Toronto Math Forum

APM346-2016F => APM346--Tests => TT2 => Topic started by: Victor Ivrii on November 17, 2016, 03:25:28 AM

Title: TT2-P3
Post by: Victor Ivrii on November 17, 2016, 03:25:28 AM
Using Fourier method find eigenvalues and eigenfunctions of Laplacian in the rectangle $\{0<x<a,\, 0<y<b\}$ with Neumann boundary conditions:
\begin{align}
&u_{xx}+u_{yy}=-\lambda u\qquad 0<x<a,\ 0<y<b,\label{3-1}\\[3pt]
&u_x|_{x=0}=u_x|_{x=a}=u_y|_{y=0}=u_y|_{y=b}=0.\label{3-2}
\end{align}
Title: Re: TT2-P3
Post by: XinYu Zheng on November 17, 2016, 08:07:51 AM
Introduce $u=X(x)Y(y)$ then applying separation of variables yields
$$\underbrace{\frac{X''}{X}}_{-\lambda_1}+\underbrace{\frac{Y''}{Y}}_{-\lambda_2}=-\lambda$$
Now we have two ODEs, namely $X''+\lambda_1X=0$ and $Y''+\lambda_2Y=0$ both with Neumann B.C. at $0,a$ and $0,b$ respectively. We know the solution to these problems:
$$\begin{cases} \lambda_{1,0}=0& X_0=\frac{1}{2}\\ \lambda_{1,n}=\frac{n^2\pi^2}{a^2}& X_{n}=\cos(n\pi x/a)\\ \lambda_{2,0}=0& Y_0=\frac{1}{2}\\ \lambda_{2,n}=\frac{m^2\pi^2}{b^2}& Y_{m}=\cos(m\pi x/b)\\ \end{cases}$$
Where $n,m\geq 1$. So the eigenfunctions are
$$u_{n,m}(x,y)=\cos(n\pi x/a)\cos(m\pi x/b)$$
With eigenvalues $\lambda_{m,n}=\lambda_{1,n}+\lambda_{2,n}=\pi^2(n^2/a^2+m^2/b^2)$. We can absorb the case where $m=0$ or $n=0$ by allowing us to plug $m,n=0$ in this equation.