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Math Help - PDE

  1. #1
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    PDE

    Can someone help me solve this PDE:

    ut = uxx - (B^2)u, 0 < x < L, t > 0
    u(0,t) = U1, u(L,t) = U2, u(x,0) = f(x)
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    Can someone help me solve this PDE:

    ut = uxx - (B^2)u, 0 < x < L, t > 0
    u(0,t) = U1, u(L,t) = U2, u(x,0) = f(x)
    I will assume that U1 and U2 are constants. First you need to move them to zero. So let u = v + ax + b and let  U1 = 0 + a 0 + b,\;\;\; U2 = 0 + aL + b which gives  a\, \text{and}\, b thus

    u = v + \frac{U2-U1}{L} x + U1

    The problem now changes

    v_t = v_{xx} - B^2\left( v + \frac{U2-U1}{L} x + U1 \right)

    subject to v(0,t) = 0,\;v(L,t) = 0,\; v(x,0) = f(x) - \frac{U2-U1}{L} x - U1

    If there was no source term then a separation of variables would lead to


    u = \sum_{n=1}^{\infty} b_n e^{\omega^2 t} \sin \omega x,\;\;\; \omega = \frac{n \pi}{L}

    but because of the source term we try

    u = \sum_{n=1}^{\infty} T_n (t) \sin \omega x,\;\;\; \omega = \frac{n \pi}{L}

    sub. into the PDE and group terms accordiong to \sin w x. You should get an ODE for T_n which you can solve. Remember, you'll need a sine series for the B^2\left(\frac{U2-U1}{L} x + U1 \right)
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  3. #3
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    I'm having trouble solving the ODE for Tn. Does anyone know how to do it?
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