Periodic impulsive fractional differential equations

Lemma 2.1.

Under assumptions (i) and (ii), there holds

for Θ = ∏ k = 0 N L k + 1 ⁢ E q ⁢ ( K ⁢ ( t k + 1 - t k ) q ) , where E q is the Mittag-Leffler function (see [8, p. 40]).

Proof.

On all intervals ( t k , t k + 1 ) , k = 0 , 1 , … , N , equation (2.1) is equivalent to

Consider two solutions x and y of (2.1) with x ⁢ ( 0 ) = x 0 and y ⁢ ( 0 ) = y 0 , respectively. Then by (2.2), we get

Applying a Gronwall fractional inequality [17, Corollary 2] to (2.3), we obtain

Then using (2.4) with (ii), we get

which implies

as desired. ∎

Theorem 2.2.

Suppose (i)–(iii) are satisfied. If Θ < 1 , then (2.1) has a unique T-periodic solution, which is in addition asymptotically stable.

Proof.

Inequality (2.5) implies that P : ℝ m → ℝ m is a contraction, so applying the Banach fixed point theorem yields that P has a unique fixed point x 0 . Moreover, since

for any y 0 ∈ ℝ m , we see that the corresponding periodic solution is asymptotically stable. ∎

Theorem 2.3.

Suppose assumptions (i) and (C1). If there is a simple zero x 0 ∈ R m of M, i.e., M ⁢ ( x 0 ) = 0 and det ⁡ D ⁢ M ⁢ ( x 0 ) ≠ 0 , then (2.6) has a unique T-periodic solution located near x 0 for any ε ≠ 0 small. Moreover, if ℜ ⁡ σ ⁢ ( D ⁢ M ⁢ ( x 0 ) ) < 0 then it is asymptotically stable, and if ℜ ⁡ σ ⁢ ( D ⁢ M ⁢ ( x 0 ) ) ∩ ( 0 , ∞ ) ≠ ∅ , then it is unstable. If M ⁢ ( x 0 ) ≠ 0 for any x 0 ∈ R m , then (2.6) has no T-periodic solution for ε ≠ 0 small.

Proof.

To find a T-periodic solution of (2.6), we need to solve P ⁢ ( ε , x 0 ) = x 0 , which by (2.7) is equivalent to

If there is a simple zero x 0 of M, then (2.8) can be solved by the implicit function theorem to get its solution x 0 ⁢ ( ε ) with x 0 ⁢ ( 0 ) = x 0 . Moreover, D ⁢ P ⁢ ( ε , x 0 ⁢ ( ε ) ) = 1 + ε ⁢ D ⁢ M ⁢ ( x 0 ) + O ⁢ ( ε 2 ) . Thus stability and instability results follow directly by the known arguments (see [10]), so we omit details. ∎

Lemma 2.4.

Under assumptions (I)–(III), there holds

and

for t ∈ I and r ∈ N , where P r is the rth iteration of P.

Proof.

On all intervals ( k ⁢ h , ( k + 1 ) ⁢ h ) , k ∈ ℕ 0 , equation (2.9) is equivalent to

which implies

for t ∈ ( n ⁢ h , ( n + 1 ) ⁢ h ) . This implies (2.10) since P r ⁢ ( x 0 ) = x ⁢ ( x 0 , r ⁢ T + ) . Moreover, we have

for t ∈ I . This implies (2.11). ∎

Theorem 2.5.

Suppose assumptions (I)–(III). If

then (2.9) has no rT-periodic solution for any r ∈ N .

Proof.

We need to solve P r ⁢ ( x 0 ) = x 0 , which is equivalent to

This gives

This contradicts (2.12). ∎

Example 2.6.

We consider the Lorenz system

which for q = 0.995 , a = 10 , b = 120 , c = 8 3 , and without impulses presents a chaotic attractor (see Fig. 1). The utilized numerical scheme to integrate Lorenz’s system is the predictor-corrector Adams–Bashforth–Moulton method for FDEs [3].

Take N = 1 and t k = k ⁢ h , k ∈ ℕ 0 . Let

which embeds Lorenz’s attractor. To apply Theorem 2.5, we need to find

Let Δ k = ( Δ k ⁢ 1 , Δ k ⁢ 2 , Δ k ⁢ 3 ) . For Δ 11 = Δ 21 = Δ ¯ 1 , Δ 12 = 0 , Δ 22 = Δ ¯ 2 , and Δ 31 = Δ 32 = Δ ¯ 3 , condition (2.12) has the form

for h = 0.004 . Note that T = ( N + 1 ) ⁢ h = 0.008 . Summarizing, we obtain the following proposition.

Theorem 2.8.

Suppose (I)–(III) and, in addition, there are constants f ± i ∈ R , i = 1 , … , m , such that either

uniformly with respect to t ∈ I and x ^ i ∈ R m - 1 for all i = 1 , … , m , or

uniformly with respect to t ∈ I and x ^ i ∈ R m - 1 for all i = 1 , … , m . If

for all i = 1 , … , m , then (2.9) has a T-periodic solution.

Proof.

We need to solve (2.13). For simplicity, we set

So we need to solve

Let F ⁢ ( x 0 ) = ( F 1 ⁢ ( x 0 ) , … , F m ⁢ ( x 0 ) ) . By (2.11), there holds lim x 0 ⁢ i → ± ∞ ⁡ x i ⁢ ( x 0 , t ) = ± ∞ uniformly with respect to t ∈ I and x ^ 0 ⁢ i ∈ ℝ m - 1 , where x 0 = ( x 01 , … , x 0 ⁢ m ) and x ⁢ ( x 0 , t ) = ( x 1 ⁢ ( x 0 , t ) , … , x m ⁢ ( x 0 , t ) ) . So by the first possibility (C1), for any ε > 0 , there is m ε > 0 such that

uniformly with respect to s ∈ I and x ^ 0 ⁢ i ∈ ℝ m - 1 for all i = 1 , … , m . This implies

for any x 0 ⁢ i ≥ m ε and x ^ 0 ⁢ i ∈ ℝ m - 1 , ∥ x ^ 0 ⁢ i ∥ ≤ m ε , while

for any x 0 ⁢ i ≤ - m ε , x ^ 0 ⁢ i ∈ ℝ m - 1 , ∥ x ^ 0 ⁢ i ∥ ≤ m ε , and for all i = 1 , … , m . Further, let us take G : ℝ m → ℝ m and H : [ 0 , 1 ] × ℝ m → ℝ m given by

Note that H ⁢ ( 0 , x 0 ) = F ⁢ ( x 0 ) and H ⁢ ( 1 , x 0 ) = G ⁢ ( x 0 ) . It is easy to check that by fixing ε > 0 sufficiently small, (2.15) implies

for any x 0 ⁢ i = ± m ε , x ^ 0 ⁢ i ∈ ℝ m - 1 , ∥ x ^ 0 ⁢ i ∥ ≤ m ε . Similarly, we verify that (2.17) holds also in the second possibility (C2). Consequently, we derive

for any λ ∈ [ 0 , 1 ] and x 0 ∈ ∂ ⁡ Ω for Ω = { x 0 ∈ ℝ m ∣ ∥ x 0 ∥ ≤ m ε } , where ∂ ⁡ Ω is the border of Ω. This gives

where deg is the Brouwer topological degree. On the other hand, a linear equation G ⁢ ( x 0 ) = - ∑ k = 1 N + 1 Δ k has the only solution x ¯ 0 , which by (2.15) is located in Ω. Moreover, the linearization D ⁢ G ⁢ ( x ¯ 0 ) is a nonsingular matrix, so deg ⁡ ( G , Ω , - ∑ k = 1 N + 1 Δ k ) = ± 1 , and thus deg ⁡ ( F , Ω , - ∑ k = 1 N + 1 Δ k ) = ± 1 ≠ 0 . Summarizing, we see that (2.16) has a solution in Ω, and so (2.13) is solvable. ∎

Theorem 2.9.

Suppose (I)–(III) and, in addition, there are constants f ± i ∈ R , i = 1 , … , m and a permutation σ : { 1 , … , m } → { 1 , … , m } such that

(C3)

uniformly with respect to t ∈ I and x ^ i ∈ R m - 1 for all i = 1 , … , m . If (2.15) holds for all i = 1 , … , m , then (2.9) has a T-periodic solution.

Example 2.10.

Consider a system

for A i > B i > 0 and all i = 1 , … , m . Then we take

Hence, assumption (2.12) has the form

and assumption (2.15) has the form

for all i = 1 , … , m , respectively. Consequently, Theorem 2.8 can be applied.