# 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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Pritha 2021 年 7 月 1 日
コメント済み: Pritha 2021 年 7 月 8 日
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.
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Pritha 2021 年 7 月 8 日
Thanks for your reply. But it shows problrm in 5th line. It will be helpful for me if you elaborate the process.

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### 回答 (2 件)

Alex Sha 2021 年 7 月 3 日
Hi, the result below should be one solution
x1: -52.5485799573576
x2: 12.1032014174617
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Pritha 2021 年 7 月 8 日
I have got another set of aswers like,
x1: 18.2653
x2: -98.5331 , for different initial condition.

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Pritha 2021 年 7 月 8 日
I have got another set of aswers like,
x1: 18.2653
x2: -98.5331 , for different initial condition.
##### 2 件のコメント表示非表示 1 件の古いコメント
Pritha 2021 年 7 月 8 日
"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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