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Thread: 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 $\displaystyle 0 < x < L$ and $\displaystyle t > 0$ that the PDE

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

    with boundary conditions

    $\displaystyle u(0,t) = u_x (L,t) = 0$

    has the general solution

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

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

    $\displaystyle u(0,t) = 0 $, $\displaystyle u_x(L,t) = 0$

    $\displaystyle u(x,t) = F(x) G(t)$

    $\displaystyle u_x(x,t) = F'(x) G(t)$

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

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

    Substitute back into PDE

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

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

    Both must equal so constant $\displaystyle k = -p^2$

    This gives 2 ODE's:

    $\displaystyle F'' + p^2 F = 0$

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

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

    $\displaystyle u(0,t) = F(0) G(t) = 0$

    $\displaystyle \Rightarrow \ F(0) = 0$

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

    $\displaystyle F(x) = B \sin (px)$

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

    $\displaystyle u_x(L,t) = F'(L) G(t) = 0$

    $\displaystyle \Rightarrow \ F'(L) = 0$

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

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

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

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

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

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

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

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

    Then $\displaystyle u_n(x,t) = F_n(x) G_n(t)$

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

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

    Therefore

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