How do I improve solving 18 second order system of algebraic equations in fsolve ?

Hi All

trying to solve 18 nonlinear (2nd order)equations using fsolve I get : possibility of inacuaracy in the results. while I have used the options like : 1e6 iterations plus 1e-5 solution tolerance.

this is the options setting :

options = optimoptions(@fsolve,'Display','iter','MaxFunEvals',1e6,'MaxIter',1e6,'TolFun',1e-6,'DerivativeCheck','on','Diagnostics','on');

and this is the error I get

Equation solved, inaccuracy possible.

The vector of function values is near zero, as measured by the default value
of the function tolerance. However, the last step was ineffective.

<stopping criteria details>

is there a way to improve it ?

[Walter Roberson]

fsolve is not appropriate solving differential equations. Use dsolve() or one of the ode*() routines or one of the boundary value problem routines, or perhaps even the PDE Toolbox would be appropriate?

[farzad]

I don't have differential equations, they are only second order equations

[farzad]

They are only algebraic equations

[Walter Roberson]

Do you mean equations that have variables that might be squared?

[John D'Errico]

Just playing with the max number of iterations and the tolerance will not help.

You don't say how many unknowns you are solving for. Remember that there MAY be dozens of solutions. The problem MAY be ill-conditioned. So the starting value will be far more important than those parameters, but it may not be sufficient if the problem is ill-conditioned.

[farzad]

@Walter : yes they are squared.

@John : I have 18 unknowns , equal to the equations number.

[farzad]

I updated my question and put the error and the options. I don't think that it's ill conditions but there is inaccuracy

[Walter Roberson]

Do you have the Symbolic Toolbox? Sometimes vpasolve() is better suited when accuracy is a problem.

Anyhow, you should make the solution tolerence smaller.

In some cases, when the coefficients are large enough, the gradient can end up so steep that you have to switch to absolute tolerences instead of relative tolerances.

Remember that a warning of inaccuracy does not mean that the solution returned is too inaccurate for your purposes -- it just means that when the solution was cross-checked, the residue was larger than a tolerance, so it is warning that the solution might not be so good. But if you are asking for tolerences of 1e-5 on a system with 18 variables, then chances are that you are not all that concerned about exact locations.

In the context of polynomials of degree 2, the warning would be valid for the situation where the polynomial has a minima or maxima that is close to zero rather than a zero crossing. The equation (x+a)^2+delta with positive delta could potentially be determined to have a root at -a if delta is close enouugh to zero even though the mathematical root is complex valued.

[farzad]

dear Walter, well with the current inaccuracy, my results are not super good and cause large errors in the next post processing on the results and give bigger deviations later. so I would need a close enough result. I thought 10^-5 is the difference of calculated with respect to the expected. I reduced it but nothing chaged. i need to make an executable later, so I dont know if vpasolve could be useful

[Walter Roberson]

Can you post the equations, together with representative values of any constants they require?