Conservation Laws and Symmetry Reductions of the Hunter-Saxton Equation via the Double Reduction Method

1. Introduction

In this research article, we focus on the Hunter-Saxton equation [1], which is a significant mathematical model in the theoretical study of nematic liquid crystals. This equation, given by

$${\left(u_t+u{u}_x\right)}_x=\frac{1}{2}u{x}^2,$$

(1)

has garnered considerable attention from researchers. Notably, Zhang et al. [2] investigated the well-posedness and global existence of solutions for the equation, shedding light on its mathematical behavior. Xu and Guo [3] explored solitary wave solutions using the modified Hirota’s bilinear method, contributing to our understanding of the equation’s soliton solutions and their properties.

Several studies on the Hunter-Saxton equation have utilized Lie symmetry analysis to study properties of the equations and to find solutions, and our study follows a similar approach. Zhdanov et al. [4] and Mikodynski and Trynkiewicz [5] determined Lie symmetries of the Hunter-Saxton equation and then used the symmetries to find conservation laws for the equation. Deng et al. [6] employed Lie symmetry analysis to study the stability of travelling wave solutions of the Hunter-Saxton equation. Mikodynski [7] found a family of exact solutions to the Hunter-Saxton equation using Lie symmetry analysis. Barakat and El-Khodary [8] discovered a new class of exact solutions, while Liu et al. [9] obtained a family of exact solutions for a modified version of the equation. Wang et al. [10] found a new class of exact solutions with nonlocal terms using Lie symmetry analysis. Johnpillai and Khalique [11] also utilized Lie symmetry analysis to find a new class of exact solutions for a generalized version of the Hunter-Saxton equation.

Our study focuses on the symmetry reductions of the Hunter-Saxton equation using the double reduction method. We determine Lie point symmetries of the equation, construct conservation laws using the multiplier method, and employ the double reduction method to perform symmetry reductions. This research complements existing studies on the Hunter-Saxton equation and contributes to the application of the double reduction method for finding solutions of partial differential equations (PDEs).

The double reduction method, introduced by Sjöberg [12,13], is a technique for solving PDEs based on the use of Lie symmetries and conservation laws. For a $(1+1)$ PDE of order q, the double reduction theory allows for the reduction of the PDE to an ODE of order $q-1$, provided that the PDE possesses a conservation law and an associated symmetry. Generalizations of the double reduction method have been proposed to handle higher-dimensional PDEs and systems of PDEs [14,15,16]. Anco and Gandarias [17] have introduced a further generalization of the double reduction method to handle partial differential equations (PDEs) with $n\ge 2$ independent variables and a symmetry algebra of dimension at least $n-1$. In their work [17], they present an algorithm for identifying all symmetry-invariant conservation laws that reduce to first integrals for the corresponding ordinary differential equation (ODE) governing symmetry-invariant solutions of the PDE.

Moreover, Anco and Gandarias [17] propose an improved formulation for assessing the symmetry-invariance of conservation laws by utilizing multipliers. This refined formulation enables the direct derivation of symmetry-invariant conservation laws, eliminating the need to first obtain conservation laws and subsequently verify their invariance.

The subsequent sections of this paper are structured as follows: Section 2 provides an overview of the necessary preliminaries and outlines the fundamental principles of the double reduction theorem. In Section 3, we calculate the Lie point symmetries and conservation laws for the Hunter-Saxton equation, determining which conservation laws are associated with symmetries. Section 4 focuses on executing symmetry reductions for the Hunter-Saxton equation. Finally, in Section 5, we present our concluding remarks.

2. Fundamentals of the Double Reduction Theorem

In this section, we present the double reduction routine for a qth-order ($q\ge 1$) partial differential equation with n independent variables $x=(x^1,x^2,\dots ,x^n)$ and one dependent variable $u=u\left(x\right)$, namely

$$F(x,u,\dots ,u_{\left(q\right)})=0,$$

(2)

where $u_{\left(q\right)}$ denotes the collection $\left\{u_q\right\}$ of qth-order partial derivatives. In this connection, we firstly present the following well-known definitions and results (see, e.g., [18,19,20,21])

1.

The total derivative operator with respect to $x^i$ is

$$D_i=\frac{\partial }{\partial x^i}+u_i\frac{\partial }{\partial u}+u_{ij}\frac{\partial }{\partial u_j}+\cdots ,i=1,2,\dots ,n,$$

(3)

where $u_i$ denotes the derivative of u with respect to $x^i$. Similarly $u_{ij}$ denotes the derivative of u with respect to $x^i$ and $x^j$.

2.

An n-tuple $T=\left(T^1,T^2,\dots ,T^n\right),i=1,2\cdots n,$ such that

$$D_i{T}^i=0$$

(4)

holds for all solutions of (2) is known as a conservation law of (2).

3.

A multiplier $\Lambda$ for Equation (2) is a non-singular function on the solution space of (2) with the property

$$D_i{T}^i=\Lambda E$$

(5)

for arbitrary function $u\left(x^1,x^2,\dots ,x^n\right).$

4.

The determining equations for multipliers are obtained by taking the variational derivative:

$$\frac{\delta }{\delta u}(\Lambda E)=0,$$

(6)

where the Euler operator $\delta /\delta u$ is defined by

$$\frac{\delta }{\delta u}=\frac{\partial }{\partial u}-D_i\frac{\partial }{\partial u_i}+D_{ij}\frac{\partial }{\partial u_{ij}}-D_{ijk}\frac{\partial }{\partial u_{ijk}}+\cdots .$$

(7)

5.

A Lie symmetry of (2) with infinitesimal generator $X={\xi }_i\partial x_i+\eta \partial u$ is said to be associated with a conserved law (4) if the symmetry and the conservation law satisfy the relations [22]

$$\left[T^i,X\right]=X\left(T^i\right)+T^i{D}_j{\xi }^j-T^j{D}_j{\xi }^i,i=1,\dots ,n.$$

(8)

Suppose that the PDE (2) admits a Lie point symmetry with infinitesimal generator $X={\xi }_i\partial x_i+\eta \partial u$ that is associated with a conservation law $D_i{T}^i=0$. The following steps constitute the routine of the double reduction method:

I. 

Find similarity variables ${\tilde{x}}_i,i=1,2,\dots ,n$ and w

$$\begin{array}{ccc}\hfill {\tilde{x}}_i& =& {\tilde{x}}_i\left(x^1,x^2,\dots ,x^n\right),i=1,2,\dots ,n\hfill \\ \hfill w\left({\tilde{x}}_1,\dots ,{\tilde{x}}_{n-1}\right)& =& \omega \left(x^1,x^2,\dots ,x^n\right)u\hfill \end{array}$$

such that in these variables $X={\displaystyle \frac{\partial }{\partial {\tilde{x}}_n}}$.

II. 

Find inverse canonical coordinates

$$\begin{array}{ccc}\hfill x^i& =& x^i\left({\tilde{x}}_1,{\tilde{x}}_2,\dots ,{\tilde{x}}_n\right),i=1,2,\dots ,n\hfill \\ \hfill u\left(x^1,x^2,\dots ,x^n\right)& =& \psi \left({\tilde{x}}_1,{\tilde{x}}_2,\dots ,{\tilde{x}}_n\right)w\hfill \end{array}$$

III.

Write partial derivatives of u in terms of the similarity variables.

IV.

Construct matrices A and $A^{-1}$ as follows:

$$A=\left(\begin{array}{cccc}{\tilde{D}}_1{x}_1& {\tilde{D}}_1{x}_2& \dots & {\tilde{D}}_1{x}_n\\ {\tilde{D}}_2{x}_1& {\tilde{D}}_2{x}_2& \dots & {\tilde{D}}_2{x}_n\\ ⋮& ⋮& ⋮& ⋮\\ {\tilde{D}}_n{x}_1& {\tilde{D}}_n{x}_2& \dots & {\tilde{D}}_n{x}_n\end{array}\right),A^{-1}=\left(\begin{array}{cccc}{D}_1{\tilde{x}}_1& D_1{\tilde{x}}_2& \dots & D_1{\tilde{x}}_n\\ D_2{\tilde{x}}_1& D_2{\tilde{x}}_2& \dots & D_2{\tilde{x}}_n\\ ⋮& ⋮& ⋮& ⋮\\ D_n{\tilde{x}}_1& D_n{\tilde{x}}_2& \dots & D_n{\tilde{x}}_n\end{array}\right)$$

V.

Write components $T^i$ of the conserved vector in terms of the similarity variables as follows:

$$\left(\begin{array}{c}{\tilde{T}}^1\\ {\tilde{T}}^2\\ ⋮\\ {\tilde{T}}^n\end{array}\right)=J{\left(A^{-1}\right)}^T\left(\begin{array}{c}{T}^1\\ T^2\\ ⋮\\ T^n\end{array}\right),$$

(9)

where $J=\det \left(A\right).$ Note that $T^1,\dots T^n$ in (9) are easily expressed in terms of the similarity variables in light of II and III.

VI.

The reduced conservation law becomes

$$D_1{\tilde{T}}^1+D_2{\tilde{T}}^2+\dots +D_{n-1}{\tilde{T}}^{n-1}=0.$$

(10)

3. Symmetries and Conservation Laws of the Hunter-Saxton Equation

The Hunter-Saxton equation (1) is a $(1+1)$ PDE with two independent variables $x=(x^1,x^2)=(t,x)$ and one dependent variable $u=u(t,x).$ It admits the following four symmetries:

$$\begin{array}{cccc}\hfill X_1& =x\frac{\partial }{\partial x}+u\frac{\partial }{\partial u}\hfill & \hfill X_2& =\frac{\partial }{\partial t}\hfill \\ \hfill X_3& =t\frac{\partial }{\partial t}+x\frac{\partial }{\partial x}\hfill & \hfill X_4& =t^2\frac{\partial }{\partial t}+2tx\frac{\partial }{\partial x}+2x\frac{\partial }{\partial u}.\hfill \end{array}$$

(11)

The symmetries are easily computed using MathLie, the symmetry-finding package for Mathematica [23] developed by G. Baumann [24]. We will use the multiplier approach to derive conservation laws for the Hunter-Saxton equation (1). We seek first-order multipliers

$$\Lambda =\Lambda (x,t,u,u_x,u_t)$$

(12)

of (1), for which the determining equation according to (6) is

$$\frac{\delta }{\delta u}\left[\Lambda \left({\left(u_t+u{u}_x\right)}_x-\frac{1}{2}u{x}^2\right)\right]=0,$$

(13)

where the standard Euler operator $\delta /\delta u$, as defined in (7), is

$$\frac{\delta }{\delta u}=\frac{\partial }{\partial u}-D_t\frac{\partial }{\partial u_t}-D_x\frac{\partial }{\partial u_x}+D_t^2\frac{\partial }{\partial u_{tt}}+D_x^2\frac{\partial }{\partial u_{xx}}+D_x{D}_t\frac{\partial }{\partial u_{tx}}-\cdots ,$$

(14)

and total derivative operators $D_t$ and $D_x$ using (3) are

$$\begin{array}{cc}\hfill D_t& =\frac{\partial }{\partial t}+u_t\frac{\partial }{\partial u}+u_{tt}\frac{\partial }{\partial u_t}+u_{tx}\frac{\partial }{\partial u_x}+\cdots ,\hfill \\ \hfill D_x& =\frac{\partial }{\partial x}+u_x\frac{\partial }{\partial u}+u_{xx}\frac{\partial }{\partial u_x}+u_{tx}\frac{\partial }{\partial u_t}+\cdots .\hfill \end{array}$$

The determining equation for the multiplier $\Lambda$ after expansion takes the following form:

$${\Omega }_0+u_{tt}{\Omega }_1+u_{tx}{\Omega }_2+{\left(u_{tx}\right)}^2{\Omega }_3+u_{xx}{\Omega }_4+u_{xx}{u}_{tt}{\Omega }_5=0,$$

(15)

where

$$\begin{array}{ccc}\hfill {\Omega }_0& =& u_x{\Lambda }_{tu}-\frac{1}{2}{u}_x^2{\Lambda }_{t{u}_t}+{\Lambda }_{tx}-\frac{1}{2}{u}_x^3{\Lambda }_{u{u}_x}-\frac{1}{2}{u}_x^2{u}_t{\Lambda }_{u{u}_t}+u{u}_x^2{\Lambda }_{uu}+u_x{u}_t{\Lambda }_{uu}\hfill \\ & & +2u{u}_x{\Lambda }_{xu}+u_t{\Lambda }_{xu}+u{\Lambda }_{xx}-\frac{1}{2}{u}_x^2{\Lambda }_{x{u}_x}+\frac{3u_x^2{\Lambda }_u}{2}+u_x{\Lambda }_x\hfill \\ \hfill {\Omega }_1& =& u_x{\Lambda }_{u{u}_t}-\frac{1}{2}{u}_x^2{\Lambda }_{u_t{u}_t}+{\Lambda }_{x{u}_t}\hfill \\ \hfill {\Omega }_2& =& 2u{u}_x{\Lambda }_{u{u}_t}+2u{\Lambda }_{x{u}_t}-u_x^2{\Lambda }_{u_t{u}_x}+2{\Lambda }_u\hfill \\ \hfill {\Omega }_3& =& u{\Lambda }_{u_t{u}_t}-{\Lambda }_{u_t{u}_x}\hfill \\ \hfill {\Omega }_4& =& {\Lambda }_{t{u}_x}-u{\Lambda }_{t{u}_t}+u_t{\Lambda }_{u{u}_x}+u{\Lambda }_{x{u}_x}+u{u}_x{\Lambda }_{u{u}_x}-u{u}_t{\Lambda }_{u{u}_t}-\frac{1}{2}{u}_x^2{\Lambda }_{u_x{u}_x}\hfill \\ & & +2u{\Lambda }_u-u_x{\Lambda }_{u_x}-u_t{\Lambda }_{u_t}+\Lambda \hfill \\ \hfill {\Omega }_5& =& {\Lambda }_{u_t{u}_x}-u{\Lambda }_{u_t{u}_t}\hfill \end{array}$$

The multiplier determining equation (15) splits with respect to different combinations of the derivatives $u_{xx},$$u_{tx}$ and $u_{tt}$ yielding an overdetermined linear system of equations for the multiplier. The system of equations was solved using Mathematica [23] to obtain

$$\Lambda =u_t\left({\delta }_2+{\delta }_{3}t-\frac{{\delta }_1{t}^2}{2}\right)+u_{x}x({\delta }_3-{\delta }_{1}t)+{\delta }_{1}x+{\delta }_4{u}_x+\frac{{\delta }_5}{u_x^2},$$

(16)

where ${\delta }_i$ are arbitrary constants. From (5) and (16), we have

$$\begin{array}{ccc}\hfill & & \left[{\left(u_t+u{u}_x\right)}_x-\frac{1}{2}u{x}^2\right][u_t\left({\delta }_2+{\delta }_{3}t-\frac{{\delta }_1{t}^2}{2}\right)+u_{x}x({\delta }_3-{\delta }_{1}t)\hfill \\ & & +{\delta }_{1}x+{\delta }_4{u}_x+\frac{{\delta }_5}{u_x^2}]=D_t{T}^t+D_x{T}^x,\hfill \end{array}$$

(17)

where

$$\begin{array}{ccc}\hfill T^t& =& u_x^2\left(u\left(\frac{{\delta }_1{t}^2}{4}-\frac{{\delta }_2}{2}-\frac{{\delta }_{3}t}{2}\right)+x\left(\frac{{\delta }_3}{2}-\frac{{\delta }_{1}t}{2}\right)+\frac{{\delta }_4}{2}\right)-\frac{{\delta }_5}{u_x}+{\varphi }_2\left(x\right)\hfill \\ & & +u_x({\delta }_{1}x-{\delta }_{1}tu+{\varphi }_1\left(u\right))\hfill \\ \hfill T^x& =& u_t^2\left(\frac{{\delta }_2}{2}+\frac{{\delta }_{3}t}{2}-\frac{{\delta }_1{t}^2}{4}\right)+u{u}_x^2\left(x\left(\frac{{\delta }_3}{2}-\frac{{\delta }_{1}t}{2}\right)+\frac{{\delta }_4}{2}\right)-\frac{{\delta }_{5}u}{u_x}+\frac{3{\delta }_{5}x}{2}\hfill \\ & & +u_x\left(u{u}_t\left({\delta }_2+{\delta }_{3}t-\frac{{\delta }_1{t}^2}{2}\right)+{\delta }_{1}ux\right)+u_t({\delta }_{1}tu-{\varphi }_1\left(u\right))+{\varphi }_3\left(t\right)\hfill \end{array}$$

for arbitrary functions $u(t,x)$. When $u(t,x)$ is a solution of Eq. (1), the left hand side of (17) vanishes and we obtain conservation laws of Hunter-Saxton equation (1) for which the conserved vectors $\left(T_i^1,T_i^2\right)$ are given by:

$$\begin{array}{ccc}\hfill T_1^1& =& u_x\left(ux-\frac{1}{2}{t}^{2}u{u}_t\right)-\frac{t^2{u}_t^2}{4}-\frac{1}{2}tu{u}_x^{2}x+u_t(tu-{\varphi }_1\left(u\right))+{\varphi }_3\left(t\right)\hfill \\ \hfill T_1^2& =& u_x^2\left(\frac{t^{2}u}{4}-\frac{tx}{2}\right)+u_x(x-tu+{\varphi }_1\left(u\right))+{\varphi }_2\left(x\right),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill T_2^1& =& {\varphi }_3\left(t\right)+u{u}_x{u}_t-u_t{\varphi }_1\left(u\right)+\frac{u_t^2}{2}\hfill \\ \hfill T_2^2& =& u_x{\varphi }_1\left(u\right)+{\varphi }_2\left(x\right)-\frac{u{u}_x^2}{2}\hfill \end{array}$$

$$\begin{array}{ccc}\hfill T_3^1& =& tu{u}_x{u}_t+\frac{t{u}_t^2}{2}+{\varphi }_3\left(t\right)+\frac{1}{2}u{u}_x^{2}x-u_t{\varphi }_1\left(u\right)\hfill \\ \hfill T_3^2& =& u_x^2\left(\frac{x}{2}-\frac{tu}{2}\right)+u_x{\varphi }_1\left(u\right)+{\varphi }_2\left(x\right),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill T_4^1& =& {\varphi }_3\left(t\right)-u_t{\varphi }_1\left(u\right)+\frac{u{u}_x^2}{2}\hfill \\ \hfill T_4^2& =& {\varphi }_2\left(x\right)+u_x{\varphi }_1\left(u\right)+\frac{u_x^2}{2},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill T_5^1& =& {\varphi }_3\left(t\right)-u_t{\varphi }_1\left(u\right)-\frac{u}{u_x}+\frac{3x}{2}\hfill \\ \hfill T_5^2& =& {\varphi }_2\left(x\right)+u_x{\varphi }_1\left(u\right)-\frac{1}{u_x}.\hfill \end{array}$$

According to (8), a symmetry X is associated with a conservation law $D_t{T}^t+D_x{T}^x=0$ if the following formula is satisfied:

$$X\left(\begin{array}{c}{T}^t\hfill \\ T^x\hfill \end{array}\right)-\left(\begin{array}{cc}{D}_t{\xi }^t& D_x{\xi }^t\\ D_t{\xi }^x& D_x{\xi }^x\end{array}\right)\left(\begin{array}{c}{T}^t\hfill \\ T^x\hfill \end{array}\right)+\left(D_t{\xi }^t+D_x{\xi }^x\right)\left(\begin{array}{c}{T}^t\hfill \\ T^x\hfill \end{array}\right)=0.$$

It turns out that the association of symmetries and conservation laws of (1) is obtained in the following cases:

$$\begin{array}{cc}\hfill {\kappa }_1(X_1+2X_3)+{\kappa }_2{X}_2& \to \left\{\begin{array}{c}{T}_2^1=\frac{u_t^2}{2}-\frac{{\delta }_1{u}_t}{u}+u{u}_x{u}_t\hfill \\ T_2^2=\frac{{\delta }_1{u}_x}{u}+\frac{{\delta }_3}{x}-\frac{u{u}_x^2}{2}\hfill \end{array}\right.\hfill \\ \hfill {\kappa }_1(X_1+X_3)+{\kappa }_2{X}_2& \to \left\{\begin{array}{c}{T}_4^1=\frac{u{u}_x^2}{2}-\frac{{\delta }_1{u}_t}{u}\hfill \\ T_4^2=\frac{{\delta }_1{u}_x}{u}+\frac{{\delta }_3}{x}+\frac{u_x^2}{2}\hfill \end{array}\right.\hfill \\ \hfill {\kappa }_1\left(X_1-\frac{X_3}{2}\right)+{\kappa }_2{X}_2& \to \left\{\begin{array}{c}{T}_5^1=\frac{{\delta }_2}{2{\kappa }_2-{\kappa }_{1}t}-\frac{{\delta }_1{u}_t}{u}-\frac{u}{u_x}+\frac{3x}{2}\hfill \\ T_5^2=\frac{{\delta }_1{u}_x}{u}+\frac{{\delta }_3}{x}-\frac{1}{u_x}\hfill \end{array}\right.\hfill \\ \hfill X_3& \to \left\{\begin{array}{c}{T}_3^1=\frac{{\delta }_1}{t}+tu{u}_x{u}_t+\frac{t{u}_t^2}{2}+\frac{1}{2}u{u}_x^{2}x-u_t{\varphi }_1\left(u\right)\hfill \\ T_3^2=\frac{{\delta }_2}{x}+u_x^2\left(\frac{x}{2}-\frac{tu}{2}\right)+u_x{\varphi }_1\left(u\right)\hfill \end{array}\right.\hfill \end{array}$$

4. Double Reduction of the Hunter-Saxton Equation

4.1. Double Reduction of (1) by $〈{\kappa }_1(X_1+2X_3)+{\kappa }_2{X}_2〉$

We transform the generator $Z={\kappa }_1(X_1+2X_3)+{\kappa }_2{X}_2$ to its canonical form

$Y=0\frac{\partial }{\partial r}+\frac{\partial }{\partial s}+0\frac{\partial }{\partial w}.$ Therefore canonical coordinates $r=r(t,x),$$s=s(t,x)$ and

$w=w(t,x,u)$ must be found such that $Z\left(r\right)=0,Z\left(s\right)=1$ and $Z\left(w\right)=0$. While the coordinates r and w are obtained from invariants of $Z,$ the coordinate s may be determined by inspection. More systematically, it can be obtained from an invariant $J=v-s(x,y)$ of the extended operator $Z+{\partial }_v$, where v is an auxiliary variable [21]. We obtain

$$r=\frac{x}{{(2{\kappa }_{1}t+{\kappa }_2)}^{3/2}},s=\frac{lnx}{3{\kappa }_1},w=\frac{u}{\sqrt{2{\kappa }_{1}t+{\kappa }_2}},{\kappa }_1\ne 0,$$

(18)

where $w=w\left(r\right)$. Inverse canonical coordinates follow from (18) and are given by

$$t=\frac{e^{2{\kappa }_{1}s}-{\kappa }_2{r}^{2/3}}{2{\kappa }_1{r}^{2/3}},x=e^{3{\kappa }_{1}s},u=\frac{w{e}^{{\kappa }_{1}s}}{r^{1/3}}.$$

(19)

Computing A and ${\left(A^{-1}\right)}^T$, we obtain

$$A=\left(\begin{array}{cc}{D}_{r}t\hfill & D_{r}x\hfill \\ D_{s}t\hfill & D_{s}x\hfill \end{array}\right)=\left(\begin{array}{cc}-\frac{e^{2{\kappa }_{1}s}}{3{\kappa }_1{r}^{5/3}}& 0\\ \frac{e^{2{\kappa }_{1}s}}{r^{2/3}}& 3e^{3{\kappa }_{1}s}{\kappa }_1\end{array}\right)$$

and

$${\left(A^{-1}\right)}^T=\left(\begin{array}{cc}{D}_{t}r& D_{x}r\\ D_{t}s& D_{x}s\end{array}\right)=\left(\begin{array}{cc}-3e^{-2{\kappa }_{1}s}{\kappa }_1{r}^{5/3}& e^{-3{\kappa }_{1}s}r\\ 0& \frac{e^{-3{\kappa }_{1}s}}{3{\kappa }_1}\end{array}\right).$$

The partial derivatives of u from (19) are given by

$$\begin{array}{cc}\hfill & u_t={\kappa }_1\sqrt[3]{r}{e}^{-{\kappa }_{1}s}(w-3r{w}_r),u_x=r^{2/3}{w}_r{e}^{-2{\kappa }_{1}s},\hfill \\ \hfill & u_{tx}=-{\kappa }_1{r}^{4/3}{e}^{-4{\kappa }_{1}s}(3r{w}_{rr}+2w_r),\hfill \\ \hfill & u_{xx}=r^{5/3}{w}_{rr}{e}^{-5{\kappa }_{1}s}.\hfill \end{array}$$

(20)

The reduced conserved form is given by

$$\left(\begin{array}{c}{T}_2^r\\ T_2^s\end{array}\right)=J{\left(A^{-1}\right)}^T\left(\begin{array}{c}{T}_2^t\\ T_2^x\end{array}\right),$$

(21)

where $J=det\left(A\right)=-\frac{e^{5{\kappa }_{1}s}}{r^{5/3}}$. By substituting (19) and (20) into (21), we obtain

$$\begin{array}{cc}\hfill T_2^r& ={\delta }_1{\kappa }_1+3{\delta }_3{\kappa }_1+3{\kappa }_1^{2}rw{w}_r-\frac{9}{2}{\kappa }_1^2{r}^2{w}_r^2-\frac{{\kappa }_1^2{w}^2}{2}+\frac{3}{2}{\kappa }_{1}rw{w}_r^2-{\kappa }_1{w}^2{w}_r,\hfill \\ \hfill T_2^s& =w_r\left({\kappa }_{1}w-\frac{{\delta }_1}{w}-\frac{w^2}{3r}\right)+\frac{{\delta }_1}{3r}-\frac{{\kappa }_1{w}^2}{6r}+w_r^2\left(w-\frac{3{\kappa }_{1}r}{2}\right),\hfill \end{array}$$

(22)

where the reduced conserved form satisfies

$$D_r{T}_2^r=0.$$

(23)

From (22) and (23), we have

$$3{\kappa }_1^{2}rw{w}_r-\frac{9}{2}{\kappa }_1^2{r}^2{w}_r^2-\frac{{\kappa }_1^2{w}^2}{2}+\frac{3}{2}{\kappa }_{1}rw{w}_r^2-{\kappa }_1{w}^2{w}_r=k,$$

where k is an arbitrary constant.

4.2. Double Reduction of (1) by $〈{\kappa }_1(X_1+X_3)+{\kappa }_2{X}_2〉$

Canonical coordinated determined from $〈{\kappa }_1(X_1+X_3)+{\kappa }_2{X}_2〉$ are:

$$r=\frac{x}{{({\kappa }_{1}t+{\kappa }_2)}^2},s=\frac{lnx}{2{\kappa }_1},w=\frac{u}{\sqrt{x}},{\kappa }_1\ne 0,$$

(24)

where $w=w\left(r\right)$, and the inverse canonical coordinates are given by

$$t=-\frac{{\kappa }_2\sqrt{r}-e^{{\kappa }_{1}s}}{{\kappa }_1\sqrt{r}},x=e^{2{\kappa }_{1}s}u=w{e}^{{\kappa }_{1}s}.$$

(25)

Therefore, the partial derivatives of u from (25) are given by

$$\begin{array}{cc}\hfill & u_t=-2{\kappa }_1{r}^{3/2}{w}_r,u_x=\frac{1}{2}{e}^{-{\kappa }_{1}s}(2r{w}_r+w),\hfill \\ \hfill & u_{tx}=-e^{-2{\kappa }_{1}s}{\kappa }_1{r}^{3/2}(2r{w}_{rr}+3w_r,\hfill \\ \hfill & u_{xx}=-\frac{1}{4}{e}^{-3{\kappa }_{1}s}(w-4r(r{w}_{rr}+w_r)).\hfill \end{array}$$

(26)

As for A and ${\left(A^{-1}\right)}^T$, we obtain

$$A=\left(\begin{array}{cc}{D}_{r}t\hfill & D_{r}x\hfill \\ D_{s}t\hfill & D_{s}x\hfill \end{array}\right)=\left(\begin{array}{cc}-\frac{e^{{\kappa }_{1}s}}{2{\kappa }_1{r}^{3/2}}& 0\\ \frac{e^{{\kappa }_{1}s}}{\sqrt{r}}& 2e^{2{\kappa }_{1}s}{\kappa }_1\end{array}\right),$$

and

$${\left(A^{-1}\right)}^T=\left(\begin{array}{cc}{D}_{t}r& D_{x}r\\ D_{t}s& D_{x}s\end{array}\right)=\left(\begin{array}{cc}-2e^{-{\kappa }_{1}s}{\kappa }_1{r}^{3/2}& e^{-2{\kappa }_{1}s}r\\ 0& \frac{e^{-2{\kappa }_{1}s}}{2{\kappa }_1}\end{array}\right).$$

Therefore, from

$$\left(\begin{array}{c}{T}_4^r\\ T_4^s\end{array}\right)=J{\left(A^{-1}\right)}^T\left(\begin{array}{c}{T}_4^t\\ T_4^x\end{array}\right),$$

(27)

where $J=\det \left(A\right)=-\frac{e^{3{\kappa }_{1}s}}{r^{3/2}},$ we obtain

$$\begin{array}{cc}\hfill T_4^r& ={\delta }_1{\kappa }_1+2{\delta }_3{\kappa }_1+{\kappa }_1{r}^2{w}_r^2+{\kappa }_{1}rw{w}_r+\frac{{\kappa }_1{w}^2}{4}-\frac{1}{2}{r}^{3/2}w{w}_r^2-\frac{w^3}{8\sqrt{r}}-\frac{1}{2}\sqrt{r}{w}^2{w}_r,\hfill \\ \hfill T_4^s& =-\frac{{\delta }_1{w}_r}{w}-\frac{w^3}{16{\kappa }_1{r}^{3/2}}-\frac{w^2{w}_r}{4{\kappa }_1\sqrt{r}}-\frac{\sqrt{r}w{w}_r^2}{4{\kappa }_1}.\hfill \end{array}$$

(28)

From the reduced conservation law $D_r{T}_4^r=0$, we obtain

$${\kappa }_1{r}^2{w}_r^2+{\kappa }_{1}rw{w}_r+\frac{{\kappa }_1{w}^2}{4}-\frac{1}{2}{r}^{3/2}w{w}_r^2-\frac{w^3}{8\sqrt{r}}-\frac{1}{2}\sqrt{r}{w}^2{w}_r=k,$$

where k is an arbitrary constant.

4.3. Double Reduction of (1) by $〈{\kappa }_1\left(X_1-\frac{X_3}{2}\right)+{\kappa }_2{X}_2〉$

Canonical coordinated determined from $〈{\kappa }_1\left(X_1-\frac{X_3}{2}\right)+{\kappa }_2{X}_2〉$ are:

$$r=x(2{\kappa }_2-{\kappa }_{1}t),s=\frac{2lnx}{{\kappa }_1},w=\frac{u}{x^2},{\kappa }_1\ne 0$$

(29)

where $w=w\left(r\right)$, and the inverse canonical coordinates are given by

$$t=\frac{2{\kappa }_2-r{e}^{-\frac{1}{2}{\kappa }_{1}s}}{{\kappa }_1},x=e^{\frac{{\kappa }_{1}s}{2}},u=w{e}^{{\kappa }_{1}s}$$

(30)

Therefore, the partial derivatives of u from (30) are given by

$$\begin{array}{cc}\hfill & u_t=-{\kappa }_1{w}_r{e}^{\frac{3{\kappa }_{1}s}{2}},u_x=e^{\frac{{\kappa }_{1}s}{2}}(r{w}_r+2w),\hfill \\ \hfill & u_{tx}=-{\kappa }_1{e}^{{\kappa }_{1}s}(r{w}_{rr}+3w_r),\hfill \\ \hfill & u_{xx}=r(r{w}_{rr}+4w_r)+2w\hfill \end{array}$$

(31)

Therefore,

$$A=\left(\begin{array}{cc}{D}_{r}t\hfill & D_{r}x\hfill \\ D_{s}t\hfill & D_{s}x\hfill \end{array}\right)=\left(\begin{array}{cc}-\frac{e^{-\frac{1}{2}{\kappa }_{1}s}}{{\kappa }_1}& 0\\ \frac{1}{2}{e}^{-\frac{1}{2}{\kappa }_{1}s}r& \frac{1}{2}{e}^{\frac{{\kappa }_{1}s}{2}}{\kappa }_1\end{array}\right)$$

and

$${\left(A^{-1}\right)}^T=\left(\begin{array}{cc}{D}_{t}r& D_{x}r\\ D_{t}s& D_{x}s\end{array}\right)=\left(\begin{array}{cc}-e^{\frac{{\kappa }_{1}s}{2}}{\kappa }_1& e^{-\frac{1}{2}{\kappa }_{1}s}r\\ 0& \frac{2e^{-\frac{1}{2}{\kappa }_{1}s}}{{\kappa }_1}\end{array}\right).$$

Therefore, from

$$\left(\begin{array}{c}{T}_5^r\\ T_5^s\end{array}\right)=J{\left(A^{-1}\right)}^T\left(\begin{array}{c}{T}_5^t\\ T_5^x\end{array}\right),$$

(32)

where $J=\det \left(A\right)=-\frac{1}{2},$ we obtain

$$\begin{array}{cc}\hfill T_5^r& =\frac{2{\kappa }_1(2{\delta }_{1}r{w}_r+4{\delta }_{1}w+{\delta }_{3}r{w}_r+2{\delta }_{3}w-1)-2{\delta }_2(r{w}_r+2w)-r(3r{w}_r+4w)}{4r{w}_r+8w},\hfill \\ \hfill T_5^s& =-\frac{2{\delta }_1{\kappa }_1{r}^2{w}_r^2+4{\delta }_1{\kappa }_{1}rw{w}_r+2{\delta }_{2}rw{w}_r+4{\delta }_2{w}^2+3r^{2}w{w}_r+4r{w}^2}{2{\kappa }_1{r}^{2}w{w}_r+4{\kappa }_{1}r{w}^2}.\hfill \end{array}$$

(33)

From the reduced conservation law $D_r{T}_5^r=0$, we obtain

$$\frac{2{\kappa }_1(2{\delta }_{1}r{w}_r+4{\delta }_{1}w+{\delta }_{3}r{w}_r+2{\delta }_{3}w-1)-2{\delta }_2(r{w}_r+2w)-r(3r{w}_r+4w)}{4r{w}_r+8w}=k,$$

where k is an arbitrary constant.

4.4. Double Reduction of (1) by $〈X_3〉$

Canonical coordinated determined from $X_3$ are:

$$r=\frac{x}{t},s=lnxw=u,$$

(34)

where $w=w\left(r\right)$, and the inverse canonical coordinates are given by

$$t=\frac{e^s}{r},x=e^{s}u=w.$$

(35)

Therefore, the partial derivatives of u from (35) are given by

$$\begin{array}{cc}\hfill & u_t=-r^2{e}^{-s}{w}_r,u_x=r{e}^{-s}{w}_r,\hfill \\ \hfill & u_{tx}=-r^2{e}^{-2s}(r{w}_{rr}+w_r),\hfill \\ \hfill & u_{xx}=r^2{e}^{-2s}{w}_{rr}\hfill \end{array}$$

(36)

As for A and ${\left(A^{-1}\right)}^T,$ we obtain

$$A=\left(\begin{array}{cc}{D}_{r}t\hfill & D_{r}x\hfill \\ D_{s}t\hfill & D_{s}x\hfill \end{array}\right)=\left(\begin{array}{cc}-\frac{e^s}{r^2}& 0\\ \frac{e^s}{r}& e^s\end{array}\right),$$

and

$${\left(A^{-1}\right)}^T=\left(\begin{array}{cc}{D}_{t}r& D_{x}r\\ D_{t}s& D_{x}s\end{array}\right)=\left(\begin{array}{cc}-e^{-s}{r}^2& e^{-s}r\\ 0& e^{-s}\end{array}\right).$$

Therefore, from

$$\left(\begin{array}{c}{T}_3^r\\ T_3^s\end{array}\right)=J{\left(A^{-1}\right)}^T\left(\begin{array}{c}{T}_3^t\\ T_3^x\end{array}\right),$$

(37)

where $J=\det \left(A\right)=-\frac{e^{2s}}{r^2},$ we obtain

$$\begin{array}{cc}\hfill T_3^r& ={\delta }_2-{\delta }_1,\hfill \\ \hfill T_3^s& =\frac{1}{2}{w}_r^2(w-r)-\frac{{\delta }_1}{r}-w_r{\varphi }_1\left(w\right).\hfill \end{array}$$

(38)

It is remarkable that in this case, because $T_3^r$ in (38) is simply a constant, the reduced conservation law $D_r{T}_3^r=0$ does not result in an ODE that can be solved for w. Therefore, no invariant solution will arise via the double reduction method from the association of $X_3$ and the conservation law $T_3.$

5. Concluding Remarks

In this paper, a study of the Hunter-Saxton equation using Lie symmetry analysis was presented. Symmetry reductions of the equation were carried out by employing the generalised approach to double reduction theory proposed by Bokhari et al [14]. By utilising the multiplier method, nontrivial conservation laws for the Hunter-Saxton equation were derived. These conservation laws, along with the Lie point symmetries of the equation, were employed to perform symmetry reductions via the double reduction method.

Through the analysis, a set of first-order ODEs was obtained, whose solutions represent invariant solutions for the Hunter-Saxton equation. Out of the five nontrivial conservation laws constructed, it was observed that only four had associated Lie point symmetries according to the definition provided by Kara and Mahomed [18]. The conservation law $T_1$ did not have any linear combination of symmetries associated with it. Additionally, it is noteworthy that despite the conservation law $T_3$ having an associated Lie point symmetry, $X_3,$ the application of the double reduction method in this case did not yield a symmetry reduction of the Hunter-Saxton equation. This outcome could be attributed to the “collapse" of the first integral, which was expected to represent a reduced ODE for the PDE but instead resulted in a constant value.