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

PDE with oscillating boundary conditions

  1. Aug 22, 2012 #1
    Hi,

    Say I have this pde:

    [itex]u_t=\alpha u_{xx}[/itex]
    [itex]u(0,t)=\sin{x}+\sin{2x}[/itex]
    [itex]u(L,t)=0[/itex]

    I know the solution for the pde below is v(x,t):

    [itex]v_t=\alpha v_{xx}[/itex]
    [itex]v(0,t)=\sin{x}[/itex]
    [itex]v(L,t)=0[/itex]

    And I know the solution for the pde below is w(x,t)

    [itex]w_t=\alpha w_{xx}[/itex]
    [itex]w(0,t)=\sin{2x}[/itex]
    [itex]w(L,t)=0[/itex]

    Would the complete solution be u(x,t)=v(x,t)+w(x,t)?
     
  2. jcsd
  3. Aug 22, 2012 #2
    Your first boundary conditions are screwed up: the left side is a function of t and the right side is a function of x. Assuming you fix that, then yes, functions satisfying pieces of the boundary conditions can be added together just as you've done. You can confirm thus just by substituting [itex]u[/itex] into the original PDE and using the linearity of differentiation.
     
  4. Aug 22, 2012 #3
    Thanks!

    I was worried I was doing it wrong. I meant to write [itex]u(0,t)=\sin{t}+\sin{2t}[/itex] , ...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: PDE with oscillating boundary conditions
Loading...