1 Introduction

Fujimoto–Watanabe, in a remarkable paper [1] decades ago, derived a complete list of third-order polynomial evolution equations of not normal type [2] with nontrivial Lie–Bäcklund symmetries. It is possible to construct chains of differential substitutions which connect some of the equations to the KdV, Sawada–Kotera and Kaup equations [3]. Most of the equations possess a recursion operator and the Lie algebras are commutative, namely the four equations of interest are

$$\begin{aligned} u_t= & {} u_x^3 u_{xxx}+\alpha u_x^4, \end{aligned}$$
(1)
$$\begin{aligned} u_t= & {} u^3 u_{xxx}+3u^2 u_x u_{xx}+\alpha (u^3 u_{xx}+u^2 u_x^2)+\frac{2}{9}\alpha ^2 u^3 u_x, \end{aligned}$$
(2)
$$\begin{aligned} u_t= & {} u^3 u_{xxx}+3 u^2 u_{x} u_{xx}+4\alpha u^3 u_x, \end{aligned}$$
(3)
$$\begin{aligned} u_t= & {} u^3 u_{xxx}+\frac{3}{2}u^2 u_x u_{xx}+\alpha (u^3 u_{xx}+u^2 u_x^2)+\frac{2}{9}\alpha ^2 u^3 u_x, \end{aligned}$$
(4)

where \(\alpha \ne 0\) is an arbitrary constant.

In the literature, some Fujimoto–Watanabe equations have been studied, but not necessarily the ones listed above. Liu [4] used a complete discrimination system to obtain the classifications of traveling wave solutions of a Fujimoto–Watanabe equation. By the method of dynamical systems, the bifurcations and dynamics of traveling wave solutions of a Fujimoto–Watanabe equation were analysed [5, 6], and also phase portraits and periodic solutions were described. In [7], a Fujimoto–Watanabe equation (not listed above) was studied from the perspective of Lie symmetries.

Symmetry methods and conservation laws are deeply intertwined in the analysis of partial differential equations [8,9,10]. The theory relating to symmetries is well known and is best studied from the book [11]. In this paper, we explore how symmetry methods combine with power series to arrive at solutions. The resulting equations obtained through Lie reductions are difficult to solve, and most other techniques will not work. Power series are a useful tool to find approximate solutions, and the solutions may be made exact by finding a recurrence relation of the series.

The plan of the paper follows. In Sect. 2, we present the symmetry details possessed by the four equations, together with the commutator, adjoint and group properties. Section 3, contains the reductions and power series solutions, while the next section demonstrates the convergence of a solution. The conserved vector components are given in Sect. 5, and a conclusion follows in Sect. 6.

2 Symmetries and groups

First of all, let us consider a one-parameter Lie group of infinitesimal transformations:

$$\begin{aligned} x\rightarrow x+\varepsilon \xi (x,t,u), \end{aligned}$$
(5)
$$\begin{aligned} t\rightarrow t+\varepsilon \tau (x,t,u), \end{aligned}$$
(6)
$$\begin{aligned} u\rightarrow u+\varepsilon \phi (x,t,u), \end{aligned}$$
(7)

with the infinitesimal \(\varepsilon \ll 1\). The vector field associated with the above transformations can be written as

$$\begin{aligned} X=\xi (x,t,u){\frac{\partial }{\partial {x}}}+\tau (x,t,u){\frac{\partial }{\partial {t}}}+\phi (x,t,u){\frac{\partial }{\partial {u}}}. \end{aligned}$$
(8)

The symmetry group of the equation will be generated by the vector field of the above form. Applying the prolongation to the given equations, we find the coefficient functions \(\xi (x,t,u),~\tau (x,t,u)\) and \(\phi (x,t,u)\) that must satisfy the Lie symmetry condition.

The Lie point symmetries of Eqs. (1)–(4) are

$$\begin{aligned} X_1={\frac{\partial }{\partial {x}}},~X_2={\frac{\partial }{\partial {t}}},~X_3=-3t{\frac{\partial }{\partial {t}}}+{u}{\frac{\partial }{\partial {u}}}, \end{aligned}$$
(9)

with Lie brackets \( [X_i,X_j]=X_iX_j-X_jX_i\), given in Table 1.

Table 1 Lie brackets
Full size table

Thus, we have the corresponding one-parameter groups \(G_i~(i=1,2,3)\) generated by the symmetries \(X_i\) of the four equations, as

$$\begin{aligned} G_1:(x,t,u)\mapsto & {} (x+\varepsilon ,t,u), \end{aligned}$$
(10)
$$\begin{aligned} G_2:(x,t,u)\mapsto & {} (x,t+\varepsilon ,u), \end{aligned}$$
(11)
$$\begin{aligned} G_3:(x,t,u)\mapsto & {} (x,e^{-3\varepsilon }t,e^{\varepsilon }u). \end{aligned}$$
(12)

We can see that \(G_1\) is a space translation, \(G_2\) is a time translation, and \(G_3\) is a scaling transformation.

The adjoints corresponding to (9) are given in Table 2,

Table 2 Adjoint representation of subalgebras
Full size table

where we require the adjoints to calculate the one-dimensional subalgebras. The approach for the one-dimensional optimal system is provided in [12]. That is, take a general element from the Lie algebra and reduce it to its simplest like form by applying carefully chosen adjoint transformations

$$\begin{aligned} \mathrm{Ad(exp}(\epsilon X_i))X_j=X_j-\epsilon [X_i,X_j]+\frac{1}{2}\,\epsilon ^2[X_i,[X_i,X_j]]- \cdots . \end{aligned}$$

Thus, for the class of equations (1)–(4), we obtain the optimal system of one-dimensional subalgebras

$$\begin{aligned} X_1,~X_3+aX_1,~X_2+\epsilon X_1, \end{aligned}$$

where a is a constant and \(\epsilon =0,\pm 1\).

3 Solutions via power series

In this section, we draw on the optimal system of one-dimensional subalgebras to reduce Eqs. (1)–(4) to ordinary differential equations. Some of the subsequent ordinary differential equations are highly nonlinear and may be solved using the power series method [13], or see [14] for an application.

Reducing Eqs. (1)–(4) by the generator \(X_1\) gives

$$\begin{aligned} y'(t)=0, \end{aligned}$$
(13)

where \(u(x,t)=y(t)\). Solving (13) gives

$$\begin{aligned} y(t)= c_1, \end{aligned}$$
(14)

where \(c_1\) is an arbitrary constant. Hence, each of the four equations possess the constant solution \(u(x,t)=c_1\).

3.1 The case of Eq. (1)

Reducing equation (1) by \(X_2+\epsilon X_1\) gives the travelling wave reduction

$$\begin{aligned} -3 p(s)^2 p{'''}(s)+4 \alpha \epsilon ^3 p(s)^3+\epsilon ^4=0, \end{aligned}$$
(15)

where \(u(x,t)=p(s)\) and \(s=\frac{t \epsilon -x}{\epsilon }\).

A reduction of (1) by \(X_3+a X_1\) gives

$$\begin{aligned}{} & {} \Bigg ((3+4a^{3}\alpha )y(z)^{4}-a^{4}y'(z)+81z^{2}y(z)^{2}y^{(1)}(z)(7y'(z)+3z(4y''(z)+z y'''(z))\Bigg )\nonumber \\{} & {} \quad +3z y(z)^{3}\Bigg ((66+4a^{3}\alpha )y'(z)+27z(4y''(z)+z y'''(z))\Bigg )=0, \end{aligned}$$
(16)

where \(u(x,t)=e^{\frac{x}{a}}y(z)\) and \(z=e^{\frac{3x}{a}}t\).

Let us consider the reduced equation (15) with \(\epsilon =1\). A substitution of the power series

$$\begin{aligned} p(s)=\displaystyle {\sum _{r{_1}=0}^{\infty }} a_{r{_1}}s^{r{_1}}, \end{aligned}$$
(17)

with \(a_{r{_1}}\) as unknown constants to be determined, into (15) provides

$$\begin{aligned}{} & {} 1+ 4\alpha \Bigg (a_0^3+3sa_0^2a_1+3s^2(a_0a_1^2+a_0^2a_2)+\displaystyle {\sum _{r{_1}=3}^{\infty }\sum _{r{_2}=0}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}}a_{r{_1}-r{_2}}a_{r{_2}-r{_3}}a_{r{_3}} \Bigg )\nonumber \\{} & {} \quad -3\displaystyle {\sum _{r{_1}=3}^{\infty }\sum _{r{_2}=0}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}} \left( r_1-r_2+3 \right) \left( r_1-r_2+2 \right) \left( r_1-r_2+1 \right) a_{r{_1}-r{_2}+3}a_{r{_2}-r{_3}}a_{r{_3}}=0. \end{aligned}$$
(18)

From (18) comparing coefficients of the power of s, we find that \(a_0,a_1,a_2\) are arbitrary,

$$\begin{aligned} a_{3}=\frac{ 4 \alpha a_0^3+1}{18 a_0^2},\quad a_{4}= \frac{\alpha a_0 a_1-\frac{a_1 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}}{6 a_0}, \end{aligned}$$
(19)

and

$$\begin{aligned} a_{5}= \frac{2 \alpha a_2 a_0^2+2 \alpha a_1^2 a_0-4 a_1 \left( \alpha a_0 a_1-\frac{a_1 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}\right) -\frac{a_2 \left( 4 \alpha a_0^3+1\right) }{3 a_0}-\frac{a_1^2 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}}{30 a_0^2}. \end{aligned}$$
(20)

In general for \(r{_1}\ge 3\), we have

$$\begin{aligned} a_{r{_1}+3}{} & {} = \frac{-3\displaystyle {\sum \nolimits _{r{_2}=1}^{r{_1}}\sum \nolimits _{r{_3}=0}^{r{_2}}}(r{_1}-r{_2}+3)(r{_1}-r{_2}+2)(r{_1}-r{_2}+1)a_{r{_1}-r{_2}+3}a_{r{_2}-r{_3}}a_{r{_3}}}{3(r{_1}+3)(r{_1}+2)(r{_1}+1)a_0^2}\nonumber \\{} & {} \quad +\frac{4\alpha \displaystyle {\sum \nolimits _{r{_2}=0}^{r{_1}}\sum \nolimits _{r{_3}=0}^{r{_2}}}a_{r{_1}-r{_2}}a_{r{_2}-r{_3}}a_{r{_3}} }{3(r{_1}+3)(r{_1}+2)(r{_1}+1)a_0^2}. \end{aligned}$$
(21)

Therefore, from (19)–(21), for arbitrary constants \(a_{0}\ne 0, a_1, a_2\) we obtain the power series solution for equation (15), viz.

$$\begin{aligned} p(s)= & {} a_0+ a_1 s+ a_2 s^2+\frac{s^3 \left( 4 \alpha a_0^3+1\right) }{18 a_0^2}+ \frac{s^4 \left( \alpha a_0 a_1-\frac{a_1 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}\right) }{6 a_0}\nonumber \\{} & {} +\frac{s^5}{30 a_0^2} \Bigg (2 \alpha a_2 a_0^2+2 \alpha a_1^2 a_0-4 a_1 \left( \alpha a_0 a_1-\frac{a_1 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}\right) -\frac{a_2 \left( 4 \alpha a_0^3+1\right) }{3 a_0}\nonumber \\{} & {} -\frac{a_1^2 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}\Bigg )\nonumber \\{} & {} +\ldots \end{aligned}$$
(22)

An inversion of the transformation gives the solution to the PDE (1), namely

$$\begin{aligned} u(x,t)= & {} a_0+ a_1 (t-x)+ a_2 (t-x)^2+\frac{(t-x)^3 \left( 4 \alpha a_0^3+1\right) }{18 a_0^2}\nonumber \\ {}{} & {} \quad +\frac{(t-x)^4 \left( \alpha a_0 a_1-\frac{a_1 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}\right) }{6 a_0} +\frac{(t-x)^5}{30 a_0^2} \Bigg (2 \alpha a_2 a_0^2+2 \alpha a_1^2 a_0\nonumber \\ {}{} & {} \quad -4 a_1 \left( \alpha a_0 a_1-\frac{a_1 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}\right) -\frac{a_2 \left( 4 \alpha a_0^3+1\right) }{3 a_0}\nonumber \\ {}{} & {} \quad -\frac{a_1^2 \left( 4 \alpha a_0^3+1\right) }{6 a_0^2}\Bigg )+\ldots \end{aligned}$$
(23)

Similarly, and very tediously, we can construct power series solutions to all reduced equations in this paper. We omit the details for all other cases. The proof of convergence of the solution is discussed later.

A solution to (16) is found from substituting

$$\begin{aligned} y(z)=\sum _{{r_1}=0}^{\infty } a_{r_1} z^{r_1}, \end{aligned}$$

into (16) and similar to the above discussion, we get

$$\begin{aligned} y(z)= & {} a_0+a_1 z+ 7 z^2 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) \nonumber \\{} & {} \quad +\frac{1}{3} z^3 \Big (280 \alpha a_0^3 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) +7392 a_0^3 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) \nonumber \\{} & {} \quad +60 \alpha a_1^2 a_0^2+1179 a_1^2 a_0^2\Big )\frac{1}{2} z^4 \Big (\frac{26}{3} \alpha a_0^3 \big (280 \alpha a_0^3 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) \nonumber \\{} & {} \quad +7392 a_0^3 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) +60 \alpha a_1^2 a_0^2+1179 a_1^2 a_0^2\big )\nonumber \\{} & {} \quad +506 a_0^3 \big (280 \alpha a_0^3 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) \nonumber \\{} & {} \quad +7392 a_0^3 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) +60 \alpha a_1^2 a_0^2+1179 a_1^2 a_0^2\big )\nonumber \\{} & {} \quad +546 \alpha a_1 a_0^2 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) \nonumber \\{} & {} \quad +27909 a_1 a_0^2 \left( 2 \alpha a_1 a_0^3+15 a_1 a_0^3\right) +26 \alpha a_1^3 a_0+870 a_1^3 a_0\Big )\nonumber \\{} & {} \quad +\ldots \end{aligned}$$
(24)

3.2 The case of Eq. (2)

In the same vein, reducing equation (2) by \(X_2+\epsilon X_1\) gives

$$\begin{aligned} p'(s)(9\epsilon ^{3}+2\alpha (9+\alpha )\epsilon ^{2}p(s)^{3}+27p(s)^{2} p''(s))+9p(s)^3 (-\alpha \epsilon p''(s)+p'''(s))=0, \end{aligned}$$
(25)

while \(X_3+a X_1\) leads to

$$\begin{aligned}{} & {} -e^{\frac{4x}{a}}\big ((36+9 a\alpha +2a^{2}\alpha (9+\alpha )\big )y(z)^{4}-a^{3}y'(z)\nonumber \\{} & {} \quad +243 z^{2}y(z)^{2}y'(z)(5y'(z)+3z y''(z))\nonumber \\{} & {} \quad +3z y(z)^{3}\Bigg ((351+45a\alpha +2a^{2}\alpha (9+\alpha )y'(z)\nonumber \\{} & {} \quad +27z((15+a\alpha )y''(z)+3z y'''(z))\Bigg )=0. \end{aligned}$$
(26)

For this case, we find the solutions

$$\begin{aligned} p(s)= & {} a_0+ a_1 s+ a_2 s^2\nonumber \\ {}{} & {} +\frac{s^3 \left( -2 \alpha ^2 a_1 a_0^3-18 \alpha a_1 a_0^3+18 \alpha a_2 a_0^3-54 a_1 a_2 a_0^2-9 a_1\right) }{54 a_0^3}\nonumber \\ {}{} & {} -\frac{s^4}{108 a_0^3} \Bigg (2 \alpha ^2 a_2 a_0^3+3 \alpha ^2 a_1^2 a_0^2-\frac{1}{2} \alpha \Big (-2 \alpha ^2 a_1 a_0^3-18 \alpha a_1 a_0^3\nonumber \\{} & {} +18 \alpha a_2 a_0^3-54 a_1 a_2 a_0^2-9 a_1\Big )\nonumber \\ {}{} & {} +\frac{3 a_1 \left( -2 \alpha ^2 a_1 a_0^3-18 \alpha a_1 a_0^3+18 \alpha a_2 a_0^3-54 a_1 a_2 a_0^2-9 a_1\right) }{a_0}\nonumber \\{} & {} +18 \alpha a_2 a_0^3+27 \alpha a_1^2 a_0^2 -27 \alpha a_1 a_2 a_0^2+54 a_2^2 a_0^2+54 a_1^2 a_2 a_0+9 a_2\Bigg )+\ldots , \end{aligned}$$
(27)

and

$$\begin{aligned} y(z)= & {} a_0+\frac{1}{9} \left( 2 \alpha ^2+27 \alpha +36\right) a_0^4 z\nonumber \\{} & {} \quad +\frac{1}{18} z^2 \Big (\frac{14}{9} \alpha ^2 \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7+33 \alpha \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7\nonumber \\{} & {} \quad +133 \left( 2 \alpha ^2+27 \alpha +36\Big ) a_0^7\right) +\frac{1}{27} z^3 \Bigg (\frac{10}{27} \alpha ^2 \left( 2 \alpha ^2+27 \alpha +36\right) ^2 a_0^{10}\nonumber \\{} & {} \quad +\frac{170}{3} \left( 2 \alpha ^2+27 \alpha +36\right) ^2 a_0^{10}+9 \alpha \left( 2 \alpha ^2+27 \alpha +36\right) ^2 a_0^{10}\nonumber \\{} & {} \quad +\frac{10}{9} \alpha ^2 a_0^3 \Big (\frac{14}{9} \alpha ^2 \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7+33 \alpha \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7\nonumber \\{} & {} \quad +133 \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7\Big )+36 \alpha a_0^3 \Big (\frac{14}{9} \alpha ^2 \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7\nonumber \\{} & {} \quad +33 \alpha \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7+133 \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7\Big )\nonumber \\{} & {} \quad +260 a_0^3 \Big (\frac{14}{9} \alpha ^2 \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7+33 \alpha \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7\nonumber \\{} & {} \quad +133 \left( 2 \alpha ^2+27 \alpha +36\right) a_0^7\Big )\Bigg )+\ldots , \end{aligned}$$
(28)

respectively.

3.3 The case of Eq. (3)

Reducing the equation by \(X_2+\epsilon X_1\) yields

$$\begin{aligned} p'(s)\Bigg (\epsilon ^{3}+4\alpha \epsilon ^{2}p(s)^{3}+3p(s)^{2}p''(s)\Bigg )+p(s)^{3}p'''(s)=0, \end{aligned}$$
(29)

and \(X_3+a X_1\) gives

$$\begin{aligned}{} & {} (-4-4a^{2}\alpha )y(z)^{4}+a^{3}y'(z)-27z^{2}y(z)^{2}y'(z)(5y'(z)+3z y''(z)\nonumber \\{} & {} \quad -3z y(z)^{3}((39+4a^{2}\alpha )y'(z)+9z(5y''(z)+z y'''(z))=0. \end{aligned}$$
(30)

In this case, we obtain

$$\begin{aligned} p(s)= & {} a_0+ a_1 s+ a_2 s^2\nonumber \\{} & {} +\frac{s^3 \left( -4 \alpha a_1 a_0^3-6 a_1 a_2 a_0^2-a_1\right) }{6 a_0^3}\nonumber \\{} & {} \frac{s^4}{12 a_0^3} \Bigg (-4 \alpha a_2 a_0^3-6 \alpha a_1^2 a_0^2-\frac{3 a_1 \left( -4 \alpha a_1 a_0^3-6 a_1 a_2 a_0^2-a_1\right) }{a_0}\nonumber \\ {}{} & {} -6 a_2^2 a_0^2-6 a_1^2 a_2 a_0-a_2\Bigg )\nonumber \\{} & {} +\ldots , \end{aligned}$$
(31)

and

$$\begin{aligned} y(z)= & {} \frac{1}{2} z^2 \left( 16 \alpha (\alpha +1) a_0^7+532 (\alpha +1) a_0^7\right) +4 (\alpha +1) a_0^4 z+a_0\nonumber \\{} & {} -\frac{2}{3} z^3 \bigg (96 \alpha (\alpha +1)^2 a_0^{10}-4080 (\alpha +1)^2 a_0^{10}+\nonumber \\{} & {} +2 \alpha a_0^3 \big (16 \alpha (\alpha +1) a_0^7+532 (\alpha +1) a_0^7\big )\nonumber \\{} & {} -130 a_0^3 \left( 16 \alpha (\alpha +1) a_0^7+532 (\alpha +1) a_0^7\right) \bigg )+\nonumber \\{} & {} +\frac{1}{4} z^4 \Bigg (-1280 \alpha (\alpha +1)^3 a_0^{13}+40768 (\alpha +1)^3 a_0^{13}\nonumber \\{} & {} -120 \alpha (\alpha +1) a_0^6 \big (16 \alpha (\alpha +1) a_0^7\nonumber \\{} & {} +532 (\alpha +1) a_0^7\big )+5226 (\alpha +1) a_0^6 \left( 16 \alpha (\alpha +1) a_0^7+532 (\alpha +1) a_0^7\right) \nonumber \\{} & {} -\frac{2678}{3} a_0^3 \bigg (96 \alpha (\alpha +1)^2 a_0^{10}-4080 (\alpha +1)^2 a_0^{10}+2 \alpha a_0^3 \big (16 \alpha (\alpha +1) a_0^7\nonumber \\{} & {} +532 (\alpha +1) a_0^7\big )-130 a_0^3 \left( 16 \alpha (\alpha +1) a_0^7+532 (\alpha +1) a_0^7\right) \bigg )\nonumber \\{} & {} +\frac{40}{3} \alpha a_0^3 \bigg (96 \alpha (\alpha +1)^2 a_0^{10}-4080 (\alpha +1)^2 a_0^{10}+2 \alpha a_0^3 \big (16 \alpha (\alpha +1) a_0^7\nonumber \\{} & {} +532 (\alpha +1) a_0^7\big )+-130 a_0^3 \big (16 \alpha (\alpha +1) a_0^7+532 (\alpha +1) a_0^7\big )\bigg )\Bigg )\nonumber \\{} & {} +\ldots \end{aligned}$$
(32)

for the above ODEs.

3.4 The case of Eq. (4)

Similarly, for Eq. (4), \(X_2+\epsilon X_1\), the travelling wave reduction produces

$$\begin{aligned} 18\epsilon ^{3}-4\alpha (9+\alpha )\epsilon ^{2}p(s)^{3}-27p(s)^{2}p''(s))+18p(s)^{3}(\alpha \epsilon p''(s)+p'''(s)=0, \end{aligned}$$
(33)

and \(X_3+a X_1\) gives

$$\begin{aligned}{} & {} \Big ((45+18 a\alpha +4a^{2}\alpha (9+\alpha )\Big )y(z)^{4}-18a^{3}y'(z)\nonumber \\{} & {} \quad +243 z^{2}y(z)^{2}y'(z)(5y'(z)+3z y''(z))\nonumber \\{} & {} \quad +3z y(z)^{3}\Bigg (540+90a\alpha +4a^{2}\alpha (9+\alpha )\Bigg )y'(z)\nonumber \\{} & {} \quad +27z \Bigg ((27+2a\alpha )y''(z)+6z y'''(z))\Bigg )=0, \end{aligned}$$
(34)

with corresponding solutions

$$\begin{aligned} p(s)= & {} a_0+ a_1 s+ a_2 s^2\nonumber \\{} & {} +\frac{s^3 \left( -2 \alpha ^2 a_1 a_0^3-18 \alpha a_1 a_0^3+18 \alpha a_2 a_0^3-27 a_1 a_2 a_0^2-9 a_1\right) }{54 a_0^3}\nonumber \\{} & {} -\frac{s^4}{216 a_0^3} \Big (4 \alpha ^2 a_2 a_0^3+6 \alpha ^2 a_1^2 a_0^2-\alpha \big (-2 \alpha ^2 a_1 a_0^3-18 \alpha a_1 a_0^3+18 \alpha a_2 a_0^3-9 a_1\nonumber \\{} & {} -27 a_1 a_2 a_0^2\big ) +\frac{9 a_1 \left( -2 \alpha ^2 a_1 a_0^3-18 \alpha a_1 a_0^3+18 \alpha a_2 a_0^3-27 a_1 a_2 a_0^2-9 a_1\right) }{2 a_0}\nonumber \\{} & {} +36 \alpha a_2 a_0^3+54 \alpha a_1^2 a_0^2 -54 \alpha a_1 a_2 a_0^2+54 a_2^2 a_0^2+54 a_1^2 a_2 a_0+18 a_2\Big )\nonumber \\{} & {} +\ldots , \end{aligned}$$
(35)

and finally,

$$\begin{aligned} y(z)= & {} a_0+\frac{1}{18} \left( 4 \alpha ^2+54 \alpha +45\right) a_0^4 z+\frac{1}{18} z^2 \Big (\frac{7}{9} \alpha ^2 \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7\nonumber \\{} & {} +\frac{33}{2} \alpha \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7+50 \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7\Big )\nonumber \\{} & {} +\frac{1}{54} z^3 \Bigg (\frac{5}{27} \alpha ^2 \big (4 \alpha ^2+54 \alpha +45\big )^2 a_0^{10}+\frac{235}{12} \left( 4 \alpha ^2+54 \alpha +45\right) ^2 a_0^{10}\nonumber \\{} & {} +\frac{9}{2} \alpha \left( 4 \alpha ^2+54 \alpha +45\right) ^2 a_0^{10}+\frac{20}{9} \alpha ^2 a_0^3 \bigg (\frac{7}{9} \alpha ^2 \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7\nonumber \\{} & {} +\frac{33}{2} \alpha \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7+50 \big (4 \alpha ^2+54 \alpha +45\big ) a_0^7\bigg )\nonumber \\{} & {} +72 \alpha a_0^3 \Big (\frac{7}{9} \alpha ^2 \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7+\frac{33}{2} \alpha \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7\nonumber \\{} & {} +50 \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7\Big )+433 a_0^3 \Big (\frac{7}{9} \alpha ^2 \big (4 \alpha ^2+54 \alpha \nonumber \\{} & {} +45\big ) a_0^7+\frac{33}{2} \alpha \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7+50 \left( 4 \alpha ^2+54 \alpha +45\right) a_0^7\Big )\Bigg )\nonumber \\{} & {} +\ldots . \end{aligned}$$
(36)

4 Convergence—exact power series

Next we prove the convergence of the power series solution (17). The rest of the series solutions above may be tested for convergence in the same manner. Consider, the case of Eq. (1), the first reduction. From (21) we obtain

$$\begin{aligned} |a_{r_1+3}|\le M\left( \sum _{r{_2}=1}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}|a_{r{_1}-r{_2}+3}||a_{r{_2}-r{_3}}||a_{r{_3}}| +\sum _{r{_2}=0}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}|a_{r{_1}-r{_2}}||a_{r{_2}-r{_3}}||a_{r{_3}}| \right) , \end{aligned}$$
(37)

where \(r_1=3,4,5 \ldots \), and \(M=max\left\{ \frac{1}{a^{2}_{0}},\frac{4|\alpha |}{3a^{2}_{0}}\right\} .\)

Suppose we have the power series \(\mu =R(s)=\sum ^{\infty }_{{r_1}=0}p_{{r_1}}s^{{r_1}}\), where

$$\begin{aligned} p_{k}=|a_{k}|~ k=0,1,\ldots ,5 \end{aligned}$$
(38)

and

$$\begin{aligned} p_{r_1+3}= M\left( \sum _{r{_2}=1}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}p_{r{_1}-r{_2}+3}p_{r{_2}-r{_3}}p_{r{_3}} +\sum _{r{_2}=0}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}p_{r{_1}-r{_2}}p_{r{_2}-r{_3}}p_{r{_3}} \right) , \end{aligned}$$
(39)

for \(r_1=3,4,5 \ldots \). Hence,

$$\begin{aligned} |a_{r_1}|\le p_{r_1}, ~r_1=0,1,2\ldots \end{aligned}$$
(40)

Next we prove that \(\mu \) is convergent in a neighborhood of a point. In fact, \(\mu \) is a majorant series of equation (17) and it can be written as follows:

$$\begin{aligned} R(s)= & {} p_{0}+p_{1}s+p_{2}s^{2}+p_{3}s^{3}+p_{4}s^{4}+p_{5}s^{5}+\sum _{r{_1}=3}^{\infty }p_{r_1+3}s^{r_1+3}\nonumber \\= & {} p_{0}+p_{1}s+p_{2}s^{2}+p_{3}s^{3}+p_{4}s^{4}+p_{5}s^{5}\nonumber \\{} & {} + M\sum _{r{_1}=3}^{\infty }\left( \sum _{r{_2}=1}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}p_{r{_1}-r{_2}+3}p_{r{_2}-r{_3}}p_{r{_3}} +\sum _{r{_2}=0}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}p_{r{_1}-r{_2}}p_{r{_2}-r{_3}}p_{r{_3}} \right) s^{r_1+3}\nonumber \\= & {} p_{0}+p_{1}s+p_{2}s^{2}+p_{3}s^{3}+p_{4}s^{4}+p_{5}s^{5}\nonumber \\{} & {} +M\left( \sum _{r{_1}=3}^{\infty } \sum _{r{_2}=1}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}p_{r{_1}-r{_2}+3}p_{r{_2}-r{_3}}p_{r{_3}}s^{r_1+3} +\sum _{r{_1}=3}^{\infty } \sum _{r{_2}=0}^{r{_1}}\sum _{r{_3}=0}^{r{_2}}p_{r{_1}-r{_2}}p_{r{_2}-r{_3}}p_{r{_3}} s^{r_1+3}\right) \nonumber \\= & {} p_{0}+p_{1}s+p_{2}s^{2}+p_{3}s^{3}+p_{4}s^{4}+p_{5}s^{5}\nonumber \\{} & {} +M[R^{3}(s)(s^3+1)-p^{2}_{0}R-R^{2}(p_1s+p_1s^2)+\nu (s)], \end{aligned}$$
(41)

where \(\nu (s)\) is a polynomial with respect to s, that is,

\(\nu (s)=-3 p_0 p_1^2\,s^5-3 p_0^2 p_2 s^5-p_1^2 p_3 s^5-4 p_0 p_2 p_3 s^5-4 p_0 p_1 p_4 s^5-2 p_0^2 p_5 s^5-p_0 p_2^2\,s^4-3 p_0^2 p_1 s^4-4 p_0 p_1 p_3 s^4-2 p_0^2 p_4 s^4-2 p_0 p_1 p_2 s^3-2 p_0^2 p_3 s^3-p_0 p_1^2\,s^2-p_0^2 p_2 s^2-p_0^3\,s^3-p_0^2 p_1 s\).

Let implicit function equation be:

$$\begin{aligned} F(s,\mu )= & {} \mu -p_{0}-p_{1}s-p_{2}s^{2}-p_{3}s^{3}-p_{4}s^{4}-p_{5}s^{5}\nonumber \\{} & {} -M[R^{3}(s)(s^3+1)-p^{2}_{0}R(s)-R^{2}(s)(p_1s+p_1s^2)+\nu (s)], \end{aligned}$$
(42)

where we obtain that \(F(0,p_{0})=0\) and \(F_{\mu }(0,r_{0})=1-2Mp_0^2\ne 0\). By virtue of the implicit function theorem [15], \(\mu =R(s)\) is analytic and convergent in a neighbourhood of the point \((0,p_{0})\) in the plane and with a positive radius. Then the power series solution (17) is convergent in the neighbourhood of a point \((0,p_{0})\)

Therefore, (17) can be written as

$$\begin{aligned} p(s)= & {} a_{0}+a_{1}s+a_{2}s^{2}+a_{3}s^{3}+a_{4}s^{4}+a_{5}s^{5}+\sum ^{\infty }_{r_1=3}a_{r_1+3}s^{r_1+3}\nonumber \\= & {} a_{0}+a_{1}s+a_{2}s^{2}+a_{3}s^{4}+a_{4}s^{4}+a_{5}s^{5}\nonumber \\{} & {} +\sum ^{\infty }_{r_1=3}\Bigg (\frac{-3\displaystyle {\sum \nolimits _{r{_2}=1}^{r{_1}}\sum \nolimits _{r{_3}=0}^{r{_2}}}(r{_1}-r{_2}+3)(r{_1}-r{_2}+2)(r{_1}-r{_2}+1)a_{r{_1}-r{_2}+3}a_{r{_2}-r{_3}}a_{r{_3}}}{3(r{_1}+3)(r{_1}+2)(r{_1}+1)a_0^2}\nonumber \\{} & {} +\frac{4\alpha \displaystyle {\sum \nolimits _{r{_2}=0}^{r{_1}}\sum \nolimits _{r{_3}=0}^{r{_2}}}a_{r{_1}-r{_2}}a_{r{_2}-r{_3}}a_{r{_3}} }{3(r{_1}+3)(r{_1}+2)(r{_1}+1)a_0^2}\Bigg )s^{r_1+3}. \end{aligned}$$
(43)

Thus, the exact power series solution of equation (1) can be written as

$$\begin{aligned} u(x,t)= & {} a_{0}+a_{1}(t-x)+a_{2}(t-x)^{2}+a_{3}(t-x)^{3}+a_{4}(t-x)^{4}+a_{5}(t-x)^{5}\nonumber \\{} & {} +\sum ^{\infty }_{r_1=3}a_{r_1+3}(t-x)^{r_1+3}\nonumber \\= & {} a_{0}+a_{1}(t-x)+a_{2}(t-x)^{2}+a_{3}(t-x)^{4}+a_{4}(t-x)^{4}+a_{5}(t-x)^{5}\nonumber \\{} & {} +\sum ^{\infty }_{r_1=3}\Bigg (\frac{-3\displaystyle {\sum \nolimits _{r{_2}=1}^{r{_1}}\sum \nolimits _{r{_3}=0}^{r{_2}}}(r{_1}-r{_2}+3)(r{_1}-r{_2}+2)(r{_1}-r{_2}+1)a_{r{_1}-r{_2}+3}a_{r{_2}-r{_3}}a_{r{_3}}}{3(r{_1}+3)(r{_1}+2)(r{_1}+1)a_0^2}\nonumber \\{} & {} +\frac{4\alpha \displaystyle {\sum \nolimits _{r{_2}=0}^{r{_1}}\sum \nolimits _{r{_3}=0}^{r{_2}}}a_{r{_1}-r{_2}}a_{r{_2}-r{_3}}a_{r{_3}} }{3(r{_1}+3)(r{_1}+2)(r{_1}+1)a_0^2}\Bigg )(t-x)^{r_1+3}, \end{aligned}$$
(44)

where \(a_{0}\ne 0,a_1,a_{2},\) are arbitrary constants, \(a_3,a_4,a_5\) are given in (19)–(20) and the rest of the constants are to be determined by (21).

5 Conservation laws

In this section, we derive the conservation laws associated with Eqs. (1)–(4). A current \({ T}=(T^1,\ldots ,T^n)\) is conserved if it satisfies

$$\begin{aligned} D_i \, T^i=0, \end{aligned}$$
(45)

along the solutions of a given differential system, say \(G_\alpha (x,u^{(k)})=0\). Multipliers or integrating factors, that produce conservation laws, are called \(\Lambda ^\alpha \) [16]. They satisfy the relation

$$\begin{aligned} D_i \, T^i=\Lambda ^\alpha G^\alpha , \end{aligned}$$
(46)

for all functions \(u^\alpha \) and the overdetermined equations for \(\Lambda ^\alpha \) are

$$\begin{aligned} {{\delta }\over {\delta u^\alpha }} \Big [\Lambda ^\alpha G^\alpha \Big ]=0, \end{aligned}$$
(47)

where \({{\delta }\over {\delta u^\alpha }}\) is the Euler operator. In the Tables 3 and 4, we list the conservation laws for each of our four cases.

Table 3 Conserved components Eqs. (1)–(2)
Full size table
Table 4 Conserved components Eq. (3)- (4)
Full size table

6 Conclusion

A special class of third-order evolutionary equations were investigated. Several distinct ordinary differential equations were obtained by reduction with the optimal system. In particular, these equations are highly nonlinear with solutions best found by power series. Through the results obtained in this paper, it can be found that the power series method is effective in solving ordinary differential equations admitted by the Lie group reduction method. A standard technique for testing the convergence of such solutions is demonstrated. Finally, the conservation laws may be used for further studies of gaining insights into the given class of equations.