Instability modulation for the (2+1)-dimension paraxial wave equation and its new optical soliton solutions in Kerr media

The dispersal of ultrashort pulses of electromagnetic radiation into a nonlinear medium is a multidimensional phenomenon. The interaction between different physical procedures such as dispersion, material dispersion, diffraction, and nonlinear response impacts the pulse dynamics. Due to the interaction of dispersion, diffraction and nonlinearity, a non-dispersive and non-diffractive wave packet called soliton (light bullet) is created. Solitons have many applications in optical microscopy, optical information storage, laser-induced particle acceleration, Bose–Einstein condensation, and high-resolution signal transmission [1].

The ubiquitous phenomenon that originates from the interplay of linear dispersion or diffraction and the nonlinear self-interaction of wave areas is called modulational instability (MI). This impact was first theoretically recognized by Benjamin and Feir in 1967 for deep-water waves [2]. MI studies are increasingly interested in nonlinear optics, fluid dynamics, Bose–Einstein condensate, physics of plasma and other areas.

Many methods are used to find and analyze solutions of nonlinear differential equations, such as shooting with Runga–Kutta fourth-order technique [3–6], finite difference method [7, 8], homotopy perturbation method [9], Adams–Bashforth–Moulton method [10], Adomian decomposition method [11], trial equation method [12], the modified Darboux transformation technique [13], the Bäcklund transformations method [14], the simple equation method [15], sine-Gordon expansion method [16], lie symmetries along with $\left({}^{G^{\prime} }/{}_{G}\right)$ expansion method [17], the the bilinear method [18, 19], extended trial equation method [20], the extended sinh-Gordon expansion method [21–23], improved Bernoulli sub-equation function method [24, 25], the multiplier approach [26], modified simple equation method [27], exp $\left(-\varphi \left(\xi \right)\right)$ expansion method [28], method of undetermined coefficients [29], couple of integration schemes [30], improved tan $\left({}^{\phi \left(\xi \right)}/{}_{2}\right)$-expansion method [31], tanh function method [32], the modified tanh-function method [33], Jacobi elliptic function anzätz method [34], the modified kudryashov method [35], and inverse mapping method [36]. In [37], authors extended the variable coefficient Jacobian elliptic function method to solve nonlinear differential equation. The balance between different-order nonlinearities and high-order dispersion/diffraction in parity-time symmetric potentials was used to construct three-dimensional optical solitons [38]. In [39, 40], exact vector multipole and vortex solitons of nonlinear Schrödinger equation were also investigated. Moreover, many powerful methods have been used and also extended to find new properties of mathematical models symbolizing serious real world problems [41–49].

In this paper, we use the modified auxiliary expansion method to seek novel soliton solutions of the paraxial nonlinear Schrödinger equation. The new solutions are presented in terms of the family solution and expressed in hyperbolic, trigonometric and exponential functions. Finally, the instability modulation of the paraxial wave equation is also presented.

Suppose that, we have the following nonlinear partial differential equation

$$\begin{eqnarray}&&P\left(u,{u}_{x},{u}^{2}{u}_{x},{u}_{t},{u}_{{tt}},\ldots \right)=0.\end{eqnarray} \tag{ 1 }$$

To find the explicit exact solutions of equation (1), we use the following transformation

$$\begin{eqnarray}&&u\left(x,y,t\right)=U\left(\xi \right),\ \ \xi =x-\nu \,t,\end{eqnarray} \tag{ 2 }$$

where ν is arbitrary constant and ξ is the symbol of the wave variable. Substituting equation (2) to equation (1), the result is a nonlinear ordinary differential equation as follows

$$\begin{eqnarray}&&N\left(U,{U}^{2},U^{\prime} ,{U}^{{\prime} ^{\prime} },\ldots \right)=0.\end{eqnarray} \tag{ 3 }$$

Now the trial equation of solution for equation (3) is defined as

$$\begin{eqnarray}&&U\left(\xi \right)={a}_{0}+\displaystyle \sum _{i=1}^{n}{a}_{i}{K}^{i{\rm{\Phi }}\left(\xi \right)}+\displaystyle \sum _{i=1}^{n}{b}_{i}{K}^{-i{\rm{\Phi }}\left(\xi \right)},\end{eqnarray} \tag{ 4 }$$

where a_i and b_i, $\left(1\leqslant i\leqslant n\right)$ are non-zero constants and ${\rm{\Phi }}\left(\xi \right)$ is the auxiliary ODE given by

$$\begin{eqnarray}&&{\rm{\Phi }}^{\prime} \left(\xi \right)=\displaystyle \frac{{K}^{-{\rm{\Phi }}\left(\xi \right)}+\mu \,{K}^{{\rm{\Phi }}\left(\xi \right)}+\lambda }{\mathrm{ln}\left(K\right)},\end{eqnarray} \tag{ 5 }$$

where μ, λ are constants and $K\gt 0,K\ne 1$. The auxiliary ODE has the general solution as follows:

• (i)  
  When λ² − 4μ > 0, then $f\left(\xi \right)={\mathrm{log}}_{K}$ $\left(-\lambda -\sqrt{{\lambda }^{2}-4\mu }\tanh \right.$ $\left.\left(\tfrac{1}{2}\sqrt{{\lambda }^{2}-4\mu }\left(C+\xi \right)\right)\right).$
• (ii)  
  When λ² − 4μ < 0, then $f\left(\xi \right)={\mathrm{log}}_{K}$ $\left(-\lambda +\sqrt{-{\lambda }^{2}+4\mu }\tan \right.$ $\left.\left(\tfrac{1}{2}\sqrt{-{\lambda }^{2}+4\mu }\left(C+\xi \right)\right)\right).$
• (iii)  
  When ${\lambda }^{2}-4\mu \ne 0$, λ = 0 and μ < 0, then $f\left(\xi \right)={\mathrm{log}}_{K}$ $\left(\sqrt{-4\mu }\coth \right.$ $\left.\left(\tfrac{1}{2}\sqrt{-4\mu }\left(C+\xi \right)\right)\right).$
• (iv)  
  When ${\lambda }^{2}-4\mu \ne 0$, λ = 0 and μ > 0, then $f\left(\xi \right)={\mathrm{log}}_{K}$ $\left(\sqrt{4\mu }\cot \right.$ $\left.\left(\tfrac{1}{2}\sqrt{4\mu }\left(C+\xi \right)\right)\right).$
• (v)  
  When λ² − 4μ > 0 and μ = 0, then $f\left(\xi \right)={\mathrm{log}}_{K}$ $\left(\tfrac{\lambda }{-1+\cosh \left(\lambda \left(\epsilon +\xi \right)\right)+\sinh \left(\lambda \left(\epsilon +\xi \right)\right)}\right).$
• (vi)  
  When λ² − 4μ = 0, $\lambda \ne 0$ and $\mu \ne 0$, then $f\left(\xi \right)={\mathrm{log}}_{K}$ $\left(-\tfrac{2\lambda \left(\xi +\epsilon \right)+4}{{\lambda }^{2}\left(\xi +\epsilon \right)}\right).$
• (vii)  
  When λ² − 4μ = 0, λ = 0 and μ = 0, then $f\left(\xi \right)={\mathrm{log}}_{K}\left(\xi +\epsilon \right).$

The paraxial NLSE in Kerr media is given by

$$\begin{eqnarray}&&{{\rm{i}}{u}}_{y}+\displaystyle \frac{\alpha }{2}{u}_{{tt}}+\displaystyle \frac{\beta }{2}{u}_{{xx}}+\gamma {\left|u\right|}^{2}u=0,\end{eqnarray} \tag{ 6 }$$

where $u=u\left(x,y,t\right)$ is the complex wave envelope function. The constants α , β and γ are the symbols of the dispersion, diffraction and Kerr nonlinearity, respectively. In equation (1) if we get elliptic nonlinear Schrödinger equation and if α β < 0, equation (1) becomes hyperbolic nonlinear Schrödinger equation. Now assume the following wave transformations:

$$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & U\left(\xi \right){{\rm{e}}}^{{\rm{i}}\theta },\,\,\,\xi =x+y-{ct},\\ \theta & = & \kappa x+\omega y-\nu t,\end{array}\end{eqnarray} \tag{ 7 }$$

Inserting equation (7) into equation (6), and separate the result into the real and imaginary part, we get

$$\begin{eqnarray}&&-\left({c}^{2}\alpha +\beta \right)U^{\prime\prime} +\left(\beta {\kappa }^{2}+\alpha {\nu }^{2}+2\omega \right)U-2\gamma {U}^{3}=0\end{eqnarray} \tag{ 8 }$$

and

$$\begin{eqnarray}&&\left(1+\beta \kappa +c\alpha \nu \right)U^{\prime} =0.\end{eqnarray} \tag{ 9 }$$

Now, we know that $U^{\prime} \ne 0$, therefore

$$\begin{eqnarray}&&\beta =\displaystyle \frac{-1-c\alpha \nu }{\kappa }.\end{eqnarray} \tag{ 10 }$$

Putting equation (10) into equation (8) to get the closed solution, we get

$$\begin{eqnarray}&&\begin{array}{l}\left(1-{c}^{2}\alpha \kappa +c\alpha \nu \right)U^{\prime\prime} -\kappa \left(\kappa +c\alpha \kappa \nu -\alpha {\nu }^{2}\right.\\ \ \ \left.-\ 2\omega \right)U-2\gamma \kappa {U}^{3}=0.\end{array}\end{eqnarray} \tag{ 11 }$$

Finding the homogeneous principal balance between $U^{\prime\prime}$ and U³, we get n = 1. Putting the value of into equation (4), the equation (4) can be written as the following

$$\begin{eqnarray}&&U\left(\xi \right)={a}_{0}+{a}_{1}{K}^{\phi \left(\xi \right)}+{b}_{1}{K}^{-\phi \left(\xi \right)}.\end{eqnarray} \tag{ 12 }$$

Using equation (12) and its second derivative with equation (11), we analyze the following cases and solutions:

Case 1. When ${a}_{0}\,=\tfrac{{\rm{i}}\lambda \sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }}{2\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}},$ ${b}_{1}\,=\tfrac{{\rm{i}}\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }}{\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}}$, a₁ = 0, $c\,=\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}-4\mu }+\tfrac{1}{\kappa -2\omega }\right)$ and $\alpha =\tfrac{{\left(\kappa -2\omega \right)}^{2}}{2{\nu }^{2}\left(\kappa -\omega \right)}$ we get the following family solution:

• Family 1. When λ² − 4μ > 0, $\lambda \ne 0$, $\mu \ne 0$ and γ > 0, then

  $$\begin{eqnarray}\begin{array}{rcl}{u}\left(x,y,t\right) & = & \displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }}{2\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}}\\ & & \left(\displaystyle \frac{{\lambda }^{2}-4\mu +\lambda \sqrt{{\lambda }^{2}-4\mu }\tanh \left(\tfrac{1}{2}\sqrt{{\lambda }^{2}-4\mu }\left(x+y+\epsilon -t\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}-4\mu }+\tfrac{1}{\kappa -2\omega }\right)\right)\right)}{\left(\lambda +\sqrt{{\lambda }^{2}-4\mu }\tanh \left(\tfrac{1}{2}\sqrt{{\lambda }^{2}-4\mu }\left(x+y+\epsilon -t\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}-4\mu }+\tfrac{1}{\kappa -2\omega }\right)\right)\right)\right)}\right.,\end{array}\end{eqnarray} \tag{ 13 }$$

  which is dark soliton solution of equation (6).
• Family 2. When λ² − 4μ < 0, $\lambda \ne 0$, $\mu \ne 0$ and γ < 0, then

  $$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & \displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }\,}{2\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}}\\ & & \left(\lambda -\displaystyle \frac{4\mu }{\lambda -\sqrt{-{\lambda }^{2}+4\mu }\tan \left[\tfrac{1}{2}\sqrt{-{\lambda }^{2}+4\mu }\left(x+y+\epsilon -t\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}-4\mu }+\tfrac{1}{\kappa -2\omega }\right)\right)\right]}\right),\end{array}\end{eqnarray} \tag{ 14 }$$

  which is singular solution of equation (6).
• Family 3.When λ = 0, μ < 0 and γ > 0, then

  $$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & \displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{-\mu }\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}-12\kappa \mu -4{\kappa }^{2}\omega +8\mu \omega }}{4\sqrt{\gamma {\mu }^{2}\left(\kappa -\omega \right)}}\\ & & \tanh \left(\sqrt{-\mu }\left(x+y+\epsilon -t\nu \left(\displaystyle \frac{1}{\kappa }-\displaystyle \frac{\kappa }{2\mu }+\displaystyle \frac{1}{\kappa -2\omega }\right)\right)\right),\end{array}\end{eqnarray} \tag{ 15 }$$

  which is dark soliton solution of equation (6).
• Family 4. When λ = 0, μ > 0 and γ < 0, then

  $$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & \displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\mu }\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}-12\kappa \mu -4{\kappa }^{2}\omega +8\mu \omega }}{4\sqrt{\gamma {\mu }^{2}\left(\kappa -\omega \right)}}\\ & & \tan \left(\sqrt{\mu }\left(x+y+\epsilon -t\nu \left(\displaystyle \frac{1}{\kappa }-\displaystyle \frac{\kappa }{2\mu }+\displaystyle \frac{1}{\kappa -2\omega }\right)\right)\right),\end{array}\end{eqnarray} \tag{ 16 }$$

  which is singular solution of equation (6).
• Family 5. When λ² − 4μ > 0, μ = 0 and γ > 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\lambda \sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa {\lambda }^{2}-4{\kappa }^{2}\omega -2{\lambda }^{2}\omega }\coth \left(\tfrac{1}{2}\lambda \left(x+y+\epsilon -t\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}}+\tfrac{1}{\kappa -2\omega }\right)\right)\right)}{2\sqrt{\gamma {\lambda }^{4}\left(\kappa -\omega \right)}},\end{eqnarray} \tag{ 17 }$$

  which is singular solution of equation (6).

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

Case 2. When ${a}_{0}=\tfrac{{\rm{i}}\lambda \sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }}{2\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}}$, ${a}_{1}=\tfrac{{\rm{i}}\mu \sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }}{\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}}$, b₁ = 0, $c=\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}-4\mu }+\tfrac{1}{\kappa -2\omega }\right)$ and $\alpha =\tfrac{{\left(\kappa -2\omega \right)}^{2}}{2{\nu }^{2}\left(\kappa -\omega \right)}$ we get the following family solution:

• Family 1. When λ² − 4μ > 0, $\lambda \ne 0$, $\mu \ne 0$ and γ > 0, then

  $$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & -\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{{\lambda }^{2}-4\mu }\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }}{2\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}}\\ & & \tanh \left(\displaystyle \frac{1}{2}\sqrt{{\lambda }^{2}-4\mu }\left(x+y+\epsilon -t\nu \left(\displaystyle \frac{1}{\kappa }+\displaystyle \frac{2\kappa }{{\lambda }^{2}-4\mu }+\displaystyle \frac{1}{\kappa -2\omega }\right)\right)\right).\end{array}\end{eqnarray} \tag{ 18 }$$
• Family 2. When λ² − 4μ < 0 , $\lambda \ne 0$, $\mu \ne 0$ and γ > 0, then

  $$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & \displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa \left({\lambda }^{2}-4\mu \right)-4{\kappa }^{2}\omega -2\left({\lambda }^{2}-4\mu \right)\omega }}{2\sqrt{\gamma {\left({\lambda }^{2}-4\mu \right)}^{2}\left(\kappa -\omega \right)}}\\ & & \left(\lambda -\displaystyle \frac{4\mu }{\lambda -\sqrt{-{\lambda }^{2}+4\mu }\tan \left(\tfrac{1}{2}\sqrt{-{\lambda }^{2}+4\mu }\left(x+y+\epsilon -t\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}-4\mu }+\tfrac{1}{\kappa -2\omega }\right)\right)\right)}\right).\end{array}\end{eqnarray} \tag{ 19 }$$
• Family 3. When λ = 0, μ < 0 and γ > 0, then

  $$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & \displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{-\mu }\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}-12\kappa \mu -4{\kappa }^{2}\omega +8\mu \omega }}{4\sqrt{\gamma {\mu }^{2}\left(\kappa -\omega \right)}}\\ & & \tanh \left(\sqrt{-\mu }\left(x+y+\epsilon -t\nu \left(\displaystyle \frac{1}{\kappa }-\displaystyle \frac{\kappa }{2\mu }+\displaystyle \frac{1}{\kappa -2\omega }\right)\right)\right).\end{array}\end{eqnarray} \tag{ 20 }$$
• Family 4. When λ = 0 and μ > 0, then

  $$\begin{eqnarray}\begin{array}{rcl}u\left(x,y,t\right) & = & \displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\mu }\sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}-12\kappa \mu -4{\kappa }^{2}\omega +8\mu \omega }}{4\sqrt{\gamma {\mu }^{2}\left(\kappa -\omega \right)}}\\ & & \tan \left(\sqrt{\mu }\left(x+y+\epsilon -t\nu \left(\displaystyle \frac{1}{\kappa }-\displaystyle \frac{\kappa }{2\mu }+\displaystyle \frac{1}{\kappa -2\omega }\right)\right)\right).\end{array}\end{eqnarray} \tag{ 21 }$$
• Family 5. λ² − 4μ > 0 and μ = 0. then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\lambda \sqrt{\kappa -2\omega }\sqrt{2{\kappa }^{3}+3\kappa {\lambda }^{2}-4{\kappa }^{2}\omega -2{\lambda }^{2}\omega }\coth \left(\tfrac{1}{2}\lambda \left(x+y+\epsilon -t\nu \left(\tfrac{1}{\kappa }+\tfrac{2\kappa }{{\lambda }^{2}}+\tfrac{1}{\kappa -2\omega }\right)\right)\right)}{2\sqrt{\gamma {\lambda }^{4}\left(\kappa -\omega \right)}}.\end{eqnarray} \tag{ 22 }$$

Zoom In Zoom Out Reset image size

Case 3. When ${a}_{0}=-\tfrac{{\rm{i}}\lambda \sqrt{c\kappa +\nu -2c\omega }}{\sqrt{2c\gamma {\lambda }^{2}-8c\gamma \mu -4\gamma \kappa \nu }}$, ${a}_{1}=-\tfrac{2{\rm{i}}\mu \sqrt{c\kappa +\nu -2c\omega }}{\sqrt{2c\gamma {\lambda }^{2}-8c\gamma \mu -4\gamma \kappa \nu }}$,b₁ = 0 and $\alpha =\tfrac{{\lambda }^{2}-4\mu +2\kappa \left(\kappa -2\omega \right)}{\left(c\kappa -\nu \right)\left(c\left({\lambda }^{2}-4\mu \right)-2\kappa \nu \right)},$ we get the following family solutions:

• Family 1. When λ² − 4μ > 0, $\lambda \ne 0$ and $\mu \ne 0$, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{{\lambda }^{2}-4\mu }\sqrt{\nu +c\left(\kappa -2\omega \right)}\tanh \left(\tfrac{1}{2}\left(-{ct}+x+y+\epsilon \right)\sqrt{{\lambda }^{2}-4\mu }\right)}{\sqrt{2}\sqrt{\gamma \left(c\left({\lambda }^{2}-4\mu \right)-2\kappa \nu \right)}}.\end{eqnarray} \tag{ 23 }$$
• Family 2. When λ² − 4μ < 0, $\lambda \ne 0$ and $\mu \ne 0$, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{-\tfrac{{\lambda }^{2}}{2}+2\mu }\sqrt{\nu +c\left(\kappa -2\omega \right)}\tan \left(\tfrac{1}{2}\left(-{ct}+x+y+\epsilon \right)\sqrt{-{\lambda }^{2}+4\mu }\right)}{\sqrt{\gamma \left(c\left({\lambda }^{2}-4\mu \right)-2\kappa \nu \right)}}.\end{eqnarray} \tag{ 24 }$$
• Family 3. When λ = 0 and μ < 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{-\mu }\sqrt{\nu +c\left(\kappa -2\omega \right)}\coth \left(\left(-{ct}+x+y+\epsilon \right)\sqrt{-\mu }\right)}{\sqrt{-\gamma \left(2c\mu +\kappa \nu \right)}}.\end{eqnarray} \tag{ 25 }$$
• Family 4. When λ = 0 and μ > 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\mu }\sqrt{\nu +c\left(\kappa -2\omega \right)}\cot \left[\left(-{ct}+x+y+\epsilon \right)\sqrt{\mu }\right]}{\sqrt{-\gamma \left(2c\mu +\kappa \nu \right)}}.\end{eqnarray} \tag{ 26 }$$
• Family 5. When λ² − 4μ > 0 and μ = 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\rm{i}}\lambda \sqrt{c\kappa +\nu -2c\omega }}{\sqrt{2c\gamma {\lambda }^{2}-4\gamma \kappa \nu }}.\end{eqnarray} \tag{ 27 }$$
• Family 6. When λ² − 4μ = 0, $\lambda \ne 0$ and $\mu \ne 0$, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\nu +c\left(\kappa -2\omega \right)}}{\left(1-{ct}+x+y+\epsilon \right)\sqrt{-\gamma \kappa \nu }}.\end{eqnarray} \tag{ 28 }$$
• Family 7. When λ² − 4μ = 0, λ = 0 and μ = 0,

  $$\begin{eqnarray}&&u\left(x,y,t\right)=0.\end{eqnarray} \tag{ 29 }$$

Zoom In Zoom Out Reset image size

Case 4. When ${a}_{0}=-\tfrac{{\rm{i}}\lambda \sqrt{c\kappa +\nu -2c\omega }}{\sqrt{2c\gamma {\lambda }^{2}-8c\gamma \mu -4\gamma \kappa \nu }}$,a₁ = 0, ${b}_{1}=-\tfrac{2{\rm{i}}\sqrt{c\kappa +\nu -2c\omega }}{\sqrt{2c\gamma {\lambda }^{2}-8c\gamma \mu -4\gamma \kappa \nu }}$ and $\alpha =\tfrac{{\lambda }^{2}-4\mu +2\kappa \left(\kappa -2\omega \right)}{\left(c\kappa -\nu \right)\left(c\left({\lambda }^{2}-4\mu \right)-2\kappa \nu \right)}$ we get the following family solution:

• Family 1. When λ² − 4μ > 0, $\lambda \ne 0$ and $\mu \ne 0$, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\nu +c\left(\kappa -2\omega \right)}\left({\lambda }^{2}-4\mu +\lambda \sqrt{{\lambda }^{2}-4\mu }\tanh \left(\tfrac{1}{2}\left(-{ct}+x+y+\epsilon \right)\sqrt{{\lambda }^{2}-4\mu }\right)\right)}{\sqrt{2}\sqrt{\gamma \left(c\left({\lambda }^{2}-4\mu \right)-2\kappa \nu \right)}\left(\lambda +\sqrt{{\lambda }^{2}-4\mu }\tanh \left(\tfrac{1}{2}\left(-{ct}+x+y+\epsilon \right)\sqrt{{\lambda }^{2}-4\mu }\right)\right)}.\end{eqnarray} \tag{ 30 }$$
• Family 2. When λ² − 4μ < 0, $\lambda \ne 0$ and $\mu \ne 0$, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\nu +c\left(\kappa -2\omega \right)}\left(\lambda -\tfrac{4\mu }{\lambda -\sqrt{-{\lambda }^{2}+4\mu }\tan \left[\tfrac{1}{2}\left(-{ct}+x+y+\epsilon \right)\sqrt{-{\lambda }^{2}+4\mu }\right]}\right)}{\sqrt{2}\sqrt{\gamma \left(c\left({\lambda }^{2}-4\mu \right)-2\kappa \nu \right)}}\end{eqnarray} \tag{ 31 }$$
• Family 3. When λ = 0 and μ < 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{-\mu }\sqrt{\nu +c\left(\kappa -2\omega \right)}\tanh \left(\left(-{ct}+x+y+\epsilon \right)\sqrt{-\mu }\right)}{\sqrt{-\gamma \left(2c\mu +\kappa \nu \right)}}.\end{eqnarray} \tag{ 32 }$$
• Family 4. When λ = 0 and μ > 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\mu }\sqrt{\nu +c\left(\kappa -2\omega \right)}\tan \left(\left(-{ct}+x+y+\epsilon \right)\sqrt{\mu }\right)}{\sqrt{-\gamma \left(2c\mu +\kappa \nu \right)}}.\end{eqnarray} \tag{ 33 }$$
• Family 5. When λ² − 4μ > 0 and μ = 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\lambda \sqrt{\nu +c\left(\kappa -2\omega \right)}\coth \left(\tfrac{1}{2}\left(-{ct}+x+y+\epsilon \right)\lambda \right)}{\sqrt{2}\sqrt{\gamma \left(c{\lambda }^{2}-2\kappa \nu \right)}}.\end{eqnarray} \tag{ 34 }$$
• Family 6. When λ² − 4μ = 0, $\lambda \ne 0$ and $\mu \ne 0$, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\nu +c\left(\kappa -2\omega \right)}}{\left(-{ct}+x+y+\epsilon \right)\sqrt{-\gamma \kappa \nu }}.\end{eqnarray} \tag{ 35 }$$
• Family 7. When λ² − 4μ = 0, λ = 0 and μ = 0, then

  $$\begin{eqnarray}&&u\left(x,y,t\right)=-\displaystyle \frac{{\mathrm{ie}}^{{\rm{i}}\left(x\kappa -t\nu +y\omega \right)}\sqrt{\nu +c\left(\kappa -2\omega \right)}}{\left(-{ct}+x+y+\epsilon \right)\sqrt{-\gamma \kappa \nu }}\end{eqnarray} \tag{ 36 }$$

In this section, we analyze the modulation instability (MI) of the stationary solutions of equation (6) by utilizing the virtue of linear stability analysis [50–52]. The MI may consist of the exponential growth of small disturbances in the amplitude or optical wave phase. It is essential to observe the MI in the nonlinear physics of the wave. Assume that equation (6) have the following stationary solutions [53, 54]:

$$\begin{eqnarray}&&u\left(x,y,t\right)=a{{\rm{e}}}^{{\rm{i}}\psi t},\end{eqnarray} \tag{ 37 }$$

where a is arbitrary real constants. Inserting equation (37) into equation (6), we get $\psi =\sqrt{\tfrac{2{a}^{2}\gamma }{\alpha }}$. Suppose that the perturbed stationary solution is given by:

$$\begin{eqnarray}&&u\left(x,y,t\right)=\left(a+\varepsilon U\left(x,y,t\right)\right){{\rm{e}}}^{{\rm{i}}\sqrt{\tfrac{2{a}^{2}\gamma }{\alpha }}t},\end{eqnarray} \tag{ 38 }$$

here $U\left(x,y,t\right)$ is a complex function. Using equations (38) and (6), the outcomes satisfy the following linear equations

$$\begin{eqnarray}&&a\gamma \left(U+{U}^{* }\right)+\alpha \,{U}_{{tt}}+2{\rm{i}}\,{U}_{y}+\beta \,{U}_{{xx}}=0.\end{eqnarray} \tag{ 39 }$$

Where U* is the conjugate function. therefore, equation (39) can be defined as

$$\begin{eqnarray}&&U\left(x,y,t\right)={u}_{1}{{\rm{e}}}^{{\rm{i}}\left({Mx}+{Ny}-{Wt}\right)}+{u}_{2}{{\rm{e}}}^{-{\rm{i}}\left({Mx}+{Ny}-{Wt}\right)},\end{eqnarray} \tag{ 40 }$$

where W denotes the complex frequency, M, N are real disturbance wave-numbers, and u₁, u₂ are the coefficients of the linear combination. Using equation (40) and putting into equation (39), we get the following homogeneous equations

$$\begin{eqnarray}&&\begin{array}{l}\left({a}^{2}\gamma -2N-{W}^{2}\alpha -{M}^{2}\beta \right){U}_{1}+\gamma {a}^{2}{U}_{2}=0,\\ \gamma {a}^{2}{U}_{1}+\left(-{W}^{2}\alpha +{M}^{2}\beta +{a}^{2}\gamma +2N\right){U}_{2}=0.\end{array}\end{eqnarray} \tag{ 41 }$$

Evaluating the determinant, we get the following relationship:

$$\begin{eqnarray}&&-4{N}^{2}+{W}^{4}{\alpha }^{2}-4{M}^{2}N\beta -{M}^{4}{\beta }^{2}-2{a}^{2}{W}^{2}\alpha \gamma =0.\end{eqnarray} \tag{ 42 }$$

Due to equation (42), we can discuss the following cases of the MI [53, 54] for equation (6) as follows:Case 1. If

$$\begin{eqnarray}&&M=\mp \sqrt{-\displaystyle \frac{2N}{\beta }-\displaystyle \frac{\sqrt{{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)}}{{\beta }^{2}}},\end{eqnarray} \tag{ 43 }$$

we observe that the MI of the equation (6) occur if satisfy the following inequalities

$$\begin{eqnarray*}&&{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)\lt 0\end{eqnarray*}$$

or

$$\begin{eqnarray*}&&-\displaystyle \frac{2N}{\beta }-\displaystyle \frac{\sqrt{{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)}}{{\beta }^{2}}\lt 0,\end{eqnarray*}$$

when $\beta \ne 0$.

Case 2. If

$$\begin{eqnarray}&&M=\mp \sqrt{-\displaystyle \frac{2N}{\beta }+\displaystyle \frac{\sqrt{{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)}}{{\beta }^{2}}},\end{eqnarray} \tag{ 44 }$$

the MI of the equation (6) occur if satisfy the following inequalities

$$\begin{eqnarray*}&&{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)\lt 0\end{eqnarray*}$$

or

$$\begin{eqnarray*}&&-\displaystyle \frac{2N}{\beta }+\displaystyle \frac{\sqrt{{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)}}{{\beta }^{2}}\lt 0,\end{eqnarray*}$$

when $\beta \ne 0$. Now we investigate MI gain spectrum as

$$\begin{eqnarray}&&{g}\left(W\right)=2\mathrm{Im}\left(M\right)=2\sqrt{-\displaystyle \frac{2N}{\beta }-\displaystyle \frac{\sqrt{{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)}}{{\beta }^{2}}},\end{eqnarray} \tag{ 45 }$$

$$\begin{eqnarray}&&{g}\left(W\right)=2\mathrm{Im}\left(M\right)=2\sqrt{-\displaystyle \frac{2N}{\beta }+\displaystyle \frac{\sqrt{{W}^{2}\alpha {\beta }^{2}\left({W}^{2}\alpha -2{a}^{2}\gamma \right)}}{{\beta }^{2}}},\end{eqnarray} \tag{ 46 }$$

which can be observe that gain the MI gain is significantly affected by and that represents dispersion, diffraction of equation (6).

In this paper, we used the modified auxiliary expansion method to construct some novel soliton solutions of the (2+1)-dimension paraxial nonlinear Schrödinger equation. We presented a new solution in terms of hyperbolic, trigonometric and exponential functions. The instability modulation of the paraxial wave equation is also presented and analyzed in two cases. According to MI, the MI gain spectrum in the normal-GVD and anomalous-GVD for both cases are studied and illustrated graphically. The affection of all parameters are also illustrated. All our solutions are new, satisfy main paraxial wave equation and might be useful and applicable in the optical fiber industry. Figure 1 and figure 3 represent the dark solution, figure 2 and figure 4 are dark-singular solutionand figure 5 is a singular solution of equation (6). After considering simulations, figure 1 and figure 3 represent the dark solution, figure 2 and figure (4) are dark-singular solution and figure 5 is a singular solution of equation (6).

From figure 6, we conclude that the MI gain spectrum in the normal-GVD regime is increasing via increasing the values of Kerr nonlinearity (γ), real amplitude (a) and real disturbance wave-number (N) while we observe contrary affection of diffraction values β. Also from figure 7, the MI gain spectrum in the anomalous-GVD is decreased by increasing the values of diffraction, real amplitude, and real disturbance wave-number while we see opposite direction of affection for values of real disturbance wave-number.

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

From figure 8, the MI gain spectrum in the normal-GVD is decreased by increasing the values of diffraction, real amplitude, and real disturbance wave-number while we see opposite direction of affection for values of real disturbance wave-number, which is the same effect in the anomals-GVD at equation (45). In another figure, we observe from figure 9 the same affection in the normal-GVD in equation (45). So, from these, we conclude that the normal-GVD obtained from equation (45) have the same characteristics of anomalous-GVD obtained from equation (46), Also anomalous-GVD obtained from equation (45) is the same as normal-GVD obtained from equation (46).

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size