Chapter 6 Laplace equation In this chapter we consider Laplace equation in d-dimensions given by u + u + ··· + u = 0. (6.1) x1 x1 x2 x2 xd xd We study Laplace equation in d = 2 throughout this chapter (excepting Section 6.2), and most of the ideas can be generalized to general space dimensions d > 2. When d = 2, = ( ) the independent variables x1, x2 are denoted by x, y, and write x x, y . Thus Laplace equation in two independent variables is + = uxx uyy 0. (6.2) The non-homogeneous problem + = uxx uyy f , (6.3) where f is a function of the independent variables x, y only is called the Poisson equation. 6.1 Green’s identities Green’s Identities play an important role in the analysis of Laplace equation. They are derived from divergence theorem. Let us recall the divergence theorem now. Theorem 6.1 (Divergence theorem). Let Ω ⊆ R2 be a bounded piecewise smooth domain. Ψ Ω ! R2 Ψ = ( ) 2 1(Ω)\ (Ω) = Let : be a function. Let 1, 2 where i C C for i 1,2. Then Z Z r.Ψ(x, y) d x d y = Ψ.n dσ, (6.4) Ω @ Ω where n is the unit outward normal to @ Ω. Theorem 6.2 (Green’s identities). Let Ω ⊆ R2 be a bounded piecewise smooth domain, and let n denote the unit outward normal to @ Ω. For u, v 2 C 2(Ω) \ C 1(Ω), we have (i) (Green’s Identity-I) Z Z ∆ ( ) = @ σ u x, y d x d y nu d , (6.5) Ω @ Ω 147 148 Chapter 6. Laplace equation @ = r where nu : u.n is called the normal derivative of u; as it represents the directional derivative of u in the dirction of the unit outward normal n. (ii) (Green’s Identity-II) Z Z ∆ − ∆ ( ) = @ − @ σ v u u v x, y d x d y v nu u nv d (6.6) Ω @ Ω (iii) (Green’s Identity-III) Z Z Z r r = @ σ − ∆ u. v d x d y v nu d v u d x d y. (6.7) Ω @ Ω Ω Proof. Applying (6.4) with Ψ = ru, Ψ = vru − urv, and Ψ = vru yield Green’s identities I, II, and III respectively. 6.2 Fundamental solutions in Rd Since Lapalce equation is invariant under translations, and rotations (see Exercise 6.4), we look for solutions to Laplace equation having such symmetric properties. d Let us fix ξ 2 R , and look for solutions to ∆u = 0 having the form vξ (x) = (r ), where v u uXd t = k − ξ k = ( − ξ )2 r x xi i . i=1 Substituting the formula for vξ in the equation, we get 00 d − 1 0 ∆vξ (x) = (r ) + (r ) = 0. r Solving the last equation, we get 0 − (r ) = C r 1 d (6.8) Integrating we get C ln r if d = 2, ( ) = r C 2−d ≥ (6.9) 2−d r if d 3. Theorem 6.3. Let K(x,ξ ) denote the function defined by K(x,ξ ) := (kx − ξ k). (i) Let ξ be a point of Ω. For u 2 C 2(Ω) the following identity holds Z Z (ξ ) = ( ξ )∆ − ( ( ξ )@ − @ ( ξ )) σ u K x, u d x K x, nu u nK x, d . (6.10) Ω @ Ω (ii) The following equality holds in the sense of distributions on Rd : ∆K(x,ξ ) = δξ . ' 2 1(Rd ) i.e., for every C0 the following equality holds. Z '(ξ ) = K(x,ξ )∆'(x) d x. (6.11) Rd Sivaji IIT Bombay 6.2. Fundamental solutions in Rd 149 (iii) If u 2 C 2(Ω) and harmonic in Ω (i.e., ∆u = 0 in Ω), then for ξ 2 Ω we get Z (ξ ) = − ( ( ξ )@ − @ ( ξ )) σ u K x, nu u nK x, d . (6.12) @ Ω Proof. Step 1: Proof of (i): Let u 2 C 2(Ω) and ξ be a point of Ω. Note that we cannot apply Green’s identity II (6.6) with v = vξ (x) = (kx − ξ k) since vξ is singular at x = ξ . Thus we cut out a ball B(ξ ,ρ) from Ω along with its boundary, and then apply Green’s identity II. Note that the domain Ωρ;= Ω n B[ξ ,ρ] is bounded by @ Ω and S(ξ ,ρ). Since ∆vξ = 0 in Ωρ, we have Z Z Z ∆ = @ − @ σ + @ − @ σ vξ u d x vξ nu u nvξ d vξ nu u nvξ d . (6.13) Ωρ @ Ω S(ξ ,ρ) Let us now compute the second term on RHS of the equation (6.13). Z Z Z @ − @ σ = @ σ − @ σ vξ nu u nvξ d vξ nu d u nvξ d (6.14) S(ξ ,ρ) S(ξ ,ρ) S(ξ ,ρ) Note that on the circle S(ξ ,ρ), we have vξ (x) = (kx − ξ k) = (ρ). Using this information, and divergence theorem, we get Z Z Z @ σ = (ρ) @ σ = − (ρ) ∆ vξ nu d nu d u d x. (6.15) S(ξ ,ρ) S(ξ ,ρ) B(ξ ,ρ) Note that the outward unit normal n on the circle S(ξ ,ρ) points towards its center ξ . @ = − 0(ρ) (ξ ρ) Also nvξ holds at points on the circle S , . Thus we get Z Z @ σ = − ρ1−d σ u nvξ d C u d . (6.16) S(ξ ,ρ) S(ξ ,ρ) Since both u and ∆u are continuous at ξ , we have the following convergences as ρ ! 0: Z Z − (ρ) ∆ ! ρ1 d σ ! ! (ξ ) u d x 0, u d d u , (6.17) B(ξ ,ρ) S(ξ ,ρ) ! Rd where d denotes the surface area of the unit sphere in . Now passing to the limit as ρ ! 0 in the equation (6.13), we get Z Z ∆ = @ − @ σ + ! (ξ ) vξ u d x vξ nu u nvξ d C d u (6.18) Ω @ Ω in view of (6.15), (6.16), (6.17). = 1 Choosing C ! in the formulae (6.9), the equation (6.18) yields (6.10). d = ' 2 1(Ω) Step 2: Proof of (ii): In the equation (6.10) if we take u C0 , then we get (6.11). Step 3: Proof of (iii): Equation (6.12) is an immediate consequence of (6.10). October 4, 2015 Sivaji 150 Chapter 6. Laplace equation 6.3 Boundary value problems associated to Laplace equation We can study Laplace equation or Poisson equation on any open domain Ω ⊆ R2. These problems will always have an infinite number of solutions. Thus it is interesting to know if there are solutions to these equations if some constraints are placed on the unknown function. When Ω is a bounded domain with piecewise smooth boundary, there are atleast three different kinds of restrictions are placed on the unknown function and its deriva- tives along the boundary @ Ω of the domain Ω. Such restrictions are called boundary conditions. Let us now describe three boundary value problems (BVP) for Poisson equation. They differ in the nature of boundary conditions. We also define the notions of solutions to these boundary value problems. 1. For given functions f , g, Dirichlet Problem consists of solving the boundary value problem ∆u = f (x) in Ω, (6.19a) u = g on @ Ω. (6.19b) Definition 6.4 (Solution to Dirichlet BVP). Let f 2 C (Ω), and g 2 C (@ Ω).A function ' 2 C 2(Ω) \ C (Ω) is said to be a solution to Dirichlet BVP if (i) the function ' is a solution to Poisson equation (6.19a), i.e., for each x 2 Ω, ∆'(x) = f (x) holds, and (ii) for each x 2 @ Ω, the equality '(x) = g(x) holds. 2. For given functions f , g, Neumann Problem consists of solving the boundary value problem ∆u = f (x) in Ω, (6.20a) @ = @ Ω nu g on . (6.20b) Definition 6.5 (Solution to Neumann BVP). Let f 2 C (Ω), and g 2 C (@ Ω).A function ' 2 C 2(Ω) \ C 1(Ω) is said to be a solution to Neumann BVP if (i) the function ' is a solution to Poisson equation (6.20a), i.e., for each x 2 Ω, ∆'(x) = f (x) holds, and 2 @ Ω @ '( ) = ( ) (ii) for each x , the equality n x g x holds. 3. For given functions f , g, and α 2 R, Third Boundary Value Problem a.k.a. Robin Problem consists of solving the boundary value problem ∆u = f (x) in Ω, (6.21a) + α@ = @ Ω u nu g on . (6.21b) Definition 6.6 (Solution to Robin BVP). Let f 2 C (Ω), and g 2 C (@ Ω). A func- tion ' 2 C 2(Ω) \ C 1(Ω) is said to be a solution to Robin BVP if Sivaji IIT Bombay 6.3. Boundary value problems associated to Laplace equation 151 (i) the function ' is a solution to Poisson equation (6.21a), i.e., for each x 2 Ω, ∆'(x) = f (x) holds, and 2 @ Ω '( ) + α@ '( ) = ( ) (ii) for each x , the equality x n x g x holds. Remark 6.7. (i) If Dirichlet BVP is posed on a bounded domain Ω, then the corresponding Dirich- let problem is called an interior Dirichlet problem. If Ω is the complement of a bounded domain, then corresponding Dirichlet problem is called exterior Dirich- let problem. (ii) There are other kinds of boundary value problems possible. For example, on a part of the boundary @ Ω one of the three boundary value problems described above is imposed and on the remaining part another one of the above three.