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Extremal Polygonal Cacti for General Sombor Index***This paper was supported by the National Natural Science Foundation of China [No. 11971406].

Jiachang Ye, Jianguo QianCorresponding author. E-mail: [email protected] (J.G. Qian)
School of Mathematical Sciences, Xiamen University,
Xiamen, 361005, P.R. China
Abstract

The Sombor index of a graph GG was recently introduced by Gutman from the geometric point of view, defined as SO(G)=uvE(G)d(u)2+d(v)2SO(G)=\sum_{uv\in E(G)}\sqrt{d(u)^{2}+d(v)^{2}}, where d(u)d(u) is the degree of a vertex uu. For two real numbers α\alpha and β\beta, the α\alpha-Sombor index and general Sombor index of GG are two generalized forms of the Sombor index defined as SOα(G)=uvE(G)(d(u)α+d(v)α)1/αSO_{\alpha}(G)=\sum_{uv\in E(G)}(d(u)^{\alpha}+d(v)^{\alpha})^{1/\alpha} and SOα(G;β)=uvE(G)(d(u)α+d(v)α)βSO_{\alpha}(G;\beta)=\sum_{uv\in E(G)}(d(u)^{\alpha}+d(v)^{\alpha})^{\beta}, respectively. A kk-polygonal cactus is a connected graph in which every block is a cycle of length kk. In this paper, we establish a lower bound on α\alpha-Sombor index for kk-polygonal cacti and show that the bound is attained only by chemical kk-polygonal cacti. The extremal kk-polygonal cacti for SOα(G;β)SO_{\alpha}(G;\beta) with some particular α\alpha and β\beta are also considered.

Keywords: general Sombor index, polygonal cactus, extremal problem

Mathematics Subject Classification: 05C07, 05C09, 05C90

1 Introduction

We consider only connected simple graphs. For a graph GG, we denote by V(G)V(G) and E(G)E(G) the vertex set and edge set of GG, respectively. For a vertex vV(G)v\in V(G), we denote by dG(v)d_{G}(v), or d(v)d(v) if no confusion can occur, the degree of vv. A vertex vv is called a cut vertex of GG if GvG-v is not connected.

In mathematical chemistry, particularly in QSPR/QSAR investigation, a large number of topological indices were introduced in an attempt to characterize the physical-chemical properties of molecules. Among these indices, the vertex-degree-based indices play important roles [4, 7, 8]. Probably the most studied are, for examples, the Randić connectivity index R(G)R(G) [18], the first and second Zagreb indices M1(G)M_{1}(G) and M2(G)M_{2}(G) [9], which were introduced for the total π\pi-energy of alternant hydrocarbons.

A vertex-degree-based index of a graph GG can be generally represented as the sum of a real function f(d(u),d(v))f(d(u),d(v)) associated with the edges of GG [10], i.e.,

If(G)=uvE(G)f(d(u),d(v)),I_{f}(G)=\sum\limits_{uv\in E(G)}f(d(u),d(v)),

where f(s,t)=f(t,s)f(s,t)=f(t,s). In the literature, If(G)I_{f}(G) is also called the connectivity function [22] or bond incident degree index [1, 21, 23].

Recently, Gutman [10] introduced an idea to view an edge e=uve=uv as a geometric point, namely the degree-point, that is, to view the ordered pair (d(u),d(v))(d(u),d(v)) as the coordinate of ee. Therefore, it is interesting to consider the function f(s,t)f(s,t) from the geometric point of view. A natural considering is to define f(s,t)f(s,t) as a geometric distance from the degree-point (s,t)(s,t) to the origin. In this sense, the first Zagreb index, i.e., M1(G)=uvE(G)(d(u)+d(v))=uvE(G)(|d(u)|+|d(v)|)M_{1}(G)=\sum_{uv\in E(G)}(d(u)+d(v))=\sum_{uv\in E(G)}(|d(u)|+|d(v)|), is exactly the index defined on the Manhattan distance. Along this direction, a more natural considering would be to define f(s,t)f(s,t) as the Euclidean distance, i.e., f(s,t)=s2+t2f(s,t)=\sqrt{s^{2}+t^{2}}. Indeed, based on this idea, Gutman [10] introduced the Somber index defined by

SO(G)=uvE(G)d(u)2+d(v)2SO(G)=\sum_{uv\in E(G)}\sqrt{d(u)^{2}+d(v)^{2}}

and further determined the extremal trees for the index. In [3], Das et al. established some bounds on the Sombor index and some relations between Sombor index and the Zagreb indices and, in [19], Redžpović studied chemical applicability of the Sombor index. Further, Cruz et al. [2] characterized the extremal chemical graphs and hexagonal systems for the Sombor index.

More recently, for positive real number α\alpha, Réti et al. [20] defined the α\alpha-Sombor index as

SOα(G)=uvE(G)(d(u)α+d(v)α)1/α,SO_{\alpha}(G)=\sum_{uv\in E(G)}(d(u)^{\alpha}+d(v)^{\alpha})^{1/\alpha},

which could be viewed as the one based on Minkowski distance. In the same paper, they also characterized the extremal graphs with few cycles for α\alpha-Sombor index.

In this paper we consider a more generalized form of Sombor index defined as

SOα(G;β)=uvE(G)(d(u)α+d(v)α)β,SO_{\alpha}(G;\beta)=\sum_{uv\in E(G)}(d(u)^{\alpha}+d(v)^{\alpha})^{\beta},

where α,β\alpha,\beta are real numbers. We note that this form is a natural generalization of the Sombor index, which was also introduced elsewhere, e.g., the first (α,β)KA(\alpha,\beta)-KA index in [12] and the general Sombor index in [11]. In addition to the first Zagreb, Sombor and the α\alpha-Sombor index listed above, the general Sombor index also includes many other known indices, e.g., the modified first Zagreb index (α=3,β=1\alpha=-3,\beta=1) [17], forgotten index (α=2,β=1\alpha=2,\beta=1) [6], inverse degree index (α=2,β=1\alpha=-2,\beta=1) [5], modified Sombor index (α=2,β=1/2\alpha=2,\beta=-1/2) [13], first Banhatti-Sombor index (α=2,β=1/2\alpha=-2,\beta=1/2) [15] and general sum-connectivity index (α=1,β\alpha=1,\beta\in\mathbb{R}) [24].

A block in a graph is a cut edge or a maximal 2-connected component. A cactus is a connected graph in which every block is a cut edge or a cycle. Equivalently, a cactus has no edge lies in more than one cycle. In the following, we call a kk-cycle (a cycle of length kk) a kk-polygon. If each block of a cactus GG is a kk-polygon, then GG is called a kk-polygonal cactus or polygonal cactus with no confusion.

In this paper, we consider the extremal kk-polygonal cacti for SOα(G;β)SO_{\alpha}(G;\beta). In the following section we establish a lower bound on α\alpha-Sombor index for kk-polygonal cacti and show that the bound is attained only by chemical kk-polygonal cacti. In the third section we characterize the extremal polygonal cactus with maximum SOα(G;β)SO_{\alpha}(G;\beta) for (i) α>1\alpha>1 and β1\beta\geq 1; and (ii) 1/2α<11/2\leq\alpha<1 and β=2\beta=2, respectively. In the fourth section, we characterize the extremal polygonal cacti with minimum SOα(G;β)SO_{\alpha}(G;\beta) for α>1\alpha>1 and β1\beta\geq 1.

2 Polygonal cacti with minimum α\alpha-Sombor index

For convenience, in what follows we denote rα(s,t;β)=(sα+tα)βr_{\alpha}(s,t;\beta)=(s^{\alpha}+t^{\alpha})^{\beta}, rα(s,t)=(sα+tα)1/αr_{\alpha}(s,t)=(s^{\alpha}+t^{\alpha})^{1/\alpha} and r(s,t)=s2+t2r(s,t)=\sqrt{s^{2}+t^{2}}, where s>0s>0, t>0t>0, α,β\alpha,\beta\in\mathbb{R} and α0\alpha\not=0. For integers nn with n1n\geq 1 and kk with k3k\geq 3, we denote by 𝒢n,k\mathcal{G}_{n,k} the class of kk-polygonal cacti with nn polygons.

In this section we consider the α\alpha-Sombor index SOα(G)SO_{\alpha}(G), i.e., SOα(G;1/α)SO_{\alpha}(G;1/\alpha). For G𝒢n,kG\in\mathcal{G}_{n,k}, it is clear that |V(G)|=nkn+1,|E(G)|=nk|V(G)|=nk-n+1,|E(G)|=nk and every vertex of GG has even degree no more than 2n2n. Further, it is clear that vv is a cut vertex of GG if and only if vv has degree no less than 4, i.e., dG(v)4d_{G}(v)\geq 4. A polygon is called a pendent polygon if it contains exactly one cut-vertex of GG. For 2st2n2\leq s\leq t\leq 2n, we denote by ns,t=ns,t(G)n_{s,t}=n_{s,t}(G) the number of edges in GG that join two vertices of degrees ss and tt. Let X={(s,t):s,t{2,4,,2n},st}X=\{(s,t):s,t\in\{2,4,\ldots,2n\},s\leq t\} and Y=X{(2,2),(2,4),(4,4)}.Y=X\setminus\{(2,2),(2,4),(4,4)\}.

Definition 2.1.

[16] Let π=(w1,w2,,wn)\pi=\big{(}w_{1},w_{2},\ldots,w_{n}\big{)} and π=(w1,w2,,wn)\pi^{\prime}=\big{(}w^{\prime}_{1},w^{\prime}_{2},\ldots,w^{\prime}_{n}\big{)} be two non-increasing sequences of nonnegative real numbers. We write ππ\pi\lhd\pi^{\prime} if and only if ππ\pi\neq\pi^{\prime}, i=1nwi=i=1nwi\sum_{i=1}^{n}w_{i}=\sum_{i=1}^{n}w^{\prime}_{i}, and i=1jwii=1jwi\sum_{i=1}^{j}w_{i}\leq\sum_{i=1}^{j}w^{\prime}_{i} for all j=1, 2,,nj=1,\,2,\,\ldots,\,n.

A function ζ(x)\zeta(x) defined on a convex set XX is called strictly convex if

ζ(μx1+(1μ)x2)<μζ(x1)+(1μ)ζ(x2)\zeta\big{(}\mu x_{1}+(1-\mu)x_{2}\big{)}<\mu\zeta(x_{1})+(1-\mu)\zeta(x_{2}) (1)

for any 0<μ<10<\mu<1 and x1,x2Xx_{1},x_{2}\in X with x1x2x_{1}\neq x_{2}.

Lemma 2.1.

[16] Let π=(w1,w2,,wn)\pi=\big{(}w_{1},w_{2},\ldots,w_{n}\big{)} and π=(w1,w2,,wn)\pi^{\prime}=\big{(}w^{\prime}_{1},w^{\prime}_{2},\ldots,w^{\prime}_{n}\big{)} be two non-increasing sequences of nonnegative real numbers. If ππ\pi\lhd\pi^{\prime}, then for any strictly convex function ζ(x)\zeta(x), we have i=1nζ(wi)<i=1nζ(wi)\sum_{i=1}^{n}\zeta{(w_{i})}<\sum_{i=1}^{n}\zeta{(w^{\prime}_{i})}.

Lemma 2.2.

Let α>1\alpha>1 and n3n\geq 3. Then
(i). rα(2n,2)rα(2n2,4)>0;r_{\alpha}(2n,2)-r_{\alpha}(2n-2,4)>0;
(ii). rα(6,2)+rα(2,2)2rα(4,2)>0.r_{\alpha}(6,2)+r_{\alpha}(2,2)-2r_{\alpha}(4,2)>0.

Proof.

Since α>1\alpha>1 and n3n\geq 3, then by Lemma 2.1, we have (2n)α+2α>(2n2)α+4α(2n)^{\alpha}+2^{\alpha}>(2n-2)^{\alpha}+4^{\alpha}. Hence (i) holds clearly. Let g(x)=rα(x,2)=(xα+2α)1/αg(x)=r_{\alpha}(x,2)=(x^{\alpha}+2^{\alpha})^{1/\alpha}, where x>0x>0 and α>1\alpha>1. Since g′′(x)=(α1)2αxα2(xα+2α)21/α>0g^{{}^{\prime\prime}}(x)=\frac{(\alpha-1)2^{\alpha}x^{\alpha-2}}{(x^{\alpha}+2^{\alpha})^{2-1/\alpha}}>0, then g(x)g(x) is strictly convex. Then by Lemma 2.1, g(6)+g(2)>2g(4)g(6)+g(2)>2g(4). Hence (ii) also holds. ∎

Lemma 2.3.

Let α0\alpha\neq 0 and G𝒢n,kG\in\mathcal{G}_{n,k}, where n1n\geq 1 and k3k\geq 3. Then

SOα(G)=(4n4)(2α+4α)1/α+2(nk4n+4)21/αSO_{\alpha}(G)=(4n-4)(2^{\alpha}+4^{\alpha})^{1/\alpha}+2(nk-4n+4)2^{1/\alpha}
+(6×21/α2(2α+4α)1/α)n4,4+(s,t)Yη(s,t;α)ns,t,~{}~{}~{}~{}~{}~{}+\left(6\times 2^{1/\alpha}-2(2^{\alpha}+4^{\alpha})^{1/\alpha}\right)n_{4,4}+\sum_{(s,t)\in Y}\eta(s,t;\alpha)n_{s,t},

where η(s,t;α)=(sα+tα)1/α2(1s+1t)21/α\eta(s,t;\alpha)=(s^{\alpha}+t^{\alpha})^{1/\alpha}-2\left(\frac{1}{s}+\frac{1}{t}\right)2^{1/\alpha}.

Proof.

By the definition of ns,tn_{s,t}, it is not difficult to see that

{nkn+1=(s,t)X(1s+1t)ns,t,nk=(s,t)Xns,t\left\{\begin{array}[]{ccl}nk-n+1&=&\sum\limits_{(s,t)\in X}\left(\frac{1}{s}+\frac{1}{t}\right)n_{s,t},\\ nk&=&\sum\limits_{(s,t)\in X}n_{s,t}\end{array}\right. (2)

because each vertex of GG contributes 1 to each of the two sides in the first equation and each edge of GG contributes 1 to each of the two sides in the second equation. Write (2) as

{4n2,2+3n2,4=4(nkn+1)2n4,44(s,t)Y(1s+1t)ns,t,n2,2+n2,4=nkn4,4(s,t)Y(1s+1t)ns,t.\left\{\begin{array}[]{ccl}4n_{2,2}+3n_{2,4}&=&4(nk-n+1)-2n_{4,4}-4\sum\limits_{(s,t)\in Y}\left(\frac{1}{s}+\frac{1}{t}\right)n_{s,t},\\ n_{2,2}+n_{2,4}&=&nk-n_{4,4}-\sum\limits_{(s,t)\in Y}\left(\frac{1}{s}+\frac{1}{t}\right)n_{s,t}.\end{array}\right. (3)

Therefore,

{n2,4=4n42n4,4,n2,2=nk4n+4+n4,4(s,t)Y(1s+1t)ns,t.\left\{\begin{array}[]{ccl}n_{2,4}&=&4n-4-2n_{4,4},\\ n_{2,2}&=&nk-4n+4+n_{4,4}-\sum\limits_{(s,t)\in Y}\left(\frac{1}{s}+\frac{1}{t}\right)n_{s,t}.\end{array}\right. (4)

Consequently, by (4) we have

SOα(G)\displaystyle SO_{\alpha}(G) =\displaystyle= (4α+4α)1/αn4,4+(2α+4α)1/αn2,4+(2α+2α)1/αn2,2\displaystyle(4^{\alpha}+4^{\alpha})^{1/\alpha}n_{4,4}+(2^{\alpha}+4^{\alpha})^{1/\alpha}n_{2,4}+(2^{\alpha}+2^{\alpha})^{1/\alpha}n_{2,2}
+\displaystyle~{}~{}+ (s,t)Y(sα+tα)1/αns,t\displaystyle\sum_{(s,t)\in Y}(s^{\alpha}+t^{\alpha})^{1/\alpha}n_{s,t}
=\displaystyle= (4n4)(2α+4α)1/α+2(nk4n+4)21/α\displaystyle(4n-4)(2^{\alpha}+4^{\alpha})^{1/\alpha}+2(nk-4n+4)2^{1/\alpha}
+\displaystyle~{}~{}+ (6×21/α2(2α+4α)1/α)n4,4\displaystyle\left(6\times 2^{1/\alpha}-2(2^{\alpha}+4^{\alpha})^{1/\alpha}\right)n_{4,4}
+\displaystyle~{}~{}+ (s,t)Y((sα+tα)1/α2(1s+1t)21/α)ns,t.\displaystyle\sum_{(s,t)\in Y}\left((s^{\alpha}+t^{\alpha})^{1/\alpha}-2\left(\frac{1}{s}+\frac{1}{t}\right)2^{1/\alpha}\right)n_{s,t}.

For α0\alpha\neq 0 and positive integer pp, let δα,p(s,t)=((s+p)α+tα)1/α(sα+tα)1/α\delta_{\alpha,p}(s,t)=\left((s+p)^{\alpha}+t^{\alpha}\right)^{1/\alpha}-\left(s^{\alpha}+t^{\alpha}\right)^{1/\alpha}, where s,t>0s,t>0.

Lemma 2.4.

If s,t>0s,t>0 and pp is an arbitrary positive integer, then
(i). rα(s,t;β)r_{\alpha}(s,t;\beta) strictly increases in ss for fixed tt, and in tt for fixed ss when α,β>0\alpha,\beta>0;
(ii). δα,p(s,t)>0\delta_{\alpha,p}(s,t)>0 and δα,p(s,t)\delta_{\alpha,p}(s,t) strictly decreases in tt for fixed ss when α>1\alpha>1;
(iii). δα,p(s,t)\delta_{\alpha,p}(s,t) strictly increases in ss for fixed tt when α>1\alpha>1.

Proof.

(i) follows directly since α,β>0\alpha,\beta>0.

Since 1<1/α1<0-1<1/\alpha-1<0 when α>1\alpha>1,

δα,1(s,t)t=tα1(((s+1)α+tα)1/α1(sα+tα)1/α1)<0.\frac{\partial\delta_{\alpha,1}(s,t)}{\partial t}=t^{\alpha-1}\big{(}\left((s+1)^{\alpha}+t^{\alpha}\right)^{1/\alpha-1}-\left(s^{\alpha}+t^{\alpha}\right)^{1/\alpha-1}\big{)}<0.

We also note that δα,1(s,t)>0\delta_{\alpha,1}(s,t)>0, hence (ii) follows as δα,p(s,t)=i=0p1δα,1(s+i,t)\delta_{\alpha,p}(s,t)=\sum^{p-1}_{i=0}\delta_{\alpha,1}(s+i,t).

Finally, since 11/α>01-1/\alpha>0 when α>1\alpha>1,

δα,1(s,t)s=((s+1)αsα+(s+1)αtα)11/α((s+1)αsα+sαtα)11/α(((s+1)α+tα)(sα+tα))11/α>0.\frac{\partial\delta_{\alpha,1}(s,t)}{\partial s}=\frac{\big{(}(s+1)^{\alpha}s^{\alpha}+(s+1)^{\alpha}t^{\alpha}\big{)}^{1-1/\alpha}-\big{(}(s+1)^{\alpha}s^{\alpha}+s^{\alpha}t^{\alpha}\big{)}^{1-1/\alpha}}{\big{(}((s+1)^{\alpha}+t^{\alpha})(s^{\alpha}+t^{\alpha})\big{)}^{1-1/\alpha}}>0.

Hence (iii) follows as δα,p(s,t)=i=0p1δα,1(s+i,t)\delta_{\alpha,p}(s,t)=\sum^{p-1}_{i=0}\delta_{\alpha,1}(s+i,t). ∎

The distance dG(u,v)d_{G}(u,v) between two vertices uu and vv of a connected graph GG is defined as usual as the length of a shortest path that connects uu and vv. In general, for two subgraphs G1G_{1} and G2G_{2} of GG, we define the distance between G1G_{1} and G2G_{2} by dG(G1,G2)=min{dG(u,v):uV(G1),vV(G2)}d_{G}(G_{1},G_{2})=\min\{d_{G}(u,v):u\in V(G_{1}),v\in V(G_{2})\}. For n2n\geq 2, a star-like cactus Sn,kS_{n,k} is defined intuitively as a kk-polygonal cactus such that all polygons have a vertex in common. It is clear that Sn,kS_{n,k} is unique and contains exactly one vertex of degree 2n2n while all other vertices have degree two.

Lemma 2.5.

Let α>1\alpha>1 and G𝒢n,kG\in\mathcal{G}_{n,k}, where n3n\geq 3 and k3k\geq 3. If GG contains a vertex of degree at least 6, then SOα(G)SO_{\alpha}(G) is not minimum in 𝒢n,k\mathcal{G}_{n,k}.

Proof.

Suppose to the contrary that GG contains qq (q1q\geq 1) vertices of degree at least 6 and SOα(G)SO_{\alpha}(G) is minimum in 𝒢n,k\mathcal{G}_{n,k}.
Case 1. q2q\geq 2.

Let u1u_{1} and ww be two vertices of degree at least 6 such that dG(u1,w)d_{G}(u_{1},w) is maximum and let PP be a shortest path connecting u1u_{1} and ww. Since dG(u1)6d_{G}(u_{1})\geq 6, u1u_{1} is contained in at least three polygons, exactly one of which, say CC, has at least two common vertices with PP. Let C1=u1u2uku1C_{1}=u_{1}u_{2}\cdots u_{k}u_{1} and C2=u1z2zku1C_{2}=u_{1}z_{2}\cdots z_{k}u_{1} be two polygons other than CC that contain u1u_{1} as a common vertex. Since dG(u1,w)d_{G}(u_{1},w) is maximum, we have dG(v)4d_{G}(v)\leq 4 for every v{u2,uk,z2,zk}v\in\{u_{2},u_{k},z_{2},z_{k}\}. Let C3=v1v2vkv1C_{3}=v_{1}v_{2}\cdots v_{k}v_{1} be a pendent polygon that lies in the same component with ww in Gu1G-u_{1} and the distance dG(u1,C3)d_{G}(u_{1},C_{3}) is as large as possible, where dG(v1)=2a4d_{G}(v_{1})=2a\geq 4 and dG(v2)=dG(v3)=2d_{G}(v_{2})=d_{G}(v_{3})=2.

Without loss of generality, assume dG(u1)=2bdG(w)6d_{G}(u_{1})=2b\geq d_{G}(w)\geq 6. Let G=Gu1u2u1uk+v2u2+v2ukG^{\prime}=G-u_{1}u_{2}-u_{1}u_{k}+v_{2}u_{2}+v_{2}u_{k}. Then by Lemma 2.4, we have

SOα(G)SOα(G)\displaystyle SO_{\alpha}(G)-SO_{\alpha}(G^{\prime}) >\displaystyle> (rα(2b,d(u2))rα(4,d(u2))+(rα(2b,d(uk))rα(4,d(uk))\displaystyle\big{(}r_{\alpha}(2b,d(u_{2}))-r_{\alpha}(4,d(u_{2})\big{)}+\big{(}r_{\alpha}(2b,d(u_{k}))-r_{\alpha}(4,d(u_{k})\big{)}
+(rα(2,2)rα(4,2))+(rα(2,2a)rα(4,2a))\displaystyle+\big{(}r_{\alpha}(2,2)-r_{\alpha}(4,2)\big{)}+\big{(}r_{\alpha}(2,2a)-r_{\alpha}(4,2a)\big{)}
+(rα(2b,d(z2))rα(2b2,d(z2)))\displaystyle+\big{(}r_{\alpha}(2b,d(z_{2}))-r_{\alpha}(2b-2,d(z_{2}))\big{)}
+(rα(2b,d(zk))rα(2b2,d(zk)))\displaystyle+\big{(}r_{\alpha}(2b,d(z_{k}))-r_{\alpha}(2b-2,d(z_{k}))\big{)}
>\displaystyle> (rα(6,4)rα(4,4))+(rα(6,4)rα(4,4))\displaystyle\big{(}r_{\alpha}(6,4)-r_{\alpha}(4,4)\big{)}+\big{(}r_{\alpha}(6,4)-r_{\alpha}(4,4)\big{)}
+(rα(2,2)rα(4,2))+(rα(2,4)rα(4,4))\displaystyle+\big{(}r_{\alpha}(2,2)-r_{\alpha}(4,2)\big{)}+\big{(}r_{\alpha}(2,4)-r_{\alpha}(4,4)\big{)}
+(rα(6,4)rα(4,4))+(rα(6,4)rα(4,4))\displaystyle+\big{(}r_{\alpha}(6,4)-r_{\alpha}(4,4)\big{)}+\big{(}r_{\alpha}(6,4)-r_{\alpha}(4,4)\big{)}
=\displaystyle= 8(3α+2α)1/α18×21/α\displaystyle 8(3^{\alpha}+2^{\alpha})^{1/\alpha}-18\times 2^{1/\alpha}
>\displaystyle> 86×21/α18×21/α\displaystyle 8\sqrt{6}\times 2^{1/\alpha}-18\times 2^{1/\alpha}
>\displaystyle> 0,\displaystyle 0,

which contradicts the minimality of GG.

Case 2. q=1q=1.

If G≇Sn,kG\not\cong S_{n,k}, then the discussion for this case is similar to that for Case 1 by choosing u1u_{1} to be the vertex with degree at least 6 and C3=v1v2vkv1C_{3}=v_{1}v_{2}\cdots v_{k}v_{1} to be a pendent polygon such that dG(u1,C3)d_{G}(u_{1},C_{3}) is maximum. Otherwise, GSn,kG\cong S_{n,k}. Let u1u_{1} be the vertex with degree 2n62n\geq 6, C1=u1u2uku1C_{1}=u_{1}u_{2}\cdots u_{k}u_{1} and C2=u1z2zku1C_{2}=u_{1}z_{2}\cdots z_{k}u_{1} be two pendent polygons. Let G=Gu1u2u1uk+z2u2+z2ukG^{\prime}=G-u_{1}u_{2}-u_{1}u_{k}+z_{2}u_{2}+z_{2}u_{k}. Then by Lemma 2.4 and Lemma 2.2, we have

SOα(G)SOα(G)\displaystyle SO_{\alpha}(G)-SO_{\alpha}(G^{\prime}) =\displaystyle= (2n×rα(2n,2)+(nk2n)×rα(2,2))\displaystyle\big{(}2n\times r_{\alpha}(2n,2)+(nk-2n)\times r_{\alpha}(2,2)\big{)}
((2n3)×rα(2n2,2)+rα(2n2,4)\displaystyle-\big{(}(2n-3)\times r_{\alpha}(2n-2,2)+r_{\alpha}(2n-2,4)
+3rα(4,2)+(nk2n1)rα(2,2))\displaystyle+3r_{\alpha}(4,2)+(nk-2n-1)r_{\alpha}(2,2)\big{)}
=\displaystyle= 2n×rα(2n,2)+rα(2,2)(2n3)×rα(2n2,2)\displaystyle 2n\times r_{\alpha}(2n,2)+r_{\alpha}(2,2)-(2n-3)\times r_{\alpha}(2n-2,2)
rα(2n2,4)3rα(4,2)\displaystyle-r_{\alpha}(2n-2,4)-3r_{\alpha}(4,2)
>\displaystyle> 3rα(2n,2)+rα(2,2)rα(2n2,4)3rα(4,2)\displaystyle 3r_{\alpha}(2n,2)+r_{\alpha}(2,2)-r_{\alpha}(2n-2,4)-3r_{\alpha}(4,2)
>\displaystyle> 2rα(2n,2)+rα(2,2)3rα(4,2)\displaystyle 2r_{\alpha}(2n,2)+r_{\alpha}(2,2)-3r_{\alpha}(4,2)
>\displaystyle> rα(6,2)+rα(2,2)2rα(4,2)\displaystyle r_{\alpha}(6,2)+r_{\alpha}(2,2)-2r_{\alpha}(4,2)
>\displaystyle> 0,\displaystyle 0,

which contradicts the minimality of GG.

Recall that a graph is called a chemical graph if it has no vertex of degree more than 4. For G𝒢n,kG\in\mathcal{G}_{n,k}, we call GG a chemical (n,k)(n,k)-cactus, or chemical cactus for short, if GG has no vertex of degree greater than 4. It is clear that every cut vertex in a chemical cactus has degree 4, which connects exactly two polygons. The following corollary follows directly from Lemma 2.3 and Lemma 2.5, which shows that the minimum value of SOα(G)SO_{\alpha}(G) among all cacti in 𝒢n,k\mathcal{G}_{n,k} is attained only by chemical cacti.

Corollary 2.1.

For α>1\alpha>1, n3,k3n\geq 3,k\geq 3 and G𝒢n,kG\in\mathcal{G}_{n,k}, if GG attains the minimum value of SOα(G)SO_{\alpha}(G), then GG is a chemical cactus and

SOα(G)=(4n4)(2α+4α)1/α+2(nk4n+4)21/α+(6×21/α2(2α+4α)1/α)n4,4(G).SO_{\alpha}(G)=(4n-4)(2^{\alpha}+4^{\alpha})^{1/\alpha}+2(nk-4n+4)2^{1/\alpha}+\left(6\times 2^{1/\alpha}-2(2^{\alpha}+4^{\alpha})^{1/\alpha}\right)n_{4,4}(G).

In the following we will determine the minimum value of SOα(G)SO_{\alpha}(G) among all chemical cacti. By Corollary 2.1, this is equivalent to determine the maximum value of n4,4(G)n_{4,4}(G) as 6×21/α2(2α+4α)1/α<6×21/α2(2×3α)1/α=06\times 2^{1/\alpha}-2(2^{\alpha}+4^{\alpha})^{1/\alpha}<6\times 2^{1/\alpha}-2(2\times 3^{\alpha})^{1/\alpha}=0 by Lemma 2.1. For a chemical cactus HH, we call a polygon CC in HH a saturated polygon if every vertex on CC is a cut vertex, i.e., a vertex of degree 4. Further, we call a chemical cactus HH nice-saturated if the following two conditions hold:
1). HH has as many as possible saturated polygons;
2). the cut vertices on each polygon of HH are successively arranged.

For a chemical cactus HH, let T(H)T(H) be the tree whose vertices are the polygons in HH and two vertices are adjacent provided their corresponding polygons has a common vertex. It is clear that T(H)T(H) is a tree with maximum vertex degree no more than kk. Let pp be the number of the vertices of degree kk in T(H)T(H), and let d1,d2,,dsd_{1},d_{2},\ldots,d_{s} be the degrees of all the vertices in HH that are neither of degree 1 nor of degree kk, i.e., 1<di<k1<d_{i}<k for each i{1,2,,s}i\in\{1,2,\ldots,s\}. Since T(H)T(H) is a tree, we have

kp+d1+d2++ds+(nps)=2n2kp+d_{1}+d_{2}+\cdots+d_{s}+(n-p-s)=2n-2 (5)

and every saturated polygon in HH corresponds to a vertex of degree kk in T(H)T(H). Further, T(H)T(H) has as many as possible vertices of degree kk if and only if d1+d2++dss<k1d_{1}+d_{2}+\cdots+d_{s}-s<k-1. This implies that

n2k11<p=n2(d1+d2++dss)k1n2k1.\frac{n-2}{k-1}-1<p=\frac{n-2-(d_{1}+d_{2}+\cdots+d_{s}-s)}{k-1}\leq\frac{n-2}{k-1}. (6)

That is, if HH is nice-saturated then HH has exactly n2k1\left\lfloor\frac{n-2}{k-1}\right\rfloor saturated cycles. As an example, a chemical (6,4)(6,4)-cactus with p<n2k1=1p<\left\lfloor\frac{n-2}{k-1}\right\rfloor=1, a chemical (6,4)(6,4)-cactus in which the cut vertices on some polygon are not successively arranged, and a nice-saturated (6,4)(6,4)-cactus are illustrated as (a), (b) and (c), respectively, in Figure 1.

Refer to caption
Figure 1: (a). p=0p=0; (b). The cut vertices on the kk-cycle CC are not successively arranged; (c). A nice-saturated (6,4)(6,4)-cactus.
Lemma 2.6.

A chemical cactus HH attains the maximum value of n4,4(H)n_{4,4}(H) if and only if HH is nice-saturated.

Proof.

Let pp and d1,d2,,dsd_{1},d_{2},\ldots,d_{s} be defined as above. By a simple calculation, we have

n4,4(H)kp+vV(G),d(v)<k(d(v)1)=kp+(d11)+(d21)++(ds1)n_{4,4}(H)\leq kp+\sum_{v\in V(G),d(v)<k}(d(v)-1)=kp+(d_{1}-1)+(d_{2}-1)+\cdots+(d_{s}-1) (7)

and the equality holds if and only if the cut vertices on each polygon of HH are successively arranged.

Suppose p<n2k1p<\left\lfloor\frac{n-2}{k-1}\right\rfloor. Then by the pervious analysis, we have d1+d2++dssk1d_{1}+d_{2}+\cdots+d_{s}-s\geq k-1. Let d1,d2,,dsd^{\prime}_{1},d^{\prime}_{2},\ldots,d^{\prime}_{s} be a sequence satisfying d1=k,1didid^{\prime}_{1}=k,1\leq d^{\prime}_{i}\leq d_{i} for i{2,3,,s}i\in\{2,3,\cdots,s\} and i=1sdi=i=1sdi\sum_{i=1}^{s}d^{\prime}_{i}=\sum_{i=1}^{s}d_{i}. Let S be the sequence obtained from the degree sequence of T(H)T(H) by replacing d1,d2,,dsd_{1},d_{2},\ldots,d_{s} by d1,d2,,dsd^{\prime}_{1},d^{\prime}_{2},\ldots,d^{\prime}_{s}, respectively. It is clear that S is still a degree sequence of a tree with maximum degree not greater than kk. Let HH^{\prime} be a cactus such that T(H)T(H^{\prime}) has degree sequence S and the cut vertices on each polygon of HH^{\prime} are successively arranged. Then by (7) and a direct calculation, we have n4,4(H)=n4,4(H)+1n_{4,4}(H^{\prime})=n_{4,4}(H)+1. That is, HH does not attain the maximum value of n4,4(H)n_{4,4}(H), which completes our proof. ∎

Theorem 2.1.

Let G𝒢n,kG\in\mathcal{G}_{n,k}, where n3n\geq 3 and k3k\geq 3. Then

SOα(G)(4n4)(2α+4α)1/α+2(nk4n+4)21/αSO_{\alpha}(G)\geq(4n-4)(2^{\alpha}+4^{\alpha})^{1/\alpha}+2(nk-4n+4)2^{1/\alpha}
+(6×21/α2(2α+4α)1/α)(n2+n2k1),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+\left(6\times 2^{1/\alpha}-2(2^{\alpha}+4^{\alpha})^{1/\alpha}\right)\big{(}n-2+\left\lfloor\frac{n-2}{k-1}\right\rfloor\big{)},

and the equality holds if and only if GG is a nice-saturated chemical cactus.

Proof.

By (7), (5) and (6), if GG is minimum, then

n4,4(G)=kp+(d11)+(d21)++(ds1)=n2+p=n2+n2k1.n_{4,4}(G)=kp+(d_{1}-1)+(d_{2}-1)+\cdots+(d_{s}-1)=n-2+p=n-2+\left\lfloor\frac{n-2}{k-1}\right\rfloor.

Hence, the theorem follows directly from Corollary 2.1 and Lemma 2.6. ∎

3 Polygonal cactus with maximum general Sombor index

In this section we will characterize the polygonal cactus with maximum general Sombor index for the two cases α1,β>1\alpha\geq 1,\beta>1; and α=2,1/2β<1\alpha=2,1/2\leq\beta<1, respectively.

Lemma 3.1.

Let ΔABM\Delta ABM be a triangle in Euclidean space and OO the midpoint of the triangle side ABAB. Then |MA|2β+|MB|2β>2|MO|2β|MA|^{2\beta}+|MB|^{2\beta}>2|MO|^{2\beta} for any real number β12\beta\geq\frac{1}{2}, where |MA||MA| is the length of the side MAMA.

Proof.

Let |MA|=a|MA|=a, |MB|=b|MB|=b and |MO|=d|MO|=d. When a=ba=b, the lemma follows directly. Without loss of generality, we now assume a>b>0a>b>0. By the triangle inequality, d<a+b2<ad<\frac{a+b}{2}<a and so (a,b)(a+b2,a+b2)(a,b)\rhd\left(\frac{a+b}{2},\frac{a+b}{2}\right). Hence, by Lemma 2.1, a2β+b2β2(a+b2)2β>2d2βa^{2\beta}+b^{2\beta}\geq 2\left(\frac{a+b}{2}\right)^{2\beta}>2d^{2\beta} when β12.\beta\geq\frac{1}{2}.

Lemma 3.2.

Let s>2s>2 and t>2t>2. Then
(i). rα(s+2,2;β)rα(s2,2;β)>0r_{\alpha}(s+2,2;\beta)-r_{\alpha}(s-2,2;\beta)>0 for any α>0\alpha>0 and β>0\beta>0;
(ii). rα(s+2,t;β)+rα(s2,t;β)2rα(s,t;β)r_{\alpha}(s+2,t;\beta)+r_{\alpha}(s-2,t;\beta)\geq 2r_{\alpha}(s,t;\beta) for any α1\alpha\geq 1 and β>1\beta>1;
(iii). rα(s+2,t2;β)+rα(s2,t+2;β)2rα(s,t;β)r_{\alpha}(s+2,t-2;\beta)+r_{\alpha}(s-2,t+2;\beta)\geq 2r_{\alpha}(s,t;\beta) for any α1\alpha\geq 1 and β>1\beta>1.

Proof.

(i) follows directly.

For (ii), by Lemma 2.1 and the monotonicity of rα(s,t;β)r_{\alpha}(s,t;\beta), we have

rα(s+2,t;β)+rα(s2,t;β)\displaystyle r_{\alpha}(s+2,t;\beta)+r_{\alpha}(s-2,t;\beta) =\displaystyle= ((s+2)α+tα)β+((s2)α+tα)β\displaystyle\left((s+2)^{\alpha}+t^{\alpha}\right)^{\beta}+\left((s-2)^{\alpha}+t^{\alpha}\right)^{\beta}
2((s+2)α+(s2)α2+tα)β\displaystyle\geq 2\left(\frac{(s+2)^{\alpha}+(s-2)^{\alpha}}{2}+t^{\alpha}\right)^{\beta}
2(sα+tα)β\displaystyle\geq 2(s^{\alpha}+t^{\alpha})^{\beta}
=2rα(s,t;β).\displaystyle=2r_{\alpha}(s,t;\beta).

Hence, (ii) holds.

The discussion for (iii) is analogous to that for (ii). ∎

Theorem 3.1.

Let n3,k3n\geq 3,k\geq 3 and G𝒢n,kG\in\mathcal{G}_{n,k}. If α1\alpha\geq 1 and β>1\beta>1 ; or α=2\alpha=2 and 12β<1\frac{1}{2}\leq\beta<1, then

SOα(G;β)2n((2n)α+2α)β+n(k2)(2α+1)βSO_{\alpha}(G;\beta)\leq 2n((2n)^{\alpha}+2^{\alpha})^{\beta}+n(k-2)(2^{\alpha+1})^{\beta}

and the equality holds if and only if GSn,kG\cong S_{n,k}.

Proof.

We first assume that α1\alpha\geq 1 and β>1\beta>1.

Let GG be such that SOα(G;β)SO_{\alpha}(G;\beta) is as large as possible. Further, let C1=z1z2zkz1C_{1}=z_{1}z_{2}\cdots z_{k}z_{1} and C2=v1v2vkv1C_{2}=v_{1}v_{2}\cdots v_{k}v_{1} be two pendent polygons such that dG(C1,C2)d_{G}(C_{1},C_{2}) is as large as possible, where z1z_{1} and v1v_{1} are the cut-vertices of C1C_{1} and C2C_{2}, respectively.

If GSn,kG\cong S_{n,k}, then the theorem follows directly. We now assume G≇Sn,kG\not\cong S_{n,k}. Then, z1v1z_{1}\neq v_{1}. Let G1=Gv1v2v1vk+z1v2+z1vkG_{1}=G-v_{1}v_{2}-v_{1}v_{k}+z_{1}v_{2}+z_{1}v_{k} and G2=Gz1z2z1zk+v1z2+v1zkG_{2}=G-z_{1}z_{2}-z_{1}z_{k}+v_{1}z_{2}+v_{1}z_{k}. We consider the following two cases:

Case 1. z1z_{1} and v1v_{1} are adjacent in GG.

In this case, we have SOα(G1;β)SOα(G;β)=SO_{\alpha}(G_{1};\beta)-SO_{\alpha}(G;\beta)=

vNG(v1){z1}(rα(dG(v1)2,dG(v);β)rα(dG(v1),dG(v);β))\displaystyle\sum_{v\in N_{G}(v_{1})\setminus\{z_{1}\}}\left(r_{\alpha}\big{(}d_{G}(v_{1})-2,d_{G}(v);\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(v);\beta\big{)}\right)
+zNG(z1){v1}(rα(dG(z1)+2,dG(z);β)rα(dG(z1),dG(z);β))\displaystyle+\sum_{z\in N_{G}(z_{1})\setminus\{v_{1}\}}\left(r_{\alpha}\big{(}d_{G}(z_{1})+2,d_{G}(z);\beta\big{)}-r_{\alpha}\big{(}d_{G}(z_{1}),d_{G}(z);\beta\big{)}\right)
+2rα(dG(z1)+2,2;β)2rα(dG(v1)2,2;β)\displaystyle+2r_{\alpha}\big{(}d_{G}(z_{1})+2,2;\beta\big{)}-2r_{\alpha}\big{(}d_{G}(v_{1})-2,2;\beta\big{)}
+rα(dG(v1)2,dG(z1)+2;β)rα(dG(v1),dG(z1);β),and\displaystyle+r_{\alpha}\big{(}d_{G}(v_{1})-2,d_{G}(z_{1})+2;\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(z_{1});\beta\big{)},\,\,\text{and}

SOα(G2;β)SOα(G;β)=SO_{\alpha}(G_{2};\beta)-SO_{\alpha}(G;\beta)=

vNG(v1){z1}(rα(dG(v1)+2,dG(v);β)rα(dG(v1),dG(v);β))\displaystyle\sum_{v\in N_{G}(v_{1})\setminus\{z_{1}\}}\left(r_{\alpha}\big{(}d_{G}(v_{1})+2,d_{G}(v);\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(v);\beta\big{)}\right)
+zNG(z1){v1}(rα(dG(z1)2,dG(w);β)rα(dG(z1),dG(z);β))\displaystyle+\sum_{z\in N_{G}(z_{1})\setminus\{v_{1}\}}\left(r_{\alpha}\big{(}d_{G}(z_{1})-2,d_{G}(w);\beta\big{)}-r_{\alpha}\big{(}d_{G}(z_{1}),d_{G}(z);\beta\big{)}\right)
+2rα(dG(v1)+2,2;β)2rα(dG(z1)2,2;β)\displaystyle+2r_{\alpha}\big{(}d_{G}(v_{1})+2,2;\beta\big{)}-2r_{\alpha}\big{(}d_{G}(z_{1})-2,2;\beta\big{)}
+rα(dG(v1)+2,dG(z1)2;β)rα(dG(v1),dG(z1);β).\displaystyle+r_{\alpha}\big{(}d_{G}(v_{1})+2,d_{G}(z_{1})-2;\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(z_{1});\beta\big{)}.

Recall that z1z_{1} and v1v_{1} are the cut-vertices of C1C_{1} and C2C_{2}, respectively. Therefore, dG(z1)4d_{G}(z_{1})\geq 4 and dG(v1)4d_{G}(v_{1})\geq 4. Combining with Lemma 3.2, we have

rα(dG(v1)+2,dG(v);β)+rα(dG(v1)2,dG(v);β)>2rα(dG(v1),dG(v);β),\displaystyle r_{\alpha}\big{(}d_{G}(v_{1})+2,d_{G}(v);\beta\big{)}+r_{\alpha}\big{(}d_{G}(v_{1})-2,d_{G}(v);\beta\big{)}>2r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(v);\beta\big{)},
rα(dG(z1)+2,dG(z);β)+rα(dG(z1)2,dG(z);β)>2rα(dG(z1),dG(z);β),\displaystyle r_{\alpha}\big{(}d_{G}(z_{1})+2,d_{G}(z);\beta\big{)}+r_{\alpha}\big{(}d_{G}(z_{1})-2,d_{G}(z);\beta\big{)}>2r_{\alpha}\big{(}d_{G}(z_{1}),d_{G}(z);\beta\big{)},
rα(dG(v1)+2,dG(z1)2;β)+rα(dG(v1)2,dG(z1)+2;β)>2rα(dG(v1),dG(z1);β),\displaystyle r_{\alpha}\big{(}d_{G}(v_{1})+2,d_{G}(z_{1})-2;\beta\big{)}+r_{\alpha}\big{(}d_{G}(v_{1})-2,d_{G}(z_{1})+2;\beta\big{)}>2r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(z_{1});\beta\big{)},
rα(dG(z1)+2,2;β)rα(dG(z1)2,2;β)>0, and\displaystyle r_{\alpha}\big{(}d_{G}(z_{1})+2,2;\beta\big{)}-r_{\alpha}\big{(}d_{G}(z_{1})-2,2;\beta\big{)}>0,\text{\quad and}
rα(dG(v1)+2,2;β)rα(dG(v1)2,2;β)>0.\displaystyle r_{\alpha}\big{(}d_{G}(v_{1})+2,2;\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1})-2,2;\beta\big{)}>0.

This means that SOα(G1;β)>SOα(G;β)SO_{\alpha}(G_{1};\beta)>SO_{\alpha}(G;\beta) or SOα(G2;β)>SOα(G;β)SO_{\alpha}(G_{2};\beta)>SO_{\alpha}(G;\beta), a contradiction.

Case 2. z1z_{1} and v1v_{1} are not adjacent in GG.

In this case, we have

SOα(G1;β)SOα(G;β)\displaystyle SO_{\alpha}(G_{1};\beta)-SO_{\alpha}(G;\beta) =\displaystyle= vNG(v1)(rα(dG(v1)2,dG(v);β)rα(dG(v1),dG(v);β))\displaystyle\sum_{v\in N_{G}(v_{1})}\left(r_{\alpha}\big{(}d_{G}(v_{1})-2,d_{G}(v);\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(v);\beta\big{)}\right)
+zNG(z1)(rα(dG(z1)+2,dG(z);β)rα(dG(z1),dG(z);β))\displaystyle+\sum_{z\in N_{G}(z_{1})}\left(r_{\alpha}\big{(}d_{G}(z_{1})+2,d_{G}(z);\beta\big{)}-r_{\alpha}\big{(}d_{G}(z_{1}),d_{G}(z);\beta\big{)}\right)
+2rα(dG(z1)+2,2;β)2rα(dG(v1)2,2;β),and\displaystyle+2r_{\alpha}\big{(}d_{G}(z_{1})+2,2;\beta\big{)}-2r_{\alpha}\big{(}d_{G}(v_{1})-2,2;\beta\big{)},\,\,\text{and}
SOα(G2;β)SOα(G;β)\displaystyle SO_{\alpha}(G_{2};\beta)-SO_{\alpha}(G;\beta) =\displaystyle= vNG(v1)(rα(dG(v1)+2,dG(v);β)rα(dG(v1),dG(v);β))\displaystyle\sum_{v\in N_{G}(v_{1})}\left(r_{\alpha}\big{(}d_{G}(v_{1})+2,d_{G}(v);\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(v);\beta\big{)}\right)
+zNG(z1)(rα(dG(z1)2,dG(z);β)rα(dG(z1),dG(z);β))\displaystyle+\sum_{z\in N_{G}(z_{1})}\left(r_{\alpha}\big{(}d_{G}(z_{1})-2,d_{G}(z);\beta\big{)}-r_{\alpha}\big{(}d_{G}(z_{1}),d_{G}(z);\beta\big{)}\right)
+2rα(dG(v1)+2,2;β)2rα(dG(z1)2,2;β).\displaystyle+2r_{\alpha}\big{(}d_{G}(v_{1})+2,2;\beta\big{)}-2r_{\alpha}\big{(}d_{G}(z_{1})-2,2;\beta\big{)}.

Recall that dG(z1)4d_{G}(z_{1})\geq 4 and dG(v1)4d_{G}(v_{1})\geq 4. Similar to Case 1, by Lemma 3.2 , we have

rα(dG(v1)+2,dG(v);β)+rα(dG(v1)2,dG(v);β)>2rα(dG(v1),dG(v);β),\displaystyle r_{\alpha}\big{(}d_{G}(v_{1})+2,d_{G}(v);\beta\big{)}+r_{\alpha}\big{(}d_{G}(v_{1})-2,d_{G}(v);\beta\big{)}>2r_{\alpha}\big{(}d_{G}(v_{1}),d_{G}(v);\beta\big{)},
rα(dG(z1)+2,dG(z);β)+rα(dG(z1)2,dG(z);β)>2rα(dG(z1),dG(z);β),\displaystyle r_{\alpha}\big{(}d_{G}(z_{1})+2,d_{G}(z);\beta\big{)}+r_{\alpha}\big{(}d_{G}(z_{1})-2,d_{G}(z);\beta\big{)}>2r_{\alpha}\big{(}d_{G}(z_{1}),d_{G}(z);\beta\big{)},
rα(dG(z1)+2,2;β)rα(dG(z1)2,2;β)>0 and\displaystyle r_{\alpha}\big{(}d_{G}(z_{1})+2,2;\beta\big{)}-r_{\alpha}\big{(}d_{G}(z_{1})-2,2;\beta\big{)}>0\text{\quad and}
rα(dG(v1)+2,2;β)rα(dG(v1)2,2;β)>0,\displaystyle r_{\alpha}\big{(}d_{G}(v_{1})+2,2;\beta\big{)}-r_{\alpha}\big{(}d_{G}(v_{1})-2,2;\beta\big{)}>0,

which means that SOα(G1;β)>SOα(G;β)SO_{\alpha}(G_{1};\beta)>SO_{\alpha}(G;\beta) or SOα(G2;β)>SOα(G;β)SO_{\alpha}(G_{2};\beta)>SO_{\alpha}(G;\beta), a contradiction.

Therefore, Sn,kS_{n,k} is the unique maximal polygonal cactus. Further, we have

SOα(Sn,k;β)=2nrα(2n,2;β)+n(k2)rα(2,2;β)=2n((2n)α+2α)β+n(k2)(2α+1)β.SO_{\alpha}(S_{n,k};\beta)=2nr_{\alpha}(2n,2;\beta)+n(k-2)r_{\alpha}(2,2;\beta)=2n((2n)^{\alpha}+2^{\alpha})^{\beta}+n(k-2)(2^{\alpha+1})^{\beta}.

The discussion for the case that α=2\alpha=2 and 12β<1\frac{1}{2}\leq\beta<1 is analogous by Lemma 3.2 (i) and Lemma 3.1.∎

4 Polygonal cacti with minimum general Sombor index

A symmetric function φ(s,t)\varphi(s,t) defined on positive real numbers is called escalating [22] if

φ(s1,s2)+φ(t1,t2)φ(s2,t1)+φ(s1,t2)\displaystyle\varphi(s_{1},s_{2})+\varphi(t_{1},t_{2})\geq\varphi(s_{2},t_{1})+\varphi(s_{1},t_{2}) (8)

for any s1t1>0s_{1}\geq t_{1}>0 and s2t2>0,s_{2}\geq t_{2}>0, and the inequality holds if s1>t1>0s_{1}>t_{1}>0 and s2>t2>0s_{2}>t_{2}>0. Further, an escalating function φ(s,t)\varphi(s,t) is called special escalating [23] if

4φ(2l,2)φ(2l2,4)φ(2l2,2)φ(4,2)φ(4,4)0\displaystyle 4\varphi(2l,2)-\varphi(2l-2,4)-\varphi(2l-2,2)-\varphi(4,2)-\varphi(4,4)\geq 0 (9)

for l3l\geq 3 and

φ(s1,s2)φ(t1,t2)0\displaystyle\varphi(s_{1},s_{2})-\varphi(t_{1},t_{2})\geq 0 (10)

for any s1t12s_{1}\geq t_{1}\geq 2 and s2t22s_{2}\geq t_{2}\geq 2.

Lemma 4.1.

If s,t>0s,t>0, then rα(s,t;β)=(sα+tα)βr_{\alpha}(s,t;\beta)=\left(s^{\alpha}+t^{\alpha}\right)^{\beta} is special escalating for α1\alpha\geq 1 and β>1\beta>1.

Proof.

Set φ(s,t)=(sα+tα)β\varphi(s,t)=\left(s^{\alpha}+t^{\alpha}\right)^{\beta}. Since α1\alpha\geq 1 and β>1\beta>1, we have (s1α+s2α,t1α+t2α)(s2α+t1α,s1α+t2α)\big{(}s_{1}^{\alpha}+s_{2}^{\alpha},t_{1}^{\alpha}+t_{2}^{\alpha}\big{)}\rhd\big{(}s_{2}^{\alpha}+t_{1}^{\alpha},s_{1}^{\alpha}+t_{2}^{\alpha}\big{)} when s1>t1>0s_{1}>t_{1}>0 and s2>t2>0s_{2}>t_{2}>0. Then by Lemma 2.1, the inequality in (8) strictly holds. Further, it is clear that the equality in (8) holds when s1=t1>0s_{1}=t_{1}>0 or s2=t2>0s_{2}=t_{2}>0. This means that (sα+tα)β\left(s^{\alpha}+t^{\alpha}\right)^{\beta} is escalating.

In addition, by Lemma 2.1 and the monotonicity of (sα+tα)β\left(s^{\alpha}+t^{\alpha}\right)^{\beta}, if l3l\geq 3 and α1\alpha\geq 1 then (2l)α+2α(2l2)α+4α>(2l2)α+2α(2l)^{\alpha}+2^{\alpha}\geq(2l-2)^{\alpha}+4^{\alpha}>(2l-2)^{\alpha}+2^{\alpha}, (2l)α+2α>4α+2α(2l)^{\alpha}+2^{\alpha}>4^{\alpha}+2^{\alpha} and (2l)α+2α6α+2α4α+4α(2l)^{\alpha}+2^{\alpha}\geq 6^{\alpha}+2^{\alpha}\geq 4^{\alpha}+4^{\alpha}. Hence, (9) follows directly as β>1\beta>1.

Finally, it is easy to see that (10) holds when α1\alpha\geq 1 and β>1\beta>1 by the monotonicity of (sα+tα)β\left(s^{\alpha}+t^{\alpha}\right)^{\beta}. Therefore, (sα+tα)β\left(s^{\alpha}+t^{\alpha}\right)^{\beta} is special escalating. ∎

A kk-polygonal cactus GG is called a cactus chain if each polygon in GG has at most two cut-vertices and each cut-vertex is the common vertex of exactly two polygons. It is clear that each cactus chain has exactly n2n-2 non-pendent polygons and two pendent polygons for n2n\geq 2. We denote by 𝒜n,k\mathcal{A}_{n,k} the class consisting of those cactus chains such that each pair of cut-vertices that lies in the same polygon of GG are adjacent. In contrast, we denote by n,k\mathcal{B}_{n,k} the class consisting of those cactus chains such that each pair of cut-vertices that lies in the same polygon of GG are not adjacent. It can be seen that 𝒜n,k\mathcal{A}_{n,k} is unique for k3k\geq 3 and n,3=\mathcal{B}_{n,3}=\emptyset.

Theorem 4.1.

[23] Let f(s,t)f(s,t) be a special escalating function and GG be a cactus of 𝒢n,k\mathcal{G}_{n,k}, where n3n\geq 3 and k3k\geq 3.
(i). If k=3k=3, then

If(G)2f(2,2)+2nf(4,2)+(n2)f(4,4)I_{f}(G)\geq 2f(2,2)+2nf(4,2)+(n-2)f(4,4)

with equality holding if and only if G𝒜n,3G\in\mathcal{A}_{n,3}.
(ii). If k4k\geq 4, then

If(G)(kn4n+4)f(2,2)+(4n4)f(4,2),I_{f}(G)\geq(kn-4n+4)f(2,2)+(4n-4)f(4,2),

where the equality holds if Gn,kG\in\mathcal{B}_{n,k}. Furthermore, if k{4,5}k\in\{4,5\}, then the equality holds if and only if Gn,k.G\in\mathcal{B}_{n,k}.

Corollary 4.1.

Let n3,k3,α1,β>1n\geq 3,k\geq 3,\alpha\geq 1,\beta>1 and G𝒢n,kG\in\mathcal{G}_{n,k}.
(i). If k=3k=3, then SOα(G;β)2(2α+1)β+2n(4α+2α)β+(n2)(24α)βSO_{\alpha}(G;\beta)\geq 2\big{(}2^{\alpha+1}\big{)}^{\beta}+2n\big{(}4^{\alpha}+2^{\alpha}\big{)}^{\beta}+(n-2)\big{(}2\cdot 4^{\alpha}\big{)}^{\beta}, where the equality holds if and only if G𝒜n,3G\in\mathcal{A}_{n,3}.
(ii). If k4k\geq 4, then SOα(G;β)(kn4n+4)(2α+1)β+(4n4)(4α+2α)βSO_{\alpha}(G;\beta)\geq(kn-4n+4)\big{(}2^{\alpha+1}\big{)}^{\beta}+(4n-4)\big{(}4^{\alpha}+2^{\alpha}\big{)}^{\beta}, where the equality holds if Gn,kG\in\mathcal{B}_{n,k}. Furthermore, if k{4,5}k\in\{4,5\}, then the equality holds if and only if Gn,k.G\in\mathcal{B}_{n,k}.

Proof.

In Theorem 4.1, set f(s,t)=rα(s,t;β)f(s,t)=r_{\alpha}(s,t;\beta). Then the corollary follows immediately by Lemma 4.1 and a simple calculation. ∎

5 Acknowledgement

This work was supported by the National Natural Science Foundation of China [Grant number: 11971406].

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