**Appendix D. Algebraic System for the** *e−R***(***ξ***)-Expansion Method**

In the algorithm of the *e*−*R*(*ξ*)-expansion method, the algorithm needs to solve the following system of equations for the *a*0, *a*1, and *ω*; here we present the system for each power of *e*−*R*(*ξ*) from 0 to 4.

$$\begin{aligned} c^0: \quad &a\_1\mu \Big( (\alpha - 3)a\_1\mu + \lambda^2 + 2\mu + 1 \Big) + a\_0((\alpha - 4)a\_1\lambda\mu + \omega) \\ &+ a\_0^2((6 - 4\alpha)a\_1\mu - 1) + (\alpha - 1)a\_0^4 = 0, \end{aligned}$$

$$\begin{aligned} e^{-\overline{R}(z)}: \quad &aa\_0a\_1\lambda^2 + 3aa\_1^2\lambda\mu - 4aa\_0^2a\_1\lambda - 8aa\_0a\_1^2\mu + 2aa\_0a\_1\mu \\ &+ 4aa\_0^3a\_1 + a\_1\lambda^3 - 4a\_0a\_1\lambda^2 - 10a\_1^2\lambda\mu + 8a\_1\lambda\mu + 6a\_0^2a\_1\lambda \\ &+ a\_1\lambda + 12a\_0a\_1^2\mu - 8a\_0a\_1\mu + a\_1\omega - 4a\_0^3a\_1 - 2a\_0a\_1 = 0,\end{aligned}$$

$$\begin{aligned} e^{-2\mathcal{R}(z)}: \quad 2a a\_1^2 \lambda^2 - 8a a\_0 a\_1^2 \lambda + 3a a\_0 a\_1 \lambda - 4a a\_1^3 \mu + 4a a\_1^2 \mu \\ + 6a a\_0^2 a\_1^2 - 4a a\_0^2 a\_1 - 7 a\_1^2 \lambda^2 + 7 a\_1 \lambda^2 + 12 a\_0 a\_1^2 \lambda - 12 a\_0 a\_1 \lambda \\ + 6a\_1^3 \mu - 14 a\_1^2 \mu + 8a\_1 \mu - 6 a\_0^2 a\_1^2 - a\_1^2 + 6 a\_0^2 a\_1 + a\_1 = 0, \end{aligned} \tag{A28}$$

$$\begin{aligned} \left(e^{-3R(z)}\right): & \quad -4\alpha a\_1^3 \lambda + 5\alpha a\_1^2 \lambda + 4\alpha a\_0 a\_1^3 - 8\alpha a\_0 a\_1^2 + 2\alpha a\_0 a\_1 \\ & \quad + 6a\_1^3 \lambda - 18a\_1^2 \lambda + 12a\_1 \lambda - 4a\_0 a\_1^3 + 12a\_0 a\_1^2 - 8a\_0 a\_1 = 0, \end{aligned}$$

$$e^{-4R(z)}: \quad \mathfrak{a}a\_1^4 - 4\mathfrak{a}a\_1^3 + 3\mathfrak{a}a\_1^2 - a\_1^4 + 6a\_1^3 - 11a\_1^2 + 6a\_1 = 0.$$

#### **Appendix E. System of Equations for Exponential Function Method**

To obtain solution through the exponential function method, the following nonlinear algebraic system needs to be solved:

$$x^0: \quad a\_2 \left(-4a\_2b\_1 + a\_2^2 + 3b\_1^2\right) \left((a-1)a\_2 + 2b\_1\right) = 0\_\prime$$

$$\begin{aligned} e^{\mathbb{Q}(z)}: \quad 4aa\_1a\_2^3 - 4aa\_2^3b\_1\lambda + 5aa\_2^2b\_1^2\lambda - 8aa\_2^3b\_0 - 8aa\_1a\_2^2b\_1 + 10aa\_2^2b\_0b\_1 + 2aa\_1a\_2b\_1^2 + 6a\_2^3b\_1\lambda \\ -18a\_2^2b\_1^2\lambda + 12a\_2b\_1^3\lambda + 12a\_2^3b\_0 + 12a\_1a\_2^2b\_1 - 36a\_2^2b\_0b\_1 - 8a\_1a\_2b\_1^2 + 24a\_2b\_0b\_1^2 - 4a\_1a\_2^3 = 0, \end{aligned}$$

$$\begin{cases} \begin{aligned} e^{2Q(z)}: \quad 4aa\_0a\_2^3 + 6aa\_1^2a\_2^2 + 2aa\_2^2b\_1^2\lambda^2 - 8aa\_2^3b\_0\lambda - 8aa\_1a\_2^2b\_1\lambda + 17aa\_2^2b\_0b\_1\lambda + 3aa\_1a\_2b\_1^2\lambda \\ \quad - 4aa\_2^3b\_1\mu + 4aa\_2^2b\_1^2\mu + 10aa\_2^2b\_0^2 - 20aa\_1a\_2^2b\_0 - 4aa\_0a\_2^2b\_1 - 4aa\_1^2a\_2b\_1 + 8aa\_1a\_2b\_0b\_1 \\ \quad - 7a\_2^2b\_1^2\lambda^2 + 7a\_2b\_1^3\lambda^2 + 12a\_2^3b\_0\lambda + 12a\_1a\_2^2b\_1\lambda - 60a\_2^2b\_0b\_1\lambda - 12a\_1a\_2b\_1^2\lambda + 48a\_2b\_0b\_1^2\lambda \\ \quad + 6a\_2^3b\_1\mu - 14a\_2^2b\_1^2\mu + 8a\_2b\_1^3\mu - 36a\_2^2b\_0^2 - a\_2^2b\_1^2 + 30a\_1a\_2^2b\_0 + 6a\_0a\_2^2b\_1 + a\_2b\_1^3 - 2a\_0a\_2b\_1^2 \\ \quad + 6a\_1^2a\_2b\_1 + 36a\_2b\_0^2b\_1 - 30a\_1a\_2b\_0b\_1 - 4a\_0a\_2^3 - 6a\_1^2a\_2^2 = 0, \end{aligned}$$

*e Q*(*z*) : 4*αa*2*a*<sup>3</sup> + <sup>12</sup>*αa*0*a*<sup>2</sup> *a*<sup>1</sup> + *<sup>α</sup>a*2*a*1*b*<sup>2</sup> *λ*<sup>2</sup> + <sup>7</sup>*αa*<sup>2</sup> *b*0*b*1*λ*<sup>2</sup> + <sup>3</sup>*αa*<sup>2</sup> *b*2 *λμ* <sup>−</sup> <sup>4</sup>*αa*2*a*<sup>2</sup> *b*1*λ* <sup>−</sup> <sup>20</sup>*αa*<sup>2</sup> *a*1*b*0*<sup>λ</sup>* + <sup>12</sup>*αa*2*a*1*b*0*b*1*<sup>λ</sup>* + <sup>18</sup>*αa*<sup>2</sup> *b*2 *<sup>λ</sup>* <sup>−</sup> <sup>4</sup>*αa*0*a*<sup>2</sup> *b*1*<sup>λ</sup>* + <sup>2</sup>*αa*2*a*1*b*<sup>2</sup> *μ* <sup>−</sup> <sup>8</sup>*αa*<sup>2</sup> *a*1*b*1*<sup>μ</sup>* <sup>−</sup> <sup>8</sup>*αa*<sup>3</sup> *b*0*<sup>μ</sup>* + <sup>14</sup>*αa*<sup>2</sup> *b*0*b*1*<sup>μ</sup>* <sup>−</sup> <sup>16</sup>*αa*2*a*<sup>2</sup> *b*<sup>0</sup> + <sup>12</sup>*αa*2*a*1*b*<sup>2</sup> <sup>−</sup> <sup>16</sup>*αa*0*a*<sup>2</sup> *b*<sup>0</sup> + *a*2*b*<sup>3</sup> *λ*<sup>3</sup> <sup>−</sup> <sup>4</sup>*a*2*a*1*b*<sup>2</sup> *λ*<sup>2</sup> + <sup>28</sup>*a*2*b*0*b*<sup>2</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>24</sup>*a*<sup>2</sup> *b*0*b*1*λ*<sup>2</sup> + <sup>8</sup>*a*2*b*<sup>3</sup> *λμ* <sup>−</sup> <sup>10</sup>*a*<sup>2</sup> *b*2 *λμ* + 6*a*2*a*<sup>2</sup> *b*1*<sup>λ</sup>* + <sup>30</sup>*a*<sup>2</sup> *a*1*b*0*<sup>λ</sup>* <sup>−</sup> <sup>44</sup>*a*2*a*1*b*0*b*1*<sup>λ</sup>* <sup>+</sup> *<sup>a</sup>*2*b*<sup>3</sup> *<sup>λ</sup>* <sup>−</sup> <sup>64</sup>*a*<sup>2</sup> *b*2 *<sup>λ</sup>* <sup>−</sup> <sup>4</sup>*a*0*a*2*b*<sup>2</sup> *<sup>λ</sup>* + <sup>6</sup>*a*0*a*<sup>2</sup> *b*1*λ* + 72*a*2*b*<sup>2</sup> *b*1*<sup>λ</sup>* <sup>−</sup> <sup>8</sup>*a*2*a*1*b*<sup>2</sup> *<sup>μ</sup>* + <sup>12</sup>*a*<sup>2</sup> *a*1*b*1*<sup>μ</sup>* + <sup>32</sup>*a*2*b*0*b*<sup>2</sup> *<sup>μ</sup>* + <sup>12</sup>*a*<sup>3</sup> *b*0*<sup>μ</sup>* <sup>−</sup> <sup>48</sup>*a*<sup>2</sup> *b*0*b*1*μ* + *a*2*b*<sup>3</sup> *<sup>ω</sup>* + <sup>24</sup>*a*2*a*<sup>2</sup> *b*<sup>0</sup> <sup>−</sup> <sup>44</sup>*a*2*a*1*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*a*2*a*1*b*<sup>2</sup> + <sup>24</sup>*a*2*b*<sup>3</sup> + <sup>4</sup>*a*2*b*0*b*<sup>2</sup> + 24*a*0*a*<sup>2</sup> *b*<sup>0</sup> <sup>−</sup> <sup>2</sup>*a*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>4</sup>*a*0*a*2*b*0*b*<sup>1</sup> <sup>−</sup> <sup>4</sup>*a*2*a*<sup>3</sup> <sup>−</sup> <sup>12</sup>*a*0*a*<sup>2</sup> *a*<sup>1</sup> = 0,

*e Q*(*z*) : *αa*<sup>4</sup> <sup>−</sup> *<sup>a</sup>*<sup>4</sup> <sup>−</sup> <sup>4</sup>*αb*0*a*<sup>3</sup> + <sup>6</sup>*b*0*a*<sup>3</sup> + <sup>3</sup>*αb*<sup>2</sup> *a*2 <sup>−</sup> <sup>11</sup>*b*<sup>2</sup> *a*2 <sup>−</sup> *<sup>b</sup>*<sup>2</sup> *a*2 + <sup>12</sup>*αa*0*a*2*a*<sup>2</sup> <sup>−</sup> <sup>12</sup>*a*0*a*2*a*<sup>2</sup> <sup>−</sup> <sup>16</sup>*αλa*2*b*0*a*<sup>2</sup> + <sup>24</sup>*λa*2*b*0*a*<sup>2</sup> + <sup>4</sup>*αa*0*b*1*a*<sup>2</sup> <sup>−</sup> <sup>6</sup>*a*0*b*1*a*<sup>2</sup> <sup>−</sup> <sup>4</sup>*αμa*2*b*1*a*<sup>2</sup> + <sup>6</sup>*μa*2*b*1*a*<sup>2</sup> <sup>−</sup> *αλb*0*b*1*a*<sup>2</sup> + <sup>4</sup>*λb*0*b*1*a*<sup>2</sup> + <sup>6</sup>*b*<sup>3</sup> *a*<sup>1</sup> + *<sup>ω</sup>b*<sup>3</sup> *a*<sup>1</sup> + <sup>21</sup>*αλa*2*b*<sup>2</sup> *a*<sup>1</sup> <sup>−</sup> <sup>76</sup>*λa*2*b*<sup>2</sup> *a*<sup>1</sup> + *αλa*0*b*<sup>2</sup> *a*<sup>1</sup> <sup>−</sup> <sup>4</sup>*λa*0*b*<sup>2</sup> *a*<sup>1</sup> + *αλμa*2*b*<sup>2</sup> *a*<sup>1</sup> <sup>−</sup> <sup>4</sup>*λμa*2*b*<sup>2</sup> *a*<sup>1</sup> + *<sup>λ</sup>*2*b*0*b*<sup>2</sup> *a*<sup>1</sup> + <sup>2</sup>*μb*0*b*<sup>2</sup> *a*<sup>1</sup> + *b*0*b*<sup>2</sup> *a*<sup>1</sup> <sup>−</sup> <sup>20</sup>*αμa*<sup>2</sup> *b*0*a*<sup>1</sup> + <sup>30</sup>*μa*<sup>2</sup> *b*0*a*<sup>1</sup> <sup>−</sup> <sup>24</sup>*αa*0*a*2*b*0*a*<sup>1</sup> <sup>+</sup> <sup>36</sup>*a*0*a*2*b*0*a*<sup>1</sup> <sup>−</sup> <sup>6</sup>*λb*<sup>2</sup> *b*1*a*<sup>1</sup> <sup>−</sup> <sup>4</sup>*αa*0*b*0*b*1*a*<sup>1</sup> <sup>+</sup> <sup>14</sup>*a*0*b*0*b*1*a*<sup>1</sup> <sup>+</sup> <sup>4</sup>*αλ*2*a*2*b*0*b*1*a*<sup>1</sup> <sup>−</sup> <sup>14</sup>*λ*2*a*2*b*0*b*1*a*<sup>1</sup> <sup>+</sup> <sup>8</sup>*αμa*2*b*0*b*1*a*<sup>1</sup> <sup>−</sup> <sup>28</sup>*μa*2*b*0*b*1*a*<sup>1</sup> <sup>−</sup> <sup>4</sup>*a*2*b*0*b*1*a*<sup>1</sup> <sup>+</sup> <sup>54</sup>*λa*2*b*<sup>3</sup> <sup>−</sup> *<sup>λ</sup>*2*a*0*b*<sup>3</sup> <sup>−</sup> <sup>2</sup>*μa*0*b*<sup>3</sup> <sup>−</sup> *<sup>a</sup>*0*b*<sup>3</sup> + <sup>2</sup>*μ*2*a*2*b*<sup>3</sup> + *λ*2*μa*2*b*<sup>3</sup> + *<sup>μ</sup>a*2*b*<sup>3</sup> + <sup>6</sup>*αa*<sup>2</sup> *a*2 <sup>−</sup> <sup>6</sup>*a*<sup>2</sup> *a*2 + <sup>8</sup>*αλ*2*a*<sup>2</sup> *b*2 <sup>−</sup> <sup>28</sup>*λ*2*a*<sup>2</sup> *b*2 + <sup>16</sup>*αμa*<sup>2</sup> *b*2 <sup>−</sup> <sup>56</sup>*μa*<sup>2</sup> *b*2 <sup>−</sup> *<sup>a</sup>*<sup>2</sup> *b*2 + <sup>6</sup>*αa*0*a*2*b*<sup>2</sup> <sup>−</sup> <sup>24</sup>*a*0*a*2*b*<sup>2</sup> + *<sup>α</sup>a*<sup>2</sup> *b*2 <sup>−</sup> <sup>3</sup>*a*<sup>2</sup> *b*2 + *αμ*2*a*<sup>2</sup> *b*2 <sup>−</sup> <sup>3</sup>*μ*2*a*<sup>2</sup> *b*2 <sup>−</sup> <sup>2</sup>*λ*2*a*0*a*2*b*<sup>2</sup> <sup>−</sup> <sup>4</sup>*μa*0*a*2*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*a*0*a*2*b*<sup>2</sup> + <sup>6</sup>*λa*0*b*0*b*<sup>2</sup> + <sup>4</sup>*λ*3*a*2*b*0*b*<sup>2</sup> + 4*λa*2*b*0*b*<sup>2</sup> + <sup>32</sup>*λμa*2*b*0*b*<sup>2</sup> + <sup>3</sup>*ωa*2*b*0*b*<sup>2</sup> <sup>−</sup> <sup>16</sup>*αλa*0*a*<sup>2</sup> *b*<sup>0</sup> + <sup>24</sup>*λa*0*a*<sup>2</sup> *b*<sup>0</sup> <sup>−</sup> <sup>4</sup>*αμa*0*a*<sup>2</sup> *b*<sup>1</sup> + <sup>6</sup>*μa*0*a*<sup>2</sup> *b*<sup>1</sup> <sup>−</sup> <sup>6</sup>*a*0*b*<sup>2</sup> *b*<sup>1</sup> + <sup>41</sup>*λ*2*a*2*b*<sup>2</sup> *b*<sup>1</sup> + <sup>46</sup>*μa*2*b*<sup>2</sup> *b*<sup>1</sup> + <sup>5</sup>*a*2*b*<sup>2</sup> *b*<sup>1</sup> + 4*αa*<sup>2</sup> *a*2*b*<sup>1</sup> <sup>−</sup> <sup>6</sup>*a*<sup>2</sup> *a*2*b*<sup>1</sup> + <sup>11</sup>*αλμa*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>36</sup>*λμa*<sup>2</sup> *b*0*b*<sup>1</sup> − 2*αλa*0*a*2*b*0*b*<sup>1</sup> = 0,

*e Q*(*z*) : <sup>−</sup> *<sup>a</sup>*0*b*<sup>3</sup> *λ*<sup>3</sup> + *<sup>a</sup>*1*b*0*b*<sup>2</sup> *λ*<sup>3</sup> + <sup>5</sup>*a*2*b*<sup>2</sup> *b*1*λ*<sup>3</sup> + <sup>38</sup>*a*2*b*<sup>3</sup> *λ*<sup>2</sup> + <sup>9</sup>*αa*1*a*2*b*<sup>2</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>32</sup>*a*1*a*2*b*<sup>2</sup> *λ*<sup>2</sup> + *αa*0*a*1*b*<sup>2</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>4</sup>*a*0*a*1*b*<sup>2</sup> *λ*<sup>2</sup> + <sup>10</sup>*a*0*b*0*b*<sup>2</sup> *λ*<sup>2</sup> + <sup>4</sup>*μa*2*b*0*b*<sup>2</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>10</sup>*a*1*b*<sup>2</sup> *b*1*λ*<sup>2</sup> <sup>−</sup> *<sup>α</sup>a*<sup>2</sup> *b*0*b*1*λ*<sup>2</sup> + <sup>4</sup>*a*<sup>2</sup> *b*0*b*1*λ*<sup>2</sup> <sup>−</sup> <sup>2</sup>*αa*0*a*2*b*0*b*1*λ*<sup>2</sup> <sup>+</sup> <sup>4</sup>*a*0*a*2*b*0*b*1*λ*<sup>2</sup> <sup>+</sup> <sup>12</sup>*a*1*b*<sup>3</sup> *λ* <sup>−</sup> <sup>8</sup>*μa*0*b*<sup>3</sup> *<sup>λ</sup>* <sup>−</sup> *<sup>a</sup>*0*b*<sup>3</sup> *<sup>λ</sup>* + <sup>5</sup>*αa*<sup>2</sup> *b*2 *<sup>λ</sup>* <sup>−</sup> <sup>18</sup>*a*<sup>2</sup> *b*2 *<sup>λ</sup>* + <sup>14</sup>*αμa*<sup>2</sup> *b*2 *<sup>λ</sup>* <sup>−</sup> <sup>48</sup>*μa*<sup>2</sup> *b*2 *λ* + 10*αa*0*a*2*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>40</sup>*a*0*a*2*b*<sup>2</sup> *<sup>λ</sup>* + <sup>3</sup>*αa*<sup>2</sup> *b*2 *<sup>λ</sup>* <sup>−</sup> <sup>10</sup>*a*<sup>2</sup> *b*2 *<sup>λ</sup>* <sup>−</sup> <sup>4</sup>*μa*0*a*2*b*<sup>2</sup> *<sup>λ</sup>* + <sup>8</sup>*μa*1*b*0*b*<sup>2</sup> *λ* + *a*1*b*0*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>4</sup>*αa*<sup>3</sup> *b*0*<sup>λ</sup>* + <sup>6</sup>*a*<sup>3</sup> *b*0*<sup>λ</sup>* <sup>−</sup> <sup>24</sup>*αa*0*a*1*a*2*b*0*<sup>λ</sup>* <sup>+</sup> <sup>36</sup>*a*0*a*1*a*2*b*0*<sup>λ</sup>* <sup>+</sup> <sup>4</sup>*αa*0*a*<sup>2</sup> *b*1*λ* <sup>−</sup> <sup>6</sup>*a*0*a*<sup>2</sup> *b*1*<sup>λ</sup>* <sup>−</sup> <sup>12</sup>*a*0*b*<sup>2</sup> *b*1*<sup>λ</sup>* + <sup>40</sup>*μa*2*b*<sup>2</sup> *b*1*<sup>λ</sup>* + <sup>5</sup>*a*2*b*<sup>2</sup> *b*1*<sup>λ</sup>* + <sup>4</sup>*αa*<sup>2</sup> *a*2*b*1*<sup>λ</sup>* <sup>−</sup> <sup>6</sup>*a*<sup>2</sup> *a*2*b*1*λ* <sup>−</sup> <sup>8</sup>*αa*0*a*1*b*0*b*1*<sup>λ</sup>* <sup>+</sup> <sup>28</sup>*a*0*a*1*b*0*b*1*<sup>λ</sup>* <sup>+</sup> <sup>4</sup>*αμa*1*a*2*b*0*b*1*<sup>λ</sup>* <sup>−</sup> <sup>12</sup>*μa*1*a*2*b*0*b*1*<sup>λ</sup>* <sup>+</sup> <sup>4</sup>*αa*0*a*<sup>3</sup> <sup>−</sup> <sup>4</sup>*a*0*a*<sup>3</sup> + <sup>40</sup>*μa*2*b*<sup>3</sup> + <sup>2</sup>*a*2*b*<sup>3</sup> + *<sup>ω</sup>a*0*b*<sup>3</sup> + <sup>2</sup>*αa*0*a*1*b*<sup>2</sup> <sup>−</sup> <sup>8</sup>*a*0*a*1*b*<sup>2</sup> + <sup>18</sup>*αμa*1*a*2*b*<sup>2</sup> <sup>−</sup> <sup>64</sup>*μa*1*a*2*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*a*1*a*2*b*<sup>2</sup> + <sup>2</sup>*αμa*0*a*1*b*<sup>2</sup> <sup>−</sup> <sup>8</sup>*μa*0*a*1*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*a*0*a*1*b*<sup>2</sup> + <sup>8</sup>*μa*0*b*0*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*a*0*b*0*b*<sup>2</sup> + 3*ωa*1*b*0*b*<sup>2</sup> + <sup>8</sup>*μ*2*a*2*b*0*b*<sup>2</sup> + <sup>4</sup>*μa*2*b*0*b*<sup>2</sup> + <sup>12</sup>*αa*<sup>2</sup> *a*1*a*<sup>2</sup> <sup>−</sup> <sup>12</sup>*a*<sup>2</sup> *a*1*a*<sup>2</sup> <sup>−</sup> <sup>8</sup>*αa*0*a*<sup>2</sup> *b*<sup>0</sup> + 12*a*0*a*<sup>2</sup> *b*<sup>0</sup> <sup>−</sup> <sup>16</sup>*αμa*0*a*<sup>2</sup> *b*<sup>0</sup> + <sup>24</sup>*μa*0*a*<sup>2</sup> *b*<sup>0</sup> <sup>−</sup> <sup>8</sup>*αa*<sup>2</sup> *a*2*b*<sup>0</sup> + <sup>12</sup>*a*<sup>2</sup> *a*2*b*<sup>0</sup> <sup>−</sup> <sup>16</sup>*αμa*<sup>2</sup> *a*2*b*<sup>0</sup> + 24*μa*<sup>2</sup> *a*2*b*<sup>0</sup> <sup>−</sup> <sup>8</sup>*μa*1*b*<sup>2</sup> *b*<sup>1</sup> + <sup>2</sup>*a*1*b*<sup>2</sup> *b*<sup>1</sup> + <sup>3</sup>*ωa*2*b*<sup>2</sup> *b*<sup>1</sup> + <sup>8</sup>*αa*<sup>2</sup> *a*1*b*<sup>1</sup> <sup>−</sup> <sup>12</sup>*a*<sup>2</sup> *a*1*b*<sup>1</sup> <sup>−</sup> <sup>2</sup>*αa*<sup>2</sup> *b*0*b*<sup>1</sup> + <sup>8</sup>*a*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>2</sup>*αμa*<sup>2</sup> *b*0*b*<sup>1</sup> + <sup>8</sup>*μa*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>2</sup>*a*<sup>2</sup> *b*0*b*<sup>1</sup> + <sup>4</sup>*αμ*2*a*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>12</sup>*μ*2*a*<sup>2</sup> *b*0*b*<sup>1</sup> − 4*αμa*0*a*2*b*0*b*<sup>1</sup> + 8*μa*0*a*2*b*0*b*<sup>1</sup> − 4*a*0*a*2*b*0*b*<sup>1</sup> = 0, (A29)

*e Q*(*z*) : 8*a*2*b*<sup>3</sup> *λ*<sup>3</sup> + <sup>4</sup>*a*0*b*0*b*<sup>2</sup> *λ*<sup>3</sup> <sup>−</sup> <sup>4</sup>*a*1*b*<sup>2</sup> *b*1*λ*<sup>3</sup> + <sup>7</sup>*a*1*b*<sup>3</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>7</sup>*μa*0*b*<sup>3</sup> *λ*<sup>2</sup> + <sup>2</sup>*αa*<sup>2</sup> *b*2 *λ*<sup>2</sup> <sup>−</sup> <sup>7</sup>*a*<sup>2</sup> *b*2 *λ*<sup>2</sup> + 4*αa*0*a*2*b*<sup>2</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>16</sup>*a*0*a*2*b*<sup>2</sup> *λ*<sup>2</sup> + <sup>2</sup>*αa*<sup>2</sup> *b*2 *λ*<sup>2</sup> <sup>−</sup> <sup>7</sup>*a*<sup>2</sup> *b*2 *λ*<sup>2</sup> + <sup>7</sup>*μa*1*b*0*b*<sup>2</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>7</sup>*a*0*b*<sup>2</sup> *b*1*λ*<sup>2</sup> <sup>−</sup> *<sup>μ</sup>a*2*b*<sup>2</sup> *b*1*λ*<sup>2</sup> <sup>−</sup> <sup>4</sup>*αa*0*a*1*b*0*b*1*λ*<sup>2</sup> <sup>+</sup> <sup>14</sup>*a*0*a*1*b*0*b*1*λ*<sup>2</sup> <sup>+</sup> <sup>52</sup>*μa*2*b*<sup>3</sup> *<sup>λ</sup>* + <sup>2</sup>*a*2*b*<sup>3</sup> *<sup>λ</sup>* + <sup>3</sup>*αa*0*a*1*b*<sup>2</sup> *λ* <sup>−</sup> <sup>12</sup>*a*0*a*1*b*<sup>2</sup> *<sup>λ</sup>* + <sup>15</sup>*αμa*1*a*2*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>52</sup>*μa*1*a*2*b*<sup>2</sup> *<sup>λ</sup>* + <sup>3</sup>*αμa*0*a*1*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>12</sup>*μa*0*a*1*b*<sup>2</sup> *λ* + 20*μa*0*b*0*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>2</sup>*a*0*b*0*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>8</sup>*αa*0*a*<sup>2</sup> *b*0*<sup>λ</sup>* + <sup>12</sup>*a*0*a*<sup>2</sup> *b*0*<sup>λ</sup>* <sup>−</sup> <sup>8</sup>*αa*<sup>2</sup> *a*2*b*0*<sup>λ</sup>* + <sup>12</sup>*a*<sup>2</sup> *a*2*b*0*λ* <sup>−</sup> <sup>20</sup>*μa*1*b*<sup>2</sup> *b*1*<sup>λ</sup>* + <sup>2</sup>*a*1*b*<sup>2</sup> *b*1*<sup>λ</sup>* + <sup>8</sup>*αa*<sup>2</sup> *a*1*b*1*<sup>λ</sup>* <sup>−</sup> <sup>12</sup>*a*<sup>2</sup> *a*1*b*1*<sup>λ</sup>* <sup>−</sup> <sup>3</sup>*αa*<sup>2</sup> *b*0*b*1*<sup>λ</sup>* + <sup>12</sup>*a*<sup>2</sup> *b*0*b*1*λ* <sup>−</sup> <sup>3</sup>*αμa*<sup>2</sup> *b*0*b*1*<sup>λ</sup>* + <sup>12</sup>*μa*<sup>2</sup> *b*0*b*1*<sup>λ</sup>* <sup>−</sup> <sup>6</sup>*αμa*0*a*2*b*0*b*1*<sup>λ</sup>* <sup>+</sup> <sup>16</sup>*μa*0*a*2*b*0*b*1*<sup>λ</sup>* <sup>+</sup> <sup>8</sup>*μa*1*b*<sup>3</sup> + *<sup>a</sup>*1*b*<sup>3</sup> + *ωa*2*b*<sup>3</sup> <sup>−</sup> <sup>8</sup>*μ*2*a*0*b*<sup>3</sup> <sup>−</sup> *<sup>μ</sup>a*0*b*<sup>3</sup> + <sup>6</sup>*αa*<sup>2</sup> *a*2 <sup>−</sup> <sup>6</sup>*a*<sup>2</sup> *a*2 + <sup>4</sup>*αμa*<sup>2</sup> *b*2 <sup>−</sup> <sup>14</sup>*μa*<sup>2</sup> *b*2 <sup>−</sup> *<sup>a</sup>*<sup>2</sup> *b*2 + 6*αμ*2*a*<sup>2</sup> *b*2 <sup>−</sup> <sup>20</sup>*μ*2*a*<sup>2</sup> *b*2 + <sup>8</sup>*αμa*0*a*2*b*<sup>2</sup> <sup>−</sup> <sup>32</sup>*μa*0*a*2*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*a*0*a*2*b*<sup>2</sup> + <sup>4</sup>*αμa*<sup>2</sup> *b*2 <sup>−</sup> <sup>14</sup>*μa*<sup>2</sup> *b*2 <sup>−</sup> *<sup>a</sup>*<sup>2</sup> *b*2 <sup>−</sup> <sup>2</sup>*μ*2*a*0*a*2*b*<sup>2</sup> + <sup>3</sup>*ωa*0*b*0*b*<sup>2</sup> + <sup>8</sup>*μ*2*a*1*b*0*b*<sup>2</sup> + *<sup>μ</sup>a*1*b*0*b*<sup>2</sup> + <sup>4</sup>*αa*<sup>3</sup> *a*<sup>2</sup> <sup>−</sup> <sup>4</sup>*a*<sup>3</sup> *a*<sup>2</sup> <sup>−</sup> <sup>4</sup>*αμa*<sup>3</sup> *b*<sup>0</sup> + 6*μa*<sup>3</sup> *b*<sup>0</sup> <sup>−</sup> <sup>4</sup>*αa*<sup>2</sup> *a*1*b*<sup>0</sup> + <sup>6</sup>*a*<sup>2</sup> *a*1*b*<sup>0</sup> <sup>−</sup> <sup>24</sup>*αμa*0*a*1*a*2*b*<sup>0</sup> <sup>+</sup> <sup>36</sup>*μa*0*a*1*a*2*b*<sup>0</sup> <sup>+</sup> <sup>4</sup>*αa*<sup>3</sup> *b*<sup>1</sup> <sup>−</sup> <sup>6</sup>*a*<sup>3</sup> *b*<sup>1</sup> + 4*αμa*0*a*<sup>2</sup> *b*<sup>1</sup> <sup>−</sup> <sup>6</sup>*μa*0*a*<sup>2</sup> *b*<sup>1</sup> <sup>−</sup> <sup>8</sup>*μa*0*b*<sup>2</sup> *b*<sup>1</sup> <sup>−</sup> *<sup>a</sup>*0*b*<sup>2</sup> *b*<sup>1</sup> + <sup>3</sup>*ωa*1*b*<sup>2</sup> *b*<sup>1</sup> + <sup>4</sup>*μ*2*a*2*b*<sup>2</sup> *b*<sup>1</sup> + 5*μa*2*b*<sup>2</sup> *b*<sup>1</sup> + <sup>4</sup>*αμa*<sup>2</sup> *a*2*b*<sup>1</sup> <sup>−</sup> <sup>6</sup>*μa*<sup>2</sup> *a*2*b*<sup>1</sup> − 8*αμa*0*a*1*b*0*b*<sup>1</sup> <sup>+</sup> <sup>28</sup>*μa*0*a*1*b*0*b*<sup>1</sup> <sup>−</sup> <sup>4</sup>*a*0*a*1*b*0*b*<sup>1</sup> <sup>+</sup> <sup>2</sup>*μ*2*a*1*a*2*b*0*b*<sup>1</sup> <sup>=</sup> 0,

*e Q*(*z*) : *a*1*b*<sup>3</sup> *λ*<sup>3</sup> <sup>−</sup> *<sup>a</sup>*0*b*<sup>2</sup> *b*1*λ*<sup>3</sup> + <sup>14</sup>*μa*2*b*<sup>3</sup> *λ*<sup>2</sup> + *<sup>α</sup>a*0*a*1*b*<sup>2</sup> *λ*<sup>2</sup> <sup>−</sup> <sup>4</sup>*a*0*a*1*b*<sup>2</sup> *λ*<sup>2</sup> + <sup>10</sup>*μa*0*b*0*b*<sup>2</sup> *λ*2 <sup>−</sup> <sup>10</sup>*μa*1*b*<sup>2</sup> *b*1*λ*<sup>2</sup> <sup>−</sup> *<sup>α</sup>a*<sup>2</sup> *b*0*b*1*λ*<sup>2</sup> + <sup>4</sup>*a*<sup>2</sup> *b*0*b*1*λ*<sup>2</sup> + <sup>8</sup>*μa*1*b*<sup>3</sup> *<sup>λ</sup>* + *<sup>a</sup>*1*b*<sup>3</sup> *<sup>λ</sup>* <sup>−</sup> <sup>12</sup>*μ*2*a*0*b*<sup>3</sup> *λ* + 3*αμa*<sup>2</sup> *b*2 *<sup>λ</sup>* <sup>−</sup> <sup>10</sup>*μa*<sup>2</sup> *b*2 *<sup>λ</sup>* + <sup>6</sup>*αμa*0*a*2*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>24</sup>*μa*0*a*2*b*<sup>2</sup> *<sup>λ</sup>* + <sup>5</sup>*αμa*<sup>2</sup> *b*2 *λ* <sup>−</sup> <sup>18</sup>*μa*<sup>2</sup> *b*2 *<sup>λ</sup>* + <sup>12</sup>*μ*2*a*1*b*0*b*<sup>2</sup> *<sup>λ</sup>* <sup>−</sup> <sup>4</sup>*αa*<sup>2</sup> *a*1*b*0*<sup>λ</sup>* + <sup>6</sup>*a*<sup>2</sup> *a*1*b*0*<sup>λ</sup>* + <sup>4</sup>*αa*<sup>3</sup> *b*1*<sup>λ</sup>* <sup>−</sup> <sup>6</sup>*a*<sup>3</sup> *b*1*λ* <sup>−</sup> <sup>8</sup>*μa*0*b*<sup>2</sup> *b*1*<sup>λ</sup>* <sup>−</sup> *<sup>a</sup>*0*b*<sup>2</sup> *b*1*<sup>λ</sup>* <sup>−</sup> <sup>12</sup>*μ*2*a*2*b*<sup>2</sup> *b*1*λ* − 8*αμa*0*a*1*b*0*b*1*λ* + 28*μa*0*a*1*b*0*b*1*λ* + *ωa*1*b*<sup>3</sup> + <sup>16</sup>*μ*2*a*2*b*<sup>3</sup> + <sup>2</sup>*μa*2*b*<sup>3</sup> + <sup>2</sup>*αμa*0*a*1*b*<sup>2</sup> <sup>−</sup> <sup>8</sup>*μa*0*a*1*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*a*0*a*1*b*<sup>2</sup> + <sup>6</sup>*αμ*2*a*1*a*2*b*<sup>2</sup> <sup>−</sup> <sup>20</sup>*μ*2*a*1*a*2*b*<sup>2</sup> + <sup>2</sup>*αμ*2*a*0*a*1*b*<sup>2</sup> <sup>−</sup> <sup>8</sup>*μ*2*a*0*a*1*b*<sup>2</sup> + <sup>8</sup>*μ*2*a*0*b*0*b*<sup>2</sup> <sup>−</sup> <sup>2</sup>*μa*0*b*0*b*<sup>2</sup> + <sup>4</sup>*αa*<sup>3</sup> *a*<sup>1</sup> <sup>−</sup> <sup>4</sup>*a*<sup>3</sup> *a*<sup>1</sup> <sup>−</sup> <sup>8</sup>*αμa*0*a*<sup>2</sup> *b*<sup>0</sup> + <sup>12</sup>*μa*0*a*<sup>2</sup> *b*<sup>0</sup> <sup>−</sup> <sup>8</sup>*αμa*<sup>2</sup> *a*2*b*<sup>0</sup> + <sup>12</sup>*μa*<sup>2</sup> *a*2*b*<sup>0</sup> + <sup>3</sup>*ωa*0*b*<sup>2</sup> *b*<sup>1</sup> <sup>−</sup> <sup>8</sup>*μ*2*a*1*b*<sup>2</sup> *b*<sup>1</sup> + <sup>2</sup>*μa*1*b*<sup>2</sup> *b*<sup>1</sup> + <sup>8</sup>*αμa*<sup>2</sup> *a*1*b*<sup>1</sup> <sup>−</sup> <sup>12</sup>*μa*<sup>2</sup> *a*1*b*<sup>1</sup> <sup>−</sup> <sup>2</sup>*αμa*<sup>2</sup> *b*0*b*<sup>1</sup> + <sup>8</sup>*μa*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>2</sup>*a*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>2</sup>*αμ*2*a*<sup>2</sup> *b*0*b*<sup>1</sup> + <sup>8</sup>*μ*2*a*<sup>2</sup> *b*0*b*<sup>1</sup> <sup>−</sup> <sup>4</sup>*αμ*2*a*0*a*2*b*0*b*<sup>1</sup> <sup>+</sup> <sup>12</sup>*μ*2*a*0*a*2*b*0*b*<sup>1</sup> <sup>=</sup> 0,

$$\begin{split} e^{8Q(z)}: \quad aa\_0^4 - aa\_0^2b\_0b\_1\lambda\mu + aa\_1a\_0b\_0^2\lambda\mu + 3aa\_0^2b\_1^2\mu^2 + 2aa\_2a\_0b\_0^2\mu^2 - 4aa\_1a\_0b\_0b\_1\mu^2 \\ \quad + aa\_1^2b\_0^2\mu^2 + 4aa\_0^3b\_1\mu - 4aa\_1a\_0^2b\_0\mu - a\_0b\_0^2b\_1\lambda^2\mu + a\_1b\_0^3\lambda^2\mu + 6a\_0b\_0b\_1^2\lambda\mu^2 \\ \quad + 6a\_2b\_0^3\lambda\mu^2 - 6a\_1b\_0^2b\_1\lambda\mu^2 + 4a\_0^2b\_0b\_1\lambda\mu - 4a\_1a\_0b\_0^2\lambda\mu - 6a\_0b\_1^3\mu^3 + 6a\_1b\_0b\_1^2\mu^3 \\ \quad - 6a\_2b\_0^2b\_1\mu^3 - 11a\_0^2b\_1^2\mu^2 - 8a\_2a\_0b\_0^2\mu^2 - 2a\_0b\_0^2b\_1\mu^2 + 14a\_1a\_0b\_0b\_1\mu^2 + 2a\_1b\_0^3\mu^2 \\ \quad - 3a\_1^2b\_0^2\mu^2 - 6a\_0^3b\_1\mu + 6a\_1a\_0^2b\_0\mu - a\_0b\_0^2b\_1\mu + a\_1b\_0^3\mu + a\_0b\_0^3\omega - a\_0^2b\_0^2 - a\_0^4 = 0. \end{split}$$

#### **References**


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**Jinyang Liu 1,2, Boping Tian 1, Deqi Wang 2,\*, Jiaxin Tang <sup>2</sup> and Yujin Wu <sup>3</sup>**


**Abstract:** In this paper, we propose a user-friendly integral inequality to study the coupled parabolic chemotaxis system with singular sensitivity under the Neumann boundary condition. Under a low diffusion rate, the classical solution of this system is uniformly bounded. Our proof replies on the construction of the energy functional containing Ω |*v*| 4 *<sup>v</sup>*<sup>2</sup> with *<sup>v</sup>* <sup>&</sup>gt; 0. It is noteworthy that the inequality used in the paper may be applied to study other chemotaxis systems.

**Keywords:** chemotaxis model; energy functional; integral inequality; global uniform boundedness

**MSC:** 35A01; 35A02

#### **1. Introduction**

Our work considers the coupled parabolic chemotaxis system with singular sensitivity

$$\begin{cases} u\_t = \nabla \cdot (\nabla u - \chi \frac{\mu}{v} \nabla v), & \quad \mathbf{x} \in \Omega, t > 0, \\ v\_t = k \Delta v - v + u, & \quad \mathbf{x} \in \Omega, t > 0, \\ \partial\_\nu u = \partial\_\nu v = 0, & \quad \mathbf{x} \in \partial \Omega, t > 0, \\ u(\mathbf{x}, 0) = u\_{0\prime} \quad v(\mathbf{x}, 0) = v\_{0\prime} & \mathbf{x} \in \Omega, \end{cases} \tag{1}$$

for parameters *<sup>χ</sup>*, *<sup>k</sup>* > 0 with the Neumann boundary condition, where <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup><sup>2</sup> is a bounded domain with a smooth boundary. *u* and *v* are the cell density and concentration of chemical stimulus with respect to time *t* and *x*, respectively. *k* represents the diffusion rate of the chemical signal. The initial functions *<sup>u</sup>*<sup>0</sup> <sup>∈</sup> *<sup>C</sup>*0(Ω¯ ) and *<sup>v</sup>*<sup>0</sup> <sup>∈</sup> *<sup>W</sup>*1,∞(Ω) satisfy *<sup>u</sup>*<sup>0</sup> <sup>≥</sup> <sup>0</sup> and *<sup>v</sup>*<sup>0</sup> <sup>&</sup>gt; 0.

In 1970, Keller and Segel [1] originally introduced the system

$$\begin{cases} \boldsymbol{u}\_{t} = \nabla \cdot (\nabla \boldsymbol{u} - \boldsymbol{u} \chi(\boldsymbol{v}) \nabla \boldsymbol{v}), & \boldsymbol{x} \in \Omega, t > 0, \\\ \boldsymbol{\tau} \boldsymbol{v}\_{t} = \boldsymbol{k} \Delta \boldsymbol{v} - \boldsymbol{a} \boldsymbol{v} + \beta \boldsymbol{u}, & \boldsymbol{x} \in \Omega, t > 0, \end{cases} \tag{2}$$

to describe chemotaxis, the oriented movement of cells in response to the concentration of chemical signal produced by themselves and self-diffusion, where *<sup>τ</sup>*, *<sup>k</sup>*, *<sup>α</sup>*, *<sup>β</sup>* > 0 are parameters. The chemical signal experiences random diffusion and decay. Particular cases and derivatives of chemotaixs models have been developed extensively, such as the parabolic– elliptic case [2–5], the fully parabolic case [6–10] and other extensive versions [11–13]. Some studies have focused on the problem of whether the solution to the respective model undergoes a chemotactic collapse in the sense that the cell density becomes unbounded in finite or infinite time [3,6,7,12]. Given the initial conditions *<sup>u</sup>*<sup>0</sup> <sup>≥</sup> 0, *<sup>v</sup>*<sup>0</sup> <sup>&</sup>gt; 0 and the Neumann boundary conditions, others have concentrated on the aggregation effect of the chemotactic sensitivity *χ*(*v*).

**Citation:** Liu, J.; Tian, B.; Wang, D.; Tang, J.; Wu, Y. Global Boundedness in a Logarithmic Keller–Segel System. *Mathematics* **2023**, *11*, 2743. https:// doi.org/10.3390/math11122743

Academic Editors: Patricia J. Y. Wong, Takashi Suzuki and Hovik Matevossian

Received: 24 March 2023 Revised: 7 June 2023 Accepted: 12 June 2023 Published: 16 June 2023

**Copyright:** © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

If *χ*(*v*) = *χ* with *τ* = *k* = *α* = *β* = 1, Osaki and Yagi [14] showed the global boundedness of solutions to (2) for *n* = 1 and Nagai et al. [15] proved the results if <sup>Ω</sup> *<sup>u</sup>*<sup>0</sup> <sup>&</sup>lt; <sup>4</sup>*<sup>π</sup>* for *<sup>n</sup>* <sup>=</sup> 2. For *<sup>n</sup>* <sup>≥</sup> 3, if *u*0*<sup>L</sup> n* <sup>2</sup> (Ω) is small enough, there exist global weak solutions [16]. Another form of sensitivity function is

$$\chi(v) = \frac{\chi\_0}{(c + \alpha v)^k}$$

for *<sup>c</sup>*, *<sup>χ</sup>*<sup>0</sup> <sup>&</sup>gt; 0, *<sup>k</sup>* <sup>&</sup>gt; 1 and *<sup>α</sup>* <sup>&</sup>gt; 0, which is non-singular. In this case, the global existence is established for *k* = 2, *c* = 1 by [17] and for *k* = 1, *α* = 1 by [12]. Furthermore, if *χ*(*v*) = *<sup>χ</sup>*<sup>0</sup> *vk* for *<sup>k</sup>* <sup>&</sup>gt; 1, *<sup>χ</sup>*<sup>0</sup> <sup>&</sup>gt; 0, there exist global classical solutions to (2) [18].

The logarithmic sensitivity function *χ*(*v*) = *<sup>χ</sup> <sup>v</sup>* with *<sup>χ</sup>* <sup>&</sup>gt; 0 is commonly considered because it is in compliance with the Weber–Fechner law [19]. Taking this form with *τ* = *k* = *α* = *β* = 1, the chemotaxis model becomes the classical version:

$$\begin{cases} u\_t = \nabla \cdot (\nabla u - \chi \mu \frac{\nabla \upsilon}{v}), & \quad \mathfrak{x} \in \Omega, t > 0, \\ v\_t = \Delta v - v + u\_\prime & \quad \mathfrak{x} \in \Omega, t > 0, \\ \partial\_\upsilon u = \partial\_\upsilon v = 0, & \quad \mathfrak{x} \in \partial \Omega, t > 0, \\ u(\mathfrak{x}, 0) = u\_0, \quad v(\mathfrak{x}, 0) = v\_0, & \quad \mathfrak{x} \in \Omega. \end{cases} \tag{3}$$

Global bounded solutions to (3) are provided by Osaki and Yagi [14] in a one-dimensional case. As for *n* = 2, Lankeit [7] introduced an energy functional and proved that the solutions are uniform bounded in a convex domain with the range of *χ* extending to slightly more than one. Moreover, Winkler [20] proved that there exist global classical solutions if 0 < *<sup>χ</sup>* < <sup>2</sup> *<sup>n</sup>* , and Fujie [6] showed the solutions are uniformly time bounded. In [21], global bounded solutions are constructed under the the condition of *<sup>χ</sup>* <sup>≤</sup> <sup>4</sup> *<sup>n</sup>* with <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup>*<sup>n</sup>* being the convex domain. Furthermore, (3) employs global weak solutions when *χ* < *<sup>n</sup>*+<sup>2</sup> <sup>3</sup>*n*−<sup>4</sup> [20]. In the radially symmetric setting, weak solutions are constructed by [22] under the condition *<sup>χ</sup>* < *<sup>n</sup> <sup>n</sup>*−<sup>2</sup> . These results imply that there is a balance between *<sup>χ</sup>* and dimension *n* for the establishment of global solutions to classic models (3). The work to extend both *χ* and *n* is laborious without giving any condition of (3). Lankeit and Winkler [23] extended the range of *χ* to

$$\chi < \begin{cases} \infty & \text{if } n = 2 \\ \sqrt{8} & \text{if } n = 3 \\ \frac{n}{n-2} & \text{if } n \ge 4 \end{cases}$$

under the definition of the generalized solution, which is constructed on the basis of the global weak solution.

There are also other results established on the changing of parameters, referring to [9,24]. Indeed, the parameters in (2) have an impact on the aggregation of cell density. Xiangdong [25] constructed global solutions to (1) with *n* ≤ 8 under some conditions, where the relationship between *k* and *χ* is established. However, if *n* = 2, the diffusion rate of the concentration of chemicals *k* does not work, since *χ* is still less than one, as in [25].

In [26], the estimates containing <sup>Ω</sup> |∇*v*| <sup>2</sup> are established to study the system where the chemotactic sensitivity is a constant and the source of the signal is modeled by *v*. In the work of Winkler [27], the only evident global quasi-dissipative structure involving Ω |∇*v*| 2 *<sup>v</sup>*<sup>2</sup> ,(*<sup>v</sup>* <sup>&</sup>gt; <sup>0</sup>) is established to address the difficulty brought about by the nonlinear source of signal. However, the system with logarithmic sensitivity presents more challenges, and the structure of <sup>Ω</sup> *f*(*v*)|∇*v*| *<sup>n</sup>* (*n* is even) is essential to the estimates. Hence, motivated by Lankeit [7] and Nagai [15], we establish an energy-type functional containing

$$\int\_{\Omega} \frac{|\nabla v|^4}{v^2}.$$

The fractional term of *v* in the energy-type functional may alleviate the difficulty of preventing the aggregation caused by nonlinear kinetics in some derivate systems such as [27,28], where the source of the signal is modeled by *uv*.

In this paper, the global existence and uniform boundedness of the classical solutions of (1) are established as follows:

**Theorem 1.** *Let* <sup>Ω</sup> *be a bounded domain with a smooth boundary <sup>∂</sup>*<sup>Ω</sup> *on* <sup>R</sup>2*, initial data <sup>v</sup>*<sup>0</sup> <sup>&</sup>gt; <sup>0</sup> *and <sup>u</sup>*<sup>0</sup> <sup>≥</sup>, ≡ <sup>0</sup> *in* <sup>Ω</sup> *with <sup>u</sup>*<sup>0</sup> <sup>∈</sup> *<sup>C</sup>*0(Ω¯ ) *and <sup>v</sup>*<sup>0</sup> <sup>∈</sup> *<sup>W</sup>*1,∞(Ω)*. For all <sup>χ</sup>* <sup>&</sup>gt; <sup>0</sup>*, there exists a constant Ck that depends on u*0, *v*0, Ω *and χ, such that whenever*

$$k \ge \mathbf{C}\_{k\nu}$$

*then (1) admits a unique classical solution* (*u*, *<sup>v</sup>*) <sup>∈</sup> *<sup>C</sup>*0(Ω¯ <sup>×</sup> [0, <sup>∞</sup>)) : *<sup>C</sup>*2,1(Ω¯ <sup>×</sup> (0, <sup>∞</sup>))*. Moreover, there exist constants <sup>δ</sup>*, <sup>C</sup> > <sup>0</sup> *such that <sup>δ</sup>* <sup>≤</sup> *<sup>v</sup>* < <sup>C</sup> *and* <sup>0</sup> <sup>≤</sup> *<sup>u</sup>* < <sup>C</sup> *for all t* <sup>∈</sup> (0, <sup>∞</sup>)*.*

Intuitively, this shows that the large diffusion rate of chemical signals can prevent the aggregation of cell density resulting from a large *χ*.

In the paper, we first demonstrate the local existence of and recall some inequalities in the preliminaries. Then, we prove our key integral inequality in the Section 3 and give some useful a priori estimates in the Section 4. Finally, we prove the uniform boundedness of the solutions.

#### **2. Preliminaries**

*2.1. Local Existence*

The local existence of classical solutions to chemotaxis systems has been well-established using the methods of standard parabolic regularity theory and an appropriate fixed-point framework, which is shown in the following. Details of proof can be seen in Theorem 2.1 of [7] or [20].

**Proposition 1.** *Let* <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup>*<sup>n</sup> be a bounded domain with a smooth boundary, and <sup>u</sup>*<sup>0</sup> <sup>∈</sup> *<sup>C</sup>*0(Ω¯ ) *and <sup>v</sup>*<sup>0</sup> <sup>∈</sup> *<sup>W</sup>*1,*q*(Ω), *<sup>q</sup>* <sup>&</sup>gt; *<sup>n</sup>* <sup>≥</sup> <sup>1</sup> *are non-negative; then, for any <sup>k</sup>*, *<sup>χ</sup>* <sup>&</sup>gt; <sup>0</sup>*, there exists <sup>T</sup>*max <sup>∈</sup> (0, <sup>∞</sup>] *and a pair of unique non-negative solutions satisfying*

$$\left\{ \begin{array}{c} u \in \mathcal{C}^{0}(\boldsymbol{\Omega} \times [0, T\_{\text{max}})) \cap \mathbb{C}^{2,1}(\boldsymbol{\Omega} \times (0, T\_{\text{max}})), \\ v \in \mathcal{C}^{0}(\boldsymbol{\Omega} \times [0, T\_{\text{max}})) \cap \mathbb{C}^{2,1}(\boldsymbol{\Omega} \times (0, T\_{\text{max}})) \cap L\_{\text{loc}}^{\infty}([0, T\_{\text{max}}); W^{1,q}(\boldsymbol{\Omega})), \end{array} \right\}$$

*such that* (*u*, *<sup>v</sup>*) *solves (1) classically in* <sup>Ω</sup> <sup>×</sup> [0, *<sup>T</sup>*max) *and, moreover, if <sup>T</sup>*max <sup>&</sup>lt; <sup>∞</sup>*, then* lim *t*→*T*max *u*(·, *t*)*L*∞(Ω) + *v*(·, *t*)*W*1,*q*(Ω) = ∞*.*

#### *2.2. The Positive Lower Boundedness of v*

In order to prove the lower boundedness of *v* in (1), we first prove the boundedness of *uL*<sup>1</sup> and *vL*<sup>1</sup> . Integrating the first and the second PDE in (1), we have the mass identities

$$\int\_{\Omega} u = \int\_{\Omega} u\_0 =: m, \ t > 0$$

and

$$\int\_{\Omega} v = \int\_{\Omega} \mu\_0 + \left(\int\_{\Omega} v\_0 - \int\_{\Omega} \mu\_0\right) \cdot \varepsilon^{-t}, \ t > 0.1$$

Based on these facts, one can deduce the non-negative lower boundedness of *v* from the abstract representation formula of the *v* equation. Copying Lemma 2.2 of [7], we write it as follows:

**Lemma 1.** *Let* (*u*, *<sup>v</sup>*) *satisfy Proposition 1; then, there exists <sup>T</sup>*max <sup>&</sup>gt; <sup>0</sup> *and a positive constant <sup>δ</sup> depending on v*<sup>0</sup> *such that*

$$w(\mathbf{x}, t) \ge \delta > 0, \forall (\mathbf{x}, t) \in \bar{\Omega} \times [0, T\_{\text{max}}). \tag{4}$$

**Proof.** Firstly, by the comparison principle and the fact of *<sup>v</sup>*<sup>0</sup> <sup>&</sup>gt; 0 on <sup>Ω</sup>¯ , we have for a small *<sup>t</sup>*

$$v(x,t) \ge \min\_{x \in \Omega} v\_0 \cdot e^{-t} > 0.$$

Let us fix *τ* = *τ*(*u*0, *v*0). Then, it follows that

$$
v(x,t) \ge \min\_{x \in \Omega} v\_0 \cdot e^{-\tau} := \delta\_1 > 0, \forall t \in [0,\tau).
$$

Now, from the well-known Neumann heat semigroup estimate for *et*<sup>Δ</sup> (see Lemma 1.3 in [29] and Lemma 2.2 in [20]), we denote by *<sup>d</sup>* the diameter of the <sup>Ω</sup> and have for <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup><sup>2</sup> that

$$\ell(e^{t\Delta}\omega) \ge \frac{1}{4\pi t} e^{-\frac{\zeta^2}{4t}} \cdot \int\_{\Omega} \omega > 0, \,\omega \in \mathbb{C}^0(\Omega).$$

Then, the abstract representation formula of *v* shows

$$\begin{split} v(\cdot, t) &= e^{t(\Delta - 1)} v\_0 + \int\_{\Omega} e^{(t - s)(\Delta - 1)} u(\cdot, t) ds \\ &\ge \int\_0^t \frac{1}{4\pi (t - s)} e^{-\left((t - s) + \frac{\theta^2}{4(t - s)}\right)} (\int\_{\Omega} u(\cdot, t)) ds \\ &\ge m \int\_0^t \frac{1}{4\pi r} e^{-\left(r + \frac{\theta^2}{4r}\right)} dr := \delta\_2 > 0, \forall t \in [\pi, \infty), \end{split} \tag{5}$$

where *r* := *t* − *s*. Choosing *δ* = min{*δ*1, *δ*2}, we deduce (4).

#### *2.3. Recall of Useful Theorems*

The well-known general Young's inequality [30] is recalled.

**Lemma 2.** *Let f* , *<sup>g</sup>* <sup>≥</sup> <sup>0</sup> *be the continuous function with p*, *<sup>q</sup>* > <sup>0</sup> *satisfying* <sup>1</sup> *<sup>p</sup>* <sup>+</sup> <sup>1</sup> *<sup>q</sup>* = 1*, then*

$$
\epsilon f \mathfrak{g} \le \epsilon f^p + \frac{1}{q} (\epsilon p)^{-\frac{q}{p}} \mathfrak{g}^q \tag{6}
$$

*holds for all* <sup>&</sup>gt; <sup>0</sup>*. Moreover, for continuous <sup>h</sup>* <sup>&</sup>gt; <sup>0</sup> *and any* 1, <sup>2</sup> <sup>&</sup>gt; <sup>0</sup>*, taking <sup>p</sup>* <sup>=</sup> 2, *<sup>q</sup>* <sup>=</sup> 3,*<sup>r</sup>* <sup>=</sup> <sup>6</sup> *such that* <sup>1</sup> *<sup>p</sup>* <sup>+</sup> <sup>1</sup> *<sup>q</sup>* <sup>+</sup> <sup>1</sup> *<sup>r</sup>* = 1*, we have*

$$fgh \le \epsilon\_1 f^2 + \frac{\epsilon\_2}{4\epsilon\_1} g^3 + \frac{\sqrt{6}}{36\epsilon\_1 \sqrt{\epsilon\_2}} h^6. \tag{7}$$

**Proof.** In (7) is given the result of the straightforward calculation of the well-known inequality (6).

**Lemma 3.** *Let* <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup>*n, n* <sup>≥</sup> <sup>1</sup> *be a smooth bounded domain. Any function f* <sup>∈</sup> *<sup>C</sup>*2(Ω) *satisfies*

$$i.\ \nabla|\nabla f|^2 = 2\nabla f \cdot \mathcal{D}^2 f,\tag{8}$$

$$\text{a.ii.}\ (\Delta f)^2 \le n|D^2f|^2,\tag{9}$$

$$
\langle \vec{u} \vec{v}, \nabla f \cdot \nabla \Delta f = \frac{1}{2} \Delta |\nabla f|^2 - |D^2 f|^2. \tag{10}
$$

All the identities and inequalities in the above lemma can be obtained from straightforward calculation. One can see [7,31] and Lemma 3.1 in [8] for their application. We could not find a precise reference in the literature that covers all that is necessary for our purpose; therefore, we conclude with a short lemma here.

#### **3. A User-Friendly Integral Inequality**

The proof of Theorem 1 is based on the extension and application of an integral inequality, which is generated within one dimension by Q. Wang [28]. The following theorem has a multidimensional form. It is worth noting that the integral inequality connects the fraction of the gradient and the second derivative. A similar inequality can be found in [7]. Furthermore, the explicit coefficient in the integral inequality is easy to use for readers.

**Theorem 2.** *Let* <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup>*<sup>n</sup> be a smooth bounded domain with <sup>w</sup>* > <sup>0</sup> *satisfying <sup>w</sup>* <sup>∈</sup> *<sup>C</sup>*2(Ω¯ ) *and ∂w ∂ν* = 0 *on ∂*Ω*. Then,*

$$\int\_{\Omega} \frac{|\nabla w|^{2p+2}}{w^{q+2}} \le \frac{n+4p\epsilon}{2q+1-\frac{p}{\epsilon}} \int\_{\Omega} \frac{|D^2 w|^2 |\nabla w|^{2p-2}}{w^q} \tag{11}$$

*for all p* <sup>≥</sup> <sup>1</sup>*, q* > <sup>−</sup><sup>1</sup> <sup>2</sup> *and* <sup>&</sup>gt; *<sup>p</sup>* <sup>2</sup>*q*+<sup>1</sup> <sup>&</sup>gt; <sup>0</sup>*.*

**Proof.** Let *J* := <sup>Ω</sup> |Δ log *w*| 2 |∇*w*| 2*p*−2 *<sup>w</sup>q*−<sup>2</sup> <sup>&</sup>gt; 0 for *<sup>p</sup>* <sup>≥</sup> 1. Directly calculating <sup>|</sup><sup>Δ</sup> log *<sup>w</sup>*<sup>|</sup> <sup>2</sup> leads to

$$J = \int\_{\Omega} \frac{|\Delta w|^2 |\nabla w|^{2p-2}}{w^q} \widetilde{-2 \int\_{\Omega} \frac{|\nabla w|^{2p} \Delta w}{w^{q+1}}} + \int\_{\Omega} \frac{|\nabla w|^{2p+2}}{w^{q+2}}.\tag{12}$$

Since *<sup>∂</sup><sup>w</sup> ∂ν* = 0 on *∂*Ω, integration by parts gives

$$\begin{split} J\_{0} &= 2 \int\_{\Omega} \frac{\nabla |\nabla w|^{2p} \cdot \nabla w}{w^{q+1}} - 2(q+1) \int\_{\Omega} \frac{|\nabla w|^{2p+2}}{w^{q+2}} \\ &= 2p \int\_{\Omega} \frac{|\nabla w|^{2p-2} \nabla |\nabla w|^{2} \cdot \nabla w}{w^{q+1}} - 2(q+1) \int\_{\Omega} \frac{|\nabla w|^{2p+2}}{w^{q+2}}. \end{split}$$

By (8) of Lemma 3 and (6), we have for <sup>&</sup>gt; 0 that

$$\begin{split} \mathcal{J}\_{0} &= 4p \int\_{\Omega} \frac{|\nabla w|^{2p} \cdot D^{2}w}{w^{q+1}} - 2(q+1) \int\_{\Omega} \frac{|\nabla w|^{2p+2}}{w^{q+2}} \\ &\leq 4p\epsilon \int\_{\Omega} \frac{|\nabla w|^{2p-2} |D^{2}w|^{2}}{w^{q}} - \left(2(q+1) - \frac{p}{\epsilon}\right) \int\_{\Omega} \frac{|\nabla w|^{2p+2}}{w^{q+2}}. \end{split} \tag{13}$$

By (9), substituting (13) into (12) gives

$$J \le (n + 4p\epsilon) \int\_{\Omega} \frac{|\nabla w|^{2p-2} |D^2 w|^2}{w^q} - \left( (2q + 1) - \frac{p}{\epsilon} \right) \int\_{\Omega} \frac{|\nabla w|^{2p+2}}{w^{q+2}}.$$

Due to *<sup>q</sup>* <sup>&</sup>gt; <sup>−</sup><sup>1</sup> <sup>2</sup> , <sup>&</sup>gt; *<sup>p</sup>* <sup>2</sup>*q*+<sup>1</sup> <sup>&</sup>gt; 0; thus, (2*<sup>q</sup>* <sup>+</sup> <sup>1</sup>) <sup>−</sup> *<sup>p</sup>* <sup>&</sup>gt; 0, and we conclude with (11). **Remark 1.** *Letting* <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup><sup>2</sup> *and taking <sup>q</sup>* <sup>=</sup> *<sup>p</sup>* <sup>=</sup> 2, > <sup>2</sup> <sup>5</sup> *, then <sup>n</sup>*+4*p* <sup>2</sup>*q*+1<sup>−</sup> *<sup>p</sup>* = <sup>2</sup>+8 <sup>5</sup><sup>−</sup> <sup>2</sup> *. Note that* <sup>2</sup>+8 <sup>5</sup><sup>−</sup> <sup>2</sup> *achieves its global minimum over* ( <sup>2</sup> <sup>5</sup> , <sup>∞</sup>) *at* <sup>=</sup> <sup>4</sup><sup>+</sup> <sup>√</sup><sup>26</sup> <sup>10</sup> (≈*0.9099) with the value* 2 21−4 <sup>√</sup><sup>26</sup> (≈*3.3117). Therefore,*

$$\int\_{\Omega} \frac{|\nabla w|^6}{w^4} \le \frac{2}{21 - 4\sqrt{26}} \int\_{\Omega} \frac{|D^2 w|^2 |\nabla w|^2}{w^2}. \tag{14}$$

#### **4. Some Useful A Priori Estimates**

Let us first give an inequality to estimate the boundary integration.

**Lemma 4.** *Let* <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup><sup>2</sup> *be a bounded smooth domain. If <sup>v</sup>* <sup>∈</sup> *<sup>C</sup>*2(Ω¯ ) *satisfies <sup>∂</sup><sup>v</sup> ∂ν* = 0*, the for any* <sup>ˆ</sup> > <sup>0</sup>*, there exists C*(ˆ) *depending on* <sup>Ω</sup> *such that*

$$\int\_{\partial\Omega} \frac{|\nabla v|^2}{v^2} \frac{\partial (|\nabla v|^2)}{\partial \nu} \le \hat{\epsilon} \int\_{\Omega} \frac{|\nabla v|^2 |D^2 v|^2}{v^2} + \mathcal{C}(\hat{\epsilon}) \tag{15}$$

*for all t* ∈ (0, *T*max) *and n* ≥ 1*.*

**Proof.** Firstly, we show that

$$\int\_{\partial\Omega} \frac{|\nabla v|^2}{v^2} \frac{\partial (|\nabla v|^2)}{\partial \nu} = 16 \int\_{\partial\Omega} |\nabla \sqrt{v}|^2 \frac{\partial (|\nabla \sqrt{v}|^2)}{\partial \nu}. \tag{16}$$

From the Neumann boundary condition, we calculate the right-hand side, obtaining

$$\begin{split} 16 \int\_{\partial\Omega} |\nabla\sqrt{v}|^2 \frac{\partial(|\nabla\sqrt{v}|^2)}{\partial v} &= \int\_{\partial\Omega} \frac{|\nabla v|^2}{v} \frac{\partial \frac{|\nabla v|^2}{v}}{\partial v} \\ = \int\_{\partial\Omega} \frac{|\nabla v|^2}{v} \frac{\frac{\partial |\nabla v|^2}{\partial v} v - \frac{\partial v}{\partial v} |\nabla v|^2}{v^2} &= \int\_{\partial\Omega} \frac{|\nabla v|^2}{v^2} \frac{\partial(|\nabla v|^2)}{\partial v} \end{split}$$

for all *t* ∈ (0, *T*max).

Now, according to (3.17) in [11], we have for any *<sup>ε</sup>* <sup>&</sup>gt; 0 and constant *<sup>C</sup><sup>ε</sup>* <sup>&</sup>gt; 0 depending on Ω that

$$\varepsilon \int\_{\partial \Omega} |\nabla \sqrt{v}|^2 \frac{\partial (|\nabla \sqrt{v}|^2)}{\partial v} \le \varepsilon \int\_{\Omega} \left| \nabla |\nabla \sqrt{v}|^2 \right|^2 + \mathbb{C}\_{\varepsilon} \quad \text{for all } t \in (0, T\_{\text{max}}). \tag{17}$$

By straightforward calculation, we have

$$\begin{split} \int\_{\Omega} \left| \nabla |\nabla \sqrt{v}|^2 \right|^2 &= \frac{1}{16} \int\_{\Omega} \left| \nabla \left( \frac{|\nabla v|^2}{v} \right) \right|^2 = \frac{1}{16} \int\_{\Omega} \left| \frac{\nabla (|\nabla v|^2)}{v} - \frac{\nabla v |\nabla v|^2}{v^2} \right|^2 \\ &= \frac{1}{16} \int\_{\Omega} \left( \frac{|\nabla |\nabla v|^2|^2}{v^2} - 2 \frac{\nabla |\nabla v|^2 \cdot \nabla v |\nabla v|^2}{v^3} + \frac{|\nabla v|^6}{v^4} \right) \\ &\leq \frac{1}{16} \int\_{\Omega} \left( (2\varepsilon\_1 + 1) \frac{\left| \nabla |\nabla v|^2 \right|^2}{v^2} + (\frac{1}{2\varepsilon\_1} + 1) \frac{|\nabla v|^6}{v^4} \right) \end{split}$$

for *<sup>ε</sup>*<sup>1</sup> <sup>&</sup>gt; 0. Then, we have from (14) that

$$\int\_{\Omega} \left| \nabla |\nabla \sqrt{v}|^2 \right|^2 \leq \frac{1}{16} \left( 2\varepsilon\_1 + 1 + \frac{\varepsilon}{8\varepsilon\_1} + \frac{\varepsilon}{4} \right) \int\_{\Omega} \frac{\left| \nabla |\nabla v|^2 \right|^2}{v^2} \tag{18}$$

for all *<sup>t</sup>* <sup>∈</sup> (0, *<sup>T</sup>*max), where ˜ <sup>=</sup> <sup>2</sup> 21−4 <sup>√</sup><sup>26</sup> for simplicity. Combining (18) and (17) with (16), we can obtain that

$$\int\_{\partial\Omega} \frac{|\nabla v|^2}{v^2} \frac{\partial (|\nabla v|^2)}{\partial \nu} \le \varepsilon \left(2\varepsilon\_1 + 1 + \frac{\tilde{\epsilon}}{8\varepsilon\_1} + \frac{\tilde{\epsilon}}{4}\right) \int\_{\Omega} \frac{\left|\nabla |\nabla v|^2\right|^2}{v^2} + 16\mathbb{C}\_{\varepsilon}.$$

Denoting ˆ = *ε*(2*ε*<sup>1</sup> + 1 + ˜ <sup>8</sup>*ε*<sup>1</sup> <sup>+</sup> ˜ <sup>4</sup> ) and *<sup>C</sup>*(ˆ) = <sup>16</sup>*Cε*, we prove (15) for any <sup>ˆ</sup> <sup>&</sup>gt; 0.

In preparation for the construction and estimation of energy-type functionals, some important *a priori* estimates are provided and collected into two lemmas in the following.

**Lemma 5.** *Let <sup>k</sup>* > <sup>0</sup> *and* (*u*, *<sup>v</sup>*) *be the solutions of (1) satisfying Proposition 1. Then, we have for any* <sup>ˆ</sup> > <sup>0</sup> *that*

$$\begin{split} \frac{d}{dt} \int\_{\Omega} \frac{|\nabla v|^4}{v^2} + \int\_{\Omega} \frac{|\nabla v|^4}{v^2} \le -\left(\frac{4k}{3} - 2k\mathfrak{k}\right) \int\_{\Omega} \frac{|\nabla v|^2 |D^2 v|^2}{v^2} - 2 \int\_{\Omega} \frac{|\nabla v|^4 u}{v^3} \\ + 4 \int\_{\Omega} \frac{|\nabla v|^2 \nabla v \cdot \nabla u}{v^2} + \mathcal{C}(\mathfrak{k}). \end{split} \tag{19}$$

**Proof.** Through straightforward calculation, we can show

$$\begin{split} \frac{d}{dt} \int\_{\Omega} \frac{|\nabla v|^4}{v^2} = 4 \int\_{\Omega} \frac{|\nabla v|^2 \nabla v \cdot \nabla v\_t}{v^2} - 2 \int\_{\Omega} \frac{|\nabla v|^4 v\_t}{v^3} \\ = 4k \int\_{\Omega} \underbrace{\frac{|\nabla v|^2 \nabla v \cdot \nabla \Delta v}{v^2}}\_{-2k \int\_{\Omega} \frac{l\_2}{v^2}} - 2 \int\_{\Omega} \underbrace{\frac{|\nabla v|^4}{v^2} + 4 \int\_{\Omega} \frac{|\nabla v|^2 \nabla v \nabla u}{v^2}}\_{-2k \int\_{\Omega} \frac{|\nabla v|^4 \Delta v}{v^3}} \\ - 2k \int\_{\Omega} \frac{|\nabla v|^4 \Delta v}{v^3} - 2 \int\_{\Omega} \frac{|\nabla v|^4 u}{v^3} .\end{split} \tag{20}$$

In light of (10), we have from (15) that

$$\begin{split} I\_{1} &= 2k \int\_{\Omega} \frac{|\nabla v|^{2} \Delta |\nabla v|^{2}}{v^{2}} - 4k \int\_{\Omega} \frac{|\nabla v|^{2} |D^{2}v|^{2}}{v^{2}} \\ &= 2k \int\_{\partial\Omega} \frac{|\nabla v|^{2}}{v^{2}} \frac{\partial (|\nabla v|^{2})}{\partial v} - 2k \int\_{\Omega} \nabla (\frac{|\nabla v|^{2}}{v^{2}}) \cdot \nabla |\nabla v|^{2} - 4k \int\_{\Omega} \frac{|\nabla v|^{2} |D^{2}v|^{2}}{v^{2}} \\ &= 2k \int\_{\partial\Omega} \frac{|\nabla v|^{2}}{v^{2}} \frac{\partial (|\nabla v|^{2})}{\partial v} - 2k \int\_{\Omega} \frac{(\nabla |\nabla v|^{2})^{2}}{v^{2}} \\ &\quad + 4k \int\_{\Omega} \frac{|\nabla v|^{2} \nabla v \cdot \nabla |\nabla v|^{2}}{v^{3}} - 4k \int\_{\Omega} \frac{|\nabla v|^{2} |D^{2}v|^{2}}{v^{2}} \\ &\leq - (12k - 2k\mathfrak{\epsilon}) \int\_{\Omega} \frac{|\nabla v|^{2} |D^{2}v|^{2}}{v^{2}} + 4k \int\_{\Omega} \frac{\nabla v |\nabla v|^{2} \cdot \nabla |\nabla v|^{2}}{v^{3}} + \mathcal{C}(\mathfrak{\epsilon}). \end{split} \tag{21}$$

Similarly, we calculate that

$$I\_2 = 2k \int\_{\Omega} \nabla (\frac{|\nabla v|^4}{v^3}) \nabla v = 4k \int\_{\Omega} \frac{|\nabla v|^2 \nabla v \cdot \nabla |\nabla v|^2}{v^3} - 6k \int\_{\Omega} \frac{|\nabla v|^6}{v^4}.$$

Given by the sum of *I*<sup>2</sup> and *I*<sup>3</sup> and taking = <sup>1</sup> <sup>3</sup> , (6) implies that

$$\begin{split} I\_2 + I\_3 &= 8k \int\_{\Omega} \frac{|\nabla v|^2 \nabla v \cdot \nabla |\nabla v|^2}{v^3} - 6k \int\_{\Omega} \frac{|\nabla v|^6}{v^4} \\ &\le 8k\epsilon \int\_{\Omega} \frac{(\nabla |\nabla v|^2)^2}{v^2} + (\frac{2k}{\epsilon} - 6k) \int\_{\Omega} \frac{|\nabla v|^6}{v^4} \\ &= \frac{32k}{3} \int\_{\Omega} \frac{|\nabla v|^2 |D^2 v|^2}{v^2} .\end{split} \tag{22}$$

Substituting (22) and (21) into (20), we finish the proof by taking the first identity of Lemma 3.

**Lemma 6.** *Supposing that* (*u*, *v*) *solves (1) and all conditions of Proposition 1 hold, then there exist small* 1, <sup>2</sup> <sup>&</sup>gt; <sup>0</sup> *and <sup>δ</sup>* <sup>&</sup>gt; <sup>0</sup> *such that*

$$\frac{1}{2}\frac{d}{dt}\int\_{\Omega}u^2 \le -(1-\chi\epsilon\_1\delta^\*)\int\_{\Omega}|\nabla u|^2 + \frac{\chi\epsilon\_2}{4\epsilon\_1}\int\_{\Omega}u^3 + \frac{\chi\sqrt{6}}{36\epsilon\_1\sqrt{\epsilon\_2}}\int\_{\Omega}\frac{|\nabla v|^6}{v^4}.\tag{23}$$

**Proof.** In light of the *u* equation of (1) and integration by parts, we can show that

$$\frac{1}{2}\frac{d}{dt}\int\_{\Omega}u^2 = \int\_{\Omega}u\nabla \cdot (\nabla u - \chi\frac{\nabla v}{v}u) = -\int\_{\Omega}|\nabla u|^2 + \int\_{\Omega}\chi u \frac{\nabla u \cdot \nabla v}{v}.\tag{24}$$

The employment of (7) implies

$$\int\_{\Omega} \chi u \frac{\nabla u \cdot \nabla v}{v} \leq \chi \epsilon\_1 \int\_{\Omega} \frac{|\nabla u|^2}{v^{\frac{2}{5}}} + \frac{\chi \epsilon\_2}{4\epsilon\_1} \int\_{\Omega} u^3 + \frac{\chi \sqrt{6}}{36\epsilon\_1 \sqrt{\epsilon\_2}} \int\_{\Omega} \frac{|\nabla v|^6}{v^4} \tag{25}$$

for small 1, <sup>2</sup> <sup>&</sup>gt; 0. Note that *<sup>v</sup>* has the lower bound for any *<sup>t</sup>* <sup>&</sup>gt; 0. Let *<sup>δ</sup>*<sup>∗</sup> :<sup>=</sup> *<sup>δ</sup>*<sup>−</sup> <sup>2</sup> <sup>3</sup> be the upper bound of *v*<sup>−</sup> <sup>2</sup> <sup>3</sup> and substitute (25) into (24) to obtain (23).

#### **5. Uniform Boundedness**

In this section, we shall finish the proof of Theorem 1. Firstly, we construct the energy functional and prove that each item of the functional is uniform bounded.

**Theorem 3.** *For <sup>α</sup>* > <sup>0</sup>*, let* <sup>F</sup>*α*(*u*, *<sup>v</sup>*) *take the following form:*

$$\mathcal{F}\_{\mathfrak{a}}(\mathfrak{u}, \mathfrak{v}) = \mathfrak{a} \int\_{\Omega} \mathfrak{u}^2 + \int\_{\Omega} \frac{|\nabla \mathfrak{v}|^4}{\mathfrak{v}^2}.$$

*Then, for* <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup><sup>2</sup> *and any <sup>χ</sup>* > <sup>0</sup>*, there exists a constant Ck*(*u*0, *<sup>v</sup>*0, <sup>Ω</sup>, *<sup>χ</sup>*) > <sup>0</sup> *such that if <sup>k</sup>* > *Ck*(*u*0, *<sup>v</sup>*0, <sup>Ω</sup>, *<sup>χ</sup>*)*, then for some C* > <sup>0</sup>

$$\frac{d}{dt}\mathcal{F}\_{\mathfrak{a}}(\mathfrak{u},\mathfrak{v}) + \mathcal{F}\_{\mathfrak{a}}(\mathfrak{u},\mathfrak{v}) < \mathbb{C} \text{ for all } \mathfrak{t} \in (0, T\_{\text{max}}).\tag{26}$$

**Proof.** Combining (19) and (23), we achieve

$$\begin{split} &\frac{d}{dt}\Big(a\int\_{\Omega}u^{2}+\int\_{\Omega}\frac{|\nabla v|^{4}}{v^{2}}\Big)+\left(a\int\_{\Omega}u^{2}+\int\_{\Omega}\frac{|\nabla v|^{4}}{v^{2}}\right) \\ &\leq -\left(2a-2a\chi\epsilon\_{1}\delta^{\*}\right)\int\_{\Omega}|\nabla u|^{2}+\dfrac{\frac{\imath\_{1}}{8\varepsilon\_{1}\varepsilon\_{2}}\int\_{\Omega}u^{3}}{2\epsilon\_{1}}+\dfrac{\frac{a\chi\sqrt{6}}{18\varepsilon\_{1}\sqrt{\varepsilon\_{2}}}\int\_{\Omega}\frac{|\nabla v|^{6}}{v^{4}}}{\int\_{\Omega}u^{3}} \\ &-\left(\frac{4k}{3}-2k\varepsilon\right)\int\_{\Omega}\frac{|\nabla v|^{2}|D^{2}v|^{2}}{v^{2}}-2\int\_{\Omega}\frac{|\nabla v|^{4}u}{v^{3}}+\underbrace{4\int\_{\Omega}\frac{|\nabla v|^{2}\nabla v\cdot\nabla u}{v^{2}}}\_{\in} \\ &+a\int\_{\Omega}u^{2}+\mathcal{C}(\mathfrak{k}). \end{split} \tag{27}$$

The Gagliardo–Nirenberg inequality and the boundedness of *uL*1(Ω) imply that there exists *<sup>C</sup>* > 0 depending on *u*0*L*1(Ω), <sup>Ω</sup> such that

$$\int\_{\Omega} \boldsymbol{u}^2 \leq \eta \int\_{\Omega} |\nabla \boldsymbol{u}|^2 + C$$

for some small *<sup>η</sup>* > 0 and

$$I\_1 = \frac{a\chi\epsilon\_2}{2\epsilon\_1} \|u\|\_{L^3(\Omega)}^3 \le \frac{a\chi\epsilon\_2}{2\epsilon\_1} (\mathbb{C}\_1 \|\nabla u\|\_{L^2(\Omega)}^2 + \mathbb{C}\_2),\tag{28}$$

where *<sup>C</sup>*1, *<sup>C</sup>*<sup>2</sup> <sup>&</sup>gt; 0, depending on *u*0*L*1(Ω) and <sup>Ω</sup>. For *<sup>I</sup>*3, we employ Lemma 2 to obtain

$$I\_3 \le \frac{1}{\epsilon\_3} \int\_{\Omega} \frac{|\nabla v|^6}{v^4} + 4\epsilon\_3 \int\_{\Omega} |\nabla u|^2 \tag{29}$$

for any <sup>3</sup> <sup>&</sup>gt; 0. Combining the first item of (29) with *<sup>I</sup>*<sup>2</sup> and employing (14), we have

$$(\frac{a\chi\sqrt{6}}{18\epsilon\_1\sqrt{\epsilon\_2}} + \frac{1}{\epsilon\_3})\int\_{\Omega} \frac{|\nabla v|^6}{v^4} \le (\frac{a\chi\sqrt{6}}{18\epsilon\_1\sqrt{\epsilon\_2}} + \frac{1}{\epsilon\_3})\varepsilon \int\_{\Omega} \frac{|\nabla v|^2 |D^2 v|^2}{v^2},\tag{30}$$

where we denote ˜ = <sup>2</sup> 21−4 <sup>√</sup><sup>26</sup> for simplicity. Thus, substituting (28)–(30) into (27) gives

$$\begin{split} \frac{d}{dt} (\boldsymbol{a} \int\_{\Omega} \boldsymbol{u}^{2} + \int\_{\Omega} \frac{|\nabla \boldsymbol{v}|^{4}}{\boldsymbol{v}^{2}} ) &+ (\boldsymbol{a} \int\_{\Omega} \boldsymbol{u}^{2} + \int\_{\Omega} \frac{|\nabla \boldsymbol{v}|^{4}}{\boldsymbol{v}^{2}}) \\ \leq &- \overbrace{(2\boldsymbol{a} (1 - \chi \boldsymbol{\epsilon}\_{1} \boldsymbol{\delta}^{\*} - \frac{\chi \boldsymbol{\epsilon}\_{2}}{4\boldsymbol{\epsilon}\_{1}} \mathbf{C}\_{1} - \eta) - 4\boldsymbol{\epsilon}\_{3}}^{\kappa\_{2}} \int\_{\Omega} |\nabla \boldsymbol{u}|^{2} \\ \quad - \overbrace{(\frac{4\boldsymbol{k}}{3} - 2\boldsymbol{k}\hat{\mathbf{\epsilon}} - \frac{\boldsymbol{a} \chi \sqrt{\boldsymbol{\delta}}}{18\boldsymbol{\epsilon}\_{1} \sqrt{\boldsymbol{\epsilon}\_{2}}} \tilde{\boldsymbol{\epsilon}} - \frac{1}{\boldsymbol{\epsilon}\_{3}} \tilde{\boldsymbol{\epsilon}}}^{\kappa\_{3}} \int\_{\Omega} \frac{|\nabla \boldsymbol{v}|^{2} |D^{2} \boldsymbol{v}|^{2}}{\boldsymbol{v}^{2}} + \mathrm{C}. \end{split} \tag{31}$$

Let 1, <sup>2</sup> and <sup>3</sup> be small, such that *κ*<sup>1</sup> = 0. Then, taking a small ˆ such that <sup>2</sup> <sup>3</sup> <sup>&</sup>gt; ˆ, we denote

$$\mathbb{C}\_{\left(\mathfrak{c}\_{1},\mathfrak{c}\_{2},\mathfrak{c}\_{3}\right)} := \frac{a\chi\sqrt{6}}{6\varepsilon\_{1}\sqrt{\varepsilon\_{2}}(4-6\mathfrak{c})}\mathfrak{c} + \frac{3}{\varepsilon\_{3}(4-6\mathfrak{c})}\mathfrak{c} > 0,$$

and let *Ck* depending on *u*0, *v*0, Ω, *χ* be the lower bound of *C*(1,2,3) provided *κ*<sup>1</sup> = 0. Therefore, for any *<sup>k</sup>* <sup>≥</sup> *Ck* <sup>&</sup>gt; 0, we have *<sup>κ</sup>*<sup>2</sup> <sup>≥</sup> 0 and can then deduce (26).

**Theorem 4.** *Let (u,v) be the solutions of (1) satisfying all conditions in Proposition 1. Then,*

$$\int\_{\Omega} u^2(\cdot, t) \le \mathbb{C} \quad \text{and} \quad \int\_{\Omega} |\nabla v(\cdot, t)|^2 \le \mathbb{C},\tag{32}$$

*with t* ∈ (0, *T*max)*.*

**Proof.** According to (26), there is *<sup>C</sup>* > 0 such that

$$\int\_{\Omega} u^2(\cdot, t) \le \mathbb{C} \quad \text{and} \quad \int\_{\Omega} \frac{|\nabla v|^4}{v^2}(\cdot, t) \le \mathbb{C}.$$

for all *t* ∈ (0, *T*max). From Young's inequality and the Gagliardo–Nirenberg inequality, there exist 4, *GN* <sup>&</sup>gt; 0 and *<sup>C</sup>* <sup>&</sup>gt; 0 such that

$$\begin{split} \int\_{\Omega} |\nabla v(\cdot, t)|^2 &\leq \mathbb{C}\_{\mathfrak{c}\_4} \int\_{\Omega} \frac{|\nabla v|^4}{v^2}(\cdot, t) + \mathfrak{e}\_4 \int\_{\Omega} v^2(\cdot, t) \\ &\leq \mathbb{C}\_{\mathfrak{c}\_4} \int\_{\Omega} \frac{|\nabla v|^4}{v^2}(\cdot, t) + \mathfrak{e}\_4 \mathfrak{e}\_{GN} \int\_{\Omega} \left| \nabla v(\cdot, t) \right|^2 + \mathbb{C} \end{split} \tag{33}$$

for all *<sup>t</sup>* <sup>∈</sup> (0, *<sup>T</sup>*max). Taking <sup>4</sup> <sup>&</sup>lt; <sup>1</sup> <sup>2</sup>*GN* , then we have 4*GN* <sup>&</sup>lt; <sup>1</sup> <sup>2</sup> and prove (32).

**Proof of Theorem 1.** Using the well-known Moser's technique [32], the *L*<sup>∞</sup> boundedness of *u* follows from Theorem 4. Indeed, one can follow the estimates of Nagai [15] or directly employ Lemma 2.3 in [7] to prove the theorem.

#### **6. Conclusions**

Our paper proves the uniform boundedness of solutions of the chemotaxis system with singular sensitivity under a small diffusion rate of the chemical signal. We prove a user-friendly inequality that has certain parameters, and construct a new energy functional that is applicable to the double Keller–Segel model with nonlinear sources.

**Author Contributions:** Conceptualization, J.L., B.T. and D.W.; Methodology, D.W.; Validation, J.L., B.T. and J.T.; Formal analysis, J.L., J.T. and Y.W.; Writing—original draft, B.T. and D.W.; Writing review & editing, J.L., J.T. and Y.W. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Data Availability Statement:** Not applicable.

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

#### **References**


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## *Article* **Lump-Type Solutions, Lump Solutions, and Mixed Rogue Waves for Coupled Nonlinear Generalized Zakharov Equations**

**Aly R. Seadawy 1,\*, Syed T. R. Rizvi <sup>2</sup> and Hanadi Zahed <sup>1</sup>**

<sup>2</sup> Department of Mathematics, COMSATS University Islamabad, Lahore Campus, Islamabad 45550, Pakistan

**\*** Correspondence: aabdelalim@taibahu.edu.sa

**Abstract:** This article studies diverse forms of lump-type solutions for coupled nonlinear generalized Zakharov equations (CNL-GZEs) in plasma physics through an appropriate transformation approach and bilinear equations. By utilizing the positive quadratic assumption in the bilinear equation, the lump-type solutions are derived. Similarly, by employing a single exponential transformation in the bilinear equation, the lump one-soliton solutions are derived. Furthermore, by choosing the double exponential ansatz in the bilinear equation, the lump two-soliton solutions are found. Interaction behaviors are observed and we also establish a few new solutions in various dimensions (3D and contour). Furthermore, we compute rogue-wave solutions and lump periodic solutions by employing proper hyperbolic and trigonometric functions.

**Keywords:** CNL-GZE; lump-type solitons; rogue wave; appropriate transformation technique

**MSC:** 35J05; 35J10; 35K05; 35L05

#### **1. Introduction**

The study of partial differential equations (PDEs) occurs in various fields such as theoretical physics, applied mathematics, biological sciences, and engineering sciences. These PDEs play a crucial role in explaining key scientific phenomena. For instance, the Korteweg–de Vries equation governs shallow water wave dynamics near ocean shores and beaches, and the nonlinear Schrödinger's equation governs the propagation of solitons through optical fibers. Some examples of PDEs and their applications can be found in [1–8].

Although the above-mentioned PDEs are scalar, a large number of PDEs are coupled. Some of them are two-coupled PDEs such as the Gear–Grimshaw equation, whereas others are three-coupled PDEs. An example of a three-coupled PDE is the Wu–Zhang equation. These coupled PDEs are also calculated in distinct areas of theoretical physics. In this paper, we will study CNL-GZE used in plasmas.

Lump waves (LWs), as superior nonlinear wave phenomena, have been visualized in various fields. LWs are theoretically viewed as a limited type of soliton and move with higher propagating energy compared to general solitons. Consequently, LWs can be destructive and even catastrophic in certain systems, such as in the ocean and finance. It is important to be able to find and anticipate LWs in practical applications. In recent years, studies on lump solutions have increased, leading to more specialized investigations. Therefore, theoretical investigations of LWs are instrumental in enhancing our understanding and predicting possible extremes in nonlinear systems [9–13].

Finding the lump solutions of PDEs has become a primary focus in recent years. As a result, several mathematical experts have developed important schemes in order to solve PDEs [14–16].

**Citation:** Seadawy, A.R.; Rizvi, S.T.R.; Zahed, H. Lump-Type Solutions, Lump Solutions, and Mixed Rogue Waves for Coupled Nonlinear Generalized Zakharov Equations. *Mathematics* **2023**, *11*, 2856. https:// doi.org/10.3390/math11132856

Academic Editors: Hovik Matevossian and Patricia J. Y. Wong

Received: 27 February 2023 Revised: 5 May 2023 Accepted: 20 June 2023 Published: 26 June 2023

**Copyright:** © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

In this article, we consider the CNL-GZE for the complex envelope *u*(*x*, *t*) of the high-frequency wave and the real low-frequency field *v*(*x*, *t*), as follows [17]:

$$\begin{cases} i\hbar\_1 \psi\_t + \psi\_{xx} - 2\hbar\_2 |\phi|^2 \psi + 2\psi \phi = 0, \\ \phi\_{tt} - \psi\_{xx} - (|\phi|^2)\_{xx} = 0. \end{cases} \tag{1}$$

where *h*<sup>1</sup> and *h*<sup>1</sup> are real constants. The cubic term in Equation (1) represents the nonlinear self-interaction in the high-frequency subsystem, which corresponds to a self-focusing effect in plasma physics.

Several researchers have worked on the stated model. For instance, Wang et al. evaluated periodic wave solutions for GZEs using the extended F-expansion method [17]. Zheng et al. performed a numerical simulation of a GZ system [18]. Bao et al. developed numerical schemes for a GZ system [19]. Bhrawy et al. constructed an efficient Jacobi pseudospectral approximation for a nonlinear complex GZ system [20]. Zhang et al. studied solitary wave solutions through a variational approach [21]. Similarly, Yildirim et al. studied some newly discovered soliton solutions of GZEs by applying He's variational approach [22]. Li et al. computed additional exact solutions of GZEs through the Expfunction method [23]. Buhe et al. studied symmetry reductions, conservation laws, and exact solutions for GZEs [24]. Lin et al. constructed some additional exact solutions for GZEs through the Exp-function method [23]. Wu et al. studied exact solutions for GZEs using a variational approach [25]. However, in this paper, we will explore lump, lump-type, lump one-strip, and lump two-strip solutions for CNL-GZEs through appropriate transformation methods and bilinear equations. We compute the lump solutions by choosing the appropriate polynomial function. In addition, we compute lump-periodic and rogue-wave solutions by using logarithmic transformation.

This article is organized as follows. In Section 2, we form bilinear equations and evaluate lump solutions for the coupled nonlinear generalized Zakharov equations in plasma physics through appropriate transformation approaches. The solutions are presented along with with their corresponding graphs. The mixed solutions of soliton and lump waves are provided in Section 3. We evaluate the lump one-strip and lump two-strip solutions using suitable profiles in Section 3. By employing a trigonometric ansatz in the bilinear equation, we compute lump periodic solutions in Section 4. By utilizing a hyperbolic ansatz in the bilinear equation, we explore rogue-wave solutions in Section 5. Section 6 discusses the results of the obtained solutions, and finally, in Section 7, we present some concluding remarks.

#### **2. Lump Solution**

For the lump solutions of Equation (1), we apply the following ansatz: [26–30],

$$\Psi(\mathbf{x},t) = \frac{h\_3 e^{(ict)} p(\mathbf{x},t)}{q(\mathbf{x},t)}, \ \ \Phi(\mathbf{x},t) = 2[\ln q(\mathbf{x},t)]\_x - c,\tag{2}$$

then, we obtain the bilinear equations,

$$\begin{aligned} 2h\_2h\_3^2p^3 + 2ch\_3pqt^2 + ch\_1h\_3pq^2 - ih\_1h\_3q^2 + p\_t + ih\_1h\_3pqq\_t - 4h\_3pqq\_x \\ + 2h\_3qp\_xq\_x - 2h\_3pq\_x^2 - h\_3q^2p\_{xx} + h\_3pqq\_{xx} = 0, \end{aligned} \tag{3}$$

and

$$\begin{aligned} 4h\_3^2 q^2 p\_x^2 q\_t^2 q\_x - q^2 q\_{tt} q\_x - 4h\_3^2 p q p\_x q\_x + 3h\_3^3 p^2 q\_x^2 - 2q q\_x^3 - 2q^2 q\_t q\_{xt} + q^3 q\_{xtt} \\ + h\_3^2 p q^2 p\_{xx} - h\_3^2 p^2 q q\_{xx} + 3q^2 q\_x q\_{xx} - q^3 q\_{xxx} = 0, \end{aligned} \tag{4}$$

respectively.

Now, to obtain the LP solution, the functions *p* and *q* in Equations (3) and (4) are assumed to be [27,28],

$$p = \mathfrak{J}\_1^2 + \mathfrak{J}\_2^2 + a\_2 \quad q = \mathfrak{J}\_1^2 + \mathfrak{J}\_2^2 + a\_3 \tag{5}$$

where *ξ*<sup>1</sup> = *a*0*x* + *t*, *ξ*<sup>2</sup> = *a*1*x* + *t*.

In addition, *ai*(1 ≤ *i* ≤ 3) are specific constants. Now, by substituting Equation (5) into Equations (3) and (4) and solving the equations obtained from the coefficients of *x* and *t*, we obtain:

**Set I.** The values of unknowns for Equations (3) and (4), respectively, are as follows:

$$\begin{cases} a\_0 = \frac{-1 + i\sqrt{3}}{2}, \; h\_1 = -\frac{2\left(\frac{1}{2}h\_3^2 + \varepsilon\right)}{\varepsilon}, \; a\_2 = a\_2, \; a\_3 = a\_3, \; a\_0 = a\_0. \\\ and \\\ a\_0 = \frac{1 - i}{2}, \; a\_1 = \frac{1 + i}{2}, \; h\_3 = 0, \; a\_2 = a\_2, \; a\_3 = a\_3. \end{cases} \tag{6}$$

Then, the values in Equation (6) generate the required solutions for Equations (3) and (4), which are, respectively,

$$\begin{cases} \Psi\_{1,1} = -\frac{2e^{it}\left(c + h\_2h\_3^2\right)\left(a\_2 + \left(t + \frac{\left(-1 + i\sqrt{3}\right)}{2}x\right)^2 + \left(t + a\_1x\right)^2\right)}{c\left(a\_3 + \left(t + \frac{\left(-1 + i\sqrt{3}\right)x}{2}\right)^2 + \left(t + a\_1x\right)^2\right)},\\ and \\ \Phi\_{1,1} = \frac{2\left(-1 + i\sqrt{3}\right)\left(\left(t + \frac{\left(-1 + i\sqrt{3}\right)^2}{2}x\right) + 2a\_1(t + a\_1x)\right)}{a\_3 + \left(t + \frac{\left(-1 + i\sqrt{3}\right)x}{2}\right)^2 + \left(t + a\_1x\right)^2} - c. \end{cases} \tag{7}$$

and

$$\begin{cases} \boldsymbol{\Psi}\_{1,2} = \frac{\boldsymbol{\epsilon}^{i\boldsymbol{t}}h\_{1}\left(\boldsymbol{\epsilon}\_{2} + \left(t + \left(\frac{1-\bar{i}}{2}\right)\boldsymbol{x}\right)^{2} + \left(t + \left(\frac{1+\bar{i}}{2}\right)\boldsymbol{x}\right)^{2}\right)}{\boldsymbol{a}\boldsymbol{y} + \left(t + \left(\frac{1-\bar{i}}{2}\right)\boldsymbol{x}\right)^{2} + \left(t + \left(\frac{1+\bar{i}}{2}\right)\boldsymbol{x}\right)^{2}},\\ \text{and} \\ \boldsymbol{\Phi}\_{1,2} = \frac{2\left((1-i)\left(t + \left(\frac{1-\bar{i}}{2}\right)\boldsymbol{x}\right) + (1+i)\left(t + \left(\frac{1+\bar{i}}{2}\right)\boldsymbol{x}\right)\right)}{\boldsymbol{a}\boldsymbol{y} + \left(t + \left(\frac{1-\bar{i}}{2}\right)\boldsymbol{x}\right)^{2} + \left(t + \left(\frac{1+\bar{i}}{2}\right)\boldsymbol{x}\right)^{2}} - \boldsymbol{c}. \end{cases} \tag{8}$$

**Set II.** The values of the parameters in Equations (3) and (4) are, respectively,

$$\begin{cases} a\_0 = \frac{-3+3i}{4}, a\_1 = \frac{3+3i}{4}, h\_1 = -2, a\_2 = 0, a\_3 = a\_3 \\ and \\ a\_0 = 1, a\_1 = 1, h\_3 = h\_3, a\_2 = a\_2, a\_3 = a\_3. \end{cases} \tag{9}$$

Then, the values in Equation (9) generate the required solutions for Equations (3) and (4), which are, respectively,

$$\begin{cases} \boldsymbol{\Psi}\_{2,1} = -\frac{2\boldsymbol{\epsilon}^{ict}\left(\left(t - \left(\frac{3-3i}{4}\right)\boldsymbol{x}\right)^{2} + \left(t + \left(\frac{3+3i}{4}\right)\boldsymbol{x}\right)^{2}\right)}{a\_{3} + \left(t - \left(\frac{3-3i}{4}\right)\boldsymbol{x}\right)^{2} + \left(t + \left(\frac{3+3i}{4}\right)\boldsymbol{x}\right)^{2}},\\ \text{and} \\ \boldsymbol{\Phi}\_{2,1} = \frac{2\left(\left(-\frac{3+3i}{2}\right)\left(t - \left(\frac{3-3i}{4}\right)\boldsymbol{x}\right) + \left(\frac{3+3i}{2}\right)\left(t + \left(\frac{3+3i}{4}\right)\boldsymbol{x}\right)\right)}{a\_{3} + \left(t - \left(\frac{3-3i}{4}\right)\boldsymbol{x}\right)^{2} + \left(t + \left(\frac{3+3i}{4}\right)\boldsymbol{x}\right)^{2}} - \boldsymbol{c}. \end{cases} \tag{10}$$

and

$$\begin{cases} \psi\_{2,2} = \frac{c^{ict} h\_1 \left(a\_2 + 2\left(t + x\right)^2\right)}{a\_3 + 2\left(t + x\right)^2}, \\ and \\ \phi\_{2,2} = -c + \frac{8\left(t + x\right)}{a\_3 + 2\left(t + x\right)^2}. \end{cases} \tag{11}$$

#### **3. Mixed Solutions of Soliton and Lump Waves**

In this section, we study the interaction of a lump soliton with a single kink wave and the interaction of a lump soliton with double kink waves.

#### *3.1. Lump One-Strip Soliton Interaction Solution*

To obtain the lump one-strip solution, we use the transformations given in Equations (3) and (4) [22,27–30]:

$$p = \mathfrak{J}\_1^2 + \mathfrak{J}\_2^2 + a\_2 + b\_0 e^{k\_1 x + k\_2 t}, \quad q = \mathfrak{J}\_1^2 + \mathfrak{J}\_2^2 + a\_3 + b\_0 e^{k\_1 x + k\_2 t}, \tag{12}$$

where *ξ*<sup>1</sup> = *a*0*x* + *t*, *ξ*<sup>2</sup> = *a*1*x* + *t*, and *ai*(1 ≤ *i* ≤ 3), *k*1, *k*2, and *b*<sup>0</sup> are any constants. Now, from Equations (12) and (4), we obtain the coefficients of *x* and *t* and solve the equations as follows:

**Set I.** The values of the parameters in Equations (3) and (4) are, respectively,

$$\begin{cases} c = \frac{-18h\_2h\_3^2 + 8 + 4\sqrt{-9h\_1h\_2 - 72h\_2h\_3^2 + 4}}{9h\_1 + 18}, \; h\_1 = \frac{2 + \sqrt{-9h\_1h\_2 - 72h\_2h\_3^2 + 4}}{3}, \; a\_0 = ia\_1, \; h\_3 = h\_3, \; a\_2 = a\_2, \\\text{and} \\\ a\_0 = \frac{1 - 2i\sqrt{5}}{3}, a\_1 = \frac{1 + 2i\sqrt{5}}{3}, a\_2 = -\frac{6930}{19h\_3^4}, a\_3 = -\frac{62100}{19h\_3^4}, b\_1 = -\frac{19}{90}h\_3^2, b\_2 = -\frac{19}{90}h\_3^2. \end{cases} \tag{13}$$

Then, the values in Equation (13) generate the required results for Equations (3) and (4), which are, respectively,

⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ *ψ*3,1 = *e i* <sup>−</sup>18*h*2*h*<sup>2</sup> <sup>3</sup>+8+<sup>4</sup> √−<sup>9</sup>*ih*1*k*2−72*h*2*h*<sup>2</sup> <sup>3</sup>+<sup>4</sup> *t* <sup>9</sup>*h*1+<sup>18</sup> *<sup>h</sup>*<sup>1</sup> ⎛ ⎜⎜⎝*<sup>a</sup>*2+*b*0*<sup>e</sup> k*2*t*+ 2+ √−<sup>9</sup>*ih*1*k*2−72*h*2*h*<sup>2</sup> <sup>3</sup>+<sup>4</sup> *x* <sup>3</sup> +(*t*+*ia*1*x*) <sup>2</sup>+(*t*+*a*1*x*) 2 ⎞ ⎟⎟⎠ *a*3+*b*0*e k*2*t*+ 2+ √−<sup>9</sup>*ih*1*k*2−72*h*2*h*<sup>2</sup> <sup>3</sup>+<sup>4</sup> *x* <sup>3</sup> +(*t*+*ia*1*x*) <sup>2</sup>+(*t*+*a*1*x*) 2 , *and <sup>φ</sup>*3,1 <sup>=</sup> <sup>−</sup>−18*h*2*h*<sup>2</sup> <sup>3</sup>+8+4 √−<sup>9</sup>*ih*1*k*2−72*h*2*h*<sup>2</sup> <sup>3</sup>+4 <sup>9</sup>*h*1+<sup>18</sup> + 2 ⎛ ⎜⎜⎝ 1 <sup>3</sup> *b*0*e k*2*t*+ 2+ √−<sup>9</sup>*ih*1*k*2−72*h*2*h*<sup>2</sup> <sup>3</sup>+<sup>4</sup> *x* <sup>3</sup> Π<sup>1</sup> ⎞ ⎟⎟⎠ *a*3+*b*0*e k*2*t*+ 2+ √−<sup>9</sup>*ih*1*k*2−72*h*2*h*<sup>2</sup> <sup>3</sup>+<sup>4</sup> *x* <sup>3</sup> +(*t*+*ia*1*x*) <sup>2</sup>+(*t*+*a*1*x*) 2 , Π<sup>1</sup> = 2 + −9*ih*1*k*<sup>2</sup> − <sup>72</sup>*h*2*h*<sup>2</sup> <sup>3</sup> + 4 + 2*ia*1(*t* + *ia*1*x*) + 2*a*1(*t* + *a*1*x*). (14)

and

$$\begin{cases} \begin{aligned} \Psi\_{4,1} &= \frac{i^{4i}t\_{1}\left(\frac{19\theta\_{2}^{2}t}{90} - \frac{19\theta\_{2}^{2}t}{90} - \frac{6930}{194\theta\_{2}^{2}}\right)^{2} + \left(t + \frac{(1 - 2i\sqrt{\pi})x}{3}\right)^{2}}{100\frac{\theta\_{2}^{2}}{90} - \frac{19\theta\_{2}^{2}t}{90} - \frac{6930}{194\theta\_{2}^{2}} + \left(t + \frac{(1 - 2i\sqrt{\pi})x}{3}\right)^{2}}, \\quad \text{and} \\\\ \Phi\_{4,1} &= \frac{2\left(-\frac{19\theta\_{2}^{2}t}{90} - \frac{19\theta\_{2}^{2}x}{90}\right)^{2} + \frac{2\left((1 - 2i\sqrt{\pi})x}{3}\right) + \left(t + \frac{(1 - 2i\sqrt{\pi})x}{3}\right)^{2}}{100\frac{\theta\_{2}^{2}}{90} - \frac{6930}{190} + \left(t + \frac{(1 - 2i\sqrt{\pi})x}{3}\right)^{2} + \left(t + \frac{(1 - 2i\sqrt{\pi})x}{3}\right)^{2}} - c\_{r} \\\\ \Pi\_{2} &= \frac{2\left(1 + 2i\sqrt{\pi}\right)}{3} \left(t + \frac{(1 - 2i\sqrt{\pi})x}{3}\right). \end{aligned} \tag{15}$$

**Set II.** The values of the parameters in Equations (3) and (4) are, respectively,

$$\begin{cases} a\_0 = ia\_1, \; k\_1 = \frac{6h\_2h\_3^2 + 3ch\_1 + 6c}{4}, \; a\_3 = a\_3, \; h\_3 = h\_3, \; a\_2 = a\_2. \\\ and \\ a\_0 = \frac{-1 - 2i\sqrt{5}}{3}, a\_1 = \frac{-1 + 2i\sqrt{5}}{3}, a\_2 = -\frac{69300}{19h\_3^4}, a\_3 = -\frac{62100}{19h\_3^4}, k\_1 = -\frac{19}{90}h\_3^2, k\_2 = -\frac{19}{90}h\_3^2. \end{cases} \tag{16}$$

Then, the values in Equation (16) generate the required results for Equations (3) and (4), which are, respectively,

$$\begin{cases} \begin{aligned} \Psi\_{5,1} &= \frac{\epsilon^{ict}h\_{1}\left(a\_{2}+b\_{0}e^{\frac{k\_{2}t}{4}}+\frac{\left(6h\_{2}h\_{3}^{2}+3ch\_{1}+6c\right)x}{4}+\left(t+ia\_{1}x\right)^{2}+\left(t+ia\_{1}x\right)^{2}\right)}{4t+b\_{0}e^{\frac{k\_{2}t}{4}}+\frac{\left(6h\_{2}h\_{3}^{2}+3ch\_{1}+6c\right)x}{4}+\left(t+ia\_{1}x\right)^{2}+\left(t+ia\_{1}x\right)^{2}} \\ \text{and} \\ \Phi\_{5,1} &= \frac{2\left(\frac{1}{9}b\_{0}e^{\frac{k\_{2}t}{4}+\frac{\left(6h\_{2}h\_{3}^{2}+3ch\_{1}+6c\right)x}{4}}+\left(6h\_{2}h\_{3}^{2}+3ch\_{1}+6c\right)+2ia\_{1}\left(t+ia\_{1}x\right)^{2}+2a\_{1}\left(t+a\_{1}x\right)^{2}\right)}{\left(t+a\_{0}x\right)^{2}+\left(t+ia\_{1}x\right)^{2}+\left(t+ia\_{1}x\right)^{2}} - \mathcal{L} \end{aligned} \end{cases} \tag{17}$$

and

⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ *ψ*5,2 = *eicth*<sup>1</sup> ⎛ ⎝*b*0*e* 19*h*2 3*t* <sup>90</sup> <sup>−</sup> <sup>19</sup>*h*<sup>2</sup> 3*x* <sup>90</sup> <sup>−</sup> <sup>69300</sup> 19*h*4 3 + *<sup>t</sup>*<sup>+</sup> (−1−2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 2 + *<sup>t</sup>*<sup>+</sup> (−1+2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 2 ⎞ ⎠ *b*0*e* 19*h*2 3*t* <sup>90</sup> <sup>−</sup> <sup>19</sup>*h*<sup>2</sup> 3*x* <sup>90</sup> <sup>−</sup> <sup>69300</sup> 19*h*4 3 + *<sup>t</sup>*<sup>+</sup> (−1−2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 2 + *<sup>t</sup>*<sup>+</sup> (−1+2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 <sup>2</sup> , *and φ*5,2 = 2 ⎛ <sup>⎝</sup><sup>−</sup> <sup>19</sup> <sup>90</sup> *b*0*e* 19*h*2 3*t* <sup>90</sup> <sup>−</sup> <sup>19</sup>*h*<sup>2</sup> 3*x* <sup>90</sup> *h*<sup>2</sup> <sup>3</sup><sup>+</sup> <sup>2</sup>(−1−2*<sup>i</sup>* <sup>√</sup>5) 3 *<sup>t</sup>*<sup>+</sup> (−1−2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 +Π<sup>3</sup> ⎞ ⎠ *b*0*e* 19*h*2 3*t* <sup>90</sup> <sup>−</sup> <sup>19</sup>*h*<sup>2</sup> 3*x* <sup>90</sup> <sup>−</sup> <sup>62100</sup> 19*h*4 3 + *<sup>t</sup>*<sup>+</sup> (−1−2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 2 + *<sup>t</sup>*<sup>+</sup> (−1+2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 <sup>2</sup> <sup>−</sup> *<sup>c</sup>*, <sup>Π</sup><sup>3</sup> <sup>=</sup> <sup>2</sup>(−1+2*<sup>i</sup>* <sup>√</sup>5) 3 *<sup>t</sup>* <sup>+</sup> (−1−2*<sup>i</sup>* <sup>√</sup>5)*<sup>x</sup>* 3 . (18)

#### *3.2. Lump Double-Strip Soliton Interaction Solution*

To obtain the lump two-strip solution, we assume the following transformation [22,27–30]:

$$p = \bigwedge\_{1}^{2} + \bigwedge\_{1}^{2} + a\_3 + m\_1 e^{k\_1 x + k\_2 t + k\_3} + m\_2 e^{k\_2 x + k\_3 t + k\_6}, \quad q = \bigwedge\_{1}^{2} + \bigwedge\_{2}^{2} + a\_4 + m\_1 e^{k\_1 x + k\_2 t + k\_3} + m\_2 e^{k\_4 x + k\_5 t + k\_6}, \tag{19}$$

where < <sup>1</sup> = *a*1*x* + *a*2*t*, < <sup>2</sup> = *a*1*x* + *a*2*t*, and *ai*(1 ≤ *i* ≤ 4), *ki*(1 ≤ *i* ≤ 6), *m*1, and *m*<sup>2</sup> are specific real parameters. Now, from Equation (19) and Equation (4), we obtain the coefficients of *x*, *t*, and *exp* and solve these equations as follows:

**Set I.** When *k*<sup>5</sup> = *k*<sup>4</sup> = *a*<sup>1</sup> = 0 for Equation (3) and *k*<sup>3</sup> = *k*<sup>6</sup> = *a*<sup>1</sup> = 0 for Equation (4), the values of the parameters are, respectively,

$$\begin{cases} a\_4 = -\frac{i\left(9h\_2h\_3^2h\_1^2 - 3a\_2h\_3h\_2^2 + \text{i}\_2^2a\_3\right)}{(3ih\_3h\_3^2 + a\_2)a\_2}, & k\_2 = \frac{2k\_1\left(3h\_2h\_3^2 - 2a\_2\right)}{3h\_2h\_3^2}, \ m\_1 = -\frac{a\_2m\_2\left(3h\_2h\_3^2 + 2a\_2\right)}{-a\_2 + 3h\_2h\_3^2} \\\ and \\ a\_2 = \frac{\sqrt{6}h\_3^2}{60}, & k\_1 = -\frac{\sqrt{6}h\_4^2 - \frac{2}{3}\sqrt{6}h\_5^2 + 3h\_4h\_5}{5\sqrt{\frac{2}{3}\sqrt{6}h\_4^2k\_5 - \frac{2}{3}h\_5^2 - \frac{1}{3}h\_5^2}}, k\_2 = \sqrt{\frac{2}{5}\sqrt{6}k\_4k\_5 - \frac{6}{5}h\_4^2 - \frac{1}{3}h\_5^2}, a\_4 = 0. \end{cases} \tag{20}$$

Then, the values in Equation (20) generate the required results for Equations (3) and (4), which are, respectively,

⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ *ψ*5,1 = *eicth*<sup>1</sup> ⎛ <sup>⎝</sup>*a*3+*k*<sup>2</sup> 6+*ea*2*<sup>t</sup> <sup>m</sup>*2<sup>−</sup> *<sup>a</sup>*2*m*2*<sup>e</sup> a*2*t* (<sup>3</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>*m*2+2*a*2) <sup>−</sup>*a*2+3*ih*2*h*<sup>2</sup> 3 + *<sup>k</sup>*3<sup>+</sup> <sup>2</sup>*k*1(<sup>3</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>−2*a*2)*<sup>t</sup>* <sup>3</sup>*h*2*h*<sup>2</sup> 3 +*k*1*x* 2 ⎞ ⎠ 2*ea*2*<sup>t</sup>* <sup>−</sup> *<sup>i</sup>*(<sup>9</sup>*ih*2*h*<sup>2</sup> 3*k*2 1−3*a*2*a*3*h*2*h*<sup>2</sup> 3+*ia*<sup>2</sup> <sup>2</sup>*a*3) (<sup>3</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>+*a*2)*<sup>a</sup>*<sup>2</sup> +*k*<sup>2</sup> 6+ *<sup>k</sup>*3<sup>+</sup> <sup>2</sup>*k*1(<sup>3</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>−2*a*2)*<sup>t</sup>* <sup>3</sup>*h*2*h*<sup>2</sup> 3 +*k*1*x* <sup>2</sup> , *φ*5,1 = 4*k*<sup>1</sup> *<sup>k</sup>*3<sup>+</sup> <sup>2</sup>*k*1(<sup>3</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>−2*a*2)*<sup>t</sup>* <sup>3</sup>*h*2*h*<sup>2</sup> 3 +*k*1*x* 2*ea*2*<sup>t</sup>* <sup>−</sup> *<sup>i</sup>*(<sup>9</sup>*ih*2*h*<sup>2</sup> 3*k*2 1−3*a*2*a*3*h*2*h*<sup>2</sup> 3+*ia*<sup>2</sup> <sup>2</sup>*a*3) (<sup>3</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>+*a*2)*<sup>a</sup>*<sup>2</sup> +*k*<sup>2</sup> 6+ *<sup>k</sup>*3<sup>+</sup> <sup>2</sup>*k*1(<sup>3</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>−2*a*2)*<sup>t</sup>* <sup>3</sup>*h*2*h*<sup>2</sup> 3 +*k*1*x* <sup>2</sup> − *c*. (21)

and

⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ *ψ*6,1 = *eicth*<sup>1</sup> ⎛ ⎝Δ1+ <sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 <sup>5</sup>*t*<sup>−</sup> ( <sup>√</sup>6*k*<sup>2</sup> <sup>4</sup><sup>−</sup> <sup>2</sup> 3 <sup>√</sup>6*k*<sup>2</sup> <sup>5</sup>+3*k*4*k*5)*<sup>x</sup>* 5 √<sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 5 2 ⎞ ⎠ 2*e h*2 3*t* <sup>10</sup>√<sup>6</sup> +(*k*5*t*+*k*4*x*) 2+ <sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 <sup>5</sup>*t*<sup>−</sup> ( <sup>√</sup>6*k*<sup>2</sup> <sup>4</sup><sup>−</sup> <sup>2</sup> 3 <sup>√</sup>6*k*<sup>2</sup> <sup>5</sup>+3*k*4*k*5)*<sup>x</sup>* 5 √<sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 5 2 , Δ<sup>1</sup> = *a*<sup>3</sup> + *e h*2 3*t* <sup>10</sup>√<sup>6</sup> *m*<sup>1</sup> + *e h*2 3*t* <sup>10</sup>√<sup>6</sup> *m*<sup>2</sup> + (*k*5*t* + *k*4*x*) 2 . *and φ*6,1 = 2 ⎛ ⎜⎜⎜⎜⎝ Δ2− ( <sup>√</sup>6*k*<sup>2</sup> <sup>4</sup><sup>−</sup> <sup>2</sup> 3 <sup>√</sup>6*k*<sup>2</sup> <sup>5</sup>+3*k*4*k*5) ⎛ ⎝ √<sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 5*t*− ( <sup>√</sup>6*k*<sup>2</sup> <sup>4</sup><sup>−</sup> <sup>2</sup> 3 <sup>√</sup>6*k*<sup>2</sup> <sup>5</sup>+3*k*4*k*5)*<sup>x</sup>* 5 √<sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 5 ⎞ ⎠ 5 √<sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 5 2⎞ ⎟⎟⎟⎟⎠ 2*e h*2 3*t* <sup>10</sup>√<sup>6</sup> +(*k*5*t*+*k*4*x*) 2+ <sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 <sup>5</sup>*t*<sup>−</sup> ( <sup>√</sup>6*k*<sup>2</sup> <sup>4</sup><sup>−</sup> <sup>2</sup> 3 <sup>√</sup>6*k*<sup>2</sup> <sup>5</sup>+3*k*4*k*5)*<sup>x</sup>* 5 √<sup>2</sup> 5 <sup>√</sup>6*k*4*k*5<sup>−</sup> <sup>6</sup> 5 *k*2 <sup>4</sup><sup>−</sup> <sup>1</sup> 5 *k*2 5 <sup>2</sup> <sup>−</sup> *<sup>c</sup>*, Δ<sup>2</sup> = 2*k*4(*k*5*t* + *k*4*x*). (22)

**Set II.** When *k*<sup>5</sup> = *k*<sup>4</sup> = *a*<sup>1</sup> = 0 for Equation (3) and *k*<sup>3</sup> = *k*<sup>6</sup> = *a*<sup>1</sup> = 0 for Equation (4), the values of the parameters are, respectively,

$$\begin{cases} a\_2 = \frac{4il\_2h\_3^2c(a\_3 - a\_4)}{-10a\_3h\_2h\_3^2 + 10a\_4h\_2h\_3^2 + 2ac - 2a\_4c + 9k\_1^2}, \; h\_1 = -\frac{-10a\_3h\_2h\_3^2 + 10a\_4h\_2h\_3^2 + 2a\_3c - 2a\_4c + 9k\_1^2}{c(a\_3 - a\_4)},\\ k\_2 = -\frac{\frac{4}{3}ic\_1(a\_3 - a\_4)}{-10a\_3h\_2h\_3^2 + 10a\_4h\_2h\_3^2 + 2a\_3c - 2a\_4c + 9k\_1^2}, \; m\_1 = -m\_2 - 4. \\\ \text{and} \\ m\_1 = -\frac{5a\_3m\_2 - 4a\_4m\_2 - 8a\_3 + 6a\_4}{5a\_3 - 4a\_4}, k\_1 = ik\_4, k\_2 = ik\_5, a\_2 = 0. \end{cases} \tag{23}$$

Then, the values in Equation (23) generate the required results for Equations (3) and (4), which are, respectively,

⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ *ψ*7,1 = − ⎛ ⎜⎜⎝*eictD*<sup>1</sup> ⎛ ⎜⎜⎝*<sup>a</sup>*3+*k*<sup>2</sup> <sup>6</sup>+(−*m*2−4)*e* <sup>4</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>*c*(*a*3−*a*4)*<sup>t</sup>* <sup>−</sup>10*a*3*h*2*h*<sup>2</sup> 3+10*a*4*h*2*h*<sup>2</sup> 3+2*a*3*c*−2*a*4*c*+9*k*<sup>2</sup> <sup>1</sup> +*m*2*e* <sup>4</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>*c*(*a*3−*a*4)*<sup>t</sup>* <sup>−</sup>10*a*3*h*2*h*<sup>2</sup> 3+10*a*4*h*2*h*<sup>2</sup> 3+2*a*3*c*−2*a*4*c*+9*k*<sup>2</sup> <sup>1</sup> *D*<sup>2</sup> ⎞ ⎟⎟⎠ ⎞ ⎟⎟⎠ (*a*3−*a*4)*c* ⎛ ⎜⎜⎝*<sup>a</sup>*4+*k*<sup>2</sup> <sup>6</sup>+2*e* <sup>4</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>*c*(*a*3−*a*4)*<sup>t</sup>* <sup>−</sup>10*a*3*h*2*h*<sup>2</sup> 3+10*a*4*h*2*h*<sup>2</sup> 3+2*a*3*c*−2*a*4*c*+9*k*<sup>2</sup> <sup>1</sup> *D*<sup>2</sup> ⎞ ⎟⎟⎠ , *<sup>D</sup>*<sup>1</sup> <sup>=</sup> <sup>−</sup>10*a*3*h*2*h*<sup>2</sup> <sup>3</sup> + <sup>10</sup>*a*4*h*2*h*<sup>2</sup> <sup>3</sup> <sup>+</sup> <sup>2</sup>*a*3*<sup>c</sup>* <sup>−</sup> <sup>2</sup>*a*4*<sup>c</sup>* <sup>+</sup> <sup>9</sup>*k*<sup>2</sup> 1, *D*<sup>2</sup> = *<sup>k</sup>*<sup>3</sup> <sup>−</sup> <sup>4</sup>*ik*1*c*(*a*3−*a*4)*<sup>t</sup>* <sup>3</sup>(−10*a*3*h*2*h*<sup>2</sup> 3+10*a*4*h*2*h*<sup>2</sup> 3+2*a*3*c*−2*a*4*c*+9*k*<sup>2</sup> <sup>1</sup>) <sup>+</sup> *<sup>k</sup>*1*<sup>x</sup>* 2 , *and φ*7,1 = −*c* + 4*k*<sup>1</sup> *<sup>k</sup>*3<sup>−</sup> <sup>4</sup>*ik*1*c*(*a*3−*a*4)*<sup>t</sup>* <sup>3</sup>(−10*a*3*h*2*h*<sup>2</sup> 3+10*a*4*h*2*h*<sup>2</sup> 3+2*a*3*c*−2*a*4*c*+9*k*<sup>2</sup> 1) +*k*1*x* ⎛ ⎜⎜⎝*<sup>a</sup>*4+*k*<sup>2</sup> <sup>6</sup>+2*e* <sup>4</sup>*ih*2*h*<sup>2</sup> <sup>3</sup>*c*(*a*3−*a*4)*<sup>t</sup>* <sup>−</sup>10*a*3*h*2*h*<sup>2</sup> 3+10*a*4*h*2*h*<sup>2</sup> 3+2*a*3*c*−2*a*4*c*+9*k*<sup>2</sup> <sup>1</sup> *D*<sup>2</sup> ⎞ ⎟⎟⎠ . (24)

and

$$\begin{cases} \psi\_{8,1} = \frac{c^{ic}h\_1\left(a\_3 + m\_2 - \frac{5a\_3m\_2 - 4a\_4m\_2 - 8a\_5 + 6a\_4}{5a\_3 - 4a\_4} + (ik\_5t + ik\_4x)^2 + (k\_5t + k\_4x)^2\right)}{2 + a\_4 + (ik\_5t + ik\_4x)^2 + (k\_5t + k\_4x)^2} \\ and \\ \phi\_{8,1} = \frac{2\left(2ik\_4(ik\_5t + ik\_4x)^2 + 2k\_4(k\_5t + k\_4x)^2\right)}{2 + a\_4 + (ik\_5t + ik\_4x)^2 + (k\_5t + k\_4x)^2} - c. \end{cases} \tag{25}$$

#### **4. Lump Periodic Soliton Solution**

To compute the LPS solution, we use the following supposition in Equations (3) and (4) [22,27–30]:

$$p = \bigwedge\_{1}^{2} + \bigwedge\_{2}^{2} + a\_2 + a\_3 \cos(n\_1 \mathbf{x} + t), \quad q = \bigwedge\_{1}^{2} + \bigwedge\_{2}^{2} + a\_4 + a\_5 \cos(n\_1 \mathbf{x} + t) \tag{26}$$

where < <sup>1</sup> = *B*0*x* + *t*, < <sup>2</sup> = *B*1*x* + *t*. In addition, *ai*(1 ≤ *i* ≤ 5) and *n*<sup>1</sup> are various parameters to be determined. Now, by substituting Equation (26) into Equations (3) and (4) and then examining the coefficients of *x*, cos function, and *t*, we obtain the following:

**Set I.** The values of the parameters for Equations (3) and (4) are, respectively,

$$\begin{cases} n\_1 = -\frac{\frac{1}{4}il\_1(a\_4 - a\_5)}{a\_4 + a\_5}, \ a\_0 = -a\_1, \ c = c, \ a\_4 = a\_4. \\ and \\ n\_1 = -\frac{4\left(a\_0^2 + a\_1^2\right)}{(a\_1 + a\_0)\left(3a\_0^2 + 3a\_1^2 - 2\right)}, \ a\_0 = a\_0, \ c = c, \ a\_4 = a\_4, \ a\_3 = a\_3. \end{cases} \tag{27}$$

Then, the values in Equation (27) generate the required results for Equations (3) and (4), which are, respectively,

$$\begin{cases} \begin{aligned} \Phi\_{9,1} &= \frac{e^{it}h\_{1}\left(a\_{2}+(t+a\_{0}\mathbf{x})^{2}+(t+a\_{1}\mathbf{x})^{2}+a\_{4}\cos\left(t-\frac{i(4\_{4}-4\mathbf{s}\_{1})h\_{1}\mathbf{x}}{4\left(4\_{4}+4\mathbf{s}\_{1}\right)}\right)\right)}{\left(a\_{3}+(t+a\_{0}\mathbf{x})^{2}+(t+a\_{1}\mathbf{x})^{2}+a\_{5}\cos\left(t-\frac{i(4\_{4}-4\mathbf{s}\_{1})h\_{1}\mathbf{x}}{4\left(4\_{4}+4\mathbf{s}\_{1}\right)}\right)\right)},\\ \text{and} \\\\ \Phi\_{9,1} &= -c+\frac{2\left(2a\_{0}(t+a\_{0}\mathbf{x})+2a\_{1}(t+a\_{1}\mathbf{x})+\frac{i\left(4\_{4}-4\mathbf{s}\_{1}\right)a\_{3}\sin\left(t-\frac{i\left(4\_{4}-4\mathbf{s}\_{1}\right)h\_{1}\mathbf{x}}{4\left(4\_{4}+4\mathbf{s}\_{1}\right)}\right)}{4\left(4\_{4}+4\mathbf{s}\_{1}\right)}\right)}. \end{aligned} \end{cases} \tag{28}$$

and

$$\begin{cases} \begin{aligned} \Phi\_{10,1} &= \frac{t^{4s}h\_{1}\left(a\_{2}+(t+a\_{0}x)^{2}+(t+a\_{1}x)^{2}+a\_{4}\cos\left(t+\frac{4\left(x\_{0}^{2}+a\_{1}^{2}\right)x}{\left(a\_{0}+4a\_{0}\right)\left(3a\_{0}^{2}+3a\_{1}^{2}-2\right)}\right)\right)}{\left(a\_{3}+(t+a\_{0}x)^{2}+(t+a\_{1}x)^{2}+a\_{5}\cos\left(t+\frac{4\left(a\_{0}^{2}+a\_{1}^{2}\right)x}{\left(a\_{1}+4a\_{0}\right)\left(3a\_{0}^{2}+3a\_{1}^{2}-2\right)}\right)\right)},\\quad\text{and}\\ \Phi\_{10,1} &= -\mathfrak{c}+\frac{4\left(a\_{0}^{2}+a\_{1}^{2}\right)a\_{5}\sin\left(t+\frac{4\left(a\_{0}^{2}+a\_{1}^{2}\right)x}{\left(a\_{1}+4a\_{0}\right)\left(3a\_{0}^{2}+3a\_{1}^{2}-2\right)}\right)}{\left(a\_{3}+\left(t+a\_{0}x\right)^{2}+\left(t+a\_{1}x\right)^{2}+a\_{5}\cos\left(t+\frac{4\left(a\_{0}^{2}+a\_{1}^{2}\right)x}{\left(a\_{1}+4a\_{0}\right)\left(3a\_{0}^{2}+3a\_{1}^{2}-2\right)}\right)\right)}.\end{aligned} \end{cases}\tag{29}$$

#### **5. Rogue-Wave Solutions**

To compute the LPS solution, we use the following supposition in Equations (3) and (4) [22,27–30]:

$$p = \bigwedge\_{1}^{2} + \bigwedge\_{2}^{2} + a\_2 + a\_3 \cosh(n\_1 \mathbf{x} + t), \quad q = \bigwedge\_{1}^{2} + \bigwedge\_{2}^{2} + a\_4 + a\_5 \cosh(n\_1 \mathbf{x} + t) \tag{30}$$

where < <sup>1</sup> = *a*0*x* + *t*, < <sup>2</sup> = *a*1*x* + *t*. In addition, *ai*(1 ≤ *i* ≤ 5) and *n*<sup>1</sup> are various parameters to be determined. Now, by substituting Equation (26) into Equations (3) and (4) and then examining the coefficients of *x*, cos function, and *t*, we obtain the following:

**Set I.** The values of the parameters for Equations (3) and (4), are, respectively,

$$\begin{cases} a\_4 = -\frac{a\_5(4in\_1+b\_1)}{(4in\_1-b\_1)}, \; a\_0 = -a\_1, \; a\_2 = a\_2, \; a\_4 = a\_4, \; a\_3 = a\_3. \\\; \text{and} \\\; a\_1 = ia\_0, \; a\_4 = 0, \; a\_1 = 1, \; a\_3 = a\_3, \; a\_5 = a\_5. \end{cases} \tag{31}$$

Then, the values in Equation (31) generate the solutions for Equations (3) and (4), which are, respectively,

$$\begin{cases} \begin{aligned} \psi\_{11,1} &= \frac{e^{ict}h\_{1}\left(a\_{2}+(t-a\_{1}x)^{2}+(t+a\_{1}x)^{2}-\frac{\mathfrak{s}\_{5}\left(4i\mathfrak{u}\_{1}+b\_{1}\right)\cosh\left(t+a\_{1}x\right)}{\left(4\mathfrak{u}\_{1}-\mathfrak{s}\_{1}\right)}\right)}{\left(a\_{3}+(t-a\_{1}x)^{2}+(t+a\_{1}x)^{2}+a\_{5}\cosh\left(t+a\_{2}x\right)\right)},\\ \text{and} \\ \Phi\_{11,1} &= \frac{2\left(-2a\_{1}(t-a\_{1}x)+2a\_{1}(t+a\_{1}x)+a\_{5}x\sinh(t+a\_{2}x)\right)}{\left(a\_{3}+(t-a\_{1}x)^{2}+(t+a\_{1}x)^{2}+a\_{5}\cosh(t+a\_{2}x)\right)} - c. \end{aligned} \end{cases} \end{cases} \tag{32}$$

and

$$\begin{cases} \psi\_{12,1} = \frac{c^{ict} h\_1 \left( a\_2 + (t + ia\_0 x)^2 + (t + a\_1 x)^2 \right)}{\left( a\_3 + (t + ia\_0 x)^2 + (t + a\_1 x)^2 + a\_5 \cosh(t + a\_2 x) \right)},\\ and \\ \Phi\_{12,1} = -c + \frac{2(2ia\_0 (t + ia\_0 x) + 2a\_0 (t + a\_0 x) + a\_5 n\_2 \sinh(t + n\_2 x))}{\left( a\_3 + (t + ia\_0 x)^2 + (t + a\_1 x)^2 + a\_5 \cosh(t + n\_2 x) \right)}. \end{cases} \tag{33}$$

#### **6. Results and Discussion**

We observed that the solution *ψ*1,1(*x*, *t*) in Equation (7) with *a*<sup>1</sup> = 10, *h*<sup>2</sup> = −2, *h*<sup>3</sup> = 2, *a*<sup>3</sup> = 2, and *c* = 3 formed two lump waves (LWs) known as upper-bright and lower-dark LWs, and that the bright and dark LWs were symmetrical about the coordinate plane. As *a*<sup>2</sup> varied from a minimum to a maximum number, the two LWs rotated counterclockwise. When *a*<sup>2</sup> = 0, the LW disappeared, but at *a*<sup>2</sup> = 5, the LW gradually reappeared (see Figure 1). The contour lump-wave profiles for *ψ*1,1(*x*, *t*) are plotted for *a*<sup>1</sup> = 10, *h*<sup>2</sup> = −2, *h*<sup>3</sup> = 2, *a*<sup>3</sup> = 2, and *c* = 3 in Figure 2. The mixed solutions of soliton and lump waves were successfully obtained. Notice that our solution *φ*3,1(*x*, *t*) in Equation (14) with

*h*<sup>1</sup> = 10, *b*<sup>0</sup> = 10, and *c* = 5 formed lump one-strip waves (LSWs) known as upper-bright LSWs. The lump one-strip wave profiles for *φ*2,1(*x*, *t*) are depicted for *h*<sup>1</sup> = 10, *b*<sup>0</sup> = 10, and *c* = 5 in Figures 3 and 4. The lump double-strip wave profiles for *φ*5,1(*x*, *t*) are plotted for *k*<sup>3</sup> = 4, *h*<sup>2</sup> = 2, *h*<sup>1</sup> = 4, *h*<sup>3</sup> = 3, *a*<sup>2</sup> = 20, *a*<sup>3</sup> = 5, *k*<sup>6</sup> = 2, *m*<sup>2</sup> = 2, and *c* = 5 in Figures 3, 5 and 6. By utilizing the assumption of the cosine function in bilinear equations in Equations (3) and (4), we have obtained the lump periodic solutions. We have successfully obtained the lump periodic graphs for *φ*9,1(*x*, *t*), which are plotted for *a*<sup>0</sup> = 10, *a*<sup>1</sup> = 5, *a*<sup>2</sup> = 4, *a*<sup>3</sup> = 2, *a*<sup>4</sup> = 3, *a*<sup>5</sup> = 5, and *h*<sup>1</sup> = 20 in Figure 7. The lump periodic contour graphs for *φ*9,1(*x*, *t*) are plotted for *a*<sup>0</sup> = 10, *a*<sup>1</sup> = 5, *a*<sup>2</sup> = 4, *a*<sup>3</sup> = 2, *a*<sup>4</sup> = 3, *a*<sup>5</sup> = 5, and *h*<sup>1</sup> = 20 in Figure 8. By utilizing the assumption of cosine hyperbolic functions in bilinear equations in Equations (3) and (4), we have obtained the lump periodic solutions. As *a*<sup>1</sup> varied from −10 to 10, the rogue wave rotated, and its behavior can be seen for *ψ*11,1(*x*, *t*) for *h*<sup>1</sup> = 4, *a*<sup>2</sup> = 3, *a*<sup>3</sup> = 1.5, *a*<sup>5</sup> = 5, *n*<sup>1</sup> = 3, *n*<sup>2</sup> = 4, and *c* = 5 in Figure 9.

**Figure 1.** Lump-wave profiles for *ψ*1,1(*x*, *t*) are plotted for *a*<sup>1</sup> = 10, *h*<sup>2</sup> = −2, *h*<sup>3</sup> = 2, *a*<sup>3</sup> = 2, *c* = 3.

**Figure 2.** Contour lump-wave profiles for *ψ*1,1(*x*, *t*) are plotted for *a*<sup>1</sup> = 10, *h*<sup>2</sup> = −2, *h*<sup>3</sup> = 2, *a*<sup>3</sup> = 2, *c* = 3.

**Figure 3.** *Cont.*

**Figure 3.** Lump one-strip wave profiles for *φ*3,1(*x*, *t*) are plotted for *h*<sup>1</sup> = 10, *b*<sup>0</sup> = 10, *c* = 5.

**Figure 4.** *Cont.*

**Figure 4.** Contour lump one-strip wave profiles for *φ*3,1(*x*, *t*) are plotted for *h*<sup>1</sup> = 10, *b*<sup>0</sup> = 10, *c* = 5.

**Figure 5.** Lump double-strip wave profiles for *φ*5,1(*x*, *t*) are plotted for *k*<sup>3</sup> = 4, *h*<sup>2</sup> = 2, *h*<sup>1</sup> = 4, *h*<sup>3</sup> = 3, *a*<sup>2</sup> = 20, *a*<sup>3</sup> = 5, *k*<sup>6</sup> = 2, *m*<sup>2</sup> = 2, *c* = 5.

**Figure 6.** Contour profiles for Figure 5.

**Figure 7.** *Cont.*

**Figure 7.** Lump periodic graphs for *φ*9,1(*x*, *t*) are plotted for *a*<sup>0</sup> = 10, *a*<sup>1</sup> = 5, *a*<sup>2</sup> = 4, *a*<sup>3</sup> = 2, *a*<sup>4</sup> = 3, *a*<sup>5</sup> = 5, *h*<sup>1</sup> = 20.

**Figure 8.** *Cont.*

**Figure 8.** Lump periodic contour graphs for *φ*9,1(*x*, *t*) are plotted for *a*<sup>0</sup> = 10, *a*<sup>1</sup> = 5, *a*<sup>2</sup> = 4, *a*<sup>3</sup> = 2, *a*<sup>4</sup> = 3, *a*<sup>5</sup> = 5, *h*<sup>1</sup> = 20.

**Figure 9.** Rogue-wave profiles for *ψ*11,1(*x*, *t*) are plotted for *h*<sup>1</sup> = 4, *a*<sup>2</sup> = 3, *a*<sup>3</sup> = 1.5, *a*<sup>5</sup> = 5, *n*<sup>1</sup> = 3, *n*<sup>2</sup> = 4, *c* = 5.

#### **7. Concluding Remarks**

In this paper, we have studied multiple forms of lump solutions for CNL-GZEs in plasma physics using appropriate transformation approaches, bilinear equations, and symbolic computations. By utilizing the positive quadratic assumption in the bilinear equation, we have derived the lump-type solutions. We have evaluated the lump onesoliton solutions through a single exponential function transformation in the bilinear

equation. Similarly, we have computed the lump two-soliton solutions using a double exponential function transformation in the bilinear equation. Mixed solutions of lump waves and solitons have been successfully evaluated. Furthermore, we have computed rogue-wave solutions and lump periodic solutions by utilizing appropriate hyperbolic and trigonometric functions. We have identified certain constraint values throughout the derivation of the solutions that must hold for the soliton solution to exist. The presented solutions have valuable uses in plasma physics.

**Author Contributions:** Methodology, Methodology and Writing—review & editing, S.T.R.R.; Formal analysis, H.Z.; Supervision, A.R.S. All authors have read and agreed to the published version of the manuscript.

**Funding:** The Deputyship for Research and Innovation in the Ministry of Education in Saudi Arabia for funding this research work under project number 141/442.

**Data Availability Statement:** Not applicable.

**Acknowledgments:** The authors extend their appreciation to the Deputyship for Research and Innovation in the Ministry of Education in Saudi Arabia for funding this research work under project number 141/442. Furthermore, the authors would like to extend their appreciation to Taibah University for its supervisory support.

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

#### **References**


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