Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 48, 2006
EXPLICIT BOUNDS ON SOME NONLINEAR INTEGRAL INEQUALITIES
YOUNG-HO KIM
D EPARTMENT OF A PPLIED M ATHEMATICS ,
C HANGWON NATIONAL U NIVERSITY,
C HANGWON 641-773, KOREA
yhkim@sarim.changwon.ac.kr
Received 24 September, 2005; accepted 09 November, 2005
Communicated by D. Hinton
A BSTRACT. In this paper, some new nonlinear integral inequalities involving functions of one
and two independent variables which provide explicit bounds on unknown functions are established. We also present some of its applications to the qualitative study of retarded differential
equations.
Key words and phrases: Integral inequalities, Retarded integral inequality, Retarded differential equations, Partial delay differential equations.
2000 Mathematics Subject Classification. 26D10, 26D15, 34K10, 35B35.
1. I NTRODUCTION
Nonlinear differential equations whose solutions cannot be found explicitly arise in essentially every branch of modern science, engineering and mathematics. One of the most useful
methods available for studying a nonlinear system of ordinary differential equations is to compare it with a single first-order equation derived naturally from some bounds on the system.
However, the bounds provided by the comparison method are sometimes difficult or impossible to calculate explicitly. In fact, in many applications explicit bounds are more useful while
studying the behavior of solutions of such systems. Another basic tool, which is typical among
investigations on this subject, is the use of nonlinear integral inequalities which provide explicit
bounds on the unknown functions. Over the last scores of years several new nonlinear integral
inequalities have been developed in order to study the behavior of solutions of such systems.
One of the most useful inequalities in the development of the theory of differential equations
is given in the following theorem. If u, f are nonnegative continuous functions on R+ = [0, ∞),
u0 ≥ 0 is a constant and
Z t
2
2
u (t) ≤ u0 + 2
f (s)u(s)ds
0
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
This research is financially supported by Changwon National University in 2005.
288-05
2
YOUNG -H O K IM
for t ∈ R+ , then
t
Z
u(t) ≤ u0
t ∈ R+ .
f (s)ds,
0
It appears that this inequality was first considered by Ou-Iang [5], while investigating the
boundedness of certain solutions of certain second-order differential equations.
In the past few years this inequality given in [5] has been used considerably in the study of
qualitative properties of the solutions of certain abstract differential, integral and partial differential equations.
Recently, Pachpatte in [9] obtained a useful upper bound on the following inequality:
n Z αi (t)
X
p
(1.1)
u (t) ≤ c + p
[ai (s)up (s) + bi (s)u(s)]ds,
i=1
αi (t0 )
and its variants, under some suitable conditions on the functions involved in (1.1), including
the constant p > 1. In fact, the results given in [9] are generalized versions of the inequalities
established by Lipovan in [4], Qu-Iang in [5] and Pachpatte in [6].
The main purpose of this paper is to obtain explicit bounds on the following retarded integral
inequality
"Z
#
Z αi (t)
n
t
X
(1.2)
u(t) ≤ c +
ai (s)up (s) +
bi (s)up (s)ds ,
i=1
t0
αi (t0 )
and its variants, under some suitable conditions on the functions involved in (1.2), including the
constant p ≥ 0, p 6= 1, or p > 1. We also prove the two independent variable generalization
of the result and present some applications of those to the global existence of solutions of
differential equations with time delay.
2. T HE I NTEGRAL I NEQUALITIES
We shall introduce some notation, R denotes the set of real numbers and R+ = [0, ∞),
I = [t0 , T ) are the given subsets of R. The first order derivative of a function z(t) with respect
to t will be denoted by z 0 (t). Let C(M, N ) denote the class of continuous functions from the
set M to the set N. In the following theorems we prove some nonlinear integral inequalities.
Theorem 2.1. Let u, ai , bi ∈ C(I, R+ ), αi ∈ C 1 (I, I) be nondecreasing with αi (t) ≤ t on I,
for i = 1, 2, . . . , n. Let p ≥ 0, p 6= 1, and c ≥ 0 be constants. If
#
"Z
Z αi (t)
n
t
X
(2.1)
u(t) ≤ c +
ai (s)up (s) ds +
bi (s)up (s) ds
i=1
t0
αi (t0 )
for t ∈ I, then
"
(2.2)
u(t) ≤ cq + q
Z
n
X
i=1
t
t0
Z
αi (t)
!# 1q
bi (s) ds
ai (s) ds +
αi (t0 )
for t ∈ [t0 , T1 ), where q = 1 − p and T1 is chosen so that the expression inside [. . . ] is positive
on the subinterval [t0 , T1 ).
Proof. From the hypotheses we observe that α0 (t) ≥ 0 for t ∈ I. Let c ≥ 0 and define a
function z(t) by the right-hand side of (2.1). Then, z(t0 ) = c, z(t) is nondecreasing for t ∈ I,
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
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E XPLICIT B OUNDS O N SOME N ONLINEAR I NTEGRAL I NEQUALITIES
3
u(t) ≤ z(t), and
z 0 (t) =
n
X
[ai (t)up (t) + bi (αi (t))up (αi (t))αi0 (t)]
i=1
≤
n
X
[ai (t)z p (t) + bi (αi (t))z p (αi (t))αi0 (t)]
i=1
≤
n
X
[ai (t) + bi (αi (t))αi0 (t)][z(t)]p .
i=1
By making the constant q = 1 − p and using the function z1 (t) = z q (t)/q we get
n
X
0
(2.3)
z1 (t) ≤
[ai (t) + bi (αi (t))αi0 (t)].
i=1
By taking t = s in (2.3) and integrating it with respect to s from t0 to t, t ∈ I, we obtain
Z t
Z t
n Z t
X
0
0
(2.4)
z1 (s) ds ≤
ai (s) ds +
bi (αi (s))αi (s) ds .
t0
t0
i=1
t0
Integrating, making the change of the function on the left side in (2.4) and rewriting yields
"Z
#
Z αi (t)
n
t
cq X
z q (t)
≤
+
ai (s) ds +
bi (s) ds
q
q
t0
αi (t0 )
i=1
for t ∈ I. It follows that
"
(2.5)
z(t) ≤ cq + q
Z
n
X
i=1
t
Z
ai (s) ds +
t0
!# 1q
αi (t)
bi (s) ds
αi (t0 )
for t ∈ [t0 , T1 ), where T1 is chosen so that the expression inside [. . . ] is positive on the subinterval [t0 , T1 ). Using (2.5) in u(t) ≤ z(t) we get the inequality in (2.2).
Corollary 2.2. Let u, bi ∈ C(I, R+ ), αi ∈ C 1 (I, I) be nondecreasing with αi (t) ≤ t on I for
i = 1, . . . , n, and let p ≥ 0, p 6= 1, and c ≥ 0 be constants. If
n Z αi (t)
X
u(t) ≤ c +
bi (s)up (s) ds
i=1
αi (t0 )
for t ∈ I, then
"
u(t) ≤ cq + q
n Z
X
i=1
# 1q
αi (t)
bi (s) ds
αi (t0 )
for t ∈ [t0 , T1 ), where q = 1 − p and T1 is chosen so that the expression inside [. . . ] is positive
on the subinterval [t0 , T1 ).
The following theorem deals with the constant 1 < p < ∞ versions of the inequalities
established in Theorem 2.1.
Theorem 2.3. Let u, a, bi , ci ∈ C(I, R+ ), αi ∈ C 1 (I, I) be nondecreasing with αi (t) ≤ t on I,
for i = 1, 2, . . . , n. Also, let a(t) be nondecreasing in t, t ∈ I and p > 1 be constants. If
"Z
#
Z αi (t)
n
t
X
(2.6)
u(t) ≤ a(t) +
bi (s)up (s) ds +
ci (s)up (s) ds
i=1
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
t0
αi (t0 )
http://jipam.vu.edu.au/
4
YOUNG -H O K IM
for t ∈ I, then
"
(2.7)
Z
n
X
u(t) ≤ a(t) 1 − (p − 1)ap−1 (t)
T = sup t ∈ I : (p − 1)ap−1 (t)
Z
n
X
i=1
Z
1
!# 1−p
αi (t)
bi (s) ds +
ci (s) ds
t0
i=1
for t ∈ [t0 , T ), where
(
t
αi (t0 )
t
Z
bi (s) ds +
t0
!
αi (t)
ci (s) ds
)
<1 .
αi (t0 )
Proof. From the hypotheses we observe that α0 (t) ≥ 0 for t ∈ I. Define a function v(t) by
"Z
#
Z αi (t)
n
t
X
v(t) =
bi (s)up (s) ds +
ci (s)up (s) ds .
t0
i=1
αi (t0 )
Then, v(t0 ) = 0, v(t) is nondecreasing for t ∈ I, u(t) ≤ a(t) + v(t), and
n
X
0
v (t) ≤
[bi (t) + ci (αi (t))αi0 (t)][a(t) + z(t)]p
i=1
≤ R(t)[a(t) + v(t)],
where
R(t) =
n
X
[bi (t) + ci (αi (t))αi0 (t)][a(t) + z(t)]p−1 .
i=1
Now by the comparison result for linear differential inequalities(see [1, Lemma 1.1., p. 2]), this
implies that
Z t
Z t
R(τ ) dτ ds
v(t) ≤
R(s)a(s) exp
t0
s
Z t
Z t
(2.8)
≤ a(t)
R(s) exp
R(τ ) dτ ds
t0
s
for s ≥ t0 . By integrating on the right hand side in (2.8) we get
Z t
(2.9)
v(t) + a(t) ≤ a(t) exp
R(τ ) dτ .
t0
From (2.9) we successively obtain
p−1
≤a
[v(t) + a(t)]
p−1
Z
t
(p − 1)R(τ ) dτ
(t) exp
,
t0
R(t) ≤ a
p−1
Z
t
(p − 1)R(τ ) dτ
(t) exp
t0
X
n
[bi (t) + ci (αi (t))αi0 (t)]
i=1
and
A(t) = (p − 1)R(t)
≤ (p − 1)a
p−1
Z
t
A(τ ) dτ
(t) exp
t0
X
n
[bi (t) + ci (αi (t))αi0 (t)].
i=1
Consequently, we get
Z t
n
X
p−1
A(t) exp −
A(τ ) dτ ≤ (p − 1)a (t)
[bi (t) + ci (αi (t))αi0 (t)],
t0
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
i=1
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E XPLICIT B OUNDS O N SOME N ONLINEAR I NTEGRAL I NEQUALITIES
5
or
(2.10)
Z t
n
X
d
p−1
− exp −
A(τ ) dτ
≤ (p − 1)a (t)
[bi (t) + ci (αi (t))αi0 (t)],
dt
t0
i=1
By taking t = s in (2.10) and integrating it with respect to s from t0 to t, t ∈ I, we obtain
Z t
(2.11)
1 − exp −
A(τ ) dτ
t0
≤ (p − 1)a
p−1
(t)
n Z
X
t
Z
ci (αi (s))αi0 (s) ds
bi (s) ds +
t0
i=1
t
.
t0
Making the change of the variables on the right side in (2.11) and rewriting yields
1
!# 1−p
Z t
"
Z t
Z αi (t)
n
X
exp
R(τ ) dτ ≤ 1 − (p − 1)ap−1 (t)
bi (s) ds +
ci (s) ds
t0
t0
i=1
αi (t0 )
for t ∈ [t0 , T ), where T is chosen so that the expression inside [. . . ] is positive in the subinterval
[t0 , T ). This, together with (2.9) and u(t) ≤ a(t) + v(t), gives the inequality in (2.7).
In the following theorem we establish two independent-variable versions of Theorem 2.1
which can be used in the qualitative analysis of hyperbolic partial differential equations with
retarded arguments. Let 4 = I1 × I2 , where I1 = [x0 , X), I2 = [y0 , Y ) are the given subsets of
the real numbers, R. The first order partial derivatives of a function z(x, y) defined for x, y ∈ R
with respect to x and y are denoted by ∂z(x, y)/∂x and ∂z(x, y)/∂y respectively.
Theorem 2.4. Let u, ai , bi ∈ C(4, R+ ), αi ∈ C 1 (I1 , I1 ), βi ∈ C 1 (I2 , I2 ) be nondecreasing
with αi (x) ≤ x on I1 , βi (y) ≤ y on I2 , for i = 1, 2, . . . , n. Let p ≥ 0, p 6= 1, and c ≥ 0 be
constants. If
n Z x Z y
X
(2.12)
u(x, y) ≤ c +
ai (s, t)up (s, t) dt ds
i=1
x0
y0
Z
αi (x)
bi (s, t)up (s, t) dt ds
+
αi (x0 )
for (x, y) ∈ I1 × I2 , then
"
(2.13)
u(x, y) ≤ cq + q
n Z
X
i=1
x
x0
Z
y
!
βi (y)
Z
βi (y0 )
Z
αi (x)
Z
bi (s, t) dt ds
ai (s, t) dt ds +
y0
!# 1q
βi (y)
αi (x0 )
βi (y0 )
for (x, y) ∈ [x0 , X) × [y0 , Y ), where q = 1 − p and X, Y are chosen so that the expression
inside [. . . ] is positive on the subintervals [x0 , X) and [y0 , Y ).
Proof. The details of the proof of Theorem 2.4 follow by an argument similar to that in the
proof of Theorem 2.1 with suitable changes. From the hypotheses we observe that α0 (x) ≥ 0
for x ∈ I1 and β 0 (y) ≥ 0 for y ∈ I2 . Let c ≥ 0 and define a function z(x, y) by the righthand side of (2.12). Then, z(x0 , y) = z(x, y0 ) = c, z(x, y) is nondecreasing for (x, y) ∈ 4,
u(x, y) ≤ z(x, y), and
"Z
#
Z βi (y)
n
y
X
∂
z(x, y) ≤
ai (x, t) dt +
bi (αi (x), t)αi0 (x) dt [z(x, y)]p .
∂x
y0
βi (y0 )
i=1
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
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6
YOUNG -H O K IM
By making the constant q = 1 − p and using the function v(x, y) = z q (x, y)/q we get
"Z
#
Z βi (y)
n
y
X
∂
(2.14)
v(x, y) ≤
ai (x, t) dt +
bi (αi (x), t)αi0 (x) dt .
∂x
y0
βi (y0 )
i=1
By taking x = s in (2.14) and integrating it with respect to s from x0 to x, x ∈ I1 , we obtain
#
"Z Z
Z x
Z αi (x) Z βi (y)
n
x
y
X
∂
(2.15)
v(s, y) ds ≤
ai (s, t) dt ds +
bi (s, t) dt ds .
x0 ∂s
x
y
α
(x
)
β
(y
)
0
0
0
0
i
i
i=1
Integrating with respect to s from x0 to x, making the change of the function on the left side in
(2.15) and rewriting yields
"Z Z
#
Z αi (x) Z βi (y)
n
x
y
z q (x, y)
cq X
≤
+
ai (s, t) dt ds +
bi (s, t) dt ds
q
q
x0 y 0
αi (x0 ) βi (y0 )
i=1
for (x, y) ∈ 4. This implies
"
(2.16)
z(x, y) ≤ cq + q
n Z
X
x
y
Z
αi (x)
Z
!# 1q
βi (y)
ai (s, t) dt ds +
x0
i=1
Z
bi (s, t) dt ds
y0
αi (x0 )
βi (y0 )
for (x, y) ∈ [x0 , X) × [y0 , Y ), where X, Y are chosen so that the expression inside [. . . ] is
positive on the subintervals [x0 , X) and [y0 , Y ). Using (2.16) in u(x, y) ≤ z(x, y) we get the
inequality in (2.13).
The following theorem deals with the constant 1 < p < ∞ versions of the inequalities
established in Theorem 2.4.
Theorem 2.5. Let u, ai , bi ∈ C(4, R+ ), αi ∈ C 1 (I1 , I1 ), βi ∈ C 1 (I2 , I2 ) be nondecreasing
with αi (x) ≤ x on I1 , βi (y) ≤ y on I2 , for i = 1, 2, . . . , n. Let a(x, y) be nondecreasing in
(x, y) ∈ 4 and 1 < p < ∞ be constant. If
n Z x Z y
X
(2.17) u(x, y) ≤ a(x, y) +
bi (s, t)up (s, t) dt ds
x0
i=1
y0
Z
αi (x)
Z
+
αi (x0 )
#
βi (y)
ci (s, t)up (s, t) dt ds
βi (y0 )
for (x, y) ∈ I1 × I2 , then
"
(2.18)
u(x, y) ≤ a(x, y) 1 − (p − 1)ap−1 (x, y)
n
X
1
# 1−p
Ψi (x, y)
i=1
for (x, y) ∈ 41 , where
Z
x
Z
y
Ψi (x, y) =
Z
αi (x)
Z
βi (y)
bi (s, t) dt ds +
x0
y0
ci (s, t) dt ds
αi (x0 )
βi (y0 )
and
(
41 = sup (x, y) ∈ 4 : (p − 1)ap−1 (x, y)
n
X
)
ξi (x, y) < 1 .
i=1
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
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E XPLICIT B OUNDS O N SOME N ONLINEAR I NTEGRAL I NEQUALITIES
7
Proof. The details of the proof of Theorem 2.5 follows by an argument similar to that in the
proof of Theorem 2.3 with suitable changes. From the hypotheses we observe that α0 (x) ≥ 0
for x ∈ I1 and β 0 (y) ≥ 0 for y ∈ I2 . Define a function v(x, y) by
#
Z αi (x) Z βi (y)
n Z x Z y
X
ci (s, t)up (s, t) dt ds .
v(x, y) =
bi (s, t)up (s, t) dt ds +
i=1
x0
αi (x0 )
y0
βi (y0 )
Then, v(x0 , y) = v(x, y0 ) = 0, v(x, y) is nondecreasing for (x, y) ∈ 4, u(x, y) ≤ a(x, y) +
v(x, y), and
∂
v(x, y) ≤ R(x, y)[a(x, y) + v(x, y)],
∂x
where
"Z
#
Z βi (y)
n
y
X
R(x, y) =
bi (x, t) dt +
ci (α(x), t)α0 (x) dt [a(x, y) + v(x, y)]p−1 .
i=1
y0
βi (y0 )
Now by keeping y fixed and using the comparison result for linear differential inequalities (see
[1, Lemma 1.1., p. 2]), this implies that
Z x
Z x
(2.19)
v(x, y) ≤
R(s, y)ai (s, y) exp
R(τ, y) dτ ds
x0
s
for s ≥ x0 . By integrating on the right hand side in (2.19) we get
Z x
(2.20)
v(x, y) + a(x, y) ≤ a(x, y) exp
R(τ, y) dτ .
x0
From (2.20) we successively obtain
p−1
[v(x, y) + a(x, y)]
p−1
≤a
Z
x
(p − 1)R(τ, y) dτ
(x, y) exp
,
x0
A(x, y) = (p − 1)R(x, y)
≤ (p − 1)a
p−1
x
Z
(x, y) exp
R(τ, y) dτ
X
n
x0
ψi (x, y),
i=1
where
y
Z
ψi (x, y) =
Z
βi (y)
bi (x, t) dt +
y0
ci (α(x), t)α0 (x) dt.
βi (y0 )
Consequently, we have
Z x
n
X
∂
(2.21)
− exp −
A(τ, y) dτ
≤ (p − 1)ap−1 (x, y)
ψi (x, y).
∂x
x0
i=1
By taking x = s in (2.21) and integrating it with respect to s from x0 to x, x ∈ I1 , we obtain
Z x
n
X
p−1
(2.22)
1 − exp −
A(τ, y) dτ ≤ (p − 1)a (x, y)
Ψi (x, y),
x0
i=1
where
Z
x
Z
y
Ψi (x, y) =
Z
αi (x)
Z
βi (y)
bi (s, t) dt ds +
x0
y0
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
ci (s, t) dt ds.
αi (x0 )
βi (y0 )
http://jipam.vu.edu.au/
8
YOUNG -H O K IM
Making the change of the function on the inequality (2.22) and rewriting yields
1
# 1−p
"
Z x
n
X
R(τ, y) dτ ≤ 1 − (p − 1)ap−1 (x, y)
Ψi (x, y)
exp
x0
i=1
for (x, y) ∈ 41 , where 41 is chosen so that the expression inside [. . . ] is positive in the subinterval 41 . This, together with (2.20) and u(x, y) ≤ a(x, y) + v(x, y), gives the inequality in
(2.18).
3. A PPLICATIONS
In this section we will show that our results are useful in proving the global existence of solutions to certain differential equations with time delay. First consider the functional differential
equation involving several retarded arguments with the initial condition
( 0
x (t) = F (t, x(t), x(t − h1 (t)), . . . , x(t − hn (t)), t ∈ I,
(3.1)
x(t0 ) = x0 ,
where x0 is constant, F ∈ C(I × Rn+1 , R) and for i = 1, . . . , n, let hi ∈ C 1 (I, R+ ) be
nonincreasing and such that t − hi (t) ≥ 0, x − hi (t) ∈ C 1 (I, I), h0i (t) < 1, and h(t0 ) = 0. The
following theorem deals with a bound on the solution of the problem (3.1).
Theorem 3.1. Assume that F : I × Rn+1 → R is a continuous function for which there exist
continuous nonnegative functions ai (t), bi (t) for t ∈ I such that
n
X
(3.2)
|F (t, v, u1 , . . . , un )| ≤
[ai (t) |v|p + bi (t) |ui |p ],
i=1
where p ≥ 0, p 6= 1 is constant, and let
1
,
i = 1, . . . , n.
t∈I 1 − h0i (x)
If x(t) is any solution of the problem (3.1), then
!# 1q
"
Z t
Z t−hi (t)
n
X
|x(t)| ≤ |x0 |q + q
ai (σ) dσ +
bi (σ) dσ
Mi = max
(3.3)
i=1
t0
t0
for t ∈ I, where bi (σ) = Mi bi (σ + hi (s)), σ, s ∈ I.
Proof. The solution x(t) of the problem (3.1) can be written as
Z t
(3.4)
x(t) = x0 +
F (s, x(s), x(s − h1 (s)), . . . , x(s − hn (s)) ds.
t0
From (3.2), (3.3), (3.4) and making the change of variables we have
Z t
n Z t
X
p
p
|x(t)| ≤ |x0 | +
ai (s)|x(s)| ds +
bi (s)|x(s − hi (s))| ds
(3.5)
≤ |x0 | +
i=1
n
X
i=1
t0
Z
t0
t
t0
ai (s)|x(s)|p ds +
Z
!
t−hi (t)
bi (σ)|x(σ)|p dσ
t0
for t ∈ I, where bi (σ) = Mi bi (σ + hi (s)), σ, s ∈ I. Now a suitable application of the inequality
in Theorem 2.1 to (3.5) yields the result.
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
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E XPLICIT B OUNDS O N SOME N ONLINEAR I NTEGRAL I NEQUALITIES
9
Remark 3.2.
(i) For 1 < p < ∞, a suitable application of the inequality in Theorem 2.3 to (3.5) yields
the following result
1
"
!# 1−p
Z t
Z t−hi (t)
n
X
|x(t)| ≤ |x0 | − (p − 1)|x0 |p
ai (s) ds +
.
bi (σ) dσ
i=1
t0
t0
This shows in particular that the solution x(t) is bounded on [t0 , T ), where T is chosen
so that the expression inside [. . . ] is positive in the subinterval [t0 , T ).
(ii) Consider the functional differential equation
(
x0 (t) = F (t, x(t − h1 (t)), . . . , x(t − hn (t)), t ∈ I,
(3.6)
x(t0 ) = x0 .
Assume that F : I×Rn+1 → R is a continuous function for which there exist continuous
nonnegative functions bi (t) for t ∈ I such that
(3.7)
|F (t, u1 , . . . , un )| ≤
n
X
bi (t) |ui | ,
i=1
(3.8)
where p ≥ 0, p 6= 1 is constant. Let Mi be a function defined by (3.3). If x(t) is any
solution of (3.6), the solution x(t) can be written as
Z t
x(t) = x0 +
F (s, x(s − h1 (s)), . . . , x(s − hn (s)) ds.
t0
(3.9)
From (3.7), (3.8) and making the change of variables we have
n Z t−hi (t)
X
bi (σ)|x(σ)|p dσ
|x(t)| ≤ |x0 | + p
i=1
t0
for t ∈ I, where bi (σ) = Mi bi (σ + hi (s)), σ, s ∈ I. Now a suitable application of the
inequality in Corollary 2.2 to (3.8) yields
"
# 1q
n Z t−hi (t)
X
|x(t)| ≤ |x0 |q + q
bi (σ) dσ
i=1
t0
for t ∈ I, where q = 1 − p, bi (σ) = Mi bi (σ + hi (s)), σ, s ∈ I.
In the following we present an application of the inequality given in Section 2 to study the
boundedness of the solutions of the initial boundary value problem for hyperbolic partial delay
differential equations of the form
 ∂
∂

z(x, y)

∂y
∂x

= F (x, y, z(x, y), z(x − h1 (x), y − g1 (y)), . . . , z(x − hn (x), y − gn (y)),
(3.10)



z(x, y0 ) = e1 (x), z(x0 , y) = e2 (y), e1 (x0 ) = e2 (y0 ) = 0,
where p > 0, p 6= 1 is constant, F ∈ C(4 × Rn+1 , R), e1 ∈ C 1 (I1 , R), e2 ∈ C 1 (I2 , R), and
hi ∈ C 1 (I1 , R+ ), gi ∈ C 1 (I2 , R+ ) are nonincreasing and such that x − hi (x) ≥ 0, x − hi (x) ∈
C 1 (I1 , I1 ), y − gi (y) ≥ 0, y − hi (y) ∈ C 1 (I2 , I2 ), h0i (t) < 1, gi0 (t) < 1, and hi (x0 ) = gi (y0 ) = 0
for i = 1, . . . , n.
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
http://jipam.vu.edu.au/
10
YOUNG -H O K IM
Theorem 3.3. Assume that F : 4 × Rn+1 → R is a continuous function for which there exists
continuous nonnegative functions ai (x, y), bi (x, y) for i = 1, . . . , n; x ∈ I1 , y ∈ I2 such that
(
P
|F (x, y, v, u1 , . . . , un )| ≤ ni=1 [ai (x, y) |v|p + bi (x, y) |ui |p ],
(3.11)
|e1 (x) + e2 (y)| ≤ c
for c ≥ 0, p ≥ 0, p 6= 1, and let
(3.12)
Mi = max
x∈I1
1
,
1 − h0i (x)
Ni = max
y∈I2
1
,
1 − gi0 (y)
If z(x, y) is any solution of the problem (3.10), then
"
Z
n Z x Z y
X
q
|z(x, y)| ≤ c + q
ai (σ, τ ) dτ dσ+
x0
i=1
y0
φ(x)
i = 1, . . . , n.
Z
!# 1q
ψ(y)
bi (σ, τ )] dτ dσ
x0
y0
for (x, y) ∈ 41 , where q = 1 − p, φ(x) = x − hi (x), ψ(y) = y − gi (y) and b(σ, τ ) =
Mi Ni bi (σ + hi (s), τ + gi (t)), σ, s ∈ I1 , τ, t ∈ I2 .
Proof. It is easy to see that the solution z(x, y) of the problem (3.10) satisfies the equivalent
integral equation
Z xZ y
(3.13) z(x, y) = e1 (x) + e2 (y) +
F (s, t, z(s, t), z(s − h1 (s), t − g1 (t)),
x0
y0
. . . , z(s − hn (s), t − gn (t)) dt ds.
From (3.11), (3.13) and making the change of variables, we have
Z xZ yX
n
(3.14)
|z(x, y)| ≤ c +
ai |z(s, t)|p + bi |z(s − hi (s), t − gi (t))|p ds dt.
x0
y0 i=1
Now a suitable application of the inequality given in Theorem 2.4 to (3.14) yields the desired
result.
Remark 3.4. For 1 < p < ∞, a suitable application of the inequality in Theorem 2.5 to (3.14)
yields the following result
1
"
# 1−p
n
X
,
|z(x, y)| ≤ c − cp (p − 1)
Ψi (x, y)
i=1
where
Z
x
Z
y
Ψi (x, y) =
Z
φ(x)
Z
ψ(y)
ai (σ, τ ) dτ dσ +
x0
y0
bi (σ, τ ) dτ dσ.
x0
y0
R EFERENCES
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Timisoara, No. 29, 1987.
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Kraqujevac, Sec. Math. Tech. Sci., 13 (1992), 23–28.
[4] O. LIPOVAN, A retarded integral inequality and its applications, J. Math. Anal. Appl., 285 (2003),
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E XPLICIT B OUNDS O N SOME N ONLINEAR I NTEGRAL I NEQUALITIES
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[5] L. QU-IANG, The boundedness of solutions of linear differential equations y 00 + A(t)y = 0, Shuxue
Jinzhan, 3 (1957), 409–415.
[6] B.G. PACHPATTE, On some new inequalities related to certain inequalities in the theory of differential equations, J. Math. Anal. Appl., 189 (1995), 128–144.
[7] B.G. PACHPATTE, Inequalities for Differential and Integral Equations, Academic Press, New York,
1998.
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Math., 5(3) (2004), Art. 80. [ONLINE: http://jipam.vu.edu.au/article.php?sid=
436].
J. Inequal. Pure and Appl. Math., 7(2) Art. 48, 2006
http://jipam.vu.edu.au/