I'm interested in finding functions $f: U \to \mathbb{R}$ satisfying $$f(x+1, y) = af(x,y), $$ $$f(x, y+1) = bf(x,y), $$ where $U \subset \mathbb{R}^2$ and $a,b \in \mathbb{R}$ are arbitrary real numbers. The function doesn't have to be globally defined (it would be more interesting to me if it isn't). This is obviously trivial when $a = b = 1$, since these are just periodic functions. In the generic case it seems to me that the solutions should be some kind of generalised spirals, I'm not sure how else to describe them.

I've tried the 1D case $f(x+1) = af(x)$, writing $f(x) = \sum_i c_i x^i$ and assuming $a \neq 1$ gives us equations for the coefficients: $$ (a-1)c_0 = c_1 + c_2 + \dots$$ $$ (a-1)c_1 = 2c_2 + 3c_3 + \dots$$ with the generic term, $$(a-1) c_n = \sum_{i > n} {{i}\choose{i-n}} c_i. $$ This series seems vaguely familiar but I haven't been able to find it or solve it.

My other thought was to try writing $z = x+iy$ and find functions $f : U \to \mathbb{R}$ with $$f(z+1) = af(z), $$ $$f(z+i) = bf(z), $$ with $U \subset \mathbb{C}$.

I imagine that if such functions exist they already have a name, but I obviously haven't been searching for the right thing.

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    $\begingroup$ If $a,b>0$, the function $f(x,y)=K\cdot a^xb^y$ satisfies your criterion $\endgroup$ – MPW Mar 1 '18 at 11:24
  • $\begingroup$ Thanks. I can't believe I didn't see the obvious solution! I'm still interested in whether these are the only solutions or not. $\endgroup$ – Joe Mar 1 '18 at 11:27
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    $\begingroup$ $g(x,y)\cdot a^xb^y$, where $g(x,y)$ is an arbitrary periodic function, will do as well. $\endgroup$ – Ivan Neretin Mar 1 '18 at 11:54

Unfortunately, I doubt these functions have a name.

As pointed out by Ivan Neretin in a comment, functions of the form $g(x,y)\cdot a^xb^y$ would work where $g$ is periodic in both directions. Conversely, assume $f(x,y)$ works and define $g(x,y)=\dfrac{f(x,y)}{a^xb^y}$. Then substituting into the equations (and assuming $ab\ne0$) reveals that $g(x,y)$ must be periodic in both directions.

At the end of the OP, it's mentioned that we could interpret these functions as functions on the complex plane, which could be good for $g$, but not as much for $f$ since we'd have $f(z)=a^{\mathrm{Re}\,z}b^{\mathrm{Im}\,z}g(z)$ which isn't really the nicest sort of function on the complex plane.

For functions like $g$ which are periodic in two directions on the complex plane, they are called doubly-periodic functions, and a very famous class of examples are Weierstrass's elliptic functions $\wp$, which are related to elliptic curves. For this special case where the periods are $1$ and $i$, we can write $\wp(z)=\dfrac1{z^2}+{\displaystyle\sum_{(m,n)\ne(0,0)}}\dfrac{1}{(z+m+ni)^2}-\dfrac{1}{(m+ni)^2}$.

  • $\begingroup$ Thanks, this really helps. $\endgroup$ – Joe Mar 1 '18 at 14:11

Let $a$, $b\in{\mathbb C}^*$. Then there are $p$, $q\in{\mathbb C}$ with $a=e^p$, $b=e^q$. Consider the function $$g(x,y):=f(x,y)\exp(-px-qy)\ .$$ It is then easy to see that $$g(x+1,y)=g(x,y+1)=g(x,y)\qquad\forall x,\>\forall y\in{\rm dom}(g)\ ,$$ hence $g$ is doubly periodic. It follows that $f$ can be written in the form

$$f(x,y)=g(x,y)\exp(px+qy)$$ with a doubly periodic function $g$, and $g$ may be chosen arbitrarily.


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