I need to evaluate the limit of $f(x)$ as $x\to0$, where the function $f$ solves the following equation: $$ f(x)=\left\{ \begin{array}{ll} g(x) & \text{if } x\geq \frac{1}{2};\\ \frac{1}{2} f(\alpha x) + \frac{1}{4} (1-x) f\left(\frac{\alpha x}{1-x}\right) + \frac{1}{4} (1+x) f\left(\frac{\alpha x}{1+x}\right) & \text{otherwise } (0<x<\frac{1}{2}), \end{array} \right. $$

where $\alpha\in (1,\frac{3}{2})$ is a given constant, and the function $g$ is known over the interval $[\frac{1}{2},\alpha]$.

I wonder if there is a nice numerical trick to compute $\lim_{x\to 0} f(x)$ ? Or maybe even a formula ?

I tried to solve the recurrence equation by discretizing $f$ over $[0,\frac{1}{2}]$ (which reduces to solving a linear system of size $1/2\epsilon$ for a discretization size of $\epsilon$ ), but the solution looks "fractal-like", so I don't know if my computation can be trusted, and if there are not numerical issues around $x=0$...

  • $\begingroup$ just an observation: taking derivatives and then sending $x\rightarrow 0$ I find $f^{(n)}(0)=\alpha^n f^{(n)}(0)$, hence $f^{(n)}(0)=0$ for any $n\geq 1$. $\endgroup$ Aug 14 '19 at 19:55
  • $\begingroup$ Since g is known on [1/2,α], we can determine f(α/3) putting x=1/2 in the functional equation. But then, which other values can be found? Putting x=α/3 again we may find f(α/(3+α)), unless α<√(3/2), which is not excluded. In any case, I don't see how f(x) can possibly be determined for x< α-1... $\endgroup$ Aug 14 '19 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.