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Heat equation with periodic BC

  1. Oct 13, 2009 #1
    I got the following PDE.

    [tex]
    {\frac {\partial }{\partial t}}u \left( y,t \right) = \nu \cdot {\frac {\partial
    ^{2}}{\partial {y}^{2}}}u \left( y,t \right)[/tex]

    With boundary conditions:
    [tex]y=0: u(0,t)=0[/tex]
    [tex]y=h: u(h,t)=U_0 \cdot cos(\omega \cdot t)[/tex]

    Now I need to show by using substitution that,
    [tex]\frac{u}{U_0} = \left( A\cos \left( \omega\,t \right) +B\sin \left( \omega\,t
    \right) \right) \sinh \left( {\frac {\alpha\,y}{h}} \right) \cos
    \left( {\frac {\alpha\,y}{h}} \right) + \left( C\cos \left( \omega\,t
    \right) + \left( D \right) \sin \left( \omega\,t \right) \right)
    \sin \left( {\frac {\alpha\,y}{h}} \right) \cosh \left( {\frac {\alpha
    \,y}{h}} \right)[/tex]
    , is a solution of the above equation.

    Here:
    [tex]\alpha = \sqrt{\frac{\Omega}{2}}[/tex] where [tex]\Omega = \frac{\omega \cdot h^2}{\nu}[/tex]

    Next to that I need to find the expressions to calculate the coefficients for a given [tex]\alpha[/tex].

    Attempt at a solution:
    Starting with the hint I substitute the expression back into the PDE.
    [tex] \left( -A\sin \left( \omega\,t \right) \omega+B\cos \left( \omega\,t
    \right) \omega \right) \sinh \left( {\frac {\alpha\,y}{h}} \right)
    \cos \left( {\frac {\alpha\,y}{h}} \right) + \left( -C\sin \left(
    \omega\,t \right) \omega+ \left( D \right) \cos \left( \omega\,t
    \right) \omega \right) \sin \left( {\frac {\alpha\,y}{h}} \right)
    \cosh \left( {\frac {\alpha\,y}{h}} \right)[/tex]
    =
    [tex]-2\, \left( A\cos \left(
    \omega\,t \right) +B\sin \left( \omega\,t \right) \right) \cosh
    \left( {\frac {\alpha\,y}{h}} \right) {\alpha}^{2}\sin \left( {\frac
    {\alpha\,y}{h}} \right) {h}^{-2}+2\, \left( C\cos \left( \omega\,t
    \right) + \left( D \right) \sin \left( \omega\,t \right) \right)
    \cos \left( {\frac {\alpha\,y}{h}} \right) {\alpha}^{2}\sinh \left( {
    \frac {\alpha\,y}{h}} \right) {h}^{-2}
    [/tex]

    This gives me that A = D and B = C. Now using this information I continue with the BC's. The first BC is automatically satisfied by the sine and hyperbolic sine term. Next I continue with the second BC.

    [tex]\left( A\cos \left( \omega\,t \right) +B\sin \left( \omega\,t
    \right) \right) \sinh \left( {\frac {\alpha\,y}{h}} \right) \cos
    \left( {\frac {\alpha\,y}{h}} \right) + \left( B\cos \left( \omega\,t
    \right) + \left( A \right) \sin \left( \omega\,t \right) \right)
    \sin \left( {\frac {\alpha\,y}{h}} \right) \cosh \left( {\frac {\alpha
    \,y}{h}} \right) =\cos \left( \omega\,t \right)[/tex]

    Here I don't see how to continue. Maple gives a rather lengthy solution. Doing it by hand it seems I miss information, but my professor said this should be enough. I have been looking at this problem for a while and maybe I am overlooking things. So I need some assistance from someone with a fresh look.
     
    Last edited: Oct 13, 2009
  2. jcsd
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