Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple-looking PDE

  1. Dec 29, 2011 #1
    Dear All,

    I got some trobule in solving the following simple-looking PDE's. Can anyone give a hint about how to solve it? thanks a lot! I guess the solution is of the form [itex]y(u,v)=A[\cos(k(u-f(v))-B]\cosh(v)-C[/itex]. But I don't know a formal way to solve.



    [itex]\frac{\partial^4y(u,v)}{\partial u^2 \partial v^2} +k^2 \frac{\partial^2y(u,v)}{\partial v^2}=0[/itex]

    [itex]\frac{\partial^2y(u,v)}{\partial u^2} +k^2 y(u,v) +b=0[/itex]

    Boundary condition:

    [itex]\frac{\partial y(u,v)}{\partial v}|_{v=0}=0[/itex]
    [itex]\frac{\partial y(u,v)}{\partial u}|_{u=0,v=0}=0[/itex]
     
  2. jcsd
  3. Dec 30, 2011 #2
    If you differentiate the second equation twice with respect to v, you will see that it is basically the same equation as the first equation. Just integrate the first equation twice to get rid of the partial derivatives wrt v and you end up with two almost identical second order ODE's.

    And in the second equation, the fact that y depends on v is not important, so to simplify things, you could start by solving y" + k^2y+b=0
    When you have found the solution, it is trivial to extend it to the case where y depends on v.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple-looking PDE
  1. Simple PDE (Replies: 3)

  2. Simple PDE (Replies: 2)

  3. Simple PDE (Replies: 3)

  4. It looks simple but (Replies: 7)

Loading...