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Wave equation PDE, can't match initial conditions

  1. Nov 19, 2007 #1
    First post, hooray! Undergrad nuke engineer here, trying to figure out a really annoying PDE. My notation for U_xx = 2nd partial of U with respect to x, U_tt = 2nd partial of U with respect to t, etc.

    1. The problem statement, all variables and given/known data
    I'm working a nonhomogenous PDE with homogeneous initial and boundary conditions. My prof told me to first solve it as if it were a homogeneous PDE and then do some other stuff. I can't even get that done, usually I have no problem.

    Given U_xx + sin(x)cos(t) = U_tt, 0<x<pi, t>0
    BC: U(0,t)=0 U(pi,t)=0
    IC:U(x,0)=0 U_t(x,0)=0

    3. The attempt at a solution
    I solved the PDE as if it were homogeneous (U_xx = U_tt), used U(x,t)=X(x)T(t), set X"/X = T"/T = -lambda^2.
    My Sturm-Louisville problems were X"+lambda^2 * X = 0 and T" + lambda^2 * T=0

    I started solving for X(x). lambda^2=0 yielded only the trivial solutions. Lambda^2>0 gave X(x)=Acos(lambda*x) + Bsin(lambda*x). The initial condition u(0,t) = 0 made A=0, then u(pi,t)=0 gave lambda = n; n =1, 2,3 ...

    The trouble started when I tried to solve T(t). Using lambda^2 >0 , I got T=Ccos(lambda*t)+Dsin(lambda*t). The initial condition u(x,0)=0 gave C=0.

    However, U_t(x,0)=0 messed me up. T'(t)=D*lambda*cos(lambda*t). I already had lambda = n; n=1,2,3 from solving X(x). Those eigenvalues do not make T'(t) = 0. I would need n/2; n=1,3,5 to make T'(t)=0 and satisfy the Neumann.

    Any ideas?
  2. jcsd
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