Results 1 to 4 of 4

Math Help - A very difficult problem for me

  1. #1
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85

    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  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.
    Last edited by Maccaman; October 18th 2008 at 09:23 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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  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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    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.)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. difficult problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 30th 2010, 04:41 AM
  2. Difficult problem
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: May 10th 2010, 07:23 PM
  3. Aah difficult problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 15th 2009, 06:14 PM
  4. a difficult problem
    Posted in the Number Theory Forum
    Replies: 13
    Last Post: June 24th 2009, 03:25 PM
  5. Difficult problem
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: November 10th 2008, 11:50 PM

Search Tags


/mathhelpforum @mathhelpforum