AN IMPULSIVE NONLINEAR SINGULAR VERSION OF THE GRONWALL-BIHARI INEQUALITY

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AN IMPULSIVE NONLINEAR SINGULAR VERSION OF
THE GRONWALL-BIHARI INEQUALITY
NASSER-EDDINE TATAR
Received 11 August 2005; Revised 18 October 2005; Accepted 20 October 2005
We find bounds for a Gronwall-Bihari type inequality for piecewise continuous functions.
Unlike works in the prior literature, here we consider inequalities involving singular kernels in addition to functions with delays.
Copyright © 2006 Nasser-Eddine Tatar. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In this paper, we are concerned with the following impulsive integral inequality:
u(t) ≤ a(t) + b(t)
t
+ c(t)
0
t
0
k1 (t,s)um (s)ds
k2 (t,s)un (s − τ)ds + d(t)
u(t) ≤ ϕ(t),
ηk u tk ,
t ≥ 0,
(1.1)
0<tk <t
t ∈ [−τ,0], τ > 0,
where a(t), b(t), c(t), and d(t) are nonnegative continuous functions, m,n > 1, ηk ≥ 0, the
points tk (called “instants of impulse effect”) are in the increasing order, and limk→∞ tk =
+∞. The kernels ki (t,s), i = 1,2, are of the form
ki (t,s) = (t − s)βi −1 sγi Fi (s),
i = 1,2,
(1.2)
where βi > 0, γi > −1, Fi (t), i = 1,2, and ϕ(t) are nonnegative continuous functions. For
this reason, we say that we are in the presence of an impulsive nonlinear singular version
of the Gronwall inequality with delay.
We would like to find bounds for solutions to this inequality in the space of piecewise
continuous functions u : X → Y (X ⊂ R,Y ⊂ RN ), with points of discontinuity of the first
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 84561, Pages 1–12
DOI 10.1155/JIA/2006/84561
2
An impulsive Gronwall-Bihari type inequality
kind at the points tk ∈ X. Our functions will also be assumed to be left continuous at the
points tk . This space will be denoted by PC(X,Y ).
Integral inequalities are an important tool to investigate some qualitative and quantitative properties of solutions to differential equations such as existence, uniqueness,
boundedness, and stability. Among these integral inequalities, we cite the famous Gronwall inequality and its different generalizations (see [3, 13]).
Impulsive integral equations, impulsive integro-differential equations, and impulsive
differential equations arise naturally in various fields such as population dynamics and
optimal control (see the monographs [2, 9, 15]). It seems that the first treatment of impulsive systems goes back to the monograph by Krylov and Bogolyubov [8].
The following impulsive integral inequality:
u(t) ≤ a +
t
c
b(s)u(s)ds +
ηk u tk ,
t ≥ 0,
(1.3)
c<tk <t
has been first used by Samoilenko and Perestyuk [14] to investigate problems of the form
x = f (t,x),
t = tk ,
Δx = Ik (x),
t = tk .
(1.4)
Then, a similar inequality with constant delay was considered by Bainov and Hristova in
[1]. Recently, Hristova in [5] treated a more general inequality with nonlinear functions
in u. However, in all previous works, the functions (kernels) involved in the integrals are
regular, even in the case of integrals of convolution or nonconvolution types (see [3, 13]).
In this work, we consider the case of singular kernels of the form (1.2). The type of
inequalities we are going to discuss arise for instance when we study impulsive evolution
problems of the form
du
+ Au = f t,u,ut , t > 0, t = tk ,
dt
u(0) = u0 ∈ X,
+
−
Δu tk = u tk − u tk ,
(1.5)
k = 1,2,...,
where A is a sectorial operator (see, for instance, [17] where the case without delay and
with globally Lipschitzian right-hand side is treated).
We point out here that nonlinear singular versions of the Gronwall-Bihari inequality
have been already considered by the present author in [6, 7, 10, 16] and Medved in [11,
12] to investigate problems of the form (1.5) and perturbed problems of (1.5) but without
impulse effects.
The plan of the paper is as follows. In the next section we present some lemmas and
notation which will be needed in the proof of our result. Section 3 contains the statement
and proof of our theorem. It is ended with some important remarks.
2. Preliminaries
In this section, we prepare some lemmas and notation which we will use in the next
section.
Nasser-Eddine Tatar 3
Lemma 2.1. For all β > 0 and γ > −1,
t
0
(t − s)β−1 sγ ds = Ct β+γ ,
t ≥ 0,
(2.1)
where C = C(β,γ) = Γ(β)Γ(γ + 1)/Γ(β + γ + 1).
Lemma 2.2. If β,γ,δ > 0, then for any t > 0,
t 1−β
t
0
(t − s)β−1 sγ−1 e−δs ds ≤ C,
(2.2)
where C = C(β,γ,δ) is a positive constant independent of t. In fact,
C = max 1,21−β Γ(γ) 1 +
γ −γ
δ .
β
(2.3)
See [6] for the proof.
Lemma 2.3. Let a, b, K, ψ be nonnegative continuous functions on the interval I = (0,T)
(0 < T ≤ ∞), let ω : (0, ∞) → R be a continuous, nonnegative, and nondecreasing function
with ω(0) = 0 and ω(u)>0 for u>0, and let A(t):= max0≤s≤t a(s) and B(t):= max0≤s≤t b(s).
Assume that
ψ(t) ≤ a(t) + b(t)
t
0
K(s)ω ψ(s) ds,
t ∈ I.
(2.4)
Then
ψ(t) ≤ H −1 H A(t) + B(t)
t
0
K(s)ds ,
t ∈ 0,T1 ,
(2.5)
v
where H(v) := v0 dτ/ω(τ) (v ≥ v0 > 0), H −1 is the inverse of H, and T1 > 0 is such that
t
H(A(t)) + B(t) 0 K(s)ds ∈ D(H −1 ) for all t ∈ (0,T1 ).
See [4] or [3, 13].
In order to lighten the statement of our result, we adopt the following notation. Let
τ
V (τ) := 1 + 0 F22 (s)ϕ2n (s − τ)ds, r := max{m,n} > 1, t0 := 0.
For p and q such that 1/ p + 1/q = 1, we define
f p (t) := sup aq (t),C q/ p pβ1 − p + 1, pγ1 bq (t)t q(β1 +γ1 )−1 ,
C q/ p pβ2 − p + 1, pγ2 cq (t)t q(β2 +γ2 )−1 ,d q (t) ,
(2.6)
with C(pβ1 − p + 1, pγ1 ) and C(pβ2 − p + 1, pγ2 ) the constants from Lemma 2.1, and T p
4
An impulsive Gronwall-Bihari type inequality
to be the sup of all values of t for which
k ti
i=1 ti−1
(i + 2)(q−1)r
×
×
i−1
r
q
1 + ( j + 2)q−1 η j f t j
j =1
q
q
m
F1 (s) f (s) + F2 (s) f n (s − τ) ds + (k + 3)(q−1)r
k
q
r
1 + ( j + 2)q−1 η j f t j
j =1
t
tk
q
q
F1 (s) f m (s) + F2 (s) f n (s − τ) ds <
V (τ)1−r
.
(r − 1)
(2.7)
If p = q = 2, put f (t) := f2 (t) and T := T2 .
3. The bounds
Without loss of generality, we will suppose that the tk are such that τ < tk+1 − tk ≤ 2τ,
k = 0,1,2,.... For the general case, see Remark 3.2 below.
Theorem 3.1. Let the above assumptions on the different parameters and functions hold.
Suppose that u is in PC([−τ,+∞],[0,+∞]) and satisfies (1.1), then
(a) if βi > 1/2 and γi > −1/2, i = 1,2, it holds that for t ∈ (tk ,t+1 ],
u(t) ≤ (k + 3) f (t)
l=1
1/q
k
1 + (k + 2)ηl2 f tl
× V (τ)1−r − (r − 1)
×
×
k ti
i=1 ti−1
i−1
(i + 2)r
j =1
F12 (s) f m (s) + F22 (s) f n (s − τ)
k
j =1
r
1 + ( j + 2)η2j f t j
t
tk
r
1 + ( j + 2)η2j f t j
ds − (r − 1)(k + 3)r
1/2(1−r)
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
(3.1)
as long as the expression between the second brackets is positive, that is, on (0,T);
(b) if 0 < βi ≤ 1/2 and −1 < γi ≤ −1/2, then it holds that for t ∈ (tk ,t+1 ],
u(t) ≤ (k + 3)q−1 f p (t)
k
l=1
× V (τ)
×
×
1−r
1/q
q
1 + (k + 2)q−1 ηl f tl
− (r − 1)
k ti
(i + 2)(q−1)r
i−1
i=1 ti−1
j =1
q
q
m
n
F1 (s) f p (s) + F2 (s) f p (s − τ) ds − (r
k
j =1
q
r
1 + ( j + 2)q−1 η j f t j
t
tk
q
r
1 + ( j + 2)q−1 η j f t j
q
− 1)(k + 3)(q−1)r
1/q(1−r)
F1 (s) f pm (s) + F22 (s) f pn (s − τ) ds
(3.2)
as long as the expression between the second brackets is positive, that is, on (0,T p ).
Nasser-Eddine Tatar 5
Proof. We will use a mathematical induction.
(a) Step 1. We start by proving the validity of (3.1) in the interval [0,t1 ] (in fact, the
argument we present is valid within the interval (0, T), this fact will be mentioned in
every occasion by indicating the right interval over which the estimate is valid). For t ∈
[0,τ] ⊂ [0,t1 ] (see assumptions on tk ), we have
u(t) ≤ a(t) + b(t)
t
+ c(t)
0
t
0
(t − s)β1 −1 sγ1 F1 (s)um (s)ds
(3.3)
(t − s)β2 −1 sγ2 F2 (s)un (s − τ)ds.
If βi > 1/2 and γi > −1/2, i = 1,2, then by the Cauchy-Schwarz inequality and Lemma 2.1,
we obtain
t
u(t) ≤ a(t) + C 1/2 2β1 − 1,2γ1 b(t)t β1 +γ1 −1/2
+C
1/2
2β2 − 1,2γ2 c(t)t
β2 +γ2 −1/2
t
0
0
1/2
F12 (s)u2m (s)ds
(3.4)
1/2
F22 (s)u2n (s − τ)ds
,
where C 2β1 − 1,2γ1 and C(2β2 − 1,2γ2 ) are the constants from Lemma 2.1. Squaring
both sides of (3.4), we find
u2 (t) ≤ 3a2 (t) + 3C 2β1 − 1,2γ1 b2 (t)t 2(β1 +γ1 )−1
2
+ 3C 2β2 − 1,2γ2 c (t)t
2(β2 +γ2 )−1
t
0
t
0
F12 (s)u2m (s)ds
(3.5)
F22 (s)u2n (s − τ)ds.
Therefore
u2 (t) ≤ 3 f (t) 1 +
≤ 3 f (t) 1 +
t
0
t
0
F12 (s)u2m (s)ds +
F12 (s)u2m (s)ds +
t
0
F22 (s)u2n (s − τ)ds
τ
0
(3.6)
F22 (s)ϕ2n (s − τ)ds .
Putting
v1 (t) := 1 +
τ
0
F22 (s)ϕ2n (s − τ)ds +
t
0
F12 (s)u2m (s)ds,
(3.7)
we see that v1 (t) is a nondecreasing positive differentiable function on [0, τ],v1 (0) =
τ
1 + 0 F22 (s)ϕ2n (s − τ)ds =: V (τ),
u2 (t) ≤ 3 f (t)v1 (t),
(3.8)
v1 (t) = F12 (t)u2m (t) ≤ 3m F12 (t) f m (t)v1m (t) ≤ 3r F12 (t) f m (t)v1r (t).
(3.9)
An integration of (3.9) (or using Lemma 2.3 directly) leads to
v1 (t) ≤ V (τ)1−r − 3r (r − 1)
t
0
1/(1−r)
F12 (s) f m (s)ds
(3.10)
6
An impulsive Gronwall-Bihari type inequality
as long as
t
2
m
1−r /3r (r
0 F1 (s) f (s)ds < V (τ)
− 1). Therefore, for t ∈ [0,τ],
u(t) ≤ 3 f (t) V (τ)1−r − 3r (r − 1)
t
0
1/2(1−r)
F12 (s) f m (s)ds
(3.11)
t
as long as 0 F12 (s) f m (s)ds < V (τ)1−r /3r (r − 1).
Let t ∈ (τ,t1 ]. Then, from (3.6) and (3.7), we have
u2 (t) ≤ 3 f (t) v1 (τ) +
t
τ
F12 (s)u2m (s)ds +
t
τ
F22 (s)u2n (s − τ)ds .
(3.12)
Let us designate
w1 (t) := v1 (τ) +
t
τ
F12 (s)u2m (s)ds +
t
τ
F22 (s)u2n (s − τ)ds.
(3.13)
Then w1 (t) is a nondecreasing positive differentiable function on (τ,t1 ],
w1 (τ) = v1 (τ) ≤ w1 (t),
u2 (t) ≤ 3 f (t)w1 (t),
(3.14)
w1 (t) = F12 (t)u2m (t) + F22 (t)u2n (t − τ).
(3.15)
Since 0 < t − τ ≤ τ (see Remark 3.2) and from (3.7), (3.8), (3.14), and (3.15),
u2 (t − τ) ≤ 3 f (t − τ)v1 (t − τ) ≤ 3 f (t − τ)v1 (τ) ≤ 3 f (t − τ)w1 (t),
(3.16)
and we can write that
w1 (t) ≤ F12 (t) 3 f (t)w1 (t)
m
+ F22 (t) 3 f (t − τ)w1 (t)
n
≤ 3r F12 (t) f m (t) + F22 (t) f n (t − τ) w1r (t).
(3.17)
Integrating (3.17) from τ to t and using (3.10), we obtain
w1 (t) ≤ w1 (τ)1−r − 3r (r − 1)
≤ V (τ)1−r − 3r (r − 1)
− 3r (r − 1)
t
τ
t
τ
τ
0
1/(1−r)
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
F12 (s) f m (s)ds
(3.18)
1/(1−r)
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
≤ V (τ)1−r − 3r (r − 1)
t
1/(1−r)
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
0
and hence, for t ∈ (τ,t1 ],
u(t) ≤ 3 f (t) V 1−r − 3r (r − 1)
t
0
1/2(1−r)
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
(3.19)
Nasser-Eddine Tatar 7
as long as
t
0
F12 (s) f m (s) + F22 (s) f n (s − τ) ds <
V 1−r
3r (r
− 1)
.
(3.20)
We define the function ψ1 : [0,t1 ] → R by
⎧
⎨v1 (t),
ψ1 (t) := ⎩
t ∈ [0,τ],
t ∈ τ,t1 .
w1 (t),
(3.21)
It can be easily seen that (3.1) in the statement of the theorem is satisfied over [0,t1 ]
(recall that t0 := 0).
Step 2. Let t ∈ (t1 ,t2 ]. If t ∈ (t1 ,t1 + τ], then
u(t) ≤ a(t) + b(t)
t
+ c(t)
0
t
0
(t − s)β1 −1 sγ1 F1 (s)um (s)ds
(3.22)
(t − s)β2 −1 sγ2 F2 (s)un (s − τ)ds + η1 d(t)u t1 .
Squaring both sides of (3.22) after applying the Cauchy-Schwarz inequality and
Lemma 2.1, as in the previous steps from (3.4) to (3.6), we find
u2 (t) ≤ 4 f (t) 1 +
t
0
F12 (s)u2m (s)ds +
≤ 4 f (t) v1 (τ) +
t
+
t1
t1
τ
t
0
F22 (s)u2n (s − τ)ds + η12 u2 t1
F12 (s)u2m (s)ds +
F12 (s)u2m (s)ds +
t
t1
t1
τ
F22 (s)u2n (s − τ)ds
(3.23)
F22 (s)u2n (s − τ)ds + η12 u2 t1
.
Note here that we have used definition (3.7) of v1 (t). Thanks to (3.13) and (3.14), we
entail that
u2 (t) ≤ 4 f (t) w1 t1 +
t
t1
F12 (s)u2m (s)ds +
w1 (t1 ) +
≤ 4 f (t) 1 + 3η12 f t1
t
t1
t
t1
F22 (s)u2n (s − τ)ds + 3η12 f t1 w1 t1
F12 (s)u2m (s)ds +
t
t1
F22 (s)u2n (s − τ)ds .
(3.24)
We define
v2 (t) := w1 t1 +
t
t1
F12 (s)u2m (s)ds +
t
t1
F22 (s)u2n (s − τ)ds.
(3.25)
8
An impulsive Gronwall-Bihari type inequality
It is clear that v2 (t) is a nondecreasing positive differentiable function on (t1 ,t1 + τ],
u2 (t) ≤ 4 f (t) 1 + 3η12 f t1 v2 (t).
v2 t1 = w1 t1 ≤ v2 (t),
(3.26)
Since t − τ ≤ t1 , by (3.6), (3.12), (3.13), and (3.25), we see that
u2 (t − τ) ≤ 3 f (t − τ)ψ1 (t − τ) ≤ 3 f (t − τ)w1 t1 ≤ 3 f (t − τ)v2 (t),
(3.27)
and thus from this estimation, (3.25) and (3.26), we get
v2 (t) = F12 (t)u2m (t) + F22 (t)u2n (t − τ)
m m
≤ 4m 1 + 3η12 f t1
f (t)F12 (t) + 3n F22 (t) f n (t − τ) v2r (t).
(3.28)
An integration of (3.28) from t1 to t together with (3.18) leads to
1−r
v2 (t) ≤ v2 t1
×
t
t1
− (r − 1)
m
4m 1 + 3η12 f t1
≤ V (τ)1−r − 3r (r − 1)
×
t
t1
4
m
t1
1/(1−r)
f m (s)F12 (s) + 3n F22 (s) f n (s − τ) ds
F12 (s) f m (s) + F22 (s) f n (s − τ) ds − (r − 1)
0
m
1 + 3η12 f t1
1/(1−r)
f m (s)F12 (s) + 3n F22 (s) f n (s − τ) ds
(3.29)
and hence, for t ∈ (t1 ,t1 + τ], we have
u(t) ≤ 2
1 + 3η12 f t1
f (t)
× V (τ)1−r − 3r (r − 1)
− (r − 1)
t
4
t1
m
t1
0
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
m
1 + 3η12 f t1
1/2(1−r)
f m (s)F12 (s) + 3n F22 (s) f n (s − τ) ds
(3.30)
as long as
3
r
t1
0
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
t
+
t1
m
4m 1 + 3η12 f t1
f m (s)F12 (s) + 3n F22 (s) f n (s − τ) ds ≤
V 1−r
.
r −1
(3.31)
Now let t ∈ (t1 + τ,t2 ], then from (3.7), (3.13), (3.14), (3.25), and
u2 (t) ≤ 4 f (t) 1 +
t
0
F12 (s)u2m (s)ds +
t
0
F22 (s)u2n (s − τ)ds + η12 u2 t1
,
(3.32)
Nasser-Eddine Tatar 9
we deduce that
u2 (t) ≤ 4 f (t) v2 t1 +τ +
t
t1+τ
F12 (s)u2m (s)ds+
≤ 4 f (t) 1 + 3η12 f t1
t
v2 t1 + τ +
t1 +τ
t
t1+τ
F22 (s)u2n (s − τ)ds + 3η12 f t1 v2 t1 +τ
F12 (s)u2m (s)ds+
t
t1 +τ
F22 (s)u2n (s − τ)ds
(3.33)
because w1 (t1 ) ≤ v2 (t1 ) ≤ v2 (t1 + τ). At this stage, we denote
w2 (t) := v2 t1 + τ +
t
t1 +τ
F12 (s)u2m (s)ds +
t
t1 +τ
F22 (s)u2n (s − τ)ds.
(3.34)
Then, clearly w2 (t) is a nondecreasing positive differentiable function on (t1 + τ,t2 ],
w2 (t1 + τ) = v2 (t1 + τ) ≤ w2 (t), and
w2 (t) = F12 (t)u2m (t) + F22 (t)u2n (t − τ).
(3.35)
Observe that by (3.33) and (3.34), we have the estimates
u2 (t) ≤ 4 f (t) 1 + 3η12 f t1 w2 (t),
(3.36)
and since t1 < t − τ < t1 + τ, it follows from (3.24) that
u2 (t − τ) ≤ 4 f (t − τ) 1 + 3η12 f t1 v2 (t − τ)
≤ 4 f (t − τ) 1 + 3η12 f t1 v2 t1 + τ
≤ 4 f (t − τ) 1 + 3η12 f t1 w2 (t).
(3.37)
Consequently,
r w2 (t) ≤ 4r 1 + 3η12 f t1
f m (t)F12 (t) + f n (t − τ)F22 (t) w2r (t).
(3.38)
Again by an integration of (3.38), we end up with
r
w2 (t) ≤ w21−r t1 + τ − 4r (r − 1) 1 + 3η12 f t1
t1
t
t1+τ
1/(1−r)
f m (s)F12 (s)+ f n (s − τ)F22 (s) ds
≤ V (τ)1−r − 3r (r − 1)
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
0
t1 +τ m m
− (r − 1)
4m 1 + 3η12 f t1
f (s)F12 (s) + 3n F22 (s) f n (s − τ) ds
t1
r
− 4r (r − 1) 1 + 3η12 f t1
t
t1 +τ
1/(1−r)
f m (s)F12 (s) + f n (s − τ)F22 (s) ds
(3.39)
10
An impulsive Gronwall-Bihari type inequality
or simply
w2 (t) ≤ V (τ)1−r − 3r (r − 1)
t1
0
F12 (s) f m (s) + F22 (s) f n (s − τ) ds
r
− 4r (r − 1) 1 + 3η12 f t1
t
t1
1/(1−r)
f m (s)F12 (s) + f n (s − τ)F22 (s) ds
.
(3.40)
Hence,
u(t) ≤ 2 f (t) 1 + 3η12 f t1
t1
2
F1 (s) f m (s) + F22 (s) f n (s − τ) ds
× V (τ)1−r − 3r (r − 1)
0
r
− 4r (r − 1) 1 + 3η12 f t1
t
t1
1/2(1−r)
f m (s)F12 (s) + f n (s − τ)F22 (s) ds
(3.41)
provided that the expression between brackets is positive. We define ψ2 : (t1 ,t2 ] → R by
⎧
⎨v2 (t),
ψ2 (t) := ⎩
w2 (t),
t ∈ t1 ,t1 + τ ,
(3.42)
t ∈ t1 + τ,t2 .
It is clear that (3.1) holds on (t1 ,t2 ].
Step 3. Finally, suppose that (3.1) is valid over (tk ,tk+1 ], then if t ∈ (tk+1 ,tk+2 ], we define
⎧
⎨vk+2 (t),
ψk+2 (t) := ⎩
wk+2 (t),
t ∈ tk+1 ,tk+1 + τ ,
t ∈ tk+1 + τ,tk+2 ,
(3.43)
with
vk+2 (t) := wk+1 (tk+1 ) +
wk+2 (t) := vk+2 tk+1 + τ +
t
tk+1
F12 (s)u2m (s)ds +
t
tk+1 +τ
t
tk+1
F12 (s)u2m (s)ds +
F22 (s)u2n (s − τ)ds,
t
tk+1 +τ
(3.44)
F22 (s)u2n (s − τ)ds.
In a similar manner as in Steps 1 and 2, we can see that (3.1) is valid over (tk+1 ,tk+2 ].
(b) If 0 < βi ≤ 1/2 and −1 < γi ≤ −1/2, then instead of the Cauchy-Schwarz inequality
we use the Hölder inequality with
1
1
, − , i = 1,2 ,
1 < p < min
1 − βi γi
(3.45)
Nasser-Eddine Tatar 11
and q > 1 such that 1/ p + 1/q = 1. We find
u(t) ≤ a(t) + C 1/ p pβ1 − p + 1, pγ1 b(t)t β1 +γ1 −1/q
+C
1/ p
pβ2 − p + 1, pγ2 c(t)t
β2 +γ2 −1/q
t
0
t
0
1/q
q
F1 (s)uqm (s)ds
(3.46)
1/q
q
F2 (s)uqn (s − τ)ds
.
Then, we raise both sides to the power q and we use the inequality
n
i=1
r
ai
≤ n r −1
n
i =1
ari ,
n ∈ N∗ , r,ai ∈ R+ , i = 1,...,n.
The rest of the proof remains the same. The proof is now complete.
(3.47)
Remark 3.2. Apart from the case treated in the proof, that is, when τ < tk+1 − tk ≤ 2τ,
k = 0,1,2,..., there are several other cases, but each and every one of them can fit in the
one considered above or one of the following cases.
Case 1. There exists an nk > 1 such that tk+1 − tk ≥ nk τ, that is, tk < tk + nk τ ≤ tk+1 . In this
case, we argue in a similar fashion over (tk ,tk + τ], (tk + τ,tk + 2τ],...,(tk + (nk − 1)τ,tk +
nk τ] and then over (tk + nk τ,t] with t > tk + nk τ. Therefore the function ψk+1 (t) will have
nk + 1 components.
Case 2. There exists k0 ≥ 1 such that (tk0 ,tk0 +1 ] does not contain any tk0 + nτ, n = 1,2,...,
that is, tk0 +1 − tk0 < τ. Here we deal with this interval in a single step using only a function
of the form vk0 +1 (t), that is, ψk0 +1 (t) := vk0 +1 (t).
/ (0,t1 ] as in Case 1. Again, in this situation,
Case 3. τ ∈ (tk1 ,tk1 +1 ] with k1 > 0, that is, τ ∈
we consider only functions of the form vk (t) until we reach the interval (tk1 ,tk1 +1 ] where
we consider both vk1 +1 (t) and wk1 +1 (t).
Remark 3.3. Obviously, if ki (t,s) = (t − s)βi −1 sγi e−δi s Fi (s), δi > 0, i = 1,2..., the proof still
works. However, using Lemma 2.2 instead of Lemma 2.1 throughout the proof, we can
have much larger intervals over which the estimations are valid.
Remark 3.4. It is clear that our result can be easily extended to other nonlinearities than
the polynomial ones, iterated integrals and the case of several variables. One may use the
Gronwall-Bihari lemma (Lemma 2.3) in case of a nondecreasing nonlinearity. See also
[3, 13] for other classes of nonlinearities.
Acknowledgments
The author is very grateful for the financial support and the facilities provided by King
Fahd University of Petroleum and Minerals. Thanks are due also to the anonymous referees.
12
An impulsive Gronwall-Bihari type inequality
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Nasser-Eddine Tatar: Department of Mathematical Sciences, King Fahd University of
Petroleum and Minerals, Dhahran 31261, Saudi Arabia
E-mail address: tatarn@kfupm.edu.sa
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