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My question is:

How do you use separation of variables on a PDE that has more than one constant in it?

All the examples I can find in my book/online only have one constant in it, like $$ \frac{\partial u}{\partial t}=k\frac{\partial^2 u}{\partial x^2}+k\frac{\partial^2 u}{\partial y^2}$$

Which then means only one of the resultant ODEs has $k$ in it. If there is more than one constant, what do you do? Still use separation of variables? My attempt is below the question.

The problem I am working on is:

Solve

$\displaystyle \frac{\partial u}{\partial t}=k_1\frac{\partial^2 u}{\partial x^2}+k_2\frac{\partial^2 u}{\partial y^2}$

on a rectangle $(0 \lt x \lt L, 0 \lt y \lt H)$ subject to

$\displaystyle u(0,y,t)=0, \frac{\partial u}{\partial y}(x,0,t)=0$

$\displaystyle u(0,y,0)=\alpha(x,y)$

$\displaystyle u(L,y,t)=0, \frac{\partial u}{\partial y}(x,H,t)=0$

What I have done so far:

Let $u(x,t)=X(x)Y(y)T(t)$. Then using separation of variables we have:

(1) $\displaystyle \frac{d T}{d t}=-\lambda(k_1+k_2)T, \text{ IC: }T(0)=\alpha(x,y)$

(2) $\displaystyle \frac{d^2X}{dx^2}=-\mu \frac{(k_1+k_2)X}{k_1}, \text{ BCs: }X(0)=0, X(L)=0$

(3) $\displaystyle \frac{d^2Y}{dy^2}=-(\lambda - \mu) \frac{(k_1+k_2)Y}{k_2}, \text{ BCs: }Y'(0)=0, Y'(H)=0$

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    $\begingroup$ What if you now solve these ODE's? The first gives an exponential, the second and third gives sines and cosines. That gives you $T,X$ and $Y$. And then check if $XYT$ is a solution? $\endgroup$ – Jens Wagemaker Mar 11 at 20:56
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You could also change the order of the assignment of variables. In $$ \frac{T'(t)}{T(t)}=k_1\frac{X''(x)}{X(x)}+k_2\frac{Y''(y)}{Y(y)} $$ all terms must be constants, so that you can set $\frac{X''(x)}{X(x)}=-λ$, $\frac{Y''(y)}{Y(y)}=-μ$ and thus in consequence $\frac{T'(t)}{T(t)}=-(k_1λ+k_2μ)$. I think the expressions for the constants in this form are less complex.

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  • $\begingroup$ So using this method, the three ODEs to solve become: (1) $\displaystyle X''+\lambda X=0, \text{ BCs: }X(0)=0, X(L)=0$ (2) $\displaystyle Y''+\mu Y=0, \text{ BCs: }Y'(0)=0, Y'(H)=0$ (3) $\displaystyle T'+(k_1 \lambda +k_2 \mu)T=0, \text{ IC: }T(0)=\alpha(x,y)$ ? $\endgroup$ – LovesPeanutButter Mar 11 at 21:48
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    $\begingroup$ Yes up to the last condition. The IC for $T$ is the Fourier coefficient corresponding to $(λ,μ)$ in the expansion of the initial function $α$ in the eigenbasis. $\endgroup$ – LutzL Mar 11 at 23:22

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