# Separation of variables PDE

1. Oct 21, 2015

### whatisreality

1. The problem statement, all variables and given/known data
The wave equation for ψ=ψ(t,x,y) is given by

$\frac{\partial ^2 \phi}{\partial t^2} - \frac{\partial ^2 \phi}{\partial x^2} - \frac{\partial ^2 \phi}{\partial y^2}$

Use separation of variables to separate the equation into 3 ODEs for T, X and Y. Use the separation constants
$-k_{x}^{2}X$ and $-k_{y}^{2}Y$

Do not introduce any more separation constants for T.
2. Relevant equations

3. The attempt at a solution
I'm fairly sure I know how to start.
Ansatz ψ(t,x,y) = T(t)X(x)Y(y). Sub the derivatives of this into the the wave equation:

$XY\frac{\partial ^2 T}{\partial t^2} - TY\frac{\partial ^2 X}{\partial x^2} - TX\frac{\partial ^2 Y}{\partial y^2}$=0

Then divide by TXY:
$\frac{1}{T}\frac{\partial ^2 T}{\partial t^2} - \frac{1}{X}\frac{\partial ^2 X}{\partial x^2} - \frac{1}{Y}\frac{\partial ^2 Y}{\partial y^2}$=0

Rearrange:
$\frac{1}{T}\frac{\partial ^2 T}{\partial t^2} = \frac{1}{X}\frac{\partial ^2 X}{\partial x^2} + \frac{1}{Y}\frac{\partial ^2 Y}{\partial y^2}$
Which is only possible if LHS = RHS = constant, I think? But then I would get
$\frac{1}{X}\frac{\partial ^2 X}{\partial x^2} + \frac{1}{Y}\frac{\partial ^2 Y}{\partial y^2}$ = constant, and how do I separate that further? Why would I need more than one constant of separation?
Because if
$\frac{1}{X}\frac{\partial ^2 X}{\partial x^2} + \frac{1}{Y}\frac{\partial ^2 Y}{\partial y^2}$ = constant

Then surely each individual term must also be a constant, and I can just write
$\frac{1}{X}\frac{\partial ^2 X}{\partial x^2}$ = $-k_{x}^{2}X$
$\frac{1}{Y}\frac{\partial ^2 Y}{\partial y^2}$ = $-k_{y}^{2}X$

And why wouldn't I introduce a new constant for T? What do I write instead,

$\frac{1}{T}\frac{\partial ^2 T}{\partial t^2}$ = constant?

2. Oct 21, 2015

### pasmith

You are missing the "= 0" which would turn that expression into an equation.

You mean
$$\frac 1X \frac{\partial^2 X}{\partial x^2} = -k_x^2 \\ \frac 1Y \frac{\partial^2 Y}{\partial x^2} = -k_y^2$$

Because you need $$\frac{1}{T}\frac{\partial ^2 T}{\partial t^2} - \frac{1}{X}\frac{\partial ^2 X}{\partial x^2} - \frac{1}{Y}\frac{\partial ^2 Y}{\partial y^2} = \frac{1}{T}\frac{\partial ^2 T}{\partial t^2} + k_x^2 + k_y^2 =0$$

3. Oct 21, 2015

### whatisreality

Oh yes, I did mean that. An yep, get the bit about not introducing a new constant too! Thank you :)