Consider a following recursion relation: \begin{equation} a_n = 1 + \sum\limits_{i=1}^{n-1} f(n-i) \cdot a_i \end{equation} subject to $a_1=1$. This is a generalization of Solving recurrence relation $a_n=1 + \sum\limits_{i=1}^{n-1}ia_{n-i}$ with $a_1=1$ . We call the function $f()$ a kernel function. Since this recursion contains a convolution it can be solved by Z transforms. Define $A(z) := \sum\limits_{n=1}^\infty a_n z^n$ and $F(z) := \sum\limits_{n=1}^\infty f(n) z^n$. Multiplying both sides by $z^n$ and summing over $n=1,2,\cdots,\infty$ and then solving for the Z transform of the sequence in question yields: \begin{equation} A(z) = \frac{z}{1-z} \cdot \frac{1}{1-F(z)} \end{equation} Now consider some particular form of the kernel function, i.e. $f(n-i) := \alpha^{n-i}-1$ where $\alpha$ is termed the memory parameter and $\alpha \in (0,1)$. Then $F(z) = z (\alpha-1)/((1-\alpha z)(1-z))$ and: \begin{equation} A(z) = \frac{z (1- \alpha z)}{\alpha z^2 - 2 \alpha z +1} = -1+ \frac{2}{\alpha} Re[\frac{1}{z-z_-} \cdot \frac{z_-(1-2 \alpha) + 1}{-z_++z_-}] \end{equation} where $z_\pm = 1/\sqrt{\alpha} \exp(\pm \imath \phi)$ where $\phi:= \arccos(\sqrt{\alpha})$. Nowthat we have decomposed the Z transform into simple fractions the inversion is straightforward and reads: \begin{equation} a_n = -\delta_{n,0} + (\sqrt{\alpha})^{n-1} \cdot \cos\left( \phi \cdot (n+1) + \theta\right) \end{equation} where $\cos(\theta) = -1+2 \alpha$ and $\sin(\theta) = \sqrt{\alpha}(2 \alpha-2)/\sqrt{1-\alpha}$. Therefore we see that the sequence $\left\{ a_n \right\}_{n=1}^\infty$ is a boring phase-shifted waveform with a constant frequency $\phi$ and an exponentially decaying amplitude. This is not a surprise because the memory parameter (which determines the frequency) does not depend on time.
Having said all this my question is the following . What happens if we allow the memory parameter to depend on time? To be precise let us assume that: \begin{equation} a_n = 1 + \sum\limits_{i=1}^{n-1} f(n-i,i) \cdot a_i \end{equation} where $a_1=1$ and \begin{equation} f(n-i,i) := \left( \alpha - i \alpha_1 \right)^{n-i} -1 \end{equation} where $0 < \alpha_1 < \alpha/n$ and $\alpha < 1$. Clearly now the memory parameter decreases with time and therefore (from the definition of $\phi$) the frequency increases with time and as such we should be expecting an interesting waveform with increasing frequency. The way the amplitude depends on time should also change in this case. My question is now how do we solve this model? Clearly the recursion in question is not a convolution anymore so Z-transform methods do not work (or do they??).
This is not a whole solution to this problem but at least a step in the right direction.Since the recursion relation is not a convolution anymore the corresponding Z-transform does not satisfy an algebraic equation but instead it satisfies a certain differential equation. Multiplying both sides of the recurrence by $z^n$ and summing over $n\ge1$ we establish, in a pretty straightforward manner, that the following identity holds: \begin{equation} A(z) = \left( \frac{z (1- \alpha z)}{1-2 \alpha z + \alpha z^2} \right) + \sum\limits_{p=1}^\infty (-\alpha_1)^p \frac{z^p}{(1-\alpha z)^{p}} \frac{(1-z)}{(1-2 \alpha z + \alpha z^2)} \cdot {\mathcal D}_z^{p} A(z) \end{equation} subject to $A(0)=0$, $A^{'}(0)=1$, $A^{''}(0)=\alpha-\alpha_1$ and so on. Here ${\mathcal D}_z := z d/d z$.
Since the parameter $\alpha_1$ is "small" we seek a solution in terms of a series expansion in this parameter. In other words we have: \begin{equation} A(z) = A_0(z) + \sum\limits_{q=1}^\infty \alpha_1^q A_q(z) \end{equation} We term $A_0(z)$ the basic solution and the functions $A_q(z)$ higher order corrections.
Inserting the above ansatz into the differential equation and comparing coefficients at consecutive powers of $\alpha_1$ on both sides gives the following recurrence relations for the higher order corrections. \begin{eqnarray} A_0(z) &=& \frac{z (1- \alpha z)}{1-2 \alpha z + \alpha z^2} \\ A_n(z) &=& \left( \frac{1-z}{1-2 \alpha z+\alpha z^2} \right)\cdot\sum\limits_{p=1}^n \left( \frac{-z}{1-\alpha z} \right)^p {\mathcal D}_z^p A_{n-p}(z) \end{eqnarray} for $n \ge 1$. Using symbolic calculation software like Mathematica we can easily generate a couple of functions above and we easily recognize the pattern: \begin{equation} A_n(z) = \frac{(1-z) z^{n+1} \cdot P_n^{(5n-3)}(z)}{(1- \alpha z)^{2 n-1} (1-2 \alpha z + \alpha z^2)^{2 n+1}} \end{equation} for $n\ge 1$. Here $P_n^{(5n-3)}(z)$ is some polynomial in $z$ of order $5n-3$. This means that the quantities in question are rational functions with the degree of the numerator always being strictly smaller than that of the denominator. This allows for a following simple fraction decomposition: \begin{equation} A_n(z) = \sum\limits_{j=1}^{2 n-1} \frac{{\mathfrak A}_j^{(n)}}{(1- \alpha z)^j} + \sum\limits_{j=1}^{2 n+1} \left( \frac{{\mathfrak B}_j^{(n)}}{(z - \frac{e^{\imath \phi}}{\sqrt{\alpha}})^j} + c.c \right) \end{equation} where the coefficients $\left\{ {\mathfrak A}_j^{(n)} \right\}_{j=1}^{2 n-1}$ and $\left\{ {\mathfrak B}_j^{(n)} \right\}_{j=1}^{2 n+1}$ depend on $\alpha$ only. We will specify the recursion relations being satisfied by those coefficients and hopefully solve the relations as well. Work on this is currently underway.