Mathematical Methods in the Applied Sciences
Volume 48, Issue 12 pp. 12427-12439
RESEARCH ARTICLE
Open Access

Analysis of the Generalized Ostrovsky Equation in the Propagation of Surface and Internal Waves in Rotating Fluids

Sol Sáez, 

Corresponding Author

Sol Sáez

Departamento de Matemáticas, Universidad de Cádiz, Cádiz, Spain

Correspondence:

Sol Sáez ([email protected])

Contribution: ​Investigation, Writing - original draft

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First published: 06 May 2025
https://doi.org/10.1002/mma.11036

Funding: The support of University of Cádiz is gratefully acknowledged by the author.

ABSTRACT

The Ostrovsky equation models long, weakly nonlinear waves, explaining the propagation of surface and internal waves in a rotating fluid. The study focuses on the generalized Ostrovsky equation. Introduced by Levandosky and Liu, this equation demonstrates the existence of solitary waves through variational methods. This paper investigates the generalized Ostrovsky equation using Lie symmetry group method and low local conservation laws, essential for analyzing differential equations and describing conserved physical and chemical processes. Specific cases reduce it to the Ostrovsky or generalized Korteweg–de Vries (KdV) equations. Detailed calculations of local conservation laws, classical point symmetries, and symmetry reductions are provided, offering invariant solutions and Lie symmetry groups. This research advances the understanding of differential equations and their applications in modeling scientific phenomena.

1 Introduction

The study of nonlinear internal waves in a rotating ocean has been an important area of research in oceanography and fluid dynamics. Ostrosvsky equation was introduced by Ostrovsky in [1] as a model for long waves which are weakly nonlinear, to explain the propagation of the surface and internal waves in a fluid of reference which describe a rotating movement. This paper aims to discuss the generalized Ostrovsky equation by using Lie symmetry group method and low local conservation laws for this equation. Conservation laws are highly significant in the analysis of differential equations because they describe chemical and physical processes with conserved quantities. We study conservation laws for these equations and we resort to the invariance and multiplier perspective by using the Euler–Lagrange operator. In this paper, we consider the generalized Ostrovsky equation.
u t x − β u x x x x + g ( u ) x x = α f ( u ) $$ {u}_{tx}-\beta {u}_{xxxx}+{\left(g{(u)}_x\right)}_x=\alpha f(u) $$ (1)
where α $$ \alpha $$ and β $$ \beta $$ are the real dispersion coefficients and the functions f $$ f $$ and g $$ g $$ are C 2 $$ {C}^2 $$ functions. This equation was introduced by Levandosky and Liu in [2] when f $$ f $$ is the identity function and they proved the existence of solitary waves, called ground states, by employing variational methods. Specifically, in the case g ( u ) = u 2 $$ g(u)={u}^2 $$ and f ( u ) = u $$ f(u)=u $$ , we obtain the Ostrovsky equation
u t x − β u x x x x + ( u 2 ) x x = α u $$ {u}_{tx}-\beta {u}_{xxxx}+{\left({\left({u}^2\right)}_x\right)}_x=\alpha u $$ (2)
The function u ( t , x ) $$ u\left(t,x\right) $$ denotes the free surface of a liquid, α ∈ ℝ $$ \alpha \in \mathbb{R} $$ is a measure of rotational effects due to Coriolis force, and β ∈ ℝ $$ \beta \in \mathbb{R} $$ determines the dispersion. Some special cases of the Ostrovsky equation has been investigated by several authors [3-7]. In [8], the authors prove that solutions of this equation converge to solutions of the Korteweg–de Vries (KdV) equation. In [9], Varlamov and Liu proved that (1) has special characteristics in the space X s $$ {X}_s $$ for s > 3 / 2 $$ s>3/2 $$ . Tsugawa [10] extended the results of Varlamov and Liu for s > − 3 / 4 $$ s>-3/4 $$ . In the absence of rotation, that is, when α = 0 $$ \alpha =0 $$ , we integrate and we obtain the generalized KdV equation:
u t − β u x x x + g ( u ) x = 0 $$ {u}_t-\beta {u}_{xxx}+{\left(g(u)\right)}_x=0 $$ (3)

The original Ostrovsky equation models the propagation of nonlinear waves in rotating media, such as oceans and the atmosphere. However, this formulation may be insufficient to describe scenarios where dispersion and nonlinearity vary more complexly. The generalized version allows for additional terms that can represent more realistic physical effects, such as spatial and temporal variations in dispersion or more sophisticated wave interactions. Moreover, studying this version expands knowledge of nonlinear differential equations and their exact solutions, which can be applied in both numerical and theoretical fluid dynamics models.

In the current paper, we have studied (1) using Lie symmetry reductions, symmetry group, and a lot of conservative laws for them [11-13]. Partial differential equations describe several scientific processes, such as the heat equation [14] and other chemical or physical processes [15, 16]. Numerous solutions of differential equations have been obtained by several methods [17], such as the direct method [18], simplest equation method [19], the tanh method [20], or the homogeneous balanced method [21]. An important and efficient method to study differential equations is the Lie group method [22, 23]. In Olver [24] and Bluman and Kumei [25], we find a precise description about symmetry groups and Lie's theory. Symmetries help us to obtain invariable solutions of differential equations which have been reduced to other equations with fewer independent variables [26, 27]. By means of symmetry groups, we can obtain solutions of partial differential equations through other solutions.

This paper is ordered as follows: Firstly, we calculate local conservation laws of low order admitted by (1) by employing the multipliers theory, and we obtain a wide variety of low-order conservation laws for Equation (1). After that, we obtain the classical point symmetries admitted by Equation (1) in Section 3. By means of the point symmetries, we calculate symmetry reductions of the equation in the following section. In addition, we obtain other invariant solutions of the generalized equation, and we obtain the Lie symmetry groups. Finally, we present some concluding comments.

2 Conservation Laws

Conservations laws have been utilized to obtain solutions of partial differential equations and in the expansion of different numerical methods [28-30]. They are used in proving the existence and uniqueness of solutions. To evaluate conserved fluxes and densities, we study conservation laws for Equation (1), and we have used the invariance and multiplier perspective with the use of the Euler–Lagrange operator. Then, we have obtained conservation laws for Equation (1), namely, the following continuity equation:
D t T + D x X | ε = 0 , $$ {\left.\left({D}_tT+{D}_xX\right)\right|}_{\varepsilon }=0, $$ (4)
in terms of the total derivative operators holding for the solutions of (1). The spatial flux is noted by X $$ X $$ , and the conserved density is noted by T $$ T $$ , depending of u , x , t $$ u,x,t $$ and some derivatives of the function u $$ u $$ . Also, D x $$ {D}_x $$ and D t $$ {D}_t $$ represent the total derivative functions with respect x $$ x $$ and t $$ t $$ , respectively.
A multiplier Q $$ Q $$ is a function Q ( t , x , u , u t , u x ,   … ) $$ Q\left(t,x,u,{u}_t,{u}_x,\dots \right) $$ that satisfies that, for solutions of Equation (1),
( u t x − β u x x x x + g ( u ) x x − α f ( u ) ) Q $$ \left({u}_{tx}-\beta {u}_{xxxx}+{\left(g{(u)}_x\right)}_x-\alpha f(u)\right)Q $$
is a divergence expression. Every significant conservation law emerges from multipliers. If we move away of the group of solutions of Equation (1), all significant conservation laws (4) are equivalent to conservation laws with the following characteristic form:
D t T ^ + D x X ^ = ( u t x − β u x x x x + g ( u ) x x − α f ( u ) ) Q , $$ \left({D}_t\hat{T}+{D}_x\hat{X}\right)=\left({u}_{tx}-\beta {u}_{xxxx}+{\left(g{(u)}_x\right)}_x-\alpha f(u)\right)Q, $$ (5)
where ( T ^ , X ^ ) $$ \left(\hat{T},\hat{X}\right) $$ varies from ( T , X ) $$ \left(T,X\right) $$ by a banal conserved current.
We solve the following determining equation to find the set of multipliers:
δ δ u ( u t x − β u x x x x + g ( u ) x x − α f ( u ) ) Q = 0 , $$ \frac{\delta }{\delta u}\left({u}_{tx}-\beta {u}_{xxxx}+{\left(g{(u)}_x\right)}_x-\alpha f(u)\right)Q=0, $$ (6)
where δ δ u $$ \frac{\delta }{\delta u} $$ represents the Euler–Lagrange factor Ê [ u ] $$ \hat{E}\left[u\right] $$ defined by
Ê [ u ] : = ∂ ∂ u + ∑ s ≥ 1 ( − 1 ) s D i 1 ⋯ D i s ∂ ∂ u i 1 i 2 … i s . $$ \hat{E}\left[u\right]\kern0.5em := \kern0.5em \frac{\partial }{\partial u}+\sum \limits_{s\ge 1}{\left(-1\right)}^s{D}_{i_1}\cdots {D}_{i_s}\frac{\partial }{\partial {u}_{i_1{i}_2\dots {i}_s}}. $$ (7)

By considering (5-7), we derive the following local conservation laws for Equation (1):

Theorem 1.Under some arbitrary constants, the generalized Ostrovsky equation (1) admits the following local conservation laws of low order.

  • Case 1. In the case f ( u ) = c 1 u + c 2 $$ f(u)={c}_1u+{c}_2 $$ , with f ≠ 0 $$ f\ne 0 $$ , we have the conservation law multiplier Q = g ′ ( u ) u x − β u x x x + u t $$ Q={g}^{\prime }(u){u}_x-\beta {u}_{xxx}+{u}_t $$ , and we get the following local conservation law:
    D t c 1 α 2 u 2 + α c 2 u + D x 1 2 u t 2 + β − 1 2 c 1 u x 2 α + ( c 1 u + c 2 ) α u x x − u t u x x x + β 2 u x x x 2 − α ∫ ( c 1 u + c 2 ) g ′ ( u ) d u + 1 2 g ′ ( u ) 2 u x 2 − u x g ′ ( u ) ( β u x x x − u t ) = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(\frac{c_1\alpha }{2}{u}^2+\alpha {c}_2u\right)\hfill \\ {}\hfill & +{D}_x\left(\frac{1}{2}{u}_t^2+\beta \left(-\frac{1}{2}{c}_1{u}_x^2\alpha +\left({c}_1u+{c}_2\right)\alpha {u}_{xx}-{u}_t{u}_{xx x}\right)\right.\hfill \\ {}\hfill \kern1em & +\frac{\beta }{2}{u}_{xx x}^2-\alpha \int \left({c}_1u+{c}_2\right){g}^{\prime }(u) du\hfill \\ {}\hfill & \left.+\frac{1}{2}{g}^{\prime }{(u)}^2{u}_x^2-{u}_x{g}^{\prime }(u)\left(\beta {u}_{xx x}-{u}_t\right)\right)=0.\hfill \end{array}} $$ (8)
  • Case 2.

    In the case f ≡ 0 $$ f\equiv 0 $$ , we obtain local conservation laws under the following conservation law multipliers:

    • a.

      Q 1 = x $$ {Q}_1=x $$ , give us the following local conservation law:

      D t x u x + D x x g ′ ( u ) u x − g ( u ) + β ( − x u x x x + u x x ) = 0 . $$ {D}_t\left(x{u}_x\right)+{D}_x\left(x{g}^{\prime }(u){u}_x-g(u)+\beta \left(-x{u}_{xx x}+{u}_{xx}\right)\right)=0. $$ (9)

    • b.

      Q 2 = F ( t ) $$ {Q}_2=F(t) $$ , with the conservation law:

      D t F ( t ) u x + D x − F ( t ) β u x x x + F ( t ) g ′ ( u ) u x − u F ′ ( t ) = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(F(t){u}_x\right)\hfill \\ {}\hfill & +{D}_x\left(-F(t)\beta {u}_{xxx}+F(t){g}^{\prime }(u){u}_x-u{F}^{\prime }(t)\right)=0.\hfill \end{array}} $$ (10)

    • c.

      Q 3 = u x x x − 1 β u t − 1 β u x g ′ ( u ) , β ≠ 0 $$ {Q}_3={u}_{xxx}-\frac{1}{\beta }{u}_t-\frac{1}{\beta }{u}_x{g}^{\prime }(u),\kern0.3em \beta \ne 0 $$ with the conservation law:

      D t 0 + D x − 1 2 β β u x x x − g ′ ( u ) u x − u t 2 = 0 . $$ {D}_t(0)+{D}_x\left(-\frac{1}{2\beta }{\left(\beta {u}_{xxx}-{g}^{\prime }(u){u}_x-{u}_t\right)}^2\right)=0. $$ (11)

  • Case 3.

    In the case g ( u ) = c 1 u + c 2 $$ g(u)={c}_1u+{c}_2 $$ , with c 1 , c 2 ∈ ℝ $$ {c}_1,{c}_2\in \mathbb{R} $$ , we obtain local conservation laws under the following conservation law multipliers:

    • a.

      Q 1 = u x $$ {Q}_1={u}_x $$ and the local conservation law:

      D t 1 2 u x 2 + D x − u x β u x x x − α ∫ f ( u ) d u + 1 2 ( β u x x 2 + c 1 u x 2 ) = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(\frac{1}{2}{u}_x^2\right)+{D}_x\hfill \\ {}\hfill & \left(-{u}_x\beta {u}_{xx x}-\alpha \int f(u) du+\frac{1}{2}\left(\beta {u}_{xx}^2+{c}_1{u}_x^2\right)\right)=0.\hfill \end{array}} $$ (12)

    • b.

      Q 2 = u t $$ {Q}_2={u}_t $$ , with the conservation law:

      D t − 1 2 ( β u x x x + u x 2 c 1 ) − α ∫ f ( u ) d u + D x − u t β u x x x + β u t x u x x + 1 2 u t 2 + c 1 u t u x = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(-\frac{1}{2}\left(\beta {u}_{xx x}+{u}_x^2{c}_1\right)-\alpha \int f(u) du\right)\hfill \\ {}\hfill & +{D}_x\left(-{u}_t\beta {u}_{xx x}+\beta {u}_{tx}{u}_{xx}+\frac{1}{2}{u}_t^2+{c}_1{u}_t{u}_x\right)=0.\hfill \end{array}} $$ (13)

    • c.

      Q 3 = u x x x $$ {Q}_3={u}_{xxx} $$ , in the case f ≡ c 3 ∈ ℝ $$ f\equiv {c}_3\in \mathbb{R} $$ , with the conservation law:

      D t − 1 2 u x x 2 + D x − 1 2 β u x x x 2 − c 3 α u x x 1 2 c 1 u x x 2 + u t x u x x = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(-\frac{1}{2}{u}_{xx}^2\right)\hfill \\ {}\hfill & +{D}_x\left(-\frac{1}{2}\beta {u}_{xx x}^2-{c}_3\alpha {u}_{xx}\frac{1}{2}{c}_1{u}_{xx}^2+{u}_{tx}{u}_{xx}\right)=0.\hfill \end{array}} $$ (14)

    • d.

      Q 4 = β u x x x + u t $$ {Q}_4=\beta {u}_{xxx}+{u}_t $$ , in the case f ≡ c 3 ∈ ℝ $$ f\equiv {c}_3\in \mathbb{R} $$ , which give us the following local conservation law:

      D t − β u x x 2 − 1 2 u x 2 − c 3 α u + D x − β 2 2 u x x x 2 + β 2 ( u x x 2 + 4 u t x u x x − 2 u t u x x x ) + 1 2 u t 2 + u t u x − c 3 α β u x x = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(-\beta {u}_{xx}^2-\frac{1}{2}{u}_x^2-{c}_3\alpha u\right)\hfill \\ {}\hfill & +{D}_x\left(-\frac{\beta^2}{2}{u}_{xx x}^2+\frac{\beta }{2}\left({u}_{xx}^2+4{u}_{tx}{u}_{xx}-2{u}_t{u}_{xx x}\right)\right.\hfill \\ {}\hfill & +\left.\frac{1}{2}{u}_t^2+{u}_t{u}_x-{c}_3\alpha \beta {u}_{xx}\right)=0.\hfill \end{array}} $$ (15)

  • Case 4. In the case g ( u ) = 1 n + 1 u n + 1 , n ≠ 0 , n ≠ − 1 , α ≠ 0 , f ( u ) = c 3 $$ g(u)=\frac{1}{n+1}{u}^{n+1},\kern0.3em n\ne 0,\kern0.3em n\ne -1,\kern0.3em \alpha \ne 0,\kern0.3em f(u)={c}_3 $$ , with c 3 ≠ 0 $$ {c}_3\ne 0 $$ , under the conservation law multiplier Q = F 2 ( t , γ ) + F 1 ( − γ ) $$ Q={F}_2\left(t,\gamma \right)+{F}_1\left(-\gamma \right) $$ , with γ = u n u x − β u x x x + u t $$ \gamma ={u}^n{u}_x-\beta {u}_{xxx}+{u}_t $$ , we obtain the following local conservations law:
    D t ( 0 ) + D x ∫ − β ( F 2 ( t , γ ) + F 1 ( − γ ) ) d u x x x = 0 . $$ {D}_t(0)+{D}_x\left(\int -\beta \left({F}_2\left(t,\gamma \right)+{F}_1\left(-\gamma \right)\right)d{u}_{xxx}\right)=0. $$ (16)
  • Case 5.

    In the case g ( u ) = 1 n + 1 u n + 1 , n ≠ 0 , n ≠ − 1 , α ≠ 0 $$ g(u)=\frac{1}{n+1}{u}^{n+1},\kern0.3em n\ne 0,\kern0.3em n\ne -1,\kern0.3em \alpha \ne 0 $$ , and f ( u ) = c 1 u n + 1 + c 2 u + c 3 $$ f(u)={c}_1{u}^{n+1}+{c}_2u+{c}_3 $$ with c 1 ≠ 0 $$ {c}_1\ne 0 $$ , we obtain local conservation laws under the following conservation law multipliers:

    • a.

      Q 1 = ( n + 1 ) α c 1 u x + u x x x − 1 β u n u x − 1 β u t , β ≠ 0 $$ {Q}_1=\left(n+1\right)\alpha {c}_1{u}_x+{u}_{xxx}-\frac{1}{\beta }{u}^n{u}_x-\frac{1}{\beta }{u}_t,\kern0.3em \beta \ne 0 $$ give us the following local conservation law:

      D t α β n + 1 2 c 1 u x 2 β + 1 n c 1 u n + 1 2 c 2 u 2 + c 3 u + D x − ( n + 1 ) α β c 1 u x u x x x − 1 2 β u x x x 2 + u n u x u x x x + u t u x x x + n + 1 2 α β c 1 u x x 2 − α c 1 u n + 1 u x x − α c 2 u u x x − α c 3 u x x + ( n + 1 ) α c 1 u n u x 2 + 1 2 α c 2 u x 2 − 1 2 β u 2 n u x 2 − 1 β u n u t u x − 1 2 β u t 2 − n + 1 n + 2 α 2 c 1 2 u n + 2 − n + 1 2 α 2 c 1 c 2 u 2 − ( n + 1 ) α 2 c 1 c 3 u + α c 1 2 β ( n + 1 ) u 2 ( n + 1 ) + α c 2 β ( n + 2 ) u n + 2 + α c 3 β ( n + 1 ) u n + 1 = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(\frac{\alpha }{\beta}\left(\frac{n+1}{2}{c}_1{u}_x^2\beta +\frac{1}{n}{c}_1{u}^n+\frac{1}{2}{c}_2{u}^2+{c}_3u\right)\right)\hfill \\ {}\hfill & +{D}_x\left(-\left(n+1\right)\alpha \beta {c}_1{u}_x{u}_{xx x}-\frac{1}{2}\beta {u}_{xx x}^2+{u}^n{u}_x{u}_{xx x}\right.\hfill \\ {}\hfill & +{u}_t{u}_{xx x}+\frac{n+1}{2}\alpha \beta {c}_1{u}_{xx}^2-\alpha {c}_1{u}^{n+1}{u}_{xx}-\alpha {c}_2u{u}_{xx}\hfill \\ {}\hfill & -\alpha {c}_3{u}_{xx}+\left(n+1\right)\alpha {c}_1{u}^n{u}_x^2+\frac{1}{2}\alpha {c}_2{u}_x^2\hfill \\ {}\hfill & -\frac{1}{2\beta }{u}^{2n}{u}_x^2-\frac{1}{\beta }{u}^n{u}_t{u}_x-\frac{1}{2\beta }{u}_t^2\hfill \\ {}\hfill & -\frac{n+1}{n+2}{\alpha}^2{c}_1^2{u}^{n+2}-\left(\frac{n+1}{2}\right){\alpha}^2{c}_1{c}_2{u}^2\hfill \\ {}\hfill & -\left(n+1\right){\alpha}^2{c}_1{c}_3u+\frac{\alpha {c}_1}{2\beta \left(n+1\right)}{u}^{2\left(n+1\right)}\hfill \\ {}\hfill & +\left.\frac{\alpha {c}_2}{\beta \left(n+2\right)}{u}^{n+2}+\frac{\alpha {c}_3}{\beta \left(n+1\right)}{u}^{n+1}\right)=0.\hfill \end{array}} $$ (17)

    • b.

      Q 2 = e n + 1 α 3 / 2 c 1 3 / 2 β t n + 1 e x ( n + 1 ) α c 1 e n + 1 α c 2 t ( n + 1 ) c 1 $$ {Q}_2={\left({e}^{\sqrt{n+1}}{\alpha}^{3/2}{c}_1^{3/2}\beta t\right)}^{n+1}{e}^{x\sqrt{\left(n+1\right)\alpha {c}_1}}{e}^{\frac{\sqrt{n+1}\sqrt{\alpha }{c}_2t}{\left(n+1\right)\sqrt{c_1}}} $$ , with the conservation law:

      D t u x e ( n + 1 ) α ( n + 1 ) c 1 ( ( n + 1 ) 2 α β c 1 2 t + ( n + 1 ) c 1 x + c 2 t ) + D x − n + 1 c 1 e ( n + 1 ) α ( n + 1 ) c 1 ( ( n + 1 ) 2 α β c 1 2 t + ( n + 1 ) c 1 x + c 2 t ) · c 1 3 / 2 α β u x + c 1 n + 1 ( β u x x x − u x u n ) − α ( n + 1 ) n + 1 c 1 β u x x − c 1 u n + 1 n + 1 − c 2 u n + 1 = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left({u}_x{e}^{\frac{\sqrt{\left(n+1\right)\alpha }}{\left(n+1\right)\sqrt{c_1}}\left({\left(n+1\right)}^2\alpha \beta {c}_1^2t+\left(n+1\right){c}_1x+{c}_2t\right)}\right)\hfill \\ {}\hfill & +{D}_x\left(-\frac{n+1}{\sqrt{c_1}}{e}^{\frac{\sqrt{\left(n+1\right)\alpha }}{\left(n+1\right)\sqrt{c_1}}\left({\left(n+1\right)}^2\alpha \beta {c}_1^2t+\left(n+1\right){c}_1x+{c}_2t\right)}\right.\hfill \\ {}\hfill & \cdotp \left({c}_1^{3/2}\alpha \beta {u}_x+\frac{\sqrt{c_1}}{n+1}\left(\beta {u}_{xx x}-{u}_x{u}^n\right)\right.\hfill \\ {}\hfill & -\left.\left.\frac{\sqrt{\alpha \left(n+1\right)}}{n+1}\left({c}_1\beta {u}_{xx}-\frac{c_1{u}^{n+1}}{n+1}-\frac{c_2u}{n+1}\right)\right)\right)=0.\hfill \end{array}} $$ (18)

    • c.

      Q 3 = e − n + 1 α 3 / 2 c 1 3 / 2 β t n + 1 e − x ( n + 1 ) α c 1 e − n + 1 α c 2 t ( n + 1 ) c 1 $$ {Q}_3={\left({e}^{-\sqrt{n+1}}{\alpha}^{3/2}{c}_1^{3/2}\beta t\right)}^{n+1}{e}^{-x\sqrt{\left(n+1\right)\alpha {c}_1}}{e}^{-\frac{\sqrt{n+1}\sqrt{\alpha }{c}_2t}{\left(n+1\right)\sqrt{c_1}}} $$ , with the conservation law:

      D t u x e − ( n + 1 ) α ( n + 1 ) c 1 ( ( n + 1 ) 2 α β c 1 2 t + ( n + 1 ) c 1 x + c 2 t ) + D x − n + 1 c 1 e − ( n + 1 ) α ( n + 1 ) c 1 ( ( n + 1 ) 2 α β c 1 2 t + ( n + 1 ) c 1 x + c 2 t ) · c 1 3 / 2 α β u x + c 1 n + 1 ( β u x x x − u x u n ) − α ( n + 1 ) n + 1 ( c 1 β u x x − c 1 u n + 1 n + 1 − c 2 u n + 1 ) = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left({u}_x{e}^{-\frac{\sqrt{\left(n+1\right)\alpha }}{\left(n+1\right)\sqrt{c_1}}\left({\left(n+1\right)}^2\alpha \beta {c}_1^2t+\left(n+1\right){c}_1x+{c}_2t\right)}\right)\hfill \\ {}\hfill & +{D}_x\left(-\frac{n+1}{\sqrt{c_1}}{e}^{-\frac{\sqrt{\left(n+1\right)\alpha }}{\left(n+1\right)\sqrt{c_1}}\left({\left(n+1\right)}^2\alpha \beta {c}_1^2t+\left(n+1\right){c}_1x+{c}_2t\right)}\right.\hfill \\ {}\hfill & \cdotp \left({c}_1^{3/2}\alpha \beta {u}_x+\frac{\sqrt{c_1}}{n+1}\left(\beta {u}_{xx x}-{u}_x{u}^n\right)\right.\hfill \\ {}\hfill & -\left.\left.\frac{\sqrt{\alpha \left(n+1\right)}}{n+1}\left({c}_1\beta {u}_{xx}-\frac{c_1{u}^{n+1}}{n+1}-\frac{c_2u}{n+1}\right)\right)\right)=0.\hfill \end{array}} $$ (19)

  • Case 6. In the case g ( u ) = 1 n + 1 u n + 1 , n ≠ 0 , n ≠ − 1 , α ≠ 0 $$ g(u)=\frac{1}{n+1}{u}^{n+1},\kern0.3em n\ne 0,\kern0.3em n\ne -1,\kern0.3em \alpha \ne 0 $$ , and f ( u ) = c 2 u + c 3 $$ f(u)={c}_2u+{c}_3 $$ , with c 2 ≠ 0 $$ {c}_2\ne 0 $$ , under the conservation law multiplier Q = − u n u x + β u x x x − u t $$ Q=-{u}^n{u}_x+\beta {u}_{xxx}-{u}_t $$ , we obtain one local conservation law:
    D t c 2 α 2 u 2 + α c 3 u + D x − 1 2 u t 2 + β 1 2 c 2 u x 2 α − c 3 α u x x + u t u x x x − β 2 u x x x 2 − α β c 2 u u x x + u x ( β u x x x − u t ) u n + α c 3 n + 1 u n + 1 + α c 2 n + 1 u n + 2 − 1 2 u x 2 u 2 n = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(\frac{c_2\alpha }{2}{u}^2+\alpha {c}_3u\right)\hfill \\ {}\hfill & +{D}_x\left(-\frac{1}{2}{u}_t^2+\beta \left(\frac{1}{2}{c}_2{u}_x^2\alpha -{c}_3\alpha {u}_{xx}+{u}_t{u}_{xx x}\right)-\frac{\beta }{2}{u}_{xx x}^2\right.\hfill \\ {}\hfill & -\alpha \beta {c}_2u{u}_{xx}+{u}_x\left(\beta {u}_{xx x}-{u}_t\right){u}^n+\frac{\alpha {c}_3}{n+1}{u}^{n+1}+\frac{\alpha {c}_2}{n+1}{u}^{n+2}\hfill \\ {}\hfill & -\left.\frac{1}{2}{u}_x^2{u}^{2n}\right)=0.\hfill \end{array}} $$ (20)
  • Case 7. In the case g ( u ) = 1 n + 1 u n + 1 , n ≠ 0 , n ≠ − 1 , α ≠ 0 , f ( u ) = c 3 $$ g(u)=\frac{1}{n+1}{u}^{n+1},\kern0.3em n\ne 0,\kern0.3em n\ne -1,\kern0.3em \alpha \ne 0,\kern0.3em f(u)={c}_3 $$ , with c 3 ≠ 0 $$ {c}_3\ne 0 $$ , under the conservation law multiplier Q = F 2 ( t , γ ) + F 1 ( − γ ) $$ Q={F}_2\left(t,\gamma \right)+{F}_1\left(-\gamma \right) $$ , with γ = u n u x − β u x x x + u t $$ \gamma ={u}^n{u}_x-\beta {u}_{xxx}+{u}_t $$ , we obtain the following local conservations law:
    D t ( 0 ) + D x ∫ − β ( F 2 ( t , γ ) + F 1 ( − γ ) ) d u x x x = 0 . $$ {D}_t(0)+{D}_x\left(\int -\beta \left({F}_2\left(t,\gamma \right)+{F}_1\left(-\gamma \right)\right)d{u}_{xxx}\right)=0. $$ (21)
  • Case 8. In the case g ( u ) = 1 n + 1 u n + 1 , n ≠ 0 , n ≠ − 1 $$ g(u)=\frac{1}{n+1}{u}^{n+1},\kern0.3em n\ne 0,\kern0.3em n\ne -1 $$ , and α = 0 $$ \alpha =0 $$ , under the conservation law multipliers Q 1 = x , Q 2 = F ( t ) $$ {Q}_1=x,\kern0.3em {Q}_2=F(t) $$ , and Q 3 = u x x x − 1 b u t − 1 b u n u x $$ {Q}_3={u}_{xxx}-\frac{1}{b}{u}_t-\frac{1}{b}{u}^n{u}_x $$ , we obtain the following local conservations laws, respectively:
    D t ( − u ) + D x x u n u x − β x u x x x + β u x x + x u t − 1 n + 1 u n + 1 = 0 , D t F ( t ) u x + D x − F ( t ) β u x x x + F ( t ) u n u x − F ′ ( t ) u = 0 , D t ( 0 ) + D x − 1 2 β ( − u n u x + β u x x x − u t ) 2 = 0 , $$ {\displaystyle \begin{array}{cc}\hfill {D}_t\left(-u\right)& +{D}_x\left(x{u}^n{u}_x-\beta x{u}_{xx x}+\beta {u}_{xx}\right.\hfill \\ {}\hfill & +\left.x{u}_t-\frac{1}{n+1}{u}^{n+1}\right)=0,\hfill \\ {}\hfill {D}_t\left(F(t){u}_x\right)& +{D}_x\left(-F(t)\beta {u}_{xx x}+F(t){u}^n{u}_x-{F}^{\prime }(t)u\right)=0,\hfill \\ {}\hfill {D}_t(0)& +{D}_x\left(-\frac{1}{2\beta }{\left(-{u}^n{u}_x+\beta {u}_{xx x}-{u}_t\right)}^2\right)=0,\hfill \end{array}} $$ (22)
    and the multiplier Q 4 = 1 $$ {Q}_4=1 $$ with the local conservation laws:
    D t ( u x ) + D x − β u x x x + u n u x = 0 , D t ( 0 ) + D x − β u x x x + u n u x + u t = 0 . $$ {\displaystyle \begin{array}{c}\hfill {D}_t\left({u}_x\right)+{D}_x\left(-\beta {u}_{xxx}+{u}^n{u}_x\right)=0,\\ {}\hfill {D}_t(0)+{D}_x\left(-\beta {u}_{xxx}+{u}^n{u}_x+{u}_t\right)=0.\end{array}} $$ (23)
  • Case 9.

    In the case g ( u ) = u , α = 0 $$ g(u)=u,\kern0.3em \alpha =0 $$ , we obtain extra local conservation laws under the following conservation law multipliers:

    • a.

      Q 1 = x $$ {Q}_1=x $$ , give us the following local conservation law:

      D t x u x + D x − β x u x x x + β u x x + x u x − u = 0 . $$ {D}_t\left(x{u}_x\right)+{D}_x\left(-\beta x{u}_{xx x}+\beta {u}_{xx}+x{u}_x-u\right)=0. $$ (24)

    • b.

      Q 2 = 1 2 x 2 − t x $$ {Q}_2=\frac{1}{2}{x}^2- tx $$ , with the conservation law:

      D t − 1 2 ( x ( 2 t − x ) u x ) + D x β 2 ( − u x x x x 2 + 2 x ( t u x x x + u x x ) − 2 t u x x − 2 u x ) − t x u x + t u = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(-\frac{1}{2}\left(x\left(2t-x\right){u}_x\right)\right)\hfill \\ {}\hfill & +{D}_x\left(\frac{\beta }{2}\Big(-{u}_{xx x}{x}^2+2x\left(t{u}_{xx x}+{u}_{xx}\right)\right.\hfill \\ {}\hfill & -\left.2t{u}_{xx}-2{u}_x\Big)- tx{u}_x+ tu\right)=0.\hfill \end{array}} $$ (25)

    • c.

      Q 3 = − x 6 ( 3 t 2 − 3 t x + x 2 ) $$ {Q}_3=-\frac{x}{6}\left(3{t}^2-3 tx+{x}^2\right) $$ , with the conservation law:

      D t − x 6 ( 3 t 2 − 3 t x + x 2 ) u x + D x β 6 ( u x x x x 3 + x 2 ( − 3 t u x x x − 3 u x x ) + x ( 3 t 2 u x x x + 6 t u x x + 6 u x ) ) + β − 1 2 u x x t 2 − t u x − u − 1 2 u x t 2 x + 1 2 u x t x 2 − 1 6 u x x 3 + 1 2 u t 2 = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(-\frac{x}{6}\left(3{t}^2-3 tx+{x}^2\right){u}_x\right)\hfill \\ {}\hfill & +{D}_x\left(\frac{\beta }{6}\Big({u}_{xx x}{x}^3+{x}^2\left(-3t{u}_{xx x}-3{u}_{xx}\right)\right.\hfill \\ {}\hfill & +x\left(3{t}^2{u}_{xx x}+6t{u}_{xx}+6{u}_x\right)\Big)\hfill \\ {}\hfill & +\beta \left(-\frac{1}{2}{u}_{xx}{t}^2-t{u}_x-u\right)-\frac{1}{2}{u}_x{t}^2x\hfill \\ {}\hfill & +\left.\frac{1}{2}{u}_xt{x}^2-\frac{1}{6}{u}_x{x}^3+\frac{1}{2}u{t}^2\right)=0.\hfill \end{array}} $$ (26)

    • d.

      Q 4 = 2 3 x u x + t β u x x x + u t + 1 3 u x $$ {Q}_4=\frac{2}{3}x{u}_x+ t\beta {u}_{xxx}+{u}_t+\frac{1}{3}{u}_x $$ , with the conservation law:

      D t 1 3 x u x 2 + 1 3 t ( − 3 β u x x 2 − u x 2 ) + D x t 6 − 3 β 2 u x x x 2 + β − 2 u x u x x x − 6 u t u x x x − 16 − u x x 4 − 3 4 u t x u x x ) + u x 2 + 6 u t u x + 3 u t 2 ) = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(\frac{1}{3}x{u}_x^2+\frac{1}{3}t\left(-3\beta {u}_{xx}^2-{u}_x^2\right)\right)\hfill \\ {}\hfill & +{D}_x\left(\frac{t}{6}\left(-3{\beta}^2{u}_{xx x}^2+\beta \left(-2{u}_x{u}_{xx x}-6{u}_t{u}_{xx x}\right.\right.\right.\hfill \\ {}\hfill & \left.-16\left(-\frac{u_{xx}}{4}-\frac{3}{4}{u}_{tx}\right){u}_{xx}\left)+{u}_x^2+6{u}_t{u}_x+3{u}_t^2\right)\right)=0.\hfill \end{array}} $$ (27)

    • e.

      Q 5 = F 2 ( t , δ ) + F 1 ( t ) $$ {Q}_5={F}_2\left(t,\delta \right)+{F}_1(t) $$ , with δ = − β u x x x + u t + u x $$ \delta =-\beta {u}_{xxx}+{u}_t+{u}_x $$ , give us the conservation law:

      D t F 1 ( t ) u x + D x ∫ − β ( F 2 ( t , δ ) + F 1 ( t ) ) d u x x x + F 1 ( t ) u x − F 1 ′ ( t ) u = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left({F}_1(t){u}_x\right)+{D}_x\hfill \\ {}\hfill & \left(\int -\beta \left({F}_2\left(t,\delta \right)+{F}_1(t)\right)d{u}_{xxx}+{F}_1(t){u}_x-{F}_1^{\prime }(t)u\right)=0.\hfill \end{array}} $$ (28)

    • f.

      Q 6 = 1 3 ( 2 t + x ) u x + t u t $$ {Q}_6=\frac{1}{3}\left(2t+x\right){u}_x+t{u}_t $$ , with the conservation law:

      D t 1 6 ( x − t ) u x 2 − 1 2 β t u x x 2 + D x 1 6 β t ( − 6 u t u x x x ) − 4 u x u x x x + 6 u t x u x x + 2 u x x 2 ) + 1 6 β ( ( − 2 x u x x x + 2 u x x ) u x + x u x x 2 ) + t u t 2 2 + u t u x + u x 2 3 + u x 2 x 6 = 0 . $$ {\displaystyle \begin{array}{cc}\hfill & {D}_t\left(\frac{1}{6}\left(x-t\right){u}_x^2-\frac{1}{2}\beta t{u}_{xx}^2\right)\hfill \\ {}\hfill & +{D}_x\left(\frac{1}{6}\beta t\left(-6{u}_t{u}_{xx x}\right)-4{u}_x{u}_{xx x}+6{u}_{tx}{u}_{xx}+2{u}_{xx}^2\Big)\right.\hfill \\ {}\hfill & +\frac{1}{6}\beta \left(\left(-2x{u}_{xx x}+2{u}_{xx}\right){u}_x+x{u}_{xx}^2\right)\hfill \\ {}\hfill & +\left.t\left(\frac{u_t^2}{2}+{u}_t{u}_x+\frac{u_x^2}{3}\right)+\frac{u_x^2x}{6}\right)=0.\hfill \end{array}} $$ (29)

They are the absolute collection of local conservation laws accepted by Equation (1) under the constants α $$ \alpha $$ and β $$ \beta $$ .

Proof.Equation (4) is satisfied when the generalized Ostrovsky equation (1) holds. The usual form of low-order multipliers of Equation (1) is represented by

Q ( t , x , u , u t , u x , u x x u x x x ) . $$ Q\left(t,x,u,{u}_t,{u}_x,{u}_{xx}{u}_{xx x}\right). $$
The determining equation (6) gives us determining systems. We solve them and we obtain the previous solution multipliers given in each case, which provide us conserved fluxes and densities of Equation (1).

3 Point Symmetries of the Generalized Ostrovsky Equation

A Lie symmetry of determined partial differential equations is an operator that transforms solutions into other solutions. The mathematician, Sophus Lie, developed a theory in the 1870s to study the symmetries of partial differential equations, using what are now known as Lie groups. Now, we apply the classical Lie method to calculate symmetry reductions of the generalized Ostrovsky equation (1), and we consider the following one-parameter group:
t ^ = t + ε τ ( t , x , u ) + Θ ( ε 2 ) , x ^ = x + ε ξ ( t , x , u ) + Θ ( ε 2 ) , û = u + ε η ( t , x , u ) + Θ ( ε 2 ) , $$ {\displaystyle \begin{array}{cc}\hfill \hat{t}=& \kern0.2em t+\varepsilon \tau \left(t,x,u\right)+\Theta \left({\varepsilon}^2\right),\hfill \\ {}\hfill \hat{x}=& \kern0.2em x+\varepsilon \xi \left(t,x,u\right)+\Theta \left({\varepsilon}^2\right),\hfill \\ {}\hfill \hat{u}=& \kern0.2em u+\varepsilon \eta \left(t,x,u\right)+\Theta \left({\varepsilon}^2\right),\hfill \end{array}} $$
∂ û ∂ t ^ = ∂ u ∂ t + ε η t ( t , x , u ) + Θ ( ε 2 ) , ∂ û ∂ x ^ = ∂ u ∂ x + ε η x ( t , x , u ) + Θ ( ε 2 ) , ∂ 2 û ∂ x ^ 2 = ∂ 2 u ∂ x 2 + ε η x x ( t , x , u ) + Θ ( ε 2 ) , ⋮ $$ {\displaystyle \begin{array}{cc}\hfill \frac{\partial \hat{u}}{\partial \hat{t}}=& \kern0.2em \frac{\partial u}{\partial t}+\varepsilon {\eta}^t\left(t,x,u\right)+\Theta \left({\varepsilon}^2\right),\hfill \\ {}\hfill \frac{\partial \hat{u}}{\partial \hat{x}}=& \kern0.2em \frac{\partial u}{\partial x}+\varepsilon {\eta}^x\left(t,x,u\right)+\Theta \left({\varepsilon}^2\right),\hfill \\ {}\hfill \frac{\partial^2\hat{u}}{\partial {\hat{x}}^2}=& \kern0.2em \frac{\partial^2u}{\partial {x}^2}+\varepsilon {\eta}^{xx}\left(t,x,u\right)+\Theta \left({\varepsilon}^2\right),\hfill \\ {}\hfill & \vdots \hfill \end{array}} $$ (30)
where ε $$ \varepsilon $$ is a small group parameter and η , τ $$ \eta, \kern0.3em \tau $$ , and ξ $$ \xi $$ represent the infinitesimals of symmetry transformations, corresponding to the dependent and independent variables, respectively, where
η t = D t ( η ) − u x D t ( ξ ) − u t D t ( τ ) , η x = D x ( η ) − u x D x ( ξ ) − u t D x ( τ ) , η x x = D x ( η x − u x x D x ( ξ ) ) − u t x D x ( τ ) , η x x x = D x ( η x x ) − u x x x D x ( ξ ) − u t x x D x ( τ ) , η x x x x = D x ( η x x x ) − u x x x x D x ( ξ ) − u t x x x D x ( τ ) , ⋮ $$ {\displaystyle \begin{array}{cc}\hfill {\eta}^t=& \kern0.2em {D}_t\left(\eta \right)-{u}_x{D}_t\left(\xi \right)-{u}_t{D}_t\left(\tau \right),\hfill \\ {}\hfill {\eta}^x=& \kern0.2em {D}_x\left(\eta \right)-{u}_x{D}_x\left(\xi \right)-{u}_t{D}_x\left(\tau \right),\hfill \\ {}\hfill {\eta}^{xx}=& \kern0.2em {D}_x\left({\eta}^x-{u}_{xx}{D}_x\left(\xi \right)\right)-{u}_{tx}{D}_x\left(\tau \right),\hfill \\ {}\hfill {\eta}^{xx x}=& \kern0.2em {D}_x\left({\eta}^{xx}\right)-{u}_{xx x}{D}_x\left(\xi \right)-{u}_{tx x}{D}_x\left(\tau \right),\hfill \\ {}\hfill {\eta}^{xx x x}=& \kern0.2em {D}_x\left({\eta}^{xx x}\right)-{u}_{xx x x}{D}_x\left(\xi \right)-{u}_{tx x x}{D}_x\left(\tau \right),\hfill \\ {}\hfill & \vdots \hfill \end{array}} $$
The operators D t $$ {D}_t $$ and D x $$ {D}_x $$ represent the total derivative functions with respect t $$ t $$ and x $$ x $$ , respectively, which are defined as
D t = ∂ ∂ t + u t ∂ ∂ u + u t x ∂ ∂ u x + u t t ∂ ∂ u t + … D x = ∂ ∂ x + u x ∂ ∂ u + u x t ∂ ∂ u t + u x x ∂ ∂ u x + … . $$ {\displaystyle \begin{array}{cc}\hfill {D}_t=& \kern0.2em \frac{\partial }{\partial t}+{u}_t\frac{\partial }{\partial u}+{u}_{tx}\frac{\partial }{\partial {u}_x}+{u}_{tt}\frac{\partial }{\partial {u}_t}+\dots \hfill \\ {}\hfill {D}_x=& \kern0.2em \frac{\partial }{\partial x}+{u}_x\frac{\partial }{\partial u}+{u}_{xt}\frac{\partial }{\partial {u}_t}+{u}_{xx}\frac{\partial }{\partial {u}_x}+\dots .\hfill \end{array}} $$ (31)
The generators associated to the Lie algebra are given by the generator X $$ X $$ , represented by
X = τ ( t , x , u ) ∂ ∂ t + ξ ( t , x , u ) ∂ ∂ x + η ( t , x , u ) ∂ ∂ u $$ X=\tau \left(t,x,u\right)\frac{\partial }{\partial t}+\xi \left(t,x,u\right)\frac{\partial }{\partial x}+\eta \left(t,x,u\right)\frac{\partial }{\partial u} $$ (32)
where
τ = d t ^ d ε ε = 0 , ξ = d x ^ d ε ε = 0 , η = d û d ε ε = 0 . $$ \tau ={\left.\frac{d\hat{t}}{d\varepsilon}\right|}_{\varepsilon =0},\kern1em \xi ={\left.\frac{d\hat{x}}{d\varepsilon}\right|}_{\varepsilon =0},\kern1em \eta ={\left.\frac{d\hat{u}}{d\varepsilon}\right|}_{\varepsilon =0}. $$
If the generator (32) gives us a symmetry of (1), then X $$ X $$ satisfies the symmetry condition
p r ( 4 ) X ( Δ ) | ε = 0 = 0 $$ {\left.p{r}^{(4)}X\left(\Delta \right)\right|}_{\varepsilon =0}=0 $$ (33)
where Δ = u t x − β u x x x x + ( g ( u ) ) x x − α f ( u ) $$ \Delta ={u}_{tx}-\beta {u}_{xxxx}+{\left({\left(g(u)\right)}_x\right)}_x-\alpha f(u) $$ and p r ( 4 ) X $$ p{r}^{(4)}X $$ represents the fourth prolongation of (32)
p r ( 4 ) X = X + ∑ J ϕ J ( t , x , u ( 4 ) ) ∂ ∂ u J $$ p{r}^{(4)}X=X+\sum \limits_J{\phi}^J\left(t,x,{u}^{(4)}\right)\frac{\partial }{\partial {u}_J} $$ (34)
where ϕ J ( t , x , u ( 4 ) ) = D J ( ϕ − τ u t − ξ x ) + ξ u J x + τ u J t $$ {\phi}^J\left(t,x,{u}^{(4)}\right)={D}_J\left(\phi -\tau {u}_t-\xi x\right)+\xi {u}_{Jx}+\tau {u}_{Jt} $$ , and J = ( j 1 , … , j k ) $$ J=\left({j}_1,\dots, {j}_k\right) $$ , for 1 ≤ j k ≤ 2 $$ 1\le {j}_k\le 2 $$ and 1 ≤ k ≤ 4 $$ 1\le k\le 4 $$ .
Precisely, the fourth prolongation of the vector field X $$ X $$ is given by
p r ( 4 ) X = X + ϕ x ∂ ∂ u x + ϕ x x ∂ ∂ u x x + ϕ t x ∂ ∂ u t x + ϕ x x x x ∂ ∂ u x x x x $$ p{r}^{(4)}X=X+{\phi}^x\frac{\partial }{\partial {u}_x}+{\phi}^{xx}\frac{\partial }{\partial {u}_{xx}}+{\phi}^{tx}\frac{\partial }{\partial {u}_{tx}}+{\phi}^{xx xx}\frac{\partial }{\partial {u}_{xx xx}} $$ (35)
where ϕ x , ϕ x x , ϕ t x , ϕ x x x x $$ {\phi}^x,\kern0.3em {\phi}^{xx},\kern0.3em {\phi}^{tx},\kern0.3em {\phi}^{xx xx} $$ are expressed as a function of τ , ξ , η $$ \tau, \kern0.3em \xi, \kern0.3em \eta $$ and the derivatives of η $$ \eta $$ . By using (33), we obtain the infinitesimals τ ( t , x , u ) , ξ ( t , x , u ) $$ \tau \left(t,x,u\right),\kern0.3em \xi \left(t,x,u\right) $$ , and η ( t , x , u ) $$ \eta \left(t,x,u\right) $$ . From (33) and (35), the invariance condition reads as
( g ′ ′ ( u ) u x x + g ′ ′ ′ ( u ) u x 2 − α f ′ ( u ) ) η + 2 g ′ ′ ( u ) u x ϕ x + g ′ ( u ) ϕ x x + ϕ t x − β ϕ x x x x = 0 , $$ {\displaystyle \begin{array}{cc}\hfill \Big({g}^{\prime \prime }(u){u}_{xx}& +{g^{\prime \prime}}^{\prime }(u){u}_x^2-\alpha {f}^{\prime }(u)\Big)\eta +2{g}^{\prime \prime }(u){u}_x{\phi}^x\hfill \\ {}\hfill & +{g}^{\prime }(u){\phi}^{xx}+{\phi}^{tx}-\beta {\phi}^{xx xx}=0,\hfill \end{array}} $$ (36)
where
ϕ x = D x η − u t D x τ − u x D x ξ , ϕ t x = D x ϕ t − u t t D x τ − u t x D x ξ , ϕ x x = D x ϕ x − u t x D x τ − u x x D x ξ , ϕ x x x x = D x ϕ x x x − u t x x x D x τ − u x x x x D x ξ , $$ {\displaystyle \begin{array}{cc}\hfill {\phi}^x=& \kern0.2em {D}_x\eta -{u}_t{D}_x\tau -{u}_x{D}_x\xi, \hfill \\ {}\hfill {\phi}^{tx}=& \kern0.2em {D}_x{\phi}^t-{u}_{tt}{D}_x\tau -{u}_{tx}{D}_x\xi, \hfill \\ {}\hfill {\phi}^{xx}=& \kern0.2em {D}_x{\phi}^x-{u}_{tx}{D}_x\tau -{u}_{xx}{D}_x\xi, \hfill \\ {}\hfill {\phi}^{xx x x}=& \kern0.2em {D}_x{\phi}^{xx x}-{u}_{tx x x}{D}_x\tau -{u}_{xx x x}{D}_x\xi, \hfill \end{array}} $$ (37)
and D x $$ {D}_x $$ and D t $$ {D}_t $$ are the total differential operators with respect to x $$ x $$ and t $$ t $$ , respectively, given by (31).
By using (33), we calculate the point symmetries of Equation (1). A point symmetry of Equation (1) is a one-parameter group of transformations depending on ( t , x , u ) $$ \left(t,x,u\right) $$ with generator (32). The prolongation of the Lie group leaves invariant Equation (1). The vector field (32) yields a point symmetry of Equation (1) when it satisfies the Lie symmetry condition (33), which gives rise to the following linear system of determining differential equations for the infinitesimals ξ ( t , x , u ) , $$ \xi \left(t,x,u\right),\kern0.3em $$ τ ( t , x , u ) , $$ \tau \left(t,x,u\right),\kern0.3em $$ η ( t , x , u ) $$ \eta \left(t,x,u\right) $$ , the functions g ( u ) , f ( u ) $$ g(u),\kern0.3em f(u) $$ , and the real parameters α $$ \alpha $$ and β $$ \beta $$ :
τ x = 0 , τ u = 0 , η x u = 0 , η u u = 0 , ξ u = 0 , ξ x x = 0 , 3 ξ x − τ t = 0 , 2 ξ x g ′ ( u ) + η g ′ ( u ) − ξ t = 0 , η t u − ξ t x + 2 η x g ′ ′ ( u ) = 0 , η u g ′ ′ ( u ) + 2 ξ x g ′ ′ ( u ) + η g ″ ′ ( u ) = 0 , − η t x + α η f ′ ( u ) − η x x g ′ ( u ) − α η u f ( u ) + 4 α ξ x f ( u ) + β η x x x x = 0 , $$ {\displaystyle \begin{array}{cc}\hfill & {\tau}_x=0,\kern1em {\tau}_u=0,\kern1em {\eta}_{xu}=0,\kern1em {\eta}_{uu}=0,\kern1em {\xi}_u=0,\hfill \\ {}\hfill & {\xi}_{xx}=0,\kern1em 3{\xi}_x-{\tau}_t=0,\kern1em 2{\xi}_x{g}^{\prime }(u)+\eta {g}^{\prime }(u)-{\xi}_t=0,\hfill \\ {}\hfill & {\eta}_{tu}-{\xi}_{tx}+2{\eta}_x{g}^{\prime \prime }(u)=0,\kern1em {\eta}_u{g}^{\prime \prime }(u)+2{\xi}_x{g}^{\prime \prime }(u)+\eta {g}^{{\prime\prime} \prime }(u)=0,\hfill \\ {}\hfill & -{\eta}_{tx}+\alpha \eta {f}^{\prime }(u)-{\eta}_{xx}{g}^{\prime }(u)-\alpha {\eta}_uf(u)+4\alpha {\xi}_xf(u)+\beta {\eta}_{xx xx}=0,\hfill \end{array}} $$ (38)

Solving (38), we obtain some conditions for the functions and parameters, and we have the following theorem with the infinitesimal generators of the generalized Ostrovsky equation (1):

Theorem 2.The point symmetries of Equation (1) are defined by the following independent operators:

  • Case 1. In the case g ( u ) = 1 n + 1 u n + 1 $$ g(u)=\frac{1}{n+1}{u}^{n+1} $$ and f ( u ) = c u 2 n + 1 $$ f(u)=c{u}^{2n+1} $$ with n ≠ 0 , n ≠ − 1 $$ n\ne 0,\kern0.3em n\ne -1 $$ , the point symmetries of Equation (1) are defined by the generators
    V 1 = ∂ t , V 2 = ∂ x , V 3 = 3 t ∂ t + x ∂ x − 2 n u ∂ u $$ {V}_1={\partial}_t,\kern1em {V}_2={\partial}_x,\kern1em {V}_3=3t{\partial}_t+x{\partial}_x-\frac{2}{n}u{\partial}_u $$ (39)
  • Case 2. In the case g ( u ) = 1 n + 1 u n + 1 $$ g(u)=\frac{1}{n+1}{u}^{n+1} $$ and f ( u ) = c u $$ f(u)= cu $$ , with n ≠ 0 , n ≠ − 1 , α ≠ 0 $$ n\ne 0,\kern0.3em n\ne -1,\kern0.3em \alpha \ne 0 $$ , the point symmetries of Equation (1) are defined by the generators V 1 = ∂ t $$ {V}_1={\partial}_t $$ and V 2 = ∂ x $$ {V}_2={\partial}_x $$ .
  • Case 3. In the case g ( u ) = u $$ g(u)=u $$ , there are point symmetries of Equation (1) in the case α ≠ 0 $$ \alpha \ne 0 $$ and f ( u ) = c u $$ f(u)= cu $$ , or in the case α = 0 $$ \alpha =0 $$ , and the generators are given by
    V 1 = ∂ t , V 2 = ∂ x , V 4 = u ∂ u $$ {V}_1={\partial}_t,\kern1em {V}_2={\partial}_x,\kern1em {V}_4=u{\partial}_u $$ (40)
  • Case 4. In the case g ( u ) = c 1 e u $$ g(u)={c}_1{e}^u $$ and f ( u ) = c 2 e 2 u $$ f(u)={c}_2{e}^{2u} $$ , the point symmetries are given by
    V 1 = ∂ t , V 2 = ∂ x , V 5 = 3 t ∂ t + x ∂ x − 2 ∂ u $$ {V}_1={\partial}_t,\kern1em {V}_2={\partial}_x,\kern1em {V}_5=3t\partial t+x\partial x-2{\partial}_u $$ (41)
  • Case 5. In the general case for g ( u ) $$ g(u) $$ and f ( u ) $$ f(u) $$ , the point symmetries are given by V 1 $$ {V}_1 $$ and V 2 $$ {V}_2 $$ .

We can easily check that the generators (39-41) are closed under the Lie bracket.

Now, we calculate the adjoint table for V i $$ {V}_i $$ and V j $$ {V}_j $$ , for i , j = 1 , … , 5 $$ i,j=1,\dots, 5 $$ , given by the following expression:
Ad ( exp ( ε V ) ) W 0 = ∑ n = 0 ∞ ε n n ! ( com V ) n ( W 0 ) = W 0 − ε [ V , W 0 ] + ε 2 2 [ V , [ V , W 0 ] ] − … . $$ {\displaystyle \begin{array}{cc}\hfill \mathrm{Ad}\left(\exp \left(\varepsilon V\right)\right){W}_0=& \kern0.2em {\sum}_{n=0}^{\infty}\frac{\varepsilon^n}{n!}{\left(\mathrm{com}V\right)}^n\left({W}_0\right)\hfill \\ {}\hfill =& {W}_0-\varepsilon \left[V,{W}_0\right]+\frac{\varepsilon^2}{2}\left[V,\left[V,{W}_0\right]\right]-\dots .\hfill \end{array}} $$
By using the adjoint representation, we obtain conjugated subalgebras through equivalence classes, which represent every subalgebra of the Lie algebra. The optimal systems in each case are given, respectively, by
{ μ V 1 + λ V 2 ,   V 3 } , { μ V 1 + λ V 2 } , { μ V 1 + λ V 2 + γ V 4 } , { μ V 1 + λ V 2 , V 5 } a n d { μ V 1 + λ V 2 } . $$ {\displaystyle \begin{array}{cc}\hfill & \left\{\mu {V}_1+\lambda {V}_2,{V}_3\right\},\left\{\mu {V}_1+\lambda {V}_2\right\},\kern0.3em \left\{\mu {V}_1+\lambda {V}_2+\gamma {V}_4\right\},\hfill \\ {}\hfill & \left\{\mu {V}_1+\lambda {V}_2,{V}_5\right\} and\left\{\mu {V}_1+\lambda {V}_2\right\}.\hfill \end{array}} $$
Now, we use (39-41) to determinate the symmetry groups g ( t , x , u ) $$ g\left(t,x,u\right) $$ associated to the generator (32) and calculate new solutions for Equation (1) associated with them. To obtain the symmetry group, we use the following system of initial problems:
∂ t ^ ∂ ϵ = τ ( t , x , u ) , ∂ x ^ ∂ ϵ = ξ ( t , x , u ) , ∂ û ∂ ϵ = η ( t , x , u ) $$ \frac{\partial \hat{t}}{\partial \epsilon }=\tau \left(t,x,u\right),\frac{\partial \hat{x}}{\partial \epsilon }=\xi \left(t,x,u\right),\frac{\partial \hat{u}}{\partial \epsilon }=\eta \left(t,x,u\right) $$ (42)
and
( t ^ , x ^ , û ) | ϵ = 0 = ( t , x , u ) $$ {\left.\left(\hat{t},\hat{x},\hat{u}\right)\right|}_{\epsilon =0}=\left(t,x,u\right) $$ (43)
where g ( t , x , u ) = ( t ^ , x ^ , û ) $$ g\left(t,x,u\right)=\left(\hat{t},\hat{x},\hat{u}\right) $$ . From (42) and (43), we get the following theorem with the corresponding Lie symmetry group.

Theorem 3.The Lie symmetry groups G i , i = 1 , … , 5 $$ {G}_i,\kern0.3em i=1,\dots, 5 $$ , generated by V i , i = 1 , … , 5 $$ {V}_i,\kern0.3em i=1,\dots, 5 $$ , are specified by

G 1 ( t , x , u ) = ( t + ϵ , x , u ) time translation $$ {G}_1\left(t,x,u\right)=\left(t+\epsilon, x,u\right)\kern0.60em time\ translation $$ (44)
G 2 ( t , x , u ) = ( t , x + ϵ , u ) space translation across the x axis $$ {G}_2\left(t,x,u\right)=\left(t,x+\epsilon, u\right)\kern0.3em space\ translation\ across\ the\ x\ axis $$ (45)
G 3 ( t , x , u ) = ( e 3 ϵ t , e ϵ x , e − 2 ϵ n u ) scaling group $$ {G}_3\left(t,x,u\right)=\left({e}^{3\epsilon }t,{e}^{\epsilon }x,{e}^{-\frac{2\epsilon }{n}}u\right)\kern1em scaling\ group $$ (46)
G 4 ( t , x , u ) = ( t , x , e ϵ u ) exponential dilation $$ {G}_4\left(t,x,u\right)=\left(t,x,{e}^{\epsilon }u\right)\kern0.6em exponential\ dilation $$ (47)
G 5 ( t , x , u ) = ( e 3 ϵ t , e ϵ x , u − 2 ϵ ) shift , $$ {G}_5\left(t,x,u\right)=\left({e}^{3\epsilon }t,{e}^{\epsilon }x,u-2\epsilon \right)\kern1em shift, $$ (48)
where ϵ $$ \epsilon $$ is the parameter of the group.

Solutions of Equation (1) are transformed into solution through the use of symmetry groups. So, on the assumption that u = f ( t , x ) $$ u=f\left(t,x\right) $$ is a solution of the generalized equation (1), we obtain new solutions for Equation (1) by means of diverse symmetry groups. Then, by using the previous groups G i , i = 1 , … 5 $$ {G}_i,\kern0.3em i=1,\dots 5 $$ , we calculate the appropriate new solutions:
û 1 = f ( t − ϵ , x ) $$ {\hat{u}}_1=f\left(t-\epsilon, x\right) $$ (49)
û 2 = f ( t , x − ϵ ) $$ {\hat{u}}_2=f\left(t,x-\epsilon \right) $$ (50)
û 3 = f ( t e − 3 ϵ , x e − ϵ ) e − 2 ϵ n $$ {\hat{u}}_3=f\left(t{e}^{-3\epsilon },x{e}^{-\epsilon}\right){e}^{-\frac{2\epsilon }{n}} $$ (51)
û 4 = f ( t e − ϵ , x ) e ϵ $$ {\hat{u}}_4=f\left(t{e}^{-\epsilon },x\right){e}^{\epsilon } $$ (52)
û 5 = f ( t e − 3 ϵ , x e − ϵ ) − 2 ϵ $$ {\hat{u}}_5=f\left(t{e}^{-3\epsilon },x{e}^{-\epsilon}\right)-2\epsilon $$ (53)

4 Symmetry Reductions and Exact Solutions

In this part, we use the optimal system computed in the anterior subsection, and we get the following symmetry reductions of Equation (1).
  • Case 1a.

    In the case g ( u ) = 1 n + 1 u n + 1 $$ g(u)=\frac{1}{n+1}{u}^{n+1} $$ and f ( u ) = c u 2 n + 1 $$ f(u)=c{u}^{2n+1} $$ with n ≠ 0 , n ≠ − 1 $$ n\ne 0,\kern0.3em n\ne -1 $$ , we use the generator V 3 $$ {V}_3 $$ , and we get

    u = h ( w ) t − 2 3 n $$ u=h(w){t}^{-\frac{2}{3n}} $$ (54)
    where w = x t − 1 3 $$ w=x{t}^{-\frac{1}{3}} $$ . By considering (54) and (1), we have a reduced ordinary reduced equation given by
    − 1 3 w h ′ ′ + n h n − 1 ( h ′ ) 2 − β h i v ) + h n h ′ ′ − 2 + n 3 n h ′ = 0 $$ -\frac{1}{3}w{h}^{\prime \prime }+n{h}^{n-1}{\left({h}^{\prime}\right)}^2-\beta {h}^{iv\Big)}+{h}^n{h}^{\prime \prime }-\frac{2+n}{3n}{h}^{\prime }=0 $$ (55)

  • Case 1b.

    In the case g ( u ) = 1 n + 1 u n + 1 $$ g(u)=\frac{1}{n+1}{u}^{n+1} $$ and f ( u ) = c u 2 n + 1 $$ f(u)=c{u}^{2n+1} $$ with n ≠ 0 , n ≠ − 1 $$ n\ne 0,\kern0.3em n\ne -1 $$ , from the generator λ V 1 + V 2 $$ \lambda {V}_1+{V}_2 $$ , we have

    u = h ( w ) $$ \kern1em u=h(w) $$ (56)
    where w = x − λ t $$ w=x-\lambda t $$ . By the substitution of (56) into (1), we have an ordinary differential equation given by
    − α h 2 n + 1 − λ h ′ ′ + n h n − 1 ( h ′ ) 2 − β h i v ) + h n h ′ ′ = 0 $$ -\alpha {h}^{2n+1}-\lambda {h}^{\prime \prime }+n{h}^{n-1}{\left({h}^{\prime}\right)}^2-\beta {h}^{iv\Big)}+{h}^n{h}^{\prime \prime }=0 $$ (57)

    We integrate with respect to w $$ w $$ twice, multiply by h ′ $$ {h}^{\prime } $$ , and integrate again with respect to w $$ w $$ for α = 0 $$ \alpha =0 $$ . We consider zero the constants of integration, and we have the following ordinary differential equation:

    λ 2 ( n + 1 ) ( n + 2 ) h 2 − h n + 2 + β ( n + 1 ) ( n + 2 ) ( h ′ ) 2 = 0 $$ \frac{\lambda }{2}\left(n+1\right)\left(n+2\right){h}^2-{h}^{n+2}+\beta \left(n+1\right)\left(n+2\right){\left({h}^{\prime}\right)}^2=0 $$ (58)
    with solution for n = 1 $$ n=1 $$
    h ( w ) = 3 λ 1 + tan ( k − w ) 18 b λ 12 b 2 $$ h(w)=3\lambda \left(1+\tan {\left(\left(k-w\right)\frac{\sqrt{18 b\lambda}}{12b}\right)}^2\right) $$ (59)
    where k ∈ ℝ $$ k\in \mathbb{R} $$ . This give us the following solution for Equation (1):
    u ( t , x ) = 3 λ 1 + tan ( k − x + λ t ) 18 b λ 12 b 2 $$ u\left(t,x\right)=3\lambda \left(1+\tan {\left(\left(k-x+\lambda t\right)\frac{\sqrt{18 b\lambda}}{12b}\right)}^2\right) $$ (60)

    In Figure 1, we consider (60) with λ = − 1 , k = 1 $$ \lambda =-1,k=1 $$ , and b = 1 $$ b=1 $$ .

    The solution for n = 2 $$ n=2 $$ is given by

    h ( w ) = − 24 b λ e − − 2 b λ 2 b ( k − w ) 24 b λ + e − 2 b λ b ( k − w ) $$ h(w)=\frac{-24 b\lambda {e}^{-\frac{-2 b\lambda}{2b}\left(k-w\right)}}{24 b\lambda +{e}^{\frac{\sqrt{-2 b\lambda}}{b}\left(k-w\right)}} $$ (61)
    and
    h ( w ) = − 24 b λ e − − 2 b λ 2 b ( k − w ) 24 b λ + e − 2 b λ b ( w − k ) $$ h(w)=\frac{-24 b\lambda {e}^{-\frac{-2 b\lambda}{2b}\left(k-w\right)}}{24 b\lambda +{e}^{\frac{\sqrt{-2 b\lambda}}{b}\left(w-k\right)}} $$ (62)
    where k ∈ ℝ $$ k\in \mathbb{R} $$ . This give us the following solution for Equation (1):
    u ( t , x ) = − 24 b λ e − − 2 b λ 2 b ( k − x + λ t ) 24 b λ + e − 2 b λ b ( k − x + λ t ) $$ u\left(t,x\right)=\frac{-24 b\lambda {e}^{-\frac{-2 b\lambda}{2b}\left(k-x+\lambda t\right)}}{24 b\lambda +{e}^{\frac{\sqrt{-2 b\lambda}}{b}\left(k-x+\lambda t\right)}} $$ (63)
    and
    u ( t , x ) = − 24 b λ e − − 2 b λ 2 b ( k − x + λ t ) 24 b λ + e − 2 b λ b ( − k + x − λ t ) $$ u\left(t,x\right)=\frac{-24 b\lambda {e}^{-\frac{-2 b\lambda}{2b}\left(k-x+\lambda t\right)}}{24 b\lambda +{e}^{\frac{\sqrt{-2 b\lambda}}{b}\left(-k+x-\lambda t\right)}} $$ (64)

    In Figure 2, we consider solution (63) with λ = − 1 , k = 1 $$ \lambda =-1,k=1 $$ , and b = 3 $$ b=3 $$ and solution (64) with λ = − 2 , k = 1 $$ \lambda =-2,k=1 $$ , and b = 1 $$ b=1 $$ , respectively.

  • Case 2.

    In the case g ( u ) = 1 n + 1 u n + 1 $$ g(u)=\frac{1}{n+1}{u}^{n+1} $$ and f ( u ) = c u $$ f(u)= cu $$ , with n ≠ 0 , n ≠ − 1 , α ≠ 0 $$ n\ne 0,\kern0.3em n\ne -1,\kern0.3em \alpha \ne 0 $$ , by using the generator λ V 1 + V 2 $$ \lambda {V}_1+{V}_2 $$ , we get the similarity variables

    w = x − λ t , u = h ( w ) , $$ w=x-\lambda t,\kern1em u=h(w), $$ (65)
    and the ODE
    − λ h ′ ′ + n h n − 1 ( h ′ ) 2 − β h i v ) + h n h ′ ′ − α h = 0 $$ -\lambda {h}^{\prime \prime }+n{h}^{n-1}{\left({h}^{\prime}\right)}^2-\beta {h}^{iv\Big)}+{h}^n{h}^{\prime \prime }-\alpha h=0 $$ (66)

  • Case 3.

    In the case g ( u ) = u $$ g(u)=u $$ and f ( u ) = c u $$ f(u)= cu $$ , by using the generator V 1 + λ 2 V 2 + λ 4 V 4 $$ {V}_1+{\lambda}_2{V}_2+{\lambda}_4{V}_4 $$ , we obtain

    w = x − λ 2 t , u = h ( w ) e λ 4 t , $$ w=x-{\lambda}_2t,\kern1em u=h(w){e}^{\lambda_4t}, $$ (67)
    and the ordinary differential equation
    λ 4 h ′ − λ 2 h ′ ′ − β h i v ) + h ′ ′ − α h = 0 $$ {\lambda}_4{h}^{\prime }-{\lambda}_2{h}^{\prime \prime }-\beta {h}^{iv\Big)}+{h}^{\prime \prime }-\alpha h=0 $$ (68)
    • ∘

      We have the following solution of (68) in the general case of α , β , λ 2 $$ \alpha, \kern0.3em \beta, \kern0.3em {\lambda}_2 $$ , and λ 4 $$ {\lambda}_4 $$

      h ( w ) = ∑ k = 1 4 C k e R o o t O f ( β Z 4 + ( λ 2 − 1 ) Z 2 − λ 4 Z + α , i n d e x = k ) w $$ h(w)=\sum \limits_{k=1}^4{C}_k{e}^{RootOf\left(\beta {Z}^4+\left({\lambda}_2-1\right){Z}^2-{\lambda}_4Z+\alpha, \kern0.3em index=k\right)w} $$ (69)
      where C k ∈ ℝ , k = 1 … 4 $$ {C}_k\in \mathbb{R},\kern0.3em k=1\dots 4 $$ , and the corresponding solution of (1)
      u ( t , x ) = ∑ k = 1 4 C k e R o o t O f ( β Z 4 + ( λ 2 − 1 ) Z 2 − λ 4 Z + α , i n d e x = k ) ( x − λ 2 t ) $$ u\left(t,x\right)=\sum \limits_{k=1}^4{C}_k{e}^{RootOf\left(\beta {Z}^4+\left({\lambda}_2-1\right){Z}^2-{\lambda}_4Z+\alpha, \kern0.3em index=k\right)\left(x-{\lambda}_2t\right)} $$ (70)
      where C k ∈ ℝ , k = 1 … 4 $$ {C}_k\in \mathbb{R},\kern0.3em k=1\dots 4 $$ .

    • ∘

      In the case α + 16 β + 4 λ 2 − 4 + 2 λ 4 = 0 $$ \alpha +16\beta +4{\lambda}_2-4+2{\lambda}_4=0 $$ , we obtain the following solutions of (68):

      h ( w ) = c 1 s e c h ( w ) 2 ( 1 + t a n h ( w ) ) 2 $$ h(w)=\frac{c_1\mathit{\operatorname{sech}}{(w)}^2}{{\left(1+\mathit{\tanh}(w)\right)}^2} $$ (71)
      h ( w ) = c 2 c s c h ( w ) 2 ( 1 + c o t h ( w ) ) 2 $$ h(w)=\frac{c_2\mathit{\operatorname{csch}}{(w)}^2}{{\left(1+\mathit{\coth}(w)\right)}^2} $$ (72)
      and these give us the following solutions for Equation (1), respectively,
      u ( t , x ) = c 1 s e c h ( x − λ 2 t ) 2 ( 1 + t a n h ( x − λ 2 t ) ) 2 $$ u\left(t,x\right)=\frac{c_1\mathit{\operatorname{sech}}{\left(x-{\lambda}_2t\right)}^2}{{\left(1+\mathit{\tanh}\left(x-{\lambda}_2t\right)\right)}^2} $$ (73)
      u ( t , x ) = c 2 c s c h ( x − λ 2 t ) 2 ( 1 + c o t h ( x − λ 2 t ) ) 2 $$ u\left(t,x\right)=\frac{c_2\mathit{\operatorname{csch}}{\left(x-{\lambda}_2t\right)}^2}{{\left(1+\mathit{\coth}\left(x-{\lambda}_2t\right)\right)}^2} $$ (74)
      where c 1 , c 2 ∈ ℝ $$ {c}_1,{c}_2\in \mathbb{R} $$ . In Figure 3, we consider solution (73) for c 1 = 0 . 5 , λ 2 = 1 $$ {c}_1=0.5,{\lambda}_2=1 $$ and solution (74) for c 2 = 1 , λ 2 = 0 . 001 $$ {c}_2=1,{\lambda}_2=0.001 $$ , respectively.

    • ∘

      In the case λ 4 = 0 $$ {\lambda}_4=0 $$ , we obtain the following general solution of (68):

      h ( w ) = C 1 e − − 2 β λ 2 − 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 w 2 b + C 2 e − 2 β λ 2 − 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 w 2 b + C 3 e − − 2 β − λ 2 + 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 w 2 b + C 4 e − 2 β − λ 2 + 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 w 2 b , $$ {\displaystyle \begin{array}{cc}\hfill h(w)=& \kern0.2em {C}_1{e}^{-\frac{\sqrt{-2\beta \left({\lambda}_2-1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}w}{2b}}\hfill \\ {}\hfill & +{C}_2{e}^{\frac{\sqrt{-2\beta \left({\lambda}_2-1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}w}{2b}}\hfill \\ {}\hfill & +{C}_3{e}^{-\frac{\sqrt{-2\beta \left(-{\lambda}_2+1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}w}{2b}}\hfill \\ {}\hfill & +{C}_4{e}^{\frac{\sqrt{-2\beta \left(-{\lambda}_2+1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}w}{2b}},\hfill \end{array}} $$ (75)
      where c i ∈ ℝ , i = 1 … 3 $$ {c}_i\in \mathbb{R},\kern0.3em i=1\dots 3 $$ .

      This give us the following solution for Equation (1):

      u ( t , x ) = C 1 e − − 2 β λ 2 − 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 ( x − λ 2 t ) 2 b + C 2 e − 2 β λ 2 − 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 ( x − λ 2 t ) 2 b + C 3 e − − 2 β − λ 2 + 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 ( x − λ 2 t ) 2 b + C 4 e − 2 β − λ 2 + 1 + − 4 α β + λ 2 2 − 2 λ 2 + 1 ( x − λ 2 t ) 2 b . $$ {\displaystyle \begin{array}{cc}\hfill u\left(t,x\right)=& \kern0.2em {C}_1{e}^{-\frac{\sqrt{-2\beta \left({\lambda}_2-1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}\left(x-{\lambda}_2t\right)}{2b}}\hfill \\ {}\hfill & +{C}_2{e}^{\frac{\sqrt{-2\beta \left({\lambda}_2-1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}\left(x-{\lambda}_2t\right)}{2b}}\hfill \\ {}\hfill & +{C}_3{e}^{-\frac{\sqrt{-2\beta \left(-{\lambda}_2+1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}\left(x-{\lambda}_2t\right)}{2b}}\hfill \\ {}\hfill & +{C}_4{e}^{\frac{\sqrt{-2\beta \left(-{\lambda}_2+1+\sqrt{-4\alpha \beta +{\lambda}_2^2-2{\lambda}_2+1}\right)}\left(x-{\lambda}_2t\right)}{2b}}.\hfill \end{array}} $$ (76)

      In Figure 4, we consider solution (76) in the cases

      C i = 1 , i = 1 … 4 , λ 2 = − 1 , β = 1 , α = − 1 , C i = 1 , i = 1 … 4 , λ 2 = 1 , β = 2 , α = 3 , C i = 1 , i = 1 … 4 , λ 2 = 70 , β = 1 , α = 1 , C 1 = C 2 = 1 , C 3 = − 0 . 5 , C 4 = 1 , λ 2 = − 50 , β = 4 , α = − 4 , $$ {\displaystyle \begin{array}{cc}\hfill {C}_i=& \kern0.2em 1,i=1\dots 4,{\lambda}_2=-1,\beta =1,\alpha =-1,\hfill \\ {}\hfill {C}_i=& \kern0.2em 1,i=1\dots 4,{\lambda}_2=1,\beta =2,\alpha =3,\hfill \\ {}\hfill {C}_i=& \kern0.2em 1,i=1\dots 4,{\lambda}_2=70,\beta =1,\alpha =1,\hfill \\ {}\hfill {C}_1=& \kern0.2em {C}_2=1,{C}_3=-0.5,{C}_4=1,{\lambda}_2=-50,\beta =4,\alpha =-4,\hfill \end{array}} $$
      respectively.

    • ∘

      In the case λ 4 = 0 $$ {\lambda}_4=0 $$ , we obtain the following solution of (68):

      h ( w ) = c 1 sinh ( c 2 w ) + c 3 cosh ( c 2 ) w $$ h(w)={c}_1\sinh \left({c}_2w\right)+{c}_3\cosh \left({c}_2\right)w $$ (77)
      where c 2 2 λ 2 + β c 2 4 − c 2 2 + α = 0 $$ {c}_2^2{\lambda}_2+\beta {c}_2^4-{c}_2^2+\alpha =0 $$ , with the following solution for Equation (1):
      u ( t , x ) = c 1 sinh ( c 2 ( x − λ 2 t ) ) + c 3 cosh ( c 2 ) ( x − λ 2 t ) $$ u\left(t,x\right)={c}_1\sinh \left({c}_2\left(x-{\lambda}_2t\right)\right)+{c}_3\cosh \left({c}_2\right)\left(x-{\lambda}_2t\right) $$ (78)

      In Figure 5, we consider solution (78) in the case c 1 = c 2 = 1 , c 3 = 0 , λ 2 = 1 , β = 1 $$ {c}_1={c}_2=1,{c}_3=0,{\lambda}_2=1,\beta =1 $$ ; the case c 1 = 0 , c 2 = 1 , c 3 = 1 , λ 2 = 1 , β = 1 $$ {c}_1=0,{c}_2=1,{c}_3=1,{\lambda}_2=1,\beta =1 $$ ; and the case c i = 1 , i = 1 … 3 , λ 2 = 1 , β = 1 $$ {c}_i=1,i=1\dots 3,{\lambda}_2=1,\beta =1 $$ , respectively.

    • ∘

      In the case λ 4 = 0 $$ {\lambda}_4=0 $$ , we obtain the following solution of (68)

      h ( w ) = c 1 sin ( c 2 w ) + c 3 cos ( c 2 w ) $$ h(w)={c}_1\sin \left({c}_2w\right)+{c}_3\cos \left({c}_2w\right) $$ (79)
      where c 2 2 λ 2 + β c 2 4 − c 2 2 − α = 0 $$ {c}_2^2{\lambda}_2+\beta {c}_2^4-{c}_2^2-\alpha =0 $$ , with the following solution for Equation (1):
      u ( t , x ) = c 1 sin ( c 2 ( x − λ 2 t ) ) + c 3 cos ( c 2 ( x − λ 2 t ) ) $$ u\left(t,x\right)={c}_1\sin \left({c}_2\left(x-{\lambda}_2t\right)\right)+{c}_3\cos \left({c}_2\left(x-{\lambda}_2t\right)\right) $$ (80)

      In Figure 6, we consider solution (80) in the case c 1 = c 2 = 1 , c 3 = 1 , λ 2 = 10 , β = 1 $$ {c}_1={c}_2=1,{c}_3=1,{\lambda}_2=10,\beta =1 $$ ; the case c 1 = 0 , c 2 = c 3 = 1 , λ 2 = 0 . 5 , β = 2 $$ {c}_1=0,{c}_2={c}_3=1,{\lambda}_2=0.5,\beta =2 $$ ; and the case c 1 = 100 , c 2 = 1 , c 3 = 0 , λ 2 = 0 , β = 1 $$ {c}_1=100,{c}_2=1,{c}_3=0,{\lambda}_2=0,\beta =1 $$ , respectively.

    • ∘

      In the case α = λ 4 = 0 $$ \alpha ={\lambda}_4=0 $$ , we obtain the following solutions if b ( λ 2 − 1 ) $$ b\left({\lambda}_2-1\right) $$ is a positive, negative, or zero real number, respectively,

      h ( w ) = k 1 sin λ 2 − 1 w b + k 2 cos λ 2 − 1 w b $$ h(w)={k}_1\sin \left(\frac{\sqrt{\lambda_2-1}w}{\sqrt{b}}\right)+{k}_2\cos \left(\frac{\sqrt{\lambda_2-1}w}{\sqrt{b}}\right) $$ (81)
      h ( w ) = k 1 e λ 2 − 1 i w b + k 2 e λ 2 − 1 i w b $$ h(w)={k}_1{e}^{\left(\frac{\sqrt{\lambda_2-1} iw}{\sqrt{b}}\right)}+{k}_2{e}^{\left(\frac{\sqrt{\lambda_2-1} iw}{\sqrt{b}}\right)} $$ (82)
      h ( w ) = k 1 w + k 2 $$ h(w)={k}_1w+k2 $$ (83)
      and these give us the following solutions for Equation (1), respectively,
      u ( t , x ) = k 1 sin λ 2 − 1 ( x − λ 2 t ) b + k 2 cos λ 2 − 1 ( x − λ 2 t ) b , $$ {\displaystyle \begin{array}{cc}\hfill u\left(t,x\right)=& \kern0.2em {k}_1\sin \left(\frac{\sqrt{\lambda_2-1}\left(x-{\lambda}_2t\right)}{\sqrt{b}}\right)\hfill \\ {}\hfill & +{k}_2\cos \left(\frac{\sqrt{\lambda_2-1}\left(x-{\lambda}_2t\right)}{\sqrt{b}}\right),\hfill \end{array}} $$ (84)
      u ( t , x ) = k 1 e λ 2 − 1 i ( x − λ 2 t ) b + k 2 e λ 2 − 1 i ( x − λ 2 t ) b $$ u\left(t,x\right)={k}_1{e}^{\left(\frac{\sqrt{\lambda_2-1}i\left(x-{\lambda}_2t\right)}{\sqrt{b}}\right)}+{k}_2{e}^{\left(\frac{\sqrt{\lambda_2-1}i\left(x-{\lambda}_2t\right)}{\sqrt{b}}\right)} $$ (85)
      u ( t , x ) = k 1 ( x − t ) + k 2 $$ u\left(t,x\right)={k}_1\left(x-t\right)+k2 $$ (86)

      In Figure 7, we consider solution (84) with k 1 = k 2 = 1 , λ 2 = 2 , b = 1 $$ {k}_1={k}_2=1,{\lambda}_2=2,b=1 $$ ; solution (85) with k 1 = 1 , k 2 = 1 , λ 2 = − 2 , b = 1 $$ {k}_1=1,{k}_2=1,{\lambda}_2=-2,b=1 $$ ; and solution (86) with k 1 = 1 , k 2 = 1 , λ 2 = 1 $$ {k}_1=1,{k}_2=1,{\lambda}_2=1 $$ , respectively.

  • Case 4a.

    In the case g ( u ) = c 1 e u $$ g(u)={c}_1{e}^u $$ and f ( u ) = c 2 e 2 u $$ f(u)={c}_2{e}^{2u} $$ , by using the generator V 3 $$ {V}_3 $$ , we have

    u = h ( w ) − 2 3 l n ( t ) $$ u=h(w)-\frac{2}{3}\mathit{\ln}(t) $$ (87)
    where w = x t − 1 3 $$ w=x{t}^{-\frac{1}{3}} $$ , and we have the ordinary reduced equation given by
    − 1 3 w h ′ ′ + c 1 h ′ ′ e h + c 1 ( h ′ ) 2 e h − α c 2 e 2 h − β h i v ) − 1 3 h ′ = 0 $$ {\displaystyle \begin{array}{cc}\hfill & -\frac{1}{3}w{h}^{\prime \prime }+{c}_1{h}^{\prime \prime }{e}^h+{c}_1{\left({h}^{\prime}\right)}^2{e}^h\\ {}\hfill & -\alpha {c}_2{e}^{2h}-\beta {h}^{iv\Big)}-\frac{1}{3}{h}^{\prime }=0\end{array}} $$ (88)

  • Case 4b.

    In the case g ( u ) = c 1 e u $$ g(u)={c}_1{e}^u $$ and f ( u ) = c 2 e 2 u $$ f(u)={c}_2{e}^{2u} $$ from the generator λ V 1 + V 2 $$ \lambda {V}_1+{V}_2 $$ , one has

    u = h ( w ) $$ \kern1em u=h(w) $$ (89)
    where w = x − λ t $$ w=x-\lambda t $$ . By substituting (89) into (1), we have the ordinary differential equation given by
    − λ h ′ ′ − β h i v ) + c 1 h ′ ′ e h + c 1 ( h ′ ) 2 e h − α c 2 e 2 h = 0 $$ -\lambda {h}^{\prime \prime }-\beta {h}^{iv\Big)}+{c}_1{h}^{\prime \prime }{e}^h+{c}_1{\left({h}^{\prime}\right)}^2{e}^h-\alpha {c}_2{e}^{2h}=0 $$ (90)

    For c 2 = 0 $$ {c}_2=0 $$ , we have the following solutions of (90) where λ $$ \lambda $$ is positive, negative, or λ = 0 $$ \lambda =0 $$ , respectively,

    h ( w ) = k 1 sin w λ β + k 2 cos w λ β $$ h(w)={k}_1\sin \left(w\sqrt{\frac{\lambda }{\beta }}\right)+{k}_2\cos \left(w\sqrt{\frac{\lambda }{\beta }}\right) $$ (91)
    h ( w ) = k 1 e i w λ β + k 2 e − i w λ β $$ h(w)={k}_1{e}^{iw\sqrt{\frac{\lambda }{\beta }}}+{k}_2{e}^{- iw\sqrt{\frac{\lambda }{\beta }}} $$ (92)
    h ( w ) = k 1 + k 2 w $$ h(w)={k}_1+{k}_2w $$ (93)
    and we get the following solutions of Equation (1):
    u ( t , x ) = k 1 sin ( x − λ t ) λ β + k 2 cos ( x − λ t ) λ β $$ u\left(t,x\right)={k}_1\sin \left(\left(x-\lambda t\right)\sqrt{\frac{\lambda }{\beta }}\right)+{k}_2\cos \left(\left(x-\lambda t\right)\sqrt{\frac{\lambda }{\beta }}\right) $$ (94)
    u ( t , x ) = k 1 e i ( x − λ t ) λ β + k 2 e − i ( x − λ t ) λ β $$ u\left(t,x\right)={k}_1{e}^{i\left(x-\lambda t\right)\sqrt{\frac{\lambda }{\beta }}}+{k}_2{e}^{-i\left(x-\lambda t\right)\sqrt{\frac{\lambda }{\beta }}} $$ (95)
    u ( t , x ) = k 1 + k 2 x $$ u\left(t,x\right)={k}_1+{k}_2x $$ (96)

  • Case 5.

    In the general case for g ( u ) $$ g(u) $$ and f ( u ) $$ f(u) $$ , from the generator λ V 1 + V 2 $$ \lambda {V}_1+{V}_2 $$ , we have u = h ( w ) , w = x − λ t $$ u=h(w),\kern0.3em w=x-\lambda t $$ , and we have the reduced differential equation given by

    − λ h ′ ′ − β h i v ) + g ′ ′ ( h ′ ) 2 + g ′ h ′ ′ − α f = 0 $$ -\lambda {h}^{\prime \prime }-\beta {h}^{iv\Big)}+{g}^{\prime \prime }{\left({h}^{\prime}\right)}^2+{g}^{\prime }{h}^{\prime \prime }-\alpha f=0 $$ (97)
    and integrating twice respect w $$ w $$ , we obtain
    − λ h − β h ′ ′ + g − α ∫ ∫ f d w d w = 0 $$ -\lambda h-\beta {h}^{\prime \prime }+g-\alpha \int \left(\int fdw\right) dw=0 $$ (98)

    For f , g $$ f,g $$ identity functions or constant functions, we have similar solutions to (94-96).

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FIGURE 1
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Exact solution of (60) for λ = − 1 , k = 1 $$ \lambda =-1,k=1 $$ , and b = 1 $$ b=1 $$ . [Colour figure can be viewed at wileyonlinelibrary.com]
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FIGURE 2
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Exact solutions of (63) and (64). [Colour figure can be viewed at wileyonlinelibrary.com]
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FIGURE 3
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Exact solution (85) and (86). [Colour figure can be viewed at wileyonlinelibrary.com]
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FIGURE 4
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Exact solution of Equation (1) given by Equation (76). [Colour figure can be viewed at wileyonlinelibrary.com]
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FIGURE 5
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Exact solution of Equation (1) given by Equation (78). [Colour figure can be viewed at wileyonlinelibrary.com]
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FIGURE 6
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Exact solution of Equation (1) given by Equation (80). [Colour figure can be viewed at wileyonlinelibrary.com]
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FIGURE 7
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Exact solution (84-86). [Colour figure can be viewed at wileyonlinelibrary.com]

5 Concluding Remarks

We have studied the generalized Ostrovsky equation (1) with real dispersion coefficients, from the point of view of symmetry analysis. First of all, we have derived conservation laws for the subjacent equation through multiplier approach conservation theorem, and we have recourse to the invariance and multiplier perspective by using the Euler–Lagrange operator. We have obtained several local conservation laws for Equation (1). Secondly, we have calculated Lie point symmetries of the equation, and subsequently, we have performed symmetry reductions. Also, we have obtained traveling wave solutions of significance importance by using the vector fields. Finally, we have obtained Lie symmetry groups generated by means of the vector field and new solutions for Equation (1) associated to them. The results obtained in this study contribute to the understanding of wave propagation in rotating fluids. The derivation of exact solutions and conservation laws allows for a better characterization of wave regimes, facilitating the development of more efficient and accurate numerical models. These solutions can serve as benchmark tests for computational simulations, helping to assess the accuracy of numerical schemes used in oceanic or atmospheric models. Furthermore, the identification of symmetries and conservation laws provides useful tools for simplifying complex equations and exploring new problem formulations in fluid dynamics.

Author Contributions

Sol Sáez: investigation; writing – original draft.

Acknowledgments

The support of University of Cádiz is gratefully acknowledged by the author.

    Conflicts of Interest

    The author declares no conflicts of interest.

    Data Availability Statement

    Data sharing is not applicable—no new data generated, or the article describes entirely theoretical research.

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