fsolve result is not desirable even giving a close starting point

I have three nonlinear equations with three unknowns and I used fsolve function.

I know that x=[0.3173;0.3173;3.6590] is a good enough solution for the three equations. However, using fsolve, I could not get to this solution even for a very close starting point as shown below. Instead the result is (upon plugging these results into the 3 equations, the values are not close to 0):

x =
   0.999999999999936
   1.000000000000067
   1.000000000000036

How can this be solved? Thanks!

CODE:

x0=[0.31; 0.31; 3.5];

options=optimset('Display','iter'); 

[x,fval]=fsolve(@myfun1,x0,options)

function F=myfun(x)

F=[13*cos(x(3)) - 5*x(1)*x(3)^2 - 5*x(2)*x(3)^2 + 13*x(1)*x(3)*sin(x(3)) + 13*x(2)*x(3)*sin(x(3)) + 10*x(1)*x(2)*x(3)^2 - 13*x(1)*x(2)*x(3)^2*cos(x(3)) + 40;

     5*x(3) - 13*sin(x(3)) + 30*x(1)*x(3) + 13*x(1)*x(3)*cos(x(3)) + 13*x(2)*x(3)*cos(x(3)) - 5*x(1)*x(2)*x(3)^3 + 13*x(1)*x(2)*x(3)^2*sin(x(3));

     390*x(3)^2*sin(x(3)) + 300*x(1)*x(3)^3 - 25*x(1)*x(3)^5 + 25*x(2)*x(3)^5 - 150*x(3)^3 - 260*x(1)*x(3)^3*cos(x(3)) - 130*x(2)*x(3)^3*cos(x(3)) + 130*x(1)*x(3)^4*sin(x(3)) - 130*x(2)*x(3)^4*sin(x(3)) - 169*x(1)*x(3)^3*cos(x(3))^2 + 169*x(2)*x(3)^3*cos(x(3))^2 - 169*x(1)*x(3)^3*sin(x(3))^2 + 169*x(2)*x(3)^3*sin(x(3))^2
     ];

[Matt J]

In future, please highlight your code and apply the

formatting button, as I have just done for you now.

[Qingbin]

Thanks, I will do this later.

Answer 1

[Matt J]

Possibly because you're passing @myfun1 to fsolve instead of @myfun? When I make this change and run your code, I get the solution you expect.

[Qingbin]

Thanks! Sorry for the mistake. One question to follow up is that how can I give a good estimate of the starting point. In the above problem, I want to solve for the max{x(3)} and min{x(3)}.

[Matt J]

There is no science to the initial guess. If you want solutions with the max/min possible x(3) value, try an initial guess with parameters well above/below what x(3) would naturally be.

Answer 2

[Alex Sha]

For Qingbin's equations, although it is a problem that has passed a long time, it is worth and interesting to have a try, there are multi-solutions:

The first one:

x1: -12.7613758329431
x2: -12.7613758329431
x3: -0.755039591199301

The second one:

x1: 0.101047394240414
x2: -3.71598808815167
x3: -1.63192533266968

The third one:

x1: -0.282297628581163
x2: 0.369875309676247
x3: 3.09470113469319

The fourth one:

x1: 0.317262018077202
x2: 0.317262018077202
x3: 3.65895757009226