The Lie symmetry group of the general Lienard-type equation

TL;DR

This paper characterizes the conditions under which a general Lienard-type differential equation admits additional Lie symmetries beyond the obvious time translation symmetry, extending previous results to higher-order cases.

Contribution

It provides a complete characterization and verifiable conditions for extra Lie symmetries in Lienard-type equations for all orders except three.

Findings

Identifies when additional Lie symmetries occur

Provides an easily checkable condition on the functions f_k

Extends symmetry analysis to higher-order Lienard equations

Abstract

We consider the general Lienard-type equation $\ddot{u} = \sum_{k=0}^n f_k \dot{u}^k$ for $n\geq 4$. This equation naturally admits the Lie symmetry $\frac{\partial}{\partial t}$. We completely characterize when this equation admits another Lie symmetry, and give an easily verifiable condition for this on the functions $f_0, \dots , f_n$. Moreover, we give an equivalent characterization of this condition. Similar results have already been obtained previously in the cases $n=1$ or $n=2$. That is, this paper handles all remaining cases except for $n=3$.

Equations

$$\ddot{u}=\sum _{k=0}^n{f}_k(u){\dot{u}}^k,$$

$$\ddot{u}=f_1(u)\dot{u}+f_0(u),$$

$$\ddot{u}=f_2(u){\dot{u}}^2+f_1(u)\dot{u}+f_0(u),$$

$$X=\xi (t,u)\frac{\partial }{\partial t}+\eta (t,u)\frac{\partial }{\partial u}.$$

$$X^{1}=\xi\frac{\partial}{\partial t}+\eta\frac{\partial}{\partial u}+(D\eta-\dot{u}D\xi)\frac{\partial}{\partial\dot{u}}=\xi\frac{\partial}{\partial t}+\eta\frac{\partial}{\partial u}+\Bigl{(}\eta_{t}+\left(\eta_{u}-\xi_{t}\right)\dot{u}-\xi_{u}\dot{u}^{2}\Bigr{)}\frac{\partial}{\partial\dot{u}}.$$

$$S^1=\frac{\partial }{\partial t}+\dot{u}\frac{\partial }{\partial u}+\left(\sum _k{f}_k{\dot{u}}^k\right)\frac{\partial }{\partial \dot{u}},$$

$$[X^1-\xi S^1,S^1]=0$$

$$0$$

$$\displaystyle=\Bigl{(}\left(\eta-\xi\dot{u}\right)\Bigl{(}\sum_{k}f_{k}\dot{u}^{k}\Bigr{)}_{u}+\Bigl{(}\eta_{t}+\left(\eta_{u}-\xi_{t}\right)\dot{u}-\xi_{u}\dot{u}^{2}-\xi\bigl{(}\sum_{k}f_{k}\dot{u}^{k}\bigr{)}\Bigr{)}\Bigl{(}\sum_{k}f_{k}\dot{u}^{k}\Bigr{)}_{\dot{u}}$$

$$\displaystyle\qquad-\Bigl{(}\eta_{t}+\left(\eta_{u}-\xi_{t}\right)\dot{u}-\xi_{u}\dot{u}^{2}-\xi\Bigl{(}\sum_{k}f_{k}\dot{u}^{k}\Bigr{)}\Bigr{)}_{t}-\dot{u}\Bigl{(}\eta_{t}+\left(\eta_{u}-\xi_{t}\right)\dot{u}-\xi_{u}\dot{u}^{2}-\xi\Bigl{(}\sum_{k}f_{k}\dot{u}^{k}\Bigr{)}\Bigr{)}_{u}$$

$$\displaystyle\qquad-\Bigl{(}\sum_{k}f_{k}\dot{u}^{k}\Bigr{)}\Bigl{(}\eta_{t}+\left(\eta_{u}-\xi_{t}\right)\dot{u}-\xi_{u}\dot{u}^{2}-\xi\Bigl{(}\sum_{k}f_{k}\dot{u}^{k}\Bigr{)}\Bigr{)}_{\dot{u}}\Bigr{)}\frac{\partial}{\partial\dot{u}}$$

$$\displaystyle=\Bigl{(}-\eta_{tt}+\left(\xi_{tt}-2\eta_{tu}\right)\dot{u}+\left(2\xi_{ut}-\eta_{uu}\right)\dot{u}^{2}+\xi_{uu}\dot{u}^{3}$$

$$\displaystyle\qquad+\sum_{k}\left(f_{k}^{\prime}\eta+(k+1)f_{k+1}\eta_{t}+(k-1)f_{k}\eta_{u}+(2-k)f_{k}\xi_{t}+(4-k)f_{k-1}\xi_{u}\right)\dot{u}^{k}\Bigr{)}\frac{\partial}{\partial\dot{u}},$$

$$-{\eta }_{tt}+f_0^{\prime }\eta +f_1{\eta }_t-f_0{\eta }_u+2f_0{\xi }_t$$

$$\displaystyle+\bigl{(}\xi_{tt}-2\eta_{tu}+f_{1}^{\prime}\eta+2f_{2}\eta_{t}+f_{1}\xi_{t}+3f_{0}\xi_{u}\bigr{)}\cdot\dot{u}$$

$$\displaystyle+\bigl{(}2\xi_{tu}-\eta_{uu}+f_{2}^{\prime}\eta+3f_{3}\eta_{t}+f_{2}\eta_{u}+2f_{1}\xi_{u}\bigr{)}\cdot\dot{u}$$

$$\displaystyle+\bigl{(}\xi_{uu}+f_{3}^{\prime}\eta+4f_{4}\eta_{t}+2f_{3}\eta_{u}-f_{3}\xi_{t}+f_{2}\xi_{u}\bigr{)}\cdot\dot{u}$$

$$\displaystyle+\textstyle{\sum_{\,k=4}^{\,n-1}}\bigl{(}f_{k}^{\prime}\eta+(k+1)f_{k+1}\eta_{t}+(k-1)f_{k}\eta_{u}+(2-k)f_{k}\xi_{t}+(4-k)f_{k-1}\xi_{u}\bigr{)}\cdot\dot{u}$$

$$\displaystyle+\bigl{(}f_{n}^{\prime}\eta+(n-1)f_{n}\eta_{u}+(2-n)f_{n}\xi_{t}+(4-n)f_{n-1}\xi_{u}\bigr{)}\cdot\dot{u}$$

$$+(3-n)f_n{\xi }_u\cdot \dot{u}$$

$$f_n(u)\ne 0.$$

$$F(u)\ne 0,$$

$$F^{\prime }(u)={\left|f_n(u)\right|}^{\frac{1}{n-1}},$$

$$-a^{2}g+f_0^{\prime }g+a{f}_{1}g+(-1)f_0{g}^{\prime }+2f_0$$

$$a\left(1-2g^{\prime }\right)+f_1^{\prime }g+2a{f}_{2}g+f_1$$

$$-g^{\prime \prime }+f_2^{\prime }g+3a{f}_{3}g+f_2{g}^{\prime }$$

$$f_k^{\prime }g+(k+1)a{f}_{k+1}g+(k-1)f_k{g}^{\prime }+(2-k)f_k$$

$${\xi }_u=0.$$

$$-{\eta }_{tt}+f_0^{\prime }\eta +f_1{\eta }_t+(-1)f_0{\eta }_u+2f_0{\xi }_t$$

$$\left({\xi }_{tt}-2{\eta }_{tu}\right)+f_1^{\prime }\eta +2f_2{\eta }_t+f_1{\xi }_t$$

$$-{\eta }_{uu}+f_2^{\prime }\eta +3f_3{\eta }_t+f_2{\eta }_u$$

$$f_k^{\prime }\eta +(k+1)f_{k+1}{\eta }_t+(k-1)f_k{\eta }_u+(2-k)f_k{\xi }_t$$

$$f_n^{\prime }\eta +(n-1)f_n{\eta }_u+(2-n)f_n{\xi }_t$$

$${\left|f_n\right|}^{\prime }\eta +(n-1)\left|f_n\right|{\eta }_u+(2-n)\left|f_n\right|{\xi }_t=0.$$

$$\frac{{\left|f_n\right|}^{\prime }\cdot {\left|f_n\right|}^{\frac{2-n}{n-1}}}{(n-1)}\eta +{\left|f_n\right|}^{\frac{1}{n-1}}{\eta }_u=\frac{n-2}{n-1}{\left|f_n\right|}^{\frac{1}{n-1}}{\xi }_t.$$

$${\left({\left|f_n\right|}^{\frac{1}{n-1}}\eta \right)}_u$$

$${\left|f_n\right|}^{\frac{1}{n-1}}\eta$$

$$\eta$$

$$F^{\prime }(u)$$

$$g(u)$$

$$\eta =g(u)\cdot {\xi }_t(t).$$

$$-(k+1)f_{k+1}g{\xi }_{tt}=\left(f_k^{\prime }g+(k-1)f_k{g}^{\prime }+(2-k)f_k\right){\xi }_t.$$

Full text

The research was supported by the European Union’s Seventh Framework Programme (FP7/2007-2013) under grant agreement no. 318202. The first, third and fourth authors were partially supported by the European Union’s Seventh Framework Programme (FP7/2007-2013) under grant agreement no. 317721. The second author was partially supported by the Hungarian Scientific Research Fund (OTKA) grant no. K109185 and grant no. FK124814. The first and fourth authors were partially supported by the EFOP-3.6.1-16-2016-00022 project. This projects have been supported by the European Union, co-financed by the European Social Fund.

1. Introduction

In this paper we consider the Liénard-type second order ordinary differential equation

[TABLE]

where the dot denotes differentiating by the independent variable $t$ representing the time. Equation (1) is a special case of the Levinson–Smith-type equation $\ddot{u}=g_{1}\left(u,\dot{u}\right)\dot{u}+g_{0}(u)$ [11, G.3, p. 198–199], for which existence and uniqueness of a limit cycle have been established under certain conditions [9, 10].

Equation (1) is a common generalization of the Rayleigh-type equation $\ddot{u}+F(\dot{u})+u=0$ when $F$ is a polynomial, the classical Liénard-type equation $\ddot{u}=f_{1}(u)\dot{u}+f_{0}(u)$, and the quadratic Liénard-type equation $\ddot{u}=f_{2}(u)\dot{u}^{2}+f_{1}(u)\dot{u}+f_{0}(u)$. These equations come up quite often in Physics or Biology. Rayleigh-type systems play an important role in the theory of sound [21] or in the theory of non-linear oscillations [20, Chapter 2.2.4]. Classical Liénard-type equations arise in the model of the van der Pol oscillator applied in physical and biological sciences [25], but electric activity of the heart rate [26] or nerve impulses are modelled by a Liénard-type model, as well [4, 12] or [6, Chapter 7]. In [15] the population of Easter Island is modelled, and the system of differential equations is then reduced to a second order quadratic Liénard-type equation. One can even find applications in economy [3, 5, 7, 8].

Symmetry analysis is a very useful tool developed to understand and solve differential equations. Several examples come from Physics (see e.g. [1, 16] for comprehensive studies on the topic), and an increasing number of examples from Biology (see e.g. [2, 13, 14, 15]). Finding some symmetries for a differential equation can be used to derive an appropriate change of coordinates which then helps to eliminate the independent variables or to decrease the order of the system.

In many cases mentioned above (e.g. the Fitzhugh–Nagumo model [4, 12] or the model for the population of Easter Island [15]) the model is based on a first order system of two equations equivalent to a second order Liénard equation. In such a case, one might benefit to consider the equivalent second order system, which would only admit a finite dimensional Lie symmetry group instead of an infinite one. If this Lie group is at least two-dimensional, then pulling the symmetries back to the original system could yield two independent symmetries of the original system, and solutions can be determined by quadratures. This method has been applied successfully in several situations in the past (see e.g. [2, 14, 15] for some recent examples in Biology). This motivates to study the Lie symmetries of (1).

Pandey, Bindu, Senthilvelan and Lakshmanan [18, 19] considered the classical Liénard equation

[TABLE]

where $f_{1}$ and $f_{0}$ are arbitrary, infinitely many times differentiable functions. They classified when (2) has a 1, 2, 3, or 8 dimensional Lie symmetry group depending on $f_{0}$ and $f_{1}$. Then Tiwari, Pandey, Senthilvelan and Lakshmanan [22, 23] classified the dimension of the Lie symmetry group of quadratic Liénard-type equations without $\dot{u}$ term, and then more generally [24] the mixed quadratic Liénard-type equation

[TABLE]

where $f_{0}$, $f_{1}$ and $f_{2}$ are arbitrary, infinitely many times differentiable functions. Further, Paliathanasis and Leach [17] showed how one can simplify (3) by removing $f_{2}$ from (3) in the case $f_{1}=0$. The question naturally arises: what are the Lie symmetries if the right-hand side of (3) is a higher order polynomial in $\dot{u}$?

In this paper we consider (1) for $n\geq 4$ and for differentiable functions $f_{k}$ depending only on $u$, and not on $t$. Note, that equation (1) is autonomous, therefore the tangential Lie algebra $\mathcal{L}$ of the Lie group of all its symmetries always contains the $1$-dimensional subalgebra generated by the vector field $\frac{\partial}{\partial t}$. Determining another generator of $\mathcal{L}$ would then lead to a solution by quadratures of (1), and of any first order system equivalent to it. In Theorem 1 (see Section 3 for details) we completely characterize the case when (1) admits a more than 1 (in fact, 2) dimensional symmetry group. In particular, we give conditions (7–10) such that the symmetry group is 2 dimensional if and only if these conditions hold.

Here, conditions (7–9) are natural, but the meaning of the system (10) seems less intuitive, even though the system (10) is easily verifiable for a particular choice of $F$. In Theorem 3 (see Section 4 for details) we provide a necessary and sufficient condition for $f_{0},\dots,f_{n}$ to satisfy (10). It turns out that $f_{0},\dots,f_{n}$ satisfy (10) if and only if they are expressible by $F$ and some constants.

2. The symmetry condition

We formulate the symmetry condition for (1) in this section. Consider equation (1) on the plane $(t,u)$, where $t$ is the independent variable, and $u$ is the dependent variable. Further, the computations will be slightly easier if we consider the right-hand side as an infinite sum $\sum_{k}f_{k}\dot{u}^{k}=\sum_{k=-\infty}^{\infty}f_{k}\dot{u}^{k}$, where $f_{k}=0$ if $k<0$ or $k>n$.

The general form of an infinitesimal generator of a symmetry of (1) has the form

[TABLE]

Let $D$ denote the total derivation by $t$, that is $D\xi=\xi_{t}+\dot{u}\xi_{u}$, $D\eta=\eta_{t}+\dot{u}\eta_{u}$. We use the convention of writing partial derivatives into the lower right index. Then the first prolongation of $X$ is

[TABLE]

Further, let

[TABLE]

be the spray corresponding to the differential equation (1). The vector field (4) is an infinitesimal symmetry of (1) if and only if its first prolongation $X^{1}$ satisfies the Lie bracket condition

[TABLE]

on the space $(t,u,\dot{u})$ (cf. [1, Chapter 4, §3]). Substituting $X^{1}$ and $S^{1}$ into (5) we obtain

[TABLE]

therefore the symmetry condition is

[TABLE]

3. Lie symmetry algebra

We consider (1) for $n\geq 4$ and for differentiable functions $f_{k}$ depending only on $u$, and not on $t$. In Theorem 1 we completely characterize the case when (1) admits more then 1 (in fact, 2) dimensional symmetry group. We prove following

Theorem 1**.**

Consider the equation (1) for some $n\geq 4$ and for $f_{0},\dots,f_{n}\colon I\to\mathbb{R}$ being differentiable functions such that $f_{n}$ is not constant zero on the open interval $I\subseteq\mathbb{R}$. Then the Lie symmetry algebra $\mathcal{L}$ of (1) is exactly 1 dimensional, unless there exists a constant $a\in\mathbb{R}$, an open interval $U\subseteq I$, and a three-times differentiable function $F\colon U\to\mathbb{R}$ such that for all $u\in U$ we have

[TABLE]

and with the notation $g(u)=\frac{(n-2)F(u)}{(n-1)F^{\prime}(u)}$ the following hold:

[TABLE]

Further, if both $F_{1}$ and $F_{2}$ satisfy conditions (7–10), then $F_{1}=F_{2}$.

Remark 2**.**

(Generator of the symmetry algebra $\mathcal{L}$ of (1).) In case conditions (7–10)

1. (1)

do not hold, then the symmetry algebra $\mathcal{L}$ is generated by $\frac{\partial}{\partial t}$, 2. (2)

hold, then the 2 dimensional Lie symmetry algebra $\mathcal{L}$ is generated

1. i)

by $\frac{\partial}{\partial t}$ and $t\frac{\partial}{\partial t}+g(u)\frac{\partial}{\partial u}$ if $a=0$, or 2. ii)

by $\frac{\partial}{\partial t}$ and $e^{at}\frac{\partial}{\partial t}+ae^{at}g(u)\frac{\partial}{\partial u}$ if $a\neq 0$.

Proof.

The left-hand side of (6) has to be zero for all $(t,u,\dot{u})$. As $\xi$, $\eta$, $f_{k}$ ($0\leq k\leq n$) do not depend on $\dot{u}$, (6) is a polynomial in $\dot{u}$. Thus, (6) holds if and only if each of its coefficients is zero. Since $f_{n}$ is not constant 0 on the interval $I$, there exists an open interval $U^{\prime}\subseteq I$ such that $f_{n}(u)\neq 0$ for $u\in U^{\prime}$. Thus, from the coefficient of $\dot{u}^{n+1}$ by $n\geq 4$ we obtain

[TABLE]

In particular, $\xi$ only depends on $t$ and not on $u$. Substituting $\xi_{u}=0$ into (6) and considering the coefficients, we obtain that

[TABLE]

In the following we analyze the system (11). Note, that $\xi=c$, $\eta=0$ (for any $c\in\mathbb{R}$) satisfies (11). Further, if $\eta=0$ then from (11e) we have $\xi_{t}=0$ and $\xi=c$ for some $c\in\mathbb{R}$. Thus, in the following we assume that $\eta$ is not constant 0.

Consider (11e) first. Now, $f_{n}$ is nonzero on the open interval $U^{\prime}$, therefore either $f_{n}(u)>0$ for all $u\in U^{\prime}$ or $f_{n}(u)<0$ for all $u\in U^{\prime}$. Choose $\varepsilon\in\left\{\,1,-1\,\right\}$ such that $\varepsilon f_{n}(u)>0$ for all $u\in U^{\prime}$, that is $\left\lvert f_{n}\right\rvert=\varepsilon f_{n}$. Then we have

[TABLE]

Multiplying by $\frac{1}{n-1}\left\lvert f_{n}\right\rvert^{\frac{2-n}{n-1}}$ yields

[TABLE]

Here, the left-hand side is the $u$-derivative of $\left\lvert f_{n}\right\rvert^{\frac{1}{n-1}}\eta$, hence

[TABLE]

Let $F\colon U^{\prime}\to\mathbb{R}$ be a function defined as in (9), that is

[TABLE]

Thus we obtained that

[TABLE]

Note, that $g(u)$ cannot be constant 0 on $U^{\prime}$ (otherwise both $F$ and $F^{\prime}=\left\lvert f_{n}\right\rvert^{\frac{1}{n-1}}$ were constant 0), thus there exists an open interval $U\subseteq U^{\prime}$ such that $F(u)\neq 0$ for all $u\in U$, and hence $g(u)\neq 0$ ($u\in U$). By substituting (12) into (11d) we have

[TABLE]

In particular, for $k=n-1$ we have $f_{n}(u)g(u)\neq 0$ for all $u\in U$, thus

[TABLE]

Now, $f_{n}$, $f_{n-1}$, and $g$ only depend on $u$, and $\xi_{tt}$ and $\xi_{t}$ only depend on $t$. Therefore there exists $a\in\mathbb{R}$ such that

[TABLE]

for some $c\in\mathbb{R}$. Substituting (15) and (16) into (11), one obtains

[TABLE]

Thus, either $f_{0},f_{1},f_{2},\dots f_{n},g,F$ satisfy the conditions (7–10), or else $c=0$, implying $\xi_{t}=0$ and $\eta=0$, a contradiction.

Finally, assume that $F_{1}$ and $F_{2}$ both satisfy conditions (8–9) on $U$. Then let $g_{1}$ and $g_{2}$ be defined from $F_{1}$ and $F_{2}$, and let $a_{1}$ and $a_{2}$ be defined using (13). Let $\xi_{1}$, $\xi_{2}$ be such that $(\xi_{i})_{t}=e^{a_{i}t}$, let $\eta_{i}=e^{a_{i}t}g_{i}(u)$, and let $X_{i}=\xi_{i}\frac{\partial}{\partial t}+\eta_{i}\frac{\partial}{\partial u}$ for $i=1,2$. Further, let $\xi=\xi_{1}-\xi_{2}$, $\eta=\eta_{1}-\eta_{2}$, and $X=\xi\frac{\partial}{\partial t}+\eta\frac{\partial}{\partial u}$. Now, if $F_{1}$ and $F_{2}$ both satisfy conditions (8–9) and (10), then both $X_{1}$ and $X_{2}$ are elements of the Lie symmetry algebra $\mathcal{L}$, and thus $X=X_{1}-X_{2}=\left(\xi_{1}-\xi_{2}\right)\frac{\partial}{\partial t}+\left(\eta_{1}-\eta_{2}\right)\frac{\partial}{\partial u}\in\mathcal{L}$, as well. However, we have proved that if either $\xi_{t}\neq 0$ or $\eta\neq 0$, then they are of the form (15) and (16). Thus, $e^{a_{1}t}-e^{a_{2}t}$ is of the form $ce^{at}$ for some $a,c\in\mathbb{R}$, that is $a_{1}=a_{2}$ and $c=0$. Then $\xi_{t}=0$, implying $\eta=0$ by (12), which yields $g_{1}=g_{2}$. Finally, by $F_{1}^{\prime}=F_{2}^{\prime}$, the definition of $g$ immediately implies

[TABLE]

and hence $F_{1}=F_{2}$. This finishes the proof of Theorem 1. ∎

4. Equivalent description of the conditions

In this paragraph we provide a necessary and sufficient condition for $f_{0},\dots,f_{n}$ to satisfy (10). We have the following

Theorem 3**.**

Let $U\subseteq\mathbb{R}$ be an open interval, $f_{0},\dots,f_{n},F\colon U\to\mathbb{R}$ be functions satisfying (7–9), $g$ as introduced in Theorem 1 and let $\varepsilon,\nu\in\left\{\,1,-1\,\right\}$, $a\in\mathbb{R}$ be real constants such that $\left\lvert f_{n}\right\rvert=\varepsilon f_{n}$, $\left\lvert F\right\rvert=\nu F$. Then $f_{0},\dots,f_{n},F,g$ satisfy (10) if and only if there exist real constants $b_{k}$ and functions $A_{k},B_{k}\colon U\to\mathbb{R}$ ($0\leq k\leq n$) where $b_{n}=\varepsilon\left(\frac{n-1}{n-2}\right)^{1-n}$, $A_{n}$ is constant 0,

[TABLE]

and

[TABLE]

such that $f_{0},\dots,f_{n}$ are of the form

[TABLE]

In particular, if $a=0$, then $B_{k}(u)=0$ for all $u\in U$ ($0\leq k\leq n$), and thus $f_{0},\dots,f_{n},F,g$ satisfy (10) if and only if there exist real constants $b_{k}^{\prime}$ ($0\leq k\leq n$) such that

[TABLE]

Further, if $F$ is positive on $U$, then

[TABLE]

First, in Section 4.1 we show that $A_{k}$ ($0\leq k\leq n$) defined by (17) satisfy a recursive system of differential equations. The details are contained in Lemma 4. Then in Section 4.2 we consider the case $a=0$, when (10) results in homogeneous equations for $f_{k}$. Finally, in Section 4.3 we prove Theorem 3 by considering the general case $a\neq 0$ and applying the method of variation of parameters.

4.1. Auxiliary functions

Let us use the notations of Theorem 3.

Lemma 4**.**

Let $A_{n}(u)=0$. For all $0\leq k\leq n-1$ a particular solution of the ordinary differential equation

[TABLE]

where $b_{k+1}$ is an arbitrary real constant, is the function $A_{k}$ defined by (17) in Theorem 3.

Proof.

We prove (17) by induction on $m=n-k$. For $m=1$ we have

[TABLE]

This proves (17) for $k=n-1$. Assume now, that (17) holds for an integer $m=n-k$, $1\leq m\leq n$, that is

[TABLE]

Putting this into (22) for $k=n-(m+1)$ one obtains

[TABLE]

Hence, (17) holds for $k=n-(m+1)$ and by induction it holds for all integers $0\leq k\leq n-1$. ∎

4.2. The homogeneous case

Assume $a=0$. Now, (10) takes the form

[TABLE]

Note, that (23d) for $k=0,1$ gives (23a) and (23b). Now, $g(u)\neq 0$ for $u\in U$, hence the solution of (23d) is

[TABLE]

for some $b_{2}\in\mathbb{R}$. This proves Theorem 3 in the case $a=0$ by selecting $b_{k}^{\prime}=b_{k}\cdot\left(\frac{n-1}{n-2}\right)^{k-1}$ ($0\leq k\leq n,k\neq 2$) and $b_{2}^{\prime}=1+b_{2}\cdot\frac{n-1}{n-1}$.

4.3. The general (inhomogeneous) case

Proof of Theorem 3.

If $a\neq 0$, then (10d) is an inhomogeneous linear differential equation for $f_{k}$, and by (24) its general solution (by variation of parameters) is

[TABLE]

for some function $B_{k}=B_{k}(u)$ and constant $b_{k}\in\mathbb{R}$. Write $f_{k}$ in the form $f_{k}=(b_{k}+B_{k})h_{k}$, where

[TABLE]

Putting (26) into (10d) we obtain

[TABLE]

Now, $h_{k}$ is a particular solution of the homogeneous differential equation (23d), thus

[TABLE]

Since $h_{k}(u)\neq 0$ for all $u\in U$, $B_{k}$ is a particular solution of the differential equation

[TABLE]

We prove by induction on $m=n-k$ that $B_{k}=A_{k}$ ($3\leq k\leq n-1$) by showing that $B_{k}$ satisfies the recursive system of differential equations (22) of Lemma 4. Let $m=1$, that is $k=n-1$. From (9) we have $f_{n}=\varepsilon(F^{\prime})^{n-1}$. Applying (27) we obtain

[TABLE]

Comparing (28) and (22) for $m=1$, we find $B^{\prime}_{n-1}=A^{\prime}_{n-1}$ by choosing $b_{n}=\varepsilon\left(\frac{n-1}{n-2}\right)^{1-n}$. Hence, a particular solution $B_{n-1}$ of the differential equation $B^{\prime}_{n-1}=A^{\prime}_{n-1}$ is $A_{n-1}$. Therefore, for $k=n-1$ we have $B_{k}=A_{k}$.

Assume that for an integer $4\leq k\leq n-1$ $B_{k}=A_{k}$ holds, thus from (26) for $k=n-m+1$ we have

[TABLE]

Putting this into (27) for $k=n-m$ we obtain

[TABLE]

Comparing (22) for $k=n-m$ and (29) we see that $B^{\prime}_{n-m}=A^{\prime}_{n-m}$. Hence a particular solution $B_{n-m}$ of the differential equation $B^{\prime}_{n-m}=A^{\prime}_{n-m}$ is $A_{n-m}$. Therefore for $k=n-m$ one has $B_{k}=A_{k}$. By induction, for all $3\leq k\leq n-1$ one has $B_{k}=A_{k}$. Thus (19a) holds for $3\leq k\leq n-1$, and (18a) is proved.

We continue by proving (19b) and (18b), that is we show the condition on $f_{k}$ for $k=2$. Now, $f_{2}$ is the solution of the inhomogeneous linear differential equation (10c). The general solution of (10c) by (25) and by variation of parameters has the form

[TABLE]

for some function $B_{2}=B_{2}(u)$ and constant $b_{2}\in\mathbb{R}$. Putting (30) into (10c), then using $\left(\frac{1}{g}\right)^{\prime}g=-\frac{g^{\prime}}{g}$, and the fact that $\frac{g^{\prime}+b_{2}}{g}$ is the solution to the homogeneous differential equation (23c), one has

[TABLE]

that is $B_{2}^{\prime}=-3af_{3}g$. Using the form of $f_{3}$ given by (26) and the definition of $g$ we obtain

[TABLE]

Now, $A_{3}=B_{3}$ by (18a), thus comparing (31) and (22) for $k=2$ yields $B_{2}^{\prime}=\nu A_{2}^{\prime}$. Hence, a particular solution $B_{2}$ of the differential equation $B^{\prime}_{2}=\nu A^{\prime}_{2}$ has the form $B_{2}=\nu A_{2}$. Thus, (19b) and (18b) hold.

Now, we obtain the condition on $f_{k}$ for $k=1$. The function $f_{1}$ is the solution of the inhomogeneous linear differential equation (10b). The general solution of (10b) by (24) and by variation of parameters is

[TABLE]

for some function $B_{1}=B_{1}(u)$ and constant $b_{1}\in\mathbb{R}$. Putting (32) into (10b), and using the fact that $\left\lvert F\right\rvert^{\frac{1-n}{n-2}}$ is the solution to the homogeneous differential equation (23b), one obtains

[TABLE]

As $f_{2}$ has the form (30) and $B_{2}=\nu A_{2}$ by (18b), we obtain that

[TABLE]

Comparing (33) and (22) for $k=1$ we obtain

[TABLE]

Hence a particular solution $B_{1}$ of the differential equation (34) has the form

[TABLE]

Therefore, (19a) holds for $k=1$, and (18c) is proved.

Finally, we prove that (19a) holds for $k=0$. For $k=0$ the function $f_{0}$ is the solution of the inhomogeneous linear differential equation (10a). The general solution of (10a) by (24) and by variation of parameters is

[TABLE]

for some function $B_{0}=B_{0}(u)$ and constant $b_{0}\in\mathbb{R}$. Putting (36) into (10a), and using the fact that $\left(\frac{n-2}{n-1}\right)\left\lvert F\right\rvert^{\frac{-n}{n-2}}\left(F^{\prime}\right)^{-1}$ is the solution to the homogeneous differential equation (23a), we obtain

[TABLE]

As $f_{1}$ has the form (32), and $B_{1}$ has the form (35), we obtain

[TABLE]

Comparing (37) and (22) for $k=0$ we have

[TABLE]

Therefore, a particular solution $B_{0}$ of (38) is

[TABLE]

Hence, (19a) holds for $k=0$, and (18d) is proved. This finishes the proof of Theorem 3 ∎

5. Open problems

Several questions arise after determining the symmetries of (1). Indeed, if the Lie group of symmetries is at least two dimensional, then one can apply the two-dimensional solvable Lie group to obtain the solutions of (1).

Problem 1**.**

Determine the solutions of (1) provided $f_{k}$ ($0\leq k\leq n$, $n\geq 4$) satisfy the conditions of Theorem 1.

The only remaining case for (1) not covered by Theorem 1 or by [17, 18, 19, 22, 23, 24] is when $n=3$. Then one cannot immediately conclude $\xi_{u}=0$ from (6), because the $\dot{u}^{4}$ term of (6) is identically 0. In fact, for $n=3$ the symmetry condition translates to

[TABLE]

A potential simplification of the system (39) might be to eliminate $f_{2}$ from (39) by using a coordinate change $v=G(u)$, for some bijective, two-times differentiable $G$, for which $G^{\prime\prime}(u)=G^{\prime}(u)f_{2}(u)$ is satisfied (see e.g. [17]). This, however, still does not give an immediate answer as to what the solutions of (39) are.

Problem 2**.**

Determine all symmetries (and solutions) of the autonomous differential equation

[TABLE]

where $f_{0},f_{1},f_{2},f_{3}$ are arbitrary continuous functions in $u$.