A very difficult problem for me

**Maccaman** · October 18th 2008, 10:12 AM

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 $u = u_0$ (const.) until a steady temperature u(x,y) is reached in the cylinder. Show that

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

for 0 < x < a, 0 < y < b.

**ThePerfectHacker** · October 19th 2008, 08:09 AM

Originally Posted by Maccaman

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 $u = u_0$ (const.) until a steady temperature u(x,y) is reached in the cylinder. Show that

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

for 0 < x < a, 0 < y < b.

You are solving $u_{xx}+u_{yy} = 0$ .

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

Now it remains to use seperation of variables.

**Maccaman** · October 20th 2008, 07:48 AM

I am lost with what to do for the boundary conditions $u(x,0)=u_0$ and $u_y(x,b)=0$ . All the 2- D Laplace equations I have done have had boundary conditions like $u(x,b)=0, u(x,2) = f(x)$ , ect.

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

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

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

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

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

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

For the PDE

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

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

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

So

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

thats all ive been able to do w/out knowing what to do with the other BC's. Can anyone please show me?

**Opalg** · October 20th 2008, 08:51 AM

That's a good start. Next thing is to look at what you know about the function F(x). You have $F''(x) = kF(x)$ and F(a)=F(0)=0. That's a standard simple harmonic motion equation, and you should be able to deduce that $F(x) = \sin\left(\tfrac{n\pi x}a\right)$ is a solution for every natural number n. Therefore $k = -\tfrac{n^2\pi^2}{a^2}$ .

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

Putting that information together, we know that

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

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

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

whenever $0<x<a$ . 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 $\textstyle f(x) = \begin{cases}u_0&(0<x<a)\\ -u_0&(-a<x<0)\end{cases}$ 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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