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Math Help - Pde

  1. #1
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    Pde

    The displacement for the transverse vibrations of a beam satisfies the differential equation
    utt + (b^2)uxxxx = 0
    Find u(x,t) if u(0,t) = 0, u(L,t) = 0, uxx(0,t) = 0, uxx(L,t) = 0, u(x,0) = f(x), ut(x,0) = g(x)

    I've never encountered a differential equation like this before. How do you solve it?
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  2. #2
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    Quote Originally Posted by JimmyT View Post
    The displacement for the transverse vibrations of a beam satisfies the differential equation
    utt + (b^2)uxxxx = 0
    Find u(x,t) if u(0,t) = 0, u(L,t) = 0, uxx(0,t) = 0, uxx(L,t) = 0, u(x,0) = f(x), ut(x,0) = g(x)

    I've never encountered a differential equation like this before. How do you solve it?
    Same way as with u_t = u_{xx} and u_{xx} + u_{yy} = 0 - separation of variables.
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  3. #3
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    So I get d^2T/dt^2 = (b^2)T*lambda and d^4X/dx^4 = X*lambda

    How do I solve it from here?
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  4. #4
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    Quote Originally Posted by JimmyT View Post
    So I get d^2T/dt^2 = (b^2)T*lambda and d^4X/dx^4 = X*lambda

    How do I solve it from here?
    Pick X^{(4)} = \lambda X first. Because of your boundary conditions \lambda = \omega^4 so X^{(4)} - \omega^4 X = 0. The characteristic equation m^4 - \omega^4 = 0\;\Rightarrow\; m = \pm i, \pm 1 so the solutions are

    X = c_1 \sin \omega x + c_2 \cos \omega x + c_3 \sinh \omega x + c_4 \cosh \omega x .

    Your bc's will determine the constants and the \omega. Then go the the T equation.
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