A functional power series equation: $(1-w-zw)F(z,w)+zw^nF\left(z,w^2\right)=z-zw$

Question

I am interested in solving the following functional equation:

$$(1-w-zw)F(z,w)+zw^nF(z,w^2)=z-zw\,.$$

Here $n\geq 2$ is a fixed integer and $F(z,w)$ is a power series in two variables with complex coefficients, which you may assume that converges in a neighborhood of the origin in $\mathbb C^2$.

My thoughts:

Defining $G(j)=F(z,w^j)$ for $j\geq 1$ and using the functional equation we obtain

$$G(j)-\underbrace{\,\frac{-zw^{jn}}{1-w^j-zw^j}\,\\[4mm]}_{a_j}\,G(2j)=\underbrace{\,\frac{z-zw^j}{1-w^j-zw^j}\,\\[4mm]}_{b_j}\,.$$

Dividing both sides by $\prod_{k=0}^\infty a_{2^kj}=c_j$ and defining $H(j)=G(j)/c_j$ we get

$$H(j)-H(2j)=\frac{b_j}{c_j}\,,$$

which implies

$$\begin{align*} F(z,w)=&\,G(1)=c_1H(1)\\ =&\,c_1\sum_{r=0}^\infty\bigl[H(2^r)-H(2^{r+1})\bigr]+c_1\lim_{r\to\infty}H(2^r)\\ =&\,\sum_{r=0}^\infty b_{2^r}\frac{c_1}{c_{2^r}}+\lim_{r\to\infty}\frac{c_1}{c_{2^r}}G(2^r)\\ %=&\,\sum_{r=0}^\infty \frac{z-zw^{2^r}}{1-w^{2^r}-zw^{2^r}}\,\prod_{k=0}^{r-1}\frac{zw^{2^kn}}{1-w^{2^k}-zw^{2^k}}+\lim_{r\to\infty}\frac{c_1}{c_{2^r}}G(2^r)\\ \end{align*}$$

Since $F$ is holomorphic, then for $|w|<1$ we have $\lim\limits_{r\to\infty}G(2^r)=F(z,0)=z$ by the functional equation. Defining

$$d_r=\frac{c_1}{c_{2^r}}=\prod_{k=0}^{r-1}\frac{-zw^{2^kn}}{1-w^{2^k}-zw^{2^k}}$$

we arrive to the following "explicit" formula, provided that the sequence $\boldsymbol{(d_r)_{r\geq1}}$ converges for $\boldsymbol{|z|,|w|}$ small:

$$F(z,w)=\sum_{r=0}^\infty b_{2^r}d_r+z\lim_{r\to\infty}d_r\,.$$

I personally believe that the sequence $(d_r)_{r\geq1}$ indeed converges. Ironically, I put a great effort into the partial solution above, but I am not feeling like solving the system of infinite linear equations satisfied by the coefficients of $F$. Who knows? some seasoned complex analyst can help me and obtain an explicit formula for both the series and the infinite product above.

Answer 1

In this answer, we make some observations that OP might find useful in his search for an explicit solution.

1. Let us rewrite the double power series $$F(z,w)~=~\sum_{j,k=0}^{\infty}a_{jk}z^jw^k ~=~\sum_{k=0}^{\infty}F_k(z)w^k \tag{A}$$ in terms of coefficient functions $F_k(z)$. OP's functional equation leads to a recursion relation for the coefficient functions $$ F_k(z) ~=~(1+z) F_{k-1}(z) - z F_{\frac{k-n}{2}}(z) + z \delta_k^0 - z \delta_k^1, \qquad k~\in~\mathbb{N}_0.\tag{B} $$ [It is implicitly understood in eq. (B) that $F_{\ell}(z)\equiv 0$ if $\ell\notin\mathbb{N}_0$.] Hence the first few coefficient functions read $$ F_0(z)~=~z, \qquad F_1(z)~=~z^2, \qquad F_2(z)~=~z^3 + (1-\delta_n^2)z^2,\qquad \ldots.\tag{C} $$ It is not hard to see that the $k$'th coefficient function $F_k(z)$ is a $k\!+\!1$ order polynomial of the form $$F_k(z)~\stackrel{(B)}{=}~\sum_{j=1}^{k+1}a_{jk}z^j~\stackrel{(B)}{=}~~z^{k+1}+ \text{lower-order terms}.\tag{D}$$

2. It becomes clear that there exists a unique solution to OP's functional equation.

3. It follows that the power series $F(z,w)$ is absolutely convergent for at least $$|z|, |w|~<~\frac{1}{2},\tag{E}$$ thereby agreeing with OP's comment about the existence of an absolutely convergent neighborhood around the origin $(z,w)=(0,0)$. To prove this convergence claim, define recursively non-negative majorant coefficient functions $$ G_0(z)~:=~|z|, \qquad G_1(z)~:=~|z|^2,$$ $$ G_k(z)~:=~(1+|z|) G_{k-1}(z) + |z| G_{\frac{k-n}{2}}(z) , \qquad k~\in~\mathbb{N}\backslash\{0,1\}. \tag{F} $$ and note that by induction $$ |z|~<~\frac{1}{2}\qquad \stackrel{(B)+(F)}{\Rightarrow} \qquad \sum_{j=1}^{k+1} |a_{jk}||z|^j~\leq~|G_k(z)|~<~2^{k-1}. \tag{G}$$

Answer 2

We begin with defining a new function $$ f(z,q,r) = z \sum_{k\ge 0} (zq)^k a_k(q,r)$$ by power series where we define $$ a_0(q,r) = 1, a_{k+1}(q,r) = (a_k(q,r) - q^kra_k(q^2,r^2))/(1-q). $$ It is not obvious from the recursive definition that if $n\ge0$ then $a_k(q,q^n)$ is a polynomial in $q$ with positive integer coefficients. The connection with the original function is $$ F(z,w) = f(z,w,w^{n-1})$$ while the new function satisfies $$ (1-q-zq)f(z,q,r) + zqrf(z,q^2,r^2) = z(1-q). $$ As special cases of $F$ we have $F(0,w) = 0$ and also if $n=1$ or $w=0$ then $F(z,w) = z$ because if $k>0$ then $a_k(w,1) = 0$ or $(zw)^k = 0$ respectively.

Another special case is if $n=2$ and $w=1$, then $F(z,1) = z A(z)$ where $A(z)$ is the generating function of sequence A008934. If we look at the power series coefficients for $n\ge1$ with $w=1$ they form the table in sequence A093729.

Answer 3

Let us write ${\Bbb D}(0,r)=\{z\in {\Bbb C}: |z|<r\}$ for the open disk of radius $r>0$ centered at the origin in the complex plan. I claim that the functional equation defines a meromorphic function in the domain $(z,w)\in {\Bbb C} \times {\Bbb D}(0,1)$.

It is slightly easier to work with a different function than $F$. Note that since $F(0,w)\equiv 0$ if $F$ is analytic in a neighborhood of the origin it has the form $F(z,w)=z H(z,w)$ again with $H$ analytic and verifying $$ H(z,w) = \frac{1-w}{1-w(1+z)} - \frac{z w^n}{1-w(1+z)} H(z,w^2).$$

To find solutions of the latter consider for $0<\delta<1$, $0<R$ the polydisk $\Omega=\Omega_{R,\delta}= {\Bbb D}(0,R) \times {\Bbb D}(0,\delta) \subset {\Bbb C}^2$ and the space $$ A=A_{R,\delta} = C^\omega( \Omega_{R,\delta}) \cap C(\overline{\Omega}_{R,\delta}) $$ of analytic functions on $\Omega$ having a continuous continuation to the boundary. It is a Banach space under the sup norm $\|H\| = \sup_{(z,w)\in \Omega} |H(z,w)|$.

We will look at the map: $$ \Phi(H) (z,w) = \frac{1-w}{1-w(1+z)} - \frac{z w^n}{1-w(1+z)} H(z,w^2), \ \ \ H \in A$$ When $1-\delta(1+R)>0$ then $\Phi$ maps $A$ into $A$ since $(z,w)\in \Omega$ implies $(z,w^2)\in \Omega$ (and the RHS is analytic). If in addition the contraction condition $$ \eta=\frac{R \delta^n}{1 - \delta(1+R)} <1 $$ holds, then $$ \|\Phi (H_1) - \Phi(H_2) \| \leq \eta \|H_1 - H_2\|, \ \ \ H_1,H_2\in A.$$ This shows that $\Phi$ is a Lipschitz contraction, whence by the fixed point theorem has a unique fixed point, again denoted by $H$. As mentioned above $F(z,w)=zH(z,w)$ then solves the posed problem on the domain $\Omega$. Note that given any $0<|z|=R<+\infty$ you may find $\delta=\delta_R>0$ which satifies the contraction condition. Furthermore, when $|w|<1$ then iterating the map $\Phi$, $p$ times we get meromorphic factors times $H(z,w^{2^p})$ and for $p$ large enough we have $|w|^{2^p}< \delta_R$ so that $H(z,w^{2^p})$ is analytic. In other words, $H(z,w)$ is meromorphic (e.g. by viewing it for fixed $w$ as a function of $w$, or the other way around). Singularities appear precisely when $1-w^{2^k}(1+z)=0$, $k\geq 0$. We conclude that the functional equation defines a function which is analytic on the domain: $$ \{ (z,w)\in {\Bbb C}^2: |w|<1 \ \ \mbox{and} \ \ \forall k\geq 0 : w^{2^k}(1+z)\neq 1\}$$

The above map $\Phi$ also gives good power series developments in $\Omega$ (assuming the contraction condition). However, when $|w|=1$, the iteration fails, and I imagine that this constitutes a natural boundary for a meromorphic extension of $H^*$ (not sure about this, and anyway wouldn't know how to prove this off hand). The reason is that if $w=\exp(2\pi i u)$ with $u$ irrational the map $w\mapsto w^2$ becomes a rather erratic (ergodic) map of the circle, which probably induces a rather bad behaviour of any possible extension. This also suggests that there are no explicit solution to the problem, other than developping the infinite product/sum obtained by iteration.