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Thread: Another form of diffusion equation

  1. #1
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    Another form of diffusion equation

    now by defining several dimensionless variables, we got another form of diffusion equation:
    $\displaystyle U_{T}=U_{XX} \ \ \ \ T>0, \ \ 0<X<1 $

    X,T,U here are all dimensionless.

    U with initial conditions:
    $\displaystyle U(X,0)=0, \ \ 0\leq X\leq1 $ and boundary conditions :
    $\displaystyle U(0,T)=T, \ \ U(1,T)=0, \ \ T\geq0 $

    $\displaystyle U(X,T)=T(1-X)+V(X,T) $

    find the analytic solution of this problem.
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    what i got

    what i got here,

    $\displaystyle V(X,0)=V(0,T)=V(1,T)=0 \ \ \ \ (1-X)+V_{T}=V_{XX} $


    but i have no idea how to solve.
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  3. #3
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    Quote Originally Posted by silversand View Post
    what i got here,

    $\displaystyle V(X,0)=V(0,T)=V(1,T)=0 \ \ \ \ (1-X)+V_{T}=V_{XX} $


    but i have no idea how to solve.
    If you were to solve

    $\displaystyle v_t = v_{xx}$ subject to $\displaystyle v(0,t) = v(1,t) = 0$

    you would use a separation of variables and ultimately lead to

    $\displaystyle v = \sum_{n=1}^\infty T_n(t) \sin n \pi x $

    where

    $\displaystyle T_n(t) = c_n e^{- n^2 \pi^2 t}$

    Since you have a source term in your new equation

    $\displaystyle v_t = v_{xx} + \underbrace{x-1}_{source}$

    then find a Fourier sine series for the source term of the form

    $\displaystyle x - 1 = \sum_{n=1}^\infty k_n \sin n \pi x $

    and seek a solution of your porblem in the form

    $\displaystyle v = \sum_{n=1}^\infty T_n(t) \sin n \pi x $

    where $\displaystyle T_n(t)$ is to be determined (this should give rise to an ODE for $\displaystyle T_n$).
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    how

    how to determine the coefficients : $\displaystyle c_{n}, k_{n} $ ?
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  5. #5
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    Quote Originally Posted by silversand View Post
    how to determine the coefficients : $\displaystyle c_{n}, k_{n} $ ?
    The $\displaystyle c_n$ is found from the initial condition. For the $\displaystyle k_n $, it's $\displaystyle k_n = \frac{2}{1} \int_0^1 (x-1) \sin n \pi x\, dx$.
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