Answer to Question #149900 in Differential Equations for Hicks

Question

Solve Laplace’s equation for u(x, y):

d2y/dx2 + d2u/dy2 = 0
0 < x < 2 , 0 < y < 1
u(x, 0) = sin πx , u(x, 1) = x/2
u(0, y) = sin πy , u(2, y) = y

Answer 1

$\displaystyle \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\\ \textsf{Let}\,\, u(x, y) = X(x)Y(y)\\ X''Y+ XY'' = 0,\,\, X''Y= -XY''\\ X''Y= -XY'' = K\\ \frac{X''}{X} = -\frac{Y''}{Y} = K\\ \frac{X''}{X} = K,\,\, \frac{Y''}{Y} = -K\\ \textsf{If}\,\, K\,\, \textsf{is chosen to be positive}\\ \textsf{say}\,\, K = \lambda^2,\,\,\textsf{then}\,\,\\ X'' = \lambda^2 X\\ X = Ae^{\lambda x} + Be^{-\lambda x}\\ Y'' = -\lambda^2 Y\\ Y = C\cos(\lambda y) + D\sin(\lambda y)\\ u(x,y) = \left(Ae^{\lambda x} + Be^{-\lambda x}\right)\left(C\cos(\lambda y) + D\sin(\lambda y)\right)\\ u(x,0) = C\left(Ae^{\lambda x} + Be^{-\lambda x}\right) = \sin(\pi x)\\ u(0,0) = C\left(A + B\right) = 0\\ u(0,y) = \left(A + B\right)\left(C\cos(\lambda y) + D\sin(\lambda y)\right) = \sin(\pi y)\\ u(0,0) = C\left(A + B\right) = 0\\ C = 0,\,\, A = -B\\ u(x,1) = \left(Ae^{\lambda x} + Be^{-\lambda x}\right)\left(C\cos(\lambda) + D\sin(\lambda)\right) = \frac{x}{2}\\ u(0,\,1) = \left(A + B\right)\left(C\cos(\lambda) + D\sin(\lambda)\right) = 0\\ \left(A + B\right)\left(D\sin(\lambda)\right) = 0\\ \textsf{We would not want}\,\, D \,\,\textsf{to be zero},\\ \textsf{so we can force}\,\, \lambda \,\, \textsf{to produce a 0 term}\\ \textsf{Choosing}\,\, D = 0 \,\, \textsf{would make}\,\,u = 0 \\ \textsf{so we must force}\,\, \sin(\lambda)\,\, \textsf{to be zero}\\ \textsf{i.e. choose} \,\, \lambda = n\pi \\ \textsf{where}\,\, n\,\, \textsf{is a positive integer}\\ \sin(\lambda) = 0, \lambda = n\pi\\ u(2,y) = \left(Ae^{2\lambda} + Be^{-2\lambda}\right)\left(C\cos(\lambda y) + D\sin(\lambda y)\right) = y\\ u(2,0) = C\left(Ae^{2\lambda} + Be^{-2\lambda}\right) = 0\\ \begin{aligned} u(x,y) &= \left(Ae^{\lambda x} - Ae^{-\lambda x}\right)\left(D\sin(\lambda y)\right)\\ &= AD\left(e^{n\pi x} - e^{-n\pi x}\right)\sin(\lambda y) = 2AD\sinh\left(n\pi x\right)\sin(n\pi y) \\&= E\sinh\left(n\pi x\right)\sin(n\pi y) \end{aligned}\\ u(x,y)\,\,\textsf{can be superposed as;}\\ \begin{aligned} u(x,y) &=\sum_{n = 1}^{\infty} E_n\sinh\left(n\pi x\right)\sin(n\pi y)\\ \end{aligned}\\ u(2,y) =y\\ \begin{aligned} u(2,y) &=\sum_{n = 1}^{\infty} E_n\sinh\left(2 n\pi\right)\sin(n\pi y) \\&=\sum_{n = 1}^{\infty} b_n\sin(n\pi y) \end{aligned}\\ \textsf{Where}\,\, b_n = E_n\sinh\left(2 n\pi\right)\\ \textsf{We have here a half-range Fourier sine}\\ \textsf{series representation of a function}\,\, u(2,y)\\ \textsf{defined over}\,\ 0< y < 1, \,\,\textsf{Extending}\,\, u(2,y)\,\, \textsf{as an}\\ \textsf{even periodic function with period}\,\, 2\,\, \textsf{and}\\ \textsf{using standard Fourier series theory gives}\\ \begin{aligned} b_n &= 2\int_0^1 y \sin(n\pi y)\, \mathrm{d}y \\&= 2\int_0^1 y\, \mathrm{d}\left(\frac{-\cos(n\pi y)}{n\pi}\right) \\&= \frac{-2y\cos(n\pi y)}{n\pi}\biggr\vert_0^1 + \frac{2}{n\pi}\int_0^1 \cos(n\pi y)\,\mathrm{d}y \\&= \frac{-2\cos(n\pi)}{n\pi}+ \frac{2}{n\pi}\cdot\frac{\sin(n\pi y)}{n\pi}\biggr\vert_0^1 \\&= \frac{-2\cos(n\pi)}{n\pi}+ \frac{2\sin(n\pi)}{n^2\pi^2} \end{aligned}\\ \textsf{But}\,\, \sin(n\pi) = 0\,\, \&\,\, \cos(n\pi) = (-1)^n \,\,\forall n \in \mathbb{Z}^{+} \\ b_n = -\frac{2(-1)^n}{n\pi} \\ E_n\sinh\left(2 n\pi\right) = -\frac{2(-1)^n}{n\pi}\\ \therefore E_n = -\frac{2(-1)^n}{n\pi\left(\sinh\left(2 n\pi\right)\right)}\\ \begin{aligned} \therefore u(x,y) &= \sum_{n = 1}^{\infty} \frac{2(-1)^{n - 1}}{n\pi\left(\sinh\left(2 n\pi\right)\right)}\sinh\left(n\pi x\right)\sin(n\pi y) \end{aligned}$