Multiplicative Sombor index of graphs

Discrete Math. Lett. 9 (2022) 80–85 DOI: 10.47443/dml.2021.s213 Discrete Mathematics Letters www.dmlett.com Research Article Multiplicative Sombor index of graphs Hechao Liu∗ School of Mathematical Sciences, South China Normal University, Guangzhou, P. R. China (Received: 26 February 2022. Received in revised form: 3 March 2022. Accepted: 19 March 2022. Published online: 28 March 2022.) c 2022 the author. This is an open access article under the CC BY (International 4.0) license (www.creativecommons.org/licenses/by/4.0/). Abstract p P The Sombor index of a graph G is defined as SO(G) = uv∈E(G) d2G (u) + d2G (v), where dG (u) denotes the degree of the p Q Q vertex u of G. Accordingly, the multiplicative Sombor index of G can be defined as SO (G) = uv∈E(G) d2G (u) + d2G (v). In this article, some graph transformations which increase or decrease the multiplicative Sombor index are first introduced. Then by using these transformations, extremal values of the multiplicative Sombor index of trees and unicyclic graphs are determined. Keywords: multiplicative Sombor index; extremal value. 2020 Mathematics Subject Classification: 05C09, 05C92. 1. Introduction The term “chemical graph theory” was coined by Nenad Trinajstić and it was used as the title of his seminal book [33]. In the new AMS subject classification 2020, the subject number 05C92 in “05C graph theory” is assigned to chemical graph theory. Gutman said that it is a major success of all those colleagues who over several decades worked and/or are working in “chemical topology”, a field of research considered by many as worthless. We only consider simple connected graph G with the vertex set V (G) and edge set E(G). Denote by NG (u) the set of the vertices that are neighbors of the vertex u ∈ V (G). Then |NG (u)| is the degree of the vertex u, denoted by dG (u) or d(u). We call the vertex u as a pendent vertex if d(u) = 1. We call a path P = u1 u2 · · · uk in G as a pendent path if d(u1 ) ≥ 3, d(uk ) = 1 and d(ui ) = 2 for 2 ≤ i ≤ k − 1. The girth of G is the length of a shortest cycle in G. Denote by Pn and Sn the path and star graphs with n vertices, respectively. All notations and terminology used, but not defined here, can be found in the textbook [3]. The Sombor index [15] was proposed by Gutman, which is defined as X q SO(G) = d2G (u) + d2G (v). uv∈E(G) Since the publication of [15], the Sombor index has attracted much attention of researchers. For the mathematical properties and chemical applications on the Sombor index or its variants, see [1, 2, 4–14, 16–32, 34, 35] and the references cited therein. According to the definition of the Sombor index, it is natural to consider the multiplicative version of the Sombor index, defined as Y Y q (G) = d2G (u) + d2G (v). SO uv∈E(G) The aim of this paper is to begin the research on mathematical properties of the multiplicative Sombor index. 2. Transformations We first introduce some transformations which will be useful in the proof of main theorems. ∗ E-mail address: H. Liu / Discrete Math. Lett. 9 (2022) 80–85 81 Lemma 2.1. Let G be a connected graph. Q Q (1) if uv ∈ E(G), then SO (G) > SO (G − uv); Q Q (2) if uv ∈ / E(G), then SO (G) < SO (G + uv). Figure 1: Transformation A. Lemma 2.2. Let G and G∗ be the graphs shown in Figure 1. We allow u = v. If uu1 u2 · · · uk and vv1 v2 · · · vl are two pendent Q Q path in G, and G∗ = G − uu1 + u1 vl , then SO (G∗ ) < SO (G). Proof. We consider two cases. Case 1. u = v. Let dG0 (u) = t ≥ 1 and NG0 (u) = {x1 , x2 , · · · , xt }. If k ≥ 2, l ≥ 2, then Y SO (G) − Y SO (G∗ ) = 5 · ((t + 2)2 + 4) · 8 k+l−4 2 · t p Y (dG0 (xi ) + 2)2 + t2 i=1 t p Y p k+l−2 − 5 · (t + 1)2 + 4 · 8 2 · (dG0 (xi ) + 1)2 + t2 √ i=1 > 8 k+l−4 2 i √ p · 5((t + 2) + 4) − 8 5 (t + 1)2 + 4 h 2 > 0. Similarly, if k = l = 1 or k = 1, l ≥ 2 or k ≥ 2, l = 1, then we have Y Y (G) − (G∗ ) > 0. SO SO Case 2. u 6= v. Q (G∗ ) QSO SO (G) p = d2G (u1 ) + 4 p Y d2G (vl−1 ) + 4 p p d2G (u1 ) + d2G (u) d2G (vl−1 ) + 1 u ∈N (u)\{u } i G 1 p d2G (ui ) + (dG (u) − 1)2 p . d2G (ui ) + d2G (u) If l ≥ 2, then dG (vl−1 ) = 2, dG (u1 ) = 2 or 1. Since dG (u) ≥ 3, one has Y Y (G∗ ) < (G). SO SO If l = 1, then dG (vl−1 ) ≥ 3. Since dG (u) ≥ 3, dG (u1 ) = 2 or 1. p √ Q d2 (u1 ) + 4 (G∗ ) 13 SO Q < √ p 2 G < 1. 10 dG (u1 ) + d2G (u) SO (G) Lemma 2.3. Let G and G∗ be the graphs as depicted in Figure 2, and G∗ = G − {uw1 , uw2 , · · · , uwt } + {vw1 , vw2 , · · · , vwt }. Q Q Then SO (G) < SO (G∗ ). H. Liu / Discrete Math. Lett. 9 (2022) 80–85 82 Figure 2: Transformation B. Proof. Let dG0 (v) = k ≥ 1 and NG0 (v) = {v1 , v2 , · · · , vk }. Then Y SO (G∗ ) − Y SO (G) ≥ t+1 ((k + t + 1)2 + 1) 2 · k q Y d2G0 (vi ) + (k + t + 1)2 i=1 k q Y p t − (k + 1)2 + (t + 1)2 · ((t + 1)2 + 1) 2 · d2G0 (vi ) + (k + 1)2 i=1 t+1 2 > p t ((k + t + 1) + 1) − (k + 1)2 + (t + 1)2 · ((t + 1)2 + 1) 2 p p (k + t + 1)2 + 1 − (k + 1)2 + (t + 1)2 > 0. > 2 Figure 3: Transformation C. Lemma 2.4. Let G and G∗ be the graphs as shown in Figure 3. If G∗ is the graph obtained from G by identifying the vertices u and v to a new vertex w and adding a pendent vertex w0 to the vertex w, then Y Y (G) < (G∗ ). SO SO Proof. Let NG (u) = {v, u1 , u2 , · · · , uk } and NG (v) = {u, v1 , v2 , · · · , vl }. Then Y SO (G∗ ) − Y SO (G) ≥ k q l q Y Y p (k + t + 1)2 + 1 · d2G (ui ) + (k + l + 1)2 · d2G (vj ) + (k + l + 1)2 i=1 j=1 k q l q Y Y p − (k + 1)2 + (l + 1)2 · d2G (ui ) + (k + 1)2 · d2G (vj ) + (l + 1)2 i=1 > p (k + t + 1)2 + 1 − j=1 p (k + 1)2 + (l + 1)2 > 0. Lemma 2.5. Let G, G∗ be the graphs in Figure 4. Note that P = v1 v2 · · · vt vt+1 is a pendent path, NG (v1 ) = {v2 , u, w} and G∗ = G − uv1 + uvt+1 . Then Y Y (G) > (G∗ ). SO SO H. Liu / Discrete Math. Lett. 9 (2022) 80–85 83 Figure 4: Transformation D. Proof. If dG (u) = 1, then by Lemma 2.2, the conclusion holds. Thus, in the following, we suppose that dG (u) ≥ 2. If t ≥ 2, then q q q q Y Y √ √ t−2 t 4 + 9 · 1 + 4 · 8 2 · d2G (u) + 9 · d2G (w) + 9 − 8 2 · d2G (u) + 4 · d2G (w) + 4 (G) − (G∗ ) ≥ SO SO = t−2 8 2   q q q √ q 65 d2G (u) + 9 · d2G (w) + 9 − 8 d2G (u) + 4 · d2G (w) + 4 > 0. If t = 1, then Y SO (G) − Y SO (G∗ ) ≥ √ 1+9 q q q q √ d2G (u) + 9 · d2G (w) + 9 − 4 + 4 d2G (u) + 4 · d2G (w) + 4 > 0. Qm p x Lemma 2.6. Let f (x) = ((x + m)2 + 1) 2 · i=1 (x + m)2 + (d(i) )2 , where d(i) (i = 1, 2, · · · , m) are all nonnegative integers. Then for any positive integers s, t, the inequality f (s + t)f (0) > f (s)f (t) holds. Proof. We prove that ln f (s + t) + ln f (0) > ln f (s) + ln f (t). For this, let g(x) = ln f (x) + ln f (0) − ln f (x1 ) − ln f (x − x1 ). Then, g 0 (x) = , m X 1 x(x + m) x+m ln((x + m)2 + 1) + + 2 (x + m)2 + 1 i=1 (x + m)2 + (d(i) )2 # " m 1 (x − x1 )(x − x1 + m) X x − x1 + m 2 − ln((x − x1 + m) + 1) + + 2 (x − x1 + m)2 + 1 (x − x1 + m)2 + (d(i) )2 i=1 h(x) − h(x − x1 ), where h(x) = m X 1 x(x + m) x+m ln((x + m)2 + 1) + + . 2 2 (x + m) + (...truncated)