I want to solve 2 equations with 2 variables but it cant be solved and the command window shows fsolve stopped because the last step was ineffective

function main1

p=.153;

x0 = [-3000, -3000];

options = optimoptions('fsolve','display', 'iter');

[sol,fval,exitflag, output] = fsolve(@(x)fun(x,p),x0, options);

if exitflag <= 0

return

end

end

function xres = fun(x,p)

k0 = 2.536;

k1 = 1.354;

k2 = -4.775;

k3 = -2.772;

k4 = -0.235;

gsN = 10.567;

gzN = -0.467;

f_pi = 93.3;

d = 0.064;

X0 = 409.769;

f_k = 122.143;

m_pi = 139;

m_k = 498;

k = (3 * p* pi^2)^(1/3);

gsN = 10.567;

gzN = -0.467;

m1 = -(gsN*x(1) + gzN*x(2));

E = sqrt(k^2 +m1^2);

rho_s = (m1/pi^2)* (k*E- m1^2 * log((k+E)/m1));

F(1) = (k0*x(1)*X0^2)-(4*k1*x(1)*(x(1)^2 + x(2)^2))-(2*k2*x(1)^3)-(2*k3*X0*x(1)*x(2))-(2*d*X0^4/(3*x(1)))+(m_pi^2*f_pi) - (gsN*rho_s);

F(2) = (k0*x(2)*X0^2)-(4*k1*x(2)*(x(1)^2 + x(2)^2))-(2*k2*x(2)^3)-(k3*X0*x(1)^2)-(d*X0^4/(3*x(2)))+((sqrt(2)*m_k^2*f_k)-(f_pi*m_pi^2/sqrt(2))) - (gzN*rho_s);

xres = F;

end

* here i want to solve x(1) and x(2)

command window

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.

<stopping criteria details>

fsolve stopped because the sum of squared function values, r, changed by 9.991357e-14

relative to its initial value; this is less than max(options.FunctionTolerance^2,eps) = 1.000000e-12.

However, r = 2.189392e+13, exceeds sqrt(options.FunctionTolerance) = 1.000000e-03.

[Amit Bhowmick]

Check your function and plot the residue, try to observe roots from the graph.

use this code

[X,Y]=meshgrid(-3000:50:-1,-3000:50:-1);

x=X(:);

y=Y(:);

for ii=1:numel(x)

res=myfun(x(ii),y(ii),p);

y1(ii)=res(1);

y2(ii)=res(2);

end

figure(2)

plot(y1)

hold

plot(y2)

[Pritha]

Thanks for your reply. But it shows problrm in 5th line. It will be helpful for me if you elaborate the process.

Answer 1

[Alex Sha]

Hi, the result below should be one solution

x1: -52.5485799573576
x2: 12.1032014174617

[Pritha]

I have got another set of aswers like,

x1: 18.2653

x2: -98.5331 , for different initial condition.

so how can i conclude that which answer is correct? please help me. also for diffent p value answers are not changing. Can you please give me hint to handle this?

Answer 2

[Pritha]

I have got another set of aswers like,

x1: 18.2653

x2: -98.5331 , for different initial condition.

so how can i conclude that which answer is correct? please help me. also for diffent p value answers are not changing. Can you please help me out?

[Amit Bhowmick]

A set of nonlinear equation can have more than one solution. However a set of linear eqution can have only one solution. For an example two straight line (Linear curve) can have only one point of intersection but two curve line can have more than one intersection. So both answer is correct unless you have any extra constraint on dependent varriable x1 and x2.

[Pritha]

Thanks sir for your answer. I understand. But in the matlab command window alongwith the answers,

"Equation solved, solver stalled.

fsolve stopped because the relative size of the current step is less than the

value of the step size tolerance squared and the vector of function values

is near zero as measured by the value of the function tolerance. fsolve stopped because the relative norm of the current step, 8.122955e-17, is less than

max(options.StepTolerance^2,eps) = 2.220446e-16. The sum of squared function values,

r = 5.297954e-16, is less than sqrt(options.FunctionTolerance) = 1.000000e-15." is coming.

so what does it mean?

and also if i change the input p the result should change but it is not happening here. please help sir.