3 Laplace's Equation
3 Laplace’s Equation We now turn to studying Laplace’s equation ∆u = 0 and its inhomogeneous version, Poisson’s equation, ¡∆u = f: We say a function u satisfying Laplace’s equation is a harmonic function. 3.1 The Fundamental Solution Consider Laplace’s equation in Rn, ∆u = 0 x 2 Rn: Clearly, there are a lot of functions u which satisfy this equation. In particular, any constant function is harmonic. In addition, any function of the form u(x) = a1x1 +:::+anxn for constants ai is also a solution. Of course, we can list a number of others. Here, however, we are interested in finding a particular solution of Laplace’s equation which will allow us to solve Poisson’s equation. Given the symmetric nature of Laplace’s equation, we look for a radial solution. That is, we look for a harmonic function u on Rn such that u(x) = v(jxj). In addition, to being a natural choice due to the symmetry of Laplace’s equation, radial solutions are natural to look for because they reduce a PDE to an ODE, which is generally easier to solve. Therefore, we look for a radial solution. If u(x) = v(jxj), then x u = i v0(jxj) jxj 6= 0; xi jxj which implies 1 x2 x2 u = v0(jxj) ¡ i v0(jxj) + i v00(jxj) jxj 6= 0: xixi jxj jxj3 jxj2 Therefore, n ¡ 1 ∆u = v0(jxj) + v00(jxj): jxj Letting r = jxj, we see that u(x) = v(jxj) is a radial solution of Laplace’s equation implies v satisfies n ¡ 1 v0(r) + v00(r) = 0: r Therefore, 1 ¡ n v00 = v0 r v00 1 ¡ n =) = v0 r =) ln v0 = (1 ¡ n) ln r + C C =) v0(r) = ; rn¡1 1 which implies ½ c1 ln r + c2 n = 2 v(r) = c1 (2¡n)rn¡2 + c2 n ¸ 3: From these calculations, we see that for any constants c1; c2, the function ½ c1 ln jxj + c2 n = 2 u(x) ´ c1 (3.1) (2¡n)jxjn¡2 + c2 n ¸ 3: for x 2 Rn, jxj 6= 0 is a solution of Laplace’s equation in Rn ¡ f0g.
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