He–Laplace method for nonlinear vibration in shallow water waves

Abstract

This study is based on an analytical approach for nonlinear vibration of nonlinear evaluation equations which arises in shallow water waves. This approach is, coupled with variational iteration method and Laplace transformation, known as He–Laplace method. In order to show the capability of the proposed method, He’s polynomials are employed to handle nonlinear terms in illustrated problems due to specific limit of Laplace transformation. Significant results and their graphical representations demonstrated that He–Laplace method is most suitable for nonlinear problems, and it is enormously active for nonlinear vibrations and nonlinear waves.

Keywords

He–Laplace method nonlinear vibration modified Boussinesq equation cubic Boussinesq equation He’s polynomials

Introduction

Mathematical modeling of many physical system leads to nonlinear ordinary or partial differential equations (PDEs) in numerous fields of physics and engineering.^1–4 It is well known that the investigation of the exact solutions of nonlinear PDEs plays an important role in the study of nonlinear physical phenomena. The Boussinesq equation appears in studying the transverse motion and nonlinearity in acoustic waves on elastic rods with circular cross section. This equation also arises in other physical applications such as nonlinear lattice waves,⁵ iron sound waves in plasma,⁶ and vibrations in a nonlinear string.⁷ Kaya⁸ investigated the explicit solution of generalized Boussinesq equation using Adomian decomposition scheme. Yu et al.⁹ estimated multiple soliton solutions of the Boussinesq equation in the light of some suitable transformations. Wazwaz^10–13 investigated different approaches for the solution of Boussinesq equation and showed reliable results to nonlinear evolution equation. Jawad¹⁴ obtained exact solution for cubic Boussinesq and modified Boussinesq using tanh method, sech method, and sine-cosine method equations. Mousa and Kaltayev¹⁵ constructed approximate and exact solutions for Boussinesq equations using homotopy perturbation Padé technique.

The idea of variational iteration method was developed by He^16–18 and was introduced as a powerful tool in various kinds of nonlinear problems.^19,20 Another efficient iterative technique was introduced to solve nonlinear differential equations, known as homotopy perturbation method.^21–24 Recently, Khuri and Sayfy²⁵ developed a technique for the solution of differential equations applying Laplace variational iteration strategy and many researchers^26–29 demonstrated the efficiency of this method with various kinds of ordinary and PDEs but all these techniques have some limitations.

In this paper, we are interested in applying He–Laplace method to obtain approximate, analytical, and exact solutions of the cubic Boussinesq and modified Boussinesq equations. The significant results show that He–Laplace method is suitable for nonlinear vibration in shallow water waves.

Basic idea of homotopy perturbation method

To illustrate the basic concept of homotopy perturbation method, consider the following nonlinear functional equation

A (u) - f (r) = 0, r \in Ω

(1)

with boundary conditions

B (u, \frac{\partial u}{\partial n}) = 0, r \in Γ

(2)

where A is a general functional operator, B is a boundary operator, f(r) is a known analytic function, and Γ is the boundary of the domain Ω. The operator A can be generally divided into two operators, L and N, where L is a linear, while N is a nonlinear operator. Therefore, equation (1) can be written as follows

L (u) + N (u) - f (r) = 0

(3)

Using the homotopy technique, we construct a homotopy $v (r, p) : Ω \times [0, 1] \to ℝ$ which satisfies

H (v, p) = (1 - p) [L (v) - L (u_{0})] + p [L (v) - N (v) - f (r)]

(4)

H (v, p) = L (v) - L (u_{0}) + p L (u_{0}) + p [N (v) - f (r)] = 0

(5)

where

p \in [0, 1]

is called homotopy parameter, and u₀ is an initial approximation for the solution of equation (1), which satisfies the boundary conditions. According to HPM, we can use p as a small parameter, and assume that the solution of equation (5) can be written as a power series in p

v = v_{0} + p v_{1} + p^{2} v_{2} + \dots = \sum_{i = 0}^{\infty} p^{i} v_{i}

(6)

Considering p = 1, the approximate solution of equation (1) will be obtained as follows

u = \lim_{p \to 1} v = v_{0} + v_{1} + v_{2} + v_{3} + \dots

(7)

Construction of He–Laplace method

In order to construct He–Laplace method, we may take the Laplace transform of equation (3) and obtain

L [L (u) + N (u) - f (r)] = 0

(8)

Multiplying the above equation with Lagrange multiplier, say $λ (s)$ , we get

λ L [L (u) + N (u) - f (r)] = 0

operating with Laplace transform

λ {L [L (u)] + L [N (u) - f (r)]} = 0

therefore, the recurrence relation becomes

u_{n + 1} (x, s) = u_{n} (x, s) + λ {L [L ({\tilde{u}}_{n})] + L [N ({\tilde{u}}_{n}) - f (r)]}

(9)

This recurrence relation is known as the Laplace variational method, taking the variation of equation (9)

δ u_{n + 1} (x, s) = δ u_{n} (x, s) + λ δ {L [L ({\tilde{u}}_{n})] + L [N ({\tilde{u}}_{n}) - f (r)]}

To identify the value of Lagrange multiplier $λ (s)$ with the help of Laplace transform, we reveal that ${\tilde{u}}_{n}$ is a restricted variable, i.e. $δ {\tilde{u}}_{n} = 0$ and

\frac{δ u_{n + 1} (x, s)}{δ u_{n} (x, s)} = 0

Now, taking the inverse Laplace transform of equation (9) as follows

u_{n + 1} (x, t) = u_{n} (x, t) + L^{- 1} [λ (s) {L [L (u_{n})] + L [N (u_{n}) - f (r)]}]

(10)

Finally, the homotopy perturbation method is implemented to investigate the approximate solution, obtained by substituting the calculated coefficients from equation (7).

Analysis of the method

In this section, He–Laplace method is applied for solving modified Boussinesq equation and cubic Boussinesq equation. The results show that He–Laplace method is very effective and valid for nonlinear evolution equations. Figures 1 and 3 show the behavior of illustrated problems, while Figures 2 and 4 show the error between exact solution and approximate solution.

Figure 1.

Behavior of u(x, t) with $- 10 \leq x \leq 10$ and $- 5 \leq t \leq 5$ .

Figure 2.

Error between exact solution and approximate solution.

Figure 3.

Behavior of u(x, t) with $- 40 \leq x \leq 40$ and $- 5 \leq t \leq 5$ .

Figure 4.

Error between exact solution and approximate solution.

Example 1

Consider modified Boussinesq equation

u_{t t} - u_{x x x x} - {(u^{3})}_{x x} = 0

(11)

with initial conditions

u (x, 0) = \sqrt{2} sech (x), u_{t} (x, o) = \sqrt{2} sech (x) \tanh (x)

(12)

Now taking the Laplace transform of equation (11)

L [\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{4} u}{\partial x^{4}} - \frac{\partial^{2} u^{3}}{\partial x^{2}}] = 0

(13)

According to He–Laplace method, the recurrence relation of equation (13) takes the form

u_{n + 1} (x, s) = u_{n} (x, s) + λ (s) L [\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{4} \tilde{u}}{\partial x^{4}} - \frac{\partial^{2} {\tilde{u}}^{3}}{\partial x^{2}}]

(14)

Now, taking the variation and applying Laplace transform on equation (14), we get

δ u_{n + 1} (x, s) = δ u_{n} (x, s) + λ (s) δ [{s^{2} u_{n} (x, s) - s {\tilde{u}}_{n} (x, 0) - {\tilde{u}}_{n}' (x, 0)} - L {\frac{\partial^{4} \tilde{u}}{\partial x^{4}} + \frac{\partial^{2} {\tilde{u}}^{3}}{\partial x^{2}}}]

which implies

δ u_{n + 1} (x, s) = δ u_{n} (x, s) + s^{2} λ δ u_{n} (x, s)

this in turn gives

λ (s) = - \frac{1}{s^{2}}

Notice that ${\tilde{u}}_{n}$ is a restricted variable, i.e. $δ {\tilde{u}}_{n} = 0$ and

\frac{δ {\tilde{u}}_{n + 1} (x, s)}{δ {\tilde{u}}_{n} (x, s)} = 0

Using the value of $λ (s)$ in equation (14)

u_{n + 1} (x, s) = u_{n} (x, s) - \frac{1}{s^{2}} L [\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{4} u}{\partial x^{4}} - \frac{\partial^{2} u^{3}}{\partial x^{2}}]

(15)

Now, taking inverse Laplace of equation (15)

u_{n + 1} (x, t) = u_{n} (x, t) - L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{4} u}{\partial x^{4}} - \frac{\partial^{2} u^{3}}{\partial x^{2}}}]

and applying He’s polynomials

\begin{array}{l} u_{0} + p u_{1} + p^{2} u_{2} + p^{3} u_{3} + p^{4} u_{4} + \dots = u_{n} (x, t) + p L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{4} u_{n}}{\partial x^{4}} + \frac{\partial^{2} u_{n}^{3}}{\partial x^{2}}}] \\ = u_{n} (x, t) + p L^{- 1} [\frac{1}{s^{2}} L {(\frac{\partial^{4} u_{0}}{\partial t^{4}} + \frac{\partial^{2} u_{0}^{3}}{\partial x^{2}}) \\ + p (\frac{\partial^{4} u_{0}}{\partial t^{4}} + 3 \frac{\partial^{2} (u_{0}^{2} u_{1})}{\partial x^{2}}) + p^{2} (\frac{\partial^{4} u_{0}}{\partial t^{4}} + \frac{\partial^{2} (3 u_{0} u_{1}^{2} + 3 u_{0}^{2} u_{2})}{\partial x^{2}}) + \dots}] \end{array}

By comparing highest powers of p

\begin{array}{l} p^{0} : u_{0} = u_{0} (x, t) + t u_{0 t} (x, t), \\ p^{1} : u_{1} = - L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{4} u_{0}}{\partial t^{4}} + \frac{\partial^{2} u_{0}^{3}}{\partial x^{2}}}], \\ p^{2} : u_{2} = - L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{4} u_{0}}{\partial t^{4}} + 3 \frac{\partial^{2} (u_{0}^{2} u_{1})}{\partial x^{2}}}], \\ p^{3} : u_{3} = - L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{4} u_{0}}{\partial t^{4}} + \frac{\partial^{2} (3 u_{0} u_{1}^{2} + 3 u_{0}^{2} u_{2})}{\partial x^{2}}}], \\ ⋮ \end{array}

so, we get

\begin{array}{l} u_{0} = \sqrt{2} sech (x) + t (\sqrt{2} sech (x) \tanh (x)), \\ u_{1} = \frac{1}{2} t^{2} (- \sqrt{2} {sech}^{5} (x) + \sqrt{2} sech (x) \tanh^{4} (x)) \\ + \frac{1}{6} t^{3} (- 5 \sqrt{2} {sech}^{5} (x) \tanh (x) - 4 \sqrt{2} {sech}^{3} (x) \tanh^{3} (x) + \sqrt{2} sech (x) \tanh^{5} (x)) \\ + \frac{1}{2} t^{4} (2 \sqrt{2} {sech}^{7} (x) - 19 \sqrt{2} {sech}^{5} (x) \tanh^{2} (x) + 9 \sqrt{2} {sech}^{3} (x) \tanh^{4} (x)) \\ + \frac{3}{10} t^{5} (2 \sqrt{2} {sech}^{7} (x) \tanh (x) - 9 \sqrt{2} {sech}^{5} (x) \tanh^{3} (x) + 3 \sqrt{2} {sech}^{3} (x) \tanh^{5} (x)), \\ u_{2} = \frac{1}{24} t^{4} (- 19 \sqrt{2} {sech}^{9} (x) + 196 \sqrt{2} {sech}^{7} (x) \tanh^{2} (x) + 90 \sqrt{2} {sech}^{5} (x) \tanh^{4} (x) \\ - 124 \sqrt{2} {sech}^{3} (x) \tanh^{6} (x) + \sqrt{2} sech (x) \tanh^{8} (x)) \\ + \frac{1}{120} t^{5} (- 11 \sqrt{2} {sech}^{9} (x) \tanh (x) + 316 \sqrt{2} {sech}^{7} (x) \tanh^{3} (x) + 162 \sqrt{2} {sech}^{5} (x) \tanh^{5} (x) \\ - 164 \sqrt{2} {sech}^{3} (x) \tanh^{7} (x) + \sqrt{2} sech (x) \tanh^{9} (x)) \\ + \frac{1}{60} t^{6} (159 \sqrt{2} {sech}^{11} (x) - 36327 \sqrt{2} {sech}^{9} (x) \tanh^{2} (x) \\ + 72819 \sqrt{2} {sech}^{7} (x) \tanh^{4} (x) - 23441 \sqrt{2} {sech}^{5} (x) \tanh^{6} (x) + 819 \sqrt{2} {sech}^{3} (x) \tanh^{8} (x)) \\ + \frac{1}{140} t^{7} (1850 \sqrt{2} {sech}^{11} (x) \tanh (x) - 11473 \sqrt{2} {sech}^{9} (x) \tanh^{3} (x) + 16021 \sqrt{2} {sech}^{7} (x) \tanh^{5} (x) \\ - 6023 \sqrt{2} {sech}^{5} (x) \tanh^{7} (x) + 273 \sqrt{2} {sech}^{3} (x) \tanh^{9} (x)) \\ + \frac{3}{280} t^{8} (44 \sqrt{2} {sech}^{13} (x) - 2866 \sqrt{2} {sech}^{11} (x) \tanh^{2} (x) + 14251 \sqrt{2} {sech}^{9} (x) \tanh^{4} (x) \\ - 12152 \sqrt{2} {sech}^{7} (x) \tanh^{6} (x) + 1575 \sqrt{2} {sech}^{5} (x) \tanh^{8} (x)) \\ + \frac{3}{40} t^{9} (4 \sqrt{2} {sech}^{13} (x) \tanh (x) - 106 \sqrt{2} {sech}^{11} (x) \tanh^{3} (x) + 357 \sqrt{2} {sech}^{9} (x) \tanh^{5} (x) \\ - 236 \sqrt{2} {sech}^{7} (x) \tanh^{7} (x) + 25 \sqrt{2} {sech}^{5} (x) \tanh^{9} (x)), \\ ⋮ \end{array}

hence the solution can be expressed as

u_{n} (x, t) = u_{0} + u_{1} + u_{2} + u_{3} + u_{4} + \dots

Using Taylor’s series, the exact solution converges to

u_{n} (x, t) = \sqrt{2} \sec h (x - t)

Example 2

Consider cubic Boussinesq equation

u_{t t} - u_{x x} + 2 {(u^{3})}_{x x} - u_{x x x x} = 0

(16)

with initial conditions

u (x, 0) = \frac{1}{x}, u_{t} (x, o) = - \frac{1}{x^{2}}

(17)

Now taking the Laplace transform of equation (16)

L [\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} + 2 \frac{\partial^{2} u^{3}}{\partial x^{2}} - \frac{\partial^{4} u}{\partial x^{4}}] = 0

(18)

According to He–Laplace method, the recurrence relation of equation (18) takes the form

u_{n + 1} (x, s) = u_{n} (x, s) + λ (s) L [\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} + 2 \frac{\partial^{2} u^{3}}{\partial x^{2}} - \frac{\partial^{4} u}{\partial x^{4}}]

(19)

Now, taking the variation and applying Laplace transform in equation (19), we get

\begin{matrix} δ u_{n + 1} (x, s) = δ u_{n} (x, s) + λ (s) δ [{s^{2} ω_{n} (x, s) - s {\tilde{u}}_{n} (x, 0) - {\tilde{u}}_{n}' (x, 0)} \\ - L {\frac{\partial^{2} u}{\partial x^{2}} - 2 \frac{\partial^{2} u^{3}}{\partial x^{2}} + \frac{\partial^{4} u}{\partial x^{4}}}] \end{matrix}

this implies

δ u_{n + 1} (x, s) = δ u_{n} (x, s) + s^{2} λ δ u_{n} (x, s)

this in turn gives

λ (s) = - \frac{1}{s^{2}}

Notice that ${\tilde{u}}_{n}$ is a restricted variable, i.e. $δ {\tilde{u}}_{n} = 0$ and

\frac{δ {\tilde{u}}_{n + 1} (x, s)}{δ {\tilde{u}}_{n} (x, s)} = 0

Using the value of $λ (s)$ in equation (19)

u_{n + 1} (x, s) = u_{n} (x, s) - \frac{1}{s^{2}} L [\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} + 2 \frac{\partial^{2} u^{3}}{\partial x^{2}} - \frac{\partial^{4} u}{\partial x^{4}}]

(20)

Now, taking inverse Laplace of equation (20)

u_{n + 1} (x, t) = u_{n} (x, t) - L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} + 2 \frac{\partial^{2} u^{3}}{\partial x^{2}} - \frac{\partial^{4} u}{\partial x^{4}}}]

Applying He’s polynomial formula

\begin{array}{l} u_{0} + p u_{1} + p^{2} u_{2} + p^{3} u_{3} + p^{4} u_{4} + \dots = u_{n} (x, t) + p L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{2} u}{\partial x^{2}} - 2 \frac{\partial^{2} u^{3}}{\partial x^{2}} + \frac{\partial^{4} u}{\partial x^{4}}}] \\ = u_{n} (x, t) + p L^{- 1} [\frac{1}{s^{2}} L {(\frac{\partial^{2} u_{0}}{\partial t^{2}} - 2 \frac{\partial^{2} u_{0}^{3}}{\partial x^{2}} + \frac{\partial^{4} u_{0}}{\partial t^{4}}) \\ + p (\frac{\partial^{2} u_{0}}{\partial t^{2}} - 6 \frac{\partial^{2} (u_{0}^{2} u_{1})}{\partial x^{2}} + \frac{\partial^{4} u_{0}}{\partial t^{4}}) \\ + p^{2} (\frac{\partial^{2} u_{0}}{\partial t^{2}} - 2 \frac{\partial^{2} (3 u_{0} u_{1}^{2} + 3 u_{0}^{2} u_{2})}{\partial x^{2}} + \frac{\partial^{4} u_{0}}{\partial t^{4}}) + \dots}] \end{array}

By comparing the highest powers of p

\begin{array}{l} p^{0} : u_{0} = u_{0} (x, t) + t u_{0 t} (x, t), \\ p^{1} : ω_{1} = L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{2} u_{0}}{\partial t^{2}} - 2 \frac{\partial^{2} u_{0}^{3}}{\partial x^{2}} + \frac{\partial^{4} u_{0}}{\partial t^{4}}}], \\ p^{2} : u_{2} = L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{2} u_{0}}{\partial t^{2}} - 6 \frac{\partial^{2} (u_{0}^{2} u_{1})}{\partial x^{2}} + \frac{\partial^{4} u_{0}}{\partial t^{4}}}], \\ p^{3} : u_{3} = L^{- 1} [\frac{1}{s^{2}} L {\frac{\partial^{2} u_{0}}{\partial t^{2}} - 2 \frac{\partial^{2} (3 u_{0} u_{1}^{2} + 3 u_{0}^{2} u_{2})}{\partial x^{2}} + \frac{\partial^{4} u_{0}}{\partial t^{4}}}], \\ ⋮ \end{array}

so, we get

\begin{array}{l} u_{0} = \frac{1}{x} - \frac{t}{x^{2}}, \\ u_{1} = \frac{t^{2}}{x^{3}} - \frac{t^{3}}{x^{4}}, \\ u_{2} = \frac{t^{4}}{x^{5}} - \frac{t^{5}}{x^{6}}, \\ u_{3} = \frac{t^{6}}{x^{7}} - \frac{t^{7}}{x^{8}}, \\ ⋮ \end{array}

hence the solution can be expressed as

u_{n} (x, t) = u_{0} + u_{1} + u_{2} + u_{3} + u_{4} + \dots

u_{n} (x, t) = \frac{1}{x} - \frac{t}{x^{2}} + \frac{t^{2}}{x^{3}} - \frac{t^{3}}{x^{4}} + \frac{t^{4}}{x^{5}} - \frac{t^{5}}{x^{6}} + \frac{t^{6}}{x^{7}} - \frac{t^{7}}{x^{8}} + \dots

Using Taylor’s series, the exact solution converges to

ω_{n} (x, t) = \frac{1}{x + t}

Conclusions and future work

In this article, He–Laplace method is provided for strongly nonlinear vibration of nonlinear evolution equations. The following conclusions are made:

This method is applied in a direct way without using linearization, transformation, discretization, or restrictive assumptions.

The suggested He–Laplace method is a key solution for nonlinear issues in the field of nonlinear vibration and nonlinear waves.

The recommended scheme is quick and influential for analytical solution of nonlinear problems.

It is an open problem for readers that He–Laplace method can be employed for fractional heat- and wave-like equations.

Footnotes

Declaration of conflicting interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: This work is supported by National Natural Science Foundation of China (Grant No. 11571057).

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