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Solving a system of nonlinear equations with unorganized unknowns?

Asked by Devanjith Fonseka on 4 Oct 2019
Latest activity Edited by Matt J
on 6 Oct 2019
Ideally when using fsolve, my entire solution vector will be unknown. However my unknowns are broken down into 2 vectors x and r such that some values of my x vector are always fixed as they are boundary conditions but the summation of the length of x which are unknown, and r gives me the total number of unknowns. Is there a way I can use fsolve to pick the indices I would like it to solve leaving the boundary conditions the way they are?.


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1 Answer

Answer by Matt J
on 4 Oct 2019
Edited by Matt J
on 4 Oct 2019

If you use lsqnonlin instead of fsolve, you can set a vector of upper and lower bounds ub and lb on the solution. Just set ub(i)=lb(i) for those x(i) elements you want constrained to something known, e.g.,
p0=[x0(:);r0(:)]; %Initial guess
[lb,ub]=deal(-inf(N,1), +inf(N,1));
ub(1:3)=10; lb(1:3)=10; %constrain x(1:3) to equal 10


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This works for a system of nonlinear equations and yourFunction is defined the same way as I would for fsolve right?
Why isn't p0 defined as p0=[x0(:),r0(:)]?
It doesn't matter. The solver will reshape p0 into a column vector anyway, as described here
I implemented it and I get this error:
The Levenberg-Marquardt algorithm does not handle bound constraints and the trust-region-reflective algorithm requires at
least as many equations as variables; aborting.
I expected this to occur considering that even though I constrain my vector to take a specific value for my known variables, I would be left with too many variables.
Is there any work around to this?
If you have fewer equations than unknowns then it is not clear what you hope for as a result. There will normally be infinite solutions in that situation, and the problem will be ill-posed.

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