Liu Keqiang, Zhao Biao

(College of Mathematics & System Sciences,Xinjiang University,Urumqi 830046,Xinjiang,China)

0 Introduction

The chemical applicability of theGA1index was examined documented in detail in the paper [8] and the reviews [9-10]. Upper and lower bounds forGA1index of general graphs have been given in [4,11-12]. In this paper, the sharp upper and lower bounds onGA1index of line graph are obtained.

LetE′={(u1v,vu2)|u1v,vu2∈E(G)}, call graph (E(G),E′) as line graph ofG, denotedL(G). Denote the edge ofL(G) by (u1,v,u2), both (u1,v,u2) and (u1,v,u2) demonstrate the same edge. LetP={(u1,v,u2)∈E(L(G))|u1v,vu2∈E(G),d(u1)=1,d(v)=2} be the set of pendant edges,Γ={H|u1,v,u2∈V(H) andu1v,vu2∈E(H) such thatd(u1)=1,d(v)=2,d(u2)=Δ=δ1} be the class of graphH.

1 Some lemmas

Lemma1[13]Let (a1,a2,…,an) be the positiven-tuples such that there exist positive numbersaandAsatisfying 0

(1)

Lemma4[14]LetGbe a graph withnvertices andmedges,anddvdenote the degree of a vertexvinG，then

ii) the degree inL(G) of an edgeuvofGisd(u)+d(v)-2.

2 Bounds for GA1 index of line graph

Theorem1LetGbe a simple connected graph of ordernwithmedges, maximum vertex degreeΔand minimum vertex degreeδ,δ≥2， then

GA1(L(G))≥GA1(G),

(2)

with equality holding if and only ifG?Cn.

ProofDefine the index

For 2≤d(u1),d(v),d(u2)≤Δ, we have

(3)

with equality holding if and only ifd(u2)=d(v)=Δ=2.Thus

(4)

Forv∈V(G), we have

(5)

Thus

(6)

Now suppose that equality holds in (2),then all inequalities in the above argument must be equalities, from equality (5) and (6), we have forv∈V(G),d(v)=2, that isG?Cn.

Conversely, we can see easily that the equality holds in (2) forG?Cn.

Theorem2LetGbe a simple connected graph of ordernwithmedges,ppendent edges, maximum vertex degreeΔ, minimum vertex degreeδ1of non-pendent edge, (d1,d2,…,dm) be a degree sequence ofL(G). Then

(7)

with equality holding if and only ifGis isomorphic to a regular graph orG∈Γ.

ProofDefine the index

For brevity, write

Forδ1≤d(u1),d(v),d(u2), we have

This implies

and finally we get

(8)

Now since

Note that |E′-P|=e(L(G))-p, using (1), we get

(9)

BecauseP={(u1,v,u2)|u1v,u2v∈E(G),d(u1)=1,d(v)=2}， then |P|=p, we have

(10)

Thus

(11)

(12)

with equality holding if and only ifΔ=δ1(for allv∈V″, thend(v)=(2m-3p)/(n-2p)).

Forδ1≤d(u2)≤Δ, we have

(13)

From (10)-(13), we get (7).

Suppose that equality holds in (7), then all inequalities in the above argument must be equalities. Now we consider two case:

i) ifp=0, from equality in (8), we getΔ=δ1. ThenGis isomorphic to a regular graph.

ii) ifp>0, from equality in (10) and (13), we getd(u1)=1,d(v)=2,d(u2)=Δ=δ1, thusG∈Γ.

Conversely, one can easily see that the equality holds in (7) for regular graph orG∈Γ.

Theorem3LetGbe a simple connected graph of ordernwithmedges,ppendant edge. Then

(14)

with equality holding in (14) if and only ifGis isomorphic to the complete graphK3orG?P3orG?P4.

ProofFor each pendant edge (u1,v,u2)∈E′,we have eitherd(u1)=1,d(v)=2 ord(u2)=1,d(v)=2.Thus

and finally we get

(15)

with equality holds in (15) if and only ifd(u1)=1,d(v)=2,d(u2)=n-2.

For each non-pendant edge (u1,v,u2)∈E′， we have 2≤d(u1),d(v),d(u2)≤n-1. Thus

then

(16)

with equality holds in (16) if and only ifd(u1)=2,d(v)=2,d(u2)=n-1.Thus

Suppose that equality holds in (14), then all inequalities in the above argument must be equalities. Now we consider two case

i) ifp=0, from equality (16), we getd(u1)=2,d(v)=2,d(u2)=n-1. Then we have a common neighborv. ThusGis isomorphic to the complete graphK3.

ii) ifp>0, from equality (16), we getd(u1)=1,d(v)=2,d(u2)=n-2. Firstly, assume thate(L(G)) =p, then all the edges are pendant edges inL(G), and henceGis isomorphic toP3orP4. Next, assume thate(L(G))>p. In this case, the maximum degree vertex， sayu2，by equality in (16) has degreen-1. There exists at least one non-pendant edge inG, and hence two vertices， sayviandvj， are adjacent to vertexu2. By equality in (16), for the non-pendant edgevivj∈E(G), eitherdi=n-1 ordj=n-1. Thus do not have any pendant vertex inGasdi=n-1, that is a contradiction.

Conversely, we can easily see that the equality holds in (14) for the complete graphK3orG?P3orG?P4.

Theorem4LetGbe a simple connected graph withnvertices,medges,ppendant edges,maximum vertex degreeΔand minimum vertex degreeδ1of non-pendant edge,(d1,d2,…,dm) be a degree sequence ofL(G).Then

(17)

with equality holding in (17) if and only ifGis isomorphic to a regular graph orG∈Γ.

ProofFor each pendant edge (u1,v,u2)∈E′,we have eitherd(u1)=1,d(v)=2 ord(u2)=1,d(v)=2.By Lemma 1,we have

(18)

with equality holds in (18) if and only ifd(u1)=1,d(v)=2,d(u2)=δ1.

Sincee(L(G))-pis the number of non-pendant edges inL(G). By Cauchy-Schwarz Inequality, we get

(19)

Note that |V″|=n-2p, by lemma 4, we have

(20)

with equality holds in (20) if and only if for ?v∈V″,d(v)=Δ.

From (18),(19) and (20), we get (17).

Suppose that equality holds in (17), then all inequalities in the above argument must be equalities. Now we consider two case：

i) ifp=0, from equality (19),(20), we getd(u1)=d(v)=d(u2)=Δ=δ1.That is，Gis isomorphic to a regular graph.

ii) ifp>0, from equality (18)-(20), we getd(u1)=1,d(v)=2,d(u2)=Δ=δ1. That is，G∈Γ.

Conversely, one can easily see that the equality holds in (17) for a regular graph orG∈Γ.

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