Perturbation of the Boundary in Boundary-Value Problems of Partial Differential Equations

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Perturbation of the Boundary in Boundary-Value Problems of Partial Differential Equations

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London Mathematical Society Lecture Note Series. 318

Perturbation of the Boundary in Boundary-Value Problems of Partial Differential Equations Dan Henry with editorial assistance from Jack Hale and Antˆonio Luiz Pereira

   Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge  , UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521574914 © Cambridge University Press 2005 This book is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2005 - -

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Contents

Introduction

page 1

1 1.1

Geometrical Preliminaries Some notation

2

Differential Calculus of Boundary Perturbations

18

3 3.1 3.2 3.3 3.4 3.5 3.6

Examples Using the Implicit Function Theorem Torsional Rigidity Simple Eigenvalues of the Dirichlet Problem for the Laplacian Capacity Green’s Function Simple Eigenvalues of Robin’s Problem Simple Eigenvalue of a General Dirichlet Problem

27 27 32 36 39 40 44

4 4.1

Bifurcation Problems Multiple Eigenvalues of the Dirichlet Problem for the Laplacian Variation of a Turning Point A Bifurcation Problem wit Two-Dimensional Kernel Generic Simplicity of Eigenvalues of a Self-Adjoint 2m-Order Dirichlet Problem

46 46 51 53

5

The Transversality Theorem

60

6 6.1

Generic Perturbation of the Boundary Generic Simplicity of Eigenvalues of the Dirichlet Problem for


79

4.2 4.3 4.4

3 3

vii

55

80

viii 6.2 6.3 6.4 6.5 6.6 6.7 6.8

7 7.1 7.2 7.3 7.4 7.5 7.6 8 8.1 8.2 8.3 8.4 8.5

Contents Simplicity of Eigenvalues with a Reflection-symmetry Constraint Simplicity of Real Eigenvalues of a General Second-Order Dirichlet Problem Generic Simplicity of Eigenvalues for the Neumann Problem for
Generic Simplicity of
u + f (x, u, ∇u) = 0 in , u = 0 on ∂ Generic Simplicity of Eigenvalues of Robin’s Problem Generic Simplicity of Solutions of
u + f (u, x) = 0 in , ∂u/∂ N = g(k, u) on ∂ Generic Simplicity of Complex Eigenvalues of a Dirichlet Problem Boundary Operators for Second-Order Elliptic Equations Weakly Singular Integral Operators with C∗r (α) Kernels Integral Equations with C∗r (α) Kernels Fourier Transform and Composition of Weakly Singular Kernels Limits of Singular Integrals Fundamental Solutions Calculation of some boundary Operators The Method of Rapidly-Oscillating Solutions Introduction Formal Asymptotic Solutions Exact Solutions Example 6.8 Revisited: Generic Simplicity of Complex Eigenvalues Generic Simplicity of Solutions of a System

Appendix 1 Appendix 2

Eigenvalues of the Laplacian in the Presence of Symmetry On Micheletti’s Metric Space

References Index

83 85 88 89 94 100 106

114 116 122 126 135 139 141 152 152 154 158 161 168

183 188 199 203

Introduction

Perturbation of the boundary (or of the domain of definition of a boundary value problem) is a rather neglected mathematical topic, though it has attracted occasional interest: Rayleigh [33] in 1877 (the first edition), Hadamard [9] in 1908, Courant and Hilbert [5] in the German edition of 1937, Polya and Sz¨ego [31] in 1951, Garabedian and Schiffer [7] in 1952, and some more recent work [3, 14, 6, 13, 26, 21–23, 42, 35, 32, 27, 10, 11]. The list is far from complete, but is notably sparse. There seem to be two related reasons for this neglect: (1) the subject is too easy; (2) it is too difficult. If you are interested only in Fundamental Questions, this is certainly a trivial topic. One perturbs the region by applying a diffeomorphism near the identity; but you can change variables via this diffeomorphism to keep the region fixed, and are then only perturbing the coefficients in a fixed region. It is simply the chain rule. However, if you try to carry out this trivial change of variables, you will become mired in such long and difficult calculations that you’ll be tempted to quit. If you persist, and are fortunate, and are extremely careful, there may be a miraculous simplification at the end. On experiencing this miracle for the second time I became suspicious – theorems 2.2, 2.4 show how to go directly to the “miracle,” bypassing the computational morass. (Peetre [27] also found part of this result, and it is implicit in Courant and Hilbert [5: vol. 1 p. 260] for variational problems.) It is, at the end, merely the chain rule, and we may then apply standard tools (implicit function theorem, Liapunov, Schmidt method, transversality theorem) to problems of perturbation of the boundary. But the standard tools are not enough – new problems arise requiring, for example, a more general form of the transversality theorem for problems with Fredholm index – ∞ (ch. 5). Open problems abound. The calculus developed in chapter 2 applies – in principle – to almost any boundary or initial boundary-value problem, but often leads immediately to difficult unsolved problems. To avoid excessive depression of author and reader, we will 1

2

Introduction

concentrate on questions we can answer – generally boundary-value problems for scalar second-order elliptic equations. The subject is not, after all, entirely trivial. I have worked on perturbation of the boundary sporadically since about 1973, after reading Joseph’s article [13]. The formulas of theorems 2.2, 2.4 date from 1975 – in a more complicated version – and most of the examples of chapters 3 and 4 (and a few from chapter 6) were developed in 1975–1981 and exposed in seminars at the University of Kentucky, Brown University and the University of S˜ao Paulo. In 1982, I had the opportunity to develop this topic at some length [11] at the University of Brasilia. Later that year I generalized the transversality theorem (lectures at S˜ao Carlos, September 1982), which solved some problems left open in [11] and raised a host of new open problems. These demanded some tedious calculation – such as those in chapter 7 – which were only completed recently, and then only in special cases: much remains to be done. If the word “I” seems to appear excessively here, it is because very few other people have worked on these problems in the past twelve years with a comparable approach, and my work has been independent of these few (aside from some of the examples cited below). There are, of course, other notions of a “small change in the domain” besides “image under diffeomorphism near the identity” (see, for example, [3, 26, 10]); the advantage of using such regular perturbations will hopefully become more clear as we proceed. In the past five years, my work has been supported by FAPESP, and of course by IME-USP. It would have been nice to have this ready for the fiftieth anniversary of the University of S˜ao Paulo, but as usual I missed the deadline. Still, better late than never: Happy 51st birthday! Most of the following (Chapters 1 to 7) was written in 1985. But at the end of that year, I found a way to circumvent Horrible Chapter Seven, the method of rapidly oscillating solutions. My course was clear: everything should be rewritten with the new method! Unfortunately, I couldn’t seem to find the time and energy needed for the task. Finally, Jack Hale and Antonio Luiz Pereira persuaded me to publish it as it stands, with only minor corrections. (Except that the correction of Theorem 7.6.10 is not so minor, and this required rewriting Example 6.8.) A new chapter was added, on the method of rapidly oscillating solutions, with a new example (8.5: generic simplicity of solutions of a system) to show the power of the method. And, with a few years perspective, Chapter 7 does not seem so horrible. Thanks Jack and Antonio Luiz for getting it moving!

Chapter 1 Geometrical Preliminaries

In this chapter we introduce some notation, find convenient representations for regions in Rn with smooth boundary (1.2–1.7) and develop some tools needed later: approximation by smooth regions (1.8), smooth extension of functions (1.9), smooth deformations of regions (1.10), differentiation of integrals (1.11), and differential operators on a hypersurface (1.12). In short, a mixed bag of topics which will be needed later, and it may be wise to omit details until the need becomes apparent. We treat only smoothly-bounded regions. Initial-boundary-value problems lead naturally to the study of regions with corners; but our examples will be elliptic boundary-value problems, and smooth regions provide sufficient variety.

1.1 Some Notation For a function f defined near x ∈ Rn , the m th derivative at x, D m f (x), may be considered as a homogeneous polynomial of degree m (h → D m f (x)h m ) on Rn , or as a symmetric m-linear form, or as the collection of partial derivatives
 α  ∂ D m f (x) = f (x) : |α| = m , ∂x depending on convenience. Then the norm |D m f (x)| may denote  α   ∂   max  f (x) or max |D m f (x)h m |. |α|=m |h|≤1 ∂x The last version has the advantage of being independent of rotation of coordinates in Rn , but the norms are equivalent. If  is an open set in Rn and m ≥ 0 an integer, C m () is the space of m-times continuously and bounded differentiable functions on  whose derivatives 3

4

Chapter 1. Geometrical Preliminaries

extend continuously to the closure , with the usual norm φC m () = max sup |D j φ(x)|. 0≤ j≤m x∈

The space of values is some normed linear space E, and is not clear which “E” is meant, we may write C m (, E). r C m () is the closed surface of C m () consisting of functions whose m th unif derivative is uniformly continuous; if  is bounded, this is C m (). r C m,α () is the subspace of C m () consisting of functions whose m th unif derivative is H¨older continuous with exponent α (0 < α ≤ 1), provided with the norm   φC m,α () = max φC m () , Hα (D m φ)

where Hα ( f ) = sup{| f (x) − f (y)|/|x − y|α : x = y ∈ } (This space is “boundedly closed” in C 0 (); that is, a bounded sequence in C m,α () which converges uniformly (in C 0 ()) has its limits in C m,α ().) r C m,α+ () is the closed subspace of C m,α (), 0 < α < 1, consisting of functions φ ∈ C m,α () such that |D m φ(x) − D m φ(y)|/|x − y|α → 0

as x − y → 0

(x, y ∈ ),

provided with the same C m,α () norm. It is sometimes convenient to write C m,0 for C m ,

m C m,0+ for Cunif

m,α m,α+ so we allow 0 ≤ α ≤ 1 in C m,α and 0 ≤ α < 1 in C m,α+ . Cloc ()[Cloc ()] is the space of functions whose restrictions to any 0 (with 0 compact in ) are in C m,α (0 )[C m,α+ (0 ), respectively].   ∞ ∞ C ∞ () = C m (), Cloc () = Cloc () m

m

C ω () = analytic functions defined on an open set of Rn containing  (hence, extending to analytic functions on an open set in Cn , when we consider Rn as {z ∈ Cn : Im z = 0}.) Definition 1.2. An open set  ⊂ Rn has C m -regular boundary [or C m,α or m,α m,α+ ∞ or Cloc or C ∞ or Cloc or C ω ; regular boundary], if there exists C m,α+ or Cloc 1 m n m,α m,α+ or C or . . . ] which is at least in Cunif (Rn , R) such φ ∈ C (R , R) [or C that  = {x ∈ Rn : φ(x) > 0}

1.1 Some Notation

5

and φ(x) = 0 implies |gradφ(x)| ≥ 1. We may also say  is a C m region or  is C m regular [ or C m,α or . . . ]. Theorem 1.3.  ⊂ Rn has C m -regular boundary (or C m,α orC m,α+ ) – at least 1 Cuni f – if and only if there are positive constants r, M such that, given any open ball B ⊂ Rn of radius r, after appropriate rotation and translation of coordinates, we have ˆ ∩ B,  ∩ B = {x ∈ Rn |xn > ψ(x)}

xˆ = (x1 , . . . , xn−1 )

ˆ ∩B ∂ ∩ B = {x ∈ R |xn = ψ(x)} n

for some ψ ∈ C m (Rn−1 , R) (or C m,α or C m,α+ ) with norm ≤ M. Note the conditions are trivial if B ∩ ∂ = ∅. Remark. This implies easily that our definition of C m -regular boundary is equivalent to that used by F. Browder and Agmon-Douglis-Nirenberg in their studies of elliptic boundary value problems. We will see that Def. 1.2 is very convenient for discussing perturbations of the boundary (as in 1.8 below). Our Definition 1.2 applied only when ∂ is at least uniformly C 1 . The condition of Thm. 1.3 is more general, in that we may permit (for example) ψ to be merely Lipschitz continuous (in C 0,1 (Rn−1 )) which gives the “minimally smooth” domains of Stein’s extension theorem [38, Sec. 6.3]. Some of our results apply to regions with convex corners, transversal intersections of smooth regions, as described in the remark following Thm. 1.9. Proof. Suppose  = {x | φ(x) > 0} is C m (or C m,α or C m,α+ ) regular, φ(x) = 0 ⇒ |grad φ(x)| ≥ 1, L = sup |grad φ|, and choose r > 0 so |x − y| ≤ 6Lr ⇒ |Dφ(x) − Dφ(y)| < 1/2. Let B be a ball of radius r in Rn which meets ∂; we may assume the center of the ball is 0. If 0 ∈ ∂, choose the positive xn -axis along the inward normal (i.e., grad φ (0)). Otherwise let p be a point of ∂ ∩ B closest to 0 and choose the xn - axis to contain p and be directed into . Then p = (0, pn ), | pn | < r , ( p) = |Dφ( p)| ≥ 1 (possibly p = 0). Also φ (0, s) has the same sign and ∂∂φ xn (x) > 1/2. as s − pn in −r < s < r, and for |x − p| ≤ 6Lr we have ∂∂φ xn n−1 ˆ ≤ 2r ; then |φ(x, ˆ xn ) − φ(0, xn )| ≤ 2Lr and ±φ(x, ˆ pn ± Let xˆ ∈ R , |x| ˆ ∈ ( pn − 4Lr ) ≥ ±φ(0, pn ± 4Lr ) − 2Lr > 0 so there exists unique ψ(x) ˆ ψ(x)) ˆ = 0. By the implicit function theorem, ψ is C m 4Lr, pn + 4Lr ) with φ(x, m,a m,α+ ˆ = |−(∂φ/∂ x)/(∂φ/∂ ˆ (or C or C , respectively) and |Dψ(x)| xn )| ≤ 2M ∞ ˆ ≤ 2r . Choose some (fixed) C θ : R → [0, 1] with θ(t) = 1 for t ≤ 1, for |x| ˆ = θ(|x|/r ˆ )ψ(x), ˆ or zero for |x| ˆ ≥ 3r/2. θ(t) = 0 for t ≥ 3/2, and let ψ0 (x) Then ψ0 ∈ C m (Rn−1 ) (or C m,α or C m,α+ ) with norm bounded by a multiple

6

Chapter 1. Geometrical Preliminaries

(depending only on n, m, α, r) of the norm of φ, and ˆ  ∩ B = {x ∈ B | xn > ψ0 (x)},

ˆ ∂ ∩ B = {x ∈ B | xn = ψ(x)}.

For the converse, we need the following √ Lemma 1.4. Let σ > n, r > 0, and for each k = (k1 , . . . , kn ) ∈ Zn , let Bk 1/2 be the open ball in Rn with radius r and center (r/σ )k, while Bk is the n concentric ball with radius r/2. Then every point of R is contained in some 1/2 Bk , and no point is contained in more than (2σ + 1)n of the balls Bk . Proof of the Lemma. The result is invariant under a homothety (x → cx, c = constant > 0) so it suffices to treat the case r = σ . If x ∈ Rn there exists k ∈ Zn √ 1/2 so x − k ∈ [−1/2 , 1/2 )n , hence |x − k| ≤ n/2 < σ/2 so x ∈ Bk . Suppose x ∈ Bk ; then for each j = 1, . . . n, |x j − k j | < σ so k j is an integer in (x j − σ, x j + σ ). But (x j − σ, x j + σ ) contains no more than 2σ + 1 integers, so there are at most (2σ + 1)n choices of k ∈ Zn such that x ∈ Bk . Completion of Proof of (1.3). Assume r, M given satisfying the requirements of the theorem; we must find φ : Rn → R which satisfies the conditions √ (1.2). Choose σ = 2n + 1 in the lemma. There is a C ∞ partition of unity {φk }k∈Zn 1/2 supp φk ⊂ Bk , φk ≥ 0, φk (x) ≡ 1 k

with φk C m (or φk C m,α ) uniformly ≤ K = K (n, r, m, α). If Bk ∩ ∂ = φ, there is a function Sk (x) of class C m (or C m,α or C m,α+ ) − ˆ after rotation of coordinates – such that Sk (x) = xn − ψ(x)  ∩ Bk = {x ∈ Bk : Sk (x) > 0}, ∂ ∩ Bk = {x ∈ Bk : Sk (x) = 0}, with Sk C m (or C m,α ) ≤ M and Sk (x) = 0 ⇒ |DSk (x)| ≥ 1. If Bk ⊂ , let Sk = 1; if Bk ∩  = ∅, let Sk = −1. Define ψ(x) =

m n m,α or C m,α+ ], ψ > 0 in , ψ = k φk (x)Sk (x). Then ψ is in C (R , R) [or C 0 on ∂, ψ < 0 outside . If x ∈ ∂ then φk (x)Sk (x) = 0 for each k and at x, grad (φk Sk ) = φk grad Sk which either vanishes or has the direction of the

inward normal so |grad ψ(x) | = k φk (x)| grad Sk (x)| ≥ 1. The following “normal coordinates” are sometimes useful.

1.1 Some Notation

7

Theorem 1.5. Let  ⊂ Rn have C m -regular boundary (or C m,α or C m,α+ , 2 ≤ m ≤ ∞). There exists r > 0 so that if Br (∂) = {x : dist(x, ∂) < r } π (x) = the point of ∂ nearest to x t(x) = ±dist(x, ∂)

(“+” outside, “−” inside)

then t(·) : Br (∂) → (−r, r ), π(·) : Br (∂) → ∂ are well-defined, π is a C m−1 (or C m−1,α or C m−1,α+ ) retraction onto ∂ (π(x) = x when x ∈ ∂) and t has the same smoothness as ∂ (C m or C m,α or C m,α+ ). Further x → (t(x), π(x)) : Br (∂) → (−r, r ) × ∂ is a C m−1 (or C m−1,α or C m−1,α+ ) diffeomorphism with inverse (t, ξ ) → ξ + t N (ξ ) : (−r, r ) × ∂ → Br (∂) where N (ξ ) is the unit outward normal to ∂ at ξ . t(·) is the unique solution of |∇t(x)| = 1 in Br (∂), with t = 0 on ∂, ∂t/∂ N > 0 on ∂. Extending the normal field N to a neighborhood of ∂ by N (ξ + t N (ξ )) = N (ξ ) −r < t < r, we have N (x) = grad t(x) on Br (∂). Also K (x) = D N (x) = D 2 t(x), restricted to the tangent space at x ∈ ∂, is the curvature of ∂. It is sometimes convenient to call K (x) the curvature, though it is degenerate (K (x)N (x) = 0) in the normal direction. Remark. The fact that t(·) has the same smoothness as ∂ – does not lose a derivative, as happens with π(·) – seems to have been noted first by Gilbarg and Trudinger [8]. The best (largest) choice of r is r = 1/ max |k|, where k is the sectional curvature of the boundary in any (tangent) direction at any point of ∂. Corollary 1.6. A C m -regular region  ⊂ Rn , m ≥ 2, may be represented by {x|φ(x) > 0} where φ is C m and |∇φ(x)| ≡ 1 on a neighborhood of ∂. In this case, φ is unique on a neighborhood of ∂. Proof of (1.5). By the inverse function theorem, the C m−1 map (t, ξ ) → ξ + t N (ξ ) = x has a C m−1 inverse t = t(x), ξ = π(x), on some neighborhood of ∂. On the other hand, for each x ∈ Rn , ξ → 1/2 |x − ξ |2 (ξ ∈ ∂) has a minimum and if ξ is a minimizing point then x − ξ ⊥ Tξ (∂) so x = ξ + t N (ξ )

8

Chapter 1. Geometrical Preliminaries

for some real t with t = ±dist(x, ∂). We show, for some r > 0, that x = ξ + t N (ξ ), t = ±dist(x, ∂) has a unique solution ξ whenever x ∈ Br (∂) so ξ = π (x) is the (unique) nearest points. Suppose  = {x : φ(x) > 0}, φ(x) = 0 ⇒ |∇φ(x)| ≥ 1 and r = 1/ sup |D 2 φ(x)|. If dist(x, ∂) < r but there are two “nearest points” ξ1 = ξ2 , x = ξ j + t N (ξ j ), then t 2 = |ξ1 − ξ2 + t N (ξ1 )|2 = |ξ1 − ξ2 |2 +  12t(ξ1 − ξ2 ). N (ξ1 ) + t 2 . Now φ(ξ2 ) = φ(ξ1 ) = 0 = ∇φ(ξ1 ) · (ξ2 − ξ1 ) + 1/2 0 D 2 φ(θ ξ2 + (1 − θ ξ1 )) · (ξ2 − ξ1 )2 dθ and N (ξ1 ) = −∇φ(ξ1 )/|∇φ(ξ1 )|, so for |t| < r 0 < |ξ1 − ξ2 |2 = 2t N (ξ1 ) · (ξ2 − ξ1 ) ≤ |t| sup |D 2 φ| |ξ1 − ξ2 |2 < |ξ1 − ξ2 |2 , a contradiction. Thus t(·), π(·) are well-defined and C m−1 on Br (∂). Extend N to be constant on normal lines, so N (x) = N (π (x)) is C m−1 on Br (∂). It is clear that t(x + s N (x)) = t(x) + s when t(x) and t(x) + s are in (−r, r ), so ∂t(x)/∂ x j = ∂ j t(x + s N (x)) + s

n ∂k t(x + s N (x))∂ j Nk (x). k=1

Let s = −t(x), so x + s N (x) = π(x) ∈ ∂, and note ∇t(π(x)) = N (π (x)) = N (x). Then on Br (∂) ∂ j t(x) = N j (x) − t(x)

m

Nk (x)∂ j Nk (x) = N j (x)

k=1

or N (x) = grad t(x). (Observe that K (x)N (x) = ∂ N /∂ N (x) = 0.) Clearly N is C m−1 so t(·) is C m . To identify D N (ξ ) = D 2 t(ξ ) as the curvature of ∂ at ξ , we may choose ˆ where ψ(0) = 0, Dψ(0) = 0 and K ij = coordinates so  is locally {xn > ψ(x)} 2 ∂ ψ/∂ xi ∂ x j (0) (1 ≤ i, j ≤ n − 1) is the curvature matrix (on the tangent plane ˆ 0) + o(|x|) as x → 0 so Rn−1 × 0). It is easy to show N (x) = (0, −1) + (K x, D N (0) = ( 0K 00 ), and the restriction in the tangent plane is the curvature K of ∂ at 0. Remark. If S is a C 2 hypersurface, x0 ∈ S, the usual existence theory [5], [6] for first order scalar PDEs says there is a unique C 1 solution of |∇t| = 1 near x0 with t = 0, ∂t/∂ N > 0 on S near x0 . If S is not C 2 (or C 1,1 ) there may be no C 1 solution of this problem. For example if 1 < p < 2, S = {x : x2 = |x1 | p}, there are two “nearest points” to x when x1 = 0, x2 > 0 (near 0) and the gradient of dist(x, S) has a discontinuous jump as x1 crosses 0 with x2 > 0. Theorem 1.7. Let  be a compact subgroup of the orthogonal group O(n) and let  ⊂ Rn be a C m -regular region such that γ () =  for all γ ∈ . Then there exists C m φ : Rn → R such that for  = {x : φ(x) > 0},

1.1 Some Notation

9

φ(x) = 0 ⇒ |∇φ(x)| ≥ 1, and φ(γ ·x) = φ(x) for all γ ∈ , x ∈ Rn . If ∂ is at least C 2 , we may choose such φ with |∇φ| = 1 on a neighborhood of ∂. Proof. There exists a function φ0 satisfying all requirements except perhaps -invariance. Let φ(x) be the average of γ → φ0 (γ x) with respect to Haar measure in ; then φ is certainly C m and -invariant. Further x ∈  ⇒ γ x ∈  ⇒ φ0 (γ x) > 0 for all γ ∈  so φ(x) > 0 in ; and similarly φ(x) = 0 on ∂, φ(x) < 0 outside . On ∂, N (x) = −∇φ0 (x)/|∇φ0 (x)| and it follows easily that N (γ x) = γ N (x) for x ∈ ∂, γ ∈ , so grad(φ0 (γ x)) = γ (grad φ0 (γ x) = −N (x)|grad φ0 (γ x)| and (averaging with respect to γ ) |grad φ(x)| = averageγ |grad φ0 (γ x)| ≥ 1 for x ∈ ∂. If in fact |∇φ0 (x)| = 1 near ∂, then φ(x) = φ0 (x) = ±dist(x, ∂) near ∂ so |∇φ(x)| = 1 near ∂. A common technique in analysis is to approximate a given function by a smooth function, to facilitate calculations, and take limits only at the end. Similarly we may approximate a given region by smooth regions. If  = {x : m region, and ψ is C m -close to φ, we show {x : ψ(x) > 0} is φ(x) > 0} is a Cunif m C -close to , that is, it is diffeomorphic to  by a diffeomorphism C m -close to the identity. Theorem 1.8. Let φ : Rn → R be (at least) uniformly C 1 , φ(x) = 0 ⇒ |grad φ(x)| ≥ 1,  = {x : φ(x) > 0}, and for some a > 0, 
1 dist(x, ∂), a for all x. |φ(x)| ≥ min 2 The last condition may always be achieved by modifying φ away from ∂. Also, let r0 > 0. Then there exists 0 > 0 such that, if ψ − φC 1 (Rn ) < 0 , there is a diffeomorphism h(·, ψ) : Rn → Rn supported in Br0 (∂) – i.e., h(x, ψ) = x when dist (x, ∂) ≥ r0 – such that h(; ψ) = {x : ψ(x) > 0} and h(·, ψ) − idRn c1 (Rn ) → 0 as ψ − φc1 (Rn ) → 0. If ψ is of class C m,α [or C m,α+ or C ∞ or C ω ] then {x : ψ(x) > 0} is a C m,α [or C m,α+ or C ∞ or C ω ] regular region, assuming ψ − φC 1 < 0 . If φ, ψ are C m,α+ then h(·, ψ) is a C m,α+ diffeomorphism, h(·, ψ) − idRn C m,α → 0 as ψ − φC m,α → 0 and (x, ψ) → h(x, ψ) : Rn × C m,α+ (Rn ) → Rn is C m,α+ . If φ, ψ are C m,α then h(·, ψ) is a C m,α diffeomorphism and (x, ψ) → h(x, ψ) : Rn × C m,α → Rn is C m,α . As ψ − φC 1 → 0 with ψC m,α bounded, h(·, ψ) remains bounded in C m,α and converges to idRn in C k,β when k + β < m + α.

10

Chapter 1. Geometrical Preliminaries

Remark. The hypothesis ψ − φC m,α → 0 does not yield C m,α convergence of h(·, ψ) → id, if φ is not C m,α+ . Proof. Choose a C ∞ vector field M(x) on a neighborhood of ∂, uniformly close to grad φ: for some r1 > 0 and C ≥ 2 1 ≤ |M(x)| ≤ C 2 and 1 |M(x)|2 if dist(x, ∂) ≤ r1 . 2 Let r2 = min{a, r0 , 1/2r1 } and choose C ∞ θ : Rn → [0, 1] so that θ = 1 on Br2 /2 (∂), θ = 0 outside Br2 (∂). We will show that, if ψ − φC 1 is small and x ∈ Br2 (∂), there is a unique t = t(x; ψ) near 0 such that M(x) · grad φ(x) ≥

ψ(x + t M(x)) = φ(x). Then define h by h(x; ψ) =



x + θ(x)t(x; ψ)

M(x) in Br2 (∂)

x

outside Br2 (∂)

and the conditions of the theorem are satisfied. First choose s0 > 0 so small that s0 ≤ 1/2r2 and |Dφ(x) − Dφ(y)| ≤ 1/8C when |x − y| ≤ C S0 . Suppose ψ − φC 0 = sup |ψ − φ| ≤ s0 /32 and sup |Dψ − Dφ| ≤ 1/8C. Then for −s0 ≤ t ≤ s0 , dist(x, ∂) ≤ r2 , ∂ (ψ(x + t M(x)) − φ(x)) = M(x) · Dψ(x + t M(x)) ∂t = M(x) · Dφ(x) + M(x) · (Dψ(x + t M(x)) − Dφ(x))   1 1 1 1 2 + ≥ ≥ |M| − |M| 2 8C 8C 16 while for t = ±s0 t(ψ(x + t M(x)) − φ(x)) ≥ t(φ(x + t M(x)) − φ(x)) − s0 ψ − φC 0   s2 s2 2 1 2 |M| − |M|/8c − 0 ≥ 0 > 0. ≥ s0 2 32 32 Thus there is a unique t = t(x, ψ) in (−s0 , s0 ) with ψ(x + t M(x)) = φ(x), and it is easy to see ∂ t(x; ψ) → 0 ∂x → 0. (Here we again use the uniform

|t(x, ψ)| ≤ 16ψ − φC 0 uniformly on Br2 (∂) as ψ − φC 1 continuity of Dψ.)

and

1.1 Some Notation

11

The evaluation map (ψ, z) → ψ(z) : C m,α × Rn → R [or C m,α+ × Rn → R] is of class C m,α [or C m,α+ ]. The proof is similar to that of the case C m in [1]. Thus, by the implicit function theorem, if φ is C m,α [or C m,α+ ], (x, ψ) → t(x; ψ) : Br2 (∂) × C m,α [or C m,α+ ] → R is of class C m,α [or C m,α+ ]. It is useful to define an extension operator E  for open  ⊂ Rn such that given u :  → R (or R N or C N ), E  (u) is defined on all Rn with the same smoothness as u and E  (u) = u on . Stein [38] defines such an operator for regions whose boundary is merely Lipschitzian. When the boundary is smooth, there is a simpler construction which also applies (with some modifications, noted below) when  has convex corners. Theorem 1.9. Let  ⊂ Rn have C m,α (or C m,α+ ) regular boundary, 1 ≤ m < ∞. There is a linear operator E  which carries functions u :  → R to functions E  (u) : Rn → R such that E  (u) = u on  and for may integer k in 0 ≤ k ≤ m and any p, β in 1 ≤ p ≤ ∞, 0 ≤ β ≤ 1, with k + β ≤ m + α, we have uW k, p () ≤ E  (u)W k, p (Rn ) ≤ Cm, uW k, p () uC k,β () ≤ E  (u)C k,β (Rn ) ≤ Cm, uC k,β () where Cm, is a constant depending only on m and . If ∂ is C k,β+ and u is C k,β+ in , then E  (u) is C k,β+ in Rn . For any r0 > 0 we may choose E  such that the support of E  (u) is in an r0 -neighborhood of the support of u (⊂ ), and such that E  (u) outside  depends only on the values of u in an r0 -neighborhood of ∂. In this case, the constant Cm, depends also on r0 . The constants Cm,h() are bounded when h varies in a C m,α -small neighborhood of ∞ and u = U | for some U ∈ Cc∞ (Rn ) the inclusion i  :  → Rn . If ∂ is Cloc ∞ n then E  (u) ∈ Cc (R ). Proof. First we consider a half-space, R × Rn−1 . Let a j , b j ≤ −1 ( j =



k 1, 2, . . .) be real such that a j → 0 rapidly, |a j b j |k < ∞ and ∞ 1 ajbj = 1 for every k = 0, 1, 2, . . . . Then if B j (t) = t for t ≥ 0, B j (t) = b j t for t ≤ 0, v → V defined by V (t, ξ ) =

∞

a j v(B j (t), ξ )

(t ∈ R, ξ ∈ Rn−1 )

j=1

is the desired extension operator for  = R+ × Rn−1 . Specifically let b j =

j 1 − 2 j (as in [35]) and define the a j by ∞ 1 a j z = A(z)  ∞ ∞

A(z) = z (1 − z/2k ) (1 − 1/2k ). k=1

1

12

Chapter 1. Geometrical Preliminaries

Then A is an entire analytic function of order 0, A(1) = 1, A(2 j ) = 0 for j = 1, 2, 3, . . . and so ∞ j=1

a j bkj

=

k   k i=0

i

(−1)i A(2i ) = 1 for each k = 0, 1, 2, . . .

In the general case, let M : Rn → Rn be C ∞ (each derivative bounded on Rn ) which is uniformly close to the outward unit normal on ∂, recall ∂ is at least uniformly C 1 (Def. 1.1). Then there exists r1 > 0 so that (t, ξ ) → ξ + t M(ξ ) : (−r1 , r1 ) × ∂ → Rn is a C m,α (or C m,α+ ) diffeomorphism onto a neighborhood V1 of ∂, and V1 ⊃ Br2 (∂) for some r2 > 0. Chose 0 < r3 < r2 and some C ∞ θ : Rn → [0, 1] such that θ = 1 in Br3 (∂), θ = 0 outside Br2 (∂). Define  in   u(x)

∞ E  (u)(x) = θ(x) k=1 ak (θu)(ξ + bk t M(ξ )) for x = ξ + t M(ξ ),   0 outside  ∪ V1 . Writing ∂/∂ M for (∂/∂t)ξ =const in the coordinates x = ξ + t M(ξ ) near ∂, it follows (∂/∂ M) j E  u(ξ ± 0 · M(ξ )) = (∂/∂ M) j u(ξ − 0 · M(ξ )), ξ ∈ ∂, for each j and E  is the desire extension operator. Remarks. We may treat similarly regions with convex corners, of the form  = {x : φ j (x) > 0 for j = 1, . . . , ν} where each φ j is C m,α (or C m,α+ ) and for each subset { j1 , . . . , jk } ⊂ {1, . . . , v}, (0, . . . , 0) is a regular value of (φ j1 , . . . , φ jk ). There are also some uniformity conditions, which we avoid by assuming ∂ is compact. Then for each point x of , there is a neighborhood of x in , C m,α k (or C m,α+ ) diffeomorphic to an open set in R+ × Rn−k (and taking x to 0), for some integer k, 0 ≤ k ≤ min(n, ν). We say x is a k th order corner (k = 0 for an interior point) and exactly k of the φ j vanish at x. We may define an extension operator for Rk+ × Rn−k by extending one variable at a time as above. The result is V (t, ξ ) = ai1 · · · aik v(Bi1 (t1 ), . . . , Bik (tk ), ξ ) i 1 ,...,i k ≥1

for t ∈ Rk , ξ ∈ Rn−k . A partition of unity and the usual tricks give the extension operator for . Another type of extension problem occurs when we have a function defined initially only on ∂. As above this reduces to extension from Rn−1 × 0 to Rn , supposing ∂ is smooth. (When ∂ has corners, there will also be some compatibility conditions to satisfy.) Such extensions relative to the Sobolev

1.1 Some Notation

13

spaces are fairly complicated, but fortunately in our applications, this extension problem occurs only in the trivial case C m . Corollary 1.10. Let  ⊂ Rn have C m,α [or C m,α+ ] boundary (m ≥ 1) at least 1 = C 1 . Let h ∈ C m,α ((−r, r ) × , Rn ) [or C m,α+ ] for some r > 0 and Cunif h(0, x) = x for x ∈ , i.e., h(0, ·) is the inclusion i  of  in Rn and h(t, ·) :  → Rn is an imbedding for small t. Then there exists H ∈ C m,α ((−r, r ) × Rn , Rn ) [or C m,α+ ] such that H (t, x) = h(t, x) for x ∈ , H (0, ·) = identity on Rn , and for some r0 > 0, H (t, ·) is a diffeomorphism on Rn with inverse H −1 (t, ·) for −r0 < t < r0 and H −1 ∈ C m,α ((−r0 , r0 ) × Rn ) [or C m,α+ , respectively]. Similar results hold when t → h(t, ·) ∈ C m,α is C k,β with (t, x) are in C m,α [or C m,α+ ] k + β ≤ m + α. If both (t, x) → h(t, x), ∂h ∂t on (−r, r ) × , h(0, ·) = i  , there exists V ∈ C m,α ((−r0 , r0 ) × Rn , Rn ) [or C m,α+ ] such that ∂ h(t, x) = V (t, h(t, x)) for x ∈ , −r0 < t < r0 ∂t h(0, x) = x. Conversely, given such V , the function h defined by this differential equation (t, x) of class C m,α [or C m,α+ , respectively]. has (t, x) → h(t, x), ∂h ∂t Remark. This corollary provides alternative representations of a “smoth curve of regions t → (t),” (t) = h(t, ). We will ordinarily use the first form (a curve of imbeddings of  in Rn ), but the other representations are also convenient in some cases. In light of Th. 1.8, we may equally use (t) = {x : φ(x, t) > 0} or (as in the proof of that theorem) ∂(t) = {ξ + S(ξ, t)M(ξ ) | ξ ∈ ∂}, a graph over ∂, given a smooth vector field M transverse to ∂, where S(ξ, t) is a smooth real-valued function on ∂ × (−r, r ) which vanishes when t = 0. The last representation has the advantage of uniqueness given M; the others are degenerate, with many imbeddings (or diffeomorphisms or . . .) yielding the same geometric result. However, this degeneracy does not seem to cause problems. Some analogous results occur as lemmas in Examples 3.1, 3.2 below. Ana Maria Micheletti [22] shows the set of regions C m -diffeomorphic to a given bounded C m region  ⊂ Rn can be considered a complete metric space relative to the “Courant distance:” d(1 , 2 ) = inf

N 

    f j − idC m +  f j−1 − idC m

j=1

considered over all N ≥ 1 and f j ∈ C m (Rn , Rn ) (1 ≤ j ≤ N ) which are diffeomorphisms with compact support ( f j (x) = x outside a compact set) and such

14

Chapter 1. Geometrical Preliminaries

that f 1 ◦ f 2 ◦ . . . ◦ f N carries 1 onto 2 . We will not use this metric space explicitly, but our representations above (for α = 0: recall  is bounded here) all yield continuous curves t → (t) in this metric. Proof of the Corollary. Let E  be the extension operator provided by Thm. 1.9, and define H (t, ·) = idRn + E  (h(t, ·) − i  ). Since E  is linear, we also have j ∂ti H (t, ·) = E  ∂t h(t, ·) for i ≤ j ≤ m and it follows that H ∈ C m,α ((−r, r ) × Rn , Rn ) [or C m,α+ ] with H (0, x) = x on Rn . Since supx | ∂∂x H (x, t) − I | < 1 for t near 0, the inverse H −1 (t, ·) exists and is mostly easily studied as (t, x) → (t, H −1 (t, ·)(x)), which is the inverse of (t, x) → (t, H (t, x)). The inversefunction theorem in the class C m,α [or C m,α+ ] shows (t, x) → H −1 (t, ·)(x) (t, x) is C m,α [or C m,α+ ] is C m,α [or C m,α+ ] for t near 0. If also (t, x) → ∂h ∂t ∂ −1 then (t, x) → H (t, ·)(x), ∂t H (t, x) and V (t, x) = (∂t H )(t, H −1 (t, ·)(x)) are C m,α [or C m,α+ ], and ∂t∂ H (t, x) = V (t, H (t, x)) for t near 0, x ∈ Rn . 1 1 region and h ∈ Cunif ((−r, r ) × , Rn ) Theorem 1.11. Let  ⊂ Rn be a Cunif with h(0, ·) = i  [h(0, x) = x for x ∈ ]. Also let f : R × Rn → R and ∂ f /∂t be continuous (at least near t = 0), with compact support. Then if (t) = h(t, ), t → (t) f (t, x)d x is C 1 near t = 0 and

d dt



 (t)

f (t, x)d x =

(t)

∂f (t, x)d x + ∂t

 ∂(t)

f (t, x)V · N d A x

where V is the velocity field (so ∂h/∂t(t, x) = V (t, h(t, x))) and N the unit outward normal. Let  be a C 2 region and assume in addition to the previous hypotheses, ∂ f /∂ x is continuous ( f is C 1 ). that ∂ 2 h/∂ x 2 and ∂ 2 h/∂t∂  x are continuous and 1 Then for t near 0, t → ∂(t) f (t, x)dA x is C and d dt

∂(t)





 f (t, x)d A x =

∂(t)

 ∂f ∂f + V · NN · + HV · N f dA ∂t ∂x

where, as before, V is the velocity field, N the outward unit normal, and H = divN is the mean curvature of ∂(t) (the sum of the principal curvatures). Proof. It suffices to prove these results when f and h are smooth (say, C 2 ) and to compute the derivative at t = 0. Then h(t, x) = x + t V (x) + O(t 2 ), ∂h (t, x) = I + t V (x) + O(t 2 ) ∂x

1.1 Some Notation and

 (t)

 =

 ∂h f (t, h(t, x))det (t, x) d x ∂x 

 f (t, y)dy =

15





{ f (0, x) + t( f t + f x · V + f div V ) + O(t 2 )} d x

   so the derivative at t = 0 is  ( f t + div( f V )) =  f t + ∂ f V · N . For the second case, we may suppose the normal field N (t, x) for ∂(t) is extended as a C 1 unit-vector field near ∂(t), and then extended in any C 1 manner throughout Rn . Then   f (t, x)d A x = div( f (t, x)N (t, x))d x ∂(t)

(t)

and we may apply the first case, merely recalling that ∂ N /∂t · N = 0 on ∂(t). We will sometimes need to use differential operators (gradient, divergence and Laplacian) in a hypersurface S ⊂ Rn . The following definitions are all equivalent to the corresponding formulas of Riemannian geometry or tensor analysis, in the metric induced in S from (Euclidean) Rn . These formulas are (of course) intrisic to S and say nothing about the surrounding Rn ; but our interest is precisely in this relation to a neighborhood of S (see Th. 1.13). Definition 1.12. Let S be a C 1 hypersurface in Rn and let ϕ : S → R be C 1 (so it may be extended to be C 1 on a neighborhood of S); then ∇ S ϕ is the tangent vector field on S such that, for each C 1 curve t → x(t) ∈ S, we have d ˙ ϕ(x(t)) = ∇ S ϕ(x(t)) · x(t). dt Let S be a C 2 hypersurface in Rn and let a : S → Rn be a C 1 vector field – it is a tangent-vector field if a · N ≡ 0 on S, where N is a unit normal field on S. Then div S a : S → R is the continuous function such that, for every C 1 ϕ : S → R with support compact in S,   (div S a)ϕ = − a · ∇ S ϕ. S

S

Note div S a depends only on the tangential component of a|S. Finally, if u : S → R is C 2 , then  S u = div S (∇s u) or equivalently, for all C 1 ϕ : S → R with compact support,   ϕ S u = − ∇ S ϕ · ∇ S u S

S

16

Chapter 1. Geometrical Preliminaries

Theorem 1.13. (i) If S is a C 1 hypersurface and ϕ : Rn → R is C 1 on a neighborhood of S, then on S ∇ S ϕ(x) = the component of grad ϕ(x) tangent to S at x = ∇ϕ(x) − N (x)∂ϕ/∂ N (x) where N is an unit-normal field on S. ∇ S ϕ depends only on the restriction ϕ|S. (ii) If S is a C 2 hypersurface, a : Rn → Rn is C 1 on a neighborhood of S, N : Rn → Rn is a C 1 unit-vector field on a neighborhood of S which is a normal field at points of S near x0 ∈ S and H = div N is the mean curvature of S (near x0 ), then ∂ (a · N ) div S a = div a − H a · N − ∂N

n ∂a j on S (near x0 ), where div a = i=1 ∂ x j . div S a depends only on the tangential component of a at point of S. (iii) If S is a C 2 hypersurface, u : Rn → R is C 2 on a neighborhood of S, and N is a normal-vector field for S (near x0 ) in the sense of (ii) above, then  S u = u − H

∂ 2u ∂u ∂N − + ∇S u · 2 ∂N ∂N ∂N

on S near x 0 . We may choose N so that ∂∂ NN = 0 on S and then the final term is omitted.  S u depends only on the restriction u|S. Proof. (i) is trivial; we prove (ii), and (iii) may be proved in the same way (or see [11]). Use the “normal coordinates” of Th. 1.5, and let St be the normal translation of S, St = {ξ + t N (ξ ) | ξ ∈ S}; we may suppose ϕ has small support, so we need only work with a small piece of S. Let (S, St ) = {ξ + θ N (ξ ) | 0 < θ < t, ξ ∈ S} and by the divergence theorem    (ϕ div a + ∇ϕ · a) = ϕa · N − ϕa · N . (S,St )

St

The derivative at t = 0 is then   (ϕ div a + ∇ϕ · a) = S

so

S

S

∂ (ϕa · N ) + H ϕa · N ∂N

 ∂ (a · N )) = 0 a · ∇ S ϕ + ϕ(div a − H a · N − ∂N


S

for all such ϕ (with small support). This is the formula claimed for a particular choice of the normal field (with ∂ N /∂ N = 0), and the general case follows easily.

1.1 Some Notation

17

Remark.  S u may also be determined from 

2  S u(x) + o( 2 ) as → 0+ . udA/|S ∩ B (x)| = u(x) + 2(n + 1) S∩B (x) Alternatively if S is written as a graph over the tangent plane Tx (S), S (near x) is {x + ξ − νx (ξ )N (x) | ξ ∈ Tx (S) near 0} where ∇x (·) is C 2 , νx (ξ ) = O(|ξ |2 ) as ξ → 0, and if ξ denotes the Laplacian relative to (induced) Cartesian coordinates in Tx (∂), then ξ u(x + ξ − νx (ξ )N (x))|ξ =0 =  S u(x).

Chapter 2 Differential Calculus of Boundary Perturbations

We will develop a kind of differential calculus in which the independent variable is the domain of definition of a boundary-value problem. There are conceptual difficulties in this task which have already been encountered in continuum mechanics. There are two customary ways to describe the motion (or deformation of the region): (i) the Lagrangian description, in which we label each “particle,” for example by giving its position at some time t0 ; (ii) the Eulerian description, in which we write the velocity and other variables as functions of time and position in a fixed coordinate system. time t

time t0

x2 V

p

x

p

time t (Lagrange)

(Euler)

x1

For example, in the Eulerian form we have a velocity function V (x, t) which is the velocity a particle would have if it were at position x at time t. The particle p occupies position x(t, p) at time t, so ∂ x(t, p) = V (x(t, p), t). ∂t The Eulerian form is frequently (not always) more natural and simpler for computations, but we use the Lagrangian form to prove theorems. In fact, we use both methods – they are essentially equivalent – and the results of this chapter show how to pass from one to the other. We will use an artificial “time” t to parameterize the regions; when there is a natural time-dependence in the problem this could cause confusion and another 18

Chapter 2. Differential Calculus of Boundary Perturbations letter (
= t) should be used. Consider a formal non-linear differential operator u → v,
 ∂u ∂u ∂ 2 u ∂ 2 u v(y) = f y, u(y), ,..., , 2, ,... , ∂ y1 ∂ yn ∂ y1 ∂ y1 ∂ y2

19

y ∈ Rn

where f is a given function and u, f might have several components. For the sake of less cumbersome notation, we define a constant matrix coefficient differential operator L 
∂u ∂ 2u ∂u (y), . . . , (y), 2 (y), . . . Lu(y) = u(y), ∂ y1 ∂ yn ∂ y1 with as many terms as are needed, so our nonlinear operator becomes u → f (·, Lu(·)). More precisely, suppose Lu(·) has values in R p and f (y, λ) is defined for (y, λ) in some open set G ⊂ Rn × R p . For subsets  ⊂ Rn define F by F (u)(y) = f (y, Lu(y)),

y∈

for sufficiently smooth functions u on  such that (y, Lu(y)) ∈ G for all y ∈ . For example, if f is continuous,  is bounded and L involves derivatives of order ≤ m, the domain of F is an open set – perhaps empty – in C m (), and the values of F are in C 0 (). [Other function spaces could be used, with the obvious modifications.] If  is bounded and the domain of F is nonempty then also the domain of Fh() is nonempty in C m (h()) for all h :  → Rn in some neighborhood of the inclusion i  . We will usually work with bounded domains, but this is not essential; in Example 4 (Capacity) below, we consider deformations of Rn \ where  is bounded, applying diffeomorphisms with compact support (equal to the identity at large distances). There may be some technical problems when ∂ is unbounded, but for many purposes it is sufficient to restrict attention to diffeomorphisms with compact support. Let h :  → Rn be a C m imbedding, i.e., a C m diffeomorphism to its image h(). We define the composition map (or pull-back) h ∗ by h ∗ u(x) = u(h(x)),

x ∈

when u is a given function on h(), Then h ∗ is an isomorphism of C m (h()) onto C m (), with inverse h ∗−1 = (h −1 )∗ , and we use the same notation for the pullback in other function spaces, H¨older spaces h ∗ : C k,β (h()) → C k,β (0 ≤ k + β ≤ m), Sobolev spaces h ∗ : W k, p (h()) → W k, p () (0 ≤ k ≤ m, 1 ≤ p ≤ ∞), etc. If u = col(u 1 , . . . , u q ) then h ∗ u = col(h ∗ u 1 , . . . , h ∗ u 1 ). An appropriate pull-back for the normal-vector field N (actually, covector) will be

20

Chapter 2. Differential Calculus of Boundary Perturbations

discussed below (2.3); our problems do not have sufficient geometric structure to make it worthwhile burdening the pull-back h ∗ with more responsibilities. For such an imbedding h of a bounded region , Fh() has an open domain D(Fh() ⊂ C m (h()) and Fh() : D(Fh() ) ⊂ C m (h() → C 0 (h()) is what we term the Eulerian form of the formal nonlinear differential operator v → f (·, Lv(·)) on h(), while h ∗ Fh() h ∗−1 : h ∗ D(Fh() ) ⊂ C m () → C 0 () is the Lagrangian form; the same terms are used relative to other function spaces. The advantage of the Lagrangian form is that it acts in spaces which don’t depend on h, facilitating use of the implicit function theorem (for example). But then we need to know the smoothness of (h, u) → h ∗ Fh() h ∗−1 (u) and we must be able to calculate derivatives with respect to h. (Since h ∗ is linear, the derivative with respect to u presents no problems.) Smoothness. First note (with v = h ∗−1 u, y = h(x), x ∈ ) h ∗ Fh() h ∗−1 u(x) = Fh() v(h(x)) = f (y, Lv(y)) = f (h(x), h ∗ Lh ∗−1 u(x)). A typical constant coefficient differential operator is (∂/∂ y)α =

n 

(∂/∂ y j )α j ,

i=1

α = (α1 , . . . , αn ), α j = integer ≥ 0. From the chain rule  α j
α
α n n    −1  ∂ ∂ ∂ ∗ ∗−1 h hx i j h u(x) = v(y) = u(x). ∂y ∂y ∂ xi j=1 i=1 Thus, for example (h, u) → (h ∗ Lh ∗−1 )u : Diff m () × C m () → C 0 () is analytic (in fact, rational), where Diff m () is the open subset of maps in C m (, Rn ) which are diffeomorphisms to their images (= imbeddings), supposing  is a bounded C m region. If in addition f is C k or analytic on G,

Chapter 2. Differential Calculus of Boundary Perturbations

21

then (h, u) → h ∗ Fh() h ∗−1 (u) = f (h(·), h ∗ Lh ∗−1 u) : Diff m () × C m () → C 0 () is C k or analytic, respectively, on its (open) domain of definition. Other function spaces may be (and will be) employed, and smoothness results are equally simple. Calculation of Derivative. We will compute the Gˆateaux derivative of h → h ∗ Fh() h ∗−1 (u), i,e., the t-derivative along a C 1 curve t → h(t, ·) of imbeddings. In the natural Eulerian form, we would compute ∂ ∂ F(t) (v)(y) = f (y, Lv(y)) ∂t ∂t with y fixed in (t) = h(t, ); but y = h(t, x) for some x ∈  = (0), and to keep y fixed, x must move. If U (x, t) = h −1 x h t and x(t) solves the differential equation d x/dt = −U (x, t), then dtd h(t, x(t)) = 0. Thus the partial derivative in t in the Eulerian form with y = h(x, t) fixed, corresponds to the anticonvective derivative Dt in the reference region  (Lagrangian form) Dt =

∂ ∂ − U (x, t) · , ∂t ∂x

U = h −1 x ht .

More precisely, we have the following. Lemma 2.1. Suppose ψ : Rn × R → R is C 1 , h : Rn × R → Rn is C 1 and det h x (x, t)
= 0 in the region of interest; then Dt (h ∗ (t)ψ(·, t)) = h ∗ (t)

∂ψ (·, t) ∂t

where h ∗ (t) is the pull-back by h(·, t). Proof. If y = h(x, t), by the chain rule,   n  ∂ ∗ Dt (h (t)ψ(·, t))(x) = ∂t − Uj ψ(h(x, t), t) ∂x j 1   n n  ∂ψ ∂h k  ∂h k ∂ψ (y, t) + − = Uj ∂t ∂ yk ∂t ∂x j 1 j=1 =

∂ψ ∂ψ (y, t) = h(t)∗ (·, t). ∂t ∂t

Remark. The above lemma (and the results below) are stated “pointwise,” but will later be interpreted as equalities of curves (parametrized by t) with values

22

Chapter 2. Differential Calculus of Boundary Perturbations

in certain function spaces. The interpretation is immediate for the spaces C m , but in general we argue by continuity, starting from C m approximations. Theorem 2.2. Suppose f (t, y, λ) is C 1 on an open set in R × Rn × R p , L is a constant-coefficient differential operator of order ≤ m with Lv(y) ∈ R p (where defined) and for open sets Q ⊂ Rn and C m functions v on Q, let FQ (t)v be the function y → f (t, y, Lv(y)),

y ∈ Q,

where defined. Suppose t → h(t, ·) is a curve of imbeddings of an open set  in Rn , j (t) = h(t, ) and for | j| ≤ m, |k| ≤ m + 1, (t, x) → ∂t ∂x h(t, x), ∂xk h(t, x), j ∂t ∂x u(t, x), ∂xk u(t, x) are continuous on R ×  near t = 0, and h(t, ·)∗−1 u(t, ·) is in the domain of F(t) (t). Then at points of ,      

Dt h ∗ F(t) (t)h ∗−1 (u) = h ∗ F˙(t) (t)h ∗−1 (u) + h ∗ F(t) (t)h ∗−1 (u)Dt u where Dt is the anti-convective derivative defined above, ∂f F˙ Q (t)v(y) = (t, y, Lv(y)) ∂t and ∂f (t, y, Lv(y))Lw(y), ∂λ is the linearization of v →  FQ (t)v. FQ (t)v · w(y) =

y ∈ Q,

Proof. Define v = h ∗−1 u, i.e., u(t, x) = v(t, h(t, x)); by the lemma, Dt (h ∗ Lh ∗−1 )(u) = Dt (h ∗ Lv) = h ∗ L

∂v = (h ∗ Lh ∗−1 )Dt u ∂t

so   Dt h ∗ F(t) (t)h ∗−1 u = Dt h ∗ f (t, ·, Lv)
 ∂f ∂v ∗ ∂f (t, ·, Lv) + (t, ·, Lv)L =h ∂t ∂λ ∂t which is the result claimed. Remark. Suppose we deal with a linear operator
α  ∂ aα (y) , A= ∂y |α|≤m

Chapter 2. Differential Calculus of Boundary Perturbations

23

not explicitly dependent on t, and h(x, t) = x + t V (x) + o(t) as t → 0. Then at t = 0  ∂ ∗ h Ah() h ∗−1 u t=0 = V · ∇(Au) + A(∂u/∂t − V · ∇u) ∂t ∂u + [V · ∇, A]u, =A ∂t The commutator of V · ∇ and A is still of order m – assuming smooth coefficients – and this is the result we would obtain by first changing variables (x in place of h(t, x) = y) and then taking the derivative at t = 0. If we applied the chain rule directly, we would never see a derivative of order (m + 1) and probably not notice the simple commutator structure, especially if (as usual) we compute the derivative at a solution u of Au = 0. This is the reason for the superiority of our general approach compared to a “bare-hands” change of variable (and the source of the ”miracle” mentioned in the Introduction). This point was also noted by Peetre [27], and related to a Lie derivative. For operators in variational form, Courant [5, Vol. 1, p. 260] gives an equivalent formula. We must also treat boundary conditions, and a quite general form of boundary condition is b(t, y, Lv(y), M N(t) (y)) = 0

for y ∈ ∂(t),

where L,M are constant-coefficient differential operators and N(t) (y) is the outward unit normal for y ∈ ∂(t), extended smoothly as a unit vector field on a neighborhood of ∂(t). For example, the Neumann problem requires N(t) · grad v = 0 on ∂(t), while some boundary conditions of the theory of elasticity involve the curvature of the boundary which may be expressed using the derivative of the normal. In these cases – and any “natural” boundary conditions – the particular extension of N(t) away from the boundary is irrelevant. Rather than hypothesizing this, we choose some extension of N in the reference region and then define N(t) = Nh(t,) by T −1 h ∗ Nh() (x) = Nh() (h(x)) = Th −1 x N (x)/ h x N (x)

(2.1)

for x near ∂, where Th −1 x is the inverse-transpose (“contra-gradient”) of the Jacobian matrix h x = [∂h i /∂ x j ]i,n j=1 and . is the Euclidean norm. This is the extension understood in the above boundary condition: b(t, y, Lv(y), M N(t) (y)) is defined for y ∈  near ∂ and has limit zero (in some sense, depending on the function spaces employed) as y → ∂(t). Note that h → h ∗ Nh() (x) is analytic for each x, but the smoothness in x depends on the smoothness of ∂ (and N and h). If  = {x : ϕ(x) > 0} and

24

Chapter 2. Differential Calculus of Boundary Perturbations

N = − grad ϕ(x)/ grad ϕ(x) near ∂, and if ψh (h(x)) = ϕ(x)(ψh = h ∗−1 ϕ) then according to the extension (2.1) above Nh ()(y) = − grad ψh (y)/|| grad ψh (y)||

for y near ∂h().

At points of ∂h(), the unit normal vector is determined geometrically, compatibly with (2.1). Lemma 2.3. Let  be a C 2 -regular region, N (·) a C 1 unit-vector field defined on a neighborhood of ∂ which is the outward normal on ∂, and for C 2 imbeddings h :  → Rn define Nh() on a neighborhood of h(∂) = ∂h() by (2.1) above. Suppose h(t, ·) is an imbedding for each t, defined by ∂ h(t, x) = V (t, h(t, x)) ∂t

for x ∈ , h(0, x) = x,

(t, y) → V (t, y) is C 2 and (t) = h(t, ), N(t) = Nh(t,) . Then for x near ∂ ∂, y = h(t, x) near ∂(t), we may compute the derivative ( ) y=const. and, if ∂t y ∈ ∂(t),
 ∂ N(t) (y) = Dt (H ∗ Nh() )(x) ∂t = −(∇∂(t) σ + σ ∂ N(t) /∂ N(t) )(y) σ = V · N(t) is the normal velocity and ∇∂(t) σ = ∇σ − N(t) ∂ N∂σ(t) is the component of the gradient tangent to ∂(t). Remark. We may, if desired, choose N so ∂ N /∂ N ≡ 0 near ∂ (as in Th. 1.5); but in general ∂ N(t) /∂ N(t) will be a nontrivial vector field orthogonal to N(t) . The formula is claimed only for y ∈ ∂(t), but to make sense of the derivative we compute initially for u near ∂(t). Proof. It suffices to find the derivative at t = 0 (since we can transfer the “origin”  of (2.1) to any (t); see below). Thus let h(t, x) = x + t V (x) + o(t), V (x) = V (0, x) so N(t) (h(t, x)) has i th component Ni − t

 ∂Vj j

∂ xi

N j + t Ni

 j,k

N j Nk

∂Vj + o(t) ∂ xk

and Dt (h ∗ N(t) )|t=0 = −V j ∂ Ni /∂ x j − N j ∂ V j /∂ xi + Ni N j Nk ∂ V j /∂ xk , summing over repeated indices. If V = σ N + τ, τ · N ≡ 0, we find Dt (h ∗ N(t) )i |t=0 = −∂σ/∂ xi + Ni Nk ∂σ/∂ xk − σ N j ∂ Ni /∂ x j + qi

Chapter 2. Differential Calculus of Boundary Perturbations

25

with qi = −τ j ∂ Ni /∂ x j + τ j ∂ N j /∂ xi − τ j Ni Nk ∂ N j /∂ xk and we have used τ · N ≡ 0, N · N ≡ 1. Now qi Ni ≡ 0 and for any ρ ⊥ N we have ρi qi = ρi τ j (∂ N j /∂ xi − ∂ Ni /∂ x j ) which vanishes on ∂: this depends only on N |∂, not on the extension, and for the extension of Th. 1.5, [∂ Ni /∂ x j ] is a symmetric matrix. Thus q = 0 on ∂, which proves the result. Theorem 2.4. Let b(t, y, λ, µ) be a C 1 function on an open set of R × Rn × R p × Rq and let L , M be constant-coefficient differential operators (and order ≤ m) of appropriate dimensions so b(t, y, Lv(y), M N (y)) makes sense. Assume  is a C m+1 region, N (x) is a C m unit-vector field near ∂ which is the outward unit normal on ∂, and define Nh() by T −1 Nh() (h(x)) =Th −1 x N (x)/ h x N (x)

when h :  → Rn is a C m+1 smooth imbedding. Also define Bh() (t) by   Bh() (t)V (y) = b t, y, Lv(y), M Nh() (y) for y ∈ h() near ∂h(). If t → h(t, ·) is a curve of C m+1 imbeddings of  and for | j| ≤ m, |k| ≤ j j m + 1, (t, x) → (∂t ∂x h, ∂xk h, ∂t ∂x u, ∂xk u)(t, x) are continuous on R ×  near t = 0, then at points of  near ∂      

Dt h ∗ Bh() (t)h ∗−1 (u) = h ∗ B˙ h() h ∗−1 (u) + h ∗ Bh() h ∗−1 (u) · Dt u      + h ∗ ∂ Bh() /∂ N h ∗−1 (u) · Dt h ∗ Nh()

where h = h(t, ·); B˙ h() , Bh() are defined as in Th. 2.2,

  ∂h  t, y, Lv(y), M Nh() (y) · Mn(y) ∂ Bh() /∂ N (v) · n(y) = ∂µ



and Dt (h ∗ Nh() )|∂ is computed in the previous lemma. Proof. The same calculation as in Th. 2.2 Change of Origin. In the above, the “origin” or reference region is , but we may easily transfer the origin to any 1 diffeomorphic to . Let H1 :  → 1 be the diffeomorphism and for every imbedding h :  → Rn define the imbedding h 1 = h ◦ H1−1 : 1 → Rn . Similarly define x1 = −1 N (x)/ · · · and then H1 (x), u 1 = H ∗−1 u, N1 (x1 ) = N H1 () (H1 x) =TH1,x h() = h 1 (1 ), h ∗ Fh() h ∗−1 u(x) = h ∗1 Fh 1 (1 ) h ∗−1 1 u 1 (x 1 ) h ∗ Bh() h ∗−1 u(x) = h ∗1 Bh 1 (1 ) h ∗−1 1 u 1 (x 1 )

26

Chapter 2. Differential Calculus of Boundary Perturbations

using the normal T −1 Nh 1 (1 ) (h 1 (x1 )) = Th −1 1,x1 N1 (x 1 )/ · · · = h x N (x)/ · · ·

= Nh() (h(x)). This “change of origin” is used frequently in the sequel, as it permits us to compute derivatives in h at h = i  , where the formulas are simpler.

Chapter 3 Examples Using the Implicit Function Theorem

We prove differentiability with respect to the domain for various quantities (e.g., eigenvalues) associated with a boundary-value problem, and compute the derivative, and sometimes a second derivative, using the formulas of Chapter 2. Examples 3.1, 3.2 and 3.5 are treated in some detail, to show the calculations are not merely formal. Example 3.1. Torsional Rigidity. The resistance to torsion of a cylindrical rod depends not only on the elastic constants of the material, but also on the geometry of the cross-section
⊂ R2 , through the torsional rigidity R(
). Ω


R(
) =

Ω




|∇u|2 where u :
→ R is the solution of

u = −1 in
,

It is convenient to use R(
) =




u = 0 on ∂
.

u, noting that




u=




−uu =




|∇u|2 .

Let 0 < α < 1 and let
⊂ R2 be a bounded C 2,α -regular region. Diff2,α (
) is the open set of C 2,α (
, R2 ) consisting of imbeddings; if h ∈ Diff2,α (
), then h(
) is also C 2,α -regular. Suppose v : h(
) → R solves v = −1 in h(
), v = 0 on ∂h(
); then R(h(
)) = h(
) v. Also u = h ∗ v solves (h ∗ h ∗−1 )u = −1 in
, and


˙ = R(h(
))

u = 0 on ∂
,

u(x)|det h  (x)|d x.




27

28

Chapter 3. Examples Using the Implicit Function Theorem

Consider the map  : (u, h) → (h ∗ h ∗−1 )u from C 2,α (
)0 × Diff2,α (
) to C (
), where C 2,α (
)0 = {u ∈ C 2,α (
)|u = 0 on ∂
}.  is an analytic map, and we apply the implicit function theorem for h near i
(= the inclusion of
in R2 ). Let u 0 be the solution of u 0 = −1 in
, u 0 = 0 on ∂
, so (u 0 , i
) = −1. Now u → (u, i
) = u : C 2,α (
)0 → C 0,α (
) is an isomorphism, by familiar results of elliptic theory. So for every h in a C 2,α (
, R2 )-neighborhood of i
, there is a unique solution u h ∈ C 2,α (
)0 of (u h , h) = −1 and h → u h is analytic, with u h = u 0 when h = i
. It follows that h → R(h(
)) =
u h (x)det h (x)d x is also analytic and we will compute its derivative at h = i
. It is convenient to have more smoothness for the initial calculations. If
is C m,α regular (m ≥ 2), the same argument shows h → (u h , R(h(
)) : C m,α (
, R2 ) → C m,α (
)0 × R is analytic on a neighborhood of i
. Suppose t → h(t, ·) ∈ C m,α (
, R2 ) is a C 1 curve near t = 0 with h(t, x) = x + t V (x) + c(t) as t → 0. Then t → u h(t,·) ∈ C m,α (
)0 is also C 1 near t = 0 and, say, 0,α

u h(t,·) = u 0 + tu 1 + o(t) as t → 0, so d R(h(t,
))|t=0 = dt





(u 1 + u 0 div V ).

Now t → Dt u h(t,·) ∈ C m−1,α (
) is continuous and (if m ≥ 3)   Dt h ∗ h ∗−1 u h(t,·) = (h ∗ h ∗−1 )Dt u h(t,·) is an equation of continuous functions of t with values in C m−3,α (
). The equation is valid if everything is smooth (Theorem 2.2) and by continuity, is valid along our solution curve t → u h(t,·) (if m ≥ 3). At t = 0 we have v1 = 0, where v1 = Dt u h(t,·) |t=0 = u 1 − V · ∇u 0 . When x ∈ ∂
, u h(t,·) (x) = 0 so u 1 = u 0 = 0 on ∂
. Also

d R(h(t,
))|t=0 = (u 1 − V · ∇u 0 + div (V u 0 )) = v1 dt
0


∂u 0 ∂v1 + u0 = −v1 u 0 + u 0 v1 = −v1 ∂ N ∂N
∂
2 
∂u 0 = V·N . ∂N ∂
This was proved for the case m ≥ 3, but we know the derivative exists in the setting of C 2,α data; we prove by continuity that the final equation

Chapter 3. Examples Using the Implicit Function Theorem d R(h(t,
))|t=0 = dt




∂


V·N

∂u 0 ∂N

29

2

also holds for C 2,α data. We may assume h(t,
) = {x | ϕ(t, x) > 0} for some ϕ such that t → ϕ(t, ·) ∈ C 2,α (R2 ) is C 1 , ϕ(t, x) = 0 ⇒ |ϕx (t, x)| ≥ 1. There exists ϕ(s, t, x) which is C ∞ for s = 0, ϕ(0, t, x) = ϕ(t, x), and (s, t) → ϕ(s, t, ·), ∂t ϕ(s, t, ·) ∈ C 2,α (R2 ) are continuous near (0,0). Let
(s, t) = {x|ϕ(s, t; x) > 0}, and let u(s, t, ·) be the solution of x u = −1 in
(s, t), u = 0 on ∂
(s, t); then for s = 0 ∂ R(
(s, t)) = ∂t




∂
(s,t)

V · N
(s,t)

∂u ∂N

2

and we want the limit as s → 0. By Theorem 1.8,
(s, t) converges to h(t,
) in C 2,α as s → 0, so the implicit function argument above proves R(
(s, t)) → R(h(t,
)) as s → 0. To see the limit of the integral has the expected form, we write it as an integral over ∂
. By Theorem 1.8,
(s, t) = h(s, t,
) for an appropriate C 2,α diffeomorphism h(s, t, ·) : Rn → Rn , and u(s, t, h(s, t, x)) ≡ u h(s,t,·) (x) for x ∈
[u h is the function provided by the previous argu|2 = |∇u(s, t, ·)|2 ; and at h(s, t, x), x ∈ ∂
, this ment]. On ∂
(s, t), | ∂u(s,t,·) ∂N ∗ ∗ −1 is |h(s, t, ·) ∇h (s, t, ·) u h(s,t,·) (x)|2 . Further V · N
(s,t) = |ϕϕxt | (s, t, h(s, t, x)) and the corresponding elements of area for ∂
(s, t), ∂
are in the proportion d A(s,t) = d A(0,0)



det(TE Thx h x E) det(TE E)

1/2

where E = [e1 , . . . , en−1 ] is an n × (n − 1) matrix whose columns are a basis for the tangent space to ∂
(for example e j = ∂ x(ξ )/∂ξ j , when ∂
is locally the image of ξ → x(ξ ) : Rn−1 → Rn ). Substitution in the integral shows we may allow s → 0 and then ∂ R(
(t)) = ∂t




∂
(t)

V · N
(t)

∂u ∂N

2 dA

where u = −1 in
(t), u = 0 on ∂
(t), and
(t) = h(t,
), t → h(t, ·) is a C 1 -curve of C 2,α -imbeddings of the C 2,α region
. The Fr´echet derivative (at h = i
) is 


h˙ →

∂


h˙ · N


∂u 0 ∂ N


2 d A : C 2,α (
, R2 ) → R.

30

Chapter 3. Examples Using the Implicit Function Theorem  Observe that dtd (area
(t)) = ∂
(t) V · N
(t) d A (by Theorem 1.11 with f ≡ 1) so
is a critical point of R(
), with fixed area, if and only if (∂u 0 /∂ N
)2 = constant on ∂
. Since u 0 > 0 in
, by the maximum principle, this says u 0 = −1 in
,

u 0 = 0 on ∂
,

∂u 0 = constant < 0 on ∂
∂N

if and only if
is a critical point of the torsional rigidity with fixed area. This is evidently the case when
is a disc in R2 , since u 0 is radially symmetric. Conversely, by a result of Serrin [36], the only (connected, bounded, C 2 ) regions for which such a solution u 0 exists are discs, so the disc is the only critical point. In fact, the disc maximizes the torsional rigidity for given area. This may be shown by an argument like that of Faber and Krahn (as given in Garabedian [6, p. 413–416]), or see the article of P´olya [30]. The argument above is in Lagrangian form – as it must be to use the implicit function theorem. But once we know a (sufficiently smooth) solution exists, subsequence calculations are simpler in the Eulerian form. Suppose h(t, x), ∂h (t, x) are C m,α on R ×
(near t = 0) with h(0, x) = x, m ≥ 2; there is (by ∂t Corollary 1.10) an extension H to R × Rn with (t, x) → H (t, x), ∂∂tH (t, x) of class C m,α , H (0, x) = x for all x ∈ Rn , H (t, x) = h(t, x) for x ∈
, t near 0 and ∂t∂ H (t, x) = V (t, H (t, x)) where (t, y) → V (t, y) is C m,α . Let u h(t,·) be the function provided by the implicit function argument above, and define U (t, x) = (E
u h(t,·) )(x) where E
is the extension operator (Theorem 1.9); then (t, x) → U (t, x) is C m,α on R × Rn (near t = 0). Finally, define v : R × Rn → R (near 0 × Rn ) by v(t, H (t, x)) = U (t, x) or v(t, y) = U (t, H (t, ·)−1 (y)), so v is also C m,α . It follows that (t, y) →  y v(t, y) is C m−2,α , and   y v(t, y) = −1 in
(t) = h(t,
) v(t, y) = 0

on ∂
(t) = h(t, ∂
).

(To prove directly the smoothness of v seems a very difficult problem.) In this  form, the calculations are quite straightforward. Thus R(
(t)) =
(t) v(t, y)dy so (recalling v = 0 on the boundary)
∂v d R(
(t)) = . dt
(t) ∂t But  y ( ∂v ) = 0 in
(t) and so (assuming m ≥ 3, so ∂v (t, ·) is at least C 2 ) ∂t ∂t  
 ∂v d ∂v R(
(t)) = − y v + y v dt ∂t ∂t
(t)
∂v ∂v − = ∂t ∂ N y ∂
(t)

Chapter 3. Examples Using the Implicit Function Theorem

31

Now if y = y(t) solves y˙ (t) = V (t, y(t)) and y(t0 ) ∈ ∂
(t0 ), for some t0 , it + V ∂v = 0 on follows y(t) ∈ ∂
(t) for all t so v(t, y(t)) ≡ 0; therefore ∂v ∂t ∂y ∂v ∂v ∂
(t). Also v(t, ·) = 0 on ∂
(t) implies ∂ y = ∂ N N
(t) hence  
∂v 2 d R(
(t)) = V · N
(t) . dt ∂ N
t ∂
(t) The calculations in this form are so simple that we will go on to compute the second derivative. Writing σ = V · N
(t) (defined on a neighborhood of ∂
(t), for each t) and recalling |∇v|2 = (∂v/∂ N
(t) )2 on ∂
(t), we have (for m ≥ 4) 
∂ d2 ∂ 2 2 2 2 (σ |∇v| (σ |∇v| R(
(t)) = ) + σ ) + H σ |∇v| dt 2 ∂N ∂
(t) ∂t where H = div N
(t) is the mean curvature (Theorem 1.11). Writing v˙ for this becomes   
∂v 2 ∂σ ∂σ d2 2 +σ + Hσ R(
(t)) = dt 2 ∂t ∂N ∂
(t) ∂ N  2 
∂ v ∂v ∂ v˙ + σ 2σ + ∂N ∂N2 ∂N ∂
(t)

∂v , ∂t

(v˙ = 0 in
(t), v˙ = σ ∂∂vN on ∂
(t); σ = V · N
(t) ) and we observe  
d2 ∂σ ∂σ 2 + σ + H σ area
(t) = . dt 2 ∂t ∂N ∂
(t) Suppose
(0) is the unit disc in R2 and keep area
(t) = constant. At

ikθ and V · N = σ = ∞ on t = 0, v = 14 (1 − r 2 ) in polar coordinates −∞ σk e  1 1 d ∂
(0), with σ0 = 0 since σ0 = 2π ∂
(0) V · N = 2π dt area
(t)|t=0 = 0. Also ˙ θ) = at t = 0, v˙ is the solution of v˙ = 0 in r < 1, v˙ = 12 σ on r = 1, or v(r, 1 ∞ 1 |k| ikθ σ r e , while ∂v/∂ N = ∂v/∂r = − on r = 1. Substitution in the −∞ k 2 2 formula above yields  
2π  1 2 ∂ v˙  d2 σ −σ  R(
(t))|t=0 = dθ dt 2 2 ∂r r =1 0 ∞ = −2π |σk |2 (k − 1). 2

This is independent of σ1 (or σ−1 = σ 1 ) which is reasonable as 


 d ∓iθ e σ = (x ∓ i y) d x d y  2π σ±1 = dt ·
(t) ∂
(0) t=0 gives the velocity of the center of mass at t = 0 and does not affect the shape of
(t). The other coefficients are all negative – hardly surprising, as the disc maximizes R(
) with fixed area.

32

Chapter 3. Examples Using the Implicit Function Theorem

Example 3.2. Simple Eigenvalues of the Dirichlet Problem for the Laplacian. Suppose
⊂ Rn is a C 2 -regular bounded region and λ0 is a simple eigenvalue of the Dirichlet problem in
, u + λu = 0 in
,

u = 0 on ∂
,



u ≡ 0

and let u 0 be the corresponding eigenfunction with
u 20 = 1. For all h ∈ C 2 (
, Rn ) in some C 2 -neighborhood of i
, h(
) is also a C 2 region and there is a unique simple eigenvalue λ(h(
)) near λ0 , h → λ(h(
)) is analytic, and its derivative at i
is  2
˙h · N
∂u 0 . ˙h → − ∂ N
∂
The derivative was computed formally, in some special cases, by Rayleigh [33, p. 338, eq. 11] and for general two-dimensional regions by Hadamard [9], but the first adequate proof is probably that of Garabedian and Schiffer [7]. We work in the Sobolev spaces H m (
) and H01 (
) = {u ∈ H 1 (
) | u = 0 on ∂
}. Define F : H 2 ∩ H01 (
) × R × Diff2 (
) → L 2 (
) × R by  
F(u, λ, h) = h ∗ ( + λ)h ∗−1 u, u 2 det h  .


Then F(u 0 , λ0 , i
) = (0, 1) and whenever F(u, λ, h) = (0, 1), v = h ∗−1 u : h(
) → R satisfies
v + λv = 0 in h(
), v = 0 on ∂h(
), v 2 = 1. h(
)

Also F is analytic and the derivative

 
∂F (u 0 ; λ0 , i
) : (v, µ) → ( + λ0 )v + µu 0 , 2 u 0 v ∂(u, λ)


is an isomorphism. Indeed, since λ0 is a simple eigenvalue, the image ( + λ0 )(H 2 ∩ H01 (
)) = {u 0 }⊥ , so for any ( f, g) ∈ L 2 (
) × R
( + λ0 )v + µu 0 = f, 2 u0v = g


has a unique solution (v, µ) ∈ H ∩
u 0 f, µ= 2




H01 (
)

× R:

1 v = v0 + gu 0 2

where ( + λ0 )v0 = f − µu 0 ,

v0 ⊥ u 0 ,

v0 ∈ H 2 ∩ H01 (
).

Chapter 3. Examples Using the Implicit Function Theorem

33

Thus the implicit function theorem applies and there are analytic functions h → u h , λh , for h − i
C 2 small, such that F(u h , λh , h) = (0, 1). If ∂
is C m regular (m ≥ 2), the same argument applies to F : H m ∩ H01 (
) × R × Diff m (
) → H m−2 (
) × R to give analytic h → (u h , λh ) ∈ H m ∩ H01 (
) × R for h − i
C m (
) small. If E
is the extension map of Theorem 1.9, h → E
(u h ) ∈ H m (Rn ) is analytic, as is h → H = idRn + E
(h − i
) ∈ C m (Rn , Rn ), and each H −1 is C m , for h − i
C m small. Let vh = (H −1 )∗ E
(u h ) ∈ H m (Rn ); then vh + λh vh ∈ H m−2 (Rn ) and vanishes in h(
), vh = 0 on ∂h(
), and h(
) vh2 = 1. We show h → vh ∈ H i (Rn ) is C m−i near h = i
for 0 ≤ j ≤ m. Lemma. Suppose m ≥ 1 and let X be a convex set in C m (Rn , Rn ) consisting of C m maps H with H (x) = x for |x| ≥ R0 (some fixed R0 ) which are diffeomorphisms with D j H (x), D j H −1 (x) (0 ≤ j ≤ m) uniformly bounded. For each j, 0 ≤ j ≤ m, H → H −1 : X ⊂ C m → C m− j (Rn , Rn ) is a C j map and for 1 ≤ p < ∞, ( f, H ) → f ◦ H −1 : W m, p (Rn ) × X → W m− j, p (Rn ) is also C j . Proof. The evaluation map (H, x) → H (x) : X × Rn → Rn is C m [1] so (H, x) → (H, H (x)) is C m and has C m inverse (H, y) → (H, H −1 (y)). We are concerned only with a compact set {|y| ≤ R0 } so (H, y) → H −1 (y) of class C m implies H → H −1 |B R0 : X → C m− j (B R0 , Rn ) is C j , by the converse Taylor theorem [1]. Again by the converse Taylor theorem, it is enough to prove: If f ν → f in L p (Rn ), gν → g uniformly on Rn and gν , g and C 1 diffeomorphisms with |Dgν |, |D(gν−1 )| uniformly bounded and gν , g equal to the identity outside a fixed compact set, then f ν ◦ gν → f ◦ g in L p (Rn ). But f ν → f in measure, and our hypotheses ensure f ν ◦ gν − f ◦ gν → 0 in measure and also f ◦ gν → f ◦ g in measure so f ν ◦ gν → f ◦ g in measure.  Further E | f ν |p → 0 as measure E → 0, uniformly in ν, which implies E | f ν ◦ gν | p ≤ const. gν (E) | f ν | p → 0 as meas E → 0, and the proof is complete. Thus if
is C 3 regular, t → h(t, ·) ∈ C 3 (
, Rn ) is C 1 with h(0, ·) = i
then t → vh(t,·) = v(t, ·) ∈ H 2 (Rn ) is C 1 , t → λh(t,·) = λ(t) is C 1 v(t, ·) + λ(t)v(t, ·) = 0 in
(t) = h(t,
)
v(t, ·) = 0 on ∂
(t), v(t, ·)2 = 1
(t)

34

Chapter 3. Examples Using the Implicit Function Theorem

with
(0) =
, v(0, ·) = u 0 , λ(0) = λ0 . Let v˙ = ˙ ˙ + h˙ · ∇u 0 = 0 on ∂
so λ(0)u 0 = 0 in
, v

∂ v| ∂t t=0

and then ( + λ0 )v˙ +


{−u 0 ( + λ0 )v˙ + ( + λ0 )u 0 v} ˙
 ∂u 0 ∂ v˙ + v˙ − u0 = ∂N ∂N ∂
 

∂u 0 2 ∂u 0 ˙ ˙ h·N −h · ∇u 0 = =− ∂N ∂N ∂
∂


˙ λ(0) =

where h˙ =



∂h  . ∂t t=0




More generally d λ(
(t)) = dt




∂
(t)

V · N
(t)

∂v(t, ·) ∂ N
(t)

2

where ∂h (t, x) = V (t, h(t, x)),
(t) = h(t,
) and we deal with sufficiently ∂t smooth (C 3 ) deformation of a C 3 domain
. But, as in the previous example, we may approximate C 2 regions by C 3 (or C ∞ or analytic) regions and then take limits to conclude this is the expression for the derivative also in the C 2 setting. Note that, when t → h(t, ·) ∈ C 2 (
, Rn ) is C 1 (or C ∞ ), our previous argument says t → v(t, ·) ∈ H 2 (Rn ) is continuous, though it is differentiable into H 1 (Rn ), and the calculation above would then involve distributions in H −1 (Rn ). We avoid working with distributions by only taking limits in the equation for dtd λ(
(t)). Now observe that
→ λ(
) has a critical point (with vol (
) fixed) if and only if the corresponding eigenfunction v
satisfies (∂v
/∂ N
)2 = constant = 0 on ∂
. If we are considering the first (principal) eigenvalue λ0 and
is connected, then v
does not change sign in
, so the condition is: ∂v
/∂ N
= constant = 0 on ∂
. By arguments of Serrin and Gidas-Ni-Nirenberg, it can be shown [11] that this holds if and only if
is a ball. The ball is the only critical point of
→ λ0 (
), the principal eigenvalue of bounded connected C 2 regions
with given volume. If fact, the ball minimizes the principal eigenvalue for given volume. This was conjectured by Rayleigh [33] on the basis of various examples and a formal calculation of the second derivative (see below) and proved by Faber and Krahn, whose argument is in [6, p. 413–416]. We now calculate the second derivative of
→ λ(
) at the point
= unit ball of Rn , for volume-preserving deformations. Specifically let h : R ×
0 → Rn be smooth near t = 0 (
0 = unit ball) and such that h(0, ·) = i
0 and vol h(t,
0 ) = vol
0 . Writing ∂h (t, x) = V (t, h(t, x)),
(t) = h(t,
), ∂t

Chapter 3. Examples Using the Implicit Function Theorem

35

etc., we differentiate d λ(
(t)) = dt




∂
(t)

V · N
(t)

∂v ∂ N
(t)

2

as in Example 3.1 to conclude d2 λ(
(t))|t=0 = − dt 2 where σ = V · N
and v˙ =


2σ ∂
0

∂ v| ∂t t=0

( + λ0 )v˙ = 0 in
0 ,

  ∂u 0 ∂u 0 ∂ v˙ −σH ∂N ∂N ∂N

satisfies v˙ + h˙ · ∇u 0 = 0 on ∂
0

 √ and
0 vu ˙ 0 = 0 (recall dtd λ(
(t))|t=0 = 0). Now u 0 (x) = kn (x)−ν Jν ( λ0 |x|) √ where ν = (n − 2)/2, Jν is the ν-order Bessel function, λ0 > 0 with Jν ( λ0 ) = 0 > 0 (say) on ∂
0 , so ∂u 0 /∂ N = 0 and kn is chosen so
0 u 20 = 1 and ∂u ∂N n−1 2λ0 /|S |. [This λ0 is a simple eigenvalue for the unit ball, which depends on a nontrivial fact (“Bourget’s hypothesis”) about Bessel functions: see Watson, Theory of Bessel Functions, Cambridge U. Press.] Expand σ in spher

ical harmonics: σ (x) = ∞ 0 P j (x) when |x| = 1, where P j (x) is a homogeneous polynomial of degree j with P j (x) = 0. (See Stein [38] for basic properties of harmonic polynomials.) Since the volume is fixed

d σ = P0 , 0 = vol
(t)|t=0 = dt ∂
0 ∂
0

√ √

x ˙ = − 2λ0 /|S n−1 | ∞ so P0 = 0. Then v(x) 1 f j ( λ0 |x|)/ f j ( λ0 )P j ( |x| ) −ν where f j (z) = z Jν+ j (z), and substitution (with H = n − 1) yields   ∞ 
 4λ0 d2 2  λ(
(t)) = n−1 Pj Q j dt 2 |S | j=1 ∂
0 t=0

     Q j = λ0 Jν+ λ0 Jν+ j λ0 + ν + 1. j Since Jν (λ0 ) = 0, it follows (by the recursion relations for J ) that Q 1 = 0 so the second derivative does not depend on P1 . This is understandable as P1 gives the initial velocity of the center of mass, and says nothing about change of shape. Rayleigh [33] computed this derivative formally when n = 2, and showed (if λ0 is the principal eigenvalue) that Q j > 0 for all j ≥ 2. The same argument works for 2 ≤ n ≤ 5, but in general we refer to the argument of Polya and

36

Chapter 3. Examples Using the Implicit Function Theorem

Sz¨ego [31, p. 135], who also treated n = 2, but whose argument shows, for every n ≥ 2, that Q j > 0 for all j ≥ 2. As a more specific example, suppose
(t) is the ellipsoid with semi-axes e K 1 t , . . . , e K n t , i.e.,   n n −K j t 2
(t) = x ∈ R | (x j e ) 0, by the maximum principle) and then K (
) = ∂
σ is the total charge on ∂
(or in
). It is also easy to show 
2 1 n |∇ψ| : ψ ∈ H (R ), ψ = 1 on
K (
) = min Rn

and this is often taken as the definition of capacity when
is not smooth. However we assume
is smooth (C 2 ) and show K (
) is an analytic function of
, for C 2 deformations, and compute the derivative: ∂ K (h(
))|h=i h˙ = ∂h




∂


h˙ · N


∂φ ∂ N


2 ,

h˙ ∈ C 2 (Rn , Rn ).

(As usual, N
is the unit outward normal from
, although the capacitary potential φ is defined in Rn \
.) Query: Is the ball the only critical point of the capacity, with given volume? Working in an unbounded region raises some technical problems. For this example, we could avoid these problems by using inversion in a sphere to reduce to a bounded domain. But we will use a method which generalizes. Choose p in 3 < p < ∞ (or more generally, p > n/(n − 2)) so |x|2−n will be p-power integrable near ∞. Also choose some large R so that
(and any

38

Chapter 3. Examples Using the Implicit Function Theorem

relevant perturbations) are in B R = {x : |x| < R}, and choose C ∞ ψ0 so that ψ0 = 1 on ∂
and ψ0 = 0 outside B R . Define Diff2R (Rn ) = {all C 2 diffeomorphisms h : Rn → Rn such that h(x) = x for |x| ≥ R} 1, p

X R = {ψ ∈ W 2, p ∩ W0 (Rn \
)|ψ = 0 outside B R } Y R = { f ∈ L p (Rn \
)| f = 0 outside B R }, with the induced norms. X R , Y R are Banach spaces while Diff2R is the translation of of an open set in a Banach space. Note for h ∈ Diff2R (Rn ) that (h ∗ h ∗−1 )ψ(x) = ψ(x) when |x| > R, and also note ψ → ψ : X R → Y R is an isomorphism. Apply the implicit function theorem to the analytic map (h, ψ) → ∗ (h h ∗−1 )(ψ + ψ0 ) : Diff2R × X R → Y R to prove, for each h ∈ Diff2R in a C 2 neighborhood of idRn , there is a unique ψh ∈ X R with (h ∗ h ∗−1 )(ψh + ψ0 ) = 0 in Rn \
, and h → ψh is analytic. Then φh = h ∗−1 (ψh + ψ0 ) : Rn \h(
) → R is a harmonic function which vanishes at ∞ and φh = 1 on ∂h(
), so φh is the capacitary potential for h(
) and



K (h(
)) =

|∇φh | = 2

Rn \h(
)

Rn \


|h ∗ ∇h ∗−1 (ψh + ψ0 )|2 det h 

is also analytic in h. (We could write this as an integral over B R \
, plus another over ∂ B R ; to avoid convergence problems at ∞.) Assuming a bit more smoothness, say C 3 , we compute the derivative. Omitting details, say h(t, x) = x + t V (x) + o(t), φh(t,·) = φ0 + tφ1 + o(t) is the capacitary potential for ω(t) = h(t,
), so φ0 = 0, φ1 = 0 in Rn \
; φ0 , φ1 vanish at ∞, O(|x|2−n ); φ0 = 1 and φ1 + V · ∇φ0 = 0 on ∂
. Then  
  d  d  ∇φh(t,·) 2 K (
(t) =   dt t=0 dt Rn \
(t)
t=0
= 2∇φ0 · ∇φ1 + −V · N (∇φ0 )2 Rn \


∂


  ∂φ0 2 ∂φ0 +V ·N =− 2φ1 ∂N ∂N ∂
2 
∂φ0 V·N . = ∂N ∂





Chapter 3. Examples Using the Implicit Function Theorem

39

Alternatively,  
d  d  K (
(t)) = −  N
(t) · ∇φh(t,·) dt t=0 dt t=0 ∂
(t) 
 ∂ 2 φ0 ∂φ0 N˙ · ∇φ0 + N · ∇φ1 + V · N =− + H V · N ∂N2 ∂N
∂
∂φ1 =− ∂
∂ N  


∂φ0 2 ∂φ1 ∂φ0 =− = φ0 φ1 V·N , =− ∂N ∂N ∂N ∂
∂
∂
using Theorems 1.11, 1.12 and Lemma 2.3. Example 3.4. Green’s Function. The first general treatment of perturbation of the boundary is due to Hadamard [9] who took as the fundamental problem calculation of the change in Green’s function. This is also the basis of the work of Schiffer and Garabedian [7], among others. Our method allows a more direct approach to Green’s function (and Neumann’s and other corresponding kernels). We treat only the case of the Dirichlet problem for a second-order uniformly elliptic operator L L=

n

ai j (x)

i, j=1

n ∂2 ∂ + b j (x) + C(x), ∂ xi ∂ x j ∂x j j=1

but other cases may be treated similarly [11]. See Peetre [27] for another approach. We assume everything is smooth, and uniqueness holds for the Dirichlet problem for L in
⊂ Rn ,
smooth and bounded, n ≥ 3. Then the Green function G
(x, y) exists and may be defined by L ∗y G
(x, y) = 0 for x = y in
, G
(x, y) = 0 for y ∈ ∂
,  (2−n)/2 jk G
(x, y) = Cn (x) a (x)(y j − x j )(yk − xk ) + O(|x − y|3−n ) j,k

as y → x(x ∈
), where [a jk ] is the inverse of the symmetric positive-definite matrix [a jk ] and

1 = (2 − n)|S n−1 | det[a jk (x)]. Cn (x) Then for any C 2 φ :
→ R and x ∈
,

 ∂G 
(x, y). d A y φ(y) φ(x) = G
(x, y)Lφ(y)dy + a jk (x)N j Nk ∂ N (y)
∂


40

Chapter 3. Examples Using the Implicit Function Theorem

Say f ∈ C ∞ (Rn , R) and h(t, x) = x + t V (x) + o(t) is a smooth curve of smooth imbeddings of
. For each t near 0 there is a solution u(t, ·) of Lu(t, ·) = f in
(t) = h(t,
), u(t, ·) = 0 on ∂
(t), and we may consider u(t, x) extended smoothly to a neighborhood of {0} ×
. Calculating in the Eulerian mode, with u˙ = ∂u/∂t and ∂h/∂t = V (t, h), we find L u˙ = 0 in
(t), u˙ + V · ∇u = 0 on ∂
(t) so (if G
(t) is the appropriate Green function and x ∈
(t)),  
∂u(t, y) ∂G
(t) (x, y) ˙ x) = − V · N
(t) (t, y) u(t, d Ay a jk N j Nk . ∂ N
(t) (y) ∂ N
(t) ∂
(t) j,k Also u(t, y) =


(t)

˙ x) = u(t,

d dt

G
(t) (y, z) f (z)dz, so




(t)

G
(t) (x, z) f (z)dz =


(t)

G˙
(t) (x, z) f (z)dz

where we define (for x, z in
)
  ∂G
(t) (x, y) G˙
(t) (x, z) = − d Ay a jk (y)N j Nk V ·N
(t) ∂ N (y) ∂
(t) ∂G
(t) (y, z) . × ∂ N (y) By the implicit function theorem, we may easily prove differentiability of t → G
(t) (x, z) for x = z in
(t), so in fact ∂ G˙
(t) (x, z) = G
(t) (x, z), ∂t

even for x = z.

Example 3.5. Simple Eigenvalues of u + (λ + C(x))u = 0 in
, ∂u/∂ N + β(x)u = 0 on ∂
. The previous examples used the Dirichlet boundary condition, but we may also treat other boundary conditions using Theorem 2.4. Suppose β, C : Rn → R are C 2 functions and
⊂ Rn is a bounded C 2 region. Assume λ0 is a simple eigenvalue of the above (Robin) problem in
, with eigenfunction u 0 ,
u 20 = 1. We show for every h :
→ Rn C 2 close to i
, the corresponding problem in h(
) has a unique eigenvalue λh near λ0 , λh is a simple eigenvalue, h → λh is differentiable and the derivative at h = i
is   
˙h · N
|∇∂
u 0 |2 − λ0 + c + β 2 − Hβ − 2 ∂β u 20 ˙h → ∂ N
∂
where H = div N
is the mean curvature of ∂
and ∇∂
u 0 is the tangential component of the gradient of u 0 .

Chapter 3. Examples Using the Implicit Function Theorem

41

Let H 1/2 (∂
) be the space of restrictions to ∂
of functions in H 1 (∂
), i.e., the quotient H 1 (∂
)/H01 (∂
). Consider the map  
∗ ∗−1 ∗ ∗−1 2 F : (u, λ, h) → h ( + c + λ)h u, h (Nh(
) · ∇ + β)h u|∂
, u


: H 2 (
) × R × Diff2 (
) → L 2 (
) × H 1/2 (∂
) × R. Then F(u 0 , λ0 , i
) = (0, 0, 1), and F(u, λ, h) = (0, 0, 1) implies v ≡ h ∗−1 u ∈ H 2 (h(
)) satisfies ( + c + λ)v = 0 in h(
),
h(
)

∂v + βv = 0 on ∂h(
), ∂N

v2 = 1, |det h  |

so λ is an eigenvalue of the Robin problem in h(
). Now (u, λ, h) → h ∗ h ∗−1 u + λu + (c ◦ h) · u is C 2 in the relevant spaces, while (h, u) → h ∗ (Nh(
) · ∇ + β)h ∗−1 u|∂
= h ∗ Nh(
) · h ∗ ∇h ∗−1 u|∂
+ (β ◦ h)u|∂
is C 1 , for (h, u) → (β ◦ h)u : Diff2 × H 2 → H 1 (
) is C 1 . Thus F is a C 1 map and we could have assumed c is C 1 . The smoothness of F is limited only by that of c and β: if c is C r , β is C r +1 , then F is C r . Also, if ∂
is C m (m ≥ 2) we could work with the corresponding F on H m × R × Diff m → H m−2 × H m−3/2 × R, which is a C r map provided c is C r +m−2 , β is C r +m−1 . Since λ0 is assumed to be a simple eigenvalue, ∂F ˙ → (( + c + λ0 )u˙ + λu ˙ 0, ˙ λ) (u 0 , λ0 , i
) : (u, ∂(u, λ)  
∂ ˙ 2 (u 0 u)) ˙ + β u, ∂N
: H 2 (
) × R → L 2 (
) × H 1/2 (∂
) × R is an isomorphism, so the implicit function theorem says F(u, λ, h) = (0, 0, 1) has a unique solution u = u h , λ = λh near u 0 , λ0 for every h C 2 -near i
, and h → (u h , λh ) is C 1 . To compute the derivative, we assume more smoothness, say
is C 3 , β is 3 C , c is C 2 , and we work in H 3 (
). The same argument gives h → (u h , λh ) : Diff3 (
) → H 3 (
) × R of class C 1 . We will calculate in the Lagrangian mode, for a change, which (in this problem) is about the same level of difficulty as the Eulerian.

42

Chapter 3. Examples Using the Implicit Function Theorem

Let u(t, ·) = u h(t,·) , λ(t) = λh(t,·) , where h(t, x) is smooth with h(t, x) = x + t V (x) + o(t) as t → ∞. Now for t near 0, h ∗ ( + c + λ(t))h ∗−1 u(t, ·) = 0 in
and ˙ Dt h ∗ ( + c + λ(t))h ∗−1 u(t, ·) = λ(t)u(t, ·) + h ∗ ( + c + λ)h ∗−1 Dt u holds by continuity (a continuous function of t with values in L 2 (
)), so at t = 0, with u˙ = ∂t∂ u(t, ·)|t=0 , ˙ ˙ · V · ∇u 0 ) in
. 0 = λ(0)u 0 + ( + c + λ0 )(u Also by Theorem 2.4     Dt h ∗ Nh(
) · ∇ + β h ∗−1 u(t, ·) = h ∗ Nh(
) · ∇ + β h ∗−1 Dt u(t, ·)   + Dt h ∗ Nh(
) · h ∗ ∇h ∗−1 u(t, ·) holds near ∂
, when everything is smooth, and by continuity also holds for our solution curve as an equality of continuous functions of t with values in H 1 on a neighborhood of ∂
. Taking t → 0 and restricting to ∂
, we find     ∂u 0 ∂ + βu 0 = + β (u˙ − V · ∇u 0 ) −V · ∇ ∂N ∂N   ∂N σ = V · N
, − ∇∂
σ + σ · ∇u 0 , ∂N by Lemma 2.3. Since ∂u 0 /∂ N + βu 0 = 0 on ∂
, the left-side of this equation involves only the normal derivative, and we have   ∂ + β (u˙ − V · ∇u 0 ) ∂N     ∂N ∂ ∂u 0 + βu 0 + ∇∂
+ σ = −σ · ∇u 0 ∂N ∂N ∂N   ∂β = div∂
(σ ∇∂
u 0 ) + σ u 0 λ0 + c + β 2 − β H − ∂N on ∂
, were we have used Theorem 1.12 and the equations satisfied by u 0 . Now multiplying the earlier equation by u 0 and integrating over
,

˙ u 20 + ( + c + λ0 )(u˙ − V · ∇u 0 ) 0 = λ(0)
 

∂ ˙ + β (u˙ − V · ∇u 0 ) u0 = λ(0) + ∂N ∂
 

∂β 2 2 ˙ + σ u 0 λ0 + c + β − β H − u 0 div∂
(σ ∇∂
u 0 ). = λ(0) + ∂N ∂
∂


Chapter 3. Examples Using the Implicit Function Theorem 43  The last integral is − ∂
σ |∇∂
u 0 |2 , which gives the result claimed. We may take limits in the final expression – as in Example 3.1 – to see the result also holds if ∂
is C 2 , c is C 1 and β is C 2 . Now suppose n = 2 and β, c are constant – we may suppose c = 0. We will compute the second derivative at t = 0 of a simple eigenvalue λ(t) of the Robin problem in the ellipse
(t) with semi-axes et , e−t . If β = 0 we always have the eigenvalue zero; but we assume λ(0) = 0 in case β = 0. It is clear ˙ that λ(−t) = λ(t), (as in Example 3.2) so λ(0) = 0, and t → λ(t) is analytic. In √ √ polar coordinates, at t = 0, the eigenvalue λ(0) = λ0 is a root of λ0 J0 ( λ0 ) + √ √ is u 0 (r ) = c J0 ( λ0r ) with c chosen so β J0 ( 2λ0 ) = 0 and the eigenfunction 2 2 u = 1, hence π u (1) = λ /(λ 0 0 0 + β ). If u(t, ·) is the eigenfunction in
(0) 0
(t), working in the Eulerian mode, and u˙ = ∂t∂ u(t, ·)|t=0 we have ( + λ0 )u˙ = 0 in the unit disc
(0) = {r < 1} and 

   ∂ ∂u 0 ˙ + β u˙ + N · ∇u 0 + V · ∇ + βu 0 = 0 ∂N ∂N

on r = 1.

Now V (t, y) = col(y1 , −y2 ) so V · N
(0) = cos 2θ on r = 1. Since u 0 is radial and N˙ · N = 0, the boundary condition is 

 ∂ ˙ r =1 = − cos 2θ(u 0 (1) + βu 0 (1)) = µu 0 (1) cos 2θ + β u| ∂r

 where µ = λ0 + β 2 − β. Further, assuming
(t) u(t, ·)2 ≡ 1, we have  √ ˙ = 0 ( ∂
V · N = 0) so u˙ = B J2 ( λ0r ) cos 2θ where the constant B
(0) u 0 u is given by

u 0 (1)µ(2β − λ0 ) . B J2 ( λ0 ) = 2(µ − β) Let σ = V · N
(t) , Q(t, ·) = |∇u(t, ·)|2 + u(t, ·)2 (βdiv N
(t,·) − 2β 2 − λ(t)) so dtd λ(t) = ∂
(t) Q σ and d2 λ(t) = dt 2




∂
(t)


∂σ ∂Q ∂σ ∂Q 2 +σ + Hσ Q + +σ . σ2 ∂t ∂N ∂ N ∂t ∂
(t)

At t = 0, Q = u 0 (r )2 + u 0 (r )2 (β/r − 2β 2 − λ0 ) is constant on {r = 1} = ∂
(0)

44

Chapter 3. Examples Using the Implicit Function Theorem

so the first integral vanishes (area
(t) = constant). Also ∂ Q/∂ N = ∂ Q/∂r |r =1 = β(4µ − 1)u 0 (1)2 at t = 0, div N˙ |t=0 = 4 cos 2θ,  ∂ Q  ˙ − 2β 2 − λ0 ) + βu 20 div N = 2∇u 0 · ∇ u˙ + 2u 0 u(β ∂t  r =1,t=0

= u(1)2 cos 2θ{4β + µ(λ0 µ − 4βµ + 2β 2 )/(µ − β)}. Substitution yields ¨ λ(0) =

µ2

λ0 [λ0 µ2 + µ(3β − 2β 2 ) − 3β 2 ], − β2

µ = λ0 + β 2 − β,

which (a minor miracle) is the same expression obtained in [11] by a different method. Note that, as β → ±∞ with λ0 bounded, this approaches √ λ0 (λ0 − 2) and J0 ( λ0 ) → 0, so we recover the result for the Dirichlet problem (Example 3.2). Example 3.6. Simple Eigenvalue of a General Dirichlet Problem. Suppose

Q ⊂ Rn is open and for multi-indices α = (α1 , . . . , αn ) with |α| = n1 α j ≤

2m we have aα : Q → C of class C |α| , while a0 is C 1 , | |α|=2m aα (x)ξ α | ≥

c0 |ξ |2m for x ∈ Q, ξ ∈ Rn and some constant c0 > 0, and |α|=2m aα (∂/∂ x)α is properly elliptic (true, for example, if n ≥ 3 or if the aα are real when |α| = 2m). Let
be a bounded C 2m -regular region with
⊂ Q, and suppose λ = λ0 is a simple eigenvalue of  α ∂ aα (x) u = λu in
, ∂ x |α|≤2m (∗)
∂ m−1 u ∂u = ··· = u= = 0 on ∂
∂N ∂ N m−1 with eigenfunction u 0 . Then u 0 is unique, up to a complex-constant multiplier, and there is an eigenfunction v0 (equally unique) of the adjoint problem  ∂ α (aα (x)v) = λv in
, − ∂x |α|≤m ∂ m−1 v ∂v = ··· = = 0 on ∂
∂N ∂ N m−1   with
u 0 v0 = 0. We may and shall assume
u 0 v0 = 1. Then if t → h(t, ·) ∈ C 2m (
, Rn ) is a C 1 function of the form h(t, x) = x + t V (x) + o(t) as t → 0,
(t) = h(t,
), there is a unique eigenvalue λ(t) near λ0 of the corresponding problem (∗ )
(t) for t near 0, λ(0) = λ0 , t → λ(t) v=

Chapter 3. Examples Using the Implicit Function Theorem is C 1 , and d λ(t)|t=0 = (−1)m+1 dt




∂






|α|=2m

aα N
α V · N


45

∂ m u 0 ∂ m v0 . ∂Nm ∂Nm

The existence and differentiability of the eigenvalue and eigenfunction are proved by applying the implicit function theorem to    
∗ α ∗−1 aα ∂ h u − λu, uv0 (u, λ, h) → h :H

2m

∩

α H0m (
, C)




× C × Diff (
) → L 2 (
, C) × C. 2m

We will compute the derivative, assuming all convenient smoothness.

From h ∗ ( α aα ∂ α − λ)h ∗ u = 0 and Theorem 2.2, we find ˙ aα ∂ α (Dt u)t=0 = λ(0)u in
. 0 α

Also w ≡ Dt u|t=0 satisfies ∂ j w/∂ N j = 0 on ∂
for 0 ≤ j ≤ m − 2 while ∂ m−1 w/∂ N m−1 = −V · N
∂ m u 0 /∂ N m on ∂
. Therefore 

α α ˙λ(0) = λ(0) ˙ u 0 v0 = aα ∂ w − (−∂) (aα v0 )w v0





= (−1)m




α



∂
|α|=2m

α

aα N α

∂ m v0 ∂ m−1 w ∂ N m ∂ N m−1

which is the result claimed. We show that a simple eigenvalue may be moved by perturbing the boundary,

assuming uniqueness in the Cauchy problem for α aα ∂ α and its adjoint – which certainly holds if the aα are analytic (Holmgren) or if m = 1 (see, for example, [2] or [12, Theorem 8.9.1]). Suppose to the contrary that λ0 is a simple eigenvalue of (∗ )
which does not ˙ move or, more generally, that
is a critical point of the eigenvalue so λ(0) =0 m m for every choice of V in the expression above. Then ∂∂ Num0 ∂∂ Nvm0 = 0 on ∂
, so on some non-empty open set in ∂
, at least one of these (say u 0 ) vanishes. By the adjacent component uniqueness in the Cauchy problem, u 0 ≡ 0 throughout  of
; in fact u 0 v0 = 0 throughout
. But
u 0 v0 = 1, a contradiction. If
is connected, we can move a simple eigenvalue by small perturbations of ∂
restricted to a small neighborhood of a given point. Of course if uniqueness fails, one might possibly have an eigenfunction (u 0 or v0 or both) which vanishes throughout a neighborhood of ∂
, and then the eigenvalue would not move. (Pli˘s gave an example of a fourth order elliptic equation with C ∞ coefficients on R3 which has nontrivial kernel containing C ∞ functions with compact support.)

Chapter 4 Bifurcation Problems

The following examples are again based on the implicit function theorem, but joined with the method of Liapunov-Schmidt, which gives them a slightly different flavor. We use the domain of definition as a bifurcation parameter. An introduction – and much more – to bifurcation theory can be found in Methods of Bifurcation Theory, S. N. Chow and J. K. Hale (Springer-Verlag). Example 4.1. Multiple Eigenvalues of the Dirichlet Problem for the Laplacian. Let
be a bounded C 2 region in Rn and let λ = λ0 be an m-fold eigenvalue of u + λu = 0 in
,

u = 0 on ∂
,

u
≡ 0,

with m > 1. For regions C 2 -close to
, there will be m eigenvalues (counting multiplicity) close to λ0 and we study their dependence on the region. Suppose {φ j }mj=1 is an orthonormal basis of the eigenfunctions in
with eigenvalue λ0 . if h is near i
in C 2 (
, Rn ) and λ is an eigenvalue of the Dirichlet problem in h(
), there is a solution v
≡ 0 of v + λv = 0 in h(
), v = 0 on ∂h(
), so h ∗ v = u ∈ H 2 ∩ H01 (
) satisfies h ∗ ( + λ)h ∗−1 u = 0 in
. Define the (real) analytic map F : (u, λ, h) → h ∗ ( + λ)h ∗−1 u : H 2 ∩ H01 (
) × R × Diff2 (
) → L 2 (
). (Recall the eigenvalues are necessarily real, so we may restrict attention to real
eigenfunctions.) Let P = m 1 φ j ⊗ φ j be the orthogonal projection onto span {φ1 , . . . , φm }  m  Pψ = φj φ j ψ.


1

46

Chapter 4. Bifurcation Problems

47

We seek λ (near λ0 ) and u
= 0 such that F(u, λ, h) = 0 or equivalently u = v + w
= 0,

v = Pu,

w = (I − P)u ∈ N (P),

such that P F(v + w, λ, h) = 0 and (I − P)F(v + w, λ, h) = 0. The second equation may be written 0 = ( + λ)w + (I − P)(h ∗ h ∗−1 − )(v + w) ∈ N (P). Since ( + λ)|N (P) : N (P) ∩ H 2 ∩ H01 (
) → N (P) ⊂ L 2 (
) is an isomorphism for λ = λ0 , it is also an isomorphism for λ near λ0 and has inverse Q λ which is an analytic function of λ. Then the above equation becomes w = −Q λ (I − P)(h ∗ h ∗−1 − )(v + w),

λ near λ0 ,

which is solvable for w if also h − i
C 2 is small. The solution w = S(λ, h)v is linear in v ∈ R(P) and analytic in (λ, h) near (λ0 , i
) with

S(λ, h) L(R(P),H 2 ∩H01 ) = O( h − i
C 2 ) as h → i
. There remains another equation, in R(P). Introducting coordinates in this m-dimensional space, v =
m
m 1 c j φ j for some c j ∈ R, and we conclude: u = 1 c j (φ j + S(λ, h)φ j )
= 0 and h ∗−1 u is an eigenfunction of the problem in h(
) with eigenvalue λ (λ near
λ0 , h near i
) if and only if the c j are not all zero and m k=1 M jk (λ, h)ck = 0 for 1 ≤ j ≤ m, where M jk (λ, h) =
φ j (h ∗ h ∗−1 + λ)(φk + S(λ, h)φk ). Thus λ is an eigenvalue if and only if det [M jk (λ, h))]mj,k=1 = 0.  Now (λ, h) → M jk (λ, h) is analytic near (λ0 , i
) and M jk (λ,mi
) =
φ j ( + λ)(φk + 0) = (λ − λ0 )δ jk so det [M jk (λ, i
)] = (λ − λ0 ) . By Rouch´e’s theorem, there are m roots λ (counting multiplicity) of det [M jk (λ, h)] = 0 near λ0 whenever h − i
C 2 is sufficiently small. (We see later that “most” choices of h split this m-fold root λ0 into m simple roots, or m simple eigenvalues; see Example 4.4 or 6.1.) Suppose h(0, ·) = i
and t → h(t, ·) ∈ C 2 (
, Rn ) is real-analytic near t = 0; for example h(t, x) = x + t V (x), for some V ∈ C 2 . Then (λ, t) → det [M jk (λ, h(t, ·))]mj,k=1 is analytic near t = 0, λ = λ0 , and by Puiseux’s theorem the roots λ = λ(t) may be expressed as analytic functions of some fractional power of t, t 1/k for some integer k ≥ 1. However we know λ(t) is real for every (small) real t. If at j/k and a(eiπ t) j/k are both real with t > 0, j = integer, then a is real and eigher a = 0 or j/k = integer; thus the “fraction” 1/k = 1[14] and the eigenvalues λ(t) are actually analytic in t, as are the eigenfunctions u(t, ·), F(u(t, ·), λ(t), h(t, ·)) = 0. We will calculate the derivatives of λ(t) at t = 0 on such a branch (there will ordinarily be m branches). As usual, it is easier to calculate in the Eulerian mode, and for this it is convenient to assume more smoothness. If
is C m,α , for example, all the arguments above may be carried out in the C m,α setting and t → (λ(t, ·), (u(t, ·)) ∈ R × C m,α (
)0 is

48

Chapter 4. Bifurcation Problems

analytic near t = 0 if t → h(t, ·) ∈ C m,α (
, Rn ) is analytic. We simply assume “sufficient smoothness,” and m = 3, 0 < α < 1, appears to be sufficient. Suppose therefore
(t) = h(t,
), and λ = λ(t), v = v(t, x) are smooth functions such that v + λv = 0 in
(t),

v = 0 on ∂
(t),

with v(0, ·)
≡ 0, and let v˙ = ∂v/∂t; then ˙ = 0 in
(t), ( + λ)v˙ + λv

v˙ + V · ∇v = 0 on ∂
(t)
where ∂h/∂t = V (t, h), h(0, ·) = i
. When t = 0, v = v(0, ·) = m 1 c j φ j for some c1 , . . . , cm , not all zero and   m  ∂φ j ∂φk ˙λ(0)ck = ˙λ(0)φk v = − · cj V·N ∂N ∂N
∂
j=1  ∂φ k Let M 0jk = ∂
V · N
∂ Nj ∂φ at t = 0, c = col(c1 , . . . , cm ) and it follows ∂N ˙λ(0)c + M 0 c = 0, c
= 0, so λ(0) is an eigenvalue of −M 0 . In fact, it is easy to prove that for each simple eigenvalue µ0 of M 0 , there is a corresponding (unique) branch of eigenvalues t → λ(t) of the form λ(t) = λ0 − tµ0 + o(t) as t → 0. λ(t) is a simple eigenvalue of the Dirichlet problem in
(t), for small ˙ t
= 0. But the eigenvalue −λ(0) of interest may not be simple, so we continue. To compute the second derivative at t = 0, we must know v| ˙ t=0 . Now there is a unique φ˙ j ∈ H 2 (
) orthogonal to φ1 , . . . , φm , such that ( + λ0 )φ˙ j ∈ span
˙ ˙ t=0 = m {φ1 , . . . , φm } and φ˙ j = −V · ∇φ j on ∂
. It follows that v| 1 (c j φ j + c˙ j φ j ) for some c˙t , . . . , c˙m (the c j are the same as before). Then at t = 0 ˙ v˙ + λ(0)v ¨ = 0 in
( + λ0 )v¨ + 2λ(0) 2 ˙ v¨ + 2V · ∇ v˙ + (V · ∇) v + V · ∇v = 0 on ∂
 ¨ ∂φk ˙ ¨ and so λ(0)c k + 2λ(0)c˙k = ∂
V ∂ N , 1 ≤ k ≤ m. This gives the equation m in R ¨ ˙ 0 = (λ(0)I + M˙ 0 )c + 2(λ(0)I + M 0 )c˙ ˙ together with 0 = (λ(0)I + M 0 )c, c
= 0, where M˙ 0 = [ M˙ 0jk ]mj,k=1 is  ∂φk ((V · ∇)2 φ j + 2V · ∇ φ˙ j + V˙ · ∇φ j ). M˙ 0jk = ∂
∂ N ˙ ¨ ¨ If λ(0) is a simple eigenvalue of −M 0 then λ(0) must be chosen so (λ(0) + 0 0 ˙ ˙ M )c is the image of λ(0) + M , i.e., orthogonal to c, i.e., ¨ λ(0) = −c M˙ 0 c/c · c

Chapter 4. Bifurcation Problems

49

˙ ¨ Another case of interest is when M 0 = 0; then λ(0) = 0 and λ(0) is an eigenvalue of − M˙ 0 with eigenvector c. Now we study in more detail the case when
= unit disc in R2 ,
(t) = the t −t ellipse with √semi-axes e , e . When t = 0 we have a double eigenvalue λ > 0 where J p ( λ) = 0 for some integer p ≥ 1, and φ1 , φ2 are given by √ φ1 + iφ2 = k J p ( λr )ei pθ  √  in polar coordinates, with k = π2 /J p ( λ), so
φ j φk = δ jk . For t near 0 there is a pair of eigenvalues λ1 (t), λ2 (t) near λ, analytic functions of t. Since
(−t) is geometrically the same as
(t), the pair {λ1 (t), λ2 (t)} is an even function of t, so either λ1 (t), λ2 (t) are both even or λ2 (t) = λ1 (−t). We have y = h(t, x) = col(x1 et , x2 e−t ), V (t, y) = col(y1 , −y2 ), V ◦ ∇ = y1 ∂∂y1 − y2 ∂∂y2 = cos 2θr ∂r∂ − sin 2θ ∂θ∂ and the matrix M 0 is    cos2 pθ cos pθ sin pθ 2λ 2π 0 M = cos 2θ π 0 sin pθ sin2 pθ which is zero when p ≥ 2, but for p = 1  λ 0 0 M = . 0 −λ √ Thus for p = 1, the double eigenvalue λ [J1 ( λ = 0] splits into √ λ1 (t) = λ − tλ + O(t 2 ) for k J1 ( λr ) cos θ + O(t), √ λ2 (t) = λ + tλ + O(t 2 ) for k J1 ( λr ) sin θ + O(t) with λ2 (t) = λ1 (−t). One may compute the second derivative as well: λ2 (t) = λ + tλ + t 2

λ2 + O(t 3 ). 8

˙ Now suppose p ≥ 2 so M 0 = 0, λ(0) = 0. We have ( + λ)φ˙ j ∈ ∂φ j ˙ sp[φ1 , φ2 ], φ j = − cos 2θ ∂r so

λ  i( p+2)θ e φ˙ 1 + i φ˙ 2 |r =1 = − + ei( p−2)θ 2π and when r = 1  

√ √ J p+2 ( λr ) i( p+2)θ J p−2 ( λr ) i( p−2)θ λ ˙ ˙ + , φ1 + i φ 2 = − √ e √ e 2π J p+2 ( λ) J p−2 ( λ)

50

Chapter 4. Bifurcation Problems

since p + 2 > p > p − 2 > − p for p ≥ 2. Then with √  √ λJm ( λ) Rm = , √ Jm ( λ) we have



2V · ∇(φ˙ 1 + i φ˙ 2 )|r =1 = −

(V · ∇)2 (φ1 + iφ2 )|r =1

λ i pθ e (R p+2 + R p−2 + e4iθ (R p+2 − p − 2) 2π

+ e−4iθ (R p−2 + p − 2)).

λ i pθ e (−2 + e4iθ ( p + 1) + e−4iθ (− p + 1)). =− 2π

Thus



− M˙ 0 = λ(R p+2 + R p−2 + 2)

1 0

0 1



+ δ p2 λ(R p−2 − 1)

0 . −1

1 0

√ ˙ = 0, If p = 2 we find J2 ( λ) = 0, λ(0)  √ λ    − (18 + 5λ) for k J2 ( λr ) cos 2θ + O(t) 6 ¨ λ(0) = √  λ   (λ − 6) for k J2 ( λr ) sin 2θ + O(t) 6 In case p = 2, the eigenvalues must be even functions of t. If p ≥ 3, both eigenvalues have the form  λt 2 λ λ1,2 (t) = λ − 2+ 2 + O(t 3 ) 2 p −1

but presumably split in higher-order terms (though this is not proved). t ? λ (J0)

(J1)

t=0

+

t≠0

−+ 2

1

(J2)

(J0)

(J3)

+−+ −

− + − 3

? 4

5

6

7

8

Reyleigh [33] found the nodal-sets for the case p = 1 by a variational argument.

Chapter 4. Bifurcation Problems

51

Example 4.2. Variation of a Turning Point. Mignon, Murat and Puel [24] studied the variation of the turning point of u + λeu = 0

in
,

u = 0 on ∂


u

(λ*, u*)

λ*

λ

when the domain
changes. In a neighborhood of the turning point (λ∗ , u ∗ ), the solutions lie on a curve of the form λ = λ∗ + kτ 2 + O(τ 3 ),

u = u ∗ + τ φ ∗ + O(τ 3 )

where k < 0 and φ ∗ satisfies ∗

φ ∗ + λ∗ eu φ ∗ = 0 in
,

φ ∗ = 0 on ∂


with φ ∗ > 0 in
. We will treat a more general case:
is a bounded C 2 -regular region in Rn , f : Rn × R → R is C 2 and there is a solution (λ, u) = (λ0 , u 0 ) of (∗)


u + λ f (x, u) = 0 in
,

u = 0 on ∂


such that the linearization at this point L 0 : φ → φ + λ0 f  (·, u 0 (·))φ : H 2 ∩ H01 (
) → L 2 (
) has a one-dimensional kernel, all multiples of φo . (Here f  (x, s) = ∂s∂ f (x, s).) We also assume   φ0 f (·, u 0 (·))
= 0, φ03 f  (·, u 0 (·))
= 0. λ0
= 0,





All these hypotheses hold for the example of [24]: the turning point is at the end of the stable branch; so 0 is the principal eigenvalue of the linearization, a simple eigenvalue with a positive eigenfunction. We prove, for each h C 2 -close i
, there is a unique turning point near (λ0 , u 0 ) and the parameter value λh is a differentiable function of h with derivative   ∂u 0 ∂φ0 d ˙ ˙ λh |h=i
: h → − h · N
φ0 f (·, u 0 (·)). dh ∂N ∂N ∂



52

Chapter 4. Bifurcation Problems

First note, by standard bifurcation arguments (
fixed), that the solutions (u, λ) of (∗)
near (u 0 , λ0 ) lie on a curve λ = λ0 + kτ 2 + o(τ 2 ),

u = u 0 + τ φ0 + O(τ 2 ),

τ small,

  where k = − 12 λ0
φ03 f  (·, u 0 (·))/
φ0 f (·, u 0 (·))
= 0. (In the case of [24], it is clear k < 0.) For any small perturbation h(
) of
, if we find a solution (λh , u h ) near (λ0 , u 0 ) of (∗)h(
) which is not simple – so the linearization has eigenvalue zero – it follows by continuity that the kernel is one-dimensional and the corresponding inequalities hold to prove (λh , u h ) is a turning point for (∗)h(
) . Choose p in n/2 < p < ∞, so W 2, p (
) ⊂ C 0 (
), and define 

∗

G : (λ, v, φ, h) → h ( + λ f )h

∗−1

∗



u, h ( + λ f )h

∗−1



 (u, φ),




φ

2

1, p

: R × W 2, p ∩ W0 (
) × H 2 ∩ H01 (
) × Diff2 (
) → L p (
) × L 2 (
) × R on a neighborhood of (λ0 ,u 0 , φ0 , i
). Then G is continuously differentiable, ∂G (λ0 , u 0 , φ0 , i
) is G(λ0 , u 0 , φ0 , i
) = (0, 0,
φ02 ), and ∂(λ,v,φ)  ˙ v, ˙ → (λ, ˙ φ)

L 0 v˙ + λ˙ f (u 0 ), L 0 φ˙ + λ˙ f  (u 0 )φ0 + λ0 f  (u 0 )φ0 v, ˙ 2




φ0 φ˙ ,

which is an isomorphism. Indeed, given (a, b, c) ∈ L p × L 2 × R,  the inverse ˙ v, ˙ is uniquely determined as follows: Choose λ˙ so λ˙ image ( λ, ˙ φ)
φ0 f (u 0 ) =  ˙ ˙ = a − λ f (u 0 ) ⊥ φ0 determines v˙ to within a multiple
aφ0 , and then L 0 v of φ0 ; this multiple is chosen so that λ˙ f  (u 0 )φ0 + λ0 f  (u 0 )φ0 v˙ ⊥ φ0 (recall to within a multiple of φ0 ; and this λ0
f  (u 0 )φ03
= 0), hence φ˙ is determined  multiple is chosen by requiring 2
φ0 φ˙ = c. By the implicit function there is a C 1 solution h → (λh , vh , φh )  theorem, 2 of G(λ, v, φ, h) = (0, 0,
φ0 ) with (λh , vh , φh ) = (λ0 , u 0 , φ0 ) when h = i
. Also h ∗−1 vh ≡ u h , h ∗−1 φh ≡ ψh satisfy u h + λh f (·, u h ) = 0 in h(
),

u h = 0 on ∂h(
)

( + λh f  (·, u h ))ψh = 0 in h(
),

ψh = 0 on ∂h(
)

  and ψh
= 0 since h(
) ψh2 /(det h  ) =
φ02 > 0. Thus (λh , u h ) is the (unique) turning-point near (λ0 , u 0 ) for the problem in h(
), h near i
. The calculation of the derivative, in the Eulerian mode, is simple.

Chapter 4. Bifurcation Problems

53

Example 4.3. A Bifurcation Problem with Two-Dimensional Kernel. We seek small solutions u of u + f (x, u) + λu = α(x) + β(x)u in
,

u = 0 on ∂


where f (x, u) = O(u 2 ) as u → 0, f and λ are fixed ( f is C 3 , λ is constant), and (α,√ β,
) are parameters near (0, 0, unit disc of R2 ). We suppose λ > 0 satisfies J1 ( λ) = 0, so the linearization at α = 0, β = 0,
= D (= √ unit disc) has two-dimensional kernel with basis {φ1 , φ2 }, φ1 + iφ2 = J1 ( λr )eiθ in polar 2 coordinates. We impose a “non-degeneracy” condition on m(x) ≡ ∂∂u 2f (x, 0), satisfied for most choices of m, namely: For some complex constants A, B, C  m(x)(φ1 + iφ2 )(ξ1 φ1 + ξ2 φ2 )2 = Aξ12 + Bξ1 ξ2 + Cξ22 , ξ ∈ R2 D

and we require Aξ12 + Bξ1 ξ2 + Cξ22
= 0 whenever (ξ1 , ξ2 )
= (0, 0). In this case {A cos2 θ + B cos θ sin θ + C sin2 θ | 0 ≤ θ ≤ 2π } is an ellipse in the complex plane which does not pass through 0, although it may have one semi-axis zero. We suppose it does not reduce to a point (A = C, B = 0) nor does this occur after a non-singular linear change of ξ -variables. Analytically, we suppose either (A+C)2 (A − C) B¯ is not real, or it is real but {(A−C) 2 +B 2 } does not belong to [1, ∞] ⊂ C. If the ellipse encloses 0, we have an elliptic umbilic i if 0 is outside the ellipse, a hyperbolic umbilic. We prove that as (α,
) vary on a neighborhood of (0, D) with β fixed (small), or as (α, β) vary near (0, 0) with
fixed (near D), or various combinations, we produce the universal unfolding of this umbilic, and there may be 0, 2 or 4 solutions, depending on the case. 0 2

2 4

4

α

hyperbolic umbilic

4 α

Ω, β

β, Ω

elliptic umbilic

(The case α ≡ 0 is quite different, as u = 0 is then always a solution.) The solutions u are critical points of  u   1 2 1 1 2 2 f (x, ·) d x u → − |∇u| + λu + α(x)u + β(x)u + 2 2 2 0
so it should not be surprising that catastrophe theory applies.

54

Chapter 4. Bifurcation Problems

Example: m(x1 , x2 ) = m 0 + m 1 x1 + m 2 x2 +

2  0

j 2− j

p j x1 x2

+

3 

j 3− j

q j x1 x2

0

for some real constants m 0 , m 1 , . . . , q3 . The corresponding quadratic form does not depend on m 0 or the p j , and is   Aξ12 + Bξ1 ξ2 + Cξ22 = Q 1 2(m 1 + im 2 ) ξ12 + ξ22  + (m 1 − im 2 )(ξ1 + iξ2 )2  + Q 2 ξ12 + ξ22 (3q0 − iq1 + q2 − 3iq3 )  + 2 ξ12 − ξ22 (q0 + iq3 )  − 2iξ1 ξ2 (2q0 + iq1 + q2 + 2iq3 ) 1 √ 1 √ where Q 1 = π4 0 J1 ( λr )3r 2 dr , Q 2 = π8 0 J1 ( λr )3r 4 dr . Suppose Q 1 , Q 2
= 0; this √ certainly holds if λ is the first (double) eigenvalue or if λ is large (J1 ( λ) = 0), and numerical integration suggests that it holds in every case. Then depending on the m 1 , m 2 , q0 , . . . , q3 , we could be in the hyperbolic case (e.g., m(x) = m 1 x1 + m 2 x2
≡ 0) or the elliptic case (e.g., m(x) =
j 2− j x23 − 3x12 x2 ) or the degenerate case (e.g., m(x) = m 0 + 20 p j x1 x2 , when A = B = C = 0). Note we do not allow m(x) = constant. Define F
(u; α, β) = u + λu + f (·, u) − α(·) − β(·)u, so F
: H 2 ∩ H01 (
) × C 3 (R2 ) × C 3 (R2 ) → L 2 (
) is C 3 . Also (u, α, β, h) → h ∗ Fh(D) h ∗−1 (u; α, β) is C 3 from H 2 ∩ H01 (D) × C 3 × C 3 × Diff2 to L 2 (D). Using the method of Liapunov-Schmidt (but supressing details) if h = i D + g, g C 2 (D) small u = ξ1 φ1 + ξ2 φ2 + ψ = φ + ψ with ψ ⊥ {φ1 , φ2 } and ξ1 , ξ2 , ψ small, and α C 3 + β C 3 is small, the equation becomes 0 = h ∗ ( + λ)h ∗−1 (φ + ψ) − α ◦ h − β ◦ h(φ + ψ) + f (h(·), φ + ψ) 1 = ( + λ)ψ + [g · ∇,  + λ](φ + ψ) + α + β(φ + ψ) + m(φ + ψ)2 + · · · 2 and we solve the component in {φ1 , φ2 }⊥ for ψ = ψ(ξ, α, β, g) = O( α + g (|ξ | + α + β ) + |ξ | β + g 2 + |ξ |3 ).

Chapter 4. Bifurcation Problems

55

Substitution in the remainder of the equation gives the bifurcation equation E(ξ ; α, β, g) = 0 in R2 , (E 1 + i E 2 )(ξ, α, β, g)  = (φ1 + iφ2 )h ∗ Fh(D) (·, α, β)h ∗−1 (ξ1 φ1 + ξ2 φ2 + ψ(ξ, α, β, g)) D

= α1 + iα2 + ξ1 (β11 + iβ21 ) + ξ2 (β12 + iβ22 ) + ξ1 (γ11 + iγ21 ) + ξ2 (γ12 + iγ22 ) 1 + Aξ12 + Bξ1 ξ2 + Cξ22 + (higher order terms) 2 where       2 φ1 φ2 β11 β12 α1 φ1 φ1 = , = , α(x) β(x) α2 φ2 β21 β22 φ1 φ2 φ22 D D    γ11 γ12 cos2 θ sin θ cos θ g·N = dθ γ21 γ22 sin θ cos θ sin2 θ ∂D  π √  √ 2 p0 + p1 p2 λJ1 λ = p2 p0 − p1 2 when g· N |∂ D = p0 + p1 cos 2θ + p2 sin 2θ + {terms orthogonal to 1, cos 2θ, sin 2θ}. To obtain the universal unfolding of the umbilic, ( αα12 ) should vary over a full neighborhood of ( 00 ) in R2 and   π √  √ 2 p0 + p1 p2 β11 β12 λJ1 λ + β21 β22 p2 p0 − p1 2 should include all (small) symmetric 2 × 2 matrices. The last condition could be achieved with β = 0 (or small), considering only ellipses
centered at the origin but with arbitrary orientation and arbitrary semi-axes (near 1). If we restricted attention to ellipses with a fixed area, then p0 = 0 and we should also allow β to vary (e.g., β an arbitrary small constant). Example 4.4. Generic Simplicity of Eigenvalues of a Self-Adjoint 2m-Order Dirichlet Problem. Following Micheletti [23], grosso modo, we show that for most C 2m -regular bounded regions
⊂ Rn (in the sense of Baire category), a self-adjoint strongly elliptic operator with the unique continuation property has only simple eigenvalues for the Dirichlet problem. I am indebted to M. Saut for bringing Micheletti’s work to my attention. Specifically, let Q be an open set in Rn and suppose aαβ = aβα : Q → C is of class C s , s = max(|α|, |β|) + 1, when the multi-indices α, β have |α| ≤ m,

56

Chapter 4. Bifurcation Problems

|β| ≤ m, and for some constant c0 > 0  aαβ (x)ξ α+β ≥ c0 |ξ |2m Re

for x ∈ Q, ξ ∈ Rn .

|α|=|β|=m

Finally assume “unique continuation” holds, in the sense: If λ is real, u ∈ H 2m (Q) with support (u) compact in Q and
α β |α|,|β|≤m D (aαβ D u) = λu as distributions in Q then u ≡ 0 in Q. (Here α1 α αn D = D1 · · · Dn , D j = 1i ∂∂x j .) This is certainly true if the aα,β are analytic (Holmgren’s theorem) or if m = 1 (see [2] or [12, Thm. 8.9.1]). Let
be a bounded C 2m -regular region with closure
⊂ Q. Then for an ample (= residual = category II) set of h in a C 2m (
, Rn )-neighborhood of i
, the Dirichlet problem in h(
)  D α (aαβ D β u) = λu in h(
), u ∈ H 2m ∩ H0m (h(
)), u
≡ 0 |α|,|β|≤m

has only simple eigenvalues. In fact the argument is more general. Without assuming unique continuation, for most choices of h near i
the eigenvalues will all have minimal multiplicity; and if λ is such an eigenvalue whose minimal multiplicity is p ≥ 2, there is a basis {φ1 , . . . , φ p } for the eigenfunctions all of which vanish throughout a neighborhood of ∂h(
). Of course such an eigenvalue must remain of multiplicity p ≥ 2 for any small perturbation of h(
); and, of course, this cannot occur (i.e., the eigenvalues must be simple) when unique continuation holds. Assume λ is an eigenvalue of multiplicity p for the Dirichlet problem in some
, and that it has minimal multiplicity with respect to small perturbations of ∂
; that is, for any  > 0, if h − i
C 2m is sufficiently small, the problem in h(
) has an eigenvalue λh of multiplicity ≥ p in |λh − λ| < . (In this case, as we see below, λh is the only eigenvalue close to λ, and in fact λh = λ.) It will be sufficient to work with a restricted class of perturbations so let h(t, x) = x + t V (x) for small real t and any given V ∈ C 2m (
, Rn ). Let
(t) = h(t,
) and define  D α (aαβ D β v) : H 2m ∩ H0m (
(t) → L 2 (
(t)) A
(t) : v → |α|,|β|≤m

and B(t) = h(t, ·)∗ A
(t) h(t, ·)∗−1 : H 2m ∩ H0m (
) → L 2 (
). Then B(t) may be considered a self-adjoint operator with respect to L 2 (
; det h x (·, t)d x) for each small t. When t = 0,
(0) =
, B(0) = A
1 2m and  there exist φ1 , . . . , φ p ∈ H ∩ H0 (
), a basis for N (A
− λ), such that ¯
φ j φ k = δ jk (1 ≤ j, k ≤ p). We apply Gram-Schmidt orthogonalization with

Chapter 4. Bifurcation Problems

57

respect to the inner product of L 2 (
; det h x(·, t)d x) to produce another basis {φ1 (t, ·), . . . , φ p (t, ·)} for N (A
− λ) with
φ j (t, ·)φk (t, ·)det h x (t, ·) = δ jk and φ j (0, ·) = φ j . Note that t → φ j (t, ·) is C 1 (in fact, rational). Define
p P(t) = 1 φ j (t) ⊗ φ j (t), the orthogonal projection (in this inner product) onto N (A
− λ). By the implicit function theorem, there is a C 1 solution (t, µ) → w j (t, µ) ∈ H 2m ∩ H0m (
) of  (I − P(t)) h ∗ A
(t) h ∗−1 − λ0 − µ (φ j (t) + w j (t, µ)) = 0 P(t)w j (t, µ) = 0 for small real t, µ, with w j (0, µ) = 0. In fact µ → w j (t, µ) is analytic. By hypothesis, for each t near 0 (h(t, ·) near i
) there is an eigenvalue λ0 + µ near λ0 of the Dirichlet problem in
(t), having multiplicity ≥ p. Thus there exists nontrivial ψ in H 2m ∩ H0m (
) with B(t)ψ = (λ0 + µ)ψ, so, for
some constants c1 , . . . , c p , not all zero, ψ = j c j (φ j (t, ·) + w j (t, µ, ·)); note the c j may depend on t, µ. For each k = 1, . . . , p  µck =




φ¯ k (B(t)ψ − λ0 ψ) det h x (t, ·) d x =

p 

c j p jk (t, µ)

j=1

where

 p jk (t, µ) =




φ¯ k (B(t) − λ0 )(φ j (t, ·) + w j (t, µ, ·)),

and p

(t, µ) = det [µδ jk − p jk (t, µ)] jk=1 = 0. Note the p jk (t, ·) are analytic functions of µ near (t, µ) = (0, 0), and C 1 in both variables. When t = 0, p jk (0, µ) = 0 and (0, µ) = µ p ; by Rouch´e’s theorem, there are p roots µ near 0 counting multiplicity, for each small t. But p by hypothesis, there is some real root µ near 0 such that [µδ jk − p jk (t, µ)] j,k=1 has at least p independent null-vectors, so (since p jk = pk j ) p jk (t, µ) = µδ jk ,

1 ≤ j, k ≤ p,

for some small µ = µ(t). For small  > 0 at t sufficiently small  ζ dζ ∂ 1 (t, ζ ) pµ(t) = 2πi |ζ |= (t, ζ ) ∂ζ so t → µ(t) is C 1 near t = 0.

58

Chapter 4. Bifurcation Problems Now

    ∂ p jk (0, 0) = φ¯ k Dt h ∗ A
(t) − λ h ∗−1 (φ j + w j ) t=0 ∂t µ=0
 + φ¯ k V · ∇(A
− λ)φ j
 = φ¯ k (A
− λ)z j ,

p˙ jk ≡




z j = Dt (φt + w j (t, 0, ·))|t=0 , ∂m φ

where z j ∈ H 2m−1 ∩ H0m−1 (
) with ( ∂∂N )m−1 z j = −V · N ∂ N mj on ∂
, and we must interpret (A
− λ)z j as a distribution in H −1 (
). Integration by parts gives  p˙ jk = φ¯ k A
z j − z j A
φk
    ∂ m φ j ∂ m φk =− aαβ N α N β V · N ∂Nm ∂Nm ∂
|α|=|β|=m = p˙ jk . Now µ(0) ˙ is an eigenvalue of [ p˙ jk ], and in fact the only eigenvalue: otherwise the multiplicity would be reduced for t
= 0. But a Hermitian matrix with only one eigenvalue is a multiple of the identity, so p˙ jk = 0 and p˙ j j = p˙ kk whenever j
= k. This must hold for every choice of V , so on ∂
 m 2  m 2  ∂ φj   ∂ m φ j ∂ m φk  =  ∂ φk  if j
= k.  = 0,    m m m m ∂N ∂N ∂N ∂N  Since the multiplicity p ≥ 2, we conclude ∂ m φ j /∂ N m = 0 on ∂
, i.e., φ j ∈ ˙ = 0. H 2m ∩ H0m+1 (
) for each j = 1, . . . , p. Also we have µ(0) Summarizing: If λ = λ(
) is a p-fold eigenvalue of the problem in
whose multiplicity cannot be reduced by small perturbations of
, then the eigenfunc˜ near
, there is a unique eigentions belong to H 2m ∩ H0m+1 (
) and, for every
˜ value λ(
) near λ(
), which also has multiplicity p, depends differentiably on ˜ and has derivative zero when
˜ =
. Thus, in fact, λ(
) ˜ = λ(
) = λ is
, constant. We next show that we may choose a basis for the eigenfunctions which does not depend on small perturbations of
. Returning to the previous notation (h(t, x) = x + t V (x),
(t) = h(t,
)) p we now know µ ≡ 0, i.e., p jk (t, 0) ≡ 0 for small t, and {ψ j (t, ·)} j=1 , defined by h(t, ·)∗ ψ j (t, ·) = φ j (t, ·) + w j (t, 0, ·), is a basis for the eigenfunctions of A
(t) for the eigenvalue λ; thus we also have the ψ j (t, ·) ∈ H 2m ∩ H0m+1 (
(t)). Define ψ˙ j (t, ·) = ∂t∂ ψ j (t, ·); then h(t, ·)∗ ψ˙ j (t, ·) = Dt (φ j (t, ·) + h(t, 0,
)) is

Chapter 4. Bifurcation Problems

59

in H 2m−1 ∩ H0m (
) so ψ˙ j (t, ·) ∈ H 2m−1 ∩ H0m (
(t)). But also A
(t) ψ˙ j (t, ·) = λψ˙ j (t, ·) as distributions in
(t), so in fact ψ˙ j (t, ·) ∈ H 2m ∩ H0m (
(t)). Since p

{ψ j (t, ·)} j=1 is a basis for N (A
(t) − λ) there is a continuous matrix B(t) = [b jk (t)] so ψ˙ j (t, ·) =

p 

b jk (t)ψk (t, ·).

k=1

˙ + C(t)B(t) = 0, with C(0) = I , and Define C(t) as the matrix solution of C(t)
p let θ j (t, ·) = k=1 C jk (t)ψk (t, ·). This is another basis for the eigenfunctions of N (A
(t) − λ), with the additional property ∂θ j (t, ·) ≡ 0, 1 ≤ j ≤ p. ∂t If y ∈
(s) for all s between 0 and t, we have θ j (t, y) = θ j (0, y) = φ j (y). Choose V close to N
and on ∂
; then for small t < 0,
(t) ⊂
and ∪−δ 0. For any y in this neighborhood, y ∈ ∂
(t) for some, t −δ < t ≤ 0, so θ j (t, y) = 0 = θ j (0, y) = φ j (y). Thus the eigenfunctions φ j must all vanish on a neighborhood of ∂
, as was to be proved. For each j = 1, 2, . . ., there is an open dense set among the C 2m -regular bounded regions
with closure in Q such that every eigenvalue λ in − j ≤ λ ≤ j has minimal multiplicity: λ is either simple, or all the eigenfunctions corresponding to λ vanish on a neighborhood of ∂
. This condition is obviously open, and the argument above shows it is dense. Taking the intersection over all j = 1, 2, . . . gives the result claimed at the beginning. Remark. For the Laplacian, Karen Uhlenbeck [41] proved this result independently, using the transversality theorem; this argument will be our introductory example in Chapter 6. Note that t → h(t, ·)∗ A
(t) h(t, ·)∗−1 is not analytic in general, unless the coefficients are analytic. Thus we cannot a priori expect to find a basis for the eigenfunctions continuous as t → 0 [14, p. 111]; hence the careful construction of a basis.

Chapter 5 The Transversality Theorem

The transversality theorem (or transversal density theorem) of Thom and Abraham is a tool useful in many branches of geometry and analysis. The usual form [1, 32, 35, 42] makes unnecessarily strong hypotheses for the case of negative Fredholm index. We will prove some generalizations, aimed at infinite-dimensional applications, with special attention to the case of negative index. (We will see many examples with index −∞ in Ch. 6). Our terminology is that of Lang [17] and Abraham and Robbin [1] except for “codimension,” defined in 5.15, and “σ -proper,” defined in 5.4(3). See 5.13–14 for semi-Fredholm operators. This chapter is logically self-contained except for Sard’s theorem. We will use only Theorem 5.4 in the sequel, but other variations on the theme of transversality seemed of sufficient interest to be worth recording. Theorem 5.1. Sard’s Theorem (Brown-Morse-Sard). Let A ⊂ Rn be an open set and f : A → Rm a C k map, where k is a positive integer and k > n − m. Then the set of critical values of f has measure zero in Rm and is meager (= first Baire category). Proof. See [1] or [37]. Recall that x is a regular point of f if the derivative f  (x) is surjective; otherwise x is a critical point. A critical value of f is the image of some critical point; all other points of the codomain are regular values of f , including all points outside the image of f . The only modification in infinite dimensions is that, for x to be a regular point of f , we require not only that f  (x) is surjective but also that its kernel splits – which always holds in our applications, as the kernel is finite dimensional. When n < m, the set of critical values of f : A ⊂ Rn → Rm is simply the image of f , and to treat the image we don’t need differentiability. 60

Chapter 5. The Transversality Theorem

61

Theorem 5.2. Suppose 1 ≤ n < m, A is any subset in Rn and f : A → Rm is of class C µ+ , µ = n/m, that is | f (x) − f (y)| →0 |x − y|µ

as x − y → 0

uniformly for x = y in bounded subsets of A. Then f (A) has measure zero and is meager in Rm . Proof. It suffices to show meas f (C ∩ A) = 0 for every compact cube C < Rn . Let  > 0 and choose δ > 0 so | f (x) − f (y) ≤ |x − y|µ when |x − y| ≤ δ in C ∩ A. Divide C into N n disjoint congruent cubes {C  }, with N so large that diam C  = N1 diam C < δ. Then for each such C  , diam f (C  ∩ A) ≤  (diam C/N )µ so f (C  ∩ A) is contained in a cube of edge ≤ 2 (diam C/N )µ and volume ≤ (2)m (diam C/N )n . Therefore meas f (C ∩ A) ≤ (2)m (diam C)n , with  > 0 arbitrary. Theorem 5.3 (i) Let n > m ≥ 1,  > 0; there is a function f : Rn → Rm of class C n−m+1− with compact support whose critical values include [0, 1]m . (The set of critical points includes a set Rm−1 × W where W ⊂ Rn−m+1 is homeomorphic to [0, 1] (but has Hausdorff dimension n − m + 1 − ).) (ii) Let 1 ≤ n < m,  > 0; there is a function f : Rn → Rm of class C (n−)/m with compact support whose image contains [0, 1]m . Remark. This shows the smoothness conditions of (5.1) for n ≥ m and (5.2) for n < m, are nearly optimal. If n = m, (5.1) only requires continuous differentiability, and we need one derivative to define “critical value.” Proof. (i) Whitney [43] showed how to construct a curve K q ⊂ [0, 1]q (for each q ≥ 2), homeomorphic to [0, 1], and a function Wq : K q → [0, 1] such that Wq (K q ) = [0, 1] and |Wq (x) − Wq (x  )| = 0(|x − x  |q− ) for all x, x  ∈ K q . (To treat arbitrarily small  > 0, we must choose the scaling factor of Whitney’s construction close to 1/2; Whitney used the factor 1/3.) Then by Whitney’s λL

L

0 < λ = scale factor < 1/2

62

Chapter 5. The Transversality Theorem

extension theorem – in the more general version of E. Stein [38] – there is a C q− (Rn ) extension, still denoted Wq , such that (∂/∂ x) j Wq (x) = 0 for all x ∈ K q and 1 ≤ | j| ≤ q − 1. If m = 1, multiplying Wn by an appropriate “cutoff” gives the required example. In general, let q = n − m + 1 ≥ 2 and then x → (x1 , . . . , xn−1 , Wq (xm , . . . , xn )) is C q− with Rm−1 × [0, 1] among its critical values. (ii) There is a C 1/m Peano curve Pm : [0, 1] onto [0, 1]m for each m ≥ 2; the construction generalizes the method of Moore [25] for m = 2, by induction on m, as shown in the figure (for m = 3).

OUT

IDE

SIDE

0 ≤ x3 ≤ 1/3

OUTS

SIDE

OUT

1/ ≤ x3 ≤ 2/ 3 3

SIDE

OUT

2/ ≤ x3 ≤ 1 3

If 1 < n < m, consider φ = Pm ◦ Wn |K n : K n ⊂ Rn → Rm , where Wn is the C n− Whitney function. For x, x  ∈ K n , 

|φ(x) − φ(x  )| = O |Wn (x) − W (n(x  )|1/m = O |x − x  |(n−)/m and there is a C (n−)/m extension of φ which satisfies the requirements of the theorem – in particular, the image contains Pm (Wn (K n )) = Pm [0, 1] = [0, 1]m . Kupka [16] gave an example of a C ∞ map from 2 to R with critical values = [0, 1]. As modified by Dovady and Bonic, the example is a cubic polynomial f : 2 → R f (x) =

∞  n=1

2−n λ(nxn ),

x = (xn )θn=1

with

∞ 

|xn |2 < ∞,

1

where λ(t) = 3t 2 − 2t 3 . The critical points are {x ∈ 2 |nxn = 0 or 1, for all n}  −n and the critical values { ∞ n=1 2 n |n = 0 or 1, for all n} = [0, 1]. Thus smoothness is not enough; but Smale [37] showed we can generalize Sard’s theorem to infinite dimensions provided we restrict attention to smooth Fredholm maps (the derivative at each point is a Fredholm linear operator) of separable spaces. Quinn [32] noted that left-Fredholm and σ -proper is sufficient. (Kupka’s example is right-Fredholm with index +∞). It is customary [1, 32] to first prove Smale’s theorem, and then obtain the transversality theorem from that by a neat geometric argument. I have sinned against geometry in order to allow weaker hypotheses in the case of negative index. We will prove the transversality theorem directly, using a Liapunov-Schmidt argument (as in

Chapter 5. The Transversality Theorem

63

[37]) to reduce to finite dimensional problem, and then appealing to Sard’s theorem. Smale’s theorem is then an immediate corollary. We treat here only the case of regular values – see 5.10 for more general transversality. Theorem 5.4. Transversality Theorem. Suppose given positive integers k, m; Banach manifolds X, Y, Z of class C k ; an open set A ⊂ X × Y ; a C k map of f : A → Z ; and a point ζ ∈ Z . Assume for each (x, y) ∈ f −1 (ζ ) that: (1) ∂∂ xf (x, y) : Tx X → Tζ Z is semi-Fredholm with index < k. (2) Either  (α) D f (x, y) = or

∂f ∂f , ∂x ∂y

 : Tx X × Ty Y → Tζ Z is surjective

     ∂f ∂f (x, y) ≥ m + dim N (x, y) . (β) dim R(D f (x, y))/R ∂x ∂x

Further assume: (3) (x, y) → y : f −1 (ζ ) → Y is σ -proper, that is f −1 (ζ ) = ∞ j=1 M j is a countable union of sets M j such that (x, y) → u : M j → Y is a proper map, for each j. [Given (xν , yν ) ∈ M j such that {yν } converges in Y , there is a convergent subsequence (or subnet) of {(xν , yν )} with limit in M j .] We note that (3) holds if f −1 (ζ ) is Lindel¨of [every open cover has a countable subcover], or more specifically, if f −1 (ζ ) is a separable metric space or if X, Y are separable metric spaces. Y A y

(Ay, y) X

Let A y = {x | (x, y) ∈ A} and Y˙crit = {y | ζ

is a critical value of

f (·, y) : A y → Z }.

Then Ycrit is a meager set in Y and, if (x, y) → y : f −1 (ζ ) → Y is proper, Ycrit is also closed. If ind ∂ f /∂ x ≤ −m < 0 on f −1 (ζ ), then 2(α) implies 2 (β) and Ycrit = {y | ζ ∈ f (A y , y)}

64

Chapter 5. The Transversality Theorem

has codimension ≥ m in Y . [See (5.15) for definition of codimension; note Ycrit is meager iff codim Ycrit ≥ 1.] Remark. The usual hypothesis is that ζ is a regular value of f , so 2(α) always holds. If 2(β) holds at some point then index ∂ f /∂ x ≤ −m at this point, since     R(D f ) ∂f ≥ dim . codim R ∂x R(∂ f /∂ x) If ind (∂ f /∂ x) ≤ −m and 2(α) holds. Then 2(β) also holds. Thus 2(β) is more general for the case of negative index (and we’ll see examples with index −∞). It is customary to assume A = X × Y . Our generalization A ⊂ X × Y , while mathematically trivial, is very convenient in some problems. Terminological quibble. I have deliberately avoided the use of “residual”. Category I = “meager” is more or less accepted, and I would suggest “ample” for the complement. “Category II” is merely uninformative; “residual” is misleading. Lemma 5.5. Let k, m = 1, X, Y, Z , A, f, ζ be gives as above, (x0 , y0 ) ∈ f −1 (ζ ), and assume hypotheses (1) and (2) – (α) or (β) – hold at (x0 , y0 ). Then there exists open neighborhoods U of x0 , V of y0 , and an open dense subset V 0 ⊂ V , such that U × V ⊂ A and ζ is a regular value of f (·, y) | U whenever y ∈ V 0 . (If ind(∂ f /∂ x) < 0, this means ζ ∈ f (U × V 0 ).) Sublemma 5.6. Suppose f (x0 , y0 ) = ζ, ∂∂ xf (x0 , y0 ) is left-Fredholm and f is continuously differentiable on a neighborhood W of (x0 , y0 ). Then there is a neighborhood W of (x0 , y0 ) such that W ⊂ W0 and (x, y) → y : f −1 (ζ ) ∩ W → Y is proper. Proof of Sublemma. The result is local, so we may assume X, Y, Z are Banach spaces. Now L ≡ ∂∂ xf (x0 , y0 ) is left-Fredholm so X 1 = N (L) is finite dimensional and splits X = X 1 ⊕ X 2 , and the restriction of L is an isomorphism from X 2 onto R(L), with a continuous inverse (R(L)) is closed. There exists c0 > 0 so |L x2 | ≥ c0 |x2 | for all x2 ∈ X 2 . Also, if B = 1 +  ∂∂ yf (x0 , y0 ), there is a bounded neighborhood W of (x0 , y0 ), so small that W ⊂ W0 and | f (x, y) − f (x  , y  ) − L(x − x  )| ≤

c0 |x − x  | + B|y − y  | 2

for (x, y) and (x  , y  ) in W . Now suppose {(xn , yn )}n≥1 is a sequence in f −1 (ζ ) ∩ W such that {yn } converges. Then xn1 (= the components of xn in X 1 ) is bounded in a finite-dimensional space and has a convergent subsequence – we suppose, in fact, that {xn1 } converges.

Chapter 5. The Transversality Theorem

65

Now




c0 c0 xn2 − xm2 ≤ L xn2 − xm2 = |L(xn − xm )| ≤ |xn − xm | + B|yn − ym | 2 |yn − ym | → 0 as n, m → ∞. Thus {xn } conso |xn2 − xm2 | ≤ |xn1 − xm1 | + 2B c0 verges, which proves Sublemma 5.6. Next we show f −1 (ζ ) Lindel¨of implies (3). Indeed, by Sublemma 5.6, each point of f −1 (ζ ) has an open neighborhood W ⊂ W ⊂ A such that (x, y) → y : f −1 (ζ ) ∩ W → Y is proper. By hypothesis, there is a countable subcover −1 (ζ ) ∩ W j . { f −1 (ζ ) ∩ W j }∞ j=1 so (3) holds with M j = f Assuming the Lemma 5.5 and that (x, y) → y : f −1 (ζ ) → Y is proper, we prove {y ∈ Y | ζ is a critical value of f (·, y)} is closed and without interior. Let {yn }n≥1 be a sequence in this set which converges in Y ; for each n, there exists xn ∈ X so (xn , yn ) ∈ f −1 (ζ ) and xn is a critical point of f (·, yn ). By hypothesis, we may suppose (xn , yn ) → (x, y), so (x, y) ∈ f −1 (ζ ). If ∂∂ xf (x, y) were onto, then so would be ∂∂ xf (xn , yn ) for n large; hence x is a critical point of f (·, y) and closedness is proed. It remains to show that for each y ∈ Y , there exists y  arbitrarily close to y so ζ is a regular value of f (·, y  ). Let K y = {x| f (x, y) = ζ }; by the properness assumption, this is a compact set. By Lemma 5.5, for each x ∈ K y there are open sts Ux  x, Vx  y and Vx0 – an open dense subset of Vx – such that U x × V x ⊂ A and f (·, y  )|Ux has ζ as a regular value for all y  ∈ Vx0 . Choose a finite subcover Ux1 , . . . , Ux N for K y and   let U˜ = Ux1 | ∪ · · · ∪ Ux N , V˜ = 1N Vx j , V˜ 0 = 1N Vx01 ; V˜ 0 is open and dense in V˜ , V˜ and U˜ are open, y ∈ V˜ , K y ⊂ U˜ , and ζ is a regular value of f (·, y  )|U˜ when y  ∈ V˜ 0 . In fact, ζ is a regular value of f (·, y  ) for all y  ∈ V˜0 sufficiently close to y. Otherwise there would exist yn → y, yn ∈ V˜ 0 , and critical points xn of f (·, yn ) with (xn , yn ) ∈ f −1 (ζ ), such that limn→∞ xn = x exists; then (x, y) ∈ f −1 (ζ ), x ∈ K y , and xn ∈ U˜ , yn ∈ V˜ 0 for n large, so xn is not a critical point of f (·, yn ), a contradiction. If Lemma 5.5 and (3) hold, the same argument shows {y ∈ Y | there is a critical point x of f (·, y) with (x, y) ∈ M j ⊂ f −1 (ζ )} is closed and nowhere dense for each j = 1, 2, . . .; hence the union of these, {y ∈ Y | there is a critical point x of f (·, y) with (x, y) ∈ f −1 (ζ )} is meager. This completes the first step of the demonstration of the theorem. Now we show the case m > 1 of Theorem 5.4 may be reduced to the case m = 1, by change of variables. Suppose therefore m > 1, k = 1 and 2(β) holds and let X˜ = X × S m−1 , Y˜ = C 1 (S m−1 , Y ), A˜ = {(x, t; y˜ ) ∈ X˜ × Y˜ |(x, y˜ (t)) ∈ A} and f˜ : A˜ → Z (= Z˜ ) : (x, t; y˜ ) → f (x, y˜ (t)). Then f˜ is C 1 and the problem “tilde” satisfies the same hypotheses as the original problem, except that “m” is replaced by “1.” If 3(β) [or 3(γ )] holds for the original problem, it also holds for

66

Chapter 5. The Transversality Theorem

“tilde,” and we recall 3(α) ⇒ 3(γ ). If f (x, y) = ζ, y = y˜ (t), so f˜ (x, t; y˜ ) = ζ , q we choose a maximal subset {t˙1 , . . . , t˙q } ⊂ Tt (S m−1 ) so { ∂∂ yf y˜  (t)·t j } j=1 are independent relative to R(∂ f /∂ x); then dim N (∂ f /∂(x, t)) = dim N (∂ f /∂ x) + p m − 1 − q. Then we choose {η1 , . . . , η p } ⊂ Ty Y so { ∂∂ xf (x, y) y˙ k}k=1 are in∂f  ˜ dependent relative to R(∂ f /∂(x, t)) = R(∂ f /∂(x) + R( ∂ y y˜ (t)); by 2(β), we may assume p + q ≥ m + dim N (∂ f /∂ x), that is       R(D f˜ ) ∂ f˜ ∂f dim = 1 + dim N . ≥ p ≥ m − q + dim N ∂x ∂(x, t) R(∂ f˜ /∂(x, t)) Thus assuming the theorem is valid for m = 1 (the problem “tilde”), we see { y˜ ∈ C 1 (S m−1 , Y ) | ζ = f (x, y˜ (t))

for some x, t with (x, y˜ (t) ∈ A}

is a meager set in C1 (S m−1 , Y ), which means C 1 -codimen {y ∈ Y |ζ ∈ f (A y , y)} > m − 1, the C 1 -codimension is ≥ m and so the codimension is ≥ m. (See the Definition 5.15 and property (1) of codimension below.) It remains only to prove Lemma 5.5. Proof of Lemma 5.5. Since the result is local – near x0 , y0 ∈ X × Y and ζ ∈ Z – we may assume X, Y, Z are Banach spaces, x0 = 0, y0 = 0, ζ = 0, f is C k on a neighborhood of (0, 0) ∈ X × Y, f (x, y) = L x + M y + o(|x| + |y|), L is semi-Fredholm with index < k, and either (α) R(L , M) = {L x + M y| for all x, y} = Z or (β) dim{R(L , M)/R(L)} > dim N (L). Since (α) ⇒ (β) for negative index, it is enough to prove the result in case (α) when ind L ≥ 0 and in case (β) when ind L < 0. Case (α). ind L ≥ 0 (or L is Fredholm), R(L , M) = Z . L is Fredholm so X = X 1 ⊕ X 2 , Z = Z 1 ⊕ Z 2 , X 1 = N (L), Z 2 = R(L), L 2 ≡ L|X 2 : X 2 → Z 2 is an isomorphism, and ind L = dim X 1 − dim Z 1 . The complement Z 1 to R(L) is not unique and we may choose Z 1 ⊂ R(M). Then there is a subspace Y1 ⊂ Y so M1 ≡ M | Y1 : Y1 → Z 1 is an isomorphism; defining Y2 = M −1 Z 2 , it follows that Y = Y1 ⊕ Y2 . Writing f in terms of its components in these spaces, f : (x, y) = (x1 , x2 , y1 , y2 ) → (M1 y1 + g(x, y), L 2 x2 + h(x, y)) where g, h are C k and g, gx , g y , h, h x all vanish at (0,0), but perhaps h y = 0. By the implicit function theorem, we may solve f (x, y) = (0, 0) for y1 = φ(x1 , y2 ) and x2 = ψ(x1 , y2 ), with φ, ψ of class C k near (0,0). In matrix form, ∂∂ xf (x, y) is       gx2 gx1 1 p 1 0  0 = h x1 L 2 + h x2 0 1 q 1 0 L 2 + h x2 for appropriate operators p, q, where  = gx1 − gx2 (L 2 + h x2 )−1 h x1 . Then ∂ f /∂ x is surjective if and only if  = X 1 → Z 1 is surjective. Now by the

Chapter 5. The Transversality Theorem

67

definition of φ and ψ, we have


−1 M1 + g y1 − gx2 L 2 + h x2 h y1 φx1 +  = 0,

and the coefficient of φx1 is an isomorphism when we are close to (0,0). Thus in a neighborhood max(|y1 |, |y2 |) < δ of y = 0, 0 is a regular value of f (·, y)| max(|x1 |, |x2 |) < , with y = y1 + y2 , if and only if y1 is a regular value of φ(·, y2 )|{|x1 | ind L = dim X 1 − dim Y1 , Sard’s theorem says, for every small y2 (|y2 | < δ), there is a dense set of y1 in |y1 | < δ such that zero is a regular value of f (·, y1 + y2 ) on {|x1 | < , |x2 | < }. This proves denseness, while openness follows from the Sublemma 5.6. Case (β): indL < 0, k = 1, dim{R(L , M)/R(L)} > dim N (L). Let n = dim N (L). There exist { f 1 , . . . , f n+1 } in R(L , M) independent  c j f j ∈ R(L) implies all c j = 0. Then f j = L x j + relative to R(L), i.e., n+1 1 M y j where {y1 , . . . , yn+1 } are linearly independent in Y , a basis for a subspace Y1 ⊂ Y such that MY1 ∩ R(L) = {0} and M is injective on Y1 . Let Z 1 = MY1 and choose Z 2 ⊃ R(L) so that Z 1 ⊕ Z 2 = Z . Let Y2 = M −1 Z 2 ; then Y = Y1 ⊕ Y2 . Also let X 1 = N (L), X = X 1 ⊕ X 2 , so n = dim X 1 < dim Z 1 = dim Y1 . Now f has the form (x, y) = (x1 , x2 , y1 , y2 ) → (M1 y1 + g(x, y), L 2 x2 + h(x, y)) where M1 = M | Y1 : Y1 → Z 1 is an isomorphism, and L 2 = L | X 2 : X 2 → Z 2 is injective with closed image. Further, g, h are C 1 and at (0,0), g, gx , g y , h, h x all vanish. We may solve M1 y2 + g(x, y) = 0 for y1 = ψ(x, y2 ) a C 1 function with ψ = 0 and ψx = 0 at the origin. Choose small δ > 0. Fix y2 ∈ Y2 , |y2 | < δ, and let S(y2 ) = {x | x = (x1 , x2 ), |x1 | ≤ δ, |x2 | ≤ δ, f (x1 , x2 , ψ(x, y2 ), y2 ) = 0} Also let p1 : X 1 × X 2 → X 1 be the projection on X 1 and π (y2 ) = p1 |S(y2 ) : S(y2 ) → X 1 . If δ is small, π(y2 ) is injective, with Lipschitz inverse; assuming this, ψ(S(y2 ), y2 ) = ψ(·, y2 ) ◦ π(y2 )−1 (π(y2 )S(y2 )) ⊂ Y1 is the Lipschitz image of a set in X 1 , and dim X 1 < dim Y1 so ψ(S(y2 ), y2 ) has measure zero in Y1 . Thus given any |y1 | < δ, |y2 | < δ, there exists y1 arbitrarily close to y1 but outside ψ(S(y2 ), y2 ), hence f (x1 , x2 , y1 , y2 ) = 0 for all |x1 | < δ, |x2 | < δ. Openness follows from (5.6), so it only remains to show π (y2 ) has Lipschitz inverse.

68

Chapter 5. The Transversality Theorem

Now |L x2 | ≥ c0 |x2 | for all x2 ∈ X 2 and some constant c0 > 0. Since ¯ ≤ δ, |y2 | ≤ δ ψx (0, 0) = 0, for sufficiently small δ > 0 and |x| ≤ δ, |x| c0 ¯ ¯ ψ(x, ¯ y2 ), y2 ) − L(x − x)| ¯ ≤ |x − x|. | f (x, ψ(x, y2 ), y2 ) − f (x, 2 If also x, x¯ ∈ S(y2 ) then c0 c0 ¯ ≤ (|x1 − x¯1 | + |x2 − x¯2 |) ¯ ≤ |x − x| c0 |x2 − x¯2 | ≤ |L(x − x)| 2 2 ¯ |x − x| ¯ ≤ 2|π x − π x|, ¯ which comso |x2 − x¯2 | ≤ |x1 − x¯1 | = |π x − π x|, pletes the proof. Warning: The remainder of this chapter (except the “Fredholm facts” at the end) will not be used in the sequel. Proceed at your own risk. Example. As a simple application, suppose E is a Banach space and M ⊂ E is a C 2 compact submanifold, of dimension M. We show there is an open dense set in L(E, R2m+1 ) such that, for each L in this set, L|M : M → R2m+1 is an embedding: L x = L x  when x = x  in M and Lv = 0 when v = 0 in Tx M, x ∈ M. Openness is easily proved, since M is compact. Let M (2) = {(x, x  ) ∈ M × M|x = x  } and f : M (2) × L(E, R2n+1 ) → R2m+1 : (x, x  , L) → L(x − x  ). Then f is C 2 and (with X = M (2) , Y = L(E, R2m+1 ), ζ = 0 ∈ R2m+1 ), hypothesis (1) holds (index = 2m − (2m + 1) < 0), and also 2(α) holds. In fact, ˙ − x  ) | for all L} = R2m+1 so if f (x, x  , L) = 0, then x − x  = 0 and { L(x  ∗ D f (x, x , L) is surjective. If E is separable, then (3) holds; in fact (3) holds in very case (write M (2) as a countable union of compact sets). Thus {L ∈ L(E, R2m+1 ) | L|M is not injective} is a meager set, and also closed. Similarly, by consideration of (x, v, L) → Lv : (T M)0 × L(E, R2m+1 ) → R2m+1 , where (T M)0 = {(x, v) ∈ T M ⊂ E × E|v = 0}, we see {L ∈ L(E, R2m+1 ) | L|Tx M is not injective, for some x ∈ M} is a meager, hence, closed without interior. In the open dense complement of these sets, L|M is an embedding. Professor W. M. Oliva asked me if there is a corresponding result for C 1 manifolds. The short answer is No!, but a more extended response is in Corollary 5.12 below. If 1 > µ > 0, a C 1+µ m-dimensional submanifold of E may be embedded by “linear projection” in R N provided N is sufficiently large; but we must allow N → ∞ as µ → 0.

Chapter 5. The Transversality Theorem

69

So far, we have considered only regular values, a special case of transversality. To justify the title of this chapter, we treat briefly the general case, transversality to submanifold which need not be a single point. Definition 5.7. (1) Let F be a Banach space and F1 a closed linear subspace of F; we say F1 splits in F (or F1 has a closed complementary space) if there is a closed subspace F2 ⊂ F such that F1 ⊕ F2 = F, i.e., F1 + F2 = F and F1 ∩ F2 = {0}. (2) Let E, F be Banach spaces, F1 a closed subspace of F which splits, and L ∈ L(E, F) a continuous linear map. Then L is transverse to F1 (in F), L  F1 , if (i) L −1 (F1 ) splits in E and (ii) R(L) + F1 = F. (3) Let X, Y be C 1 Banach manifolds and Z ⊂ Y a C 1 submanifold [so q ∈ Z implies Tq Z splits in Tq Y ], and suppose f : X → Y is C 1 . Then f is transverse to Z at p ∈ X , f   Z if either F( p) ∈ / Z or f ( p) = q ∈ Z  Tq Z . We say f is transverse to Z, f   Z , if f   Z for every and f  ( p)  p ∈ X. Remark. This definition is equivalent to that of Lang and, what is less obvious, to that of Abraham and Robbin [1] and the notes of R. Palais. These authors require – instead of (2)(ii) for L   F1 – that the image of L contain a closed complement of F1 . This certainly implies 2(ii), and the following lemma (part (c)) shows that such a complement must exist under our definition. Lemma 5.8. (Eine kleine Linearwissenschaft) (1) Let F1 be a closed subspace of a Banach space F, if F1 is finite dimensional or has finite codimension, then F1 splits in F. If F is a Hilbert space, every closed linear subspace splits in F. (2) If every closed linear subspace of F splits in F, then F is a Hilbert space, i.e., there is an inner product in F which defines an equivalent norm. (3) Suppose F1 splits in F and F2 , F2 are closed complements: F = F1 ⊕ F2 = F1 ⊕ F2 . Then there is an isomorphism L : F → F such that L|F1 is the identity in F1 and L(F2 ) = F2 . (4) Let E, F be Banach spaces and F1 ⊂ F a closed splitting subspace. Then L ∈ L(E, F) is transverse to F1 if and only if there exist splittings E = E 1 ⊕ E 2 , F = F1 ⊕ F2 , such that the corresponding matrix form of L is       L 11 L 12 E1 F1 : → 0 L 22 E2 F2

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where L 22 : E 2 → F2 is an isomorphism. In this case, E 1 = L −1 (F1 ) and L(E 2 ) is a closed complement to F1 . If we choose F2 = L(E 2 ) then in the matrix form  F1 } is open in L(E, F). L 12 = 0. The set {L ∈ L(E, F)|L  Proof. (1) is straightforward and (2) is a theorem of Lindenstrauss and Tzafiri [19]. (3) Let P, P  ∈ L(E) be the projections onto F1 whose kernels are F2 , F2 respectively. Then for real t, P(t) = (1 − t)P + t P  is a projection onto F1 . Choose an integer N > 2P − P  (1 + P + P  ) and define P j = P( j/N ), W j = P j P j+1 + (I − P j )(I − P j+1 ). Then W j − I  < 1, W j is an isomorphism, W j |F1 is the identity in F1 , P j+1 = W j−1 P j W j , and L = W0 W1 · · · W N −1 is the desired isomorphism. In (4), suppose L has the indicated matrix form; then L −1 (F1 ) = E 1 splits  F1 . Conversely suppose L   F1 , F2 and R(L) + F1 = F1 + L(E 2 ) = F so L  −1 is a closed complement to F1 , E 1 = L (F1 ) and E 2 is a closed complement to E 1 . In the matrix form for L, L 21 : E 1 → F2 is zero since L E 1 ⊂ F1 ; L 22 is injective, since E 1 = L −1 (F1 ); and L 22 is surjective, since F1 + R(L) = F. It is easy to see that F1 ⊕ L(E 2 ) = F, and that L(E 2 ) is closed (a graph over F2 = L 22 E 2 ). Openness of transverality also follows easily in the matrix form. Remark. Transversality of a map to a submanifold, as in (5.7(3)) seems to be the most useful form, but it is enlightening to contemplate the more geometric version, transversality of two submanifolds. We cite only the linear case. If E, F are closed splitting subspaces of a Banach space G, then E, F have transversal intersection (E   F or F   E) when one of the following equivalent conditions holds: (i) the inclusion E ⊂ G is transverse to F; (ii) the inclusion F ⊂ G is transverse to E; (iii) E + F = G and E ∩ F splits in E or in F or in G, in which case it splits in all three. (iv) There are closed subspaces E 1 ⊂ E, F2 ⊂ F so that E = E 1 ⊕ (E ∩ F),

F = (E ∩ F) ⊕ F2

and G = E 1 ⊕ (E ∩ F) ⊕ F2 . F2

F

E∩F E E1

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71

When dim G < ∞, it is easy to see E + F = G implies (x + E) ∩ (y + F) is nonempty with the same dimension for all x, y, while E + F = G implies (x + E) ∩ (y + F) is empty for almost all x, y. Lemma 5.9. Let X, Y be C k Banach manifolds (k ≥ 1) and let Z be a C k submanifold of Y , and let f : X → Y be C k . Given q ∈ Z there exists an open neighborhood V of q in Y , a Banach space E and a C k map g : V → E such that 0 is regular value of g and g −1 (0) = V ∩ Z . If U is an open set in X such that F(U ) ⊂ V , then the restriction f |U : U → Y is transverse to Z if and only if 0 is a regular value of g ◦ f |U : U → E. Proof. This follows immediately from the definitions. We only recall that a submanifold of Z (following [1, 17]) is what is sometimes called a split-embedded submanifold, and at a regular point, the kernel of the derivative splits. The above lemma shows that transversality conditions reduce (locally) to conditions on regular values. The following theorem is, then, merely a translation of Theorem 5.4 to more general language. Theorem 5.10. Suppose given positive integers k, m; Banach manifolds X, Y, Z , W of class C k , with W a C k submanifold of Z 1 ; A an open set in X × Y ; and a C k map f : A → Z . Assume the following: (1) Either (α) W is finite dimensional and σ -compact, and for (x, y) ∈ f −1 (W ), ∂f (x, y) is semi-Freholdm with ind ∂∂ xf (x, y) + dim W < k; or ∂x (β) X is finite dimensional and σ -compact, W is σ -closed, and dim X − codim W < k. [W is σ -closed if Z is metrizable.] (2) For each (x, y) ∈ f −1 (W ), with q = f (x, y), either (α) R(D f (x, y)) + Tq W = Tq Z 1 or (β) dim{(R(D f (x, y)) + Tq W )/(R(∂ f /∂ x) + Tq W )} ≥ m + dim N ( ∂∂ xf (x, y)) + dim (Tq W ∩ R( ∂∂ xf (x, y))). (3) (x, y) → ( f (x, y), y) : f −1 (W ) → W × Y is σ -proper which is true, for example, if X is σ -compact and W σ -closed, or if f −1 (W ) is Lindel¨of. Then Ycrit ≡ {y ∈ Y | f (·, y) is not transverse to W } is meager in Y . Ycrit is closed if X is compact and W and f −1 (W ) are closed, or if W is compact and (x, y) → ( f (x, y), y) : f −1 (W ) → W × Y is proper. If dim W < ∞ and ind (∂ f /∂ x) + dim W ≤ −m on f −1 (W ), or if dim X < ∞ and dim X − codimen W ≤ −m, then 2(α) implies 2(β) and Ycrit = {y ∈ Y | f (A y , y) meets W } has codimension ≥ m. Remark. Theorem 5.4 is the case W = {ζ }. To obtain the semi-Fredholm property, it seems necessary to suppose at least one of W, X is finite dimensional.

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If both X and Z have finite dimension,  ind

∂f ∂x

 + dim W = dim X − dim Z + dim W = dim X − codim W.

Proof. The “σ -compactness” in (1) and “σ -properness” in (3) permit reduction to a local problem: given (x0 , y0 ) ∈ f −1 (W ), there are open neighborhoods U of x0 , V of y0 and an open dense V 0 ⊂ V such that U × V ⊂ A and f (·, y)|U  W for y ∈ V 0 . By Lemma 5.9, this local problem reduces to a problem on  regular values to which Theorem 5.4 applies. The details are omitted. Another variation on this theme, using the method rather than the result of Theorem 5.4, gives a “non-differentiable” transversality theorem. Theorem 5.11. Let Y, Z , W be C 1 Banach manifolds, W a σ -closed submanifold of Z , and let (X, d) be a metric space, σ -compact, with finite Hausdorff dimension dim X . Let A be an open set in X × Y and assume f : A → Z is locally of class C µ (0 < µ < 1). Suppose each point of f −1 (W ) has a neighborhood where y → f (x, y) is differentiable and (x, y) → ∂∂ yf (x, y) is continuous. Further suppose f (x, y) = z ∈ W implies dim

  

∂f R ∂ y (x, y) + Tz W Tz W

>

1 dim X. µ

 W for each x. This holds if µ > dim X/codim W and f (x, ·)  Then {y | f (X × {y} ∩ A) meets W } is a meager set in Y , and it is closed if X is compact and f −1 (W ) is closed, or if (x, y) → y : f −1 (W ) → Y is proper. Remark. Both dim X and the H¨older exponent µ depend on the metric chosen for X . Proof. We reduce to a local problem as before: each (x0 , y0 ) ∈ f (W ) has an open neighborhood U × V ⊂ X × Y , with U × V ⊂ A, and an open dense V 0 ⊂ V , such that f (U, V 0 ) does not meet W . For the local problem with X compact, we may suppose Y, Z , W are Banach spaces, y0 = 0, f (x0 , 0) = 0 and there is a subspace Y1 ⊂ Y such that MY1 ∩ W = {0} [M = ∂∂ yf (x0 , 0)], M|Y1 is injective and ∞ > dim Y1 > 1 dim X . There is a projection P of Z onto MY1 and a complement Y2 , Y = µ Y1 ⊕ Y2 . If f (x, y) ∈ W then P f (x, y) = 0 and near (x0 , 0) we may solve the last equation for y1 = ψ(x, y2 ), y = y1 + y2 , where ψ is C µ . It follows that dim ψ(X, y2 ) ≤ µ1 dim X < dim Y1 , so ψ(X, y2 ) has measure zero in Y1 . For any (small) y = y1 + y2 we may choose y1 arbitrarily near y1 so y1 ∈ ψ(X, y2 ), hence f (X, y1 + y2 ) does not meet W .

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73

onto

Example. Let Pm : [0, 1]−→[0, 1]m be a C 1/m Peano curve and let X = [0, 1], Y = Z = Rm , W = {0} and f (x, y) = Pm (x) − y. Then f is C µ with µ = m1 = dim X/codim W , but all other hypotheses are satisfied – in particular ∂f (x, y) is surjective. And the conclusion fails: 0 ∈ f ([0, 1], y) for all y ∈ ∂y [0, 1]m . Corollary 5.12. Let E be a finite or infinite-dimensional Banach space, 0 < µ < 1, and M ⊂ E a compact C 1+µ submanifold of dimension m. (We could allow M to have a boundary or corners.) If N ≥ 2m + 1 and N > m(1 + 1/µ) − 1, then there is an open dense set of L ∈ L(E, R N ) such that the restriction L|M : M → R N is an embedding. For each d = 2, 3, 4, . . .there exists a curve K d ⊂ Rd+1 , which is a submanifold of class C 1+1/d and dimension m = 1, such that; L|K d : K d → Rd is never a C 1 embedding for any L ∈ L(Rd+1 , Rd ), although it is a topological embedding for most L if d ≥ 3. Note d = m(1 + 1/µ) − 1 when m = 1, µ = 1/d. Proof. Openness is clear, and denseness reduces to a local problem: for each p ∈ M, there is a neighborhood U of p and an open dense S p ⊂ L(E, R N ) so L|M ∩ U is an embedding for L ∈ S p . Since N ≥ 2m + 1, L|M is injective for most L, as noted earlier. For the local problem, suppose p = 0, E = Rm × E 2 and M ∩ (neighborhood of 0) = {(ξ, ψ(ξ ) | ξ ∈ Rm near 0} where ψ : Rm → E 2 is C 1+µ near 0, ψ(0) = 0, ψ  (0) = 0. A linear map from Rm × E 2 to R N has the form (ξ, η) → Pξ + Qη (P ∈ R N ×m , Q ∈ L(E 2 , R N )) and we want P + Qψ  (ξ ) injective (rank m) for all (small) ξ , for most choices of P, Q. (Apply the theorem above to f : (ξ, P, Q) → P + Qψ  (ξ ) with X = {neighborhood of 0 in Rm }, Y = R N ×m × L(E 2 , R N ), Z = R N ×m and W = Wr the set of N × m matrices of rank r (for each 0 ≤ r ≤ m − 1). Note codim Wr = (m − r )(N − r ) ≥ N − m + 1 > m/µ. onto To construct the curves K d , first choose a C 1 map q : [0, 1]d −→S d and onto d let qd = q ◦ Pd : [0, 1]−→S , of class C 1/d . For any φ in C 1 ([0, 1], R), t let Vφ (t) = 0 φ(s)qd (s)ds, 0 ≤ t ≤ 1. Applying Theorem 5.4 to (t, t  , φ) →  Vφ (t) − Vφ (t )(t = t  ) we see Vφ (·) is injective for most choices of φ. Choose such φ with ϕ(t) > 0 for all t. Then t → Vφ (t) : [0, 1] → Rd+1 is a C 1+1/d embedding, whose image is K d ; and it is easy to show for each L : Rd+1 → Rd , there exists t ∈ (0, 1) with L dtd Vφ (t) = 0, so L|K d is not a C 1 embedding. Review of Fredholm and Semi-Fredholm Operators As a general reference for this topic, see Kato [14].

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Definition 5.13. Let X, Y be Banach spaces and T : X → Y a continuous linear operator. Then T is Fredholm [or semi-Fredholm] if R(T ) is closed and both [or, at least one] of dim N (T ), codim R(T ) are finite; in this case, the index of T is ind(T ) = dim N (T ) − codimR(T ). which is an integer or +∞ or −∞. A semi-Fredholm operator with finite index is Fredholm. A semi-Fredholm operator is left-Fredholm if its index is not +∞ (dim N (T ) finite), right-Fredholm if its index is not −∞ (codim R(T ) finite). Examples (1) If X, Y are finite dimensional, every linear operator T : X → Y is Fredholm with index = dim X − dim Y . In fact, suppose T has rank r ; then dim N (T ) = dim X − r , codim R(T ) = dim Y − r and ind(T ) does not depend on r or T . If at least one of X, Y is finite-dimensional, every T ∈ L(X, Y ) is semi-Fredholm and ind(T ) = dim X − dim Y . If both X, Y are infinite dimensional the zero operator is not semi-Fredholm, and any compact operator from X to Y is not semi-Fredholm. (2) Let X = Y = 2 , the usual sequence space. The identity in 2 (or any isomorphism) is Fredholm with index 0. The shift S : (x1 , x2 , . . . ) → (x2 , x3 , . . . ) is Fredholm with index 1 and S m is Fredholm with index m (m = 1, 2, 3, . . . ). The adjoint S ∗ : (x1 , x2 , . . . ) → (0, x2 , x2 , . . . ) is Fredholm with index −1 and (S ∗ )m has index −m(m = 1, 2, 3, . . . ). S n (S ∗ )m and (S ∗ )m S n have index n − m. L : (x1 , x2 , . . . ) → (x1 , 0, x2 , 0, x3 , 0, . . . ) is left-Fredholm with index −∞, and its adjoint L ∗ : (x1 , x2 , . . . ) → (x1 , x3 , x5 , . . . ) is right-Fredholm with index +∞. Note L ∗ L is the indentity but L L ∗ is not semi-Fredholm. (3) If there exists a Fredholm operator from X to Y with (finite) index m, then X × Rm is isomorphic to Y (if m ≤ 0) or X is isomorphic to Y × Rm (if m ≥ 0), so the spaces are “nearly the same.” Our familiar ∞-dimensional Banach spaces E have E × R isomorphic to E, but it is an open problem whether this holds generally. In any case, many properties (separable, reflexive, uniformly convex, . . . ) must be the same in both X and Y . 5.14. Fredholm Facts (1) T ∈ L(X, Y ) is semi-Fredholm iff T ∗ ∈ L(Y ∗ , X ∗ ) is semi-Fredholm, and in this case ind T ∗ = −ind T (see Example 2 above).

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75

(2) If T ∈ L(X, Y ) is semi-Fredholm, there exists δ > 0 such that, whenever S − T L(X,Y ) < δ, S is semi-Fredholm and ind S = ind T . The set of semi-Fredholm operators with given index is open in L(X, Y ). (3) If T ∈ L(X, Y ) is semi-Fredholm and K ∈ L(X, Y ) is compact, then T + K is semi-Fredholm and ind (T + K ) = ind T . (If T is an isomorphism, this result contains a substantial fraction of the Riesz-Schauder theory of compact operators.) (4) If T ∈ L(X, Y ) is semi-Fredholm, there exists F ∈ L(X, Y ) of finite rank such that, for all real t = 0, T + t F is surjective (if ind T ≥ 0) or injective (if ind T ≤ 0), or an isomorphism (if ind T = 0). (5) If T ∈ L(X, Y ), S ∈ L(Y, Z ) are both semi-Fredholm and their indices may be summed (excluding only the case ∞ − ∞), then ST ∈ L(X, Z ) is semi-Fredholm and ind(ST ) = ind S + ind T . In the case “∞ − ∞,” ST need not be semi-Fredholm, and if it is, the index may take any value in Z ∪ {+∞, −∞} (see Example 2 above). (6) For each finite i and integer j ≥ 0, the set   T ∈ L(X, Y ) | T Fredholm, ind T = 1 and       dim N (T ) = j (if i ≤ 0)   or     codim R(T ) = j (if i ≥ 0). is either empty or an analytic submanifold of L(X, Y ) with codimension j( j + |i|), and is an open set when j = 0. (7) If H1 , H2 are Hilbert spaces, the set of semi-Fredholm operators is open and dense in L(H1 , H2 ). In fact {T ∈ L(H1 , H2 ) | T is injective with closed image or T is surjective} = {T ∈ L(H1 , H2 ) | T has a left-inverse or a right-inverse} is open and dense. (See Halmos, A Hilbert Space Problem Book, #109.) The set of Fredholm operators is not dense unless both H1 and H2 are finite dimensional. (8) Let L j ∈ L(X, Yj ) ( j = 1, . . . , n) and suppose L 1 is left Fredholm. Then x → (L 1 x, . . . , L n x) is left-Fredholm, and it has index −∞ if some Y j ( j > 1) has infinite dimension. This simple fact is used frequently in Chapter 6. Codimension of Sets Definition 5.15. Let E be a topological Baire space. For any closed or σ -closed (= countable union of closed sets = Fσ ) subset F ⊂ E and integer m ≥ 0, codim F > m means {φ ∈ C(I m , E)|φ(I m ) meets F} is meager in C(I m , E).

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Chapter 5. The Transversality Theorem

Here I m = [0, 1]m , I 0 = {0}, and C(I m , E) has the obvious compact-open m topology. In place of I m , we could use (for example) B , the closed unit ball m+1 in Rm , or S m = ∂ B , which give equivalent conditions. Remark. If E is a Banach space or C k Banach manifold, we may also define C k -codim F > m to mean {φ ∈ C k (S m , E) | φ(S m ) meets F} is meager in C k (S m ; E). It is clear that C k -codim F > m implies C j -codim F > m for 0 ≤ j ≤ k, and C 0 -codim F > m is equivalent to codim F > m. Most of the properties below hold also for C k -codimension (in particular (7)), but we will use only C 0 -codim = codim. Note that, if F is closed, codim F > m is equivalent to saying {φ ∈ C k (S m , E) | φ(S m ) meets F} is nowhere dense in the C 0 topology; thus we need not worry about Peano curves. It is easy to prove codim(F1 ∪ F2 ∪ . . . ) > m iff codim F j > m for every j = 1, 2, . . . and m > 0, codim F > m imply codim F > m − 1. By the first, we need only consider closed sets F; and by the second, we may define codimension (not only the phrase “codim F > m”). Definition. (Continued) codim F = m means codim F ≯ m and either m = 0 or codim F > m − 1. codim F = ∞ means codim F > m for every m ≥ 0. Then for every closed or σ -closed F ⊂ E there is a well-defined m ∈ {0, 1, 2, . . . , ∞} such that codim F = m, and we call m the codimension of F. Obviously the codimension of F depends on E, so we sometimes specify “codimension of F in E.” 5.16. Simple Properties of Codimension (1) F1 ⊂ F2 implies codim F1 ≤ codim F2 . (2) codim F > 0 if and only if F is meager in E. (If E were not a Baire space, one could have codim E > 0.) (3) codim (F1 ∪ F2 ∪ . . . ) = min{codimF j | j = 1, 2, . . . , } (4) codim φ = ∞. If E = Rn , any nonempty F ⊂ E has codim F ≤ n. Antoine’s necklace [Hocking and Young, Topology, p. 177] is a compact totally-disconnected set in R3 , so it has topological dimension zero. But its codimension is not 3; R3 − (Antoine’s necklace) is not simply connected, so the codimension is ≤ 2, by (5) below. In fact the codimension equals 2 : it is greater than 1, since any continuous curve in R3 may be slightly perturbed so it misses the necklace, since it misses the (sufficiently small) approximating tori. (One may conjecture that generally dim F + codim F ≤ dim E, using topological dimension. What is the Hausdorff dimension? For a particular (economical) construction it is no more than 1+ log2 /log3 , which might be exact. If dim A > 1 we would

Chapter 5. The Transversality Theorem

77

have the amusing result top-dim A + codim A < dim E < H − dim A + codim A for A = Antoine’s necklace C E = R3 (5) Suppose E is a Banach space, F ⊂ E is closed; then codim F > 1 implies E\F is connected, codim F > 2 implies E\F is simply connected, codim F > m + 1 implies every map in C 0 (S m , E\E) is homotopic to a constant. (6) If h : E → E 1 is homeomorphism and F1 = h(F), then codim F (in E) = codim F1 (in E 1 ). (7) Suppose E is a C 1 Banach manifold and F is C 1 submanifold with finite codimension m in the usual sense, and F has a countable base. Then codim F = m according to our definition. The proofs are obvious, except possibly for (7). In (7), since F has a countable basis for its topology, we may work locally and (applying a C 1 diffeomorphism) reduce to the case: E = Rm × E 2 , F = {0} × F2 where F2 is the closure of a neighborhood of 0 in E 2 . Given any C 1 φ : S m−1 → E, we have φ = (φ1 , φ2 ), φ j : S m−1 → E j , and φ(S m−1 ) meets F only if 0 ∈ φ1 (S m−1 ) ⊂ Rm . But a C 1 -small perturbation φ˜ 1 of φ1 makes 0 ∈ φ˜ 1 (S m−1 ) [apply transversality to (t, ψ) → ψ(t)] hence C 1 -codim F > m − 1, so also codim F > m − 1 (see remark above). To prove the codimension is not greater than m, let φ(t) = (t, 0), φ : B m → E = Rm × E 2 . Any φ˜ = (φ˜ 1 , φ˜ 2 ) ∈ C(B m , E) uni˜ m ) ∩ F nonempty, i.e., φ˜ 1 (t) = 0 for some t; otherwise formly close to φ has φ(B we could construct a retraction of B m on ∂B m .

Interlude When condition (2.β) of Theorem 5.4 fails, we typically conclude that a certain complicated operator has finite rank. Chapters 7 and 8 provide different ways to obtain a more concrete condition (G(x, y) = 0, in certain Banach spaces) from the “finite-rank” hypothesis. Both methods are difficult, so we pause to show the importance of these results in a fairly general context. We start with some smooth f : X + Y → Z such that f x = ∂ f /∂ x is Fredholm, and hope to show 0 ∈ Z is a regular value of f (x, y) for most choices of y ∈ Y – at least, a dense set. We argue by contradiction: assume for every y near y0 ∈ Y , these exists x ∈ X with f (x, y) = 0 and f x (x, y) not onto. Then in fact D f (x, y) = ( f x , f y ) is not onto, by the transversality theorem, and we derive from this an additional infinite-dimensional condition ∂(x, y) = 0 ∈ W1 (dim W1 = ∞), hoping for a contradiction. But perhaps it is not contradictory, or not obviously so; we continue.

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For each y near y0 , there exists x so (x, y) is a critical point of f and ( f (x, y), g(x, y)) = (0, 0) ∈ Z × W1 . Since dim W1 = −∞, the x-derivative of ( f, g) is semi-Fredholm with index −∞, and condition (2.β) of Theorem 5.4 must fail. It follows that, for each y˙ ∈ Ty Y , there exists x˙ ∈ Tx X ˙ gx x˙ + g y y) ˙ ∈ E, for a fixed finite-dimensional subspace so that ( f x x˙ + f y y, E ⊂ T0 Z × T0 W1 . Since f x is Fredholm, it has an “inverse” [ f x−1 ], module finite rank operators. Then we can “solve” for x˙ = −[ f x−1 ] f y y˙ (module finite rank) and conclude g y − gx [ f x−1 ] f y : Ty y → T0 W1 has finite rank. The methods of Chapters 7 and 8 show that this (in typical problems) implies yet another infinite-dimensional condition : h(x, y) = 0 ∈ W2 (dim W2 = ∞). Thus for each y near y0 there is a solution X of the even-more-overdetermined problem: ( f (x, y), g(x, y), h(x, y)) = (0, 0, 0) ∈ Z × W1 × W2 , D F(x, y) not onto, If this is not yet clearly contradictory, we may repeat the process. Or we may get tired, return to the beginning and modify our hypotheses so that we do have a contradiction. Essential steps in this cyclic argument come from condition (2.β) of Theorem 5.4 and the methods of Chapters 7 and 8.

Chapter 6 Generic Perturbation of the Boundary

One of the compensations for the difficulty of PDEs, compared to ordinary differential equations, is that the regions in which we work can have almost any shape. This may seem only an added difficulty, but it turns into an advantage if we restrict attention to properties which are generic with respect to perturbation of the boundary. This is quite a strong condition, as the class of perturbations is infinite-dimensional (but it is finite-dimensional for ODEs). In many problems it is also reasonable to require “genericity” with respect to certain coefficients, etc. (See Uhlenbeck [42] and Saut and Teman [34] for some results of this kind.) But some problems are very rigid in this regard (e.g., the Navier-Stokes equation). Perturbation of the boundary is almost always reasonable, if we allow for occasional symmetry constraints as in Example 6.2 below. We will in any event perturb only the boundary – as a way of testing the strength of our tools, as a first step in more general “genericity” arguments, and for the intrinsic interest of the problems. In this chapter, “generic” properties are those that hold for most domains
, in the sense of Baire category, or for most domains in a certain stated class C (C m -regular, bounded, perhaps connected, perhaps contained in a certain open set or the boundary is contained in a certain open set, or the boundary meets a certain open set, or . . .). More precisely, given any region
0 of the class C, if h ∈ Diff m (
0 ) is outside a certain meager subset F ⊂ Diff m (
0 ) and if h(
0 ) is in the class C, then the property (supposed generic) holds. In all our applications, the excluded set of embeddings F is defined by properties of the image of the embedding, so it is invariant under composition with C m -diffeomorphisms
0 →
0 . This implies (though we don’t prove it here) that the image F(
0 ) in Micheletti’s metric space [22] of regions C m -diffeomorphisms to
0 is also meager, and in fact has the same codimension as F (in the sense of Definition 5.15).

79

80

Chapter 6. Generic Perturbation of the Boundary

Frank Morgan (“Measures on spaces of surfaces,” Arch. Rat. Mech. Anal. 78 (1982)) has defined a class of measures of smooth embeddings – generalizing Wiener measure for continuous curves – which induces a notion of “measurezero” for surfaces (images of the embeddings). He obtained interesting results holding “almost-everywhere” in this sense. Some work has been done on generic geometry without direct mention of PDEs – though many of the results are motivated by applications to PDEs. See, for example, “Generic caustics by reflection,” Amer. Math. Mo., V. I. Arnold, “Singularities of systems of rays,” Russ. Math. Surveys 38 (1983), V. M. Petkov and L. Stojanov, “Periodic geodesics of generic nonconvex domains,” Univ. Fed. Pernambuco, preprint (1985). Some of our results (as the lemma of Example 6.8) might also be classed as generic geometry. The main tool in this chapter is the transversality Theorem 5.4. Our first example was proved by Micheletti [21] and Uhlenbeck [41] and is also contained in Example 4.4 above; but it is a convenient example to introduce the use of transversality. Example 6.1. Generic Simplicity of Eigenvalues of the Dirichlet Problem for . We prove, for most C 2 regular bounded regions
⊂ Rn , that all eigenvalues λ of u + λu = 0 in
;

u = 0 on ∂
,

u ≡ 0

are simple. First suppose
0 is connected, as well as bounded and C 2 . Consider the map F : (u, λ, h) → h ∗ ( + λ)h ∗−1 u 
: H 2 ∩ H01 (
)\{0} × R × Diff2 (
0 ) → L 2 (
0 ). Suppose, for some such h, (u, λ) → F(u, λ, h) has zero as a regular value; then with
= h(
0 ), 0 ∈ L 2 (
) is a regular value of 
(v, λ) → v + λv : H 2 ∩ H01 (
)\{0} × R → L 2 (
). This says, whenever v + λv = 0, v = 0, then ˙ → ( + λ)v˙ + λv ˙ : H 2 ∩ H01 (
) × R → L 2 (
) (v, ˙ λ) is surjective, hence ( + λ) | H 2 ∩ H01 (
) has (closed) image of codimension ≤ 1, and in fact codimension = 1, since λ is an eigenvalue; therefore λ is a simple eigenvalue. Thus all eigenvalues of the Dirichlet problem in h(
0 ) are simple. Therefore it is sufficient to show zero is a regular value of F(·, ·, h)

Chapter 6. Generic Perturbation of the Boundary

81

for most h, or (since F is analytic, F(·, ·, h) is Fredholm for each h and the spaces are separable) to show zero is a regular value of F, verifying 2(α) of Theorem 5.4. This is possible, but raises some technical questions of smoothness which we may avoid by a slight reformulation of the problem. For each integer j (= 1, 2, 3, . . .), the set of h ∈ Diff2 (
0 ) such that all eigenvalues λ ≤ j are simple (for the problem in h(
0 )) is an open set in Diff2 (
0 ): there are only finitely many such eigenvalues and each depends continuously (analytically!) on h, according to Example 3.2. We show the set is also dense, for each j, and then taking the intersection over all j gives the result desired. Thus proof of denseness is the key step, and for this we may use stronger norms. By Theorem 1.8, we may approximate
0 (in the C 2 sense) by a C 3 (or C ∞ or analytic) region and then show denseness in the stronger C 3 topology. Suppose therefore
0 is in fact C 3 ; we show zero is a regular value of (u, λ, h) → F(u, λ, h) for h ∈ Diff 3 (
0 ). If that were false, there would exist a critical point (u 0 , λ0 , h 0 ) such that F(u 0 , λ0 , h 0 ) = 0. We transfer the 1 2 “origin” from
0 to
= h 0 (
0 ) defining v0 = h ∗−1 0 u 0 ∈ H ∩ H0 (
)\{0} and F˜ : (v, λ, h) → h ∗ ( + λ)h ∗−1 v
 : H 2 ∩ H01 (
)\{0} × R × Diff3 (
) → L 2 (
) ˜ λ, h) = h ∗−1 F(h ∗ v, λ, h ◦ h 0 ) has a critical point (v0 , λ0 , i
) with so F(v, 0 0 ˜ F(v0 , λ0 , i
) = 0 (recall h ∗0 is an isomorphism). (Such “changes of origin” may be made later without comment.) ˜ 0 , λ0 , i
) By hypothesis, the derivative D F(v ˙ → ( + λ0 )v˙ + [h˙ · ∇,  + λ0 ]v0 + λv ˙ h) ˙ 0 (v, ˙ λ, ˙ 0 = ( + λ0 )(v˙ − h˙ · ∇v0 ) + λv is not surjective. The image is closed in L 2 (
), since the image of {v˙ ∈ H 2 ∩ H01 (
), λ˙ = 0, h˙ = 0} is closed with finite codimension, so there ex˙ we see ists  ψ = 0 in L 2 (
) orthogonal  to the image. Keeping v˙ = 0,2 h = 0, 1 ˙ ψv = 0; keeping h = 0, ψ( + λ ) v ˙ = 0 for all v ˙ ∈ H ∩ H (
) says 0 0 0

ψ is a weak (hence, strong) solution of ( + λ0 )ψ = 0 in
, ψ = 0 on ∂
, so ψ ∈ H 2 ∩ H01 (
). In fact, since ∂
is C 3 , both v0 and ψ are in C 2,α (
) ∩ W 3, p (
) for any α < 1, p < ∞. Now allowing h˙ to vary in C 3 (
, Rn )

82

Chapter 6. Generic Perturbation of the Boundary

we have   0= ψ( + λ0 )h˙ · ∇v0 = {ψ( + λ0 )h˙ · ∇v0 − ( + λ0 )ψ h˙ · ∇v0 }
 
  ∂ψ ∂v0 ∂ψ ˙ ∂ ˙ h˙ · N = h · ∇v0 − h · ∇v0 = − ψ ∂N ∂N ∂N ∂N ∂
∂
for all h˙ ∈ C 3 (
, Rn ). [Note v0 ∈ W 3, p (
) so h˙ · ∇v0 ∈ W 2, p (
) − which is why we chose to work in a C 3 domain.] Therefore our hypothesis (which we hope to contradict) implies the existence of a non-zero C 2 solution ψ 0 = 0 everywhere of ( + λ0 )ψ = 0 in
, ψ = 0 on ∂
, such that ∂∂ψN ∂v ∂N on ∂
. 0 0 = 0; then ∂v = 0 on a neighborhood Suppose at some x0 ∈ ∂
, that ∂v ∂N ∂N ∂ψ of this point so ∂ N = 0 and ψ = 0 on a neighborhood (in ∂
) of x0 , while ψ + λψ = 0 in
. By uniqueness in the Cauchy problem for second-order elliptic equations (see [2] or [12, Th. 8.9.1], or the statement of the result at the end of this section), it follows ψ ≡ 0 in
, since
is connected; but this is false. 0 = 0 at every point of ∂
− but this implies v0 ≡ 0 in
, which Therefore ∂v ∂N is also false. Thus we have proved that zero is a regular value of F (restricted to C 3 deformations), so there is a C 3 -dense set of h ∈ Diff3 (
0 ) such that all eigenvalues of the Dirichlet problem in h(
0 ) are simple, and the argument is complete for connected (bounded, C 2 ) regions. Further, given a simple eigenvalue λ and corresponding eigenfunction u for the problem in
(connected), if V ∈ C 2 (
, Rn ) and
(t) = {x + t V (x) | x ∈
} for t near 0, there is unique eigenvalue λ(t) near λ of the problem in
(t), λ(0) = λ, and 
2  V · N
∂∂u  d ∂
N
 λ(t) = − , 2 dt t=0
u | does not vanish by Example 3.2. By uniqueness in the Cauchy problem, ∂∂u N
∂
2 n identically ( Q∩∂
( ∂∂u ) > 0 for any open Q ⊂ R which meets ∂
) so we N
can move the eigenvalue λ by perturbing
. In the general case, suppose
0 has several components
1 , . . . ,
N [finitely many, since
0 is bounded and C 2 -regular]. We may first perturb each
i (1 ≤ i ≤ N ) separately, to ensure all eigenvalues of the problem restricted to any
i and with modulus ≤ j, are simple; a further slight perturbation will separate any eigenvalues which occur simultaneously for two or more components. For an open dense set of embedding h ∈ Diff2 (
0 ), all eigenvalues λ ≤ j are simple ( j = 1, 2, . . .); taking the intersection, we see all eigenvalues are simple in most bounded C 2 regions.

Chapter 6. Generic Perturbation of the Boundary

83

Remark. We will often have occassion to cite the theorem on uniqueness in the Cauchy problem, so we state the result in more detail here. The proof is in H¨ormander [12, Theorem 8.9.1] and – more accessible, but assuming ai j are C 2 – in Agmon [2].

Uniqueness in the Cauchy Problem for Second-Order Elliptic Equations Suppose Q is a connected open set in Rn , B is a ball which meets ∂ Q in a C 2 hypersurface B ∩ ∂ Q, ai j = a ji : Q → R are C 1 functions with  2 n i, j ai j (x)ξi ξ j ≥ c0 |ξ | for x ∈ Q, ξ ∈ R and some constant c0 > 0. Assume 2 u ∈ H (Q)and for some constant K

B

  ∂ 2u  a (x) i j  ∂x ∂x i

j

   ≤ K (|∇u(x)| + |u(x)|) a.e. in Q 

and a.e. on ∂ Q ∩ B we have u = 0, ∂∂uN = 0. [More exactly,     ∂u 2  =0 |u|2 +  ∂N  ∂ Q∩B so, if we set u = 0 in B\Q, then u ∈ H 2 (B ∪ Q).] Then u = 0 a.e. in Q. Example 6.2. Simplicity of Eigenvalues with a Reflection-Symmetry Constraint. Sometimes one imposes geometrical symmetry conditions on the regions
, of the form: γ (
) =
for all γ ∈ , where is a given subgroup of the orthogonal group O(n). We treat here the simplest example, when = {I, J }, J = I, J 2 = I so J = J −1 = J . We may rotate coordinates so J has the form J = ( 0I p 0−I ) ( p + q = n, q > 0), a reflection. (If q is even, J q may also be considered a rotation.) We will prove, for most bounded C 2 regions
⊂ Rn such that J
=
, all eigenvalues of the Dirichlet problem for the Laplacian in
are simple, and so every eigenfunction φ is either even (φ ◦ J = φ) or odd (φ ◦ J = −φ). As was pointed out to me by A. L. Pereira, for a third-order symmetry (J 3 = I, J = I ), one may have double eigenvalues which cannot be split by perturbing the boundary unless the symmetry is violated.

84

Chapter 6. Generic Perturbation of the Boundary

J

J2 = I

J3 = I

To simplify, we assume
is connected. For the involution J (J 2 = I, J = I ) we may write any function f on
as f = f even + f odd where f even = 1 1 2 ( f + f ◦ J ) = f even ◦ J, f odd2= 2( f −2 f ◦ J) = 2− f odd ◦ J , and observe
(even). (odd) = 0 so
f =
f even +
f odd when f ∈ L 2 (
). Thus odd L 2 (
) = L even 2 (
) ⊕ L 2 (
) as an orthogonal sum, and the Laplacian preserves eveness or oddness. We restrict attention to one of these (even or odd) subspaces, and prove, within this subspace, all eigenvalues are generically simple; then we show how to split an eigenvalue which happens to occur simultaneously in both spaces (has both an even and odd eigenfunction). Lemma. Let J 2 = I, J = I, J (
) =
and let φ : ∂
→ R be a continuous even function (φ ◦ J = φ). Then there exists V : Rn → Rn of class C ∞ with V (J x) = J V (x) everywhere and such that V · N
| ∂
is uniformly arbitrarily close to φ· (If φ ∈ L p (∂
), 1 ≤ p < ∞, and is even, we may approximate in L p (∂
).) Proof. Choose W : Rn → Rn , C ∞ , uniformly close to φ N
on ∂
, and define V (x) = 12 (W (x) + J −1 W (J x)). Then V is C ∞ , V (J x) = J V x and on ∂
, V (x) is close to 12 φ(x)(N
(x) + J N
(J x)) = φ(x)N
(x)). Remark. The same “symmetrizing” argument shows any J -invariant region C 2 -close to
is of the form h(
) for some h near i
with h(J x) = J h(x) for all x ∈
. Let Diff2J
= {h ∈ Diff2 (
) | h(J x) = J h(x) for all x ∈
} so, for every such h, h(
) is also J -invariant. By the same calculation as in Example 6.1, above applied to
even × Diff2J (
) → L 2 (
)even , (u, λ, h) → h ∗ ( + λ)h ∗−1 u : H 2 ∩ H01 (
)  we must show the impossibility of ∂
h˙ · N
∂∂uN ∂∂ψN = 0 for all h˙ ∈ C 2 (
, Rn ) ˙ x) = J h(x), ˙ with h(J when u, ψ are both even eigenfunctions. Similarly in the ∂ψ is even on ∂
and, by the lemma, odd subspace. In either case the product ∂u N ∂N ∞ we may find C V so that V ◦ J = J · V and V · N
is close to sgn ( ∂∂uN ∂∂ψN ) (in L 1 (∂
)). But if h(t, x) = x + t V (x) for small real t, then h(t, ·) ∈ Diff2J
so V · N
= h˙ · N
| t=0 is orthogonal to ∂∂uN ∂∂ψN . It follows that ∂∂uN ∂∂ψN = 0 on

Chapter 6. Generic Perturbation of the Boundary

85

∂
, violating uniqueness in the Cauchy problem. Thus, in each subspace, the eigenvalues are generically simple. Now suppose λ0 is an eigenvalue in
for both even and odd subspaces, which is simple in each space separately. There are eigenfunctions   u 0 , v0 corresponding to λ0 , with u 0 ◦ J = u 0 , v0 ◦ J = −v0 ,
u 20 = 1 =
v02 . A small perturbation h(t, x) = x + t V (x) (with V ◦ J = J · V ) moves the eigenvalues to

 λeven (t) = λ0 − t  λodd (t) = λ0 − t

∂


∂


V ◦ N


V ◦ N


∂u 0 ∂N ∂v0 ∂N

2 + O(t 2 ) 2 + O(t 2 )

Thus we can split λ0 into two simple eigenvalues by (symmetric) perturbation of
, unless perhaps

∂u 0 ∂N



2 =

∂v0 ∂N

2 on ∂
.

   2 =
u 20 +
v02 = Let w ± = u 0 ± v0 ; w + , w − are eigenfunctions with
w ± + ∂w − 0 2 0 2 = ( ∂u ) − ( ∂v ) , which cannot vanish 2, w + = w − = 0 on ∂
, and ∂w ∂N ∂N ∂N ∂N a.e. in ∂
by uniqueness in the Cauchy problem. Thus, the eigenvalues may always be split into simple eigenvalues, by small (symmetric) perturbations of the boundary. For an open dense set of h ∈ Diff2J (
), all eigenvalues with modulus ≤ k for h(
) are simple (k = 1, 2, 3, . . .); so for an ample set of such h, all eigenvalues for h(
) are simple. Remark. According to Theorem 1.7, a J -invariant C m -regular region
may be represented in the form {x : φ(x) > 0} where φ is C m , φ(x) = 0 ⇒ |∇φ(x)| ≥ 1 and φ(J x) = φ(x) for all x. To approximate
by C ∞ (or analytic) J -invariant ˜ regions, it suffices to approximate φ by appropriately smooth ψ, define ψ(x) = 1 m ˜ (ψ(x) + ψ(J x)), and then {x : ψ(x) > 0} is a smooth J -invariant region C 2 near
. (Theorem 1.6). Thus, we could carry out the calculation above under the hypothesis that
is C m , for any large m; then conclude all eigenvalues with modulus ≤ k are simple for an open dense set of J -invariant C 2 -regular regions (k = 1, 2, . . .). Example 6.3. Simplicity of Real Eigenvalues of a General Second-Order Dirichlet Problem. We study elliptic operators of the form A(λ) =

n i, j=1

ai j (x)

n ∂2 ∂ + b j (x, λ) + c(x, λ) ∂ xi ∂ x j ∂ xj j=1

86

Chapter 6. Generic Perturbation of the Boundary

where ai j (x) = a ji (x) is C 2 , and b j (x, λ), c(x, λ) are polynomials in λ with C 2 coefficients, all real-valued when λ is real. We also assume that  2 n i, j ai j (x)ξi ξ j ≥ c0 |ξ | for all x, ξ in R and some constant c0 > 0. Given 2 n bounded C -regular
⊂ R , λ is an eigenvalue if there is a solution u ≡ 0 of A(λ)u = 0 in
,

u = 0 on ∂
;

λ is a simple eigenvalue if the kernal of A(λ) | H 2 ∩ H01 (
) is one-dimensional d A(λ))u, (so the image has codimension one) and the image does not contain ( dλ for u = 0 in the kernel. This definition ensures that simple eigenvalues may be studied without trouble using the implicit function theorem – as in the usual case b(x, λ) = b(x), c(x, λ) = c(x) + λ. Such polynomials occur, for example, when we seek exponential solutions u(t, x) = eλt v(x) ≡ 0 of the wave equation u tt + r (x)u t = u in
,

u = 0 on ∂
,

so v − (λ2 + λr (x))v = 0 in
, v = 0 on ∂
, v ≡ 0. It is important to know also the complex eigenvalues, but these are more difficult to handle; they will be discussed in Example 6.8 below. The important question is whether the coefficients are real or complex – thus in the example, if r (x) ≡ 0, we could treat λ on the imaginary axis by the methods of this section. (Example 4.4 treats self-adjoint problems, possibly complex-valued, with the more usual form of eigenvalue parameter.) Returning to the general problem above, let A
(λ) be the restriction of A(λ) to a map on H 2 ∩ H01 (
) into L 2 (
) (the real spaces, when λ is real). It is then easy to show zero is a regular value of (λ, u) → A
(λ)u : R × (H 2 ∩ H01 (
)\{0}) → L 2 (
) if and only if all real eigenvalues λ are simple. We will prove, for most bounded C 2 regions
⊂ Rn , that all real eigenvalues are simple, and that all simple real eigenvalues can be moved by small perturbations of
. Our hypotheses allow the case when A(λ) does not depend on λ – in that case, there can be no “simple eigenvalues,” so generically A
is an isomorphism. Suppose
is connected. Consider the map F : (u, λ, h) → h ∗ Ah(
) (λ)h ∗−1 (u) : (H 2 ∩ H01 (
)\{0}) × R × Diff2 (
) → L 2 (
), which is of class C 2 . For fixed h, (u, λ) → F(u, λ, h) is Fredholm with index 1 (index 0, if we keep λ fixed). The spaces are separable, and we only need to prove zero is a regular value of F. Suppose to the contrary that F has a critical point (u, λ, h) with F(u, λ, h) = 0; we may suppose h = i
. Then A
(λ)u = 0 in
, u = 0 on ∂
, u ≡ 0, and

Chapter 6. Generic Perturbation of the Boundary the derivative of F at (u, λ, i
) ˙ → A
(λ)(u˙ − h˙ · ∇u) + A
(λ)u λ˙ ˙ h) ˙ λ, (u,



A
(λ)

87

d A
(λ) = dλ

is not surjective, so there exists ψ = 0 in L 2 (
) orthogonal to the image. It  follows that ψ ∈ H 2 ∩ H01 (
), A∗
(λ)ψ = 0 in
,
ψ A
(λ)u = 0, and for all h˙ ∈ C 2 (
, Rn )   0= ψ A
(λ)h˙ · ∇u = {ψ A
(λ)h˙ · ∇u − A∗
(λ)ψ h˙ · ∇u}

∂ψ ∂u − ai j Ni N j h˙ · N = ∂ N ∂ N i, j ∂
hence ∂∂ψN ∂∂uN = 0 on ∂
. But this violates uniqueness in the Cauchy problem, so zero is not a critical value of F. Thus, generically in
, all real eigenvalues are simple. Now suppose λ is a simple eigenvalue for the problem in
. Let φ, ψ ∈ H 2 ∩ 1 H0 (
)\{0} with A
(λ)φ = 0, A∗
(λ)ψ = 0; φ, ψ are unique (up to constant real multiples) and
ψ A
(λ)φ = 0, by definition of “simplicity.” For any C 2 V : Rn → Rn , let h(t, x) = x + t V (x),
(t) = h(t,
) for small real t. A slight variant of the argument in Example 3.2 (or 3.6) shows there is a unique eigenvalue λ(t) near λ for the problem in
(t), λ(t) is a simple eigenvalue and differentiable in t and     − ∂
V · N
( ai j Ni N j ) ∂∂φN ∂∂ψN d   λ(t) = .  dt t=0
ψ A
(λ)φ Since ∂∂φN ∂∂ψN |∂
≡ 0 (in fact, it is nonzero a.e. in ∂
) we may choose V so the derivative is nonzero. Thus every simple eigenvalue can be moved by small perturbations of ∂
, and indeed by small perturbations restricted to a small neighborhood of a given point of ∂
. So what’s wrong with complex coefficients? The problem with complex coefficients is that we cannot work entirely in the complex field: the embeddings h are necessarily real. If we take λ, u in complex spaces, the image of the derivative of F will be a real-linear subspace of L 2 (
, C), but need not be linear over the complex field. Following the ˙ =0 ˙ h) ˙ λ, argument above, there exists ψ = 0 so that Re
ψ D F(u, λ, i
)(u, ∂ψ ∂u ˙ and we wind up with the condition: Re ( ˙ h) ˙ λ, ) = 0 on ∂
. It for all (u, ∂N ∂N is not at all clear that this is impossible. We return to the complex case later (Example 6.8). Although complex coefficients cause trouble, we can easily treat more general real dependence on λ. Examining the argument above, we see that when 

88

Chapter 6. Generic Perturbation of the Boundary

is a C m+1 m-dimensional manifold and (x, λ) → b j (x, λ), c(x, λ) : Rn ×  → R are C m+1 , then for most bounded C m+1 -regular regions
⊂ Rn , (u, λ) → A
(λ)u : (H 2 ∩ H01 (
)\{0}) ×  → L 2 (
) has zero as a regular value. Thus whenever A
(λ)u = 0 for some u = 0, it follows that A
(λ) (H 2 ∩ H01 (
)) + d A
(λ)u)Tλ  = L 2 (
). This implies dim N (A
(λ)) ≤ m = dim , and ( dλ provides at least a first step in the study of multi-parameter eigenvalue problems. Example 6.4. Generic Simplicity of Eigenvalues for the Neumann Problem for . The Neumann eigenvalue problem in
⊂ Rn is u + λu = 0 in
,

∂u = 0 on ∂
, ∂N

u ≡ 0

and always has the eigenvalue λ = 0. The corresponding eigenfunctions are constant in each component of
, and the multiplicity of the zero eigenvalue is the number of components of
. We prove that, for most bounded C 2 -regular regions
⊂ Rn , all non-zero eigenvalues are simple, and may be moved by small perturbations of
. As usual, it is enough to do this for connected regions
. Given a C 2 -regular bounded region
, the Neumann problem in
has all non-zero eigenvalues simple if and only if (0, 0) is a regular value of  

∂u  F
: (u, λ) → u + λu, ∂ N ∂
: (H 2 (
)\{0}) × (R\{0}) → L 2 (
) × H 1/2 (∂
). Suppose in fact
is C 3 -regular connected and bounded and consider the map G : (u, λ, h) → h ∗ Fh(
) h ∗−1 (u, λ) with F as above and h ∈ Diff3 (
), G is an analytic map of separable spaces, (u, λ) → G(u, λ, h) is Fredholm (index 1) for each H , and we wish to prove (0, 0) is a regular value of G. Suppose to the contrary there is a critical point (u, λ, h) of G with G(u, λ, h) = (0, 0); we may suppose h = i
. Then u + λu = 0 in
, ∂u/∂ N = 0 on ∂
, u ≡ 0, λ = 0 and there exists non-zero (ψ, θ) ∈ L 2 (
) × H −1/2 (∂
) orthogonal to the image of the derivative.  ˙ ψ{( + λ)(u˙ − h˙ · ∇u ) + λu} 0=




 ∂u ∂ ˙ ˙ ˙ (u˙ − h · ∇u) + N · ∇u + h · ∇ + θ, ∂N ∂ N ∂
where ·, ·∂
indicates the duality of H −1/2 (∂
) × H 1/2 (∂
). Keeping h˙ = 0, λ˙ = 0, it follows that ψ ∈ H 2 (
), ψ + λψ = 0 in
, ∂ψ/∂ N = 0 on ∂


Chapter 6. Generic Perturbation of the Boundary 89   and θ, g∂
= − ∂
ψg for all g ∈ H 1/2 (∂
). As λ˙ varies, we see
ψu = 0; and when h˙ varies, we find (supposing, as we may, ∂ N /∂ N = 0 on ∂
) 



∂ 2u 0= ψ −∇∂
σ · ∇u + σ ∂N2 ∂


 for all

σ = h˙ · N ∈ C 3 (∂
)

since N˙ = −∇∂
σ (2.3). Now ∂
is C 3 so, u, ψ are in the fact C 2,α functions for any α < 1, and −ψ∇∂
σ · ∇u = −div∂
(σ ψ∇∂
u) + σ (∇∂
ψ · ∇∂
u + ψ∂
u) so 0 = ∇∂
ψ · ∇∂
u + ψ(∂
u + ∂ 2 u/∂ N 2 ) on ∂
. Since ∂u/∂ N = 0 on ∂
and u = −λu, this (with 1.12) says ∇∂
ψ · ∇∂
u = λuψ on ∂
, which we show is impossible. By uniqueness in the Cauchy problem uψ = 0 on a dense open set of ∂
; choose some x0 ∈ ∂
where uψ = 0. Define a curve t → x(t) ∈ ∂
by x(0) = x0 , d x/dt = ∇∂
u(x); ∇∂
u(·) is a C 1 tangent-vector field on ∂
, and ∂
is compact, so this curve is defined on −∞ < t < ∞. Further, dtd u(x(t)) =  t |∇∂
u(x)|2 ≥ 0 and ψ(x(t)) = ψ(x0 ) exp(λ 0 u(x(s))ds)  t for all t. If u(x0 ) > 0 then u(x(t)) ≥ u(x0 ) > 0 for all t ≥ 0, and λ > 0, so λ 0 u(x(s))ds → +∞ as t → +∞. This is impossible, since ψ is bounded. But if u(x0 ) < 0, the same argument applies at t → −∞; in any case, we have a contradiction. Uhlenbeck [41] gives a similar argument. Thus (0, 0) is a regular value of G, and in most C 3 (bounded, connected) regions
⊂ Rn , all eigenvalues of the Neumann problem are simple. The set of (bounded, connected) C 2 -regions
, such that the eigenvalues of modulus ≤ k are simple, is open (by Example 3.5) and dense (by the C 3 case). This holds for each k = 1, 2, 3, . . . , and taking the intersection gives the general (connected) case.  If λ is a non-zero simple eigenvalue with eigenfunction u,
u 2 = 1, then according to Example 3.5 we can move λ by small perturbations of
, except possibly in the case where |∇∂
u|2 = λu 2 everywhere on ∂
. In this case, let u(x0 ) = max u | ∂
; then ∇∂
u(x0 ) = 0 so u(x0 ) = 0 and similarly min u | ∂
= 0, so u ≡ 0 on ∂
, u ≡ 0 in
, which false. (Since ∂
is C 2 , u ∈ W 2, p (
) for any p < ∞ so u is at least C 1 .) Example 6.5. Generic Simplicity of Solutions of u + f (x, y, ∇u) = 0 in
, u ≥ 0 on ∂
. Let Q be an open set in Rn × R × Rn and let f : Q → R be locally C 2 . We prove, for most bounded C 2 -regular regions
⊂ Rn , all solutions u of u + f (x, u, grad u) = 0 in
,

u = 0 on ∂


90

Chapter 6. Generic Perturbation of the Boundary

are simple, i.e., the linearization ∂f ∂f (·, u, ∇u) · ∇φ + (·, u, ∇u)φ ∂p ∂u : H 2 ∩ H01 (
) → L 2 (
)

φ → φ +

is an isomorphism. (One of the requirements for a C 1 function u to be a solution ¯ It follows that the set of soluis that (x, u(x), grad u(x)) ∈ Q for all x ∈
). tions is (generically) discrete, and if f is bounded (for example), this set is finite. Thus u + sin u = 0, in
, u = 0 on ∂
, has only finitely many solutions for most
. The same result holds with any uniformly elliptic second-order operator in place of the Laplacian. Saut and Teman [34] proves this for u + f (x, u) = 0, under the hypothesis – inadvertantly omitted from [34] – that f (x, 0) ≡ 0. (The author had independently proved this case). The general form stated above seems to require the extension of the transversality theorem given in Chapter 5 (Theorem 5.4.). Indeed this is one of the examples which motivated study of this generalization. We prove the result first under the hypothesis that f is everywhere defined and bounded (as well as locally C 2 ), and
is C 3 regular and connected. Consider the map F : (u, h) → h ∗ ( + f )h ∗−1 (u) : W 2, p ∩ W0 (
) × Diff3 (
) → L p (
) 1, p

for some p in n < p < ∞ (so W 2, p (
) ⊂ C 1 (
)). If zero is a regular value of u → F(u, h), it is easily shown that all solutions of the problem in h(
) are simple. As usual, the only difficulty is to know whether zero is a regular value of F. We will try to prove this by contradiction – and fail. For certain f and
, zero may be a critical value – but then the critical point has very special properties. Using conditions 2(β) of Theorem 5.4, we show there is an open dense subset D0 ⊂ Diff 3 (
) such that the restriction of F to D0 has zero as a regular value, and then there is an open dense D1 ⊂ D0 such that, if h ∈ D1 , the problem in h(
) has only finitely-many simple solutions. The usual tricks then give the general result. Suppose now (u, i
) is a critical point of F with F(u, i
) = 0. There exists ψ = 0 in L p (
) (1/ p + 1/ p  = 1) orthogonal to the image of the derivative,  1, p ψ L(u˙ − h˙ · ∇u) = 0 for all u˙ ∈ W 2, p ∩ W0 (
), h˙ ∈ C 3 (
)


 where L =  + n1 ∂∂pfk (·, u(·), ∇u(·)) ∂∂xk + ∂∂uf (·, u(·), ∇u(·)). Since ∂
is C 3 , u is C 2,α for any α < 1, so the coefficients of L are C 1 . Fixing h˙ = 0, we see L ∗ ψ = 0 in
, ψ = 0 on ∂
and ψ ∈ W 2,q (
) for any q < ∞. Allowing

Chapter 6. Generic Perturbation of the Boundary 91  ˙ so ∂ψ ∂u ≡ 0 on ∂
. Since h˙ to vary, we see 0 = ∂
h˙ · N ∂∂ψN ∂∂uN for all h, ∂N ∂N ∂u ψ ≡ 0 it follows that ∂ N ≡ 0 on ∂
. If f (x, 0, 0) ≡ 0, it follows by uniqueness in the Cauchy problem that u ≡ 0. But the zero solution may be studied separately; by Example 6.3 (in the case where the coefficients don’t depend on λ), for most regions
, the zero solution 1, p is simple. Applying the above argument for u ∈ W 2, p ∩ W0 (
)\{0}, we see th nonzero solutions are also generically simple. (This is the case treated in [34].) More generally, suppose f (x, 0, 0) vanishes on some neighborhood of ∂
; then also u vanishes near ∂
, so we have a special solution u = u 0 which does not depend on
(for small perturbations), and the same argument applies. u 0 1, p is generically simple, by Example 6.3, and solutions in W 2, p ∩ W0 (
)\{u 0 } are generically simple. It is easy to find examples u, f,
, such that f (x, 0, 0) = 0 on a dense set and u is a solution with u = 0, ∂u/∂ N = 0 on ∂
. (Hint: First choose u.) But we want to show this is an unusual property of
, given f . For an open dense set of (bounded, connected, C 3 -regular) regions
, either f (x, 0, 0) ≡ 0 on a neighborhood of ∂
, or f (x, 0, 0) = 0 at some point of ∂
. The first case is treated above, and in second case we show there is an open dense subset of a C 3 -neighborhood of i
such that, for any h in this set, there is no solution u of the problem in h(
) with both u = 0 and ∂u/∂ N = 0 on ∂h(
). (We assume here n ≥ 2; the case n = 1 is easy, but requires separate treatment.) When n ≥ 2, the excluded set of embeddings has infinite codimension. Choose δ0 > 0 so h − i
C 3 < δ0 implies h is an embedding and f (x, 0, 0) = 0 somewhere on ∂h(
). Define G : (h, u) → h ∗ ( + f )h ∗−1 (u) : BδC0 (i
) × W0 (
) → L p (
). 3

2, p

2, p

(Note u ∈ W 2, p (
) is in W0 (
) if and only if u = 0 and ∂u/∂ N = 0 on ∂
.) Now G is of class C 1 and G(·, h) is semi-Fredholm with index −∞ for each h. We verify condition 2(β) of Theorem 5.4 at (u, i
), supposing G(u, i
) = 2, p ∂ G(u, i
) = L|W0 (
) is injective, by uniqueness in the 0. Observe that ∂u Cauchy problem, though we don’t use this fact. Let us suppose

  ∂ G(u, i
) 0). ∂ G(u, i
)) is infinite-dimensional and condition Thus R(DG(u, i
))/R( ∂u 2(β) holds. (It holds at any point (u, h) ∈ G −1 (0), by usual change of origin.) It is also easy to verify that (u, h) → h : G −1 (0) → Bδ0 is proper (recall f is bounded). Thus excluding only a closed subset of infinite codimension, for any 2, p h with h − i
C 3 < δ0 , G(u, h) = 0 for all u ∈ W0 (
). Thus there is an open dense set of h ∈ Diff3 (
)such that: either f (x, 0, 0) ≡ 2, p 0 on a neighborhood of ∂h(
) and the solution (if any) in W0 (h(
)) is simple, or f (x, 0, 0) = 0 somewhere on ∂h(
) and the there are no solutions in 2, p W0 (h(
)). This is the set D0 claimed above, so the proof (for n ≥ 2,
of class C 3 and f bounded) is complete. With f bounded, the set of C 3 -regular regions having all solutions simple, is open (by the implicit function theorem: there are finitely many solutions) and dense, since we may approximate by C 3 regions. Now suppose n = 1: u  + f (x, u, u  ) = 0

in

a < x < b,

u(a) = u(b) = 0.

Perturbation of
= (a, b) means merely changing a and b, but the argument above applies until the last step. We must still show there are no solutions u with u  (a) = 0, u  (b) = 0, for most choices a, b. Let U (x; a) be the solution

Chapter 6. Generic Perturbation of the Boundary

93

of Ux x + f (x, U, Ux ) = 0 with U = 0, Ux = 0 when x = a. Assume for some a < b that f (b, 0, 0) = 0 and U (b; a) = 0, Ux (b; a) = 0. Since Ux x (b; a) = − f (b, 0, 0) = 0 we may solve Ux (x; α) = 0 for x = β(α) near b when α is near 2, p a. Then there are no solutions of our problem in W0 (α, β) (α near a, β near b) unless β = β(α) and U (β(α); α) = 0, a condition at least of codimension 1. Thus we have the result when f is bounded and defined everywhere. For the general case, choose open sets Q 1 ⊂ Q 2 ⊂ · · · ⊂ Q whose union is Q with Q j compact in Q j+1 , and choose f j ∈ C 2 (Rn × R × Rn ) such that f i = f on Q j . For each f j there is an open-dense set D j ⊂ Diff2 (
) of “good” embeddings. Let h ∈ ∩∞ 1 D j and let u be any solution of our nonlinear Dirichlet problem in h(
); {(x, u(x), grad u(x)) | x ∈
} is a compact subset of Q, so it is contained in some Q j . Thus u also satisfies the problem with f j in place of f , and the linearization L j is an isomorphism (h ∈ D j ). But L j = L, and the proof is complete.

Interlude: On the Need for Chapter 7 The last example is the simplest case of a problem we encounter with full force in the succeeding examples. We wish to apply the transversality theorem to some f (x, y), with ∂∂ xf (x, y) Fredholm, by showing 0 is a regular value of f . We suppose, for contradiction, that there exists (x, y) with f (x, y) = 0 and D f (x, y) not surjective, and deduce some other infinite dimensional condition g(x, y) = 0, hoping for a contradiction. But it is not contradictory, or at least not obviously so; we continue. Hoping to show it is unusual, if not contradictory, we apply transversality again to (x, y) → ( f (x, y), g(x, y)). Now we have index −∞ (see 5.14, #8). Trying to verify 2(β) of Theorem 5.4 by contradiction, we suppose the quotient space has finite dimension n 0 , so for appropriate {θ1 , . . . , θn 0 }: For any y˙ there exist x˙ and constants c1 , . . . , cn 0 such that n0 ˙ gx x) ˙ + ( f y y˙ , g y y˙ ) = cjθj. ( f x x, 1

The operator f x is Fredholm, so we can almost solve the first component   n0 
c j θ 1j , θ j = θ 1j , θ 2j f x x˙ = − f y y˙ + 1

˙ modulo operators of finite rank; then substitute in the second component for x, to conclude   : y˙ → g y y˙ − gx f x−1 f y y˙ is of finite rank. (The “inverse” should not be taken too seriously.) In the last example, we could reduce this to a simple form; but in general is a complicated abstract gadget which might (who knows?) be of finite rank. To show it is

94

Chapter 6. Generic Perturbation of the Boundary

not of finite rank to obtain the longed-for contradiction, we need quite concrete representation of (especially, [ f x−1 ]), and this is the purpose of Chapter 7. In the remainder of this chapter, we merely quote the results needed from Chapter 7, hopefully fortifying our spirits for the trials to come. Example 6.6. Generic Simplicity of Eigenvalues of Robin’s Problem. Given real-valued c(x), β(x) of class C 2 , C 3 respectively on Rn , consider the eigenvalue problem (∗)


u + (λ + c(x))u = 0 in (
),

∂u + β(x)u = 0 on ∂
, ∂N

u ≡ 0

for bounded C 2 -regions
⊂ Rn . In the special case β ≡ 0, c ≡ constant (Neumann), λ = −c is an eigenvalue for any choice of
, with multiplicity equal to the number of components of
, and locally-constant eigenfunctions. All other eigenvalues are simple for most choices of
, and move when we perturb
: Example 6.4. We will prove, for some bounded C 2 -regions
, all eigenvalues of (∗)
are simple, and may be moved by perturbation of
, except possibly in the case; β ≡ 0 on a neighborhood of
and c ≡ constant on a neighborhood of ∂
i , where
i is some component of
. In the exceptional case, let c0 be the constant value of c; then λ = −c0 may be an eigenvalue which does not move when
is slightly perturbed; whose eigenfunctions are locally constant on a neighborhood of ∂
i (if
i is a component of
such that c(x) ≡ c0 near ∂
i ) and vanish in any component
k such that c(x) ≡ c0 on a neighborhood of ∂
k ; and the multiplicity of λ = −c0 is at most the number of components of the first kind. Any eigenvalues not of this special type are simple (generically) and move when we perturb ∂
. We will assume
is connected; then (generically) all eigenvalues are simple, and all may be moved by small perturbation of
expect perhaps λ = −c0 in case c ≡ c0 and β ≡ 0 on a neighborhood of ∂
with an eigenfunction locally constant near ∂
.

Ω

0 12

3

Chapter 6. Generic Perturbation of the Boundary

95

Example. Choose 0 < R0 < R1 < R2 < R3 with J0 (R1 ) > 0 > J0 (R2 ), J0 is the zero-order Bessel function. Choose φ(t) of class C ∞ so that φ(t) = J0 (t) on R1 ≤ t ≤ R2 , φ(t) = 0 outside [R1 , R2 ], φ(t) = 1 for t ≤ R0 , φ(t) = −1 for t ≥ R3 . Let u(x) = φ(|x|), β(x) ≡ 0, c(x) = −u(x)/u(x) for x ∈ R2 , so u, c are C ∞ and c(x) has support in {R0 ≤ | x| ≤ R1 } ∪ {R2 ≤ |x| ≤ R3 }. For any annular region
⊂ R2 with inner boundary in {|x| < R0 } and outer boundary in {|x| > R3 }, zero is an eigenvalue with eigenfunction u(x) which is locally constant (but not constant) near ∂
, and also not of constant sign. For most such
, all eigenvalues (including 0) are simple. If J0 has k roots in (R1 , R2 ), then zero is at least the (k + 1)st eigenvalues. Each of the following conditions is open with respect to perturbation of
: (i) β(x) = 0 somewhere on ∂
; (ii) β(x) ≡ 0 on a neighborhood of ∂
, and c | ∂
is not constant; (iii) β(x) ≡ 0 and c(x) ≡ constant on a neighborhood of ∂
. [We assume
is connected, but ∂
may not be connected.] Further, given any (bounded, connected, C 2 -regular) region
0 ⊂ Rn , some arbitrarily small C 2 perturbation
of
0 will satisfy at least one of these conditions. We may thus assume we are in one of these cases, and restrict ourselves to small perturbations so we always remain in that case. In this, as in most genericity arguments, the hard part is to prove denseness. Assume for contradiction that there is a bounded connected C 2 region
0 , satisfying one of (i), (ii) or (iii) above, such that, for every
C 2 -close to
0 , (∗)
has a multiple eigenvalue. We will show we must be in Case (iii), and then examine this case more closely. Choose such
(close to
0 ) and define
F(u, λ, h) = h ∗ ( + λ + c)h ∗−1 u,  h ∗ (Nh(
) · ∇ + β)h ∗−1 u ∈ L 2 (
) × H 1/2 (∂
) for u = 0 in H 2 (
), λ ∈ R, and h near i
in C 2 (
, Rn ). If (u, λ) → F(u, λ, h) has (0, 0) as a regular value, all eigenvalues λ of (∗)h(
) would be simple – which is false when h is close to i
. By the transversality theorem, (0, 0) must be a critical value of F and in fact there is a solution (u, λ) of (∗)
with u = 0 so (u, λ, i
) is a critical point of F. It follows – as in Example 6.4 – that there exists ψ = 0 in H 2 (
) which also satisfies (∗)
for the same λ and further (∗∗)
∇∂
ψ · ∇∂
u = ψu(λ + c + β 2 − β div N
− ∂β/∂ N
) on ∂
 and
ψu = 0. [We will not use this last condition; the argument below only says (λ, u) and (λ, ψ) satisfy (∗)
with u ≡ 0, ψ ≡ 0, and (λ, u, ψ) satisfies (∗∗)
.]

96

Chapter 6. Generic Perturbation of the Boundary

If the coefficient of ψu in (∗∗)
is always positive on ∂
– for example, if λ is large – the argument of Example 6.4 applies to give a contradiction. But in general, (∗∗)
is not contradictory. Example.
is the unit ball of Rn , n ≥ 2, u(x) = x1 , ψ(x) = x2 , λ = 0, c ≡ 0, and on ∂
, β = −1 and ∂β/∂ N = n + 1; then (∗∗)
holds. [If n = 1, the eigenvalue are always simple.] It is, however, unusual. Unless we are in Case (iii) – with β ≡ 0, c(x) ≡ constant near ∂
– there is an open dense set of embeddings h in a C 2 -neighborhood of i
such that, if u, ψ are both non-trivial solutions of (∗)h(
) for some λ, then (∗∗)h(
) fails. Even in Case (iii), the condition (∗∗)h(
) will fail for most h near i
unless λ + c ≡ 0 and |∇u∇ψ| ≡ 0 on a neighborhood of ∂
. (This shows that, in Cases (i) and (ii), all eigenvalues are generically simple; and in Case (iii), all are simple except perhaps λ = −c|∂
.) To prove this, consider G(u, ψ, λ, h) = F(u, λ, h), F(ψ, λ, h), h ∗ K h(
) h ∗−1 (u, ψ, λ)|∂
) with F as above, and K
(u, ψ, λ) = ∇u · ∇ψ − (λ + Q)uψ, Q = c + 2β 2 − β div N
− N
· ∇β, with values in (say) L 1 (∂
). Assume for every h near i
, there exist u, ψ in H 2 (
)\{0} and λ ∈ R such that G(u, ψ, λ, h) = (0, 0, 0, 0, 0) ∈ (L 2 (
) × H 1/2 (∂
))2 × L 1 (∂
). By Theorem 5.4 [with x = (u, ψ, λ), y = h, ζ = 0], this can only happen if 2(β) fails, and in particular, for the solution of G(u, ψ, λ, i
) = 0, the quotient space R(DG(u, ψ, λ, i
))/R(∂G/∂(u, ψ, λ)) must have finite dimension – say, dimension n 0 . We may choose θ1 , . . . , θn 0 in (L 2 (
) × H 1/2 (∂(
))2 × L 1 (∂
) such that: given h˙ ∈ C 2 (
) there exist ˙ λ) ˙ ∈ H 2 × H 2 × R and real numbers c1 , . . . , cn 0 so ˙ ψ, (u, ˙ = ˙ λ, ˙ h) ˙ ψ, DG(u, ψ, λ, i˙
)(u,

n0

cjθj.

1

This is a collection of five equations. Consider the first pair. ˙ = ( + λ + c)/(u˙ − h˙ · ∇u) + λu





n0

c j θ 1j ∈
,

1

∂u ∂ + β (u˙ − h˙ · ∇u) + N˙ · ∇u + h˙ · ∇ + βu ∂N ∂N

 =

n0

c j θ 2j ∈ ∂
,

1

where θ j = (θ 1j , . . . , θ 5j ). We can almost solve for u˙ − h˙ · ∇u, modulo finite dimensions. Specifically, let {φ1 , . . . , φ p } be a basis for the eigenfunctions of

Chapter 6. Generic Perturbation of the Boundary

97

(∗)
, with eigenvalue λ; there are continuous linear operators C : H 1/2 (∂
) → H 2 (
), A : L 2 (
) → H 2 (
), such that v ≡ A f + Cg is the solution of ( + λ + c)v − f |
∈ span {φ1 , . . . , φ p } ∂v + βv = g on ∂
∂ N 


vφ j = 0

for 1 ≤ j ≤ p.

Then u˙ − h˙ · ∇u = −C( N˙ · ∇u + h˙ · ∇( ∂∂uN + βu) + (finite) where “(finite)” denotes some unspecified element of a specific finite-dimensional space – ˙ in this case, having dimension ≤ n 0 + p + 1, as we vary λ, c1 , . . . , cn 0 and ˙ ˙ φ ( u − h · ∇u) for 1 ≤ j ≤ p.
j Similarly from the second pair of equations, we have 



∂ψ ˙ ˙ ˙ ˙ + βψ + (finite). ψ − h · ∇ψ = −C N · ∇ψ + h · ∇ ∂N Substitution in the final equation n0

c j θ 5j = h˙ · ∇ K
(u, ψ, λ) + uψ(β div N˙
+ N˙
· ∇β)

1

+ (∇u · ∇ − (λ + Q)u)(ψ˙ − h˙ · ∇ψ) + (∇ψ · ∇ − (λ + Q)ψ)(u˙ − h˙ · ∇u)

shows the operator ∂ K
+ uψ(−β∂
σ − ∇∂
σ · ∇β) ∂N 



∂ ∂ψ + βψ + (∇u ·∇ − (λ + Q)u) C ∇∂
σ ·∇ψ − σ ∂N ∂N



 ∂u ∂ + (∇ψ ·∇ − (λ + Q)ψ) C ∇∂
σ ·∇u − σ + βu ∂N ∂N

: σ → σ

has finite rank. Here σ = h˙ · N , and we have assumed ∂ N
/∂ N
= 0 on ∂
, so N˙
= −∇∂
σ . Case (i): β(x) = 0 somewhere on ∂
. The operator is of order 2 and σ → σ + βuψ∂
σ is continuous from (say) H 3/2 (∂
) to L 1 (∂
). Since βuψ ≡ 0 on ∂
, by uniqueness in the Cauchy problem, it follows cannot be of finite rank, so we cannot be in Case (i). In more detail, suppose x0 ∈ ∂
and βuψ = 0 at x0 hence, on a neighborhood of x0 . Choose a smooth function σ0 supported in a small neighborhood of x0 , σ0 (x0 ) = 0, and a smooth θ : ∂
→ R with ∇∂
θ = 0 on supp σ0 . For large

98

Chapter 6. Generic Perturbation of the Boundary

real k, let σk (x) = k −2 σ0 (x) exp(ikθ(x)); then σk  H 3/2 (∂
) → 0 as k → ∞ and σk (x) = βuψσ0 (x) · |∇∂
θ(x)|2 exp(ikθ(x)) + o(1) so is not of finite rank. Case (ii) or (iii): β ≡ 0 near ∂
. The condition (∗∗)
becomes ∇u · ∇ψ = (λ + c)uψ on ∂
, and Q = c in the definition of K
. Let  S(x, y) be a fundamental solution for  + λ + c. Then Cg(x) = ˜ where g˜ solves −2 ∂
S(x, y)g(y)dAy  ∂S ˜ ˜ g(x) = g(x) −2 (x, y)g(y)dAy, x ∈ ∂
. ∂
∂ N x (We disregard operator of finite rank.) If E n (x − y) is the fundamental solution of the Laplacian, it follows (see 7.6) that   E n (x − y)g(y) + R(x, y)g(y), x ∈ ∂
, Cg (x) = −2 ∂


∂


where the “remainder” R is a kernel of class C∗1 (2) (in the notation of Chapter 7), defining an integral operator of order −2 on ∂
, while C is of order −1. We use this representation of C to find the principal part of , namely the operator order 1.  (y − x) ⊗ ∇∂
σ (y) 2 P.V. d Ay : M(x) σ → wn |y − x|n ∂
with M(x) = ∇∂
u(x) ⊗ ∇∂
ψ(x) + ∇∂
ψ(x) ⊗ ∇∂
u(x), w n = |S n−1 | and we use the notation 1 ≤ i, j ≤ n (a ⊗ b)i j = ai b j , Pi j Q i j . P:Q= i, j

The difference between and the operator above is continuous (for example) from H 1/2 (∂
) to L 1 (∂
). Since has finite rank, it is necessary that (τ1 ⊗ τ2 ) : M(x) = 0 for all x ∈ ∂
and τ1 , τ2 ∈ Tx (∂
) ∂u (x) ∂ψ (x) ∂τ ∂τ

= 0 for x ∈ ∂
, τ ⊥ N
(x) (see 7.4). Suppose ∇∂
u(x) = 0 at some point x ∈ ∂
; then for almost all tangent di(x) = 0 so ∂ψ (x) = 0, so ∇∂
ψ(x) = 0. Similarly on interchangrections τ, ∂u ∂τ ∂τ ing u, ψ; thus |∇∂
u∇∂
ψ| ≡ 0 on ∂
. By (∗∗)
, (λ + c)uψ ≡ 0 on ∂
so in fact λ + c ≡ 0 on ∂
. Thus we cannot be in Case (ii), so we must be in Case (iii), with β ≡ 0, c ≡ constant = −λ near ∂
, and also |∇u∇ψ| = 0 on ∂
. On a neighborhood of ∂
, u and ψ are harmonic functions; on a neighborhood of a given component of ∂
, at least one of u, ψ must be constant, by uniqueness in the Cauchy problem, and so |∇u∇ψ| = 0 near (as well as on) ∂
. or

Chapter 6. Generic Perturbation of the Boundary

99

Now suppose we are in Case (iii), β ≡ 0 and c(x) ≡ constant near ∂
– we may suppose c(x) ≡ 0 near ∂
. We want to prove 0 is either not an eigenvalue, or is a simple eigenvalue, for most regions near
. Suppose 0 is a p-fold eigenvalue ( p ≥ 1) for
and {φ1 , . . . , φ p } is an orthonormal basis for the eigenfunctions. Applying the Liapunov-Schmidt method, for h near i
, 0 is an eigenvalue for h(
) only if there is a solution q ∈ R p \{0} of M(h)q = 0 p (and then h ∗−1 ( j=1 q j (φ j + S j (h))) is an eigenfunction) where S j (h) is the solution of 0 = ( + c)v + (h ∗ ( + c)h ∗−1 −  − c)(φ j + v) (const.) φk




 ∂v ∂ ∗ ∂ ∗ 0= + h h − (φ j + v) on ∂
∂N ∂N ∂N

k

vφk = 0 for 1 ≤ k ≤ p, p and M(h) = [M jk (h)] j,k=1 is  φk (h ∗ ( + c)h ∗−1 −  − c)(φ j + S j (h)) M jk (h) =


 ∂ ∗−1 ∂ h (φ j + S j (h)). φk h ∗ − − ∂N ∂N ∂


For any V ∈ C 2 (
, Rn ),

p  d M(i
+ t V )|t=0 = − V · N ∇φ j · ∇φk dt ∂
j,k=1

We could reduce the multiplicity of the zero eigenvalue by small perturbations of
unless ∇φ j · ∇φk = 0 on ∂
for all j, k (If p = 1, “reduce multiplicity” means 0 is not an eigenvalue.) Thus if the zero eigenvalue is of minimal multiplicity p ≥ 1, then in fact p = 1. For the eigenfunctions φ j are locallyconstant on ∂
(and also near ∂
, being harmonic there), so they would be linearly dependent if we had p > 1. This completes the proof of generic simplicity, and we now consider moving eigenvalues by perturbing
. In the case just discussed, the eigenvalue (and corresponding eigenfunction) are independent of small perturbations of
; but this is the only  exception. In general, let λ be a simple eigenvalue with eigenfunction φ,
φ 2 = 1. By Example 3.5, there is a unique eigenvalue λ(t) near λ for the problem in
(t) = {x + t V (x) | x ∈
}, t small, and  d λ(t)|t=0 = V · N
{|∇∂
φ|2 − (λ + c + β 2 − β div N
− ∂β/∂ N
)φ 2 }. dt ∂
Thus the derivative is nonzero for some V unless K
(φ, φ, λ) = 0 on ∂
, in the previous notation. But our argument above shows – for most choices of

100

Chapter 6. Generic Perturbation of the Boundary


– there are no solutions (u, ψ, λ) of u + (λ + c)u = 0 = ψ + (λ + c)ψ in
, u ≡ 0 ∂u ∂ψ + βu = 0 = + βψ on ∂
, ψ ≡ 0 ∂N ∂N K
(u, ψ, λ) = 0 on ∂
except in the special Case (iii) just discussed. The argument above assumed n ≥ 2, in the “finite-rank” discussion; but for n = 1 all eigenvalues are always simple. Suppose some eigenvalue λ of u  + (λ + c) · u = 0 in(a, b),

−u + β1 u = 0 at a,

u  + β2 u = 0 at b

cannot be moved by perturbing a, bi then β1 + β12 + λ + c = 0 near a, −β2 + β22 + λc ≡ 0 near b. We may change variables by v = e−ϕ u where ϕ(x) is chosen so ϕ 1 = β1 near a, ϕ 1 = −β2 near bi the problem becomes v  + 2ϕ  (x)v  + (λ + c˜ (x))v = 0 in (a, b) v  = 0 at x = a, x = b. In this form, λ cannot be moved by perturbing a, b only if λ + c˜ ≡ 0 near a, b, and the eigenfunction v is locally constant near ∂(a, b). At least after changing variables, we get a result similar to the case n ≥ 2. Example 6.7. Generic Simplicity of Solutions of u + f (u, x) = 0 in
, ∂u/∂ N = g(x, u) on ∂
. We wish to prove (0, 0) is a regular value of 

∂u − g(·, u)|∂
F
: u → u + f (·, u), ∂ N
for most choices of
⊂ Rn (in appropriate spaces), which will follow – under mild smoothness hypotheses – if (0, 0) is a regular value of (h, u) → h ∗ Fh(
) h ∗−1 (u). Under some plausible “special hypotheses” for f, g, described below, we show that – restricting h to an open dense subset – (0, 0) will be a regular value of (h, u) → h ∗ Fh(
) h ∗−1 (u), so the solutions will all be simple for most choices of
. As an example (suggested by population genetics theory) suppose r : R → R is C 2 and has zero as a regular value (r (c) = 0 ⇒ r  (c) = 0) and S : Rn → R is C 2 such that and s : Rn → R is ξ 2 with S(x) = 0 on a dense set. Then for most bounded connected C 2 regions
⊂ Rn , all solutions u of u + r (u)S(x) = 0 in
,

∂u = 0 on ∂
∂N

are simple. Note that, if c ∈ r −1 (0), u(x) ≡ c is a solution for any choice of
; it is simple for most choices of
. Another case of interest is when

Chapter 6. Generic Perturbation of the Boundary

101

f, g are independent of x, we require f to be C 2 , g to be C 3 and f (c) = g(c) = 0 ⇒ | f  (c)| + |g  (c)| = 0. Then the solutions of u + f (u) = 0 in
, ∂u/∂ N = g(u) on ∂
, are all simple for most choices of
. We are assuming n ≥ 2; some analogous results for n = 1 are given in Theorem 11 of [D. Henry, Some infinite-dimensional Morse-Smale systems, J. Diff. Eq. 58 (1985)] but the case n = 1 is still essentially unsolved. [Indeed the case n ≥ 2 is not completely satisfactory.] If we impose the boundary condition ∂u/∂ N = g(x, u) on certain components of ∂
while u = 0 on other components (at least one!) a mild variant of the argument in Example 6.5 – restricting our perturbations to the “Dirichlet” components of ∂
shows the solutions are generically simple. We will work in an anti-logical fashion, starting only with smoothness assumptions ( f is C 2 , g is C 3 ) and discovering appropriate additional hypotheses in the course of the calculation. This is the way the assumptions for this and other examples were found. But if you prefer a logical order, you may turn directly to the statement of the result at the end of this section. we always suppose n ≥ 2 and ∂
is connected, bounded and (in the initial calculation) C 3 . Suppose for contradiction that there exists a solution u of F
(u) = 0 such that (u, i
) is a critical point of (v, h) → h ∗ Fh(
) h ∗−1 (v), and further, that this holds for every
in some (C 3 ) neighborhood of a given bounded, C 3 , connected region
0 . It follows (as in Examples 6.4, 6.6) that there exists ψ = 0 in the kernel of the linearization with K
(u, ψ) = 0 on ∂
, K
(u, ψ) = ∇∂
ψ · ∇∂
u − ψ( f (·, u) + g(·, u) div N
+ N
· gx (·, u) + gg  ) where g  = ∂g/∂u, gx = ∂g/∂ x, evaluated at (x, u(x)), x ∈ ∂
. Thus for every small perturbation
of
0 , there is a solution (u, ψ) of the over-determined problem  ∂u   u + f (·, u) = 0 in
, = g(·, u) on ∂
  ∂ N   ∂ψ ψ + f  (·, u)ψ = 0 in
, = g  (·, u)ψ on ∂
, ψ ≡ 0,   ∂ N     K
(u, ψ) = 0 on ∂
. The first four equations (with ψ ≡ 0) define a Fredholm map for (u, ψ) of index zero; including the fifth, we have a semi-Fredholm map of index −∞. Let (u, ψ) → G
(u, ψ) denote the resulting map from W 2, p (
) × (W 2, p (
)\{0}) to (L p (
) × W 1/ p, p (∂
))2 × L 1 (∂
), for some n < p < ∞ (so W 2, p ⊂ C 1 ). By hypothesis, h ∗ G h(
) h ∗−1 (u, ψ) = (0, 0, 0, 0, 0) has a solution (u, ψ) for every h in a C 3 -neighborhood of i
, so assumption 2(β) of Theorem 5.4 must fail at (u, ψ, i
). In particular, the quotient space has finite dimension (say, dimension n 0 ) and for some θ1 , . . . , θn 0 the following holds: given any

102

Chapter 6. Generic Perturbation of the Boundary

˙ ψ˙ in W 2. p (
) and real numbers c1 , . . . , cn 0 so h˙ ∈ C 3 (
, Rn ) there exist u, DG
(u, ψ)(u˙ − h˙ · ∇u, ψ˙ − h˙ · ∇ψ) + h˙ · ∇G
(u, ψ) n0 ∂G
+ cjθj. (u, ψ) N˙
= ∂N 1 Let θ j = (θ 1j , . . . , θ 5j ), and the first two components of this equation are: L u (u˙ − h˙ · ∇u) =

n0

c j θ 1j in
,

1

Mu (u˙ − h˙ · ∇u) =

n0 1

c j θ 2j − N˙ · ∇u − h˙ · N

∂ ∂N



 ∂u − g(·, u) on ∂
∂N



where L u =  + f (·, u), Mu = ∂/∂ N
− g  (·, u). Then 



∂ ∂u ˙ ˙ ˙ u˙ − h · ∇u = −C L ,M N · ∇u + h · N − g(·, u) + (finite), ∂N ∂N where C L ,M is the appropriate solution operator (studied in 7.6). Similarly the second pair of equations yields

 ∂  ˙ ˙ ˙ ˙ ˙ (Mu ψ) − g (·, u)ψ(u˙ − h · ∇u) ψ − h · ∇ψ = −C L ,M N · ∇ψ + h · N ∂N ˙ · ∇u)) + (finite), = −A L ,M ( f  (·, u)ψ(u˙ − h) and we substitute the earlier expression for u˙ − h˙ · ∇u on the right-hand side. Then we substitute these in the fifth component equation, to conclude: σ → σ is an operator of finite rank, where σ = h˙ · N
, N˙
= −∇∂
σ (i.e., we suppose ∂ N
/∂ N
= 0) and ∂ K
(u, ψ) σ = ψg∂
σ + ψ∇∂
σ · gx + σ ∂ N 

∂ (Mu ψ) + (∇u · ∇ − Q)C L ,M ∇∂
σ · ∇ψ − σ ∂N



 ∇ψ · ∇ − ψ Q  +(∇u · ∇ − Q)C L ,M g  ψ + C L ,M (∇∂
σ · ∇u −(∇u · ∇ − Q)A L ,M f  ψ 

∂ ∂u −σ −g ∂N ∂N K
(u, ψ) = ∇u · ∇ψ − ψ Q(u) (= 0 on ∂
) Q(u) = f (·, u) + g(·, u) div N + 2g  g + N · gx , ∂ Q(u). Q  (u) = ∂u The principal part of is σ → ψg∂
σ , of second order, while the remainder is first order (C L ,M is order −1, A L ,M is order −2). Thus will not be

Chapter 6. Generic Perturbation of the Boundary

103

of finite rank unless ψg(·, u) ≡ 0 on ∂
. Since ψ ≡ 0, this says g(·, u) ≡ 0 on ∂
. Thus u satisfies the over-determined problem ∂u = 0 and g(·, u) = 0 on ∂
. ∂N

u + f (·, u) = 0 in
,

We study such over-determined problems in the lemma below. By hypothesis, a solution u exists for every
near
0 , hence by the lemma, there is a solution u which also satisfies gx (·, u) = 0 and g  (·, u)∇∂
u = 0 on ∂
. But this information may be useless – for example, if g ≡ 0 (the Neumann problem) – so we continue. Knowing that g(·, u) = 0, gx (·, u) = 0 and g  (·, u)∇∂
u = 0 on ∂
we see K
(u, ψ) = ∇u · ∇ψ − ψ f (·, u) = 0 on ∂
and the principal part of is of order 1: σ → ∇u · ∇C L ,M (∇∂
σ · ∇ψ) + ∇ψ · ∇C L ,M (∇∂
σ · ∇u) = ∇∂
u · ∇∂
C L ,M (∇∂
σ · ∇∂
ψ) + ∇∂
ψ · ∇∂
C L ,M (∇∂
σ · ∇∂
u) or, more explicitly, σ (x) =

2 P.V. wn

 ∂


(y − x) ⊗ ∇∂
σ (y) : M(x) + (order zero) |x − y|n

with M(x) = ∇∂
u ⊗ ∇∂
ψ + ∇∂
ψ ⊗ ∇∂
u at x ∈ ∂
. For this to be of fi∂ψ(x) = 0 for every x ∈ ∂
, τ ⊥ N
(x). nite rank, it is necessary that ∂u(x) ∂τ ∂τ This implies |∇∂
u||∇∂
ψ| = 0 on ∂
, and so (from K
(u, ψ) = 0, ψ ≡ 0) it follows f (·, u) ≡ 0 on ∂
. Applying the lemma to f , we have also that f x (·, u) = 0 and f  (·, u)∇∂
u = 0 on ∂
. Applying the lemma again to f x , gx , and then to gx x (since g is C 3 ) we conclude: For every
C 3 -close to
0 , there is a solution (u, ψ) of u + f (·, u) = 0 in
, ∂u = 0 on ∂
, ∂N

ψ + f  (·, u)ψ = 0 in
, ∂ψ = g  (·, u)ψ on ∂
, ∂N

ψ ≡ 0,

and on ∂
, gx (·, u) = 0,

g(·, u) = 0,

f (·, u) = 0, 

g (·, u)∇∂
u = 0, f  (·, u)∇∂
u = 0,

gx x (·, u) = 0, f x (·, u) = 0,

gx (·, u)∇∂
u f x (·, u)∇∂
u = 0,

gx x x (·, u) = 0,

f x x (·, u) = 0, gx x (·, u)∇∂
u = 0,

= 0, and

|∇∂
u||∇∂
ψ| = 0.

104

Chapter 6. Generic Perturbation of the Boundary

With all these conditions, we should be able to find a contradiction under mild additional assumptions on f, g. But first, we prove the lemma – in a more general case then we need here. Lemma. Suppose  is an open set in Rq , V is open in Rn which meets ∂
0 , W0 is open in W 2, p (
0 ), f (x, u, p, λ) and g(x, u, λ) are C 2 and φ(x, u, λ) is C 1 , and for every
= h(
0 ) in a C 3 neighborhood of
0 there is a solution (u, λ) of u + f (x, u, ∇u, λ) = 0 in
,

∂u = g(x, u, λ) on ∂
, ∂N

with h ∗ u ∈ W0 , λ ∈ , such that φ(x, u, λ) = 0 on ∂
∩ V. Then there is such a solution (u, λ) which also satisfies φ  (x, u, λ)∇∂
u = 0 and

φx (x, u, λ) + N
(x)φ  (x, u, λ)g(x, u, λ) = 0

on ∂
∩ V. Proof. Condition 2(β) of Theorem 5.4 must fail so we conclude



  ∂u ∂ ∂  φ(·, u, λ) + φ (·, u, λ) C L ,M ∇∂
σ · ∇u − σ − g  σ → σ ∂N ∂N ∂N ∂
∩V is of finite rank. The principal part is of order zero, namely (after integration by parts). σ → σ (x)(N (x) · φx + φ  g)  − 2φ  (x, u(x), λ) P.V.

∂


  (y − x) · ∇∂
u(x) σ (y) , n w n |x − y| x∈∂
∩V

with the remainder an operator of order −1. Thus on ∂
∩ V , N · φx + φ  g = 0 and φ  ∇∂
u = 0. Since φ(x, u(x), λ) ≡ 0 on ∂
∩ V , differentiation in a tangential direction τ gives τ · φx (·, u(·), λ) = 0 which (with the equation N · φx = −φ  g) gives the result claimed. Now we will try to formulate plausible sufficient conditions on f, g to ensure generic simplicity of solutions. Special Hypotheses on f, g f (x, u) is locally C 2 , g(x, u) is locally C 3 , on Rn × R. (1) There is a discrete set {c j } ⊂ R, possible empty, such that f (x, c j ) ≡ 0, g(x, c j ) ≡ 0 for all x, but | f  (x, c j )| + |g  (x, c j )| = 0 on a dense set of Rn . (2) For any c ∈ R\{c j }, the set {x ∈ Rn | at (x, c) we have g = 0, gx = 0, gx x = 0, gx x x = 0, f = 0, f x = 0, and f x x = 0} does not contain a C 2

Chapter 6. Generic Perturbation of the Boundary

105

hypersurface; e.g., has dimension < n − 1. [If f, g are independent of x, this says | f (c)| + |g(c)| = 0 when c ∈ {c j }.] (3) {(x, u) ∈ Rn × R | u ∈ {c j } and at (x, u) we have g = 0, gx = 0, gx x = 0, gx x x = 0, g  = 0, gx = 0, gx x = 0, f = 0, f x = 0, f x x = 0, f  = 0, and f x = 0, does not contain a subject which is a C 1 -graph over a C 2 hypersurface in Rn , e.g., has dimension < n − 1. Remark. These conditions are far from optimal, but seem adequate for most applications. (In the genetics problem, they do not allow the extreme cases “dominant” or “recessive;” but these always have non-simple solutions. Hypothesis (1) says there may be trivial constant solutions u(x) ≡ c j , but these will be simple for most choices of
by Example 6.6. If we omit hypothesis (1), the conclusion is that all solutions other than the constant solutions {u ≡ c j } will be simple, in most regions
. Theorem. If f, g satisfy the “special hypotheses” above, then for most bounded C 2 regions
⊂ Rn , all solutions u of u + f (x, u) = 0 in
,

∂u = g(x, u) on ∂
, ∂N

are simple. Proof. It is enough to treat the case where
is connected and f, g are uniformly bounded; then instead of “most regions” we can say “open, dense set in Diff2 (
).” Let F
(u) = (u + f (·, u), ∂u/∂ N
− g(·, u)|∂
), F
: W 2, p (
) → L p (
) × W 1/ p, p (∂
). Then F
is C 2 , Fredholm with index zero, and (h, u) → h ∗ Fh(
) h ∗−1 (u) is C 1 on Diff2 (
) × W 2, p (
). Suppose h ν → h in Diff2 (
) and there exists u ν ∈ W 2, p (
) so h ∗ν Fh ν (
) h ∗−1 ν (u ν ) = (0, 0) and (u ν , h ν ) is a critical point. Then the u ν are bounded in W 2, p (
) so we may assume they converge in C 1 to some u, and it follows that u ∈ W 2, p (
), h ∗ Fh(
) h ∗−1 (u) = (0, 0) and (u, h) is a critical point. Thus the set h such that there is a non-simple solution in h(
), is closed and its complement (all solutions simple) is open in Diff2 (
). We prove denseness by contradiction. Suppose for every
in some C 2 neighborhood of a given
0 that there is a non-simple solution. This implies the weaker claim with C 3 in place of C 2 (supposing, as we may,
0 is C 3 ). Then we have the hypothesis of the previous calculation: there exists u so F
(u) = (0, 0) and (u, i
) is a critical point of (u, h) → h ∗ Fh(
) h ∗−1 u [h ∈ Diff3 (
)], and this holds for every region
C 3 -close to
0 . Then there exist (u, ψ) with F
(u) = (0, 0), L u ψ = 0 in
, Mu ψ = 0 on ∂
, ψ ≡ 0, and all the conditions listed above (before the “special hypothesis”) must hold. And again, all this is still true for any regions near
; so we may assume, for a start, that all the constant solutions {c j } are simple. Suppose u|∂
is constant on a neighborhood

106

Chapter 6. Generic Perturbation of the Boundary

of some x0 ∈ ∂
, say u(x) ≡ c. If c ∈ {c j }, we contradict hypothesis (2); if c = c j for some j, then by uniqueness in the Cauchy problem, u(x) ≡ c j in
, which is false since u is not a simple solution. Thus, |∇∂
u| = 0 on a dense open set of ∂
, in particular on some ∂
∩ B with B an open ball. Then {(x, u(x)) | x ∈ ∂
∩ B} is a set of the kind prohibited by (3). In every case, we have a contradiction, so the proof is complete. Example 6.8. Generic Simplicity of Complex Eigenvalues of a Dirichlet Problem. Let b j (x, λ), c(x, λ) ( j = 1, . . . , n) be polynomials in λ ∈ C with coefficients C 2 functions of x ∈ Rn possibly complex valued. [c need only be C 1 .] Define A(λ) =  +

n

b j (x, λ)

j=1

∂ + C(x, λ) =  + b · ∇ + c ∂x j

and consider the eigenvalue problem A(λ)u = 0 in
,

u = 0 on ∂
,

u ≡ 0

(with u ∈ H 2 ∩ H01 (
, C)). An eigenvalue is any λ ∈ C for which a solution u ≡ 0 exists; and the eigenvalue λ is simple if the kernel of the operator d A(λ))u is not A(λ) | H 2 ∩ H01 (
, C) is one-dimensional (= C · u) and if ( dλ contained in the range of the operator. This implies λ and u vary smoothly with perturbations of the coefficients or the domain
, λ remaining simple, and reduces to the usual definition in the usual case: b = b(x), c = c(x) + λ. λ is a simple eigenvalue of A(λ) if and only if it is a simple eigenvalue of A∗ (λ) or of e−φ Aeφ , for any smooth scalar φ(x, λ). We will assume n ≥ 2, but the one-dimensional case should be mentioned. Let U (·; α, λ) be the solution of Ux x + bUx + cU = 0 such that U = 0, Ux = 1 for x = α. Then, for the interval (α, β), λ is an eigenvalue when U (β; α, λ) = 0 ∂ U (β; α, α) = 0. This notion of “simplicity” and it is simple when, in addition, ∂λ is equivalent to that above when n = 1. An important function associated with L =  + b · ∇ + C is D(L) = C −

n n ∂b j 1 1 1 1 div b − b2 = C − − b2 . 2 4 2 1 ∂x j 4 1 j

(I call this the “discriminant” of L, but perhaps another name has priority; surely it has been noticed before.) In case n = 1, if b, c are constants, the corresponding quadratic (in the form suggested by Fourier transformation) is √ (i z)2 + b(i z) + c, with roots z = ib/2 ± D(L). In general, D(L ∗ ) = D(L) where L ∗ =  − b · ∇ + c − div b. (For the complex sesquilinear inner product, we should take conjugates: then D(L ∗ ) = D(L).) For any smooth scalar

Chapter 6. Generic Perturbation of the Boundary

107

function φ(x, λ), D(eφ Le−φ ) = D(L) eφ Lc−φ =  + (b − 2∇φ) · ∇ + c − b · ∇φ + (∇φ)2 − φ. D(L) also appears in the fundamental solution for L: in the expansion of the singularity (see Theorem 7.5.1) when ‘c’ first appears, it is in the combination D(L). Similarly, D(L) appear in ∂∂N B L (g) – see Theorem 7.6.4 (In these cases, the b j , c may even be matrix-valued!) For the Dirichlet problem in
– assuming the b j , c and
are smooth – L is self-adjoint in L 2 (
, C) if and only if Re b ≡ 0 and D(L) is real-valued in
. Here we use the customary inner product of L 2 , but if we include a weight (measure e2φ d x instead of d x, for some smooth real-valued φ) the condition is: Re b ≡ 2∇φ and D(L) is real valued. We could change variables (u → ueφ ) to get an operator eφ Le−φ which is self-adjoint in the usual sense. If D(L) is real and Im b = 2∇φ is a gradient, we could change variables (u → ueiφ ) to get an operator eiφ Le−iφ with real coefficients. We treated the real case in Example 6.3. In Example 4.3, we considered a self-adjoint case with complex coefficients but with simple dependence on the (real) eigenvalue parameter. The general second-order self-adjoint case – b purely imaginary and c − 1/2 div b real, with b, c polynomials in λ – remains open. (I managed to treat the case: b independent of λ, ∂c/∂λ > 0 on a dense set for each real λ.) The non-real and non-self-adjoint case is: D(L) is not real, or it is real but neither Reb nor Imb is a gradient. Here we only use the first condition. Im D(L) = 0, but will give a more general result in 8.4. We will assume: for certain given open sets  ⊂ C, V ⊂ Rn , when λ ∈  (∗) Im D(A(λ)) = Im(c − 1/2 div b − 1/4 b2 ) = 0 on a dense set of x ∈ V We will conclude, for most bounded C 2 open
⊂ Rn such that ∂
meets V , all eigenvalues in  are simple and “freely moveable,” i.e., they can be moved in any complex direction by perturbation of ∂
. Thus, if the imaginary axis is in , we may also say, generically, all eigenvalues have real part different from zero. (If the b j , c are real-valued when λ is real, any real simple eigenvalue is moveable by perturbation of ∂
but it is not freely moveable, since it must remain on the real line. In this case, the real line cannot meet .) Here we assume n ≥ 2. The case n = 1 always needs special treatment since the space of domains (intervals) is finite dimensional. For n = 1, the coefficient of du/dx is always a gradient, and may be removed; and the kernel is always one dimensional, for any eigenvalue; but whether eigenvalues are generically simple, for most intervals, is not known. (Except in the self-adjoint case, which coincides with the real case.)

108

Chapter 6. Generic Perturbation of the Boundary

Example. Suppose m, k j , r, b j , c are smooth real-valued functions of x ∈ Rn and we seek solutions of the form u(x, t) = v(x)eλt ≡ 0 for mu tt + k · ∇u t + ru t = u + b · ∇u + cu in ∂
with u = 0 on ∂
, i.e., A(λ)v ≡ v + (b − λk) · ∇v + (c − mλ2 − r λ)v = 0 in
, with v = 0 on ∂
, v ≡ 0. Then Im D(A(λ)) = −Im λ{M Re λ + R} with M = 2m + |k|2 , R = r − 1/2 div k − 1/2 b · k. Consider three cases, for some open V ⊂ Rn (perhaps the whole space): (i) M ≡ 0, R = 0 on a dense set of V , with  = C\R; (ii) M =  0 on a dense set and R/M is not cosntant on any open set in V , with  = C\R; (iii) M = 0 on a dense set and R/M = constant = µ0 in V , with  = C\{R ∪ (−µ0 + iR)}. In each case, hypothesis (∗) holds so, when ∂
meets V , eigenvalues in  are (generically) simple and freely moveable. When λ is real, we can apply 6,3: generically in
, the real eigenvalues are simple and moveable, though not ¯ ⊂ V and k ≡ 0, for eigenvalues of freely moveable. In case (iii), supposing
the form λ = −µ0 + iσ (σ real), the equation becomes v + b · ∇v + (c + mµ20 + mσ 2 )v = 0 with real coefficients, so these eigenvalues are also generically simple and moveable. (If k ≡ 0, or k is not a gradient, these eigenvalues are still not understood.) If zero is a regular value of (u, λ) → A
(λ)u : (H 2 ∩ H01 (
, C)\{0}) ×  → L 2 (
, C), (for a given open  ⊂ C) then all eigenvalues in  are simple. This holds for most choices of
= h(
0 ) provided zero is a regular value of (u, λ, h) → h ∗ Ah(
) (λ)h ∗−1 u. If 0 is any compact subset of , the set of
(near a certain
0 ) where this

Chapter 6. Generic Perturbation of the Boundary

109

last condition fails (with λ ∈ 0 ) is closed, consisting of
such that  A(λ)u = 0 in
, u = 0 on ∂
, u ≡ 0 (∗∗) A∗ (λ)ψ = 0 in
, ψ = 0 on ∂
, ψ ≡ 0 and Re(∂ψ/∂ N ∂u/∂ N ) = 0 on ∂
has a solution (u, ψ, λ) with λ ∈ 0 . We show it has no interior. Suppose, for contradiction, it has interior: we may suppose
0 is in the interior and ∂
0 is C 4 , and (∗∗) is solvable with λ ∈  for every
in some C 4 -neighborhood of
0 . (Note that u, ψ are then in C 3,α (
) for any α < 1.) We will impose the final condition only in ∂
∩ V , for some given open V ⊂ Rn – which may be only a subset of the V in hypothesis (∗) − and we suppose V meets ∂
0 , so it also meets any ∂
for
near
0 . Then apply the transversality theorem to F(u, ψ, λ, h) = (h ∗ Ah(
) (λ)h ∗−1 u, h ∗ A∗h(
) (λ)h ∗−1 ψ, h ∗ K h(
) h ∗−1 (u, ψ)) where K
(u, ψ) = Re(∇u · ∇ψ|∂
∩ V ) ∈ L 1 (∂
∩ V ). (Since u = ψ = 0 on ∂
, K
(u, ψ) = 0 on ∂
.) For each h near i
(and
near
0 ) there is a solution u, ψ, λ of F(u, ψ, λ, h) = (0, 0, 0) ∈ L 2 (
, C)2 × L 1 (∂
∩ V ) with λ ∈  so condition (2.β) of Theorem 5.4 must fail. Thus we have solutions (u, ψ, λ) of (∗∗) which also satisfy: is of finite rank, where    ∂ ∂ ∗ B A(λ) (σ uN ) + uN B A (λ) (σ ψN )  , (σ ) = Re ψN ∂N ∂N ∂
∩v σ = h˙ N is the normal velocity of the perturbation, uN = ∂u/∂N, ψN = ∂/∂N , and B A , B A∗ are the solution operators for the Dirichlet problem (see 7.6). We apply Theorems 2, 3 and 4 of Section 7.6. [An alternative argument is in Section 8.3.] Many terms vanish because Re(u n ψn ) ≡ 0 on ∂
, hence Re (ψN (x)uN (y) + ψN (y)uN (x))   2ψN (x)uN (y) + (y − x)D(uN ψN )(x)  = Re  +1/2 (y − x)2 : D 2 (uN ψN )(x) 3− −(y − x) · DuN (x)(y − x) · DψN (x) + O(|x − y| ) = − Re((y − x) · DuN (x)(y − x) · DψN (x)) + O(|x − y|3− ) = O(|x − y|2 ),

110

Chapter 6. Generic Perturbation of the Boundary

recalling uN , ψN are C 3− , for any ∈> 0. The part remaining is  2σ (y)d A y (σ )(x) = Re(τ · ∇∂
uN (x) + 1/2 b · τ uN (x)) n−2 ∂
ωn (x − y) × (τ · ∇∂
ψN (x) − 1/2 b · τ ψN (x))  + 2σ (y)d A y E n (y − x) ∂


× Re(uN (y)ψN (y)(c − 1/2 div b − 1/4 b2 )(y, λ))  + {O(|x − y|3−−n ) + (smooth kernel)} σ (y)d A y , ∂


y−x for x ∈ ∂
∩ V and any smooth real σ . We have written τ for |y−x| in the first intergral, which is nearly a unit tangent vector for y near x. In case n = 2, the kernel in the first integral is smooth, but the coefficient of 1 log(y − x) must vanish. Since uN ψN is purely imaginary and E λ (y − x) = 2π non-zero on a dense set, we conclude Im D(A(λ)) = 0 on ∂
∩ V . In case n ≥ 3, at each x ∈ ∂
∩ V and for each τ ⊥ N x with |τ | = 1,



 ∂uN 1 ∂ψN 1 + /2 b · τ uN − /2 b · τ ψN Re ∂τ ∂τ 1 Re(uN ψN D(A)) = n−2

(see 7.3.7). Consider a point of ∂
∩ V where uN = 0 and ψN = 0 – almost any point. We may define (locally) real functions R, S, φ by uN = Re iφ ,

ψN = iSe−iφ

and then the equation becomes (τ · ∇∂
φ + 1/2 Im b · τ )(τ · ∇∂
log |S/R| − Re b · τ ) =

1 Im D(A). n−2

At any point of ∂
∩ V , we may choose τ ⊥ N so also τ ⊥ (∇∂
φ + 1/2 Im btan ), since n ≥ 3, to conclude: Im D(A(λ)) = 0 on ∂
∩ V,

when n ≥ 2.

(We also have, for n ≥ 3, where uN ψN = 0 in ∂
∩ V , 0 = |Im btan + ∇∂
(2 arg uN )| · |Re btan − ∇∂
log |ψN /uN ||. We will use this condition in 8.3.) This is not yet a contradiction to (∗): for each
near
0 there exists a solution (u, ψ, λ) of (∗∗) with λ ∈  which satisfies also Im D(A(λ, ·)) = Im (c − 1/2 div b − 1/4 b2 ) = 0 on ∂
∩ V . But the lemma below shows there

Chapter 6. Generic Perturbation of the Boundary

111

is a nonempty open V0 ⊂ V − we may suppose it is near ∂
0 − and some λ0 ∈ ∧ such that Im D(A(λ0, ·)) ≡ 0 in V0 . This contradiction proves (∗∗) has not solutions for most
; in fact, restricting λ to a compact set of , there are no solutions except perhaps in a closed set of infinite codimension. (In Theorem 5.4, m is arbitrary.) The same condition – nonsolvability of (∗∗) – implies the eigenvalue λ, assumed simple, is freely moveable. Indeed consider h(x, t) = x + t V (x) for any smooth V : Rn → Rn ; the eigenvalue λ(t) near λ, for the Dirichlet problem in h(
, t) is well-defined for small t and 

  d d λ(t)|t=0 = A(λ) u, σ = V · N. σ uN ψN / ψ dt dλ ∂

The denominator is non-zero, since λ is simple, and the set of possible “velocities,” for various choices of V or σ , is a real-linear subpace of C. It fails to be the whole complex plane only if uN ψN |∂
has constant argument, i.e., for some real constant θ, Re(e−iθ uN ψN )|∂
≡ 0. But this implies (u, e−iθ ψ, λ) solves (∗∗), which has no solutions. Thus λ is freely moveable. It remains only to prove the lemma. Lemma. Let V be an open set in Rn , n ≥ 2, and M any set in Rm , and suppose  F(x, µ) = |α|≤k Fα (x)µα is a polynomial in µ ∈ Rm with coefficients Fα : V → R p continuous. Let
0 be a C r domain (r ≥ 1) in Rn whose boundary ∂
0 meets V and assume: for every
in some C r -neighborhood of
0 such that
\V =
0 \V (i.e., perturbed only inside V ) there exists µ = µ(
) in M such that F(x, µ) = 0

for all x ∈ ∂
∩ V.

Then there exists µ0 ∈ M and a nonempty open set V0 ⊂ V such that F(x, µ0 ) = 0 for all x ∈ V0 . Remark. The result is entirely reasonable, but the hypothesis is not. Our argument is algebraic – hence, the polynomial dependence – but it seems to miss the point: it should be a result of topological dimension theory. We satisfy an infinite-dimensional condition (F(·, µ)|∂
∩ V ≡ 0) by choice of a finitedimensional parameter µ, and this is only possible if it is trivially satisfied for some µ0 . But I haven’t managed to prove it by topological dimension theory. (Our argument probably also applies with only analytic dependence on µ, but this still seems excessive.) If we apply the transversality theorem directly, assuming F and M are smooth, we conclude that the hypothesis still holds for (x, µ) → (F(x, µ), ∂∂x1 F(x, µ), . . . , ∂∂xn F(x, µ)) in place of F. If x → F(x, µ) were

112

Chapter 6. Generic Perturbation of the Boundary

analytic for each µ with all (∂/∂ x)α F(x, µ) continuous, we could continue to arbitrarily high-order derivatives and eventually find µ0 ∈ M with F(·, µ0 ) ≡ 0 in V , supposing V is connected. But analyticity in x is even less acceptable than analyticity in µ, for typical applications; and analyticity seems totally irrelevant. Proof. Though not essential, it is convenient to reduce to the case of linear dependence. Let N be the number of multi-indices α = (α1 , . . . , αm ) with |α| ≤ k, so µ → (µα )|α|≤k : Rm → R N defines an imbedding of M onto A ⊂ R N \{0},  and instead of F(x, µ) we use |α|≤k aα Fα (x) = a · F(x) for a ∈ A. We suppose, for contradiction, that a · F(x) = 0 on a dense set of V for every choice of a ∈ A ⊂ R N \{0}. Choose a1 ∈ A; then a1 · F(x1 ) = 0 for some x1 in V arbitrarily near ∂
0 . Choose such x1 , so that there exists
1 sufficiently near
0 whose boundary contains x1 . (“Sufficiently near” means being inside the C r -neighborhood of
0 in our hypothesis.) Define L 1 ⊂ R N by L 1 = {a ∈ R N : a · F(x1 ) = 0} so a1 ∈ / L 1 but L 1 meets A: for some a1 ∈ A, a1 · F ≡ 0 on ∂
1 ∩ V , in particular at x1 . Choose a2 in A ∩ L 1 and xˆ = x1 on ∂
1 ∩ V ; a2 · F = 0 on a dense set so ˆ Choose such x2 and
2 near
1 (hence, a2 · F(x2 ) = 0 at points x2 near x. near
0 ) such that ∂
2 contains both x1 and x2 . Then define L 2 = {a ∈ R N : a · F(x1 ) = 0 and a · F(x2 ) = 0}. We have L 1 ⊃ L 2 since a2 ∈ L 1 \L 2 , and also =

L 2 meets A: for some a2 ∈ A, a2 · F = 0 on ∂
2 ∩ V ⊃ {x1 , x2 }. We may continue thus indefinitely, in particular for N steps: R N ⊃ L 1 ⊃ = = L 2 ⊃ · · · ⊃ L N , with each L k a linear space which meets A. But N > dim L k > = = dim L k+1 for each k ≥ 1 so we find a contradiction: L N = {0} meets A ⊂ R N \{0}.

Some Obvious Questions In Example 6.5, are the solutions generically hyperbolic? What about a system of equations, not just one? How about the Navier-Stokes equation! If not a system, why not a fourth-order scalar problem? Introduce some parameters; as the parameters change, we can expect bifurcation of solutions, but is the bifurcation “generic” for most
? In Example 6.7 can we allow dependence on ∇u in f ? If so, are the resulting solutions hyperbolic? And what about systems? Higher-order equations? Generic bifurcation? Mixed boundary conditions?

Chapter 6. Generic Perturbation of the Boundary

113

In Example 6.2, how about other symmetries? Continuous symmetries? Symmetries in nonlinear problems? Bifurcation with symmetry? Rework the whole chapter with symmetry! Why not add “genericity” of coefficients to generic domains? Surely this will give much stronger theorems! What are generic properties of multi-parameter eigenvalue problems? In Example 6.6, does the result hold for non-self-adjoint problems? Can one prove results like Examples 6.3 or 6.8 with other boundary conditions? Or for systems? Or higher-order equations? In the self-adjoint problem Example 4.4 can we allow polynomial dependence on λ, as in Examples 6.3 and 6.8? Why not study some time-dependent problems? Say periodic solutions in time-dependent domains, or fixed generic domains. Are they generically simply. Hyperbolic? How do they bifurcate? Are the stable and unstable manifolds transversal, generically with respect to the domain? How about control theory with the domain as parameter? Is this enough for local controllability? If we can move only a small part of the boundary, how much does this accomplish? Can we use perturbation of the boundary to supplement other control parameters, perhaps closing reachable sets? Or extending them in new directions? Can we optimize? Or will this lead to very irregular regions? Author’s Note: I only claimed to have questions, not answers.

Chapter 7 Boundary Operators for Second-Order Elliptic Equations

For the Laplace operator
in a smooth bounded region  ⊂ Rn , among the operators of interest are ∂∂N B
|∂ , C
|∂ , ∂∂N A
B
|∂ where u = A
f is the solution of
u = f in , v = B
g is the solution of
v = 0 in , w = C
h is the solution of
w = constant in ,

u = 0 on ∂, v = g on ∂, ∂w = h on ∂, ∂N



1 with  w = 0 (the constant is || ∂ h). Corresponding operators may be defined for other second-order elliptic equations (and for higher-order equations and systems). For applications to generic perturbation of the boundary, we need fairly explicit representations of these operators. One may write, for example,   ∂G  (x, y)g(y)dA y G  (x, y) f (y)dy, B
g(x) = A
f (x) =  ∂ ∂ N (y) where G is Green’s function for the Dirichlet problem in ; but this is useless without fairly precise information about the singularity of G near the boundary. P. Levy [18] gives some information for the Laplacian in R3 : he computes, in effect, the principal and subprincipal parts of ∂∂N B
, by subtracting singularities of known harmonic functions. The method seems more difficult than ours when we attempt the next step, and we need more detail and more generality. In any case, this reference was found too late to be used here, though it gives a valuable verification for part of Theorem 7.6.7. Assuming everything is C ∞ one may use the “projection on the Cauchy data” of the theory of pseudodifferential operators [4, 37]. The principal symbols have been computed, but we need much more detail. This is presumably the best way to attack the problem – especially for more general elliptic systems – but I have not managed to complete the calculation with this method. (I did manage the calculation 114

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

115

for a special type of rapidly-oscillating solution in 1985, after completing this manuscript. Chapter 8 presents the results. I hope Chapter 7 – with an important correction to Thm. 7.6.4 – is still of interest, presenting another method for a difficult problem.) We will use the more concrete (although more clumsy) method of integral operators. We can avoid singular integrals – until the final results – by integration by parts, so the detailed calculations use only weaklysingular (absolutely integrable) kernels. In the first three sections, we study weakly-singular kernels with emphasis on differentiability properties (but without going to the C ∞ extreme). Then we treat limits of singular integrals and fundamental solutions of second-order equations having the Laplacian as principal part. We obtain integral equations for the operators of interest and “solve” these (to second or third order) to obtain fairly explicit representations of the operators. For example if  is a bounded C 4 region in Rn (n ≥ 2) and E(z) is the fundamental solution of the Laplacian |z|2−n (2 − n)ωn 1 log |z| E(z) = 2π

E(z) =

for n = 2, for n = 2,

where ωn = |δ n−1 | = 2π n/2 / (n/2), then for x ∈ ∂ and g ∈ C 2 (∂),  ∂ 2 E(x − y) n − 2 H (x) ∂ B
g(x) = 2 P.V. g(x) (g(y) − g(x))dA y − ∂N ∂ N ∂ N n−1 2 x y ∂  n−1 (y − x). Kˆ (x)(y − x) − P.V. g(y)dA y 2ωn−1 |y − x|n+1 ∂     y−x 2−n + |x − y| Q 1 x, R1 (x, y)g(y). g(y) + |y − x| ∂ ∂ Here K (x) = D N (x) is the curvature of ∂ at x (at least when restricted to the tangent space Tx (∂); see Th. 1.5), H (x) = trace K(x) is the mean (x) ˆ P(x) has trace zero (P(x) = I − N (x)⊗ N (x) is curvature, K(x) = K(x) − Hn−1 the projection on the tangent space), σ → Q 1 (x, σ ) is a fourth-order polynomial computed explicitly below (Section 7.6), and the “remainder” R1 is a kernel of class C∗1 (2) in ∂, in the notation of Section 7.1 below (or C∗r −3 (2), if ∂ is C r ). “P.V.” indicates the Cauchy principal value when the integrand is singular at x,   P.V. · · · = lim ···. ∂

→0 ∂\B (x)

If n = 2 then ∂ is a finite union of curves and (for definiteness) suppose it is a single curve of length L. Functions on ∂ may be written as functions of

116

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

the arc length, extended as L-periodic functions on the real-line and then the operator has a simple form:   ∂ B
g(t) = λ(t − s)g

(s)ds + R1 (t, s)g(s)ds ∂N ∂ ∂   λ (t − s)g (s) + R1 (t, s)g(s) = P.V. ∂  ∂

λ (t − s)(g(s) − g(t)) + R1 (t, s)g(s) = P.V. ∂

where λ(t) is L-periodic, C λ(t) =

∞

1 log |t|, π

∂

except at {0, ±L , ±2L 1 , . . .}, with λ (t) =

1 , πt

λ

(t) =

−1 π +2

near t = 0, while R1 (t, s) is L-periodic in each variable and of class C 1− (or C r −3− , if ∂ is C r ) for any > 0. ( ∂∂N B
g = H g , where H is the Hilbert transform associated with the curve ∂.)

7.1 Weakly-Singular Integral Operators with C∗r (α) Kernels We study integral operators  u → K (x, y)u(y)d M y ,

x ∈ M,

M

where M is an m-dimensional differentiable manifold without boundary (m ≥ 1) having a positive measure d M locally equivalent to m-dimensional Lebesgue measure, and K (x, ·) is absolutely integrable. (In our applications M = ∂ for some  ⊂ Rn , m = n − 1, and d M is the surface-area measure d A in ∂ induced from n-dimensional Lebesgue measure in Rn .) We study only a special class of kernels K (x, y) (class C∗r (α), r = integer ≥ 0, α > 0, defined below) r -times differentiable for x = y with certain growth-restrictions on the derivatives as x − y → 0. Standard tricks – partition of unity and local change of variables – permit us to work in Euclidean Rm , and we do so until Assumptions 7.1.4. We will see later that fundamental solutions of second order elliptic equations are kernels of this class, as are the kernels of many integral operators associated with elliptic boundary value problems. The space of value of the kernels K (x, y) has little importance here – we will take the values to be real or complex matrices of given dimensions, and whenever products occur, there is the implicit assumption that the matrices are compatible. The transpose of a matrix B is written  B.

7.1 Weakly-Singular Integral Operators with C∗r (α) Kernels

117

Definition 7.1.1. Let A be an open set in Rm ; a function K (x, y) defined for x = y in A is a kernel of class C∗r (α) in A (r = integer ≥ 0, α > 0) if it is C r for x = y and for any δ > 0 and |i| + | j| + |k| ≤ r   ∂xi ∂ yj (∂x + ∂ y )k K (x, y) = O 1 + |x − y|α−m−|i|−| j|−δ uniformly for x = y in compact subsets of A. If α > m + |i| + | j|, we require j also that ∂xi ∂ y (∂x + ∂ y )k K (x, y) extend continuously to {x = y}; if m > 1 and |i| + | j| < r , this condition is superfluous, since A × A is not disconnected by the diagonal. We say the kernel K has exponent α, and we will sometimes use “operator of order −α” to mean an integral operator with kernel of exponent α. (See Theorems 7.1.2 and 7.1.3 for partial justification of this usage.) Remark. If φ : Rm \{0} → R is differentiable, (∂x + ∂ y )φ(x, y) = 0. Thus if φ is C r and homogeneous of degree α − m on Rm \{0}, φ(x − y) is a C∗r kernel. K (x, y) is of class C∗r (α) if and only if ∂x K (x, y) and ∂ y K (x, y) are of class C∗r −1 (α − 1) and (∂x + ∂ y ) K (x, y) is C∗r −1 (α). If α > r + m, a “kernel of class C∗r (α)” simply means a C r function. If K (x, y) is a C∗r (α) kernel in A ⊂ Rm , h : A0 ⊂ Rm → A is a C r +1 diffeomorphism and ψ : A0 × A0 → R is of class C r , then (x, y) → ψ(x, y)K (h(x), h(y)) is a C∗r (α) kernel in A0 . This permits definition of C∗r (α) kernels on C r +1 manifolds (Assumptions 7.1.4 below). Example. Let Q(x, y, σ ) be C r on (Rm )3 , α > 0, j = integer ≥ 0 and   y−x α−m j (log |x − y|) Q x, y, K (x, y) = |x − y| for x = y; |y − x| then K is a C∗r (α) kernel in Rm . We will see many kernels of this kind later. Here we have used the Euclidean norm | · | but we have the same conclusion with | · |x,y in place of | · |, where |v|x,y = (v · G(x, y)v)1/2 , G(x, y) is a C r function whose values are symmetric positive matrices. For this case, in a neighborhood of a given x0 ∈ Rm , let |v|0 = |v|x0 ,x0 and write y−x (log |x − y|0 )k Q ik (x, y, |y−x| ) K (x, y) as a finite sum of kernels |x − y|α+i−m 0 0 [where i ≥ 0, 0 ≤ k ≤ j, Q ik (x, y, σ ) is C r ], plus a C r function of x, y. Theorem 7.1.2. Let K be a C∗r (α) kernel in Rm with compact support and values in Ca×b (= the a × b complex matrices), and let K˜ denote the corresponding integral operator  K (x, y)u(y)dy, x ∈ Rm . K˜ u(x) = Rm

118

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

Then K˜ is a compact operator in any of the following pairs of spaces (with j, k integers in 0 ≤ j ≤ k ≤ r ; 1 ≤ p, q ≤ ∞; 0 ≤ σ, τ ≤ 1), (1) (2) (3) (4)

W j, p (Rm , Cb ) → W k,q (Rm , Ca ) if j − m/p + α > k − m/q C j,σ (Rm , Cb ) → W k,q (Rm , Ca ) if j + σ + α > k − m/q W j, p (Rm , Cb ) → C k,τ (Rm , Ca ) if j − m/p + α > k + τ, k < r C j,σ (Rm , Cb ) → C k,τ (Rm , Ca ) if j + σ + α > k + τ, k < r, k < j + α.

Recalling that j − m/p is the “net smoothness” of functions in W j, p (Rm ), an approximate summary is: for 0 < α ≤ r, K˜ is smoothing of order α, or an operator of order −α. Proof. The identities   ∂ (when α > 1, r ≥ 1) K (x, y)u(y)dy = ∂x K (x, y)u(y) ∂x  = (∂x + ∂ y )K (x, y)u(y)  + K (x, y)∂ y u(y) (when α > 0, r ≥ 1) allow us to reduce the case (α, r ; j, k) to (α − 1, r − 1; j, k − 1) when α > 1, r ≥ 1, or to (α, r − 1; j, k − 1) and (α, r ; j − 1, k − 1) when α > 0, r ≥ 1. Using this and the Sobolev embeddings (in the cases α > r ; r ≥ α > k; r ≥ k ≥ α), we may reduce to one of the following cases: (1) (2) (3)

K˜ : L p → L q is compact if α > 0, r ≥ 0, α − m/p > −m/q, p ≤ q; K˜ : L p → C τ is compact if 0 < α ≤ 1 = r, α − m/p > τ ; K˜ : C σ → C τ is compact if 0 < α ≤ 1 = r, α + σ > τ . Here C σ = C 0,σ or C 0,σ + for σ < 1, and for σ = 1 may denote C 1 or C 0,1 (= Lipschitz), whichever gives the stronger conclusion.

In cases (2) and (3), it suffices to prove continuity; compactness follows from the compactness of the support of K and the compactness of the inclusion C σ + ⊂ C σ ( > 0). (1) Continuity follows immediately from Young’s inequality ( f ∗ g L q ≤

f L s g L p when 1 ≤ p, q, s ≤ ∞ and 1/q = 1/ p + 1/s for the conjugate exponents). If θ : Rn → R is C ∞ with compact support, θ ≡ 1 ))K (x, y) for > 0, then K˜ is near 0, and K (x, y) = (1 − θ( x−y certainly compact for any > 0 and Young’s inequality shows

K˜ − K˜ L(L p ,L q ) → 0 as → 0, so K˜ is compact.

7.1 Weakly-Singular Integral Operators with C∗r (α) Kernels

119

(2) For λ < α − m (arbitrarily close) we have |K (x , y) − K (x, y)| ≤ C min{|x − y|λ + |x − y|λ , |x − x|(|x − y|λ−1 + |x − y|λ−1 } ≤ C |x − x |θ (|x − y|λ−θ + |x − y|λ−θ ) for 0 ≤ θ ≤ 1 (constants C, C ). Take θ = τ and (2) follows from H¨older’s inequality. (3) Say K (x, y) = 0 if |x| > R or |y| > R. For |x| ≤ R, |x + h| ≤ R  K˜ u(x + h) − K˜ u(x) =

|y| 1 this gives continuity into C 0,1 (= Lipschitz); but it is easy to see K˜ (C 1 ) ⊂ C 1 so in fact K˜ (C σ ) ⊂ C 1 . Theorem 7.1.3. Let K , L be kernels with compact support of class C∗r (α), C∗r (β) respectively in Rm ; we suppose their values can be multiplied. Then K ◦ L defined by  K (x, z)L(z, y)dz K ◦ L(x, y) = Rm

is a kernel of class C∗r (α + β) in Rm with compact support. If α + β > r + m, K ◦ L is a C r function on Rm × Rm . Proof. The case r = 0 is well-known (see, for example, W. Pogorzelski, Integral Equations and Their Applications, vol. I (Pergamon Press, 1966)). Suppose r ≥ 1, α > 1; we have  (∂x + ∂ y )K ◦ L(x, y) = (∂x + ∂ y )K (x, z)L(z, y)  + K (x, z)(∂z + ∂ y )L(z, y)  ∂x K ◦ L(x, y) = ∂x K (x, z)L(z, y)   ∂ y K ◦ L(x, y) = ∂z K (x, z)L(z, y) + K (x, z)(∂z + ∂ y )L(z, y)

120

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

so the case (r, α, β) reduces to (r − 1, α − 1, β), when r ≥ 1, α > 1. Similarly (r, α, β) reduces to (r − 1, α, β − 1) when r ≥ 1, β > 1. Thus we reduce eventually to either r = 0 – which is known – or to r ≥ 1 with 0 < α, β ≤ 1, which we now examine. Let θ : Rn → R be C ∞ , θ (x) = 0 for |x| ≤ 1, θ(x) = 1 for |x| ≥ 2, and for any > 0 define      y−x y−x ˜ K (x, y), K (x, y) = 1 − θ K (x, y), K (x, y) = θ and similarly define L , L˜ . We estimate the derivatives of K ◦ L near a given (x0 , y0 ), x0 = y0 . For any > 0, K ◦ L = K ◦ L + K ◦ L˜ + K˜ ◦ L + K˜ ◦ L˜ and we choose = 15 (x0 − y0 ). For (x, y) in a neighborhood of (x0 , y0 ) we have 4 < |x − y| < 6 so K˜ ◦ L (x, y) = 0; and since K , L are C r funcj tions, we see K ◦ L is C r near (x0 , y0 ). Now for |i| + | j| + |k| ≤ r, ∂xi ∂ y (∂x + ∂ y )k K˜ ◦ L (x, y) is a sum of terms       x−z y−z



1−θ θ (∂x + ∂z )k +i K (x, z) const.





× ∂zi ∂ yj (∂ y + ∂z )k L(y, z) with k + k

= k, i + i

= i. The terms containing some derivative of ) vanish, since this gives disjoint supports. Since 0 < α, β ≤ m and θ( y−z |y − z| > , |x − z| < 2 on the support of the integrand, we may estimate the integral by  |x − z|α−m−δ β−m−i− j−δ dz = O( α+β−m−i− j−2δ ) const. |x−z| 0. Similarly for K ◦ L˜ . In the case K ◦ L , we have a sum of terms      x−z z−y i



const. θ θ ∂x (∂x + ∂z )k K (x, z)∂ yj (∂z + ∂ y )k L(z, y) (k + k

= k), in addition to integrals where one or more of the derivatives ∂xi ∂ y falls on θ( x−z ) or θ( z−y ). But in the latter, |x − z| or |z − y| is comparable to in the support of the integrand, and the estimates are like the previous case. In the remaining integrals we have |x − z| ≥ , |y − z| ≥ and we write the j

7.1 Weakly-Singular Integral Operators with C∗r (α) Kernels

integrals as I + I I = 0 and p ∈ Hk (Rm ), |x|α−m p(x/|x|) or more precisely  φ→

(7.3.1)

Rm

|x|α−m p(x/|x|)φ(x)d x,

φ ∈ S(Rm ),

is a tempered distribution which is an analytic function of α and has analytic continuation for all complex α with α ∈ {−k, −k − 2, −k − 4, · · · }. In fact, for any integer N ≥ 0 

|x|α−m p(x/|x|)φ(x)   = |x|α−m p(x/|x|)φ(x) + |x|α−m p(x/|x|) |x|>1 |x| 0, ξ ∈ Rm ,  exp (−λ|x|2 − iξ · x) p(x)d x Rm  √ = dz exp(−z 2 − ξ 2 /4λ)λ−(m+k)/2 p(z − iξ/2 λ) Rm m/2 −m/2−k −ξ 2 /4λ

λ

= π

e

p(−iξ/2).

Let φ ∈ Cc∞ (Rm \{0}); then for 0 < δ < (m + k)/2,  ˆ (δ)|x|−2δ p(x)φ(x)d x Rm  ∞  2 ˆ = λδ−1 dλ e−λx p(x)φ(x)d x 0

= π m/2

 Rm

Rm



∞

λδ− 2 −k−1 e−ξ m

2

/4λ

 dλ p(−iξ/2)φ(ξ )dξ,

0

and we have used the fact that |ξ | ≥ const. > 0 on supp φ. If also δ > k/2, the last integral converges and the equation holds for all φ ∈ S(Rm ), by continuity. Now let δ = 12 (m + k − α); k/2 < δ < (m + k)/2 when 0 < α < m, and this gives the result claimed for 0 < α < m. Analytic continuation gives the general case.

7.3 Fourier Transform and Composition of Weakly-Singular Kernels

129

Now we examine the transform for α near {m + k, m + k + 2, . . .}. If α = m + k + 2 j ( j = integer ≥ 0), |x|α−m p(x/|x|) = |x|2 j p(x) is a polynomial and the transform is well-known: |x|2 j p(x)}∧ (ξ ), φ(ξ ) = (2π )m − (−
ξ ) j p(−i∂ξ ) φ (ξ )|ξ =0 so the transform is a derivative of Dirac’s δ, with support {0}. We compute the derivative with respect to α: {|x|2 j p(x) log |x|}∧ (ξ ), φ(ξ ) 

  −α

= lim γα,k log |ξ |−1 + γα,k |ξ | p(ξ/|ξ |), φ(ξ ) . α→m+k+2 j

Using formula (7.3.2) – or its derivative with respect to α – we see that this is γ f.p.| ξ|−m−k−2 j p(ξ/|ξ |), φ(ξ ) + 

 P(ω)ω2 |ω|=1

lim



α→m+k+2 j

γα,k

mk + 2 j − α

+

|S|=k+2 j

1 S ∂ φ(0) S! 

γα,k (m + k + 2 j − α)2

∂



where γα,k = ∂α γα,k , and γ is γα,k for α = m + k + 2 j. If α = m + k + 2 j + δ, the final coefficient above is

γ + δγ

+ O(δ 2 ) δγ + 12 δ 2 γ

+ O(δ 3 ) 1 + → − γ

. 2 −δ δ 2 Thus there is a polynomial q(x), homogeneous of degree k + 2 j, such that (7.3.4) and (7.3.5)

{|x|2 j p(x) log |x| + q(x)} ∧ (ξ ) = γ f.p. (|ξ |−m−k−2 j p(ξ/|ξ |))   k+α ∂ j+1 −k α−1 m/2 γα,k |α=m+k+2 j = j!(−1) i 2 π  γ = = 0. ∂α 2

In fact, q(x) = (γ

/2γ )|x|2 j p(x) – but we do not need to know q explicitly. Differentiating j times with respect to α gives  ∧  j   x j (s) α−m j (log |x|) p (ξ ) = (log |ξ |−1 ) j−s |ξ |−α p(ξ/|ξ |) |x| γα,k s |x| S=0 (s) ∂ s where γα,k = ( ∂α ) γα,k , so we may handle integral powers of log |x| with equal ease (or difficulty, depending on α). We may also treat more general “direction functions” φ(x/|x|) by expansion in harmonic polynomials. Choose an orthonormal basis for L 2 (S m−1 ),

130

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

{ p j |S m−1 }∞ j=0 , with each p j a homogeneous harmonic polynomial ( p0 = constant) and j → d j = deg p j non-decreasing. (We assume m > 1; m = 1 j)1/m−1 as j → ∞,
S pj = −d j (d j + m − 2) p j is trivial.) Then d j ≈ ( (m−1)! 2 m−1 on S , and the Sobolev spaces on S m−1 may be written σ

H (S

m−1

) = D(1 −
s)

σ/2

  ∞ ∞  2 σ  = Cj pj |C j | (1 + d j (d j + m − 2)) < ∞ 0

0

for all σ ≥ 0. If α is not an integer and φ = 

α−m

|x|

(7.3.6)

 φ

x |x|

∧

∞ 0

(ξ ) = |ξ |

−α

φ j p j ∈ H σ (S m−1 ) then  Tα φ

ξ |ξ |



 m−1 where Tα φ(ω) = ∞ . If α is an integer, there may 0 φ j γα,d j p j (ω), ω ∈ S be finitely many j for which γα,d j is zero or ∞ (the troublesome terms have d j ≤ max(α − m, −α)); such terms are treated separately, and we define Tα only in the orthogonal complement. Since |γα,d | ≈ (2π)m d α−m/2 as d → ∞, it follows that Tα : H σ (S m−1 ) → H σ −α+m/2 (S m−1 ) [modulo, perhaps, a finitedimensional subspace] is an isomorphism. For example, if α, β and α + β are not integers (to simplify) and σ > −1/2, φ ∈ H σ +α (S m−1 ), ψ ∈ H σ +β (S m−1 ), then         x x x α−m β−m α+β−m φ ψ θ |x| ∗ |x| = |x| |x| |x| |x| −1 where θ = Tα+β (Tα φ · Tβ ψ) ∈ H σ +α+β (S m−1 ). Here we use the fact that σ m−1 ) is an algebra under pointwise multiplication when σ > (m − 1)/2. H (S Fortunately our explicit calculations involve only polynomials, hence finite sums. To facilitate these calculations, we include a table of values of γα,k and introduce n = m + 1 (which occurs naturally in our calculations).

n = m + 1, γα,k

ω = ωn ,

ω¯ = ωn−1 = ωm = 2π m/2 / (m/2).

k=0

k=1

k=2

k=3

k=4

/2 (n − 2)ω

−i ω¯

−1/2 ω

2i ω(n ¯ − 1)

3ω/2n

(n − 3)ω¯

−1/2 (n − 2)iω

−2ω¯

/2 iω

8ω/(n ¯ − 1)

α = 3 /2 (n − 2)(n − 4)ω

−2(n − 3)i ω¯

− /2 (n − 2)ω

α = 4 2(n − 3)(n − 5)ω¯

− /2 (n − 2)(n − 4)iω

−8(n − 3)ω¯

α=1

1

α=2 1

3

3

8i ω¯

3

/2 (n − 2)iω

15

/2 iω

15

48ω¯

When γα,k = 0, this table cannot be used to find the inverse transform; but in ∂

= ∂α γα,k = 0 and we can use (7.3.5). We complement this table that case γα,k

7.3 Fourier Transform and Composition of Weakly-Singular Kernels

131

for the points where γα,k = 0. by giving the derivative γα,k

n=2 (m=1)

k=0 1 2 3 γα,k α = 1 −π πi α=2 r r 3π α = 3 12π r r −36πi −15πi α=4

n=4

k=0 α=1 2 3 −2π 2 r 4

1

6π 2 i

n=3

n=5

k=0 1 2 α=1 2 −2π 4πi 3 r r 16π 4 8π

k=0 α=1 2 3 4 −16π 2

Another simpler method seems to always work, which we illustrate with 1 1 |x|3−n |x|2−m = − ωn−1 ; and an example. If n = 3, (F.T.)−1 (|ξ |−2 ) = (n−3)ω 3−n n−1 the result for n = 3 is obtained by replacing |x|3−n /(3 − n) by log |x|, 1 log |x|. More precisely, (F.T.)−1 (f.p.|ξ |−2 ) is i.e., (F.T.)−1 (|ξ |−2 ) = − 2π 1 − 2π log |x| plus a constant, when n = 3 (m = 2). The Fourier transform shows the action of an integral operator on rapidly oscillating functions. Suppose, for example,  is a bounded C r +1 region in Rn , K (x, y) is a C∗r (α) kernel in ∂, 0 < α ≤ r , φ is C r on ∂ and ψ is C r +1 and real-valued with ∇∂ ψ(y) = 0 on supp φ, and consider  (7.3.7) J (t, x) ≡ K (x, y)φ(y) exp(it ψ(y))d Ay, x ∈ ∂ ∂

as t → +∞. If x is outside supp φ, it is easy to see J (t, x) = O(t −r ), and for every x (and > 0) we show J (t, x) = O(t −α+ ) as t → +∞. It is enough to treat the case when supp φ is small, and then we may write ∂ (in supp φ) as {h(ξ ) | ξ ∈ V open ⊂ Rn−1 }, h : V → ∂ ⊂ Rn is a C r +1 embedding, and   J (t, h(ξ )) = K (h(ξ ), h(η))φ(h(η)) exp(itψ(h(η))) det  h η h η dη. V

We may choose new coordinates ζ in Rn−1 so that ζ1 = ψ(h(η)) and the problem reduces to: if K˜ is a C∗r (α) kernel with compact support in Rm (m = n − 1)  K˜ (x, y)eit y1 dy = O(t −α+ ) as t → +∞. Rm

132

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

If r ≥ α > 1 then Rm K˜ (x, y)eit y1 dy = − it1 Rm ∂ y1 K˜ (x, y)eit y1 dy and we eventually reduce to the case 0 < α ≤ 1 ≤ r . But in this case, if e1 = (1, 0, . . . , 0),   K˜ (x, y)eit y1 dy = − K˜ (x, y + πe1 /t)eit y1 dy m m R R 1 = /2 ( K˜ (x, y) − K˜ (x, y + π e1 /t))eit y1 dy Rm

so

   

  it y1 ˜ K (x, y)e dy  Rm  ≤ Const. (|y − x|σ + |y + πe1 /t − x|σ ) 5 |y−x|< /t  t −1 (|y − x|σ −1 + |y + π e1 /t − x|σ −1 ) + Const. R≥|y−x|≥5/t

  ≤ O t −(σ +m) ) = O(t −α+ , where σ = α − m − . Suppose K (x, y) − K 0 (x, y − x) is a C∗S (β) kernel on ∂ with r ≥ s ≥ β > α, K 0 (x, z) depends only on the component of z in the tangent space transform Kˆ 0 (x, ·) of K 0 (x, ·)|Tx (∂) is computable. Tx (∂),
and the Fourier itψ(y) d Ay = Kˆ 0 (x, −t∇∂ ψ(x))φ(x)eitψ(x) + o(t −α ) as Then ∂ K (x, y)φ(y)e t → +∞. If z → K 0 (x, z) is homogeneous of degree α − m, and if K˜ is of finite rank, it follows that K 0 (x, z) = 0 for all z ∈ Tx (∂). Now consider a bounded open set  ⊂ Rn with C r +1 boundary ∂ (r ≥ 2) and a C∗r (α) kernel K (x, y) on ∂ of the form   y−x α−m j K (x, y) = |x − y| (log(x − y)) Q x, y, |y − x| where m = n − 1 = dim ∂, α > 0 and Q(x, y, ω) is a C r function. For y ∈ ∂ near x, y = x + η − νx (η)N x , η ∈ Tx (∂); i.e., we may write ∂ as a graph over the tangent plane at x, η = Px (y − x) is the orthogonal projection of y − x onto Tx (∂), and νx (η) = 1/2 K x (η2 ) + O(|η|3 ) as η → 0, where K x is the curvature at x. Then |x − y| = |η|(1 + νx (η)2 /|η|2 )1/2 = |η|(1 + O(|η|2 )) and K (x, x + η − νx (η)N x ) = |η|α−m (log |η|) j Q(x, x, η/|η|) + |η|α+1−m (log |η|) j (ωQ y (x, y, ω) −1/2 K x ω2 N x Q ω (x, y, ω))| y=x,ω=η/|η| + · · · N −1 K˜ j (x, η) + K˜ N (x, η) = j=0

7.3 Fourier Transform and Composition of Weakly-Singular Kernels

133

 r− j or K (x, y) = 0N K j (x, y), K j (x, y) = K˜ j (x, Px (y − x)), where K j is C∗ (α + j) (0 ≤ j ≤ N ) and the Fourier transforms of η → K˜ j (x, η) (0 ≤ j < N ) are explicitly computable when ω → Q(x, y, ω) is a polynomial. Suppose L is another such C∗r (β) kernel on ∂, and we compute the composition K ◦ L,  K (x, y)L(y, z)d A y . K ◦ L(x, z) = ∂

Let θ : Rn → R be C ∞ with small support, but θ ≡ 1 near 0 : a “cutoff ” function. If K , L are C∗r (α), C∗r (β) respectively, K θ (x, y) = K (x, y)θ(y − x) and L θ (y, z) = L(y, z)θ(z − y), then K − K θ , L − L θ are C r kernels and K ◦ L − K θ ◦ L θ is also a C r kernel. Without loss of generality, we shall assume K and L are supported in a small neighborhood of the diagonal of ∂ × ∂. Define h x : Tx (∂) → ∂ : η → x + η − νx (η)N x (η small) and then for z close to x   K (x, h x (η)) 1 + |νx (η)|2 · L(h z (η), ˜ z)|h z (η)=h dη. K ◦ L(x, z) = ˜ x (η) Tx (∂)

We can solve h z (η) ˜ = h x (η) for η˜ ∈ Tz (∂) as a function of x, z and η ∈ Tx (∂). In fact, let z = h x (ζ ), ζ ∈ Tx (∂) near 0, and define Jx,z = I − N x ⊗ νx (ζ ) ∈ Rn×n , where the gradient νx (ζ ) is considered as a vector in Tx (∂) ⊥ N x . Jx,z is an isomorphism of Rn and it carries Tx (∂) onto Tz (∂): given a curve ζ = ζ (t) in Tx (∂) we have h x (ζ (t)) ∈ ∂ and d h x (ζ (t)) = Jx,z ζ˙ (t) ∈ Tz (∂) dt if h x (ζ (t)) = z. Now h z (η) ˜ = h x (η) with η˜ = Jx,z σ, z = h x (ζ ), (σ, ζ ∈ Tx (∂), becomes η − ζ − (νx (η) − νx (ζ ))N x = Jx,z σ − νz (Jx,z σ )Nz . Since Px Jx z σ = σ for σ ∈ Tx (∂), we have η − ζ = σ − νz (Jx z σ )Px Nz . But τ = σ − νz (Jx z σ )Px Nz (σ, τ ∈ Tx (∂)) is solvable for σ = Fx,z (τ ) = τ + ˜ = h x (η) is solvable for η˜ = Jx z Fx z (η − ζ ) and the integral O(|τ |2 |ζ |) so h z (η) above becomes   K (x, h x (η)) 1 + |νx (η)|2 L(h z (Jx z Fx z (η − ζ )), z)dη, K ◦ L(x, z) = Tx (∂)

134

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

which is almost a convolution. Let  N K˜ j (x, η), K (x, h x (η)) 1 + |νx (η)|2 = 0

(7.3.8)

K˜ j (x, η) = O(|η|

α+ j−m−δ

)

as η → 0 in Tx (∂), for any δ > 0, and N     L˜ j (x, z, τ ), L h z Jx z Fx z Jx−1 τ , z = z j=0

(7.3.9)

L˜ j (x, z, τ ) = O(|τ |

β+ j−m−δ

)

as τ → 0 in Tz (∂), for any δ > 0. We suppose η → K˜ j (x, η) ( j < N ) on Tx (∂) and τ → L˜ k (x, z, τ ) (k < N ) on Tz (∂) have explicitly-computable Fourier transforms. Then we can compute the sum of convolutions  K˜ j (x, η) L˜ k (x, z, Jx z (η − ζ ))dη (7.3.10) j+k0 |ω|=1

Q(x, x, ω)/ω0 N x d Aω



= lim+ →0



ω·N x > |ω|=1

Q(x, x, ω)/ω · N x .

[When ω → Q(x, x, ω) is a polynomial, we may compute q(x) using (7.4.3) below.]

136

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

Example. (n = 2) ∂ = Y ⊂ R2 ≈ C and using complex notation,  g(ζ ) dζ for z ∈ , J (z) = ζ −z γ where γ is given the usual orientation. Then τζ = dζ /ds is the “positive” unit tangent vector at ζ ∈ γ
and −iτζ is the outward normal, and with −z ) |ζds . The “unit sphere” of Tζ γ Q(z, ζ, ω) = ωτζ g(ζ ), J (z) = γ Q(z, ζ, |ζζ −z| −z| has two points, ±τζ , and Q(ζ, ζ, τζ ) + Q(ζ, ζ, −τζ ) = 0. Also  Q(ζ, ζ, ω) q(ζ ) = lim ω·N > →0 ω·N |ω|=1  π− dθ g(ζ )(cos θ + i sin θ) = iπg(ζ ) = lim →0 sin θ so

 lim z→z 0 z∈R

g(ζ ) dζ = iπg(z 0 ) + P.V. ζ −z

γ

 γ

g(ζ ) dζ ζ − z0

for z 0 ∈ γ ,

one of the formulas of Plemelj. We first treat a simpler case. Lemma 7.4.2. Let m ≥ 1 and suppose Q(x, t; ξ, τ ) is Lipschitzian on a neighborhood of (x, t) = (0, 0) ∈ Rm × R and (ξ, τ ) ∈ S+m , S+m = {(ξ, τ ) ∈ Rm × R : |ξ |2 + τ 2 = 1, τ ≥ 0}. Also assume Q(x, t; ξ, τ ) = 0 when |x| ≥ r0 and  Q(0, 0; ξ, 0)d Aξ = 0. S m−1

Define J (t) = then


Rm

Q(x, t; x/R , t/R )R −m d x, R =

lim+ J (t) = q + P.V.

t→0

where



 Rm



|x|2 + t 2 , for small t > 0;

  Q x, 0; x/|x| , 0 |x|−m d x

1 {Q(0, 0; ξ, τ ) − Q(0, 0; ξ/|ξ |, 0)}d Aξ,τ S+m τ  1 Q(0, 0; ξ, τ )d Aξ,τ (if we integrate first with respect to ξ ) = S+m τ  π/2  (sin θ)m−1 = dθ Q(0, 0; σ sin θ, cos θ)d Aσ cos θ 0 S m−1

q=

7.4 Limits of Singular Integrals Proof. Let q(t) =


|x| 0). j=1 f j (x) ⊗ g j (y) for some f j , g j ∈ C

7.6 Calculation of Some Boundary Operators

141

Let S1 (x, y) = E n (z){I + β(x) · z + 1/2 z · B(x)z + K n γ (x)|z|2 (log |z|)σ },   z = y − x, where β(x) · z = n1 β j (x)z j , z · Bz = nj,k=1 B jk z j z k , β j (x) = 1 /2 b j (x), B jk (x) = Bk j (x) = 1/2 (β j βk + βk β j + ∂β j /∂ xk + ∂βk /∂ x j ), γ (x) =   1 1 /2 (C(x) − n1 ∂β j /∂ x j − n1 β 2j ), kn = n−4 and σ = 0 when n = 4; kn = −1, σ = 1 when n = 4. Then S(x, y) − E n (x − y)I is C∗r +2 (3), S(x, y) − E n (x − y){I + β(x) · (y − x)} is C∗r +1 (4) and S(x, y) − S1 (x, y) is C∗r (5).  n Proof. Let R1 (x, y) be defined by L ∗ y S1 (x, y) = R1 (x, y), x = y in R ; then R1 is a kernel of class C∗r (3) on Rn . Let B be a large ball in Rn such that Q 0 ⊂ B and let {g1 , . . . , g N } be an L 2 (B)-orthonormal basis for {g | L g = 0 in B, g = 0 on ∂ B}; we may suppose the g j ∈ C r +3− (B), for any > 0, since the coefficients are C r +1 . We seek a solution y → S2 (x, y) of L ∗ y S2 (x, y) = N   − R1 (x, y) + 1 g j (y ⊗ f j (x) (y ∈ B) with S2 (x, y) = 0 for y ∈ ∂ B and n (for n > 3), some appropriate f j . Since R1 (x, ·) ∈ L p (B) when 1 < p < n−3 1, p 2, p or any 1 < p < ∞
(for n = 2, 3), there is a solution S2 (x, ·) ∈ W ∩ W0 (B) provided f j (x) = B R1 (x, y)g j (y)dy (x ∈ B, 1 ≤ j ≤ N ), which is at least C r +2 on B. If G(x, y) is the Green function for the Laplacian in B, we may write an integral equation for S2

 S2 (x, y) =

G(y, z)(F(x, z) + σ (x, z)) 

B

σ (x, y) =

G 1 (y, z)(F(x, z) + σ (x, z)) B

where F(x, y) = −R1 (x, y) + # G 1 (y, z) =

N

f j (x) ⊗ g j (y)

1

n 1

$ n ∂ b j (y) − c(y) + ∂ j b j (y) G(y, z). ∂yj 1

Now G is C∗∞ (2) kernel while G 1 is C∗r +1 (1), F is C∗r
(3). Let  be a resolvent for kernel for G 1 , class C∗r +1 (1), and then σ (x, y) = B G 2 (y,
z)F(x, z) (plus, perhaps, something from the kernel), G 2 (y, z) = G 1 (y, z) + B (y, ·)G 1 (·, z) is C∗r +1 (1), so σ is C∗r (4). Substitution shows S2 is C∗r (5). The other claims are proved similarly.

7.6 Calculation of Some Boundary Operators Let b j : Rn → C p× p be locally C r +2 (1 ≤ j ≤ n), C : Rn → C p× p locally C r +1 and let S(x, y) be an approximate fundamental solution (Th. 7.5.1)

142

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

for Lx =
+

n 1

b j (x)

∂ + c(x), ∂x j


=

n ∂2 . 2 1 ∂x j

We will obtain representations of various solution operators associated with L (in a given C r +4 smooth region ) such as: 1) A L f + B L g = u is a solution of Lu − f | ∈ {finite},

u|∂ = g

where {finite} is some unspecified element of a specific finite-dimensional space – in this case, complementary to the image of L | H 2 ∩ H01 (, C p ). We also require that u belongs to a given complement to the kernel of L | H 2 ∩ H01 . 2) A L ,M f + C L ,Mg = v is a solution of Lv − f | ∈ {finite}, where Mv =

∂v ∂ N

Mv|∂ = g

+ A(x)v, given A : Rn → C p× p of class C r +2 .

For our applications, operators of finite rank are negligible and will be simply omitted, or indicated by a term “{finite}.” The first step is to obtain integral equations for the operators. 1) Let u = A L f + B Lg as above; by (7.5.4)     ∂S S(x, y) f (y)dy + (x, y)g(y) − S(x, y)b, N (y)g(y) u(x) =  ∂ ∂ N y  ∂u − (y) + {finite}, for x ∈  S(x, y) ∂ N ∂ so

∂u | ∂ N ∂

≡ u˙ satisfies

 ∂S ˙ ˙ + /2 u(x) (x, y)u(y)dA y ∂ ∂ N x  ∂S 1 = (x, y) f (y)dy + b · N (x)g(x) 2  ∂ Nx  ∂S − (x, y) b · N (y)g(y)dA y ∂ ∂ N x  ∂ 2 S(x, y) (7.6.1) + lim g(y) + {finite} x→∂ ∂ ∂ N x ∂ N y
2 for x ∈ ∂. The term ∂ ∂∂ NS(x,y) g(y) dA y has a limit as x → ∂ (from the x ∂ Ny interior) provided g is C 2 , since all the other limits exist, we examine this in more detail below. 1

7.6 Calculation of Some Boundary Operators

143

2) Let v = A L ,M f + C L ,M g; by (7.5.4)   v(x) = S(x, y) f (y)dy − S(x, y)g(y)dA y  ∂    ∂S + (x, y) + S(x, y)(A(y) − b · N (y)) v(y)dA y + {finite} ∂ ∂ N y for x ∈ , and v|∂ ≡ v0 satisfies  1 /2 v0 (x) − K (x, y)v0 (y)dA y ∂   (7.6.2) = S(x, y) f (y)dy − S(x, y)g(y) + {finite} 

∂

∂S (x, ∂ Ny

for x ∈ ∂, where K (x, y) =

y) + S(x, y)(A(y)b · N (y))|∂×∂ .

Suppose R(x, y) is the resolvent kernel for 2K (x, y) provided by Th. 7.2.3 – ˜ modulo operators of finite rank. Since K is a kernal of (I − 2 K˜ )−1 = I + R, class C∗r +1 (1) on ∂(K (x, y) = O(|x − y|2−n ) = O(|x − y|1−m ) on ∂, m = n − 1 = dim ∂), and ∂ is C r +2 (in fact, C r +4 ), it follows that R is C∗r +1 (1) and the solution of (7.6.2) is     S(x, y) f (y) + 2 R(x, z) S(z, y) f (y) d A z v0 (x) = 2  ∂      −2 S(x, y)g(y) − 2 R(x, z)S(z, y)d A z g(y) + {finite}. ∂

∂

∂

In particular  (7.6.3)

C L ,M g|∂ (x) = −2

where

∂

C(x, y)g(y)dA y + {finite}

 C(x, y) = S(x, y) +

∂

R(x, z)S(z, y)d A z

= E n (x − y)I + (kernel of class C∗r +1 (2) on ∂) To obtain more detail, note C(x, y) = E n (x − y)I + E n (x − y)β(x) · (y − x)  + R0 (x, z)E n (z − y)d A z + (class C∗r (3)) ∂

2x{ ∂ En∂(x−y) Ny

where R0 (x,
y) = + E n (x − y)(a(x) − 3/2 b · N (x))} so we want
∂ En (x−z) to compute ∂ E n (x − z)E n (z − y) and ∂ ∂ Nz E n (z − y) – or at least,

144

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

their leading terms. Computation of the leading terms is easy, since we merely 1−m find the convolution in the tangent space. Thus E n (x − y) = |x−y| so, from (1−m)ωn the table in 7.3,  ∂

E n (x − z)E n (z − y)d A z ≈

2 |x − y|2−m γ1,0 |x − y|3−n = , (2 − n)2 ωn2 γ2,0 4ωn−1 (n − 3)

−1 log |x − y|; if n = 2, − 1/8 |x − y| at least for n > 3; if n = 3, we have 8π 0 (which is the result obtained with n = 2 in the general expression above). With p the convention that “ |z|p ” means log |z| when p = 0, we have

 ∂

(7.6.4)

E n (x − z)E n (z − y)d A z = −

|x − y|3−n 4ωn−1 (3 − n)

+ (kernel of class C∗r +3 (3)).

Now ∂ E n (x − z) Nz · (z − x) ζ · K (x)ζ = = + O(|ζ |3−n ) ∂ Nz ωn |z − x|n 2ωn |ζ |n where ζ = z − x, and K (x) = D N (x) is the (degenerate) curvature matrix (x) which we write K (x) = Kˆ (x) + Hn−1 P(x), P(x) = I − N (x) ⊗ N (x), so trace Kˆ (x) = 0. In the tangent space at x, this is |ζ |1−m ζ · K (x)ζ = 2ωn |ζ |n 2ωn



ζ H (x) ζ · Kˆ (x) + |ζ | |ζ | n − 1



with Fourier transform |ξ |−1 2ωn

#

$ ξ · Kˆ ξ H (x) + γ1,0 γ1,2 , |ξ |2 n−1

so the leading term of the convolution is 

∂ En E n (z − y)d A z ∂ ∂ N x 

|x − y|3−n (y − x) · Kˆ (x)(y − x) 2(n − 2) H (x) =− + 16ωn−1 |y − x|2 (n − 1)(n − 3)

(7.6.5)

+ (kernel of class C∗r +1 (3))

7.6 Calculation of Some Boundary Operators

145

Thus the kernel C(x, y) of (7.6.3) is, more exactly, C(x, y) = E n (x − y)I + E n (x − y)β(x) · (y − x) $ # 2(n − 2) |x − y|3−n (y − x) · Kˆ (x)(y − x) H (x) · I + + 16ωn−1 |y − x|2 (n − 1)(n − 3) −

1 |x − y|3−n + (kernel of class C∗r (3)) (a(x) − 3/2 b · N (x)) 2ωn−1 3−n

(7.6.6) The calculation for the “Dirichlet case,” in the integral equation (7.6.1), are similar but more complicated. We first study the case L =
. Theorem 7.6.1. If  is a bounded C r +4 -regular region in Rn (n ≥ 2, r ≥ 0) and g ∈ C 2 (∂) then  ∂ 2 E n (x − y) g(y)dA y lim x→∂ ∂ Nx ∂ N y ∂ x∈  ∂ 2 E(x − y) = P.V. (g(y) − g(x)) ∂ ∂ N x ∂ N y   (x − y) · ∇∂ g(y) dA y + Q 0 (x, y)g(y) = P.V. ωn |x − y|n ∂ ∂   = E n (x − y)
∂ g(y)dA y + Q 0 (x, y)g(y) ∂

where



Q 0 (x, y) =

∂

∂2 E ∂E ∂2 E + + H (y) ∂ Nx ∂ N y ∂ N y2 ∂ Ny

    

∂×∂

1 = (|K (x)z|2 + H (x)z · K (x)z − n(z · k(x)z)2 /|z|2 ) 2ωn |z|n + O(|z|3−n ), z = y − x K (x) = D N (x) is the (degenerate) curvature matrix (see Th. 1.5). H = trace K is the mean curvature, and Q 0 is a kernel of class C∗r +2 (1) on ∂. If n = 2, Q 0 is a C r +2 function.
dA y = χ (x) so for x ∈  Proof. ∂ ∂ En∂(x−y) Ny   ∂ 2 E(x − y) ∂ 2 E(x − y) g(y) = (g(y) − g(x)), ∂ ∂ N x ∂ N y ∂ ∂ N x ∂ N y to which we apply Th. 7.4.1. We have ∂ 2 E(x − y) Dg(x) · z (g(y) − g(x)) = ∂ Nx ∂ N y ωn |z|n

  (N (x) · z)2 −1 + n + O(|z|2−n ) |z|2

146 so, if

Chapter 7. Boundary Operators for Second-Order Elliptic Equation z |z|

= ξ + t N (x)(ξ ⊥ N (x)) the first term becomes

Dg(x) · ξ + t ∂∂gN (x) (−1 + nt 2 ), ωn and (7.4.3) gives q(x) = 0 and the first equality claimed above. Now for x ∈ , y ∈ ∂, |z|1−n


y E n (x − y) = 0 =
∂ y E(x − y) + H (y) so



∂ E(x − y) ∂ 2 E(x − y) + ∂ Ny ∂ N y2



∂2 E ∂E ∂2 E + + H (y) 2 ∂ Ny ∂ N y2 ∂ ∂ N x ∂ N y   E(x − y)
∂ g(y) + =− ∂

∂

 g(y) ∂2 E g(y) ∂ Nx ∂ N y

for x ∈ . Another application of Th. 7.4.1 (with q = 0 again) gives the third equality of the theorem, which implies the second on integration by parts. To obtain the form claimed for Q 0 , write ∂ near x as a graph over the tangent 3 plane: y = x + η − νx (η)N x with νx (η) = 1/2 η · K (x)η  + O(|η| ) as η → 0



2 (η ∈ Tx (∂) ⊥ N (x)). Then N (y) = (νx (η) + N x )/ 1 + |νx | , considering ∂ νx (η) ∈ Tx (∂), N x˙ (y − x) = − 1/2 η· K (x)η + O(|η|3 ), N y˙ (y − x) = νx (η) = ∂η  1 /2 η · K (x)η + O(|η|3 ), N x˙ N y = 1/ 1 + |νx |2 = 1 − 1/2 |K (x)η|2 + O(|η|3 ). ∂ νx (x) is C r +3 . Note νx (·) is C r +4 for fixed x and (x, n) → ∂η The two-dimensional case is sufficiently simple to deserve sparate treatment. Theorem 7.6.2. For n = 2, ∂ is a finite-union of closed curves and for definiteness we suppose it is a single curve of length L. We write functions on ∂ as functions of the arc-length, extended as L-periodic functions on R. Let λ : R → R be L-periodic, C ∞ on R\|[0, ±L , ±2L , . . .] with λ(t) =

1 log |t|, π

λ (t) =

1 , πt

λ

(t) = −

1 πt 2

near t = 0.

Then there exists R : R × R → R of class C r +1− (any > 0), L-periodic in each argument, so that   ∂

B
g(t) = λ(t − s)g (s)ds + R(t, s)g(s)ds ∂N ∂ ∂   + P.V. λ (t − s)g (s)ds + R(t, s)g(s)ds ∂   ∂

λ (t − s)(g(s) − g(t))ds + R(t, s)g(s)ds = P.V. ∂

∂

7.6 Calculation of Some Boundary Operators

147

for any C 2 g. The middle-form is closest to the usual representation of the Hilbert transform H, ∂∂N B
g = H g , where φ + i H φ is
the restriction to ∂ of an analytic function of x + i y for (x, y) ∈ , with ∂ H φ = 0 to fix the constant. Proof. The kernel Q 0 of the previous theorem is C r +2 on ∂ × ∂, x (x−y) x (η) = π (η2ν+ν where y = x + η − νx (η)N x , as is 2 ∂ N∂ x E(x − y) = π1 N|x−y| 2 2 x (η) ) η ∈ Tx (∂) ≈ R. ; then from Th. 7.6.1 and Let R1 (x, y) be the resolvent kernel for −2 ∂ E(x−y) ∂ Nx (7.6.1) – with S = E n – we find ∂ 1 B
g(t) = ∂N π

 ∂

log |t − s|g

(s)ds +

 ∂

R2 (t, s)g(s)ds

where 

1 ∂2 R2 (t, s) = 2Q 0 (t, s) + 2 R1 (t, ·)Q 0 (·, s) + π ∂s 2 ∂

 ∂

R1 (t, s) log |σ − s|dσ.

Since Q 0 , R1 are C r +2 while log |σ − s| is a kernel of class C∗r +3 (1) on ∂, R2 (t, s) is at least C r, and it is easily shown to be C r +1− ( > 0). Theorem 7.6.3. Let  ⊂ Rn be bounded and C r +4 (n ≥ 3, r ≥ 0). For g ∈ C 2 (∂),  ∂ 2 E n (x − y) n − 2 H (x) ∂ B
g(x) = 2 lim g(x) g(y) − x→∂ ∂ ∂N ∂ Nx ∂ N y n−1 2  n−1 (y − x) · Kˆ (x)(y − x) − P.V. g(y) 2ωn−1 |y − x|n+1 ∂    y−x + |x − y|2−n Q 1 x, g(y) |y − x| ∂ R1 (x, y)g(y), x ∈ ∂, + ∂

where R1 (x, y) is a kernel of class C∗r (2) on ∂ and   1 1 n−1 ζ · D 2 N (x)ζ 2 + ζ · ∇∂ H (x) Q 1 (x, ζ ) = − ωn−1 4 4  1 n 2 n−2 + (ζ · Kˆ (x)ζ )2 − | Kˆ (x)ζ |2 + ζ · Kˆ (x)ζ ωn 6 3 (n − 1)  n−2 1 2 2 ˆ trace( K (x) ) + H (x) + 2(n − 1)2 3(n − 2)

148

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

(x) K (x) = D N (x),H (x) = trace K (x), Kˆ (x) = K (x) − Hn−1 P(x), P(x) = I − N x ⊗ N x . Note K (x) = D 2 t(x),D 2 N (x) = D 3 t(x) where t(x) is the “normal” coordinate of Th. 1.5.

Remark. If n = 2 then Kˆ = 0, D 2 N (x)ζ 3 = ζ · ∇∂ H (x) for ζ ∈ Tx (∂), |ζ | = 1, and the above reduces to the previous theorem; but here we shall assume n > 2. The first term above is treated in Th. 7.6.1. The singular integral (P.V.) above exists since Kˆ (x) | Tx (∂) has trace zero.
Proof. Let Pg(x) = limx→∂ ∂ (∂
2 E/∂ N x ∂ N y )g(y)dA y , the operator treated γ (y)dA y so the integral equain (7.6.7) above, and Mγ (x) = ∂ ∂ E(x−y) ∂ Nx tion (7.6.1) for γ = ∂∂N B
g becomes 1/2 γ + Mγ = Pg. Let γ0 = 2Pg − 4M Pg + 8M 2 Pg and γ1 = γ − γ0 ; then 1/2 γ1 + Mγ1 = −8M 3 Pg. We calculate M P, M 2 P with sufficient accuracy, and see in particular that M 2 P is an integral operator with kernel of class C∗r (1), so M 3 P has kernel of class C∗r (2) and may be absorbed in the remainder.
Now P = P0 + Q˜ 0 , P0 g(x) = ∂ E n (x − y)
∂ g(y) and M P0 has C∗r +2 (2) kernel – which we calculate to within a remainder C∗r +2 (4), so that integration by parts still leaves a remainder C∗r (2). The kernel of M P0 is  K 1 (x, y) =

∂

∂ E(x − z) E(z − y)d A z ∂ Nx

which is clearly C r +3 for x = y and we are concerned only with the behavior near x = y. If z = x + ζ − νx (ζ )N x then (D 2 N )x ζ 2 = νx

(0)ζ 2 − N x |K (x)ζ |2 + O(|ζ |3 ) so N x (x − z) νx (ζ ) ∂ E(x − z) = = n 2 ∂ Nx ωn |x − z| ωn (|ζ | + νx (ζ )2 )n/2   1 1 1 1 2 2 4−n ζ · (D = / ζ · K (x)ζ + N ζ )) + O(|ζ | ) 2 x ωn |ζ |n ωn (ζ )n 6 We calculate the Fourier transform (of the first terms) as before: ξ · Kˆ ξ n−2 1 H |ξ |−1 + − |ξ |−1 4 |ξ |2 4(n − 1)   ξ · ∇∂ H 3i −2 1 1 2 2 ξ · (D N ξ ) − + |ξ | 2 6 2(n + 1) |ξ | −

ξ · ∇∂ H n−2 i|ξ |−2 . 4(n + 1) |ξ |

7.6 Calculation of Some Boundary Operators

149

Similarly for E n ; multiply, and take the inverse transform to conclude  |η|3−n 1 n−2 |η|3−n η · Kˆ η H K 1 (x, y) = + − 2 ωn−1 16 |η| 8(n − 1) 3 − n   2 2 η · ∇H 3 4−n 1 η · D N η 1 − |η| − 32 6 |η|3 2(n + 1) |η|  3−n n − 2 |η| η · ∇∂ H + (kernel of class C∗r +1 (4)) + 16(n + 1) 3 − n Similarly (details omitted)  ∂ E(x − z) K 2 (x, y) = K 1 (z, y)d A z ∂ Nx ∂  |η|4−n n (η · Kˆ η)2 1 | Kˆ η|2 n − 2 η · Kˆ η = − + 4 2 ωn 48 |η| 12 |η| 8(n − 1) |η|2  n−2 1 2 ˆ 2) tr( K + H + 16(n − 1)2 24(n − 2) + (kernel of class C∗r +1 (4)) Now

 K 1 (x, y)
∂ g(y) = lim K 1 (x, y)
∂ g(y) →0 ∂\B (x) ∂  n − 2 H (x) g(x) = P.V.
∂ y K 1 (x, y)g(y) + n−1 8 ∂  M 2 P0 g(x) =
∂ y K 2 (x, y)g(y). 

M P0 g(x) =

∂

∂ ∂ To compute
∂ y , note that this is
η + O(|η|2 ∂η 2 ) + O(|η| ∂η ) where y = x + η − νx (η)N x , η ∈ Tx (∂) near 0. Straightforward calculation completes the proof.  Now we consider the general case: L =
+ n1 b j (x) ∂∂x j + c(x). By Thm. 7.4.1   ∂ 2 (S − E) ∂ 2 (S − E) g(y)dA y = −β · N (x)g(x) + P.V. g(y) lim x→∂ ∂ ∂ N x ∂ N y ∂ ∂ N x ∂ N y  z · β(x) g(y) = −β · N (x)g(x) − P.V. n ∂ ωn |z|    g(y) z Q 2 x, + |z| |z|n−2 ∂  + (C∗r (2) kernel)g(y), z = y − x, 2

∂

150

Chapter 7. Boundary Operators for Second-Order Elliptic Equation

with Q 2 (x, ζ ) = − 2ω1 n ζ · B(x)ζ + |ζ |n−2 E n (ζ ){ ∂∂βN · N − N · B N + γ }. The singular integral above may be written   z · β(x) g(y) = E(y − x)div∂ (β(y)g(y)) −P.V. n ∂ ωn |z| ∂      z 2−n r + (C∗ (2) kernel) g(y) + Q 3 x, |z| |z| ∂ where Q 3 (x, ζ ) = ω1n (ζ · (Dβ(x)ζ ) − 1/2 β · N ζ · K (x)ζ ) and we recall  β · N = nj=1 β j N j is p × p matrix. Thus   ∂ 2 S(x, y) lim g(y) = E n (x − y)(
∂ g(y) + div∂ (β(y)g(y)) x→∂ ∂ ∂ N x ∂ N y ∂      z 2−n r + C∗ (2) kernel g(y) + Q 4 x, |z| |z| ∂ for some (computable) Q 4 , and the calculation is now similar to that in Th. 7.6.3. Without more detail, we record the conclusion. Theorem 7.6.4. If b j : Rn → C p×P are C r +2 (1 ≤ j ≤ n), C : Rn → C p× p is C r +1 ,  is a bounded C r +4 region in Rn and g ∈ C 2 (∂, C p ) then for  x ∈ ∂, L =
+ n1 b j (x) ∂∂x j + c(x) ∂ ∂ B L g(x) − B
g(x) ∂N ∂N  1 b(x) · (y − x) = − b · N (x)g(x) − P.V. g(y) n 2 ∂ ωn |y − x|       1 y−x y−x 1 + Q x, Q 3 x, + n−2 2 r ωn−1r n−2 r ∂ ωn r   ˜ + E n (y − x)Q 4 (x) g(y) + R(x, y)g(y) ∂

˜ where R(x, y) is a

C∗r (2)

kernel in ∂,

n ∂b j 1 1 1 ζ j ζk b j bk − + ζ · Kˆ (x)ζ b · N (x) ζ j ζk 4 j,k=1 2 ∂ xk 2 ∂b j 1 1 Q 3 (x, ζ ) = (b · N b · ζ − b · ζ b · N ) + (ζ j Nk − ζk N j ) 8 4 j,k ∂ xk

Q 2 (x, ζ ) = −

n−1 ζ · Kˆ (x)ζ b(x) · ζ 4 1 ∂b j 1 2 Q 4 (x, ζ ) = c(x) − − b . 2 j ∂x j 4 j j −

7.6 Calculation of Some Boundary Operators

151

In case n = 2, this simplifies slightly: taking ∂ as a curve of length L (as in Th. 7.6.2), and writing “log|t|,” “sgn t” and “1/ t” for L-periodic functions having this form near t = 0, we have    ∂ ds 1 1 1 B L g(t) = P.V. g (s) + btan (t)g(s) − b · N (t)g(t) ∂N π t − s 2 2 ∂     1 1 ∂b1 ∂b2 (b2 b1 − b1 b2 ) + + − sgn(t − s)g(s) 16 8 ∂ x2 ∂ x1 ∂     1 1 1 log(t − s)g(s) + c(t) − div b(t) − b12 + b22 2 4 2π ∂  ˜ s)g(s) + R(t, ∂

˜ s) is a C∗r (2) kernel (in particular, C 1 ) b · N = b1 N1 + b2 N2 , where R(t, btan = −N2 b1 + N1 b2 .

Chapter 8 The Method of Rapidly-Oscillating Solutions

8.1 Introduction The theory of pseudo-differential operators (abbreviated ψDOp ) shows the essential behavior of many operators appears most clearly when applied to rapidly oscillating functions, studying the behavior as the frequency tends to infinity. This provides a more flexible and transparent approach to our “finite-rank” problems than the method of integral operators of Chapter 7. It is also often simpler, for simple problems – but for this reason, we are encouraged to attack harder problems. (If you are not familiar with ψDOp , don’t worry; our argument is direct. Here, the intention is only to show why we don’t cite results from this theory.) We have a ψDOp which, by hypothesis – a hypothesis we hope to contradict – has finite rank. A ψDOp of finite rank is trivial: its symbol is identically zero. So all we have to do is compute the symbol. But it must be computed in some detail. Often the (apparent) principal and subprincipal symbols both vanish identically and uselessly – we must go to the third stage, sometimes further. This is not surprising since we start with second order operators, so the coefficient of the zero-order part only begins to play a role at the third stage. We also deal with quite degenerate problems; after all, we expect to find a contradiction. The theory of ψDOp serves only as inspiration, since it ordinarily computes explicitly only the principal symbol. One shows the other terms are, in principle, computable, but there is rarely any need to compute them – except in our problems. Consider an example: the operator 
∂ ∂ r B A (σ u N ) + u N B A∗ (σ ψ N ) σ
→ (σ ) = Re ψ N ∂N ∂N was studied is 6.8 above, and will be treated by our new method in 8.4 below. (This calculation in the 1985 manuscript was incorrect, revealing an error in 152

8.1. Introduction

153

the earlier version of Theorem 7.6.10 – an error which only became clear on comparing with the results of the method of rapidly oscillating solutions.) Apparently,  is first-order; it would be first-order for most choices of u N , ψ N . But we treat the case where Re(u N ψ N ) ≡ 0 on the boundary, so the first-order and zero-order parts vanish and we are left with an operator of order −1: (γ cos ωθ ) = ω−1 γ C(θ) cos ωθ + O(ω−2 )

as ω → +∞,

where the coefficient C(θ) is computed explicitly in 8.4 (We assume γ , θ are smooth real functions on ∂ and |∇∂ θ| ≡ 1 on the support of γ .)  is assumed to have finite rank and in consequence γ C(θ) = 0 on ∂, for all permitted choices of γ , θ, which gives us new information about u N , ψ N and the coefficients of the elliptic operator A appearing in . Here we use the simple Lemma. Suppose S is a C 1 manifold; A, B ∈ L 2 (S) with compact support; θ is real-valued and C 1 on S with ∇s θ = 0 in supp A ∪ supp B; E is a finite dimensional subspace of L 2 (S); and u(ω) ∈ E for all large real ω, satisfying u(ω) = A cos ωθ + B sin ωθ + o(1) in L 2 (S) as ω → +∞. Then A = 0, B = 0. Proof. Let {e1 , . . . , e N } be an orthonormal basis for E; then u(ω) =

N  k=1

 Ck (ω)ek ,

Ck (ω) =

e¯k (A cos ωθ + B sin ωθ + o(1)) S

so each Ck (ω) → 0 as ω → +∞i by the Riemann-Lebesgue lemma, and

u(ω) L 2 (S) → 0. But  1 (|A|2 + |B|2 ) as ω → +∞.

u(ω) 2L 2 (S) → S 2 In Section 2, we find formal asymptotic expansions for, among others, B L (geiωθ ) as ω → +∞, i.e., the solution u of Lu = u + b r ∇u + cu = 0 in ,

u = geiωθ on ∂,

where |∇∂ θ| = 1 in supp G. We give a summary of the formulas at the end of this chapter, for reference. In Section 3, we show the formal solutions are close to exact solutions. In Section 4, we consider the operator  mentioned above (already treated in 6.8), to illustrate our method. We extend the argument of 6.8 to allow a much weaker hypothesis when n ≥ 3.

154

Chapter 8. The Method of Rapidly-Oscillating Solutions

In Section 5, we treat a new problem: proving generic simplicity of solutions of a system u j + f i (x, u 1 , u 2 , . . . , u p ) = 0 in ,

u j = 0 on ∂ (1 ≤ j ≤ p).

Most of these results were stated, without proof, in “Generic properties of equilibrium solutions by perturbation of the boundary,” D. B. Henry, in Dynamics of Infinite Dimensional Systems, ed. S. N. Chow and J. K. Hale, NATO ASI Series (Springer-Verlag, 1987).

8.2 Formal Asymptotic Solutions Results will be summarized at the end of the chapter. We study the differential operator L = I +

n 

b j (x)

j=1

∂ + c(x), ∂x j

x ∈ Rn ,

where I is the p × p identity matrix and b j (x), c(x) are smooth functions with values in C p× p (the p × p complex matrices); we will be more precise about smoothness conditions in 8.3. We work in an open set  ⊂ Rn with smooth boundary – typically, we will work in a small neighborhood of a given point of the boundary. We seek a formal solution u of Lu = f = 2ωeωS F in ,  Fk (x) F= , (2ω)k k≥0

u = g = eiωθ G on ∂  G k (x) G= (2ω)k k≥0

∂S |∂ > 0 (θ|∂ given) ∂N  and u has the form u = eωS r k≥0 Uk (x)/(2ω)k . Note Re S < 0 inside  so f and u both tend rapidly to zero as ω → +∞, except on or very near ∂. Setting U−1 ≡ 0, we find on substitution S|∂ = iθ,

0 = ω2 (∇ S)2

∞ 

Re

Uk /(2ω)k

0

∞  + (2ω)−k ( Uk + LUk−1 − Fk ) in , as ω → ∞ 0

with Uk | ∂ = G k , k ≥ 0, where

= ∇ S r ∇ + 1/2( S + b r ∇ S). (Scalar operators, like ∇ S r ∇, are implictly multiplied by an identity matrix,  and b r ∇ S(x) = nj=1 b j (x)∂ S/∂ x j ∈ C p× p .)

8.2. Formal Asymptotic Solutions

155

We choose the complex-valued S so (∇ S)2 =

n  (∂ S/∂ x j )2 = 0 1

and choose the Uk inductively, solving 

Uk + LUk−1 = Fk

(k ≥ 0)

Uk | ∂ = G k with U−1 ≡ 0 Well . . . not quite. There ordinarily are no exact solutions, but we only need that (∇ S)2 and the ∇Uk + LUk−1 − Fk tend to zero rapidly as x → ∂. (See 8.3 for exact conditions.) We find the approximate solutions using the “normal coordinates” near ∂ treated in Theorem 1.5, and we need more details about these.

Normal Coordinates (1) (y, t)
→ x = y + t N (y) is a diffeomorphism of (y, t) ∈ ∂ × (−r, r ) onto a neighborhood Br (∂) of ∂ in Rn , for small r > 0, when ∂ is at least C 2 , with inverse x
→ (y, t): r y = the point of ∂ closest to x (= π(x), in notation of Thm. 1.5)
r t = ± distance from x to ∂, sign +, outside  −, inside . r We extend the unit normal N by: N (y + t N (y)) = N (y), −r < t < r, y ∈ ∂. (2) d x = (I + t K  (y))dy + N (y)dy = (I + t K  (y))(N dt + dy) where K = D N is the (degenerate) curvature matrix, K (x) = D 2 t(x), with K N = 0, and “dy” is tangent to ∂ at y. (3) vol(d x) = det(I + t K  (y))dt dA y = eλ dt dA y where λ(t, y) =  (−1)m−1 m log det(I + t K (y)) = ∞ t Hm (y) for small t, Hm = trace K m , m=1 m trace K is the mean curvature. ˜ t) = u(y + t N (y)) for small t; then (4) Let u(x) be C 1 near  and define u(y, du = ∇u r d x = u˜ t dt + u˜ y r dy so ∇u(y + t N (y)) = (I + t K (y))−1 u˜ y + N (y)u˜ t . (We don’t always distinguish u˜ from u, and sometimes write ∂u/∂ N for u˜ t and ∇∂ u for u˜ y .) 2 (5) If  u, v are C near ∂ and v has small support, we change variables in 0 = (v u + ∇v r ∇u)vol(d x) by (3), (4) and integrate by parts to obtain: u(y + t N (y)) = u˜ tt + λt u˜ t + (I + t K )−2 λ y r u˜ y + div∂ ((I + t K )−2 u˜ y ).

156

Chapter 8. The Method of Rapidly-Oscillating Solutions (On ∂, at t = 0, this reduces to u = u N N + H u N + ∂ u.)

Calculation of S S(y + t N (y)) =



t k Sk (y)/k!,

k≥0

at least asymptotically, as t → 0 with y ∈ ∂ (in the relevant portion of ∂, which is usually small). On ∂ we have S = S0 = iθ(y)

(θ( r ) given, smooth, real)

and ∂ S/∂ N |∂ = S1 has Re S1 > 0, and we want (∇ S)2 (y + t N (y)) = 0 (asymptotically, as t → 0). Now ∇ S(y + t N (y)) = N (S1 + t S2 + t 2 /2 S3 + . . .) + (I − t K + 2 2 t K − . . .)(S0 + t S1 + t 2 /2 S2 + . . .), writing φ  for φ y = ∇∂ φ, so (∇ S)2 = 0 = (S0 )2 + S12 + t(2S1 S2 + 2S0 r S1 − 2S0 r K S0 ) + . . . . The zero-order term gives S1 =

 −(S0 )2 = |∇∂ θ|,

since Re S1 > 0; we assume |∇∂ θ| ≡ 1 in the region of interest, which makes things much simpler. In this case S0 = iθ,

S1 = 1,

S2 = −∇∂ θ r K ∇∂ θ,

and we may compute as many terms as desired – assuming ∂ and θ are sufficiently smooth. (Sk , k ≤ 4, are given explicitly in the summary at the end of this chapter.) We note, for later use, that each Sk is the sum of a real part, even in θ, and a purely-imaginary part, odd in θ. The hypothesis |∇∂ θ| ≡ 1 is very helpful in our problems (since the toporder operator is the Laplacian), so we pause to discuss this equation. For n = 2, ∂ is a curve (or several) and θ is arc-length along the curve. For n ≥ 3, there will be many choices of θ. Let x0 ∈ ∂ and let  be a C 2 hypersurface containing x0 and transverse to ∂. Define θ = ± {distance in ∂ from x to ∂ ∩ }, changing sign as we cross ∂ ∩ . Then for various choices of  and various choices of the real constant C, θ = C ± θ (x) yields all the C 2 solutions of |∇∂ θ| ≡ 1 near x0 . If  and ∂ are C m (m ≥ 2) then θ is C m . In particular, given τ ⊥ N (x0 ), |τ | = 1, we may choose a C m solution of |∇∂ θ| = 1 near x0 such that ∇∂ θ(x0 ) = τ. (This may be proved directly, or we may appeal to the theory of first-order scalar PDEs [5, 6].) For n = 2, θ is essentially unique, so this case often differs from n ≥ 3.

8.2. Formal Asymptotic Solutions

157

Calculation of Λ

= ∇ S r ∇ + 1/2σ,

σ = S + b r ∇ S

and we will write these in normal coordinates. ∇ S r ∇ = (I + t K )−2 Sy r ∂ y + St ∂t = iθ  r ∂ y + ∂t + t(−2i K θ  r ∂ y + S2 ∂t )

+ t 2 3i K 2 θ  r ∂ y + 1/2 S2 r ∂ y + 1/2 S3 ∂t + · · ·

S = S2 + t S3 + t 2 /2 S3 + · · · + (H − t H2 + · · ·) 1 + t S2 + t 2 /2 S3 + · · ·



+ (I − 2t K + · · ·) t H  − t 2 /2 H2 + · · · iθ  + t 2 /2 S2 + · · ·

+ div∂ ((I − 2t K + 3t 2 K 2 − · · ·) iθ  + t 2 /2 S2 + · · · = S2 + H + i ∂ θ + t(S3 + H S2 − H2 + iθ  r H  − 2idiv∂ (K θ  )) + t 2 /2 S4 + H S3 − 2H2 S2 + 2H3 − 4i K θ  r H  − iθ  r H2

+ 3idiv∂ (K 2 θ  ) + 1/2 ∂ S2 + O(t 3 ) ¨ + ··· ˙ + t 2 /2 b(y) Writing b(y + t N (y)) = b(y) + t b(y)   ˙ + N ) + b r (−i K θ  + N S2 )) + · · · b r ∇ S(y + t N (y)) = b r (iθ + N ) + i(b(iθ σ = S + b r ∇ S = σ0 (y) + tσ1 (y) + t 2 /2 σ2 (y)) + · · ·.

Calculation of U0

U0 = F0 (y + t N (y)) = F0 (y) + t F˙o (y) + t 2 /2 F¨0 (y)) + · · · with U0 | ∂ = G 0 ,

U0 (y + t N (y)) = G 0 (y) + t U˙ 0 (y) + t 2 /2 U¨ 0 (y) + · · ·

and U0 = F0 becomes F0 + t F˙0 + t 2 /2 F¨0 + · · · = U˙ 0 + iθ  r ∂ y G 0 + 1/2σ0 G 0

+ t U¨ 0 + iθ  r ∂ y U˙ 0 + 1/2σ0 + S2 U˙ 0

+ −2i K θ  r ∂ y G 0 + 1/2σ1 G 0

˙¨0 + iθ  r ∂ y U¨ 0 + 1/2σ0 + 2S2 U¨ 0 + t 2 /2 U + (−4i K θ  r ∂ y U˙ 0 + σ1 U˙ 0 )

+ 6i K 2 θ  r ∂ y G 0 + S2 r ∂ y G 0 + 1/2σ2 G 0 + O(t 3 ) ˙¨0 on ∂. (These all vanish off the support which determines U0 = G 0 , U˙ 0 , U¨ 0 , U of G 0 , where F0 , F˙0 , F¨0 vanish.)

158

Chapter 8. The Method of Rapidly-Oscillating Solutions

Calculation of U1 , U2

U1 = F1 − LU0 , U1 | ∂ = G 1 so we must find LU0 = U0 + b r ∇U0 + cU LUo (y + t N (y)) = U¨0 + (H + b r N )U˙ 0 + ∂ G 0 + b r ∇∂ G 0 + cG 0 ˙¨0 + (H + b r N )U¨0 + ∂ U˙ 0 + b r ∇∂ U˙ 0 + t{(U + cU˙ 0 + (b˙ r N U˙ 0 − H2 ) − 2 div∂ (K G 0 ) ˙ 0 )} + O(t 2 ), + (b˙ − bK + H ) r G 0 + cG and substitution gives U˙ 1 , F¨1 on ∂ (U1 | ∂ = G 1 ). From these we compute LU1 | ∂, hence U˙ 2 | ∂. Thus we formally solve the Dirichlet problem to any desired order, assuming sufficient smoothness (and patience). If instead, we are given, for example,  1 1 ∂U + βU |∂ = eiωθ 2ω H0 + H1 + H + . . . 2 ∂N 2ω (2ω)2 then (supposing β is independent of ω) we may determine the corresponding G’s from H0 = 1/2G 0 ,

Hk = U˙ k−1 + 1/2G k + βG k−1

(k ≤ 1),

or G 0 = 2H0 ,

G 1 = 2H1 − 4β H0 − 2F0 + 4iθ  r ∂ y H0 + 2σ0 H0 ,

and so on. We summarize our results at the end of the chapter for ease of reference.

8.3 Exact Solutions Suppose the integer N ≥ 0 and for k = 0, 1, . . . , N , F(x) is C N +1−k near ∂ and G k is C N +2−k on ∂, both with compact support. Also suppose b j (x) ∈ C p× p is C N +1 and c(x) ∈ C p× p is C N . For our “generic” problems, we may always suppose ∂ is as smooth as desired, but for this calculation, class C N +3 suffices, and we suppose θ has the same smoothness (as always, suppose |∇∂ θ| ≡ 1). We may then choose S(y + t N (y)) of class C N +3 such that

as d(x) = dist. (x, ∂) → 0. (∇ S(x))2 = O d(x) N +2 We choose Uk (x) of class C N +2−k so that, for 0 ≤ k ≤ N ,

Uk + LUk−1 − Fk (x) = O(d(x) N +1−k )

8.3. Exact Solutions Uk |∂ = G k

159

(U−1 ≡ 0)

Choose a C ∞ “cutoff ” χ : χ ≡ 1 near ∂ ∩ ∪k supp Fk ∪ supp G k , but χ is supported near this set. We compare the formal solution u form. = χ r eωS

N 

Uk /(2ω)k

k=0

with the exact solution u = B L (g) + A L ( f ) as ω → +∞, where

g − eiωθ

N 

G k /(2ω)k C 2 (∂) = o(ω−N )

k=0

f − χ r 2ωeωS

N 

Fk /(2ω)k L p() = o(ω−N )

k=0

and we show u − u form W p2 () = o(ω−N ) as ω → +∞. (We might use the norm 2−1/ p

(∂) instead of C 2 (∂) for the boundary values.) Here 1 < P < ∞ of W p is fixed and will ordinarily be chosen with p > n so W p2 ⊂ C 1 and, in particular,  

N N   ∂u

iωθ k k ˙ r ω Uk /(2ω) = o(ω−N ) −χ e G k /(2ω) + ∂ N ∂ 0 0 uniformly on ∂ as ω → +∞, or even in C α (∂) provided 0 < α < 1 − n/ p. Sometimes we need a stronger norm, such as W p2 (); the calculations are similar, assuming a bit more smoothness for the b j , c: one more derivative, for W p3 . There are various ways to define B L , A L , differing by operators of finite rank, but the above results holds in any case. Suppose L | H 2 ∩ H01 () has m-dimensional kernel, so its image is closed, with codimension m. Let {ψ1 , . . . , ψm } be a basis for any complement to the image; then if {σ1 , . . . , σm } is a basis for the kernel of L ∗ | H 2 ∩ H01 (), L ∗ = − b r ∇ + c − div b, we have det [  σ j r ψk]mj,k=1 = 0. Also choose a complement to the kernel of L , in the form {u :  u r φ j = 0 (1 ≤ j ≤ m)}. We could suppose the {φ¯ j } are a basis for the kernel of L , but anything nearby would work as well. The solutions σk of the adjoint equation are C N +2− , any  > 0, and we suppose the ψk , φ j have similar smoothness. Then u = B L (g) + A L ( f )

160

Chapter 8. The Method of Rapidly-Oscillating Solutions

is the solution of Lu = u + b r ∇u + c = f +

m 

α j ψ j in 

j=1

for some choice of the α j ∈ C, such that  u = g on ∂ and u r φ j = 0, 

1 ≤ j ≤ m.

The α j are uniquely determined, since for each 1 ≤ k ≤ m   m  αj σk r ψ j = σk r (Lu − f ) j=1









=−

σk f − r



∂

∂σk r g, ∂N

1 ≤ k ≤ m,

with an invertible matrix. Now   N N   Fk Uk /(2ω)k − 2ωeωS L eωS (2ωk ) 0 0  N  r Uk eωS 2+N ∇ S ∇ S + LU N (2dω) = N N +2 (2ω) 4d (2ω)k 0  N  ( U + LU − F ) k k−1 k + (2ω)1−k+N d N +1−k k=0 and Re S(x) < −1/2d(x) in  near ∂. Since χ ≡ 1 near ∂ − at least, the part where the G k and Fk , hence Uk , are nonzero – we see, for some constant C, |Lu form. − χ r 2ωeωS

N 

Fk /(2ω)k | ≤ Cω−N e−ωd/4

0

in , uniformly as ω → + ∞. Let z ∈ W p2 () be defined by z = u form. − u −

m 

βk τ k

k=1

where {τ1 , . . . , τm} is a basis for the kernel of L|H 2 ∩ H01 (), with βk ∈ C r chosen to ensure   z φmj = 0 for all j = 1, . . . , m. r Note det [  τk φ j ] j,k=1 = 0; this certainly holds when the {φ¯ j } form a basis for the kernel, and it still is true when they are nearby. Thus the βk are well-defined.

8.4. Example 6.8 Revisited

161

In  |L z | = |Lu form − Lu| ≤ | f − χ r 2ωeωS

N 

Fk /(2ω)k | + Cω−N e−ωd/4 +

0

m 

|α j | sup |ψ j |

j=1

and on ∂, ω N z = ω N (u form. − u) = ω N ( iωθ 0N G k /(2ω)k − g)|∂ which tends to zero in C 2 (∂) as ω → +∞. Also 

 

σk r f +

∂

N  ∂σk r g = o(ω−N ) + (2ω)1− j ∂N j=0

 N  −j + (2ω) j=0

∂

 

σk r F j eωS

σk r Gj eiωθ = o(ω−N )

as ω → +∞,

by the Riemann-Lebesgue lemma, so the |α j | = o(ω−N ). Similarly m  k=1

 βk

 

τk

r

φj =



uform

r

  N φj = (2ω)−i χUi  i=0

r

φ j eωS = o(ω−N )

so the |βk | = o(ω−N ) as ω → ∞. Thus ||Lz|| L p () = o(ω−N ),

|| z | ∂||C 2 (∂) = o(ω−N ),

so ||z||W p2 () = o(ω−N ). Then m



βk τk

W 2 () = o(ω−N ) ||u form − u||W p2 () ≤ ||z||W p2 () +

1

p

and the claim is proved.

8.4 Example 6.8 Revisited: Generic Simplicity of Complex Eigenvalues In the example, bj (x, λ) and c(x, λ) are polynomials in λ with smooth complexvalued coefficients. (It suffices that the coefficients of the b j are C 2 and those of c are C 1 , but we won’t worry about smoothness in the calculation.) We often suppress the λ-dependence and treat A = + b r ∇ + c,

A∗ = A − b r ∇u + c − div b.

162

Chapter 8. The Method of Rapidly-Oscillating Solutions

As in 6.8, our hypothesis for contradiction is that, for each  near 0 , there exists λ ∈ and solutions u, ψ of A∗ ψ = 0 in ,

Au = 0 in ,

u ≡ 0,

u = 0 = ψ on ∂

ψ ≡ 0

with Re(u N ψ N ) ≡ 0 on ∂. (u N = ∂u/∂ N , ψ N = ∂ψ/∂ N .) We have u N ψ N = 0, purely imaginary, on a dense set of ∂. Solvability of this problem for each  near 0 implies we have a solution (λ, u, ψ) which also satisfies:  ∂ ∂ B A (σ u N ) + u N B A∗ (σ ψ N ) σ →
(σ ) = Re ψ N ∂N ∂N


has finite rank. Instead of appealing to Chapter 7, we will show: if γ , θ are smooth, real functions on ∂ with |∇∂ θ| = 1 in supp γ , then uniformly on ∂, (γ cos ωθ) =

γ cos ωθ C(θ) + o(ω−1 ) ω

as ω → ∞

where 



 C(θ) = Re D θ u N + 1/2bθ u N r D θ ψ N − 1/2bθ ψ N − qu N ψ N in supp γ , D θ = ∇∂ − ∇∂ θ ∂θ∂ , bθ = b − N b r N − ∇∂ θb r ∇∂ θ and q = c − 1/2 div b − 1/4b2 (= D(A(λ))). The lemma of the introduction (8.1) then shows C(θ) = 0 for all such θ; we examine this condition later. Note when n = 2, D θ = 0 and bθ = 0 so Im q = 0 on ∂. We construct formal asymptotic solutions  1 U1 + · · · B A (γ u N eiωθ ) = eωS U0 + 2ω  1 iωθ ωS B A∗ (γ ψ N e ) = e 1 + · · · 0 + 2ω with the same “S” since this depends only on the highest derivative, but we write σ = S + b r ∇ S for A, σ ∗ = S − b r ∇ S for A∗ . It is convenient to use the complex exponential: (γ cos ωθ) = Re 1/2(0 (γ eiωθ ) + 0 (γ e−iωθ )) = Re r even 0 (γ eiωθ )

8.4. Example 6.8 Revisited

163

where ∂ ∂ B A (γ u N eiωθ ) + u N B A∗ · (γ ψ N eiωθ ) ∂N ∂N
˙0 = eiωθ 2ωγ u N ψ N + ψ N U˙ 0 + u N 

0 (γ iωθ ) = ψ N

 1 −1 ˙ ˙ (ψ N U1 + u N 1 ) + o(ω ) + 2ω Thus the coefficient of “ω” in  (γ cos ωθ) is Re r even eiωθ 2γ u N ψ N = γ cos ωθ Re (u N ψ N ) ≡ 0. The coefficient of “ω0 ” is Re r even eiωθ (ψ N U˙ 0 + u N ˙0 ) = − cos ωθγ (q − H ) Re u N ψ N + sin ωθ(∂θ γ + γ ∂ θ) Re u N ψ N + γ ∂θ (Re u N ψ N ) ≡ 0 The coefficient of ω−1 does not vanish trivially, but there is a lot of cancellation; we pause to organize the calculations. We distinguish terms of type R or type I : r type R = (real-valued, even in θ) + (pure imaginary, odd in θ) r type I = (real, odd) + (imaginary, even).

Note the products: R × R = R, R × I = I, I × I = R. Type I is also called “negligible” since Re r even r eiωθ (R + I ) = Re r even eiωθ R. (Indeed eiωθ = cos ωθ + i sin ωθ is type R so eiωθ I is type I with real-evenpart zero.) For example u N ψ N is type I , as is iγ ∂θ (u N ψ N ) or iγ ∂ θu N ψ N , or ˙ 0 , where S and every Sk is type R. Also note ∇∂ , i ∂ , ∇ S r ∇ + ψ N U˙ 0 + u N  ∂θ 1/ S preserve the type. 2 Some terms, containing b or c, may have parts of both types; we will simply carry those along in the calculation but terms which are clearly negligible will be collected in “+ (negl.)”. The coefficient of (2ω)−1 is ˙ 1 = −(ψ N U¨ 0 + (H + b r N )ψ N U˙ 0 + ψ N ∂ γ u N ψ N U˙ 1 + u N  ¨ 0 + (H − b r N )u N  ˙0 + ψ N b r ∇∂ γ u N + cγ u N ψ N + u N  + u N ∂ γ ψ N − u N b r ∇∂ γ ψ N + (c − div b)γ u N ψ N ).

164

Chapter 8. The Method of Rapidly-Oscillating Solutions

We have ψ N ∂ γ u N + u N ∂ γ ψ N = 2u N ψ N ∂ γ + ∇∂ γ r ∇∂ (u N ψ N ) + γ (ψ N ∂ u N + u N ∂ ψ N ) = −2γ ∇∂ ψ N r ∇∂ u N + (negl.), so ˙ 1 = − (ψ N U¨ 0 + u N  ¨ 0 ) − b r N (ψ N U˙ 0 − u N  ˙ 0) ψ N U˙ 1 + u N  + 2γ ∇∂ ψ N ∇∂ u N − γ b r (ψ N ∇∂ u N − u N ∇∂ ψ N ) − (2c − div b)γ u N ψ N + (negl.). Also



¨ 0 ) = ψ N i∂θ U˙ 0 + 1/2σ0 − q U˙ 0 −(ψ N U¨ 0 + u N 



+ − 2ikθ  r ∇∂ + 1/2σ1 γ u N

˙0 ˙ 0 + 1/2σ0∗ − q  + u N i∂θ 



 r ∗ + − 2ikθ ∇∂ + 1/2σ1 γ ψ N ˙ 0 ) − i∂θ ψ N U˙ 0 − i∂θu N  ˙0 = i∂θ (ψ N U˙ 0 + u N  ˙ 0) + 1/4(σ0 + σ0∗ − 2q)(ψ N U˙ 0 + u N  ˙ 0) + 1/2b r (N + iθ  )(ψ N U˙ 0 − u N  + 1/2(σ1 + σ1∗ )γ u N ψ N + (negl.),

with some negligible terms underlined. Further ˙0 −i∂θ ψ N U˙ 0 − i∂θ u N  i = − 2γ ∂θ ψ N ∂θ u N + γ b r (N + iθ  )(u N ∂θ ψ N − ψ N ∂θ u N ) + (negl.), 2 and more such calculations yield finally: ˙ 1 = 2γ ∇∂ ψ N r ∇∂ u N − γ b r (ψ N ∇∂ u N − u N ∇∂ ψ N ) ψ N U˙ 1 + u N  − 2γ ∂θ ψ N ∂θ u N + γ b r θ  (ψ N ∂θ u N − u N ∂θ ψ N ) − 2γ {c − 1/2 divb − 1/4(b r N )2 − 1/4(b r θ  )2 }u N ψ N + (negl.) Writing Dθ for ∇∂ − (∇∂ θ) ∂θ∂ and bθ = b − N b r N − ∇∂ θb r θ  . This becomes



2γ Dθ ψ N − 1/2bθ ψ N r Dθ u N + 1/2bθ u N   − 2γ c − 1/2 div b − 1/4b2 u N ψ N + (negl.).

8.4. Example 6.8 Revisited

165

The non-negligible part is even in θ, so we obtain (γ cos ωθ) =



γ cos ωθ  Re Dθ ψ N − 1/2bθ ψ N r Dθ u N + 1/2bθ u N ω

 − c − 1/2 div b − 1/4b2 u N ψ N + o(ω−1 )

as ω → ∞,

−1

and the coefficient of ω must vanish for all allowed choices of γ , θ. If n = 2, Dθ = 0 and bθ = 0 so we obtain Im D(A(λ)) = 0 on ∂ ∩ V . Let A = ∇∂ ψ N − 1/2btan ψ N , B = ∇∂ u N + 1/2btan u N , and C = c − 1/2 div b − 1/4b2 (= D(A(λ))). At each x ∈ V ∩ ∂, for each τ ⊥ N (x) with |τ | = 1, we can choose θ with |∇∂ θ| = 1 near x, ∇∂ θ(x) = τ, γ supported near x with γ (x) = 0, and conclude Re{(A − τ A r τ ) r (B − τ B r τ ) − C} = 0 or Re(A r τ B r τ ) = Re(A r B − C) for all τ ⊥ N , |τ | = 1. 1 Re A r B, which The average value of the left-side, over all such τ , is n−1 implies (n − 2) Re A r τ B r τ = Re C when |τ | = 1, τ ⊥ N , or   ∂u N 1 ∂ψ N 1 − /2b r τ ψ N + /2b r τ u N (n − 2) Re ∂τ ∂τ = Re D(A(λ))u N ψ N ,

the same equation obtained in 6.8 above. As in 6.8, we conclude: for n ≥ 2,

Im D(A(λ)) = 0 on ∂ ∩ V ;

and for n ≥ 3, where u N ψ N = 0 in ∂ ∩ V , |Im btan − ∇∂ (2 arg u N )| r |Re btan − ∇∂ log |ψ N /u N = 0. We show this contradicts the hypothesis (for n ≥ 3): for each λ ∈ , if c − 1/2 div b − 1/4 b2 is real everywhere in a neighborhood of some x ∈ V , then neither Re b not Im b is a gradient on any neighborhood of x. Differently stated: for λ ∈

|Im D(A(λ))| +



n 

β=Re b,Im b i, j,=1

on a dense set of V.

|∂βi /∂ x j − ∂β j /∂ xi | > 0

166

Chapter 8. The Method of Rapidly-Oscillating Solutions

The lemma of 6.8 shows Im D(A(λ)) = 0 in some nonempty open V0 ⊂ V , so (Re b), (Im b) are both non-zero on a dense set of V0 , where (β) = [∂βi /∂ x j − ∂β j /∂ xi ]i,n j=1 is an anti-symmetric matrix at each x. (Recall b generally also depends on λ.) But we also know (for n ≥ 3) on a dense open set in V ∩ ∂, each point has a neighborhood where Re btan or Im btan is a gradient on ∂; this will give a contradiction. (Here we use some simple topology: if two continuous real-valued functions f 1 , f 2 have product f 1 r f 2 ≡ 0, then interior f 1−1 (0)∪ interior f 2−1 (0) is dense in the domain.) (1) Suppose β(x) = (β1 (x), . . . , βn (x)) ∈ Rn is C 1 and ∂ is C 2 , and in some neighborhood of x0 ∈ ∂ there is C 1 function φ such that βtan = ∇∂ φ on ∂ in this neighborhood. Then  = (β) = [∂βi /∂ x j − ∂β j /∂ xi ]i,n j=1 satisfies  =  N ⊗ N − N ⊗  N on ∂, in this neighborhood. In fact ∇∂ φ = βtan is C 1 so we may modify φ, if necessary, so that φ is locally C 2 . Further, ∂ = {x : ψ(x) = 0} for some ψ ∈ C 2 with |∇ψ(x)| ≥ 1 where ψ(x) = 0, and φ + (constant) r ψ could be used in place of φ, so we may assume ∂φ/∂ N = 0 in the neighborhood of interest. Now consider β(x) − ∇φ(x), a C1 function on a full (Rn )-neighborhood of x0 , such that β − ∇φ = µN on ∂ near x0 , for some non-zero C 1 function µ. We, may extend N , µ with |N | ≡ 1, µ = 0, so this equation still holds off ∂, and then i j = ∂i µN j − ∂ j µNi + µ(∂i N j − ∂ j Ni ) on this neighborhood; so, on ∂ near x0 ,  = τ ⊗ N − N ⊗ τ with τ = ∇∂ µ − µ∂ N /∂ N ⊥ N . It follows that  N = τ so  = ( N ) ⊗ N − N ⊗ . (This condition would be trivial for n = 2, but we assume n ≥ 3.) Remark. We will use the notation “a ⊗ b” more generally below. It may be defined for any linear spaces E, F : if e ∈ E, f ∈ F, e ⊗ f is the bilinear form on the product of the duals E  × F  given by (e ⊗ f )(φ, ψ) = φ(e) r ψ( f )

for φ ∈ E  ,

ψ ∈ F .

Note e ⊗ f = 0 if and only if e = 0 or f = 0. In each E = E  = Rn , F = F  = Rm , we may write e ⊗ f as a matrix: e = (e1 , . . . , en ), f = ( f 1 , . . . , f m ), (e ⊗ f )i j = ei f j so (e ⊗ f ) (x, y) = x r((e ⊗ f )y) = x r e f r y for x ∈ Rn , y ∈ Rm . (2) Let 1 = (Re b), 2 = (Im b), Q j (N ) =  j + N ⊗  j N − ( j N ) ⊗ N ( j = 1, 2). On a dense set of ∂ ∪ V , either Re btan or Im btan is a gradient on ∂ near given any point, so Q 1 (N ) or Q 2 (N ) vanishes, or Q 1 (N ) ⊗ Q 2 (N ) = 0 on ∂ ∩ V . We note, for later use, if n ≥ 3 and at some point of ∂ ∩ V, Q j (N ) = 0 and  Q j (N )τ = 0 for all τ ⊥ N , then  j = 0 at this point. In fact, Q j (N )τ = 0 says τ ⊗  j N + N ⊗  j τ =  j τ ⊗ N +  j N ⊗ τ (for all τ ⊥ N ). Applying this to

8.5. Example 6.8 Revisited

167

τ yields  j N = τ ( j N ) r τ. If  j N = 0, then for all τ ⊥ N , τ has the direction of  j N , which is false for n ≥ 3. Thus,  j N = 0 so (since Q j = 0)  j = 0. (3) So far,  (and the corresponding λ) have been kept fixed. Allowing  to vary near 0 there exists λ ∈ so that Im D(A(λ)) = 0 and Q 1 (N ) ⊗ Q 2 (N ) = 0 on ∂ ∩ V . By the transversality theorem, whenever τ ⊥ N , we also have Q 1 (N )τ ⊗ Q 2 (N ) + Q 1 (N ) ⊗ Q 2 (N )τ = 0. Assuming n ≥ 3, if Q 1 (N ) = 0 at some point of ∂ ∩ V , then Q 2 (N ) = 0 and Q 2 (N )τ = 0 for all τ ⊥ N , hence 2 = 0 at that point. Similarly Q 2 (N ) = 0 implies 1 = 0. If 1 ⊗ 2 = 0, then both  j are nonzero so both Q j (N ) = 0. Thus we have, on ∂ ∩ V . Im D(A(λ)) = 0

and

(1 ⊗ ) ⊗ (Q 1 (N ), Q 2 (N )) = 0.

Applying the transversality theorem to this condition, we see also (1 ⊗ 2 ) ⊗ (Q 1 (N )τ, Q 2 (N )τ ) = 0 for all τ ⊥ N . Suppose 1 ⊗ 2 = 0 at some point of ∂ ∩ V ; then for both j = 1, 2, Q j (N ) = 0 and Q j (N )τ = 0 when τ ⊥ N , so 1 = 2 = 0, a contradiction. Hence at each point of ∂ ∩ V , Im D(A(λ)) = 0 and 1 ⊗ 2 = 0 (or |1 2 | = 0). By the lemma of 6.8, we obtain the long-sought contradiction. We summarize Theorem. Assume n ≥ 3 and let b j (x, λ), c(x, λ) be polynomials in λ with x
→ b j (x, λ) ∈ C of class C 2 , x
→ c(x, λ) ∈ C of class C 1 ( j = 1, 2, r , n). Let ⊂ C and V ⊂ Rn be open sets such that: for any λ ∈ , if (c − 1/2div b − 1/4b2 ) is real everywhere in a neighborhood of some point of ∇, then neither Re b( r , λ) nor Im b( r , λ) is a gradient in any neighborhood of that point. Then for most C 2 bounded regions  ⊂ Rn such that ∂ meets V, all eigenvalues λ ∈ of u + b

r

∇u + cu = 0 in ,

u = 0 on ∂,

u ≡ 0,

are simple and freely moveable. (For n = 2, we have the same conclusion assuming Im(c − 1/2 div b − = 0 on a dense set of V , for each λ ∈ ; see Example 6.8. Whether the above weaker hypothesis suffices for n = 2 is not known.)

1/ b2 ) 4

Remark. If the b j , c are real-valued for real λ, then for real λ, c − 1/2 div b − is real and Im b( r , λ) ≡ 0, so cannot meet the real axis.

1/ b2 4

168

Chapter 8. The Method of Rapidly-Oscillating Solutions

8.5 Generic Simplicity of Solutions of a System We will study solutions of u j + f j (x, u 1 , u 2 , . . . , u p ) = 0 in , u j = 0 on ∂ ( j = 1, 2, . . . , p) for p ≥ 2, n ≥ 2, where f : Rn × R p → R p is a given smooth function, and the region  is assumed open, bounded and C 2 (or better). We assume  is connected during the argument, but this is not important – we may perturb each comp ponent separately. We write f  (x, u) for the p × p matrix [∂ f j (x, u)/∂u k ] j,k=1 . If zero is a regular value of u
→ u + f ( r , u), from Wq2 ∩ Wq1 (, R p )0 to L q (, R p ), n < q < ∞, then all solutions u of u + f ( r , u) = 0 in , u = 0 on ∂, are simple. This would hold for most  if zero were a regular value of (h, u)
→ h ∗ ( + f )h ∗−1 u. We prove this by contradiction, under appropriate hypotheses on f . Effective hypotheses are far from obvious, so we start merely assuming reasonable smoothness and that, for every  near a certain 0 , there is a solution u which is not simple. This will imply a sequence of additional conditions, and solvability of ever more over-determined problems. Then it will be clear how to select our hypotheses for an argument by contradiction. Roughly speaking, we treat the cases (1) f (x, 0) ≡ 0; (2) f (x, 0) ≡ 0 and f  (x, 0) ≡ 0, depending on x; (3) f (u) is independent of x, f (0) = 0, and f  (0) has only simple eigenvalues (and p is not too large); (4) p = 2, f : R2 → R2 is independent of x ∈ Rn , f (0) = 0, f  (0) = λI for some real λ, f  (0) = 0. (The exact conditions are more complicated: see Theorem 8.5.1 for (1), (2); Thm. 8.5.3 for (3); Thm. 8.5.4 for (4).) In each case, if f is fairly smooth, we conclude all solutions of u + f (x, u) = 0 in , u = 0 on ∂, are simple, for most choices of  ⊂ Rn . As in the scalar case (6.5), if there is a non-simple solution u for every  near 0 , then there is a solution (u, ψ) of
(∗)

u + f (x, u) = 0 in , ψ + T f  (x, u)ψ = 0 in ,

u = 0 on ∂ ψ = 0 on ∂,

ψ ≡ 0

with ∂∂ψN r ∂∂uN ≡ 0 on ∂. Also, by uniqueness in the Cauchy problem, ∂ψ/∂ N = 0 on a dense set of ∂; but since p ≥ 2, we cannot conclude that ∂u/∂ N vanishes.

8.5. Generic Simplicity of Solutions of a System

169

Applying the transversality theorem, we find: for every  near 0 there is a solution (u, ψ) of (∗) such that ψ N r u N = 0 and also σ
→

∂ {B L ∗ r (σ ψ N ) − A L ∗ r Tf  ( r , u)ψB L (σ u N )} r u N ∂N ∂ B L (σ u N ) + σ ψ N r f 0 + ψN r ∂N

has finite rank, L = + f  ( r , u), L ∗ = + T f  ( r , u), f 0 = ( r , 0). (We have used the facts ψ N N + H ψ N = 0, u N N + H u n + f 0 = 0 on ∂.) Taking σ = γ eiωθ , |∇∂ θ| = 1 in supp γ , as usual, we find ˙ 0 r u N + ψ N r U˙ 0 + γ ψ N 

r

f 0 = 0,

˙ k r u N + ψ N r U˙ k = 0 for k ≥ 1 

where




Uk + LUk−1 = 0

(U−1 = 0),

k + L ∗ k−1 = −Tf  (u)ψUk−1 ,

Uk | ∂ =

γ uN 0


k | ∂ =

γ ψN 0

if k = 0 if k > 0 if k = 0 if k > 0

and = ∇ S r ∇ + 1/2 S is the same for L or L ∗ . It is convenient to treat a more general case, so assume: for every  near 0 , there is a solution (u, ψ) of (∗) such that ψ N r A = 0 on ∂

for all a ∈ A

and ψ N r Mu N = 0 on ∂

for all M ∈ M

where A ⊂ C(Rn , R p ), M ⊂ C(Rn , R p× p ) are given sets (or linear spaces). This certainly holds for A = {0}, M = {I }, the constant functions. The same argument as above yields: ˙ 0 r Mu N + ψ N r M U˙ 0 − σ ∂ N (ψn r Mu k ) = 0 ˙ 0 r a − γ ∂ N (ψ N r a) = 0   and ˙k ra = 0 

˙ k r Mu N + ψ N r M U˙ k = 0, 

for k ≥ 1.

Step 1. Take k = 0: if a, M are C 1 , 0 = iγ ∂θ ψ N r a + γ ∂ N (ψ N r a), 0 = iγ (∂θ ψ N r Mu N + ψ N r M∂θ u N ) + γ ∂ N (ψ N r Mu N ).

170

Chapter 8. The Method of Rapidly-Oscillating Solutions

The first of these (using ψ N N + H ψ N = Aψ | ∂ = 0) yields ψ N r ∂a/∂ x j = 0 on ∂

(1 ≤ j ≤ n),

a ∈ A ∩ C 1.

The second yields (since u N N + H u N + f 0 = 0 on ∂, f 0 = f ( r , 0)) ψ N r ∂x j Mu N − N j ψ N r M f 0 = 0 on ∂ (1 ≤ j ≤ n). This is not quite the form treated before, but a similar transversality argument yields (in the principal symbol, for a change) 0 = ψ N r ∂x j Mu N − ie j r ∇∂ θ ψ N r M f 0 where e j is the unit vector on the x j -axis. This shows each term vanishes separately: for M ∈ M ∩ C 1 , on ∂, ψ N r ∂x j Mu N = 0

(1 ≤ j ≤ n),

ψ N r M f 0 = 0.

˙1 ra = Step 2. Next consider k = 1, with a, M in C 2 . Writing f 0 = f  ( r , 0),  0 yields ¨ 0 + H ˙ 0 + A∂ γ ψ N + T f 0 (γ ψ N ) r a. 0 = ( We have 0 = ∂ (γ ψ N r a) so, by previous results, 0 = ∂ (γ ψ N ) r a, and ¨ 0 r a = −i∂θ  ˙ 0 r ∂θ a = 0 so ˙ 0 r a. Finally  ˙ 0 r a = i clearly 0 =  ψN

r

f 0 a = 0.

˙ 1 r Mu N + ψ N r M U˙ 1 yields Similarly, 0 =  0 = ψ N r ( f 0 M + M f 0 )u N + ∂ ψ N r Mu N + ψ N r M ∂ u N − ∂θ2 ψ N r Mu n − ψ N r M∂θ2 u N . If n = 2, θ is arclength in the curve ∂ and ∂ = ∂ 2 /∂θ 2 so 0 = ψ N r ( f M + M f 0 )u N . We will show this also holds for n ≥ 3, by more calculation. Step 3. From ∂θ2 (ψ N r Mu N ) = 0, we see

− ∂θ2 ψ N r Mu N + ψ N r M∂θ2 u N = 2∂θ ψ N r M∂θ u N ,

8.5. Generic Simplicity of Solutions of a System

171

so ∂r ψ N r M∂r u N = A = constant for all directions τ with τ ⊥ N , |r | = 1, so (in component form) A=

p 

  Mαβ arg τ r ∇∂ ψ Nα τ r ∇∂ u βn

α,β=1

=

1  β Mαβ ∇∂ ψ Nα r ∇∂ u N n − 1 α,β

=−

1 ( ∂ ψ N r Mu N + ψ N r M ∂ u N ). 2(n − 1)

Then for τ ⊥ N , 2(n − 2)∂τ ψ N r M∂τ u N = |τ |2 ψ N r ( f 0 M + M f 0 )u N . Further, previous results imply ∂τ ψ N r Mu N N + ψ N N r M∂τ u N = 0, ψ N N Mu N N = 0 so, for any z ∈ Rn , with ∂z = z r ∇,

r

2(n − 2)∂z ψ N r M∂z u N = (|z|2 − (z r N )2 )ψ N r ( f 0 M + M f 0 )u N . We apply transversality again (M ∈ M ∩ C 2 ) with this condition to conclude σ
→ 2(n − 2){∂z (∂ N (δψ) − N˙ r ∇ψ) r M∂z u N + ∂z ψ N r M∂z (∂ N (δu) − N˙ r ∇u)} − 2z r N z r N˙ ψ N r M1 u N − |z tan |2 ((δψ) N r M1 u N + ψ N r M1 (δu) N ) − σ ∂ N (2(n − 2)∂z ψ N r M∂z u N − |z tan |2 ψ N r M1 u N ) has finite rank, where M1 = f 0 M + M f 0 and δu + f  (u)δu = (finite),

δu|∂ = σ u N

δψ + T f  (u)δψ + T f  (u, ψ)δu = (finite),

δψ|∂ = σ ψ N .

Since ψ = 0, on ∂, also, N˙ r ∇ψ = 0 on ∂ and we find ∂z ( N˙ r ∇ψ) = z r N N˙ r ∇ψ N = −z r N ∇∂ σ r ∇∂ ψ N (assuming ∂ N /∂ N = 0 on ∂); and similarly for u. We take σ = γ eiωθ as usual, and as usual, the principal part (coefficient of ω2 ) says nothing new. But the coefficient of ω is 0 = 2(n − 2)γ {(−z r (N + iθ  )i∂θ ψ N + z tan r ∇∂ψ N ) r M∂z u N + ∂z ψ N r M(−z r (N + iθ  )i∂θ u N + z tan r ∇∂ u N )} + (2z r N z r θ  iγ − |z tan |2 γ )ψ N r M1 u N .

172

Chapter 8. The Method of Rapidly-Oscillating Solutions

The part odd in θ must vanish, but this tells us nothing new. The even part also vanishes, so taking z = τ, θ  = σ , unit tangent vectors (⊥ N ), we obtain 0 = (n − 2)γ τ r σ (∂τ ψ N r M∂σ u N + ∂σ ψ N r M∂τ u N ). Choosing τ = σ we see (n − 2)∂τ ψ N r M∂τ u N = 0 = ψ N r M1 u N so ψ N r (M f 0 + f 0 M)u n = 0 on ∂ for n ≥ 2. Thus the hypothesis. (A, M): For all  near 0 , there is a solution (u, ψ) of (∗) such that on ∂, ψ N r a = 0 for all a ∈ A, ψ N r Mu N = 0 for all M ∈ M, implies this still holds with A1 , M1 in place of A, M,
 ∂a A1 = A ∪ ( j = 1, . . . , n) for a ∈ A ∩ C 1 ∂x j ∪ {M f 0 , M ∈ M ∩ C 1 } ∪ { f 0 a, a ∈ A ∩ C 2 } 
∂M M1 = M ∪ ( j = 1, . . . , n) for M ∈ M ∩ C 1 ∂x j ∪ { f 0 M + M f 0 , M ∈ M ∩ C 2 }. We apply this, starting from A = {0}, M = {I }, to obtain first A = { f 0 }, M = {I, 2 f 0 }; then
   ∂ M = I, 2 f 0 , 2∂x j f 0 , 4 f 02 ; f 0 , f 0 f 0 , A = f0, ∂x j etc. Theorem 8.5.1. Suppose n ≥ 2, p ≥ 2, and let f : Rn × R p → R p , f  : Rn × R p → R p× p be at least C 1 and f 0 = f ( r , 0), f 0 = f  ( r , 0) at least C 2 . Define linear spaces A ⊂ C(Rn , R p ), M ⊂ C(Rn , R p× p ) by   A= Aj M= Mj j≥0

A0 = {0}, and for j ≥ 0, A j+1

j≥0

M0 = {I },


∂a = span A j ; (1 ≤ k ≤ n, a ∈ A j ∩ C 1 ); f 0 a (a ∈ A j ∩ C 2 ); ∂ xk  1 M f 0 (M ∈ M j ∩ C )

M j+1

8.5. Generic Simplicity of Solutions of a System
∂M = span M j ; (1 ≤ k ≤ n, M ∈ M j ∩ C 1 ); ∂ xk    2 f 0 M + M f 0 (M ∈ M j ∩ C ) .

173

M contains (at least) all polynomials in f 0 and A contains (polynomial in f 0 ) × f 0 . (1) Assume, for a dense set of x ∈ Rn , that {a(x),

all a ∈ A} = R p .

Then for most C 2 bounded regions  ⊂ Rn , all solutions of u + f (x, u) = 0 in , u = 0 on ∂, are simple. (2) In case f (x, 0) ≡ 0, we have A = {0}, but assume M is large so that: for a dense set of x ∈ Rn , whenever β = 0 in R p , {M(x)β, all M ∈ M} = R p . Then we have generic simplicity of solutions, as in case (1). Remark. An important case is when f (u) is independent of x and f (0) = 0;  p−1 then A = {0} and M = { k=0 ck f  (0)k (ck ∈ R)}, and hypothesis (2) ordinarily fails. (The only exception is when p = 2 and f  (0) has complex eigenvalues.) We will treat this case below. Proof. The difficult point, as usual, is density. We suppose this fails: for every  near some 0 (connected, bounded, smooth), suppose there is a non-simple solution. Then there is a solution (u, ψ) of (∗) such that ψ N r a = 0, ψ N r Mu N = 0 on ∂ for all a ∈ A, M ∈ M. In case (1), we may suppose {a(x) all a ∈ A} = R p for a dense (open) set in ∂ and conclude ψ N = 0 on ∂, hence ψ = 0 in , a contradiction. (It suffices to consider a finite subset of A, depending on 0 .) In case (2), we may suppose the dense (open) set of the hypothesis is also dense in ∂. Since ψ N = 0 on a dense set of ∂, we conclude u N ≡ 0 on ∂. By uniqueness in the Cauchy problem, u ≡ 0. Thus for all  near 0 , there is a nontrivial solution φ of φ + f 0 φ = 0 in , φ = 0 on ∂, and it is not simple. This is a particular (linear) case of our initial hypothesis, and M is unchanged when we replace f (x, u) by f  (x, 0)u, so we obtain the same conclusion: φ = 0, φ N = 0 on ∂, so φ ≡ 0 in , contrary to hypothesis. Now suppose f (u) is independent of x and f (0) = 0. Our hypothesis for contradiction is that, for any  near 0 , there is a solution (u, ψ) of
u + f (u) = 0 in , u = 0 on ∂ (∗) ψ + T f  (u)ψ = 0 in , ψ = 0 on ∂, ψ ≡ 0

174

Chapter 8. The Method of Rapidly-Oscillating Solutions

such that ψ N r f  (0) j u N = 0 on ∂ for all j ≥ 0 (or 0 ≤ j ≤ p − 1). For any , u ≡ 0 solves the first equation, but the zero solution is usually simple. In fact, it fails to be simple if and only if some eigenvalue λ of the matrix f  (0) is also an eigenvalue of the scalar Dirichlet problem φ + λφ = 0 in ,

φ = 0 on ∂,

φ ≡ 0.

For a dense open set of , we avoid the eigenvalues of f  (0) so assume henceforth: the zero solution is simple. Then our hypothesis above includes the condition u ≡ 0. Consider the case where all eigenvalues of f  (0) are simple (distinct). Say  f (0) p j = λ j p j , T f  (0)q j = λ j q j , with q j r p j = 1, q j r pk = 0 for j = k (1 ≤ j, k ≤ p). Then the condition ψ N r f  (0) j u N = 0 for all j ≥ 0 becomes  j λk ψ N r pk qk r u N = 0 for all j ≥ 0, k

hence ψ N pk qk r u N = 0 for k = 1, 2, . . . , p. Suppose we can prove qk r u N = 0 on a dense set of ∂, for every k; then ψ N r pk = 0 for every k, ψ N = 0 on ∂, ψ = 0 in , contrary to hypothesis.  The argument is a bit more general: 0 1 if each eigenvalue of f (0) has only one  r eigenvector (as in the case f (0) = 0 0 ) and qk u N = 0 densely ∂ for every eigenvector qk of T f  (0), we again conclude ψ ≡ 0. In the following lemma we make no hypothesis about f  (0). r

Lemma 8.5.2. Assume f : R p → R p is C 4 with f (0) = 0. Let Q ⊂ C p be any subspace = {0} and let V ⊂ Rn be open and assume: for every  near 0 , there is a solution u of u + f (u) = 0 in ,

u = 0 on ∂,

u ≡ 0,

such that q r ∂u/∂ N = 0 on ∂ ∩ V for every q ∈ Q. Then, for every  near 0 , there is a solution u ≡ 0 such that for all q ∈ span {T f  (0) j Q : j ≥ 0}, q r u N = 0,

q r f  (0)u 2N = 0,

q r f  (0)(u N , f  (0)u N ) = 0,

q r f  (0)u 3N = 0,

and q r f iv (0)u 4N = 0

on ∂ ∩ V. Proof. The transversality theorem says there is a solution u ≡ 0 such that, for q ∈ Q, q r u N = 0 on ∂ ∩ V and also σ
→ q r ∂ N B L (σ u N )|∂∩V ,

L =  + f  (u)

8.5. Generic Simplicity of Solutions of a System

175

has finite rank. (It suffices to use a finite set of q in Q.) Taking σ = iθ γ e , |∇∂ θ| = 1 on supp γ , we conclude q r U˙ k = 0 on ∂ ∩ V for k ≥ 0

(at least, 0 ≤ k ≤ 4.)

Now U0 | ∂ = γ u N and Uk | ∂ = 0 for k ≥ 0, so also q r Uk = 0 on ∂ ∩ V , or q r Uk (y + t N (y)) = O(t 2 ) as t → 0 with y ∈ ∂ ∩ V, k ≥ 0. Since = ∇ S r ∇ + 1/2 S is a scalar operator, acting separately on each component, and q is constant, we have q r U0 = q r U0 = 0 with q r U0 |∂ = 0, and so q r U0 (y + t N (y)) = o(|t| N ) as t → 0, y ∈ ∂, for arbitrary N – assuming ∂, θ, γ are smooth, as we may. For k ≥ 1

q r Uk + (q r Uk−1 ) + q r f  (u)Uk−1 ∼ = 0,

q r Uk = O(t 2 )

Define ck by q r f  (u)U0 (y + t N (y)) = c0 + tc1 + t 2 c2 + t 3 c3 + o(t 3 ), recalling f  is C 3 , and then

q r U1 = −(c0 + c1 t + c2 t 2 + c3 t 3 ) + o(t 3 ) as t → 0, in ∂ ∩ V. At t = 0, q r U˙ 1 = −c0 = 0 or q r f  (0)γ u N = 0 for q ∈ Q.Thus our hypothesis holds also with span {Q,T f  (0)Q} in place of Q. We may, and shall, assume T  f (0)Q ⊂ Q. Since c0 = 0, q r U1 = −c1 t 2 /2 + O(t 3 ) so (q r U1 )|∂ = −c2 so 0 =

q r U2 + q r U1 + q r f  (u)U1 |t=0 = q r U˙ 2 − c1 , so c1 = 0. (We compute c1 , c2 , c3 below.) Since c1 = 0, q r U1 = −c2 t 3 /3 + O(t 4 ) so 0 = q r U2 + (−2c2 t + O(t 2 )) + q r f  (0)t U˙ 1 + O(t 2 ), so q r U2 = c2 t 2 + O(|t|3 ) (recalling q r f  (0)U˙ 1 = 0). Then 0 = q r U3 + q r U2 + q r f  (u)U2 |∂ = q r U˙ 3 + 2c2 so also c2 =0. Now q r U1 = −c3 t 4 /4 + o(t 4 ), and this step is more delicate. We have q r 2 f  (u)(t U˙ 1 + t2 U¨ 1 + · · ·) = tq r f  (0)U˙ 1 + t 2 (q r f  (0)u N U˙ 1 + 1/2 q r f  (0)U¨ 1 ) + o(t 2 ), and the t-coefficient vanishes, so q r U2 = at 3 + o(t 3 ) where a = c3 −   1/ (q r f  (0)u U 1 ˙ ¨ ˙ r r r N 1 + /2 q f (0)U1 ). Then 0 = q U3 + 6at + tq f (0)U2 + 3 2 2 o(t) so q r U3 = −3at + o(t ), and 0 = q r U˙ 4 − 6a, or a = 0. In fact q r U1 = O(t 4 ) for any q ∈ Q so also q r f  (0)U1 = O(t 4 ) and q r f  (0)U¨ 1 = 0 in ∂ ∩ ∇. Thus we have q r f  (u)U0 = c0 + c1 t + c2 t 2 + c3 t 3 + o(t 3 ) t3 = (q r f  (0)U˙ 1 ) + o(t 3 ) 3 = q r ( f  (u) − f  (0))U0 + o(t 3 )

176

Chapter 8. The Method of Rapidly-Oscillating Solutions  t2 t3  r = q f (0) tu N + u N N + u N N N + · · · 2 6  2 t ¨ ˙ r γ U N + t U0 + U0 + · · · 2  2 t2  1 r + /2 q f (0) tu N + u N N + · · · 2 ˙ r (γ U N + t U0 + · · ·) +1/6 q r f iv (0)(tu N + · · ·)3 (γ U N + · · ·) + o(t 3 ).

The coefficient of t vanishes, so q r f  (0)u 2N = 0 in ∂ ∩ V, which implies 0 = ∂τ (q r f  (0)u 2N ) = 2q r f  (0)u N ∂τ u N for any τ ⊥ N , so q r f  (0)u N U˙ 0 = 0. Thus the coefficient of t 2 is γ γ 0 = q r f  (0)u N u N N + q r f  (0)u 3N 2 2 since u N N + H u N = 0 (= u + f (u)|∂) we have q r f  (0)u 3N = 0 on ∂ ∩ V . Omitting some terms clearly zero, we find from the coefficient of t 3, γ (q

f iv (0)u 4N + q r f  (0)u N u N N N ) = 2q r f  (0)u N U˙ 1 − 3q r f  (0)u N U¨ 0 r

= −5q r f  (0)u N U¨ 0 − 2q r f  (0)u N (H U˙ 0 + ∂ γ u N + f  (0)γ u N ) = 5γ q r f  (0)u N ∂θ2 u N − 2γ q r f  (0)u N ( ∂ u N + f  (0)u N ). Since 0 = ∂ N ( u + f (u))|∂ = u N N N − (H 2 + H2 )u N + ∂ u N + f  (0)u N , we have, in supp γ ⊂ ∂ ∩ V , 5q r f  (0)u N ∂θ2 u N = q r f iv (0)u 4N + q r f  (0)u N ( ∂ u N + f  (0)u N ). Since ∂θ2 (q r f  (0)u 2N ) = 0, the left-side is −5q r f  (0)(∂τ u N )2 when ∇∂ θ = τ ⊥ N , |τ | = 1, and this is independent of the choice of τ . Thus q r f  (0)(∂τ u N )2 = average q r f  (0)(τ r ∇∂ u N )2 τ ⊥ N ,|τ |=1

1 q r f  (0)(∇∂ u N )2 n−1 1 q r f  (0)u N ∂ u N =− n−1 =

(The penultimate term is ambiguous, but clear in component form.) Then (n − 6)q r f  (0)(∂τ u N )2 = Q(u N )

for |τ | = 1,

where Q(u N ) = q r f iv (0)u 4N + q r f  (0)(u N , f  (0)u N ).

τ ⊥ N,

8.5. Generic Simplicity of Solutions of a System

177

A similar condition occurred above, just before Theorem 5.1, and this is treated the same way. Since q r f  (0)∂τ u N u N N = 0 = q r f  (0)u 2N N (u N N = −H u N ) for z ∈ Rn , (n − 6)q r f  (0)(∂z u N )2 − |z|2tan Q(u N ) = 0 on ∂ ∩ V. We apply the transversality theorem, and as usual the principal part says nothing new, but the sub-principal part gives (with z = τ ⊥ N , ∇∂ θ = σ ⊥ N , |σ | = |τ | = 1) 2(n − 6)q r f  (0)(∂τ u N )(∂τ u N + τ r σ ∂σ u N ) = Q  (u N )u N = 2{Q(u N ) + (τ r σ )2 Q(u N )}. Choose τ = σ: then Q  (u N )u N = 4Q(u N ), so the quadratic part vanishes, q r f  (0)(u N , f  (0)u N ) = 0. if n ≥ 3, we may choose τ ⊥ σ , so also q r f iv (0)u 4N = 0. This completes the proof of the lemma. Theorem 8.5.3. Let f : R p → R p be C 4 with f (0) = 0. Assume f  (0) has only simple eigenvalues, or more generally, each eigenvalue of f  (0) has only one eigenvector. Further assume, whenever β = 0 in R p , that R p = span f  (0) j {β, f  (0)β 2 , f  (0)β 3 , f  (0)(β  , f (0)β), f iv (0)β 4 }. j≥o

Then for most bounded C 2 regions  ⊂ Rn , all solutions of u + f (u) = 0 in  ⊂ Rn ,

u = 0 on ∂,

are simple. Proof. Let {q1 , . . . , qm } be a basic for the eigenfunctions of T f  (0) and let {V1 , V2 , . . .} be a countable basis for the topology of Rn – for example, all open balls of rational radius whose center has rational coordinates. By the lemma, for each j ∈ {1, 2, . . . , m} and each k ∈ {1, 2, . . .}, there is an ample (= category II) set of regions  such that, for any solution u ≡ 0, we have q j r ∂u/∂ N = 0 at some point of ∂ ∩ Vk (unless the intersection is empty). In the countable intersection, which is still ample, any solution u ≡ 0, satisfies q j r ∂u/∂ N = 0 on a dense set of ∂, for all j ∈ {1, . . . , m}. As note above (before the lemma), this implies denseness of the property “only simple solutions.” Example. ( p = 2) f : R2 → R2 is C 4 with  Q 1 (u 2 ) + K 1 (u 3 ) + L 1 (u 4 ) + o(|u|4 ) f (u) = f  (0)u + Q 2 (u 2 ) + K 2 (u 3 ) + L 2 (u 4 ) + o(|u|4 )

178

Chapter 8. The Method of Rapidly-Oscillating Solutions

where Q 1 , Q 2 are (real homogeneous) quadratic polynomials; K 1 , K 2 are cubics; L 1 , L 2 are quartics; and we consider the cases (i) f  (0) has two complex eigenvalues 0 (ii) f  (0) = ( λ 0 µ ) has real distinct eigenvalues 1 (iii) f  (0) = ( λ 0 λ ) has one eigenvalue and only one eigenvector

0 (iv) f  (0) = ( λ 0 λ ) has one real eigenvalue with two eigenvactors.

(We may always change the u-variable to put f  (0) in real Jordan canonical form.) Case (i): Theorem 8.5.1, part (2), applies. Say f  (0)( p + iq) = (α + iβ) ( p + iq) where α, β ∈ R and p, q ∈ R2 , β = 0 and ( p, q) = (0, 0). It is easy to see p, q are independent when β = 0, and verify {γ , f  (0)γ } are independent for any γ = 0 in R2 . Case (ii): Let e1 = col(1, 0), e2 = col(0, 1): then assuming







0 = Q 1 e22 + K 1 e23 + L 1 e24 and







0 = Q 1 e12 + K 2 e13 + L 2 e14 ,

the hypotheses of Theorem 8.5.3 hold. Case (iii): If α r f  (0) j β = 0 for j = 0, 1, and α = 0, β = 0 in R2 , then α1 = 0, β2 = 0, and we may suppose without loss of generality α = e2 , β = e1 . The hypotheses of Theorem 8.5.3 hold if







0 = Q 2 e12 + K 2 e13 + L 2 e14 . Case (iv): This case is not included above, but will be permitted in the next theorem provided for some γ ∈ R2   γ1 Q 1 (γ 2 ) det  0 ≡ γ2 Q 2 (γ 2 ) and for all γ = 0 in R2 , the 2 × 4 matrix [γ , Q(γ 2 ), K (γ 3 ), L(γ 4 )] has rank 2. Remark. Many cases remain open – for example Case (iv) when f is odd, f (−u) = − f (u), or f  (0) = λI with n ≥ 3.

8.5. Generic Simplicity of Solutions of a System Theorem 8.5.4. Let f : R2 → R2 be C 4 , f (0) = 0, f  (0) = ( λ0 real λ, and assume

179 1 ) λ

for some

(i) for some γ = 0 in R2 , f  (0)γ 2 is not a real multiple of γ , or det[γ , f  (0)γ 2 ] ≡ 0; (ii) for every γ = 0 in R2 , the 2 × 4 matrix [γ , f  (0)γ 2 , f  (0)γ 3 , f iv (0)γ 4 ] has rank 2. Then for most C 2 bounded regions  ⊂ R2 (n ≥ 2), all solutions u of u + f (u) = 0 in , u = 0 on ∂ are simple. Proof. As noted earlier, we may assume the zero solution is simple. If there is a non-simple solution for every  near some 0 , then there is a solution (u, ψ) of (∗) with u ≡ 0, ψ ≡ 0 and ψ N r u N = 0 on ∂. We apply the transversality theorem to conclude that also (in usual notation) ˙ k r u N + ψ N r U˙ k = 0 on ∂, 

for k ≥ 0.

The equation for k = 0 says nothing new; consider k = 1. ˙ 1 r u N − ψ N r U˙ 1 0 = − ¨ 0 + H ˙ 0 + ∂ γ ψ N + f  (0)γ ψ N ) r u N = ( + ψ N r (U¨ 0 + H U˙ 0 + ∂ γ u N + f  (0)γ u N ), which (with f  (0) = λI ) simplifies to

γ ( ∂ ψ N r u N + ψ N r ∂ u N ) = γ ∂θ2 ψ N r u N + ψ N r ∂θ2 u N . If n = 2, then ∂ = ∂θ2 so we have nothing new; but assuming n ≥ 3, it follows (as above) that, where γ = 0, ∂θ2 ψ N r u N + ψ N r ∂θ2 u N = −2∂θ ψ N r ∂θ u N =

1 (ψ N r ∂ u N + ∂ ψ N r u N ) n−1

so ∂τ ψ N r ∂τ u N = 0 for any τ ⊥ N . For any tangent vector field σ, Dσ = (scalar) σ r ∇∂ + (scalar) is a general first-order operator, and a second-order

180

Chapter 8. The Method of Rapidly-Oscillating Solutions

(tangential) operator is a sum of terms Dσ1 Dσ2 + Dσ3 , and we have ψ N r u N = 0,

Dσ ψ N r u N + ψ N r Dσ u N = 0,

Dσ1 ψ N r Dσ2 u N + Dσ2 ψ N r Dσ1 u N = 0 and, for any second-order tangential operator D 2 , D 2 ψ N r u N + ψ N r D 2 u N = 0. (For example, ∂ (γ ψ N ) r u N + ψ N r ∂ (γ u N ) = 0.) This greatly simplifies the calculation for k = 2 – in fact, almost everything ˙ 2 r u N reduces to 0 = 2γ ψ N r f  (0)u 2N . vanishes and 0 = ψ N r U˙ 2 +  In case n = 2, the equations for k = 0, k = 1, tell us nothing. We must go ˙ 2 r u N = 0, knowing only ψ N r u N = 0 and n = 2. Fordirectly to ψ N r U˙ 2 +  tunately, the calculation simplifies for n = 2 since q = H = κ (the scalar curvature of ∂), θ is arc-length along ∂, ∂ = ∂ 2 /∂θ 2 , and σ0 = −q + H + i ∂ θ = 0. Again everything cancels but one term, though the calculations are more difficult and we conclude, for n ≥ 2, ψ N r f  (0)u 2N = 0 on ∂. Now let {β1 . . . , βm } be the finite set of β ∈ R2 with |β| = 1 and det[β, f  (0)β 2 ] = 0. Choose γk = βk⊥ ∈ R2 , βk rotated 90◦ . By Lemma 8.5.2 and hypothesis (ii), we may suppose any non-zero solution u satisfies γk r u N = 0 on a dense set of ∂ for each k = 1, . . . , m, so {u N , f  (0)u 2N } is a basis for Rn . But ψ N r u N = 0 and ψ N r f  (0)u 2N = 0 on ∂, so ψ N ≡ 0 on ∂, ψ ≡ 0 in , contrary to hypothesis.

Summary of 8.2 (1) |∇∂ θ| ≡ 1, ∇ S r ∇ S = 0, S|∂ = iθ, Re ∂∂sN > 0 S(y + t N (y)) =

 t k Sk (y) k≥0

k!

t3 qt 2 + (−q 2 + 3r + i∂θ q) = iθ(y) + t − 2! 3! t4 + (−i∂θ S3 + 3q S3 − 12s − 6i∇∂q r K ∇∂ θ) + O(t 5 ) 4! where q = θ  r K θ  , r = θ  r K 2 θ  , S = θ  r K 3 θ  , θ  = ∇∂ θ, ∂θ = θ  r ∇∂ . ˙ + (2) = ∇ S r ∇ + 1/2 σ, σ = S + b r ∇ S, b(y + t N (y)) = b(y) + t b(y) t2 ¨ b(y) + · · · 2! t2 σ (y + t N (y)) = σ0 (y) + tσ1 (y) + σ2 (y) + · · · 2! = −q + H + i ∂ θ + b r (N + iθ  ) + t{S3 − q H − H2 + i∂θ H − 2idiv∂ (K θ  )

8.5. Generic Simplicity of Solutions of a System

181

+ b˙ r (N + iθ  ) + b r (−q N − i K θ  )} t2 {S4 + q H S3 + 2q H2 + 2H3 − i∂θ H2 2! − 4i(K θ  ) r H  + b¨ r (N + iθ  ) + 2b˙ r (−q N − i K θ  ) +

+ 6i div∂ (K θ  ) − ∂ q + b r (S3 N − q  + 2i K 2 θ  )} + O(t 3 ) Hm = trace K m , H = H1 , K = D N = curvature. ∇ S r ∇ = ∂t + i∂θ + t(−q∂t − 2i K θ  r ∂ y ) +

t2 (S3 ∂t + 6i K 2 θ  r ∂ y − q  r ∂ y ) + · · · 2

=L    (3) ( + b r ∇ + c) eωS (U0 + 1 F + · · ·) 2ω 1 U j |∂ = G j ,

U0 = F0 ,

1 U 2ω 1

+

1 U (2ω)2 2

+ · · ·) = 2ωeωS (F0 +

U j + LU j−1 = F j (U−1 = 0)

t2 U0 (y + t N (y)) = G 0 (y) + t U˙ 0 (y) + U¨ 0 (y) + · · · 2!

At t = 0: F0 = U˙ 0 + i∂θ G 0 + 1/2σ0 G 0



F˙0 = U¨ 0 + i∂θ + 1/2σ0 − q U˙ 0 + − 2i K θ  r ∂ y + 1/2σ1 G 0



˙¨ 0 + i∂θ + 1/2σ0 − 2q U¨ 0 + (−4i K θ  r ∂ y + σ1 + S3 U˙ 0 F¨0 = U

+ 6i K 2 θ  r ∂ y − q  r ∂ y + 1/2σ2 G 0

U1 + LU0 = F1 ,

U1 |∂ = G 1 ,

U1 = G 1 + t U˙ 1 +

t2 ¨ U1 + · · · 2

At t = 0:

F1 = U˙ 1 + i∂θ + 1/2σ0 G 1 + U¨ 0 + (H + b r N )U˙ 0

+ ∂ G 0 + b r ∇∂ G 0 + cG 0



F˙1 = U¨ 1 + i∂θ + 1/2σ0 − q U˙ 1 + −2i K θ  r ∂ y + 1/2σ1 G 1 ˙¨0 + (H + b r N )U¨ 0 + ∂ U˙ 0 + b r ∇∂ U˙ 0 + cU˙ 0 +U

182

Chapter 8. The Method of Rapidly-Oscillating Solutions + (b˙ r N − H2 )U˙ 0 − 2 div∂ (K G 0 ) ˙ 0 + (b˙ + H  ) r G 0 − b r K G 0 + cG

U2 + LU1 = F1

At t = 0:

U2 |∂ = G 2

F2 = U˙ 2 + i∂θ + 1/2σ0 G 2 + U¨ 1 + (H + b r N )U˙ 1 + ∂ G 1 + b r ∇∂ G 1 + cG 1

Appendix 1 Eigenvalues of the Laplacian in the Presence of Symmetry

A. L. Pereira, in his doctoral thesis investigated more general symmetry conditions than the one discussed in 6.2. We describe some of his results (and sketch some of the arguments) that appear with more details in [28], which is the general reference for this section. We proved (Example 6.1) that, for most C 2 regular bounded regions
⊂ Rn , all eigenvalues of u + λu = 0 in
,

u = 0 in


are simple. Given this result, it may be surprising to find higher multiplicity in many specific examples. This is well-known in the case of the disc or the rectangle in R2 . Another example is the equilateral triangle, where eigenvalues with arbitrarily high multiplicity exist (see Pinsky [29]). It seems that higher multiplicity appears whenever a detailed analysis can be done. Now, a common characteristic in these cases is the symmetry of the regions. In fact, it is the symmetry that makes a detailed analysis possible. To be more precise, we start with some definitions. Let G be a subgroup of the orthogonal group O(n) on Rn . We say that
is G-symmetric (or G-invariant) if g(
) =
for every g ∈ G. A map h :
→ Rn is (G-) equivariant if h ◦ g = g ◦ h for all g ∈ G, so h(
) is G-symmetric when
is G-symmetric. A function φ on Rn is even or G-invariant if φ ◦ g = φ for all g ∈ G. If
is G-symmetric and u is an eigenfunction then u ◦ g −1 is also an eigenfunction for every g ∈ G, so the multiplicity of λ is at least the dimension of span {u ◦ g −1 | g ∈ G}. In fact, suppose there exists a point x ∈
such that {g ∈ G : gx = x} = {Id} (such a point is called a free point under the natural action of G in
; there is always a free point in
if G is, for example, finite or commutative). Then, if some irreducible representation of G has real dimension m, there are infinitely many eigenvalues of the Laplacian with multiplicity ≥ m in any G-symmetric bounded open set. 183

184

Appendix 1. Eigenvalues of the Laplacian in the Presence

If we want only simple eigenvalues, the irreducible representations of G must have real dimension one, i.e., G is isomorphic to Z 2 ⊕ Z 1 ⊕ · · · ⊕ Z 2 (m times) or G is generated by m commuting reflections. In any other case the symmetry forces higher multiplicity. Suppose
is G-symmetric, λ is an eigenvalue of the Laplacian in
with an eigenfunction u; if span {u ◦ g −1 : g ∈ G} = N (λ + ); i.e., the orbit of some eigenfunction generates all eigenfunctions corresponding to λ, we say that λ is a G-simple eigenvalue. [If we use complex-valued functions, we should also ¯ Equivalently, λ is G-simple if the natural include the G-orbit of the conjugate u.] action of G(u → u ◦ g −1 ) on N ( + λ) is irreducible. Suppose
is a G-symmetric bounded open set in Rn which is C 2 regular, and λ is a G-simple eigenvalue of multiplicity m. Then, for every
, C 2 , G˜ near
there is a unique eigenvalue λ(
) ˜ for the Laplacian equivariant region
˜ ˜ in
which is close to λ = λ(
), λ(
) is also G-simple of multiplicity m, and ˜ → λ(
) ˜ is analytic.
Thus a G-simple eigenvalue in a G-symmetric region behaves like a simple eigenvalue when no symmetry is imposed. The most we could hope for then is: in generic G-symmetric C 2 -domains, all eigenvalues of the Laplacian are G-simple. This is proved in certain cases – for finite commutative groups (argument sketched below) and for finite subgroups of O(2) ⊂ O(n) (such as the non-commutative dihedral groups) along with partial results for some other cases (see [28]). The commutative case is comparatively simple, since the irreducible representations all have real or complex dimension one it suffices to use characters. A character of G is a map χ : G → C with |χ (g)| = 1, χ(Id ) = 1, χ (g1 • g2 ) = χ(g1 ) • χ (g2 ) for g, g1 , g2 ∈ G. The character is real if it only has values in {−1, 1}. If χ is a character, define Mχ = {u ∈ L 2 (
, C) | u ◦ g −1 = χ (g)u, g ∈ G} If χ1 , χ2 are distinct characters, then Mχ1 ⊥ Mχ2 . In fact, if u ∈ Mχ1 , v ∈ Mχ2 , then






u v¯ =




u ◦ g −1 v¯ ◦ g −1 = χ1 (g)χ2 (g)





u v, ¯

 for all g ∈ G, so if
u v¯ = 0, we must have χ1 (g) = χ2 (g) for all g ∈ G, or χ1 = χ2 . Since G is commutative L 2 (
, C) = ⊕ Mχ (see [20] p. 51). Further χ (H 2 ∩ H01 (
, C) ∩ Mχ ) ⊂ Mχ for every character χ and G-symmetric
. It suffices then to prove the following:

Appendix 1. Eigenvalues of the Laplacian in the Presence

185

(A) For every character χ and integer k, there is a dense open set of G-invariant C 3 regions
such that the restriction  | H 2 ∩ H01 (
, C) ∩ Mχ → Mχ has only simple eigenvalues λ with |λ| ≤ k. (B) For an integer k and every pair of characters with χ1 = χ2 and χ1 = χ¯ 2 , there is a dense open set of G-invariant C 3 regions
such that there is no eigenvalue λ with |λ| ≤ k with both an eigenfunction in Mχ1 and one in Mχ2 . For real characters the argument is like in Example 6.1. In any case, openness is easy to obtain; to obtain density we use transversality arguments. For (A) with a complex χ , given a C 3 , G-invariant
0 ⊂ Rn consider the map F (u, λ, h) −→ h ∗ ( + λ)h ∗−1 u ∈ Mχ with u ∈ H 2 ∩ H01 (
, C) ∩ Mχ \{0}, λ ∈ C, |λ| ≤ k, h ∈ E 3 where E 3 = {C 3 G-equivariant imbeddings of
0 ⊂ Rn }. It is enough to show that 0 is a regular value of F. If this is not true, we find that there exists λ ∈ C, |λ| ≤ k and u, ψ ∈ H 2 ∩ H01 (
, C) ∩ Mχ \ {0} such that   u + λu = 0 in
, ψ + λψ = 0 in
 ¯  (∗)  ∂ ψ ∂u ¯ = 0,  ψu = 0 on ∂
. Re
∂N ∂N

 ¯ = 0). (Note that ψ = iku, k ∈ R\{0} satisfies all conditions except that
ψu To see this is not possible for most
(i.e., except in a closed set with empty interior), we consider the map  G (u, ψ, λ, h) −→ h ∗ ( + λ)h ∗−1 u, h ∗ ( + λ)h ∗−1 ψ,
¯ h ∗ Bh ∗−1 (u, ψ)|∂
ψu,


¯ u, ψ ∈ H 2 ∩ H01 (
, C) ∩ Mχ \{0}, λ ∈ C, |λ| ≤ where B(u, v) = Re
u •
v, 3 k, h ∈ E and the values are in Mχ × Mχ × C × L 1 (
, C). Again, it is easy to see that G −1 (0, 0, 0, 0) is closed. If it is (0, 0, 0, 0) for some choice of (u, ψ, λ) for every h in an open set, we may suppose i
0 is in its interior and then, condition (2.β) of Theorem 5.4 must fail, so the quotient R(DG(u, ψ, λ, h))   ∂G R ∂(u,ψ,λ)

186

Appendix 1. Eigenvalues of the Laplacian in the Presence

must be finite dimensional. This implies that for every
, C 3-close to
0 , there is a solution (u, ψ, λ) of (∗) such that

σ −→ Re{u N ∂ N B+λ (σ ψ¯ N ) + ψ N ∂ N B+λ (σ u N )} with σ even, has finite rank. Here B+λ is given by v = B+λ g, where B+λ is defined by

( + λ)v ∈ N ( + λ) v ⊥ N ( + λ) v|∂
= g, The method of rapidly oscillating solutions shows, as ω → ∞

(γ cos ωθ) = ω−1 γ cos ωθ Re(∇∂
u N

•

∇∂
ψ¯ N − ∂θ u N ∂θ ψ¯ N ) + O(ω−2 )

where γ , θ are even |∇∂
θ| = 1 in supp γ and γ is supported in a small neighborhood of {Gx} for some x ∈ ∂
– recall G is finite. Then Re(∂τ u N ∂τ ψ N ) = Re(∇∂
u N • ∇∂
ψ¯ N ) for all τ ⊥ N , |τ | = 1 which implies Re(∂τ u N ∂τ ψ¯ N ) =

1 Re(∇∂
u N n−1

•

∇∂
ψ¯ N ).

If n ≥ 3 we conclude Re(∂τ u N ∂τ u¯ N ) = 0 for all τ ⊥ N . If n = 2, more calculation is needed – we only quote the result:

(γ cos ωθ) = −ω−3 λγ Re(∂θ ψ¯ N ∂θ u N ) + O(ω−4 ). Thus in every case, on ∂
Re(u N , ψ¯ N ) = 0

and

Re(u n ψ¯ N ) = 0

where  indicates any tangential derivative. Consider a point where u N = 0, ψ N = 0 u N = q1 + iq2

ψ N = r1 + ir2

(q j , r j real),

then q1r1 + q2r2 = 0, q1 r1 + q2 r2 = 0 so, for some α, r1 = αq2 , r2 = −αq1 and 0 = α  (q2 q1 − q1 q2 ). If ∇∂
α = 0, almost all tangential directions have α 1 = 0 so q1 /q2 (or q2 /q1 ) is locally constant. It follows, for some z ∈ C − {0}, that u/z is real valued in
. But this is impossible in Mχ \{0}, when χ is a complex character. If α is locally constant, for some  iαu N on a  α ∈ R, ψ N = nonempty open set of ∂
, so ψ = −iαu in
, 0 =
ψ¯ • u = iα
|u|2 . Thus α = 0 so ψ = 0 which is false. This contradiction shows, for an open dense set of E 3 (∗) is not solvable; restricting F to this set, we obtain (A).

Appendix 1. Eigenvalues of the Laplacian in the Presence

187

(B) Suppose, for every G, C 3 -near
0 , there exists u, v, λ with  u ∈ Mχ1 , v ∈ Mχ2 , u, v in H 2 ∩ H01 (
, C)\{0} |λ| ≤ k (∗∗) u + λu = 0 in
, v + λv = 0 in
We may assume λ is a simple eigenvalue for the restriction of − in Mχ1 ˜ G-symmetric near
there is a unique eigenvalue λ j (
) ˜ and in Mχ2 . For
˜ ˜ near λ for the restriction of  to Mχ j and λ1 (
) = λ2 (
). Thus  the derivative ˜ − λ2 (
) ˜ vanishes at
˜ =
so (assuming |u|2 = |v|2 = 1) we of λ1 (
)





also have | ∂∂uN |2 = | ∂∂vN |2 on ∂
. Thus, for every h ∈ E 3 near i
0 , there exists u, v ∈ H 2 ∩ H01 (
, C)\{0} with u ∈ Mχ1 , v ∈ Mχ2 , and λ ∈ C, with |λ| ≤ k and (h ∗ ( + λ)h ∗−1 u, h ∗ ( + λ)h ∗−1 v, h ∗ Ch ∗−1 (u, v)|∂
) = (0, 0, 0)

where C(u, v) = |∇u|2 − |∇v|2 ; the values are in Mχ1 × Mχ2 × L 1 (∂
). Theorem 5.4 (2.β) then implies for every G-symmetric region
, C 3 -near  
0 , 2 there is a solution (u, v, λ) of (∗∗) which also satisfies
|u| = 1 =
|v|2 , |u N |2 = |v N |2 on ∂
and  σ −→ Re(u¯ N ∂ N (B+λ (σ u N ) + v¯ N ∂ N B+λ (σ v N )) has finite rank. This implies (with γ , θ even and |∇∂
θ| = 1 is supp γ ) (γ cos ωθ) = 2ω−1 γ cos ωθ (|∇∂
u N |2 − |∇∂
v N |2 − |∂θ u N |2 + |∂θ v N |2 ) + O(ω−2 ) If n = 2, we have λ cos ωθ(|∂θ u N |2 − |∂θ v N |2 ) + O(ω−4 ) 2 Then – since G is finite – we find |∂τ u N | = |∂τ v N | on ∂
for all τ ⊥ N . Let u N = Reiρ , v N = Reiσ , then (γ cos ωθ) = −ω−3

|∂τ u N | = |Rτ + i Rρτ | = |Rτ + i Rστ | so ρτ2 = στ2 , ∂τ (ρ − σ )∂τ (ρ + σ ) = 0 when τ ⊥ N . Either ρ − σ (and u N /v N ) is locally constant or ρ + σ (and u N /v¯ N ) is locally constant, thus either u, v are linearly dependent (so χ1 = χ2 ) or u, v¯ are linearly dependent (so χ1 = χ¯ 2 ). Therefore we have a contradiction and (B) follows. With (A) and (B) we conclude: for some G-symmetric C 2 regions
⊂ n R (n ≥ 2), all eigenvalues are G-simple: simple eigenvalues when the eigenfunction corresponds to a real character, double when it corresponds to a complex character with independent eigenfunctions u, u¯ (or Re, u, Im u).

Appendix 2 On Micheletti’s Metric Space

Anna Maria Micheletti [22] proved the set Mm (
)(1 ≤ m < ∞) of regions in Rn which are C m -diffeomorphic to a given set
⊂ Rn is metrizable using the “Courant distance” dm : for
1 ,
2 ∈ Mm (
), dm (
1 ,
2 )
 N   −1  = inf || f k − Id||C m +  f k − IdC m | f 1 ◦ f 2 ◦ · · · ◦ f N (
1 ) =
2 k=1

where we consider all N ≥ 1 and all C m diffeomorphisms f k : Rn → Rn such that D j ( f k (x) − x) → 0 as |x| → ∞ for 0 ≤ j ≤ m. Further, the metric space (Mm (
), dm ) is complete and separable. (In fact, we only consider regions diffeomorphic to
by diffeomorphisms which are near the identity at infinity. Diffeomorphic images of
could be “knotted” in different ways; we ignore such complications, so Mm (
) does not usually include all diffeomorphs of ˜ ∈ Mm (
) contains a neigh
.) In the metric topology, a neighborhood of
n n ˜ borhood of the form {F (
) | F : R → R a C m map with ||F − Id||C m <  and D j (F(x) − x) → 0 as |x| → ∞, 0 ≤ j ≤ m} for small  > 0; and such ˜ < δ} for some δ > 0. These a neighborhood contains a ball {
 : dm (
 ,
) results are proved below (A.9), in the manner of Micheletti (who considered m = 3). If
is bounded and C m -regular, Mm (
) may be considered a C 0 Banach manifold, modeled on the Banach space C m (∂
, R) (see (A.10)). Given a C m vector field V : Rn → Rn transverse to ∂
and (small) σ ∈ C m (∂
, R) we define
(σ ) near
by ∂
(σ ) = {x + σ (x)V (x) : x ∈ ∂
}, which defines a homeomorphism σ →
(σ ) from C m (∂
, R) [near zero] onto a neighborhood of
in Mm (
); various such choices of coordinates are C 0 -compatible. 188

Appendix 2. On Micheletti’s Metric Space

189

This coordinate system has the advantage of uniqueness – given V – but is otherwise difficult to use. We will usually employ the set of C m imbeddings h :
→ Rn with h(
) ∈ Mm (
), so the restriction h :
→ h(
) ⊂ Rn is a diffeomorphism. This is an open set Diff m (
) in the Banach space C m (
, Rn ) and naturally inherits the structure of an analytic Banach manifold. The map h → h(
) : Diff m (
) → Mm (
) is continuous, surjective and open, with a local right-inverse near any point of Mm (
). [We use the above (C 0 ) coordinate system to define the right inverse; it is assumed
is bounded and C m -regular.] We call “σ -closed” a countable union of closed sets, also termed an “Fσ ”. In our genericity problems, we consider sets F ⊂ Diff m (
) defined by properties of the image F(
) = {h(
) : h ∈ F} ⊂ Mm (
), so F = {h : h(
) ∈ F(
)}. Assuming this, F is closed [or σ -closed, or meager] in Diff m (
) if and only if F(
) is closed [or σ -closed, or meager] in Mm (
) (see (A.12)). More precisely, when F or F(
) is closed, the codimension of F in Diff m (
) equals the codimension of F(
) in Mm (
). (“Codimension” is defined in 5.16; positive codimension is equivalent to meagerness.) We begin with the formula of Fa´a de Bruno: for k ≥ 2, if f, g are C k ,  k! f ( j) (g(x))(g  (x)y)α1 ( f ◦ g)(k) (x)y k = α! 1≤ j=|α|≤k |α  |=k



1  g (x)y 2 2!

α2

 ···

1 (k) g (x)y k k!

αk (A.1)

= f (k) (g(x))(g  (x)y)k + f  (g(x))g (k) (x)y k  + ··· 2≤ j=|α|≤k−1 |α  |=k αi =0 for i≥k

where α = (α1 , α2 , α3 , . . . ), α  = (α1 , 2α2 , 3α3 , . . . , jα j , . . . ), the α j are integers ≥ 0. |α| = α1 + α2 + α3 + . . . , α! = α1 !α2 !α3 ! . . . , and the second equality of (A.1) isolates the terms with a highest-order (k th ) derivative. It is enough to prove this when f, g are polynomials, and then it becomes the multinomial formula: g ( j) (x)y j g(x + y) = g(x) + γ1 + γ2 + . . . , γ j = j! and

 1 ( f ◦ g)(k) (x)y k = f (g(x) + γ1 + γ2 + . . . ) k! k≥0  1 f ( j) (g(x))(γ1 + γ2 + . . . ) j = j!  1 f ( j) (g(x))γ1α1 γ2α2 . . . . = α! |α|= j≥0

f (g(x + y)) =

190

Appendix 2. On Micheletti’s Metric Space

Counting the powers of y in γ1α1 γ2α2 , . . . gives α1 + 2α2 + 3α3 + . . . = |α j |. Define |ψ ( j) (x)| = sup|y|≤1 |ψ ( j) (x)y j |, ||ψ ( j) || = supx |ψ ( j) (x)|, and [ψ]k,t = k t j ( j) j=2 j! ||ψ || for k ≥ 2, [ψ]k,t = 0 for k < 2. Lemma A.2. For f, g of class C m , we have || f ◦ g|| ≤ || f ||,

||( f ◦ g) || ≤ || f  ||||g  ||

and if 2 ≤ k ≤ m, t > 0, [ f ◦ g]k,t ≤ [ f ]k,t||g || + || f  ||[g]k,t + [ f ]k−1,eTk−1 (g) where Tk−1 (g) = t||g  || + [g]k−1,t .

For 2 ≤ k ≤ m, from (A.1) [ f ◦ g]k,t =

k   tj  ( f ◦ g)( j)  ≤ [ f ]k,t||g || + || f  ||[g]k,t j! j=2 αi k−1  i   ej   t  (i)  ( j)    + f g α! i! 2≤|α|= j≤k−1 i=1 2≤|α  |≤k

 i ej ( j) and the final sum is k−1 ||( 1k−1 i!t ||g ( j) ||) j . We used the fact about j=2 j! || f symmetric j-linear maps, such as f ( j) (x), that ( j) f (x)h 1 h 2 . . . h j hi = ≤ e j f ( j) (x) ,

0, (|h 1 ||h 2 | . . . |h j |) proved in a stronger form (with j j /j! in place of e j ) by the polarization identity in, for example, H. Cartan, Differential Calculus. Now we generalize the fundamental lemma of Micheletti [22]. Lemma A.3. Let each φ j : Rn → Rn be C m -bounded and write ||φ||C m = 0m k!1 ||φ (k) ||, ||φ1(k) || + ||φ2(k) || + . . . + ||φ N(k) || ≤ σk (0 ≤ k ≤ m); then || (Id + φ N ) ◦ (Id + φ N −1 ) ◦ . . . ◦ (Id + φ1 ) − Id||C m m  σk Pm (eσ1 , σ2 , . . . , σm−1 ) ≤ k! k=0 for a certain polynomial Pm with positive coefficients, dependent only on m. Observe that the number of factors N is not mentioned in the estimate.

Appendix 2. On Micheletti’s Metric Space

191

Proof. Let 0 = Id,  j = (Id + φ j ) ◦ . . . ◦ (Id + φ1 ) = (Id + φ j ) ◦  j−1 for j ≥ 1, so || j − Id|| = || j−1 − Id + φ j ◦  j−1 || ≤ || j−1 − Id|| + ||φ j || ≤

j 

||φi || ≤ σ0 .

1

Also ||j || ≤ (1 + ||φ j ||)||j−1 || ≤

j 

(1 + ||φi ||) ≤ eσ1

1

||j − Id|| ≤ −1 +

j 

(1 + ||φ j ||) ≤ eσ1 − 1 ≤ σ1 eσ1 .

1

From m ≥ 2, t > 0 [ j ]m,t = [ j − Id]m,t ≤ [ j−1 ]m,t + [φ j ◦  j−1 ]m,t ≤ (1 + ||φ j ||)[ j−1 ]m,t + [φ j ]m,teσ1 + [φ j ]m−1,eTm−1 (t, j−1 ) so [k ]m,t ≤

k  k 

(1 + ||φi ||)([φ j ]m,teσ1 + [φ j ]m−1,eTm−1 ( j−1 )

j=1 j+1

Define the polynomials Q k : Rk−1 → R by: Q 1 ≡ 1, Q k (τ2 , . . . , τk ) = 1 +

k  τi 2

i!

+

k−1  i   e τi 2

i!

(Q k−1 (τ2 . . . τk−1 ))i .

Each Q k (0) = 1 and Q k is a polynomial with coefficients ≥ 0. By induction on m ≥ 2, we have with z = teσ1 , j ≥ 1. Tm−1 (t,  j−1 ) =

m−1  i i=1

 t  (i)  j−1 i!

≤ z Q m−1 (zeσ1 σ2 , z 2 eσ1 σ3 , . . . , z m−2 eσ1 σm−1 ). (For the induction, let τk = eσ1 (teσ1 )k−1 σk in the expression for Q m .) Then taking t = 1, k ≥ 0, k − Id C m ≤ σ0 + σ1 eσ1 + eσ1 (Q m (τ2 , . . . , τm ) − 1))

with τ j = e jσ1 σ j .

192

Appendix 2. On Micheletti’s Metric Space m τi Q m,i (τ2 , . . . , τm ) for polynomials Q m,i We have Q m (τ2 , . . . , τm ) − 1 = i=2 with positive coefficients and thus obtain the form claimed. Let C0m = { f : Rn → Rn of class C m with f (k) (x) → 0 as |x| → ∞, 0 ≤ k ≤ m} which is a Banach space in the norm f C m . Let Diff m (Rn ) be the translate of an open set in C0m , with the induced topology, 

Diff m (Rn ) = F : Rn → Rn diffeomorphism with F − Id ∈ C0m . (A.4) It is easy to show that F, G ∈ Diff m (Rn ) imply that F ◦ G and F −1 are in Diff m (Rn ). Further, since each F ∈ Diffm (Rn ) has x → D j F(x) (0 ≤ j ≤ m) uniformly continuous, (F, G) → F ◦ G and F → F −1 are continuous (in the induced topology): Diff m (Rn ) is a topological group. Define δm : Diff m (Rn ) → R by  N   δm (F) = inf f k − Id C m +  f k−1 − IdC m : N ≥ 1, k=1

f k ∈ Diff m and F = f N ◦ f N −1 ◦ . . . ◦ f 1

(A.5)

Lemma A.6. δm (F) = δm (F −1 ) ≥ 0, δm (F ◦ G) ≤ δm (F) + δm (G), and for F, G ∈ Diff m (Rn ) and there exist r1 > 0, r2 > 0, K > 1 so that F − Id C m ≤ r1 implies δm (F) ≤ r2 and δm (F) ≤ r2 implies K −1 ≤

δm (F) ≤ K. F − Id C m

In particular, δm (F) = 0 only when F = Id.

Proof. Let F ∈ Diff m (m ≥ 1) and define G = F −1 ; if y = F(x), G  (y) = F  (x)−1 and for 2 ≤ k ≤ m, the k th derivative of G(F(x) = x gives 0 = G (k) (y)h k + F −1 (x)F (k) (x)(F −1 (x)h)k αi k−1   k!  1 (i) G ( j) (y) F (x)(F −1 (x)h)i + α! i! 2≤ j=|α|≤k−1 i=1 |α  |=k

(This shows G − Id ∈ C0m so F −1 = G ∈ Diff m and also shows F → F −1 is continuous.) It follows, for an increasing polynomial Pm (a, b), that F −1 − Id C m ≤ F − Id C m · Pm ( (F  )−1 , F − Id C m )

Appendix 2. On Micheletti’s Metric Space

193

so δm (F) ≤ F − Id C m + F −1 − Id C m ≤ K F − Id C m for F − Id C m small (so, for example, |F  (x) − I | ≤ 1/2, |F −1 (x)| ≤ 2) with K > 1 + Pm (2, 0). N fk − If F = f N ◦ f N −1 ◦ . . . ◦ f 1 with all f j ∈ Dif m (Rn ) and S = k=1 Id C m we have, by Lemma A.3, F − Id C m ≤ S · Q m (S) for an increasing polynomial Q m . We may choose the N and f j so S ≤ 2δm (F) and then F − Id C m ≤ 2δm (F) · Q m (2δm (F)) to obtain the inequality K −1 ≤

δm (F) F−I d C m

≤ K for F near Id.

For F, G ∈ Diff m (Rn ) define dm (F, G) = δm (F ◦ G −1 )

(A.7)

with δm defined in (A.5) above. Lemma. dm is a distance (metric) on Diff m (Rn ), giving the induced topology from Id + C0m and in this metric, (Diff m (Rn ), dm ) is separable and complete. (This is also a topological group under composition (F, G) → F ◦ G.) Proof. From (A.6) it is clear we have a metric which is topologically equivalent to the (translated-norm) metric of C0m + Id. Since C0m is separable, so is Diff m (Rn ). Let {Fk }k≥1 be a Cauchy sequence in (Diff m (Rn ), dm ); it is enough to show some subsequence converges, or equally, to treat the case   dm (Fk+1 , Fk ) = δm Fk+1 ◦ Fk−1 ≤ r · 2−k (k = 1, 2, . . . ) for some fixed r > 0. We choose r = 12 min{r2 , 1/4k} where r2 , k are the constants of (A.6). Since δm (Fk ◦ F j−1 ) ≤ 2r · 2− j < r2 when k ≥ j ≥ 1, we have Fk ◦ FJ−1 − Id C m ≤ 2K r · 2− j . Now Fk+1 − Fk C m = Fk+1 ◦ Fk−1 − Id) ◦ (Fk ◦ F1−1 ) ◦ F1 C m ≤ O(2−k ) for all k ≥ 1, so {Fk − Id}k≥1 is a Cauchy sequence in C0m , with limit f = −1 − limk→∞ Fk − Id. Similarly, Fk−1 − Id C m is bounded and Fk ◦ Fk+1 −1 −k −k Id C m < K r · 2 so Fk ◦ F j − Id C m ≤ 2kr · 2 for j ≥ k ≥ 1     −1    F − F −1  0 =  F −1 ◦ Fk ◦ F −1 − F −1 ◦ Id 0 ≤ O(2−k ) k k k k+1 k+1 C C since m ≥ 1, so Fk−1 − Id → g uniformly (in C 0 ) and (Id + f ) ◦ (Id + g) = Id = (Id + g) ◦ (Id + f ). It follows that Id + f is a bijection as well as locally C m diffeomorphism (implicit function theorem) so Id + f ∈ Diff m (Rn ). Since Fk → Id + f as k → ∞, also Fk−1 → (Id + f )−1 in Diff m as k → ∞.

194

Appendix 2. On Micheletti’s Metric Space

Now define Mm (
) = {F(
) : F ∈ Diff m (Rn )} for any fixed open set
⊂ Rn . For
1 ,
2 in Mm (
), define dm (
1 ,
2 ) = inf{δm (F) : F ∈ Diff m (Rn ), F(
1 ) =
2 }  N   = inf f j − Id C m +  f j−1 − IdC m : N ≥ 1, j=1

(A.8) f j ∈ Diff m (Rn ) and f n ◦ . . . ◦ f 1 (
1 ) =
2 Theorem A.9. (Mm (
), dm ) is a complete separable metric space. Proof. Since Diff m (Rn ) is separable, so is Mm (
). Let {
j } j≥1 be a sequence in Mm (
) with dm (
j+1 ,
j ) < 2− j . There exists F j ∈ Diff m (Rn ) with F j (
j ) =
j+1 ; δm (F j ) < 2− j . Let G j = F j ◦ F j−1 ◦ . . . ◦ F1 so G j ∈ Diff m (Rn ), G j (
1 ) =
j and for k ≥ j,   ≤ δm (Fk ) + . . . + δm (F j+1 ) < 2− j . dm (G k , G j ) = δm G k ◦ G −1 j Since {G k }k≥1 is a Cauchy sequence in Diff m (Rn ), there exists G ∈ Diff m (Rn ) with dm (G j , G) → 0. Therefore G(
1 ) ∈ Mm (
) and dm (
j , G(
1 )) = dm (G j (
1 ), G(
1 )) ≤ δm (G j ◦ G −1 1 ) → 0 as j → +∞, or
j → G(
1 ) in Mm (
). Remark. Micheletti defines the closed subgroup Z m (
) ⊂ Diff m (Rn ) by Z m (
) = {F ∈ Diff m (Rm ) : F(
) =
} and obtains Mm (
) as the quotient space Diff m (
)/Z m (
) with the quotient norm dm (G 1 (
), G 2 (
)) = inf{dm (G 1 ◦ H1 , G 2 ◦ H2 ) : H j ∈ Z m (
)}. This is isometric with our definition. Theorem A.10. Suppose
is a bounded C m -regular region in Rm (1 ≤ m < ∞); then Mm (
) is a C 0 Banach manifold modeled on C m (∂
, R). (A coordinate system was mentioned in the introduction; here we give details). ˜ ∈ Proof. First we define local coordinates on a neighborhood of
. (If
˜ so coordinates are similarly defined near Mm (
), then Mm (
) = Mm (
); any point of Mm (
).) Let V : Rn → Rn be a C m vector field transverse to ∂
, i.e., V · N = 0 on ∂
where N is the outward unit normal. (V need only be defined near ∂
.) By the inverse function theorem, for small r > 0, (x, t) → x + t V (x) : ∂
× (−r, r ) → Rn

Appendix 2. On Micheletti’s Metric Space

195

is a C m imbedding of ∂
× (−r, r ) onto a neighborhood U of ∂
, and there is a C m inverse y → (π (y), τ (y)) : U → ∂
× (−r, r ), y = x + t V (x) if and only if x = π(y), t = τ (y) for y ∈ U, x ∈ ∂
, −r < t < r . ˜
) < ] we have ˜ in a small C 1 -neighborhood of
[d1 (
, For any
˜ ˜ ˜ ˜ ∂
⊂ U and V transverse to ∂
so π | ∂
: ∂
→ ∂
is a diffeomorphism and ˜ = {π(y) + τ (y)V (π(y)) : y ∈ ∂
} ˜ = {x + σ (x,
)V ˜ (x) : x ∈ ∂
} where ∂
−1 ˜ → σ (x,
) ˜ later. ˜ = τ ◦ (π|∂
) ˜ . We prove the continuity of
σ (· ,
) Choose C ∞ θ : Rn → [0, 1] with θ ≡ 1 near ∂
, θ ≡ 0 outside U and define H (σ ) : Rn → Rn by x + θ(x)σ (π(x))V (x), x ∈ U H (σ )(x) = x, x ∈ U for σ ∈ C m (∂
, R). H (0) = Id and H (σ ) ∈ Diff m (Rn ) for small σ C m , with σ → H (σ ) : C m (∂
, R) [near zero] → Diff m (Rn ) continuous. Since (σ, x) → H (σ )(x) is C m , by smoothness of the evaluation map. We define
(σ ) = H (σ )(
) ∈ Mm (
), and σ →
(σ ) ∈ Mm (
) is continuous; we show it is a homeomorphism of C m (∂
, R) [near zero] onto a neighborhood of
. ˜ → σ (· ,
) ˜ mentioned above: The inverse map is
˜ =
˜ and σ (· ,
(τ )) = τ.
(σ (· ,
)) ˜
). ˜ near
,
˜ = F(
) for F ∈ Diff m (Rn ) near Id, say δn (F) ≤ 2dm (
, For
m Now π ◦ F|∂
: ∂
→ ∂
is a C diffeomorphism, depending continuously on F (near the identity) and σ (· , F(
)) = τ ◦ F ◦ (π ◦ F | ∂
)−1 ∈ C m (∂
, R) ˜ in Mm (
), we may choose ˜ν →
also depends continuously on F. If
m n m n ˜ ν = Fν (
),
˜ = F(
), so Fν ∈ Diff (R ) with Fν → F in Diff (R ),
m ˜ in C (∂
, R). ˜ n ) → σ (· ,
) σ (· ,
We note, for later use, that h → h(
) [h ∈ C m (
, Rn ) near the inclusion i
:
⊂ Rn ] has a continuous local right-inverse p. In fact, let p(
(σ )) = H (σ )|
; then p(
(σ ))(
) = H (σ )(
) =
(σ ) for small σ ∈ C m (∂
, R). Such a local right-inverse exists near each point of Mm (
), and shows h → h(
) is an open map (restricted to C m imbeddings h). Diff m (
) = {h :
→ Rn C m

imbedding with h(
) ∈ Mm (
)}

= {H|
: H ∈ Diff (Rn )}, m

(A.11)

196

Appendix 2. On Micheletti’s Metric Space

which is an open set in C m (
, Rn ) if
is bounded, with the induced metric and Banach manifold structures. As noted above h → h(
) : Diff m (
) → Mm (
) is continuous, surjective and an open map with a local right-inverse near each point provided
is bounded and C m -regular. Theorem A.12. Let
⊂ Rn be a bounded open set with ∂
C m-regular, and let F ⊂ Diff m (
) be a set defined by its image F(
) = {h(
) : h ∈ F}, i.e., F = {h ∈ Diff m (
) : h(
) ∈ F(
)}. Then F is closed [or σ -closed, or meager] in Diff m (
) if and only if F(
) is closed [or σ -closed, or meager] in Mm (
). More precisely, if F and F(
) are σ -closed, codim F (in Diff m (
)) = codim F(
) (in Mm (
)). Remark. “Codimension” is defined in 5.15, and positive codimension is equivalent to meagerness. Proof. According to 5.16 (3), codim{F1 ∪ F2 ∪ . . . } = min{codimF j : j ≥ 1} so it suffices to treat closed sets F. For the same reason, we may suppose ˜ ∈ Mm (
), there exists r > 0 and F(
) has small diameter. In fact, for each
m ˜ → Diff (
)[ p(
 )(
) =
 for dm (
 ,
) ˜ < a local right-inverse p : Br (
) ˜ :
˜ ∈ Mm (
)} cover Mm (
) which is separable, so a r ]. The sets {Br/2 (
) ˜ We countable subcover suffices, and it is enough to treat each F(
) ∩ B¯ r/2 (
). ¯ ˜ thus suppose F is closed, F(
) ⊂ Br/2 (
) and there is a right-inverse p defined ˜ (Since h → h(
) is continuous and open, F is closed if and only if on Br (
). F(
) is closed.) Suppose codim F > j, so on an open dense set of V ∈ C(I j , Diff m (
)) we have γ (I j ) ∩ F = ∅. Given continuous ψ : I j → Mm (
) we show – by arbitrarily small changes in ψ – we can achieve ψ(I j ) ∩ F(
) = ∅. This will prove codim F(
) ≥ codim F. ˜ this is easy: p ◦ ψ ∈ C(I j , Diff m (
)) and there exists If ψ(I j ) ⊂ Br (
), γ near p ◦ ψ with γ (I j ) ∩ F = ∅, so γ (t)(
) is near p(ψ(t))(
) = ψ(t) for t ∈ I j and γ (I j )(
) ∩ F(
) = ∅. In general, divide I j in small congruent cubes {Ci } with disjoint interiors, with diameter so small that diam ψ(Ci ) < r/2. For each i, either ψ(Ci ) is out˜ (hence, disjoint from F(
)), or inside Br (
) ˜ when the previous side B¯ r/2 (
) argument applies. In modifying ψ | Ci to avoid F(
), we make sufficiently

Appendix 2. On Micheletti’s Metric Space

197

small changes so we don’t upset previous changes in other sub-cubes – which is possible since F(
) is closed. Thus codim F(
) ≥ codim F in every case. If codim F = ∞, codim F(
) = ∞. If codim F = 0, then F has interior so also F(
) has interior, codim F(
) = 0. Suppose k is a positive integer and codim F = k; we show codim F(
) ≤ k, i.e., there exists ψ0 ∈ C(I k , F(
)) so that for any ψ near ψ0 , ψ(t) ∈ F(
) for some t ∈ I k . We know codim F = k, so there exists γ0 ∈ C(I k , F) so that any nearby γ has γ (I k ) ∩ F = ∅, and we may suppose γ0 (I k ) ∩ F = ∅ [dim ∂ I k < k]. ˜ this is again simple. Define γ˜0 (t) = p(γ0 (t)(
)), If γ0 (I k )(
) ⊂ B2r/3 (
), k t ∈ I ; we have γ˜0 (t)(
) = γ0 (t)(
) so there is a (unique) map H (t) :
→
with γ˜0 (t) ◦ H (t) = γ0 (t), and (by the implicit function theorem) it is a C m diffeomorphism with t → H (t) ∈ C m continuous. Then γ near γ˜0 in C(I k , Diff m (
)) implies γ ◦ H is near γ˜0 ◦ H = γ0 so γ (t) ◦ H (t) ∈ F and γ (t) ∈ F for some t ∈ I k . Choose ψ ∈ C(I k , Mm (
)) near ψ0 = ˜ so p ◦ ψ(t) is well-defined and near {t → γ˜0 (t)(
)}; then ψ(I k ) ⊂ Br (
) k p(γ˜0 (t)(
)) = γ˜0 (t) for all t ∈ I , so p ◦ ψ(t) ∈ F for some t, hence ψ(t) ∈ F(
). In general, we may choose a small k-dimensional cube C ⊂ I k so that ˜ and any γ | C near γ0 | C has γ (C) ∩ F = ∅, so the γ0 (C)(
) ⊂ B2r/3 (
), above argument applies and the proof is complete.

Estimates of

∞

n=1 (1

+ z/2n ) and Extension Operator

Working with the extension operator E (for a half-space) of Thm. 1.9, we   sometimes want estimates of k≥1 |ak ||bk | p or k≥1 |ak |2kp (bk = 1 − 2k ) for  ∞ κ large p. Let f (z) = 1 (1 + z/2n ) so A(z) = z f (−z)/ f (−i) = ∞ 1 ak z and ∞ max|z|=r |A(z)| = r f (r )/ f (−1) = 1 |ak |r k . (1) f (z) = +



zk 1≤n 1