Stability analysis for discrete time abstract fractional differential equations

Definition 2.1

Let α > 0, the α-th order fractional sum operator is defined by

Definition 2.2

Let α ∈ (0, 1], the α-th order fractional difference operator (in the sense of Riemann-Liouville-like) is defined by

where Δ denotes the forward Euler operator by Δu(n) := u(n + 1) − u(n), n ∈ ℕ₀.

Definition 2.3

Let α > 0 and let A be a closed linear operator with domain D(A) defined on a Banach space X. An operator-valued sequence {Sα(n)}n∈ℕ0⊂ℬ(X) is called an α-resolvent sequence generated by A if it satisfies the following conditions:

1. S_α(n)x ∈ D(A) and AS_α(n)x = S_α(n)Ax, for all n ∈ ℕ₀ and x ∈ D(A);

2. S_α(n)x = k^α(n)x + A(k^α ∗ S_α)(n)x, for all n ∈ ℕ₀ and x ∈ X.

Lemma 2.1

([1]) Letρ(A) be the resolvent set of operatorAand let{Sα(n)}n∈ℕ0be anα-resolvent sequence generated byA. Then

1. 1 ∈ ρ(A), and for allx ∈ Xwe have thatS_α(0)x = (I − A)⁻¹x.

2. For allx ∈ Xwe have thatS_α(0)x ∈ D(A) andS_α(n)x ∈ D(A²) for alln ∈ ℕ.

Lemma 2.2

([10]) Let 0 < α ≤ 1 and letAbe the generator of a boundedC₀-semigroup {T(t)}_t≥0defined on a Banach spaceX. Then, Agenerates anα-resolvent sequence{Sα(n)}n∈ℕ0given by

wherep_n(t) := e^−ttⁿ/n! (n ∈ ℕ₀, t ≥ 0) is the Poisson distribution and functionf_{α, s}(t) denotes the stable Lévy process given by

in which the branch ofz^αis taken such thatRe(z^α) > 0 forRe(z) > 0.

Lemma 2.3

([10]) Let 0 < α ≤ 1 and letAbe the generator of a boundedC₀-semigroup {T(t)}_t≥0defined on a Banach spaceX. Then,

Furthermore, ifAis the generator of a compactC₀-semigroup {T(t)}_t>0. Then, Agenerates a compactα-resolvent sequence{Sα(n)}n∈ℕ0.

Remark 2.1

Noting that if X is a finite dimension space, we see that the semigroup T(t) can be rewritten by e^At, and if A is the generator of a C₀-semigroup {e^At}_t≥0 with respect to ‖A‖ℬ≤1. Then, A generates an α-resolvent sequence if and only if {Sα(n)}n∈ℕ0 is given by

In fact, by the properties of stable Lévy process (iii) and Mittag-Leffler functions, we get for n∈ℕ0

Conversely, it is easy to check that S_α(n) satisfies the Definition 2.3 for every n∈ℕ0.

Definition 3.1

Let 0 < α < 1 and A be the generator of an α-resolvent sequence {Sα(n)}n∈ℕ0. We say that u ∈ l^∞(ℕ₀; D(A)) is a solution of (1.1) if u satisfies u(0) = u₀ ∈ D(A) and

Lemma 3.1

LetU ⊂ l^∞(ℕ;X) satisfy

1. The setHn(U)={u(n): u∈U}is relatively compact inX, for alln∈ℕ.

2. limn→∞supu∈U‖u(n)‖=0, that is, for eachε > 0, there is aN > 0 such that‖u(n)‖<ε, for eachn ≥ Nand for allu ∈ U.

ThenUis relatively compact inl^∞(ℕ;X).

Proof

Let {um}m=1∞ be a sequence in U, then by (a), for any given n∈ℕ, there exists a convergent subsequence {umk}k=1∞⊂{um}m=1∞ such that limk→∞umk(n)=u(n), i.e., for each ε > 0, there exists a constant N^∗ = N^∗(n, ε) > 0, such that

From the assumption (b), for each ε > 0, there exists a constant N′ > 0, such that for n = N′

Let N′ be fixed. For each 1 ≤ n < N′, then for j, k > N = N(N′, ε), we have

Therefore, one has ‖umk−umj‖∞<ε. This means that {umk}k=1∞ is a Cauchy subsequence in l^∞ (ℕ; X). We thus derive that {umk}k=1∞ has a convergence element u ∈ l^∞(ℕ; X), which implies that U is relatively compact in l^∞(ℕ;X).

Theorem 3.1

Assume that operatorAsatisfies (H1) andfsatisfies (H2)-(H3). Then, the problem (1.1) withu₀ ∈ D(A) has at least one stable solution.

Proof

Let us define the map P:l∞(ℕ0;D(A))→l∞(ℕ0;D(A))as follows

and (Pu)(0)=u0. We first show that P is well defined. In fact, let u ∈ l^∞(ℕ₀;D(A)) be given, it follows from (H2) that

additionally, in view of Lemma 2.3 and observing that 1 + α − β ∈ (0, 1] for 0 < α < β < 1, using the semigroup relationship (k^α ∗ k^1−β)(n − 1) = k^1+α−β(n − 1) for each n∈ℕ we obtain

where we also use the fact that 0 < k^s(n) ≤ k^s(0) = 1 owing to the nonincreasing property of k^s(n) for each 0 < s ≤ 1 and n∈ℕ. This proves that P is well defined. Now, we show that P is continuous. Let {um}m=1∞⊂l∞(ℕ0;D(A)) be a sequence such that u_m → u as m → ∞ in the norm topology of l^∞(ℕ₀;D(A)). First, we have

and using the semigroup property of k^α(n) for all n∈ℕ0, we get

Therefore, for all n∈ℕ, be virtue of the property of series, it is easy to check that

as m → ∞, which implies that ‖Pum−Pu‖∞→0 as m → ∞. Therefore P is continuous.

Since T(t) is compact for t > 0, then from Lemma 2.3, we know that the sequence of operators {Sα(n)}n∈ℕ0 is compact. Let r > 0 be given. We define a set by

Clearly, Sr is a bounded, closed and convex subset of l^∞(ℕ; D(A)). In view of (H2), we can deduce that P maps Sr into itself. Thus, it remains to show that P is a compact operator.

In order to prove that U:=PSr is relatively compact, we will use Lemma 3.1. We check that the conditions in this lemma are satisfied, and now we check that U satisfies all the assumptions:

1. Let v=Pu for any u∈Sr. We have

   vε(n)=(Pεu)(n)=∑j=0n−1Sαε(j)f(n−1−j,u(n−1−j)), n∈ℕ,

   where

   Sαε(j)x:=∫0∞∫ε∞pj(t)fs,α(t)T(s)x dsdt=T(ε)∫0∞∫ε∞pj(t)fs,α(t)T(s−ε)x dsdt, x∈X,

   where we use the semigroup property of T(t). Hence, it remains to prove that (Q*f)(n−1) is bounded, where

   Q(j)x=∫0∞∫ε∞pj(t)fs,α(t)T(s−ε)x dsdt.

   In fact, noting that from the properties of stable Lévy process (ii), we have the following identity

   ∫0∞∫0∞pj(t)fs,α(t)dsdt=kα(j), j∈ℕ0,

   it is easy to check

   ‖∑j=0n−1Q(j)f(n−1−j,u(n−1−j))‖≤∑j=0n−1∫0∞∫ε∞pj(t)fs,α(t)‖T(s−ε)f(n−1−j,u(n−1−j))‖ dsdt≤MLf∑j=0n−1kα(j)k1−β(n−1−j)r≤MLfr.

   For any fixed n^∈ℕ and for all n≥n^+1, since

   ∑j=n^n−1kα(j)k1−β(n−1−j)≤k1+α−β(n−1),

   and by (2.1) we have

   k1+α−β(n−1)=1Γ(1+α−β)(n−1)α−β[1+O(1n−1)],

   for n large enough, it follows that for any δ > 0 there is n*∈ℕ large enough such that n_∗ + 1 ≤ n with n large enough and

   ∑j=n*n−1kα(j)k1−β(n−1−j)<δ2MLfr.

   Therefore, one has

   ∑j=n*n−1‖(Sα(j)−Sαε(j))f(n−1−j,u(n−1−j))‖≤2MLf∑j=n*n−1kα(j)k1−β(n−1−j)‖u(n−1−j)‖<δ.

   For all n∈ℕ, by the proof of [10, Corollary 3.2.], we know ‖(Sα(n)−Sαε(n))x‖≤ε for all n∈ℕ0 and for x ∈ X. Hence, we get

   ∑j=0n*−1‖(Sα(j)−Sαε(j))f(n−1−j,u(n−1−j))‖≤MLfrεn*.

   Together the above arguments, we see that

   ‖v(n)−vε(n)‖≤∑j=0n*−1‖(Sα(j)−Sαε(j))f(n−1−j,u(n−1−j))‖+∑j=n*n−1‖(Sα(j)−Sαε(j))f(n−1−j,u(n−1−j))‖≤MLfrεn*+δ.

   For the arbitrariness of δ, it yields that ‖v(n) − v^ε(n)‖ → 0 as ε → 0. We thus conclude that the set H_n(U) is relatively compact in X for all n∈ℕ.

2. Let u∈Sr and v=Pu. For each n∈ℕ we have

   ‖v(n)‖≤MLf∑j=0n−1kα(n−1−j)k1−β(j)‖u(j)‖≤MLf‖u‖∞kα+1−β(n−1),

   which implies that lim_{n → ∞}‖v(n)‖ = 0 independently of u∈Sr. Therefore, U=PSr is relatively compact in l^∞(ℕ;D(A)) from Lemma 3.1, and by applying the continuity of operator P, we conclude that P is a completely continuous operator. Thus, the Schauder’s fixed point theorem shows that P has at least one fixed point u ∈ l^∞(ℕ₀;D(A)).

   Additionally, let u^∗ be a solution of problem (1.1) in l^∞(ℕ₀; D(A)), which means that there is a constant C > 0 such that ‖u^∗‖_∞ ≤ C and

   u*(n)=Sα(n)(I−A)u0+∑j=0n−1Sα(n−1−j)f(j,u*(j)), n∈ℕ,

   moreover, in view of (2.1) we have

   ‖u*(n)‖≤‖Sα(n)(I−A)u0‖+∑j=0n−1‖Sα(n−1−j)f(j,u*(j))‖≤Mkα(n)‖(I−A)u0‖+MLf∑j=0n−1kα(n−1−j)k1−β(j)‖u*(j)‖≤2Mkα(n)‖u0‖A+MLfCk1+α−β(n−1)→0, as n→∞.

   Thus, u^∗ is a stable solution. The proof is completed.

Remark 3.1

Theorem 3.1 shows that it is not necessary to use the Lipschitz condition to establish the existence for problem (1.1), and this is a general result of the paper [10].

Example 3.1

Let Ω = [0, π] and X = L²(Ω). We consider the following discrete abstract Cauchy problem

where Δ^α is the Riemann-Liouville-like fractional difference operator of order 0 < α < β < 1.

Let us consider the operator A:D(A) ⊆ X → X defined by

Clearly A is closed densely defined in X and it is well known that A generates a compact, uniformly bounded and analytic C₀-semigroup {T(t)}_t>0. Furthermore, A has a discrete spectrum with eigenvalues of the form −m², m∈ℕ, and corresponding normalized eigenfunctions given by ϕm(z)=2/πsin(mz). In addition, {ϕm}m∈ℕ is an orthogonal basis for X, and

Hence, by applying Lemmas 2.2-2.3, we get the discrete compact α-resolvent family {Sα(n)}n∈ℕ0 as follows

Let

since the inequalities |E_{α, α}(−m²t^α)| ≤ 1/Γ(α) for all m∈ℕ, t∈ℝ+ and |Sm(n)|≤kα(n) for n∈ℕ0, it follows that Sm(n) tend to zero as n → ∞ for all m∈ℕ. Thus, we have

Therefore, let f(n, u(n)) = k^1−β(n)u(n), the problem (3.1) possesses a stable solution by Theorem 3.1 and its expression form is given by

Definition 4.1

If u(n) satisfies

where ϑ(n) ≥ 0 for all n∈ℕ0, and there exist a solution v(n) of the problem (1.1) and a constant C > 0 independent of u(n) and v(n) with

for all n∈ℕ0, then problem (1.1) is called the Ulam-Hyers-Rassias stability. In particular, if ϑ(n) is substituted for a constant in the above inequalities, then problem (1.1) is called the Ulam-Hyers stability.

Remark 4.1

Obviously, v solves (4.1) if and only if there exists g:ℕ0→X satisfying

such that

Furthermore, if v∈l∞(ℕ0,X) is a solution of inequality (4.1), there exists a constant C > 0 such that v is a solution of the following inequality

Remark 4.2

It is hard to get the Ulam-Hyers stability of problem (1.1), because if we substitute the sequence ϑ(n) for a constant, then from Remark 4.1, we see that

in which ∑j=0∞kα(j) is divergent according to the Raabe’s discriminant, hence the above inequality does not make sense and we can not find a suitable stability in the sense of Ulam-Hyers.

Theorem 4.1

Assume that (H4) holds. Letϑ(n):ℕ0→ℝ+be an increasing sequence such that∑j=0n−1ϑ(j)≤ϑ(n)and letu∈l∞(ℕ0,D(A))be a solution of inequality (4.1), then problem (1.1) is Ulam-Hyers-Rassias stable.

Proof

Let v∈l∞(ℕ0,D(A)) be a solution of inequality (4.1). By Remark 4.1, from the property of 0 < k^α(n) ≤ 1 for α ∈ (0, 1), n∈ℕ0, we have

Let us denote by u∈l∞(ℕ0,D(A)) the unique solution of the Cauchy problem

The solution u of above equation satisfies

therefore, it follows that

On the other hand, let b(n) = ‖u(n) − v(n)‖, from 0≤b(n)≤a(n)+∑j=0n−1L(j)b(j) with respect to a increasing sequence a(n) for all n∈ℕ0, we get

In fact, in view of b(0) = 0, we have for n = 1 that b(1) ≤ a(1); for n = 2, we get that b(2) ≤ a(2) + a(1)L(1) ≤ a(2)(1 + L(1)). Assume that it is true for some n=k∈ℕ2. Let n = k + 1, then the induction implies

which implies the desired inequality. Since ∑j=1∞L(j) is convergent absolutely, it follows that ∏j=1∞(1+L(j)) is convergent absolutely and then there exists a constant M_∗ > 0 such that

Thus, let a(n) = Mϑ(n), there exists a constant C := MM_∗ > 0 such that

Therefore, we conclude the desired result. The proof is completed.

Example 4.1

For any 0 < λ < 1 and 0 < α < 1, let us consider the following fractional difference equation

where g(n) is a bounded sequence on l∞(ℕ0) with n²|g(n)| ≤ 1, parameter ν > 0. Clearly, λ > 0 is the generator of the exponentially bounded C₀-semigroup T(t) = e^−λt for t ≥ 0. Hence, (H1) holds. Let f(n, u) = νg(n) sin (u), it is easy to check the condition (H4) and if ϑ(n) = 2ⁿ for all n∈ℕ0 and the inequality

holds, then (4.2) is the Ulam-Hyers-Rassias stable by Theorem 4.1.