Solve PDE with separation of variables

  • #1

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?
 

Answers and Replies

  • #2
Orodruin
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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)?
 
  • #3
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.
 
  • #4
Orodruin
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No, it asks you to show there is a solution of that form.
 
  • #5
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.
 
  • #6
Orodruin
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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
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?
 
  • #8
Or am I being silly, and I just let B=0? Would I need to justify that?
 
  • #9
Orodruin
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Or am I being silly, and I just let B=0? Would I need to justify that?
Your solution for T is a solution independent of what the values of A and B are. In particular, B=0 is a solution.
 

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