The (distance) signless Laplacian spectral radius of digraphs with given arc connectivity

Linear Algebra and its Applications 581 (2019) 85–111

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Linear Algebra and its Applications www.elsevier.com/locate/laa

The (distance) signless Laplacian spectral radius of digraphs with given arc connectivity ✩ Weige Xi a,b , Wasin So c , Ligong Wang a,b,∗ a

Department of Applied Mathematics, School of Science, Northwestern Polytechnical University, Xi’an, Shaanxi 710072, China b Xi’an-Budapest Joint Research Center for Combinatorics, Northwestern Polytechnical University, Xi’an, Shaanxi 710129, China c Department of Mathematics and Statistics, San Jose State University, San Jose, CA 95192-0103, USA

a r t i c l e

i n f o

Article history: Received 25 October 2018 Accepted 5 July 2019 Available online 11 July 2019 Submitted by R. Brualdi MSC: 05C50 15A18 Keywords: Strongly connected Signless Laplacian spectral radius Distance signless Laplacian spectral radius Arc connectivity

a b s t r a c t Let G n,k denote the set of strongly connected digraphs with ∗ order n and arc connectivity k, and let G n,k denote the set of digraphs in G n,k with all vertices having outdegree and indegree greater than k. In this paper, we determine the unique digraph with the maximum signless Laplacian spectral radius among all digraphs in G n,k . We also determine the unique one with the maximum signless Laplacian spectral ∗ radius among all digraphs in G n,k with k = 1, 2. For the general case, we propose a conjecture on the maximum ∗ signless Laplacian spectral radius among all digraphs in G n,k . Moreover, we characterize the extremal digraph achieving the minimum distance signless Laplacian spectral radius among all digraphs in G n,k . We also characterize the extremal digraph achieving the minimum distance signless Laplacian ∗ spectral radius among all digraphs in G n,k with k = 1, 2. For the general case, we propose a conjecture on the minimum

✩ Supported by the National Natural Science Foundation of China (No. 11871398), the Natural Science Basic Research Plan in Shaanxi Province of China (Program No. 2018JM1032) and China Scholarship Council (No. 201706290182). * Corresponding author at: Department of Applied Mathematics, School of Science, Northwestern Polytechnical University, Xi’an, Shaanxi 710072, China. E-mail addresses: [email protected] (W. Xi), [email protected] (W. So), [email protected], [email protected] (L. Wang).

https://doi.org/10.1016/j.laa.2019.07.010 0024-3795/© 2019 Elsevier Inc. All rights reserved.

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distance signless Laplacian spectral radius among all digraphs ∗ in G n,k . © 2019 Elsevier Inc. All rights reserved.

1. Introduction Let G be a digraph with vertex set V (G) = {v1 , v2 , . . . , vn } and arc set E(G). A digraph is simple if it has no loops and multiple arcs. A digraph is strongly connected if for every pair of vertices vi , vj ∈ V (G), there exists a directed path from vi to vj . For a digraph G, if there is an arc from vi to vj , then we indicate this by writing e = (vi , vj ) ∈ E(G), and e is said to be out-incident to vi and in-incident to vj ; vi is said to be out-adjacent to vj and vj is said to be in-adjacent to vi . For any vertex vi ∈ V (G), Ni+ = Nv+i (G) = {vj : (vi , vj ) ∈ E(G)} and Ni− = Nv−i (G) = {vj : (vj , vi ) ∈ E(G)} are + called the sets of out-neighbors and in-neighbors of vi , respectively. Let d+ i (G) = di = − − |Ni+ | denote the outdegree of vertex vi , and d− i (G) = di = |Ni | denote the indegree of + + − vertex vi in the digraph G. We use δ , Δ and δ to denote the minimum outdegree, maximum outdegree and minimum indegree, respectively. A digraph is r-regular if all vertices have outdegree r and indegree r. Throughout this paper, we only consider finite simple strongly connected digraphs. ←→

Let Kn denote the complete digraph on n vertices in which for two arbitrary distinct ←→

←→

vertices vi , vj ∈ V (Kn ), there are arcs (vi , vj ), (vj , vi ) ∈ E(Kn ). Let G1  G2 denote the digraph obtained from two disjoint digraphs G1 , G2 with vertex set V = V (G1 ) ∪ V (G2 ) and arc set E = E(G1 )∪E(G2 )∪{(u, v), (v, u) : u ∈ V (G1 ), v ∈ V (G2 )}. Let 1 ≤ k ≤ n−2 ←→

←→

←→

and 1 ≤ m ≤ n − k − 1, K(n, k, m) denote the digraph Kk  (Kn −k−m ∪ Km ) ∪ E, where ←→

←→

E = {(u, v) : u ∈ V (Km ), v ∈ V (Kn −k−m )}. For a strongly connected digraph G, a set of arcs S ⊂ E(G) is an arc cut set if G − S is not strongly connected. The arc connectivity κ(G) of G, is the minimum number of arcs whose deletion yields the resulting digraph not-strongly connected. Let G n,k denote the set of strongly connected digraphs with order n and arc connec←→

tivity κ(G) = k ≥ 1. If k = n − 1, then G n,n−1 = {Kn }. So we only consider the cases 1 ≤ k ≤ n − 2. Obviously, the vertex connectivity of K(n, k, m) is k, and K(n, k, m) has ∗ arc connectivity k if and only if m = 1 or m = n − k − 1. Let G n,k denote the set of strongly connected digraphs with order n and arc connectivity k ≥ 1 whose vertices have ∗ both outdegree and indegree greater than k. Obviously, G n,k  G n,k . Note that G n,k is nonempty, we list some examples below. The first example of digraphs in G n,k are K(n, k, 1) and K(n, k, n − k − 1). Now the second example of digraphs in G n,k are Cayley digraphs Cay(Zn , S), which are digraphs constructed as follows. Let Zn = {0, 1, 2, . . . , n − 1} denote the additive group of integers modulo n, and S = {s1 , s2 , . . . , sk } ⊂ Zn be a set of group elements

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such that 0 ∈ / S and gcd(n, s1 , s2 , . . . , sk ) = 1. The Cayley digraph of Zn with respect to S, denoted by Cay(Zn , S), is defined as follows: the vertices are the elements of Zn , for any two vertices i, j ∈ Zn , if there is an arc from i to j if and only if (j − i) ∈ S, where (j − i) modulo n. It can be checked that Cay(Zn , S) is k-regular and κ(Cay(Zn , S)) = k, i.e., Cay(Zn , S) ∈ G n,k . Let A(G) = (aij ) be the adjacency matrix of a digraph G, where aij = 1 if (vi , vj ) ∈ + + E(G) and aij = 0 otherwise. Let Diag(G) = diag(d+ 1 , d2 , . . . , dn ) be the diagonal matrix with outdegrees of the vertices of G. Then we denote Q(G) = Diag(G)+A(G) the signless Laplacian matrix of G. The eigenvalue of Q(G) with the largest modulus is called the signless Laplacian spectral radius of the digraph G, denoted by q(G). Let G be a strongly connected digraph with vertex set V (G) = {v1 , v2 , . . . , vn }. For vi , vj ∈ V (G), the distance from vi to vj , denoted by dG (vi , vj ) or dij , is the length of the shortest directed path from vi to vj in the digraph G. For vi ∈ V (G), the transmission of vertex vi in G is the sum of distances from vi to all other vertices of G, denoted by T rG (vi ). The distance matrix of G is an n × n matrix D(G) = (dij ). In fact, for 1 ≤ i ≤ n, the transmission of vertex vi , T rG (vi ) is just the i-th row sum of D(G). Let T r(G) = diag(T rG (v1 ), T rG (v2 ), · · · , T rG (vn )) be the diagonal matrix with vertex transmissions of G. The distance signless Laplacian matrix of G is an n × n matrix defined as DQ (G) = T r(G) + D(G). The eigenvalue of DQ (G) with the largest modulus is called the distance signless Laplacian spectral radius of G, denoted by λ(G). If G is a strongly connected digraph, then it follows from the Perron Frobenius Theorem [5] that q(G) (resp., λ(G)) is an eigenvalue of Q(G) (resp., DQ (G)), and there is a unique positive unit eigenvector corresponding to q(G) (resp., λ(G)). The positive unit eigenvector corresponding to q(G) (resp., λ(G)) is called the Perron vector of Q(G) (resp., DQ (G)). The investigation of matrices related to various graphical structures is a very large and growing area of research. In particular, the spectral radius and distance spectral radius of digraphs have been well studied in the literature, see [3,4,7,9,15–17,19,18]. Recently, there are many articles on the (distance) signless Laplacian spectral radius of digraphs. The bounds of (distance) signless Laplacian spectral radius can be found in [2, 6,8,21,23,25,26]. Especially, there are articles on maximizing or minimizing the (distance) signless Laplacian spectral radius among a given set of digraphs. In [6], Hong and You determined the digraph which achieves the minimum (or maximum) signless Laplacian spectral radius among all strongly connected digraphs with some given parameters such as clique number, girth or vertex connectivity. In [20], Xi and Wang determined the extremal digraph which attains the maximum (or minimum) signless Laplacian spectral radius among all strongly connected bicyclic digraphs. In [10], Li et al. determined the extremal digraph which has the minimum distance signless Laplacian spectral radius among all strongly connected digraphs with given dichromatic number. In [22], Xi and Wang determined the extremal digraph with the maximum signless Laplacian spectral radius among all strongly connected digraphs with given dichromatic number. In [11] and [22], the authors independently determined the extremal digraph with the minimum

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distance signless Laplacian spectral radius among all strongly connected digraphs with given vertex connectivity. In [13], Li et al. determined the unique digraph which attains the maximum (or minimum) signless Laplacian spectral radius among generalized  ∞  and θ-digraphs. However, there is no work on the relation between the (distance) signless Laplacian spectral radius and arc connectivity of a digraph. Previous work has been done on maximizing the adjacency spectral radius of digraphs with given arc connectivity [14]. In [17], Lin and Shu determined the extremal digraph with the minimum distance spectral radius with given arc connectivity. It is interesting to consider the problems of maximizing the signless Laplacian spectral radius and minimizing the distance signless Laplacian spectral radius of digraphs subject to fixed arc connectivity. In [24], Ye et al. determined the extremal graphs with the maximum signless Laplacian spectral radius among all connected graphs of order n and edge connectivity r. In [12], Li et al. determined the extremal graph with the minimum distance spectral radius among all connected graphs of order n and edge connectivity r. In this paper, we study the (distance) signless Laplacian spectral radius of strongly connected digraphs. We try to use the similar methods in [24] and [12] to determine the extremal digraph which has the maximum signless Laplacian spectral radius and the minimum distance signless Laplacian spectral radius among all strongly connected digraphs with given order and arc connectivity, respectively. But it does not work. It forces us to use a new approach to determine the digraph which has the maximum signless Laplacian spectral radius and the minimum distance signless Laplacian spectral radius among all strongly connected digraphs in G n,k . Moreover, we also determine the unique ∗ digraph with the maximum signless Laplacian spectral radius among all digraphs in G n,k with k = 1, 2. For the general case, we propose a conjecture on the maximum signless ∗ Laplacian spectral radius among all digraphs in G n,k . In addition, we also characterize the extremal digraph achieving the minimum distance signless Laplacian spectral radius ∗ among all digraphs in G n,k with k = 1, 2. For the general case, we propose a conjecture ∗ on the minimum distance signless Laplacian spectral radius among all digraphs in G n,k . 2. Preliminaries In this section, we give some lemmas which will be used in later sections. Lemma 2.1. ([5]) Let M = (mij ) be an n × n nonnegative matrix with spectral radius n  ρ(M ), and let Ri (M ) be the i-th row sum of M , i.e., Ri (M ) = mij (1 ≤ i ≤ n). j=1

Then min{Ri (M ) : 1 ≤ i ≤ n} ≤ ρ(M ) ≤ max{Ri (M ) : 1 ≤ i ≤ n}. Moreover, if M is irreducible, then either one equality holds if and only if R1 (M ) = R2 (M ) = . . . = Rn (M ).

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Definition 2.2. ([1,5]) Let A = (aij ), B = (bij ) be n × n matrices. If aij ≤ bij for all i and j, then A ≤ B. If A ≤ B and A = B, then A < B. If aij < bij for all i and j, then A B. Lemma 2.3. ([1,5]) Let A, B be nonnegative matrices. If 0 ≤ A ≤ B, then ρ(A) ≤ ρ(B), where ρ(A) is the spectral radius of A. Furthermore, if B is irreducible and 0 ≤ A < B, then ρ(A) < ρ(B). By Lemma 2.3, we have the following results in terms of digraphs. Corollary 2.4. Let G be a digraph and H be a spanning subdigraph of G. (i) q(G) ≥ q(H). (ii) If G is strongly connected, and H is a proper subdigraph of G, then q(G) > q(H). Corollary 2.5. Let G be a digraph and H be a spanning subdigraph of G. (i) If G and H are strongly connected, then λ(H) ≥ λ(G). (ii) If G is strongly connected, and H is a proper strongly connected subdigraph of G, λ(H) > λ(G). Lemma 2.6. ([6]) Let G be a strongly connected digraph on n vertices, u, v, w be three distinct vertices of V (G), (u, v) ∈ E(G) and X = (x1 , x2 , . . . , xn )T be the unique positive unit eigenvector corresponding to the signless Laplacian spectral radius q(G), where xi corresponds to the vertex i. Let H = G−{(u, v)}+{(u, w)}. (Noting that if (u, w) ∈ E(G), then H has multiple arc (u, w).) If xw ≥ xv , then q(H) ≥ q(G). Furthermore, if H is strongly connected and xw > xv , then q(H) > q(G). Lemma 2.7. ([10]) Let G be a strongly connected digraph with u, v ∈ V (G) and (u, v) ∈ / E(G). Then λ(G) > λ(G + {(u, v)}). Lemma 2.8. ([22]) Let n, k, m be positive integers with 1 ≤ k ≤ n − 2 and 1 ≤ m ≤ √ 3n+m−4+ n2 −7m2 +6mn−8mk . n − k − 1. Then λ(K(n, k, m)) = 2 Lemma 2.9. ([6]) Let n, k be positive integers with 1 ≤ k ≤ n − 2, and G be a strongly connected digraph with order n and vertex connectivity k. Then q(G) ≤ q(K(n, k, n − k − 1)) with equality if and only if G ∼ = K(n, k, n − k − 1). Lemma 2.10. ([22]) Let n, k be positive integers with 1 ≤ k ≤ n − 2, and G be a strongly connected digraph with order n and vertex connectivity k. Then λ(G) ≥ λ(K(n, k, 1)) with equality if and only if G ∼ = K(n, k, 1). Lemma 2.11. Let G be a strongly connected digraph with order n and arc connectivity k ≥ 1, and Ek be an arc cut set of G of size k such that G − Ek has exactly two strongly connected components, say G1 and G2 with |V (G1 )| = n1 and |V (G2 )| = n2 , where

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− n1 + n2 = n. If d+ v > k and dv > k for each vertex v ∈ V (G), then n1 ≥ k + 2, n2 ≥ k + 2.

Proof. Without loss of generality, we assume that there are no arcs from G1 to G2 in G − Ek . If n1 ≤ k, then there exists a vertex w of G1 such that d+ w ≤ n1 − 1 +

k k k ≤ (n1 − 1) × + = k, n1 n1 n1

a contradiction. If n1 = k +1, then there exists a vertex u of G1 which is not out-incident with any arc of Ek . This implies that d+ u ≤ n1 − 1 = k, also a contradiction. If n2 ≤ k, then there exists a vertex v of G2 such that d− v ≤ n2 − 1 +

k k k ≤ (n2 − 1) × + = k, n2 n2 n2

a contradiction. If n2 = k + 1, then there exists a vertex z of G2 which is not in-incident with any arc of Ek . This implies that d− z ≤ n2 − 1 = k, also a contradiction. So the result follows. 2 3. The maximum signless Laplacian spectral radius of digraphs with given arc connectivity In this section, we investigate the maximum signless Laplacian spectral radius of strongly connected digraphs with given arc connectivity. Lemma 3.1. Let G ∈ G n,k . If G contains a vertex of outdegree k, then q(G) ≤ q(K(n, k, n − k − 1)). Proof. Let w be a vertex of G such that d+ w = k. Then the arcs out-incident to w form an arc cut set of size k. Adding all possible arcs from G\{w} to G\{w} ∪ {w}, we obtain a digraph H, which is isomorphic to K(n, k, n − k − 1), the arc connectivity of H remains equal to k. If G = K(n, k, n − k − 1), by Corollary 2.4, the addition of any such arc will give q(G) < q(K(n, k, n − k − 1)). So the result follows. 2 Lemma 3.2. Let G ∈ G n,k . If G contains a vertex of indegree k, then q(G) ≤ q(K(n, k, 1)). Proof. Let w be a vertex of G such that d− w = k. Then the arcs in-incident to w form an arc cut set of size k. Adding all possible arcs from w to G\{w}, and all possible arcs from G\{w} to G\{w}, we obtain a digraph H  , which is isomorphic to K(n, k, 1), the arc connectivity of H  remains equal to k. If G = K(n, k, 1), by Corollary 2.4, the addition of any such arc will give q(G) < q(K(n, k, 1)). So the result follows. 2

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Let δ 0 (G) = min{δ + (G), δ − (G)}. The following result shows that the digraph K(n, k, n − k − 1) maximizes the signless Laplacian spectral radius among all digraphs in G n,k when κ(G) = δ 0 (G) = k ≥ 1. Theorem 3.3. For each 1 ≤ k ≤ n − 2, among all digraphs in G n,k with δ 0 (G) = k, the digraph K(n, k, n − k − 1) maximizes the signless Laplacian spectral radius. Proof. Let G be a digraph with arc connectivity k, and δ 0 (G) = k. If δ 0 (G) = δ + (G), then there exists a vertex of outdegree k. So by Lemma 3.1, K(n, k, n−k−1) maximizes the signless Laplacian spectral radius. If δ 0 (G) = δ − (G), then there exists a vertex of indegree k. So by Lemma 3.2, K(n, k, 1) maximizes the signless Laplacian spectral radius. Moreover, by Lemma 2.9, q(K(n, k, n −k −1)) > q(K(n, k, 1)). Therefore, we complete the proof. 2 Theorem 3.4. For each 1 ≤ k ≤ n − 2, the digraph K(n, k, n − k − 1) maximizes the signless Laplacian spectral radius among all digraphs in G n,k . Proof. Let G be a digraph in G n,k . Note that each vertex in the digraph G has outdegree at least k and indegree at least k, otherwise G ∈ / G n,k . Then, we discuss the following two cases. − Case 1. If there exists a vertex u of G with d+ u = k or du = k, by Theorem 3.3, q(G) ≤ q(K(n, k, n − k − 1)) and the equality holds if and only if G ∼ = K(n, k, n − k − 1). − Case 2. For arbitrary vertex u of G, d+ > k and d > k. Let S be an arc cut set of u u G containing k arcs, then G − S consists of exactly two strongly connected components G1 , G2 , respectively, of orders a, b and a + b = n. Without loss of generality, we may assume that there are no arcs from G1 to G2 in G − S. By Lemma 2.11, a ≥ k + 2, b = n − a ≥ k + 2, then k + 2 ≤ a ≤ n − k − 2, n ≥ a + k + 2 ≥ 2k + 4. Next we create a new digraph H by adding to G any possible arcs from G2 to G1 ∪ G2 or any possible arcs from G1 to G1 that were not present in G. Obviously, the arc connectivity of H remains equal to k and all vertices of H have outdegree greater than k and indegree still greater than k. By Corollary 2.4, the addition of any such arc will give q(H) > q(G). For ←→

←→

k + 2 ≤ a ≤ n − k − 2, let H  = Ka ∪ Kn −a , U = {u1 , u2 , · · · , uk } be a set of k vertices in ←→

←→

V ( Ka ) and W = {v1 , v2 , · · · , vk } be a set of k vertices in V (Kn −a ). Let H4 be a digraph obtained from H  by adding all possible arcs from U to W , and adding all possible arcs ←→

←→

from Kn −a to Ka . Then, H be a spanning subdigraph of H4 , therefore, by Corollary 2.4, q(H) ≤ q(H4 ). Hence, in order to prove the digraph K(n, k, n − k − 1) maximizes the signless Laplacian spectral radius in G n,k , it suffices to prove q(H4 ) < q(K(n, k, n−k−1)). However, we can know the vertex connectivity of H4 is k and H4  K(n, k, n − k − 1). Hence, by Lemma 2.9, we know that q(H4 ) < q(K(n, k, n − k − 1)). Therefore, q(G) ≤ q(H) ≤ q(H4 ) < q(K(n, k, n − k − 1)).

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Fig. 1. Three digraphs G1 , G2 and G3 , and each undirected edge for digraphs G1 , G2 , G3 denotes a pair of oppositely directed arcs.

Hence, combining the above two cases, we obtain that the digraph K(n, k, n − k − 1) is the unique digraph which has the maximum signless Laplacian spectral radius among all digraphs in G n,k . 2 It is natural to ask: what are digraphs in G n,k whose signless Laplacian spectral radius is minimum for each 1 ≤ k ≤ n − 2? Theorem 3.5. For each 1 ≤ k ≤ n − 2, and G ∈ G n,k , q(G) ≥ 2k. Equality holds if and only if G is k-regular. Proof. By Lemma 2.1, for any strongly connected digraph H, 2δ + ≤ q(H) ≤ 2Δ+ , with either equality if and only if the outdegrees of all vertices in H are equal. Since G ∈ G n,k , then δ + ≥ k, δ − ≥ k. So we obtain that q(G) ≥ 2k with the equality if and only if the outdegrees of all vertices in G is k, and the indegrees of all vertices in G is also k because δ − ≥ k. That is q(G) ≥ 2k with the equality if and only if G is a k-regular digraph. 2 The Cayley digraph Cay(Zn , S) defined in the first section is k-regular digraph with arc connectivity k. Here, we give some examples of k-regular digraphs with arc connectivity k. Example 3.6. Let G1 = Cay(Z6 , S1 ), where S1 = {1, 2, 4, 5}, G2 = Cay(Z7 , S2 ), where S2 = {2, 4, 6}, G3 = Cay(Z8 , S3 ), where S3 = {1, 2, 3, 6, 7} be the strongly connected digraphs of orders 6, 7 and 8 respectively, as shown in Fig. 1. Note that G1 is a 4-regular digraph with arc connectivity 4, G2 is a 3-regular digraph with arc connectivity 3, and G3 is a 5-regular digraph with arc connectivity 5. ∗

4. The digraph with maximum signless Laplacian spectral radius in G n,k In Section 3, we show that the digraph K(n, k, n − k − 1) is the unique digraph which has the maximum signless Laplacian spectral radius among all digraphs in G n,k . Since ∗ K(n, k, n − k − 1) contains a vertex of outdegree k, we have K(n, k, n − k − 1) ∈ / G n,k . We

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want to find the digraph with the maximum signless Laplacian spectral radius among all ∗ ∗ digraphs in G n,k . Moreover, by Lemma 2.11, if G ∈ G n,k , then n ≥ 2k + 4. Furthermore, n−4 we deduce 1 ≤ k ≤ n−4 2 . So in this section, we only consider the cases 1 ≤ k ≤ 2 . We start with special cases with k = 1, 2. ←→

←→

Let G(n, a) be a digraph obtained from Ka ∪ Kn −a by adding 1 arc from one vertex

←→

←→

←→

←→

in Ka to one vertex in Kn −a and adding all possible arcs from Kn −a to Ka . Obviously, ∗ for 3 ≤ a ≤ n − 3, G(n, a) ∈ G n,1 ∗

Theorem 4.1. Let G ∈ G n,1 . Then q(G) ≤ q(G(n, 3)) < 2n − 4 with the first equality if and only if G ∼ = G(n, 3). Moreover, q(G(n, 3)) is the largest real root of the equation x3 − (2n + 1)x2 + (12n − 24)x − 14n + 36 = 0. ∗

Proof. Since G ∈ G n,1 , all vertices of G have outdegree greater than 1 and indegree greater than 1. Let E1 be an arc cut set of G containing 1 arc, then G − E1 consists of exactly two strongly connected components G1 , G2 , respectively, of orders a, b and a + b = n. Without loss of generality, we may assume that there are no arcs from G1 to G2 in G − E1 . By Lemma 2.11, a ≥ 1 + 2 = 3, b = n − a ≥ 1 + 2 = 3, then 3 ≤ a ≤ n − 3, n ≥ a + 3 ≥ 6. Next we create a new digraph H by adding to G any possible arcs from G2 to G1 ∪ G2 or any possible arcs from G1 to G1 that were not present in G. Obviously, the arc connectivity of H remains equal to 1 and all vertices of H have outdegree greater than 1 and indegree still greater than 1, and H ∼ = G(n, a). By Corollary 2.4, the addition of any such arc will give q(H) > q(G). Let v0 be the unique vertex in G1 out-adjacent to one vertex in G2 . Then the signless Laplacian matrix of G(n, a) has the form ⎛

a

⎜ 1 ⎜ (a−1)×1 ⎜ ⎝ 1 1(n−a−1)×1

11×(a−1) (a − 2)Ia−1 + Ja−1 11×(a−1) J(n−a−1)×(a−1)

1 0(a−1)×1 n−1 1(n−a−1)×1

01×(n−a−1)

⎞

⎟ 0(a−1)×(n−a−1) ⎟ ⎟. ⎠ 11×(n−a−1) (n − 2)In−a−1 + Jn−a−1

Let X be the Perron vector of Q(G(n, a)), the coordinate corresponding to v0 is x0 . It is easy to know that all coordinates of X corresponding to vertices of V (G1 )\{v0 } are equal, say x1 , all coordinates corresponding to vertices of V (G2 ) are equal, say x2 . For convenience, we use q to denote q(G(n, a)). Therefore, we have ⎧ ⎪ ⎪ ⎨ax0 + (a − 1)x1 + x2 = qx0 , x0 + (a − 1)x1 + (a − 2)x1 = qx1 , ⎪ ⎪ ⎩x + (a − 1)x + (n − 1)x + (n − a − 1)x = qx . 0 1 2 2 2 Or equivalently

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⎛

a a−1 ⎜ ⎝ 1 2a − 3 1 a−1

⎞⎛ ⎞ ⎛ ⎞ 1 x0 x0 ⎟⎜ ⎟ ⎜ ⎟ 0 ⎠ ⎝ x1 ⎠ = q ⎝ x1 ⎠ . 2n − a − 2 x2 x2

It is easy to know that the characteristic polynomial of the above matrix is x3 − (2n − 5 + 2a)x2 + (6an − 6n − a2 − 7a + 6)x + 8an − 4a2 n + 2a3 − 6a − 2n, where 3 ≤ a ≤ n − 3. In order to find the digraph which has the maximum signless Laplacian spectral radius in G(n, a), we partition into two cases. Case 1. a = 3, then H ∼ = G(n, 3). In the following, we want to prove that 2n − 5 < q(G(n, 3)) < 2n − 4. Take f (x) = x3 − (2n + 1)x2 + (12n − 24)x − 14n + 36, then q(G(n, 3)) is the largest real root of f (x) = 0. Then f  (x) = 3x2 − 2(2n + 1)x + 12n − 24, and the discriminant of f  (x) is 4(2n + 1)2 − 4 × 3 × (12n − 24) = 4(4n2 − 32n + √ 2 73) > 0 for n ≥ 6. Hence, f  (x) > 0 when x > 2n+1+ 4n3 −32n+73 . Since n ≥ 6, √ 2n+1+ 4n2 −32n+73 < 2n+1+2n−2 = 4n−1 < 2n −4. Therefore, f  (x) > 0 for all x ≥ 2n −4. 3 3 3 Furthermore, we have f (x) is increasing for all x ≥ 2n − 4, i.e., for all x ≥ 2n − 4, f (x) ≥ f (2n − 4) = 4n2 − 30n + 52 > 0 since n ≥ 6. Hence q(G(n, 3)) < 2n − 4. On the other hand, we can know that q(G(n, 3)) > 2n − a − 2 = 2n − 5. Therefore, if a = 3, 2n − 5 < q(H) = q(G(n, 3)) < 2n − 4. Case 2. 4 ≤ a ≤ n − 3, take g(x) = x3 − (2n − 5 + 2a)x2 + (6an − 6n − a2 − 7a + 6)x + 8an − 4a2 n + 2a3 − 6a − 2n, then q(G(n, a)) is the largest real root of g(x) = 0. In the following, we want to prove that q(G(n, a)) < 2n − 5 when 4 ≤ a ≤ n − 3. Since g  (x) = 3x2 − 2(2n − 5 + 2a)x + 6an − 6n − a2 − 7a + 6, g  (x) = 6x − 2(2n − 5 + 2a). We can know that g  (x) > 0 for all x ≥ 2n − 5. Therefore g  (x) ≥ g  (2n − 5) = 4n2 − 26n + 31 − 2an + 13a − a2 ≥ 4n2 − 26n + 31 − 2(n − 3)n + 13a − (n − 3)2 ≥ n2 − 14n + 22 + 13 × 4 = n2 − 14n + 74 > 0 (for n ≥ 7). So g(x) ≥ g(2n − 5) = 2a3 − (6n − 5)a2 + (4n2 + 4n − 21)a − 12n2 + 40n − 30 for all x ≥ 2n − 5. Take h(a) = 2a3 − (6n − 5)a2 + (4n2 + 4n − 21)a − 12n2 + 40n − 30, where 4 ≤ a ≤ n − 3. Then h (a) = 12a − (12n − 10) < 0. Thus, for fixed n, the minimal value of h(a) must be taken at either a = 4 or a = n − 3. Hence, for all x ≥ 2n − 5,

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g(x) ≥ g(2n −5) = h(a) ≥ min{h(4), h(n −3)} = min{3n2 −23n +24, 4n2 −40n +94} > 0 since n ≥ 7. Therefore q(G(n, a)) < 2n − 5 < q(G(n, 3)). Thus combining the above two cases, we have q(G) ≤ q(G(n, 3)) < 2n − 4 with the first equality if and only if G ∼ = G(n, 3). Moreover, q(G(n, 3)) is the largest real root of 3 2 the equation x − (2n + 1)x + (12n − 24)x − 14n + 36 = 0. 2 ←→

←→

←→

For 4 ≤ a ≤ n − 4, let G = Ka ∪ Kn −a , u, v be two vertices in V ( Ka ) and w, z be two ←→

vertices in V (Kn −a ). Let H1 (n, a) be a digraph obtained from G by adding 2 arcs (u, w) ←→

←→

and (u, z), and adding all possible arcs from Kn −a to Ka . Let H2 (n, a) be a digraph obtained from G by adding 2 arcs (u, w) and (v, w), and adding all possible arcs from ←→ Kn −a

←→

to Ka . Let H3 (n, a) be a digraph obtained from G by adding 2 arcs (u, w) and ←→

←→

(v, z), and adding all possible arcs from Kn −a to Ka . Obviously, for 4 ≤ a ≤ n − 4, ∗ Hi (n, a) ∈ G n,2 , where i = 1, 2, 3. ∗

Theorem 4.2. Let G ∈ G n,2 . Then q(G) ≤ q(H1 (n, 4)) < 2n − 5 with the first equality if and only if G ∼ = H1 (n, 4). Moreover, q(H1 (n, 4)) is the largest real root of the equation x3 − (2n + 4)x2 + (20n − 40)x − 44n + 136 = 0. ∗

Proof. Since G ∈ G n,2 , all vertices of G have outdegree greater than 2 and indegree greater than 2. Let E2 be an arc cut set of G containing 2 arcs, then G − E2 consists of exactly two strongly connected components G1 , G2 , respectively, of orders a, b and a + b = n. Without loss of generality, we may assume that there are no arcs from G1 to G2 in G − E2 . By Lemma 2.11, a ≥ 2 + 2 = 4, b = n − a ≥ 2 + 2 = 4, then 4 ≤ a ≤ n − 4, n ≥ a + 4 ≥ 8. Next we create a new digraph H by adding to G any possible arcs from G2 to G1 ∪ G2 or any possible arcs from G1 to G1 that were not present in G. Obviously, the arc connectivity of H remains equal to 2 and all vertices of H have outdegree greater than 2 and indegree still greater than 2, and H ∼ = H1 (n, a), ∼ ∼ or H = H2 (n, a), or H = H3 (n, a). By Corollary 2.4, the addition of any such arc will give q(Hi (n, a)) > q(G) for i = 1, 2, 3. By Lemma 2.6, it is easy to know that q(H2 (n, a)) = q(H3 (n, a)). Therefore, in the following, it suffices to find the digraph in H1 (n, a) and H2 (n, a) which has the maximum signless Laplacian spectral radius. Case 1. H ∼ = H1 (n, a). Then the signless Laplacian matrix of H1 (n, a) has the form ⎛

a+1

⎜ 1 ⎜ (a−1)×1 ⎜ ⎝ 12×1 1(n−a−2)×1

11×(a−1) (a − 2)Ia−1 + Ja−1 J2×(a−1) J(n−a−2)×(a−1)

11×2 0(a−1)×2 (n − 2)I2 + J2 J(n−a−2)×2

01×(n−a−2)

⎞

⎟ 0(a−1)×(n−a−2) ⎟ ⎟. ⎠ J2×(n−a−2) (n − 2)In−a−2 + Jn−a−2

Let X be the Perron vector of Q(H1 (n, a)), the coordinate corresponding to u is x0 . It is easy to know that all coordinates of X corresponding to vertices of V (G1 )\{u} are equal, say x1 , all coordinates corresponding to vertices of V (G2 ) are equal, say x2 . Therefore, we have

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⎧ ⎪ ⎪ ⎨(a + 1)x0 + (a − 1)x1 + 2x2 = q(H1 (n, a))x0 , x0 + (a − 1)x1 + (a − 2)x1 = q(H1 (n, a))x1 , ⎪ ⎪ ⎩x + (a − 1)x + (n − 1)x + (n − a − 1)x = q(H (n, a))x . 0 1 2 2 1 2 Or equivalently ⎛ a+1 ⎜ ⎝ 1 1

a−1 2a − 3 a−1

⎛ ⎞ ⎞⎛ ⎞ x0 2 x0 ⎜ ⎟ ⎟⎜ ⎟ 0 ⎠ ⎝ x1 ⎠ = q(H1 (n, a)) ⎝ x1 ⎠ . 2n − a − 2 x2 x2

It is easy to know that the characteristic polynomial of the above matrix is x3 −(2n −4 + 2a)x2 + (6an − a2 − 6a − 4n)x − 4a2 n + 2a3 + 2a2 + 4an − 4a + 4n − 8, where 4 ≤ a ≤ n − 4. In order to find the digraph which has the maximum signless Laplacian spectral radius in H1 (n, a), we partition into two cases. Subcase 1. a = 4. Take f (x) = x3 − (2n + 4)x2 + (20n − 40)x − 44n + 136, then q(H1 (n, 4)) is the largest real root of f (x) = 0. In the following, we want to prove that 2n − 6 < q(H1 (n, 4)) < 2n − 5. Because f  (x) = 3x2 − 2(2n + 4)x + 20n − 40, and the discriminant of f  (x) is 4(2n + 4)2 − 4 × 3 × (20n − 40) = 4(4n2 − 44n + √ 2n+4+ 4n2 −44n+136  136) > 0 for n ≥ 8. Hence, f (x) > 0 when x > . Since n ≥ 8, 3 √ 2n+4+ 4n2 −44n+136 2n+4+2n−4 4n  < = 3 < 2n − 5. Therefore, f (x) > 0 for all x ≥ 2n − 5. 3 3 Furthermore, we have f (x) is increasing for all x ≥ 2n − 5, i.e., for all x ≥ 2n − 5, f (x) ≥ f (2n − 5) = 4n2 − 44n + 111 > 0 for n ≥ 8. Hence q(H1 (n, 4)) < 2n − 5. On the other hand, we can know that q(H1 (n, 4)) > 2n − a − 2 = 2n − 6. Therefore, if a = 4, 2n − 6 < q(H1 (n, 4)) < 2n − 5. Subcase 2. 5 ≤ a ≤ n − 4. Take g(x) = x3 − (2n − 4 + 2a)x2 + (6an − a2 − 6a − 4n)x − 2 4a n + 2a3 + 2a2 + 4an − 4a + 4n − 8, then q(H1 (n, a)) is the largest real root of g(x) = 0. We want to prove that q(H1 (n, a)) < 2n − 6 when 5 ≤ a ≤ n − 4. Because g  (x) = 3x2 − 2(2n − 4 + 2a)x + 6an − a2 − 6a − 4n, g  (x) = 6x − 2(2n − 4 + 2a). We can know that g  (x) > 0 for all x ≥ 2n − 6. Therefore g  (x) ≥ g  (2n − 6) = 4n2 − 36n + 60 − 2an + 18a − a2 ≥ 4n2 − 36n + 60 − 2(n − 4)n + 18a − (n − 4)2 ≥ n2 − 20n + 44 + 18 × 5 = n2 − 20n + 134 > 0 (for n ≥ 9).

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So g(x) ≥ g(2n − 6) = 2a3 − (6n − 8)a2 + (4n2 + 4n − 40)a − 16n2 + 76n − 80 for all x ≥ 2n − 6. Take h(a) = 2a3 − (6n − 8)a2 + (4n2 + 4n − 40)a − 16n2 + 76n − 80, where 5 ≤ a ≤ n − 4. Then h (a) = 12a − (12n − 16) < 0. Thus, for fixed n, the minimal value of h(a) must be taken at either a = 5 or a = n − 4. Hence, for all x ≥ 2n − 6, g(x) ≥ g(2n −6) = h(a) ≥ min{h(5), h(n −4)} = min{4n2 −54n +170, 4n2 −44n +80} > 0 for n ≥ 9. Therefore, if 5 ≤ a ≤ n − 4, q(H1 (n, a)) < 2n − 6 < q(H1 (n, 4)). Therefore, if H ∼ = H1 (n, a), combining the above two subcases, we know that H1 (n, a) has maximal signless Laplacian spectral radius when a = 4. Moreover, q(H1 (n, 4)) is the largest real root of the equation x3 − (2n + 4)x2 + (20n − 40)x − 44n + 136 = 0. Case 2. H ∼ = H2 (n, a). Then the signless Laplacian matrix of H2 (n, a) has the form ⎛

(a − 1)I2 + J2 ⎜ 1 ⎜ (a−2)×2 ⎜ ⎝ 11×2 J(n−a−1)×2

J2×(a−2) (a − 2)Ia−2 + Ja−2 11×(a−2) J(n−a−1)×(a−2)

12×1 0(a−2)×1 n−1 1(n−a−1)×1

02×(n−a−1)

⎞

⎟ 0(a−2)×(n−a−1) ⎟ ⎟. ⎠ 11×(n−a−1) (n − 2)In−a−1 + Jn−a−1

Let X be the Perron vector of Q(H2 (n, a)). It is easy to know that the coordinates of X corresponding to u, v are equal, say x0 , all coordinates corresponding to vertices of V (G1 )\{u, v} are equal, say x1 , all coordinates corresponding to vertices of V (G2 ) are equal, say x2 . Therefore, we have ⎧ ⎪ ⎪ ⎨ ax0 + x0 + (a − 2)x1 + x2 = q(H2 (n, a))x0 , 2x0 + (a − 1)x1 + (a − 3)x1 = q(H2 (n, a))x1 , ⎪ ⎪ ⎩ 2x + (a − 2)x + (n − 1)x + (n − a − 1)x = q(H (n, a))x . 0 1 2 2 2 2 Or equivalently ⎛

⎞⎛ ⎞ ⎛ ⎞ a+1 a−2 1 x0 x0 ⎜ ⎟ ⎜ ⎟⎜ ⎟ 2a − 4 0 ⎝ 2 ⎠ ⎝ x1 ⎠ = q(H2 (n, a)) ⎝ x1 ⎠ . 2 a − 2 2n − a − 2 x2 x2 It is easy to know that the characteristic polynomial of the above matrix is x3 − (2n − 5 + 2a)x2 + (6an − a2 − 7a − 6n + 4)x − 4a2 n + 2a3 + 8an − 6a − 4, where 4 ≤ a ≤ n − 4. In order to find the digraph which has the maximum signless Laplacian spectral radius in H2 (n, a), we partition into two cases. Subcase 1. a = 4. Take α(x) = x3 − (2n + 3)x2 + (18n − 40)x − 32n + 100, then q(H2 (n, 4)) is the largest real root of α(x) = 0. In the following, we want to prove that 2n − 6 < q(H2 (n, 4)) < 2n − 5. Because α (x) = 3x2 − 2(2n + 3)x + 18n − 40,

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and the discriminant of α (x) is 4(2n + 3)2 − 4 × 3 × (18n − 40) = 4(4n2 − 42n + √ 2 129) > 0 for n ≥ 8. Hence, α (x) > 0 when x > 2n+3+ 4n3 −42n+129 . Since n ≥ 8, √ 2n+3+ 4n2 −42n+129  < 2n+3+2n−3 = 4n 3 3 3 < 2n − 5. Therefore, α (x) > 0 for all x ≥ 2n − 5. Furthermore, we have α(x) is increasing for all x ≥ 2n − 5, i.e., for all x ≥ 2n − 5, α(x) ≥ α(2n − 5) = 4n2 − 42n + 100 > 0 for n ≥ 8. Hence q(H2 (n, 4)) < 2n − 5. On the other hand, we can know that q(H2 (n, 4)) > 2n − a − 2 = 2n − 6. Therefore, if a = 4, 2n − 6 < q(H2 (n, 4)) < 2n − 5. Subcase 2. 5 ≤ a ≤ n − 4. Take β(x) = x3 − (2n − 5 + 2a)x2 + (6an − a2 − 7a − 6n + 4)x − 4a2 n + 2a3 + 8an − 6a − 4, then q(H2 (n, a)) is the largest real root of β(x) = 0. In the following, we want to prove that q(H2 (n, a)) < 2n − 6 when 5 ≤ a ≤ n − 4. Because β  (x) = 3x2 − 2(2n − 5 + 2a)x + 6an − a2 − 7a − 6n + 4, β  (x) = 6x − 2(2n − 5 + 2a). We can know that β  (x) > 0 for all x ≥ 2n − 6. Therefore β  (x) ≥ β  (2n − 6) = 4n2 − 34n + 52 − 2an + 17a − a2 ≥ 4n2 − 34n + 52 − 2(n − 4)n + 17a − (n − 4)2 ≥ n2 − 18n + 36 + 17 × 5 = n2 − 18n + 121 > 0 (for n ≥ 9). So β(x) ≥ β(2n − 6) = 2a3 − (6n − 8)a2 + (4n2 + 6n − 36)a − 16n2 + 68n − 64 for all x ≥ 2n − 6. Take θ(a) = 2a3 − (6n − 8)a2 + (4n2 + 6n − 36)a − 16n2 + 68n − 64, where 5 ≤ a ≤ n − 4. Then θ (a) = 12a − (12n − 16) < 0. Thus, for fixed n, the minimal value of θ(a) must be taken at either a = 5 or a = n − 4. Hence, for all x ≥ 2n − 6, β(x) ≥ β(2n −6) = θ(a) ≥ min{θ(5), θ(n −4)} = min{4n2 −52n +206, 6n2 −56n +80} > 0 for n ≥ 9. Therefore, if 5 ≤ a ≤ n − 4, q(H2 (n, a)) < 2n − 6 < q(H2 (n, 4)). Therefore, if H ∼ = H2 (n, a), combining the above two subcases, we know that H2 (n, a) has maximal signless Laplacian spectral radius when a = 4, and q(H2 (n, 4)) is the largest real root of the equation x3 − (2n + 3)x2 + (18n − 40)x − 32n + 100 = 0. Thus, if H ∼ = H1 (n, a), H1 (n, a) has maximal signless Laplacian spectral radius when a = 4, and q(H1 (n, 4)) > 2n − 6 is the largest real root of the equation x3 − (2n + 4)x2 + (20n − 40)x − 44n + 136 = 0; if H ∼ = H2 (n, a), H2 (n, a) has maximal signless Laplacian spectral radius when a = 4, and q(H2 (n, 4)) > 2n − 6 is the largest real root of the equation x3 − (2n + 3)x2 + (18n − 40)x − 32n + 100 = 0. Therefore, in order to find the ∗ digraph which has the maximum signless Laplacian spectral radius in G n,2 , it suffices to compare the signless Laplacian spectral radii of H1 (n, 4) and H2 (n, 4). Let η(x) = x3 − (2n + 4)x2 + (20n − 40)x − 44n + 136, φ(x) = x3 − (2n + 3)x2 + (18n − 40)x − 32n + 100. Then from the above proof, we can know that q(H1 (n, 4)) is the largest real root of η(x) = 0 and q(H2 (n, 4)) is the largest real root of φ(x) = 0. Hence,

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φ(x) − η(x) = x2 − 2nx + 12n − 36, which is increasing for all x > 2n − 6. Therefore, for all x > 2n − 6, φ(x) − η(x) = x2 − 2nx + 12n − 36 > (2n − 6)2 − 2n(2n − 6) + 12n − 36 = 0. Then we have q(H1 (n, 4)) > q(H2 (n, 4)). Thus, we have q(G) ≤ q(H1 (n, 4)) < 2n − 5 with the first equality if and only if ∼ G = H1 (n, 4). Moreover, q(H1 (n, 4)) is the largest real root of the equation x3 − (2n + 4)x2 + (20n − 40)x − 44n + 136 = 0. 2 For general 1 ≤ k ≤ examples.

n−4 2 ,

we propose the following conjecture based on numerical ∗

Conjecture 4.3. Let G be a digraph in G n,k with the maximum signless Laplacian spectral radius. Then G is obtained from ←→ Kk +2

←→

←→ Kk +2

←→

∪ Kn −k−2 by adding k arcs from one vertex in ←→

←→

to k vertices in Kn −k−2 and adding all possible arcs from Kn −k−2 to Kk +2 .

The special cases with k = 1, 2 are proved in Theorem 4.1 and Theorem 4.2, respectively. 5. The minimum distance signless Laplacian spectral radius of digraphs with given arc connectivity In this section, we discuss the minimum distance signless Laplacian spectral radius among all digraphs in G n,k , where 1 ≤ k ≤ n − 2. Lemma 5.1. Let G ∈ G n,k , which contains a vertex of outdegree k. Then λ(G) ≥ λ(K(n, k, n − k − 1)). Proof. Let w be a vertex of G such that d+ w = k. Then the arcs out-incident to w form an arc cut set of size k. Adding all possible arcs from G\{w} to G\{w}∪{w}, we will arrive at a digraph H, which is isomorphic to K(n, k, n − k − 1), the arc connectivity of H remains equal to k. By Lemma 2.7, if G = K(n, k, n − k − 1), then λ(G) > λ(K(n, k, n − k − 1)). So the result follows. 2 Lemma 5.2. Let G ∈ G n,k , which contains a vertex of indegree k. Then λ(G) ≥ λ(K(n, k, 1)). Proof. Let w be a vertex of G such that d− w = k. Then the arcs in-incident to w form an arc cut set of size k. Adding all possible arcs from w to G\{w}, and all possible arcs from G\{w} to G\{w}, we will arrive at a digraph H  , which is isomorphic to K(n, k, 1),

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the arc connectivity of H  remains equal to k. By Lemma 2.7, if G = K(n, k, 1), then λ(G) > λ(K(n, k, 1)). So the result follows. 2 Theorem 5.3. Let G ∈ G n,k with 1 ≤ k ≤ n − 2. Then λ(G) ≥

3n − 3 +

√

n2 + 6n − 8k − 7 , 2

equality holds if and only if G ∼ = K(n, k, 1). Proof. Let G be a digraph in G n,k . Note that each vertex in the digraph G has outdegree no less than k and indegree no less than k, otherwise G ∈ / G n,k . Then we discuss the following two cases. − + Case 1. If there exists a vertex u of G with d+ u = k or du = k. If du = k, by Lemma 5.1, λ(G) ≥ λ(K(n, k, n−k−1)). If d− However, by u = k, by Lemma 5.2, λ(G) ≥ λ(K(n, k, 1)). √ 3n−3+ n2 +6n−8k−7 Lemmas 2.10 and 2.8, we have λ(K(n, k, n−k−1)) > λ(K(n, k, 1)) = . 2 So if there exists a vertex u of G with outdegree or indegree k, the digraph K(n, k, 1) is the unique digraph which has the minimum distance signless Laplacian spectral radius among all digraphs in G n,k . The result follows in this case. Case 2. We suppose that all vertices of G have outdegree greater than k and indegree greater than k. Let S be an arc cut set of G containing k arcs, then G − S consists of exactly two strongly connected components G1 , G2 , with order n1 , n2 , respectively. Without loss of generality, we may assume that there are no arcs from G1 to G2 in G − S. Next we create a new digraph G by adding to G any possible arcs from G2 to G1 ∪ G2 or any possible arcs from G1 to G1 that were not present in G. Obviously, the arc connectivity of G remains equal to k and all vertices of G have outdegree greater than k and indegree still greater than k. By Lemma 2.7, the addition of any such arc ←→

←→

will give λ(G) > λ(G ). Let H  = Kn1 ∪ Kn2 , U = {u1 , u2 , · · · , uk } be a set of k vertices ←→

←→

in V (Kn1 ) and W = {v1 , v2 , · · · , vk } be a set of k vertices in V (Kn2 ). Let H0 be a digraph obtained from H  by adding all possible arcs from U to W , and adding all ←→

←→

possible arcs from Kn2 to Kn1 . Note that G is a spanning subdigraph of H0 , therefore, by Corollary 2.5, λ(G ) ≥ λ(H0 ). However, we can know the vertex connectivity of H0 is k and H0  K(n, k, 1). Hence, by Lemma 2.10 we know that λ(H0 ) > λ(K(n, k, 1)). Therefore, λ(G) ≥ λ(G ) ≥ λ(H0 ) > λ(K(n, k, 1)). Consequently, K(n, k, 1) is the unique digraph which has the minimum distance signless Laplacian spectral radius among all digraphs in G n,k . That is to say, for any G ∈ G n,k with 1 ≤ k ≤ n − 2, λ(G) ≥

3n − 3 +

√

n2 + 6n − 8k − 7 , 2

equality holds if and only if G ∼ = K(n, k, 1). This completes the proof. 2

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D. Li et al. in [10] proved that Cn has the maximum distance signless Laplacian spectral radius among all strongly connected digraphs on n vertices, where Cn denotes the directed cycle on n vertices. It is not difficult to see that Cn has the maximum distance signless Laplacian spectral radius in G n,1 because of κ(Cn ) = 1. If κ(G) = ←→

←→

n − 1, then the complete digraph Kn is the unique digraph in G n,n−1 , therefore Kn has the maximum distance signless Laplacian spectral radius in G n,n−1 . So we propose the following question. Question 5.4. What is the digraph G in G n,k whose distance signless Laplacian spectral radius is maximum for each k = 2, 3, . . . , n − 2? ∗

6. The digraph with minimum distance signless Laplacian spectral radius in G n,k In Section 5, we prove that the digraph K(n, k, 1) is the unique digraph with the minimum distance signless Laplacian spectral among all digraphs in G n,k . Since K(n, k, 1) ∗ / G n,k . Therefore, we want to know which has a vertex of indegree k, we get K(n, k, 1) ∈ digraph has the minimum distance signless Laplacian spectral radius among all digraphs ∗ ∗ in G n,k . Moreover, by Lemma 2.11, if G ∈ G n,k , then n ≥ 2k +4. Furthermore, we deduce n−4 1 ≤ k ≤ n−4 2 . So in this section, we only consider the cases 1 ≤ k ≤ 2 . Firstly, we start with some special cases when k = 1, 2. Let 3 ≤ a ≤ n − 3, and G(n, a) be a digraph defined as in Section 4. ∗

Theorem 6.1. Let G ∈ G n,1 . Then λ(G) ≥ λ(G(n, n − 3)) with the equality if and only if G∼ = G(n, n − 3). Moreover, λ(G(n, n − 3)) is the largest root of the equation x3 − (4n + 1)x2 + (5n2 − 5n + 30)x − 2n3 + 6n2 − 30n + 12 = 0. ∗

Proof. Since G ∈ G n,1 , all vertices of G have outdegree greater than 1 and indegree greater than 1. Let E1 be an arc cut set of G containing one arc, then G − E1 consists of exactly two strongly connected components G1 , G2 with order a, b and a + b = n, respectively. Without loss of generality, we may assume that there are no arcs from G1 to G2 in G − E1 . By Lemma 2.11, a ≥ 1 + 2 = 3, b = n − a ≥ 1 + 2 = 3, then 3 ≤ a ≤ n − 3, n ≥ a + 3 ≥ 6. Next we get a new digraph G by adding to G any possible arcs from G2 to G1 ∪ G2 or any possible arcs from G1 to G1 that were not present in G. Obviously, G ∼ = G(n, a), and the arc connectivity of G remains equal to  1 and all vertices of G have outdegree greater than one and indegree still greater than one. By Lemma 2.7, the addition of any such arc will give λ(G) > λ(G ) = λ(G(n, a)). ←→

←→

Let v0 be the unique vertex in Ka out-adjacent to one vertex in Kn −a . Let X be the Perron vector of DQ (G(n, a)), the coordinate corresponding to v0 is x0 . It is easy to ←→

know that all coordinates of X corresponding to vertices of V ( Ka )\{v0 } are equal, say ←→

x1 , all coordinates corresponding to vertices of V (Kn −a ) are equal, say x2 . Therefore, we get

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⎧ ⎪ ⎪ ⎨(2n − a − 2)x0 + (a − 1)x1 + x2 + 2(n − a − 1)x2 = λ(G(n, a))x0 , (3n − 2a − 2)x1 + x0 + (a − 2)x1 + 2x2 + 3(n − a − 1)x2 = λ(G(n, a))x1 , ⎪ ⎪ ⎩x + (a − 1)x + (n − 1)x + (n − a − 1)x = λ(G(n, a))x . 0

1

2

2

2

Or equivalently ⎛

2n − a − 2 ⎜ 1 ⎝ 1

⎞⎛ ⎞ ⎛ ⎞ a−1 2n − 2a − 1 x0 x0 ⎟⎜ ⎟ ⎜ ⎟ 3n − a − 4 3n − 3a − 1 ⎠ ⎝ x1 ⎠ = λ(G(n, a)) ⎝ x1 ⎠ . x2 x2 a−1 2n − a − 2

It is easy to know that λ(G(n, a)) is the largest real root of the equation x3 + (−7n + 3a + 8)x2 + (−17na + 6a2 − 35n + 15a + 16n2 + 21)x + 18 + 40n2 − 12n3 + 4a3 + 20a − 16na2 − 46na + 18a2 + 22n2 a − 44n, where 3 ≤ a ≤ n − 3. If 6 ≤ n ≤ 8, then we can easily check that G(n, a) is minimal when a = n − 3, and λ(G(n, n − 3)) is the largest real root of the equation x3 − (4n + 1)x2 + (5n2 − 5n + 30)x − 2n3 + 6n2 − 30n + 12 = 0. Therefore, in the following, we assume that n ≥ 9. In order to find the digraph which has the minimum distance signless Laplacian spectral radius in G(n, a), we discuss the following two cases. Case 1. n2 < a ≤ n − 3. If a < n − 3, then n2 < a ≤ n − 4. In the following, we want to compare λ(G(n, a)) and λ(G(n, a + 1)). Let fa (x) = x3 + (−7n + 3a + 8)x2 + (−17na + 6a2 − 35n + 15a + 16n2 + 21)x + 18 + 40n2 − 12n3 + 4a3 + 20a − 16na2 − 46na + 18a2 + 22n2 a − 44n. λ(G(n, a)) is the largest real root of fa (x) = 0. Then we get fa+1 (x) − fa (x) = x3 + (−7n + 3(a + 1) + 8)x2 + (−17n(a + 1) + 6(a + 1)2 − 35n + 15(a + 1) + 16n2 + 21)x + 18 + 40n2 − 12n3 + 4(a + 1)3 + 20(a + 1) − 16n(a + 1)2 − 46n(a + 1) + 18(a + 1)2 + 22n2 (a + 1) − 44n − [x3 + (−7n + 3a + 8)x2 + (−17na + 6a2 − 35n + 15a + 16n2 + 21)x + 18 + 40n2 − 12n3 + 4a3 + 20a − 16na2 − 46na + 18a2 + 22n2 a − 44n] = 3x2 + (12a − 17n + 21)x + 42 − 32na + 12a2 + 22n2 − 62n + 48a, with axis of symmetry x =

17n − 12a − 21 < 3n − a − 4. 6

On the other hand, from λ(G(n, a))x1 = (3n − a − 4)x1 + x0 + (3n − 3a − 1)x2 > (3n − a − 4)x1 , we get λ(G(n, a)) > 3n − a − 4 > x . Note that fa (λ(G(n, a))) = 0. Then for all x ≥ λ(G(n, a)),

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fa+1 (x) − fa (x) = 3x2 + (12a − 17n + 21)x + 42 − 32na + 12a2 + 22n2 − 62n + 48a 2

≥ 3λ(G(n, a)) + (12a − 17n + 21)λ(G(n, a)) + 42 − 32na + 12a2 + 22n2 − 62n + 48a > 3(3n − a − 4)2 + (12a − 17n + 21)(3n − a − 4) + 42 − 32na + 12a2 + 22n2 − 62n + 48a = −2n2 + 3na − 3n + 3a2 + 3a + 6  n 2  n n n + 3 × + 6 for a ≥ > −2n2 + 3n × − 3n + 3 × 2 2 2 2 1 = (n2 − 6n + 24) > 0. 4 Obviously, λ(G(n, a + 1)) is the largest real root of fa+1 (x) = 0, together with the fact that fa+1 (x) ≥ 0 = fa+1 (λ(G(n, a + 1))) for all x ≥ λ(G(n, a + 1)). Hence λ(G(n, a)) > λ(G(n, a + 1)). Therefore, if n2 < a ≤ n − 3, we get λ(G(n, a)) is minimal when a = n − 3, and λ(G(n, n − 3)) is the largest real root of the equation x3 − (4n + 1)x2 + (5n2 − 5n + 30)x − 2n3 + 6n2 − 30n + 12 = 0. Take g(x) = x3 − (4n + 1)x2 + (5n2 − 5n + 30)x − 2n3 + 6n2 − 30n + 12, then λ(G(n, n − 3)) is the largest real root of g(x) = 0. In the following, we want to prove λ(G(n, n − 3)) < 3n − 5. Since g  (x) = 6x − 2(4n + 1) ≥ 6(3n − 5) − 2(4n + 1) = 10n − 32 > 0 for all x ≥ 3n − 5. So g  (x) is increasing for all x ≥ 3n − 5, g  (x) ≥ g  (3n − 5) = 3(3n − 5)2 − 2 × (4n + 1) × (3n − 5) + 5n2 − 5n + 30 = 8n2 − 61n + 115 > 0 for n ≥ 9. Furthermore, we get g(x) is increasing for all x ≥ 3n − 5, i.e., for all x ≥ 3n − 5, g(x) ≥ g(3n − 5) = 4n3 − 58n2 + 240n − 288 > 0 for n ≥ 9. Hence, we get λ(G(n, n − 3)) < 3n − 5. That is, if n 2 < a ≤ n − 3, we get λ(G(n, a)) ≥ λ(G(n, n − 3)) with equality if and only if a = n − 3 and λ(G(n, n − 3)) < 3n − 5. Case 2. 3 ≤ a ≤ n2 . In the following, we want to prove λ(G(n, a)) > 3n − 5 when 3 ≤ a ≤ n2 . Let fa (x) = x3 + (−7n + 3a + 8)x2 + (−17na + 6a2 − 35n + 15a + 16n2 + 21)x + 18 + 40n2 − 12n3 + 4a3 + 20a − 16na2 − 46na + 18a2 + 22n2 a − 44n. λ(G(n, a)) is the largest real root of fa (x) = 0. Take l(a) = fa (3n − 5) = 4a3 + (2n − 12)a2 + (−6n2 − 20n + 20)a + 2n2 + 4n − 12. Then l (a) = 24a + 2(2n − 12) > 0. Thus, for fixed n, the maximal value of l(a) must be taken at either a = 3 or a = n2 . Hence, fa (3n − 5) = l(a) ≤ max{l(3), l( n2 )} = max{−4n2 + 4n + 48, −4n2 + 14n − 12} < 0 for n ≥ 9. Hence λ(G(n, a)) > 3n − 5 > λ(G(n, n − 3)) for all 3 ≤ a ≤ n2 . Thus combining the above two cases, we get λ(G) ≥ λ(G(n, n − 3)) with the first equality if and only if G ∼ = G(n, n − 3). Moreover, λ(G(n, n − 3)) is the largest real root of the equation x3 − (4n + 1)x2 + (5n2 − 5n + 30)x − 2n3 + 6n2 − 30n + 12 = 0. 2 For 4 ≤ a ≤ n − 4, let Hi (n, a) (i = 1, 2, 3) be a digraph defined as in Section 4.

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∗

Theorem 6.2. Let G ∈ G n,2 . Then λ(G) ≥ λ(H3 (n, n − 4)) with the equality if and only if G ∼ = H3 (n, n − 4). Moreover, λ(H3 (n, n − 4)) is the largest real root of the equation x3 − (4n + 3)x2 + (5n2 − n + 54)x − 2n3 + 4n2 − 46n − 16 = 0. ∗

Proof. Since G ∈ G n,2 , all vertices of G have outdegree greater than 2 and indegree greater than 2. Let E2 be an arc cut set of G containing 2 arcs, then G − E2 consists of exactly two strongly connected components G1 , G2 , respectively, of orders a, b and a + b = n. Without loss of generality, we may assume that there are no arcs from G1 to G2 in G − E2 . By Lemma 2.11, a ≥ 2 + 2 = 4, b = n − a ≥ 2 + 2 = 4, then 4 ≤ a ≤ n − 4,  n ≥ a +4 ≥ 8. Next we get a new digraph G by adding to G any possible arcs from G2 to G1 ∪G2 or any possible arcs from G1 to G1 that were not present in G. Obviously, the arc   connectivity of G remains equal to 2 and all vertices of G have outdegree greater than   2 and indegree still greater than 2. Then, we have G ∼ = H1 (n, a), or G ∼ = H2 (n, a), or  ∼ G = H3 (n, a). By Lemma 2.7, the addition of any such arc will give λ(G) > λ(Hi (n, a)) for i = 1, 2, 3.  If G ∼ = H1 (n, a). Let X1 be the Perron vector of DQ (H1 (n, a)), the coordinate corresponding to u is x0 . It is easy to know that all coordinates of X1 corresponding to ←→

vertices of V ( Ka )\{u} are equal, say x1 , all coordinates corresponding to vertices of ←→

V (Kn −a ) are equal, say x2 . Therefore, we get ⎧ ⎪ ⎪ ⎨(2n − a − 3)x0 + (a − 1)x1 + 2x2 + 2(n − a − 2)x2 = λ(H1 (n, a))x0 ,

x0 + (3n − 2a − 3)x1 + (a − 2)x1 + 2 × 2x2 + 3(n − a − 2)x2 = λ(H1 (n, a))x1 , ⎪ ⎪ ⎩x + (a − 1)x + (n − 1)x + (n − a − 1)x = λ(H (n, a))x . 0

1

2

2

1

2

Or equivalently ⎛

2n − a − 3 ⎜ 1 ⎝ 1

a−1 3n − a − 5 a−1

⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 2n − 2a − 2 x0 x0 x0 ⎜ ⎟ ⎟ ⎜ ⎟ def ⎜ ⎟ 3n − 3a − 2 ⎠ ⎝ x1 ⎠ = B1 ⎝ x1 ⎠ = λ(H1 (n, a)) ⎝ x1 ⎠ . x2 x2 x2 2n − a − 2

Then λ(H1 (n, a)) is the largest eigenvalue or the spectral radius of the matrix B1 .  Similarly, if G ∼ = H2 (n, a). Let Y be the Perron vector of DQ (H2 (n, a)). It is easy to know that the coordinates corresponding to u, v are equal, say, y0 , all coordinates of Y ←→

corresponding to vertices of V ( Ka )\{u, v} are equal, say y1 , all coordinates corresponding ←→

to vertices of V (Kn −a ) are equal, say y2 . Therefore, we get ⎛

⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 2n − a − 1 a−2 2n − 2a − 1 y0 y0 y0 ⎜ ⎟ ⎜ ⎟ def ⎜ ⎟ ⎜ ⎟ 2 3n − a − 5 3n − 3a − 1 ⎠ ⎝ y1 ⎠ = B2 ⎝ y1 ⎠ = λ(H2 (n, a)) ⎝ y1 ⎠ . ⎝ y2 y2 y2 2 a−2 2n − a − 2 Then λ(H2 (n, a)) is the largest eigenvalue or the spectral radius of the matrix B2 .

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If G ∼ = H3 (n, a). Let Z be the Perron vector of DQ (H3 (n, a)). It is easy to know, the coordinates corresponding to u, v are equal, say, z0 , all coordinates of Z corresponding ←→

to vertices of V ( Ka )\{u, v} are equal, say z1 , all coordinates corresponding to vertices ←→

of V (Kn −a ) are equal, say z2 . Therefore, we get ⎛

2n − a − 1 ⎜ 2 ⎝ 2

⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ z0 a−2 2n − 2a − 1 z0 z0 ⎟ ⎜ ⎟ def ⎜ ⎟ ⎜ ⎟ 3n − a − 6 3n − 3a − 2 ⎠ ⎝ z1 ⎠ = B3 ⎝ z1 ⎠ = λ(H3 (n, a)) ⎝ z1 ⎠ . a−2 2n − a − 2 z2 z2 z2

Then λ(H3 (n, a)) is the largest eigenvalue or the spectral radius of the matrix B3 . Because 0 < B3 < B2 and B2 irreducible, then from Lemma 2.3, we have λ(H3 (n, a)) < λ(H2 (n, a)). Therefore, in the following, it suffices to find the digraph which has the minimum distance signless Laplacian spectral radius in H1 (n, a) and H3 (n, a).  Case 1. G ∼ = H1 (n, a). It is easy to know that λ(H1 (n, a)) is the largest real root of the equation x3 + (−7n + 3a + 10)x2 + (−17na + 6a2 − 44n + 20a + 16n2 + 32)x + 32 + 36a − 68n − 16na2 − 60na + 24a2 + 50n2 + 22n2 a − 12n3 + 4a3 = 0, where 4 ≤ a ≤ n − 4. If 8 ≤ n ≤ 11, then we can easily check that H1 (n, a) is minimal when a = n − 4, and λ(H1 (n, n − 4)) is the largest real root of the equation x3 − (4n + 2)x2 + (5n2 − 4n + 48)x − 2n3 + 6n2 − 48n + 16 = 0. Therefore, in the following, we assume that n ≥ 12. In order to find the digraph which has the minimum distance signless Laplacian spectral radius in H1 (n, a), we discuss the following two cases. Subcase 1. n2 < a ≤ n − 4. If a < n − 4, then n2 < a ≤ n − 5. In the following, we want to compare λ(H1 (n, a)) and λ(H1 (n, a + 1)). Let θa (x) = x3 + (−7n + 3a + 10)x2 + (−17an + 6a2 + 16n2 + 32 − 44n + 20a)x + 32 − 60an + 24a2 + 22an2 − 16na2 + 50n2 + 4a3 + 36a − 12n3 − 68n. λ(H1 (n, a)) is the largest real root of θa (x) = 0. Then we have θa+1 (x) − θa (x) = x3 + (−7n + 3(a + 1) + 8)x2 + (−17n(a + 1) + 6(a + 1)2 + 16n2 + 32 − 44n + 20(a + 1))x + 32 − 60n(a + 1) + 24(a + 1)2 + 22n2 (a + 1) − 16n(a + 1)2 + 50n2 + 4(a + 1)3 + 36(a + 1) − 12n3 − 68n − [x3 + (−7n + 3a + 10)x2 + (−17an + 6a2 + 16n2 + 32 − 44n + 20a)x + 32 − 60an + 24a2 + 22an2 − 16na2 + 50n2 + 4a3 + 36a − 12n3 − 68n] = 3x2 + (12a − 17n + 26)x + 64 − 32na + 12a2 + 22n2 − 76n + 60a, with axis of symmetry x =

17n − 12a − 26 < 3n − a − 5. 6

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On the other hand, from λ(H1 (n, a))x1 = x0 + (3n − 2a − 3)x1 + (a − 2)x1 + 2 × 2x2 + 3(n − a − 2)x2 > (3n − a − 5)x1 , we get λ(H1 (n, a)) > 3n − a − 5 > x . Note that θa (λ(H1 (n, a))) = 0. Then for all x ≥ λ(H1 (n, a)), θa+1 (x) − θa (x) = 3x2 + (12a − 17n + 26)x + 64 − 32na + 12a2 + 22n2 − 76n + 60a 2

≥ 3λ(H1 (n, a)) + (12a − 17n + 26)λ(H1 (n, a)) + 64 − 32na + 12a2 + 22n2 − 76n + 60a > 3(3n − a − 5)2 + (12a − 17n + 26)(3n − a − 5) + 64 − 32na + 12a2 + 22n2 − 76n + 60a = −2n2 + 3na − 3n + 3a2 + 4a + 9  n 2  n n n ≥ −2n2 + 3n × − 3n + 3 × + 4 × + 9 for a ≥ 2 2 2 2 1 = (n2 − 4n + 36) > 0. 4 Obviously, λ(H1 (n, a + 1)) is the largest real root of θa+1 (x) = 0, together with the fact that θa+1 (x) ≥ 0 = θa+1 (λ(H1 (n, a + 1))) for all x ≥ λ(H1 (n, a + 1)). Hence λ(H1 (n, a)) > λ(H1 (n, a + 1)). Therefore, if n2 < a ≤ n − 4, we get λ(H1 (n, a)) is minimal when a = n − 4, and λ(H1 (n, n − 4)) is the largest real root of the equation x3 − (4n + 2)x2 + (5n2 − 4n + 48)x − 2n3 + 6n2 − 48n + 16 = 0. Take h(x) = x3 − (4n + 2)x2 + (5n2 − 4n + 48)x − 2n3 + 6n2 − 48n + 16, then λ(H1 (n, n − 4)) is the largest real root of h(x) = 0. In the following, we want to prove λ(H1 (n, n − 4)) < 3n − 5. Since h (x) = 6x − 2(4n + 2) ≥ 6(3n − 5) − 2(4n + 2) = 10n − 34 > 0 for all x ≥ 3n − 5. So h (x) is increasing for all x ≥ 3n − 5, h (x) ≥ h (3n − 5) = 3(3n − 5)2 − 2 × (4n + 2) × (3n − 5) + 5n2 − 4n + 48 = 8n2 − 66n + 143 > 0 for n ≥ 12. Furthermore, we get h(x) is increasing for all x ≥ 3n − 5, i.e., for all x ≥ 3n − 5, h(x) ≥ h(3n − 5) = 4n3 − 64n2 + 301n − 399 > 0 for n ≥ 12. Hence, we get λ(H1 (n, n − 4)) < 3n − 5. That is, if n2 < a ≤ n − 4, we get λ(H1 (n, a)) ≥ λ(H1 (n, n − 4)) with equality if and only if a = n − 4 and λ(H1 (n, n − 4)) < 3n − 5. Subcase 2. 4 ≤ a ≤ n2 . In the following, we want to prove λ(H1 (n, a)) > 3n − 5 when 4 ≤ a ≤ n2 . Let θa (x) = x3 +(−7n+3a+10)x2 +(−17an+6a2 +16n2 +32−44n+20a)x+32−60an+ 24a2 + 22an2 − 16na2 + 50n2 + 4a3 + 36a − 12n3 − 68n. λ(H1 (n, a)) is the largest real root of θa (x) = 0. Take p(a) = θa (3n −5) = 4a3 +(2n −6)a2 +(−2n2 −5n +11)a +3n2 −2n −3. Then p (a) = 24a + 2(2n − 6) > 0. Thus, for fixed n, the maximal value of p(a) must be taken at either a = 4 or a = n2 . Hence, θa (3n − 5) = p(a) ≤ max{p(4), p( n2 )} = max{−5n2 + 10n + 201, −n2 + 72 n − 3} < 0 for n ≥ 12. Hence λ(H1 (n, a)) > 3n − 5 > λ(H1 (n, n − 4)) for all 4 ≤ a ≤ n2 .

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Thus combining the above two subcases, we get λ(H1 (n, a)) is minimal when a = n −4. Moreover, λ(H1 (n, n − 4)) is the largest real root of the equation x3 − (4n + 2)x2 + (5n2 − 4n + 48)x − 2n3 + 6n2 − 48n + 16 = 0.  Case 2. G ∼ = H3 (n, a). It is easy to know that λ(H3 (n, a)) is the largest real root of the 3 equation x +(−7n +3a +9)x2 +(−17an +6a2 +16n2 +22 −37n +16a)x +16 −50an +20a2 + 22an2 − 16na2 + 42n2 + 4a3 + 24a − 12n3 − 46n, where 4 ≤ a ≤ n − 4. If 8 ≤ n ≤ 11, then we can easily check that H3 (n, a) is minimal when a = n − 4, then λ(H3 (n, n − 4)) is the largest real root of the equation x3 −(4n +3)x2 +(5n2 −n +54)x −2n3 +4n2 −46n −16 = 0. Therefore, in the following, we assume that n ≥ 12. In order to find the digraph which has the minimum distance signless Laplacian spectral radius in H3 (n, a), we discuss the following two cases. n+2 Subcase 1. n+2 2 < a ≤ n − 4. If a < n − 4, then 2 < a ≤ n − 5. In the following, we want to compare λ(H3 (n, a)) and λ(H3 (n, a + 1)). Let φa (x) = x3 + (−7n + 3a + 9)x2 + (−17an + 6a2 + 16n2 + 22 − 37n + 16a)x + 16 − 50an + 20a2 + 22an2 − 16na2 + 42n2 + 4a3 + 24a − 12n3 − 46n. λ(H3 (n, a)) is the largest real root of φa (x) = 0. Then we get φa+1 (x) − φa (x) = x3 + (−7n + 3(a + 1) + 9)x2 + (−17n(a + 1) + 6(a + 1)2 + 16n2 + 22 − 37n + 16(a + 1))x + 16 − 50n(a + 1) + 20(a + 1)2 + 22n2 (a + 1) − 16n(a + 1)2 + 42n2 + 4(a + 1)3 + 24(a + 1) − 12n3 − 46n − [x3 + (−7n + 3a + 9)x2 + (−17an + 6a2 + 16n2 + 22 − 37n + 16a)x + 16 − 50an + 20a2 + 22an2 − 16na2 + 42n2 + 4a3 + 24a − 12n3 − 46n] = 3x2 + (12a − 17n + 22)x + 48 − 32na + 12a2 + 22n2 − 66n + 52a, with axis of symmetry x =

17n − 12a − 22 < 3n − a − 6. 6

On the other hand, from λ(H3 (n, a))z1 = 2z0 + (3n − 2a − 3)z1 + (a − 3)z1 + 2 × 2z2 + 3(n − a − 2)z2 > (3n − a − 6)z1 , we get λ(H3 (n, a)) > 3n − a − 6 > x . Note that φa (λ(H3 (n, a))) = 0. Then for all x ≥ λ(H3 (n, a)), φa+1 (x) − φa (x) = 3x2 + (12a − 17n + 22)x + 48 − 32na + 12a2 + 22n2 − 66n + 52a 2

≥ 3λ(H3 (n, a)) + (12a − 17n + 22)λ(H3 (n, a)) + 48 − 32na + 12a2 + 22n2 − 66n + 52a > 3(3n − a − 6)2 + (12a − 17n + 22)(3n − a − 6) + 48 − 32na + 12a2 + 22n2 − 66n + 52a = −2n2 + 3na − 6n + 3a2 − 6a + 24.

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Take ν(a) = −2n2 + 3na − 6n + 3a2 − 6a + 24 = 3a2 + (3n − 6)a − 2n2 − 6n + 24, with axis of symmetry  a = 6−3n < n+2 6 2 < a. Then for all x ≥ λ(H3 (n, a)), φa+1 (x) − φa (x) > ν(a) 2    n+2 n+2 n+2 + (3n − 6) × − 2n2 − 6n + 24 for a > >3× 2 2 2 =

1 2 (n − 12n + 84) > 0. 4

Obviously, λ(H3 (n, a + 1)) is the largest root of φa+1 (x) = 0, together with the fact that φa+1 (x) ≥ 0 = φa+1 (λ(H3 (n, a + 1))) for all x ≥ λ(H3 (n, a + 1)). Hence λ(H3 (n, a)) > λ(H3 (n, a + 1)). Therefore, if n+2 2 < a ≤ n − 4, we get λ(H3 (n, a)) is minimal when a = n − 4, and λ(H3 (n, n − 4)) is the largest real root of the equation x3 − (4n + 3)x2 + (5n2 − n + 54)x − 2n3 +4n2 −46n−16 = 0. Take α(x) = x3 −(4n+3)x2 +(5n2 −n+54)x−2n3 +4n2 −46n−16, then λ(H3 (n, n − 4)) is the real largest root of α(x) = 0. In the following, we want to prove λ(H3 (n, n − 4)) < 3n − 7. Since α (x) = 6x − 2(4n + 3) ≥ 6(3n − 7) − 2(4n + 3) = 10n − 48 > 0 for all x ≥ 3n − 7. So α (x) is increasing for all x ≥ 3n − 7, α (x) ≥ α (3n − 7) = 3(3n − 7)2 − 2 × (4n + 3) × (3n − 7) + 5n2 − n + 54 = 8n2 − 89n + 135 > 0 for n ≥ 12. Furthermore, we get α(x) is increasing for all x ≥ 3n − 7, i.e., for all x ≥ 3n − 7, α(x) ≥ α(3n − 7) = 4n3 − 82n2 + 494n − 884 > 0 for n ≥ 12. Hence, we get λ(H3 (n, n − 4)) < 3n − 7. That is, if n+2 2 < a ≤ n − 4, we get λ(H3 (n, a)) ≥ λ(H3 (n, n − 4)) with equality if and only if a = n − 4 and λ(H3 (n, n − 4)) < 3n − 7. Subcase 2. 4 ≤ a ≤ n+2 2 . In the following, we want to prove λ(H3 (n, a)) > 3n − 7 when 4 ≤ a ≤ n+2 2 Let φa (x) = x3 + (−7n + 3a + 9)x2 + (−17an + 6a2 + 16n2 + 22 − 37n + 16a)x + 16 − 50an + 20a2 + 22an2 − 16na2 + 42n2 + 4a3 + 24a − 12n3 − 46n. λ(H3 (n, a)) is the largest real root of φa (x) = 0. Take ϕ(a) = φa (3n − 7) = 4a3 + (2n − 22)a2 + (−2n2 − 9n + 59)a + 5n2 − n − 40. Then ϕ (a) = 24a + 2(2n − 22) > 0. Thus, for fixed n, the maximal value of ϕ(a) must be taken at either a = 4 or a = n+2 2 . Hence, n+2 11 2 2 φa (3n − 7) = ϕ(a) ≤ max{ϕ(4), ϕ( 2 )} = max{−3n − 5n + 100, −2n + 2 n + 1} < 0 for n ≥ 12. Hence λ(H3 (n, a)) > 3n − 7 > λ(H3 (n, n − 4)) for all 4 ≤ a ≤ n+2 2 . Thus, combining the above two subcases, we get λ(H3 (n, a)) is minimal when a = n−4. Moreover, λ(H3 (n, n − 4)) is the largest real root of the equation x3 − (4n + 3)x2 + (5n2 − n + 54)x − 2n3 + 4n2 − 46n − 16 = 0.  Hence, if G ∼ = H1 (n, a), λ(H1 (n, a)) ≥ λ(H1 (n, n − 4)) with the equality if and only if  a = n − 4; if G ∼ = H3 (n, a), λ(H3 (n, a)) ≥ λ(H3 (n, n − 4)) with the equality if and only if a = n − 4. Therefore, in order to find the digraph which has the minimum distance ∗ signless Laplacian spectral radius among all digraphs in G n,2 , it suffices to compare the distance signless Laplacian spectral radius of H1 (n, n − 4) and H3 (n, n − 4).

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Let h(x) = x3 − (4n + 2)x2 + (5n2 − 4n + 48)x − 2n3 + 6n2 − 48n + 16 and α(x) = x3 − (4n + 3)x2 + (5n2 − n + 54)x − 2n3 + 4n2 − 46n − 16. Moreover, from the above proof, we can know λ(H1 (n, n − 4)) is the real largest root of h(x) = 0 and λ(H3 (n, n − 4)) is the real largest root of α(x) = 0. For 8 ≤ n ≤ 11, we can easily check that λ(H1 (n, n − 4)) > λ(H3 (n, n − 4)). Then, for 8 ≤ n ≤ 11, λ(G) ≥ λ(H3 (n, n − 4)) with the equality if and only if G ∼ = H3 (n, n − 4). Moreover, λ(H3 (n, n − 4)) is the largest real root of the equation x3 − (4n + 3)x2 + (5n2 − n + 54)x − 2n3 + 4n2 − 46n − 16 = 0. So the result follows in this case. Next we assume that n ≥ 12. Let η(x) = h(x) − α(x) = x3 − (4n + 2)x2 + (5n2 − 4n + 48)x − 2n3 + 6n2 − 48n + 16   − x3 − (4n + 3)x2 + (5n2 − n + 54)x − 2n3 + 4n2 − 46n − 16 = x2 − (3n + 6)x + 2n2 − 2n + 32. √

2

√

2

It is easy to know that η(x) < 0 when 3n+6− n2 +44n−92 < x < 3n+6+ n2 +44n−92 . √ √ 2 2 For 12 ≤ n ≤ 17, 3n+6− n2 +44n−92 < n, and 3n+6+ n2 +44n−92 > 2n + 9. In the following, we want to prove λ(H3 (n, n − 4)) < 2n + 9. Since α (x) = 6x − 2(4n + 3) ≥ 6(2n + 9) − 2(4n + 3) = 4n + 48 > 0 for all x ≥ 2n + 9. So α (x) is increasing for all x ≥ 2n + 9, α (x) ≥ α (2n + 9) = 3(2n + 9)2 − 2 × (4n + 3) × (2n + 9) + 5n2 − n + 54 = n2 + 23n + 243 > 0 for all x ≥ 2n + 9. Furthermore, we get α(x) is increasing for all x ≥ 2n + 9, i.e., for all x ≥ 2n + 9, α(x) ≥ α(2n + 9) = −n2 + 107n + 956 > 0 for 12 ≤ n ≤ 17. Hence, we get λ(H3 (n, n − 4)) < 2n + 9 for 12 ≤ n ≤ 17. Furthermore, by Lemma 2.1, we get λ(H3 (n, n − 4)) > 2n − 2. Therefore, for 12 ≤ √ √ 2 2 n ≤ 17, 3n+6− n2 +44n−92 < n < 2n − 2 < λ(H3 (n, n − 4)) < 2n + 9 < 3n+6+ n2 +44n−92 . Thus η(λ(H3 (n, n − 4))) = h(λ(H3 (n, n − 4))) − α(λ(H3 (n, n − 4))) < 0. Note that α(λ(H3 (n, n − 4))) = 0. Hence, h(λ(H3 (n, n − 4))) = η(λ(H3 (n, n − 4))) < 0, together with the fact that h(x) ≥ 0 = h(λ(H1 (n, n − 4))) for all x ≥ λ(H1 (n, n − 4)), which implies that λ(H (n, n − 4)) > λ(H3 (n, n − 4)). So the result follows. √ 1 √ 2 2 For n ≥ 18, 3n+6− n2 +44n−92 < n, and 3n+6+ n2 +44n−92 ≥ 2n + 10. In the following, we want to prove λ(H3 (n, n − 4)) < 2n + 10. Since α (x) = 6x − 2(4n + 3) ≥ 6(2n + 10) − 2(4n + 3) = 4n + 54 > 0 for all x ≥ 2n + 10. So α (x) is increasing for all x ≥ 2n + 10, α (x) ≥ α (2n + 10) = 3(2n + 10)2 − 2 × (4n + 3) × (2n + 10) + 5n2 − n + 54 = n2 + 27n + 294 > 0 for all x ≥ 2n + 10. Furthermore, we get α(x) is increasing for all x ≥ 2n + 10, i.e., for all x ≥ 2n + 10, α(x) ≥ α(2n +10) = 132n +1224 > 0 for n ≥ 18. Hence, we get λ(H3 (n, n −4)) < 2n +10 for n ≥ 18.

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Therefore, for n ≥ 18, √ 3n−6+ n+ 44n−92 . Thus 2

√ 3n−6− n+ 44n−92 2

< n < 2n − 2 < λ(H3 (n, n − 4)) < 2n + 10 ≤

η(λ(H3 (n, n − 4))) = h(λ(H3 (n, n − 4))) − α(λ(H3 (n, n − 4))) < 0. Note that α(λ(H3 (n, n − 4))) = 0. Hence, h(λ(H3 (n, n − 4))) = η(λ(H3 (n, n − 4))) < 0, together with the fact that h(x) ≥ 0 = h(λ(H1 (n, n − 4))) for all x ≥ λ(H1 (n, n − 4)), which implies that λ(H1 (n, n − 4)) > λ(H3 (n, n − 4)). So the result follows. ∗ Therefore, for any G ∈ G n,2 , we have λ(G) > λ(H3 (n, n − 4)) with the equality if and only if G ∼ = H3 (n, n − 4). Moreover, λ(H3 (n, n − 4)) is the largest real root of the equation x3 − (4n + 3)x2 + (5n2 − n + 54)x − 2n3 + 4n2 − 46n − 16 = 0. 2 For general 1 ≤ k ≤ examples.

n−4 2 ,

we propose the following conjecture based on some numerical ∗

Conjecture 6.3. Let G be a digraph in G n,k with the minimum distance signless Laplacian ←→

←→

spectral radius. Then G is obtained from Kk +2 ∪ Kn −k−2 by adding k arcs from k vertices ←→

←→

←→

←→

in Kn −k−2 to k vertices in Kk +2 and adding all possible arcs from Kk +2 to Kn −k−2 . Declaration of Competing Interest No potential conflict of interest was reported by the authors. Acknowledgements The authors would like to thank the anonymous referee very much for his or her corrections and suggestions, which are very helpful for improving the presentation of the manuscript. References [1] A. Berman, R.J. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Academic Press, New York, 1979. [2] Ş.B. Bozkurt, D. Bozkurt, On the signless Laplacian spectral radius of digraphs, Ars Combin. 108 (2013) 193–200. [3] Ş.B. Bozkurt, D. Bozkurt, X.D. Zhang, On the spectral radius and the energy of a digraph, Linear Multilinear Algebra 63 (2015) 2009–2016. [4] S.W. Drury, H.Q. Lin, Colorings and spectral radius of digraphs, Discrete Math. 339 (2016) 327–332. [5] R.A. Horn, C.R. Johnson, Matrix Analysis, Cambridge University Press, New York, 1985. [6] W.X. Hong, L.H. You, Spectral radius and signless Laplacian spectral radius of strongly connected digraphs, Linear Algebra Appl. 457 (2014) 93–113. [7] Y.L. Jin, X.D. Zhang, On the spectral radius of simple digraphs with prescribed number of arcs, Discrete Math. 338 (2015) 1555–1564. [8] W.W. Lang, L.G. Wang, Sharp bounds for the signless Laplacian spectral radius of digraphs, Appl. Math. Comput. 238 (2014) 43–49.

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