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Solution to Homogenous PDE

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the solution of:

    utt-uxx = sin(∏x) for 0<x<1
    u(x,0)=0 for 0<=x<=1
    ut(x,0)=0 for 0<=x<=1
    u(0,t)=0
    u(1,t)=0


    2. Relevant equations

    utt-uxx = sin(∏x) for 0<x<1
    u(x,0)=0 for 0≤x≤1
    ut(x,0)=0 for 0≤x≤1
    u(0,t)=0
    u(1,t)=0

    3. The attempt at a solution

    I was thinking that I would need to somehow make a change of variables, converting the PDE to a homogenous PDE and continuing from there. I don't really know how I can do this or solve it at all without some assistance!
     
  2. jcsd
  3. Sep 26, 2013 #2

    pasmith

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    The strategy is the same as for linear ODEs with constant coefficients: Find a particular solution which satisfies the PDE but not necessarily the boundary conditions. Then add a complementary solution (a solution of [itex]u_{tt} - u_{xx} = 0[/itex]) in order to satisfy the boundary conditions.
     
  4. Sep 26, 2013 #3
    So basically something like Autt-Bxx= sin∏x and Cutt-Duxx=0 and then sum the up into a solution u(x,t)?
     
  5. Sep 26, 2013 #4

    pasmith

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    No. You have [itex]u(x,t) = u_p(x,t) + u_c(x,t)[/itex] where [itex]u_p[/itex] is any solution of
    [tex]
    u_{tt} - u_{xx} = \sin \pi x
    [/tex]
    and [itex]u_c[/itex] is the solution of
    [tex]
    u_{tt} - u_{xx} = 0
    [/tex]
    subject to [itex]u_c(x,t) = -u_p(x,t)[/itex] on the boundary.

    But here it is actually simplest to try a solution of the form [itex]u(x,t) = f(t)\sin \pi x[/itex], given that [itex](\sin(\pi x))'' = -\pi^2 \sin(\pi x)[/itex]. This reduces the problem to an ODE for f(t).
     
  6. Sep 27, 2013 #5
    So you're saying that I need to set u(x,t)= u_tt-u_xx = f(t)sin(∏x)?
    How do I find the function f(t)?

    What I did was this:
    found u_xx, u_tt and plugged them into u_tt - u_xx = sin(∏x). What do I do with the boundary conditions and the initial conditions (how do they factor in)?
     
  7. Sep 27, 2013 #6

    pasmith

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    The conditions [itex]u(0,t) = f(t) \sin 0 = 0[/itex] and [itex]u(1,t) = f(t) \sin \pi = 0[/itex] are satisfied automatically.

    The remaining two conditions require that
    [tex]u(x,0) = f(0) \sin\pi x = 0[/tex]
    for all [itex]0 < x < 1[/itex] and
    [tex]u_t(x,0) = f'(0) \sin \pi x = 0[/tex]
    for all [itex]0 < x < 1[/itex]. This gives you initial conditions for [itex]f[/itex] and [itex]f'[/itex].
     
  8. Sep 29, 2013 #7
    Great! I got it!
    How would I solve this problem using a technique like changing variables instead of separation of variables?
     
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