THE DIRICHLET PROBLEM FOR THE LAPLACE OPERATOR

Transcription

1 THE DIRICHLET PROBLEM FOR THE LAPLACE OPERATOR Stefano Meda Università di Milano-Bicocca c Stefano Meda 2013

2 ii A Francesco

3 Contents I The Dirichlet problem via Perron s method 1 1 The classical Dirichlet problem Introduction Separation of variables Background and preliminary results Subharmonic functions Maximum principle and uniqueness Green s function The Poisson kernel for the half space The Poisson kernel for the ball The two dimensional case The higher dimensional case Perron s method The Lebesgue spine II The Dirichlet problem via integral equations 55 2 Linear operators on Banach spaces Basic definitions Bounded operators The spectrum of a linear operator The adjoint of a bounded operator iii

4 iv CONTENTS 2.5 Bilinear forms Unitary operators and projections Compact operators The spectra of compact operators The spectral theorem Dirichlet via integral equations Kernels of type α The double layer potential The single layer potential Solvability Full proofs for generic C 2 hypersurfaces III The Dirichlet problem via L 2 methods Introduction to Dirichlet s principle Why weak solutions? Plan of Part III Details left behind Distributions and Sobolev spaces Continuous functions and measures Smooth functions and distributions Derivatives of distributions Fundamental solutions Sobolev spaces Dirichlet L Poincare s inequality Solution to the minimization problem The dual of H0(Ω) Weak solutions and regularity

5 Part I The Dirichlet problem via Perron s method 1

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7 Chapter 1 The classical Dirichlet problem 1.1 Introduction Suppose that n 2, and that Ω is a bounded open set in R n. We denote by the Laplace operator, which acts on a function f in C 2 (Ω) by f(x) = n j 2 f(x) x Ω. j=1 The classical Dirichlet problem is the following: given a continuous function g on Ω, find a function u in C 2 (Ω) C(Ω) such that { u = 0 in Ω u Ω = g. This is a very challenging problem, which has been considered by many outstanding mathematicians in the past two centuries. The reader is referred to the interesting article [G] of L. Garding for an historical account of the research on the Dirichlet problem until the first half of the twentieth century. As we shall see, it is comparatively easy to prove that if a solution to the Dirichlet problem exists, then it is unique. By contrast, it is entirely nontrivial to prove that, under suitable assumptions on the domain Ω, the Dirichlet problem is solvable. In some textbooks, mainly bounded domains with C 2 boundary are considered. From the point of view of applications, this assumption is far inadequate. Indeed, Dirichlet problems in a square, or in domains of the plane with polygonal boundary are quite common, if not paradigmatic. Thus, it seems 3

8 4 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM reasonable to focus on theories which at least cover domains with Lipschitz boundary. In this chapter, we take up the classical theory of harmonic functions on a very general class of domains, which include Lipschitz domains. Then we will illustrate Perron s method of solving the Dirichlet problem and give a characterisation of the domains in R n where the Dirichlet problem is solvable. We mainly follow [GT, Chapter 2]. There are many problems in Mathematics and in Physics that lead to consider the Dirichlet problem. Here we recall two of them. The first is concerned with the heat diffusion in a body Ω R n (n = 2, 3 are the most important cases). Assume that a fixed temperature distribution at the boundary is maintained by a heating and refrigeration system. By suitably normalising the physical constants involved, we may assume that the temperature u(x, t) of the point x Ω at time t satisfies the following equation (known as the heat equation) t u(x, t) u(x, t) = 0 x Ω t > 0, (1.1.1) where the Laplacian acts on the x variable. Denote by g(x) the temperature at which the heating and refrigeration system keeps the point x Ω. Then we must have u(x, t) = g(x) x Ω t > 0. (1.1.2) Of course, the body has an initial temperature u(x, 0) at each point x Ω. Given that the boundary temperature is kept at a steady state, it is plausible that the system will evolve towards an equilibrium state u(x), that is u(x) = lim t u(x, t). Since for all T > 0 the function u T (x, t) := u(x, T + t) satisfies the boundary value problem (1.1.1) (1.1.2), it is reasonable to expect that the same will happen to u. Since u does not depend on t, u will satisfy { u = 0 in Ω u Ω = g. The second problem leading to the Dirichlet boundary value problem is internal to Mathematics. Riemann tackled the problem of constructing a conformal mapping ϕ between a proper simply connected domain Ω in

9 1.2. SEPARATION OF VARIABLES 5 the complex plane and the unit disc D := {z C : z < 1}. By this we mean that ϕ is a holomorphic bijection between Ω and D. Suppose that Ω has smooth boundary. We look for a homeomorphism ϕ : Ω D, which, restricted to Ω, is a biholomorphism between Ω and D. We may assume that 0 Ω. By possibly composing ϕ with a Möbius transformation, we may assume that ϕ(0) = 0. If such a conformal mapping ϕ exists, then z ϕ(z)/z is nonvanishing (recall that ϕ (z) 0 for all z Ω, for ϕ is conformal), whence there exists a holomorphic function f on Ω such that ϕ(z) z = e f(z) z Ω. Since the left hand side extends to a continuous map on Ω that necessarily maps Ω onto D, we have that ϕ(z) = 1 for every z Ω. Therefore log ϕ(z) = log z + Re f(z) z Ω and Re f(z) = log z z Ω. Clearly Re f is harmonic in Ω, for it is the real part of a holomorphic function, and it is a solution to the boundary value problem { u(z) = 0 in Ω u(z) Ω = log z. 1.2 Separation of variables It is quite rare to be able to find explicit solutions to Dirichlet problems. However, there are a few exceptions. In particular, when the domain Ω has suitable geometric features, the method of separation of variables, due to Daniel Bernoulli, may be successfully employed. Here we consider just a basic example, which however, illustrates quite clearly the gist of the method. Denote by R the strip {(x, y) R 2 : 0 x π, y 0}. Assume that the stationary temperature u(x, y) is continuous and bounded and satisfies the following boundary conditions u(0, y) = 0 = u(π, y) y > 0, u(x, 0) = f(x) x [0, π], (1.2.1)

10 6 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM where f is an assigned continuous function. We seek functions of the form u(x, y) = X(x) Y (y) that satisfy the Laplace equation and the boundary conditions on the vertical edges of the strip. We substitute u in the Laplace equation, and obtain X = Y X Y. Since the left hand side depends only on x and the right hand side depends only on y, neither can depend on either. Hence both are equal to a constant, which we write as c, where c is, for the time being, a complex number. Then, we are led to solve the ordinary differential equations X + c X = 0 and Y c Y = 0. Furthermore, X must be a solution to the following boundary value problem X + c X = 0 and X(0) = 0 = X(π). (1.2.2) It is straightforward to check that this problem has a nontrivial solution only if c is of the form k 2, where k is a positive integer (prove this!). The solutions of (1.2.2) are then all the multiples of sin(kx). The function Y must be a solution to the following problem Y k 2 Y = 0 and Y bounded on [0, ). (1.2.3) All the solutions to this problem are multiples of e ky. Thus, we are led to consider the functions They satisfy u k (x, y) := e ky sin(kx). u k = 0, u k (0, y) = 0 = u k (π, y) y > 0. Unless the datum f is one of the functions sin(kx), none of the functions u k will match the boundary condition u k (x, 0) = f(x). The idea to circumvent this difficulty is to consider superpositions of the functions u k, i.e., to see whether functions of the form c k e ky sin(kx) (1.2.4) k=1 satisfy the given Dirichlet problem. Of course, this idea is suggested by the fact that the Laplace operator v v is linear. At least formally, if the

11 1.3. BACKGROUND AND PRELIMINARY RESULTS 7 function above satisfies the Dirichlet problem, then its value at (x, 0) must be equal to f(x). In other words, f should have the following Fourier sine series expansion f(x) = c k sin(kx). k=1 It is a nontrivial fact, proved long after the appearance of Fourier s original paper (1822), that the sequence { (2/π) 1/2 sin(kx) } is a complete orthonormal system in L 2 ((0, π)). We shall come back to this fact later. This will force c k to be the k th Fourier coefficient of the sine expansion of the function f. Note that if f is the restriction to [0, π] of a π periodic function in C 2 (R) such that f(0) = 0, then c k = O(k 2 ), and the series in (1.2.4) converges uniformly on [0, π] [0, ). Therefore its sum is a continuous function on R that matches the boundary values. Furthermore, the sum of the series is infinitely many times differentiable at all the points in R with y > 0, and satisfies the Laplace equation therein. Exercise Prove all the assertions above on Fourier sine series, except for the completeness of the system { (2/π) 1/2 sin(kx) }. Exercise Show that the method of separation of variables for the Dirichlet problem discussed above leads, in the case where f(x) = 1, to the solution u(x, y) := 4 [ e y sin x + 1 π 3 e 3y sin(3x) + 1 ] 5 e 5y sin(5x) + = 2 sin x arctan π sinh y. Show that u satisfies the boundary conditions except at the corners. Draw the isothermals for small x and y. Hint. To find the sum of the series within square brackets, write z = x +iy and observe that the series is the imaginary part of a power series in the variable e iz, whose sum is (1/2) log[(1+e iz )/(1 e iz )]. 1.3 Background and preliminary results We briefly review some definitions and results from advanced calculus. A subset of R n is a domain if it is an open connected set.

12 8 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM A subset S of R n is a hypersurface of class C k if for every x 0 S there exists an open subset V of R n containing x 0 and a real valued function φ C k (V ) such that φ does not vanish on S V and S V = { x V : φ(x) = 0 }. For the sake of definiteness, suppose that n φ does not vanish on S V. Then, by the implicit function theorem, there exists a C k function ψ such that x n = ψ ( x 1,..., x n 1 ) for all ( x 1,..., x n 1, x n ) in S V. For convenience, denote by x the point (x 1,..., x n 1 ) in R n 1. Note that the map (x, x n ) ( x, x n ψ(x ) ) maps V S onto a neighbourhood of the point x 0 of the hyperplane x n = 0. The inverse of the map x ( x, ψ(x ) ), which is defined in a suitable neighbourhood of x 0, gives a local chart of S around x 0. Now, for every x S, the vector φ(x) is orthogonal to S (i.e., it is orthogonal to the hyperplane tangent to S at x 0 ). We shall assume that S is oriented, i.e. there is a choice of a unit normal vector ν(x) at each point x of S that varies continuously with x. In particular ν(x) = ± φ(x) φ(x). This formula shows that the normal field x ν(x) is of class C k 1 on S. Definition We say that a domain Ω has C k boundary, or that Ω is of class C k, if Ω is a hypersurface of class C k. Definition Suppose that Ω is an open set, and that k is a positive integer. We denote by C k (Ω) the vector space of all functions f in C k (Ω) such that D α f extends to a continuous function on Ω for all multiindices of length k. We recall the classical divergence theorem.

13 1.3. BACKGROUND AND PRELIMINARY RESULTS 9 Theorem Suppose that Ω is a bounded domain with C 1 boundary, and denote by ν the unit outward normal to Ω. Suppose that w C 1 (Ω). Then div w dv = w ν dσ, where σ denotes the surface measure of Ω. Ω In the applications, the boundary does not always satisfy the assumptions of Theorem There is a generalisation of the divergence theorem that covers domains with a very general boundary. It is beyond the scope of these notes to deal with such generalisations, for which the reader is referred to the book of Evans and Gariepy [EG]. However, as explained in the Introduction, it is important to deal with a class of domains that include, at least, cubes, cylinders and other simple domains with edges. We shall focus on the so called Lipschitz domains. Recall that a map ϕ : R n 1 R is Lipschitz if there exists a constant L such that ϕ(x) ϕ(y) L x y x, y R n 1. Ω The infimum of all the constants L such that the above inequality holds is called the Lipschitz constant of ϕ. Lipschitz functions are possibly not differentiable at some points (for instance, x x is Lipschitz with constant 1, but it is not differentiable at the origin), but the set where it is not differentiable is small in the measure theoretic sense. In fact, the following nontrivial result holds. For its proof we refer the reader to [EG]. Theorem (Rademacher) Suppose that ϕ is Lipschitz. Then the set of points where ϕ is not differentiable is of null measure. Suppose that y is a point in the n-dimensional Euclidean space. A cylinder Γ centred at a point y is a set, which, in a suitable system of coordinates (x, x n ) (x is an (n 1)-dimensional vector) centred at y, is of the form B I, where B is a ball with centre 0 in the hyperplane {(x, x n ) R n : x n = 0}, and I is an interval of the real line {(x, x n ) R n : x = 0}. We say that the coordinate system (x, x n ) above is adapted to the cylinder Γ. Definition property: A Lipschitz domain is a domain Ω with the following

14 10 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM for each y Ω there exists a cylinder Γ centred at y, which, may be written as B I with respect to a suitable system of coordinates (x, x n ) centred at y, and a Lipschitz map ϕ : B R, with ϕ(0 ) = 0, such that (i) Ω Γ = {(x, x n ) Γ : x n = ϕ(x ) for all x in B } (ii) Ω Γ = {(x, x n ) Γ : x n < ϕ(x ) for all x in B }. A consequence of the definition and of Rademacher s theorem is that a Lipschitz domain Ω admits a well defined outward unit normal ν at almost every point of Ω (with respect to the surface measure). The surface measure of Ω has the following expression in terms of the local coordinates (x, ϕ(x )) in a neighbourhood of the point y (see Definition for the notation) dσ(x, ϕ(x )) = 1 + ϕ(x ) 2 dx. The following generalised divergence theorem holds. Theorem Suppose that Ω is a bounded domain with Lipschitz boundary, and denote by ν the unit outward normal to Ω, which is defined at almost every point of the boundary. Suppose that w C 1 (Ω). Then Ω div w dv = where σ denotes the surface element of Ω. Ω w ν dσ, In some cases the assumption that w be in C 1 (Ω) is too strong. We shall need the following corollary, whose proof follows a strategy which occurs in many other similar situations. Given a domain Ω, denote by Cb 1 (Ω) the subspace of C 1 (Ω) of all functions which are bounded and have bounded first order partial derivatives. Corollary Suppose that Ω and ν are as in the theorem above, and that w Cb 1(Ω) C(Ω). Then div w dv = w ν dσ, Ω Ω where σ denotes the surface element of Ω.

15 1.3. BACKGROUND AND PRELIMINARY RESULTS 11 Proof. To each y Ω we associate a cylinder Γ(y) with centre in y. In particular, if y Ω, we choose Γ(y) of the form B I, so that the part of the boundary in Γ(y) is the graph of a Lipschitz function defined on B. Clearly {Γ(y) : y Ω} is a convering of Ω. Since Ω is compact, we may extract a finite subcovering Γ(y 1 ),..., Γ(y M ) of Ω. Another compactness argument shows that we may obtain a finite cover of Ω by Γ(y 1 ),..., Γ(y M ), and, possibly, of other cylinders Γ(y M+1 ),..., Γ(y N ), which may be chosen so that they have empty intersection with Ω. We denote by ψ 1,..., ψ N a smooth partition of unity associated to the chosen covering. Thus, N ψ j = 1 on Ω. j=1 As a consequence, div w dv = Ω = N div(ψ j w) dv j=1 N Ω j=1 Ω Γ(y j ) div(ψ j w) dv. The last equality follows from the fact that the support of ψ j is contained in Γ(y j ). We distinguish two cases, according to the fact that Γ(y j ) has empty or nonempty intersection with Ω, equivalently M + 1 j N or 1 j M. In the first case, Ω Γ(y j ) = Γ(y j ), which is, of course, a Lipschitz domain. Furthermore, ψ j w C 1 (Γ(y j )), and it vanishes on Γ(y j ). Hence, by Theorem 1.3.6, div(ψ j w) dv = 0. Γ(y j ) In the second case, we argue as follows. Fix j, consider the coordinate system with centre y j, adapted to the cylinder Γ(y j ), and denote by ϕ j the corresponding Lipschitz map. Then Ω Γ(y j ) div(ψ j w) dv = B j = lim ε 0 B j ϕj (x n) dx div(ψ j w)(x, x n ) dx n ϕj (1.3.1) (x n) ε dx div(ψ j w)(x, x n ) dx n ;

16 12 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM the last equality is a consequence of the dominated convergence theorem, which may be applied because div(ψ j v) is bounded, hence integrable, on Γ(y j ). For ε small, denote by A ε j the set {(x, x n ) Ω Γ(y j ) : x n < ϕ j (x ) ε}. Observe that ϕj (x n) ε dx div(ψ j w)(x, x n ) dx n = B j A ε j div(ψ j w) dv. (1.3.2) Clearly, A ε j is a Lipschitz domain, and ψ j w C 1 (A ε j ). Then, by Theorem 1.3.6, div(ψ j w) dv = ψ j w ν dσ A ε j A ε j ( ) ψ j w ν dσ Ω Γ(y j ) (1.3.3) as ε tends to 0, because ψ j w is continuous on Ω, hence on Ω Γ(y j ). Observe that ( ) ψ j w ν dσ = ( ) ψ j w ν dσ, Ω Γ(y j ) Ω Γ(y j ) because ψ j vanishes on the boundary of Γ(y j ). By combining this, (1.3.1), (1.3.2) and (1.3.3), and summing over j, we obtain the desired formula. The following corollary of the divergence theorem will be extensively used in the sequel. Corollary Suppose that Ω is a bounded Lipschitz domain, and that u and v are in C 2 (Ω). The following hold: (i) (ii) ( first Green s identity) v u dv + Ω Ω u dv = (iii) ( second Green s identity) (v u u v) dv = Ω Ω Ω ν u dσ; v u dv = Ω Ω v ν u dσ; ( v ν u u ν v ) dσ;

17 1.3. BACKGROUND AND PRELIMINARY RESULTS 13 (iv) ( integration by parts) v j u dv + Ω Ω u j v dv = Ω u v ν j dσ. Proof. To prove (i), just take w = u in the divergence theorem. To prove (ii), just take w = v u in the divergence theorem. By interchanging the role of u and v in (ii), and subtracting the resulting equalities, we get (iii). Finally, (iv) follows from the divergence theorem by taking w to be a vector field, all of whose components vanish but the j th, which is equal to uv. Observe that statements (i)-(iv) above sometimes hold under less restrictive assumptions that u, v C 2 (Ω). For instance, by using Corollary 1.3.7, we see that (iv) holds under the assumption that u, v Cb 1 (Ω) C(Ω). We leave to the reader the task of reformulating similar conditions for statements (i)-(iii). Denote by ω n the surface measure of the sphere B 1 (0) in R n. Various formulae will involve ω n and the Lebesgue measure of the ball B 1 (0). Some of the exercises below describe an elegant way to compute these quantities. We recall the following formula of integration in polar coordinates. Proposition Suppose that f is an integrable function on R n. Then f dv = dr r n 1 f(rω) dσ(ω) R n 0 S n 1 = dr f(ω ) dσ(ω ). 0 B r Exercise Prove that the measure of the unit ball in R n is ω n /n. Hint: use the divergence theorem for an appropriate vector field. Exercise By following the steps below, prove that the measure of the unit ball in R n is π n/2 /Γ(n/2 + 1) (here Γ denotes Euler s Gamma function): (i) show that if a > 0, then R n e a x 2 dx = ( π/a ) n/2 ;

18 14 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM (ii) compute by computing I := R p ( θ 1,..., θ n 1 ) dθ1 dθ n 1, R n e a x 2 with (i), and using polar coordinates; (iii) compute V ( B 1 (0) ), by integrating the characteristic function of B 1 (0) in polar coordinates; (iv) conclude that the surface measure of B 1 (0) is 2π n/2 /Γ(n/2). Exercise Suppose that Ω is a domain in R n. Prove that the divergence operator is the adjoint of the gradient, in the sense that for every smooth vector field w and every smooth function ϕ with compact support contained in Ω ϕ div w dv = ϕ w dv. Ω Exercise Suppose that φ is a C 2 diffeomorphism between the domains Ω and φ(ω), that f is a function in C 2 (φ(ω)), and that F C 2 (φ(ω); R n ). We denote by x the variable in Ω, by y that in φ(ω), and by J(x) the determinant of the differential map φ (x). Prove the following change of variables formulae (in these formulae y = φ(x), and the gradient of a scalar function is a row vector): (i) y f(y) = x ( f φ ) (x) φ (x) 1 ; Ω dx (ii) div y F(y) = 1 J(x) div [ x J (F φ) [(φ ) t ] 1] (x); (iii) y f(y) = 1 J(x) div [ x J x (f φ) ( (φ ) t φ ) 1 ] (x); (iv) if φ is an orthogonal transformation, then the columns of φ (x) are of the form a j u j, where u 1,..., u n are orthonormal vectors. Then φ (x) t φ (x) = diag(a 2 1,..., a 2 n) and J(x) = a 1 a n. Prove that y f(y) = 1 a 1 a n div x ( a2 a n a 1 1 (f φ),..., a 1 a n 1 a n ) n (f φ) (x);

19 1.3. BACKGROUND AND PRELIMINARY RESULTS 15 (v) prove that polar coordinates in R 2 are associated to an orthogonal transformation, and compute the Laplacian in polar coordinates by using (iii) above, or by specialising the formula found in (iv); (vi) suppose that φ is a conformal mapping, i.e., that φ (x) = ϱ(x) U(x), where ϱ(x) is a scalar factor and U(x) is an orthogonal matrix. Prove that y f(y) = 1 ϱ(x) div [ n x ϱ n 2 x (f φ) ] (x); (vii) prove that the inversion map y = x/ x 2 is a conformal mapping between appropriate domains, with conformal factor x 2. Compute the Laplace operator with respect to the new coordinates; (viii) suppose further that Ω is a domain in C and that φ : Ω φ(ω) is a one-to-one holomorphic function. Prove that where w = φ(z). f(w) = φ (z) 2 (f φ)(z), Hint: Exercise may be helpful to prove (ii). Exercise Suppose that u is a smooth function on R 2 \ {0}. Prove that u(r, θ) = r 2 u + 1 r ru + 1 r 2 2 θu. Then find all radial solutions to the equation u = 0 in R 2 \ {0}. Exercise Suppose that u is a smooth function on R 3 \ {0}. Prove that u(r, θ, ψ) = r 2 u + 2 r ru + 1 [ 1 ] r 2 (sin ψ) 2 2 θu + ψu 2 + cot ψ ψ u. Then find all radial solutions to the equation u = 0 in R 3 \ {0}. Prove also that the differential operator within square brackets in the formula above is the Laplace Beltrami operator on the sphere S 2 with respect to the Riemannian metric induced by the standard Euclidean metric of R 3. Exercise Show that commutes with all the rotations of R n. Deduce that if u is a smooth radial function in R n, then so is u. Exercise Prove that if P (D) is a differential polynomial that commutes with all rotations of R n, then P (D) is, in fact, a polynomial in.

20 16 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM 1.4 Subharmonic, superharmonic and harmonic functions Observe that preserves the class of real functions and of purely imaginary functions. Definition Suppose that Ω is a domain in R n. A real valued function u C 2 (Ω) is subharmonic (superharmonic) if u 0 ( 0) in Ω. A function u in C 2 (Ω) is harmonic if u = 0. Note that u is harmonic if and only if the real and the imaginary part of u are harmonic. Exercise Prove that the real and the imaginary parts of a holomorphic function in a domain Ω are harmonic in Ω. Prove that the functions x 2 y 2 and xy are harmonic in R 2, and so are the functions r j cos(jθ) and r j sin(jθ) for each nonnegative integer j. Exercise Suppose that u is harmonic in R n \ {0}. Prove that the function u (x) = x 2 n u ( x/ x 2) is harmonic in R n \ {0}. The map u u is called the Kelvin transform. Hint: Exercise (vii) may be helpful. Exercise Suppose that Ω is a domain in the complex plane, that φ : Ω φ(ω) is a one-to-one holomorphic mapping, and that f is harmonic in φ(ω). Prove that the function f φ is is harmonic in φ(ω). Hint: Exercise (viii) may be useful. Exercise Prove that if u is harmonic in Ω, and Ω Ω, then ν u dσ = 0. Ω Conversely, show that if u C 2 (Ω) satisfies ν u dσ = 0 for every ball B Ω, then u is harmonic in Ω. B

21 1.4. SUBHARMONIC FUNCTIONS 17 Exercise For which values of α is the function x x α subharmonic? Answer: α(α + n 2) 0. This shows that if n 2, then a convex function need not be subharmonic. Exercise In this exercise we discuss subharmonic functions on intervals of the real line. (i) Find all harmonic functions on an interval I of R; (ii) give a characterisation of subharmonic functions on I in terms of convexity; (iii) prove that a continuous real function ϕ on an open interval I is convex if and only if ( x + y ) ϕ ϕ(x) + 1 ϕ(y) x, y I. (1.4.1) 2 (iv) enlarge the definition of subharmonic function, by requiring that f is subharmonic in I if and only if f is continuous in I and satisfies the mean value inequality on every interval [a, b] I, i.e. if (1.4.1) holds. Conclude that a continuous function u on I is subharmonic if and only if for every interval [a, b] I u(x) H a,b (x) x [a, b], where H a,b is the harmonic function on [a, b] such that H a,b (a) = u(a) and H a,b (b) = u(b). The next result relates subharmonic, superharmonic and harmonic functions with their spherical and solid means over balls. In fact, these kind of functions may be characterised by the behaviour of their means, as we shall see later. For a ball B, we denote by σ( B) the surface measure of B and by V (B) the Lebesgue measure of B. Theorem (Mean value inequalities) Suppose that Ω is a domain. The following hold: (i) for every subharmonic function u and every ball B Ω u(y) 1 u dσ and u(y) 1 σ( B) V (B) B B u dv ;

22 18 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM (ii) for every superharmonic function u and every ball B Ω u(y) 1 u dσ and u(y) 1 u dv ; σ( B) V (B) B (iii) for every harmonic function u and every ball B Ω u(y) = 1 u dσ and u(y) = 1 σ( B) V (B) B B B u dv. Proof. We shall prove (i). The proof of (ii) is similar and is omitted. Part (iii) is a straightforward consequence of (i) and (ii). To prove (i), suppose that B = B R (y), that 0 < ρ < R, and observe that, by Corollary (i), 0 ν u dσ B ρ(y) = u(y + ω ω ) B ρ(0) ω dσ(ω ) = ρ n 1 u(y + ρω) ω dσ(ω) B 1 (0) = ρ n 1 d [ ] u(y + ρω) dσ(ω) B 1 (0) dρ = ρ n 1 d u(y + ρω) dσ(ω) dρ B 1 (0) = ρ n 1 d [ 1 ] u dσ. dρ ρ n 1 B ρ(y) Now we divide both sides by ρ n 1, integrate with respect to ρ between ε and R, and obtain 0 1 u dσ 1 u dσ. R n 1 B R (y) ε n 1 B ε(y) Since u is continuous in y, lim ε 0 1 ε n 1 B ε(y) u dσ = ω n u(y). By combining the last two formulae, we obtain that 1 u(y) u dσ, ω n R n 1 B R (y)

23 1.4. SUBHARMONIC FUNCTIONS 19 which is equivalent to the first formula in (i). To prove the second formula in (i), we multiply by R n 1 both sides of the inequality above and integrate with respect to R between 0 and r. We obtain r n n u(y) 1 r dr u dσ, ω n which is equivalent to the required inequality. 0 B R (y) It is an important fact that a converse of Theorem holds. In its proof we shall use the fact that if φ is a smooth compactly supported function, and f is locally integrable in R n, then the convolution φ f of φ and f, defined by φ f(x) = R n φ(x y) f(y) dy x R n, is a smooth function. For a proof of this fact, and further properties of convolution, the reader may consult [Me, Section 2.1]. Theorem Suppose that u is a continuous function in the domain Ω and that for every ball B Ω the following holds: u(c B ) = 1 u dσ, σ( B) where c B denotes the centre of B. Then u is harmonic in Ω. Proof. Denote by φ a smooth radial function with support contained in the ball B 1 (0) such that φ dv = 1, and denote by ψ its profile, i.e., ψ( x ) = φ(x). For every ε > 0, set φ ε (x) = ε n φ(x/ε). Clearly the support of φ ε is contained in B ε (0). If x belongs to B Ω ε := {x : B ε (x) Ω}, then the support of y φ ε (x y) is contained in Ω. Observe that u(y) φ ε (x y) dy = u(x εy) φ(y) dy R n y <1 1 (polar coordinates) = u(x εry ) ψ(r) r n 1 dσ(y ) dr (by assumption) (polar coordinates) 0 y =1 1 = ω n u(x) ψ(r) r n 1 dr 0 = u(x) φ dv R n = u(x).

24 20 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Now, the left hand side, being the convolution of u with the smooth function φ ε, can be differentiated infinitely many times, whence u is in C (Ω ε ). Since ε is arbitrary, u is in C (Ω). It remains to show that u is harmonic. Suppose that B r (x) Ω. By Corollary (i), u dv = ν u dσ B r(x) B r(x) ω = u(x + ω) ω =r ω dσ(ω) = r n 1 u(x + rω ) ω dσ(ω ) ω =1 = r n 1 d u(x + rω ) dσ(ω ). dr The last integral is equal to ω n 1 σ( B r (x)) ω =1 B r(x) u dσ, which, in turn is equal to ω n u(x), for u possesses the mean value property by assumption. Thus, we have proved that u dv = ω n r n 1 d dr u(x) = 0 B r(x) for every ball B r (x) such that B r (x) Ω. vanishes at every point in Ω, as required. Since u is continuous, u Exercise Show that if u is a function in C 2 (Ω) and x 0 Ω, then 2n [ 1 ] u(x 0 ) = lim u(x r 0 r 2 0 ) u(x σ(s n r ω) dσ(ω). ) S n 1 Deduce that if u satisfies the mean value inequality of Theorem (i), then u is subharmonic. 1.5 Maximum principle and uniqueness for the Dirichlet problem In this section we derive some consequences of the mean value inequalities proved in the previous section. In particular, we shall prove the maximum

25 1.5. MAXIMUM PRINCIPLE AND UNIQUENESS 21 and the minimum principles for harmonic functions. As a corollary we shall obtain that the classical Dirichlet problem on a domain Ω has, at most, one solution. The problem of the existence of solutions to the Dirichlet problem will be addressed later. Theorem Suppose that Ω is a domain in R n, and that u C 2 (Ω). The following hold: (i) (strong maximum principle) if u is subharmonic and there exists a point y Ω such that u(y) = sup Ω u, then u is constant; (ii) (strong minimum principle) if u is superharmonic and there exists a point y Ω such that u(y) = inf Ω u, then u is constant; (iii) if u is harmonic, then u cannot assume an interior maximum or minimum, unless it is constant. Proof. We prove (i). The proof of (ii) is similar and is omitted. Part (iii) is a direct consequence of (i) and (ii). Set M := sup Ω u, and Ω M := {x Ω : u(x) = M}. Clearly Ω M contains y, and it is closed, for u is continuous. We shall prove that Ω M is open. It will follow that Ω M = Ω, for Ω is connected by assumption, i.e. u = M in Ω, as required. To prove that Ω M is open, choose z in Ω M. Observe that u M is subharmonic ( (u M) = u), so that, by the mean value inequality, 0 = u(z) M 1 (u M) dv 0 V (B) for every ball B with radius small enough. Therefore (u M) dv = 0. B If u were strictly less than M on an open subset of B, then the integral above would be strictly negative. Therefore u = M on B, and Ω M is open, as required. B Corollary Suppose that Ω is a bounded domain in R n, and that u C 2 (Ω) C(Ω). The following hold:

26 22 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM (i) (weak maximum principle) if u is subharmonic, then sup Ω u = sup u; Ω (ii) (weak minimum principle) if u is superharmonic, then (iii) if u is harmonic, then inf Ω u = inf Ω u; inf u u sup u. Ω Ω Proof. To prove (i), observe that sup Ω u = sup Ω u, for u is continuous in Ω. Now, if sup Ω u > sup Ω u, then there exists a point y in Ω for which u(y) = sup Ω u. Consequently u is constant in Ω by Theorem (i). Hence sup Ω u = sup Ω u, thereby contradicting the assumption. The proof of (ii) is almost verbatim the same as the proof of (i), and is omitted. Part (iii) follows directly from (i) and (ii). Exercise Suppose that u and v are in C 2 (Ω) C(Ω) and that u Ω = v Ω. Prove that if u is harmonic and v is subharmonic, then v u on Ω. This result justifies the term subharmonic. State and prove a corresponding result for superharmonic functions. A noteworthy consequence of Corollary is the following uniqueness result for the Dirichlet problem. Theorem Suppose that v and w are functions in C 2 (Ω) C(Ω) that solve the Dirichlet problem { u = f in Ω Then v = w. u Ω = g.

27 1.6. GREEN S FUNCTION 23 Proof. Set u = v w. Clearly u solves { u = 0 u Ω = 0. in Ω In particular, u is a harmonic function in C 2 (Ω) C(Ω) which vanishes on Ω. By Corollary (iii), u = 0 on Ω, i.e. v = w, as required. We wish to emphasize the fact that no claims are made concerning the existence of a solution to the Dirichlet problem. 1.6 Green s function In this section we establish an important representation formula for solutions of the Dirichlet problem, under the assumption that a solution exists. Definition The Newtonian potential in R n is the function N : R n \ {0} R, defined by 1 log x if n = 2 N(x) = 2π 1 (1.6.1) x 2 n if n 3. (2 n) ω n Given a point x in R n, the Newtonian potential with pole x is the function N(x ). Note that if n 3, then N is a negative function. By contrast, if n = 2, then N has not a definite sign. This simple fact will have far reaching consequences. The reason for the normalisation of N will be clarified below (see Definition 5.4.2). It will be proved later that N is a fundamental solution of the Laplace operator: it will play an important role in finding distributional solutions to the Poisson equation u = f, where f is a given datum. All this will be discussed in Section 5.4. Exercise Prove, by direct calculation, that N is harmonic in R n \ {0}.

28 24 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Exercise Prove that B r(x) ν N(x y) dσ(y) = 1, Exercise Recall that the Laplacian in R n may be written in polar coordinates as follows = 2 r + n 1 r r + 1 r 2 S n 1, where S n 1 is the Laplace Beltrami operator on the sphere S n 1 (it is a second order differential operator in the angular variables only). Find all radial harmonic functions in R n. We now establish Green s representation formula. Theorem (Green s representation formula) Suppose that Ω is a bounded Lipschitz domain and that u C 2 b (Ω) C1 (Ω). The following hold: (i) for every x in Ω [ ] u(x) = u(y) ν N(x y) ν u(y) N(x y) dσ(y) Ω + N(x y) u(y) dv (y); Ω (1.6.2) (ii) if the support of u is a compact set in Ω, then u(x) = N(x y) u(y) dv (y) x Ω; Ω (iii) if u is harmonic in Ω, then [ ] u(x) = u(y) ν N(x y) ν u(y) N(x y) dσ(y) x Ω; Ω (iv) suppose that x Ω and that there exists a function h x C 1 (Ω) which is harmonic on Ω and satisfies (h x ) Ω = N(x ) Ω. Denote by G(x, ) the function h x + N(x ). Then u(x) = u(y) ν G(x, y) dσ(y) + G(x, y) u(y) dv (y). Ω The normal derivative of G in the integral above is taken with respect to the variable y. Ω

29 1.6. GREEN S FUNCTION 25 Proof. We prove the theorem in the case where n 3. The case n = 2 is left to the reader. First we prove (i). Consider the domain Ω \ B r (x), for r small. Apply the second Green s identity (see Corollary (iii) and the comments at the end of the proof thereof) with N(x ) in place of v and Ω \ B r (x) in place of Ω. We obtain N(x y) u(y) dv (y) Ω\B r(x) [ = N(x y) ν u(y) u(y) ν N(x y) ] dσ(y) Ω [ + N(x y) ν u(y) u(y) ν N(x y) ] dσ(y). B r(x) The required formula will follow from this and the dominated convergence theorem once we prove that and that lim N(x y) ν u(y) dσ(y) = 0 (1.6.3) r 0 B r(x) lim u(y) ν N(x y) dσ(y) = u(x). (1.6.4) r 0 B r(x) To prove (1.6.3), note that N(x y) C r 2 n and that y B r (x) ν u(y) max u(y), y Ω which is finite, because u C 1 (Ω). Thus, N(x y) ν u(y) dσ(y) C max u(y) r 2 n σ ( B r (x) ), B r(x) which tends to 0 as r tends to 0, for σ ( B r (x) ) r n 1, as required. y Ω To prove (1.6.4), write the integral in (1.6.4) as [ ] u(y) u(x) ν N(x y) dσ(y) + u(x) B r(x) Since u is smooth, B r(x) ν N(x y) dσ(y). u(y) u(x) r max u Ω y B r (x).

30 26 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Recall that N is homogeneous of degree 2 n, hence its partial derivatives are homogeneous of degre 1 n. Thus, ν N(x y) C r 1 n y B r (x). Therefore B r(x) [ u(y) u(x) ] ν N(x y) dσ(y) C r r 1 n σ ( B r (y) ), which tends to 0 as r tends to 0. Finally, by Exercise 1.6.3, ν N(x y) dσ(y) = 1, B r(x) which completes the proof of (i). Note that (ii) and (iii) are direct consequences of the representation formula established in (i). It remains to prove (iv). Write N(x y) = G(x, y) h x (y) in (1.6.2). We obtain [ ] u(x) = u ν G(x, ) ν u G(x, ) dσ + G(x, y) u(y) dv (y) Ω Ω + ν u h x dσ u ν h x dσ h x u dv. Ω Ω Note that, by the second Green s identity and the fact that h x is harmonic, ν u h x dσ = u ν h x dσ + h x u dv, Ω and the required formula follows. Ω The last part of the theorem above suggests that the function G defined therein may play an important role in the theory. This justifies the following definition. Definition Suppose that Ω is a bounded domain in R n. Assume that for every x Ω the Dirichlet problem { u = 0 in Ω u Ω = N(x ) is solvable, and that the solution h x is in C 1 (Ω). The function G : Ω Ω \ {(z, z) : z Ω}, defined by G(x, y) := h x (y) + N(x y), is called the Green function for the domain Ω. Ω Ω

31 1.6. GREEN S FUNCTION 27 Note that we do not assert the existence of a Green s function for a generic domain Ω. In fact, there are domains which do not admit a Green s function. Exercise Prove that the Green s function for a bounded domain Ω, if it exists, is unique. We explicitly state a straightforward but important consequence of Theorem (iv): a representation formula for the solution to the Dirichlet problem { u = 0 in Ω u Ω = g, under the assumption that a solution u exists and it is of class C 1 (Ω). Corollary Suppose that Ω is a bounded Lipschitz domain, which admits a Green s function G. If u C 1 (Ω) is a solution to the Dirichlet problem above, then u(x) = g(y) ν G(x y) dσ(y) x Ω. Ω Exercise Prove that Corollary holds under the weaker assumption that u C(Ω). Hint: imitate the strategy of the proof of Corollary Definition Suppose that G is the Green function for the domain Ω. The function P : Ω Ω R, defined by P (x, Y ) := ν(y ) G(x, Y ) is called the Poisson kernel for the domain Ω. Corollary indicates that it is reasonable to produce efforts to determine the Poisson kernel of a given domain Ω. However, the Poisson kernel may be explicitly found only in a few, albeit important, cases, where the domain Ω has a suitable shape. The next section is devoted to the case of the upper half space in R n. Exercise Suppose that G is the Green s function of the bounded domain Ω. Prove the following: (i) G(x, y) = G(y, x) for every x, y in Ω, x y;

32 28 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM (ii) G(x, y) < 0 for every x, y in Ω, x y; (iii) prove that G(x, y) N(x y) for every x and y in Ω; (iv) if f is bounded in Ω, then G(x, y) f(y) dv (y) 0 as x Ω. Ω Hints: (i) for every z in Ω and every (small) ε > 0, denote by Ω ε (z) the set Ω\ B ε (z). Then write the harmonic function G(x, ) by using the representation formula in Theorem (iii) with Ω ε (x) in place of Ω and similarly write the harmonic function G(y, ) by using the representation formula with Ω ε (y) in place of Ω. Then use second Green s identity and the fact that G(x, ) = N(x ) + h x ( ); (ii) note that G(x, y) tends to if y tends to x. Therefore, G < 0 near x. Apply the maximum principle to the domain Ω \ B ε (x); (iii) observe that, by the minimum principle, h x > 0 in Ω and use (ii) and the definition of G; (iv) use (iii) and the Lebesgue dominated convergence theorem, away from Ω, and the extra-integrability of the Newtonian potential near Ω. 1.7 The Poisson kernel for the half space Our aim is to compute the Poisson kernel for the upper half space R n + in R n, defined as follows R n + := {(x, x n ) R n 1 R : x n > 0}. First we need to compute the Green s function for R n +. Fix a point x = (x, x n ) in the upper half space, and consider the Newtonian potential N(x ) with pole x. We must find a function h x in C 1 (R n +) that is harmonic on R n + and such that N(x ) + h x vanishes identically when x n = 0. Observe that the Newtonian potential is, up to a constant, the potential generated by a unit negative charge placed at the point x := (x, x n ). By symmetry, its values on the hyperplane x n = 0 are the same of those of a Newtonian potential generated by a unit negative charge placed at the point x. Thus, the Green s function G of R n + is given by 1 x y log if n = 2 G(x, y) = 2π x y [ 1 (2 n) ω n x y 2 n x y 2 n] if n 3. (1.7.1)

33 1.7. THE POISSON KERNEL FOR THE HALF SPACE 29 Exercise Check that the function G defined above is the Green s function for the upper half space. First we consider the case where n 3. By definition, for every x R n + and y R n 1 the Poisson kernel P (x, y ) is then given by P (x, y ) = ν G(x, y ) = yn G(x, y ) = 1 [ yn x n ω n x y y n + x n. n x y ] y n n=0 Observe that x y = x y when y n = 0. Hence P (x, y ) = 2 ω n x n x y n = 2 ω n x n ( x y 2 + x 2 n) n/2 x n > 0, x, y R n 1. (1.7.2) A similar computation shows that the formula above holds also in the case where n = 2. Exercise Compute the Poisson kernel for a quadrant of the plane. Hint: refine the method of images illustrated above for the half plane. Note that, given a function g on R n 1 the Dirichlet problem { u = 0 in R n + u R n + = g may have more than one solution. In particular, if g vanishes identically, then the null function and the function u(x, x n ) = x n both solve the Dirichlet problem above. Thus, if we want to recover uniqueness, we must impose further restriction on the solution. A typical statement which holds in this case is the following. Theorem Suppose that g is continuous and bounded on R n 1. Then there is a unique bounded continuous function u on R n + which solves the Dirichlet problem { u = 0 in R n + u R n + = g.

34 30 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Furthermore u(x) = g(y ) P (x, y ) dy R n 1 = 2 ω n R n 1 g(y ) x n ( x y 2 + x 2 n) n/2 dy x R n +, (1.7.3) and u(x, 0) = g(x ) for every x R n 1. Remark Set p(x ) := 2 ω n 1 ( x ) n/2 x R n 1. Note that p is in L 1 (R n 1 ) and that P (x, y ) = p xn (x y ), where, as customary, we write p xn to be the dilated and normalised version of p in L 1 (R n 1 ). Explicitly, p xn (w ) = x 1 n n p ( ) w /x n w R n 1 x n > 0. Observe that p xn 1 = p 1, thereby justifying the terminology. With this notation, formula (1.7.3) may be rewritten as follows where stands for convolution on R n 1. u(x) = g p xn (x ) x R n +, (1.7.4) Exercise Prove that (1.7.3) gives, indeed, a solution to the Dirichlet problem on the upper half space with boundary datum g. By using a celebrated theorem of Liouville and the Schwartz reflection principle, we may prove that if g is continuous and bounded, then (1.7.4) gives the unique bounded solution to the Dirichlet problem in the upper half space. 1.8 The Poisson kernel for the ball The aim of this section is to provide the solution to the classical Dirichlet problem in the ball B R (0). The theory developed here is important at least for two reasons: (i) it leads to an explicit formula for the solution; (ii) the fact that the classical Dirichlet problem in every ball is solvable is a key step towards the solution of the classical Dirichlet problem in more general domains (see Definition and the proof of Theorem ).

35 1.8. THE POISSON KERNEL FOR THE BALL The two dimensional case We first look at the two dimensional case, where the theory of functions of a complex variable is of valuable help. Suppose that ϕ is a conformal map between the domains Ω C and ϕ(ω). In particular, ϕ is a bijection between Ω and ϕ(ω) and ϕ (z) 0 for every z in Ω. Proposition Suppose that Ω and ϕ are as above, that ϕ extends to a biholomorphic map between Ω and ϕ(ω) and assume that ϕ(ω) admits a Green s function G ϕ(ω). Then Ω admits a Green s function G Ω and G Ω (z, ζ) = G ϕ(ω) (ϕ(z), ϕ(ζ) ) (z, ζ) Ω Ω \ {(ω, ω) : ω Ω}. Proof. The assumptions on ϕ imply that ζ Ω if and only if ϕ(ζ) (ϕ(ω)). Then for each z Ω and ζ Ω G Ω (z, ζ) = G ϕ(ω) (ϕ(z), ϕ(ζ) ) = 0, because G ϕ(ω) is the Green s function for ϕ(ω). For the same reason, and the fact that ϕ is conformal, 0 = w G ϕ(ω) (w, ϕ(ζ)) = ϕ (z) 2 z [ G ϕ(ω)( ϕ(z), ϕ(ζ) )] = ϕ (z) 2 z G Ω (z, ζ); we have used (1.3.13) (viii) in the second equality above. Hence G Ω (, ζ) is harmonic for every ζ in Ω. Finally, we need to prove that G Ω (z, ζ) (2π) 1 log z ζ is harmonic in Ω. Clearly this function is harmonic in Ω \ {z}. In order to prove that it extends to a harmonic function on Ω, it suffices to show that G Ω (z, ζ) (2π) 1 log z ζ is bounded in a neighbourhood of z. Notice that G Ω (z, ζ) 1 log z ζ 2π = G ϕ(ω) (ϕ(z), ϕ(ζ) ) 1 log z ζ 2π = 1 2π log ϕ(z) ϕ(ζ) + h ( ) 1 ϕ(z) ϕ(ζ) 2π = 1 2π log ϕ(z) ϕ(ζ) ( ) + h ϕ(z) ϕ(ζ) z ζ log z ζ

36 32 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Observe that [ϕ(z) ϕ(ζ)]/[z ζ] tends to ϕ (z) as ζ tends to z, and recall that ϕ (z) 0, for ϕ is conformal. Therefore the function ζ 1 2π log ϕ(z) ϕ(ζ) ( ) + h ϕ(z) ϕ(ζ) (z ζ) is harmonic in Ω \ {z} and bounded in a neighbourhood of z, hence it is harmonic in Ω. We have proved that G Ω (z, ) may be written as the sum of the Newtonian potential with pole z and a function harmonic in Ω. This completes the proof of the proposition. We would like to apply this result to the case where the domain Ω is just the unit disc D := {z C : z < 1} and ϕ(z) = i 1 + z z D. 1 z It is not hard to check that ϕ maps D conformally onto the upper half plane Π, and D onto R. However, notice that ϕ does not extend to a biholomorphic mapping from D to Π, so that, strictly speaking, the result above does not apply. Nonetheless, we conjecture that for z D { G D G Π( ϕ(z), ϕ(ζ) ) if ζ 1 (z, ζ) = 0 if ζ = 1. Recall that in the last section we have proved that the Green s function for the upper half plane is given by G Π (w, ω) := 1 2π Denote by z and ζ the points in D such that log w ω w ω. ϕ(z) = w and ϕ(ζ) = ω. It is straightforward to check that w = ϕ(1/z), i.e., w is the image under ϕ of the point obtained from z by the mapping z 1 z, which is the inversion with respect to the circle { z = 1}. Note that G Π( ϕ(z), ϕ(ζ) ) = 1 2π log ϕ(z) ϕ(ζ) ϕ(1/z) ϕ(ζ) = 1 2π = 1 2π 1+z 1 z log 1+ζ 1 ζ 1+(1/z) 1+ζ 1 (1/z) log z ζ 1 z ζ. 1 ζ

37 1.8. THE POISSON KERNEL FOR THE BALL 33 Clearly, for each z D, the function ζ G Π( ϕ(z), ϕ(ζ) ) is harmonic in D, vanishes on D and is of class C 1 (D \ {z}. Thus, G D (z, ζ) = G Π( ϕ(z), ϕ(ζ) ) z D ζ D \ {z}. (1.8.1) Now we write z = r e iφ and ζ = s e iθ, and obtain The Poisson kernel is then G D (z, ζ) = 1 2π log r s ei(θ φ) 1 rs e i(θ φ) = 1 4π log r2 2rs cos(θ φ) + s 2 1 2rs cos(θ φ) + r 2 s 2. s G D (r e iφ, s e iθ ) s=1 = 1 [ 2r cos(θ φ) + 2 4π r 2 2r cos(θ φ) + 1 = 1 1 r 2 2π 2r cos(θ φ) + ] 2r2 1 2r cos(θ φ) + r 2 1 2r cos(θ φ) + r 2. (1.8.2) Therefore, if a solution u C 1 (D) to the Dirichlet problem exists, then u(r e iφ ) = 1 2π 2π 0 { u = 0 u D = g, in D 1 r 2 1 2r cos(θ φ) + r 2 g(eiθ ) dθ. (1.8.3) Furthermore, by arguing much as in Proposition below, one can prove that if g is a continuous function on D, then the function u defined in D as in (1.8.3) and equal to g on D is a classical solution to the Dirichlet problem in D with datum g. The aim of the following series of exercises is to determine the Green s function, hence the Poisson kernel, of certain domains in the complex plane that are conformally equivalent to the unit disc. Exercise Compute the Green s function and the Poisson kernel of the half disc F := {z C : z < 1, Im(z) > 0}.

38 34 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Exercise Prove that the map z e z is a conformal map of the strip S := {z C : 0 < Im(z) < π} onto the upper half plane. Then compute the Green s function and the Poisson kernel of the strip S. Exercise Prove that the map z z α/π is a conformal map of the upper half plane onto the sector Γ := {z C : 0 < arg(z) < α}. Then compute the Green s function and the Poisson kernel of the sector Γ The higher dimensional case We now consider the case where n 3. First, we deal with the case where R = 1. The idea is that the potential on the unit sphere B 1 (0) induced by a unit charge at x 0 is equal to the potential on B 1 (0) induced by a charge of magnitude x 2 n at the point x/ x 2, obtained by inversion with respect to B 1 (0) of x. Lemma Suppose that Y B 1 (0) and that x B 1 (0) \ {0}. Then Y x = x Y x. x Proof. This lemma is geometrically obvious (just draw a picture). For an analytic proof, just square both sides of the formula above and make the required computation. Proposition Suppose that R > 0. The following hold: (i) the function, defined for every x B 1 (0) and for every y B 1 (0) by G 1 (x, y) = N(x y) x 2 n N ( x/ x 2 y ) = is the Green s function of B 1 (0); 1 (2 n) ω n [ x y 2 n x 1 x x y 2 n ],

39 1.8. THE POISSON KERNEL FOR THE BALL 35 (ii) the Poisson kernel for the ball B R (0) is given by P R (x, Y ) = R2 x 2 ω n R x Y n x B R (0) Y B R (0). (1.8.4) Proof. First we prove (i). Recall that y N(x y) is harmonic in R n \ {x}. The first equation in (i) then shows that G 1 (x, ) is harmonic in B 1 (0) \ {x}. By Lemma 1.8.5, the second proves that G 1 (x, y) = N(x y) + h x (y), where h x is harmonic in B 1 (0) and h x (Y ) = N(x Y ) when Y B 1 (0). Since the Green s function, if it exists, is unique, G 1 is the Green s function of B 1. Next we prove (ii). It is straightforward to check that the function G R (x, y) = 1 R n 2 G 1(x/R, y/r) is the Green s function for B R (0). Indeed, by Lemma 1.8.5, if Y B R (0) and x B R (0), then Y x = Y x R R x. (1.8.5) x Clearly G R (x, ) is harmonic in B R (0) \ {x}. Furthermore, if Y B R (0), then 1 [ G R (x, Y ) = R 2 n x (2 n) ω n R Y 2 n R x R x R x Y 2 n ] R R 1 [ = x Y 2 n R (2 n) ω n x x x 2 n] R Y = 0, where the last equality follows from (1.8.5). Denote by ν the outward normal of B R (0) at the point Y. By the definition of Poisson kernel, as required. P R (x, Y ) = ν G R (x, Y ) = 1 [ Y x ω n x Y Y n Y x R = R2 x 2 ω n R x Y n, x R Y R x x R x x Y n x R Y ] Y A direct consequence of Corollary and Definition is the following representation formula for solutions to the Dirichlet problem on B R (0).