New Function Solutions of Ablowitz-Kaup-Newell-Segur Water Wave Equation via Power Index Method

In this article, the applications of the Power Index Method (PIM) are presented to ascertain function solutions of the AKNS equation. We ﬁ nd out adequate explicit general solutions for the AKNS in the form of exponential, trigonometric, and logarithmic functions which have their importance in the physical sciences. We have used function transformations in which indexes of independent variables are nonrestricted parameters. During our investigation, we have seen that the nonlinear AKNS can be reduced to linear ODE by using PIM. All the solutions contain independent variables with parameters, and they are graphically represented by choosing suitable values of parameters.


Introduction
The exact solutions of the partial differential equation have their own consequence in the numerous branches of physical sciences, applied sciences, and other areas of engineering to comprehend their physical interpretation. Therefore, the search for an exact solution of nonlinear systems has been an interesting and valuable topic for mathematicians and physicists in applied science. A transformation of variables can sometimes be found that transforms a nonlinear PDE into a nonlinear ODE such as Lie Symmetry Transformations [1,2], the traveling wave transformation, and the Cole-Hopf transformation method [3]. In the current year, many effective methods including the simple and modified equation methods [4,5] and the sine-cosine method [6] have been employed to attain exact solutions of the equation.
In models of partial differential equations (PDEs), the AKNS equations play a central role and have many applications in applied physics. Various methods have been established for obtaining analytic solutions of the AKNS equations, for instance, the Exp-function method [7], multilinear variable separation approach method, variable coefficient method, and extended auxiliary equation method [8,9]. The truncated Painlevé analysis of the AKNS equation with their symmetries [10] and the complex combined darkbright solutions of the AKNS equation using SGEM [11] are obtained. Some transformations like the inverse-scattering [12][13][14][15] Backlund transformations [16] and the Darboux transformation [17] have been applied to find solutions of the AKNS equation.
The paper preparation is as follows: in Section 2, we describe briefly the Power Index Method. In Section 3, we apply this method to the nonlinear AKNS equation by using function transformation. In this method, by using transformations like (8) and (12), we have obtained linear ODEs. Our exact solutions are an explicit general form and contain independent variables with nonrestricted parameters. Our findings are a more general form; see for details (60) and (61).

Power Index Method
Step 1. Consider the partial differential equation [18] 4 where α is a perturbation parameter with α > 0. We have employed the Power Index Method on (1) to obtain its exact solutions.
We introduce the new variable and the travelling transformation Some forms of the new variable are Figure 1 shows the procedure of transformation of PDE to ODE.
Step 2. Express the nonlinear PDE into a mixed algebraic form where p t = a 1 n 1 + a 2 n 2 + a 3 n 3 + a 4 , Step 3. We choose the highest power index term of one variable in the algebraic equation equal with the other highest power index term of the same variable p t ðn 1 , n 2 , n 3 Þ = q t ðn 1 , n 2 , n 3 Þ. Similarly, we get other relations of power indexes so that we can choose particular values of power indexes. We continue the same process for other variables. Now, we have selected a new variable and transformation with adjustable indexes. Next, each term is replaced by a new variable and transformation, and we have reduced PDE into ODE.
Step 4. Solve the obtained ODE by using computerized symbolic packages. The exact solution of PDE can be obtained from the analytic solution of ODE and the new variable transformation. Figure 2 demonstrates the exact solution of (10).

Exact Solutions of AKND by Using Function Transformations
Case 1. Let us start with an elementary exponential function and introduce a new variable as Using (7), equation (1) can be expressed into ODE of the form  Journal of Function Spaces and n 1 = 1, then we have the following analytic solution of (8).
From (7) and (9), the analytic solution of the PDE is as follows: If we choose m 1 = 1 and n 1 = −1, then we have the following exact solution of the PDE (1). Figure 3 displays the graphical illustration of the analytical solution of (11).
We have observed that for the valuesm 1 andn 1 with the same sign or opposite signs, the ODE (8) has the same analytic solution.
Case 2. Now, we consider another exponential function of the form The transformation (12) reduces (1) to ODE in the form With the help of Symbolic Packages, the analytic solution of (13) is given as follows: By (12), (14), and (15), we get the exact solution of the PDE (1) as follows: We note that the solutions u 21 and u 22 contain parameters n 2 and m 2 . For different values of n 2 and m 2 , we can find more solutions of (1). Figure 4 contrasts with (1) which is given in (16). For n 2 = 0 in (13), we get the following ODE.
The analytic solution of the ODE (18) is We obtain the exact solution of (1) in the term of x: We note that the exponential function in terms of y and t has no ODE reduction form.

Journal of Function Spaces
Case 3. Next, we consider a monomial of the form Using (21), the PDE (1) is reduced in the form The analytic solution of the ODE (22) is The exact solution of the PDE (1) is obtained from (23) and transformation (21) Using (25), the PDE (1) is reduced in the form The analytic solution of the ODE (26) are Figures 6 and 7 show the performance of the solution of (1) which is given in (24).
If we replace y by t in (25), the new transformation becomes Using (31), the PDE (1) is reduced in the form The analytic solutions of the ODE (32) are   Figure 8 acts as the solution of (1) which is given in (30).

Case 5.
We define a new variable as x m 5 y 2s 5 +2r 5 −1 In (35), two terms contain only the y variable, so if we want to make this variable vanish in these terms, we have to take 2s 5 + r 5 − 1 = s 5 + r 5 . This implies s 5 = 1. Further, the term that contain x m 5 must have the r 5 power of y. From this, we get 2s 5 + 2r 5 − 1 = r 5 or s 5 + 2r 5 = r 5 ; these imply r 5 = −1. First, we take m 5 = 1, r 5 = −1, and s 5 = 1 in ((34)) and get the following ODE: The analytic solution of (36) is given as follows: Now, we take r 5 = 1, m 5 = 1, and s 5 = 1 in (35) and get the following ODE.
The analytic solution of (41) is given as follows:
To find the suitable value of m 6 and n 6 , we can take different combinations of the y-index terms, and finally, we get m 6 = 0 and n 6 = 1. We have the following final form of the transformation (44).
and the corresponding ODE is given as follows: The analytic solution of (47) is given as follows: The exact solution of (1) is given as follows: Case 7. In this case, we have considered the logarithmic function of the form The algebraic form of the PDE (1) has the following y-index terms:  Journal of Function Spaces To choose the suitable index value of y, we take 2 n 7 − 1 = n 7 , and we get n 7 = 1. Finally, we have the following transformation: and the corresponding ODE is obtained when the PDE (1) is transformed. The analytic solution of (55) is given by The exact solution of (1) has the form This transformation transformed the nonlinear PDE (1) to a product of two linear ODEs.
Using (62), (65), and (66), the exact solutions of (1) are Solving these ODEs by using Maple, we get the same solutions of (1) which are given in (67) and (68). Figure 17 displays the graphical illustration of the analytical solution of (61).