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Math Help - General Solution of PDE using Fourier's method

  1. #1
    Member Maccaman's Avatar
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    General Solution of PDE using Fourier's method

    Hello,
    Can someone please have a look at this problem and tell me if I have answered it correctly (I do get the solution so I'm more concerned about the process being correct).

    Show that using the Fourier's method for 0 < x < L and t > 0 that the PDE

    u_t(x,t) = c^2u_{xx}(x,t)

    with boundary conditions

    u(0,t) = u_x (L,t) = 0

    has the general solution

    u(x,t) = \sum_{n=1}^\infty A_ne^{-(\frac{(2n+1)\pi c}{2L})^2t} \sin (\frac{(2n+1)\pi x}{2L})
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  2. #2
    Member Maccaman's Avatar
    Joined
    Sep 2008
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    Here is my answer:

    u_t(x,t) = c^2u_{xx}(x,t)

    u(0,t) = 0 , u_x(L,t) = 0

    u(x,t) = F(x) G(t)

    u_x(x,t) = F'(x) G(t)

    u_{xx}(x,t) = F''(x) G(t)

    u_t(x,t) = F(x) \dot{G}(t)

    Substitute back into PDE

    F\dot{G} = c^2 F''G

    \Rightarrow \ \frac{\dot{G}}{c^2G} = \frac{F''}{F}

    Both must equal so constant k = -p^2

    This gives 2 ODE's:

    F'' + p^2 F = 0

    \dot{G} + c^2 p^2 G = 0

    F(x) = A \cos (px) + B \sin (px)

    u(0,t) = F(0) G(t) = 0

    \Rightarrow \ F(0) = 0

    F(0) = A \ \Rightarrow \ A=0

    F(x) = B \sin (px)

    F'(x) = Bp \cos (px)

    u_x(L,t) = F'(L) G(t) = 0

    \Rightarrow \ F'(L) = 0

    F'(L) = Bp \cos(pL)

    \Rightarrow \ p = \frac{z \pi}{2L} for some z = 2n+1 for n = 1,2,3,....

    F_n(x) = \sin (\frac{(2n+1)\pi x}{2L}) , (Let B = 1)

    Let \lambda_n^2 = c^2 p^2

    = \frac{c^2 (2n+1)^2 \pi}{4L^2}

    \Rightarrow \ \lambda_n = \frac{c(2n+1) \pi}{2L}

    \dot{G} + \lambda_n^2 G = 0

    \Rightarrow G_n(t) = A_n e^{-\lambda_n^2 t} where A_n is some co-efficient

    Then u_n(x,t) = F_n(x) G_n(t)

    = A_n e^{-\lambda_n^2 t} \sin (\frac{(2n+1)\pi x}{2L})

    = A_n e^{-(\frac{(2n+1)c \pi}{2L})^2 t} \ \sin (\frac{(2n+1)\pi x}{2L})

    Therefore

    u(x,t) = \sum_{n=1}^\infty u_n(x,t) = \sum_{n=1}^\infty A_n e^{-(\frac{(2n+1)c \pi}{2L})^2 t} \ \sin (\frac{(2n+1)\pi x}{2L})
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