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1. Assume first that u, v and 0 have compact supports, and so belong to L 1 (R). Since 1 l p + l /q + 1/r ; = 2, if we put ilr (x) (-x), then by Exercises 3.2 and 3.3, I(u * v,4)I = I(u * v * c)(O)I j, 1
Xr (x)
and put uj = X1 u, vj = Xjv and ¢j = Xj 4,. Since (I1J I * It; 1, I4) I) -< IIuj IIL,,(R") III; IILq(R")II4); IIL,.(R")
< IIuIILp(R")IIVIILq(R")II4IIL,.(R")
for all j, the monotone convergence theorem implies that I(u * v, 4))I _< (lul * Ivl, I4I) R v(y)P' d) 1/p
< (fly-xlE,
r
J
*,(x) dx = 1.
These properties of 'VE mean that (`YE * u)(x) is a kind of local average of u around x, and for this reason 'YE * u ti u when c is small.
Differentiation
63
Theorem 3.4 Let 1 < p < oo. If `/. E is as above, and if u E L p (R" ), then
and II'/'E * U - uII L,(R") < co,,(, Eu),
11 *c * uII Lp(1R") < IIUIIL,,(R")
u in Lp(R") as e 4. 0.
so VE * u
Proof. Since 11 *E
IIL,(R") =
1, we see from Theorem 3.1 that
IIi *UIILn(R") < II1IreIIL,(R")IIUIILp(R") = IIUIIL,,(R")
Also,
'- * u(x) - u(x) = u * 'VE(x) - u(x)
0,., 0,., (Y) dY R"
and so, for any 0 E L p. (W'), Holder's inequality implies that
I(*E*u-u,0)I
=I fYISe
R*E(Y)
f [u(x - y) - u(x)]O(x) dx dy
I
J
'Y E(Y)Wp(E,
u)IIOIILr,+(R^) dy = Wp(E,
Y1 <E
and the desired estimate for ikE * u - u follows.
Corollary 3.5 If I < p < oo, then C,,,,,p(S2) is dense in Lp(SZ).
Proof. We choose any K j C S2 such that KI S; K2 e ... and 0 = U1 K); for example,
Kj = {x E S2 : dist(x, R" \ SZ) > 1/j and Ixj < j).
Let U E L(), define U j E L(R) by u (x) =
u(x) if x E Kj,
ifx E1R' \Kj,
0
and consider the function uj,E =
YkE * u j.
Theorem 3.3 shows that U j., E
C a ,p(R"), Exercise 3.4 shows that supp u j,E C Q
if E < dist(K1, R" \ 0),
and by Theorem 3.4 Iluj.E - UIILE,(c2) S IIUE.j - uj IIL,(R1') + IIuJ - uII L,,(s )
< Wp(e, U j) + IIuIIL,,(Q\KJ)
64
Sobolev Spaces
Since II U II L,,(s,\K) --). 0 as j -+ oo, and since cop (E, uj) -* 0 as c -* 0 for each fixed j, the result follows. We can also use convolution to smooth out the characteristic function of a set.
Theorem 3.6 Let F be a closed subset of R". For each E > 0, there exists XE E C°°(R') satisfying
0 < X, ,(x) < 1 and I aaXE (x) I < CE-IaI
if x E F, if 0 < dist(x, F) < E,
XE (x) = 0
if dist(x, F) > E.
XE (x) = 1
Proof Define vE E L°O(]R") by J1
VE(x)
0
if dist(x, F) < E, if dist(x, F) > E,
so that vE = I on a neighbourhood of F, and choose 1/r E C mp(R) satisfying (3.9). With the help of Exercise 3.4, one sees that the function XE _ *E/a * VE/2 has the required properties.
Schwartz Distributions
A (measurable) function u : 0 -* C is said to be locally integrable if u is absolutely integrable on every compact subset of Q. We denote the set of all such functions by L1,1oc(52). The following observation is the starting point for the theory of distributions.
Theorem 3.7 If u, v E L1,10 (0) satisfy
1 u(x)q(x) dx = J v(x)o(x) dx for all O E C mp(S2), then u = v almost everywhere on Q. Proof. Let K C= 52, and choose an open set 521 such that K C= S21 c S21 Cc Q.
We define f E L1(R") by
f(x)=
J
u(x) - v(x) if x E 521, 0
ifxER"\521,
'YE be as in Theorem 3.4. There is an co > 0 such that if x E K and 0 < E < co then (x - -) E C mp (521) and therefore (Vi, * f) (x) = (f, *E(x - .))tt,, = (u - v, 'YE(x - .))a = 0. Since '/FE * f -* f in L1(R"),
and let
Yk E
Schwartz Distributions
65
it follows that f = 0 almost everywhere on K, i.e., u = v almost everywhere on K. 0 Theorem 3.7 shows that a locally integrable function u is uniquely determined by its associated linear functional 4) H (u, 0) a. We wish to introduce a larger
class of linear functionals on C,'O.P (0). Following the notation introduced by Schwartz [92], put
E(Q) =
C°O(S2),
DK(S2) = CK (S2),
V(S2) = C mP(S2),
for any K C= Q. Since we want our functionals to be continuous in an appropriate sense, we now define convergence of sequences in each of these function spaces. No deeper properties of the underlying locally convex topologies will be used; cf. [106]. Let (4j) J _ , be a sequence in E (S2). We write
cj -+ 0 in S(Q) if, for each compact set K and for each multi-index a,
8'Oj -* 0 uniformly on K. When, for a fixed K, this condition holds for all a, and in addition supp c! c K for all j, we write
cj - 0 in DK (S2). 0 in D(7) means that 4j -+ 0 in DK(S2) for some K C= 0. Convergence to a non-zero function is then defined in the obvious way. For instance, cj --* ¢ in DK (0) means ¢ E DK (0), 4j E DK (0) for all j, and ¢j - 4) -3 0
Finally, 4j
in DK (S2).
Consider an abstract linear functional f : D(52) - C. For the moment, we write e(0) to denote the value off at ¢ E D(S2). If f is sequentially continuous, i.e., if for every sequence ¢j in D(S2), c1 -* 0 in D(S2)
implies t(4 f) -+ 0,
then t is called a (Schwartz) distribution on Q. In this context, the elements of D(Q) are referred to as test functions on 0. The set of all distributions on S2 is denoted by D*(0). Associated with each u E L 1,1 (0) is the linear functional to defined by (Lu)(4)) = (u, 4))c
for 0 E D(S2).
(3.11)
66
Sobolev Spaces
It is clear that to : D(7) -f C is sequentially continuous, and hence a distribution on 92. Furthermore, Theorem 3.7 shows that the linear map i : L 1,1"C (0) --+
D*(S2) is one-one. Hence, we may identify u with tu, and thereby make L1,1oc(92) into a subspace of D*(92). Those distributions that are not locally integrable can then be viewed as generalised functions.
As an example, fix x E R", and let E E D*(92) be the associated point evaluation functional defined by
f(0) = 0(x) for 0 E D(R"). It is not possible to represent this e by a locally integrable function. To see why,
suppose for a contradiction that £ = to for some u E L1.1oc(R"). It follows from Theorem 3.7 that u = 0 almost everywhere on R" \ (x}, and since the set (x) has measure zero, this means that u = 0 almost everywhere on R". Hence, ¢(x) = (u, 4)) = 0 for all 0 E D(R"), a contradiction. Following the convention introduced by Dirac, we denote the point evaluation functional for x by S., or in the case x = 0, just by S. In keeping with the philosophy that distributions are generalised functions, we henceforth write (u, 0) n for the value of u E 'D* (0) at 0 E D(92), whether or not u is locally integrable on 0. If 92 = R", then we usually omit the subscript, and just write (u, 0); for instance,
(S. ,0) =0(x) for0 E D(R"). Suppose that 921 is an open subset of 0, and for any ¢ E D(921) let E D(92) denote the extension of ¢ by zero. For any distribution u E D*(92) the restriction u I n, E D* (921) is defined by
(uIn,, 0)si, = (u, )st for 0 E D(9Z1). We say that u = 0 on S21 if u I a, = 0, and define supp u to be the largest relatively closed subset of Q such that u = 0 on n \ supp U. If U E L1,1,, (92), then the distributional support of u is the same as its essential support as a function, i.e., supp u is the largest closed set such that u = 0 almost everywhere on S2 \ supp u. We define E*(92) in the obvious way, i.e., a linear functional u : E(Q) -+ C
belongs to E* (Q) if it is sequentially continuous: (u, 4 )n -+ 0 whenever ¢J -} 0 in E(92). Exercise 3.8 shows that the inclusion D(92) g E(S2) is continuous and dense, so E*(92) S; D*(92). In fact, we can characterise E*(92) as follows.
Schwartz Distributions
67
Theorem 3.8 The space E* (S2) coincides with the space of distributions having compact support, i.e., E* (Q) = [U E D* (0) : supp u C= S2).
Proof Suppose that u E D*(S2) and supp u C= S2. By Theorem 3.6, there is
a X E D(l) with X = 1 on a neighbourhood of supp u. We define a linear functional ii on E(52) by putting for ¢ E E(SZ),
and claim that
(i) O; -+ 0 in E(S2) implies (u, c;)Q --). (u, O)g; (ii) (il, ¢)o = (u, O)n for ¢ E D(S2). Together, these facts mean that u = u E E* (S2) when E* (S2) is viewed as a subspace of D*(S2). We remark that u does not depend on the choice of X. To prove (i), assume Oj --+ 0 in E(0). It follows at once that XOj XO
in D(cz), so (u, oi)n = (u, XOi) - (u, xO)si = (u, 0)cz. To prove (ii), assume ¢ E D(S2). Since (1 - X)0 = 0 on a neighbourhood of supp u, we have (u, (I - X)0)a = 0 and therefore (u, 0) _ (u, 0) 52 (u, (1 - x)O)si _ (u, A2. Conversely, let u E E*(S2), and suppose for a contradiction that supp u is not compact. Choose an increasing sequence of compact sets KI c K2 c .. . with S2 = U,° , K j, then u I a\KJ # 0 for all j. Thus, we can find 4 E V (S2) such that (u, cj)n = 1 and supp4J Cc n \ K3. It follows that t -+ 0 in E(S2), 0, a contradiction. so (u, /j)n
The restriction map u r-+ u In, is just one of many linear operations that can be extended from functions to distributions. For instance, to define partial differentiation of distributions one formally integrates by parts:
(ac'u, cn = (-1)I0" (u, aacc
for u E D*(S2) and 0 E D(0).
Here, the sequential continuity of 8au follows at once from the fact that if 0 in D(S2), then 8a0j ->- 0 in D(S2). Also, we define the complex ¢j conjugate u E D*(S2) of U E D*(S2) by
(u, 95) = (u, 0)
for 0 E D(S2),
Sobolev Spaces
68
and we generalise the meaning of the inner product in L2(0),
(u, On =
Jsi
u(x)v(x)dx,
by putting
(u, 4)n = (u, ¢)n
for u E D*(0) and 0 E D(S2).
When S2 = l(8", we just write (u, 0). The convolution of a distribution with a test function is defined in the obvious way,
(u * 0)(x) = (u, i0(x - ))
forx E R1, U E
D*(R")
and 0 E D(R"). (3.12)
Thus, in particular, S * 0 _ 0, or formally fyt. 8(x - y)o (y) dy = 0(x), so we can think of 8(x - y) as the continuous analogue of Kronecker's Stk. Finally, multiplication of a distribution u E D*(S2) by a smooth function * E COO(Q) is defined by
(ifu, q )n = (u, *0)j2
for0 E D(S2).
In each of the above examples, Theorem 3.7 guarantees that the generalised concept is consistent with the classical one. For instance, if the classical partial derivative aj u exists and is locally integrable on 0, then 8i (cu) = c (a; u), where i is the imbedding defined in (3.11). Distributions are a powerful conceptual tool for the study of partial differential equations, and provide, in particular, a very effective system of notation. Fortunately, we shall require few technical results from the theory of distributions. However, the following fact will be used.
Theorem 3.9 Suppose that u E D* (0) and X E Q. If supp u C (x }, then for some m > 0,
u = E ac a"3,, on 0, 1" I I. Define XE(Y) = X(E-,(y -X)) so that XE = 1 on B,/2 (x), and xE = 0 outside BE(x). Hence, (1 0 on a neighbourhood of suppu = {x}, so (u, (1 - XE)42) = 0 and therefore Since laaXEI < CE-I"l and (u,102) = (u, [x + (1 - XE)14)2) = (u, 18014611
C is L -periodic if
u(x + kL) = u(x)
for x E R" and k E V,
or in other words, if u is L-periodic in each of its n variables. We can think of
70
Sobolev Spaces
such a function as being defined on the additive quotient group TL = R" / (L7G" ),
and introduce the L2 inner product
(u, v)T =
u(x)v(x) dx,
J
where integration over TL just means integration over any translate of the cube (0, L)". Using separation of variables, we can easily find the normalised L-periodic eigenfunctions of the Laplacian (1.1): for k E Z", the function 'bk(x) = satisfies
-Aok = (27rIkI)2k
on TL,
and II0kIIL2(rL) = 1. Since -0 is self-adjoint, we expect from Theorem 2.37 that the Ok will form a complete orthonormal system in L2(TL). Thus, for a general L-periodic function u, we should have
u(x) = 1: (Ok, u)r bk(x) =
1L"
uL(kIL)e'22r(k/c).X,
(3.14)
kEZ
kEZ"
where
UL( ) =
JaI
e-ibrg-xu(x)dx.
The analogous expansion for a non-periodic function can be viewed as arising in the limit as L -+ 00. If U E L 1(1R' ), then we define its Fourier transform u =
2u by
u(t;) _ x {u(x)) =
fp
dx fort E
!lam"
and expect from (3.14) that, under appropriate conditions, u = .F*u, where.F* is the adjoint of the integral operator T, i.e.,
u(x) _
-,x{u(t)} = f
(3.15)
for x E W. In fact, we can readily prove the following.
Theorem 3.10 If both u and u = 'Fu belong to Li(R"), then the Fourier inversion formula (3.15) is valid at every point x where u is continuous.
Fourier Transforms
71
Proof. Let *(x) = e-"1"12 and `YE (x) = E-"* (e -'x). Exercise 3.11 shows that 1/r is invariant under the Fourier transform: F* = 1/r = -7-1*. Therefore, by and .F*1i E _ 'YE In other words, the inversion Exercise 3.12, formula is valid for *E, implying that
j
J
d
d =
fR. Cfa
R11
l.
d dY
= fit" u(Y) J
f,u(Y)*E(x - Y)dy 1 as c y 0 for each 1f Since 7' dominated convergence theorem that
fRna(')ei2'
E IR", we deduce from the
d = lim(1/*E * u)(x).
(3.16)
E10
Assuming u is continuous at x, let co > 0 and choose So > 0 such that Iu(x - y) - u(x)I < co for IYI < So. Observe that fR YE(Y) dy = E(0) = 1, so for 0 < E < co,
f
[u (x - y) - u(x)]*E(Y)dy
Iu(x - y) - u(x)IfE(Y)dy lyl-So
< Eo f *e (Y) dY
+
(f Iu(x - y) - u(x)I dy) \ at^
sup *E(Y) lyl?ao
"(60/E)'
Eo +2IIuIIL,(R^)E-"e
and the result follows at once from (3.16).
0
Corollary 3.11 If both u and u are continuous everywhere and belong to L 1(IR!1), then.F*.Fu = u = .F.F* u.
72
Sobolev Spaces
We now consider the actions of .7 and 1* on the Schwartz space of rapidly decreasing, C°° functions, 8(W1)
E C°°(R") : sup Ix"afl¢(x)I < oo XER"
for all multi-indices a and
}.
Sequential convergence in this space is defined by interpreting the statement
¢j -+ 0 in S(R't) to mean that, for all multi-indices a and 0,
x"0O(x) -+0 uniformly forx E R". Elementary calculations show that if 0 E S(R"), then and
F:,,g[(-i27rx)"O(x)} (3.17)
so the Fourier transform defines a (sequentially) continuous linear operator
.F : S(R") -* S(R"). Moreover, by Corollary 3.11, this operator has a continuous inverse, namely
the adjoint.1'' : S(R") -* S(R"). By Exercise 3.8, the inclusions D(R") C S(RI) C E(R") are continuous with dense image, so we have
E*(1(g") c S*(R") c D*(
),
and the elements of S*(R"), i.e., the continuous linear functionals on S(IR'1), are called temperate distributions. A sufficient condition for a function u E LI,,,,(R") to be a temperate distribution is that it is slowly growing: u(x) _ O (Ix I'') as Ix I -+ oo, for some r. The formulae
(J u, 0) = (u, TO) and
(.F*u, ') = (u, F 4 )
are obviously valid if both u and 0 belong to S(R"), and serve to define extensions
S* (R") -). S*(R")
and
J'" : S* (R") -> S*(k8")
We also have the following result, known as Plancherel's theorem.
Sobolev Spaces - First Definition
73
Theorem 3.12 The Fourier transform and its adjoint determine bounded linear operators
.F: L2(R") -a L2(R") and F* : L2(R") -+ L2(
n
with .F-' = .F". Furthermore, these operators are unitary: (.Fu, .Fv) _ (u, v) = (.F*u, .F*v) for u, v E L2(Il8").
Proof. If u, v E S(R"), then (.Fu,.Fv) = (.F*.Fu, v), and .F*.Fu = u by Corollary 3.11, because u E S(W) c LI(W), so (.Fu, Fv) = (u, v). In particular, taking v = u, we see that II.FuIIL2(Rn) = IIuIIL2(RI) for U E S(Rn). Corollary 3.5 implies that S(R") is dense in L2 (R"), so F has a unique extension from S(R") to a unitary operator on L2(Illi"). Furthermore, this extension satisfies (.Fu, 0) _ (u,.Fo) for all ¢ E S(lR"), consistent with the definition of Fu as a temperate distribution. In other words, the extension from S(R") agrees with the restriction from S*(If8"). Similar arguments yield the same reO
sults for .F*.
Corollary 3.13 The Fourier transform preserves the L2-norm: III fIL,(R'") _ IIuIIL2(1R")
a
Another important fact about the Fourier transform is its effect on convolutions: if u, v E L (Ilk") then
V) W) =
J
J
u(x - y)v(Y)dydx
f f e-1br(x-y)-4u(x - y)
aw
J
dy
= h()v() Sobolev Spaces - First Definition Suppose I < p < oo, and let 0 be a non-empty open subset of R". The Sobolev space WP' (n) of order r based on Ln(S2) is defined by
Wp(E2) = {u E Lp(Q) : 8Yu E L,(c2) for Iaj < r}. Here, of course, 8au is viewed as a distribution on n, so the condition 8"u E L p (Q) means that there exists a function g,, E L p (n) such that (u, 8"O) s- = (-1)I"I (ga, O)n for all 0 E D(S2), or equivalently 8'u = tga where t : L1j,(S2)
Sobolev Spaces
74
-+ D*(0) is the imbedding defined by (3.11). Such a function ga is often described as a weak partial derivative of u. The completeness of L p (S2) implies that WP (0) becomes a Banach space on putting
Ilullw;,(12) _
(tr1PdX)
1/p
To define Sobolev spaces of fractional order, we denote the Slobodeckil seminorm by
lu(x) - u(y)Ip lulu,p>sa =
CJsi J
oo
I x- y I"+P"L
dx dy
)1/p
for 0 < µ < 1.
(3.18)
Notice that the integrand is the pth power of lu(x) - u(y)I/lx - yliL+"/p, so for p = oo we get the usual Holder seminorm. For s = r + µ, we define
Wp(l) = {u E W,(2): Iaaulµ.p.n < oo for lal = r}, and equip this space with the norm
Ilullw;(n) =
E Iaaulµ.P", Ia1=r
For any integer r > 1, the negative-order space WT (S2) is defined to be the space of distributions u E D* (Q) that admit a representation
u = E as fa
with f, E L p(S2).
(3.19)
lal 0.
Sobolev Spaces - Second Definition In this section, we introduce a second family of Sobolev spaces, which later will turn out to be equivalent to the one given in the preceding section. For S E R, we define a continuous linear operator ,75 : S(RI) S(lR"), called the Bessel potential of order s, by
d for x E R.
J5u(x) = fR11 (1 +
In this way,
x-. (,75u(x)} _ (1 + I I2)s"2u( ),
(3.20)
so under Fourier transformation the action of ?S is to multiply u by a function that is D(I IS) for large . We can therefore think of ?SEas a kind of differential
operator of orders; cf. (3.17). Notice that for all s, t
Js+' = JS Jt ,
R,
= J--, Jo = identity operator.
It follows from (3.20) that
(Yu, v) = (u, ,75v)
and
(.75u, v) _ (u, .7sv)
for all u, v E S (R"), giving a natural extension of the Bessel potential to a linear operator ?5 : S* (W) -+ S* (WI) on the space of temperate distributions.
Sobolev Spaces
76
For any s E R, we define HS(R"), the Sobolev space of order s on R", by Hs(R??) = {u E S*(JRf) : JSu E L2(R")),
and equip this space with the inner product (u, v)H=(R") = (JSu, 3Sv)
and the induced norm (3.21)
(u, u)H3(R") =
Notice that the Bessel potential
,7S : HS(R") - L2(R") is a unitary isomorphism, and in particular, since J°u = u,
H°(R") =L2(R"). Several facts about HS(R") follow immediately from standard properties of L2(R ). For instance, HS(R") is a separable Hilbert space, and D(R") is
dense in HS(R") because ,75[S(R')] = S(R") is dense in L2(R"), and the inclusion D(R") c S(W) is continuous with dense image. Also, one sees from (3.2) and (3.3), with p = 2, that H-S(R") is an isometric realisation of the dual space of HS (RII ), i.e.,
H-S(R") = [HS(R")]*
fors E R,
(3.22)
and sup
I(u, v)1
OOVEH3(R") IIuIIH=(R")
=
sup
1(u, v)I
o#vEH=(R") IIVIIHI(R")
for U E H-'(RI). Plancherel's theorem (Theorem 3.12) and (3.20) imply that IIUIIH=(R,t) =
j(1 + 12u2d, "
so if s < t then IIuIIH=(R") < IIuIlH,(R") and hence H'(R") c HS(R"). This inclusion is continuous with dense image. For any closed set F C R", we define the associated Sobolev space of order s by
HF.=(uEHS(R"):suppucF),
Sobolev Spaces - Second Definition
77
whereas for any non-empty open set 7 c Rn we define
HS(S2) _ {u E D*(S2) : u = UIn for some U E HI (R")). We see at once that H. is a closed subspace of HS (RI), and is therefore a Hilbert
space when equipped with the restriction of the inner product of H''(R"). A Hilbert structure for HS (0) is defined with the help of the orthogonal projection
P=P,,n: HS(R")-f which satisfies
PUI n = 0 and (I - P)UI n = UI n
for all U E HS(R").
Noting that if U I n = 0 then P U = U because U E HI \n, we see that a well-defined inner product on H' (S2) arises by putting
(u, u)HI(n) = ((1 - P)U, (I - P)V)HX(R,,)
if u = U112 and v = VI
for U, V E HS (R" ). Notice that the induced norm satisfies IIuIIH=(n) =
(u, u)H°(n)
= vin
uEHt(R»)
IIUIIHS(R,I),
(3.23)
because if U In = u then IIUIIHs(R") = IIPUIIHa(R") + II(1- P)UIIH
(R
II(I - P)UIIHt(R") =
The map U r-* (1 - P)Ul n is a unitary isomorphism from the orthogonal complement of onto HS(c2). Therefore, H4(S2) is a separable Hilbert space. Also, (3.23) shows that the restriction operator U H UIn is continuous from HS (R") to HS (0), and thus the space
D(7)=(u:u=Ulnforsome UED
11)
I
is dense in HI (Q) because D(R") is dense in H' (R'). We also define two other Sobolev spaces on 0, HS(S2) = closure of V(S2) in HS(R"), Ho (S2) = closure of D(S2) in HS(S2),
which we make into Hilbert spaces in the obvious way, by restriction of the
78
Sobolev Spaces
inner products in HS (R") and in HS (0), respectively. It is clear that HS (S2) c
H
and
H" (S2) c Ho (S2),
and later we shall establish the reverse inclusions subject to conditions on 92 and s. Note that an element of H is a distribution on R", but, provided the n-dimensional Lebesgue measure of the boundary of n is zero, the restriction operator u H uIn defines an imbedding
H&cL2(0) fors?0.
(3.24)
(If U E H and uIa = 0, then suppu c a2 = SZ \ S2, implying u = 0 on R".) In general, if s < -Z, then Hazy {0} no matter how smooth the boundary of 0, so we cannot imbed H. in a space of distributions on S2; see Lemma 3.39. The necessity of introducing more than one kind of Sobolev space on S2 can be seen already from the next theorem, which extends our earlier observation
that H-'(R") is an isometric realisation of the dual space of H`(R"); see also Theorem 3.30. Theorem 3.14 Let S2 be a non-empty open subset of R", and let s E R.
(i) If fl` (Q) = HI'r, and if we define (u, v) a = (u, V) for U E H-1(S2) and v = V IQ with V E HI (R"), then H''(0) is an isometric realisation of HS (0)*. (ii) If HS (S2) = Ham, and if we define
(u, v)n = (U, v) foru = UIn with U E H-'(R"), and V E H5(S2), then H-5(n) is an isometric realisation of H5(S2)*.
Proof First note that
(u, V) = 0 if u E D(Q), V E HS(R") and VIsi = O, so (u, v)c is well defined for u E H-s(S2) and V E H'(S2). We claim that (lu)(v) _ (u, v)Q defines an isometric isomorphism t : H-S(S2) -* HS(S2)*. Indeed,
I(lu)(v)I = I(u, V)I
IIUIIH-S(R,1)IIVIIH-(R")
whenever v = VIQ,
so I(lu)(v)I < IIUIIH-r(n)IIVIIH'(a) and hence IItuIIH-(Q)» < IIuIIH-=(O). Fur-
thermore, given a functional f E H3(S2)*, the map V H f(V In) is bounded
Equivalence of the Norms
79
on HS (R") because the restriction map V i-3 V I n is bounded from H3 (RI)
onto H'(Q). We know already that H-3(R!) is an isometric realisation of [HS (R")]*, so there exists u E
H-S (R") satisfying
f(Vlc) = (u, V) for all V E HS(R"), and
It(Vlra)I o#VEH'(R")
II V II Hs(a^)
If V E D(R \ n), then V I = 0, so (u, V) = 0, showing that suppu c S2, i.e., u E His. Assume now that H-S(S2) = H 3. In this case, u E and we see from
f(v) = t(Vlg) = (u, V) = (tu)(v)
for v = Vlo and V E HS(R'l)
that f = tu, sot is onto. Finally,
It(Vln)I < implying that IIuIIH-,(n) = IIulIH-'(a°) 0
l2
dco dt.
I u I1,,.
in
80
Sobolev Spaces
Proof. Define the forward difference operator Sh by
Shu(x) = u(x + h) - u(x), and consider the Fourier transform
I)i ( ).
.FShu( ) _
(3.25)
By making the substitution h = y - x, applying Plancherel's theorem and then reversing the order of integration, one finds that
f
2=
IIShUIIL_(Rn)
l
Ihl'-u+n
FC"
f
f
dh = R
IhI2µ+n
We transform the inner integral to polar coordinates, letting h = pw, where p = I hl and co = h/IhI. Since dh = pi-1 dp dw,
f
112
'H
Ihl2u+n
A=f
p-2µ-1 >o
lei2rrpl;."
- 112 dcodp = au l
I2µ,
0
where we used the substitution p = I
that f
f
1-1 t
and exploited radial symmetry. Note
lei2'
(,,j_1 ' - 112 dw is O(t2) as t J. 0, and is 0(1) as t Too, so that aµ is a finite, positive real number for 0 < µ < 1.
Theorem 3.16 Ifs ? 0, then W'(R') = H' (RI) with equivalent norms. Proof. Let r be a non-negative integer, and let 0 < p < 1. In view of (3.17), Plancherel's theorem gives IIu11w'.(R")
_
j br( )lu( )led
1:
lal 0, if there exists a continuous linear operator E : Ws (0) -> Ws (R") such that Eu I n = u
for all u E Ws (0), then
H'(Q) = W3(c) with equivalent norms.
Proof. If U E Ws(S2), then is = Uln for U = Eu E Ws(R") = Hs(R"), so u E Hs(S2) and IIuIIH,'(n) _< IIUIIH=(R") - IlEullwS(R,,) < Cllullws(j2),
giving a continuous inclusion W1 (S2) c H3(0). No extension operator is needed to establish the corresponding result when s is a negative integer.
Theorem 3.19 For any non-empty open set 7 C R", and for any integer r > 0,
H-r(2) = W-r(2) with equivalent norms.
Proof First consider the case 0 =1R' , and recall that [H'(R"))* = H-r (Rn) We define a Banach space isomorphism J : Wr(W") --+ H -1 (WI) by (Ju, v) = (u, v)Wr(R,.)
foru, v E Wr(R") = Hr(Rn),
Sobolev Spaces
82
and introduce another inner product and norm for H-r (R),
((u, v))-r = (J-'u, J-'v)wr Obviously, Illu III-r
and
((u, u))-r =
IIIuIII-r =
IIJ-iullwr(tt").
flu II H-' (>a")
If U E H-r(R") and V E H" (R), then
(u, v) _
(J-lu,
where fa = (-1)Ialaa J-1u E W-r(R") with IIuIIW-r(R^)
(aaJ-lu, aav) _
v)wr(R") =
(aafa, v),
IaI1 for S subordinate to W. Moreover, the 4j can be chosen in such a way that supp .ij is compact for each j.
Proof Put 0 = U W. We begin by constructing open n-cubes Q 1, Q2.... with the following properties: (i) the family { Q j l j> 1 is an open cover of n;
(ii) for each j there exists a set W E W such that Q j C W; (iii) each point of 9 has a neighbourhood that intersects only finitely many of the Q p Let 01 C 02 C SZ3 C . be a strictly increasing sequence of bounded open sets whose union is 0 and that satisfy SZj C= S2 j+1 for each j > 1. For convenience,
we put 0-1 = S20 = 0, and then define the compact sets K j = S2 j \ S2 j-1 for j > 1. Since K j does not intersect S2 j_2, given x E K j we can find an open cube G centred at x with G j,s C= W \ S2 j-2 for some W E W. The family {G j.X : x E K j } is an open cover for K j, so by compactness we can extract a finite subcover Q p After relabelling the cubes, we obtain a countable family Uj> 1 Qj = (Q1, Q2, ... } with the required properties (i)-(iii). (The set Q j is disjoint from each cube in Q i+2 U Q j+3 U .... and so intersects only cubes in the finite family Q1 U U Qj+1.) For each j > 1, we can use Exercise 3.6 to construct a function 1/rj E C ,p (]Il;") satisfying i/ij > 0 on Q j, and ij = 0 on R" \ Q p. Property (iii) of the Q j implies that the sum 11(x) = > j> 1 *j (x) defines a function E CO° (S2), and property (i) implies that > 0 on Q. Hence, we obtain the desired partition of unity by defining q 5j (x) = *j (x) / ' (x) for x E 0, and O j (x) = 0 otherwise.
Corollary 3.22 Given any countable open cover {W1, W2, ...} of a set S C R", there exists a partition of unity (P1, 02, ... for S having the property that supp q5j C Wj for each j > 1.
Density and Imbedding Theorems
85
Proof. Let 01, 02, ... be a partition of unity for S subordinate to the given open cover, define the index sets 11 = {k > 1 : supp 0k C= WI) and
Ij = {k > 1 : supp Ok C= W3 and k f I, U ... U Ij _ 1) for j > 2, and then put O! (x) = >kEI; Ok(x) for j > 1. We will now show that the Sobolev spaces on ]R" are invariant under sufficiently regular changes of coordinates; u o K denotes the composite function defined by (u o K)(x) = u[K(x)].
Theorem 3.23 Suppose that K : IR" -* ]R" is a bijective mapping and r is a positive integer, such that a"K and FK-1 exist and are (uniformly) Lipschitz on R" for J a I < r - 1. For 1 - r < s < r, we have u E H'(1R") if and only if u o K E HS(IR"), in which case 11U
0 K11 H-
(RI-)
11U11
S(R")
Proof. It suffices to show that for 1 - r < s < r, 11U 0KIIii
(R1-) < Cr11UIIH'(R-')
If s = r, then the estimate follows directly from the chain rule, because H" (IR") = W r (]R" ). The same estimate holds for s =1- r because H 1-r (]R") [Hr-1 (R'l)]* and
(u o K, v)
= (u, (v o K-1)I det(K
The case 1 - r < s < r then follows by interpolation, using Theorem B.7.
Density and Imbedding Theorems We saw earlier that D(SZ) is dense in H8(S2), but it is easy to find examples where D(S2) is not dense in Ws (0); see Exercise 3.18. However, we have the following theorem of Meyers and Serrin [64]. The proof relies on a technical lemma. Lemma 3.24 Lets E ]R and c > 0. For each u E HS (I8") there exists v E D(W' ) satisfying 11U
- v1IH-I(tt'-) < E
and
supp u c {x E ]R" : dist(x, suppu) < E).
Proof. See Exercises 3.14 and 3.17.
Sobolev Spaces
86
Theorem 3.25 For any open set 0 and any real s > 0, the set W5(S2) fl g (o) is dense in Ws(9).
Proof. Define a strictly increasing sequence of bounded open sets W1 C W2C ... by Wj = {x E S2 : Ix1 < j and dist(x,R" \ 0) > 1/j), and choose a partition of unity 01, 02, ... as in Corollary 3.22. Let U E WS (S2) and c > 0. For each j, the function 4j belongs to D(S2), so Oju E WI (R-1) = HI (W) and we can apply Lemma 3.24 to obtain a sequence (v)1 of functions in D(S2) satisfying
IItju - vj11wT(n) :
2jj
and
suppvj c Wj+i.
(Here, we use the fact that 110j u - vj II w=(n) = II /. ju - vj 11 w'(ttn).) Define v(x) _ F_1 v j (x), and note that this sum is finite for x in any compact subset
ofS2,sovEE(c2)and 00
Ilu - v11w(n) = T(Oju
00
j=1
E
1
- vj) W (SZ)
j=1
0 Next, we prove the Sobolev imbedding theorem, which shows that if s is large enough then the elements of HS (1[8") can be thought of as continuous functions, and the elements of Ho (S2) as functions that vanish on the boundary of n.
Theorem 3.26 Suppose 0 < tt < 1. If U E H"t 2+u (R` ), then u is (almost everywhere equal to) a Holder-continuous function. In fact, IUWI 0, we say that the set (3.26) is a Ck hypograph if the function : lR' -+ ]f8 is Ck, and if 81 is bounded for lad < k. In the obvious way, we then define a Ck domain by substituting "Ck" for "Lipschitz" throughout Definition 3.28. Likewise, for 0 < It < 1, we define a domain by adding the requirement that the kth-order partial derivatives of be Holder-continuous with exponent µ, i.e.,
18' (x') - d' (Y') i < Mix' - A,
f o r all x', y' E Rn-1 and IaI = k.
Hence, a Lipschitz domain is the same thing as a CO, 1 domain. Notice that in the definition of a Ck or domain, we can assume if we want that has compact support, because r is always assumed to be compact. The class of Lipschitz domains is broad enough to cover most cases that arise in applications of partial differential equations. For instance, if k > I and r, is a compact, (n -1)-dimensional Ck submanifold of R", then 0 is a Ck domain and hence also a Lipschitz domain. Furthermore, any polygon in RI or polyhedron in RI is a Lipschitz domain. One can construct many other examples using the fact that if K : R" - * R" is a C 1 diffeomorphism and if 0 is a Lipschitz domain, then the set K(S2) is again a Lipschitz domain. Figure 2 shows some examples of open sets that fail to be Lipschitz domains: (i) is disqualified because of the cusp at the point A; (ii) because of the crack B C (a Lipschitz domain cannot be on both sides of its boundary); and (iii) because in any neighbourhood of the point D it is impossible to represent r as the graph of a function. For a Lipschitz domain, in fact even for a CO domain, a much stronger density result than Theorem 3.25 holds.
Lipschitz Domains
91
(i)
0
Figure 2. Examples of regions that fail to be Lipschitz domains.
Theorem 3.29 If 0 is a CO domain, then (i) D(S2) is dense in WS (Q) for s > 0; NO D(S2) is dense in H or in other words HS (S2) = Hn for S E R. Proof. Suppose to begin with that n is of the form (3.26) for some continuous function : ]R"_' - ]R having compact support.
Lets > 0, u E WI (0) and e > 0. For S > 0, we define
ua(x)=u(x',x"-S)
and
528=(x ER" :x,,
so that ua E WS(Qs). Since a"us = (a' u)8, we can choose 3 small enough so that Ilu-uslsalIW$(si) < 2
and then choose a cutoff function X E E(]R") satisfying X = 1 on 0 and X = 0 on ]R" \ 52(8/2), so that Xus E WS (]R"). Hence, by Theorem 3.16, there exists V E D(]R") such that E
Ilxua-VIIWs(atn) 0. This time we put us (x) = u (x', x,, + 8), and observe that us E HI (R") and suppus c {x E R" : x" (x') - 8}. Choose S small enough to ensure IIu -US IIH.(ttl')
0, define
v(x)
1
x
u(y)dy f'[u(x) -u(y)]dy=u(x)-J x X
o
and
w(x) = r°° X
V
dY Y
Since v'(x) = u'(x) - x-1v(x) and w'(x) = -x-1v(x), we see that
u'(x) = v'(x) - w'(x) forx > 0. Furthermore, u has compact support, and both v(x) and w(x) tend to zero as x tends to infinity, so
u(x)=v(x)-w(x) forx>0. 'By the Cauchy-Schwarz inequality, Iv(x)12
0. If S2 is a Lipschitz domain, then
P(Q) = fu E L2(52) : u E H'(R")) C Hp(S2), where u denotes the extension of u by zero:
u(x) =
u(x)
if x E 92,
0
ifxER"\S2.
In fact,
H's(Q) = Ho(S2)
provideds 0 {2, 2, 2,
}
Proof. For the moment, we think of the elements of H' (S2) as distributions
on R. If U E Hs(S2), then the restriction v = ulst belongs to L2(S2), and u = v as a distribution on R", so D E H'(R"). Conversely, if u E L2(S2) and u E H' (R"), then supp u c S2, so u E H = HS (S2). We have already seen that HS (S2) c _Ho (S2).
Now view HI(Q) as a subspace of L2(c2), and let u E D(S2). For any integer r > 0, Theorems 3.16 and 3.30 give IIuIIHr(p) = IIuIIHr(R")
IlaauIILz(R") Ial5r
=
IIUI1Hr(si),
-
Ila"uIIL,(n) IaI Q(X")),
where x" = (x1, ... , xn_2); for n = 2, the function q reduces to a constant. In the obvious way, we extend the notion of a Lipschitz dissection to the case when S2 is the image under a rigid motion of a Lipschitz hypograph.
Next, suppose that n is a Lipschitz domain. We say that (3.30) is a Lipschitz dissection of r if, in the notation of Definition 3.28, there are Lipschitz dissections 8 S2, = r,, u l1, U r2, such that
W,nr1 = W,nr1,,
W,nfI= W,nII,, W,nr2= W,nr2,,
for all j. We remark that, in this case, the subsets r 1 and 1`2 are not necessarily
connected. LetD(r1) = 1-0 E D(r) : supp0 c r1}; by defining
Hs(r1) = {U1r, : U E Hs(r)), Hs(171) = closure of D(171) in HI (r),
Ho (r) = closure of D(rI) in Hs(ri), the properties of Sobolev spaces on Lipschitz domains in R carry over to Sobolev spaces on r1, subject to the condition that Is I < 1, or Is I < k if (3.30) is Ck-1.1 in the obvious sense.
Sobolev Spaces
100
The Trace Operator In studying boundary value problems, we shall need to make sense of the restriction u I r as an element of a Sobolev space on r when u belongs to a Sobolev space on 7. The main idea is contained in the following lemma.
Lemma 3.35 Define the trace operator y : D(IR") -* D(IR"-') by
yu(x) = u(x', 0) forx' E
IR
n-
Ifs > 2, then y has a unique extension to a bounded linear operator
y : H$(1R") -4 H" '/2(1R"-'). Proof. For u E D(1R"), the Fourier inversion formula (3.15) gives
f
Yu(x') = f,'
u(',n)
(f
dt = J
f
"
oo
and so 00
Yu(')=J 0000 u(', n)dSn=J
(1+1 12)-S/2(1+1 12)s'%u 00
Applying the Cauchy-Schwarz inequality, we obtain the bound
f
W
(1 +
where, using the substitution t;,, = (1 + oo
Ms( ') =
1W1I2)'/2t,
f
00
dtn
00 (1+I 112+I n12)S
(1 +I
112)S-1/2
The integral with respect to t converges because s >
dt (1+t2)S.
oo
so if we write MS =
MS (0) then
(1 +
MS
f
(I +
do
00
Integrating over ' E 1R" gives MSIIUIIHs(Rn),
and since D(1R") is dense in H5 (R"), we obtain a unique continuous extension
for y.
The Trace Operator
101
The lemma above is sharp in the sense that
Hs-l/2(R"-') = {yu : u E HS(R")) for s > 1, because y has a continuous right inverse rio, as we now show.
Lemma 3.36 For each integer j > 0, there exists a linear operator
77j:S(1R"-')-* S(R") satisfying
a" (rij u) (x', 0) =
8",u(x')
if a = j,
0
if
for x' E Ri-', U E S(R"-)) and any multi-index a = (a', a"). Moreover, rij has a unique extension to a bounded linear operator i7 j :
Hs-j-1/2(Rn-i) -+ H' (W') for S E R.
Proof Choose a function O j E D (R) satisfying 9 j (y) = yj /j! for I y j < 1, and define z1
u(x) =
dl;' forx ER".
(1 +
JR1
Since 9j(k) (0) = S jk, we see that
aa(77ju)(x', 0) =
J
j
gives
as required. The substitution x = (1 +
11 u() _
(1
+ ,i )j/2
a"'u(x')s j"
ei2nt' X'
f4 (
L(1 + I
I)
x] dxn
j[(1 +
_
(1
gives
and the substitution 4n = (1 +
4-1 X
J
I
,12)1+l
[(1+1 12)-i/2 n]I2d ndS 00
Sobolev Spaces
102
where 00
CS = J 0 (1 +t2)SIdj(t)I2dt < oo,
for all s E R. For Sobolev spaces on domains, we can now prove the following.
Theorem 3.37 Define the trace operator y : D(S2) -+ D(F) by
Yu = uIr If SZ is a
Ck-1,1 domain,
and if 2 < s < k, then y has a unique extension to a
bounded linear operator
y : H$(Q) -+
(3.31)
HS-112(r),
and this extension has a continuous right inverse.
Proof Since H5(]R") is invariant under a Ck-1.1 change of coordinates if 1 - k < s < k, one sees, via a partition of unity and a local flattening of the boundary, that if z < s < k then IIYUIIHs-1/2(r) 1,
11uI1E5 < CIIu11Hs (R.")
and we claim
IIu(,
CIIuflE,
In fact, the substitution n = (1 + d
f7a, (
" = C5(1 + )
1
,12)-(s-1/2),
for z < s < Z.
yields where CS =
f
00 00
t -2(l+ t t2)'
104
Sobolev Spaces
with C, < oo for 2 < s < 2, so, applying the Cauchy-Schwarz inequality,
Ilu(,
0)II2
Hs-1i2(Ra-1)
f
J R°-r
(1
J - -l
(1 +
I
hI2)s-I/2I f
2
00
u( '> n) dSn 00
f (J-00 as ()
)
00 a,()II ( )12dn)
d'
00
=Csf as()Iu()12d =C'11U112Es 1'
Thus, we see that Ilut(, 0)IIHf-II2(RI-r) -< CIIul1El < C11U11El < CIIUIIHS(R")
for 1 < s < 2
which, combined with Theorem 3.37, shows that the trace operator (3.31) is
bounded for 1 < s < 2 . The next lemma is a version of a standard fact [41, p. 47] about distributions supported by a hyperplane, and will allow us to characterise Ho (S2) using the trace operator. The symbol ® means the tensor product of distributions, so, formally, (vj ® SWjW)(x) = vj(x')SWD(x"). In the proof, we use the notation R+ = {x E R'r : x" > 0) for the upper half space.
Lemma 3.39 Consider the hyperplane F = {x E R" : x = 0).
(i) Ifs > - 2, then HF = {0). (ii) If s < - 2, then HF is the set of distributions on R" having the form
u = E uj ®S(j)
with vj E
H''+j+l/2(R'r-t).
(3.32)
0:5j - ? . Suppose now that u E H F . with -k - 2 < s < -k - f o r an integer k > 0, z and assume 0 < j < k. Let ri; be as in Lemma 3.36, and define v; E D* (][8"- ) by
(v1,4) _ (-1)j(u, rl;0) for d E D(R't-'). Observe that I(v;, 0)1 Z, and let E = {u E HS(7) : y(8au) = 0 for (aI < }, noting that this definition makes sense because y o 8' H' (Q) -z HS-111(F) for kal < s - 2. Let f E E* satisfy £(o) = 0 for all ¢ E D(Q). By the Hahn-Banach theorem, there is a w E HS (S2)* = H-S (S2) such that f(o) = (w, gyp) for all 0 E E. Since W E HFS, part (ii) of Lemma 3.39
s-
shows that w = Eo<j<s-1/2 Vi ® 3(j) with vj E
H-s+j+1/2(R"-1)
Hence, for
every q5 E E,
> (-l)J(vj, Y(3
)) = 0,
o<j<s-1/2
and so D(S2) is dense in E, proving part (ii).
0
Vector-Valued Functions Thus far, the present chapter has dealt only with spaces of scalar-valued (generalised) functions. The results obtained extend in a straightforward manner to spaces of vector-valued functions
u: S2-+ c", and this final section does no more than establish some notational conventions. We denote the space of compactly supported, C"'-valued, C°O test functions by
D(Q)," = D(S2; C').
The (sequentially) continuous linear functionals on D(O)"' are then the ("valued distributions on 0, and we view these objects as generalised C"'-valued functions, by writing
(u, v)n = 1 u(x) v(x) dx,
Jn
(3.34)
where the dot denotes the bilinear form on cm whose restriction to R" coincides
Exercises
107
with the standard Euclidean inner product, i.e., "1
U.V=EuJVJ. j=1
The set of all Cm -valued distributions on SZ is denoted by D* (0)"' = D* (S2; C'").
We think of u as a column vector or m x I matrix, and let u* denote the row vector or 1 x m matrix obtained by transposing the complex conjugate u. Using matrix multiplication, we may then write the standard unitary inner product in CC' as m
u*V=U.V=Eujvp I=1
In this way, the sesquilinear form associated with the bilinear pairing (3.34) is given by
(u, On =
u(x)*v(x) dx. fn
Of course, if u and v are square-integrable functions from S2 to CC', then (u, v)n is their inner product in L2 (S2)m = L2 (S2; C"'). The definitions of the vector Sobolev spaces on S2, W'(S2)' = Wp (S2; (C'" ),
H3 (S2)" = HS (S2; (C'"),
HS (S2)m = HS (S2; (Cm),
should now be obvious. Likewise for the vector Sobolev spaces on r. Occasionally, we shall encounter normed spaces of matrix-valued functions, such as L,, (S2)mx,n = L,,(12; C">°"), whose meaning should also be obvious.
Exercises 3.1 Suppose that u E Lp(S2) satisfies I(u, v)nl < MIIvIILp.(n)
for all V E Lp.(S2).
(i) Show that if 1 < p < oo, then II u 11 L,,(sa) < M. [Hint: take v = ] sign(u) lulp-1
(ii) Show that if p = oo, then for every measurable set E C 0 with I E l > 0, the mean value of l u I over E is bounded by M, i.e., E1
I11JElu(x)Idx<M.
Deduce that 11U IIL (n) < M. [Hint: take v = sign(u) XE, where XE is
the characteristic function of E.]
Sobolev Spaces
108
3.2 Prove that convolution is associative:
(u*v)*w=u*(v*w) foru,v,wEL1(R") 3.3 For 1 < j < k, let f j E L 1(R") be a compactly supported, non-negative 1, and let 0 < X j < 1. Fix X E 1I8", and
function satisfying II fj II L,
define
g(A)=(fi' *...* f,R)(x) forA=(At,...,Ak). (Take f to be identically 1 outside as well as inside supp f3.) ° that g (e j) = 1 where ee j E Rk is the vector with compo(i) Show nents (ej )1 = 1 - S jl. (ii) Use the fact that, for any positive a j and any non-negative A j and µj, k
k
k
j=1
j=I
j=1
H ajt-r)xj+ruj = exp (1 - t) L log a^' + t L log aµ' to show that the function g : [0, I ]k --> [0, oo) is convex. (iii) Deduce that g(A) < 1 if At +...+Ak = k -1. [Hint: A Aj)ej.] (iv) Hence show that k
(1-
1
if E-=k-1. j=1;pj 3.4 Show that supp(u * v) c supp u+ supp v = {x + y : x E supp u and y E supp U).
3.5 Recall the notation (3.7). (i) Show that
(d)k dt
u(x + ty) = u(k)(x + ty; y).
[Hint: k!/a! equals the number of permutations of k = IaI objects when there are a j objects of type j, for I < j < n, and it is assumed that objects of different types are distinguishable, but objects of the same type are indistinguishable.] (ii) Use integration by parts to verify Taylor's formula for a function of one variable: k
f (s) =
f(j)(0)
E j j=o
1
Sk+t
Si +
k1
1
J
(1 - t)k f(k+I)(ts) dt.
Exercises
109
(iii) By taking f (s) = u(x + sy), derive Taylor's formula (3.8) for a function of n variables.
3.6 Define f : R -+ R by
f(t) =
e-' 1'
10
if t > 0,
ift 0. (ii) Deduce that f E C°°(R).
(iii) Construct a CO0 function g : IR - R with g > 0 on (-1, 1) and with supp g = [-1, 1]. (iv) Construct a COG function * E C mP(R") satisfying (3.9). 3.7 Let S2 = (0, 1), choose any 0 E D(S2) not identically zero, and define
¢j E D(S2) by 0, (x) = 0 (j -' x). Show that -Oj -+ 0 in E(Q), -but not in D(S2).
3.8 Establish that the inclusions D(S2) C E(S2) and D(R'1) C S(R") E(1E8") are continuous with dense image, by proving each of the following statements:
(i) If ¢j - 0 in D(S2), then qj -+ 0 in E(S2).
(ii) If 4i -+ 0 in D(R), then ¢f - 0 in S(R"). (iii) If 4j -* 0 in S(R"), then Oj - 0 in E(R"). (iv) Let Ki C K2 C be an increasing sequence of compact sets whose union is S2, and let Xi E D(S2) satisfy Xi = 1 on Kj. If 0 E E(S2), then Xicb -+ 4) in E(S2).
(v) Let X E D(W) satisfy X (x) = 1 for IxI < 1, and define Xf E D(R") for each positive integer j by X,i (x) = X (j _'X). If 0 E S (W), then
XjO -* 0 in S(W). 3.9 Consider a linear functional f : D(S2) -+ C. Show that a is sequentially continuous (and hence a distribution on S2) if and only if for each compact set K C S2 there exists an integer m > 0 such that
It(4))I -S CK,," E sup laaol for all 4) E DK(S2). Ia1 n/2. 3.19 Let U E V* (Q) and S E (i) Suppose W is an open set and X E E) (W). Show that if u E HI (w n 0), then XU E HI (0) and 11Xu1IH,(92) < Cx,s1Iu1IHs(wns ).
(ii) Suppose (¢i )
is a partition of unity of the type used in the proof of Theorem 3.29. Show that u E H-'(9) if and only if 95i u E Hs (Wi n o)
for 0 < j < J, in which case I IIUIIH=(n)
^' L .i=o
3.20 Prove the following inequalities, due to Hardy: for a > 0 and 1 < p < oo,
dY\ dx LJ'(x-" 0
Jo
X
If(Y)I Y
I/P
"0.I.
3.22 Consider the half space Q = {x E R,1 : x" < 01. Let U E D(S2), and define U(X)
U (X) =
if x" < 0,
0
(i) Show that Ck."(1 + It'I)-k(1 + 14"1) - I for every k > 0. (ii) Hence show that U E Hi = HS (S2) for s < 1
(iii) Show that if a,u(x', 0) = 0 for 0 < k < j, then U E Hr (S2) for s < j + 3 .
4
Strongly Elliptic Systems
We are now ready to begin our study of boundary value problems for linear elliptic systems of second-order partial differential equations. The first task is to explain how, via the first Green identity, such problems fit into the abstract scheme treated in Chapter 2. We then define the class of strongly elliptic operators, and investigate when such operators are coercive. After that, an existence and uniqueness theorem for weak solutions in H' (S2)" is given, expressed in the form of the Fredholm alternative. Next, we prove regularity of the solution on the interior and up to the boundary, under appropriate assumptions on the data and the domain. We also prove the transmission property, which will be used later to show regularity at the boundary of surface potentials for smooth domains. The final section of the chapter presents some rather technical estimates relating the H1-norm of the trace and the L2-norm of the conormal of
derivative. These estimates will allow us to prove some limited regularity of surface potentials for Lipschitz domains.
The First and Second Green Identities Suppose that Q is a non-empty open, possibly unbounded subset of ll8", and consider a linear second-order partial differential operator P of the form n
it
It
Pu=->>8J(Ajkaku)+j:AjBJu+Au onc2,
(4.1)
j=1
j=1 k=1
where the coefficients
Ajk = [apgJ,
AJ
= `a',q],
A = [apgl
are functions from S2 into C""", the space of complex m x m matrices. Thus,
I < p < m and 1 < q < m, and P acts on a (column) vector-valued function 113
Strongly Elliptic Systems
114
U : Q -- C'" to give a vector-valued function Pu : cZ -,, Cm, whose compo-
nents are M
M
(POP
- L+ L Lr nn
nn
aJ (ap4aku4) +
j=1 k=1 R=1
nn
j=1 q=1
M
ap9u9
ap9aju4 + 9=1
fort
cIIUIIHI(n)s.
- CIIuIIi,(0),0I
for U E V.
Obviously, in this context L2 (S2) acts as the pivot space for V. When seeking to determine whether or not a given differential operator is coercive, we can ignore the lower-order terms, and consider just the sesquilinear form corresponding to the principal part, n
4>0(U, v) =
E(Ajkaku)*8jvdx.
Jj=.1
k=1
Lemma 4.5 The differential operator P is coercive on V if and only if its principal part Po is coercive on V.
Proof. For any 6 > 0, we have I'D (u, v) -'Do(u, v)I -< C11U11HI (n),,, IIv11L,(n)" < C(EIIU1121
so if Po is coercive, i.e., if Re co (u, u) > c II u 11 H
+E-'IIVI1L2(s2)
Al" - C II u II L2(2)"' , then
Re4D(u,u) > and by choosing E sufficiently small, we see that P is coercive. The converse is proved in the same way. 0
Strongly Elliptic Operators
119
The differential operator P is said to be strongly elliptic on S2 if "
It
Re>2>2 [Ajk(x)
for all x E 0, l; E R" and >) E C"'.
k 1 *Sj77 >
j=1 k=1
(4.7)
Depending on the subspace V and the regularity of S2, this purely algebraic condition on the leading coefficients is often necessary and sufficient for P to be coercive.
Theorem 4.6 Assume that the coefficients Ajk are (bounded and) uniformly continuous on Q. The differential operator P is strongly elliptic if and only if it is coercive on Ho (S2)1'1.
Proof. Suppose that P is strongly elliptic. First we consider the special case when the leading coefficients A jk are constant. Let U E Ho (S2)'" = H' (S2)"',
i.e., let u E H1(R")"' with supp C S2. Since .F,,k[aju(x)} Plancherel's theorem implies that
f (Ajkaku)*aju dx = (27r )2 R11
)J* ju(S) dS,
JR"
so, taking ri = u(l) in (4.7), Re(Do(u, u)
(27r
)2
f
c>2
g
f. Iajul2dx
j=1
CIIUI12'(2)u' -CIIUIIL2(s2)"
By Lemma 4.5, we see that P is coercive on Ho (S2)"' To handle the general case, let c > 0 and choose 8 > 0 such that
max IA jk(x) - Ajk(Y)I < E j,k
for Ix - yI < S.
(4.8)
Cover SZ with a locally finite family of open balls B1, B2, B3, ... , each of radius S. (If 0 is bounded, then the family of balls will be finite.) Since the diameters of the balls are bounded away from zero, we can assume that for each d > 0, there is a number Nd such that any given set of diameter less than d intersects at most Nd balls. By Corollary 3.22 and Exercise 4.6, we can find real-valued functions 01, 02, 03, ... in C,1,,mp(W) with 01 > 0 and supp,0) c B1, such that 01(x)2 = 1, 1>1
>2cbi(x) < C, and >2I8j0,(x)I < C, 1?1
!?1
forx E S2.
Strongly Elliptic Systems
120
Note that the number of non-zero terms in each sum is finite, and is bounded independently of x. Since [Ajkak(4lu))*aj(Oly) = 01 (Ajkaku)*ajll + (akOl)0!(Ajku)*ajv + O1(aj0l)(Ajkaku)*v + (akOl)(ajO,)(Ajku)*v, we have 0o(0lu, 01u) < CIIuIfHI(s2)'n IIu1IHI(2)n, 1> 1
and also Re (Do(u, u)
>
Re4 o(01u, 01u) - CIIu1IHI(n)n' IIuIIL,(9),1 . 1>1
Let fio denote the sesquilinear form obtained from (Do by freezing the coefficients Ajk (x) at x = x1, the centre of the ball BI, and observe that
4'o(0tu, 01u) - Co(01u, 0tu) = f {[Ajk(x)
- Ajk(x!)Jak(01u)}*aj(olu) dx.
From the special case considered earlier, we know that (Do is coercive on Ho (S2),
with constants independent of 1, and by (4.8),
IAjk(x) - Ajk(x!)I < E forx E BI, so
Re to(Otu, 01u)
Re (Do(01u, 01u)
-
6110,UI12
> (c -E)IIfiluIIHa(n),n -CII01uIIL,(n)"1, and
Re Do (u,u)> 1>1
-C
1101U112
),n -
CIIuIIHo(S2)n, IIu11L2(szyn
1>1
Since the 0,2 form a partition of unity,
II0IUIIL2(n) = f E0l(x)2lu(x)I2dx = IIUIIL2(n),n !>i
l>1
Strongly Elliptic Operators
121
and
Il018;u + (a;o,)ull2L2(),H IIajul1L2(n)N, - C11
uIIHo(Q),,,11u11L2(n)"'-
Using the inequality ab < (E'a2 + b2/E'), we see that 2
/
Re(Do(u, u) > (c - E - E')IIUIIHo(n)m - Cf l +
1
E,
)IIU112,(si)u'
so P is coercive on Ho (S2).
To prove the converse, take a real-valued cutoff function >/r E C mp(R") satisfying
*>0onR", Jr=0forlxl>1,
*(x)2dx=1. fRII
Let xo E 0, and put *E (x) _ E-"j21/r(E-l (x small,
- xo)), so that for E sufficiently
E C mp(S2) with
r 'VE >
0 for Ix - xol > e,
0 on n,
J
/rE(x)2 dx = 1.
Thus, '1/!E (x)2 converges to 8 (x - x0) as c 4. 0. Consider the function uE(x) _ 1E(x)e`t.XI?. Since
IIUEIlL2(n)"' = 1171,
and ajtfE = E-'(aj*)E, we have
and since ajuE =
Now,
(AjkakuE)*ajuE =
'YE (Arjk
+i1IE(ak /E)(Ajk17)* J
-',,/E(aj'YE)(Ajk k?I)*17 + (ak
so if we define
A'k
=
f1/Ie(x)2Ajk(x)dx,
t
E)(aj*E)(AjkY1 !)*TI,
Strongly Elliptic Systems
122 then
ReE
C(E-2
j11 > ReIo(uE, uE) -
j=1 k=1
If we now assume that P is coercive on Ho (0)', then Re'Do(uE, uE) _> c11 uEII HacWN - CIIuEllc2(si),,, >-
c(E-2 +
CIrjJ2,
implying that n
Re
C(l + E-2 +
kr1)* j17 ?
it
j=1 k=1
Now replace by tl; where t > 0, divide through by t2, and send t -+ 00 to obtain (4.7) with A)k in place of A,k. Since Ask -+ Ajk(xo) as c y 0, we conclude that P is strongly elliptic. 0
For scalar problems, i.e., when m = 1, the strong ellipticity condition (4.7) simplifies to n
Re
1: 1: Ajk(x)44k4'j > it
for allx E S2 and
E R",
j=1 k=1
and the next result is usually sufficient for establishing that the differential operator is coercive on the whole of H 1(S2), not just on Ho (S2); cf. Exercise 4.1.
Theorem 4.7 Assume that P has scalar coefficients (i.e., m = 1), and that P is strongly elliptic on 0. If the leading coefficients satisfy Ak j = A jk
for all j and k,
on S2,
then P is coercive on H 1(S2).
Proof Define F : 0 x C" --* C by !Y
n
F(x,
Ajk(x) k `;j. j=1 k=1
The symmetry condition on the leading coefficients implies that
F(x, + ir1) = F(x, ) + F(x, >)) for x,17 E R",
Strongly Elliptic Operators
123
and so by strong ellipticity, Re F(x, l;) > cI:;12 for all l; E C" (not just for L"). Hence, for all u E H' (S2),
E
cgradull?.().
Re (DO (u, u) = j Re F(x, gradu) dx z
El
In a similar fashion, when m > lit is easy to see that P is coercive on H' (S2)" if Akj = Ask and n
n
n
E[A,k(x)k]*;
Re
j=1 k=I
>c
lei l2
j=1
for all x E cZ and 1, ... , l; E R'",
(4.9)
but this assumption excludes some strongly elliptic operators with important applications; see Exercise 10.3. Using an approach due to Ne6as [72, pp. 187-195], we shall prove a sufficient condition for coercivity on H' (S2)"` in the case when the leading coefficients can be split into sums of Hermitian rank-1 matrices, i.e.,
t Ajk =
brj r=1
where the blj are (column) vectors in C". It follows that Po must be formally self-adjoint, and that L
(Do(u, v) =
J
2
It
uNvdx
where Nu = E b* 8j u.
(4.10)
J=I
1=1
Note that the first-order differential operator N acts on a vector-valued function u to produce a scalar-valued function Nu, and that L
b0(u, u) =
QNulli2cn) >_0
for u E H' (p)'".
r=1
An important example of a strongly elliptic operator of this type is described in Chapter 10; see in particular Theorem 10.2. The proof of coercivity is based on the following technical lemma, whose proof turns out to be surprisingly difficult.
Strongly Elliptic Systems
124
Lemma 4.8 If l is a Lipschitz domain, then for any integers p > 0 and q > 1, and for any u E D(S2), IIuIIH-P(s2) 0, let
'NE be as in (3.9) and (3.10), introduce the C°° function f (y', c) = (`YE * ) (y'), and define
KE (Y) = (Y', f (Y', -Ey) + y,1) Since grad
E L,,. (R"- 1), we find that a
(
for y < 0.
(ayn-1)«,,-, (aE
I ) «I ...
ay,
C )a,r f (Y', E) < EIaI-1
for laI > 1.
(4.11)
Thus, a,(KE)n(y) = 1 - Ea. f (y', -Ey11) = 1 + O(E), and we now fix c small enough so that c < 8,1(KE),1(Y) < C
for y < 0,
and write K = KE. In this way, K (y) is a strictly increasing function of y, E (-oo, 0), with K. (y) f ' (y') as y71 T 0, and so K : lR" - 0 is a C°O diffeomorphism, and it can be shown using (4.11) that Ia"K(Y)I
1.
Strongly Elliptic Operators
125
In the substitution x = K(y), we have x' = y', so the Jacobian is simply
detDK(y) =
K. (y).
Also, by differentiating the equation xn = f (x', one sees that ay,
-ajf(x', -Eyn)
axj
1 - Eanf (x', -Eyn)
yn with respect to x,
(Nu, Q10) _ 1=1
>(N1 u, Q10)a, 1=1
so L
I (anal ur, 4'Tl 1. In this case, the homogeneous adjoint problem (4.16) also has exactly p linearly independent solutions, say v1, ... , VP E H 1(Sl)"', and the inhomogeneous problem (4.13) is solvable in H 1(0)'" if and only if
for I < j < p.
(v1, f )Q + (Yv1, 9N)rN = (Evv1, 9D)rp
When the data satisfy these p conditions, the solution set is u + W, where u is any particular solution. Moreover
IIu+WIIH'(i)"lw 0, if u E H1(922)' and f E Hr (922)"' satisfy
Pu = f on 02,
Strongly Elliptic Systems
136
and if the coefficients of P belong to Cr,l (S22)mx"1, then u E Hr+I(Q I)," and
UUIIHI-+2 Ajaju - Au, j=1
(J.k)0(n.n)
so by Lemma 4.17, 82u11L2(a1)" 11
C n-I Il81ullH-(92,)'»,
C11 f IILZ(sll)N + C11 uII H'(n,)u +
(4.26)
!=1
giving the desired estimate for II U11 H2(S2, )^,
For non-zero yu E H3/2(r2)m, we use Lemma 3.36 to find w E H2 (02)' satisfying y w = y u on F2, and II w II H2($i2)m < C II Y u II H3/2(1'2) n . The differ-
ence u - w E H'(02) "' satisfies P(u - w) = f - Pw on 522, and y (u - w) = 0
Regularity of Solutions
139
on 1`2, so by the argument above,
flu - w1IH2(n,)1,, < CllU - wllHl(n,),,, +C11f -PW11L,(n,)u', and therefore
IIUIIH2(n,) 0, then v = e,,.) We denote the one-sided trace
operator for n' by y±, and the sesquilinear form and conormal derivative and L3v , respectively. On account of our convention for SZ} by '1 regarding the meaning of v, the first Green identity (Lemma 4.1) takes the form
(D+(u, v)
= (Pu, On- T (B}u, y±v)r
for u E H2( 1)"' and V E Hl(S2t)"',
(4.30)
and dually (Lemma 4.2)
+(u, v) = (u, P"v)12t : (y±u, BV Or
for u E H) (S2±)"' and VE
H2(i)m.
(4.31)
As explained in the discussion following Lemma 4.3, we extend the definitions of B} and B to make these identities valid for u, v E H 1(S2#)"' whenever in the former case Pu, or in the latter case P*v, belongs to k-1 (01)'In this setting, it is convenient to think of u as a function defined on the whole
of R", but possibly having jumps in its trace and conormal derivatives across
Strongly Elliptic Systems
142
the surface 1,. We denote these jumps by
[ulr = y+u - y-u,
[Bvul r = 13,,'u - Bvu,
[Bul r =13 u -
provided both of the corresponding one-sided quantities make sense. When the jump vanishes, the + or - superscript is redundant and will often be dropped, i.e., we shall write
yu = y+u = y-u if [u]r' = 0, and similarly for the conormal derivatives. For brevity, we sometimes write
u = uln= to emphasise that we are considering the pieces of u separately.
Lemma 4.19 Let f } E H-1(SZ±)"', define f = f + + f - E H_I (RU)"', and suppose that u E L2(R")"' with u± E H' (SZ±)'". If
Pu} = f ± on Sgt,
(4.32)
then
c+(u+, v) + 4-(u-, v) _ (f, v) - ([B,u]r, yv)r for v E H'
(]E8")"',
(4.33) and
(Pu, 0) = (f, 0) + ([ulr, BvO)r - ([Bvulr, YO)r for 0 E D(R")"'. (4.34)
Thus, Pit = f on R" if and only if [ulr = [Bvulr = 0. Proof. Equation (4.33) follows at once from (4.30). The definition of Pu as a distribution on R!' gives (Pu, 0) = (u, P*O) = (u+, P*O)Qa- + (u-, P*O)n-, and by (4.31),
(u}, P*4)n- = 4±(ut, 4) + (Y:-u, BvO)r, so (4.34) follows from (4.33).
0
We can now prove the transmission property for (4.32): if the jumps in u and its conormal derivative are smooth, and if the right hand sides f + and f - are
The Transmission Property
143
smooth up to the boundary, then the restrictions u+ and u- must be smooth up to the boundary. Jumps are also allowed in the coefficients of P, although we shall not use this fact in later chapters. The proof again uses Nirenberg's method of difference quotients. Results like the following theorem, but involving more general types of transmission conditions on F, appear in work by Schechter [90] and Roitberg and Seftel [89]. Theorem 4.20 Let G 1 and G2 be bounded open subsets of R" such that G 1 C= G2 and G 1 intersects 17, and put
S2 =G;ns2l and
for j = 1, 2.
Suppose, for an integer r > 0, that x2 is
Pu± = f±
Cr+1. t, and consider the two equations on S22 ,
r+1
where P is strongly elliptic on G2 with coefficients in C'' 1(SZ )! L2(G2)"' satisfies
u± E H1[ulr E H'+312 (x2"
If U E
[B,,u]r E H(x2)"'
,
and if f ± E Hr(S22)"', then U-1- E H''+2(Q+)'" and IIu+II Hr+2(n ),,, + II U
GIIu+IIHI(n
),,,
II
+CIIU IIHI(1
+ C II [u]r2 II
)8;
C II [Bvulr, II Hl+',Z(r,)-
+CIIf+IIH,(Q2),,, +CIIf Proof As in the proof of Theorem 4.16, it suffices to consider the case when SZ+ is the half space x" > 0. Suppose to begin with that [u]r = 0 on r2i so that u E H1(G2)r by Exercise 4.5. We fix a cutoff function X E D(G2) such that X = .l on G 1, then, as with (4.22), (x u:-)
on Q:'
where the functions fl and f satisfy II f1 IIL,(Q: (Q:)- < C11 Xf IIL,(si})r, +C1lu}IIH1(0) .
By Lemma 4.19,
4) G,(Xu, v) = (fl, V) G, -
Yu)r,
for V E Ho (G2)"',
Strongly Elliptic Systems
144
an equation eerily like (4.28). Repeating the argument from the proof of interior regularity, but with an extra term involving [B (Xu)]r, we see that < CHUIIH(GZ)" + CIIf1IIL2(G2)"
11Al.h(XU)112
+ CI ([B,, (X 01r, y A,, -h v) r2
where v = O,./,(Xu); cf. (4.23). If 1 < 1 < n - 1, then the argument leading to (4.29) shows that I([Bv(xu)]r, Yol,-hv)r, I < (II[Evu)rlIHI/2(r2)"' + IIUIIH'(G2) X Ilol.h(Xu)Ilk'(G2)1'-,
and therefore Il a,(xu)IIH'(G2)" -< CIIUIIH'(G2)" + CII[E,1u]rHIH"12(r2)"' + Cll f1IIL2(G2)
Of course, 8"(Xu) is generally not in HI(G2)'", because [8u]r # 0, but as in (4.25),
f3
on n2" ,
with
CIIftIIL2(7
)1" + CIIuIIH'(nt),,, + C> IIa,UIIH'(s2'),,,. 1=I
Hence, WE E H2(7 )'" and the desired estimate holds for r = 0.
For non-zero [ulr E H312(F2)'", we use the extension operator 770 of Lemma 3.36 to construct w E H2(G2)"' satisfying
Yw = [u]r on I'2
and
IIB,wIIH'12(r2)'" + IIwIIH2(G2)1,, < CII[u]rl!H3P(r2)'".
Consider the function
I u+
UI =
onQ
u- +w on Q2.
Since
Put = )
f
on Q21
f-+Pw oncz
,
with [u1]r = [u]r - w = 0 on I'2, and [B ullr = [Buu]r -
E
HI/2(I'2)m, the preceding argument applies, showing that ui E H2(S2 )"'.
Estimates for the Steklov-Poincare Operator
145
Therefore, u± E H2(0)"' and Ilu
Ilui IIH2cn;
IIH2co, .
Ilu1 Ilx'csa >>° + Ilwllx'csi .^
< CIIu1IIHt(G2)1" + C11[8,,ui]r1iH"r(r2)" + Cflf+IIL2c2;>^'
+Cllf +PzIIL2(n)" +Ilwllx2(n;),,,, which, because CIIwitH2(n2)ur
< CII[u]rllx3/2cr2yn,,
shows that the desired estimate for the case r = 0 holds also when [u]r 0 0. An inductive argument like the one used for part (ii) of Theorem 4.18 takes care
ofr> 1. Estimates for the Steklov-Poincare Operator Consider the semi-homogeneous Dirichlet problem,
Pu = 0
on Q,
(4.35)
yu=g onl' and the adjoint problem
P*v = 0
on cl,
yv=ci
onr.
4.36)
If the fully homogeneous problem has only the trivial solution in H' (S2)"', i.e., if g = 0 implies u = 0, then under the usual assumptions we are able to define solution operators
U:gHu and
V:OF-).v,
with
U: H'/2(r)m -)-H'(S2)m
and V: H'/2(r )ra -+ H'(P)"'.
We can also form the Steklov-Poincare operators 13,U : g H 0H that satisfy
H'12(r)n, -+ H-1/2(1')'' and
13,V :
Hi/2(r)"'
--
(4.37) and 13,V :
H-'/2(1')"'. (4.38)
Strongly Elliptic Systems
146
The purpose of this section is to prove that, under certain conditions, 13,U
are also bounded from H' (r)m to L2a fact that will be used
and
later in our study of surface potentials and boundary integral operators; see Theorem 6.12. Notice that
(Bvug, O)r = t'(Ug, V-0) = (g, B,VO)r,
(4.39)
so (B,U)* =13 V. If SZ is at least C", then the regularity estimates of Theorem 4.18 apply, and we can extend (4.37) and (4.38) as follows. C'+'.1 domain, for some integer Theorem 4.21 Assume that SZ is a bounded, If r > 0, and that P is strongly elliptic on S2, with coefficients in Cr.1 (0) the Dirichlet problem (4.35) has only the trivial solution in H 1(SZ)'" when g = 0, then the solution operator has the mapping property
U : Hs+1/2(r)m
Hs+l (S2)m
for 0 < s < r + 1,
and the Steklov-Poincare operator has the mapping property
B,U : Hs+1/2(r)m -+ Hs-1/2(r )m for -r - 1 < s < r + 1.
Proof. The case s = 0 is covered already in (4.37) and (4.38). Part (i) of Theorem 4.18 shows that U : H'+3/2(r)m -+ H'+2(Q)m, and thus 8,U: Hr+3/2(r)m -+ H'+1/2(r)m, which means that the result holds for s = r + 1. Boundedness for the range 0 < s < r + 1 now follows by interpolation, i.e., by Theorems B.8 and B. 11. The same arguments apply to V and B, V, so, in view of (4.39), we can extend X3VU in a unique way to a linear operator that is
bounded for -r - 1 < s < 0. Our task is much harder when c2 is permitted to be Lipschitz. In this case, with the help of the following integral identity. we will estimate Here, for the sake of brevity, we use the summation convention, i.e., we sum any repeated indices from 1 to n. Lemma 4.22 Assume that cZ is Lipschitz, and that the leading coefficients Ajk belong to W111 (c2)m"". For any real-valuedfunctions h 1, h2,
... , h" E W1 m),
and for' any u, v E H2(Q)m,
v,y{[(hlAJk - hJAlk - hkAjl)aku]*aiv} dx Jr
_ f {(Djkaku)*ajv +
(hk8ku)*(Pov)} dx,
Estimates for the Steklov-Poincare Operator
147
where
Djk = at(hIAjk) - (alhj)Alk - (alhk)Aji. Proof By the divergence theorem, it suffices to show that 8({[(hlAjk - h jAlk
=
(DjkakU)*ajv
- hkAjr)aku]*ajv} + (Pou)*(hjajv) + (hkaku)*(Pov).
In fact, the left-hand side expands to a sum of five terms,
[8r(hlAjk - h jAlk
- hkAjl)(aku)]*ajv
+ [(hlAjk - hkA j,)8,aku]*ajv + [ - h jAlkalaku]*ajv
+ [(h,Ajk - h jAik)aku]*alajv + [ - hkAjIaku]*a,ajv. The second term vanishes because its factor (...) is skew-symmetric in 1 and k, and likewise the fourth term vanishes because its factor (...) is skew-symmetric in I and j. The third term equals
h j [Pou + (alA1k)aku]*ajv = (Pou)*(hj81v) + [hj(atArk)aku]*ajv, and the fifth term equals
-(hkaku)*Aj*iatajv = (hkaku)*[Pov + (a,A;t)ajv] _ (hkaku)*(Pov) + [hk(a,Ajl)aku]*ajv, so we get the desired right hand side with
Djk = aI(hlAjk -hjArk -hkAji)+hj8lAlk+hkalAjl.
0
Rellich [85] used a special case of the above identity to obtain an integral representation for the Dirichlet eigenvalues of the Laplacian. Subsequently, Payne and Weinberger [81] generalised the Rellich identity to handle secondorder elliptic systems with variable coefficients, and used it to bound the errors in certain approximations to (D(u, u) and pointwise values of u, when u is the solution to a Dirichlet or Neumann problem. In what follows, we use the arguments of Necas [72, Chapitre 5]. We will use certain first-order partial differential operators of the form
Qu = Qky(aku) with Qk E L,,(P)mxm If vk Qk = 0 on I', then such a Q is said to be tangential to r.
Strongly Elliptic Systems
148
Lemma 4.23 Assume that 0 is a Lipschitz domain, and let u E CCIomP(S2)m.
(i) If Q is afirst-order, tangential differential operator, then IIQiIIL,(r)-n < CIIYullH-(r),,,.
(ii) The normal and conormal derivatives of u satisfy IIt3vuIIL2(r) n < Cllau/avlIL2cryn +
CIIYuIIHi(ryn,
and, when P is strongly elliptic on 2, I1au/a))1L,(r),n c,
(4.41)
then
da
I113" ull2Z(rro < C J
r
< CIIYajUIIL,(r)"(IIQjuIIL,(r)- + IIQ'UIIL,(r)- + IIQ';uHIL,(r) +CIIuIIH(n),., +CII7'ouliL2(s)''IIUIIHI(n)
By Lemma 4.23, IIYajuIIL,(r)"N
cllau/avjIL,tr)- +CIIYuIIH1(ryST CIIL.u1IL2(r)" + CIIYulIHI(r)-,
and so CII5vuIIL,(n-11YullH-(r)- +CIIYuIIHI(ry, +CIIuIIHI(n)w + CIIPouIIL2(n),"l111IIHI(n)Since II Pou II L, (n)., < II f 11L,(92)' + C 11U 11 HI (U)"',
the estimate in part (i) follows.
Now drop the requirement that u E H2(SZ)'", i.e., allow u E H' (c2)1, but assume y u E H 1(I')". It suffices to consider S2 of the form x" < (x'), where is Lipschitz with compact support. By Theorem 4.6, the operator P is coercive on Ho (0)'", so for A sufficiently large,
4)(u. u)
clIuIIH1(nr.
for u E Ho (l)'".
Choose a sequence Sr E CI (R"-1) such that 1.
S, -*
2. r
K is the sesquilinear form with coefficients
Ajk(Y) _ (ay Ars (x) 8xk) J(Y), \
r
s
, Aj(Y) _ (Ar(X)2)J(Y) ax,
AK(Y) = A (x)J(y).
(ii) Let PI denote the differential operator with the coefficients in part (i).
Show that Pu = f on 0 if and only if PKuK = fK on UK, where
fK(Y) = f(x)J(Y) (iii) Show that P is strongly elliptic (respectively, coercive) on Q if and only if PI is strongly elliptic (respectively, coercive) on 7K. 4.3 Show that if f (0) = 0 and a > 2, then
t-af
00
(Jo
I
1/2
(t) 12
dt)
[Hint: use Exercise 3.20.]
1/2
1
0,
on r, fort>0, on S2,
when t = 0,
where a and b are positive constants. Let 401, 02, ... and. 11, A2, ... be the eigenfunctions and eigenvalues of the stationary problem, as in Theorem 4.12, i.e.,
-a 0Oj = ,Xj¢j on 0, a0;
-}-by¢j =0
av
on!,
and derive the series representation 00
u(x, t) = > e-"(0 , uo)ccpj(x) In what sense does this sum converge?
for x E Q and t > 0.
5
Homogeneous Distributions
For a E C, a function u : R" \ (0}
u(tx) = t°u(x)
C is said to be homogeneous of degree a if
for all t > 0 and x E 1R" \ t0}.
(5.1)
To extend this concept to distributions, we introduce the linear operator Mr, defined by
MMu(x) = u(tx)
for 0 o t E'1R and x E IR",
and observe that for every u E L 1,10 (1R" ),
(Miu, 0) = Its-"(u, M1ltfi)
fort # 0 and 45 E D(1R").
(5.2)
If U E D*(IR") then (5.2) serves to define Mtu E D*(1R"), because M11, : D(1R") -* D(RI) is continuous and linear. We then say that u ED*(R) is homogeneous of degree a on IR" if M,u = t°u on 1R" in the sense of distributions, for all t > 0. This chapter develops the theory of homogeneous distributions, using Hadamard's notion of a finite-part integral to extend homogeneous functions on 1R" \ {0} to distributions on R. We consider in some detail the Fourier trans-
form of, and the change of variables formula for, such finite-part extensions. A technique used several times is to reduce the general n-dimensional case to a one-dimensional problem by transforming to polar coordinates. Most of the material that follows can be found in standard texts such as Schwartz [92], Gel'fand and Shilov [27], and Hormander [41], but the final two sections dealing with finite-part integrals on surfaces - include some less well-known results from the thesis of Kieser [48]. The results of this chapter will be applied later in our study of fundamental solutions of elliptic partial differential operators, and of boundary integral operators with non-integrable kernels. 158
Finite-Part Integrals
159
Finite-Part Integrals We begin our study of homogeneous distributions by focusing on the simplest example, namely, the one-dimensional, homogeneous function xa
10
if x > 0,
ifx -1, then x+ is locally integrable on R, and is obviously homogeneous of degree a as a distribution on R, not just as a function on R \ {0}. To deal with
the interesting case Re a < -1, we use the following concept, introduced by Hadamard [37] in the context of Cauchy's problem for hyperbolic equations.
..., bN+2 be complex numbers, with Re a1>0andaf#0 for I<j 1 and any 0 E S(R),
Ha(0) =
if Re a > -k - 1 and
(-1)kH°+k(0(k)) (a + 1)(a + 2) ... (a + k)
-1,-2,...,-k,
a
but 00
H-1(q) = -J
0'(x)logxdx 0
and
1
H-k-1 (0) =
1 O(k)(0) +
j
k! j=1
k!
1H-1
(O(k))
Proof. Integration by parts gives
_E) a
-
H + 1O,)
if a# -1.
Suppose first that Re a > -k - 1 and a ; -1, -2, ... , -k. By Taylor expansion, k-1 O(j) Ea+1O E)
=
i0)Ej+a+1
+0(6 Rea+k+l ),
j=0
so if Ha+1(q') exists, then Ha+1(0')
a+l and the first part of the lemma follows by induction. Next, integration by parts and Taylor expansion give 00
-0(E) logE -
_ -0(0) logE -
f0'(x)logxdx
J0
00 O'(x)
logx dx + O(E log,-),
implying the formula for H-1(0). Finally, when a H-k-1,, (
_-
E-ko
= -k - 1,
(E) + H-k W)
Homogeneous Distributions
162
and by Taylor expansion, k-1
(1)
1=1
J'
(k)
k
so if H_k (4') exists then (k)
+ Xk ( k
The formula for H_k_ 1(0) follows by induction.
'O
The next lemma shows that the distribution f.p. x+ is homogeneous of degree a on R, except when a is a negative integer.
Lemma 5.4 If c E S(R) and t > 0, then t-1Ha(Ml/tO) = taHa(O) fora # -1, -2, -3,
...,
t-1 H k-1(M'/to) = t-k-1 H k-1(46) + (t-k-1 log t)
(k)
but
ki0)
for any integer k > 0.
Proof If Re a > -1, then it suffices to make the substitution x = ty to obtain
t-'Ha(Mi/tcb) = t-' f OOxa¢(x/t)dx = to
f00
Ya,O (y) dy = taH.
-1, -2, ... , -k. For cf. (5.2). Now suppose that Re a > .-k - 1 and a brevity, write bk = (-1)k/[(a + 1) (a + k)]; then by Lemma 5.3, t-'Ha(Mi/t4) = t-1bkHa+k((M11 )(k)) = t-'bkHa+k (t-kMl/t4(k))
= bkt-k-lHa+k(M1/tO(k)) = bktaHO+k(O(k)) = taHa(t) However, when a = -1, we have
t-1H-l (Ml/A = -t-' J
(Ml/to)'(x) logx dx
0
_ -t-2
00
J0
0'(t-1x) l o g
dx,
Finite-Part Integrals
163
and the substitution x = ty gives 00
t-1 H-1 (MI/to) = -t-1 J
O'(Y) log(tY) dy = t-1 H_I (0) + 0 (O)t-' log t. 0
1/j, then
In general, if we let ck = (1/k!)
t-'H-k-1 (M11,O) =
1
t-'Ck(M1/iO)(k)(0) + k! k
= Ckt-k-1o(k)(O) + t kl
H_i ((M1/t0)(k))
1
H_I(M1/,0(k))
= t-k-1 (CkO(k) (0) + I H-1(O(k)))
+
0(ki 0) t-k-1
log t,
giving the second formula in the lemma. We shall also have use for the homogeneous function 0
x° = (-x)+ _
1xIa
if x > 0, if x < 0,
and its finite-part extension, E
(f.p. X!, 0 (x)) = f.P
fO -00
= f.p.
Ix JaO (x) dx
f xa0(-x)dx = Ha(M-t4b) E
It is easy to verify that
f.p.(-x)+ = f.p. x,,
(5.6)
and that f.p. xa is homogeneous of degree a on R, except when a is a negative integer. Indeed, Lemma 5.4 shows that if t > 0, then
f.p. (tx)f = to f.p. xt for a
-1, -2, -3, ... ,
but
f.p.(tx) fk-i = t-k-1 f.p.
x±k-1
+ (F1)k
t-k
log t 8(k) k
(x)
for any integer k > 0.
(5.7)
This loss of homogeneity does not occur in the case of the function x-k-1
Homogeneous Distributions
164
Indeed, the finite-part extension
x-k-10(x) dx
(f.p. x-k-1, 4b(x)) = f.p. E(O J XI>E
= H-k-1(0) + (-1)k+1H-k-I(M-10), satisfies x+k-1 + (_ l)k+1
f.p. x-k-1 = f.p.
f.p.
x_k-1
(5.8)
and so, by (5.7) and (5.6), f.p.(tx)-k-1 = t-k-1 f.p.x-k-1
if 0
(5.9)
t E IR.
One can formally integrate by parts k + 1 times to express (f.p. x-k-1, 0(x)) as a convergent integral.
Lemma 5.5 For q5 E S(R) and for any integer k > 0, 00 (f.p.x-k-1,,0(x))
r 0(k+1)(x)logIxI dx.
= k J o0
Proof. By (5.8) and Lemma 5.3,
(f.p.x-k-1, 0(x)) = H_k-1(0) + (-1)k+1H-k-I (M-1l) kk
E1
[O(k)(0) + (-1)k+I (M-10)(k)(0)]
k!J=1
(_l)k+1(M-1,)(k)),
+ k1 H-1 (.(k) + and thus, because (M_10)(k) = (-1)kM_1q5(k),
(f.p.x-k-1, q5(x)) = kI H_1 (0(k)
=
1
k1
f
°O
{
- M-I0(k)) (k+l)(x)
+.O(k+l)(-x)] logx dx,
giving the desired formula.
0
Later, when studying the Fourier transforms of f.p. xt and f.p. x-k-1, we shall encounter the distribution (x ± i0)', defined by
((x f MY', 0(x)) = li m foo (x ± iy)°4(x) dx, with the branch of z° = exp(a log z) chosen so that -ir < arg z < it.
(5.10)
Finite-Part Integrals
165
Lemma 5.6 The formula (5.10) defines a temperate distribution (x ± i0)' E S*(R), given by
(x ± i0)' = f.p. x+ + e}""` f.p. x° f o r a # -1, -2, -3, ... , and
(_ I k+l (x ± i0)-k-I = f.p. x-k-1 ± i7r
(k)
S
k
(x) for any integer k > 0.
Proof If Re a > -1, then we may apply the dominated convergence theorem to obtain the first formula. If -k - 1 < Re a < -k for some integer k > 0, so that Re(a + k) > -1, then integration by parts gives 00
(-1)k (x ± iy)a+k
foo
(a + ])(a+ 2)...(a + k)
f 00 (x + iy)aQb(x) dx =
01k) (x)
dx,
so in the limit as y 4. 0, we have (-1)k (xa+k + efin(a+k)xa+k)
((x f i0)a, fi(x)) _
(a + l)(a + 2) ... (a + k)
'
(k)
dk x+ k + (-1)kefinaxa+k dxk
and we have only to apply Exercise 5.3. However, when a = -k - 1,
L:x ± iy)14 (x) dx =
f log(x ± iy)
1(x) dx,
and since -n < arg(x ± iy) < it, yy im
log(x ± iy)
logx
ifx > 0,
log Ix I ± i7r
if x < 0.
Thus, by Lemma 5.5,
((x ±
i0)-k-1,
0(x)) =
00 flogIxI1)(x)dx
kI 1
k
_
° f(±iir)(x)dx
(f.p.x-k1, 0(x)) -
which yields the desired expression for (x ±
i0)-k-1.
ki (fiir) 0(k)(0),
Homogeneous Distributions
166
Extension from W'\{0} to 1[t" If U E L1,1oc(R" \ {0)), then we can try to define the finite-part extension f.p. u as a distribution on R' by writing
(f.p. U, 0) = f.p. f
u(x)¢(x) dx for O E D(R").
XI>E
In the special case when the finite part of the integral is just a limit, we speak of the principal value p.v. u, i.e.,
(p.v. U, ) = limJ
40 IXI>E
u(x)O (x) dx.
Suppose now that u E C°°(R' \ (0)) is homogeneous of degree a. By in-
troducing polar coordinates p = lxi and w = x/p, so that x = poi and dx = pn-1 dp dco, we find that
f
u(x)cb(x) dx = J
pat"-1-0 (pce)) dp dcv.
u(co) p>E
X I>E
This formula prompts us to define the linear operator Ra by Racb(x)
= f.p.
40 JE
po+n-1cb(px) 00
dp = (f P
p+++n-1,
(px))
for X E R" \ {0),
(5.11)
so that
(f.p. U, 0) =
JI of=1
u(c))Ra¢(co) dcv
for ¢ E S(R").
(5.12)
Here, to justify taking the finite part inside the integral with respect to co, it suffices to check that the o(1) term in the expansion of fE00 pa-n+l0(pco) dp tends to zero uniformly for lwJ = 1. As a consequence of Lemma 5.4, we are able to prove the following.
Theorem 5.7 Suppose that u E C°°(R" \ 0) is a homogeneous function of degree a.
(i) If a -n, -n - 1, -n - 2, ... , then f.p. u is the unique extension of u to a homogeneous distribution on W.
(ii) If a = -n - k for some integer k > 0, then a homogeneous extension
Extension from R" \ (0) to R"
167
exists if and only if u satisfies the orthogonality condition
whenever l a l= k.
wa u (co) d w= 0
(5.13)
In this case, the homogeneous extensions of u consist of all distributions of the form
f.p. u + Y' ca 8.8, Ial=k
with arbitrary coefficients ca E C.
Proof Consider
f
(Mr f.p. U, 0) = t-r" (f.p. U, MI/r4>) =
u(w)t-"RaMI/r4>(w) dcv.
IwI=I
Since RaMI/ra(w) = Ra4>(t-ico), we see from Exercise 5.4 that for a as in part (i), U(w)t-n(t-1)-a-lzRa4)(w)
(Mr f.p. U, 0) = f
dw = (ta f.p. u, 0).
wl=1
However, if a = -n - k, then (Mr f.p. U, 0) U(w)t-"
((t_t)kR_il_k4>(cv)
-
(t-1)klog(t-1)
Iwl=1
E aao(0)wa a
dw
lal=k
f.p. U, 0) +
t-"-k
as of
log t
a.
Ial=k
w"u(w) dw,
Iwl=1
so f.p. u is homogeneous on R" if and only if (5.13) holds. To settle the question of uniqueness, and to show the necessity of the orthogonality condition for existence if a = - n - k, let u E V* (R") be any extension of u. Since u - f.p. u = 0 on R" \ 0, Theorem 3.9 implies that
ii -f.p.u=Ecaa,3, a
where the sum is finite. The result follows because 8a8 is homogeneous of degree -n - lad; see Exercises 5.1 and 5.2.
Homogeneous Distributions
168
The next theorem complements the one above, and introduces a particularly important class of homogeneous functions satisfying the orthogonality condition (5.13).
Theorem 5.8 Suppose that u E C°O(R" \ {0)) is a homogeneous function of degree -n - k, for some integer k > 0. If u has parity opposite to k, i.e., if U (_X) = (_1)k+1 U (X) for x E R" \ {0},
(5.14)
then
(f.p. U, 0) =
p-k-1, O(pw)) dw for ¢ E S(R"),
12 f
Iwl=1
and f.p. u is the unique extension of u to a homogeneous distribution with parity opposite to k.
Proof. The parity condition (5.14) implies that (f.p. U, ci} =
f
u(-m)R-"-k.(-w) dw
Iwl=1
f
(-1)k+1u(w)R-n-k41(-w) dw,
wl=1
and by (5.6),
R-n-ko(-w) _ (f.p.
p+k-1,
(f.p.
p_k-1, O(pw)),
so we have
f =2 f
(f.p. U, ) = 1
2
u(w)[R-n-k O(w) + (-1)k+1R-n-k(-w)] dw
w l=t
u(w)(f.p. p+k-1 + (_1)k+1 f p
p- k-1, 0) dw,
wl=1
giving the desired formula; recall (5.8). The homogeneity of f.p. u follows from Theorem 5.7 because the parity condition (5.14) implies the orthogonality condition (5.13). Alternatively, one sees from (5.9) that
M, f.p. u = Itl-"t-k f.p. U
on R" fort E lR \ {0),
and in particular, f.p. u has parity opposite to k. Finally, if Ia I = k then F 8 has the same parity as k, so f.p. u is the only homogeneous extension of u having the opposite parity to k. 0
Fourier Transforms
169
The uniqueness results in Theorems 5.7 and 5.8 yield a simple proof of the following fact.
Theorem 5.9 Suppose that u E C°O(R" \ {0}) is a homogeneous function of degree a, and assume, if a = -n - k for some integer k > 0, that u has parity opposite to k. Then for any multi-index a, 8 " f.p. u = f.p.(a"u)
on I[8".
P r o o f By Exercise 5.2, if a # -1, -2, -3, ... , then a" f.p. u and f.p.(a"u) are homogeneous extensions of a"u with degree a - I a I , and must therefore coincide. If a = -n - k but u has parity opposite to k, then
a"u(-x) _(-1)k+I"I+1a"u(x) for 0 0 x E W, so 8"u is homogeneous of degree -n - (k + lal) and has parity opposite to k + la l. Thus, a" f.p. u and f.p.(a"u) are again homogeneous extensions
0
of a"u, and both have parity opposite to k, so they must coincide.
Fourier Transforms Our aim in this section is to compute the Fourier transform of the finite-part extension of a homogeneous function. Following the pattern of previous sections, the one-dimensional distributions f.p. xf and f.p. x-k-1 will be treated first. In order to state the next lemma, we require the gamma function,
'(a)
=f
00
forRea>0.
In the usual way, r is extended by means of the identity
r(a + l) = ar(a) to a meromorphic function on C having simple poles at 0, -1, -2, ... , and satisfying r (k + 1) = k ! for any integer k > 0. T:r_.4 {f.p. x' 1. If a # -1, -2, -3, ... , then
Lemma 5.10 Let IIa
}
r1Q ()
_
r(a + 1) (±i2ir)"+1
(
i0)
,
but for any integer k > 0,
IZ}k-1( ) _
(:Fi2 r
k
(1o27rll ± 12 sign() - r'(1) - E
I
j=1 I
(When k = 0, the empty sum over j is interpreted as zero.)
Homogeneous Distributions
170
Proof To begin with, suppose that Re a > -1. For any 17 > 0, the function e-2nglxlx11 belongs to LI(R), and 00
Fx,
{e-2nglxlXa l = fJ J
t
dx.
Making the substitution z = 27r (ri f ii )x, and then applying Cauchy's theorem to shift the contour of integration back to the positive real axis, we find that o
[27r(ii±i
_
(±i27r)a+l
e-zzadz
Jo
J
1
00
a1??)--1
1'(a + 1) (
Sending ri J, 0, we obtain the first part of the lemma for Re a > -1, and the
case Re a < -1 then follows by analytic continuation; see Exercise 5.6. For the second part of the lemma, consider 00
H_k-1,E
)=f
(a-i2Trl
e-ibrtxx-k-1 dx,
E
where f,.00 is interpreted as limn. fE if k = 0. Suppose l; E R}, and make ±i27rIr; Ix to obtain the substitution z = H-k-I,E(e-i2n1;,)
_
(i27rf)k
f fi00 Jti2ir I IE
e-:.z-k-1 dz.
Applying Cauchy's theorem (and Jordan's lemma, if k = 0), we see that fi0c
2,rI
i2n1? IE
f.
o00
e-zz-k-1 dz =
e-zz-k-1
dz = f 1
4
=±
e-zZ-k-I dz
- fc.,
e-zz-k-1 dz,
IE
f
e-zz-k-1
resse-zz-k-1
=
dz
(_I)_ k
o l2
(i27rj)k (H_k_I.2rIIE() T
k
i7r (-1)k 2
k!
)
Fourier Transforms
171
By Exercise 5.7 together with Lemma 5.3, we have f-
k' 0) log 27r l
P (0)
(P
k!
- (-1)k k
k
- - tog27r ICI + k! H-t 1
1
i=t J
E i=t k
1
J
- l0 g27r ICI+e-xlogxdx
I
(-i27rt)k
i7r
2 sign(t)-1
kt
(f.p. xk-', fi(x)) = f.p. f
x-k-t f
J
E
oo
,
k (1)-E _1
i=t
00
(n-k-t'
I
I
e-i2n1XO(t) dt dx
00
00 Elo
foo
where the final step is justified because the o(1) term in the expansion of can be bounded by f (E)g(t), with f (E) = o(1) as c 4. 0, and g(') having only polynomial growth as It I oo. The formula for n±k-i (t) H_k-t.E(e-i2"t')
is now established, and the one for II-k-t (t) then follows by (5.6) because the 0 Fourier transform commutes with M_ t; see Exercise 5.8.
As an immediate consequence of Lemmas 5.6 and 5.10, we obtain the formulae below; see also Exercise 5.10. Corollary 5.11 F o r a n y integer k, let 1-lk() =
.
_
1 nk(t,) = (-i2zr s'kl(t) and l1-k_, (t) _ ( )k
{f-p. xk}. If k > 0, then 1227r 7r
tk sign(t).
Turning to the general, n-dimensional setting, we require the following technical lemma.
Homogeneous Distributions
172
Lemma 5.12 Let a E C and U E D* (R" ). If u is homogeneous of degree a on R" \ {O}, i.e., if Mtu = tau on Rn \ {0} for all t > 0, then u E S*(R"). If, in addition, u is CO° on R" \ (0), then u is C°° on R' \ (0). Proof Let* E C mP (R" \f0}) be as in Exercise 5.12, and define X E C mP (R" ) by X(x)=1-J1
*(tx)
dt
forxER".
0
Put uO = Xu and ul = (1 - X)u; then uo E S* and uo E C°°(R") because uo has compact support, so it suffices to consider u I. We have
(ui, ) = (j' ir(tx)
a tu(x),
0(x)=
f '(u,
OM1
fr)
dt
for q E D(Rn),
(u, OM *) = (u, Mt (*Ml/t-O)) = (t-"Ml/tu, rMl/to)
= t-n-a(u, *M,1, 0). Let K = supp *; then there is an integer k such that
1(u, iMi,t4)I < C E max I aa(*Mltt4) I< C E t-I°`I max I a"O(t-lx)I lal n + ja I + Re a, then the function 8. [(-i2zrx)a u 1(x)] belongs to L 1(1 ' ), and we deduce from (3.17) that (i2iri; )O 8"u i
is continuous
onR R.
Lemma 5.12 shows that the Fourier transform of a homogeneous distribution always exists as a temperate distribution. Furthermore, the following holds. Theorem S.13 Let a E C. If u is a homogeneous distribution of degree a on I[2",
then its Fourier transform u is a homogeneous distribution of degree -a - n
on R. Proof. For t > 0 and 0 E D(R ),
(Mru, 0) = (Mr.Tu, ) = t-"(u, .FM1/ro), and by Exercise 5.8, t-"(u, .FM11r4) = t-"(u, t"M,.Fo) = t-"(M111u, F4) _ (t-n-au, ,F4) = (t-n-a Fu, 95),
so Mrir. = t-n-au.
If U E C°°(lf8" \ (0)) is homogeneous of degree a, then by Lemma 5.12 the finite-part extension f.p. u is a temperate distribution on 1l8", and, recalling (5.12), the Fourier transform of f.p. u is given by
(.F f.p. U, 0) = (f.p. U, ) =
dco
JIwI=1
for g5E S( (5.15)
We can express Raq in terms of the one-dimensional Fourier transform in Lemma 5.10.
Lemma 5.14 If X E 1[8" \ {0}, then
Rac(x) = (n +n-1(S
x), ( ))
for 4 E S(R").
40+11-1, (px)). To express the Proof By (5.11), we have (f.p. function p H c (px) as a one-dimensional Fourier transform, we make the substitution = i;1 + tx/Ix12, where t = i; x and thus '_r is the orthogonal
projection of onto the hyperplane normal to x. In this way,
(px) = J 0 f tl=o
t l dal dt
_
(p),
Homogeneous Distributions
174
where
0,'(0 =
1+
Ixl JX =o
IX12x
dal,
so
Rac(Px) = (f.p p++n 1, O^x (P)) = (na+n-! OS) Now see Exercise 5.11.
Together, (5.15) and Lemma 5.14 show that the Fourier transform of f.p. u is given by
f.p.u,q5)= f
))dw for
u(w)(n
Iwl=
ES(R'), (5.16)
and similarly, the inverse Fourier transform of f.p. u is given by
u(w)(II +,:-1(-x w), q5(x)) dcv for O E S(R").
(.F* f.p. U, 0) Iwl=1
(5.17)
If Re a < I - n so that
E L 1,10c (R), then we may write
(f.p. u(x)) = and
-,C{f.p. u(i )} =
L
IIa+n-1(
co)u(a)) do)
I=1
Iwsee
f
]Ia}n-1(-x w)u(w) dco;
(5.18)
Exercise 5.12 for alternative formulae that do not require any restriction on a.
Change of Variables
We wish to investigate the effect of a change of variable x = K(y), where K : S2" -+ S2 is a Cx diffeomorphism satisfying K (O) = 0,
and S2" and S2 are open neighbourhoods of 0 in R". For any E > 0, let BE={yER,':lyl<E}
Change of Variables
175
denote the open ball in 1R" with radius c and centre 0, so that if u E L 1.10c (Q \ (0} )
and 0 E D(Q), then
J
u(x)O (x) dx = J
(5.19)
Theorem 5.15 below implies the existence of the finite part of the right-hand side of (5.19) when u is homogeneous, or, more generally, when u is the sum of finitely many homogeneous functions and a remainder term that is integrable on Q.
Theorem 5.15 If U E C°°(1R" \ (0}) is homogeneous of degree a, then there exist E > 0 and functions wo, w1, w2, ... and Ro, R1, R2, ... with the following properties: (i) For every m > 0, the composite function u o K admits the expansion m
u(K(Y)) _ > wi(y) + R,. (y) for0 < IYI < E. i=o
(ii) For every j > 0, the function wj is C°O on IR" \ (0) and homogeneous of degree a + j :
wj (ty) = t'
wi (y)
fort >0 and y ER' ' \ {0}.
(iii) For every m > 0, the function R, is CO° on BE \ (0) and, for every multi-index a, IaaRm(y)1
1, we can define a C°° function f : SZ -+ IR by
(K(y))a=
1
[K'(0)]m
ifyES2\0,
ify=0,
Change of Variables
179
and so f. p. E JO
1
{K(±E)j'n
L p. EO
f (±E) =
f
(±E)m
cnn)
(0) mI
Thus,
Jk(0) =
Jk-l(0') k
and (5.22) follows by induction on k.
For c sufficiently small, the set K(BE) is approximately ellipsoidal and can be described using the function g in the next lemma. Recall that Sn-' = {w E R" : I w I = 1 } denotes the unit sphere in R". Lemma 5.17 There is an co > 0 and a C°O function g : (-EO, EO) X
n S-1
-o- IIB,
such that
K(BE)fl(pw:0 0) but u satisfies u(-x) _ (-1)k+)u(x), then f.p. Elo
f
u(x)O(x) dox \B.
= f.p. E.IA
for 0 E D(r).
f IX'1>E
u(x', (x'))O(x', (x')) 1 +
dx'
Homogeneous Distributions
182
Proof We introduce polar coordinates in R"-', writing
x' = rw,
r = Ix'I,
co = x'/r E S!,-2.
For X E F, we have IX 12 = r2 + (r(0)2, and so
x E F \ BE f r l +
E.
Since (0) = 0, there exists EO > 0 and a C°O function g : (-EO, EO) X S"-2 R such that 1' \ BE = { (x', (x')) : x' = rw, r > g (E, co) and w E Sn-2 }
for 0 < E < co,
with g(E, w) _ -g(-E, -co); cf. Lemma 5.17. The result follows by arguing as in the proof of Theorem 5.18, remembering that now the integral is over RI-I instead of R". Finite-part integrals on surfaces arise as boundary values of functions of the form
f (x) _ f u(x - y),/i(y) day for x E R" \ IF, r
(5.26)
where the integral is divergent if x E F. We shall consider u of the following type.
Assumption 5.20 The distribution u is of the form
u(x) = Jct%x[v(4))
with
P( )
where p and q are homogeneous polynomials satisfying
(degree of p) - (degree of q) = j - 1 for some j > -n + 2, and in addition
q(); 0 forall ER' \{0). Thus, V E C°O(R" \ {0)) is homogeneous of degree j - 1, and so is locally integrable on lid" because j - 1 > -n + 1. By Theorem 5.13, we see that u is a homogeneous distribution on ][l;" of degree -n + 1 - j, and by Lemma 5.12, u is C°° on R" \ {0}. Also, by Exercise 5.8,
v(-l;) = (-1)'-'v(l;)
and
u(-x) = (-1)j-lu(x).
(5.27)
Finite-Part Integrals on Surfaces
183
Figure 4. Integration contours in the definition (5.28) of f }.
To begin with, we investigate the simplest case, when I' = R' ' and so
f(x) =
f
u(x' - Y', x")ir(Y) dy'
for x,,
0.
The next lemma gives an alternative representation for f in terms of the Fourier transform of *. We denote the upper and lower complex half planes by
C+={zEC:Imz>O} and C-={zEC:Imz ro and z E C}, then we denote the integral of w over C} by
f w(z) dz =
f
cr
w(z) dz
for r >ro.
By Cauchy's theorem, this integral is independent of r.
Lemma 5.21 If u satisfies Assumption 5.20, and if i,r E S(Ri-1), then
f (x) =
forx E R±,
where
m±(', x") =
f
f
Furthermore,
t-ix,,) = tim±(',
fort >0 and i'540,
(5.28)
Homogeneous Distributions
184 and
m+(-C, -xn) = (_1)r-'m-(', xn). Proof The function f has a natural extension to a distribution on R", namely the convolution u * (i/r 0 8), where (Vr (9 8)(y) = +/r(y')S(y"). Hence, for all 0 E S(II8"),
(f, 0) = (u * (tk 0 8), .F.F*O) = (F[u * ( ® S)], f*O)
= J '*O. Suppose now that supp o c R". In this case, for each i;' the function ¢( ', ) is continuous on C± U R, is analytic on C±, and satisfies where
bounds of the form
4(01:5 CM,N(1 + WD-MO + It U-" fore' E R` I and
E C} U
Hence,
H t"
f
d'n oo
and to shift the contour of integration in the "-plane, we consider the poles of putting Z(i') = ( " E C : q('', l") = 0 ). Since q is homogeneous, we have Z(ti;') = and since the coefficients of the polynomial q(i', ) are continuous functions of ',
for
f
f
00
d ,= =
v( 00
=
f v()() d = f f f
v(
)ei2'
x0 (x) dx d "
x,:) dx.
yR 'RI,
the substitution z = ti;, gives
Finally, since
m±(t ', t-ixn) =
f fv(t)ei2ht x t
dz
trm f(i;, x),
Finite-Part Integrals on Surfaces
185
the substitution z = - gives
and since
M+(-C, -xn) =
dz
Jfc;
f
C,
by (5.27).
which equals
Since m±(k', is a C°C function of x,,, we see that the restrictions f I w. and f IR,. can be extended to C°O functions on R. We now consider the one-sided boundary values of f on the hyperplane x, = 0, given by
f.+(x') =
lim
Z--s(X'.o),ZEWI
f (z) = fR11
-I
mf(',
')e'2rtX d ' for X' E R!'- 1.
Theorem 5.22 Suppose that u satisfies Assumption 5 .20, and that i E S(Il8i-1)
(i) If j < -1, then f+ W) = .t--(x') =
f
u(x' - y', 0)f(y') dy'.
(5.29)
Rn-1
(ii) If j > 0, then
f+ (x') + f_(x') = 2 f.p. E4O
f
IX'-Y'l>E
u(x' - y', 0)i(y') dy',
and the jump in f across the hyperplane x = 0 has the form
f+(x') - f-(x') = E c«a°*(x'), IaI=i
for some coefficients Cc, E C.
Proof Using the sum and difference of m+ and m_, we define us(x')
where
ms( ') =
0) + m-( ', 0),
and
ud(x') =
where md(') = m+(,', 0) - m-(', 0),
Homogeneous Distributions
186
so that
f++f_=u5*ifr
f+-f_=ud*1/r.
and
By Lemma 5.21, m5 and and are homogeneous of degree j, and satisfy
(-1)'-lm5(4') and md(- ') = (-1)'md('), so us and Ud are homogeneous distributions on R" of degree -j - (n - 1), and satisfy
u5(-x') = (-1)1-1u5(x')
and
ud(-x') = (-1)!ud(x').
Since u(x - y) is C°° for x # y, it is easy to see that if x'
supp *, then (5.29) holds (even if j > 0), and in particular f+ (x') - f- (x') = 0. Therefore
suppud(x' - ) c {x'}. If j < -1, then Ud E L111,,(Rn-1) so Ud = 0. If j > 0, then, with the help of Theorem 3.9 and Exercise 5.1, we deduce from the homogeneity of Ud that
Ud*t/t= Ecaaa*, lal=i
for some ca E C. Furthermore, u5(x' - y') = 2u(x' - y', 0) for x'
y', because (5.29) holds when x' f supp *, so the homogeneous distributions u5 and 0) are equal on R"-1 \ {0}. Since both have degree -(n - 1) - j 2f.p. and parity j - 1, we see by Theorem 5.8 (applied in R11- 1, not R") that in fact D 0) as distributions on R' 1. us = 2 f.p. Suppose now that r is the graph of a C°° function We denote the epigraph and hypograph of by
Q' _ {x E 1R" : x" > C(x')}
: IRi-1 -+ R, as in (5.25).
S2- = {x E 1R" : x < (x')},
and
and denote the boundary values oof the function f defined in (5.26) by
ft(x) =
lim
J u(z - y)*(y) day for x E F. r
It is possible to generalise Theorem 5.22 as follows.
Theorem 5.23 As before, let r be the CO° surface (5.25), with (0) = 0. If u satisfies Assumption 5.20, and if r E D(r), then the restrictions f Iu+ and f In-
can be extended to C°O functions on 1R1. Moreover, for x = (x', (x')) E F we have
(i) if j < -1, then /'
f+(x) = f-(x) = J u(x - y)f(y) day; r
Exercises
187
(ii) if j > 0, then
u(x - y)f(y)dav f+(x)+f-(x) =2f.p.J 40 r\e, (x) and, for some coefficients ba E C°O (Rn-1),
f+(x) - f-(x) = E ba(x')8 I (x', (x')). Ia1
We shall not prove this result, because techniques from the theory of pseudodifferential operators would be required; cf. Kieser [48, Satz 4.3.6] or Chazarain and Piriou [ 10, p. 280]. In all subsequent proofs, we avoid using Theorem 5.23. It does, however, help to account for some of our results in Chapter 7.
Exercises
5.1 Show that the Dirac distribution 8 E D* (R") is homogeneous of de-
gree -n. 5.2 Show that if u E CcO(R" \ {0}) is homogeneous of degree a E C as a function on R" \ (0), then for any multi-index a the partial derivative 8"u is homogeneous of degree a - lad on R" \ {0). Show further that if a E D*(R") is homogeneous of degree a as a distribution on R", then 8"u is homogeneous of degree a - IaI on R". 5.3 Show that
d -
f.p. x} = ±a f.p. x f
1
for a
-1, -2, -3, ... ,
-
but
d f.p. x fk =- Fk-f .p. xtk-1 ± (k i)k
6(k) (x)
for any integer k > 1.
Deduce that
dx
f.p. x-k = -k f.p.
x-k-1
5.4 Recall the definition (5.11) of Ra0. Show that if 0 E D(R"), t > 0 and
xEIR"\{0},then R0¢(tx) = t-a-"RaO(x)
for a O -n, -n - 1, -n - 2, ...,
Exercises
188 but
R-n-k 0(tx) = tkR-n-k4 (x) - tk logt E
8 0i )x
I"I=k
a!
for any integer k > 0.
5.5 Show that if u is a homogeneous function in C°°(]RI \ {0}), and if * E
C' (R"), then* f.p. u = f.p.(*u). 5.6 Show that for each 0 E S(R), the function a H H,,(0) is analytic for a ¢ -1, -2, -3, ... , with all simple poles, and residues ck)(0)
res
a=-k-l Ha(4) _
kI
5.7 Show that if A > 0 and 0 E S(R), then Otki0)
f.p. H_k-l,AE(O) = H-k-1(0) -
logA
Ej0
for any integer k > 0. 5.8 Show that for t 0,
M,T= ItI-"FM111,
M,.7 r'*= ItI-"-T*MI1t,
,'Mr = ItI-"M111.P, F*Mt = ItI-"Mllt.'F*. 5.9 Show that f.p. El0
J
x°-1 a-z dx = F(a)
for a E C \ (0, -1, -2, ...}.
E
5.10 Prove Corollary 5.11 directly, i.e., without using Lemma 5.10. For the first part, use (3.17), and for the second part, show that H_A_I (e- i2a
)=
(-12Jt)k+1 2kl
k
for any integer k > 0.
sign (s)
5.11 Show that if u E L1,10c (R) and 0 E D(R"), then
u(t)oa(t)dt,
.ifR." u(a x)q5 (x) dx = FOO
where
f
0. (t) = l
al I
.,.L=O O
(x' + talla) dxl, I
Exercises
189
and dx 1 is the surface element on the (n - 1)-dimensional hyperplane
normal to a. Show further that 0a E D(R), and that if u is a distribution on R, then u(a x) makes sense as a distribution on I[8". 5.12 Derive alternative formulae to (5.12), (5.16) and (5.17), as follows. (i) Show that there exists f E C mP (0, oo) satisfying
r
00
f (t)
dt t
00
=1
and fo
f (t) log t dt = 0.
[Hint: look for f in the form f (t) = Cg (At) for appropriate constants
C>0and),>0.] (ii) Deduce that the cutoff function 1i (x) = f (Ix 1) belongs to Cmp (R" \ {0}), and satisfies
I
W
(tx) t = 1 and
f0*
(tx) logt
dlog
IxI
for x E R" \ {0}.
(iii) Show that if u E C0(R" \ {0}) is homogeneous of degree a, then (f.p. U, ¢) =
J
u(x)*(x)RQ¢(x) dx
for O E S(W").
(iv) Deduce that
TX,g{f.p.u(x)} = (n
*(x)u(x))
and
)7":(f-p-U(0) = (nt+n-1(-S . x), / (f)u( )) 5.13 Show that if u E C°°(Il8" \ (0)) is homogeneous of degree -n - k for some integer k > 0, and if u satisfies the parity condition (5.14), then
",(f.p.u( )} _
fl-k-1 (-x co)u(w)dco
-1
2 1.1=i (i2ir)k+1
4k!
J WI=i
sign(x CO) (X w)ku(w) dw.
5.14 Suppose that K E C°°(R" \ {0}) is homogeneous of degree -n, and that
K(w)dw=0. Iw1=1
Exercises
190
Define
K(x-y)u(y)dy forx ER",
KEu(x)=J ly-xl>E
whenever this integral exists and is finite, and put
Ku(x) = limKEu(x), whenever this limit exists. (i) Show that if u is Holder-continuous and has compact support, then
Ku (x) = I
K(x - y)[u(y) - u(x)] dy
Ix-yl 0 is any number such that u (y) = 0 for Ix - y I > Px (ii) Show, with the assumptions of (i), that KEu -* Ku uniformly on compact subsets of R. (iii) Show that p.v. K exists and is a homogeneous distribution of degree -n on R. [Hint: see Theorem 5.7.] (iv) Show that Ku = (p.v. K) * u. C II u II Hs (R") for -oo < s < oo, and (v) Deduce that II Ku II H (w') that I Ku IK < C I u I ,,, for 0 < .t < 1. [Hint: use Theorem 5.13 and Lemma 3.15.]
6
Surface Potentials
Following the notation of Chapter 4, we consider a second-order partial differential operator n
n
n
Pu = -TL: aj(Ajkaku) +>Ajaju + Au. J=1 k=1
J=1
From this point onwards, we shall always assume that P has C°O coefficients and is strongly elliptic on W. Thus, Alk, Aj and A are (bounded) C°O functions from R" into CmXI, with the leading coefficients satisfying n
n
Re >2>2[Ajk(x)4k71]* jrl ? CIA121,712
for x,
E R" and rl E Cm .
j=1 k=1 (6.1)
In this and subsequent chapters, we shall develop integral equation methods for solving boundary value problems involving P. Such methods require a twosided inverse for P, or, more precisely, they require a linear operator !9 with the property that
PGu = u = GPu for u E £*(R")"'.
(6.2)
Since P is a partial differential operator, it is natural to seek g in the form of an integral operator:
1
CJu(x) = f G(x, y)u(y) dy for x E W.
(6.3)
R's
The kernel G is said to be a fundamental solution for P, and the same term is also applied to the operator g, although we shall sometimes refer to the latter as a volume potential. We shall also work with a kind of approximate fundamental solution, known as a parametrix, that is generally easier to construct. 191
Surface Potentials
192
The plan of the chapter is as follows. The first two sections set out the main properties of parametrices and fundamental solutions, emphasising the simplest case when P has constant coefficients. Next, we prove the third Green identity, in which the single- and double-layer potentials arise. Following the approach of Costabel [14], we then prove the jump relations and mapping properties of these surface potentials for the case of a Lipschitz domain. The final section of the chapter establishes some relations between the surface potentials associated with P and those associated with P*.
Parametrices A smoothing operator on R" is an integral operator
)Cu(x) _ ! K(x, y)u(y)dy forx E R", whose kernel K is C°O from R" such K satisfies
into C" xm a it is easy to see that any
E(
K : E* (R")'"
Conversely, it can be shown that every continuous linear operator from E* (R") into E(IR")m has a C°° kernel; see [10, p. 28].
A linear operator G : E*(Rn)"' -+ D* (W)"' is called a parametrix for P if there exist smoothing operators K 1 and K2 such that
PGu = u - Klu and GPu = u - /C2u
for U E E* (W')'.
(6.4)
Roughly speaking, a parametrix allows us to invert P modulo smooth functions. Later, we shall write G as an integral operator as in (6.3), and refer also to its kernel G (x, y) as a parametrix for P. When P has constant coefficients, we can easily construct a parametrix via the Fourier transform. Indeed, let PO and P (t;) denote the polynomials corresponding to Po and P, respectively, i.e., n
Po(e) =
(27r)2
j Ajk k j=1 k=1
and
P( j=1
Parametrices
193
For any u E S* (R")'",
and
.7=x_{Pou(x)} = Po( )u( )
{Pu(x)} =
..F
and the strong ellipticity condition (6.1) can be written as for
Re77*Po(')'1 ?
E R" and 17 E C"'.
then Thus, if b = Re r7*b < 17711b1, giving follows that the £2 matrix norm of Po(e)-1 satisfies the bound IPo( )-11
SF,I[1-x( )]I I-2 ()I2d C II u I I H- I (W,),,,
proving the desired mapping property.
In the general case when P is permitted to have non-constant coefficients, a parametrix g can be constructed using the symbol calculus from the theory of pseudodifferential operators; see Chazarain and Piriou [10, pp. 221-224]. The mapping property of Theorem 6.1 remains valid locally, i.e., given any fixed cutoff functions X1, X2 E Cm p(R"), X1 CJX2 :
H'-'(R")
HS+1(1[8")"`
for -oo < s < oo.
(6.10)
The next lemma will help us to describe the behaviour of the kernel G (x, y).
Lemma 6.2 Suppose that v E C°O (ll8" \ {0}) is homogeneous of degree -j for
some integer j > 1. If
u(x) = T4*'.'(f.p. v()}, then the distribution u is locally integrable on 1[8", and is CO° on 118" \ {0}. Moreover,
(i) if 1 < j < n - 1, then u is homogeneous of degree j - n; (ii) if j > n, then u(x) = u1(x) +u2(x)log1xI, where u 1 and u2 are homogeneous of degree j - n and C°O on R" \ {0}, with u2 a polynomial.
Parametrices
195
Proof Part (i) follows at once from Lemma 5.12 and Theorem 5.13, because v is locally integrable on R". If j > n, then by (5.18),
II±-x w)v(w) dw,
u(x) _ ICI=1
and part (ii) follows from Lemma 5.10.
0
We state the next theorem for the general case, but give a proof only for P having constant coefficients.
Theorem 6.3 Assume that P is strongly elliptic with C°° coefficients on R". There exists a parametrix 9 for P whose kernel admits an expansion of the form N
G(x,y)Gj(x,x-y)+RN(x,y),
(6.11)
j=U
for each N > 0, where the functions Go, GI, G2, ... and Ro, RI, R2, ... have the following properties:
(i) For each j > 0, the function G j is C°D on Il8" x (Rn \ (0)), and has the same parity as j in its second argument, i.e., G j (x, -z) = (-1)j G j (x, z). (ii) If 0 < j < n - 3, then G j (x, z) is homogeneous in z of degree 2 - n + j. (iii) If j > n - 2, then G j has the form G j(x, z) = Gj1(x, z) + G j2(x, z) log Izl, where G j I (x, z) and G j2(x, z) are homogeneous in z of degree 2 - n + j, with G j2(x, z) a polynomial in z.
(iv) If0 n - 2, then RN is C2-"+N on R", and
aaRN(x, y) = O(Ix - y12-n+(N+1)--IaI log Ix _ y;) as Ix - yI
0,forIal >2-n+(N+1).
Proof As mentioned above, we assume P has constant coefficients, and consider G given by (6.9). The choice of po ensures that there exists an expansion
Surface Potentials
196
of the form
P( )-1 =
for ICI > Po, j=o
with Vj E CO0(R" \ (0))"'11 rational and homogeneous of degree -2 - j. We define
Gj(z) _ and apply Lemma 6.2 to obtain (ii) and (iii). The expansion (6.11) serves
to define RN, and we see that G j (-z) = (-1)jG j (z) because Vj (-t) _ (-1)jV1(4) Write RN(x, y) = RN,1(x - y) + RN,2(x - y) + RN,3(x - y), where N+I
RN 1 = -X E f.p. V3,
00
RN.2 = f.p. VN+I,
RN,3 = (1-X) E V3.
j=0
j=N+2
We see that RN, I is C°O on R" because RN, I has compact support. Parts (ii) and (iii) imply that RN,2(X, Y) = RN,2(x - y) = GN+1 (x - y) satisfies conditions (iv)-(v). To deal with the remaining term RN,3 (X, Y) = RN,3(x - y), we use the bounds
ITx-, {(-127rz)fla"RN,3(z)}I = la
)I
C(1 +
if -2 - (N + 2) + I al < -n - 1, and thus 8" RN,3 is continuous on I[8" if la l < 3 - n + N.
Indeed, taking fi = 0, we see that .'{8"RN,3} E
Furthermore, by summing over I Ig I = r > 0, we see that I z I' 8" RN, 3 (z) is
bounded for z E IE8" if -2 - (N + 2) + lal - r = -n - 1, i.e., if -r = 2 - n + (N + 1) - Jul < 0, and thus laaRN,3(z)l < CIzl2-n+(N+l)-Ial if
lal>2-n+(N+1). Notice in particular that the parametrix G(x, y) in Theorem 6.3 is CO0 for x # y. We can use this fact, together with the mapping property (6.10), to extend the interior regularity result of Theorem 4.16. Theorem 6.4 Let 521 and 02 be open subsets of R", such that SZ I C= 02, and
assume that P is strongly elliptic with C°° coefficients. For s, t E R, if u E H' (522)"' and f E H-'(02)' satisfy
Pu = f on 522,
Fundamental Solutions
197
then u E HS+2 (cZ 1)' and IIuIIH=+'(sz1),,, < CIIa1IHI(n2)' + C11 f
Proof. Choose an open set 2 c 02 such that SZ1 C= 0 and SZ C Q2, and then choose a cutoff function X1 E C mp(SZ2) such that x1 = 1 on Q. We have
X1U - LC2(Xiu) = GP(xiu) = G(xif) +G[Pxiu - XIPul, and thus IIuIIH+2(&2,)1r' _ IIK2(Xiu) +G(Xif) +G[PXIU - X1Pu]IIH=+2(c1)m
Since K2 is a smoothing operator, II1C2(X1u)IIHs+2(Q,)"' < CIIuIIH,(n2)"', and
it follows from (6.10) that
II(XIGXI)f IIHs+2(n,)-n
wl=
defines a fundamental solution for P. (ii) I fn > 3, then G (z) = P (l;)-' } defines a fundamental solution for P, and G is homogeneous of degree 2 - n. (iii) If n = 3, then G in (ii) has the integral representation G(z)
=
2z1
f l P(w1)-1 dwl,
where Sz = {w1 E S2 : wl z = 01 is the unit circle in the plane normal to z, and dwj is the element of arc length on Sz . Proof. We see from (6.6) that P (l;) is invertible for C°O and homogeneous of degree -2 for E R" \ {0}.
0. Thus, P (t;)-' is
Suppose n = 2. We know from Lemma 5.12 that f.p. P(l;)-' is a temperate distribution, so G(z) is well defined as the inverse Fourier transform of f.p. P(l;')-', and by Exercise 5.5, f.p.
I = .Fz,. (3(z)},
Fundamental Solutions
199
implying that G is a fundamental solution for P. Since P (-w) = P (w), we see from (5.18) that
G(z) =
12 f.1=1 [II±,(-w z) + II±1(w z)]P(co)-' dc),
II±I() = F'(1) -
sign(se),
2
giving the integral representation for G (z), and completing the proof of part (i). Part (ii) is clear from Theorem 5.13, because for n > 3 the function P ()-1 is locally integrable on R", and thus homogeneous of degree -2 as a distribution on lid".
To prove (iii), we note that by Exercise 5.12, if n = 3 then
G(z) = (IIa (-
()P( )-1)
z),
Moreover, since P(-i) =
for z E R3 \ {0}.
and since we can choose * so that
(- . z) + IIo ( z), ()P( Z(rI Observe that I1 (- z) = IIo (' z) and x+ + x° = 1, so fl + 11 G(z) =
.F{ I) = S, and therefore, if we define cp (') = 1/r () P (t) proof of Lemma 5.14, then, by Exercise 5.11,
G(z) =
_
z),()P()-1) =
Z(8( 1
21zl
t t=0
2(8, 0z) =
=
and Oz (t) as in the
2c6z(0)
P 1)-1 d l
( Y'
Introducing polar coordinates in the plane normal to z, i.e., putting l = pwl with p = and wl = l/p E Si'-, we find that
f
l Z=0
(l)P(Sl)-'d
1 = J J'1E
(Awl)P(Awl)
>O
Sl
(f o
which yields the desired formula for G(z).
Adpdwl
(Awl) dp) P(wl)-' dw1, A
Surface Potentials
200
Of course, in part (i) of Theorem 6.8 we can simply take (6.13)
zl)P(co)-1 dcv,
G(z) = Js (log IW1
because P = Po annihilates constants. We shall not prove any other general existence result for fundamental solutions, although Chapter 9 treats a particular example with A 0 0. Dieudonne [19,
pp. 253-256] discusses the history of existence proofs for fundamental solutions of general classes of partial differential operators. Gel' fand and Shilov [27, p. 122] give a reasonably simple proof for scalar elliptic operators of arbitrary order with constant coefficients, and Hormander [41, Theorem 7.3.10] considers arbitrary (not necessarily elliptic) partial differential operators with constant coefficients. Miranda [67, Theorem 19, VIII] treats second-order elliptic equations with variable coefficients.
The Third Green Identity Let us recall the notation used in our discussion of the transmission property (Theorem 4.20). The set S2- is a bounded Lipschitz domain in JE81, SZ+ is the complementary, unbounded Lipschitz domain, r = a S2+ = 8 S2-, and we have
the sesquilinear forms ct = Ogf, defined by n
n
n
cI(u, v) = f E E(A jkaku)*ajv + (A jaju)*v + (Au)*v dx; t j=1 k=1 j=1 cf. (4.2). The one-sided trace operators for SZ+ and S2- are denoted by y+ and y-, respectively, so that
y±u = (U±) I r
when u = U}1c2± for some Ut E D(1[8")"'
In the usual way, we extend y+ to a bounded linear operator
y} : Hs(c2±)m -*
Hs-1/2(r)m
for 2 < s < 2
The one-sided conormal derivatives of a function u E H2(S2±)m are defined by BV
n j=1
n
n
Vjy} EAjkaku
u
(k=1
and 13
vjy j=1
EA* k=1
with the usual generalisation via the first Green identity, as in Lemma 4.3. Remember also our convention that the unit normal v points out of S2- and into 52+.
The Third Green Identity
201
When u is defined on the whole of R, we sometimes write ut = uIn± for its restriction to Q±. To avoid redundant + and - signs, we write the onesided traces as y+u and y-u instead of y+u+ and y-u-, and similarly for the one-sided conormal derivatives. The jumps in these quantities are denoted by
[u)r=Y+u-y u, Au)r=Bvu-B-u, and we often indicate that a jump vanishes by dropping the + or - superscript; for instance, we write
yu = y+u = y-u if [u]r = 0. The first new symbols are y*, the adjoint of the two-sided trace operator, and B*, defined by
(Y*, 0) = (Vr, YO)r and
for 0 E E(R")m (6.14)
0)
By Theorem 3.38, y0 E HI-Emm for 0 < c < 2, so y*4r makes sense as a distribution on R" for any * E HE-' (F)'". Similarly, since 9.-0 E L.(1')"', makes sense as a distribution on R" for any * E Li (I')"'. we see that Obviously,
supp y*Vr c supp /r c r and
supp B* fr c supp * c r.
Using y* and B*, we can restate Lemma 4.19 as follows.
Lemma 6.9 Let f } E H-I (S2±)'" and put f = f+ + f - E
H_1 (1i$
n )n' and
suppose that u E L2(Rn)'" with u± E H1(SZt)"' If
Pu± = f } on Q±, then
Pu = f + B*[u]r -
on lR".
(6.15)
Now let C be a parametrix for P. Thus, there are smoothing operators 1C1 and 1C2 such that
P9u = u - ICiu and cPu = u - IC2u
for U E E*(Rn)"',
and of course ICI = 0 = IC2 if G is a fundamental solution. Motivated by Lemma 6.9, we define the single-layer potential SL and the double-layer
Surface Potentials
202
potential DL by
SL=Gy* and DL =GBv. Applying G to both sides of (6.15), the third Green identity follows immediately.
Theorem 6.10 If, in addition to the hypotheses of Lemma 6.9, the function u
has compact support in 1R" (and thus also f has compact support in R"), then
u = g f + DL[u]r -
K2u
on R".
The definitions above mean that
(SL V,, 0) = O1i, Yc*0)r and (DL i/r, 0)
1Z\ ifl BAG*0)r' for 0 E E( H
so by considering test functions with supp 0 C= R" \ F, and recalling from Theorem 6.3 and Corollary 6.5 that G(x, y) is C°° for x # y, one obtains the integral representations
SL*(x) =
f
G(x, y)i/i(y) dc,,
(6.16)
DL*r(x) = f [Bv.,G(x, y)*]*1/r(y) day,
(6.17)
for x E 1R" \ r. Notice also that
PSL Jr=y*Vr-ICI y*Vr and PDLKlon R, (6.18)
and hence
P SL Vr = -Ki y** and P DL
on SZ}.
(6.19)
In particular, if G is a fundamental solution, then P SL Vr = 0 = P DL Vr on 52±.
Jump Relations and Mapping Properties The surface potentials SLib and DL* are C°O on S2± because G(x, y) is C°° for x # y. We shall now investigate their behaviour at the boundary F. The results in the next two theorems, for general Lipschitz domains, are from the paper of Costabel [14].
Jump Relations and Mapping Properties
203
Theorem 6.11 Fix a cutoff function x E C o ,(R). The single-layer potential SL and the double-layer potential DL give rise to bounded linear operators
XSL: H-1/2(r )m
H1(l[$n)m,
X DL : Hl/2(r)m
HI(S2:)m,
ySL : H-t/2(r),n
H1/2(r)m,
yt DL : HI/2(r)m
HI/2(11)m,
H-1/2(r)",
B DL : H1/2(r)m
H-1/2(r),n,
B} SL : H-1/2(r)m
and satisfy the jump relations
[SL f]r = 0 and [B SL 1/rl r = -1/r
for* E H-1/2(r)m,
[DL Tf]r = Vr and [B DL *lr = 0
for* E HI/2(r)m.
and
Proof Choose a second cutoff function xt E D(W), satisfying Xt = 1 on a neighbourhood of 0- U r. For * E D(r)m and 0 E D(li8")'n, (xgxtY**,
(xSL 1/r, 4)) =
4)) = (*, Y(xiG*X)4))r,
and by Theorem 3.38 and (6.10),
y : HI (R n)m -* H1/2(r)m
and
X19*X : H-'(R)' -+
HI (Rn)nt,
(6.20) so IIY(xtg*x)4)IIH112(Rn)>-<
3, then, by Theorem 6.3, the leading term Go in the homogeneous expansion of G has degree 2 - n. If n = 2, then Go contains a logarithm. Consequently,
IG(z, y) I dvy < rnB, (X)
CE
for Z E IR" and n > 3,
CE (1 + I log E I)
fort E B, (x) and n = 2,
Boundary Integral Equations
220
and it is easy to see that if, say, * E L,, (r)n`, then
Sr(x) = J G(x, y)if(y) day and S*Vr(x) =
J
G(y, x)*l/r(y) da,, (7.8)
for x E F. Hence, S and S* are integral operators on F with weakly singular kernels. To handle the other six boundary operators, we define
,,*(x) = 2 fr T, **(x) =
\B,(x)
[ B&.yG(x, Y)*]*f(Y) day,
2f \B,(x) B,.xG(Y,x)*y+(Y)day,
TE(x) = 2 /
r\B,(x)
[Bv,yG(Y, x)]**(Y) day,
TEr(x) = 2 J
Bv.., G(x, Y)l(Y) day,
r\Be (x)
REik (x) = - f
R:f(x) = - f
\B,(x)
Bv,x[Bu,yG(x, Y)*]**(Y) day,
Bv,x[Bv,yG(Y,x)]* (Y)day; \B, (x)
cf. the integral formulae for the single- and double-layer potentials given in (7.1) and (7.2). Recalling the definition of By from (4.3) and (4.4), and the definition of By from (4.5), we see that the kernels of the last six integral operators above are given explicitly as follows:
2[B,,,G(x, y)*]* = 2[an+kG(x, Y)Akj(Y) + G(x, y)Aj(Y)]vj(y), 2Bv,xG(y, x)* = 2[an+kG(Y, x)Akj (x) + G(y, x)Aj (x)]*v j (x),
2[Bv,yG(y, x)]* = 2vj (Y)[Ajk(Y)akG(Y, x)]*,
2Bv,xG(x, y) = 2vj(x)Ajk(x)8kG(x, y), (7.9)
-Bv,x[Bv,yG(x, y)*]* _ -vj(x)[Ajk(x)akan+nnG(x, y)Anr1(Y)]u1(Y) - vj (x)[Ajk(x)akG(x, y) Al (y)] vi (y), -Bv.x[Bv,yG(y, x)]* =
x)A,,,1(x)]*vl(x)
- vj(Y)[Ajk(Y)akG(Y,
x)Al(x)]*vl(x).
Here, we have used the summation convention, and note that an+kG(x, y) = a),,. G(x, y).
Integral Representations
221
In general, the six kernels in (7.9) are all strongly singular on the (n - 1)dimensional surface F, because the leading term in the homogeneous expansion
of VG is of degree 2 - n - Ia I. To investigate what happens as e .0, suppose that SZ- is given locally
by x < (x'), and define the directional derivative (x,
d (x', h') = lim
+ t h') - (x')
tlo
t
For X E F, i.e., for x _ (x'), we shall say that r is uniformly directionally differentiable at x if
(x' + h') _ (x') +
h') +o(Ih'I)
as Ih'I - 0.
(7.10)
h') is homogeneous of degree 1, but not necessarily linear, Note that in h'. In order to state our next theorem, we define two subsets of the unit sphere §n-1 c lf8n
T+(x) = {ro E S"-' : w >
' co))
T -(x) = [co E S"-' : w
3.
' dW (7.12)
Since an+k Go (x, z) is homogeneous in z of degree 1-n, and since (7.10) implies that
ma ([ }(x) \ l E (x)l U[ 1 E (x) \ T (x)l J= 0, A
E
Integral Representations
223
we see that
lim]
[13,,,yG(x, y)*]*u(y)day =+a+(x)Vr(x),
40 S2T-naBEcx)
giving the formula for y t DL * (x). The expression for T * (x) then follows immediately from the definition of Tin (7.3). The formulae for y± DL *(X) and T >/r (x) follow by a similar argument, with the help of Exercise 6.2.
When I' is sufficiently smooth, the preceding results for T and T simplify, and we can deal with the other four boundary operators; cf. Theorem 5.23. Theorem 7.4 Let X E I' and Mfr E D(P)"'.
(i) If I' has a tangent plane at x, then
and T*(x) = limTEl/r(x).
Tif(x) = lim TE* (x) CIO
CIO
(ii) If r is C1,'` (with 0 < µ < 1) on a neighbourhood of x, then T*l/r(x) = 1imTE Vr(x)
and
T**(x) = 1imTEi/r(x).
E40
CIO
(iii) If r is C2 on a neighbourhood of x, then Ri/r(x) = f.p. RE1/r(x)
and
R*1/r(x) = f.p. RE*(x). CIO
w), and since TI(x) is given Proof. Since a"+kGo(x, -cv) = by (7.11), we see that a-(x) = a+ (x) and so TEi/r(x) -* T i/r(x). In the same way, a+(x) = ii-(x), so T,1/r(x) -+ Ti/r(x). Part (ii) follows from part (i) because, cf. (7.12), the combination
B,,,xG(y, x)* + 13,,,yG(x, y)* = [vi (x)Akj(x)* - vj (y)Aki
(y)*]
x a"+kGo(x, x - y)*
+
=
J 0(1+lloglx-yil) ifn=2, 1 O(Ix - y12-")
O(Ix
-
if n > 3,
you+1-n)
is only a weakly singular kernel on F, and [13,,,yG(x, y)*]* is the kernel of T.
224
Boundary Integral Equations
We now deal with the hypersingular operators. By Lemma 6.16,
R,f (x) = -BL DL * (x) = -B' SLBvu(x)
Em([X3v,.,G(x, )]*, Pu),f ,
where S2E = Sgt \ BE (x). Since P* [Bv x G (x, )]* = 0 on S2E , the second Green identity (Theorem 4.4) gives
+([13v,xG(x, .)]*, Pu)n
= -(Bv[Bv,xG(x, )]*, Yu)a,E *
+ ([B.,,xG(x, .)]
Bvu)a52, .
From (7.9), we see that ')]*,
-(Bv[Bv,xG(x,
yu)anz = RE1f(x)
± SEtnaBf(x)
Bv,x [Bv,yG(x,
Y)*]*u(Y)
day
and
([Bv.xG(x, )]*,Bvu)asa{
-
B,,,xG(x,Y)Bvu(Y)dory, (x)
where v(y) is the outward unit normal to BE(x). Thus,
Rilr(x) =
2
[::FBvu(x) + T*Bvu(x)] + im(REf(x) + CIO
fJ
2T
*Bvu(x)
{Bv,x [Bv.yG(x, Y)*]*u(Y) B, (x)
- A xG(x, y)Bvu(Y)} derv I, and by arguing as in the proof of Theorem 7.3 and noting that co), we find that
(7.13)
Go (x, -(0) _
B,,,xG(x, y)B,u(Y) dcy S2 naBE(x)
f
vj(x)Ajk(x)8kG(x, x + Ew)wl8lu(x + Ew)En-1 dw. (x)
Differentiating the expansion in Theorem 6.3 with respect to x, one obtains
as the leading term
x - y), and since
z) is odd and
Integral Representations
225
homogeneous of degree 1 - n as a function of z, we have Bv,xG(x, y)13vu(y) day, S2+naBf(x)
=
I
-2 1
Vj (x)Ajk(X)a.+kGo(X, w)a,u(x)wl do) + 0(E).
wI-1
Hence, taking the average of the + and - expressions for Ri/r(x) in (7.13), we are left with R,/i(x) = lim RE*(x) +
1
L-naBEx )
13V,x[ VyG(x, y)`]*it(y) day *
1
2 L+fl8B2 (x)
Bv,x[9v.YG(X,y)*]u(y)da,,
1
In view of Theorem 6.3 and (7.9), if we let u,,,l(y) = A,,,,(y)u(y) and ul(y) _ A,(y)u(y), then -,Bv,x[Bv.YG(x,
y)*]*u(y)
= vj(x)Ajk(x)[an+kan+,nGO(x, x - y)un:l(x) + akan+1GO(x, x - y)umt(X) + an+kan+mG1 (x, x - y)unsl (X)
an+ka,,+mGo(x, x - y)8pum!(X)(yp - xp)
- an+kGO(X, x - y)u!(x)]vl(y)
+ 0(jx - y12-',). For the leading term, we apply Exercise 7.2 with f (w) = Aml(x)wl to obtain
f
i naBf(x)
=E
Go (x, -w)
an+kan+mGo(x, x - y)v!(y) day
fT-(x)
8n+kam+kG0 (x, w)wl do) + 0(e).
Each of the remaining strongly singular terms in the integrand has the form
f (x, x - y), where f (x, z) is even and homogeneous of degree 1 - n as a function of z. Since
lim J e10
S2{naBE(.r)
f (x, x - y) day: =
JT'(x)
f (x, co) dw,
the contributions from T+(x) and T- (x) cancel, and part (iii) follows.
Boundary Integral Equations
226
The Dirichlet Problem We now show how the single- and double-layer potentials allow a pure Dirichlet problem to be reformulated as a boundary integral equation of the first kind with a weakly singular kernel.
Theorem 7.5 Let f E H-1(S2-)m and g E H1/2(f')"'. (i) If U E H '(Q-)' is a solution of the interior Dirichlet problem
Pu = f y-u = g
on St-,
(7.14)
on 1',
then the conormal derivative * = BV _U E H-1/2(F)m is a solution of the boundary integral equation
(g + Tg) - y9 f on r,
Sl/r =
(7.15)
2
and u has the integral representation
u=Gf -DLg+SLi/r on Q-.
(7.16)
(ii) Conversely, if ilr E H-112(1F)m is a solution of the boundary integral equation (7.15), then the formula (7.16) defines a solution u E Hi (l-)"' of the interior Dirichlet problem (7.14).
Proof As in Theorem 6.10, we view f as a distribution in H-'(R)' with supp f C 52--; cf. Theorem 3.29 (ii). Suppose that u E H1(Q-) satisfies (7.14), and define u = 0 on the exterior domain Q+. Applying Theorem 6.10, we obtain the representation formula,
u=Gf -DLy-u +SLB.u on Q-,
(7.17)
and then by (7.5),
y-u = y9 f - (-y-u + Ty-u) + SB- u i
on F.
Part (i) now follows from the boundary condition y-u = g. To prove (ii), suppose that i/i E H-112(I')m satisfies (7.15), and define u by (7.16). The mapping property (6.10) of the volume potential, together with those of the surface potentials given in Theorem 6.11, imply that G f , DL g and
SL i/r all belong to H1(Q-), so u E H1(SZ-)m. By (6.2), we have PG f = f on lR", and by (6.19), we have P DL g = P SL 0 on SZ-, so Pu = f on Q-. Finally, y-u = g by (7.5). The next theorem shows that the boundary integral equation (7.15) satisfies
the Fredholm alternative; cf. Theorem 2.33. The method of proof was first
The Dirichlet Problem
227
used by Nedelec and Planchard [76], [74], Le Roux [56], [57], and Hsiao and
Wendland [42], for the case when P is the Laplacian. These authors were all concerned with error estimates for Galerkin boundary element methods, in which context positivity up to a compact perturbation is of fundamental importance for establishing stability.
Theorem 7.6 The boundary operator S = y SL admits a decomposition
S=So+L, in which So : H-1/2 (r)m - H1/2(r')m is positive and bounded below, i.e.,
Re(So*, Or ? CII IIH-1n(r)for* E H-1/2(r)m, and in which L : H-1/2 (r')m -+ H 1/2 (r)m is a compact linear operator. Hence,
S : H-1/2(r)m -) H1/2(r')m is a Fredholm operator with index zero.
Proof. Put u = X SL i/r and v = X SL 4), where 0 E H-1/2(r)" and X E C mp(R") is a cutoff function satisfying X= 1 on a neighbourhood of S2-. Since Si/r = yu and 4 _ -[Bp,v]r, and since Pv = 0 on Q-, the first Green identity implies that
(SQL, fi)r = (yu, By v - By v)r = 'Do- (u, v) + Dn+(u, v) + (L1', 4))r, where
(L1l, 4))r = -(u, Pv)o+.
(7.18)
We have [u]r = 0 so u E H 1(R")m, and hence the strong ellipticity of P implies that Re (Do- (u, u) + Re (Dsy+ (u, u) = Re OR,. (u, u) ? C II U II h l (R,,),,, + (L2i, Or,
(7.19)
where
(L2l,10)r = -C(u, v)R,,. Furthermore, II
IIH-,ncr)y = IIBV u - By uIIH-"n(r)'" < CIIuIIHI(R),-,
(7.20)
Boundary Integral Equations
228
so if L = L 1 + L2 and So = S - L, then So is positive and bounded below, as required. By Theorem 3.27, to show that L : H-1/2(r')"' -+ H112(r')'" is compact, it suffices to show that L:
HE-1(r')"'
is bounded for 0 < E < 1. In fact,
-* H1-E (F)"'
1 has the form
L
L1ifr(x) = f K1(x, y)*(y)dory, r where K1 is C°° on a neighbourhood of r x F, because G(x, y) is C°O forx # y, and Pv has compact support in SZ+. To deal with L2, we apply the CauchySchwarz inequality and obtain I(L2f,' )rl < CIIuIILZ(R"")"IIVIILZ(R")"'
By the mapping property of the single-layer potential in Theorem 6.12, Ilu1ILZ(R")nr < CllullH'+112(Rn)", < CIIfIIH'-I(r)", so
I(L2i,')rl /rN+gD,weseefrom (7.17)that u = 9f - DL( *N + gD) + SL( *D + gN)
on Q-,
=0
and since u satisfies the boundary conditions, *D = 0 on 1'N, and on 1'D. Hence, by (7.5),
gD = y u = yGf - 2 (-gr + TDN*N + TgD) + SDD7GD + SgN
on FD,
and
gN = BV u = B-9f + RNN*N + RgD + 2
(gN
+ TND 1rD + T*gN)
Of IN,
giving
SDDY'D - 2TDNYN = (gD +TgD) - SgN - yCf 2
on rD,
TND*D + RNN1N = 2 (gN - T*gN) - RgD - 5; g f on I'N,
which is just the 2 x 2 system (7.27). This argument proves part (i).
Conversely, suppose that V'D E H1/2(rD)"' and *N E H-'/2(I'N)m satisfy (7.27). By (6.10) and Theorem 6.11, the equation (7.28) defines a function u E H' 02-)'", and obviously Pu = f on S2-. Finally, by working backwards through the calculations above, we see that y-u = gD on I'D, and ;t3; u = gN on 1'N, proving part (ii).
Mixed Boundary Conditions
233
By putting SDD
A=
IT* 2
ND
1
-2TDN
nb
[*D]
Y
*N
h=
hD hN
RNN
we can write the system (7.27) as
AO =h, and by putting (0, O)rDxrN = (1kD, OD)r0 + (*N, ON)rN,
we have (A/, 0)roxrN = (SDD*D, OD) rD - I(TDNfN, IOD)rD + 2 (TNDtD, ON)rN + (RNN1N, ON)rD.
When P* = P, a simple argument shows that the Fredholm alternative is valid for (7.27). Theorem 7.10 Let H = H-1hI2(F D)"' X H1/2(rN)"'. If 2 isformallyself-adjoint, and if P is coercive on H' (S2-)"' and on H' (S2+)', then
A=Ao+L, where AO : H -* H* is positive and bounded below, i.e.,
Re(Ao,o,'tb)rOxrN ? cIIijIH for' E H, and where L : H -+ H* is a compact linear operator. Hence,
A : H -+ H* is a Fredholm operator with index zero. Proof Let So be as in Theorem 7.6, and let Robe as in the proof of Theorem 7.8.
Thus, S=So+LsandR=Ro+LR,where Ls: H-1/2(r)"
H1/2(r)m
and LR : H1/2(r),n _, H-112(F),n
Boundary Integral Equations
234
are compact linear operators. Noting that T = T because P is formally selfadjoint, we define
A0 _ [(S01rD)Ir.D - ITDN*N W
Lzb =
and
(Ls1D)Iro (LRl N)IrN
2TNDWD -r (R0 N)I rN
In this way, A = Ao + L, the operator L : H --> H* is compact, and
(Aoi, Y')rDxrN = ((So D)Irp,
D)ro
- I(TDNZG'N, *D)rp
fN)rN + ((RokN)IrN, *N)rN.
+ Since
(TN)*o, *N)rN = (*D, TDN*N)rN = (TDN*N,'D)rp, and since supp 1D c I'D U 11 and supp 1/'N c rN U 11, it follows that
Re(Aoi, tP)roxrN = (So1D, y'D)r +
V'N)r
+cDI1NIIH,/2(r),,, = cII
IIH,
as required.
Exterior Problems Integral equation methods are particularly suited to boundary value problems posed on the exterior domain 52+. It turns out that, in general, the solution will
not belong to H1(S2+)"', but only to HI (cZ)'" for each finite p, where 0v is defined as in (7.25). Furthermore, to make use of the third Green identity on S2+ we require a somewhat stronger result than Theorem 6.10, that incorporates a suitable radiation condition. In other words, some assumption about the behaviour of the solution at infinity is needed, and here we shall adopt the approach of Costabel and Dauge [16].
Lemma 7.11 Let u E D*(S2+)"`. If Pu has compact support in Q+, then there exists a unique function .Mu E C°°(R")"' such that
Mu(x) = f G(x,
dory
- j [Ev.yG(x, y)*]*u(y) dory
(7.29)
,
for x in any bounded Lipschitz domain S2 such that S2 U supp Pu C= S2 and where r 1 = 8 SZ I .
Proof. First note that, by Theorem 6.4, the distribution u is C°O on S2+ \ supp Pu. Given X E R11, we define .M u (x) to equal the right-hand side of (7.29)
Exterior Problems
235
with r, the boundary of any ball Q- = B. centred at the origin with radius p large enough to ensure that S2- U supp Pu C- Bp and x c Bp. By applying the second Green identity over an annular region of the form Bp, \ B,o, , one sees that the definition of Mu(x) is independent of the choice of p, because Pu = 0 = P*G(x, )* on Bp, \ Bp, . Similarly, to see that (7.29) holds for x in any bounded Lipschitz domain 01 with Q- U supp Pu C 0, , we apply the second Green identity over Bp \ S2- for any p such that Q7 C Bp. Notice that PMu = 0 on R". We now give the desired version of the third Green identity for Q+. Theorem 7.12 Suppose that f E H-i (9+)m has compact support, and choose po large enough so that
T UI'C=Bpo
and
supp f C-52+x.
If U E D* (52+)m satisfies
Pu = f
on SZ+,
and if the restriction of u to 52+A belongs to H l (S2+)"', then
u = 9f + DL y+u - SL B' u + Mu on 52+.
(7.30)
Proof By Theorem 6.10 with u- identically zero, the representation formula (7.30) holds on the bounded Lipschitz domain S2+ fl &2I, with Mu(x) given by (7.29) for x E SZ
.
(Keep in mind that v is the inward unit normal to S2+ fl
0, on r, but the outward unit normal on r I.) Before considering boundary value problems on 52+, we need to know how M acts on volume and surface potentials.
Lemma 7.13 Fix Z E R". If u(y) = G(y, z), then Mu = 0 on R". Proof Let z, x E R' with x # z. We choose a bounded Lipschitz domain S2such that z E S2- and X E 52+, and define
G(x, y)*
v(y)-{0
for y E Q-
foryE52+
Since P*v = 0 on SZ±, the third Green identity (6.29) gives
v(z) = SL B,, v(z) - DL y-v(z),
Boundary Integral Equations
236 or equivalently,
G(x, z)* =
fr
G(y, z)*B,,,yG(x, y)* day -
Jr
[B,,,,G(y, z)]*G(x, y)* dQy.
Thus,
G(x, z) = J [t3.,,yG(x, y)*]*G(y, z) day -
f
G(x, y)BV.yG(y, z) day,
or equivalently,
u(x) = DL yu(x) and therefore Mu(x) = 0 by Theorem 7.12, because Pu = 0 on SZ+.
Lemma 7.14 Let f E E* (R")'°. If U = 9f, then Mu = 0 on R". Proof. Obviously, Pu = 0 on R'1 \ supp f , and since G(x, y) is C°O for x # y, one sees from Lemma 7.13 that
Mu(x) _ (M.G(x, y), f (y)) = 0 for x E R" \ supp f . Since PMu = 0 on 1R", it follows from the third Green identity that Mu = 0
onR". To state the main result for this section, it is convenient to introduce the notation H11
(Q+),,,
= {u E D* (S2+)m U I Q+ E H 1(SZp)"` for each finite
p > 0 such that Q- C BPI.
We point out that Exercise 7.4 gives some simple sufficient conditions on u for
ensuring that .Mu = 0, in the case when P = Po. Theorem 7.15 Suppose that f E H-1 (SZ+)m has compact support. (i) Let g E H 1/2(I')"'. If U E H11 (Q+)'" is a solution of the exterior Dirichlet problem
Pu = f on St+,
y+u=g onr, Mu = 0
on lR",
(7.31)
Exterior Problems
237
then the conormal derivative Mfr = B+ U E H-112(r)'" is a solution of the boundary integral equation
Si/i = y9 f - (g - Tg) on r,
(7.32)
;
and u has the integral representation
u=Gf+DLg-SL,/i one
.
(7.33)
Conversely, if ili E H-112(I')ry` is a solution of the boundary integral equation (7.32), then the formula (7.33) defines a solution u E H11 (S2+)m of the exterior Dirichlet problem (7.31).
(ii) Let g E H-112(x)"'. If U E Hi, (S2+)'" is a solution of the exterior Neumann problem
Pu = f on 52+,
B+u=g on r, Mu = 0
(7.34)
on R",
then the trace* = y+u E H 112 (r ),n is a solution of the boundary integral equation
Ri/i = By G f - i (g + T*g) on r,
(7.35)
and u has the integral representation
u = G f + DL *- SL g on Q+.
(7.36)
Conversely, if ili E H1/2(I')'" is a solution of the boundary integral equation (7.35), then the formula (7.36) defines a solution u E HIOc, (Q+)... of the exterior Neumann problem (7.34). (iii) Let gD E H1/2(r)and gN E H-1/2(I')", and define hD E H1i2(rD)"' and hN E H-1/2(rN)m by
hD = Y9 f - SgN - (gD - TgD)
on FD,
2
hN = BV +9f - RgD - (gN + T *gN) 2
on FN.
Boundary Integral Equations
238
If u E H11 (S2+)"' is a solution of the exterior mixed problem
Pu = f
on SZ+,
Y+u=gD on I'D, l3+ u = gN
on FN,
Mu=0
on R",
(7.37)
then the differences
*D = l3+ u - gN E H-1/2(rD)n'
and *N = Y+u - gD E
H112(pN)'n
satisfy
-
SDD
RNN
TND
hD
1/JD
Z TDN
j[Nj=[hN]'
(7.38)
and u has the integral representation
u = 9f + DL(i/rN + 9D) - SL(*D +,N)
on Q+.
(7.39)
Conversely, if ilrD E H-1/2 (rD)"' and *N E H1/2(rN)' satisfy the system of boundary integral equations (7.38), then the formula (7.39) defines a solution u E H11 (S2+)"' of the exterior mixed problem (7.37).
Proof. Suppose that u E Hil (S2+) is a solution to the exterior Dirichlet problem (7.31). By Theorem 7.12,
u = 9f + DL y+u - SL l3+ u on 52+,
(7.40)
and then by (7.5),
Y+u = yG f + (y+u + T y+u)
- S13+ u
on F.
2
Using the boundary condition y+u = g, and putting /r = B+ u, we arrive at the boundary integral equation (7.32) and the integral representation (7.33). To complete the proof of (i), suppose conversely that * E H-1/2(F)m satisfies (7.32), and define u by (7.33). Together, (6.10) and Theorem 6.11 im-
ply that u E H11 (S2+)"', and it follows from (6.2) and (6.19) that Pu = f on 52+. Also, y+u = g by (7.5). Finally, Lemma 7.14 shows that M9 f = 0,
M DL g = MG9vg = 0 and M SL Vr = MGy"i/r = 0, so Mu = 0, and hence u is a solution of (7.31).
Regularity Theory
239
The proof of part (ii) proceeds in the same way, except that one takes the conormal derivative of both sides of (7.40), instead of the trace, to obtain
B'u=8'Gf -Ry+u-z(-l3vu+T*B-,) on F. The proof of part (iii) is similar to that of Theorem 7.9.
Regularity Theory In Theorem 4.18, we proved local regularity up to the boundary for solutions to elliptic partial differential equations. A simple argument based on this result yields the following local regularity estimates for the boundary integral equations. Recall the definition (7.25) of the set Q+ Theorem 7.16 Let G 1 and G2 be bounded open subsets of R" such that G 1 C= G2
and G 1 intersects r. Put
Sgt-= G; n c2} and r = Gl n S2+ and suppose, for some integer r > 0, that 1`2 is
for j = 1, 2, Cr+1,1
(i) If */r E H-1/2(I')'" and f E Hr+3/2(r2)"` satisfy
Si/r= f onF2, then t/r E Hr+l/2(rl)"' and IIkIIH1+'/2(r,)< GIIiIIH-"2(r)"
+CIIfIIHr+3/2(r2)m
(ii) If P is coercive on H1(Q-)"' and on H1(c2)"', and if i// E H1/2(I')' and f E Hr+1/2(x2)"' satisfy
R* = f
on I'2,
then i/r E Hr+3/2(F1)"' and II*IIHI+3/2(r,)
< CII
IIH'/2(r), + Cll f IIH1+i/2(r2)"-
Proof. Let G3/2 be a bounded open subset of R" such that G1 C= G3/2 C G3/2 C= G2,. and put 523/2 = G3/2 n Sgt.
Boundary Integral Equations
240
In case (i), the single-layer potential u = SL * E H 1(S22 )"' satisfies
Pu=O one , yu= f on r2, so using the jump relation for I31 SL * (Theorem 6.11), together with the trace estimates (Theorem 3.37) and elliptic regularity (Theorem 4.18), we find that JIB, u
- By uIIH'+1r-(r,)
CIIuIIHr'+2(n3/,)1" + CIIuIIH'+2(niZ)m
CIIuIIH'(sz2)", +CIIu1IH'(n )", +CIIf1IHr+3r_(r2),,,.
The estimate follows because II u II H' (s2±)m < C II *II H-'/2(r)m
In case (ii), the double-layer potential u = DL i E H' (S22 )"' satisfies
Pu=0 onc22, l3,u = f on r2, so
IIfilH'+sn(r1)", = IIY+u - y-UII
CIIu and finally IIuIIH'(12
)m
< C11
IIuII
H'(Q )", + CII f 1I
IIH'/2(r)m
D
We saw in Theorem 7.2 that the mapping properties of the boundary integral operators hold for an extended range of Sobolev spaces when F is smoother than just Lipschitz. The regularity result just proved allows us also to extend the Fredholm property for R and S.
Theorem 7.17 If I' is
C''+1.1 for
some integer r > 0, then
H5+1/2(r) is Fredholm with index zerofor -r < s (i) S : Hs-1/2 (r)"' r, and ker S does not depend on s in this range; (r)m HS-1/2 (r)"' is Fredholm with index zero for -r -1 < (ii) R : s < r + 1, and ker R does not depend on s in this range.
-
Proof We know from Theorem 7.6 that S is Fredholm with index zero when
s = 0. Thus, let 01, ..., ¢p be a basis for kerS in H-1/2(r)'". In fact, by Theorem 7.16 these basis functions belong to
Hr+1/2(r)",, and we can choose
Exercises
241
them to be orthonormal in L2(r')m. The same reasoning applied to S* yields an orthonormal basis 01, ... , BP for ker S*, with each 9 j E H''+1 j2 (r)"` . We can therefore define a compact linear operator
for -r < s < r,
K : HS-1/2(r)m by P
Ki/r = E(Oj, '1` j=1
and a bounded linear operator
A = S + K : HS-112(r)r -+ H:+1/2(f
)m
for -r < s < r.
The operator A is certainly Fredholm with index zero when s = 0, and since
(0j, Silr)r = 0 and (0j, K,lr) = (0j, Or
for all * E H-1/2(r)"',
it is easy to see that the homogeneous equation A* = 0 has only the trivial H-t/2(r)m. Thus, the inhomogeneous equation A*/r = f has a solution in unique solution * E H-112(1,)m for every f E H1j2(F)"`. Furthermore, if f E Hr+112(r)m, then Si/r E Hr+1/2(x)"1 because Kt/r E Hr+112(ryn, and so * E Hi-112(r)"' by Theorem 7.16. It follows that A has a bounded inverse for s = r, and the same is true of A*. Hence, by interpolation and duality, A is invertible for -r < s < r. Therefore, since K is compact, the operator S must be Fredholm with index zero for -r < s < r. Also, ker S does not depend on s, H-r-1/2(1,)"1 satisfies Si/r = 0, then A* = Kiln E Hr+i'2(r)m because if * E A-1 Kulr E Hr-1/2(x)m. The proof of (i) is now complete. and thus i/r = Part (ii) may be proved using the same approach. The allowed range of s is larger because the basis functions for ker R belong to Hr+3/2(r)m
0
Exercises 7.1 Suppose that S2 is a C°O hypograph x < (x'), and let K (x, y) be any one of the six kernels in (7.9). Show that
K(x,y)u(y)day
f.p. J E,o r\B,(f x)
= f.p. ElO
X,-y'I>E
K(x, y', (y'))i(y', (y')) 1 + I grad (y')I2 dy'
242
Boundary Integral Equations
for x = (x') and'* E D(1')'. [Hint: apply Theorems 5.15, 5.19 and 6.3.]
7.2 Suppose that S2- is a C2 hypograph given by yn < C(y'), and assume (without loss of generality) that
(0) = 0 and
grad (0) = 0.
Put 'T'E = {w E Si-1 : Ew E S2+},
and show the following. (i) There exists a = a(E, r7) such that rE+
= IN E Sn-1 co = (r7 cos 9, sin 9), 17 E S"-2, a (E, r7) < 0 < 7r/2).
(If n = 2, then 17ES0={+1,-1}.) (ii) The function a satisfies lima (6, r7) = 0
a=
and
E
o
as ac
1
2
` n-1 n-1
(E r7 cos a).]
(iii) With S+'-' = [Co E S"-1 : (0n > 0}, n/2
f
T,+
f(w)dw =
sin0)dOdr7 10E.0 f (i?(E,nf(r7cos0, Cos 0,
5'-2
=f X
n-1 n-1
f (w) dw - 46
a, a, (0) p=1 q=1
+
f (17, 0)r7p r7q d 77 + 0 (E2). J1-2
(iv) In part (iii), the term in E vanishes if f (-w) = - f (w) for all co E Sn -1.
7.3 Show that the right-hand side of the boundary integral equation (7.23) satisfies
z (8 - T"8) - Ci- Gf, dr)r = 0
for every solution 0 E H1/2(1')'" of the homogeneous adjoint problem
Exercises
243
R*4 = 0 if and only if f and g satisfy
(g, y v)r + (f, v),- - = 0 for every solution v E H 1(Q-)"' of
P*v=0 onQ
,
onl'. [Hint: see the discussion following Theorem 7.6.] 7.4 Assume that P has no lower-order terms (i.e., P = PO), and that G (x, y) _ G(x - y) is as in Theorem 6.8. Show that if u satisfies the hypotheses of Lemma 7.11 and if, as Ix I -+ oo,
u(x) = o(l) and 13,u(x)
o([Ixl log IxI]-1)
when n = 2,
o(IxI-1)
when n > 3,
then Mu = 0 on R". Here, v is the outward unit normal to the ball Bp.
7.5 Show that if Pu = 0 on R, then Mu = u on R. [Hint: apply the third Green identity over the ball BP.] 7.6 The Calderon projection is the linear operator PC defined by
PCO =
y-(SL*2 -DL*I)
,
where
13v (SL *2 - DL *1)
_
*1 *2
Theorem 6.11 implies the mapping property
Pc : H1/2(r)n' X H-1/2(r)m -+ H1/2(r)m
X
H-I/2(r)m
and if 0 = Pct/i, then the function u = SL *2 - DL *1 satisfies
Pu = 0
on Q-,
y-u=01 on I',
B u=42 onr, so u = SL 4)2 - DL 01 by Theorem 6.10. Hence, Pc4) = 45, or in other words PC20 = Pct', demonstrating that Pc really is a projection. (i) Show that
PC=
2(I - T)
S
ER
2
(I + T'*)
Boundary Integral Equations
244
(ii) Deduce that
SR = 4(I - T2), ST* = TS, RT = T*R, RS = 4[I - (T*)2].. 7.7 Recall the definition of the Steklov-Poincare operators 13 V from (4.38). Show that if S : H-'t2(I')'" -+ H1/2(P)"' is invertible, then we have the representations
B,U = R+ 4(I +T*)S-'(1 +T) = 2(I +T*)S-1 and
B,,V = R* + (1 + T *)(S*)-1(1 + T) = (I + T *)(S*)-' 2
4
7.8 Let S2± = R1, and think of r as R"-'. Also, assume P = Pa with constant coefficients, so that is a homogeneous quadratic polynomial, and let G(x, y) = G(x - y) be the fundamental solution given by Theorem 6.8. (i) By applying Lemma 5.21, show that if n > 3, then
S*(x') = J
dal
ms(
for x' E R"-',
where, with the notation (5.28),
The function ms is called the symbol of S. (ii) In the same fashion, show that for n > 2,
R*(x') = J
mR( ')
(')e12ngxd",
where n
MR( O =
J
it
E1: j=1 k=1 n
n
_ -(2X)2 E E Anj (J
ff
SjSkP(S', n)-'
j=1 k=1
(iii) From the homogeneity properties
ms(t') =
and mR(t:;') = tms(1;')
fort > 0,
Exercises
245
deduce that S:
Hs-1/2(RR-1)m
Hs+1/2(R"-I)m
Hs+1/2(wn-1)m
Hs-1/2(Rn-1)m
and
R:
for all s E R. (iv) Let X E C mP(lR") satisfy X = I on a neighbourhood of zero, and consider
u(x) = Ti-1s1l1 - X{OJv( )). where v is rational and homogeneous of degree j - 1, as in Assump-
tion 5.20, but with j < -1 + n. (Thus, v is not locally integrable on RI.) Modify Lemma 5.21 accordingly, and hence show that the mapping property of S in part (iii) holds also when n = 2.
The Laplace Equation
Our development of the general theory of elliptic systems and boundary integral equations is now complete. In this and the remaining two chapters, we concentrate on three specific examples of elliptic operators that are important in applications. This chapter deals with the Laplace operator in R", denoted by If
A
a2 J=1
The Laplacian constitutes the simplest example of an elliptic partial differential operator, and its historical role was discussed already in Chapter 1. After deriving the fundamental solution for the Laplacian, we shall introduce a classical tool from potential theory: spherical harmonics. These functions turn out to be eigenfunctions of the boundary integral operators associated with the Laplace
equation on the unit ball. They are also useful for studying the behaviour of harmonic functions at infinity, leading to simple radiation conditions. The final section of the chapter investigates ker S and ker R, and the sense in which S and R are positive-definite. The operator P = -A has the form (4.1) with constant, scalar coefficients
Ajk = Sjk,
Aj = 0,
A = 0.
Associated with -A is the Dirichlet form (Pn (u, v) =
f grad u grad v dx,
and the conormal derivative (4.4) is simply the normal derivative,
=
au a
. V
Obviously, -A is formally self-adjoint and strongly elliptic; see (4.6) and (4.7).
246
Fundamental Solutions
247
Fundamental Solutions The Fourier transform of -Du is P(4)u(4) where P(l;) = (27r )2 1 1; 12, so by Theorem 6.8 a fundamental solution for -A is
G(x,y)=G(x-y),
where G(x)=4
1
whenn=3.
Exercise 8.1 is the corresponding calculation for n = 2. To give a fundamental
solution for a general n, we denote the surface area of S"-', the unit sphere in W, by
r
r
cia-
7rn/2
dcv=2
Iwl=1
r(n/2)
;
(8.1)
see Exercise 8.2.
Theorem 8.1 A fundamental solution for the operator -A is given by
G(x) =
when n > 3,
1
1
(n - 2)T
IxIn-2
and, for any constant r > 0, by
G(x) = Zn log Ixl
when n = 2.
Proof The Laplacian is radially symmetric (see Exercise 8.3), so it is natural to seek G in the form G(x) = w(p) where p = Ix 1. Since AG = 0 on R" \ (0), Exercise 8.4 shows that w must satisfy
Ld
PIT-t dp p
-l dw
dp = 0
for p > 0,
so
w(p) =
an
1
+ bn
when n > 3,
n - 2 pn-2
or 1
w(p) = a2 109 - + b2 p
when n = 2,
for some constants an and bn. The choice of b is arbitrary, but an is fixed by the requirement that G satisfy (6.12), i.e., by the requirement that -AG = 8
248
The Laplace Equation
on R", or in other words
-(G, AO) _ O(0) for O E D(1R").
(8.2)
Any test function 0 E D(R") has compact support, so we can apply the second Green identity (1.9) over the unbounded domain {x :IxI > c), and arrive at the formula
-
f
G(x)Oq(x) dx =
f
0(x)a,G(x) da,r - J
(x) do-,
X I=E
X I>E
(8.3)
where yr = -x/E. Since grad G(x) = dp IXI = - IXI"
for n > 2,
(8.4)
we have 8,G(x) = -(x/E) grad G(x) = a,, -0` for IxI = E. Thus, by the mean-value theorem for integrals,
f
Ej J
fi(x) dax =
(x)8vG(x) dox = a"
IxI=E
xI=E
for some xE satisfying IxE I = E, whereas
O(E)
JIxI=E
G(x)8vO(x) da,, = 10(cllog,-I)
if n > 3, if n = 2.
Thus, if a" = 1/Ta, then (8.2) follows from (8.3) after sending f 0. An alternative method of determining a" is to apply the third Green identity
to the constant function u = I over the unit ball. One obtains the formula DL 1(0) = -1, from which it again follows that a" = 1 / T . 0 Throughout the remainder of this chapter, G(x, y) = G(x - y) will always denote the fundamental solution from Theorem 8.1. Recalling the definitions of the boundary integral operators S, R and T given in (7.3), we see first that by (7.8), Si/r(x) =
*(Y)
1
(n - 2)Tn Jr Ix
-
yI"-2
d6y
when n > 3,
and
S* (x) =
1* 2,
r
(y) log Ix
day YI
when n = 2,
249
Fundamental Solutions for x E I". By (7.9) and (8.4), the kernel of T is
2a,,,yc(x, y) -
2 v,, (x - y)
T.
IX - Y111
and the kernel of R is 1
.yG(x,y)=
Ix-yln+2
'
then
for n > 2. Notice that if r is
0, let P,,, (]R") denote the set of homogeneous polynomials of degree m in n variables, i.e., the set of functions u of the form
u(x) = E aaxa for x E R,
(8.5)
IaI=m
with coefficients as E C. A solid spherical harmonic of degree m is an element of the subspace
xm(R")= (uEPm(1R"):Au =0on W). Apart from the results involving the boundary integral operators, our general approach to the study of spherical harmonics is essentially that of Miiller [70]. Let
M(n, m) = dim Pm(R") and N(n, m) = dimf,"(IR") for n > I and m > 0. (8.6) By a standard combinatorial argument, the number of non-negative integer + a" = m is solutions al, . . . , a" to the equation al +
M(n, m) _
1
1
(m+n- ), n
(8.7)
Spherical Harmonics
251
and since each u E P," (R") has a unique representation m
u(x) _
>2Vk(xI)Xn -k
E Pk(R"-'),
(8.8)
forn > 2 and m > 0.
(8.9)
with Uk
k=0
we see at once that
M(n, m) _ >2 M(n - 1, k) k=0
Also, Po = 71o is just the space of constant functions, and P1 = WI is just the space of homogeneous linear functions, so
M(n, 0) = N(n, 0) = 1 and M(n,1)=N(n,l)=n
forn> 1.
Taking the Laplacian of (8.8), we find after some simple manipulations that m
Au (x) =
[A'vk(x') + (m - k + 2) (m - k + 1)uk_2(x')]Xnl-k k=2
forn>2 and m>2, where A' is the Laplacian on IRn-1. Thus, U E 7-(R") if and only if vk-2(X')
-A'vk (x')
(m-k+2) (m-k+1) fort2 and m> 1, ( 8.
11 )
and so it follows from (8.9) that m
N(n,m)=>2N(n-1,k)
forn> 1 and m> 1.
(8.12)
k=0
Furthermore, since (
ifm=Oorl,
1
N(1, m) = t 0 ifm > 2, we have
N(2, m) _
1
2
ifm = 0, if m> 1,
(8.13)
The Laplace Equation
252
and in view of (8.7) and (8.11),
N(n , m) _
2m + n2 2
n-
forn > 3 and m > 0.
(m n+ n 3 3l
(8.14)
In particular, N(3, m) = 2m + 1. Put
7-l," (&'-') = {
: * = u lso-t for some u E H. (R")I.
Corollary 8.3 implies that the restriction map u H u Is- i is one-one, and hence is an isomorphism from 7I,,, (R") onto 7l", (S"- ), so
dimfl,,,(S"-') = N(n, m) forn > 2 and m > O. An element of 7-1,,, (S' ') is called a surface spherical harmonic of degree m.
We now show that 7-l": (S'-') is an eigenspace of each of the four boundary integral operators on S". Later, in Theorem 8.17, we shall see that the spherical harmonics account for all of the eigenfunctions of these operators.
Theorem 8.4 If T = S"-i forn > 3, then T = T * _ - (n - 2) S, Ri/r =
m(m+n-2) /r and SiJr =
2m+n-2
1
2m+n-2
fori/r E 7L (8.15)
Proof. Observe that
forx,yEF,
vy.=y and
(8.16)
so the kernel of T is
2 vy (x-y) =-(n-2)G(x,y), r" Ix - yl" and therefore T = T * = - (n - 2) S. Now suppose that Jr = y u E 71,,, ; with u E 7-l, (Rn). Euler's relation for homogeneous functions, u(y) = mu(y), implies that au
av
=mi/r on IF,
-1) y; a;
(8.17)
and therefore, applying Theorems 7.5 and 7.7,
S(mi/r) =
Ti/r) 2
and
R1* r = 2
T*(mi/r)}.
(8.18)
Spherical Harmonics
253
Since T = T* = -(n - 2) S, it follows that mSi/r = [i/r - (n - 2)Si/r]
Rir = 2m[i/r + (n - 2)S*],
and
Z
0
giving (8.15).
The same method of proof yields the following result in two dimensions.
Theorem 8.5 If F = S', then
T* = T** =
27r
fr k(y) dcr for*
E
L2(F),
(8.19)
and when'i/r E R,, (S'), Ri/r = 0,
Ri/r= 2ir,
Si/r = (log r)i t,
Si/r= 2ir, 1
T ilr = T *ifr = -,lr for m = 0;
Ti/r=T*1/r=0 form> 1.
Proof The relations (8.16) remain valid for x, y E S" kernel of T is now I
1
it Ix-y12
27r
(8.20)
(8.21)
when n = 2, but the
forx,yEF,
implying (8.19). To prove (8.20), suppose that m = 0. We see at once from (8.18) that R +' L= 0. Furthermore, i/r is constant, so by symmetry Si/r is also constant,
and then uniqueness for the solution of the interior Dirichlet problem shows that SL * is constant on the disc Q-. At the origin, we have
SL*(0) =
27r
Ivl=i
log
IYI*(Y)day = (logy)
.
so SL i/r = (log r)ilr on SZ-, and hence Sii = y- SL a/r = (log r)i/ on r. If m > 1, then using (8.17) and the divergence theorem,
T*
T*v/r
27rm, r
8vdc
2rrmfst-Dudx=O.
Thus, (8.21) follows at once from (8.18). We now consider spherical harmonics that are invariant under rotation about the nth coordinate axis.
254
The Laplace Equation
Lemma 8.6 Given m > 0, there exists a unique function u satisfying (i) U E H. (R"); (ii) if A E R"" is an orthogonal matrix satisfying Ae = e, then
u(Ax) = u(x) forx E R"; (iii) U(en) = 1. In fact,
u(x) =
1
ri,-I
(x + ix'
j7)... drl.
(8.22)
fs"-2
Proof. One easily verifies that (8.22) defines a function u satisfying (i), (ii) and (iii). To show uniqueness, suppose that A E IR""" satisfies the assumptions of (ii). It follows that A has the block structure
A= where A' E
1[l;("-l) "t"-ti is orthogonal. Thus, with uk as in (8.8),
u(Ax) _ E vk(A'x')x, -k, k=0
and so the conclusion of (ii) means that vk (A'x') = vk (x') for all x' E IEI;n-1 which in turn means that vk (x') depends only on Ix'j. Hence, every it E P (R")
satisfying (ii) has the form (8.8) with vk(x') = dklxIk, where dk = 0 if k is odd. Exercise 8.4 shows that
O'Ix'Ir = r(n - r
- 3)Ix'jr-2,
so the condition (8.10) for u E T(m (R") holds if and only if the coefficients satisfy
-k(n + k - 3) dk-2 = (m - k + 2)(m, - k + 1) dk
for2 2, by
1 - t2 + ten, for -1 < t < 1 and r?E Sii-2,
where to =
Pm(t) = u(w),
1 and u (-x) _
noting that u (w) is independent of q. Observe that since u
(-1)nlu (x), the Legendre polynomials satisfy
form>0 and n>2.
Pm(1)=1 and P,n(-t)=(-1)mPm(t) Also, we have the explicit representation nl
Pm(n, t) = >dk(n, m)(1 _
t2)k/2tm-k,
k=O
where the coefficients dk = dk(n, m) are determined by the recurrence relation (8.23) with the starting values do = 1 and d1 = 0.
When n = 2, the integral (8.22) becomes just the sum over n E S° _ (-1, +1), with To = 2, so we find that
1 - t2)m + (t - i 1 - t2)m].
P. (2, t) = 2 [(t + i
(8.24)
If t = cos 0, then t ± i-,,/1 - t2 = e±'O, so P. (2, cos 0) = cos m46,
and therefore Pm (2, t) is the mth Chebyshev polynomial of the first kind. Exercise 8.8 shows that Pm(3, t) is the usual Legendre polynomial of degree m. The fundamental solution for the Laplacian can be expanded in terms of these polynomials, as follows. Theorem 8.7 Let bm = bm (n) be the coefficients in the Taylor expansion 00
(1 _ 1z)n-2
=
bnzm for I z I < 1.
(8.25)
m=O
If 0 < IxI < IYI and x y = Ix IIyI cos0, then 1
IX
00
> bm Pm (n, cos 0)
- yln-2 = n1=0
Ix lIn lyln-2+n1
for n > 3,
(8.26)
The Laplace Equation
256
and m
00
log Ix
= log
1
A
IYI
E
+ in=1
M
Pm(2, cos 0)
.
IYI-
Proof. By Taylor expansion about x = 0, we have 00 1
Ix - y In-2
_
Fm(x, y)
(8.27)
for IxI < IYI,
M=O
where
F.(x,Y) = > a, (y)x"
and
a. (y) =
1
IaI
1
lal=m
Note that F. (x, y) is homogeneous of degree m in x, and of degree - (n -2) -m in y. Taking the Laplacian of (8.27) with respect to x, we see that 00
EOxFm(x,Y)=0 for IxI < IYI, ,n=0
so by uniqueness of the coefficients in a Taylor expansion, Fm
y) E 7-1m (ill;")
for each integer in > 0. Moreover, if A E R" ,n is an orthogonal matrix, then
jAx - Ayl=Ix - Yl,and so Fm (Ax, Ay) = F,,, (x, y).
In particular, Fn, (Ax, y) = Fm (x, y) if Ay = y, and therefore by Lemma 8.6,
Fm (w, C) = b,n P,,, (co ) for co,
E S" -1,
and for some constant b,,,. Thus, Fm (x, Y) = Ixln
IYI-(,r-2)-m
Fm (ti' IyI l
brPm(COS0)
In-2+m '
(8.28)
and by choosing x and y so that I y I = 1 and cos 0 = 1, we have Ix - yI = (Ix12 - 2IxI + 1)1h/2 = 1 - IxI, so the b,n are as in (8.25). The proof of (8.26) is now complete. When n = 2, we proceed in the same way, except that now Ial
a,(Y)_ (ai
8y logIYI'
Spherical Harmonics
257
so if m > 1, then F,n (x, y) is homogeneous of degree m in x, and of degree -m in y. It follows that for some constants bn m
1
Fo(x, y) = log
and F. (x, y) = b,n Pm (2, Cos 0) Ix
lYl
l
form > 1,
iyitm
and by choosing lyl = 1 and cos0 = 1, we see that b,n = 1/m because log
I
lxl = -log(l - lxl)
m 00 m=1
m
for lxl < 1.
The next theorem gives some expansions of general harmonic functions in terms of spherical harmonics. Recall the definition of Mu given in Lemma 7.11.
Theorem 8.8 Write x = pco, where p = Ixl and w = x/p. (i) If /u(x) = O for p < po, then there exist
u(x) = EP "'
(w)
E fn,(Sa-1) such that
for p < P0.
m=0
(ii) If Du (x) = 0 for p > P0, and if M u = 0 on R", then there exist
E
such that
U(X) = EP 00 2
(w)
for p > Po,
when n > 3,
,n=0
and 00
U (X) = (log P)1o(w) +
p-I"Yin. (w)
for p > Po,
when n = 2.
m=1
Proof Let S2- = Bp,,, the open ball of radius po and centre 0, and consider the special case when u = SL 0 for some 0 E L 1(I'). By (8.27), we get the desired expansion for p < po, with ,n (w) _
(n - 2)Tn
Fm(we Y)O(Y) day.
If n > 3, then by interchanging x and y in (8.27) we see that 00
Ix - yin-2
= L F,n(y, x) m=0
for Ixi > IYI,
The Laplace Equation
258
and by (8.28),
F
X)
2+2,,
yl
(XI)
so u has the desired expansion for p > po, with
f
IYIn-2+2mFm(w, y).0 (Y)day. 1 ..(w) = (n - 2)Tn r When n = 2, the only essential change is to the terms with m = 0, which are
Jr
Fo(x, y)o (Y) day = - J (log IYI)O(Y) dcry
if p < po,
Fo(y, x)o (Y) day = -(log p) J 0 (Y) day r
if p > po.
r
and
Jr
Likewise, any double-layer potential u = DL 0 has expansions of the desired form, with *o = 0 in part (ii), because if n > 3, then a
=0 E ava F(x,y)
1
avy IX
M=0
forlxl 0, with Mv°° =0 on R".
The Laplace Equation
260
To prove part (i), let n > 3 and write w = x/Ixl = xd/I xd I. By Theorem 8.8, there are surface spherical harmonics *,0, and of degree m, such that 00
v°(xd)
= Y Ixd Im *° ((O)
for Ixd I < P0,
,n=0
and 00
v°O(xd) _
(w)
Ix0
for Ixal > 0.
,n=0
Suppose that u(x) = as I x I -+ oo. This assumption means that ud is bounded at zero, and thus r10 must be identically zero for all m > 0. Hence, O(Ixi2-n)
ud=v° and u(x)
(A) n-2 ud(xa) _ E Po-2+2,n 00
x
(w)
for IxI > p0.
,n=0
We conclude that 8"u(x) = O(Ixl2-"-1,I) for all a, and so Mu = 0 on R" by Exercise 7.4. Conversely, if Mu = 0 on R" then u(x) = O(Ixi2-") by Theorem 7.12.
Now suppose that n = 2. By arguing as above, it is easy to see that if u(x) =bi log lxi +b2 + O(Ixl-1) as lxl -+ oo, then the function 5(x)= u (x) - b1 log Ix I - b2 satisfies Mu = 0, and therefore Mu = b2M 1= b2 by Lemma 7.13 and Exercise 7.5. The converse again follows by Theorem 7.12.
0 Solvability for the Dirichlet Problem
We know from Theorems 7.6 and 7.8 that, for any bounded Lipschitz domain Q-, the boundary integral operators
S : H-1/2(F) - H1/2(1') and R : H'/2(f') -+
H-1/2(1.)
are Fredholm with index zero. The following uniqueness theorem for the exterior Dirichlet problem will help us to investigate ker S. We shall see that complications arise when n = 2. Theorem 8.10 A function u E H11 (Q+) satisfies
Au = 0
on Q+,
y+u = 0 u(x) = if and only if u = 0 on Q+.
on r, O(Ix12-")
as IxI -* 00
(8.31)
Solvability for the Diriehlet Problem
261
Proof. Suppose that (8.31) holds. Applying the first Green identity over S2p = 7+ fl Bp for p sufficiently large, we have (DS2P (u, u) _ -(av u, Y+u)r
=
- fa
da B,,
(-f'_-u(Pw))u(Pw)P"_' dw. dp
Theorems 8.8 and 8.9 imply that
u(pw) = O(p2-n) and
-d u(pw) _ dp
0(p")
if n > 3,
O(p-2)
if n = 2,
so
fzP0(p-')
Igrad ul2dx=fiS2+(u,u)
O(p2-n)
if n > 3,
ifn = 2. (
Sending p -+ oo, we deduce that grad u = 0 on Q+, and thus u is constant on each component of 52+. Since y+u = 0, it follows that u = 0 on 52+. The converse is obvious. 0
Corollary 8.11 Let* E H-1"2(F) satisfy S* = 0 on r. (i) Ifn > 3, then i/i = 0.
(ii) Ifn=2and(l,+/i)r=0,then*/r=0. Proof. The single-layer potential u = SL * satisfies
Au=0 on Q-'-, y±u = 0 on l', and as IxI -+ oo, we have u(x) = O(Ixl2-") when n > 3, but
it(x)=-- (1,OrlogIxl+O(Ixl ') whenn=2. Thus, provided we assume that (1, Or = 0 when n = 2, it follows from Theorem 8.10 that it is identically zero, and hence . =
0.
For the Laplacian, we can prove a stronger version of Theorem 7.6.
Theorem 8.12 Let
V = H-'/2(F)
ifn > 3,
262
The Laplace Equation
and
V =[ * E H-112(1') : (1, lr')r = 0 } if n = 2. The boundary operator S satisfies
(Si,r, Or = J grad SL 1k grad SL 4, dx for r1i E V and O E H-1/2(r), (8.32)'
and is strictly positive-definite on V, i.e.,
(S*, *) r > 0 for all * E V \ (0). Proof. Let 1lr E V and 0 E H-112(r). If p is sufficiently large, and if we put
u = SL * and v = SL 0, then (as in the proof of Theorem 7.6) the jump relations and the first Green identity imply that
(SL, 4,)r = (yu, av v - av v)r = n-(u, v) + ('n+ (u, v) + I Bp BP
The integral over aBp is O(p2-") if n > 3, and is O(p-1) if n = 2. Thus, in either case, we obtain (8.32) after sending p -+ oo.
It is obvious from (8.32) that (S*, Or > 0. Moreover, if (S*, Or = 0,
0
then grad
Corollary 8.13 If n > 3, then S is positive and bounded below on H-1/2(r), i.e.,
(Si, Or ? CII
IIH-iIa(r)
for all',/r E H-112(x).
(8.33)
Proof. By Theorem7.6,S:H-1/2(I') -* H"2 (F) is Fredhoim with zero index, and since S is strictly positive-definite, ker S = (0}. Hence, S has a bounded inverse. Since S-1 is self-adjoint, and since the inclusion H1/2(r) c_ L2(r) is compact by Theorem 3.27, the result follows from Corollary 2.38 with A = S-1.
0 By modifying the boundary integral equation Silr = f and adding a side condition, we obtain a system that is always uniquely solvable, even when n = 2.
Lemma 8.14 Given any f E H1/2(r) and b E C, the system of equations
S*+a= f and (1,*)r=b, has a unique solution ,/r E H-1i2(r) and a E C.
Solvability for the Dirichlet Problem
263
Proof Introduce the Hilbert space H = H-1/2(1') x C, identify the dual space H* with H1"2(I') x C by writing ((i/r, a), (0, b))
0)r + ab,
and define a bounded linear operator A : H --* H* by
A(*, a) = (S* + a, (1, if) r). In this way, A is self-adjoint, and we now show that A has a bounded inverse.
Let So and L be as in Theorem 7.6, so that S = So + L with So invertible and L compact as operators from H-112(f) to H1/2(I'). We define Ao(,b, a) = (So,/r, a)
and
K(,f, a) = (a + L+/r, (l, i/r)r
- a),
H* comso that A = AO + K, with AO : H H* invertible, and K : H pact. By Theorem 2.26, A is invertible if the homogeneous system A (*, a) _ (0, 0) has only the trivial solution. In fact, if
S* + a =0 and
(1,
')r = 0,
then (Si/r, Or = (-a, *)r = -a(1, *)r = 0, so (r = 0 by Theorem 8.12,
and inturna= -S* = 0. Theorem 8.15 There exists a unique distribution,/req E
H-1/2
(F) such that S*eq
is constant on r, and (1, t/req)r = 1. If n > 3, then S*eq > 0.
Proof Let
9'eq be the solution of the system in Lemma 8.14 when f = 0 and b = 1. Thus, Slfeq = -a is constant on r, and by Theorem 8.12, if n > 3, then
-a = -a(1, ifeq)r = (S*eq, lkeq)r > 0. The distribution *eq is real-valued, and is called the equilibrium density for F. If n > 3, then the reciprocal of the positive constant S*eq is called the capacity of r, a quantity we denote by Capr, so that 1
Capr
= Sz/req
when n > 3.
This terminology has its origins in electrostatics: if an isolated conductor carries
a charge Q in equilibrium, so that the potential V is constant throughout the conductor, then the ratio Q/ V does not depend on Q, and is called the capacitance. Mutual repulsion causes all of the charge to lie on the boundary of the conductor, so (with appropriately normalised units) the electrostatic potential is
The Laplace Equation
264
SL 1/r, where i/r is the surface charge density. Thus, Q = ,fr l/r da and V = Si,
and in the case of a unit charge Q = 1, we have /r = *eq, so the capacitance is the reciprocal of S*eq. Now consider the case n = 2, and write S = S, to indicate the dependence on the choice of the parameter r in the fundamental solution from Theorem 8.1. The equilibrium density 1/req is the same for all r, but not so the constant Sr1/req. Since Srl/eq is not always positive, one introduces the logarithmic capacity,
Capr = e2'', so that 1
2n
1
log Capr
_
S1 *,q
when n = 2.
Notice that Sri//
Si* +
(127r, Or
log.1,r,
and hence S,.1/req = 2I log
r CaPr
In particular, Sr1leq = 0 if and only if r = Capr.
Theorem 8.16 Consider S,.: H-1/2(I') -3 H1/2(F) when n = 2. (i) The operator S,. is positive and bounded below on the whole of H-1/2(F)
if and only if r > Capr. (ii) The operator S,. has a bounded inverse if and only if r 0 Capr. Proof. For brevity, put ar = Sr1/req = (27r)-l log(r/Capr). Let Vr E H-1/2(F),
define *0 = * - (1, *)r*eq, and observe that
i/I = *o + (1,1)rieq,
(1, *o)r = 0
and Sr.
/ = Sr'Yo + a,.(1, ifr)r
Also, since (Sri/ro, *eq) _ (*o, Srl, eq)r = 0, we have
(Sri, *)r = (Sr*o, *o) r +a,(1/r. 1)r(1, if)r.
(8.34)
If r < Capr, then (Sr 1/req, +feq) r = ar < 0. To complete the proof of (i), suppose
that r > Capr, or equivalently, ar > 0. By Theorem 8.12, both terms on the right-hand side of (8.34) are non-negative, and the first is zero if and only if *o = 0. Thus, (Sr*, Or 0, with equality if and only if *o = 0 and (1, Or = 0, i.e., if and only if /r = 0. Hence, Sr is strictly positive-definite on
Solvability for the Dirichlet Problem
265
the whole of H-112(F). Arguing as in the proof of Corollary 8.13, we conclude that Sr is positive and bounded below on H-1/2(r).
Turning to part (ii), we note that if r = Capr, then Sr cannot be invertible because Sr lyeq = 0. Thus, suppose that r
Capr and S,. ly = 0. We have Sr ly0 = -ar(l, *)r, hence (S,.1/ro, fo)r = 0, and therefore ly0 = 0 by Theorem 8.12.
In turn, (1,1/r)r = 0 because ar 0 0, giving 1/r = 0. Thus, the homogeneous equation has only the trivial solution, and Sr is invertible.
In the case of the unit sphere r = Sn-1, it is clear from symmetry that 1/req takes the constant value I/ T,,, and in view of Theorems 8.4 and 8.5,
ifn = 2,
1
{(n-2)T,, ifn>3. Further properties of Capr are given in Exercises 8.10 and 8.11, and in the books of Hille [40, pp. 280-289] and Landkof [52]. We conclude this section with an interesting application of Theorem 2.36.
Theorem 8.17 If m 0 1, then 11,,, (Si -1) and f-l1 (Sn-1) are orthogonal to L2(Sn-1). Furthermore, the orthogonal direct sum each other as subspaces of ®O0 (Si-1) is dense in 7-f m=0 m L2(S"-1)
Proof. Recall from Theorems 8.4 and 8.5 that l,n (Sn-1) is an eigenspace of S. Hence, the orthogonality of 7-l n, (Sn-1) and Iii (Sn-1) follows at once from the
fact that S is self-adjoint. We may assume, by choosing r > 1 if n = 2, that S is strictly positive-definite on
H-112(Sn-1).
Since the inclusion
L2(Sn-1) c
L2(Sn-1) is compact H-1/2(Sn-1) is compact, the operator S: L2(Si-1) -+ with ker S = (0). Hence, the eigenfunctions of S span a dense subspace of L2(Si_1),
and to complete the proof it suffices to show that every eigenfunction of S is a spherical harmonic. Suppose for a contradiction that ly E L2(r) is a non-trivial solution of Sly =
µ1y on Si-1 for some (necessarily positive) µ, and that * 1 7-l,n(Si-1) for every m > 0. It follows from Theorem 7.2 that * E C°°(Si-1), so by Theorem 6.13 the single-layer potential u = SL * is C°O up to the boundary of the unit ball. If ly,,, E 7-1,,, (S-1) is as in part (i) of Theorem 8.8, then for 0 < p < 1,
f
- 1=1
00
SL*(pw)1/r(w)dw = 11 p,n m=1
and so, sending p f 1, we see that (S*, l tradiction.
f
,n(W)ly(co)do) = 0,
fJwI=1
0, implying, ly = 0, a con-
The Laplace Equation
266
Solvability for the Neumann Problem Solutions of the Neumann problem for the Laplacian are unique only modulo constants; more precisely, the following holds.
Theorem 8.18 A function u E H' (Q-) satisfies
Au = 0 on cZ-,
(8.35)
a-u=0 onr
if and only if u is constant on each component of 12-. Likewise, u e Hloc(S2+) satisfies
Au=0 avu=0 u(x) = if and only if u is constant on each component of S2+ and, when n > 3, is zero on the unbounded component of 52+.
Proof The first part of the theorem is a special case of Corollary 8.3. The exterior problem is handled in a similar fashion, by applying the first Green identity over S2p , and arguing as in the proof of Theorem 8.10.
Let Q, , ... , Op be the components (i.e., the maximal connected subsets) of S2-, and define
of =
1
on S2 ,
0
on S2- \ S2i ,
for 1 < j < p. Thus, the functions v1, ..., vi,, form a basis for the solution space of the homogeneous interior Neumann problem (8.35).
Theorem 8.19 Let f E H(S2-) and g E H-1/2(p). The interior Neumann problem
-Au = f
a,u=g
on Q-,
(8.36)
onF,
has a solution u E H1(S2-) if and only if the data f and g satisfy
Jsz,
f dx +
f
asp,
gda=0 for t<j 0. Moreover, if(Ri/i, r =0, then grad u = 0 on R' \ T, so u is constant on each component of R" \ r, implying that * = [ulr is constant on each component of F. Hence, by Theorem 8.20, (Rill, ill) r = 0 implies Ri/i = 0. Therefore, the coercive, self-adjoint operator R is strictly positive-definite on the orthogonal complement of ker R, and we may appeal to part (iii) of Exercise 2.17.
Exercises 8.1
Let G (x, y) = G (x -y) be the fundamental solution given by Theorem 6.8
when n = 2 and P(4) = (2ir)21'I2. (i) Show that
G(x) =
1
2ir
1
log lxI -F
I"(1) -log27r 27r
- (2,r)2 J 1
2ir
log I sin BI dB.
(ii) Use the substitution 0 = 20 to show that 2n
I 8.2
Deduce (8.1) from
(L: 8.3
log I sin 0 I d9 = -27r log 2.
= Tf
re`dt) ePep"dp = 0
Show that if A E Rnxn is an orthogonal matrix, and if v(x) = u(Ax), then Lv(x) = (Du)(Ax). In particular, when u is harmonic, so is v.
Exercises 8.4
269
Show that if u(x) = w(p), where p = IxI, then
Au(x)dew+n-ldwp
dpe 8.5 8.6
dp \
dp /
T. Jr
(x yl1y) [.
(y)
- *(x)] da,.
Deduce from (8.12) that the numbers N(n, m) have the generating function 00
E N(n,
1+z
m)z,,,
Z)11-1
,,,=o
8.8
p'
dp
d
Show that T1 = -1 and T*+//eq = -*eq. Assume the hypotheses of Theorem 7.3, take P =-A, and let X E r. (i) Show that a:: (x) = -a [T+(x)]/T,,. (ii) Show that limEl.o TE 1(x) _ -1 - [a+ (x) - a-(x)] = -2a+ (x). (iii) Deduce that
T,l/(x) = -*(x) + 8.7
1
for IzI < 1.
Let b,,, = b,,, (n) be as in Theorem 8.7. Show that when n > 3, the Legendre polynomials have the generating function 00
Ebm(n)P.(n,t)zm = m =0
1,
1
(1 - 2tz + Z2)(n-2)/2
and by differentiating both sides with respect to z, derive the three-term recurrence
Po(t) = 1,
b1 P1(t) = (n - 2)t,
(m + 1)b,,,+1 P,n+1(t) - (2m + n - 2)t b,,, P,,, (t)
+(m + n - 3)b,,,-1'n-1 (t)=0 form > 1. Show likewise that when n = 2, 00
,,,=1
1m P»,(2, t)zm = log
1
1 - 2tz + ze
for IzI < 1
and
Po(2, t) = 1 ,
P1(2, t) = t,
Pm+1(2, t) - 2t P,,, (2, t) + Pm-1(2, t) = 0
form > 1.
The Laplace Equation
270
8.9
Recall the definition (8.29) of the inverse point xd with respect to the sphere a Ba, .
(i) Show that
_
axk
po) l2
axk
x;xk (s;k - 2IX11
CIXI
and
" axi ax;
E
4
=
PO
aXJ axk
1=1
Ixl
8jk'
(ii) Consider two curves x =x(t) and y = y(t) that intersect at a point a E II8n when t = 0. Show that
4dx dy
dxO
dy=
PO
dt
dt
(Ia)
dt
dt
when t = 0,
and deduce that the mapping x i--> xI is conformal, i.e., anglepreserving. (iii) Show that if E is a plane or sphere, then so is Ed. [Hint: consider the
equation a Ix 12 + b x + c = 0, where the coefficients a and c are scalars, but b is a vector.] (iv) Think of = xx as a system of orthogonal curvilinear coordinates for x = , and deduce that -2,q
Au =
ICI)
[(PO)2j,-4 au
n
j=j
Next, establish the identity
P, 2-n a
2n-4 au
a; (ICI)
(ICI)
(po) n 2
au a
-2a; a;
a2
+ a; (ICI)
ICI n-2
a2
ICI) PO
n-2
uA
n-2
a2
u
(Iii) PO
and finally conclude that
aupo)
n+2 n
ICI
1
n-2
a2
n+2
n
po
a; (ICI) u =(IXI)
where u° is the Kelvin transform (8.30).
1
a2ud
a;'
Exercises
271
8.10 Let n > 3, and consider the exterior Dirichlet problem
Au = 0
on T2+,
y+u=1
on I', u(x)=O(Ix12-n)
aslxl -+ oo.
Show that the unique solution is u = Capr SL *eq, and deduce that
Capr = -
f
8+ u dcr.
This result can sometimes be used to compute Capr; see Landkof [52, p. 165]. 8.11 Suppose that I' is a simple, closed curve in the complex z-plane, and let w = f (z) define a conformal mapping of S2+ U r onto I w I > 1. Since f is one-one on cZ+, it must have a simple pole at oo; see Markushevich [63, pp. 90-91]. Thus, there is a constant pr such that
f(z)= Z +O(1) asz -* oo. Pr
Moreover, we can assume that pr is real, because the domain I w I ? 1 is invariant under rotation, i.e., under multiplication by e`9 for any angle 0. The constant pr, is then known as the external conformal radius. (i) Show that the real-valued function u defined by f = eh'+", satisfies
Au =0
on Q+,
y+u = 0 u(z) = log
on I', IZI
Pr
+0(l) as z - oo.
(ii) Hence show that
u = log
r
Capr
- 2n SL *eq,
and deduce that pr = Capr. (iii) Suppose that a > b > 0, and let r be the ellipse
Iz-cl+lz+cl=2a, where c= a2-b2. Thus, a and b are the semimajor and semiminor axes, respectively,
The Laplace Equation
272
and c/a is the eccentricity. Verify that the formula
z=
a+b 2
w+
a-b 2w
defines a conformal mapping w l-4 z of the region I w > 1 onto 52+, and deduce that Capr = 1(a + b). For further examples, see Landkof [52, p. 172]. 8.12 Prove the (crude) bound
Capr
ifn = 2,
diam(1') (n - 2)T,,
_-
diam(I')"-2
if n > 3.
8.13 For any a > 0, we write ar = {ax : x E 17). By expressing the equilibrium density for al' in terms of the equilibrium density for r, show that
ifn = 2,
a Capr Caper = t&3_2Capr
if n > 3.
8.14 Derive the following variational characterisation of the capacity: min (Si, ,/,EH-'12(r). (l.*)r=l
Or = (Sieq, *eq)r
12n
log
Cap,
r
Capr
ifn = 2,
ifn>3.
LI
[Hint: if (1, *)r = 1, then fr = *o + *eq with (1, fo)r = 0.] 8.15 Let Q+ = C \ [-1, 1]. The formula
- 1Iw+w Z-2 1
defines a conformal mapping w H z of the region I w I> 1 onto Q+. [This mapping is a degenerate case of the one in Exercise 8.11 (iii) above.] Define
fo(z) = log
wr
and f,n(Z) =
2m w'"
form > 1,
and let un, (x, y) = Re fm (z),
where z = x + iy.
Exercises
(i) Show that w = z + ru
273
z2 - 1, and that
z2 - 1 = ::+-i
fl - x2
for-1 <x < 1.
(ii) Hence show that if -1 < x < 1, then 8:':U",
(x, 0) =
vlimp ifmz) T1
=
1 -x2
x
1
when in = 0,
2(x±i 1 -x2 -'n
when m > 1.
(iii) Let r = (-1, 1), and show that u,,, = SL 1/r,,,, where Vr,,, = -[8vu,,, Jr is given by
fo()
2
and d *,,,(x) =
1 - xz
P. (2-, x) 1
form > 1.
xz
Here, P(2, x) is the Chebyshev polynomial (8.24). (iv) Deduce that for x E r,
r
1
2rrlog\ Ix-YI/ (v) Assume r
if m = 0,
2 log 2r
Pm (2, y)
1-y2 dy
1
l)m(2,X)
if m > 1.
Z, and express the solution of Si/ = f as a series
involving the Fourier-Chebyshev coefficients of f . 8.16 Let c2+, r and w H z be as in Exercise 8.15, but now put
um(x, y) = Re(2iwm+i
)
form > 1.
(i) Show that um = DLi/r,,,, where 1/r,,, = [u,,,Jr is given by *.(x) = Q,,, (x) VI-1- -
x2 and Qm (cos 8) =
sin(m + 1)0' sin 8
(The function Q,,, is the mth Chebyshev polynomial of the second kind.) (ii) Deduce that f.p.
27r EyO
Jc-,,l)\B,(x) (x - y)
2
Qm(Y)
1 -y2dy =
m-
Q,n(x)
2
for-1<x2(X,, *)rXj, i=1
are as in Theorem 8.20. (i) Show that R, is self-adjoint, and that
where X1,
, Xq
(R1r, i)r ? cIII/FI12hIp()
for all *E H1/2(1').
(ii) Show that TXJ = -Xj, and deduce that
(Xi,h)r=±
f
fdx+J gdQ fort < j 3,
and the result for a general n follows. In the particular case n = 3, we can use Exercise 9.1.
The Sommerfeld Radiation Condition When dealing with exterior problems for the Helmholtz equation, it is convenient to introduce the Hankel functions of the first and second kind, Hu1)(z) = JN,(z) +iYN,(z)
and
Hµ)(z) = Jµ(z) - iY,.(z),
The Sommerfeld Radiation Condition
281
which form an alternative basis for the solution space of Bessel's equation. We also put h
(z) = Jm (z) + 1Ym (z)
and
(Z) = Jm (Z) - lYm (Z),
writing h(1) (n, z) and h(2) (n, z) when it is desirable to indicate the dimension n explicitly. If n = 3, then and h,(nt) and h,(n2) are known as spherical Hankel
functions.
From the standard asymptotic expansions for the Bessel functions - see Abramowitz and Stegun [1, p. 364] or Gradshteyn and Ryzhik [29, pp. 961962] - we find that [e'tz-(2'+"- 1).,/4)] +
1
h,(11) (Z) =
z
OC
(n-1)/2
1)],
{\
Z1
(9.10) Z
(n-1)/2 Ce-i[z-(2m+n-1)n/4)] + O Z
hm2) (z) =
L
1\\
J,
J
as z - oo with -rr < arg z < 7r. By Lemma 9.3, the function u(x) = hI')(kp)i/r(w)
forx = pw, *
E'H.(Sn-1)
(9.11)
is a solution of the Helmholtz equation, and the corresponding time-harmonic solution of the wave equation (9.1) satisfies Re
[e-tar h(;) (kp)
* (w)}
= pcos[kp - At - (2m + n - 1)nr/4] + O(p-(n+1)/2) as p - * oo. Physically, this solution is an outgoing or radiating wave; if h ;2j is used in place of h;,;), then we obtain an incoming wave.
Definition 9.5 Write x = pcw with p = I x I and co E Sn', and let u(x) be a solution of the Helmholtz equation for p sufficiently large. We call u a radiating solution if it satisfies the Sommerfeld radiation condition
um
au
p(n-1)/2
C
uniformly in co.
p
- iku) = 0,
The Helmholtz Equation
282
The derivatives of h;,;) and h(2) have the asymptotic behaviour
d l) dhn,z
=
_
1
z (n-1)/2
-i z(n-l)/2
dz
2nn-1),r/4)1
[e'
+
o\(
i, (9.12)
[e_2m+n1)a/4)1 +
p(n-l)/2 (au ap
_ iku) = o 1 ),
\p
/
(9.13)
and is therefore a radiating solution. Also, by taking a = i in (9.8), we obtain a radiating fundamental solution, kn-2
klxl),
G(x) =
(9.14)
and, as a special case, eikI.t
G(x) =
47rIxI
when n = 3.
(9.15)
Recalling our assumption (9.3), we note that if Im k > 0 then G (x) has exponential decay at infinity. The following theorem reveals the connection between Definition 9.5 and our earlier treatment of radiation conditions for general elliptic equations; recall the definition of the operator M given in Lemma 7.11. Theorem 9.6 Let u be a solution of the exterior Helmholtz equation
-Au-k2u=0 onQ+, and suppose that M is defined using the radiating fundamental solution (9.14).
(i) If Mu = 0, then u satisfies (9.13). (ii) If u satisfies lim
P-'a% aBr,
then Mu = 0.
au
-ap- iku
2
dcr = 0,
(9.16)
The Sommerfeld Radiation Condition
283
Hence, the requirement Mu = 0 is equivalent to the Sommerfeld radiation condition.
Proof. By enlarging S2- if necessary, we can assume that u is C°O on SZ+. By
Theorem 7.12, if Mu = 0, then u = DL y+u - SL 8v u on Q+, so part (i) follows from Exercise 9.4. Assume now that u satisfies (9.16). We claim that
Iul2dcr=O(1) asp-+ oo.
(9.17)
LBp
In fact,
-au ap
2
2
8u
iku
I
ap
+ IkI2IU12 +2Im1
au
kavul,
and by applying the first Green identity over the bounded domain SZp = S2+ n Bp, we see that
f(gradu12-k21uI2)dx=fBpavuda- f a-vuda.
(9.18)
(In the first integral on the right, v is the outward unit normal to S2p+, but in the second integral v is the inward unit normal.) Multiplying (9.18) by k, and taking the imaginary part, we obtain
- f Im(kaa u) da = Im(k) -
Ik12Iu12) dx
JBPI
m(kavu) da
and so
'
2
Im(k) f12 (Igradu12 + Ik12Iu12) dx + j 2
- f Iml k a 4 uJ da
as p
\I aU I
+
Ik121u12 I dQ
(9.19)
oo.
The claim (9.17) now follows from our assumptions that k # 0 and Im k > 0. To complete the proof of (ii), we simply write
Mu(x) =
G(x,
Ja BPp
-JB
A
iku(y)] day
[a,,. G(x, y) - ikG(x, y)]u(y) day
284
The Helmholtz Equation
for p sufficiently large, and then apply the Cauchy-Schwarz inequality, noting that the radiating fundamental solution G (x, y) = G (x - y) satisfies
f IG(x, Y) 12 day < C sP
JaBP
Ix -
yl-(n-u day .5 C
and
f Ia,i,yG(x, y) - ikG(x, Y)I2 day < C J aP
Ix -
YI-(n+1) dory
< Cp-'-
aBP
0 We are now in a position to give an expansion of the fundamental solution in spherical harmonics or Legendre polynomials.
Theorem 9.7 For m > 0, let {*n,p : 1 < p < N(n, m)} be an orthonormal basis for The radiating fundamental solution G(x, y) = G(x - y) given by (9.14) has the expansion oo N(n,m)
G(x, y)
=ikn-2
hm)(klxl)srnep(xllxl)jm(k!Y!)/mp(Y/IYI) M=O p=1
ikn-2
T.
00
N (n, m) h(,) (k l x l)jm (kl y l) P, (n, cos 0) m=0
forlxI>IYI>0, where
cos 8 = IxIIYI
Proof Let
E 1-1,,, (Si-1), and put
u(y) = j,n(kp)i/r(co) and v(y) = h;;)(kp)*(w)
for y = pa.
By Lemma 9.3, we have -Au - k2u = 0 on W, and -Av - k2v = 0 on R' \ {0). Hence, Mu = u by Exercise 7.5, and therefore by Theorem 7.12,
f
[a",yG(x, y)u(y) - G(x,
day = 0 for IxI > p,
(9.20)
aP
whereas My = 0 by Theorem 9.6, so
f [a,,,yG(x y)v(y) - G(x, aP
day = v(x)
for lxl > p.
(9.21)
The Sommerfeld Radiation Condition
285
It follows from Exercise 9.3 that cu
m) = dzt m z
j,n(z)ddz
x (9.20) - j,,,(kp) x (9.21) gives the equation
so the combination
f
_
for lxl > p, P
or equivalently,
ik,=-2j».(kp)h;)(klxl)*(x/Ixl)
G(x, pW)/(w) da) =
for lxl > p.
By Theorem 8.17, we have 00 N(n.m)
G(x, y) = G(x, pw) = E E f m=o p=1
.
, G(x, prl)*,np(n) dri *,np(o)),
implying the first expansion, and the second then follows by Corollary C.2. However, so far we have proved only convergence in the sense of It is easy to see from the definition of the Legendre polynomial P," (n, t) immediately following Lemma 8.6 that
L2(S"-I).
I P. (n, t) I < 1
for -1 < t < 1,
so the mth term in the second expansion is bounded by N(n, m)Ih(;)(klxl)j,,, (k I y 1)1. Using the standard large-argument approximations [ 1, p. 365),
Jµ(z) = Yp (z)
(?-)[1+o1] µ \ ez
and
} [I + 0(1)] asµ -+ 00
(with fixed z), we find after some calculation that
iy,n(klxl)jm(klyl) [1 +o(1)]
-i
(k1
In 1)
2m+n-2(klxl)
[I + 0(1)]
asm -±oo.
Since (8.14) shows that N(n, m) = O(m"-2) as in -+ oo with n fixed, the expansions converge pointwise for Ix I > I y 1.
286
The Helmholtz Equation
Uniqueness and Existence of Solutions Theorems 4.12 and 8.2 imply that for each Lipschitz dissection r = I'D U 11 U rN there exist interior eigenvalues 0 < Al < A2 < , and corresponding interior eigenfunctions 01, -02, ... in H1 (S2-), satisfying
-L\/b1 =,,joj on S2-,
Y-0j = 0
on 1'D,
aOj=0
On PN,
with O j not identically zero. Therefore, by Theorem 4.10, the interior problem
-Du -k2u = f
on n-,
Y -U = gD
on FD,
au u = gN
on rN,
(9.22)
has a unique solution u E H1(S2-) for each f E H-1(S2-), gD E H1/2(rD) and gN E H-1/2(FN) if and only if k2 is not an interior eigenvalue. Otherwise, a solution exists if and only if the data satisfy
(4j, Do- + (Y-0j, gN)rN = (a, 4'j, gD)rp for all j such that Aj = k2. Notice that since X j > 0, if Im k > 0 then k2 cannot be an interior eigenvalue, and so (9.22) is uniquely solvable. The following result of Rellich [86] will help us to prove uniqueness for exterior problems.
Lemma 9.8 For any real wave number k > 0, if
-Au-k2u=0 on R"\BPo and if lim
P-+oo
f
I u (x) I2 do. = 0,
(9.23)
xI=P
then u = 0 on R" \ BPo. Proof. Let (1//mp : 1 < p < N(n, m)) be an orthonormal basis for Theorem 6.4 shows that u (x) is C°° for Ix I > po, so
7-lm(S"-1)
oo N(n,m)
u(x) = E E fmp(kP)hI/mp(co) M=O p=1
for x = pw, p > po and w E S)'-
Uniqueness and Existence of Solutions
287
where
fmp(t) = J u(k-1 tw)rmp(w) dw for t > kpp. n-I
The sum converges in L2(Sn-1), and 00 N(n,m)
Iu(x)I2 dax Ixl=p
=j
Iu(Aw)I2An-1
n -i
dw = E
An-1
Ifmp(kP)I2.
M=O P=1
Since u satisfies the Helmholtz equation, the function fmp is a solution of (9.5), and hence fmp(t) = amlphin)(t) + amphm)(t) for some constants amp and am2p. By (9.10),
A"-1I fmp(kP)I2 =
lan pe2i(kp-(2m+n-1)a/4] +amPl2
+ O(A-1),
so the assumption (9.23) implies that a npeie +am2p = 0 for all real 0. Therefore, amp = a, ;,p = 0, which means that fn p is identically zero. Lemma 9.9 Suppose that u E H1oc (S2+) is a radiating solution of the Helmholtz equation, i.e., suppose
-Au-k2u=0 on Q+ and
lim A(n-1)l2 au
- iku) = 0.
(9.24)
A
If
Im(k f (8v u)u de) > 0, r
then u = 0 on 0+. Proof If Im k > 0, then we see from (9.19) that fsi+ I u (x) 12 dx 0 as p oo, and thus u must be identically zero on 52+. If Im k = 0, then we can apply Lemma 9.8 because (9.19) shows that f xl=p I u (x) I2 d cx -* 0 as p oo, and k > 0 by our assumption (9.3). The desired uniqueness theorem follows at once.
288
The Helmholtz Equation
Theorem 9.10 I fu E H11 (S2+) is a solution of the homogeneous exterior mixed problem
-Au - k2u = 0 y+u = 0
on 52+,
on I'D,
a+u=0 onrN, and if u satisfies the Sommerfeld radiation condition (9.24), then u = 0 on
W.
It is now possible to deduce existence results for exterior problems using boundary integral equations. For brevity, we treat only the pure Dirichlet problem; but see also Exercise 9.5. Theorem 9.11 If f E H-' (S2+) has compact support, and if g E H 1/2 (I'), then the exterior Dirichlet problem for the Helmholtz equation,
-Au - k2u = f on 52+, y+u = g
on r,
has a unique radiating solution u E HioC(SZ+).
Proof. We have already proved uniqueness. By Theorems 7.15 and 9.6, a solution exists if and only if there exists i/r E H-1/2(r) satisfying
Si/r = y9 f - 2(g - Tg) on F.
(9.25)
(In the usual way, to define 9f we view f as a distribution on Rwith supp f c 52+.) Theorem 7.6 shows that the Fredholm alternative is valid for this boundary
integral equation, and by Theorem 7.5 the set ker S* consists of all functions of the form a. v where v E H' (S2-) is a solution of the interior homogeneous adjoint problem
-Lv-k2v=0 on Q-, y-v=0 on T. Using (7.5) and the second Green identity, we find that for all such v,
YGf - 2(g - Tg))r =
Y (Gf +DLg))r
= (y v, a. (9f +DLg))r _((_A_ k2)v, gf + DLg)o_ + (v, (-A - k2) (g f + DL g)) n_ .
A Boundary Integral Identity
289
Each of the three terms on the right vanishes, because y-v = 0 on F, (-A k2)v = O on S2-, and (-t - k2)(G f +DL g) = f = 0 on S2-. Thus, * exists, as required.
We remark that the solution fr of the boundary integral equation (9.25) is unique if and only if k2 is not an interior Dirichlet eigenvalue for -A.
A Boundary Integral Identity In this section, we derive a remarkable identity connecting the hypersingular boundary integral operator R with the weakly singular operator S, associated with the Helmholtz equation in 1[83. This identity, together with analogous ones for other elliptic equations, was introduced by Nedelec [75] as a way of avoiding the evaluation of hypersingular integrals in Galerkin boundary element methods
involving R. It will be convenient in what follows to work with the bilinear instead of the sesquilinear form form For any scalar test function w E D(I83) and any vector-valued test function W E D(I83)3, we have the identities
div(uW) = (grad u) W + u div W, div(wF) = F F. grad w + (div F)w,
div(F x W) _ (curl F) W - F curl W, if, say, u : JR3 -* C and F : R3 -± C3 are C'. Consequently, the divergence theorem implies that
(grad u, W) = -
J
u div W dx,
arty
(div F, w) = -
Ja3
F grad w dx,
(curl F, W) = f F curl W dx,
Jrt;
so for distributions u E D*(R3) and F E D* (R3)3,
(gradu, W) = -(u, divW), (div F, w) = - (F, grad w), (curl F, W) = (F, curl W). Given a scalar test function
E D(I83), we shall write Or = Ojr, and define
290
The Helmholtz Equation
yt and at by
(Or, fr)r and (8v0r,
(YtOr,
for * E D(R3);
(Or, 8v5v )r
cf. (6.14). The following identities hold in the sense of distributions.
Lemma 9.12 If 0, * E D(R3), then div y`(Orv)
-a'. Or and curl yt(Orv)
-y`(v x y grad 0)
on R3.
Proof We have (div YL(Orv), w) = -(Yt(Orv), grad w) = -(Orv, y grad w) r
_ -(Or, avw)r = (-8.'Or, w), which proves the first part of the lemma. Next,
(curl y`(4rv), W) = (ryt(Orv), curl W) = (Or v, y curl W)r
=
Jr
v y (0 curl W) da,
so by the divergence theorem,
(curl y`(Orv), W) = F
div(cb curl W) dx. JL
From the identities
div (0 curl W) = grad 0 curl W + 0 div curl W = grad
curl W + 0
and
div (grad 0 x W) = (curl grad ) W - grad
curl W = 0 - grad ¢ curl W,
we have div(O curl W) = -div(grad 0 x W). Thus,
(curl y`(Orv), W) = f
Jsrzt
_- J
div(grad 0 x W) dx
r
which proves the second part of the lemma.
fr
v
(grad o x W) da
A Boundary Integral Identity
291
Now fix a 0 E D(R3), and define
u = DL Or on R3
and Ft = grad u} on
using, in the double-layer potential, any fundamental solution of the form (9.9). We also construct a locally integrable vector field F : R3 _+ C3 by putting
F = I F+
F-
on 52+,
on S2-.
In this way, the support of the distribution F - grad u is a subset of 1'. Two further technical lemmas are required. Lemma 9.13 As distributions on R3,
u = -div SL(Or v),
div F = -k2u,
grad u = F + y`(Or v),
curl F = yt(v x y grad 0).
Proof. Using Lemma 9.12, we find that since 8j commutes with the convolution operator G,
u = DL Or = 9(8vOr) _ c(-div Y`(Or v))
= -divc(y`(Orv)) _ -divSL(Orv). Next, since
(F-grad u,W) =J3(F.W+udivW)dx and
f(F. W+udivW)dx =f } f div(u±W)dx = F zt
r
v y}(uW)da,
it follows by the jump relation for the double-layer potential, [u] r = [DL 'r ] r = Or, that
(F-gradu, W) =- fr v -(cbrYW)do = -(Orv,YW)r = (-Y`(Orv),W), so grad u = F + y`(tr v). Another application of Lemma 9.12 gives
div F = div[grad u - y`(Orv)] = Au + 8tbr,
292
The Helmholtz Equation
and since -Au - k2u = (-A - k2)GB,t,¢r = a,`,Or, we see that div F = -k2u. Finally, since curl grad u = 0, the second part of Lemma 9.12 implies that
curl F = curl[gradu - yt(Orv)] = yt(v x y grad 0). Lemma 9.14 The vector potential
A = g(curl F) = SL(v x y grad 0) satisfies
div A = 0 and curl A = F - k2 SL(Orv)
on R3.
Proof Since div curl F=O, we have div A = div g (curl F) = 9 (div curl F) 0, whereas since
curl curl F = grad div F - A F = grad(-k2u) -AF
= -k2[F + Yt(cbrv)] -AF = (-OF - k2F) - k2yt(Orv), we have
(curl A, W) = (curl g(curl F), W) = (9 (curl curl F), W)
= ((-A - k2) F - k2yt(Or v), G W) = (F, (-Lx - k2)gW) - (k2c(YtOrv), W)
= (F - k2 SL(4,rv), W), as claimed.
We can now prove the main result for this section; see also Exercises 8.18 and 9.6.
Theorem 9.15 Let G (x, y) = G (x - y) where G is given by (9.9). If 0, i/r E D(R3), then
(ROr, +/rr)r = (S(v x Y grad-0), v x y grad *) r - k2(S(Orv), frv)r Proof. Using the definition of R and the first Green identity, we see that
(Ror, fr)r = (-av u, fr)r = +
Jet
(grad u . grad * - k2ui/i) dx,
Exercises
293
and on 01 we have
grad u grad aJr - k2ui/r = [curl A + k2 SL(Orv)] grad * + k2 div SL(4rv)* = curl A grad * + k2 { SL(0r v) grad + [div SL(Or v)]i/r }
= div[A x grad 1/r + k2 SL(¢rv)*]. Hence, the divergence theorem gives
(Ror, *r)r = - J v y[A x grad
dQ
r _ (yA, v x y grad *)r - k2(S(Orv),1/r'rv)r,
0
and finally y A = S(v x y grad 0).
Exercises 9.1 Show from the series definitions of JI12 (z) and Y112 (z) that the zero-order spherical Bessel functions may be written as
Jo(3, z) _
sin z z
and
yo (3, z) _
-cos z z
9.2 Prove Theorem 9.4 by showing that as Ix I -+ 0,
a;G(x) = a;Go(x) + +
I O(Ixl log lxi) O(Ix13-n)
if n = 2, if n > 3,
and then arguing as in the proof of Theorem 8.1. 9.3 Show that if f and f2 are solutions of the differential equation (9.5), then their Wronskian
W = W (fl, f2) _
fi
f2
fl'
f2
is a solution of
dW +n-1W=0. dz
z
Deduce that W = const/z"-', and in particular W
(h(,'), (2)) n, h n?
= -2i
Z"-I
[Hint: use (9.10) and (9.12).]
and W Um, ym) _
294
The Helmholtz Equation
9.4 Let G(x, y) = G(x - y) be the radiating fundamental solution given by (9.14), and write x = pco with p = Ix I and w E
Sn-1
(i) Show that Ix - yI = IxI - w y + O(IxI-1) as IxI
oo, uniformly
foryEF. (ii) Use (9.10) and (9.12) to show that k(n-3)/2e-i(n-3)Yr/4
SL 1Jr(x) =
2(27r)(n-1)/2
x
eikp
p(n-1)/2
(f(f
e-ikmy*(y)day+O(p-I)
and
k(n-3)/2e-i(n-3)n/4 DL 11f (x) =
2(2.7r)(':-1)/2
xC
eikp
p(n-1)/2
f (a,,ve-ikw,y)i(y)
r
day + O(p-1)l
as Ix I -* oo, uniformly in w. (iii) Show that SL f and DL f satisfy the Sommerfeld radiation condition. (iv) Deduce that if u E Him (Q+) is a radiating solution of the -Au -k2 u = 0 on 52+, then there exists a unique function uc,. E CO°(Si-I) such that eikp
p(n-1)/2 [uoo(w) + O(p-1)]
as p -+ oo,
uniformly in w. The function u,,. is called the far field pattern of u. [Hint: use Theorem 7.12.] (v) Show that if u... = 0 on Sn-1, then u = 0 on Sa+. [Hint: use Lemma 9.8.]
9.5 Show that if f E H-I (S2+) has compact support, and if g E H-1/2(I'), then the exterior Neumann problem for the Helmholtz equation,
-Au - k2u = f on 52+, 8v u = g
on l',
has a unique radiating solution u E HIa, (S2+). [Hint: reformulate the
Exercises
295
problem as a boundary integral equation involving the hypersingular operator R, and apply the Fredholm alternative as in the proof of Theorem 9.11.] 9.6 Show that, in two dimensions, the identity of Theorem 9.15 takes the form
(ROr, *r)r = (Sai0, atf)r - k2(S(Orv), irrv)r for 0,' E D(R2) where r = (-V2, vi) is the tangent vector to I' satisfying v x r = e3, and
a,0 = r
grad 0 denotes the tangential derivative of 0.
10
Linear Elasticity
In the preceding two chapters, we considered the simplest and most important examples of scalar elliptic equations. Now we turn to the best-known example of an elliptic system, namely, the equilibrium equations of linear elasticity. For the history of these equations, we refer to the article by Cross in [30, pp. 10231033], the introduction of the textbook by Love [60], and the collection of essays by Truesdell [101]. Our aim in what follows is simply to show how the elasticity equations fit into the general theory developed in earlier chapters. Necas and Hlavd6ek [73] give a much more extensive but still accessible treatment of these equations, without, however, discussing boundary integral formulations. Let u denote the displacement field of an elastic medium. Mathematically, u : 7 -+ C', so m = n in our usual notation, and physically u is R"-valued and the dimension n equals 3. In Cartesian coordinates, the components of the (infinitesimal) strain tensor are given by
Ejk(u)=Z(ajuk+akuj) forj,kE(l,2,...,n), and we denote the components of the stress tensor by E jk. Thus, using the summation convention, the kth component of the traction over r is vi E jk. (The
traction is the force per unit area acting on S2 through the surface T.) If f is the body force density, then in equilibrium we have aJ Eik + fk = 0.
(10.1)
For a linear homogeneous and isotropic elastic medium, the stress-strain relation is E jk (u) = 2µE jk (u) + A(diV u)s jk,
where the Lame coefficients µ and X are real constants. We can write the 296
Korn's Inequality
297
equilibrium equations (10.1) in our standard form Pu = f by putting
Pu = -a3Bju
and
(13ju)k = Ejk(u).
Notice that the conormal derivative has a direct physical meaning:
B, u = traction on r, and since
aj Ejk =,1 aj(ajuk + akU j) + a.ak(div u) = µajajuk + (µ + ).)ak(diV U), the second-order partial differential operator P can be written in the form
Pu = -µ/u - (µ + A) grad(div u).
(10.2)
This chapter begins with a proof of Korn's inequality, thereby establishing that P is coercive on H 1(S2)" . After that, we derive the standard two- and three-dimensional fundamental solutions. The third and final section discusses uniqueness theorems and the positivity of the boundary integral operators, but only for the three-dimensional case.
Korn's Inequality We see from (10.2) that the Fourier transform of Pu is P u ( ), where P ( is the homogeneous, R"'-valued quadratic polynomial with jk-entry
)
Pjk( ) = (27r or, letting I, denote the n x n identity matrix,
(202[µI I21" + (µ + A)
(10.3)
Thus,
fori ElR' andnEV, and therefore P is strongly elliptic if and only if
µ > 0 and 2µ +.l > 0;
(10.4)
cf. (6.5). Landau and Lifshitz [54, p. 11] explain the physical significance of (10.4). Since
(BjU)*ajV = Ejk(u)aJVk =2pEjk(it)ajVk+X(dlvu)akVk = 21.LE jk (u)Ejk (v) + A(div u) (div v),
Linear Elasticity
298
the sesquilinear form associated with P is
(pc(u, v) = f [21LEjk(u)Ejk(v)+A(divu)(divv)]dx. For a physical displacement field u : 0 -+ 1R3, the quadratic form
2Osz(u, u) = 2 f Ejk(u)Ejk(u)dx z
is the free energy of the elastic medium within S2; see Landau and Lifshitz [53, p. 12]. It will be convenient to let grad u denote then x n matrix whose jk-entry entry is ajuk, and to write
IIE(u)Ili2(sz)"xn = f Ejk(u)Ejk(u)dx and z
J
8juk8jukdx.
Notice that
cl z(u, u) =
21L11E(u)IIi2(U)"xn
+ I1divu11t2(12).
(10.5)
If the Lame coefficients satisfy (10.4), so that P is strongly elliptic, then we know from Theorem 4.6 that (D is coercive on Ho (S2)". In fact, a stronger result known as Korn's first inequality holds. Recall that, by Theorem 3.33, Hi (Q)" = Hp (S2)" if S2 is Lipschitz.
Theorem 10.1 If 0 is an open subset of R", then IIE(u)112 (n)1xn
grad u11L2foru E
>_
(Q) n.
all
Proof. It suffices to consider u E D(O)". As in the first part of the proof of Theorem 4.6, we apply Plancherel's theorem to obtain
JR" Ejk(u)Ejk(u)dx = fR#l
=j 2
kuj)(17T)(5juk+} kuj)d
I . ul2) d "
f
2
fR- Iaju12dx. j=1
The result follows, because we can replace R" by 0 in both of the integrals with respect to x.
0
Fundamental Solutions
299
A much deeper result is Korn's second inequality (or just Korn's inequality). Theorem 10.2 If S2 is a Lipschitz domain, then IIE(u)IlL,()nxn ? cllgradUHL,(f)nxn - CIIuIIL2ISZ),7
foru E H'(SZ)n.
Proof. The left-hand side has the form (4.10) with L = n2, and for convenience we change the index set, writing N1ku = EJk (u) instead of N u. Since
.AIku = 2(a;uk + aku;) = 2(ekaj +ei ak)u, we have
(Nfku (x) } = Njk (i2rc )u (l; ), where NJk (
) = 2 (ek1 + e Sk).
The desired inequality holds because the hypotheses of Theorem 4.9 are sat-
isfied, with q,. = 1 for 1 < r < m = n. For instance, when n = 3 and
r=1, =2el,
1N11(e)T =tiel,
-6 N22
2NI1(S)T = 12e1,
-1N23(e)T +3N12()T +2N,3( )T = 2e3e1,
t3 N, ]( )T = 1e3e1,
-1N33( )T +3N13()T + 3N31( )T = t3e1,
)T
and the other cases can be handled in the same way.
In view of (10.5), Korn's second inequality implies the following result, which allows us to apply the general theory of elliptic systems. Necas and Hlavacek [73, pp. 46-49] give simple physical arguments showing that both Lames coefficients should be positive for typical elastic materials.
Theorem 10.3 For any Lipschitz domain S2, the elasticity operator (10.2) is coercive on H I (S2)" if µ > 0 and X > 0.
Fundamental Solutions Since the polynomial (10.3) is homogeneous, we may obtain a fundamental solution for the elasticity operator using Theorem 6.8; see Landau and Lifshitz [54, p. 30] for an alternative approach.
Linear Elasticity
300
Theorem 10.4 If µ # 0 and 2µ + A # 0, then a fundamental solution G (x, y) G(x - y) for the elasticity operator (10.2) is given by T
G(x)
8nµ(2µ + .X)
C(3µ + X)
1 13 + (µ +,X) IxI3) I
when n = 3, (10.6)
and by G(x)
=
1
4gµ.(2µ +X)
((3p' + .) log
1 I2 + (µ + A) xxTI2) 1
I
when n = 2.
IX
Proof By (10.3), if CO E S", then
P(w) = (2n)2[µ1 + (µ +,l)WWT ] and so by Exercise 10.1, z
p(w)-I =
(27r)
µ(2µ + ),)
[(2µ + ),)I - (µ +))WWT ].
Assume that n = 3, and choose an orthonormal basis ?71, 712 E JR3 for the plane normal to x, so that the unit circle in this plane has the parametric representation SY : w1 = (cos 0)x1i + (sin 0) 712
for -7r < B < n.
By Theorem 6.8(iii), z
G(x)
- I µ(2µ + X) J [(2µ
+ ),)I3 - (µ + )A)wlwl] dw1,
and since
fs
xx T
+ (sine 0)r1271i] dO = 7r (7JI 17T + 712711) = 7r (13
- IxI2 )'
the formula for G(x) follows. Suppose now that n = 2; by (6.13), the formula
G(x) =
-(2n)
z
µ(2µ +,X)
J((log Iw sI
xI)[(2µ + ),)12
- (µ +)-)wwT ] dw
Uniqueness Results
301
defines a fundamental solution. The vectors
ni =
x
x,
1
=
Ixl - Ixl
X,21
and
X2
J7,
Ixl L
form an orthonormal basis for R`, and by putting co = (cos 9)r71 + (sin 0) 112 we see that
Js
log Iw xI dco =
T
_n
log(Ix cos01) dO = 2.7r log IxI + const
1
and
(log 1w x l )WWT dco
Js'
=J log(Ixcos01)[(cos20)17,riT +(co59sin9)(ri,riz +rl2rli) + (sin2 9)7721i2 ] d9
J
"
1 +cos 20 2
1og(IxCOSel)
T
,+
1 - cos 20
T1 dB.
2
2 J/
n
An integration by parts gives ;r
f
log (Ix cos 91) cos 20 d0
=
-,
rr
,
IT
sin O sin 20 dO = Cos 9 2
sine 0 dO = rr, n
and one easily verifies that
xxT
and
'?Irli +rl2riz =12
ntrli -7722 =2Ixle -12,
so
f
1
(log lc) xl)wwTdco= 1
r
log(IxcosOl)dO12+
/ 2xxT
I 7rI
1x12
12)
= rr log Ix 112 + it x 12 + const,
leading to the stated formula for G(x).
U
Uniqueness Results Throughout this section, we assume that the components of u are real, and treat only three-dimensional problems.
Linear Elasticity
302
To apply the Fredholm alternative (Theorem 4.10) to the mixed boundary value problem in linear elasticity, we must determine all solutions of the homogeneous problem. As a first step, we show that the only strain-free displacement fields are the infinitesimal rigid motions, and that such displacement fields are also stress-free.
Lemma 10.5 Let S2 be a connected open subset of R3. A distribution u E D*(52;R3) satisfies E(u) = 0 on S2 if and only if there exist constant vectors a, b E R3 such that
u(x)=a+bxx forxES2.
(10.7)
Moreover, in this case E(u) = 0 on S2.
Proof. Let B = [b,3] E 83x3 denote the skew-symmetric matrix defined by Bx = b x x, i.e.,
B=
0
-b3
b2
b3
0
-b1
-b2
b1
0
Ifu(x)=a+b xx,then ajuk =bkj,so Ejk(u) = 1 (bkj + bjk) = 0. To prove the converse, assume that E(u) = 0 on S2. Since, with the notation of (3.9) and (3.10),
E(*,*u)='YE*E(u) on(xES2:dist(x,I')>e}, we can assume that u E C`O(S2)3. The diagonal entries of the strain tensor are
just Ejj(u) = ajitj (no sum over j), so a1u1 = a2u2 = 83113 = 0
on 0.
Since the off-diagonal entries of the strain tensor also vanish, we can show that
a2u1=83u1=0,
82
a1u3=a2u3=0
on Q;
for instance, a uk = al (2Elk (u) - aku 1) = -ak (al u 1) = 0. Therefore, Fu = 0 on 0 if (a I > 3, implying that u is a quadratic polynomial in x. In fact, the vanishing of the partial derivatives listed above shows that u must have the form ul(x) = al + b12X2 + b13x3 + C1X2X3,
u2(X) = a2 + b21X1+ b23X3 + C2X1X3. u3 (X) = a3 + b31x1 + b32x2 + c3X1X2,
Uniqueness Results
303
for some constants aj, bik and cj. Since E(u) equals 0
b12+b2,
b13+b31
b,2 + b2,
0
b23 + b32
b, 3 + b3,
b23 + b32
0
1
1
+ 2
0
(C) + C2)X3
(Cl + C3)X2
(Cl + C2)x3
0
(C2 + C3)x1
(Cl + C3)X2
(C2 + C3)xl
0
we conclude that bfk = -bk! and c! = 0 for all j, k. Finally, if u has the form (10.7), then E(u) = 2AE(u) + A(divu)13 = 0 because E(u) = 0 and div u = E!j (u) = 0. O Theorem 10.6 Assume that 0 is a bounded, connected Lipschitz domain in and that the Lame coefficients satisfy
A>0 and A>0.
(10.8)
Let W denote the set of solutions in H' (S2; R3) to the homogeneous, mixed boundary value problem
-p Au - (µ + A) grad(div u) = 0 on 0,
yu =0
on I'D,
0
on F.
(10.9)
(i) If I'D # 0, then W = {0}, i.e., (10.9) has only the trivial solution. (ii) If I'D = 0, so that (10.9) is a pure Neumann problem, then W consists of all functions of the form (10.7) for a, b E 1R3.
Proof. If U E H' (Q)3 is a solution of (10.9), then c(u, v) = 0 for all v E H1(Q)3 such that yv = 0 on I'D. In particular, by taking v = u and recalling (10.5), we see that 2 tjIE(u)IIL,(Q)3x3 +AIIdIvuIIL,(g2) = 0.
Our assumptions on a and A then imply that E(u) = 0 on Q, and so u has the form (10.7). If FD # 0, then, because rD is relatively open in r, we can find x, y, z E rD such that x - y and z - y are linearly independent. From the three equations
a+bxx=0, a+bxy=0, a+bxz=0,
Linear Elasticity
304
it follows that b x (x - y) = 0 and b x (z - y) = 0, and we conclude that b = 0. (Otherwise, x - y and z - y would both be scalar multiples of b.) In turn, a = 0, and part (i) is proved. If I'D = 0 and u has the form (10.7), then E (u) = 0 on S2, by Lemma 10.5, and so u is a solution of the homogeneous Neumann problem, i.e., u E W. 0 To conclude this section, we consider the boundary integral operators S : H-'t2(F; 1183)
H112(F; 1183)
and
R : H't2(r; R3) -+
H-1/2(1,; R 3)
associated with the three-dimensional elasticity operator (10.2), and defined using the standard fundamental solution (10.6). Theorem 10.7 Assume that the Lame coefficients satisfy (10.8).
(i) The weakly singular boundary integral operator is positive and bounded below on the whole of its domain, i.e.,
(Si, Or ? cIIrf1H-,r-(r)3 for all Ir E H-1/2(r; R3). In particular, ker S = (0). (ii) The hypersingular boundary integral operator is positive and bounded below on the orthogonal complement of its null space. Indeed, if SZ- is connected, then the six functions Xi : I' -+ 1183 (1 < j < 6) defined by
forxEI'and1 1, an extension operator Ek : WP (S2) -3 WP (R) that is bounded for 1 < p < oo. Using a different method, Stein [96, p. 181] obtained an extension operator E : W PI (S2) -+ WP (118"), not depending on k > 0, and bounded
for 1 < p < oo. In the case when 0 is smooth, there is a simpler construction, due to Seeley [93]; see Exercise A.3. In the main result of this appendix, Theorem A.4, we shall use a modified version of Calderon's extension. Suppose that S2 is a hypograph, S2= (x
x _ 1), and the function
: 1R11- I -- R is Lipschitz:
Mix' - y'i for x', Y' E R"-'.
(A.2)
A crude extension operator is obtained simply by reflection in the boundary of S2.
Theorem A.1 If 0 is the Lipschitz hypograph (A.1), and if
Eou(x) =
for x E S2,
U (X)
1u(x',2 (x')-x") forx c- W' \S2,
then Eo : W p (S2) --* Wp (IR") is bounded for 0 < s < 1 and I < p < oo. 309
Extension Operators for Sobolev Spaces
310
Proof Puti = (x',
xn). One easily verifies that x E 0 if and only if x E R" \ 0, and vice versa, with x = x. Moreover, the Jacobian determinant of the transformation x H x is identically equal to -1. Thus, II Eou II L,,(W'\n) _ IIUIILP(n), and IIEouIIL,,(R11) = 211uIILn(12) Also, if x E R" \ 0, then
for 1 < j < n - 1, for j = n,
8;u(2)
ajEou(x)
-anu(2)
implying that II8;UIILP(s2) +2MIIanUIIL,,(52) { I1
fort < j < n - 1, f j = n.
u11 z,,(9)or
8
Hence, Eo : W, (0) --+ WP (R") is bounded if s = 0 or 1. Assume now that 0 < s < 1 and 1 < p < oo. Recalling the definition (3.18) of the Slobodeckii seminorm, we write IEouIf pR,,
=lulsps2+I(+12+I3,
where
I,
lu(x) - u(Y)IP
Ix -
fL >x').
Yln+ps
dxdy ,
n
Iu(2)-u(y)Ipdxd
I213 = Since I2, that
Y,
n+ps
Iu(x) fL,t(Y')
v,
Ix - Y 112+p$
dxdY.
/f 4M2 IX - y1, we see
- 5 I < 21(x') - (y')I + Ixn -
(y'), then I Eou(x) - Eou(y)I = Iu(z) - u(Y)I < Iuls.oo.nlx - YI'`
< [2V-1 + M2lsluls,oo,szlx - yls.
If x > (x') and y < (y'), then Eou(x) - Eou(y)I = Iu(i) - u(y)l < Iuls,..nIi - yls < CIuls,.,S21x - yls, and similarly if x < (x') and y > (y'), then I Eou (x) - Eou (Y) I < C l u (s,oo,5
Ix - yIs. To obtain an extension operator for s > 1, we shall use the Sobolev representation formula; recall the notation of (3.7). Lemma A. 2 For each integer k > 1, if i/r E C°°(S"-') satisfies
f
f(co) dco =
-1=1
(-1)k (k - 1)!'
(A.3)
and if u E C o ,(R" ), then
u(x)=f t^
* \ I_
u(k) (x
I'
+ y; Y) dy for x E R".
Proof By Exercises A.1 and 3.5,
(k_I()( 00
u(x)
(k -11)!
dp
°o
k
(k- 1)!
x + pco) dp
pk-lu(k)(x+pco;co)dp.
10
312
Extension Operators for Sobolev Spaces
Multiplying both sides by * (w) and integrating with respect to co, we see that 00
u(x) = f
*(c)) f
uI=1
p-flu(x + pco; pw) p"-1 dp dm.
0
We use the abbreviation I u 1 µ for the Slobodeckil seminorm on R" with p = 2.
Lemma A.3 If K E C u E D(R"), then
\ (0}) is homogeneous of degree 1 - n, and if
(
for-oo<so
K(Pw)X'(P)wjpn-1
dpdw,
which, using the homogeneity of aj K and K, simplifies to
f
0
00
X(P)dp f P
ajK(co)dco=-f' X'(p)dp
f
K(w)wjdo.
I"I_1
IWI=1
Since ff ° X (p) dp/p = 1 and fo x'(p) dp = -X (0) = 0, the function Kj satisfies (A.4), as required.
We are now in a position to obtain the desired extension operator.
Theorem A.4 Assume that 0 is a Lipschitz domain. For each integer k > 0, there exists an extension operator Ek : WZ (Q) -* W2 (R") that is bounded
forks 1, if f : [0, oo) -+ C is a Ck function with compact support, then
_ .f (0) =
(k
k
-11) Jo
o0
tk-1 fck) (t)
dt.
o
A.2 In Exercise A.3, we shall need a sequence (,lk)k° satisfying 00
E 2'klk = (-1)' for all j > 0.
(A.5)
k--O
(i) Let ao, ..., aN-1 be distinct complex numbers, let b be any complex number, and consider the N x N linear system N-1
E(ak)'xk=b' for0< j 0.
I k=° (i) Show that E : D(Q) -+ D(R"). (ii) Show that E : Wn (S2) -+ WP (R") is bounded for s > 0 and I< p 0, and choose a cutoff function X E C mp[0, co) satisfying x (y) = 1 for 0 < y < 1. Define the C°'µ epigraph n = {(x, Y) E R2 : y > IxI'`}, and the function
u(x, Y) =
y'-Ex
(y)
for (x, y) E Q.
2(µ-' - 1). (i) Show that u E W2 (S2) if E < (ii) Show that u l C°'- (Q) if I - E < X < 1. (iii) Deduce that no extension operator from W2 (S2) to W2(llt2) exists. [Hint: use Theorem 3.26.]
Appendix B Interpolation Spaces
Suppose that Xo and X 1 are normed spaces, and that both are subspaces of some
larger (not necessarily normed) vector space. In this case, X0 and X1 are said to form a compatible pair X = (Xo, X1), and we equip the subspaces Xo n X1 and Xo + X 1 with the norms IIuIIX,)1/z
IIulIxonx, = (IIuIIXa + and
IIuIlxa+x, = inf {(IIuoIIX0 +
11U1 11X21)1/2
u = uo + ul where uo E Xo and u1 E X1 }.
Notice that for j = 0 and 1,
X0 nX1 cXj cX0 +X1, and these inclusions are continuous because IIuIlxa+x, < II U II x1 < II ulIxonx,
If XI c Xc,then XonX, =X1 andXc+X1 =X0. In this appendix, we present a general method for constructing, from any given compatible pair X, a family of normed spaces Xe,q = (Xo, X 1)e.q
for 0 < 9 < 1 and I < q < oo,
each of which is intermediate with respect to Xo and X1, in the sense that Xo n X1 c Xe,q c Xo + X1. 317
(B.1)
Interpolation Spaces
318
Moreover, we shall see that XB,q has the following interpolation property. Take
a second compatible pair of normed spaces Y = (Yo, Y1), and two bounded linear operators Ao : Xo -+ Yo
and
A, : X 1 -+ Y1.
If Aou = Alu
for U E Xo n X1,
then AO and A 1 are said to be compatible, and there is a unique bounded linear operator AB : XB,q -* YY,q
such that
ABU =Aou = Alu foruEXonXl.
(B.2)
We will also show that if X0 and X1 are Sobolev spaces based on L2, then so is X8.2
For technical reasons, it is convenient to construct Xe,q in two different ways. Thus, we shall define two spaces, Ko,q (X) and Jo,q (X), and show X B,q = KB,q (X) = Je.q (X ),
with equivalent norms. The K-method will be used to prove the interpolation properties of HS (c2), after which the interpolation properties of HS (Q) follow by a duality argument that relies on the J-method. We conclude by considering the interpolation properties of HI (F). For more on the theory of interpolation spaces, see Bergh and Lofstrom [5].
The K-Method The K -functional is defined for t > 0 and u E Xo + X 1 by K(t, u) = inf {(IIuoIIX0 + t2IIu111X,)112
u = uo + u l where uo E X0 and U1 E X1 }. When necessary, we write K (t, u; X) to show explicitly the choice of the compatible pair X = (Xo, X1). For fixed t > 0, the K-functional is an equivalent
The K-Method
319
norm onX0+X1:
K(t, u + v) < K(t, u) + K(t, v)
K(t,.Xu) = IXIK(t, u), and
min(l, t)Ilullxo+x, 0
This weighted norm has an important dilatation property, namely,
Ilt H f(at)Ile,q =aellflle,q fora > 0.
(B.3)
Now define
Ko,q(X) = (u E Xo + X1 :
II
u)IIe,q < oo},
and put II u11 K., (x) =
u) III,?,
where the constant Ne,q > 0 may be any desired normalisation factor. As the default value, we take
ifl q 1 K(t, 01 u; X1)2 < K(t, u; X)2. It follows that ¢IU E Ke,2(XI) _ HI (rl) for each 1, with IIuIIH= (r)
IIp1uIIKe.2(x,) -
to IIuIIK.,4(x). [Hint: use (B.6).] B.3 Give a direct proof of the interpolation property for Je,q (X), i.e., show that
Theorem B.2 holds if Ke,q is replaced by Jo,q. [Hint: J(t, Af (t); Y) < MoJ(at, f (t); X), where a = M1 /Mo.] B.4 Fix real numbers Ao > 0 and A 1 > 0, and a complex number z. Show that min (AoIzo12 + A1Iz112) =
z=zo+zi
AOA1
Ao + Al
Iz12,
and that the minimum is achieved whenAozo = A1z1-= AoA1z/(Ao+A1). B.5 Use contour integration to show that cc t 1-26
,r
1 + t2 dt
for 0 < 9 < 1.
2 sin r9
0
B.6 Show that if Xo and X 1 are complete, then so are Xo fl X 1, X0 + X 1 and
Xe,q for 0 < 9 < 1 and 1 < q < oo. [Hint: use Exercise 2.1.] B.7 Show that (X1, Xo)B,q = (Xo, X1)I-o,q for0 < 9 < 1 and I < q < oo. B.8 Assume that H, V and A satisfy the hypotheses of Corollary 2.38, and equip V with the energy norm II II A. By arguing as in the proof of Theorem B.7, show
K(t, u; H, V )2
= E 1+?
t2I(0;,u)12,
and deduce that for the normalisation (B.9), CO
IIUIIK9.z(H.v) =
E'Xjl(4),,
j=1
u)12 = IIAB"2uIl2.
Appendix C Further Properties of Spherical Harmonics
We shall prove a result (Corollary C.2) used in Chapter 9, and also construct the classical spherical harmonics, which form an orthogonal basis for xm Recall the definition of the Legendre polynomial Pn, (n, t) in the discussion following Lemma 8.6. (S2).
Theorem C.1 The orthogonal projection Q,n : L2 (5"-1) -> xm (S' -1) is given by the formula Q,n
,/ (CO) = N(n, m)
Pm (n, w 1) (r1) d?l
T. 4-1 form > 0, w E S' ' and it E
L2(S"-1).
Proof. Let (rmp : 1 < p < N(n, m)) be an orthonormal basis for
xm(S'
)
By part (i) of Exercise C.1, Qm 1F' (w) =
f
_
K(w, n)i(n) dn,
where
N(n,m)
'I, 'I' K(w, n) = E Y'mp(w) lmp(n) p=1
If A E
nxn
is an orthogonal matrix, then
J
*(Aw) dco =
J
*(w) dco for r E L1(S"-1),
and, by Exercise 8.3, the function co 1-* u(Aco) belongs to Rm(S") whenever u E xm (Sn-1). It follows by part (iii) of Exercise C. 1 that
K(Aco, An) = K(w, n) for co, n E Sr-1, 334
Further Properties of Spherical Harmonics
335
and in particular, when Aen = e we have K(Aw, en) = K(cv,
Since
w H K (co, e,,) belongs to ln, (S"-' ), if u is as in Lemma 8.6 then
K(w, en) = au(w) = aP.. (n, w en)
for w E S"_ 1,
where a = 1/K(en, en). Given r! E Sn-1, choose an orthogonal matrix A E Rnxn such that Ail = en; then K(w, rl) = K(Aw, en) = aP.,(n, Acv en)
Finally, since K(w, w) = aP,n(n, 1) = a, we see that N(n,m)
aTn=
f ,-
K(w,co)dw= E Il*mpllL,
,
p=1
giving a = N(n, m)/Tn.
O
The proof above also establishes the addition theorem.
-
Corollary C.2 If (if,np : 1 < p < N(n, m)) is an orthonormal basis for xm (Sn-1), then N(n,m)
'
for w, )l E Sn-'
(CO) Y',np (11)
P=I
The next result is known as the Funk-Hecke formula.
Theorem C.3 Let f : [-1, 1] -+ C be a continuous function. If m > 0, then
f (w rl)(n) drf = A(w) for* E 71 fS.-I
where
= Tn-1
f
f (t)Pm(n, t)(1 -
t2)(n-3)/2
dt.
(C.1)
Proof We begin by showing that f (l; w) Pm (n, rl co) d w = A Pm (n, l; rl)
fore, rl E Sn-' .
(C.2)
rl) to be the left-hand side of (C.2), and observe that for a fixed l;, Define Also, if A E RI"" is an the function >) r-# F(l;, r7) belongs to
Further Properties of Spherical Harmonics
336
orthogonal matrix, then F(A4, Arl) = F(4, rl), so by arguing as in the proof of Theorem C.1 we see that ,XP,,, (n, rl) for some constant .l. Since
Pm(n,1)=1, I = XPm(n, 1) = F(en, en) =
f
f(en w)P,n(n, en W) do),
S`n-I
and the formula (C.1) follows by Exercise C.2. Thus, (C.2) holds, and upon multiplying both sides of this equation by *(q), where 1/r E xm (Sn- t ), and integrating with respect to q, we have
f
S.-I
f(
. w)
\
Sn
-
Pm(n, n
w)*(rl) drl dw
I
fs.-I
P. (n,
rl)*(rl) dil.
Applying Theorem C.l, the result follows at once. We now begin our construction of an explicit orthogonal basis for H. (S! I- ),
starting with the case n = 2. Remember that 1-1o(Sn-1) consists of just the constant functions on S", so it suffices to deal with m > 1. Lemma C.4 Let 9 be the polar angle in the usual parametric representation of the unit circle S1, co = (cos9, sin 9).
If m > 1, then N(2, m) = 2 and the functions cosm9
1rm1(w) _
and
*,n2 (W)
1/7-r
r7r
sin m9
form an orthonormal basis for Hm (S1)
Proof. We have seen already in (8.13) that N(2, m) = 2 form > 1. Define two real-valued solid spherical harmonics u 1, u2 E Nm (R2) by u1(x) + iu2(x) = I (x1 + ix2)m,
and observe that 1lrn,1 and i/r,,i2 are the corresponding surface spherical harmonics in 11m (S1), i.e., 1/ m p = u p is, for p = 1 and 2. One readily verifies that *,,l and
1/rm2 are orthonormal in L2(S1).
Any non-trivial function Amj (n, t) satisfying the conclusion of the next theorem is called an associated Legendre function of degree m and order j for the dimension n.
Further Properties of Spherical Harmonics
337
Theorem C.5 Assume n > 3, let 0 < j < m, and define
A,nj(n, t) = (1 - t2)jj2Pm-j(n +2j, t).
(C.3)
If i/r E ?-l j (Si-2) and
where co =
W(co) = A,nj (n, t)i/r(r1),
then
1 - t2 n + ten and rl E gi-2,
E fm(Sn-1)
Proof For any 1(r E L i (Sn-2), the formula
u(x) = f-2(xn + ix' )'n
(
) dt
defines a solid spherical harmonic of degree m, whose restriction to S"-' may be written as
u(w)=f (t+i 1-t2 r! If we now assume that and Exercise C.5,
E
?j j (S, -2), then by Theorem C.3, Exercise C.4(ii)
t
u(w) = Tn-2*(r1) f (t + i l - t2 s)mPj(n - 1, s)(1 = cl (17) (1 - t2)'/2
J
(t + i l - t2 1
s)m-I (1
-
s2)(n-4)/2
ds
s2)r+(n-4)/2
ds
= c2i(11)(1 - t2)j/2Pm- j(n + 2j, t) = c2`1(w), where the constants cl and c2 depend on n, m and j. Thus, 41 E ?-lm claimed.
(S"), as
Combining Theorem C.5 and Exercise C.6, and recalling (8.12), we see how to construct an orthogonal basis for W n (S"-1) by recursion on the dimension n. Corollary C.6 If {7(rjp
x j (Sn-2), and if
:
I < p < N(n - 1, j)) is an orthogonal basis for
q'mjp(w) = Amj(n, t)'Yjp(q),
where w =
1 - t2 q + ten and r1 E Sn-2,
then {Wmjp:O < j < m and 1 < p < N(n - 1, j)) is an orthogonal basis
Further Properties of Spherical Harmonics
338
for H,n (Sn-1), and
N(n + 2j, in - j) T +2l! 1 II*jP ,(S,,-2) In particular, taking n = 3 and using a basis for fl3(S1) of the type in Lemma C.4, we arrive at the classical spherical harmonics. Theorem C.7 Use the standard parametric representation for the unit sphere S2,
w = (sin 0 cos 0, sin 0 sin 0, cos 0) for 0 < 0 < 2ir and 0 < 0 < rr.
If in > 1, then N(3, m) = 2m + 1 and the functions *mo (w) = Pn, (3, cos 0),
*mj (w) = (sin Ql )' Pm_j (3 + 2j, cos 0) cos j O for 1 < j < m, YIm.m+i (w) = (sin 1)i
Pm_j (3 + 2j, cos ¢) sin jO for 1 < j < m,
form an orthogonal basis for lm (52), With II1f,no II L2(s2) = 4n' and T3+2i
It
'I, II
n=; II L2cs2) = IIY'm.m+J HL,(S2) =
N(3 + 2j, m - j) T2+2j
for 1 < j < m.
Exercises C.1 Suppose that (S, p) is a measure space, and let Q be the orthogonal projection from L2(S, µ) onto a finite-dimensional subspace V. (i) Show that if {¢p)n 1 is an orthonormal basis for V, and if we define N
K(x, y)
LOp(x)Op(y), p=1
then
Qu(x) =
Js
K(x, y)u(y) dµy
for x E S and u E L2(S, µ).
(ii) Show that the kernel K does not depend on the choice of the orthonormal basis for V.
(iii) Suppose that a group G acts on S, and that µ and V are invariant under G, i.e., if g E G, then
u(gx) d,i = J f u(x) d/.L s
s
for u E L1 (S, A),
Exercises
339
and
x H u (gx) belongs to V whenever u E V.
Show that the kernel K is invariant under G, i.e., if g E G then K(gx, gy) = K(x, y) for x, y E S. C.2 Show that if f is, say, continuous on Si-1, then
f Jg f
(co)
f
dw =
I I
Jg,-2
f
1 - t2 >7 + te,i) d>) (1 - t2)(n-3)/2 dt.
C.3 Prove the orthogonality property of the Legendre polynomials:
f
T.
1
1
P., (n, t)PI(n, t)(1 - t2)(n-3)l2dt =
N(n, m) Tn_1
1
smI
form >Oandl >0. [Hint: use Theorems C.1 and C.3.]
C.4 For n > 2 and m > 0, let m
P n
(t) = (1 -
t2)-(n-3)/2-d
dtm (1
-
t2)m+(n-3)/2
(i) Show that pm is a polynomial of degree m. (ii) Use integration by parts to show that if, say, f E C'n [-1, 1], then
f
f(t)Pm(t)(1 - t2)(n-3)/2dt 1
j f ('n) (t) (1 -
t2)m+(n-3)/2
dt.
(iii) Deduce the orthogonality property fJ
Pm(t)PI(t)(1 -
t2)3/2 dt = 0
if m
(iv) Find pm(1), and conclude from Exercise C.3 that
(-),n
P(n, t) _ (n+2m-3)(n+2m-5)...(nresult known as the Rodrigues formula.
1)
C5 Show that
P.(n,t)=
1
Tn-t
g,-z
(t+i
foralla)
ES"-2,
Further Properties of Spherical Harmonics
340
and then derive the Laplace representation,
P. (n, t) =
Tn-1
j (t + i
1 - t2 s)m (1 -
s2)(n-4)12 ds
for n > 3.
[Hint: use Theorem C.3 with f (s) = (t + i 1 - t2 s)m and 1/r = 1 E ? jO(Sn-2).)
C.6 Suppose n > 3, and let Am j (n, t) be the associated Legendre function (C.3).
Show that for *1, *2 E C(Sn-2), if and
IPi(W) = Amj(n, where W =
42(W) = Amj(n,
1 - t2 + ten, then
('P1, W2)L20°-i) =
1
N(n + 2j, m - j)
(*I,
2)
Tn+2j-1L2(5-z).
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[1] M. Abramowitz and I. Stegun, Handbook of Mathematical Functions, Dover, New York, 1970. [2] R. A. Adams, Sobolev Spaces, Academic Press, 1975. [3] K. E. Atkinson, The Numerical Solution of Integral Equations of the Second Kind, Cambridge University Press, 1997. [4] A. Beer, Allgemeine Methode zur Bestimmung der elektrischen and magnetischen Induction, Ann. Phys. Chem. 98 (1856), 137-142. [5] J. Bergh and J. Lbfstrbm, Interpolation Spaces - An Introduction, Springer, 1976. [6] Yu. D. Burago and V. G. Maz'ya, Potential theory and function theory for irregular regions, Seminars in Mathematics, Volume 3, V. A. Steklov Institute, Leningrad. (English translation: Consultants Bureau, New York, 1969.) [7] G. Birkhoff and U. Merzbach, A Source Book in Classical Analysis, Harvard University Press, 1973. [8] A. P. Calderbn, Lebesgue spaces of differentiable functions and distributions, Partial Differential Equations, Proc. Sympos. Pure Math. 4 (1961), 33-49. [9] A. P. Calderdn, Cauchy integrals on Lipschitz curves and related operators, Proc. Nat. Acad. Sci. U.S.A. 74 (1977), 1324-1327. [10] J. Chazarain and A. Piriou, Introduction to the Theory of Linear Partial Differential Equations, North-Holland, 1982. [11] R. R. Coifman, A. McIntosh and Y Meyer, L'integrale de Cauchy definit un operateur borne sur L2 pour les courbes lipschitziennes, Ann, of Math. 116 (1982), 361-387. [12] D. Colton and R. Kress, Integral Equation Methods in Scattering Theory, Wiley, 1983.
[13] D. Colton and R. Kress, Inverse Acoustic and Electromagnetic Scattering Theory, Springer, 1992. [14] M. Costabel, Boundary integral operators on Lipschitz domains: elementary results, SIAM J. Math. Anal. 19 (1988), 613-626. [15] M. Costabel and W. L. Wendland, Strong ellipticity of boundary integral operators, J. Reine Angew. Math. 372 (1986), 34-63. [16] M. Costabel and M. Dauge, On representation formulas and radiation conditions, Math. Methods Appl. Sci. 20 (1997), 133-150. [17] B. E. J. Dahlberg, C. E. Kenig and G. C. Verchota, Boundary value problems for the systems of elastostatics in Lipschitz domains, Duke Math. J. 57 (1988), 795-818.
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Index
Adjoint (formal) of a differential operator, 116 of an abstract linear operator, 37, 43 of conormal derivative operator, 201 of trace operator, 201 Annihilator of a subspace, 23 Arzela-Ascoli theorem, 28 Associated Legendre function, 336, 340 Beer, A., 11 Beltrami operator, 277 Bessel function, 278 spherical, 279. 293 Bessel potential, 75 Bochner integral, 321 Boundary integral equation and logarithmic capacity, 264, 275 for Dirichlet problem, 226 for exterior problem, 236 for problem with mixed boundary conditions, 231 for Neumann problem, 229,242 side condition for, 262. 274 Boundary integral operator (Schwartz) kernel of, 220, 223 adjoint of, 218 arising from a self-adjoint differential operator, 218 arising from the Laplacian, 248, 249 definition of, 218, 233 mapping properties of, 219, 245 on an open surface, 275 symbol of, 244, 275
Calderdn, A. P., 309 Calderbn projection, 243 Capacity, 263 and exterior Dirichlet problem, 271 behaviour under dilatation, 272
logarithmic, 264 and conformal mapping, 271 of a line segment, 275 of an ellipse, 272 of an open surface, 275 variational characterisation of, 272 Cauchy-Riemann equations, 2 Cauchy-Schwarz inequality, 54 Chebyshev polynomials, 255 explicit solutions in terms of, 272, 273 Codimension, 18 Coercivity abstract definition of, 44 for differential operators, 118 change of coordinates, 156 on H 1(12), 122, 126 on Ha (0), 119 for elasticity operator, 298, 299 for hypersingular integral operator, 230 Compact linear operator, 28 properties of, 54 transpose of, 29 Compact subsets of LN, 28 Compatible pair of normed spaces, 317 Completely continuous linear operator, see compact linear operator Conjugate exponent, 58 Conjugation, 37 Conormal derivative, 114 L2 estimates, 149 generalised, 117 relative to the formal adjoint, 116 Convergence in D(S2), 65 in E(S2), 65
in E(S2) but not in D(P), 109 in S(]R' ), 72 Convex set, 38
347
348
Index
Convolution, 58 and approximation, 63, 111 and differentiation, 62. 109 and Fourier transform, 73 associativity of, 108 support of, 108 Cooling-off problem, 157 Costabel. M., 102, 202 Cutoff function, 83
d'Alembert, J., 2 Density of Cm p(S2) in L p (c ). 63 of D(S2) in E(S2), 109
of D(Q) in Hs(9), 77 of D(R") in S(R"), 109 of D(S2) in H'(Q), 111 of S(R11) in E(W), 109 of D(S2) in W" (Q), 91 of Ws (0) fl E(92) in Ws(S2), 86 Dirac delta function(al), 66 convolution with, 68 homogeneity of, 187 Dirichlet form, 246 Dirichlet problem, 4 solution operator for, 145 Dirichlet's principle, 8 abstract form, 55 Dirichlet, P., 8 Distribution, 65, 109 multiplication with a smooth function, 68
of form u(a x), 189 partial derivative of, 67 temperate, 72 with compact support, 67 Divergence theorem, 97 Domain Lipschitz, see Lipschitz domain 89 of class Ck, 90 of class 90 Double-layer potential, see surface potential du Bois-Reymond, P., 12 Dual space, 20 of Lo, 58, 107 of a Sobolev space, see Sobolev spaces, duality relations realisation of, 27
Eigenfunction expansions, see spectral theory Eigenvalue, 45 Elasticity operator, 297 Energy inner product, 44 example from linear elasticity, 298 Epigraph, 186, 316 Equicontinuous set of functions in C(X), 28 in LP, 28
Equilibrium density, 263 Euler, L., 2 Extension operator, 81, 309, 313 non-existence of, 316 Seeley, 316
External conformal radius, 271 Far-field pattern, 294 Finite part, 159 Finite-part extension differentiation of, 169 homogeneity of, 162, 166, 168 in n dimensions, 166
ofx'k-1, 164 of x+, 160 Finite part integral change of variables formula, 177, 180 hypersingular integral operator. 223 on a surface, 181 Fourier transform, 70 inversion theorem, 70 of a homogeneous distribution, 172, 173, 189
of a temperate distribution, 72 of partial derivatives, 72 of f.p. u, 174 Fourier, J.-BA., 4 Fredholm alternative, 14, 37, 43 for boundary integral equations, 226, 228, 229, 240 for coercive operators, 44 for the mixed boundary value problem, 128 for the third boundary value problem, 131 relation to eigensystem, 51 Fredholm equation of the second kind, 13 abstract theory, 30, 35 Fredholm operator, 32 Fredholm, 1., 13 Fundamental solution, 191, 197 for elasticity operator, 300 for Laplacian, 2, 11, 247, 268 for the Helmholtz operator, 279 radiating, 282 integral formula for, 198 series expansion of, 255, 284 Funk-Hecke formula, 335
Gamma function, 169, 188 Gauss, C. F., 7 Generalised function, 66 Green identity first, 4, 114, 116, 1 1 8, 141
first, dual version of, 115, 118, 141 second, 4, 118 third, 5, 202
dual version of, 211 with radiation condition, 235
Index Green's function, 5 symmetry of, 16 Green, G., 4, 8
Hadamard, J., 159 Hahn-Banach theorem, 20 Hankel function, 280 spherical, 281, 293 Hardy's inequalities, l 11, 112 Harmonic analysis techniques, 209 Heat equation, 4 Helmholtz equation, 276 radiating solution of, 281 Hermitian sesquilinear form, 43, 116 Hilbert space, 38 best-approximation properties, 38, 39 dual of, 41, 42 weak sequential compactness in, 42 Hilbert, D., 14 Holder's inequality, 58 Homogeneous distribution, 158 derivatives of, 187 Homogeneous function, 158 and change of variables, 175 derivatives of, 187 orthogonality condition for, 167 parity condition for, 168, 169, 175 Hypograph, 186 of class C't, 90 Lipschitz, see Lipschitz hypograph Image, 18 Imbedding of S* (W) in D* (W), 72
of L11(2) in D`(S2), 66 Index of a Fredholm operator, 33 compact perturbation, 36 homotopy, 54 small perturbation, 54 Inner product, 38 Interpolation of normed spaces, 318 duality properties, 324, 326
J-method, 322 K-method, 319 reiteration theorem, 327 Inverse point with respect to a sphere, 259, 270 J-functional, 322 Jump relations, see surface potential K-functional, 318 determined by a positive-definite, self-adjoint operator, 333 for Sobolev spaces, 329 Kelvin transform, 259, 270 Kelvin, Lord, see Thomson, W.
349
Kernel (null space), 18 of boundary integral operator, 240 Helmholtz equation, 288 Laplacian, 267 linear elasticity, 304 Kom's inequality first, 298 second, 299, 305 Lagrange, J.-L., 2 Lame coefficients, 296 Laplace equation, 1 Laplace operator, 246 eigenvalues of, 249 rotational invariance of, 268 Laplace, P. S., 3 Lax-Milgram lemma, 43 Le Roux, J., 12 Legendre polynomials, 255 generating function for, 269 Laplace representation, 340 orthogonality property, 339 recurrence relation for, 269 Rodrigues formula for, 339 Liouville, J., 6-8 Lipschitz dissection, 99 Lipschitz domain definition, 89 non-examples of, 90 outward unit normal to, 96 surface element for, 96 Lipschitz hypograph, 89 Locally integrable functions. 64
Meyers-Serrin theorem, 85 Modulus of continuity, 60, 110 Mollifier, see convolution Multi-index, see partial derivative Ne&as, J., 123, 126, 147 Nedelec, J. C., 289 Neumann, C., 10, 12 Newtonian potential, see volume potential Nirenberg, L., 133 Nitsche, J. A., 305 Noether, F., 33
Open mapping theorem, 19 Orthogonal complement, 40 Orthogonal projection, see projection, orthogonal
Parametrix, 192 adjoint of, 197, 211 behaviour of kernel, 195 mapping property for, 193, 197
350
Index
Partial derivative, 61 weak, 74 Partition of unity, 83 and Sobolev norm, 111, 331 Peetre's inequality, 88, 110 Pivot space, 44 interpolation property of, 331 use of L2, 118 Plancherel's theorem, 73 Poincard, H., 10, 13, 145 Poisson integral formula, 5 Poisson's equation, 15
Poisson, S: D., 3-6 Positive and bounded below, 43 boundary integral operator, 262, 264, 267, 275 Potential electrostatic, 3, 5, 263 gravitational, 3 surface, see surface potential vector, 292 volume, see volume potential Principal part, 114 and coercivity, 118 Principal value, 166, 190 Projection, 20 orthogonal, 40, 54
Quotient norm, 19 Quotient space, 18 Radiation condition, 234, 243 for the Laplacian, 259 Sommerfeld, 281, 283, 294 Reflexive Banach space, 22, 37 Regulariser, 35 Regularity theory for boundary integral equations, 239 interior, 135,196 up to the boundary, 137 Relatively compact set, 27 Rellich-Payne-Weinberger identity, 146
Rellich, F., 87, 147, 286 Riesz representation theorem, 40 Riesz, F., 15 Rigid motion (infinitesimal), 302 Scalar wave equation, 276 Self-adjoint operator, see adjoin Separation of variables, 4, 277 Sequential compactness, 27 Sesquilinear form, 42 arising from a boundary integral operator,
261,275 arising from an elliptic differential operator, 114
arising from the Helmholtz operator, 276 arising from the Laplacian, 246 Single-layer potential, see surface potential Singular integral operator, 190, 312 Slobodeckii seminorm, 74, 79 and Fourier transform, 79 Smoothing operator, 192 Sobolev imbedding theorem, 86 Sobolev representation formula, 311 Sobolev spaces compact imbeddings, 87 definition via Bessel potentials, 76 definition via weak derivatives, 74 density theorems, see density duality relations, 76, 78, 92, 98 equivalent norms for, 96, 110
Hs(1R") = W(W), 80 Hs(Q) = Ws (0), 81, 92 Hs (S2) = Ho (S2), 95,112
Hs(S2)=H' 91 interpolation properties of, 329-331 invariance under change of coordinates, 85 of negative order, 74 of vector-valued (generalised) functions, 106
on the boundary of a domain, 98, 99 Sommerfeld radiation condition, see radiation condition, Sommerfeld Spectral radius, 55 Spectral theory for coercive self-adjoin[ operators, 49 for compact self-adjoint operators, 47, 55 for self-adjoint elliptic differential operators, 132 Spectrum of a linear operator, 45 Spherical harmonics, 250, 252 addition theorem for, 335 and boundary integral operators, 252 and the Helmholtz equation, 279 classical, 338 eigenfunctions of the Beltrami operator, 278 for the circle, 336 orthogonal basis for, 337 orthogonal projection onto, 334 L2(S"-1), orthogonality in 265 series expansion in, 257 Stein, E. M., 309 Steklov-Poincare operators, 145 estimates for a C"+1 , 1 domain, 146 estimates for a Lipschitz domain, 155 representations in terms of boundary integral operators, 244 Strain tensor, 296 Stress tensor, 296 Stress-strain relation, 296 Strictly positive-definite operator, 44
Index Strongly elliptic differential operator, 119 change of coordinates, 156 linear elasticity, 297 with constant coefficients, 193 Sturm-Liouville problem, 6 Sturm, C. F., 6 Successive approximations, method of, 11 Support essential, 66 of a distribution, 66 of a function, 61 Surface area of unit sphere, 247, 268 Surface potential double layer, 10, 202 traces of, 221 duality relations, 212, 213 for self-adjoint differential operator, 212, 218 jump relations for, 3, 11, 186, 203, 215 mapping properties of, 203, 205, 210 single layer, 3, 201 traces and conormal derivatives of, 218 Tangential differential operator, 147 Taylor expansion, 61 Test function, 65 Thomson, W., 6, 8 Totally bounded set, 27 Trace operator, 100 and Ho (f ), 105
351
and surface potentials, 209 for Ct- 1, 1 domains, 102 for Lipschitz domains, 102 112 from Wk (R") to W; one-sided, 141 right inverse for, 101 Traction, 296 alternative formula for, 308 as conormal derivative, 297 Transmission property, 142, 143 for surface potentials, 183, 186, 210 Transpose of a linear operator, 22 inverse of, 53 Uniformly directionally differentiable surface, 221
Uniqueness theorem for the Helmholtz equation, 288 for the Laplace equation, 260 mixed boundary conditions, 250 Neumann problem, 266 Volterra, V., 13 Volume potential, 2, 191 behaviour at the boundary, 216
Wave number, 276 Weak convergence, 42, 55 Weber, H., 9 WeierstraB, K., 10
Index of Notation
Functional Analysis A*
A/ At
dist(u, W) im A
(g, u) )A II
11A
ker A
£(X, Y) ®
(g, u)
u1v
uIW ti spec(A) Wa
av
uj - u (X0, X1)e,q X*
adjoint of A, 37 induced map on cosets modulo ker A, 18 transpose of A, 22 distance from point u to set W, 21 image (range) of linear operator A, 18 same as (g, u), 37 energy inner product for A, 44 energy norm for A, 44 kernel (null space) of linear operator A, 18 space of bounded linear operators from X to Y, 18 direct sum, 20 value of functional g E X* at u E X, 20 u is orthogonal to v, 39 u is orthogonal to the set W, 39 equivalence of norms, 17 spectrum of A, 45 subspace of X* that annihilates W c X, 23 subspace of X that annihilates V C X*, 23 uj converges weakly to u, 42 interpolation space, 318 dual space of X, 20
Theory of Distributions Ccomp°O
(Q)
Comp(c2)
space of C°O functions with compact support in 0, 61 space of Cr functions with compact support in 0, 61 353
Index of Notation
354
D(S2) D(S2)
space of infinitely differentiable functions on 0, 61 space of functions in CO°(S2) having support in K, 61 space of r times continuously differentiable functions on 0, 61 space of functions in C` (S2) having support in K, 61 C mp(Q) with sequential convergence defined, 65 space of restrictions to 7 of functions in D(IR"), 77
S
same as So, 66
CO0 (Q)
CK (S2)
C' (n) CK (S2)
S(R")
Dirac delta function(al) at x, 66 CK (S2) with sequential convergence defined, 65 space of Schwartz distributions on S2, 65 space of distributions with compact support in 92, 66 C°°(0) with sequential convergence defined, 65 finite-part extension of u, 166 principal value of u, 166 finite-part integral of xa i (x) over the half line x > 0, 160 abbreviation for (u, v)g when S2 = R", 68 same as (u, v)s2, 68, 107 space of locally integrable functions on 0, 64 dilatation operator, 158 tensor product of functions or distributions, 104 abbreviation for (u, v)n when 0 =1R", 66 integral (generalised, if necessary) of u v over S2, 58, 66, 106 Fourier transform of f.p. xt, 169 pa+'t-(x) over p > 0, 166 finite-part integral of Schwartz class of rapidly decreasing Coo functions on 1R", 72
x+
xa if x > 0, but 0 if x < 0, 159
f.p. X"
f.p. x-k-1
finite-part extension of x+, 160 (x l" if x < 0, but 0 if x > 0, 163 finite-part extension of x° , 163 finite-part extension of x-k_1 for an integer k > 0, 164
(x ± i0),
164
SX
DK (S2)
-D. (0)
E. (0) E(S2) f.p. U P.V. U H a ((a)
(u, v)
(u, On L1.1°c(0)
A ® (u, v)
(u, v)c
n () Ra0
X0
f.p. X'.
Sobolev Spaces HF HS (R") HS (I-) HS (S2)
HS (S2)'"
space of distributions in HI (RI) with support in F, 76 Sobolev space on 1R" (definition via Bessel potential), 76 Sobolev space on 17, 98 space of restrictions to S2 of distributions in HS (R" ), 77 space of HS functions on Q with values in Cm, 107
Sobolev Spaces
Ho (0) H` (Q)
closure of D(S2) in HS (S2), 77
HS (52)"'
space of HS functions on S2 with values in Cm, 107 Bessel potential of order s, 75 Slobodeckil seminorm, 74
is
HA.P,0
355
closure of D(2) in HS (I8"), 77
I
Wp(S2) Wp(S2)m
abbreviation for W2 (S2), 75 Sobolev space of orders > 0 based on LP(0), 73, 74 space of WP functions on S2 with values in C"', 107
Differential and Integral Operators AS"-1
B; B,,
BV 13*
B; B,,
DL DL Eik(u) G (x, y) G
G(x - y) Gj(x, x - y)
[u]r
P PO
(P = On
4)t R
M S
Elk
Beltrami operator on the unit sphere, 277 ft h component of generalised flux or traction, 114 conormal derivative, 114 conormal derivative from S21, 141 adjoint of B,,, 201 dual version of B;, 115 dual version of 8,,, 115 double-layer potential, 10, 202 dual version of DL, 211 strain tensor, 296 fundamental solution or parametrix, 2, 191 volume potential, 191 G(x, y) in translation-invariant case, 193 jth term in homogeneous expansion of G(x, y), 195 jump in across r, 142
jump in BA across P, 142 jump in u across I', 142 Laplace operator, 1 second-order differential operator, usually strongly elliptic, 113 principal part of P, 114 sesquilinear form arising from P for the domain £2, 114 abbreviation for (Dnt, 141 boundary integral operator, conormal derivative of DL, 218 operator arising in radiation conditions, 234 boundary integral operator, trace of SL, 7, 218 stress tensor, 296
Index of Notation
356 S§L_ L
T T
U V
single-layer potential, 3, 202 dual version of SL, 211 boundary integral operator, sum of one-sided traces of DL, 11, 218 dual version of T, 218 solution operator for the Dirichlet problem, 145 solution operator for the adjoint Dirichlet problem, 145
Other Symbols a!
IaI
8"u ya C+
CCapr u*v A1,h
dQ
u = ,'Fu u = .P*u
t y
factorial of the multi-index a, 61 order of the partial derivative determined by a, 61 partial derivative of u determined by the multi-index a, 61 monomial determined by the multi-index a, 61 complex upper half plane Im z > 0, 183 complex lower half plane Im z < 0, 183 capacity of IF, 263 convolution of u and v, 58 difference quotient in lth variable with step size h, 62 element of surface area on 1, 1, 97 Fourier transform of u, 70 inverse Fourier transform of u, 70 (common) boundary of S2 = St- and of 52+, 1, 89, 141 trace operator for S2, 100, 102
Yadjoint of y, 201 Y}
rD FN hml), h(2)
Ju j," LP (S2)
M(n, m) N(n, m)
trace operator for Sgt, 1, 141 portion of r with Dirichlet boundary condition, 128 portion of r with Neumann boundary condition, 128 spherical Hankel functions, 281 Bessel function of the first kind, 278 spherical Bessel function of the first kind, 279 Lebesgue space of pth-power-integrable functions on 0, 58 dimension of P,, (R"), 250
v
dimension of R. (R"), 250 Euclidean norm in R" or unitary norm in C'", 1 outward unit normal to 9 = S2-, 1, 97, 141
0
domain in 1R", 1
Sgt p*
interior (-) and exterior (+) domains, 1, 141 conjugate exponent to p, 58
1X I
Other Symbols
generalised Legendre polynomial of degree m for the dimension n, 255 space of homogeneous polynomials of degree m, 250
P. (n, t)
P. (R")
upper half space x > 0, 183 lower half space x < 0, 183
]R+
]R"
xm (W)
f. (S'
I)
u*
uv utkI(x; y)
T. T}(x) xa Yu ym
F(a) HM H(2)
f
space of solid spherical harmonics of degree m, 250 space of surface spherical harmonics of degree m, 252 transpose of the complex conjugate of the vector u, 107 dot product of vectors u and v, 107 kth Fr6chet derivative of u, 61 Kelvin transform of u, 259 surface area of S"- 1, 247 221
inverse point of x with respect to a sphere, 258 Bessel function of the second kind, 278 spherical Bessel function of the second kind, 279 gamma function, 169 Hankel functions, 280 special contour integral, 183
357
