[FreakLegion]

> all nonlinear equation systems are nonconvex

Maybe you have something more particular in mind when you say "systems", but not all nonlinear functions are non-convex. Least squares, for example, is nonlinear and convex.

Also note that IPOPT, while wonderful, is a local solver. It may not be limited to convex problems, but those are the only ones it's guaranteed to solve to optimality.

[wenc]

I was talking in terms of convex optimization. The criteria for convex optimization is convex objective, convex inequality constraints and convex feasible region, and linear (not just convex) equality constraints.

I’m aware that ipopt is a local solver.

[FreakLegion]

> all nonlinear equation systems are nonconvex

> I was talking in terms of convex optimization

I'm not sure what subtlety you're pointing to here. As far as non-convex problems, though, my point is that IPOPT isn't special in this regard. Any convex solver can be a non-convex solver if you don't care about global optimality.

[wenc]

> Any convex solver can be a non-convex solver if you don't care about global optimality.

Aside: Structurally I’m not sure how this would be true.

Convex solvers have very specific forms. For instance a QP solver requires a very particular form and does not admit any arbitrary non convex form except for one: the non-PSD Hessian which is the concave problem.

My point is that all NLEs power inside an optimization problem gives rise to a non convex optimization problem with no guarantee of a global solution. So convex optimization is not applicable here.

[Bimos]

The feasible region {x | f(x) = 0} is nonconvex no matter whether f is convex.

[graycat]

Consider claim:

> The feasible region {x | f(x) = 0} is nonconvex no matter whether f is convex.

For some positive integer n and for the set of real numbers R, consider closed (in the usual topology for R^n), convex set A a subset of R^n. Define function f: R^n --> R so that f(x) = 0 for all x in A, f(x) > 0 for all x not in A, and for all x in R^n f infinitely differentiable at x. It is a theorem that such an f exists.

Then { x | f(x) = 0 } = A and is both closed and convex in contradiction to the claim.

[ubj]

I'm assuming you're referring to nonlinear f(x) because this statement is trivially false for linear f(x).

But consider the function f(x) = max(0, g(x) - c) where the following holds:

* g(x) is nonlinear, positive definite in x, and convex.

* c > 0

Then f(x) is nonlinear and convex (it's the pointwise maximum of convex functions), and the set {x : f(x) = 0} is a convex set.