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

Authors: Khalil Ahmad; Khudija Bibi
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this article, the applications of the Power Index Method (PIM) are presented to ascertain function solutions of the AKNS equation. We find 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.


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## Body

## 1. 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 dark-bright solutions of the AKNS equation using SGEM [11] are obtained. Some transformations like the inverse-scattering [12–15] 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 Section2, 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).


## 2. Power Index Method

Step 1.
Consider the partial differential equation [18]
(1)4∂2w∂x∂t+∂4w∂x3∂t+8∂w∂x∂2w∂x∂y+4∂w∂y∂2w∂x2−α∂2w∂x2=0,
whereα 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(2)ξ=qt,x,y,and the travelling transformation
(3)wt,x,y=q2t,x,yhξ+q3t,x,y.
Some forms of the new variable are(4)ξ1=a1xm1tm2+a2xm3ym4+a3tm5,ξ2=b1em1x+n1t+b2em2x+n2y,ξ3=c1cosx+y+c2cosx+t+c3tanx+y+c4tanx+t.Figure1 shows the procedure of transformation of PDE to ODE.Step 2.
Express the nonlinear PDE into a mixed algebraic form(5)Ft,x,y≔tptxpxypyφξ,gξ,g′ξ,g′′ξ,⋯,where
(6)pt=a1n1+a2n2+a3n3+a4,px=b1m1+b2m2+b3m3+b4,py=c1r1+c2r2+c3r3+c4.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 variableptn1,n2,n3=qtn1,n2,n3. 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 1
Flow chart of solution of the PDE (1).Figure2 demonstrates the exact solution of (10).Figure 2
3D plot of (10) for α=1.


## 3. Exact Solutions of AKND by Using Function Transformations

Case 1.
Let us start with an elementary exponential function and introduce a new variable as(7)ξ1t,x=em1x+n1t,u1t,x,y=h1ξ1.Using (7), equation (1) can be expressed into ODE of the form
(8)m12n1ξ13h1′′′′ξ1+6m12n1ξ12h1′′′ξ1+ξ17m12n1−m1α+4n1h1′′ξ1+m12n1−m1α+4n1h1′ξ1=0.Equation (8) shows that for each real value of m1 and n1, the PDE (1) can be reduced to ODE (8). If we choose m1=1 and n1=1, then we have the following analytic solution of (8).
(9)h1ξ1=C1+C2lnξ1+C3Sin4−αlnξ1+C4Cos4−αlnξ1.From (7) and (9), the analytic solution of the PDE is as follows:
(10)u1t,x==C1+C2x+t+C3Sin4−αx+t+C4Cos4−αx+t.If we choosem1=1 and n1=−1, then we have the following exact solution of the PDE (1).
(11)u1t,x,y==C1+C2x+t+C3Sin4+αx−t+C4Cos4+αx−t.Figure3 displays the graphical illustration of the analytical solution of (11).Figure 3
3D plot of (11) for α=1.We have observed that for the valuesm1andn1with 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(12)ξ2x,y=em2x+n2y,u2t,x,y=h2ξ2.
The transformation (12) reduces (1) to ODE in the form
(13)−12n2ξ22h2′ξ2h2′′ξ2+12n2ξ2h2′2ξ2−αξ2h2′′ξ2−αh2′ξ2=0.
With the help of Symbolic Packages, the analytic solution of (13) is given as follows:
(14)h2ξ2=α12n2lnξ2−α2−24n2C112n2lnξ2+C2,(15)h2ξ2=α12n2lnξ2+α2−24n2C112n2lnξ2+C2.
By (12), (14), and (15), we get the exact solution of the PDE (1) as follows:
(16)u21x,y=α12n2m2x+n2y−α2−24n2C112n2m2x+n2y+C2,(17)u22x,y=α12n2m2x+n2y+α2−24n2C112n2m2x+n2y+C2.We note that the solutionsu21 and u22 contain parameters n2 and m2. For different values of n2 and m2, we can find more solutions of (1).Figure4 contrasts with (1) which is given in (16).Figure 4
Plot of (16) for α=5.Forn2=0 in (13), we get the following ODE.
(18)ξ2h2′′ξ2+h2′ξ2=0.The analytic solution of the ODE (18) is
(19)h2ξ2=C2lnξ2+C1.We obtain the exact solution of (1) in the term of x:
(20)u23x=C2m2x+C1.We note that the exponential function in terms ofy and t has no ODE reduction form.Case 3.
Next, we consider a monomial of the form(21)ξ3x=x2m3,u3t,x,y=tn3+yr3+h3ξ3.
Using (21), the PDE (1) is reduced in the form
(22)m3ξ3h3′′ξ3+m3−12h3′ξ3=0.
The analytic solution of the ODE (22) is
(23)h3ξ3=C1+C2ξ31/2m3.
The exact solution of the PDE (1) is obtained from (23) and transformation (21)
(24)u3t,x,y=C1+C2x+tn3+yr3.Figure5 presents the solution of (1) which is given in (20).Case 4.
In this case, we have considered the polynomial of the form(25)ξ4x,y=cosx+y,ux,y=h4ξ4.
Using (25), the PDE (1) is reduced in the form
(26)ξ42−1h4′′ξ4+ξ4h4′ξ4121−ξ42h4′ξ4+α=0.
The analytic solution of the ODE (26) are
(27)h41ξ4=C1+C2∫1ξ42−11/24dξ4,(28)h42ξ4=C1−αarcsinξ412,(29)w41x,y=C1+C2∫1cosx+y2−11/24secx+ydx+dy,(30)w42x,y=C1−αarcsincosx+y12.Figure 5
Plot of (20) for m2=−5,−3,−1,1,3,5.Figures6 and 7 show the performance of the solution of (1) which is given in (24).Figure 6
3D plot of (24) for x=1.Figure 7
3D plot of (24) for t=10.If we replacey by t in (25), the new transformation becomes
(31)ξ4x,y=Cosx+y,ux,y=h4ξ4.Using (31), the PDE (1) is reduced in the form
(32)ξ42−12h4′′′′ξ4+α+3ξ42−αh4′′ξ4+ξ46ξ42−1h4′′′ξ4+α−3h4′ξ4=0.The analytic solutions of the ODE (32) are
(33)h43ξ4=C1+C2lnξ4+ξ42−1+C3∫ξ4+ξ42−1α−4ξ42−1+C4∫ξ4+ξ42−1−α−4ξ42−1dξ4,w43x,t=C1+C2lncosx+t+cosx+t2−1+C3∫cosx+t+cosx+t2−1α−4cosx+t2−1secx+tdx+dt.+C4∫cosx+t+cosx+t2−1−α−4cosx+t2−1secx+tdx+dt.Figure8 acts as the solution of (1) which is given in (30).Case 5.
We define a new variable as(34)ξ5x,y=xm5yr5,u1t,x,y=ys5h5ξ5,(35)xm5y2s5+2r5−12s5+3r5m5−r5h5’ξ52+m5s5h5ξ5h5”ξ5−12m5r5x2m5y2s5+3r5−1h5’ξ5h5”ξ5−4m5−1s5y2s5+r5−1h5ξ5h5’ξ5+m5xm5ys5+2r5−1h5”ξ5+m5−1ys5+r5h5’ξ5=0.
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 2s5+r5−1=s5+r5. This implies s5=1. Further, the term that contain xm5 must have the r5 power of y. From this, we get 2s5+2r5−1=r5 or s5+2r5=r5; these imply r5=−1. First, we take m5=1, r5=−1, and s5=1 in ((34)) and get the following ODE:
(36)h5′′ξ53ξ5h5ξ5+14α−4h5ξ5=0.
The analytic solution of (36) is given as follows:
(37)h51ξ5=α4+C1ξ51/3,(38)h52ξ5=C1ξ5+C2,(39)u51ξ5=α4+C1x1/3y2/3,(40)u52ξ5=C1x+C2.Figure 8
3D plot of (30) for α=10.Now, we taker5=1, m5=1, and s5=1 in (35) and get the following ODE.
(41)ξ5h5′ξ5+14α−4h5ξ5=0.The analytic solution of (41) is given as follows:
(42)h53ξ5=α4+C11ξ5.(43)w53y=α4y+C1y2.Case 6.
In this case, we have considered a polynomial with a logarithmic function of the form(44)ξ6x,y=ym6lnx,w6t,x,y=yn6h6ξ6.
The algebraic form of the PDE (1) has the following y-index terms:
(45)py1=2m6+2n6−1,py2=3m6+2n6−1,py3=m6+2n6−1,py4=m6+n6,py5=2m6+n6.Figure9 acts as the solution of (1) which is given in (39). Figure 10 entertains the solution of (1) which is given in (43).Figure 9
3D plot of (39) for α=1.Figure 10
Plot of (43) for α=1,10.To find the suitable value ofm6 and n6, we can take different combinations of the y-index terms, and finally, we get m6=0 and n6=1. We have the following final form of the transformation (44).
(46)ξ6x,y=lnx,w6t,x,y=yh6ξ6,and the corresponding ODE is given as follows:
(47)8h6′′ξ6−4h6ξ6+αh6′ξ6+h6′′ξ64h6ξ6−α=0.The analytic solution of (47) is given as follows:
(48)h61ξ6=12C1expξ6+12C21/34+α4,h62ξ6=−12C1expξ6+12C21/38+312C1expξ6+12C21/38i+α4,h63ξ6=−12C1expξ6+12C21/38−312C1expξ6+12C21/38i+α4.The exact solution of (1) is given as follows:
(49)w61x,y=y12C1x+12C21/34+αy4,(50)w62x,y=−y12C1x+12C21/38+3y12C1x+12C21/38i+αy4,(51)w63x,y=−y12C1x+12C21/38−3y12C1x+12C21/38i+αy4.Case 7.
In this case, we have considered the logarithmic function of the form(52)ξ7x,y=lnxm7,w7t,x,y=lnxm7yn7h7ξ7.
The algebraic form of the PDE (1) has the following y-index terms:
(53)py1=2n7−1,py2=n7.To choose the suitable index value ofy, we take 2n7−1=n7, and we get n7=1. Finally, we have the following transformation:
(54)ξ7x,y=lnxm7,w7t,x,y=lnxm7yh7ξ7,and the corresponding ODE is obtained when the PDE (1) is transformed.
(55)42m7−ξ7h7ξ72+h7ξ74ξ76m7−ξ7h7′ξ7+4m7ξ72h7′′ξ7+α+8m7ξ72h7′ξ62+αξ7−2m7h7′ξ6−m7αh7′ξ6h7′′ξ7=0.Figures11, 12, and 13 present the graphical diagrams of the analytical solutions of (49), (50), and (51), respectively.Figure 11
3D plot of (49) for α=1.Figure 12
3D complex plot of (50) for α=1.Figure 13
3D complex plot of (51) for α=1.The analytic solution of (55) is given by
(56)h71ξ7=−12m7C1expξ7/m7+12m7C2−α31/38ξ7+α4ξ7,(57)h72ξ7=−−12m7C1expξ7/m7+12m7C2−α31/38ξ7−312m7C1expξ7/m7+12m7C2−α31/38ξ7i+α4ξ7,(58)h73ξ7=−−12m7C1expξ7/m7+12m7C2−α31/38ξ7+3−12m7C1expξ7/m7+12m7C2−α31/38ξ7i+α4ξ7,(59)w71x=−12m7C1x+12m7C2−α31/38lnxm7+α4lnxm7,(60)W72x=−−12m7C1x+12m7C2−α31/38lnxm7+3−12m7C1x+12m7C2−α31/38lnxm7i+α4lnxm7.The exact solution of (1) has the form
(61)W73x=−−12m7C1x+12m7C2−α31/38lnxm7−3−12m7C1x+12m7C2−α31/38lnxm7i+α4lnxm7.Case 8.
In this case, we have considered a trigonometric function of the form(62)ξ8x,y=tanm8x+n8y,w8t,x,y=h8ξ8.
This transformation transformed the nonlinear PDE (1) to a product of two linear ODEs.
(63)α−12n81+ξ82h8′ξ8=0,(64)2ξ8h8′ξ8+1+ξ82h8′′ξ8=0,(65)h81ξ8=α12n8tan−1ξ8+C,(66)h82ξ8=C1tan−1ξ8+C2.
Using (62), (65), and (66), the exact solutions of (1) are
(67)w81x,y=α12n8m8x+n8y+C,(68)w82x,y=C1m8x+n8y+C2.Figures14, 15, and 16 present the graphical diagrams of the analytical solutions of (59), (60), and (61), respectively.Figure 14
3D complex plot of (59) for α=1 and m7=1.Figure 15
3D complex plot of (60) for α=1 and m7=1.Figure 16
3D complex plot of (61) for α=1 and m7=1.If we choose following transformations,(69)ξ8x,y=tan1/2m8x+n8y,w8t,x,y=h8ξ8,(70)ξ8x,y=tan1/4m8x+n8y,w8t,x,y=h8ξ8,then we get the first two ODEs by using (69) and the next two ODEs by using (70).
(71)αξ8−6n8+ξ84h8′ξ8=0,3ξ84−1h8′ξ8+ξ81+ξ84h8′′ξ8=0,αξ83−3n81+ξ88h8′ξ8=0,5ξ88−3h8′ξ8+ξ81+ξ88h8′′ξ8=0.Solving these ODEs by using Maple, we get the same solutions of (1) which are given in (67) and (68).Figure17 displays the graphical illustration of the analytical solution of (61).Case 9.
In this case, we have considered the trigonometric function of the form(72)ξ9x,y=tanx+t,w9t,x,y=h9ξ9.
This transformation reduces the nonlinear PDE (1) to linear ODE.
(73)1+ξ926h9′′′′ξ9+12ξ91+ξ922h9′′′ξ9+12−α+36ξ921+ξ92h9′′ξ9−2ξ9α−121+ξ92h9′ξ9=0.
The analytic solution of (73) has the form
(74)h9ξ9=C1+C2tan−1ξ9+C3ξ9+ii−ξ9−α−4/2+C4ξ9+ii−ξ9α−4/2.
The closed form function solution of (1) is given by
(75)w9x+t=C1+C2x+t+C3x+t+ii−x−t−α−4/2+C4x+t+ii−x−tα−4/2.Figure 17
3D complex plot of (61) for α=1 and m7=1.


## 4. Conclusion

The intention of this research paper is to find function solutions using exponential, logarithmic, trigonometric, and rational functions which are attained from the AKNS equation via PIM. Our attained solutions are an explicit form, and they contain independent variables with parameter indexes. We have seen that for some of our transformations which are applied to nonlinear AKNS, we get linear ODEs. All the found solutions of the equation are new, and some solutions are more general (see (59), (60), and (61)). The exact solutions of AKNS which we have obtained in this article are useful to understand the physical phenomenon in many aspects. They are explicitly characterized by choosing appropriate values of parameters. This method has been efficaciously employed to get exact solutions of the equation.



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