I am trying to prove stability and get a non-asymptotic upper bound on the convergence rate of a nonlinear discrete-time dynamical system, whose dynamics are stated in terms of the (non-elementary) Lambert $W$ function.
Let $\psi: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be defined by $$ \psi(b, x) = xW(\exp(b + 1/x + \log(1/x))), $$ where $W$ denotes the principal branch of the Lambert $W$ function, i.e. $W(x) \exp(W(x)) = x$ for $x\geq 0$.
Further let $c>0$ and $\mathbf{g} \in \mathbb{R}^M$ with $g_i \in (0, 1]$, and define $\mathbf{r}: \mathbb{R}^M \rightarrow \mathbb{R}^M$ as $$ r_i(\mathbf{x}) = \frac{1}{g_i} \log(\psi(c g_i^2, x_i)). $$ Finally, let $\mathbf{a} = (a_1, \ldots, a_M) \in \mathbb{R}^M$ with $a_i \geq 0$, and let $\mathbf{w}_1, \ldots, \mathbf{w}_M$ be unit-norm vectors in $\mathbb{R}^d$ satisfying $ \left\langle a_1\mathbf{w}_1 + \cdots + a_M\mathbf{w}_M, \mathbf{w}_i \right\rangle > 0$ for all $i$.
Now, define a sequene in $\mathbb{R}^M$ by $$\begin{aligned} \mathbf{b}_0 &= \mathbf{1}, \\ \mathbf{b}_{t+1} &= \mathbf{b}_t \odot \exp \left( -\frac{1}{M} \mathbf{g} \odot (G \cdot \mathbf{r}(\mathbf{b}_t)) \right), \end{aligned}$$ where $\odot$ denotes the element-wise product, and $G \in \mathbb{R}^{M \times M}$ is the Gram matrix of the $\mathbf{w}_1, \dots, \mathbf{w}_M$.
Question: Does $\mathbf{b}$ go to zero? If so, can we upper bound $\|\mathbf{b}_t\|_1$ in terms of $t$ and the parameters $M, c, \mathbf{g}, G$?
Some immediate observations about the system:
- The projection $\psi(b, \cdot)$ is positive, increasing, concave, and $\psi(b, x) \rightarrow 1$ as $x \rightarrow 0^+$. Consequently, $r_m$ is positive, increasing, concave (in terms of $x_m$) and $r_m(\mathbf{x}) \rightarrow 0$ as $x_m \rightarrow 0^+$.
- Our sequence $\mathbf{b}_t$ stays inside the positive orthant.
- The only stationary point in the positive orthant is $\mathbf{0}$. To see this, let $\mathbf{w}_* = \sum_i a_i\mathbf{w}_i$ and note that any stationary point $\mathbf{x}$ needs to satisfy $G \cdot \mathbf{r}(\mathbf{x}) = 0$. Thus we have $$\begin{aligned} \langle G \cdot \mathbf{r}(\mathbf{x}), \mathbf{a} \rangle &= \sum_{i=1}^M (G \cdot \mathbf{r}(\mathbf{x}))_i a_i \\ &= \sum_{i=1}^M \left( \sum_{j=1}^N G_{ij} r_j(\mathbf{x}) \right) a_i \\ &= \sum_{i=1}^M \sum_{j=1}^N \langle \mathbf{w}_i, \mathbf{w}_j \rangle r_j(\mathbf{x}) a_i \\ &= \sum_{j=1}^M \left\langle \sum_{i=1}^M a_i \mathbf{w}_i, \mathbf{w}_j \right\rangle r_j(\mathbf{x}) \\ &= \sum_{j=1}^M \left\langle \mathbf{w}_*, \mathbf{w}_j \right\rangle r_j(\mathbf{x}) \\ &> 0, \end{aligned}$$ where the last line uses the fact that $\langle \mathbf{w}_*, \mathbf{w}_j \rangle > 0$ for all $j$ and $r_j(\mathbf{x}) > 0$ whenever $\mathbf{x} > 0$.
In the special case that $M=2$, the function $L(\mathbf{x}) = \max(r_1(\mathbf{x}), r_2(\mathbf{x}))$ is a Lyapunov function for $\mathbf{b}_t$. In this case, I can show that $L(\mathbf{b}_t) \leq \mathcal{O}(1/t)$ and consequently that $\|\mathbf{b}_t\| \leq \mathcal{O}(1/t)$, where $\mathcal{O}$ omits dependence on $M, c, \mathbf{g}$, and $\mathbf{G}$. The proof is too long to include here, but I want to mention it because it seems promising to try to generalize this Lyapunov function when $M \geq 2$.
