How to use Fsolve with multiple variables

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Sina 2013-10-22

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https://ww2.mathworks.cn/matlabcentral/answers/91108-how-to-use-fsolve-with-multiple-variables

评论： Walter Roberson 2022-1-24

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Hi. I am trying to solve two non-linear equations using fsolve. Here are my two codes:

-------------

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.

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LAKKIMSETTI SUNANDA 2021-2-1

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 2022-1-21

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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Walter Roberson 2013-10-22

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[x,fval] = fsolve(@(x) Teth3(x, P, H, L), x0);

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Walter Roberson 2016-11-18

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That cannot be determined using fsolve(). You would need the Symbolic toolbox to get started.

There is no closed form solution for your x values. The values can be expressed in non-linear terms as formulae involving a quantity which for the moment I will call R.

R = RootOf( H^2 * Z^2 * exp(Z) - L^2 * Z^2 * exp(Z) + p^2 * (exp(Z))^2 - 2 * p^2 * exp(Z) + p^2, Z)

That is, R is the set of value, Z, such that the expression inside the RootOf() becomes 0. There is no closed form solution for that expression. There may be multiple solutions, especially complex-valued. 0 is a solution, but in the overall expression that leads to division by 0, so it is not a correct solution to the overall equations.

Once you have a value for R, then

x(1) = -R*exp(R)*p*(H-L)*(H+L)*(exp(R)-1)*(R-ln(-p*(exp(R)-1)/(R*(H-L))))/(((H^2-L^2)*R^2+p^2)*(exp(R))^2-2*exp(R)*p^2+p^2)
x(2) = (H-L)*exp(R)*R*(H+L)*mu*p*(exp(R)-1)/(((H^2-L^2)*R^2+p^2)*(exp(R))^2-2*exp(R)*p^2+p^2)

which is not pretty but at least can be simply mechanically applied once you have an R.

With the values I randomly selected for p, H, L above, the solutions to R are all imaginary (except for R = 0, a false solution.)

Walter Roberson 2016-11-21

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Alek Yawfimetz 2020-4-18

I appreciate the help Walter, I had a similar issue and this helped me solve it instantly. Thanks.

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Manuela Gräfe 2017-4-24

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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.

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safi58 2016-11-21

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https://ww2.mathworks.cn/matlabcentral/answers/91108-how-to-use-fsolve-with-multiple-variables#answer_244058

06104164.pdf

in this paper on page 3284, they have used fsolve to solve those equations. see upper right side.

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safi58 2016-12-4

编辑：Walter Roberson 2016-12-5

在 MATLAB Online 中打开

Hi Walter,

Sorry to disturb you again and again.

But I am still having problem.

I have done some changes in equations.

Can you please tell me what should be done now?

f =@(x) [x(1)+[[x(1)-(1/x(4))+1]*cos(x(3))+x(2)*sin(x(3))-1]*cos(k*(m-x(3)))+(1/k)*[[-x(1)+(1/x(4))-1]*sin(x(3))+x(2)*cos(x(3))]*sin(k*(m-x(3)))+(1/x(4));
x(2)+[[-x(1)+(1/x(4))-1]*xos(x(3))-x(2)*sin(x(3))+1]*k*sin(k*(m-x(3)))+[[-x(1)+(1/x(4))-1]*sin(x(3))+x(2)*cos(x(3))]*cos(k*(m-x(3)));
       [-x(1)+(1/x(4))-1]*sin(x(3))+x(2)*cos(x(3))-x(2)-n*x(3);
       [[[-x(1)+1/x(4)-1]*(1-cos(x(3)))+x(2)*sin(x(3))-x(2)*x(3)+(n*x(3)^2)/2]*((pri^2*R)/m*sec^2*Z)]-1];

for these equations it is showing

Undefined function 'xos' for input arguments of type 'double'.
Error in dcmb (line 10)
f =
[x(1)+[[x(1)-(1/x(4))+1]*cos(x(3))+x(2)*sin(x(3))-1]*cos(k*(m-x(3)))+(1/k)*[[-x(1)+(1/x(4))-1]*sin(x(3))+x(2)*cos(x(3))]*sin(k*(m-x(3)))+(1/x(4));
Error in fsolve (line 217)
            fuser = feval(funfcn{3},x,varargin{:});
Error in dcmb_driver (line 4)
X = fsolve(@dcmb, x0, optimset('MaxFunEvals',2000,'MaxIter',2000));
Caused by:
    Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue.

safi58 2016-12-5

solved it!!!!!!!!

Manuela Gräfe 2017-4-24

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.

请先登录，再进行评论。

safi58 2016-12-6

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/91108-how-to-use-fsolve-with-multiple-variables#answer_246114

06104164.pdf

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

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GUANGHE HUO 2022-1-21

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https://ww2.mathworks.cn/matlabcentral/answers/91108-how-to-use-fsolve-with-multiple-variables#answer_878905

编辑：Walter Roberson 2022-1-21

在 MATLAB Online 中打开

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?

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Walter Roberson 2022-1-22

在 MATLAB Online 中打开

clearvars
load K_mesh_sp.mat K_mesh_sp
load K_mesh_rp.mat K_mesh_rp
load Theta_s.mat Theta_s
load alpha_tspd.mat alpha_tspd
load alpha_trpd.mat alpha_trpd
load beta_d.mat beta_d
load K_mesh_sp.mat R_pd  %not in file R_pd.mat !
load R_sd.mat R_sd
load T_in.mat T_in
syms X [1 6]
Rb_sd=0.122867515153017;
position=270;
delta=zeros(position,6);
delta_0=[1e-6 1e-6 1e-6 1e-6 1e-6 1e-6];    % Initial value of delta   
delta(1,:)=delta_0;
for i=1:1:position 
    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');
    T1 = F(X);
    sol = solve(T1([1:3 5:7]), X)
    T2 = subs(T1(4), sol)
    assert(isAlways(T2 == 0, 'unknown', 'true'), 'seventh equation fail')
    delta(i+1,:)=fsolve(F,x0,options);
end

You will get seventh equation fail in the first iteration.

The last three equations are of the form Xi - Xj which forces Xi and Xj to be the same. Substitute those equalities into the first three equations and you get equations that are all linear in a single variable and the right hand side of the equation is 0. When a linear multiple of a variable must be equal to 0, then it forces the variable to be equal to 0. So the first three variables must be 0 and the last 3 variables must be the same as the first three variables and so must equal 0 as well. This leaves you with one equation out of the 7, and if you substitute in all 0 for the variables there, you get something that cannot possibly be 0.

So, at least for the first round, the equations are inconsistent.

Walter Roberson 2022-1-24

在 MATLAB Online 中打开

        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 2022-1-24

在 MATLAB Online 中打开

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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How to use Fsolve with multiple variables

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