On the sum of the two largest Laplacian eigenvalues of unicyclic graphs

Abstract

Let G be a simple graph and $S_{2}(G)$ be the sum of the two largest Laplacian eigenvalues of G. Guan et al. (J. Inequal. Appl. 2014:242, 2014) determined the largest value for $S_{2}(T)$ among all trees of order n. They also conjectured that among all connected graphs of order n with m ($n \leq m\leq2n -3$) edges, $G_{m,n}$ is the unique graph which has maximal value of $S_{2}(G)$, where $G_{m,n}$ is a graph of order n with m edges which has $m-n+1$ triangles with a common edge and $2n-m-3$ pendent edges incident with one end vertex of the common edge. In this paper, we confirm the conjecture with $m=n$.

1 Introduction

Let $G=(V,E)$ be a simple connected graph with vertex set $V(G)$ and edge set $E(G)$. Its order is $|V(G)|$, denoted by $n(G)$ (or n for short), and its size is $|E(G)|$, denoted by $m(G)$ (or m for short). For a vertex $v \in V(G)$, let $N(v)$ be the set of all neighbors of v in G and let $d(v)=|N(v)|$ be the degree of v. Particularly, denote by $\Delta(G)$ the maximum degree of G. A pendent vertex is a vertex with degree one. The diameter of a connected graph G, denoted by $d(G)$, is the maximum distance among all pairs of vertices in G. Let $S_{n}$ and $P_{n}$ be the star and the path of order n, respectively. Let $S^{k}_{a,b}$ be the tree of order n obtained from two stars $S_{a+1}$, $S_{b+1}$ by joining a path of length k between their central vertices (see Figure 2). For all other notions and definitions, not given here, see, for example, [1] or [2] (for graph spectra).

Let $A(G)$ and $D(G)$ be the adjacency matrix and the diagonal matrix of vertex degrees of G, respectively. The matrix $L(G)=D(G)-A(G)$ is called Laplacian matrix of G. The Laplacian matrix is an important topic in the theory of graph spectra. We use the notation $I_{n}$ for the identity matrix of order n and denote by $\phi(G, x) = \det(xI_{n}-L(G))$ the Laplacian characteristic polynomial of G. It is well known that $L(G)$ is positive semidefinite symmetric and singular. Denote its eigenvalues by $\mu_{1}(G) \geq\mu_{2}(G)\geq\cdots\geq \mu_{n}(G)$ (or simply $\mu_{1} \geq\mu_{2} \geq\cdots\geq \mu_{n}$ sometimes for convenience) which are always enumerated in non-increasing order and repeated according to their multiplicity. Note that each row sum of $L(G)$ is 0 and, therefore, $\mu_{n}(G)=0$. Fiedler [3] showed that the second smallest eigenvalue $\mu_{n-1}(G)$ of $L(G)$ is 0 if and only if G is disconnected. Thus the second smallest eigenvalue of $L(G)$ is popularly known as the algebraic connectivity of G. The largest eigenvalue $\mu_{1}(G)$ of $L(G)$ is usually called the Laplacian spectral radius of the graph G. Recently, some of the research has been focused on $\mu_{1}$, $\mu_{2}$ or $u_{n-1}$ (see [3–9]).

Let $S_{k}(G)= \sum_{i=1}^{i=k}\mu_{i}(G)$ be the sum of the k largest Laplacian eigenvalues of G. Brouwer conjectured that $S_{k}(G) \leq m(G)+ \binom{k+1}{2}$ for $k=1,2,\ldots, n$. The conjecture is still open, but some advances on the conjecture have been achieved (see [10–14]). Specially, for $k = 2$, Haemers et al. [12] proved that $S_{2}(G) \leq m(G)+3$ for any graph G. When G is a tree, Fritscher et al. [15] improved this bound by giving $S_{2}(T) \leq m(T)+3-\frac{2}{n(T)}$, which implies that Haemers’ bound is always not attainable for trees. Therefore, it is interesting to determine which tree has maximal value of $S_{2}(T)$ among all trees of order n. Guan et al. [11] proved that $S_{2}(T)\leq S_{2}(T^{1}_{\lceil\frac{n-2}{2}\rceil, \lfloor\frac{n-2}{2} \rfloor})$ for any tree of order $n \geq4$, and the equality holds if and only if $T \cong T^{1}_{\lceil\frac {n-2}{2}\rceil, \lfloor\frac{n-2}{2} \rfloor}$. For any graph G of order n with m edges, note that $\mu_{1}(G)\leq n(G)$. Then Haemers’ bound is clearly not attainable when $2n(G)< m(G)+3$. For $m(G)+3 \leq2n(G)$, Guan et al. [11] showed that $S_{2}(G_{m,n})=m(G_{m,n})+3$, where $G_{m,n}$ is a graph of order n size m which has $m-n+1$ triangles with a common edge and $2n-m-3$ pendent edges incident with one end vertex of the common edge ($G_{n+1,n}$ is illustrated in Figure 1). This indicates that Haemers’ bound is always sharp for connected graphs ($m \leq2n-3$) other than trees. The following conjecture on the uniqueness of the extremal graph is also presented in [11].

Figure 1

$\pmb{G_{n+1,n}}$ . $G_{m,n}$ is a graph of order n with m edges which has $m-n+1$ triangles with a common edge and $2n-m-3$ pendent edges incident with one end vertex of the common edge.

Conjecture 1.1

[11]

Among all connected graphs of order n with m edges ($n \leq m\leq2n -3$), $G_{m,n}$ is the unique graph with maximal value of $S_{2}(G)$, that is, $S_{2}(G_{m,n})=m(G_{m,n})+3$.

In this paper, we confirm Conjecture 1.1 with $m=n$.

2 Preliminaries

In this section, we present some lemmas which will be useful in the subsequent sections. For $\mu_{1}(G)$, the following results are well known.

Lemma 2.1

Let G be a connected graph of order n, $d_{i}=d(v_{i})$ and $m_{i}=\sum_{v_{j} \in N(v_{i})}d_{j}/d_{i}$. Then

• (1) [4] $\mu_{1}(G) \leq n(G)$ with equality if and only if the complement of G is disconnected;

• (2) [8] $\mu_{1}(G) \leq \max\{d_{i}+m_{i}|v_{i} \in V(G)\}$.

Lemma 2.2

[7]

Let G be a connected graph of order $n \geq4$ with m edges. Then

$$\mu_{1}(G) < \max \biggl\{ \Delta(G), m-\frac{n-1}{2} \biggr\} +2. $$

Lemma 2.3

[9]

Let T be a tree of order n with $d(T)\geq3$. Then $\mu_{1}(T) < n-0.5$.

Lemma 2.4

Let T be a tree of order n with $d(T)\geq4$. Then $\mu_{1}(T) < n-1$.

Proof

For any tree T of order n with $d(T)\geq4$, it follows that $n \geq5$ and $\Delta(T)\leq n-3$. That is, $\Delta(T)+2 \leq n-1$ and $m-\frac{n-1}{2} +2=\frac{n-1}{2} +2 \leq n-1$. Then the result follows from Lemma 2.2. □

The following theorem from matrix theory plays a key role in our proofs. We denote the eigenvalues of a symmetric matrix M of order n by $\lambda_{1}(M)\geq\lambda_{2}(M)\geq\cdots \geq\lambda_{n}(M)$.

Theorem 2.5

[16]

Let A and B be two real symmetric matrices of size n. Then, for any $1\leq k \leq n$,

$$\sum_{i=1}^{k}\lambda_{i}(A+B) \leq\sum_{i=1}^{k}\lambda_{i}(A)+ \sum_{i=1}^{k}\lambda_{i}(B). $$

From Theorem 2.5, the following lemma is immediate.

Lemma 2.6

Let $G_{1}, \ldots, G_{r}$ be some edge disjoint graphs. Then, for any k,

$$S_{k}(G_{1}\cup\cdots\cup G_{r})\leq \sum _{i=1}^{r}S_{k}(G_{i}). $$

The following lemma can be found in [2] and is well known as the interlacing theorem of Laplacian eigenvalues.

Lemma 2.7

[2]

Let G be a graph of order n and let $G^{\prime}$ be the graph obtained from G by inserting a new edge into G. Then the Laplacian eigenvalues of G and $G^{\prime}$ interlace, that is,

$$\mu_{1}\bigl(G^{\prime}\bigr) \geq\mu_{1}(G) \geq\cdots \mu_{n}\bigl(G^{\prime}\bigr) \geq\mu_{n}(G)=0. $$

Lemma 2.8

[2]

Let A be a symmetric matrix of order n with eigenvalues $\lambda _{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{n}$ and B be a principal submatrix of A of order m with eigenvalues $\mu_{1} \geq\mu_{2} \geq\cdots\geq\mu_{m}$. Then the eigenvalues of B interlace the eigenvalues of A, that is, $\lambda_{i} \geq\mu_{i} \geq\lambda _{n-m+i}$ for $i=1, \ldots, m$. Specially, for $v \in V(G)$, let $L_{v}(G)$ be the principal submatrix of $L(G)$ formed by deleting the row and column corresponding to vertex v, then the eigenvalues of $L_{v}(G)$ interlace the eigenvalues of $L(G)$.

Lemma 2.9

For any tree T of order n, if there exists an edge $e\in E(T)$ such that $\min\{e(T_{1}), e(T_{2})\} \geq2$ and $\min\{d(T_{1}), d(T_{2})\} \geq3$ or $\max\{d(T_{1}), d(T_{2})\} \geq4$, then $S_{2}(T) < n(T)+1$, where $T_{1}$, $T_{2}$ are the two components of $T-e$.

Proof

Let $T-e=T_{1} \cup T_{2}$. By Lemma 2.6, it suffices to show that $S_{2}(T_{1} \cup T_{2}) < n(T)-1$. If $\mu_{1}(T)= \mu_{1}(T_{1})$ and $\mu_{2}(T)= \mu_{2}(T_{1})$ (or $\mu_{1}(T)= \mu_{1}(T_{2})$ and $\mu_{2}(T)= \mu_{2}(T_{2})$), then the result follows since $S_{2}(T_{1} \cup T_{2})=S_{2}(T_{i}) < m(T_{i})+3 \leq m(T)=n(T)-1$ for $i=1, 2$. In what follows, we assume that $S_{2}(T_{1} \cup T_{2})=\mu_{1}(T_{1})+\mu_{1}(T_{2})$. If $\min\{d(T_{1}), d(T_{2})\} \geq3$, then Lemma 2.3 implies that $S_{2}(T_{1} \cup T_{2}) < (n(T_{1})-0.5)+ (n(T_{2})-0.5) = n(T)-1$. If $\max\{d(T_{1}), d(T_{2})\} \geq4$, say $d(T_{2}) \geq4$, then Lemmas 2.1 and 2.4 imply that $S_{2}(T_{1} \cup T_{2}) < n(T_{1})+ (n(T_{2})-1)=n(T)-1$. This completes the proof. □

Lemma 2.10

For any tree T of order n with $d(T)=5$, $S_{2}(T)< n(T)+1$.

Proof

Without loss of generality, we assume that $v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}$ is a path of length 5 in T. We now consider the following three cases.

Case 1. $\min\{d(v_{3}), d(v_{4})\} \geq3$.

Let $T_{1}$, $T_{2}$ be the two components of $T-v_{3}v_{4}$. Then the result follows from Lemma 2.9.

Case 2. $d (v_{3}) = d (v_{4}) = 2$.

Note that $d(T)=5$. Then T is isomorphic to $S^{3}_{a,b}$ (see Figure 2), where $a, b \geq1$ and $a+b+4=n$. By an elementary calculation, we have $\phi(S^{3}_{a,b},\lambda)=\lambda(\lambda-1)^{n-5}f_{1}(\lambda)$, where $f_{1}(\lambda)= \lambda^{4}-(n+3)\lambda^{3}+(5n+ab-4)\lambda^{2}-(6n+3ab-10)\lambda+n$. Denoted by $\lambda_{1} \geq\lambda_{2} \geq\lambda_{3} \geq \lambda_{4}>0$ are the four roots of $f_{1}(\lambda)=0$. Then we have $\lambda_{1} + \lambda_{2} + \lambda_{3} + \lambda_{4}=n+3$ since $\sum_{i=1}^{i=n}\mu_{i}=2m$. Note that $S^{3}_{a,b}$ contains $P_{6}$ as a subgraph. Then Lemma 2.7 implies that $\lambda_{3} \geq\mu_{3}(P_{6})=2$. Thus, $S_{2}(T)=\lambda_{1} + \lambda_{2}=n+3-(\lambda_{3} + \lambda_{4})< n+3-\lambda_{3}< n-1$, as desired.

Figure 2

Two trees of order n .

Case 3. $d(v_{3}) \geq3$ and $d(v_{4}) = 2$ (or $d(v_{4}) \geq3$ and $d(v_{3}) = 2$).

Without loss of generality, we assume that $d(v_{3}) \geq3$ and $d(v_{4}) = 2$. If $d(v_{2}) \geq3$, let $T-v_{2}v_{3}= T_{1} \cup T_{2}$; if there is $P_{3}=uvv_{3}$ attached to $v_{3}$, let $T-v_{3}v_{4}= T_{1} \cup T_{2}$, where $u, v \neq v_{i}$ ($i=1,2,\ldots,6$). Then the result follows from Lemma 2.9. We now assume that $d(v_{2})=d(v_{4})=2$ and all the neighbors of $v_{3}$ except for $v_{2}$ and $v_{4}$ are pendent vertices. That is, T is isomorphic to $H_{a,b}$, where $H_{a,b}$ (see Figure 2) is the tree of order n obtained from $v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}$ by attaching a and $b-1$ pendent vertices to $v_{3}$ and $v_{5}$, respectively, where $a, b \geq1$ and $a+b+5=n$. Note that the matrix $1\cdot I_{n}-L(H_{a,b})$ has a and b different identical rows. Then the multiplicity of eigenvalue 1 is at least $n-7$. Let $\lambda_{1} \geq\lambda_{2} \geq\lambda_{3} \geq \lambda_{4}\geq\lambda_{5} \geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues. Then $\lambda_{1} + \lambda_{2} + \lambda_{3} + \lambda_{4}+\lambda_{5} + \lambda_{6}+ \lambda_{7}=n+5$ since $\sum_{i=1}^{i=n}\mu_{i}=2m$. For $a, b\geq 2$, $H_{a,b}$ contains $H_{2,2}$ as a subgraph. Then by Lemma 2.7 we have $\lambda_{3} \geq\mu_{3}(H_{2,2})=2.44$ and $\lambda_{4} \geq\mu_{4}(H_{2,2})=1.59$. Therefore, $S_{2}(T)=\lambda_{1} + \lambda_{2} < n+5-(\lambda_{3}+\lambda_{4})< n+1$, as required. If $a=1$, then by Lemmas 2.1 and 2.8 we have $\mu_{1}(H_{1,b}) \leq(n-5)+\frac{n-4}{n-5}$ and $\mu_{2}(H_{1,b}) \leq\mu_{1}(L_{v_{5}}(H_{1,b}))=4.26$. That is, $S_{2}(T)=S_{2}(H_{1,b})=\mu_{1}(H_{1,b})+\mu_{2}(H_{1,b})< n+1$, as required. Similarly, if $b=1$, then by Lemmas 2.1 and 2.8 we have $\mu_{1}(H_{a,1}) \leq(n-4)+\frac{n-2}{n-4}$ and $\mu_{2}(H_{a,1}) \leq\mu_{1}(L_{v_{3}}(H_{a,1}))=3.0$. It follows that $S_{2}(T)=S_{2}(H_{a,1})=\mu_{1}(H_{a,1})+\mu _{2}(H_{a,1})< n+1$, as required.

From the discussion above, the proof is completed. □

Lemma 2.11

Let T be a tree of order n with $d(T)\geq6$. Then $S_{2}(T)< n(T)+1$.

Proof

We now consider the following two cases.

Case 1. $d(T)\geq7$.

Let $v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{7}v_{8}$ be a path of length 7 in T and $T-v_{4}v_{5}=T_{1} \cup T_{2}$. Then the result follows from Lemma 2.9.

Case 2. $d(T)=6$.

If $T= P_{7}$, then the result follows since $S_{2}(P_{7}) < 8$. If $T\neq P_{7}$, let $v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{7}$ be a path of length 6 in T. If $d(v_{4})\geq3$, let $T-v_{3}v_{4}=T_{1} \cup T_{2}$; if $d(v_{3})\geq3$ (or $d(v_{5})\geq3$), let $T-v_{3}v_{4}=T_{1} \cup T_{2}$ (or $T-v_{4}v_{5}=T_{1} \cup T_{2}$); if $d(v_{2})\geq3$ (or $d(v_{6})\geq3$), let $T-v_{2}v_{3}=T_{1} \cup T_{2}$ (or $T-v_{5}v_{6}=T_{1} \cup T_{2}$). In each of the above cases, by Lemma 2.9, we have $S_{2}(T)< n(T)+1$. This completes the proof. □

A firefly graph $F_{s,t,n-2s-2t-1}$ ($s \geq0$, $t \geq0$ and $n-2s-2t-1 \geq0$) is a graph with n vertices that consists of s triangles, t pendent paths of length 2 and $n-2s-2t-1$ pendent edges, sharing a common vertex. An example of a firefly graph $F_{2,3,4}$ is illustrated in Figure 3. Clearly $F_{0,0,n-1} \cong S_{n}$ and $F_{1,0,n-3} \cong G_{n,n}$.

Figure 3

An example of firefly graph.

Lemma 2.12

[6] The second largest Laplacian eigenvalue of $F_{1, t, n-2t-3}$ satisfies $\mu_{2}(F_{1, t, n-2t-3})=3$.

Note that for $t \geq1$, the complement of $F_{1, t, n-2t-3}$ is connected. Hence Lemma 2.1 implies that $\mu_{1}(F_{1, t, n-2t-3})< n(F_{1, t, n-2t-3})$. This together with Lemma 2.12 implies the following lemma.

Lemma 2.13

For $t \geq1$, $S_{2}(F_{1, t, n-2t-3}) < n(F_{1, t, n-2t-3})+3$.

3 Main result

A unicyclic graph is a connected graph whose number of edges m is equal to the number of vertices n. It is easy to see that each unicyclic graph can be obtained by attaching rooted trees to the vertices of a cycle $C_{k}$ for some k. Thus if $R_{1},\ldots, R_{k}$ are k rooted trees (of orders $n_{1},\ldots, n_{k}$, say), then we adopt the notation $U(R_{1},\ldots, R_{k})$ to denote the unicyclic graph G (of order $n=n_{1}+\cdots+n_{k}$) obtained by attaching the rooted tree $R_{i}$ to the vertex $v_{i}$ of a cycle $C_{k}=v_{1}v_{2} \cdots v_{k}v_{1}$ (i.e., by identifying the root of $R_{i}$ with the vertex $v_{i}$ for $i=1,\ldots, k$). Denote by $e(v_{i})$ the maximum distance between $v_{i}$ and any vertex of $R_{i}$. In the special case when $R_{i}$ is a rooted star $K_{1,a_{i}}$ with the center of the star as its root (that is, $e(v_{i})=1$), we will simplify the notation by replacing $R_{i}$ by the number $a_{i}$.

The following lemma is immediate from Lemmas 2.6, 2.10 and 2.11.

Lemma 3.1

For any unicyclic graph G of order n with m edges, if there exists an edge $e\in E(G)$ such that $G-e$ is a tree with $d(G-e) \geq5$, then $S_{2}(G)< m(G)+3$.

Let $uvwu$ be a triangle and $T_{a,b,c}$ be the graph obtained by attaching a, b, c pendent vertices to u, v, w, respectively, where $a+b+c=n-3$ and $a \geq b \geq c \geq0$. Note that $T_{n-3,0,0} \cong G_{n,n}$. Denote by $Q_{a,b}$ the graph obtained by attaching a and b pendent vertices to two non-adjacent vertices of a quadrangle, respectively, where $a+b=n-4$ and $a \geq b \geq0$. $T_{a,b,c}$ and $Q_{a,b}$ are illustrated in Figure 4.

Figure 4

Two unicyclic graphs of order n . $T_{a,b,c}$ is a unicyclic graph of order n obtained by attaching a, b, c pendent vertices to three vertices of a triangle $uvw$, respectively, and $Q_{a,b}$ is a unicyclic graph of order n obtained by attaching a and b pendent vertices to two non-adjacent vertices of a quadrangle, respectively.

Lemma 3.2

For $T_{a,b,c}$, $S_{2}(T_{a,b,c}) \leq m(T_{a,b,c})+3$ with equality if and only if $a=n-3$ (that is, $T_{a,b,c} \cong G_{n,n}$).

Proof

Note that the matrix $1\cdot I_{n}-L(T_{a,b,c})$ has a, b and c different identical rows. So the multiplicity of eigenvalue 1 is at least $n-6$. Let $\lambda_{1} \geq\lambda_{2} \geq\lambda_{3} \geq\lambda_{4} \geq \lambda_{5} > \lambda_{6}=0$ be the other six eigenvalues of $L(T_{a,b,c})$. Then we have $\lambda_{1} + \lambda_{2} + \lambda_{3} + \lambda_{4}+\lambda_{5} =n+6$ since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. For $c \geq2$, $T_{a,b,c}$ contains $T_{2,2,2}$ as a subgraph. Then by Lemma 2.7 we have $\lambda_{3} \geq\mu_{3}(T_{2,2,2})=3.0$. Therefore, $S_{2}(T_{a,b,c})=\lambda_{1} + \lambda_{2} =n+6-( \lambda_{3} + \lambda_{4}+\lambda_{5})< n+6-\lambda_{3} \leq n+3 = m(T_{a,b,c})+3$, as required. For $c=1$ and $b \geq4$, $T_{a,b,c}$ contains $T_{4,4,1}$ as a subgraph. Then by Lemma 2.7 we have $\lambda_{3} \geq\mu_{3}(T_{4,4,1})=3.0$. That is, $S_{2}(T_{a,b,1}) = n+6-( \lambda_{3} + \lambda_{4}+\lambda _{5})< n+6-\lambda_{3} \leq n+3 = m(T_{a,b,c})+3$, as required. If $c=1$ and $b=3$, then by Lemmas 2.1 and 2.8, we have $\mu_{1}(T_{a,3,1}) \leq n-4+\frac{6}{n-5}$ and $\mu_{2}(T_{a,3,1}) \leq\mu_{1}(L_{u}(T_{a,3,1}))=5.88$. It follows that $S_{2}(T_{a,3,1})= \mu_{1}(T_{a,3,1})+ \mu_{2}(T_{a,3,1}) < m(T_{a,3,1})+3$, as required. If $c=1$ and $b=2$, then by Lemmas 2.1 and 2.8, we have $\mu_{1}(T_{a,2,1}) \leq n-3+\frac {5}{n-4}$ and $\mu_{2}(T_{a,2,1}) \leq\mu_{1}(L_{u}(T_{a,2,1}))=5.05$. That is, $S_{2}(T_{a,2,1})= \mu_{1}(T_{a,2,1})+ \mu_{2}(T_{a,2,1}) < m(T_{a,2,1})+3$ for $n \geq10$ (that is, $a \geq4$), as required. A direct calculation shows that $S_{2}(T_{3,2,1}) < m(T_{3,2,1})+3$ (or $S_{2}(T_{2,2,1}) < m(T_{2,2,1})+3$). Similarly, if $c=1$ and $b=1$, then by Lemmas 2.1 and 2.8, we have $\mu_{1}(T_{a,1,1}) \leq n-2+\frac{4}{n-3}$ and $\mu_{2}(T_{a,1,1}) \leq\mu _{1}(L_{u}(T_{a,1,1}))=4.30$. That is, $S_{2}(T_{a,1,1})= \mu _{1}(T_{a,1,1})+ \mu_{2}(T_{a,1,1} )< m(T_{a,1,1})+3$, for $n \geq9$ (that is, $a \geq4$), as required. A direct calculation shows that $S_{2}(T_{3,1,1}) < m(T_{3,1,1})+3$ (or $S_{2}(T_{2,1,1}) < m(T_{2,1,1})+3$ or $S_{2}(T_{1,1,1}) < m(T_{1,1,1})+3$). Next, we assume $c=0$. By an elementary calculation, we have $\phi (T_{a,b,0},\lambda)=\lambda(\lambda-1)^{n-5}f_{2}(\lambda)$, where $f_{2}(\lambda)= \lambda^{4}-(n+5)\lambda^{3}+(5n+ab+7)\lambda ^{2}-(7n+2ab+3)\lambda+3n$. Let $x_{1}\geq x_{2} \geq x_{3}\geq x_{4}$ be the roots of $f_{2}(\lambda)=0$. Then

$$\begin{aligned}& x_{1}+ x_{2} + x_{3}+ x_{4}=n+5, \end{aligned}$$

(3.1)

$$\begin{aligned}& x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}=5n+ab+7, \end{aligned}$$

(3.2)

$$\begin{aligned}& x_{2}x_{3}x_{4}+x_{1}x_{3}x_{4}+x_{1}x_{2}x_{4}+x_{1}x_{2}x_{3}=7n+2ab+3. \end{aligned}$$

(3.3)

If

$$ x_{1}+ x_{2}=n+3, $$

(3.4)

then, by (3.1), we have

$$ x_{3}+ x_{4}=2. $$

(3.5)

From (3.2)-(3.5) it follows that

$$\begin{aligned}& x_{1}x_{2}+x_{3}x_{4}=3n+ab+1, \end{aligned}$$

(3.6)

$$\begin{aligned}& (n+3)x_{3}x_{4}+2x_{1}x_{2}=7n+2ab+3. \end{aligned}$$

(3.7)

By (3.6) and (3.7), we have

$$\begin{aligned}& x_{3}x_{4}=1. \end{aligned}$$

(3.8)

Combining (3.5) and (3.8), we have

$$ x_{3}=x_{4}=1. $$

(3.9)

Then $f_{2}(1)=-2ab=0$, which implies that $b=0$. Therefore, if $b \geq 1$, then $S_{2}(T_{a,b,0}) < m(T_{a,b,0})+3$. A direct calculation shows that $S_{2}(T_{n-3,0,0}) = m(T_{n-3,0,0})+3$. This completes the proof. □

Lemma 3.3

For $Q_{a,b}$, $S_{2}(Q_{a,b}) < m(Q_{a,b})+3$.

Proof

By a direct calculation, we have $\phi(Q_{a, b},\lambda)=\lambda(\lambda-1)^{n-6}f_{3}(\lambda)$, where $f_{3}(\lambda)=(\lambda-2)(\lambda^{4}-(n+4)\lambda ^{3}+(5n+ab+1)\lambda^{2}-(6n+2ab-2)\lambda+2n)$. Let $\lambda_{1} \geq\lambda_{2} \geq\lambda_{3} \geq\lambda_{4}\geq \lambda_{5}>0$ be the five roots of $f_{3}(\lambda)=0$. Then we have $\lambda_{1} + \lambda_{2} + \lambda_{3} + \lambda _{4}+\lambda_{5} =n+6$ since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. For $b \geq1$, $Q_{a,b}$ contains $Q_{1,1}$ as a subgraph. Then by Lemma 2.7 we have $\lambda_{3} \geq\mu_{3}(Q_{1,1})=2$ and $\lambda_{4} \geq\mu_{4}(Q_{1,1})=1.26$. Therefore, $S_{2}(Q_{a,b}) = \lambda_{1} + \lambda_{2} = n+6- \lambda _{3} - \lambda_{4} - \lambda_{5} < n+3$. In what follows, we assume $b=0$. Since $f_{3}(1)=0$, we can rewrite $\phi(Q_{a, 0},\lambda )=\lambda(\lambda-1)^{n-5}f_{4}(\lambda)$, where $f_{3}(\lambda )=(\lambda-1)f_{4}(\lambda)$. Let $\lambda_{1}' \geq\lambda_{2}' \geq\lambda_{3}' \geq\lambda _{4}'> 0$ be the four roots of $f_{4}(\lambda)=0$. Then we have $\lambda_{1}' + \lambda_{2}' + \lambda_{3}' + \lambda _{4}' = n+5$ since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. For $a \geq1$, $Q_{a, 0}$ contains $Q_{1,0}$ as a subgraph. Then by Lemma 2.7 we have $\lambda_{3}' \geq\mu_{3}(Q_{1,0})=2$. Thus, $S_{2}(Q_{a,b}) = \lambda_{1}' + \lambda_{2}'= n+5-\lambda_{3}' - \lambda_{4}' < n+3$. If $a=0$, then $S_{2}(Q_{a,b}) = S_{2}(C_{4}) < m(C_{4})+3$ by a straight calculation. This completes the proof. □

Now, we come to the main results of this paper.

Theorem 3.4

For any unicyclic graph G, $S_{2}(G) \leq m(G)+3$ with equality if and only if $G \cong T(n-3,0,0)$.

Proof

For any unicyclic graph G, we assume that $C_{k}=v_{1}v_{2} \cdots v_{k}v_{1}$ is the unique cycle in G (for some k) and G has the form $U(R_{1},\ldots, R_{k})$. For $k \geq5$, $G-v_{1}v_{2}$ is a tree with $d (G-v_{1}v_{2}) \geq 5$. Then by Lemma 3.1 we have $S_{2}(G) < m(G)+3$. We now consider the following two cases.

Case 1. $k=4$.

Let $C_{4}=v_{1}v_{2}v_{3}v_{4}v_{1}$ be the unique cycle in G. If there exist $e(v_{i}) \geq2$, say $e(v_{1})\geq2$, then $G-v_{1}v_{2}$ is a tree with $d(G-v_{1}v_{2}) \geq5$; if there are two adjacent vertices in $C_{4}=v_{1}v_{2}v_{3}v_{4}v_{1}$, say $v_{1}$ and $v_{2}$, such that $e(v_{1}) \geq1$ and $e(v_{2}) \geq1$, then $G-v_{1}v_{2}$ is a tree with $d (G-v_{1}v_{2}) \geq5$. Then by Lemma 3.1 we have $S_{2}(G) < m(G)+3$. We now assume $G \cong Q_{a, b}$ (see Figure 4). Then the result follows from Lemma 3.3.

Case 2. $k=3$.

If $\max \{e(v_{1}), e(v_{2}), e(v_{3}) \} \geq3$, say $e(v_{1})\geq 3$, then $G-v_{1}v_{2}$ is a tree with $d(G-v_{1}v_{2}) \geq5$; if there are two vertices in $C_{3}=v_{1}v_{2}v_{3}v_{1}$, say $v_{1}$ and $v_{2}$, such that $e(v_{1})=2$ and $e(v_{2}) \geq1$, then $G-v_{1}v_{2}$ is a tree with $d (G-v_{1}v_{2}) \geq5$. Therefore, Lemma 3.1 implies that $S_{2}(G) < m(G)+3$. We now assume $G \cong F_{1,t,n-2t-3}$ ($t \geq1$) or $G \cong T_{a,b,c}$. Then the result follows from Lemma 2.13 or Lemma 3.2.

This completes the proof. □