*3.1. The Cubic-Quartic Nonlinear Schrödinger Equation*

To solve Equation (3), by the *m* + *<sup>G</sup>*- *G* expansion method, we use the following transformation:

$$\mu \left( \mathbf{x}, t \right) = \mathcal{U}(\xi) e^{i\theta}, \qquad \xi = \mathbf{x} - \nu t, \qquad \theta = -\kappa \mathbf{x} + \omega t. \tag{17}$$

In the above equation, *θ* (*x*, *t*) symbolize the phase component of the soliton, *κ* represent the soliton frequency, while *ω* denote the wave number, and *ν* symbolize the velocity of the soliton. Substitute wave transformation into Equation (3), and separate the outcome equation into real and imaginary parts. We can write the real part as follows:

$$-\left(a\kappa^3 - \beta\kappa^4 + \omega\right)lI + c\_1lI^3 + c\_2lI^5 + 3a\kappa lI^{\prime\prime} - 6\beta\kappa^2lI^{\prime\prime} + \beta lI^{(4)} = 0,\tag{18}$$

and the imaginary part can be written as:

$$\left(3\alpha \kappa^2 - 4\beta \kappa^3 + \nu\right) \mathcal{U}' - \left(\kappa - 4\beta \kappa\right) \mathcal{U}^{(3)} = 0. \tag{19}$$

From Equation (19) *U*- = 0 and *U*---= 0, then:

$$
\lambda \nu = 4\beta \kappa^3 - 3a\kappa^2, \quad a = 4\beta \kappa. \tag{20}
$$

Hence, Equation (18) can be rewritten as:

$$\left(3\beta\kappa^4 + \omega\right)\mathcal{U} - c\_1\mathcal{U}^3 - c\_2\mathcal{U}^5 - 12\beta\kappa^2\mathcal{U}'' + 6\beta\kappa^2\mathcal{U}'' - \beta\mathcal{U}^{(4)} = 0. \tag{21}$$

Multiplying both sides of Equation (21) by *U*and taking its integration with respect to *ξ*, we get:

$$\times \left(-12\left(\mathcal{U}''\right)^2 + 24\mathcal{U}'''\mathcal{U}'\right) + 6c\_1\mathcal{U}^4 + 4c\_2\mathcal{U}^6 + 72\beta\kappa^2 \left(\mathcal{U}'\right)^2 - \left(36\beta\kappa^4 + 12\omega\right)\mathcal{U}^2 = 0.\tag{22}$$

Finding the balance, we gain *n* = 1. Replacing this value of balance into Equation (8), we get:

$$\mathcal{U}\left(\mathcal{J}\right) = a\_{-1}(m+F)^{-1} + a\_0 + a\_1\left(m+F\right). \tag{23}$$

By substituting Equation (23) into Equation (3) by using Equation (9), we get the following solutions:

Case 1. When *a*<sup>0</sup> = *<sup>λ</sup>a*<sup>1</sup> <sup>2</sup> , *κ* = ∓ (2*m*+*λ*) <sup>2</sup>−4*<sup>μ</sup>* <sup>√</sup><sup>6</sup> , *<sup>c</sup>*<sup>1</sup> <sup>=</sup> <sup>8</sup>*β*((2*m*+*λ*) <sup>2</sup>−4*μ*) *a*2 1 , *<sup>c</sup>*<sup>2</sup> <sup>=</sup> <sup>−</sup><sup>24</sup> *<sup>β</sup> a*4 1 , *a*−<sup>1</sup> = 0, and Δ = (*λ* + 2*m*) <sup>2</sup> <sup>−</sup> <sup>4</sup>*μ*, we get an exponential function solution as follows:

$$\begin{split} u(x,t) &= \mathrm{e}^{\mathrm{i}\left(\sqrt{\frac{\Delta}{8}}x + \frac{5}{12}\beta\Delta^{2}t\right)}\\ &\quad \left(\frac{\lambda a\_{1}}{2} + a\_{1}\left(m + \frac{1}{2}\left(-2m + \left(1 - \frac{2A\_{1}}{A\_{1} + A\_{2}\mathrm{e}^{\sqrt{\Delta}\left(x - \frac{2}{3}\sqrt{\frac{3}{3}\beta\beta\Delta^{3}t}\right)}}\right)\sqrt{\Delta} - \lambda\right)\right)\right), \end{split} \tag{24}$$

which is a dark solution, as shown in Figure 1, *A*<sup>1</sup> and *A*<sup>2</sup> are non-zero numbers, and Δ > 0. Figure 1 shows that Equation (24) is a dark soliton under the suitable values of parameters.

**Figure 1.** 3D surface of Equation (24), which is a dark optical soliton solution plotted when *A*<sup>1</sup> = 1, *A*<sup>2</sup> = 0.3, *β* = 0.2, *a*<sup>1</sup> = 0.4, *λ* = 1, *m* = 1, *μ* = −1, and *t* = 2 for 2D.

*Appl. Sci.* **2020**, *10*, 219

Case 2. When *<sup>a</sup>*<sup>0</sup> <sup>=</sup> <sup>−</sup> *<sup>λ</sup>a*−<sup>1</sup> <sup>2</sup>*m*(*m*+*λ*)−2*<sup>μ</sup>* , *<sup>a</sup>*<sup>1</sup> <sup>=</sup> 0, *<sup>a</sup>*<sup>2</sup> <sup>=</sup> <sup>12</sup>*<sup>ω</sup>* 5((2*m*+*λ*) <sup>2</sup>−4*μ*) <sup>2</sup> , *κ* = (2*m*+*λ*) <sup>2</sup>−4*<sup>μ</sup>* <sup>√</sup><sup>6</sup> , *<sup>c</sup>*<sup>1</sup> <sup>=</sup> 96(−*m*(*m*+*λ*)+*μ*) 2*ω* 5((2*m*+*λ*) <sup>2</sup>−4*μ*)*a*<sup>2</sup> −1 , *<sup>c</sup>*<sup>2</sup> <sup>=</sup> <sup>−</sup>288(−*m*(*m*+*λ*)+*μ*) 4*ω* 5((2*m*+*λ*) <sup>2</sup>−4*μ*) 2 *a*4 −1 , and Δ = (*λ* + 2*m*) <sup>2</sup> <sup>−</sup> <sup>4</sup>*μ*, we obtain an exponential function solution:

$$u\left(x,t\right) = \frac{\operatorname{sinc}\left(-\frac{z\sqrt{\left(2m+\lambda\right)^{2}-4\mu}}{\sqrt{6}}+t\omega\right)}{m+\frac{1}{2}\left(-2m+\left(1-\frac{2A\_{1}}{\sqrt{\pi\left(\frac{s}{\lambda}+\frac{\sqrt{\lambda}}{8\sqrt{\lambda}}\tau\right)}}\right)\sqrt{\Delta}-\lambda\right)} + \frac{\lambda a\_{-1}\operatorname{sinc}\left(-\frac{z\sqrt{\left(2m+\lambda\right)^{2}-4\mu}}{\sqrt{6}}+t\omega\right)}{2m\left(m+\lambda\right)-2\mu},\tag{25}$$

which is a soliton solution, as shown in Figure 2, *A*<sup>1</sup> and *A*<sup>2</sup> are non-zero numbers, and Δ > 0.

With the suitable values, Figure 2 presents that Equation (25) is a singular soliton.

**Figure 2.** 3D surface of Equation (25), which is a singular soliton solution plotted when *A*<sup>1</sup> = 2, *A*<sup>2</sup> = 3, *<sup>β</sup>* = 6, *<sup>a</sup>*−<sup>1</sup> = 6, *<sup>λ</sup>* = 1, *<sup>m</sup>* = 1, *<sup>μ</sup>* = −1, and *<sup>t</sup>* = 2 for 2D.

Case 3. When *<sup>a</sup>*−<sup>1</sup> <sup>=</sup> <sup>−</sup> <sup>i</sup> <sup>√</sup><sup>3</sup> *<sup>c</sup>*1(*m*(*m*+*λ*)−*μ*) *c*2((2*m*+*λ*) <sup>2</sup>−4*μ*) , *a*<sup>0</sup> = <sup>i</sup> <sup>√</sup><sup>3</sup> *<sup>c</sup>*1*<sup>λ</sup>* 2 *c*2((2*m*+*λ*) <sup>2</sup>−4*μ*) , *<sup>a</sup>*<sup>1</sup> <sup>=</sup> 0, *<sup>ω</sup>* <sup>=</sup> <sup>−</sup>5*c*<sup>1</sup> 2 32*c*<sup>2</sup> , *κ* = ∓ (2*m*+*λ*) <sup>2</sup>−4*<sup>μ</sup>* <sup>√</sup><sup>6</sup> , *<sup>γ</sup>* <sup>=</sup> <sup>−</sup> <sup>3</sup>*c*<sup>1</sup> 2 8*c*2((2*m*+*λ*) <sup>2</sup>−4*μ*) <sup>2</sup> , and Δ = (*λ* + 2*m*) <sup>2</sup> <sup>−</sup> <sup>4</sup>*μ*, we have an exponential

function solution:

$$\begin{aligned} \mathbf{u} \ (x, t) &= \mathbf{e}^{\frac{1}{2} \left( -\frac{5z\_1^2}{3\sqrt{c\_1}} t + \frac{\sqrt{3}}{\sqrt{c\_1}} x \right)} \\ &\quad \left( \frac{\mathbf{i}\sqrt{3}\sqrt{c\_1}\lambda}{2\sqrt{c\_2}\Delta} - \frac{\mathbf{i}\sqrt{3}\sqrt{c\_1} \left( m(m+\lambda) - \mu \right)}{\left( m + \frac{1}{2} \left( -2m + \left( 1 - \frac{2A\_1}{A\_{1+2\varepsilon}\frac{z\_1^2}{2\sqrt{c\_2}\Delta}} \right) \sqrt{\Delta} - \lambda \right) \right) \sqrt{c\_2}\Delta} \right), \end{aligned} \tag{26}$$

which is a soliton solution, as shown in Figure 3, *A*<sup>1</sup> and *A*<sup>2</sup> are non-zero numbers, and Δ > 0. Considering some values of parameters, Figure 3 shows singular soliton solution.

**Figure 3.** 3D surface of Equation (26), which is a singular soliton solution plotted when *A*<sup>1</sup> = 0.3, *A*<sup>2</sup> = 2, *c*<sup>1</sup> = 0.3, *c*<sup>2</sup> = 2, *λ* = 1, *m* = 1, *μ* = −1, and *t* = 2 for 2D.
