On the Maximum Sombor Index of Unicyclic Graphs with a Fixed Girth

Let G be a graph having the set of edges E ( G ) . Represent by d G ( u ) the degree of a vertex u of G . Te Sombor (SO) index of G is defned as SO ( G ) � 􏽐 uv ∈ E ( G )


Introduction
In this paper, we are concerned with only connected and fnite graphs. Tus, throughout this study, the term "graph" means a connected and fnite graph. Te graph-theoretical terminology utilized in this work without giving their definitions may be found in the books [1][2][3].
For a graph G, represent by E(G) and V(G) the sets of edges and vertices, respectively. For a vertex u ∈ V(G), the set N G (u) consists of all those vertices of G that are adjacent with u. Te degree of a vertex u ∈ V(G) is represented by d G (u), which is equal to the number of elements of the set N G (u). If d G (u) � 1, then u is called a pendent vertex, and if d G (u) ≥ 3, then u is called a branching vertex. Denote by G − w the graph deduced from G by removing the vertex w ∈ V(G) as well as its incident edges. Te length of a shortest cycle in a graph G is known as the girth of G. A connected graph with the same order and size is usually referred to as a unicyclic graph. In other words, a connected cyclic graph having cyclomatic number one is known as a unicyclic graph [4].
Te Sombor (SO) index for a graph G is defned [5] as Te SO index, although proposed recently [5], has received a lot of attention from academics and led to numerous articles; for instance, see the surveys [6,7] and associated references are given therein. Possible chemical applications of the SO index may be found in the papers [8,9].
In the present article, we are concerned with the solution to an extremal problem involving the SO index for unicyclic graphs with a fxed girth and order. Te SO index of unicyclic graphs has been studied in several papers. Te graphs possessing the least and largest values of the SO index among all unicyclic graphs with a fxed order were found independently in [10,11]. Liu [12] (respectively, Alidadi et al. [13]) reported the graphs attained the highest value (respectively, least value) of the SO index among all unicyclic graphs with a fxed diameter and order. Te graphs attaining the minimum (respectively, maximum) SO index over the class of all unicyclic graphs with a fxed maximum degree (respectively, matching number) and order were characterized in [14] (respectively, in [15]). In this paper, we give the characterization of the graphs possessing the frsttwo maximum values of the SO index in the class of all unicyclic graphs with a fxed girth and order.

Main Results
We start this section with an elementary lemma.
is strictly decreasing.
Next, we provide a lemma that is crucial for proving our frst main result.

Lemma 2.
Let G be a connected graph of order at least 4. Let u and v be adjacent vertices of degrees at least 2 such that they do not have any neighbor in common. Let H be the graph deduced from G by dropping all the edges, except uv, incident with v and inserting the edges ux for all as desired. Te next result gives the characterization of the graphs possessing the frst maximum value of the SO index in the class of all unicyclic graphs with a fxed girth and order. □ Theorem 1. If G is a unicyclic graph with order n and girth k, then with equality if and only if G � C k,n− k , where C k,n− k is the graph formed by attaching n − k pendent vertex/vertices to exactly one vertex of the cycle graph C k of order k.
Proof. In [11], it was proved that C 3,n− 3 is the unique graph having the maximum Sombor index among all unicyclic graphs of order n; thus, our theorem follows from this result when k � 3. In the rest of the proof, we assume that 4 ≤ k ≤ n. Also, suppose that SO(G) is as large as possible. By Lemma 2, every vertex that lies on the cycle of G has exactly two nonpendent neighbors. Note that n can be written in terms of k; that is, n � k, k + 1, k + 2, . . .. Now, we apply induction on n.
Te result trivially holds for n ∈ k, k + 1 { }, because there is only one graph in either of these two cases. Tence, the induction starts. Let w ∈ V(G) be a pendent vertex having the neighbor v.
Observe that the graph G − w has order n − 1 and girth k.
By Lemma 1, one has where the equality sign in (5) holds if and only if each of the two nonpendent neighbors of v has degree 2. Note that the function ψ defned by is strictly increasing for x ≥ 3 because its derivative is positive for x ≥ 3 as the inequalities and hence, from (6), it follows that with equality if and only if each of the two nonpendent neighbors of v has degree 2 and n − k + 2 � q. Now, by using the inductive hypothesis in (11), one arrives at with equality if and only if each of the two nonpendent neighbors of v has degree 2, n − k + 2 � q, and G − w � C k,n− k− 1 . Tis completes the induction and thence the proof. In order to prove our second main result, which gives the characterization of the graphs possessing the second-maximum value of the SO index in the class of all unicyclic graphs with a fxed girth and order, we prove a series of lemmas. □ Lemma . Let G be a connected graph of order at least 5. Let uv be an edge of G such that each of its end vertices (i) has exactly two nonpendent neighbors and (ii) contains at least one pendent neighbor provided that be the graph obtained from G by removing all the pendent neighbors of v and then attaching them to the vertex u. It holds that SO(G) < SO(G ′ ).

Proof. Let v, w
{ } be the set of nonpendent neighbors of u with d G (w) � s ≥ 2. Let u, w ′ be the set of nonpendent neighbors of (HTML translation failed) with d G (w ′ ) � t ≥ 2. Also, assume that d G (u) � p and d G (v) � q. Ten, p ≥ q ≥ 3.

Lemma 5. Let G be a connected graph of order at least 6. Let u and v be nonadjacent vertices of G such that each of them (i) has at least one pendent neighbor and (ii) contains exactly two nonpendent neighbors such that each of them has degree 2 and d G (u) ≥ d G (v) ≥ 3. Let G ′ be the graph obtained from G by removing a pendent neighbor of v and then attaching it to u. It holds that SO(G) < SO(G ′ ).
Proof. If d G (u) � p and d G (v) � q, then p ≥ q ≥ 3, and hence, we have (17) □ Lemma 6. Let G be the graph deduced from the cycle C k of order k by attaching p pendent vertices to a vertex u ∈ V(C k ) and attaching q pendent vertices to another vertex with equality if and only if p � n − 4.
with equality if and only if p � n − k − 1 and uv ∉ E(G).
Proof. (i) Tis part follows directly from Lemma 4 because uv ∈ E(G) when k � 3.
(ii) If uv ∉ E(G), then the desired conclusion follows from Lemma 5. In the rest of the proof, assume that uv ∈ E(G). By Lemma 4, we have with equality if and only if p � n − k − 1. Let Here, it holds that Journal of Mathematics √ .

(22)
Taking p � n − k − 1, we have for p ≥ 1. □ Lemma 7. For n − k − 1 ≥ q ≥ 1 and k ≤ n − 2, if G is the graph deduced from the unicyclic graph C k,n− k− q (see Teorem 1) by attaching q pendent vertices to exactly one pendent vertex of C k,n− k− q , then with equality if and only if q � 1.
Proof. Te result trivially holds for k � n − 2 because there is only one graph in this case. Tus, we suppose that k ≤ n − 3.
Here, we have Te substitution n − k � α in equation (25) yields where α ≥ q + 1 ≥ 2. By a computer program, it is checked that the maximum value of the right-hand-sided expression of (26) under the given constraints is achieved if and only if q � 1. Now, we determine the graph attaining the secondmaximum value of the Sombor index in the class of all unicyclic graphs with a fxed order and girth. □ Theorem 2. Let G be a unicyclic graph with order n and girth g such that G ≠ C k,n− k (see Teorem 1).
with equality if and only if G is the graph deduced from C 3,n− 4 by attaching a pendent vertex to one of its vertices of degree 2.
(ii) If 4 ≤ k ≤ n − 2, then with equality if and only if G is a graph deduced from C k,n− k− 1 by attaching a pendent vertex to a vertex w ∈ V(C k,n− k− 1 ), where d C k,n− k− 1 (w) � 2 and w is not adjacent to the branching neighbor.
If the set v 1 , v 2 , . . . , v k contains at least three branching vertices, then by using Lemmas 2, 3 (or 5), and 6, we fnd a unicyclic graph G ′ with order n and girth g such that G ′ ≠ C k,n− k and satisfying SO(G) < SO(G ′ ) ≤ a.
If the set v 1 , v 2 , . . . , v k contains exactly two branching vertices, then by using Lemmas 2, 4 (or 5), and 6, we fnd a unicyclic graph G ″ with order n and girth g such that G ″ ≠ C k,n− k and satisfying SO(G) ≤ SO(G ″ ) ≤ a, where both the equality signs hold if and only if G is one of the two extremal graphs defned in the statement of the theorem.
In the remaining proof, we assume that v 1 , v 2 , . . . , v k contains exactly one branching vertex; let v 1 be the branching vertex. Since G ≠ C k,n− k , the vertex v 1 has at least three nonpendent neighbors. If v 1 has either at least four nonpendent neighbors or it has only three nonpendent neighbors, each of which has at least two nonpendent neighbor, then by using Lemmas 2 and 7, we fnd a unicyclic graph G ‴ with order n and girth g such that G ‴ ≠ C k,n− k and satisfying SO(G) < SO(G ‴ ) ≤ b, where b is the upper bound on SO given in Lemma 7, that is, At the end of the proof, we will prove that b − a < 0.
If v 1 has only three nonpendent neighbors such that one of them has only one nonpendent neighbor, then from Lemma 7, it follows that SO(G) ≤ b with equality if and only if q � 1.
In order to complete the proof, in the following, we prove that b − a < 0. First, we assume that k � 3. Ten, Now, we suppose that 4 ≤ k ≤ n − 2. Ten, for x � n − k + 1 ≥ 3.

Data Availability
Te data used to support the fndings of this study are available from the corresponding author upon request.

Conflicts of Interest
Te authors declare that they have no conficts of interest.