How to use Fsolve with multiple variables

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Hi. I am trying to solve two non-linear equations using fsolve. Here are my two codes:
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function G=Teth3(x,p,H,L);
g=9.82;
mu=0.0141*g;
G=[H-(x(2)/mu).*(cosh(mu.*(x(1)+p)/x(2))-cosh(mu.*x(1)/x(2))); %function G defines a vector of two functions to be solved
L-(x(2)/mu).*(sinh(mu.*(x(1)+p)/x(2))-sinh(mu.*x(1)/x(2)))];
end
-------------
function F = tethsolve3(p,H,L);
x0=[2;2];
g=9.82;
mu=0.0141*g;
[x,fval]=fsolve(@Teth3,x0,p,H,L);
end
-------------
I want the second one (tethsolve3) to get three inputs p, H, L (which will be used as constants) and solve the first function (Teth3) which is a function of x(vector of 2 variables) and p,H,L. I don't know what format I should use for the Fsolve. Please advise. Thank you very much.
  2 Comments
LAKKIMSETTI SUNANDA
LAKKIMSETTI SUNANDA on 1 Feb 2021
Hi. I also have a problem in solving the equations. I have four equations and four unknowns and I have to find those 4 unknown variables. The equations are as follows:
2*Y1*tan(t1_1)+Y2*tan(t2_1)+Y3*tan(t3_1)==0;
2*Y1*tan(t1_3)+Y2*tan(t2_3)+Y3*tan(t3_3)==0;
b1== (t1_1*Y1)+(t2_1*(Y2/2)*((sec(t2_1)^2)/(sec(t1_1)^2)))+(t3_1*(Y3/2)*((sec(t3_1)^2)/(sec(t1_1)^2)));
b3== (t1_3*Y1)+(t2_3*(Y2/2)*((sec(t2_3)^2)/(sec(t1_3)^2)))+(t3_3*(Y3/2)*((sec(t3_3)^2)/(sec(t1_3)^2)));
Here Y1,b1,b3 are known values and the unknown variables are Y2,Y3,l2,l3. The above t1,t2,t3 are in terms of l2 and l3.
t2_1=(2*pi/lambda(1))*l2;
t3_1=(2*pi/lambda(1))*l3;
t2_3=(2*pi/lambda(3))*l2;
t3_3=(2*pi/lambda(3))*l3;
t1_1=(2*pi/lambda(1))*l1;
t1_3=(2*pi/lambda(3))*l1;
Please help me with this. Thank you.
Alan Weiss
Alan Weiss on 21 Jan 2022
I suggest that you ask your question in a new topic rather than reopening an old topic.
That said, have you tried the Problem-Based Workflow for Solving Equations? It is a very natural way of solving equations.
Alan Weiss
MATLAB mathematical toolbox documentation

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Accepted Answer

Walter Roberson
Walter Roberson on 22 Oct 2013
[x,fval] = fsolve(@(x) Teth3(x, P, H, L), x0);
  23 Comments
Walter Roberson
Walter Roberson on 21 Nov 2016
You can send messages through my profile.
Alek Yawfimetz
Alek Yawfimetz on 18 Apr 2020
I appreciate the help Walter, I had a similar issue and this helped me solve it instantly. Thanks.

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More Answers (4)

Manuela Gräfe
Manuela Gräfe on 24 Apr 2017
Hello umme mumtahina,
please send me an personal message. I am also interested in the solutions of your questions.
Sometimes you just write: "Got it!", but you don't give the final solution. Due to the fact, that this is a public community, you should provide the corresponding answers to your questions.
So please send me ASAP a personal message.

safi58
safi58 on 21 Nov 2016
in this paper on page 3284, they have used fsolve to solve those equations. see upper right side.
  10 Comments
Manuela Gräfe
Manuela Gräfe on 24 Apr 2017
Hello umme mumtahina,
please send me an personal message. I am also interested in the solutions of your questions.
Sometimes you just write: "Got it!", but you don't give the final solution. Due to the fact, that this is a public community, you should provide the corresponding answers to your questions.
So please send me ASAP a personal message.

Sign in to comment.


safi58
safi58 on 6 Dec 2016
Hi Walter, after solving these equations i have found the initial condition. Can you please tell me how to draw these waveforms? see page 3247

GUANGHE HUO
GUANGHE HUO on 21 Jan 2022
Edited: Walter Roberson on 21 Jan 2022
for i=1:1:position
global alpha_tspd alpha_trpd beta_d R_pd R_sd Rb_sd T_in
K_sp1=K_mesh_sp(i,1);
K_sp2=K_mesh_sp(i,2);
K_sp3=K_mesh_sp(i,3);
K_rp1=K_mesh_rp(i,1);
K_rp2=K_mesh_rp(i,2);
K_rp3=K_mesh_sp(i,3);
theta_s=Theta_s(i,1);
F=@(x)[ (K_sp1*x(1)*cos(alpha_tspd)-K_rp1*x(4)*cos(alpha_trpd))*cos(beta_d)*R_pd;
(K_sp2*x(2)*cos(alpha_tspd)-K_rp2*x(5)*cos(alpha_trpd))*cos(beta_d)*R_pd;
(K_sp3*x(3)*cos(alpha_tspd)-K_rp3*x(6)*cos(alpha_trpd))*cos(beta_d)*R_pd;
(K_sp1*x(1)+K_sp2*x(2)+K_sp3*x(3))*cos(beta_d)*cos(alpha_tspd)*R_sd-T_in;
x(4)-theta_s*Rb_sd-x(1);
x(5)-theta_s*Rb_sd-x(2);
x(6)-theta_s*Rb_sd-x(3) ];
x0=delta(i,:);
options=optimoptions('fsolve','Algorithm','levenberg-marquardt');
delta(i+1,:)=fsolve(F,x0,options);
end
Hi, above are my code, every time I need to change some variables, but when I run, it shows that no solutions 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.
I do not know why, can someone help me?
  8 Comments
Walter Roberson
Walter Roberson on 24 Jan 2022
x(4)-theta_s*Rb_sd-x(1);
x(5)-theta_s*Rb_sd-x(2);
x(6)-theta_s*Rb_sd-x(3) ];
If theta_s happened to be 0, then those force equality relationships.
If theta_s were non-zero then those force simple linear relationships that you can substitute into the first four equations.
Walter Roberson
Walter Roberson on 24 Jan 2022
syms x [1 6]
syms alpha_trpd alpha_tspd beta_d K_rp1 K_rp2 K_rp3 K_sp1 K_sp2 K_sp3 R_pd R_sd Rb_sd T_in theta_s
eqn = [ (K_sp1*x(1)*cos(alpha_tspd)-K_rp1*x(4)*cos(alpha_trpd))*cos(beta_d)*R_pd;
(K_sp2*x(2)*cos(alpha_tspd)-K_rp2*x(5)*cos(alpha_trpd))*cos(beta_d)*R_pd;
(K_sp3*x(3)*cos(alpha_tspd)-K_rp3*x(6)*cos(alpha_trpd))*cos(beta_d)*R_pd;
(K_sp1*x(1)+K_sp2*x(2)+K_sp3*x(3))*cos(beta_d)*cos(alpha_tspd)*R_sd-T_in;
x(4)-theta_s*Rb_sd-x(1);
x(5)-theta_s*Rb_sd-x(2);
x(6)-theta_s*Rb_sd-x(3) ]
eqn = 
sol = solve(eqn([1:3 5:end]), x)
sol = struct with fields:
x1: -(K_rp1*Rb_sd*theta_s*cos(alpha_trpd))/(K_rp1*cos(alpha_trpd) - K_sp1*cos(alpha_tspd)) x2: -(K_rp2*Rb_sd*theta_s*cos(alpha_trpd))/(K_rp2*cos(alpha_trpd) - K_sp2*cos(alpha_tspd)) x3: -(K_rp3*Rb_sd*theta_s*cos(alpha_trpd))/(K_rp3*cos(alpha_trpd) - K_sp3*cos(alpha_tspd)) x4: -(K_sp1*Rb_sd*theta_s*cos(alpha_tspd))/(K_rp1*cos(alpha_trpd) - K_sp1*cos(alpha_tspd)) x5: -(K_sp2*Rb_sd*theta_s*cos(alpha_tspd))/(K_rp2*cos(alpha_trpd) - K_sp2*cos(alpha_tspd)) x6: -(K_sp3*Rb_sd*theta_s*cos(alpha_tspd))/(K_rp3*cos(alpha_trpd) - K_sp3*cos(alpha_tspd))
eqn4 = subs(eqn(4), sol)
eqn4 = 
In order for the system of equations to hold, that eqn4 would have to hold. It is independent of the x values: if it holds then there is a solution to the system, and if it does not hold then there is no solution to the system.

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