1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)
    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


    User Avatar
    Homework Helper

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

    Let's try looking for a separable solution [itex]u(x,t) = X(x)T(t)[/itex]. Substituting that into the PDE, we get
    T' X - T X'' = e^{-2t} \sin\frac{\pi x}L
    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


    User Avatar
    Science Advisor
    Homework Helper
    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 ?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted