Application of the Generalized Double Reduction Method to the (1+1)-Dimensional Kaup-Boussinesq (K-B) System: Exploiting Lie Symmetries and Conservation Laws

1. Fundamental Operators, Definitions and Theorems

This section presents the well-known definitions and theorems in the literature (see [21,25,30,31,32,33]) which will be used later in this study. Let us consider a $k\mathrm{th}$ order system of r partial differential equations of n independent variables $x=$$\left(x^1,x^2,\dots ,x^n\right)$ and m dependent variables $u=\left(u^1,u^2,\dots ,u^m\right)$

$$E^{\sigma }\left(x,u,u_{\left(1\right)},u_{\left(2\right)},\dots ,u_{\left(k\right)}\right)=0,\sigma =1,\dots ,r.$$

(2)

Collections of all first, second, ..., $kth$-order partial derivatives are denoted by $u_{\left(1\right)},u_{\left(2\right)},\dots ,u_{\left(k\right)}$ respectively, that is

$$u_i^{\alpha }=D_i\left(u^{\alpha }\right),u_{ij}^{\alpha }=D_j{D}_i\left(u^{\alpha }\right),\dots ,$$

(3)

with the total differentiation operator with respect to $x^i$ given by,

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

(4)

A Lie-Bäcklund or generalized operator is defined by

$$X={\xi }^i{\displaystyle \frac{\partial }{\partial x^i}}+{\eta }^{\alpha }{\displaystyle \frac{\partial }{\partial u^{\alpha }}}+{\zeta }_i^{\alpha }{\displaystyle \frac{\partial }{\partial u_i^{\alpha }}}+\sum _{s\ge 1}{\zeta }_{i_1\cdots i_s}^{\alpha }{\displaystyle \frac{\partial }{\partial u_{i_1\cdots i_s}^{\alpha }}},{\xi }^i,{\eta }^{\alpha }\in \mathcal{A}$$

(5)

where $\mathcal{A}$ is the universal space of differential functions and the additional coefficients are determined uniquely by the following formulas:

$$\begin{array}{cc}\hfill {\zeta }_i^{\alpha }& =D_i\left({\eta }^{\alpha }\right)-u_j^{\alpha }{D}_i\left({\xi }^j\right),\hfill \\ \hfill {\zeta }_{i_1\cdots i_s}^{\alpha }& =D_{i_s}\left({\zeta }_{i_1\cdots i_{s-1}}^{\alpha }\right)-u_{j{i}_1\cdots i_{s-1}}^{\alpha }{D}_{i_s}\left({\xi }^j\right),s>1.\hfill \end{array}$$

(6)

The Lie point symmetry of equation (2) is an operator X of the form (5) that satisfies

$$X^{\left[k\right]}{E}^{\sigma }{|}_{\left(\right)}=0,$$

(7)

where $X^{\left[k\right]}$ is the kth prolongation of X defined by

$$X^{\left[k\right]}={\xi }^i{\displaystyle \frac{\partial }{\partial x^i}}+{\eta }^{\alpha }{\displaystyle \frac{\partial }{\partial u^{\alpha }}}+{\zeta }^{\alpha }{\displaystyle \frac{\partial }{\partial u_i^{\alpha }}}+\dots +{\zeta }_{i_1\dots i_k}^{\alpha }{\displaystyle \frac{\partial }{\partial u_{i_1\dots i_k}^{\alpha }}}.$$

(8)

This means that equation (2) is invariant under the action of the generator X.

A conserved vector of (2) is n-tuple $T=\left(T^1,T^2\right.$, $\left.\dots ,T^n\right)$ satisfying the relation

$${\left.D_i{T}^i\right|}_{\left(\right)}=0$$

(9)

where $T^i=T^i(x,u,u_{\left(1\right)},u_{\left(2\right)},\dots ,u_{\left(q\right)})\in \mathcal{A},i=$$1,2,\dots ,n$.

A conservation law can be expressed in characteristic form [34] as

$$D_i{T}^i={\Lambda }_{\sigma }{E}^{\sigma },\sigma =1,2,\dots ,r,$$

(10)

where ${\Lambda }_{\sigma }={\Lambda }_{\sigma }(x,u,u_{\left(1\right)},\dots ,u_{\left(q\right)})$ are the characteristics or multipliers for the PDE system (2). The determining equations for multipliers are obtained by taking the variational derivative

$${\displaystyle \frac{\delta }{\delta u^{\alpha }}}\left[\sum _{\sigma =1}^p{\Lambda }_{\sigma }{E}^{\sigma }\right]=0,\alpha =1,2,\dots ,m,$$

(11)

where the Euler operator ${\displaystyle \frac{\delta }{\delta u^{\alpha }}}$ is defined by

$${\displaystyle \frac{\delta }{\delta u^{\alpha }}}={\displaystyle \frac{\partial }{\partial u^{\alpha }}}+\sum _{s\ge 1}{(-1)}^s{D}_{i_1}\cdots D_{i_s}{\displaystyle \frac{\partial }{\partial u_{i_1\cdots i_s}^{\alpha }}},\alpha =1,\dots ,m.$$

(12)

Definition 1.

A Lie-Bäcklund symmetry generator X of the form (5) is associated with a conserved vector T of the system (2) if X and T satisfy the relations

$$X\left(T^i\right)+T^i{D}_j{\xi }^j-T^j{D}_j{\xi }^i,=0,i=1,\dots ,n.$$

(13)

Theorem 1.

Suppose $D_i{T}^i=0$ is a conservation law of the $PDE$ system (2). Then under a similarity transformation of a symmetry X of the form (5) for the PDE, there exist functions ${\tilde{T}}^i$ such that X is still symmetry for the PDE ${\tilde{D}}_i{\tilde{T}}^i=0$, where ${\tilde{T}}^i$ is given by

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

(14)

where

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

and $J=\det \left(A\right).$

Corollary 1.(The necessary and sufficient condition for reduced conserved form [21]). The conserved form $D_i{T}^i=$ 0 of the PDE system (2) can be reduced under a similarity transformation of a symmetry X to a reduced conserved form ${\tilde{D}}_i{\tilde{T}}^i=0$ if and only if X is associated with the conservation law T.

Corollary 2.(see [21]). A nonlinear system of qth-order PDEs with n independent and m dependent variables which admits a nontrivial conserved form that has at least one associated symmetry in every reduction from the n reductions (the first step of double reduction) can be reduced to a $(q-1)$ th-order nonlinear system of ODEs.

2. Symmetries and Conservation laws of K-B System (1)

The KB system (1) admits the following four Lie point symmetries [35]

$$\begin{array}{cc}\hfill X_1& ={\displaystyle \frac{\partial }{\partial x}},\hfill \\ \hfill X_2& ={\displaystyle \frac{\partial }{\partial t}},\hfill \\ \hfill X_3& ={\displaystyle \frac{\partial }{\partial v}}-2t{\displaystyle \frac{\partial }{\partial x}},\hfill \\ \hfill X_4& =v{\displaystyle \frac{\partial }{\partial v}}-2t{\displaystyle \frac{\partial }{\partial t}}+2u{\displaystyle \frac{\partial }{\partial u}}-x{\displaystyle \frac{\partial }{\partial x}}.\hfill \end{array}$$

(15)

To construct conservation laws for (1) we employ the multiplier method [36,37,38] and look for first-order multipliers of the form

$$\begin{array}{cc}\hfill & {\Lambda }_1={\Lambda }_1\left(x,t,u,v,u_x,v_x\right),\hfill \\ \hfill & {\Lambda }_2={\Lambda }_2\left(x,t,u,v,u_x,v_x\right).\hfill \end{array}$$

The determining equations for multipliers ${\Lambda }_1$ and ${\Lambda }_2$ become

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\delta }{\delta u}}\left({\Lambda }_1\left(v_t-u_x-2v{v}_x\right)+{\Lambda }_2\left(u_t-v_{xxx}-2v{u}_x-2u{v}_x\right)\right)\equiv 0,\hfill \\ \hfill & {\displaystyle \frac{\delta }{\delta v}}\left({\Lambda }_1\left(v_t-u_x-2v{v}_x\right)+{\Lambda }_2\left(u_t-v_{xxx}-2v{u}_x-2u{v}_x\right)\right)\equiv 0\hfill \end{array}$$

(16)

where the Euler operators ${\displaystyle \frac{\delta }{\delta u}}$ and ${\displaystyle \frac{\delta }{\delta v}}$ are given by

$$\begin{array}{cc}\hfill {\displaystyle \frac{\delta }{\delta u}}& ={\displaystyle \frac{\partial }{\partial u}}-D_t{\displaystyle \frac{\partial }{\partial u_t}}-D_x{\displaystyle \frac{\partial }{\partial u_x}}+D_x^2{\displaystyle \frac{\partial }{\partial u_{xx}}}\cdots ,\hfill \\ \hfill {\displaystyle \frac{\delta }{\delta v}}& ={\displaystyle \frac{\partial }{\partial v}}-D_t{\displaystyle \frac{\partial }{\partial v_t}}-D_x{\displaystyle \frac{\partial }{\partial v_x}}+D_x^2{\displaystyle \frac{\partial }{\partial v_{xx}}}+\cdots ,\hfill \end{array}$$

(17)

and total derivative operators $D_t$ and $D_x$ are

$$\begin{array}{cc}\hfill D_t& ={\displaystyle \frac{\partial }{\partial t}}+u_t{\displaystyle \frac{\partial }{\partial u}}+v_t{\displaystyle \frac{\partial }{\partial v}}+u_{tt}{\displaystyle \frac{\partial }{\partial u_t}}+v_{tt}{\displaystyle \frac{\partial }{\partial v_t}}+u_{tx}{\displaystyle \frac{\partial }{\partial u_x}}+v_{tx}{\displaystyle \frac{\partial }{\partial v_x}}+\cdots \hfill \\ \hfill D_x& ={\displaystyle \frac{\partial }{\partial x}}+u_x{\displaystyle \frac{\partial }{\partial u}}+v_x{\displaystyle \frac{\partial }{\partial v}}+u_{xx}{\displaystyle \frac{\partial }{\partial u_x}}+v_{xx}{\displaystyle \frac{\partial }{\partial v_x}}+u_{tx}{\displaystyle \frac{\partial }{\partial u_t}}+v_{tx}{\displaystyle \frac{\partial }{\partial v_t}}+\cdots \hfill \end{array}$$

(18)

The system (16) after expansion and splitting with respect to the derivatives of u and v, gives the determining equations:

$$\begin{array}{cccccc}\hfill {\Lambda }_{2_{xx}}& =0,\hfill & \hfill {\Lambda }_{2_{vx}}& =0,\hfill & \hfill {\Lambda }_{2_{vv}}& =0,\hfill \\ \hfill {\Lambda }_{1_t}-2u{\Lambda }_{2_x}& =0,\hfill & \hfill {\Lambda }_{2_t}-2v{\Lambda }_{2_x}& =0,\hfill & \hfill {\Lambda }_{1_x}& =0,\hfill \\ \hfill {\Lambda }_{1_u}-{\Lambda }_{2_v}& =0,\hfill & \hfill {\Lambda }_{2_u}& =0,\hfill & \hfill {\Lambda }_{1_v}& =0,\hfill \\ \hfill {\Lambda }_{1_{u_x}}& =0,\hfill & \hfill {\Lambda }_{2_{u_x}}& =0,\hfill & \hfill {\Lambda }_{1_{v_x}}& =0,\hfill \\ \hfill {\Lambda }_{2_{v_x}}& =0.\hfill \end{array}$$

(19)

The system of determining equations (19) is solved and we obtain

$$\begin{array}{cc}\hfill & {\Lambda }_1=u(2C_{1}t+C_2)+C_4,\hfill \\ \hfill & {\Lambda }_2=C_1(2tv+x)+C_{2}v+C_3,\hfill \end{array}$$

(20)

where $C_i,i=1,2,\dots ,4$ are arbitrary constants. The multipliers of the KB system satisfy formula (10), i.e,

$${\Lambda }_1(v_t-u_x-2v{v}_x)+{\Lambda }_2(u_t-v_{xxx}-2v{u}_x-2u{v}_x)=D_t{T}^t+D_x{T}^x,$$

(21)

for all functions $u(t,x)$ and $v(t,x)$. From (20) and (21), we obtain four conserved vectors for (1)

$$\begin{array}{ccc}\hfill T_1& :& \left\{\begin{array}{cc}\hfill T_1^t& {\displaystyle =2tuv+ux,}\hfill \\ \hfill T_1^x& {\displaystyle =-t{u}^2-4tu{v}^2-2tv{v}_{xx}+t{v}_x^2-2uvx+v_x-x{v}_{xx},}\hfill \end{array}\right.\hfill \end{array}$$

(22)

$$\begin{array}{ccc}\hfill T_2& :& \left\{\begin{array}{cc}\hfill T_2^t& {\displaystyle =uv,}\hfill \\ \hfill T_2^x& {\displaystyle =-\frac{u^2}{2}-2u{v}^2-v{v}_{xx}+\frac{v_x^2}{2},}\hfill \end{array}\right.\hfill \end{array}$$

(23)

$$\begin{array}{ccc}\hfill T_3& :& \left\{\begin{array}{cc}\hfill T_3^t& {\displaystyle =u,}\hfill \\ \hfill T_3^x& {\displaystyle =-2uv-v_{xx},}\hfill \end{array}\right.\hfill \end{array}$$

(24)

$$\begin{array}{ccc}\hfill T_4& :& \left\{\begin{array}{cc}\hfill T_4^t& {\displaystyle =v,}\hfill \\ \hfill T_4^x& {\displaystyle =-u-v^2.}\hfill \end{array}\right.\hfill \end{array}$$

(25)

3. Double Reduction of K-B System (1)

We are now applying the double reduction theorem based on Lie symmetries and conservation laws of (1) to find the reductions and exact solutions. For two independent variables t and x, the formula (13) yields

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

(26)

Using (26), we establish association of the symmetries (15) and the conserved vectors (22) - (25). The results are presented in the Table 1.

3.1. Reduction of (1) Using $〈X_3〉$

The generator $X_3$ takes canonical form $X=\partial /\partial s$ when

$${\displaystyle \frac{dt}{0}}={\displaystyle \frac{dx}{-2t}}={\displaystyle \frac{du}{0}}={\displaystyle \frac{dv}{1}}={\displaystyle \frac{dr}{0}}={\displaystyle \frac{ds}{1}}={\displaystyle \frac{dw}{0}}={\displaystyle \frac{dq}{0}},$$

(27)

and from (27) we get the the canonical coordinates

$$r=t,s=-{\displaystyle \frac{x}{2t}},w=u,q={\displaystyle \frac{2tv+x}{2t}},$$

(28)

where $w=w\left(r\right)$ and $q=q\left(r\right)$. From (28), the inverse canonical coordinates are given by

$$t=r\gamma ,x=-2rs,u=w,v=q+s.$$

(29)

The partial derivatives of v from (29) are

$$\begin{array}{cccc}\hfill v_x& =-{\displaystyle \frac{1}{2r}}\hfill & \hfill v_{xx}& =0.\hfill \end{array}$$

(30)

For two independent variables t and x, the formula (14) reduces to the conserved form

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

(31)

where

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

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

and $J=\det \left(A\right)=2r$. Thus, the conserved vectors $T_1$ and $T_3$ reduce to

$$\begin{array}{ccc}\hfill {\tilde{T}}_1& :& \left\{\begin{array}{cc}\hfill T_1^r& {\displaystyle =-4q{r}^{2}w,}\hfill \\ \hfill T_1^s& {\displaystyle =-rw\left(4q^2+w\right)-\frac{1}{4r},}\hfill \end{array}\right.\hfill \end{array}$$

(32)

$$\begin{array}{ccc}\hfill {\tilde{T}}_3& :& \left\{\begin{array}{cc}\hfill T_3^r& {\displaystyle =-2rw,}\hfill \\ \hfill T_3^s& {\displaystyle =-2qw,}\hfill \end{array}\right.\hfill \end{array}$$

(33)

and the conserved vectors ${\tilde{T}}_1$ and ${\tilde{T}}_3$ satisfy the reduced conserved form

$$D_r{T}_i^r=0,i=1,3.$$

Consequently, we obtain

$$\begin{array}{ccc}\hfill q{r}^{2}w& =& {\kappa }_1,\hfill \end{array}$$

(34)

$$\begin{array}{ccc}\hfill rw& =& {\kappa }_2,\hfill \end{array}$$

(35)

where ${\kappa }_1$ and ${\kappa }_2$ are arbitrary constants. Solving (34) and (35) simultaneously for w and p and using (28) leads to a solution

$$\begin{array}{ccc}\hfill u(t,x)& =& {\displaystyle \frac{{\kappa }_2}{t}},\hfill \end{array}$$

(36)

$$\begin{array}{ccc}\hfill v(t,x)& =& {\displaystyle \frac{2{\kappa }_1-{\kappa }_{2}x}{2{\kappa }_{2}t}},\hfill \end{array}$$

(37)

for the system (1).

3.2. Reduction of (1) Using $〈X_1+\gamma X_2〉$

According to Table 1, the symmetries $X_1$ and $X_2$ are both associated with the conserved vectors $T_2,T_3$, and $T_4$. For the reduction process involving $X_1$ and $X_2$, we will use the linear combination $X_1+\kappa X_2$, where $\kappa$ is a parameter. The generator $X_1+\kappa X_2$ has the canonical form $X=\partial /\partial s$ when

$${\displaystyle \frac{dt}{\gamma }}={\displaystyle \frac{dx}{1}}={\displaystyle \frac{du}{0}}={\displaystyle \frac{dv}{0}}={\displaystyle \frac{dr}{0}}={\displaystyle \frac{ds}{1}}={\displaystyle \frac{dw}{0}}={\displaystyle \frac{dq}{0}},$$

(38)

which results in canonical coordinates

$$r={\displaystyle \frac{\gamma x-t}{\gamma }},s={\displaystyle \frac{t}{\gamma }},w=u,q=v,\gamma \ne 0$$

(39)

where $w=w\left(r\right)$ and $q=q\left(r\right)$. From (39), the inverse canonical coordinates are given by

$$t=s\gamma ,x=r+s,u=w,v=q.$$

(40)

The partial derivatives of v from (40) are

$$\begin{array}{cccc}\hfill v_x& =q_r\hfill & \hfill v_{xx}& =q_{rr}.\hfill \end{array}$$

(41)

Again, using formula (14) the reduced conserved form is given by

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

(42)

where

$$A=\left(\begin{array}{cc}{D}_{r}t\hfill & D_{r}x\hfill \\ D_{s}t\hfill & D_{s}x\hfill \end{array}\right)=\left(\begin{array}{cc}0& 1\\ \gamma & 1\end{array}\right),$$

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

and $J=\det \left(A\right)=-\gamma$.

This reduces the conserved vectors $T_2$, $T_3$, and $T_4$ to

$$\begin{array}{ccc}\hfill {\tilde{T}}_2& :& \left\{\begin{array}{cc}\hfill T_2^r& {\displaystyle =2\gamma q^{2}w+q(\gamma q_{rr}+w)+\frac{1}{2}\gamma \left(w^2-q_r^2\right),}\hfill \\ \hfill T_2^s& {\displaystyle =-qw,}\hfill \end{array}\right.\hfill \end{array}$$

(43)

$$\begin{array}{ccc}\hfill {\tilde{T}}_3& :& \left\{\begin{array}{cc}\hfill T_3^r& {\displaystyle =2\gamma qw+\gamma q_{rr}+w,}\hfill \\ \hfill T_3^s& {\displaystyle =-w,}\hfill \end{array}\right.\hfill \end{array}$$

(44)

$$\begin{array}{ccc}\hfill {\tilde{T}}_4& :& \left\{\begin{array}{cc}\hfill T_4^r& {\displaystyle =\gamma q^2+q+\gamma w,}\hfill \\ \hfill T_4^s& {\displaystyle =-q,}\hfill \end{array}\right.\hfill \end{array}$$

(45)

where the conserved vectors ${\tilde{T}}_2$, ${\tilde{T}}_3$, and ${\tilde{T}}_4$ satisfy the reduced conserved form

$$D_r{T}_i^r=0,i=2,3,4.$$

Therefore, this leads to the system of equations

$$\begin{array}{cc}\hfill 2\gamma q^{2}w+q(\gamma q_{rr}+w)+{\displaystyle \frac{1}{2}}\gamma \left(w^2-q_r^2\right)=& {\kappa }_1,\hfill \\ \hfill 2\gamma qw+\gamma q_{rr}+w=& {\kappa }_2,\hfill \\ \hfill \gamma q^2+q+\gamma w=& {\kappa }_3,\hfill \end{array}$$

(46)

where ${\kappa }_1$, ${\kappa }_2$, and ${\kappa }_3$ are arbitrary constants.

Any pair of equations (46) can be solved to obtain a solutions of the K-B system (1). However, directly solving the system (46) can pose significant difficulties. Nevertheless, a helpful simplification arises by setting ${\kappa }_3=0$ and solving for w in terms of q and r in the third equation of (46) to get

$$w\left(r\right)={\displaystyle \frac{-\gamma q^2-q}{\gamma }}.$$

(47)

Substituting (47) in the first equation of (46), we get the second-order ODE

$$2\gamma q{q}_{rr}-\gamma q_r^2-3\gamma q^4-{\displaystyle \frac{q^2}{\gamma }}-4q^3={\kappa }_1.$$

(48)

Similarly, substituting (47) into the second equation of (46) results in the second-order ODE

$$\gamma q_{rr}-2\gamma q^3-{\displaystyle \frac{q}{\gamma }}-3q^2={\kappa }_2.$$

(49)

Now, expressing (49) as

$$q_{rr}={\displaystyle \frac{\gamma {\kappa }_2+2{\gamma }^2{q}^3+3\gamma q^2+q}{{\gamma }^2}}$$

(50)

and replace $q_{rr}$ in (48) by the right-hand side of (50) we obtain the first-order ODE

$${\kappa }_1-\gamma q_r^2+\gamma q^4+{\displaystyle \frac{q^2}{\gamma }}+2{\kappa }_{2}q+2q^3=0.$$

(51)

By observing that (51) admits the translational symmetry $\Gamma =\partial /\partial r$, we apply the method of canonical variables [39] to solve (51) in the case ${\kappa }_1=0={\kappa }_2$. This yields the solution

$$q\left(r\right)={\displaystyle \frac{e^{r/\gamma }}{M-\gamma e^{r/\gamma }}},$$

(52)

where $M=e^{J/\gamma }$ and J is the constant of integration.

Finally, the equations (52) and (47), using (39) lead to the following solution for the K-B system (1)

$$\begin{array}{ccc}\hfill u(t,x)& =& -{\displaystyle \frac{M{e}^{(t+\gamma x)/{\gamma }^2}}{\gamma {\left(M{e}^{t/{\gamma }^2}-\gamma e^{x/\gamma }\right)}^2}},\hfill \end{array}$$

(53)

$$\begin{array}{ccc}\hfill v(t,x)& =& {\displaystyle \frac{e^{x/\gamma }}{M{e}^{t/{\gamma }^2}-\gamma e^{x/\gamma }}},\hfill \end{array}$$

(54)

arising from $X_1+\gamma X_2$ via $T_2$, $T_3$, and $T_4$.

3.3. Reduction of (1) Using $〈X_4〉$

The generator $X_4$ takes canonical form $X=\partial /\partial s$ when

$${\displaystyle \frac{dt}{-2t}}={\displaystyle \frac{dx}{-x}}={\displaystyle \frac{du}{2u}}={\displaystyle \frac{dv}{v}}={\displaystyle \frac{dr}{0}}={\displaystyle \frac{ds}{1}}={\displaystyle \frac{dw}{0}}={\displaystyle \frac{dq}{0}},$$

(55)

and from (55) we get the the canonical coordinates

$$r={\displaystyle \frac{x}{\sqrt{t}}},s=-{\displaystyle \frac{log\left(t\right)}{2}},w=tu,q=\sqrt{t}v,$$

(56)

where $w=w\left(r\right)$ and $q=q\left(r\right)$. From (56), the inverse canonical coordinates are given by

$$t=e^{-2s},x=e^{-s},u=e^{2s}w,v=-e^{s}q.$$

(57)

The partial derivatives of v from (57) are

$$\begin{array}{cccc}\hfill v_x& =-e^{2s}{q}_r\hfill & \hfill v_{xx}& =-e^{3s}{q}_{rr}.\hfill \end{array}$$

(58)

Again, the formula (14) reduces to the conserved form

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

(59)

where

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

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

and $J=\det \left(A\right)=2e^{-3s}$.

The conserved vectors $T_1$ and $T_4$ so reduce to

$$\begin{array}{ccc}\hfill {\tilde{T}}_1& :& \left\{\begin{array}{cc}\hfill T_1^r& {\displaystyle =-8q^{2}w-4q{q}_{rr}+6qrw+2q_r^2-2q_r+2r{q}_{rr}-r^{2}w-2w^2,}\hfill \\ \hfill T_1^s& {\displaystyle =w(2q-r),}\hfill \end{array}\right.\hfill \end{array}$$

(60)

$$\begin{array}{ccc}\hfill {\tilde{T}}_4& :& \left\{\begin{array}{cc}\hfill T_4^r& {\displaystyle =-2q^2+qr-2w,}\hfill \\ \hfill T_4^s& {\displaystyle =q,}\hfill \end{array}\right.\hfill \end{array}$$

(61)

where the conserved vectors ${\tilde{T}}_1$ and ${\tilde{T}}_4$ satisfy the reduced conserved form

$$D_r{T}_i^r=0,i=1,4.$$

Thus, we obtain the system of equations

$$\begin{array}{cc}\hfill -8q^{2}w-4q{q}_{rr}+6qrw+2q_r^2-2q_r+2r{q}_{rr}-r^{2}w-2w^2& ={\kappa }_1,\hfill \\ \hfill -2q^2+qr-2w& ={\kappa }_2,\hfill \end{array}$$

(62)

where ${\kappa }_1$ and ${\kappa }_2$ are arbitrary constants. Solving the system (62) directly can be quite challenging. However, a useful simplification is to solve for w in terms of q and r from the second equation of (62), and setting ${\kappa }_2=0$, to obtain

$$w\left(r\right)={\displaystyle \frac{1}{2}}qr-q^2.$$

(63)

Substituting the expression for w from (63) into the first equation of (62) yields a second-order ODE

$$4r{q}_{rr}-8q{q}_{rr}+4q_r^2-4q_r-r^{3}q+7r^2{q}^2+12q^4-16r{q}^3={\kappa }_1.$$

(64)

Thus, the solution of the K-B system arising from $X_4$ via $T_1$ and $T_4$ is given by

$$\begin{array}{cc}\hfill & u(t,x)={\displaystyle \frac{1}{t}}w\left(r\right),\hfill \\ \hfill & v(t,x)={\displaystyle \frac{1}{\sqrt{t}}}q\left(r\right),\hfill \end{array}$$

(65)

where q is the solution of the ODE (64), w is given by (63), and $r=x/\sqrt{t}$.

4. Concluding Remarks

This paper illustrates the application of the generalized double reduction method by analyzing the (1+1)-dimensional Kaup-Boussinesq (K-B) system, which serves as a model for nonlinear wave propagation. This method takes advantage of the association of Lie point symmetries and conservation laws, and has proven to be an effective tool for simplifying and finding solutions to nonlinear systems. In this study, we began by computing Lie point symmetries for the K-B system and constructed four non-trivial conservation laws using the multiplier method. Using the generalized double reduction method, we successfully reduced the K-B system to second-order ordinary differential equations and algebraic equations. This reduction yielded two distinct exact solutions: one derived from the algebraic equations and another from the second-order ODEs, where further exploitation of Lie point symmetries allowed us to utilize the method of canonical variables for the solution process. The generalized double reduction method is versatile and can be applied to other nonlinear systems that are rich in symmetries and conservation laws, in which case exact solutions may be discovered. Future research will explore further applications of this methods to different PDE models, potentially contributing to advancements in the study of nonlinear wave equations and other related phenomena.