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Ordinary differential equations and BVP

  1. Sep 28, 2013 #1
    Solve BVP by separating variables and using eigenfunction expansion method




    PDE:Ut-Uxx=e-2tsin(pi x/L) U=U(x,t),x(0,L)
    BC1:U(0,t)=0
    BC2:U(L,t)=0
    IC:U(x,0)=sin(pi x/L)
    U(x,t)=X(x)T(t),X''=(lambda) X ,lambda is the separation of parameter.

    I have calculated the basis functions as Xn=root(2/L)sin((n pi x) /L) for n=1,2 and labdan=-((n pi)/L)2

    Please help to solve the problem
     
  2. jcsd
  3. Sep 28, 2013 #2

    pasmith

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    Homework Helper

    Your PDE is not homogeneous. You have:
    [tex]
    u_{t} - u_{xx} = e^{-2t} \sin\frac{\pi x}L
    [/tex]

    Let's try looking for a separable solution [itex]u(x,t) = X(x)T(t)[/itex]. Substituting that into the PDE, we get
    [tex]
    T' X - T X'' = e^{-2t} \sin\frac{\pi x}L
    [/tex]
    Oh dear. Dividing by [itex]XT[/itex] isn't necessarily going to give us terms which are functions of a single variable and whose sum is a constant, so the standard method doesn't work (or rather, we are forced to make particular choices for X and T in order to do that, and those choices don't lead to a solution of the PDE).

    But if we look at the initial and boundary conditions, we see that the initial condition is a constant multiple of [itex]\sin \frac{\pi x}L[/itex] and the boundary conditions require that [itex]u[/itex] vanish at points where [itex]\sin \frac{\pi x}L[/itex] vanishes. This suggests that we should try [itex]X(x) = \sin \frac{\pi x}L[/itex].
     
  4. Sep 28, 2013 #3

    arildno

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    Science Advisor
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    Gold Member
    Dearly Missed

    Make a trial solution e^(-2t)*F(t)*G(x)

    That will remove one part of your problem.
     
  5. Sep 30, 2013 #4
    I am a little unsure if I have the correct approach but here goes

    U(x,0) = X(x)T(0) = sin(pi*x/L) (the initial condition)

    Let a=pi/L then X(x) = sin(ax)/T(0), Here, X(0)T(t) = X(L)T(0) = 0 (the boundary conditions)

    Substituting this back in the original equation gives:

    d/dt[T(t)]*sin(ax) + T(t)*sin(ax)*a^2 = exp(-2t)*sin(ax)*T(0)

    Solving this gives:

    T(t) = c0*exp(-a^2*t) + T(0)*exp(-2t)/(a^2-4)

    U(x,t) = X(x)*T(t) = [ c1*sin(ax)*exp(-a^2*t) + sin(ax)*exp(-2t)/(a^2-4) ]

    Is the answer correct? How to use eigen function expansion method ?
     
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