Solving symbolically a system of 8 non-linear equations

Hi,

I have got a system of 8 equations that are non-linear as below (unknowns are x, y and z).

I have tried to solve with the Matlab function solve, but this results in an empty sym 0 by 1. When, I only solve the first 6, I have got a parameterised solution, I would have thought that having 8 equations with only 3 unknowns would be easy to solve. Could you help?

Best,

FD

clear all; close all;
%knowns and their interval
syms A B
assume([A, B], 'real') % A and B are real
assume([A, B], 'positive') % A and B are >0
assume(0.2 <= A <= 0.4) % A takes values between 0.2 and 0.4
syms ang D
assume([ang, D], 'real')  % angle and D are real
assume([ang, D] > 0) % angle and D are >0
assume(0.2 <= D <= 1) % D takes values between 0.2 and 1
assume(40 <= ang <= 90) % angle takes values between 40deg and 90deg
%unknowns to determine
syms x y z; assume([x, y, z], 'real'); assume([x, y, z] > 0) % unknowns are real and >0
%LHS terms of equality
p2_z11   = B/A;
p2_z12   = -(B*exp(sym('1i')*deg2rad(ang)))/D;
p2_z22   = (B*A*(D+exp(2*sym('1i')*deg2rad(ang))))/(D*D*(1-A));
p2_det   = p2_z11 * p2_z22 - p2_z12 * p2_z12;
p2z2p    = [p2_z11, p2_z12; p2_z12, p2_z22]

p2z2p = 

p2abcd_A = p2_z11 / p2_z12;
p2abcd_B = p2_det / p2_z12;
p2abcd_C = 1 / p2_z12;
p2abcd_D = p2_z22 / p2_z12;
p2abcd   = [p2abcd_A, p2abcd_B; p2abcd_C, p2abcd_D]

p2abcd = 

%RHS terms of equality
x_A        = cosd(ang);
x_B        = sym('1i')*x*sind(ang);
x_C        = sym('1i')*(1/x)*sind(ang);
x_D        = cosd(ang);
x_ABCD     = [x_A, x_B ; x_C, x_D]

x_ABCD = 

z_A        = 1;
z_B        = 0;
z_C        = 1/z;
z_D        = 1;
z_ABCD     = [z_A, z_B ; z_C, z_D]

z_ABCD = 

y_A        = cosd(2*ang);
y_B        = sym('1i')*y*sind(2*ang);
y_C        = sym('1i')*(1/y)*sind(2*ang);
y_D        = cosd(2*ang);
y_ABCD     = [y_A, y_B ; y_C, y_D]

y_ABCD = 

p3abcd     = x_ABCD * z_ABCD * y_ABCD

p3abcd = 

p3abcd_A   = p3abcd(1,1);
p3abcd_B   = p3abcd(1,2);
p3abcd_C   = p3abcd(2,1);
p3abcd_D   = p3abcd(2,2);
%Solving equations: equating real and imag term to term LHS to RHS
equations = [real(p2abcd_A) == real(p3abcd_A),...
             imag(p2abcd_A) == imag(p3abcd_A),...
             real(p2abcd_B) == real(p3abcd_B),...
             imag(p2abcd_B) == imag(p3abcd_B),...
             real(p2abcd_C) == real(p3abcd_C),...
             imag(p2abcd_C) == imag(p3abcd_C),...
             real(p2abcd_D) == real(p3abcd_D),...
             imag(p2abcd_D) == imag(p3abcd_D),...
             ];
equations'

ans = 

sol_1shot = solve(equations, [x, z, y ], 'ReturnConditions',true)

sol = struct with fields:
             x: [0×1 sym]
             z: [0×1 sym]
             y: [0×1 sym]
    parameters: [1×0 sym]
    conditions: [0×1 sym]

sol_6first = solve(equations(1:6), [x, z, y ], 'ReturnConditions',true)

[Torsten]

What do you expect as solution for 8 equations in 3 unknowns ? The usual result in the linear case is that the system is overdetermined and thus the solution is empty.

[dave]

I would expect some of the equations (like three of them) used for determining the 3 unknowns and the others ending up being knowns in terms of knowns if the expressions of the unknowns are injected in those as per substitution, these can be used for verifying the consistency of the solutions. I wonder if Matlab has a function that verify the consistency of overdetermined system?

[Torsten]

If you suspect your equations are consistent, proceed as follows:

Take 3 equations out of the 8 such that the Jacobian with respect to the unknown variables is regular.

Solve the three equations in the three unknowns.

Insert the result in the remaining 5 equations and check whether they are satisfied.

[dave]

Many thanks @Torsten !

Do you mean by consistent that for example the system of equations describe a physical problem for which at least one solution exists? If so, I have redifined the system of equations below which indeed describes a physical problem and for which there is at least one solution (found by fixing some variables and verified by having 0 relative error when fitting to the measured data).

The system is defined with 8 equations and there are 5 variables.

When calculating the Jacobian, I have got some conditions since the symbolic expressions.

Proceeding forward to solving the first 5 equations, I obtain parametrised solutions. Could you please advise?

I have tried to reduce the number of equations by trying to find if some are linearly proportional to others, but without success! However, the equations show some similar paterns in some parts. Could you please advise if these can in any way be used ?

Best,

Dave

close all; clear all;
syms x1 x2 y1 y2 t % unknowns
assume([x1, x2, y1, y2, t], 'real');
assume([y1, y2, t], 'positive');

syms ar ai br bi cr ci dr di % knowns, are all real
assume([ar, ai, br, bi, cr, ci, dr, di], 'real');

eq1 = ar == cos(x1)*cos(x2) - (y1/y2)*(sin(x1)*sin(x2))
eq2 = ai == (y1*sin(x1)*cos(x2))/t
eq3 = br == - (y1*y2*sin(x1)*sin(x2))/t
eq4 = bi == y1 * cos (x2) * sin(x1) + y2 * cos(x1) * sin(x2)
eq5 = cr == (cos(x1)*cos(x2))/t
eq6 = ci == ((cos(x2)*sin(x1))/y1)+((cos(x1)*sin(x2))/y2)
eq7 = dr == cos(x1)*cos(x2) - (y2/y1)*sin(x1)*sin(x2)
eq8 = di == (y2 * cos(x1) * sin(x2))/t

equations = [eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8]
equations.'

J1to5 = jacobian(equations(1:5), [x1, x2, y1, y2, t]);
det_J1to5 = simplify(det(J1to5))

sol_5first = solve(equations(1:5), [x1, x2, y1, y2, t], 'ReturnConditions',true)

[Torsten]

What is your final aim ? t, y1 and y2 arbitrarily chosen, x1 an odd multiple of pi/2 and x2 a multiple of pi give an infinite number of solutions. But I'm not sure if you can really expect "solve" to return this result. And I don't know if there are other solutions.

I cannot execute the code because it takes too long with MATLAB online. Which parametrized solution does it return ?

[dave]

My final aim is to have {x1, x2, y1, y2, t} = function (ar, ai, br, bi, cr, ci, dr, di).

Could you explain why "t, y1 and y2 arbitrarily chosen, x1 an odd multiple of pi/2 and x2 a multiple of pi give an infinite number of solutions." ?

What if I would like to impose not having x1 and x2 taking those values, do the code lines below impose the condition to "solve"?

close all; clear all;
syms x1 x2 y1 y2 t % unknowns
assume([x1, x2, y1, y2, t], 'real');
assume([y1, y2, t], 'positive');
syms ar ai br bi cr ci dr di % knowns, are all real
assume([ar, ai, br, bi, cr, ci, dr, di], 'real');

syms n
assume(n, 'integer')

cond_on_x1 = (x1*(2*n+1) ~= (pi/2)) 
assume (cond_on_x1)

cond_on_x2 = (x2/sym('pi')) ~= 'integer'
assume (cond_on_x2)

eq1 = ar == cos(x1)*cos(x2) - (y1/y2)*(sin(x1)*sin(x2))
eq2 = ai == (y1*sin(x1)*cos(x2))/t
eq3 = br == - (y1*y2*sin(x1)*sin(x2))/t
eq4 = bi == y1 * cos (x2) * sin(x1) + y2 * cos(x1) * sin(x2)
eq5 = cr == (cos(x1)*cos(x2))/t
eq6 = ci == ((cos(x2)*sin(x1))/y1)+((cos(x1)*sin(x2))/y2)
eq7 = dr == cos(x1)*cos(x2) - (y2/y1)*sin(x1)*sin(x2)
eq8 = di == (y2 * cos(x1) * sin(x2))/t
equations = [eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8]
equations.'

J1to5 = jacobian(equations(1:5), [x1, x2, y1, y2, t]);
det_J1to5 = simplify(det(J1to5))
sol_5first = solve(equations(1:5), [x1, x2, y1, y2, t], 'ReturnConditions',true)

Which parametrized solution does it return ? See attached

Thanks,

Dave

[Torsten]

Could you explain why "t, y1 and y2 arbitrarily chosen, x1 an odd multiple of pi/2 and x2 a multiple of pi give an infinite number of solutions." ?

Choose arbitrary values for t, y1 and y2, choose x1 an odd multiple of pi/2 and x2 a multiple of pi, insert these values into your equations and you will see that all 8 are satisfied.

What if I would like to impose not having x1 and x2 taking those values, do the code lines below impose the condition to "solve"?

The condition on x1 is x1 ~= (2*n+1)*pi/2, not x1*(2*n+1) ~= pi/2. But "solve" will most probably only repeat your equations together with your assumptions. If "solve" without the option "'ReturnConditions',true returns an empty solution (I think this is the case here, isn't it ?), you cannot expect an even more complicated solution with this option.

[dave]

Many thanks @Torsten!

• Having x1 an odd multiple of pi/2 and x2 a multiple of pi giving an infinite number of solutions is insightful.
• It seems that empty solution is all I can get with solve as the system of equation is too complicated for a one-or-two-solve-shot.

Best,

Dave