Solving the difference equation $h(\vec x)\cdot A(\vec x)=\sum_{i=1}^m A(\vec x - \vec e_i)$

Question

I am trying to solve the following difference equation:

$$h(x_1,x_2,x_3,x_4) \cdot A(x_1,x_2,x_3,x_4)=A(x_1-1,x_2,x_3,x_4)+A(x_1,x_2-1,x_3,x_4)+A(x_1,x_2,x_3-1,x_4)+A(x_1,x_2,x_3,x_4-1)$$

with bounday constraints: $A(0,0,0,0)=1$ and $A(0,*,*,*)=\dots =A(*,\dots ,*,0)=0$ otherwise

Or more general: $$h(\vec x)\cdot A(\vec x)=\sum_{i=1}^m A(\vec x - \vec e_i)$$

with the same boundary constraints. ( $A(\vec 0) = 1$, $A(0,*)=\dots = A(*,0)=0$

I would like to know:

1) Is there a closed form for this recurrence?

2) Is it possible to find the generating function?

3) Is this the discretization of any well known PDE?

Answer 1

The equation $$h(\vec x)\cdot A(\vec x)=\sum_{i=1}^m A(\vec x - \vec e_i)$$ represents a general translation on the function $A(\vec x)$ and can be stated introducing the operator $$h(\vec x)=\exp\left(-\sum_{i=1}^m{\vec e_i}\partial_i\right)$$ so that $$\sum_{i=1}^m A(\vec x - \vec e_i)=\exp\left(-\sum_{i=1}^m{\vec e_i}\partial_i\right)A(\vec x).$$

You can always introduce a set of eigenvalues of the operators $-i\partial_i$ that we can call $p_i$. You have a bounded set and so what you will get is discrete set of eigenvalues $\{\vec p_n,n\in\mathbb{Z}\}$ and eigenfunctions $\{\phi_n^{(i)}(x_i),n\in\mathbb{Z}\land i\in[1\ldots m]\}$ for $\vec p$. Assuming you can expand $A(\vec x)$ using these eigenfunctions, you will get $$h(\vec x)\cdot A(\vec x)=\sum_{n\in\mathbb{Z}}e^{-i\sum_{l=1}^m{\vec p}_n\cdot ({\vec x}-{\vec e}_i) }\tilde A_n.$$ In this way, your problem is simply reduced to solving a set of algebraic equations involving the eigenvalues $\vec p_n$.

Your equation states that a sum of translations on $A(\vec x)$ reduces to a multiplicative factor. I do not know if a general solution exists to this kind of problem.