In this paper, the fixed/predefined-time generalized synchronization problem of stochastic complex dynamical networks with delays is studied for the first time. First, based on the feedback controller without linear terms, the results show that the controlled system has strong stability. Second, stochastic analysis methods, inequality techniques, and an extension of the existing fixed/predefined-time stability lemma (η range extension) are used to make the results of this paper more general. The sufficient conditions for generalized synchronization are established, and the settling time independent of the initial values are given. To illustrate the theoretical results, a numerical example is given.
Citation: Qike Zhang, Tao Xie, Wenxiang Fang. Fixed/predefined-time generalized synchronization for stochastic complex dynamical networks with delays[J]. AIMS Mathematics, 2024, 9(3): 5482-5500. doi: 10.3934/math.2024266
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In this paper, the fixed/predefined-time generalized synchronization problem of stochastic complex dynamical networks with delays is studied for the first time. First, based on the feedback controller without linear terms, the results show that the controlled system has strong stability. Second, stochastic analysis methods, inequality techniques, and an extension of the existing fixed/predefined-time stability lemma (η range extension) are used to make the results of this paper more general. The sufficient conditions for generalized synchronization are established, and the settling time independent of the initial values are given. To illustrate the theoretical results, a numerical example is given.
Consider an infinite, locally finite, and connected graph G=(V(G),E(G),o), where V(G) denotes the vertex set, E(G) the edge set, and o is a designated root. Consider Markov chain. There is a stationary measure π(⋅) such that for any two adjacent vertices x and y, π(x)p(x,y)=π(y)p(y,x), where p(x,y) is the transition probability. For the edge joining vertices x and y, assign a weight
c(x,y)=π(x)p(x,y). |
Now we call the weights of the edges conductance and their reciprocals resistance. In this paper, we study the spectral radius of irreducible Markov chains on weighted graphs. More precisely, we focus on the spectral radius of a biased random walk, which is defined as follows:
Fix a root o in a graph G. Let |x| be the graph distance between x and o for any vertex x of G. Write N for the set of natural numbers, and let Z+=N∪{0}. For any n∈Z+, we define two subsets of vertices:
The ball of radius n centered at o is denoted by BG(n)={x∈V(G): |x|≤n}.
The boundary of this ball is ∂BG(n)={x∈V(G): |x|=n}. Fix any λ∈[0,∞). If edge e is at distance n from o, then let its conductance be λ−n. Denote by RWλ the random walk associated the above conductances, and we call RWλ a biased random walk. Recall that a random walk on an infinite connected network is transient iff the effective conductance from any of its vertices to infinity is positive [19].
The motivation for introducing RWλ is to design a Monte-Carlo algorithm for self-avoiding walks by Berretti and Sokal [5]. See [13,20,21] for refinements of this idea. Due to interesting phenomenology and similarities to concrete physical systems ([7,9,10,12,23,24,25]), biased random walks and biased diffusions in disordered media have attracted much attention in the mathematical and physics communities since the 1980s.
In the following, we assume G is transitive. Let Mn=|∂BG(n)| be the cardinality of ∂BG(n) for any n∈Z+. Define the growth rate of G as
gr(G)=lim infn→∞n√Mn. |
Since the sequence {Mn}∞n=0 is submultiplicative, the limit gr(G)=limn→∞n√Mn exists indeed. R. Lyons [15] showed that the critical parameter for RWλ on a general tree is exactly the exponent of the Hausdorff dimension of the tree boundary. Moreover, R. Lyons [16] proved that for Cayley graphs and degree-bounded transitive graphs, the growth rate is exactly the critical parameter of the RWλ.
Let d be the vertex degree of G. For any vertex v of G except o, denote by d−v the number of edges connecting v to ∂BG(|v|−1). For the definition of RWλ (Xn)∞n=0 starting at o, the transition probability from v to an adjacent vertex u is
p(v,u)={1/dif v=o,λd+(λ−1)d−vif u∈∂BG(|v|−1) and v≠o,1d+(λ−1)d−votherwise. |
Note that RW1 is just a simple random walk on G.
We write
p(n)(x,y)=p(n)(x,y,λ)=Px(Xn=y). |
The Green function is defined by
G(x,y|z)=∞∑n=0p(n)(x,y)zn,x,y∈V(G),z∈C. |
Define
τy=τy(λ)=inf{n≥1|Xn=y},f(n)(x,y)=Px(τy=n), |
the associated generating function
U(x,y|z)=∞∑n=1f(n)(x,y)zn,x,y∈V(G),z∈C. |
Given any function g(z), let us denote the radius of convergence by Rg. By the Cauchy-Hadamard criterion,
RG=RG(λ)=1lim supn→∞n√pn(x,y). |
Recall [19, Exercise 1.2] that RG is independent of x and y due to its irreducibility. Define
ρ=ρG(λ)=ρ(λ)=1RG=lim supn→∞n√pn(x,x). |
ρ is called the spectral radius of the biased random walk. The reason for this name we can refer to [14,19] for more information on the spectral radius. Moreover,
ρ=lim supn→∞n√pn(o,o). |
Define the speed of RWλ, (Xn)∞n=0, as the limit of |Xn|n as n→∞, if it exists almost surely (or in probability).
There are many deep and important questions related to how the spectral radius and the speed depend on the bias parameter λ (see [14], Chapter 9). In [17,p.2005 Questions], R. Lyons, R. Pemantle, and Y. Peres raised the Lyons-Pemantle-Peres monotonicity problem. For a Cayley graph G with gr(G)>1, does the speed of RWλ exist and be positive for all λ∈(1,gr(G))? For more information, readers can refer to [1,4].
Moreover, R. Lyons, R. Pemantle and Y. Peres [18] conjectured that the speed of RWλ on the supercritical Galton-Watson tree without leaves is strictly decreasing. The conjecture has been confirmed for λ lying in some regions (see [2,4,26]). For Galton-Watson trees without leaves, the Lyons-Pemantle-Peres monotonicity problem was answered positively for λ≤m11160 by Ben Arous, Fribergh and Sidoravicius [4], where m1 is the minimal degree of the Galton-Watson tree. And Aïdékon [1] improved the just-mentioned result to λ≤12 by a completely different approach. In [3], Ben Arous, Hu, Olla, and Zeitouni obtained the Einstein relation for RWλ on Galton-Watson trees, which implies the conjecture holds in a neighborhood of m.
We are interested in the continuity of the spectral radius of RWλ. For λ>gr(G), RWλ is recurrent, ρ=1. When λ=0 RWλ is not irreducible.
Problem 1.1. For a Cayley graph G, is the spectral radius of RWλ continuous for all λ∈(0,gr(G)]?
In this paper, our focus is centered on addressing Problem 1.1 within the realm of a distinctive class of block-free product groups. To ensure a comprehensive understanding, the precise definition and pertinent details regarding free products will be meticulously introduced in Section 2.
Theorem 1.2. Let graph G be a free product of complete graphs. Then the spectral radius ρ(λ) of RWλ on G is continuous in λ∈(0,gr(G)].
To prove Theorem 1.2, the primary technical obstacle lies in establishing the generating function and subsequently demonstrating that the Cheeger constant is indeed positive.
Theorem 1.2 affirmatively resolves Problem 1.1, specifically for the scenario involving the free product of complete graphs. A key observation here is that the d-regular tree Td, represents a particular instance of such free products of complete graphs. Consequently, it is deduced from the theorem that the spectral radius, ρ(λ) of RWλ defined on Td, exhibits a characteristic of continuity over the interval (0, gr(Td)], where gr(Td) signifies the growth rate of the d-regular tree. This finding underscores the robustness of the spectral property with respect to variations in the parameter λ.
Intuitively, a "free product" of finite Cayley graphs Gi is a rule to construct a new Cayley graph G by gluing these m cells at "common vertices" without edge intersection, step by step. Concretely, we construct the Cayley graph G of H1∗H2∗⋯∗Hr by the following steps: Here, ∗ denotes a free product.
Step 1. Glue each i-cell (1≤i≤r) at a common vertex o such that any two of the r cells only have one common vertex o. View o as the birth root of these m cells. Usually o is chosen to be the identity element 1. Denote the obtained graph as G(1), and mark any vertex x in the j-cell as [j].
Step 2. For any x∈G(1)∖{o}, it must be in some cell, say an i-cell, then we glue each j-cell (1≤j≠i≤r) at x such that any of the r−1 cells has only one common vertex x with G(1) and any two of the r−1 cells only have one common vertex x. View x as the birth root of just added r−1 cells. For any distinctive two vertices x and y of G(1)∖{o}, we require that any cell glued at x is disjoint from any cell glued at y. Let G(2) be the resulting graph. And for any 1≤j≤r, also mark a vertex x of G(2)∖G(1) in the j-cell as [j].
Step 3. For all x∈G(2)∖G(1), according to ⟨x⟩, we can determine which type of cell x belongs to, and glue other type cells as Step 2. And then mark all new added vertices y and define ⟨y⟩ and [y] as Step 2. Denote by G(3) the obtained graph. And so on, we obtain G with a type function [⋅] in its vertices, where for convenience we let [o]=0.
Now let G be the Cayley graph of H1∗H2∗⋯∗Hr with root o, where each Hi (1≤i≤r) is a finite group whose Cayley graph is the complete graph Kmi+1 on mi+1 vertices. Let ∑ri=1mi=m. Thus, the transition probability from v to an adjacent vertex u is
p(v,u)={1/mif v=o,λm+λ−1if u∈∂BG(|v|−1) and v≠o,1m+λ−1otherwise. |
Let f(n)i(o,o) (i=1,2,⋯,r) be the probability of the biased random walk on G starting at o and τo=n, which does visit a vertex of Kmi+1. Define
Ui(o,o|z)=∞∑n=1f(n)i(x,y)zn. |
Hence,
U(o,o|z)=r∑i=1Ui(o,o|z). | (2.1) |
Note the tree-like structure of G. To compute f(n)i(o,o), a biased random walk must reach an edge in Kmi+1 from o and return o by an edge in Kmi+1 at last step. Each vertex of Kmi+1 glues a copy of Kmj+1 (j≠i). By the symmetry of Kmi+1, we can regard Kmi+1 as an edge with a cycle and glue the same structure as in G. Thus
Ui(o,o|z)=mimzλm+λ−1z∞∑n=0(i−1∑j=0Mj(z)+˜Mi(z)+r∑k=i+1Mk(z))n. |
Here Mj(z) denotes the generating function associated with the hitting probability of Kmi+1, which starts at a vertex of Kmj+1 (j≠i). Note that the probability from o to verteices of Kmj+1 is mim and from a vertex of Kmi+1 to the vertices of Kmj+1 is mim+λ−1. And the other steps obey the same law as Uj(o,o|z). Thus
Mj(z)=mm+λ−1Uj(o,o|z),˜Mi(z)=mi−1m+λ−1z. |
Therefore, in the domain {z∈C||z|<RU},
Ui(o,o|z)=λmi(m+λ−1)2z211−(mm+λ−1r∑i=1Uj(o,o|z)−mm+λ−1Ui(o,o|z)+˜Mi(z))=λmi(m+λ−1)2z211−(mm+λ−1U(o,o|z)−mm+λ−1Ui(o,o|z)+˜Mi(z)). |
Hence,
Ui(o,o|z)=−((m+λ−1)−mU(o,o|z)−(mi−1)z)2m+√((m+λ−1)−mU(o,o|z)−(mi−1)z)2+4λmiz22m. | (2.2) |
Drawing upon the results established in Eqs 2.1 and 2.2, we are able to deduce that
U(o,o|z)=r∑i=1Ui(o,o|z)=r∑i=1−((m+λ−1)−mU(o,o|z)−(mi−1)z)2m+r∑i=1√((m+λ−1)−mU(o,o|z)−(mi−1)z)2+4λmiz22m. |
Let
∂BGi(n)={x∈V(Gi): |x|=n}. |
Define
Si(z)=∑n≥1|∂BGi(n)|zn. |
Lemma 2.1. [8]
gr(G)=1zS, |
where zS is the unique real number with
r∑i=1Si(zS)1+Si(zS)=1. | (2.3) |
In the work of E. Candellero, L. A. Gilch, and S. Müller [8], they derive an upper bound for the upper box-counting dimension and a complementary lower bound for the Hausdorff dimension of the geometric endpoint boundary of the trace. This analysis is instrumental in establishing Lemma 2.1. Our approach subsequently leverages Lemma 2.1 as a pivotal step to demonstrate that the growth rate of the free product of complete graphs is indeed positive.
Since the Cayley graph of Hi (1≤i≤r) is a complete graph on mi+1 vertices,
Si(z)=miz. |
Recall Lemma 2.1,
gr(G)=1zS, |
where zS is the unique real positive number with
r∑i=1mizS1+mizS=1. |
Let Γi=(Vi,Ei) generated by the vertex set
Vi={x∈V(G),the geodesic betweenoandxstarting atoto visitKmi+1}. |
Clearly, for any n∈Z+, 1≤i≤r,
|BΓi(n)|=mi∑j≠i|BΓj(n−1)|. |
Hence, for n large enough and 1≤i≠j≤r,
|∂BΓi(n)|∼|∂BΓj(n)|∼gr(G)n. |
To continue, we introduce some preliminaries about the Cheeger constant. For a weighted graph H (with weight c(x,y) for the edge joining vertices x and y), we say that H satisfies the isoperimetric inequality, briefly IP∞, if there exists a κ>0 such that C(∂E(S))≥κC(S) for any finite connected subset S. Here C(S)=∑x∈SCx and C(∂E(S))=∑x∈S,y∉Sc(x,y), where Cx=∑y∼xc(x,y). The largest possible κ is the: Kesten-Cheeger-Dodziuk-Mohar theorem, see [14] Chapter 7, Theorem 7.3. A weighted graph can also be regarded as a network. Readers can refer to [19] for details.
Theorem 2.2. For any connected infinite network H, the following are equivalent:
(1) H satisfies IP∞ with κ>0.
(2) 0<ρ<1.
In fact, κ2/2≤1−ρ≤κ.
In the subsequent discussion, we will undertake the task of proving that the Cheeger constant of G is indeed positive. This assertion is facilitated by invoking the Kesten-Cheeger-Dodziuk-Mohar theorem (Theorem 2.2), which leads us directly to the derivation of the following lemma.
Lemma 2.3. For any λ∈(0,gr(G)), 0<ρ(λ)<1.
Proof. To return o, a path can hit a vertex x of type [i] with probability 1m, then move to another neighbour of x with type [j] with probability mj−1m+λ−1. Until the n−th step, the random walk runs away from o. From the (n+1)−th until the 2n−th the random walk returns o with probability λm+λ−1 for every step. Hence, for any n≥2
p(2n)(o,o)≥r∑i=11m(min1≤i≤rmim+λ−1)n−1(λm+λ−1)n. |
Thus
ρ(λ)≥min1≤i≤r√miλm+λ−1>0. |
Let S be any connected, finite subgraph of G. Assume that S is between ∂BG(n) and ∂BG(n+m). Note that
S=∪n+mj=n(S∩∂BG(j))=∪n+mj=nSj,|S|=|∂S|+|So|, |
where Sj denotes S∩∂BG(j) and So denotes the interior points of S.
For any x∈Sn, denote the subgraph of x as Sx=(V(Sx),E(G)). Here
V(Sx)={y∈V(G),the geodesic betweenyandovisitsx}, |
E(Sx)={xy∈E(G),x,y∈V(Sx)}, |
Sx(n)={y∈V(Sx):the graph distance ofxandy≤n}. |
Given any fixed ϵ>0, from the definition of gr(G), {there exists n0 such that, for any k>n0,
|∂BG(k)|≥(gr(G)−ϵ)k. |
Define
n1=inf{k>n,there existsy∈∂S∩Sx(k−n)}, |
which means that at ∂Sx(k−n), we can find at least one vertex belonging to ∂S. Hence,
C(∂Sx∩Sx(n1−n))C(Sx(n1−n))≥λ−n1mn1−n(m+λ−1)λ−n=λn−n1mn1−n(m+λ−1). |
If n1−n≤n0, then
C(∂Sx∩Sx(n1−n))C(Sx(n1−n))≥λ−n0mn0(m+λ−1)>0. |
Now assume that, n1−n>n0. Notice that every vertex x≠o has m−mi neighbors in BG(|x|+1) and only one neighbour in BG(|x|−1) if x∈[i]. Let Vij be the set of all vertices of type [i] in So∩Sj. It is easy to see that
r∑i=1(m−mi)|Vij|≤|Sj+1|. |
Notice that ∑ri=1(m−mi)|Vij| denotes the growth way of So∩Sj. Since n1−n>n0, for n0≤j≤n1−n,
|So∩∂Sx(j)|(gr(G)−ϵ)≤r∑i=1(m−mi)|Vij|≤|∂Sx(j+1)|. |
Write c=gr(G)−ϵ. Hence, for any n0≤j≤n1−n,
|So∩∂Sx(j)|≤1c|∂Sx(j+1)|=1c|So∩∂Sx(j+1)|+1c|∂S∩∂Sx(j+1)|≤1c(1c|So∩∂Sx(j+2)|+1c|∂S∩∂Sx(j+2)|)+1c|∂S∩∂Sx(j+1)|=(1c)2|So∩∂Sx(j+2)|+(1c)2|∂S∩∂Sx(j+2)|≤(1c)n1−n−j|So∩∂Sx(n1−n)|+(1c)n1−n−j|∂S∩∂Sx(n1−n)|. |
In the case when |∂S∩∂Sx(n1−n)|≥ϵ|∂Sx(n1−n)|,
|So∩∂Sx(n1−n)|≤(1−ϵ)|∂Sx(n1−n)|≤1−ϵϵ|∂S∩∂Sx(n1−n)|. |
Therefore, for any n0≤j≤n1−n,
|So∩Sx(j)|≤1ϵ(1c)n1−n−j|∂S∩∂Sx(n1−n)|. |
Notice that every vertex in So∩Sx(j) has weight λ−j(m+λ−1). Therefore,
C(So∩∂Sx(n0+1))≤m+λ−1λn+n0+11ϵ(1c)n1−n−n0−1|∂S∩∂Sx(n1−n)|,C(So∩∂Sx(n0+2))≤m+λ−1λn+n0+21ϵ(1c)n1−n−n0−2|∂S∩∂Sx(n1−n)|,C(So∩∂Sx(n1−n−1))≤m+λ−1λn1−11ϵ1c|∂S∩∂Sx(n1−n)|. |
Combining with that
C(So∩Sx(n1−n))=n1−n∑i=1C(So∩∂Sx(i)),C(∂S∩Sx(n1−n))=n1−n∑i=0C(∂S∩∂Sx(i)), |
we have that
C(So∩Sx(n1−n))≤C(S0∩Sx(n0))+n1−n−n0−1∑i=1m+λ−1ϵ(λc)iλ−n1|∂S∩∂Sx(n1−n)|. |
Since λ<gr(G)−ϵ,
n1−n−n0−1∑i=1(λc))i≤λc1−λc. |
Hence, we obtain that,
C(So∩Sx(n1−n))≤C(S0∩Sx(n0))+m+λ−1ϵλc1−λcλ−n1|∂S∩∂Sx(n1−n)|. |
Notice that
λ−n1|∂S∩∂Sx(n1−n)|≤C(∂Sx∩(Sx(n1−n)∖Sx(n0)). |
Thus,
C(∂Sx∩Sx(n1−n))C(Sx(n1−n))C(So∩Sx(n0))+C(So∩(Sx(n1−n)∖Sx(n0))C(∂S∩Sx(n0))+C(∂S∩(Sx(n1−n)∖Sx(n0))≥min{λ−n0mn0(m+λ−1),1m+λ−1ϵλc1−λc+1}>0. |
Notice that the left case is |∂S∩∂Sx(n1−n)|<ϵ|∂Sx(n1−n)|. However, we will discuss the case in the following ways. If n1−n≤n0, for x1∈∂Sx(n1), define
n2(x1)=inf{k>n1,there existsy∈∂S∩Sx1(k−n1)}. |
Then, similar to the proof of Sx(n1−n), we can discuss the case of Sx1(n2−n1). If n1−n>n0, define
n2=inf{n2(y),y∈∂Sx(n1)∩So}. |
Moreover, if |∂S∩∂Sx(n2−n)|≥ϵ|∂Sx(n2−n)|, then for n1≤j≤n2,
|So∩∂Sx(j)|(gr(G)−2ϵ)≤r∑i=1(m−mi)|Vij|≤|∂Sx(j+1)|. |
Similarly, as above, we have
C(∂S∩Sx(n2−n))C(Sx(n2−n))≥min{λ−n0mn0(m+λ−1),1m+λ−1ϵλc11−λc1+1}>0. |
Here c1=gr(G)−2ϵ. The left case is |∂S∩∂Sx(n1−n)|<ϵ|∂Sx(n1−n)| and |∂S∩∂Sx(n2−n1)|<ϵ|∂Sx(n2−n1)|. Notice that S is a finite graph; then there exists K>0 such that nK=n+m. And
∂S∩∂Sx(nK−n)=∂S∩∂Sx(m)=∂Sx(nK−n). |
Hence,
|∂S∩∂Sx(nK−n)|>ϵ|∂Sx(nK−n)|. |
Whatever, from the above discussion, we can divide Sx into several parts. S1x,S2x,⋯ such that every part satisfies C(∂S∩Six)C(Six)>0 (i∈N). Therefore,
C(∂S∩∂Sx)C(∂Sx)≥min{λ−n0mn0(m+λ−1),1m+λ−1ϵλcK−11−λcK−1+1}>0. |
Here cK−1=gr(G)−Kϵ. Hence, for λ∈(0,gr(G)−Kϵ),
C(∂S)C(S)>0. |
Note that K is decreasing as ϵ↓0. Thus, we prove that G is a positive Cheeger constant for λ∈(0,gr(G)), which completes the proof by Theorem 2.2.
Lemma 2.4. G(o,o|RG)<∞, U(o,o|RG)<1, and RG=RU.
Proof. Note that for any z>0
G(o,o|z)=1+∞∑i=1(U(o,o|z))i. |
It is easy to see that RG≤RU, and in |z|<RG.
G(o,o|z)=11−U(o,o|z). |
So U(o,o|RG)≤1.
Recall the following: Pringsheim's Theorem: If f(z)=∑∞n=0anzn with an≥0, then the radius of the convergence is the smallest positive singularity of f(z).
Hence, the smallest positive singularity RG of G(o,o|z) is either one of the radius of convergence RU of U(o,o|z) or the smallest positive number z1 with U(o,o|z1)=1. Whatever, notice that U(o,o|z) is strictly increasing for z. Therefore, z1 is the unique positive number satisfying U(o,o|z)=1. Thus, to prove the theorem, we only need to prove U(o,o|RG)<1.
Assume that U(o,o|RG)=1. We exclude the trivial case where mi=1 for 1≤i≤r. Notice that RG≥1. If RG=1, then U(o,o|1)<1 by transience. So the left-case is RG>1. Firstly, let us consider the case of 0<λ≤1. Recall that
U(o,o|z)=r∑i=1Ui(o,o|z)=r∑i=1−((m+λ−1)−mU(o,o|z)−(mi−1)z)2m+√((m+λ−1)−mU(o,o|z)−(mi−1)z)2+4λmiz22m. |
Thus
1=r∑i=1−((λ−1)−(mi−1)RG)+√((λ−1)−(mi−1)RG)2+4λmiR2G2m=r∑i=1((1−λ)+(mi−1)RG)+√((1−λ)+(mi−1)RG)2+4λmiR2G2m. |
Notice (1−λ)+(mi−1)RG≥(1−λ)+(mi−1). Since there is at least one mi>1, which implies (1−λ)+(mi−1)RG>(1−λ)+(mi−1), we have
1>r∑i=1((1−λ)+(mi−1))+√((1−λ)+(mi−1))2+4λmi2m=r∑i=1mi−λ+mi+λ2m=1. |
It is a contradiction. That is for 0<λ≤1, U(o,o|RG)<1.
Consider the case when λ>1. Notice that
[λ−1−(mi−1)z]2+4λmiz2=[λ−1+(mi+1)z]2+4λmiz2−4miz2−4(λ−1)miz. |
For λ>1,
4λmiR2G−4miR2G−4(λ−1)miRG=4RG(λmiRG−miRG−(λ−1)mi)=4RG((λ−1)miRG−(λ−1)mi)=4RG(λ−1)mi(RG−1)≥0. |
Since RG>1, we have
4λmiR2G−4miR2G−4(λ−1)miRG>0. |
Thus, we get the following contradiction:
1=r∑i=1−((λ−1)−(mi−1)RG)+√((λ−1)−(mi−1)RG)2+4λmiR2G2m=r∑i=1−((λ−1)−(mi−1)RG)2m+r∑i=1√((λ−1)−(mi+1)RG)2+4λmiR2G−4R2G+4(λ−1)miRG2m>r∑i=1−((λ−1)−(mi−1)RG)+((λ−1)+(mi+1)RG)2m=RG>1. |
Hence, for λ≥1, U(o,o|RG)<1.
Suppose that RG<RU. There exists ˜z with RG<˜z<RU such that U(o,o|˜z)<1 since U(o,o|RG)<1. Therefore, G(o,o|˜z)<∞. It is in contradiction with RG, which is the convergence radius of G(o,o|z).
Therefore, we complete the proof.
Lemma 2.5. For any λ0∈(0,gr(G)), limλ→λ0RG(λ)=RG(λ0).
Proof. We will employ proof by contradiction to establish the aforementioned theorem.
Fix a sequence {λk}k≥1 with λk↑λ0. Suppose lim supλ→λ0RG(λ)>RG(λ0)=z∗. Then we can find a subsequence nk with limk→∞RG(λnk)>z∗. Without loss of generality, we can assume z′=limk→∞RG(λk)>z∗. For a large enough k,
1>U(RG(λk),λk)=∞∑n=0f(n)(o,o,λk)RG(λk)n. |
Applying Fatou's lemma,
1>lim infk→∞U(RG(λk),λk)=lim infk→∞∞∑n=0f(n)(o,o,λk)RG(λk)n=∞∑n=0f(n)(o,o,λ0)z′n=∞. |
It is a contradiction. Hence, lim supλ→λ0RG(λ)≤RG(λ0), and especially lim infλ→λ0ρ(λ)≤ρ(λ0).
Specially lim supλ→gr(G)RG(λ)≤RG(gr(G))=1. Notice that for any λ, RG(λ)≥1. So limλ→gr(G)RG(λ)=RG(gr(G))=1.
Here we use f(n)(λ) to denote f(n)(o,o) and U(o,o,λ) to U(o,o|z). Let
Πn={all the paths with lengthnandτo=n}. |
Thus
f(n)(λ)=∑γ∈ΠnP(γ,λ). |
Here P(γ,λ)=∏ni=0p(wi,wi+1) for γ=w0w1⋯wn. Note that
p(v,u)={1/mif v=o,λm+λ−1if u∈∂BG(|v|−1) and v≠o,1m+λ−1otherwise. |
Given λ0∈(0,gr(G)) and z0. For any ϵ>0, RG(λ),RG(λ0), z<z0 with |z−z0|2+|λ−λ0|2<ϵ, there exists a 0<δ<√ϵ such that P(γ,λ)≤(1+δ)nP(γ,λ0). Hence, f(n)(λ)≤(1+δ)nf(n)(λ0). And there exists a δ1>0 such that
(1+δ)zz0≤(1+δ1)<Rλ0z0. |
Therefore,
U(o,o,λ)=Σ∞n=0f(n)(λ)zn≤Σ∞n=0(1+δ1)nf(n)(λ0)zn0<∞. |
If lim supλ→λ0RG(λ)<RG(λ0). We can find a subsequence nk such that limλnk→λ0RG(λ)<RG(λ0). Thus for any ϵ>0, let z=Rλ0−ϵ2. Since limλ→λ0RG(λ)<RG(λ0),
U(o,o,z)=∞. |
It is impossible. It means that limλ→λ0RG(λ)=RG(λ0).
For any vertex set A and Z, let τA=inf{n≥0|Xn∈A}. If RWλ starts at a vertex in A, then τA=0. Write τ+A=inf{n>0|Xn∈A}. τ+A is different from τA only when RWλ starts in A. Consider the probability of a RWλ starting at a vertex x that visits A before its visit Z:
Px(A→Z,λ)=Px(τZ<τ+A). |
For λ∈[0,λc], define θ(λ)=Po(τ+o(λ)<∞). Clearly, θ(λ)=U(o,o|1)=∑∞n=1f(n)(o,o,λ), and θ(0)=0. Suppose A={o} and (Gn)n≥1 be any sequence of finite subgraphs of G that exhaust G. That is Gn⊆Gn+1 and G=⋃Gn. And let Zn be the set of vertices in G∖Gn. So limn→∞Po(o→Zn,λ) is the probability of never returning to o. And limn→∞Po(o→Zn,λ) is independent of (Zn)n≥1 see [19] Exercise 2.4. We may regard the entire circuit between o and Zn as a single conductor of effective conductance Cc(o↔Zn). Recall [19] Chapter 2.2, Cc(o↔Zn)=π(o)Po(o→Zn,λ). Hence,
θ(λ)=1−limn→∞Po(o→Zn,λ). |
Recall the following: Rayleigh's monotonicity principle.
Theorem 3.1. Let H be a finite graph and A and Z be two disjoint subsets of its vertices. If c and c′ are two assignments of conductances on H with c≤c′, then Cc(A↔Z)≤Cc′(A↔Z).
Notice that c(x,y)=c(x,y,λ) is the edge weight of edge xy. And recall the definition of RWλ, c(x,y) is decreasing of λ. By Theorem 3.1, for λ1≤λ2,
Cc(λ1)(o↔Zn)≥Cc(λ2)(o↔Zn). |
Whatever, for o, π(o) can be any positive constant that is independent of λ. In fact, for any vertex x∈G, we can choose π(x)=∑y∼xc(x,y). Therefore, for λ1≤λ2,
limn→∞Po(o→Zn,λ1)≥limn→∞Po(o→Zn,λ2). |
And
θ(λ1)≤θ(λ2). |
The number of visits to vertex o prior to escape is modeled by a geometric random variable, which has an expected value or mean (1−θ(λ))−1. Note that the mean of τ+o(λ),
Eo(τ+o(λ))=∞∑n=1nf(n)(o,o,λ). |
Recall the Varopoulos-Carne bound ([19] Chapter 13.2, Theorem 13.4) of n-step transition probability. For any x, y,
p(n)(x,y,λ)≤2√π(y)/π(x)ρn. |
Clearly, if ρ<1, then
Eo(τ+o(λ))=∞∑n=1nf(n)(o,o,λ)<∞. |
Hence, for large enough n, by the Strong Law of Large Numbers,
♯{the number of visits o before timen}∼nEo(τ+o(λ)). |
Problem 3.2. Is Eo(τ+o(λ)) increasing for λ∈[0,gr(G)]?
lim supn→∞p(n)(o,o,λ)=Po(τ+o(λ)<∞)1Eo(τ+o(λ))? |
If both problems have a positive answer, then the spectral radius ρ is increasing with λ. And for λ∈[0,gr(G)), if Eo(τ+o(λ))<∞, then ρ<1.
Moreover,
Problem 3.3. For RWλ on the free product of complete graphs with λ∈[0,λc], is θ(λ) continuous and strictly increasing?
Recall the following result: [19] Chapter 6, Proposition 6.6.
Proposition 3.4. Consider a graph H with an upper exponent growth rate b>1. For a reversible Markov chain (Xn)n≥0 starting at o on its vertex set with reversible measure π(⋅) is bounded and π(o)>0 and ρ<1. Then, in the graph metric,
lim infn→∞|Xn|n>−lnρlnb. |
Hence, for RWλ on G with ρ<1, lim infn→∞|Xn|n>0. So the key, to answer Lyons-Pemantle-Peres monotonicity problem, is to prove the existence of the speed. Furthermore, R. Lyons, R. Pemantle, and Y. Peres [17] proved that on the lamplighter group Z⋉∑x∈ZZ2, which has a growth rate of (1+√5)/2, the speed of RWλ is 0 at λ=1, and is strictly positive when 1<λ<(1+√5)/2. We wonder whether the spectral radius ρ has a similar property or not. That is
Problem 3.5. For λ∈(0,1], ρ(λ)=1? And for 1<λ<(1+√5)/2, ρ(λ)<1?
Moreover, does a similar result hold for amenable groups with exponential growth?
Citing Chapter 9 of [14], it is demonstrated that the n-step transition probability is governed by the spectral radius ρ(λ), with the following upper bound holding:
p(n)(x,y)≤2√π(y)π(x)ρn. |
This inequality underscores the influence of the spectral radius; as the spectral radius decreases, the probability of transition between any two states x and y after n steps becomes more restricted. Given that the speed, a measure of how rapidly the random walk explores the graph, is inherently tied to the transition probabilities, a positive speed necessitates that p(n)(x,y) decreases over time, implicating a requirement for ρ(λ) to be sufficiently small. Based on these considerations, we posit that the existence of a positive speed aligns with the premise that the spectral radius ρ(λ) must be adequately restrained.
In the paper, we establish the continuity of the spectral radius ρ(λ) of RWλ on the free product of complete graphs, as a function of the parameter λ within the interval (0,gr(G)].
He Song, Longmin Wang, Kainan Xiang contributed to the conception of the study; He Song, Longmin Wang, Kainan Xiang, Qingpei Zang contributed significantly to analysis and manuscript preparation; He Song, Longmin Wang, Kainan Xiang, Qingpei Zang contributed to the writing of the manuscript. He Song, Longmin Wang, Kainan Xiang, Qingpei Zang helped perform the analysis with constructive discussions. All authors have read and approved the final version of the manuscript for publication.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors would like to thank the anonymous referees, the editors for their constructive comments that improved the quality of this paper.
KX's research is supported partially by National Natural Science Foundation of China (Nos. 11671216, 11871032) and by Hu Xiang, Gao Ceng, Ci Ren, Cai Ju, Jiao Gong, Cheng Chuang Xin, Ren Cai (No. 2019RS1057).
HS's research is supported partially by the Natural Science Foundation of Jiangsu Higher Education Institutions of China (No. 21KJB110003), Huai'an City Science and Technology Project (HAB202357) and by Xiang Yu, Ying Cai (No. 31SH002).
QZ's research is supported partially by National Natural Science Foundation of China (Nos. 11971197).
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1. | He Song, Longmin Wang, Kainan Xiang, The speed of a biased walk on a Galton–Watson tree without leaves is monotonic for low values of bias, 2025, 0021-9002, 1, 10.1017/jpr.2024.113 |