Proof.
Let X=(x1,x2,…,xn)T be the Perron vector of G. From Lemma 12 and by direct calculations, we have, for μ≥3, q(G1)>q(B(3,1,3))≈5.5615>3+5. So, in the following, we only consider those graphs, which have signless Laplacian spectral radius greater than q(G)>3+5.

Choose G*∈Bn(2μ) such that q(G*) is as large as possible. Then G* consists of a subgraph H which is one of graphs B(p,1,q), B(p,l,q), and P(p,l,q) (see Figures 1 and 2).

Let T be a tree attached at some vertex, say, z, of H; |V(T)| is the number of vertices of T including the vertex z. In the following, we prove that tree T is formed by attaching at most one path of length 1 and other paths of length 2 at z.

Suppose P:v0v1⋯vk is a pendant path of G* and vk is a pendant vertex. If k≥3, let H1=G*-v2v3+v0v3. From Corollary 8, we have H1∈Bn(2μ) and q(H1)>q(G*), which is a contradiction.

For each vertex u∈V(T-z), we prove that d(u)≤2. Otherwise, there must exist some vertex u0 of T-z such that d(z,u0)=max{d(z,v)∣v∈V(T)-z,d(v)≥3}. From the above proof, we have the pendant paths attached u0 which have length of at most 2. Obviously, there exists an internal path between u0 and some vertex w of G*, denoted by P¯:u0w1⋯wm (wm=w). If m≥2, let H2 be the graph obtained from G* - u0w1 - w1w2 by amalgamating u0, w1, and w2 to form a new vertex s1 together with attaching a new pendant path s1s2s3 of length 2 at s1. From Corollary 5, we have H2∈Bn(2μ) and q(H2)>q(G*), which is a contradiction. If m=1, by Lemma 14 and Corollary 3, we can get a new graph H3 such that H3∈Bn(2μ) and q(H3)>q(G*), which is a contradiction.

From the proof as above, we have the tree T which is obtained by attaching some pendant paths of length 2 and at most one pendant path of length 1 at z.

From Corollary 2, we have all the pendant paths of length 2 in G* which must be attached at the same vertex of H.

In the following, we prove that G* is isomorphic to one of graphs G1,G2,…,G6 (see Figure 3). We distinguish the following two cases:

Case 1 (
G
*
∈
B
n
+
(
2
μ
)
). We prove that G* is isomorphic to one of graphs G1, G2, and G3.

Assume that there exists some cycle Cp of G* with length of at least 4. From Corollary 5, we have each internal path of G*, which is not a triangle, has length 1. Note that all the pendant paths of length 2 in G* must be attached at the same vertex, then there must exist edges v1v2∈E(G*), v1v3∈E(Cp), and v1v4∈E(Cp) and d(v1)=3, d(v2)=1, d(v3)≥3, and d(v4)≥3. Let H4 (H5) be the graph obtained from G*-v1v3 (G*-v1v4) by amalgamating v1 and v3(v4) to form a new vertex y1(y3) together with subdividing the edge y1v2 (y3v2) with a new vertex y2 (y4). From Lemma 13, we have Hi∈Bn+(2μ) (i=4,5) and either q(H4)>q(G*) or q(H5)>q(G*), which is a contradiction. Then for each cycle Cg of G*, we have g=3.

Assume that l≥4. If there exists an internal path P¯*:vivi+1⋯vm (1≤i<m≤l) with length greater than 1 in G*. Then, by Corollary 5, we can get a new graph H6 such that q(H6)>q(G*) and H6∈Bn+(2μ), which is a contradiction. Thus, d(vi)≥3 (i=1,2,…,l) and either d(v2)=3 or d(v3)=3. By Lemma 13, we can also get a new graph H7 such that q(H7)>q(G*) and H7∈Bn+(2μ), which is a contradiction. Hence, l≤3.

We distinguish the following three subcases:

Subcase 1.1 (
l
=
1
). Then G* is the graph obtained by attaching all the pendant paths of length 2 at the same vertex of G¯, where G¯ is one of graphs G¯1,…,G¯5 (see Figure 4).

Assume that G¯=G¯2. If xu≥xv, let H8=G*-rv-sv+ru+su; if xv≥xu, let H9=G*-ut+tv. Obviously, Hi∈Bn+(2μ) (i=8,9) and either q(H8)>q(G*) or q(H9)>q(G*) by Lemma 1, which is a contradiction. By similar reasoning, we have also G¯≠G¯3.

Subcase 1.2 (
l
=
2
). Then G* is the graph obtained by attaching all the pendant paths of length 2 at the same vertex of G¯, where G¯ is one of graphs G¯6,…,G¯14 (see Figure 4).

Assume that G¯=G¯6. If xv1≥xv2, let H10=G*-v2u+v1u; if xv2≥xv1, let H11=G*-v1r+v2r. Obviously, Hi∈Bn+(2μ) (i=10,11) and either q(H10)>q(G*) or q(H11)>q(G*) by Lemma 1, which is a contradiction. By similar reasoning, we have also G¯≠G¯j (j=6,…,14).

Subcase 1.3 (
l
=
3
). Then G* is the graph obtained by attaching all the pendant paths of length 2 at the same vertex of G¯, where G¯ is one of graphs G¯15,…,G¯20 (see Figure 4).

Assume that G¯=G¯15. If xv1≥xv2, let H12=G*-v2v3+v1v3; if xv2≥xv1, let H13=G*-v1z1+v2z1. Obviously, Hi∈Bn+(2μ) (i=12,13) and either q(H12)>q(G*) or q(H13)>q(G*) by Lemma 1, a contradiction. By similar reasoning, we have also G¯≠G¯j (j=15,…,20).

Thus, G¯ is isomorphic to one of the graphs G¯1, G¯4 and G¯5. In the following, we prove that G* is isomorphic to one of graphs G1, G2 and G3.

Assume that G* is obtained by attaching all the pendant paths of length 2 at vertex y4 of G¯1. If xv1≥xy4, let H14 be the graph obtained from G¯1 by attaching μ-3 pendant paths of length 2 at v1. If xy4≥xv1, let H15=G*-v1y3-v1y1-v1y2+y4y3+y4y1+y4y2. Obviously, H14=H15=G1 and q(G1)>q(G*) by Lemma 1, a contradiction. Then G*=G1. By similar reasoning, the result follows.

Case 2 (
G
*
∈
B
n
+
+
(
2
μ
)
). By similar reasoning as that of Case 1, we have G* is the graph obtained by attaching all the pendant paths of length 2 at the same vertex of G¯, where G¯ is one of graphs G¯21,…,G¯24 (see Figure 4).

From Lemma 1, it is easy to prove that G¯≠G¯22 and all the pendant paths of length 2 are attached at the vertex of degree 3 of G¯21 or of degree 4 of G¯i (i=23,24). Thus, G* is isomorphic to one of graphs G4, G5 and G6 (see Figure 3).

So, G* is isomorphic to one of graphs G1,…,G6. From Lemma 15, we know q(G1)>q(Gi), (i=2,3,…,6). Thus, G*=G1.