The preceding theorems will be used to estimate remainders, but here we show how to calculate explicitly compositions of weakly-singular kernels of the form
|x−y|α−m(log|x−y|)jQ
x,y, y−x
|y−x|
.
These compositions are (approximately) convolutions, so the Fourier transform is a convenient tool. We define the transform f → fˆby
fˆ(ξ)=
Rme−i xãξ f(x)d x, f(x)=(2π)−m
Rmei xãξfˆ(ξ)dξ for “nice” f, extended to tempered distributionsT ∈S(Rm) by: ˆT ∈S(Rm) satisfies
Tˆ, φ = T,φˆ for allφ∈S(Rm).
7.3 Fourier Transform and Composition of Weakly-Singular Kernels 127 [S(Rm) is the Schwartz space ofC∞functionsφonRmsuch thatxj∂xkφ(x) is bounded onRmfor all j,kandS(Rm) is the dual space.]
LetHk(Rm) denote the space of harmonic polynomials onRm(p(x)≡0) which are homogeneous of degree k, the “solid spherical harmonics.” The restrictions to Sm−1 of the harmonic polynomials give all polynomials on Sm−1; given any homogeneous k-order polynomial p(x) on Rm, we have p(x)=
0≤j≤k/2|x|2jpj(x) for somepj ∈Hk−2j(Rm). (See Stein [38] for this and other properties of harmonic polynomials.) If p∈Hj(Rm),q ∈Hk(Rm) and j =k, then
Sm−1p(x)q(x)d Ax =0.
Forα >0 andp∈Hk(Rm),|x|α−mp(x/|x|) or more precisely φ→
Rm|x|α−mp(x/|x|)φ(x)d x, φ∈S(Rm), (7.3.1)
is a tempered distribution which is an analytic function ofαand has analytic continuation for all complexαwithα∈ {−k,−k−2,−k−4,ã ã ã }. In fact, for any integerN ≥0
Rm|x|α−mp(x/|x|)φ(x)
=
|x|>1|x|α−mp(x/|x|)φ(x)+
|x|<1|x|α−mp(x/|x|)
×
φ(x)−
|j|≤N
1
j!xj∂jφ(0)
+
|j|≤N
1 j!
∂jφ(0)
|j| +α
Sm−1p(ω)ωjd Aω (7.3.2)
and
Sm−1p(ω)ωj =0 unless|j| ∈ {k,k+2,k+4, . . .}. This formula gives
|x|α−mp(x/|x|),φ(x)for anyα∈C\{−k,−k,−2, . . .}, provided we choose N large enough. Ifα= −k−2i for some integeri ≥0 the terms with|j| = k+2iin the sum are not defined; we merely throw away these terms, and the rest is (by definition) thefinite part,f.p.(|x|−m−k−2ip(x/|x|), φ(x). We see that, whenα= −k−2i+
|x|α−mp(x/|x|), φ(x) − 1
|j|=k+2i
1 j!∂jφ(0)
|ω|=1p(ω)ωj (7.3.3)
converges tof.p.(|x|−m−k−2ip(x/|x|)), φ(x)as →0.
Now we find the Fourier transform (following Stein [38]).
128 Chapter 7. Boundary Operators for Second-Order Elliptic Equation Theorem 7.3.1. {|x|α−mp(x/|x|)}∧(ξ)=γα,κ|ξ|−αp(ξ/|ξ|), where
γα,k=i−kπm/22α k+α
2
m+k−α 2
p∈Hk(Rm)andαis complex with
α∈ {−k,−k−2,−k−4, . . .;m+k,m+k+2, . . .}
Remark. If 0< Reα <m, both distributions make “classical” sense; and if 0<Reα <m/2, then|x|α−mp(x/|x|) is the sum of a function in L1(Rm) – supported in (ξ)≤1 – and a function in L2(Rm), so the Fourier transform also makes sense (in L∞+L2). But in general, we must interpret these as distributions, obtained by analytic continuation outside the “classical strip,”
0< Reα <m.
The left-side of the equation is also analytic forα=m+k,m+k+2, . . .; we study the behavior of the transform near these points, after this proof.
Proof. By the mean-value property of harmonic functions, for anyx ∈Rm, the average value of p(x+y) on any sphere|y| =const.is p(x), therefore
Rme−|y|2p(x+h)d y=p(x)
e−y2d y=πm/2p(x). By analytic continuation, this holds for allx∈Cn. Thus for anyλ >0, ξ∈Rm,
Rm exp (−λ|x|2−iξãx)p(x)d x
=
Rmd zexp(−z2−ξ2/4λ)λ−(m+k)/2p(z−iξ/2√ λ)
= πm/2λ−m/2−ke−ξ2/4λp(−iξ/2).
Letφ∈Cc∞(Rm\{0}); then for 0< δ <(m+k)/2,
Rm(δ)|x|−2δp(x) ˆφ(x)d x
= ∞
0
λδ−1dλ
Rme−λx2p(x) ˆφ(x)d x
= πm/2
Rm
∞
0
λδ−m2−k−1e−ξ2/4λdλ
p(−iξ/2)φ(ξ)dξ, and we have used the fact that|ξ| ≥const. >0 on suppφ. If alsoδ >k/2, the last integral converges and the equation holds for allφ∈S(Rm), by continuity.
Now letδ= 12(m+k−α);k/2< δ <(m+k)/2 when 0< α <m, and this gives the result claimed for 0< α <m.Analytic continuation gives the general case.
7.3 Fourier Transform and Composition of Weakly-Singular Kernels 129 Now we examine the transform for α near {m+k,m+k+2, . . .}. If α=m+k+2j(j =integer≥0), |x|α−mp(x/|x|)= |x|2jp(x) is a polyno- mial and the transform is well-known:
|x|2jp(x)}∧(ξ), φ(ξ) =(2π)m−(−ξ)jp(−i∂ξ)φ(ξ)|ξ=0
so the transform is a derivative of Dirac’sδ, with support{0}. We compute the derivative with respect toα:
{|x|2jp(x) log|x|}∧(ξ), φ(ξ)
= lim
α→m+k+2j
γα,klog|ξ|−1+γα,k
|ξ|−αp(ξ/|ξ|), φ(ξ) . Using formula (7.3.2) – or its derivative with respect toα– we see that this is
γf.p.|ξ|−m−k−2jp(ξ/|ξ|), φ(ξ) + lim
α→m+k+2j
|S|=k+2j
1 S!∂Sφ(0)
|ω|=1
P(ω)ω2
γα,k
mk+2j−α+ γα,k
(m+k+2j−α)2
where γα,k= ∂α∂ γα,k, and γ is γα,k for α=m+k+2j. If α=m+k+ 2j+δ, the final coefficient above is
γ+δγ+O(δ2)
−δ +δγ+12δ2γ+O(δ3)
δ2 → −1
2γ. Thus there is a polynomialq(x), homogeneous of degreek+2j, such that
{|x|2jp(x) log|x| +q(x)}∧(ξ)=γf.p.(|ξ|−m−k−2jp(ξ/|ξ|)) (7.3.4)
and
γ= ∂
∂αγα,k|α=m+k+2j = j!(−1)j+1i−k2α−1πm/2 k+α
2
=0.
(7.3.5)
In fact,q(x)=(γ/2γ)|x|2jp(x) – but we do not need to knowqexplicitly.
Differentiating jtimes with respect toαgives
|x|α−m(log|x|)jp x
|x| ∧
(ξ)=
j
S=0
j s
γα,(s)k(log|ξ|−1)j−s|ξ|−αp(ξ/|ξ|) whereγα,(s)k=(∂α∂ )sγα,k, so we may handle integral powers of log|x|with equal ease (or difficulty, depending onα).
We may also treat more general “direction functions”φ(x/|x|) by expan- sion in harmonic polynomials. Choose an orthonormal basis for L2(Sm−1),
130 Chapter 7. Boundary Operators for Second-Order Elliptic Equation {pj|Sm−1}∞j=0, with each pj a homogeneous harmonic polynomial (p0= constant) and j →dj =degpj non-decreasing. (We assume m>1;m=1 is trivial.) Thendj ≈((m−21)!j)1/m−1as j → ∞,Spj = −dj(dj+m−2)pj
onSm−1, and the Sobolev spaces onSm−1may be written Hσ(Sm−1)=D(1−s)σ/2=
Cjpj ∞
0
∞ 0
|Cj|2(1+dj(dj+m−2))σ<∞
for allσ ≥0. Ifαis not an integer andφ=∞
0 φjpj ∈ Hσ(Sm−1) then
|x|α−mφ x
|x|
∧
(ξ)= |ξ|−αTαφ ξ
|ξ| (7.3.6)
where Tαφ(ω)=∞
0 φjγα,djpj(ω), ω∈ Sm−1. If α is an integer, there may be finitely many j for whichγα,dj is zero or∞(the troublesome terms have dj≤ max(α−m,−α)); such terms are treated separately, and we define Tα only in the orthogonal complement. Since|γα,d| ≈(2π)mdα−m/2 asd → ∞, it follows thatTα:Hσ(Sm−1)→ Hσ−α+m/2(Sm−1) [modulo, perhaps, a finite- dimensional subspace] is an isomorphism. For example, ifα, βandα+βare not integers (to simplify) andσ >−1/2,φ∈Hσ+α(Sm−1),ψ∈ Hσ+β(Sm−1), then
|x|α−mφ x
|x|
∗
|x|β−mψ x
|x|
= |x|α+β−mθ x
|x|
where θ=Tα+β−1 (TαφãTβψ)∈Hσ+α+β(Sm−1). Here we use the fact that Hσ(Sm−1) is an algebra under pointwise multiplication whenσ >(m−1)/2.
Fortunately our explicit calculations involve only polynomials, hence finite sums. To facilitate these calculations, we include a table of values ofγα,kand introducen=m+1 (which occurs naturally in our calculations).
n=m+1, ω=ωn, ω¯ =ωn−1=ωm=2πm/2/ (m/2).
γα,k k=0 k=1 k=2 k=3 k=4
α=1 1/2(n−2)ω −iω¯ −1/2ω 2iω¯(n−1) 3ω/2n α=2 (n−3) ¯ω −1/2(n−2)iω −2 ¯ω 3/2iω 8 ¯ω/(n−1) α=31/2(n−2)(n−4)ω −2(n−3)iω¯ −3/2(n−2)ω 8iω¯ 15/2iω α=4 2(n−3)(n−5) ¯ω −3/2(n−2)(n−4)iω −8(n−3) ¯ω 15/2(n−2)iω 48 ¯ω
Whenγα,k=0, this table cannot be used to find the inverse transform; but in that caseγα,k= ∂α∂ γα,k=0 and we can use (7.3.5). We complement this table
7.3 Fourier Transform and Composition of Weakly-Singular Kernels 131 by giving the derivativeγα,kfor the points whereγα,k=0.
(mn=2=1) γα,k k=0 1 2 3 α=1 −π
α=2 r πi
α=3 12π r 3π
α=4 r −36πi r −15πi
n=3 k=0 1 2
α=1 2 −2π
3 r 4πi
4 8π r 16π
n=4 k=0 1
α=1 2 3−2π2 4 r 6π2i
n=5 k=0
α=1 2 3 4−16π2
Another simpler method seems to always work, which we illustrate with an example. If n =3,(F.T.)−1(|ξ|−2)= (n−3)1ω
n−1|x|2−m= −ωn−11 |x3|−3−nn ; and the result for n=3 is obtained by replacing |x|3−n/(3−n) by log|x|, i.e., (F.T.)−1(|ξ|−2)= −21πlog|x|. More precisely, (F.T.)−1(f.p.|ξ|−2) is
−2π1 log|x|plus a constant, whenn =3 (m=2).
The Fourier transform shows the action of an integral operator on rapidly oscillating functions. Suppose, for example,is a boundedCr+1region inRn, K(x,y) is aC∗r(α) kernel in∂, 0< α≤r,φisCron∂andψisCr+1and real-valued with∇∂ψ(y)=0 on suppφ, and consider
J(t,x)≡
∂K(x,y)φ(y) exp(itψ(y))d Ay, x∈∂ (7.3.7)
ast→ +∞.Ifxis outside suppφ, it is easy to seeJ(t,x)=O(t−r), and for everyx(and >0) we show J(t,x)=O(t−α+ ) ast → +∞.It is enough to treat the case when suppφis small, and then we may write∂(in suppφ) as {h(ξ)|ξ ∈Vopen⊂Rn−1},h:V →∂⊂Rn is aCr+1embedding, and
J(t,h(ξ))=
V
K(h(ξ),h(η))φ(h(η)) exp(i tψ(h(η)))
dethηhηdη.
We may choose new coordinates ζ in Rn−1 so that ζ1=ψ(h(η)) and the problem reduces to: if ˜K is a C∗r(α) kernel with compact support in Rm (m=n−1)
Rm
K˜(x,y)ei t y1d y=O(t−α+ ) ast → +∞.
132 Chapter 7. Boundary Operators for Second-Order Elliptic Equation If r ≥α >1 then
Rm K˜(x,y)ei t y1d y= −i t1
Rm∂y1K˜(x,y)ei t y1d y and we eventually reduce to the case 0< α≤1≤r. But in this case, if e1= (1,0, . . . ,0),
Rm
K˜(x,y)ei t y1d y= −
Rm
K˜(x,y+πe1/t)ei t y1d y
=1/2
Rm( ˜K(x,y)−K˜(x,y+πe1/t))ei t y1d y
so
Rm
K˜(x,y)ei t y1d y
≤Const.
|y−x|<5/t
(|y−x|σ+ |y+πe1/t−x|σ) +Const.
R≥|y−x|≥5/t
t−1(|y−x|σ−1+ |y+πe1/t−x|σ−1)
≤ O
t−(σ+m))=O(t−α+
, whereσ =α−m− .
Suppose K(x,y)−K0(x,y−x) is a C∗S(β) kernel on ∂with r≥s≥ β > α,K0(x,z) depends only on the component of z in the tangent space Tx(∂), and the Fourier transform ˆK0(x,ã) of K0(x,ã)|Tx(∂) is computable.
Then
∂K(x,y)φ(y)ei tψ(y)d Ay=Kˆ0(x,−t∇∂ψ(x))φ(x)ei tψ(x)+o(t−α) as t → +∞. Ifz→ K0(x,z) is homogeneous of degreeα−m, and if ˜K is of finite rank, it follows thatK0(x,z)=0 for allz∈Tx(∂).
Now consider a bounded open set⊂RnwithCr+1boundary∂(r ≥2) and aC∗r(α) kernelK(x,y) on∂of the form
K(x,y)= |x−y|α−m(log(x−y))jQ
x,y, y−x
|y−x|
wherem=n−1=dim∂, α >0 and Q(x,y, ω) is aCr function. For y∈
∂nearx,y=x+η−νx(η)Nx, η∈Tx(∂); i.e., we may write∂as a graph over the tangent plane atx, η=Px(y−x) is the orthogonal projection ofy− x ontoTx(∂), andνx(η)=1/2Kx(η2)+O(|η|3) asη→0, where Kx is the curvature atx. Then|x−y| = |η|(1+νx(η)2/|η|2)1/2= |η|(1+O(|η|2)) and
K(x,x+η−νx(η)Nx)= |η|α−m(log|η|)jQ(x,x, η/|η|) + |η|α+1−m(log|η|)j(ωQy(x,y, ω)
−1/2Kxω2NxQω(x,y, ω))|y=x,ω=η/|η|+ ã ã ã
=
N−1
j=0
K˜j(x, η)+K˜N(x, η)
7.3 Fourier Transform and Composition of Weakly-Singular Kernels 133 or K(x,y)=N
0 Kj(x,y), Kj(x,y)=K˜j(x,Px(y−x)), whereKjisC∗r−j (α+j) (0≤ j ≤N) and the Fourier transforms ofη→K˜j(x, η) (0≤ j <N) are explicitly computable whenω→ Q(x,y, ω) is a polynomial.
SupposeL is another suchCr∗(β) kernel on∂, and we compute the com- positionK ◦L,
K◦L(x,z)=
∂K(x,y)L(y,z)d Ay.
Let θ:Rn →R be C∞ with small support, but θ≡1 near 0 : a “cutoff ” function. If K,L areC∗r(α),Cr∗(β) respectively,Kθ(x,y)=K(x,y)θ(y−x) and Lθ(y,z)=L(y,z)θ(z−y), then K −Kθ,L−Lθ are Cr kernels and K ◦L−Kθ◦Lθis also aCrkernel. Without loss of generality, we shall assume K andL are supported in a small neighborhood of the diagonal of∂×∂.
Definehx :Tx(∂)→∂:η→x+η−νx(η)Nx(ηsmall) and then forz close tox
K◦L(x,z)=
Tx(∂)K(x,hx(η))
1+ |νx(η)|2ãL(hz( ˜η),z)|hz( ˜η)=hx(η)dη.
We can solve hz( ˜η)=hx(η) for ˜η∈Tz(∂) as a function of x,z and η∈ Tx(∂). In fact, let z=hx(ζ), ζ ∈Tx(∂) near 0, and define Jx,z =I− Nx⊗νx(ζ)∈Rn×n, where the gradient νx(ζ) is considered as a vector in Tx(∂)⊥Nx.Jx,zis an isomorphism ofRnand it carriesTx(∂) ontoTz(∂):
given a curveζ =ζ(t) inTx(∂) we havehx(ζ(t))∈∂and d
dthx(ζ(t))= Jx,zζ˙(t)∈Tz(∂)
if hx(ζ(t))=z. Now hz( ˜η)=hx(η) with η˜=Jx,zσ,z=hx(ζ),(σ, ζ ∈ Tx(∂), becomes
η−ζ −(νx(η)−νx(ζ))Nx =Jx,zσ−νz(Jx,zσ)Nz. Since PxJx zσ =σ forσ ∈Tx(∂), we have
η−ζ =σ−νz(Jx zσ)PxNz.
Butτ =σ−νz(Jx zσ)PxNz(σ, τ ∈Tx(∂)) is solvable forσ =Fx,z(τ)=τ + O(|τ|2|ζ|) sohz( ˜η)=hx(η) is solvable for ˜η=Jx zFx z(η−ζ) and the integral above becomes
K◦L(x,z)=
Tx(∂)
K(x,hx(η))
1+ |νx(η)|2L(hz(Jx zFx z(η−ζ)),z)dη,
134 Chapter 7. Boundary Operators for Second-Order Elliptic Equation which is almost a convolution. Let
K(x,hx(η))
1+ |νx(η)|2 =
N
0
K˜j(x, η), (7.3.8)
K˜j(x, η)=O(|η|α+j−m−δ) asη→0 inTx(∂), for anyδ >0, and
L hz
Jx zFx zJx z−1τ ,z
=
N
j=0
L˜j(x,z, τ), (7.3.9)
L˜j(x,z, τ)=O(|τ|β+j−m−δ)
as τ →0 in Tz(∂), for anyδ >0. We suppose η→K˜j(x, η) (j <N) on Tx(∂) andτ →L˜k(x,z, τ) (k<N) on Tz(∂) have explicitly-computable Fourier transforms. Then we can compute the sum of convolutions
j+k<N
Tx(∂)
K˜j(x, η) ˜Lk(x,z,Jx z(η−ζ))dη (7.3.10)
(where x,z are treated as parameters, and only at the end do we recall that ζ =Px(z−x)). The difference between this sum andK◦Lis a kernel of class Cr∗−N(α+N).
Ifφ∈S(Tx(∂))we have
Tx(∂)
K˜j(x, η) ˜Lk(x,z,Jx z(η−ζ))dη,( ˆφζ)
=
Tx(∂)(F.T.K˜j(x,ã)) (ξ)ãF.T.( ˜Lk(x,z,−Jx zã)) (ξ)φ(ξ)dξ.
Further
F.T.( ˜Lk(x,z,−Jx zã)) (ξ)=
Tx(∂)e−iξãτL˜k(x,z,−Jx zτ)dτ
=
Tz(∂)
eiξãσL˜k(x,z, σ)dσ
1+ |νx(ζ)|2, (7.3.11)
using the fact that ξãJx zτ =ξãτ for τ, ξ ∈Tx(∂). Thus we may com- pute the transform of ˜Lk(x,z,ã)|Tz(∂), using the earlier formulas, substitute –
Pzξ for ξ and divide by
1+ |νx(ζ)|2 to get the transform of L˜k(x,z,−Jx z)|Tx(∂). If L(y,z)= |y−z|β−m(log|y−z|)jR(y,z,|zz−−yy|) and R(y,z, ω) is a polynomial in ω, the transform of ˜Lk(x,z,−Jx z)|Tx(∂) will be a sum of terms|Pzξ|−β−k(log|Pzξ|)iSi k(x,z,Pzξ/|Pzξ|) withSi k(x,z, ω) a polynomial inω. Writingωforξ/|ξ|,|Pzξ| = |ξ|(1+1+|ν(ωãνx(ζ))2
x(ζ)|2)1/2and so