Numerical Solution to System of Non-linear Equations

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SUMMARY

The discussion centers on solving a system of non-linear equations using Matlab and Mathematica. The user reports discrepancies between the solutions obtained from Matlab's fsolve() and Mathematica's NSolve[], with Mathematica indicating an infinite solution set. The equations involve independent variables a, b, c, d, e, and f, and the analysis reveals that the system does not have a unique solution. A re-scaled approach using Maple provides two distinct solution types, confirming the non-uniqueness of the solution.

PREREQUISITES
  • Understanding of non-linear equations and their solutions.
  • Familiarity with Matlab's fsolve() function.
  • Knowledge of Mathematica's NSolve[] function.
  • Experience with Maple for solving polynomial equations.
NEXT STEPS
  • Explore advanced techniques for solving non-linear equations in Matlab.
  • Learn about the implications of infinite solutions in Mathematica.
  • Investigate the use of Maple for polynomial root finding.
  • Study the methods for rescaling variables in complex systems of equations.
USEFUL FOR

Mathematicians, engineers, and researchers dealing with non-linear systems, as well as students seeking to understand numerical methods for solving equations in Matlab, Mathematica, and Maple.

Maximilien
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Homework Statement



I am having problems solving this system of non-linear equations in Matlab and Mathematica. a,b,c,d,e,f are all independent variables.

In Matlab I use fsolve() and get one solution, in Mathematica I use NSolve[] and get another totally different solution. Mathematica also reports this error: "NSolve::infsolns: Infinite solution set has dimension at least 1."

What is the problem here? Is it that there is no unique solution to this set of equations?

Homework Equations



<br /> -\frac{21 a b c}{200000}+\frac{21 a^2 d}{400000}-\frac{21 b^2 d}{400000}+\frac{e}{100}=0<br />
<br /> \frac{21 a^2 c}{400000}-\frac{21 b^2 c}{400000}+\frac{21 a b d}{200000}+\frac{f}{100}=0<br />
<br /> \frac{c}{100}-\frac{19 a b e}{200000}+\frac{19 a^2 f}{400000}-\frac{19 b^2 f}{400000}=0<br />
<br /> \frac{d}{100}+\frac{19 a^2 e}{400000}-\frac{19 b^2 e}{400000}+\frac{19 a b f}{200000}=0<br />
<br /> \frac{a}{100}+\frac{b c e}{10000}-\frac{a d e}{10000}-\frac{a c f}{10000}-\frac{b d f}{10000}-\frac{1}{5 \sqrt{2}}=0<br />
<br /> \frac{b}{100}+\frac{a c e}{10000}+\frac{b d e}{10000}+\frac{b c f}{10000}-\frac{a d f}{10000}=0<br />
 
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Maximilien said:

Homework Statement



I am having problems solving this system of non-linear equations in Matlab and Mathematica. a,b,c,d,e,f are all independent variables.

In Matlab I use fsolve() and get one solution, in Mathematica I use NSolve[] and get another totally different solution. Mathematica also reports this error: "NSolve::infsolns: Infinite solution set has dimension at least 1."

What is the problem here? Is it that there is no unique solution to this set of equations?

Homework Equations



<br /> -\frac{21 a b c}{200000}+\frac{21 a^2 d}{400000}-\frac{21 b^2 d}{400000}+\frac{e}{100}=0<br />
<br /> \frac{21 a^2 c}{400000}-\frac{21 b^2 c}{400000}+\frac{21 a b d}{200000}+\frac{f}{100}=0<br />
<br /> \frac{c}{100}-\frac{19 a b e}{200000}+\frac{19 a^2 f}{400000}-\frac{19 b^2 f}{400000}=0<br />
<br /> \frac{d}{100}+\frac{19 a^2 e}{400000}-\frac{19 b^2 e}{400000}+\frac{19 a b f}{200000}=0<br />
<br /> \frac{a}{100}+\frac{b c e}{10000}-\frac{a d e}{10000}-\frac{a c f}{10000}-\frac{b d f}{10000}-\frac{1}{5 \sqrt{2}}=0<br />
<br /> \frac{b}{100}+\frac{a c e}{10000}+\frac{b d e}{10000}+\frac{b c f}{10000}-\frac{a d f}{10000}=0<br />

I used a re-scaled system (which you should always do, as a matter of course): I set x = 100*X for x = a,b,c,d,e,f and X = A,B,C,D,E,F, then solved the system exactly in Maple. Here is the solution:


> sol:=%;

sol := {A = %1, B = 0, C = C,

2 2 2 1/2
D = 1/20 RootOf(_Z + 400 C + 190 %1 - 19 %1 2 ),

2 2 2 2 1/2
E = -21/8 %1 RootOf(_Z + 400 C + 190 %1 - 19 %1 2 ),

2
F = -105/2 %1 C},

1/2
2
{A = ----, B = 0, C = 0, D = 0, E = 0, F = 0}, {A = %1, B = 0,
10

2 2 1/2
C = 1/20 RootOf(_Z + 190 %1 - 19 %1 2 ), D = 0, E = 0,

2 2 2 1/2
F = -21/8 %1 RootOf(_Z + 190 %1 - 19 %1 2 )}

4
%1 := RootOf(9975 _Z - 4)

It is a bit hard to read here, but there are two distinct solution types, one being
{A = .14150988295122837392, B = 0., C = .24393682254305351184e-2*I, D = 0., E = 0., F = -.25645443231646279921e-2*I} (where I = sqrt(-1)) and the other contained in the first curly brackets. In this "other" solution, C is arbitrary and all the other variables are determined as functions of C, but of a complicated kind: C appears in the coefficients of 4th degree polynomials, and A, B, etc., involve roots of these polynomials. So, no, the solution is certainly not unique.

RGV
 

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