Lie point symmetries classification of the mixed Liénard-type equation

Abstract

In this paper we develop a systematic and self-consistent procedure based on a set of compatibility conditions for identifying all maximal (eight parameter) and non-maximal (one and two parameter) symmetry groups associated with the mixed quadratic-linear Liénard-type equation, $\ddot{x} + f(x){\dot{x}}^{2} + g(x)\dot{x}+h(x)= 0$, where $f(x),\,g(x)$ and h(x) are arbitrary functions of x. With the help of this procedure we show that a symmetry function b(t) is zero for non-maximal cases, whereas it is not so for the maximal case. On the basis of this result the symmetry analysis gets divided into two cases, (i) the maximal symmetry group $(b\ne 0)$ and (ii) non-maximal symmetry groups $(b=0)$. We then identify the most general form of the mixed quadratic linear Liénard-type equation in each of these cases. In the case of eight-parameter symmetry group, the identified general equation becomes linearizable. In the case of non-maximal symmetry groups the identified equations are all integrable. The integrability of all the equations is proved either by providing the general solution or by constructing time-independent Hamiltonians.

Appendices

Appendix 1: Relationship among the functions $f,\,g$ and h

Integrating Eqs. (7)–(10) consistently one can obtain the infinitesimals $\xi$ and $\eta$ in terms of the arbitrary functions $f,\,g$ and h. Even though Eq. (7) can be integrated to yield $\xi$, with two unknown arbitrary functions, the explicit form of these two unknown arbitrary functions as well as the other infinitesimal $\eta$ are difficult to obtain explicitly from the rest of the determining equations. To determine the infinitesimals $\xi$ and $\eta$ uniquely from the determining Eqs. (7)–(10), we formulate two integrability conditions by imposing the compatibility between these four equations. By considering each one of the integrability conditions separately, we then integrate the determining Eqs. (7)–(10) and explore the Lie point symmetries of (2).

$(\mathrm{i})$ First auxiliary equation

To begin with we seek the compatibility between Eqs. (8) and (9). We then simplify this equation appropriately so that the resultant equation (which we call as first auxiliary equation) involves the infinitesimals $\xi$ and $\eta$ and their first derivative only.

Differentiating Eq. (8) with respect to t,

$$\begin{aligned} \eta _{xxt}+f \eta _{xt}+f_{x}\eta _{t}-2 \xi _{txt}+2 g \xi _{xt}=0. \end{aligned}$$

(52)

Next, replacing the term $\eta _{xt}$ in the resultant equation by substituting (9) in it, one gets

$$\begin{aligned}&2\eta _{xxt}-f \eta \, g_{x}-3f h\,\xi _{x}+2\,\left( f_x-f^2\right) \eta _{t}-f g\,\xi _{t} \nonumber \\&\quad +\,4g\,\xi _{xt}+f\xi _{tt}-4\xi _{txt}=0. \end{aligned}$$

(53)

Differentiating Eq. (9) with respect to x once, we get

$$\begin{aligned}&2\eta _{txx}+\eta g_{xx}+g_x \eta _x+2 f_x \eta _t+2 f \eta _{tx}+g_x \xi _t \nonumber \\&\quad +\,g \xi _{tx} -\xi _{ttx}+3 h_x \xi _x+3 h \xi _{xx}=0. \end{aligned}$$

(54)

Substituting Eqs. (7) and (9) in Eq. (54) for $\xi _{xx}$ and $\eta _{xt}$, respectively, and simplifying the resultant equation, we obtain the relation

$$\begin{aligned}&2\eta _{txx}+\eta \left( g_{xx}-fg_x\right) +g_x\eta _x+3\,h_x\,\xi _x \nonumber \\&\quad +\,2\left( f_x-f^2\right) \eta _t \nonumber \\&\quad +\left( g_x-fg\right) \xi _t+g\,\xi _{tx}+f\xi _{tt}-\xi _{ttx}=0. \end{aligned}$$

(55)

Eliminating $\eta _{xxt}$ from (53) and (55), we arrive at

$$\begin{aligned}&\eta g_{xx}+g_x \eta _x+3\left( h_x+f h\right) \xi _x+g_x \xi _t-3 g \xi _{tx} \nonumber \\&\quad +\,3 \xi _{ttx}=0. \end{aligned}$$

(56)

The expression which we obtained from the compatibility between Eqs. (8) and (9) contains the terms $\xi _{tx}$ and $\xi _{ttx}$. Now, we express these two terms in terms of first derivatives. To achieve this we differentiate (52) with respect to x and obtain

$$\begin{aligned}&\eta _{xxtx}+f_{xx}\eta _t+f_x\eta _{xt}+f \eta _{xtx}+f_x\eta _{tx}-2 \xi _{txtx} \nonumber \\&\quad +\,2 g_x \xi _{xt}+2 g \xi _{xtx}=0. \end{aligned}$$

(57)

We again replace the terms $\xi _{xxt}$ and $\xi _{txtx}$ which appear in (57) in terms of their lower order derivatives by using Eq. (7), and then replace $\eta _{xt}$ by (9) and finally $\eta _{xtx}$ by (53) so that Eq. (57) now simplifies to

$$\begin{aligned}&2\eta _{xxtx}-\eta \,g_x \left( 2f_x-f^2\right) -3\,h\,\left( 2 f_x-f^2\right) \nonumber \\&\quad \xi _x+2(f_{xx} -3 f f_x+f^3) \eta _t-g\left( 2f_x-f^2\right) \xi _t \nonumber \\&\quad +\,4 g_x \xi _{xt} +\left( 2f_x-f^2\right) \xi _{tt}=0. \end{aligned}$$

(58)

Differentiating (54) with respect to x, we get

$$\begin{aligned}&2\eta _{txxx}+\eta g_{xxx}+2 g_{xx}\eta _x+3 h_{xx}\xi _x+g_x\eta _{xx} \nonumber \\&\quad +\,6 h_x\xi _{xx} +3 h\xi _{xxx}+2 f_{xx}\eta _t \nonumber \\&\quad +\,g_{xx}\xi _t+4 f_x\eta _{tx}+2 g_x\xi _{tx} \nonumber \\&\quad +\,2 f\eta _{txx}+g\xi _{txx}-\xi _{ttxx}=0. \end{aligned}$$

(59)

Substituting Eqs. (7)–(10) and (55) in the above Eq. (59) and simplifying the later, we find

$$\begin{aligned}&2\eta _{txxx}+\eta \left( g_{xxx}-f g_{xx}-3f_x g_x+f^2 g_x\right) \nonumber \\&\quad +\,(3h_{xx} -3 h f_x+3f h_x-2g g_x+3f^2 h) \xi _x \nonumber \\&\quad +\,2\left( g_{xx}-f g_x\right) \eta _x +2\left( f_{xx}-3 f f_x+f^3\right) \eta _t \nonumber \\&\quad +\,(g_{xx}-2g f_x-f g_x +f^2 g) \xi _t+4 g_x \xi _{tx} \nonumber \\&\quad +\,\left( 2f_x-f^2\right) \xi _{tt}=0. \end{aligned}$$

(60)

Eliminating the variable $\eta _{xxtx}$ in (58) with the help of (60), we arrive at

$$\begin{aligned}&\eta \left( g_{xxx}-f g_{xx}-f_x g_x\right) +2\left( g_{xx}-f g_x\right) \eta _x+(3 h_{xx}\nonumber \\&\quad +\,3 h f_x+3 f h_x-2 g g_x) \xi _x+\left( g_{xx}-f g_x\right) \xi _t=0.\nonumber \\ \end{aligned}$$

(61)

Note that Eq. (61) which comes out from the compatibility of the Eqs. (8) and (9), with the help of appropriate usage of the other two Eqs. (7) and (10) in them, involves the infinitesimals $\xi ,\,\eta$ and their first derivatives $\eta _x,\,\xi _x$ and $\xi _t$ alone. The coefficients of the derivatives of $\xi$ and $\eta$ now turn out to be functions of $f,\,g,\,h$ and their derivatives.

To proceed further, we rewrite Eq. (61) in the form

$$\begin{aligned} PL_1+ \xi _xL_2+\eta L_{1x} =0, \end{aligned}$$

(62)

where

$$\begin{aligned} L_1= & {} g_{xx}-f g_x, \end{aligned}$$

(63)

$$\begin{aligned} L_2= & {} 3 h_{xx}+3 h f_x+3 f h_x-2 g g_x, \end{aligned}$$

(64)

with $P=2\eta _x+\xi _t$. We call Eq. (62) as the first auxiliary equation.

$(\mathrm{ii})$ Second auxiliary equation

We construct the second auxiliary equation by considering the compatibility between Eqs. (8) and (10). We then simplify this equation augmenting another equation which comes out from the compatibility of Eqs. (9) and (10) so that the resultant expression involves only $\xi$ and $\eta$ and does not contain any second derivatives.

Now, let us differentiate Eq. (10) once with respect to x, which in turn yields

$$\begin{aligned}&\eta _{ttx}+\eta h_{xx}-h \eta _{xx}+g_x \eta _t+2 h_x \xi _t+g \eta _{tx} \nonumber \\&\quad +\,2 h \xi _{tx}=0. \end{aligned}$$

(65)

Repeating the differentiation one more time, we find

$$\begin{aligned}&\eta _{ttxx}+\eta h_{xxx}+h_{xx} \eta _x-h_x \eta _{xx}-h \eta _{xxx}\nonumber \\&\quad +g_{xx} \eta _t+2 h_{xx} \xi _t+\,2 g_x \eta _{tx} \nonumber \\&\quad +\,4 h_x \xi _{tx}+g \eta _{txx}+2 h \xi _{txx}=0. \end{aligned}$$

(66)

Substituting Eqs. (7)–(10) in (66) and then replacing $\eta _{xxt}$ by (53) and simplifying the resultant equation, we obtain

$$\begin{aligned}&2\eta _{ttxx}+\eta \left( 2h_{xxx}+2h f_{xx}+2f_x h_x-2f h f_x+ f g g_x -2g_x^2\right) \nonumber \\&\quad +\,\left( 4 h_{xx}-2g g_x+f g^2\right) \xi _t \nonumber \\&\quad +\,(4 g h_x-2h g_x +3 f g h) \xi _x \nonumber \\&\quad +\,2\left( g_{xx}-g f_x-2 f g_x+f^2 g\right) \eta _t +4\left( h_x + f h-g^2\right) \xi _{tx} \nonumber \\&\quad +\,2(h_{xx}+2 h f_x+f h_x -f^2 h) \eta _x \nonumber \\&\quad +\left( 2g_x -f g\right) \xi _{tt}+4 g \xi _{txt}=0. \end{aligned}$$

(67)

Next, we consider Eq. (52) and differentiate it with respect to t, so that one has

$$\begin{aligned} \eta _{xxtt}+f_x \eta _{tt}+f \eta _{xtt}+2 g \xi _{xtt}-2 \xi _{txtt}=0. \end{aligned}$$

(68)

Now let us replace the term $\eta _{tt}$ in the above Eq. (68) by (10) and $\eta _{xtt}$ by (65). As a consequence Eq. (68) can now be brought to the form

$$\begin{aligned}&2\eta _{xxtt}-\eta \left( 2f h_{xx}+2f_x h_x+2 h f f_x- f g g_x\right) \nonumber \\&\quad -\, f \,g\, h\, \xi _x -2\left( g f_x+f g_x-f^2 g\right) \eta _t \nonumber \\&\quad +\,2\,h\,\left( f_x-f^2\right) \eta _x-(4 h f_x +4 f h_x-f g^2) \xi _t \nonumber \\&\quad -\, f g \xi _{tt}+4 g \xi _{xtt}-4 \xi _{txtt}=0. \end{aligned}$$

(69)

The fourth derivative of $\eta$, namely, $\eta _{ttxx}$, can be eliminated from (67) with the help of (69). The result shows that

$$\begin{aligned}&\eta \left( h_{xxx}+h f_{xx}+f h_{xx}+2 f_x h_x-g_x^2\right) \nonumber \\&\quad +\,(2g(h_x+fh) -h g_x) \xi _x+2\left( h_x+f h-g^2\right) \xi _{tx} \nonumber \\&\quad +\,2 \xi _{txtt}+(h_{xx} +h f_x+f h_x) \eta _x \nonumber \\&\quad +\left( 2 h_{xx}+2 h f_x+2 f h_x-g g_x\right) \xi _t +\left( g_{xx}-f g_x\right) \eta _t \nonumber \\&\quad +\,g_x \xi _{tt}=0. \end{aligned}$$

(70)

The equation which arises from the compatibility of (8) and (10) contains both second and higher derivative terms. As in the previous case here also, we try to formulate an auxiliary equation which involves first derivatives only. To achieve this task, we express the terms $\xi _{tt},\,\xi _{tx}$ and $\xi _{txtt}$ which appear in (70) in terms of their lower order derivatives. We differentiate Eq. (54) with respect to t and obtain an expression

$$\begin{aligned}&2\eta _{txxt}+ g_{xx} \eta _t+g_x \eta _{xt}+3 h_x \xi _{xt}+3 h \xi _{xxt} \nonumber \\&\quad +\,2 f_x \eta _{tt} +g_x \xi _{tt}+2 f \eta _{txt}+g \xi _{txt}-\xi _{ttxt}=0. \end{aligned}$$

(71)

Substituting Eqs. (7)–(9) and (65) in (71) and simplifying the later equation, we get

$$\begin{aligned}&4 \eta _{txxt}-\eta (4f h_{xx}+4f_x\, h_x+4f\, h\, f_{x}-2 f\, g \,g_{x}+{g_{x}}^2) \nonumber \\&\quad +\,4\left( h f_x-f^2 h\right) \eta _x+2 g \xi _{txt} \nonumber \\&\quad +\,2(g_{xx}-3 f g_x-2g f_x +2f^2 g) \eta _t \nonumber \\&\quad -\,\left( 8 f h_x+8 h f_x+ g g_x-2 f g^2\right) \xi _t -2 \xi _{ttxt} \nonumber \\&\quad -\,h\,\left( 3 g_x+2 f g\right) \xi _x+6\left( h_x+f h\right) \xi _{tx} \nonumber \\&\quad +\,\left( 3 g_x-2fg\right) \xi _{tt}=0. \end{aligned}$$

(72)

To eliminate the term $\eta _{ttxx}$ in (72) we substitute (67) in the former. The result yields

$$\begin{aligned}&\eta \left( 4 h_{xxx}+4 f h_{xx}+4 h f_{xx}+8 f_x h_x-3 {g_x}^2\right) \nonumber \\&\quad +\,(8 h_{xx} +8h f_x+8f h_x-3 g g_x) \xi _t \nonumber \\&\quad +\,(8 g h_x+8 f g h-h g_x) \xi _x +2( g_{xx}-f g_x) \eta _t \nonumber \\&\quad +\,4(h_{xx}+h f_x+f h_x) \eta _x \nonumber \\&\quad +\,2(h_x +f h-4 g^2) \xi _{tx}+g_x \xi _{tt}+6 g \xi _{txt} \nonumber \\&\quad +\,2 \xi _{ttxt}=0. \end{aligned}$$

(73)

Now, subtracting (70) from (73) we can remove the terms $\xi _{tt}$ and $\xi _{txtt}$ but the resultant equation involves the terms $\xi _{txt}$. To eliminate this term we multiply (56) by 2g and add the later one to the subtracted equation which in turn cancels the term $\xi _{txt}$. The final expression obtained in this process reads

$$\begin{aligned}&\eta \left( 3 h_{xxx}+3 f h_{xx}+3 h f_{xx}+6 f_x h_x-2 {g_x}^2-2 g g_{xx}\right) \nonumber \\&\quad +\,(g_{xx}-f g_x) \eta _t+(3 h_{xx} \nonumber \\&\quad +\,3 h f_x+3 f h_x-2 g g_x) \eta _x+2\left( 3 h_{xx} \right. \nonumber \\&\quad \left. +\,3 h f_x+3 f h_x-2 g g_x\right) \xi _t=0. \end{aligned}$$

(74)

Equation (73) can be rewritten in a compact form

$$\begin{aligned} \eta _tL_ 1 + QL_2 + \eta L_ {2x} = 0, \end{aligned}$$

(75)

where $L_1$ and $L_2$ are given in (63) and (64) and $Q=\eta _x + 2\xi _t$. We call Eq. (75) as the second auxiliary equation. This auxiliary equation which arises from the compatibility of both (8) and (10) as well as (9) and (10) involves only the infinitesimals $\xi$ and $\eta$ and their first derivatives. Interestingly, we observe that the compatibility conditions have now been written in a neater form and more importantly with two common coefficients, namely $L_1$ and $L_2$. Now, we can solve these two auxiliary equations and formulate the integrability conditions for the determining equations.

(iii) Integrability conditions

We consider the auxiliary equations as two algebraic equations for the unknowns $L_1$ and $L_2$. In other words we try to remove the terms $L_{1x}$ and $L_{2x}$ which appear in these two equations. We then solve the resultant algebraic equations and establish the integrability condition for the existence of symmetries.

Differentiating (62) with respect to t, we get

$$\begin{aligned} P_{t}L_1+\xi _{tx} L_2+\eta _{t} L_{1x}=0, \end{aligned}$$

(76)

and replacing the term $L_{1x}$ by using (62) again, Eq. (76) can be brought to the form

$$\begin{aligned} (P_t\eta -P\eta _t)L_1+(\xi _{tx}\eta -\xi _x\eta _t)L_2=0. \end{aligned}$$

(77)

Similarly, differentiating Eq. (76) with respect to t,

$$\begin{aligned} P_{tt}L_1+\xi _{ttx} L_2+\eta _{tt} L_{1x}=0, \end{aligned}$$

(78)

and then replacing $L_{1x}$ in the resultant equation by (62), we obtain another equation which involves only $L_1$ and $L_2$ that is

$$\begin{aligned} \left( P_{tt} \eta -P\eta _{tt}\right) L_1+\left( \xi _{ttx} \eta -\xi _x \eta _{tt}\right) L_2=0. \end{aligned}$$

(79)

Upon solving Eqs. (77) and (79) for $L_1$ and $L_2$, we find that solution exists under the following four conditions:

$$\begin{aligned}&(\mathrm{i})\,L_1 = 0,\quad L_2=0,\quad M\ne 0, \nonumber \\&(\mathrm{ii})\,L_1\ne 0,\quad L_2\ne 0,\quad M=0, \nonumber \\&(\mathrm{iii})\,L_1 = 0,\quad L_2\ne 0,\quad M=0, \nonumber \\&(\mathrm{iv})\,L_1\ne 0,\quad L_2=0,\quad M=0, \end{aligned}$$

(80)

where

$$\begin{aligned} M&=\left( P_{tt} \eta -P\eta _{tt}\right) \left( \xi _{tx}\eta -\xi _x\eta _{t}\right) \nonumber \\&\quad -\,\left( P_{t} \eta -P\eta _{t}\right) \left( \xi _{ttx}\eta -\xi _{tx}\eta _{tt}\right) . \end{aligned}$$

(81)

By recalling the second auxiliary equation (vide Eq. 75) and proceeding in the same manner as we did above for Eq. (62), we can derive the following two equations from (75), namely

$$\begin{aligned}&\left( \eta _{tt}\eta -\eta _{t}^2\right) L_1+\left( Q_{t}\eta -Q\eta _{t}\right) L_2=0, \end{aligned}$$

(82)

$$\begin{aligned}&\left( \eta _{ttt}\eta _t-\eta _{tt}^2\right) L_1+\left( Q_{tt}\eta _t-Q_t\eta _{tt}\right) L_2=0. \end{aligned}$$

(83)

Upon solving these two Eqs. (82) and (83), we find that the following four conditions should be fulfilled in order to have non-trivial solutions, namely

$$\begin{aligned}&(i)\,L_1=0,\quad L_2=0,\quad N\ne 0, \nonumber \\&(ii)\,L_1\ne 0,\quad L_2\ne 0,\quad N=0,\nonumber \\&(iii)\,L_1=0,\quad L_2\ne 0,\quad N=0, \nonumber \\&(iv)\,L_1\ne 0,\quad L_2=0,\quad N=0, \end{aligned}$$

(84)

where N is given by

$$\begin{aligned} N= & {} \left( \eta _{ttt}\eta _t-\eta _{tt}^2\right) \left( Q_{t}\eta -Q\eta _{t}\right) \nonumber \\&-\,\left( Q_{tt}\eta _t-Q_t\eta _{tt}\right) \left( \eta _{tt}\eta -\eta _{t}^2\right) . \end{aligned}$$

(85)

Analyzing Eqs. (80) and (84), we conclude that one can integrate the determining Eqs. (7)–(10) and obtain nontrivial symmetries for the ODE (2) under the following four conditions, namely:

$$\begin{aligned} (\mathrm{i})\,L_1= & {} 0,\,L_2=0,\quad M\ne 0,\quad N\ne 0, \end{aligned}$$

(86)

$$\begin{aligned} (\mathrm{ii})\,L_1\ne & {} 0,\quad L_2\ne 0,\quad M=0,\quad N=0, \end{aligned}$$

(87)

$$\begin{aligned} (\mathrm{iii})\,L_1= & {} 0,\quad L_2\ne 0,\quad M=0,\quad N=0, \end{aligned}$$

(88)

and

$$\begin{aligned} (\mathrm{iv})\,L_1\ne 0,\quad L_2=0,\quad M=0,\quad N=0. \end{aligned}$$

(89)

We call the above four conditions as the integrability conditions for the system of determining equations. The first condition says that both the expressions which involve arbitrary functions f, g and h (say $L_1$ and $L_2$) should vanish, whereas none of the expressions (81, 85) which interconnect the infinitesimals $\xi$ and $\eta$ should vanish. From the second condition we infer that this should be in the opposite way. The expressions (M and N) which involve the infinitesimals or the infinitesimal $\eta$ alone should vanish whereas the expression which involves the arbitrary functions in Eq. (2) should not vanish. Finally, in the third and fourth conditions one of the functions involving $f,\,g$ and h ($L_1$ or $L_2$) are zero along with the expressions M and N.

In the following, we analyze these four conditions in some detail.

Appendix 2: The case $L_1=0,\,L_2=0$ and $M, N\ne 0$

In Sect. 3, we have shown that the case $L_1=0,\,L_2=0,\,M,\,N\ne 0$ corresponds to Eq. (17) admitting the maximal symmetry group, namely eight parameters. We have also shown in Sect. 4 that this equation is linearizable.

It has been proved that the linearization of a scalar second-order nonlinear ODE, $\ddot{x}+f(t,x,\dot{x})=0$, by point transformation will result in a cubic polynomial in first derivative [26],

$$\begin{aligned} \ddot{x}=A(t,x)\dot{x}^{3}+B(t,x)\dot{x}^{2}+C(t,x)\dot{x}+D(t,x), \end{aligned}$$

(90)

where the coefficients $A,\,B,\,C$ and D should satisfy the following two conditions, namely,

$$\begin{aligned}&3A_{tt}+3A_{t}C-3A_{x}D+3AC_{t}+C_{xx}-6AD_{x} \nonumber \\&\quad +\,BC_{x} -2BB_{t}-2B_{tx}=0, \end{aligned}$$

(91)

$$\begin{aligned}&6A_{t}D-3B_{x}D+3AD_{t}+B_{tt}-2C_{tx}-3BD_{x} \nonumber \\&\quad +\,3D_{xx} +2CC_{x}-CB_{t}=0, \end{aligned}$$

(92)

where the suffix refers to partial derivatives. By comparing our case (2) with (90), we find that the linearizability criteria (91) and (92) reduce to

$$\begin{aligned}&g_{xx}-fg_{x}=0, \end{aligned}$$

(93)

$$\begin{aligned}&3h_{xx}+3 h f_x+3 f h_x-2 g g_x=0, \end{aligned}$$

(94)

which is exactly equivalent to $L_1=0$ and $L_2=0$ as shown in Sect. 3. Thus we observe that the first integrability condition which we have derived in the previous section is nothing but the linearizability condition for the present nonlinear ODE (2) under point transformation. Thus we conclude that whenever the given functions f g and h satisfy the relations (93) and (94) the symmetry determining equations (7)–(10) can be integrated to give maximal number of symmetries. This is also confirmed by the non-vanishing of M and N.

Appendix 3: The case $L_1\ne 0,\,L_2\ne 0$ and $M=0,\,N=0$

Now, we analyze the second condition, that is, $L_1\ne 0,L_2\ne 0$ and $M=0,N=0$. Since we have $L_1\ne 0$ and $L_2\ne 0$ one may preassume that the determining equations can be integrated to yield lesser parameter Lie point symmetries only. However, here we are interested in understanding the implications of the conditions $M=0$ and $N=0$. For this purpose we investigate these two conditions in detail.

To begin with let us consider the condition $M=0$ which is equivalent to

$$\begin{aligned}&\left( P_{tt} \eta -P\eta _{tt}\right) \left( \xi _{tx}\eta -\xi _x\eta _{t}\right) \nonumber \\&\quad -\,\left( P_{t} \eta -P\eta _{t}\right) \left( \xi _{ttx}\eta -\xi _{tx}\eta _{tt}\right) =0, \end{aligned}$$

(95)

where $P=2\eta _x+\xi _t$. Integrating (95) and substituting the form of P in it, we get

$$\begin{aligned} 2\eta _x+\xi _t=c_1(x)\xi _x+c_2(x)\eta , \end{aligned}$$

(96)

where $c_1$ and $c_2$ are arbitrary functions of x.

Repeating the calculation for $N=0$, in the same manner, we end up at

$$\begin{aligned} \eta _t=c_3(x)(\eta _x+2\xi _t)+c_4(x)\eta , \end{aligned}$$

(97)

where $c_3$ and $c_4$ are functions of x. With the help of (96) the above Eq. (97) can be rewritten as

$$\begin{aligned} \eta _t=\eta \left( \frac{1}{2} c_2 c_3+c_4\right) +\frac{1}{2} c_1 c_3 \xi _x +\frac{3}{2} c_3 \xi _t. \end{aligned}$$

(98)

Now, we seek a compatibility between Eqs. (96) and (98) so that the resultant equation admits only one variable, say $\xi$ [which can be determined from (18)]. Doing so, we find

$$\begin{aligned}&\left( c_2 c_{3x}+c_3 c_{2x}+2 c_{4x}\right) \eta +\left( c_1c_4+c_1 c_{3x}+ c_3c_{1x}\right) \xi _x \nonumber \\&\quad +\left( 3c_{3x} -2c_2 c_3 -c_4\right) \xi _t \nonumber \\&\quad +\, c_1 c_3 \xi _{xx}+(3 c_3 - c_1) \xi _{tx}+ \xi _{tt}=0. \end{aligned}$$

(99)

Differentiating Eq. (99) with respect to x, we obtain an equation of the form

$$\begin{aligned}&A_{1}\eta +A_2\xi _x+A_3\xi _{xx}+A_4\xi _{xxx} \nonumber \\&\quad +A_5\xi _t+A_6\xi _{tx} +A_7\xi _{txx}+2\xi _{ttx}=0, \end{aligned}$$

(100)

where the coefficients ${A_i}^{'}s,\,i=1,2,\ldots ,7$, are functions of x alone whose explicit expressions can be found straightforwardly.

Differentiating (100) one more time with respect to x

$$\begin{aligned}&\eta (A_1 c_2+2 A_{1x})+(A_1 c_1+2 A_{2x}) \xi _x+2(A_2+A_{3x})\xi _{xx} \nonumber \\&\quad +\,2(A_3+A_{4x})\xi _{xxx}+2A_4\xi _{xxxx} \nonumber \\&\quad +\,(2A_{5x}-A_1)\xi _t +2(A_5+A_{6x})\xi _{tx} \nonumber \\&\quad +\,2(A_6+A_{7x}t)\xi _{txx}+2 A_7 \xi _{txxx} +4\xi _{ttxx}=0. \end{aligned}$$

(101)

and eliminating $\xi _t$ from (101) by using (100), we find

$$\begin{aligned}&\eta B_1+B_2 \xi _x+B_3 \xi _{xx}+B_4 \xi _{xxx}+B_5 \xi _{xxxx}+B_6 \xi _{tx} \nonumber \\&\quad +\,B_7\xi _{txx}+B_8 \xi _{txxx}+B_9 \xi _{ttx}+B_{10}\xi _{ttxx}=0,\nonumber \\ \end{aligned}$$

(102)

where the functions ${B_i}^{'}s,\,i=1,2,\ldots ,10,$ are functions of x alone whose explicit expressions again can be fixed unambiguously.

Differentiate Eq. (102) with respect to x, we get

$$\begin{aligned}&C_1\eta +C_2\xi _x+C_3\xi _{xx}+C_4\xi _{xxx}+C_5\xi _{xxxx} \nonumber \\&\quad +\,C_6\xi _{} +C_7\xi _{tx}+C_8\xi _{txx} \nonumber \\&\quad +\,C_9\xi _{txxx}+C_{10}\xi _{txxxx}+C_{11}\xi _{ttx} \nonumber \\&\quad +\,C_{12}\xi _{ttxx}+C_{13}\xi _{ttxxx}=0, \end{aligned}$$

(103)

where ${C_i}^{'}s,\,i=1,2,\ldots ,13,$ are functions of x alone and are connected to the earlier functions.

We have now two equations with $\eta$, vide Eqs. (102) and (103). Eliminating $\eta$ from these two equations and substituting the explicit form of $\xi (=b(t)\mathfrak {I}(x) +a(t))$ where $\mathfrak {I}=\int {F(x)\mathrm{d}x}$ and $F(x)=\mathrm{e}^{\int {f\mathrm{d}x}})$ in the resultant equation, we arrive at

$$\begin{aligned}&\left( -B_2C_1+B_1C_2\right) bF+\left( -B_3C_1+B_1C_3\right) bfF \nonumber \\&\quad +\,(B_1C_5 -B_5C_1)(f_{xx}+3f_xf+f^3)bF \nonumber \\&\quad +\left( -B_4C_1+B_1C_4\right) (f_x +f^2)bF \nonumber \\&\quad +\,B_1C_6 (f_{xxx}+4ff_{xx}+3f_x^2+6f^2f_x+f^4)bF \nonumber \\&\quad +\,B_1C_{10}(f_{xx}+3f_xf+f^3)\dot{b}F \nonumber \\&\quad +\left( -B_9C_1+B_1C_{11}\right) \ddot{b}F +\left( -B_6C_1+B_1C_7\right) \dot{b}F \nonumber \\&\quad +\left( -B_7C_1+B_1C_8\right) \dot{b}fF +\left( -B_8C_1+B_1C_9\right) \nonumber \\&\quad \times \,(f_x+f^2)\dot{b}F+(-B_{10}C_1\nonumber \\&\quad +\,B_1C_{12}) \ddot{b}fF+B_1C_{13}(f_x+f^2)\ddot{b}F=0, \end{aligned}$$

(104)

where overdot denotes differentiation with respect to t. Note that in (104) each and every term inside the brackets are functions of x only. With this observation we rewrite Eq. (104) in a more compact form

$$\begin{aligned} D_1(x)b(t)+D_2(x)\dot{b}(t)+D_3(x)\ddot{b}(t)=0, \end{aligned}$$

(105)

where $D_1(x),\, D_2(x)$ and $D_3(x)$ are arbitrary functions of x and are defined to be

$$\begin{aligned} D_1&=F (B_1 C_2-C_1B_2+f (B_1 C_3-C_1B_3) \nonumber \\&\quad +(B_1 C_5 -C_1B_5)(f^3+3ff_x+f_{xx}) \nonumber \\&\quad +(B_1 C_4-C_1B_4)(f_x+f^2) \nonumber \\&\quad +\,B_1 C_6(f^4+6f^2f_x+3 f_x^2+4ff_{xx}+f_{xxx})), \end{aligned}$$

(106)

$$\begin{aligned} D_2&=F \left( B_1 C_7-C_1B_6+f\left( B_1C_8-C_1B_7\right) +(B_1C_9 \right. \nonumber \\&\quad \left. -\,C_1B_8)\left( f_x+f^2\right) +B_1C_{10}\left( f^3+3ff_x+f_{xx}\right) \right) , \end{aligned}$$

(107)

$$\begin{aligned} D_3&=F (B_1 C_{11}-C_1B_9+f\left( B_1 C_{12}-C_1B_{10}\right) \nonumber \\&\quad +\,B_1 C_{13}\left( f_x+f^2\right) ). \end{aligned}$$

(108)

Since $D_1(x),\, D_2(x)$ and $D_3(x)$ are arbitrary integration constants which are functions of x and do not vanish in general. We conclude that the left-hand side can become zero if $\ddot{b}=0,\,\,\,\dot{b}=0,\,\,\,b=0$. Consequently, the arbitrary function b(t) which is present in one of the infinitesimals, $\xi$, Eq. (18), is zero, implying lesser parameter symmetries. However, there is one exceptional case to the above: It corresponds to the case where all the three functions $D_1(x),\, D_2(x)$ and $D_3(x)$ become arbitrary constants. In this case let us rewrite Eq. (105) in the form

$$\begin{aligned} \ddot{b}-k_2\dot{b}+k_3b=0. \end{aligned}$$

(109)

where $-k_2=\frac{D_2}{D_3}$ and $k_3=\frac{D_1}{D_3}$. Comparing Eq. (109) with Eq. (20) with $k_1=0$ we observe that the associated ODE will be a subcase of the general linearizable Eq. (17). From this result we conclude that while integrating the determining Eqs. (7)–(10) with $L_1\ne 0,L_2\ne 0$ and $M=0,N=0$ one will always get lesser parameter Lie point symmetries if $b=0$. Otherwise one obtains a special case of the eight-parameter symmetry case (linearizable case).

Appendix 4: The case $L_1=0,\,L_2\ne 0,\,M=0,\,N=0$

In the previous section we have shown that for the case $L_1\ne 0,L_2\ne 0$ and $M=0,N=0$ one will always get lesser parameter Lie point symmetries provided $b=0$. Proceeding in the same manner, one can also show that even for the case $L_1=0,\,L_2\ne 0,\,M=0,\,N=0$ one will get lesser parameter symmetries if $b=0$, whereas for $b\ne 0$ we will get a subcase of the linearizable equation.

However, in this section we will consider this case in an alternate way. To study this case we fix $L_1=0$ and look for the possible combinations of the infinitesimals $\xi ,\eta$ and their derivatives which make M and N zero. All the possible combinations and their corresponding equations along with the symmetries are given in the following Table 1 ($\mathfrak {I}(x)=\int {F\mathrm{d}x}$ and $F=\mathrm{e}^{\int {f\mathrm{d}x}}$).

Table 1 The case $L_1=0,\,L_2\ne 0,\,M=0,\,N=0$

All the cases given in Table 1 correspond to one-parameter symmetries except for the case 5, that is $\xi _x=0,\,\eta _t=0$. This particular case corresponds to two-parameter symmetries. It is to be noted that this equation can be transformed to the modified Emden equation ($\ddot{X}+g_1 X\dot{X}+h_1 X^3=0$) with the help of the transformation $X=\mathfrak {I}(x)+\lambda _2$, where $\lambda _2=\frac{g_2}{g_1}$.

Appendix 5: The case $L_1\ne 0,\,L_2=0,\,M=0,\,N=0$

Now, in this section, we again look for the possible combinations of the infinitesimals $\xi ,\eta$ and their derivatives that make M and N to be zero. The equations under all the possible combinations of $\xi ,\eta$ and their derivatives are shown in the following Table 2.

Table 2 The case $L_1\ne 0,\,L_2=0,\,M=0,\,N=0$

The cases 2 and 3 listed in Table 2 belong to one-parameter symmetry case. However, the cases 1 and 4 are subcases of the maximal symmetry case (17) and case 5 belongs to two-parameter symmetries case which is again a subcase of Eq. (41).