Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2010, Article ID 984141, 14 pages
doi:10.1155/2010/984141
Research Article
Some Nonlinear Integral Inequalities in
Two Independent Variables
Wei Nian Li1, 2
1
2
Department of Mathematics, Binzhou University, Shandong 256603, China
School of Mathematics Science, Qufu Normal University, Shandong 273165, China
Correspondence should be addressed to Wei Nian Li, wnli@263.net
Received 14 January 2010; Accepted 29 May 2010
Academic Editor: Binggen Zhang
Copyright q 2010 Wei Nian Li. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
We investigate some new nonlinear integral inequalities in two independent variables. The
inequalities given here can be used as tools in the qualitative theory of certain nonlinear partial
differential equations.
1. Introduction
It is well known that the integral inequalities involving functions of one and more than
one independent variables which provide explicit bounds on unknown functions play a
fundamental role in the development of the theory of differential equations. In the past
few years, a number of integral inequalities had been established by many scholars, which
are motivated by certain applications. For details, we refer to literatures 1–10 and the
references therein. In this paper we investigate some new nonlinear integral inequalities in
two independent variables, which can be used as tools in the qualitative theory of certain
partial differential equations.
2. Main Results
In what follows, R denotes the set of real numbers and R 0, ∞ is the given subset of R.
The first-order partial derivatives of a zx, y defined for x, y ∈ R with respect to x and y are
denoted by zx x, y, and zy x, y respectively. Throughout this paper, all the functions which
appear in the inequalities are assumed to be real-valued and all the integrals involved exist
on the respective domains of their definitions, CM, S denotes the class of all continuous
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functions defined on set M with range in the set S, p and q are constants, and p ≥ 1,
0 < q ≤ p.
We firstly introduce two lemmas, which are useful in our main results.
Lemma 2.1 Bernoulli’s inequality 11. Let 0 < α ≤ 1 and x > −1. Then 1 xα ≤ 1 αx.
Lemma 2.2 see 7. Let ut, at, and bt be nonnegative and continuous functions defined for
t ∈ R .
i Assume that at is nondecreasing for t ∈ R . If
ut ≤ at t
2.1
bsusds,
0
for t ∈ R , then
ut ≤ at exp
t
2.2
bsds ,
0
for t ∈ R .
ii Assume that at is nonincreasing for t ∈ R . If
ut ≤ at ∞
2.3
bsusds,
t
for t ∈ R , then
ut ≤ at exp
∞
bsds ,
2.4
t
for t ∈ R .
Next, we establish our main results.
Theorem 2.3. Let ux, y, ax, y, bx, y, fx, y, gx, y ∈ CR2 , R and ax, y > 0.
i If
up x, y ≤ a x, y b x, y
x ∞
0
fs, tup s, t gs, tuq s, t dt ds,
x, y ∈ R ,
y
E1
then
1
u x, y ≤ a1/p x, y a1/p−1 x, y b x, y h x, y exp
p
x ∞
Fs, tdt ds ,
0
x, y ∈ R ,
y
2.5
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3
where
x ∞
h x, y 0 y fs, tas, t gs, taq/p s, t dt ds,
2.6
q
F x, y b x, y f x, y aq/p−1 x, y g x, y .
p
2.7
ii If
u x, y ≤ a x, y b x, y
p
∞ ∞
x
fs, tup s, t gs, tuq s, t dt ds,
x, y ∈ R ,
y
E 1
then
1
u x, y ≤ a1/p x, y a1/p−1 x, y b x, y h x, y exp
p
∞ ∞
Fs, tdt ds ,
x
y
x, y ∈ R ,
2.5 where
h x, y ∞ ∞
fs, tas, t gs, taq/p s, t dt ds,
x
y
2.6 and Fx, y is defined by 2.7.
Proof. We only give the proof of i. The proof of ii can be completed by following the proof
of i.
i Define a function zx, y by
z x, y x ∞
0
fs, tup s, t gs, tuq s, t dt ds,
x, y ∈ R .
y
2.8
Then E1 can be restated as
b x, y z x, y
.
u x, y ≤ a x, y b x, y z x, y a x, y 1 a x, y
p
2.9
Using Lemma 2.1, from 2.9, we easily obtain
1
u x, y ≤ a1/p x, y a1/p−1 x, y b x, y z x, y ,
p
2.10
q
uq x, y ≤ aq/p x, y aq/p−1 x, y b x, y z x, y .
p
2.11
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Combining 2.8, 2.9, and 2.11, we have
z x, y ≤
x ∞
fs, tas, t bs, tzs, t
0
y
q
gs, t aq/p s, t aq/p−1 s, tbs, tzs, t dt ds
p
x ∞
Fs, tzs, tdt ds, x, y ∈ R ,
h x, y 0
2.12
y
where hx, y and Fx, y are defined by 2.6 and 2.7, respectively. Obviously, hx, y is
nonnegative, continuous, nondecreasing in x, and nonincreasing in y for x, y ∈ R .
Firstly, we assume that hx, y > 0 for x, y ∈ R . From 2.12, we easily observe that
x ∞
z x, y
Fs, tzs, t
dt ds.
≤1
hs, t
h x, y
0 y
2.13
Letting
v x, y 1 x ∞
0
y
Fs, tzs, t
dt ds,
hs, t
2.14
we easily see that vx, y is nonincreasing in y, y ∈ R , and
z x, y ≤ h x, y v x, y ,
x, y ∈ R .
2.15
Therefore,
vx x, y ∞
y
≤
Fx, tzx, t
dt
hx, t
∞
2.16
Fx, tvx, tdt
y
≤ v x, y
∞
Fx, tdt.
y
Treating y, y ∈ R , fixed in 2.16, dividing both sides of 2.16 by vx, y, setting x s, and
integrating the resulting inequality from 0 to x, x ∈ R , we have
v x, y ≤ exp
x ∞
Fs, t dt ds .
0
y
2.17
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It follows from 2.15 and 2.17 that
z x, y ≤ h x, y exp
x ∞
2.18
Fs, t dt ds .
0
y
Therefore, the desired inequality 2.5 follows from 2.10 and 2.18.
If hx, y is nonnegative, we carry out the above procedure with hx, y ε instead
of hx, y, where ε > 0 is an arbitrary small constant, and subsequently pass to the limit as
ε → 0 to obtain 2.5. This completes the proof.
Theorem 2.4. Assume that ux, y, ax, y, bx, y, fx, y, gx, y ∈ CR2 , R , and L ∈
CR3 , R . Let ax, y > 0, and
0 ≤ Ls, t, u − Ls, t, v ≤ Ks, t, vu − v,
2.19
for u ≥ v ≥ 0, where K ∈ CR3 , R .
i If
up x, y ≤ a x, y b x, y
x ∞
0
fs, tuq s, t Ls, t, us, t dt ds,
x, y ∈ R ,
y
E2
then
x ∞
1 1/p−1 1/p
u x, y ≤ a
Hs, tdt ds ,
x, y a
x, y b x, y G x, y exp
p
0 y
x, y ∈ R ,
2.20
where
G x, y x ∞
fs, taq/p s, t L s, t, a1/p s, t dt ds,
0
2.21
y
1 1/p−1 q H x, y f x, y aq/p−1 x, y K x, y, a1/p x, y
a
x, y b x, y .
p
p
2.22
ii If
up x, y ≤ a x, y b x, y
∞ ∞
x
y
fs, tuq s, t Ls, t, us, t dt ds,
x, y ∈ R ,
E 2
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then
1
u x, y ≤ a1/p x, y a1/p−1 x, y b x, y G x, y
p
∞ ∞
× exp
Hs, tdt ds , x, y ∈ R ,
x
2.20 y
where
G x, y ∞ ∞
fs, taq/p s, t L s, t, a1/p s, t dt ds,
x
y
2.21 and Hx, y is defined by 2.22.
Proof. We only prove the part i. The proof of ii can be completed by following the proof of
i.
i Define a function zx, y by
z x, y x ∞
0
fs, tuq s, t Ls, t, us, t dt ds.
2.23
y
Then, as in the proof of Theorem 2.3, we obtain 2.9–2.11. Therefore, we have
x ∞
0
y
x ∞
0
x ∞
fs, tuq s, tdt ds ≤
y
0
y
q
fs, t aq/p s, t aq/p−1 s, tbs, tzs, t dt ds,
p
2.24
x ∞ 1 1/p−1
1/p
Ls, t, us, tdt ds ≤
L s, t, a s, t a
s, tbs, tzs, t
p
0 y
− L s, t, a1/p s, t L s, t, a1/p s, t dt ds
≤
x ∞ L s, t, a1/p s, t dt ds
0
y
x ∞
0
y
1
K s, t, a1/p s, t a1/p−1 s, tbs, tzs, tdt ds.
p
2.25
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It follows from 2.23–2.25 that
z x, y ≤
x ∞
0
fs, taq/p s, t L s, t, a1/p s, t dt ds
y
x ∞
0
y
q
fs, taq/p−1 s, tbs, t
p
K s, t, a
G x, y 1/p
2.26
1
s, t a1/p−1 s, tbs, t zs, tdt ds
p
x ∞
Hs, tzs, tdt ds,
0
y
where Gx, y and Hx, y are defined by 2.21 and 2.22, respectively.
It is obvious that Gx, y is nonnegative, continuous, nondecreasing in x, and
nonincreasing in y for x, y ∈ R . By following the proof of Theorem 2.3, from 2.26, we
have
z x, y ≤ G x, y exp
x ∞
2.27
Hs, t dt ds .
0
y
Combining 2.10 and 2.27, we obtain the desired inequality 2.20. The proof is complete.
Theorem 2.5. Let ax, y, ux, y, Ls, t, u, and Ks, t, u be the same as in Theorem 2.4, and
rx, y ∈ CR2 , R .
i Assume that ax, y is nondecreasing in x, x ∈ R , and the condition 2.19 holds. If
up x, y ≤ a x, y x
0
r s, y up s, y ds x ∞
0
Ls, t, uq s, tdt ds,
x, y ∈ R ,
y
E3
then
u x, y ≤ B1/p x, y
x ∞
1 1/p−1 1/p
x, y a
x, y J x, y exp
Ms, tdt ds ,
× a
p
0 y
x, y ∈ R ,
2.28
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where
B x, y exp
x
r s, y ds ,
2.29
0
J x, y x ∞ L s, t, Bq/p s, taq/p s, t dt ds,
0
2.30
y
q
M x, y K x, y, Bq/p x, y aq/p x, y Bq/p x, y aq/p−1 x, y .
p
2.31
ii Assume that ax, y is nonincreasing in x, x ∈ R , and the condition 2.19 holds. If
up x, y ≤ a x, y ∞
r s, y up s, y ds x
∞ ∞
x
Ls, t, uq s, tdt ds,
x, y ∈ R ,
y
E 3
then
u x, y ≤ B1/p x, y
∞ ∞
1
t dt ds ,
× a1/p x, y a1/p−1 x, y J x, y exp
Ms,
p
x
y
x, y ∈ R ,
2.28 where
B x, y exp
∞
r s, y ds ,
2.29 x
J x, y ∞ ∞ L s, t, Bq/p s, taq/p s, t dt ds,
x
2.30 y
x, y K x, y, Bq/p x, y aq/p x, y Bq/p x, y q aq/p−1 x, y .
M
p
2.31 Proof. i Define a function zx, y by
z x, y a x, y v x, y ,
2.32
where
v x, y x ∞
0
Ls, t, uq s, tdt ds.
2.33
y
Then E3 can be restated as
u x, y ≤ z x, y p
x
0
r s, y up s, y ds.
2.34
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Noting the assumption that ax, y is nondecreasing in x, x ∈ R , we easily see that
zx, y is a nonnegative and nondecreasing function in x, x ∈ R . Therefore, treating y, y ∈
R , fixed in ?? and using part i of Lemma 2.2 to ??, we get
up x, y ≤ B x, y z x, y .
2.35
that is,
v x, y
u x, y ≤ B x, y a x, y v x, y B x, y a x, y 1 ,
a x, y
p
2.36
where Bx, y is defined by 2.29. Using Lemma 2.1, from 2.35 we have
1
u x, y ≤ B 1/p x, y a1/p x, y a1/p−1 x, y v x, y ,
p
q
uq x, y ≤ Bq/p x, y aq/p x, y aq/p−1 x, y v x, y .
p
2.37
2.38
Combining 2.33 and 2.37, and noting the hypotheses 2.19, we obtain
x ∞ q
L s, t, Bq/p s, t aq/p s, t aq/p−1 s, tvs, t
p
0 y
q/p
q/p
q/p
q/p
−L s, t, B s, ta s, t L s, t, B s, ta s, t dt ds
x ∞
≤ J x, y Ms, tvs, tdt ds,
v x, y ≤
0
2.39
y
where Jx, y and Mx, y are defined by 2.30 and 2.31, respectively.
It is obvious that Jx, y is nonnegative, continuous, nondecreasing in x and
nonincreasing in y for x, y ∈ R . By following the proof of Theorem 2.3, from 2.38, we
obtain
v x, y ≤ J x, y exp
x ∞
Ms, tdt ds ,
0
x, y ∈ R .
2.40
y
Obviously, the desired inequality 2.28 follows from 2.36 and 2.39.
ii Noting the assumption that ax, y is nonincreasing in x, x ∈ R , and using the
part ii of Lemma 2.2, we can complete the proof by following the proof of i with suitable
changes. Therefore, the details are omitted here.
By using the ideas of the proofs of Theorems 2.5 and 2.3, we easily prove the following
theorem.
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Theorem 2.6. Let ax, y, ux, y, rx, y, fx, y, gx, y ∈ CR2 , R , and ax, y > 0.
i Assume that ax, y is nondecreasing in x, x ∈ R . If
up x, y ≤ a x, y x ∞
0
x
r s, y up s, y ds
0
fs, tup s, t gs, tuq s, t dt ds,
E4
x, y ∈ R ,
y
then
u x, y ≤ B1/p x, y
x ∞
1 1/p−1 1/p
P s, tdt ds ,
× a
x, y a
x, y H x, y exp
p
0 y
x, y ∈ R ,
2.41
where
H x, y x ∞
fs, tBs, tas, t gs, tBq/p s, taq/p s, t dt ds,
0
y
q
P x, y f x, y B x, y g x, y Bq/p x, y aq/p−1 x, y ,
p
2.42
and Bx, y is defined by 2.29.
ii Assume that ax, y is nonincreasing in x, x ∈ R . If
u x, y ≤ a x, y p
∞ ∞
x
∞
r s, y up s, y ds
x
fs, tup s, t gs, tuq s, t dt ds,
E 4
x, y ∈ R ,
y
then
u x, y ≤ B1/p x, y
∞ ∞
1 1/p−1 1/p
x, y a
x, y H x, y exp
× a
P s, tdt ds ,
p
x
y
x, y ∈ R ,
2.40 Advances in Difference Equations
11
where
tas, t gs, tBq/p s, taq/p s, t dt ds,
x, y ∞ ∞ fs, tBs,
H
x y
q
P x, y f x, y B x, y g x, y Bq/p x, y aq/p−1 x, y ,
p
2.41 y is defined by 25 .
and Bx,
Remark 2.7. Noting that p and q are constants, and p ≥ 1, 0 < q ≤ p, we can obtain many
special integral inequalities by using our main results. For example, let p 1, q 1/4, and
p q 2, respectively; from Theorem 2.3, we obtain the following corollaries.
Corollary 2.8. Let ux, y, ax, y, bx, y, fx, y, gx, y ∈ CR2 , R and ax, y > 0.
i If
u x, y ≤ a x, y b x, y
x ∞
0
fs, tus, t gs, tu1/4 s, t dt ds,
x, y ∈ R ,
y
E5
then
x ∞
x, y exp
tdt ds ,
u x, y ≤ a x, y b x, y h
Fs,
0
x, y ∈ R ,
2.43
y
where
x, y h
x ∞
fs, tas, t gs, ta1/4 s, t dt ds,
0
2.44
y
1 −3/4 x, y g x, y .
F x, y b x, y f x, y a
4
2.45
ii If
u x, y ≤ a x, y b x, y
∞ ∞
x
fs, tus, t gs, tu1/4 s, t dt ds,
x, y ∈ R ,
y
E 5
then
x, y exp
u x, y ≤ a x, y b x, y h
x ∞
0
y
Fs, tdt ds ,
x, y ∈ R ,
2.42 12
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where
x, y h
∞ ∞
fs, tas, t gs, ta1/4 s, t dt ds,
x
y
2.43 y is defined by 2.44.
and Fx,
Corollary 2.9. Let ux, y, ax, y, bx, y, gx, y ∈ CR2 , R and ax, y > 0.
i If
u x, y ≤ a x, y b x, y
2
x ∞
0
gs, tu2 s, tdt ds,
x, y ∈ R ,
y
E6
then
1
u x, y ≤ a1/2 x, y a−1/2 x, y b x, y m x, y
2
x ∞
× exp
bs, tgs, tdt ds , x, y ∈ R ,
0
2.46
y
where
m x, y x ∞
2.47
gs, tas, tdt ds.
0
y
ii If
u x, y ≤ a x, y b x, y
2
∞ ∞
x
gs, tu2 s, tdt ds,
x, y ∈ R ,
y
E 6
then
1
x, y
u x, y ≤ a1/2 x, y a−1/2 x, y b x, y m
2
x ∞
bs, tgs, tdt ds , x, y ∈ R ,
× exp
0
2.45 y
where
m
x, y ∞ ∞
gs, tas, tdt ds.
x
y
2.46 Advances in Difference Equations
13
Remark 2.10. If we add ax, y > 0 to the assumptions of 7, Theorems 2.2–2.4, then we easily
see that 7, Theorems 2.2–2.4 are special cases of Theorems 2.3, 2.5, and 2.6, respectively.
Therefore, our paper gives some extensions of the results of 7 in a sense.
3. An Application
In this section, using Theorem 2.3, we obtain the bound on the solution of a nonlinear
differential equation.
Example 3.1. Consider the partial differential equation:
pup−1 x, y uxy x, y p p − 1 up−2 x, y ux x, y uy x, y h x, y, u x, y r x, y ,
ux, ∞ σx,
u ∞, y τ y ,
u∞, ∞ d,
3.1
where h ∈ CR2 × R, R, r ∈ CR2 , R, σ, τ ∈ CR , R, and d is a real constant, and p ≥ 1 is a
constant.
Suppose that
h x, y, u x, y ≤ f x, y u x, y p g x, y u x, y q ,
∞ ∞
rs, tdt ds ≤ a x, y ,
σx τ y − d x
y
3.2
where ax, y, fx, y, gx, y ∈ CR2 , R and ax, y > 0 for x, y ∈ R , and 0 < q ≤ p is a
constant. Let ux, y be a solution of 3.1 for x, y ∈ R ; then
u x, y ≤ a1/p x, y 1 a1/p−1 x, y E x, y exp
p
∞ ∞
Ws, tdt ds ,
x
x, y ∈ R ,
y
3.3
where
E x, y ∞ ∞
fs, tas, t gs, taq/p s, t dt ds,
x
y
q
W x, y f x, y aq/p−1 x, y g x, y .
p
3.4
In fact, if ux, y is a solution of 3.1, then it can be written as see 1, page 80
p
σx τ y − d u x, y
∞ ∞
x
for x, y ∈ R .
y
hs, t, us, t rs, tdt ds,
3.5
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It follows from 3.2 and 3.5 that
u x, y p ≤ a x, y ∞ ∞
x
y
fs, t|us, t|p gs, t|us, t|q dt ds.
3.6
Now, a suitable application of part ii of Theorem 2.3 to 3.6 yields the required
estimate in 3.3.
Acknowledgment
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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