1 Introduction↩︎

Graphs discussed in this paper are simple and undirected. Let \(G\) be a graph with vertex set \(V(G)=\{v_1,v_2,\ldots,v_n\}\) and edge set \(E(G)\). The order of \(G\) is denoted by \(|V(G)|=n\). A graph \(G\) is called trivial if \(n=1\). The diagonal degree matrix of \(G\) is denoted by \(D(G)=diag(d_1,d_2,\ldots,d_n)\), where \(d_i\) denotes the degree of a vertex \(v_i\) in \(G\). The adjacency matrix of \(G\) is defined to be the \((0,1)\)-matrix \(A(G)=(a_{ij})_{n\times n}\), where \(a_{ij}=1\) if and only if \(v_i\) and \(v_j\) are adjacent in \(G\). Let \(Q(G)=D(G)+A(G)\) be the signless Laplacian matrix of \(G\). The largest eigenvalues of \(A(G)\) and \(Q(G)\), denoted by \(\rho(G)\) and \(q(G)\), are called the spectral radius and the signless Laplacian spectral radius of \(G\), respectively.

For a vertex subset \(S\) of \(V(G)\), let \(G[S]\) denote the subgraph of \(G\) induced by \(S\), and let \(G-S\) be the subgraph derived from \(G\) by deleting the vertices in \(S\) together with their incident edges. Let \(G_1\) and \(G_2\) be two vertex disjoint graphs. We denote by \(G_1\cup G_2\) the disjoint union of \(G_1\) and \(G_2\), and by \(G_1\vee G_2\) the join of \(G_1\) and \(G_2\). Let \(K_n\) denote the complete graph of order \(n\).

For two integers \(a\) and \(b\) with \(1\leq a\leq b\), a spanning subgraph \(F\) of \(G\) is called an \([a,b]\)-factor of \(G\) if \(a\leq d_F(v)\leq b\) for every \(v\in V(G)\). If \(a=1\) and \(b=k\), then an \([a,b]\)-factor is a \([1,k]\)-factor. If \(n\) is even and \(k=1\), then a \([1,k]\)-factor is a 1-factor (or a perfect matching). A spanning tree of \(G\) is called a spanning \(k\)-tree if its maximum degree is at most \(k\), where \(k\geq2\) is an integer. It is obvious that a graph \(G\) admits a spanning \(k\)-tree if and only if \(G\) contains a connected \([1,k]\)-factor.

Spanning \(k\)-trees and \([a,b]\)-factors have attracted many researchers’ attention. Many sufficient conditions for graphs to possess \([1,2]\)-factors were derived by Johnson, Paulusma and Wood [1], Kelmans [2], Kawarabayashi, Matsuda, Oda and Ota [3], Gao and Wang [4], Wu [5], Zhou [6], [7], Zhou, Zhang and Sun [8], Zhou, Sun and Bian [9], Zhou, Wu and Bian [10], Zhou, Sun and Liu [11], [12], Wang and Zhang [13], Liu and Pan [14]. Kim, O, Park and Ree [15], Wang and Zhang [16] provided some sufficient conditions for the existence of \([1,b]\)-factors in graphs. Zhou and Liu [17] investigated the connections between the spectral radius of a connected graph and its odd \([1,b]\)-factors, and provided a lower bound on the existence of odd \([1,b]\)-factors by virtue of the spectral radius. Matsuda [18], Furuya and Yashima [19], Wu [20], Zhou [21], Zhou, Pan and Xu [22], Zhou, Zhang and Liu [23], Zhou, Pan and Xu [24], Zhou and Zhang [25], Wang and Zhang [26], [27] studied the properties of \([a,b]\)-factors in graphs, and showed some graphic parameter conditions for graphs to possess \([a,b]\)-factors. Ding, Johnson and Seymour [28], Kyaw [29], Neumann and Eduardo [30] provided some sufficient conditions for connected graphs to possess spanning trees. Broersma and Tuinstra [31], Tsugaki and Yamashita [32] established the connections between degree conditions and spanning trees in connected graphs. Kyaw [33] presented a degree and neighborhood condition for a connected graph to possess a spanning \(k\)-tree. Neumann-Lara and Rivera-Campo [34] gave an independence number condition to guarantee that a connected graph contains a spanning \(k\)-tree. Win [35] create a connection between toughness and the existence of a spanning \(k\)-tree of a connected graph. Very recently, Fan, Goryainov, Huang and Lin [36] established a lower bound on a spanning \(k\)-tree of a connected graph by virtue of its spectral radius.

Kaneko [37] introduced the concept of leaf degree of a spanning tree. Let \(T\) be a spanning tree of a connected graph \(G\). The leaf degree of a vertex \(v\in V(T)\) is defined as the number of leaves adjacent to \(v\) in \(T\). Furthermore, the leaf degree of \(T\) is the maximum leaf degree among all the vertices of \(T\). Kaneko [37] obtained a criterion for the existence of a spanning tree with leaf degree at most \(k\) in a connected graph.

Motivated by [37], [38] directly, it is natural and interesting to put forward a sufficient condition to guarantee that a graph contains a spanning tree with leaf degree at most \(k\). In this paper, we study the relation between the signless Laplacian spectral radius of a connected graph and its spanning tree with leaf degree at most \(k\), and obtain a signless Laplacian spectral radius condition for the existence of a spanning tree with leaf degree at most \(k\) in a connected graph, which are shown in the following.

Let \(k\) be a positive integer, and let \(G\) be a connected graph of order \(n\geq k+2\).

() For \((k,n)\notin\{(1,6),(2,8),(3,10),(4,12)\}\), if \(q(G)>\theta(k,n)\), then \(G\) has a spanning tree with leaf degree at most \(k\), where \(\theta(k,n)\) is the largest root of \(x^{3}-(3n-2k-5)x^{2}+n(2n-2k-5)x-2(n-k-3)(n-k-2)=0\).

() For \((k,n)=(1,6)\), if \(q(G)>4+2\sqrt{3}\), then \(G\) has a spanning tree with leaf degree at most 1.

() For \((k,n)=(2,8)\), if \(q(G)>5+\sqrt{21}\), then \(G\) has a spanning tree with leaf degree at most 2.

() For \((k,n)=(3,10)\), if \(q(G)>6+4\sqrt{2}\), then \(G\) has a spanning tree with leaf degree at most 3.

() For \((k,n)=(4,12)\), if \(q(G)>7+3\sqrt{5}\), then \(G\) has a spanning tree with leaf degree at most 4.

The proof of Theorem 1.1 will be given in Section 3.

2 Some preliminaries↩︎

In this section, we showed some necessary preliminary lemmas, which are very useful in the proof of Theorem 1.1. Kaneko [37] provided a criterion for the existence of a spanning tree with leaf degree at most \(k\) in a connected graph.

([37]). Let \(k\) be a positive integer and \(G\) be a connected graph. Then \(G\) contains a spanning tree with leaf degree at most \(k\) if and only if \[i(G-S)<(k+1)|S|\] for every nonempty subset \(S\) of \(V(G)\), where \(i(G-S)\) denotes the number of those isolated vertices in \(G-S\).

Cvetković and Simić [39] obtained the signless Laplacian spectral radius of a complete graph.

([39]). Let \(K_n\) be a complete graph of order \(n\), where \(n\geq2\) is an integer. Then \(q(K_n)=2n-2\).

Shen, You, Zhang and Li [40] posed a result on the signless Laplacian spectral radius of a connected graph.

([40]). Let \(G\) be a connected graph. If \(H\) is a subgraph of \(G\), then \(q(H)\leq q(G)\). If \(H\) is a proper subgraph of \(G\), then \(q(H)<q(G)\).

Let \(M\) be a real symmetric matrix whose rows and columns are indexed by \(V=\{1,2,\cdots,n\}\). Assume that \(M\), with respect to the partition \(\pi: V=V_1\cup V_2\cup\cdots\cup V_t\), can be written as \[\begin{align} M=\left( \begin{array}{ccc} M_{11} & \cdots & M_{1t}\\ \vdots & \ddots & \vdots\\ M_{t1} & \cdots & M_{tt}\\ \end{array} \right), \end{align}\] where \(M_{ij}\) denotes the submatrix (block) of \(M\) formed by rows in \(V_i\) and columns in \(V_j\). Let \(q_{ij}\) denote the average row sum of \(M_{ij}\). Then matrix \(M_{\pi}=(q_{ij})\) is called the quotient matrix of \(M\). If the row sum of every block \(M_{ij}\) is a constant, then the partition is equitable.

([41]). Let \(M\) be a real symmetric matrix with an equitable partition \(\pi\), and let \(M_{\pi}\) be the corresponding quotient matrix. Then every eigenvalue of \(M_{\pi}\) is an eigenvalue of \(M\). Furthermore, if \(M\) is nonnegative, then the largest eigenvalues of \(M\) and \(M_{\pi}\) are equal.

3 The proof of Theorem 1.1↩︎

In this section, we prove Theorem 1.1.

Let \(\varphi(x)=x^{3}-(3n-2k-5)x^{2}+n(2n-2k-5)x-2(n-k-3)(n-k-2)\) and let \(\theta(k,n)\) is the largest root of \(\varphi(x)=0\). Suppose that \(G\) has no spanning tree with leaf degree at most \(k\). It follows from Lemma 2.1 that \[i(G-S)\geq(k+1)|S|\] for some nonempty subset \(S\) of \(V(G)\). Choose such a connected graph \(G\) so that its signless Laplacian spectral radius is as large as possible. In light of Lemma 2.3 and the choice of \(G\), the induced subgraph \(G[S]\) and each connected component of \(G-S\) are complete graphs, respectively. Furthermore, \(G\) is just the graph \(G[S]\vee(G-S)\).

We may see that \(G-S\) has at most one non-trivial connected component. Otherwise, we can add edges among all non-trivial connected components to get a non-trivial connected component of larger size, which gives a contradiction (based on Lemma 2.3). For convenience, let \(i(G-S)=i\) and \(|S|=s\). Thus, we possess \[\begin{align} \label{eq:3461} i\geq(k+1)s. \end{align}\tag{1}\] In what follows, we proceed by considering two possible cases.

\(G-S\) admits just one non-trivial connected component.

In this case, it is obvious that \(G=K_s\vee(K_{n_1}\cup iK_1)\), where \(n_1\geq2\) is an integer. We are to show \(i=(k+1)s\). Assume \(i\geq(k+1)s+1\), then we construct a new graph \(H_1\) derived from \(G\) by joining each vertex of \(K_{n_1}\) with one vertex in \(iK_1\) by an edge. Then \(i(H_1-S)\geq(k+1)s\) and \(G\) is a proper spanning subgraph of \(H_1\). Utilizing Lemma 2.3, we have \(q(G)<q(H_1)\), which is a contradiction to the choice of \(G\). Hence, \(i\leq(k+1)s\). Combining this with (1 ), we get \(i=(k+1)s\), and so \(G=K_s\vee(K_{n_1}\cup(k+1)sK_1)\). Obviously, \(n=(k+2)s+n_1\geq(k+2)s+2\geq k+4\). In view of the equitable partition \(V(G)=V(K_s)\cup V(K_{n_1})\cup V((k+1)sK_1)\), the quotient matrix of \(Q(G)\) equals \[\begin{align} M_1=\left( \begin{array}{ccc} n+s-2 & n-(k+2)s & (k+1)s\\ s & 2n-(2k+3)s-2 & 0\\ s & 0 & s\\ \end{array} \right), \end{align}\] whose characteristic polynomial is \[\begin{align} f_1(x)=&x^{3}-(3n-(2k+1)s-4)x^{2}+(2n^{2}-(2k-1)sn-6n-4(k+1)s^{2}+4ks+4)x\\ &-s(2n^{2}-4(k+1)sn-6n+2(k+1)^{2}s^{2}+6(k+1)s+4). \end{align}\] According to Lemma 2.4, the largest root, say \(\theta_1\), of \(f_1(x)=0\) equals \(q(G)\).

Note that \(K_s\vee(n-s)K_1\) is a proper subgraph of \(G\). In terms of the equitable partition \(V(K_s\vee(n-s)K_1)=V(K_s)\cup V((n-s)K_1)\), the quotient matrix of \(Q(K_s\vee(n-s)K_1)\) is \[\begin{align} M_2=\left( \begin{array}{ccc} n+s-2 & n-s \\ s & s \\ \end{array} \right). \end{align}\] Then the characteristic polynomial of the matrix \(M_2\) is given as \[f_2(x)=x^{2}-(n+2s-2)x+2s(s-1).\] Using Lemma 2.4, the largest root, say \(\theta_2\), of \(f_2(x)=0\) equals \(q(K_s\vee(n-s)K_1)\). Thus, we get \[\begin{align} \label{eq:3462} q(K_s\vee(n-s)K_1)=\theta_2=\frac{n+2s-2+\sqrt{(n+2s-2)^{2}-8s(s-1)}}{2}. \end{align}\tag{2}\] Together with Lemma 2.3, we infer \[\begin{align} \label{eq:3463} \theta_1=q(G)>q(K_s\vee(n-s)K_1)=\frac{n+2s-2+\sqrt{(n+2s-2)^{2}-8s(s-1)}}{2}. \end{align}\tag{3}\]

To verify \(q(G)\leq\theta(k,n)\), it suffices to prove \(\varphi(\theta_1)\leq0\). Recall that \(f_1(\theta_1)=0\). Hence, \[\begin{align} \label{eq:3464} \varphi(\theta_1)=&\varphi(\theta_1)-f_1(\theta_1)\nonumber\\ =&(s-1)(-(2k+1)\theta_1^{2}+((2k-1)n+4(k+1)s+4)\theta_1\nonumber\\ &+2(n-k-3)(n-k-2)-4(k+1)sn+2(k+1)^{2}s(s+1)+6(k+1)s)\nonumber\\ =&(s-1)g_1(\theta_1), \end{align}\tag{4}\] where \(g_1(\theta_1)=-(2k+1)\theta_1^{2}+((2k-1)n+4(k+1)s+4)\theta_1+2(n-k-3)(n-k-2)-4(k+1)sn+2(k+1)^{2}s(s+1)+6(k+1)s\). According to (3 ) and \(n\geq(k+2)s+2\), we possess \[\frac{(2k-1)n+4(k+1)s+4}{2(2k+1)}<n+s-2<\frac{n+2s-2+\sqrt{(n+2s-2)^{2}-8s(s-1)}}{2}<\theta_1.\] Thus, we derive \[\begin{align} \label{eq:3465} g_1(\theta_1)<&g_1\left(\frac{n+2s-2+\sqrt{(n+2s-2)^{2}-8s(s-1)}}{2}\right)\nonumber\\ =&n^{2}-(4k+5)sn-(2k+5)n+2(k+1)(k+3)s^{2}+(2k^{2}+10k+10)s\nonumber\\ &+2k^{2}+6k+6-(n-s-2k-3)\sqrt{(n+2s-2)^{2}-8s(s-1)}. \end{align}\tag{5}\] Recall that \(n\geq(k+2)s+2\geq3s+2\). For \(s\geq2\), we deduce \[n-s-2k-3\geq(k+2)s+2-s-2k-3=(k+1)s-2k-1\geq2(k+1)-2k-1>0\] and \[(n+2s-2)^{2}-8s(s-1)-(n+s-1)^{2}=(s-1)(2n-5s-3)>0.\] Together with (5 ), \(s\geq2\) and \(n\geq(k+2)s+2\), we obtain \[\begin{align} \label{eq:3466} g_1(\theta_1)<&n^{2}-(4k+5)sn-(2k+5)n+2(k+1)(k+3)s^{2}+(2k^{2}+10k+10)s\nonumber\\ &+2k^{2}+6k+6-(n-s-2k-3)\sqrt{(n+2s-2)^{2}-8s(s-1)}\nonumber\\ <&n^{2}-(4k+5)sn-(2k+5)n+2(k+1)(k+3)s^{2}+(2k^{2}+10k+10)s\nonumber\\ &+2k^{2}+6k+6-(n-s-2k-3)(n+s-1)\nonumber\\ =&-(4k+5)sn-n+(2k^{2}+8k+7)s^{2}+(2k^{2}+12k+12)s+2k^{2}+4k+3\nonumber\\ \leq&-(4k+5)s((k+2)s+2)-((k+2)s+2)+(2k^{2}+8k+7)s^{2}\nonumber\\ &+(2k^{2}+12k+12)s+2k^{2}+4k+3\nonumber\\ =&-(2k^{2}+5k+3)s^{2}+(2k^{2}+3k)s+2k^{2}+4k+1\nonumber\\ \leq&-2(2k^{2}+5k+3)s+(2k^{2}+3k)s+2k^{2}+4k+1\nonumber\\ =&-(2k^{2}+7k+6)s+2k^{2}+4k+1\nonumber\\ \leq&-2(2k^{2}+7k+6)+2k^{2}+4k+1\nonumber\\ =&-2k^{2}-10k-11\nonumber\\ <&0. \end{align}\tag{6}\] It follows from (4 ), (6 ) and \(s\geq1\) that \[\varphi(\theta_1)=(s-1)g_1(\theta_1)\leq0,\] which yields \(q(G)=\theta_1\leq\theta(k,n)\), which contradicts \(q(G)>\theta(k,n)\) for \(n\geq k+2\) and \((k,n)\notin\{(1,6),(2,8),(3,10),(4,12)\}\). As for \((k,n)=(1,6)\), we possess \(q(G)=\theta_1\leq\theta(1,6)<4+2\sqrt{3}\), which is a contradiction to \(q(G)>4+2\sqrt{3}\) for \((k,n)=(1,6)\). As for \((k,n)=(2,8)\), we have \(q(G)=\theta_1\leq\theta(2,8)<5+\sqrt{21}\), which contradicts \(q(G)>5+\sqrt{21}\) for \((k,n)=(2,8)\). As for \((k,n)=(3,10)\), we obtain \(q(G)=\theta_1\leq\theta(3,10)<6+4\sqrt{2}\), which contradicts \(q(G)>6+4\sqrt{2}\) for \((k,n)=(3,10)\). As for \((k,n)=(4,12)\), we derive \(q(G)=\theta_1\leq\theta(4,12)<7+3\sqrt{5}\), which is a contradiction to \(q(G)>7+3\sqrt{5}\) for \((k,n)=(4,12)\).

\(G-S\) has no non-trivial connected component.

In this case, it is clear that \(G=K_s\vee iK_1\). We are to show \(i\leq(k+1)s+1\). Assume that \(i\geq(k+1)s+2\). Then we can construct a new graph \(H_2\) obtained from \(G\) by adding an edge in \(iK_1\). Thus, we see \(i(H_2-S)\geq(k+1)s\) and \(G\) is a proper subgraph of \(H_2\). Using Lemma 2.3, we have \(q(G)<q(H_2)\), which contradicts the choice of \(G\). Therefore, \(i\leq(k+1)s+1\). Together with (1 ), it suffices to consider \(i=(k+1)s\) and \(i=(k+1)s+1\).

\(i=(k+1)s\).

In this subcase, we have \(G=K_s\vee(k+1)sK_1=K_s\vee(n-s)K_1\), where \(n=(k+2)s\geq k+2\). In terms of (2 ), we get \[q(G)=q(K_s\vee(n-s)K_1)=\theta_2=\frac{n+2s-2+\sqrt{(n+2s-2)^{2}-8s(s-1)}}{2}.\] Note that \(f_2(\theta_2)=0\). Then \[\begin{align} \label{eq:3467} \varphi(\theta_2)=&\varphi(\theta_2)-\theta_2f_2(\theta_2)\nonumber\\ =&-(2n-2s-2k-3)\theta_2^{2}+(2n^{2}-2kn-5n-2s^{2}+2s)\theta_2-2(n-k-3)(n-k-2)\nonumber\\ =&-(s+1)n^{2}+(3s+2k+2)sn+(2k+5)n-(2s^{2}+4k+4)s-2k^{2}-6k-6\nonumber\\ &+(-sn+n+s^{2}+2ks+2s-2k-3)\sqrt{(n+2s-2)^{2}-8s(s-1)}. \end{align}\tag{7}\]

\(s=1\).

Obviously, \(n=k+2\) and \(q(G)=n=\theta(k,k+2)\), which contradicts \(q(G)>\theta(k,n)\) for \(n=k+2\).

\(s=2\).

Note that \(n=2k+4\). If \(k=1\), then \(n=6\) and \(q(G)=4+2\sqrt{3}\), a contradiction. If \(k=2\), then \(n=8\) and \(q(G)=5+\sqrt{21}\), a contradiction. If \(k=3\), then \(n=10\) and \(q(G)=6+4\sqrt{2}\), a contradiction. If \(k=4\), then \(n=12\) and \(q(G)=7+3\sqrt{5}\), a contradiction. If \(k\geq5\), then it follows from \(n=2k+4\) and (7 ) that \(\varphi(\theta_2)=-2k^{2}+4k+6+2\sqrt{(k+1)(k+5)}<0\), which implies that \(q(G)=\theta_2<\theta(k,n)\) for \(s=2\), a contradiction.

\(s\geq3\).

Note that \(n=(k+2)s\). We easily deduce \[\begin{align} &-sn+n+s^{2}+2ks+2s-2k-3\\ =&-(k+2)s^{2}+(k+2)s+s^{2}+2ks+2s-2k-3\\ =&-(k+1)s^{2}+(3k+4)s-2k-3\\ <&0 \end{align}\] and \[(n+2s-2)^{2}-8s(s-1)-(n+s-1)^{2}=(s-1)(2n-5s-3)\geq0\] for \(s\geq3\). Together with (7 ) and \(n=(k+2)s\), we get \[\begin{align} \label{eq:3468} \varphi(\theta_2)\leq&-(s+1)n^{2}+(3s+2k+2)sn+(2k+5)n-(2s^{2}+4k+4)s-2k^{2}-6k-6\nonumber\\ &+(-sn+n+s^{2}+2ks+2s-2k-3)(n+s-1)\nonumber\\ =&-(2k^{2}+5k+3)s^{3}+(4k^{2}+16k+13)s^{2}-7(k+1)s-2k^{2}-4k-3\nonumber\\ :=&p(s). \end{align}\tag{8}\] Then we see that \(p'(s)=-3(2k^{2}+5k+3)s^{2}+2(4k^{2}+16k+13)s-7(k+1)\) and \(p''(s)=-6(2k^{2}+5k+3)s+2(4k^{2}+16k+13)\). For \(s\geq3\), we infer \[\begin{align} p''(s)=&-6(2k^{2}+5k+3)s+2(4k^{2}+16k+13)\\ \leq&-18(2k^{2}+5k+3)+2(4k^{2}+16k+13)\\ =&-28k^{2}-58k-28\\ <&0. \end{align}\] Hence, \(p'(s)\) is decreasing in the interval \([3,+\infty)\). Then we have \[\begin{align} p'(s)\leq&p'(3)\\ =&-27(2k^{2}+5k+3)+6(4k^{2}+16k+13)-7(k+1)\\ =&-30k^{2}-46k-10\\ <&0, \end{align}\] which implies that \(p(s)\) is decreasing in the interval \([3,+\infty)\). Thus, we get \[\begin{align} p(s)\leq&p(3)\\ =&-27(2k^{2}+5k+3)+9(4k^{2}+16k+13)-21(k+1)-2k^{2}-4k-3\\ =&-20k^{2}-16k+12\\ <&0. \end{align}\] Together with (8 ), we obtain \(\varphi(\theta_2)\leq p(s)<0\) for \(s\geq3\), which implies \(q(G)=\theta_2<\theta(k,n)\), a contradiction.

\(i=(k+1)s+1\).

In this subcase, it is obvious that \(G=K_s\vee((k+1)s+1)K_1\) and \(n=(k+2)s+1\geq k+3\). According to the equitable partition \(V(G)=V(K_s)\cup V(((k+1)s+1)K_1)\), the quotient matrix of \(Q(G)\) equals \[\begin{align} M_3=\left( \begin{array}{ccc} (k+3)s-1 & (k+1)s+1\\ s & s \\ \end{array} \right). \end{align}\] Thus we derive its characteristic polynomial as \[f_3(x)=x^{2}-((k+4)s-1)x+2s(s-1).\] By virtue of Lemma 2.4, the largest root, say \(\theta_3\), of \(f_3(x)=0\) equals \(q(G)\). Thus, we possess \[q(G)=\theta_3=\frac{(k+4)s-1+\sqrt{(k^{2}+8k+8)s^{2}-2ks+1}}{2}.\] Note that \(n=(k+2)s+1\) and \(f_3(\theta_3)=0\). Then \[\begin{align} \label{eq:3469} \varphi(\theta_3)=&\varphi(\theta_3)-\theta_3f_3(\theta_3)\nonumber\\ =&-(2(k+1)s-2k-1)\theta_3^{2}+(2(k+1)(k+3)s^{2}-k(2k+5)s-2k-3)\theta_3\nonumber\\ &-2(k+2)^{2}s^{2}+2(k+2)(2k+3)s-2(k+1)(k+2)\nonumber\\ =&-k(k+1)s^{3}+(k^{2}-1)s^{2}+(2k^{2}+5k+3)s-2(k+1)^{2}\nonumber\\ &-(k+1)(s-1)(s-2)\sqrt{(k^{2}+8k+8)s^{2}-2ks+1}\nonumber\\ \leq&-k(k+1)s^{3}+(k^{2}-1)s^{2}+(2k^{2}+5k+3)s-2(k+1)^{2}\nonumber\\ &-(k+1)(s-1)(s-2)(k+3)s. \end{align}\tag{9}\]

If \(s=1\), then \(n=k+3\). Together with (9 ), we get \(q(G)=\theta_3=\theta(k,k+3)\), which contradicts \(q(G)>\theta(k,n)\) for \(n=k+3\). If \(s=2\), then \(n=2k+5\). Combining this with (9 ), we infer \[\begin{align} \varphi(\theta_3)=&-8k(k+1)+4(k^{2}-1)+2(2k^{2}+5k+3)-2(k+1)^{2}\\ =&-2k^{2}-2k\\ <&0, \end{align}\] which implies \(q(G)=\theta_3<\theta(k,2k+5)\), which contradicts \(q(G)>\theta(k,n)\) for \(n=2k+5\). If \(s\geq3\), then it follows from (9 ) that \[\begin{align} \varphi(\theta_3)\leq&-k(k+1)s^{3}+(k^{2}-1)s^{2}+(2k^{2}+5k+3)s-2(k+1)^{2}\\ &-(k+1)(s-1)(s-2)(k+3)s\\ <&-3k(k+1)s^{2}+(k^{2}-1)s^{2}+(2k^{2}+5k+3)s-2(k+1)^{2}\\ =&-(2k^{2}+3k+1)s^{2}+(2k^{2}+5k+3)s-2(k+1)^{2}\\ \leq&-3(2k^{2}+3k+1)s+(2k^{2}+5k+3)s-2(k+1)^{2}\\ =&-4k(k+1)s-2(k+1)^{2}\\ <&0, \end{align}\] which yields \(q(G)=\theta_3<\theta(k,n)\), a contradiction. This completes the proof of Theorem 1.1. \(\Box\)

4 Extremal graphs↩︎

Finally, we establish several graphs to claim that the bounds obtained in Theorem 1.1 are sharp.

Let \(k\) and \(n\) be two positive integers with \(n\geq k+2\), and let \(\theta(k,n)\) be the largest root of \(x^{3}-(3n-2k-5)x^{2}+n(2n-2k-5)x-2(n-k-3)(n-k-2)=0\). If \((k,n)\notin\{(1,6),(2,8),(3,10),(4,12)\}\), then \(q(K_1\vee(K_{n-k-2}\cup(k+1)K_1))=\theta(k,n)\) and \(K_1\vee(K_{n-k-2}\cup(k+1)K_1)\) contains no spanning tree with leaf degree at most \(k\). If \((k,n)=(1,6)\), then \(q(K_2\vee4K_1)=4+2\sqrt{3}\) and \(K_2\vee4K_1\) contains no spanning tree with leaf degree at most 1. If \((k,n)=(2,8)\), then \(q(K_2\vee6K_1)=5+\sqrt{21}\) and \(K_2\vee6K_1\) contains no spanning tree with leaf degree at most 2. If \((k,n)=(3,10)\), then \(q(K_2\vee8K_1)=6+4\sqrt{2}\) and \(K_2\vee8K_1\) contains no spanning tree with leaf degree at most 3. If \((k,n)=(4,12)\), then \(q(K_2\vee10K_1)=7+3\sqrt{5}\) and \(K_2\vee10K_1\) contains no spanning tree with leaf degree at most 4.

In terms of the equitable partition \(V(K_1\vee(K_{n-k-2}\cup(k+1)K_1))=V(K_1)\cup V(K_{n-k-2})\cup V((k+1)K_1))\), the quotient matrix of \(Q(K_1\vee(K_{n-k-2}\cup(k+1)K_1))\) equals \[\begin{align} M(K_1\vee(K_{n-k-2}\cup(k+1)K_1))=\left( \begin{array}{ccc} n-1 & n-k-2 & k+1\\ 1 & 2n-2k-5 & 0\\ 1 & 0 & 1\\ \end{array} \right), \end{align}\] whose characteristic polynomial is given as \(x^{3}-(3n-2k-5)x^{2}+n(2n-2k-5)x-2(n-k-3)(n-k-2)\). In view of Lemma 2.4, the largest root of \(x^{3}-(3n-2k-5)x^{2}+n(2n-2k-5)x-2(n-k-3)(n-k-2)=0\), \(\theta(k,n)\) equals \(q(K_1\vee(K_{n-k-2}\cup(k+1)K_1))\). Namely, \(q(K_1\vee(K_{n-k-2}\cup(k+1)K_1))=\theta(k,n)\). Set \(S=V(K_1)\). Then \(|S|=1\) and \(i(K_1\vee(K_{n-k-2}\cup(k+1)K_1)-S)=i((K_{n-k-2}\cup(k+1)K_1))=k+1=(k+1)|S|\). According to Lemma 2.1, \(K_1\vee(K_{n-k-2}\cup(k+1)K_1)\) has no spanning tree with leaf degree at most \(k\).

By virtue of the equitable partition \(V(K_2\vee4K_1)=V(K_2)\cup V(4K_1)\), the quotient matrix of \(Q(K_2\vee4K_1)\) equals \[\begin{align} M(K_2\vee4K_1)=\left( \begin{array}{ccc} 6 & 4\\ 2 & 2 \\ \end{array} \right). \end{align}\] Then its characteristic polynomial is \(x^{2}-8x+4\). By means of Lemma 2.4, the largest root of \(x^{2}-8x+4=0\) equals \(q(K_2\vee4K_1)\). Namely, \(q(K_2\vee4K_1)=4+2\sqrt{3}\). Write \(S=V(K_2)\). Then \(|S|=2\) and \(i(K_2\vee4K_1-S)=4=2|S|\). Applying Lemma 2.1, \(K_2\vee4K_1\) has no spanning tree with leaf degree at most 1.

According to the equitable partition \(V(K_2\vee6K_1)=V(K_2)\cup V(6K_1)\), the quotient matrix of \(Q(K_2\vee6K_1)\) equals \[\begin{align} M(K_2\vee6K_1)=\left( \begin{array}{ccc} 8 & 6\\ 2 & 2 \\ \end{array} \right). \end{align}\] Then its characteristic polynomial is \(x^{2}-10x+4\). Using Lemma 2.4, the largest root of \(x^{2}-10x+4=0\) equals \(q(K_2\vee6K_1)\). That is, \(q(K_2\vee6K_1)=5+\sqrt{21}\). Let \(S=V(K_2)\). Then \(|S|=2\) and \(i(K_2\vee6K_1-S)=6>4=2|S|\). In light of Lemma 2.1, \(K_2\vee6K_1\) has no spanning tree with leaf degree at most 2.

By the equitable partition \(V(K_2\vee8K_1)=V(K_2)\cup V(8K_1)\), the quotient matrix of \(Q(K_2\vee8K_1)\) is \[\begin{align} M(K_2\vee8K_1)=\left( \begin{array}{ccc} 10 & 8\\ 2 & 2 \\ \end{array} \right), \end{align}\] whose characteristic polynomial is given as \(x^{2}-12x+4\). According to Lemma 2.4, the largest root of \(x^{2}-12x+4=0\) equals \(q(K_2\vee8K_1)\). That is, \(q(K_2\vee8K_1)=6+4\sqrt{2}\). Let \(S=V(K_2)\). Then \(|S|=2\) and \(i(K_2\vee8K_1-S)=8>4=2|S|\). Utilizing Lemma 2.1, \(K_2\vee8K_1\) has no spanning tree with leaf degree at most 3.

In view of the equitable partition \(V(K_2\vee10K_1)=V(K_2)\cup V(10K_1)\), the quotient matrix of \(Q(K_2\vee10K_1)\) is \[\begin{align} M(K_2\vee10K_1)=\left( \begin{array}{ccc} 12 & 10\\ 2 & 2 \\ \end{array} \right), \end{align}\] whose characteristic polynomial is given as \(x^{2}-14x+4\). By Lemma 2.4, the largest root of \(x^{2}-14x+4=0\) equals \(q(K_2\vee10K_1)\). That is, \(q(K_2\vee10K_1)=7+3\sqrt{5}\). Put \(S=V(K_2)\). Then \(|S|=2\) and \(i(K_2\vee10K_1-S)=10>4=2|S|\). By means of Lemma 2.1, \(K_2\vee10K_1\) has no spanning tree with leaf degree at most 4. \(\Box\)

Declaration of competing interest↩︎

The authors declares that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Data availability↩︎

No data was used for the research described in the article.

References↩︎