Journal of Inequalities in Pure and
Applied Mathematics
EXPLICIT BOUNDS ON SOME NONLINEAR
INTEGRAL INEQUALITIES
volume 7, issue 2, article 48,
2006.
YOUNG-HO KIM
Department of Applied Mathematics
Changwon National University
Changwon 641-773, Korea.
Received 24 September, 2005;
accepted 09 November, 2005.
Communicated by: D. Hinton
EMail: yhkim@sarim.changwon.ac.kr
Abstract
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c 2000 Victoria University
ISSN (electronic): 1443-5756
288-05
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Abstract
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.
2000 Mathematics Subject Classification: 26D10, 26D15, 34K10, 35B35.
Key words: Integral inequalities, Retarded integral inequality, Retarded differential
equations, Partial delay differential equations.
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
This research is financially supported by Changwon National University in 2005.
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Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3
Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
References
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1.
Introduction
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
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
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for t ∈ R+ , then
Z
u(t) ≤ u0
t
f (s)ds,
t ∈ R+ .
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.
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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:
(1.1)
p
u (t) ≤ c + p
n Z
X
i=1
αi (t)
[ai (s)up (s) + bi (s)u(s)]ds,
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 )
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
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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.
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2.
The Integral Inequalities
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
Young-Ho Kim
αi (t0 )
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for t ∈ I, then
Contents
"
(2.2)
Explicit Bounds On some
Nonlinear Integral Inequalities
q
u(t) ≤ c + q
Z
n
X
i=1
t
t0
Z
αi (t)
ai (s) ds +
!# 1q
bi (s) ds
αi (t0 )
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for t ∈ [t0 , T1 ), where q = 1 − p and T1 is chosen so that the expression inside
[. . . ] is positive on the subinterval [t0 , T1 ).
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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
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nondecreasing for t ∈ I, 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
z10 (t) ≤
(2.3)
n
X
[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
i=1
t0
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
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
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for t ∈ I. It follows that
"
(2.5)
z(t) ≤ cq + q
Z
n
X
t
Z
ai (s) ds +
t0
i=1
!# 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
u(t) ≤ c +
n Z
X
i=1
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
αi (t)
bi (s)up (s) ds
αi (t0 )
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for t ∈ I, then
"
q
u(t) ≤ c + 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
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and p > 1 be constants. If
(2.6)
u(t) ≤ a(t) +
"Z
n
X
i=1
t
bi (s)up (s) ds +
t0
Z
#
αi (t)
ci (s)up (s) ds
αi (t0 )
for t ∈ I, then
"
(2.7) u(t) ≤ a(t) 1 − (p − 1)ap−1 (t)
×
Z
n
X
T = sup t ∈ I : (p − 1)a
p−1
(t)
Z
n
X
i=1
Z
1
!# 1−p
αi (t)
bi (s) ds +
Young-Ho Kim
αi (t0 )
t
Z
!
αi (t)
bi (s) ds +
t0
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 .
i=1
t0
Explicit Bounds On some
Nonlinear Integral Inequalities
ci (s) ds
t0
i=1
for t ∈ [t0 , T ), where
(
t
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
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i=1
≤ R(t)[a(t) + v(t)],
J. Ineq. Pure and Appl. Math. 7(2) Art. 48, 2006
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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
v(t) ≤
R(s)a(s) exp
R(τ ) dτ ds
t0
s
Z t
Z t
(2.8)
≤ a(t)
R(s) exp
R(τ ) dτ ds
t0
Explicit Bounds On some
Nonlinear Integral Inequalities
s
Young-Ho Kim
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 .
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t0
Contents
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) ≤ ap−1 (t) exp
Z
t
(p − 1)R(τ ) dτ
t0
X
n
[bi (t) + ci (αi (t))αi0 (t)]
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i=1
and
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A(t) = (p − 1)R(t)
≤ (p − 1)ap−1 (t) exp
Page 9 of 23
Z
t
A(τ ) dτ
t0
X
n
i=1
[bi (t) + ci (αi (t))αi0 (t)].
J. Ineq. Pure and Appl. Math. 7(2) Art. 48, 2006
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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
i=1
or
Z t
d
− exp −
A(τ ) dτ
(2.10)
dt
t0
≤ (p − 1)a
p−1
(t)
n
X
[bi (t) + ci (αi (t))αi0 (t)],
Explicit Bounds On some
Nonlinear Integral Inequalities
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
i=1
t
Z
t
ci (αi (s))αi0 (s) ds
bi (s) ds +
t0
.
t0
Making the change of the variables on the right side in (2.11) and rewriting
yields
Z t
exp
R(τ ) dτ
t0
"
≤ 1 − (p − 1)ap−1 (t)
n
X
i=1
Z
t
Z
bi (s) ds +
t0
1
!# 1−p
αi (t)
ci (s) ds
αi (t0 )
Young-Ho Kim
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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
(2.12) u(x, y) ≤ c +
n Z
X
x
x0
i=1
Z
y
y0
αi (x) Z
αi (x0 )
!
βi (y)
bi (s, t)up (s, t) dt ds
+
Young-Ho Kim
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ai (s, t)up (s, t) dt ds
Z
Explicit Bounds On some
Nonlinear Integral Inequalities
βi (y0 )
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for (x, y) ∈ I1 × I2 , then
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(2.13) u(x, y)
"
≤ cq + q
n Z x
X
i=1
x0
Z
y
Z
αi (x)
Z
ai (s, t) dt ds +
y0
!# 1q
βi (y)
bi (s, t) dt ds
αi (x0 )
βi (y0 )
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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 right-hand 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
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
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
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Z
x
(2.15)
x0
∂
v(s, y) ds
∂s
"Z
n
X
≤
i=1
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x
x0
Z
y
Z
αi (x)
Z
βi (y)
ai (s, t) dt ds +
y0
#
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bi (s, t) dt ds .
αi (x0 )
βi (y0 )
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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
x
y
α
(x
)
β
(y
)
0
0
0
0
i
i
i=1
for (x, y) ∈ 4. This implies
(2.16) z(x, y)
"
q
≤ c +q
n Z
X
i=1
x
x0
Z
y
Z
αi (x)
Z
ai (s, t) dt ds +
y0
!# 1q
βi (y)
bi (s, t) dt ds
αi (x0 )
βi (y0 )
Young-Ho Kim
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
i=1
x0
Explicit Bounds On some
Nonlinear Integral Inequalities
y0
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Z
αi (x)
Z
+
αi (x0 )
#
βi (y)
βi (y0 )
ci (s, t)up (s, t) dt ds
J. Ineq. Pure and Appl. Math. 7(2) Art. 48, 2006
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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 xZ
Ψi (x, y) =
x0
y
y0
Z
αi (x)
Z
βi (y)
bi (s, t) dt ds +
ci (s, t) dt ds
αi (x0 )
Explicit Bounds On some
Nonlinear Integral Inequalities
βi (y0 )
and
Young-Ho Kim
(
41 = sup (x, y) ∈ 4 : (p − 1)ap−1 (x, y)
n
X
)
ξi (x, y) < 1 .
i=1
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
v(x, y) =
bi (s, t)up (s, t) dt ds +
ci (s, t)up (s, t) dt ds .
i=1
x0
y0
αi (x0 )
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βi (y0 )
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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
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where
R(x, y) =
"Z
n
X
i=1
y
Z
βi (y)
bi (x, t) dt +
y0
#
ci (α(x), t)α0 (x) dt [a(x, y)+v(x, y)]p−1 .
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 .
Young-Ho Kim
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x0
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From (2.20) we successively obtain
p−1
[v(x, y) + a(x, y)]
≤a
p−1
Z
x
(p − 1)R(τ, y) dτ
(x, y) exp
x0
≤ (p − 1)a
Z
x
(x, y) exp
R(τ, y) dτ
x0
where
Z
y
ψi (x, y) =
Z
βi (y)
bi (x, t) dt +
y0
,
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A(x, y) = (p − 1)R(x, y)
p−1
Explicit Bounds On some
Nonlinear Integral Inequalities
βi (y0 )
X
n
Close
ψi (x, y),
i=1
ci (α(x), t)α0 (x) dt.
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Consequently, we have
(2.21)
Z x
n
X
∂
p−1
− exp −
A(τ, y) dτ
≤ (p − 1)a (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
x
Z
1 − exp −
(2.22)
A(τ, y) dτ
≤ (p − 1)a
p−1
(x, y)
x0
n
X
Ψi (x, y),
Explicit Bounds On some
Nonlinear Integral Inequalities
i=1
Young-Ho Kim
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 )
Contents
βi (y0 )
Making the change of the function on the inequality (2.22) and rewriting yields
Z
x
exp
R(τ, y) dτ
x0
"
≤ 1 − (p − 1)ap−1 (x, y)
n
X
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1
# 1−p
Ψi (x, y)
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).
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3.
Applications
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
(3.2)
|F (t, v, u1 , . . . , un )| ≤
n
X
[ai (t) |v|p + bi (t) |ui |p ],
where p ≥ 0, p 6= 1 is constant, and let
Mi = max
t∈I
1
,
1 − h0i (x)
Young-Ho Kim
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i=1
(3.3)
Explicit Bounds On some
Nonlinear Integral Inequalities
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i = 1, . . . , n.
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If x(t) is any solution of the problem (3.1), then
"
|x(t)| ≤ |x0 |q + q
Z
n
X
i=1
t
!# 1q
t−hi (t)
Z
ai (σ) dσ +
bi (σ) dσ
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.
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
t0
From (3.2), (3.3), (3.4) and making the change of variables we have
Title Page
|x(t)| ≤ |x0 | +
(3.5)
≤ |x0 | +
n Z
X
i=1
n
X
i=1
t
ai (s)|x(s)|p ds +
t0
Z
bi (s)|x(s − hi (s))|p ds
t0
t
t0
t
Z
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.
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Remark 1.
(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
bi (σ) dσ
|x(t)| ≤ |x0 | − (p − 1)|x0 |p
ai (s) ds +
.
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)),
(3.6)
x(t0 ) = x0 .
t ∈ I,
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
Young-Ho Kim
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bi (t) |ui | ,
i=1
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
(3.8)
x(t) = x0 +
F (s, x(s − h1 (s)), . . . , x(s − hn (s)) ds.
t0
Explicit Bounds On some
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From (3.7), (3.8) and making the change of variables we have
(3.9)
|x(t)| ≤ |x0 | + p
n Z
X
i=1
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 Corollary 2.2 to (3.8) yields
"
|x(t)| ≤ |x0 |q + q
n Z
X
i=1
# 1q
t−hi (t)
bi (σ) dσ
t0
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
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)
= F (x, y, z(x, y), z(x − h1 (x), y − g1 (y)), . . . ,


 ∂y ∂x
z(x − hn (x), y − gn (y)),
(3.10)



z(x, y0 ) = e1 (x), z(x0 , y) = e2 (y), e1 (x0 ) = e2 (y0 ) = 0,
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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.
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Theorem 3.2. 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)
x0
Explicit Bounds On some
Nonlinear Integral Inequalities
i = 1, . . . , n.
Young-Ho Kim
Z
!# 1q
ψ(y)
bi (σ, τ )] dτ dσ
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
(3.13) z(x, y) = e1 (x) + e2 (y)
Z xZ y
+
F (s, t, z(s, t), z(s − h1 (s), t − g1 (t)),
x0
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y0
. . . , z(s − hn (s), t − gn (t)) dt ds.
J. Ineq. Pure and Appl. Math. 7(2) Art. 48, 2006
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From (3.11), (3.13) and making the change of variables, we have
(3.14)
|z(x, y)|
Z
≤c+
x
y
Z
x0
n
X
ai |z(s, t)|p + bi |z(s − hi (s), t − gi (t))|p ds dt.
y0 i=1
Now a suitable application of the inequality given in Theorem 2.4 to (3.14)
yields the desired result.
Remark 2. For 1 < p < ∞, a suitable application of the inequality in Theorem
2.5 to (3.14) yields the following result
"
|z(x, y)| ≤ c − cp (p − 1)
n
X
Ψi (x, y)
,
y
Ψi (x, y) =
Z
φ(x) Z
ψ(y)
ai (σ, τ ) dτ dσ +
x0
y0
bi (σ, τ ) dτ dσ.
x0
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where
xZ
Young-Ho Kim
1
# 1−p
i=1
Z
Explicit Bounds On some
Nonlinear Integral Inequalities
y0
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References
[1] D. BAINOV AND P. SIMEONOV, Integral Inequalities and Applications,
Kluwer Academic Publishers, Dordrecht, 1992.
[2] S.S. DRAGOMIR,
The Gronwall Type Lemmas and Applications,Monografii Mathematica, Univ. Timisoara, No. 29, 1987.
[3] S.S. DRAGOMIR, On some nonlinear generalizations of Gronwall’s inequality, Coll. Sci. Fac. Sci. 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), 336–443.
[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.
[8] B.G. PACHPATTE, Explicit bounds on certain integral inequalities, J. Math.
Anal. Appl., 267 (2002), 48–61.
[9] B.G. PACHPATTE, On some new nonlinear retarded integral inequalities,
J. Inequal. Pure and Appl. Math., 5(3) (2004), Art. 80. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=436].
Explicit Bounds On some
Nonlinear Integral Inequalities
Young-Ho Kim
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