Homework Help: Numerical Solution to System of Non-linear Equations

1. Jul 23, 2012

Maximilien

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

2. Jul 23, 2012

Ray Vickson

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