# Solve PDE with separation of variables

1. Oct 22, 2015

### whatisreality

1. The problem statement, all variables and given/known data
The wave equation for ψ(t, x) in 3D is

$\frac{\partial ^2 \psi}{\partial t^2}$ - Δ $\psi =0$
Let ϒ(x) satisfy Δϒ = λϒ where λ<0.

The x is in bold presumably to indicate it is in 3D, so represents also y and z?

Show there is a solution of the form ψ(t, x) = sin(ωt)ϒ(x) and find ω.

The attempt at a solution
I'm going to stop putting x in bold now, though it still represents x, y and z.

I tried using separation of variables, so letting ψ = ϒ(x)T(t). That would mean

$\frac{\partial ^2 \psi}{\partial t^2} =$ ϒ(x)$\frac{\partial ^2 T(t)}{\partial t^2}$

Sub this into the given wave equation:
ϒ(x)$\frac{\partial ^2 T(t)}{\partial t^2} -$Δ$\psi =0$

This is most probably the bit that's wrong, but I think
Δ$\psi$ = T Δϒ(x)
Which, subbing in the given expression for Δϒ, gives

Δ$\psi$ = Tλϒ. Putting that all into the wave equation:

ϒ(x)$\frac{\partial ^2 T(t)}{\partial t^2}-$Tλϒ = 0

But ϒ vanishes from this equation, and if I carry on to solve the remaining equation for T:
$\frac{\partial ^2 T(t)}{\partial t^2}-T \lambda$ = 0
Given $\lambda$ is negative, -T$\lambda$ becomes +T x (magnitude of lambda)

The solution to which is T = Asin($\sqrt{\lambda}t$) + Bcos($\sqrt{\lambda}t$)
Nowhere near the solution I'm after! Where did that go wrong?

2. Oct 22, 2015

### Orodruin

Staff Emeritus
What is the solution that you are after? Why do you think yours is wrong (apart from the fact that you should have the absolute value of $\lambda$ under the square roots)?

3. Oct 22, 2015

### whatisreality

The question asks me to show that the solution is of the above form.

4. Oct 22, 2015

### Orodruin

Staff Emeritus
No, it asks you to show there is a solution of that form.

5. Oct 22, 2015

### whatisreality

Isn't my solution of a different form to the one asked for though? I haven't shown that there is a solution of that form.

6. Oct 22, 2015

### Orodruin

Staff Emeritus
You have just quoted what you get for $T(t)$. Try multiplying it with $\Upsilon(\vec x)$ and select suitable constants $A$ and $B$.

7. Oct 22, 2015

### whatisreality

Oh, right. So I've said the solution is ψ = T(t)ϒ, which subbing in the solution for T is

$\psi =$ϒ$Asin(t\sqrt{\lambda})$+ϒ$Bcos(t\sqrt{\lambda})$ (Where $\lambda$ represents the magnitude of lambda)
So are both terms individually solutions? Because the sin terms does look like the solution I'm asked for. But I thought only both terms added together constituted a solution?

8. Oct 22, 2015

### whatisreality

Or am I being silly, and I just let B=0? Would I need to justify that?

9. Oct 22, 2015

### Orodruin

Staff Emeritus
Your solution for T is a solution independent of what the values of A and B are. In particular, B=0 is a solution.