- #1
whatisreality
- 290
- 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?