# Some New Upper Bounds for theY-Index of Graphs

Authors: Durbar Maji; Ganesh Ghorai; Faria Ahmed Shami
Journal: Journal of Mathematics
 (2022)

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## Abstract

In mathematical chemistry, the topological indices with highly correlation factor play a leading role specifically for developing crucial information in QSPR/QSAR analysis. Recently, there exists a new graph invariant, namely,Y-index of graph proposed by Alameri as the sum of the fourth power of each and every vertex degree of that graph. The approximate range of the descriptors is determined by obtaining the bounds for the topological indices of graphs. In this paper, firstly, some upper bounds for the Y-index on trees with several types of domination number are studied. Secondly, some new bounds are also presented for this index of graphs in terms of relevant parameters with other topological indices. Additionally, a new idea on bounds for the Y-index by applying binary graph operations is computed.


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## Body

## 1. Introduction

In this paper, we only consider the molecular graphsMG [1], which are simple and connected. In chemical graph theory, molecules or molecular compounds are often modelled by chemical structure as MG. The atoms of a molecular compound are to be represented as the vertices of the MG, whereas the edges represent the chemical bonds. Let G=GV,E be a MG with VG=μ1,μ2,…,μn as the vertex set and EG=e1,e2,…,em as the edge set, such that VG=n and EG=m. The degree of μ∈VG, denoted by ξμ/G, is the total number of edges, which are associated with μ. Obviously, 0≤δ≤ξμ/G≤Δ≤n−1, where δ=minξμ/G|μ∈G and Δ=maxξν/G|ν∈G. A set S⊆VG satisfying the condition ∀μ∈VG∖S, NGμ∩S≠∅ is called a dominating set of G. When ℛ⊆Vℛ satisfies the condition ∀μ∈Vℛ∖ℛ, ξℛμ,ν≤κ for some ν∈ℛ, where ξℛμ,ν denotes the distance between μ and ν, is said to be a distance κ-domination DDk set of ℛ. The DDk number ([2, 3]) of G, denoted by ζκ, is the minimum cardinality among all ℛ sets. The notations diamG=maxεGμ|μ∈VG and εGμ=maxξGμ,ν|ν∈VG denote the diameter of G and the eccentricity of μ, respectively. A path P is called a diameter path DP of G when the length of P is equal to diamG. We follow the book [4] for the graph theoretical definitions and notations.The graph invariantGI is a number that is uniquely determined by a graph. The subset of GIs is topological indices, which are used to predict several properties such as physical, chemical, pharmaceutical, and biological activities of chemical species. In 1947, the great chemist holder Wiener initiated a first-time idea about the topological index TI. He presented the first TI, namely, Wiener index [5] to search the boiling points of alkanes. After long years, Gutman and Trinajstic [6] investigated two oldest GIs. They are defined as M1G=∑μ∈VGξ2μ/G and M2G=∑μν∈EGξμ/Gξν/G, respectively. The concept of first general ZI was considered by Li and Zheng [7]. It is defined as(1)M1λG=∑μ∈VGξλμG=∑μν∈EGξλ−1μG+ξλ−1νG,where λ∈ℜ−0,1. For λ=3, it becomes forgotten topological index proposed by Furtula et al. [8]. It is presented as(2)FG=∑μ∈VGξ3μG=∑μν∈EGξ2μG+ξ2νGG.Liu et al. ([9]) introduced the reformulated F-index of G as follows:(3)RFG=∑e∈EGξ3eG=∑e∼f∈EGξ2eG+ξ2fG.In [10], Milicevic et al. introduced the first reformulated Zagreb index of a graph G. It is defined as(4)EM1G=∑e∈EGξ2e,whereξe=ξμ/G+ξν/G−2. Recently, Alameri et al. [11] introduced a GI named Y-index YI and defined as(5)YG=∑μ∈VGξ4μG=∑μν∈EGξ3μG+ξ3νG.TheYI is the special case of the first general Zagreb index for λ=4.In this study, we obtain some new upper boundsUB for the YI in terms of different graph parameters, on ζκT for tree of vertex n and TIs. We arrange the remaining work as follows: Section 2 contains the UB for the YI on trees with ζκ. Section 3 contains some UB for YI in behavior of some relevant parameters. Section 4 collects some UB for YI under several graph operations. Finally, Section 5 presents the conclusions of the obtained results. To know more related to this field, readers are referred to [12–17].


## 2. Preliminaries

To establish the main results, the following lemmas are required.Lemma 1.
[18]). Let T be a n>3 vertex tree with e=p1p2∈ET, a nonpendant edge. Suppose the union of T1 and T2 is equal to T−p1p2, where pi∈Ti for i∈1,2. Let T be a new tree obtained by taking an edge joining transformation (EJT) of T on e. It is attained by identifying p1∈T1 with p2∈T2 and also joining a pendent vertex s to the p=p1=p2. In short, we denote T´=ϕT,p1p2. Then, we get YT´<YT.Lemma 2.
[19]). If G is an n vertex MG graph with n=κ+1, then ζκG=⌊n/κ+1⌋.Lemma 3.
[20]). Let ℋ and T be two trees with n and κ+1n vertices, respectively. Then, ζκT=n holds iff at least one of following conditions is satisfied:(1)
T is any κ+1-vertex tree.(2)
T is equal to ℋ°κ obtained by taking ℋ and n copies of Pκ−1 and then link with the jth vertex of ℋ to exactly one end vertex in the jth copy of Pκ−1.Lemma 4.
[2]). If G contains the maximum value of the ZI s among all MG s of n-vertices with ζκG and SG=μ∈VG|ξμ/G=1,ζκG−μ=ζκG. If SG≠∅, then NGSG=1.Lemma 5.
[2]). Suppose μ and ν be two vertices in G such that p1,p2,…,pr and q1,q2,…,qt pendent vertices adjacent to p and q respectively. Define G′=G−qq1,qq2,…,qt+pq1,pq2,…,pqt and G′′=G−pp1,pp2,…,ppr+qp1,qp2,…,qpr. Then either MiG′ is greater than MiG or MiG′′ is greater than MiG,i=1,2.Lemma 6.
[2]). Let T be a tree of order n with Δ and ζκ≥2. Then κζκT≤n−ΔT.Lemma 7.
[21]). Suppose μν is any edge of G with n vertices. Then, for any integer t≥2(i)
ξμ/G,ξν/GSG,t=nt−2n−ξμ/G−ξν/G+⊳μ,ν(ii)
ξμ/G,ξν/G+1SG,t=nt−2ξμ/G−⊳μ,ν−βnt−2ξμ/G(iii)
ξμ/G+1,dν/GSG,t=nt−2ξμ/G−⊳μ,ν−βnt−2ξν/G(iv)
ξμ/G+1,ξν/G+1SG,t=nt−2⊳μ,ν+1+βnt−2ξμ/G+ξν/G+1.Lemma 8.
[22]). (Radon’s inequality) Let x=xii=1n and y=yii=1n be two sequences of positive real numbers. For any α≥0,(6)∑i=1nxiα+1yiα≥∑i=1nxiα+1∑i=1nyiα.where the equality occurs for xi=pyi for some constant p, for all i=1,2,…,n.


## 3. Main Results and Discussions

### 3.1. SomeUB for the YI on Trees with DDκ Number

In this section, we establish some sharpUB for the YI of graphs on the trees as to the DDκ number, ζκ. The set of all n vertex trees with ζκ and the star of order n−κζ+1 with u1,u2,…,un−κζκ pendent vertices are denoted as Tn,κ,ζκ and Sn−κζκ+1, respectively.Theorem 1.
LetTbe a tree of ordernand it containsζκT=2; then theUBofYTcan be expressed asYT≤n−2κ4+n−2κ+1+16κ+1+16κ−15. The equality holds forT⇔Tn,κ,2j, wherej∈1,2,…,k.Proof.
LetT∈Tn,κ,2 be a tree with a DP such that P:u0,u1,…,ud. For d≤2κ, there exists a DDκ set μ⌊d/2⌋ of T, a contradiction. In case d≥2κ+2, also denoted by T´=ϕT,uiui+1 the tree obtained from T by EJT (Lemma 1) on the edge uiui+1 for some i∈1,2,…,d−2, then ζκT´=2; therefore, T´∈Tn,κ,2. But YT´>YT, a contradiction. Thus, it is only s for d=2κ+1.
In this case, we consider a treeTn,κ,2α∈Tn,κ,2 obtained from the path P2κ+2=w0w1…w2κ+1 by attaching n−2κ+1 pendent vertices to wα, where α∈1,…,2κ. Moreover, Tn,κ,2α≅Tn,κ,2d−α for κ+1≤α≤d−1 and also YTn,κ,2α=YTn,κ,2β for 1≤α≠β≤d−1. Consequently, T≅Tn,κ,2α, for some α∈1,2,…,κ. Therefore, the YI for the tree T can be directly computed as YT=YTn,κ,2α=n−2κ4+n−2κ+1+16κ+1+16κ−15.Theorem 2.
Consider annvertex treeTthat belongs toTn,κ,3. Then,YT≤n−3κ4+n−3κ+1+216κ+1+16κ−15. The equality occurs asT⇔Tn,κ,3.Proof.
Given thatT∈Tn,κ,3. Obviously n≥κ+1ζκ, by Lemma 2. Also, the equality n=κ+1ζκ of Lemma 3 holds the results. Now, we proceed to prove the theorem by induction hypothesis IH on n. Assume that n>3κ+1 and the statement is true for n−1. Our main goal is to reach T≅Tn,κ,3.
LetP=u0u1…ud and D be a DP and minimum DDκ set of T, respectively. Actually, we have to prove d≥2κ+2. Otherwise, uκ,uκ+1 is a DDκ set, a contradiction. Let us assume that uκ,ud−κ⊆D such that ∪r=0κVTr\uκ∩D=∅ and ∪r=d−κdVTr\ud−κ∩D=∅.
Chooseu0=α1,u2,…,ud=αt as the pendent vertices of T and also ζT=αi|ζκT−αi=ζκT for 1≤i≤t. If ζT=∅, then ζT=αi|ζκT−αi=ζκT−1 for 1≤i≤t.
Whenξui/T≥3, then α2,…,αt−1∩VTi≠∅. Taking uκ,ud−κ∈D, so ζκT−x=ζκT for x∈α2,…,αt−1∩VTi, a contradiction. Therefore ξui/T=2 for i∈1,2,…,κ,d−κ,…,d−1. Clearly ζκT−u0=ζκT−1, since ξu1/T=2.
Remark thatξTu1,uκ+1=κ and ∪r=0κVTrvκ∩D=∅. So, uκ+1∈D. Likewise, ud−κ−1∈D. For d>2κ+2, the vertices uκ,uκ+1,ud−κ−1,ud−κ are distinguished, a contradiction. So, d=2κ+2 and D=uκ,uκ+1,ud−κ.
On the other side, ifξuκ+1/T=2, then T≅P2κ+3 and uκ,ud−κ is a DDκ set, which is an inconsistency. Therefore, ξuκ+1/T≥3 and also ζκ=3≤m. When m>3, then ζκT−αi=ζκT for some i∈1,…,m, an impropriety. So, m=3. Thus, Tκ+1 is a path of which ended vertices are uκ+1 and α3. That is, ξuκ+1,α3=κ. Hence, VT=3κ+1, which contradicts n>3κ+1.
Assume thatv is a unique vertex αi which is a pendent vertex with ζκT−αi=ζκT. Note that ξv/T≤Δ≤n−3κ, by Lemma 6. So, by the IH and the definition of YT, we get(7)YT=YT−αi+4ξ3vT−6ξ2vT+4ξvT≤n−1−3κ4+n−1−3κ+1+216κ+1+16κ−15+4n−3κ3−6n−3κ2+4n−3κ=n−3κ4+n−3κ+1+216κ+1+16κ−15
Therefore, the equality arrives if and only ifT−αi≅Tn−1,κ,3 and ξv/T=Δ=n−3κ, that is, T⇔Tn,κ,3.Theorem 3.
LetTbe a tree havingnvertices andζkT≥3. Ifn=κ+1ζk, we haveYT≤ζκ−14+4ζκ−13+ζκ−16ζκ+5+28κ−3ζκ−8.The equality is attained when T⇔Tn,κ,ζκ.Proof.
Given thatn=κ+1ζκ for the tree T of n vertices with DDκ number, ζκ≥3. By Lemma 3, we get T=G°κ for some tree G on ζκ vertices. Let us consider VG=u1,u2,…,uκ. Then, ξui/G=ξvi/T−1. Therefore, ∑i=1ζκξui/G=2ζκ−1 since for every tree (assume T′) containing n-vertices with vertex set x1,x2,…,xn occurs ∑i=1ζκξxi/T′=2n−1. By the definition of the YI, we can express(8)YT=∑i=1ζκξ4uiG+∑vi∈VT∖VGξ4viT=∑i=1ζκξuiT−14+4∑i=1ζκξuiT−13+6∑i=1ζκξuiT−12+4∑i=1ζκξuiT−1+ζκ+16κ−1ζκ+ζκ=YG+4FG+6M1G+8ζκ−1+2ζκ+16κ−1ζκ≤YSζκ+4FSζκ+6M1ζκ+16κζκ−6ζκ−8=ζκ−14+4ζκ−13+ζκ−16ζκ+5+28κ−3ζκ−8for equalities G⇔Sζκ that imply T⇔Tn,κ,ζκ.Theorem 4.
ConsiderTas ann-vertex tree whoseDDκnumber isζκ≥3. Then,YT≤n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15.
The equality occurs forT⇔Tn,κ,ζκ.Proof.
LetT⇔Tn,κ,ζκ be a tree containing a DP such that P=u0u1…ud that maximized the YI of graphs. The main goal is to establish the maximization of YT with respect to T≅Tn,κ,ζκ. Let us consider D to be a minimum DDκ set of T and also define ζT=x∈VT|ξx/T=1andζκT−x=ζκT. If ΓT=∅, then ζκT−ui=ζκT−1fori=0,d. Also, for ΓT≠∅, by Lemma 4, NTΓT=1. In case, u0,ud∈ΓT, as d−1>1, we get u1,ud−1⊆NTΓT that implies NTΓT>1, a contradiction. Therefore, we consider that ζκT−u0=ζκT−1, and thus uκ,uκ+1,ud−κ⊆D, from Theorem 2.
By Lemma1, applying EJT of T on any nonpendent edge of Tα repeatedly for α=1,…,κ, it is to be constructed a tree T´ from T such that T´α≅SVT´α, where T´α is the component of T´−uα−1uα,uαuα+1 having uα, for α=1,…,κ. Then, we have T´∈Tn,κ,ζκ and also YT≤YT´, where the equality holds T⇔T´.
Now letT∗=T´−∪α∈1,…,κ∖αruαw|w∈NT´uα∖uα−1,uα+1+∪αr∈1,…,κ∖αruαw||w∈NT´uα∖uα−1,uα+1 for some αr∈1,…,κ.
Then, by Lemma5, we get YT´≤T∗ with equality if and only if T´≅YT∗.
Again, define byT⋆=T∗−uαrw|w∈NT∗uαr∖uαr−1,uαr+1+uκ+1w|w∈NT∗uαr∖uαr−1,uαr+1. In fact, let NT∗uαr∖uαr−1,uαr+1=p,p≥0.
Then,ξuα/T⋆=2 for α=1,…,κ and also D will be the minimum DDκ set of T⋆. It implies that all the vertices in ∪α=0kNT⋆αuκ∖u0,…,uκ can be determined by uκ+1∈D. Therefore, D∖uκ will be a DDκ set of T⋆−u0,…,uκ. Suppose that PNκ,Dy is the set of all private κ-neighbors of y upon D in T⋆. Then, PNκ,Duκ+1⊆VT⋆∖u0,…,uκ. Thus, D∖uκ will be a minimum DDκ set of the tree T⋆−u0,…,uκ. Therefore, ζκT⋆−u0,…,uκ=ζκ−1=ζκT⋆−u0,…,uκ−1.
So, from the definition ofYI, we have(9)YT⋆−YT⋆=duκ+1T∗+p4+24−ξ4uκ+1T∗−p+24=2p2ξ3uκ+1T∗+3pξ2uκ+1T∗+2p2ξuκ+1T∗−4p2−12p−16≥0
It means thatYT⋆≥YT⋆, where the equality holds iff either p=0 i.e. T∗≅T⋆ or ξuκ+1/T∗=2.
So far, we have provedYT⋆≤n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15 with equality iff T⋆≅Tn,κ,ζκ by induction on ζk. We have from Theorem 2 that the assertion is mathematics for n≥κ+1ζκ as well as ζk=3.
Now let us consider the affirmation contains forζk−1 and all the vertices n≥κ+1ζκ−1.
Because ofζkT⋆−u0,…,uκ and VT⋆−u0,…,uκ=n−κ−1≥κ+1ζκ−1, we get by the IH(10)YT⋆=YT∗−u0,…,uκ+4ξ3uκ+1−6ξ2uκ+1+4ξuκ+1−1+∑α=0κξT∗4uα=YTn−κ−1,κ,ζκ−1+4n−κζκ3−6n−κζκ2+4n−κζκ+16κ=n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15,where the equality holds iff T⋆−u0,…,uκ≅Tn−κ−1,κ,ζκ−1 and also ξT⋆uκ+1=Δ=n−κζκ and otherwise T⋆≅Tn,κ,ζκ with ξT⋆uα=2 for α=1,…,κ. Therefore, we can conclude thatYT≤YT´≤YT∗≤YT∗≤n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15 with either equality iff T≅T´≅T∗≅T⋆≅Tn,κ,ζκ or T≅T´≅T∗ with ξT∗uκ+1=2. Besides, T⋆≅Tn,κ,ζκ.Here, we determine someUB on the YI of trees containing n-vertices with domination number ζ ([23]). The DDκ number of a graph is said to be the domination number of that graph if κ=1. If T is an n-vertex tree containing a DP such that P:u0u1…ud, then denote by Ti the component of T−ui−1ui,uiui+1 containing ui,i=1,2,…,d−1. To compute our main outcome, at first, we will focus on the following definition.ConsiderTn,ζ to be a tree constructed from a star K1,n−1 with involvement of a pendant edge to its ζ−1 pendant vertices. Note that Tn,ζ∈Tn,ζ, a class of n vertex trees and domination number ζ. Also, ζ=1 occurs iff T≅T1,n−1.Corollary 1.
IfT∈Tn,ζ, thenYT≤n−ζ4+n−2ζ+1+17ζ−1 with equality holding for T⇔Tn,ζ.Proof.
ForΔ=2, it occurs that T≅Pnn≥2. The equality holds for T≅T2,1≡P2, T≅T3,1≡P3 and T≅T4,2≡P4. But for n≥5, the above inequality is strict. Now, we consider a diameter path P=u0,u1,…,ud and a minimum dominating set D of T with Δ≥3. To prove the theorem, we will take the way of IH on n. Let us consider that Theorem 1 is true for n−1 and also the statement is to be proved as well as truth by replacing n+1 from n.When ζT−u0=ζT, then by the IH we have(11)YT=YT−u0+4ξ3u1T−6ξ2u1T+4ξu1T,≤n−1−ζ4+n−2ζ+17ζ−1+4n−ζ3−6n−ζ2+4ξn−ζ (since ξu/T≤n−ζ, by Lemma 6) =n−ζ4+n−2ζ+1+17ζ−1. The equality holds iff the pendant vertex u0 is adjacent to the vertex u1 of degree Δ=n−ζ, that is, T≅Tn,ζ.
Otherwise, assume thatζT−u0=ζT−1. So, it will be ξu1/T=2 which also implies that u1 belongs to every minimum dominating set, i.e., ζT−u0=ζT. Therefore, we can obtain by the IH(12)YT=YT−u0+4ξ3u1/T−6ξ2u1/T+4ξu1/T≤n−ζ+14+n−2ζ+3+17ζ−2+16=n−ζ4+n−2ζ+1+17ζ−1,where the equality holds iffT−u0≅Tn−1,ζ−1 and pendant vertex u0 is adjacent to the vertex u1 of degree 2, that is, T≅Tn,ζ.


### 3.2. SomeUB for the YI of Graphs with respect to Some Standard Parameters and Others TI

In this section, we establish the some sharpUB for the YI of G⋅w⋅r to some graph parameters such as n,m,δ,Δ and others TI such as M1G,M2G,FG,EFG,EM1G. Let DG=ξu1/G,ξu1/G,…,ξun/G. If DG=r, then G is said to be r-regular. If DG=r,s, then G is r,s biregular and so on. Motivating the proof technique as in [24], we obtain an UB for YI in the following theorem.Theorem 5.
ConsiderGto be an,mgraph, i.e.,Gcontainsnvertices andmedges. ThenYG≤2mδ+Δδ2+Δ2−nδΔδ2+δΔ+Δ2+δ−tΔ3+δΔ2+δΔ−tδ2+tδ+t2, wheretis the integer defined by relation2m−nδ≡t−δmodΔ−δ,δ≤t≤Δ−1and the equality holds iff at most one vertex ofGhas different degree fromδandΔ.Proof.
Considerxi as the number of vertices of degree i in G. From the definition of YI, we can write(13)YG=∑i=δΔi4xi.
Obviously,(14)∑i=δΔixi=2m,∑i=δΔxi=n.
After calculation, we get(15)xδ=1Δ−δnΔ−2m+∑i=δ+1Δ+1i−Δxi,(16)xΔ=1Δ−δ2m−nδ+∑i=δ+1Δ+1δ−ixi.
Using (15) and (16), we have(17)YG=δ4xδ+Δ4xΔ+∑i=δ+1Δ−1i4xi=δ4Δ−δnΔ−2m+∑i=δ+1Δ+1i−Δxi+Δ4Δ−δ2m−nδ+∑i=δ+1Δ+1δ−ixi+∑i=δ+1Δ−1i4xi.=1Δ−δδ4nΔ−2m+Δ42m−nδ+1Δ−δ∑i=δ+1Δ−1δ4i−Δ+Δ4δ−i+i4Δ−δxi.
Actually, the termδ4i−Δ+Δ4δ−i+i4Δ−δ will be negative for δ+1≤i≤Δ−1. So, the YG will be maximum if xi=0 for i=δ+1,…,Δ−1. Therefore, (15) and (16) become to xδ=nΔ−2m/Δ−δ and xΔ=2m−nδ/Δ−δ. These two equations require that(18)2m−nδ≡0modΔ−δ.
If the requirement is not true, we choosexi such that nt=1 and xi=0 for all i=δ+1,…,Δ−1, except for i=t. Then, (15) and (16) become xδ=nΔ−2m+k−Δ/Δ−δ and xΔ=2m−n−1δ/Δ−δ which satisfy conditions 3 and 4. In [25], there survives a n vertex graph G of which one vertex with degree 0 when n and δ are both odd. If we add the edges to this graph, the vertex degrees increase one at a time up to Δ. There occurs 2m−nδ≡t−δmodΔ−δ that implies that the degree of one more vertex may be increased up to t. Therefore, there exists a graph of order n and size m along with a unique vertex of degree t that different from δ and Δ.
Suppose now that the graphG contains two vertices of degrees i and k for δ+1≤i≤k≤Δ−1. If the sum of vertex degrees remains the same by reducing the first vertex degree by 1 and increasing the second vertex degree by 1, the value of the YI is replaced by(19)δ−i−1Δ3+δΔ2+δ2Δ−i−1δ2+i−1δ+i−12−δ−iΔ3+δΔ2+δ2Δ−iδ2+iδ+i2+δ−k+1Δ3+δΔ2+δ2Δ−k+1δ2+k+1δ+k+12−δ−kΔ3+δΔ2+δ2Δ−kδ2+kδ+k2=k+13−i−13+3k3−i3+3k2+i2+k−i>0.
It means that condition 8 is not true, and it will be the optimal choice of the quantitiesxi=0 for δ+1≤i≤Δ−1 such that xt=1 (except for i=t). Therefore, we can conclude from (17) that(20)YG≤2mδ+Δδ2+Δ2−nδΔδ2+δΔ+Δ2+δ−tΔ3+δΔ2+δΔ−tδ2+tδ+t2.Theorem 6.
IfGis an,mgraph, then(21)YG≤EFG+6EM1G+12M1G−3M12Gn−16m,where the equality holds iff either G is regular or semiregular bipartite graph.Proof.
Using Lemma8, for α=1, setting xi=ξj/G+ξk/G and yi=1/ξj/G+1/ξk/G for the graph G in (1), we have(22)∑jk∈EGξj/G+ξk/G21/ξj/G+1/ξk/G≥∑jk∈EGξj/G+ξk/G2∑jk∈EG1/ξj/G+1/ξk/G,(23)i.e.∑jk∈EGξjGξkGξjG+kG≥M12Gn.Also,(24)∑jk∈EGξjGξkGξjG+ξkG=13∑jk∈EGξjG+ξkG3−∑jk∈EGξ3jG+ξ3kG=13∑e∈EGξeG+23−YG;whereξe=ijG=ξiG+ξjG−2.=13∑e∈EGξ3eG+2∑e∈EGξ2eG+4∑e∈EGξeG+83m−13∑jk∈EGξ3jG+ξ3kG=13EFG+2EM1G+4M1G−2m+83m−13YG.since ∑e∈EGξe/G=∑i∈VLGξi/LG=M1G−2m.
From (23) and (24), M12G/n≤1/3EFG+2EM1G+4M1G−2m+8/3m−1/3YG.Theorem 7.
Suppose a graphGthat containsnvertices andmedges. Then,YG≤M1GFG−M2G, with equalities G⇔P2.Proof.
Ifa, b, …, l are positive numbers sets with m elements in each set and p,q,…,t are positive numbers such that p+q+⋯+t>1, then by Jensen’s theorem ∑i=1maipbiq…lit≤∑i=1maip∑i=1mbiq…∑i=1mlit. We know (25)YG=∑uv∈EGξ3uG+ξ3vG=∑uv∈EGξuG+ξvGξ2uG+ξ2vG−ξuGξvG≤∑uv∈EGξuG+ξvG∑uv∈EGξ2uG+ξ2vG−ξuGξvG.
Settingai=ξu/G+ξv/G and bi=ξ2u/G+ξ2v/G−ξu/Gξv/G and p=q=1, then by Jensen’s theorem =M1GFG−M2G.Theorem 8.
LetGbe a graph ofnorder andmsize. Then,YG≤16m4/n1631/4Δ/δ16/3+4/3δ/Δ4. The equality occurred whenGis regular graph.Proof.
We prove the theorem using the following inequalities.
If1<x,y<∞, pi,qi≥0 and ϕqiy≤pix≤φqiy for 1≤i≤n, then(26)∑i=1npix1/x∑i=1nqiy1/y≤cxϕ,φ∑i=1npiqi,where cxϕ,φ=max1/xϕ/φ1/y+1/yφ/ϕ1/x,1/xφ/ϕ1/y+1/yϕ/φ1/x is a constant with some positive constants ϕ,φ. If pi>0 for some 1≤i≤n, then the equality holds if and only if ϕ=φ and pix=ϕqiy for every 1≤i≤n. Setting pi=ξui/G, qi=1 and x=4,y=4/3 and also ϕ=δ4, φ=Δ4, we have(27)∑i=1nξ4uiG1/4∑i=1n14/3≤max14δ4Δ44/3+43Δ4δ41/4,14Δ4δ44/3+43δ4Δ41/4∑i=1nξuiGso YG1/4n4/3≤2m1/4Δ/δ16/3+4/3δ/Δ, i.e., YG≤16m4/n1631/4Δ/δ16/3+4/3δ/Δ4. This completes the proof.Theorem 9.
For annvertex graphG, we haveYG≤FG4/3. The equality is satisfied whenGis regular.Proof.
Letx1,x2,…,xn be n positive real numbers, and let s,t be positive rational numbers. Then, by Jensen’s inequality ([26]) ∑i=1nxit1/t≤∑i=1nxis1/s if t>s>0. The equality holds iff x1=x2=⋯=xn. Considering xi=ξui/G for t=4,s=3, then we have ∑i=1nξ4ui/G1/4≤∑i=1nξ3ui/G1/3, that is, YG≤FG4/3.Theorem 10.
LetGbe a (n, m) graph. Then,YG≤δ2+Δ2M1G−nδ2Δ2. The equality is attained whenGis regular.Proof.
Supposexi,yi,h and H are the positive real numbers such that hxi≤yi≤Hxi for i=1,2,…,n. Then, by Diaz-Metacalf inequality [27], ∑i=1nyi2+hH∑i=1n≤h+H∑i=1nxiyi ant the equality is attained if and only if yi=hxi and yi=Hai. Now taking xi=1 and yi=ξ2ui/G and h=δ2, H=Δ, we get(28)∑i=1nξ4uiG+δ2Δ2∑i=1n1≤δ2+Δ2∑i=1nξ2uiG.Thus, YG≤δ2+Δ2M1G−nδ2Δ2.Theorem 11.
LetGbe a graph whose number of vertices isnand edgesm. Then,(29)YG≤βnΔ−δ2Δ2+Δδ+δ2+2mFGn.
The equality is attained iffx1=x2=⋯=xnandz1=z2=⋯=znand alsoβn=n⌈n/2⌉1−1/n⌈n/2⌉, where⌈x⌉is the largest integer greater than or equal tox.Proof.
Letxi and zi be positive real numbers for which there exist real constants x,z,X and Z such that x≤xi≤X and z≤zi≤Z for 1≤i≤n, respectively. Then, we have (discrete) Gruss inequality ([27]) n∑i=1nxizi−∑i=1nxi∑i=1nzi≤βnX−xY−y.
The equality controls iffx1=x2=⋯=xn and z1=z2=⋯=zn.
By settingxi=ξ3u/G and zi=ξu/G for every i=1,2,…,n, we have X=Δ3 and x=δ3.
Then, the inequality becomes(30)n∑i=1nξ4uiG−∑i=1nξ3uiG∑i=1nξuiG≤βnΔ3−δ3Δ−δ.So, YG≤βnΔ−δ2Δ2+Δδ+δ2+2mFG/n. This completes our claim.Corollary 2.
Sinceβn≤n2/4, thereforeYG≤n2Δ−δ2Δ2+Δδ+δ2+8mFG/4n.Theorem 12.
LetGbe an,mgraph. Then,(31)YG≤3Δ+δFG−n−1Δ−δ+3δ2−3δΔM1G−n−12m−nΔδ2−4n−2mδΔ+2mΔ3+6mδΔ2−δΔ3nwith equality holds if and only if G is Δ,δ biregular.Proof.
We have from [28] that YG=n−1FG−Y¯G, where Y¯G be the Y-coindex of G. From [29], FG≤M1GΔ+2δ−δ22m−nΔ−4mδΔ. Define by XG=n−1∑u∈VGξu/G−Δ2ξu/G−δ−∑u∈VGξu/G−Δ3ξu/G−δ. Since XG≥0, we have Y¯G≥n−1δ+2Δ+3Δ2+3δΔM1G−3Δ+δFG−2mΔ2+δΔ2−4mδΔ−2mΔ3−6mδΔ2+δΔ3n.
After simplification, we get the required result. □Theorem 13.
LetGbe an,mgraph, we have(32)YG≤2x+Δ+δFG−x2+δΔM1G+2mx2δΔ+xδ+Δ−x2δΔ.
The equality holds whenG be a Δ,δ biregular graph and also δ≤x≤Δ, where x be a positive real number.Proof.
Define byF1G=∑u∈VGξu/G−x2ξu/G−yξu/G−z.Setting δ≤x≤Δ, y=Δ and z=δ, then F1G≤0. Thus, (33)F1G=YG−2x+Δ+δFG+x2+δΔM1G−2mx2δΔ+xδ+Δ+x2δΔ≤0.Theorem 14.
LetGbe an,mgraph. Then,(34)YG≤4Δ−6FG−ΔΔ−1+2Δ−12Δ−5+Δ−2Δ−3M1G+2mΔ−2Δ−32Δ−1+ΔΔ−12Δ−5−nwxyz.
The equality occurs whenG is a tetra-regular graph.Proof.
Suppose that(35)F2G=∑u∈VGξuG−wξuG−xξuG−yξuG−z=YG−w+x+y+zFG+xw+w+xy+z+yzM1G−2mw+xyz+y+zxw+wxyz,where w,x,y,z are the positive real numbers.Setting w=Δ,x=Δ−1,y=Δ−2 and z=Δ−3, then F2G≤0. Therefore, we get the required result. The equality is satisfied when G is a tetra-regular graph.Corollary 3.
LetGbe a graph withnvertices andmedges. Then,YG≤3Δ+δFG−3ΔΔ+δM1G+2mΔ2Δ+3δ−Δ3δn and also YG≤3δ+ΔFG−3δδ+ΔM1G+2mδ2δ+3Δ−δ3Δn with equality holding when G is a Δ,δ biregular graph.Proof.
Consider an auxiliary functionF3G=∑u∈VGξu/G−x3ξu/G−y, where x and y are the real numbers. Thus,(36)F3G=∑u∈VGξ4uG−3x+yξ3uG+3xx+yξ2uG−x2x+3yξuG+x3y=YG−3x+yFG+3xx+3yM1G−2mx2x+3y+x3yn.Taking x=Δ, y=δ then F3G≤0 and YG≤3Δ+δFG−3ΔΔ+δM1G+2mΔ2Δ+3δ−Δ3δn. Also for x=δ and y=Δ, we have F3G≤0. Thus, YG≤3δ+ΔFG−3δδ+ΔM1G+2mδ2δ+3Δ−δ3Δn.Corollary 4.
LetGbe a graph of ordernand sizem. Then,(37)Y¯G≤n−13Δ−1+3Δ2Δ−1M1G−4Δ−1FG−2mn−1Δ3Δ−2−2mΔ24Δ−3+Δ2Δ−1n+Δ−1nwhere the equality is satisfied iff G is a Δ,Δ−1 biregular graph.Proof.
Define byF4G=n−1∑u∈VGξu/G−Δ2ξu/G−Δ−1−∑u∈VGξu/G−Δ3ξu/G−Δ−1(38)F4G=n−1∑u∈VGξuG−Δ2ξuG−Δ−1−∑u∈VGξuG−Δ3ξuG−Δ−1≤n−1FG−3Δ−1M1G+2mΔ3Δ−2−Δ2Δ−1n−YG+4Δ−1FG−3Δ2Δ−1M1G+2mΔ24Δ−3−Δ3Δ−1n.
SinceF4G≤0(39)Y¯G≤n−13Δ−1+3Δ2Δ−1M1G−4Δ−1FG−2mn−1Δ3Δ−2−2mΔ24Δ−3+Δ2Δ−1n+Δ−1nCorollary 5.
IfGis a graph withnvertices andmedges, the upper bounds of theY¯Gare given by3Δn+2Δ−1M1G−4ΔFG+nΔ2n+Δ−1−2mΔ23n+4Δ−3. The equality holds ifGis a regular graph.Proof.
Similarly, it is to be proved by definingF5G=n−1∑u∈VGξu/G−Δ3−∑u∈VGξu/G−Δ4. Obviously, F5G≤0.In 2005, Klavzar et al. [30] introduced the generalized Sierpinski graph gSG,t. It is obtained from SG,t by adding a new vertex u, called the special vertex of gSG,t, and edges joining u with all extreme vertices of SG,t.Theorem 15.
LetGbe a graph of ordernand sizemand letgSG,tbe its generalized Sierpinski graph with dimensiont≥2. Then, theYIofgSG,tis given by(40)YgSG,t≤2mΔ3nt−1+4nt−2+4βnt−2+2mnt−2+βnt−26Δ2+4Δ+1.
The upper bound is achieved iffG is a Δ-regular graph.Proof.
TheYI of gSG,t can be defined as(41)YgSG,t=∑uv∈EG∑i,j=01ξuG+i,ξvG+jSG,tξuG+i3+ξuG+j3.
By applying Lemma7, we have(42)=∑uv∈EGnt−2n−ξuG−ξvG+⊳u,vξ3uG+ξ3vG+nt−2ξvG−⊳u,v−βnt−2ξuGξ3uG+ξvG+13+nt−2ξuG−⊳u,v−βnt−2ξvGξuG+13+ξ3vG+nt−2⊳u,v+1+βnt−2ξuG+ξvG+1ξuG+13+ξvG+13≤∑uv∈EGnt−2n−2Δ+⊳u,v2Δ3+2nt−2Δ−⊳u,v−βnt−2ΔΔ3+Δ+13+nt−2⊳u,v+1+βnt−22Δ+12Δ+13=2mΔ3nt−1+4nt−2+4βnt−2+2mnt−2+βnt−26Δ2+4Δ+1.


## 3.1. SomeUB for the YI on Trees with DDκ Number

In this section, we establish some sharpUB for the YI of graphs on the trees as to the DDκ number, ζκ. The set of all n vertex trees with ζκ and the star of order n−κζ+1 with u1,u2,…,un−κζκ pendent vertices are denoted as Tn,κ,ζκ and Sn−κζκ+1, respectively.Theorem 1.
LetTbe a tree of ordernand it containsζκT=2; then theUBofYTcan be expressed asYT≤n−2κ4+n−2κ+1+16κ+1+16κ−15. The equality holds forT⇔Tn,κ,2j, wherej∈1,2,…,k.Proof.
LetT∈Tn,κ,2 be a tree with a DP such that P:u0,u1,…,ud. For d≤2κ, there exists a DDκ set μ⌊d/2⌋ of T, a contradiction. In case d≥2κ+2, also denoted by T´=ϕT,uiui+1 the tree obtained from T by EJT (Lemma 1) on the edge uiui+1 for some i∈1,2,…,d−2, then ζκT´=2; therefore, T´∈Tn,κ,2. But YT´>YT, a contradiction. Thus, it is only s for d=2κ+1.
In this case, we consider a treeTn,κ,2α∈Tn,κ,2 obtained from the path P2κ+2=w0w1…w2κ+1 by attaching n−2κ+1 pendent vertices to wα, where α∈1,…,2κ. Moreover, Tn,κ,2α≅Tn,κ,2d−α for κ+1≤α≤d−1 and also YTn,κ,2α=YTn,κ,2β for 1≤α≠β≤d−1. Consequently, T≅Tn,κ,2α, for some α∈1,2,…,κ. Therefore, the YI for the tree T can be directly computed as YT=YTn,κ,2α=n−2κ4+n−2κ+1+16κ+1+16κ−15.Theorem 2.
Consider annvertex treeTthat belongs toTn,κ,3. Then,YT≤n−3κ4+n−3κ+1+216κ+1+16κ−15. The equality occurs asT⇔Tn,κ,3.Proof.
Given thatT∈Tn,κ,3. Obviously n≥κ+1ζκ, by Lemma 2. Also, the equality n=κ+1ζκ of Lemma 3 holds the results. Now, we proceed to prove the theorem by induction hypothesis IH on n. Assume that n>3κ+1 and the statement is true for n−1. Our main goal is to reach T≅Tn,κ,3.
LetP=u0u1…ud and D be a DP and minimum DDκ set of T, respectively. Actually, we have to prove d≥2κ+2. Otherwise, uκ,uκ+1 is a DDκ set, a contradiction. Let us assume that uκ,ud−κ⊆D such that ∪r=0κVTr\uκ∩D=∅ and ∪r=d−κdVTr\ud−κ∩D=∅.
Chooseu0=α1,u2,…,ud=αt as the pendent vertices of T and also ζT=αi|ζκT−αi=ζκT for 1≤i≤t. If ζT=∅, then ζT=αi|ζκT−αi=ζκT−1 for 1≤i≤t.
Whenξui/T≥3, then α2,…,αt−1∩VTi≠∅. Taking uκ,ud−κ∈D, so ζκT−x=ζκT for x∈α2,…,αt−1∩VTi, a contradiction. Therefore ξui/T=2 for i∈1,2,…,κ,d−κ,…,d−1. Clearly ζκT−u0=ζκT−1, since ξu1/T=2.
Remark thatξTu1,uκ+1=κ and ∪r=0κVTrvκ∩D=∅. So, uκ+1∈D. Likewise, ud−κ−1∈D. For d>2κ+2, the vertices uκ,uκ+1,ud−κ−1,ud−κ are distinguished, a contradiction. So, d=2κ+2 and D=uκ,uκ+1,ud−κ.
On the other side, ifξuκ+1/T=2, then T≅P2κ+3 and uκ,ud−κ is a DDκ set, which is an inconsistency. Therefore, ξuκ+1/T≥3 and also ζκ=3≤m. When m>3, then ζκT−αi=ζκT for some i∈1,…,m, an impropriety. So, m=3. Thus, Tκ+1 is a path of which ended vertices are uκ+1 and α3. That is, ξuκ+1,α3=κ. Hence, VT=3κ+1, which contradicts n>3κ+1.
Assume thatv is a unique vertex αi which is a pendent vertex with ζκT−αi=ζκT. Note that ξv/T≤Δ≤n−3κ, by Lemma 6. So, by the IH and the definition of YT, we get(7)YT=YT−αi+4ξ3vT−6ξ2vT+4ξvT≤n−1−3κ4+n−1−3κ+1+216κ+1+16κ−15+4n−3κ3−6n−3κ2+4n−3κ=n−3κ4+n−3κ+1+216κ+1+16κ−15
Therefore, the equality arrives if and only ifT−αi≅Tn−1,κ,3 and ξv/T=Δ=n−3κ, that is, T⇔Tn,κ,3.Theorem 3.
LetTbe a tree havingnvertices andζkT≥3. Ifn=κ+1ζk, we haveYT≤ζκ−14+4ζκ−13+ζκ−16ζκ+5+28κ−3ζκ−8.The equality is attained when T⇔Tn,κ,ζκ.Proof.
Given thatn=κ+1ζκ for the tree T of n vertices with DDκ number, ζκ≥3. By Lemma 3, we get T=G°κ for some tree G on ζκ vertices. Let us consider VG=u1,u2,…,uκ. Then, ξui/G=ξvi/T−1. Therefore, ∑i=1ζκξui/G=2ζκ−1 since for every tree (assume T′) containing n-vertices with vertex set x1,x2,…,xn occurs ∑i=1ζκξxi/T′=2n−1. By the definition of the YI, we can express(8)YT=∑i=1ζκξ4uiG+∑vi∈VT∖VGξ4viT=∑i=1ζκξuiT−14+4∑i=1ζκξuiT−13+6∑i=1ζκξuiT−12+4∑i=1ζκξuiT−1+ζκ+16κ−1ζκ+ζκ=YG+4FG+6M1G+8ζκ−1+2ζκ+16κ−1ζκ≤YSζκ+4FSζκ+6M1ζκ+16κζκ−6ζκ−8=ζκ−14+4ζκ−13+ζκ−16ζκ+5+28κ−3ζκ−8for equalities G⇔Sζκ that imply T⇔Tn,κ,ζκ.Theorem 4.
ConsiderTas ann-vertex tree whoseDDκnumber isζκ≥3. Then,YT≤n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15.
The equality occurs forT⇔Tn,κ,ζκ.Proof.
LetT⇔Tn,κ,ζκ be a tree containing a DP such that P=u0u1…ud that maximized the YI of graphs. The main goal is to establish the maximization of YT with respect to T≅Tn,κ,ζκ. Let us consider D to be a minimum DDκ set of T and also define ζT=x∈VT|ξx/T=1andζκT−x=ζκT. If ΓT=∅, then ζκT−ui=ζκT−1fori=0,d. Also, for ΓT≠∅, by Lemma 4, NTΓT=1. In case, u0,ud∈ΓT, as d−1>1, we get u1,ud−1⊆NTΓT that implies NTΓT>1, a contradiction. Therefore, we consider that ζκT−u0=ζκT−1, and thus uκ,uκ+1,ud−κ⊆D, from Theorem 2.
By Lemma1, applying EJT of T on any nonpendent edge of Tα repeatedly for α=1,…,κ, it is to be constructed a tree T´ from T such that T´α≅SVT´α, where T´α is the component of T´−uα−1uα,uαuα+1 having uα, for α=1,…,κ. Then, we have T´∈Tn,κ,ζκ and also YT≤YT´, where the equality holds T⇔T´.
Now letT∗=T´−∪α∈1,…,κ∖αruαw|w∈NT´uα∖uα−1,uα+1+∪αr∈1,…,κ∖αruαw||w∈NT´uα∖uα−1,uα+1 for some αr∈1,…,κ.
Then, by Lemma5, we get YT´≤T∗ with equality if and only if T´≅YT∗.
Again, define byT⋆=T∗−uαrw|w∈NT∗uαr∖uαr−1,uαr+1+uκ+1w|w∈NT∗uαr∖uαr−1,uαr+1. In fact, let NT∗uαr∖uαr−1,uαr+1=p,p≥0.
Then,ξuα/T⋆=2 for α=1,…,κ and also D will be the minimum DDκ set of T⋆. It implies that all the vertices in ∪α=0kNT⋆αuκ∖u0,…,uκ can be determined by uκ+1∈D. Therefore, D∖uκ will be a DDκ set of T⋆−u0,…,uκ. Suppose that PNκ,Dy is the set of all private κ-neighbors of y upon D in T⋆. Then, PNκ,Duκ+1⊆VT⋆∖u0,…,uκ. Thus, D∖uκ will be a minimum DDκ set of the tree T⋆−u0,…,uκ. Therefore, ζκT⋆−u0,…,uκ=ζκ−1=ζκT⋆−u0,…,uκ−1.
So, from the definition ofYI, we have(9)YT⋆−YT⋆=duκ+1T∗+p4+24−ξ4uκ+1T∗−p+24=2p2ξ3uκ+1T∗+3pξ2uκ+1T∗+2p2ξuκ+1T∗−4p2−12p−16≥0
It means thatYT⋆≥YT⋆, where the equality holds iff either p=0 i.e. T∗≅T⋆ or ξuκ+1/T∗=2.
So far, we have provedYT⋆≤n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15 with equality iff T⋆≅Tn,κ,ζκ by induction on ζk. We have from Theorem 2 that the assertion is mathematics for n≥κ+1ζκ as well as ζk=3.
Now let us consider the affirmation contains forζk−1 and all the vertices n≥κ+1ζκ−1.
Because ofζkT⋆−u0,…,uκ and VT⋆−u0,…,uκ=n−κ−1≥κ+1ζκ−1, we get by the IH(10)YT⋆=YT∗−u0,…,uκ+4ξ3uκ+1−6ξ2uκ+1+4ξuκ+1−1+∑α=0κξT∗4uα=YTn−κ−1,κ,ζκ−1+4n−κζκ3−6n−κζκ2+4n−κζκ+16κ=n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15,where the equality holds iff T⋆−u0,…,uκ≅Tn−κ−1,κ,ζκ−1 and also ξT⋆uκ+1=Δ=n−κζκ and otherwise T⋆≅Tn,κ,ζκ with ξT⋆uα=2 for α=1,…,κ. Therefore, we can conclude thatYT≤YT´≤YT∗≤YT∗≤n−κζκ4+n−κ+1ζκ+16κ+1ζκ−1+16κ−15 with either equality iff T≅T´≅T∗≅T⋆≅Tn,κ,ζκ or T≅T´≅T∗ with ξT∗uκ+1=2. Besides, T⋆≅Tn,κ,ζκ.Here, we determine someUB on the YI of trees containing n-vertices with domination number ζ ([23]). The DDκ number of a graph is said to be the domination number of that graph if κ=1. If T is an n-vertex tree containing a DP such that P:u0u1…ud, then denote by Ti the component of T−ui−1ui,uiui+1 containing ui,i=1,2,…,d−1. To compute our main outcome, at first, we will focus on the following definition.ConsiderTn,ζ to be a tree constructed from a star K1,n−1 with involvement of a pendant edge to its ζ−1 pendant vertices. Note that Tn,ζ∈Tn,ζ, a class of n vertex trees and domination number ζ. Also, ζ=1 occurs iff T≅T1,n−1.Corollary 1.
IfT∈Tn,ζ, thenYT≤n−ζ4+n−2ζ+1+17ζ−1 with equality holding for T⇔Tn,ζ.Proof.
ForΔ=2, it occurs that T≅Pnn≥2. The equality holds for T≅T2,1≡P2, T≅T3,1≡P3 and T≅T4,2≡P4. But for n≥5, the above inequality is strict. Now, we consider a diameter path P=u0,u1,…,ud and a minimum dominating set D of T with Δ≥3. To prove the theorem, we will take the way of IH on n. Let us consider that Theorem 1 is true for n−1 and also the statement is to be proved as well as truth by replacing n+1 from n.When ζT−u0=ζT, then by the IH we have(11)YT=YT−u0+4ξ3u1T−6ξ2u1T+4ξu1T,≤n−1−ζ4+n−2ζ+17ζ−1+4n−ζ3−6n−ζ2+4ξn−ζ (since ξu/T≤n−ζ, by Lemma 6) =n−ζ4+n−2ζ+1+17ζ−1. The equality holds iff the pendant vertex u0 is adjacent to the vertex u1 of degree Δ=n−ζ, that is, T≅Tn,ζ.
Otherwise, assume thatζT−u0=ζT−1. So, it will be ξu1/T=2 which also implies that u1 belongs to every minimum dominating set, i.e., ζT−u0=ζT. Therefore, we can obtain by the IH(12)YT=YT−u0+4ξ3u1/T−6ξ2u1/T+4ξu1/T≤n−ζ+14+n−2ζ+3+17ζ−2+16=n−ζ4+n−2ζ+1+17ζ−1,where the equality holds iffT−u0≅Tn−1,ζ−1 and pendant vertex u0 is adjacent to the vertex u1 of degree 2, that is, T≅Tn,ζ.


## 3.2. SomeUB for the YI of Graphs with respect to Some Standard Parameters and Others TI

In this section, we establish the some sharpUB for the YI of G⋅w⋅r to some graph parameters such as n,m,δ,Δ and others TI such as M1G,M2G,FG,EFG,EM1G. Let DG=ξu1/G,ξu1/G,…,ξun/G. If DG=r, then G is said to be r-regular. If DG=r,s, then G is r,s biregular and so on. Motivating the proof technique as in [24], we obtain an UB for YI in the following theorem.Theorem 5.
ConsiderGto be an,mgraph, i.e.,Gcontainsnvertices andmedges. ThenYG≤2mδ+Δδ2+Δ2−nδΔδ2+δΔ+Δ2+δ−tΔ3+δΔ2+δΔ−tδ2+tδ+t2, wheretis the integer defined by relation2m−nδ≡t−δmodΔ−δ,δ≤t≤Δ−1and the equality holds iff at most one vertex ofGhas different degree fromδandΔ.Proof.
Considerxi as the number of vertices of degree i in G. From the definition of YI, we can write(13)YG=∑i=δΔi4xi.
Obviously,(14)∑i=δΔixi=2m,∑i=δΔxi=n.
After calculation, we get(15)xδ=1Δ−δnΔ−2m+∑i=δ+1Δ+1i−Δxi,(16)xΔ=1Δ−δ2m−nδ+∑i=δ+1Δ+1δ−ixi.
Using (15) and (16), we have(17)YG=δ4xδ+Δ4xΔ+∑i=δ+1Δ−1i4xi=δ4Δ−δnΔ−2m+∑i=δ+1Δ+1i−Δxi+Δ4Δ−δ2m−nδ+∑i=δ+1Δ+1δ−ixi+∑i=δ+1Δ−1i4xi.=1Δ−δδ4nΔ−2m+Δ42m−nδ+1Δ−δ∑i=δ+1Δ−1δ4i−Δ+Δ4δ−i+i4Δ−δxi.
Actually, the termδ4i−Δ+Δ4δ−i+i4Δ−δ will be negative for δ+1≤i≤Δ−1. So, the YG will be maximum if xi=0 for i=δ+1,…,Δ−1. Therefore, (15) and (16) become to xδ=nΔ−2m/Δ−δ and xΔ=2m−nδ/Δ−δ. These two equations require that(18)2m−nδ≡0modΔ−δ.
If the requirement is not true, we choosexi such that nt=1 and xi=0 for all i=δ+1,…,Δ−1, except for i=t. Then, (15) and (16) become xδ=nΔ−2m+k−Δ/Δ−δ and xΔ=2m−n−1δ/Δ−δ which satisfy conditions 3 and 4. In [25], there survives a n vertex graph G of which one vertex with degree 0 when n and δ are both odd. If we add the edges to this graph, the vertex degrees increase one at a time up to Δ. There occurs 2m−nδ≡t−δmodΔ−δ that implies that the degree of one more vertex may be increased up to t. Therefore, there exists a graph of order n and size m along with a unique vertex of degree t that different from δ and Δ.
Suppose now that the graphG contains two vertices of degrees i and k for δ+1≤i≤k≤Δ−1. If the sum of vertex degrees remains the same by reducing the first vertex degree by 1 and increasing the second vertex degree by 1, the value of the YI is replaced by(19)δ−i−1Δ3+δΔ2+δ2Δ−i−1δ2+i−1δ+i−12−δ−iΔ3+δΔ2+δ2Δ−iδ2+iδ+i2+δ−k+1Δ3+δΔ2+δ2Δ−k+1δ2+k+1δ+k+12−δ−kΔ3+δΔ2+δ2Δ−kδ2+kδ+k2=k+13−i−13+3k3−i3+3k2+i2+k−i>0.
It means that condition 8 is not true, and it will be the optimal choice of the quantitiesxi=0 for δ+1≤i≤Δ−1 such that xt=1 (except for i=t). Therefore, we can conclude from (17) that(20)YG≤2mδ+Δδ2+Δ2−nδΔδ2+δΔ+Δ2+δ−tΔ3+δΔ2+δΔ−tδ2+tδ+t2.Theorem 6.
IfGis an,mgraph, then(21)YG≤EFG+6EM1G+12M1G−3M12Gn−16m,where the equality holds iff either G is regular or semiregular bipartite graph.Proof.
Using Lemma8, for α=1, setting xi=ξj/G+ξk/G and yi=1/ξj/G+1/ξk/G for the graph G in (1), we have(22)∑jk∈EGξj/G+ξk/G21/ξj/G+1/ξk/G≥∑jk∈EGξj/G+ξk/G2∑jk∈EG1/ξj/G+1/ξk/G,(23)i.e.∑jk∈EGξjGξkGξjG+kG≥M12Gn.Also,(24)∑jk∈EGξjGξkGξjG+ξkG=13∑jk∈EGξjG+ξkG3−∑jk∈EGξ3jG+ξ3kG=13∑e∈EGξeG+23−YG;whereξe=ijG=ξiG+ξjG−2.=13∑e∈EGξ3eG+2∑e∈EGξ2eG+4∑e∈EGξeG+83m−13∑jk∈EGξ3jG+ξ3kG=13EFG+2EM1G+4M1G−2m+83m−13YG.since ∑e∈EGξe/G=∑i∈VLGξi/LG=M1G−2m.
From (23) and (24), M12G/n≤1/3EFG+2EM1G+4M1G−2m+8/3m−1/3YG.Theorem 7.
Suppose a graphGthat containsnvertices andmedges. Then,YG≤M1GFG−M2G, with equalities G⇔P2.Proof.
Ifa, b, …, l are positive numbers sets with m elements in each set and p,q,…,t are positive numbers such that p+q+⋯+t>1, then by Jensen’s theorem ∑i=1maipbiq…lit≤∑i=1maip∑i=1mbiq…∑i=1mlit. We know (25)YG=∑uv∈EGξ3uG+ξ3vG=∑uv∈EGξuG+ξvGξ2uG+ξ2vG−ξuGξvG≤∑uv∈EGξuG+ξvG∑uv∈EGξ2uG+ξ2vG−ξuGξvG.
Settingai=ξu/G+ξv/G and bi=ξ2u/G+ξ2v/G−ξu/Gξv/G and p=q=1, then by Jensen’s theorem =M1GFG−M2G.Theorem 8.
LetGbe a graph ofnorder andmsize. Then,YG≤16m4/n1631/4Δ/δ16/3+4/3δ/Δ4. The equality occurred whenGis regular graph.Proof.
We prove the theorem using the following inequalities.
If1<x,y<∞, pi,qi≥0 and ϕqiy≤pix≤φqiy for 1≤i≤n, then(26)∑i=1npix1/x∑i=1nqiy1/y≤cxϕ,φ∑i=1npiqi,where cxϕ,φ=max1/xϕ/φ1/y+1/yφ/ϕ1/x,1/xφ/ϕ1/y+1/yϕ/φ1/x is a constant with some positive constants ϕ,φ. If pi>0 for some 1≤i≤n, then the equality holds if and only if ϕ=φ and pix=ϕqiy for every 1≤i≤n. Setting pi=ξui/G, qi=1 and x=4,y=4/3 and also ϕ=δ4, φ=Δ4, we have(27)∑i=1nξ4uiG1/4∑i=1n14/3≤max14δ4Δ44/3+43Δ4δ41/4,14Δ4δ44/3+43δ4Δ41/4∑i=1nξuiGso YG1/4n4/3≤2m1/4Δ/δ16/3+4/3δ/Δ, i.e., YG≤16m4/n1631/4Δ/δ16/3+4/3δ/Δ4. This completes the proof.Theorem 9.
For annvertex graphG, we haveYG≤FG4/3. The equality is satisfied whenGis regular.Proof.
Letx1,x2,…,xn be n positive real numbers, and let s,t be positive rational numbers. Then, by Jensen’s inequality ([26]) ∑i=1nxit1/t≤∑i=1nxis1/s if t>s>0. The equality holds iff x1=x2=⋯=xn. Considering xi=ξui/G for t=4,s=3, then we have ∑i=1nξ4ui/G1/4≤∑i=1nξ3ui/G1/3, that is, YG≤FG4/3.Theorem 10.
LetGbe a (n, m) graph. Then,YG≤δ2+Δ2M1G−nδ2Δ2. The equality is attained whenGis regular.Proof.
Supposexi,yi,h and H are the positive real numbers such that hxi≤yi≤Hxi for i=1,2,…,n. Then, by Diaz-Metacalf inequality [27], ∑i=1nyi2+hH∑i=1n≤h+H∑i=1nxiyi ant the equality is attained if and only if yi=hxi and yi=Hai. Now taking xi=1 and yi=ξ2ui/G and h=δ2, H=Δ, we get(28)∑i=1nξ4uiG+δ2Δ2∑i=1n1≤δ2+Δ2∑i=1nξ2uiG.Thus, YG≤δ2+Δ2M1G−nδ2Δ2.Theorem 11.
LetGbe a graph whose number of vertices isnand edgesm. Then,(29)YG≤βnΔ−δ2Δ2+Δδ+δ2+2mFGn.
The equality is attained iffx1=x2=⋯=xnandz1=z2=⋯=znand alsoβn=n⌈n/2⌉1−1/n⌈n/2⌉, where⌈x⌉is the largest integer greater than or equal tox.Proof.
Letxi and zi be positive real numbers for which there exist real constants x,z,X and Z such that x≤xi≤X and z≤zi≤Z for 1≤i≤n, respectively. Then, we have (discrete) Gruss inequality ([27]) n∑i=1nxizi−∑i=1nxi∑i=1nzi≤βnX−xY−y.
The equality controls iffx1=x2=⋯=xn and z1=z2=⋯=zn.
By settingxi=ξ3u/G and zi=ξu/G for every i=1,2,…,n, we have X=Δ3 and x=δ3.
Then, the inequality becomes(30)n∑i=1nξ4uiG−∑i=1nξ3uiG∑i=1nξuiG≤βnΔ3−δ3Δ−δ.So, YG≤βnΔ−δ2Δ2+Δδ+δ2+2mFG/n. This completes our claim.Corollary 2.
Sinceβn≤n2/4, thereforeYG≤n2Δ−δ2Δ2+Δδ+δ2+8mFG/4n.Theorem 12.
LetGbe an,mgraph. Then,(31)YG≤3Δ+δFG−n−1Δ−δ+3δ2−3δΔM1G−n−12m−nΔδ2−4n−2mδΔ+2mΔ3+6mδΔ2−δΔ3nwith equality holds if and only if G is Δ,δ biregular.Proof.
We have from [28] that YG=n−1FG−Y¯G, where Y¯G be the Y-coindex of G. From [29], FG≤M1GΔ+2δ−δ22m−nΔ−4mδΔ. Define by XG=n−1∑u∈VGξu/G−Δ2ξu/G−δ−∑u∈VGξu/G−Δ3ξu/G−δ. Since XG≥0, we have Y¯G≥n−1δ+2Δ+3Δ2+3δΔM1G−3Δ+δFG−2mΔ2+δΔ2−4mδΔ−2mΔ3−6mδΔ2+δΔ3n.
After simplification, we get the required result. □Theorem 13.
LetGbe an,mgraph, we have(32)YG≤2x+Δ+δFG−x2+δΔM1G+2mx2δΔ+xδ+Δ−x2δΔ.
The equality holds whenG be a Δ,δ biregular graph and also δ≤x≤Δ, where x be a positive real number.Proof.
Define byF1G=∑u∈VGξu/G−x2ξu/G−yξu/G−z.Setting δ≤x≤Δ, y=Δ and z=δ, then F1G≤0. Thus, (33)F1G=YG−2x+Δ+δFG+x2+δΔM1G−2mx2δΔ+xδ+Δ+x2δΔ≤0.Theorem 14.
LetGbe an,mgraph. Then,(34)YG≤4Δ−6FG−ΔΔ−1+2Δ−12Δ−5+Δ−2Δ−3M1G+2mΔ−2Δ−32Δ−1+ΔΔ−12Δ−5−nwxyz.
The equality occurs whenG is a tetra-regular graph.Proof.
Suppose that(35)F2G=∑u∈VGξuG−wξuG−xξuG−yξuG−z=YG−w+x+y+zFG+xw+w+xy+z+yzM1G−2mw+xyz+y+zxw+wxyz,where w,x,y,z are the positive real numbers.Setting w=Δ,x=Δ−1,y=Δ−2 and z=Δ−3, then F2G≤0. Therefore, we get the required result. The equality is satisfied when G is a tetra-regular graph.Corollary 3.
LetGbe a graph withnvertices andmedges. Then,YG≤3Δ+δFG−3ΔΔ+δM1G+2mΔ2Δ+3δ−Δ3δn and also YG≤3δ+ΔFG−3δδ+ΔM1G+2mδ2δ+3Δ−δ3Δn with equality holding when G is a Δ,δ biregular graph.Proof.
Consider an auxiliary functionF3G=∑u∈VGξu/G−x3ξu/G−y, where x and y are the real numbers. Thus,(36)F3G=∑u∈VGξ4uG−3x+yξ3uG+3xx+yξ2uG−x2x+3yξuG+x3y=YG−3x+yFG+3xx+3yM1G−2mx2x+3y+x3yn.Taking x=Δ, y=δ then F3G≤0 and YG≤3Δ+δFG−3ΔΔ+δM1G+2mΔ2Δ+3δ−Δ3δn. Also for x=δ and y=Δ, we have F3G≤0. Thus, YG≤3δ+ΔFG−3δδ+ΔM1G+2mδ2δ+3Δ−δ3Δn.Corollary 4.
LetGbe a graph of ordernand sizem. Then,(37)Y¯G≤n−13Δ−1+3Δ2Δ−1M1G−4Δ−1FG−2mn−1Δ3Δ−2−2mΔ24Δ−3+Δ2Δ−1n+Δ−1nwhere the equality is satisfied iff G is a Δ,Δ−1 biregular graph.Proof.
Define byF4G=n−1∑u∈VGξu/G−Δ2ξu/G−Δ−1−∑u∈VGξu/G−Δ3ξu/G−Δ−1(38)F4G=n−1∑u∈VGξuG−Δ2ξuG−Δ−1−∑u∈VGξuG−Δ3ξuG−Δ−1≤n−1FG−3Δ−1M1G+2mΔ3Δ−2−Δ2Δ−1n−YG+4Δ−1FG−3Δ2Δ−1M1G+2mΔ24Δ−3−Δ3Δ−1n.
SinceF4G≤0(39)Y¯G≤n−13Δ−1+3Δ2Δ−1M1G−4Δ−1FG−2mn−1Δ3Δ−2−2mΔ24Δ−3+Δ2Δ−1n+Δ−1nCorollary 5.
IfGis a graph withnvertices andmedges, the upper bounds of theY¯Gare given by3Δn+2Δ−1M1G−4ΔFG+nΔ2n+Δ−1−2mΔ23n+4Δ−3. The equality holds ifGis a regular graph.Proof.
Similarly, it is to be proved by definingF5G=n−1∑u∈VGξu/G−Δ3−∑u∈VGξu/G−Δ4. Obviously, F5G≤0.In 2005, Klavzar et al. [30] introduced the generalized Sierpinski graph gSG,t. It is obtained from SG,t by adding a new vertex u, called the special vertex of gSG,t, and edges joining u with all extreme vertices of SG,t.Theorem 15.
LetGbe a graph of ordernand sizemand letgSG,tbe its generalized Sierpinski graph with dimensiont≥2. Then, theYIofgSG,tis given by(40)YgSG,t≤2mΔ3nt−1+4nt−2+4βnt−2+2mnt−2+βnt−26Δ2+4Δ+1.
The upper bound is achieved iffG is a Δ-regular graph.Proof.
TheYI of gSG,t can be defined as(41)YgSG,t=∑uv∈EG∑i,j=01ξuG+i,ξvG+jSG,tξuG+i3+ξuG+j3.
By applying Lemma7, we have(42)=∑uv∈EGnt−2n−ξuG−ξvG+⊳u,vξ3uG+ξ3vG+nt−2ξvG−⊳u,v−βnt−2ξuGξ3uG+ξvG+13+nt−2ξuG−⊳u,v−βnt−2ξvGξuG+13+ξ3vG+nt−2⊳u,v+1+βnt−2ξuG+ξvG+1ξuG+13+ξvG+13≤∑uv∈EGnt−2n−2Δ+⊳u,v2Δ3+2nt−2Δ−⊳u,v−βnt−2ΔΔ3+Δ+13+nt−2⊳u,v+1+βnt−22Δ+12Δ+13=2mΔ3nt−1+4nt−2+4βnt−2+2mnt−2+βnt−26Δ2+4Δ+1.


## 4. SomeUB for YI under Some Graph Operations

In this section, we derive someUB for YI under several graph operations. Let Gi be a graph with the vertex set VGi=ni and the edge set EGi=mi for i=1,2. For each u∈VG1 and v∈VG2, we get ξu/G1≤ΔG1 and ξv/G2≤ΔG2.


### 4.1. Cartesian Product

The Cartesian product ([31]) of G1 and G2, denoted by G1⊗G2, is the graph with vertex set VG1⊗G2=VG1×VG2 and its degree distribution is ξu,v/G1⊗G2=ξu/G1+ξv/G2.Theorem 16.
TheYIofG1⊗G2satisfies the following inequality:YG1⊗G2≤n1n2Δ4G1+Δ4G2+4Δ3G1ΔG2+4ΔG1Δ3G2+6Δ2G1Δ2G2with equality occurring whenG1andG2are regular graphs.Proof.
By the definition ofY-index, we have(43)YG1⊗G2=∑u,v∈VG1×G2ξu,vG1×G24=∑u∈VG1∑v∈VG2ξuG1+ξvG24≤n1n2Δ4G1+Δ4G2+4Δ3G1ΔG2+4ΔG1Δ3G2+6Δ2G1Δ2G2.The inequality must be equality if ξu1/G1+ξv1/G2=ξu2/G1+ξv2/G2 for any u1,u2∈VG1 and v1,v2∈VG2.


### 4.2. Join

The degree of a vertexu for the join [32] of G1 and G2, denoted by G1+G2, is given by(44)ξuG1+G2=ξuG1+n2ifu∈VG1ξuG2+n1ifu∈VG2.Theorem 17.
TheUBon theY-index of two graphsG1andG2for join is given by(45)YG1+G2≤2m1ΔG1+n23+2m2ΔG2+n13+n1n2ΔG1+n23+ΔG2+n13.
The equality holds whenG1 and G2 are regular graphs.Proof.
By the definition of theYI, we get(46)YG1+G2=∑uv∈EG1ξuG1+n23+ξvG1+n23+∑uv∈EG2ξuG2+n13+ξvG2+n13+∑u∈VG1∑v∈VG2ξuG1+n23+ξvG2+n13≤2m1ΔG1+n23+2m2ΔG2+n13+n1n2ΔG1+n23+ΔG2+n13.


### 4.3. Composition

For the compositionG1G2 of two graphs G1 and G2 [11], the degree of a vertex u,v∈VG1G2 is given by du,v/G1G2=n2ξu/G1+ξv/G2.Theorem 18.
TheUBofYIforG1G2is given byYG1G2≤n25Δ4G1+n1n2Δ4G2+6n1n23Δ2G1Δ2G2+4n1n24Δ3G1ΔG2+4n1n22ΔG1Δ3. The equality carries for G1 and G2 regular graphs.Proof.
From the definition ofY-index, we have(47)YG1G2=∑u∈VG1∑v∈VG2n2ξuG1+ξvG24≤n25Δ4G1+n1n2Δ4G2+6n1n23Δ2G1Δ2G2+4n1n24Δ3G1ΔG2+4n1n22ΔG1Δ3.


### 4.4. Corona Product

For Corona product [33] of G1 and G2, denoted by G1⋄G2, the degree of a vertex u∈G1⋄G2 is given by(48)ξuG1⋄G2=ξuG1+n2ifu∈VG1ξuG2+1ifu∈VG2,i,i=1,2,…,n1,,where G2,i is the i-th copy of the graph G2.Theorem 19.
LetG=G1⋄G2be the corona product ofG1andG2.YGsatisfies the following inequalitiesYG≤n1ΔG1+n24+n1n2ΔG2+14. The equality holds when G1 and G2 are regular.Proof.
From definition ofYI, we have(49)YG1⋄G2=∑u∈VG1ξuG1+n24+n1∑u∈VG2ξuG2+14≤n1Δ4G1+4n2Δ3G1+6n22Δ2G1+4n23ΔG1+n24+n1n2Δ4G2+4Δ3G2+6ΔG2+4ΔG2+1.


### 4.5. Strong Product

Considerξu,v/G1∗G2=ξu/G1+ξv/G2+ξu/G1v/G2 as a degree distribution of a vertex u,v in the strong product [11] G1∗G2.Theorem 20.
The sharpUB of YI for G1∗G2 is given by(50)YG1∗G2≤n1n2Δ4G1+Δ4G2+Δ4G1Δ4G2+6Δ2G1Δ2G2Δ2G1+Δ2G2+1+4ΔG1ΔG2Δ3G1+Δ3G2+Δ3G1Δ2G2+Δ2G1Δ3G2+Δ2G1+Δ2G2+12Δ2G1Δ2G2ΔG1+ΔG2+ΔG1ΔG2.
The equality occurs whenG1 and G2 are regular.


## 4.1. Cartesian Product

The Cartesian product ([31]) of G1 and G2, denoted by G1⊗G2, is the graph with vertex set VG1⊗G2=VG1×VG2 and its degree distribution is ξu,v/G1⊗G2=ξu/G1+ξv/G2.Theorem 16.
TheYIofG1⊗G2satisfies the following inequality:YG1⊗G2≤n1n2Δ4G1+Δ4G2+4Δ3G1ΔG2+4ΔG1Δ3G2+6Δ2G1Δ2G2with equality occurring whenG1andG2are regular graphs.Proof.
By the definition ofY-index, we have(43)YG1⊗G2=∑u,v∈VG1×G2ξu,vG1×G24=∑u∈VG1∑v∈VG2ξuG1+ξvG24≤n1n2Δ4G1+Δ4G2+4Δ3G1ΔG2+4ΔG1Δ3G2+6Δ2G1Δ2G2.The inequality must be equality if ξu1/G1+ξv1/G2=ξu2/G1+ξv2/G2 for any u1,u2∈VG1 and v1,v2∈VG2.


## 4.2. Join

The degree of a vertexu for the join [32] of G1 and G2, denoted by G1+G2, is given by(44)ξuG1+G2=ξuG1+n2ifu∈VG1ξuG2+n1ifu∈VG2.Theorem 17.
TheUBon theY-index of two graphsG1andG2for join is given by(45)YG1+G2≤2m1ΔG1+n23+2m2ΔG2+n13+n1n2ΔG1+n23+ΔG2+n13.
The equality holds whenG1 and G2 are regular graphs.Proof.
By the definition of theYI, we get(46)YG1+G2=∑uv∈EG1ξuG1+n23+ξvG1+n23+∑uv∈EG2ξuG2+n13+ξvG2+n13+∑u∈VG1∑v∈VG2ξuG1+n23+ξvG2+n13≤2m1ΔG1+n23+2m2ΔG2+n13+n1n2ΔG1+n23+ΔG2+n13.


## 4.3. Composition

For the compositionG1G2 of two graphs G1 and G2 [11], the degree of a vertex u,v∈VG1G2 is given by du,v/G1G2=n2ξu/G1+ξv/G2.Theorem 18.
TheUBofYIforG1G2is given byYG1G2≤n25Δ4G1+n1n2Δ4G2+6n1n23Δ2G1Δ2G2+4n1n24Δ3G1ΔG2+4n1n22ΔG1Δ3. The equality carries for G1 and G2 regular graphs.Proof.
From the definition ofY-index, we have(47)YG1G2=∑u∈VG1∑v∈VG2n2ξuG1+ξvG24≤n25Δ4G1+n1n2Δ4G2+6n1n23Δ2G1Δ2G2+4n1n24Δ3G1ΔG2+4n1n22ΔG1Δ3.


## 4.4. Corona Product

For Corona product [33] of G1 and G2, denoted by G1⋄G2, the degree of a vertex u∈G1⋄G2 is given by(48)ξuG1⋄G2=ξuG1+n2ifu∈VG1ξuG2+1ifu∈VG2,i,i=1,2,…,n1,,where G2,i is the i-th copy of the graph G2.Theorem 19.
LetG=G1⋄G2be the corona product ofG1andG2.YGsatisfies the following inequalitiesYG≤n1ΔG1+n24+n1n2ΔG2+14. The equality holds when G1 and G2 are regular.Proof.
From definition ofYI, we have(49)YG1⋄G2=∑u∈VG1ξuG1+n24+n1∑u∈VG2ξuG2+14≤n1Δ4G1+4n2Δ3G1+6n22Δ2G1+4n23ΔG1+n24+n1n2Δ4G2+4Δ3G2+6ΔG2+4ΔG2+1.


## 4.5. Strong Product

Considerξu,v/G1∗G2=ξu/G1+ξv/G2+ξu/G1v/G2 as a degree distribution of a vertex u,v in the strong product [11] G1∗G2.Theorem 20.
The sharpUB of YI for G1∗G2 is given by(50)YG1∗G2≤n1n2Δ4G1+Δ4G2+Δ4G1Δ4G2+6Δ2G1Δ2G2Δ2G1+Δ2G2+1+4ΔG1ΔG2Δ3G1+Δ3G2+Δ3G1Δ2G2+Δ2G1Δ3G2+Δ2G1+Δ2G2+12Δ2G1Δ2G2ΔG1+ΔG2+ΔG1ΔG2.
The equality occurs whenG1 and G2 are regular.


## 5. Application

As an application, we compute theYI of C80 Fullerene, by using Theorem 11. Fullerenes are the molecules such as cage-like polyhedra, containing solely carbon atoms. Fullerenes contain the networks of pentagons and hexagons. Here, we consider the fullerene C80 such that molecules made up entirely of n (natural number) carbon atoms contain twelve pentagonal sides and n/2−10 hexagonal faces, where n≠22≥20. For the graph representing fullerene C80 which is given in [34], we have FC80=2160. The number of edges(m) in fullerene C80 is m=nr/2=80×3/2=120. Thus, by Theorem 11, YG≤βnΔ−δ2Δ2+Δδ+δ2+2mFG/n =  2×120×2160/80=6480.


## 6. Conclusion

TheYI is one of the new chemical descriptors, which passes the test of having a highly correlation with the physiochemical properties it claims to describe in [11]. It comes as no surprise. Then, we determine some new UBs for the YI using various parameters such as order, size, maximum degree, minimum degree, distance κ-domination number, and some other topological indices. Furthermore, some sharp UB for the YI based on graph binary operations are obtained. At last, we consider an application for YI index of C80 Fullerene. The appeal of computing the UB is of course their generality and simple proofs. Along in this line, determining new lower bounds for YI is considered to be studied in the future.



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