An. Şt. Univ. Ovidius Constanţa
Vol. 18(2), 2010, 149–160
SOME NONLINEAR DELAY INTEGRAL
INEQUALITIES AND THEIR DISCRETE
ANALOGUES
Wei Nian Li
Abstract
In this paper, we investigate some nonlinear delay integral inequalities and their analogues which provide explicit bounds on unknown functions. The inequalities given here can be used as tools in the qualitative
theory of certain delay differential equations, delay integral equations
and delay difference equations.
1
Introduction
The integral inequalities and the finite difference inequalities play a fundamental role in the development of the theory of differential equations, integral
equations and difference equations. During the past few years, many such inequalities have been discovered, which are motivated by certain applications.
For example, see the monographes[1, 2, 9, 10], papers[3–7, 11, 12] and the references therein. However, in the qualitative analysis of some classes of delay
differential equations, delay integral equations and delay difference equations,
the bounds provided by the earlier inequalities are inadequate and it is necessary to seek some new integral inequalities and their discrete analogues in
order to achieve a diversity of desired goals. In this paper, we investigate
some nonlinear delay integral inequalities and their discrete analogues which
provide explicit bounds on unknown functions.
Key Words: Integral inequality, discrete inequality, delay, differential equation, integral
equation, difference equation
2010 Mathematics Subject Classification: 26D15, 26D10, 34C11, 39A12
Received: September, 2009
Accepted: September, 2010
149
150
2
Wei Nian Li
Formulation of the Problem
In what follows, R denotes the set of real numbers, R+ = [0, ∞) is the given
subset of R, C(M, S) denotes the class of all continuous functions defined
on set M with range in the set S, and N0 = {0, 1, 2, · · · } denotes the set
of nonnegative integers. We use the usual conventions that empty sums and
products are taken to be 0 and 1 respectively. Throughout this paper, all the
functions which appear in the inequalities are assumed to be real-valued and
all the involved sums exist on the respective domains of their definitions.
In this paper, on the one hand, we study the following nonlinear delay
integral inequalities
Z t
xp (t) ≤ a(t) + b(t)
[f (s)xp (s − τ ) + g(s)x(s) + h(s)]ds, t ∈ R+ , (E1)
0
and
Z
t
xp (t) ≤ a(t) + b(t)
L(s, x(s − τ ))ds, t ∈ R+ ,
(E2)
0
with the initial condition
½
x(t) = ϕ(t), t ∈ [−τ, 0],
ϕ(t − τ ) ≤ (a(t))1/p for t ∈ R+ with t − τ ≤ 0,
(I)
where p > 1 and τ ∈ R+ are constants, ϕ(t) ∈ C([−τ, 0], R+ ), and L ∈
2
C(R+
, R+ ).
On the other hand, we also investigate the following discrete analogues of
(E1) and (E2)
xp (n) ≤ a(n) + b(n)
n−1
X
[f (s)xp (s − σ) + g(s)x(s) + h(s)], n ∈ N0 ,
(E 0 1)
s=0
and
xp (n) ≤ a(n) + b(n)
n−1
X
V (s, x(s − σ)), n ∈ N0 ,
(E 0 2)
s=0
with the initial condition
½
x(n) = ψ(n), n ∈ {−σ, · · · , −1, 0},
ψ(n − σ) ≤ (a(n))1/p for n ∈ N0 with n − σ ≤ 0,
(I 0 )
where p > 1, σ ∈ N0 are constants, ψ(n) ∈ R+ , n ∈ {−σ, · · · , −1, 0}, and
V : N0 × R+ → R+ .
SOME NONLINEAR DELAY INTEGRAL INEQUALITIES
3
151
Main Results
The following lemmas are useful in our main results.
Lemma 1[8]. Assume that
1 1
+ = 1 with p > 1. Then
p q
1
1
xp yq ≤
x y
+ ,
p
q
(1)
for x, y ∈ R+ .
Lemma 2[11]. (i) Assume that u(t), a(t), b(t) ∈ C(R+ , R+ ), and a(t) is
nondecreasing for t ∈ R+ . If
Z t
u(t) ≤ a(t) +
b(s)u(s)ds,
0
for t ∈ R+ , then
µZ
u(t) ≤ a(t) exp
t
¶
b(s)ds ,
0
for t ∈ R+ .
(ii) Assume that u(n), a(n), b(n) are nonnegative functions defined for n ∈
N0 , and a(n) is nondecreasing for n ∈ N0 . If
u(n) ≤ a(n) +
n−1
X
b(s)u(s), n ∈ N0 ,
s=0
then
u(n) ≤ a(n)
n−1
Y
[1 + b(s)], n ∈ N0 .
s=0
Theorem 1. Assume that x(t), a(t), b(t), f (t), g(t), h(t) ∈ C(R+ , R+ ). If a(t)
and b(t) are nondecreasing in R+ , then the inequality (E1) with the initial
condition (I) implies
½
x(t) ≤
µZ
a(t) + b(t)B(t) exp
0
t
µ
¶ ¶¾ p1
g(s)
b(s) f (s) +
ds
,
p
(2)
for t ∈ R+ , where
Z t·
B(t) =
f (s)a(s) +
0
¸
a(s) + p − 1
g(s) + h(s) ds.
p
(3)
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Wei Nian Li
Proof. Fixing any positive number T , we define a function z(t) by
½
z(t) =
Z
t
a(T ) + b(t)
¾ p1
p
[f (s)x (s − τ ) + g(s)x(s) + h(s)]ds
, t ∈ [0, T ]. (4)
0
It is easy to see that z(t) is a nonnegative and nondecreasing function, and
x(t) ≤ z(t), t ∈ [0, T ].
Therefore,
x(t − τ ) ≤ z(t − τ ) ≤ z(t), t − τ ≥ 0, t ∈ [0, T ].
(5)
Using the initial condition (I), we have
x(t − τ ) = ϕ(t − τ ) ≤ (a(t))1/p ≤ (a(T ))1/p ≤ z(t), t − τ ≤ 0, t ∈ [0, T ]. (6)
(5) and (6) guarantee
x(t − τ ) ≤ z(t), t ∈ [0, T ].
It follows from (4) and (7) that
Z t
z p (t) ≤ a(T ) + b(t)
[f (s)z p (s) + g(s)z(s) + h(s)]ds, t ∈ [0, T ].
(7)
(8)
0
Taking t = T in (8), we obtain
Z
z p (T ) ≤ a(T ) + b(T )
T
[f (s)z p (s) + g(s)z(s) + h(s)]ds.
(9)
0
Noting that T ∈ R+ is arbitrary, from (9), we have
Z t
z p (t) ≤ a(t) + b(t)
[f (s)z p (s) + g(s)z(s) + h(s)]ds, t ∈ R+ .
(10)
0
Similarly, we obtain
x(t) ≤ z(t), t ∈ R+ .
Define a function u(t) by
Z t
u(t) =
[f (s)z p (s) + g(s)z(s) + h(s)]ds, t ∈ R+ .
(11)
(12)
0
Then (10) can be restated as
z p (t) ≤ a(t) + b(t)u(t), t ∈ R+ .
(13)
SOME NONLINEAR DELAY INTEGRAL INEQUALITIES
153
Using Lemma 1, from (13), we easily obtain
1
z(t) ≤ (a(t) + b(t)u(t)) p (1)
p−1
p
≤
a(t) b(t)
p−1
+
u(t) +
, t ∈ R+ .
p
p
p
(14)
Combining (12)–(14), we get
µ
¶
·
¸
0
g(t)
a(t) + p − 1
u (t) ≤ b(t) f (t) +
u(t) + f (t)a(t) +
g(t) + h(t) ,
p
p
i.e.
Z
t
u(t) ≤ B(t) +
0
µ
¶
g(s)
b(s) f (s) +
u(s)ds, t ∈ R+ ,
p
(15)
where B(t) is defined by (3). Using the Part (i) of Lemma 2, from (15), we
have
µZ t
µ
¶ ¶
g(s)
u(t) ≤ B(t) exp
b(s) f (s) +
ds , t ∈ R+ .
(16)
p
0
Clearly, the desired inequality (2) follows from (11), (13) and (16). The
proof is complete.
Theorem 2. Assume that x(t), a(t), b(t) ∈ C(R+ , R+ ), a(t) and b(t) are
nondecreasing in R+ . If
0 ≤ L(t, x) − L(t, y) ≤ K(t, y)(x − y),
(17)
2
for x ≥ y ≥ 0, where K ∈ C(R+
, R+ ), then the inequality (E2) with the initial
condition (I) implies
½
x(t) ≤
µZ
t
a(t) + b(t)E(t) exp
0
µ
¶
¶¾ p1
a(s) + p − 1 b(s)
K s,
ds
,
p
p
(18)
for t ∈ R+ , where
Z
t
E(t) =
0
µ
¶
a(s) + p − 1
L s,
ds.
p
(19)
Proof. Fixing any positive number T , we define a function z(t) by
½
z(t) =
Z
t
a(T ) + b(t)
¾ p1
L(s, x(s − τ ))ds , t ∈ [0, T ].
0
Using a similar way in the proof of Theorem 1 and noting the condition
(17), we easily obtain that z(t) is a nonnegative and nondecreasing function,
and
x(t) ≤ z(t), t ∈ R+ ,
(20)
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Wei Nian Li
and
Z
t
p
z (t) ≤ a(t) + b(t)
L(s, z(s))ds, t ∈ R+ .
(21)
0
Define a function u(t) by
Z
t
u(t) =
L(s, z(s))ds.
(22)
0
Then (21) can be restated as
z p (t) ≤ a(t) + b(t)u(t), t ∈ R+ .
(23)
As in the proof of Theorem 1, from (23), we obtain (14). Noting the
condition (17), from (22) and (14), we have
0
u (t) =
≤
≤
L(t,
µ z(t))
¶
µ
¶
a(t) b(t)
p−1
a(t) p − 1
L t,
+
u(t) +
− L t,
+
p
p
p
p
µ p
¶
a(t) p − 1
+L t,
+
p
p ¶
µ
µ
¶
a(t) p − 1 b(t)
a(t) p − 1
K t,
+
u(t) + L t,
+
,
p
p
p
p
p
i.e.
Z
t
u(t) ≤ E(t) +
0
µ
¶
a(s) + p − 1 b(s)
K s,
u(s)ds, t ∈ R+ ,
p
p
(24)
where E(t) is defined by (19). Using the Part (i) of Lemma 2, it follows from
(24) that
µZ
t
u(t) ≤ E(t) exp
0
µ
¶
¶
a(s) + p − 1 b(s)
K s,
ds , t ∈ R+ ,
p
p
(25)
We easily see that the desired inequality (18) follows from (20), (23) and
(25). This completes the proof of Theorem 2.
Theorem 3. Assume x(n), a(n), b(n), f (n), g(n), h(n) be nonnegative functions defined for n ∈ N0 . If a(n) and b(n) are nondecreasing in N0 , then the
inequality (E 0 1) with the initial condition (I 0 ) implies
½
x(n) ≤
a(n) + b(n)G(n)
n−1
Y·
s=0
µ
¶¸¾ p1
g(s)
1 + b(s) f (s) +
,
p
(26)
155
SOME NONLINEAR DELAY INTEGRAL INEQUALITIES
for n ∈ N0 , where
G(n) =
n−1
X·
f (s)a(s) +
s=0
¸
a(s) + p − 1
g(s) + h(s) .
p
(27)
Proof. Fixing any positive integer M , we define a function z(n) by
½
z(n) =
a(M )+b(n)
n−1
X·
¸¾ p1
p
f (s)x (s−σ)+g(s)x(s)+h(s)
, n ∈ NM , (28)
s=0
where NM = {0, 1, · · · , M }. It is easy to see that z(n) is a nonnegative and
nondecreasing function, and
x(n) ≤ z(n), n ∈ NM .
Therefore, for n ∈ N0 with n − σ ≥ 0, we have
x(n − σ) ≤ z(n − σ) ≤ z(n), n ∈ NM .
(29)
Using the initial condition (I 0 ), for n ∈ N0 with n − σ ≤ 0, we have
x(n − σ) = ϕ(n − σ) ≤ (a(n))1/p ≤ (a(M ))1/p ≤ z(n), n ∈ NM .
(30)
Combining (29) and (30), we obtain
x(n − σ) ≤ z(n), n ∈ NM .
(31)
Therefore,
z p (n) ≤ a(M ) + b(n)
n−1
X
[f (s)z p (s) + g(s)z(s) + h(s)], n ∈ NM .
(32)
s=0
Taking n = M in (32), we obtain
z p (M ) ≤ a(M ) + b(M )
M
−1
X
[f (s)z p (s) + g(s)z(s) + h(s)].
(33)
s=0
Noting that M ∈ N0 is arbitrary, from (33), we observe that
z p (n) ≤ a(n) + b(n)
n−1
X
s=0
[f (s)z p (s) + g(s)z(s) + h(s)], n ∈ N0 .
(34)
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Wei Nian Li
Using a Similar way, we obtain
x(n) ≤ z(n), n ∈ N0 .
(35)
Define a function u(n) by
u(n) =
n−1
X
[f (s)z p (s) + g(s)z(s) + h(s)], n ∈ N0 .
(36)
s=0
Then (34) can be restated as
z p (n) ≤ a(n) + b(n)u(n), n ∈ N0 .
(37)
Using Lemma 1, from (37), we easily obtain
a(n) + p − 1
b(n)
u(n) +
, n ∈ N0 .
p
p
z(n) ≤
(38)
Therefore,
u(n + 1) − u(n) ≤
·
¸
a(n) + p − 1
f (n)a(n) +
g(n) + h(n)
p¶
µ
g(n)
+b(n) f (n) +
u(n), n ∈ N0 .
p
(39)
Substituting n = s and taking the sum over s from 0 to n − 1, it follows
from (39) that
u(n) ≤ G(n) +
n−1
X·
s=0
µ
¶¸
g(s)
1 + b(s) f (s) +
u(s), n ∈ N0 ,
p
(40)
where G(n) is defined by (27). Using the Part (ii) of Lemma 2, we easily see
that (40) guarantees
u(n) ≤ G(n)
n−1
Y·
s=0
µ
¶¸
g(s)
1 + b(s) f (s) +
, n ∈ N0 .
p
(41)
It is easy to see that the desired inequality (26) follows from (35), (37) and
(41). This completes the proof of Theorem 3.
Theorem 4. Let x(n), a(n), b(n) be nonnegative functions for n ∈ N0 , a(n)
and b(n) be nondecreasing in N0 . If
0 ≤ V (n, x) − V (n, y) ≤ W (n, y)(x − y),
(42)
SOME NONLINEAR DELAY INTEGRAL INEQUALITIES
157
for x ≥ y ≥ 0, where W : N0 × R+ → R+ , then the inequality (E 0 2) with the
initial condition (I 0 ) implies
½
µ
¶
¸¾ p1
n−1
Y·
a(s) + p − 1 b(s)
x(n) ≤ a(n) + b(n)F (n)
1 + W s,
,
(43)
p
p
s=0
for n ∈ N0 , where
F (n) =
µ
¶
a(s) + p − 1
V s,
.
p
s=0
n−1
X
(44)
Proof. Fixing any positive integer M , we define a function z(n) by
½
z(n) =
a(M ) + b(n)
n−1
X
¾ p1
V (s, x(s − σ)) , n ∈ NM .
s=0
Using a similar way in the proof of Theorem 3 and noting the condition
(42), we easily obtain that z(n) is a nonnegative and nondecreasing function,
and
x(n) ≤ z(n), n ∈ N0 ,
(45)
and
z p (n) ≤ a(n) + b(n)
n−1
X
V (s, z(s)), n ∈ N0 .
(46)
s=0
Define a function u(n) by
u(n) =
n−1
X
V (s, z(s)), n ∈ N0 .
(47)
s=0
Then (46) can be restated as
z p (n) ≤ a(n) + b(n)u(n), n ∈ N0 .
(48)
As in the proof of Theorem 3, from (48), we obtain (38). Noting the
condition (42), from (47) and (38), we have
u(n + 1) − u(n) = V µ
(n, z(n))
¶
a(n) b(n)
p−1
≤ V n,
+
u(n) +
p
p
µ p
¶
µ
¶
a(n) p − 1
a(n) p − 1
−V n,
+
+ V n,
+
p
p ¶
p
p
µ
a(n) p − 1 b(n)
+
u(n)
≤ W n,
p ¶p
µ p
a(n) p − 1
+V n,
+
, n ∈ N0 .
p
p
(49)
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Wei Nian Li
Substituting n = s and taking the sum over s from 0 to n − 1, it follows from
(49) that
n−1
X µ a(s) + p − 1 ¶ b(s)
u(s), n ∈ N0 ,
(50)
u(n) ≤ F (n) +
W s,
p
p
s=0
where F (n) is defined by (44). Using the Part (ii) of Lemma 2, from (50), we
have
¶
¸
µ
n−1
Y·
a(s) + p − 1 b(s)
, n ∈ N0 ,
(51)
u(n) ≤ F (n)
1 + W s,
p
p
s=0
The desired inequality (43) follows from (45), (48) and (51). This completes
the proof.
Finally, we present an application of Theorem 1.
Example. Consider the delay differential equation
0
(xp (t)) = P (t, x(t), x(t − τ )), t ∈ R+ ,
with the initial condition
½
x(t) = φ(t), t ∈ [−τ, 0],
1
φ(t − τ ) ≤ |C| p for t ∈ R+ with t − τ ≤ 0,
(52)
(53)
where P ∈ C(R+ × R2 , R), C = xp (0), p > 1, τ ∈ R+ are constants, and
φ ∈ C([−τ, 0], R).
Assume that
|P (t, x(t, ), x(t − τ ))| ≤ f (t)|xp (t − τ )| + g(t)|x(t)| + h(t),
(54)
where f (t), g(t), h(t) are as defined in Theorem 1. If x(t) is a solution of the
equation (52) satisfying the initial condition (53), then
½
µZ t
µ
¶ ¶¾ p1
g(s)
e
|x(t)| ≤ |C| + b(t)B(t) exp
b(s) f (s) +
,
(55)
ds
p
0
for t ∈ R+ , where
e =
B(t)
Z t·
|C|f (s) +
0
¸
|C| + p − 1
g(s) + h(s) ds.
p
(56)
In fact, the solution x(t) of equation (52) satisfying the initial condition
(53) satisfies the equivalent delay integral equation
Z t
p
x (t) = C +
P (s, x(s), x(s − τ ))ds, t ∈ R+ ,
(57)
0
SOME NONLINEAR DELAY INTEGRAL INEQUALITIES
159
with the initial condition (53). Noting the assumption (54), we have
Z
|xp (t)| ≤ |C| +
t
[f (s)|xp (s − τ )| + g(s)|x(s)| + h(s)]ds
(58)
0
with the initial condition (53). Now a suitable application of Theorem 1 to
(58) yields (55).
Remark 1. The right–hand side of (55) gives us the bound on the solution
x(t) of the equation (52) satisfying the initial condition (53) in terms of the
known functions for t ∈ R+ .
Remark 2. We can present some applications of Theorems 2–4. Due to
limited space, their statements are omitted here.
Acknowledgments
This work is supported by the National Natural Science Foundation of
China (10971018), the Natural Science Foundation of Shandong Province
(ZR2009AM005), China Postdoctoral Science Foundation Funded Project (20080440633),
Shanghai Postdoctoral Scientific Program (09R21415200), the Project of Science and Technology of the Education Department of Shandong Province
(J08LI52), and the Doctoral Foundation of Binzhou University (2006Y01).
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Department of Mathematics,
Binzhou University, Shandong, 256603, P.R.China
e-mail: wnli@263.net