Periodic BVP for a class of nonlinear differential equation ... - SciELO

CUBO A Mathematical Journal Vol.17, N¯o 01, (11–27). March 2015

Periodic BVP for a class of nonlinear differential equation with a deviated argument and integrable impulses Alka Chadha and Dwijendra N Pandey Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee-247667 [email protected], [email protected] ABSTRACT This paper deals with periodic BVP for integer/fractional order differential equations with a deviated argument and integrable impulses in arbitrary Banach space X for which the impulses are not instantaneous. By utilizing fixed point theorems, we firstly establish the existence and uniqueness of the mild solution for the integer order differential system and secondly obtain the existence results for the mild solution to the fractional order differential system. Also at the end, we present some examples to show the effectiveness of the discussed abstract theory. RESUMEN Este art´ıculo estudia las ecuaciones diferenciales de orden entero/fraccional con condiciones de frontera peri´odicas con un argumento desviado e impulsos integrables en espacios de Banach arbitrarios X donde los pulsos no son instant´ aneos. Utilizando teoremas de punto fijo, establecemos la existencia y unicidad de soluciones temperadas para los sistemas diferenciales de orden entero, y luego obtenemos resultados de existencia para soluciones temperadas del sistema diferencial de orden fraccional. Adem´as, presentamos un ejemplo para mostrar la efectividad de la teor´ıa abstracta discutida. Keywords and Phrases: Deviating arguments, Fixed point theorem, Impulsive differential equation, Periodic BVP, Fractional calculus. 2010 AMS Mathematics Subject Classification: 34G20, 34K37, 34K45, 35R12, 45J05.

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1

Alka Chadha & Dwijendra N Pandey

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Introduction:

In a few decades, impulsive differential equations have received much attention of researchers mainly due to its demonstrated applications in widespread fields of science and engineering. Impulsive differential equations have played an important role in real world problems for describing a process which is characterized by the development of a sudden change in system’s state. Such processes are investigated in various fields such as biology, physics, control theory, population dynamics, medicine and many others. Impulsive differential equations are an appropriate model to hereditary phenomena for which a delay argument arises in modelling equations. For more details on impulsive differential equation, we refer to the monographs [1],[2] and papers [3]-[12] and references given therein. A differential equation with boundary conditions arise in many areas of applied sciences, for example, chemical engineering, blood flow problems, thermoelasticity, population models, underground water flow and many others. For more details on differential equation with integral boundary conditions, we refer to [10, 17, 18, 19, 23, 27] and references given therein. On the other hand, fractional calculus have many applications in various areas of sciences and engineering for example, fluid dynamics, like fractal theory, diffusion in porous media and fractional biological neurons, traffic flow, polymer rheology. The fractional differential equation is an important tool to describe nonlinear oscillation of earthquake. For more study on fractional calclus, we refer to books [13]-[16]. In this work, we consider the periodic boundary value problems for integer order nonlinear differential equations in a Banach space X of the form with non-instantaneous integrable impulses u ′ (t) = u(t) =

f(t, u(t), u([h(u(t), t)])), t ∈ (sm , tm+1 ], m = 0, 1, 2, · · · , δ, Zt Gm (s, u(s))ds, t ∈ (tm , sm ], m = 1, 2, · · · , δ, δ ∈ N

(1.1) (1.2)

tm

u(0)

=

u(T ).

(1.3)

Next, we consider the periodic boundary value problems for nonlinear fractional differential equations in a Banach space X of the form with non-instantaneous integrable impulses c

Dq 0,t u(t) u(t)

= I2−q f(t, u(t), u([h(u(t), t)])), t ∈ (sm , tm+1 ], 0 < q < 1, t Zt Gm (s, u(s))ds, t ∈ (tm , sm ], m = 0, 1, 2, · · · , δ, δ ∈ N, =

(1.4) (1.5)

tm

u(0)

= u(T ),

(1.6)

where 0 < T < ∞, c Dq 0,t represents the Caputo fractional derivative of the order q with lower limit 0, 0 = t0 = s0 < t1 ≤ s1 ≤ t2 < · · · < tδ ≤ sδ ≤ tδ+1 = T are fixed numbers, Gm : (tm , sm ] × X → X, m = 1, · · · , δ. The nonlinear X-valued functions f and h are appropriate functions and satisfy some suitable conditions to be stated later. In this system of equations

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Periodic BVP for a class of nonlinear differential equation . . .

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(1.1)-(1.3) and (1.4)-(1.6), the impulses begin all of a sudden at the points ti and continue their proceeding on a finite interval [ti , si ]. According to the authors in [4]-[5], there are many different inspirations for consideration of the problem (1.1)-(1.3) and (1.4)-(1.6). The hemodynamical equilibrium of a person is an example of such systems. One can prescribe some intravenous drugs (insulin) in the case of a decompensation (for example, low or high level of glucose). Since the introduction of the drugs in the bloodstream and the consequent absorption of the body are successive and continuous processes, we can describe this situation as an impulsive action which start suddenly and stays active on a finite time interval. The organization of the paper is as follows: In section 2, we give some basic definitions, assumptions, lemmas and theorems as preliminaries which will be used for proving our main results. In section 3, we prove the existence of a mild solution to the problem (1.1)-(1.3) and problem (1.4)-(1.6). Some examples are also presented at the end of the paper.

2

Preliminaries and Assumptions

In this section, we discuss some basic definitions, preliminaries, theorem and lemmas which will be used for proving the required result. Let (X, k · k) be a Banach space. Let C(J; X), where J = [0, T ] denotes the space of all continuous X-valued functions on interval J which is a Banach space with the norm k ukC = supt∈J k u(t)k. The space of all Bochner integrable functions u : (0, T ) → X represented by L1 ((0, T ); X), is a RT Banach space with norm k uk1 = 0 k u(t)kdt. The Br (x, X) denotes the closed ball with center at x and radius r in X. To study the impulsive differential equation, we introduce the following space PC([0, T ]; X)

=

{u : [0, T ] → X; u ∈ C((tj , tj+1 ]; X),

∃

u(t+ j )

and

u(t− j ), j

j = 0, 1, · · · , m, and

= 1, · · · , m exist with u(t− j ) = u(tj )}.

It is clear that PC([0, T ]; X) is a Banach space with the norm kukPC = max k u(t)k. t∈[0,T ]

For a function u ∈ PC([0, T ]; X) and j ∈ {0, 1, · · · , m}, we define the function uej ∈ C([tj , tj+1 ]; X) such that u(t), for t ∈ (tj , tj+1 ], uej (t) = (2.1) u(t+ j ), for t = tj . fj = {uej : u ∈ B} and we have following For B ⊂ PC([0, T ]; X) and j ∈ {0, 1, · · · , m}, we have B Accoli-Arzel` a type criteria.

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Lemma 2.1. [4]. A set B ⊂ PC([0, T ]; X) is relatively compact in PC([0, T ]; X) if and only if each set f Bj (j = 1, 2, · · · , m) is relatively compact in C([tj , tj+1 ], X)(j = 0, 1, · · · , m). Now, we recall some basic definition. Definition 2.1. The Riemann-Liouville fractional integral of f with order q defined by Iq 0,t f(t) =

1 Γ (q)

Zt

(t − s)q−1 f(s)ds.

(2.2)

0

Definition 2.2. The fractional derivative of function f : [0, ∞) → R in the Riemann-Liouville sense with order q is defined by Dq 0,t f(t) =

dn 1 n dt Γ (n − q)

Zt

(t − s)n−q−1 f(s)ds,

t > 0,

n − 1 < q < n.

(2.3)

0

Definition 2.3. The fractional derivative of function f : [0, ∞) → R in the Caputo sense of order q is defined by c

Dq 0,t f(t)

=

1 Γ (n − q)

Zt

(t − s)n−q−1 fn (s)ds,

(2.4)

0

for n − 1 < q < n, n ∈ N, t > 0, with the following property: c

q Dq 0,t (I0,t f(t)) =

f(t) −

n−1 X k=1

tk k f (0). k!

(2.5)

Before expressing and demonstrating the required main result, we present the following definition of mild solution to the system (1.1)-(1.3) and (1.4)-(1.6). Lemma 2.2. For given continuous function f : [0, T ] → X and Gm ∈ C([tm , sm ], X), a function u ∈ PC([0, T ]; X) is a mild solution for the impulsive periodic boundary value problem u ′ (t) = u(t)

=

f(t), t ∈ (sm , tm+1 ], m = 0, 1, 2, · · · , δ, δ ∈ N Zt Gm (s)ds, t ∈ (tm , sm ], m = 1, 2, · · · , δ,

(2.6) (2.7)

tm

u(0)

=

u(T ),

(2.8)

if and only if u(·) satisfies the following Rsδ RT Rt  G (s)ds + sδ f(s)ds + 0 f(s)ds, t ∈ [0, t1 ],   R tδ δ Rt sm u(t) = t ∈ (sm , tm+1 ], Gm (s)ds + sm f(s)ds, tm   R t t ∈ (tm , sm ], tm Gm (s)ds,

for each m = 1, · · · , δ.

(2.9)

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Lemma 2.3. For the continuous function f : [0, T ] → X and Gm ∈ C([tm , sm ], X), a function u ∈ PC([0, T ]; X) is said to be a mild solution for the system c

Dq 0,t u(t)

=

u(t)

=

I2−q f(t), t ∈ (sm , tm+1 ], 0 < q < 1, t Zt Gm (s)ds, t ∈ (tm , sm ], m = 0, 1, 2, · · · , δ, δ ∈ N,

(2.10) (2.11)

tm

u(0)

=

u(T ),

if and only if u(0) = u(T ), u(t) = following integral equations

Rt

tm

(2.12)

Gm (t), ∀ t ∈ (tm , sm ], m = 1, · · · , δ and u(·) satisfies the

R Rsδ RT sδ    tδR Gδ (s)ds − 0 (sδ − s)f(s)ds + 0 (T − s)f(s)ds t t ∈ [0, t1 ], u(t) = + 0 (t − s)f(s)ds,  R R  s s  m G (s)ds − m (s − s)f(s)ds + Rt (t − s)f(s)ds, t ∈ (s , t m m m m+1 ], tm 0 0

(2.13)

for each m = 1, · · · , δ.

Further, we list the following assumption which will be used to establish the main result. Assumptions on f, h and Gm , (m = 1, · · · , δ) : (A1) The function f : [0, T ] × X × X → X is continuous and there exist a positive constant Lf and 0 < γ1 ≤ 1 such that kf(t1 , u1 , v1 ) − f(t2 , u2 , v2 )k ≤ Lf [|t1 − t2 |γ1 + ku1 − u2 kX + kv1 − v2 kX ],

(2.14)

for all (tj , uj , vj ) ∈ [0, T ] × X × X, j = 1, 2. (A2) h : X × [0, T ] → [0, T ] is continuous function and there exist positive constants Lh and 0 < γ2 ≤ 1 such that |h(u1 , t1 ) − h(u2 , t2 )| ≤ Lh [ku1 − u2 kX + |t1 − t2 |γ2 ],

(2.15)

for each (uj , tj ) ∈ X × [0, T ], for j = 1, 2. (A3) Gm : [0, T ] × X → X, m = 1, 2, · · · , δ, are continuous functions and there exist constants LGm > 0 such that k Gm (t, x) − Gm (t, y)k ≤ kGm(t, u(t))k

≤

LGm k x − yk,

(2.16)

Km ,

(2.17)

for all (t, x), (s, y) ∈ [0, T ] × X, u ∈ X and Km > 0, m = 1, · · · , δ are constants.

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Existence Result

In this section, we establish the existence of a mild solutions for the systems (1.1)-(1.3) and (1.4)(1.6) by using fixed point theorems. Let Y0 = PC(J; X)

=

{y ∈ PC(J; X) : y ∈ C((tm , tm+1 ], X), m = 0, 1, · · · , δ + and y(t− m ) = y(tm ), y(tm ) exist}.

(3.1)

and Y1 = {y ∈ Y0 : ky(t) − y(s)k ≤ L|t − s|, ∀ t ∈ [tm , tm+1 ], m = 0, 1, · · · , δ}.

(3.2)

Where L is an appropriate positive constant to be defined later.

3.1

Integer Order case

Theorem 3.1. We assume that assumptions (A1) − (A3) are satisfied. If Θ

=

sup{ max [LGm (sm − tm ) + Lf (1 + Lh L)(tm+1 − sm )], m=1,··· ,δ

LGδ (sδ − tδ ) + Lf (1 + Lh L)(T − sδ + t1 )} < 1.

(3.3)

Then, the system (1.1)-(1.3) has a unique mild solution on the interval J. Proof. In order to transform the system (1.1)-(1.3) into a fixed point problem, we consider the map Q : S → S defined by R t G (s, u(s))ds, t ∈ (tm , sm ], m = 1, · · · , δ,  tm m    Rsδ G (s, u(s))ds + RT f(s, u(s), u([h(u(s), s)]))ds δ tδ sδ Rt (3.4) Qu(t) =  f(s, u(s), u([h(u(s), s)]))ds, t ∈ [0, t1 ], +  0   Rt Rsδ tδ Gm (s, u(s))ds + sm f(s, u(s), u([h(u(s), s)]))ds, t ∈ (sm , tm+1 ], where S = {u ∈ Y0 ∩Y1 : kukPC ≤ R}. Clearly, S is a closed and bounded subset of Y1 and complete metric space. It is not difficult to show that Qu ∈ Y0 . Now, it remains to show that Qu ∈ Y1 . For u ∈ S and τ2 , τ1 ∈ [0, t1 ] with τ1 < τ2 , Z τ2 kf(s, u(s), u([h(u(s), s)]))kds, kQu(τ2 ) − Qu(τ1 )k ≤ τ1

≤

(3.5)

H|τ2 − τ1 |.

where H = supt∈[0,T ] kf(t, u(t), u([h(u(t), t)]))k. Similarly, τ2 , τ1 ∈ (tm , sm ], m = 1, · · · , δ Z τ1 Z τ2 Gm (s, u(s))dsk Gm (s, u(s))ds − kQu(τ2 ) − Qu(τ1 )k ≤ k tm

≤ Km |τ2 − τ1 |,

tm

(3.6)

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17

and for τ2 , τ1 ∈ (sm , tm+1 ], m = 1, · · · , δ kQu(τ2 ) − Qu(τ1 )k ≤ H|τ2 − τ1 |.

(3.7)

Therefore, we conclude that Qu ∈ Y1 with suitable constant L = min{ max

m=1,··· ,δ

Km , H}.

Now, we show that Q(S) ⊆ S. For t ∈ [0, t1 ] and u ∈ S, we get ZT Z sδ Gδ (s, u(s))dsk + kf(s, u(s), u([h(u(s), s)]))kds kQu(t)k ≤ k tδ

Zt

+

sδ

kf(s, u(s), u([h(u(s), s)]))kds,

0

≤ Kδ (sδ − tδ ) + H(T − sδ + t1 ) ≤ Kδ T + HT.

(3.8)

For t ∈ [sm , tm+1 ], m = 1, · · · , δ, kQu(t)k ≤

Km (sm − tm ) + H(tm+1 − sm ) ≤ Km T + HT,

(3.9)

and for t ∈ (sm , tm ], we have that kQu(t)k ≤ Km T . We choose R = max[Kδ T +HT,

sup

{Km T +

m=1,··· ,δ

HT }]. Thus, we get that Q(S) ⊆ S. In the next step, we prove that Q is a contraction map. For w1 , w2 ∈ S and t ∈ [0, t1 ], we get kQw1 (t) − Qw2 (t)k

≤ [LGδ (sδ − tδ ) + Lf (1 + Lh L)(T − sδ + t1 )] ×kw1 − w2 kPC .

(3.10)

For t ∈ [sm , tm+1 ], m = 1, · · · , δ kQw1 (t) − Qw2 (t)k

≤ ≤

[LGm (sm − tm ) + Lf (1 + Lh L)(tm+1 − sm )]kw1 − w2 kPC , max [LGm (sm − tm ) + Lf (1 + Lh L)(tm+1 − sm )]

m=1,··· ,δ

×kw1 − w2 kPC ,

(3.11)

and for t ∈ (tm , sm ], we obtain that kQw1 (t) − Qw2 (t)k ≤

max

m=1,··· ,δ

LGm (sm − tm ) × kw1 − w2 kPC .

(3.12)

From the inequalities (3.10)-(3.12), we get kQw1 − Qw2 kPC ≤ Θkw1 − w2 kPC .

(3.13)

Thus, by the inequality (3.3), we conclude that Q is a contraction on S and there exists a unique fixed point u ∈ S of the map Q. It is obvious that u is a mild solution for the system (1.1)-(1.3). Our second existence result is based on Krasnoselskii’s theorem. The statement of the theorem is given as:

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Theorem 3.2. Let F ⊂ X be a closed convex and nonempty subset of X, where X is a Banach space. Let P1 and P2 be the operator such that (a) P1 w1 + P2 w2 ∈ F, whenever, w1 , w2 ∈ F, (b) P1 is a contraction, (c) P2 is compact and continuous. Then, the map P = P1 + P2 has a fixed point x ∈ F i.e., x = P1 x + P2 x. Theorem 3.3. Assume that (A1) − (A3) are satisfied. Then, there exists a mild solution for the system (1.1)-(1.3) on J provided that Ξ = max{KGm (sm − tm ); m = 1, · · · , δ} < 1.

(3.14)

Proof. We define the following operators Q1 : S → S which is decomposition of operator Q, by Rsδ  G (s, u(s))ds, t ∈ [0, t1 ],   R tδ δ t Q1 u(t) = (3.15) t ∈ (tm , sm ], m = 1, · · · , δ, sm Gm (s, u(s))ds,   Rsm Gm (sm , u(sm )), t ∈ (sm , tm+1 ] m = 1, · · · , δ. tm

and Q2 : S → S by R T Rt  f(s, u(s), u(h(u(s), s)))ds + 0 f(s, u(s), u(h(u(s), s)))ds, t ∈ [0, t1 ],   sδ (3.16) Q2 u(t) = 0, t ∈ (tm , sm ], m = 1, · · · , δ,   R t sm f(s, u(s), u(h(u(s), s)))ds, t ∈ (sm , tm+1 ] i = 1, · · · , δ. We choose r such that

max{ max (Km + H)T, (Kδ + H)T } < r.

(3.17)

Br = {u ∈ Y0 ∩ Y1 : kukPC ≤ r}.

(3.18)

m=1,··· ,δ

Consider It is clear that the mappings Q1 and Q2 are well-defined. Now, we show the result in several steps. Step 1. For u, v ∈ Br and t ∈ [0, t1 ], we have ZT Z sδ kf(s, v(s), v(h(v(s), s)))kds Gδ (s, u(s))dsk + k(Q1 u + Q2 v)(t)k ≤ k sδ

tδ

+

Zt

kf(s, v(s), v(h(v(s), s)))kds,

0

≤ Kδ (sδ − tδ ) + H[T − sδ − t1 ] ≤ Kδ T + HT.

(3.19)

For t ∈ (sm , tm+1 ], m = 1, · · · , δ, k(Q1 u + Q2 v)(t)k

≤ k

Z sm tm

Gm (s, u(s))dsk +

Zt

kf(s, u(s), u(h(u(s), s)))kds,

sm

≤ Km (sm − tm ) + H(tm+1 − sm ) ≤ (Km + H)T,

(3.20)

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19

and for t ∈ [tm , sm ], we have k(Q1 u + Q2 v)(t)k ≤ Km T . Thus, by the choice of r, we get that k(Q1 u + Q2 v)kPC ≤ r, for all t ∈ [0, T ].

(3.21)

Hence, Q1 u + Q2 v ∈ Br . Step 2. We show that Q1 is contraction map. For w1 , w2 ∈ Br and t ∈ [0, t1 ], kQ1 w1 (t) − Q2 w2 (t)k

≤ KGδ kw1 (sδ ) − w2 (sδ )k × |sδ − tδ | ≤ KGδ (sδ − tδ )kw1 − w2 kPC .

(3.22)

For t ∈ (tm , sm ], m = 1, · · · , δ, we get kQ1 w1 (t) − Q2 w2 (t)k ≤ KGm (sm − tm )kw1 − w2 kPC ,

(3.23)

and t ∈ (sm , tm+1 ], m = 1, · · · , δ kQ1 w1 (t) − Q2 w2 (t)k ≤ KGm (sm − tm )kw1 − w2 kPC .

(3.24)

From the above inequalities, we conclude that kQ1 w1 − Q2 w2 kPC ≤ Ξkw1 − w2 kPC ,

(3.25)

which gives that Q1 is a contraction. Step 3. Q2 is continuous map. Let {zp }∞ p=1 be a sequence such that zp → z ∈ Br . For t ∈ [0, t1 ], kQ2 zp (t) − Q2 z(t)k

ZT

≤

kf(s, zp (s), zp (h(zp (s), s))) − f(s, z(s), z(h(z(s), s)))kds

sδ

+

Zt

kf(s, zp (s), zp (h(zp (s), s))) − f(s, z(s), z(h(z(s), s)))kds,

0

by the continuity of f and h, we have that s ∈ [0, t] f(s, zp (s), zp (h(zp (s), s))) → h(zp (s), s) →

f(s, z(s), z(h(z(s), s))), as p → ∞,

h(z(s), s),

From the dominated convergence theorem, we get

as p → ∞,

kQ2 zp − Q2 zkPC → 0, as p → ∞,

For t ∈ (tm , sm ], m = 1, · · · , δ,

kQ2 zp (t) − Q2 z(t)k = 0. Similarly, for t ∈ (sm , tm+1 ], m = 1, · · · , δ kQ2 zp (t) − Q2 z(t)k

≤

Zt

sm

kf(s, zp (s), zp (h(zp (s), s))) − f(s, z(s), z(h(z(s), s)))kds,

(3.26) (3.27)

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CUBO 17, 1 (2015)

by the continuity of f, h and the dominated convergence theorem, we deduce that kQ2 zp − Q2 zkPC → 0,

as p → ∞.

Step 3. Q2 is compact. Since f is continuous map and k(Q2 u)(t)k ≤ 2HT < r. This implies that Q2 is uniformly bounded on Br . Now, we show that Q2 maps bounded set into equicontinuous set of Br . For τ2 > τ1 ∈ [0, t1 ] and u ∈ Br , we have kQ2 u(τ2 ) − Q2 u(τ1 )k

≤

LF (τ2 − τ1 ).

(3.28)

For τ2 > τ1 ∈ (tm , sm ], we have kQ2 u(τ2 ) − Q2 u(τ1 )k = 0. For τ2 > τ1 ∈ (sm , tm+1 ], m = 1, · · · , δ and u ∈ Br , we have kQ2 u(τ2 ) − Q2 u(τ1 )k

≤

LF (τ2 − τ1 ).

(3.29)

Thus, we conclude that kQ2 u(τ2 ) − Q2 u(τ1 )k → 0 as τ2 → τ1 . Hence Q2 is equicontinuous. By the Steps (3) − (4) and Arzela-Ascoli theorem, we deduce that Q2 : Br → Br is continuous and compact i.e. completely continuous. Since Q1 is contraction and Q2 is completely continuous operator. Thus, Q = Q1 + Q2 has a fixed point by using Krasnoselskiis fixed point theorem which is just a mild solution for the system (1.1)-(1.3). The proof of the theorem is finished.

3.2

Fractional order case

Now, we obtain the existence results for the problem (1.4)-(1.6) via fixed points theorems, the first existence result of the mild solution for problem (1.4)-(1.6) is obtained by using Banach fixed point theorem and second existence results is obtained by using Krasnoselskii’s fixed point theorem. Theorem 3.4. Assume that hypotheses (A1) − (A3) are fulfilled and Lf (1 + LLh )(t2m+1 + s2m ) ], max (sm − tm )LGm , m=1,··· ,δ m=1,··· .δ 2 Lf (1 + LLh )(T 2 + s2δ + t21 LGδ (sδ − tδ ) + } < 1. (3.30) 2 Then, the problem (1.4)-(1.6) has at least one mild solution on [0, T ]. Λ = sup{ max [LGm (sm − tm ) +

Proof. We firstly define the operator Q : S → S by Rsδ Rsδ   tδR Gδ (s, u(s))ds − 0 (sδ − s)f(s, u(s), u([h(u(s), s)]))ds  T   + 0 (T − s)f(s, u(s), u([h(u(s), s)]))ds   R   + t (t − s)f(s, u(s), u([h(u(s), s)]))ds, t ∈ [0, t1 ], 0 (Qu)(t) = Rt  G (s, u(s))ds, t ∈ (tm , sm ], m = 1, · · · , δ,  tm m   R Rs  s m   Gm (s, u(s))ds − 0m (sm − s)f(s, u(s), u([h(u(s), s)]))ds  tm  R  t + 0 (t − s)f(s, u(s), u([h(u(s), s)]))ds, t ∈ (sm , tm+1 ], m = 1, · · · , δ.

(3.31)

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21

It is clear that Qu ∈ Y0 . So it remains to show that Qu ∈ Y1 . For u ∈ S and τ2 , τ1 ∈ [0, t1 ] with τ1 ≤ τ2 , we get k(Qu)(τ2 ) − (Qu)(τ1 )k =

k

≤ k

Z τ2

Z0τ1

(τ2 − s)f(s, u(s), u([h(u(s), s)]))ds −

Z τ1

(τ1 − s)f(s, u(s), u([h(u(s), s)]))dsk,

[(τ2 − s) − (τ1 − s)]f(s, u(s), u([h(u(s), s)]))dsk

0

+k

Z τ2

(τ2 − s)f(s, u(s), u([h(u(s), s)]))dsk,

τ1

≤ H(τ2 − τ1 )2 + H

(τ2 − τ1 )2 , 2

≤ 2HT |τ2 − τ1 |,

(3.32)

Similarly, for τ2 , τ1 ∈ (sm , tm+1 ], m = 1, · · · , δ, k(Qu)(τ2 ) − (Qu)(τ1 )k ≤ HT |τ2 − τ1 |

(3.33)

and for τ2 , τ1 ∈ (tm , sm ], k(Qu)(τ2 ) − (Qu)(τ1 )k ≤ Km |τ2 − τ1 |. Thus, from (3.32)-(3.34), we conclude that Qu ∈ Y1 with L = min{ max

m=1,··· ,δ

(3.34) Km , 2HT, HT }. Hence

Q is well defined on S. Next we show that Q(S) ⊆ S. For u ∈ S and t ∈ [0, t1 ], we get kQu(t)k ≤

k

Z sδ

+ +

tδ ZT 0 Zt

Gδ (s, u(s))dsk +

Z sδ

(sδ − s)kf(s, u(s), u([h(u(s), s)]))kds

0

(T − s)kf(s, u(s), u([h(u(s), s)]))kds

(t − s)f(s, u(s), u([h(u(s), s)]))ds,

0

≤

Kδ (sδ − tδ ) +

3HT 2 H(T 2 + s2δ + t21 ) ≤ Kδ T + . 2 2

(3.35)

Similarly, for t ∈ (sm , tm+1 ], m = 1, · · · , δ, kQu(t)k

≤ Km T + T 2 H,

(3.36)

and for t ∈ (tm , sm ], we get kQu(t)k ≤ Km T. 2

(3.37)

We choose R = max{Kδ T + 3T2 H , supm=1,··· ,δ Km T +T 2 H} such that kQu(t)k ≤ R, for all t ∈ [0, T ]. We now show that Q is a contraction map on S. For u∗ , u∗∗ ∈ S and t ∈ [0, t1 ], we get

22


CUBO 17, 1 (2015)

k(Qu∗ )(t) − (Qu∗∗ )(t)k Z sδ [Gδ (s, u∗ (s)) − Gδ (s, u∗∗ (s))]dsk ≤ k tδ Z sδ (sδ − s)kf(s, u∗ (s), u∗ ([h(u∗ (s), s)])) − f(s, u∗∗ (s), u∗∗ ([h(u∗∗ (s), s)]))kds + 0

+ +

ZT

0 Zt

(T − s)kf(s, u∗ (s), u∗ ([h(u∗ (s), s)])) − f(s, u∗∗ (s), u∗∗ ([h(u∗∗ (s), s)]))kds

(t − s)kf(s, u∗ (s), u∗ ([h(u∗ (s), s)])) − f(s, u∗∗ (s), u∗∗ ([h(u∗∗ (s), s)]))kds,

0

Lf (1 + LLh )(T 2 + s2δ + t21 ]ku∗ − u∗∗ kPC . (3.38) 2 Similarly, for t ∈ (sm , tm+1 ], m = 1, · · · , δ k(Qu∗ )(t) − (Qu∗∗ )(t)k Z sm [Gm (s, u∗ (s)) − Gm (s, u∗∗ (s))]dsk ≤ k t Zm sm (sm − s)kf(s, u∗ (s), u∗ ([h(u∗ (s), s)])) − f(s, u∗∗ (s), u∗∗ ([h(u∗∗ (s), s)]))kds + ≤

[LGδ (sδ − tδ ) +

0

+

Zt

(t − s)kf(s, u∗ (s), u∗ ([h(u∗ (s), s)])) − f(s, u∗∗ (s), u∗∗ ([h(u∗∗ (s), s)]))kds,

0

≤

max [LGm (sm − tm ) +

m=1,··· ,δ

Lf (1 + LLh )(t2m+1 + s2m ) ]ku∗ − u∗∗ kPC , 2

(3.39)

and for t ∈ (tm , sm ], we get k(Qu∗ )(t) − (Qu∗∗ )(t)k ≤

max

m=1,··· .δ

LGm (sm − tm )ku∗ − u∗∗ kPC .

(3.40)

From the inequalities (3.38)-(3.40), we obtain k(Qu∗ )(t) − (Qu∗∗ )(t)k ≤ Λku∗ − u∗∗ kPC .

(3.41)

Thus, by the inequality (3.30), we conclude that Q is a contraction on S i.e., there exists a unique fixed point of the map u ∈ S such that Qu(t) = u(t) for all t ∈ [0, T ]. Hence problem (1.4)-(1.6) has a unique mild solution on [0, T ]. Theorem 3.5. Assume that (A1)-(A3) are fulfilled and Ξ = max{LGm |sm − tm |; m = 1, · · · , δ} < 1.

(3.42)

Then, problem (1.4)-(1.6) has at least one mild solution on [0, T ]. Proof. We consider the operators Q1 and Q2 on Bq,r = {u ∈ Y0 ∩ Y1 : kukPC ≤ r} defined by R t  G (s, u(s))ds, t ∈ (tm , sm ],   R tm m sδ Q1 u(t) = (3.43) tδ Gδ (s, u(s))ds, t ∈ [0, t1 ]   Rsm Gm (s, u(s))ds, t ∈ (sm , tm+1 ], m = 1, · · · , δ, tm

CUBO


17, 1 (2015)

23

and

Q2 u(t) =

R T  (T − s)f(s, u(s), u([h(u(s), s)]))ds   0Rsδ    − R0 (sδ − s)f(s, u(s), u([h(u(s), s)]))ds   + t (t − s)f(s, u(s), u([h(u(s), s)]))ds, t ∈ [0, t1 ] 0

 0, t ∈ (tm , sm ], m = 1, · · · , δ,    Rsm    − (sm − s)f(s, u(s), u([h(u(s), s)]))ds    R0t + 0 (t − s)f(s, u(s), u([h(u(s), s)]))ds, t ∈ (sm , tm+1 ], m = 1, · · · , δ.

(3.44)

where r is a positive constant such that max{

sup

H(t2m+1 + s2m ) H(T 2 + s2δ + t21 ) , Kδ (sδ − tδ ) + } ≤ r. 2 2

Km (sm − tm ) +

m=1,··· ,δ

(3.45)

For the purpose of convenience, we separate the proof into a few steps. Step 1. We show that Q1 u + Q2 u ∈ Bq,r for each u ∈ Bq,r . For t ∈ [0, t1 ], we have kQ1 u(t) + Q2 u(t)k ≤ k

Z sδ

Gδ (s, u(s))dsk + k

Z sδ

(T − s)f(s, u(s), u([h(u(s), s)]))dsk

0

tδ

+k

ZT

(sδ − s)f(s, u(s), u([h(u(s), s)]))dsk + k

0

Zt

(t − s)f(s, u(s), u([h(u(s), s)]))dsk

0

≤ Kδ (sδ − tδ ) +

H(T 2 + s2δ + t21 ) , 2

(3.46)

where H = supt∈[0,T ] kf(t, u(t), u([h(u(t), t)]))k. Similarly, for t ∈ (sm , tm+1 ], m = 1, · · · , δ kQ1 u(t) + Q2 u(t)k Z sm Z sm (sm − s)f(s, u(s), u([h(u(s), s)]))dsk Gm (tm , u(tm ))k + k ≤ k tm Zt

+k

0

(t − s)f(s, u(s), u([h(u(s), s)]))dsk,

0

≤

Km (sm − tm ) +

H(t2m+1 + s2m ) , 2

(3.47)

and for t ∈ (tm , sm ], m = 1, · · · , δ, kQ1 u(t) + Q2 u(t)k ≤ Km (sm − tm ).

(3.48)

By inequality (3.45), we get kQ1 u(t) + Q2 u(t)k ≤ r for all t ∈ [0, T ]. Hence, Q1 u + Q2 u ∈ Bq,r . Step 2. The map Q1 is contraction on Bq,r . From the step 2 of Theorem 3.3, we have that Q1 is a contraction on Bq,r . Step 3. The map Q2 is continuous on Bq,r . Let {up }∞ p=1 be a sequence in Bq,r such that limp→∞ up = u ∈ Bq,r . For t ∈ (tm , sm ], m =

24


CUBO 17, 1 (2015)

1, · · · , δ, it is obvious since Q2 up (t) = 0. For t ∈ [0, t1 ], we get k(Q2 up )(t) − (Q2 u)(t)k ≤

ZT

(T − s)kf(s, up (s), up ([h(up (s), s)])) − f(s, u(s), u([h(u(s), s)]))kds Z sδ (sδ − s)kf(s, up (s), up ([h(up (s), s)])) − f(s, u(s), u([h(u(s), s)]))kds + 0

0

+

Zt

(t − s)kf(s, up (s), up ([h(up (s), s)])) − f(s, u(s), u([h(u(s), s)]))dskds,

0

by the continuity of f and Lebesgue dominated convergence theorem, we estimate k(Q2 up )(t) − (Q2 u)(t)k → 0, as p → ∞.

(3.49)

Similarly, t ∈ (sm , tm+1 ], m = 1, · · · , δ, k(Q2 up )(t) − (Q2 u)(t)k ≤

Z sm

(sm − s)kf(s, up (s), up ([h(up (s), s)])) − f(s, u(s), u([h(u(s), s)]))kds

+

(t − s)kf(s, up (s), up ([h(up (s), s)])) − f(s, u(s), u([h(u(s), s)]))kds

0

Zt 0

by the continuity of f and Lebesgue dominated convergence theorem, we estimate k(Q2 up )(t) − (Q2 u)(t)k → 0, ∀ t ∈∈ (sm , tm+1 ] as p → ∞.

(3.50)

Hence, Q2 is continuous map on Bq,r . Step 4. Q2 is compact. Q2 is firstly uniformly bounded on Bq,r , since kQ2 ukPC ≤ r. We now prove that Q2 maps bounded set into equicontinuous set of Bq,r . For t ∈ (tm , sm ], m = 1, · · · , δ, it is obvious. For τ2 , τ1 ∈ [0, t1 ] with τ1 < τ2 , we have kQ2 (τ2 ) − Q2 (τ1 )k ≤ H(τ2 − τ1 )2 + H

(τ2 − τ1 )2 . 2

(3.51)

For τ2 , τ1 ∈ (sm , tm+1 ], m = 1, · · · , δ with τ2 > τ1 , kQ2 (τ2 ) − Q2 (τ1 )k ≤ H(τ2 − τ1 )2 + H

(τ2 − τ1 )2 . 2

(3.52)

The right hand side of inequalities (3.51)-(3.52) tend to zero as τ2 → τ1 . Thus, Q2 (Bq,r ) is equicontinuous. By Arzela-Ascoli theorem, we conclude that Q2 is completely continuous. Therefore, from the Krasnoselskiis fixed point theorem, we deduce that Q = Q1 + Q2 has a fixed point which is just a mild solution for the problem (1.4)-(1.6).

CUBO 17, 1 (2015)

4


25

Examples

For illustrating the application of the theory, we consider the following examples. Let us consider the following impulsive nonlinear Cauchy problems with boundary conditions as 1/2

u ′ (t)(or c D0,t u(t))

=

|u( 13 u(t))| |u(t)| 1 ] [ + 3 + t3 6(1 + |u(t)|) 1 + |u( 31 u(t))| 3/2

or (I0,t

|u( 13 u(t))| 1 |u(t)| [ ]), + 3 3 + t 6(1 + |u(t)|) 1 + |u( 31 u(t))|

t ∈ (0, 1] ∪ (2, 3]

(4.1) u(t)

=

u(0)

=

Zt

|u(s)| ds, t ∈ (1, 2], (9s + 1)(1 + |u(s)|) 1 u(3),

(4.2) (4.3)

where 0 = s0 < t1 = 1 < s1 = 2 < t2 = 3 and J = [0, 3] and u ∈ C1 ([0, 3], [0, 3]). Then, u ∈ CL ([0, 3], [0, 3]). Here CL ([0, 3], [0, 3]) = {u ∈ C([0, 3], [0, 3]) : |u(t) − u(s)|L ≤ L|t − s|, ∀ t, s ∈ [0, 3]}

(4.4)

and f(t, u(t), u(h(u(t), t)))

=

G1 (t, u(t)) =

|u( 31 u(t))| 1 |u(t)| [ ], + 3 + t3 6(1 + |u(t)|) 1 + |u( 31 u(t))|

(4.5)

|u(t)| . (9t + 1)(1 + |u(t)|)

(4.6)

It is easy to show that f and g satisfy the following condition kf(t, u1 , v1 ) − f(t, u2 , v2 )k

≤

kG1 (t, u1 ) − G1 (t, u2 )k

≤

kG1 (t, u)k

≤

Lf [ku1 − u2 k + kv1 − v2 kL ], u1 , u2 ∈ [0, 3], 1 ku1 − u2 k, 10 1 1 = K1 ≤ 9t + 1 10

(4.7) (4.8) (4.9)

Thus all the assumptions of Theorem 3.1/3.3 or 3.4/3.5 are fulfilled. Hence, there exists a mild solution for the problem (4.1). Acknowledgement. The authors would like to thank the referee for valuable comments and suggestions. The work of the first author is supported by the UGC (University Grants Commission, India) under Grant No (6405 − 11 − 061) and Indian Institute of Technology, Roorkee. nt. Received:

December 2014.

Accepted:

January 2015.

26


CUBO 17, 1 (2015)

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