# Extremal Values of Variable Sum Exdeg Index for Conjugated Bicyclic Graphs

Authors: Muhammad Rizwan; Akhlaq Ahmad Bhatti; Muhammad Javaid; Ebenezer Bonyah
Journal: Journal of Chemistry
 (2021)

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## Abstract

A connected graphGV,E in which the number of edges is one more than its number of vertices is called a bicyclic graph. A perfect matching of a graph is a matching in which every vertex of the graph is incident to exactly one edge of the matching set such that the number of vertices is two times its matching number. In this paper, we investigated maximum and minimum values of variable sum exdeg index, SEIa for bicyclic graphs with perfect matching for k≥5and a>1.


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## Body

## 1. Introduction

LetG=VG,EG be a simple, finite, and connected graph, where the set of vertices and the set of edges are denoted by VG and EG, respectively. For any vertex t∈VG, dt or dtG is defined as the degree of vertex t. The vertices of degree 1 are called pendant vertices, whereas the vertices of degree greater than 2 are called branching vertices. A graph GV,E is said to be a tree and a bicyclic graph if the number of edges is one less than its vertices and one more than its vertices, respectively. Note that NGv represents the set which contains all the neighbouring vertices of v, whereas NGv=NGv∪v. The maximum degree and minimum degree of a graph G are denoted by ΔG and δG, respectively. An edge e which is obtained by joining the vertices r and s is denoted by e=rs. If r and s are nonadjacent vertices in a graph G, then the expression G+rs shows increment of an edge in G between vertices r and s. An edge with a pendent vertex is said to be pendent edge. If B′⊆EG and B⊆VG, then the subgraphs G−B′ and G−B have been obtained by removing the edges and vertices from G, respectively.A subsetMG⊂EG is said to be m-matching in G if no two elements in MG are adjacent, where |MG|=m. A perfect matching of a graph is a matching in which every vertex of the graph is incident to exactly one edge of the matching set such that n=2m. If a vertex u associates with an edge e∈MG, then we say that vertex u is saturated by MG. The following are the notations to denote the path graph, cycle graph, and star graph on n vertices Pn, Cn, and Sn, respectively. Let Tn,m denote the collection of all n-vertex trees with m matching. When we attach m−1 vertices of degree 1 to the noncentral vertices of Sn−m+1, the new graph is expressed by T0n,m. It is clear that m≤n+1/2. For m≤n/2, we have a tree T0n,m with m-matching, whereas for m=n+1/2 we have a tree T0n,m with m−1-matching. For any n-vertex tree, if n=2m, then it means tree has perfect matching. We assume that G∗,w1 and G∗∗,w2 are the graphs rooted at w1 and w2, respectively. Then G∗,w1⋓G∗∗,w2 is obtained by identifying w1 and w2 as the same vertex. Let G′ be a resulting graph when we apply some graph transformation on a graph G; that is, VG=VG′. When such graphs will be discussed, we will always consider dx as the degree of a vertex x in G. To read about the expressions and definitions related to this paper, we refer the readers to [1].In chemical graph theory, usually we can represent the molecular structure by graph, and, in these graphs, vertices correspond to atoms, whereas edges correspond to bonds [2]. To better understand physicochemical properties of a chemical graph, nowadays researchers are using topological indices. 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 to determine the extremal values of chemical structure because these extremal values of the chemical structure give information about the physicochemical properties of the related molecule. In this way, the study of the topological indices about chemical structure can provide useful information about a chemical compound without paying expensive experiments [4].There exist many topological indices in literature which have their application in theoretical chemistry, particularly in quantitative structure-activity relationship (QSAR) and quantitative structure-property relationship (QSPR) researches [5, 6]. Here we will discuss a few of them which are the oldest and most used topological indices in literature, for example, Wiener index, Platt index, and first Zagreb index.The great chemist Wiener first of all used the topological index in the year 1947 in the field of chemical graph theory. This index was a distance-based index which was named Wiener index. This index was proposed to predict boiling points of paraffins [7]. In 1952, the Platt index was proposed for predicting paraffin properties [8]. In 1972, for a molecule to calculate the total π-electron energy, the technique of sum of the squares of degrees (valencies) of the vertices (nodes) was used. After that, this topological index was named Zagreb index and became very famous [9]. Regarding extremal graph theory in literature, upper and lower bounds for cyclic graphs have been investigated for many well-swotted molecular structure descriptors such as hyper-Zagreb index [10], variable sum exdeg index [11], and Zagreb indices [12]. For more details about the topological indices, we refer the readers to [5, 13] and [14].For a graphG, Vukicevic [14] proposed variable sum exdeg index as(1)SEIaG=∑uv∈EGadu+adv=∑u∈VGduadu,where a>0 but a≠1. This topological index is correlated well with octane-water partition coefficient [14] and is employed for the study of octane isomers; see [15]. The role of this index in nanoscience can be seen in [16]. The authors investigated some bounds on SEIa for conjugated unicyclic graphs with respect to the length of its cycle [17]. We refer the readers to [13, 14] to see chemical application of this index.


## 2. Some Known Results

In the subsequent section, some lemmas will be listed and these lemmas will be used in proving our main theorems.Lemma 1 (see [17]).
Letm≥1, n≥4, and a>1 be integers and if T∈T2m,m, then SEIaT≤m⋅am+2m−1a2+am, and the sign of equality holds when T≅T02m,m.Lemma 2 (see [17]).
Letm≥1, n≥4, and a>1 be integers. If T∈T2m,m, then SEIaT≥2a+22m−2a2, and the sign of equality holds when T≅P2m.Lemma 3 (see [17]).
Suppose thatm≥1,a>1 and T∈T2m+1,m; then SEIaT≥SEIaP2m+1, and the sign of equality holds when T≅P2m+1.Lemma 4 (see [17]).
LetT∈T2m+1,m; then SEIaT≤SEIaT02m+1,m, and the sign of equality holds when T≅T02m+1,m, where m≥1,a>1.Lemma 5 (see [18]).
Suppose thatG∈Bk2m,m, k<2m; then, for every xi∈A, Txi∈Tni,ni/2 or Txi∈Tni,ni−1/2.Proposition 1.
: Ifa>1 and x>1, then fx=x⋅ax is an increasing function.Proof.
The proof is easy to understand from the definition ofSEIa.
From [19], we know that any graph GV,E with two cycles, say C1 and C2, can be depicted in three ways. The first possible way is that there is no common vertex between cycles C1 and C2 as shown in Figure 1(a): G1. The second one is that there is a single common vertex between cycles C1 and C2 as shown in Figure 1(b): G2. The last possible way is that there is a common path between cycles C1 and C2 as shown in Figure 1(c): G3. Let Bk12m,mrespectivelyBk22m,m,Bk32m,m be the collection of all 2m-vertex conjugated bicyclic graphs containing G1respectivelyG2,G3 as their subgraphs such that VG1=VG2=VG3=k. We also assume that Bk2m,m=Bk12m,m∪Bk22m,m∪Bk32m,m.
LetG be a bicyclic graph that have Gi,i=1,2,3, as a subgraph, where Gi=k. Here we define A=xi∈VGi:dxi≥3 and Txi is a connected component of G−NGixi, containing xi, where ni=VTxi.


## 3. Extremal Values of Variable Sum Exdeg Index,SEIa, for Conjugated Bicyclic Graphs

In this section, conjugated bicyclic graphs will be characterized with extreme values ofSEIa. First, we put up some lemmas which will be used in proving our main results.Lemma 6.
LetG1, G2, and G3 be the graphs as defined above such that VG1=VG2=VG3=n; then SEIaG1=SEIaG3<SEIaG2, where a>1.Proof.
From the structures ofG1, G2, and G3, it is easy to calculate the following differences:(2)SEIaG1−SEIaG3=2.3a3+n−22a2−2.3a3+n−22a2=0.
(3)SEIaG2−SEIaG1=4a4+n−12a2−2.3a3+n−22a2=4a4−3a3−3a3−2a2=aμ21+μ2lna−aμ11+μ1lna.
From the inequalitiesa>1, 2<μ1<3<μ2<4, and μ2>μ1, we have SEIaG2>SEIaG1.Lemma 7.
LetG∈Bk2m,m, k<2m, be a graph having minimum SEIa, where a>1; then, for every xi∈A, we have Txi=Pni, where xi is the end vertex of Txi.Proof.
LetG∈Bk2m,m be a graph having minimum SEIa, where a>1, and k<2m . We suppose on the contrary that Txi≠Pni; then there must exist a vertex, say v∈VTxi, such that dvTxi≥3. It is easy to see that there exist at least two pendent vertices, say v1 and v2, in G such that P1 and P2 are two distinct paths which are expressed by v−v1 and v−v2, respectively. We assume that x1 and x2 are the neighbouring vertices of v in paths P1 and P2, respectively. Since G∈Bk2m,m, there must exist at least one of the two vx1,vx2 edges, which does not belong to the matching set. We assume that x1v∉MG. We define G′=G−x1v+x1v2. We have(4)SEIaG−SEIaG′=dv⋅adv−dv−1⋅adv−1−dv2+1⋅adv2+1−dv2⋅adv2=aμ21+μ2lna−aμ11+μ1lna.
Fromμ1∈dv2,dv2+1,μ2∈dv−1,dv,μ2>μ1, and a>1, we have SEIaG>SEIaG′, and this contradicts with our choice about G.
There are two possible options for vertexxi related to its position. The fist possibility is that vertex xi is a pendent vertex in Txi and the second one is that vertex xi is not an end vertex of Txi. If xi is an end vertex of Txi, then the above portion of the proof is enough for proving the existing lemma; otherwise, we make the following proof.
Now we assume thatTxi=Pni and xi is not an end vertex in Txi. So dxiTxi=2 and Txi has two vertices of degree 1, say z1 and z2. Let u′ and u′′ be the neighbouring vertices of xi. We denote the path Pni=z1…u′xiu′′…z2. Since G∈Bk2m,m, at least one edge u′xi or u′′xi does not belong to matching set. Without loss of generality, we assume that xiu′′∉MG. We define G′=G−xiu′′+u′′z1, and we have(5)SEIaG′−SEIaG=dz1+1⋅adz1+1−dz1⋅adz1−dxi⋅adxi−dxi−1⋅adxi−1=aμ11+μ1lna−aμ21+μ2lna.
Fromμ1∈dz1,dz1+1,μ2∈dxi−1,dxi,μ2>μ1, and a>1, we have SEIaG>SEIaG′; this contradicts with our assumption. This completes the proof.Lemma 8.
LetG∈Bkj2m,m be a graph with k<2m and a>1 such that SEIaG is maximum, and j=1,2,3 for each xi∈A; we have the following:(1)
IfVTxi=ni≡0mod2, then Txi≅T0ni,ni/2(2)
IfVTxi=ni≡1mod2, then Txi≅T0ni,ni+1/2Proof.
LetG∈Bkj2m,m, k<2m, be a graph having maximum SEIaG for j=1,2,3. Let MG denote the perfect matching of G. In the proof of existing lemma, we will just show that results (1) and (2) are true for Bk22m,m. By this proof, we can also say that the above results are also true for G∈Bk12m,m and G∈Bk32m,m. For A≥1 and xi∈A, we make the following cases:Case 1: ifxi∉VC1∩VC2.We suppose thatxi∉VC1∩VC2. Let xi−1 and xi+1 be the neighbouring vertices of xi along cycle C1 or C2. So we can assume the following expression:(6)Q=dxiadxi−dxi−2adxi−2+dxi−1adxi−1−dxi−1−1adxi−1−1+dxi+1adxi+1−dxi+1−1adxi+1−1.LetG∗=G−xixi−1−xixi+1 be a connected component and it does not contain xi, so we have SEIaG=SEIaG∗+Q+SEIaTxi. By Lemma 5, xi∈A, Txi∈Tni,ni/2, or Txi∈Tni,ni−1/2, and we observe dxi−2≤△Txi. So we make the following subcases:Subcase 1.1: whenni is even.Ifni is even, then △Txi≤ni/2. By Lemma 5, Lemma 1, and Proposition 1, we have SEIaG≤SEIaG∗+Q+SEIaT0ni,ni/2, and the sign of equality holds when Txi≅T0ni,ni/2. Hence, we have the desired result.Subcase 1.2: whenni is odd.Ifni is odd, then △Txi≤ni−1/2+1. It is easy to check that if dxi−2=△Txi=ni−1/2+1, then graph G has no perfect matching. So dxi−2<△Txi or dxi−2=△Txi<ni−1/2+1. If we choose dxi−2<△Txi, then △Txi≥2 and ni≥3. Here we consider du=△Txi=△.Subcase 1.2.1: if△Txi=2 and ni=3, then Txi≅P3≅T03,2 and the result holds.Subcase 1.2.2: if we consider that the maximum degree ofTxi is 2 but ni≥5 or Txi≅Pni, where xi is the pendent vertex of Txi, let Txi=xi…pqr. We define H1=G−pq+qxi and clearly H1∈Bk22m,m. We have(7)SEIaH1−SEIaG=dxi+1adxi+1−dxiadxi−2a2−a=aμ21+μ2lna−aμ11+μ1lna.Fromμ1∈du−1,du,μ2∈dxi,dxi+1,μ2>μ1 and a>1, we have SEIaH1>SEIaG; this contradicts with our choice about G.Subcase 1.2.3: if△Txi≥3, then there exist △−2 vertices, say y1,y2,….y△−2, each of which does not exist on a path whose end points are xi and u, where yiu∉MG for i=1,2,…△−2.We defineH2=G−∑t=1△−2uyt+∑t=1△−2xiyt and clearly H2∈Bk22m,m.Whendxi≥du, we have(8)SEIaH2−SEIaG=dxi+du−2adxi+du−2−dxiadxi−duadu−2a2=du−2aμ21+μ2lna−aμ11+μ1lna.Fromμ1∈2,du,μ2∈dxi,dxi+du−2,μ2>μ1, and a>1, we conclude that SEIaH2>SEIaG; this contradicts with our assumption.Whendxi<du, we have(9)SEIaH2−SEIaG=dxi+du−2adxi+du−2−duadu−dxiadxi−2a2=dxi−2aμ21+μ2lna−aμ11+μ1lna.Fromμ1∈2,dxi,μ2∈du,dxi+du−2,μ2>μ1, and a>1, we find that SEIaH2>SEIaG; again this contradicts with our assumption.Ifdxi−2=△Txi=ni−1/2, then Txi≅T0ni,ni+1/2 and dxi−2=△T0ni,ni+1/2 as required.Ifdxi−2=△Txi<ni−1/2, then we make the two following subcases:Subcase 1.2.3.1: we consider a vertexw∈VTxi−xi with dw≥3. As we know that dw≥3, we assume that w1∈Nw such that w1 does not exist on the path whose end vertices are xi and w and ww1∉MG. We define H3=G−ww1+w1xi. Obviously H3∈Bk22m,m. We have(10)SEIaH3−SEIaG=dxi+1adxi+1−dxiadxi−dwadw−dw−1adw−1=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈dw−1,dw,μ2∈dxi,dxi+1,μ2>μ1, for a>1SEIaH3>SEIaG. Again, we have a contradiction.Subcase 1.2.3.2: for all the verticesw∈VTxi−xi, dw≤2. In this subcase, we have two possible cases. Either Txi−xi is a connected component, say J, such that J≅Pl≥4, or there exist at least two connected components in Txi−xi, say J1 and J2, such that J1≅P3≅J2. We consider the former because in case the latter happens, it becomes easy to check G∉Bk22m,m (because G does not have perfect matching). We consider a pendent vertex w1 in J such that w1 is not adjacent to xi in G, where w2∈Nw1. We define H4=G−w1−w2+w2xi+w1w2 and clearly H4∈Bk22m,m. We have(11)SEIaH4−SEIaG=dxi+1adxi+1−dxiadxi−2a2−a=aμ21+μ2lna−aμ11+μ1lna,whereμ1∈1,2,μ2∈dxi,dxi+1,μ2>μ1, where a>1SEIaH4>SEIaG; a contradiction arises.Case 2:xi∈VC1∩VC2.Suppose thatNxi∩NC1=x1,x2 and Nxi∩NC2=x3,x4. We consider the expression Q=∑t=14dxt−dxt−1+dxiadxi−dxi−4adxi−4. Suppose that G∗∗ and G∗∗∗ are the connected components of graph G−xixj, where j=1,2,3,4 and i≠j. Graphs G∗∗ and G∗∗∗ do not contain vertex xi but have vertices x1,x2 and x3,x4, respectively. So we have SEIaG=SEIaG∗∗+SEIaG∗∗∗+Q+SEIaTxi. According to the same arguments to the proof of case 1, the result holds when G∈Bk22m,m. Hence, the proof of Lemma 8 is accomplished.
Now, by using Lemmas1–8, we are going to prove our main results.Theorem 1.
IfG∈Bk12m,m with k<2m, then SEIaG≥3⋅3a3+2m−42⋅a2+a and the sign of equality holds when G≅G1,xj⋓P2m−k+1,xj, where dxjG1=2, a>1, and G1 is a graph as shown in Figure 1.Figure 1
(a)G1, (b) G2, and (c) G3.
(a)(b)(c)Proof.
LetG∈Bk12m,m be a graph with k<2m, which minimizes SEIa, where a>1. For the choice of G, Lemma 7 ensures that, for every vertex xi∈A, Txi is isomorphic to Pni, where Pni is a pendent path attached to xi. In the current situation, set A will not contain any vertex with degree greater than or equal to 4; otherwise, we will find a graph in Bk12m,m whose SEIa value is minimum compared to G. If A=3, then we have the desired result. Now we assume that A≥4 and dw=3 for all w∈A. Obviously Txj≅Pnjj=1,2…A. We represent Txj=v0jv1j…vtjjtj≥1, where v0j=xjj=1,2,….A. We define G′ by applying some transformation on G as shown in Figure 2, such that G′=G−v02v12−v03v13−…−v0Av1A+vt11v12+vt22v13+…+vt|A|−1|A|−1v1A. We have(12)SEIaG′−SEIaG=A−12a2−a−3a3−2a2=A−1aμ11+μ1lna−aμ21+μ2lna.
Fromμ1∈1,2,μ2∈2,3,μ2>μ1, and a>1, we have SEIaG′<SEIaG; this contradicts with our choice about G.Figure 2
G and G′ in Theorem 1.Theorem 2.
LetG∈Bk32m,m with k<2m; then SEIaG≥3⋅3a3+2m−42⋅a2+a and equality holds when G≅G3,xj⋓P2m−k+1,xj, where dxjG3=2, a>1, and G3 is graph shown in Figure 1.Proof.
The proof can be drawn from the proof of Theorem1.Theorem 3.
LetG∈Bk22m,m with k<2m; then SEIaG≥4a4+3a3+2m−32⋅a2+a and the sign of equality holds when G≅G2,xj⋓P2m−k+1,xj, where dxjG2=2, a>1, and G2 is a graph shown in Figure 1.Proof.
LetG∈Bk22m,m be a graph with k<2m, which minimizes SEIa, a>1. For the choice of G, Lemma 7 ensures that, for every xi∈A, Txi is isomorphic to Pni, where Pni is a pendent path attached to xi. In the current situation, set A will not have any vertex with degree greater than or equal to 5; otherwise, we will find a graph in Bk22m,m whose SEIa value is minimum compared to G. If A=2, then we have the desired result. Now we assume that A≥3 and dz=3 for all z∈A−x, where dxG2=4. According to the same reasoning defined in Theorem 1, we obtain a contradiction and we complete the proof of Theorem 3.Theorem 4.
LetG∈Bk2m,m with a>1; then the following holds:(i)
If2m=k, then SEIa≥2⋅3a3+2m−22⋅a2 and the sign of equality holds when G≅G3 or G≅G1, where G1 and G3 are shown in Figure 1 with VG1=2m=VG3(ii)
If2m≥k+1, then SEIa≥3⋅3a3+2m−42⋅a2+a and the equality holds when G≅Gi,xj⋓P2m−k+1,xj, where dxjGi=2 and Gi is a graph shown in Figure 1 with i=1,3Proof.
(i)
According to Lemma6, we have SEIaG≥SEIaG1=SEIaG3=2⋅3a3+2m−22⋅a2 which is desired.(ii)
Whenk<2m, we choose graphs H1, H2, and H3 in Bk12m,m, Bk22m,m, and Bk32m,m, respectively, where SEIaH1, SEIaH2, and SEIaH3 are minimum. According to Theorem 1, Theorem 2, and Theorem 3, respectively, we obtainSEIaH1=SEIaH3=3⋅3a3+2m−42⋅a2+aSEIaH2=4a4+3a3+2m−32⋅a2+a
ForSEIa, we have(13)SEIaH1−SEIaH2=3.3a3+2m−42.a2+a−4a4+3a3+2m−32.a2+a=3a3−2.a2−4a4−3a3=aμ11+μ1lna−aμ21+μ2lna.
Fromμ1∈2,3,μ2∈3,4,μ2>μ1, and a>1, we have SEIaH1<SEIaH2.
Here we define a graphAk which is obtained from C2r+1,u0⋓C2t,v0 by attaching one pendent edge to v0 and one pendent edge to v2i−1, respectively, i∈1,2,…t,2r+2t=k, as shown in Figure 3.Figure 3
GraphC2r+1,u0⋓C2t,v0, r≥1, t≥2.Theorem 5.
LetG∈Bk22m,m be a graph with 5≤k<2m and a>1. Then the following results will hold:(1)
If2m=k+1, then SEIaG≤5⋅a5+2m−22a2+a and the equality holds if G≅G∗, where graph G∗ is acquired from G2 when we join a pendent edge to the vertex of degree 4(2)
If2m=k+2, then SEIaG≤2a+2m−42a2+3⋅a3+5⋅a5 and the equality holds if and only if G∈Ak(3)
If2m≥k+3, where k is odd, then SEIaG≤m−k−1/2a+m+k−3/22a2+m−k−9/2m−k−9/2 and the sign of equality holds if G≅G2,xi⋓T02m−k+1,2m−k+1/2,xi, where dxiG2=4 and dxi−4=ΔT02m−k+1,2m−k+1/2(4)
If2m≥k+3, where k is even, then SEIaG≤m−k−2/2a+m+k−6/22a2+3a3+m−k−8/2am−k−8/2 and the sign of equality holds if G≅G2∗,xi⋓T02m−k,2m−k/2,xi, where dxiG2=4 and dxi−4=ΔT02m−k,2m−k/2Proof.
For any graphG∈Bk22m,m with 2m=k+1, we can construct two types of graphs from Bk22m,m, say S1 and S2, whereas S1 and S2 are obtained from G2 by attaching a pendent edge to a vertex of degree 4 and degree 2, respectively. We have(14)SEIaS1−SEIaS2=5a5−4a4−3a3−2a2=aμ21+μ2lna−aμ11+μ1lna,>0,where μ1∈2,3,μ2∈4,5,μ2>μ1, and a>1, and we have SEIaS1>SEIaS2. Thus, claim (1) is true.
(2) For2m=k+2, first we find SEIa values of all graphs which are shown in Figure 4. We have(15)SEIaD1=5a5+2m−22⋅a2+a,SEIaD2=4a4+3a3+2m−32⋅a2+a,SEIaD3=5a5+3a3+2m−42⋅a2+2a,SEIaD4=SEIaD5=4a4+2⋅3a3+2m−52⋅a2+2a.
It becomes easy to calculate from Figure4 that MaxSEIaD1,SEIaD2,SEIaD3,SEIaD4,SEIaD5=SEIaD3. Hence, the proof of (2) is finished.
(3) For2m≥k+3, we assume that G∈Bk22m,m such that SEIaG is maximum. According to Lemma 8, Txi≅T0ni,ni/2 or Txi≅T0ni,ni+1/2. For every vertex xi∈A with VTxi=ni being odd or even, here we assume that x∈VC1∩VC2. Here we choose that MG is a perfect matching in G. We consider the following cases:Case 1: whenA=1IfA=1, according to Lemma 8, proof of (3) is straightforwardCase 2: whenA=2.If|A|=2, then we can assume that A=x,x1Subcase 2.1:n0=VTx=1.We haven0=VTx=1 and we know that 2m≥k+3 so n1=VTx1≥4 and here we choose that n1 is even. Since G has maximum SEIa, Tx1≅T0n1,n1/2 and dx1≥4. There must exist dx1−3 vertices, say z1,z2,…zdx1−3, in Nx1∩VTx1; that is, x1zi∉MG, where i=1,2,…dx1−3. We define G1=G−∑t=1dx1−3x1zt+∑t=1dx1−3xzt. Clearly G1∈Bk22m,m, so we have(16)SEIaG1−SEIaG=dx1+1adx1+1−dx1adx1−4a4−3a3=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈3,4,μ2∈dx1,dx1+1,μ2>μ1, and a>1, and SEIaG1>SEIaG; this contradicts with our choice about G.Subcase 2.2: letn0=VTx=2 and VTx1=n1≥3, where n1 is odd. Since we have assumed that G has maximum SEIa, Tx1≅T0n1,n1+1/2 and dx1≥3. Let Nx1∩VTx1=z1,z2,…zdx1−2,x1zi∉MG, where i=1,2,…dx1−2. We define G2=G−∑t=1dx1−2x1zt+∑t=1dx1−2xzt. Clearly G2∈Bk22m,m.Whendx1≥5,(17)SEIaG2−SEIaG=5+dx1−2a5+dx1−2−dx1adx1−5a5−2a2=3⋅aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈2,5,μ2∈dx1,dx1+3,μ2>μ1, and a>1SEIaG2>SEIaG; this contradicts with our choice about G.Whendx1<5, we have(18)SEIaG2−SEIaG=5+dx1−2a5+dx1−2−5a5−dx1adx1−2a2=dx1−2aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈2,dx1,μ2∈5,dx1+3,μ2>μ1, and a>1SEIaG2>SEIaG; this contradicts with our choice about G.Subcase 2.3: letVTx=n0≥3 and VTx1=n1=2; here we assume that n0 is odd. According to Lemma 8, Tx≅T0n0,n0+1/2. Since G∈Bk22m,m, there exists a vertex w∈Nx∩VG2, where xw∈MG. Let Nw−x=t. Nx1=w1,w2,z1, where dz1=1. We define G3=G−wt−w1x1−w2x1+xx1+z1t+w1w2. Clearly G3∈Bk22m,m.(19)SEIaG3−SEIaG=dx1+1adx1+1−dxadx−3a3−2a2=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈2,3,μ2∈dx,dx+1,μ2>μ1, and a>1SEIaG3>SEIaG, a contradiction to our assumption.Subcase 2.4: letn0=VTx and VTx1=n1≥3. Without loss of generality, we assume that dx≥dx1. Obviously, there will be a path P∗=x1z1z2 in Tx1 such that dz1=2 and dz2=1. We define G4=G−x1z1+xz1 and clearly G4∈Bk22m,m.(20)SEIaG4−SEIaG=dx+1adx+1−dxadx−dx1adx1−dx1−1adx1−1=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈dx1−1,dx1,μ2∈dx,dx+1,μ2>μ1, where a>1SEIaG4>SEIaG, a contradiction to our assumption.Case 3: letn0=VTx and VTxi=ni, where xi∈A−x, i=1,2,…A−1. Here we consider the following cases:Subcase 3.1: we assume that there exists at least one vertex, sayx1∈A−x, such that n1≥3.Subcase 3.1.1: letn1=3; this implies that Tx1≅P3, where x1 is the pendent vertex in P3. Suppose that Tx1=x1t1t2, where dt1=2 and dt2=1. We define G5=G−x1t1+xt1. Clearly G5∈Bk22m,m.(21)SEIaG5−SEIaG=dx+1adx+1−dxadx−3.a3−2a2=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈2,3,μ2∈dx,dx+1,μ2>μ1, where a>1SEIaG5>SEIaG, a contradiction to our assumption.Subcase 3.1.2: letn1≥4, so there must exist at least dx1−3 vertices, say z1,z2,…zdx1−3, in Nx1∩VTx1; that is, dzj=2, j=1,2,…dx1−3. We define G6=G−∑t=1dx1−3x1zt+∑t=1dx1−3xzt. Clearly G6∈Bk22m,m.Whendx1≤dx,(22)SEIaG6−SEIaG=dx+dx1−3adx+dx1−3−dxadx−dx1.adx1−dx1−dx1+3adx1−dx1+3=dx1−3aμ21+μ2lna−aμ11+μ1lna,fromμ1∈3,dx1,μ2∈dx,dx+dx1−3,μ2>μ1, and a>1 we have SEIaG6>SEIaG; this contradicts with our choice about G.Whendx1>dx,(23)SEIaG6−SEIaG=dx+dx1−3adx+dx1−3−dx1.adx1−dxadx−3a3=dx−3aμ21+μ2lna−aμ11+μ1lna,fromμ1∈3,dx,μ2∈dx1,dx+dx1−3,μ2>μ1, and a>1, we have SEIaG6>SEIaG; this contradicts with our choice about G.fromμ1∈3,dx,μ2∈dx1,dx+dx1−3,μ2>μ1, and a>1, we have SEIaG6>SEIaG; this contradicts with our choice about G.Subcase 3.2: letni=2 for every vertex xi∈A−x, i=1,2,…A−1. Let Nxi∩VTxi=zi and Nxi−zi=wi1,wi2, i=1,2. We define G7=G−x1z1−x2z2+xz1+z1z2. Clearly G7∈Bk22m,m.
We have(24)SEIaG7−SEIaG=dx+1adx+1+dx1−1adx1−1+dx2−1adx2−1+dz1+1adz1+1−dxadx+dx1adx1+dx2adx2+dz1adz1,where a>1, and we have SEIaG7>SEIaG; here we obtain a contradiction. This completes the proof of (3). By using the same arguments of proof of (3), the proof of (4) can be solved.Figure 4
D1,D2, D3,D4,D5.Theorem 6.
LetG∈Bk12m,m∪Bk32m,m be a graph with 5≤k<2m, and a>1. Then the following results will hold:(1)
If2m=k+1, then SEIa≤4⋅a4+3⋅a3+2m−32a2+a and the sign of equality holds if and only if G≅Gj∗′, j=1,3, where graph Gj∗′ is acquired from Gj when we join a pendent edge to the degree 3 vertex.(2)
If2m=k+2, then SEIa≤2⋅4a4+2m−42a2+2a and the sign of equality holds if and only if G is obtained from Gj,j=1,3, when each vertex of degree 3 has a pendent edge(3)
If2m≥k+3, where k is odd, then SEIaG≤m−k−1/2a+m+k−5/22a2+3.a3+m−k−7/2am−k−7/2 and the sign of equality holds if G≅Gj,xi⋓T02m−k+1,2m−k+1/2,xi, where dxiGj=3 and dxi−3=ΔT02m−k+1,2m−k+1/2, j=1,3(4)
If2m≥k+3, where k is even, then SEIaG≤m−k−2/2a+m+k−6/22a2+4⋅a4+m−k−6/2am−k−6/2 and the sign of equality holds if G≅Gj∗,xi⋓T02m−k,2m−k/2,xi, where dxiGj∗=3 and dxi−3=ΔT02m−k,2m−k/2, j=1,3Theorem 7.
LetG∈Bk2m,m be a graph with 5≤k, and a>1. Then the following results will hold:(1)
If2m=k, then SEIaG≤4⋅a4+2m−12a2+a and the sign of equality holds if and only if G≅G2(2)
If2m=k+1, then SEIaG≤5a5+2m−22a2+a and the sign of equality holds if and only if G≅G∗, where G∗ is obtained by attaching a pendent edge to the vertex of degree 4(3)
If2m=k+2, then SEIaG≤5a5+3a3+2m−42a2+2a and the sign of equality holds if and only if G≅Ak(4)
If2m≥k+3, where k is odd, then SEIaG≤m−k−1/2a+m+k−3/22a2+m−k−9/2am−k−9/2 and the equality holds if G≅G2,xi⋓T02m−k+1,2m−k+1/2,xi, where dxiG2=4 and dxi−4=ΔT02m−k+1,2m−k+1/2(5)
If2m≥k+3, where k is even, then SEIaG≤m−k−2/2a+m+k−6/22a2+3⋅a3+m−k−8/2am−k−8/2 and the sign of equality holds if G≅G2∗,xi⋓T02m−k,2m−k/2,xi, where dxiG2j=4 and dxi−4=ΔT02m−k,2m−k/2Proof.
(1)
According to Lemma6, for 2m=k, we have SEIaG≤SEIaG2=4⋅≤a4+2m−12⋅a2.(2)
For2m=k+1, according to Theorems 5 and 6, we obtain G∗ and G∗′, where SEIaG∗=5a5+2m−22a2+a; SEIaG∗′=4a4+3a3+2m−32a2+a.(25)SEIaG∗−SEIaG∗′=5a5+2m−22a2+a−4a4+3a3+2m−32a2+a=5a5−4a4−3a3−2a2=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈2,3,μ2∈4,5,μ2>μ1, and a>1SEIaG∗>SEIaG∗′.(3)
For2m=k+2, according to Theorems 5 and 6, we obtain G∗ and G∗′, where G∗ is obtained when we attach one pendent edge to the vertex of degree 4 and one with degree 2, whereas G∗′ is obtained from G1 or G3 by attaching one pendent edge to each vertex of degree 3. SEIaG∗=5a5+3.a3+2m−42a2+2a; SEIaG∗′=2.4a4+2m−42a2+2a.We calculate the difference:(26)SEIaG∗−SEIaG∗′=5a5+3.a3+2m−42a2+2a−2.4a4+2m−42a2+2a=5a5−4a4−4a4−3a3=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈3,4,μ2∈4,5,μ2>μ1, and a>1, so we have SEIaG∗>SEIaG∗′.(4)
For2m≥k+3, where k is odd, according to Theorems 5 and 6, we have two graphs E1 and E2, respectively. SEIaE1=m−k−1/2a+m+k−3/22a2+m−k−9/2am−k−9/2; SEIaE2=m−k−1/2a+m+k−5/22a2+3a3+m−k−7/2am−k−7/2.We calculate the difference:(27)SEIaE1−SEIaE2=m−k−92am−k−9/2−m−k−72am−k−7/2−3a3−2a2=aμ21+μ2lna−aμ11+μ1lna,>0,whereμ1∈2,3,μ2∈m−k−7/2,m−k−9/2,μ2>μ1, for a>1SEIaE1>SEIaE2. Hence, result (4) satisfied.(5)
For2m≥k+3, where k is even, according to Theorems 5 and 6, we have two graphs E3 and E4, respectively. SEIaE3=m−k−2/2a+m+k−6/22a2+3a3+m−k−8/2am−k−8/2; SEIaE4=m−k−2/2a+m+k−6/22a2+4a4+m−k−6/2am−k−6/2.
We calculate the difference:(28)SEIaSEIaE3−SEIaSEIaE4=m−k−82am−k−8/2−m−k−62am−k−6/2−4a4−3a3=aμ21+μ2lna−aμ11+μ1lna,>0,where μ1∈3,4,μ2∈m−k−6/2,m−k−8/2,μ2>μ1, a>1SEIaE3>SEIaE4, and, hence, result (5) satisfied. Hence, Theorem 7 is finished.


## 4. Conclusion

Determining the sharp bounds on any topological index regarding various graph parameters is a significant job. In this paper, for conjugated bicyclic graphs, extremal values ofSEIa have been explored. Extremal graphs for each sharp bound have also been examined. The following are the main points of conclusion:Maximum and minimum values ofSEIa for conjugated bicyclic graphs from Bk12m,m, Bk22m,m, and Bk32m,m have been found. Lower and upper bounds of SEIa for Bk2m,m have been also investigated [20, 21].



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