Re: solving simultaneous nonlinear equation
- From: Alan Weiss <aweiss@xxxxxxxxxxxxx>
- Date: Wed, 22 Apr 2009 08:50:39 -0400
Flo Felis wrote:
Hie, I would like to seek for advice in solving simultaneous nonlinear equations. I have 3 parameters that are required to determined via three equations, and this is not possible to be solved by hand due to the complexity. Thus, i used the "syms" function in Matlab. However, MatLab was not able to generate the answers also. May I know if there is anything I have done wrong? Below are my codes;Flo, your problem is not well-posed.
syms a b c
eq1='16.8 = (18.6*exp(-0.1/(a/b))+c*6.4974*(1-exp(-0.1/(a/b))))'
eq2='10.3 = (18.6*exp(-72/(a/b))+c*6.4974*(1-exp(-72/(a/b))))'
eq3='7.4 = (18.6*exp(-534/(a/b))+c*6.4974*(1-exp(-534/(a/b))))'
[a,b,c]=solve(eq1,eq2,eq3,a,b,c)
Thanks and Regards; Flo
1. You have only two free parameters, namely c and a/b. Thus there is no reason to believe that your problem has a solution.
2. Call R = exp(-0.1/(a/b)), and C = c*6.4974. You can write your equations as follows:
16.8 = (18.6*R + C*(1-R))
10.3 = (18.6*R^720 + C*(1-R^720))
7.4 = (18.6*R^5340 + C*(1-R^5340))
3. R is between 0 and 1. It must be very close to 1, in fact.
The first and second equations show that R^720 must be something like 2/3 or so. Therefore R is on the order of 1 - 1/720.
4. The third equation then shows that C is close to 7.4, since R^5340 is something like 1/20.
5. But the first equation shows that C is about 16.8 - 18.6, since R is so close to 1. Therefore, as I suspected, there is no solution.
Alan Weiss
MATLAB mathematical toolbox documentation
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