Difference equation satisfied by discrete harmonic functions on square lattice

A function $$f:\mathbb Z^2 \rightarrow \mathbb R$$ is said to be discrete harmonic if it satisfies the discrete Laplacian equation $$\Delta f(m,n) = f(m+1,n)+f(m-1,n)+f(m,n+1)+f(m,n-1)-4f(m,n) = 0~.$$ Let $$\mathcal F$$ be the set of all discrete harmonic functions on the square lattice.

Question: Is there another difference equation with integer (equiv. rational) coefficients that is satisfied by all discrete harmonic functions?

If $$D$$ is a difference operator with integer coefficients, then any $$f\in \mathcal F$$ trivially satisfies $$(D\circ\Delta)f(m,n)=0~.$$ My question is if there is a difference equation with integer coefficients satisfied by all $$f\in\mathcal F$$ that cannot be written as $$D\circ \Delta$$ for any $$D$$.

I can rephrase the above question as follows. Any function $$f:\mathbb Z^2\rightarrow\mathbb R$$ can be uniquely represented as $$\hat f(x,y) = \sum_{m,n\in\mathbb Z} f(m,n) x^m y^n~.$$ The discrete Laplacian operator $$\Delta$$ can be associated with the Laurent polynomial $$p(x,y) = x + x^{-1} + y + y^{-1} - 4 \in \mathbb Z[x,x^{-1},y,y^{-1}]~.$$ Then, any $$f\in\mathcal F$$ satisfies $$p(x,y) \hat f(x,y) = 0~.$$ My question can be rephrased as follows:

Is there a $$q(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$$ that is not a multiple of $$p(x,y)$$, and $$q(x,y)\hat f(x,y)=0~,$$ for all $$f\in\mathcal F$$?

Here, by a multiple of $$p(x,y)$$, I mean something of the form $$r(x,y)p(x,y)$$ where $$r(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$$.

$$\DeclareMathOperator\u{\mathbf u}\DeclareMathOperator\e{\mathbf e}$$I think I know how to answer my question. First, note that $$p(x,y) = x^{-1}y^{-1} \tilde p(x,y)$$, where $$\tilde p(x,y) = (x-1)^2y+x(y-1)^2 \in \mathbb Z[x,y]$$. More generally, any $$q(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$$ can be written as $$q(x,y) = x^a y^b \tilde q(x,y)$$ for some $$a,b\in\mathbb Z$$ such that $$\tilde q(x,y)\in\mathbb Z[x,y]$$. So, my question can be rephrased as follows:

Is there a $$\tilde q(x,y)\in\mathbb Z[x,y]$$ that is not divisible by $$\tilde p(x,y)$$ such that $$\tilde q(x,y) \hat f(x,y)=0$$ for any $$f\in\mathcal F$$?

I will show that the answer is no, i.e., any $$\tilde q(x,y)$$ that satisfies $$\tilde q(x,y) \hat f(x,y)=0$$ for any $$f\in\mathcal F$$ must be divisible by $$\tilde p(x,y)$$.

Choosing a lexicographic monomial order with $$x\succ y$$ for multivariate division, any $$\tilde q(x,y)$$ can be uniquely written as $$\tilde q(x,y) = x^2\alpha(x) + x \beta(y) + \gamma(y) + \tilde r(x,y)\tilde p(x,y)~,\tag{1}\label{1}$$ where $$\tilde r(x,y)\in\mathbb Z[x,y]$$ and $$\alpha(X),\beta(X),\gamma(X)\in\mathbb Z[X]$$. Let $$u$$, $$v$$, and $$w$$ be the degrees of $$\alpha(X)$$, $$\beta(X)$$, and $$\gamma(X)$$ respectively, and let $$\alpha(X) = \sum_{i=0}^u a_i X^i~,\qquad \beta(X)=\sum_{j=0}^v b_j X^j~,\qquad \gamma(X)=\sum_{k=0}^w c_k X^k~.$$ I will show that $$a_i = b_j = c_k = 0$$ if $$\tilde q(x,y) \hat f(x,y)=0$$ for all $$f\in\mathcal F$$.

Consider the function $$f(m,n;t) = (-1)^m t^{-m+n} \left(\frac{1+t}{1-t}\right)^{-m-n}~.$$ It is easy to check that it is a discrete harmonic function on a square lattice for any $$t\ne 0,\pm 1$$ (see page 13 of this note for a related function). Now, using \eqref{1}, the equation $$\tilde q(x,y) \hat f(x,y;t) = 0$$ becomes $$x_t^2 \alpha(x_t) + x_t \beta(y_t) + \gamma(y_t) = 0~,$$ where $$x_t := t\left(\frac{1+t}{1-t}\right)$$ and $$y_t := t\left(\frac{1-t}{1+t}\right)$$. Let $$\nu := \max(v-1,w)$$. Multiplying by $$(1-t)^{u+2} (1+t)^\nu$$ gives a polynomial in $$t$$ given by $$\sum_{i=0}^u a_i t^{i+2} (1+t)^{\nu+i+2} (1-t)^{u-i} + \sum_{j=0}^v b_j t^{j+1} (1+t)^{\nu-j+1} (1-t)^{u+j+1} + \sum_{k=0}^w c_k t^k (1+t)^{\nu-k} (1-t)^{u+k+2} = 0~.\tag{2}\label{2}$$ Since this equation holds for any $$t\ne0,\pm 1$$, the polynomial in $$t$$ must vanish identically (even at $$t=0,\pm1$$).

We can write \eqref{2} as $$\sum_{i=0}^u a_i \alpha_i(t;\u) + \sum_{j=0}^v b_j \beta_j(t;\u) + \sum_{k=0}^w c_k \gamma_k(t;\u) = 0~.\tag{3}\label{3}$$ where $$\u:=(u,v,w)$$. I will proceed by induction to show that the set $$S(\u)$$ of polynomials $$\alpha_i(t;\u)$$'s, $$\beta_j(t;\u)$$'s, and $$\gamma_k(t;\u)$$'s is linearly independent for any $$\u$$:

• Base cases: For $$\u=\mathbf 0$$, $$S(\mathbf 0)$$ contains $$\alpha_0(t;\mathbf 0) = t^2(1+t)^2~, \qquad \beta_0(t;\mathbf 0) = t(1-t^2)~,\qquad \gamma_0(t;\mathbf 0) = (1-t)^2~,$$ which are clearly linearly independent. Similarly, for $$\u=(0,1,0)$$, $$S(0,1,0)$$ contains $$\alpha_0(t;\u) = t^2(1+t)^2~, \qquad \beta_0(t;\u) = t(1-t^2)~, \qquad \gamma_0(t;\u) = (1-t)^2~, \\ \beta_1(t;\u) = t^2(1-t)^2~,$$ which are also linearly independent.

• Induction step: Assume that $$S(\u)$$ is linearly independent.

1. For $$\u+\e_1$$, where $$\e_1:=(1,0,0)$$, $$S(\u+\e_1)$$ contains $$\alpha_i(t;\u+\e_1) = \begin{cases} (1-t)\alpha_i(t;\u)~,&\text{for } i = 0,\ldots,u~, \\ t^{u+3} (1+t)^{\nu+u+3}~,&\text{for } i=u+1~, \end{cases} \\ \beta_j(t;\u+\e_1) = (1-t) \beta_j(t;\u)~,\quad\text{for } j = 0,\ldots,v~, \\ \gamma_k(t;\u+\e_1) = (1-t) \gamma_k(t;\u)~,\quad\text{for } k = 0,\ldots,w~.$$ Except for $$\alpha_{u+1}(t;\u+\e_1)$$, all the other polynomials are linearly independent because they are all $$(1-t)$$ times polynomials that are linearly independent by hypothesis. On the other hand, $$\alpha_{u+1}(1;\u+\e_1)\ne0$$, so it is linearly independent of the other polynomials, which vanish at $$t=1$$. Therefore, $$S(\u+\e_1)$$ is linearly independent.

2. For $$\u+\e_2$$ and $$\u+\e_3$$, the above argument is a bit subtle. Instead, we can proceed as follows. Say $$S(u,v_0,v_0-1)$$ is a linearly independent set. Then $$S(u,v_0,w)$$ with $$w, and $$S(u,v,v_0-1)$$ with $$v are also linearly independent sets because $$S(u,v,v_0-1)\subset S(u,v_0,v_0-1)$$ and $$S(u,v_0,w)\subset S(u,v_0,v_0-1)$$. So, all I need to show is that if $$S(u,v_0,v_0-1)$$ is linearly independent, then $$S(u,v_0+1,v_0)$$ is also linearly independent. This can be shown using an argument similar to the one above (it's only slightly more complicated).

Since the polynomials $$\alpha_i(t;\u)$$'s, $$\beta_j(t;\u)$$'s, and $$\gamma_k(t;\u)$$'s are all linearly independent for any $$\u$$, it follows from \eqref{3} that $$a_i = b_j = c_k = 0$$. Therefore, $$\tilde q(x,y)$$ is divisible by $$\tilde p(x,y)$$.