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Homework Help: PDE: separation of variables problem

  1. Nov 18, 2005 #1
    I am to reduce the following PDE to 2 ODEs and find only the particular solutions:

    u_tt - u_xx - u = 0; u_t(x,0) = 0; u(0,t) = u(1,t) = 0

    I guess u = X(x)T(t), and plug u_tt, u_xx into PDE and divide by u to get:

    T''/T = X''/X + 1 = K

    I solve X'' + (1-K)X = 0 first.

    From characteristic equation, r = +- sqrt(1-K)

    Becausse of boundary conditions, we must have 1-K < 0

    So r = +- i sqrt(K-1)

    (I think I make an error somewhere around here...)

    ==> u = c1 cos(sqrt(K-1)x) + c2 sin(sqrt(K-1)x)

    From boundary conditions, c1 =0, and sin(sqrt(K-1)) = 0 => sqrt(K-1) = n*pi

    => K-1 = n^2*pi^2
    => K = n^2*pi^2 + 1

    Correct so far?

    Now for T:

    T'' - KT = 0

    T'' - (n^2*pi^2 + 1)T = 0

    r = +- sqrt(n^2*pi^2+1)

    I think my K = n^2*pi^2 + 1 is wrong because it is strictly positive and I don't think an e^t solution will satisfy the initial condition.

    Please help.

    The book's answer is cos( sqrt(n^2*pi^2-1)t) * sin(n*pi*x)
  2. jcsd
  3. Nov 18, 2005 #2


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    Homework Helper

    When you separate variables and equate to a separation constant, usually need to check the boundary conditions against a positive, negative, and zero separation constant. For now, these types of equations just have a negative separation constant. That is:

    [tex]\frac{T^{''}}{T}=\frac{X^{''}}{X}+1=K=-\lambda;\quad \lambda>0[/tex]

    So, need to solve:




    Solving for the roots:

    [tex]m=\pm \sqrt{-(1+\lambda)}[/tex]

    Thus have:


    Substituting the boundary conditions yield:

    [tex]\lambda=(n\pi)^2-1;\quad n>0[/tex]


    [tex]X(x)=A_2Sin(n\pi x)[/tex]

    Can you do T now?

    Edit: Also, I think you need another constraint to obtain a particular solution, you know, wave equations usually specify the initial velocity as well.
    Last edited: Nov 18, 2005
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