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I would like to solve the following algebraic linear q-difference equation:

\begin{equation} a\left(x\right)f\left(x\right)=f\left(qx\right) \end{equation}

The parameter $q$ is real, positive and smaller than $1$, while the functions $a\left(x\right)$ and $f\left(x\right)$ are $\mathbb{R}\rightarrow\mathbb{R}$. Moreover $a\left(x\right)$ is known, continuous and such that $a\left(0\right)=0$ (this is an important constraint), and I would like to calculate $f\left(x\right)$. So for example we can consider the functional equation $xf\left(x\right)=f\left(qx\right)$.

Do you know how to calculate $f\left(x\right)$?

Thanks in advance for your help!

EDIT:

Please observe that $f\left(0\right)=0$ and that the solution for $x<0$ doesn't depend on the solution for $x>0$, so for example we can have $f\left(x\right)=0$ only for $x\leq0$ and not for $x>0$, or viceversa.

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Putting for simplicity $x=e^t, q=e^{-b}, F(t)=f(e^t), A(t)=a(e^t)$, we obtain $$A(t)F(t)=F(t-b).$$ Now you can assign $F$ arbitrarily on any interval of length $b$, for example on $(0,b)$, and this formula defines you a solution everywhere left of this interval. If you want a continuous function, you want $A(b)F(b)=F(0)$, otherwise $F$ is arbitrary on $(0,b)$. Similarly, if you want it smooth etc. If $A(t)\neq 0$ for all $t$, you can also extend your solution to the right.

The answer on further questions depends on what is exactly known about $a$, and what properties you want $f$ to have.

Edit. For example, if $b=1$, $a(x)=x$, we obtain $$e^tF(t)=F(t-1).$$ Taking logs, $$t+\phi(t)=\phi(t-1),\quad\phi(t)=\log F(t).$$ Differentiating twice gives that $\phi^{\prime\prime}$ is an (arbitrary) periodic function. The simplest solution is obtained when $\phi^{\prime\prime}={\mathrm{const}}$. Then $\phi(t)=-(1/2)(t^2+t)$, and the solution of your original equation is $$f(x)=\exp\left(-(1/2)(\log^2x+\log x)\right),$$ which satisfies $f(x/e)=xf(x)$.

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  • $\begingroup$ I think it should be added at this link. $\endgroup$ – user2983638 Jan 10 '14 at 23:13
  • $\begingroup$ user2983638: OK, go ahead:-) Functional equations is in fact an enormous subject, and one cannot cover the whole of it in a short article. $\endgroup$ – Alexandre Eremenko Jan 11 '14 at 16:20

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