Abstract

We identify graphs with the maximal Laplacian spectral radius among all unicyclic graphs with vertices and diameter .

1. Introduction

Following [1], letbe a simple undirected graph onvertices and edges (sois its order and is its size). For,ordenotes the degree ofanddenotes the set of all neighbors of vertex. A pendant vertex is a vertex of degree 1 and a pendant edge is an edge incident with a pendant vertex. Let. For two verticesand(), the distance between and is the number of edges in the shortest path joining and . The diameter of a graph is the maximum distance between any two vertices of. Let () be a path ofwith(unless). If,, then we call an internal path of ; ifand, then we calla pendant path of; if the subgraph induced byinisitself, that is,, then we callan induced path. Obviously, the shortest path between any two distinct vertices ofis an induced path. We will use,to denote the graph obtained fromby deleting a vertex, or an edge, respectively (this notation is naturally extended if more than one vertex, or edge, is deleted).

Denote byandthe cycle and the path with vertices, respectively. We calla unicyclic graph if , where is the number of vertices andis the number of edges. We will useto denote the sets of all unicyclic graphs withvertices and diameter. Letbe a graph of orderobtained from the cycleby attachingpendant edges and a path of lengthat one vertex of the cycle, and a path of lengthto another nonadjacent vertex of the cycle respectively, where .

Letbe the Laplacian matrix, whereis the diagonal matrix andis the adjacency matrix. The matrixis real symmetric and positive semidefinite; the eigenvalues of can be arranged as, where the largest eigenvalueis called the Laplacian spectral radius of.

The investigation on the Laplacian spectral radius of graphs is an important topic in the theory of graph spectra. Recently, the problem concerning graphs with maximal Laplacian spectral radius of a given class of graphs has been studied extensively. Li et al. [2] determined those graphs which maximized Laplacian spectral radius among all bipartite graphs with (edge-) connectivity at mostand characterized graphs of orderwithcut-edges, having Laplacian spectral radius equal to. X. L. Zhang and H. P. Zhang [3] studied the largest Laplacian spectral radius of the bipartite graphs with vertices andcut edges and the bicyclic bipartite graphs, respectively. The Laplacian spectral radius of unicyclic graphs has been studied by many authors (see [4–6]). Liu et al. [7] determined the graphs with the largest Laplacian spectral radii among all unicyclic graphs and bicyclic graphs with vertices andpendant vertices. Hua et al. [8] determined extremal graphs with maximal Laplacian spectral radius among all unicyclic graphs with given order and given pendant vertices number.

In 2007, Liu et al. [9] determined graphs with the maximal spectral radius among all unicyclic graphs with vertices and diameter. In 2012, He and Li [6] identified graphs with the maximal signless Laplacian spectral radius among all unicyclic graphs with vertices of diameter. Next, Guo [4] considered the Laplacian spectral radius of unicyclic graphs with fixed diameter and proposed Conjecture 1.

In this paper, we prove the conjecture as Theorem 1.

Theorem 1. Letbe a graph in,. Consider the following.(i)Ifis odd,and equality holds if and only if.(ii)Ifis even and,and equality holds if and only if.(iii)Ifis even and,and equality holds if and only if.

The rest of this paper is organized as follows. In Section 2, we present some notations and lemmas which will be used later on. In Section 3, we determine graphs with the largest Laplacian spectral radius among all unicyclic graphs withvertices and diameter.

2. Lemmas

In this section, we list some lemmas which will be used to prove our main results.

Lemma 2 (see [10]). Suppose thatare two distinct vertices of a connected graph. Letbe the graph obtained from by attachingnew paths at . Let , where corresponds to the vertex, be a unit eigenvector of corresponding to . Let If, then. Further, if, then.

Lemma 3 (see [10]). Letbe a pendant edge of a connected graphwithvertices and let be a pendant vertex. Letbedisjoint connected graphs and let be a vertex of . Letbe the graph obtained by addingnew edges among . Let
(i)If, then.(ii)If, then, with equality if and only if eitheror there exists somesuch that.

Letbe a vertex of a connected graphwith at least two vertices. Let () be the graph obtained fromby attaching two new paths and of length and , respectively, at , where and are distinct new vertices. Let .

Lemma 4 (see [11]). Letbe a connected graph onvertices andbe a vertex of. Letbe the graph defined as previously mentioned. If, then, with equality if and only if there exists a unit eigenvector ofcorresponding totaking the valueon vertex.

Lemma 5 (see [1]). Letbe a graph obtained by deleting an edge from the graph. Then , .

Letbe a graph obtained from the cycleby attaching pendant edges at one vertex of the cycle.

Lemma 6 (see [5]). Letbe a unicyclic graph onvertices; then; when, the equality holds if and only if; when, the equality holds if and only if,.

Lemma 7. Letbe a connected graph with at least one edge, let be its maximal degree, and let be the degree of vertexand; then ; the equality holds if and only if [12]; ; the equality holds if and only ifis regular or semiregular bipartite graph [13].

Letbe the principal submatrix obtained fromby deleting the corresponding row and column of. Generally, letbe the principal submatrix obtained fromby deleting the corresponding rows and columns of all vertices of . For any square matrix , denote by the characteristic polynomial of . In particular, if , we write by for convenience. If , then suppose that .

Letbe the graph obtained by joining the vertexof the graphto the vertex ofof the graphby an edge. We calla connected sum ofatandat.

Lemma 8. Letandbe two graphs. If for , then . (In general, let and be polynomials with positive leading coefficients. If for , then , where and are the largest roots of and , resp.)

Proof. If, then,, a contradiction.

Lemma 9 (see [14]). Letbe a connected sum ofatandat; then

Lemma 10 (see [14]). Letbe a connected graph withvertices which consists of a subgraphand distinct pendant edges (not in ) attaching to a vertexin. Then

Lemma 11 (see [15]). Letbe the matrix obtained fromby deleting the rows and columns corresponding to two pendant vertices of; suppose that,; then(i);(ii);(iii), ;(iv).

From Lemma 11(i), all eigenvalues ofare, where. Other characterizations ofcan be shown below.

Lemma 12. Letbe the matrix as above. Consider the following.(i)If, thenwhen.(ii)If, thenwhen.(iii), where.

Proof. From Lemma 11(ii), it is easy to prove (i) and (ii) by introduction on.
By Lemma 11(iii), we have as desired.

Lemma 13. Suppose thatare two adjacent vertices of the cycle, whereis even. Let() be the graph obtained fromby attaching two new paths and of length and at and , respectively, where and are distinct new vertices. Let . Then .

Proof. Using Lemma 9, we have where
From Lemmas 11(ii) and 11(iv), (6) becomes which is greater than 0 whenby Lemma 12(i). And follows from Lemma 7(i). Thusholds by Lemma 8.

For, we haveand . If , then . If , then , or . By Lemma 7, has the largest Laplacian spectral radius.

Therefore, in the following, we assume that .

Letbe the unicyclic graph of ordershown in Figure 1. Letbe a graph of orderobtained fromby attachingpendant vertices to each, respectively, wherewhenor. Denote that

(a)
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Figure 1 

Lemma 14. Let. Then there is a graph such that .

Proof. Letand let be a unit eigenvector of , where corresponds to the vertex(). Let. Then. Let, . Assume, without loss of generality, that. Let. Let By Lemma 2, we have. Note thatforandfor. If , then we will useto repeat the above step until the cardinality of, being nonzero, is only one. So we haveand. Note that, and hence the lemma holds.

Lemma 15. For any,, where; the equality holds if and only if.

Proof. Suppose that. If , the result is obvious.
If, by Lemma 7, we have
Case 1..
When , let. Let Then, in all cases,. Thus by Lemma 2,.
Next, we show that for. Because, we can getin which the rows and columns correspond to vertices as the ordering . Furthermore, letbe a square matrix of order, whereandwheneverand; letbe a square matrix of order, whereandwheneverand. Then Hence, In order to simplify the notation, we denoteandby and, respectively. Similarly, In general, by Lemma 11(ii), we have
Hence, by Lemma 12(iii), From (17) and Lemma 10, holds for.
Thusfollows from Lemma 8.
Case 2. .
First note that.
Next, since, we can derivein which the rows and columns correspond to vertices as the ordering. Then
Combining Lemma 10 with (15) and (19), we get which is greater than 0 whenby Lemma 12(i). Thus, by Lemma 8, holds. Hence, the proof is completed.

Letbe a graph of orderobtained from a triangle by attachingpendant edges and a path of length at one vertex of the triangle, and a path of length to another vertex of the triangle, respectively, where.

Lemma 16. , where.

Proof. Suppose that; by Lemma 7(i),. Let; by Lemma 9, where By Lemmas 10 and 11(ii) and (15) and (21), for.
Sofollows from Lemma 8.

In view of Lemma 16, the next corollary is obvious.

Corollary 17. , where; the equality holds if and only if.

Let be the unicyclic graph of order shown in Figure 1. Letbe a graph of orderobtained fromby attachingpendant vertices to each, respectively. Denote

Lemma 18. Let. Then there is a graphsuch that.

Proof. The proof is similar to that of Lemma 14.

Lemma 19. For any,, where; the equality holds if and only if.

Proof. Suppose that. If, the result is trivial.
If, by Lemma 7, we have
Case 1. .
If ,.
If, we can obtainin which the rows and columns correspond to vertices as the ordering. Then by Lemma 11(ii), Similarly, Combining the two equations above with Lemmas 10 and 12(iii), we get for.
From Lemma 8,holds.
Case 2. , .
From Lemma 7, when, we have
For, we can obtainin which the rows and columns correspond to vertices as the ordering. Then where By (26) and (30), From Lemmas 11(ii) and 12(i), when, hold (sincefor).
By Lemma 9 and (32), when, From Lemma 8,holds.
Hence, we complete the proof.

Letbe a graph of orderobtained from the cycleby attaching pendant edges and a path of lengthat one vertex of the cycle and a path of lengthto another nonadjacent vertex of the cycle, respectively, where.

Lemma 20. Letbe a graph defined as above; then(i)if, then, where;(ii)ifi.e., , then(a)when, or,;(b)when,.

Proof. Suppose that; by Lemma 7(i),holds. Let; by Lemma 9, where By Lemmas 10 and 11(ii) and (26) and (35), where.
When, by Lemma 11(ii), we get Denote bythe largest root of. Then Hence, by Lemma 12(i), (37) is greater thanwhen. Sofollows from Lemma 8.
When, (37) becomes
If, letandbe two graphs with. Through Maple 15, the largest root ofis (up to four decimal places), which is less thanand. From Lemma 5,andhold when. So, (40) is greater than 0 for . By Lemma 8,holds.
If, by Lemmas 11(ii) and 12(iii), where (one may refer to (52) in Lemma 24).
By Lemma 12(ii),when. And by derivative, when.
Thus, (41) is greater thanfor. From Lemma 8,holds. Putinto (40), whose right side is less than 0. So. We complete the proof.

Form Lemma 20, the below corollary holds.

Corollary 21. When,(i)ifis odd, then; the equality holds if and only if;(ii)ifis even, then(a)when,; the equality holds if and only if;(b)when,; the equality holds if and only if.

Letbe the unicyclic graph of ordershown in Figure 2. Letbe a graph of orderobtained fromby attachingpendant vertices to each, respectively. When,and. Denote that

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Figure 2 

Lemma 22. Let. Then there is a graphsuch that.

Proof. The proof is similar to that of Lemma 14.

Lemma 23. For any,, where; the equality holds if and only if.

Proof. Suppose that. If, the result is obvious.
If, by Lemma 7, we have Hence, the lemma holds.

3. Main Results and Their Proofs

In this section, we first show that.

Lemma 24. .

Proof. Suppose that; by Lemma 7,, holds. We distinguish the following two cases.
Case 1. ().
Letandbe two graphs on the left of Figure 3. If, denoteandbyand, respectively. Letand. By Lemma 9, where Note thatand.
Combining the equations above with Lemma 10, we get Hence, by Lemmas 11(ii) and 12(i), when, where Soholds by Lemma 8.
Case 2. ().
Letandbe two graphs on the right of Figure 3. If, denoteandbyand , respectively. By similar computations as Case 1, we have
Hence, by Lemmas 11(ii), 12(i), and 12(iii), when, where Sofollows from Lemma 8.

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Figure 3 

Letbe a graph of orderobtained from a triangle by attachingpendant edges, a path of lengthand a path of lengthat one vertex of the triangle, where.

Lemma 25. .

Proof. Suppose that; by Lemma 7(i),holds. We distinguish the following two cases.
Case 1. .
Letbe a graph on the left of Figure 3. If, denoteby. Let. By Lemma 9, where By Lemma 10, we get By (47), we have Hence,
When, (59)holds for. Sofollows from Lemma 8.
When,also holds (the proof will be given in Case 2 of the lemma).
Case 2.().
Letbe a graph on the right of Figure 3. By similar computations as in Case, we have By (51), we get Hence,
Letand the largest root ofis denoted by, where.
We first show thatis strictly increasing.
We use the induction on. Clearly, holds. Generally, assume that, then by Lemmas 11(ii) and 12(iii), Putinto (63), whose right side is less than 0. So.
Furthermore,has the upper bound.
Denote that the largest root ofis. If there exists somesuch that , we substitutewithand put into (64). So, the right side of it is less than and equal toand the left side is greater than, a contradiction.
Letand     .
By (60) and (62), where.
By Lemma 7(i), is the largest root of . If , then and , a contradiction. So .
By derivative, when, From Lemma 8,holds. Furthermore, and hold by Lemma 5. Hence, (62) is greater than 0 for. From Lemma 8, we getas desired.

Letbe the unicyclic graph of ordershown in Figure 2.

Lemma 26. .

Proof. Note that by Lemma 7(i),.
Case 1. ).
By Lemma 9, we have where.
By (47), we get Hence, by Lemmas 11(ii), 12(ii), and 12(iii), when, So follows from Lemma 8.
Case 2. ().
By a similar proof as of Case 1,holds.

Next we give the proof of Theorem 1, which is the most important result.

Proof of Theorem 1. Letandbe a unit eigenvector of, wherecorresponds to the vertex().
Choosesuch that the Laplacian spectral radius ofis as large as possible. Then, by Lemma 6, we can assume that. Letbe the induced path of lengthand letbe the only cycle in. Since, we have, say. We first show some claims.
Claim 1. .
Proof of Claim 1. Otherwise, sinceis connected, there exists an only pathconnectingand, where,and.
For each, letbe a rooted tree (withas its root) attached at(), where the order ofis. We assume that all trees, but, are kept fixed, while(along with its root) can be changed. Suppose that. Letbe a vertex belonging tochosen so thatand that(the distance betweenand) is the largest. By Lemma 4 (applied in the reverse direction), the Laplacian spectral radius is increased when any hanging path atis replaced by a hanging star (namely, edges of a hanging path now become the hanging edges at). If the same is repeated for other hanging paths at, we get one star attached at(its central vertex is identified with) whose size is equal to the sum of the lengths of the aforementioned paths. Letbe a vertex in, adjacent to, and belonging to the (unique) path betweenand. By Lemma 3, the Laplacian spectral radius is increased when all hanging edges atbecome the hanging edges at. Note also that. By repeating the same procedure (for any other vertex as), we arrive at, where the rooted treebecomes a star, so that.
By the same way to other rooted trees, we arrive at, where every rooted tree(), and. From, applying Lemma 2, we arrive at, where only a rooted treeis attached at some vertex. So.
From, by Lemma 3, the Laplacian spectral radius increased when all vertices adjacent tobecome adjacent vertices to. By repeating the same procedure (for any other vertex ofas), we arrive at, where, and.
Hence, we have, a contradiction.
By Claim 1,. Denote that (), where and .
Claim 2. for.
Proof of Claim 2. Consider other rooted trees attached atand, respectively. By a similar proof as Claim 1 (the procedure until), we can getand, where only a rooted tree is attached at(), a contradiction.
Claim 3. and.
Proof of Claim 3. Denote that and .
Case 1. . By Lemma 7, for().
When(),. By Lemma 23,for any, where .
Hence, by Lemma 4,has the larger Laplacian spectral radius. Furthermore, from Lemmas 24 and 25, holds.
Case 2. .
Subcase 2.1. . By Lemma 7, for().
When(),, By Lemma 15,for any, where.
Subcase 2.2. . By Lemma 7, for().
When(), by Lemma 13,. Furthermore, from Lemma 26,.
When(),.
Hence, in view of Subcases 2.1 and 2.2, by Corollary 17 and Lemma 24, holds.
Case 3. . By Lemma 7, for().
When(), , by Lemma 19,for any, where.
Hence, in view of Cases 1, 2, and 3 and Corollary 21, ifis odd,has the largest Laplacian spectral radius; ifis even and,has the largest Laplacian spectral radius; If is even and,has the largest Laplacian spectral radius, a contradiction.
By Claims 1, 2, and 3, Theorem 1 follows immediately.

Acknowledgments

The author would like to express sincere gratitude to the referees for a very careful reading of the paper and for all their insightful comments and valuable suggestions, which led to improving this paper. The research of the author is supported by the National Natural Science Foundation of China (nos. 11371078 and 61303020) and the National Natural Science Foundation of Shanxi Province (no. 2012011019-2).