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Math Help - Wave Equation problem

  1. #1
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    Wave Equation problem

    Solve the wave equation subject to the boundary conditions of u(0,t) - 0 for t>=0, and u(L,t)=0, for t>=0.

    u(x,0) = f(x) for 0<=x<=L
    du/dt (x,0) = g(x)

    f(x) = sin (2pix/L) , g(x) = 0

    Note: Sorry but I don't even understand what the problem is asking, I have a really bad professor for this course (Applied Math), and to be honest, I had him before and I had to learn everything on my own. But then was Calculus I, and now this is difficult, I don't know even where to start, please help!

    Thank you.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Solve the wave equation subject to the boundary conditions of u(0,t) - 0 for t>=0, and u(L,t)=0, for t>=0.

    u(x,0) = f(x) for 0<=x<=L
    du/dt (x,0) = g(x)

    f(x) = sin (2pix/L) , g(x) = 0

    Note: Sorry but I don't even understand what the problem is asking, I have a really bad professor for this course (Applied Math), and to be honest, I had him before and I had to learn everything on my own. But then was Calculus I, and now this is difficult, I don't know even where to start, please help!

    Thank you.
    The problem is to solve:
    u_{tt} - c^2u_{xx} = 0
    given
    u(0, t) = 0 for t \geq 0
    and
    u(L, t) = 0 for t \geq 0
    and
    u(x, 0) = sin \left ( \frac{2 \pi x}{L} \right ) for 0 \leq x \leq L
    and
    u_t(x, 0) = 0

    The general solution of the wave equation is
    u(x, t) = \frac{f(x - ct) + f(x + ct)}{2} + \frac{1}{2c} \int_{x - ct}^{x + ct} g(s) ds
    where
    u(x, 0) = f(x)
    and
    u_t(x, 0) = g(x)

    So the solution will be of the form:
    u(x, t) = \frac{1}{2} sin \left ( \frac{2 \pi (x - ct)}{L} \right ) + \frac{1}{2} sin \left ( \frac{2 \pi (x + ct)}{L} \right )

    You may check that this solution indeed satisfies the boundary conditions.

    -Dan
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  3. #3
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    Another way which works nicely here is to use seperation of variables. Since \cos \frac{2\pi x}{L} is orthogonal to sines and cosine the infinite series is actually finite. Try it.

    I need to leave now. When I come back.
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  4. #4
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    I'm still quite lost. So far we still haven't learn this general solution of <br />
u(x, t) = \frac{f(x - ct) + f(x + ct)}{2} + \frac{1}{2c} \int_{x - ct}^{x + ct} g(s) ds<br />

    Is this given or you actually compute it?

    My professor did one example, and man, he didn't even get to finish it because he tried to do it in the last few seconds of the class. He has f(x) = 0, g(x) = sin(pi*x/L). Then he took the du/dt, and then he stopped there...

    So is that a way to solve it without using the general solution? Like the separation of variables as the last post suggested?

    Thank you.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tttcomrader View Post
    I'm still quite lost. So far we still haven't learn this general solution of <br />
u(x, t) = \frac{f(x - ct) + f(x + ct)}{2} + \frac{1}{2c} \int_{x - ct}^{x + ct} g(s) ds<br />

    Is this given or you actually compute it?

    My professor did one example, and man, he didn't even get to finish it because he tried to do it in the last few seconds of the class. He has f(x) = 0, g(x) = sin(pi*x/L). Then he took the du/dt, and then he stopped there...

    So is that a way to solve it without using the general solution? Like the separation of variables as the last post suggested?

    Thank you.
    You might find this helpful.

    -Dan
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  6. #6
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    Thanks! Now I understand how to get the general solution and how to use it.

    I presume that the second part of the general solution, the integral of g(s) will go to zero as g(x) = 0? (1st fundamental theorem of cal?)
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tttcomrader View Post
    Thanks! Now I understand how to get the general solution and how to use it.

    I presume that the second part of the general solution, the integral of g(s) will go to zero as g(x) = 0? (1st fundamental theorem of cal?)
    Yup!

    -Dan
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  8. #8
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    Quote Originally Posted by tttcomrader View Post
    Solve the wave equation subject to the boundary conditions of u(0,t) - 0 for t>=0, and u(L,t)=0, for t>=0.

    u(x,0) = f(x) for 0<=x<=L
    du/dt (x,0) = g(x)

    f(x) = sin (2pix/L) , g(x) = 0
    Given the equation,
    \frac{\partial ^ 2 u}{\partial t^2} - c^2 \frac{\partial ^2 u}{\partial x^2} = 0 .

    With homogenous boundary value problem,
    u(0,t)=u(L,t)=0

    And initial value problems,
    u(x,0)=f(x) \mbox{ and }u_t(x,0)=g(x)

    The solution is,
    u(x,t) = \sum_{n=1}^{\infty} \left( A_n \cos \frac{\pi n c t}{L} + B_n \sin \frac{\pi n c t}{L}\right) \sin \frac{\pi n x}{L}

    Where,
    A_n  = \frac{2}{L} \int_0^L f(x) \sin \frac{\pi n x}{L} dx
    B_n = \frac{2}{\pi n c}\int_0^L g(x) \sin \frac{\pi n x}{L} dx

    In this problem g(x) = 0 \mbox{ on }[0,L] thus, B_n = 0 for all n\geq 1.

    And, f(x) = \sin \frac{2\pi x}{L}.
    Thus, A_n = \frac{2}{L}\int_0^L \sin \frac{2\pi x}{L} \cdot \sin \frac{\pi n x}{L} dx


    Use the following result:
    Theorem: The set \left\{ \sin \frac{\pi n x}{L} \right\} is orthogonal on [-L,L].
    And furthermore,
    \int_{-L}^L \sin \frac{\pi n x}{L} \sin \frac{\pi m x}{L} dx = L \delta_{nm}.
    (Where \delta_{nm} is the Kronecker Delta.)

    Which means,
    \int_0^L \sin \frac{2\pi x}{L} \sin \frac{\pi n x}{L} = \frac{L}{2} \cdot \delta_{2n} = \left\{ \begin{array}{c}\frac{L}{2} \mbox{ if }n=2 \\ 0 \mbox{ if }n\not = 2 \end{array} \right.

    This tells us that A_1=0,A_2=1,A_3=0,A_4=0,....

    Hence the solution is,
    u(x,t) = \cos \frac{2\pi ct}{L} \sin \frac{2\pi x}{L}


    NOTE: Topsquark used something called d'Alembert's General Solution.
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