Extremal Trees for the General Randić Index with a Given Domination Number

Abstract

For a graph G with vertex set $V_G$ and edge set $E_G$, the general Randić index $R_{\alpha }(G)$ is defined as the sum of items $(d_ud_v)^{\alpha }$ of all edges $uv\in E_G$, where $d_u$ and $d_v$ denote the degrees of vertices $u,v\in V_G$, respectively, and $\alpha$ is an arbitrary real number. In this paper, we determine the upper bound on $R_{\alpha }(G)$ with $\alpha \in [\alpha _1,0)$ of trees in terms of domination number, where $\alpha _1\approx -0.5287$ is the unique nonzero root of the equation $27^\alpha -6^\alpha -4^\alpha +2^\alpha =0$. Moreover, the lower bound on $R_{\alpha }(G)$ with $\alpha \in [\alpha _2,0)$ of trees in terms of domination number is also obtained, where $\alpha _2\approx -0.5696$ is the unique nonzero root of the equation $2\cdot 4^\alpha +2^{\alpha +1}-3^\alpha -2\cdot 6^\alpha -1=0$. Finally, the corresponding extremal trees are characterized.

1 Introduction

The Randić index, also known as the connectivity index or the branching index, was introduced by the chemist Randić [22] in 1975, which has been extensively utilized in applications for chemistry, biology, and complex network [8, 12, 13, 23]. In addition, various mathematical, physical, and chemical properties of the Randić index have been established in decades [7, 14, 18, 24, 27]. The formula of the Randić index for a graph $G=(V_{G},E_{G})$ is given by

$$\begin{aligned} R=R(G)=\sum _{uv\in E_G}(d_ud_v)^{-\frac{1}{2}}, \end{aligned}$$

where $d_u$ and $d_v$ denote the degrees of vertices $u,v\in V_G$, respectively.

Ballobás and Erdös [2] generalized the Randić index by replacing $-\frac{1}{2}$ with a real number $\alpha$ and named it the general Randić index, which is defined as

$$\begin{aligned} R_{\alpha }=R_{\alpha }(G)=\sum _{uv\in E_G}(d_ud_v)^{\alpha }. \end{aligned}$$

A key turning point involving the mathematical investigation of the general Randić index appeared in the second half of the 1990s, when a significant and ever growing research on this matter started, causing numerous publications (see [4, 9, 10, 14, 20] and the reference therein).

Regarding the problem of extremal trees on the Randić index and the general Randić index, researchers have tried to find precise bounds of R(G) and $R_{\alpha }(G)$ with some given graph parameters, such as the matching number [16, 21], the number of pendent vertices [5, 28], and the degree sequence [25]. Recently, the relationship between topological indices and domination number has become an interesting research topic and has attracted lots of attention [3, 6, 11, 15, 17, 19, 26]. Bermudo et al. [1] got the extremal trees of Randić index with a given domination number. Naturally, we intend to generalize their results by considering the general Randić index.

First, some terminology and notations need to be introduced. A set $D\subseteq V_G$ is called a dominating set in G if every vertex in $V_G{\setminus } D$ is adjacent to at least one vertex in D. The domination number of G, denoted by $\gamma (G)$, is defined as the minimum cardinality among all dominating sets D of G, i.e., $\gamma (G)=\min _{D\subseteq V_G}\lbrace |D| \rbrace$. A vertex $v\in V_{G}$ is called a pendent vertex if $d_v=1$, while it is called a support vertex if v is a neighbor of a pendent vertex. The set of neighbors of v is $N(v)=\lbrace u\in V_G|~uv\in E_G \rbrace$. The diameter of a tree is the longest path between two pendent vertices. Besides, $G-\lbrace v\rbrace$ denote the graph obtained from G by deleting the vertex v together with its incident edges.

In this paper, we determine the upper bound of $R_{\alpha }(G)$ with $\alpha \in [\alpha _1,0)$ of trees in terms of domination number, where $\alpha _1\approx -0.5287$ is the unique nonzero root of the equation $27^\alpha -6^\alpha -4^\alpha +2^\alpha =0$. Moreover, the lower bound of $R_{\alpha }(G)$ with $\alpha \in [\alpha _2,0)$ of trees in terms of domination number is also obtained, where $\alpha _2\approx -0.5696$ is the unique nonzero root of the equation $2\cdot 4^\alpha +2^{\alpha +1}-3^\alpha -2\cdot 6^\alpha -1=0$. Finally, we characterize the corresponding extremal trees.

2 The Upper Bound on the $R_{\alpha }(G)$ with $\alpha \in [\alpha _1,0)$ of Trees in Terms of Domination Number

To characterize extremal n-vertex trees of the Randić index with a given domination number $\gamma$, Bermudo et al. [1] constructed two graph families $\mathcal {T}_{1}$ and $\mathcal {T}_{2}$, which are described in Definitions 2.1 and 3.1, respectively. In this section, we determined the upper bound on the general Randić index $R_{\alpha }$ of trees with a given domination number $\gamma$ for $\alpha \in [\alpha _1,0)$, where $\alpha _1\approx -0.5287$ is the unique nonzero root of the equation $27^\alpha -6^\alpha -4^\alpha +2^\alpha =0$. Then, we prove that the corresponding extremal trees also belong to $\mathcal {T}_{1}$.

Definition 2.1

[1] Let $\mathcal {T}_{1}$ be a set of trees and the path with 3t vertices ($P_{3t}$) all belong to $\mathcal {T}_{1}$, where t is a positive integer number. The following are two ways to construct a new graph that belongs to the graph family $\mathcal {T}_{1}$:

• (i) If there exist a tree $T'\in \mathcal {T}_{1}$ and a pendent vertex $v\in V_{T'}$, we take a path $P_{3t}$, whose consecutive vertices are $w_1,w_2,\ldots , w_{3t}$, then the tree T, such that $V_{T}=V_{T'}\cup V_{P_{3t}}$ and $E_T=E_{T'}\cup E_{P_{3t}}\cup \lbrace vw_1\rbrace$, belongs to $\mathcal {T}_{1}$ (see Fig. 1a).

• (ii) If $T'\in \mathcal {T}_{1}$ satisfies that there exists a vertex $v\in D(T')$ such that $N(v)=\lbrace u_1,u_2\rbrace$ and $d_{u_1}=d_{u_2}=2$, where $D(T')$ is a minimum dominating set in $T'$, we take a path $P_{3t+1}$, whose consecutive vertices are $w_1,w_2,\ldots , w_{3t+1}$, then the tree T, such that $V_{T}=V_{T'}\cup V_{P_{3t+1}}$ and $E_T=E_{T'}\cup E_{P_{3t+1}}\cup \lbrace vw_1\rbrace$, belongs to $\mathcal {T}_{1}$ (see Fig. 1b).

Fig. 1

Two trees in the graph family $\mathcal {T}_{1}$

To simplify the proof of the following theorem giving upper bound on $R_{\alpha }(T)$, we first give a lemma to show the monotonicity of two functions.

Lemma 2.2

Let $\psi _1(z)=z^{\alpha }(z-1)$ and $\phi _1(z)=\psi _1(z)-\psi _1(z-1)=z^{\alpha }(z-1)-(z-1)^{\alpha }(z-2)$. If $\alpha \in (-1,0)$, the function $\psi _1(z)$ increases monotonically for $z\in (0,\infty )$ and the function $\phi _1(z)$ decreases monotonically for $z\in (1,\infty )$.

Proof

Since $\psi '_1(z)=(\alpha +1)z^{\alpha }-\alpha z^{\alpha -1}>0$ for $\alpha \in (-1,0)$ and $z>0$, we obtain that $\psi _1(z)$ is a monotonically increasing function for $z\in (0,\infty )$ if $\alpha \in (-1,0)$. Note that $\psi ''_1(z)=\alpha z^{\alpha -2}[(\alpha +1)z-(\alpha -1)]<0$ for $\alpha \in (-1,0)$ and $z>0$. Suppose that $\alpha \in (-1,0)$ and there exist $z_1$ and $z_2$ such that $z_2>z_1>1$, then by Lagrange mean value theorem, we get

$$\begin{aligned} \phi _1(z_2)-\phi _1(z_1)&=\phi '_1(\xi _1)(z_2-z_1)\\&=[\psi '_1(\xi _1)-\psi '_1(\xi _1-1)](z_2-z_1)\\&=\psi ''_1(\eta _1)(z_2-z_1)<0, \end{aligned}$$

where $1<z_1<\xi _1<z_2$ and $0<\xi _1<\eta _1<\xi _1$. Hence, if $\alpha \in (-1,0)$, $\phi _1(z)$ is a monotonically decreasing function for $z\in (1,\infty )$.

This complete the proof. $\square$

Theorem 2.3

Let T be an n-vertex tree with domination number $\gamma$. Then,

$$\begin{aligned} {R_{\alpha }(T)}&\le \frac{4n+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma )\nonumber \\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) , \quad \mathrm{for}\quad \alpha \in [\alpha _1,0), \end{aligned}$$

(2.1)

with the equality holding if and only if $T\in \mathcal {T}_{1}$, where $\alpha _1\approx -0.5287$ is the unique nonzero root of the equation $27^\alpha -6^\alpha -4^\alpha +2^\alpha =0$.

For convenience, we denote $f(n,\gamma ,\alpha )=\frac{4n+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma )\cdot (3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha })$. Next, we introduce a necessary lemma.

Lemma 2.4

Let T be an n-vertex tree with domination number $\gamma$. Suppose that there exists a vertex $v\in V_T$ such that $d_v\ge 3$, $N(v)=\lbrace u_1, u_2,\ldots ,u_{d_v} \rbrace$, $d_{u_1}\ge 2$ and $d_{u_i}=1$ for $i=2,3,\ldots ,d_v$. Take $T_{-1}=T-\lbrace u_{d_v}\rbrace$, then

• (i) If $d_{v}\ge 4$, $\alpha \in (-1,0)$ and $R_{\alpha }(T_{-1})\le f(n-1,\gamma ,\alpha )$, then $R_{\alpha }(T)<f(n,\gamma ,\alpha )$.

• (ii) If $d_v=3$, $d_{u_1}=2$, $\alpha \in (-1,0)$ and $R_{\alpha }(T_{-1})\le f(n-1,\gamma ,\alpha )$, then $R_{\alpha }(T)<f(n,\gamma ,\alpha )$

Proof

(i) By calculation, we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-1})-(d_v-2)\cdot (d_v-1)^{\alpha }-[d_{u_1}\cdot (d_v-1)]^{\alpha } +(d_v-1)\cdot (d_v)^{\alpha }\\&\quad +(d_{u_1}\cdot d_v)^{\alpha }\\&=R_{\alpha }(T_{-1})+\left[ (d_v)^{\alpha }-(d_v-1)^{\alpha }\right] \cdot [(d_v-2)+(d_{u_1})^{\alpha }]+(d_v)^{\alpha }\\&\le \frac{4(n-1)+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma -1)\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad +\left[ (d_v)^{\alpha }-(d_v-1)^{\alpha }\right] \cdot [(d_v-2)+(d_{u_1})^{\alpha }]+(d_v)^{\alpha }\\&= f(n,\gamma ,\alpha )+3\cdot 4^{\alpha }-3\cdot 6^{\alpha }-2^{\alpha } +\left[ (d_v)^{\alpha }-(d_v-1)^{\alpha }\right] \\&\quad \cdot [(d_v-2)+(d_{u_1})^{\alpha }] +(d_v)^{\alpha }\\&\le f(n,\gamma ,\alpha )+3\cdot 4^{\alpha }-3\cdot 6^{\alpha }-2^{\alpha }+(d_v)^{\alpha }(d_v-1)-(d_v-1)^{\alpha }(d_v-2). \end{aligned}$$

From Lemma 2.2, we get $3\cdot 4^{\alpha }-3\cdot 6^{\alpha }-2^{\alpha }+(d_v)^{\alpha }(d_v-1)-(d_v-1)^{\alpha }(d_v-2)=3\cdot 4^{\alpha }-3\cdot 6^{\alpha }-2^{\alpha }+\phi _1(d_v)\le 3\cdot 4^{\alpha }-3\cdot 6^{\alpha }-2^{\alpha }+\phi _1(4)=6\cdot 4^\alpha -3\cdot 6^\alpha -2\cdot 3^\alpha -2^\alpha$ for any $d_v\ge 4$.

It can be checked, using computer program, that $6\cdot 4^\alpha -3\cdot 6^\alpha -2\cdot 3^\alpha -2^\alpha <0$ for every $\alpha \in (-1,0)$ (see Fig. 2a), i.e., $R_{\alpha }(T)<f(n,\gamma ,\alpha )$.

(ii) If $d_v=3$, $d_{u_1}=2$, we have $3\cdot 4^{\alpha }-3\cdot 6^{\alpha }-2^{\alpha }+\left[ (d_v)^{\alpha }-(d_v-1)^{\alpha }\right] \cdot [(d_v-2)+(d_{u_1})^{\alpha }]+(d_v)^{\alpha }=3\cdot 4^{\alpha }-3\cdot 6^{\alpha }-2^{\alpha }+(3^{\alpha }-2^{\alpha })\cdot (1+2^{\alpha })+3^{\alpha }=2\cdot (2^\alpha -1)\cdot (2^\alpha -3^\alpha )$. It can be checked that $2^\alpha -1<0$ and $2^\alpha -3^\alpha >0$ for $\alpha \in (-1,0)$, which implies that $2(2^\alpha -1)(2^\alpha -3^\alpha )<0$ for every $\alpha \in (-1,0)$ (see Fig. 2b). Thus, we get $R_{\alpha }(T)<f(n,\gamma ,\alpha )$.

Fig. 2

The two functions of $\alpha$ in Lemma 2.4

This proves the lemma. $\square$

Proof of Theorem 2.3

For any tree with 4 vertices ($S_4$ or $P_4$), we have $R_{\alpha }(S_4)=3\cdot 3^{\alpha }<f(4,1,\alpha )$, and $R_{\alpha }(P_4)=4^{\alpha }+2\cdot 2^{\alpha }<f(4,2,\alpha )$ if $\alpha \in (-1,0)$. Suppose that the inequality (2.1) holds for any $(n-1)$-vertex trees and then we consider a tree with n vertices. We take a diameter path of T, denoted by $v_1v_2\cdots v_d$. By Lemma 2.4, we can suppose that $d_{v_2}\le 3$. Then, we discuss the following two cases.

Case 1 $d_{v_2}=3$.

From Lemma 2.4, one can assume that $d_{v_3}\ge 3$. Denote $N(v_2)=\lbrace v_1,v_3,u_1 \rbrace$ and $N(v_3)=\lbrace v_2,v_4,w_1,w_2,\ldots ,w_{d_{v_3}-2} \rbrace$. Then, we take $T_{-3}=T-\lbrace v_1,v_2,u_1 \rbrace$. There exists a dominating set D(T) such that $v_2\in D(T)$ and $v_3\in N[D{\setminus }\lbrace v_2 \rbrace ]$, implying $\gamma (T)=\gamma (T_{-3})+1$. Note that $d_{w_i}\le 3$ for $i=1,2,\ldots ,d_{v_3}-2$. By the above consideration, we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-3})-[d_{v_4}\cdot (d_{v_3}-1)]^{\alpha }-(d_{v_3}-1)^{\alpha }\cdot \sum _{i=1}^{d_{v_3}-2}(d_{w_i})^{\alpha }+(d_{v_3}\cdot d_{v_4})^{\alpha }\\&\quad +(d_{v_3})^{\alpha }\cdot \sum _{i=1}^{d_{v_3}-2}(d_{w_i})^{\alpha }+(3\cdot d_{v_3})^{\alpha }+2\cdot 3^{\alpha }\\&\le \frac{4(n-3)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-3-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [(d_{v_4})^{\alpha }+\sum _{i=1}^{d_{v_3}-2}(d_{w_i})^{\alpha }]+[(d_{v_3})^{\alpha }+2]\cdot 3^{\alpha }\\&\le f(n,\gamma ,\alpha )-[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [(d_{v_3}-2)\cdot 3^{\alpha }]+[(d_{v_3})^{\alpha }+2]\\&\quad \cdot 3^{\alpha }-3\cdot 4^{\alpha }\\&=f(n,\gamma ,\alpha )+[(d_{v_3}-1)\cdot (d_{v_3})^{\alpha }-(d_{v_3}-2)\cdot (d_{v_3}-1)^{\alpha }+2]\\&\quad \cdot 3^{\alpha }-3\cdot 4^{\alpha }. \end{aligned}$$

By Lemma 2.2, we calculate that $[(d_{v_3}-1)\cdot (d_{v_3})^{\alpha }-(d_{v_3}-2)\cdot (d_{v_3}-1)^{\alpha }+2]\cdot 3^{\alpha }-3\cdot 4^{\alpha }=\left[ \phi _1(d_{v_3})+2\right] \cdot 3^{\alpha }-3\cdot 4^{\alpha } \le \left[ \phi _1(3)+2\right] \cdot 3^{\alpha }-3\cdot 4^{\alpha }=2\cdot 9^\alpha -6^\alpha +2\cdot 3^\alpha -3\cdot 4^\alpha <0$ for any $d_{v_3}\ge 3$ and $\alpha \in (-1,0)$ (see Fig. 3a). Thus, for $\alpha \in (-1,0)$, $R_{\alpha }(T)<f(n,\gamma ,\alpha )$.

Case 2 For any diameter path of T, $d_{v_2}=2$.

Case 2.1 $d_{v_3}\ge 3$. Denote $N(v_3)=\lbrace v_2,v_4,w_1,w_2,\ldots ,w_{d_{v_3}-2} \rbrace$. Then, we take $T_{-2}=T-\lbrace v_1,v_2\rbrace$. Obviously, there exists a dominating set D(T) such that $v_2\in D(T)$ and $v_3\in N[D{\setminus }\lbrace v_2 \rbrace ]$, implying $\gamma (T)=\gamma (T_{-2})+1$. By calculation, we get

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-2})-[d_{v_4}\cdot (d_{v_3}-1)]^{\alpha }-(d_{v_3}-1)^{\alpha }\cdot \sum _{i=1}^{d_{v_3}-2}(d_{w_i})^{\alpha }+(d_{v_4}\cdot d_{v_3})^{\alpha }\\&\quad +(d_{v_3})^{\alpha }\cdot \sum _{i=1}^{d_{v_3}-2}(d_{w_i})^{\alpha }+(2\cdot d_{v_3})^{\alpha }+2^{\alpha }\\&\le \frac{4(n-2)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-2-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot \left[ (d_{v_4})^{\alpha }+\sum _{i=1}^{d_{v_3}-2}(d_{w_i})^{\alpha }\right] +[(d_{v_3})^{\alpha }+1]\cdot 2^{\alpha }\\&\le f(n,\gamma ,\alpha )-[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [(d_{v_3}-2)\cdot 2^{\alpha }]+[(d_{v_3})^{\alpha }+1]\cdot 2^{\alpha }\\&\quad -6\cdot 4^{\alpha }+3\cdot 6^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )+[(d_{v_3}-1)(d_{v_3})^{\alpha }-(d_{v_3}-2)(d_{v_3}-1)^{\alpha }]\cdot 2^{\alpha }\\&\quad -6\cdot 4^{\alpha }+3\cdot 6^{\alpha }+2\cdot 2^{\alpha } \end{aligned}$$

Combining Lemma 2.2, we calculate that $[(d_{v_3}-1)(d_{v_3})^{\alpha }-(d_{v_3}-2)(d_{v_3}-1)^{\alpha }]\cdot 2^{\alpha }-6\cdot 4^{\alpha }+3\cdot 6^{\alpha }+2^{\alpha +1}=\phi _1(d_{v_3})\cdot 2^\alpha -6\cdot 4^{\alpha }+3\cdot 6^{\alpha }+2^{\alpha +1}\le \phi _1(3)\cdot 2^\alpha -6\cdot 4^{\alpha }+3\cdot 6^{\alpha }+2^{\alpha +1}= 5\cdot 6^\alpha -7\cdot 4^\alpha +2^{\alpha +1}<0$ for $d_{v_3}\ge 3$ and $\alpha \in [-0.68,0)$ (see Fig. 3b). Hence, for $\alpha \in [-0.68,0)$, $R_{\alpha }(T)<f(n,\gamma ,\alpha )$.

Fig. 3

The two functions of $\alpha$ in Case 1 and Case 2.1 of Theorem 2.3

Case 2.2 $d_{v_3}=2$. Denote $N(v_4)=\lbrace v_3,v_5,x_1,x_2,\ldots ,x_{d_{v_4}-2}\rbrace$. For any $x_i$ ($i=1,2,\ldots ,d_{v_4}-2$), if there exist two vertices $y_1$, $y_2$ such that $y_2\in N(x_i)$ and $y_1\in N(y_2)$, then we get a new diameter path of T, that is $y_1,y_2,x_i,v_4,v_5,\ldots ,v_d$, then by the above cases, we can assume that $d_{x_i}=d_{y_2}=2$. Moreover, if $x_i$ is a support vertex, from Lemma 2.4, we can suppose that $d_{x_i}=2$ or $d_{x_i}=3$.

Case 2.2.1 $d_{v_4}\ge 4$. Take $T_{-3}=T-\lbrace v_1,v_2,v_3\rbrace$, then we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-3})-[(d_{v_4}-1)\cdot d_{v_5}]^{\alpha }-(d_{v_4}-1)^{\alpha }\cdot \sum _{i=1}^{d_{v_4}-2}(d_{x_i})^{\alpha }\\&\quad +(d_{v_4}\cdot d_{v_5})^{\alpha }+(d_{v_4})^{\alpha }\cdot \sum _{i=1}^{d_{v_4}-2}(d_{x_i})^{\alpha }+(2\cdot d_{v_4})^{\alpha }+4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-3)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-3-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -[(d_{v_4}-1)^{\alpha }-(d_{v_4})^{\alpha }]\cdot \left[ (d_{v_5})^{\alpha }+\sum _{i=1}^{d_{v_4}-2}(d_{x_i})^{\alpha }\right] +(2\cdot d_{v_4})^{\alpha }+4^{\alpha }+2^{\alpha }\\&\le f(n,\gamma ,\alpha )-[(d_{v_4}-1)^{\alpha }-(d_{v_4})^{\alpha }]\cdot [(d_{v_4}-2)\cdot 3^{\alpha }]\\&\quad +(2\cdot d_{v_4})^{\alpha }-2\cdot 4^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )+[(d_{v_4}-1)(d_{v_4})^{\alpha }-(d_{v_4}-2)(d_{v_4}-1)^{\alpha }]\cdot 3^{\alpha }\\&\quad +(2^\alpha -3^\alpha )(d_{v_4})^{\alpha }-2\cdot 4^{\alpha }+2^{\alpha }. \end{aligned}$$

If $\alpha <0$, the function $(2^\alpha -3^\alpha )(d_{v_4})^\alpha$ monotonically decreases for $d_{v_4}\ge 0$. Combining Lemma 2.2, we calculate that $[(d_{v_4}-1)(d_{v_4})^{\alpha }-(d_{v_4}-2)(d_{v_4}-1)^{\alpha }]\cdot 3^{\alpha }+(2^\alpha -3^\alpha )(d_{v_4})^{\alpha }-2\cdot 4^{\alpha }+2^{\alpha }=\phi _1(d_{v_4})\cdot 3^\alpha +(2^\alpha -3^\alpha )(d_{v_4})^{\alpha }-2\cdot 4^{\alpha }+2^{\alpha }\le \phi _1(4)\cdot 3^\alpha +(2^\alpha -3^\alpha )\cdot 4^{\alpha }-2\cdot 4^{\alpha }+2^{\alpha }= 2\cdot 12^\alpha -2\cdot 9^\alpha +8^\alpha -2\cdot 4^\alpha +2^\alpha <0$ for $d_{v_4}\ge 4$ and $\alpha \in [-0.64,0)$ (see Fig. 4a). Hence, for $\alpha \in [-0.64,0)$, $R_{\alpha }(T)<f(n,\gamma ,\alpha )$.

Case 2.2.2 $d_{v_4}\le 3$. If there exists a minimum dominating set D(T) such that $v_4\in D(T)$, then we take $T_{-2}=T-\lbrace v_1,v_2 \rbrace$. By calculation, we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-2})-(d_{v_4})^{\alpha } +(2\cdot d_{v_4})^{\alpha }+4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-2)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha } +[n-2-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) -(d_{v_4})^{\alpha }+(2\cdot d_{v_4})^{\alpha }+4^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )-(d_{v_4})^{\alpha }+(2\cdot d_{v_4})^{\alpha }-5\cdot 4^{\alpha }+2\cdot 2^{\alpha }+3\cdot 6^{\alpha }. \end{aligned}$$

Note that $-(d_{v_4})^{\alpha }+(2\cdot d_{v_4})^{\alpha }-5\cdot 4^{\alpha }+2\cdot 2^{\alpha }+3\cdot 6^{\alpha }\le -5\cdot 4^{\alpha }+2\cdot 2^{\alpha }+4\cdot 6^{\alpha }-3^\alpha <0$ for any $d_{v_4}\le 3$ and $\alpha \in [-0.6395,0)$ (see Fig. 4b), then we have $R_{\alpha }(T)<f(n,\gamma ,\alpha )$ for $\alpha \in [-0.6395,0)$.

Fig. 4

The two functions of $\alpha$ in Case 2.2.1 and Case 2.2.2 of Theorem 2.3

Next, we suppose that $v_4\notin D(T)$ for any minimum dominating set D in T.

Case 2.2.2.1 $d_{v_4}=3$. Denote $N(v_5)=\lbrace v_4,a_1,\ldots ,a_{d_{v_5}-1} \rbrace$, then we study the following three cases, which are depicted in Fig. 5.

Fig. 5

Three trees for Case 2.2.2.1

(I) If $d_{v_5}\ge 9$, we take $T_{-7}=T-\lbrace v_1,v_2,v_3,v_4,x_1,y_1,y_2 \rbrace$. One can check that $\gamma (T_{-7})=\gamma (T)-2$ and

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-7})-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }] \cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }\\&\quad +2\cdot 6^{\alpha }+2\cdot 4^{\alpha }+2\cdot 2^{\alpha }\\&\le \frac{4(n-7)+3(\gamma -2)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-7-3(\gamma -2)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+2\cdot 4^{\alpha }+2\cdot 2^{\alpha }\\&= f(n,\gamma ,\alpha )\!-\![(d_{v_5}-1)^{\alpha }\!-\!(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }-6^{\alpha }-4^{\alpha }\!+\!2^{\alpha }. \end{aligned}$$

Easily, it can be checked that $(3\cdot d_{v_5})^{\alpha }-6^{\alpha }-4^{\alpha }+2^{\alpha }\le 27^\alpha -6^\alpha -4^\alpha +2^\alpha <0$ for any $d_{v_5}\ge 9$ and $\alpha \in [-0.5287,0)$ (see Fig. 6a). Hence, for $\alpha \in [-0.5287,0)$, $R_{\alpha }(T)<f(n,\gamma ,\alpha )$. For $d_{v_5}\le 8$, we take $T_{-3}=T-\lbrace v_1,v_2,v_3 \rbrace$. One can see $\gamma (T_{-3})=\gamma (T)-1$ and

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }({T_{-3}})-4^{\alpha }-(2\cdot d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+(3\cdot d_{v_5})^{\alpha }+4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-3)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-3-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -4^{\alpha }-(2\cdot d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+(3\cdot d_{v_5})^{\alpha }+4^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )-3\cdot 4^{\alpha }+(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+2^{\alpha }. \end{aligned}$$

Since the inequality $-3\cdot 4^{\alpha }+(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+2^{\alpha }\le 24^\alpha -16^\alpha -3\cdot 4^\alpha +2\cdot 6^\alpha +2^\alpha <0$ holds for any $d_{v_5}\le 8$ and $\alpha \in [-0.6241,0)$ (see Fig. 6b), we have $R_{\alpha }(T)<f(n,\gamma ,\alpha )$ for $\alpha \in [-0.6241,0)$.

Fig. 6

The two functions of $\alpha$ in Case 2.2.2.1 (I) of Theorem 2.3

(II) Assume that $d_{v_5}\ge 4$, then we take $T_{-7}=T-\lbrace v_1,v_2,v_3,v_4,x_1,y_1,y_2 \rbrace$. Similarly, we get

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-7})-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }+9^{\alpha }\\&\quad +2\cdot 3^{\alpha }+6^{\alpha }+4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-7)+3(\gamma -2)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-7-3(\gamma -2)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) -[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha \\&\quad +(3\cdot d_{v_5})^{\alpha }+9^{\alpha }+2\cdot 3^{\alpha }+6^{\alpha }+4^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1} (d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }+9^{\alpha }\\&\quad +2\cdot 3^{\alpha }-2\cdot 6^{\alpha }-2\cdot 4^{\alpha }. \end{aligned}$$

It can be compute that $(3\cdot d_{v_5})^{\alpha }+9^{\alpha }+2\cdot 3^{\alpha }-2\cdot 6^{\alpha }-2\cdot 4^{\alpha }\le 12^\alpha +9^\alpha +2\cdot 3^\alpha -2\cdot 6^\alpha -2\cdot 4^\alpha$ is always negative for $d_{v_5}\ge 4$ and $\alpha \in [-0.7773,0)$ (see Fig. 7a). Thus, for $\alpha \in [-0.7773,0)$, $R_{\alpha }(T)<f(n,\gamma ,\alpha )$. If $d_{v_5}\le 3$, we take $T_{-3}=T-\lbrace v_1,v_2,v_3\rbrace$. Then,

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-3})-(2\cdot d_{v_5})^{\alpha }-6^{\alpha }+(3\cdot d_{v_5})^{\alpha }+9^{\alpha }+6^{\alpha }+4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-3)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-3-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad +(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+9^{\alpha }+4^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )-2\cdot 4^{\alpha }+(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+9^{\alpha }+2^{\alpha }. \end{aligned}$$

Since $-2\cdot 4^{\alpha }+(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+9^{\alpha }+2^{\alpha }\le 2\cdot 9^\alpha -6^\alpha -2\cdot 4^\alpha +2^\alpha <0$ for any $d_{v_5}\le 3$ and $\alpha \in [-0.6894,0)$ (see Fig. 7b), we have for $\alpha \in [-0.6894,0)$, $R_{\alpha }(T)<f(n,\gamma ,\alpha )$.

Fig. 7

The two functions of $\alpha$ in Case 2.2.2.1 (II) of Theorem 2.3

(III) In this case, we assume that $d_{v_5}\ge 5$. Let $T_{-6}=T-\lbrace v_1,v_2,v_3,v_4,x_1,y_1\rbrace$, then we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-6})-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }] \cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }\\&\quad +2\cdot 6^{\alpha }+4^{\alpha }+2\cdot 2^{\alpha }\\&\le \frac{4(n-6)+3(\gamma -2)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-6-3(\gamma -2)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+4^{\alpha }+2\cdot 2^{\alpha }\\&=f(n,\gamma ,\alpha )-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(3\cdot d_{v_5})^{\alpha }\\&\quad +2\cdot 6^{\alpha }-5\cdot 4^{\alpha }+2\cdot 2^{\alpha }. \end{aligned}$$

Since the inequality $(3\cdot d_{v_5})^{\alpha }+2\cdot 6^{\alpha }-5\cdot 4^{\alpha }+2\cdot 2^{\alpha }\le 15^\alpha +2\cdot 6^{\alpha }-5\cdot 4^{\alpha }+2\cdot 2^{\alpha }<0$ holds for any $d_{v_5}\ge 5$ and $\alpha \in [-0.5336,0)$ (see Fig. 8a), we have $R_{\alpha }(T)<f(n,\gamma ,\alpha )$ for $\alpha \in [-0.5336,0)$. For $d_{v_5}\le 4$, we take $T_{-3}=T-\lbrace v_1,v_2,v_3 \rbrace$. Then,

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-3})-(2\cdot d_{v_5})^{\alpha }-4^{\alpha }+(3\cdot d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-3)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-3-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad +(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )+(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+2^{\alpha }-3\cdot 4^{\alpha }. \end{aligned}$$

Since $(3^{\alpha }-2^{\alpha })\cdot (d_{v_5})^{\alpha }+2\cdot 6^{\alpha }+2^{\alpha }-3\cdot 4^{\alpha }\le 12^\alpha -8^\alpha +2\cdot 6^{\alpha }+2^{\alpha }-3\cdot 4^{\alpha }$ is negative for $d_{v_5}\le 4$ and $\alpha \in [-0.747,0)$ (see Fig. 8b), we get $R_{\alpha }(T)<f(n,\gamma ,\alpha )$ for $\alpha \in [-0.747,0)$.

Fig. 8

The two functions of $\alpha$ in Case 2.2.2.1 (III) of Theorem 2.3

Case 2.2.2.2 $d_{v_4}=2$. The corresponding tree is depicted in Fig. 9. We denote $N(v_5)=\lbrace v_4,a_1,a_2,\ldots ,a_{d_{v_5}-1}\rbrace$ and consider the following two cases:

Fig. 9

The tree for Case 2.2.2.2

Case 2.2.2.2.1 $d_{v_5}=2$. Let $T_{-3}=T-\lbrace v_1,v_2,v_3 \rbrace$, then we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-3})-2^{\alpha }+3\cdot 4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-3)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-3-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) +3\cdot 4^{\alpha }\\&=\frac{4n+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma )\cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&=f(n,\gamma ,\alpha ). \end{aligned}$$

Case 2.2.2.2.2 $d_{v_5}\ge 3$. If $d_{a_i}\le 2$ with $i=1,2,\ldots ,d_{v_5}-1$, we take $T_{-4}=T-\lbrace v_1,v_2,v_3,v_4\rbrace$, then

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-4})-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \sum _{i=1}^{d_{v_5}-1}(d_{a_i})^\alpha +(2\cdot d_{v_5})^{\alpha }+2\cdot 4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-4)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n-4-3(\gamma -1)]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) -[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot \left( d_{v_5} -1\right) \\&\quad \cdot 2^{\alpha } +2\cdot 4^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )+{[(d_{v_5})^{\alpha +1}-(d_{v_5}-1)^{\alpha +1}]\cdot 2^\alpha +2\cdot 4^{\alpha }-3\cdot 6^{\alpha }}. \end{aligned}$$

One can check that ${[(d_{v_5})^{\alpha +1}-(d_{v_5}-1)^{\alpha +1}]\cdot 2^\alpha +2\cdot 4^{\alpha }-3\cdot 6^{\alpha }}\le \left( 3^{\alpha +1}-2^{\alpha +1}\right) \cdot 2^\alpha +2\cdot 4^{\alpha }-3\cdot 6^{\alpha }=0$ for any $d_{v_5}\ge 3$. Hence, $R_{\alpha }(T)\le f(n,\gamma ,\alpha )$.

Based on the above discussion, we suppose that $d_{a_1}=\max _{1\le i\le d_{v_5}-1}d_{a_i}\ge 3$. Denote $N(a_1)=\lbrace v_5,b_1,b_2,\ldots ,b_{d_{a_1}-1}\rbrace$, then by all above cases and Lemma 2.4, one can see that $1\le d_{b_j}\le 3$ and all vertex in ${N(b_j){\setminus } \lbrace a_1\rbrace }$ are pendent vertices. For convenience, the number of vertices in $\lbrace b_1,b_2,\ldots ,b_{d_{a_1}-1} \rbrace$ with degree i is written by $n_i$; then, we obtain $n_1+n_2+n_3=d_{a_1}-1$. Next, we construct two subgraphs $T'$, $T''$ of T such that $V_{T}=V_{T'}\cup V_{T''}$ and $E_{T}=E_{T'}\cup E_{T''}\cup \lbrace v_5a_1\rbrace$. By calculation, we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T')+R_{\alpha }(T'')-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\\&\quad \cdot \left[ 2^{\alpha }+\sum _{i=2}^{d_{v_5}-1}(d_{a_i})^\alpha \right] -n_1\cdot (d_{a_1}-1)^{\alpha }-n_2\cdot [2(d_{a_1}-1)]^{\alpha }\\&\quad -n_3\cdot [3(d_{a_1}-1)]^{\alpha }+n_1\cdot (d_{a_1})^{\alpha } +n_2\cdot (2\cdot d_{a_1})^{\alpha }+n_3\cdot (3\cdot d_{a_1})^{\alpha }\\&\quad +(d_{a_1}\cdot d_{v_5})^{\alpha }\\&\le \frac{4n+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma ) \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad +2\cdot 2^{\alpha }-3\cdot 4^{\alpha }+(d_{a_1}\cdot d_{v_5})^{\alpha }\\&\quad -[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot [2^{\alpha }+(d_{v_5}-2) \cdot (d_{a_1})^{\alpha }]\\&\quad -[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }] \cdot (n_1+n_2+n_3)\cdot 3^{\alpha }\\&\le f(n,\gamma ,\alpha )-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot [2^{\alpha } +(d_{a_1})^{\alpha }]\\&\quad -[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha }\\&\quad +2\cdot 2^{\alpha }-3\cdot 4^{\alpha }+(d_{a_1}\cdot d_{v_5})^{\alpha } \end{aligned}$$

Now, we consider the function $g(z)=-[(z-1)^{\alpha }-z^{\alpha }]\cdot [2^{\alpha }+(d_{a_1})^{\alpha }]+(d_{a_1}\cdot z)^{\alpha }$. Since

$$\begin{aligned} g'(z)&=\alpha \left\{ \left[ z^{\alpha -1}-(z-1)^{\alpha -1} \right] \left[ 2^{\alpha }+(d_{a_1})^{\alpha } \right] + (d_{a_1})^{\alpha }\cdot z^{\alpha -1} \right\} \ge 0 \\&\Leftrightarrow \left[ z^{\alpha -1}-(z-1)^{\alpha -1} \right] \left[ 2^{\alpha }+(d_{a_1})^{\alpha } \right] + (d_{a_1})^{\alpha }\cdot z^{\alpha -1}\le 0\\&\Leftrightarrow z^{\alpha -1}\left[ 2^{\alpha }+2\cdot (d_{a_1})^{\alpha } \right] \le {(z-1)^{\alpha -1}\left[ 2^{\alpha }+(d_{a_1})^{\alpha } \right] }\\&\Leftrightarrow \left( \frac{z}{z-1} \right) ^{\alpha -1}\le \frac{2^{\alpha }+(d_{a_1})^{\alpha }}{2^{\alpha }+2\cdot (d_{a_1})^{\alpha }} \\&\Leftrightarrow \left( \frac{z}{z-1} \right) ^{1-\alpha }\ge \frac{2^{\alpha }+2\cdot (d_{a_1})^{\alpha }}{2^{\alpha }+(d_{a_1})^{\alpha }} \\&\Leftrightarrow \frac{z}{z-1}\ge \left[ \frac{2^{\alpha }+2\cdot (d_{a_1})^{\alpha }}{2^{\alpha }+(d_{a_1})^{\alpha }} \right] ^{\frac{1}{1-\alpha }}\\&\Leftrightarrow z\le \frac{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}}{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1}, \end{aligned}$$

we have that g(z) attains its maximum value if $z=\frac{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}}{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1}$. Hence,

$$\begin{aligned}&-[(d_{v_5}-1)^{\alpha }-(d_{v_5})^{\alpha }]\cdot [2^{\alpha }+(d_{a_1})^{\alpha }]-[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha }+2\cdot 2^{\alpha }-3\cdot 4^{\alpha }\\&\qquad +(d_{a_1}\cdot d_{v_5})^{\alpha }\\&\quad \le 2\cdot 2^{\alpha }-3\cdot 4^{\alpha }-\left( \left\{ \frac{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}}{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1}-1\right\} ^{\alpha }-\left\{ \frac{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}}{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1}\right\} ^{\alpha } \right) \\&\qquad \cdot [2^{\alpha }+(d_{a_1})^{\alpha }]\\&\qquad -[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha }+(d_{a_1})^{\alpha }\cdot \left\{ \frac{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}}{\left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1}\right\} ^{\alpha }\\&\quad =-[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha }+2\cdot 2^{\alpha }-3\cdot 4^{\alpha }\\&\qquad -\frac{1}{\left( \left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1\right) ^\alpha } \left\{ [2^{\alpha }+(d_{a_1})^{\alpha }]-\left[ 2^\alpha +2\cdot (d_{a_1})^{\alpha }\right] \cdot \left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{\alpha }{1-\alpha }}\right\} \\&\quad =-[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha } +2\cdot 2^{\alpha }-3\cdot 4^{\alpha }\\&\qquad +\frac{2^{\alpha }+(d_{a_1})^{\alpha }}{\left( \left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1\right) ^\alpha }\left( \left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1\right) \\&\quad =-[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha } +2\cdot 2^{\alpha }-3\cdot 4^{\alpha }\\&\qquad +[2^{\alpha }+(d_{a_1})^{\alpha }]\left( \left[ \frac{2\cdot (d_{a_1})^{\alpha }+2^{\alpha }}{(d_{a_1})^{\alpha }+2^{\alpha }}\right] ^{\frac{1}{1-\alpha }}-1\right) ^{1-\alpha }\\&\quad = -[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha }+2\cdot 2^{\alpha }-3\cdot 4^{\alpha }\\&\qquad +[2^{\alpha }+(d_{a_1})^{\alpha }]\left( \left[ 1+\frac{2^{-\alpha }}{(d_{a_1})^{-\alpha }+2^{-\alpha }}\right] ^{\frac{1}{1-\alpha }}-1 \right) ^{1-\alpha }. \end{aligned}$$

Note that the function $\varphi (z)=2\cdot 2^{\alpha }-3\cdot 4^{\alpha }+(2^{\alpha }+z^\alpha )\left( \left[ 1+\frac{2^{-\alpha }}{z^{-\alpha }+2^{-\alpha }} \right] ^{\frac{1}{1-\alpha }}-1 \right) ^{1-\alpha }$ always decreases for $z>0$ if $\alpha <0$. By calculation, we obtain $\varphi (16)<0$ for $\alpha \in [-0.5287,0)$ (see Fig. 10a), which implies that the function $\omega (d_{a_1})=-[(d_{a_1}-1)^{\alpha }-(d_{a_1})^{\alpha }]\cdot (d_{a_1}-1)\cdot 3^{\alpha }+2\cdot 2^{\alpha }-3\cdot 4^{\alpha }+[2^{\alpha }+(d_{a_1})^{\alpha }]\left( \left[ 1+\frac{2^{-\alpha }}{(d_{a_1})^{-\alpha }+2^{-\alpha }}\right] ^{\frac{1}{1-\alpha }}-1 \right) ^{1-\alpha }$ is always negative for any $d_{a_1}\ge 16$ and $\alpha \in [-0.5287,0)$. Therefore, we only check the value of $\omega (d_{a_1})$ for $3\le d_{a_1}\le 15$ (see Fig. 10b, c). So, we have $R_{\alpha }(T)<f(n,\gamma ,\alpha )$ for $\alpha \in [-0.5287,0)$.

Fig. 10

Several functions of $\alpha$ in Case 2.2.2.2.2 of Theorem 2.3

Next, we will prove that $R_{\alpha }(T)=f(n(T),\gamma (T),\alpha )$ if and only if $T\in \mathcal {T}_{1}.$ For any path $P_{3t}$, we have $R_{\alpha }(P_{3t})=f(3k,k,\alpha )$. If $T'\in \mathcal {T}_{1}$ such that $R_{\alpha }(T')=f(n(T'),\gamma (T'),\alpha )$, we take a pendent vertex $v\in T'$ and a path $P_{3t}=w_1w_2\cdots w_{3t}$, and consider a new tree T such that $V_{T}=V_{T'}\cup V_{P_{3t}}$ and $E_T=E_{T'}\cup E_{P_{3t}}\cup \lbrace vw_1\rbrace$ (Definition 2.1 (i)). By calculation, we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T')+R_{\alpha }(P_{3t})-2\cdot 2^{\alpha }+3\cdot 4^{\alpha }\\&=\frac{4n(T')+3\gamma (T')-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n(T')-3\gamma (T')]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad +2\cdot 2^{\alpha }+(3t-3)\cdot 4^{\alpha }-2\cdot 2^{\alpha }+3\cdot 4^{\alpha }\\&=\frac{4n+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n(T)-3\gamma (T))\cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&=f(n(T),\gamma (T),\alpha ). \end{aligned}$$

Suppose that there exists a tree $T'\in \mathcal {T}_{1}$ such that $R_{\alpha }(T')=f(n(T'),\gamma (T'),\alpha )$ and a vertex $v\in V_{T'}$ satisfies that $N(v)=\lbrace u_1,u_2\rbrace$, $d_{u_1}=d_{u_2}=2$, and $v\in D(T')$, where D is a minimum dominating set in $T'$. Then, we construct a new tree $T\in \mathcal {T}_{1}$. Let $P_{3t+1}=w_1w_2\cdots w_{3t+1}$, $V_T=V_{T'}\cup V_{P_{3t+1}}$ and $E_T=E_{T'}\cup E_{P_{3t+1}}\cup \lbrace vw_1\rbrace$ (Definition 2.1 (ii)). Hence,

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T')+R_{\alpha }(P_{3t+1})-2\cdot 4^{\alpha }-2^{\alpha }+3\cdot 6^{\alpha }+4^{\alpha }\\&=\frac{4n(T')+3\gamma (T')-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+[n(T')-3\gamma (T')]\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -2\cdot 2^{\alpha }+(3t-2)\cdot 4^{\alpha }-2\cdot 4^{\alpha }-2^{\alpha }+3\cdot 6^{\alpha }+4^{\alpha }\\&=\frac{4n+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n(T)-3\gamma (T))\cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&=f(n(T),\gamma (T),\alpha ). \end{aligned}$$

Finally, we need to prove that any tree T will belong to the family $\mathcal {T}_{1}$ if $R_{\alpha }(T)=f(n(T),\gamma (T),\alpha )$. Suppose that there exists a tree T such that $R_{\alpha }(T)=f(n(T),\gamma (T),\alpha )$, and $T\notin \mathcal {T}_{1}$. Let T be the tree with the minimum number of vertices, which satisfies the above assumptions. If $v_1v_2\cdots v_d$ is a diameter path of T, by above proofs, we suppose that $d_{v_2}=d_{v_3}=d_{v_4}=2$, $2\le d_{v_5}\le 3$, $d_{a_1}\le 2$ and $d_{a_2}\le 2$. Then, we investigate the following cases.

If $d_{v_5}=3$ and $d_{a_2}\le 2$, we let $T_{-4}=T-\lbrace v_1,v_2,v_3,v_4 \rbrace$. Then,

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-4})-(2^{\alpha }-3^{\alpha })\cdot [(d_{a_1})^{\alpha }+(d_{a_2})^{\alpha }]+6^{\alpha }+2\cdot 4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-4)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma -1)\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&\quad -(2^{\alpha }-3^{\alpha })\cdot [(d_{a_1})^{\alpha }+(d_{a_2})^{\alpha }]+6^{\alpha }+2\cdot 4^{\alpha }+2^{\alpha }\\&=f(n,\gamma ,\alpha )-(2^{\alpha }-3^{\alpha })\cdot [(d_{a_1})^{\alpha }+(d_{a_2})^{\alpha }]-2\cdot 6^{\alpha }+2\cdot 4^{\alpha }. \end{aligned}$$

If $d_{a_1}=1$ or $d_{a_2}=1$, then we get $-(2^{\alpha }-3^{\alpha })\cdot [(d_{a_1})^{\alpha }+(d_{a_2})^{\alpha }]-2\cdot 6^{\alpha }+2\cdot 4^{\alpha }<0$ for $\alpha <0$, a contradiction. If $d_{a_1}=d_{a_2}=2$, then $-(2^{\alpha }-3^{\alpha })\cdot [(d_{a_1})^{\alpha }+(d_{a_2})^{\alpha }]-2\cdot 6^{\alpha }+2\cdot 4^{\alpha }=0$ and $R_{\alpha }(T_{-4})=f(n(T_{-4}),\gamma (T_{-4}),\alpha )$. If $T_{-4}\in \mathcal {T}_{1}$, based on the definition of $T_{-4}$, one can check that $T\in \mathcal {T}_{1}$, a contradiction. If $T_{-4}\notin \mathcal {T}_{1}$, we have $|V_{T_{-4}}|<|V_{T}|$, a contradiction.

If $d_{v_5}=2$, we take $T_{-3}=T-\lbrace v_1,v_2,v_3 \rbrace$. Then, we get

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-3})-2^{\alpha }+3\cdot 4^{\alpha }+2^{\alpha }\\&\le \frac{4(n-3)+3(\gamma -1)-15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma )\\&\quad \cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) +3\cdot 4^{\alpha }\\&=\frac{4n+3\gamma -15}{5}\cdot 4^{\alpha }+2\cdot 2^{\alpha }+(n-3\gamma )\cdot \left( 3\cdot 6^{\alpha }+2^{\alpha }-\frac{19}{5}\cdot 4^{\alpha }\right) \\&=f(n,\gamma ,\alpha ). \end{aligned}$$

Note that $R_{\alpha }(T_{-3})=f(n(T_{-3}),\gamma (T_{-3}),\alpha )$. Analogously, we can assume that $T_{-3}\in \mathcal {T}_{1}$. Since $v_4$ is a pendent vertex in $T_{-3}$, we can conclude that $T\in \mathcal {T}_{1}$. If $T_{-3}\notin \mathcal {T}_{1}$, we obtain a contradiction with minimality of T.

This completes the proof. $\square$

3 The Lower Bound on the $R_{\alpha }$ with $\alpha \in [\alpha _2,0)$ of Trees in Terms of Domination Number

In this section, for $\alpha \in [\alpha _2,0)$, the lower bound on the general Randić index $R_{\alpha }$ of trees using not only the order, but also the order and the domination number, is obtained, where $\alpha _2\approx -0.5696$ is the unique nonzero root of the equation $2\cdot 4^\alpha +2^{\alpha +1}-3^\alpha -2\cdot 6^\alpha -1=0$. Moreover, we prove the corresponding extremal trees belong to $\mathcal {T}_{2}$.

Definition 3.1

[1] Let $\mathcal {T}_{2}$ be a set of trees with order n and domination number $\gamma$, which are obtained from the star $S_{n-\gamma +1}$ by attaching a pendent edge to its $\gamma -1$ pendent vertices (see Fig. 11).

Theorem 3.2

Let T be an n-vertex tree with a domination number $\gamma$. Then,

$$\begin{aligned} R_{\alpha }(T)\ge (n-2\gamma +1)(n-\gamma )^{\alpha }+2^{\alpha }(\gamma -1)[1+(n-\gamma )^{\alpha }],\quad \mathrm{for}\quad \alpha \in [\alpha _2,0), \end{aligned}$$

(3.1)

with the equality holding if and only if $T\in \mathcal {T}_{2}$, where $\alpha _2\approx -0.5696$ is the unique nonzero root of the equation $2\cdot 4^\alpha +2^{\alpha +1}-3^\alpha -2\cdot 6^\alpha -1=0$.

Fig. 11

Several extremal trees in graph family $\mathcal {T}_2$

Proof

To simplify the computations, we denote $h(n,\gamma ,\alpha )=(n-2\gamma +1)(n-\gamma )^{\alpha }+2^{\alpha }(\gamma -1)[1+(n-\gamma )^{\alpha }]$. For any 4-vertex tree, one can see that $R_{\alpha }(S_4)=3\cdot 3^{\alpha }=h(4,1,\alpha )$, and $R_{\alpha }(P_4)=4^{\alpha }+2\cdot 2^{\alpha }=h(4,2,\alpha )$. We suppose that the inequality shown in (3.1) is always true for any tree with $n-1$ vertices. Next, we will consider an n-vertex trees. We take a diameter path $v_1v_2\cdots v_d$ in T, and denote $N(v_2)=\lbrace v_1,v_3,u_1,u_2,\ldots ,u_{d_{v_2}-2} \rbrace$, $N(v_3)=\lbrace v_2,v_4,w_1,w_2,\ldots ,w_{d_{v_3}-2}\rbrace$. Let $T_{-1}=T-\lbrace v_1 \rbrace$, then we investigate the following two cases.

Case 1 If $\gamma (T_{-1})=\gamma (T)$, we obtain

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-1})+(d_{v_2}-1)\cdot (d_{v_2})^{\alpha } +(d_{v_2}\cdot d_{v_3})^{\alpha }-(d_{v_2}-2)\cdot (d_{v_2}-1)^{\alpha }\\&\quad -[(d_{v_2}-1)\cdot d_{v_3}]^{\alpha }\\&\ge (n-2\gamma )\cdot (n-\gamma -1)^{\alpha }+(\gamma -1)\cdot [2\cdot (n-\gamma -1)]^{\alpha }+(\gamma -1)\cdot 2^{\alpha }\\&\quad +(d_{v_2}-1)\cdot (d_{v_2})^{\alpha }+(d_{v_2}\cdot d_{v_3})^{\alpha }-(d_{v_2}-2)\cdot (d_{v_2}-1)^{\alpha }\\&\quad -[(d_{v_2}-1)\cdot d_{v_3}]^{\alpha }\\&=h(n,\gamma ,\alpha )+[(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha }\\&\quad -(n-\gamma )^{\alpha }]-(n-\gamma -1)^{\alpha }\\&\quad -[(d_{v_2}-1)^{\alpha }-(d_{v_2})^{\alpha }]\cdot [(d_{v_2}-1)+(d_{v_3})^{\alpha }]+(d_{v_2}-1)^{\alpha }. \end{aligned}$$

Since $T\cong S_{n}$ if $n=d_{v_2}+1$ and $T\in \mathcal {T}_{2}$ if $n=d_{v_2}+2$, we only need to study the case that $n\ge d_{v_2}+3$. Consider the following function:

$$\begin{aligned} \kappa (\gamma )&=[(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha }-(n-\gamma )^{\alpha }]-(n-\gamma -1)^{\alpha }\\&=[(n-2\gamma )+(\gamma -1)\cdot 2^{\alpha }](n-\gamma -1)^{\alpha }-[(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]\\&\qquad (n-\gamma )^{\alpha }. \end{aligned}$$

By calculation, we get

$$\begin{aligned} \kappa '(\gamma )&=-\alpha (n-\gamma -1)^{\alpha -1}[(n-2\gamma )+(\gamma -1)\cdot 2^{\alpha }]+(n-\gamma -1)^{\alpha }(2^\alpha -2)\\&\quad +\alpha (n-\gamma )^{\alpha -1}[(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]-(n-\gamma )^{\alpha }(2^\alpha -2)\\&=(n-\gamma -1)^{\alpha -1}\left\{ -\alpha [(n-2\gamma )+(\gamma -1)\cdot 2^{\alpha }]+(n-\gamma -1)(2^\alpha -2) \right\} \\&\quad -(n-\gamma )^{\alpha -1}\left\{ -\alpha [(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]+(n-\gamma )(2^\alpha -2) \right\} . \end{aligned}$$

Let $\vartheta _1(\gamma )=-\alpha [(n-2\gamma )+(\gamma -1)\cdot 2^{\alpha }]+(n-\gamma -1)(2^\alpha -2)$ and $\vartheta _2(\gamma )=-\alpha [(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]+(n-\gamma )(2^\alpha -2)$. It follows that $\vartheta '_1(\gamma )=\vartheta '_2(\gamma )=-\alpha (2^\alpha -2)+(2-2^\alpha )=(1+\alpha )(2-2^\alpha )>0$ for $\alpha \in (-1,0)$. Hence, $\vartheta _1(\gamma )\le \vartheta _1(\frac{n}{2})=-\alpha (\frac{n}{2}-1)2^{\alpha }+(\frac{n}{2}-1)(2^\alpha -2)=(\frac{n}{2}-1)[(1-\alpha )2^\alpha -2]<0$ for $\alpha \in (-1,0)$, and $\vartheta _2(\gamma )\le \vartheta _2(\frac{n}{2})=-\alpha [1+(\frac{n}{2}-1)2^{\alpha }]+\frac{n}{2}(2^\alpha -2)=(\frac{n}{2}-1)[(1-\alpha )2^\alpha -2]+(2^\alpha -2-\alpha )<0$ for $\alpha \in (-1,0)$.

Suppose that $\alpha \in (-1,0)$, then we consider the inequality $\kappa '(\gamma )=(n-\gamma -1)^{\alpha -1}\vartheta _1(\gamma )-(n-\gamma )^{\alpha -1}\vartheta _2(\gamma )<0$. Since

$$\begin{aligned}&(n-\gamma -1)^{\alpha -1}\vartheta _1(\gamma )-(n-\gamma )^{\alpha -1}\vartheta _2(\gamma )<0\nonumber \\&\quad \Leftrightarrow \frac{\vartheta _1(\gamma )}{\vartheta _2(\gamma )}>\left( \frac{n-\gamma }{n-\gamma -1}\right) ^{\alpha -1}\nonumber \\&\quad \Leftrightarrow \frac{-\alpha [(n-2\gamma )+(\gamma -1)\cdot 2^{\alpha }]+(n-\gamma -1)(2^\alpha -2)}{-\alpha [(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]+(n-\gamma )(2^\alpha -2)}>\left( \frac{n-\gamma -1}{n-\gamma }\right) ^{1-\alpha }\nonumber \\&\quad \Leftrightarrow 1-\frac{2-2^\alpha +\alpha }{(n-\gamma )(2-2^\alpha +\alpha )+(1-\gamma )(\alpha -\alpha \cdot 2^\alpha )}>\left( 1-\frac{1}{n-\gamma } \right) ^{1-\alpha }\nonumber \\&\quad \Leftrightarrow 1-\frac{1}{(n-\gamma )+\frac{\alpha (1-\gamma )(1-2^\alpha )}{2-2^\alpha +\alpha }}>\left( 1-\frac{1}{n-\gamma } \right) ^{1-\alpha }. \end{aligned}$$

(3.2)

Note that the inequality $\frac{\alpha (1-\gamma )(1-2^\alpha )}{2-2^\alpha +\alpha }\ge 0$ holds for $\alpha \in (-1,0)$ and $\gamma >1$; one can check that

$$\begin{aligned} 1-\frac{1}{(n-\gamma )+\frac{\alpha (1-\gamma )(1-2^\alpha )}{2-2^\alpha +\alpha }}>1-\frac{1}{n-\gamma }>\left( 1-\frac{1}{n-\gamma } \right) ^{1-\alpha }, \end{aligned}$$

for $\alpha \in (-1,0)$ and $\gamma >1$. So the function $\kappa '(\gamma )$ is always negative if $\alpha \in (-1,0)$ and $\gamma >1$, which implies that if $\alpha \in (-1,0)$, the function $\kappa (\gamma )$ monotonically decreases for $\gamma >1$.

Note that $\gamma \le \frac{n-d_{v_2}+2}{2}$ and $d_{v_3}\ge 2$, then we have $\kappa (\gamma )\ge \kappa \left( \frac{n-d_{v_2}+2}{2}\right)$ and

$$\begin{aligned}&[(n-2\gamma +1)+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha }-(n-\gamma )^{\alpha }]-(n-\gamma -1)^{\alpha }\\&\qquad -[(d_{v_2}-1)^{\alpha }-(d_{v_2})^{\alpha }]\cdot [(d_{v_2}-1)+(d_{v_3})^{\alpha }]+(d_{v_2}-1)^{\alpha }\\&\quad \ge \left[ \left( d_{v_2}-\frac{1}{2}\right) ^{\alpha }-\left( d_{v_2}+\frac{1}{2}\right) ^{\alpha }\right] \cdot \left( d_{v_2}-1+\frac{3}{2}\cdot 2^{\alpha }\right) \\&\qquad +[(d_{v_2})^{\alpha }-(d_{v_2}-1)^{\alpha }]\cdot (d_{v_2}+2^{\alpha }-1)+(d_{v_2}-1)^{\alpha }-\left( d_{v_2}-\frac{1}{2}\right) ^{\alpha }\\&\quad =\left\{ \left[ \left( d_{v_2}-\frac{1}{2}\right) ^{\alpha }-\left( d_{v_2}+\frac{1}{2}\right) ^{\alpha }\right] -[(d_{v_2}-1)^{\alpha }-(d_{v_2})^{\alpha }] \right\} (d_{v_2}+2^{\alpha }-1)\\&\qquad +2^{\alpha -1}\left[ \left( d_{v_2}-\frac{1}{2}\right) ^{\alpha }-\left( d_{v_2}+\frac{1}{2}\right) ^{\alpha }\right] +(d_{v_2}-1)^{\alpha }-\left( d_{v_2}-\frac{1}{2}\right) ^{\alpha }. \end{aligned}$$

Consider functions $\psi _2(z)=z^{\alpha }$ and $\phi _2(z)=\psi _2(z-1)-\psi _2(z)=(z-1)^{\alpha }-z^{\alpha }$.

Note that $\psi '_2(z)=\alpha z^{\alpha -1}<0$ and $\psi ''_2(z)=\alpha (\alpha -1) z^{\alpha -2}>0$. Suppose that $\alpha <0$ and there exist $z_2>z_1>1$, then we obtain

$$\begin{aligned} \phi _2(z_2)-\phi _2(z_1)&=\phi '_2(\xi _2)(z_2-z_1)\\&=[\psi '_2(\xi _2-1)-\psi '_2(\xi _2)](z_2-z_1)\\&=-{\psi ''_2(\eta _2)}(z_2-z_1)<0, \end{aligned}$$

where $1<z_1<\xi _2<z_2$ and $0<\xi _2-1<\eta _2<\xi _2$. Thus, if $\alpha <0$, $\phi _2(z)$ is a monotonically decreasing function for $z\in (1,\infty )$. One can see that for any $d_{v_2}\ge 2$ and $\alpha \in (-1,0)$, $\phi _2(d_{v_2}+\frac{1}{2})-\phi _2(d_{v_2}+1)>0$, $\psi _2(d_{v_2}-\frac{1}{2})-\psi _2(d_{v_2}+\frac{1}{2})>0$ and $\psi _2(d_{v_2}-1)-\psi _2(d_{v_2}-\frac{1}{2})>0$. Obviously, if $d_{v_2}\ge 2$ and $\alpha \in (-1,0)$, the inequality $\left\{ [(d_{v_2}-\frac{1}{2})^{\alpha }-(d_{v_2}+\frac{1}{2})^{\alpha }]-[(d_{v_2}-1)^{\alpha }-(d_{v_2})^{\alpha }] \right\} (d_{v_2}+2^{\alpha }-1)+2^{\alpha -1}[(d_{v_2}-\frac{1}{2})^{\alpha }-(d_{v_2}+\frac{1}{2})^{\alpha }]+(d_{v_2}-1)^{\alpha }-(d_{v_2}-\frac{1}{2})^{\alpha }>0$ holds, i.e., $R_{\alpha }(T)> h(n,\gamma ,\alpha )$.

Case 2 Suppose that $\gamma (T_{-1})=\gamma (T)-1$. Obviously, there exists a minimum dominating set D(T) such that $v_3\in D(T)$ and $d_{v_2}=2$. Then,

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-1})-(d_{v_3})^{\alpha }+(2\cdot d_{v_3})^{\alpha }+2^{\alpha }\\&\ge [n-1-2(\gamma -1)+1]\cdot (n-1-\gamma +1)^{\alpha }+(\gamma -2)\cdot 2^{\alpha }\\&\quad \cdot (n-1-\gamma +1)^{\alpha }+(\gamma -2)\cdot 2^{\alpha }\\&\quad -(d_{v_3})^{\alpha }+(2d_{v_3})^{\alpha }+2^{\alpha }\\&=h(n,\gamma ,\alpha )+(n-\gamma )^{\alpha }-[2\cdot (n-\gamma )]^{\alpha }-(d_{v_3})^{\alpha }+(2\cdot d_{v_3})^{\alpha }. \end{aligned}$$

It is easy to see that $(n-\gamma )^{\alpha }-[2(n-\gamma )]^{\alpha }-(d_{v_3})^{\alpha }+(2 d_{v_3})^{\alpha }\ge 0$ if $d_{v_3}\ge n-\gamma$. However, this only happens when $d_{v_3}=n-\gamma$ which implies that $T\in \mathcal {T}_{2}$.

If $d_{v_3}\le n-\gamma -1$, we denote $N(v_3)=\lbrace v_2,v_4,w_1,w_2,\ldots ,w_{d_{v_3}-2} \rbrace$. By the above investigations, we suppose that $w_i$ ($i=1,2,\ldots ,d_{v_3}-2$) is a pendent vertex or a support vertex with degree 2. Besides, if $v_4$ is a pendent vertex or a support vertex with degree 2, we can conclude that $T\in \mathcal {T}_{2}$. Without loss of generality, we let $d_{w_1}=d_{w_2}=\cdots =d_{w_{l_1}}=1$ and $d_{w_{l_1+1}}=\cdots =d_{w_{l_1+l_2}}=2$, where $l_1+l_2=d_{v_3}-2$. Then, we discuss the following two cases.

Case 2.1 $l_1\ge 2$. We take $T_{-1}=T-\lbrace w_1\rbrace$, implying $\gamma (T_{-1})=\gamma (T)$. Note that $\gamma -(2+l_2)\le \frac{n-(d_{v_3}+1+l_2)}{2}$, i.e., $\gamma \le \frac{n-l_1+1}{2}$, we obtain

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-1})-(d_{v_3}-1)^{\alpha }\cdot [(d_{v_4})^{\alpha } +(l_1-1)+(l_2+1)\cdot 2^{\alpha }]+(d_{v_3})^{\alpha }\\&\quad \cdot [(d_{v_4})^{\alpha } +l_1+(l_2+1)\cdot 2^{\alpha }]\\&\ge (n-2\gamma )\cdot (n-\gamma -1)^{\alpha }+(\gamma -1)\cdot 2^{\alpha }\cdot (n-\gamma -1)^{\alpha }+(\gamma -1)\cdot 2^{\alpha }\\&\quad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [(d_{v_4})^{\alpha }+l_1+(l_2+1)\cdot 2^{\alpha }]+(d_{v_3}-1)^{\alpha }\\&= h(n,\gamma ,\alpha )+[n-2\gamma +1+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha }\\&\quad -(n-\gamma )^{\alpha }]-(n-\gamma -1)^{\alpha }\\&\quad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [l_1+(l_2+2)\cdot 2^{\alpha }]+(d_{v_3}-1)^{\alpha } \end{aligned}$$

There exists $l\ge 0$ such that $n=l_1+2l_2+5+l$, then we have $\gamma \le \frac{l_1+2l_2+5+l-l_1+1}{2}=l_2+3+\frac{l}{2}$. By the above consideration, we get

$$\begin{aligned}&[n-2\gamma +1+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha }-(n-\gamma )^{\alpha }]-(n-\gamma -1)^{\alpha }\\&\qquad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [l_1+(l_2+2)\cdot 2^{\alpha }]+(d_{v_3}-1)^{\alpha }\\&\quad \ge \left[ l_1+\left( l_2+2+\frac{l}{2}\right) \cdot 2^{\alpha }\right] \cdot \left[ \left( d_{v_3}+\frac{l}{2}-1\right) ^{\alpha }-\left( d_{v_3}+\frac{l}{2}\right) ^{\alpha }\right] \\&\qquad -\left( d_{v_3}+\frac{l}{2}-1\right) ^{\alpha }\\&\qquad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [l_1+(l_2+2)\cdot 2^{\alpha }]+(d_{v_3}-1)^{\alpha }\\&\quad =\left[ \left( d_{v_3}+\frac{l}{2}\right) \cdot 2^{\alpha }+(1-2^{\alpha })l_1\right] \cdot \left[ \left( d_{v_3}+\frac{l}{2}-1\right) ^{\alpha }-\left( d_{v_3}+\frac{l}{2}\right) ^{\alpha }\right] \\&\qquad -\left( d_{v_3}+\frac{l}{2}-1\right) ^{\alpha }\\&\qquad -[d_{v_3}\cdot 2^{\alpha }+(1-2^{\alpha })l_1]\cdot [(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]+(d_{v_3}-1)^{\alpha }. \end{aligned}$$

Next, we consider the function $r(z)=[2^{\alpha }z+(1-2^{\alpha })l_1]\cdot [(z-1)^{\alpha }-z^{\alpha }]-(z-1)^{\alpha }$. If $(1-2^{\alpha })l_1\le 2^{\alpha -1}z$ and $\alpha \in (-1,0)$, we have

$$\begin{aligned} r'(z)&=(-\alpha )\cdot \left\{ -\frac{2^{\alpha }}{\alpha }\cdot [(z-1)^{\alpha }-z^{\alpha }] +(z-1)^{\alpha -1}-[(z-1)^{\alpha -1}-z^{\alpha -1}]\right. \\&\quad \cdot [2^{\alpha }z-(2^{\alpha }-1)l_1]\bigg \}\\&\ge (-\alpha )\cdot \left\{ -\frac{2^{\alpha }}{\alpha }\cdot [(z-1)^{\alpha }-z^{\alpha }]+(z-1)^{\alpha -1}-[(z-1)^{\alpha -1}-z^{\alpha -1}]\right. \\&\quad \cdot \left. \left( \frac{3}{2}\cdot 2^{\alpha }z\right) \right\} \\&=(-\alpha )\cdot \left\{ \left( 3\cdot 2^{\alpha -1}+\frac{2^{\alpha }}{\alpha }\right) [z^{\alpha }-(z-1)^{\alpha }]+(1-3\cdot 2^{\alpha -1})(z-1)^{\alpha -1} \right\} \\&=(-\alpha )\cdot \left\{ \left( 3\cdot 2^{\alpha -1}+\frac{2^{\alpha }}{\alpha }\right) [\psi _2(z)-\psi _2(z-1)]+(1-3\cdot 2^{\alpha -1})\right. \\&\qquad \cdot \left. \frac{1}{\alpha }\psi '_2(z-1) \right\} \end{aligned}$$

Suppose that $z\ge 2$ and $\alpha \in (-\frac{2}{3},0)$, we consider the following inequality:

$$\begin{aligned}&\left( 3\cdot 2^{\alpha -1}+\frac{2^{\alpha }}{\alpha }\right) [\psi _2(z)-\psi _2(z-1)]+\frac{1-3\cdot 2^{\alpha -1}}{\alpha }\psi '_2(z-1)>0\\&\quad \Leftrightarrow \left( 3\cdot 2^{\alpha -1}+\frac{2^{\alpha }}{\alpha }\right) \cdot \psi '_2(z-1)\cdot {\frac{\psi _2(z)-\psi _2(z-1)}{\psi '_2(z-1)}}\\&\qquad +\frac{1-3\cdot 2^{\alpha -1}}{\alpha }\psi '_2(z-1)>0\\&\quad \Leftrightarrow {\frac{\psi _2(z)-\psi _2(z-1)}{\psi '_2(z-1)}}>\frac{3\cdot 2^{\alpha -1}-1}{3\alpha \cdot 2^{\alpha -1}+2^{\alpha }}. \end{aligned}$$

Let $t(z)=\frac{\psi _2(z)-\psi _2(z-1)}{\psi '_2(z-1)}=\frac{z^{\alpha }-(z-1)^{\alpha }}{\alpha (z-1)^{\alpha -1}}$. Consider the inequality $t'(z)=\left( \frac{z}{z-1}\right) ^{\alpha -1}+\frac{1-\alpha }{\alpha }\left( \frac{z}{z-1}\right) ^{\alpha }-\frac{1}{\alpha }>0$, then we get

$$\begin{aligned}&\left( \frac{z}{z-1}\right) ^{\alpha -1}+\frac{1-\alpha }{\alpha }\left( \frac{z}{z-1}\right) ^{\alpha }-\frac{1}{\alpha }>0\\&\quad \Leftrightarrow \alpha \left( \frac{z}{z-1}\right) ^{\alpha -1}+(1-\alpha )\left( \frac{z}{z-1}\right) ^{\alpha }-1<0\\&\quad \Leftrightarrow \alpha +(1-\alpha )\frac{z}{z-1}<\left( \frac{z}{z-1}\right) ^{1-\alpha } \\&\quad \Leftrightarrow \left( 1+\frac{1}{z-1}\right) ^{1-\alpha }-(1-\alpha )\frac{1}{z-1}>1. \end{aligned}$$

Let $\rho (s)=(1+s)^{1-\alpha }-(1-\alpha )s$, where $s\in (0,1]$ and $\alpha \in (-\frac{2}{3},0)$. Since $\rho '(s)=(1-\alpha )\left[ (1+s)^{-\alpha }-1\right] >0$ for $s\in (0,1]$ and $\rho '(0)=0$, we have $\rho (s)>\rho (0)=1$ for $s\in (0,1]$, which implies that the inequality $\left( 1+\frac{1}{z-1}\right) ^{1-\alpha }-(1-\alpha )\frac{1}{z-1}>1$ holds for $z\ge 2$.

Based on the above discussion, one can see that t(z) is a monotonically increasing function for $z\ge 2$ if $\alpha \in (-\frac{2}{3},0)$. We calculate that $t(2)=\frac{2^{\alpha }-1}{\alpha }>\frac{3\cdot 2^{\alpha -1}-1}{3\alpha \cdot 2^{\alpha -1}+2^{\alpha }}$ for any $\alpha \in (-\frac{2}{3},0)$ (the function $\frac{2^{\alpha }-1}{\alpha }-\frac{3\cdot 2^{\alpha -1}-1}{3\alpha \cdot 2^{\alpha -1}+2^{\alpha }}$ is depicted in Fig. 12a), what implies that $r'(z)>0$ for any $z\ge 2$ and $\alpha \in (-\frac{2}{3},0)$. If $l=0$, one can check that $T\in \mathcal {T}_{2}$. Moreover, $d_{v_3}+\frac{l}{2}\ge d_{v_3}\ge l_1+2\ge l_1(2^{1-\alpha }-2)$ with $l>0$, and thus, we get $r(d_{v_3}+\frac{l}{2})>r(d_{v_3})$, which implies that

$$\begin{aligned}&[n-2\gamma +1+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha }-(n-\gamma )^{\alpha }]-(n-\gamma -1)^{\alpha }\\&\quad >[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [l_1+(l_2+2)\cdot 2^{\alpha }]+(d_{v_3}-1)^{\alpha }. \end{aligned}$$

It is easy to conclude that $R_{\alpha }(T)>h(n,\gamma ,\alpha )$ for $\alpha \in \left( -\frac{2}{3},0\right)$.

Case 2.2 $l_1\le 1$. Let $T_{-2}=T-\lbrace v_1,v_2 \rbrace$, then we have

$$\begin{aligned} R_{\alpha }(T)&=R_{\alpha }(T_{-2})-[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [(d_{v_4})^{\alpha }+1+(d_{v_3}-3)\cdot 2^{\alpha }]\\&\quad +(2\cdot d_{v_3})^{\alpha }+2^{\alpha }\\&\ge [n-2-2(\gamma -1)+1]\cdot (n-2-\gamma +1)^{\alpha }+(\gamma -2)\cdot 2^{\alpha }\\&\quad \cdot (n-2-\gamma +1)^{\alpha }+(\gamma -2)\cdot 2^{\alpha }\\&\quad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [(d_{v_4})^{\alpha }+1+(d_{v_3}-3)\cdot 2^{\alpha }]+(2\cdot d_{v_3})^{\alpha }+2^{\alpha }\\&= h(n,\gamma ,\alpha )+[n-2\gamma +1+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha } -(n-\gamma )^{\alpha }]\\&\quad -2^{\alpha }\cdot (n-\gamma -1)^{\alpha }\\&\quad -[(d_{v_3}-1)^{\alpha }-(d_{v_3})^{\alpha }]\cdot [1+(d_{v_3}-2)\cdot 2^{\alpha }]+(2\cdot d_{v_3})^{\alpha } \end{aligned}$$

Note that $\gamma \le \frac{n}{2}$ for any graph, we get

$$\begin{aligned}&[n-2\gamma +1+(\gamma -1)\cdot 2^{\alpha }]\cdot [(n-\gamma -1)^{\alpha }-(n-\gamma )^{\alpha }]-2^{\alpha }\cdot (n-\gamma -1)^{\alpha }\\&\quad \ge -2^{\alpha }\cdot \left( \frac{n}{2}-1\right) ^{\alpha }-\left[ \left( \frac{n}{2}\right) ^{\alpha }-\left( \frac{n}{2}-1\right) ^{\alpha }\right] \cdot \left[ 2^{\alpha }\cdot \left( \frac{n}{2}-1\right) +1\right] \\&\quad =\left[ 2^\alpha \cdot \left( \frac{n}{2}-1\right) +1-2^\alpha \right] \cdot \left( \frac{n}{2}-1\right) ^{\alpha }-\left( 2^\alpha \cdot \frac{n}{2}+1-2^\alpha \right) \cdot \left( \frac{n}{2}\right) ^{\alpha } . \end{aligned}$$

Consider functions $\psi _3(z)=(2^\alpha \cdot z+1-2^\alpha )\cdot z^{\alpha }$ and $\phi _3(z)=\psi _3(z)-\psi _3(z+1)$, then we calculate that $\psi '_3(z)=[(\alpha +1)2^{\alpha }\cdot z+\alpha (1-2^\alpha )]\cdot z^{\alpha -1}$ and $\psi ''_3(z)=2^{\alpha -2}\alpha \cdot [(\alpha +1)2^{\alpha }\cdot z+(\alpha -1)(1-2^{\alpha })]$.

Fig. 12

The four functions of $\alpha$ in Theorem 3.2

Suppose that $\alpha \in (-\frac{2}{3},0)$ and $z>0$. Since $\frac{(1-\alpha )(1-2^{\alpha })}{(\alpha +1)2^{\alpha }}<3$ holds for every $\alpha \in (-\frac{2}{3},0)$ (see Fig. 12b), it can be checked that if $z\ge 3>\frac{(1-\alpha )(1-2^{\alpha })}{(\alpha +1)2^{\alpha }}$, the inequality $(\alpha +1)2^{\alpha }\cdot z+\alpha (1-2^\alpha )>0$ holds, implying $\psi '_3(z)>0$ and $\psi ''_3(z)<0$ for $\alpha \in (-\frac{2}{3},0)$ and $z\ge 3$.

Suppose that $\alpha \in (-\frac{2}{3},0)$ and there exist $z_2>z_1\ge 3$, we obtain

$$\begin{aligned} \phi _3(z_2)-\phi _3(z_1)&=\phi '_3(\xi _3)(z_2-z_1)\\&={-[\psi '_3(\xi _3+1)-\psi '_3(\xi _3)]}(z_2-z_1)\\&=-\psi ''_3(\eta _3)(z_2-z_1)>0, \end{aligned}$$

where $3\le z_1<\xi _3<z_2$ and $3<\xi _3<\eta _3<\xi _3+1$. In addition, we calculate that $\phi _3(3)-\phi _3(2)=4\cdot 6^\alpha +2\cdot 3^\alpha -3\cdot 8^\alpha -2\cdot 4^\alpha -2^\alpha >0$ for $\alpha \in [-0.65,0)$ and $\phi _3(2)-\phi _3(1)=2\cdot 4^\alpha -2^{\alpha +1}-2\cdot 6^\alpha -3^\alpha -1>0$ for $\alpha \in [-0.5696,0)$ (see Fig. 12c, d). Based on the above discussions, we conclude that $\phi _3(z)$ is a monotonically increasing function if $\alpha \in [-0.5696,0)$. Then, we get ${\phi _3(\frac{n}{2}-1)\ge \phi _3(d_{v_3}-1)}$ for $n\ge 2d_{v_3}$. The equality holds if and only if $n=2d_{v_3}$, which implies that $T\in \mathcal {T}_{2}$. On the whole, the equality in (3.1) holds for every tree in $\mathcal {T}_2$, and if $T\notin \mathcal {T}_2$, the inequality (3.1) is strict.

This completes the proof. $\square$

4 Conclusions

The extremal problem of the general Randić index of trees in terms of domination number is investigated in this paper. For any n-vertex tree T with domination number $\gamma$, we show that $R_{\alpha }(T)\le f(n,\gamma ,\alpha )$ if $\alpha \in [-0.5287,0)$ and $R_{\alpha }(T)\ge h(n,\gamma ,\alpha )$ if $\alpha \in [-0.5696,0)$. In each case, extremal graphs are also characterized.

Fig. 13

Two non-isomorphic 7-vertex trees with domination number 3

For $P_4$ and $P_5$, it can be calculated that $R_{\alpha }(P_4)>f(4,2,\alpha )$ and $R_{\alpha }(P_5)>f(5,2,\alpha )$ if $\alpha <-1$. Moreover, we construct a 7-vertex tree with domination number 3, written as $T'(7,3)$ (depicted in Fig. 13b), and show that $R_{\alpha }(T'(7,3))=9^{\alpha }+6^{\alpha }+3\cdot 3^{\alpha }+2^{\alpha }<h(7,3,\alpha )$ if $\alpha \le -1.3568$. By observing the above simple examples, it can easily see that only when $\alpha$ is in a small interval, Theorems 2.3 and 3.2 hold.

The two ranges of the parameter $\alpha$ obtained in this paper may not be very accurate, and two lower bounds on $\alpha$ ($\alpha _1$ and $\alpha _2$) are quite near $-\frac{1}{2}$. It is hoped that in the future, one can discover new approaches to establish more extremal properties of the general Randić index and provide a more exact range of the parameter $\alpha$ about this issue.