Global Stability of Dynamic Systems of High Order

This paper deals with global asymptotic stability of prolongations of flows induced by specific vector fields and their prolongations. The method used is based on various estimates of the flows.


Introduction
Global stability of dynamic systems is a vast domain in ordinary differential equations and it is one of its main topics. Many works have been done in this context, we list some of them: [3,4,5,6,7,8]. However, little is known in the stability of high order (see [10] and [2]). In this paper, we are concerned with the global asymptotic stability of prolongations of flows generated by some specific vector fields and their perturbations. The method used is based on various estimates of the flows and their prolongations. To justify the study of the dynamic of prolongations of flows, we consider the Lie algebra χ(R n ) of vector fields on R n endowed with the weak topology, which is the topology of the uniform convergence of vector fields and all their derivatives on a compact sets. The Lie bracket is a fundamental operation not only in differential geometry but in many fields of mathematics, such as dynamic and control theory. The invertibility of this latter is of many uses i.e. given any vector fields X, Z find a vector field Y such that [X, Y ] = Z. In the case of vector fields X defined in a neighborhood of a point a with X(a) = 0 we have a positive answer: since in this case the vector field X is locally of the form ∂ ∂x 1 and the solution is given by In the case of singular vector fields, i.e. X(a) = 0 little is known.
Consider a singular vector field X defined in a neighborhood U of the origin 0 with X(0) = 0 and let φ t be the flow generated by X. Suppose that X is complete and consider a vector field Y defined on an open set V ⊃ φ t (U ) for all t ∈ R. The transportation of a vector field Y along the flow φ t is defined as and the derivative with respect to t is given as follows So if (φ t ) * Z converges to 0 and the integral Y = − +∞ 0 (φ s ) * Zds is convergent in the weak topology, then Y is a solution of our equation.
As applications of the right invertibility of the bracket operation on germs of vector fields at a singular point we refer the reader to the papers by the authors [1,2] (see also [10]).

Generalities
First we recall some definitions on global asymptotic stability as introduced in [9]. Let · be the Euclidean norm on R n , K ⊂ R n is a compact set and f any smooth function on R n , we put Definition 1. A point a ∈ R n is said globally asymptotically stable (in brief G.A.S.) of the flow φ t if i) a is an asymptotically stable (in brief A.S.) equilibrium of the flow φ t ; ii) for any compact set K ⊂ R n and any ε > 0 there exists T K > 0 such that for any Definition 2. The point a ∈ R n is said globally asymptotically stable of order r (1 ≤ r ≤ ∞) for the flow φ t if i) a is a G.A.S. point for the flow φ t ; ii) for any compact set K ⊂ R n and where I denotes the identity map.
A vector field X will be called semi-complete if the X-flow φ t = exp(tX) is defined for all t ≥ 0.
First we quote the following proposition which characterizes the uniform asymptotic stability, for a proof see the book of W. Hahn [5].
Let (φ) t denote a flow defined on R n . Proposition 1. The origin 0 in R n is G.A.S. point for the flow φ t if for any ball B(0, ρ), centered at 0 and of radius ρ > 0, there exist t 0 ≥ 0 and functions a, b such that with a a continuous function on B(0, ρ) monotonously increasing such that a(0) = 0 and b is a continuous function defined for any t ≥ t 0 monotonously decreasing such that lim t→+∞ b(t) = 0.

Estimates of prolongations of f lows
We start with some perturbations of linear vector fields.

Perturbation of linear vector fields
Consider the following linear vector field where the coefficients α i ∈ [a, b] ⊂ R and are not all 0. The X 1 -flow, ψ 1 t = exp(tX 1 ) is then and its estimates are given by Consider now a perturbation of the vector field X 1 of the form Y 1 = X 1 + Z 1 , where Z 1 is a smooth vector field globally Lipschitzian on R n . The explicit form of the Y 1 -flow is then Lemma 1. If the perturbation Z 1 fulfills then the vector field Y 1 is complete and the Y 1 -flow satisfies the estimates Proof . Clearly the Y 1 -flow ψ 1 t is bounded for any t ∈ [0, T ] with T < +∞ and any x ∈ R n . The same is true if we replace t by −t. Then ψ 1 t is complete. Consider now the equation Letting y = ψ 1 t (x) , we deduce and by integrating we obtain Let B(0, 1) be the open unit ball centered at the origin 0.
Lemma 2. If the perturbation Z 1 fulfills the estimates then Y 1 is complete and the Y 1 -flow fulfills the following estimates for ant t ≥ 0 Proof . Taking account of the explicit form of the flow (5) and the estimates (8), we deduce that and putting b 0 = b + c 0 , a 0 = a − c 0 , we deduce the following estimates x e a 0 t ≤ y ≤ x e b 0 t for any t ≥ 0.
The same is also true in the on R n \ B (0, 1).

Lemma 3.
Suppose that all the coefficients α i are negative, a ≤ α i ≤ b < 0.
If the perturbation Z 1 fulfills the estimates for any x ∈ R n and any integer m ≥ 1, then the vector field Y 1 is semi-complete and the Y 1 -flow satisfies the estimates for any t ≥ 0 Proof . By the relation (5) and the estimates (10), we deduce that the vector field Y 1 is semicomplete. Letting y = ψ 1 t (x) and taking into account the equation (7) and the estimates (10) we deduce that and by integration we have Example 1. Let the vector field such that all the coefficients fulfilling and all the exponents m i are even positive integers with 0 < m ′ 0 ≤ m i ≤ m 0 . The associated flow φ 3 t = exp(tX 3 ) is the solution of the dynamic system This latter is a Bernoulli type equation and its solution is given by The X 3 -flow φ 3 t = exp (tX 3 ) then has the explicit form and the following estimates are true, ∀ t ≥ 0 3.2 Estimation of the k th prolongation of the Y 1 -flow where ν ∈ R n , the first derivative with respect to x of the Y 1 -flow, solution of the dynamic system Lemma 4. If the perturbation Z 1 fulfills the estimate then the derivative of the Y 1 -flow is complete and has the following estimates, for any t ≥ 0 Proof . Consider as in previous lemmas the following equation and put z = η 1 and then ν e a 1 t ≤ z ≤ ν e b 1 t for any t ≥ 0 and ν ∈ R n .
Lemma 5. If the perturbation Z 1 fulfils the estimates with l = 0, 1, then the first derivative of the Y 1 -flow is complete and is estimated by, for any By the same arguments as in previous lemmas we get the estimates (18).
If the perturbation Z 1 fulfills the estimates x m for all x ∈ R n and any integers m ≥ 1.
Then the estimates of the first derivation of the Y 1 -flow are as follows, for any t ≥ 0 and taking into account the estimates given by the relation (11), we obtain Example 2. We consider the same vector field as in Example 1. Denote by ξ 1 3 (t, x, ν) = Dφ 3 t (x)ν, ∀ ν ∈ R n , the first derivation of the X 3 -flow. In coordinates, we have for any i, j = 1, . . . , n, and by the estimates (13) we get The second derivative is Consequently, for l = 1, 2 and any x ∈ B (0, ρ) with ρ > 0 arbitrary fixed, there are constants M l > 0 such that

Perturbation of a nonlinear vector field
Consider the nonlinear vector field The explicit form of the X 2 -flow is then given by for any t ≥ 0 in the sense Lemma 7. If the following assumptions are true i) all the coefficients β i are non positive, −a ′ ≤ β i ≤ −b ′ ≤ 0 ii) all the exponents m i are even positive integers; 0 < m 0 ≤m i ≤ m ′ 0 . Then the vector field X 2 is semi-complete and the X 2 -flow satisfies the estimates Proof . Clearly the flow φ 2 t = exp(tX 2 ) given by (19) is semi-complete i.e. defined for all t ≥ 0. Consider the equation and we get the estimates given in (20).
3.4 Estimation of the k th order derivation of the X 2 -flow where i, j = 1, . . . , n. Consequently To get the estimates of the second derivative, we put where a 2 1 and a 2 2 are real constants. Let ρ > 0 be any arbitrary and fixed real number, then for any x ∈ B(0, ρ) and any t ≥ t 0 > 0 and l = 1, 2 there is M l > 0 such that Suppose that for l = 1, . . . , k − 1, with fixed k, there exist constants a l j and M l > 0 such that where a l j are real constants and For the estimates of the k th derivative, we compute where a k j are real constants. So we resume Proposition 2. Suppose that i) all the coefficients satisfy β i ≤ 0, −a ′ ≤ β i ≤ −b ′ , ii) the exponents m i are even natural numbers such that 0 < m 0 ≤ m i ≤ m ′ 0 . Let ρ > 0 be any arbitrary fixed real number. For any x ∈ B(0, ρ), for any t ≥ t 0 > 0 and ∀ k ≥ 1 there exist a constant M k > 0 such that the perturbation of the nonlinear vector field X 2 and denote by ψ 2 t = exp(tY 2 ) the solution of the dynamic system d dt In coordinates we have, i = 1, . . . , n, Putting we get The Cauchy problem reads as and has the following solution so we have the explicit form of the Y 2 -flow Now we will estimate the Y 2 -flow.
3) let ρ > 0 and t 0 > 0 be fixed, then for any x ∈ B (0, ρ) and any t ≥ t 0 > 0 there is a constant M 0 > 0 such that hence the vector Y 2 is semi-complete, i.e. defined for all t ≥ 0. Consider the equation we get y i = ψ 2 t (x) i and y i (0) = |x i |, so we deduce We put b 0 = b ′ − c 0 and a 0 = a ′ + c 0 , the solutions are estimated as Hence, we have the estimate (25). Now, we estimate the first derivation of the Y 2 -flow. Let η 1 2 (t, x, ν) = Dψ 2 t (x)ν, ∀ ν ∈ R n the solution of the dynamic system Lemma 9. Suppose that i) the coefficients are such that β i ≤ 0, −a ′ ≤ β i ≤ −b ′ ; ii) the coefficients m i are even natural numbers, Then the first derivation of the Y 2 -flow has the following estimates, for any t > 0 Let ρ > 0 be arbitrary and fixed for any x ∈ B(0, ρ), and any t ≥ t 0 > 0 there is a constant M 1 > 0 such that Proof . Let x ∈ B (0, 1), for l = 0, 1 we have Let c l = max {c ′ l , c ′′ l } then for x ∈ R n one has Consider the equation and put z(t) = η 1 2 (t, x, ν) with z(0) = ν , then 1 2 The solutions fulfill the following estimates Consequently the solutions satisfy Then there are constants m 0 > 0 and m ′ 0 > 0 such that ∀ ν ∈ R n and for any t > 0.

Perturbation of binomial vector fields
Let , be the perturbation of the binomial vector field X 3 and let ψ 3 t = exp(tY 3 ) be the Y 3 -flow which is the solution of the dynamic system and in coordinates, we get which is a Bernoulli type equation and by the same method as in the proof of previous lemmas and with putting we get the solution and the implicit form of the Y 3 -flow reads as

Estimation of the Y 3 -flow
By the same arguments as in the previous, we get the following estimates of the Y 3 -flow.
there exist constants m > 0 and m ′ > 0 such that the Y 3 −flow has the estimates, ∀ t ≥ 0 2) for any t > 0 there are positive constants c 1 and c 2 such that 3) the vector field Y 3 is semi-complete.
By similar calculations as in previous lemmas, we get the following estimates to the first derivative of the Y 3 -flow.
Lemma 11. Suppose that i) all the coefficients α i are negative, −a ≤ α i ≤ −b < 0; ii) all the coefficients β i are non positive, −a ′ ≤ β i ≤ −b ′ ; iii) the exponents m i are even natural numbers such that 0 < m 0 ≤ m i ≤ m ′ 0 ; iv) Then there exist constants m > 0 and m ′ > 0 such that for any t ≥ 0 and for any t ≥ 0, there is a constant M 1 > 0 such that

Global stability of prolongations of f lows
With notations of the previous sections, we will give global stability of some flows.

Global stability of the Y 1 -flow
Lemma 12. Let the vector fields with the following assumptions i) all the coefficients are negative, −a ≤ α i ≤ −b < 0; ii) Then the origin 0 is a globally asymptotically stable equilibrium to the Y 1 -flow ψ 1 t on R n .
Proof . Let ψ 1 t = exp(tY 1 ) be the Y 1 -flow, then by the assumptions and the estimates given by Lemma 2 we get that and by Proposition 1, the origin 0 is G.A.S. for ψ 1 t on R n .
Example 3. We consider the vector field is then given by Let ρ > 0 be arbitrary and fixed real number. By the estimates (13), we have for any x ∈ B(0, ρ) and any t ≥ t 0 ≥ 0 By Proposition1 the origin 0 is a G.A.S. for the flow φ 3 t on R n .

Global stability of the first prolongation of the Y 1 -flow
Lemma 13. With the same assumptions as in Lemma 12 and the following conditions Then the origin 0 is a globally asymptotically stable for the first prolongation of the Y 1 -flow ψ 1 t on R n . Proof . By the estimates (18) and the hypothesis we deduce that and by Proposition 1,we obtain that the origin 0 is a G.A.S. equilibrium on R n for η 1

Global stability of the k th prolongation of the Y 1 -flow
Suppose that i) all the coefficients are negative, −a ≤ α i ≤ −b < 0; ii) for any l = 1, . . . , k − 1 D l Z 1 (x) ≤ c ′ l x 1−l+m for any x ∈ B (0, 1) and for any integer m ≥ l − 1, where ν ∈ R n . Since by Lemmas 12 and 13 the origin 0 is an G.A.S. equilibrium for η l 1 , with l = 0, 1, on R n , we suppose that this property remains true for l = 0, 1, . . . , k − 1 with k ≥ 2 i.e. for any ρ > 0 and any x ∈ B(0, ρ) there exist constants M l > 0 such that for any t ≥ t 0 > 0 We will show that the origin 0 is a G.A.S. equilibrium for η k 1 on R n . η k 1 (t, x, ν, . . . , ν) = D k ψ 1 t (x)ν k is solution of the dynamic system d dt Consequently we get The integral is well defined at s = 0, since and there exist constants A l > 0 such that We will show that it converges uniformly with respect to x as t + ∞. Put there are constants b l > 0 such that ∀ y ∈ R n , D l y Y 1 (y) ≤ b l and by the assumption of recurrence there exist constants M l > 0 such that We deduce that there is a constant C k > 0 such that So for any x ∈ R n one has M l b l ν l and the integral I k is uniformly convergent with respect to x ∈ R n as t → +∞. Consequently and there is a constant M ′ k > 0 such that This show by Proposition 1 that the origin 0 is a G.A.S. equilibrium to η k 1 on R n . We formulate our proving as follows Proposition 3. Let k ≥ 0 be any integer. The origin 0 is a G.A.S. equilibrium of order k for the Y 1 -flow and there is a constant M k > 0 such that ∀ t > 0 5 Global stability of a f low generated by nonlinear perturbed vector f ields First we will start with monomial vector fields.

Global stability of the X 2 -flow
Let (ii) all the exponents m i are even natural integers with 0 < m 0 ≤m i ≤ m ′ 0 . Let φ 2 t = exp (tX 2 ) be the X 2 -flow. By the estimations (19) we obtain Let ρ > 0 be arbitrary fixed, for any x ∈ B(0, ρ) and any t ≥ t 0 > 0 there is a constant M 0 > 0 such that By Proposition 1, the origin is a globally asymptotically stable equilibrium to the flow φ 2 t on R n . Let l = 1, 2, . . . any positive integer. By Proposition 2, we have: for any fixed ρ > 0, and all x ∈ B(0, ρ) and t ≥ t 0 > 0, there exist constants M l > 0 and M ′ l > 0 such that So the origin 0 is a G.A.S. equilibrium for D l φ 2 t (x) on R n . Resuming our proving, we get Proposition 4. Let k ≥ 0 be any integer. Under the above conditions (i) and (ii), the origin 0 is a G.A.S. of order k for the X 2 -flow on R n .

Global stability of high order of the
be a smooth vector field on R n such that i) all the coefficients β i ≤ 0 are non negative with −a ′ ≤ β i ≤ −b ′ ; ii) m i are even natural numbers with 0 < m 0 ≤m i ≤ m ′ 0 ; iii) for k = 0, . . . , 1 + m i is well defined at s = 0 and s = t, since lim s→0 + Dψ 2 t−s (ψ 2 s (x)) = Dψ 2 t (x).
By the recurrent assumption D l ψ 2 0 (x) are bounded and there exist constants A l > 0 such that In the same way lim s→t − Dψ 2 t−s (ψ 2 s (x)) = identity. Now, we have to show that converges uniformly on any compact set K ⊂ R n as t → 0. Let x ∈ K, by the relations (26) and (28) we get for all t ≥ 0 So y = ψ 2 t (x) ≤ x and Dψ 2 t−s (ψ 2 s (x)) is bounded. Since for any x ∈ R n and any l = 1, . . . , 1+ m i , D l Z 2i (x) ≤ c l |x i | 1−l+m i then D l y Y 2 (y) are bounded. Now by the assumption of recurrence there exist constants M l > 0 such that for any t > 0 The integral I 2 k converges uniformly on any compact K ⊂ R n as t → +∞. Now since the integral is well defined at s = 0, then lim t→0 η k 2 (t, x, ν, . . . , ν) ≤ lim t→0 Dψ 2 t (x)ν = ν hence there is a constant M ′ k > 0 such that In the same way as above the integral t 0 Dψ 2 t−s (ψ 2 s (x)) G k 2 (s, x, ν) ds is well defined and putting τ = s t we obtain η k 2 (t, x, ν, . . . , ν) = Dψ 2 t (x)ν + t 1 0 Dψ 2 t(1−τ ) (ψ 2 tτ (x))G k 2 (tτ, x, ν)dτ.
Since b 1 = b ′ (1 + m 0 ) − c 1 > a 0 m ′ 0 , by the estimates (26) and (28), we deduce the existence of a constant M k > 0 such that η k 2 (t, x, ν, . . . , ν) ≤ Dψ 2 t (x)ν + t 1 0 G k 2 (tτ, x, ν) dτ Which shows that the origin 0 is a G.A.S. equilibrium for η k 2 on R n . We formulate this fact as Proposition 5. Let k ≥ 0 be any integer. Under the above conditions (i), (ii), (iii), (iv) and (v), the origin 0 is a G.A.S. of order k on R n for the Y 2 -flow and there is a constant M k > 0 such that for any t ≥ t 0 > 0 (33)

Global stability of prolongations of the Y 3 -flow
Let ii) all the coefficients β i ≤ 0 and −a ′ ≤ β i ≤ −b ′ ; iii) the exponents m i are even natural numbers with 0 < m 0 ≤ m i ≤ m ′ 0 ; iv) For any k = 0, . . . , 1 + m i Remark 2. If x ∈ B (0, 1) then D k Z 3i (x) ≤ c ′ k |x i | 2−k+m i ≤ c ′ k |x i | 1−k+m i . Let c l = max {c ′ l , c ′′ l }, for any x ∈ R n one has D k Z 3i (x) ≤ c k |x i | 1−k+m i .