Unique solution to an arbitrary monotonic non-linear system

[dmytro]

Quick version:

I have a vector field $f:\mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^n$ of two arguments $x \in \mathbb{R}^n, y \in \mathbb{R}^m$, which has the following properties:

1. The jacobian matrix of $f$ wrt to the first argument $\frac{\partial f}{\partial x}: \mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^{n\times n}$ 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, but I should probably relax it later to monotonous w.r.t to y) $f$ is linear in the second argument, i.e. $\frac{\partial f}{\partial y} = \text{const}$ is a constant $n\times m$ matrix. Or, even, $f(x,y)=g(x)+Gy$, where $G$ is some (known) matrix
3. $f$ is sufficiently differentiable, nice and all

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

Some details:

I need this to show that a linear feedback control stabilizes my system, that's where $W^Tx$ comes from. This question arose as a generalization to a 1d case, which has a nice solution, as described below. However, I don't know which tools can I use to study the 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 "the system has unique solution for all $w$, 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.

One more thing: for stable fixed points, 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

 

[dmytro]

edit:

$f:\mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^n$ should be $f:\mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}^n$

$\frac{\partial f}{\partial x}: \mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^{n\times n}$ should be $\frac{\partial f}{\partial x}: \mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}^{n\times n}$

 

[Stephen Tashi]

One thought is to see if the Banach fixed point theorem helps.. Let $T$ to be the mapping of $\mathbb{R}^n$ into itself defined by $T(x) = f(x, W^T x)$.

http://en.wikipedia.org/wiki/Banach_fixed-point_theorem

Correction: Let $T$ to be the mapping of $\mathbb{R}^n$ into itself defined by $T(x) = f(x, W^T x) + x$. so the fixed point will be relevant to the solution of your equation.

 

[dmytro]

One thought is to see if the Banach fixed point theorem helps

Thanks for the clues. I can't seem to figure out how to make use of that theorem though...

However, I found a solution for a special case of $W$. If one picks $W^T$ with only first non-zero column, then the product $\Gamma = \frac{\partial f}{\partial y}W^T$ also has only first column not equal to zero, that is, a special case of a triangular matrix. Therefore, to make sure that the full derivative of $f$ is a stable triangular matrix, one should only require that the $(1, 1)$ elememnt of $\Gamma$ is always negative, that is:
$$sign(W_{1i}) = -sign(\frac{\partial}{\partial y_i}f_1(x, y)), \forall i, x, y$$
This condition is easy to fulfill, if one assumes that $f_1$ is monotonic in the second argument.

For the full-rank $W$, the question is still open, and that's exactly the case I need :(

 

[blue_raver22]

I find this problem to be quite interesting and challenging. The fact that the jacobian matrix of f with respect to x is lower triangular with negative elements on the main diagonal suggests that the system is strictly decreasing in each component of x. This is a strong assumption and gives us a lot of information about the behavior of the system.

Additionally, the linearity of f with respect to y is also a useful property that can potentially simplify the problem. However, the question of finding constraints on the matrix W to ensure a unique solution to the equation f(x, W^Tx)=0 is not an easy one. It requires a deep understanding of the system and its dynamics.

One approach to solving this problem could be to use techniques from nonlinear control theory, such as Lyapunov stability analysis, to analyze the stability of the system. This could potentially lead to finding constraints on W that ensure a unique solution.

Another approach could be to use numerical methods to solve the equation f(x, W^Tx)=0 for different values of W and see if there are any patterns or relationships between the solutions and the matrix W. This could give us some insight into the problem and potentially lead to finding constraints on W.

In conclusion, this is a complex and interesting problem that requires a combination of theoretical and numerical approaches to solve. I would recommend consulting with experts in nonlinear control theory and dynamical systems to get more insights and possible solutions to this problem.