II. The Potential Equation 1. Introduction. Our objective here is to show that the Dirichlet boundary value problem is well- posed for Poisson’s equation (1) −∆u(x) = f(x) Pn which contains the Laplace operator ∆u ≡ i=1 uxixi . That is, in a domain G n in R we seek a solution of (1) which takes on prescribed values u|∂G = g on the boundary ∂G. Thus, the data in the Dirichlet problem consists of the two functions, f and g, and the domain G. It is this last condition, the domain G, which makes the problem difficult, since a solution which agrees with g on the entire boundary ∂G is demanded. In particular, it is rather easy to find solutions of the partial differential equation itself, but the real problem is to satisfy the boundary conditions. Although local solutions of (1) might seem interesting, or even a single global solution on all of G, they usually represent very little progress towards solving the boundary value problem. In Section 2 we shall derive a representation for solutions of (1) in terms of f on G and the function u and its normal derivative on ∂G. Various properties of solutions of Laplace’s equation (2) −∆u(x) = 0 will be derived in Section 3 from this representation. This representation will be refined in Section 4 to construct a candidate for the solution of the Dirichlet problem, and we shall use this to solve explicitly the Dirichlet problem for very special domains. After examining some of the consequences of solvability on the sphere, we turn finally to prove existence of a solution on general domains. 2. A Fundamental Representation. Let G denote a normal domain in Rn.A harmonic function on G is a function u ∈ C2(G) which satisfies Laplace’s equation (1) ∆u = 0 in G. As is often the case with linear equations, certain special solutions which depend only on the distance from some point can be useful in the construction of other solutions. In particular, solutions of (1) of the form u(x) = w(r), where n 1 P 2 2 r = kx − ξk = ( i=1(xi − ξi) ) is the distance from x to ξ, are given by 1 ln( r ) , n = 2 , u(x) = 1 rn−2 , n ≥ 3 . Each of these functions has a singularity at ξ characterized by the dimension n. It will be convenient to define the singular solution of (1) by ( 1 1 2π ln( r ) , n = 2 , s(x, ξ) ≡ 1 r = kx − ξk , n−2 , n ≥ 3 , (n−2)ωnr 1 2 n where ωn denotes the surface area of the unit sphere in R . In particular, ω2 = 2π and ω3 = 4π. The coefficients are chosen to normalize the singularity for calculations to follow below. Recall that the Divergence Theorem Z Z ∇ · F~ dx = F~ · ~νdS G ∂G 1 holds on G when F~ = (F1,F2,...,Fn) has components in C(G¯) ∩ C (G) and ~ν is the unit outward normal on the boundary, ∂G. If u, v ∈ C1(G¯) ∩ C2(G), then we can set Fj = u vxj in the divergence theorem to obtain the First Green’s Identity Z Z ∂v (2) (u∆v + ∇u · ∇v) dx = u dS G ∂G ∂ν in which ∇v = (vx1 , vx2 , . , vxn ) is the gradient of v and ∂v = ∇v · ~ν ∂ν is the normal derivative, the directional derivative in the normal direction, ν.A corresponding result holds with u and v interchanged, and by subtracting these we obtain the Second Green’s Identity Z Z ∂v ∂u (3) (u∆v − v∆u) dx = (u − v ) dS . G ∂G ∂ν ∂ν From (2) we obtain our first uniqueness result for Laplace’s equation. If u = v is a harmonic function in G and belongs to C1(G¯), then from (2) follows Z Z ∂u (k∇uk2) dx = u dS , G ∂G ∂ν so if additionally u = 0 on ∂G, it follows that u is constant on G, hence, identically zero. This shows that there is at most one such solution in C1(G¯) ∩ C2(G) of the boundary value problem for Poisson’s equation −∆u(x) = f(x), x ∈ G, (4) u(x) = g(x), x ∈ ∂G . We shall show later that there is at most one solution of (4) in the larger class C(G¯) ∩ C2(G). Note also that by setting v = 1 in (3) we obtain Gauss’ law Z Z ∂u (5) ∆u dx = dS . G ∂G ∂ν This is related to the conservation of mass in many applicatons, and it will be useful in some computations below. We want now to obtain a representation of a solution of the boundary value problem (4) in terms of integrals over G and around ∂G. The identity (3) is our starting point. We shall apply it to a smooth function u and choose v(x) = s(x, ξ), 3 a singular solution of Laplace’s equation (1) with singularity at ξ ∈ G. Because of the singularity we cannot apply (3) directly to the region G, but we shall instead apply it to the region Gε obtained from G by deleting the sphere Sε of radius ε > 0 centered at ξ. Now since ∆s(·, ξ) = 0 in Gε, ∂Gε = ∂G − ∂Sε, and at x ∈ ∂Sε the x−ξ unit outward normal to ∂Gε is given by − ε , we obtain for n ≥ 3 Z Z ∂s(x, ξ) ∂u(x) − ∆u(x)s(x, ξ) dx = u(x) − s(x, ξ) dSx Gε ∂G ∂ν ∂ν Z 1 1 ∂u(x) (6) − u(x) − n−1 − n−2 dSx . ∂Sε ωnε (n − 2)ωnε ∂ν We consider the behaviour of each term as ε → 0. From (5) we get the estimate Z Z n ∂u(x) ωnε | dSx| = | ∆u(x) dx| ≤ max |∆u(x)| x∈Sε ∂Sε ∂ν Sε n in which the coefficient on the last term is the volume of the sphere Sε. This gives the first limit 1 Z ∂u(x) lim{ dSx} = 0 . ε→0 n−2 ε ∂Sε ∂ν n−1 Since the surface area of Sε is given by ωnε , we obtain Z Z n−1 | u(x) dSx − ωnε u(ξ)| = | (u(x) − u(ξ)) dSx| ∂Sε ∂Sε n−1 ≤ ωnε max |u(x) − u(ξ)| . x∈∂Sε The continuity of the function u then yields the limit 1 Z lim{ u(x) dSx} = u(ξ) . ε→0 n−1 ωnε ∂Sε Thus, we have shown that the right side of (6) converges as ε → 0, and so it follows that the left side does also, and we have obtained the integral representation Z Z ∂u(x) ∂s(x, ξ) (7) u(ξ) = s(x, ξ) − u(x) dSx − ∆u(x)s(x, ξ) dx . ∂G ∂ν ∂ν G This identity expresses the value of a function u ∈ C1(G¯) ∩ C2(G) at a point ∂u ξ ∈ G in terms of ∆u in the interior G and both of u and ∂ν on the boundary ∂G. Thus, the first term in (7) containing the normal derivative of u on the boundary is a defect in the representation of a solution of the boundary value problem (4). However the identity (7) will be very useful in the following discussion of properties of harmonic functions, and we shall return in Section 4 to make the appropriate modifications to eliminate the defect in the representation, i.e., to eliminate the ∂u term involving ∂ν . 4 Exercises. 1. If u is a function whose value at x depends only on the distance r = kx − ξk 1−n n−1 from some point ξ, show that ∆u = r (r ur)r. Then show that such a function is harmonic if and only if it is of the form c1+c2s(x, ξ), where c1 and c2 are constants and s(x, ξ) is the singularity function. n 2.a If ωn denotes the surface area of the unit sphere in R , then the surface area n−1 of the sphere of radius ε is ωnε , and its volume is given by Z ε n n−1 ωnε vn(ε) = ωnε dε = . 0 n 2.b Show that Z 1 p 2 vn+1(1) = 2 vn( 1 − x ) dx . 0 ω2 4π Note that v2(1) = π = 2 , so ω2 = 2π. Also v3(1) = 3 , so ω3 = 4π. Show that 2 ω4 = 2π . 2.c Note that π Z 1 Z 2 ωn 2 n ωn n+1 vn+1(1) = 2 (1 − x ) 2 dx = 2 cos (θ) dθ , 0 n n 0 hence, π Z 2 n + 1 n+1 ωn+1 = 2ωn cos (θ) dθ . n 0 Derive the equality π π Z 2 n Z 2 cosn+1(θ) dθ = cosn−1(θ) dθ 0 n + 1 0 from which we get the recursive formula n − 2 ωnωn−1 ωn+1 = n ≥ 3 n − 1 ωn−2 2.d Show by induction that n 2π 2 ωn = n , n ≥ 1, Γ( 2 ) where Γ(·) is the Gamma function. 3. Show that if u is harmonic in G and u ∈ C1(G¯), then Z ∂u dSx = 0 . ∂G ∂ν 4.a State and prove a uniquenes result for the Neumann problem which asks for a solution of −∆u(x) = f(x), x ∈ G, ∂u = g(x), x ∈ ∂G . ∂ν 5 In particular, show that any solution is determined up to a constant and that this is the best that can be done. 4.b Show that a necessary condition for the Neumann problem to have a solution is that the compatibility condition Z Z f(x) dx + g(x) dSx = 0 G ∂G hold.