Solving system of quadratic equations

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Alessandro Arduino
Alessandro Arduino 2022-4-11
评论： Alex Sha 2022-4-12
Hi guys!
I'm trying to solve a set of quadratic equations for a code I'm working on. I've tried to use vpasolve and solve but the code doesn't bring any solution. The equations are correct and I'm sure there are solutions to it as I can solve them with Mathematica but I'd like to be able to solve them in matlab so that I can write my code in there instead of Mathematica.
The code is something like this:
syms y1 y2 y3 y4 z1 z2
depd = [y1 y2 y3 y4 z1 z2];
% Assign the independent variables
x1 = sqrt(2/3);
x2 = sqrt(1/6);
x3 = sqrt(1/2);
x4 = sqrt(1/2);
z3 = sqrt(1/2);
z4 = sqrt(1/2);
prev = [-0.438450, -0.505030, -0.076748, -0.048791, 0.455646, 0.989215];
% write constraint equations
eq1 = (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 - 1/3 == 0;
eq2 = (x3 - x1)^2 + (y3 - y1)^2 + (z3 - z1)^2 - tan(pi/12) == 0;
eq3 = (x3 - x2)^2 + (y3 - y2)^2 + (z3 - z2)^2 - tan(pi/12) == 0;
eq4 = (y4 - x4)^2 + (z4 - y4)^2 + (x4 - z4)^2 - 1 == 0;
eq5 = (y4 - x1)^2 + (z4 - y1)^2 + (x4 - z1)^2 - 2 == 0;
eq6 = (z3 - x2)^2 + (x3 + y2)^2 + (y3 + z2)^2 - 0.84529946 == 0;
eqs = [eq1, eq2, eq3, eq4, eq5, eq6];
sol = vpasolve(eqs, depd, prev);
I don't need a precise solution but rather a numerical approximation. Is there something that can provide that in Matlab?
  2 个评论
Matt J
Matt J 2022-4-11
Are you sure the first equation shouldn't be,
eq1 = (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 - 1/3 == 0;
Alessandro Arduino
Alessandro Arduino 2022-4-11
Yes I just updated it as you commented, regardless the problem is still there

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采纳的回答

Davide Masiello
Davide Masiello 2022-4-11
Ran in:
I am not really confident with Simulink, but it does work on Matlab using the fsolve function.
In the equation, I changed z1 and z2 to y(5) and y(6) respectively so to make possible the indexing of the dependent variable.
clear,clc
y0 = [-0.438450, -0.505030, -0.076748, -0.048791, 0.455646, 0.989215];
y = fsolve(@eqSystem,y0)
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
y = 1×6
-0.4416 -0.3699 0.0505 0.0000 0.5894 0.7509
function out = eqSystem(y)
x1 = sqrt(2/3);
x2 = sqrt(1/6);
x3 = sqrt(1/2);
x4 = sqrt(1/2);
z3 = sqrt(1/2);
z4 = sqrt(1/2);
out = [ (x2 - x1)^2 + (y(2) - y(1))^2 + (y(6) - y(5)) - 1/3;...
(x3 - x1)^2 + (y(3) - y(1))^2 + (z3 - y(5))^2 - tan(pi/12);...
(x3 - x2)^2 + (y(3) - y(2))^2 + (z3 - y(6))^2 - tan(pi/12);...
(y(4) - x4)^2 + (z4 - y(4))^2 + (x4 - z4)^2 - 1;...
(y(4) - x1)^2 + (z4 - y(1))^2 + (x4 - y(5))^2 - 2;...
(z3 - x2)^2 + (x3 + y(2))^2 + (y(3) + y(6))^2 - 0.84529946;...
];
end
  2 个评论
Alex Sha
Alex Sha 2022-4-12
There are four solutions:
No. y1 y2 y3 y4 z1 z2
1 -0.408248290310322 -0.408248290883343 -2.65940075527358E-10 5.66489492618543E-16 0.408248289890841 0.816496580354705
2 -0.561883068548743 -0.453705322086167 -0.0890364837212022 1.4142135623731 0.527109082977269 0.9207640611459
3 -0.332314192176063 -0.0311689186229293 -0.387501968689092 5.89828382789581E-16 1.21003557681 0.934394159596848
4 -0.473380938970007 -0.109067418601998 -0.390919363116206 1.4142135623731 1.20628926550608 1.02205483456621

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