, Volume 1, Issue 4, pp 227–232

# Constructing new wave solutions to the $$(2 + 1)$$-dimensional Davey–Stewartson equation (DSE) which arises in fluid dynamics

Letters

## Abstract

In this paper, we utilize the uniform algebraic method and construct some new wave solutions to the $$(2 + 1)$$-dimensional Davey–Stewartson equation which arises in fluid dynamics. We successfully obtain some hyperbolic and trigonometric function solutions to this equation. Subsequently, We have also built the Lagrangian and the Hamiltonian for the second-order nonlinear ODE corresponding to the traveling wave reduction of the (DSE)-equation. Moreover, we plot 2D and 3D surfaces representing the obtained solutions by considering some suitable values of the parameters. The unique (2 + 1)-dimensional dynamics of the obtained exact solutions of the (DSE) equation may help to understand the formation and key properties of traveling waves in many other physical settings. Before the end of this paper, we compare our results with the existing results in the literature by presenting comprehensive conclusions.

## Keywords

(2+1)-Davey–Stewartson equation Uniform algebraic method Complex hyperbolic function solution

## Mathematics Subject Classification

35Q53 35Q51 37K10 35B

## 1 Introduction

In fluid dynamics, the Davey–Stewartson equation (DSE) was introduced in a paper by Davey and Stewartson  to describe the evolution of a three-dimensional wave-packet on water of finite depth. It is a system of partial differential equations for a complex (wave-amplitude) field $$q = \,q(x, y,t)$$ and a real (mean-flow) field $$r =\, r(x, y,t)$$:
\begin{aligned} iq_{t}+\frac{1}{2}\sigma ^{2} (q_{xx}+\sigma ^{2}q_{yy})+\lambda |q|^{2}q- qr_{x} & = 0,\nonumber \\ r_{xx}-\sigma ^{2} r_{yy}-2\lambda (|q|^{2})_{xx} & = 0. \end{aligned}
(1)
where where q and r are the dependent variables while x and y are the slow, horizontal scales parallel and perpendicular to the fast oscillations, respectively, whereas t is the slow time in the group velocity frame. As usual, irrotational flow of an inviscid fluid is studied, and q is connected with the velocity potential. Of particular significance is the parameter $$\sigma ^{2}$$, and $$\sigma = 1$$ and $$\sigma = -i$$ is called the DS I and DS II equations, respectively. The constant $$\lambda$$ measures the cubic nonlinearity. Furthermore, DS equations, a pair of coupled nonlinear equations in two dependent variables, can be reduced to the (1+1)-dimensional nonlinear Schrödinger (NLS) equation by carrying out a suitable dimensional reduction.

Much work had been done in the past few decades to improve this equation. Especially, in 1988, Boiti et al.  considered the DS equation and constructed solutions that decay exponentially in all directions. Using inverse scattering method, Fokas and Santini [3, 4] derived such solutions and they named these objects as dromion. Hietarinta and Hirota [5, 6] and Radha and Lakshmanan  gained multidromion solutions or one-dromion solutions to DS equation by bilinear method.

Ohta et al.  have investigated the dynamics of rogue waves for DSE. They have submitted to the literature the some basic properties of DSE such as fundamental rogue waves, multi-rogue waves, higher order rogue waves, exploding rogue waves. Gurefe et al.  have obtained the new exact solutions of DSE by using various methods. Zedan et al.  have successfully applied the sine-cosine method to the DSE. Shen et al.  have conducted a new construction procedure based on nonlinear variable separation method for DSE. Recently, numerical and analytical solutions of DS equations are proposed by many authors. Mirzazadeh  used the trial equation method and ansatz approach method. Bhrawy et al.  presented extended Jacobian elliptic function expansion method. Jafari et al.  used first integral method double. The Davey–Stewartson equations are also reduced to Hamiltonian ODEs , and so exact solutions could be furnished by the integrability  of finite dimensional Hamiltonian systems. Their works developed the DS equation in an interesting way. The study of this paper gives rise to some new results for Eq. (1).

One of the most effective direct methods to develop the traveling wave solution of nonlinear PDEs is the uniform algebraic method . This method has been successfully applied to obtain exact solutions for a variety of nonlinear PDEs [17, 18, 19, 20].

The rest of this paper is organized as follows. In Sect. 2, we present some properties of the uniform algebraic method. In Sect. 3, we obtain new families of soliton solutions, and periodic solutions of Davey–Stewartson equation. Graphics of some solutions in Sect. 4. We give the integrals of motion corresponding to the second-order nonlinear ODE in Sect. 5. Section 6 gives out the summary and results of this paper. Finally, Sect. 7 ends the paper with some conclusions.

## 2 The uniform algebraic method

Step 1: Considering a nonlinear evolution equation, with a physical field u, and two independent variables x and t as
\begin{aligned} P(\psi , \psi _{t}, \psi _{x},\psi _{tx},\psi _{tt},\psi _{xx},\ldots ), \end{aligned}
(2)
Then we seek traveling wave solutions to (1) in the form $$\psi (x, t)= \psi (\xi ),\;\xi =k(x-ct)$$, which leads to be a nonlinear ordinary differential equation
\begin{aligned} F(\psi ,\psi ^{\prime},\psi ^{\prime},\ldots ), \end{aligned}
(3)
where the prime denotes $$\frac{{\text{d}}}{{\text{d}}\xi }$$.
Step 2: By introducing $$f (\varpi ),\; g(\varpi )$$ and parameter of $$\lambda$$,  derived a uniform frame for soliton solutions and periodic solutions:
\begin{aligned} \left\{ \begin{array}{ll} {\psi (\xi )=\sum ^{i=N}_{i=-N}a_{i}f^{i}(\varpi ) + \sum ^{i=N}_{i=1}b_{i}f^{i-1}(\varpi )g(\varpi )}}+{\sum ^{i=-1}_{i=-N} c_{i}f^{i}(\varpi )g(\varpi )}}, \end{array} \right. \end{aligned
(4)
With $$\frac{\varpi }{{\text {d}}\xi }=g(\varpi ),\;(f^{i}(\varpi ))^{2}-(g^{i}(\varpi ))^{2} = \lambda ,\;\frac{{\text {d}}f(\varpi )}{{\text {d}}\varpi }=g(\varpi ),\;\frac{{\text {d}}g(\varpi )}{{\text {d}}\varpi }=f(\varpi )$$ (much more generally, $$\frac{{\text {d}}^{2k+1}f(\varpi )}{{\text {d}}\varpi ^{2k+1}}=g(\varpi ),\; \frac{{\text {d}}^{2k}f(\varpi )}{{\text {d}}\varpi ^{2k}}=g(\varpi ),\; \frac{{\text {d}}^{2k+1}g(\varpi )}{{\text {d}}\varpi ^{2k+1}}=f(\varpi )$$ and $$\frac{{\text {d}}^{2k}g(\varpi )}{{\text {d}}\varpi ^{2k}}=f(\varpi ),\;k\in \mathbf {Z} ^{+})$$. and $$\lambda ^{2}=1$$, where $$a_{0},\;a_{\pm 1}, \ldots a_{\pm N},\;b_{1} \ldots b_{N}$$, $$c_{-1} \ldots c_{-N},k,\;c$$ are constants to be determined later. N can be determined by balancing the highest degree linear term with the nonlinear term in (4). Namely,
\begin{aligned}&\left\{ \begin{array}{ll} \text{ when } \; \lambda =1, f(\varpi )=\cosh (\varpi ), \text { and}\; g(\varpi ) =\sinh (\varpi ),\\ \psi (\xi )={\sum ^{i=N}_{i=-N}}}a_{i}\big (-\coth (\xi )\big )^{i} -{\sum ^{i=N}_{i=1}}}b_{i}\big (-\coth (\xi )\big )^{i-1}{\text {csc}}h (\xi ) \\ -{\sum ^{i=-1}_{i=-N}}}c_{i}\big (-\coth (\xi )\big )^{i}{\text {csc}}h (\xi ), \end{array} \right. \end{aligned
(5)
\begin{aligned}&\left\{ \begin{array}{ll} \text { when}\; \lambda =-1, f(\varpi )=\sinh (\varpi ), \text{ and }\; g(\varpi ) =\cosh (\varpi ),\\ \psi (\xi )={\sum ^{i=N}_{i=-N}}}a_{i}\big (-\cot (\xi )\big )^{i} +{\sum ^{i=N}_{i=1}}}b_{i}\big (-\cot (\xi )\big )^{i-1}\csc (\xi )\\ +{\sum ^{i=-1}_{i=-N}}}c_{i}\big (-\cot (\xi )\big )^{i}\csc (\xi ), \end{array} \right. \end{aligned
(6)
Step 3: Substituting (4) into (3), we obtain a finite series of $$f^{k}(\varpi )g^{i}(\varpi )\;(k=0,\;i=0,\;1,...m),$$

Step 4: Set the coefficients of $$f^{k}(\varpi )g^{i}(\varpi )$$ to zero, and we get a set of algebraic equations with respect to the unknown $$a_{i},\; b_{i},\; c_{i}, \;k$$ and c,

Step 5: Solve the equations. Then substitute $$a_{i},\; b_{i},\; c_{i}, \;k$$ and c into (5) and (10).

## 3 The exact solutions to Davey–Stewartson

Assume that Eq. (1) has an exact solution in the form
\begin{aligned} q(x,y,t)= \,& {} \exp (i\theta )\psi (\xi ),\; r(x,y,t)=h(\xi ), \nonumber \\ \xi= & {} x+y-\nu t,\; \theta =px+sy+\omega t \end{aligned},
(7)
where $$\psi (x,y,t)$$ is a real function and $$p,\;s$$ are constants. Substituting (7) into (1) and separating the real and imaginary parts, we obtain
\begin{aligned} \left\{ \begin{array}{ll} \nu =\sigma ^{2}(p+\sigma ^{2}s),\\ \sigma ^{2}(1+\sigma ^{2})\psi ^{\prime}+2\psi ^{3}-2h\psi -(2\omega + \sigma ^{2}(p^{2}+\sigma ^{2}s^{2}))\psi =0,\\ (1-\sigma ^{2})h^{\prime}=2\lambda (\psi ^{2})^{'}. \end{array} \right. \end{aligned}
(8)
Integrating (8) with respect to $$\xi$$ and setting the constant of integration to zero, we find
\begin{aligned} h^{'}=\frac{2\lambda }{1-\sigma ^{2}} \psi ^{2} \end{aligned}.
(9)
Substituting (9) in Eq. (8), we have
\begin{aligned}&\sigma ^{2}(1-\sigma ^{4})\psi ^{\prime}+2\lambda (1+\sigma ^{2})\psi ^{3}\nonumber \\&\quad - (\sigma ^{2}-1)(2\omega +\sigma ^{2}(p^{2}+\sigma ^{2}s^{2}))\psi =0, \end{aligned}
(10)
then Eq. (10) takes the form
\begin{aligned} A\psi ^{\prime}(\xi )+B\psi +C\psi ^{3}=0, \end{aligned}
(11)
where $$A=\sigma ^{2}(1-\sigma ^{4})$$, $$B=(\sigma ^{2}-1)(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2}))$$ and $$C=2\lambda (1+\sigma ^{2}).$$
Substituting $$N = 1$$ into Eq. (4), we obtain
\begin{aligned} \psi (\xi )=a_{0}+a_{1}f(\varpi )+\frac{a_{-1}}{f(\varpi )} +b_{1}g(\varpi )+\frac{c_{-1}g(\varpi )}{f(\varpi )}. \end{aligned}
(12)
On substituting (12) into (11), the left-hand sides of (11) are converted into a finite power series of $$f^{k}(\varpi )g^{i}(\varpi )$$. If we set each coefficient of powers to zero, there is a set of algebraic equations for $$a_{-1}, \;a_{0},\; a_{1},\; b_{1}, \;c_{-1},\; A, \;B$$ and C, we omit the equations. Solving the algebraic equations, we can obtain:
• Case 1:$${A=-\lambda B,\;a_{-1}=0,\;a_{0}=0,\;a_{1}=0, \;b_{1}=0},\; c_{-1}=\sqrt{-\frac{2B}{C}},$$

• Case 2:$${A=2B\lambda ,\;a_{-1}=0,\;a_{0}=\sqrt{-\frac{B}{C}}},\;a_{1}=0, \;b_{1}=0,\;c_{-1}=0,$$

• Case 3:$${A=\frac{B}{2\lambda } ,\;a_{-1}=\sqrt{-\frac{B\lambda }{C}}},\;a_{0}=,\;a_{1}=0, \;b_{1}=0,\;c_{-1}=0,$$

• Case 4:$${A=-\frac{B}{\lambda } ,\;a_{-1}=0,\;a_{0}=\sqrt{-\frac{B}{C}}},\;a_{1}=00, \;b_{1}=0,\;c_{-1}=0,$$

• Case 5:$${A=-\lambda B,\;a_{-1}=0,\;a_{0}=0,\;a_{1}=0,\;b_{1}=\sqrt{\frac{2B}{C\lambda }}},\;c_{-1}=0.$$

Substitute $$\lambda =1$$ into Case 1–Case 5, therefore
\begin{aligned} \psi (\xi )= \,& {} a_{0}-a_{1}\coth (\xi )-a_{-1}{\text {tan}}h (\xi ) \nonumber \\&-b_{1}{\text {csc}}h (\xi )+c_{-1}{\text {sec}}h(\xi ), \end{aligned}
(13)
which can be expressed as follows:
\begin{aligned} \left\{ \begin{array}{ll} \psi _{1}(\xi )=i\sqrt{\frac{2B}{C}} {\text {sec}}h (\xi ),\\ \psi _{2}(\xi )=-i\sqrt{\frac{B}{C}}{\text {tan}}h(\xi ), \\ \psi _{3}(\xi )=\sqrt{\frac{2B}{C}} {\text {csc}}h (\xi ).\\ \end{array} \right. \end{aligned}
(14)
So according to (7), we have
\begin{aligned} \left\{ \begin{array}{ll} q_{1}(\xi )=i\sqrt{\frac{(\sigma ^{2}-1)(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2}))}{\lambda (1+\sigma ^{2})}}\times {\text {sec}}h (\xi )\times \exp (i\theta ),\\ \;r_{1}(\xi )=\frac{2(2\omega +\sigma ^{2}(p^{2}+\sigma ^{2}s^{2})}{(1+\sigma ^{2})}\times {\text {tan}}h(\xi ) \end{array}\right. \end{aligned}
(15)
\begin{aligned} \left\{ \begin{array}{ll} q_{2}(\xi )=-i\sqrt{\frac{(\sigma ^{2}-1)(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2}))}{2\lambda (1+\sigma ^{2})}}\times {\text {tan}}h(\xi )\times \exp (i\theta ),\\ \;r_{2}(\xi )=\frac{(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2})}{(1+\sigma ^{2})}\times ((\xi )-{\text {tan}}h^{2}(\xi )) \end{array}\right. \end{aligned}
(16)
\begin{aligned} \left\{ \begin{array}{ll} q_{3}(\xi )=\sqrt{\frac{(\sigma ^{2}-1)(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2}))}{\lambda (1+\sigma ^{2})}}\times {\text {csc}}h (\xi )\times \exp (i\theta ),\\ \;r_{3}(\xi )=-\frac{2(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2})}{(1+\sigma ^{2})}\times \coth (\xi ). \end{array} \right. \end{aligned}
(17)
with Case 1–Case 5, we can obtain
\begin{aligned} \left\{ \begin{array}{ll} \psi _{4}(\xi )=\sqrt{\frac{B}{C}}\tan (\xi ),\; \\ \psi _{5}(\xi )=-i\sqrt{\frac{2B}{C}}\csc (\xi ).\\ \end{array} \right. \end{aligned}
(18)
Similarly, for the cases of (18), we have another exact solution to (1) that can be written as:
\begin{aligned} \left\{ \begin{array}{ll} q_{4}(\xi )=\sqrt{\frac{(\sigma ^{2}-1)(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2}))}{2\lambda (1+\sigma ^{2})}}\times \tan (\xi )\times \exp (i\theta ), \\ r_{4}(\xi )\frac{(2\omega +\sigma ^{2}(p^{2}+\sigma ^{2}s^{2})}{(1+\sigma ^{2})} \times (-(\xi )+\tan (\xi )), \end{array} \right. \end{aligned}
(19)
\begin{aligned} \left\{ \begin{array}{ll} q_{5}(\xi )=-i\sqrt{\frac{(\sigma ^{2}-1)(2\omega +\sigma ^{2}(p^{2} +\sigma ^{2}s^{2}))}{\lambda (1+\sigma ^{2})}}\times \csc (\xi )\times \exp (i\theta ).\\ r_{5}(\xi )\frac{2(2\omega +\sigma ^{2}(p^{2}+\sigma ^{2}s^{2})}{(1+\sigma ^{2})} \times (-\cot ^{2}(\xi )) \end{array} \right. \end{aligned}
(20)
The solutions we obtained here, are more abundant than those by means of the trial equation method , which only constructed exact soliton solutions.

## 4 Graphs for the solutions

See Figs. 123, and 4. Fig. 1The 3D surfaces of Eq. (15) by considering the values $$B=C,\;p=1,\;\omega =-1,\;y=0,\;2\lambda =\sigma ^{2}-1$$, $${\nu =1}$$, $$s=1$$, $$-2\le x \le 4, \; t >0$$ Fig. 2The 2D surfaces of Eq. (15) by considering the values $$B=C,\;p=1,\;\omega =-1,\;y=0,\;2\lambda =\sigma ^{2}-1$$, $$\nu =1$$, $$s=1$$, $$t=1$$, $$-2\le x \le 4$$ Fig. 3The 3D surfaces of Eq. (20) by considering the values $$p=1,\;\omega =-1,\;y=0,\;2\lambda =\sigma ^{2}-1,\;B=C$$, $$\nu =1$$, $$s=1$$, $$-5\le x \le 6,\;t>0$$ Fig. 4The 2D surfaces of Eq. (20) by considering the values $$p=1,\;\omega =-1,\;y=0,\;2\lambda =\sigma ^{2}-1,\;B=C$$, $$\nu =1$$, $$s=1,\;t=1$$, $$-5\le x \le 6$$

## 5 Integrals of motion

Since Eq. (11) is a motion-type, we can write the corresponding Lagrangian
\begin{aligned} L=\frac{1}{2}\,\psi _{\xi }^2+\frac{B}{2A}\,\psi ^2 +\frac{C}{4A}\,\psi ^4-D,\;A\ne 0 \end{aligned}
(21)
and, the Hamiltonian $$H=\psi _{\xi } P_\psi -L$$ reads
\begin{aligned} H(\psi ,P_\psi ,\xi )=\frac{1}{2}\left[ P_\psi ^2 -\left( \frac{B}{A}\,\psi ^2 +\frac{C}{2A}\,\psi ^4-2D\right) \right], \end{aligned}
(22)
where the canonical momentum is
\begin{aligned} P_\psi =\frac{\partial L}{\partial \psi _\xi }=\psi _\xi . \end{aligned}
(23)
The independent variable $$\xi$$ does not appear explicitly in (22), then H is a constant of motion, $$H=E,$$ with
\begin{aligned} E=\frac{1}{2}\left[ \left( \frac{{\text {d}}\psi }{{\text {d}}\xi }\right) ^{2} -\left( \frac{B}{A}\,\psi ^2 +\frac{C}{2A}\,\psi ^4-2D\right) \right] . \end{aligned}
(24)
Equation (24) can also be expressed as a product of two independent constant of motions
\begin{aligned} 2E=E_{\text {Total}}\,= I_{1}\,I_{2}, \end{aligned}
(25)
where
\begin{aligned} I_{1}(\xi )=\left( \psi _\xi + \sqrt{\frac{B}{A}\,\psi ^2+\frac{C}{2A}\,\psi ^4-2D}\,\right) \,e^{ S(\xi )}, \end{aligned}
(26)
and
\begin{aligned} I_{2}(\xi )=\left( \psi _\xi - \sqrt{\frac{B}{A}\,\psi ^2+\frac{C}{2A}\,\psi ^4-2D}\,\right) \,e^{ S(\xi )}, \end{aligned}
(27)
the phase $$S(\xi )$$ is chosen in such a way that $$I_1,\;I_{2}$$ be constants of motion ($${\text {d}}I_{i}(\xi )/{\text {d}}\xi =0,\;i=1,2$$)
\begin{aligned} S(\xi )=\int \frac{\frac{B}{A}\,\psi +\frac{C}{A}\,\psi ^3}{\sqrt{\frac{B}{A}\,\psi ^2 +\frac{C}{2A}\,\psi ^4-2D}}\,{\text {d}}\xi . \end{aligned}
(28)
We have expressed the energy (24) as a product of two independent constant of motions (25). Then, we have seen that these factors are related with first-order ODEs that allow us to get the solutions of the nonlinear second-order ODE. In fact, whenever such autonomous systems arise, the factorization procedure is simplified due to the constant of motion. Observe that the Lagrangian underlying the nonlinear system also permits us to get solutions. There are some interesting papers showing the way to obtain traveling wave solutions of some nonlinear equations starting from the Lagrangian [21, 22, 23].

## 6 Results and discussion

In , the trial equation method and ansatz approach were developed and have been utilized in solving the 2D Davey–Stewartson equation and various solutions in hyperbolic functions form were obtained. Secondly, the well-known extended tanh method  has been employed to this equation and some exact hyperbolic and trigonometric function were obtained. We observe that our results are new, different trigonometric and hyperbolic function solutions when compared with the existing results obtained by using these two methods [12, 24]. We observe that these new solutions include some significant various types of solutions such as: periodic wave solution and soliton wave solutions. In Figs. 12, we plot 2D and 3D surfaces of $$q_{1}(\xi )$$ and $$r_{1}(\xi )$$ in Eq. (15), which explain the vitality of solutions with appropriate parameters. In Figs. 34, we draw 2D and 3D surfaces of $$q_{5}(\xi )$$ and $$r_{5}(\xi )$$ in Eq. (20), which show the dynamics of solutions with proper parameters. Moreover, such hyperbolic functions are of different physical meanings as well. We observed that the obtained solutions in this study have some important physical meanings, for instance; tangent hyperbolic function arises in the calculation of magnetic moment and rapidity of special relativity, the hyperbolic cosine function arises in the shape of a hanging cable . It is estimated that these newly obtained hyperbolic function solutions in this paper may help to explain some complex physical aspects in the nonlinear physical sciences and are related to such physical properties (Figs. 1234).

## 7 Conclusion

The uniform algebraic method is used to investigate an 2D Davey–Stewartson equation. The solitary wave solutions and periodic wave solutions of the equation are obtained under different circumstances. Moreover, three-dimensional and two-dimensional graphics of these exact solutions have been plotted. Hence, we can conclude that extended uniform algebraic method has an important role, to find analytical solutions of nonlinear differential equations. Finally, it is worthwhile to mention that we can apply this method to many other nonlinear like equations such as generalized-Zakharov equations and Kadomtsev–Petviashvili equation. This is our task in a future work.

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