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  1. #1
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    I'm having trouble with this one

    A semi-infinite cylinder of metal lies along the positive x-axis with its sides and the face at $\displaystyle x = 0$ insulated. The initial temperature distribution is given by $\displaystyle u(x,0) = u_0 e^{\frac{-x}{L}}$ where $\displaystyle u_0$ and $\displaystyle L$ are positive constants. Show that the temperature in the cylinder is given for $\displaystyle t > 0$ and $\displaystyle x > 0$ by

    $\displaystyle u(x,t) = \frac{u_0}{2} e^{\frac{c^2 t}{L^2}} $ $\displaystyle \Biggl( \ e^{-\frac{x}{L}} \Biggl[ 1 + \mbox{erf}$ $\displaystyle \ \ \Biggl( \frac{x - \frac{2c^2t}{L}}{\sqrt{(4c^2t)}} \Biggr) \Biggr]$ $\displaystyle + \ e^{\frac{x}{L}} $ $\displaystyle \Biggl[1 - \mbox{erf}$ $\displaystyle \ \ \Biggl( \frac{x - \frac{2c^2t}{L}}{\sqrt{(4c^2t)}} \Biggr) \Biggr] \Biggr)$
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  2. #2
    Super Member Rebesques's Avatar
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    Faces insulated, initial temperature distribution $\displaystyle u(x,0) = u_0 e^{\frac{-x}{L}}$.


    So, the solution is given by

    $\displaystyle u(x,t)=\int H(x-y,t)u(y,0)dy$, where $\displaystyle H$ is the heat kernel $\displaystyle H(x,t)=\frac{1}{2c\sqrt{\pi t}}\exp\left(\frac{-x^2}{4c^2t}\right)$. Have fun calculating
    Last edited by Rebesques; Jan 18th 2010 at 02:40 PM.
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