Applying the Lie symmetry analysis, we can get the symmetry of system (2):
$$ X_{1}=\frac{\partial }{\partial t},\qquad X_{2}=\frac{\partial }{ \partial x},\qquad X_{3}= \biggl( \frac{1}{2}x+\beta t \biggr)\frac{ \partial }{\partial x}+t\frac{\partial }{\partial t}- \frac{1}{2}v\frac{ \partial }{\partial v}-w\frac{\partial }{\partial w}. $$
(4)
The vector field \(X_{3}\) has the following characteristic equation:
$$ \frac{dt}{t}=\frac{dx}{\frac{1}{2}x+\beta t}= \frac{dv}{-\frac{1}{2}v}= \frac{dw}{-w}, $$
(5)
which gives rise to
$$ \biggl(\beta t+\frac{1}{2}x\biggr)\,dt-t\,dx=0. $$
(6)
One integration factor of (6) is given by
$$ \mu =e^{-\int \frac{3}{2t}\,dt}=t^{-\frac{3}{2}}, $$
which transfers (6) to the complete integration equation
$$ \beta t^{-\frac{1}{2}}\,dt+d \bigl(-t^{-\frac{1}{2}}x \bigr)=0, $$
from which we have the invariant variable \(\xi =2\beta t^{\frac{1}{2}}-t ^{-\frac{1}{2}}x\). In terms of Eq. (5), we have the formal invariants
$$ v=t^{-\frac{1}{2}}f(\xi ),\qquad w=t^{-1}g(\xi ), $$
(7)
where \(f(\xi )\) and \(g(\xi )\) are arbitrary smooth functions of ξ. Substituting (7) into system (2) yields the ordinary differential system
$$ \textstyle\begin{cases} -\frac{1}{2}f(\xi )+\frac{1}{2}xt^{-\frac{1}{2}}f'(\xi )=\frac{\alpha }{2}[f''(\xi )+2f(\xi )f'(\xi )+2g'(\xi )], \\ -g(\xi )+\frac{1}{2}xt^{-\frac{1}{2}}g'(\xi )=-\frac{\alpha }{2}[g''( \xi )-2g'(\xi )f(\xi )-2g(\xi )f'(\xi )]. \end{cases} $$
(8)
Let \(\beta =0\), Then system (8) reduces to
$$ \textstyle\begin{cases} f(\tau )+\tau f'(\tau )=-\alpha [f''(\tau )+2f(\tau )f'(\tau )+2g'( \tau )], \\ g(\tau )+\frac{1}{2}\tau g'(\tau )=\frac{\alpha }{2}[g''(\tau )-2g'( \tau )f(\tau )-2g(\tau )f'(\tau )], \end{cases} $$
(9)
where \(\tau =-t^{-\frac{1}{2}}x\), which is a reduction of ξ. In facr, system (9) is an ordinary differential system corresponding to the BK system (1).
The group-invariant transformations of the generalized BK system are as follows:
$$ \textstyle\begin{cases} g_{1}:(x,t,v,w)\rightarrow (x,t+\epsilon ,v,w), \\ g_{2}:(x,t,v,w)\rightarrow (x+\epsilon ,t,v,w), \\ g_{3}:(x,t,v,w)\rightarrow (2\beta te^{\epsilon }+(x-2\beta t)e^{ \frac{1}{2}\epsilon },te^{\epsilon },e^{-\frac{1}{2}\epsilon }v,we ^{-\epsilon }). \end{cases} $$
(10)
In what follows, we consider solutions to the BK system. Set \(v=V(\rho )\), \(w=W(\rho )\), \(\rho =x+lt\). Then (2) becomes
$$ \textstyle\begin{cases} lV'=\frac{\alpha }{2}(V''-2VV'-2W')-\beta V', \\ lW'=-\frac{\alpha }{2}(W''+2W'V+2WV')-\beta W', \end{cases} $$
from which we have
$$ \textstyle\begin{cases} lV-\frac{\alpha }{2}(V'-V^{2}-2W)+\beta V=c_{1}, \\ lW+\frac{\alpha }{2}(W'+2WV)-\beta W'=c_{2}. \end{cases} $$
(11)
A special solution to (11) is given by
$$ V=\frac{1}{c+\xi },\qquad W=\frac{1}{(c+\xi )^{2}} $$
(12)
in the case of \(l=-\beta \), \(c_{1}=c_{2}=0\). Hence we get a set of solutions to the generalized BK system (2):
$$ v=\frac{1}{c+x-\beta t},\qquad w=\frac{1}{(c+x-\beta t)^{2}}. $$
(13)
Applying the group-invariant transformation (10), we can deduce some other new solutions to system (2):
$$ \textstyle\begin{cases} g_{1}:\quad v=\frac{1}{c+x-\beta (t+\epsilon )},\qquad w=\frac{1}{[c+x- \beta (t+\epsilon )]^{2}}, \\ g_{2}:\quad v=\frac{1}{c+x-\beta t+\epsilon },\qquad w=\frac{1}{(c+x-\beta t+ \epsilon )^{2}}, \\ g_{3}:\quad v=\frac{e^{-\frac{1}{2}\epsilon }}{c+\beta te^{\epsilon }+(x-2 \beta t)e^{\frac{1}{2}\epsilon }},\qquad w=\frac{e^{-\epsilon }}{[c+ \beta te^{\epsilon }+(x-2\beta t)e^{\frac{1}{2}\epsilon }]^{2}}. \end{cases} $$
Taking \(\beta =0\), we can obtain group-invariant solutions to the BK system (1). In particular, we can get the series solutions to the BK system. Indeed, let
$$ f(\tau )=\sum_{n=0}^{\infty }c_{n} \tau ^{n},\qquad g(\tau )=\sum_{m=0} ^{\infty }c_{m}\tau ^{m}, $$
(14)
and substituting into (9), we have that
$$\begin{aligned}& \begin{gathered} c_{0}+\sum_{n=1}^{\infty }c_{n} \tau ^{n}+\tau c_{1}+\sum_{n=1}^{ \infty }(n+1)c_{n+1} \tau ^{n+1}\\ \quad =-\alpha \Biggl[2c_{2}+\sum _{n=1}^{\infty }(n+2) (n+1)c _{n+2}\tau ^{n}\Biggr] +2\Biggl(c_{0}+\sum _{n=1}^{\infty }c_{n}\tau ^{n}\Biggr) \Biggl(c_{1}+ \sum _{n=1}^{\infty }(n+1)c_{n+1}\tau ^{n}\Biggr)\\ \qquad {}+2d_{1} +2\sum_{m=1}^{\infty }(m+1)d_{m+1} \tau ^{m}, \end{gathered} \\& d_{0}+\sum_{m=1}^{\infty }d_{m} \tau ^{m}+\frac{1}{2}\tau d_{1}+ \frac{1}{2}\sum_{m=1}^{\infty }(m+1)d_{m+1} \tau ^{m+1} \\& \quad = \frac{\alpha }{2}[2d_{2}+\sum _{m=1}^{\infty }(m+2) (m+1)d_{m+2}\tau ^{m}-2\Biggl(c_{0} +\sum_{n=1}^{\infty }c_{n} \tau ^{n}\Biggr) \Biggl(d_{1}+\sum _{m=1}^{ \infty }(m+1)d_{m+1}\tau ^{m}\Biggr)\\& \qquad {}-2\Biggl(d_{0}+\sum _{m=1}^{\infty }d_{m}\tau ^{m}\Biggr) \Biggl(c_{1}+\sum _{n=1}^{\infty }(n+1)c_{n+1}\tau ^{n}\Biggr), \end{aligned}$$
from which we infer that
$$\begin{aligned}& c_{2}=-c_{0}c_{1}-\frac{1}{2\alpha }c_{0}-d_{1}, \\& d_{2}=c_{0}d_{1}+d_{0}c_{1}+ \frac{1}{\alpha }d_{0}, \\& c_{3}=-\frac{1}{3\alpha }c_{1}- \frac{2}{3}c_{0}c_{2}-\frac{1}{6}c _{1}^{2}-\frac{2}{3}d_{1}, \\& d_{3}=\frac{1}{2\alpha }d_{1}+\frac{2}{3}c_{0}d_{2}- \frac{1}{3}d_{1}c _{1}+\frac{2}{3}d_{0}c_{2}, \\& \cdots \cdots , \\& c_{n+2}=\frac{1}{\alpha (n+1)(n+2)}\Biggl[-c_{n}-2\alpha c_{0}(n+1)c_{n+1}-2 \alpha c_{1}c_{n}\\& \hphantom{c_{n+2}=}{}- \alpha \sum_{i,j=2}^{n}c_{i}c_{j+1}(j+1) \tau ^{i+j}-2 \alpha (n+1)d_{n+1}\Biggr], \\& d_{n+2}=\frac{1}{(n+1)(n+2)}\Biggl[\frac{2}{\alpha }d_{n}+2(n+1)c_{0}d_{n+1} +2d_{1}c_{n}+2\sum_{i,j=2}^{n}c_{i}d_{j+1} \tau ^{i+j}\\& \hphantom{d_{n+2}=}{}+2(n+1)d_{0}c_{n+1}+2c _{1}d_{n} +2\sum_{i,j=2}^{n}(j+1)d_{i}c_{j+1} \tau ^{i+j}\Biggr], \end{aligned}$$
where \(c_{0}\), \(d_{0}\), \(c_{1}\), \(d_{1}\) are arbitrary parameters. Inserting these expressions into (14), we get the series solutions of the BK system. The second equation of system (9) can be reduced to
$$ g''(\tau )-\frac{1}{\alpha } \tau g'(\tau )-\frac{2}{\alpha }g(\tau )=0 $$
(15)
under the condition
$$ (fg)'=0\quad \Rightarrow\quad fg=c. $$
(16)
As long as the solution of (15) is obtained, we can get the solution \(f(\tau )\) from (16). If \(g_{1}(\tau )\) is the known solution of (15), then we assume that \(g(\tau )=u(\tau )g_{1}(\tau )\). If \(u(\tau )\) is known, then the solution \(g(\tau )\) to Eq. (15) can be presented. It is easy to see that
$$ g''(\tau )=g_{1}( \tau )u''(\tau )+2u'(\tau )g_{1}'(\tau )+u(\tau )g _{1}''( \tau ). $$
(17)
Substituting (17) into Eq. (15) yields
$$\begin{aligned}& g_{1}(\tau )u''(\tau )+ \biggl(2g_{1}'(\tau )-\frac{1}{\alpha } \tau g_{1}(\tau ) \biggr)u'(\tau )+ \biggl(g_{1}''( \tau )-\frac{1}{ \alpha }\tau g_{1}'(\tau )- \frac{2}{\alpha } g_{1}(\tau ) \biggr)u( \tau )=0. \end{aligned}$$
Since
$$ g_{1}''(\tau )-\frac{1}{\alpha } \tau g_{1}'(\tau )-\frac{2}{\alpha }g _{1}(\tau )=0, $$
we have
$$ g_{1}(\tau )u''(\tau )+ \biggl(2g_{1}'(\tau )-\frac{1}{\alpha }\tau g _{1}(\tau ) \biggr)u'(\tau )=0. $$
(18)
Assume that \(u'(\tau )=z(\tau )\). Then Eq. (18) becomes
$$ g_{1}(\tau )z'(\tau )+ \biggl(2g_{1}'( \tau )-\frac{1}{\alpha }\tau g _{1}(\tau ) \biggr)z(\tau )=0, $$
which has the solution
$$ z=\frac{c}{g_{1}^{2}(\tau )}e^{\int \frac{1}{\alpha }\tau\, d\tau }=\frac{c}{g _{1}^{2}(\tau )}e^{\frac{1}{2\alpha }\tau ^{2}}, $$
where c is a constant. Thus we have
$$\begin{aligned}& \begin{gathered} u(\tau )=c \int^{\tau }\frac{1}{g_{1}^{2}(\tau )}e^{\frac{\tau ^{2}}{2 \alpha }}\,d \tau +\bar{c}, \\ g(\tau )=g_{1}(\tau ) \biggl[c \int^{\tau } \frac{1}{g_{1}^{2}(\tau )}e^{\frac{\tau ^{2}}{2\alpha }}\,d \tau +\bar{c} \biggr]. \end{gathered} \end{aligned}$$
(19)
Substituting (19) into Eq. (16), we can get \(f(\tau )\). Thus a type of special solutions to system (9) can be obtained.