MATLAB Answers

fsolve and "Equation solved, fsolve stalled"

33 views (last 30 days)
Federico MegaMan
Federico MegaMan on 20 Nov 2019
Edited: Matt J on 21 Nov 2019
Hello,
when I run my fsolve it works, but appears this message:
"Equation solved, fsolve stalled.
fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance and the vector of function values
is near zero as measured by the default value of the function tolerance.
<stopping criteria details>".
What does it mean precisely?
clear all
clc
close all
kT21=0;
kT22=pi/2;%87°
kr23=3.1623;
kT23=4.391248;%248.5°
kT31=0;%0.226;%13°
kr32=3.1623;
kT32=1.8919369;%106°
kT33=1.5*pi;
kT11=1.106538;%62
kT12=0;
kT13=0;%01.5*pi+1.517
kT14=0;%0.226;
kT15=5.176646;%316°
%x0=[kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,jT21];
j=0;
xsol=zeros(13,51);
T21=zeros(1,51);
for t=0:0.1:5
j=j+1;
T11(j)=0.073*cos(t)+1.08;
end
for k=1:numel(T11)
xsol(:,k)=fsolve(@(x) funz2(x), [T11(k),kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,kT21]);
kxsol=xsol(:,k);
kT13=kxsol(3);
kT21=kxsol(13);
kT12=kxsol(2);
kT14=kxsol(4);
kT15=kxsol(5);
kT22=kxsol(6);
kT23=kxsol(7);
kr23=kxsol(8);
kT31=kxsol(9);
kT32=kxsol(10);
kT33=kxsol(11);
kr32=kxsol(12);
T13(k)=kxsol(3);
T21(k)=kxsol(13);
T12(k)=kxsol(2);
T14(k)=kxsol(4);
T15(k)=kxsol(5);
T22(k)=kxsol(6);
T23(k)=kxsol(7);
r23(k)=kxsol(8);
T31(k)=kxsol(9);
T32(k)=kxsol(10);
T33(k)=kxsol(11);
r32(k)=kxsol(12);
end
figure;
plot(1:1:51, xsol);
title('xsol in funzione di t');
xlabe1=('t');
ylabe1=('xsol');
grid on;
figure;
plot(1:1:51, T11*180/pi);
title('T11 in funzione di t');
xlabe3=('t');
ylabe3=('T11');
grid on;
figure;
plot(1:1:51, T12*180/pi);
title('T12 in funzione di t');
xlabe4=('t');
ylabe4=('T12');
grid on;
figure;
plot(1:1:51, T13*180/pi);
title('T13 in funzione di t');
xlabe4=('t');
ylabe4=('T13');
grid on;
figure;
plot(1:1:51, T21*180/pi);
title('T21 in funzione di t');
xlabe2=('t');
ylabe2=('T21');
grid on;
figure;
plot(1:1:51, T23*180/pi);
title('T23 in funzione di t');
xlabe5=('t');
ylabe5=('T23');
grid on;
figure;
plot(1:1:51, r23);
title('r23 in funzione di t');
xlabe5=('t');
ylabe5=('r23');
grid on;
figure;
plot(1:1:51, T31*180/pi);
title('T31 in funzione di t');
xlabe6=('t');
ylabe6=('T31');
grid on;
figure;
plot(1:1:51, r32);
title('r32 in funzione di t');
xlabe6=('t');
ylabe6=('r32');
grid on;
function F=funz2(x)
T11=x(1);
T12=x(2);
T13=x(3);
T14=x(4);
T15=x(5);
T22=x(6);
T23=x(7);
r23=x(8);
T31=x(9);
T32=x(10);
T33=x(11);
r32=x(12);
T21=x(13);
F(1)=T12-T21;
F(2)=T31-T14;
F(3)=T13+T22-pi/2;
F(4)=T33-T22-pi;
F(5)=r23-(1+9-6*cos(T12+T13-1.5*pi))^(0.5);
F(6)=r32-(1+9-6*cos(pi-T14+T33))^(0.5);
F(7)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;
F(8)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);
F(9)=cos(T21)+3*cos(T22)+r23*cos(T23);
F(10)=sin(T21)+3*sin(T22)+r23*sin(T23);
F(11)=cos(T31)+r32*cos(T32)+3*cos(T33);
F(12)=sin(T31)+r32*sin(T32)+3*sin(T33);
end

  0 Comments

Sign in to comment.

Answers (1)

Matt J
Matt J on 20 Nov 2019
It means stopping criteria for the iterations were met. Fsolve cannot find a reason to continue.

  2 Comments

Federico MegaMan
Federico MegaMan on 21 Nov 2019
Okay, thank you.
Another thing, the results aren't the expected, is there some reason for this? Maybe because there are cos and sin?
Matt J
Matt J on 21 Nov 2019
Perhaps your expectations (whatever those are) are incorrect.
Or, perhaps you have incorrectly implemented your equation function.
Or, perhaps you have given an insufficiently accurate initial guess.

Sign in to comment.

Sign in to answer this question.


Translated by