Answer to Question #131393 in Differential Equations for Meenu Rani

Question

Apply the method of separation of variables to solve the following Dirichlet boundary value problem
uxx +uyy = 0, 0 <x<a, 0<y <b,
u(x,0) = f(x), u(x,b) = 0, 0 ≤ x ≤ a,
u(0,y) = 0, u(a,y) = 0, 0 ≤ y ≤ b.

Answer 1

$u_{xx}+u_{yy}=0, x\in(0,a), y\in(0,b)\\ u(0,y)=0,u_x(a,y)=0,y\in[0,b]\\ u(x,0)=f(x), u(x,b)=0, x\in[0,a]\\ u(x,y)=X(x)Y(y)\\ u_x=X'Y, u_{xx}=X''Y\\ u_y=XY',u_{yy}=XY''\\$

the separation of variables and used

$u(0,y)=0,u_x(a,y)=0,y\in[0,b]$

we have

$X''(x)+\lambda X(x)=0, 0<x<a, X(0)=X'(a)=0$

eigenvalues

$\lambda_k=(\frac{2k+1}{a}\frac{\pi}{2})^2, k\in Z_+$

eigenfunctions

$X_k(x)=\sin\frac{2k+1}{a}\frac{\pi}{2}x, k\in Z_+$

$Y$ find from

$Y''_k-\lambda_kY_k=0, k\in Z_+\\ Y_k(y)=A_ke^{\sqrt{\lambda_k}y}+B_ke^{-\sqrt{\lambda_k}y}, k\in Z_+$

solution of problem is

$u(x,y)=\sum\limits_{k=0}^{\infty}(A_ke^{\frac{2k+1}{a}\frac{\pi}{2}y}+B_ke^{-\frac{2k+1}{a}\frac{\pi}{2}y})\sin\frac{2k+1}{a}\frac{\pi}{2}x$

$u(x,y)\to u(x,0)=f(x), u(x,b)=0, x\in[0,a]$

$\left\{\begin{matrix} \sum\limits_{k=0}^{\infty}(A_k+B_k)\sin\frac{2k+1}{a}\frac{\pi}{2}x=f(x)\\ \sum\limits_{k=0}^{\infty}(A_ke^{\frac{2k+1}{a}\frac{\pi}{2}b}+B_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b})\sin\frac{2k+1}{a}\frac{\pi}{2}x=0, \end{matrix}\right.\\ x\in[0,a]\\ \left\{\begin{matrix} A_k+B_k=f_k\\ A_ke^{\frac{2k+1}{a}\frac{\pi}{2}b}+B_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b}=0, k\in Z_+ \end{matrix}\right.$

where

$f_k=\frac{2}{a}\int^{a}_{0}f(z)\sin\frac{2k+1}{a}\frac{\pi}{2}zdz, k\in Z_+$

Then

$A_k=\frac{-f_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b}}{2sh\frac{2k+1}{a}\frac{\pi}{2}b}\\ B_k=\frac{f_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b}}{2sh\frac{2k+1}{a}\frac{\pi}{2}b}, k\in Z_+$

$u(x,y)=\sum ^{\infty}_{k=0}\frac{f_ksh{\frac{2k+1}{a}\frac{\pi}{2}(b-y)}}{sh\frac{2k+1}{a}\frac{\pi}{2}y}\sin\frac{(2k+1)\pi}{2a}x, \\ x\in(0,a), y\in(0,b)$