The bipartite unicyclic graphs with the first ⌊n−34⌋ largest matching energies

Applied Mathematics and Computation 268 (2015) 644–656

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

The bipartite unicyclic graphs with the first  n−3  largest 4 matching energies Lin Chen, Jinfeng Liu∗ Center for Combinatorics and LPMC-TJKLC, Nankai University, Tianjin 300071, PR China

a r t i c l e

i n f o

Keywords: Matching energy Bipartite unicyclic graphs Quasi-order Coulson integral formula

a b s t r a c t The theory of matching energy of graphs since been proposed by Gutman and Wagner in 2012, has attracted more and more attention. It is denoted by Bn,m, the class of bipartite graphs with order n and size m. In particular, Bn,n denotes the set of bipartite unicyclic graphs, which is an interesting class of graphs. In this paper, for odd n, we characterize the bipartite unicyclic graphs with the first  n−3  largest matching energies. There is an interesting correspondence: 4 we conclude that the graph with the second maximal matching energy in Bn,n for odd n ≥ 11 is Pn6 , which is the only graph attaining the maximum value of the energy among all the (bipartite) unicyclic graphs for n ≥ 16. © 2015 Elsevier Inc. All rights reserved.

1. Introduction In theoretical chemistry and biology, molecular structure descriptors are used for modeling physical– chemical, toxicologic, pharmacologic, biological and other properties of chemical compounds. These descriptors are mainly divided into three types: degree-based indices, distance-based indices and spectrum-based indices. Degree-based indices [55] contain (general) Randic´ index [43,44], (general) zeroth order Randic´ index [31,32], Zagreb index [1,24,29,38,50,57,59], connective eccentricity index [63] and so on. Distance-based indices [61] include the Balaban index [14], the Wiener index [19,30,39,48,49,56] and Wiener polarity index [51], the Szeged index [2,20], ABC index [54], the Kirchhoff index [41], the Harary index [4]. Eigenvalues of graphs, various of graph energies [6–8,15,16,21,26,52], HOMO– LUMO index [45,53] belong to spectrum-based indices. Actually, there are also some topological indices defined based on both degrees and distances, such as degree distance [18], graph entropies [9]. In 1977, Gutman [22] proposed the concept of graph energy. The energy of a simple graph G is defined as the sum of the absolute values of its eigenvalues, namely,

E (G) =

n 

|λi |,

i=1

where λ1 , λ2 , . . . , λn denote the eigenvalues of G. The graph energy has been rather widely studied by theoretical chemists and mathematicians. For details, we refer the book on graph energy [46] and some new recent references [33,34,47]. A matching in a graph G is a set of pairwise nonadjacent edges. A matching M is called a k-matching if the size of M is k. Let m(G, k) denote the number of k-matchings of G, where m(G, 1) = m and m(G, k) = 0 for k >  n2  or k < 0. In addition, define

∗

Corresponding author. Tel.: +8615620601023. E-mail addresses: [email protected] (L. Chen), [email protected] (J. Liu).

http://dx.doi.org/10.1016/j.amc.2015.06.115 0096-3003/© 2015 Elsevier Inc. All rights reserved.

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

645

m(G, 0) = 1. Then the matching polynomial of the graph G is defined as

α(G) = α(G, μ) =



( − 1)k m(G, k)μn−2k .

k≥0

Similar to graph energy, in [28], Gutman and Wagner proposed the concept of matching energy. They defined the matching energy of a graph G as

ME(G) =

n 

|μi |,

i=1

where μi (i = 1, 2, . . . , n) are the roots of α(G, μ) = 0. Besides, Gutman and Wagner also gave the following equivalent definition of matching energy. Definition 1 ([28]). Let G be a simple graph, and let m(G, k) be the number of its k-matchings, k = 0, 1, 2, . . .. The matching energy of G is

ME = ME(G) =

2

π



∞ 0





 1 ln m(G, k)x2k dx. x2

(1)

k≥0

Formula (1) is called the Coulson integral formula of matching energy. Obviously, by the monotonicity of the logarithm function, this formula implies that the matching energy of a graph G is a monotonically increasing function of any m(G, k). Particularly, if G1 and G2 are two graphs for which m(G1 , k) ≥ m(G2 , k) holds for all k ≥ 0, then ME(G1 ) ≥ ME(G2 ). If, in addition, m(G1 , k) > m(G2 , k) for at least one k, then ME(G1 ) > ME(G2 ). Thus, we can define a quasi-order  as follows: If G1 and G2 are two graphs, then

G1  G2 ⇐⇒ m(G1 , k) ≥ m(G2 , k)

f or all

k.

(2)

And if G1  G2 we say that G1 is m-greater than G2 or G2 is m-smaller than G1 , which is also denoted by G2 G1 . If G1  G2 and G2  G1 , the graphs G1 and G2 are said to be m-equivalent, denote it by G1 ∼ G2 . If G1  G2 , but the graphs G1 and G2 are not m-equivalent (i.e., there exists some k such that m(G1 , k) > m(G2 , k)), then we say that G1 is strictly m-greater than G2 , write G1 G2 . If neither G1  G2 nor G2  G1 , the two graphs G1 and G2 are said to be m-incomparable and we denote this by G1 #G2 . According to Eq. (1) and Eq. (2), we get G1  G2 ⇒ ME(G1 ) ≥ ME(G2 ) and G1 G2 ⇒ ME(G1 ) > ME(G2 ) directly. In [28], Gutman and Wagner pointed out that the matching energy is a quantity of relevance for chemical applications. They arrived at the simple relation:

TRE(G) = E (G) − ME(G). Where TRE(G) is the so-called “topological resonance energy” of G. About the chemical applications of matching energy, for more details see [27]. As the research of extremal energy is an amusing work, the study on extremal matching energy is also interesting. In [28], the authors gave some elementary results on the matching energy and obtained that ME(Sn+ ) ≤ ME(G) ≤ ME(Cn ) for any unicyclic graph G of order n, where Sn+ is the graph obtained by adding a new edge to the star Sn . In [37], Ji et al. characterized the graphs with the extremal matching energy among all bicyclic graphs, while Chen and Shi [10] proved the same extremal results for tricyclic graphs. In [11], Chen et al. characterized the graphs with minimal matching energy among all unicyclic and bicyclic graphs with a given diameter d. For some more extremal results on matching energy of graphs see [13,42,58,60,62]. All graphs considered here are simple, finite and undirected. We follow the book [5] for all the notations and terminology not defined here. By convention, denote by Pn , Cn , Sn the path, cycle, star of order n. Tn denotes the set of trees with n vertices. Refer to graphs with no odd cycles as even-cycle graphs, i.e., the bipartite graphs. Denote by Bn the class of even-cycle graphs with order n, and Bn,m the class of graphs in Bn with m edges. Especially, we call the graphs in Bn,n as bipartite unicyclic graphs, which is an interesting class of graphs. What’s more, we introduce some new notations appearing in [28] and [40]. The sun graph, denoted by Cl (Ps1 +1 , . . . , Psl +1 ), is one obtained from the cycle Cl = v1 v2 . . . vl v1 by identifying one pendent vertex of path Psi +1 with vertex vi

for i = 1, . . . , l. Note that Cl (Pn−l+1 , P1 . . . , P1 ) is also called lollipop graph and abbreviated as Pnl . Let G be a connected graph with at least two vertices, and let u be one of its vertices. Denote by P(n, k, G, u) the graph obtained by identifying u with the vertex l , u) for examples, as vk of a simple path P = v1 v2 . . . vn . We in the following give the graphs Cl (Ps1 +1 , . . . , Psl +1 ) and P (s, k, Pn−s+1 shown in Fig 1. In [12], the authors determined the graphs with the second through the fourth maximal matching energies in On,n when n is odd, where On,n is the set of unicyclic odd-cycle graphs. Inspired by this, we investigate the extremal values of matching energy of bipartite unicyclic graphs. We characterize in this paper the bipartite unicyclic graphs with the first  n−3 4  largest matching energies when n is odd. One of the most interesting things is that the extremal graphs for matching energy in this paper are Pnl for some values of l, which are related to the extremal graph (i.e., Pn6 ) having maximal energy among all the bipartite unicyclic graphs for n ≥ 16 (see [36]). In fact, when n ≥ 11, with regard to matching energy, the graph Pn6 is precisely the second maximal graph in Bn,n for n being odd.

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

sl

646

s1 v vl−1 l v1 Cl v 2

u

v2

s2

Cl

v3

Cl (Ps1 +1, . . . , Psl +1)

v1

u = vk n−l−s

vs−1 vs

l P (s, k, Pn−s+1 , u)

l Fig. 1. The graphs Cl (Ps1 +1 , . . . , Psl +1 ) and P (s, k, Pn−s+1 , u).

2. Preliminaries In this section, we list several known results at first. Then some useful lemmas are shown, which play the key roles in proving our main results. Lemma 1 ([17,23]). Let G be a simple graph, e = uv be an edge of G, and N(u) = {v1 ( = v), v2 , . . . , v j } be the set of all neighbors of u in G. Then we have

m(G, k) = m(G − uv, k) + m(G − u − v, k − 1), m(G, k) = m(G − u, k) +

j 

m(G − u − vi , k − 1).

(3) (4)

i=1

From Lemma 1, we know that m(P1 ∪ G, k) = m(G, k). And one can also obtain that Lemma 2 ([11]). Let G be a simple graph and H be a subgraph (resp. proper subgraph) of G. Then G  H (resp. H). Lemma 3 ([25]). Now H1 and H2 are two graphs. If H1 H2 , then H1 ∪ G H2 ∪ G, where G is an arbitrary graph. Lemma 4 ([40]). Let n, l be positive integers, n > l ≥ 3. Denote by Ul,n the set of unicyclic graphs with n vertices and a cycle of length l, then for any graph G ∈ Ul,n , we have

ME (Pnl ) ≥ ME (G), with equality if and only if G ∼ = Pnl . Actually, the authors in [40] proved that Pnl G for any G ∈ Ul,n \ {Pnl }. Lemma 5 ([22,37]). In regard to the quasi-order , we have the following ordering:

Pn P2 ∪ Pn−2 P4 ∪ Pn−4 · · · P3 ∪ Pn−3 P1 ∪ Pn−1 . Lemma 6 ([22,28]). If F is a forest with n(n ≥ 6) vertices, then F Pn , with F ∼ Pn if and only if F ∼ = Pn . Lemma 7 ([28]). Let G be a connected graph with at least two vertices, and let u be one of its vertices. Denote by P(n, k, G, u) the graph obtained by identifying u with the vertex vk of a simple path v1 , v2 , . . . , vn . Write n = 4p + i, i ∈ {1, 2, 3, 4}, and l = (i − 1)/2. Then the inequalities

ME (P (n, 2, G, u)) < ME (P (n, 4, G, u)) < · · · < ME (P (n, 2p + 2l, G, u)) < ME (P (n, 2p + 1, G, u)) < · · · < ME (P (n, 3, G, u)) < ME (P (n, 1, G, u)) hold. Lemma 8 ([28]). Suppose that G is a connected graph and T an induced subgraph of G such that T is a tree and T is connected to the rest of G only by a cut vertex v. If T is replaced by a star of the same order, centered at v, then the matching energy decreases (unless T is already such a star). If T is replaced by a path, with one end at v, then the matching energy increases (unless T is already such a path). Lemma 9. Let n be odd and l be even with 4 ≤ l ≤ n − 3. Then Pnl Pnl+2 . Proof. For 0 ≤ k ≤

m(

)

n−1 2 ,

by Eq. (3),

(

)

(

∪ Pn−l , k − 1), ( ∪ Pn−l−2 , k − 1).

Pnl , k = m Pn , k + m Pl−2 Pnl+2 , k = m Pn , k + m Pl

m(

)

(

)

Since n is odd but l is even, then l − 2 and l are even, while n − l and n − l − 2 are odd. Applying Lemma 5, we know Pl−2 ∪ Pn−l Pl ∪ Pn−l−2 . This deduces that Pnl Pnl+2 . 

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

x Cl

y

s1

x

w z

n−

Cl

l−

s w t

y

s1

z

n−

Cl (Ps1 +1, Pn−l−s1+1 , P1, . . . , P1 )

647

l−

s

Cl (Ps+1 , P1, . . . , P1 , Pn−l−s+1, P1, . . . , P1 ) t

Fig. 2. The graphs Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ) and Cl (Ps+1 , P1 , . . . , P1 , Pn−l−s+1 , P1 , . . . , P1 ).

  t

Remark 1. Through simple calculations, it is easy to derive Pnl ∼ Pnn−l+2 . Then by the conclusion above, we get Pnn−l+2 Pnn−l . Notice that both n − l + 2 and n − l are odd, hence this result is in accordance with Lemma 8 in [12]; Remark 2. Taking advantage of the lemma above, for n being odd, we obtain Pn4 Pn6 Pn8 · · · Pnn−5 Pnn−3 Pnn−1 directly. Lemma 10. Let G1 and G2 be two vertex-disjoint graphs. Then

α(G1 ∪ G2 , x) = α(G1 , x) · α(G2 , x). Proof. Set n1 = |V (G1 )|, n2 = |V (G2 )|, with n1 + n2 = n. Then

α(G1 ∪ G2 , x) =



( − 1)k m(G1 ∪ G2 , k)xn−2k

k≥0

=



( − 1)k

k≥0

 =



k 

 m(G1 , j)m(G2 , k − j) xn−2k

j=0

    ( − 1) j m(G1 , j)xn1 −2 j · ( − 1)k− j m(G2 , k − j)xn2 −2(k− j)

j≥0

=

k− j≥0

α(G1 , x) · α(G2 , x).

The proof is thus completed.  Let GࣇCn be a connected graph in Bn,n . Denote the unique cycle of G by Cl . We call each maximal tree outside Cl with one vertex attached to some vertex of Cl a “branch” of G, namely, any two branches of G have no common vertices. It is both consistent and convenient to define a vertex in Cl with no neighbor outside Cl also as a branch of G. Any branch with just one vertex is referred to as trivial. All other branches are nontrivial. If G ࣇ Cn is a graph in Bn,n with l branches outside the unique cycle Cl , the ith branch has si + 1 (si ≥ 0) vertices for i = 1, . . . , l, where s1 + s2 + . . . + sl = n − l. Then according to Lemma 8, the matching energy of G increases when each of the branches becomes a path (unless G is already such a graph). Thus ME(G) ≤ ME(Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 )), with equality if and only if G ∼ = Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ). In the following, we will show that Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ). Lemma 11. The graphs Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) and Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ) are shown in Fig 1 and Fig 2, respectively, where l is even with 4 ≤ l ≤ n − 2, 1 ≤ s1 ≤ n − l − 1. Then Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ), with Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) ∼ Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ) if and only if Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) ∼ = Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ). Proof. For 0 ≤ k ≤  n2 , by Lemmas 1, 4 and 6, we have

m(Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ), k) = m(Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) − uv1 , k) + m(Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) − u − v1 , k − 1) l ≤ m(Pn−s ∪ Ps1 , k) + m(Pn−s1 −1 ∪ Ps1 −1 , k − 1) 1

= m(Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ) − xw, k)

+ m(Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ) − x − w, k − 1)

= m(Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ), k).

which yields that Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ). The equality holds for all k if and only if l ∪ Ps1 , meanwhile Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) − u − v1 ∼ Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) − uv1 ∼ = Pn−s = Pn−s1 −1 ∪ Ps1 −1 . That is, if and only 1 if Cl (Ps1 +1 , Ps2 +1 , . . . , Psl +1 ) ∼ = Cl (Ps1 +1 , Pn−l−s1 +1 , P1 , . . . , P1 ). 

648

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

s

x Cl

y

x Cl

n−

l−

y

n−l−

2

w

s

Cl (Ps+1, Pn−l−s+1, P1 , . . . , P1 )

Cl (Pn−l−1, P3 , P1 , . . . , P1 )

Fig. 3. The graphs used in Lemma 12.

w u0

v2

v1

u1 u2 u3

· · · u4 un−6

Cl u 0

v2

v1 ··

n− l

P (n − 4, 3, P54, u1)

Cl u 0

· · ·

u1 u2 · · ·

ui

v2

v1

−

ui−1 u1 u2 · · · u i

P (n − l + 1, 3, Cl , u0)

n

Cl u 0

·

−2

l− i− 2 l P (n − l − i + 1, 3, Pl+i , ui )

G

Fig. 4. Some graphs needed in our paper.

As a special case of Lemma 11, for the graphs Cl (Ps+1 , P1 , . . . , P1 , Pn−l−s+1 , P1 , . . . , P1 ), where l is even with 4 ≤ l ≤ n − 2, 1 ≤

  t

s ≤ n − l − 1. Let the two 3-degree vertices be x and y, if the number of vertices in the unique cycle between x and y is t with 1 ≤ t ≤ l−2 2 (as shown in Fig 2). Then it follows immediately that Cl (Ps+1 , P1 , . . . , P1 , Pn−l−s+1 , P1 , . . . , P1 ) ≺ Cl (Ps+1 , Pn−l−s+1 , P1 , . . . , P1 ).

  t

Lemma 12. Let l be even with 4 ≤ l ≤ n − 2, s be an integer with 1 ≤ s ≤ n − l − 1. Then Cl (Ps+1 , Pn−l−s+1 , P1 , . . . , P1 )

Cl (Pn−l−1 , P3 , P1 , . . . , P1 ) (these two graphs are shown in Fig 3), with Cl (Ps+1 , Pn−l−s+1 , P1 , . . . , P1 ) ∼ Cl (Pn−l−1 , P3 , P1 , . . . , P1 ) if and only if s = n − l − 2 or s = 2. Proof. For 0 ≤ k ≤  2n , we have

m(Cl (Ps+1 , Pn−l−s+1 , P1 , . . . , P1 ), k) = m(Cl (Ps+1 , Pn−l−s+1 , P1 , . . . , P1 ) − xy, k) + m(Cl (Ps+1 , Pn−l−s+1 , P1 , . . . , P1 ) − x − y, k − 1) = m(Pn , k) + m(Pl−2 ∪ Ps ∪ Pn−l−s , k − 1), m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ), k) = m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ) − xy, k) + m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ) − x − y, k − 1) = m(Pn , k) + m(Pl−2 ∪ Pn−l−2 ∪ P2 , k − 1). Since Ps ∪ Pn−l−s Pn−l−2 ∪ P2 by Lemma 5, then m(Pl−2 ∪ Ps ∪ Pn−l−s , k − 1) ≤ m(Pl−2 ∪ Pn−l−2 ∪ P2 , k − 1), which yields that m(Cl (Ps+1 , Pn−l−s+1 , P1 , . . . , P1 ), k) ≤ m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ), k). The equality holds for all k if and only if Ps ∪ Pn−l−s ∼ = Pn−l−2 ∪ P2 , namely, s = n − l − 2 or s = 2. Hence, the result holds.  Lemma 13. Let n be odd with n ≥ 9, l be even with 4 ≤ l ≤ n − 3. Then P (n − 4, 3, P54 , u1 ) Cl (Pn−l−1 , P3 , P1 , . . . , P1 ). Where the graph P (n − 4, 3, P54 , u1 ) is shown in Fig 4. Proof. For 0 ≤ k ≤

n−1 2 ,

Eq. (3) leads to

m(P (n − 4, 3, P54 , u1 ), k) = m(P (n − 4, 3, P54 , u1 ) − u1 v2 , k) + m(P (n − 4, 3, P54 , u1 ) − u1 − v2 , k − 1) 4 = m(P2 ∪ Pn−2 , k) + m(C4 ∪ Pn−7 , k − 1) 4 = m(P2 ∪ Pn−2 , k) + m(P4 ∪ Pn−7 , k − 1) + m(P2 ∪ Pn−7 , k − 2),

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

w Cl u 0

u1 u2 · · ·

v2

v1

un−l−4

l P (5, 3, Pn−4 , un−l−4)

w Cl u 0

u1

649

un−l−3 · · · u

n−l−2

l P (3, 2, Pn−2 , un−l−2)

l l Fig. 5. The graphs P (5, 3, Pn−4 , un−l−4 ) and P (3, 2, Pn−2 , un−l−2 ).

m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ), k) = m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ) − yw, k) + m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ) − y − w, k − 1) l = m(P2 ∪ Pn−2 , k) + m(Pn−3 , k − 1) l = m(P2 ∪ Pn−2 , k) + m(P4 ∪ Pn−7 , k − 1) + m(P3 ∪ Pn−8 , k − 2).

Since n is odd with n ≥ 9 and 4 ≤ l ≤ n − 3, then by Lemmas 3, 5 and Remark 2 after Lemma 9, we get P2 ∪ Pn−7 4 l  P2 ∪ Pn−2 . Hence m(P (n − 4, 3, P54 , u1 ), k) ≥ m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ), k). It follows from P2 ∪ Pn−7 P3 ∪ Pn−8 and P2 ∪ Pn−2 P3 ∪ Pn−8 that there exists some k0 such that m(P2 ∪ Pn−7 , k0 ) > m(P3 ∪ Pn−8 , k0 ), which deduces m(P (n − 4, 3, P54 , u1 ), k0 + 2) > m(Cl (Pn−l−1 , P3 , P1 , . . . , P1 ), k0 + 2). Therefore, P (n − 4, 3, P54 , u1 ) Cl (Pn−l−1 , P3 , P1 , . . . , P1 ).  Summarizing the above analysis, the graphs in Bn,n with more than one nontrivial branch have been discussed. Now consider the graph G in Bn,n with just one nontrivial branch. This branch is connected to the unique cycle Cl of G by a cut vertex, say u0 . The n − l vertices outside Cl are denoted by u1 , u2 , . . . , un−l . Suppose G  Pnl . If d(u0 ) ≥ 4, then by Lemma 7 and Lemma 8, we know that ME(G) ≤ ME(P (n − l + 1, 3, Cl , u0 )), with equality if and only if G ∼ = P (n − l + 1, 3, Cl , u0 ). If d(u0 ) = 3, since G  Pnl , there exists some vertex in {u1 , u2 , . . . , un−l } having degree not less than 3. Assume that d(ui ) ≥ 3 with 1 ≤ i ≤ n − l − 2, and l , u )), with equality if and only d(u j ) = 2 for 1 ≤ j ≤ i − 1 (see Fig 4). Then similarly, we have ME(G) ≤ ME(P (n − l − i + 1, 3, Pl+i i l l ∼ if G = P(n − l − i + 1, 3, P , ui ). The graphs P (n − l + 1, 3, Cl , u0 ) and P (n − l − i + 1, 3, P , ui ) with 1 ≤ i ≤ n − l − 3 are shown l+i

in Fig 4.

l+i

Lemma 14. Let n be odd with n ≥ 9, l be even with 4 ≤ l ≤ n − 3. Then we can obtain P (n − 4, 3, P54 , u1 ) P (n − l + 1, 3, Cl , u0 ). Proof. For 0 ≤ k ≤

n−1 2 ,

one can check that

m(P (n − 4, 3, P54 , u1 ), k) = m(P (n − 4, 3, P54 , u1 ) − u1 v2 , k) + m(P (n − 4, 3, P54 , u1 ) − u1 − v2 , k − 1) 4 = m(P2 ∪ Pn−2 , k) + m(C4 ∪ Pn−7 , k − 1) 4 ≥ m(P2 ∪ Pn−2 , k) + m(P4 ∪ Pn−7 , k − 1),

m(P (n − l + 1, 3, Cl , u0 ), k) = m(P (n − l + 1, 3, Cl , u0 ) − u0 v2 , k) + m(P (n − l + 1, 3, Cl , u0 ) − u0 − v2 , k − 1) l = m(P2 ∪ Pn−2 , k) + m(Pl−1 ∪ Pn−l−2 , k − 1).

4 l Since n is odd with n ≥ 9 and l is even with 4 ≤ l ≤ n − 3, then P4 ∪ Pn−7 Pl−1 ∪ Pn−l−2 and P2 ∪ Pn−2  P2 ∪ Pn−2 . Hence 4 4 m(P (n − 4, 3, P5 , u1 ), k) ≥ m(P (n − l + 1, 3, Cl , u0 ), k). Moreover, m(P (n − 4, 3, P5 , u1 ), 2) > m(P (n − l + 1, 3, Cl , u0 ), 2). Hence P(n − 4, 3, P54 , u1 ) P (n − l + 1, 3, Cl , u0 ).  l , u ) P (5, 3, P l , un−l−4 ), Lemma 15. Let n be odd, l be even with 4 ≤ l ≤ n − 5, 2 ≤ i ≤ n − l − 3. Then P(n − l − i + 1, 3, Pl+i i n−4 l , u ) ∼ P (5, 3, P l l , u ) ∼ P (5, 3, P l , u ) if and only if P ( n − l − i + 1, 3, P , u ) (i.e., i=n− with P(n − l − i + 1, 3, Pl+i i i = n−l−4 n−l−4 n−4 n−4 l+i l l − 4). Where the graph P(5, 3, Pn−4 , un−l−4 ) is shown in Fig 5.

Proof. For 0 ≤ k ≤

n−1 2 ,

l l m(P (n − l − i + 1, 3, Pl+i , ui ), k) = m(P (n − l − i + 1, 3, Pl+i , ui ) − v2 ui , k) l + m(P (n − l − i + 1, 3, Pl+i , ui ) − v2 − ui , k − 1) l l = m(Pn−2 ∪ P2 , k) + m(Pl+i−1 ∪ Pn−l−i−2 , k − 1) l = m(Pn−2 ∪ P2 , k) + m(Pl+i−1 ∪ Pn−l−i−2 , k − 1)

+ m(Pl−2 ∪ Pi−1 ∪ Pn−l−i−2 , k − 2), m(P (

l 5, 3, Pn−4 , un−l−4

l ), k) = m(P(5, 3, Pn−4 , un−l−4 ) − v2 un−l−4 , k) l + m(P (5, 3, Pn−4 , un−l−4 ) − v2 − un−l−4 , k − 1) l l = m(Pn−2 ∪ P2 , k) + m(Pn−5 ∪ P2 , k − 1) l = m(Pn−2 ∪ P2 , k) + m(Pn−5 ∪ P2 , k − 1) + m(Pl−2 ∪ Pn−l−5 ∪ P2 , k − 2).

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L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

w u0 u1

v2 · · ·

w v2

v1 Cl u 0

un−8

4 P (5, 3, Pn−4 , un−8)

v1

u1

· · · n−l−3

l , u1 ) P (n − l, 3, Pl+1

4 l Fig. 6. The graphs P (5, 3, Pn−4 , un−8 ) and P (n − l, 3, Pl+1 , u1 ).

Since 2 ≤ i ≤ n − l − 3, then Pl+i−1 ∪ Pn−l−i−2 Pn−5 ∪ P2 and also Pl−2 ∪ Pi−1 ∪ Pn−l−i−2 Pl−2 ∪ Pn−l−5 ∪ P2 . Which mean m(P (n − l , u ), k) ≤ m(P (5, 3, P l l − i + 1, 3, Pl+i , un−l−4 ), k). The equality holds for all k if and only if i = n − l − 4. It follows that the proof i n−4 is completed.  l , un−l−2 ) (as shown in Fig. 5). In this case, we can also arrive at Furthermore, if i = n − l − 2 and G  Pnl , then G ∼ = P (3, 2, Pn−2

l l , un−l−2 ) ≺ P (5, 3, Pn−4 , un−l−4 ). P (3, 2, Pn−2

l l Lemma 16. For even l with 4 ≤ l ≤ n − 5, we have P (3, 2, Pn−2 , un−l−2 ) ≺ P (5, 3, Pn−4 , un−l−4 ).

Proof. For all 0 ≤ k ≤  n2 , since 4 ≤ l ≤ n − 5, l l m(P (3, 2, Pn−2 , un−l−2 ), k) = m(P (3, 2, Pn−2 , un−l−2 ) − un−l−3 un−l−2 , k) l + m(P (3, 2, Pn−2 , un−l−2 ) − un−l−3 − un−l−2 , k − 1) l l = m(Pn−3 ∪ P3 , k) + m(Pn−4 , k − 1)

= m(Pn−3 ∪ P3 , k) + m(Pl−2 ∪ Pn−l−3 ∪ P3 , k − 1) + m(Pn−4 , k − 1) + m(Pl−2 ∪ Pn−l−4 , k − 2), m(P (

l 5, 3, Pn−4 , un−l−4

l ), k) = m(P(5, 3, Pn−4 , un−l−4 ) − v2 un−l−4 , k) l + m(P (5, 3, Pn−4 , un−l−4 ) − v2 − un−l−4 , k − 1) l l = m(Pn−2 ∪ P2 , k) + m(Pn−5 ∪ P2 , k − 1) = m(Pn−2 ∪ P2 , k) + m(Pl−2 ∪ Pn−l−2 ∪ P2 , k − 1) + m(Pn−5 ∪ P2 , k − 1) + m(Pl−2 ∪ Pn−l−5 ∪ P2 , k − 2).

Clearly, Pn−3 ∪ P3 ≺ Pn−2 ∪ P2 , Pl−2 ∪ Pn−l−3 ∪ P3 Pl−2 ∪ Pn−l−2 ∪ P2 , Pn−4 ≺ Pn−5 ∪ P2 and Pl−2 ∪ Pn−l−4 Pl−2 ∪ Pn−l−5 ∪ P2 . Which l l , un−l−2 ) ≺ P (5, 3, Pn−4 , un−l−4 ).  imply that P(3, 2, Pn−2 4 ,u l Lemma 17. Let n ≥ 13 be odd, l be even with 6 ≤ l ≤ n − 7, then P(5, 3, Pn−4 n−8 ) P (5, 3, Pn−4 , un−l−4 ). Where the graph 4 P (5, 3, Pn−4 , un−8 ) is shown in Fig 6.

Proof. Since l is even and 6 ≤ l ≤ n − 7, then by Lemma 5, we have P2 ∪ Pn−6 Pl−2 ∪ Pn−l−2 , P2 ∪ Pn−9  Pl−2 ∪ Pn−l−5 . For k ≥ 0, by Lemma 1, we can obtain that 4 4 4 m(P (5, 3, Pn−4 , un−8 ), k) = m(P (5, 3, Pn−4 , un−8 ) − u0 w, k) + m(P (5, 3, Pn−4 , un−8 ) − u0 − w, k − 1) 4 = m(P (5, 3, Pn−4 , un−8 ) − u0 w, k) 4 + m(P (5, 3, Pn−4 , un−8 ) − u0 − w − un−8 v2 , k − 1) 4 + m(P (5, 3, Pn−4 , un−8 ) − u0 − w − un−8 − v2 , k − 2) 4 = m(P (5, 3, Pn−4 , un−8 ) − u0 w, k) + m(P2 ∪ P2 ∪ Pn−6 , k − 1) + m(P2 ∪ P2 ∪ Pn−9 , k − 2) l ≥ m(P (5, 3, Pn−4 , un−l−4 ) − u0 w, k) + m(P2 ∪ Pl−2 ∪ Pn−l−2 , k − 1)

+ m(P2 ∪ Pl−2 ∪ Pn−l−5 , k − 2) l = m(P (5, 3, Pn−4 , un−l−4 ), k).

Since P2 ∪ Pn−6 Pl−2 ∪ Pn−l−2 , by Lemma 3, P2 ∪ P2 ∪ Pn−6 P2 ∪ Pl−2 ∪ Pn−l−2 . Thus there exists some k0 such that 4 ,u l m(P2 ∪ P2 ∪ Pn−6 , k0 ) > m(P2 ∪ Pl−2 ∪ Pn−l−2 , k0 ). So m(P (5, 3, Pn−4 n−8 ), k0 + 1) > m(P (5, 3, Pn−4 , un−l−4 ), k0 + 1). Therefore, 4 ,u l P (5, 3, Pn−4 n−8 ) P (5, 3, Pn−4 , un−l−4 ).



4 ,u Lemma 18. Let n be odd with n ≥ 11, then we have P (n − 4, 3, P54 , u1 ) P (5, 3, Pn−4 n−8 ).

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

651

Proof. For all k ≥ 0, applying Lemma 1, one can get

m(P (n − 4, 3, P54 , u1 ), k) = m(P (n − 4, 3, P54 , u1 ) − u1 v2 , k) + m(P (n − 4, 3, P54 , u1 ) − u1 − v2 , k − 1) 4 = m(P2 ∪ Pn−2 , k) + m(P (n − 4, 3, P54 , u1 ) − u1 − v2 − u3 u4 , k − 1)

+ m(P (n − 4, 3, P54 , u1 ) − u1 − v2 − u3 − u4 , k − 2) 4 = m(P2 ∪ Pn−2 , k) + m(C4 ∪ P2 ∪ Pn−9 , k − 1) + m(C4 ∪ Pn−10 , k − 2),

m(P (

4 5, 3, Pn−4 , un−8

4 4 ), k) = m(P(5, 3, Pn−4 , un−8 ) − un−8 v2 , k) + m(P (5, 3, Pn−4 , un−8 ) − un−8 − v2 , k − 1) 4 4 = m(P2 ∪ Pn−2 , k) + m(P (5, 3, Pn−4 , un−8 ) − un−8 − v2 − u0 u1 , k − 1) 4 + m(P (5, 3, Pn−4 , un−8 ) − un−8 − v2 − u0 − u1 , k − 2) 4 = m(P2 ∪ Pn−2 , k) + m(C4 ∪ P2 ∪ Pn−9 , k − 1) + m(P2 ∪ P3 ∪ Pn−10 , k − 2).

Obviously, C4 P2 ∪ P3 , by Lemma 3, we get C4 ∪ Pn−10 P2 ∪ P3 ∪ Pn−10 . Thus m(C4 ∪ Pn−10 , k − 2) ≥ m(P2 ∪ P3 ∪ Pn−10 , k − 2), and 4 ,u 4 4 then m(P (n − 4, 3, P54 , u1 ), k) ≥ m(P (5, 3, Pn−4 n−8 ), k). Moreover, m(P (n − 4, 3, P5 , u1 ), 3) > m(P (5, 3, Pn−4 , un−8 ), 3). Hence 4 4 P(n − 4, 3, P5 , u1 ) P (5, 3, Pn−4 , un−8 ). The proof is finished.  l , u ). Where the graph P (n − Lemma 19. Let n ≥ 13 be odd, l be even with 6 ≤ l ≤ n − 7, then P(n − 4, 3, P54 , u1 ) P (n − l, 3, Pl+1 1

l , u ) is shown in Fig 6. l, 3, Pl+1 1

Proof. For all k ≥ 0, it follows from Lemma 1 that

m(P (n − 4, 3, P54 , u1 ), k) = m(P (n − 4, 3, P54 , u1 ) − u0 w, k) + m(P (n − 4, 3, P54 , u1 ) − u0 − w, k − 1) = m(P (n − 4, 3, P54 , u1 ) − u0 w − u1 v2 , k) + m(P (n − 4, 3, P54 , u1 ) − u0 w − u1 − v2 , k − 1) + m(P2 ∪ Pn−4 , k − 1) = m(P2 ∪ Pn−2 , k) + m(P4 ∪ Pn−7 , k − 1) + m(P2 ∪ Pn−4 , k − 1), m(P (n −

l l, 3, Pl+1 , u1

l l ), k) = m(P(n − l, 3, Pl+1 , u1 ) − u0 w, k) + m(P (n − l, 3, Pl+1 , u1 ) − u0 − w, k − 1) l = m(P (n − l, 3, Pl+1 , u1 ) − u0 w − u1 v2 , k) l + m(P (n − l, 3, Pl+1 , u1 ) − u0 w − u1 − v2 , k − 1) + m(Pl−2 ∪ Pn−l , k − 1)

= m(P2 ∪ Pn−2 , k) + m(Pl ∪ Pn−l−3 , k − 1) + m(Pl−2 ∪ Pn−l , k − 1).

Since l is even and 6 ≤ l ≤ n − 7, then P4 ∪ Pn−7  Pl ∪ Pn−l−3 , P2 ∪ Pn−4 Pl−2 ∪ Pn−l . Hence m(P (n − 4, 3, P54 , u1 ), k) ≥ m(P (n − l , u ), k). In particular, there exists some k such that m(P ∪ P l, 3, Pl+1 1 2 n−4 , k0 ) > m(Pl−2 ∪ Pn−l , k0 ), which implies that m(P (n − 0 l , u ), k + 1). Consequently, P (n − 4, 3, P 4 , u ) P (n − l, 3, P l , u ). 4, 3, P54 , u1 ), k0 + 1) > m(P (n − l, 3, Pl+1 1 0 1 1 5 l+1



So far, what is remaining to be discussed is the comparison of ME(P (n − 4, 3, P54 , u1 )) and ME(Pnl ). By utilizing the Coulson integral formula of matching energy, as well as the help of computer, we will show ME(P (n − 4, 3, P54 , u1 )) < ME(Pnl ) for 4 ≤ l ≤ n+1 2 in the next section. 3. Main results Let G be a simple graph, e be an edge of G connecting the vertices vr and vs . By G(e/j) we denote the graph obtained by inserting j(j ≥ 0) new vertices (of degree two) on the edge e. On the number of k-matchings of the graph G(e/j), the property that m(G(e/ j + 2), k) = m(G(e/ j + 1), k) + m(G(e/ j), k − 1) for all j ≥ 0 was given in [25]. In addition, on the matching polynomial of G(e/j), we have shown that α(G(e/ j + 2), x) = xα(G(e/ j + 1), x) − α(G(e/ j), x) in [10]. Lemma 20. For 3 ≤ l ≤ n − 1, the matching polynomials of Pn and Pnl have the following forms:

α(Pn , x) = A1 (x)(Y1 (x))n + A2 (x)(Y2 (x))n ; α(Pnl , x) = B1 (x)(Y1 (x))n + B2 (x)(Y2 (x))n . Where Y1 (x) =

√

x+

x2 −4 , Y2 (x) 2

=

√

x−

x2 −4 . 2

3 Proof. By the definition of G(e/j), Pn = P2 (e1 /n − 2) and Pnl = Pn−l+3 (e2 /l − 3), where e1 is the unique edge of P2 , e2 is one of the 3 edges of the triangle in Pn−l+3 . Hence both α (Pn , x) and α(Pnl , x) satisfy the recursive formula

f (n, x) = x f (n − 1, x) − f (n − 2, x). The general solution of this linear homogeneous recurrence relation is

f (n, x) = C1 (x)(Y1 (x))n + C2 (x)(Y2 (x))n ,

652

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

√ √ 2 2 where Y1 (x) = x+ 2x −4 , Y2 (x) = x− 2x −4 , with Y1 (x) + Y2 (x) = x and Y1 (x)Y2 (x) = 1. Take the initial values as α(P2 , x) = x2 − 1 3 and α(P3 , x) = x − 2x. We then get

α(Pn , x) = A1 (x)(Y1 (x))n + A2 (x)(Y2 (x))n , where A1 (x) = For 3 ≤ l ≤

Y1 (x)α(P3 ,x)−α(P2 ,x) Y (x)α(P ,x)−α(P ,x) , A2 (x) = 2(Y (x))34 −(Y (x))22 . (Y1 (x))4 −(Y1 (x))2 2 2 n − 1, m(Pnl , k) = m(Pn , k) + m(Pl−2 ∪ Pn−l , k − 1).

α(Pnl , x) =



So

( − 1)k m(Pnl , k)xn−2k

k≥0

=



( − 1)k (m(Pn , k) + m(Pl−2 ∪ Pn−l , k − 1))xn−2k

k≥0

=



( − 1)k m(Pn , k)xn−2k +

k≥0



( − 1)k m(Pl−2 ∪ Pn−l , k − 1)xn−2k

k≥0

α(Pn , x) − α(Pl−2 ∪ Pn−l , x) = α(Pn , x) − α(Pl−2 , x) · α(Pn−l , x) = A1 (x)(Y1 (x))n + A2 (x)(Y2 (x))n − (A1 (x)(Y1 (x))l−2 + A2 (x)(Y2 (x))l−2 ) · (A1 (x)(Y1 (x))n−l + A2 (x)(Y2 (x))n−l ) = A1 (x)(Y1 (x))n − (A1 (x))2 (Y1 (x))n−2 − A1 (x)A2 (x)(Y1 (x))l−2 (Y2 (x))n−l + A2 (x)(Y2 (x))n − (A2 (x))2 (Y2 (x))n−2 − A1 (x)A2 (x)(Y1 (x))n−l (Y2 (x))l−2 . =

Therefore, α(Pnl , x) = B1 (x)(Y1 (x))n + B2 (x)(Y2 (x))n . Where

B1 (x) = A1 (x) − (A1 (x))2 (Y2 (x))2 − A1 (x)A2 (x)(Y2 (x))2l−2 , B2 (x) = A2 (x) − (A2 (x))2 (Y1 (x))2 − A1 (x)A2 (x)(Y1 (x))2l−2 for 3 ≤ l ≤

n+2 2 ;

B1 (x) = A1 (x) − (A1 (x))2 (Y2 (x))2 − A1 (x)A2 (x)(Y2 (x))2n−2l+2 , B2 (x) = A2 (x) − (A2 (x))2 (Y1 (x))2 − A1 (x)A2 (x)(Y1 (x))2n−2l+2 for

n+2 2 < l ≤ n − 1. We complete the proof. 

Lemma 21. Let n(n ≥ 9) be odd and 4 ≤ l ≤

n+1 2

be even. Then ME (P (n − 4, 3, P54 , u1 )) < ME (Pnl ).

Proof. If l = 4, then by Lemma 4, we get ME(P (n − 4, 3, P54 , u1 )) < ME(Pn4 ) directly. If n = 9, then l = 4, this is the case just discussed. Hence in the following we assume that n ≥ 11 and l ≥ 6. Obviously, P (n − 4, 3, P54 , u1 ) = P (4, 3, P54 , u1 )(e/n − 8), where e is the pendent edge incident with u1 in P (4, 3, P54 , u1 ). Similarly,

α(P(n − 4, 3, P54 , u1 ), x) = C1 (x)(Y1 (x))n + C2 (x)(Y2 (x))n with Y1 (x) =

√

x+

x2 −4 , Y2 (x) 2

=

√

x2 −4 . 2

x−

The initial values can be chosen as:

α(P(4, 3, P54 , u1 ), x) = C1 (x)(Y1 (x))8 + C2 (x)(Y2 (x))8 = x8 − 8x6 + 18x4 − 12x2 + 2;

α(P(

5, 3, P54 , u1

), x) = C1 (x)(Y1 (x))9 + C2 (x)(Y2 (x))9 = x9 − 9x7 + 25x5 − 25x3 + 8x.

Solving the above two equations, we get

C1 (x) =

Y1 (x)α(P (5, 3, P54 , u1 ), x) − α(P (4, 3, P54 , u1 ), x) , (Y1 (x))10 − (Y1 (x))8

C2 (x) =

Y2 (x)α(P (5, 3, P54 , u1 ), x) − α(P (4, 3, P54 , u1 ), x) . (Y2 (x))10 − (Y2 (x))8

Set Z1 (x) = −iY1 (ix) =

√

x+

x2 +4 , 2

Z2 (x) = −iY2 (ix) =

Z1 (x) · Z2 (x) = −1, Z1 (x) + Z2 (x) = x, Z1 (x) − Z2 (x) =

√

x−

x2 +4

2

x2

, where i2 = −1. Then we have Y1 (ix) = iZ1 (x), Y2 (ix) = iZ2 (x),

+ 4. Besides, set

f1 = −α(P2 , ix) = x + 1; 2

f2 = iα(P3 , ix) = x3 + 2x; g1 = α(P (4, 3, P54 , u1 ), ix) = x8 + 8x6 + 18x4 + 12x2 + 2; g2 = −iα(P (5, 3, P54 , u1 ), ix) = x9 + 9x7 + 25x5 + 25x3 + 8x.



L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

For 4 ≤ l ≤

n+1 2 ,

653

according to Lemma 20 as well as the results got above,

A1 (ix) = A2 (ix) =

Y1 (ix)α(P3 , ix) − α(P2 , ix) Z1 (x) f2 + f1 = ; (Y1 (ix))4 − (Y1 (ix))2 (Z1 (x))4 + (Z1 (x))2 Y2 (ix)α(P3 , ix) − α(P2 , ix) Z2 (x) f2 + f1 = ; (Y2 (ix))4 − (Y2 (ix))2 (Z2 (x))4 + (Z2 (x))2

B1 (ix) = A1 (ix) − (A1 (ix))2 (Y2 (ix))2 − A1 (ix)A2 (ix)(Y2 (ix))2l−2 =

Z1 (x) f2 + f1 (Z1 (x) f2 + f1 )2 + + (Z1 (x))4 + (Z1 (x))2 (Z1 (x))2 ((Z1 (x))4 + (Z1 (x))2 )2

(Z1 (x) f2 + f1 )(Z2 (x) f2 + f1 )(Z2 (x))2l−2 ; ((Z1 (x))4 + (Z1 (x))2 )((Z2 (x))4 + (Z2 (x))2 )

B2 (ix) = A2 (ix) − (A2 (ix))2 (Y1 (ix))2 − A1 (ix)A2 (ix)(Y1 (ix))2l−2 = C1 (ix) =

Z2 (x) f2 + f1 (Z2 (x) f2 + f1 )2 + + (Z2 (x))4 + (Z2 (x))2 (Z2 (x))2 ((Z2 (x))4 + (Z2 (x))2 )2

(Z1 (x) f2 + f1 )(Z2 (x) f2 + f1 )(Z1 (x))2l−2 ; ((Z1 (x))4 + (Z1 (x))2 )((Z2 (x))4 + (Z2 (x))2 )

Y1 (ix)α(P (5, 3, P54 , u1 ), ix) − α(P (4, 3, P54 , u1 ), ix) (Y1 (ix))10 − (Y1 (ix))8

Z1 (x)g2 + g1 ; (Z1 (x))10 + (Z1 (x))8 Y2 (ix)α(P (5, 3, P54 , u1 ), ix) − α(P (4, 3, P54 , u1 ), ix) C2 (ix) = (Y2 (ix))10 − (Y2 (ix))8 Z2 (x)g2 + g1 = . (Z2 (x))10 + (Z2 (x))8 =

And then

ME(P (n − 4, 3, P54 , u1 )) − ME(Pnl ) =



2

π −

∞ 0

2





 1 ln m(P (n − 4, 3, P54 , u1 ), k)x2k dx 2 x k≥0



π

0



∞

∞





 1 ln m(Pnl , k)x2k dx 2 x k≥0

α(P(n − 4, 3, P54 , u1 ), ix) = ln dx π 0 α(Pnl , ix)  ∞ 2 C1 (ix)(Y1 (ix))n + C2 (ix)(Y2 (ix))n = ln dx. π 0 B1 (ix)(Y1 (ix))n + B2 (ix)(Y2 (ix))n 2

Since n is odd,





C1 (ix)(Y1 (ix))n+2 + C2 (ix)(Y2 (ix))n+2 C1 (ix)(Y1 (ix))n + C2 (ix)(Y2 (ix))n K0 (x) ln − ln = ln 1 + . B1 (ix)(Y1 (ix))n+2 + B2 (ix)(Y2 (ix))n+2 B1 (ix)(Y1 (ix))n + B2 (ix)(Y2 (ix))n H0 (n, x) Where

K0 (x) = (C1 (ix)(Y1 (ix))n+2 + C2 (ix)(Y2 (ix))n+2 ) · (B1 (ix)(Y1 (ix))n + B2 (ix)(Y2 (ix))n ) − (B1 (ix)(Y1 (ix))n+2 + B2 (ix)(Y2 (ix))n+2 ) · (C1 (ix)(Y1 (ix))n + C2 (ix)(Y2 (ix))n ) = (C1 (ix)B2 (ix) − C2 (ix)B1 (ix))((Y1 (ix))2 − (Y2 (ix))2 )



= (C1 (ix)B2 (ix) − C2 (ix)B1 (ix))( − x x2 + 4); H0 (n, x) = (B1 (ix)(Y1 (ix))n+2 + B2 (ix)(Y2 (ix))n+2 ) · (C1 (ix)(Y1 (ix))n + C2 (ix)(Y2 (ix))n ) l α(Pn+2 , ix) · α(P (n − 4, 3, P54 , u1 ), ix)



   l = in+2 m(Pn+2 , k)xn+2−2k · in m(P (n − 4, 3, P54 , u1 ), k)xn−2k

=

k≥0

= i2n+2

=





l m(Pn+2 , k)xn+2−2k

k≥0

 k≥0

l m(Pn+2 , k)xn+2−2k



·

k≥0

 k≥0





m(P (n − 4, 3, P54 , u1 ), k)xn−2k



m(P (n − 4, 3, P54 , u1 ), k)xn−2k .

k≥0

l Apparently, since x > 0, meanwhile, m(Pn+2 , k) ≥ 0 and m(P (n − 4, 3, P54 , u1 ), k) ≥ 0 hold for all k ≥ 0, then H0 (n, x) > 0. Next, we shall verify C1 (ix)B2 (ix) − C2 (ix)B1 (ix) > 0.

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L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

According to the expressions of Cj (ix), B j (ix)( j = 1, 2), together with Z2 (x) f2 + f1 = (Z2 (x))4 and Z1 (x) f2 + f1 = (Z1 (x))4 , through a series of calculations, we derive

C1 (ix)B2 (ix) − C2 (ix)B1 (ix) = ((Z2 (x))8 (1 + (Z2 (x))2 )2 (Z1 (x)g2 + g1 ) + (Z2 (x))10 (1 + (Z1 (x))2 )

(Z1 (x)g2 + g1 ) + (Z1 (x))2l−10 (1 + (Z2 (x))2 )(Z1 (x)g2 + g1 ) − (Z1 (x))8 (1 + (Z1 (x))2 )2 (Z2 (x)g2 + g1 ) − (Z1 (x))10 (1 + (Z2 (x))2 )(Z2 (x)g2 + g1 ) − (Z2 (x))2l−10 (1 + (Z1 (x))2 )(Z2 (x)g2 + g1 ))/(x2 + 4)2 . By the help of the computer, we get

Z (x) = (Z2 (x))8 (1 + (Z2 (x))2 )2 (Z1 (x)g2 + g1 ) + (Z2 (x))10 (1 + (Z1 (x))2 ) =

(Z1 (x)g2 + g1 ) − (Z1 (x))8 (1 + (Z1 (x))2 )2 (Z2 (x)g2 + g1 ) − (Z1 (x))10 (1 + (Z2 (x))2 )(Z2 (x)g2 + g1 )

x2 + 4x13 + 13 x2 + 4x11 + 63 x2 + 4x9 + 139 x2 + 4x7 + 131 x2 + 4x5 + 28 x2 + 4x3 − 10 x2 + 4x.

Let H0 (l, x) = (Z2 (x))2l−10 (1 + (Z1 (x))2 )(Z2 (x)g2 + g1 ) − (Z1 (x))2l−10 (1 + (Z2 (x))2 )(Z1 (x)g2 + g1 ). It suffices to show Z(x) > H0 (l, x) holds for x > 0. Take the derivative of H0 (l, x) with respect to l, let H0 (l, x) denote the derived function. We claim that H0 (l, x) is decreasing on l. Claim. For 6 ≤ l ≤ n+1 2 and any given x with x > 0, the function H0 (l, x) is decreasing on l. Proof. Clearly, Z1 (x) > 1 and Z2 (x) < 0. Moreover, since Z1 (x) · Z2 (x) = −1, then ln (Z1 (x)) + ln ( − Z2 (x)) = ln (Z1 (x) · ( − Z2 (x))) = 0, which implies that ln ( − Z2 (x)) = − ln (Z1 (x)). Accordingly,

H0 (l, x) = 2(1 + (Z1 (x))2 )(Z2 (x)g2 + g1 )(Z2 (x))2l−10 ln ( − Z2 (x)) − 2(1 + (Z2 (x))2 )(Z1 (x)g2 + g1 )(Z1 (x))2l−10 ln (Z1 (x)) = −2 ln (Z1 (x))((1 + (Z2 (x))2 )(Z1 (x)g2 + g1 )(Z1 (x))2l−10 − (1 + (Z1 (x))2 )( − Z2 (x)g2 − g1 )(Z2 (x))2l−10 ). Make full use of the computer, we obtain

1 10 1 2 11 8 9 2 41 6 25 2 x + x + x + x + 4x9 + x + 4x7 + x + 4x5 2 2 2 2 2 2 61 4 25 2 x + x + 4x3 + 16x2 + 4 x2 + 4x + 2 > 0; + 2 2 (1 + (Z2 (x))2 )(Z1 (x)g2 + g1 ) − (1 + (Z1 (x))2 )( − Z2 (x)g2 − g1 ) Z1 (x)g2 + g1 =

= x10 + 12x8 + 50x6 + 84x4 + 50x2 + 8 > 0. Namely, (1 + (Z2 (x))2 )(Z1 (x)g2 + g1 ) > 0 and (1 + (Z2 (x))2 )(Z1 (x)g2 + g1 ) > (1 + (Z1 (x))2 )( − Z2 (x)g2 − g1 ). On the other hand, since Z1 (x) > |Z2 (x)| > 0, then (Z1 (x))2l−10 > (Z2 (x))2l−10 > 0 for l ≥ 6. Consequently, we always have (1 + (Z2 (x))2 )(Z1 (x)g2 + g1 )(Z1 (x))2l−10 − (1 + (Z1 (x))2 )( − Z2 (x)g2 − g1 )(Z2 (x))2l−10 > 0. Hence H0 (l, x) < 0. That is, H0 (l, x) is decreasing on l. It follows from the claim that H0 (l, x) ≤ H0 (6, x). As Z (x) − H0 (6, x) =















x2 + 4x13 + 14 x2 + 4x11 + 75 x2 + 4x9 +

190 x2 + 4x7 + 224 x2 + 4x5 + 98 x2 + 4x3 + 8 x2 + 4x > 0 for all x > 0, we demonstrate that Z(x) > H0 (6, x) ≥ H0 (l, x). Therefore, C1 (ix)B2 (ix) − C2 (ix)B1 (ix) = (Z (x) − H0 (l, x))/(x2 + 4)2 > 0. K (x) Up to now, we have established that C1 (ix)B2 (ix) − C2 (ix)B1 (ix) > 0, which indicates that K0 (x) < 0. Hence ln (1 + H 0(n,x) ) < ln 1 = 0. Namely, we have ln

ln

C1 (ix)(Y1 (ix))n+2 +C2 (ix)(Y2 (ix))n+2 B1 (ix)(Y1 (ix))n+2 +B2 (ix)(Y2 (ix))n+2

< ln

C1 (ix)(Y1 (ix))n +C2 (ix)(Y2 (ix))n . B1 (ix)(Y1 (ix))n +B2 (ix)(Y2 (ix))n

0

Thus

C1 (ix)(Y1 (ix))n + C2 (ix)(Y2 (ix))n C1 (ix)(Y1 (ix))11 + C2 (ix)(Y2 (ix))11 ≤ ln n n B1 (ix)(Y1 (ix)) + B2 (ix)(Y2 (ix)) B1 (ix)(Y1 (ix))11 + B2 (ix)(Y2 (ix))11

for n ≥ 11. This yields that for 6 ≤ l ≤

n+1 2 ,

ME(P (n − 4, 3, P54 , u1 )) − ME(Pnl ) = ≤

2

π 2

π



ln

C1 (ix)(Y1 (ix))n + C2 (ix)(Y2 (ix))n dx B1 (ix)(Y1 (ix))n + B2 (ix)(Y2 (ix))n

ln

C1 (ix)(Y1 (ix))11 + C2 (ix)(Y2 (ix))11 dx B1 (ix)(Y1 (ix))11 + B2 (ix)(Y2 (ix))11

∞

0



0

∞

l = ME(P (7, 3, P54 , u1 )) − ME(P11 ).

6 ) = 13.77695. When n = 11, then l = 6. By computer-aided calculations, we arrive at ME(P (7, 3, P54 , u1 )) = 13.75635, ME(P11 4 l 4 6 4 Hence ME(P (n − 4, 3, P5 , u1 )) − ME(Pn ) ≤ ME(P (7, 3, P5 , u1 )) − ME(P11 ) < 0, i.e., ME(P (n − 4, 3, P5 , u1 )) < ME(Pnl ). The proof is thus completed. 

Based on the lemmas we established, we can now state our main results.

L. Chen, J. Liu / Applied Mathematics and Computation 268 (2015) 644–656

655

Theorem 1. Let n ≥ 9 be odd and l be even. If G is an arbitrary graph in Bn,n other than the graphs Pnl (4 ≤ l ≤ ME (Pnl ).

n+1 2 ),

then ME (G) <

Proof. Let G be an arbitrary graph in Bn,n other than the graphs Pnl (4 ≤ l ≤ n+1 2 ). Suppose the girth of G is g = g(G). g If n = 9, then l = 4, by Lemma 4 and Remark 2 after Lemma 9, it’s easy to obtain, for such a graph G, that ME(G) ≤ ME(P9 ) ≤ 4 4 ME(P9 ). Furthermore, the equalities cannot hold simultaneously. Hence ME(G) < ME(P9 ). 4 ) > ME(P 6 ), it suffices to show ME(G) < ME(P 6 ). If g ≥ 6, then according to Lemma 4 If n = 11, then l = 4, 6. Since ME(P11 11 11 and Remark 2 after Lemma 9, there is no need to elaborate. If g = 4, then we should only consider the graph P (7, 3, P54 , u1 ) on the 6 ). basis of the Lemmas 11, 12, 13, 14, 15, 16, 17, 18, 19. Applying Lemma 21 directly, we get ME(P (7, 3, P54 , u1 )) < ME(P11 If n ≥ 13, for g > n+1

n+1 2 ,

n+1

we have ME(G) ≤ ME(Pn ) < ME(Pn 2 ) ≤ ME(Pnl ). For g = g

n+1 2 ,

n+1

since G  Pn 2 , we have ME(G) <

ME(Pn 2 ) ≤ ME(Pnl ). For g < n+1 2 ≤ n − 5, putting Lemmas 11, 12, 13, 14, 15, 16, 17, 18, 19 together with Lemma 21, we can show ME(G) < ME(Pnl ). The theorem is thus proved.  Combining Theorem 1 with Remark 2 after Lemma 9, it’s not difficult to obtain the key point of our paper. Theorem 2. Let n ≥ 9 be odd. Then we have n+1

(i) If n ≡ 3 (mod 4), Pn4 , Pn6 , . . . , Pn 2 are the graphs in Bn,n with the first (ii) If n ≡ 1 (mod 4),

Pn4 , Pn6 ,

n−1 2

. . . , Pn

are the graphs in Bn,n with the first

n−3 4

largest matching energies;

n−5 4

largest matching energies.

4. Conclusion l In this paper, we established the graphs in Bn,n with the first  n−3 4  largest matching energies. They all have the form of Pn for some l. Among these graphs, the graph Pn6 plays an important role in unicyclic graphs. In [36], the authors determined that Pn6 is the only graph which attains the maximum value of the energy among all the bipartite unicyclic graphs for n ≥ 16. Furthermore, it’s the graph having maximal energy among all unicyclic graphs (see [3] and [35]). While in this paper, for odd n, we conclude that Pn6 has the second maximal matching energy in Bn,n when n ≥ 11.

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