Download Laplace Equation on Spheres: Solutions and Legendre Polynomials and more Study notes Differential Equations and Transforms in PDF only on Docsity! Notes on the Laplace equation for spheres §1. Laplacian in spherical coordinates Let (r, φ, θ) be the spherical coordinates, related to the Cartesian coordinates by x = r sinφ cos θ, y = r sinφ sin θ, z = r cosφ . In polar coordinates, the Laplacian ∆ = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 becomes ∆u = 1 r2 ∂ ∂r ( r2∂u ∂r ) + 1 r2 sinφ ∂ ∂φ ( sinφ ∂u ∂φ ) + 1 r2 sin2 φ ∂2u ∂θ2 (1) Although it is possible to verify (1) directly using the chain rule, we give a simpler proof using the formula ∇u · ∇f = ∂u ∂r ∂f ∂r + 1 r2 ∂u ∂φ ∂f ∂φ + 1 r2 sin2 φ ∂u ∂θ ∂f ∂θ (2) for the inner product of the gradients of two functions, which is quite easy to check. The proof of (1) uses (2), the change of variable formula, and integration by parts, without much calculation: For each smooth function f(x, y, z) which vanishes outside of a bounded region in R3, we have∫∫∫ f ∆u r2 sinφ dr dφ dθ = ∫∫∫ f ∆u dx dy dz change of variables = − ∫∫∫ ∇u · ∇f dx dy dz integration by parts = − ∫∫∫ ( ∂u ∂r ∂f ∂r + 1 r2 ∂u ∂φ ∂f ∂φ + 1 r2 sin2 φ ∂u ∂θ ∂f ∂θ ) r2 sinφ dr dφ dθ by (2) = ∫∫∫ f [ ∂ ∂r ( r2 ∂u ∂r ) sinφ+ ∂ ∂φ ( sinφ ∂u ∂φ ) + ∂2u ∂θ2 ] dr dφ dθ integration by parts = ∫∫∫ f [ 1 r2 ∂ ∂r ( r2 ∂u ∂r ) + 1 r2 sinφ ∂ ∂φ ( sinφ∂u ∂φ ) + 1 r2 sin2 φ ∂2u ∂θ2 ] r2 sinφ dr dr dφ dθ Since the above equality holds for every smooth function f vanishing outside some bounded region, (1) follows. §2. Solutions invariant under rotations How does one find a function u(r, φ θ) defined on a sphere 0 ≤ r ≤ R of radius R such that ∆u = 0, u(R, φ, θ) = f(φ) (3) where f(φ) is a (given) function on the boundary of the sphere r = R, expressed in spherical coordinates, which is symmetric about the z-axis? The general procedure of separation of variables says that one should first find harmonic functions of the form F (r)G(φ); these “eigenfunctions” will be parametrized by some discrete parameters. 1 Separation of variables leads to the system of ordinary equations d dr ( r2 dF dr ) − k F (r) = 0, (4) and 1 sinφ d dφ ( sinφ dG dφ ) + k G(φ) = 0 (5) for some k ∈ R. The equation (4) has solutions F (r) = rn, with n ∈ R≥0 and k = n(n+ 1). The equation (5) will look a lot better if we use the variable w = cosφ, with 0 ≤ w ≤ 1. Let H(w) be the function in w corresponding to G(φ), so that H(cosφ) = G(φ), then the equation (5) becomes the standard Legendre equation with parameter n: d dw ( (1− w2) dH dw ) + n(n+ 1)H(w) = 0 (6) Although one does not see any restriction on k at this stage other than n ≥ 0, it turns out that n has to be an integer, otherwise F (r)G(φ) cannot be a well-defined harmonic function on the sphere. This follows from a careful analysis of solutions of (6). §3. Legendre polynomials With n being a non-negative integer, the solutions of (6) that behave well at w = ±1 are all multiples of the Legendre polynomial Pn(w), defined by Pn(w) = 1 2n n! dn dwn [(w2 − 1)n] n ∈ Z≥0 . (7) Besides being the solution of the Legendre differential equation (6), the Legendre polynomials satisfy the orthogonality relations:∫ 1 −1 Pm(w)Pn(w) dw = 2 2m+ 1 δm,n (8) where δm,n is equal to 0 if m 6= n, and δm,n = 1 if m = n. To check (8), one uses integration by parts repeatedly. Another way to define the Legendre polynomials is a succinct expression of its generating function: (1− wt+ t2)− 1 2 = ∞∑ n=0 tn Pn(w) . (9) We have P0(w) = 1, P1(w) = w, P2(w) = 3w2−1 2 , P3(w) = 1 2 (5w3−3w),etc. In case you like to have a good reference for the special functions you encountered in this course, I recommend A Course of Modern Analysis by Whittaker a Watson, an absolute classic. 2
