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:


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

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

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.