How to use options in fsolve

Hello,

I'm trying to use the options in fsolve but i didn't understand how i can do...

I know that i have to do fsolve(@(x) myfunz,x0,options) but I don't know what i have to write, option=..?

This is my code:

j=0;
xsol=zeros(13,28);
T21=zeros(1,28);
for t=0:0.1:2.7
    j=j+1;
    T21(j)=0.333*cos(t)-0.245;
    jT21=T21(j);
%jT21=0;
jT22=pi/2;%87°
jr23=3.1623;
jT23=4.391248;%248.5°

jT31=0;%0.226;%13°
jr32=3.1623;
jT32=1.8919369;%106°
jT33=1.5*pi;

jT11=1.106538;%62
jT12=0;
jT13=0;%01.5*pi+1.517
jT14=0;%0.226;
jT15=5.176646;%316°


for k=1:numel(T21)
    
x0=[jT11,jT12,jT13,jT14,jT15,jT22,jT23,jr23,jT31,jT32,jT33,jr32,jT21];

xsol=fsolve(@(x) funzz(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,jT21]);
kT11=jxsol(1);
kT12=jxsol(2);
kT13=jxsol(3);
kT14=jxsol(4);
kT15=jxsol(5);
kT22=jxsol(6);
kT23=jxsol(7);
kr23=jxsol(8);
kT31=jxsol(9);
kT32=jxsol(10);
kT33=jxsol(11);
kr32=jxsol(12);
end
end
function F=funzz(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)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;
F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);
F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);
F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);
F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);
F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);
F(7)=T12-T21;
F(8)=T31-T14;
F(9)=T13+T22-pi/2;
F(10)=T33-T22-pi;
F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);
F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);

end

But when I run it, it says:

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>

Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt

algorithm instead.

> In fsolve (line 310)

In wsadfg (line 30)

I saw that Levenberg-Marquardt is linked to options(is it right?), but i didn't understand how i can do.

Another thing, how can i make xsol a xsol(k)? Because matlab give me error if i put it.

Thank you in advance.

Answer 1

[Matt J]

I know that i have to do fsolve(@(x) myfunz,x0,options)

I don't think so. The exit message says "Equation solved".

Another thing, how can i make xsol a xsol(k)?

for k=1:numel(T21)
    

xsol(:,k)=fsolve(@(x) funzz(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,jT21]);

end

[Federico MegaMan]

Thank you for the answer.

So do you think the code is right? Because the output are all costant...

I did some changes for T21 <0 and T21>0 and add the plot, but the output are all costant...

%jT21=0;
kT22=pi/2;%87°
kr23=3.1623;
kT23=4.391248;%248.5°

kT31=0;
kr32=3.1623;
kT32=1.8919369;%106°
kT33=1.5*pi;

kT11=1.106538;%62
kT12=0;
kT13=0;
kT14=0;
kT15=5.176646;%316°
    

j=0;
xsol=zeros(13,28);
T21=zeros(1,28);
for t=0:0.1:2.7
    j=j+1;
    T21(j)=0.333*cos(t)-0.245;
    jT21=T21(j);
end

x0=[kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,jT21];


for k=1:numel(T21)
    
    if T21<=0
xsol(:,k)=fsolve(@(x) funz1(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,T21(k)]);

    else
        xsol(:,k)=fsolve(@(x) funz2(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,T21(k)]);
    end
        jxsol=xsol(:,k);

kT11=jxsol(1);
kT12=jxsol(2);
kT13=jxsol(3);
kT14=jxsol(4);
kT15=jxsol(5);
kT22=jxsol(6);
kT23=jxsol(7);
kr23=jxsol(8);
kT31=jxsol(9);
kT32=jxsol(10);
kT33=jxsol(11);
kr32=jxsol(12);
end


figure;
plot(1:1:28, xsol);
title('xsol in funzione di t');
xlabe1=('t');
ylabe1=('xsol');
grid on;



function F=funz1(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)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;
F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);
F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);
F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);
F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);
F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);
F(7)=T12-T21;
F(8)=T31-T14;
F(9)=T13+T22-pi/2;
F(10)=T33-T22-pi;
F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);
F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);

end



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)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;
F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);
F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);
F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);
F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);
F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);
F(7)=T12-T21;
F(8)=T31-T14;
F(9)=T13+T22-3*pi/2;
F(10)=T33-T22-pi;
F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);
F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);

end

So can i ignorate this "Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt

algorithm instead. "?

[Matt J]

So do you think the code is right? Because the output are all costant...

Of course it is constant. funz1 and T21 aren't changing, so you are solving the same equations over and over again.

So can i ignorate this "Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt

Yes.

[Federico MegaMan]

Oh, wait I want they changes. Doesn't T21 vary with for? Anyway how can i let them change?

[Matt J]

By doing something in the loop that makes them changes with k...? Currently, nothing is done with them.

[Federico MegaMan]

I did it putting jxsol=xsol(:,k), and after T11=jxsol(1), T12=jxsol(2), ecc... I did it to changes their value every iteration.

[Matt J]

Perhaps you meant to write

if T21(k)<=0  %add indexing