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Convergence of a sandpile on a triangular lattice under rescaling A. A. Alieva, N. S. Kalininb a Department of Mathematics and Computer Science, Saint Petersburg State University, St. Petersburg, Russia b Guangdong Technion Israel Institute of Technology, Shantou, Guangdong Province, P. R. China
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     We begin by stating the problem in § 1, and in § 2 we present a plan of the proof, together with a survey of related results; in § 3 and § 4 we find bounds for the order of growth of the domain occupied by the sand. In § 5 and § 6 we show the existence of a limit in the weak-$^*$ topology. Section 7 is devoted to recalling various facts about discrete harmonic functions on lattices. § 1. Introduction and notationConsider the following dynamic model, known as the sandpile model or the dollar game. Let $G=(V,E)$ be an infinite regular connected graph without loops and multiple edges, and let the degree (valency) $\mathrm{deg}(z)$ of every vertex $z\in V$ be equal to $d=6$. The state of the sandpile model on $G$ is a function $\phi\colon V\to\mathbb Z_{\geqslant 0}$. The function $\phi(z)$ can be thought of as the number of grains of sand at the vertex $z$. If $\phi(z)\geqslant d$, then a toppling at a vertex $z$ is allowed, which means transferring a grain of sand from $z$ to every vertex adjacent to $z$. Formally, we build a new state $\phi'=T_z\phi$ according to the following rule:
Definition 1.1. Let $\triangle$ be a triangular lattice on a plane which is composed of regular triangles with side $1$. We view it as a graph in which each vertex has degree $6$. Consider the state $n\delta_0$, that is, the state equal to $n\in\mathbb N$ at the point $(0,0)$ and to zero at all other points. We begin producing topplings for this state; the process of producing topplings as long as possible is called relaxation. It is easy to show that any relaxation of the state $n\delta_0$ is finite, and its result $(n\delta_0)^\circ$ does not depend on the order in which topplings at particular points were carried out (see [2] and [3]). Definition 1.2. Let $u_n\colon \triangle\to\mathbb{Z}_{\geqslant 0}$ denote the number of topplings at the corresponding vertex during a relaxation of the state $n\delta_0$ on $\triangle$. Then $u_n$ is called the toppling function (odometer) of $n\delta_0$. Definition 1.3. Let $T_{n}=\{ z\mid u_{n}(z)>0\} \subset \triangle$ be the set of lattice points at which at least one toppling occurred during a relaxation of $n\delta_{0}$ on $\triangle$. The aim of this paper is to estimate the growth of $T_n$ with $n$ and to prove the existence of a limit of the cardinality of the sets $T_n$ under rescaling. That is, if $T_n$ is squeezed with coefficient $\sqrt n$, then Figure 1 has a limit in the weak-$^*$ topology, that is, for almost all points the average amount of sand in a sufficiently small vicinity of the point stabilizes as $n\to\infty$. We note that pointwise convergence cannot be expected: the number of grains of sand at each vertex is equal to $0$, $1$, $2$, $3$, $4$, or $5$, and in Figure 1 one can see repeating patterns that persist under rescaling.  Figure 1.A triangular lattice $\triangle$, the state $(n\delta_0)^\circ$: (a) $n=2^{13}$; (b) $n=2^{18}$. The white colour corresponds to the absence of sand, and the black colour corresponds to the maximum number of grains of sand (namely, $5$) at the vertex in a stable state, and shades of gray correspond to intermediate quantities of grains of sand. Triangular areas paved with some periodic patterns can be seen near the boundary of the domain. Some domains are subdivided into parts by linear graphs that are also made up of repeating patterns. The figure has been borrowed from [1]. Our methods are parallel to those used in [4] but differ in details, and at some places we present a more detailed description. There is virtually no literature in Russian about sandpile models, and our paper is intended to fill this gap partially. We note that, despite the fact that the limit of rescaled sets $T_n$ exists, very little is known about the boundary of the resulting set (see Figure 1); it is only known that it is Lipschitz (see [5]), and the proof is based on the monotonicity of the toppling function in any fixed direction ‘from the origin’. In this paper we consider aspects of the sandpile model related to rescaling, but there also are other major topics in the study of the sandpile model. For example, the sandpile dynamics on finite graphs is related to the so-called sandpile group (see [6]), which is also called the critical group of the graph (see [7]), or the Jacobian of a graph (by analogy with Riemann surfaces; see [8]). The order of this group is equal to the number of spanning trees in the graph. A recent survey on this topic is [9]. Questions relating to the behaviour of the sandpile group under graph rescaling were discussed in [10]–[15]. § 2. Plan of the proof and a survey of related resultsFor $u\colon \triangle \to \mathbb{R}$ we define the discrete Laplace operator by Instead of studying the (complicated) sandpile dynamics, it is more convenient to study the so-called toppling function. We note that applying $\Delta$ to the toppling function calculates the balance of the grains of sand incoming and outgoing during the relaxation; this is especially clear if we write A toppling at a vertex adjacent to $x$ adds a grain of sand to $x$; in a toppling at $x$ the vertex $x$ loses $d$ (the degree of the vertex) grains of sand. We obtain The so-called the least action principle says that the toppling function is a pointwise minimum integer-valued nonnegative function with this property (see [16]). We note that $T_n$ is exactly the support of the toppling function of the state $n\delta_0$.For a square lattice (every integer point is connected by edges with four adjacent points) it can be shown that, if there was a toppling at a point $(k,0)$ upon relaxation of $n\delta_0$, then there also were topplings at all vertices $(i,j)$ such that $|i|+|j|\leqslant k$, and if $(i,j)$ underwent a toppling, then so also did $(i,0)$. In this way the following theorem is proved. Theorem 2.1 (see [17]). On a square lattice, the set of vertices $T_n$ satisfies the following condition: there exists $r>0$ such that where $J_r=\{(x,y)\mid |x|+|y|\leqslant r\}$ and $I_r=\{(x,y)\mid \max(|x|,|y|)\leqslant r\}$. Now it can readily be shown that the growth rate of $r$ has the order $\sqrt n$. Lemma 2.2 (see [17]). If $\Omega$ is the set of vertices at which a toppling occurred during a relaxation, then the number of grains of sand remaining in $\Omega$ is not less than the number of internal edges of $\Omega$ (that is, edges with both ends in $\Omega$). The proof of the lemma is based on the fact that to every internal edge of $\Omega$ one can uniquely assign the last grain of sand ‘transferred’ along this edge. Thus, there has been a toppling at every vertex in $J_{r}$. It follows from Lemma 2.2 that $n$ (the initial number of grains of sand at the origin) is not less than the number of internal edges in $J_{r}$, that is, $n\geqslant 4r^2$ and $r\leqslant\sqrt n/2$. On the other hand, at the final state $(n\delta_{0})^\circ$ there are at most three grains of sand at every vertex, and therefore $n \leqslant 3 (2r+1)^2$ (tree times the number of vertices in $I_{r+1}$). We obtain
$$
\begin{equation*}
\frac{\sqrt n}{\sqrt{12}}-1\leqslant r\leqslant \frac{\sqrt n}{2}.
\end{equation*}
\notag
$$
It is easy to generalize these arguments to the case of a triangular lattice, but the bounds thus obtained give one a very rough idea about the shape of $T_n$ (although they show that $T_n$ grows at a rate of $\sqrt n$). We claim that the boundary of the support of $u_n$ (the toppling function for the state $n\delta_0$) lies between two circles of close radii $\approx0.23\sqrt n$ and $\approx0.303\sqrt n$. The main idea is in the two-sided bound
$$
\begin{equation}
c_1 |x|^2-c'n\log|x|+C_1\leqslant u(x)\leqslant c_2 |x|^2-c'n\log|x|+C_2
\end{equation}
\tag{2}
$$
on the set $T_n$. The involvement of $\log |x|$ is explained as follows. There is a discrete Green’s function on each lattice (square, triangular, hexagonal, …), that is, a function $g$ such that $\Delta g=\delta_0$. Such a function grows as $c'\log|x|$ (where $c'$ is a constant, specific for each lattice). We have and it is natural to assume that $u_n+ng$ grows quadratically. A lower bound for $u_n$ is obtained as follows. We consider sand to be divisible and, instead of causing a toppling, distribute the excess of grains of sand over $d-1$ to the adjacent vertices (one can imagine a wafer on which condensed milk is poured and when the amount of milk in a cell exceeds $d-1$, the excess flows to adjacent cells; see [18]). The relaxed sandpile of $n$ grains of sand must obviously be larger than a puddle formed by $n$ drops of condensed milk spread around. Analytically, this corresponds to choosing the constant $c_1$ in (2) in such a way that $\Delta c_1|x|^2=d-1$. It is more difficult to obtain an upper bound for $u_n$, because even inside $T_n$ there are vertices at which $\Delta u_n$ is zero. However, in the mean the value of $\Delta u_n$ (equal to the number of grains of sand in the final state after relaxation) is bounded away from zero by Lemma 2.2, and it is equal to the average number of edges per vertex, that is, $d/2$. That is, in the mean, there are at least three grains of sand per vertex inside the relaxed sandpile. To reveal this phenomenon we average all functions over discs of sufficiently large radius and choose $c_2=3=d/2$ in (2). The key element of the proof is the discrete maximum principle (Proposition 7.4): a discrete function $f$ that is subharmonic (that is, $\Delta f\geqslant 0$) on a finite subset of the lattice attains its maximum at the ‘boundary’ of this subset, that is, at a vertex one of whose neighbours lies outside the subset under consideration. In addition, in both bounds one must evaluate carefully what occurs near the boundary $T_n$ (about which, we recall, nothing is known in general). The first half of this paper is devoted to careful estimates following these ideas (borrowed from [18], where they were applied to a square lattice). The constants $3=d/2$ and $5=d-1$ occur in the statement of Theorem 3.7 and have been explained above. The other half of the paper follows [4] and is devoted to the problem of convergence of the toppling functions In § 5, using arguments of the type of the Arzelà-Ascoli theorem, we prove that there is a convergent subsequence of these functions (our functions are bounded and have a bounded discrete Laplacian as the step of the lattice tends to zero). Then, using standard methods of functional analysis, we show that the limit function is twice differentiable almost everywhere.Then we show in § 6 that only one limit function is possible, that is, the limits for all subsequences coincide. If there were at least two limit functions, then it would be possible to take functions approximating them on lattices and reduce (truncate) one of them using the other, which would contradict the principle of least action. The main idea here is that, if the limit function is twice differentiable at a point, then there is an integer-valued function with almost the same Hessian ‘in the mean’ on the whole lattice, and one can make the ‘truncation’ mentioned above using just this function. The required integer-valued function is constructed as the minimum of suitably transferred functions in a neighbourhood of a point at which the Hessian of the limit is known. By implementing these ideas for a square lattice Pegden and Smart, this time jointly with Levine, classified exactly which $2\times 2$ matrices can be realized as the Hessians of the limit function, that is, described the possible ‘quadratic’ patterns encountered during the relaxation of a large sandpile at the origin (see [19] and [20]). It turns out that the patterns are encoded by circles in the Apollonian packaging of the plane by circles (subsequently, such considerations were used to describe the neutral element of the sandpile group for some ellipses; see [21]). However, the following question remains open: what part of the picture is described by these patterns? In [22] Pegden and Smart showed that the dimension of the complement to the patterns is strictly less than two. As can be seen in Figure 1, the complement to the patterns is actually one-dimensional (a graph formed by line segments). This has not been proved yet. It is extremely difficult to obtain similar results for lattices other than square ones. The graph $\Gamma$ encoding the structure of emerging patterns looks much more complicated (see [23]). However, this was done for $F$-lattices by Bou-Rabee in [24]. Instead of Apollonius circles, in the case of an arbitrary lattice sandpile patterns are probably encoded by Kleinian bugs (see [25]). A result related to rescaling was obtained by Bou-Rabee in [26]: cross-sections of the neutral element of the sandpile group for a cube of large dimension are similar to the neutral element of the sandpile group of a cube of lower dimension, that is, the structure of patterns is stable not only under rescaling but also under changes of the dimension of the problem. A description of the linear patterns mentioned above was given in the series of papers [27]–[30], where it was shown that linear patterns in sandpile models converge to tropical curves after rescaling. Without technical details, the main ideas of those papers were described in [31]. Multidimensional analogues of the linear patterns mentioned above were described in [32] and [33]. § 3. The order of growth of the domain occupied by sand: a lower boundLemma 3.1 (see [34]–[36]). There is a discrete Green’s function (a function with the properties $\Delta g(x)=\delta_{0}$ and $g(0,0)=0$) for the triangular lattice $g\colon \triangle \to \mathbb{R}$ that has the following asymptotic expansion for some constant $C_0$: Let us represent $\triangle$ as a set of integer combinations of the vectors $(1,0)$ and $(1/2,\sqrt 3/2)$. Proposition 3.2. Consider the function $x\mapsto|x|^2$ as a function on $\triangle$. Then $\Delta |x|^2=6$. Below we follow [18]. Fix $c$ and consider the function It is bell shaped: there is a local maximum at zero, then the function decreases with distance from zero, and then it increases. Let $r_0$ be such that $n=4\pi\sqrt3\, cr_0^2$. The circle of radius $r_0$ turns out to be the ‘base’ of the bell described above. Set
$$
\begin{equation}
\gamma_{n}(x)=\widetilde{\gamma}_{n}(x)-\widetilde{\gamma}_{n}(\lfloor r_0 \rfloor,0).
\end{equation}
\tag{5}
$$
The paper [18] contains a similar lemma, but for $\varepsilon=0$. Proof of Lemma 3.3. Consider the leading part of $\widetilde{\gamma}_n$: Then
$$
\begin{equation}
\widetilde{\gamma}_{n}(x)=\Psi_{n}(|x|)-C_0n+O(n|x|^{-2}).
\end{equation}
\tag{8}
$$
We expand $\Phi_{n}$ in a Taylor series around $r=r_0$:
$$
\begin{equation}
\Psi_{n}(r)=\Psi_{n}(r_0)+\Psi_{n}'(r_0)(r-r_0)+\frac{1}{2}\Psi_{n}''(t)(r-r_0)^2
\end{equation}
\tag{9}
$$
for some $t\in [r_0,r]$. We note that it follows from the relation $n=4\sqrt3\, \pi c r_0^2$ that $\Psi_{n}'(r_0)=0$ ($r_0$ was chosen precisely from these considerations). Then for some $t_{1}$ between $r=|x|$ and $r_0$ and $t_{2}$ between $r_0$ and $\lfloor r_0 \rfloor$ we have
$$
\begin{equation}
\begin{aligned} \, \notag \gamma_{n}(x) &=\widetilde{\gamma}_{n}(x)-\widetilde{\gamma}_{n}(\lfloor r_0 \rfloor,0) =\Psi_{n}(|x|)-\Psi_{n}(\lfloor r_0 \rfloor )+O(n|x|^{-2}) \\ \notag &=\biggl(c+\frac{n}{2\sqrt3\, \pi t_{1}^{2}}\biggr)(r-r_0)^2 -\biggl(c+\frac{n}{2\sqrt3\, \pi t_{2}^{2}}\biggr)(r_0-\lfloor r_0 \rfloor)^2 +O(n|x|^{-2}) \\ &\geqslant c(r_0-|x|)^2+O(n|x|^{-2})-\biggl(c+\frac{n}{2\sqrt3\, \pi t_{2}^{2}}\biggr)(r_0-\lfloor r_0\rfloor)^2. \end{aligned}
\end{equation}
\tag{10}
$$
This completes the proof of the lemma. We denote by $B_{R}$ the ball of radius $R$ centred at zero in the Euclidean metric on the plane. Lemma 3.4. There is an absolute constant $C$ such that for fixed $c$, $n=4\sqrt3\, \pi c r_0^2$ and any $x\in B_{r_0+1}\setminus B_{r_0-1}$ The proof follows from the expansion (10). Proof. Since $\Delta (\gamma_{n}(x)-c|x|^2)\leqslant0$, it follows that the minimum of $\gamma_{n}(x)-c|x|^2$ in $B_{r_0/4}$ is attained on the boundary of this ball at a point $z\in \partial B_{r_0/4}$. That is, for any $x\in B_{r_0/4}$, by Lemma 3.3 we have This completes the proof of the lemma. Lemma 3.6. There is an absolute constant $C$ (independent of $n$) such that $\gamma_{n}\geqslant C$ everywhere for each $n$. Proof. Lemma 3.3 implies the existence of such $C$ outside any ball with centre at zero. Lemma 3.5 shows that a constant can also be found in a neighbourhood of zero. This completes the proof of the lemma. Theorem 3.7. Let $n=({2\pi}/{\sqrt3}) r^2$. Then for any $\varepsilon>0$ there exists $c_{\triangle}$ independent of $n$ such that where $k_{1}=\sqrt{1/5}$ and $k_{2}=\sqrt{1/3}+\varepsilon$. Proof. The line of reasoning follows [18]. The choice of $r$ comes from the choice of $c=1/6$ in Lemma 3.3. The renormalization for $c=5/6$ yields the constant $k_{1}=\sqrt{1/5}$.
Set and
$$
\begin{equation}
\xi_n(x) :=\widetilde{\xi}_n(x)-\widetilde{\xi}_n(\lfloor {k_{1}r}\rfloor,0).
\end{equation}
\tag{15}
$$
By Lemma 3.4, for all $x \in \partial B_{k_{1}r}$ we have
$$
\begin{equation}
u_{n}(x)-\widetilde{\xi}_{n}(x)\geqslant-\widetilde{\xi}_{n}(x)\geqslant -5C
\end{equation}
\tag{16}
$$
for some $C$ independent of $n$. We have $\Delta(u_{n}-\xi_{n})=\Delta u_{n}+n\delta_{0}-5\leqslant 0$; therefore, it follows from the discrete maximum principle that $u_{n}-\xi_{n}\geqslant-5 C$ in all the $B_{k_{1}r}$. Further, by Lemma 3.3, for $x \in B_{k_{1}r}$ we have
$$
\begin{equation}
u_n(x)\geqslant\xi_{n}(x)-5C\geqslant 5\biggl(\frac{1}{6}(k_{1}r-|x|)^{2} -C\frac{(k_{1}r)^2}{|x|^2}-C-\varepsilon\biggr)
\end{equation}
\tag{17}
$$
for some absolute constant $C$, some $\varepsilon$, and for sufficiently large $n$ and $r$. Now let ${k_{1}r}/{4}\leqslant|x|\leqslant k_{1}r-c_{2}$. We choose $c_{2}$ in such a way that
$$
\begin{equation}
u_n(x)\geqslant5\biggl(\frac{1}{6}\,c_{2}^{2}-\frac{C}{16}-C-\varepsilon\biggr)>0.
\end{equation}
\tag{18}
$$
For $|x|\leqslant{k_{1}r}/{4}$, by Lemma 3.5 we have Thus, we have proved the inclusion $B_{k_{1}r-c}\subset T_{n}$.
§ 4. The order of growth of the domain occupied by the sand: an upper boundWe continue the proof of Theorem 3.7. Let us prove the following simple and useful lemma. Lemma 4.1. For any vertex $x \in T_n$ adjacent to a vertex outside $T_n$ there is a path $x=x_{0} \thicksim x_{1} \thicksim x_{2} \thicksim\dots \thicksim x_{m}=0$ from $x$ to the origin in $T_{n}$ on which the inequality $u_{n}(x_{i+1})\geqslant u_{n}(x_{i})+1$ holds for all $i=0,\dots, m-1$. Proof. This is similar to the proof of Lemma 4.2 in [18]. We construct a sequence of vertices that ends at the origin. We note that for $y\neq 0$ we have $\Delta u_n(y)\geqslant 0$. We recall that $u_n$ is the toppling function and $n\delta_0+\Delta u_n$ is the final state after the relaxation of the sandpile at zero.
Let $x_{0}:=x\in T_{n}$ be adjacent to a vertex outside $T_n$, that is, there exists $x_{-1}\notin T_{n}$ such that $x_{0}\thicksim x_{-1}$. If $x_{0}=(0,0)$, then there is nothing to prove. We note that $u_{n}(x_{0}) > 0=u_{n}(x_{-1})$. It is also known that and therefore there exists $z\thicksim x$ such that $u_{n}(z)>u_{n}(x_{0})$. We have $u_{n}(z)>0$, and so $z\in T_{n}$. Set $x_{1}:=z$; if $x_{1}=0$, then the proof is over. Otherwise we repeat the procedure. Thus, at the $k$th step we have a path $x_{0} \thicksim x_{1} \thicksim x_{2} \thicksim \dots \thicksim x_{k}$ in $T_{n}$ on which $u_{n}(x_{i})>u_{n}(x_{i-1})$ for $i=1,\dots,k$. We note that all points constructed in this way are different, and there are finitely many points in $T_{n}$. This means that we reach the point $(0,0)$ at some step. This completes the proof of the lemma.For $x\in \triangle$ we define $Q_{k}(x)$ as the set of vertices in $\triangle$ lying at a distance (here the distance means the length of the shortest path in $\triangle$) at most $k$ from $x$, and of the edges in $\triangle$ between these vertices. Thus, $Q_{k}(x)$ contains $3k^{2}+3k+1$ vertices and $9k^2+3k$ edges. We set
$$
\begin{equation}
u^{(k)}_{n}(x):=\frac{1}{3k^{2}+3k+1}\sum_{y\in Q_{k}(x)}u_{n}(y).
\end{equation}
\tag{20}
$$
Then it can readily be seen that the total number of grains of sand in $Q_{k}(x)$ after relaxation is equal to
$$
\begin{equation}
(3k^{2}+3k+1)\Delta u^{(k)}_{n}(x)+n\chi_{Q_{k}(0)}(x),
\end{equation}
\tag{21}
$$
where $\chi_{Q_{k}(0)}(x)$ denotes the characteristic function of the set $Q_k(0)$, that is, it is equal to $1$ if $x\in Q_k(0)$ and to zero otherwise. Thus, by Lemma 2.2 we have
$$
\begin{equation}
\Delta u^{(k)}_{n}(x)\,{\geqslant}\, \frac{9k^2+3k}{3k^{2}+3k+1} -\frac{n}{3k^{2}+3k+1}\chi_{Q_{k}(0)}(x) \quad \forall\, x\,{\in}\, T^{(k)}_{n}\,{:=}\,\{x\mid Q_{k}(x)\,{\subset}\, T_{n}\}.
\end{equation}
\tag{22}
$$
Next, we continue to do the same as in [18]: we set We note that $c_k$ tends to $\sqrt{1/3}<k_2$ as $k\to\infty$. We write
$$
\begin{equation}
\widehat{\psi}^{(k)}_{n}(x) :=c_{k}^{-2}\frac{1}{6}|x|^2-ng(x),
\end{equation}
\tag{24}
$$
$$
\begin{equation}
\widetilde{\psi}^{(k)}_{n}(x) :=\frac{1}{3k^{2}+3k+1}\sum_{y\in Q_{k}(x)}\widehat{\psi}^{(k)}_{n}(y)
\end{equation}
\tag{25}
$$
and
$$
\begin{equation}
\psi^{(k)}_{n}(x) :=\widetilde{\psi}^{(k)}_{n}(x) -\widetilde{\psi}^{(k)}_{n}(\lfloor c_{k}r\rfloor,0).
\end{equation}
\tag{26}
$$
It follows from (22) that $\Delta (u^{(k)}_{n}-\psi^{(k)}_{n})\geqslant0$. Substituting $c_{k}^{2}n$ into Lemma 3.6 we see that there exists $a_{k}$ for which
$$
\begin{equation}
\widehat{\psi}^{(k)}_{n}(x)-\widehat{\psi}^{(k)}_{n}(\lfloor c_{k}r\rfloor,0)\geqslant-a_{k}.
\end{equation}
\tag{27}
$$
That is, Lemma 4.2. If the set $Q_{k}(x)\setminus T_{n}$ is not empty, then $u^{(k)}_{n}(x)\leqslant l_k$ for some constant $l_{k}$ independent of $n$. Proof. Let $y\in Q_{k}(x)\setminus T_{n}$. Then there has been no toppling at the point $y$, that is, at most five grains of sand have come to this point from its neighbours in $Q_{k}(x)$; then at most 20 grains of sand have come to each of these neighbours, and so on. Applying this reasoning sufficiently many times (depending only on $k$) we obtain the required constant (which is quite large). This completes the proof of the lemma. We continue the proof of Theorem 3.7. From Lemma 4.2 we know that
$$
\begin{equation}
u^{(k)}_{n}(x)-\psi^{(k)}_{n}(x)\leqslant l_{k}+a_{k} \quad \text{for } x\in\partial T^{(k)}_{n}.
\end{equation}
\tag{29}
$$
That is, from the discrete maximum principle we see that
$$
\begin{equation}
u^{(k)}_{n}(x)-\psi^{(k)}_{n}(x)\leqslant l_{k}+a_{k} \quad \text{for } x\in T^{(k)}_{n}.
\end{equation}
\tag{30}
$$
Now, by Lemma 3.4,
$$
\begin{equation}
u^{(k)}_{n}(x)\leqslant\psi^{(k)}_{n}(x)+l_{k}+a_{k}\leqslant \widetilde{c_2} \quad \text{for } c_{k}r-1\leqslant x \leqslant c_{k}r;
\end{equation}
\tag{31}
$$
then $u_{n}(x)\leqslant c_{2}':=c_{2}(3k^{2}+3k+1)$. Thus, by Lemma 4.1 we have $T_{n}\subset B_{c_{k}r+c_{2}'}$. Choosing a sufficiently large $k$ completes the proof of Theorem 3.7.
§ 5. Auxiliary assertions for the proof of convergenceBelow we follow [4] in our considerations. Notation 5.1. For $x\in \mathbb{R}^2$ let $\lfloor x \rceil$ denote the point closest to $x$ in $\triangle$. If there are several points of this type, then we select any of them. The inequality $|\lfloor x \rceil - x|\leqslant {1}/{\sqrt{3}}$ holds. Every function on a lattice can be extended to a function on the plane by using nearest neighbours: Let $s_n \colon \triangle \to \mathbb{Z}_{\geqslant 0}$ denote the final configuration after the relaxation of $n\delta_{0}$ on the triangular lattice. We extend $s_n$ to a function on the whole plane by using nearest neighbours. The following proposition is a direct consequence of Theorem 3.7. Proposition 5.2. For some $R>0$ and all positive integers $n$ the inclusion $\{{s_{n}>0}\}\subset B_{R n^{1/2}}$ holds. Recall that the function $u_{n}\colon \triangle\to\mathbb{Z}_{\geqslant 0}$ is the number of topplings at every vertex for the entire relaxation time of $n\delta_{0}$. We extend $u_{n}$ to a function on the whole plane by using nearest neighbours. Definition 5.3. For $h := n^{-1/2}$ set $y \thicksim^{h} x$ if $|y - x| = h$ and $y - x \in h\triangle$, and set and Proposition 5.5. The equality $s_{n}(x)=\Delta^{1} u_{n} (x)+n\delta_{0}$ holds, and $0\leqslant \Delta^{1} u_{n} (x)+n\delta_{0}\leqslant 5$. The toppling function $u_n$ is a pointwise minimum function providing a stable state; this can be proved using induction on the number of topplings. Proposition 5.6 (principle of least action; see [16]). Let $u\colon\triangle \to \mathbb{Z}_{\geqslant 0}$, and let the inequality $\Delta^{1} u (x)+n\delta_{0}\leqslant 5$ hold; then $u_{n}\leqslant u$. Proposition 5.7. Let $u\in C^{\infty}$. Then it follows from the Taylor series expansion of $u$ that $\Delta^{h} u \to \frac{3}{2}\Delta u$ locally uniformly (that is, uniformly on every compact set) as $h\to 0$. Proposition 5.8. If $u, v\colon h\triangle\to\mathbb{R}$, $w :=\min\{u, v\}$ and $w(x)=u(x)$, then $\Delta^{h}w(x) \leqslant \Delta^{h}u(x)$. Proposition 5.9. For $h=n^{-1/2}$ the Laplacian on a triangular lattice has a fundamental solution $\Phi_{n}\colon h\triangle \to \mathbb{R}$: In addition, $\Phi_{n} \to (2\sqrt{3}\,\pi)^{-1}\log|x|$ in $C^{\infty}(\mathbb{R}^{2}\setminus{0})$ locally uniformly. We take $\Phi_{n}(x):=g(\sqrt n x)-C_0$ for the discrete Green’s function $g$. Definition 5.10. A function $u\in C(\mathbb{R}^{2})$ is twice differentiable at a point $x\in\mathbb{R}^2$ if there is a vector $Du(x)\in \mathbb{R}^2$ and a symmetric $2 \times 2$ matrix $D^2u(x)\in\mathbb{R}^{2 \times 2}$ such that Definition 5.11. Let $s\in L^\infty(B_{1})$. We say that $u\in L^\infty(B_{1})$ is a weak solution of the equation $\Delta u=s$ in $B_{1}$ if for any $\phi \in C_{0}^{\infty}(B_{1})$. We write $\overline{\omega}_{n}:=\overline{u}_{n}+\Phi_{n}$. Then $\Delta^{h}\overline{\omega}_{n}=\overline{s}_{n}$. Proposition 5.12. Let $\overline{\omega}_{n}\colon h\triangle=n^{-1/2}\triangle\to\mathbb{R}$ be a sequence of functions satisfying the conditions Then every sequence $n_{k}\to\infty$ contains a subsequence $n_{k_{j}}\to\infty$ such that $\overline{\omega}_{n_{k_{j}}}\to \omega$ locally uniformly in $B_{1}$ for some $\omega\in C(B_{1})$. Proof. Replacing $\overline{\omega}_{n}$ by $\overline{\omega}_{n}+C$ if necessary we can assume that all sequences $\overline{\omega}_{n}$ are nonnegative. We extend $\overline{\omega}_{n}$ to continuous functions as follows: to every triangle of the lattice we continue the function by linearity. The rest of the proof consists in verifying the fact that the extended functions $\overline{\omega}_{n}$ satisfy the conditions of the Arzelà-Ascoli theorem. The uniform boundedness is ensured by the assumption. We verify equicontinuity.
We set $\widetilde{\omega}_{n}(x):=\overline{\omega}_{n}(h^{-1}x)$. Then for $x\in\triangle$ we have
$$
\begin{equation*}
|\Delta^{1} \widetilde{\omega}_{n}(x)|=h^2|\Delta^{h} \overline{\omega}_{n}(h^{-1}x)|\leqslant \frac{C}{n}.
\end{equation*}
\notag
$$
Let $x_{1},y_{1}\in B_{1}$ and $|x_{1}-y_{1}|<\delta/2$. We assume that $n$ is sufficiently large so that $|x-y|<\delta$ for some $x,y\in B_{1}\cap h\triangle$. Then $|h^{-1}x-h^{-1}y|<\delta n^{1/2}$. Thus, according to Lemma 7.15, we have
$$
\begin{equation*}
\begin{aligned} \, |\overline{\omega}_{n}(x)-\overline{\omega}_{n}(y)| &=|\widetilde{\omega}_{n}(h^{-1}x)-\widetilde{\omega}_{n}(h^{-1}y)| \\ &\leqslant\Bigl |\max_{B_{\delta n^{1/2}}(h^{-1}x)}\widetilde{\omega}_{n} -\min_{B_{\delta n^{1/2}}(h^{-1}x)}\widetilde{\omega}_{n}\Bigr| \leqslant \frac{C_{1}\delta^2 n}{n}=C_{1}\delta^2. \end{aligned}
\end{equation*}
\notag
$$
This implies readily that we can indeed choose $\delta$ required for equicontinuity. This completes the proof of the proposition. Proposition 5.13. Let $u \in C(B_{4})$ be a weak solution of the equation $\Delta u=s$ for some $s\in L^{\infty}(B_{4})$. Then the function $u$ is twice differentiable almost everywhere in $B_{1}$. Proof. Since $s\,{\in}\, L^{p}(B_{4})$ for all $p\,{>}\,1$, we have $u \,{\in}\, W^{2,p}(B_{2})$ for any $p>1$ (we recall that $W^{2,p}(B_2)$ is the space of functions in $L^p$ whose weak first and second derivatives also lie in $L^p$). Therefore, $Du\in W^{1,p}(B_{1})$. Then we find from Theorem 5 in [37], § 5.8.3, that the function $Du$ is differentiable almost everywhere, which means that $u$ is twice differentiable almost everywhere. This completes the proof of the proposition. Proposition 5.14. Let $s\,{\in}\, L^{\infty}(B_1)$, $f\,{\in}\, C(B_{1\,{+}\,\varepsilon})$ and $f(x)\,{=}\,0$ for $|x|\,{\geqslant}\,1$. If $f$ is a weak solution of the equation $\Delta f=s\geqslant 0$ and $\sup_{B_{1}}f>0$, then the set has a positive measure. Proof. In Theorem 3.2 of [38]1 set $d=\lambda=\Lambda=1$ and $f\equiv c$ for sufficiently large $c$ and take the function $-f$ as $u$. Let $\Gamma_{u}$ denote the convex hull of the function $-u^{-}$ on $B_{2}$. Since $\sup u^{-}>0$, it follows from the positivity of the integral in Theorem 3.2 of [38] that the set $\{u= \Gamma_{u}\}$ has a positive measure. Further, it is obvious that $u(x)<0$ for $x\in\{u=\Gamma_{u}\}$. It follows from the proof of Theorem 3.2 of [38] that Lemma 3.5 of [38] holds for $u$. That is, $D^2\Gamma_{u}>0$ on a subset $A\subset\{u=\Gamma_{u}\}$ of positive measure. Further, since $u$ is twice differentiable almost everywhere, we see that $D^2u\geqslant D^2\Gamma_{u}>0$ for almost all $x\in A$. This completes the proof of the proposition.
§ 6. Proof of convergenceWe follow the line of reasoning in [4]. First we prove convergence along subsequences and then the uniqueness of the limit. Lemma 6.1. In every sequence $n_{k}\to\infty$ there is a subsequence $n_{k_{j}}\to\infty$ such that $\overline{\omega}_{n_{k_{j}}}\to u$ locally uniformly for some $u\in C(\mathbb{R}^2)$, and $\overline{s}_{n_{k_{j}}}\to s$ converges weak-$^*$ in $L^{\infty}(\mathbb{R}^2)$ for some $s$. Here $\omega$ is a weak solution of the equation $\Delta\omega=\frac{2}{3}s$. Proof. By Proposition 5.2 we have $\{\overline{u}_{n}>0\}\subset B_{R}$; in addition, $\overline{u}_{n}\geqslant0$. Therefore,
$$
\begin{equation}
\overline{\omega}_{n}\geqslant \Phi_{n} \quad\text{for } x\in h\triangle,
\end{equation}
\tag{35}
$$
and Further, let and where $\partial^{h}E$ is the set $(B_{R+1}\setminus B_{R-1})\cap h\triangle$. Then we have $\phi\leqslant\overline{\omega}_{n}\leqslant\psi$ on $\partial^{h}E$, and $6=\Delta^{h}\phi\geqslant\Delta^{h}\overline{\omega}_{n}\geqslant\Delta^{h}\psi=0$ (since $\Delta^{h}\overline{\omega}_{n}=\overline{s}_{n}$ and $0\leqslant\overline{s}_{n}\leqslant 5$).
According to the maximum principle,
$$
\begin{equation}
|x|^2-(R+h)^2+\inf_{\partial^{h}E}\Phi_{n} \leqslant\overline{\omega}_{n}\leqslant\sup_{\partial^{h}E}\Phi_{n} \quad\text{for } x\in B_{R}.
\end{equation}
\tag{40}
$$
For every $R'>R$ we know that $\Phi_{n}\to\Phi$ uniformly in $B_{R'}\setminus B_{R}$ (where $\Phi$ is from Proposition 5.9). Thus, we obtain
$$
\begin{equation}
\|\overline{\omega}_{n}\|_{L^{\infty}(R')}\leqslant C(R').
\end{equation}
\tag{41}
$$
In addition, we know that $|\Delta^{h}\overline{\omega}_{n}|\leqslant d-1$. Therefore, having previously chosen a sufficiently large $R$, we conclude from Proposition 5.12 that there is a sequence $n_{j}$ and a function $\omega\in C(\mathbb{R}^{2})$ such that $\overline{\omega}_{n_{j}} \to \omega$ locally uniformly. Since every uniformly bounded sequence of functions has a weak-$^*$ convergent subsequence (a fact from functional analysis), we can select another subsequence $n_{k_{j}}$ and a function $s\in L^{\infty}(\mathbb{R}^2)$ in such a way that we have the weak-$^*$ convergence $s_{n_{k_{j}}}\to s$. We denote this sequence by $s_{n_{j}}$ again to avoid overloading the notation.
Then for $\phi\in C_{0}^{\infty}(\mathbb{R}^2)$ we have
$$
\begin{equation}
\int_{\mathbb{R}^2}\phi \overline{s}_{n_{j}}\,dx =\int_{\mathbb{R}^2}\phi \Delta^{h_{j}}\overline{\omega}_{n_{j}}\,dx =\int_{\mathbb{R}^2}(\Delta^{h_{j}}\phi) \overline{\omega}_{n_{j}}\,dx,
\end{equation}
\tag{42}
$$
where the first equality follows from the definition, and the other from the fact that $\overline{\omega}_{n_{j}}$ is constant inside regular hexagons with vertices at the centres of lattice triangles. Since $\overline{\omega}_{n_{j}}\to \omega$ locally uniformly n $\mathbb{R}^2$, the convergence is uniform on any compact set, in particular, on the closure of some neighbourhood of the support of $\phi$. Moreover, $\Delta^{h_{j}} \phi \to \frac{3}{2}\Delta \phi$ uniformly. That is, the right-hand side of (42) converges to $\displaystyle\frac{3}{2}\int \Delta\phi\cdot\omega \,dx$. The left-hand side converges to $\displaystyle\int \phi s\, dx$ by the definition of weak-$^*$ convergence. Then $\frac{2}{3}s=\Delta\omega$ by definition. This completes the proof of the lemma. An immediate consequence is as follows: $\overline{v}_{n_{j}}\to v:=\omega-\Phi$ in $\mathbb{R}^{2}\setminus \{0\}$ locally uniformly. Let us now construct an integer-valued function on the lattice whose Hessian is equal to $D^{2}v(x_{0})$ the hessian of the limit function $v$ at the point $x_0$. Such a function is constructed as follows. We consider an integer-valued function on the lattice which is close to the limit function, take a neighbourhood where its Hessian is close to the limiting Hessian $D^{2}v(x_{0})$, and by means of parallel translations extend this function to the whole lattice, each time adding a linear part in such a way that the graphs of the transferred functions lie near the graph of the quadratic function with Hessian $D^{2}v(x_{0})$. Next we prove that the pointwise minimum of the functions transferred lies close to the quadratic function mentioned above. By using these functions, we bypass the difficulty that subsequences with different limits in Lemma 6.1 can be defined on different lattices. Lemma 6.2. Let $v_{n}\colon \triangle\to\mathbb{Z}$ be an arbitrary sequence of functions such that ${\Delta v_{n}\leqslant K\in \mathbb{Z}}$. Let $v \in C(B_{r}(x_{0}))$ for some $r>0$, and let $\overline{v}_{n}(x):=h^2 v_{n }(h^{-1}x)$ converge uniformly to $v$ along some sequence $n_{k}\to\infty$. If $v$ is twice differentiable at $x_{0}$, then for any $\varepsilon$ there is a function $u\colon \triangle \to\mathbb{Z}$ such that for $x\in\triangle$ Proof. We proceed almost verbatim in the same way as in [4]. If necessary, we replace $v_{n}$ by
$$
\begin{equation*}
v_{n}'(x):=v_{n}(x+\lfloor h^{-1}x_{0}\rceil)-v_{n}(\lfloor h^{-1}x_{0}\rceil) -\lfloor h^{-1}Dv(x_{0})\rceil x,
\end{equation*}
\notag
$$
to assume that $x_{0}=0$, $v(0)=0$, $Dv(0)=0$ and $v_{n}(0)=0$. We write $\phi(x)=\frac{1}{2}x^{\top}D^{2}v(0)x$. The function $v$ is twice differentiable at zero, so we can reduce $r$ and choose a large $n=n_{k}$:
$$
\begin{equation}
\sup_{B_{r}}|\overline{v}_{n}-\phi|\leqslant \varepsilon r^2.
\end{equation}
\tag{44}
$$
After rescaling this transforms into
$$
\begin{equation}
\sup_{B_{h^{-1}r}}|v_{n}-\phi|\leqslant\varepsilon h^{-2}r^2.
\end{equation}
\tag{45}
$$
We write
$$
\begin{equation}
\psi(x):=\frac{1}{2}x^{\top}(D^2v(0)-\varepsilon I)x.
\end{equation}
\tag{46}
$$
For $x\in B_{h^{-1}r/4}$ we have
$$
\begin{equation}
v_{n}(x)\leqslant\phi+\varepsilon h^{-2}r^2=\psi(x)+\frac{\varepsilon}{2}|x|^2 +\varepsilon h^{-2}r^{2}\leqslant \psi(x)+\biggl(\frac{1}{32} +1\biggr)\varepsilon h^{-2}r^{2}.
\end{equation}
\tag{47}
$$
For $x\in B_{h^{-1}r}\setminus B_{h^{-1}r/2}$ we have
$$
\begin{equation}
v_{n}(x)\geqslant\phi+\varepsilon h^{-2}r^2=\psi(x)+\frac{\varepsilon}{2}|x|^2 +\varepsilon h^{-2}r^{2}\geqslant \psi(x)+\biggl(\frac{1}{8} +1\biggr)\varepsilon h^{-2}r^{2}.
\end{equation}
\tag{48}
$$
We set
$$
\begin{equation}
v_{n,y}(x):=v_{n}(x-y)+\lfloor D\psi(y)\rceil(x-y)+\lfloor\psi(y)\rceil
\end{equation}
\tag{49}
$$
and
$$
\begin{equation}
u(x):=\min\{v_{n,y}(x)\mid y\in\triangle\cap B_{h^{-1}r}(x)\}.
\end{equation}
\tag{50}
$$
Further, for $x\in B_{h^{-1}r/4}(y)$ we have
$$
\begin{equation*}
\begin{aligned} \, v_{n,y}(x) &=v_{n}(x-y)+\lfloor D\psi(y)\rceil(x-y)+\lfloor\psi(y)\rceil \\ & \leqslant \psi(x-y)+\biggl(\frac{1}{32}+1\biggr)\varepsilon h^{-2}r^2 +\lfloor D\psi(y)\rceil(x-y)+\lfloor\psi(y)\rceil \\ &\leqslant\psi(x-y)+\biggl(\frac{1}{32}+1\biggr)\varepsilon h^{-2}r^2+ D\psi(y)(x-y)+\psi(y)+\frac{h^{-1}r}{4\sqrt{3}}+\frac{1}{\sqrt{3}} \\ &\leqslant \psi(x)+\biggl(\frac{1}{32}+1\biggr)\varepsilon h^{-2}r^{2} +\frac{h^{-1}r}{4\sqrt{3}}+\frac{1}{\sqrt{3}}. \end{aligned}
\end{equation*}
\notag
$$
For $x\in B_{h^{-1}r}(y)\setminus B_{h^{-1}r/2}(y)$ we have
$$
\begin{equation*}
\begin{aligned} \, v_{n,y}(x) &=v_{n}(x-y)+\lfloor D\psi(y)\rceil(x-y)+\lfloor\psi(y)\rceil \\ &\geqslant \psi(x-y)+\biggl(\frac{1}{8}+1\biggr)\varepsilon h^{-2}r^2 +\lfloor D\psi(y)\rceil(x-y)+\lfloor\psi(y)\rceil \\ &\geqslant\psi(x-y)+\biggl(\frac{1}{8}+1\biggr)\varepsilon h^{-2}r^2+ D\psi(y)(x-y)+\psi(y)-\frac{h^{-1}r}{\sqrt{3}}-\frac{1}{\sqrt{3}} \\ &\geqslant \psi(x)+\biggl(\frac{1}{8}+1\biggr)\varepsilon h^{-2}r^{2} -\frac{h^{-1}r}{\sqrt{3}}-\frac{1}{\sqrt{3}}. \end{aligned}
\end{equation*}
\notag
$$
Now we choose $n$ sufficiently large so that
$$
\begin{equation}
\biggl(\frac{1}{8}+1\biggr)\varepsilon h^{-2}r^{2} -\frac{h^{-1}r}{\sqrt{3}}-\frac{1}{\sqrt{3}} > \biggl(\frac{1}{32}+1\biggr)\varepsilon h^{-2}r^{2} +\frac{h^{-1}r}{4\sqrt{3}}+\frac{1}{\sqrt{3}},
\end{equation}
\tag{51}
$$
It follows from (51) and (52) that
$$
\begin{equation}
u(x)=\min\{v_{n,y}(x)\mid y \in \triangle \cap B_{h^{-1}r/2}(x)\}
\end{equation}
\tag{53}
$$
is well defined. Indeed, since ${h^{-1}r}/{4}>1$, we can choose $y_{1}\in\triangle\cap B_{h^{-1}r/4}(x) $, and for any $y_{2}\in B_{h^{-1}r}(x)\setminus B_{h^{-1}r/2}(x)$, by the previous calculations and (51) we have $v_{n,y_{1}}(x)<v_{n,y_{2}}(x)$. Further, for $z\in B_{h^-1r/2}(x)$ we have $B_{h^{-1}r/2}(z)\subset B_{3h^{-1}r/4 }(x)\subset B_{h^{-1}r}(z)$. Therefore, we have
$$
\begin{equation}
u=\min\{v_{n,y}\mid y \in \triangle \cap B_{3h^{-1}r/4}(x)\} \quad\text{in}\ B_{h^{-1}r/4}(x).
\end{equation}
\tag{54}
$$
Further, since $\Delta^{1}v_{n,y}\leqslant K$ and ${h^{-1}r}/{4}>1$, Proposition 5.8 enables us to conclude that $\Delta^{1}u(x)\leqslant K$.
Now we estimate $u$ from below. Fix a point $x\in\triangle$. Then for some $y\in B_{h^{-1}r/2}(x)$ we have
$$
\begin{equation}
\begin{aligned} \, u(x) \notag &=v_{n,y}(x)=v_{n}(x-y)+\lfloor D\psi(y)\rceil(x-y)+\lfloor\psi(y)\rceil \\ &\geqslant \psi(x)-\frac{1}{\sqrt{3}}-\frac{h^{-1}r}{2\sqrt{3}}+\min_{B_{h^{-1}r/2}}v_{n} \geqslant \psi(x)-C \end{aligned}
\end{equation}
\tag{55}
$$
for some $C\in\mathbb{Z}$. We can take $u+C$ for the required function. This completes the proof of Lemma 6.2. Theorem 6.3. The functions $\overline s_{n}(x):=s_{n}(n^{1/2}x)$ converge weak-$^*$ to a function $s \in L^{\infty}(\mathbb{R}^{2})$. In addition, the limit satisfies the following conditions: $\displaystyle\int_{\mathbb{R}^{2}}s=1$, $0\leqslant s \leqslant d-1=5$ and $s=0$ in $\mathbb{R}^{2}\setminus B_{R}$ for some $R>0$. We prove a slightly stronger assertion below. Theorem 6.4. There are functions $s \in L^{\infty}(\mathbb{R}^{2})$ and $\omega\in C(\mathbb{R}^2)$ such that the sequence $\overline s_{ n}(x)$ converges weak-$^*$ to $s$, the sequence $\overline{\omega}_{n}$ converges to $\omega$ locally uniformly in $\mathbb{R}^ 2$, and $\omega$ is a weak solution of the equation $\Delta\omega=\frac{2}{3}s$. In addition, $\displaystyle\int_{\mathbb{R}^{2}}s=1$, $0\leqslant s\leqslant d-1=5$, $s=0$ in $\mathbb{R} ^{2}\setminus B_{R}$ for some $R>0$. Proof. By Lemma 6.1 there is a convergence along subsequences, so it is sufficient to prove the uniqueness of $s$ and $\omega$. It follows from Proposition 5.13 that $\Delta \omega= \frac{2}{3}s$ almost everywhere, so it is sufficient to prove the uniqueness of $\omega$. We prove it by contradiction.
Let $\overline{\omega}_{n_{k}}\to\omega$ and $\overline{\omega}_{n'_{k}}\to\omega'$ be two different limits. Since $\omega=\omega'=\Phi$ outside $B_{R}$ for some $R>0$, we can assume that
$$
\begin{equation*}
\sup_{B_R}(\omega - \omega') > 0=\sup_{\partial B_{R}}(\omega-\omega').
\end{equation*}
\notag
$$
By Proposition 5.14 we can choose a point $a\in\mathbb{R}^2\setminus 0$ such that $\omega$ and $\omega'$ are twice differentiable at $a$ and $ D^2\omega(a)\leqslant D^2\omega'(a)-2\varepsilon I$ for some $\varepsilon>0$. Set $v=\omega - \Phi$ and $v'=\omega' - \Phi$. Since $a\neq 0$, we have $\overline{v}_{n'_{k}}\to v'$ uniformly in a small neighbourhood of $a$. Moreover, $v'$ is twice differentiable at $a$, and therefore by Lemma 6.2 we can find $u\colon \triangle\to\mathbb{Z}$ such that, for all $x \in\triangle$, Since $D^2 v(a)+\varepsilon I\leqslant D^2 v'(a)-\varepsilon I$, it follows that
Using the following lemma we arrive at a contradiction and complete the proof of Theorem 6.3. Lemma 6.7. For small $r>0$ and large $n_{k}$ the function $\widetilde{v}:=\min\{v_{n_{k}},u_{n_{k}}\} $ contradicts the principle of least action for $v_{n_{k}}$. Proof. First of all, Therefore, now we are to select $r$ and $n_{k}$ to ensure two facts: the nonnegativity of $\widetilde{v}$ and the condition $n_{k}\delta_{0}+\Delta^{1}\widetilde{v }\leqslant 5$. Then the application of the principle of least action completes the proof.
Nonnegativity. It suffices to prove that $u_{n_{k}}\geqslant0$ in $B_{h_{k}^{-1}r}(a_{k})$. The function $v$ is twice differentiable at $a$, $\overline{v}_{n_{k}}\to v$ in $B_{r}(a)$ uniformly, and $\omega(a)>\omega'(a)$. Therefore, for small $r>0$, large $n_{k}$ and $x\in B_{r}(a)$ we have
$$
\begin{equation}
\overline{v}_{n_{k}}(a)+Dv(a)(x-a)+\frac{1}{2}(x-a)^{\top}D^2v(a)(x-a)\geqslant \frac{v(a)}{2}>\frac{v'(a)}{2}\geqslant 0.
\end{equation}
\tag{59}
$$
Then for $x\in B_{h_{k}^{-1}r}(a_{k})$ we calculate
$$
\begin{equation*}
\begin{aligned} \, &u_{n_{k}}(x)=u(x-a_{k})+\lfloor h_{k}^{-1}Dv(a)\rceil(x-a_{k})-u(0) +v_{n_{k}}(a_{k})-1 \\ &\geqslant \frac{1}{2}(x-a_{k})^\top D^2v(a)(x-a_{k}) +h_{k}^{-1}Dv(a)(x-a_{k})\,{-}\,u(0)\,{+}\,v_{n_{k}}(a_{k})\,{-}\,1\,{-}\, \frac{1}{\sqrt{3}}h_{k}^{-1}r \\ &\geqslant h_{k}^{-2}\frac{v(a)}{2}-u(0)-1-\frac{1}{\sqrt{3}}h_{k}^{-1}r-O(1)\geqslant 0 \end{aligned}
\end{equation*}
\notag
$$
for large $n_{k}$ and small $r$. The first inequality in the above chain follows from (57), and the penultimate one from (59).
The inequality $n_{k}\delta_{0}+\Delta^{1}\widetilde{v}\leqslant 5$. To prove it, according to Proposition 5.8, it suffices to show that $v_{n_{k}}\leqslant u_{n_{k}}$ in $B_{h_{k}^{-1}r}(a_{k})\setminus B_{h_{k}^{-1}r/2}(a_{k})$ and $0\notin B_{h_{k}^{-1}r/2}(a_{k}) $ for large $n_{k}$ and small $r>0$. Since $v$ is twice differentiable at $a$, for large $n_{k}$ and small $r>0$ we have
$$
\begin{equation*}
\overline{v}_{n_{k}}(x)\leqslant \overline{v}_{n_{k}}(a)+Dv(a)(x-a)+\frac{1}{2}(x-a)^{\top}D^2v(a)(x-a)+\frac{\varepsilon}{8}r^2
\end{equation*}
\notag
$$
for $x\in B_{r}(a)$. Rescaling, we obtain
$$
\begin{equation*}
v_{n_{k}}(x)\leqslant v_{n_{k}}(a_{k})+h_{k}^{-1}Dv(a)(x-a) +\frac{1}{2}(x-a)^{\top}D^2v(a)(x-a)+h_{k}^{-2}\frac{\varepsilon}{8}r^2
\end{equation*}
\notag
$$
for $x\in B_{h_{k}^{-1}r}(a_{k})$. Since
$$
\begin{equation*}
u(x)\geqslant \frac{1}{2}x^{\top}D^2v(a)x+\frac{\varepsilon}{4}h_{k}^{-2}r^{2}
\end{equation*}
\notag
$$
for $x\in B_{h_{k}^{-1}r}(a_{k})\setminus B_{h_{k}^{-1}r/2}(a_{k})$, we conclude that for $x\in B_{h_{k}^{-1}r}(a_{k})\setminus B_{h_{k}^{-1}r/2}(a_{k})$. The correction term is positive for large $n_{k}$, from which we conclude that for large $n_{k}$ and small $r>0$, which completes the proof. § 7. Application: harmonic functions on a triangular latticeRecall the following well-known results concerning discrete harmonic functions, discrete Harnack’s principle and so on. This is necessary to prove that a positive function with bounded discrete Laplacian on $\triangle$ grows no more rapidly than a quadratic function (Lemma 7.15). All functions in this section are defined only on lattices, and all sets are subsets of the lattice $\triangle$. Definition 7.1. Let $\Omega\subset\triangle$ be a finite set. Let $\partial_{\mathrm{out}}\Omega:=\{e_{\mathrm{out}}\mid e\in\partial\Omega\}$ and $\overline{\Omega}:=\Omega\cup\partial_{\mathrm{out}}\Omega$. Definition 7.2. A function $u\colon \triangle\to\mathbb{R}$ is said to be harmonic in $\Omega$ if $\Delta u \equiv 0$ in $\Omega$. This is proved by direct calculation. Proposition 7.4 (discrete maximum principle). Let $\Omega$ be a finite subset of $\triangle$, and let the function $u$ satisfy $\Delta u\geqslant 0$ on $\Omega$. Then $\max_{\partial \Omega}u\geqslant\max_{\Omega}u$. In fact, if the maximum $M=\max u$ is attained at an interior vertex $x$, then the values of $u$ at neighbours of $x$ must also be equal to $M$ by discrete subharmonicity. Repeating this argument, we find that $u$ is a constant function. Proposition 7.5. For $f\colon \partial_{\mathrm{out}}\Omega\to\mathbb{R}$ the Dirichlet problem, that is, the system of equations has a unique solution. Proof. The problem is to verify the solvability of the system of $|\overline\Omega|$ linear equations with $|\overline\Omega|$ unknowns. It follows from the maximum principle that for $f\equiv 0$ the solution is unique (and is identically equal to zero), which means that for any $f$ there is at most one solution. It follows from the coincidence of the number of equations and unknowns that for any $f$ a solution exists and is unique. This completes the proof of the proposition. Definition 7.6. Given $A\subset\partial_{\mathrm{out}}\Omega$, we let $\mathrm{hm}(\cdot, A)$ be the solution of the Dirichlet problem for $f=\mathbb{I}_{ A}$. Proposition 7.7. Let $h$ be a harmonic function in $\Omega$. Then the following representation holds: Indeed, both parts are harmonic functions and coincide at the boundary. According to the maximum principle, this means that they coincide everywhere. Lemma 7.8. Given $x\in\Omega$, there is a unique function $G_{\Omega}(\cdot,x)\colon \overline{\Omega}\to\mathbb{R }$ satisfying $G_{\Omega}(y,x)\equiv0$ for $y\in\partial_{\mathrm{out}}\Omega$ and Moreover, $G_{\Omega}(y,x)=G_{\Omega}(x,y)$. The existence and uniqueness are proved in the same way as in the Dirichlet problem. The symmetry follows from Green’s formula. The values of the left- and right-hand sides on the boundary coincide, and thus we have to verify the harmonicity of the right-hand side. This is a direct calculation. Lemma 7.11. There is a positive constant $C$ such that if the function $h$ is harmonic in $B_{R}(z),$ then Proof. It suffices to prove the lemma for $z=0$ and sufficiently large $R$, since for any fixed $R$ we can take $C=10R^2$ and use the maximum principle to obtain the necessary estimate. We let
$$
\begin{equation*}
H(x):=g(x)+\frac{R^2-|x|^2}{4\sqrt{3}\,\pi R^2}-C_{0}-\frac{1}{2\sqrt{3}\,\pi}\log R,
\end{equation*}
\notag
$$
where $C_0$ is the constant from the asymptotic expansion of the Green function $g(v)$. We note that if $x=e_{\mathrm{out}}$ or $x=e_{\mathrm{in}}$ for $e\in \partial B_{R}(0)$, then $|R- |x||\leqslant 1$, and therefore
$$
\begin{equation*}
H(x)=\frac{1}{2\sqrt{3}\,\pi}\log\biggl(1+\frac{|x|-R}{R}\biggr) +\frac{1}{2\sqrt{3}\,\pi}\frac{R-|x|}{R}-\frac{(R-|x|)^2}{4\sqrt{3}\,\pi R^2}+O(R^{-2})=O(R^{-2}).
\end{equation*}
\notag
$$
The application of Proposition 7.3 to $f_1=h$ and $f_2=H$ completes the proof. Lemma 7.12. There exist positive constants $C_{1}$ and $C_{2}$ such that for any ${x\in\partial_{\mathrm{out}}B_{R}(0)}$ Proof. If $h$ is harmonic and nonnegative, then $h(x)\geqslant \frac{1}{6}h(y)$ for any $y \thicksim x$. Therefore, we can assume that $R$ is sufficiently large. By Lemma 7.9 it suffices to prove that
$$
\begin{equation*}
\frac{C_{1}}{R}\leqslant -G_{B_{R}(0)}(0,x)\leqslant\frac{C_{2}}{R}
\end{equation*}
\notag
$$
for every $x$ such that $R-5<|x|<R-3$. We have
$$
\begin{equation*}
\begin{aligned} \, &-G_{B_{R}(0)}(0,x) =\biggl(-g(x)+C_0+\frac{1}{2\sqrt{3}\,\pi}\log R\biggr) \\ &\qquad +\biggl(g(x)-C_0-\frac{1}{2\sqrt{3}\,\pi}\log R -G_{B_{R}(0)}(0,x)\biggr) :=H_{1}(x)+H_{2}(x). \end{aligned}
\end{equation*}
\notag
$$
We know from the asymptotic expression for $g$ that for $x$ lying at a bounded distance from the boundary we have
$$
\begin{equation*}
H_{1}(x)=-\frac{1}{2\sqrt{3}\,\pi}\log\biggl(1+\frac{|x|-R}{R}\biggr)+O(R^{-2}) =-\frac{1}{2\sqrt{3}\,\pi}\frac{|x|-R}{R}+O(R^{-2}).
\end{equation*}
\notag
$$
Further, $H_{2}$ is harmonic, and $H_{1}\equiv H_{2}$ on $\partial_{\mathrm{out}}B_{R}(0)$. Therefore, by the maximum principle for $x\in B_{R}(0)$ we have Hence which completes the proof. Lemma 7.13. There is a positive constant $C$ such that if $h\geqslant 0$ is harmonic in the disc $B_{R+1}(z)$, then Proof. By Lemmas 7.11 and 7.12 we have
$$
\begin{equation*}
\begin{aligned} \, |h(z)-M_{R}h(z)| &\leqslant \frac{C}{R^2}\sum_{e\in\partial B_{R}(z)}(h(e_{\mathrm{out}})+h(e_{\mathrm{in}})) \leqslant 7 \frac{C}{R^2} \sum_{x\in\partial_{\mathrm{out}} B_{R}(z)}h(x) \\ &\leqslant \frac{7{C}}{R}\sum_{x\in\partial_{\mathrm{out}} B_{R}(z)}h(x)\mathrm{hm}_{B_{R}(z)}(z,\{x\})=\frac{7C}{R}h(z). \end{aligned}
\end{equation*}
\notag
$$
The constant $7$ arises here from the inequality $h(e_{\mathrm{out}})+h(e_{\mathrm{in}})\leqslant 7h(e_{\mathrm{out}})$. This completes the proof of the lemma. Lemma 7.14. There are positive constants $C_{1}$ and $C_{2}$ such that if $h\geqslant 0$ is discretely harmonic in $B_{R}(z)$ and $|z-w|\leqslant R/4$, then $C_{1}h(z)\leqslant h(w)\leqslant C_{2}h(z)$. Proof. We have
$$
\begin{equation*}
M_{R}h(z)=\frac{\sqrt{3}}{2}\frac{1}{\pi R^{2}}\sum_{w\in B_{R}(z)}h(w)\geqslant \frac{\sqrt{3}}{2}\frac{1}{\pi R^{2}}\sum_{w\in B_{R/2}(z)}h(w)=4M_{R/2}h(w).
\end{equation*}
\notag
$$
Then by Lemma 7.13,
$$
\begin{equation*}
h(z)\biggl(1-\frac{C}{R}\biggr)\leqslant M_{R/4}h(z) \leqslant\frac{1}{4}M_{R/2}h(w)\leqslant\frac{1}{4}h(w)\biggl(1+\frac{C}{R}\biggr).
\end{equation*}
\notag
$$
That is, we have obtained the first inequality for sufficiently large $R$. For small $R$ it follows from Lemma 7.11. The other inequality is obtained similarly from the relation $M_{R/2}h(z)\geqslant 4M_{R/4}h(w)$. This completes the proof of the lemma. Lemma 7.15. Let $u\colon \triangle\to \mathbb{R}_{+}$, $\max_{B_{10R}(0)}|\Delta f|\leqslant \lambda$ and $f(0)=0 $. Then there exists a constant $C$ independent of $f$ and $R$ such that Proof. This repeats the proof of Lemma 2.17 in [39]. Consider the functions We know from the assumptions of the lemma that $\Delta f_{+}\geqslant 0$ and $\Delta f_{-}\leqslant 0$. Fix $x\in B_{R}(0)$, and let $r= 9|x|$. Next, let $h_{\pm}$ be the harmonic functions in the ball $B=B_{r-1}(0)$ whose values at the boundary coincide with $f_{\pm}$. Then, first of all, $h_{+}\geqslant0$ on $B$; in addition, in $B$ we have
By Lemma 7.14 we know that there exists $C_{2}$ such that $h_{+}(y)\leqslant C_{2}h_{+}(0)$ for $y\in B_{(r-1)/4}(0)$. Then, since $f_{-}(0)=f(0)=0$, we have Thus,
$$
\begin{equation*}
f(x)\leqslant h_{+}(x)-\lambda |x|^2\leqslant (162C_{2}-\lambda)|x|^2.
\end{equation*}
\notag
$$
This completes the proof of the lemma.
Citation in format AMSBIB
\Bibitem{AliKal23} |
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