using fsolve when the function handle takes in 2 variables that are both vectors

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Purnjay Peshawaria
Purnjay Peshawaria 2020-4-18
评论： John D'Errico 2021-10-3
I'm trying to solve system of non-linear equations. My function handle fun takes in 2 variables ( x and y) that are both vectors.Each vector has 2 components. I want fsolve to return me 2 vectors with 2 components each. However, fsolve can only return one vector. How do I find an appropriate solution for my problem? Any help appreciated.
fun=@(x,y) [exp(-exp(-(x+y))) - x.*(1+x.^2); x.*cos(y) + y.*sin(x) - [0.5;0.5]];
x0=[1;1]; % initial guess for x vector
y0=[0;1]; % initial guess for y vector
[result1,result2]=fsolve(fun,x0,y0);

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dpb
dpb 2020-4-19
Recast as 4-vector w/ v(1:2) -- x; v(3:4) -- y:
fun2=@(x) [exp(-exp(-(x(1:2)+x(3:4)))) - x(1:2).*(1+x(1:2).^2); ...
x(1:2).*cos(x(3:4)) + x(3:4).*sin(x(1:2)) - [0.5;0.5]];
>> fsolve(fun2,[x0;,y0])
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
ans =
0.4627
0.4627
0.0876
0.0876
>> fun2(ans)
ans =
1.0e-11 *
-0.2900
-0.0679
-0.7623
-0.0910
>>
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mary
mary 2021-10-3
@dpb could you please answer this question: https://nl.mathworks.com/matlabcentral/answers/1464779-solve-a-set-of-two-linear-equations
It's very similar to this one. With the differnce that I want to keep f1 and f2 as they are. And for the sake of clarity, I dont want to concatenate x and y into a vector.
John D'Errico
John D'Errico 2021-10-3
Ran in:
@mary - what you want to do, and what you CAN do are very different things. We can't always get what we want. A fellow named Mick eloquently told us that many years ago.
Optimizers are not set up to optimize functions where the parameters are scattered among the inputs. Surely you understand that cannot happen. So even if the variables of interest are the only variables, it cannot be done, at least not directly by a tool like fsolve.
Can you create a helper function, that will solve the problem? Well, yes. For example, I want to optimize the objective f(x,y), and you refuse to reformulate it as essentialy f(xy), where xy is now a vector of length 2? I'll use fminunc here as an example...
Fun = @(x,y) (x-1).^2 + (x - y).^2 - x + y;
varsplit = @(XY) Fun(XY(1),XY(2));
fminunc(varsplit,[2,2])
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
ans = 1×2
1.0000 0.5000
So fminunc found the minimum of Fun, as a function of two variables.

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