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How to solve the q-shift difference equation $W(x)=(x-a)W(xq^{-2})$? Here $W(x)$ is the unknown function. Thank you very much.

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Have you tried using that equation like a fixed-point equation, iterating and trying to see if the process converges? –  Mariano Suárez-Alvarez Jan 17 '12 at 19:21
@Mariano, the process will run forever. –  LJR Jan 17 '12 at 19:27
That is exactly why I suggested to see of the process converges. I very much doubt there is a solution to the equation in finite terms! –  Mariano Suárez-Alvarez Jan 17 '12 at 19:30
Do ypu understand what 'see if the process converges' means? –  Mariano Suárez-Alvarez Jan 17 '12 at 19:35
No, that's not what I mean. You should review —in a calculus book— the properties of infinite products (it usually comes right after those of infinite series). The little book by Knopp titled Theory and application of infinite series is a great source. –  Mariano Suárez-Alvarez Jan 17 '12 at 19:48

1 Answer 1

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Assume $a\neq 0$ and $q>1$. We can define $G(u) = a^{-u} f(u) W(q^{2u})$, where $f(u)$ is a function satisfying $f(u+1)=-f(u)$. Then $$ \begin{eqnarray} G(u) &=& a^{-u}f(u)W(q^{2u}) \\ &=& a^{-u}f(u)(q^{2u} - a)W(q^{2u}q^{-2}) \\ &=& a^{-(u-1)}f(u-1)(1-a^{-1}q^{2u}) W(q^{2(u-1)}) \\ &=& (1-a^{-1}q^{2u})G(u-1). \end{eqnarray} $$ This gives $G(u)$ as a convergent infinite product times a scale factor (equal to $\lim_{u\rightarrow-\infty}G(u)$), which we will take to be $1$: $$ G(u) = (1-a^{-1}q^{2u})(1-a^{-1}q^{2(u-1)})\cdots = (a^{-1}q^{2u};q^{-2})_{\infty}, $$ where $(\alpha;\beta)_\infty$ is the q-Pochhammer symbol. Now, taking $x=q^{2u}$, we can invert this to obtain $$ \begin{eqnarray} W(x) = \frac{a^{u}}{f(u)}\left(\frac{x}{a};q^{-2}\right)_{\infty}=e^{(i\pi+\ln a)u}\left(\frac{x}{a};q^{-2}\right)_{\infty}, \end{eqnarray} $$ choosing $f(u)=e^{-i\pi u}$, or $$ W(x) = \exp\left({\frac{i\pi + \ln a}{2 \ln q}\cdot\ln x}\right)\left(\frac{x}{a};q^{-2}\right)_{\infty}. $$ This is a particular solution, given by a particular choice of $f$. To obtain the general solution, multiply $W(x)$ by any function of $u=\ln x / (2 \ln q)$ that is periodic with period $1$.

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thank you very much. Why do you know how to define $G(u)$ to solve this problem? –  LJR Jan 18 '12 at 16:37
The change of variable $x=q^{2u}$ turns the multiplicative shift into an additive shift. You then have an infinite-looking product, but its terms approach $(-a)$ instead of $1$ (which they need to approach for the product to converge). Factoring out the $(-a)$, in the form of $a^{-u}f(u)$, leaves the remaining product convergent. –  mjqxxxx Jan 18 '12 at 16:54

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