In this section, we consider LEL of line graph, subdivision graph, paraline graph, and total graph of a regular graph. We begin with the case of line graphs.
Theorem 3.1
(see [4])
Let
G
be a regular graph of degree
r
with
n
vertices. Then
$$\mathit{LEL}\bigl(\mathcal{L}(G)\bigr)=\mathit{LEL}(G)+\frac{n(r2)}{2} \sqrt{2r}. $$
By Theorem 3.1 and (5), we have the following corollary directly.
Corollary 3.2
Let
G
be a regular graph of degree
r
with
n
vertices. Then
$$\frac{nr}{\sqrt{r+1}}+\frac{n(r2)}{2}\sqrt{2r} \leq\mathit{LEL}\bigl( \mathcal{L}(G)\bigr) \leq\sqrt{r+1}+\sqrt{(n2) (nrr1)}+\frac{n(r2)}{2} \sqrt{2r}, $$
with both equalities holding if and only if
\(G\cong K_{n}\).
Remark 1
In [17], Wang and Luo proved that if G is a regular graph of n vertices and of degree r, then
$$ \frac{n(r2)}{2}\sqrt{2r} < \mathit{LEL}\bigl(\mathcal{L}(G)\bigr) \leq \sqrt{(n1)nr}+\frac{n(r2)}{2}\sqrt{2r}, $$
(6)
with the right equality if and only if \(G\cong K_{n}\). Evidently, the lower bound in Corollary 3.2 is better than that in (6). For the upper bound, our bound is also better than that in (6). To see this, we only need to show that
$$ \sqrt{r+1}+\sqrt{(n2) (nrr1)}\leq\sqrt{(n1)nr}, $$
(7)
that is,
$$(n2) (nrr1)\leq \bigl(\sqrt{(n1)nr}\sqrt{r+1} \bigr)^{2}, $$
that is,
$$2\sqrt{(n1) (r+1)nr}\leq nr+(n1) (r+1)= (\sqrt{nr} )^{2}+ \bigl( \sqrt{(n1) (r+1)} \bigr)^{2}, $$
which is clearly obeyed.
We now consider the case of subdivision graphs.
Theorem 3.3
Let
G
be a regular graph of degree
r
with
n
vertices. Then

(i)
\(\mathit{LEL}(\mathcal{S}(G))\leq(n1)\sqrt{r+2+\frac{2}{n1}\mathit {LEL}(G)} +\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2}\), with the equality if and only if
\(G\cong K_{n}\);

(ii)
\(\mathit{LEL}(\mathcal{S}(G))>(n1)\sqrt{r+\frac{3}{2}+\frac {2}{n1}\mathit{LEL}(G)} +\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2}\).
Proof
Suppose that \(\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{n1}\geq\mu_{n}=0\) are the Laplacian eigenvalues of G. Then from (1) and Lemma 2.2, it follows that
$$\begin{aligned} \mathit{LEL}\bigl(\mathcal{S}(G)\bigr) =&\sum_{i=1}^{n1} \biggl(\sqrt{\frac {r+2+\sqrt {(r+2)^{2}4\mu_{i}}}{2}} +\sqrt{\frac{r+2\sqrt{(r+2)^{2}4\mu_{i}}}{2}} \biggr) \\ &{}+\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\ =&\sum_{i=1}^{n1}\sqrt{ \biggl(\sqrt{ \frac{r+2+\sqrt{(r+2)^{2}4\mu_{i}}}{2}} +\sqrt{\frac{r+2\sqrt{(r+2)^{2}4\mu_{i}}}{2}} \biggr)^{2}} \\ &{}+\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\ =&\sum_{i=1}^{n1}\sqrt{r+2+2\sqrt{ \mu_{i}}}+\sqrt{r+2}+\frac{\sqrt {2}n(r2)}{2}. \end{aligned}$$
(8)
Further, by the CauchySchwarz inequality, we obtain
$$\begin{aligned} \mathit{LEL}\bigl(\mathcal{S}(G)\bigr) \leq&\sqrt{(n1)\sum _{i=1}^{n1}(r+2+2\sqrt {\mu _{i}})}+ \sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\ =&(n1)\sqrt{r+2+\frac{2}{n1}\mathit{LEL}(G)}+\sqrt{r+2}+ \frac{\sqrt {2}n(r2)}{2}. \end{aligned}$$
Moreover, it is easy to see that the equality holds if and only if \(\mu_{1}=\mu_{2}=\cdots=\mu_{n1}\), which, by Lemma 2.7, is equivalent to \(G\cong K_{n}\). Hence, (i) follows.
We now prove (ii). Take \(a_{i}=\sqrt{r+2+2\sqrt{\mu_{i}}}\) and \(b_{i}=1\), \(i=1,2,\ldots,n1\). Set \(A=\sqrt{r+2+2\sqrt{2r}}\), \(a=\sqrt{r+2}\), and \(B=b=1\). Clearly, \(0< a\leq a_{i}\leq A\) (as \(0\leq\mu_{n1}\leq\cdots\leq\mu _{1}\leq2r\)), and \(0< b\leq b_{i}\leq B\). Furthermore, we get
$$(ABab)^{2}=\frac{8r}{ (\sqrt{r+2+2\sqrt{2r}}+\sqrt{r+2} )^{2}} < \frac{8r}{4(r+2)}< 2. $$
Then by Lemma 2.5, it follows from (8) that
$$\begin{aligned} \mathit{LEL}\bigl(\mathcal{S}(G)\bigr) \geq& \sqrt{(n1)\sum _{i=1}^{n1}(r+2+2\sqrt{\mu_{i}}) \frac {1}{4}(n1)^{2}(ABab)^{2}} \\ &{}+\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\ >&(n1)\sqrt{r+\frac{3}{2}+\frac{2}{n1}\mathit{LEL}(G)}+\sqrt {r+2}+\frac{\sqrt {2}n(r2)}{2}, \end{aligned}$$
as desired, completing the proof. □
Theorem 3.3, together with (5), would yield the next immediate corollary.
Corollary 3.4
Let
G
be a regular graph of degree
r
with
n
vertices. Then

(i)
\(\mathit{LEL}(\mathcal{S}(G))\leq(n1)\sqrt{r+2+\frac{2 (\sqrt {r+1}+\sqrt{(n2)(nrr1)} )}{n1}} +\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2}\), with the equality if and only if
\(G\cong K_{n}\);

(ii)
\(\mathit{LEL}(\mathcal{S}(G))>(n1)\sqrt{r+\frac{3}{2}+\frac {2nr}{(n1)\sqrt{r+1}}} +\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2}\).
Remark 2
It was showed in [17] that if G is a regular graph of n vertices and of degree r, then
$$ \frac{\sqrt{2}n(r2)}{2}+n\sqrt{r+2} < \mathit{LEL}\bigl(\mathcal{S}(G)\bigr) \leq(n1) \sqrt{r}+\sqrt{r+2}+\frac{\sqrt{2}(nr2)}{2}, $$
(9)
the right equality holds if and only if \(G\cong K_{2}\). Note that the lower and upper bounds in Corollary 3.2 are better than those in (9), respectively. Indeed, for the lower bound, since \(\sqrt{\frac{r}{r+1}}\geq\frac{\sqrt{2}}{2}\) and \(n>n1\), we have
$$\begin{aligned}& \sqrt{r+\frac{3}{2}+\frac{2nr}{(n1)\sqrt{r+1}}} +\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\& \quad>(n1)\sqrt{r+\frac{3}{2}+\sqrt{2r}}+\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} >n \sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2}. \end{aligned}$$
For the upper bound, using (7) and the fact that \(\frac {n}{n1}\leq2\), we obtain
$$\begin{aligned}& (n1)\sqrt{r+2+\frac{2 (\sqrt{r+1}+\sqrt{(n2)(nrr1)} )}{n1}} +\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\& \quad\leq(n1)\sqrt{r+2+\frac{2\sqrt{(n1)nr}}{n1}} +\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\& \quad\leq(n1)\sqrt{r+2+2\sqrt{2r}}+\sqrt{r+2}+\frac{\sqrt{2}n(r2)}{2} \\& \quad=(n1)\sqrt{r}+\sqrt{r+2}+\frac{\sqrt{2}(nr2)}{2}. \end{aligned}$$
Next, we turn our attention to the case of paraline graphs. Notice that if G is rregular, then \(\mathcal{C}(G)\) is also rregular, and hence it follows from Lemma 2.3 that the Laplacian eigenvalues of \(\mathcal{C}(G)\) are \(r+2\) (with multiplicity \(n(r2)/2\)), r (with multiplicity \(n(r2)/2\)), and
$$\frac{r+2\pm\sqrt{(r+2)^{2}4\mu_{i}}}{2}\quad (i=1,2,\ldots,n), $$
where \(\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{n}\) are the Laplacian eigenvalues of G. Using the same argument as the proof of Theorem 3.3, we may arrive at the following result.
Theorem 3.5
Let
G
be a regular graph of degree
r
with
n
vertices. Then

(i)
\(\mathit{LEL}(\mathcal{C}(G))\leq(n1)\sqrt{r+2+\frac{2}{n1}\mathit {LEL}(G)} +(\frac{n(r2)}{2}+1)\sqrt{r+2}+\frac{n(r2)}{2}\sqrt{r}\), with the equality if and only if
\(G\cong K_{n}\);

(ii)
\(\mathit{LEL}(\mathcal{C}(G))>(n1)\sqrt{r+\frac{3}{2}+\frac {2}{n1}\mathit{LEL}(G)} +(\frac{n(r2)}{2}+1)\sqrt{r+2}+\frac{n(r2)}{2}\sqrt{r}\).
Likewise, the next corollary follows evidently from Theorem 3.5 and (5).
Corollary 3.6
Let
G
be a regular graph of degree
r
with
n
vertices. Then

(i)
\(\mathit{LEL}(\mathcal{C}(G))\leq(n1)\sqrt{r+2+\frac{2 (\sqrt {r+1}+\sqrt{(n2)(nrr1)} )}{n1}} +(\frac{n(r2)}{2}+1)\sqrt{r+2}+\frac{n(r2)}{2}\sqrt{r}\), with the equality if and only if
\(G\cong K_{n}\);

(ii)
\(\mathit{LEL}(\mathcal{C}(G))>\sqrt{r+\frac{3}{2}+\frac {2nr}{(n1)\sqrt{r+1}}} +(\frac{n(r2)}{2}+1)\sqrt{r+2}+\frac{n(r2)}{2}\sqrt{r}\).
We finally consider the case of total graphs.
Theorem 3.7
Let
G
be a regular graph of degree
r
with
n
vertices. Then

(i)
\(\mathit{LEL}(\mathcal{T}(G))\leq n\sqrt{r+2}+\sqrt{2(n1)nr}+\frac {n(r2)}{2}\sqrt{2r+2}\), with the equality if and only if
\(G\cong K_{n}\);

(ii)
\(\mathit{LEL}(\mathcal{T}(G))>(n1)\sqrt{4r+2}+\sqrt{r+2}+\frac {n(r2)}{2}\sqrt{2r+2}\).
Proof
Note first that \(\mathcal{T}(G)\) is a regular graph of degree 2r with \(\frac{n(r+2)}{2}\) vertices. Then from Lemma 2.4, it follows that \(\mathcal{T}(G)\) has \(\frac{n(r2)}{2}\) Laplacian eigenvalues equal to \(2r+2\) and the following 2n Laplacian eigenvalues:
$$\frac{r+2+2\mu_{i}\pm\sqrt{(r+2)^{2}4\mu_{i}}}{2}\quad (i=1,2,\ldots,n), $$
where \(\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{n}=0\) are the Laplacian eigenvalues of G. Then by (1) and some computation, we get
$$ \mathit{LEL}\bigl(\mathcal{T}(G)\bigr)=\sum_{i=1}^{n1} \sqrt{r+2+2\mu_{i}+2\sqrt{\mu _{i}^{2}+(r+3) \mu_{i}}}+\sqrt{r+2}+\frac{n(r2)}{2}\sqrt{2r+2}. $$
(10)
On the other hand, by using the CauchySchwarz inequality and the fact that \(\sum_{i=1}^{n1}\mu_{i}=nr\) and \(\sum_{i=1}^{n1}\mu_{i}^{2}=nr^{2}+nr\), we have
$$\begin{aligned}& \sum_{i=1}^{n1}\sqrt{r+2+2 \mu_{i}+2\sqrt{\mu _{i}^{2}+(r+3) \mu_{i}}} \\& \quad\leq\sqrt{(n1)\sum_{i=1}^{n1} \bigl(r+2+2\mu_{i}+2\sqrt{\mu _{i}^{2}+(r+3) \mu_{i}} \bigr)} \\& \quad\leq(n1)\sqrt{r+2+\frac{2nr}{n1}+2\sqrt{\frac{1}{n1}\sum _{i=1}^{n1} \bigl(\mu_{i}^{2}+(r+3) \mu_{i} \bigr)}} \\& \quad=(n1)\sqrt{r+2+\frac{2nr}{n1}+2\sqrt{\frac{2nr}{n1}(r+2)}} \\& \quad=(n1)\sqrt{r+2}+\sqrt{2(n1)nr}, \end{aligned}$$
which, together with (10), yields the required upper bound. For the sharpness of this bound, it is not difficult to check that if \(G\cong K_{n}\) then the equality holds. Conversely, we assume that the equality holds. Then the above two inequalities should be equalities. Thus, it follows from the second inequality that, for any \(1\leq i,j\leq n1\) and \(i\neq j\), \(\sqrt{\mu_{i}^{2}+(r+3)\mu_{i}}=\sqrt{\mu_{j}^{2}+(r+3)\mu_{j}}\), that is, \((\mu_{i}\mu_{j})(\mu_{i}+\mu_{j}+r+3)=0\), which implies that \(\mu_{i}=\mu_{j}\). In other words, we have \(\mu_{1}=\mu_{2}=\cdots=\mu_{n1}\). Note that in this case, the first inequality is also an equality. Eventually, by Lemma 2.7, we get \(G\cong K_{n}\).
We next prove (ii). Since \(\mu_{i}\geq0\), \(i=1,2,\ldots,n1\), by (10) we get
$$ \mathit{LEL}\bigl(\mathcal{T}(G)\bigr) \geq\sum_{i=1}^{n1} \sqrt{r+2+2\mu_{i}+2\sqrt{(r+3)\mu_{i}}}+\sqrt {r+2}+ \frac{n(r2)}{2}\sqrt{2r+2}. $$
(11)
We now take \(a_{i}=\sqrt{r+2+2\mu_{i}+2\sqrt{(r+3)\mu_{i}}}\) and \(b_{i}=1\), \(i=1,2,\ldots,n1\). Set \(A=\sqrt{5r+2+2\sqrt{2(r+3)r}}\), \(a=\sqrt{r+2}\), and \(B=b=1\). It is easy to see that \(0< a\leq a_{i}\leq A\), and \(0< b\leq b_{i}\leq B\). Furthermore, if \(r\geq3\), then \(A=\sqrt{5r+2+2\sqrt{2(r+3)r}}\leq\sqrt {9r+2}\) and hence,
$$\begin{aligned} (ABab)^{2} \leq& (\sqrt{9r+2}\sqrt{r+2} )^{2} = \frac{(8r)^{2}}{10r+4+2\sqrt{(9r+2)(r+2)}} \\ < &\frac{64r^{2}}{16r}=4r; \end{aligned}$$
if \(r=2\), then by a direct calculation, we also obtain \((ABab)^{2}<4r\). Thus by Lemma 2.5 and (5), we get
$$\begin{aligned}& \sum_{i=1}^{n1}\sqrt{r+2+2 \mu_{i}+2\sqrt{(r+3)\mu_{i}}} \\ & \quad\geq\sqrt{(n1)\sum_{i=1}^{n1} \bigl(r+2+2\mu_{i}+2\sqrt{(r+3)\mu _{i}} \bigr) \frac{1}{4}(n1)^{2}(ABab)^{2}} \\ & \quad>(n1)\sqrt{\frac{2nr}{n1}+\frac{2\sqrt{r+3}}{n1}\mathit {LEL}(G)+2} \\ & \quad>(n1)\sqrt{ \biggl(\sqrt{\frac{r+3}{r+1}}+1 \biggr)\frac {2nr}{n1}+2} \\ & \quad>(n1) \sqrt{4r+2}, \end{aligned}$$
which, together with (11), would yield the required lower bound.
This completes the proof. □
Remark 3
It was proved in [17] that if G is a regular graph of n vertices and of degree r, then
$$\begin{aligned}& \mathit{LEL}\bigl(\mathcal{T}(G)\bigr) > \frac{n(r2)}{2}\sqrt{2r+2}+n\sqrt {r+2}, \quad\mbox{and} \end{aligned}$$
(12)
$$\begin{aligned}& \mathit{LEL}\bigl(\mathcal{T}(G)\bigr) \leq (n1)\sqrt{3r}+\frac{(nr2)}{2} \sqrt {2r+2}+\sqrt{r+2}, \end{aligned}$$
(13)
with the equality if and only if \(G\cong K_{2}\). Obviously, the lower bound in Theorem 3.7 is better than that in (12). For the upper bound, it is easy to check that, for \(n\geq2\) and \(r>1\),
$$\sqrt{r+2}+\sqrt{\frac{2nr}{n1}}\leq\sqrt{r+2}+2\sqrt{r}\leq\sqrt {3r}+ \sqrt{2r+2}, $$
implying that
$$(n1)\sqrt{r+2}+\sqrt{2(n1)nr}\leq(n1) (\sqrt{3r}+\sqrt {2r+2} ), $$
from which we would find that the upper bound in Theorem 3.7 is better than that in (13).