Algebraic Properties of First Integrals for Scalar Linear Third-Order ODEs of Maximal Symmetry

Abstract

By use of the Lie symmetry group methods we analyze the relationship between the first integrals of the simplest linear third-order ordinary differential equations (ODEs) and their point symmetries. It is well known that there are three classes of linear third-order ODEs for maximal cases of point symmetries which are 4, 5, and 7. The simplest scalar linear third-order equation has seven-point symmetries. We obtain the classifying relation between the symmetry and the first integral for the simplest equation. It is shown that the maximal Lie algebra of a first integral for the simplest equation 𝑦^′′′=0 is unique and four-dimensional. Moreover, we show that the Lie algebra of the simplest linear third-order equation is generated by the symmetries of the two basic integrals. We also obtain counting theorems of the symmetry properties of the first integrals for such linear third-order ODEs. Furthermore, we provide insights into the manner in which one can generate the full Lie algebra of higher-order ODEs of maximal symmetry from two of their basic integrals.

1. Introduction

Ordinary differential equations (ODEs) are a fertile area of study, especially the Lie algebraic properties of such equations and their first integrals have great importance. Initial investigations since the works of Lie [1] were motivated by physical problems, such as the free particle, one-dimensional harmonic oscillator, and Emden-Fowler equations and classification. Indeed, most of the earlier works dealt with second-order linear equations (see, e.g., [2–8]). In the investigation by Mahomed and Leach [9], they found that the Lie point symmetries of the maximal cases of scalar linear 𝑛th-order ODEs, 𝑛≥3, are 𝑛+1, 𝑛+2, and 𝑛+4. Thus for scalar linear third-order equations these correspond to 4, 5, and 7 symmetries. Moreover for scalar third-order linear ODEs, Govinder and Leach [10] provided the algebraic structure of the basic first integrals for such equations. They showed that the three equivalence classes each has certain first integrals with a specific number of point symmetries. They followed on the initial investigation of Leach and Mahomed [11] who considered the point symmetries of the basic first integrals of linear second-order ODEs. Then in the work [12] Flessas et al. attempted the symmetry structure for the first integrals of higher-order equations of maximal symmetry.

The subject of the present paper is the investigation of the Lie algebraic properties of first integrals of scalar linear third-order ODEs of the maximal class which is represented by 𝑦^′′′=0. We remind the reader that for the simplest class there has been some analysis made in Flessas et al. [12]. This is in regards to the maximal algebra possessed by an integral of 𝑦^′′′=0 which is listed in Table  II of [12]. However, this is incomplete. We extend this study and provide a complete analysis on the Lie point symmetries and first integrals for the simplest third-order ODE including the maximal algebra case. We firstly deduce the classifying relation between the point symmetries and first integrals for this simple class. Then we use this to study the point symmetry properties of the first integrals of 𝑦^′′′=0 which also represents all linearizable by point transformations third-order ODEs that reduce to this class.

We begin by noting the condition for symmetries of the first integrals of scalar linear ODEs of order one. Then for completeness we review briefly the results of the paper by Mahomed and Momoniat [13] which discusses the relationship between the point symmetries of the first integrals of scalar linear second-order ODEs. These two cases are shown to be distinct in terms of their algebraic properties of their integrals when compared to higher-order ODEs of maximal symmetry.

1.1. Linear First-Order Equations

Consider the simplest first-order ODE 𝑦^′=0. (1) It is easy to see that 𝜕𝑋=𝜉𝑥,𝑦𝜕𝜕𝑥+𝜂𝑥,𝑦𝜕𝑦 (2) is a point symmetry generator of (1) if 𝑋^[1]𝑦^′_𝑦′=0=0, (3) where 𝑋^[1]𝜕=𝜉𝜕𝜕𝑥+𝜂𝜕𝑦+𝜁_𝑥𝜕𝜕𝑦^′ (4) with 𝜁_𝑥=𝐷_𝑥𝜂−𝑦^′𝐷_𝑥(𝜉), (5) in which 𝐷_𝑥 is the total differentiation operator and 𝑋^[1] is the first prolongation of the generator 𝑋. We quickly see that 𝜂_𝑥=0,𝜂=𝜂𝑦, (6) where 𝜂 is an arbitrary function of 𝑦. Therefore, 𝜕𝑋=𝜉𝑥,𝑦𝑦𝜕𝜕𝑥+𝜂.𝜕𝑦 (7) Thus there is an infinite number of point symmetries. We now show that only 𝑋=𝜉(𝑥,𝑦)𝜕/𝜕𝑥 are symmetries of the first integral.

This forms an infinite-dimensional subalgebra of the Lie algebra of (1).

Now 𝐼=𝑦 is a first integral of (1). It has point symmetry 𝑋 as in (2) if 𝑋𝐼=0. (8) This implies that 𝜂=0 which immediately results in 𝜕𝑋=𝜉𝑥,𝑦.𝜕𝑥 (9) There is an infinite number of symmetries of the first integral 𝐼=𝑦 of (1).

Let 𝐹 be an arbitrary function of 𝐼, namely, 𝐹=𝐹(𝐼). The symmetry of this general function of the first integral is 𝑋𝐹=𝑋𝐼𝜕𝐹𝜕𝐼=0. (10) Therefore 𝑋 as in (9) is a symmetry of 𝐼=𝑦 and also any function of 𝐹(𝑦). Since any scalar first-order ODE is equivalent to the simplest one (1), this means that a first integral of a nonlinear first-order ODE has infinitely many symmetries too.

As an example, we consider the nonlinear first-order Riccati equation 𝑦^′+𝑦²=0, (11) the first integral of which is 1𝐼=𝑦−𝑥. (12) A symmetry of the first integral (12) is 𝜕𝑋=𝜕𝑥−𝑦²𝜕.𝜕𝑦 (13) In fact, we have an infinite number of symmetries given by 𝜕𝑋=𝜉𝑥,𝑦𝜕𝑥−𝑦²𝜕,𝜕𝑦 (14) where 𝜉 is an arbitrary function in its arguments.

Therefore we note here that the symmetries of the first integrals of a first-order equation form a proper subalgebra of the Lie algebra of the equation itself. We cannot generate the full algebra as is the case for linear scalar second-order ODEs [11] by use of the algebra of the basic integral alone.

The symmetries of the first integrals of scalar linear second-order ODEs have interesting properties [11, 13]. The first integrals of such linear equations can have 0, 1, 2, and the maximum 3 symmetries [13]. The Lie algebra of the maximum case is unique. Peculiar to such equations is the other remarkable property that their Lie algebra is generated by the symmetry properties of the basic integrals and their quotient [11].

Below we study the symmetry properties of first integral for the simplest scalar linear third-order ODEs of maximal point symmetry. In the case of the basic first integrals, the algebraic properties are known from the work [10]. Here we pursue the relationship between symmetries and first integrals of scalar linear third-order ODEs for the simplest and maximal class. We obtain the classifying relation for this class and invoke this to arrive at counting theorems and the result on the maximal case of symmetries of the first integrals.

In the following we look at algebraic properties of first integrals for the seven point symmetry case by deriving the classifying relation between the symmetries and their first integrals. We use this relation to arrive at interesting properties which appear for the first time in the literature.

2. Algebraic Properties of the First Integrals of 𝑦^′′′=0

We consider the simplest third-order ODE 𝑦^′′′=0, (15) which as is well known has the seven symmetries (Lie [1] and e.g., [4]) 𝑋₁=𝜕𝜕𝑦,𝑋₂𝜕=𝑥,𝑋𝜕𝑦₃=𝑥²𝜕𝜕𝑦,𝑋₄𝜕=𝑦𝜕𝑦,𝑋₅=𝜕,𝑋𝜕𝑥₆𝜕=𝑥𝜕𝜕𝑥+𝑦𝜕𝑦,𝑋₇=𝑥²𝜕𝜕𝜕𝑥+2𝑥𝑦.𝜕𝑦 (16) We have listed the symmetries in the order of the solution symmetries being first, then the homogeneity symmetry, and the remaining three which form the algebra 𝑠𝑙(2,𝑅). This ODE (15) also represents all linearizable third-order ODEs reducible to it via point transformation. The order in which the symmetries appear in (16) is used in what follows. It is obvious that (15) has three functionally independent first integrals𝐼₁=𝑦^′′,𝐼₂=𝑥𝑦^′′−𝑦^′,𝐼₃=12𝑥²𝑦^′′−𝑥𝑦^′+𝑦. (17a)(17b)(17c)We use the ordering of the integrals as given in [10]. The first integral (17a) has four symmetries [10] 𝑋₁=𝜕,𝑋𝜕𝑥₂=𝜕,𝑋𝜕𝑦₃𝜕=𝑥,𝑋𝜕𝑦₄𝜕=𝑥𝜕𝜕𝑥+2𝑦,𝜕𝑦 (18) from which we observe that there are two solution symmetries, one translation in 𝑥 symmetry and a scaling symmetry. The translation in 𝑥 symmetry is a subset of the 𝑠𝑙(2,𝑅) symmetries with 𝑋₄ being a combination of the uniform scaling symmetry in both variables contained in the 𝑠𝑙(2,𝑅) symmetries together with the homogeneity symmetry. Part of this fact was also noted in [12]. The second first integral (17b) has three symmetries [10] 𝑌₁=𝜕,𝑌𝜕𝑦₂=𝑥²𝜕,𝑌𝜕𝑦₃𝜕=𝑥𝜕𝜕𝑥+𝑦,𝜕𝑦 (19) with two solution symmetries and 𝑌₃ being part of the 𝑠𝑙(2,𝑅) symmetries. The third first integral (17c) also has four symmetries [10] 𝐺₁𝜕=𝑥,𝐺𝜕𝑥₂𝜕=𝑥,𝐺𝜕𝑦₃=𝑥²𝜕,𝐺𝜕𝑦₄=𝑥²𝜕𝜕𝜕𝑥+2𝑥𝑦.𝜕𝑦 (20) Again one can see the solution symmetries, scaling, and the symmetry 𝐺₄ which is contained in the 𝑠𝑙(2,𝑅) symmetries. Note further that the symmetries in (20) are found by multiplying those of (18) by the factor 𝑥. In fact these two sets are equivalent via a point transformation [10]. The other important properties are discussed in the next section.

Below we obtain the classifying relation.

2.1. Classifying Relation for the Symmetries and Integrals

Now let 𝐹 be an arbitrary function of the integrals (17a), (17b), and (17c), 𝐼₁, 𝐼₂, and 𝐼₃, namely, 𝐹=𝐹𝐼₁,𝐼₂,𝐼₃. (21) The symmetry of this general function of the first integrals is 𝑋^[2]𝐹=𝑋^[2]𝐼₁𝜕𝐹𝜕𝐼₁+𝑋^[2]𝐼₂𝜕𝐹𝜕𝐼₂+𝑋^[2]𝐼₃𝜕𝐹𝜕𝐼₃=0, (22) where 𝑋^[2]𝐼₁=𝜉𝜕𝜕𝜕𝑥+𝜂𝜕𝑦+𝜁_𝑥𝜕𝜕𝑦^′+𝜁_𝑥𝑥𝜕𝜕𝑦^′′𝑦^′′=𝜁_𝑥𝑥,𝑋^[2]𝐼₂=𝜉𝜕𝜕𝜕𝑥+𝜂𝜕𝑦+𝜁_𝑥𝜕𝜕𝑦^′+𝜁_𝑥𝑥𝜕𝜕𝑦^′′𝑥𝑦^′′−𝑦^′=𝜉𝑦^′′−𝜁_𝑥+𝑥𝜁_𝑥𝑥,𝑋^[2]𝐼₃=𝜉𝜕𝜕𝜕𝑥+𝜂𝜕𝑦+𝜁_𝑥𝜕𝜕𝑦^′+𝜁_𝑥𝑥𝜕𝜕𝑦^′′×12𝑥²𝑦^′′−𝑥𝑦^′+𝑦=𝜉𝑥𝑦^′′−𝑦^′+𝜂−𝑥𝜁_𝑥+12𝑥²𝜁_𝑥𝑥. (23) The coefficient functions 𝜉, 𝜂, 𝜁_𝑥, and 𝜁_𝑥𝑥 are 𝜉=𝑎₅+𝑥𝑎₆+𝑥²𝑎₇,𝜂=𝑎₁+𝑥𝑎₂+𝑥²𝑎₃+𝑦𝑎₄+𝑦𝑎₆+2𝑥𝑦𝑎₇,𝜁_𝑥=𝑎₂+2𝑥𝑎₃+𝑦^′𝑎₄+2𝑦𝑎₇,𝜁_𝑥𝑥=2𝑎₃+𝑦^′′𝑎₄−𝑦^′′𝑎₆+2𝑦^′−2𝑥𝑦^′′𝑎₇. (24) These are obtained by setting 𝑋^[2]=⁷_𝑖=1𝑎_𝑖𝑋_𝑖^[2], (25) where the 𝑋_𝑖 are the symmetry generators as given in (16) and the 𝑎_𝑖 are constants. The reason being that the symmetries of the first integrals are always the symmetries of the equation (see [14]).

After substitution of the values of 𝑋^[2]𝐼₁, 𝑋^[2]𝐼₂, and 𝑋^[2]𝐼₃ as in (23), with 𝜉, 𝜂, 𝜁_𝑥, and 𝜁_𝑥𝑥 given in (24), as well as by use of the first integrals 𝐼₁=𝑦^′′, 𝐼₂=𝑥𝑦^′′−𝑦^′, and 𝐼₃=(1/2)𝑥²𝑦^′′−𝑥𝑦^′+𝑦 in (22), we arrive at the classifying relation 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂𝜕𝐹𝜕𝐼₁+−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃𝜕𝐹𝜕𝐼₂+𝑎₁+𝑎₄+𝑎₆𝐼₃+𝑎₅𝐼₂𝜕𝐹𝜕𝐼₃=0. (26) The relation (26) explicitly provides the relationship between the symmetries and the first integrals of the simple third-order equation (15). We invoke this to classify the first integrals according to their symmetries in what follows.

We use the classifying relation (26) to establish the number and property of symmetries possessed by the first integrals of the simplest third-order equation (15).

There arise five cases. We deal with each below.

Case 1 (no symmetry). If 𝐹 is any arbitrary function of 𝐼₁, 𝐼₂, and 𝐼₃, then 𝐹_𝐼1, 𝐹_𝐼2, and 𝐹_𝐼3 are not related to each other. In this case we have from (26) that 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂=0,−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃𝑎=0,₁+𝑎₄+𝑎₆𝐼₃+𝑎₅𝐼₂=0. (27) It is easy to see from (27) that all the 𝑎’s are zero. Therefore there exists no symmetry for this case.
As an illustrative example, if we take 𝐹=𝐼₁𝐼₂ln𝐼₃, then (26) straightforwardly yields 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂𝐼₂𝐼₃ln𝐼₃+−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃𝐼₁𝐼₃ln𝐼₃+𝑎₁+𝑎₄+𝑎₆𝐼₃+𝑎₅𝐼₂𝐼₁𝐼₂=0. (28) This easily results in all the 𝑎’s being zero.

Case 2 (one symmetry). If 𝐹 satisfies the relation (26), then there exists one symmetry. For the simple symmetries of (15) one obtains further symmetries except for 𝑋₆ which we consider below.
If we take 𝐹=𝐼₁𝐼₂𝐼₃ or any function of this product, then the relation (26) becomes 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂𝐼₂𝐼₃+−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃𝐼₁𝐼₃+𝑎₁+𝑎₄+𝑎₆𝐼₃+𝑎₅𝐼₂𝐼₁𝐼₂=0. (29) In (29), 𝑎₁ to 𝑎₇ are zeros except 𝑎₆ which gives the one symmetry 𝑋₆𝜕=𝑥𝜕𝜕𝑥+𝑦.𝜕𝑦 (30) Consider 𝐽=𝐼₁𝐼₃−(1/2)𝐼²₂ and 𝐼₂=𝑥𝑦^′′−𝑦^′. Now let 𝐹=𝐽/𝐼₂ (cf. [12]). Then relation (26) becomes 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂2𝐼₂𝐼₃+−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃−2𝐼₁𝐼₃−𝐼²₂+𝑎₁+𝑎₄+𝑎₆𝐼₃+𝑎₅𝐼₂2𝐼₁𝐼₂=0. (31) We see here that all the 𝑎’s are zero except 𝑎₆. Therefore there exists one symmetry which again is (30).
In fact similar to the free particle equation [13], there are many one symmetry cases.

Case 3 (two symmetries). Here there are many cases as well. We begin by utilizing the Lie table for the classification of the two-dimensional algebras in the plane which are given, for example, in [6]. They are 𝑌₁=𝜕𝜕𝑦,𝑌₂=𝜕,𝑌𝜕𝑥₁=𝜕𝜕𝑦,𝑌₂𝜕=𝑥,𝑌𝜕𝑦₁=𝜕𝜕𝑦,𝑌₂𝜕=𝑥𝜕𝜕𝑥+𝑦,𝑌𝜕𝑦₁=𝜕𝜕𝑦,𝑌₂𝜕=𝑦.𝜕𝑦 (32) These form subalgebras of the Lie algebra of symmetries of (15) as can clearly be observed.
We take the first realization listed above. If 𝑎₁ is arbitrary in (26), then 𝑋₁=𝜕/𝜕𝑦 implies that 𝐹 is independent of 𝐼₃. Further 𝑋₅=𝜕/𝜕𝑥 yields that 𝐹 does not depend on 𝐼₂ as well. Since we require that 𝜕𝐹/𝜕𝐼₁≠0, we have 2𝑎₃+𝐼₁𝑎₄−𝑎₆−2𝑎₇𝐼₂=0, (33) from which it follows that 𝑎₃=𝑎₇=0 and 𝑎₄=𝑎₆. Thus we end up with two more symmetries 𝑋₂ and 𝑋₄+𝑋₆. These turn out to be the four symmetries of the integral 𝐼₁ given in (18).
Likewise for the second realization we obtain (18) again. For the third realization listed above we find the symmetries of 𝐼₂ as in (19).
Hence the first three realizations listed do not provide maximal two symmetries for the first integrals of (15).
In fact the fourth realization results in a two symmetry case as 𝑎₁ and 𝑎₄ are arbitrary and give rise to 𝜕𝐹𝜕𝐼₃=0,𝐼₁𝜕𝐹𝜕𝐼₁+𝐼₂𝜕𝐹𝜕𝐼₂=0, (34) which has the solution 𝐼𝐹=𝐻²𝐼₁. (35) The further substitution of this form into the relation (26) constrains all the 𝑎’s to be zero except for 𝑎₁ and 𝑎₄. This result prompts the following simple products and quotients that do give two symmetries.
If 𝐹=𝐼₁𝐼₂, then relation (26) yields 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂𝐼₂+−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃𝐼₁=0. (36) Here we observe that 𝑎₂, 𝑎₃, 𝑎₅, and 𝑎₇ are zeros whereas 𝑎₁ is arbitrary and 𝑎₆=2𝑎₄, and therefore we obtain the two symmetries 𝑋₁=𝜕,𝜕𝜕𝑦𝑌=2𝑥𝜕𝜕𝑥+3𝑦,𝜕𝑦 (37) which form a two-dimensional algebra with 𝑋₁,𝑌=3𝑋₁. (38) If we set 𝐹=𝐼₁𝐼₃, then we end up getting 𝑎₁, 𝑎₃, 𝑎₄, 𝑎₅, and 𝑎₇ equal to zero. Since 𝑎₂ and 𝑎₆ are arbitrary so they result in two symmetries 𝑋₂𝜕=𝑥,𝑋𝜕𝑦₆𝜕=𝑥𝜕𝜕𝑥+𝑦,𝜕𝑦 (39) with Lie bracket 𝑋₂,𝑋₆=0. (40) If we take 𝐹=𝐼₂𝐼₃, then we see that 𝑎₃ is arbitrary and 𝑎₆=−2𝑎₄ which then give rise to the two symmetries 𝑋₃=𝑥²𝜕,𝜕𝜕𝑦𝑌=2𝑥𝜕𝜕𝑥+𝑦,𝜕𝑦 (41) with 𝑋₃,𝑌=−3𝑋₃. (42) Consider now 𝐹=𝐼₃/𝐼₁. This shows that 𝑎₂ and 𝑎₄ are arbitrary, and the resulting two symmetries are 𝑋₂𝜕=𝑥,𝑋𝜕𝑦₄𝜕=𝑦,𝜕𝑦 (43) with commutation relation 𝑋₂,𝑋₄=𝑋₂. (44) If we let 𝐹=𝐼₃/𝐼₂, then here 𝑎₃ and 𝑎₄ are arbitrary, and therefore the two symmetries are 𝑋₃=𝑥²𝜕,𝑋𝜕𝑦₄𝜕=𝑦,𝜕𝑦 (45) with 𝑋₃,𝑋₄=𝑋₃. (46) If 𝐽=𝐼₁𝐼₃−(1/2)𝐼²₂, 𝐼₁=𝑦^′′, and 𝐹=𝐽/𝐼₁ (cf. [12]), then we have from the relation (26) 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂12𝐼²₂+−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃−𝐼₁𝐼₂+𝑎₁+𝑎₄+𝑎₆𝐼₃+𝑎₅𝐼₂𝐼²₁=0. (47) This results in 𝑎₁, 𝑎₂, 𝑎₃, and 𝑎₇ being zero whereas 𝑎₅ is arbitrary, and 𝑎₄=−𝑎₆, which give rise to the two symmetries 𝑋₅=𝜕,𝜕𝜕𝑥𝑌=𝑥,𝜕𝑥 (48) with 𝑋₅,𝑌=𝑋₅. (49) If 𝐽=𝐼₁𝐼₃−(1/2)𝐼²₂, 𝐼₃=(1/2)𝑥²𝑦^′′−𝑥𝑦^′+𝑦, and 𝐹=𝐽/𝐼₃ (cf. [12]), then relation (26) yields 2𝑎₃+𝑎₄−𝑎₆𝐼₁−2𝑎₇𝐼₂2𝐼²₃+−𝑎₂+𝑎₄𝐼₂+𝑎₅𝐼₁−2𝑎₇𝐼₃−2𝐼₃𝐼₂+𝑎₁+𝑎₄+𝑎₆𝐼₃+𝑎₅𝐼₂𝐼²₂=0. (50) The above relation shows that 𝑎₁, 𝑎₂, 𝑎₃, and 𝑎₅ are zeros, 𝑎₇ is arbitrary, and 𝑎₄=𝑎₆. This imply two symmetries 𝑋₇=𝑥²𝜕𝜕𝜕𝑥+2𝑥𝑦,𝜕𝜕𝑦𝑌=𝑥𝜕𝜕𝑥+2𝑦,𝜕𝑦 (51) together with 𝑋₇,𝑌=−𝑋₇. (52) There are thus many two symmetry cases. One could obtain more. Also they could arise as different combinations of the seven symmetries (16). So one concludes that the two-dimensional algebra cases are not unique. We have seen the occurrence of both Abelian and non-Abelian Lie algebras.

Case 4 (three symmetries). Here we use the three-dimensional real realizations of Mahomed and Leach [7]. The notation used is that given in [4]. Since we adapt these realizations as symmetries of third-order equations, the entries for the first entry 𝐿_3;1 and the nonsolvable algebra 𝐿^I_3;8 are those which are symmetries of such equations.
Note that for 𝐿^II_3;8, 𝐿^III_3;8, and 𝐿^IV_3;8 one can use the realizations as given in Table 1 or the ones obtained by interchanging 𝑥 and 𝑦 in the realizations given. The reason for this is that one still obtains third-order representative equations for the latter realizations.
As 𝐿_3;1 is the first three-dimensional algebra in Table 1, we start with that. We want this Abelian algebra to be admitted by a first integral of (15). We utilize the classifying relation (26). Therefore 𝑎₁, 𝑎₂, and 𝑎₃ are arbitrary which imply that 𝐹 is constant. Hence this algebra is not admitted by any first integral of (15) although, if it is admitted by a nonlinear third-order ODE, it implies linearization (see Mahomed and Leach [6]).

Table 1 

Realizations of three-dimensional algebras in the real plane.

It is not difficult to deduce that the same applies to the algebras 𝐿_3;2, 𝐿^I_3;3, 𝐿^II_3;3, 𝐿^I_3;4, 𝐿^II_3;4, 𝐿^I_3;5, 𝐿^II_3;5, and 𝐿^I_3;6 when 𝑎≠2 and 𝐿^II_3;6 when 𝑎≠1/2,−1. If 𝑎=2 for 𝐿^I_3;6, then one ends up with four symmetries of 𝐹=𝐹(𝐼₁) which are (18). The same applies to 𝐿^II_3;6, 𝑎=1/2. For both these cases, the algebras 𝐿^I_3;6, 𝑎=2 and 𝐿^II_3;6, 𝑎=1/2 are admitted by a first integral 𝐹=𝐹(𝐼₁) but these are not maximal and contained in a four-dimensional algebra, spanned by operators (18).

Now we focus on the three symmetry case which is admitted by 𝐹=𝐹(𝐼₂). These three symmetries are given in (19). The Lie algebra of the generators (19) has nonzero commutators 𝑋₁,𝑋₃=𝑋₁,𝑋₂,𝑋₃=−𝑋₂, (53) with the elements thus forming the Lie algebra 𝐿^II_3;6, 𝑎=−1. In fact the transformation 𝑋=𝑥²,𝑌=𝑥+𝑦, (54) maps it to the canonical form of Table 1, namely, 𝑋₁=𝜕,𝜕𝑌𝑋₂𝜕=𝑋,𝜕𝑌𝑋₃𝜕=2𝑋𝜕𝜕𝑋+𝑌.𝜕𝑌 (55) Therefore the symmetries of 𝐹=𝐹(𝐼₂) has the Lie algebra 𝐿^II_3;6, 𝑎=−1.

There is yet another three-dimensional algebra which is admitted by a first integral of (15). This occurs for 𝐿^I_3;8. We use the classifying relation (26). Here 𝑎₅, 𝑎₆, and 𝑎₇ are arbitrary. Making these constants one at a time unity and the rest zero yield is 𝐼₁𝜕𝐹𝜕𝐼₂+𝐼₂𝜕𝐹𝜕𝐼₃=0,−𝐼₁𝜕𝐹𝜕𝐼₁+𝐼₃𝜕𝐹𝜕𝐼₃𝐼=0,₂𝜕𝐹𝜕𝐼₁+𝐼₃𝜕𝐹𝜕𝐼₂=0. (56) The solution to this system (56) yields 𝐹=𝐹𝐼₁𝐼₃−12𝐼²₂. (57) Thus the basic first integral 𝐽=𝐼₁𝐼₃−(1/2)𝐼²₂ has the algebra 𝐿^I_3;8.

The Lie algebras 𝐿^II_3;8, 𝐿^III_3;8, and 𝐿^IV_3;8 are not admitted by any first integral of (15) as these are subalgebras of the maximal six-dimensional algebras (see Ibragimov and Mahomed [3]) which are admitted by nonlinear third-order ODEs not reducible to the simplest equation (15).

In the case of the algebra 𝐿_3;9 one has the situation that this algebra is not a subalgebra of the seven-dimensional algebra of (15) (Wafo Soh and Mahomed [15]).

In conclusion of this discussion, we have two three-dimensional algebras admitted by a first integral of (15) which are 𝐿^I_3;8 and 𝐿^II_3;6, 𝑎=−1.

We state the following theorem.

Theorem 1. If a first integral of the simplest third-order ODE, 𝑦^′′′=0, admits a three-dimensional algebra, then it is one of the two three-dimensional algebras 𝐿^II_3;6, 𝑎=−1, or 𝐿^I_3;8.

The proof follows from the previous discussions. Note here that the three-dimensional algebra admitted is not unique.

Case 5 (four symmetries). In the four symmetry case we have that only 𝐼₁ and 𝐼₃ possess four symmetries. They are given by (18) and (20). Both are similar by a point transformation [10]. The Lie algebra of (18) is 𝐿^I_4;9 (see [12]). This can be seen by interchanging 𝑋₁ and 𝑋₂ in (18). Thus the nonzero commutation relations are 𝑋₂,𝑋₃=𝑋₁,𝑋₁,𝑋₄=2𝑋₁,𝑋₂,𝑋₄=𝑋₂,𝑋₃,𝑋₄=𝑋₃. (58) We have the following theorem.

Theorem 2. The maximum dimension of the Lie algebra admitted by any first integral of the simplest third-order ODE, 𝑦^′′′=0, or any third-order ODE linearizable by point transformation to the simplest ODE is 𝐿^I_4;9.

Proof. Any first integral of 𝑦^′′′=0 or third-order ODE reducible to the simplest ODE by point transformation cannot admit a maximal algebra of dimension greater than four since the basic integrals 𝐼₁ and 𝐼₃ have the unique four dimension algebra 𝐿^I_4;9. The other integrals have lower dimensional Lie algebras in the classification obtained above.

Finally, we have the following counting theorem.

Theorem 3. A first integral of the simplest third-order ODE, 𝑦^′′′=0, or any linearizable third-order ODE by point transformation to the simplest ODE can have 0, 1, 2, 3, or the maximum 4 symmetries. The four symmetry case is unique.

3. Symmetry Properties of First Integrals of Higher-Order ODEs: Some Remarks

In the case of symmetries of the simplest first-order ODE 𝑦^′=0 we have seen that the algebra of any first integral constitutes a proper subalgebra of the equation itself. One cannot generate the full algebra of 𝑦^′=0 via the algebras of any integral. This result also applies to any scalar first-order ODE due to equivalence of this with the simplest equation.

What occurs for scalar linear second-order ODEs is very different to the first-order ODE case. Here as has been shown in [11], the Lie algebra of 𝑦^′′=0 which represents any linear or linearizable second-order ODE can be generated by the three-dimensional algebras of the triplets of the basic integrals and their quotient which are isomorphic to each other. Thus in this case one requires not only the basic integrals, say 𝐼₁ and 𝐼₂, but also a functionally dependent quotient integral 𝐽=𝐼₂/𝐼₁.

Another important point to make is that the full Lie algebra of the simplest third-order equation (15) is generated by the four symmetries (18) as well as the three symmetries 𝐺₂, 𝐺₃, and 𝐺₄ of (20). Hence one requires only the symmetries of the basic integrals 𝐼₁ and 𝐼₃ to generate the full algebra of our equation (15). One should contrast this with what happens for 𝑦^′=0 and 𝑦^′′=0 discussed above. So one has the seven symmetries of our equation (15) being generated by four symmetries of 𝐼₁ together with three symmetries of 𝐼₃. The natural question then is as follows: what occurs for higher-order ODEs of maximal symmetry? Patterns emerge, some of which are discussed in the three propositions in the paper of Flessas et al. [12]. We discuss another important property, namely, that of generation of the full algebra via integrals now.

Consider the 𝑛th-order ODE of maximal symmetry 𝑦^(𝑛)=0,𝑛≥3. (59) This ODE (59) has 𝑛+4 symmetries as is well known. The 𝑛 first integrals of (59) are easily constructible and we focus on the first and last which are 𝐼₁=𝑦^(𝑛−1),𝐼_𝑛=^𝑛_𝑖=1(−1)^𝑖−1()!𝑥𝑛−𝑖^𝑛−𝑖𝑦^{(𝑛−𝑖)}. (60)(61) The first integral (60) has 𝑛+1 symmetries which are 𝑋_𝑖=𝑥^𝑖−1𝜕𝑋𝜕𝑦,𝑖=1,…,𝑛−1,_𝑛=𝜕,𝑋𝜕𝑥_𝑛+1𝜕=𝑥+()𝑦𝜕𝜕𝑥𝑛−1.𝜕𝑦 (62) This forms an 𝑛+1-dimensional subalgebra of (59). Now the first integral (61) has symmetries 𝑌_𝑖=𝑥^𝑖𝜕𝑌𝜕𝑦,𝑖=1,…,𝑛−1,_𝑛𝜕=𝑥,𝑌𝜕𝑥_𝑛+1=𝑥²𝜕+()𝜕𝜕𝑥𝑛−1𝑥𝑦,𝜕𝑦 (63) which one can see comes from multiplying the symmetries of (62) by 𝑥. We can observe from these two sets (62) and (63) that the full Lie algebra of our equation (59) is generated from the 𝑛+1 symmetries of (62) and 3 symmetries of (63), namely, 𝑌_𝑛−1, 𝑌_𝑛, and 𝑌_𝑛+1. Further the two sets (62) and (63) are equivalent to each other by means of the point transformation 1𝑥=𝑥,𝑦𝑦=𝑥^𝑛−1. (64) This is an extension of the transformation given in [10] and is for higher-order ODEs. In [10] it was given for third-order ODEs.

We have the following theorem, the proof of which is evident from the above.

Theorem 4. The full Lie algebra of the 𝑛th-order ODE 𝑦^(𝑛)=0, 𝑛≥3, is generated by two subalgebras, namely, the 𝑛+1-dimensional algebra ⟨𝑋_𝑗∶𝑗=1,…,𝑛+1⟩ of 𝐼₁ and the three-dimensional subalgebra ⟨𝑌_𝑛−1,𝑌_𝑛,𝑌_𝑛+1⟩ of 𝐼_𝑛.

Hence the picture is quite distinct for the manner in which the full Lie algebra is generated for the ODEs 𝑦^′=0, 𝑦^′′=0 and, 𝑦^(𝑛)=0, 𝑛≥3. This is also consistent with the properties of their symmetry algebra which are different (see, e.g., [4]).

Acknowledgments

K. S. Mahomed is thankful to the University of the Witwatersrand and the NRF (South Africa) for financial support. E. Momoniat thanks the NRF (South Africa) for financial support.