Author Topic: Web Bonus Problem 5  (Read 459 times)

Victor Ivrii

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Web Bonus Problem 5
« on: February 02, 2015, 03:06:05 PM »
Consider
\begin{align}
&u_{xx}-u_{yy}=0\qquad &&0<x<a,\; 0<y<b\label{eq-1}\\[3pt]
&u|_{x=0}=u|_{x=a}=0&& 0<y<b,\label{eq-2}\\[3pt]
&u|_{y=0}=u|_{y=b}=0&&0<x<a.\label{eq-3}
\end{align}
Find the relationship between $a$ and $b$ when this problem has a non–trivial (that means which is not identically $0$) solution $u(x,y)$.

Hints

(a) Apply method of separation of variables.

(b) Alternatively, apply the method of characteristics (which is look for $u(x,y)=f(x+y)+g(x-y)$).

« Last Edit: February 02, 2015, 03:07:58 PM by Victor Ivrii »

Zacharie Leger

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Re: Web Bonus Problem 5
« Reply #1 on: April 14, 2015, 01:35:22 PM »
Following hint (a) we let $u(x,y)=X(x)Y(y)$ to obtain
$$\frac{X''(x)}{X(x)}=\frac{Y''(y)}{Y(y)}=-\lambda$$
from equation (1). For $\lambda=\omega^2, \omega>0$ we obtain solutions of the form
$$X(x)=A\cos(\omega _x x)+B\sin(\omega _x x)$$
Applying BCs we see that
$$X(0)=A=0 \Rightarrow X(x)=B\sin(\omega _x x) \Rightarrow X(a)=B\sin(\omega _x a)=0 \Rightarrow \omega _x =\frac{n_x\pi}{a} $$
Similarly for $Y(y)$ we can get that
$$Y(y)=C\cos(\omega _y y)+D\sin(\omega _y y)$$
$$Y(0)=C=0 \Rightarrow Y(y)=D\sin(\omega _y y) \Rightarrow y(b)=B\sin(\omega _y b)=0 \Rightarrow \omega _y =\frac{n_y\pi}{b} $$
Note that $\omega _x=\frac{n_x\pi}{a}=\frac{n_y\pi}{b}=\omega _y$. Hence, $\frac{a}{b}=\frac{n_x}{n_y}$ so
$$  \frac{a}{b} \in \mathrm{Q}$$
« Last Edit: April 14, 2015, 02:46:52 PM by Zacharie Leger »