给定片段数的树、单圈图和双圈图的极值p-谱半径.pdf

第 43 卷第 3 期2023 年 9 月数学理论与应用MATHEMATICAL THEORY AND APPLICATIONSVol.43No.3Sep.2023The Extremalpspectral Radii of Trees,Unicyclic and Bicyclic Graphswith Given Number of SegmentsQiu MairongHe Xiaocong(School of Mathematics and Statistics,Central South University,Changsha 410083,China)AbstractLet G be a finite and simple graph.A walk S is called a segment of G if the endpoints(not necessarilydistinct)of S are of degree 1 or at least 3,and each of the rest vertices is of degree 2 in G.In this paper,we determinethe graphs that maximize the pspectral radius for p 1 among trees,unicyclic and bicyclic graphs with given orderand number of segments,respectively.Key wordspspectral radiusTreeUnicyclic graphBicyclic graphSegment给定片段数的树、单圈图和双圈图的极值p谱半径邱买容贺晓聪*(中南大学数学与统计学院,长沙,410083)摘要设 G 是一个有限简单图.S 是 G 的一条途径.如果 S 的端点(可以相同)在 G 中的度是 1 或者至少是 3,且其他的顶点在 G 中的度都是 2,则称 S 为 G 的一个片段.本文对大于 1 的实数 p,分别确定固定阶数和片段数的树、单圈图和双圈图的最大 p谱半径,并刻画对应的极图.关键词p谱半径树单圈图双圈图片段doi:10.3969/j.issn.1006 8074.2023.03.0031IntroductionAssume all graphs in this paper are finite and simple.Let G be a graph with vertex set V(G)=v1,v2,vn and edge set E(G).We follow the notation and terminologies in 1 except otherwisestated.If two vertices viand vjare adjacent,we write vi vj,otherwise,write vi vj.The adjacencymatrix of G is defined as A(G)=(aij)nnwhere aij=1 if vi vj,and aij=0 otherwise.Let dG(vi)be the degree of viand NG(vi)be the set of neighbors of viin G.As usual,denote by(G)the maximumdegree in G.The distance between two vertices u and v,denoted by d(u,v),is the length of a shortestpath between u and v.If H is a subgraph of G,let d(u,H)=mind(u,v)|v V(H).Denote by Pnand Cnthe path and the cycle on n vertices,respectively.k paths Pl1,Pl2,Plkaresaid to have almost equal lengths if|li lj|1 for 1 i,j k.A vertex of degree 1 in a graph is saidThisworkissupportedbytheFundamentalResearchFundsfortheCentralUniversitiesofCentralSouthUniversity(No.2021zzts0034)Corresponding author:He Xiaocong Email:收稿日期:2022 年 10 月 31 日62数学理论与应用to be a pendant vertex and a vertex of degree at least 3 is called a branching vertex.An induced path withvertices u1,u2,utis called a pendant path of G,if dG(u2)=dG(ut1)=2 and dG(ut)=1(note that there is no requirement on the degree of u1).A walk is a sequence of vertices u1,u2,ukofG such that ui1 uifor i=2,k,where the first vertex u1and last vertex ukare called endpoints.A walk S is a segment of a graph if the endpoints(not necessarily distinct)of S are of degree 1 or at least3,and each of the rest vertices is of degree 2 in the graph.Let m=|E(G)|.If G is connected and m=n+c1,then G is called a ccyclic graph.Especially,if c=0,1 and 2,then G is called a tree,unicyclic graph and bicyclic graph,respectively.For c 1,Thebase of a ccyclic graph G,denoted byG,is the minimal ccyclic subgraph of G.It is easy to know thatG is the unique ccyclic subgraph of G containing no pendant vertices,while G can be obtained fromGby attaching trees to some vertices ofG.The pspectral radius was introduced for uniform hypergraphs by Keevash,Lenz and Mubayi in 10andsubsequentlystudiedin5,8,17,18.ForasimplegraphG,letx=(xvi)Rn,wherexvicorrespondstothevertexvi.ThequadraticformofGisdefinedasPG(x):=2vivjE(G)xvixvj=xTA(G)x.Usingthe definition of quadratic form of graphs,for any real number p 1,the pspectral radius of G is definedas(p)(G)=maxxp=1PG(x)=maxxp=12vivjE(G)xvixvj.A vector x Rnsatisfying xp=1 and(p)(G)=PG(x)is called an eigenvector to(p)(G).Forany p 1,if x is a vector such that xp=(ni=1|xvi|p)1p=1 and(p)(G)=PG(x),then thevector x=(|xvi|)also satisfies xp=1,and so(p)(G)=PG(x)PG(x)(p)(G)implies that(p)(G)=PG(x).Therefore,there is always a nonegative vector x such that xp=1 and(p)(G)=PG(x).Let x be an eigenvector to(p)(G).If p 1,by Lagranges method,one can show that(p)(G)xp1vk=viNG(vk)xvifor k=1,2,n,which is called the eigenequation of(p)(G)for the vertex vk.The pspectral radius(p)(G)is a multifaceted parameter,as(1)(G)/2 is the Lagrangian of G(see 19).Especially,(2)(G)is the usual spectral radius introduced by Cooper and Dutle 4,andlimp(p)(G)=2|E(G)|(see 17).The investigation on the spectral radius of graphs is an importanttopic in the theory of graph spectra.Recently,the problem concerning graphs with maximal or minimalspectral radius of a given class of graphs has been studied extensively.Brualdi and Solheid 2 studiedthe spectral radius of connected graphs.Guo 7 determined graphs with the largest spectral radius amongall the unicyclic graphs and all bicyclic graphs on n vertices and k pendant vertices,respectively.Lin andZhou 13 obtained the structure of trees that maximize the distance spectral radius among trees with givenorder and number of segments.Lin and Song 16 considered the family of all trees with given number ofsegment sequences and proved that the generalized star minimizes the Wiener index.It should be notedthat extremal problems for(p)(G)mainly focus on hypergraphs(see 4,5,8,9,11,17,18).Kang,Liu,给定片段数的树、单圈图和双圈图的极值 p谱半径63Lu and Wang 8 characterized the 3uniform hypergraphs with maximum pspectral radius for p 1among the BergeG hypergraphs when G is a path,a cycle or a star.For more advances in research onspectral radius of graphs,we refer the readers to 3,6,7,12 15.Let k be the number of segments in G and be the number of vertices of degree 2 in G.If S is asegment of G,we write S G.Denote the length of a segment S by lSand the number of vertices ofdegree 2 in S by S.It is easy to know thatSGlS=m(G)and S=lS 1.Then=SGS=SG(lS 1)=SGlS k=m(G)k.Therefore,in the class of ccyclic nvertex graphs,as long as the number of segments of graphs is given,the number of vertices of degree 2 is determined.Based on this observation,in this paper,we determinethe structure of trees,unicyclic and bicyclic graphs attaining the maximum pspectral radius with givennumber of segments,respectively.Let G(n,c,k),T(n,k),U(n,k)and B(n,k)be the set of ccyclic graphs,trees,unicyclic graphs andbicyclic graphs with n vertices and k segments,respectively.Obviously,T(n,1)=Pnand T(n,0)=T(n,2)=U(n,0)=Cnand U(n,1)=B(n,0)=B(n,1)=.Thus,we only need to considertrees in T(n,k)for 3 k n1,unicyclic graphs in U(n,k)for 2 k n,bicyclic graphs in B(n,k)for 2 k n+1.Let Tn,kbe an nvertex tree with exactly one branching vertex formed by identifying one endvertexof each of the k paths with almost equal lengths(see Figure 1 for it and all the following graphs).Let(i,j,k)be an nvertex unicyclic graph obtained from C3by attaching i,j and k pendant vertices to eachvertex of C3,respectively.We write knfor a unicyclic graph on n vertices obtained from C3by attachingk paths of almost equal lengths to one vertex.Let Bbe a bicyclic graph with 4 vertices and Bbe abicyclic graph with 5 vertices and exactly one branching vertex.Let Bkbe a bicyclic graph obtained byattaching k pendant paths of almost equal lengths to one vertex of degree 3 in B.Let Bkbe a bicyclicgraph obtained by attaching k pendant paths of almost equal lengths to the vertex of degree 4 in B.LetH1be attained from Bby attaching k7 pendant vertices to one vertex of degree 3 of B,and attachinga pendant vertex to the vertices of degree 2 in B,respectively.Let H2be attained from Bby attachingk 5 pendant vertices to one vertex of degree 3 of B,and attaching a pendant vertex to one vertex ofdegree 2 in B.We can now state our main results.Theorem 1.1Let p 1,3 k n 1 and T T(n,k).Then(p)(T)1.For any U U(n,k),we have(1)if n=k and k 6,then(p)(U)(p)(k 5,1,1),unless U=(k 5,1,1)(2)if n=k+1 and k 4,then(p)(U)(p)(k 3,1,0),unless U=(k 3,1,0)(3)if n k+2 and k 2,then(p)(U)1.For any B B(n,k),we have(1)if n=k 1 and k 7,then(p)(B)(p)(H1),unless B=H1(2)if n=k and k 5,then(p)(B)(p)(H2),unless B=H2(3)if n=k+1 or k+2 and k 4,then(p)(B)(p)(Bk3),unless B=Bk3(4)if n k+3 and k 4,then(p)(B)max(p)(Bk2),(p)(Bk3),unless either B=Bk2or B=Bk3(5)if n 6 and k=3,then(p)(B)(p)(B1),unless B=B1(6)ifn 5andk=2,then(p)(B)1 and x Sn1p,+satisfies the equationsxp1vk=viNG(vk)xvi,k=1,2,nfor =(p)(G),then xv1,xv2,xvnare positive(b)If p 2,then there is a unique positive eigenvector x to(p)(G)(c)If p 2 and a vector x Sn1p,+satisfies the above equations for some,then =(p)(G).给定片段数的树、单圈图和双圈图的极值 p谱半径65Lemma 2.2(8)Let p 1 and G be a connected graph.Let u,v be two vertices of G andv1,v2,vs N(v)N(u)(1 s dG(v).Let x=(x1,x2,xn)Tbe a nonegative eigenvectorof(p)(G)with xp=1,where xicorresponds to the vertex vi(i=1,2,n).Let Gbe the graphobtained from G by deleting the edges vviand adding the edges uvi,1 i s.If xu xv,then(p)(G)(p)(G).Furthermore,if p 1,then(p)(G)(p)(G(u;k+1,s 1)for p 1.Suppose a,b are two adjacent vertices of G with dG(a)2 and dG(b)2.Let G(1)(k,s)be thegraph obtained from G by attaching a pendant path of length k at vertex a and attaching a pendant path oflength s at vertex b,respectively.Similar to the result of Lemma 2.3,we haveLemma 2.4Let G be a connected graph.If k s 1,then(p)(G(1)(k,s)(p)(G(1)(k+1,s 1)for p 1.Proof Suppose(p)(G(1)(k,s)(p)(G(1)(k+1,s 1).Let x be a positive eigenvector withxp=1 corresponding to(p)(G(1)(k+1,s 1).Let au1ukuk+1and bv1vs2vs1be twopendant paths at a and b of lengths k+1 and s 1 in G(1)(k+1,s 1),respectively.For convenience,let u0=a,v0=b.We will prove by induction thatxuki xvsi1,i=0,1,s 1.For i=0,if xuk xvs1,let G1=G(1)(k+1,s 1)ukuk+1+vs1uk+1.Note thatG1=G(1)(k,s).By Lemma 2.2,we have(p)(G(1)(k,s)=(p)(G1)(p)(G(1)(k+1,s 1),which is a contradiction.Hence,xuk xvs1.For i 1,suppose xuk(i1)xvs1(i1).We will show that xuki xvs1i.Let G2=G(1)(k+1,s1)ukiuk(i1),vsi1vsi+vsi1uk(i1),ukivsi.Clearly,G2=G(1)(k,s).Since0 (p)(G(1)(k,s)(p)(G(1)(k+1,s 1)=(p)(G2)(p)(G(1)(k+1,s 1)2vwE(G2)xvxw 2vwE(G(1)(k+1,s1)xvxw=2(xuk(i1)xvsi)(xvsi1 xuki),66数学理论与应用by the induction assumption,we obtain xuki xvs1i.If xuki=xvs1i,then(p)(G2)=(p)(G(1)(k+1,s 1)and x is also an eigenvector of(p)(G2).Using the eigenequations for uki,we have0=(p)(G2)(p)(G(1)(k+1,s 1)xp1uki=xvsi xuk(i1)xvs1i.So,xuki xvs1ifor i=0,1,s 1.In particular,xuk(s1)xv0=xb.Case 1 k s 1.Note that uks+1 a.Let G3=G(1)(k+1,s 1)ua|u NG(a)b ub|u NG(b)a+uuks|u NG(a)b+uuks+1|u NG(b)a.Clearly,G3=G(1)(k,s).For convenience,let =uNG(a)bxu,and =uNG(b)axu.By calculation,we have0 (p)(G(1)(k,s)(p)(G(1)(k+1,s 1)=(p)(G3)(p)(G(1)(k+1,s 1)2(xuks+1 xb)+(xuks xa).Since 0,0 and(xuks+1 xb)0,we obtain xuks xa xuks.Let G4=G(1)(k+1,s1)uksuks+1,abub|u NG(b)a+uks+1a,uksb+uuks+1|u NG(b)a.Clearly,G4=G(1)(k,s).We have0 (p)(G(1)(k,s)(p)(G(1)(k+1,s 1)=(p)(G4)(p)(G(1)(k+1,s 1)2(xaxuks+1+xbxuks xaxb xuksxuks+1)+2(xuks+1 xb)2(xaxuks+1+xbxuks xaxb xuksxuks+1)=2(xuks+1 xb)(xa xuks)0,which is a contradiction.Case 2 k=s.Note that xuks+1=xu1 xb.Let G5=G(1)(k+1,s1)ub|u NG(b)a+uu1|u NG(b)a.Clearly,G5=G(1)(k,s).By Lemma 2.2,we have(p)(G(1)(k,s)=(p)(G5)(p)(G(1)(k+1,s 1),which is a contradiction.This completes the proof.Suppose a,b,w are three vertices of G such that dG(a)2,dG(b)2,dG(w)=2,and a w,b w,a b.Let G(2)(k,s)be the graph obtained from G by attaching a pendant path of length k atvertex a and attaching a pendant path of length s at vertex b,respectively.给定片段数的树、单圈图和双圈图的极值 p谱半径67Lemma 2.5Let G be a connected graph.If k1 s 1,then(p)(G(2)(k,s)(p)(G(2)(k+1,s 1)for p 1.Proof Suppose(p)(G(2)(k,s)(p)(G(2)(k+1,s 1).Let x be a positive eigenvector withxp=1 corresponding to(p)(G(2)(k+1,s 1).Let au1ukuk+1and bv1vs2vs1be twopendant paths at a and b of lengths k+1 and s 1 in G(2)(k+1,s 1),respectively.For convenience,let u0=a,v0=b.Similarly,as the proof of Lemma 2.4,we have xuks+1 xv0=xb.Case 1 k s 2.Note that uks a.Let G1=G(2)(k+1,s 1)ua|u NG(a)w ub|u NG(b)w+uuks1|u NG(a)w+uuks+1|u NG(b)w.Clearly,G1=G(2)(k,s).For convenience,let =uNG(a)wxuand =uNG(b)wxu.By calculation,we have0 (p)(G(2)(k,s)(p)(G(2)(k+1,s 1)=(p)(G1)(p)(G(2)(k+1,s 1)2(xuks1 xa)+2(xuks+1 xb).Since 0,0,we have xa xuks1.We proceed by considering the following two subcases.Subcase 1.1 xw xuks.Let G2=G(2)(k+1,s1)uksuks+1,wbub|u NG(b)w+uks+1w,uksb+uuks+1|u NG(b)w.Clearly,G2=G(2)(k,s).However,(p)(G2)(p)(G(2)(k+1,s 1)2(xuks+1 xb)+2(xuks+1 xb)(xw xuks)0,which contradicts to our assumption.Subcase 1.2 xw 0,which also contradicts to our assumption.Case 2 k=s+1.Note that xu2=xuks+1 xb.Let G4=G(2)(k+1,s1)ub|u NG(b)w+uu2|u NG(b)w.Clearly,G4=G(2)(k,s).By Lemma 2.2,we have(p)(G(2)(k,s)=(p)(G4)(p)(G(2)(k+1,s 1),which is a contradiction.This completes the proof.Lemma 2.6Let G be a connected graph with uv E(G)and dG(u),dG(v)2.Suppose u and vhave no common neighbors.Let Gbe the graph obtained from G by deleting the edge uv,identifying uand v as a new vertex w and attaching a pendant vertex to w.Then for p 1(p)(G)(p)(G).68数学理论与应用Proof LetG1=G zv|z NG(v)u+zu|z NG(v)u,if xu xvG zu|z NG(u)v+zv|z NG(u)v,if xu(p)(G).In the following proofs,we will frequently use the following Lemma.Lemma 2.7Let G be a connected graph such that u1v1 E(G),u2v2 E(G),u1u2/E(G),v1v2/E(G),p 1.Let G=Gu1v1,u2v2+v1v2,u1u2.Let x be a positive eigenvectorof(p)(G)with xp=1.If xv1 xu2,xv2 xu1,then(p)(G)(p)(G).Moreover,if one of the two inequalities is strict,then(p)(G)(p)(G).Proof By(p)(G)=maxyp=1PG(y)PG(x),we have(p)(G)(p)(G)PG(x)PG(x)=2(xv1xv2+xu1xu2 xu1xv1 xu2xv2)=2(xv1 xu2)(xv2 xu1)0,i.e.,(p)(G)(p)(G).If(p)(G)=(p)(G),then x must be an eigenvector of(p)(G).From the eigenequations of(p)(G)in vertices v1and v2,we have(p)(G)xp1v1=uNG(v1)xu=xu1+uNG(v1)v2,u1xu xv2+uNG(v1)v2,u1xu=uNG(v1)xu=(p)(G)xp1v1,and(p)(G)xp1v2=uNG(v2)xu=xu2+uNG(v2)v1,u2xu xv1+uNG(v2)v1,u2xu=uNG(v2)xu=(p)(G)xp1v2.Therefore,xv1=xu2,xv2=xu1.This completes the proof.给定片段数的树、单圈图和双圈图的极值 p谱半径69Lemma 2.8Let G be a graph among G(n,c,k)obtaining the maximum pspectral radius,wherec 0,p 1.Let u and v be two vertices of G.Then we have(a)if dG(u)dG(v),then xu xv(b)if xu xv,then dG(u)dG(v)(c)if xu=xv,then dG(u)=dG(v).Proof(a)Suppose to the contrary that xu xv.Note that there exists a shortest path Puvbetweenu and v.Let dG(u)dG(v)=r.Notice that r 0.Then there must exist w1,wr NG(u)NG(v)and w1,wr/V(Puv).Let G1=G wiu|i=1,r+wiv|i=1,r.Clearly,G1 G(n,c,k).By Lemma 2.2,we have(p)(G1)(p)(G),which is a contradiction.Hence(a)holds.The statement of(c)is immediately from(a).Now,we prove(b).Otherwise,if dG(v)dG(u)=r 0,by the statement of(a),we have xv xu,which is a contradiction.Hence(b)holds.Definition 2.1Let G be a ccyclic graph with c 1 and u V(G).A closer vertex v of u is avertex adjacent to u satisfying d(v,G)=d(u,G)1.A further vertex w of u is a vertex adjacent to usatisfying d(w,G)=d(u,G)+1.Lemma 2.9Let G be a ccyclic graph among G(n,c,k)obtaining the maximum pspectral radius,where c 1.Then dG(v)2 for any v V(G)V(G).Proof If the conclusion is not ture,then there is a vertex v V(G)V(G)such that dG(v)3.Let u be the closest vertex to v in V(G).Let Puvbe the shortest path connecting u and v.Clearly,V(Puv)V(G)=u.Note that dG(u)3.Let x be a positive eigenvector of(p)(G).LetG1=G vw|w NG(v)V(Puv)+uw|w NG(v)V(Puv),if xu xvG uw|w NG(u)V(Puv)+vw|w NG(u)V(Puv),if xv xu.Obviously,G1 G(n,c,k).By Lemma 2.2,(p)(G1)(p)(G),which is a contradiction.Lemma 2.10Let G be a ccyclic graph among G(n,c,k)obtaining the maximum pspectral radius,wherec 1.LetP=v0v1vqbeapendantpat