Existence results of noninstantaneous impulsive fractional integro-differential equation

Definition 2.1

[35] The Riemann-Liouville fractional integral operator of β>0, of function h∈L1(ℝ+) is defined as

provided the integral on the right-hand side exists, where Γ(⋅) is the gamma function.

Definition 2.2

[36] The Caputo fractional derivative of order β>0, n−1<β<n, n∈ℕ, is defined as

where the function h(t) has absolutely continuous derivatives up to order (n−1).

Theorem 2.1

(Banach fixed point theorem) [37] Let E be a closed subset of a Banach space(X,||⋅||)and letT:E→Econtraction, then T has unique fixed point in E.

Theorem 2.2

(Krasnoselskii’s fixed point theorem) [37] Let E be a closed convex nonempty subset of a Banach space(X,||⋅||)and P and Q are two operators on E satisfying:

1. Pu+Qv∈E, wheneveru,v∈E,

2. P is contraction,

3. Q is completely continuous,

then the equationPu+Qu=uhas unique solution.

Definition 2.3

(Completely continuous operator) [38] Let X and Y be Banach spaces. Then the operator T:D⊂X→Y is called completely continuous if it is continuous and maps any bounded subset of D to relatively compact subset of Y.

Definition 3.1

The function u(t) is called mild solution of the impulsive fractional equation (3.1) over the interval if u(t) satisfies the integral equation

where

are the linear operators defined on U. Here, ζα(θ) is the probability density function over the interval [0,∞) defined by

and the operator S(t) is the semi-group generated by evolution operator A.

Assumptions 3.1

Assumptions for the existence and uniqueness of the mild solution of fractional evolution equation with noninstantaneous impulses.

(A1) The evolution operator A generates C0 semigroup S(t) for all t∈[0,T].

(A2) The function f:[0,T]×U×U→U is continuous with respect to t and there exist positive constants f1⁎ and f2⁎ such that ||f(t,u1,v1)−f(t,u2,v2)||≤f1⁎||u1−u2||+f2⁎||v1−v2|| for u1,v1,u2,v2∈Br0={u∈U;||u||≤r0} for some r0.

(A3) The operator K:[0,T]×U→U is continuous and there exist a constant k⁎ such that ||Ku−Kv||≤k⁎||u−v|| for u,v∈Br0.

(A4) The functions gk:[tk,sk]×U are continuous and there exist positive constants 0<gk⁎<1 such that ||gk(t,u(t))−gk(t,v(t))||≤gk⁎||u−v||.

Lemma 3.1

[10] If the evolution operator A generatesC0semigroupS(t), then the operatorsU(t)andV(t)are strongly continuous and bounded. This means there exist positive constant M such that||U(t)u||≤M||u||and||V(t)u||≤MΓ(α)||u||for allt∈[0,T].

Theorem 3.2

If Assumptions (A1)–(A4) are satisfied, then the semilinear fractional integro-differential equation with noninstantaneous impulses (3.1) has unique mild solution.

Proof

Define the operator ℱ on U by

where ℱ1, ℱ2k and ℱ3k are

for all k=1,2,…p.

In view of this operator ℱ, equation (3.2) has unique solution if and only if the operator equation u(t)=ℱu(t) has unique solution. This is possible if and only if each of u(t)=ℱ1u(t), u(t)=ℱ2ku(t) and u(t)=ℱ3ku(t) has unique solution over the interval [0,t1), [tk,sk) and [sk,tk+1) for all k=1,2,…,p, respectively, as let u1(t),u2k(t) and u3k(t) be the solutions of u(t)=ℱ1u(t), u(t)=ℱ2ku(t) and u(t)=ℱ3ku(t), respectively. Defining,

then one can easily show that u(t) is a unique solution of u(t)=ℱu(t).

For all t∈[0,t1) and u,v∈Br0,

By applying Assumptions (A1)–(A3) and Lemma 3.1, we get

Considering supremum over interval [0,t1) we get ||ℱ1(n)u−ℱ1(n)v|≤c⁎||u−v||→0 for fixed t1. Therefore, there exist m such that ℱ1(m) is contraction on Br0. Thus, by general Banach contraction theorem the operator equation u(t)=ℱ1u(t) has unique solution over the interval [0,t1).

For all k=1,2,…,p, t∈[tk,sk) and u,v∈U and assuming (A4)

Then ℱ2k is contraction and by the Banach fixed point theorem the operator equation u(t)=ℱ2ku(t) has unique solution for the interval [tk,sk) for all k=1,2,…,p. This means for all k=1,2,…,p, u(t)=gk(t,u(t)) has unique solution for all t∈[tk,sk). Lipschitz continuity of gk leads to uniqueness of the solution at point sk also.

For all k=1,2,…,p, t∈[sk,tk+1) and u,v∈Br−0,

Applying Assumptions (A1)–(A3) and Lemma 3.1, we get

Considering supremum over interval [sk,tk+1) we get ||ℱ3k(n)u−ℱ3k(n)v|≤c⁎||u−v||→0 for fixed sub-interval [sk,tk+1) for all k=1,2,…,p. Therefore, there exist m such that ℱ3k(m) is contraction on Br0. Thus, by general Banach contraction theorem the operator equation u(t)=ℱ3ku(t) has unique solution over the interval [sk,tk+1) for all k=1,2,…,p.

Hence, the operator equation u(t)=ℱu(t) has unique solution over the interval [0,T], which is nothing but mild solution of Eq. (3.1).□

Example 3.2.1

The fractional order integro-differential equation:

over the interval [0,1] with initial condition u(0,x)=u0(x) and boundary condition u(t,0)=u(t,1)=0. Equation (3.3) can be reformulated as the fractional order abstract equation in U=L2([0,1],ℝ) as:

over the interval [0,1] by defining z(t)=u(t,⋅), operator Au=u″ (second-order derivative with respect to x). The functions f and g over respected domains are defined as f(t,z(t),Kz(t))=(z2(t))′/2+∫0te−z(s)ds and g(t,z(t))=z(t)2(1+z(t)), respectively.

1. The linear operator A over the domain D(A)={u∈U;u″ exists and continuous with u(0)=u(1)=0} is self-adjoint, with compact resolvent and is the infinitesimal generator of C0 semigroup S(t) over the interval [0,1] given by

   (3.5)S(t)u=∑n=1∞exp(−n2π2t)<u,ϕn>ϕn,

   where ϕn(s)=2sin(nπs) for all n=1,2,… is the orthogonal basis for the space X.

2. The function K:[0,1]×[0,1]×X→X is continuous with respect to t and differentiable with respect to z for all z and hence K is Lipschitz continuous with respect to z. This means that there exist positive constant k⁎ such that ||K(t,z1)−K(t,z2)||≤k⁎||z1−z2||.

3. The function f:[0,1]×X×X→X is continuous with respect to t and is differential with respect to argument z and Kz. Therefore, there exist positive constants f1⁎ and f2⁎ such that ||f(t,z1,Kz1)−f(t,z2,Kz2)||≤f1⁎||z1−z2||+f2⁎||Kz1−Kz2||, z1,z2∈Br0 for some r0.

4. The impulse g is continuous with respect to t and Lipchitz continuous with respect to z with Lipschitz constant g⁎=1/2<1.

Therefore, by Theorem 3.2, equation (3.4) has unique solution over [0,1]. Hence, equation (3.3) has unique solution over the interval [0,1].

Definition 4.1

The function u(t) is called mild solution of the impulsive fractional equation (3.1) over the interval if u(t) satisfies the integral equation

where

are the linear operators defined on U. Here, ζα(θ) is the probability density function over the interval [0,∞) defined by

and the operator S(t) is the semi-group generated by evolution operator A.

Assumptions 4.1

Assumptions for the existence of the mild solution of fractional evolution equation with noninstantaneous impulses.

(B1) The evolution operator A generates C0 semigroup S(t) for all t∈[0,T].

(B2) The function f(t,⋅,⋅) is continuous and f(⋅,u,v) is measurable on [0,T]. Also, there exist β∈(0,α) with mf∈L1β([0,T],ℝ)su such that |f(t,u,v)|≤mf(t) for all u,v∈U.

(B3) The operator K:[0,T]×U→U is continuous and there exists a constant k⁎ such that ||Ku−Kv||≤k⁎||u−v||.

(B4) The operator h:U→U is Lipschitz continuous with respect to u with Lipschitz constant 0<h⁎≤1.

(B5) The functions gk:[tk,sk]×U are continuous and there exist positive constants 0<gk⁎<1 such that ||gk(t,u(t))−gk(t,v(t))||≤gk⁎||u−v||.

Theorem 4.1

(Existence theorem) If Assumptions (B1)–(B5) are satisfied, then the nonlocal semi-linear fractional order integro-differential equation (4.2) has mild solution providedMh⁎<1andMg⁎<1.

Proof

From Lemma 3.1, ||U(t)||≤M for all u∈Bk={u∈U:||u||≤k} for any positive constant k. Therefore,

According to (B2), f(⋅,u,v) is measurable on [0,T] and one can easily show that (t−s)α−1∈L11−β[0,t] for all t∈[0,T] and β∈(0,α). Let

By Holder’s inequality and Assumption (B2), for t∈[0,T],

For t∈[0,t1) and for positive r we define F1 and F2 on Br as,

then u(t) is the mild solution of the semilinear fractional integro-differential equation if and only if the operator equation u=F1u+F2u has solution for u∈Br for some r. Therefore, the existence of a mild solution of (3.1) over the interval [0,t1) is equivalent to determining a positive constant r0, such that F1+F2 has a fixed point on Br0.

Step 1:||F1u+F2v||≤r0 for some positive r0.

Let u,v∈Br0, choose

and consider

Therefore, ||F1u+F2v||≤r0 for every pair u,v∈Br0.

Step 2:F1 is contraction on Br0.

For any u,v∈Br0 and t∈[0,t1), we have |F1u(t)−F1v(t)|≤Mh⁎||u−v||. Taking supremum over [0,t1), ||F1u−F1v||≤Mh⁎||u−v||. Since Mh⁎<1, F1 is contraction.

Step 3:F2 is a completely continuous operator on Br0.

Let {un} be the sequence in Br0 converging to u∈Br0 and consider,

which implies

Continuity of f and K leads to ||F2un−F2u||→0 as n→∞. Thus, F2 is continuous.

To show {F2u(t),u∈Br0} is relatively compact it is sufficient to show that the family of functions {F2u,u∈Br0} is uniformly bounded and equicontinuous, and for any t∈[0,t1), {F2u(t),u∈Br0} is relatively compact in U.

Clearly for any u∈Br0, ||F2u||≤r0, which means that the family {F2u(t),u∈Br0} is uniformly bounded.

For any u∈Br0 and 0≤τ1<τ2<t1,