# Solve PDE with separation of variables

## Homework Statement

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?

Orodruin
Staff Emeritus
Homework Helper
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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)?

Show there is a solution of the form ψ(t, x) = sin(ωt)ϒ(x) and find ω.
The question asks me to show that the solution is of the above form.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
No, it asks you to show there is a solution of that form.

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

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

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

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

Orodruin
Staff Emeritus