**3. The Lower Bound of the Distance Laplacian Spectral Radius of Graphs among** B*n*,4

If *G* ∈ B*n*,*d*, then there exists a partition {*V*0, *V*1, ... , *Vd*} of *V*(*G*) such that |*V*0| = 1 and *d*(*u*, *v*) = *i* for *u* ∈ *V*<sup>0</sup> and *v* ∈ *Vi* (*i* = 1, 2, . . . , *d*).

**Lemma 6** (Lemma 2.1, [12])**.** *Let G* ∈ B*n*,*<sup>d</sup> with a vertex partition described as above. Then G*[*Vi*] *induces an empty graph (i.e., containing no edge) for each i* ∈ {0, 1, . . . , *d*}*.*

**Lemma 7.** *Let d* ≥ 3 *and G* ∈ B*n*,*d. If d*(*G* + *e*) < *d when any edge e is added to G, then* |*Vd*| = 1 *and the induced subgraph G*[*Vi*−<sup>1</sup> ∪ *Vi*] (*i* = 1, 2, . . . , *d*) *is a complete bipartite graph.*

**Proof.** From Lemma 6, it is clear that *G*[*Vi*−<sup>1</sup> ∪ *Vi*] (*i* = 1, 2, ... , *d*) is a complete bipartite graph. Moreover, let *u* ∈ *Vd* and *v* ∈ *Vd*−3. Assume, on the contrary, that |*Vd*| ≥ 2, then the graph *G* + *euv* ∈ B*n*,*d*, a contradiction.

**Remark 1.** *Denote a subset of* B*n*,*<sup>d</sup> by* B *<sup>n</sup>*,*d, consisting of all the graphs satisfying Lemma 7. For instance, if G* ∈ B *<sup>n</sup>*,4*, then <sup>G</sup> is of the form shown in Figure 1. Then the partition of <sup>V</sup>*(*G*) *can be written as V*<sup>0</sup> = {*w*}*, V*<sup>1</sup> = {*v*1, ... , *vs*}*, V*<sup>2</sup> = {*u*1, ... , *ut*}*, V*<sup>3</sup> = {*y*1, ... , *yk*} *and V*<sup>4</sup> = {*z*}*, where s* + *t* + *k* + 2 = *n and s*, *t*, *k* ≥ 1*.*

**Figure 1.** A graph *G* ∈ B *<sup>n</sup>*,4.

Before giving the main conclusion of this section, we first investigate the properties of the eigenvector corresponding to *ρ*L(*G*) for *G* ∈ B *<sup>n</sup>*,4.

Let *G* ∈ B *<sup>n</sup>*,4 and the partition of *<sup>V</sup>*(*G*) be arranged as in Remark 1. Without loss of generality, suppose |*V*3|≥|*V*1| ≥ 1 (i.e., *k* ≥ *s* ≥ 1).

**Lemma 8.** *Let the eigenvector corresponding to ρ*L(*G*) *be x. Then*

$$\begin{cases} \mathfrak{x}(v\_i) = \mathfrak{x}(v\_j) \ (1 \le i, j \le s), \\ \mathfrak{x}(u\_i) = \mathfrak{x}(u\_j) \ (1 \le i, j \le t), \\ \mathfrak{x}(y\_i) = \mathfrak{x}(y\_j) \ (1 \le i, j \le k). \end{cases}$$

**Proof.** Since the proofs of the three results are parallel, here we only give the first one. As the vertices of *V*<sup>1</sup> are twin points (if *s* > 1), *d*(*v*, *vi*) = *d*(*v*, *vj*) for each *v* ∈ *V*(*G*)\{*vi*, *vj*}, and thus *Tr*(*vi*) = *Tr*(*vj*) = 2*s* + *t* + 2*k* + 2. Considering the characteristic equations indexed by *vi* and *vj*, it is obtained that

$$\begin{cases} \rho\_{\mathcal{L}}(G) \cdot \mathfrak{x}(v\_{i}) = \sum\_{\boldsymbol{v} \in V(G)} d(v\_{i\boldsymbol{\nu}} \boldsymbol{v}) (\mathfrak{x}(v\_{i}) - \mathfrak{x}(v)) \\ \rho\_{\mathcal{L}}(G) \cdot \mathfrak{x}(v\_{j}) = \sum\_{\boldsymbol{v} \in V(G)} d(v\_{j\boldsymbol{\nu}} \boldsymbol{v}) (\mathfrak{x}(v\_{j}) - \mathfrak{x}(v)) .\end{cases}$$

Then it follows that *ρ*L(*G*) · (*x*(*vi*) − *x*(*vj*)) = (*Tr*(*vi*) + 2)(*x*(*vi*) − *x*(*vj*)). From Lemma 4, we easily obtain

$$\begin{array}{rcl} \rho\_{\mathcal{L}}(G) & \geq & Tr\_{\text{max}} = Tr(w) = s + 2t + 3k + 4 \\ & > & 2s + t + 2k + 4 = Tr(v\_i) + 2. \end{array}$$

Thus, *x*(*vi*) = *x*(*vj*) follows.

For the eigenvector *x* in Lemma 8, suppose *x*(*w*) = *x*0, *x*(*vi*) = *x*<sup>1</sup> (1 ≤ *i* ≤ *s*), *x*(*ui*) = *x*<sup>2</sup> (1 ≤ *i* ≤ *t*), *x*(*yi*) = *x*<sup>3</sup> (1 ≤ *i* ≤ *k*) and *x*(*z*) = *x*4. Then *x* can be written as

$$\boldsymbol{\pi} = (\boldsymbol{\pi}\_{0}, \underbrace{\boldsymbol{\pi}\_{1}, \dots, \boldsymbol{\pi}\_{1}}\_{s}, \underbrace{\boldsymbol{\pi}\_{2}, \dots, \boldsymbol{\pi}\_{2}}\_{t}, \underbrace{\boldsymbol{\pi}\_{3}, \dots, \boldsymbol{\pi}\_{3}}\_{k}, \boldsymbol{\pi}\_{4})^{T} \boldsymbol{\dots}$$

**Lemma 9.** *Let x be as just described. If* |*V*1| = |*V*3| *(i.e., s* = *k), then*

$$\begin{cases} \mathfrak{x}\_0 = -\mathfrak{x}\_4 \neq 0, \\ \mathfrak{x}\_1 = -\mathfrak{x}\_3 \neq 0, \\ \mathfrak{x}\_2 = 0. \end{cases}$$

**Proof.** Applying Lemma 8, the characteristic equation L(*G*) · *x* = *ρ*L(*G*) · *x* can be simplified in the conventional form as follows:

$$\begin{cases} m\_0 \cdot \mathbf{x}\_0 - \mathbf{s} \cdot \mathbf{x}\_1 - 2t \cdot \mathbf{x}\_2 - 3k \cdot \mathbf{x}\_3 - 4\mathbf{x}\_4 = \rho\_\mathcal{L}(\mathbf{G}) \cdot \mathbf{x}\_0 \\ -\mathbf{x}\_0 + m\_1 \cdot \mathbf{x}\_1 - t \cdot \mathbf{x}\_2 - 2k \cdot \mathbf{x}\_3 - 3\mathbf{x}\_4 = \rho\_\mathcal{L}(\mathbf{G}) \cdot \mathbf{x}\_1 \\ -2\mathbf{x}\_0 - s \cdot \mathbf{x}\_1 + m\_2 \cdot \mathbf{x}\_2 - k \cdot \mathbf{x}\_3 - 2\mathbf{x}\_4 = \rho\_\mathcal{L}(\mathbf{G}) \cdot \mathbf{x}\_2 \\ -3\mathbf{x}\_0 - 2s \cdot \mathbf{x}\_1 - t \cdot \mathbf{x}\_2 + m\_3 \cdot \mathbf{x}\_3 - \mathbf{x}\_4 = \rho\_\mathcal{L}(\mathbf{G}) \cdot \mathbf{x}\_3 \\ -4\mathbf{x}\_0 - 3s \cdot \mathbf{x}\_1 - 2t \cdot \mathbf{x}\_2 - k \cdot \mathbf{x}\_3 + m\_4 \cdot \mathbf{x}\_4 = \rho\_\mathcal{L}(\mathbf{G}) \cdot \mathbf{x}\_4 \end{cases} \tag{1}$$

where *m*<sup>0</sup> = *Tr*(*w*), *m*<sup>1</sup> = *Tr*(*vi*) − 2(*s* − 1), *m*<sup>2</sup> = *Tr*(*ui*) − 2(*t* − 1), *m*<sup>3</sup> = *Tr*(*yi*) − 2(*k* − 1) and *m*<sup>4</sup> = *Tr*(*z*).

The sum of the first equality and the fifth one gives

$$(\mathbf{x} + \mathbf{t} + 2\mathbf{k} + 2)\mathbf{x}\_0 - 4(\mathbf{x}\_0 + s\mathbf{x}\_1 + t\mathbf{x}\_2 + k\mathbf{x}\_3 + \mathbf{x}\_4) + (\mathbf{n} + \mathbf{t} + 2\mathbf{s} + 2)\mathbf{x}\_4 = \rho\_\mathcal{L}(\mathbf{G}) \cdot (\mathbf{x}\_0 + \mathbf{x}\_4). \tag{2}$$
 
$$\text{Since } \mathbf{s} = \mathbf{k} \text{ and } \mathbf{x}\_0 + s\mathbf{x}\_1 + t\mathbf{x}\_2 + k\mathbf{x}\_3 + \mathbf{x}\_4 = \mathbf{0} \text{, we have}$$

$$
\mathcal{L}n(\mathbf{x}\_0 + \mathbf{x}\_4) = \rho\_{\mathcal{L}}(G) \cdot (\mathbf{x}\_0 + \mathbf{x}\_4). \tag{3}
$$

Take a 2 × 2 principal submatrix *M* of L(*G*), where *M* = *Tr*(*w*) −4 −4 *Tr*(*z*) ! . Note that *Tr*(*w*) = *Tr*(*z*) = 2*n* for *s* = *k*. Then, applying Lemma 4,

$$\rho\_{\mathcal{L}}(G) \ge \lambda\_1(M) = Tr(w) + 4 = 2n + 4.$$

Thus, we obtain *x*<sup>0</sup> + *x*<sup>4</sup> = 0, *i*.*e*., *x*<sup>0</sup> = −*x*<sup>4</sup> from (3). Similarly, from the second and the fourth equalities in (1), it follows that

$$-2(\varkappa\_0 + \varkappa\_4) + (n+2)(\varkappa\_1 + \varkappa\_3) = \rho\_\mathcal{L}(G) \cdot (\varkappa\_1 + \varkappa\_3).$$

Since *x*<sup>0</sup> + *x*<sup>4</sup> = 0 and *ρ*L(*G*) ≥ 2*n* + 4 > *n* + 2, *x*<sup>1</sup> + *x*<sup>3</sup> = 0, *i*.*e*., *x*<sup>1</sup> = −*x*3. The fourth equality in (1) minus the second one indicates that

$$\begin{array}{rcl} \mathbf{2}(\mathbf{x}\_{4}-\mathbf{x}\_{0}) &=& [\rho\_{\mathcal{L}}(\mathbf{G}) - (\mathbf{2}s + t + 2k + 4)](\mathbf{x}\_{3} - \mathbf{x}\_{1}) \\ &=& (\rho\_{\mathcal{L}}(\mathbf{G}) - 2n + t)(\mathbf{x}\_{3} - \mathbf{x}\_{1}). \end{array} \tag{4}$$

If *x*<sup>0</sup> = 0, then *x*<sup>4</sup> = −*x*<sup>0</sup> = 0. Further, from (4), it follows that *x*<sup>1</sup> = −*x*<sup>3</sup> = 0 (note that *ρ*L(*G*) ≥ 2*n* + 4). Recalling that *x*<sup>0</sup> + *sx*<sup>1</sup> + *tx*<sup>2</sup> + *kx*<sup>3</sup> + *x*<sup>4</sup> = 0, we know *x*<sup>2</sup> = 0, and thus *x* is a zero vector, a contradiction. Hence, *x*<sup>0</sup> = −*x*<sup>4</sup> = 0. Similarly, *x*<sup>1</sup> = −*x*<sup>3</sup> = 0, which implies *x*<sup>2</sup> = 0. The proof is complete.

For convenience, denote the graph *G* ∈ B *<sup>n</sup>*,4 by *<sup>G</sup>*(1,*s*, *<sup>t</sup>*, *<sup>k</sup>*, 1) with vertex partition shown in Remark 1. We next determine the unique extremal graph minimizing the distance Laplacian spectral radius among B *<sup>n</sup>*,4.

**Theorem 1.** *The graph G*(1, 1, *n* − 4, 1, 1) *in Figure 2 is the unique graph with minimum distance Laplacian spectral radius among* B *<sup>n</sup>*,4*.*

**Figure 2.** The graph *G*(1, 1, *n* − 4, 1, 1) ∈ B *<sup>n</sup>*,4 and graph *<sup>G</sup>* .

**Proof.** Let *G*<sup>0</sup> = *G*(1,*s*, *t*, *k*, 1) ∈ B *<sup>n</sup>*,4 with *<sup>s</sup>* <sup>+</sup> *<sup>t</sup>* <sup>+</sup> *<sup>k</sup>* <sup>+</sup> <sup>2</sup> <sup>=</sup> *<sup>n</sup>* and *<sup>s</sup>*, *<sup>t</sup>*, *<sup>k</sup>* <sup>≥</sup> 1. Without loss of generality, assume that *s* ≤ *k*. We proceed by proving the following three claims, which will imply the conclusion.

**Claim 1.** If *s* ≥ 2 in graph *G*0, then let *G*<sup>1</sup> = *G*(1,*s* − 1, *t*, *k* + 1, 1). We claim that *ρ*L(*G*0) < *ρ*L(*G*1).

In graph *G*0, let *V*(*G*0) = {*V*0, ... , *V*4} and *Vi* be expressed as that in Remark 1. Then we easily obtain

$$\begin{cases} Tr(w) = s + 2t + 3k + 4, & Tr(v\_i) = Tr(y\_i) = 2s + t + 2k + 2, \\ Tr(u\_i) = s + 2t + k + 2, & Tr(z) = 3s + 2t + k + 4, \end{cases} \tag{5}$$

and the distance Laplacian matrix of *G*<sup>0</sup> is

$$
\mathcal{L}(\mathsf{G}\_{0}) = \begin{pmatrix}
\operatorname{Tr}(\boldsymbol{w}) & -\mathsf{I}\_{1\times s} & -2\mathsf{I}\_{1\times t} & -3\mathsf{I}\_{1\times k} & -4 \\
\end{pmatrix}.
$$

Further, we have

$$|\lambda I\_{\rm II} - \mathcal{L}(\mathsf{G}\_{\rm 0})| = (\lambda - \mathrm{Tr}(v\_{\rm i}) - 2)^{s-1} (\lambda - \mathrm{Tr}(u\_{\rm i}) - 2)^{t-1} (\lambda - \mathrm{Tr}(y\_{\rm i}) - 2)^{k-1} \cdot |\lambda I\_{\rm S} - \mathcal{R}(\mathsf{G}\_{\rm 0})|,\tag{6}$$
 where

$$R(\mathbb{C}\_0) = \begin{pmatrix} Tr(w) & -s & -2t & -3k & -4 \\ -1 & Tr(v\_i) - 2(s-1) & -t & -2k & -3 \\ -2 & -s & Tr(u\_i) - 2(t-1) & -k & -2 \\ -3 & -2s & -t & Tr(y\_i) - 2(k-1) & -1 \\ -4 & -3s & -2t & -k & Tr(z) \end{pmatrix} . \tag{7}$$

From the above, we say that the largest eigenvalue of *R*(*G*0) is the spectral radius of *G*0. In fact, *λ*1(*R*(*G*0)) ≥ *Tr*(*w*) from Lemma 3, and *Tr*(*vi*) + 2, *Tr*(*yi*) + 2 and *Tr*(*ui*) + 2 are the eigenvalues of L(*G*0) apart from those of *R*(*G*0) from (6). Furthermore, *Tr*(*w*) > *Tr*(*vi*) + 2 = *Tr*(*yi*) + 2 and *Tr*(*w*) > *Tr*(*ui*) + 2 by (5) clearly. Thus, *λ*1(*R*(*G*0)) = *ρ*L(*G*0) holds.

For graph *G*1, we obtain the matrices L(*G*1) and *R*(*G*1) by substituting *s* − 1 and *k* + 1 for *s* and *k* in L(*G*0) and *R*(*G*0), respectively. Analogously, we have *λ*1(*R*(*G*1)) = *ρ*L(*G*1). Denote the characteristic polynomials of *R*(*G*0) and *R*(*G*1) by *ψ*0(*λ*) and *ψ*1(*λ*), respectively. Next, we are aimed at proving

$$
\psi\_1(\rho\_{\mathcal{L}}(G\_0)) < 0. \tag{8}
$$

By using MATLAB, we obtain

$$
\psi\_1(\lambda) - \psi\_0(\lambda) = 4(s - k - 1)\lambda \left(\lambda^2 - s\lambda - k\lambda - 2n\lambda - 4\lambda + 6n + sn + kn + n^2\right). \tag{9}
$$

Let *g*(*λ*) = *λ* (*λ*<sup>2</sup> − *sλ* − *kλ* − 2*nλ* − 4*λ* + 6*n* + *sn* + *kn* + *n*2). Then the derivative of *g*(*λ*) is

$$g'(\lambda) = 3\lambda^2 - (2s + 2k + 4n + 8)\lambda + n^2 + sn + kn + 6n^2$$

with symmetry axis ( *<sup>λ</sup>* <sup>=</sup> <sup>2</sup>*n*+*s*+*k*+<sup>4</sup> <sup>3</sup> . Since *<sup>s</sup>* <sup>≤</sup> *<sup>k</sup>*, *Trmax* <sup>=</sup> *Tr*(*w*) = <sup>2</sup>*<sup>n</sup>* <sup>−</sup> *<sup>s</sup>* <sup>+</sup> *<sup>k</sup>* in graph *<sup>G</sup>*0, and thus *ρ*L(*G*0) ≥ *Trmax* = 2*n* − *s* + *k* from Lemma 4. By simple calculation, we obtain ( *λ* < 2*n* − *s* + *k*, and since *n* ≥ *s* + *k* + 3, we have

$$\,\_3\mathrm{g}'(2n-s+k) = 5n^2 + (5k-11s-10)n + 5s^2 - 6sk + 8s + k^2 - 8k > 0.$$

We now say that *g*(*λ*) is strictly increasing for *λ* ≥ 2*n* − *s* + *k*. Moreover, from *n* ≥ *s* + *k* + 3, it follows that

$$\log(2n - s + k) = (2n - s + k)(n^2 + kn - 3sn - 2n + 4s - 4k - 2sk + 2s^2) > 0.$$

Note that *s* − *k* − 1 < 0 in (9). Then we have

$$\begin{array}{rcl} \psi\_1(\rho\_{\mathcal{L}}(\mathcal{G}\_0)) &=& \psi\_1(\rho\_{\mathcal{L}}(\mathcal{G}\_0)) - \psi\_0(\rho\_{\mathcal{L}}(\mathcal{G}\_0)) < \psi\_1(2n - s + k) - \psi\_0(2n - s + k) \\ &=& 4(s - k - 1) \cdot \lg(2n - s + k) < 0, \end{array}$$

which establishes (8).

Applying (8), we can easily prove that *ρ*L(*G*0) < *ρ*L(*G*1). Assume on the contrary that *ρ*L(*G*0) > *ρ*L(*G*1) (noting that *ρ*L(*G*0) = *ρ*L(*G*1) since *ψ*1(*ρ*L(*G*0)) < 0 and *ψ*1(*ρ*L(*G*1)) = 0). Observing that *ψ*1(*λ*) tends to infinity when *λ* tends to infinity (as the leading coefficient of *ψ*1(*λ*) is 1), we can find a sufficiently large *q* > *ρ*L(*G*0) such that *ψ*1(*q*) > 0. As *ψ*1(*λ*) is a continuous function, from *ψ*1(*ρ*L(*G*0)) < 0 and *ψ*1(*q*) > 0, it follows that *ψ*1(*p*) = 0 for a positive number *p* between *ρ*L(*G*0) and *q*, which is a contradiction to the fact that *ρ*L(*G*1) is the largest root of *ψ*1(*λ*) = 0. Therefore, *ρ*L(*G*0) < *ρ*L(*G*1).

**Claim 2.** Assume *G*<sup>2</sup> = *G*(1,*s* − 1, *t* + 2, *k* − 1, 1), where *k* ≥ *s* ≥ 2 and *t* ≥ 1. Then we claim that *ρ*L(*G*0) > *ρ*L(*G*2).

Let the unit eigenvector corresponding to *ρ*L(*G*2) be

$$\mathbf{x} = (\mathbf{x}\_0, \underbrace{\mathbf{x}\_1, \dots, \mathbf{x}\_1}\_{s-1}, \underbrace{\mathbf{x}\_2, \dots, \mathbf{x}\_2}\_{t+2}, \underbrace{\mathbf{x}\_3, \dots, \mathbf{x}\_3}\_{k-1}, \mathbf{x}\_4)^T.$$

By Rayleigh's principle,

$$\begin{array}{rcl} \rho\_{\mathcal{L}}(\mathcal{G}\_{0}) - \rho\_{\mathcal{L}}(\mathcal{G}\_{2}) & \geq & \mathbf{x}^{T} \cdot (\mathcal{L}(\mathcal{G}\_{0}) - \mathcal{L}(\mathcal{G}\_{2})) \cdot \mathbf{x} \\ &=& \sum\_{\{u,v\} \subseteq V(\mathcal{G}\_{0})} (d\_{\mathcal{G}\_{0}}(u,v) - d\_{\mathcal{G}\_{2}}(u,v)) (\mathbf{x}(u) - \mathbf{x}(v))^{2} \\ &=& 2(s-1)(\mathbf{x}\_{1} - \mathbf{x}\_{2})^{2} + 2(k-1)(\mathbf{x}\_{3} - \mathbf{x}\_{2})^{2} \geq 0. \end{array} \tag{10}$$

Next, we show that *ρ*L(*G*0) − *ρ*L(*G*2) > 0. First, if *s* = *k*, then from Lemma 9, it follows that *x*<sup>1</sup> = −*x*<sup>3</sup> = 0 and *x*<sup>2</sup> = 0, and thus

$$
\rho\_{\mathcal{L}}(G\_0) - \rho\_{\mathcal{L}}(G\_2) \ge 4(s - 1)x\_1^2 > 0.
$$

On the other hand, suppose *s* < *k* and 2(*s* − 1)(*x*<sup>1</sup> − *x*2)<sup>2</sup> + 2(*k* − 1)(*x*<sup>3</sup> − *x*2)<sup>2</sup> = 0 in (10). Then *x*<sup>1</sup> = *x*<sup>2</sup> = *x*3. Substitute *t* + 2 for *t* in (4), and then *x*<sup>0</sup> = *x*<sup>4</sup> follows by applying *x*<sup>1</sup> = *x*3. In addition, by replacing *s*, *k* and *t* with *s* − 1, *k* − 1 and *t* + 2 in (2), respectively, it gives

$$
\rho\_{\mathcal{L}}(G\_2) \cdot \mathbf{x}\_0 = 2n \cdot \mathbf{x}\_{0\prime}.
$$

and hence *x*<sup>0</sup> = 0 for the reason that *ρ*L(*G*2) ≥ *TrG*<sup>2</sup> (*w*) = 2*n* + *k* − *s* > 2*n*. Recalling that *x*<sup>0</sup> + (*s* − 1)*x*<sup>1</sup> + (*t* + 2)*x*<sup>2</sup> + (*k* − 1)*x*<sup>3</sup> + *x*<sup>4</sup> = 0, we have *x*<sup>1</sup> = *x*<sup>2</sup> = *x*<sup>3</sup> = 0, and then eigenvector *x* is the zero vector, a contradiction. Hence, if *s* < *k*, then

$$2(s-1)(\mathbf{x}\_1 - \mathbf{x}\_2)^2 + 2(k-1)(\mathbf{x}\_3 - \mathbf{x}\_2)^2 > 0.5$$

In summary, *ρ*L(*G*0) − *ρ*L(*G*2) > 0 holds.

**Claim 3.** If *s* = *k* − 1(≥ 1) in graph *G*0, then let *G*<sup>3</sup> = *G*(1,*s*, *t* + 1, *k* − 1, 1). We claim that *ρ*L(*G*0) > *ρ*L(*G*3).

Let the unit eigenvector corresponding to *ρ*L(*G*3) be

$$\boldsymbol{\pi} = (\boldsymbol{\pi}\_0, \underbrace{\boldsymbol{\pi}\_1, \dots, \boldsymbol{\pi}\_1}\_s, \underbrace{\boldsymbol{\pi}\_2, \dots, \boldsymbol{\pi}\_2}\_{t+1}, \underbrace{\boldsymbol{\pi}\_3, \dots, \boldsymbol{\pi}\_3}\_{k-1=s}, \boldsymbol{\pi}\_4)^T.$$

Then by Rayleigh's principle and Lemma 9,

$$\begin{array}{rcl} \rho\_{\mathcal{L}}(\mathcal{G}) - \rho\_{\mathcal{L}}(\mathcal{G}\_{3}) & \geq & \mathbf{x}^{T} \cdot (\mathcal{L}(\mathcal{G}) - \mathcal{L}(\mathcal{G}\_{3})) \cdot \mathbf{x} \\ &=& \sum\_{\{u,v\} \subseteq V(\mathcal{G})} (d\_{\mathcal{G}}(u,v) - d\_{\mathcal{G}\_{3}}(u,v)) (\mathbf{x}(u) - \mathbf{x}(v))^{2} \\ &=& (\mathbf{x}\_{0} - \mathbf{x}\_{2})^{2} + \mathbf{s}(\mathbf{x}\_{1} - \mathbf{x}\_{2}) - t(\mathbf{x}\_{2} - \mathbf{x}\_{2})^{2} + \mathbf{s}(\mathbf{x}\_{2} - \mathbf{x}\_{3})^{2} - (\mathbf{x}\_{2} - \mathbf{x}\_{4})^{2} \\ &=& 2\mathbf{s} \cdot \mathbf{x}\_{1}^{2} > 0. \end{array}$$

Now we are in a position to complete the proof of the theorem.

For graph *G*<sup>0</sup> = *G*(1,*s*, *t*, *k*, 1) ∈ B *<sup>n</sup>*,4, suppose *<sup>k</sup>* <sup>≥</sup> *<sup>s</sup>*. If *<sup>k</sup>* <sup>≥</sup> 2 and (*<sup>k</sup>* <sup>−</sup> *<sup>s</sup>*) <sup>≡</sup> <sup>0</sup>(*mod* <sup>2</sup>), then by Claim 1,

$$\rho\_{\mathcal{L}}(G\_0) \ge \rho\_{\mathcal{L}}(G(1, s + \frac{k-s}{2}, t, k - \frac{k-s}{2}, 1)) = \rho\_{\mathcal{L}}(G(1, \frac{s+k}{2}, t, \frac{s+k}{2}, 1))$$

with equality if and only if *s* = *k*. Furthermore, from Claim 2, we have

$$
\rho\_{\mathcal{L}}(G(1, \frac{s+k}{2}, t, \frac{s+k}{2}, 1)) > \rho\_{\mathcal{L}}(G(1, 1, n-4, 1, 1)).
$$

On the other side, if *k* ≥ 3 and (*k* − *s*) ≡ 1(*mod* 2), then by Claim 1,

$$\rho\_{\mathcal{L}}(G\_0) \ge \rho\_{\mathcal{L}}(G(1, s + \frac{k-s-1}{2}, t, k - \frac{k-s-1}{2}, 1)) \\ = \rho\_{\mathcal{L}}(G(1, \frac{s+k-1}{2}, t, \frac{s+k+1}{2}, 1)),$$

with equality if and only if *k* − *s* = 1. Moreover, from Claim 3,

$$
\rho\_{\mathcal{L}}(G(1, \frac{s+k-1}{2}, t, \frac{s+k+1}{2}, 1)) > \rho\_{\mathcal{L}}(G(1, \frac{s+k-1}{2}, t+1, \frac{s+k-1}{2}, 1)).
$$

Finally, from Claim 2, it follows that

$$
\rho\_{\mathcal{L}}(G(1, \frac{s+k-1}{2}, t+1, \frac{s+k-1}{2}, 1)) > \rho\_{\mathcal{L}}(G(1, 1, n-4, 1, 1)).
$$

For the case of *s* = 1 and *k* = 2, from Claim 3, it is straightforward that

$$
\rho\_{\mathcal{L}}(G(1,1,n-5,2,1)) > \rho\_{\mathcal{L}}(G(1,1,n-4,1,1)).
$$

This completes the proof.

From Theorem 1 and Lemma 5, we indicate that if *G* ∈ B*n*,*<sup>d</sup>* is not a spanning subgraph of *G*(1, 1, *n* − 4, 1, 1), then *ρ*L(*G*) > *ρ*L(*G*(1, 1, *n* − 4, 1, 1)). In addition, if *s* = *k* = 1 and *t* = *n* − 4 in (7), then we obtain *λ*1(*R*0) = 4 + <sup>1</sup> <sup>2</sup> (3*<sup>n</sup>* <sup>+</sup> <sup>√</sup>*n*<sup>2</sup> <sup>+</sup> <sup>16</sup>) by using MATLAB, i.e.,

$$
\rho\_{\mathcal{L}}(G(1,1,n-4,1,1)) = 4 + \frac{1}{2}(3n + \sqrt{n^2 + 16})\dots
$$

Thus, we have the following theorem.

**Theorem 2.** *Let G* ∈ B*n*,*d. Then ρ*L(*G*) ≥ 4 + <sup>1</sup> <sup>2</sup> (3*<sup>n</sup>* <sup>+</sup> <sup>√</sup>*n*<sup>2</sup> <sup>+</sup> <sup>16</sup>) *with equality if and only if G* = *G*(1, 1, *n* − 4, 1, 1)*.*

**Proof.** Denote the graph *<sup>G</sup>*(1, 1, *<sup>n</sup>* − 4, 1, 1) − *evu*<sup>1</sup> by *<sup>G</sup>* (see Figure 2). First, we show that *ρ*L(*G* ) > *ρ*L(*G*(1, 1, *n* −4, 1, 1)). Take a 2×2 principal submatrix *M* = *Tr*(*w*) −4 −4 *Tr*(*z*) ! = 2*n* + 2 −4 −4 2*n* ! from L(*G* ). By simple calculation, *λ*1(*M*) > 2*n* + 5 > 4 + <sup>1</sup> <sup>2</sup> (3*n* + <sup>√</sup>*n*<sup>2</sup> <sup>+</sup> <sup>16</sup>). From Lemma 4, it follows that

$$
\rho\_{\mathcal{L}}(G') \ge \lambda\_1(M) > 2n + 5 > 4 + \frac{1}{2}(3n + \sqrt{n^2 + 16}) = \rho\_{\mathcal{L}}(G(1, 1, n - 4, 1, 1)).
$$

Thus, we say that for any spanning subgraph *H* = *G*(1, 1, *n* − 4, 1, 1) of *G*(1, 1, *n* − 4, 1, 1), *ρ*L(*H*) > *ρ*L(*G*(1, 1, *n* − 4, 1, 1)) from Lemma 5.

Hence, now, it is clear from Theorem 1 and the above result that for any graph *G* ∈ B*n*,*d*,

$$\rho\_{\mathcal{L}}(G) \ge \rho\_{\mathcal{L}}(G(1, 1, n - 4, 1, 1)) = 4 + \frac{1}{2}(3n + \sqrt{n^2 + 16})$$

with equality if and only if *G* = *G*(1, 1, *n* − 4, 1, 1).

## **4. Conclusions**

In this paper, a lower bound of the distance Laplacian spectral radius of the *n*-vertex bipartite graphs with diameter 4 is obtained. The method used here is helpful for solving the general case and we conjecture that the graph *G*(1, ... , 1, *n* − *d*, 1, ... , 1) is the unique one minimizing the distance Laplacian spectral radius among *n*-vertex bipartite graphs with even diameter *d* ≥ 4.

**Author Contributions:** Conceptualization, L.Q. and L.M.; methodology, W.Z.; formal analysis, L.L.; writing—original draft preparation, L.Q.; writing—review and editing, W.Z. and L.L.; supervision, L.M.; project administration, L.M.; funding acquisition, L.M. All authors have read and agreed to the published version of the manuscript.

**Funding:** This work was supported by the 2021 Visiting Scholar "Teacher Professional Development Project" in the University of Zhejiang Provincial Department of Education (Grant No. FX2021169).

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**

