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Thread: Eigenfunctions and PDE's

  1. #1
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    Eigenfunctions and PDE's

    Solve,

    $\displaystyle \frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=F(x,t)$

    $\displaystyle u(0,t)=0, \frac{\partial u}{\partial x}(\pi,t )=0, u(x,0)=0$

    where $\displaystyle F(x,t)$ is a given function.

    (a) Determine the x-dependent eigenfunctions appropriate to the solution of the equation with given boundary conditions.

    (b) Solve the method of eigenfunction expansions to obtain a series solution for the the problem in terms of the eigenfunctions determined in (a)

    (c) Simplify for the particular case $\displaystyle F(x,t)=1$


    Any help appreciated. Thanks
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  2. #2
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    Quote Originally Posted by featherbox View Post
    Solve,

    $\displaystyle \frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=F(x,t)$

    $\displaystyle u(0,t)=0, \frac{\partial u}{\partial x}(\pi,t )=0, u(x,0)=0$

    where $\displaystyle F(x,t)$ is a given function.

    (a) Determine the x-dependent eigenfunctions appropriate to the solution of the equation with given boundary conditions.

    (b) Solve the method of eigenfunction expansions to obtain a series solution for the the problem in terms of the eigenfunctions determined in (a)

    (c) Simplify for the particular case $\displaystyle F(x,t)=1$


    Any help appreciated. Thanks
    When you have a forced equation problem you can use Duhamel's principle
    Duhamel's principle - Wikipedia, the free encyclopedia

    So we need to solve the associated problem

    $\displaystyle \phi(x,t;s)$ such that

    $\displaystyle \phi_t-\phi_{xx}=0$

    $\displaystyle \phi(0,s)=0=\phi_{x}(\pi,s) \quad \phi(x,s)=F(x,s)$

    Separating variables as usual gives

    $\displaystyle \phi=XT \iff \frac{X''}{X}=\frac{\dot{T}}{T}=-\lambda^2$

    $\displaystyle T(t;s)=e^{-\lambda^2(t-s)} \quad X(x)=c_1\cos(\lambda x)+x_2\sin(\lambda x)$

    imposing boundary conditions gives $\displaystyle \lambda =\frac{1}{2}(2n+1)$

    $\displaystyle X(x)=\sin\left( \lambda x\right)$

    $\displaystyle \phi_n(x,t;s)=A_ne^{-\lambda^2(t-s)}\sin(\lambda)x$

    Now using the initial condition gives

    $\displaystyle F(x,s)=A_n\sin(\lambda x) \implies A_n = \frac{2}{\pi}\int_{0}^{\pi}F(x,s)\sin(\lambda x)dx$

    So phi has the form

    $\displaystyle \phi_n(x,t;s)=A_ne^{-\lambda^2(t-s)}\sin(\lambda x)$

    So now by Duhamel's Principle the final solution is

    $\displaystyle u(x,t)=\sum_{n=1}^{\infty}\int_{0}^{t}\phi_n ds$

    $\displaystyle u(x,t)=\sum_{n=1}^{\infty}\int_{0}^{t}\left( A_ne^{-\lambda^2(t-s)}\sin(\lambda x) \right)ds$
    Last edited by TheEmptySet; Jun 1st 2011 at 09:16 PM. Reason: typo
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