Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unique solution to an arbitrary monotonic non-linear system

  1. Aug 5, 2014 #1
    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:

    1. 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]
    2. (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
    3. [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

    \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 [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
    Last edited: Aug 5, 2014
  2. jcsd
  3. Aug 6, 2014 #2

    [itex]f:\mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^n[/itex] should be [itex]f:\mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}^n[/itex]

    [itex]\frac{\partial f}{\partial x}: \mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^{n\times n}[/itex] should be [itex]\frac{\partial f}{\partial x}: \mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}^{n\times n}[/itex]
    Last edited: Aug 6, 2014
  4. Aug 8, 2014 #3

    Stephen Tashi

    User Avatar
    Science Advisor

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


    Correction: Let [itex] T [/itex] to be the mapping of [itex] \mathbb{R}^n [/itex] into itself defined by [itex] T(x) = f(x, W^T x) + x [/itex]. so the fixed point will be relevant to the solution of your equation.
    Last edited: Aug 8, 2014
  5. Aug 11, 2014 #4
    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 [itex]W[/itex]. If one picks [itex]W^T[/itex] with only first non-zero column, then the product [itex]\Gamma = \frac{\partial f}{\partial y}W^T[/itex] 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 [itex]f[/itex] is a stable triangular matrix, one should only require that the [itex](1, 1)[/itex] elememnt of [itex]\Gamma[/itex] is always negative, that is:
    [tex]sign(W_{1i}) = -sign(\frac{\partial}{\partial y_i}f_1(x, y)), \forall i, x, y[/tex]
    This condition is easy to fulfill, if one assumes that [itex]f_1[/itex] is monotonic in the second argument.

    For the full-rank [itex]W[/itex], the question is still open, and that's exactly the case I need :(
    Last edited: Aug 11, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Unique solution to an arbitrary monotonic non-linear system