Algebraic and qualitative remarks about the family yy′ = (αx^m+k–1 + βx^m–k–1)y + γx^2m–2k–1

Theorem 1.1

Let X, Y be analytic functions with polynomial part containing terms of degree greater than 1. Consider the planar differential system

being the origin an isolated critical point. Assume

as the solutions of y + X(x, y) = 0 near to (0, 0). Suppose

and also

Then the following statements hold:

1. If α is even and α > 2β + 1, then (0, 0) is a saddle node.

   If α is even and α < 2β + 1 or Φ(x) ≡ 0, then the flow near of (0, 0) have two hyperbolic sectors.

2. If α is odd and a > 0, then (0, 0) is a saddle point.

3. If α is odd and a < 0, several cases can occur:

   ∙c1α>2β+1,βeven∙c2α=2β+1,βeven,b2+4a(β+1)≥0,

   in this case for b < 0 the critical point (0, 0) is a stable node, while for b > 0 the critical point (0, 0) is unstable node.

   ∙c3α>2β+1,βodd∙c4α=2β+1,βodd,b2+4a(β+1)≥0,

   in this case the flow near to the critical point (0, 0) is topologically conformed by an elliptic sector joint with an hyperbolic sector.

   ∙c5α=2β+1yb2+4a(β+1)<0∙c6α<2β+1,or,Φ(x)≡0,

   in this case the critical point (0, 0) is focus or center.

Proposition 2.1

Given the family of Liénard equations of the form:

the change of variables z = x^k and y = x^m(t + ax^k + bx^–k) allow to transform any equation of this family to a Riccati equation.

Proof

The system of equations, associated with this Liénard equation is:

Now, applying the transformation

and differentiating we have that:

then, the associated foliation has the form ydy = (f(x)y – g(x))dx, being f(x) = (a(2m + k)x^m+k–1 + b(2m – k)x^m–k–1) and g(x) = (a²mx^4k+cx^2k+b²m)x^2m–2k–1 as the Liénard equations. Now we compute each part of this equality, thus we obtain the left side as:

and the right side as

For our purpose, we organize the terms with respect to dz, that is:

Now organizing the terms again we have:

Thus, we obtain the Riccati equation:

and we conclude the proof.□

Lemma 3.1

If R = a₁x² + a₁x + a₀, S = b₁x + b₀, with a₂, b₁ ≠ 0. Then the differential equation

with y = y(x) is transformed in the equation

with

Proof

R2∂x2y+SR∂xy+Cy=0 then replacing

We divide all by a22, thus we get:

Now in R we complete the square

Then:

Assume

then

Replacing x in the polynomial S in term of τ, we get:

Then, if l1=b1a2 and l0=−b1.a12a22+b0a2, we get l₁τ + l₀ = L(τ).

Now, if λ=Ca22, the differential equations will be

where ŷ = y(x(τ)), and the transformation τ=x+a12a2 send ∂_x on ∂_τ.□

Remark 3.1

The differential equation of the form

with λ = n(n + 1 + ã + b̃) and n ∈ ℕ, is known as Jacobi equation (in general form). It is a particular case of the hypergeometric equation, but the solutions include Jacobi polynomials. If we take ã = b̃ and λ̃ = n(n + 2ã) with n ∈ ℕ, we get a Gegenbauer equation (or ultraspherical case):

Theorem 3.2

The differential equation

with a_i ∈ ℂ(x), with a_n ≠ 0, can be transformed into the differential equation

with b_i ∈ ℂ(x).

Proof

Following the Lemma 3.1 and taking the implicit transformation z = ε(t)y + μ(t) with ε(t), μ(t) ∈ ℂ(x), we compute the first equation applying the change of variable

Then, computing in general form the Leibniz rule we get:

Continuing of this form, we divide all equation by ε. Then, we use the same method of the indeterminate coefficients to calculate ε and the b_i coefficient:

If we take y^(n–1):

then:

From this differential equation we get an appropriate ε value, and with it we obtain the coefficient b_i.

If we take y^(n–2):

If we take y^(n–3):

Continuing of this form we see, that for any k ∈ ℕ the recurrent formulae will be:

□

Corollary 3.3

The differential equation

with a_i ∈ ℂ(x) can be transformed in the following equation

where b_i ∈ ℂ(x).

Example

Applying the transformation over the general second order differential equation z″ + a₁z′ + a₀z = 0:

Using Theorem 3.2, we get 2ε′ε+a1=0, then ε′=−a12ε. Now, through derivatives and dividing by ε, we get ε″=−a1′2−ε′εa12, but ε′ε=−a12. Thus, we obtain

In this way we arrive to b=−a1′2+a124, for the differential equation y″ + by = 0.

In the following theorem we recall that a Hamiltonian Change of Variable z = z(x) is a change of variable, where (z(x), z′(x)) is a solution curve of a Hamiltonian system of one degree of freedom. The new derivation is given by ∂^z=α∂z, being α = (∂_xz)², see [15, 16, 17] and references therein.

Theorem 3.4

Let Q and L be as in (3.1), with a₁ = b₁ = 0 or b0=a1b12a2. Through the Hamiltonian change of variable ξ = ∂_τ Q , the differential equation

is transformed in the equation

Owing to λ = n(n + 1 + ã + b̃), we have the Jacobi equation with ã = b̃. Moreover, if λ = n(n + 2ã) then we obtain a Gegenbauer equation.

Proof

1. If we assume a₁ = b₀ = 0, we obtain l0=0,l1=b1a2,q0=a0a2. Then:

   (τ2+q0)2∂τ2w+l1τ(τ2+q0)∂τw+λw=0.

   Now, by hypothesis ξ = ∂_τ Q , that is

   ξ=∂τQ2Q=∂τ(τ2+q0)2τ2+q0,

   thus

   ξ2=τ2+q0τ2+q0−q0τ2+q0q01−ξ2=τ2+q0τ2=q01−ξ2−q0τ=±ξq01−ξ2.

   Furthermore, due to ξ = ∂_τ Q then α we arrive to:

   α=∂τξ=∂τ2Q=1q0(1−ξ2−ξ.2ξ21−ξ2)1−ξ2=1q0(1−ξ2+ξ2)1−ξ2(1−ξ2)=(1−ξ2)3q0,

   i.e α=(1−ξ2)3q0.

   Now ∂^ξ=(1−ξ2)3q0∂ξ, and ∂^ξ2=α∂ξ2+12∂ξα∂ξ=(1−ξ2)3q0∂ξ2−3(1−ξ2)2q0∂ξ.

   We compute all elements of the Hamiltonian change of variable Q^=Q(τ(ξ))=q021−ξ2,L^=l1ξq01−ξ2, ŵ(ξ(τ)) = w(τ). Then:

   Q2∂τ2w⇝Q^2∂^ξ2w^=(q021−ξ2)2((1−ξ2)3q0∂ξ2w^−3(1−ξ2)2q0∂ξw^)=q0(1−ξ2)∂ξ2w^−3ξ∂ξw^,

   LQ∂τw⇝L^Q^∂^ξw^=(l1ξq01−ξ2)(q021−ξ2)(1−ξ2)3q0∂ξw^=l1q0ξ∂ξw^.

   If we replace, in the transformed differential equation, we obtain:

   q0(1−ξ2)∂ξ2w^+(l1q0−3)ξ∂ξw^+λw^=0.

   It is equivalent to Gegenbauer equation with λ~=λq0,−2a~+1=l1−3q0 i.e a~=3q0−l1−12 then:

   (1−ξ2)∂ξ2w^+l1−3q0ξ∂ξw^+λq0w^=0.

   Finally if λq0=n(n+3q0−l1−1) with n ∈ ℕ, then the solutions of this equation are Ultraspherical Gegenbauer polynomial.

2. If b0=a1b12a2 then:

   Q(τ)=τ+q0withq0=a0a2−(a12a2)2,L(τ)=l1~τ+l0~,l1~=b1a2and l0~=−b1a12a22+b0a2=−b1a12a22+b1a12a22=0

   i.e the initial differential equation will be:

   (τ2+q0)2∂τ2w^+l1~τ(τ2+q0)∂τw^+λw^=0

   for instance, the same differential equation of the previous case.□

Lemma 3.5

The Gegenbauer equation

is transformed into an hypergeometric equation

where a = μ – ν, b = ν + μ + 1, c = μ + 1 and z=1−x2.

Proof

For this transformation we will use a Hamiltonian change of variable, over the independent variable of the Gegenbauer equation x = 1 – 2z. Then α=∂xz=−12, that is, α=14,∂z^=12∂z and ∂z2^=14∂z2.

Substituting in the Gegenbauer equation, we obtain

Thus, we obtain the equation