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Thread: BVP

  1. #1
    ux0
    ux0 is offline
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    BVP

    For what constant value of q does the boundary value problem

    $\displaystyle u_t=u_{xx} + q$ . . . $\displaystyle 0<x<2, t>0$
    $\displaystyle u_x(0,t)=2$ . . . . .$\displaystyle t \geq 0$
    $\displaystyle u_x(2,t)=0$ . . . . .$\displaystyle t \geq 0$
    $\displaystyle u(x,o)=f(x)$ . . . . .$\displaystyle 0 \leq x \leq 2$

    have a steady state solution? Find the complete solution when

    $\displaystyle
    f(x)=\left\{\begin{matrix}
    2x \mapsto 0 \leq x \leq 1
    \\
    x \mapsto 1 \leq x \leq 2
    \end{matrix}\right.$




    So to find the steady state solution i did the following steps. (The notation i use is the notation we use in class)

    $\displaystyle 0=U''+q \Rightarrow U''=-q$

    $\displaystyle U'(x)=-qx + A$

    $\displaystyle U(x)= \frac{-q}{2}x^2 + Ax + B$

    using my endpoint conditions i get

    $\displaystyle U'(2) = 0 = -q2 + A$

    $\displaystyle A=2q$

    $\displaystyle U'(0)=2=2q \rightarrow 1=q \rightarrow A=2$

    So my steady state solution is

    $\displaystyle U(x)=\frac{-1}{2}x^2+2x+B$

    which is only valid when $\displaystyle q = 1$... If this part is right, I can't seem to find the complete solution when i have the Constant B...
    Last edited by ux0; Oct 29th 2009 at 05:54 PM.
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  2. #2
    Super Member Rebesques's Avatar
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    Take B=0, and it is a steady state solution.
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