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Math Help - Partial Differential Equations

  1. #1
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    Partial Differential Equations

    Hi I am stuck on a PDE boundary question, I think I have shown the equation seperates and understand what the boundarys mean but I am not sure how to show the general solution and then solve the boundary value problems. Please help

    Assume solution u(x,t) = X(x)T(t) to the modified diffusion equation u_t - Du_{xx} - au = 0.

    First it says show the equation seperates, I think the follwing is correct?:

    substituting the solution into the equation:
    u_t - au = Du_{xx}

     <br />
XT' - aXT = DX''T<br />

    so \frac {T'}{T} - a = D \frac {X''}{X}

    Then it wants the general solution for X(x) and T(t)

    and then assuming D>0, a>=0, L>0 solve the boundary value problem:
    u_t - Du_{xx} - au = 0 for all 0=<x=<L , t>=0

    u(0,t) = u(L,t) = 0 for all t>=0

     <br />
u(x,0) = sin(\pi x/L) + sin(2\pi x/L)<br />

    and finally characterise the differenc between solutions for a>D\pi^2 / L^2 and a<D\pi^2 / L^2

    I have found a similiar problem in Boyce's elementary Diff equations book but just cant work it out!

    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathgrad View Post
    Hi I am stuck on a PDE boundary question, I think I have shown the equation seperates and understand what the boundarys mean but I am not sure how to show the general solution and then solve the boundary value problems. Please help

    Assume solution u(x,t) = X(x)T(t) to the modified diffusion equation u_t - Du_{xx} - au = 0.

    First it says show the equation seperates, I think the follwing is correct?:

    substituting the solution into the equation:
    u_t - au = Du_{xx}

     <br />
XT' - aXT = DX''T<br />

    so \frac {T'}{T} - a = D \frac {X''}{X}
    Assume that this is correct (your job to check), than the right hand side is a function of x only and the left hand side is a function of y only, so as they are equal they are also equal to a constant, which leaves you with the two ordinary differential equations:

    {T'}= (K+a) T

    and:

    X''=\frac{K}{D}X

    for some constant K

    RonL
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  3. #3
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    Jul 2007
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    Thanks, ok I understand that but still dont know how to find the general solution and solve the boundary problem? Sorry if I'm missing the point here.
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