Numerical Solution to System of Nonlinear Equationsby Maximilien Tags: equations, mathematica 8, matlab, nonlinear, numerical 

#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

HW Helper
Thanks
P: 4,670

> 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 


Register to reply 
Related Discussions  
How to know if an infinite system of linear equations has a solution  Linear & Abstract Algebra  1  
Numerical Simultaneous Solution of NonLinear Coupled Equations  General Math  14  
Particular Solution to a System of First Order Linear Equations  Calculus & Beyond Homework  1  
Solution to system of linear equations in range of system matrix  Calculus & Beyond Homework  2  
Patterns of Solution Sets of a System of Linear Equations  Linear & Abstract Algebra  2 