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Re: Is a solution possible to this exponential equation?

At 14:48 1-05-97 -0400, Michael Hucka wrote:
>I have a pair of equations involving exponentials that I'd like to be able to
>solve for one of the variables, but I can't seem to find a way to do it.  MMA
>3.0's Solve operator complains its usual complaint about the equation
>involving transcendental functions.  Quite possibly there is no analytical
>solution, but I'd like to find out from the experts out there about what
>approaches one might try.  The equations are as follows:
>   1/2 = kc * Exp[ -a * w^2 * x^2 ]  -  ks * Exp[ -a * b^2 * w^2 * x^2]
>where a and b are constants, and kc, ks, w and x are variables.  There is an
>additional condition,
>   kc - ks = Sqrt[ 1/2 ]
>So there are 2 equations and 4 unknowns.  I'd like to solve this for w, or
>rather, to express w in terms of the other unknowns.  The problem, of course,
>is that w is in the exponent of two of the terms.
>What is the right approach to take in cases like this?  This is actually
>something I've come across before in my work, but haven't had much luck with.
>Mike Hucka    hucka at    University
> PhD to be, computational models of human visual processing (AI Lab)     of
>   UNIX systems administrator & programmer/analyst (EECS DCO)         Michigan


look what happens if you substitute 
rule1=ks->   kc -  Sqrt[ 1/2 ]
rule2= Exp[ -a * w^2 * x^2 any_:1 ]  ->  u^any
it=   1/2 ==kc * Exp[ -a * w^2 * x^2 ]  -  ks * Exp[ -a * b^2 * w^2 * x^2]

        1/2 == kc*u - (-(1/Sqrt[2]) + kc)*u^(b^2)
so you have in essence

        alfa x^n + beta x + gamma ==0

no symbolic solution to that I'm afraid. Numerics go ok though.
if b and kc are constants with a known numerical value, you can solve for u
using NSolve. The rest is simple : you can take the log and find (w * x).
To separate w and x needs further conditions.

does this help any?

Dr. Wouter L. J. MEEUSSEN
eu000949 at
w.meeussen.vdmcc at

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