Hyperbolic Boundary Valued Problem

[robby991]

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.

 

[LCKurtz]

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.

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.

 

[robby991]

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:

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

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

Yields the following:

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

where, \alpha^{2} = \frac{Vm}{KmD}

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 \frac{\partial P}{\partial x} and the 2 diffusion-based equations. When taking the limit, which part is evaluated? Thank you very much.