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In Strauss's Partial Differential Equations, the eigenvalue problem $-\Delta v=\lambda v,\qquad v\big|_{\partial \Omega}=0$ is solved by separating the $x,y,z$ variables: $v=X(x)Y(y)Z(z)$, $ \frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=-\lambda$

The separated BCs are $X(0)=X(\pi)=Y(0)=Y(\pi)=Z(0)=Z(\pi)=0$

Here are my questions:

  • How do we deduce that the solutions are $v(x,y,z)=\sin lx\sin my \sin nz$ where $l^2+m^2+n^2=\lambda\quad (1\leq m,l,n<\infty)$

  • Why the shape of $\Omega=[a,b]\times[c,d]\times[e,f]$ is needed for this kind of method?

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From \frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}=-\lambda, we get that exists $(c_1,c_2,c_3)$ such that \frac{X''(x)}{X(x)} =c_1, \quad\frac{Y''(y)}{Y(y)}=c_2, \quad\frac{Z''(z)}{Z(z)}=c_3 and $c_1+c_2+c_3=-\lambda$. You will get the result by solving those differential equations.

We have to define $X$, $Y$ and $Z$ and these functions have to be defined on intervals, which explains the shape of $\Omega$.

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    The point is that $X''(x)/X(x)$ depends only on $x$, but since this is also $-\lambda - Y''(y)/Y(y) - Z''(z)/Z(z)$ which depends only on $y$ and $z$, it must be constant. As for the shape, $X(x) Y(z) Z(z) = 0$ when one of the three factors is 0. So you get a boundary consisting of (pieces of) planes where $x$, $y$ and $z$ are constant.2011-04-22