System of nonlinear equations unique solution

This is my first longer answer on math.stack, but you asked something I'm working in.

If you are interested in algorithmic results and real equations, then one could say that for polynomials, everything is decidable and for more general functions, almost everything is undecidable. In particular, the whole first-order theory of real numbers with multiplication is decidable; so you can include polynomial equations, quantifiers, disjunctions etc.

However, once you allow to use arbitrary complicated functions containing compositions of polynomials and the sin function, then the existence of a solution is already algorithmically undecidable.

Common software packages are based on iterative methods, I think that most of them use some variants of Newton method; but the above result shows that you cannot have completeness results using such algorithms.

For compact domains and some reasonably computable input functions, there are still undecidable results, look at this question that I asked on mathoverflow.

If you have $n$ equations in $n$ variables $f(x)=0$ on a bounded box $B=[0,1]^n$ such that $f$ can be reasonably represented on a computer - for example, it is a combination of common functions - then you can always disprove the existence of a solution by interval arithmetic, if the solution doesn't exists; however, such algorithm would never terminate if the solution exists. On the other hand, you can prove the existence of a solution if $0\notin f(\partial B)$ and the degree $\deg(f,B)\neq 0$ (the degree can be computed); this is equivalent to nonextendability of $\partial B\to\mathbb{R}^n\setminus\{0\}$ to a continuous map $B\to\mathbb{R}^n\setminus\{0\}$. This is always the case if the solution is robust (i.e. resistant wrt. perturbations of $f$); for this statement and a bit of history of the problem, here is a paper on this topic. This is an analogy to the determinant, because if the determinant is nonzero, the solution is also "stable" wrt. small perturbations of the equations.

An interesting question is if, for a piece-wise linear map $f: K\to\mathbb{R}^n$ on a finite simplicial complex $K$ and a number $\alpha>0$, whether it holds that for each $\alpha$-perturbation $g$ of $f$ (that is, $\|g-f\|\leq \alpha)$, $g(x)=0$ has a solution in $K$. Surprisingly, this is decidable if $\dim K\leq 2n-3$ or $n$ is even, and undecidable for a fixed odd $n$ and arbitrary $K$. You can find the corresponding paper here.

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