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Math Help - 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:
     U_{T}=U_{XX} \ \ \ \ T>0, \ \ 0<X<1

    X,T,U here are all dimensionless.

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

     U(X,T)=T(1-X)+V(X,T)

    find the analytic solution of this problem.
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  2. #2
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    what i got

    what i got here,

     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,

     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

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

    you would use a separation of variables and ultimately lead to

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

    where

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

    Since you have a source term in your new equation

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

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

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

    and seek a solution of your porblem in the form

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

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

    how to determine the coefficients :  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 :  c_{n}, k_{n} ?
    The c_n is found from the initial condition. For the k_n , it's k_n = \frac{2}{1} \int_0^1 (x-1) \sin n \pi x\, dx.
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