fsolve with vectors x and y (not x(1) x(2))

Question

anto 2022-12-26

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1884732-fsolve-with-vectors-x-and-y-not-x-1-x-2

编辑： Torsten 2022-12-26

I'd like to use fsolve, to solve for the vectors x and y, this nonlinear system which i wrote in a function.

function F = EquationsList(x,y)
F(1) = -0.00011 + 2 * x(1) * (cos(y(1))+sin(y(1))) + 2 * x(2) * (cos(y(2))+sin(y(2))) + 2 * x(3) * (cos(y(3))+sin(y(3))); 
F(2) = -0.00013 + 2 * x(1) * (cos(y(1))+sin(y(1))) + 2 * x(2) * (cos(y(2))+sin(y(2))) + 2 * x(3) * (cos(y(3))+sin(y(3))); 
...
...
F(9) = -0.00015 + 2 * x(1) * (cos(y(1))+sin(y(1))) + 2 * x(2) * (cos(y(2))+sin(y(2))) + 2 * x(3) * (cos(y(3))+sin(y(3)));  
end

in the code above I only reported the first 2 equations. It's a large system.

the equations derive from the development of the sum shown below:

in which a,x,y are all scalars.

For using fsolve I coded this:

%defining constants:
N_mesh=3;
%define the problem for fsolve
problem.objective = @EquationsList; 
%initial point
x0=zeros(N_mesh*N_mesh,1);
y0=zeros(N_mesh*N_mesh,1);
problem.x0 = x0; 
problem.y0 = y0; 
%solver
problem.solver = 'fsolve';
%set tolerances
problem.options = optimoptions('fsolve', ...
    'MaxIter', 4000, ...
    'MaxFunEvals',4000, ...
    'StepTolerance',1e-16, ...
    'FunctionTolerance',1e-16, ...
    'OptimalityTolerance',1e-10,...
    'Algorithm','levenberg-marquardt');
%solve
[x,y] = fsolve(problem);

But it gives me the error ' Not enough input arguments. '

Does anybody know how to use fsolve with these equations? I can't use only the vector x as the problem I'm trying to resolve requires two different output: x and y as vectors.

Thanks in advance for any replies.

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anto 2022-12-26

在 MATLAB Online 中打开

Sorry, I've cutted some scalars in the LATEX equation for better clearance but you're technically right.

In reality the complete system which I'm trying to solve is including other constants which multiplies the (cos + sin) addend.

The equations are in this form, in which all parameters are scalars.

So the system of non-linear equations written in the function is something like this:

function F = MyFun(x,y)
F(1) = -0.0001121770+2*x(1)*(cos(y(1))*cos(0.0000000000)+sin(y(1))*sin(0.0000000000))*((0.0021349691)^2)*(0.1666666667)...
+2*x(2)*(cos(y(2))*cos(0.0000000000)+sin(y(2))*sin(0.0000000000))*((0.0021349690)^2)*(0.1666666667)...
+2*x(3)*(cos(y(3))*cos(0.0000000000)+sin(y(3))*sin(0.0000000000))*((0.0021349690)^2)*(0.0000000000)...
+2*x(4)*(cos(y(4))*cos(0.0000000000)+sin(y(4))*sin(0.0000000000))*((0.0021349691)^2)*(0.0000000000);
F(2)= ....
...
...
end

and so on.

I'm just trying to use a syntax which lets me use fsolve with those unknowns. Do you know how to do it?

Any help would be greatly appreciated.

Karim 2022-12-26

在 MATLAB Online 中打开

well, what you show here has 4 unknowns for x and y, hence i would guess that you need to set the inital values accordingly. Why do you take the square of the number of variables?

% ...
N_mesh = 4;
% ...
x0 = zeros(N_mesh,1);
y0 = zeros(N_mesh,1);

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Answer 1

Karim 2022-12-26

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1884732-fsolve-with-vectors-x-and-y-not-x-1-x-2#answer_1136577

在 MATLAB Online 中打开

Based on the comments, I would guess that you are looking for 8 parameters i.e.

syms x [1 4]; x

x =

syms y [1 4]; y

y =

can you try modifying your function "MyFun.m" such that:

function F = MyFun( input, numX )
% extract the variables from the input
x = input( 1:numX );
y = input( (numX+1):end );
% evaluate the equations
F(1) = -0.0001121770+2*x(1)*(cos(y(1))*cos(0.0000000000)+sin(y(1))*sin(0.0000000000))*((0.0021349691)^2)*(0.1666666667)...
    +2*x(2)*(cos(y(2))*cos(0.0000000000)+sin(y(2))*sin(0.0000000000))*((0.0021349690)^2)*(0.1666666667)...
    +2*x(3)*(cos(y(3))*cos(0.0000000000)+sin(y(3))*sin(0.0000000000))*((0.0021349690)^2)*(0.0000000000)...
    +2*x(4)*(cos(y(4))*cos(0.0000000000)+sin(y(4))*sin(0.0000000000))*((0.0021349691)^2)*(0.0000000000);
F(2) = ...
end

and then you can call fsolve as:

% define the number of parameters
numX = 4;
numY = 4;
% define the problem for fsolve
problem.objective = @MyFun(x, numX);
% set the initial values
problem.x0 = zeros( numX+numY,1); 
% set the solver
problem.solver = 'fsolve';
% set the options
problem.options = optimoptions('fsolve', ...
    'MaxIter', 4000, ...
    'MaxFunEvals',4000, ...
    'StepTolerance',1e-16, ...
    'FunctionTolerance',1e-16, ...
    'OptimalityTolerance',1e-10,...
    'Algorithm','levenberg-marquardt');
% solve the problem
solution = fsolve(problem);
% extract the variables
x = solution( 1:numX );
y = solution( (numX+1):end );

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anto 2022-12-26

Yes, that's what I've been thinking also just now:

fsolve wants only a vector of unknowns, so we just "assign" the first numX unknowns to the first vector (which i called x in the LATEX), and the following numY unknowns as the second vector (which I called y in the LATEX).

Thanks again Karim.

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Answer 2

Torsten 2022-12-26

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1884732-fsolve-with-vectors-x-and-y-not-x-1-x-2#answer_1136572

编辑：Torsten 2022-12-26

在 MATLAB Online 中打开

%defining constants:
N_mesh=3;
%define the problem for fsolve
problem.objective = @(z)EquationsList(z,Nmesh); 
%initial point
x0=zeros(1,N_mesh*N_mesh);
y0=zeros(1,N_mesh*N_mesh);
problem.x0 = [x0,y0];
%solver
problem.solver = 'fsolve';
%set tolerances
problem.options = optimoptions('fsolve', ...
    'MaxIter', 4000, ...
    'MaxFunEvals',4000, ...
    'StepTolerance',1e-16, ...
    'FunctionTolerance',1e-16, ...
    'OptimalityTolerance',1e-10,...
    'Algorithm','levenberg-marquardt');
%solve
z = fsolve(problem);
x = z(1:Nmesh*Nmesh);
y = z(Nmesh*Nmesh+1:2*Nmesh*Nmesh);
function F = MyFun(z,Nmesh)
  x = z(1:Nmesh*Nmesh);
  y = z(Nmesh*Nmesh+1:2*Nmesh*Nmesh);
  F(1) = -0.0001121770+2*x(1)*(cos(y(1))*cos(0.0000000000)+sin(y(1))*sin(0.0000000000))*((0.0021349691)^2)*(0.1666666667)...
         +2*x(2)*(cos(y(2))*cos(0.0000000000)+sin(y(2))*sin(0.0000000000))*((0.0021349690)^2)*(0.1666666667)...
         +2*x(3)*(cos(y(3))*cos(0.0000000000)+sin(y(3))*sin(0.0000000000))*((0.0021349690)^2)*(0.0000000000)...
         +2*x(4)*(cos(y(4))*cos(0.0000000000)+sin(y(4))*sin(0.0000000000))*((0.0021349691)^2)*(0.0000000000);
  F(2)= ....
  ...
  ...
end