In this section, we determine the first five maximum spectral radii of graphs in \(\mathbb{U}(n)\).
Let \(U_{n,m}\) be the unicyclic graph obtained from the cycle \(C_{m}=v_{1}v_{2}\cdots v_{m}v_{1}\) by attaching \(n-2m+1\) pendent vertices to \(v_{1}\) and a pendent vertex to each of the other vertices on \(C_{m}\). Obviously, \(U_{n,m}\in\mathbb{U}(n)\).
Lemma 7
Let
\(G\in\mathbb{U}(n)\)
with unique cycle of length
\(m \ge3\), where
\(n \ge2m\). Then
\(\lambda_{1}(G)\le\lambda_{1}(U_{n,m})\)
with equality if and only if
\(G \cong U_{n,m}\).
Proof
Let G be a graph with maximum spectral radius satisfying the given condition. Let \(C_{m}=v_{1}v_{2}\cdots v_{m}v_{1}\) be the unique cycle of G. By Lemma 2, the vertices of G outside \(C_{m}\) are all pendent vertices.
Suppose that \(G\ncong U_{n,m}\). Then we may choose two vertices, say \(v_{i}\) and \(v_{j}\), on \(C_{m}\) such that \(d_{G}(v_{i}),d_{G}(v_{j})\ge4\), where \(1\le i < j\le m\). Let x be the principal eigenvector of G. Suppose without loss of generality that \(x_{v_{i}}\ge x_{v_{j}}\). Consider \(G_{1}=G-vv_{j}+vv_{i}\), where v is a neighbor of \(v_{j}\) outside \(C_{m}\). Obviously, \(G_{1}\in\mathbb{U}(n)\) and its unique cycle is still of length m. By Lemma 1, \(\lambda_{1}(G)<\lambda_{1}(G_{1})\), which is a contradiction. Thus there is at most one vertex on \(C_{m}\) with degree at least four in G, i.e., \(G \cong U_{n,m}\). □
Lemma 8
Let
\(G\in\mathbb{U}(n)\)
with unique cycle of length
\(m \ge4\), where
\(n\ge2m\). Then
\(\lambda_{1}(G)\le\lambda_{1}(U_{n,4})\)
with equality if and only if
\(G \cong U_{n,4}\).
Proof
If \(m\ge5\), then applying Lemma 2 to \(G=U_{n,m}\) by setting uv to be an edge on the cycle incident to the vertex of maximum degree, we have \(\lambda_{1}(U_{n,m})<\lambda_{1}(U_{n,m-1})\), and thus \(\lambda_{1}(U_{n,m})<\lambda_{1}(U_{n,4})\). If \(m=4\), then by Lemma 7, \(\lambda_{1}(G)\le\lambda_{1}(U_{n,4})\) with equality if and only if \(G \cong U_{n,4}\). □
Let \(C_{3}(T_{1},T_{2},T_{3})\) be the n-vertex unicyclic graph with the triangle \(v_{1}v_{2}v_{3}v_{1}\) such that the deletion of the three edges on the triangle would result in three vertex-disjoint trees \(T_{1}\), \(T_{2}\), \(T_{3}\), where \(v_{i}\in V(T_{i})\) for \(i=1,2,3\).
Let \(S_{n}(a,b,c)=C_{3}(T_{1},T_{2},T_{3})\), where \(|V(T_{1})|=a-1\), \(|V(T_{2})|=b-1\), \(|V(T_{3})|=c-1\), \(a+b+c=n+3\), \(a\ge b\ge c\ge2\), and \(T_{1}\), \(T_{2}\), \(T_{3}\) are all stars with centers \(v_{1}\), \(v_{2}\), \(v_{3}\), respectively. Obviously, \(S_{n}(n-3,3,3) \cong U_{n,3}\).
Lemma 9
Among the graphs
\(S_{n}(a,b,c)\)
with
\(a+b+c=n+3\)
and
\(a\ge b\ge c\ge 3\), \(S_{n}(n-3,3,3) \cong U_{n,3}\)
for
\(n\ge6\), \(S_{n}(n-4,4,3)\)
for
\(n\ge 8\), and
\(S_{n}(n-5,5,3)\)
for
\(n\ge10\)
are, respectively, the unique graphs with the first, the second, and the third maximum spectral radii.
Proof
Let x be the principal eigenvector of \(S_{n}(a,b,c)\).
Suppose that \(a>b\). If \(x_{v_{1}}< x_{v_{2}}\), then by Lemma 1,
$$\lambda_{1}\bigl(S_{n}(a,b,c)\bigr)< \lambda_{1} \bigl(S_{n}(a-1,b+1,c)\bigr)< \cdots< \lambda_{1} \bigl(S_{n}(b,a,c)\bigr), $$
which is a contradiction. If \(x_{v_{1}}\ge x_{v_{2}}\), then by Lemma 1,
$$\lambda_{1}\bigl(S_{n}(a,b,c)\bigr)< \lambda_{1} \bigl(S_{n}(a+1,b-1,c)\bigr). $$
If \(a=b\), then whether \(x_{v_{1}}\ge x_{v_{2}}\) or \(x_{v_{1}}< x_{v_{2}}\), we have by Lemma 1 that
$$\lambda_{1}\bigl(S_{n}(a,b,c)\bigr)< \lambda_{1} \bigl(S_{n}(a+1,b-1,c)\bigr). $$
Then
$$\lambda_{1}\bigl(S_{n}(a,b,c)\bigr)< \lambda_{1} \bigl(S_{n}(a+1,b-1,c)\bigr) $$
for \(b\ge4\). This implies that among the graphs \(S_{n}(a,b,3)\) with \(a+b+3=n+3\) and \(a\ge b\ge3\), \(S_{n}(n-3,3,3) \cong U_{n,3}\) for \(n\ge6\), \(S_{n}(n-4,4,3)\) for \(n\ge8\), and \(S_{n}(n-5,5,3)\) for \(n\ge10\) are, respectively, the unique graphs with the first, the second, and the third maximum spectral radius.
If \(c\ge4\), then by similar arguments as above,
$$\lambda_{1}\bigl(S_{n}(a,b,c)\bigr)\le \lambda_{1}\bigl(S_{n}(n-5,4,4)\bigr)< \lambda_{1} \bigl(S_{n}(n-5,5,3)\bigr). $$
Now the result follows. □
Let \(U_{n}^{1}\) be the unicyclic graph obtained by attaching a path on two vertices to the vertex of degree \(n-5\) of \(S_{n-2}(n-5,3,3)\), where \(n\ge8\). Let \(U_{n}^{2}\) be the unicyclic graph obtained by attaching a path on two vertices to the vertex of degree two of \(S_{n-2}(n-4,3,2)\), where \(n\ge7\). Let \(U_{n}^{3}\) be the unicyclic graph obtained by attaching a path on two vertices to the vertex of degree \(n-6\) of \(S_{n-2}(n-6,4,3)\), where \(n\ge10\).
For \(u,v\in V(G)\), let \(d_{G}(u,v)\) be the distance from u to v in G.
Let \(P_{n}\) be the path on n vertices. Let \(G \cup H\) be the vertex-disjoint union of graphs G and H.
Lemma 10
Let
\(G=C_{3}(T_{1},T_{2},T_{3})\)
with
\(|V(T_{1})|=a-1\), \(|V(T_{2})|=b-1\), \(|V(T_{3})|=2\), \(a+b=n\), \(a\ge b\), \(b=3, 4\), and
\(n\ge 10\). Suppose that
\(G\ncong S_{n}(a,b,c), U_{n}^{1}, U_{n}^{2}\). Then
\(\lambda_{1}(G)<\lambda_{1}(S_{n}(n-5,5,3))\).
Proof
First suppose that \(b=3\). Let \(r=\max\{d_{G}(u,v_{1}):u\in V(T_{1})\}\). Obviously, \(r\ge2\) as \(G\ncong S_{n}(a,b,c)\).
If \(r\ge3\), then by Lemma 2, we may get a unicyclic graph with \(r=2\) with a larger spectral radius.
Suppose that \(r=2\).
If there are at least two non-pendent neighbors of \(v_{1}\) outside the triangle in G, then by Lemma 2, we may get a unicyclic graph with exactly one non-pendent neighbor of \(v_{1}\) outside the triangle with a larger spectral radius.
Suppose that there is exactly one non-pendent neighbor, say w, of \(v_{1}\) outside the triangle of G. Let \(d_{G}(v_{1})=s\) and \(d_{G}(w)=t\). Obviously, \(s\ge3\), and \(t\ge3\) as \(G\ncong U_{n}^{1}\). If \(t=3\), then G is the graph obtained by attaching a pendent vertex to the vertex of degree two in \(U_{n-1}^{1}\), which is denoted by \(L_{n}\). If \(t=n-5\), then G is the graph obtained by attaching \(n-6\) pendent vertices to a pendent vertex of \(S_{6}(3,3,3)\), which is denoted by \(H_{n}\).
Suppose that \(4\le t\le n-6\). Let x be the principal eigenvector of G. If \(x_{w}\ge x_{v_{1}}\), then by Lemma 1, \(\lambda_{1}(G)<\lambda_{1}(H_{n})\). If \(x_{w}< x_{v_{1}}\), then by Lemma 1, \(\lambda_{1}(G)<\lambda_{1}(L_{n})\). Thus
$$\lambda_{1}(G)< \max\bigl\{ \lambda_{1}(L_{n}), \lambda_{1}(H_{n})\bigr\} . $$
Now we compare \(\lambda_{1}(L_{n})\) with \(\lambda_{1}(H_{n})\). Let \(r_{n}(x)=\phi(L_{n},x)-\phi(H_{n},x)\). Applying Lemma 3 to \(G=L_{n}\) by setting v to be a pendent neighbor of \(v_{1}\),
$$\phi(L_{n},x)=x\cdot\phi(L_{n-1},x)-x^{n-9} \phi(P_{3}\cup P_{4},x), $$
and to \(G=H_{n}\) by setting v to be a pendent vertex whose unique neighbor is of degree \(n-5\),
$$\phi(H_{n},x)=x\cdot\phi(H_{n-1},x)-x^{n-7}\phi \bigl(S_{5}(3,3,2),x\bigr), $$
which implies that
$$\begin{aligned} r_{n}(x) =&x\cdot r_{n-1}(x)+x^{n-7}\phi \bigl(S_{5}(3,3,2),x\bigr)-x^{n-9}\phi (P_{3}\cup P_{4},x) \\ =&x\cdot r_{n-1}(x)+x^{n-7}\bigl(x^{5}-5x^{3}-2x^{2}+3x \bigr) \\ &{}-x^{n-9}\bigl(x^{3}-2x\bigr) \bigl(x^{4}-3x^{2}+1 \bigr) \\ =&x\cdot r_{n-1}(x)+2x^{n-8}\bigl(1-2x^{2}-x^{3} \bigr). \end{aligned}$$
Note that \(r_{8}(x)=0\) as \(L_{8}=H_{8}\), and it is easily seen that \(1-2x^{2}-x^{3}<0\) for \(x>1\), and thus \(r_{n}(x)<0\) (i.e., \(\phi(L_{n},x)<\phi(H_{n},x)\)) for \(x>1\), which implies that \(\lambda_{1}(L_{n})>\lambda_{1}(H_{n})\). Then
$$\lambda_{1}(G)\le \lambda_{1}(L_{n}) $$
if \(b=3\).
Suppose that \(b=4\). First suppose that \(d_{G}(v_{2})=3\). Note that \(G\ncong U_{n}^{2}\), then by similar arguments as above, \(\lambda_{1}(G)\le \lambda_{1}(H)\), where H is the graph obtained by attaching a path on two vertices to the vertex of degree \(n-6\) in \(U_{n-2}^{2}\). By Lemma 2, \(\lambda_{1}(H)<\lambda_{1}(U_{n}^{3})\). If \(d_{G}(v_{2})=4\), then by similar arguments as above, \(\lambda_{1}(G)\le \lambda_{1}(U_{n}^{3})\). Thus
$$\lambda_{1}(G)\le\lambda_{1}\bigl(U_{n}^{3} \bigr) $$
if \(b=4\).
Now we have shown that
$$\lambda_{1}(G)\le \max\bigl\{ \lambda_{1}(L_{n}), \lambda_{1}\bigl(U_{n}^{3}\bigr)\bigr\} $$
for \(b=3\mbox{ or }4\).
To determine \(\max\{\lambda_{1}(L_{n}),\lambda_{1}(U_{n}^{3})\}\), we compare \(\lambda_{1}(U_{n}^{3})\) with \(\lambda_{1}(L_{n})\). Let \(T^{(1)}\) be the \((n-3)\)-vertex tree obtained by attaching \(n-8\) pendent vertices to the center of the path on five vertices. Applying Lemma 3 to \(G=U_{n}^{3}\) by setting v to be a pendent neighbor of \(v_{2}\),
$$\phi\bigl(U_{n}^{3},x\bigr)=x\cdot\phi\bigl(U_{n-1}^{1},x \bigr)-x\cdot\phi\bigl(T^{(1)},x\bigr), $$
and to \(G=L_{n}\) by setting v to be a pendent vertex in \(T_{1}\) of degree three,
$$\phi(L_{n},x)=x\cdot\phi\bigl(U_{n-1}^{1},x \bigr)-x\cdot\phi\bigl(S_{n-3}(n-6,3,3),x\bigr). $$
Note that \(T^{(1)}\) is a proper spanning subgraph of \(S_{n-3}(n-6,3,3)\), and thus by Lemma 4, \(\phi(S_{n-3}(n-6,3,3),x)<\phi(T^{(1)},x)\) for \(x\ge \lambda_{1}(S_{n-3}(n-6,3,3))\), which implies that \(\phi(U_{n}^{3},x)<\phi(L_{n},x)\) for \(x\ge\lambda_{1}(S_{n-3}(n-6,3,3))\), i.e., \(\lambda_{1}(U_{n}^{3})>\lambda_{1}(L_{n})\). Thus
$$\lambda_{1}(G)\le\lambda_{1}\bigl(U_{n}^{3} \bigr) $$
for \(b=3\mbox{ or }4\).
Now we are left to compare \(\lambda_{1}(U_{n}^{3})\) with \(\lambda_{1}(S_{n}(n-5,5,3))\). Let \(T^{(2)}\) be the \((n-4)\)-vertex tree obtained by attaching \(n-7\) pendent vertices to an end vertex of the path on three vertices. Applying Lemma 3 to \(G=U_{n}^{3}\) by setting v to be a pendent vertex whose unique neighbor is of degree two,
$$\phi\bigl(U_{n}^{3},x\bigr)=x\cdot\phi\bigl(S_{n-1}(n-5,4,3),x \bigr)-\phi\bigl(S_{n-2}(n-6,4,3),x\bigr), $$
and to \(G=S_{n}(n-5,5,3)\) by setting v to be a pendent neighbor of \(v_{2}\),
$$\phi\bigl(S_{n}(n-5,5,3),x\bigr)=x\cdot\phi\bigl(S_{n-1}(n-5,4,3),x \bigr)-x^{2}\cdot\phi\bigl(T^{(2)},x\bigr). $$
Note that \(P_{1}\cup P_{1}\cup T^{(2)}\) is a proper spanning subgraph of \(S_{n-2}(n-6,4,3)\), and thus by Lemma 4, \(\phi(S_{n-2}(n-6,4,3),x)< x^{2}\cdot\phi(T^{(2)},x)\) for \(x\ge \lambda_{1}(S_{n-2}(n-6,4,3))\), which implies that \(\phi(U_{n}^{3},x)>\phi(S_{n}(n-5,5,3),x)\) for \(x\ge \lambda_{1}(S_{n-2}(n-6,4,3))\), i.e., \(\lambda_{1}(U_{n}^{3})<\lambda_{1}(S_{n}(n-5,5,3))\).
Then the result follows. □
Lemma 11
For
\(n\ge8\), \(\lambda_{1}(U_{n}^{2})<\lambda_{1}(U_{n}^{1})<\lambda_{1}(S_{n}(n-4,4,3))\), for
\(n=10\), \(\lambda_{1}(U_{n}^{2})<\lambda_{1}(U_{n}^{1})<\lambda_{1}(S_{n}(n-5,5,3))\), for
\(n=11\), \(\lambda_{1}(U_{n}^{2})<\lambda_{1}(S_{n}(n-5,5,3))<\lambda_{1}(U_{n}^{1})\), and for
\(n\ge12\), \(\lambda_{1}(S_{n}(n-5,5,3))<\lambda_{1}(U_{n}^{2})<\lambda_{1}(U_{n}^{1})\).
Proof
Let T be the \((n-3)\)-vertex tree obtained by attaching \(n-6\) pendent vertices to an end vertex of the path on three vertices.
Applying Lemma 3 to \(G=U_{n}^{1}\) by setting v to be a pendent vertex whose unique neighbor is of degree two,
$$\phi\bigl(U_{n}^{1},x\bigr)=x\cdot \phi \bigl(S_{n-1}(n-4,3,3),x\bigr)-\phi\bigl(S_{n-2}(n-5,3,3),x \bigr), $$
and to \(G=S_{n}(n-4,4,3)\) by setting v to be a pendent neighbor of \(v_{2}\),
$$\phi\bigl(S_{n}(n-4,4,3),x\bigr)=x\cdot\phi\bigl(S_{n-1}(n-4,3,3),x \bigr)-x\cdot\phi(T,x). $$
Note that \(P_{1}\cup T\) is a proper spanning graph of \(S_{n-2}(n-5,3,3)\), and thus by Lemma 4,
$$x\cdot\phi(T,x)>\phi\bigl(S_{n-2}(n-5,3,3),x\bigr) $$
for \(x\ge \lambda_{1}(S_{n-2}(n-5,3,3))\), which implies that
$$\phi\bigl(U_{n}^{1},x\bigr)>\phi\bigl(S_{n}(n-4,4,3),x \bigr) $$
for \(x\ge \lambda_{1}(S_{n-2}(n-5,3,3))\), i.e., \(\lambda_{1}(U_{n}^{1})<\lambda_{1}(S_{n}(n-4,4,3))\).
By Lemma 3, we have
$$\begin{aligned}& \phi\bigl(U_{n}^{1},x\bigr)=x^{n-8} \bigl[x^{8}-nx^{6}-2x^{5}+(4n-17)x^{4}+2x^{3}-(4n-24)x^{2}+n-7 \bigr], \\& \phi\bigl(U_{n}^{2},x\bigr)=x^{n-6} \bigl[x^{6}-nx^{4}-2x^{3}+(4n-17)x^{2}+2x-3n+17 \bigr], \\& \phi\bigl(S_{n}(n-5,5,3),x\bigr)=x^{n-6} \bigl[x^{6}-nx^{4}-2x^{3}+(5n-28)x^{2}-3n+21 \bigr], \end{aligned}$$
and thus
$$\begin{aligned}& \phi\bigl(U_{n}^{1},x\bigr)-\phi\bigl(U_{n}^{2},x \bigr)=-x^{n-8}(n-7) (x+1) (x-1), \\& \phi\bigl(U_{n}^{2},x\bigr)-\phi\bigl(S_{n}(n-5,5,3),x \bigr)=x^{n-6}(n-7)f(x), \end{aligned}$$
where
$$f(x)=(11-n)x^{2}+2x-4. $$
Obviously, \(\phi(U_{n}^{1},x)<\phi(U_{n}^{2},x)\) for \(x>1\), i.e., \(\lambda_{1}(U_{n}^{1})>\lambda_{1}(U_{n}^{2})\).
Note that \(\lambda_{1}(G)>2\) if G is a unicyclic graph different from a cycle. If \(n=10\), then
for \(x>2\). If \(n=11\), then
for \(x>2\). If \(n\ge12\), then
$$\begin{aligned} f(x) =&(11-n)x^{2}+2x-4 \\ < &(11-n)2^{2}+2\cdot2-4=4(11-n)< 0. \end{aligned}$$
Thus \(f(x)>0\) for \(x>2\) if \(n=10,11\) and \(f(x)<0\) for \(x>2\) if \(n\ge12\), which implies that \(\lambda_{1}(U_{n}^{2})<\lambda_{1}(S_{n}(n-5,5,3))\) for \(n=10,11\) and \(\lambda_{1}(U_{n}^{2})>\lambda_{1}(S_{n}(n-5,5,3))\) for \(n\ge12\).
By direct calculation, we have \(\lambda_{1}(U_{n}^{1})<\lambda_{1}(S_{n}(n-5,5,3))\) for \(n=10\) and \(\lambda_{1}(U_{n}^{1})>\lambda_{1}(S_{n}(n-5,5,3))\) for \(n=11\).
The result follows easily. □
Now we prove our main result in this section.
Theorem 1
Among the graphs in
\(\mathbb{U}(n)\),
-
(i)
\(U_{n,3} \cong S_{n}(n-3,3,3)\)
for
\(n\ge6\)
and
\(S_{n}(n-4,4,3)\)
for
\(n\ge8\)
are respectively the unique graphs with the first and the second maximum spectral radii;
-
(ii)
\(S_{n}(n-5,5,3)\)
for
\(n=10\)
and
\(U_{n}^{1}\)
for
\(n\ge11\)
are the unique graphs with the third maximum spectral radius;
-
(iii)
\(S_{n}(n-5,4,4)\)
for
\(n=10\), \(S_{n}(n-5,5,3)\)
for
\(n=11\), and
\(U_{n}^{2}\)
for
\(n\ge12\)
are the unique graphs with the fourth maximum spectral radius;
-
(iv)
\(U_{n}^{1}\)
for
\(n=10\), \(U_{n}^{2}\)
for
\(n=11\), and
\(S_{n}(n-5,5,3)\)
for
\(n\ge 12\)
are the unique graphs with the fifth maximum spectral radius,
and the spectral radii of
\(U_{n,3} \cong S_{n}(n-3,3,3)\), \(S_{n}(n-4,4,3)\), \(S_{n}(n-5,5,3)\), \(U_{n}^{1}\)
and
\(U_{n}^{2}\)
are, respectively, the largest roots of the equations on
x
as follows:
$$\begin{aligned}& x^{6}-nx^{4}-2x^{3}+(3n-12)x^{2}-n+5=0, \\& x^{6}-nx^{4}-2x^{3}+(4n-19)x^{2}-2n+12=0, \\& x^{6}-nx^{4}-2x^{3}+(5n-28)x^{2}-3n+21=0, \\& x^{8}-nx^{6}-2x^{5}+(4n-17)x^{4}+2x^{3}-(4n-24)x^{2}+n-7=0, \\& x^{6}-nx^{4}-2x^{3}+(4n-17)x^{2}+2x-3n+17=0. \end{aligned}$$
Proof
Let \(G\in\mathbb{U}(n)\) with unique cycle of length \(m\ge3\), where \(n \ge2m\).
If \(m\ge4\), then by Lemmas 8 and 2,
$$\lambda_{1}(G)\le \lambda_{1}(U_{n,4})< \lambda_{1}\bigl(S_{n}(n-5,5,3)\bigr). $$
Suppose in the following that \(m=3\). Then \(G\cong C_{3}(T_{1},T_{2},T_{3})\), where \(|V(T_{1})|=a-1\), \(|V(T_{2})|=b-1\), \(|V(T_{3})|=c-1\), \(a+b+c=n+3\), \(a\ge b\ge c\ge3\).
If \(G \cong S_{n}(a,b,c)\), then by Lemma 9, \(U_{n,3}\cong S_{n}(n-3,3,3)\), \(S_{n}(n-4,4,3)\) and \(S_{n}(n-5,5,3)\) are, respectively, the unique graphs with the first, the second and the third maximum spectral radii.
Suppose that \(G\ncong S_{n}(a,b,c)\). If \(c\ge4\), then by Lemmas 2 and 9, \(\lambda_{1}(G)<\lambda_{1}(S_{n}(n-5,5,3))\). Suppose that \(c=3\). If \(b=3\) or 4 and \(G\ncong U_{n}^{1},U_{n}^{2}\), then by Lemma 10, \(\lambda_{1}(G)< \lambda_{1}(S_{n}(n-5,5,3))\). If \(b \ge5\), then by Lemmas 2 and 9, \(\lambda_{1}(G)<\lambda_{1}(S_{n}(n-5,5,3))\). Thus
$$\lambda_{1}(G)< \lambda_{1}\bigl(S_{n}(n-5,5,3) \bigr) $$
if \(G\ncong S_{n}(n-3,3,3)\), \(S_{n}(n-4,4,3)\), \(S_{n}(n-5,5,3)\), \(U_{n}^{1}\), \(U_{n}^{2}\) for \(n\ge10\).
It follows from Lemma 11 that
$$\lambda_{1}\bigl(U_{n}^{2}\bigr)< \lambda_{1}\bigl(U_{n}^{1}\bigr)< \lambda_{1}\bigl(S_{n}(n-4,4,3)\bigr) $$
for \(n\ge8\), and thus (i) follows.
Suppose that \(G\ncong S_{n}(n-3,3,3)\), \(S_{n}(n-4,4,3)\).
Suppose that \(n=10\). If \(m\ge4\), then by Lemma 8 and direct calculation,
$$\lambda_{1}(G)\le \lambda_{1}(U_{10,4})< \lambda_{1}\bigl(U_{10}^{1}\bigr). $$
Suppose that \(m=3\). If \(G\cong S_{10}(a,b,c)\), then \(G \cong S_{10}(5,5,3)\) or \(S_{10}(5,4,4)\). Suppose that \(G\ncong S_{10}(a,b,c), U_{10}^{1}\). If \(c \ge4\), then by Lemmas 1 and 2, we may construct another graph still different from \(S_{10}(a,b,c)\) and \(U_{10}^{1}\) with a larger spectral radius. Suppose that \(c = 3\). If \(b = 3\) or 4 and \(G \ncong U_{10}^{2}\), then similar to the proof of Lemma 10 and direct calculation,
$$\lambda_{1}(G)\le \lambda_{1}\bigl(U_{10}^{3} \bigr)< \lambda_{1}\bigl(U_{10}^{2}\bigr)< \lambda_{1}\bigl(U_{10}^{1}\bigr). $$
If \(b \ge5\), then by Lemmas 1 and 2, we may construct another graph still different from \(S_{10}(a,b,c)\) and \(U_{10}^{1}\) with a larger spectral radius. So if \(G \ncong S_{10}(a,b,c), U_{10}^{1}\), then
$$\lambda_{1}(G)< \lambda_{1}\bigl(U_{10}^{1} \bigr). $$
Now by direct calculation,
$$\lambda_{1}\bigl(U_{10}^{1}\bigr)< \lambda_{1}\bigl(S_{10}(5,4,4)\bigr)< \lambda_{1} \bigl(S_{10}(5,5,3)\bigr), $$
and thus the result for \(n=10\) follows.
Suppose that \(n=11\). If \(m\ge4\), then by Lemma 8 and direct calculation,
$$\lambda_{1}(G)\le \lambda_{1}(U_{11,4})< \lambda_{1}\bigl(U_{11}^{2}\bigr). $$
Suppose that \(m=3\). If \(G \cong S_{11}(a,b,c)\), then \(G\cong S_{11}(6,5,3)\), \(S_{11}(6,4,4)\), or \(S_{11}(5,5,4)\). Suppose that \(G \ncong S_{11}(a,b,c)\), \(U_{11}^{1}\), \(U_{11}^{2}\). If \(c \ge4\), then by Lemmas 1 and 2, either \(\lambda_{1}(G)<\lambda_{1}(U_{11}^{2})\), or we may construct another graph still different from \(S_{11}(a,b,c)\), \(U_{11}^{1}\) and \(U_{11}^{2}\) with a larger spectral radius. Suppose that \(c = 3\). If \(b = 3\) or 4, then similar to the proof of Lemma 10 and direct calculation,
$$\lambda_{1}(G)\le \lambda_{1}\bigl(U_{11}^{3} \bigr)< \lambda_{1}\bigl(U_{11}^{2}\bigr)< \lambda_{1}\bigl(U_{11}^{1}\bigr). $$
If \(b \ge5\), then by Lemmas 1 and 2, either \(\lambda_{1}(G)< \lambda_{1}(U_{11}^{2})\), or we may construct another graph still different from \(S_{11}(a,b,c)\), \(U_{11}^{1}\) and \(U_{11}^{2}\) with a larger spectral radius. So if \(G \ncong S_{11}(a,b,c), U_{11}^{1}, U_{11}^{2}\), then
$$\lambda_{1}(G)< \lambda_{1}\bigl(U_{11}^{2} \bigr). $$
Now by direct calculation,
$$\lambda_{1}\bigl(S_{11}(5,5,4)\bigr)< \lambda_{1} \bigl(S_{11}(6,4,4)\bigr)< \lambda _{1}\bigl(U_{11}^{2} \bigr)< \lambda_{1}\bigl(S_{11}(6,5,3)\bigr)< \lambda_{1}\bigl(U_{11}^{1}\bigr), $$
and thus the result for \(n=11\) follows.
If \(n\ge12\), then the result follows from Lemma 11. □