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I have a map $f:\mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}^n$ of two arguments $x, y$, which has a following properties:

  1. The jacobian matrix of $f$ wrt to the first argument $\frac{\partial f}{\partial x}: \mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}^{n\times n}$ is a lower triangular matrix with negative elements on the main diagonal, for all input argument values. This assumption is equivalent to the statement that each $f_i$ is strictly decreasing wrt to $x_i$ for $i = 1 \dots n$
  2. (optional, if it helps) $f$ is linear in the second argument, i.e. $\frac{\partial f}{\partial y} = \text{const}$ is a constant $n\times m$ matrix
  3. $f$ is sufficiently differentiable, nice and all

The question is, given the assumptions above, is there any hope of finding some constrains on a matrix $W \in \mathbb{R}^{n\times m}$, such that the equation $f(x, W^Tx)=0$ has a unique solution?


Comment 1

This question arose as a generalization to a 1d case, which has a nice solution, as described below. I don't know where to look for tools to study this generalized problem.

In 1D case, i.e. $n = m=1$, we have the following:

  • Let be of the form $f(x, y) = g(x) + \gamma y$
  • then $f(x, wx) = 0 \implies \gamma wx = -g(x)$
  • g(x) is strictly decreasing (according to 1.), then a sufficient condition for $g(x) + \gamma w x = 0$ to have a unique root is that $\gamma w x$ is strictly decreasing, i.e. $\gamma w < 0$.

So, had I asked this question in 1D, the answer would be like "choose any $w$ such that $\gamma w < 0$ holds". I was hoping to get smth like that for the general case, but with no luck so far. I'd also be grateful is someone points me to the suitable mathematical apparatus to figure it out.

Comment 2

I'm actually interested in a global stable solutions, so that the original question can be equivalently restated as: find constrains on $W$, such that the matrix

$$ \frac{\partial}{\partial x}[f(x, W^Tx)] = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}W^T $$

has only negative eigenvalues, for all $x$. The equivalence can be shown using the dynamical systems theory. So an answer to this question is as welcome as to the original one

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This is not supposed to be an answer but it might provide some light on it, and it was too long for a comment.

Take the easiest case, that is $f$ is linear, so we write $f(x,y)=Ax+By$ with $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times m}$. As per condition one, $A$ is the jacobian of $g$, so it has the mentioned properties.

Next $f(x,W^Tx)=Ax+BW^Tx$. Clearly $BW^T\in\mathbb{R}^{n\times n}$. So $f(x,W^Tx)=0$ has a unique solution if and only if $(A+BW^T)$ is invertible. For a discussion of such a problem see here.


As for the nonlinear case, let's see the one dimensional problem. Let $g(x)=-ax+bx^2$, $a>0$, and $f(x,y)=-ax+bx^2+cy$. Then $f(x,wx)=-ax+bx^2+cwx=x(-a+cw+bx)$. Thus $x=0$ is naturally a solution, if you want it to be unique you require $-a+cw=0$ (in contrast to the linear case!)

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