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Thread: Solving 1-d wave equation

  1. #1
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    Solving 1-d wave equation

    Hey guys,
    i have a small question. 1D-Wave equation is given as:

    $\displaystyle
    \frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2} $

    i need to find $\displaystyle u_s(x,t)=X(x)sinwt$

    From the equation form above , U_xx=$\displaystyle X''(x)sinwt$
    U_tt=$\displaystyle -w^2X(x)sinwt$ are obtained as given as a solution.

    My question is how $\displaystyle -w^2$ is obtained from $\displaystyle u_s(x,t)=X(x)sinwt$

    thank you,
    Last edited by Ali Sura; Mar 29th 2009 at 04:20 PM.
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  2. #2
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    What are your boundary conditions? That is what determines k.
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  3. #3
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    Two Boundary conditions are:

    1)
    $\displaystyle u(0,t)=0
    $


    2)
    $\displaystyle u(1,t)=sinwt
    $


    thanks,
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  4. #4
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    If $\displaystyle u_s(x,t)= X(x)sin(\omega t)$

    Then $\displaystyle u_s(0,t)= X(0)sin(\omega t)= 0$ so X(0)= 0.

    Also $\displaystyle u_s(1,t)= X(1)sin(\omega t)= sin(\omega t)$ so X(1)= 1.

    Now, $\displaystyle u_xx= X_xx sin(\omega t)$ and $\displaystyle u_tt= -\omega^2 X sin(\omega t)$ so the partial differential equation becomes $\displaystyle X_xx sin(\omega t)= -\omega^2 X sin(\omega t)$ for all x and t and so we must have $\displaystyle X_xx= -\omega^2 X$ for all x.\

    Can you solve the ordinary differential equation $\displaystyle X_xx+ \omega^2 X$ with the boundary conditions X(0)= 0, X(1)= 1?
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