2. Lie Point Symmetries

In this section, we study the Lie point symmetries of (2). Following the classical Lie technique for calculating the symmetries of differential equations [⁶, ¹⁹, ³⁵, ³⁶], we carried out the complete group classification for (2). The corresponding result can be stated as follows.

Proposition 2.1. The Lie point symmetry group of the generalization of Kummer -Schwarz differentialequation (2)with arbitrary f, is generated by

for the special choices of f listed below (Table 1) we have,

Where

with z ₁ ,Z ₂ - linearly independent solutions of 2z" - f (x)z 0.

Proof. A general form of the one-parameter Lie group admitted by (2) is given by

where Є is the group parameter. The vector field associated with the group of transformations shown above can be written as . Applying its third prolongation

to (2), we must find the infinitesimals ξ(x, y), η(x,y) satisfying the symmetry condition

associated with (2). Here, η _[x] , η _[xx] and η _[xxx] are the coefficients in Γ ⁽³ given by

where D _x is the total derivative operator: D _x = ∂ _x + y _x ∂ _y + y _xx ∂ _yx + y _xxx ∂ _yx Replacing (11) into (10) we obtain:

If we denote f ≅ f (x) and substitute in the last expression we have

Thus, we can rearrange the last expression with respect to 1, y _x , y ² _x , y ³ _x , y ⁴ _x ,y _xx , y ² _xx , y ⁻¹ _x y _x x, y _x y _xx , y ⁻¹ _x y ² _xx , y ⁻² _x y _xx and y ⁻² _x y ² _xx and canceling some terms we obtainthe determining equations for the symmetry group of (2). That is:

Solving in (12a), we have ξ = c₁(x) and , with k ₁ , k ₂ , k ₃ as arbitrary constants and c₁(x) arbitrary function. If f is arbitrary then in (12b) we have ξ = 0 and , where and k ₃ are arbitrary constants. Therefore the group of symmetries are Π₁ = ∂ _y , Π2 = y _∂ y, Π₃ = y ² ∂ _y . This is consistent with What is presented in [¹⁸, ⁹]. If f = α # 0 where a is an arbitrary constant. Thus, we have two cases in (12b) which are α > 0 and α < 0.

Case I: If α > 0 then in (12b) we have 2αξ_x + ξ _xxx = 0 and solving for ξ we obtain where k₇ , k₈ and kg are arbitrary constants, then of (12a) we have , Therefore the group of symmetries are

Case II: If α < 0 then in (12b) we have -2αξ _x + ξ _xxx = 0 and solving for ξ we obtain where k ₁₀ ,k ₁₁ and k₁₂ are arbitrary constants, then of (12a) we have . Therefore the group of symmetries are

If f = 0 then in (12b) we have ξ _xxx = 0 and solving for ξ we obtain where k₄, k₅ and k ₆ are arbitrary constants, then of (12a) we have . Therefore the group of symmetries are

This is consistent with what is presented in [⁹, ²⁰]. Let's remember that ξ = c₁(x), then for the other cases of f in (126) the following differential equation must be satisfied c ₁ (x)f _x + 2c _1,x f + c _1¡xxx = 0, so, if the functions c₁(x) and f (x) satisfy Eq. (126) then we have the following operator, additional to (9): X _c1 = c₁(x) ∂ _x, therefore following [³] we have the solution is given in the terms of two linearly independent solutions of the associated to (126) homogeneous second-order Sturm-Liouville equation

By the existence and uniqueness theorem if f (x) Є C ⁰ then there exists a nonzero solution z₁ of Eq. (13). We look for another solution z₂ of (13) such that

that is z ₁ and z ₂ are linearly independent solutions of (13) with Wronskian = 1 . Thus therefore following [³] we have the solution c₁(x) = k₁₃z₁ ²(x) + k₁₄z₁(x)z₂(x) + k₁₅z₂(x), then it follows that for any f ЄC ⁰ there are three additional symmetries. Namely:

Thus, the demonstration of the Proposition 2.1 is finished. 0

3. Optimal Algebra

Taking into account [¹⁵, ¹⁷, ³⁵, ⁴²], we present in this section the optimal algebra associated to the symmetry group of (1), that shows a systematic way to classify the invariant solutions.

To obtain the optimal algebra, we should first calculate the corresponding commutator table, which can be obtained from the operator

where i = 1, 2, with α, β = 1, • • • , 6 and ξ _α ⁱ , ξ _βi are the corresponding coefficients of the infinitesimal operators Π _α ,Π _β , . After applying the operator (14) to the symmetry group of (1), we obtain the operators that are shown in the following table

Now, the next thing is to calculate the adjoint action representation of the symmetries of (1) and to do that, we use Table 2 and the operator

Making use of this operator, we can construct the Table 3, which shows the adjoint representation for each Π_i.

Proposition 3.1. The optimal algebra associated to the equation (1) is given by the vector fields

Proof. To calculate the optimal algebra system, we start with the generators of symmetries (1) and a generic nonzero vector. Let

The objective is to simplify as many coefficients α _i as possible, through maps adjoint to G, using Table 3.

1) Assuming a_e = 1 in (15) we have that G = α ₁Π₁+ α ₂Π₂+ α ₃Π₃+ α ₄Π₄+α ₅Π₅+Π₆.Applying the adjoint operator to (Π₁ ,G), we get

1.1) Case is eliminated, therefore G ₁ = b ₁Π₁ +α ₃Π₃ + α ₄Π₄ + α ₅Π₅ +Π₆ with . Now, applying the adjoint operator to (Π₂ ,G ₁) we don’t have any reductions, thus applying the adjoint operator (Π₃ ,G ₁) we get G ₂ = Ad (exp (λ ₂Π₃))G ₁ = b ₁Π₁+b ₁ λ ² ₂ Π₂+( α ₃+b ₁ λ ₂ ²)Π₃ + α ₄Π₄ + α ₅Π₅ + Π₆.

1.1.A) Case α ² ₂ −4 α ₁ # 0. Using with α ² ₂ −4 α ₁# 0, is eliminated Π₃, thus G ₂ = b ₁Π₁ − α ₃Π₂+ α ₄Π₄+ α ₅Π₅+Π₆. Now, applying the adjoint operatorto (Π₄ ,G ₂), we get

Using is eliminated Π₅, thus G₃ = b₁Π₁ - α₃Π₂ + b₂ Π₄ + Π₆, with . Now applying the adjoint operator to (Π₅, G ₃ ), we don't have any reduction, but applying the adjoint operator to (Π₆, G ₃) we get

1.1.A.A1) Case α ² ₅ −2α ₄ # 0. Using with α ² ₅ −2α ₄ # 0, is eliminated Π₆. Then we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

1.1.A.A ₂ ) Case α ² ₅ - 2α _4- = 0. Then we get b₂ = 0, thus G₄ = b₁Π₁ - α ₃Π₂ + Π₆. Then we have an element of the optimal algebra

This is another way how a reduction of the generic element (15) ends.

1.1.B) Case α ² ₂ - 4α₁ = 0. We get b₁ = 0, thus G ₂ = α ₃Π₃ + α ₄Π₄ + α ₅Π₅ + Π₆. Now, applying the adjoint operator to (Π₄ ,G ₂), we have

Using , then Π ₅ is eliminated, then we obtain G ₅ = α ₃Π₃ + b ₄Π₄ + Π₆, with . Now applying the adjoint operator to (Π₅ ,G ₅), we don’t have any reduction, but applying the adjoint operator to (Π₆ ,G ₅) we have G ₆ =Ad (exp (λ ₆Π₆))G ₅ = α ₃Π₃ + b ₄Π₄ + 2b ₄ λ ₆Π₅ + (2λ ² ₆ b ₄ + 1)Π₆ .

1.1.B.B1) Case α ² ₅ −4α ₄ # 0. Using with α ² ₅ −4α ₄ # 0, is eliminated Π₆. Then we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

1.1.B.B2) Case α ² ₅ − 4α ₄ = 0. We get b ₄ = 0, then we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

1.2) Case α ₃ = 0. We get, G ₁ = (α ₁ − α 2λ1)Π₁ + α ₂Π₂ + α ₄Π₄ + α ₅Π₅ + Π₆. 1.2.A) Case α ₂ # 0. Using , with α ₂ # 0, is eliminated Π1, thus we get G ₁ = α ₂Π₂ + α ₄Π₄ + α ₅Π₅ + Π₆. Now, applying the adjoint operator to (Π₂ ,G ₅) and (Π₅ ,G ₅), we don’t have any reduction, then applying the adjoint operator to (Π₃ ,G ₁) we get G ₇ = Ad (exp (λ ₇Π₃))G ₁ = α ₂Π₂ + α ₂ λ ₇Π₃ + α ₄Π₄ + α ₅Π₅ +Π₆, it is clear that we don’t have any reduction, then we have G ₇ = α ₂Π₂+b ₆Π₃+ α ₄Π₄+ α ₅Π₅ +Π₆, with b ₆ = α ₂ λ ₇. Now, applying the adjoint operator to (Π₄ ,G ₇) we get G ₈ = Ad (exp (λ ₈Π₄))G ₇ = α ₂Π₂+b ₆Π₃+( α 4− α ₅ λ ₈+λ ² ₈ )Π4+( α ₅ −2λ ₈)Π₅+Π₆. Using , is eliminated Π₅, the we have G ₈ = α ₂Π₂ + b ₆Π₃ + b ₇Π₄ + Π₆, with . Now, applying the adjoint operator to (Π₆ ,G ₈) we get G ₉ =Ad (exp (λ ₉Π₆))G ₈ = α ₂Π₂ + b ₆Π₃ + b ₇Π₄ + 2b ₇ λ ₉Π₅ + (2b ₇ λ ² ₉ + 1)Π₆.

1.2.A.A1) Case α ² ₅ −4α ₄ # 0. Using with α ² ₅ −4α ₄ # 0, is eliminated Π₆. Then we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

1.2.A.A2) Case α² ₅ - 4α _4- = 0. We get b₇ = 0, then we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

1.2.B) Case α ₂ = 0. We get G ₁ = α ₁Π₁ + α ₄Π₄ + α ₅Π₅ + Π₆. Now, applying the adjoint operator to (Π₂ ,G ₅) and (Π₅ ,G ₅), we don’t have any reduction, then applying the adjoint operator to (Π₃ ,G ₁) we get G ₁₀ = Ad (exp (λ ₁₀Π₃))G ₁ =α ₁Π₁ +2α ₁ λ ₁₀Π₂ +α ₁ λ ₁₀Π₃ +α ₄Π₄ +α ₅Π₅ +Π₆, it is clear that we don’t have any reduction, then we have G ₁₀ = α ₁Π₁ + b ₉Π₂ + b ₁₀Π₃ + α ₄Π₄ + α ₅Π₅ + Π₆, with b ₉ = 2α ₁ λ ₁₀ and b ₁₀ = α ₁ λ ₁₀.

Now, applying the adjoint operator to (Π₄ ,G ₁₀) we get G ₁₁ =Ad (exp (λ ₁₁Π₄))G ₁₁ = α ₁Π₁ + b ₉Π₂ + b ₁₀Π₃ + (α ₄ − α ₅ λ ₁₁ + λ ² ₁₁)Π⁴ + (α ₅ − 2λ ₁₁)Π₅ + Π₆. Using , is eliminated Π₅, then we have G ₁₁ = α ₁Π₁ + b ₉Π₂ + b ₁₀Π₃ + b ₁₁Π₄ + Π₆, with .

Now, applying the adjoint operator to (Π₆ ,G ₁₁) we get G ₁₂ =Ad (exp (λ ₁₂Π₆))G ₁₁ = α ₁Π₁+b ₉Π₂+b ₁₀Π₃+b ₁₁Π₄+2b ₁₁ λ ₁₂Π₅+(2b ₁₁ λ ² ₁₂+1)Π₆.

1.2.B.A ₁) Case 4α ₄ − α ² ₅ # 0. Using , is eliminated Π₆, then wehave an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

1.2.B.A ₂) Case 4α ₄ −α ² ₅ = 0. We get b ₁₁ = 0, thus we have G ₁₂ = α ₁Π₁ +b ₉Π₂ +b ₁₀Π₃ + Π₆, then we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

2) Assuming α ₆ = 0 and α ₅ = 1 in (15) we have that G = α ₁Π₁ + α ₂Π₂ + α ₃Π₃ + α ₄Π₄ + Π₅. Applying the adjoint operator to (Π₂ ,G) and (Π₅ ,G) we don’t haveany reduction, thus applying the adjoint operator to (Π₁ ,G) we get

2.1) Case α₃ # 0. Using with α₃ # 0, is eliminated Π ₂, we get . Then using , we have G ₁₃ = b ₁₃Π₁ + α ₃Π₃ + α ₄Π₄ + Π₅. Now, applying the adjoint operator (Π₃, G ₁₃), we have

2.1.A) Case 4 α ₁ α ₃ - α² ₂ =0. Using is eliminated Π₃, the we get G₁₄ = b₁₃ Π₁ + 2b₁₃λ₁₄ Π ₂ + α ₄ Π ₄ + Π ₅. Now applying the adjoint operator to (Π ₄,G₁₄), we have:

Using λ ₁₅ = α ₄, is eliminated Π₄, then G ₁₅ = b ₁₃Π₁+2b ₁₃ α ₄Π₂+Π₅ . Now applying the adjoint operator to (Π₆ ,G ₁₅), we have:

We don't have any reduction, then we have an element of the optimal algebra

With b ₁₄= λ ₁₆. This is how a reduction of the generic element (15) ends.

2.1. B) Case 4α₁α₃ - α₂ ² , =0. We get b₁₃ = 0, then G ₁₄ = α ₃Π₃ + α ₄Π₄ + Π₅. Now applying the operator to (Π₄ ,G ₁₄) we have

Using λ ₁₇ = α ₄, is eliminated Π₄, then we have G ₁₇ = α ₃Π₃ + Π₅ . Now using the operator to (Π₆ ,G ₁₇), we get:

We don't have any reduction, then using λ₁₈ = b₁₅, we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

2.2) Case α₃ = 0. We get G ₁₃ = (α ₁ − α ₂ λ ₁₃)Π₁ + α ₂Π₂ + α ₄Π₄ + Π₅.

2.2.A1) Case α₂ # 0. Using , with α ₂ 0, is eliminated Π₁, then we have G ₁₃ = α ₂Π₂ + α ₄Π₄ + Π₅. Now applying the operator to (Π₃ ,G ₁₃) we have G ₁₉ = Ad (exp (λ ₁₉Π₃))G ₁₃ = α ₂Π₂ + α ₂ λ ₁₉Π₃ + α ₄Π₄ + Π₅ . It is clear that we don’t have any reduction, then using b ₁₆ = α ₂ λ ₁₉ we get

Now applying the operator to (Π₄ ,G ₁₉) we get: G ₂₀ = Ad (exp (λ ₂₀Π₄))G ₂₀ =α ₂Π₂ + b ₁₆Π₃ + (α ₄ − λ ₂₀)Π₄ + Π₅ . Then using λ ₂₀ = a ₄, is eliminated Π₄, thus G ₁₉ = α ₂Π₂ + b ₁₆Π₃ + Π₅ . Now applying the operator (Π₆ ,G ₂₀), we get

It is clear that we don't have any reduction. The using λ ₂₁ = b ₁₇ , we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

2.2.A₂) Case α₂ = 0. We get G₁₃ = α ₁Π₁+ α ₄Π₄+Π₅. Now applying the operator to (Π₃, G ₁₃) we have

we don't have any reduction, then using b₁₈ = 2α₁λ₂₂ and b₁₉ = α₁λ² ₂₂, we obtain

Now, applying the operator to (Π₄, G ₂₂) we have

Using λ₂₃ = α₄, is eliminated II₄ then we get G₂₃ = α ₁ Π ₁ + b₁₈ Π₂ + b_1g Π₃ + Π₅. Now applying the operator to (Π₃, G₂₃) we have

It is clear that we don't have any reduction, then using λ₂₄ = b₂₀, we have an element of the optimal algebra

This is how a reduction of the generic element (15) ends.

3) Assuming α ₃ = α ₅ =0 and α ₄ = 1 in (15), we have that G = α ₁ Π ₁ + α ₂ Π₂ + α ₃Π₃ + Π₄. Applying the adjoint operator to (Π₂,G), (Π ₄ ,G) and (Π₅,G) we don't have any reduction, on the other hand applying the adjoint operator to (Π₁ , G) we get

3.1) Case α ₃ # 0. Using with α3 # 0, in (32), Π₂ is liminated, therefore G ₂₅ = b ₂₁Π₁ + α ₃Π₃ + Π₄, with b ₂₁ = α ₁ + 2λ ₂₅ + α ₃ λ ² ₂₅. Now, applying the adjoint operator to (Π₃ ,G ₂₅), we get G ₂₆ = Ad (exp (λ ₂₆Π₃))G ₂₅ = b ₂₁Π₁ +2b ₂₁ λ ₂₆Π₂ + (α ₃ + b ₂₁ λ ² ₂₆)Π₃ + Π₄. It is clear that we don’t have any reduction, then G ₂₆ = b ₂₁Π₁ +b ₂₂Π₂ +b ₂₃Π₃ +Π₄, with b ₂₂ = 2b ₂₁ λ ₂₆ and b ₂₃ = α ₃ +b ₂₁ λ ² ₂₆. Now applying the adjoint operator to (Π₆ ,G ₂₆) we have

It is clear that we don't have any reduction, then using λ₂₇ = b₂₄, we have other element of the optimal algebra

This is how other reduction of the generic element (15) ends.

3.2) Case α ₃ = 0. We get G ₂₅ = (α ₁ − 2λ ₂₅)Π₁ + α ₂Π₂ + Π₄, using , is eliminated Π₁, thus G ₂₅ = α ₂Π₂ + Π₄. Now applying the adjoint operator to (Π₃ ,G ₂₅) we get:

we don’t have any reduction, then using α ₂ λ ₂₈ = b ₂₅, we obtain G ₂₈ = α ₂Π₂ +b ₂₅Π₃ + Π₄ . Now applying the adjoint operator to (Π₆ ,G ₂₈) we have:

It is clear that we don't have any reduction, then using λ ₂₉ = b ₂₆, we have other element of the optimal algebra

4) Assuming α₆ = α₅ = α₄ = 0 and α₃= 1 in (15), we have that G = α ₁Π₁+α ₂Π₂+Π₃. Applying the adjoint operator to (Π₂ ,G), (Π₄ ,G), (Π₅ ,G) and (Π₆ ,G) we don’t have any reduction, on the other hand, applying the adjoint operator to (Π₁ ,G) we get

Using , in (35), is eliminated Π₂, therefore G₃₀ = b₂₇ Π₁ + Π₃, with . Now, applying the adjoint operator to (Π₃,G₃₀) we get:

Then, we don't have any reduction, thus using b₂₈ = 2b₂₇λ₃₁ and b₂₉ = 1 + λ² ₃₁, we have other element of the optimal algebra

This is how other reduction of the generic element (15) ends.

5) Assuming α₆ = α₅ = α₄ = α₃ = 0 and α₂ = 1 in (15), we have that G = α₁ Π₁ + Π₂. Applying the adjoint operator to (Π₂, G), (Π₄, G), (Π₅, G) and (Le, G) we don't have any reduction, on the other hand, applying the adjoint operator to (Π₁ , G) we get

Using λ ₃₂ = α₁ , is eliminated Π₁ , thus we get G ₃₂ = Π₂. Now applying the adjoint operator to (Π₃, G ₃₂), we have

We don't have any reduction, then using λ ₃₃ = b ₃₀ we have other element of the optimal algebra

This is how other reduction of the generic element (15) ends.

6) Assuming α ₆ = α ₅ = α ₄ = α ₃ = α ₂ = 0 and α ₁ = 1 in (15), we have that G = Π₁. Applying the adjoint operator to (Π₁ , G), (Π₂, G), (Π₃, G), (Π₄, G), (Π₅, G) and (Π₆ , G) we don't have any reduction, on the other hand, applying the adjoint operator to (Π₃, G) we get

It is clear that we don't have any reduction, then using λ ₃₄ = b ₃₁ we have other element of the optimal algebra

This is how other reduction of the generic element (15) ends.

4. Invariant solutions by generators of the optimal algebra

In this section, we characterize all invariant solutions taking into account some operators that generate the optimal algebra presented in Proposition 3.1. For this purpose, we use the method of invariant curve condition [¹⁵] (presented in section 4.3), which is given by the following equation

Using the element Π₂+Π₃ from Proposition 3.1, under the condition (41), we obtain that Q = η ₁ − yxξ ₁ = 0, which implies (y + y ²) − y _x (0) = 0, then solving this ODE we have , which is trivial solution for (2) with y(x) = 0 or y(x) = −1, using an analogous procedure with all of the elements of the optimal algebra (Proposition 3.1), we obtain both implicit and explicit invariant solutions that are shown in the Table 4, with c being a constant.

Remark 1. The column three in Table 4 which contain the solutions for Q(x, y,y _x ), was calculated using by symbolic software: Mathematica.

5. Classification of Lie algebra

In the theory of finite dimensional Lie algebra it is possible to classify the Lie algebra by using the Levi's theorem, which state that any finite dimensional Lie algebra can be write as a semidirect product of a semisimple Lie algebra and a Solvable Lie algebra, that means that, the classification of Lie algebras reduces to the classification of the Solvable and semisimple Lie algebra. In the case of a semisimple Lie algebra (see [¹⁶]) we can use the Cartan's criterion to state if a Lie algebra is or not semisimple.

Let g the Lie algebra related to the symmetry group of infinitesimal generators of the equation (1) as stated by the table of the commutators, it is enough to consider the no vanish brackets: [Π₁ ,Π₂] = Π₁, [Π₁ ,Π₃] = 2Π₂, [Π₂ ,Π₃] = 2Π₃, [Π₄ ,Π₅] = Π₄, [Π₄ ,Π₆] = 2Π₅, [Π₅ ,Π₆] = Π₆, Using that we calculate Cartan-Killing form K as follows.

which the determinant is not zero, and thus by Cartan’s criterion this Lie algebra is semisimple. First, We remark that in the special lineal algebra sl(2, ℝ) with basis X, Y and H, we have the relations [X, Y ] = H, [X,H] = 2X [Y,H] = −2Y . Next, notice that if we use the transformation: X ₁: = Π₁ , Y ₁: = Π₃ H ₁ = 2Π₂, X ₂: = Π₄ , Y ₂: = Π₆ H ₂ = 2Π5. From that we can see that X ₁ , Y ₁ ,H ₁ and X ₂ , Y ₂ ,H ₂ generate a three dimensional Lie algebra which are isomorphic each one to the special linear Lie algebra sl(2, ℝ). In fact since [X ₁ ,H ₁] = 2x ₁, [X1, Y ₁] = H1, [Y ₁ ,H ₁] = −2Y ₁, and [X ₂ ,H ₂] = 2x ₂, [X ₂ , Y ₂] = H ₂, [Y ₂ ,H ₂] = −2Y ₂, that is we have the same structure constants that the Lie algebra sl(2, ℝ), and since the structure constant classify the Lie algebra up isomorphism we obtain the complete classification of the Lie algebra g. In general terms we can write a element of the Lie algebra as Z = X+Y where X ∈ sl(2, ℝ) and Y ∈ sl(2, ℝ). Consequently, we have the next proposition

Proposition 5.1. The 6-dimensional Lie algebra g related to the symmetry group of the equation (1) is isomorphich with sl(2, ℝ) ⊕ sl(2, ℝ) .