Fslove don't find the right solution

I try to solve a non linear system, but Fsolve can't solve it.

 F=@(q)[sqrt((q(1)-5)^2 + q(2)^2 + q(3)^2 )-sqrt( q(1)^2 + (q(2)+10)^2 + q(3)^2)-Rez1;
    sqrt(q(1)^2 + (q(2)+10)^2 + q(3)^2 )-sqrt( q(1)^2 + (q(2)-10)^2 + q(3)^2)-Rez2;
    sqrt( q(1)^2 + (q(2)-10)^2 + q(3)^2 )-sqrt( q(1)^2 + q(2)^2 + q(3)^2)-Rez3];

 q0=[50,0,0];
 options = optimset('Display','iter','maxfunevals',500,'tolfun',1e-1,'tolx',1e-1,'tolcon',1e-2);
 fsolve(F,q0,options)

Rez1,2,3 are some variable that I calculate before that. And the response of matlab is

I try to modify the values of tolfun and tolx, but no succes. Any help?

Answer 1

[Star Strider]

We don’t have your ‘Rez’ constants, so we can’t run your code.

However, using 0 as any initial parameter estimate results in a much more difficult optimisation.

Try this instead:

q0=[50,-1,1];

[Jessie Bessel]

 v=340;

 Tsos1=1.35
 Tsos2=1.47
 Tsos3=1.47
 Tsos4=1.44

 Rez1=v*(Tsos1-Tsos2)/10

 Rez2=v*(Tsos2-Tsos3)/10

 Rez3=v*(Tsos3-Tsos4)/10

 F=@(q)[sqrt((q(1)-5)^2 + q(2)^2 + q(3)^2 )-sqrt( q(1)^2 + (q(2)+10)^2 + q(3)^2)-Rez1;
    sqrt(q(1)^2 + (q(2)+10)^2 + q(3)^2 )-sqrt( q(1)^2 + (q(2)-10)^2 + q(3)^2)-Rez2;
    sqrt( q(1)^2 + (q(2)-10)^2 + q(3)^2 )-sqrt( q(1)^2 + q(2)^2 + q(3)^2)-Rez3];
 q0=[50,0,0];
 options = optimset('Display','iter','maxfunevals',500,'tolfun',1e-1,'tolx',1e-1,'tolcon',1e-2);
 fsolve(F,q0,options)

[Star Strider]

With this initial estimate:

q0 = randi(99, 3, 1);

and assigning an output for the fsolve result:

Qv = fsolve(F,q0,options)

I get solutions that appear to converge to the same value to ‘q(1)’ and ‘q(3)’, with ‘q(2)’ less well defined:

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the selected value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

Qv =

      31.0008e+000
     103.9495e-003
      37.4620e+000

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the selected value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

Qv =

      30.5924e+000
     208.7728e-003
      36.5371e+000

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the selected value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

Qv =

      31.3360e+000
      19.2740e-003
      37.2166e+000

Answer 2

[Alex Sha]

I get the results:

1:

q1: 31.25152
q2: -3.3648581278838E-15
q3: -37.1015437789933

2:

q1: 31.2515200000001
q2: 1.37892196278414E-15
q3: 37.1015437789935