NOTES on ELLIPTIC BOUNDARY VALUE PROBLEMS for the LAPLACE OPERATOR Date

NOTES ON ELLIPTIC BOUNDARY VALUE PROBLEMS FOR THE LAPLACE OPERATOR

CHARLES EPSTEIN

Date: Feb 5, 1998 ; Run: February 5, 1998

In these notes we present the pseudodifferential approach to elliptic boundary value problems for the Laplace operator acting on functions on a smoothly bounded compact domain in a compact manifold. This is an elaboration of the classical method of multiple layer potentials. After a short discussion of this method we consider the theory of homogeneous distributions on Rn. This is useful in our subsequent discussion of boundary value problems and provides an interesting concrete complement to the rather abstract general theory developed earlier in the course. We then turn to boundary value problems. We analyze the smoothness and boundary regularity of multiple layer potentials for the Laplace equation. This allows the reduction of a boundary value problem to the solvability of a system of pseudodifferential equations on the boundary itself. After considering several different boundary value problems for smooth data we establish the Sobolev regularity properties of the single and double layer potentials. The estimates allow us to extend the existence results to data with finite differentiability and also establish the standard “elliptic estimates” for the solutions of elliptic boundary value problems for the Laplacian. This treatment is culled from material in L. Hörmander, The analysis of Linear Partial Differential Operators, III,M.Taylor,Partial Differential Equations, II and Introduction to the theory of Linear Partial Differential Equations by J. Chazarain and A. Piriou. I would finally like to thank Dara Cosgrove for the remarkable typing job.

0. Preface

A fundamental class of problems in partial diﬀerential equations is elliptic boundary value problems. The classical problems of this type are the Dirichlet and Neumann problems for the Laplace equation on a smoothly bounded domain, Ω ⊆ Rn :

Dirichlet Problem ( ∆u =0 inΩ Find a function u ∈ C2(Ω) ∩ C0(Ω) such that (D) u| bΩ = f

Neumann Problem ( ∆u =0 inΩ ∈ 2 ∩ 1 Find a function u C (Ω) C (Ω) such that (N) ∂u ∂ν = g

∂u Here ∂ν is the outward normal derivative along bΩ . We would like conditions under which (D) and (N) are solvable and relate the regularity of the solution u to the data f or g respectively. A starting point for the analysis of this problem is Green’s formula: We let G(x, y) denote the fundamental solutions of the Laplace equation in Rn, ( c2 log |x − y| n =2 G(x, y)= 2−n cn|x − y| n>2

1 2 CHARLES EPSTEIN

A simple integration by parts shows that Z Z  (G(x, y) ∂u(y) − ∂G (x, y)u(y)) dσ(y)+u(x) x ∈ Ω  ∂ν ∂νy bΩ (0.1) G(x, y)∆udy= Z  c Ω  (G(x, y) ∂u(y) − ∂G (x, y)u(y)) dσ(y) x ∈ Ω ∂ν ∂νy bΩ From this formula several things are apparent

1) We can always reduce the inhomogeneous problem, ∆u = f to a homogeneous problem by letting u = v + w where

∆v =0 Z w = G(x, y)f(y)dy Ω 2) Suppose that we can show that for f, g ∈ C∞(bΩ) the functions which are deﬁnd, a priori in bΩc by Z Sf(x)= G(x, y)f(y)dσ(y) ZbΩ Dg(x)= ∂G (x, y)g(y)dσ(y) ∂νy bΩ ∞ ∞ C have extensions as elements of C (Ω) and C (Ω ), respectively. Then Sf | bΩ , Dg | bΩ | ∂u | are given by linear operators, S f, D g.Letu0 and u1 denote u bΩ and ∂ν bΩ.Ifu is a harmonic function in Ω then Green’s formula implies that + + u0 = −S u1 + D u0 So + + (0.2) S u1 =(D − I)u0 If we could show that S+ were an invertible operator then (0.2) would imply that

+ −1 + (0.3) u1 =(S ) (D − I)u0 Using (0.3) we can solve the Dirichlet problem. − We let g =(S+) 1(D+ − I)f, and u = Df −Sg + + + −1 + So That u0 = D f − S (S ) (D − I)f = f as desired. If Sf and Dg have smooth extensions it would also follow that S ∂ν f = S1 f

D ∂ν g = D1 g so that + − + u1 = D1 u0 S1 u1 + so if D1 is invertible then + −1 + u0 =(D1 ) (I + S1 )u1 and we could therefore solve the Neumann problem. ELLIPTIC BOUNDARY VALUE PROBLEMS 3

In the next few sections we will show that Sf and Df have the smooth extension property posited above. This is established by using the Fourier representation of the Green’s function and techniques from the theory of pseudodifferential operators. In fact we show that, for each ∈ N ∈ ∗ k 0 there are ψDOs Dk , Sk Ψ (bΩ) such that kS (∂ν ) f = Sk f kD (∂ν ) f = Dk f. With this information we can give sufficient conditions for a more general boundary value problem: ( ∆u =0 (0.4) ∗ b0u0 + b1u1 = gbi ∈ Ψ (bΩ) to be Fredholm problems. This will be the case if the pseudodifferential system (D+ − I)u − S+u 0 0 1 = b0u0 + b1u1 g is elliptic. In this case (0.4) is solvable for g satisfying finitely many linear conditions. We also establish that the solution u depends on the boundary data in a specific way: for example in the Dirichlet problem k k ≤ k k 1 U Hs(Ω) Cs f s− 1 s> H 2 (bΩ) 2 and for the Neumann problem k k ≤ k k 1 U Hs(Ω) Cs g s− 3 s> . H 2 (bΩ) 2

In fact as it requires no additional eﬀort we establish these results for the ∆g deﬁned by a metric g on a smooth, compact manifold with boundary.

1. Introduction

Let M be a smooth compact manifold with nonempty boundary. Choose a smooth Rie- ∞ mannian metric g on M. This deﬁnes a second order elliptic operator, ∆g acting on C (M). If the metric is given in local coordinates, (x ,...,x )by X1 n 2 ds = gijdxidxj then 1 X √ ∆ f(x)=√ ∂x gij g∂x f(x) g g i j ij where g is the matrix inverse to gij and g =detgij. The symbol of this operator is easily seen to be | |2 σ2(∆g)(x, ξ)= ξ g and therefore the operator is elliptic. In this chapter we consider boundary value problems of the form: ( ∆ u =0 (1.1) g | ∂u | b0u bM + b1 ∂ν bM = g. ∈ ∗ ∂u Here bi Ψ (bM) are pseudodiﬀerential operators and ∂ν is the outer normal derivative of u along bM. Without loss of generality we can assume that M is a domain in a compact manifold, Mc, and that the metric g has a smooth extension to a metric on gb. c In a previous section we analyzed ∆gb on M and showed that there is an operator G0 ∈ Ψ−2(Mc) such that

∆gbG0 = G0∆gb = I − π0 4 CHARLES EPSTEIN Z where π0(f)= fdVgb. Mc Choose a point p ∈ Mc\M and deﬁne: the function G on M × M by

G(x, y)=G0(x, y) − [G0(p, y)+G0(x, p)]. A simple calculation shows that 1 1 ∆yG =∆yG(x, y)=δ − − [δ − ]=δ − δ g x g x V p V x p c ∈ ∞ where V =Vol(M). Thus if we select a function u Cc (M)then

hGx, ∆gui = u(x) .

∈ ∞ \{ } ∈ ∞ For each x M, Gx(y) is an integrable function in C (M x ). Choosing ψ Cc (Ux) such that ψ(x)=1andUx is a neighborhood of x with Ux ⊂⊂ M we see that hG ,D ui = hG , ∆ φui + hG , ∆ (1 − φ)ui x g x gZ x g

= u(x)+ Gx(y)∆g(1 − φ)udy. M We can integrate by parts in the second term to obtain Z ∂u(y) − = u(x)+ (Gx(y) ∂ν ∂νy Gx(y)u(y)) dσ(y) . bM If ∆ f = 0 then we obtain Green’s formula: g Z − ∂u (1.2) u(x)= (∂νy Gx(y)u(y) Gx(y) ∂ν ) dσ(y) . bM For (x, y) in a neighborhood of M × M, G(x, y) is the Schwarz kernel of a pseudodifferential operator of order -2 which differs from G0(x, y)byasmoothkernel,thusσ(G)=σ(G0). We use formula (1.2) to reduce the solution of the boundary value problem, (1.1) to the solution of a pseudodifferential equation on bM. At the same time we obtain conditions on b0, b1 that imply that the solution to (1.1) satisfies standard “elliptic estimates”: If s ∈ R 1 greater than 2 then k k ≤ k | k k ∂uk u Hs(M) c( u bM s− 1 + s− 3 ) . H 2 (bM) ∂ν H 2 (bM) For f ∈ C∞(bM) we define two operators: Z Sf(x)= G(x, y)f(y)dσ(y) x ∈ M bM this is called a single layer potential; Z D f(x)= ∂νy G(x, y)f(y)dσ(y) bM this is called a double layer potential. A harmonic function satisfies

u(x)=Du0(x) −Su1(x) | ∂u | where u0 = u bM and u1 = ∂ν bM . Suppose we show that Su(x) = lim Su(x) x→bM Du(x) = lim Du(x) x→bM exist. Then for a harmonic function

u0 = Du0 − Su1 . ELLIPTIC BOUNDARY VALUE PROBLEMS 5

If we could show, for example, that

(I − D)u0 = −Su1 can be solved, either for u0 or u1 then we could substitute into the boundary condition to obtain:

−1 (1.3) (b1 − b0(I − D) S)u1 = g.

−1 Thus a reasonable condition to impose on b1, b0 is that b1 − b0(I − D) S is an elliptic operator. Then a solution to (1.3) exists for g satisfying ﬁnitely many linear conditions. We will in fact show that for f ∈ C∞(bM) Sf, Df have extensions as elements of C∞(M) and show that kS | k k ∈ k−1 ∂ν f bΩ = A f, A Ψ (bM) kD | k k ∈ k ∂ν f bΩ = B f, B Ψ (bM) .

Computing the principal symbols of these operators will allow us to analyze (1.1). Finally we prove estimates kS k ≤ k k f Hs (M) Cs f s− 3 (bM) H 2 ∈ R (1.4) kD k ≤ e k k s . f Hs (M) Cs f s− 1 (bM) H 2 In addition to the general calculus of pseudodiﬀerential operators on manifolds which we have already developed, we require some specialized tools to carry through this analysis: the theory of homogeneous distributors. In the next section we present such a theory. After that we analyze the single and double layer potentials and apply this analysis to study (1.1). In the ﬁnal section we prove the estimates given in (1.4).

2. Homogeneous Distributions

∈ R ∈ ∞ Rn Suppose that some m , φ Cc ( )satisﬁes: φ(λx)=λ−mφ(x) λ ∈ (0, ∞) , we say that φ is homogeneous of degree m. Such a function certainly deﬁnes an element of ∞ Rn\{ } 0 (Cc ( 0 )) by Z

`φ(ψ)=hφ, ψi = φψdx. Rn\{U}

n If we define ψλ(x)=λ ψ(λx) then a simple change of variables shows that m (2.1) hφ, ψλi = λ hφ, ψi . This is the weak formula of homogeneity. If ` ∈ C−∞(Rn \{0}) satisfies (2.1) then we say that φ is homogeneous of degree m. In this section we consider the problem of extending `φ as a distribution in S0(Rn) and the extent to which the extended distribution can be made homogeneous. We begin with the one dimensional case. For s ∈ C we define ( es log x if x>0 (2.2) xs = + 0ifx ≤ 0.

Here log z is defined so that log x ∈ R if x>0. If <(s) > −1andψ ∈S(R)then Z ∞ Z ∞ s s s I+(ψ)= x+ψ(x)dx = x+ψ(x)dx −∞ 0 6 CHARLES EPSTEIN converges absolutely and evidently defines a homogeneous distribution of degree s.Infact s < − < I+(ψ) is an analytic function of s in (s) > 1. If (s) > 0 we can integrate by parts k–times to obtain Z ∞ s+k s x+ [k] (2.3) I+(ψ)= ··· ψ (x)dx . 0 (s +1) (s + k) s From (2.3) it follows that I+(ψ) has a meromorphic extension to the entire complex plane with simple poles at the negative integers, −N. Z ∞ ψ[k](x)dx (−1)kψ[k−1](0) lim (s + k)Is+k(ψ)=(−1)k−1 = →− + − − s k 0 (k 1)! (k 1)! For s/∈−N it is easy to verify that 7→ s ψ I+(ψ) defines a distribution which is homogeneous of degree s. Note that for <(s) > 0 d s s − 0 ( dx I+)(ψ)=I+( ψ ) Z ∞ − s 0 = s+ψ (x)dx 0 Z ∞ s−1 = s x+ ψ(x)dx 0 s−1 = sI+ (ψ) . Thus d s s−1 < (2.4) ( dx I+)(ψ)=sI+ (ψ)for(s) > 0 , as both sides of (2.4) are meromorphic the equation extends to s/∈−N0. Note also that for <(s) > −1: s s (xI+)(ψ)=I+(xψ) Z ∞ s = x+xψdx 0 Z ∞ s+1 = x+ ψdx 0 s+1 = I+ (ψ) . Therefore as above s s+1 ∈−N xI+ = I+ s/ . ∈ ∞ R\{ } Finally observe that for ψ Cc ( 0 ) Z ∞ s s ∈−N I+(ψ)= x ψ(x)dx , s / 0 s s S0 R so I+ is an extension of x as a homogeneous distribution in ( ). We are left with the cases: s ∈−N. We define an extension by subtracting the pole: − k (k−1) −k def s − ( 1) ψ (0) I+ (ψ) = lim I+(ψ) s→−k s + k (k − 1)! ! Z ∞ xs+kψ[k](x) (−1)kψ[k](x) = lim + + dx →− ··· − s k 0 (s +1) (s + k) (s + k)(k 1)!     Z − (−1)k−1 ∞ Xk 1 1 (2.5) =  (log xψ[k](x))dx +   ψ(k−1)(0) , (k − 1)! j 0 j=1 ELLIPTIC BOUNDARY VALUE PROBLEMS 7 the sum is absent if k =1. −k Let us investigate the homogeneity of I+ . A calculation using the previous formula shows that: (−1)k−1 I−k(ψ )=λk I−k(ψ) − log λψ[k−1](0) . + λ + (k − 1)! −k − In other words, I+ is not homogeneous of degree ( k)as k−1 (−1) − (2.6) I−k(ψ ) − λkI−k(ψ)=−λk log λ δ(k 1)(ψ) . + λ + (k − 1)! 0

−k In fact there is no homogeneous extension of x+ , any other extension, ` would diﬀer from −k I+ by a distribution supported at zero, hence X −k (j) ` = I+ + cjδ0 . j

But δ(j) is homogeneous of degree −(i + j) and therefore no such sum can ever remove the log term in (2.6). Observe that

(xψ)(j) = xψ(j) = xψ[j] + jψ[j−1]

−k Using this relation in the deﬁnition of I+ we easily show that

−k 1−k xI = I+ and by induction

j −k j−k (2.7) x I+ = I+ . Finally we obtain that d −k − −k 0 dx I+ = I+ (ψ ) (−1)k−1ψ(k)(0) (2.8) = −kI−1−k(ψ) − . + k −k The failure of I+ to be homogeneous is reﬂected in the failure of the Euler equation −k 6 − −k x∂xI+ = kI+ . We deﬁne another homogeneous distribution on R by setting ( ≥ s 0 x 0 x− = . |x|s x<0

h s i h s ˇi ˇ − As x−,ψ = x+, ψ where ψ(x)=ψ( x) the extension of this family follows in a straight s R\{ } forward way from our analysis of x+. Every homogeneous function of degree s on 0 is of the form s s ∈ C φ = a−x− + a+x+,a . So we see that if s/∈−N then φ has a unique extension as a homogeneous distribution in S0(R). If s = −k then φ may or may not have a homogeneous extension: − k−1 − k [k−1] −k ( 1) λ log λψ (0) k−1 hψ,ψ i−λ hφ, ψi = (a +(−1) a−) . λ (k − 1)! +

k−1 So φ has an extension as a homogeneous distribution iﬀ a+ =(−1) a− =0. For example the functions x−k have extensions as homogeneous distributions. We denote these by x−k.If 8 CHARLES EPSTEIN k =1then −1 −1 −1 x (ψ)=x (ψ) − x− (ψ) +Z Z ∞ 0 = − log xψ0(x)dx − log(−x)ψ0(x)dx −∞ 0 Z Z ∞ − = − lim log xψ0(x)dx + log(−x)ψ0(x)dx ↓0 Z −∞ Z ∞ − − |∞ − ψ(x) − |− − ψ(x) = lim log xψ(x) dx + log( x)ψ(x) −∞ dx ↓0 x x Z −∞ ∞ ψ(x) = P.V. dx + lim(ψ() − ψ(−)) log x ↓0 Z−∞ ∞ ψ(x) = P.V. dx −∞ x So Z Z ∞ ψ(x) ∞ x−1(ψ)=P.V. dx = − log |x|ψ0(x)dx , −∞ x −∞ which implies that −1 ∂x log |x| = x .

Another interesting and important family of homogeneous distributions arise from a slightly diﬀerent regularization of xs: Z h(x i0)s,ψi = lim (x i)sψ(x)dx . ↓0 h s i It turns out that the functions Ψs = (x i) ,ψ areentirefunctionsofs if >0with ↓ − uniform limits as 0. Ψx is clearly uniformly analytic in Re s > 1. We use Taylor’s formula to obtain:

Z Z Z Z −1 ∞ 1 Xk (j) j 1 s s ψ (0)x s Ψx ()= + (x i) ψ(x)dx + (x i) + (x i) rk(x)dx −∞ − j! − 1 1 j=0 1

Xk ψ(0)xj where r (x)=ψ(x) − . Clearly |r (x)| = o(|x|k) and therefore all terms but the k j! k j=0 third integral are analytic in Re s > −k.LetC = {|z| =1; Imz > 0}. By Cauchy’s theorem: Z Z 1 (x i)sxjdx = (z i)szj dz −1 C These are uniformly entire functions of s as ↓ 0. This shows that (x i0)s is well deﬁned s −s s for all s.ForRe s > −1 h(x i0) ,ψλi = λ h(x i0) ,ψi. As both sides are entire, this identity persists for all s and therefore (x i0)s is a homogeneous distribution of degree s. Using Cauchy’s theorem we see that Z Xk+1 ψ(j)(0) h[(x + i0)−k − (x − i0)−k],ψi = ( zj )z−kdz +− − j! C C j=0 ψ(k−1)(0) =(2πi) (k − 1)!

Before proceeding to the n-dimensional case, we compute the Fourier transforms of these distributions. First we make a general observation about the Fourier transform of a homogeneous distribution: ELLIPTIC BOUNDARY VALUE PROBLEMS 9

Suppose φ ∈S0(Rn)satisﬁes:

−s hφ, ψλi = λ hφ, ψi then b b hφ, ψλi = hφ, ψλi.

b b n b A simple calculation shows that ψλ(ξ)=ψ(ξ/λ)=λ ψλ(ξ). Thus

b n b n+s b hφ, ψλi = λ hφ, ψ1/λi = λ hφ, ψi Hence the Fourier transform is homogeneous of order −(n + s). s ∈−N − We now compute the Fourier transforms of x+,s / .ForRe s > 1weseethat Z ∞ xs (ξ) = lim e−ix·ξe−xxsdx + ↓ 0 0 Z ∞ = lim e−x(+iξ)xsdx. ↓ 0 0 We compute this using the Cauchy integral formula

R arg z( + iξ)=0, deﬁnes the contour, Γ,ξ Z e−z(+iξ)zsdz =0 R Γ,ξ T {| | } R | −z(+iξ) |≤ −R It is elementary to show that along z = R Γ,ξ e e and therefore as R + ∞ we obtain: Z Z ∞ ∞ y dy e−x(+iξ)xsdx = e−x( )s 0 0 + iξ ( + iξ) Γ(s +1) = ( + iξ)s+1 + πi (s+1) Γ(s +1)e 2 = (ξ − i)s+1

−πi(s+1) Γ(s +1)e 2 Thus xcs (ξ)= for Re s > −1. These two functions are meromorphic in + (ξ − i0)s+1 C and so must coincide for s/∈−N. We compute

d−1 d−1 − 1 b x+ (ξ) = limx+ (ξ) δ0(ξ) ↓0 − πi 1 = limΓ()e 2 (ξ − i0) − ↓0  + πi  Γ( +1)e 2 ξ − 1 (2.9) lim for ξ>0 ↓0 = − πi  Γ( +1)e 2 |ξ| − 1 lim for ξ<0 ↓0 πi =log|ξ| + sgn ξ +Γ0(1). 2

d−k −1 To compute x+ one simply uses (2.8). From the deﬁnition of x and (2.9) we obtain d x−1(ξ)=πi sgn ξ 10 CHARLES EPSTEIN

−k −(k+1) As ∂xx = −kx we see that d − − x−k(ξ)=k!(−1)k 1(iξ)k 1πi sgn ξ = πik!(−i)k−1ξk−1 sgn ξ.

We now consider this extension problem in dimensions greater than 1. Suppose that φ ∈ C∞(Rn\{0})andforsomem ∈ C satisfies: m (2.10) φ(λx)=λ φ(x)forλ ∈ R+ We want to extend φ to Rn as a distribution which satisfies (2.10) in so far as this is possible. For ψ ∈ C∞(Rn\{0})wehavetherelation: c Z hφ, ψi = φ(x)ψ(x)dx Z Z ∞ = φ(ω) ψ(rω)rm+n−1dr dσ(ω) Sn−1 0 If we interpret Z ∞ m+n−1 h m+n−1 i ψ(rω)r = r+ ,ψ(rω) 0 = Rm(ψ)(ω) then we can use the one dimensionalZ results to obtain the desired extension: e n (2.11) hφ, ψi = φ(ω)Rm(ψ)(ω)dσ(ω) ,ψ∈S(R ) Sn−1 m+n−1 ∈S0 R 7→ Sn−1 S R As r+ ( )andω ψ(rω) is a smooth mapping of into ( )itisclearthat ∞ n−1 e −∞ n−1 Rm(ψ)(ω) ∈ C (S )sothathφ, ψi is well defined even if φ ∈ C (S ) . It defines Sn−1 a distribution which is homogeneous so long as m ∈{−n, −n − 1,...}. This extension process has several nice properties: 1) If P (x) is a homogeneous polynomial then

(P^(x)φ)=P (x)φe for all m ∈ C . 2) If m/∈{1 − n, −n,...} then ] e (∂xiφ)=∂xiφi=1,...,n Both statements are immediate consequences of (2.11). If m = −(n + k)thenφe is not always a homogeneous distribution: with ψ(λx)=λnψ(λx) Z e hφ, ψλi = φ(ω)R−(n+k)(ψ)(ω)dσ(ω) Z h i − k k+n − ( 1) k (2.12) = φ(ω)λ R−(n+k)(ψ)(ω) log λ ∂r ψ(rω) dσ(ω) . k! r=0 Thus we see that −λk+n log λ(−1)k X hφ,e ψ i−λk+nhφ,e ψi = ∂αψ(0)M(φωα) λ k! x |α|=k where Z M(φωα)= φ(ω)ωαdσ(ω) . |ω|=1 The log-term vanishes if and only if M(φωα)=0 ∀ α with |α| = k. ELLIPTIC BOUNDARY VALUE PROBLEMS 11

As in the one dimensional case, if the extension φe is not homogeneous then it is not possible e to find a homogeneous extension. This is because any other extensionP differs from φ by a α distribution supported at 0 and therefore a finite sum of the form cα∂x δ0 .Asthisis itself a sum of homogeneous distributions no such sum can suffice to remove the log-term in (2.12). Note that if φ(−x)=−(−1)kφ(x)thenS(ωαφ)=0∀ α with |α| = k . This condition is satisfied by ratios of homogeneous polynomials in odd dimensions. The Fourier transforms of homogeneous distributions are again homogeneous distributions: be e b hφ, ψλi = hφ, ψλi −n e b = λ hφ, ψ 1 i λ = λ−(m+n)hφ,e ψbi b = λ−(m+n)hφ,e ψi . b If φe is homogeneous of degree m then φe is homogeneous of degree −(m+ n). If m = −(k + n) b and φe is not homogeneous then φe also transforms with a log-term: b b λ−k log λ(−1)k X (2.13) hφ,e ψ i = λ−khφ,e ψi− S(φωα)hxα,ψi λ k! From (2.13) we easily obtain that if φ is homogeneous of degree −(k + n)then be b φ = φ1 +log|x|p(x) b where φ1 is homogeneous of degree k and p(x) is a homogeneous polynomial of degree k. ∞ n In all cases it is easy to show that if φ ∈ C (R \{0}) is homogeneous of degree m then b e ∈ ∞ Rn ≡ φ is smooth as well. To prove this we select a function χ Cc ( )sothatχ 1in Rn\{0} a neighborhood of 0. Then b c φe = χφe +[(1− χ)φe]∧ .

The first term is analytic as χφe is a compactly supported distribution. Using the oscillatory integral definition of [(1 − χ)φe]∧ we obtain that Z e−ix·ξ (2.14) [(1 − χ)φe]∧(ξ)= ∆k[(1 − χ)φ(x)] dx . |ξ|2k − ∞ Rn Any derivatives applied to 1 χ again leads to the Fourier transform of a function in Cc ( ), the only term which is therefore not obviously smooth in Rn\{0} is: Z 1 (1 − χ)(∆kφ)e−ix·ξdx . |ξ|2k Since ∆kφ is homogeneous of order m−2k. For any fixed j there is a k so that this expression is absolutely convergent along with all derivatives of order j. This completes the proof. We close with a simple application of these ideas: X α Let P (D)= aαD be an elliptic operator, that is |α|=m X α P (ξ)= aαξ |α|=m is nonvanishing in Rn\{0}. There exists a fundamental solution E of the form

E(x, y)=e0(x − y)+log|x − y|p(x − y) 12 CHARLES EPSTEIN where e0 is homogeneous of degree m − n and p is a polynomial of degree m − n (identically b g1 zero if m

This implies that P (D)E = δ0 .

3. Elliptic boundary value problem for the Laplacian

3.1. Multiple layer potentials. As it introduces no additional complexity we consider the somewhat more general problem of a boundary value problem for the Laplace, ∆g,ofametric, g on a compact domain, Ω with smooth boundary in a compact manifold, X. In section 2 we showed that there is a fundamental solution deﬁned on a neighborhood, Ω1,ofΩ.Thisis ∞ represented by a function Q ∈ C (Ω1 × Ω1\∆) which satisﬁes: y ∈ (3.1) ∆gQ(x, y)=δx(y) x Ω1 in the distribution sense, that is Z ∀ ∈ ∞ Q(x, y)∆φ(y)dV ol(y)=φ(x) φ Cc (Ω1) .

−2 Q is the Schwartz kernel of a ψDO in Ψ (Ω1), its principal symbol, σ−2(q)satisﬁes: | |−2 (3.2) σ−2(Q)(x, ξ)= ξ g(x) . For f ∈ C∞(bΩ) we deﬁne: Z Sf(x)= Q(x, y)f(y)dσ(y) bΩ and Z ∂Q Df(x)= (x, y)f(y)dσ(y) bΩ ∂νy

Here dσ is the surface measure on bΩand∂ν is the outer normal derivative along bΩ. It is immediate from (3.1) that Sf, Df ∈ C∞(Ω) and:

∆gSf(x)=∆gDf(x)=0 forx ∈ Ω1\bΩ .

If ∆gu = 0 in Ω then Green’s formula states that

(3.3) u(x)=Du0(x) −Su1(x)

∂u D S Here u0 = u and u1 = ∂ν . We now consider the operators , and prove the following bΩ bΩ basic result: Theorem 3.1. If f ∈ C∞(bΩ) then Sf, Df have extensions as elements of C∞(Ω).For each k ∈ N there are ψDO S , D in Ψk−1(bΩ) and Ψk(bΩ) respectively such that: 0 k k

kS ∂ν f = Skf bΩ (3.4) k =0, 1,... kD ∂ν f = Dkf bΩ Moreover 1 −1 σ− (S )(x, ξ)= |ξ| 1 0 2 g ∈ ∗ (3.5) 1 (x, ξ) T bΩ . σ−1(D0)(x, ξ)= 2 ELLIPTIC BOUNDARY VALUE PROBLEMS 13

In the theorem we use the metric induced on bΩ from the metric deﬁned on X.

Proof: A more invariant description of the operators, S and D takes advantage of the fact that pseudodiﬀerential operators act on distributions. Let δ denote the distribution: Z bΩ

hφ, δbΩi = φdσ . bΩ Then the operators S and D are deﬁned for f ∈ C∞(bΩ) by:

Sf = Q(fδbΩ) (3.6) Df = −Q(∂ν (fδbΩ)) . From these formulae it is apparent that, as distributions singsupp Sf ⊆ supp f (3.7) singsupp Df ⊆ supp f. From (3.7) it is clear that to prove the theorem it suffices to work in a single local coordinate | |2 e patch. We choose coordinates to simplify the form of the metric, ξ g:LetU be a small disk e centered on a point p ∈ bΩandlet(x1,...,xn) be coordinates defined on U ∩ bΩ. We define e x0 to be the signed geodesic distance from a position U to a point on bΩ. We take x0 to be positive for points in Ω and negative for points in X\Ω, this function is smooth in some neighborhood of bΩ. In a possibly smaller neighborhood, U of p,wecanuse(x0,x1,...,xn) as coordinates. In such a coordinate patch the metric takes the form: Xn 2 2 ds = dx0 + gij(x0,x)dxidxj , i,j=1 2 | |2 the principal symbol of ∆g is then σ2(∆g)=ξ0 + ξ g whereweusethenotation: Xn | |2 ij ξ g = g ξiξj . i,j=1

Since it entails no further eﬀort and is useful for subsequent applications we will consider the limiting behavior as x0 → 0 from above and below. We use (+) to denote x0 & 0and( −)

% to denote x0 0. Recall that in our domain, Ω, the variable x0 > 0. Relative to Ω, ∂x0 x0=0 is the inward pointing unit normal vector. ThecompletesymbolofQ, in this local coordinate system takes the form: X−3 ∼ 2 | |2 −2 σ(Q)(x0,x; ξ0,ξ)=q(x0,x; ξ0,ξ) (ξ0 + ξ g) + aj(x0,x; ξ0,ξ) , j=−∞ here aj is homogeneous in (ξ0,ξ)oforderj in (ξ0,ξ). In the sequel, the statement that a function of the form k(x0,x; ξ0,ξ) ‘is homogeneous’ means that it is homogeneous as a function of (ξ0,ξ). For our applications we require a more precise statement than this: for each j there is a smooth family of homogeneous polynomials, pj (x0,x; ξ0,ξ) and an integer kj, such that:

pj (x0,x; ξ0,ξ) aj (x0,x; ξ0,ξ)= . 2kj |ξ|g

In other words, the terms in the asymptotic expansion of σ(Q) are rational functionsT in (ξ0,ξ). ∈ ∞ This is a simple consequence of the parametrix construction. For f Cc (U bΩ) we see that Z Z ∞ · · dξ0dξ S ix ξ ix0 ξ0 b f(x0,x)= e e f(ξ)q(x0,x; ξ0,ξ) n+1 Rn −∞ (2π) 14 CHARLES EPSTEIN

Inthecaseathandtheξ0-integral converges absolutely, in other cases one uses (3.6) to give such expression meaning as an oscillatory integral. If ψ(a, b) is a function in C∞(R2)such that ( 1 |a|2 + |b|2 > 1 ψ (a, b)= 1 | |2 | |2 1 0 a + b < 2 then X−3 − | | 2 | |2 −2 (3.8) rN (x0,x; ξ0,ξ)=q(x0,x; ξ0,ξ) ψ1(ξ0, ξ g) (ξ0 + ξ g) + aj j=−N belongs to S−(N+1)(Rn+1; Rn+1). We have the following simple lemma: ≥ ∈ ∞ Rn Lemma 3.1. :IfN 1 then for f Cc ( ) Z Z ∞ b iξ0 ·x0 iξ·x dξ0dξ (3.9) RN f(x0,x)= f(ξ)e e rN (x0,x; ξ0,ξ) n+1 Rn −∞ (2π)

N−1 n+1 −N n belongs to C (R ) moreover RN f(0,x)=ρN (f)(x),whereρN ∈ Ψ (R ).

Proof: Proving RN f(x0,x)has(N − 1)-derivatives is simply a matter of differentiation under the integral sign. The only degradation to the convergence of the integral arises from differentiating eiξ0x0 , but this term can be differentiated (N −1)-times leading to an integrand 1 which is 0( ξ2 ). Restricting to ξ0 =0gives: 0 Z Z ∞ b ix·ξ dξ0 dξ RN f(0,x)= f(ξ)e rN (0,x; ξ0,ξ) n −∞ 2π (2π) Thus we see that Z dξ σ(ρ )= r (0,x; ξ ,ξ) 0 N N 0 2π

| α β |≤ Cαβ Dx Dξ rN (0,x; ξ0,ξ) N+1+|β| . (1 + |ξ0| + |ξ|g) Integrating this estimate shows that e | α β |≤ Cαβ Dx Dξ σ(ρN )(x, ξ) N+|β| . (1 + |ξ|g) This completes the proof of the lemma. As a simple corollary we have: Corollary 3.1. If k

Proof: We can simply differentiate (3.9) to obtain: Z Z k b ix·ξ k ix ·ξ dξ0 dξ ∂ R f(x ,x)= f(ξ)e ∂ r (x ,x; ξ ,ξ)e 0 0 x0 N 0 x0 N 0 0 n Z Z 2π (2π) b ix·ξ k ix ·ξ dξ0 dξ = f(ξ)e r (x ,x; ξ ,ξ)(iξ ) e 0 0 + l.o.t. N 0 0 0 2π (2π)n Here l.o.t. is a sum of terms of order strictly less that k − N. The lemma, its corollary and (3.8) show that to prove the theorem it suffices to consider each term in the asymptotic expansion separately. First we analyze the principal term. We ELLIPTIC BOUNDARY VALUE PROBLEMS 15 define: ZZ b ix·ξ f(ξ)e · dξ0dξ S f(x ,x)= ψ (|ξ| )ψ (ξ , |ξ| )eix0 ξ0 . 0 0 (ξ2 + |ξ|2) 2 g 1 0 g (2π)n+1 ( 0 g 0 |a|≤1 Here ψ2(a)= . 1 |a|≥2

We evaluate the ξ0 integral, for |ξ|g > 1, using the Cauchy residue formula we obtain: Z · −| || | eix0 ξ0 dξ e x0 ξ g (3.10) 0 = 2 | |2 | | (ξ0 + ξ g) 2π 2 ξ g Thus we see that Z −|x0 ||ξ|g S b | | e ix·ξ dξ (3.11) f(x0,x)= f(ξ)ψ2( ξ g) e n 2|ξ|g (2π)

n n From (3.11) it is clear that S0f(x0,x) has smooth extensions to R ×(−∞, 0] and R ×[0, ∞) which in general do not agree across x0 =0.Ifg is independent of x0 then  Z  1 b | | | |k−1 ix·ξ dξ − 2 f(ξ)ψ2( ξ g) ξ g e (2π)n ( ) k S Z lim ∂x 0f(x0,x)= → 0  − k − · x0 0  ( 1) b | | | |k 1 ix ξ dξ 2 f(ξ)ψ2( ξ g) ξ g e (2π)n (+) ( S− f(x)(−) = k0 S+ f(x)(+) k0 ∈ k−1 Rn Note that Sk Ψ ( )andthat 0 ( 1 | |k−1 − 2 ξ g ( ) σk−1(S )= − k k0 ( 1) | |k−1 2 ξ g (+)

More generally if g does depend on x0 then it follows easily from (3.11) and the fact that ψ2(|ξ|g)|ξ|g is a symbol that lim ∂k S f(x ,x)=S f + Ef here E ∈ Ψk−2(Rn). x0 0 0 k0 k k x0→0 This completes the analysis of leading part. Now we need to consider the lower order terms in the asymptotic expansion for σ(Q). These are expressions of the form: ZZ b ix·ξ iξ ·x dξ0dξ A f(x ,x)= f(ξ)e a (x ,x ; ξ ,ξ)e 0 0 ψ (ξ , |ξ| ) j 0 j 0 i 0 1 0 g (2π)n+1 where aj is homogeneous in (ξ0,ξ)ofdegreej. In fact a somewhat more precise statement is true: j (3.12) aj (x0,x; λξ0,λξ)=λ aj(x0,xi; ξ0,ξ) λ ∈ R\{0} .

This follows because aj is a rational function in (ξ0,ξ). From this we can conclude that taking the Fourier transform of aj in the ξ0-variable will not lead to terms of the form log |x0|.Up to a smoothing term: Z Z · eiξ0 x0 (3.13) A f(0,x)= fb(ξ)eix·ξψ (|ξ| ) a (x ,x; ξ ,ξ) dξ dξ . j 2 g j 0 0 2π 0

There are two approaches to evaluating the ξ0-integral. We can use the fact that

pj(x0,x; ξ0,ξ) aj(x0,x; ξ0,ξ)= 2 | |2 kj (ξ0 + ξ g) 16 CHARLES EPSTEIN and the Cauchy integral formula to obtain that for |ξ|g > 1: for x0 ≶ 0: Z ∞ ix ξ ix ξ p (x ,x; ξ ,ξ)e 0 0 dξ i − e 0 0 p (x ,x; ξ ,ξ) j 0 0 0 = ∂kj 1 j 0 0 2 2 kj ξ0 kj −∞ (ξ + |ξ| ) 2π (kj − 1)! (ξ0 i|ξ|g) ξ0 =i sgn x0|ξ|g (3.14) 0 g −|x ||ξ| e 0 g qj(x0,x; i|ξ|g,ξ) = 2k −1 . (i|ξ|g) j

Recall that deg pj − 2kj = −j, while qj is no longer a homogeneous polynomial, a moment’s reﬂection shows that:

(3.15) deg qj − (2kj − 1) = 1 − j.

From (3.13) and the second line in (3.14) it is evident that Aj f(x0,x) again has smooth extensions to Rn × (−∞, 0] and Rn × [0, ∞). From (3.14) and (3.15) we conclude that lim ∂k A f(x ,x)=A f(x)whereA ∈ Ψk+1−j(Rn). x0 j 0 jk jk x0→0

This analysis used only that j ≤−2 and the general form of the symbols, {aj }. There is a second approach to this computation that works in greater generality. To compute Z · dξ0 (3.16) α (x ,x; ξ)= a (x ,x; ξ ,ξ)eiξ0 x0 j 0 j 0 0 2π

1+j ξ0 ξ ξ0 we rewrite aj (x0,x; ξ0,ξ)=|ξ| aj (x0,x; ,ω)whereω = .Welets = in (3.16) to g |ξ|g |ξ|g |ξ|g obtain: Z ∞ | | | |1+j isx0 ξ g αj(x0,x; ξ)= ξ g aj(x0,x; s, ω)e ds . −∞ The fact that we can allow the integral to run from −∞ to ∞ is a consequence of (3.12).

To evaluate this integral we make further usage of the homogeneity of aj : ω a (x ,x; s, ω)=sj a (x ,x;1, ) , j 0 j 0 s where again we use (3.12). We use the Taylor expansion for aj: h i X ∂α ω ω (3.17) a (x ,x; s, ω)=sj ξ a (x ,x;1, 0)( )α + r (x ,x;1, ) j 0 α! j 0 s N 0 s |α|≤N where r (x ,x;1,y)=0(|y|N+1)aty =0. N 0 ( 1 |s| < 1 Let φ(s)= then 0 |s| > 2 Z · | | ds α (x ,x; ξ)=|ξ|j a (x ,x; s, ω)φ(s)eis x0 ξ g j 0 g j 0 2π Z X α ∂ aj(x0,x;1, 0) ω + |ξ|j ξ (x ,x;1, 0)( )α + g α! 0 s |α|≤N ω | | r (x ,x;1, )sj eisx0 ξ g (1 − φ(s))ds N 0 s j n The ﬁrst term is obviously a smooth function of x0 taking values in Ψ (R ), its k-fold j+k n x0-derivative takes values in Ψ (R ). To evaluate the second term we observe that sj−|α| 1 − φ(s) = sj−|α| − φ(s)sj−|α| ELLIPTIC BOUNDARY VALUE PROBLEMS 17 where sm denotes the homogeneous distribution on R deﬁned in §2. As φ(s)sj−|α| is a compactly supported distribution, its Fourier transform is a smooth function of x0. On the other hand, we have computed the Fourier transforms of sj−|α| and it is:

\j−|α| |α|−j−1 s (ξ0)=iπ(|α|−j)!(iξ0) sgn ξ0 . Thus the sum in (3.17) contributes: X α ∂ aj(x0,x;1, 0) |ξ|j ξ ωαiπ(|α|−j)!(ix |ξ| )|α|−j−1 sgn (x ) . g α! 0 g 0 |α|≤N Recalling that j ≤−3 we see that these terms have smooth extensions to Rn × (−∞, 0] and Rn × ∞ ω j − − [0, ). The error term rN (x0,x;1, s )s (1 φ(s)) is N (j + 1)-times diﬀerentiable. As n N is arbitrary we once again conclude that Ajf(x0,x) has smooth extensions to R ×(−∞, 0] and Rn × [0, ∞). Using the various expressions used to evaluate (3.16) we again show that ∈ k+1−j Rn Ajk Ψ ( ), we leave the details of this argument to the reader. This completes the proof of the statement in Theorem 3.1 regarding Sf. A similar argument could be used to study Df, however this is not really necessary as the properties of Df are easily deduced from those of Sf. Observe that ZZ b · · f(ξ)dξ (3.18) Df(x ,x)= eix ξeix0 ξ0 (−iξ )q(x ,x; ξ ,ξ) 0 0 0 0 (2π)n+1 On the other hand: ZZ b · · f(ξ)dξ0dξ ∂ Sf(x ,x)= eix ξeix0 ξ0 (iξ )q(x ,x; ξ ,ξ) x0 0 0 0 0 (2π)n+1 ZZ (3.19) ∂q fb(ξ)dξ dξ ix·ξ ix0·ξ0 0 + e e (x0,x; ξ0,ξ) n+1 ∂x0 (2π) = S1f + S2f.

The symbol in the second term in (3.19), ∂q deﬁnes an operator with exactly the same ∂x0 properties as S itself. Of course, the symbols are diﬀerent but the orders of the operators k S k S which appear in ∂x0 2f(0 ,x) are identical to those which arose in ∂x0 f(0 ,x). So we see that D − S S (3.20) f = ∂x0 f + 2f. Thus all the required properties of Df follow in a straightforward manner from the analysis we have just presented. For example, we see that ( (−1) − 2 ( ) σ0(D )= 0 1 (+) ( 2 (3.21) −| | ξ g (−) σ (D)= 2 1 1 +|ξ|g 2 (+) using these calculations we can easily prove the following proposition:

∞ Proposition 3.1. If f ∈ C (bΩ) then Sf(x0,x) is continuous as x0 → 0 whereas (3.22) Df(0+,x) −Df(0−,x)=f. D → Similarly ∂x0 f(x0,x) is continuous as x0 0 whereas S + − S − − (3.23) ∂x0 f(0 ,x) ∂x0 f(0 ,x)= f. 18 CHARLES EPSTEIN

3.2. The Calderon projector and Dirichlet to Neumann operator. If u is a harmonic function in Ω then

u(x)=Du0 −Su1,

where u0 = u bΩ and u1 = ∂ν u. So in particular + − + u0 = D0 u0 S0 u1 (3.24) and + − + u1 = D1 u0 S1 u1 . If we let f, g be smooth functions on bΩthen

U = Df −Sg is a harmonic function defined in Ω. From Theorem 3.1 it follows that + + U0 = D f − S g (3.25) 0 0 + − + U1 = D1 f S1 g We define the operator: + − + f D0 S0 f C = + − + g D1 S1 g From (3.24) it follows that C is a projection operator, that is C2 = C. From Theorem 3.1 it follows that C is a pseudodifferential operator with principal symbol: −1 1 1 |ξ| σprin(C)= g . 2 −|ξ|g 1

2 As expected σprin(C) = σprin(C). This operator is called the Calderon projector. Another operator of considerable interest is the Dirichlet to Neumann operator. As we shall soon see, the classical Dirichlet problem, ( ∆ u =0 inΩ (D) u = f bΩ can always be solved. We deﬁned an operator

∂u N f = ∂ν bΩ + + where u is the solution to (D). Observe that S0 is an elliptic ψDO on bΩandletP0 denote a parametrix:

+ + + P0 S0 = I + e0 + ∈ −∞ where E0 Ψ (bΩ). Using (3.24) we deduce that + + − + − + u1 =(P0 D0 P0 )u0 E0 u1 . So again N is a ψDO of order 1 with |ξ| σ (N )= q . 1 2 The Dirichlet to Neumann operator plays an important role in geophysical inverse scattering problems. ELLIPTIC BOUNDARY VALUE PROBLEMS 19

3.3. Elliptic boundary value problems. Now we use the properties of the multilayer potentials to show that certain boundary value problems for the Laplace operator are solvable for data satisfying ﬁnitely many conditions. Let’s consider the simplest such problem, the Dirichlet problem. With u = Du0 −Su1 we see that − + − + (I D0 )u0 = S0 u1 .

If we can solve for u1, then setting − + −1 − + u1 = (S0 ) (I D0 )f we obtain the Cauchy data for the solution to ( ∆ u =0 inΩ

u = f, bΩ i.e. if we set D S + −1 − + (3.26) u = f + (S0 ) (I D0 )f

then ∆u =0 and

+ + + −1 − + u = D0 f + S0 (S0 ) (I D0 )f bΩ = f.

+ 1 | |−1 We need to show that S0 is invertible. As its symbol is 2 ξ g it is clear that for every s ∈ R: + s → s+1 S0 : H (bΩ) H (bΩ) + ⊆ ∞ is a Fredholm operator of index 0 and kerS0 C (bΩ). To avoid technical diﬃculties we ⊆ Rn + now consider only the case of Ω . If there is a function f0 such that S0 f0 =0thenwe deﬁne

u = Sf0 . Rn\ + This function is harmonic in Ω. As its boundary values from inside Ω equal S0 f =0 + − it follows from the maximum principle that U ≡ 0. Let U denote U .AsSf0 is Ω Rn\Ω continuous, it follows that U − = 0 as well. On the other hand we know from Proposition bΩ + − 3.1 that ∂νU − ∂ν U = −f, and therefore − ∂ν U = f. As Z f(x)dσ(y) U(x)=c n | − |n−2 bΩ x y it follows that lim U −(x) = 0. We can therefore apply the maximum principle to conclude |x|→∞ − ≡ + that U Ω 0. This implies that f = 0, hence S0 is an invertible operator. This shows that the Dirichlet problem always has a smooth solution for a smooth bounded domain, Ω ⊆ Rn and f ∈ C∞(bΩ). Using the formula (3.26) and estimates for S and D proved in the last ∈ s 1 section, we will extend this result to f H (bΩ) for s> 2 as well as obtaining a precise statement of the regularity of the solution, u. ⊆ Rn + Without restricting to Ω we could have concluded that the range of S0 is of ﬁnite codimension. Thus there are ﬁnitely many linear functionals `1,...,`m so that if − + `j ((I D0 )f)=0 j =1,...,m then equation − + − + (I D0 )f = S0 u1 20 CHARLES EPSTEIN has a solution, g. Again setting u = Df −Sg we obtain that

+ − + u bΩ = D0 f S0 g + − + = D0 f +(I D0 )f = f, as desired. As before, we will be able to apply the estimates proved in the next section to extend this existance result to boundary data with ﬁnite diﬀerentiability and also obtain precise regularity results. In fact, the Dirichlet problem is always solvable on a smooth compact manifold with boundary. Now we turn our attention to more general boundary value problems: ( ∆u =0 inΩ (Pb) b0u0 + b1u1 = f on bΩ k k−1 Here b0 ∈ Ψ (bΩ) and b1 ∈ Ψ (bΩ). We want conditions which imply that this is an “elliptic problem”. For us this will be simply the condition that the system of equations: I − D+ S+ u 0 (3.27) 0 0 0 = b0 b1 u1 f is elliptic. This is a principal symbol calculation:

Deﬁnition: The boundary value problem, Pb is elliptic if

|ξ|gσk−1(b1) − σk(b0) is elliptic. This is the determinant of the principal symbol of the operator in (3.27). In this case (3.27) is a Fredholm system, the principal symbol of the parametrix is given by: −1 1 σk−1(bg) | | 2 ξ g 6 1 ,ξ=0, D −σk(b0) 2 where D = 1 σk − 1(b ) − σk(b0) . Using the symbol calculus for Ψ∗(bΩ) we easily show that 2 1 |ξ|g there exists a matrix of operators, RS TU such that −1 RS 1 σk−1(b1) | | σ = 2 ξ g TU − 1 D σk(b0) 2 RS I − D+ S+ and 0 0 = I + E, TU b0 b1 + + I − D S RS 0 0 0 = I + E b b TU 0 1 I − D+ S+ where E,E0 ∈ Ψ−∞(bΩ; C2). This shows that 0 0 is a Fredholm map whenever b0 b1 the boundary problem is elliptic. From this we easily deduce:

Theorem 3.2. If (Pb) is an elliptic boundary value problem then it has a smooth solutions for all f ∈ C∞(bΩ) which satisfy a ﬁnite number of linear conditions of the form: Z

fgj dσ =0 j =1,...,m bΩ ∞ where gj ∈ C (bΩ). ELLIPTIC BOUNDARY VALUE PROBLEMS 21

Corollary 3.2. If (Pb) is elliptic then the set of solutions to Pb with f =0is ﬁnite dimensional.

The classical Neumann problem ( ∆u =0 inΩ (N) ∂u ∂ν = f on bΩ corresponds to b0 =0,b1 = 1, this is obviously an elliptic problem which therefore has a solution for data which satisfies finitely many conditions. For a domain in Ω ⊆ X it is a consequence of Green’s formula, Z Z ∂v ∂u u∆gv − v∆gu = u − v Ω ∂ν ∂ν applied to u, a solution of (N)andv =1that Z f =0 bΩ is a necessary condition for (N) to be solvable. In fact, for this case, this is the only condition though the proof again requires non-pseudodifferential techniques. We can consider a more general problem of this type; the oblique derivative problem: ( ∆u =0 (O ) Y ∂u + Y (u )=f. ∂ν bΩ Here Y ∈ C∞(bΩ; TbΩ) is a smooth vector field tangent to bΩ. In this case

σ0(b1)=1,σ1(b0)=ihY, ξi . Recall that (x, ξ) ∈ T ∗bΩ. This problem is elliptic if hY, ξi = < 1 ∀ ξ =06 . |ξ| This is the case if Y is a real vector field. There is a very important special case of this problem which is not elliptic. If Ω ⊆ Cn is a domain with a smooth boundary then we choose a unit vector field, N transverse to bΩ such that JN = T is tangent to bΩ. Here J is the endomorphism of T Cn which defines the standard complex structure. The complex vector ≡ field Z = N + iJN is of type (0, 1), so if u is a holomorphic function in ΩthenZu bΩ 0. There is an infinite dimensional space of such functions; these functions are also harmonic. This implies that the oblique derivative problem: ( ∆n =0 inΩ (σ ) Z Zu = f bΩ is not Fredholm: it has an infinite dimensional null space. The endomorphism, J, is orthogonal and therefore hZ,ξi hJN,ξi I = m ξ |ξ| attains the value 1 in exactly one codirection. This shows that σZ is not an elliptic problem in the sense defined above. This problem is called the ∂-Neumann problem and is of fundamental importance in the theory of holomorphic functions of several variables. What does the condition − + + (3.28) (I D0 )u0 + S0 u1 =0 22 CHARLES EPSTEIN

Rn+1 mean? To understand this we consider a simple model situation, the upper half space + = [0, ∞) × Rn. We want to solve the equation ( Rn+1 ∆u =0 in +

∂u b0u + b1 ∂ν = f. bΩ Take the Fourier transform of the equation in the Rn-factor. We obtain 2b ∂ u −| |2b (3.29) 2 (x0,ξ) ξ u(x0,ξ)=0. ∂x0 The general solution to (3.28) is of the form:

−x0|ξ| x0 |ξ| ub(x0,ξ)=a−|ξ|e + a+(ξ)e . The condition (3.28) is 1 |ξ| a−(ξ)+a (ξ) + a (ξ) − a−(ξ) =0, 2 + 2|ξ| + that is a+(ξ) = 0. In other words, the exponentially growing part is identically zero. In P.D.E. the existence of exponentially growing solutions is intimately tied to non-well posedness. If b0 and b1 are convolution operators, then the boundary condition becomes: b b b b0(ξ) −|ξ|b1(ξ) a−(ξ)=f(ξ) .

Elliptically is then the condition that we can solve for the remaining coefficient, a−(ξ)in terms of the data fb(ξ). A similar interpretation for these conditions exists in the general case: one simply replaces the operator ∆g at a boundary point (0,x) with the model problem that comes form freezing the coefficients of the principal symbol at (0,x): 2 ∂ b −| |2b 2 u(x0,ξ) ξ gu(x0,ξ)=0. ∂x0 We leave this to the interested reader, see also Hörmander, vol. 3. In the next and final section we prove that the operator S and D extend to define bounded maps between the L2–Sobolev spaces on bΩandthoseofΩ.

4. Basic elliptic estimates

In the previous sections we showed that the multiple later potentials S, D define maps from C∞(bΩ) to C∞(Ω). This shows that when an elliptic boundary value problem with C∞ boundary data is solvable, the solution is smooth up to bΩ. In this section we prove the following estimates: for each s ∈ R s s+ 3 S : H (bΩ) → H 2 (Ω) (4.1) s s+ 1 D : H (bΩ) → H 2 (Ω) . Using these estimates and the density of C∞(bΩ) in Hs(bΩ) we can use our previous analysis to solve elliptic boundary value problems with boundary data of finite differentiability.

To prove (4.1) we introduce local coordinates (x0,x) to ﬂatten out the boundary where, as before, the lines x =const are arclength parametrized geodesics orthogonal to bΩ. Using partitions of unity it evidently suﬃces to prove that s+ 3 Se s Rn → 2 Rn × ∞ : Hloc( ) Hloc [0, ) and 0 s+ 1 (4.1 ) De s Rn → 2 Rn × ∞ : Hloc( ) Hloc [0, ) where Se,De are local coordinate representations of S and D. We start with a lemma: ELLIPTIC BOUNDARY VALUE PROBLEMS 23

Lemma 4.1. Let a(x0,x; ξ0,ξ) be a homogeneous function in (ξ0,ξ) of the form:

p(x0,x; ξ0,ξ) a(x0,x; ξ0,ξ)= q(x0,x; ξ0,ξ) where p and q are homogeneous polynomials in (ξ0,ξ),withdeg p − deg q = j. Suppose that for ξ =06 q(x ,x; ξ ,ξ)=0has no real roots. Let 0 0 Z

ix0ξ0 k(x0,x; ξ)= a(x0,x; ξ0,ξ)e dξ0 for x0 > 0 and ξ =06 be deﬁned as an oscillatory integral. Then for each pair of multiindices, (α0,α), (β0,β) we have the following estimates: − β β0 α α0 ≤ | | j+1+α0 β0 (4.2) (x x0 ∂x ∂x0 k) c(1 + ξ ) .

Proof: Using the oscillatory integral definition means simply that for x0 > 0andsuffi- ciently large `, Z ∞ − · (4.3) k(x ,x; ξ)=( 1 )` ∂` a(x ,x; ξ ,ξ)eix0 ξ0 dξ . 0 ix ξ0 0 0 0 0 −∞ If ` is large enough, then (4.3) is an absolutely convergent expression. Since a is a ratio of polynomials, we can define the contour of integration to a simple closed curve, Γ, contained in the upper half plane which encloses the zeros of q(x ,x; ξ). Thus: Z 0 − · k(x ,x; ξ)=( 1 )` ∂` a(x ,x; ξ ,ξ)eix0 ξ0 dξ 0 ix0 ξ0 0 0 0 Γ (4.4) Z ξ ix0·ξ0 = a(x0,x; ξ0,ξ)e dξ0 . Γξ

The last statement follows: if f(ξ0), g(ξ0) are holomorphic functions in the neighborhood of a simple closed contour, γ,then Z Z

∂ξ0 f(ξ0)g(ξ0)dξ0 = df g γ γZ = − fdg γ by Stokes’ theorem. The estimates now follow easily: Z · xβ0 xβ∂α∂α0 k =(−i)β0 xβ ∂α0 ∂αa(x ,x; ξ ,ξ)∂β0 eix0 ξ0 dξ 0 x x0 x0 x 0 0 ξ0 0 Γ α0 Again we can integrate by parts in ξ0 and apply the Leibniz formula to compute ∂x0 .We obtain: Z Xα0 − · (4.5) (−i)β0 xβ c ∂j ∂α∂β0 a(iξ )β0 jeix0 ξ0 dξ . j x0 x ξ0 0 0 j=0 Γξ

As q is homogeneous the length of Γξ can be taken proportional to ξ| the estimates in (4.2) follow easily from this, (4.5) and the symbolic estimates satisﬁed by a.Toprovetheesti- mates (4.10) it evidently suﬃces to prove analogous results for the homogeneous terms in the asymptotic expansion of the symbols of Q and ∂Q . Handling the remainder terms is an easy ∂νy exercise which we leave to the reader. We would like to use the simpler form of the Sobolev norms available for Rn+1 (as opposed n n+1 to R × [0, ∞) . To that end we extend k(x0,x; ξ)toallofR in such a way that the estimates, (4.2), continue to hold. This can be accomplished using the Seeley extension theorem, (see Melroses notes). Choosing constants {λp} we set: X∞ e p k(x0,x; ξ)= λpk(−2 x0,x; ξ) p=1 24 CHARLES EPSTEIN where we can suppose that k is compactly supported in (x0,x). There exist constants, {Cs} such that for any s e kkkHs(Rn+1 ) ≤ CskkkHs(Rn+1 ). Let Z

b i(x0·ζ0+x·ζ)e k(ζ,ξ)= e k(x0,x; ξ)dx0dx . Rn+1 We use Lemma 4.1 to obtain estimates for kb:

Lemma 4.2. For al l a, b ∈ M there are constants, Ca,b so that: |ζ | −b (4.6) bk(ζ,ξ) ≤ C(1 + |ξ|)2(1 + |ζ|)−a 1+ 0 . 1+|ξ|

Proof: We use the previous lemma and take q>aand p>1 to conclude that α αe j+1+α −q −p |∂ D k(x ,x; ξ) ≤ C(1 + |ξ|) 0 (1 + |x|) (1 + |x |(1 + |ξ|) . x0 x 0 0 Therefore Z ∞ −p | α0 αb |≤ | | j+1+α0 | | | | ≤ | | j+α0 ξ0 ζ k(ζ,ξ) C(1 + ξ ) 1+ x0 (1 + ξ ) dx0 C(1 + ξ ) . −∞ This easily implies (4.6). Let Z ix·ξe u(x)= e k(x0,x; ξ)vb(ξ)dξ .

Because

ku kHs(Rn×[0,∞)) ≤ CkukHs(Rn+1 ) R×[0,∞) is suﬃces to show that k k ≤ k k (4.7) u s−j+ 1 C v Hs(Rn) . H 2 (Rn+1 ) Using the weak formulation of the norm we see that it suﬃces to show that

|hu, φi| ≤ Ckvkskφk 1 − , j+ 2 s for all φ ∈ C∞(Rn+1). Note that c Z b ub(y)= k(y0,y− ξ, ξ)vb(ξ)dξ and so by the Plancherel theorem: Z b b hu, φi = k(y0,y− ξ, ξ)vb(ξ)φ(−y0, −y)dy0dydξ and therefore:

|hu, φi| ≤ Ckφk− 1 kV k0 where s+j+ 2 Z b s−j− 1 V (y0,y)= |k(y0,y− ξ; ξ)|(1 + |y0| + |y|) 2 |vb(ξ)|dξ .

Thus we need to prove that

kV k0 ≤ Ckvks . We replace kb by its estimate, (4.6), and deﬁne Z − − 1 − |y | − | | j | | | | s j 2 | − | r 0 q|b | W (y0,y)= (1 + ξ ) (1 + y0 + y ) (1 + y ξ ) (1 + 1+|ξ| ) v(ξ) dξ .

Evidently |v(y0,y)|≤|w(y0,y)| and so it suﬃces to show that

kW k0 ≤ CkV ks . ELLIPTIC BOUNDARY VALUE PROBLEMS 25

y0 We let | | = ye0 and obtain that 1+ y Z 1 2 k k2 | |2 2 | | W 0 = (1 + y ) W (y, (1 + y )y0) dydy0 . We use Peetre’s inequality: 2 |s| 2 s 2 |s| (1 + |ξ| )s ≤ 2 (1 + |y| ) (1 + |ξ − y| ) to obtain that: 1 j−s s−j− 1 −r (1 + |y|) 2 (1 + |ξ|) (1 + (1 + |y|)y0 + |y|) 2 (1 + |y − ξ|) ≤ | | s−j+ 1 −r+|j−s| y0 −q C(1 + |y |) 2 (1 + |y − ξ|) (1 + ) 0 1+|y − ξ| so that Z − − 1 k k2 ≤ | |2 s j 2 k · k2 W 0 C (1 + y0 ) γ( ,y0) 0dy0 where Z q−r+|j−s| −q s γ(y, y0)= (1 + |ξ − y|) (1 + |ξ − y| + |y0|) (1 + |ξ|) |vb(ξ)|dξ . In fact by the Cauchy Schwartz inequality: k · k2 ≤ k k2 2 γ( ,y0) 0 C V gS (y0) where Z 2 q−r+|j−s| −q S (y0)= (1 + |y|) (1 + |y0| + |y|) dy .

− − − 1 | − | We choose q>n 1, q>s j + n 2 and r> j s + q to obtain that −q+n−1 S(y0) ≤ C(1 + |y0|) and therefore: Z − − 1 − − k k2 ≤ k k2 k k2 s j 2 q+n 1 W 0 C V s (1 + y0 ) dy0 ≤ ek k2 C V s . This completes the proof. The proof given here comes essentially verbatim from the book of Chazarain and Piriou. We apply the estimate (4.7) with j = −2forSe and j = −1forDe to obtain the estimates (4.10). We apply these results to prove estimates for the solution of the boundary value problems considered in the previous section.

If u = Df −Sg for f, g ∈ C∞(bΩ) then (4.1) implies that k k ≤ k k k k (4.8) u Hs(Ω) C( f s− 1 + g s− 3 ) . H 2 (bΩ) H 2 (bΩ) ∞ s− 1 s− 3 If {fn,gn}⊂C (bΩ) are sequences which converge in H 2 (bΩ) and H 2 (bΩ) to (f, g) s respectively, then un = Dfn −Sgn converges in H (Ω) to u = Df −Sg.

If s> 1 then u has a well deﬁned restriction to bΩandifs> 3 then ∂u is also well deﬁned. 2 2 ∂ν

From (2.26) we see that the solution of the Dirichlet problem: (∆u =0inΩ,u = f)is bΩ given by D S + −1 − + (4.9) u = f + (s0 ) (I D0 )f. s s+ 1 If f ∈ H (bΩ), for s>0thenu ∈ H 2 (Ω) satisﬁes ∆u =0 inΩ

u = f. bΩ 26 CHARLES EPSTEIN

The boundary condition has to be interpreted in the sense of traces, i.e. in terms of the bounded linear map

s s− 1 1 H (Ω) → H 2 (bΩ) s> . 2 Note that (4.8) and (4.9) imply that k k ≤ k k u Hs(Ω) C f s− 1 . H 2 (bΩ) For the more general elliptic boundary value problems, (P ) where we need to solve ( b ∆gu =0 inΩ (P ) b ∂u b0(u )+b1 ∂ν = f bΩ k k−1 s where b0 ∈ Ψ (bΩ),b1 ∈ Ψ (bΩ). If for f ∈ H (bΩ) there is a solution u then u ∈ s+ 1 +k H 2 (Ω). In general this type of boundary value problem makes sense for s + k>0.

Theorem 4.1. If (Pb) is an elliptic boundary value problem then there is a ﬁnite number of ∞ functions {ψ1,...,ψm}⊆C (bΩ) such that for s>−k the problem (Pb) has a solution u (in the weak sense) for all f ∈ Hs(bΩ) which satisfy

hψi,fi =0 i =1,...,m. The solution u satisﬁes the estimates: k k ≤ k k (4.10) u s+ 1 +k C f Hs (bΩ) . H 2 (Ω)

We have proven this theorem for a compact domain with smooth boundary in a manifold. The operator, ∆g is the Laplacian defined by a smooth metric. This is just an example of the sort of results which can be proven by this method for systems of elliptic differential and pseudodifferential equations on domains in manifolds. Beyond this there is a theory of 1 subelliptic boundary value problems. In this theory the 2 appearing on the left hand side of − 1 1 (4.10) is replaced by a number in the interval [ 2 , 2 ]. These results can also be proven using pseudodifferential methods, see M. Taylor, Pseudodifferential Operators.

Department of Mathematics, University of Pennsylvania

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