Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 111, 2006
ON SOME NEW RETARD INTEGRAL INEQUALITIES IN n INDEPENDENT
VARIABLES AND THEIR APPLICATIONS
XUEQIN ZHAO AND FANWEI MENG
D EPARTMENT OF M ATHEMATICS
Q UFU N ORMAL U NIVERSITY
Q UFU 273165,
P EOPLE ’ S R EPUBLIC OF C HINA .
xqzhao1972@126.com
Received 24 October, 2005; accepted 24 May, 2006
Communicated by S.S. Dragomir
A BSTRACT. In this paper, we established some retard integral inequalities in n independent
variables and by means of examples we show the usefulness of our results.
Key words and phrases: Retard integral inequalities; integral equation; Partial differential equation.
2000 Mathematics Subject Classification. 26D15, 26D10.
1. I NTRODUCTION
The study of integral inequalities involving functions of one or more independent variables
is an important tool in the study of existence, uniqueness, bounds, stability, invariant manifolds
and other qualitative properties of solutions of differential equations and integral equations.
During the past few years, many new inequalities have been discovered (see [1, 3, 4, 7, 8]). In
the qualitative analysis of some classes of partial differential equations, the bounds provided
by the earlier inequalities are inadequate and it is necessary to seek some new inequalities in
order to achieve a diversity of desired goals. Our aim in this paper is to establish some new
inequalities in n independent variables, meanwhile, some applications of our results are also
given.
2. P RELIMINARIES AND L EMMAS
In this paper, we suppose R+ = [0, ∞), is subset of real numbers R, e
0 = (0, . . . , 0), α
e(t) =
n
n
n
(α1 (t1 ), . . . , αn (tn )) ∈ R+ , t = (t1 , . . . , tn ) ∈ R+ , s = (s1 , . . . , sn ) ∈ R+ , re = (r1 , . . . , rn ),
re0 = (r10 , . . . , rn0 ), ze = (z1 , . . . , zn ), ze0 = (z10 , . . . , zn0 ), T = (T1 , . . . , Tn ) ∈ Rn+ .
If f : Rn+ → R+ , we suppose
(1) s ≤ t ⇔ si ≤ ti (i = 1, 2, . . . , n);
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
318-05
2
X UEQIN Z HAO AND FANWEI M ENG
R αe(t)
R α (t )
R α (t )
(2) e0 f (s)ds = 0 1 1 · · · 0 n n f (s1 , . . . , sn )dsn . . . ds1 ;
(3) Di = dtdi , i = 1, 2,. . . , n.
3. M AIN R ESULTS
In this part, we obtain our main results as follows:
Theorem 3.1. Let ψ ∈ C(R+ , R+ ) be a nondecreasing function with ψ(u) > 0 on (0, ∞), and
let c be a nonnegative constant. Let αi ∈ C 1 (R+ , R+ ) be nondecreasing with αi (ti ) ≤ ti on R+
(i = 1, . . . , n). If u, f ∈ C(Rn+ , R+ ) and
Z αe(t)
(3.1)
u(t) ≤ c +
f (s)ψ(u(s))ds,
e
0
for e
0 ≤ t < T , then
"
u(t) ≤ G−1 G(c) +
(3.2)
Z
#
α
e(t)
f (s)ds ,
e
0
where
Z
ze
ds
,
ze ≥ ze0 > 0.
ze0 ψ(s)
G−1 is the inverse of G, T ∈ Rn+ is chosen so that
Z αe(t)
e
(3.3)
G(c) +
f (s)ds ∈ Dom(G−1 ),
0 ≤ t < T.
G(e
z) =
e
0
Define the nondecreasing positive function z(t) and make
Z αe(t)
(3.4)
z(t) = c + ε +
f (s)ψ(u(s))ds,
e
0 ≤ t < T,
e
0
where ε is an arbitrary small positive number. We know that
u(t) ≤ z(t),
(3.5)
D1 D2 · · · Dn z(t) = f (e
α)ψ(u(e
α))α10 α20 · · · αn0 .
Using (3.5), we have
D1 D2 · · · Dn z(t)
≤ f (e
α)α10 α20 · · · αn0 .
ψ(z(t))
(3.6)
For
(3.7)
Dn
D1 D2 · · · Dn−1 z(t)
ψ(z(t))
D1 D2 · · · Dn z(t)ψ(z(t)) − D1 D2 · · · Dn−1 z(t)ψ 0 Dn z(t)
=
,
ψ 2 (z(t))
using D1 D2 · · · Dn−1 z(t) ≥ 0, ψ 0 ≥ 0, Dn z(t) ≥ 0 in (3.7), we get
D1 D2 · · · Dn−1 z(t)
D1 D2 · · · Dn z(t)
(3.8)
Dn
≤
≤ f (e
α)α10 α20 · · · αn0 .
ψ(z(t))
ψ(z(t))
Fixing t1 , . . . , tn−1 , setting tn = sn , integrating from tn to ∞, yields
Z αn (tn )
D1 D2 · · · Dn−1 z(t)
0
(3.9)
≤
f (α1 (t1 ), . . . , αn−1 (tn−1 ), sn )α10 α20 · · · αn−1
dsn .
ψ(z(t))
0
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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R ETARD I NTEGRAL I NEQUALITIES
3
Using the same method, we deduce that
Z α2 (t2 )
Z αn (tn )
D1 z(t)
(3.10)
≤
···
f (t1 , s2 , . . . , sn )α10 dsn . . . ds2 ,
ψ(z(t))
0
0
and integration on [t1 , ∞) yields
Z
(3.11) G(z(t)) ≤ G(c + ε) +
α1 (t1 )
αn (tn )
Z
···
0
f (s1 , s2 , . . . , sn )dsn . . . ds1 ,
t ∈ Rn+ .
0
From the definition of G and letting ε → 0, we can obtain inequality (3.2).
Remark 3.2. If we let G(z) → ∞, z → ∞, then condition (3.3) can be omitted.
Corollary 3.3. If we let ψ = sr , 0 < r ≤ 1 in Theorem 3.1, then for t ∈ Rn+ , we have
 h
1
i 1−r
R αe(t)

1−r

c
+
(1
−
r)
, 0 < r < 1;
f
(s)ds

e
0
(3.12)
u(t) ≤


 c exp R αe(t) f (s)ds ,
r = 1.
e
0
Remark 3.4. If we let n = 1, r = 1, α
e(t) = t, in Corollary 3.3, we obtain the Mate-Nevai
inequality.
Theorem 3.5. Let ϕ ∈ C(R+ , R+ ) be an increasing function with ϕ(∞) = ∞. Let ψ ∈
C(R+ , R+ ) be a nondecreasing function and let c be a nonnegative constant. Let αi ∈ C 1 (R+ , R+ )
be nondecreasing with αi (ti ) ≤ ti on R+ (i = 1, . . . , n). If u, f ∈ C(Rn+ , R+ ) and
Z αe(t)
(3.13)
ϕ(u(t)) ≤ c +
f (s)ψ(u(s))ds,
e
0
for e
0 ≤ t < T , then
(
(3.14)
u(t) ≤ ϕ−1
G−1 [G(c) +
Z
)
α
e(t)
f (s)ds] ,
e
0
R ze
ds
where G(e
z ) = ze0 ψ[ϕ−1
, ze ≥ ze0 > 0, ϕ−1 , G−1 are respectively the inverse of ϕ and G,
(s)]
T ∈ Rn+ is chosen so that
Z αe(t)
e
(3.15)
G(c) +
f (s)ds ∈ Dom(G−1 ),
0 ≤ t < T.
e
0
Proof. From the definition of the ϕ, we know (3.13) can be restated as
Z αe(t)
(3.16)
ϕ(u(t)) ≤ c +
f (s)ψ[ϕ−1 (ϕ(u(s)))]ds,
t ∈ Rn+ .
e
0
Now an application of Theorem 3.1 gives
"
Z
−1
(3.17)
ϕ(u(t)) ≤ G
G(c) +
α
e(t)
#
f (s)ds ,
e
0 ≤ t < T.
e
0
So,
(
(3.18)
u(t) ≤ ϕ−1
G−1 [G(c) +
Z
α
e(t)
)
f (s)ds] ,
e
0 ≤ t < T.
e
0
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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4
X UEQIN Z HAO AND FANWEI M ENG
Corollary 3.6. If we let ϕ = sp , ψ = sq , p, q are constants, and p ≥ q > 0 in Theorem 3.5, for
e
0 ≤ t < T , then
 h
1
R
i p−q
α
e(t)
1− pq
q


, when p > q;
+ 1 − p e0 f (s)ds
 c
(3.19)
u(t) ≤


 c p1 exp 1 R αe(t) f (s)ds ,
when p = q.
p e
0
Theorem 3.7. Let u, f and g be nonnegative continuous functions defined on Rn+ , let c be a
nonnegative constant. Moreover, let w1 , w2 ∈ C(R+ , R+ ) be nondecreasing functions with
wi (u) > 0 (i = 1, 2) on (0, ∞). Let αi ∈ C 1 (R+ , R+ ) be nondecreasing with αi (ti ) ≤ ti on R+
(i = 1, . . . , n). If
Z αe(t)
Z t
(3.20)
u(t) ≤ c +
f (s)w1 (u(s))ds +
g(s)w2 (u(s))ds,
e
0
e
0
for e
0 ≤ t < T , then
(i) for the case w2 (u) ≤ w1 (u),
(
u(t) ≤ G−1
1
(3.21)
Z
α
e(t)
G1 (c) +
Z
f (s)ds +
e
0
(ii) for the case w1 (u) ≤ w2 (u),
(
u(t) ≤ G−1
2
(3.22)
t
Z
G2 (c) +
)
g(s)ds ,
e
0
α
e(t)
Z
f (s)ds +
e
0
t
)
g(s)ds ,
e
0
where
Z
ze
Gi (e
z) =
(3.23)
ze0
ds
,
wi (s)
ze ≥ ze0 > 0,
(i = 1, 2)
n
and G−1
i (i = 1, 2) is the inverse of Gi , T ∈ R+ is chosen so that
Z αe(t)
Z t
(3.24)
Gi (c) +
f (s)ds +
g(s)ds ∈ Dom(G−1
(i = 1, 2),
i ),
e
0
e
0 ≤ t < T.
e
0
Proof. Define the nonincreasing positive function z(t) and make
Z αe(t)
Z t
(3.25)
z(t) = c + ε +
f (s)w1 (u(s))ds +
g(s)w2 (u(s))ds,
e
0
e
0
where ε is an arbitrary small positive number. From inequality (3.20), we know
u(t) ≤ z(t)
(3.26)
and
(3.27)
D1 D2 · · · Dn z(t) = [f (e
α)w1 (u(e
α))α10 α20 · · · αn0 + g(t)w2 (u(t))].
The rest of the proof can be completed by following the proof of Theorem 3.1 with suitable
modifications.
Theorem 3.8. Let u, f and g be nonnegative continuous functions defined on Rn+ , and let ϕ ∈
C(R+ , R+ ) be an increasing function with ϕ(∞) = ∞ and let c be a nonnegative constant.
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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R ETARD I NTEGRAL I NEQUALITIES
5
Moreover, let w1 , w2 ∈ C(R+ , R+ ) be nondecreasing functions with wi (u) > 0 (i = 1, 2) on
(0, ∞), αi ∈ C 1 (R+ , R+ ) be nondecreasing with αi (ti ) ≤ ti on R+ (i = 1, . . . , n). If
Z αe(t)
Z t
(3.28)
ϕ(u(t)) ≤ c +
f (s)w1 (u(s))ds +
g(s)w2 (u(s))ds,
e
0
e
0
for e
0 ≤ t < T , then
(i) for the case w2 (u) ≤ w1 (u),
(
"
u(t) ≤ ϕ−1
(3.29)
G−1
G1 (c) +
1
Z
α
e(t)
Z
f (s)ds +
e
0
(ii) for the case w1 (u) ≤ w2 (u),
"
(
u(t) ≤ ϕ−1
(3.30)
G2 (c) +
G−1
2
Z
#)
t
;
g(s)ds
e
0
α
e(t)
Z
f (s)ds +
e
0
#)
t
g(s)ds
,
e
0
where
Z
ze
Gi (e
z) =
ze0
ds
,
wi (ϕ−1 (s))
ze > ze0 ,
(i = 1, 2),
n
and ϕ−1 , G−1
i (i = 1, 2) are respectively the inverse of Gi , ϕ, T ∈ R+ is chosen so that
Z αe(t)
Z t
e
(3.31)
Gi (c) +
f (s)ds +
g(s)ds ∈ Dom(G−1
(i = 1, 2),
0 ≤ t < T.
i ),
e
0
e
0
Proof. From the definition of ϕ, we know (3.28) can be restated as
Z
(3.32)
ϕ(u(t)) ≤ c +
α
e(t)
e
0
f (s)w1 [ϕ−1 (ϕ(u(s)))]ds
Z t
+
g(s)w2 [ϕ−1 (ϕ(u(s)))]ds,
t ∈ Rn+ .
e
0
Now an application of Theorem 3.7 gives
(
)
Z αe(t)
Z t
ϕ(u(t)) ≤ G−1
Gi (c) +
f (s)ds +
g(s)ds ,
i
e
0
e
0 ≤ t < T,
e
0
where T satisfies (3.31). We can obtain the desired inequalities (3.29) and (3.30).
Theorem 3.9. Let u, f and g be nonnegative continuous functions defined on Rn+ and let c
be a nonnegative constant. Moreover, let ϕ ∈ C(R+ , R+ ) be an increasing function with
ϕ(∞) = ∞, ψ ∈ C(R+ , R+ ) be a nondecreasing function with ψ(u) > 0 on (0, ∞) and
αi ∈ C 1 (R+ , R+ ) be nondecreasing with αi (ti ) ≤ ti on R+ (i = 1, . . . , n). If
Z αe(t)
[f (s)u(s)ψ(u(s)) + g(s)u(s)]ds,
(3.33)
ϕ(u(t)) ≤ c +
e
0
for e
0 ≤ t < T , then
(
(3.34)
u(t) ≤ ϕ−1
"
Ω−1 G−1
Z
G[Ω(c) +
Z
g(s)ds] +
e
0
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
α
e(t)
!#)
α
e(t)
f (s)ds
,
e
0
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6
X UEQIN Z HAO AND FANWEI M ENG
where
Z
re
Ω(e
r) =
re0
Z ze
G(e
z) =
ze0
ds
ϕ−1 (s)
,
re ≥ re0 > 0 ,
ds
,
ψ{ϕ−1 [Ω−1 (s)]}
ze ≥ ze0 > 0,
Ω−1 , ϕ−1 , G−1 are respectively the inverse of Ω, ϕ, G. And T ∈ R+ is chosen so that
"
# Z
Z αe(t)
α
e(t)
e
G Ω(c) +
g(s)ds +
f (s)ds ∈ Dom(G−1 ),
0 ≤ t < T,
e
0
e
0
and
( "
G−1
Z
G Ω(c) +
#
α
e(t)
Z
g(s)ds +
e
0
)
α
e(t)
f (s)ds
∈ Dom(Ω−1 ),
e
0 ≤ t < T.
e
0
Proof. Let us first assume that c > 0. Defining the nondecreasing positive function z(t) by the
right-hand side of (3.33)
Z αe(t)
z(t) = c +
[f (s)u(s)ψ(u(s)) + g(s)u(s)]ds,
e
0
we know
u(t) ≤ ϕ−1 [z(t)]
(3.35)
and
D1 D2 · · · Dn z(t) = [f (e
α)u(e
α)ψ(u(e
α)) + g(e
α)u(e
α)]α10 α20 · · · αn0 .
(3.36)
Using (3.35), we have
D1 D2 · · · Dn z(t)
≤ [f (e
α)ψ(ϕ−1 (z(α))) + g(e
α)]α10 α20 · · · αn0 .
ϕ−1 (z(t))
(3.37)
For
(3.38) Dn
D1 D2 · · · Dn−1 z(t)
ϕ−1 (z(t))
D1 D2 · · · Dn z(t)ϕ−1 (z(t)) − D1 D2 · · · Dn−1 z(t)(ϕ−1 (z(t)))0 Dn z(t)
=
,
(ϕ−1 (z(t)))2
using D1 D2 · · · Dn−1 z(t) ≥ 0, Dn z(t) ≥ 0, (ϕ−1 (z(t)))0 ≥ 0 in (3.38), we get
D1 D2 · · · Dn−1 z(t)
D1 D2 · · · Dn z(t)
(3.39)
Dn
≤
−1
ϕ (z(t))
ϕ−1 (z(t))
≤ [f (e
α)ψ(ϕ−1 z(e
α)) + g(e
α)]α10 α20 · · · αn0 .
Fixing t1 , . . . , tn−1 , setting tn = sn , integrating from 0 to tn with respect to sn yields
(3.40)
D1 D2 · · · Dn−1 z(t)
ϕ−1 (z(t))
Z αn (tn )
≤
[f (α1 (t1 ), . . . , αn−1 (tn−1 ), sn )ψ(ϕ−1 (z(α1 (t1 ), . . . , αn−1 (tn−1 ), sn )))
0
0
+ g(α1 (t1 ), . . . , αn−1 (tn−1 ), sn )]α10 α20 · · · αn−1
dsn .
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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R ETARD I NTEGRAL I NEQUALITIES
7
Using the same method, we deduce that
Z α2 (t2 )
Z αn (tn )
D1 z(t)
≤
(3.41)
···
[f (α1 (t1 ), s2 , . . . , sn )ψ(ϕ(z(α1 (t1 ), s2 , . . . , sn )))
ϕ−1 (z(t))
0
0
+ g(α1 (t1 ), s2 , . . . , sn )]α10 dsn . . . ds2 .
Setting t1 = s1 , and integrating it from 0 to t1 with respect to s1 yields
Z αe(t)
Z αe(t)
−1
(3.42)
Ω(z(t)) ≤ Ω(c) +
f (s)ψ(ϕ (z(s))ds +
g(s)ds,
e
0
e
0
R αe(T1 )
Let T1 ≤ T be arbitrary, we denote p(T1 ) = Ω(c) + 0
g(s)ds, from (3.42), we deduce that
Z αe(t)
e
Ω(z(t)) ≤ p(T1 ) +
f (s)ψ[ϕ−1 z(s)]ds,
0 ≤ t ≤ T1 ≤ T.
0
Now an application of Theorem 3.5 gives
(
"
Z
−1
−1
G(p(T1 )) +
G
z(t) ≤ Ω
#)
α
e(t)
f (s)ds
e
0 ≤ t ≤ T1 ≤ T,
,
e
0
so
(
u(t) ≤ ϕ−1
"
Ω−1 G−1
!#)
α
e(t)
Z
G(p(T1 )) +
f (s)ds
e
0 ≤ t ≤ T1 ≤ T.
,
e
0
Taking t = T1 in the above inequality, since T1 is arbitrary, we can prove the desired inequality
(3.34).
If c = 0 we carry out the above procedure with ε > 0 instead of c and subsequently let ε → 0.
Setting f (t) = 0, n = 1, we can obtain a retarded Ou-Iang inequality.
Let u, f and g be nonnegative continuous functions defined on Rn+ and let c be a nonnegative
constant. Moreover, let ψ ∈ C(R+ , R+ ) be a nondecreasing function with ψ(u) > 0 on (0, ∞)
and αi ∈ C 1 (R+ , R+ ) be nondecreasing with αi (ti ) ≤ ti on R+ (i = 1, . . . , n). If
Z αe(t)
2
2
[f (s)u(s)ψ(u(s)) + g(s)u(s)]ds,
u (t) ≤ c +
e
0
for e
0 ≤ t < T , then
"
u(t) ≤ Ω−1
1
Ω c+
2
Z
where
Z
!
α
e(t)
g(s)ds
e
0
1
+
2
Z
#
α
e(t)
f (s)ds ,
e
0
ze
ds
ze > ze0 ,
ze0 ψ(s)
Ω−1 is the inverse of Ω, and T ∈ Rn+ is chosen so that
!
Z
Z
1 αe(t)
1 αe(t)
Ω c+
g(s)ds +
f (s)ds ∈ Dom(Ω−1 ),
2 e0
2 e0
Ω(e
z) =
e
0 ≤ t < T.
Corollary 3.10. Let u, f and g be nonnegative continuous functions defined on Rn+ and let c
be a nonnegative constant. Moreover, let p, q be positive constants with p ≥ q, p 6= 1. Let
αi ∈ C 1 (R+ , R+ ) be nondecreasing with αi (ti ) ≤ ti on R+ (i = 1, . . . , n). If
Z αe(t)
p
u (t) ≤ c +
[f (s)uq (s) + g(s)u(s)]ds,
t≥0
e
0
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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8
X UEQIN Z HAO AND FANWEI M ENG
for e
0 ≤ t < T then
 p
p−1
i
h R
R e(t)
e(t)

(1− p1 )
p−1 α
1 α

c
+
g(s)ds
f
(s)ds
,
when p = q;
exp

e
p
p e

0
0

u(t) ≤
1
p−q
p−q

p−1

R
R
1

α
e
(t)
α
e
(t)

+ p−q
, when p > q.
g(s)ds
f (s)ds
c(1− p ) + p−1

e
e
p
p
0
0
Theorem 3.11. Let u, f and g be nonnegative continuous functions defined on Rn+ , and let
ϕ ∈ C(R+ , R+ ) be an increasing function with ϕ(∞) = ∞ and let c be a nonnegative constant.
Moreover, let w1 , w2 ∈ C(R+ , R+ ) be nondecreasing functions with wi (u) > 0 (i = 1, 2) on
(0, ∞), and αi ∈ C 1 (R+ , R+ ) be nondecreasing with αi (ti ) ≤ ti (i = 1, . . . , n). If
Z αe(t)
Z t
(3.43)
ϕ(u(t)) ≤ c +
f (s)u(s)w1 (u(s))ds +
g(s)u(s)w2 (u(s))ds,
e
0
e
0
for e
0 ≤ t < T , then
(i) for the case w2 (u) ≤ w1 (u),
(
"
Ω−1 G−1
1
u(t) ≤ ϕ−1
(3.44)
Z
α
e(t)
G1 (Ω(c)) +
f (s)ds +
e
0
(ii) for the case w1 (u) ≤ w2 (u),
(
"
Ω−1 G−1
2
u(t) ≤ ϕ−1
(3.45)
Z
!#)
t
Z
g(s)ds
,
e
0
α
e(t)
G2 (Ω(c)) +
f (s)ds +
e
0
!#)
t
Z
g(s)ds
,
e
0
where
re
Z
Ω(e
r) =
re0
Z ze
Gi (e
z) =
ze0
ds
ϕ−1 (s)
,
re ≥ re0 > 0 ,
ds
wi
{ϕ−1 [Ω−1 (s)]}
,
ze ≥ ze0 > 0
(i = 1, 2)
Ω−1 , ϕ−1 , G−1 are respectively the inverse of Ω, ϕ, G, and T ∈ R+ is chosen so that
!
Z αe(t)
Z t
e
Gi Ω(c) +
f (s)ds +
g(s)ds ∈ Dom(G−1
0 ≤ t ≤ T,
i ),
e
0
e
0
and
"
G−1
Gi Ω(c) +
i
Z
α
e(t)
Z
f (s)ds +
e
0
t
!#
g(s)ds
∈ Dom(Ω−1 ),
e
0 ≤ t ≤ T.
e
0
Proof. Let c > 0 and define the nonincreasing positive function z(t) and make
Z αe(t)
Z t
(3.46)
z(t) = c +
f (s)u(s)w1 (u(s))ds +
g(s)u(s)w2 (u(s))ds.
e
0
e
0
From inequality (3.43), we know
u(t) ≤ ϕ−1 [z(t)],
(3.47)
and
(3.48)
D1 D2 · · · Dn z(t) = [f (e
α)u(e
α)w1 (u(e
α))α10 α20 · · · αn0 + g(t)u(t)w2 (u(t))].
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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R ETARD I NTEGRAL I NEQUALITIES
9
Using (3.47), we have
D1 D2 · · · Dn z(t)
≤ f (e
α)w1 (u(e
α))α10 α20 · · · αn0 + g(t)w2 (u(t)).
ϕ−1 (z(t))
(3.49)
For
(3.50) Dn
D1 D2 · · · Dn−1 z(t)
ϕ−1 (z(t))
D1 D2 · · · Dn z(t)ϕ−1 (z(t)) − D1 D2 · · · Dn−1 z(t)(ϕ−1 (z(t)))0 Dn z(t)
=
,
(ϕ−1 (z(t)))2
using D1 D2 · · · Dn−1 z(t) ≥ 0, (ϕ−1 (z(t)))0 ≥ 0, Dn z(t) ≥ 0 in (3.50), we get
D1 D2 · · · Dn−1 z(t)
D1 D2 · · · Dn z(t)
Dn
≤
ϕ−1 (z(t))
ϕ−1 (z(t))
≤ f (e
α)α10 α20 · · · αn0 w1 (ϕ−1 (e
α(t))) + g(t)w2 (ϕ−1 (t)).
Fixing t1 , . . . , tn−1 , setting tn = sn , integrating from tn to ∞, yields
D1 D2 · · · Dn−1 z(t)
ϕ−1 (z(t))
Z αn (tn )
0
≤
f (α1 (t1 ), . . . , αn−1 (tn−1 ), sn )w1 (ϕ−1 (α1 (t1 ), . . . , αn−1 (tn−1 ), sn ))α10 α20 · · · αn−1
dsn
0
Z tn
+
g(t1 , . . . , tn−1 , sn )w2 (ϕ−1 (t1 , . . . , tn−1 , sn ))dsn .
0
Deductively
D1 z(t)
ϕ−1 (z(t))
Z α2 (t2 )
Z
≤
···
(3.51)
0
0
αn (tn )
f (α1 (t1 ), s2 , . . . , sn )w1 (ϕ−1 (α1 (t1 ), s2 , . . . , sn ))α10 dsn . . . ds2
Z t2
Z tn
+
···
g(t1 , s2 , . . . , sn )w2 (ϕ−1 (t1 , s2 , . . . , sn ))dsn . . . ds2 .
0
0
Fixing t2 , . . . , tn , setting t1 = s1 , integrating from 0 to t1 with respect to s1 yields
Z αe(t)
(3.52) Ω(z(t)) ≤ Ω(c) +
f (s1 , . . . , sn )w1 (ϕ−1 (z(s))
e
0
Z t
+
g(s)w2 (ϕ−1 (z(s))ds,
t ∈ Rn+ .
e
0
From Theorem 3.8, we obtain
"
z(t) ≤ Ω−1 G−1
1
Z
G1 (Ω(c)) +
α
e(t)
Z
f (s)ds +
e
0
!#
t
g(s)ds
,
e
0
using (3.47), we get the inequality (3.44).
If c = 0 we carry out the above procedure with ε > 0 instead of c and subsequently let ε → 0.
(ii) when w1 (u) ≤ w2 (u).
The proof can be completed with suitable changes.
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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10
X UEQIN Z HAO AND FANWEI M ENG
4. S OME A PPLICATIONS
Example 4.1. Consider the integral equation:
(4.1) up (t1 , . . . , tn )
Z
= f (t1 , . . . , tn ) +
α
e(t)
K(s1 , . . . , sn )g(s1 , . . . , sn , u(s1 , . . . , sn ))ds1 . . . dsn ,
e
0
where f, K : Rn+ → R, g : Rn+ × R → R are continuous functions and p > 0 and p 6= 1is
constant, α
ei (t) ∈ C 1 (R+ , R+ ) is nondecreasing with αi (t) ≤ ti on R+ (i = 1, . . . , n). In [8]
B.G. Pachpatte studied the problem when α(t) = t, n = 1. Here we assume that every solution
under discussion exists on an interval Rn+ . We suppose that the functions f , K, g in (4.1) satisfy
the following conditions
(4.2)
|f (t1 , . . . , tn )| ≤ c1 ,
|K(t1 , . . . , tn )| ≤ c2 ,
|g(t1 , . . . , tn , u)| ≤ r(t1 , . . . , tn )|u|q + h(t1 , . . . , tn )|u|,
where c1 , c2 , are nonnegative constants, and p ≥ q > 0, and r : Rn+ → R+ , h : Rn+ → R+ are
continuous functions. From (4.1) and using (4.2), we get
Z αe(t)
p
[c2 r(s1 , . . . , sn )|u|q + c2 h(s1 , . . . , sn )|u|ds1 . . . dsn .
(4.3)
|u (t1 , . . . , tn )| ≤ c1 +
e
0
Now an application of Corollary 3.10 yields
 p−1
h R
i
p

(1− p1 )

e(t)
e(t)
c2 (p−1) R α
c2 α

c
h(s)ds
.
.
.
ds
+
exp
r(s)ds
.
.
.
ds

1
n
1
n
1
e

p
p e
0
0




when p = q,


|u(t)| ≤
1
"
# p−q

p−q
p−1

1

R
R
(1−
)

α
e(t)
α
e(t)


c1 p + c2 (p−1)
h(s)ds1 . . . dsn
+ c2 (p−q)
r(s)ds1 . . . dsn
e
e

p
p
0
0




when p > q.
If the integrals of r(s), h(s) are bounded, then we can have the bound of the solution u(t) of
(4.1). Similarly, we can obtain many other kinds of estimates.
Example 4.2. Consider the partial delay differential equation:
(4.4)
(4.5)
(4.6)
∂ 2 up (x, y)
= f (x, y, u(x, y), u(x − h1 (x), y − h2 (y))),
∂x1 ∂x2
up (x, 0) = a1 (x)
a1 (0) = a2 (0) = 0,
up (0, y) = a2 (y)
|a1 (x) + a2 (y)| ≤ c,
where f ∈ C(R+ × R+ × R2 , R), a1 ∈ C 1 (R+ , R), a2 ∈ C 1 (R+ , R), c and p are nonnegative
constants. h1 ∈ C 1 (R+ , R+ ), h2 ∈ C 1 (R+ , R+ ), such that
x − h1 (x) ≥ 0,
h01 (x) < 1,
y − h2 (y) ≥ 0,
h02 (y) < 1.
Suppose that
(4.7)
|f (x, y, u, v)| ≤ a(x, y)|v|q + b(x, y)|v|,
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
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R ETARD I NTEGRAL I NEQUALITIES
11
where a, b ∈ C(R+ × R+ , R) and let
1
,
x∈R+ 1 − h01 (x)
If u(x, y) is any solution of (4.4) – (4.7), then
(i) if p = q, we have
M1 = max
(4.8)
(4.9) |u(x, y)| ≤
1
c(1− p ) +
M1 M2 (p − 1)
p
1
.
1 − h02 (y)
M2 = max
y∈R+
Z
φ1 (x)
Z
p
! p−1
φ1 (y)
eb(σ, τ )dσdτ
0
0
"
M1 M2
× exp
p
(ii) if p > q, we have

(4.10) |u(x, y)| ≤  c
(1− p1 )
M1 M2 (p − 1)
+
p
Z
φ1 (x)
Z
Z
φ1 (x)
0
Z
0
#
φ1 (y)
e
a(σ, τ )dσdτ
p−q
! p−1
φ1 (y)
eb(σ, τ )dσdτ
0
0
M1 M2 (p − q)
+
p
Z
0
φ1 (x)
Z
0
1
# p−q
φ1 (y)
e
a(σ, τ )dσdτ
.
In which φ1 (x) = x − h1 (x), x ∈ Rn+ , φ2 (y) = y − h2 (y), y ∈ Rn+ and
eb(σ, τ ) = b(σ + h1 (s), τ + h2 (t)), e
a(σ, τ ) = a(σ + h1 (s), τ + h2 (t)),
for σ, s, τ, t ∈ Rn+ .
In fact, if u(x, y) is a solution of (4.4) – (4.7), then it satisfies the equivalent integral
equation:
Z xZ y
p
(4.11) [u(x, y)] = a1 (x) + a2 (y) +
f (s, t, u(s, t), u(s − h1 (s), t − h2 (t)))dtds.
0
(Rn+
0
Rn+ , R).
for x, y ∈
×
Using (4.5), (4.7) in (4.11) and making the change of variables, we have
Z φ1 (x) Z φ1 (y)
p
(4.12)
|u(x, y)| ≤ c + M1 M2
e
a(σ, τ )|u(σ, τ )|q + eb(σ, τ )|u(σ, τ )|dσdτ.
0
0
Now a suitable application of the inequality in Corollary 3.10 to (4.12) yields (4.9) and
(4.10).
R EFERENCES
[1] 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.
[2] B.G. PACHPATTE, Explicit bounds on certain integral inequalities, J. Math. Anal. Appl., 267 (2002),
48–61.
[3] B.G. PACHPATTE, On some new inequalities related to a certain inequality arising in the theory of
differential equations, J. Math. Anal. Appl., 251 (2000), 736–751.
[4] B.G. PACHPATTE, On a certain inequality arising in the theory of differential equations, J. Math.
Anal. Appl., 182 (1994), 143–157.
[5] I. BIHARI, A generalization of a lemma of Bellman and its application to uniqueness problems of
differential equations, Acta. Math. Acad. Sci. Hungar., 7 (1956), 71–94.
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
http://jipam.vu.edu.au/
12
X UEQIN Z HAO AND FANWEI M ENG
[6] M. MEDVED, Nonlinear singular integral inequalities for functions in two and n independent variables, J. Inequalities and Appl., 5 (2000), 287–308.
[7] O. LIPOVAN, A retarded Gronwall-like inequality and its applications, J. Math. Anal. Appl., 252
(2000), 389–401.
[8] O. LIPOVAN, A retarded integral inequality and its applications, J. Math. Anal. Appl., 285 (2003),
436–443.
J. Inequal. Pure and Appl. Math., 7(3) Art. 111, 2006
http://jipam.vu.edu.au/