- #1
dmytro
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Quick version:
I have a vector field [itex]f:\mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^n[/itex] of two arguments [itex]x \in \mathbb{R}^n, y \in \mathbb{R}^m[/itex], which has the following properties:
The question is, given the assumptions above, and setting [itex]y=W^Tx[/itex], is there any hope of finding some constrains on the matrix [itex]W \in \mathbb{R}^{n\times m}[/itex], such that the equation [itex]f(x, W^Tx)=0[/itex] has a unique solution?
Some details:
I need this to show that a linear feedback control stabilizes my system, that's where [itex]W^Tx[/itex] 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. [itex]n = m=1[/itex], we have the following:
So, had I asked this question in 1D, the answer would be like "the system has unique solution for all [itex]w[/itex], such that [itex]\gamma w < 0[/itex] 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 [itex]W[/itex], such that the matrix
[tex]
\frac{\partial}{\partial x}[f(x, W^Tx)] = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}W^T
[/tex]
has only negative eigenvalues, for all [itex]x[/itex]. The equivalence can be shown using the dynamical systems theory. So an answer to this question is as welcome as to the original one
I have a vector field [itex]f:\mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^n[/itex] of two arguments [itex]x \in \mathbb{R}^n, y \in \mathbb{R}^m[/itex], which has the following properties:
- The jacobian matrix of [itex]f[/itex] wrt to the first argument [itex]\frac{\partial f}{\partial x}: \mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^{n\times n}[/itex] 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 [itex]f_i[/itex] is strictly decreasing wrt to [itex]x_i[/itex] for [itex]i = 1 \dots n[/itex]
- (optional, if it helps, but I should probably relax it later to monotonous w.r.t to y) [itex]f[/itex] is linear in the second argument, i.e. [itex]\frac{\partial f}{\partial y} = \text{const}[/itex] is a constant [itex]n\times m[/itex] matrix. Or, even, [itex]f(x,y)=g(x)+Gy[/itex], where [itex]G[/itex] is some (known) matrix
- [itex]f[/itex] is sufficiently differentiable, nice and all
The question is, given the assumptions above, and setting [itex]y=W^Tx[/itex], is there any hope of finding some constrains on the matrix [itex]W \in \mathbb{R}^{n\times m}[/itex], such that the equation [itex]f(x, W^Tx)=0[/itex] has a unique solution?
Some details:
I need this to show that a linear feedback control stabilizes my system, that's where [itex]W^Tx[/itex] 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. [itex]n = m=1[/itex], we have the following:
- Let be of the form [itex]f(x, y) = g(x) + \gamma y[/itex]
- then [itex]f(x, wx) = 0 \implies \gamma wx = -g(x)[/itex]
- g(x) is strictly decreasing (according to 1.), then a sufficient condition for [itex]g(x) + \gamma w x = 0[/itex] to have a unique root is that [itex]\gamma w x[/itex] is strictly decreasing, i.e. [itex]\gamma w < 0[/itex]
So, had I asked this question in 1D, the answer would be like "the system has unique solution for all [itex]w[/itex], such that [itex]\gamma w < 0[/itex] 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 [itex]W[/itex], such that the matrix
[tex]
\frac{\partial}{\partial x}[f(x, W^Tx)] = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}W^T
[/tex]
has only negative eigenvalues, for all [itex]x[/itex]. 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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