The Lie symmetry group of the general Liénard-type equation

Ágota Figula; Gábor Horváth; Tamás Milkovszki; Zoltán Muzsnay

doi:10.1080/14029251.2020.1700623

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Volume 27, Issue 2, January 2020, Pages 185 - 198

The Lie symmetry group of the general Liénard-type equation

Authors

Ágota Figula, Gábor Horváth, Tamás Milkovszki, Zoltán Muzsnay

University of Debrecen, Institute of Mathematics, Pf. 400, Debrecen, 4002, Hungary,figula@science.unideb.hu,ghorvath@science.unideb.hu,milkovszki@science.unideb.hu,muzsnay@science.unideb.hu

Received 8 June 2019, Accepted 13 August 2019, Available Online 27 January 2020.

DOI: 10.1080/14029251.2020.1700623 How to use a DOI?
Keywords: second order ordinary differential equation; Liénard-type equation; Levinson–Smith-type equation; Lie group; symmetry analysis; Lie algebra of infinitesimal symmetries
Abstract: We consider the general Liénard-type equation u¨=∑k=0nfku˙k for n ≥ 4. This equation naturally admits the Lie symmetry ∂∂t. We completely characterize when this equation admits another Lie symmetry, and give an easily verifiable condition for this on the functions f₀,..., 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.
Copyright: © 2020 The Authors. Published by Atlantis and Taylor & Francis
Open Access: This is an open access article distributed under the CC BY-NC 4.0 license (http://creativecommons.org/licenses/by-nc/4.0/).

1. Introduction

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

u¨=∑k=0nfk(u)u˙k,(1.1)

where the dot denotes differentiating by the independent variable t representing the time. Eq. (1.1) is a special case of the Levinson–Smith-type equation u¨=g1(u,u˙)u˙+g0(u) [1, G.3, p. 198–199], for which existence and uniqueness of a limit cycle have been established under certain conditions [2,3].

Eq. (1.1) is a common generalization of the Rayleigh-type equation u¨+F(u˙)+u=0 when F is a polynomial, the classical Liénard-type equation u¨=f1(u)u˙+f0(u), and the quadratic Liénard-type equation u¨=f2(u)u˙2+f1(u)u˙+f0(u). These equations come up quite often in Physics or Biology. Rayleigh-type systems play an important role in the theory of sound [4] or in the theory of non-linear oscillations [5, Chapter 2.2.4]. Classical Liénard-type equations arise in the model of the van der Pol oscillator applied in physical and biological sciences [6], but electric activity of the heart rate [7] or nerve impulses are modelled by a Liénard-type model, as well [8,9] or [10, Chapter 7]. In [11] 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 [12–15].

Symmetry analysis is a very useful tool developed to understand and solve differential equations. Several examples come from Physics (see e.g. [16, 17] for comprehensive studies on the topic), and an increasing number of examples from Biology (see e.g. [11, 18–20]). 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 [8, 9] or the model for the population of Easter Island [11]) 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. [11, 18, 20] for some recent examples in Biology). This motivates to study the Lie symmetries of (1.1).

Pandey, Bindu, Senthilvelan and Lakshmanan [21,22] considered the classical Liénard equation

u¨=f1(u)u˙+f0(u),(1.2)

where f₁ and f₀ are arbitrary, infinitely many times differentiable functions. They classified when (1.2) has a 1, 2, 3, or 8 dimensional Lie symmetry group depending on f₀ and f₁. Then Tiwari, Pandey, Senthilvelan and Lakshmanan [23, 24] classified the dimension of the Lie symmetry group of quadratic Liénard-type equations without u˙ term, and then more generally [25] the mixed quadratic Liénard-type equation

u¨=f2(u)u˙2+f1(u)u˙+f0(u),(1.3)

where f₀, f₁ and f₂ are arbitrary, infinitely many times differentiable functions. Further, Paliathanasis and Leach [26] showed how one can simplify (1.3) by removing f₂ from (1.3) in the case f₁ = 0. The question naturally arises: what are the Lie symmetries if the right-hand side of (1.3) is a higher order polynomial in u˙?

In this paper we consider (1.1) for n ≥ 4 and for differentiable functions f_k depending only on u, and not on t. Note, that (1.1) is autonomous, therefore the tangential Lie algebra ℒ of the Lie group of all its symmetries always contains the 1-dimensional subalgebra generated by the vector field ∂∂t. Determining another generator of ℒ would then lead to a solution by quadratures of (1.1), and of any first order system equivalent to it. In Theorem 3.1 (see Section 3 for details) we completely characterize the case when (1.1) admits a more than 1 (in fact, 2) dimensional symmetry group. In particular, we give conditions (3.1–3.4) such that the symmetry group is 2-dimensional if and only if these conditions hold.

Here, conditions (3.1–3.3) are natural, but the meaning of the system (3.4) seems less intuitive, even though the system (3.4) is easily verifiable for a particular choice of F. In Theorem 4.1 (see Section 4 for details) we provide a necessary and sufficient condition for f₀,... , f_n to satisfy (3.4). It turns out that f₀,... , f_n satisfy (3.4) if and only if they are expressible by F and some constants.

2. The symmetry condition

We formulate the symmetry condition for (1.1) in this section. Consider (1.1) 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 ∑kfku˙k=∑k=−∞∞fku˙k, where f_k = 0 if k < 0 or k > n.

The general form of an infinitesimal generator of a symmetry of (1.1) has the form

X=ξ(t,u)∂∂t+η(t,u)∂∂u.(2.1)

Let D denote the total derivation by t, that is Dξ=ξt+u˙ξu, Dη=ηt+u˙ηu. We use the convention of writing partial derivatives into the lower right index. Then the first prolongation of X is

X1=ξ∂∂t+η∂∂u+(Dη−u˙Dξ)∂∂u˙=ξ∂∂t+η∂∂u+(ηt+(ηu−ξt)u˙−ξuu˙2)∂∂u˙.

Further, let

S1=∂∂t+u˙∂∂u+(∑kfku˙k)∂∂u˙,

be the spray corresponding to the differential equation (1.1). The vector field (2.1) is an infinitesimal symmetry of (1.1) if and only if its first prolongation X¹ satisfies the Lie bracket condition

[X1−ξS1,S1]=0(2.2)

on the space (t, u, u˙ ) (cf. [16, Chapter 4, §3]). Substituting X¹ and S¹ into (2.2) we obtain

0=[X1−ξS1,S1]=((η−ξu˙)(∑kfku˙k)u+(ηt+(ηu−ξt)u˙−ξuu˙2−ξ(∑kfku˙k))(∑kfku˙k)u˙−(ηt+(ηu−ξt)u˙−ξuu˙2−ξ(∑kfku˙k))t−u˙(ηt+(ηu−ξt)u˙−ξuu˙2−ξ(∑kfku˙k))u−(∑kfku˙k)(ηt+(ηu−ξt)u˙−ξuu˙2−ξ(∑kfku˙k))u˙)∂∂u˙=(−ηtt+(ξtt−2ηtu)u˙+(2ξut−ηuu)u˙2+ξuuu˙3+∑k(fk′η+(k+1)fk+1ηt+(k−1)fkηu+(2−k)fkξt+(4−k)fk−1ξu)u˙k)∂∂u˙,

therefore the symmetry condition is

−ηtt+f0′η+f1ηt−f0ηu+2f0ξt+(ξtt−2ηtu+f1′η+2f2ηt+f1ξt+3f0ξu)⋅u˙+(2ξtu−ηuu+f2′η+3f3ηt+f2ηu+2f1ξu)⋅u˙2+(ξuu+f3′η+4f4ηt+2f3ηu−f3ξt+f2ξu)⋅u˙3+∑k=4n−1(fk′η+(k+1)fk+1ηt+(k−1)fkηu+(2−k)fkξt+(4−k)fk−1ξu)⋅u˙k+(fn′η+(n−1)fnηu+(2−n)fnξt+(4−n)fn−1ξu)⋅u˙n+(3−n)fnξu⋅u˙n+1=0.(2.3)

3. Lie symmetry algebra

We consider (1.1) for n ≥ 4 and for differentiable functions f_k depending only on u, and not on t. In Theorem 3.1 we completely characterize the case when (1.1) admits more than 1 (in fact, 2) dimensional symmetry group. We prove following

Theorem 3.1.

Consider (1.1) for some n ≥ 4 and for f₀,... , f_n : I → ℝ being differentiable functions such that f_n is not constant zero on the open interval I ⊆ ℝ. Then the Lie symmetry algebra ℒ of (1.1) is exactly 1 dimensional, unless there exists a constant a ∈ ℝ, an open interval U ⊆ I, and a three-times differentiable function F : U → ℝ such that for all u ∈ U we have

fn(u)≠0.(3.1)

F(u)≠0,(3.2)

F′(u)=|fn(u)|1n−1,(3.3)

and with the notation g(u)=(n−2)F(u)(n−1)F′(u) the following hold:

−a2g+f0′g+af1g+(−1)f0g′+2f0=0,(3.4a)

a(1−2g′)+f1′g+2af2g+f1=0,(3.4b)

−g″+f2′g+3af3g+f2g′=0,(3.4c)

fk′g+(k+1)afk+1g+(k−1)fkg′+(2−k)fk=0, 3≤k≤n−1.(3.4d)

Further, if both F₁ and F₂ satisfy conditions (3.1–3.4), then F₁ = F₂.

Remark 3.1.

(Generator of the symmetry algebra ℒ of (1.1).) In case conditions (3.1–3.4)

1.
do not hold, then the symmetry algebra ℒ is generated by ∂∂t,
2.
hold, then the 2 dimensional Lie symmetry algebra ℒ is generated
1. (a)
  by ∂∂t and t∂∂t+g(u)∂∂u if a = 0, or
2. (b)
  by ∂∂t and eat∂∂t+aeatg(u)∂∂u if a ≠ 0.

Proof.

The left-hand side of (2.3) has to be zero for all (t, u, u˙). As ξ, η, f_k (0 ≤ k ≤ n) do not depend on u˙, (2.3) is a polynomial in u˙. Thus, (2.3) 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′ ⊆ I such that f_n(u) ≠ 0 for u ∈ U′. Thus, from the coefficient of u˙n+1 by n ≥ 4 we obtain

ξu=0.

In particular, ξ only depends on t and not on u. Substituting ξ_u = 0 into (2.3) and considering the coefficients, we obtain that

−ηtt+f0′η+f1ηt+(−1)f0ηu+2f0ξt=0,(3.5a)

(ξtt−2ηtu)+f1′η+2f2ηt+f1ξt=0,(3.5b)

−ηuu+f2′η+3f3ηt+f2ηu=0,(3.5c)

fk′η+(k+1)fk+1ηt+(k−1)fkηu+(2−k)fkξt=0, 3≤k≤n−1(3.5d)

fn′η+(n−1)fnηu+(2−n)fnξt=0.(3.5e)

In the following we analyze the system (3.5). Note, that ξ = c, η = 0 (for any c ∈ ℝ) satisfies (3.5). Further, if η = 0 then from (3.5e) we have ξ_t = 0 and ξ = c for some c ∈ ℝ. Thus, in the following we assume that η is not constant 0.

Consider (3.5e) first. Now, f_n is nonzero on the open interval U′, therefore either f_n(u) > 0 for all u ∈ U′ or f_n(u) < 0 for all u ∈ U′. Choose ε ∈ {1, −1} such that εf_n(u) > 0 for all u ∈ U′, that is | f_n| = εf_n. Then we have

|fn|′η+(n−1)|fn|ηu+(2−n)|fn|ξt=0.

Multiplying by 1n−1|fn|2−nn−1 yields

|fn|′⋅|fn|2−nn−1(n−1)η+|fn|1n−1ηu=n−2n−1|fn|1n−1ξt.

Here, the left-hand side is the u-derivative of |fn|1n−1η, hence

(|fn|1n−1η)u=n−2n−1|fn|1n−1ξt,|fn|1n−1η=n−2n−1ξt∫|fn|1n−1du,η=ξt(n−2)∫|fn|1n−1du(n−1)|fn|1n−1.

Let F : U′ → ℝ be a function defined as in (3.3), that is

F′(u):=|fn(u)|1n−1,

and let g be defined as in Theorem 3.1. that is

g(u):=(n−2)F(u)(n−1)F′(u).

Thus we obtained that

η=g(u)⋅ξt(t).(3.6)

Note, that g(u) cannot be constant 0 on U′ (otherwise both F and F′=|fn|1n−1 were constant 0), thus there exists an open interval U ⊆ U′ such that F(u) ≠ 0 for all u ∈ U, and hence g(u) ≠ 0 (u ∈ U). By substituting (3.6) into (3.5d) we have

−(k+1)fk+1gξtt=(fk′g+(k−1)fkg′+(2−k)fk)ξt.

In particular, for k = n − 1 we have f_n(u)g(u) ≠ 0 for all u ∈ U, thus

ξtt=−fn−1′g+(n−2)fn−1g′+(3−n)fn−1nfngξt.

Now, f_n, f_n₋₁, and g only depend on u, and ξ_tt and ξ_t only depend on t. Therefore there exists a ∈ ℝ such that

a=−fn−1′g+(n−2)fn−1g′+(3−n)fn−1nfng,(3.7)

ξtt=aξt,(3.8)

or else ξ_t = 0 implying η = 0 by (3.6), a contradiction. Thus,

ξt=ceat,(3.9)

ηt=ceatg(u)(3.10)

for some c ∈ ℝ. Substituting (3.9) and (3.10) into (3.5), one obtains

(−a2g+f0′g+af1g+(−1)f0g′+2f0)⋅ceat=0,(a(1−2g′)+f1′g+2af2g+f1)⋅ceat=0,(−g″+f2′g+3af3g+f2g′)⋅ceat=0,(fk′g+(k+1)afk+1g+(k−1)fkg′+(2−k)fk)⋅ceat=0,3≤k≤n−1.

Thus, either f₀, f₁, f₂,... f_n,g,F satisfy the conditions (3.1–3.4), or else c = 0, implying ξ_t = 0 and η = 0, a contradiction.

Finally, assume that F₁ and F₂ both satisfy conditions (3.2–3.3) on U. Then let g₁ and g₂ be defined from F₁ and F₂, and let a₁ and a₂ be defined using (3.7). Let ξ₁, ξ₂ be such that (ξ_i)_t = e^a_it, let η_i = e^a_itg_i(u), and let Xi=ξi∂∂t+ηi∂∂u for i = 1,2. Further, let ξ = ξ₁ − ξ₂, η = η₁ − η₂, and X=ξ∂∂t+η∂∂u. Now, if F₁ and F₂ both satisfy conditions (3.2–3.3) and (3.4), then both X₁ and X₂ are elements of the Lie symmetry algebra ℒ, and thus X=X1−X2=(ξ1−ξ2)∂∂t+(η1−η2)∂∂u∈ℒ, as well. However, we have proved that if either ξ_t ≠ 0 or η ≠ 0, then they are of the form (3.9) and (3.10). Thus, e^a₁t − e^a₂t is of the form ce^at for some a,c ∈ ℝ, that is a₁ = a₂ and c = 0. Then ξ_t = 0, implying η = 0 by (3.6), which yields g₁ = g₂. Finally, by F′₁ = F′₂, the definition of g immediately implies

0=g1−g2=n−2(n−1)F1′(F1−F2),

and hence F₁ = F₂. This finishes the proof of Theorem 3.1.

4. Equivalent description of the conditions

In this paragraph we provide a necessary and sufficient condition for f₀,... , f_n to satisfy (3.4). We have the following

Theorem 4.1.

Let U ⊆ ℝ be an open interval, f₀,... , f_n, F : U → ℝ be functions satisfying (3.1–3.3), g as introduced in Theorem 3.1 and let ε, ν ∈ {1,−1}, a ∈ ℝ be real constants such that |f_n| = εf_n, |F| = νF. Then f₀,... , f_n, F, g satisfy (3.4) if and only if there exist real constants b_k and functions A_k, B_k : U → ℝ (0 ≤ k ≤ n) where bn=ε(n−1n−2)1−n, A_n is constant 0,

Ak(u)=∑i=1n−k(−ν)i(k+ii)aibk+i|F(u)|i(n−1)(n−2), (0≤k≤n−1),(4.1)

and

Bk(u)=Ak(u), (3≤k≤n)(4.2a)

B2(u)=νA2(u),(4.2b)

B1(u)=A1(u)−a|F(u)|n−1n−2(2b2(1−ν)+1),(4.2c)

B0(u)=A0(u)+a2(1+b2(1−ν))ν|F(u)|2(n−1)n−2,(4.2d)

such that f₀,... , f_n are of the form

fk(u)=(bk+Bk(u))⋅(n−1n−2)k−1⋅|F(u)|k−nn−2⋅(F′(u))k−1, (0≤k≤n, k≠2)(4.3a)

f2(u)=(b2+B2(u))⋅n−1n−2⋅F′(u)F(u)+F′(u)F(u)−F″(u)F′(u).(4.3b)

In particular, if a = 0, then B_k(u) = 0 for all u ∈ U (0 ≤ k ≤ n), and thus f₀,... , f_n, F, g satisfy (3.4) if and only if there exist real constants b′_k (0 ≤ k ≤ n) such that

fk(u)=bk′⋅|F(u)|k−nn−2⋅(F′(u))k−1, (0≤k≤n,k≠2)f2(u)=b2′⋅F′(u)F(u)−F″(u)F′(u).

Further, if F is positive on U, then

Bk(u)=Ak(u), (2≤k≤n)B1(u)=A1(u)−a(F(u))n−1n−2,B0(u)=A0(u)+a2(F(u))2(n−1)n−2.

First, in Section 4.1 we show that A_k (0 ≤ k ≤ n) defined by (4.1) satisfy a recursive system of differential equations. The details are contained in Lemma 4.1. Then in Section 4.2 we consider the case a = 0, when (3.4) results in homogeneous equations for f_k. Finally, in Section 4.3 we prove Theorem 4.1 by considering the general case a ≠ 0 and applying the method of variation of parameters.

4.1. Auxiliary functions

Let us use the notations of Theorem 4.1.

Lemma 4.1.

Let A_n(u) = 0. For all 0 ≤ k ≤ n − 1 a particular solution of the ordinary differential equation

Ak(u)′=−(k+1)a(n−1n−2)|F(u)|1n−2F(u)′(bk+1+Ak+1(u)),(4.6)

where b_k₊₁ is an arbitrary real constant, is the function A_k defined by (4.1) in Theorem 4.1.

Proof.

We prove (4.1) by induction on m = n − k. For m = 1 we have

An−1′=−nabn(n−1n−2)|F|1n−2F′,

and a particular solution is

An−1=−νnabn|F|n−1n−2.

This proves (4.1) for k = n − 1. Assume now, that (4.1) holds for an integer m = n − k, 1 ≤ m ≤ n, that is

An−m=(∑i=1m(−ν)i(n−m+ii)aibn−m+i|F|i(n−1)n−2).

Putting this into (4.6) for k = n − (m + 1) one obtains

An−(m+1)′=−(n−m)a(n−1n−2)|F|1n−2F′⋅(bn−m+(∑i=1m(−ν)i(n−m+ii)aibn−m+i|F|i(n−1)n−2)).

By integrating, one can obtain a particular solution as

An−(m+1)=−ν(n−m)abn−m|F|n−1n−2−ν∑i=1m(−ν)in−mi+1(n−m+ii)ai+1bn−m+i|F|(i+1)(n−1)n−2=−ν(n−m)abn−m|F|n−1n−2+∑j=2m+1(−ν)jn−mj(n−m−1+jj−1)ajbn−m−1+j|F|j(n−1)n−2=∑i=1m+1(−ν)i(n−(m+1)+ii)aibn−(m+1)+i|F|i(n−1)(n−2).

Hence, (4.1) holds for k = n − (m + 1) and by induction it holds for all integers 0 ≤ k ≤ n − 1.

4.2. The homogeneous case

Assume a = 0. Now, (3.4) takes the form

f0′g+(−1)f0g′+2f0=0,(4.7a)

f1′g+f1=0,(4.7b)

−g″+f2′g+f2g′=0,(4.7c)

fk′g+(k−1)fkg′+(2−k)fk=0, (3≤k≤n−1).(4.7d)

Note, that (4.7d) for k = 0,1 gives (4.7a) and (4.7b). Now, g(u) ≠ 0 for u ∈ U, hence the solution of (4.7d) is

fk=exp(∫(1−k)g′+(k−2)g)=exp((1−k)ln|g|+(k−2)∫1g)=|g|1−k⋅exp((k−2)∫(n−1)F′(n−2)F)=|g|1−k⋅exp((k−2)(n−1)n−2∫(ln|F|)′)=bk⋅|g|1−k⋅|F|(k−2)(n−1)n−2=bk⋅(n−1n−2)k−1⋅(F′)k−1⋅|F|k−nn−2(4.8)

for some b_k ∈ ℝ (0 ≤ k ≤ n − 1, k ≠ 2). For k = 2 eq. (4.7c) has the form (−g′ + f₂g)′ = 0, thus

f2=g′+b2g=(1+b2n−1n−2)⋅F′F−F″F′(4.9)

for some b₂ ∈ ℝ. This proves Theorem 4.1 in the case a = 0 by selecting bk′=bk⋅(n−1n−2)k−1(0≤k≤n,k≠2) and b2′=1+b2n−1n−2.

4.3. The general (inhomogeneous) case

Proof.

[Proof of Theorem 4.1] If a ≠ 0, then (3.4d) is an inhomogeneous linear differential equation for f_k, and by (4.8) its general solution (by variation of parameters) is

fk=(bk+Bk)⋅(n−1n−2)k−1⋅|F|k−nn−2⋅(F′)k−1,(4.10)

for some function B_k = B_k(u) and constant b_k ∈ ℝ. Write f_k in the form f_k = (b_k + B_k)h_k, where

hk=(n−1n−2)k−1⋅|F|k−nn−2⋅(F′)k−1.

Putting (4.10) into (3.4d) we obtain

((bk+Bk)hk′g+(k−1)(bk+Bk)hkg′+(2−k)(bk+Bk)hk)+Bk′hkg+(k+1)afk+1g=0.

Now, h_k is a particular solution of the homogeneous differential equation (4.7d), thus

(bk+Bk)hk′g+(k−1)(bk+Bk)hkg′+(2−k)(bk+Bk)hk=0.

Since h_k(u) ≠ 0 for all u ∈ U, B_k is a particular solution of the differential equation

Bk′=−(k+1)afk+1hk=−(k+1)afk+1(n−1n−2)1−k|F|n−kn−2(F′)1−k, (3≤k≤n−1).(4.11)

We prove by induction on m = n − k that B_k = A_k (3 ≤ k ≤ n − 1) by showing that B_k satisfies the recursive system of differential equations (4.6) of Lemma 4.1. Let m = 1, that is k = n − 1. From (3.3) we have f_n = ε(F′)ⁿ⁻¹. Applying (4.11) we obtain

Bn−1′=−nafn(n−1n−2)2−n(F′)2−n|F|1n−2=−naε(n−1n−2)2−n|F|1n−2F′.(4.12)

Comparing (4.12) and (4.6) for m = 1, we find B′_n₋₁ = A′_n₋₁ by choosing bn=ε(n−1n−2)1−n. Hence, a particular solution B_n₋₁ of the differential equation B′_n₋₁ = A′_n₋₁ is A_n₋₁. Therefore, for k = n − 1 we have B_k = A_k.

Assume that for an integer 4 ≤ k ≤ n − 1 B_k = A_k holds, thus from (4.10) for k = n − m + 1 we have

fn−m+1=(bn−m+1+An−m+1)⋅(n−1n−2)n−m⋅|F|1−mn−2⋅(F′)n−m.

Putting this into (4.11) for k = n − m we obtain

Bn−m′(u)=−(n−m+1)afn−m+1⋅(n−1n−2)1−n+m⋅|F|mn−2⋅(F′)1−n+m=−(n−m+1)a⋅n−1n−2⋅|F|1n−2⋅(F′)(bn−m+1+An−m+1).(4.13)

Comparing (4.6) for k = n − m and (4.13) we see that B′_n₋_m = A′_n₋_m. Hence a particular solution B_n₋_m of the differential equation B′_n₋_m = A′_n₋_m is A_n₋_m. Therefore for k = n − m one has B_k = A_k. By induction, for all 3 ≤ k ≤ n − 1 one has B_k = A_k. Thus (4.3a) holds for 3 ≤ k ≤ n − 1, and (4.2a) is proved.

We continue by proving (4.3b) and (4.2b), that is we show the condition on f_k for k = 2. Now, f₂ is the solution of the inhomogeneous linear differential equation (3.4c). The general solution of (3.4c) by (4.9) and by variation of parameters has the form

f2=g′+(b2+B2)g(4.14)

for some function B₂ = B₂(u) and constant b₂ ∈ ℝ. Putting (4.14) into (3.4c), then using (1g)′g=−g′g, and the fact that g′+b2g is the solution to the homogeneous differential equation (4.7c), one has

0=−g″+(g′+b2g)′g+B2′+B2(1g)′g+3af3g+g′+b2gg′+B2gg′=−g″+(g′+b2g)′g+g′+b2gg′+B2((1g)′g+g′g)+B2′+3af3g=B2′+3af3g,

that is B′₂ = −3af₃g. Using the form of f₃ given by (4.10) and the definition of g we obtain

B2′=−3a(b3+B3)n−1n−2ν|F|1n−2F′.(4.15)

Now, A₃ = B₃ by (4.2a), thus comparing (4.15) and (4.6) for k = 2 yields B′₂ = νA′₂. Hence, a particular solution B₂ of the differential equation B′₂ = νA′₂ has the form B₂ = νA₂. Thus, (4.3b) and (4.2b) hold.

Now, we obtain the condition on f_k for k = 1. The function f₁ is the solution of the inhomogeneous linear differential equation (3.4b). The general solution of (3.4b) by (4.8) and by variation of parameters is

f1=(b1+B1)|F|1−nn−2(4.16)

for some function B₁ = B₁(u) and constant b₁ ∈ ℝ. Putting (4.16) into (3.4b), and using the fact that |F|1−nn−2 is the solution to the homogeneous differential equation (4.7b), one obtains

0=a(1−2g′)+(b1+B1)(|F|1−nn−2)′g+B1′|F|1−nn−2g+2af2g+(b1+B1)|F|1−nn−2=(b1+B1)((|F|1−nn−2)′g+|F|1−nn−2)+B1′|F|1−nn−2g+a(1−2g′)+2af2g=B1′|F|1−nn−2g+a(1−2g′)+2agf2.

As f₂ has the form (4.14) and B₂ = νA₂ by (4.2b), we obtain that

B1′=|F|n−1n−2ga(2g′−1−2(g′+b2+B2))=−a(n−1n−2)ν|F|1n−2F′(1+2b2+2νA2)=−2a(b2+A2)(n−1n−2)|F|1n−2F′−a(2b2(ν−1)+ν)(n−1n−2)|F|1n−2F′.(4.17)

Comparing (4.17) and (4.6) for k = 1 we obtain

B1′=A1′−a(2b2(ν−1)+ν)n−1n−2|F|1n−2F′.(4.18)

Hence a particular solution B₁ of the differential equation (4.18) has the form

B1=A1−a(2b2(1−ν)+1)|F|n−1n−2.(4.19)

Therefore, (4.3a) holds for k = 1, and (4.2c) is proved.

Finally, we prove that (4.3a) holds for k = 0. For k = 0 the function f₀ is the solution of the inhomogeneous linear differential equation (3.4a). The general solution of (3.4a) by (4.8) and by variation of parameters is

f0=(b0+B0)(n−2n−1)|F|−nn−2(F′)−1(4.20)

for some function B₀ = B₀(u) and constant b₀ ∈ ℝ. Putting (4.20) into (3.4a), and using the fact that (n−2n−1)|F|−nn−2(F′)−1 is the solution to the homogeneous differential equation (4.7a), we obtain

0=−a2g+(b0+B0)(n−2n−1)(|F|−nn−2(F′)−1)′g+B0′(n−2n−1)|F|−nn−2(F′)−1g+af1g−(b0+B0)(n−2n−1)|F|−nn−2(F′)−1g′+2(b0+B0)(n−2n−1)|F|−nn−2(F′)−1=B0′(n−2n−1)|F|−nn−2(F′)−1g−a2g+af1g.

As f₁ has the form (4.16), and B₁ has the form (4.19), we obtain

B0′=(a2−af1)(n−1n−2)|F|nn−2F′=(a2−a(b1+B1)|F|1−nn−2)(n−1n−2)|F|nn−2F′=(a2−a(b1+A1−a(2b2(1−ν)+1)|F|n−1n−2)|F|1−nn−2)(n−1n−2)|F|nn−2F′=−a(b1+A1)(n−1n−2)|F|1n−2F′+a2(2+2b2(1−ν))(n−1n−2)|F|nn−2F′=−a(b1+A1)(n−1n−2)|F|1n−2F′+2a2(1+b2(1−ν))(n−1n−2)|F|nn−2F′.(4.21)

Comparing (4.21) and (4.6) for k = 0 we have

B0′=A0′+2a2(1+b2(1−ν))(n−1n−2)|F|nn−2F′.(4.22)

Therefore, a particular solution B₀ of (4.22) is

B0=A0+a2(1+b2(1−ν))ν|F|2(n−1)n−2.

Hence, (4.3a) holds for k = 0, and (4.2d) is proved. This finishes the proof of Theorem 4.1

5. Open problems

Several questions arise after determining the symmetries of (1.1). 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 (1.1).

Problem 5.1.

Determine the solutions of (1.1) provided f_k (0 ≤ k ≤ n, n ≥ 4) satisfy the conditions of Theorem 3.1.

The only remaining case for (1.1) not covered by Theorem 3.1 or by [21–26] is when n = 3. Then one cannot immediately conclude ξ_u = 0 from (2.3), because the u˙4 term of (2.3) is identically 0. In fact, for n = 3 the symmetry condition translates to

−ηtt+f0′η+f1ηt−f0ηu+2f0ξt=0,(ξtt−2ηtu)+f1′η+2f2ηt+f1ξt+3f0ξu=0,(2ξtu−ηuu)+f2′η+3f3ηt+f2ηu+2f1ξu=0,ξuu+f3′η+2f3ηt−f3ξt+f2ξu=0.(5.1)

A potential simplification of the system (5.1) might be to eliminate f₂ from (5.1) by using a coordinate change v = G(u), for some bijective, two-times differentiable G, for which G″(u) = G′(u) f₂(u) is satisfied (see e.g. [26]). This, however, still does not give an immediate answer as to what the solutions of (5.1) are.

Problem 5.2.

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

u¨=f0(u)+u˙f1(u)+u˙2f2(u)+u˙3f3(u),

where f₀, f₁, f₂, f₃ are arbitrary continuous functions in u.

Acknowledgements

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. These projects have been supported by the European Union, co-financed by the European Social Fund.

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TY  - JOUR
AU  - Ágota Figula
AU  - Gábor Horváth
AU  - Tamás Milkovszki
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TI  - The Lie symmetry group of the general Liénard-type equation
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