# Encyclopedia of Complexity and Systems Science

Living Edition
| Editors: Robert A. Meyers

# Analytical Soliton Solutions for Some Nonlinear Dynamical Water Waves Models

Living reference work entry
DOI: https://doi.org/10.1007/978-3-642-27737-5_737-1

## Keywords

Modified Liouville equation The Symmetric Regularized Long Wave (SRLW) equation Fourth-order nonlinear Ablowitz-Kaup-Newell-Segur water wave (AKNS) equation Riccati equation Modified F-Expansion method Soliton solutions

## Introduction

In different areas of applied science, the investigation of exact travelling wave solutions has played a vital role for describing the character wave problems. For plentiful demonstration of dilemmas in mathematical physics, different nonlinear wave systems have been discussed such as the phenomena flow of heat, plasma physics, and optical fibers. Peripheral of solitary wave is growing daily, so essential to find for exact traveling wave solutions to NLEEs. Exact solutions assist us understand the necessary back ground of problems.

The exact solutions of nonlinear partial differential equations have been investigated by many authors [1–36]. The research of traveling wave solutions of some nonlinear evolution equations derived from such fields played an imperative role in the analysis of some phenomena such as the Bernoulli’s sub-ODE method (Wang et al. 2007), Exp(-Φ(ξ))-expansion method (Islam et al. 2015), extended simple equation method (Ali et al. 2017; Lu et al. 2017a, 2018a), modified extended direct algebraic method (Arshad et al. 2017; Seadawy et al. 2017), the extended sinh-cosh and sin-cos methods (Wang et al. 2005), the Jacobi elliptic function method (Ali 2011), the Exp-function method (He and Wu 2006; Akbar and Ali 2012; Naher et al. 2012), the ansatz method (Triki et al. 2012; Sassaman and Biswas 2009; Chowdhury and Biswas 2012; Song et al. 2013), the perturbation method (Biswas et al. 2012), modified extended mapping method (Abdullah et al. 2017), the extended tanh-function method (Zhou et al. 2017; Lu et al. 2017b; Mirzazadeh 2016), modified Kudryashov and hyperbolic function methods (Seadawy et al. 2018), and many more (Helal and Seadawy 2009, 2011, 2012; Khater et al. 2006; Seadawy 2011, 2015, 2016, 2017).

Modified Liouville and the symmetric regularized long wave equations have been useful for the description of weakly nonlinear ion acoustic and space charge waves in mathematical physics. Similarly the tremendous benefit of fourth-order nonlinear Ablowitz-Kaup-Newell-Segur water wave equation is that it can be concentrated to few eminent nonlinear evolution equations such as the KdV, the mKdV, and the sine-Gordon equations which have enormous applications in applied mathematics. Now recently Lu. D (Lu et al. 2018a) investigated exact and solitary wave solutions on modified Liouville and the symmetric regularized long wave equation by applying the modified simple equation and exp(−Ψ(ξ)) expansion methods. Furthermore, Ali. A (Ali et al. 2018) lately investigated the soliton solutions of AKNS equation by employing simple equation and modified simple equation methods. Our focus in this current entry is to demonstrate the so-called modified F-expansion method for construction of soliton solutions above illustrated three important models, having valuable advantages for handling the nonlinear wave problems in nonlinear science.

The entry is organized as follows: section “Applications,” applied modified F-expansion method to our selective three nonlinear wave models; results discussion in section “Discussion of the Results”; and finally section “Conclusion.”

## Applications

### Application of Modified Liouville Equation

The generalized form of Modified Liouville equation in Lu et al. (2018a)
$${d}_1^2{v}_{xx}-{v}_{tt}+{d}_2{e}^{\beta v}=0,$$
(1)
where d1,  β,  d2 are arbitrary constants. Let wave transformation;
$$v\left(x,\, t\right)\, =V\left(\xi \right),\quad \xi = kx+\omega t,\quad\quad V={e}^{\beta v}$$
(2)
Replacement of Eq. (2) in Eq. (1) gives
$$\left(\frac{k^2{d}_1^2}{\beta }-\frac{\omega^2}{\beta}\right){V^{\prime}}^{\prime }V-\left(\frac{k^2{d}_1^2}{\beta }-\frac{\omega^2}{\beta}\right)\;{V^{\prime}}^2+{d}_2{V}^3=0.\,$$
(3)
Suppose Eq. (3) has solution (Aasaraai 2015)
$$V={a}_0+{a}_1F\left(\xi \right)+{a}_2{F}^2\left(\xi \right)+{b}_1{F}^{-1}\left(\xi \right)+{b}_2{F}^{-2}\left(\xi \right)$$
(4)
Here F(ξ) convinces the following Riccati equation (Helal and Seadawy 2012)
$${F}^{\prime}\left(\xi \right)={H}_1+{H}_2F\left(\xi \right)+{H}_3{F}^2\left(\xi \right)$$
(5)

Submission of Eq. (4), Eq. (5) in Eq. (3), gives a large equations system with a0,  a1,  b1,  a2,  b2,  H1,  H2,  H3,  d1,  d2,  k,  ω. With help of Mathematica, have subsequent solutions categories.

#### The Soliton-Like Solutions of Eq. (1)

1. 1.

When H1 = 0,  H2 = 1,  H3 = − 1, then we have,

$${a}_0=0,\quad {a}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\, {a}_1=\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }{b}_1=0,\, {b}_2=0.\,$$
(6)
Substitute Eq. (6) in Eq. (4),
$${V}_1\left(\xi \right)=\frac{\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta}\left(1+\tanh \left(\frac{1}{2}\xi \right)\right)\left(1-\frac{1}{2}\left(1+\tanh \left(\frac{1}{2}\xi \right)\right)\right)$$
(7)
1. 2.

When H1 = 0,  H2 = − 1,  H3 = 1, we have,

$${a}_0=0,\quad {a}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\, {a}_1=\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\, {b}_1=0,\quad {b}_2=0.$$
(8)
Replace (8) in (4) (Figs. 1 and 2), Fig. 1The 3D and 2D for solution (7) with; d1 = 2.9,    d2 = 4,    k = 0.3,    β = − 1.5,    ω = − 0.05 Fig. 2Dark Solitary waves (9) are plotted with; d1 = − 1.9,    d2 = − 0.05,    k = 0.3,    β = − 1.5,    ω = 1
$${V}_2\left(\xi \right)=\frac{\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta}\left(1-\coth \left(\frac{1}{2}\xi \right)\right)\left(1-\frac{1}{2}\left(1-\coth \left(\frac{1}{2}\xi \right)\right)\right)$$
(9)
1. 3.

When $${H}_1=\frac{1}{2},\, {H}_2=0,\, {H}_3=-\frac{1}{2},$$ we have,

Family-I
$${a}_0=\frac{d_1^2{k}^2-{\omega}^2}{2{d}_2\beta },\, {a}_1=0,\quad {a}_2=0,\quad {b}_1=0,\quad {b}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }$$
(10)
Put Eq. (10) in Eq. (4),
$${V}_{31}\left(\xi \right)=\frac{d_1^2{k}^2-{\omega}^2}{2{d}_2\beta }+\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta {\left(\coth \left(\xi \right)\pm \mathit{\operatorname{csch}}\left(\xi \right)\right)}^2}$$
(11)
Family-II
$${a}_0=\frac{d_1^2{k}^2-{\omega}^2}{2{d}_2\beta },\, {a}_1=0,\quad {a}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\, {b}_1=0,\quad {b}_2=o$$
(12)
Substitute Eq. (12) in (4),
$${V}_{32}\left(\xi \right)=\frac{d_1^2{k}^2-{\omega}^2}{2{d}_2\beta }+\frac{\left({\omega}^2-{d}_1^2{k}^2\right){\left(\coth \left(\xi \right)\pm \mathit{\operatorname{csch}}\left(\xi \right)\right)}^2}{2{d}_2\beta }$$
(13)
Family-III
$${a}_0=\frac{d_1^2{k}^2-{\omega}^2}{d_2\beta },\quad {a}_1=0,\quad {a}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {b}_1=0,\quad {b}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }$$
(14)
Substituting Eq. (14) into Eq. (4), we obtained,
$${V}_{33}\left(\xi \right)=\frac{d_1^2{k}^2-{\omega}^2}{d_2\beta }+\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta}\left({\left(\coth \left(\varsigma \right)\pm \mathit{\operatorname{csch}}\left(\varsigma \right)\right)}^2+\frac{1}{{\left(\coth \left(\varsigma \right)\pm \mathit{\operatorname{csch}}\left(\varsigma \right)\right)}^2}\right)$$
(15)
1. 4.

When H1 = 1,  H2 = 0,  H3 = − 1,

Family-I
$${a}_0=\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {a}_1={a}_2={b}_1=0,\quad {b}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }$$
(16)
Substituting (16) in (4),
$${V}_{41}\left(\xi \right)=\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta {\left(\tanh \left(\xi \right)\right)}^2}$$
(17)
Family-II
$${a}_0=\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {a}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {b}_1={b}_2={a}_1=0,$$
(18)
Put (18) in (4),
$${V}_{42}\left(\xi \right)=\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right){\left(\tanh \left(\xi \right)\right)}^2}{d_2\beta }$$
(19)
Family-III
$${a}_0=\frac{4\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {a}_1=0,\quad {a}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {b}_1=0,\quad {b}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }$$
(20)
Putting Eq. (20) in Eq. (4),
$${V}_{43}\left(\xi \right)=\frac{4\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta}\left({\left(\tanh \left(\xi \right)\right)}^2+\frac{1}{{\left(\tanh \left(\xi \right)\right)}^2}\right)$$
(21)

#### The Trigonometric Function Solutions of Eq. (1)

1. 1.

When $${H}_1=\frac{1}{2},\, {H}_2=0,\, {H}_3=\frac{1}{2},$$

Family-I
$${a}_0=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {a}_1=0,\quad {a}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {b}_1=0,\quad {b}_2=o$$
(22)
Substitute Eq. (22) in Eq. (4) (Fig. 3), Fig. 3Periodic Solitary waves (23) are drawn with: d1 = 0.7,  d2 = 1,  k = 4.5,  β = − 4.5,  ω = − 5.3
$${V}_{51}\left(\xi \right)=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }+\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta }{\left(\sec \left(\xi \right)+\tan \left(\xi \right)\right)}^2$$
(23)
Family-II
$${a}_0=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {a}_1=0,\quad {a}_2=0,\quad {b}_1=0,\quad {b}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }$$
(24)
Eq. (24) into Eq. (4) gives,
$${V}_{52}\left(\xi \right)=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }+\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta }{\left(\frac{1}{\sec \left(\xi \right)+\tan \left(\xi \right)}\right)}^2$$
(25)
Family-III
$${a}_0=\frac{\omega^2-{d}_1^2{k}^2}{d_2\beta },\quad {a}_1=0,\quad {a}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {b}_1=0,\quad {b}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }$$
(26)
Substituting (26) in (4),
$${V}_{53}\left(\xi \right)=\frac{\omega^2-{d}_1^2{k}^2}{d_2\beta }+\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta}\left({\left(\sec \left(\xi \right)\right)}^2+{\left(\tan \left(\xi \right)\right)}^2+2\left(\sec \left(\xi \right)\right)\left(\tan \left(\xi \right)\right)+\frac{1}{{\left(\sec \left(\xi \right)+\tan \left(\xi \right)\right)}^2}\right)$$
(27)
1. 2.

When $${H}_1=-\frac{1}{2},\, {H}_2=0,\, {H}_3=-\frac{1}{2},$$

Family-I
$${a}_0=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {a}_1=0,\quad {a}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {b}_1=0,\quad {b}_2=0$$
(28)
Putting (28) in (4),
$${V}_{61}\left(\xi \right)=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }+\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta }{\left(\sec \left(\xi \right)-\tan \left(\xi \right)\right)}^2$$
(29)
Family-II
$${a}_0=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {a}_1=0,\quad {a}_2=0,\quad {b}_1=0,\quad {b}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }$$
(30)
Substituting (30) in (4),
$${V}_{62}\left(\xi \right)=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }-\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta {\left(\sec \left(\xi \right)-\tan \left(\xi \right)\right)}^2}$$
(31)
Family-III
$${a}_0=\frac{\omega^2-{d}_1^2{k}^2}{d_2\beta },\quad {a}_1=0,{a}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta },\quad {b}_1=0,\quad {b}_2=\frac{\omega^2-{d}_1^2{k}^2}{2{d}_2\beta }$$
(32)
Replace of Eq. (33) in Eq. (4),
$${V}_{63}\left(\xi \right)=\frac{\omega^2-{d}_1^2{k}^2}{d_2\beta }-\frac{\left({\omega}^2-{d}_1^2{k}^2\right)}{2{d}_2\beta}\left(\left(\sec \left(\xi \right)\Big){}^2+{\left(\tan \left(\xi \right)\right)}^2-2\left(\sec \left(\xi \right)\right)\left(\tan \left(\xi \right)\right)\right)-\frac{1}{{\left(\sec \left(\xi \right)-\tan \left(\xi \right)\right)}^2}\right)$$
(33)
1. 3.

When H1 = 1(−1), H2 = 0,  H3 = 1(−1),

Family-I
$${a}_0=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {a}_1={b}_1=0,{a}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {b}_2=0$$
(34)
Put Eq. (34) in Eq. (4),
$${V}_{71}\left(\xi \right)=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }{\left(\tan \left(\xi \right)\left(\cot \left(\xi \right)\right)\right)}^2$$
(35)
Family-II
$${a}_0=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {a}_1=0,\quad {a}_2=0,\quad {b}_1=0,\quad {b}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }$$
(36)
Substitute Eq. (36) in Eq. (4),
$${V}_{72}\left(\xi \right)=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta {\left(\tan \left(\xi \right)\left(\cot \left(\xi \right)\right)\right)}^2}$$
(37)
Family-III
$${a}_0=-\frac{4\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {a}_1=0,\quad {a}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta },\quad {b}_1=0,\quad {b}_2=-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }$$
(38)
Put Eq. (38) in Eq. (4),
$${V}_{73}\left(\xi \right)=-\frac{4\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta }-\frac{2\left({d}_1^2{k}^2-{\omega}^2\right)}{d_2\beta}\left({\left(\tan \left(\xi \right)\left(\cot \left(\xi \right)\right)\right)}^2+\frac{1}{{\left(\tan \left(\xi \right)\left(\cot \left(\xi \right)\right)\right)}^2}\right)$$
(39)

#### The Rational Function Solutions of Eq. (1)

1. 1.
When H1 = 0,  H2 = 0,  H3 ≠ 0,
$${a}_0={a}_1=0,\quad {a}_2=-\frac{2\left({d}_1^2{H}_3^2{k}^2-{H}_3^2{\omega}^2\right)}{d_2\beta },\quad {b}_1={b}_2=0,$$
(40)

Substitute Eq. (40) in Eq. (4),
$${V}_8\left(\xi \right)=-\frac{2\left({d}_1^2{H}_3^2{k}^2-{H}_3^2{\omega}^2\right)}{d_2\beta }{\left(\frac{1}{H_3\xi +\varepsilon \Big)}\right)}^2$$
(41)
1. 2.

When H2 = 0,  H3 = 0,

$${a}_0=0,\quad {a}_1=0,\quad {a}_2=0,\quad {b}_1=0,\quad {b}_2=-\frac{2\left({d}_1^2{H}_1^2{k}^2-{H}_1^2{\omega}^2\right)}{d_2\beta }$$
(42)
Replace Eq. (42) in Eq. (4),
$${V}_9\left(\xi \right)=-\frac{2\left({d}_1^2{H}_1^2{k}^2-{H}_1^2{\omega}^2\right)}{d_2{\beta \xi}^2}$$
(43)
1. 3.

When H1 ≠ 0,  H2 ≠ 0,  H3 = 0,

$${a}_0=0,\quad {a}_1=0,\quad {a}_2=0,\quad {b}_1=-\frac{2\left({d}_1^2{H}_1{H}_2{k}^2-{H}_1{H}_2{\omega}^2\right)}{d_2\beta },\quad {b}_2=-\frac{2\left({d}_1^2{H}_1^2{k}^2-{H}_1^2{\omega}^2\right)}{d_2\beta }$$
(44)
Putting Eq. (44) in (4) (Fig. 4), Fig. 4The 3D and 2D solitary waves (45) are dawned with: H1 = 0.5,  H2 = 0.5,  d1 = 0,  d2 = − 0.3,  k = 2.3,  β = 3,  ω = 1
$${V}_{10}\left(\xi \right)=-\frac{2\left({d}_1^2{H}_1{H}_2{k}^2-{H}_1{H}_2{\omega}^2\right)}{d_2\beta}\left(\frac{H_2}{\mathit{\exp}\left({H}_2\xi \right)-{H}_1}\right)-\frac{2\left({d}_1^2{H}_1^2{k}^2-{H}_1^2{\omega}^2\right)}{d_2\beta }{\left(\frac{H_2}{\mathit{\exp}\left({H}_2\xi \right)-{H}_1}\right)}^2$$
(45)

### Application of the Symmetric Regularized Long Wave Equation

Consider the SRLW equation (Lu et al. 2018a),
$${v}_{tt}-{v}_{xx}+{\left(\frac{v^2}{2}\right)}_{xt}-{v}_{xx tt}=0$$
(46)
Let the transformation,
$$v\left(x,\, t\right)=V\left(\xi \right),\quad \xi =x+ kt$$
(47)
Substitute of Eq. (47) in (46),
$$\left({k}^2-1\right){V}^{{\prime\prime} }-k{\left(\frac{V^2}{2}\right)}^{\prime \prime }-{k}^2{V}^{{\prime\prime} {\prime\prime} }=0$$
(48)
Taking two times integration of Eq. (48) with disregard constant of integral.
$$\left({k}^2-1\right)V-k\left(\frac{V^2}{2}\right)-{k}^2{V}^{{\prime\prime} }=0$$
(49)

Let formal solution of Eq. (49) is like Eq. (4). Revolution of (4), (5) in Eq. (49), gives varieties of solutions

#### The Soliton-Like Solutions of Eq. (46)

1. 1.

For H1 = 0,  H2 = 1,  H3 = − 1, then we have,

$${a}_0=\pm \sqrt{2},\quad {a}_2={a}_1=\pm 6\sqrt{2},\quad {b}_1=0,\, k=\pm \frac{1}{\sqrt{2}},\quad {b}_2=0$$
(50)
Substitute Eq. (50) in (4),
$${V}_{11}\left(\xi \right)=\pm \sqrt{2}\pm 3\sqrt{2}\left(1+\tanh \left(\frac{1}{2}\xi \right)\right)\pm \frac{3\sqrt{2}}{2}{\left(1+\tanh \left(\frac{1}{2}\xi \right)\right)}^2$$
(51)
1. 2.
For H1 = 0,  H2 = − 1,  H3 = 1, then we have (Fig. 5),
$${a}_0=\pm \sqrt{2},\quad {a}_2=\pm 6\sqrt{2},\quad {a}_1=\pm 6\sqrt{2},\quad {b}_1=0,\quad k=\pm \frac{1}{\sqrt{2}},\quad {b}_2=0$$
(52) Fig. 5The solitary wave of solution (51) is plotted at (a) and (b), respectively
Replace (52) in (4),
$${V}_{12}=\pm \sqrt{2}\pm 3\sqrt{2}\left(1-\coth \left(\frac{1}{2}\xi \right)\right)\pm \frac{6\sqrt{2}}{4}{\left(1-\coth \left(\frac{1}{2}\xi \right)\right)}^2$$
(53)
1. 3.

For $${H}_1=\frac{1}{2},\, {H}_2=0,\, {H}_3=-\frac{1}{2},$$

Family-I
$${a}_0=\pm \frac{1}{\sqrt{2}},\quad k=\pm \frac{1}{\sqrt{2}},\quad {a}_1={a}_2={b}_1=0,{b}_2=\pm \frac{3}{\sqrt{2}}$$
(54)
substitute Eq. (54) in Eq. (4),
$${V}_{131}=\pm \frac{1}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}\left(\frac{1}{{\left(\coth \left(\xi \right)\right)}^2+{\left(\mathit{\operatorname{csch}}\left(\xi \right)\right)}^2+2\left(\coth \left(\xi \right)\right)\pm \left(\mathit{\operatorname{csch}}\left(\xi \right)\right)}\right)$$
(55)
Family-II
$${a}_0=\pm \frac{1}{\sqrt{2}},\quad k=\pm \frac{1}{\sqrt{2}},\quad {a}_1={b}_2=0,\quad {b}_1=0,\quad {a}_2=\pm \frac{3}{\sqrt{2}}$$
(56)
Putting Eq. (56) in Eq. (4),
$${V}_{132}\left(\zeta \right)=\pm \frac{1}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}{\left(\coth \left(\xi \right)\pm \mathit{\operatorname{csch}}\left(\xi \right)\right)}^2$$
(57)
Family-III
$${a}_0=\pm \frac{2}{\sqrt{5}},\quad k=\pm \frac{1}{\sqrt{5}},\quad {a}_1=0,\quad {a}_2=\pm \frac{3}{\sqrt{5}},\quad {b}_1=0,\quad {b}_2=\pm \frac{3}{\sqrt{5}}$$
(58)
Putting Eq. (58) into Eq. (4),
$${V}_{133}=\pm \frac{2}{\sqrt{5}}\pm \frac{3}{\sqrt{5}}\left({\left(\coth \left(\xi \right)\pm \mathit{\operatorname{csch}}\left(\xi \right)\right)}^2+\frac{1}{{\left(\coth \left(\xi \right)\pm \mathit{\operatorname{csch}}\left(\xi \right)\right)}^2}\right)$$
(59)
1. 4.

For H1 = 1,  H2 = 0,  H3 = − 1,

Family-I
$${a}_0=\pm \frac{4}{\sqrt{5}},\quad k\pm \frac{1}{\sqrt{5}},\quad {a}_1=0,\quad {a}_2=0,\quad {b}_1=0,\quad {b}_2=\pm \frac{12}{\sqrt{5}}$$
(60)
Replacing Eq. (60) into Eq. (4),
$${V}_{141}=\pm \frac{4}{\sqrt{5}}\pm \frac{12}{\sqrt{5}}{\left(\frac{1}{\tanh \left(\xi \right)}\right)}^2.$$
(61)
Family-II
$${a}_0=\pm \frac{4}{\sqrt{5}},\quad k\pm \frac{1}{\sqrt{5}},\quad {a}_1=0,\quad {b}_2=0,\quad {b}_1=0,\quad {a}_2=\pm \frac{12}{\sqrt{5}}$$
(62)
After replace Eq. (62) in Eq. (4),
$${V}_{142}\left(\xi \right)=\pm \frac{4}{\sqrt{5}}\pm \frac{12}{\sqrt{5}}{\left(\tanh \left(\xi \right)\right)}^2,$$
(63)
Family-III
$${a}_0=\pm \frac{8}{\sqrt{17}},\quad k=\pm \frac{1}{\sqrt{17}},\quad {a}_1=0,{a}_2=\pm \frac{12}{\sqrt{17}},\quad {b}_1=0,\quad {b}_2=\pm \frac{12}{\sqrt{17}}$$
(64)
Substitute Eq. (64) in Eq. (4),
$${V}_{143}\left(\xi \right)=\pm \frac{8}{\sqrt{17}}\pm \frac{12}{\sqrt{17}}\left({\left(\tanh \left(\xi \right)\right)}^2+\frac{1}{{\left(\tanh \left(\xi \right)\right)}^2}\right).$$
(65)

#### The Trigonometric Function Solutions of Eq. (46)

1. 1.

For $${H}_1=\frac{1}{2},\, {H}_2=0,\, {H}_3=\frac{1}{2},$$

Family-I
$${a}_0=\pm \frac{3}{\sqrt{2}},\quad k\pm \frac{1}{\sqrt{2}},\quad {a}_1=0,\quad {b}_2=0,\quad {b}_1=0,\quad {a}_2=\pm \frac{3}{\sqrt{2}}$$
(66)
Put Eq. (66) into Eq. (4) (Fig. 6),
$${V}_{151}\left(\xi \right)=\pm \frac{3}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}{\left(\sec \left(\xi \right)+\tan \left(\xi \right)\right)}^2.$$
(67)
Family-II
$${a}_0=\pm \frac{3}{\sqrt{2}},\quad k\pm \frac{1}{\sqrt{2}},\quad {a}_1=0,\quad {a}_2=0,{b}_1=0,\quad {b}_2=\pm \frac{3}{\sqrt{2}}$$
(68)
Alternate Eq. (68) into Eq. (4)
$${V}_{152}\left(\xi \right)=\pm \frac{3}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}\left(\frac{1}{{\left(\sec \left(\xi \right)\right)}^2+{\left(\tan \left(\xi \right)\right)}^2+2\left(\sec \left(\xi \right)\right)\left(\tan \left(\xi \right)\right)}\right).$$
(69)
Family-III
$${a}_0=\pm \frac{6}{\sqrt{5}},\quad k=\frac{1}{\sqrt{5}},\quad {a}_1=0,\quad {a}_2=-\pm \frac{3}{\sqrt{5}},\quad {b}_1=0,\quad {b}_2=\pm \frac{3}{\sqrt{5}}$$
(70)
Substituting (70) into (4)
$${V}_{153}\left(\xi \right)=\pm \frac{6}{\sqrt{5}}\pm \frac{3}{\sqrt{5}}\left({\left(\sec \left(\xi \right)+\tan \left(\xi \right)\right)}^2+\frac{1}{{\left(\sec \left(\xi \right)+\tan \left(\xi \right)\right)}^2}\right).$$
(71)
1. 2.

For $${H}_1=-\frac{1}{2},\, {H}_2=0,\, {H}_3=-\frac{1}{2},$$

Family-I
$${a}_0=\pm \frac{3}{\sqrt{2}},\quad k\pm \frac{1}{\sqrt{2}},\quad {a}_1=0,\quad {b}_2=0,\quad {b}_1=0,\quad {a}_2=\pm \frac{3}{\sqrt{2}}$$
(72)
Substituting Eq. (72) into Eq. (4)
$${V}_{161}\left(\xi \right)=\pm \frac{3}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}{\left(\sec \left(\xi \right)-\tan \left(\xi \right)\right)}^2.$$
(73)
Family-II
$${a}_0=\pm \frac{3}{\sqrt{2}},\quad k\pm \frac{1}{\sqrt{2}},\quad {a}_1=0,\quad {a}_2=0,{b}_1=0,\quad {b}_2=\pm \frac{3}{\sqrt{2}}$$
(74)
Substitution of (74) with Eq. (4) gives,
$${V}_{162}\left(\xi \right)=\pm \frac{3}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}\left(\frac{1}{{\left(\sec \left(\xi \right)\right)}^2+{\left(\tan \left(\xi \right)\right)}^2-2\left(\sec \left(\xi \right)\right){\left(\tan \left(\xi \right)\right)}^2}\right).$$
(75)
Family-III
$${a}_0=\pm \frac{6}{\sqrt{5}},\quad k\pm \frac{1}{\sqrt{5}},\quad {a}_1=0,\quad {a}_2=\pm \frac{3}{\sqrt{5}},\quad {b}_1=0,\quad {b}_2=\pm \frac{3}{\sqrt{5}}$$
(76)
Substituting Eq. (76) to Eq. (4)
$${V}_{163}\left(\xi \right)=\pm \frac{6}{\sqrt{5}}\pm \frac{6}{\sqrt{5}}\left({\left(\sec \left(\xi \right)-\tan \left(\xi \right)\right)}^2+\frac{1}{{\left(\sec \left(\xi \right)-\tan \left(\xi \right)\right)}^2}\right).$$
(77)
1. 3.

For H1 = 1(−1), H2 = 0,  H3 = 1(−1),

Family-I
$${a}_0=\pm \frac{12}{\sqrt{5}},\quad k\pm \frac{1}{\sqrt{5}},\quad {a}_1=0,\quad {b}_2=0,{b}_1=0,\quad {a}_2=\pm \frac{12}{\sqrt{5}}$$
(78)
Replace Eq. (78) in Eq. (4)
$${V}_{171}\left(\xi \right)=\pm \frac{12}{\sqrt{5}}\pm \frac{12}{\sqrt{5}}{\left(\left(\tan \left(\xi \right)\left(\cot \left(\xi \right)\right)\right)\right)}^2.$$
(79)
Family-II
$${a}_0=\pm \frac{12}{\sqrt{5}},\quad k\pm \frac{1}{\sqrt{5}},\quad {a}_1={b}_2=0,\quad {a}_2=0,\quad {b}_2=\pm \frac{12}{\sqrt{5}}$$
(80)
Put Eq. (80) in Eq. (4)
$${V}_{172}\left(\xi \right)=\pm \frac{12}{\sqrt{5}}\pm \frac{12}{\sqrt{5}}{\left(\frac{1}{\tan \left(\xi \right)\Big(\cot \left(\xi \right)}\right)}^2.$$
(81)
Family-III
$${a}_0=\pm \frac{24}{\sqrt{17}},\quad k\pm \frac{1}{\sqrt{17}},\quad {a}_1=0,\quad {a}_2=\pm \frac{12}{\sqrt{17}},\quad {b}_1=0,\quad {b}_2=\pm \frac{12}{\sqrt{17}}$$
(82)
Surrogate Eq. (82) within Eq. (4),
$${V}_{173}\left(\xi \right)=\pm \frac{24}{\sqrt{17}}\pm \frac{12}{\sqrt{17}}\left(\Big(\tan \left(\xi \right){\left(\cot \left(\xi \right)\right)}^2+\frac{1}{\Big(\tan \left(\xi \right){\left(\cot \left(\xi \right)\right)}^2}\right).$$
(83)

#### The Rational Function Solution of Eq. (46)

For H1 = 0,  H2 = 0,  H3 = ≠ 0,
$${a}_0=0,\quad k=1,\quad {a}_1=0,\quad {a}_2=-12{H}_3^2,\quad {b}_1=0,\quad {b}_2=0$$
(84)
Substitute Eq. (84) inside Eq. (4)
$${V}_{18}\left(\xi \right)=12{P}_3^2{\left(\frac{1}{H_3\xi +\varepsilon \Big)}\right)}^2$$
(85)
For H2 = 0,  H3 = 0,
$${a}_0=0,\quad k=1.0,\quad {a}_1=0,\quad {a}_2=0,\quad {b}_1=0,\quad {b}_2=-12{H}_1^2$$
(86)
Transfer Eq. (86) in Eq. (4), provides
$${V}_{19}\left(\xi \right)=-\frac{12}{\xi^2}$$
(87)
For H1 ≠ 0,  H2 ≠ 0,  H3 = 0,
$${a}_0=\pm \frac{2{H}_2^2}{\sqrt{H_2^2+1}},\quad k=\pm \frac{1}{\sqrt{H_2^2+1}},\quad {a}_2=0,\quad {b}_1\pm \frac{12{H}_1{H}_2}{\sqrt{H_2^2+1}},\quad {b}_2=\pm \frac{12{H}_1^2}{\sqrt{H_2^2+1}},\quad {a}_1=0,$$
(88)
Substituting Eq. (88) into Eq. (4) (Fig. 7),
$${V}_{18}=\pm \frac{2{H}_2^2}{\sqrt{H_2^2+1}}\pm \frac{12{H}_1{H}_2}{\sqrt{H_2^2+1}}\left(\frac{H_2}{\left(\exp \left({H}_2\xi \right)-{H}_1\right)}\right)\pm \frac{12{H}_1^2}{\sqrt{H_2^2+1}}{\left(\frac{H_2}{\left(\exp \left({H}_2\xi \right)-{H}_1\right)}\right)}^2$$
(89)

### Application of the Fourth-Order Nonlinear Ablowitz-Kaup-Newell-Segur Water Wave Equation

Consider the general form of AKNS (Ali et al. 2018):
$$4{u}_{xt}+{u}_{xx xt}+8{u}_x{u}_{xy}+4{u}_{xx}{v}_y-\delta {u}_{xx}=0,$$
(90)
Consider the traveling waves transformation,
$$u\left(x,\, y,\, t\right)=V\left(\xi \right),\, \xi =x+y+\mu t,$$
(91)
Using Eq. (91) into Eq. (90), we have
$$\left(4\mu -\delta \right){V}^{\prime }+6{V^{\prime}}^2+\mu V^{\prime\prime\prime }=0$$
(92)
Balancing the order of V′2 and V′′′, Eq. (92) has solution (Aasaraai 2015)
$$V\left(\xi \right)={a}_0+{a}_1F\left(\xi \right)+{b}_1{F}^{-1}\left(\xi \right)$$
(93)

From Eq. (93), Eq. (5) and Eq. (92), we acquired several equations involving these, a0,  a1,  b1,  H1,  H2,  H3,  δ, and μ. We have the following solutions

#### The Soliton-Like Solutions of Eq. (90)

When H1 = 0,  H2 = 1,  H3 = − 1,
$$\mu =\frac{\delta }{5},\quad {a}_1=\frac{\delta }{5},\quad {b}_1=0,$$
(94)
From Eq. (94) and (93), we have
$${V}_{19}\left(\xi \right)={a}_0+\frac{\delta }{10}\left(1+\tanh \left(\frac{1}{2}\xi \right)\right)$$
(95)
When H1 = 0,  H2 = − 1,  H3 = 1,
$$\mu =\frac{\delta }{5},\quad {a}_1=-\frac{\gamma }{5},\quad {b}_1=0,$$
(96)
Substituting Eq. (96) into Eq. (93), we have
$${V}_{20}\left(\xi \right)={a}_0-\frac{\delta }{10}\left(1-\coth \left(\frac{1}{2}\xi \right)\right)$$
(97)

When $${H}_1=\frac{1}{2},\, {H}_2=0,\, {H}_3=-\frac{1}{2},$$

Family-I
$$\mu =\frac{\gamma }{3},\quad {a}_1=0,\quad {b}_1=-\frac{\delta }{6},$$
(98)
Substituting Eq. (98) into Eq. (93), we obtained (Fig. 8),
$${V}_{211}\left(\xi \right)={a}_0-\frac{\delta }{6}\left(\frac{1}{\coth \left(\xi \right)\pm \operatorname{csch}\left(\xi \right)}\right)$$
(99) Fig. 8The movement of (95) at (a) and (97) at (b) under these values of parameters a0 = 1,  δ = 0.5 and a0 = 1,  δ = 5 respectively
Family-II
$$\mu =\frac{\gamma }{3},\quad {b}_1=0,\quad {a}_1=\frac{\delta }{6}$$
(100)
Substituting Eq. (100) into Eq. (93), which gives,
$${V}_{212}\left(\zeta \right)={a}_0+\frac{\delta }{6}\left(\coth \left(\xi \right)\pm \operatorname{csch}\left(\xi \right)\right)$$
(101)

When H1 = 1,  H2 = 0,  H3 = − 1,

Family-I
$$\mu =\frac{\delta }{8},\quad {a}_1=0,\quad {b}_1=\frac{\delta }{8},$$
(102)
Substituting Eq. (106) into Eq. (93), We achieved
$${V}_{221}\left(\xi \right)={a}_0+\frac{\delta }{8}\left(\frac{1}{\tanh \left(\xi \right)}\right)$$
(103)
Family-II
$$\mu =\frac{\delta }{8},\quad {b}_1=0,\quad {a}_1=\frac{\delta }{8},$$
(104)
Substituting Eq. (104) into Eq. (93), this action gives,
$${V}_{222}\left(\xi \right)={a}_0+\frac{\delta }{8}\tanh \left(\xi \right)$$
(105)
Family-III
$$\mu =\frac{\delta }{20},\quad {b}_1=\frac{\delta }{20},\quad {a}_1=\frac{\delta }{20},$$
(106)
Substituting Eq. (108) into Eq. (93)
$${V}_{223}\left(\zeta \right)={a}_0+\frac{\gamma }{20}\left(\tanh \left(\zeta \right)+\frac{1}{\tanh \left(\zeta \right)}\right)$$
(107)

#### The Trigonometric Function Solutions of Eq. (90)

When $${A}_1=\frac{1}{2},\, {A}_2=0,\, {A}_3=\frac{1}{2},$$

Family-I
$$\mu =\frac{\delta }{3},\quad {a}_1=-\frac{\delta }{6},\quad {b}_1=0,$$
(108)
Substituting Eq. (108) into Eq. (93),
$${V}_{231}\left(\zeta \right)={a}_0-\frac{\delta }{6}\left(\sec \left(\xi \right)+\tan \left(\xi \right)\right)$$
(109)
Family-II
$$\mu =\frac{\delta }{3},\quad {b}_1=\frac{\delta }{6},\quad {a}_1=0,$$
(110)
Substituting Eq. (110) into Eq. (93),
$${V}_{232}\left(\xi \right)={a}_0+\frac{\delta }{6}\left(\frac{1}{\sec \left(\xi \right)+\tan \left(\xi \right)}\right)$$
(111)

When $${H}_1=-\frac{1}{2},\, {H}_2=0,\, {H}_3=-\frac{1}{2}$$,

Family-I
$$\mu =\frac{\delta }{3},\quad {a}_1=\frac{\delta }{6},\quad {b}_1=0,$$
(112)
Substituting Eq. (112) into Eq. (93),
$${V}_{241}\left(\xi \right)={a}_0+\frac{\gamma }{6}\left(\sec \left(\xi \right)-\tan \left(\xi \right)\right)$$
(113)
Family-II
$$\mu =\frac{\delta }{3},\quad {b}_1=-\frac{\delta }{6},\quad {a}_1=0,$$
(114)
Substituting Eq. (114) into Eq. (93),
$${V}_{242}\left(\xi \right)={a}_0-\frac{\gamma }{6}\left(\frac{1}{\sec \left(\xi \right)-\tan \left(\xi \right)}\right)$$
(115)
$${H}_1=-1,\, {H}_2=0,\, {H}_3=1\left(-1\right),$$
$$\mu =-\frac{\delta }{12},\quad {a}_1=-\frac{\delta }{12},\quad {b}_1=\frac{\delta }{12},$$
(116)
Substituting Eq. (116) into Eq. (93) (Fig. 9),
$${V}_{25}\left(\xi \right)={a}_0-\frac{\delta }{12}\left(\tan \left(\xi \right)\left(\cot \left(\xi \right)\right)-\frac{1}{\tan \left(\xi \right)\left(\cot \left(\xi \right)\right)}\right)$$
(117) Fig. 9The 3D movement of the solitary waves solution (111) on (a) and for (117) on (b), a0 = 1,  δ = 5

#### The Rational Function Solutions of Eq. (6)

When H1 = 0,  H2 = 0,  H3 ≠ 0,

Family-I
$${b}_1=\frac{4\mu -\delta }{6{H}_3},\quad {a}_1=0,$$
(118)
Substituting Eq. (118) into Eq. (93),
$${V}_{261}\left(\xi \right)={a}_0-\frac{4\mu -\delta }{6{H}_3}\left({H}_3\xi +\varepsilon \right)\Big)$$
(119)
Family-II
$$\mu =\frac{\delta }{4},\quad {a}_1=-\frac{H_3\delta }{4},\quad {b}_1=0,$$
(120)
Substituting Eq. (120) into Eq. (93),
$${V}_{262}\left(\xi \right)={a}_0+\frac{H_3\delta }{4}\left(\frac{1}{\left({H}_3\xi +\varepsilon \right)}\right)$$
(121)

When H2 = 0,  H3 = 0, then we have,

Family-I
$${a}_1=\frac{-4\mu +\delta }{6{H}_1},\quad {b}_1=0,$$
(122)
Substituting Eq. (122) into Eq. (93),
$${V}_{271}\left(\xi \right)={a}_0-\frac{4\mu +\delta }{6}\left(\xi \right)\Big)$$
(123)
Family-II
$$\mu =\frac{\delta }{4},\quad {b}_1=\frac{H_1\delta }{4},\quad {a}_1=0,$$
(124)
Substituting Eq. (124) into Eq. (93),
$${V}_{272}\left(\xi \right)={a}_0+\frac{H_1\delta }{4}\left(\frac{1}{\left(\xi \right)}\right)$$
(125)
When H1 ≠ 0,  H2 ≠ 0,  H3 = 0,
$$\mu =\frac{\delta }{4+{A}_2^2},\quad {b}_1=\frac{H_1\delta }{4+{H}_2^2},\quad {a}_1=0,$$
(126)
Substituting Eq. (126) into Eq. (93) (Fig. 10),
$${V}_{28}\left(\xi \right)={a}_0+\frac{H_1\delta }{4+{H}_2^2}\left(\frac{H_2}{\left(\exp \left({A}_2\xi \right)-{H}_1\right)}\right)$$
(127) Fig. 10Solitary waves of (125) at (a) and (127) at (b) with; ε = 1,  H3 = 1,  a0 = 1,  δ = 1 and H2 = 2,  H1 = 1,  a0 = 1,  δ = − 2, respectively

## Discussion of the Results

After successfully employing the modified F-expansion technique on our three selective models, i.e., Modified Liouville, SRLW, and AKNS equations, now we have discussed the similarities and dissimilarities of our new constructed results with other results in previous different literature. Few novel constructed results are likely similar with results in different articles with the following points.
• Our solutions (7) and (61) are approximate similar with solutions (24) and (31) in Lu et al. (2018b) respectively.

• Further our solutions of (43) and (45) are nearly same with the solutions (50) and (41) in Lu et al. (2018a), respectively.

• Moreover, our solutions (95) and (97) are likely similar with solutions (38) and (39) in Inca et al. (2017), respectively.

• In the same way, our constructed solutions (101) and (121) are same with the solutions (57) and (42) in Lu et al. (2018b), respectively.

• Fig. 3 of our solution (23) with these values of parameters d1 = 0.7, d2 = 1, k = 4.5, β = − 4.5, ω = − 5.3 are equal with the solution (51) of Fig. 5 in Lu et al. (2018a) with =0.5, = 2, = 1, a = 1, b = 1, = 1, k = 1, = 0.5.

The segment, discussion of the results shows that our this new achievement is more efficiency and fruitful to face the nonlinear problems for solving them in mathematics, physics, and engineering.

## Conclusion

Here current work is about the study of modified Liouville, the symmetric regularized long wave, and Fourth order Ablowitz-Kaup-Newell-Segur wave models through the successful implement of the modified F-expansion method, series new soliton solutions in the form of hyperbolic, trigono metric, and rational functions are obtained. The graphical movements some solutions are fruitful to comprehend the physical phenomena of these three nonlinear models. Finally our constructed new results have many potential merits to handling different types of nonlinear wave system.

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