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Martin Boundary of Killed Random Walks on Isoradial Graphs

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Abstract

We consider killed planar random walks on isoradial graphs. Contrary to the lattice case, isoradial graphs are not translation invariant, do not admit any group structure and are spatially non-homogeneous. Despite these crucial differences, we compute the asymptotics of the Martin kernel, deduce the Martin boundary and show that it is minimal. Similar results on the grid \(\mathbb {Z}^{d}\) are derived in a celebrated work of Ney and Spitzer.

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Notes

  1. This is inspired by the Darboux covering for tetrahedral hypergeometric equations of the Schwarz type (1/3, 1/3, 2/3) [53, §6.1], see also [54]. Note that the same change of variables has been used in [9, §4].

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Acknowledgments

We warmly thank Béatrice de Tilière for interesting discussions. We also thank an anonymous referee for his/her reading and comments.

Author information

Authors and Affiliations

  1. Sorbonne Université, CNRS, Laboratoire de Probabilités, Statistique et Modélisation (UMR 8001), 4 Place Jussieu, 75005, Paris, France

    Cédric Boutillier

  2. Institut Universitaire de France, Paris, France

    Cédric Boutillier

  3. CNRS & Institut Denis Poisson (UMR 7013), Université de Tours et Université d’Orléans, Parc de Grandmont, 37200, Tours, France

    Kilian Raschel

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  1. Cédric Boutillier
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Correspondence to Kilian Raschel.

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Alin Bostan authored the Appendix section of this paper.

This project has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme under the Grant Agreement No. 759702 and from the ANR DIMERS (ANR-18-CE40-0033).

Appendix: Some thoughts on a positivity question (by Alin Bostan)

Appendix: Some thoughts on a positivity question (by Alin Bostan)

We consider the following question: let \(s(k)={\sum }_{n \geqslant 0} s_{n} k^{n}\), starting

$$ s(k) =\frac{1}{2}+\frac{3}{32}k^{2}+\frac{3}{64}k^{4}+\frac{123}{4096}k^{6}+\frac{177}{8192}k^{8}+\frac{34887}{2097152}k^{10}+\cdots, $$

be the unique power series solution of the algebraic equation

$$ k^{2}s^{4}-2k^{2}s^{3}+2s-1=0. $$
(30)

We want to prove that all the coefficients of s(k) are non-negative, and more precisely that all the coefficients s2n are positive.

Note that the sequence \((s_{n})_{n \geqslant 0}\) satisfies various types of recurrences, either non-linear, obtained by extracting the coefficient of kn in both sides of Eq. (30), or even linear, obtained by using a linear differential equation (with polynomial coefficients in k) satisfied by s(k). However, these recurrences cannot be used directly to solve our positivity question. For instance, from the algebraic Eq. (30), Cockle’s algorithm (see [8] and the references therein) proves that s(k) satisfies the 3rd order linear inhomogeneous differential equation

$$ \begin{array}{@{}rcl@{}} 9 {k}^{3} \left( k-1 \right)^{2} \left( k+1 \right)^{2}&& s^{\prime\prime\prime} \left( k \right) +9 {k}^{2} \left( k-1 \right) \left( k+1 \right) \left( 5 {k}^{2}-1 \right) s^{\prime\prime} \left( k \right) \\ &&+k \left( 35 {k}^{4}-14 {k}^{2}-13 \right) s^{\prime} \left( k \right) -4 \left( {k}^{2}-2 \right) s \left( k \right) = 4-2 {k}^{2}, \end{array} $$
(31)

from which a coefficient extraction proves that \((s_{n})_{n \geqslant 0}\) satisfies the linear recurrence

$$ \begin{array}{@{}rcl@{}} \left( 3 n+10 \right) \left( 3 n+14 \right) \left( n+2 \right) s_{n+4} + n \left( 3 n+2 \right) \left( 3 n+4 \right) s_{n} \\ = 2 \left( 9 {n}^{3}+54 {n}^{2}+106 n+70 \right) s_{n+2}. \end{array} $$
(32)

It is not obvious from this recurrence how to derive a non-negativity proof because of the plus sign in front of the coefficient of sn. In principle, two quite general methods might be applied to this kind of question (alone, or combined): one relying on (effective) asymptotics for the coefficients of algebraic functions [24, p. 504–505], the other relying on the approach of [37, §4.2]. Instead, we present two different proofs.

The first one is based on the following hypergeometric expression for s(k):

$$ s(k)=\frac{1}{2}+ \frac{3}{2} \cdot \left( \frac{k}{4} \cdot {}_{2}F_{1}\left[\begin{array}{cc}\frac{1}{2} \frac{5}{6}\\\frac{5}{3} \end{array}; k^{2}\right] \right)^{2}, $$
(33)

where the Gauss�� hypergeometric series with parameters \(\frac 12, \frac 56\) and \(\frac 53\) is defined by

$$_{2}F_{1}\left[\begin{array}{cc}\frac{1}{2} \frac{5}{6}\\\frac{5}{3} \end{array}; k^{2}\right] = \sum\limits_{n=0}^{\infty} \frac{(\frac12)_{n}(\frac56)_{n}}{(\frac53)_{n}} \frac {k^{n}} {n!} = 1+{\frac{1}{4}}k+{\frac{33}{256}}{k}^{2}+{\frac{85}{1024}}{k}^{3} +\cdots , $$
(34)

and (x)n denotes the Pochhammer symbol (x)n = x(x + 1)⋯(x + n − 1) for \(n\in \mathbb {N}\).

Once found (e.g., using algorithmic tools for automated guessing [36], or for differential equation solving [28]), equality Eq. (33) can easily be proved using closure properties of D-finite functions [52]: from the 2nd order linear differential equation satisfied by \({}_{2}F_{1}\left [\begin {array}{cc}\frac {1}{2} \frac {5}{6}\\\frac {5}{3} \end {array}; k^{2}\right ]\), one computes a 3rd order differential equation satisfied by the right-hand side of Eq. (33), which appears to coincide with the differential Eq. (31) satisfied by the left-hand side of Eq. (33). Therefore, the coefficients sequences of both right-hand side and left-hand side of Eq. (33) satisfy the recurrence relation Eq. (32), hence they are equal since their initial terms coincide.

A shorter proof of equality Eq. (33), with a different (more geometric) flavor, goes as follows: on the one hand, the hypergeometric function in Eq. (34) satisfies the algebraic transformationFootnote 1

$${~}_{2}F_{1}\left[\begin{array}{cc}\frac{1}{2} \frac{5}{6}\\\frac{5}{3} \end{array}; \frac{x(x+2)^{3}}{(2x+1)^{3}}\right] =4 {\frac { \left( 2 x+1 \right)^{\frac32}}{ \left( x+2 \right)^{2}}}.$$

On the other hand, it is easy to prove, starting from the polynomial Eq. (30), that the algebraic function s(k) satisfies

$$ s \left( {\frac{x^{\frac12} \left( x+2 \right)^{\frac32}}{ \left( 2 x+1 \right)^{\frac32}}} \right) = {\frac {2 x+1}{x+2}}.$$

Denoting \(\displaystyle {{\varphi }(x) = { {x \left (x+2 \right )^{3}} / { \left (2 x+1 \right )^{3}}}}\), identity Eq. (33) becomes trivial in the new variable x:

$${\frac {2 x+1}{x+2}}=\frac12+\frac{3}{2} \cdot {\varphi}(x) \cdot \left( {\frac { \left( 2 x+1 \right)^{3/2}}{ \left( x+2 \right)^{2}}}\right)^{2}.$$

Now that Eq. (33) is proved, the positivity of the coefficients s2n of s(k) is a direct consequence of the obvious fact that the hypergeometric power series in Eq. (34) has positive coefficients.

The second proof is based on the so-called Stieltjes inversion formula [47, Lecture 2], that allows to express in some cases the n-th term of a sequence as the n-th moment of a positive density. Using this formula, one can deduce the following expression:

$$ s_{2n}= \frac{\sqrt{3}}{\pi}\cdot {{\int}_{0}^{1}} k^{n}\cdot H_{1}(k) \cdot \left( \frac{1}{{2}^{4/3}}\cdot {\frac {H_{2}(k)}{{k}^{2/3}}} -\frac{1}{{2}^{8/3}} \cdot \frac {H_{1}(k)}{{k}^{1/3}} \right)\mathrm{d} k, $$

where H1(k) and H2(k) are the hypergeometric functions

$$H_{1} = {}_{2}F_{1}\left[\begin{array}{cc}\frac{1}{6} \frac{5}{6}\\\frac{4}{3} \end{array}; k\right] = 1+{\frac{5}{48}}k+\cdots, \quad H_{2} = {}_{2}F_{1}\left[\begin{array}{cc}-\frac{1}{6} \frac{1}{2}\\\frac{2}{3} \end{array}; k\right] =1-{\frac{1}{8}}k-{\frac{3}{64}}{k}^{2}-\cdots. $$

At this point, positivity is not yet apparent. However, using the change of variables k = φ(x), and using the fact that H1(φ(x)) and H2(φ(x)) become the simple algebraic functions 2(x + 1)1/3(2x + 1)1/2/(x + 2) and (2x + 1)− 1/2, the previous expression simplifies to:

$$ s_{2n} = \frac{\sqrt {3} \cdot \sqrt [3]{2}}{\pi} \cdot {{\int}_{0}^{1}} {\frac { \left( x-1 \right)^{2} \left( \sqrt [3]{2}-\sqrt [3]{x^{2}+x} \right) \sqrt [3]{x+1}}{ \left( x+2 \right) \left( 2 x+1 \right)^{2}{x}^{2/3}}} {\varphi}(x)^{n}\mathrm{d} x. $$
(35)

Once found, equality Eq. (35) can again be proved automatically using algorithmic tools, notably the method of “creative telescoping” [2, 10, 40].

Note that Lagrange inversion is a tempting alternative to the previous moment approach; however, in our case, it gives that for all n > 0,

$$ s_{2n} = \frac{1}{2n} \cdot [z^{-1}] \left( {\frac { \left( z+1 \right)^{3} \left( 3-z \right) }{16 z}} \right)^{n} = \frac{(-1)^{n}}{{n\cdot {2}^{4n+1}}} \cdot \sum\limits_{i=0}^{n}{3 n\choose i}{n\choose i+1} \left( -3 \right)^{i+1}, $$

and positivity is not clear on any of these expressions.

In contrast, identity Eq. (35) obviously proves that the sequence \((s_{2n})_{n\geqslant 0}\) is positive. And it actually proves much more, namely that \((s_{2n})_{n\geqslant 0}\) is a Stieltjes-Hausdorff moment sequence, and in particular that it is log-convex, i.e., \( s_{2n+2}s_{2n-2} \geqslant s_{2n}^{2}\) for all \(n\geqslant 1\) [50, §2] and \({\Delta }^{k} s_{2n} \geqslant 0\) for all \(k\geqslant 0\) [51, p. 338], where Δ is the difference operator Δ(cn) = (cn − cn+ 1).

We also consider the following related question: let \(t(k)={\sum }_{n \geqslant 0} t_{n} k^{n}\), starting

$$ t(k)=1+\frac{3}{64}k^{4}+\frac{3}{64}k^{6}+\frac{711}{16384}k^{8}+\frac{327}{8192}k^{10}+\cdots, $$

be the unique power series solution of the algebraic equation

$$ -27(1-k^{2})t^{4}+18(1-k^{2})t^{2}+2(2-k^{2})^{2}t+1-k^{2}+k^{4}=0. $$
(36)

We want to prove that all the coefficients tn are non-negative.

First, we remark that \(t({\varphi }(x)^{\frac 12}) = { {({x}^{2}+x+1)}/{(1+x-2x^{2})}}\). Since \(s({\varphi }(x)^{\frac 12}) = (2x+1)/(x+2)\), this implies that t(k) = R(s(k)), where R is the rational function

$$R(z) = {\frac {{z}^{2}-z+1}{3 z \left( 1-z \right) }}. $$

As a consequence of this and of identity Eq. (33), we get that

$$ t(k) = (1+3 H^{4})/(1-9 H^{4}) = (1+3 H^{4}) \cdot \sum\limits_{n \geqslant 0} (9H^{4})^{n}, $$
(37)

where H is the hypergeometric function with non-negative coefficients

$$ H = \frac{k}{4} \cdot {}_{2}F_{1}\left[\begin{array}{cc}\frac{1}{2} \frac{5}{6}\\\frac{5}{3} \end{array}; k^{2}\right] = {\frac{1}{4}}k+{\frac{1}{16}}{k}^{3}+{\frac{33}{1024}}{k}^{5}+\cdots,$$

therefore all the coefficients of t(k) are non-negative.

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Boutillier, C., Raschel, K. Martin Boundary of Killed Random Walks on Isoradial Graphs. Potential Anal 57, 201–226 (2022). https://doi.org/10.1007/s11118-021-09912-5

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