CHAPTER PoEqP.1 The Potential EquationIn principle little will change when eigenfunction expansions are applied to potentialproblems. In practice, however, the mechanics of solving such problems will become morecomplicated because the expansion coefficients are found as solutions of boundary valueproblems which often are more difficult to solve than the initial value problems arising indiffusion and wave propagation problems.P.1.1 The Solution TechniqueA standard model for potential problems solvable with a separation of variables tech-nique is the following formulation:(P.1.1) Lu ≡ uxx+ uyy= F (x, y), (x, y) ∈ D = {(x, y) : 0 < x < a, 0 < y < b}u = g(x, y), (x, y) ∈ ∂Dwhere ∂D denotes the boundary of the rectangle D. The operatorLu ≡ uxx+ uyyis known as the two-dimensional Laplacian and is commonly denoted byLu = ∆u = ∇2u = ∇ ◦ ∇u.∆u is the preferred notation for the Laplacian in the mathematics literature and we shalladopt it here.∆u = 0is known as Laplace’s equation and∆u = Ffor non-zero F is called Poisson’s equation. Hence problem (P.1.1) calls for the solution ofPoisson’s equation subject to “Dirichlet” data which implies that the function u is given onthe boundary of the domain D.1In order to obtain zero boundary conditions it is customary to decompose (P.1.1) intotwo problems of the form(P.1.2) ∆u1= F (x, y)u1(0, y) = 0, u1(a, y) = 0u1(x, 0) = g(x, 0), u1(x, b) = g(x, b)(P.1.3) ∆u2= 0u2(x, 0) = 0, u2(x, b) = 0u2(0, y) = g(0, y), u2(a, y) = g(a, y)since u(x, y) = u1(x, y) + u2(x, y) then will solve (P.1.1). If the boundary values of (P.1.1)are continuous but not zero at the corners of D then such crude splitting of (P.1.1) willintroduce artificial discontinuities into problems (P.1.2) and (P.1.3) at the corners of D.Any eigenfunction expansion will invariably show a Gibbs phenomenon at such points. Suchdiscontinuities can be avoided if the problem is so reformulated that the boundary data arezero at the corner points of D before the splitting is carried out. This is easy to achieve forthe Dirichlet problem if we subtract from u a function v(x, y) which takes on the values ofg at the corners of D. A simple function is the polynomial solution to Laplace’s equationgiven byv(x, y) = c0+ c1x + c2y + c3xywhere the coefficients are determined such that v = g at the corners. Simple algebra showsthatv(x, y) = g(0, 0)(x − a)(y − b)ab+ g(a, b)·1 +x − aa+y − bb+(x − a)(y − b)ab¸− g(a, 0)·y − bb+(x − a)(y − b)ab¸− g(0, b)·x − aa+(x − a)(y − b)ab¸takes on the values of g at the corners of D and that w = u−v satisfies the Dirichlet problem∆w = F (x, y) in Dw = g(x, y) − v(x, y) on D.2An easier and efficient method exists if the boundary data in (P.1.1) are given by asmooth function g(x, y) which is defined on all of D. We setv(x, y) = g(x, y)and obtain forw(x, y) = u(x, y) − g(x, y)the Dirichlet problem∆w = ∆u − ∆g = F (x, y) − ∆g(x, y) in Dw = 0 on ∂D.In either case, if we now split the problem for w as above into Dirichlet problems for w1and w2then both functions have continuous boundary data on ∂D. The importance ofmaintaining continuous boundary data is dramatically demonstrated by example 1 below.Note that if v(x, y) = g(x, y) then by inspection w2≡ 0.Let us now consider the solution of (P.1.2). It is much the same as the problem for thewave equation if we identify y with t. For ease of notation we shall suppress the subscript1. The homogeneous boundary data at x = 0 and x = a suggest that we associate withuxx+ uyy= F (x, y)u(0, y) = u(a, y) = 0the eigenvalue problemφ00(x) = µφ(x)φ(0) = φ(a) = 0.Its solution isλn=nπa, φn(x) = sin λnx, µn= −λ2n, n = 1, 2, . . .The approximation to problem (P.1.2) is(P.1.4) uxx+ uyy= PNF (x, y)3u(0, y) = u(a, y) = 0u(x, 0) = PNg(x, 0)u(x, b) = PNg(x, b)wherePNF (x, y) =NXn=1γn(y)φn(x)PNg(x, 0) =NXn=1ˆαnφn(x)PNg(x, b) =NXn=1ˆβnφn(x)withγn(y) =hF (x, y), φnihφn, φniˆαn=hg(x, 0), φnihφn, φniˆβn=hg(x, b), φnihφn, φni.Problem (P.1.4) is known to have a unique solution. We shall show that for all values of yit belongs to span {φn(x)}Nn=1. If we write(P.1.5) uN(x, y) =NXn=1αn(y)φn(x)and substitute this representation into (P.1.4) then∆uN=NXn=1[αn(y)φ00n(x) + α00n(y)φn(x)]=NXn=1£αn(y)(−λ2nφn(x) + α00n(y)φn(x)¤=NXn=1γn(y)φn(x)uN(x, 0) =NXn=1αn(0)φn(x) = PNg(x, 0)uN(x, b) =NXn=1αn(b)φn(x) = PNg(x, b)4which will hold for all x if−λ2nαn(y) + α00n(y) = γn(y), n = 1, . . . , Nαn(0) = ˆαn, αn(b) =ˆβn.The solution to this problem is again a linear combination of complementary solutions plusa particular integral. The complementary solutions are exponentials eλnyand e−λny, butoften a more useful form is their combination in terms of hyperbolic functions. We shallwriteαn(y) = c1cosh λny + c2sinh λny + αnp(y)where αnp(y) is a particular integral of the differential equation. Depending on the formof γn(y) it is found by guessing, the method of undetermined coefficients or the method ofvariation of parameters. Once αnp(y) is known the coefficients c1and c2can be determinedsuch that αn(y) satisfies the boundary conditions.The problem (P.1.3) is approximated by the same process. Since u is zero at y = 0 andy = b an expansion in terms of the eigenfunctions ofφ00(y) = µφ(y)φ(0) = φ(b) = 0is used. They are˜λn=nπb, φn(y) = sin˜λny, µn= −˜λ2n.The approximate problem then isuxx+ uyy= 0u(x, 0) = u(x, b) = 0u(0, y) = Png(0, y), u(a, y) = PNg(a, y).It is solved by(P.1.6) uN(x, y) =NXn=1αn(x)φn(y)5where−λ2nαn(x) + α00n(x) = 0αn(0) =hg(0, y), φnihφn, φniαn(a) =hg(a, y), φnihφn, φni.Since there is no source term the solution αn(x) is readily found to beαn(x) = αn(0) cosh λnx −αn(a) − αn(0) cosh λnasinh λnasinh λnx.The functions (P.1.5) and (P.1.6) are added to give the solution of (P.1.1). Note that formallythey also solve (P.1.1) when the boundary function g is not zero at the corners of D. Butif, for example, g(0, 0) 6= 0 thenPN(g(x, 0)) ∈ span {sin λnx}Nn=1andPN(g(0, y)) ∈ spannsin˜λnyoNn=1will be poor approximations to g near (0, 0) since PN(g(0, 0)) = 0 for all N .Problem (P.1.1) is the simplest problem for the Laplacian on a rectangle while(P.1.7) ∆u = F (x, y) in Dg1u + g2∂u∂n= g3(x, y) on Dwith g1g2≥ 0 and g21+ g226= 0 is the most general problem we can treat. Here∂u∂nis thenormal derivative of u on ∂D in the direction of the