Bicyclic graphs with maximum sum of the two largest Laplacian eigenvalues

Abstract

Let G be a simple connected graph and $S_{2}(G)$ be the sum of the two largest Laplacian eigenvalues of G. In this paper, we determine the bicyclic graph with maximum $S_{2}(G)$ among all bicyclic graphs of order n, which confirms the conjecture of Guan et al. (J. Inequal. Appl. 2014:242, 2014) for the case of bicyclic graphs.

1 Introduction

Let $G=(V(G), E(G))$ be a simple connected graph with vertex set $V(G)=\{v_{1}, v_{2}, \ldots, v_{n}\}$. The numbers of its vertices and edges are denoted by $n(G)$ and $m(G)$ (or n and m for short). For a vertex $v \in V(G)$, let $N(v)$ be the set of all neighbors of v in G. The degree of v, denoted by $d(v)$, is the cardinality of $\vert N(v) \vert$, that is, $d(v)=\vert N(v) \vert$. A vertex with degree one is called pendant vertex. Particularly, denote by $\Delta(G)$ (or Δ for short) the maximum degree of G. The matrix $L(G)=D(G)-A(G)$ is called Laplacian matrix of G, where $A(G)$ is the adjacency matrix of G and $D(G)=\operatorname{diag}(d(v_{1}), d(v_{2}), \ldots, d(v_{n}))$ is the diagonal matrix of vertex degrees of G. 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 and so its eigenvalue are nonnegative real number. The eigenvalues of $L(G)$ are called the Laplacian eigenvalues of G and are denoted by $\mu_{1}(G) \geq\mu_{2}(G)\geq\cdots\geq\mu_{n}(G)$ (or $\mu _{1} \geq\mu_{2} \geq\cdots\geq\mu_{n}$ for short), which are always enumerated in non-increasing order and repeated according to their multiplicity. It is well known that $\sum_{i=1}^{i=n}\mu _{i}=2m(G)$. Note that each row sum of $L(G)$ is 0 and, therefore, $\mu_{n}(G)=0$. Fiedler [2] 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. The investigation of Laplacian spectra of graphs is an interesting topic on which much literature focused in the last two decades (see [2–9]).

Let $S_{k}(G)=\sum_{i=1}^{i=k}\mu_{i}(G)$ be the sum of the k largest Laplacian eigenvalues of G. Hamers in [10] mentions that Brouwer conjectured that $S_{k}(G) \leq m(G)+ \binom {k+1}{2}$ for $k=1,2,\ldots, n$. This conjecture is interesting and still open. Up to now, little progress on it has been made (see [1, 10–13]). When $k = 2$, Haemers et al. [10] proved the conjecture by showing $S_{2}(G) \leq m(G)+3$ for any graph G. Especially when G is a tree, Fritscher et al. [14] improve this bound by showing $S_{2}(T) \leq m(T)+3-\frac{2}{n(T)}$, which indicates that Haemers’ bound is always not attainable for trees. Therefore, it is interesting to determine which tree has maximum value of $S_{2}(T)$ among all trees of order n. 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. Guan et al. [1] 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 $T \cong T^{1}_{\lceil\frac {n-2}{2}\rceil, \lfloor\frac{n-2}{2} \rfloor}$. Note that $\mu_{1}(G)\leq n(G)$ for any graph G of order n. When $2n(G)< m(G)+3$, it follows that $S_{2}(G) \leq2n(G) < m(G)+3$, which means that Haemers’ bound is not attainable. When $2n(G) \geq m(G)+3$, Guan et al. [1] showed that $S_{2}(G_{m,n})=m(G_{m,n})+3$, where $G_{m,n}$ is a graph with n vertices and m edges which has $m-n+1$ triangles with a common edge and $2n-m-3$ pendant edges incident with one end vertex of the common edge (illustrated in Figure 1 are $G_{n,n}$ and $G_{n+1,n}$). This indicates that Haemers’ bound is always sharp for connected graphs $(n \leq m \leq2n-3)$. The following conjecture on the uniqueness of the extremal graph is also presented in [1].

Figure 1

$\pmb{G_{n,n}}$ and $\pmb{G_{n+1,n}}$ .

Conjecture 1.1

[1]

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

Zheng et al. [15] determined the uicyclic graph with maximum $S_{2}(G)$ among all uicyclic graphs of order n, which confirms Conjecture 1.1 for $m=n$.

Theorem 1.2

[15]

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

In this paper, we prove that Conjecture 1.1 holds for $m=n+1$. The main result is as follows.

Theorem 1.3

For any bicyclic graph G of order n, $S_{2}(G) \leq m(G)+3$ with equality if and only if $G \cong G_{n+1, n}$.

In Section 2 of this paper, we give some well known lemmas which are useful in the proof of Theorem 1.3. In Section 3, we will give the proof of Theorem 1.3.

2 Preliminaries

We first present some well-known results on $\mu_{1}(G)$.

Lemma 2.1

[3]

Let G be a connected graph of order n, then $\mu_{1}(G) \leq n(G)$ with equality if and only if the complement of G is disconnected.

Lemma 2.2

[8]

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

$$ \mu_{1}(G) \leq \max\bigl\{ d_{i}+m_{i}\mid v_{i} \in V(G)\bigr\} . $$

Lemma 2.3

[7]

Let G be a connected graph with $n \geq4$ vertices and m edges. Then

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

Let M be a real symmetric matrix of order n. Then all eigenvalues of M are real and can be denoted by $\lambda_{1}(M)\geq\lambda _{2}(M)\geq\cdots \geq\lambda_{n}(M)$ in non-increasing order. The following result in matrix theory plays a key role in our proofs.

Lemma 2.4

[16]

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

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

The next lemma follows from Lemma 2.4 immediately.

Lemma 2.5

Suppose $G_{1}, \ldots, G_{r}$ are edge disjoint graphs. Then, for any k,

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

The following lemma can be found in [17] and is known as the interlacing theorem of Laplacian eigenvalues.

Lemma 2.6

[17]

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

$$\begin{aligned} \mu_{1}(G) \geq\mu_{1}\bigl(G^{\prime}\bigr) \geq\mu _{2}(G) \geq\cdots\geq\mu_{n-1}\bigl(G^{\prime}\bigr) \geq\mu_{n}(G) \geq \mu_{n}\bigl(G^{\prime}\bigr)=0. \end{aligned}$$

Lemma 2.7

[17]

Let A be a real 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 $\lambda^{\prime}_{1} \geq\lambda^{\prime}_{2} \geq\cdots\geq \lambda^{\prime}_{m}$. Then the eigenvalues of B interlace the eigenvalues of A, that is, $\lambda_{i} \geq\lambda^{\prime}_{i} \geq\lambda_{n-m+i}$, 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 $L(G)$.

The multiplicities of an eigenvalue λ for $L(G)$ is denoted by $m_{G}(\lambda)$. For a graph G of order n, it is well known that $m_{G}(1)=n-r(I_{n}-L(G))$, where $r(I_{n}-L(G))$ is the rank of $I_{n}-L(G)$.

Lemma 2.8

Let v be a pendant vertex of graph G with n vertices and $G^{\prime}$ be the graph obtained from G by adding a new vertex $v^{\prime}$ which has a unique neighbor u, where u is the neighbor of v. (See Figure  2.) Then

$$\begin{aligned} m_{G^{\prime}}(1)= m_{G}(1)+1. \end{aligned}$$

Figure 2

Graphs $\pmb{G^{\prime}}$ and G .

Proof

Let $L(G)$ and $L(G^{\prime})$ be the Laplacian matrix of G and $G^{\prime}$, respectively. It is not difficult to check that $r(I_{n}-L(G))= r(I_{n+1}-L(G^{\prime }))$. Thus the result follows from the facts that $m_{G}(1)=n-r(I_{n}-L(G))$ and $m_{G^{\prime }}(1)=(n+1)-r(I_{n+1}-L(G^{\prime}))$. □

Let $P_{n}$ and $C_{n}$ be the path and cycle of order n, respectively. A connected graph with n vertices and $n+1$ edges is called a bicyclic graph. Let $\mathcal{B}_{n}$ be the set of bicyclic graphs of order n. There are two basic bicyclic graphs: ∞-graph and θ-graph. More concisely, an ∞-graph, denoted by $\infty(p,q,l)$-graph, is obtained from two vertex-disjoint cycles $C_{p}$ and $C_{q}$ by connecting one vertex of $C_{p}$ and one vertex of $C_{q}$ with a path $P_{l+1}$ of length l, where $q \geq p \geq3$ and $l\geq0$ (in the case of $l=0$, we identify the above two vertices). A θ-graph, denoted by $\theta(p,q,l)$, is a union of three internally disjoint paths $P_{p}$, $P_{q}$, $P_{l}$ of length $p-1$, $q-1$, $l-1$, respectively with common end vertices, where $l \geq q \geq p \geq2$ and at most one of them is 2. Observe that any bicyclic graph G is obtained from a basic bicyclic graph $\infty(p,q,l)$ or $\theta(p,q,l)$ by attaching trees to some of its vertices. For any bicyclic graph G, we call its basic bicyclic graph $\infty(p,q,l)$ or $\theta(p,q,l)$ the kernel of G. For a vertex v of the kernel of G, if there is a tree $T_{v}$ attached to it, we denote by $e(v)$ the maximum distance between v and any vertex of $T_{v}$ (that is, $e(v)=\max\{d(u,v)\mid u \in V(T_{v})\}$); if there is no tree attached to it, we define $e(v)=0$. Let $\mathcal{B}^{\infty}_{n}(p,q,l)$ and $\mathcal{B}^{\theta }_{n}(p,q,l)$ be the sets of bicyclic graphs of order n with $\infty(p,q,l)$ and $\theta(p,q,l)$ as their kernel, respectively. Clearly, $\mathcal{B}_{n}=\mathcal{B}^{\infty}_{n}(p,q,l)\cup\mathcal{B}^{\theta}_{n}(p,q,l)$. Let S be a multiset of some nonnegative integers, denote by $\Vert S \Vert$ the number of nonzero elements in S, that is, $\Vert S \Vert = \vert \{a \in S \mid a \geq1 \} \vert$, where elements are counted according to their multiplicity.

Lemma 2.9

Let G be the union of some disjoint graphs $G_{1}, G_{2}, \ldots, G_{r}$, where $G_{i}$ $(i \in\{1,\ldots,r \})$ is a tree or an unicyclic graph of order $n_{i}$ which is not isomorphic to $G_{n_{i},n_{i}}$ ($G_{n, n}$ is the unicyclic graph shown in Figure  1). Then $S_{2}(G)< m(G)+3$.

Proof

By Theorem 1.2 and the fact $S_{2}(T) < m(T)+3$ for any tree, we have $S_{2}(G_{i}) < m(G_{i})+3$ for $i \in\{1, \ldots, r \}$. If $\mu_{1}$ and $\mu_{2}$ of G attain the same component, say $G_{i_{0}}$ $(1 \leq i_{0} \leq r)$, then $S_{2}(G) = S_{2}(G_{i_{0}})< m(G_{i_{0}})+3 \leq m(G)+3$. Otherwise, $\mu_{1}$ and $\mu_{2}$ of G attain two different components, without loss of generality, we assume that $\mu_{1}$ and $\mu_{2}$ attain $G_{1}$ and $G_{2}$, respectively, that is, $\mu_{1}=\mu _{1}(G_{1})$ and $\mu_{2}=\mu_{1}(G_{2})$. Then $S_{2}(G)= \mu _{1}(G_{1})+ \mu_{1}(G_{2}) \leq n(G_{1})+ n(G_{2}) \leq(m(G_{1})+1)+ (m(G_{2})+1) \leq m(G)+2 < m(G)+3$. □

For any subgraph H of G, let $G-H$ be the graph obtained from G by deleting all edges from H.

Lemma 2.10

Let G be a simple graph with at least 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that $G-4P_{2}= \bigcup G_{i}$ (or $G-3P_{3}= \bigcup G_{i}$), where $G_{i}$ is a tree or an unicyclic graph of order $n_{i}$ which is not isomorphic to $G_{n_{i},n_{i}}$ ($G_{n, n}$ is the unicyclic graph shown in Figure  1). Then $S_{2}(G)< m(G)+3$.

Proof

By direct calculation, we have $S_{2}(4P_{2})=m(4P_{2})=4$ and $S_{2}(3P_{3})=m(3P_{3})=6$. Using Lemmas 2.5 and 2.9, we have $S_{2}(G) \leq S_{2}(G-4P_{2})+S_{2}(4P_{2}) < (m(G-4P_{2})+3)+m(4P_{2}) = m(G)+3$ (or $S_{2}(G) \leq S_{2}(G-3P_{3})+S_{2}(3P_{3}) < (m(G-3P_{3})+3)+m(3P_{3}) = m(G)+3$), this completes the proof. □

3 Proof of Theorem 1.3

First of all, using the facts that $S_{2}(G) \leq m(G)+3$ for any graph G and that $\mathcal{B}_{n}=\mathcal{B}^{\infty}_{n}(p,q,l)\cup \mathcal{B}^{\theta}_{n}(p,q,l)$, we give the main idea of the proof.

• 1. For each class, we show that $S_{2}(G) < m(G)+3$ for the majority of graphs of $\mathcal{B}^{\infty}_{n}(p,q,l)$ or $\mathcal {B}^{\theta}_{n}(p,q,l)$ by Lemma 2.10.

• 2. For the remaining graphs in $\mathcal{B}^{\infty}_{n}(p,q,l)$ or $\mathcal{B}^{\theta}_{n}(p,q,l)$ (except for $G_{n+1, n}$), we prove that $S_{2}(G) < m(G)+3$ by discussing case by case.

• 3. We show that $S_{2}(G_{n+1, n}) = m(G_{n+1, n})+3$, that is, the condition that equality holds.

Next, we discuss according to the following two subsections.

3.1 Bicyclic graphs in $\mathcal{B}^{\infty}_{n}(p,q,l)$

In this subsection, we prove that $S_{2}(G) < m(G)+3$ for $G\in \mathcal{B}^{\infty}_{n}(p,q,l)$.

Lemma 3.1

Let $G \in\mathcal{B}^{\infty}_{n}(p,q,l)$. If $(p,q,l)$ satisfies one of the following conditions:

• (1) $l\geq1$,

• (2) $(p,q,l)=(p,q,0)$ and $q \geq p \geq4$,

• (3) $(p,q,l)=(3,q,0)$ and $q \geq5$,

then $S_{2}(G)< m(G)+3$.

Proof

Direct calculation shows that $S_{2}(C_{p} \cup C_{q}) \leq m(C_{p} \cup C_{q})$, where $q \geq p \geq3$. For (1) $l \geq1$, using Lemmas 2.5 and 2.9, we have $S_{2}(G) \leq S_{2}(G-C_{p}-C_{q})+ S_{2}(C_{p} \cup C_{q}) < (m(G-C_{p}-C_{q})+3)+m(C_{p} \cup C_{q})=m(G)+3$, since each component of $G-C_{p}-C_{q}$ is tree. For both (2) and (3), it is not difficult to check that G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree. Thus the result follows from Lemma 2.10. □

By Lemma 3.1, it suffices to consider the following two cases: $G\in\mathcal{B}^{\infty}_{n}(3,4,0)$ and $G\in\mathcal{B}^{\infty }_{n}(3,3,0)$.

Lemma 3.2

For $G\in\mathcal{B}^{\infty}_{n}(3,4,0)$, we have $S_{2}(G)< m(G)+3$.

Proof

For $G\in\mathcal{B}^{\infty}_{n}(3,4,0)$, its kernel is $\infty(3,4,0)$, denoted by $G^{0}_{1}$ for short shown in Figure 3. Let $V(G^{0}_{1})= \{v_{1},\ldots,v_{6}\}$. If there exists a vertex $v_{i} \in V(G^{0}_{1})$ such that $e(v_{i}) \geq2$, then G has 4 vertex-disjoint $P_{2}$ such that each component of $G-4P_{2}$ is a tree and the result follows from Lemma 2.10. Thus it suffices to consider the case that G is isomorphic to $G_{1}$ shown in Figure 3.

Figure 3

Some bicyclic graphs with $\pmb{\infty(3, 4, 0)}$ as their kernel.

When $\Vert \{n_{1},n_{2},\ldots, n_{6}\} \Vert \geq3$, if G is not isomorphic to $G^{1}_{1}$ shown in Figure 3, then G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree or a unicyclic graph of order $n_{i}$ which is not isomorphic to $G_{n_{i},n_{i}}$. Thus the result follows from Lemma 2.10. If G is isomorphic to $G^{1}_{1}$, then we have $m_{G}(1) \geq n-9$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{8} > \lambda_{9}=0$ be the other night eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{9} =n+11$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda_{4}+\lambda _{5} \geq\mu_{3}(G^{2}_{1})+\mu_{4}(G^{2}_{1})+\mu _{5}(G^{2}_{1})=3.22+2.22+2.00>7$, since $G^{1}_{1}$ contains $G^{2}_{1}$ shown in Figure 3 as a subgraph. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

If $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert =2$, then we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$.

When $n_{3}=0$, G contains a subgraph isomorphic to $G^{i}_{1}$ $(i \in\{3,4,5\})$ shown in Figure 3. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(G^{i}_{1})+\mu_{4}(G^{i}_{1})+\mu_{5}(G^{i}_{1}) > 6$ for $i \in\{3,4,5\}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

When $n_{3} \geq1$, it can be checked that $m_{G}(1) \geq n-7$ by direct calculation and Lemma 2.8. Let $\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7}=n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda _{4} \geq\mu_{3}(G^{0}_{1})+\mu_{4}(G^{0}_{1})=5$, since $G_{1}$ contains $G^{0}_{1}$ as a subgraph. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

If $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert =1$, then we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda _{2} + \cdots+\lambda_{7}=n+9$, since $\sum_{i=1}^{i=n}\mu _{i}=2m(G)$. By Lemma 2.6 and direct calculation, we have $\lambda_{3}+\lambda_{4} \geq\mu_{3}(G^{0}_{1})+\mu _{4}(G^{0}_{1})=5$, since G contains $G^{0}_{1}$ as a subgraph. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

If $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert =0$, then G is isomorphic to $G^{0}_{1}$. Direct calculation shows that $S_{2}(G)< m(G)+3$.

The proof is completed. □

Lemma 3.3

For $G\in\mathcal{B}^{\infty}_{n}(3, 3, 0)$, we have $S_{2}(G)< m(G)+3$.

Proof

For $G\in\mathcal{B}^{\infty}_{n}(3,3,0)$, its kernel is $\infty(3,3,0)$, denoted by $G^{0}_{2}$ for short shown in Figure 4. Let $V(G^{0}_{2})= \{v_{1},\ldots,v_{5}\}$. If there exists a vertex $v_{i} \in V(G^{0}_{2})$ such that $e(v_{i}) \geq2$, then it suffices to consider the case that G is isomorphic to $G^{1}_{2}$ or $G^{2}_{2}$ shown in Figure 4, otherwise G has 4 vertex-disjoint $P_{2}$ such that each component of $G-4P_{2}$ is a tree and the result follows from Lemma 2.10.

Figure 4

Some bicyclic graphs with $\pmb{\infty(3, 3, 0)}$ as their kernel.

When G is isomorphic to $G^{i}_{2}$ $(i \in\{1,2\})$, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations we have $\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{i}_{2})+\mu _{4}(H^{i}_{2})+\mu_{5}(H^{i}_{2})>6$, where $H^{i}_{2}$ $(i \in\{ 1,2\})$ is a subgraph of $G^{i}_{2}$ shown in Figure 4. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

Now, it suffices to consider the case that G is isomorphic to $G_{2}$ shown in Figure 4.

When $\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert \geq3$, if G is isomorphic to $G^{3}_{2}$ shown in Figure 4, then G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree and the result follows from Lemma 2.10. If G is isomorphic to $G^{3}_{2}$, we have $m_{G}(1) \geq n-7$ by direct calculation and Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. We have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations we have $\lambda_{3}+\lambda_{4} \geq\mu _{3}(H^{3}_{2})+\mu_{4}(H^{3}_{2})=3.00+2.33 > 5$, where $H^{3}_{2}$ is a subgraph of G shown in Figure 4. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

When $\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert = 2$, it suffices to consider the case that G is isomorphic to $G^{i}_{2}$ $(i \in\{4,5,6\})$ shown in Figure 4. When G is isomorphic to $G^{4}_{2}$, we have $m_{G}(1) \geq n-6$ by direct calculation and Lemma 2.8. Let $\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$ be the other six eigenvalues of G. We have $\lambda_{1} + \lambda_{2} + \cdots +\lambda_{6} =n+8$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations we have $\lambda_{3}+\lambda_{4} \geq\mu_{3}(H^{4}_{2})+\mu _{4}(H^{4}_{2})=3.00+1.51 > 4$, where $H^{4}_{2}$ shown in Figure 4 is a subgraph of G. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$. When G is isomorphic to $G^{i}_{2}$ $(i \in\{5,6\})$, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq \cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations we have $\lambda _{3}+\lambda_{4} \geq\mu_{3}(H^{i}_{2})+\mu_{4}(H^{i}_{2})> 5$, where $H^{i}_{2}$ shown in Figure 4 is a subgraph of $G^{i}_{2}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

If $\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert =1$, then it suffices to consider the case that G is isomorphic to $G^{7}_{2}$ or $G^{8}_{2}$ shown in Figure 4. When G is isomorphic to $G^{7}_{2}$, we have $m_{G}(1) \geq n-6$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{5} > \lambda_{6}=0$ be the other six eigenvalues of G. We have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations we have $\lambda_{3}+\lambda_{4} \geq\mu_{3}(H^{7}_{2})+\mu _{4}(H^{7}_{2})=3.00+1.4 > 4$, where $H^{7}_{2}$ shown in Figure 4 is a subgraph of $G^{7}_{2}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$. When G is isomorphic to $G^{8}_{2}$, by an elementary calculation, we have $\phi (G^{8}_{2},\lambda)= \lambda(\lambda-1)^{n-4}(\lambda -3)^{2}(\lambda-n)$. It follows that $S_{2}(G^{8}_{2})=n+3 < m+3$.

If $\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert =0$, then G is isomorphic to $G^{0}_{2}$. Direct calculation shows that $S_{2}(G)< m(G)+3$. □

We safely come to the following result by the above discussion.

Theorem 3.4

For $G\in\mathcal{B}^{\infty}_{n}(p,q,l)$, where $q \geq p \geq3$ and $l \geq0$, we have $S_{2}(G) < m(G)+3$.

3.2 Bicyclic graphs in $\mathcal{B}^{\theta}_{n}(p,q,l)$

In this subsection, we prove that $S_{2}(G) \leq m(G)+3$ for $G\in \mathcal{B}^{\theta}_{n}(p,q,l)$ and equality holds if and only if $G\cong G_{n+1,n}$. We begin with the following lemma, which follows from Lemma 2.10 immediately.

Lemma 3.5

For $G\in\mathcal{B}^{\theta}_{n}(p,q,l)$, if $(p,q,l)$ satisfies one of the following conditions:

• (1) $4 \leq p \leq q \leq l$,

• (2) $(p,q,l)=(3, q, l)$ and $4 \leq q \leq l$,

• (3) $(p,q,l)=(3, 3, l)$ and $5 \leq l$,

then $S_{2}(G)< m(G)+3$.

By Lemma 3.5, it suffices to consider the following three cases: (1) $G\in\mathcal{B}^{\theta}_{n}(3,3,4)$, (2) $G\in\mathcal {B}^{\theta}_{n}(3,3,3)$, and (3) $G\in\mathcal{B}^{\theta }_{n}(2,q,l)$, where $l \geq q \geq3$.

First, we consider the case $G\in\mathcal{B}^{\theta}_{n}(3,3,4)$.

Lemma 3.6

For $G\in\mathcal{B}^{\theta}_{n}(3,3,4)$, we have $S_{2}(G)< m(G)+3$.

Proof

For $G\in\mathcal{B}^{\theta}_{n}(3,3,4)$, its kernel is $\theta(3,3,4)$, denoted by $G^{0}_{3}$ for short shown in Figure 5. Let $V(G^{0}_{3})= \{v_{1},\ldots,v_{6}\}$. If there exists a vertex $v_{i}$ of $G^{0}_{3}$ such that $e(v_{i}) \geq2$, then G has 4 vertex-disjoint $P_{2}$ such that each component of $G-4P_{2}$ is a tree and the result follows from Lemma 2.10. Thus it suffices to consider the case that G is isomorphic to $G_{3}$ shown in Figure 5. For $n(G) \leq8$, it is easy to check that $S_{2}(G) < m(G)+3$ by a direct calculation. Thus we assume that $n(G) \geq9$ in the following.

Figure 5

Some bicyclic graphs with $\pmb{\theta(3,3,4)}$ as their kernel.

If $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \geq3$, then G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree and the result follows from Lemma 2.10.

If $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \leq2$, then we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. When G is not isomorphic to $G_{3}^{1}$ shown in Figure 5, it contains a subgraph which is isomorphic to $G^{i}_{3}$ $(i \in\{2,3,4\})$ shown in Figure 5. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda_{4}+\lambda_{5} \geq \lambda_{3}(G^{i}_{3})+\lambda_{4}(G^{i}_{3})+\lambda_{5}(G^{i}_{3}) > 6$ for $i \in\{2,3,4\}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$. When G is isomorphic to $G_{3}^{1}$, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} = n+9$, since $\sum^{i=n}_{i=1}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(G^{0}_{3})+\mu _{4}(G^{0}_{3})+\mu_{5}(G^{0}_{3})=2.00+2.00+1.26 > 5$, since G contains $G^{0}_{3}$ as a subgraph. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

The proof is completed. □

Second, we consider the case $G\in\mathcal{B}^{\theta}_{n}(3,3,3)$.

Lemma 3.7

For $G\in\mathcal{B}^{\theta}_{n}(3,3,3)$, we have $S_{2}(G)< m(G)+3$.

Proof

For $G\in\mathcal{B}^{\theta}_{n}(3,3,3)$, its kernel is $\theta(3,3,3)$, denoted by $G^{0}_{4}$ for short shown in Figure 6. Let $V(G^{0}_{4})= \{v_{1},\ldots,v_{5}\}$. If there exists a vertex $v_{i} \in V(G^{0}_{4})$ such that $e(v_{i}) \geq2$, then by Lemma 2.10, it suffices to consider the case that G is isomorphic to $G^{1}_{4}$ or $G^{2}_{4}$ shown in Figure 6.

Figure 6

Some bicyclic graphs with $\pmb{\theta(3,3,3)}$ as their kernel.

When G is isomorphic to $G^{1}_{4}$, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} = n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculation, we have $\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{1}_{4})+\mu _{4}(H^{1}_{4})+\mu_{5}(H^{1}_{4})=3.00+2.00+2.00 > 5$, where $H^{1}_{4}$ shown in Figure 6 is a subgraph of $G^{1}_{4}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

When G is isomorphic to $G^{2}_{4}$, we have $m_{G}(1) \geq n-9$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{8} > \lambda_{9}=0$ be the other nine eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{9} = n+11$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. If $n_{4} \geq1$ and $n_{5} \geq1$, then $G^{2}_{4}$ contains $H^{2}_{4}$ shown in Figure 6 as a subgraph. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}+\lambda_{5}+\lambda_{6} \geq\mu _{3}(H^{2}_{4})+\mu_{4}(H^{2}_{4})+\mu_{5}(H^{2}_{4})+\mu _{6}(H^{2}_{4})=2.47+2.00+2.00+1.22 > 7$. Therefore $S_{2}(G)=\lambda _{1} + \lambda_{2} < n+4 = m(G)+3$. If $n_{4} = 0$ or $n_{5} = 0$, then we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} = n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculation, we have $\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{3}_{4})+\mu _{4}(H^{3}_{4})+\mu_{5}(H^{3}_{4})=2.32+2.00+2.00 > 6$, where $H^{3}_{4}$ shown in Figure 6 is a subgraph of $G^{2}_{4}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

Now, we consider the case that G is isomorphic to $G_{4}$ shown in Figure 6.

First, we consider the case that $\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert \geq3$. If G is not isomorphic to $G^{4}_{4}$ shown in Figure 6, then G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree and the result follows from Lemma 2.10. When G is isomorphic to $G^{4}_{4}$, we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculation, we have $\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{4}_{4})+\mu _{4}(H^{4}_{4})+\mu_{5}(H^{4}_{4}) > 3.08+2.00+1.29 > 6$, where $H_{4}^{4}$ shown in Figure 6 is a subgraph of $G_{4}^{4}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

When $\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert = 2$, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. Then by Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4} \geq\mu_{3}( H^{i}_{4})+\mu_{4}(H^{i}_{4}) \geq5$ $(i \in\{5, 6\})$, where $H_{4}^{i}$ $(i \in\{5, 6\})$ shown in Figure 6 is a subgraph of G. Therefore $S_{2}(G)=\lambda _{1} + \lambda_{2}< n+4 = m(G)+3$.

When $\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert = 1$, we have $m_{G}(1) \geq n-6$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$ be the other six eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculation, we have $\lambda _{3}+\lambda_{4} \geq\mu_{3}(G^{0}_{4})+\mu _{4}(G^{0}_{4})=2.00+2.00=4$, since G contains $G^{0}_{4}$ as a subgraph. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$, as required.

When $\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert = 0$, G is isomorphic to $G^{0}_{4}$. Direct calculation shows that $S_{2}(G)< m(G)+3$.

From the above discussion, we complete the proof. □

Finally, we consider the case $G\in\mathcal{B}^{\theta}_{n}(2,q,l)$, where $l \geq q \geq3$.

Lemma 3.8

If $G\in\mathcal{B}^{\theta}_{n}(2,q,l)$ with $l \geq q \geq4$, then $S_{2}(G)< m(G)+3$.

Proof

If $q \geq5$, it is obviously that G has 4 vertex-disjoint $P_{2}$ such that $G-4P_{2}$ is a forest and the result follows immediately from Lemma 2.10. When $q = 4$ and $l \geq 5$, the result follows immediately from Lemma 2.10, since G has 4 vertex-disjoint $P_{2}$ such that $G-4P_{2}$ is a forest. Thus it suffices to consider the case that $G\in\mathcal{B}^{\theta}_{n}(2,4,4)$.

For $G\in\mathcal{B}^{\theta}_{n}(2,4,4)$, its kernel is $\theta (2,4,4)$, denoted by $G^{0}_{5}$ for short shown in Figure 7. Let $V(G^{0}_{5})=\{v_{1},\ldots,v_{6} \}$. If there exists a vertex $v_{i} \in V(G^{0}_{5})$ such that $e(v_{i}) \geq2$, then G has 4 vertex-disjoint $P_{2}$ such that each component of $G-4P_{2}$ is a tree and the result follows from Lemma 2.10. Now, it suffices to consider the case that G is isomorphic to $G_{5}$ shown in Figure 7.

Figure 7

Some bicyclic graphs with $\pmb{\theta(2,4,4)}$ as their kernel.

When $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \geq3$, G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree and the result follows from Lemma 2.10.

When $1 \leq \Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \leq2$, we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. If G is not isomorphic to $G^{1}_{5}$ shown in Figure 7, then G contains a subgraph which is isomorphic to $G^{2}_{5}$ shown in Figure 7. By Lemma 2.6 and direct calculation, we have $\lambda_{3}+\lambda_{4}+\lambda_{5} \geq\mu _{3}(G^{2}_{5})+\mu_{4}(G^{2}_{5})+\mu_{5}(G^{2}_{5})=3.00+2.22+1.38 > 6$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$. When G is isomorphic to $G^{1}_{5}$, we have $m_{G}(1)=n-6$ by direct calculation and Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$ be the other six eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculation, we have $\lambda _{3}+\lambda_{4} \geq\mu_{3}(G^{0}_{5})+\mu _{4}(G^{0}_{5})=3.00+2.00> 4$, since $G^{1}_{5}$ contains $G^{0}_{5}$ as a subgraph. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

When $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert = 0$, G is isomorphic to $G^{0}_{5}$. Direct calculation shows that $S_{2}(G) < m(G)+3$.

The proof is completed. □

Lemma 3.9

If $G\in\mathcal{B}^{\theta}_{n}(2,3,l)$ with $l \geq5$, then $S_{2}(G)< m(G)+3$.

Proof

When $l \geq6$, it is obviously that G has 4 vertex-disjoint $P_{2}$ such that $G-4P_{2}$ is a forest and the result follows immediately from Lemma 2.10. Thus it suffices to consider the case that $G\in\mathcal{B}^{\theta }_{n}(2,3,5)$. For $n(G) \leq7$, it is easy to check that $S_{2}(G) < m(G)+3$ by a direct calculation. Thus we assume that $n(G) \geq8$ in the following.

For $G\in\mathcal{B}^{\theta}_{n}(2,3,5)$, its kernel is $\theta (2,3,5)$, denoted by $G^{0}_{6}$ for short shown in Figure 7. Let $V(G^{0}_{6})=\{ v_{1},\ldots,v_{6}\}$. If there exists a vertex $v_{i}$ of $G^{0}_{6}$ such that $e(v_{i}) \geq2$, then G has 4 vertex-disjoint $P_{2}$ such that $G-4P_{2}$ is a forest and the result follows from Lemma 2.10. Now, we can assume that G is isomorphic to $G_{6}$ shown in Figure 7.

If $\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \geq3$, then G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree and the result follows from Lemma 2.10.

If $\Vert \{n_{1},n_{2},\ldots, n_{6}\} \Vert \leq2$, then we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of G. We have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. Note that G contains a subgraph which is isomorphic to $G^{i}_{6}$ $(i \in\{1,2,3,4,5 \})$ shown in Figure 8. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(G^{i}_{6})+\mu _{4}(G^{i}_{6})+\mu_{5}(G^{i}_{6}) > 6$ $(i \in\{1,2,3,4,5 \})$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

Figure 8

Some bicyclic graphs with $\pmb{\theta(2,3,5)}$ as their kernel.

The proof is completed. □

Lemma 3.10

For $G\in\mathcal{B}^{\theta}_{n}(2,3,4)$, we have $S_{2}(G)< m(G)+3$.

Proof

For $G\in\mathcal{B}^{\theta}_{n}(2,3,4)$, its kernel is $\theta(2,3,4)$, denoted by $G^{0}_{7}$ for short shown in Figure 9. Let $V(G^{0}_{7})= \{v_{1},\ldots,v_{5}\}$.

Figure 9

Some bicyclic graphs with $\pmb{\theta(2,3,4)}$ as their kernel.

If there exists a vertex $v_{i}$ such that $e(v_{i}) \geq2$, then by Lemma 2.10, it suffices to consider the cases that G is isomorphic to $G^{i}_{7}$ $(i \in\{1,2,3\})$ shown in Figure 9.

When G is isomorphic to $G^{i}_{7}$ $(i \in\{1,2,3\})$ shown in Figure 9, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. We have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{i}_{7})+\mu _{4}(H^{i}_{7})+\mu_{5}(H^{i}_{7}) > 6$, where $H^{i}_{7}$ shown in Figure 9 is a subgraph of $G^{i}_{7}$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$.

Now, it suffices to consider the case that G is isomorphic to $G_{7}$ shown in Figure 9. For $n(G) \leq8$, it is easy to check that $S_{2}(G) < m(G)+3$ by a direct calculation. In the following, we assume that $n(G) \geq9$.

If $\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert \geq3$, then G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree and the result follows from Lemma 2.10.

If $\Vert \{n_{1},n_{2}, \ldots,n_{5}\} \Vert = 2$, then we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. If G is not isomorphic to $G^{4}_{7}$ shown in Figure 9, then it contains a subgraph isomorphic to $H^{i}_{7}$ $(i \in\{5,\ldots, 10\})$ shown in Figure 9. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda_{4}\geq\mu _{3}(G^{i}_{7})+\mu_{4}(G^{i}_{7}) > 5$ $(i \in\{5,\ldots, 10\})$. Therefore $S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$. When G is isomorphic to $G^{4}_{7}$, we have $m_{G}(1) \geq n-6$ by direct calculation and Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$ be the other six eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots +\lambda_{6} =n+8$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda _{4} \geq\mu_{3}(H^{4}_{7})+\mu_{4}(H^{4}_{7})> 4$, where $H^{4}_{7}$ shown in Figure 9 is a subgraph of $G^{4}_{7}$. Therefore $S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

If $\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert = 1$, then we have $m_{G}(1) \geq n-6$ by Lemma 2.8. Let $\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$ be the other six eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}\geq\mu_{3}(H^{i}_{7})+\mu_{4}(H^{i}_{7}) > 4$ $(i\in\{11, 12, 13\})$, where $H^{i}_{7}$ shown in Figure 9 is a subgraph of G. Therefore $S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

The proof is completed. □

Lemma 3.11

Let $G\in\mathcal{B}^{\theta}_{n}(2,3,3)$ and $G^{0}$ shown in Figure  10 be the kernel of G. If there exists a vertex $v_{i}$ of $G^{0}$ such that $e(v_{i})\geq3$, then $S_{2}(G)< m(G)+3$.

Figure 10

Some bicyclic graphs with $\pmb{\theta(2,3,3)}$ as their kernel.

Proof

By Lemma 2.10, it suffices to consider the case that G is isomorphic to $G_{i}$ $(i \in\{8,9,10\})$ shown in Figure 10.

When G is isomorphic $G_{i}$ $(i \in\{8,9,10\})$, we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq \cdots\geq\lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of $G_{i}$. We have $\lambda_{1} + \lambda_{2} + \cdots +\lambda_{8} =n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}+\lambda_{5}\geq\mu_{3}(G^{\prime}_{i})+\mu _{4}(G^{\prime}_{i}) +\mu_{5}(G^{\prime}_{i}) > 6$, where $G^{\prime}_{i}$ shown in Figure 10 is the subgraph of $G_{i}$. Therefore $S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

The proof is completed. □

Lemma 3.12

Let $G\in\mathcal{B}^{\theta}_{n}(2,3,3)$ and $G^{0}$ shown in Figure  10 be the kernel of G. If $\max_{i=1}^{i=4} e(v_{i}) = 2$, then $S_{2}(G)< m(G)+3$, where $v_{i}$ is the vertices of $G^{0}$.

Proof

By Lemma 2.10, it suffices to consider the case that G is isomorphic to $G_{11}$ or $G_{12}$ shown in Figure 11. Here, we only prove the case that G is isomorphic to $G_{11}$. For the case G is isomorphic to $G_{12}$, we can discuss similarly.

Figure 11

Some bicyclic graphs with $\pmb{\theta(2,3,3)}$ as their kernel.

When $\Vert \{n_{1}, n_{2}, n_{3}, n_{4}\} \Vert \geq2$, G has 4 vertex-disjoint $P_{2}$ (or 3 vertex-disjoint $P_{3}$) such that each component of $G-4P_{2}$ (or $G-3P_{3}$) is a tree and the result follows from Lemma 2.10.

When $\Vert \{n_{1}, n_{2}, n_{3}, n_{4}\} \Vert \leq1$, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. We have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda _{4}\geq\mu_{3}(G^{\prime}_{11})+\mu_{4}(G^{\prime }_{11})=3.00+2.00 = 5$, where $G^{\prime}_{11}$ shown in Figure 11 is a subgraph of $G_{11}$. Thus $S_{2}(G_{11}) < m(G_{11})+3$. □

Lemma 3.13

Let $G_{13}$ be the bicyclic graph of order n shown in Figure  12, where $a \geq b \geq1$ and $a+b+5=n$. Then $S_{2}(G_{13}) < m(G_{13})+3$.

Figure 12

Some bicyclic graphs with $\pmb{\theta(2,3,3)}$ as their kernel.

Proof

For $n(G_{13}) \leq14$, it is easy to check that $S_{2}(G_{13}) < m(G_{13})+3$ by a direct calculation. In following, we assume that $n(G_{13}) \geq15$.

If $b \geq5$, then $G_{13}$ contains $G^{\prime}_{13}$ shown in Figure 12 as a subgraph. Note that $m_{G_{13}}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots \geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of $G_{13}$. We have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}\geq\mu_{3}(G^{\prime}_{13})+\mu_{4}(G^{\prime }_{13})=3.197+1.810 > 5$. Therefore $S_{2}(G_{13})= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

If $b = 4$, then we have $\lambda_{1}(G_{13})< \triangle+2 =n-4$ by Lemma 2.3 and $\lambda_{2}(G_{13})< \lambda _{1}(L_{v_{1}}(G_{13}))=7.95$ by Lemma 2.7 and direct calculation. It follows that $S_{2}(G_{13})< n+4= m(G)+3$.

If $b = 3$, then we have $\lambda_{1}(G_{13})< \triangle+2=n-3$ by Lemma 2.3 and $\lambda_{2}(G_{13})< \lambda _{1}(L_{v_{3}}(G_{13}))=6.96$ by Lemma 2.7 and direct calculation. It follows that $S_{2}(G_{13})< n+4= m(G)+3$.

If $b =2$, then we have $\lambda_{1}(G_{13})<(n-4)+\frac{n+3}{n-4}$ by Lemma 2.2 and $\lambda_{2}(G_{3})< \lambda _{1}(L_{v_{1}}(G_{13}))=6.005<6.10$ by Lemma 2.7 and direct calculation. Therefore $S_{2}(G_{13})< n-4+\frac{n+3}{n-4}+6.10 < n+4= m(G)+3$.

If $b =1$, then we have $\lambda_{1}(G_{13})<(n-3)+\frac{n+3}{n-4}$ by Lemma 2.2 and $\lambda_{2}(G_{13})< \lambda _{1}(L_{v_{1}}(G_{13}))=5.10<5.20$ by Lemma 2.7 and direct calculation. Thus $S_{2}(G_{13})< n-3+\frac{n+3}{n-4}+5.10 < n+4= m(G)+3$. □

Lemma 3.14

Let $G_{14}$ be the bicyclic graph of order n shown in Figure  12, where $a \geq b \geq1$ and $a+b+4=n$. Then $S_{2}(G_{14}) < m+3$.

Proof

By some elementary calculations, we have $\phi(G_{14}, x)= x(x-2)(x-1)^{6}g(x)$, where $g(x)= x^{4}-(n+6)x^{3}+(6n+ab+9)x^{2}-(9n+2ab+4)x+4n$. Let $x_{1}\geq x_{2} \geq x_{3}\geq x_{4}$ be the roots of $g(x)=0$. Then

$$\begin{aligned}& x_{1}+ x_{2} + x_{3}+ x_{4}=n+6, \end{aligned}$$

(3.1)

$$\begin{aligned}& x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}=6n+ab+9, \end{aligned}$$

(3.2)

$$\begin{aligned}& x_{2}x_{3}x_{4}+x_{1}x_{3}x_{4}+x_{1}x_{2}x_{4}+x_{1}x_{2}x_{3}=9n+2ab+4. \end{aligned}$$

(3.3)

If

$$ x_{1}+ x_{2}=n+4 $$

(3.4)

then, by (3.1), we have

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

(3.5)

From (3.2)-(3.5) follows that

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

(3.6)

$$\begin{aligned}& (n+4)x_{3}x_{4}+2x_{1}x_{2}=9n+2ab+4. \end{aligned}$$

(3.7)

By (3.6) and (3.7), we have

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

(3.8)

Combining (3.5) and (3.8), we have

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

(3.9)

Then $g(1)=-2ab=0$, which implies that $b=0$. Therefore, $b \geq1$ implies that $S_{2}(G_{14}) < m(G_{14})+3$. □

Lemma 3.15

Let G be the bicyclic graph of order n shown in Figure  13, where $a \geq b \geq0$, $c \geq d \geq0$ and $a+b+c+d+4=n$. Then $S_{2}(G) \leq m+3$ with equality if and only if $a=n-4$ and $b=c=d=0$ (that is, $G \cong G_{n+1,n}$).

Figure 13

Some bicyclic graphs with $\pmb{\theta(2,3,3)}$ as their kernel.

Proof

For $n(G) \leq12$, it is easy to check that $S_{2}(G) \leq m(G)+3$ with equality if and only if $G \cong G_{n+1, n}$ by a direct calculation. In the following, we assume that $n(G) \geq13$.

When $\Vert \{a,b,c,d\} \Vert = 4$, we have $m_{G}(1) \geq n-8$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots \geq\lambda_{7} > \lambda_{8}=0$ be the other eight eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$.

If $c \geq2$, then G contains $H_{1}$ shown in Figure 12 as a subgraph. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}\geq\mu_{3}(H_{1})+\mu_{4}(H_{1})=3.94+2.14 > 6$. If $c=d=1$ and $b \geq2$, then G contains $H_{2}$ shown in Figure 12 as a subgraph. By Lemma 2.6 and direct calculations, we have $\lambda_{3}+\lambda_{4}\geq\mu _{3}(H_{2})+\mu_{4}(H_{2})=3.414+2.593 > 6$. Therefore $S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$. Now, we can assume that $b=c=d=1$. Under the assumption, G is isomorphic to $H_{3}$ shown in Figure 12. We have $\lambda_{1}(H_{3})< n-2$ by Lemma 2.3 and $\lambda _{2}(H_{3})< \lambda_{1}(L_{v_{1}}(H_{3}))=5.23$ by Lemma 2.7 and direct calculation. Therefore, $S_{2}(G)< (n-2)+5.23 < n+4 = m(G)+3$.

When $\Vert \{a,b,c,d\} \Vert = 3$, we have $m_{G}(1) \geq n-7$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots \geq\lambda_{6} > \lambda_{7}=0$ be the other seven eigenvalues of G. Then we have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. If G is isomorphic to $G_{13}$, then $S_{2}(G) < m(G)+3$ by Lemma 3.13. Otherwise it contains a subgraph isomorphic to $H_{i}$ $( i \in\{4,5,6,7\})$ shown in Figure 12. Then we have $\lambda_{3}+\lambda_{4}\geq\lambda_{3}(H_{i})+\lambda _{i}(H_{i}) > 5$ $(i\in\{4,5,6,7\})$ by Lemma 2.6 and direct calculations. Therefore $S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$.

When $\Vert \{a,b,c,d\} \Vert = 2$, G is isomorphic to $G_{14}$ shown in Figure 12 or $H_{i}$ $(i \in\{8,10\})$ shown in Figure 12. For G is isomorphic to $G_{14}$, we have $S_{2}(G_{14}) < m(G_{14})+3$ by Lemma 3.14. For G is isomorphic to $H_{8}$ (or $H_{10}$), we have $m_{G}(1) \geq n-6$ by Lemma 2.8. Let $\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda _{6}=0$ be the other six eigenvalues of $H_{8}$ (or $H_{10}$). We have $\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$, since $\sum_{i=1}^{i=n}\mu_{i}=2m(G)$. Note that $H_{8}$ contains $H_{9}$ as a subgraph ($H_{10}$ contains $H_{11}$ or $H_{12}$ as a subgraph), where $H_{i}$ $(i= 9, 11, 12)$ shown in Figure 12. By Lemma 2.6 and direct calculations, we have $\lambda _{3}+\lambda_{4}\geq\mu_{3}(H_{i})+\mu_{4}(H_{i}) > 4$ $(i=9,11,12)$. Therefore $S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$, as required. When $\Vert \{a,b,c,d\} \Vert = 1$, G is isomorphic to $H_{13}$ or $G_{n+1, n}$. We have $\lambda_{1}(H_{13}) < n$ by Lemma 2.2 and $\lambda _{2}(H_{13}) =4$ by direct calculations. It follows that $S_{2}(H_{13})< n +4$. For $G_{n+1, n}$, a direct calculation shows that $S_{2}(G_{n+1, n})=m+3$. □

From the discussion above, we safely come to the following result.

Theorem 3.16

For $G\in\mathcal{B}^{\theta}_{n}(p,q,l)$, where $l \geq q \geq p \geq2$ and at most one of them is 2, we have $S_{2}(G) \leq m(G)+3$ and the equality holds if and only if $G \cong G_{n+1, n}$.

Theorem 1.3 follows immediately from Theorems 3.4 and 3.16.