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Volume 2010, Article ID 275826, 15 pagesdoi:10.1155/2010/275826 Research Article Gronwall-OuIang-Type Integral Inequalities on Time Scales Ailian Liu1, 2 and Martin Bohner2 1 School of S

Volume 2010, Article ID 275826, 15 pages

doi:10.1155/2010/275826

Research Article

Gronwall-OuIang-Type Integral Inequalities on Time Scales

Ailian Liu1, 2 and Martin Bohner2

1 School of Statistics and Mathematics, Shandong Economic University, Jinan 250014, China

2 Department of Mathematics and Statistics, Missouri University of Science and Technology,

Rolla, MO 65409-0020, USA

Correspondence should be addressed to Martin Bohner,bohner@mst.edu

Received 20 April 2010; Accepted 3 August 2010

Academic Editor: Wing-Sum Cheung

Copyrightq 2010 A Liu and M Bohner 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 present several Gronwall-OuIang-type integral inequalities on time scales Firstly, an OuIang inequality on time scales is discussed Then we extend the Gronwall-type inequalities to multiple integrals Some special cases of our results contain continuous Gronwall-type inequalities and their discrete analogues Several examples are included to illustrate our results at the end

1 Introduction

OuIang inequalities and their generalizations have proved to be useful tools in oscillation theory, boundedness theory, stability theory, and other applications of differential and difference equations A nice introduction to continuous and discrete OuIang inequalities can

be found in1,2, and studies in 3 5 give some of their generalizations to multiple integrals and higher-dimensional spaces Like Gronwall’s inequality, OuIang’s inequality is also used

to obtain a priori bounds on unknown functions Therefore, integral inequalities of this type

The calculus on time scales has been introduced by Hilger7 in order to unify discrete

In this paper, we are concerned with Gronwall-OuIang-type integral inequalities on time scales, which unify and extend the corresponding continuous inequalities and their discrete

2 OuIang Inequality

Throughout this section, we fix t0∈ T and let T

t  {t ∈ T : t ≥ t0}

Lemma 2.1 Let y ∈ Crd, p∈ R, p t ≥ 0, for all t ∈ T

t0, and α ∈ R Then

y t ≤ α 

t

t0

implies that

y t ≤ αept, t0 ∀t ∈ T



p  qt : p t − qt

8,9

Now we will give the OuIang inequality on time scales

Theorem 2.2 Let u and v be real-valued nonnegative rd-continuous functions defined on T

t0 If

u2t ≤ c 

t

t0

where c is a positive constant, then

u t ≤√c 1

2

t

t0

Proof Let

w t 

t

t0

From2.4, we have

The definition of w gives

Dividing both sides of2.8 byc  wt and integrating from t0to t∈ T

t0, we have

t

t0

wΔτ



t

t0

2

t

t0

√

c  wΔτΔτ 

t

t0

2wΔτ



≤

t

t0

wΔτ



≤

t

t0

v τΔτ,

2.10

so



c  wt −√c≤ 1

2

t

t0

Combining2.4 and 2.11 yields 2.5 and completes the proof

Theorem 2.3 Let y and g be nonnegative rd-continuous functions on T

t0 Let α, M, N be nonnegative constants and −α ∈ R If

y2t ≤ M2y2t0  2

t

t0



then

y t ≤ Myt0e −α t, t0 

t

t0

Ng τe −α t, τΔτ ∀t ∈ T

Proof Let

z t  M2y2t0  2

t

t



zΔt  2αy2t  2Ngtyt ≤ 2αzt  2Ngtz t

≤ αz σt 

z t z σt

2.15

Hence,

zΔt



z t z σt − α



Multiplying both sides of2.16 by e −α t, t0, we have

√

Integrating2.17 from t0to t, we obtain that



z te −α t, t0 ≤ Myt0 

t

t0

completes the proof

Remark 2.4 If α  0 and N  1/2, thenTheorem 2.3reduces toTheorem 2.2

Remark 2.5 If we multiply inequality2.16 by another exponential function on time scales,

for example, e 2α t, t0, we could get another kind of inequality, which is a special case of

Theorem 3.4

3 Gronwall-OuIang-Type Inequality

Pachpatte discussed several integral inequalities arising in the theory of differential equations and difference equations 3,4 Now, we extend some of these results to time scales First, we give some notations and definitions which are used in our subsequent discussion

such that

and define the differential operators Li, 0≤ i ≤ n, by

L0x  x, L i x 1

For t∈ T

0, we set

A t, r1, , r n−1, r 

t

0

r1t1 · · ·

t n−2

0

r n−1tn−1

t n−1

0

r tnΔtnΔtn−1· · · Δt1. 3.3

Theorem 3.1 Let F and r be real-valued nonnegative rd-continuous functions on T

0, and let q > 1

be a constant If

F q t ≤ c  At, r1, , r n−1, rF  ∀t ∈ T

where c > 0 is a constant, then

F t ≤

c q−1/qq− 1

q A t, r1, , r n−1, r

1/q−1

∀t ∈ T

Proof Let

From3.6, it is easy to observe that

z 1/qΔ

q z

Δ1

0

z  μzΔh1/q−1

we have

L n z



z 1/qσ ≤ L n z

z 1/q ≤ r ≤ r  L n−1z



z 1/qΔ

z 1/q

that is,

L n−1z

z 1/q

Δ

Integrating3.10 with respect to tn from 0 to t and using the fact that Ln−1z0  0, we obtain that

L n−1z t

z 1/q t ≤

t

0

which implies that

Ln−2zΔt

r n−1tz 1/q t ≤

t

0

Ln−2zΔt



z 1/qσ

Ln−2zΔt

z 1/q t ≤ rn−1t

t

0

r tnΔtn ≤ rn−1t

t

0

r tnΔtn L n−2z tz 1/qΔ

t

z 1/q tz 1/qσ

t ,

3.13 that is,

L n−2z

z 1/q

Δ

t ≤ rn−1t

t

0

By setting t  tn−1in3.14 and integrating with respect to tn−1from 0 to t and using the fact that Ln−2z0  0, we get

L n−2z t

z 1/q t ≤

t

0

r n−1tn−1

t n−1

0

Continuing this way, we obtain that

L1z t

z 1/q t ≤

t

0

r2t2 · · ·

t n−2

0

r n−1tn−1

t n−1

0

that is,

zΔt

z 1/q t ≤ r1t

t

0

r2t2 · · ·

t n−2

0

r n−1tn−1

t n−1

0

For zΔt ≥ 0, from the chain rule in 8, Theorem 1.90,

1

1− 1/q z −1/q1

Δ

 zΔ1

0

z  hμzΔ−1/q

dh

 z −1/q zΔ

1

0



1 hμ zΔ

z

−1/q

dh

≤ z −1/q zΔ.

3.18

Letting t  t1in3.17 and integrating with respect to t1from 0 to t, we have

q

q− 1 zt q−1/q − z0 q−1/q

≤

t

0

zΔt1

z 1/q t1Δt1

≤

t

0

r1t1

t1

0

r2t2 · · ·

t n−2

0

r n−1tn−1

t n−1

0

r tnΔtnΔtn−1· · · Δt2Δt1,

3.19

which means that

F t ≤ z 1/q t ≤

c q−1/qq− 1

q A t, r1, r2, , r n−1, r

1/q−1

This completes the proof

Remark 3.2. Theorem 3.1also holds for c 0 To show this, assume 3.4 holds for c  0, that

is,

F q t ≤ At, r1, , r n−1, rF  ∀t ∈ T

Now, let d > 0 be arbitrary Then

F q t ≤ d  At, r1, , r n−1, rF  ∀t ∈ T

that is,3.4 holds for c  d ByTheorem 3.1,3.5 also holds for c  d, that is,

F t ≤

d q−1/qq− 1

q A t, r1, , r n−1, r

1/q−1

∀t ∈ T

Since3.23 holds for arbitrary d > 0, we may let d → 0in3.23 to arrive at

F t ≤

q− 1

q A t, r1, , r n−1, r

1/q−1

∀t ∈ T

that is,3.5 holds for c  0.

Theorem 3.3 Let u, v, and h j for j  1, 2, 3, 4 be real-valued nonnegative rd-continuous functions

on t∈ T

0 and let q > 1 be a constant If c1, c2, and α are nonnegative constants such that

u q t ≤ c1 At, r1, , r n−1, h1u   At, r1, , r n−1, h2v  ∀t ∈ T

v q t ≤ c2 At, r1, , r n−1, h3u   At, r1, , r n−1, h4v  ∀t ∈ T

where u  e q

α ·, 0u and v  e q α·, 0v, then for all t ∈ T

0,

u t ≤ eαt, 0

2q−1c1 c2q−1/qq− 1

q A t, r1, , r n−1, 2 q−1h

1/q−1

,

v t ≤

2q−1c1 c2q−1/qq− 1

q A t, r1, , r n−1, 2 q−1h

1/q−1

,

3.27

where h t  max{h1t  h3t, h2t  h4t}.

Proof Multiplying3.25 by e q

α t, 0 yields

e q α t, 0u q t ≤ c1e α q t, 0  At, r1, , r n−1, h1u e q α t, 0  At, r1, , r n−1, h2v e q α t, 0

≤ c1 At, r1, , r n−1, h1u   At, r1, , r n−1, h2v .

3.28 Define

d2q≤ 2q−1d q

1d q

2, where d1, d2are nonnegative reals, and also noticing3.26 and e α t, 0 ≤

1, we get

F q t ≤ 2 q−1

e q α t, 0u q t  v q t

≤ 2q−1{c1 At, r1, , r n−1, h1u   At, r1, , r n−1, h2v

c2 At, r1, , r n−1, h3u   At, r1, , r n−1, h4v}

 2q−1{c1 c2  At, r1, , r n−1, h1 h3u  At, r1, , r n−1, h2 h4v}

≤ 2q−1c1 c2  At, r1, , r n−1, 2 q−1hF

.

3.30

Now,Theorem 3.1yields

F t ≤

2q−1c1 c2q−1/qq− 1

q A t, r1, , r n−1, 2 q−1h

1/q−1

concludes the proof

Theorem 3.4 Let q > 1 and B be the set of all nonnegative real-valued rd-continuous functions

defined on 0, t ∩ T Let K and L be monotone increasing linear operators on B If there exists a

positive constant c such that, for y ∈ B,

y q t ≤ c 

t

0



qL

y q

τ  Ky

τΔτ ∀t ∈ T

then, for all t∈ T

0,

y t ≤ e 1/q qL t, 0



c q−1/q q− 1

q

t

0

1/q−1

where L  Lid, K  Kid with ids ≡ 1 for all s ∈ T.

Proof Let

z t  c 

t

0



qL

y q

τ  Ky

Hence, zs ≤ zt for all 0 ≤ s ≤ t, so that z ≤ ztid on 0, t, and thus

Hence Lzt ≤ ztLt, and therefore Lz ≤ zL Similarly, Kz 1/q  ≤ z 1/q K Using this and

3.32, we obtain that

zΔ qLy q

 Ky

≤ qLz  Kz 1/q

e qL ·, 0zΔ  qL

e qL ·, 0z  e σ

qL ·, 0zΔ

e qL ·, 0z  1 μ  qL

eqL ·, 0zΔ

 e qL ·, 0  qL

z 1 μ  qL

zΔ

≤ e qL ·, 0  qL

z 1 μ  qL qLz  Kz 1/q

 e qL ·, 0



−qL



1/q

 e qL ·, 0



−qL

qL



Kz 1/q



 e qL ·, 0 1 μ  qL

Kz 1/q

3.37

In summary,

wΔ≤ 1 μ  qL

Kw 1/q e 1/q−1

Obviously

ww σ > 0, which implies wΔ

so that the chain rule9, Theorem 2.37 yields

1

1− 1/q w −1/q1

Δ

 w −1/q wΔ

1

0



1 hμ wΔ

w

−1/q

Dividing both sides of3.38 by w 1/qprovides that

w −1/q wΔ≤ 1 μ  qL

Ke 1/q−1

Integrating both sides of3.41 from 0 to t and noticing 3.40, we find that

q

q− 1 w1−1/qt − w1−1/q0



≤

t

0

τKe 1/q−1

Substitute the expression of wt, we have

z t

e qL t, 0 ≤



c q−1/qq− 1

q

t

0

1  μτqLτKe 1/q−1 qL τ, 0Δτ

q/q−1

Remark 3.5 As in the discussion inRemark 3.2,Theorem 3.4also holds true for c 0

4 Some Applications

In this section, we indicate some applications of our results to obtain the estimates of the solutions of certain integral equations for which inequalities obtained in the literature thus

dynamic equation

yΔΔ p σ ty  y σ

Theorem 4.1 Assume that p is a differentiable positive function such that pΔ is rd-continuous If there exist t0∈ T and M > 0 such that

1



p t e |pΔ|/2p t, t0 ≤ M ∀t ∈ T



then all nonoscillatory solutions of 4.1 are bounded.

Proof Let y be a nonoscillatory solution of4.1 Without loss of generality, we assume there

exists t0∈ T such that

y t > 0 ∀t ∈ T

Then

yΔΔt  −p σ ty t  y σ t< 0 ∀t ∈ T

Hence, yΔis strictly decreasing onT

t0 Thus, either

yΔt > 0 ∀t ∈ T

or there exists t1∈ T

t0such that

yΔt < 0 ∀t ∈ T

We now claim that4.6 is impossible to hold To show this, let us assume that 4.6 is true

Then y is strictly decreasing onT

t1and

y t  yt1 

t

t1

yΔτΔτ ≤ yt1  yΔt1t − t1 ∀t ∈ T

Hence, there exists t2∈ T

t1such that

y t < 0 ∀t ∈ T

contradicting yt > 0 for all t ∈ T

t0 Similarly, we can prove that if yt < 0, then yΔΔt > 0 and yΔt ≤ 0 for t ∈ T

t1 Multiplying4.1 on both sides by yΔand taking integral from t1to t, we have

t

t

yΔτyΔΔτΔτ 

t

t

From the integration by parts in8, Theorem 1.77,

yΔt2− yΔt12−

t

t1

yΔΔτy Δσ τΔτ  pty2t − pt1y2t1 −

t

t1

pΔτy2τΔτ  0,

4.10

Thus, with c1 pt1y2t1  yΔt12> 0, we have



p tyt 2 ≤ c1

t

t1

pΔτy τ



p τ



p τy τΔτ ∀t ∈ T

Theorem 2.2gives that



p tyt

 ≤√c1 1

2

t

t1

pΔτy τ



p τ Δτ 

√

c1

t

t1

pΔτ

2pτ



p τyτ

t1.

4.12



p tyt

 ≤√c1e |pΔ|/2p t, t1 ∀t ∈ T

Hence,

y t ≤ √c1

1



p t e |pΔ|/2p t, t1 ≤

√

which completes the proof

also obtain the following results

Corollary 4.2 Let T  R If p is a continuously differentiable positive function such that p is nonnegative, then all nonoscillatory solutions of 4.1 are bounded.

Proof ForT  R, we have

1



p t e |pΔ|/2p t, 0 

1



p t e

t

0pτ/2pτdτ

p t e

1/2 lnpt/p0

p t

p t

p0

1/2

p0,

4.15

Example 4.3 Consider the nonlinear one-dimensional integral equation of the form

u q t  ft 

t

0

where f :T

0 → R, k : T

0× T

0 → R, g : T

0× R → R are rd-continuous functions, and q > 1 is

and Okrasi ´nski has studied the existence and uniqueness of solutions14

Here, we assume that every solution u of4.16 exists on the interval T

that the functions f, k, g in4.16 satisfy the conditions

From4.16 and using 4.17, it is easy to observe that

|ut| q ≤ c1

t

0

|ut| ≤



c q−1/q1  q− 1

q

t

0

c2r sΔs

1/q−1

which gives the bound on u.

f t≤ c1e q α t, 0, |kt, s| ≤ hse q

where c1 and r are as above, α > 0 is a constant, h :T

and

∞

0

From4.16 and 4.20, it is easy to observe that

|eαt, 0ut| q ≤ c1

t

0

e αt, 0|ut| ≤



c q−1/q1 q− 1

q

t

0

h srse α s, 0Δs

1/q−1

So,

|ut| ≤ c∗e α t, 0, where c∗ c q−1/q1  q− 1

q

∞

0

From4.24, we see that the solution ut of 4.16 approaches zero as t → ∞.

Acknowledgments

This work is supported by Grants 60673151 and 10571183 from NNSF of China, and by Grant 08JA910003 from Humanities and Social Sciences in Chinese Universities

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