Volume 9 (2008), Issue 3, Article 82, 10 pp.
EXPLICIT BOUNDS ON SOME NONLINEAR RETARDED INTEGRAL
INEQUALITIES
MAN-CHUN TAN AND ZHI-HONG LI
D EPARTMENT OF M ATHEMATICS
J INAN U NIVERSITY
G UANGZHOU 510632
P EOPLE ’ S R EPUBLIC OF C HINA
tanmc@jnu.edu.cn
Received 21 November, 2007; accepted 10 July, 2008
Communicated by S.S. Dragomir
A BSTRACT. In this paper some new retarded integral inequalities are established and explicit
bounds on the unknown functions are derived. The present results extend some existing ones
proved by Lipovan in [A retarded integral inequality and its applications, J. Math. Anal. Appl.
285 (2003) 436-443].
Key words and phrases: Integral inequality; retarded; nonlinear; explicit bound.
2000 Mathematics Subject Classification. 26D10, 26D15.
1. I NTRODUCTION
During the past decades, studies on integral inequalities have been greatly enriched by the
recognition of their potential applications in various applied sciences [1] – [6]. Recently, integral inequalities with delays have received much attention from researchers [7] – [12]. In
this paper, we establish some new retarded integral inequalities and derive explicit bounds on
unknown functions, the results of which improve some known ones in [9].
2. M AIN R ESULTS
Throughout the paper, R denotes the set of real numbers and R+ = [0, +∞). C(M, S)
denotes the class of all continuous functions from M to S. C 1 (M, S) denotes the class of
functions with continuous first derivative.
Theorem 2.1. Suppose that p > q ≥ 0 and c ≥ 0 are constants, and u, f, g, h ∈ C (R+ , R+ ).
Let w ∈ (R+ , R+ ) be nondecreasing with w(u) > 0 on (0, ∞), and α ∈ C 1 (R+ , R+ ) be
The research was jointly supported by grants from the National Natural Science Foundation of China (No. 50578064) and the Natural
Science Foundation of Guangdong Province, China (No.06025219).
344-07
2
M AN -C HUN TAN AND Z HI -H ONG L I
nondecreasing with α(t) ≤ t on R+ . Then the following integral inequality
Z s
Z α(t) p
2
q
q
(2.1)
u (t) ≤ c + 2
f (s)u (s)
g(τ )w(u(τ ))dτ + h(s)u (s) ds, t ∈ R+
0
0
implies for 0 ≤ t ≤ T ,
(
(2.2)
"
G−1
u(t) ≤
2(p − q)
G(ξ(t)) +
p
Z
α(t)
Z
f (s)
0
1
#) p−q
s
g(τ )dτ ds
0
holds, where
ξ(t) = c
(2.3)
2(p−q)
p
Z
2(p − q)
+
p
r
1
G(r) =
(2.4)
r0
w s
1
p−q
α(t)
Z
h(s)ds,
0
r ≥ r0 > 0,
ds,
G−1 denotes the inverse function of G, and T ∈ R+ is chosen so that
Z
Z s
2(p − q) α(t)
G(ξ(t)) +
f (s)
g(τ )dτ ds ∈ Dom G−1 , for all 0 ≤ t ≤ T.
p
0
0
Proof. The conditions α ∈ C 1 (R+ , R+ ) and α(t) ≤ t imply that α(0) = 0. Firstly we assume
that c > 0. Define the nondeceasing positive function z(t) by
Z s
Z α(t) 2
q
q
z(t) := c + 2
f (s)u (s)
g(τ )w(u(τ ))dτ + h(s)u (s) ds.
0
0
2
Then z(0) = c and by (2.1) we have
1
u(t) ≤ [z(t)] p ,
(2.5)
1
1
and consequently u(α(t)) ≤ [z(α(t))] p ≤ [z(t)] p . By differentiation we get
"
!
#
Z α(t)
z 0 (t) = 2uq (α(t)) f (α(t))
g(τ )w(u(τ ))dτ + h(α(t)) α0 (t)
0
≤ 2 [z(t)]
q
p
"
Z
!
α(t)
f (α(t))
g(τ )w(u(τ ))dτ
#
+ h(α(t)) α0 (t).
0
Hence
z 0 (t)
q
p
0
Z
≤ 2f (α(t))α (t)
α(t)
g(τ )w(u(τ ))dτ + 2h(α(t))α0 (t).
[z(t)]
0
Integrating both sides of last relation on [0, t] yields
Z α(t)
Z α(t)
Z s
p−q
p−q
p
p
p
p
[z(t)]
≤
[z(0)]
+2
h(s)ds + 2
f (s)
g(τ )w(u(τ ))dτ ds,
p−q
p−q
0
0
0
which can be rewritten as
Z α(t)
2(p−q)
p−q
2(p
−
q)
(2.6)
[z(t)] p ≤ c p +
h(s)ds
p
0
Z
Z s
2(p − q) α(t)
+
f (s)
g(τ )w(u(τ ))dτ ds.
p
0
0
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 82, 10 pp.
http://jipam.vu.edu.au/
R ETARDED I NEQUALITY
3
Let T1 (≤ T ) be an arbitrary number. For 0 ≤ t ≤ T1 , from (2.3) and (2.6) we have
Z α(t)
Z s
p−q
2(p
−
q)
(2.7)
[z(t)] p ≤ ξ(T1 ) +
f (s)
g(τ )w(u(τ ))dτ ds.
p
0
0
1
1
Denoting the right-hand side of (2.7) by m(t), we know u(t) ≤ [z(t)] p ≤ [m(t)] p−q . Since w
is nondecreasing, we obtain
h
i
h
i
h
i
1
1
1
w[u(τ )] ≤ w (z(τ )) p ≤ w (z(α(t))) p ≤ w (z(t)) p , for τ ∈ [0, α(t)].
Hence
Z α(t)
2(p − q)
0
f (α(t))α (t)
g(τ )w(u(τ ))dτ
m (t) =
p
0
Z α(t)
i
1
2(p − q) h
0
p
w (z(t)) f (α(t))α (t)
≤
g(τ )dτ
p
0
Z α(t)
i
1
2(p − q) h
0
≤
w (m(t)) p−q f (α(t))α (t)
g(τ )dτ .
p
0
0
That is
(2.8)
m0 (t)
2(p − q)
≤
f (α(t))α0 (t)
1
p
p−q
w[(m(t)) ]
Z
α(t)
g(τ )dτ .
0
Integrating both sides of the last inequality on [0, t] and using the definition (2.4), we get
Z
Z s
2(p − q) α(t)
(2.9)
G(m(t)) − G(m(0)) ≤
f (s)
g(τ )dτ ds.
p
0
0
1
Taking t = T1 in inequality (2.9) and using u(t) ≤ [m(t)] p−q , we have
1
#) p−q
(
"
Z α(T1 )
Z s
2(p
−
q)
u(T1 ) ≤ G−1 G [ξ(T1 )] +
f (s)
g(τ )dτ ds
.
p
0
0
Since T1 (≤ T ) is arbitrary, we have proved the desired inequality (2.2).
The case c = 0 can be handled by repeating the above procedure with ε > 0 instead of c and
subsequently letting ε → 0. This completes the proof.
Remark 1. If c = 0 and h(t) ≡ 0 hold, G(ξ(t)) = G(0) in (2.4) is not defined. In such a case,
the upper bound on solutions of the integral inequality (2.1) can be calculated as
1
(
"
#) p−q
Z α(t)
Z s
2(p
−
q)
u(t) ≤ lim G−1 G(ε) +
f (s)
g(τ )dτ ds
.
ε→0+
p
0
0
From Theorem 2.1, we can easily derive the following corollaries.
Corollary 2.2. Suppose that u, h ∈ C (R+ , R+ ) and c ≥ 0 is a constant. Let α ∈ C 1 (R+ , R+ )
be nondecreasing with α(t) ≤ t on R+ . Then the following inequality
Z α(t)
2
2
u (t) ≤ c + 2
h(s)u(s)ds,
0
implies
Z
u(t) ≤ c +
α(t)
h(s)ds.
0
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 82, 10 pp.
http://jipam.vu.edu.au/
4
M AN -C HUN TAN AND Z HI -H ONG L I
Remark 2. If α(t) ≡ t, from Corollary 2.2 we get the Ou-Iang inequality.
Corollary 2.3. Suppose that u, f, g, h ∈ C (R+ , R+ ), and c ≥ 0 is a constant. Let w ∈
(R+ , R+ ) be nondecreasing with w(u) > 0 on (0, ∞), and α ∈ C 1 (R+ , R+ ) be nondecreasing
with α(t) ≤ t on R+ . Then the following inequality
Z s
Z α(t) 2
2
f (s)u(s)
u (t) ≤ c + 2
g(τ )u(τ )dτ + h(s)u(s) ds
0
0
implies
α(t)
Z
u(t) ≤ ξ(t) exp
Z
f (s)
R α(t)
!
g(τ )dτ
0
where ξ(t) = c +
s
ds
0
h(s)ds.
0
Theorem 2.4. Suppose that p > q ≥ 0 and c ≥ 0 are constants, and u, f, g, h ∈ C (R+ , R+ ).
Let w ∈ (R+ , R+ ) be nondecreasing with w(u) > 0 on (0, ∞), and α ∈ C 1 (R+ , R+ ) be
nondecreasing with α(t) ≤ t on R+ . Then the following integral inequality
Z α(t) p
2
q
(2.10) u (t) ≤ c + 2
f (s)u (s) w(u(s))
0
Z s
q
+
g(τ )w(u(τ ))dτ + h(s)u (s) ds,
t ∈ R+
0
implies for 0 ≤ t ≤ T
(
(2.11)
"
G−1 G(ξ(t)) +
u(t) ≤
2(p − q)
p
α(t)
Z
Z
s
f (s) 1 +
g(τ )dτ
0
1
#) p−q
ds
,
0
where ξ(t) and G(r) are defined by (2.3) and (2.4), respectively, and T ∈ R+ is chosen so that
Z
Z s
2(p − q) α(t)
G(ξ(t)) +
f (s) 1 +
g(τ )dτ ds ∈ Dom G−1 , for all 0 ≤ t ≤ T.
p
0
0
Proof. Firstly we assume that c > 0. Define the nondeceasing positive function by
Z α(t) Z s
2
q
q
z(t) := c + 2
f (s)u (s) w(u(s))+
g(τ )w(u(τ ))dτ + h(s)u (s) ds,
0
0
2
then z(0) = c and by (2.10) we have
1
u(t) ≤ [z(t)] p ,
(2.12)
and
"
z 0 (t) = 2uq (α(t)) f (α(t)) w(u(α(t))) +
Z
!
α(t)
g(τ )w(u(τ ))dτ
#
+ h(α(t)) α0 (t)
0
≤ 2 [z(t)]
q
p
"
Z
f (α(t)) w(u(α(t))) +
!
α(t)
g(τ )w(u(τ ))dτ
#
+ h(α(t)) α0 (t).
0
Hence
z 0 (t)
[z(t)]
q
p
≤ 2h(α(t))α0 (t) + 2f (α(t))α0 (t) w(u(α(t)) +
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 82, 10 pp.
Z
!
α(t)
g(τ )w(u(τ ))dτ
.
0
http://jipam.vu.edu.au/
R ETARDED I NEQUALITY
5
Integrating both sides of the last inequality on [0, t], we get
Z α(t)
p−q
p−q
p
p
[z(t)] p ≤
[z(0)] p + 2
h(s)ds
p−q
p−q
0
Z α(t)
Z s
+2
f (s) w(u(s)) +
g(τ )w(u(τ ))dτ ds.
0
0
Using (2.3), we get
[z(t)]
p−q
p
2(p − q)
≤ ξ(t) +
p
α(t)
Z
s
Z
f (s) w(u(s)) +
g(τ )w(u(τ ))dτ
0
ds.
0
Let T1 (≤ T ) be an arbitrary number. From last inequality we know the following relation holds
for t ∈ [0, T1 ],
Z
Z s
p−q
2(p − q) α(t)
p
[z(t)]
≤ ξ(T1 ) +
f (s) w(u(s)) +
g(τ )w(u(τ ))dτ ds.
p
0
0
Letting
2(p − q)
m(t) = ξ(T1 ) +
p
(2.13)
we get [z(t)]
p−q
p
Z
α(t)
Z
s
f (s) w(u(s)) +
g(τ )w(u(τ ))dτ
0
ds,
0
≤ m(t). Since w is nondecreasing, we have
h
i
h
i
h
i
1
1
1
w[u(α(t))] ≤ w (z(α(t))) p ≤ w (z(t)) p ≤ w (m(t)) p−q
and
h
i
h
i
h
i
1
1
1
w[u(τ )] ≤ w (z(τ )) p ≤ w (z(α(t))) p ≤ w (z(t)) p ,
τ ∈ [0, α(t)].
for
From (2.13), by differentiation we obtain
!
Z α(t)
2(p − q)
m (t) =
f (α(t)) w(u(α(t))) +
g(τ )w(u(τ ))dτ α0 (t)
p
0
(
)
Z α(t)
1
1
2(p − q)
≤
f (α(t)) w [m(t)] p−q +
g(τ )w [m(t)] p−q dτ α0 (t)
p
0
!
Z α(t)
2(p − q)
1
= w [m(t)] p−q
f (α(t)) 1 +
g(τ )dτ α0 (t).
p
0
0
Hence
m0 (t)
2(p − q)
≤
f (α(t)) 1 +
1
p
w [m(t)] p−q
Z
!
α(t)
g(τ )dτ
α0 (t).
0
Integrating both sides of the last inequality on [0, t], from (2.4) we get
Z
Z s
2(p − q) α(t)
G(m(t)) ≤ G(m(0)) +
f (s) 1 +
g(τ )dτ ds.
p
0
0
Hence
"
(2.14)
m(t) ≤ G−1
2(p − q)
G(ξ(T1 )) +
p
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 82, 10 pp.
Z
α(t)
Z
f (s) 1 +
0
s
#
g(τ )dτ ds .
0
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6
M AN -C HUN TAN AND Z HI -H ONG L I
1
Taking t = T1 in inequality (2.14) and using u(t) ≤ [m(t)] p−q , we have
1
(
"
#) p−q
Z α(T1 )
Z s
2(p
−
q)
.
u(T1 ) ≤ G−1 G(ξ(T1 )) +
f (s) 1 +
g(τ )dτ ds
p
0
0
Since T1 (≤ T ) is arbitrary we have proved the desired inequality (2.11).
If c = 0, the result can be proved by repeating the above procedure with ε > 0 instead of c
and subsequently letting ε → 0. This completes the proof.
Remark 3. Theorem 2.1 of Lipovan in [9] is special case of above Theorem 2.4, under the
assumptions that p = 2, q = 1 and g(t) ≡ 0.
Theorem 2.5. Suppose that p > q ≥ 0 and c ≥ 0 are constants, and u, f, g, h ∈ C (R+ , R+ ).
Let w ∈ (R+ , R+ ) be nondecreasing with w(u) > 0 on (0, ∞), and α, β ∈ C 1 (R+ , R+ ) be
nondecreasing with α(t) ≤ t, β(t) ≤ t on R+ . Then the following integral inequality
Z α(t) Z s
p
2
q
(2.15) u (t) ≤ c + 2
f (s)u (s) w(u(s)) +
g(τ )w(u(τ ))dτ ds
0
0
β(t)
Z
h (s)uq (s) w (u (s)) ds,
+2
t ∈ R+
0
implies for 0 ≤ t ≤ T
"
(
(2.16) u(t) ≤
G
−1
G(c
2(p−q)
p
2(p − q)
)+
p
Z
α(t)
Z
f (s) 1 +
0
s
g(τ )dτ
ds
0
2(p − q)
+
p
1
#) p−q
β(t)
Z
h (s)ds
,
0
where G(r) is defined by (2.4) and T ∈ R+ is chosen so that
Z s
2(p−q) 2(p − q) Z α(t)
p
G c
f (s) 1 +
+
g(τ )dτ ds
p
0
0
Z
2(p − q) β(t)
+
h (s)ds ∈ Dom G−1 , for all 0 ≤ t ≤ T.
p
0
1
Proof. The conditions that α, β ∈ C (R+ , R+ ) are nondecreasing with α(t) ≤ t, β(t) ≤ t
imply that α(0) = 0 and β(0) = 0.
Let us first assume that c > 0. Denoting the right-hand side of (2.15) by z(t), we know z(t)
1
is nondecreasing, z(0) = c2 and u(t) ≤ [z(t)] p . Consequently we have
1
1
u(α(t)) ≤ [z(α(t))] p ≤ [z(t)] p
1
1
and u(β(t)) ≤ [z(β(t))] p ≤ [z(t)] p .
Since w is nondecreasing, we obtain
0
Z
q
!
α(t)
z (t) = 2f (α(t))u (α(t)) w(u(α(t))) +
g(τ )w(u(τ ))dτ
α0 (t)
0
+ 2h (β (t)) uq (β (t)) w (u (β (t))) β 0 (t)
!
Z α(t)
q
≤ 2 [z(t)] p [f (α (t)) w (u (α (t))) +
g (τ ) w (u (τ )) dτ α0 (t)
0
0
+ h (β (t)) w (u (β (t))) β (t)].
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 82, 10 pp.
http://jipam.vu.edu.au/
R ETARDED I NEQUALITY
7
Hence
z 0 (t)
Z
≤ 2f (α (t)) w (u (α (t))) +
q
[z(t)] p
!
α(t)
α0 (t)
g (τ ) w (u (τ ))dτ
0
+ 2h (β (t)) w (u (β (t))) β 0 (t) .
Integrating both sides on [0, t], we get
p−q
p−q
p
p
[z(t)] p ≤
[z(0)] p
p−q
p−q
Z α(t)
Z s
Z
+2
f (s) w(u(s)) +
g(τ )w(u(τ ))dτ ds + 2
0
0
β(t)
h (s)w (u (s)) ds,
0
which can be rewritten as
(2.17)
[z(t)]
p−q
p
≤c
2(p−q)
p
2(p − q)
+
p
Z
α(t)
Z
s
f (s) w(u(s)) +
0
g(τ )w(u(τ ))dτ
ds
0
2(p − q)
+
p
Z
β(t)
h (s)w (u (s)) ds.
0
p−q
Denoting the right-hand side of (2.17) by m (t), we know [z (t)] p ≤ m (t) and
!
Z α(t)
2(p
−
q)
m0 (t) =
f (α(t)) w(u(α(t))) +
g(τ )w(u(τ ))dτ α0 (t)
p
0
2 (p − q)
h (β (t)) w (u (β (t))) β 0 (t)
p
!
1
Z α(t)
1
2(p − q)
≤
f (α(t)) w z p (α (t)) +
g(τ )w z p (τ ) dτ α0 (t)
p
0
1
2 (p − q)
+
h (β (t)) w z p (β (t)) β 0 (t)
p
"
!
#
Z α(t)
1 2(p − q)
f (α(t)) 1 +
g(τ )dτ α0 (t) + h (β (t)) β 0 (t)
≤ w z p (t)
p
0
"
!
#
Z α(t)
2(p − q)
1
f (α(t)) 1 +
g(τ )dτ α0 (t) + h (β (t)) β 0 (t) .
≤ w m p−q (t)
p
0
+
The above relation gives
"
!
#
Z α(t)
2(p − q)
≤
f (α(t)) 1 +
g(τ )dτ α0 (t) + h (β (t)) β 0 (t) .
p
0
(t)
m0 (t)
1
w m p−q
Integrating both sides on [0, t] and using definition (2.4) we get
"Z
#
Z s
Z β(t)
α(t)
2 (p − q)
G (m (t)) ≤ G (m (0)) +
f (s) 1 +
g (τ ) dτ ds +
h (s) ds
p
0
0
0
"
#
Z s
Z β(t)
2(p−q) 2 (p − q) Z α(t)
+
f (s) 1 +
g (τ ) dτ ds +
h (s) ds .
≤G c p
p
0
0
0
1
1
Using the relation u(t) ≤ [z(t)] p ≤ [m(t)] p−q , we get the desired inequality (2.16).
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 82, 10 pp.
http://jipam.vu.edu.au/
8
M AN -C HUN TAN AND Z HI -H ONG L I
If c = 0, the result can be proved by repeating the above procedure with ε > 0 instead of c
and subsequently letting ε → 0. This completes the proof.
Remark 4. Theorem 2 of Lipovan in [9] is a special case of Theorem 2.5 above, under the
assumptions that p = 2, q = 1, g(t) ≡ 0 and β(t) ≡ t.
3. A PPLICATION
Example 3.1. Consider the delay integral equation
Z α(t) Z s
5
2
3
3
(3.1) x (t) = x0 + 2
x (s)M s, x(s),
N (s, τ, w(|x(τ )|))dτ + h(s)x (s) ds.
0
0
Assume that
(3.2)
|M (s, t, v)| ≤ f (s) |v| ,
|N (s, t, v)| ≤ g(t) |v| ,
where f, g, h, α and w are as defined in Theorem 2.1. From (3.1) and (3.2) we obtain
Z α(t) Z s
5
3
3
2
|x(t)| ≤ x0 + 2
|x(s)| f (s)
g(τ )w(|x(τ )|)dτ + h(s) |x(s)| ds.
0
0
Applying Theorem 2.1 to the last relation, we get an explicit bound on an unknown function
#) 12
(
"
Z α(t)
Z s
4
(3.3)
|x(t)| ≤ G−1 G(ξ(t)) +
f (s)
g(τ )dτ ds
,
5 0
0
where
q Z
5 4 α(t)
4
ξ(t) = x0 +
h(s)ds.
5 0
In particular, if ω(t) ≡ t holds in (3.1), from (2.4) we derive
Z t
Z t
Z t
√
1
1
− 12
1 ds =
(3.4)
G(t) =
ds
=
s
ds
=
2
t
1
0 ω s p−q
0 s p−q
0
and
1
G−1 (t) = t2 .
4
Substituting (3.4) and (3.5) into inequality (3.3), we get
Z
Z s
p
2 α(t)
|x (t)| ≤ ξ(t) +
f (s)
g (τ )dτ.
5 0
0
(3.5)
Example 3.2. Consider the following equation
Z α(t) 8
2
4
(3.6) x (t) = x0 + 2
x (s) M (s, x(s), w(|x(s)|))
0
Z
+
s
Z
N (s, τ, w(|x(τ )|))dτ ds + 2
0
α(t)
h(s)x4 (s) ds.
0
Assume that
(3.7)
|M (s, t, v)| ≤ f (s) |v| ,
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 82, 10 pp.
|N (s, t, v)| ≤ f (s)g(t) |v| ,
http://jipam.vu.edu.au/
R ETARDED I NEQUALITY
9
where f, g, h, α and w are as defined in Theorem 2.4. From (3.6) and (3.7) we obtain
Z α(t) 8
4
2
|x(t)| ≤ x0 + 2
|x(s)| f (s) w(|x(s)|)
0
Z s
4
+
g(τ )w(|x(τ )|)dτ + h(s) |x(s)| ds.
0
By Theorem 2.4 we get an explicit bound on an unknown function
"
(
Z
Z
α(t)
(3.8)
|x(t)| ≤
G−1 G (ξ(t)) +
s
f (s) 1 +
0
g(τ )dτ
#) 14
ds
,
0
where
α(t)
Z
ξ(t) = |x0 | +
h(s)ds.
0
In particular, if ω (t) ≡ t3 holds in (3.6), from (2.4) we obtain
Z t
Z t
Z t
3
1
1
1
1 ds =
(3.9)
G(t) =
s− 4 ds = 4t 4
3 ds =
0 ω s p−q
0 s p−q
0
and
(3.10)
G−1 (t) =
1 4
t.
256
Substituting (3.9) and (3.10) into (3.8) we get
Z
Z s
1
1 α(t)
4
|x(t)| ≤ [ξ (t)] +
f (s) 1 +
g (τ ) dτ ds.
4 0
0
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