# Some Bounds on Bond Incident Degree Indices with Some Parameters

Authors: Muhammad Rizwan; Akhlaq Ahmad Bhatti; Muhammad Javaid; Fahd Jarad
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

It is considered that there is a fascinating issue in theoretical chemistry to predict the physicochemical and structural properties of the chemical compounds in the molecular graphs. These properties of chemical compounds (boiling points, melting points, molar refraction, acentric factor, octanol-water partition coefficient, and motor octane number) are modeled by topological indices which are more applicable and well-used graph-theoretic tools for the studies of quantitative structure-property relationships (QSPRs) and quantitative structure-activity relationships (QSARs) in the subject of cheminformatics. Theπ-electron energy of a molecular graph was calculated by adding squares of degrees (valencies) of its vertices (nodes). This computational result, afterwards, was named the first Zagreb index, and in the field of molecular graph theory, it turned out to be a well-swotted topological index. In 2011, Vukicevic introduced the variable sum exdeg index which is famous for predicting the octanol-water partition coefficient of certain chemical compounds such as octane isomers, polyaromatic hydrocarbons (PAH), polychlorobiphenyls (PCB), and phenethylamines (Phenet). In this paper, we characterized the conjugated trees and conjugated unicyclic graphs for variable sum exdeg index in different intervals of real numbers. We also investigated the maximum value of SEIa for bicyclic graphs depending on a>1.


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## Body

## 1. Introduction

In chemical graph theory, molecules and macromolecules (such as organic compounds, nucleic acids, and proteins) are represented by graphs wherein vertices correspond to the atoms, whereas edges represent the bonds between atoms [1, 2]. A topological index is a numerical value associated with chemical constitution for correlation of chemical structure with various physicochemical properties [3]. Topological indices play a significant role in organic chemistry and particularly in pharmacology [4, 5]. Physicochemical properties of chemical compounds such as relative enthalpy of formation, biological activity, boiling points, melting points, molar refraction, acentric factor, octanol-water partition coefficient, and motor octane number are modeled by topological indices in quantitative structure-property relation (QSPR) and quantitative structure-activity relation (QSAR) studies [4, 6–8].In chemistry, the usage of topological index started in 1947 when the chemist Wiener developed the Wiener index (a distance-based topological index) to predict boiling points of paraffins [9]. Platt index (the oldest degree-based topological index) was proposed in 1952 for predicting paraffin properties [10]. The π-electron energy of a molecular graph was calculated by adding square of degrees (valencies) of its vertices (nodes) in the year 1972. The same computational result, afterwards, was named the first Zagreb index, and in the field of molecular graph theory, it turned out to be a well-swotted topological index [11]. For more details about the topological indices in the field of chemistry, we refer to [6, 8, 12–15].Many well-known topological indices such as hyper Zagreb index [16], variable sum exdeg index [17], and Zagreb indices [18, 19] have been used to find out sharp bounds for unicyclic, bicyclic, and tricyclic graphs. Vukicevic [15] propounded variable sum exdeg index for a graph G and defined it as(1)SEIaG=∑uv∈EGadu+adv=∑u∈VGduadu,where a is a positive integer other than 1. This topological index is correlated well with octane-water partition coefficient [15] and is employed to the study of octane isomers (see [20–22]). This topological index in the form of polynomial was proposed by Yarahmadi and Ashrafi, and they find its application in nanoscience [23]. Chemical application of this index can be seen in the papers [12, 13, 15].In this paper, we mainly targeted three main problems. First of all, we find the extremal values of variable sum exdeg index (SEIa) for conjugated trees. After that, we investigated lower and upper bounds of unicyclic conjugated graphs with respect to the length of this cycle in different intervals. At the end of this paper, we find upper bounds of SEIa for bicyclic graphs. This paper contains seven sections. In the first section, we have given introduction while in Section 2, we have given the proofs of some lemmas and preliminary results. In Section 3, we discovered the bounds of a conjugated trees and this section helps us to find out lower and upper bounds of unicyclic conjugated graphs with respect to the length of this cycle in Section 4. In Section 5, we discussed an important theorem related with conjugated unicyclic graphs. In Section 6, we discovered the upper bounds of bicyclic graphs. In the last section, we have drawn the conclusion.


## 2. Preliminary Results

All graphs under consideration in this paper will be connected, simple, and finite. SupposeG=VG,EG is a simple and finite graph, whereas set of vertices is denoted by VG and the set of edges is denoted by EG. Let v∈VG for which dv is defined as the cardinality of edges incident with the vertex v. Suppose NGv denotes the set of all vertices which are adjacent with the vertex v and NGv=NGv∪v. Note that ΔG and δG represent the maximum and minimum degree of a graph G, respectively. A pendent vertex is a vertex of degree one. An edge whose one end is a pendent vertex is called pendent edge. Let B⊆VG and B′⊆EG; then, G−B and G−B′ are subgraphs of G which are obtained by deleting the vertices and edges fromG, respectively. An edge between the vertices x and y is denoted by e=xy. If B=v and B′=xy, then G−B and G−B′ can be expressed as G−v and G−xy, respectively.In a graphG, if the vertices x and y are nonadjacent, then G+xy means there is an addition of an edge between the vertices x and y in a graph G. We use Sn,Cn, and Pn to denote the star graph, cycle graph, and path graph on n vertices, respectively. We assume that graphs G∗,w1 and G∗∗,w2 be rooted at w1 and w2, respectively. Then, G∗,w1⋓G∗∗,w2 is obtained by identifying w1 and w2 as the same vertex. A graph which has no cycle is called a tree. A graph G is said to be unicyclic graph if it has a unique cycle. A graph G is said to be bicyclic graph if G has exactly n+1 edges. Let Uln represent the collection of all those graphs which have order n and a unique cycle of length l. We denote Ul2m,m the collection of all conjugated unicyclic graphs of order n in which length of its cycle is l, whereas m is the matching number of G. Let G∈Ul2m,m be a unicyclic graph of length l and it is denoted by Cl. Let G∈Ul2m,m; if n=2m=l or n=2m=l+1, then its SEIaG is unique. That is why in this paper we will assume n=2m≥l+2. One can find terminologies and expressions “indefinito” in [24–26].Suppose thatG′ is a graph acquired from another graph G by using some graph alteration such that VG=VG′. In all sections of this paper, whenever such two graphs are under debate, we always mean the vertex degree dx the degree of the vertex x in G.Lemma 1.
LetG be a graph of order n if G contains the vertices u,v∈VG such that du=s>1,dv=t>1 and s≥t; then, there exists a graph G′ such that SEIaG′>SEIaG for a>1.Proof.
Letu,v∈VG and v1,v2,v3,...vk be the pendent vertices adjacent to the vertex v. We define a new graph G′, i.e., G′=G−v1v,v2v,…,vkv+v1u,v2u,…,vku as in Figure 1. By the definition of SEIaG, we have(2)SEIaG−SEIaG′=du.adu+dv.adv−du+k.adu+k+dv−k.adv−k=dv.adv−dv−k.adv−k−du+k.adu+k−du.adu=kaμ11+μ1lna−aμ21+μ2lna<0,where μ1∈t−k,t,μ2∈s,s+k,μ2>μ1 for a>1. Thus, the proof of the above lemma is accomplished.Figure 1
(a) G and (b) G′ is constructed from G.
(a)(b)Lemma 2.
LetG be a graph having two components G1and T1, where G1 is a cycle graph and T1 is a star graph with central vertex v. Let u∈VG1 and du=p, such that uv is an edge in G. Let v1,v2,v3,...vk be the pendent vertices adjacent with the vertex v, i.e., NGv−u=v1,v2,v3,…,vk. We define G′=G−v1v,v2v,…,vkv+v1u,v2u,…,vku such that SEIaG′>SEIaG.Proof.
LetG be a graph having two components G1and T1 where G1 is a cycle graph and T1 is a star graph with central vertex v. Let u∈VG1, du=p, such that uv is an edge in G. Let v1,v2,v3,...vk be the pendent vertices adjacent with the vertex v, i.e., NGv−u=v1,v2,v3,...vk. We define G′=G−v1v,v2v,…,vkv+v1u,v2u,…,vku as in Figure 2. By the definition of SEIa, we have(3)SEIaG−SEIaG′=du.adu+dv.adv−du+k.adu+k+dv−k.adv−k=dv.adv−dv−k.adv−k−du+k.adu+k−du.adu.
Ifp≥k+1, then(4)SEIaG−SEIaG′=dv.adv−dv−k.adv−k−du+k.adu+k−du.adu=kaμ11+μ1lna−aμ21+μ2lna<0,where μ1∈1,k+1,μ2∈p,p+k,μ2>μ1, and a>1SEIaG′>SEIaG.
Ifp≤k+1, then(5)SEIaG−SEIaG′=du.adu−dv−k.adv−k−du+k.adu+k−dv.adv=z.aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈1,p,μ2∈k+1,k+p,μ2>μ1,z=p−1 and a>1. Thus, we have SEIaG′>SEIaG.Figure 2
(a) Graph G; (b) the graph G′ is obtained from G.
(a)(b)


## 3. Extremal Values of Variable Sum Exdeg Index for Conjugated Trees

First we introduce some notations which will be used in the following lemmas and theorems. Suppose thatTn,m be the collection of all trees with n vertices and m-matching number with n≥2m. When m−1 pendent vertices are attached with each certain non-central vertices of Sn−m+1, then the resulting graph is denoted by T0n,m. If we choose n=2m, then it means every tree from Tn,m and T0n,m contains perfect matching.Lemma 3.
(see [26]). If an n-vertex tree T has perfect matching, then there must exist at least two vertices of degree one with neighbouring vertices of degree two, where n≥3.Lemma 4.
(see [26]). If an n-vertex tree T has an m-matching with n>2m, then there must exist a pendent vertex u which is not saturated by m-matching.In the following, we will find two theorems which will give extreme values ofSEIa for all trees T in T2m,m.Theorem 1.
Letm≥1, n≥4, and a>1 be integers and T∈T2m,m; then, SEIaT≤m.am+2m−1a2+am, where equality meets when T≅T02m,m.Proof.
SupposeT∈T2m,m. If the tree T is isomorphic to T02m,m, then SEIaT=SEIaT02m,m. On the other hand, if T is not isomorphic to T02m,m, then we assume that the vertex u∈VT, i.e., du=ΔT where du≥2. Lemma 3 assures that there exist vertices u1 and v1 adjacent by an edge with du1=2 and dv1=1. Let Nu1−v1=w1. We define T1=T−u1w1+u1u. It is clear that T1∈T2m,m. By the definition of SEIa, we have(6)SEIaT−SEIaT1=du.adu+dw1.adw1−du+1.adu+1+dw1−1.adw1−1=dw1.adw1−dw1−1.adw1−1−du+1.adu+1−du.adu=aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈dw1−1,dw1, μ2∈du,du+1, and μ2>μ1 for a>1.
Note thatT∗=T1−u1,v1; then, obviously T∗∈T2m−1,m−1. Then, by the construction of T∗ and keeping Lemma 3 in our mind, we can choose u2 and v2 in T∗ where du2=2 and dv2=1. It is clear that duT∗=ΔT=ΔT∗. Let Nu2−v2=w2. We set T∗∗=T∗−u2w2+u2u. Similarly, SEIaT∗∗>SEIaT∗. We define T2=T1−u2w2+u2u; then,(7)SEIaT2−SEIaT1=du+2.adu+2+dw2−1.adw2−1−du+1.adu+1+dw2.adw2=du+2.adu+2−du+1.adu+1−dw2.adw2−dw2−1.adw2−1=aμ41+μ4lna−aμ31+μ3lna>0,where μ3∈dw2−1,dw2,μ4∈du+1,du+2, and μ4>μ3 for a>1.
This implies thatSEIaT2−SEIaT1>0. We repeat the above process on the graph T again and again and we obtain a sequence of graphs T1,T2,…,Ts,… with the relation SEIaT1<SEIaT2<…SEIaTs<…
For some positive integerp, we have Tp≅Tp+1 and Tp≅T02m,m.
Hence,SEIaT<SEIaT02m,m.Theorem 2.
Suppose thatm≥1, n≥4, and a>1 be integers. If T∈T2m,m, then SEIaT≥2a+22m−2a2, where equality meets when T≅P2m.Proof.
We claim thatT≅P2m; then, SEIaT=SEIaP2m. If we apply the above-defined process (in previous Theorem 1) on T, then we will obtain the expression SEIaT1<SEIaT2<…SEIaTs<… for some positive integer p≥1SEIaTp>SEIaP2m. Hence, SEIaT≥SEIaP2m=2a+22m−2a2 equality meets when T≅P2m.


## 4. Extremal Values of Variable Sum Exdeg Index for Conjugated Unicyclic Graphs

In this portion of the paper, we will find extreme values forSEIaG among all the conjugated unicyclic graphs in Ul2m,m for a>1. In this concern, we will prove some lemmas which will support our main theorems.Lemma 5.
(see [26]). For any tree T from T2m+1,m, we find at least one vertex of degree 1 which will be adjacent with a vertex v of degree 2, i.e., dv=2.Lemma 6.
Suppose thatm≥1,a>1 and T∈T2m+1,m; then, SEIaT≥SEIaP2m+1, where sign of equality meets when T≅P2m+1.Proof.
LetT∈T2m+1,m; then, by Lemma 4, we find a pendent vertex u in T which is not saturated by an m-matching of T. Obviously, the vertices in T−u are saturated by the maximal m−matching. This implies that T−u∈T2m,m. Assume that Nu=w; then, SEIaT=SEIaT−u+dwadw+duadu−dw−1adw−1. According to Theorem 2, we have(8)SEIaT≥SEIaP2m+dwadw+a−dw−1adw−1=2a+22m−2a2+dwadw+a−dw−1adw−1=2a+22m−1a2−2a2+dwadw+a−dw−1adw−1≥2a+22m−1a2=SEIaP2m+1.
The above inequality holds ifdwadw−dw−1adw−1−2a2−a≥0.
Ifdw=2, then(9)dwadw−dw−1adw−1−2a2−a=0.
Ifdw≥3, then we have(10)aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈1,2,μ2∈dw−1,dw, and μ2>μ1 for a>1. Finally, we have SEIaT≥SEIaP2m+1.Lemma 7.
LetT∈T2m+1,m; then, SEIaT≤SEIaT02m+1,m equality meets when T≅T02m+1,m where m≥1,a>1.Proof.
LetT∈T2m+1,m; then, by Lemma 4, we find a pendent vertex u in T which is not saturated by a maximal m′-matching of T. Suppose that Nu=z1. Suppose v is a vertex in T, i.e., dv=ΔT. Define T′=T−uz1+uv; then, clearly T′−u∈T2m,m. With the help of Theorem 1, we have SEIaT′−u≤SEIaT02m,m, so we have(11)SEIaT′=SEIaT′−u+duadu+dv+1adv+1−dvadv≤a+SEIaT02m,m+dv+1adv+1−dvadv∗∗∗<a+SEIaT02m,m+m+1am+1−mam=a+mam+2a2m−1+a.m+m+1am+1−mam=a+mam+2a2m−1+a.m+m+1am+1−mam=2a2m−1+1+ma+m+1am+1=SEIaT02m+1,m.
If we show thatdv+1adv+1−dvadv<m+1am+1−mam, then it will be enough for the existence of the expression ∗∗∗. Since we know that ΔT02m,m=m, T is not isomorphic to T02m+1,m and dv≤m. If we assume dv=m, then dv+1adv+1−dvadv−m+1am+1+mam=0. If we assume dv<m, then(12)=dv+1adv+1−dvadv−m+1am+1−mam=aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈dv,dv+1,μ2∈m,m+1,μ2>μ1 and a>1.
Hence,SEIaT<SEIaT′<SEIaT02m+1,m. So, we conclude that SEIaT≤SEIaT02m+1,m, and sign of equality meets when T≅T02m+1,m.
We define a setB=xi∈VCl:dxi≥3. Remember that Txi represents the connected component having the vertex xi of the graph G−xi−1xi,xixi+1.Lemma 8.
(see [26]). Let G∈Ul2m,m; then, for every xi∈B, Txi∈Tni,ni/2 or Txi∈Tni,ni−1/2.Lemma 9.
LetG∈Ul2m,m such that SEIaG is minimum if Txi≅Pni where xi∈B,ni=nTxi,a>1 and xi is one of the pendent vertices of Pni.Proof.
SupposeG∈Ul2m,m with minimum variable sum exdeg index. We also assume that xi∈B and the vertices xi−1 and xi+1 are the neighbouring vertices of the vertex xi along Cl. Here we consider the expression(13)Q=dxiadxi−dxi−2adxi−2+dxi−1adxi−1−dxi−1−1adxi−1−1+dxi+1adxi+1−dxi+1−1adxi+1−1.
We assume thatG∗ is the connected component of G−xixi−1,xixi+1 which does not contain the vertex xi. We can write the expression, SEIaG=SEIaG∗+Q+SEIaTxi. According to Lemma 8, Txi∈Tni,ni/2 or Txi∈Tni,ni−1/2. In either situation, there exists the following relation: SEIaG≥SEIaG∗+Q+SEIaPni according to Theorem 2 and Lemma 6. Furthermore, the sign of equality meets iff Txi≅Pni. Next we will prove that the vertex xi is one of the pendent vertices of Pni such that dxi=3. We suppose that dxi≥4, so there must exist two vertices u and v, i.e., Nxi−xi−1,xi+1=u,v. Then, there must be one edge of xiu,or xiv which is not included in m-matching. Without loss of generality, suppose that xiu does not belong to the m-matching. Let Pv=v1,v2,…vq where q≥2 represents the path with v=v1 as a pendent vertex of Pv. Define G′=G−xiu+uvq; it is clear that G′∈Ul2m,m. By the definition of SEIa, we have(14)SEIaG′−SEIaG=dxi−1adxi−1+dvq+1advq+1−dxiadxi+dvqadvq=dvq+1advq+1−dvqadvq−dxiadxi−dxi−1adxi−1=aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈dvq,dvq+1,μ2∈dxi−1,dxi,μ2>μ1 and a>1.
SEIaG′<SEIaG which contradicts our choice of G.Lemma 10.
IfG∈Ul2m,m, a>1 with maximum SEIaG; then, for every vertex xi∈B, there exist Txi which will be isomorphic to T0ni,ni/2 or T0ni,ni−1/2. If Txi is isomorphic to T0ni,ni/2, then dxi−2 will be equal to ΔT0ni,ni/2. If Txi is isomorphic to T0ni,ni−1/2, then the vertex xi will be the one end vertex of T0ni,ni−1/2 and xi will be adjacent to some maximum degree vertex of T0ni,ni−1/2.Proof.
LetG∈Ul2m,m with maximum variable sum exdeg index. We also assume that xi∈B and the vertices xi−1 and xi+1 are the neighbouring vertices of the vertex xi along Cl. Here we consider the expression(15)Q=dxiadxi−dxi−2adxi−2+dxi−1adxi−1−dxi−1−1adxi−1−1+dxi+1adxi+1−dxi+1−1adxi+1−1.
We assume thatG∗ is the connected component of G−xixi−1,xixi+1 which does not contain the vertex xi. We can write the expression SEIaG=SEIaG∗+Q+SEIaTxi.
According to Theorem1, Lemma 7, and Lemma 8, we have(16)SEIaG≤SEIaG∗+Q+SEIaT0ni,ni2,or(17)SEIaG≤SEIaG∗+Q+SEIaT0ni,ni−12,for ni being even or odd, respectively. Above two inequalities hold iff Txi≅T0ni,ni/2 and Txi≅T0ni,ni−1/2, respectively. Next we will prove that(1)
IfTxi≅T0ni,ni/2, then dxi−2=ΔT0ni,ni/2.(2)
IfTxi≅T0ni,ni−1/2, then T0ni,ni−1/2 has the vertex xi as a pendent vertex that is adjacent to the vertex of maximum degree in T0ni,ni−1/2.
For the proof of (i), we assume thatdxi−2<ΔT0ni,ni/2. Let y∈VT, i.e., dy=ΔT0ni,ni/2. We define G′=G−xixi−1−xixi+1+xi−1y+xi+1y. By the definition of SEIaG,(18)SEIaG−SEIaG′=dxiadxi+dyady−dxi−2adxi−2+dy+2ady+2=dxiadxi−dxi−2adxi−2−dy+2ady+2−dyady.
Ifdxi≥dy, then(19)SEIaG−SEIaG′=dyady−dxi−2adxi−2−dy+2ady+2−dxiadxi=z.aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈dx−2,dy,μ2∈dxi,dy+2,μ2>μ1, a>1, and z=dy−dxi+2.
SEIaG′>SEIaG which contradicts our choice of G.
Ifdxi<dy, then(20)SEIaG−SEIaG′=dxiadxi−dxi−2adxi−2−dy+2ady+2−dyady=2.aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈dxi−2,dxi,μ2∈dy,dy+2,μ2>μ1, and a>1.
SEIaG′>SEIaG which contradicts our choice of G.
For the proof of (ii), we will just show thatdxi=3 and dw1=ΔTvi where w1=Nxi−xi+1,xi−1.
SinceG∈Ul2m,m and Txi is isomorphic to T0ni,ni−1/2, dxi−2<ΔT0ni,ni−1/2.
Note that any vertexw2 other than the vertex of maximum degree in T0ni,ni−1/2 has the degree 2 or 1. If dxi−2=2, this implies that in T0ni,ni−1/2, there will be a vertex which is not saturated by the maximal matching of G. Here a contradiction arises fordxi−2=1. This implies dxi=3. If we assume dw1<ΔT0ni,ni−1/2, then once again we find a vertex in T0ni,ni−1/2 which is not saturated by the maximal matching in G and again we will find a contradiction. From the above discussion, the proof is accomplished.Theorem 3.
SupposeG∈Ul2m,m; then, SEIa≥1+3a3+22m−2a2 for a>1 and the sign of equality meets when G≅Cl,xi⋓P2m−k+1,xi where xi∈Cl is a pendent vertex of P2m−k+1.Proof.
SupposeG∈Ul2m,m having minimum SEIa. According to Lemma 9, for the minimum SEIaG, Txst will be isomorphic to Pnst for every xst∈B where nst=nTxst. For B=1, the above result holds. Now we discuss the above result for B≥2. We have Txst≅Pnstt=1,2,…,B. So, we denote Txst=y0ty1t…ybttbt≥1, where y0t=xstt=1,2….B. We define G∗=G−y02y12−y03y13…−y0By1B+yb11y12+yb22y13+⋯+ybB−1B−1y1B. It is clear that G∗∈Ul2m,m; then, by the definition of SEIa,(21)SEIaG∗−SEIaG=B−12a2−a−3a3−2a2=B−1aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈1,2,μ2∈2,3,μ2>μ1, and a>1.
SEIaG∗<SEIaG which contradicts our choice of G. Hence, the proof of above theorem is finished.


## 5. Main Result

Theorem 4.
LetG∈Ul2m,m and a>1; then, the following results must hold:(1)
If2m=l+2, then SEIa≤2a+2l−2a2+3a3 and the sign of equality meets when G is not isomorphic to Cl,P3.(2)
If2m≥l+3 and l is odd, then SEIaG≤m−l−1/2a+2m+l−3/2a2+m−l−5/2am−l−5/2 sign of equality meets iff G≅Cl,xi⋓T02m−l+1,2m−l+1/2,xi.(3)
If2m≥l+3 and l is even, then SEIaG≤m−l/2a+2m+l/2−2a2+3a3+m−l/2+1am−l/2+1 sign of equality meets iff G≅Cl,xi⋓T02m−l+1,2m−l/2,xi, where dxi=3, Nxi−xi−1,xi+1=y, and y=ΔT02m−l+1,2m−l/2,xi.Proof.
LetG∈Ul2m,m, a>1 with maximum SEIaG. According to Lemma 8, we are sure that Txi is isomorphic to T0ni,ni−1/2 or T0ni,ni/2 where xi∈B.
ForB=1, we have no graph G≇Cl,P3 for which 2m=l+2. Whenever 2m=l+2, then G must be isomorphic to Cl,P3 where G∈Ul2m,m. According to Theorem 3, we can find a graph G′ such that SEIaG′>SEIaCl,P3=SEIaG for G′≠Cl,P3 which contradicts the choice of G. With the help of Lemma 10, the proof of (2) or (3) is satisfied. For |B|≥2, we may have the below cases.Case 1. Let 2m=l+2; then, for any two graphs G∗and G∗∗, both graphs are not isomorphic to Cl,P3 and SEIaG∗=SEIaG∗∗. According to Theorem 3, we have SEIaG>SEIaCl,P3 where G≇Cl,P3, and hence (1) satisfies.Case 2. For 2m≥l+3, we make the following subcases. Let xst∈B,t=1,2,…,B and nTxst=nt.Subcase 2.1. Let nt=2 for every xst∈B; then, we have the following set of vertices: VG−VCl=y1,y2,…,yB and Nyt=xst,t=1,2,…,B. Let Nxst−yt=xst−1,xst+1,t=1,2,…,B. Choose B≥3. If B=3, we define G∗=G−xs2y2−xs3y3+xs1y2+y2y3. If B≥4, we define G∗=G−xs2y2−xs3y3−xs2−1xs2−xs2xs2+1−xs3xs3+1+xs2−1xs2+1+xs2xs3+xs2xs3+1+xs1y2+y2y3In both aboveG∗, we have G∗∈Ul2m,m and we have the expression(22)SEIaG∗−SEIaG=4a4−3a3+2a2−a−23a3−2a2=4a4−3.3a3+3.2a2−a>0,which contradicts our choice ofG.Subcase 2.2. Let xsl′∈B, i.e., nl′=3; here we have B≥2; there must exist a vertex xst′ in B−xsl′. We define G∗=G−xsl′yl′+xst′yl′; then, clearly G∗∈Ul2m,m. By the definition of SEIaG, we have(23)SEIaG∗−SEIaG=dxst′+1adxst′+1+2a2−dxst′adxst′+3a3=dxst′+1adxst′+1−dxst′adxst′−3a3−2a2.Sincedxst′≥3 and a>1, this implies that SEIaG∗−SEIaG>0 which is a contradiction to the choice of G.Subcase 2.3. Let xst∈B and nt≥4. In this concern, two subcases arise.Subcase 2.3.1. Suppose that dxsl=ΔG for some xsl∈B; since we have B≥2, there exists a vertex xsr in B−xsl. According to Lemma 8, Lemma 3, and Lemma 5, there must exist some adjacent vertices say us and vs in Txsr, i.e., dus=2 and dvs=1. Let Nus−vs=ws. We define G∗=G−wsus+xslus; then, clearly G∗∈Ul2m,m and(24)SEIaG∗−SEIaG=dxsl+1adxsl+1−dws−1adws−1−dxsladxsl+dwsadws=dxsl+1adxsl+1−dxsladxsl−dwsadws−dws−1adws−1=aμ21+μ2lna−aμ11+μ1lna>0,whereμ1∈dws−1,dws,μ2∈dxsl,dxsl+1,μ2>μ1, and a>1.SEIaG∗>SEIaG which contradicts our choice of G.Subcase 2.3.2. Suppose that dxsl<ΔG for some xsl∈B. Let y be a vertex in G with dy=ΔG, so this implies that y∈Txsl where l is a positive integer. Since we have B≥2, then there exists some vertex xsr in B−xsl. According to Lemma 8, Lemma 3, and Lemma 5, there must exist some adjacent vertices say us and vs in Txsr, i.e., dus=2 and dvs=1. Let Nus−vs=ws.
Remaining portion of the under discussion subcase is similar to subcase 2.3.1 and once again we find the contradiction. According to all above discussion and argument, we follow the desired result.Theorem 5.
LetG∈Ul2m,m, a>1, and nTxi≥3 for every xi∈B; then,(1)
If2m≥l+3 and l is odd, then SEIaG≤m−l−1/2a+2m+l−3/2a2+m−l−5/2am−l−5/2 sign of equality meets iff G≅Cl,xi⋓T02m−l+1,2m−l+1/2,xi.(2)
If2m≥l+3 and l is even, then SEIaG≤m−l/2a+2m+l/2−2a2+3a3+m−l/2+1am−l/2+1 sign of equality meets iff G≅Cl,xi⋓T02m−l+1,2m−l+1/2,xi, where dxi=3, Nxi−xi−1,xi+1=y, and y=ΔT02m−l+1,2m−l/2,xi.


## 6. Extremal Values of Variable Sum Exdeg Index for Bicyclic Graphs

Here we are going to define some notations. LetGn,n+1 be the collection of bicyclic graphs or all those graphs which have n vertices and n+1 number of edges. Note that if G∈Gn,n+1, then there exist two cycles say Ci and Cj in G.(i)
Ai,j is the collection of graphs G∈Gn,n+1 in which cycles Ci and Cj share a single vertex only.(ii)
Bi,j is the collection of graphs G∈Gn,n+1 in which cycles Ci and Cj share no common vertex.(iii)
ℂi,j,l is the collection of graphs G∈Gn,n+1 in which cycles Ci and Cj share a common path of length l.


### 6.1. Extremal Graphs inAi,j

SupposeSni,j is a graph from the collection Ai,j, i.e., there are k=n−i−j+1 pendent vertices adjacent to a common vertex of Ci and Cj as shown in Figure 3.Figure 3
Sn (i, j) graphs.Lemma 11.
LetG∈Ai,j; if G≇Sni,j, then SEIaG<SEIaSni,j for a>1.Proof.
LetG∈Ai,j; then, by Lemma 2, we obtain another graph say G′ for which SEIaG′>SEIaG. Further by Lemma 1, the graph G′ can be changed into another graph say G″ in which pendent vertices will be attached with some common vertex u, of Ci and Cj. If u is a not a common vertex of Ci and Cj, then G″≇Sni,j. By the definition of SEIaG, we have(25)SEIaSni,j−SEIaG″=k+4.ak+4+2a2−k+2.ak+2+4a4.Case−1: for k=n+1−i−j=1,a>1, we have(26)SEIaSni,j−SEIaG″=k+4.ak+4−4a4−k+2.ak+2−2a2=aμ21+μ2lna−aμ11+μ1lna>0,whereμ1∈2,k+2,μ2∈4,k+4,μ2>μ1, a>1.Case−2: for k=n+1−i−j≥2,a>1, we have(27)SEIaSni,j−SEIaG″=k+4.ak+4−k+2.ak+2−4a4−2a2=k.aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈2,4,μ2∈k+2,k+4,μ2>μ1, a>1.
From the above two cases, we conclude thatSEIaSni,j>SEIaG″.Lemma 12.
LetSni,j∈Ai,j; then,aSEIaSni,j<SEIaSni−1,j, a>1, i>3.bSEIaSni,j<SEIaSni,j−1, a>1, j>3.Proof.
By the definition ofSEIaG, we have(28)SEIaSni,j−SEIaSni−1,j=2a2−a−k+5ak+5−k+4ak+4=aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈1,2,μ2∈k+4,k+5,μ2>μ1 and a>1SEIaG′>SEIaG.
Proof of (ii) is the same as proof of (i).Theorem 6.
IfG∈Ai,j, then SEIaG will be maximal if G≅Sni,j and for all i≥3,j≥3, the graph from Ai,j with maximum SEIa is Sn3,3.Proof.
Proof of this theorem can be obtained by Lemma11 and Lemma 12.


### 6.2. Extremal Graphs inBi,j

Here we define thatTnri,j is a graph which is obtained by joining Ci and Cj by a path P of length r and the remaining number of vertices k=n−i−j−r+1 are attached to the same end vertex of P as shown in Figure 4.Figure 4
Tr(i, j) graphs.Lemma 13.
LetG∈Bi,j; if G≇Tnri,j, then SEIaG<SEIaTnri,j for a>1.Proof.
LetG∈Bi,j; then, by Lemma 2, we obtain another graph say G′ for which SEIaG′>SEIaG. Further by Lemma 1, the graph G′ can be changed into another graph say G″ in which pendent edges are attached with same vertex u, i.e., SEIaG″>SEIaG′. If u is not end vertex of path P, then we will show SEIaTnr>SEIaG″. By the definition of SEIa, we have(29)SEIaTnr−SEIaG″=k+3ak+3−k+2ak+2−3a3−2a2=aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈2,3,μ2∈k+2,k+3,μ2>μ1, a>1, and k=n+1−i−j−r.Lemma 14.
LetTnr∈Bi,j; then,aSEIaTnri−1,j>SEIaTnri,j, a>1, i>3.bSEIaTnri,j−1>SEIaTnri,j, a>1, j>3.cSEIaTnr−1i,j>SEIaTnri,j, a>1, r>1.Proof.
By the definition ofSEIaG,(30)θ=SEIaTnri−1,j−SEIaTnri,j=k+4ak+4−k+3ak+3−2a2−a=aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈1,2,μ2∈k+3,k+4,μ2>μ1, a>1, and k=n+1−i−j−r. This implies that θ>0.
Proof of (ii) is the same as proof of (i).Proof.
By the definition ofSEIaG,(31)θ=SEIaTnr−1i,j−SEIaTnri,j=k+4ak+4−k+3ak+3−2a2−a=aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈1,2,μ2∈k+3,k+4,μ2>μ1, a>1. This implies that θ>0.
SEIaTnr−1i,j>SEIaTnri,j. After proving Lemma 13 and Lemma 14, we are able to present the following theorem.Theorem 7.
IfG∈Bi,j, then SEIaG will be maximal if G≅Tni,j and for all i≥3,j≥3, the graph from Bi,j with maximum SEIa is Tn3,3.


### 6.3. Extremal Graphs inℂi,j,l

Here we define thatΛnli,j is a graph which is obtained by attaching n+l+1−i−j edges to one of the vertices of degree 3 in G∈ℂi,j,l (see Figure 5). Here we define some lemmas but skip their proofs. We refer Lemma 13 and Lemma 14 for the proof of following lemmas.Figure 5
Λnli,j graphs.Lemma 15.
LetG∈ℂi,j,l; if G≇Λnli,j, then SEIaG<SEIaΛnli,j for a>1.Lemma 16.
LetΛnli,j∈ℂi,j,l; then,aSEIaΛnli−1,j>SEIaΛnli,j, i>3.bSEIaΛnli,j−1>SEIaΛnli,j, j>3.cSEIaΛnl−1i,j>SEIaΛnli,j, l>1.Theorem 8.
Fora>1 and the graph from the collection ℂi,j,l with maximum SEIaG for all i≥3, j≥3 and l>1 is Λn13,3.Theorem 9.
A graphG∈Gn,n+1 has maximum variable sum exdeg index if and only if G≅Λn13,3 for a>1.Proof.
SinceSn3,3, Tn13,3, and Λn13,3 belong to Gn,n+1. All the previous lemmas and theorems make it very clear and easy to understand that Sn3,3, Tn13,3, and Λn13,3 have maximum SEIaG, and Sn3,3, Tn13,3, and Λn13,3 belong to Ai,j, Bi,j, and ℂi,j,l, respectively, for n≥6. Now we just need to compare the SEIa of Sn3,3, Tn13,3, and Λn13,3.(32)θ1=SEIaΛn13,3−SEIaSn3,3=n−1.an−1+3.a3+2.2a2+n−4a−n−1.an−1+4.2a2+n−5a=3a3−2.2a2+n−5a>0.
This implies thatθ1>0.(33)θ2=SEIaSn3,3−SEIaTn13,3=n−1.an−1+4.2a2+n−5a−n−3.an−3+4.2a2+3.a3+n−6a=n−1an−1−n−3an−3−3a3−a=2aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈1,3,μ2∈n−3,n−1,μ2>μ1,a>1. This implies that θ2>0. From the above discussion, we conclude that Λn13,3>Sn3,3>Tn13,3.


## 6.1. Extremal Graphs inAi,j

SupposeSni,j is a graph from the collection Ai,j, i.e., there are k=n−i−j+1 pendent vertices adjacent to a common vertex of Ci and Cj as shown in Figure 3.Figure 3
Sn (i, j) graphs.Lemma 11.
LetG∈Ai,j; if G≇Sni,j, then SEIaG<SEIaSni,j for a>1.Proof.
LetG∈Ai,j; then, by Lemma 2, we obtain another graph say G′ for which SEIaG′>SEIaG. Further by Lemma 1, the graph G′ can be changed into another graph say G″ in which pendent vertices will be attached with some common vertex u, of Ci and Cj. If u is a not a common vertex of Ci and Cj, then G″≇Sni,j. By the definition of SEIaG, we have(25)SEIaSni,j−SEIaG″=k+4.ak+4+2a2−k+2.ak+2+4a4.Case−1: for k=n+1−i−j=1,a>1, we have(26)SEIaSni,j−SEIaG″=k+4.ak+4−4a4−k+2.ak+2−2a2=aμ21+μ2lna−aμ11+μ1lna>0,whereμ1∈2,k+2,μ2∈4,k+4,μ2>μ1, a>1.Case−2: for k=n+1−i−j≥2,a>1, we have(27)SEIaSni,j−SEIaG″=k+4.ak+4−k+2.ak+2−4a4−2a2=k.aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈2,4,μ2∈k+2,k+4,μ2>μ1, a>1.
From the above two cases, we conclude thatSEIaSni,j>SEIaG″.Lemma 12.
LetSni,j∈Ai,j; then,aSEIaSni,j<SEIaSni−1,j, a>1, i>3.bSEIaSni,j<SEIaSni,j−1, a>1, j>3.Proof.
By the definition ofSEIaG, we have(28)SEIaSni,j−SEIaSni−1,j=2a2−a−k+5ak+5−k+4ak+4=aμ11+μ1lna−aμ21+μ2lna<0,where μ1∈1,2,μ2∈k+4,k+5,μ2>μ1 and a>1SEIaG′>SEIaG.
Proof of (ii) is the same as proof of (i).Theorem 6.
IfG∈Ai,j, then SEIaG will be maximal if G≅Sni,j and for all i≥3,j≥3, the graph from Ai,j with maximum SEIa is Sn3,3.Proof.
Proof of this theorem can be obtained by Lemma11 and Lemma 12.


## 6.2. Extremal Graphs inBi,j

Here we define thatTnri,j is a graph which is obtained by joining Ci and Cj by a path P of length r and the remaining number of vertices k=n−i−j−r+1 are attached to the same end vertex of P as shown in Figure 4.Figure 4
Tr(i, j) graphs.Lemma 13.
LetG∈Bi,j; if G≇Tnri,j, then SEIaG<SEIaTnri,j for a>1.Proof.
LetG∈Bi,j; then, by Lemma 2, we obtain another graph say G′ for which SEIaG′>SEIaG. Further by Lemma 1, the graph G′ can be changed into another graph say G″ in which pendent edges are attached with same vertex u, i.e., SEIaG″>SEIaG′. If u is not end vertex of path P, then we will show SEIaTnr>SEIaG″. By the definition of SEIa, we have(29)SEIaTnr−SEIaG″=k+3ak+3−k+2ak+2−3a3−2a2=aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈2,3,μ2∈k+2,k+3,μ2>μ1, a>1, and k=n+1−i−j−r.Lemma 14.
LetTnr∈Bi,j; then,aSEIaTnri−1,j>SEIaTnri,j, a>1, i>3.bSEIaTnri,j−1>SEIaTnri,j, a>1, j>3.cSEIaTnr−1i,j>SEIaTnri,j, a>1, r>1.Proof.
By the definition ofSEIaG,(30)θ=SEIaTnri−1,j−SEIaTnri,j=k+4ak+4−k+3ak+3−2a2−a=aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈1,2,μ2∈k+3,k+4,μ2>μ1, a>1, and k=n+1−i−j−r. This implies that θ>0.
Proof of (ii) is the same as proof of (i).Proof.
By the definition ofSEIaG,(31)θ=SEIaTnr−1i,j−SEIaTnri,j=k+4ak+4−k+3ak+3−2a2−a=aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈1,2,μ2∈k+3,k+4,μ2>μ1, a>1. This implies that θ>0.
SEIaTnr−1i,j>SEIaTnri,j. After proving Lemma 13 and Lemma 14, we are able to present the following theorem.Theorem 7.
IfG∈Bi,j, then SEIaG will be maximal if G≅Tni,j and for all i≥3,j≥3, the graph from Bi,j with maximum SEIa is Tn3,3.


## 6.3. Extremal Graphs inℂi,j,l

Here we define thatΛnli,j is a graph which is obtained by attaching n+l+1−i−j edges to one of the vertices of degree 3 in G∈ℂi,j,l (see Figure 5). Here we define some lemmas but skip their proofs. We refer Lemma 13 and Lemma 14 for the proof of following lemmas.Figure 5
Λnli,j graphs.Lemma 15.
LetG∈ℂi,j,l; if G≇Λnli,j, then SEIaG<SEIaΛnli,j for a>1.Lemma 16.
LetΛnli,j∈ℂi,j,l; then,aSEIaΛnli−1,j>SEIaΛnli,j, i>3.bSEIaΛnli,j−1>SEIaΛnli,j, j>3.cSEIaΛnl−1i,j>SEIaΛnli,j, l>1.Theorem 8.
Fora>1 and the graph from the collection ℂi,j,l with maximum SEIaG for all i≥3, j≥3 and l>1 is Λn13,3.Theorem 9.
A graphG∈Gn,n+1 has maximum variable sum exdeg index if and only if G≅Λn13,3 for a>1.Proof.
SinceSn3,3, Tn13,3, and Λn13,3 belong to Gn,n+1. All the previous lemmas and theorems make it very clear and easy to understand that Sn3,3, Tn13,3, and Λn13,3 have maximum SEIaG, and Sn3,3, Tn13,3, and Λn13,3 belong to Ai,j, Bi,j, and ℂi,j,l, respectively, for n≥6. Now we just need to compare the SEIa of Sn3,3, Tn13,3, and Λn13,3.(32)θ1=SEIaΛn13,3−SEIaSn3,3=n−1.an−1+3.a3+2.2a2+n−4a−n−1.an−1+4.2a2+n−5a=3a3−2.2a2+n−5a>0.
This implies thatθ1>0.(33)θ2=SEIaSn3,3−SEIaTn13,3=n−1.an−1+4.2a2+n−5a−n−3.an−3+4.2a2+3.a3+n−6a=n−1an−1−n−3an−3−3a3−a=2aμ21+μ2lna−aμ11+μ1lna>0,where μ1∈1,3,μ2∈n−3,n−1,μ2>μ1,a>1. This implies that θ2>0. From the above discussion, we conclude that Λn13,3>Sn3,3>Tn13,3.


## 7. Conclusion

Ascertaining the upper and lower bounds on any molecular structure descriptor with regard to various graph parameters is a significant job. We have sought the maximum value ofSEIa for unicyclic graphs. Sharp bounds have also been investigated for conjugated trees and conjugated unicyclic graphs. We also investigated the extremal graphs for each upper and lower bounds. Following are the main points of conclusion.(i)
We have provided maximum and minimum values ofSEIa for conjugated trees.(ii)
We have also provided lower and upper bounds ofSEIa for unicyclic conjugated graphs with respect to the length of this cycle.(iii)
At the end of this paper, we have determined the maximum value ofSEIa for bicyclic graphs or n,n+1−graphs.



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