How to Solve matrix equations in Matlab

I am trying to solve the following set of matrix equations

The values of r1, r2, r3 and ψare known. The values of θ and ϕ are to be found by solving this equation in matlab. i attempt to do this using the fsolve function. However, it is not able to arrive to a solution. Is there a better way to solve this?

function F = root2d(Ficksang)

rfinal = [0,-0.101233861621737,0.365119069777688];

theta_f = Ficksang(1);
phi_f   = Ficksang(2);
psi_f   = 0;

r1      = rfinal(1);
r2      = rfinal(2);
r3      = rfinal(3);


F(1) = r1 - ( ( tan(psi_f/2) - (tan(theta_f/2) * tan(phi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );

F(2) = r2 - ( ( tan(phi_f/2) + (tan(theta_f/2) * tan(psi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) ); 
    
F(3) = r3 - ( ( tan(theta_f/2) - (tan(phi_f/2) * tan(psi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );

fun = @root2d;
x0 = [0,0];
x = fsolve(fun,x0)

Answer 1

[Matt J]

With an over-determined system (3 equations and only 2 unknowns), you can't expect an exact solution. However, the solution that fsolve does find does seem to be valid as a least squares solution, judging from the surface plot below.

fun = @root2d;
x0 = [0,0];
[x,f] = fsolve(fun,x0);

Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt algorithm instead.

No solution found.

fsolve stopped because the last step was ineffective. However, the vector of function
values is not near zero, as measured by the value of the function tolerance. 

x

x = 1×2

0.6950 -0.1785

[Theta,Phi]=ndgrid( linspace(-pi/2,pi/2,300));
fun=@(x,y)norm(root2d([x,y]));
F= arrayfun(fun,Theta,Phi);

 surf(Theta,Phi,F,'EdgeColor','none')
 xlabel 'Theta', ylabel 'Phi'
 view(60,65)

Answer 2

[Sargondjani]

Your method seems valid. But of course fsolve will only attempt to find a local solution, and it might get stuck in a place where there is locally no solution.

Did you try with other starting values? This might work in general if you know where your solutions should approximately be.

[Julian Blackthorne]

I did try with a couple of other initial values but fsolve seems to still have ineffective steps leading to no solution at all.

Answer 3

[Walter Roberson]

Optimal solutions, in the sense of smallest least-squared.

Are there other solutions? Yes: you can ask to solve eqn2 with return conditions set, and the answer will be parameterized . One of the variables of parameterization, k will add multiples of pi .

(The other one is a nuisance variable -- the expression inside the root() is pulled out into a parameterized variable and the "conditions" are that the fifth degree polynomial = 0. And then to express F2 you have to extract the expression from the conditions and wrap it with a root() yourself... Do-able, but a nuisance.)

format long g
syms F [1 2]
fun = root2d(F);
residue = sum(fun.^2)

residue = 

bestF1 = solve(diff(residue, F(1)),F(1))

bestF1 = 

eqn2 = subs(residue, F(1), bestF1)

eqn2 = 

sol2 = solve(diff(eqn2, F(2)),F(2));

F2 = sol2;
F1 = subs(bestF1, F(2), F2);

F1

F1 = 

F2

F2 = 

F1n = double(F1)

F1n = 0.6950

F2n = double(F2)

F2n = -0.1785

function F = root2d(Ficksang)
rfinal = [0,-0.101233861621737,0.365119069777688];
theta_f = Ficksang(1);
phi_f   = Ficksang(2);
psi_f   = 0;
r1      = rfinal(1);
r2      = rfinal(2);
r3      = rfinal(3);
F(1) = r1 - ( ( tan(psi_f/2) - (tan(theta_f/2) * tan(phi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );
F(2) = r2 - ( ( tan(phi_f/2) + (tan(theta_f/2) * tan(psi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) ); 
    
F(3) = r3 - ( ( tan(theta_f/2) - (tan(phi_f/2) * tan(psi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );
end

[Julian Blackthorne]

@Walter Roberson but it looks like the solution obtained using the return conditions are exactly the ones that are retured by the fsolve. So, I guess even if the exact solution is not obtained, this would be the best soliution for the system then?

[Walter Roberson]

When you use 'returnconditions' then you get forms in which you can substitute different integer values for k (an automatically generated parameter) in order to get different periods (since tan() is periodic)

A more careful analysis would probably involve using returnconditions on bestF1 as well so that the atan() were in the periodic form.

format long g
syms F [1 2]
fun = root2d(F);
residue = sum(fun.^2)

residue = 

bestF1 = solve(diff(residue, F(1)),F(1))

bestF1 = 

eqn2 = subs(residue, F(1), bestF1)

eqn2 = 

sol2 = solve(diff(eqn2, F(2)),F(2), 'returnconditions', true);

sol2.parameters

ans = 

assume(sol2.parameters(1),'integer')
z_values = vpa(cellfun(@rhs, children(simplify(sol2.conditions))));

F2 = subs(sol2.F2, sol2.parameters(2), z_values(:));
F1 = subs(bestF1, F(2), F2);

F1

F1 = 

F2

F2 = 

kvals = -2:2;
F1n = double(subs(F1, sol2.parameters(1), kvals))

F1n = 
          0.695035633270614 +                     0i          0.695035633270614 +                     0i          0.695035633270614 +                     0i          0.695035633270614 +                     0i          0.695035633270614 +                     0i
           1.44113625911814 -      3.69313313087171i           1.44113625911814 -      3.69313313087171i           1.44113625911814 -      3.69313313087171i           1.44113625911814 -      3.69313313087171i           1.44113625911814 -      3.69313313087171i
           1.44113625911814 +      3.69313313087171i           1.44113625911814 +      3.69313313087171i           1.44113625911814 +      3.69313313087171i           1.44113625911814 +      3.69313313087171i           1.44113625911814 +      3.69313313087171i
          -1.78865407575345 +      3.72460215957041i          -1.78865407575345 +      3.72460215957041i          -1.78865407575345 +      3.72460215957041i          -1.78865407575345 +      3.72460215957041i          -1.78865407575345 +      3.72460215957041i
          -1.78865407575345 -      3.72460215957041i          -1.78865407575345 -      3.72460215957041i          -1.78865407575345 -      3.72460215957041i          -1.78865407575345 -      3.72460215957041i          -1.78865407575345 -      3.72460215957041i

F2n = double(subs(F2, sol2.parameters(1), kvals))

F1n = 
          -12.7448796695526 +                     0i          -6.46169436237299 +                     0i         -0.178509055193401 +                     0i           6.10467625198619 +                     0i           12.3878615591658 +                     0i
          -14.2687797481358 -      2.40307293360451i          -7.98559444095626 -      2.40307293360451i          -1.70240913377667 -      2.40307293360451i           4.58077617340291 -      2.40307293360451i           10.8639614805825 -      2.40307293360451i
          -14.2687797481358 +      2.40307293360451i          -7.98559444095626 +      2.40307293360451i          -1.70240913377667 +      2.40307293360451i           4.58077617340291 +      2.40307293360451i           10.8639614805825 +      2.40307293360451i
          -10.7747069529858 -      2.43542709294936i          -4.49152164580621 -      2.43542709294936i           1.79166366137337 -      2.43542709294936i           8.07484896855296 -      2.43542709294936i           14.3580342757325 -      2.43542709294936i
          -10.7747069529858 +      2.43542709294936i          -4.49152164580621 +      2.43542709294936i           1.79166366137337 +      2.43542709294936i           8.07484896855296 +      2.43542709294936i           14.3580342757325 +      2.43542709294936i

F2n = 

function F = root2d(Ficksang)
rfinal = [0,-0.101233861621737,0.365119069777688];
theta_f = Ficksang(1);
phi_f   = Ficksang(2);
psi_f   = 0;
r1      = rfinal(1);
r2      = rfinal(2);
r3      = rfinal(3);
F(1) = r1 - ( ( tan(psi_f/2) - (tan(theta_f/2) * tan(phi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );
F(2) = r2 - ( ( tan(phi_f/2) + (tan(theta_f/2) * tan(psi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) ); 
    
F(3) = r3 - ( ( tan(theta_f/2) - (tan(phi_f/2) * tan(psi_f/2)) ) / ...
        ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );
end

[Julian Blackthorne]

@Walter Roberson Thank you so much for taking your time to explain this.

Answer 4

[DEENA DHAYALAN]

function F = root2d(Ficksang)
rfinal = [0,-0.101233861621737,0.365119069777688];
theta_f = Ficksang(1);
phi_f   = Ficksang(2);
psi_f   = 0;
r1      = rfinal(1);
r2      = rfinal(2);
r3      = rfinal(3);
F(1) = r1 - ( ( tan(psi_f/2) - (tan(theta_f/2) * tan(phi_f/2)) ) / ...
    ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );
F(2) = r2 - ( ( tan(phi_f/2) + (tan(theta_f/2) * tan(psi_f/2)) ) / ...
    ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) ); 

F(3) = r3 - ( ( tan(theta_f/2) - (tan(phi_f/2) * tan(psi_f/2)) ) / ...
    ( 1 + (tan(theta_f/2) * tan(phi_f/2) * tan(psi_f/2)) ) );
end