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fsolve solves systems of nonlinear equations. However, it does not allow you to include any constraints, even bound constraints. The question is, how can you solve systems of nonlinear equations when you have constraints?
The short answer is, there are no guarantees that a solution exists that satisfies your constraints. There is no guarantee that any solution exists, even one that does not satisfy your constraints. Nevertheless, there are techniques that can help you search for solutions that satisfy your constraints.
To illustrate the techniques, consider how to solve the equations
$$\begin{array}{c}{F}_{1}(x)=\left({x}_{1}+1\right)\left(10-{x}_{1}\right)\frac{1+{x}_{2}^{2}}{1+{x}_{2}^{2}+{x}_{2}}\\ {F}_{2}(x)=\left({x}_{2}+2\right)\left(20-{x}_{2}\right)\frac{1+{x}_{1}^{2}}{1+{x}_{1}^{2}+{x}_{1}},\end{array}$$ | (11-9) |
where the components of x must be nonnegative. Clearly, there are four solutions to the equations:
x = (–1,–2)
x =
(10,–2),
x = (–1,20),
x = (10,20).
There is only one solution that satisfies the constraints, namely x = (10,20).
To solve the equations numerically, first enter code to calculate F(x).
function F = fbnd(x)
F(1) = (x(1)+1)*(10-x(1))*(1+x(2)^2)/(1+x(2)^2+x(2));
F(2) = (x(2)+2)*(20-x(2))*(1+x(1)^2)/(1+x(1)^2+x(1));
Save this code as the file fbnd.m on your MATLAB^{®} path.
Generally, a system of N equations in N variables has isolated solutions, meaning each solution has no nearby neighbors that are also solutions. So one way to search for a solution that satisfies some constraints is to generate a number of initial points x0, and run fsolve starting at each x0.
For this example, to look for a solution to Equation 11-9, take 10 random points that are normally distributed with mean 0 and standard deviation 100.
rng default % for reproducibility N = 10; % try 10 random start points pts = 100*randn(N,2); % initial points are rows in pts soln = zeros(N,2); % allocate solution opts = optimoptions('fsolve','Display','off'); for k = 1:N soln(k,:) = fsolve(@fbnd,pts(k,:),opts); % find solutions end
Examine the solutions in soln, and you find several that satisfy the constraints.
There are three fsolve algorithms. Each can lead to different solutions.
For this example, take x0 = [1,9] and examine the solution each algorithm returns.
x0 = [1,9]; opts = optimoptions(@fsolve,'Display','off',... 'Algorithm','trust-region-dogleg'); x1 = fsolve(@fbnd,x0,opts)
x1 = -1.0000 -2.0000
opts.Algorithm = 'trust-region-reflective';
x2 = fsolve(@fbnd,x0,opts)
x2 = -1.0000 20.0000
opts.Algorithm = 'levenberg-marquardt';
x3 = fsolve(@fbnd,x0,opts)
x3 = 0.9523 8.9941
Here, all three algorithms find different solutions for the same initial point. In fact, x3 is not even a solution, but is simply a locally stationary point.
lsqnonlin tries to minimize the sum of squares of the components of a vector function F(x). Therefore, it attempts to solve the equation F(x) = 0. Furthermore, lsqnonlin accepts bound constraints.
Formulate the example problem for lsqnonlin and solve it.
lb = [0,0];
rng default
x0 = 100*randn(2,1);
[x res] = lsqnonlin(@fbnd,x0,lb)
x = 10.0000 20.0000 res = 2.4783e-25
You can use lsqnonlin with the Global Optimization Toolbox MultiStart solver to search over many initial points automatically. See MultiStart Using lsqcurvefit or lsqnonlin.
You can reformulate the problem and use fmincon as follows:
Give a constant objective function, such as @(x)0, which evaluates to 0 for each x.
Set the fsolve objective function as the nonlinear equality constraints in fmincon.
Give any other constraints in the usual fmincon syntax.
For this example, write a function file for the nonlinear inequality constraint.
function [c,ceq] = fminconstr(x) c = []; % no nonlinear inequality ceq = fbnd(x); % the fsolve objective is fmincon constraints
Save this code as the file fminconstr.m on your MATLAB path.
Solve the constrained problem.
lb = [0,0]; % lower bound constraint rng default % reproducible initial point x0 = 100*randn(2,1); opts = optimoptions(@fmincon,'Algorithm','interior-point','Display','off'); x = fmincon(@(x)0,x0,[],[],[],[],lb,[],@fminconstr,opts)
x = 10.0000 20.0000