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

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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.
  2 个评论
Amit Bhowmick
Amit Bhowmick 2021-7-1
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)

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回答(2 个)

Alex Sha
Alex Sha 2021-7-3
Hi, the result below should be one solution
x1: -52.5485799573576
x2: 12.1032014174617
  1 个评论
Pritha
Pritha 2021-7-8
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?

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Pritha
Pritha 2021-7-8
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?
  2 个评论
Amit Bhowmick
Amit Bhowmick 2021-7-8
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
Pritha 2021-7-8
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.

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