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Thread: A very difficult problem for me

  1. #1
    Member Maccaman's Avatar
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    A very difficult problem for me

    A long cylinder of material with a with a rectangular cross-section occupies the region 0 < x < a, 0 < y < b in space. The faces at x = 0 and x = a are held at temperature u=0 , and the face at y = b is thermally insulated. The face at y = 0 is held at temperature $\displaystyle u = u_0 $ (const.) until a steady temperature u(x,y) is reached in the cylinder. Show that

    $\displaystyle u(x,y) = \frac{4u_0}{\pi}\sum_{m=1}^{\infty} \ \frac{sin ([2m -1] \pi x/a) \ cosh ([2m - 1] \pi [y -b]/a)}{[2m -1] \ cosh([2m - 1] \pi b/a)} $

    for 0 < x < a, 0 < y < b.
    Last edited by Maccaman; Oct 18th 2008 at 08:23 PM.
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  2. #2
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    Quote Originally Posted by Maccaman View Post
    A long cylinder of material with a with a rectangular cross-section occupies the region 0 < x < a, 0 < y < b in space. The faces at x = 0 and x = a are held at temperature u=0 , and the face at y = b is thermally insulated. The face at y = 0 is held at temperature $\displaystyle u = u_0 $ (const.) until a steady temperature u(x,y) is reached in the cylinder. Show that

    $\displaystyle u(x,y) = \frac{4u_0}{\pi}\sum_{m=1}^{\infty} \ \frac{sin ([2m -1] \pi x/a) \ cosh ([2m - 1] \pi [y -b]/a)}{[2m -1] \ cosh([2m - 1] \pi b/a)} $

    for 0 < x < a, 0 < y < b.
    You are solving $\displaystyle u_{xx}+u_{yy} = 0$.

    With the conditions: $\displaystyle u(0,y)=u(a,y)=0$, $\displaystyle u(x,0)=u_0$, and $\displaystyle u_y(x,b)=0$.

    Now it remains to use seperation of variables.
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  3. #3
    Member Maccaman's Avatar
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    I am lost with what to do for the boundary conditions $\displaystyle u(x,0)=u_0 $ and $\displaystyle u_y(x,b)=0 $. All the 2- D Laplace equations I have done have had boundary conditions like $\displaystyle u(x,b)=0, u(x,2) = f(x) $, ect.

    I have never seen or done one with $\displaystyle u_y $.

    Anyway, here is all I have done so far....


    $\displaystyle u(x,y) = F(x)G(y) $

    $\displaystyle \Rightarrow \ u(0,y) = F(0)G(y) = 0 \ \Rightarrow F(0) = 0 \ [or \ G(y) = 0 \ \Rightarrow u(x,y) = 0 ]\$ (trivial)

    $\displaystyle \Rightarrow \ u(a,y) = F(a)G(y) = 0 \ \Rightarrow \ F(a) = 0$

    unsure what to do with the other to BC's....

    For the PDE

    $\displaystyle F''(x)G(y) + F(x)G''(y) = 0$

    dividing by $\displaystyle F(x)G(x) \ \Rightarrow $

    $\displaystyle \frac{F''(x)}{F(x)} = - \frac{G''(y)}{Y(y)} = k $ (separation constant)

    So

    $\displaystyle F''(x) = kF(x), \ G''(y) = -kG(y) $

    thats all ive been able to do w/out knowing what to do with the other BC's. Can anyone please show me?
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  4. #4
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    That's a good start. Next thing is to look at what you know about the function F(x). You have $\displaystyle F''(x) = kF(x)$ and F(a)=F(0)=0. That's a standard simple harmonic motion equation, and you should be able to deduce that $\displaystyle F(x) = \sin\left(\tfrac{n\pi x}a\right)$ is a solution for every natural number n. Therefore $\displaystyle k = -\tfrac{n^2\pi^2}{a^2}$.

    Now look at what you know about the function G(x). You have $\displaystyle G''(y) = \tfrac{n^2\pi^2}{a^2}G(y)$. That has a solution $\displaystyle G(y) = \cosh\left(\tfrac{n\pi}a(y-A)\right)$ for any constant A. You also have a boundary condition $\displaystyle G'(b)=0$ (from $\displaystyle u_y(x,b)=0$). But $\displaystyle G'(y) = \tfrac{n\pi}a\sinh\left(\tfrac{n\pi}a(y-A)\right)$, and this will be 0 if $\displaystyle A=b$. So we have the solution $\displaystyle G(y) = \cosh\left(\tfrac{n\pi}a(y-b)\right)$.

    Putting that information together, we know that

    . . . . . . $\displaystyle u(x,y) = \sum_{n=1}^\infty C_n\sin\left(\tfrac{n\pi x}a\right)\cosh\left(\tfrac{n\pi}a(y-b)\right)$

    is a solution, for all choices of the coefficients C_n, and we want to choose these coefficients so that the remaining boundary condition $\displaystyle u(x,0) = u_0$ holds. So we want

    . . . . . . $\displaystyle u_0 = \sum_{n=1}^\infty C_n\sin\left(\tfrac{n\pi x}a\right)\cosh\left(\tfrac{n\pi b}a\right)$ . . . . .(*)

    whenever $\displaystyle 0<x<a$. That is where the Fourier sine series will come in. Since sin is an odd function, if a Fourier sine series is u_0 when 0<x<a then it must take the value -u_0 when -a<x<0.

    So there is still some work to do. You need to calculate the Fourier series for the function $\displaystyle \textstyle f(x) = \begin{cases}u_0&(0<x<a)\\ -u_0&(-a<x<0)\end{cases}$ and compare the answer with (*), to find what the coefficients C_n should be. (From the given solution, it looks as though only the odd coefficients will be nonzero.)
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