Abstract

We determine the second to fourth largest (resp. the second smallest) signless Laplacian spectral radii and the second to fourth largest signless Laplacian spreads together with the corresponding graphs in the class of unicyclic graphs with n vertices. Moreover, we prove that one class of unicyclic graphs are determined by their signless Laplacian spectra.

1. Introduction

Throughout the paper, 𝐺=(𝑉,𝐸) is an undirected simple graph with 𝑛 vertices and 𝑚 edges. If 𝐺 is connected with 𝑚=𝑛+𝑐1, then 𝐺 is called a 𝑐-cyclic graph. Especially, if 𝑐=0 or 1, then 𝐺 is called a tree or a unicyclic graph, respectively. Let 𝕌𝑛 be the class of unicyclic graphs with 𝑛 vertices. The neighbor set of a vertex 𝑣 is denoted by 𝑁(𝑣). We write 𝑑(𝑣) for the degree of vertex 𝑣. In particular, let Δ(𝐺) and 𝛿(𝐺) be the maximum degree and minimum degree of 𝐺, respectively. Let 𝐴(𝐺) be the adjacency matrix and 𝐷(𝐺) be the diagonal matrix whose (𝑖,𝑖)-entry is 𝑑(𝑣𝑖), of 𝐺, respectively. The signless Laplacian matrix of 𝐺 is 𝑄(𝐺)=𝐷(𝐺)+𝐴(𝐺). Clearly, 𝑄(𝐺) is positive semidefinite [1] and its eigenvalues can be arranged as𝜇1(𝐺)𝜇2(𝐺)𝜇𝑛(𝐺)0.(1.1) Let 𝜇(𝐺) be the signless Laplacian spectral radius of 𝐺, namely, 𝜇(𝐺)=𝜇1(𝐺). The signless Laplacian spread of 𝐺 is defined as 𝑆𝑄(𝐺)=𝜇1(𝐺)𝜇𝑛(𝐺) [1, 2]. Let Φ(𝐺,𝑥) be the signless Laplacian characteristic polynomial of 𝐺, that is, Φ(𝐺,𝑥)=det(𝑥𝐼𝑄(𝐺)).

Recently, the research on the spectrum of 𝑄(𝐺) receive much attention. Some properties of signless Laplacian spectra of graphs and some possibilities for developing the spectral theory of graphs based on 𝑄(𝐺) are discussed in [35]. The largest signless Laplacian spectral radius and the largest signless Laplacian spread among the class of unicyclic graphs with 𝑛 vertices were determined in [6] and [1], respectively. The smallest signless Laplacian spectral radius among the class of unicyclic graphs with 𝑛 vertices was determined in [7]. In this paper, we will determine the second to fourth largest and the second smallest signless Laplacian spectral radii together with the corresponding graphs in the class of unicyclic graphs with 𝑛 vertices. Moreover, we also indentify the second to fourth largest signless Laplacian spreads together with the corresponding graphs in the class of unicyclic graphs with 𝑛 vertices. In the end of this paper, we will prove that a class of unicyclic graphs are determined by their signless Laplacian spectra.

2. The Signless Laplacian Spectral Radii of Unicyclic Graphs

As usually, let 𝐾1,𝑛1, 𝑃𝑛, and 𝐶𝑛 be the star, path, and cycle with 𝑛 vertices, respectively. In the following, let 𝑆3𝑛(𝑛4) be the unicyclic graph obtained by adding one edge to two pendant vertices of 𝐾1,𝑛1, and let 𝐹𝑛, 𝐻𝑛, 𝑆4𝑛 be the unicyclic graphs with 𝑛 vertices as shown in Figure 1.

In [6], the largest signless Laplacian spectral radius in the class of unicyclic graphs was determined, and it was proved as follows.

Theorem 2.1 (see [6]). If 𝑈𝕌𝑛, and 𝑛4, then 𝜇(𝑈)𝜇(𝑆3𝑛), where the equality holds if and only if 𝑈𝑆3𝑛.

Theorem 2.2. Suppose 𝑈𝕌𝑛, and 𝑛8. (1) If 𝑈𝑆3𝑛, then 𝜇(𝑈)𝜇(𝐹𝑛), where the equality holds if and only if 𝑈𝐹𝑛, and 𝜇(𝐹𝑛) equals the maximum root of the equation 𝑥5(𝑛+5)𝑥4+(6𝑛+3)𝑥3(9𝑛1)𝑥2+(3𝑛+8)𝑥4=0. (2) If 𝑈𝑆3𝑛 and 𝑈𝐹𝑛, then 𝜇(𝑈)𝜇(𝐻𝑛), where the equality holds if and only if 𝑈𝐻𝑛, and 𝜇(𝐻𝑛) equals the maximum root of the equation 𝑥5(𝑛+5)𝑥4+(6𝑛+4)𝑥3(10𝑛2)𝑥2+(3𝑛+12)𝑥4=0. (3) If 𝑈𝑆3𝑛, 𝑈𝐹𝑛 and 𝑈𝐻𝑛, then 𝜇(𝑈)𝜇(𝑆4𝑛), where the equality holds if and only if 𝑈𝑆4𝑛, and 𝜇(𝑆4𝑛) equals the maximum root of the equation 𝑥3(𝑛+3)𝑥2+(4𝑛2)𝑥2𝑛=0.

In order to prove Theorem 2.2, the following lemmas are needed.

Lemma 2.3 (see [8]). 𝜇(𝐺)max{𝑑(𝑣)+𝑚(𝑣),𝑣𝑉(𝐺)}, where 𝑚(𝑣)=𝑢𝑣𝑑(𝑢)/𝑑(𝑣).

Proposition 2.4. Suppose 𝑐1 and 𝐺 is a 𝑐-cyclic graph on 𝑛 vertices with Δ𝑛3. If 𝑛2𝑐+5, then 𝜇(𝐺)𝑛1.

Proof. We only need to prove that max{𝑑(𝑣)+𝑚(𝑣)𝑣𝑉}𝑛1 by Lemma 2.3. Suppose 𝑑(𝑢)+𝑚(𝑢)=max{𝑑(𝑣)+𝑚(𝑣)𝑣𝑉}. We consider the next three cases.
Case 1 (𝑑(𝑢)=1). Suppose 𝑣𝑁(𝑢). Then, 𝑑(𝑢)+𝑚(𝑢)=1+𝑑(𝑣)1+Δ𝑛2<𝑛1.
Case 2 (𝑑(𝑢)=2). Suppose that 𝑣,𝑤𝑁(𝑢). Then, 𝑑(𝑢)+𝑚(𝑢)=2+𝑑(𝑣)+𝑑(𝑤)22+2Δ2=Δ+2𝑛1.(2.1)
Case 3 (3𝑑(𝑢)𝑛3). Note that 𝐺 has 𝑛+𝑐1 edges and 3𝑑(𝑢)𝑛3. Then, 𝑑(𝑢)+𝑚(𝑢)𝑑(𝑢)+2(𝑛+𝑐1)𝑑(𝑢)2𝑑(𝑢)=𝑑(𝑢)1+2𝑛+2𝑐4𝑑(𝑢).(2.2) Next we will prove that 𝑑(𝑢)1+((2𝑛+2𝑐4)/𝑑(𝑢))𝑛1, equivalently, 𝑑(𝑢)(𝑛𝑑(𝑢))2𝑛+2𝑐4. Let 𝑓(𝑥)=(𝑛𝑥)𝑥, where 3𝑥𝑛3. Since 𝑓(𝑥)=𝑛2𝑥 and 3𝑥𝑛3, we have 𝑓(𝑥)min{𝑓(3),𝑓(𝑛3)}=3(𝑛3)2𝑛+2𝑐4.
By combining the above arguments, the result follows.

Corollary 2.5. Suppose 𝑈𝕌𝑛. If 𝑛7 and Δ𝑛3, then 𝜇(𝑈)𝑛1.

Lemma 2.6 (see [6]). If 𝐺 is a connected graph of order 𝑛4, then 𝜇(𝐺)Δ+1, where the equality holds if and only if 𝐺𝐾1,𝑛1.

Suppose 𝐵 is a square matrix, let 𝑎𝑖𝑖(𝐵) be the entry appearing in the 𝑖th row and the 𝑖th column of 𝐵. The next result gives a new method to calculate the signless Lapalacian characteristic polynomial of an 𝑛-vertex graph via the aid of computer.

Lemma 2.7 (see [9]). Let 𝐺 be a graph on 𝑛𝑘 (1𝑘𝑛2) vertices with 𝑉(𝐺)={𝑣𝑛,𝑣𝑛1,,𝑣𝑘+1}. If 𝐺 is obtained from 𝐺 by attaching 𝑘 new pendant vertices, say 𝑣1,,𝑣𝑘, to 𝑣𝑘+1, then Φ𝑄𝐺,𝑥=(𝑥1)𝑘det𝑥𝐼𝑛𝑘𝑄(𝐺)𝐵𝑛𝑘,(2.3) where 𝑎11(𝑄(𝐺)) is corresponding to the vertex 𝑣𝑘+1, and 𝐵𝑛𝑘=diag{𝑘+(𝑘/(𝑥1)),0,,0}.

Example 2.8. Let 𝐹𝑛 be the unicyclic graph as shown in Figure 1. By Lemma 2.7, we have Φ𝐹𝑛,𝑥=(𝑥1)𝑛4det(𝐵),where𝐵=𝑥(𝑛2)𝑛4𝑥11101𝑥21011𝑥31001𝑥1.(2.4) By using “Matlab”, it easily follows that Φ𝐹𝑛,𝑥=(𝑥1)𝑛5𝑥5(𝑛+5)𝑥4+(6𝑛+3)𝑥3(9𝑛1)𝑥2+(3𝑛+8)𝑥4.(2.5) With the similar method, by Lemma 2.7 we have Φ𝐻𝑛,𝑥=(𝑥1)𝑛5𝑥5(𝑛+5)𝑥4+(6𝑛+4)𝑥3(10𝑛2)𝑥2,Φ𝑆+(3𝑛+12)𝑥4(2.6)4𝑛,𝑥=𝑥(𝑥1)𝑛5𝑥(𝑥2)3(𝑛+3)𝑥2+.(4𝑛2)𝑥2𝑛(2.7)

Proof of Theorem 2.2. Note that 𝑆3𝑛 is the unique unicyclic graph with Δ=𝑛1, and 𝐹𝑛, 𝐻𝑛, 𝑆4𝑛 are all the unicyclic graphs with Δ=𝑛2. Now suppose 𝑈𝕌𝑛{𝑆3𝑛,𝐹𝑛,𝐻𝑛,𝑆4𝑛}. By Lemma 2.6 and Corollary 2.5, we have 𝜇(𝑆4𝑛)>𝑛1𝜇(𝑈) because Δ(𝑈)𝑛3.
To finish the proof of Theorem 2.2, we only need to show that 𝜇(𝑆4𝑛)<𝜇(𝐻𝑛)<𝜇(𝐹𝑛) by Theorem 2.1. By Lemma 2.6, it follows that 𝜇(𝐹𝑛)>𝑛1,𝜇(𝐻𝑛)>𝑛1, and 𝜇(𝑆4𝑛)>𝑛1. When 𝑥𝑛1 and 𝑛8, by (2.5), (2.6) and (2.7), it follows that Φ𝐻𝑛𝐹,𝑥Φ𝑛,𝑥=(𝑥1)𝑛5𝑥𝑥2Φ𝑆(𝑛1)𝑥+4>0,4𝑛𝐻,𝑥Φ𝑛=,𝑥(𝑥1)𝑛52𝑥2+(𝑛12)𝑥+4(𝑥1)𝑛5(𝑥(3𝑛14)+4)>0.(2.8) Therefore, we have 𝜇(𝐹𝑛)>𝜇(𝐻𝑛)>𝜇(𝑆4𝑛). Thus, Theorem 2.2 follows.

In [7], the smallest signless Laplacian spectral radius among all unicyclic graphs with 𝑛 vertices was determined, and that is as follows.

Theorem 2.9 (see [7]). If 𝑈𝕌𝑛, then 𝜇(𝑈)4, where the equality holds if and only if 𝑈𝐶𝑛.

The lollipop graph, denoted by 𝑊𝑛,𝑝, is obtained by appending a cycle 𝐶𝑝 to a pendant vertex of a path 𝑃𝑛𝑝. The next result extends the order of Theorem 2.9.

Theorem 2.10. For any 𝑛, if 𝑈𝕌𝑛{𝐶𝑛,𝑊𝑛,𝑛1}, then 𝜇(𝑈)>𝜇(𝑊𝑛,𝑛1)>𝜇(𝐶𝑛).

To prove Theorem 2.10, we will introduce more useful lemmas and notations.

Let 𝐺 be a connected graph, and 𝑢𝑣𝐸(𝐺). The graph 𝐺𝑢,𝑣 is obtained from 𝐺 by subdividing the edge 𝑢𝑣, that is, adding a new vertex 𝑤 and edges 𝑤𝑢,𝑤𝑣 in 𝐺𝑢𝑣. An internal path, say 𝑣1𝑣2𝑣𝑠+1(𝑠1), is a path joining 𝑣1 and 𝑣𝑠+1 (which need not be distinct) such that 𝑣1 and 𝑣𝑠+1 have degree greater than 2, while all other vertices 𝑣2,,𝑣𝑠 are of degree 2.

Lemma 2.11 (see [3, 10]). Let 𝑢𝑣 be an edge of a connected graph 𝐺. If 𝑢𝑣 belongs to an internal path of 𝐺, then 𝜇(𝐺)>𝜇(𝐺𝑢,𝑣).

By 𝐺𝐺, we mean that 𝐺 is a subgraph of 𝐺 and 𝐺𝐺.

Lemma 2.12 (see [10]). If 𝐺𝐺 and 𝐺 is a connected graph, then 𝜇(𝐺)<𝜇(𝐺).

If 𝑢𝑉(𝐺) and 𝑑(𝑢)3, then we called 𝑢 a branching point of 𝐺. Let 𝑈 be a connected unicyclic graph and 𝑇𝑣 be a tree such that 𝑇𝑣 is attached to a vertex 𝑣 of the unique cycle of 𝑈. The vertex 𝑣 is called the root of 𝑇𝑣, and 𝑇𝑣 is called a root tree of 𝑈. Throughout this paper, we assume that 𝑇𝑣 does not include the root 𝑣. Clearly, 𝑈 is obtained by attaching root trees to some vertices of the unique cycle of 𝑈.

Proof of Theorem 2.10. In the proof of this result, we assume that the unique cycle of 𝑈 is 𝐶𝑝. Now choose 𝑈𝕌𝑛{𝐶𝑛} such that 𝜇(𝑈) is as small as possible. Since 𝑈𝐶𝑛, 𝑈 has at least one branching point. We consider the next two cases.
Case 1. There are at least two branching points in 𝐶𝑝.
Now suppose 𝑢 and 𝑣 are two branching points in 𝐶𝑝 such that there does not exsit other barnching point between the shortest path in 𝐶𝑝 connected 𝑢 and 𝑣. Let 𝑢=𝑣1𝑣2𝑣𝑠=𝑣(𝑠2) be the shortest path in 𝐶𝑝 connected 𝑢 and 𝑣. Since 𝑢 is a branching point of 𝑈, there is at least one pendant vertex, say 𝑤, in 𝑇𝑢. Suppose 𝑧 is the unique neighbor vertex of 𝑤 in 𝑈 (may be 𝑧=𝑢). Let 𝑈1=𝑈𝑤𝑧 and 𝑈2=𝑈1𝑤. Then, 𝑈2𝑈, and hence 𝜇(𝑈2)<𝜇(𝑈) by Lemma 2.12. Let 𝑈3=𝑈1𝑣1𝑣2+𝑣1𝑤+𝑣2𝑤. By the hypothesis, 𝑣1𝑣2𝑣𝑠 is an internal path of 𝑈2. By Lemma 2.11, we can conclude that 𝜇(𝑈3)<𝜇(𝑈2)<𝜇(𝑈). But 𝑈3 is also a unicyclic graph with 𝑛 vertices and 𝑈3𝐶𝑛 because 𝑣 is a branching point of 𝑈3, it is a contradiction to the choice of 𝑈. Thus, Case 1 is impossible.
Case 2. There is unique branching point in 𝐶𝑝.
Subcase 1. There is at least a branching point outside 𝐶𝑝.
It can be proved analogously with Case 1.

Subcase 2. There does not exist any branching point outside 𝐶𝑝.Suppose 𝑢 is the unique brancing poing in 𝐶𝑝. By the hypothesis, 𝑢 is also the unique branching point of 𝑈. Then, 𝑈 is obtained by attaching 𝑑(𝑢)2 paths to the vertex 𝑢 of 𝐶𝑝.
If 𝑑(𝑢)4, then there are at least two paths being attaching to 𝑢. It can be proved analogously with Case 1.
If 𝑑(𝑢)=3, then 𝑈 is a lollipop graph, that is, 𝑈𝑊𝑛,𝑝. Let 𝑉(𝑊𝑛,𝑝)={𝑣1,𝑣2,,𝑣𝑛} and 𝐸(𝑊𝑛,𝑝)={𝑣𝑝𝑣1,𝑣𝑖𝑣𝑖+1,1𝑖𝑛1}. If 𝑛𝑝2, since 𝑊𝑛,𝑝𝑣𝑛1𝑣𝑛𝑊𝑛,𝑝, 𝜇(𝑊𝑛,𝑝)>𝜇(𝑊𝑛,𝑝𝑣𝑛1𝑣𝑛)=𝜇(𝑊𝑛,𝑝𝑣𝑛) by Lemma 2.12. Moreover, since 𝑊𝑛,𝑝𝑣𝑛1𝑣𝑛𝑣1𝑣𝑝+𝑣𝑝𝑣𝑛+𝑣𝑛𝑣1 is the graph obtained from 𝑊𝑛,𝑝𝑣𝑛 by subdividing the edge 𝑣1𝑣𝑝. Thus, by Lemma 2.11 it follows that 𝜇𝑊𝑛,𝑝𝑣𝑛1𝑣𝑛𝑊=𝜇𝑛,𝑝𝑣𝑛𝑊>𝜇𝑛,𝑝𝑣𝑛1𝑣𝑛𝑣1𝑣𝑝+𝑣𝑝𝑣𝑛+𝑣𝑛𝑣1𝑊=𝜇𝑛,𝑝+1.(2.9)
Therefore, 𝜇(𝑊𝑛,𝑝)>𝜇(𝑊𝑛,𝑝+1). Repeating the above process, we can conclude that 𝜇(𝑊𝑛,𝑝)>𝜇(𝑊𝑛,𝑝+1)>>𝜇(𝑊𝑛,𝑛1) holds for 𝑛𝑝2.
By combining the above arguments, 𝑈𝑊𝑛,𝑛1.

3. The Signless Laplacian Spreads of Unicyclic Graphs

In [1], the largest signless Laplacian spread among all unicyclic graphs with 𝑛 vrtices was determined, as follows.

Theorem 3.1 (see [1]). If 𝑛8 and 𝑈𝕌𝑛{𝑆3𝑛}, then 𝑆𝑄(𝑆3𝑛)>𝑆𝑄(𝑈).

The next result extends the order of Theorem 3.1 to the first four largest values.

Theorem 3.2. If 𝑛16 and 𝑈𝕌𝑛{𝑆3𝑛,𝑆4𝑛,𝐹𝑛,𝐻𝑛}, then 𝑆𝑆𝑄3𝑛𝑆>𝑆𝑄4𝑛𝐹>max𝑆𝑄𝑛𝐻,𝑆𝑄𝑛𝐹min𝑆𝑄𝑛𝐻,𝑆𝑄𝑛>𝑆𝑄(𝑈).(3.1)

Remark 3.3. With the aid of computer, we always have 𝑆𝑄(𝐹𝑛)<𝑆𝑄(𝐻𝑛). But it seems rather difficult to be proved.

To prove Theorem 3.2, we need to introduce more lemmas as follows.

Proposition 3.4. Suppose 𝑈 is a unicyclic graph on 𝑛 vertices with Δ𝑛3. If 𝑛9, then 𝑆𝑄(𝑈)𝑛1.1.

Proof. Note that 𝜇𝑛(𝑈)0 and 𝑆𝑄(𝑈)=𝜇1(𝑈)𝜇𝑛(𝑈)𝜇1(𝑈). We only need to prove max{𝑑(𝑣)+𝑚(𝑣)𝑣𝑉}𝑛1.1 by Lemma 2.3. Suppose 𝑑(𝑢)+𝑚(𝑢)=max{𝑑(𝑣)+𝑚(𝑣)𝑣𝑉}. We consider the next three cases.
Case 1 (𝑑(𝑢)=1). Suppose 𝑣𝑁(𝑢). Then, 𝑑(𝑢)+𝑚(𝑢)=1+𝑑(𝑣)1+Δ𝑛2<𝑛1.1.
Case 2 (𝑑(𝑢)=2). Suppose 𝑁(𝑢)={𝑤,𝑣}. Note that 𝑈 is a unicyclic graph. Then, |𝑁(𝑣)𝑁(𝑤)|2 and |𝑁(𝑣)𝑁(𝑤)|𝑛. Therefore, 𝑑(𝑢)+𝑚(𝑢)=2+𝑑(𝑣)+𝑑(𝑤)22+𝑛+22<𝑛1.1.(3.2)
Case 3 (3𝑑(𝑢)𝑛3). Note that 𝑈 has 𝑛 edges and 3𝑑(𝑢)𝑛3. By inequality (2.2), we have 𝑑(𝑢)+𝑚(𝑢)𝑑(𝑢)1+2𝑛2𝑑(𝑢).(3.3) Next we will prove that 𝑑(𝑢)1+((2𝑛2)/𝑑(𝑢))𝑛1.1, equivalently, 𝑑(𝑢)(𝑛𝑑(𝑢)0.1)2𝑛2. Let 𝑔(𝑥)=(𝑛𝑥0.1)𝑥, where 3𝑥𝑛3. Since 𝑔(𝑥)=𝑛0.12𝑥 and 3𝑥𝑛3, we have 𝑔(𝑥)min{𝑔(3),𝑔(𝑛3)}>2𝑛2.
By combining the above arguments, the result follows.

Lemma 3.5. If 𝑛16, then 𝑛1.1<𝑆𝑄(𝐹𝑛)<𝑛1.

Proof. Let 𝑓1(𝑥)=𝑥5(𝑛+5)𝑥4+(6𝑛+3)𝑥3(9𝑛1)𝑥2+(3𝑛+8)𝑥4. Clearly, 𝑓1112𝑛=32𝑛580𝑛556𝑛432𝑛310𝑛2.+10𝑛1(3.4) Next we will prove that 𝑓1(1/2𝑛)<0 when 𝑛16. Let 𝜓(𝑛)=80𝑛556𝑛432𝑛310𝑛2+10𝑛1. When 𝑛16, since 𝜓(𝑛)=4800𝑛21344𝑛192>0, we have 𝜓(𝑛)=1600𝑛3672𝑛2192𝑛20𝜓(16)=6378476>0, then 𝜓(𝑛)=400𝑛4224𝑛396𝑛220𝑛+10𝜓(16)=25272010>0. Thus, 𝜓(𝑛)𝜓(16)=80082591>0, and hence 𝑓1(1/2𝑛)<0.
With the similar method, we have 𝑓1(0.1)=0.2159𝑛3.18749>0,𝑓11(0.5)=𝑓32(112𝑛)<0,1(3)=9𝑛52>0,𝑓1(𝑛1)=20+21𝑛5𝑛2𝑓<0,11𝑛1+=12𝑛32𝑛516𝑛8320𝑛7+1232𝑛61728𝑛5+1192𝑛4504𝑛3+140𝑛220𝑛+1>0.(3.5)
By (2.5), we can conclude that 1/2𝑛<𝜇𝑛(𝐹𝑛)<0.1 and 𝑛1<𝜇1(𝐹𝑛)<𝑛1+1/2𝑛. Thus, 𝑛1.1<𝑆𝑄(𝐹𝑛)=𝜇1(𝐹𝑛)𝜇𝑛(𝐹𝑛)<𝑛1.

Lemma 3.6. If 𝑛16, then 𝑛1.1<𝑆𝑄(𝐻𝑛)<𝑛1.

Proof. Let 𝑓2(𝑥)=𝑥5(𝑛+5)𝑥4+(6𝑛+4)𝑥3(10𝑛2)𝑥2+(3𝑛+12)𝑥4. It is easily checked that 𝑓2112𝑛=32𝑛580𝑛5112𝑛440𝑛314𝑛2𝑓+10𝑛1<0,2(0.1)=0.2059𝑛2.77649>0,𝑓21(0.5)=𝑓32(8710𝑛)<0,2(2.8)>0.2464𝑛2.14>0,𝑓2(𝑛1)=24+25𝑛5𝑛2𝑓<0,21𝑛1+=12𝑛32𝑛516𝑛8320𝑛7+1376𝑛61888𝑛5+1288𝑛4520𝑛3+144𝑛220𝑛+1>0.(3.6) By (2.6), we can conclude that 1/2𝑛<𝜇𝑛(𝐻𝑛)<0.1 and 𝑛1<𝜇1(𝐻𝑛)<𝑛1+(1/2𝑛). Thus, 𝑛1.1<𝑆𝑄(𝐻𝑛)=𝜇1(𝐻𝑛)𝜇𝑛(𝐻𝑛)<𝑛1.

Proof of Theorem 3.2. By Lemma 2.6 and (2.7), 𝑆𝑄(𝑆4𝑛)=𝜇1(𝑆4𝑛)𝜇𝑛(𝑆4𝑛)=𝜇1(𝑆4𝑛)>𝑛1. Note that 𝑆3𝑛 is the unique unicyclic graph with Δ=𝑛1, and 𝐹𝑛, 𝐻𝑛, 𝑆4𝑛 are all the unicyclic graphs with Δ=𝑛2. Now suppose 𝑈𝕌𝑛{𝑆3𝑛,𝐹𝑛,𝐻𝑛,𝑆4𝑛}. Then, Δ(𝑈)𝑛3. By Lemmas 3.5 and 3.6, Theorem 3.1, and Proposition 3.4, we can conclude that 𝑆𝑆𝑄3𝑛𝑆>𝑆𝑄4𝑛𝐹>𝑛1>max𝑆𝑄𝑛𝐻,𝑆𝑄𝑛𝐹min𝑆𝑄𝑛𝐻,𝑆𝑄𝑛>𝑛1.1𝑆𝑄(𝑈).(3.7) This completes the proof of Theorem 3.2.

4. A Class of Unicyclic Graphs Determined by Their Signless Laplacian Spectra

A graph 𝐺 is said to be determined by its signless Laplacian spectrum if there does not exist other nonisomorphic graph 𝐻 such that 𝐻 and 𝐺 share the same signless Laplacian spectra (see [11]). Let 𝑆3(𝑛,𝑘) be the unicyclic graph on 𝑛 vertices obtained by attaching 𝑘, and 𝑛𝑘3 pendant vetrices to two vertices of 𝐶3, respectively. By the definition, 𝑆3(𝑛,𝑛3)=𝑆3𝑛. The next theorem is the main result of this section.

Theorem 4.1. For any 𝑘𝑛/21, if 8𝑘(𝑛3𝑘)9(𝑛3), then 𝑆3(𝑛,𝑘) is determined by its signless Laplacian spectrum.

To prove Theorem 4.1, we need some more lemmas as follows.

Lemma 4.2 (see [12]). If 𝐺 is a graph on 𝑛 vertices with vertex degrees 𝑑1𝑑2𝑑𝑛 and signless Laplacian eigenvalues 𝜇1𝜇2𝜇𝑛, then 𝜇2𝑑21. Moreover, if 𝜇2=𝑑21, then 𝑑1=𝑑2, and the maximum and the second maximum degree vertices are adjacent.

Lemma 4.3 (see [12]). If 𝐺 is a connected graph with 𝑛 vertices, then 𝜇𝑛(𝐺)<𝛿(𝐺).

Lemma 4.4 (see [7]). In any graph, the multiplicity of the eigenvalue 0 of the signless Laplacian matrix of 𝐺 is equal to the number of bipartite components of 𝐺.

Let 𝕌(𝑛,Δ) be the class of unicyclic graphs on 𝑛 vertices with maximum degree Δ.

Lemma 4.5 (see [13]). For any 𝑘𝑛/21, if 𝑈𝕌(𝑛,𝑘+2), then 𝜇(𝑈)𝜇(𝑆3(𝑛,𝑘)), where the equality holds if and only if 𝑈𝑆3(𝑛,𝑘).

Lemma 4.6 (see [13]). Let 𝐺 be the graph with the largest signless Laplacian spectral radius in 𝕌(𝑛,Δ). If Δ𝑛2, then there must exist some graph 𝐺1𝕌(𝑛,Δ+1) such that 𝜇(𝐺)<𝜇(𝐺1).

Proof of Theorem 4.1. By an elementary computation, we have Φ𝑆3(𝑛,𝑘),𝑥=(𝑥1)𝑛5𝑓3(𝑥),(4.1) where 𝑓3(𝑥)=𝑥5(𝑛+5)𝑥4+(𝑘𝑛+5𝑛𝑘23𝑘+7)𝑥3(2𝑘𝑛+7𝑛2𝑘26𝑘+7)𝑥2+(3𝑛+8)𝑥4. Now suppose that there exists another graph 𝐺 such that 𝐺 and 𝑆3(𝑛,𝑘) share the same signless Laplacian spectra. Next we will prove that 𝐺𝑆3(𝑛,𝑘). We only need to prove the following facts.
Fact 1. 𝐺 is a connected unicyclic graph.Proof of Fact 1. Assume that 𝐺 has exactly 𝑡 connected components, say 𝐺1,,𝐺𝑡, where 𝑡1. By Lemma 4.4, 𝐺𝑖 is not a bipartite graph for 1𝑖𝑡 because 𝑆3(𝑛,𝑘) is not a biparite graph. Thus, 𝐺𝑖 is a connected unicyclic graph for 1𝑖𝑡 because 𝐺 has 𝑛 edges (since 𝑆3(𝑛,𝑘) has 𝑛 edges). Moreover, since 8𝑘(𝑛3𝑘)9(𝑛3), we have 𝑓3(4)0. Thus, 𝐺𝑖 is not a cycle for 1𝑖𝑡 because 4 is the eigenvalue of the signless Laplacian matrix of a cycle. By the a bove arguments, we can conclude that 𝐺𝑖 is not a bipartite graph and has at least one pendant vertex. Thus, 𝐺𝑖 has at least two signless Laplacian eigenvalues being larger than 1 by Lemma 4.2, and the smallest signless Laplacian eigenvalue of 𝐺𝑖 is less than 1 by Lemma 4.3. Therefore, by (4.1) we can conclude that 𝑡1, and hence 𝐺 is connected. Clearly, 𝐺 is a connected unicyclic graph because 𝐺 has 𝑛 edges.Fact 2. Δ(𝐺)𝑘+2. Proof of Fact 2. Note that 𝑘𝑛/21. Then, 𝑛<2𝑘+3. By Lemmas 2.3 and 2.6, 𝑆Δ(𝐺)+1𝜇(𝐺)=𝜇3(𝑛,𝑘)max𝑘+2+𝑛+1𝑘+2,𝑛1𝑘+𝑛+1𝑛𝑘1,2+𝑛+12<𝑘+4.(4.2) Thus, Δ(𝐺)𝑘+2 holds.Fact 3. 𝐺𝑆3(𝑛,𝑘). Proof of Fact 3. By Fact 2, we have Δ(𝐺)𝑘+2. If Δ(𝐺)𝑘+1<Δ(𝑆3(𝑛,𝑘)), then 𝜇(𝐺)<𝜇(𝑆3(𝑛,𝑘)) by Lemmas 4.5 and 4.6, a contradiction. Thus, Δ(𝐺)=𝑘+2, and hence the result follows from Lemma 4.5 because 𝜇(𝐺)=𝜇(𝑆3(𝑛,𝑘)).
This completes the proof of Theorem 4.1.

Acknowledgments

This work is supported by the Foundation for Distinguished Young Talents in Higher Education of Guangdong, China (no. LYM10039) and NNSF of China (no. 11071088).