1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solve PDE with separation of variables

  1. Oct 22, 2015 #1
    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. jcsd
  3. Oct 22, 2015 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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)?
     
  4. Oct 22, 2015 #3
    The question asks me to show that the solution is of the above form.
     
  5. Oct 22, 2015 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, it asks you to show there is a solution of that form.
     
  6. Oct 22, 2015 #5
    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.
     
  7. Oct 22, 2015 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    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##.
     
  8. Oct 22, 2015 #7
    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?
     
  9. Oct 22, 2015 #8
    Or am I being silly, and I just let B=0? Would I need to justify that?
     
  10. Oct 22, 2015 #9

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Your solution for T is a solution independent of what the values of A and B are. In particular, B=0 is a solution.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solve PDE with separation of variables
Loading...