Volume 9 (2008), Issue 2, Article 34, 11 pp.
GENERALIZATION OF AN IMPULSIVE NONLINEAR SINGULAR
GRONWALL-BIHARI INEQUALITY WITH DELAY
SHENGFU DENG AND CARL PRATHER
D EPARTMENT OF M ATHEMATICS
V IRGINIA P OLYTECHNIC I NSTITUTE AND S TATE U NIVERSITY
B LACKSBURG , VA 24061, USA
sfdeng@vt.edu
prather@math.vt.edu
Received 25 August, 2007; accepted 23 May, 2008
Communicated by S.S. Dragomir
A BSTRACT. This paper generalizes a Tatar’s result of an impulsive nonlinear singular GronwallBihari inequality with delay [J. Inequal. Appl., 2006(2006), 1-12] to a new type of inequalities
which includes n distinct nonlinear terms.
Key words and phrases: Gronwall-Bihari inequality, Nonlinear, Impulsive.
2000 Mathematics Subject Classification. 26D15, 26D20.
1. I NTRODUCTION
In order to investigate problems of the form
x0 = f (t, x),
∆x = Ik (x),
t 6= tk ,
t = tk ,
Samoilenko and Perestyuk [6] first used the following impulsive integral inequality
Z t
X
u(t) ≤ a +
b(s)u(s)ds +
ηk u(tk ), t ≥ 0.
c
0<tk <t
Then Bainov and Hristova [2] studied a similar inequality with constant delay. In 2004, Hristova
[3] considered a more general inequality with nonlinear functions in u. All of these papers
treated the functions (kernels) involved in the integrals which are regular. Recently, Tatar [7]
investigated the following singular inequality
Z t
Z t
m
u(t) ≤ a(t) + b(t)
k1 (t, s)u (s)ds + c(t)
k2 (t, s)un (s − τ )ds
0
0
X
+d(t)
ηk u(tk ), t ≥ 0,
0<tk <t
(1.1)
279-07
u(t) ≤ ϕ(t), t ∈ [−τ, 0], τ > 0
2
S HENGFU D ENG AND C ARL P RATHER
where the kernels ki (t, s) are defined by ki (t, s) = (t − s)βi −1 sγi Fi (s) for βi > 0 and γi >
−1, i = 1, 2, the points tk (called "instants of impulse effect") are in increasing order and
limk→∞ tk = +∞. This inequality was called the impulsive nonlinear singular version of the
Gronwall inequality with delay by Tatar [7]. In this paper, we will consider an inequality
n Z bi (t)
X
u(t) ≤ a(t) +
(t − s)βi −1 sri fi (t, s)wi (u(s))ds
0
i=1
+
m+n
X
j=n+1
(1.2)
+ d(t)
Z
bj (t)
(t − s)βj −1 srj fj (t, s)wj (u(s − τ ))ds
0
X
ηL u(tL ),
t ≥ 0,
0<tL <t
u(t) ≤ ϕ(t),
t ∈ [−τ, 0], τ > 0,
where n, m are positive integers, βl > 0, rl > −1 for l = 1, . . . , n + m and ηL ≥ 0 and other
assumptions are given in Section 2. This inequality is more general than (1.1) since (1.2) has n
nonlinear terms.
2. M AIN R ESULTS
w2
Notation: Following [1] and [5], we say w1 ∝ w2 for w1 , w2 : A ⊂ R → R\{0} if w
is
1
nondecreasing on A. This concept helps us to compare the monotonicity of different functions.
Now we make the following assumptions:
(H1) all wi (i = 1, . . . , n + m) are continuous and nondecreasing on [0, ∞) and positive on
(0, ∞), and w1 ∝ w2 ∝ · · · ∝ wn
(H2) a(t) and d(t) are continuous and nonnegative on [0, ∞);
(H3) all bl : [0, ∞) → [0, ∞) are continuously differentiable and nondecreasing such that
0 ≤ bl (t) ≤ t on [0, ∞), tL ≤ bl (t) ≤ tL +τ for t ∈ [tL , tL +τ ] and tL +τ ≤ bl (t) ≤ tL+1
for t ∈ [tL + τ, tL+1 ], l = 1, . . . , n + m and L = 0, 1, 2, . . . where t0 = 0. The points tL
are called instants of impulse effect which are in increasing order, and limL→∞ tL = ∞;
(H4) all fl (t, s) (l = 1, . . . , n + m) are continuous and nonnegative functions on [0, ∞) ×
[0, ∞);
(H5) ϕ(t) is nonnegative and continuous;
(H6) u(t) is a piecewise continuous function from R → R+ = [0, ∞) with points of discontinuity of the first kind at the points tL ∈ R. It is also left continuous at the points tL .
This space is denoted by P C(R, R+ ).
Without loss of generality, we will suppose that the tL satisfy τ < tL+1 − tL ≤ 2τ , L =
0, 1, 2, . . . . As in Remark 3.2 of [7], other cases can be reduced to this one.
Theorem 2.1. Let the above assumptions hold. Suppose that u satisfies (1.2) and is in
P C([−τ, ∞), [0, ∞)). Then if βα > − p1 + 1 and rα > − p1 , it holds that

uL,0 (t), t ∈ (tL , tL + τ ],






uL,1 (t), t ∈ (tL + τ, tL+1 ],



uk,0 (t), t ∈ (tk , tk + τ ] if tk + τ ≤ T,
u(t) ≤




uk,1 (t), t ∈ (tk + τ, T ] if tk + τ < T,





uk,0 (t), t ∈ (tk , T ] if tk + τ > T,
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
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G RONWALL -B IHARI I NEQUALITY
where tk ≤ T < tk+1 and
"
uL,l (t) = Wn−1
Z
bn (t)
Wn (γL,l,n (t)) +
3
!# 1q
(n + m + L + 1)q−1 cqn (t)f˜nq (t, s)ds
,
tL +lτ
γL,l,j (t) =
−1
Wj−1
Wj−1 (γL,l,j−1 (t))
Z
bj−1 (t)
+
(n + m + L +
q
1)q−1 cqj−1 (t)f˜j−1
(t, s)ds
,
j 6= 1,
tL +lτ
"
γL,l,1 (t) = (n + m + L + 1)
q−1
q
ã (t) +
n Z
X
i=1
+
n+m
X
j=n+1
Z
bj (t)
tL +lτ
cqi (t)f˜iq (t, s)wiq (φ(s))ds
0
cqj (t)f˜jq (t, s)wjq (ψ(s − τ ))ds +
0
L
X
#
d˜q (t)ηeq uqe−1,1 (te ) ,
e=1

u (t), t ∈ (tL , tL + τ ], t ∈ (tk , tk + τ ] if tk + τ ≤ T,

 L,0
and t ∈ (tk , T ] if tk + τ > T,
φ(t) =


uL,1 (t), t ∈ (tL + τ, tL+1 ] and t ∈ (tk + τ, T ] if tk + τ < T,

ϕ(t),
t ∈ [−τ, 0],




 uL,0 (t), t ∈ (tL , tL + τ ], t ∈ (tk , tk + τ ] if tk + τ ≤ T,
ψ(t) =

and t ∈ (tk , T ] if tk + τ > T,




uL,1 (t), t ∈ (tL + τ, tL+1 ] and t ∈ (tk + τ, T ] if tk + τ < T,
˜ = max d(x),
ã(t) = max a(x), f˜α (t, s) = max fα (x, s), d(t)
0≤x≤t
0≤x≤t
Z
u
Wi (u) =
0≤x≤t
dv
u > 0, ui > 0,
1 ,
wiq (v q )
p1
1
Γ(1
+
p(β
−
1))Γ(1
+
pr
)
α
α
+β
+r
−1
α
α
,
cα (t) = t p
Γ(2 + p(βα + rα − 1))
for L = 0, 1, . . . , k − 1, α = 1, 2, . . . , n + m, l = 0, 1, and i, j = 1, . . . , n where p1 + 1q = 1 for
p > 0 and q > 1, and T is the largest number such that
Z bj (t)
Z ∞
dz
q
q−1
˜
(2.1)
Wj (γL,l,j (t)) +
(n + m + L + 1) cj (t)fj (t, s)ds ≤
q 1/q ,
)
tL +lτ
uj wj (z
ui
for all t ∈ (tL , tL + τ ], all t ∈ (tk , tk + τ ] if tk + τ ≤ T and all t ∈ (tL , T ] if tk + τ > T as
l = 0, or all t ∈ [tL + τ, tL+1 ] and all t ∈ [tk + τ, T ) if tk + τ < T as l = 1 where j = 1, . . . , n,
l = 0, 1 and L = 0, 1 . . . , k − 1.
Before the proof, we introduce a lemma which will play a very important role.
Lemma 2.2 ([1]). Suppose that
(1) all wi (i = 1, . . . , n) are continuous and nondecreasing on [0, ∞) and positive on
(0, ∞), and w1 ∝ w2 ∝ · · · ∝ wn .
(2) a(t) is continuously differentiable in t and nonnegative on [t0 , t1 ),
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
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4
S HENGFU D ENG AND C ARL P RATHER
(3) all bl are continuously differentiable and nondecreasing such that bl (t) ≤ t for t ∈
[t0 , t1 )
where t0 , t1 are constants and t0 < t1 . If u(t) is a continuous and nonnegative function on
[t0 , t1 ) satisfying
u(t) ≤ a(t) +
n Z
X
i=1
bi (t)
t0 ≤ t < t1 ,
fi (t, s)wi (u(s))ds,
bi (t0 )
then
"
u(t) ≤ W̃n−1 W̃n (γn (t)) +
bn (t)
Z
#
f˜n (t, s)ds ,
t0 ≤ t ≤ T1 ,
bn (t0 )
where
"
−1
γi (t) = W̃i−1
W̃i−1 (γi−1 (t)) +
Z
bi−1 (t)
#
f˜i−1 (t, s)ds ,
i = 2, 3, . . . , n,
bi−1 (t0 )
t
Z
|a (s)|ds,
γ1 (t) = a(t0 ) +
u
Z
0
W̃i (u) =
t0
ui
dz
,
wi (z)
ui > 0,
dz
,
wi (z)
i = 1, . . . , n.
T1 < t1 and T1 is the largest number such that
bi (T1 )
Z
W̃i (γi (T1 )) +
f˜i (T1 , s)ds ≤
bi (t0 )
Z
∞
ui
Proof of Theorem 2.1. Since βα > − p1 + 1 and rα > − p1 for α = 1, . . . , n + m, by Hölder’s
inequality we obtain
u(t) ≤ a(t) +
n Z
X
+
Z
X
Z
bi (t)
! 1q
fiq (t, s)wiq (u(s))ds
0
t
(t − s)p(βj −1) sprj ds
0
j=n+1
+
(t − s)p(βi −1) spri ds
p1
0
i=1
m+n
X
t
p1
Z
bj (t)
! 1q
fjq (t, s)wjq (u(s − τ ))ds
0
d(t)ηL u(tL )
0<tL <t
≤ a(t) +
n
X
Z
ci (t)
+
j=n+1
! 1q
fiq (t, s)wiq (u(s))ds
0
i=1
m+n
X
bi (t)
Z
cj (t)
bj (t)
! 1q
fjq (t, s)wjq (u(s − τ ))ds
0
+
X
d(t)ηL u(tL )
0<tL <t
where we use bα (t) ≤ t and the definition of cα (t). Now we use the following result [4]:
If A1 , . . . , An are nonnegative for n ∈ N, then for q > 1,
(A1 + · · · + An )q ≤ nq−1 (Aq1 + · · · + Aqn ).
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
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G RONWALL -B IHARI I NEQUALITY
Since tk ≤ t ≤ T < tk+1 , we have
"
q
q−1
u (t) ≤ (1 + n + m + k)
n
X
q
a (t) +
cqi (t)
+
cqj (t)
Z
bj (t)
fiq (t, s)wiq (u(s))ds
0
i=1
m+n
X
bi (t)
Z
5
fjq (t, s)wjq (u(s
− τ ))ds +
0
j=n+1
k
X
#
q
d
(t)ηLq uq (tL )
.
L=1
˜ ≥ d(t) and f˜α (t, s) ≥ fα (t, s) and they are continuous and
We note that ã(t) ≥ a(t), d(t)
nondecreasing in t. The above inequality becomes
"
Z tL+1
k−1
n
X
X
q
q
q−1
q
ci (t)
f˜iq (t, s)wiq (u(s))ds
u (t) ≤ (1 + n + m + k)
ã (t) +
i=1
+ cqi (t)
Z
bi (t)
tL
L=0
!
f˜iq (t, s)wiq (u(s))ds
tk
+
(2.2)
m+n
X
k−1
X
j=n+1
L=0
+ cqj (t)
Z
bj (t)
cqj (t)
Z
tL+1
f˜jq (t, s)wjq (u(s − τ ))ds
tL
!
f˜jq (t, s)wjq (u(s − τ ))ds
+
#
k
X
tk
d˜q (t)ηLq uq (tL ) .
L=1
In the following, we apply mathematical induction with respect to k.
(1) k = 0. We note that t0 = 0 and we have for any fixed t̃ ∈ [0, t1 ]
"
Z bi (t)
n
X
q
q
q−1
q
(2.3) u (t) ≤ (n + m + 1)
ã (t̃) +
c (t̃)
f˜q (t̃, s)wq (u(s))ds
i
i
i
0
i=1
m+n
X
+
cqj (t̃)
Z
bj (t)
#
f˜jq (t̃, s)wjq (u(s − τ ))ds
0
j=n+1
for t ∈ [0, t̃] since cα (t) are nondecreasing.
Now we consider t̃ ∈ [0, τ ] ⊂ [0, t1 ] and t ∈ [0, t̃]. Note that 0 ≤ bj (t) ≤ t so −τ ≤
bj (t) − τ ≤ 0 for t ∈ [0, t̃]. Since u(t) ≤ ϕ(t) for t ∈ [−τ, 0], we have
uq (t) ≤ z0,0 (t),
t ∈ [0, t̃],
where
"
(2.4) z0,0 (t) = (n + m + 1)
q−1
q
ã (t̃) +
n
X
cqi (t̃)
Z
bi (t)
f˜iq (t̃, s)wiq (u(s))ds
0
i=1
+
m+n
X
j=n+1
cqj (t̃)
Z
bj (t̃)
#
f˜jq (t̃, s)wjq (ϕ(s − τ ))ds .
0
It implies that
(2.5)
u(t) ≤ z0,0 (t)1/q ,
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
t ∈ [0, t̃].
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6
S HENGFU D ENG AND C ARL P RATHER
Thus, (2.4) becomes
"
(2.6) z0,0 (t) ≤ (n + m + 1)
q−1
q
ã (t̃) +
n
X
cqi (t̃)
bi (t)
Z
1/q
f˜iq (t̃, s)wiq (z0,0 (s))ds
0
i=1
m+n
X
+
cqj (t̃)
Z
z0,0 (t) ≤ Wn−1 Wn (γ̃0,0,n (t)) +
bn (t)
Z
#
f˜jq (t̃, s)wjq (ϕ(s − τ ))ds .
0
j=n+1
By Lemma 2.2, (2.6) and (2.1), we have
"
bj (t̃)
#
(n + m + 1)q−1 cn (t̃)f˜nq (t̃, s)ds ,
0
γ̃0,0,j (t) =
−1
Wj−1
Wj−1 (γ̃0,0,j−1 (t))
bj−1 (t)
Z
+
#
q
(n + m + 1)q−1 cj−1 (t̃)f˜j−1
(t̃, s)ds ,
j 6= 1,
0
"
γ̃0,0,1 (t) = (n + m + 1)q−1 ãq (t̃) +
n+m
X
Z
j=n+1
bj (t̃)
#
cqj (t̃)f˜jq (t̃, s)wjq (ψ(s − τ ))ds
0
since ψ(t) = ϕ(t) for t ∈ [−τ, 0].
Since (2.5) is true for any t ∈ [0, t̃] and γ̃0,0,j (t̃) = γ0,0,j (t̃), we have
u(t̃) ≤ z0,0 (t̃)1/q ≤ u0,0 (t̃).
We know that t̃ ∈ [0, τ ] is arbitrary so we replace t̃ by t and get
u(t) ≤ u0,0 (t),
(2.7)
for t ∈ [0, τ ].
This implies that the theorem is true for t ∈ [0, τ ] and k = 0.
For t ∈ [τ, t̃] and t̃ ∈ [τ, t1 ], use the assumption (H3) and then we know that bα (τ ) = τ and
τ ≤ bα (t) ≤ t1 for t ∈ [τ, t1 ] and α = 1, . . . , n + m. Thus,
0 ≤ bα (t) − τ ≤ t1 − τ ≤ τ
since τ < t1 − t0 = t1 ≤ 2τ . Using this fact, (2.3) and (2.7), we get
"
Z τ
n
X
q
q
q−1
q
u (t) ≤ (n + m + 1)
ã (t̃) +
ci (t̃)
f˜iq (t̃, s)wiq (u(s))ds
i=1
+
+
n
X
cqi (t̃)
i=1
m+n
X
+
j=n+1
0
f˜iq (t̃, s)wiq (u(s))ds
τ
τ
cqj (t̃)
Z
cqj (t̃)
Z
f˜jq (t̃, s)wjq (ψ(s − τ ))ds
0
j=n+1
m+n
X
bi (t)
Z
bj (t̃)
#
f˜jq (t̃, s)wjq (u(s − τ ))ds
τ
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
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G RONWALL -B IHARI I NEQUALITY
"
≤ (n + m + 1)
q−1
q
ã (t̃) +
n
X
cqi (t̃)
+
+
cqi (t̃)
bi (t)
Z
cqj (t̃)
0
f˜iq (t̃, s)wiq (u(s))ds
m+n
X
τ
Z
f˜jq (t̃, s)wjq (ψ(s − τ ))ds
0
j=n+1
cqj (t̃)
bj (t̃)
Z
#
f˜jq (t̃, s)wjq (u0,0 (s − τ ))ds
τ
j=n+1
"
≤ (n + m + 1)
q−1
q
ã (t̃) +
n
X
cqi (t̃)
+
+
cqi (t̃)
bi (t)
Z
τ
Z
f˜iq (t̃, s)wiq (φ(s))ds
0
i=1
n
X
f˜iq (t̃, s)wiq (u0,0 (s))ds
τ
i=1
m+n
X
+
τ
Z
i=1
n
X
7
f˜iq (t̃, s)wiq (u(s))ds
τ
i=1
m+n
X
cqj (t̃)
bj (t̃)
Z
#
f˜jq (t̃, s)wjq (ψ(s
− τ ))ds
0
j=n+1
:= z0,1 (t),
where we use the definitions of φ and ψ. Thus,
1/q
u(t) ≤ z0,1 (t),
(2.8)
t ∈ [τ, t̃].
Therefore,
"
z0,1 ≤ (n + m + 1)
q−1
q
ã (t̃) +
n
X
cqi (t̃)
+
cqi (t̃)
i=1
f˜iq (t̃, s)wiq (φ(s))ds
0
i=1
n
X
τ
Z
bi (t)
Z
1/q
f˜iq (t̃, s)wiq (z0,1 (s))ds
τ
m+n
X
+
cqj (t̃)
bj (t̃)
Z
#
f˜jq (t̃, s)wjq (ψ(s
− τ ))ds .
0
j=n+1
Using Lemma 2.2, (2.1) and bα (τ ) = τ , we obtain for t ∈ [τ, t̃]
"
#
Z bn (t)
z0,1 (t) ≤ Wn−1 Wn (γ̃0,1,n (t)) +
(n + m + 1)q−1 cqn (t̃)f˜nq (t̃, s)ds ,
τ
Z
−1
γ̃0,1,j (t) = Wj−1 Wj−1 (γ̃0,1,j−1 (t))+
#
bj−1 (t)
(n + m +
q
1)q−1 cqj−1 (t̃)f˜j−1
(t̃, s)ds
, j 6= 1,
τ
"
γ̃0,1,1 (t) = (n + m + 1)
q−1
q
ã (t̃) +
n Z
X
i=1
τ
cqi (t̃)f˜iq (t̃, s)wiq (φ(s))ds
0
+
n+m
X
j=n+1
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
Z
bj (t̃)
#
cqj (t̃)f˜jq (t̃, s)wjq (ψ(s − τ ))ds .
0
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8
S HENGFU D ENG AND C ARL P RATHER
Since (2.8) is true for any t ∈ [τ, t1 ] and γ̃0,1,1 (t̃) = γ0,1,1 (t̃), we have
1/q
u(t̃) ≤ z0,1 (t̃) ≤ u0,1 (t̃).
We know that t̃ ∈ [τ, t1 ] is arbitrary so we replace t̃ by t and get
u(t) ≤ u0,1 (t),
(2.9)
t ∈ [τ, t1 ].
This implies that the theorem is valid for t ∈ [τ, t1 ] and L = 0.
(2) L = 1. First we consider t ∈ (t1 , t̃], where t̃ ∈ (t1 , t1 +τ ] is arbitrary. Note that τ < t2 −t1 ≤
2τ . (H3) gives bα (t1 ) = t1 and t1 ≤ bα (t) ≤ t1 + τ for t ∈ (t1 , t1 + τ ] so t1 − τ ≤ bα (t) − τ ≤ t1
for t ∈ (t1 , t1 + τ ]. By (2.7) and (2.9), (2.2) can be written as
"
Z τ Z t1 n
X
q
q
q−1
q
u (t) ≤ (n + m + 2)
ã (t̃) +
c (t̃)
+
f˜q (t̃, s)wq (u(s))ds
i
i
0
i=1
+
+
n
X
bi (t)
Z
cqi (t̃)
τ
Z
cqj (t̃)
0
cqj (t̃)
t1 Z
+
j=n+1
+
f˜iq (t̃, s)wiq (u(s))ds
t1
i=1
m+n
X
n
X
bj (t̃)
Z
f˜jq (t̃, s)wjq (u(s − τ ))ds
τ
#
f˜jq (t̃, s)wjq (u(s
− τ ))ds +
"
≤ (n + m + 2)
q−1
q
ã (t̃) +
n
X
cqi (t̃)
+
cqi (t̃)
bi (t)
Z
Z
t1
f˜iq (t̃, s)wiq (φ(s))ds
0
i=1
+
d˜q (t̃)η1q uq (t1 )
t1
j=1
n
X
i
τ
f˜iq (t̃, s)wiq (u(s))ds
t1
i=1
m+n
X
cqj (t̃)
Z
bj (t̃)
#
f˜jq (t̃, s)wjq (ψ(s − τ ))ds + d˜q (t̃)η1q uq0,1 (t1 )
0
j=n+1
:=z1,0 (t),
where we use the definitions of φ and ψ so
1/q
u(t) ≤ z1,0 (t),
(2.10)
t ∈ (t1 , t̃].
Thus,
"
z1,0 (t) ≤ (n + m + 2)
q−1
q
ã (t̃) +
n
X
cqi (t̃)
+
n
X
cqi (t̃)
i=1
m+n
X
f˜iq (t̃, s)wiq (φ(s))ds
0
i=1
+
t1
Z
Z
bi (t)
1/q
f˜iq (t̃, s)wiq (z1,0 (s))ds
t1
cqj (t̃)
j=n+1
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
Z
bj (t̃)
#
f˜jq (t̃, s)wjq (ψ(s − τ ))ds + d˜q (t̃)η1q uq0,1 (t1 ) .
0
http://jipam.vu.edu.au/
G RONWALL -B IHARI I NEQUALITY
9
By Lemma 2.2, (2.1) and bα (t1 ) = t1 , we obtain for t ∈ (t1 , t̃]
"
Z
bn (t)
z1,0 (t) ≤
Wn−1
Wn (γ̃1,0,n (t)) +
(n + m +
#
2)q−1 cqn (t̃)f˜nq (t̃, s)ds
,
t1
Z
−1
γ̃1,0,j (t) = Wj−1 Wj−1 (γ̃1,0,j−1 (t))+
bj−1 (t)
#
q
(n + m + 2)q−1 cqj−1 (t̃)f˜j−1
(t̃, s)ds , j 6= 1,
t1
"
γ̃1,0,1 (t) = (n + m + 2)
q−1
n Z
X
q
ã (t̃) +
+
bj (t̃)
Z
cqi (t̃)f˜iq (t̃, s)wiq (φ(s))ds
0
i=1
n+m
X
t1
#
cqj (t̃)f˜jq (t̃, s)wjq (ψ(s − τ ))ds + d˜q (t̃)η1q uq0,1 (t1 ) .
0
j=n+1
Since (2.10) is true for any t ∈ (t1 , t̃] and γ̃1,0,1 (t̃) = γ1,0,1 (t̃), we have
1/q
u(t̃) ≤ z1,0 (t̃) ≤ u1,0 (t̃).
We know that t̃ ∈ (t1 , t1 + τ ] is arbitrary so we replace t̃ by t and get
u(t) ≤ u1,0 (t),
(2.11)
t ∈ (t1 , t1 + τ ].
This implies that the theorem is valid for t ∈ (t1 , t1 + τ ] and L = 1.
We now consider t ∈ [t1 + τ, t̃], where t̃ ∈ [t1 + τ, t2 ] is arbitrary. Again, by (H3) we have
t1 +τ ≤ bα (t) ≤ t2 for t ∈ [t1 +τ, t2 ] and bα (t1 +τ ) = t1 +τ so t1 ≤ bα (t)−τ ≤ t2 −τ ≤ t1 +τ
since τ < t2 − t1 ≤ 2τ . Obviously, by (2.7), (2.9) and (2.11), (2.2) becomes
"
Z t1 +τ
n
X
q
q
q−1
q
u (t) ≤ (n + m + 2)
ã (t̃) +
ci (t̃)
f˜iq (t̃, s)wiq (u(s))ds
0
i=1
+
+
n
X
cqi (t̃)
i=1
m+n
X
Z
+
f˜iq (t̃, s)wiq (u(s))ds
t1 +τ
cqj (t̃)
t1 +τ
Z
f˜jq (t̃, s)wjq (u(s − τ ))ds
0
j=n+1
m+n
X
bi (t)
cqj (t̃)
bj (t̃)
Z
#
f˜jq (t̃, s)wjq (u(s
t1 +τ
j=n+1
"
≤ (n + m + 2)
q−1
q
ã (t̃) +
n
X
i=1
+
+
n
X
cqi (t̃)
i=1
m+n
X
j=n+1
− τ ))ds +
d˜q (t̃)η1q uq1,0 (t1 )
Z
bi (t)
cqi (t̃)
Z
t1 +τ
f˜iq (t̃, s)wiq (φ(s))ds
0
f˜iq (t̃, s)wiq (u(s))ds
t1 +τ
cqj (t̃)
Z
bj (t̃)
#
f˜jq (t̃, s)wjq (ψ(s − τ ))ds + d˜q (t̃)η1q uq0,1 (t1 )
0
:= z1,1 (t),
that is,
(2.12)
1/q
u(t) ≤ z1,1 (t),
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 34, 11 pp.
t ∈ [t1 + τ, t̃].
http://jipam.vu.edu.au/
10
S HENGFU D ENG AND C ARL P RATHER
Thus,
"
z1,1 (t) ≤ (n + m + 2)
q−1
q
ã (t̃) +
+ cqi (t̃)
n
X
cqi (t̃)
t1 +τ
Z
f˜iq (t̃, s)wiq (φ(s))ds
0
i=1
bi (t)
Z
1/q
f˜iq (t̃, s)wiq (z1,1 (s))ds
t1 +τ
m+n
X
+
cqj (t̃)
bj (t̃)
Z
#
f˜jq (t̃, s)wjq (ψ(s − τ ))ds + d˜q (t̃)η1q uq0,1 (t1 ) .
0
j=n+1
Using Lemma 2.2, (2.1) and bα (t1 + τ ) = t1 + τ , we obtain for t ∈ (t1 , t̃]
#
"
Z bn (t)
z1,1 (t) ≤ W −1 Wn (γ̃1,1,n (t)) +
(n + m + 2)q−1 cq (t̃)f˜q (t̃, s)ds ,
n
n
n
t1 +τ
Z
h
−1
γ̃1,1,j (t) = Wj−1
Wj−1 (γ̃1,1,j−1 (t)) +
bj−1 (t)
#
q
(n + m + 2)q−1 cqj−1 (t̃)f˜j−1
(t̃, s)ds , j 6= 0,
t1 +τ
"
γ̃1,1,1 (t) = (n + m + 2)
q−1
n Z
X
q
ã (t̃) +
i=1
+
n+m
X
j=n+1
Z
bj (t̃)
t1 +τ
cqi (t̃)f˜iq (t̃, s)wiq (φ(s))ds
0
#
cqj (t̃)f˜jq (t̃, s)wjq (ψ(s − τ ))ds + d˜q (t̃)η1q uq0,1 (t1 ) .
0
Since (2.12) is true for any t ∈ (t1 , t̃] and γ̃1,1,1 (t̃) = γ1,1,1 (t̃), we have
1/q
u(t̃) ≤ z1,1 (t̃) ≤ u1,1 (t̃).
We know that t̃ ∈ [t1 + τ, t2 ] is arbitrary so we replace t̃ by t and get
u(t) ≤ u1,1 (t),
t ∈ [t1 + τ, t2 ].
This implies that the theorem is valid for t ∈ [t1 + τ, t2 ] and L = 1.
(3) Finally, suppose that the theorem is valid for k, then for k + 1 we redefine φ and ψ by
replacing k with k + 1. In a similar manner as in steps (1) and (2), we can see that the theorem
holds for t ∈ (tk+1 , T ] ⊂ (tk+1 , tk+2 ].
The proof is now completed.
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[1] R.P. AGARWAL, S. DENG AND W. ZHANG, Generalization of a retarded Gronwall-like inequality
and its applications, Appl. Math. Comput., 165 (2005), 599–612.
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[3] S.G. HRISTOVA, Nonlinear delay integral inequalities for piecewise continuous functions and applications, J. Inequal. Pure Appl. Math., 5(4) (2004), Art. 88. [ONLINE: http://jipam.vu.
edu.au/article.php?sid=441].
[4] M. MEDVED, A new approach to an analysis of Henry type integral inequalities and their Bihari
type versions, J. Math. Anal. Appl., 214 (1997), 349–366.
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G RONWALL -B IHARI I NEQUALITY
11
[5] M. PINTO, Integral inequalities of Bihari-type and applications, Funkcial. Ekvac., 33 (1990), 387–
403.
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http://jipam.vu.edu.au/