# On the Minimum Variable Connectivity Index of Unicyclic Graphs with a Given Order

Authors: Shamaila Yousaf; Akhlaq Ahmad Bhatti; Akbar Ali
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

The variable connectivity index, introduced by the chemist Milan Randić in the first quarter of 1990s, for a graphG is defined as ∑vw∈EGdv+γdw+γ−1/2, where γ is a non-negative real number and dw is the degree of a vertex w in G. We call this index as the variable Randić index and denote it by Rvγ. In this paper, we show that the graph created from the star graph of order n by adding an edge has the minimum Rvγ value among all unicyclic graphs of a fixed order n, for every n≥4 and γ≥0.


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## Body

## 1. Introduction

All graphs that we discuss in the present study are simple, connected, undirected, and finite. For a graphG=V,E, the number VG is called its order and EG is called its size. For a vertex v1∈VG, denoted by Nv1, the set of all those vertices of G are adjacent with v1. The number dv1=Nv1 is called degree of v1. If dv1=1, then v1 is called a pendent vertex or a leaf. A graph of order n is known as an n-vertex graph. As usual, the n-vertex path and star graphs are denoted by Pn and Sn, respectively. An n-vertex graph containing exactly one cycle is called a unicyclic graph. The class of all n-vertex unicyclic graphs is denoted by Un. The graph obtained from Sn by adding an edge is denoted by Sn1. For the (chemical) graph theoretical notation and terminology that are not defined in this paper, we refer the reader to some standard books, such as [1–3].To model the heteroatoms molecules, it is better to use the vertex-weighted graphs, which are the graphs whose one or more vertices are distinguished in some way from the rest of the vertices [4]. Let G be a vertex-weighted graph with the vertex set v1,v2,…,vn and the vertex weight wi of the vertex vi is for i=1,2,…,n. The augmented vertex-adjacency matrix of G is an n×n matrix denoted by avAG and is defined as avAG=ai,jn×n, where(1)ai,j=1,ifvivj∈EG,wi,ifi=j,0,otherwise.The variable connectivity index [5, 6], proposed by Randić, for the graph G is defined as(2)χ1fG=∑vivj∈EG1dvi+widvj+wj.We associate this index’s name with its inventor Randić by calling it as the variable Randić index. This index was actually introduced within the QSPR/QSAR (quantitative structure-property/activity relationship) studies of heteroatoms molecules. IfG is the molecular graph of a homoatomic molecule, then w1=w2=⋯=wn=γ (say) and hence the variable Randić index χ1fG becomes(3)χ1fG=∑vivj∈EG1dvi+γdvj+γ.In the rest of this paper, we denote this index byRvγ instead of χ1f. Clearly, if we take γ=0 then the invariant RvγG is the classical Randić index [7, 8]. Details about the chemical applications of the variable Randić index can be found in [4, 7, 9–17] and related references listed therein. In [18], a mathematical study of the variable Randić index was initiated and it was proved that the star graph has the minimum variable Randić index among all trees of a fixed order n, where n ≥ 4. It needs to be mentioned here that the variable Randić index seems to have more chemical applications than the several well-known variable indices, see, for example, the variable indices considered in the papers [19–25].For convenience, we introduce some further notation and terminology. LetAn,l,jk be the unicyclic graph constructed from a path P=u0u1…ujj≥1 by connecting k pendant vertices to u0 and a cycle Cl,l=n−j−k to uj, respectively. Let An,lk be the unicyclic graph constructed from cycle Cl,l=n−k by connecting k pendant vertices to one vertex of Cl and An,li,j,k be the unicyclic graph created from cycle Cl,n=i+j+k+l by attaching K1,i, K1,j, and K1,k to the vertices of Cl. Let D be the such class of Un whose every member has unique 3-cycle, and the vertices apart from 3-cycle are pendent vertices. Let F be the such class of Un whose every member has unique 4-cycle, and the vertices apart from 4-cycle are pendent vertices and are joined to nonadjacent vertices of the unique 4-cycle.In this paper, we characterize the collection of unicyclic graphs onn vertices that minimize variable Randić index. We further show that, for γ≥0, Sn1 has minimum variable Randić index among the collection Un and(4)RvγG≥n−31+γn−1+γ+22+γn−1+γ+12+γ2+γ,where equality holds if and only if G≅Sn1.


## 2. Transformations Which Decrease the Variable Randić Index

We introduce some transformations and prove some lemmas to establish main results.Lemma 1.
Ifγ≥0 and s,t≥1, then the function Ψ defined as(5)Ψγ,s,t=s1+γ1s+2+γ−1s+t+3+γ+t1+γ1t+2+γ−1s+t+3+γ+1s+2+γt+2+γ−11+γs+t+3+γ,is positive valued.Proof.
We note that the function∂Ψ/∂s is strictly increasing in t on the interval 1,∞ because(6)∂Ψ∂s=1γ+11s+2+γ−1s+t+3+γ+s2γ+11s+t+3+γ3/2−1s+2+γ3/2+t2γ+1s+t+3+γ3/2−12t+2+γs+2+γ3/2+12γ+1s+t+3+γ3/2,∂∂t∂Ψ∂s=141s+2+γ3/2t+2+γ3/2+s+t+9+4γ1+γs+t+3+γ5/2>0.
Also, it can be seen that the value of the function∂Ψ/∂s at t=1 is >0, which implies that the function ∂Ψ/∂s is positive valued for t>1, and hence the function Ψ is strictly increasing w.r.t. s on the interval 1,∞. In our case Ψγ,s,t=Ψγ,s,t, the function Ψ is also strictly increasing w.r.t. t on the interval 1,∞. Hence, Ψγ,s,t>0 for all s,t>1 and γ≥0.Lemma 2.
Ifγ≥0 and s,t≥1, then the function Φ defined as(7)Φγ,s,t=s1+γ1s+2+γ−1s+t+4+γ+t1+γ1t+2+γ−1s+t+4+γ+1s+2+γ2+γ−11+γs+t+4+γ+1t+2+γ2+γ−11+γs+t+4+γ,is positive valued.
The proof of Lemma2 is analogous to the proof of Lemma 1.Lemma 3.
Ifγ≥0, t>2, and r>1, it holds that(8)Ωγ,r,t=r−11+γ1r+γ−1r+t+γ+1r+γt+1+γ−11+γr+t+γ>0.Proof.
Let(9)Ωγ,r,t=r−1t−11+γr+γt+γr+t+γ×t+γr+γ+r+t−1+γ−1r+γt+γ+1+γr+t−1+γ=r−1t−11+γr+γt+γr+t−1+γ×t+γr+γt+γ+1+γr+t−1+γr+γ+r+t−1+γ)(r+γ)(t+γ+1+γ)(r+t−1+γ−r+γ+r+t−1+γr+γ+r+t−1+γr+γt+γ+1+γr+t−1+γ=r−1t−11+γr+γt+γr+t−1+γ×r+γt+γ−1+r+t−1+γ1+γt+γ−1r+γ+r+t−1+γr+γt+γ+1+γr+t−1+γ.Since γ≥0, t>2 so t+γ>0, also 1+γt+γ−1>0. Hence, Ωγ,r,t>0.Lemma 4.
Forγ≥0, it holds that(10)fγ=22+γ1γ+2−1γ+3+12+γ2+γ−11+γγ+3>0.
The proof of Lemma4 is analogous to the proof of Lemma 3.Lemma 5.
Forγ≥0, it holds that(11)fγ=11+γ1γ+3−1γ+4+22+γ1γ+3−1γ+4+12+γ2+γ−11+γγ+4>0.
The proof of Lemma5 is analogous to the proof of Lemma 3.
Now, we use three transformations, which will reduce the variable Randić index.


### 2.1. Transformation 1

LetG∈Un and uv be an edge of G such that dv=r≥2. Nv=u,w1,w2,…,wr−1, where w1,w2,…,wr−1 are leaves.DefineG´=G−vw1,vw2,…,vwr−1+uw1,uw2,…,uwr−1. G and G´ are depicted in Figure 1.Figure 1
The graphG´ (obtained using Transformation 1) from G.Lemma 6.
LetG,G´∈Un. By applying Transformation 1 and for γ≥0, we have(12)RvγG>RvγG´.Proof.
Letdu=t≥3:(13)RvγG−RvγG´>dv−11+γdv+γ+1dv+γdu+γ−dv1+γdv+du−1+γ,RvγG−RvγG´>dv−11+γ1dv+γ−1dv+du−1+γ+1dv+γdu+γ−11+γdv+du−1+γ,RvγG−RvγG´>r−11+γ1r+γ−1t+r−1+γ+1r+γt+γ−11+γt+r−1+γ>0.
Using Lemma3, one can see that inequality (13) holds.


### 2.2. Transformation 2

LetG,G´∈Un such that G=G´ and their unique cycles are r−cycle and r−1−cycle, respectively, r≥4. For both graphs, the vertices apart from the cycle are leaf vertices joined to exactly one vertex of the cycle.Lemma 7.
LetG,G´∈Un such that G=G´. By applying Transformation 2 and for γ≥0, we have(14)RvγG>RvγG´.Proof.
Since(15)RvγG=r−2γ+2γ+2+n−rγ+1n−r+γ+2+2γ+2n−r+γ+2,RvγG´=r−3γ+2γ+2+n−r+1γ+1n−r+γ+3+2γ+2n−r+γ+3,RvγG−RvγG´=n−r1+γ1n−r+γ+2−1n−r+γ+3+22+γ1n−r+γ+2−1n−r+γ+3+12+γ2+γ−11+γn−r+γ+3.
Now, there will be three cases.Case I: ifn=r, then(16)RvγG−RvγG´=22+γ1γ+2−1γ+3+12+γ2+γ−11+γγ+3.Using Lemma4, one can see that relation (16) is positive valued.Case II: ifn−r=1, then(17)RvγG−RvγG´=11+γ1γ+3−1γ+4+22+γ1γ+3−1γ+4+12+γ2+γ−11+γγ+4.Using Lemma5, one can see that relation (17) is positive valued.Case III: ifn−r≥2, then(18)RvγG−RvγG´=n−r1+γ1n−r+γ+2−1n−r+γ+3+22+γ1n−r+γ+2−1n−r+γ+3+12+γ2+γ−11+γn−r+γ+3,RvγG−RvγG´>12+γ2+γ−11+γn−r+γ+3>0.Sincen−r1+γ1n−r+γ+2−1n−r+γ+3>0and22+γ1n−r+γ+2−1n−r+γ+3>0.


### 2.3. Transformation 3

LetG∈Un with G=n and u,v∈VG are on the unique cycle of G. Pendant neighbors of u and v are ui, where 1≤i≤s, and vj, where 1≤j≤t, respectively. Assume that the path between u and v on cycle is Puv such that EPuv=r, dVPuv=2 if r≥2 and EPvu≥3. Construct G´ from G by removing the edges uui and vvj, reducing the path Puv into one vertex uv and attaching a star K1,r+s+t+1 to uv, by making sure that G´=G. G and G´ are depicted in Figure 2.Figure 2
The graphG and G´ (obtained using Transformation 3).Lemma 8.
LetG,G´∈Un such that G´=G. By applying Transformation 3 and for γ≥0, s,t≥1, we have(19)RvγG>RvγG´.Proof.
Case I: letr=1 and uv∈EG, then(20)RvγG−RvγG´>s1+γs+2+γ+t1+γt+2+γ+1s+2+γt+2+γ−s+t+11+γs+t+3+γ,RvγG−RvγG´>s1+γ1s+2+γ−1s+t+3+γ+t1+γ1t+2+γ−1s+t+3+γ+1s+2+γt+2+γ−11+γs+t+3+γ.Using Lemma1, one can see that relation (20) is positive valued.Case II: now letr≥2, then(21)RvγG−RvγG´>s1+γs+2+γ+t1+γt+2+γ+1s+2+γ2+γ+12+γt+2+γ+r−22+γ2+γ−s+t+r1+γs+t+r+2+γ.
Sincer≥2 and keeping in mind the fact that(22)Ψr,s,t=s+t+r1+γs+t+r+2+γ,is an increasing function because(23)∂Ψ∂r=s+t+r+4+2γ2γ+1s+t+r+2+γ3/2>0.
We have(24)RvγG−RvγG´>s1+γ1s+2+γ−1s+t+4+γ+t1+γ1t+2+γ−1s+t+4+γ+1s+2+γ2+γ−11+γs+t+4+γ+1t+2+γ2+γ−11+γs+t+4+γ.
Using Lemma2, one can see that relation (24) is positive valued.
LetDn,3i,j,k be the such class of D that is obtained from cycle Cl,l=3 by attaching K1,i, K1,j, and K1,k to the vertices of Cl,l=3. Let Fn,4r,s be the such class of F that is obtained from cycle Cl,l=4 by attaching K1,r and K1,s to the nonadjacent vertices of Cl,l=4.Lemma 9.
LetDn,3i,j,k, Dn,3i,j−1,k, and Dn,3i+1,j−1,k be three unicyclic graphs in D as in Figure 3. If i≥j≥1, then(25)RvγDn,3i,j,k>RvγDn,3i+1,j−1,k,for γ≥0.Figure 3
The graphDn,3i,j,k (used in Lemma 9).Proof.
LetDn,3i,j,k,Dn,3i,j−1,k,andDn,3i+1,j−1,k be three unicyclic graphs in D as in Figure 3.(26)RvγDn,3i,j,k−RvγDn,3i+1,j−1,k=j1+γj+2+γ−j−11+γj+1+γ+1i+2+γj+2+γ−1i+2+γj+1+γ+1k+2+γj+2+γ−1k+2+γj+1+γ,RvγDn,3i,j,k−RvγDn,3i+1,j−1,k=11+γj+2+γ−j+1+γ+1i+2+γ+1k+2+γ−2+γ1+γ1j+2+γ−1j+1+γ,RvγDn,3i,j,k−RvγDn,3i+1,j−1,k=11+γ12ξ1−12+1i+2+γ+1k+2+γ−2+γ1+γ−12ξ2−32,where ξ1,ξ2∈j+1,j+2. Similarly, we have(27)RvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k=i+11+γi+3+γ−i1+γi+2+γ+1i+3+γj+1+γ−1i+2+γj+1+γ+1k+2+γi+3+γ−1k+2+γi+2+γ,RvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k=11+γi+3+γ−i+2+γ+1j+1+γ+1k+2+γ−γ+21+γ1i+3+γ−1i+2+γRvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k=11+γ12ξ3−1/2+1j+1+γ+1k+2+γ−γ+21+γ−12ξ3−1/2,where ξ3,ξ4∈i+2,i+3.
Bearing in mind the fact thati≥j, one can see that(28)RvγDn,3i,j,k−RvγDn,3i,j−1,k>RvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k.
Hence,RvγDn,3i,j,k>RvγDn,3i+1,j−1,k.Lemma 10.
LetFn,4r,s,Fn,4r,s−1,andFn,4r+1,s−1 be three unicyclic graphs in F as in Figure 4. If r≥s≥1, then(29)RvγFn,4r,s>RvγFn,4r+1,s−1,for γ≥0.Figure 4
The graphFn,4r,s (used in Lemma 10).Proof.
LetΨt=t/1+γ+2/2+γ1/t+2+γ and keeping in mind Figure 4, we have(30)RvγFn,4r,s=r1+γr+2+γ+s1+γs+2+γ+22+γr+2+γ+22+γs+2+γ,RvγFn,4r,s−1=r1+γr+2+γ+s−11+γs+1+γ+22+γr+2+γ+22+γs+1+γ,RvγFn,4r+1,s−1=r+11+γr+3+γ+s−11+γs+1+γ+22+γr+3+γ+22+γs+1+γ,RvγFn,4r,s−RvγFn,4r,s−1=s1+γ+22+γ1s+2+γ−s−11+γ+22+γ1s+1+γ,RvγFn,4r,s−RvγFn,4r,s−1=Ψs−Ψs−1=Ψ′ξ1,where ξ1∈s+1,s+2. Similarly, we have(31)RvγFn,4r+1,s−1−RvγFn,4r,s−1=r+11+γ+22+γ1r+3+γ−a1+γ+22+γ1r+2+γ,RvγFn,4r+1,s−1−RvγFn,4r,s−1=Ψr+1−Ψr=Ψ′ξ2,where ξ2∈r+2,r+3.
Bearing in mind the fact thatr≥s≥1, Ψ″t=−1/2t+2+γ5/22t+2+γ/1+γ−3/2t/1+γ+2/2+γ<0 for all t>0. Also, Ψ′ξ1−Ψ′ξ2>0, which implies that RvγFn,4r,s>RvγFn,4r+1,s−1.
As a consequence of Lemmas6–10, one has the following.Theorem 1.
Among all the unicyclic graphs of ordern≥4, Sn1 has the minimum variable Randić index for γ≥0 and its value is(32)RvγG≥n−31+γn−1+γ+22+γn−1+γ+12+γ2+γ.


## 2.1. Transformation 1

LetG∈Un and uv be an edge of G such that dv=r≥2. Nv=u,w1,w2,…,wr−1, where w1,w2,…,wr−1 are leaves.DefineG´=G−vw1,vw2,…,vwr−1+uw1,uw2,…,uwr−1. G and G´ are depicted in Figure 1.Figure 1
The graphG´ (obtained using Transformation 1) from G.Lemma 6.
LetG,G´∈Un. By applying Transformation 1 and for γ≥0, we have(12)RvγG>RvγG´.Proof.
Letdu=t≥3:(13)RvγG−RvγG´>dv−11+γdv+γ+1dv+γdu+γ−dv1+γdv+du−1+γ,RvγG−RvγG´>dv−11+γ1dv+γ−1dv+du−1+γ+1dv+γdu+γ−11+γdv+du−1+γ,RvγG−RvγG´>r−11+γ1r+γ−1t+r−1+γ+1r+γt+γ−11+γt+r−1+γ>0.
Using Lemma3, one can see that inequality (13) holds.


## 2.2. Transformation 2

LetG,G´∈Un such that G=G´ and their unique cycles are r−cycle and r−1−cycle, respectively, r≥4. For both graphs, the vertices apart from the cycle are leaf vertices joined to exactly one vertex of the cycle.Lemma 7.
LetG,G´∈Un such that G=G´. By applying Transformation 2 and for γ≥0, we have(14)RvγG>RvγG´.Proof.
Since(15)RvγG=r−2γ+2γ+2+n−rγ+1n−r+γ+2+2γ+2n−r+γ+2,RvγG´=r−3γ+2γ+2+n−r+1γ+1n−r+γ+3+2γ+2n−r+γ+3,RvγG−RvγG´=n−r1+γ1n−r+γ+2−1n−r+γ+3+22+γ1n−r+γ+2−1n−r+γ+3+12+γ2+γ−11+γn−r+γ+3.
Now, there will be three cases.Case I: ifn=r, then(16)RvγG−RvγG´=22+γ1γ+2−1γ+3+12+γ2+γ−11+γγ+3.Using Lemma4, one can see that relation (16) is positive valued.Case II: ifn−r=1, then(17)RvγG−RvγG´=11+γ1γ+3−1γ+4+22+γ1γ+3−1γ+4+12+γ2+γ−11+γγ+4.Using Lemma5, one can see that relation (17) is positive valued.Case III: ifn−r≥2, then(18)RvγG−RvγG´=n−r1+γ1n−r+γ+2−1n−r+γ+3+22+γ1n−r+γ+2−1n−r+γ+3+12+γ2+γ−11+γn−r+γ+3,RvγG−RvγG´>12+γ2+γ−11+γn−r+γ+3>0.Sincen−r1+γ1n−r+γ+2−1n−r+γ+3>0and22+γ1n−r+γ+2−1n−r+γ+3>0.


## 2.3. Transformation 3

LetG∈Un with G=n and u,v∈VG are on the unique cycle of G. Pendant neighbors of u and v are ui, where 1≤i≤s, and vj, where 1≤j≤t, respectively. Assume that the path between u and v on cycle is Puv such that EPuv=r, dVPuv=2 if r≥2 and EPvu≥3. Construct G´ from G by removing the edges uui and vvj, reducing the path Puv into one vertex uv and attaching a star K1,r+s+t+1 to uv, by making sure that G´=G. G and G´ are depicted in Figure 2.Figure 2
The graphG and G´ (obtained using Transformation 3).Lemma 8.
LetG,G´∈Un such that G´=G. By applying Transformation 3 and for γ≥0, s,t≥1, we have(19)RvγG>RvγG´.Proof.
Case I: letr=1 and uv∈EG, then(20)RvγG−RvγG´>s1+γs+2+γ+t1+γt+2+γ+1s+2+γt+2+γ−s+t+11+γs+t+3+γ,RvγG−RvγG´>s1+γ1s+2+γ−1s+t+3+γ+t1+γ1t+2+γ−1s+t+3+γ+1s+2+γt+2+γ−11+γs+t+3+γ.Using Lemma1, one can see that relation (20) is positive valued.Case II: now letr≥2, then(21)RvγG−RvγG´>s1+γs+2+γ+t1+γt+2+γ+1s+2+γ2+γ+12+γt+2+γ+r−22+γ2+γ−s+t+r1+γs+t+r+2+γ.
Sincer≥2 and keeping in mind the fact that(22)Ψr,s,t=s+t+r1+γs+t+r+2+γ,is an increasing function because(23)∂Ψ∂r=s+t+r+4+2γ2γ+1s+t+r+2+γ3/2>0.
We have(24)RvγG−RvγG´>s1+γ1s+2+γ−1s+t+4+γ+t1+γ1t+2+γ−1s+t+4+γ+1s+2+γ2+γ−11+γs+t+4+γ+1t+2+γ2+γ−11+γs+t+4+γ.
Using Lemma2, one can see that relation (24) is positive valued.
LetDn,3i,j,k be the such class of D that is obtained from cycle Cl,l=3 by attaching K1,i, K1,j, and K1,k to the vertices of Cl,l=3. Let Fn,4r,s be the such class of F that is obtained from cycle Cl,l=4 by attaching K1,r and K1,s to the nonadjacent vertices of Cl,l=4.Lemma 9.
LetDn,3i,j,k, Dn,3i,j−1,k, and Dn,3i+1,j−1,k be three unicyclic graphs in D as in Figure 3. If i≥j≥1, then(25)RvγDn,3i,j,k>RvγDn,3i+1,j−1,k,for γ≥0.Figure 3
The graphDn,3i,j,k (used in Lemma 9).Proof.
LetDn,3i,j,k,Dn,3i,j−1,k,andDn,3i+1,j−1,k be three unicyclic graphs in D as in Figure 3.(26)RvγDn,3i,j,k−RvγDn,3i+1,j−1,k=j1+γj+2+γ−j−11+γj+1+γ+1i+2+γj+2+γ−1i+2+γj+1+γ+1k+2+γj+2+γ−1k+2+γj+1+γ,RvγDn,3i,j,k−RvγDn,3i+1,j−1,k=11+γj+2+γ−j+1+γ+1i+2+γ+1k+2+γ−2+γ1+γ1j+2+γ−1j+1+γ,RvγDn,3i,j,k−RvγDn,3i+1,j−1,k=11+γ12ξ1−12+1i+2+γ+1k+2+γ−2+γ1+γ−12ξ2−32,where ξ1,ξ2∈j+1,j+2. Similarly, we have(27)RvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k=i+11+γi+3+γ−i1+γi+2+γ+1i+3+γj+1+γ−1i+2+γj+1+γ+1k+2+γi+3+γ−1k+2+γi+2+γ,RvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k=11+γi+3+γ−i+2+γ+1j+1+γ+1k+2+γ−γ+21+γ1i+3+γ−1i+2+γRvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k=11+γ12ξ3−1/2+1j+1+γ+1k+2+γ−γ+21+γ−12ξ3−1/2,where ξ3,ξ4∈i+2,i+3.
Bearing in mind the fact thati≥j, one can see that(28)RvγDn,3i,j,k−RvγDn,3i,j−1,k>RvγDn,3i+1,j−1,k−RvγDn,3i,j−1,k.
Hence,RvγDn,3i,j,k>RvγDn,3i+1,j−1,k.Lemma 10.
LetFn,4r,s,Fn,4r,s−1,andFn,4r+1,s−1 be three unicyclic graphs in F as in Figure 4. If r≥s≥1, then(29)RvγFn,4r,s>RvγFn,4r+1,s−1,for γ≥0.Figure 4
The graphFn,4r,s (used in Lemma 10).Proof.
LetΨt=t/1+γ+2/2+γ1/t+2+γ and keeping in mind Figure 4, we have(30)RvγFn,4r,s=r1+γr+2+γ+s1+γs+2+γ+22+γr+2+γ+22+γs+2+γ,RvγFn,4r,s−1=r1+γr+2+γ+s−11+γs+1+γ+22+γr+2+γ+22+γs+1+γ,RvγFn,4r+1,s−1=r+11+γr+3+γ+s−11+γs+1+γ+22+γr+3+γ+22+γs+1+γ,RvγFn,4r,s−RvγFn,4r,s−1=s1+γ+22+γ1s+2+γ−s−11+γ+22+γ1s+1+γ,RvγFn,4r,s−RvγFn,4r,s−1=Ψs−Ψs−1=Ψ′ξ1,where ξ1∈s+1,s+2. Similarly, we have(31)RvγFn,4r+1,s−1−RvγFn,4r,s−1=r+11+γ+22+γ1r+3+γ−a1+γ+22+γ1r+2+γ,RvγFn,4r+1,s−1−RvγFn,4r,s−1=Ψr+1−Ψr=Ψ′ξ2,where ξ2∈r+2,r+3.
Bearing in mind the fact thatr≥s≥1, Ψ″t=−1/2t+2+γ5/22t+2+γ/1+γ−3/2t/1+γ+2/2+γ<0 for all t>0. Also, Ψ′ξ1−Ψ′ξ2>0, which implies that RvγFn,4r,s>RvγFn,4r+1,s−1.
As a consequence of Lemmas6–10, one has the following.Theorem 1.
Among all the unicyclic graphs of ordern≥4, Sn1 has the minimum variable Randić index for γ≥0 and its value is(32)RvγG≥n−31+γn−1+γ+22+γn−1+γ+12+γ2+γ.



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*Source: 1217567-2020-07-18.xml*