MATLAB: ‘solve’ results in incomplete solutions

problemsolvesolver problems

Hi,
I'm trying to find the equilibrium points of a system described by 4 equations. This is the code:
 clear all;
 clc;
 syms x1 x2 x3 x4 g m1 m2 l1 l2 pi
 eq1 = x2
 eq2 = -((g*(2*m1+m2)*sin(x1)+m2*(g*sin(x1-2*x3)+2*(l2*x4^2+l1*x2^2*cos(x1- x3))*sin(x1-x3)))/(2*l1*(m1+m2-m2*cos(x1-x3)^2)));
 eq3 = x4
 eq4 = (((m1+m2)*(l1*x2^2+g*cos(x1))+l2*m2*x4^2*cos(x1-x3))*sin(x1-x3))/(l2*(m1+m2-m2*cos(x1-x3)^2));
 eq2=subs(eq2, g, 9.81); eq2=subs(eq2, m1, 2); eq2=subs(eq2, m2, 1);   eq2=subs(eq2, l1, 2); eq2=subs(eq2, l2, 1)
 eq4=subs(eq4, g, 9.81); eq4=subs(eq4, m1, 2); eq4=subs(eq4, m2, 1);  eq4=subs(eq4, l1, 2); eq4=subs(eq4, l2, 1)
 res=solve(0==eq1,0==eq2,0==eq3,0==eq4,x1,x2,x3,x4)
 [res.x1 res.x2 res.x3 res.x4]
And the result I get is:
 [  0, 0,  0,  0]
 [  0, 0, pi,  0]
 [ pi, 0,  0,  0]
 [ z2, 0, z3, z4]
I have 2 problems with this result:
1) All the first 3 are valid solutions, however, the 4th row seems gibberish to me. What is the "z" variable?
2) Also, solve skipped one of the obvious solutions (its a double pendulum system) [pi, 0, pi, 0].
In advance, thank you for the replies.

Best Answer

The z2, z3, z4 values I believe are because of this (from the solve documentation):
  • If the solution of an equation or a system of equations contains parameters, the solver can choose one or more values of the parameters and return the results corresponding to these values. For some equations and systems, the solver returns parameterized solutions without choosing particular values. In this case, the solver also issues a warning indicating the values of parameters in the returned solutions.
Except it didn't issue any warning indicating it was making up parameter symbols when I ran your code. (It has with other code I've run in the past. I have no idea why it didn't with yours.)
Without changing anything else in your code, I added these two lines (after the respective original lines for eq2 and eq4 respectively:
eq2 = simplify(collect(expand(eq2)))
eq4 = simplify(collect(expand(eq4)))
(because the solve function usually does better after such simplifications), and these:
[X1 X2 X3 X4] = solve(0==eq1,0==eq2,0==eq3,0==eq4, x1,x2,x3,x4);
SymMtx = [X1 X2 X3 X4]
VPAMtx = vpa(SymMtx, 6)
and got this solution:
SymMtx =
[     0, 0,                                        0, 0]
[    pi, 0,                                        0, 0]
[     0, 0,                                       pi, 0]
[  pi/2, 0,    2*atan(((2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
[  pi/2, 0,  2*atan((- (2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
[  pi/2, 0,   -2*atan(((2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
[ -pi/2, 0,    2*atan(((2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
[ -pi/2, 0,   -2*atan(((2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
[  pi/2, 0, -2*atan((- (2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
[ -pi/2, 0,  2*atan((- (2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
[ -pi/2, 0, -2*atan((- (2^(1/2)*2*i)/3 - 1/3)^(1/2)), 0]
that using vpa produced:
VPAMtx =
[       0, 0,                    0, 0]
[ 3.14159, 0,                    0, 0]
[       0, 0,              3.14159, 0]
[  1.5708, 0,   1.5708 + 1.14622*i, 0]
[  1.5708, 0,   1.5708 - 1.14622*i, 0]
[  1.5708, 0, - 1.5708 - 1.14622*i, 0]
[ -1.5708, 0,   1.5708 + 1.14622*i, 0]
[ -1.5708, 0, - 1.5708 - 1.14622*i, 0]
[  1.5708, 0, - 1.5708 + 1.14622*i, 0]
[ -1.5708, 0,   1.5708 - 1.14622*i, 0]
[ -1.5708, 0, - 1.5708 + 1.14622*i, 0]
I'm not certain this was what you were anticipating (or want), but it at least managed to avoid producing a solution with z2, z3, z4, even though it still only produced three real results. If this wasn't what you were expecting, it might be worth the effort to look over your original code and check for errors.
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