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1-D Wave equation with mixed boundary conditions

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve, [tex]u_{t} = u_{xx}c^{2}[/tex]

    given the following boundary and initial conditions

    [tex]u_{x}(0,t) = 0, u(L,t) = 0[/tex]

    [tex]u(x,0) = f(x) , u_{t}(x,0) = g(x)[/tex]

    2. Relevant equations

    [tex]u(x,t) = F(x)G(t)[/tex]

    3. The attempt at a solution
    I solved it, I am just not sure if it is right.

    [tex]u(x,t) = \sum_{n=1}^\infty(a_{n}cos(\lambda_{n}t) + b_{n}sin(\lambda_{n}t))cos((n-\frac{1}{2})\frac{\pi}{L}x)

    , \lambda_{n} = (n-\frac{1}{2})\frac{\pi}{L}c [/tex]

    [tex]a_{n} = \frac{2}{L}\int_0^L f(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx,

    b_{n} = \frac{4}{(2n-1)c\pi}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx

    Can someone please verify this for me?

    Thanks in advance,
    Last edited: Dec 12, 2009
  2. jcsd
  3. Dec 13, 2009 #2


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    Science Advisor

    Why does your formula for bn[/b] involve cosine rather than sine?
  4. Dec 13, 2009 #3
    Thanks for replying HallsofIvy.

    Well to get [tex]b_{n}[/tex] I applied the initial condition

    [tex]u_{t}(x,0) = g(x).[/tex]

    So I first find [tex]u_{t}[/tex] which comes to

    [tex]u_{t} = \sum_{n=1}^\infty(-a_{n}\lambda_{n}sin(\lambda_{n}t) + b_{n}\lambda_{n}cos(\lambda_{n}t))cos((n-\frac{1}{2})\frac{\pi}{L}x) [/tex]

    now evaluating [tex]u_{t}[/tex] at t = 0 and setting it equal to g(x) we get,

    [tex]u_{t}(x,0) = \sum_{n=1}^\infty b_{n}\lambda_{n}cos((n-\frac{1}{2})\frac{\pi}{L}x) = g(x) [/tex]

    But that is just the Fourier cosine series of g(x) just with the extra [tex]\lambda_{n}[/tex] and with the zero term equal to zero

    so [tex]b_{n} = \frac{2}{\lambda_{n}}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx[/tex]

    where [tex]\lambda_{n} = (n-\frac{1}{2})\frac{\pi}{L}c

    [tex] \Rightarrow b_{n} = \frac{4}{(2n-1)c\pi}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx[/tex]

    Please let me know if I did something wrong.

    Thanks again,
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