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Numerical Solution to System of Nonlinear Equations 
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#1
Jul2312, 09:23 AM

P: 5

1. The problem statement, all variables and given/known data
I am having problems solving this system of nonlinear 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 [tex] \frac{21 a b c}{200000}+\frac{21 a^2 d}{400000}\frac{21 b^2 d}{400000}+\frac{e}{100}=0 [/tex] [tex] \frac{21 a^2 c}{400000}\frac{21 b^2 c}{400000}+\frac{21 a b d}{200000}+\frac{f}{100}=0 [/tex] [tex] \frac{c}{100}\frac{19 a b e}{200000}+\frac{19 a^2 f}{400000}\frac{19 b^2 f}{400000}=0 [/tex] [tex] \frac{d}{100}+\frac{19 a^2 e}{400000}\frac{19 b^2 e}{400000}+\frac{19 a b f}{200000}=0 [/tex] [tex] \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 [/tex] [tex] \frac{b}{100}+\frac{a c e}{10000}+\frac{b d e}{10000}+\frac{b c f}{10000}\frac{a d f}{10000}=0 [/tex] 


#2
Jul2312, 11:40 AM

Sci Advisor
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Thanks
P: 4,956

> 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 = .24393682254305351184e2*I, D = 0., E = 0., F = .25645443231646279921e2*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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