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Hyperbolic Boundary Valued Problem

  1. Dec 18, 2011 #1
    Hi, I am trying to understand solving boundary valued partial differential equations and it's relation to hyperbolic functions. In one of my problems, there is a PDE and the solution contains the hyperbolic function "cosh". I was just curious if anyone has any information for me to read up on this, I cannot find anything in any book that gives any relation of the cosh function to solution to PDEs. Thanks.
     
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  3. Dec 20, 2011 #2

    LCKurtz

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    Remember that ##\cosh(x) = \frac {e^x + e^{-x}} 2## and ##\sinh(x) = \frac {e^x - e^{-x}} 2##, so any DE for which the solutions are exponentials could have the solutions expressed as sinh or cosh functions. The hyperbolic functions come in handy because of their values of 0 or 1 when x = 0 and similarly for their derivatives. For example consider the BVP
    ##y''-\lambda^2y=0##

    ##y(0) = 0, y(a) = 0##.
    Find the solution to that using the pair {##e^x,e^{-x}##} and compare that to the work using the equivalent pair {##\sinh(x),\sinh(a-x)##}. This extra simplification can come in handy in some fourier series boundary value problems in PDE's.
     
    Last edited: Dec 20, 2011
  4. Dec 22, 2011 #3
    Thank you, I understand now. I am evaluating this problem which I cannot duplicate the solution. It is as follows:

    ##i(t) =nDF\frac{\partial P}{\partial x} @ x = 0##

    Evaluating the following limit

    I = Lim i(t) as t→∞

    by solving the following equations and boundary conditions:

    [itex]\frac{\partial S}{\partial t}=[/itex][itex]D\frac{\partial^{2}S}{\partial x^{2}}[/itex]-[itex]\frac{Vm}{Km}S[/itex]
    [itex]\frac{\partial P}{\partial t}=[/itex][itex]D\frac{\partial^{2} P}{\partial x^{2}}[/itex]+[itex]\frac{Vm}{Km}P[/itex]

    b.c. [itex]\frac{\partial S}{\partial x}[/itex] = 0 @ x = 0, P = 0 @ x = 0, S = [itex]S_{0}[/itex] @ x ≥ d, P = 0 @ x ≥ d

    Yields the following:

    I=nDF[itex]\frac{S_{0}}{d}[/itex] (1-[itex]\frac{1}{cosh(\alpha d)}[/itex])

    where, [itex]\alpha^{2}[/itex] = [itex]\frac{Vm}{KmD}[/itex]

    I am unable to produce "I" and I was wondering if someone can give me some insight. I don't completely understand the relation between [itex]\frac{\partial P}{\partial x}[/itex] and the 2 diffusion-based equations. When taking the limit, which part is evaluated? Thank you very much.
     
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