Asymptotic analysis of solutions to elliptic and parabolic problems

Linköping Studies in Science and Technology. Dissertations No. 1044 Asymptotic analysis of solutions to elliptic and p...

Author: Peter Rand.

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Linköping Studies in Science and Technology. Dissertations No. 1044

Asymptotic analysis of solutions to elliptic and parabolic problems Peter Rand

Matematiska institutionen Linköpings universitet, SE-581 83 Linköping, Sweden Linköping 2006

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ii

Asymptotic analysis of solutions to elliptic and parabolic problems c 2006 Peter Rand ° Matematiska institutionen Linköpings universitet SE-581 83 Linköping, Sweden [email protected] ISBN 91-85523-04-6 ISSN 0345-7524 Printed by UniTryck, Linköping 2006

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iii

Abstract In the thesis we consider two types of problems. In Paper 1, we study small solutions to a time-independent nonlinear elliptic partial differential equation of Emden-Fowler type in a semi-infinite cylinder. The asymptotic behaviour of these solutions at infinity is determined. First, the equation under the Neumann boundary condition is studied. We show that any solution small enough either vanishes at infinity or tends to a nonzero periodic solution to a nonlinear ordinary differential equation. Thereafter, the same equation under the Dirichlet boundary condition is studied, the nonlinear term and right-hand side now being slightly more general than in the Neumann problem. Here, an estimate of the solution in terms of the righthand side of the equation is given. If the equation is homogeneous, then every solution small enough tends to zero. Moreover, if the cross-section is star-shaped and the nonlinear term in the equation is subject to some additional constraints, then every bounded solution to the homogeneous Dirichlet problem vanishes at infinity. In Paper 2, we study asymptotics as t → ∞ of solutions to a linear, parabolic system of equations with time-dependent coefficients in Ω × (0, ∞), where Ω is a bounded domain. On ∂Ω × (0, ∞) we prescribe the homogeneous Dirichlet boundary condition. For large values of t, the coefficients in the elliptic part are close to time-independent coefficients in an integral sense which is described by a certain function κ(t). This includes in particular situations when the coefficients may take different values on different parts of Ω and the boundaries between them can move with t but stabilize as t → ∞. The main result is an asymptotic representation of solutions for large t. As a corollary, it is proved that if κ ∈ L1 (0, ∞), then the solution behaves asymptotically as the solution to a parabolic system with time-independent coefficients.

Acknowledgements I would like to thank my supervisors Vladimir Kozlov and Mikael Langer for all their support and invaluable hints during this work and Anders Björn and Jonna Gill for helping me with LATEX. Thanks also to everyone else who has helped me in some way.

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iv

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v

Contents Introduction

1

References

2

Paper 1: Asymptotic analysis of a nonlinear partial differential equation in a semicylinder 7 1 Introduction

7

2 The 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

Neumann problem Notation . . . . . . . . . . . . . . . . . . . . . . . Problem formulation and assumptions . . . . . . The main asymptotic result . . . . . . . . . . . . Corollaries of Theorem 2.1 . . . . . . . . . . . . . The corresponding problem in C . . . . . . . . . An auxiliary ordinary differential equation . . . . The equation for v . . . . . . . . . . . . . . . . . Asymptotics of small solutions of problem (2.10) End of the proof of Theorem 2.1 . . . . . . . . .

3 The 3.1 3.2 3.3 3.4 3.5 3.6

Dirichlet problem Problem formulation and assumptions . . . . . . The main asymptotic result . . . . . . . . . . . . The corresponding problem in C . . . . . . . . . End of the proof of Theorem 3.1 . . . . . . . . . The case of a star-shaped cross-section . . . . . . An estimate for solutions of a nonlinear ordinary ential equation . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . . . . . . . . . . . . . differ. . . .

A Some results from functional analysis A.1 Eigenvalues and eigenvectors of −∆ . . . . . . . . . . . A.2 Existence and uniqueness of bounded solutions of Poisson’s equation in C . . . . . . . . . . . . . . . . . . . . . A.3 A local estimate for solutions of Poisson’s equation . . .

10 10 11 11 12 14 16 24 28 31 32 32 33 33 34 35 38 40 40 43 50

Paper 2: Asymptotic analysis of solutions to parabolic systems 57 1 Introduction

57

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vi 2 Problem formulation and elementary properties 2.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Spaces not involving time . . . . . . . . . . 2.1.2 Spaces involving time . . . . . . . . . . . . 2.2 Problem formulation and assumptions . . . . . . . 2.3 An estimate for u . . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

62 62 63 64 65 68

3 Spectral splitting of the solution u

69

4 Estimating the function v 4.1 A general estimate . . . . . . . . . . . . . . . . . . . . . 4.2 Estimate for v . . . . . . . . . . . . . . . . . . . . . . . .

71 71 82

5 Norm estimates for Rkl and gk (w)

84

6 Functions hJ+1 , . . . , hM 6.1 Definition of functions v0 , v1 and v2 . . . . 6.2 Integro-differential system for hJ+1 , . . . , hM 6.3 A general estimate . . . . . . . . . . . . . . 6.4 A particular case of equation (6.8) . . . . . ˇ . . . . . . . . . . . . . . . . 6.5 Estimate for h ˇ . . . . . . . . . . . . 6.6 A representation for h

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

86 86 87 87 96 99 99

7 Functions h1 , . . . , hJ 100 ˆ existence and uniqueness results . . . . 100 7.1 Equation for h; 7.2 The homogeneous equation . . . . . . . . . . . . . . . . 104 7.3 A particular solution of (7.15) . . . . . . . . . . . . . . . 116 8 Proof of Theorem 1.1

118

9 Corollaries of Theorem 1.1

120

A Eigenfunctions of a time-independent operator

124

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1

Introduction Most differential equations and systems are not possible to solve exactly. Hence, it is important to develop other methods of analyzing the properties of solutions. One of these methods is based on asymptotic analysis. Although the solutions are unknown, it may be possible to find information about their behaviour as some variables tend to some finite value or to infinity. Asymptotic analysis is used to study time-dependent evolution problems as well as time-independent stationary problems. Frequently, one is interested in behaviour of solutions as time tends to infinity, for example in questions concerning stability, periodicity, rate of growth etc. A survey of evolution problems and a general theory of analyzing them, including asymptotic analysis, can be found in Dautray, Lions [1], [2] or Lions, Magenes [8], [9], [10]. An important class of evolution problems are reactiondiffusion problems. Such occur frequently in biology and chemistry, see for example Fife [3] or Murray [11]. Important contributions to asymptotic methods for evolution problems can be found in Friedman [4], Pazy [12] and Vishik [13]. In this thesis, we use an approach developed in Kozlov, Maz’ya, [6], [7]. Starting with a linear or nonlinear equation or system of equations, the problem is reduced to first order ordinary differential equations with operator coefficients. Then, by use of a spectral splitting, a finite dimensional system of first order ordinary differential equations perturbed by a small integro-differential term is obtained for the leading term. The main difficulty is to perform the above reduction and the study of the system of ordinary differential equations for the leading term. The main result is that the asymptotic behaviour of solutions of the initial system of equations is described by solutions of the above finite dimensional system. We use and extend this approach. In paper 1 we consider a nonlinear Emden-Fowler type time-independent partial differential equation in a semi-infinite cylinder and study the asymptotics of solutions when the

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2 unbounded coordinate tends to infinity. The equation is complemented by the Neumann or Dirichlet boundary condition. We analyze the asymptotic behaviour of a given, small solution of the problem. In the Neumann case, we obtain a nonlinear ordinary differential equation for the leading term in the asymptotics of solution. We also find an estimate for the remainder term. From this asymptotic formula it follows that the solution behaves asymptotically like a periodic solution. In the Dirichlet case we show that small solutions decrease exponentially. We also consider the case of a starshaped cross section and show that if the nonlinear term in the equation is subject to some additional constraints, then every bounded solution of the homogeneous Dirichlet problem vanishes at infinity. The use of Pohoˇzaev’s identity is essential in the proof. Paper 2 is devoted to the study of a linear parabolic system of equations in a bounded domain under Dirichlet boundary conditions and with prescribed initial values. We consider the asymptotic behaviour of solutions as time tends to infinity. The elliptic part of the system is here considered as a perturbation of time-independent coefficients. We consider a larger class of perturbations than Kozlov, Maz’ya [6]. Smallness of the perturbations is assumed only in integral sense. In particular, we include such situations when the leading coefficients may take different values on different parts of Ω and the boundaries between them can move with t but stabilize as t → ∞. Here we use another reduction than Kozlov, Maz’ya [6] to obtain the first order system of ordinary differential equations perturbed by an integro-differential term for the leading terms. Then an approach from Kozlov [5] is used to study the asymptotic behaviour of solutions to this system.

References [1] R. Dautray, J-L Lions, Mathematical Analysis and Numerical Methods for Science and Technology. Volume 5. Springer-Verlag, 1992. [2] R. Dautray, J-L Lions, Mathematical Analysis and Numerical Methods for Science and Technology. Volume 6. Springer-Verlag, 1993. [3] P. C. Fife, Mathematical Aspects of Reacting and Diffusing Systems. Lecture Notes in Biomathematics, 28. Springer-Verlag, 1979. [4] A. Friedman, Partial differential equations of parabolic type. Prentice-Hall, Inc., Englewood Cliffs, N.J., 1964. [5] V. Kozlov, Asymptotic representation of solutions to the Dirichlet problem for elliptic systems with discontinuous coefficients near the boundary. Electron. J. Differential Equations 10 (2006), 46 pp. [6] V. Kozlov, V. Maz’ya, Differential Equations with Operator Coefficients. Springer-Verlag, 1999.

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3 [7] V. Kozlov, V. Maz’ya, An asymptotic theory of higher-order operator differential equations with nonsmooth nonlinearities. Journal of Functional Analysis 217 (2004), 448–488. [8] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Problems and Applications. Volume I. Springer-Verlag, 1972. [9] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Problems and Applications. Volume II. Springer-Verlag, 1972. [10] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Problems and Applications. Volume III. Springer-Verlag, 1973. [11] J. D. Murray, Mathematical Biology. Springer-Verlag, 1993. [12] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations. Springer-Verlag, 1983. [13] M. I. Vishik, Asymptotic behaviour of solutions of evolutionary equations. Cambridge University Press, 1992.

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4

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5

Paper 1

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6

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Asymptotic analysis of a nonlinear partial differential equation in a semicylinder Peter Rand Abstract We study small solutions of a nonlinear partial differential equation in a semi-infinite cylinder. The asymptotic behaviour of these solutions at infinity is determined. First, the equation under the Neumann boundary condition is studied. We show that any solution small enough either vanishes at infinity or tends to a nonzero periodic solution of a nonlinear ordinary differential equation. Thereafter, the same equation under the Dirichlet boundary condition is studied, but now the nonlinear term and right-hand side are slightly more general than in the Neumann problem. Here, an estimate of the solution in terms of the right-hand side of the equation is given. If the equation is homogeneous, then every solution small enough tends to zero. Moreover, if the cross-section is star-shaped and the nonlinear term in the equation is subject to some additional constraints, then every bounded solution of the homogeneous Dirichlet problem vanishes at infinity. An estimate for the solution is given.

1

Introduction

Let Ω be a bounded domain in Rn−1 with C 2 -boundary. We define the semi-infinite cylinder C+ = {x = (x0 , xn ) : x0 ∈ Ω, xn > 0}. In Section 2 we study bounded solutions of the equation ∆U + q(U )U = H

in C+

(1.1)

on ∂Ω × (0, ∞).

(1.2)

under the boundary condition ∂U =0 ∂ν

Our aim is to describe the asymptotic behaviour as xn → ∞ of solutions U of problem (1.1), (1.2) subject to |U (x)| ≤ Λ

for x ∈ C+ ,

(1.3)

where Λ is a positive constant. 7

8 We assume that q(u) > 0 if u 6= 0. Moreover, q is continuous and |s|, |t| ≤ Λ ⇒ |q(s)s − q(t)t| ≤ CΛ |s − t|

(1.4)

with CΛ < λ1 . Here, λ1 is the first positive eigenvalue of the Neumann problem for the operator −∆0 = −

n−1 X k=1

∂2 ∂x2k

in Ω. We set Ct = Ω × (t, t + 1) and define Lrloc (C+ ), 1 ≤ r ≤ ∞, as the space of functions which belong to Lr (Ct ) for every t ≥ 0. We also suppose H ∈ Lploc (C+ ) and Z ∞ (1 + s)kHkLp (Cs ) ds < ∞, (1.5) 0

where

(

p > n/2 p=2

if n ≥ 4 if n = 2, 3.

(1.6)

The main result of Section 2 is Theorem 2.1, which states that one of two alternatives is valid: 1. U admits the asymptotic representation U (x) = uh (xn ) + w(x)

as xn → +∞,

where uh is a nonzero periodic solution of u00h + q(uh )uh = 0 and w → 0 as xn → ∞. An estimate for the remainder term w is given. 2. U → 0 as xn → ∞. An estimate for U is given in the theorem. If, for example, H = 0 and CΛ → 0 as Λ → 0, then Corollary 2.3 gives the following estimate for U in the second case: |U (x0 , xn )| ≤ C² e−

√ λ1 −² xn

,

where ² is an arbitrary small positive number and C² is a constant depending on ². In Sections 3.1-3.4 we study solutions U of (1.1) subject to (1.3) under the Dirichlet boundary condition U =0

on ∂Ω × (0, ∞).

September 6, 2006 (12:43)

(1.7)

9 Now we suppose that q is continuous and that |q(v)| ≤ CΛ

if |v| ≤ Λ,

where CΛ < λD . Here, λD is the first eigenvalue of the Dirichlet problem for −∆0 in Ω. We assume also that kHkLp (Ct ) , with p as in (1.6), is a bounded function of t, t ≥ 0. The main result is Theorem 3.1 which gives an explicit bound for kU kL∞ (Ct ) in terms of the function kHkLp (Ct ) . This implies in particular that kU kL∞ (Ct ) → 0 as t → ∞ if the same is valid for kHkLp (Ct ) . If H = 0 and q(0)=0, then the estimate from Theorem 3.1 implies that √ |U (x0 , xn )| ≤ C² e− λD −² xn , (1.8) where ² > 0 is arbitrary. In Section 3.5 we study all bounded solutions of (1.1), (1.7). Here we suppose additionally that n ≥ 4, that the domain Ω is star-shaped with respect to the origin, that the function q is continuous with q(0) = 0 and that q(u) is positive for u 6= 0. We also assume that Z u n−3 q(u)u2 − (n − 1) q(v)v dv ≥ ² q(u)u2 (1.9) 2 0 for some ² > 0. Then Theorem 3.4 states that every bounded solution of (1.1), (1.7) with H = 0 satisfies (1.8). Some examples of functions satisfying (1.9) are • q(u) = |u|p ,

p>

• q(u) = |u|p e|u| ,

4 n−3 ,

p>

• q(u) = |u|p (e|u| − 1),

4 n−3 ,

p>

7−n n−3 ,

• linear combinations with positive coefficients of the functions above. A natural question: under which conditions on q is it possible to remove ² in the relation (1.8)? The aim of Section 3.6 is to study a similar question for the ordinary differential equation u00 − λu + q(u)u = 0,

(1.10)

where λ > 0, q is continuous with q(0) = 0 but q(u) > 0 for u 6= 0 and Z

1

−1

q(u) du < ∞. |u|

Theorem 3.5 states that every solution u of (1.10) subject to u(t) → 0 as t → ∞ satisfies ³ √ ´ |u(t)| + |u0 (t)| = O e− λt

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10 for large positive t. The problem (1.1) under the boundary conditions (1.2) or (1.7) with q(U ) = |U |p−1 , p > 1 has been studied in Kozlov [14]. There it is shown that the restriction (1.3) is essential for Theorems 2.1 and 3.1. One of the goals of this thesis is to extend some results from [14] to the equation (1.1). The equation ∆u − a|u|q−1 u = 0 in C+ , (1.11) where q > 1, a > 0 and with the boundary condition (1.2) is considered in Kondratiev [11]. Furthermore, the problem  Lu = 0 in C+  ∂u  + a|u|q−1 u = 0 on ∂Ω × (0, ∞), ∂ν where L is an elliptic partial differential operator, a > 0 and q > 1 are constants is studied in Kondratiev [12]. In both these cases it is proved that the solutions of these problems have asymptotics of the form u(x0 , xn ) = Cx−σ n with σ > 0. This shows that the minus sign in (1.11) essentially changes the asymptotic behaviour of solutions at infinity. There is a lot of research on positive solutions of nonlinear problems in an infinite cylinder and other unbounded domains. We direct the reader to Bandle and Essén [3], Berestycki [4], Berestycki, Caffarelli and Nirenberg [5], Berestycki, Larrouturou and Roquejoffre [6], Berestycki and Nirenberg [7] and Kondratiev [13] where also further references can be found. Small global solutions of the equation ∆u + λu + f (u, ux , uy ) = 0 in a two-dimensional strip with homogeneous Dirichlet boundary conditions are studied in Amick, Toland [2] and Kirchgässner, Scheurle [10].

2 2.1

The Neumann problem Notation

The Laplace operator and the gradient in Rn are denoted by ∆ and ∇, respectively. For the corresponding operators in Rn−1 we introduce ∆0 =

n−1 X k=1

and

µ 0

∇ =

∂2 ∂x2k

∂ ∂ ,..., ∂x1 ∂xn−1

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¶ .

11 By Ω we denote a bounded domain in Rn−1 with C 2 -boundary and (n−1)dimensional Lebesgue measure |Ω|. We introduce the cylinder C = {(x0 , xn ) : x0 ∈ Ω and xn ∈ R} and the semicylinder C+ = {(x0 , xn ) : x0 ∈ Ω and xn > 0}. We let ν and ν 0 denote the outward unit normals to ∂C and ∂Ω, respectively. Thus ν ∈ Rn while ν 0 ∈ Rn−1 and ν = (ν 0 , 0). After introducing Ct = Ω × (t, t + 1), we say that a function u : Rn → R belongs to Lrloc (C) k,r or Wloc (C), 1 ≤ r ≤ ∞ and k = 0, 1, . . ., if it belongs to Lr (Ct ) or W k,r (Ct ) for every t ∈ R.

2.2

Problem formulation and assumptions

Assume that p is subject to (1.6). We study the asymptotic behaviour as 2,p xn → ∞ of solutions U ∈ Wloc (C+ ) of the problem   ∆U + q(U )U = H ∂U  =0 ∂ν

in C+ on ∂Ω × (0, ∞)

(2.1)

satisfying (1.3). We assume that q is continuous and positive for u 6= 0 and satisfies (1.4). We suppose further that H ∈ Lploc (C+ ) is subject to (1.5) In order to motivate (1.6), let us consider a bounded solution 1,2 U ∈ Wloc (C+ ) of (2.1). By Lemma A.16 in Section A.3 we get that 2,p U ∈ Wloc (C+ ). Furthermore, it follows from well-known Sobolev inequalities, see for example Theorem 5.6 in Evans [8], that, since p > n/2, there exists a positive γ such that either U ∈ C 0,γ (Ct ) or U ∈ C 1,γ (Ct ) for every t > 0. Hence it is meaningful to assume that the studied solution belongs 2,p to Wloc (C+ ) and is bounded.

2.3

The main asymptotic result

The aim of Section 2 is to prove the following theorem concerning the asymptotic behaviour of solutions of (2.1) subject to (1.3): 2,p Theorem 2.1 Suppose that U ∈ Wloc (C+ ), where p satisfies (1.6), is a solution of (2.1) subject to (1.3). Suppose also that q is continuous, q(u) > 0 if u 6= 0 and that the Lipschitz condition (1.4) is fulfilled. Finally, assume that H ∈ Lploc (C+ ) satisfies (1.5). Then one of the following alternatives is valid:

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12 1. U (x) = uh (xn ) + w(x), where uh is a nonzero periodic solution of u00h + q(uh )uh = 0 and

µZ kwkL∞ (Ct ) ≤ C Z

∞ t

t

+t

e

skHkLp (Cs ) ds

√ − λ1 −CΛ (t−s)

0

√ − λ1 −CΛ t

¶

kHkLp (Cs ) ds + te

for t ≥ 1. The right-hand side tends to 0 as t → ∞. 2. kU (·, xn )kL∞ (Ω) → 0 as xn → ∞. Furthermore, U (x) = u0 (xn ) + w(x), where µZ ∞ Z u0 (t) (u00 (t))2 + q(v)v dv ≤ C kHkLp (Cs ) ds (2.2) 2 0 t ¶ Z t √ √ + e− λ1 −CΛ (t−s) kHkLp (Cs ) ds + e− λ1 −CΛ t 0

and kwkL∞ (Ct ) ≤ C

µZ

∞

√

e−

λ1 −CΛ |t−s|

0

kHkLp (Cs ) ds + e−

√ λ1 −CΛ t

¶

(2.3)

for t ≥ 1. The proof of this theorem is contained in Sections 2.5 - 2.9. Theorem 2.1 is a generalization of Theorem 3 in Kozlov [14], where the case q(U ) = |U |p−1 , p > 1 is studied. We use the same approach in this thesis. Since most of the proofs in [14] are brief or absent, we present here complete proofs of all assertions. Our restriction (1.5) is different from the corresponding restriction in Theorem 3 [14]. This rigorous analysis of the proofs indicates that possibly (1.5) is the right assumption also in [14]. In the next section we give some corollaries of Theorem 2.1.

2.4

Corollaries of Theorem 2.1

Corollary 2.2 Suppose, in addition to the conditions in Theorem 2.1, that the nonlinear term q has the property that the constant CΛ in (1.4) tends to 0 as Λ tends to 0. Then the estimates (2.2) and (2.3) √ can be improved, √ namely, the constant λ1 − CΛ can be replaced by λ1 − ² where ² > 0 is arbitrary. In this case, the constant C appearing in (2.2) and (2.3) is dependent of ². Proof. Since kU (·, xn )kL∞ (Ω) → 0 as xn → ∞ we can apply the theorem for the semicylinder Ω × (T, ∞) where T is sufficiently large and Λ small enough.

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13 Corollary 2.3 Suppose, in addition to the conditions in Corollary 2.2, that H = 0. If case 2 in Theorem 2.1 occurs, then, for every ² ∈ (0, λ1 ), there exists a constant C² such that √

|U (x0 , xn )| ≤ C² e−

λ1 −² xn

for xn ≥ 1.

(2.4)

Proof. We begin with proving that there exists a constant A² such that 1

|U (x0 , xn )| ≤ A² e− 2

√

λ1 −² xn

for xn ≥ 1.

(2.5)

From (2.2) it follows that 1

|u00 (t)| ≤ Ce− 2

√

λ1 −CΛ t

and u0 (t) → 0 Since

as t → ∞. Z

∞

u00 (s) ds

u0 (t) = − t

we get 1

|u0 (t)| ≤ Ce− 2

√ λ1 −CΛ t

.

(2.6)

Furthermore, (2.3) gives √

kwkL∞ (Ct ) ≤ Ce−

λ1 −CΛ t

,

(2.7)

where C does not depend on t. Since U (x) = u0 (xn ) + w(x) we get from (2.6) and (2.7) that 1

|U (x0 , xn )| ≤ Ce− 2

√ λ1 −CΛ xn

.

(2.8)

Using that CΛ → 0 as Λ → 0 and considering problem (2.1) in a semicylinder Ω × (t0 , ∞), where t0 is sufficiently large, we can suppose that CΛ < ². Then the estimate (2.5) follows from (2.8). We set Ct0 = Ω × (t + 1/4, t + 3/4). Lemma A.16 in Section A.3 implies that kU kW 2,2 (Ct0 ) ≤ C(kU kL2 (Ct ) + kq(U )U kL2 (Ct ) ) so from (1.4) it follows that kU kW 2,2 (Ct0 ) ≤ CkU kL2 (Ct )

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14 and by using (2.5) we get the estimate 1

kU kW 2,2 (Ct0 ) ≤ Ce− 2

√ λ1 −² t

.

Corollary √ 6.2.5 in Kozlov and Maz’ya [15], with the parameters k− = 0, k+ = λ1 , m− = 2 and m+ = 1 together with the fact that R can be made suitably small, implies that kU kW 2,2 (Ct0 ) ≤ Ce−

√ λ1 −² t

.

(2.9)

We can now use local estimates and an iteration procedure as in the proof of Lemma A.16 in Section A.3 to obtain kU kL∞ (Ct00 ) ≤ CkU kW 2,2 (Ct0 ) , where Ct00 = Ω × (t + 3/8, t + 5/8). Combination of the last estimate and (2.9) gives kU kL∞ (Ct ) ≤ Ce−

√ λ1 −² t

if t ≥ 3/8. This implies (2.4).

2.5

The corresponding problem in C

We now begin proving Theorem 2.1. Before turning to equation (2.1), we 2,p study a solution u ∈ Wloc (C) of the problem   ∆u + q(u)u = h ∂u  =0 ∂ν

in C on ∂C

(2.10)

satisfying sup |u(x)| ≤ Λ,

(2.11)

x∈C

where Λ is the same constant as in (1.3) and p is subject to (1.6). By the Sobolev embedding theorem the solution is continuous. Thus we do not need to use essential supremum. As before, we assume that q is continuous and positive for u 6= 0 and satisfies (1.4). We also suppose that h ∈ Lploc (C) and that Z

∞ −∞

(1 + |s|)khkLp (Cs ) ds < ∞.

(2.12)

Theorem A.8 in Section A.1 states that there exists an ON-basis of L2 (Ω) consisting of eigenfunctions of the operator −∆0 for the Neumann problem in Ω. Let φ0 (·, xn ) denote the eigenfunction with L2 (Ω)-norm equal to 1 corresponding to the eigenvalue λ0 = 0, i.e. φ0 = |Ω|−1/2 . Set u

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15 to the orthogonal projection of u onto the subspace of L2 (Ω) spanned by φ0 , that is Z 1 u(xn ) = u(x0 , xn ) dx0 , |Ω| Ω and define v(x) by the equality u(x) = u(xn ) + v(x).

(2.13)

Inserting (2.13) in (2.10) and integrating over Ω, we obtain Z Z 1 1 ∆v(x0 , xn ) dx0 + f (u(x0 , xn )) dx0 = h(xn ), (2.14) u00 (xn ) + |Ω| Ω |Ω| Ω where h(xn ) =

1 |Ω|

Z h(x0 , xn ) dx0 Ω

and f (t) = q(t)t.

(2.15)

Due to the homogeneous boundary condition in (2.10), Greens formula gives Z ∆0 u dx0 = 0. Ω

Therefore 1 |Ω|

Z

µ ¶ Z Z 1 d2 1 ∆0 v dx0 + 2 v dx0 |Ω| Ω dxn |Ω| Ω Z 1 = ∆0 u dx0 = 0. |Ω| Ω

∆v dx0 = Ω

Equation (2.14) can now be written as Z 1 u00 (xn ) + f (u(x0 , xn )) dx0 = h(xn ) |Ω| Ω

(2.16)

and by defining

we get

K(u, v)(xn ) = f (u(xn )) − f (u + v)(xn )

(2.17)

u00 (xn ) + f (u(xn )) = h(xn ) + K(u, v)(xn ).

(2.18)

We will often write K(xn ) or K instead of K(u, v)(xn ). Using (2.10), the equation ∆v = h − f (u + v) − u00 is obtained which, together with (2.16), implies that   ∆v = −M (u, v) + h1 in C (2.19)  ∂v = 0 on ∂C, ∂ν

September 6, 2006 (12:43)

16 where M (u, v)(x) = f (u(xn ) + v(x)) − f (u + v)(xn )

(2.20)

h1 = h − h.

(2.21)

and The equations (2.18) and (2.19) will play a central role in the sequel.

2.6

An auxiliary ordinary differential equation

In this section we study the equation ξ 00 (t) + q(ξ(t))ξ(t) = g(t),

t ≥ t0 ,

(2.22)

where t0 ≥ 1 is given. We assume that q(v) is positive for every v ∈ R, possibly except for v = 0, and continuous. We also assume that q(v)v is Lipschitz continuous on every finite interval and that Z ∞ |sg(s)| ds < ∞. (2.23) t0

Since solutions of (2.22) with g = 0 will play an important role in the asymptotic representation of ξ as t → ∞, we will now describe them. We have thus the equation ξ 00 (t) + q(ξ(t))ξ(t) = 0,

t ≥ t0 .

(2.24)

Multiplying (2.24) by ξ 0 and integrating, we obtain Z t 1 0 1 2 (ξ (t)) + q(ξ)ξξ 0 ds = (ξ 0 (t0 ))2 . 2 2 t0 Using that

dG(ξ(t)) = q(ξ)ξξ 0 , dt Z v G(v) = q(w)w dw,

where

(2.25)

0

we get

1 0 (ξ (t))2 + G(ξ(t)) = c0 , 2

where c0 is constant. If ½Z c0 < min

Z

∞

−∞

q(s)s ds, 0

¾ q(s)s ds ,

(2.26)

0

then ξ and ξ 0 are bounded and ξ is periodic. Conversely, if we know in advance that ξ is a bounded solution, then c0 satisfies (2.26). This fact and an even more general result will be deduced in the proof of Theorem 2.4.

September 6, 2006 (12:43)

17 We also note that if the right-hand side of (2.26) is infinite, then every solution of (2.24) is periodic. In the following theorem we describe the asymptotic behaviour of bounded solutions of (2.22). Theorem 2.4 Let ξ be a bounded solution of (2.22). Then one of the two following alternatives occurs: 1. ξ(t) = ξh (t)+w(t), where ξh (t) is a nonzero periodic solution of (2.24) and µZ ∞ ¶ |w(t)| + |w0 (t)| = O

|sg(s)| ds

(2.27)

t

as t → ∞. 2. Both ξ(t) and ξ 0 (t) tend to 0 as t → ∞ and µZ ∞ ¶ Z ξ(t) (ξ 0 (t))2 + q(v)v dv = O |g(s)| ds . 2 0 t

(2.28)

The remaining part of this section is devoted to the proof of this theorem. We start with the following lemma: Lemma 2.5 Let ξ be a bounded solution of (2.22). Then µZ ∞ ¶ 1 0 2 (ξ (t)) + G(ξ(t)) = c0 + O |g(s)| ds 2 t

(2.29)

as t → ∞, where c0 is a nonnegative constant depending on t0 and ξ. Proof. We begin with proving that ξ 0 is bounded. Multiplying (2.22) by ξ 0 and integrating, we obtain Z t Z t 1 0 1 (ξ (t))2 − g(s)ξ 0 (s) ds = (ξ 0 (t0 ))2 − q(ξ(s))ξ(s)ξ 0 (s) ds. (2.30) 2 2 t0 t0 Now (2.25) implies that Z t q(ξ(s))ξ(s)ξ 0 (s) ds = G(ξ(t)) − G(ξ(t0 )), t0

where G(ξ(t)) is uniformly bounded in t due to the boundedness of ξ. This means that also the left hand side of (2.30) is bounded in t. Set CT = sup |ξ 0 (t)|. t0 ≤t≤T

We have for t0 ≤ t ≤ T Z

t

t0

Z gξ 0 ds ≤ CT

∞

|g| ds, t0

September 6, 2006 (12:43)

18 which is finite because of (2.23). Therefore, from (2.30) it follows that Z ∞ 1 0 2 (ξ (t)) − CT |g| ds ≤ C 2 t0 with a constant C independent of t and T . Taking supremum over t0 ≤ t ≤ T we get Z ∞ 1 2 CT − CT |g| ds ≤ C 2 t0 which gives an upper bound for CT independent of T . Thus ξ 0 (t) is bounded for all t ≥ t0 . The equation (2.30) is equivalent to Z t 1 0 1 (ξ (t))2 + G(ξ(t)) = (ξ 0 (t0 ))2 + G(ξ(t0 )) + gξ 0 ds. (2.31) 2 2 t0 From (2.23) and the boundedness of ξ 0 , we get µZ Z t Z ∞ 0 0 gξ ds = C1 − gξ ds = C1 + O t0

t

∞

¶ |g| ds

t

for some constant C1 . This equality applied to (2.31) finally implies (2.29). By letting t → ∞ it follows that c0 ≥ 0. Proof of Theorem 2.4: Since ξ(t) is bounded, there exists a number L such that |ξ(t)| ≤ L, t ≥ t0 . (2.32) We rewrite (2.22) as the system of first order equations ( y10 (t) = y2 (t) y20 (t) = g(t) − q(y1 (t))y1 (t), where y1 = ξ and y2 = ξ 0 . In polar coordinates, y1 = r cos φ, y2 = r sin φ, the above system takes the form ( r0 cos φ − rφ0 sin φ = r sin φ r0 sin φ + rφ0 cos φ = g − q(r cos φ)r cos φ. This implies that  0  r = g sin φ + (1 − q(r cos φ))r sin φ cos φ  φ0 = g cos φ − q(r cos φ) cos2 φ − sin2 φ. r

(2.33)

Define the function ρ as r2 sin2 φ + ρ= 2

Z

r cos φ

q(v)v dv 0

September 6, 2006 (12:43)

(2.34)

19 and observe that ρ ≥ 0. Since ρ(t) =

Z

1 0 (ξ (t))2 + 2

ξ(t)

q(v)v dv, 0

it follows from Lemma 2.5 that µZ

∞

ρ(t) = c0 + O

¶ |g(s)| ds

(2.35)

t

with c0 ≥ 0. If c0 = 0, then the second alternative in the theorem is valid. Suppose now that c0 > 0. It is more convenient to use the variables (ρ, φ) instead of (r, φ). The second equation of (2.33) becomes φ0 =

g cos φ − F (ρ, φ), r(ρ, φ)

(2.36)

where F (ρ, φ) = q(r(ρ, φ) cos φ) cos2 φ + sin2 φ.

(2.37)

For readability we will make an abuse of notation and sometimes consider r as a function of t and sometimes of ρ and φ. We now show that c0 satisfies (2.26). If both integrals in (2.26) are infinite, this is obvious. Suppose that one of them is finite. First we conclude that there exists a positive number r0 and a t1 ≥ t0 such that r(t) ≥ r0

(2.38)

for t ≥ t1 . Indeed, if r(tj ) → 0 for some sequence {tj }∞ j=2 then also ρ(tj ) → 0 which contradicts (2.35). Thus (2.38) follows. Integrating (2.36) from t1 to t, where t ≥ t1 , we obtain Z

t

φ(t) = φ(t1 ) +

Z gr−1 cos φ ds −

t1

t

Z q cos2 φ ds −

t1

t

sin2 φ ds.

(2.39)

t1

From (2.23) and (2.38) it follows that the first integral in (2.39) has a finite limit as t → ∞. Let us show that one of the last two integrals tends to infinity as t → ∞. Suppose that Z ∞ sin2 φ ds < ∞. t1

Then l({t ≥ t1 : sin2 φ(t) > 1/2}) < ∞, where l(D) denotes the Lebesgue measure of D. Thus l(E) = ∞ for E = {t ≥ t1 : cos2 φ(t) ≥ 1/2}. From (2.38) and (2.32) it then follows that √ r0 / 2 ≤ |r cos φ| ≤ L

September 6, 2006 (12:43)

20 on E. Therefore, there exists a positive constant q0 such that q(r cos φ) ≥ q0 on E, which implies that Z ∞ q0 q cos2 φ ds ≥ l(E) = ∞. 2 t1 This proves that φ(t) → −∞

(2.40)

as t → ∞. Because of (2.38) and (2.40), we can find sequences {τj }, {sj } such that τj , sj → ∞ as j → ∞ and r(τj ) sin φ(τj ) = r(sj ) sin φ(sj ) = 0, r(τj ) cos φ(τj ) < 0 but r(sj ) cos φ(sj ) > 0. Equations (2.34) and (2.32) then imply that Z r(sj ) cos φ(sj ) Z L Z ∞ ρ(sj ) = q(v)v dv ≤ q(v)v dv < q(v)v dv. 0

0

Analogously,

Z

Z

−L

ρ(τj ) ≤

0 −∞

q(v)v dv
0, we can choose M so large that the last integral is less than ²/2. The integral Z

M −∞

α

e− 2 (t−s) (1 + |s|)khkLp (Cs ) ds

is then majorized by C exp(−αt/2) (where C depends on M and h) and the expression in the right-hand side of (2.80) is less than ² if t is large enough. Hence the right-hand side of (2.68) tends to 0 as t → ∞ and the proof is complete.

2.9

End of the proof of Theorem 2.1

We are now in position to complete the proof of Theorem 2.1. Proof of Theorem 2.1: Let η(t) be a smooth function with 0 ≤ η ≤ 1, η(t) = 0 if t ≤ 1 and η(t) = 1 if t ≥ 2. We set u(x) = η(xn )U (x). Then (

∆u + q(u)u = h in C ∂ν u = 0 on ∂C,

where h = ηH + q(u)u − ηq(U )U + 2η 0 Uxn + η 00 U. Obviously, ηH ∈ Lploc (C). The functions η 00 U and χ = q(u)u − ηq(U )U are bounded and equal to 0 for xn ≤ 1 and xn ≥ 2. Furthermore, we have that the function η 0 Uxn belongs to Lp (C) and is also equal to 0 for xn ≤ 1 and xn ≥ 2. Thus, the inequality (2.12) follows from (1.5). Now we can apply Lemma 2.7 on u. In the first case we have the representation u = uh + w1 where uh and w1 are subject to (2.67) and (2.68), respectively. We set   w1 w = U − uh   0

if xn ≥ 2 if 1 ≤ xn < 2 otherwise.

September 6, 2006 (12:43)

32 Then w ∈ L∞ (C+ ) and for t > 2, µZ kwkL∞ (Ct ) ≤ C Z

2

Z

∞ t

skHkLp (Cs ) ds + t

−α(t−s)

¶ khkLp (Cs ) ds

t

2

e−α(t−s) kHkLp (Cs ) ds

+t e 0 µZ ∞ ¶ Z t ≤C skHkLp (Cs ) ds + t e−α(t−s) kHkLp (Cs ) ds + te−αt . t

0

It is easy to see that this estimate is valid also for 1 ≤ t ≤ 2 so the first case in Theorem 2.1 is proved. In the second case of Lemma 2.7, we have u = u + v, where u and v are subject to (2.69) and (2.70), respectively. We set u0 = u and   if xn ≥ 2 v w = U − u if 1 ≤ xn < 2   0 otherwise. The inequalities (2.2) and (2.3) now follow from (2.69) and (2.70) and the proof is complete. 2

3 3.1

The Dirichlet problem Problem formulation and assumptions

In Section 3 we study bounded solutions of the Dirichlet problem ( ∆U + q(U )U = H in C+ U =0 on ∂Ω × (0, ∞).

(3.1)

We assume that U fulfills sup |U (x)| ≤ Λ

(3.2)

x∈C+

and that q is continuous. Let λD be the first eigenvalue of the Dirichlet problem for −∆0 in Ω. We suppose that there is a constant CΛ < λD such that |v| ≤ Λ ⇒ |q(v)| ≤ CΛ . (3.3) Finally, we assume that H ∈ Lploc (C+ ), where p satisfies (1.6), and that kHkLp (Ct ) is a bounded function of t, t ≥ 0.

September 6, 2006 (12:43)

33

3.2

The main asymptotic result

The main result of Section 3 is the following theorem concerning the asymptotic behaviour of solutions U of (3.1). 2,p Theorem 3.1 Assume that U ∈ Wloc (C+ ), where p satisfies (1.6), is a solution of (3.1) subject to (3.2). Assume further that q is continuous and that (3.3) is satisfied. Also, assume that H ∈ Lploc (C+ ) and that kHkLp (Ct ) is a bounded function of t for t ≥ 0. Then µZ ∞ √ ¶ √ − λD −CΛ |t−s| − λD −CΛ t kU kL∞ (Ct ) ≤ C e kHkLp (Cs ) ds + e , (3.4) 0

where C is independent of t. In particular, if H = 0 and q(0) = 0, then ³ √ ´ kU kL∞ (Ct ) = O e− λD −² t , (3.5) where ² > 0 is arbitrary. Remark 3.2 If kHkLp (Ct ) → 0 as t → ∞, then, by the same reasoning as in the end of the proof of Lemma 2.7, we see that the right-hand side of (3.4) tends to 0 as t → ∞. The proof of Theorem 3.1 is similar to the proof of Theorem 2.1 but shorter. It is contained in Section 3.3 and 3.4. Theorem 3.1 is a generalization of Theorem 2(i) in Kozlov [14]. As in the Neumann problem, the case q(U ) = |U |p−1 is studied in [14].

3.3

The corresponding problem in C

As in the Neumann problem in Section 2, we first consider the problem in 2,p the whole cylinder. Suppose that u ∈ Wloc (C) is a solution of the problem ( ∆u + q(u)u = h in C (3.6) u=0 on ∂C satisfying sup |u(x)| ≤ Λ

(3.7)

x∈C

with the same Λ as in (3.2). We have the following: Lemma 3.3 Let u be a solution of (3.6) fulfilling (3.7) and suppose that h ∈ Lploc (C) with khkLp (Ct ) bounded on R. Then we have the estimate Z kukL∞ (Ct ) ≤ C

∞

√ λD −CΛ |t−s|

e−

−∞

September 6, 2006 (12:43)

khkLp (Cs ) ds.

(3.8)

34 Proof. We consider the boundary value problem ( −∆v = g in C v=0 on ∂C, where

Z

∞

−∞

√

e−

λ1 |s|

kgkL2 (Cs ) ds < ∞.

From Section A.2 it follows that it has a unique bounded solution 1,2 v ∈ Wloc (C) with Z ∞ √ 1 √ kv(·, xn )kL2 (Ω) ≤ e− λD |xn −s| kg(·, s)kL2 (Ω) ds. 2 λD −∞ Applying this estimate to the solution u of the problem (3.6), we obtain ku(·, xn )kL2 (Ω) Z ∞ √ ¡ ¢ 1 ≤ √ e− λD |xn −s| kh(·, s)kL2 (Ω) + CΛ ku(·, s)kL2 (Ω) ds. 2 λD −∞ Iterating this estimate in the same way as in the proof of Lemma 2.6 in Section 2.7, we get Z ∞ √ 1 ku(·, xn )kL2 (Ω) ≤ √ e− λD −CΛ |xn −s| kh(·, s)kL2 (Ω) ds. 2 λD − CΛ −∞ (3.9) Now use the local estimate kukL∞ (Ct0 ) ≤ C(kukL2 (Ct ) + khkLp (Ct ) ) (compare with (2.65)) and estimate the term kukL2 (Ct ) by (3.9). Doing the same calculations as in Section 2.7 we arrive at (3.8).

3.4

End of the proof of Theorem 3.1

We now complete the proof of Theorem 3.1. Proof of Theorem 3.1: We use the same smooth function η as in the proof of Theorem 2.1 (page 31) and get, for u = ηU , the problem ( ∆u + q(u)u = h in C u=0 where

on ∂C

h = ηH + q(u)u − ηq(U )U + 2η 0 Uxn + η 00 U.

Clearly, khkLp (Ct ) is bounded and the estimate (3.4) follows from (3.8).

September 6, 2006 (12:43)

35 Suppose now that H = 0 and q(0) = 0. From (3.4) it follows that ³ √ ´ kU kL∞ (Ct ) = O e− λD −CΛ t . By choosing T large enough and considering (3.1) in Ω × (T, ∞) instead of C+ , we can make Λ and, since q(u) → 0 as u → 0, also CΛ arbitrary small. From this (3.5) follows. 2

3.5

The case of a star-shaped cross-section

In this section we show that under some special assumptions on Ω and q, every bounded solution of (3.1) with H = 0 will satisfy (3.5). This is a generalization of Theorem 2(iii) in [14] where the case q(U ) = |U |p−1 is studied. Theorem 3.4 Suppose that n ≥ 4 and Ω is star-shaped with respect to the origin and has C 2 -boundary. Also assume that q is continuous with q(0) = 0, and q(u) > 0 otherwise, and that Z u n−3 2 q(u)u − (n − 1) q(v)v dv ≥ ² q(u)u2 (3.10) 2 0 for some ² > 0. Then every bounded solution of (3.1) with H = 0 is subject to (3.5). Before proving this theorem we give some examples of functions q satisfying (3.10). Let us check that all functions of the form q(u) = f (|u|)|u|a+δ

(3.11)

with a = 4/(n − 3), δ > 0 and f being a nondecreasing function satisfy (3.10). Obviously, the function q in (3.11) is even. Therefore also both sides of the inequality (3.10) are even and we can assume that u ≥ 0. We have Z u Z u f (u)ua+2+δ q(v)v dv ≤ f (u) v a+1+δ dv = a+2+δ 0 0 and by using this inequality we obtain Z u n−3 q(u)u2 − (n − 1) q(v)v dv ≥ 2 0 n−3 (n − 1)f (u)ua+2+δ f (u)ua+2+δ − = 2 a+2+δ bf (u)ua+2+δ , where b=

δ(n − 3) . 2(a + 2 + δ)

September 6, 2006 (12:43)

36 Obviously, b > 0 for every δ > 0 since n ≥ 4. Choosing ² = b, we see that (3.10) is fulfilled. Here are some examples of functions satisfying (3.11): (i) q(u) = |u|p ,

p>

(ii) q(u) = |u|p e|u| ,

4 n−3 .

p>

(iii) q(u) = |u|p (e|u| − 1),

4 n−3 .

p>

7−n n−3 .

(iv) Linear combinations with positive coefficients of functions from (i) (iii). Proof of Theorem 3.4: The function u is a solution of the problem ( ∆u + q(u)u = 0 in C+ (3.12) u=0 on ∂Ω × (0, ∞). Ru As before, we set G(u) = 0 q(v)v dv. Using Pohoˇzaev’s identity with respect to the x0 -variables, compare with Section 3.2, Kozlov [14], we get 0 = (∆0 u + f (u))(x0 · ∇0 u) + uxn xn (x0 · ∇0 u) ¶ µ 0 2 n−3 0 2 0 0 0 0 0 0 |∇ u| = div ∇ u(x · ∇ u) − x + x G(u) + |∇ u| 2 2 − (n − 1)G(u) + uxn xn (x0 · ∇0 u). This implies that ¶ µ 0 x |∇u|2 + x0 G(u) + u2xn 0 = div (∇u(x0 · ∇0 u)) + div0 − 2 n−3 + |∇u|2 − (n − 1)G(u). 2

(3.13)

Set CN = Ω × (1, N ) and ΓN = ∂Ω × (1, N ) and observe that ∇u =

∂u ν ∂ν

on ΓN .

This follows by representing the vector ∇u at a certain point on ΓN as the sum of a normal and a tangent vector. The tangent vector is then zero because of the homogeneous Dirichlet boundary condition. Using that G(u) = 0 on ΓN and integrating (3.13) over CN , we obtain ¶ Z µ n−3 2 2 |∇u| + uxn − (n − 1)G(u) dx 0= 2 CN (3.14) µ ¶2 Z 1 ∂u + (x · ν) dS + F1 (N ) − F1 (1), 2 ΓN ∂ν

September 6, 2006 (12:43)

37 where

Z

¯ ¯ uxn (x · ∇ u) dx ¯ 0

F1 (t) = Ω

0¯

0

. xn =t

Multiplying (3.12) by u and using Green’s formula, we have Z ¢ ¡ −|∇u|2 + q(u)u2 dx + F2 (N ) − F2 (1), 0=

(3.15)

CN

where

Z

¯ ¯ uxn u dx ¯ 0¯

F2 (t) = Ω

. xn =t

A linear combination of (3.14) and (3.15) gives ! ¶2 Z Ãµ ∂u n−3 2 + q(u)u − (n − 1)G(u) dx ∂xn 2 CN Z µ ¶2 ∂u 1 (x · ν) dS = F (1) − F (N ), + 2 ΓN ∂ν (3.16) where

n−3 F2 (t). 2 Let us show that F is bounded on [1, ∞). For given t > 0, choose r > n. Using the Sobolev embedding theorem, we obtain F (t) = F1 (t) +

k∇ukL∞ (Ct0 ) ≤ Ck∇ukW 1,r (Ct0 ) ≤ CkukW 2,r (Ct0 ) . Now, Lemma A.16 in Section A.3 gives kukW 2,r (Ct0 ) ≤ C(kq(u)ukLr (Ct ) + kukL2 (Ct ) ) ≤ C1 . Therefore, k∇ukL∞ (Ct0 ) is a bounded function. This implies that F1 and F2 are bounded functions on [1, ∞). Since Ω is star-shaped, it follows that x0 · ν 0 ≥ 0. This and (3.10) imply that every term in (3.16) is non-negative and, from the previous analysis, bounded in N . Thus Z q(u)u2 dx < ∞ (3.17) C+

and from (3.15) it therefore follows that Z |∇u|2 dx → 0 as t → ∞.

(3.18)

Ct

Corollary A.18 in Section A.3 gives ¡ ¢ kukL∞ (Ct0 ) ≤ C kq(u)ukLp (Ct ) + kukL2 (Ct ) .

September 6, 2006 (12:43)

(3.19)

38 From Poincaré’s inequality it follows that kukL2 (Ct ) ≤ Ck∇ukL2 (Ct ) and using Hölder’s inequality, we get kq(u)ukpLp (Ct ) ≤ kq(u)1/2 ukL2 (Ct ) kq(u)p−1/2 |u|p−1 kL2 (Ct ) . Since u is bounded and q continuous, the right-hand side is also bounded. Applying the last two inequalities for estimating the right-hand side in (3.19), we obtain ³ ´ 1/(2p) kukL∞ (Ct0 ) ≤ C kq(u)u2 kL1 (Ct ) + k∇ukL2 (Ct ) . Using (3.17) and (3.18) we see that kukL∞ (Ct ) → 0 as t → ∞. Applying Theorem 3.1 on v(x0 , xn ) = u(x0 , xn + N ) for N large enough completes the proof. 2

3.6

An estimate for solutions of a nonlinear ordinary differential equation

The estimate (3.5) contains an arbitrary small parameter ². Is it possible to remove ² from this estimate? In order to see what kind of requirements we need on the function q, we consider here the model equation u00 − λu + q(u)u = 0.

(3.20)

Theorem 3.5 Let u be a solution of (3.20) subject to the condition u(t) → 0

as t → ∞.

(3.21)

Suppose that λ > 0, that the function f (u) = q(u)u is Lipschitz continuous with f (0) = 0 and that q(u) > 0 if u 6= 0. Suppose also that Z

1

−1

Then

q(u) du < ∞. |u|

(3.22)

³ √ ´ |u(t)| + |u0 (t)| = O e− λt .

Proof. We may assume that u is not identically zero. Using (3.21), we see from (3.20) that u00 (t) → 0 as t → ∞. Since Z

t+1

0

u(t + 1) = u(t) + u (t) −

(s − t − 1)u00 (s) ds,

t

September 6, 2006 (12:43)

39 we have |u0 (t)| ≤ |u(t)| + |u(t + 1)| + Therefore also u0 (t) → 0 as t → ∞. As before, we use the notation Z G(u) =

1 sup |u00 (s)|. 2 [t,t+1]

u

q(v)v dv.

0

We multiply (3.20) by u0 and obtain ¡ 0 2 ¢0 (u ) − λu2 + 2G(u) = 0. Therefore (u0 )2 − λu2 + 2G(u) is a constant. Since the functions u and u0 vanish at ∞ we conclude that this constant is 0. Thus p u0 = ± λu2 − 2G(u). (3.23) Let us show that the function λu2 (t) − 2G(u(t)) has no zeros for t large enough. First, suppose that u(t0 ) = 0 for some t0 ∈ R. Then (3.23) implies that u0 (t0 ) = 0 and we get u(t) ≡ 0 by the uniqueness of solutions to the Cauchy problem of equation (3.20). Furthermore, define Z u 1 Q(u) = q(v)v dv. λu2 0 Since

Z 1 u q(v) dv, λ 0 v we have, because of (3.21) and (3.22), that 0 ≤ Q(u) ≤

Q(u(t)) → 0 as t → ∞. This implies that λu2 − 2G(u) is positive for large t. From (3.23) it follows that u0 (t) has no zeros for large t. Since u(t) → 0 as t → ∞, we meet here one of two possibilities: Either u > 0 and u0 < 0 or else u < 0 and u0 > 0. We consider the first possibility. The second one is considered analogously. We have √ p u0 = − λ u 1 − 2Q(u). (3.24) Power series expansion yields that there exists an ² > 0 such that ¯ ¯ ¯ ¯ ¯√ 1 ¯ ≤ 2|x| − 1 ¯ 1 − 2x ¯

(3.25)

if |x| ≤ ². Furthermore, since Q(v) → 0 as v → 0, there exists a δ > 0 such that |Q(v)| ≤ ² if |v| ≤ δ. Choose t so large that |u(t)| ≤ δ. Integrating (3.24), we obtain Z δ dv 1 √ p = t − C1 . (3.26) λ u v 1 − 2Q(v)

September 6, 2006 (12:43)

40 The left-hand side can be written as 1 √ λ

Z

δ

u

dv 1 +√ v λ

Z

δ

u

1 v

Ã p

1 1 − 2Q(v)

! −1

dv.

Furthermore, we get from (3.25) that ¯Z Ã ! ¯ Z 1 ¯ ¯ δ1 1 Q(v) ¯ ¯ p dv < ∞, − 1 dv ¯ ≤ 2 ¯ ¯ ¯ u v 1 − 2Q(v) −1 |v|

(3.27)

(3.28)

where the last inequality follows from (3.22). From (3.26), (3.27) and (3.28) it follows that √ ln u = B(t) − λ t, ³ √ ´ where B is a bounded function. Therefore u(t) = O e− λt . Using (3.24), ³ √ ´ we also obtain that u0 (t) = O e− λt .

A A.1

Some results from functional analysis Eigenvalues and eigenvectors of −∆

Here we show that if ∂Ω is smooth enough there exists an ON-basis of L2 (Ω) consisting of eigenvectors of the operator −∆. For the Dirichlet problem this result is often proved in textbooks in partial differential equations so we focus on the Neumann problem and prove it since in this case the proof is not easily found. On our way we need some lemmas which will be stated without proof. Throughout this appendix, let H denote a Hilbert space and h·, ·iH the inner product in H. If A is a bounded linear operator on H, we let σ(A) denote the spectrum of A and σp (A) the set of eigenvalues of A, i.e. the point spectrum. The following lemmas are well-known results from functional analysis. Lemma A.1 Let Ω be an open set in R. Then C0∞ (Ω) is dense in L2 (Ω). Definition A.2 If X is a normed space we say that a subset M ⊂ X is total in X if the span of M is dense in X. Lemma A.3 If M is total in X and for some x ∈ X we have that x⊥M , then x = 0. Conversely, if X is complete and x⊥M implies x = 0, then M is total in X. Lemma A.4 Suppose K is a compact linear operator on H. Then N (I−K) is finite dimensional. If dim H = ∞ then also

September 6, 2006 (12:43)

41 (i) 0 ∈ σ(K). (ii) σ(K) − {0} = σp (K) − {0}. (iii) σ(K) − {0} is either finite or countable with the unique limit point 0. Lemma A.5 Suppose A is a linear, bounded and symmetric operator on H and define m = inf{hAu, uiH : u ∈ H, kuk = 1} M = sup{hAu, uiH : u ∈ H, kuk = 1}. Then m, M ∈ σ(A) ⊂ [m, M ]. Remark A.6 Since A is symmetric, hAu, uiH is always real. Lemma A.7 Suppose H is separable and A : H → H is linear, bounded, symmetric and compact. Then there exists a countable orthonormal basis of H consisting of eigenvectors of A. Theorem A.8 Let Ω be an open, bounded region in Rn with C 1 -boundary. 2 Then there exists an ON-basis {φk }∞ k=0 of L (Ω) consisting of eigenfunctions of the operator −∆ for the Neumann problem, i.e. Z Z ∇φk · ∇v dx = λk φk v dx, ∀v ∈ H 1 (Ω), (A.1) Ω

Ω

and the eigenvalues λk are subject to 0 = λ0 < λ 1 ≤ λ2 ≤ λ3 ≤ . . . ,

λk → ∞ as k → ∞,

(A.2)

where each eigenvalue is repeated according to its multiplicity. 1 Moreover, {φk }∞ k=0 is also an orthogonal basis of H (Ω). R Proof. Given u ∈ H 1 (Ω), the functional v 7→ Ω vu dx is linear and bounded on H 1 (Ω). According to Riesz representation theorem, there exists a unique T u ∈ H 1 (Ω) such that Z vu dx = hv, T uiH 1 (Ω) , ∀v ∈ H 1 (Ω). (A.3) Ω

We therefore define the operator T : H 1 (Ω) → H 1 (Ω) by the relation (A.3). It is easy to see that T is linear and from the closed graph theorem it follows that T is bounded. Namely, let Γ = {(x, T x) : x ∈ H 1 (Ω)} be the graph of T and suppose that (xn , yn ) ∈ Γ, (xn , yn ) → (x, y)Rin H 1 (Ω) × H 1 (Ω). From (A.3) it follows that hyn − T x, yn − T xiH 1 (Ω) = Ω (yn − T x)(xn − x) which, together with Hölder’s inequality, shows that yn → T x and thus y = T x. Hence the graph is closed.

September 6, 2006 (12:43)

42 T is also symmetric, because the equality hT u, viH 1 (Ω) = hu, T viH 1 (Ω) follows by combining the fact that hT u, viH 1 (Ω) = hv, T uiH 1 (Ω) with the relation (A.3). Finally, T is compact: let i : H 1 (Ω) → L2 (Ω) be the inclusion of H 1 (Ω) in L2 (Ω) and T1 : L2 (Ω) → H 1 (Ω) be the extension of T to L2 (Ω) as defined by the left-hand side of (A.3). Since H 1 (Ω) is compactly embedded in L2 (Ω) and T1 , as T , is bounded, it follows that T = T1 ◦ i is compact. Lemma A.7 now gives that there exists a countable orthonormal ba1 sis {ψj }∞ j=0 of H (Ω) consisting of eigenfunctions of T and we will now prove that, after normalization, the same set is an ON-basis for L2 (Ω). Let R µj be the eigenvalue corresponding to ψj . From (A.3) we get that ψ ψ dx = µk hψj , ψk iH 1 (Ω) so the eigenfunctions are orthogonal to each Ω j k other also in L2 (Ω). By setting φj =

ψj , kψj kL2 (Ω)

2 we thus get an orthonormal sequence {φj }∞ j=0 in L (Ω), each φj also an eigenfunction of T with the same eigenvalue as ψj . It remains to see that each function in L2 (Ω) can be written as a (finite or infinite) linear combination of them. 1 Set M = {φj }∞ j=0 . The fact that M is a basis of H (Ω) implies that M 1 1 2 is total in H (Ω). Since H (Ω) is dense in L (Ω) according to Lemma A.1, M is also total in L2 (Ω). For f ∈ L2 (Ω) we construct

g=

∞ X

hf, φk iL2 (Ω) φk ,

k=0

which is convergent in L2 (Ω). We get immediately hf − g, φj iL2 (Ω) = 0 for every j so Lemma A.3 gives that f = g. Thus {φj }∞ j=0 is an ON-basis for L2 (Ω) and the coordinates of f are its usual Fourier coefficients. The final step is to investigate the eigenvalues of T . From (A.3) it follows that hT u, uiH 1 (Ω) = kuk2L2 (Ω) (A.4) so Lemma A.5 gives that σ(T ) ⊂ [0, 1] with 0, 1 ∈ σ(T ). (A.4) also shows that 0 is not an eigenvalue. Furthermore, if λ is an eigenvalue, Lemma A.4 shows that N (λI − T ) = N (I − λ−1 T ) is finite dimensional. But since dim H 1 (Ω) = ∞ and the eigenfunctions form a basis, we conclude that T has infinitely many eigenvalues. Lemma A.4 also gives that 1 really is an eigenvalue (with constants as eigenfunctions) and that we can label the eigenvalues in decreasing order as 1 = µ0 ≥ µ1 ≥ . . . > 0, We see that

µk → 0 as k → ∞.

Z hu, viH 1 (Ω) =

f v dx,

∀v ∈ H 1 (Ω)

Ω

September 6, 2006 (12:43)

(A.5)

43 if and only if u = T f. This shows that if φk is an eigenfunction of T with eigenvalue µk , then it is also a solution of (A.1) with λk =

1 −1 µk

and thus an eigenfunction of the operator −∆ for the Neumann problem. We also see from (A.1) that the only eigenfunctions to λ0 = 0 are the constant functions. Hence λ0 has single multiplicity and (A.2) follows from (A.5). Theorem A.9 Let Ω be a bounded domain in Rn . Then there exists an 2 1 ON-basis {φk }∞ k=1 of L (Ω) where φk ∈ H0 (Ω) are eigenfunctions of the operator −∆ for the Dirichlet problem, i.e. Z Z ∇φk · ∇v dx = λk φk v dx, ∀v ∈ H01 (Ω), Ω

Ω

and the eigenvalues λk are subject to 0 < λ 1 ≤ λ2 ≤ λ3 ≤ . . . ,

λk → ∞ as k → ∞,

where each eigenvalue is repeated according to its multiplicity. 1 Also, {φk }∞ k=0 is an orthogonal basis of H0 (Ω). Proof. We prove only the last part of the theorem because the remaining parts of the complete proof can be found in Evans [8]. For u ∈ H01 (Ω) we have the equality hu, φj iH01 (Ω) = (1 + λj )hu, φj iL2 (Ω) 1 so if u ∈ H01 (Ω) is orthogonal to {φj }∞ 1 in H0 (Ω), it is also orthogonal to 2 ∞ {φj }1 in L (Ω). From this and Lemma A.3 it now follows that {φj }∞ 1 is total in H01 (Ω) and that it in fact is an orthogonal set.

A.2

Existence and uniqueness of bounded solutions of Poisson’s equation in C

In this section we will see that under some conditions the problem   −∆u = f in C  ∂u = 0 on ∂C ∂ν has a unique bounded solution with mean value 0 over Ω and also derive a bound for its L2 (Ω)-norm. We then turn to the problem ( −∆u = f in C (A.6) u = 0 on ∂C

September 6, 2006 (12:43)

44 and state that practically the same results hold in this case. We use the theory of eigenvalues of the operator −∆0 in Ω presented in Section A.1 and construct the solution in each case as an infinite linear combination of the eigenfunctions. We denote the first positive eigenvalue by λ1 in the Neumann case and by λD in the Dirichlet case. Lemma A.10 Suppose f ∈ L2loc (C) fulfills Z ∞ √ e− λ1 |s| kf kL2 (Cs ) ds < ∞. −∞

Then also

Z

∞

e−

√ λ1 |s|

−∞

kf (·, s)kL2 (Ω) ds < ∞.

Proof. First note that if a ≥ 0 and |x − y| ≤ c for some c, then e−a|y| ≤ eac e−a|x| .

(A.7)

This follows from the fact that |x| − |y| ≤ |x − y|. We write g(t) = kf (·, t)kL2 (Ω) and observe that for a ≥ 0, Hölder’s inequality gives Z a √ kf (·, t + s)kL2 (Ω) ds = kgkL1 (t,t+a) ≤ akf kL2 (Ω×(t,t+a)) . (A.8) 0

It is easy to see that Z

∞

e

√ − λ1 |s|

−∞

Z kf (·, s)kL2 (Ω) ds =

0

1

Z

∞

√

e−

λ1 |s+t|

−∞

kf (·, s + t)kL2 (Ω) ds dt

by noting that the value of the inner integral on the right-hand side is independent of t. By applying (A.7) and (A.8) on the right-hand side we arrive at Z ∞ √ Z ∞ √ √ − λ1 |s| λ1 e kf (·, s)kL2 (Ω) ds ≤ e e− λ1 |s| kf kL2 (Cs ) ds −∞

−∞

from which the lemma now directly follows. √ Remark A.11 It follows from (A.7) with a = λ1 , c = |xn |, x = s and y = s − xn that √ √ √ e− λ1 |xn −s| ≤ e λ1 |xn | e− λ1 |s| . Under the same conditions as in Lemma A.10 we therefore also have that Z ∞ √ e− λ1 |xn −s| kf (·, s)kL2 (Ω) ds < ∞ −∞

for every xn ∈ R.

September 6, 2006 (12:43)

45 For functions v ∈ H 1 (Ω) subject to Z v dx = 0

(A.9)

Ω

the following inequality holds: k∇0 vk2L2 (Ω) ≥ λ1 kvk2L2 (Ω) .

(A.10)

2 In order to see this, let us consider the ON-basis {φj }∞ j=0 of L (Ω) consisting of eigenfunctions of the Neumann problem for −∆0 in Ω corresponding to the eigenvalues 0 = λ0 < λ1 ≤ λ2 ≤ . . . (see Theorem A.8 in Section A.1). We have that ∞ X v= vj , j=0

where vj = hv, φj iL2 (Ω) φj . Since φ0 = const, it follows from (A.9) that v0 = 0. The set {φj } also forms an orthogonal basis of H 1 (Ω) so ∇0 v =

∞ X

∇0 v j

j=1

and (A.10) now follows from Parseval’s identity together with the identity h∇0 vi , ∇0 vj iL2 (Ω) = λi hvi , vj iL2 (Ω) obtained from Greens formula. Lemma A.12 Suppose f ∈ L2loc (C) satisfies Z f (x0 , xn ) dx0 = 0 for a.e.

xn ∈ R

(A.11)

Ω

and

Z

∞

e−

√ λ1 |s|

−∞

kf kL2 (Cs ) ds < ∞.

Then the problem

  −∆u = f in C  ∂u = 0 on ∂C ∂ν 1,2 has a bounded solution u ∈ Wloc (C) with Z u(x0 , xn ) dx0 = 0 for a.e. xn ∈ R

(A.12)

(A.13)

Ω

and 1 ku(·, xn )kL2 (Ω) ≤ √ 2 λ1

Z

∞

−∞

e−

√ λ1 |xn −s|

kf (·, s)kL2 (Ω) ds,

(A.14)

the last expression being finite. This solution is unique up to a constant.

September 6, 2006 (12:43)

46 Remark A.13 Since ∂Ω ∈ C 2 , it follows from Section A.3 that if 1,2 f ∈ Lrloc (C) for some r ∈ [2, ∞), then the fact that u ∈ Wloc (C) implies 2,r that u ∈ Wloc (C), i.e. there is no difference between a weak and a strong solution of (A.12). Proof. Lemma A.10 gives that the right hand side of (A.14) is finite (see the remark). We begin by finding a solution of (A.12). According to Theorem A.8 in Section A.1 there exists an orthonormal basis {φj }∞ j=0 of L2 (Ω) consisting of eigenfunctions of the Neumann problem in Ω, i.e.  0  −∆ φj = λj φj in Ω  ∂φj = 0 on ∂Ω ∂ν 0

(A.15)

with λ0 = 0 and all other λj > 0. Defining Z f φj dx0 ,

fj =

j = 1, 2, . . . ,

Ω

we get, since f0 = 0 by (A.11), f (x0 , xn ) =

∞ X

fj (xn )φj (x0 ).

j=1

We also define 1 uj (t) = p 2 λj

Z

∞

e−

√

λj |t−s|

fj (s) ds,

j = 1, 2, . . .

−∞

and set for N ≥ 1 u(N ) (x0 , xn ) =

N X

uj (xn )φj (x0 ),

j=1

f

(N )

=

N X

fj (xn )φj (x0 ).

j=1

It then follows that −∆u(N ) = f (N ) . Then, using Parseval’s identity, Minkowski’s inequality (see for example Section 6.3, Folland [9] or Section 2.4, Lieb and Loss [16]) and Parseval

September 6, 2006 (12:43)

47 again yields

(N )

ku

(·, xn )kL2 (Ω) =

µX N 

≤

¶1/2 2

uj (xn )

j=1 N X

Ã

j=1

1 ≤ √ 2 λ1 1 ≤ √ 2 λ1

Z

1 p 2 λj  ∞

!2 µZ

∞

e−

e−2

∞

 ¶2 1/2 λj |xn −s| |fj (s)| ds 

−∞ √

λ1 |xn −s|

−∞

Z

√

N X

1/2 fj (s)2 

ds

j=1

e−

√ λ1 |xn −s|

−∞

kf (·, s)kL2 (Ω) ds

(A.16)

which shows that {u(N ) }N is a Cauchy sequence in L2 (Ω). When working with u0j instead of uj and noting that u0j (t) ≤

1 2

Z

∞

e−

√ λ1 |t−s|

|fj (s)| ds,

−∞

it can be proved in the same way that also {∂u(N ) /∂xn }N is Cauchy in L2 (Ω). Write 1/2

k∇0 u(N ) (·, xn )kL2 (Ω) = h−∆0 u(N ) (·, xn ), u(N ) (·, xn )iL2 (Ω) µX ¶1/2 N 2 = λj uj (xn ) j=1

  µZ ∞ √ ¶2 1/2 N X 1 ≤ e− λ1 |xn | |fj (s)| ds  4 −∞ j=1 and proceed as before to obtain (N )

k∇u

(·, xn )kL2 (Ω)

1 ≤ 2

Z

∞

e−

√ λ1 |xn −s|

−∞

kf (·, s)kL2 (Ω) ds.

(A.17)

This proves that also {∇0 u(N ) }N is Cauchy in L2 (Ω). The conclusion is 1,2 that u(N ) converges in W 1,2 (Ω). To show the convergence in Wloc (C), we (N ) estimate ku kW 1,2 (Ct ) for a fixed t in the same way as above. Using

September 6, 2006 (12:43)

48 (A.16), Minkowski’s inequality and (A.7), we get µZ ku(N ) kL2 (Ct ) =

t

t+1

¶1/2 ku(N ) (·, xn )k2L2 (Ω) dxn

¶1/2 Z ∞ µZ t+1 √ 1 ds ≤ √ e−2 λ1 |xn −s| kf (·, s)k2L2 (Ω) dxn 2 λ1 −∞ t µZ t+1 ¶1/2 Z ∞ √ ≤C kf (·, s)kL2 (Ω) e−2 λ1 |t−s| dxn ds −∞ t Z ∞ √ √ ≤ Ce λ1 |t| e− λ1 |s| kf (·, s)kL2 (Ω) ds. −∞

The right-hand side is finite from Lemma A.10 so {u(N ) } is Cauchy also in L2loc (C). In the same way it follows from (A.17) that {∇u(N ) } is Cauchy in 1,2 2 Lloc (C) and that u(N ) converges to some function u in Wloc (C). Regularity theory gives that u solves (A.12). φ0 is constant so due to the orthogonality of {φj } we have R Furthermore, (N ) 0 u dx = 0. Thus, the property (A.13) follows from the convergence Ω in L2 (Ω). To prove the assertion about uniqueness, we now show that the only bounded solutions of   ∆u = 0 in C  ∂u = 0 on ∂C ∂ν are the constant functions. Using the decomposition u(x) =

∞ X

uj (xn )φj (x0 )

j=0

and (A.15), we get Z Z Z 0=− φj ∆u dx0 = − u∆0 φj dx0 − φj uxn xn dx0 Ω

Ω

= λj uj (xn ) − u00j (xn ),

j = 0, 1, 2, . . . .

Ω

(A.18)

If j 6= 0, then λj > 0 and in these cases the only bounded solution of (A.18) is uj = 0. Furthermore, λ0 = 0, so in the case j = 0 we get the only bounded solutions as u0 (xn ) = A, where A is constant. Since also φ0 is constant, we get that u is a constant function. We now prove (A.14). We assume that u is non-vanishing (so that ku(·, xn )kL2 (Ω) is twice differentiable). In other case, we can replace ku(·, xn )kL2 (Ω) by (ku(·, xn )k2L2 (Ω) + ²)1/2 in the calculations below and then let ² → 0 to obtain the same result.

September 6, 2006 (12:43)

49 For clarity, we skip the index L2 (Ω) in the norm and inner product 1,2 notations. Since u ∈ Wloc (C) we get that u(·, xn ) ∈ W 1,2 (Ω) for almost every xn . When multiplying (A.12) by u, integrating over Ω and using Green’s theorem, we arrive at Z Z Z ∂2u 0 2 0 |∇ u| dx − u 2 dx = f u dx. Ω Ω ∂xn Ω It follows, by (A.10) together with Cauchy-Schwartz inequality, that ¿ 2 À ∂ u , u ≥ λ1 kuk2 − kf kkuk. (A.19) ∂x2n Differentiating ku(·, xn )k twice gives kuk

d2 kuk = dx2n

¿

° ¶ À µ¿ À2 ° ° ∂u °2 ∂2u ∂u −2 ° ° kuk2 . , u − kuk , u − ° ∂xn ° ∂x2n ∂xn

The Cauchy-Schwartz inequality gives that h∂xnu, ui2 − k∂xnuk2 kuk2 ≤ 0 so from (A.19) d2 kuk ≥ λ1 kuk − kf k. (A.20) dx2n Consider the equation −

d2 ku(·, xn )k + λ1 ku(·, xn )k = g(xn ). dx2n

(A.21)

By using a Green function and the fact that u is bounded, we get Z ∞ √ 1 ku(·, xn )k = √ e− λ1 |xn −s| g(s) ds. 2 λ1 −∞ But from (A.20) and (A.21) we have g(xn ) ≤ kf (·, xn )k so (A.14) now directly follows. We now turn to the Dirichlet problem (A.6) and have the following analogue of Lemma A.12: Lemma A.14 Suppose f ∈ L2loc (C) is subject to Z ∞ √ e− λ1 |s| kf kL2 (Cs ) ds < ∞. −∞

1,2 Then the problem (A.6) has a unique, bounded solution u ∈ Wloc (C) and Z ∞ √ 1 ku(·, xn )kL2 (Ω) ≤ √ e− λD |xn −s| kf (·, s)kL2 (Ω) ds, 2 λD −∞

where the last expression is finite.

September 6, 2006 (12:43)

50 Remark A.15 Since all eigenvalues of the Dirichlet problem are positive, we do not have the conditions that f and u are orthogonal to constants in L2 (Ω) as in the Neumann problem. Proof. The proof is similar to the proof of Lemma A.12. A difference is however that the estimate (A.10), with λ1 replaced by λD , is now valid for all v ∈ W01,2 (C).

A.3

A local estimate for solutions of Poisson’s equation

We say that D is a cylindrical type domain (CTD) if D = Ω × (a, b), where −∞ < a < b < ∞. Furthermore, if D = Ω × (a, d) and D0 = Ω × (b, c) are two CTD:s, we say that D0 is compactly contained in D in the xn -direction if a < b < c < d. 1,2 Given a solution u ∈ Wloc (C) to some of the problems   ∆u = f in C (A.22)  ∂u = 0 on ∂C ∂ν or ( ∆u = f in C (A.23) u = 0 on ∂C, 2,p where f ∈ Lploc (C), we will see that in fact u ∈ Wloc (C) and find a bound for kukW 2,p (Ct0 ) expressed in terms of kukL2 (Ct ) and kf kLp (Ct ) . 1,2 Lemma A.16 Suppose that p fulfills (1.6) and that u ∈ Wloc (C) is a solution of (A.22) or (A.23). Suppose further that E1 and E2 are CTD:s and that E1 is compactly contained in E2 in the xn -direction. Suppose also that 2,p E2 ⊂ Ct for some t ∈ R. Then u ∈ Wloc (C) and

kukW 2,p (E1 ) ≤ C0 (kukL2 (E2 ) + kf kLp (E2 ) ),

(A.24)

for some constant C0 depending on p, n, E1 and E2 but not on t. Remark A.17 The assumption that E2 ⊂ Ct for some t ∈ R is not essential for the result but indicates that the constant C0 is independent of t. Proof. Given t ∈ R, make first the transformations v(x0 , xn ) = u(x0 , xn + t), g(x0 , xn ) = f (x0 , xn + t) and note that the equation ∆u = f on Ct is equivalent to the equation ∆v = g on C0 . From Theorem 15.1” in Agmon, Douglis and Nirenberg, [1], we get the interior estimate kvkW 2,r (D0 ) ≤ C(kvkLr (D) + kgkLr (D) )

(A.25)

if r ∈ (1, ∞) and D0 ⊂⊂ D ⊂ C0 . The constant C depends only on D, D0 and r. Furthermore, suppose x is a point on ∂Ω × (0, 1). After

September 6, 2006 (12:43)

51 straightening the boundary in a neighborhood of x, Theorem 15.3 [1] and the remark immediately after it gives the estimate (A.25) for D0 and D equal to hemispheres located at x. Using a partition of unity we finally get the estimate kvkW 2,r (D1 ) ≤ C(kvkLr (D2 ) + kgkLr (D2 ) ) (A.26) for any D1 , D2 such that D1 is compactly contained in D2 ⊂ C0 in the xn -direction. The constant C depends on r, D1 and D2 but not on v and g. We can therefore revert to the functions u and f and get the inequality kukW 2,r (D1,t ) ≤ C(kukLr (D2,t ) + kf kLr (D2,t ) ),

(A.27)

where D1,t and D2,t are equal to D1 and D2 translated t steps in the xn direction and C is the same constant as in (A.26), i.e. independent of t. If n = 2 or 3, the lemma follows from (A.27) with r = 2, since p = 2 in these cases. Now suppose ω is any bounded open subset of Rn with Lipschitz boundary and introduce φ(x) = nx/(n − 2x) with inverse ψ(x) = nx/(n + 2x). A well-known Sobolev inequality states that if r < n/2 and s = φ(r), then kukLs (ω) ≤ CkukW 2,r (ω)

(A.28)

kukL∞ (ω) ≤ CkukW 2,r (ω) .

(A.29)

and if r > n/2 k Set E10 = E1 and choose a sequence {E1k }∞ k=1 such that E1 is compactly k+1 contained in E1 as well as in E2 in the xn -direction for k ≥ 0. We set p1 = p, p2 = ψ(p) and notice that p2 < n/2. Using (A.27), (A.28) and then (A.27) again we obtain

kukW 2,p (E1 ) ≤ C1 (kukLp (E11 ) + kf kLp (E2 ) ) ≤ C2 (kukW 2,p2 (E11 ) + kf kLp (E2 ) ) ≤ C3 (kukLp2 (E12 ) + kf kLp (E2 ) ).

(A.30)

We now continue the iteration indicated in (A.30) in the following way: for k ≥ 2, set pk+1 = ψ(pk ) and combine again (A.27) and (A.28) to obtain kukLpk (E1k ) ≤ C(kukLpk+1 (E k+1 ) + kf kLp (E2 ) ). 1

Obviously, p1 > p2 > . . . and eventually pK ≤ 2 for some K. We can assume pK > 1, since if pK = ψ(pK−1 ) ≤ 1, then pK−1 can be increased so that the condition pK > 1 is met. We get kukW 2,p (E1 ) ≤ C(kukLpK (E1K ) + kf kLp (E2 ) ), from which (A.24) follows.

September 6, 2006 (12:43)

52 Corollary A.18 Under the same conditions as in Lemma A.16 kukL∞ (E1 ) ≤ C(kukL2 (E2 ) + kf kLp (E2 ) ).

Proof. Since p > n/2, the corollary follows by combining Lemma A.16 with the Sobolev inequality (A.29). Remark A.19 The lemma as well as the corollary also hold with C and ∂C replaced by C+ and ∂Ω × (0, ∞), respectively, but with the additional condition t ≥ 0.

References [1] S. Agmon, A. Douglis, L. Nirenberg, Estimates Near the Boundary for Solutions of Elliptic Partial Differential Equations Satisfying General Boundary Conditions. I. Communications on pure and applied mathematics 12 (1959), 623–727. [2] C. J. Amick, J. F. Toland, Nonlinear elliptic eigenvalue problems on an infinite strip - global theory of bifurcation and asymptotic bifurcation. Math. Ann. 262 (1983), 313–342. ń, On the positive solutions of Emden equations [3] C. Bandle, M. Esse in cone-like domains. Arch. for Rat. Mech. 12 (1990), 319–338. [4] H. Berestycki, Some nonlinear PDE’s in the theory of flame propagation. ICIAM 99 Proceedings of the Fourth International Congress on Industrial and Applied Mathematics, Oxford university press (2000), 13–22. [5] H. Berestycki, L. Caffarelli, L. Nirenberg, Further Qualitative Properties for Elliptic Equations in Unbounded Domains. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 25 (1997), 69–94. [6] H. Berestycki, B. Larrouturou, J. M. Roquejoffre, Stability of Travelling Fronts in a Model for Flame Propagation, Part I: Linear Analysis. Arch. Rational Mech. Anal. 117 (1992), 97–117. [7] H. Berestycki, L. Nirenberg, Travelling fronts in cylinders. Ann. Inst. H. Poincaré, Analyse non linéaire 9 (1992), 497–572. [8] L. C. Evans, Partial Differential Equations. American Mathematical Society, 1998. [9] G. B. Folland, Real Analysis: Modern Techniques and Their Applications. Wiley-Interscience, 1999.

September 6, 2006 (12:43)

53 ¨ ssner, J. Scheurle, On the Bounded Solutions of a [10] K. Kirchga Semilinear Elliptic Equation in a Strip. Journal of differential equations 32 (1979), 119–148. [11] V. A. Kondratiev, Asymptotic behaviour of solutions of some nonlinear parabolic or elliptic equations. Asymptot. Anal. 14 (1997), 117– 156. [12] V. A. Kondratiev, On some nonlinear boundary value problems in cylindrical domains. Journal of Mathematical Sciences 85 (1997), 2385–2401. [13] V. A. Kondratiev, On the existence of positive solutions of secondorder semilinear elliptic equations in cylindrical domains. Russ. J. Math. Phys. 10 (2003), 99–108. [14] V. Kozlov, On Bounded Solutions of the Emden-Fowler Equation in a Semi-cylinder. Journal of differential equations 179 (2002), 456–478. [15] V. Kozlov, V. Maz’ya, Differential Equations with Operator Coefficients. Springer, 1999. [16] E. H. Lieb, M. Loss, Analysis. American Mathematical Society, 1997.

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55

Paper 2

September 6, 2006 (12:43)

56

September 6, 2006 (12:43)

Asymptotic analysis of solutions to parabolic systems Vladimir Kozlov

Mikael Langer

Peter Rand

Abstract We study asymptotics as t → ∞ of solutions to a linear, parabolic system of equations with time-dependent coefficients in Ω × (0, ∞), where Ω is a bounded domain. On ∂Ω × (0, ∞) we prescribe the homogeneous Dirichlet boundary condition. For large values of t, the coefficients in the elliptic part are close to time-independent coefficients in an integral sense which is described by a certain function κ(t). This includes in particular situations when the coefficients may take different values on different parts of Ω and the boundaries between them can move with t but stabilize as t → ∞. The main result is an asymptotic representation of solutions for large t. As a corollary, it is proved that if κ ∈ L1 (0, ∞), then the solution behaves asymptotically as the solution to a parabolic system with time-independent coefficients.

1

Introduction

Let Ω denote an open, bounded region in Rn with Lipschitz boundary and introduce Q = Ω × (0, ∞). By x = (x1 , . . . , xn ) we denote the variables in Ω and by t the unbounded variable. We consider the parabolic system ut −

n X

(Aij uxj )xi + Au = 0

in Q,

(1.1)

i,j=1

where u = (u1 , . . . , uN ) is a function from Q to CN and Aij , i, j = 1, . . . , n, and A are quadratic matrices of size N × N whose elements are functions from Q to C. We will assume that u satisfies the Dirichlet boundary condition u(x, t) = 0 if x ∈ ∂Ω, t > 0 (1.2) and that u(x, 0) = ψ(x), (1.3) ¡ 2 ¢N where ψ is a function from L (Ω) . For general theory of parabolic equations and systems, which include in particular solvability and uniqueness results, we refer to Ladyˇzenskaja et al [12], Dautray, Lions [1], Lions, Magenes [13] and Eidel’man [2]. Evolution 57

58 problems of the above type appear for example in biology and chemistry when studying reaction diffusion problems, see for example Murray [14] or Fife [3]. Another application can be found in multigroup diffusion in neutron physics, see Example 3, Chapter XVIIB, §3, Dautray, Lions [1]. We are concerned only with the asymptotic behaviour of solutions as t → ∞. Therefore, we suppose that the matrices Aij and A can be written as (0)

(1)

Aij (x, t) = Aij (x) + Aij (x, t)

(1.4)

A(x, t) = A(0) (x) + A(1) (x, t),

(1.5)

and (1)

where Aij and A(1) are considered as perturbations. We assume that the relation ¡ (0) ¢∗ (0) (1.6) = Aji , Aij where A∗ denotes the adjoint matrix of A, holds for every pair (i, j) and that A(0) is hermitian, i.e. ¡ (0) ¢∗ A = A(0) . (0)

The matrices Aij fulfill the two-sided inequality ν

n X

2

|ξi | ≤

i=1

n X

(0) (Aij ξj , ξi ) i,j=1

≤ν

−1

n X

|ξi |2

(1.7)

i=1

for all ξi , ξj ∈ CN and some positive constant ν. Here we use the notations (u, v) =

N X

uk vk

k=1

and |u| = (u, u)1/2 for u, v ∈ CN . The matrix A(0) is supposed to belong ¡ ¢N 2 to Lq (Ω) , where   q ∈ [n, ∞] q ∈ (2, ∞]  q ∈ [2, ∞]

if n ≥ 3 if n = 2 if n = 1.

(1.8)

Writing (0)

(0)

A(0) = A+ − A− , (0)

(0)

(1.9) (0)

where both matrices A+ and A− are positive, we require that A− is bounded. This means that there exists a constant ν1 such that (0)

kA− kL∞ (Q) ≤ ν1 .

September 6, 2006 (12:43)

(1.10)

59 Furthermore, we also assume that there exists a constant ν2 such that Z ¡ (0) ¢ A u, u dx ≤ ν2 k∇uk2L2 (Ω) (1.11) Ω

¡

¢N for every u ∈ L2 (Ω) . (0)

Under the above conditions on the matrices Aij and A(0) , there ex¡ ¢N ists an ON-basis of L2 (Ω) consisting of eigenfunctions of the timeindependent operator n X

−

(0)

(Aij uxj )xi + A(0) u

i,j=1

with the Dirichlet boundary condition. Let λk , k = 1, 2, . . . , denote the eigenvalues in increasing order and J be the multiplicity of λ1 . This means that λ1 = . . . = λJ < λJ+1 ≤ λJ+2 ≤ . . . ,

λk → ∞ as k → ∞.

Let furthermore φ1 , φ2 , . . . , φJ be a basis in the eigenspace of λ1 which is (0) orthogonal in L2 -sense. The conditions on Aij , A(0) and Ω imply that ¡ 1,p ¢N φk ∈ W (Ω) for some p > 2 and k = 1, 2, . . . , J, see Theorem A.2 in Appendix A. We also assume some similar conditions on Aij and A, see Section 2.2. The main characteristic of our perturbation is the function κ(t) =

n X

(1)

kAij kLs1 ,2 (Ct ) + kA(1) kLs2 ,1 (Ct ) ,

(1.12)

i,j=1

where Ct = Ω × (t, t + 1), s1 = and

2p , p−2

  2 s2 = s02  

2np np−2(n−p)

if n < p if n = p if n > p,

(1.13)

(1.14)

where s02 denotes an arbitrary number in (2, n]. In (1.12) we have ex(1) tended the matrices Aij , i, j = 1, . . . , n, and A(1) to Ω × R by setting (1)

Aij (x, t) = A(1) (x, t) = 0 for t < 0, so κ(t) is defined for every t ∈ R. We set κ0 = sup κ(t), (1.15) t≥0

September 6, 2006 (12:43)

60 and consider perturbations subject to κ0 ≤ κ,

(1.16) (0) Aij ,

where κ is a sufficiently small constant depending on n, N , Ω, A(0) , p, 0 s2 , ν, ν1 and ν2 . An exact value of κ is difficult to give but the requirement is that κ is so small that some inequality type conditions appearing in the (1) proof of Theorem 1.1 are satisfied. Note that κ does not depend on Aij or A(1) . ¡ 1,0;2 ¢N We define W0,loc (Q) as the space consisting of functions u vanishing on ∂Ω × (0, ∞) such that u has a weak derivate with respect to every xk ¡ ¢N and every such derivate, together with u itself, belong to L2 (Ct ) for every t ≥ 0. See further in Section 2.1.2. We let ∇ denote the gradient ¡ 2 ¢N with respect to the x-variables and define V0,loc (Q) as the subspace of ¡ 1,0;2 ¢N W0,loc (Q) consisting of functions u such that |u|Ct = ess sup ku(·, s)kL2 (Ω) + k∇ukL2 (Ct ) t<s 0, QT = Ω × (0, T ). For t ∈ R, we set Ct = Ω × (t, t + 1). ¡ ¢N Let u denote a measurable function from Ct to CN . By Lq,r (Ct ) we mean the space of all such functions with the norm ³ ´1/r  R t+1 r ku(·, s)k ds if 1 ≤ r < ∞ q L (Ω) t kukLq,r (Ct ) = (2.1) ess sup ku(·, s)k q if r = ∞ L (Ω)

t<s 0

(2.4)

and the initial condition u(x, 0) = ψ(x), (2.5) ¢ N where ψ ∈ L2 (Ω) , are valid. We suppose that the matrices Aij and A (1) can be written as (1.4) and (1.5), where Aij and A(1) are small perturbations in a sense described later. (0) Let us now give the assumptions on A(0) and Aij . We require that ¡

(0)

the symmetry condition (1.6) for matrices Aij holds and the two-sided inequality (1.7) is fulfilled. The upper inequality in (1.7) implies in particular (0) that Aij all are bounded. The matrix A(0) is supposed to be hermitian and ¡ ¢N 2 , where q is given by (1.8). We suppose also that the belong to Lq (Ω) (0) (0) matrix A admits the representation (1.9), where both matrices A+ and (0) A− are positive, and that the conditions (1.10) and (1.11) are satisfied. Let us introduce the operator L(0) u = −

n X

³ ´ (0) ∂xi Aij uxj + A(0) u

i,j=1

September 6, 2006 (12:43)

66 ¡ ¢N defined on W01,2 (Ω) . This operator has an infinite number of eigenvalues λ1 ≤ λ2 ≤ . . . ,

λk → ∞ as k → ∞,

and the corresponding eigenfunctions {φk }∞ k=1 form an ON-basis in ¡ 2 ¢N L (Ω) , see Theorem A.1 in Appendix A. We denote by p ∈ (2, ∞] some number for which ¡ ¢N φk ∈ W01,p (Ω) , k = 1, 2, . . . . (2.6) Relation (2.6) is always true for p sufficiently close to 2, see Theorem A.2 in Appendix A. We suppose that the matrices Aij and A satisfy the following conditions. The relations A∗ij = Aji , i, j = 1, . . . , n, hold and the matrix A is hermitian. With ν the same as in (1.7), we assume further that ν

n X

|ξi |2 ≤

i=1

n X

(Aij ξj , ξi ) ≤ ν −1

i,j=1

n X

|ξi |2

(2.7)

i=1

for every set of N -dimensional vectors ξ1 , . . . , ξn . We also assume that ¡ ¢N 2 A ∈ Lq,r , loc (Q) where q is the same as in (1.8) and r=

2q . 2q − n

Furthermore, writing A = A+ − A− ,

(2.8)

where A+ and A− are positive, we assume that A− is bounded by ν1 from (1.10), i.e. kA− kL∞ (Q) ≤ ν1 . (2.9) Let us show that the function κ(t) introduced in (1.12) is finite. It can be checked that 2np ≤n np − 2(n − p) for n > p ≥ 2 and since n ≤ q, it follows from (1.14) that 2 ≤ s2 ≤ q. ¡ ¢N 2 (1) Since A(1) ∈ Lq,r and Aij is bounded, the function κ(t) is finite loc (Q) for every t. We define a weak solution of (2.3) under conditions (2.4) and (2.5). We begin with studying the problem ut −

n X

(Aij uxj )xi + Au = 0

i,j=1

September 6, 2006 (12:43)

in QT ,

(2.10)

67 u(x, t) = 0

if x ∈ ∂Ω, t ∈ (0, T ),

u(x, 0) = ψ(x)

(2.12)

for some fixed, positive T . We introduce ¸ Z ·X n L1 (u, η) = (Aij uxj , ηxi ) + (Au, η) dx Ω

(2.11)

(2.13)

i,j=1

¡ ¢N and say that u ∈ V02 (QT ) is a weak solution of problem (2.10)–(2.12) if Z T Z Z ¡ ¢ − (u, ηt ) dx dt + L1 (u, η) dt = ψ(x), η(x, 0) dx QT

Ω

0

¡ ¢N for all η ∈ W01,1;2 (QT ) such that η(x, T ) = 0. It is a well-known result that problem (2.10)–(2.12) has a unique weak solution u = uT from ¡ 2 ¢N V0 (QT ) . Indeed, this result for a single equation and a system where Aij are scalars can be found in Ladyˇzenskaja et al [12], sections III.1-III.4 and VII.1. The generalization to systems of equations where Aij are not necessarily scalars is straightforward. Since this can be done for any T > 0, and the solution is unique, we obtain a unique function u defined on Q. ¡ 2 ¢N We define this function as the weak solution from V0,loc (Q) of equation (2.3) under conditions (2.4) and (2.5). Equivalently, one can say that u is a weak solution of the problem ¡ 2 ¢N (2.3)–(2.5) if u ∈ V0,loc (Q) is the unique function satisfying the relation Z Z ∞ Z ¡ ¢ − (u, ηt ) dx dt + L1 (u, η) dt = ψ(x), η(x, 0) dx (2.14) Q

0

Ω

¡

¢N for all η ∈ W01,1;2 (Q) with bounded support. The symbol C, possibly with a subscript, is frequently used to denote (0) a constant depending only on n, N , Ω, Aij , A(0) , p, s02 , ν, ν1 and ν2 . The lower-case letter c is used for constants depending on the same quantities but in situations where the symbol denotes a specific constant which may be referred to in some other part of the paper. The remaining part of the paper except Section 9 is devoted to the proof of Theorem 1.1. Without loss of generality, we can assume that λ1 = 0. Namely, if λ1 6= 0, we set U (x, t) = eλ1 t u(x, t). Then U satisfies the equation Ut −

n X

(Aij Uxj )xi + A0 U = 0

in Q

(2.15)

i,j=1

and the initial and boundary conditions (2.4), (2.5). In (2.15) we have 0 0 0 A0 = A(0) + A(1) , where A(0) = A(0) − λ1 I. Clearly, the matrix A(0) satisfies (1.10) and (1.11), possibly with other constants ν1 and ν2 .

September 6, 2006 (12:43)

68

2.3

An estimate for u

In this section, an estimate for u, to be used later in the paper, is derived. Lemma 2.1 There exists a constant a0 depending on ν, ν1 and diam Ω such that the inequality ku(·, t)kL2 (Ω) ≤ kψkL2 (Ω) ea0 t

(2.16)

is valid a.e. Proof. From (2.7) and Poincaré’s inequality it follows for an arbitrary ¡ ¢N function w ∈ W01,2 (Ω) that Z

n X

(Aij wxj , wxi ) dx ≥ νk∇wk2L2 (Ω) ≥ C1 kwk2L2 (Ω) .

(2.17)

Ω i,j=1

Furthermore, using the decomposition (2.8) and inequality (2.9) together with the fact that A+ is positive, we get (Aw, w) = (A+ w, w) − (A− w, w) ≥ −ν1 |w|2 .

(2.18)

From (2.13), (2.17) and (2.18) the inequality L1 (w, w) ≥ −a0 kwk2L2 (Ω)

(2.19)

follows, where a0 = ν1 − C1 . In the same way as in Ladyˇzenskaja et al [12] Section III.2, we can derive the equality Z Z t ¯t 1 ¯ |u(x, s)|2 dx¯ + L1 (u, u) ds = 0 (2.20) 2 Ω s=0 0 for a.e. t > 0. This is equivalent to Z t 1 1 ku(·, t)k2L2 (Ω) = kψk2L2 (Ω) − L1 (u, u) ds. (2.21) 2 2 0 ¡ 2 ¢N Since Aij is bounded and u ∈ V0,loc (Q) , we see by Hölder’s inequality that the term (Aij uxj , uxi ) occurring in the expression of L1 (u, u) is an element of L1 (Ω) for almost every t. The same is valid for (Au, u); this can be seen from the derivation of (A.3) by replacing A(0) by A. Hence we can differentiate (2.21) so that ´ 1 d ³ ku(·, t)k2L2 (Ω) = −L1 (u, u) 2 dt and it follows from (2.19) that ´ 1 d ³ ku(·, t)k2L2 (Ω) ≤ a0 ku(·, t)k2L2 (Ω) . 2 dt

September 6, 2006 (12:43)

69 Hence

´ d ³ ku(·, t)k2L2 (Ω) e−2a0 t ≤ 0 dt

so ku(·, t)k2L2 (Ω) ≤ kψk2L2 (Ω) e2a0 t . After taking the square root, inequality (2.16) follows.

3

Spectral splitting of the solution u

Let u denote the unique weak solution of (2.3) under the conditions (2.4) and (2.5) in Q as defined in Section 2.2. Setting Z ¡ ¢ hk (t) = u(·, t), φk dx, k = 1, . . . , M, Ω

we obtain the representation M X

u(x, t) =

hk (t)φk (x) + v(x, t),

(3.1)

k=1

¡ ¢N where v(·, t) is orthogonal to φk in L2 (Ω) for k = 1, 2, . . . , M . The integer M will be chosen later. Recall that it is assumed that λ1 = 0 and let J denote the multiplicity of the eigenvalue 0. This means that 0 = λ1 = . . . = λJ < λJ+1 ≤ . . . .

(3.2)

To get an equation for hk , we choose the functions η in (2.14) as η(x, t) = ξ(t)φk (x), where 1 ≤ k ≤ M and the scalar function ξ belongs Cc1 (0, ∞). This gives, after using the orthogonality between the elements φ1 , . . . , φM and v, that Z −

∞

Z

∞

ξ 0 hk dt +

ξ

0

where g R kl =

0

Z ·X n Ω

µX M

¶ g R h + g e (v) dx = 0, kl l k

l=1

¸ (Aij φl xj , φk xi ) + (Aφl , φk ) dx

i,j=1

are known functions of t and ¸ Z ·X n gek (v) = (Aij vxj , φk xi ) + (Av, φk ) dx. Ω

i,j=1

September 6, 2006 (12:43)

70 This implies that hk has a (distributional) derivate h0k and we get the system of equations h0k +

M X

g R kl hl + gek (v) = 0,

k = 1, 2, . . . , M.

(3.3)

l=1

Introducing Rkl =

Z ·X n Ω

and gk (v) =

¸ φl , φk ) dx

(3.4)

¸ (1) (Aij vxj , φk xi ) + (A(1) v, φk ) dx,

(3.5)

(1) (Aij φl xj , φk xi )

(1)

+ (A

i,j=1

Z ·X n Ω

i,j=1

we see, by using (1.4) and (1.5), that g R kl = B[φl , φk ] + Rkl and gek (v) = B[v, φk ] + gk (v), with B as given in (A.2). Since ( B[φl , φk ] =

0 λk

and

if k 6= l if k = l Z

B[v, φk ] = B[φk , v] = λk

(φk , v) dx = 0 Ω

for k = 1, . . . , M , it follows that (3.3) can be rewritten as h0k + λk hk +

M X

Rkl hl + gk (v) = 0,

k = 1, 2, . . . , M.

(3.6)

l=1

We also have the initial values Z hk (0) = (ψ, φk ) dx,

k = 1, 2, . . . , M.

(3.7)

Ω

The system of equations (3.6), (3.7) will be analyzed further in Sections 6 and 7. In order to find an equation for v we use representation (3.1) in (2.14). This time we suppose that η(·, t) is orthogonal to φk , k = 1, . . . , M , in

September 6, 2006 (12:43)

71 ¡

¢N L2 (Ω) for almost every t ∈ (0, ∞). This implies that the same orthogonality relation holds between ηt (·, t) and φk and it follows that Z

Z

−

∞

(v, ηt ) dx dt + Q

· ¸ M X L1 (v, η) + hk L1 (φk , η) dt =

0

k=1

Z

¡ ¢ ψ(x), η(x, 0) dx. (3.8)

Ω

Owing to decompositions (1.4) and (1.5), we can write (1)

L1 (φk , η) = B[φk , η] + L1 (φk , η), where (1) L1 (φk , η)

=

Z ·X n Ω

(1) (Aij φkxj , ηxi )

+ (A

(1)

¸ φk , η) dx.

i,j=1

Because of the orthogonality between φk and η, it follows from (A.4) that B[φk , η] = 0 for a.e. t and hence we can rewrite (3.8) as Z

Z

−

∞

(v, ηt ) dx dt + Q

0

¸ · M X (1) L1 (v, η) + hk L1 (φk , η) dt = k=1

Z

¡ ¢ ψ(x), η(x, 0) dx. (3.9)

Ω

4

Estimating the function v

4.1

A general estimate

¡ 2 ¢N We will now study solutions w ∈ V0,loc (Q) of the equation Z −

Z

Q

Z

∞

(w, ηt ) dx dt +

[L1 (w, η) + L2 (f , η)] dt = 0

¡ ¢ ψ(x), η(x, 0) dx

Ω

subject to the orthogonality condition Z ¡ ¢ w(x, t), φk (x) dx = 0

(4.1)

(4.2)

Ω

for k = 1, . . . , M and a.e. t. The operator L1 was defined in (2.13) and L2 (f , η) =

Z ·X n Ω

¸ (fi , ηxi ) + (f, η) dx,

i=1

September 6, 2006 (12:43)

(4.3)

72 ¡ ¢N ¡ ¢N where f ∈ L2,1 and fi ∈ L2loc (Q) , i = 1, 2, . . . , n. Equalloc (Q) ity (4.1) should be satisfied for every η from the same class of functions ¡ 1,1;2 ¢N W0 (Q) with bounded support as in (2.14) such that Z ¡ ¢ η(x, t), φk (x) dx = 0 Ω

for k = 1, . . . , M and a.e. t. In the proof of the next theorem we use the estimate Z t Z t |f (s)| ds ≤ kf kL1 (s,s+1) ds 0

(4.4)

−1

for a function f ∈ L1loc (R). This is obtained by noting that, for τ ∈ [0, 1], we have Z t Z t−τ Z 1 Z t−τ |f (s)| ds = |f (s + τ )| ds = |f (s + τ )| ds dτ 0

Z

−τ t

0

Z

1

≤

−τ t

Z

|f (s + τ )| dτ ds = −1

0

−1

kf kL1 (s,s+1) ds.

¡ ¢N ¡ ¢N Proposition 4.1 Suppose that f ∈ L2,1 and fi ∈ L2loc (Q) loc (Q) for i = 1, 2, . . . , n. Then, for any M ≥ 0, there exists a unique function ¡ 2 ¢N w ∈ V0,loc (Q) satisfying (4.1) and fulfilling (4.2) for k = 1, . . . , M and a.e. positive t. For every b > 0, there exists an integer M and a constant (0) C, both depending on b, n, N , Ω, Aij , A(0) , p, s02 , ν, ν1 and ν2 , such that the solution w is subject to the inequality µ ¶ Z t |w|Ct ≤ C kψkL2 (Ω) e−bt + e−b(t−s) χ(s) ds + χ(t) , (4.5) −1

where the norm in the left-hand side is defined by (2.2) and χ(t) =

n X

kfi kL2 (Ct ) + kf kL2,1 (Ct ) .

(4.6)

i=1

Here we have extended f and fi by 0 for negative values of t. By extending also w by 0 for t < 0, estimate (4.5) becomes valid for all t ≥ −1. The existence and uniqueness of the solution can be proved in the same way as in the case of a single equation, treated in sections III.3 and III.4 in Ladyˇzenskaja et al [12]. We confine ourself to prove inequality (4.5) under an appropriate choice of M . The proof will be divided into several steps. Proof of estimate (4.5): Step 1. Deriving a differential inequality for kw(·, t)k2L2 (Ω) . Analogously to (2.20), we can derive the equality Z Z t ¯t 1 ¯ |w(x, s)|2 dx¯ + [L1 (w, w) + L2 (f , w)] ds = 0 (4.7) 2 Ω s=t0 t0

September 6, 2006 (12:43)

73 for any t0 and t such that 0 ≤ t0 < t < ∞. This corresponds to (2.13), Chapter III in Ladyˇzenskaja et al. Since L1 (w, w) and L2 (f , w) belong to L1loc (0, ∞), it follows that kw(·, t)k2L2 (Ω) is differentiable for t > t0 and ´ 1 d ³ kw(·, t)k2L2 (Ω) = 2 dt ¸ Z ·X n n X ¡ ¢ Aij wxj , wxi + (Aw, w) + (fi , wxi ) + (f, w) dx. (4.8) − Ω

i,j=1

i=1

We now estimate the terms in the right-hand side of (4.8). The decomposition A = A+ − A− , where A+ is positive and A− is bounded, and inequality (2.9) imply that Z Z Z − (Aw, w) dx = − (A+ w, w) dx + (A− w, w) dx (4.9) Ω Ω Ω ≤ ν1 kw(·, t)k2L2 (Ω) . Together with (2.7) this yields −

Z ·X n ¡ Ω

¸ ¢ Aij wxj , wxi + (Aw, w) dx

i,j=1

≤ −νk∇w(·, t)k2L2 (Ω) + ν1 kw(·, t)k2L2 (Ω) .

(4.10)

Since w is orthogonal to φ1 , . . . , φM , we have Z ·X n ³ ´ ³ ´¸ (0) Aij wxj , wxi + A(0) w, w dx ≥ λM +1 kw(·, t)k2L2 (Ω) , Ω

i,j=1

cf. (A.5) in Appendix A. From this fact, together with the upper inequality in (1.7) and property (1.11), we obtain Z ·X n ³ ´ ³ ´¸ (0) (0) 2 k∇w(·, t)kL2 (Ω) ≥ ν Aij wxj , wxi + A w, w dx Ω

i,j=1

Z ³ ´ −ν A(0) w, w dx Ω

≥ νλM +1 kw(·, t)k2L2 (Ω) − νν2 k∇w(·, t)k2L2 (Ω) and hence

νλM +1 kw(·, t)k2L2 (Ω) . 1 + νν2 This gives together with (4.10) the estimate k∇w(·, t)k2L2 (Ω) ≥

−

Z ·X n ¡ Ω

¢

¸

Aij wxj , wxi + (Aw, w) dx

i,j=1

ν ≤ − k∇w(·, t)k2L2 (Ω) − bkw(·, t)k2L2 (Ω) , 2

September 6, 2006 (12:43)

74 where

ν 2 λM +1 − ν1 . 2(1 + νν2 )

b=

Since λM +1 → ∞ as M → ∞, the constant b can be made arbitrarily large. We continue to estimate the terms in the right-hand side of (4.8) and get µX ¶ Z X n n − (fi , wxi ) dx ≤ kfi (·, t)kL2 (Ω) k∇w(·, t)kL2 (Ω) Ω i=1

i=1

1 ν ≤ k∇w(·, t)k2L2 (Ω) + 2 2ν

µX n

¶2 kfi (·, t)kL2 (Ω)

,

i=1

(4.11) where we have used the elementary inequality ab ≤

α 2 1 2 a + b , 2 2α

α > 0,

in the last line. Also, Z − (f, w) dx ≤ kf (·, t)kL2 (Ω) kw(·, t)kL2 (Ω) Ω

and when combining all estimates with (4.8), we get 1 d (kw(·, t)k2L2 (Ω) ) ≤ −bkw(·, t)k2L2 (Ω) 2 dt 1 + kf (·, t)kL2 (Ω) kw(·, t)kL2 (Ω) + 2ν

µX n

¶2 kfi (·, t)kL2 (Ω)

. (4.12)

i=1

Choose M so large that b becomes positive and set h(t) = kw(·, t)k2L2 (Ω) and

µX n

1 ν

G(t) =

¶2 kfi (·, t)kL2 (Ω)

.

(4.13)

i=1

Then, from (4.12), we obtain the differential inequality h0 (t) + 2bh(t) ≤ 2kf (·, t)kL2 (Ω) h(t)1/2 + G(t),

t > 0.

(4.14)

Step 2. Finding an estimate for kw(·, t)k2L2 (Ω) . We set h+ (t) =2h(0)e

−2bt

µZ + 12 0

t

3 + 2

Z

t

e−2b(t−τ ) G(τ ) dτ

0

e−b(t−τ ) kf (·, τ )kL2 (Ω) dτ

September 6, 2006 (12:43)

¶2 .

(4.15)

75 Let us show that h(t) ≤ h+ (t). We begin by proving that Z t ³ ´ h+ (t) ≥ h(0)e−2bt + e−2b(t−τ ) 2kf (·, τ )kL2 (Ω) h+ (τ )1/2 + G(τ ) dτ. 0

(4.16) Let us denote the right-hand side by r(t). When inserting the expression √ √ √ for h+ in r(t) and using the inequality a + b ≤ a + b for a, b ≥ 0, we obtain · Z t p −2bt −2b(t−τ ) r(t) ≤h(0)e 2 2h(0)kf (·, τ )kL2 (Ω) e−bτ + e 0

√ + 6kf (·, τ )kL2 (Ω)

µZ

Z √ + 4 3kf (·, τ )kL2 (Ω)

τ

0 τ 0

¶1/2 e−2b(τ −s) G(s) ds ¸ e−b(τ −s) kf (·, s)kL2 (Ω) ds + G(τ ) dτ. (4.17)

We estimate the terms in (4.17) one by one, beginning with Z t p e−2b(t−τ ) kf (·, τ )kL2 (Ω) e−bτ dτ 2 2h(0) 0 Z t p −bt = 2 2h(0)e e−b(t−τ ) kf (·, τ )kL2 (Ω) dτ 0

µZ

t

≤ h(0)e−2bt + 2 0

Since τ ≤ t, we have µZ √ Z t −2b(t−τ ) 6 e kf (·, τ )kL2 (Ω) 0

≤

√

t

6

e

e

−2b(τ −s)

kf (·, τ )kL2 (Ω) dτ

e

−b(t−τ )

kf (·, τ )k

L2 (Ω)

(4.18)

t

e

−2b(t−s)

¶1/2 G(s) ds

0

1 + 2

Z 0

t

e−2b(t−s) G(s) ds (4.19)

¶ e−b(τ −s) kf (·, s)kL2 (Ω) ds dτ 0 0 µZ t ¶ √ Z t −b(t−τ ) −b(t−s) ≤4 3 e kf (·, s)kL2 (Ω) ds dτ e kf (·, τ )kL2 (Ω)

√ Z t −2b(t−τ ) e kf (·, τ )kL2 (Ω) 4 3

τ

0

0

√

µZ

dτ

.

¶1/2 G(s) ds dτ

¶ µZ ¶2

t

0

and

−b(t−τ )

0

≤3

τ

0

µZ

µZ

¶2 e−b(t−τ ) kf (·, τ )kL2 (Ω) dτ

µZ

t

=4 3 0

e−b(t−τ ) kf (·, τ )kL2 (Ω) dτ

¶2 . (4.20)

September 6, 2006 (12:43)

76 Using (4.18), (4.19) and (4.20) in (4.17), we arrive at Z 3 t −2b(t−τ ) r(t) ≤ 2h(0)e−2bt + e G(τ ) dτ 2 0 µZ t ¶2 ³ √ ´ −b(t−τ ) e + 5+4 3 kf (·, τ )kL2 (Ω) dτ 0

≤ h+ (t). This implies (4.16). We now prove that h(t) ≤ h+ (t) and conclude first that both of the functions h and h+ are continuous. For h+ , this fact follows from (4.15) since G as well as kf kL2 (Ω) belong to L1loc (0, ∞). Furthermore, the continuity of h follows from (4.7). Suppose first that h(0) > 0. Then h+ (0) = 2h(0) > h(0) and let us prove that h+ (t) > h(t) for all t ∈ [0, ∞). Namely, suppose that h(τ ) < h+ (τ ) if τ ∈ [0, t) for some t > 0 (obviously, such a t exists because of the continuity of h and h+ ). From (4.14) it follows that ¡

he2bt

¢0

≤ 2e2bt kf (·, t)kL2 (Ω) h(t)1/2 + e2bt G(t)

and hence Z h(t) ≤ h(0)e−2bt + 0

t

³ ´ e−2b(t−τ ) 2kf (·, τ )kL2 (Ω) h(τ )1/2 + G(τ ) dτ.

(4.21) There are two possibilities. Either kf (·, τ )kL2 (Ω) = 0 for almost every τ ∈ [0, t] or else kf (·, τ )kL2 (Ω) > 0 on a subset of [0, t] with positive measure. In the first case it follows from (4.21) that Z t −2bt h(t) ≤ h(0)e + e−2b(t−τ ) G(τ ) dτ 0

and when comparing this with (4.15), we see that h(t) < h+ (t). Let us now consider the case when kf (·, τ )kL2 (Ω) is not identically 0 on [0, t], still assuming that h(0) > 0 and h < h+ on [0, t). From (4.21) and (4.16) we get that Z t ´ ³ h+ (t) − h(t) ≥ 2 e−2b(t−τ ) kf (·, τ )kL2 (Ω) h+ (τ )1/2 − h(τ )1/2 dτ > 0, 0

so also in this case h(t) < h+ (t). The conclusion is that the set {t ≥ 0 : h(τ ) < h+ (τ ) for τ ∈ [0, t]} is non-empty, open (because of the continuity of h and h+ ) and closed in [0, ∞). The last statement follows from the preceding analysis. Hence,

September 6, 2006 (12:43)

77 it must be equal to [0, ∞) and we have proved that if h(0) > 0, then h(t) < h+ (t) for every t ≥ 0. We now consider the case h(0) = 0. For a given ² > 0, introduce h² (t) = h(t) + ²e−2bt and h²+ (t) =2h² (0)e−2bt + µZ

t

+ 12

e

Z

3 2

t

e−2b(t−τ ) G(τ ) dτ

0

−b(t−τ )

0

¶2 kf (·, τ )kL2 (Ω) dτ

.

(4.22)

Since h1/2 ≤ (h² )1/2 , it follows from (4.21) that Z t ³ ´ h² (t) ≤ h² (0)e−2bt + e−2b(t−τ ) 2kf (·, τ )kL2 (Ω) h² (τ )1/2 + G(τ ) dτ. 0

(4.23) The relations (4.22) and (4.23) are the same as (4.15) and (4.21) with the only difference that h and h+ are replaced by h² and h²+ , respectively and also (4.16) holds with the same substitutions. Obviously, h² (0) = ² > 0, so the previous case h(0) > 0 shows that h² (t) < h²+ (t) for all t ≥ 0. The last inequality says that h(t) + ²e−2bt < h+ (t) + 2²e−2bt and by letting ² tend to 0 we see that h(t) ≤ h+ (t) for every t ≥ 0. We have thus obtained the inequality Z 3 t −2b(t−s) 2 2 −2bt kw(·, t)kL2 (Ω) ≤2kψkL2 (Ω) e + e G(s) ds 2 0 µZ t ¶2 −b(t−s) 2 + 12 e kf (·, s)kL (Ω) ds . (4.24) 0

Step 3. Finding an estimate for kw(·, t)kL2 (Ω) . Our next aim is to find an estimate for kw(·, t)kL2 (Ω) . This will be done departing from (4.24). p Before doing this, we set γ(t) = G(t) and prove that the inequality µZ

t

e 0

−2b(t−s)

¶1/2 Z b γ(s) ds ≤e 2

t

−1

e−b(t−s) kγkL2 (s,s+1) ds,

(4.25)

where γ has been extended by 0 for t < 0, holds. It is readily verified that if τ ∈ [0, 1], then µZ 0

t

¶1/2 Ã X dte Z −2b(t−s) 2 e γ(s) ds ≤ k=0

k+1−τ k−τ

September 6, 2006 (12:43)

!1/2 e−2b(t−s) γ(s)2 ds

,

78 where dte denotes the least integer larger than or equal to t. Since the left-hand side is independent of τ , we integrate the inequality from 0 to 1 with respect to τ and obtain µZ

t

e

−2b(t−s)

¶1/2 Z γ(s) ds ≤

1

2

0

Ã

0

dte Z X

!1/2

k+1−τ

e

−2b(t−s)

2

γ(s) ds

dτ.

k−τ

k=0

√ √ √ By using the inequality a + b ≤ a + b for a, b ≥ 0, we can move the sum outside the parentheses. This gives the inequality µZ

¶1/2

t

e

−2b(t−s)

2

≤ eb

γ(s) ds

0

dte Z X

1

0

k=0

e−b(t+τ −k) kγkL2 (k−τ,k−τ +1) dτ. (4.26)

In the last integral we make the substitution s = k − τ . This yields dte Z X 0

k=0

=

1

dte Z X

e−b(t+τ −k) kγkL2 (k−τ,k−τ +1) dτ k

k−1

k=0

Z e−b(t−s) kγkL2 (s,s+1) ds =

dte

e−b(t−s) kγkL2 (s,s+1) ds. (4.27)

−1

The left-hand side of (4.25) shows that we can set γ(s) = 0 for s > t. Thus, (4.25) follows from (4.26) and (4.27). In the remaining part of the proof, we let C denote a constant which (0) may depend on n, N , Ω, Aij , A(0) , p, s02 , ν, ν1 and ν2 and, additionally, b. We use (4.24) and get the estimate r µZ t ¶1/2 3 e−2b(t−s) G(s) ds 2 0

√ kw(·, t)kL2 (Ω) ≤ 2kψkL2 (Ω) e−bt +

√ Z t −b(t−s) +2 3 e kf (·, s)kL2 (Ω) ds. 0

An application of (4.25) yields, after using (4.13), that µ n Z X kw(·, t)kL2 (Ω) ≤ C kψkL2 (Ω) e−bt + Z +

i=1

t

e

−b(t−s)

0

t

−1

e−b(t−s) kfi kL2 (Cs ) ds

¶ kf (·, s)kL2 (Ω) ds .

(4.28)

By using inequality (4.4), we obtain Z 0

t

Z e−b(t−s) kf (·, s)kL2 (Ω) ds ≤ eb

t

−1

e−b(t−s) kf kL2,1 (Cs ) ds

September 6, 2006 (12:43)

(4.29)

79 and (4.28) becomes

µ Z kw(·, t)kL2 (Ω) ≤ C kψkL2 (Ω) e−bt +

t

¶ e−b(t−s) χ(s) ds ,

(4.30)

−1

with χ as defined in (4.6). The convention that w(x, t) = 0 for t < 0 makes (4.30) valid for all t ≥ −1. Step 4. Completing the proof of (4.5). We now go on with deriving the estimate (4.5). Fix a value of t > −1 and introduce the matrices ( Aij (x, s) if s ≤ t + 1 A˜ij (x, s) = 0 if s > t + 1 and

( ˜ s) = A(x, s) if s ≤ t + 1 A(x, 0 if s > t + 1.

(0) (1) The matrices A˜ij , A˜ij , A˜(0) , A˜(1) and the functions f˜i , f˜ are defined ˜ f˜i , f˜ in (4.1) and analogously. By replacing Aij , A, fi , f with A˜ij , A, (4.3), we see that the solution w is unchanged on the interval [0, t + 1]. Introduce n X χ(s) ˜ = kf˜i kL2 (Cs ) + kf˜kL2,1 (Cs ) . i=1

For ² ∈ [0, 1], inequality (4.30) yields that µ Z kw(·, t + ²)kL2 (Ω) ≤ C kψkL2 (Ω) e−bt + Z

t

e−b(t−s) χ(s) ˜ ds

−1

t+²

+

¶ e−b(t−s) χ(s) ˜ ds .

t

Since obviously χ(s) ˜ ≤ χ(t) for t ≤ s ≤ t + 1, it follows that µ ¶ Z t ess sup kw(·, s)kL2 (Ω) ≤ C kψkL2 (Ω) e−bt + e−b(t−s) χ(s) ds + χ(t) . t<s

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