Recently I have faced a problem of the following general form:

Find function f such that $f(x, y) = \alpha f(a_1x + b_1y + c_1, a_2x + b_2y + c_2)$ for all $0<x<1$, $0<y<1$, where $a_1$, $a_2$, $b_1$, $b_2$, $c_1$, $c_2$ and $\alpha$ are constants. Note that the value of the constants ensure that $0<a_ix+b_iy+c_i<1$. In other words, $f$ is defined on a $(0, 1) \times (0, 1)$ square.

I searched a bit and it seems that this problem has no well-established method of solving. It would be great if I find some insight here.


1 Answer 1


This is a fine, but difficult problem. The solution depends on the dynamics produced by the iterates of the map $$T:\quad Q\to Q,\qquad(x,y)\mapsto\left\{\eqalign{x'&:=a_1x+b_1y+c_1\cr y'&:=a_2x+b_2y+c_2\cr}\right.\quad.$$ About $T$ we are only told that it maps the open square $Q:=\ ]0,1[\, ^2$ into itself.

There is a large number of morphologically different cases to consider. E.g., $T$ might or might not have an attractive fixed point in $Q$, etcetera.

  • $\begingroup$ Thank you for your prompt reply, Christian. The original problem is a 3D one. However, I found it difficult to solve the 3D problem, so I started with the 2D form of the problem. That is why the question is posed on a square. The original problem is: $2f(x, -x+1, -2x+1) = f(0, -x+1, -x+1) ; 0<x<frac{1}{2}$ To solve it, first I tried to guess the solution! Next, I Taylor expanded the right-hand side around the point $(x, -x+1, -2x+1)$. The resulting equation is a PDE. However, it contains all possible orders of derivatives (up to $n$-th derivative, where $n\rightarrow \infty$). $\endgroup$ Commented Oct 22, 2016 at 15:12
  • $\begingroup$ Truncating the PDE to the terms with first or second derivatives does not work as the residual of the truncated terms is not negligible. So, it seems all orders of derivatives shall be considered; not an easy task. That is why I am looking for other methods. I see your point. As $x=0$ is not in the domain, there is no fixed point. I don’t know if this type of problem has even a specific name in the scientific community. It would be great if you suggest some references or keywords toward the solution of this problem? $\endgroup$ Commented Oct 22, 2016 at 15:13
  • $\begingroup$ The "3D-problem" you sketch in your comment only refers to the values of $f$ on two selected lines, and poses no conditions whatsoever on the values of $f$ in the rest of the envisaged cube. $\endgroup$ Commented Oct 22, 2016 at 15:45
  • $\begingroup$ Generally, the domain of $f$ is the cube. The aforementioned relation between the values of $f$ on the two lines is the only condition that must be met in determining $f$. You raised a good point, indeed, as I am not sure if this condition would result in a unique solution. I'd expect a class of solutions. $\endgroup$ Commented Oct 22, 2016 at 18:29

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