Gronwall-Bellman Type Inequalities On Time

WSEAS TRANSACTIONS on MATHEMATICS

Qinghua Feng, Fanwei Meng

Gronwall-Bellman Type Inequalities On Time Scales And Their Applications Qinghua Fenga,b,∗ Fanwei Meng University of Technology Qufu Normal University School of Science School of Mathematical Sciences Zhangzhou Road 12, Zibo, 255049 Jingxuan western Road 57, Qufu, 273165 China China b Qufu Normal University [email protected] School of Mathematical Sciences Jingxuan western Road 57, Qufu, 273165 China [email protected] a Shandong

Abstract: In this work, we investigate some new Gronwall-Bellman type dynamic inequalities on time scales in two independent variables, which provide a handy tool in deriving explicit bounds on unknown functions in certain dynamic equations on time scales. The established results generalize the main results on integral inequalities for continuous functions in [1] and their corresponding discrete analysis in [2]. Key–Words: Dynamic inequality; Gronwall-Bellman inequality; Time scales; Dynamic equation; Bounded

1

Introduction

wo independent variables on time scales containing integration on infinite intervals, which generalize the main results in [1] and [2], and present some applications for them. First we will give some preliminaries on calculus of time scales and some universal symbols used in this paper. Throughout the paper, R denotes the set of real numbers and R+ = [0, ∞). Z denotes the set of integers. For two given sets G, H, we denote the set of maps from G to H by (G, H). A time scale is an arbitrary nonempty closed subset of the real numbers. In this paper, T denotes an arbitrary time scale. On T we define the forward and backward jump operators σ ∈ (T, T) and ρ ∈ (T, T) such that σ(t) = inf {s ∈ T, s > t}, ρ(t) = sup{s ∈ T, s < t}.

During the past decades, with the development of the theory of differential and integral equations, a lot of integral and difference inequalities, for example, [3-12] and the references therein, have been discovered, which play an important role in the research of boundedness, global existence, stability of solutions of differential and integral equations as well as difference equations. In these inequalities, Gronwall-Bellman type inequalities and their generations have been paid much attention by many authors (for example, see [3-8]). On the other hand, Hilger [13] initiated the theory of time scales, and one of the purposes of the theory of time scales is to unify continuous and discrete analysis. Since then, many integral inequalities on time scales have been established(for example, see [14-20] and the references therein), which on one hand provide a handy tool in the study of qualitative as well as quantitative properties of solutions of certain dynamic equations on time scales, on the other hand unify continuous and discrete analysis to some extent. But to our knowledge, Gronwall-Bellman type inequalities in two independent on time scales containing integration on infinite intervals have been paid little attention in the literature so far. In this paper, we will establish some new Gronwall-Bellman type dynamic inequalities in t-

ISSN: 1109-2769

Definition 1 The graininess µ(t) ∈ (T, R+ ) is defined by µ(t) = σ(t) − t. Remark 2 Obviously, µ(t) = 0 if T = R while µ(t) = 1 if T = Z. Definition 3 A point t ∈ T is said to be leftdense if ρ(t) = t and t ̸= inf T, right-dense if σ(t) = t and t ̸= sup T, left-scattered if ρ(t) < t and right-scattered if σ(t) > t. Definition 4 The set Tκ is defined to be T if T does not have a left-scattered maximum, otherwise it is T without the left-scattered maximum. 239

Issue 7, Volume 10, July 2011



Definition 5 A function f (t) ∈ (T, R) is called rd-continuous if it is continuous in right-dense points and if the left-sided limits exist in left-dense points, while f is called regressive if 1+µ(t)f (t) ̸= 0. Crd denotes the set of rd-continuous functions, while R denotes the set of all regressive and rd-continuous functions, and R+ = {f |f ∈ R, 1 + µ(t)f (t) > 0, ∀t ∈ T}.

(i)

a [f∫(x, y) + g(x, y)]∆x b + a g(x, y)∆x,

∫b

(ii)

a (αf )(x, y)∆t

∫b

(iii) (iv)

Definition 6 For some t ∈ Tκ , and a function f ∈ (T, R), the delta derivative of f at t is denoted by f ∆ (t) (provided it exists) with the property such that for every ε > 0 there exists a neighborhood U of t satisfying

(v)

a

=α

f (x, y)∆x = −

∫b a

∫a b

=

∫b a

f (x, y)∆x

f (x, y)∆x,

f (x, y)∆x,

∫b

∫c f (x, y)∆x = a f (x, y)∆x ∫b + c f (x, y)∆x, a

∫a a

f (x, y)∆x = 0,

(vi) if f (x, y) ≥ 0 for all a ≤ x ≤ b, then ∫b a f (x, y)∆x ≥ 0.

|f (σ(t)) − f (s) − f ∆ (t)(σ(t) − s)| ≤ ε|σ(t) − s| for all s ∈ U.

Remark 10 If b = ∞, then all the conclusions of Theorem 1.1 still hold.

Similarly, for some x ∈ Tκ , and a function f (x, y) ∈ (T × T, R), the partial delta derivative of f (x, y) with respect to x is denoted by (f (x, y))∆ x , and satisfies

Definition 11 The cylinder transformation ξh is defined by  

|f (σ(x), y) − f (s, y) − (f (x, y))∆ x (σ(x) − s)| ≤ ε|σ(x) − s|, for ∀ε > 0,

ξh (z) =

where s ∈ U, and U is a neighborhood of x. The function f (x, y) is called partial delta dif f erential with respect to x on Tκ .

Log(1+hz) , h



if h ̸= 0 (f or z ̸= − h1 ),

z,

if h = 0,

where Log is the principal logarithm function. Definition 12 For p(x, y) ∈ R with respect to x, the exponential f unction is defined by ∫ x ξµ(τ ) (p(τ, y))∆τ ) ep (x, s) = exp(

Remark 7 If T = R, then f ∆ (t) becomes the usual derivative f ′ (t), while f ∆ (t) = f (t+1)−f (t) if T = Z, which represents the forward difference.

s

Definition 8 If F ∆ (t) = f (t), t ∈ Tκ , then F is called an antiderivative of f , and the Cauchy integral of f is defined by ∫b a f (t)∆t = F (b) − F (a), where a, b ∈ T.

for s, x ∈ T. Remark 13 If T = R, then for x ∈ R the following formula holds ∫ x ep (x, s) = exp( p(τ, y)dτ )

Similarly, for a, b ∈ T and a function f (x, y) : T×T → R, the Cauchy partial integral of f (x, y) with respect to x is defined by ∫b a f (x, y)∆x = F (b, y) − F (a, y), κ where (F (x, y))∆ x = f (x, y), x ∈ T .

s

f or s ∈ T. If T = Z, then for t ∈ Z, ep (x, s) =

The following theorem includes some important properties for Cauchy partial integral on time scales.

x−1 ∏

[1 + p(τ, y)]

τ =s

f or s ∈ T and s < x.

Theorem 9 If a, b, c ∈ T, α ∈ R, and f (x, y), g(x, y) ∈ Crd (T × T, R), then

ISSN: 1109-2769

∫b

The following two theorems include some known properties on the exponentialf unction.

240




Theorem 14 If p(x, y) ∈ R with respect to x, then the following conclusions hold (i) ep (x, x) ≡ 1, and e0 (x, s) ≡ 1, (ii) ep (σ(x), s) = (1 + µ(x)p(x, y))ep (x, s), (iii) If p ∈ R+ with respect to x, then ep (x, s) > 0 for ∀s, x ∈ T, (iv) If p ∈ R+ with respect to x, then ⊖p ∈ + R , 1 (v) ep (x, s) = ep (s,x) = e⊖p (s, x), where

The proof for Lemma 18 is similar to [21, Theorem 5.6], and we omit it here. Lemma 19 Under the conditions of Lemma 18, and furthermore assume a(x, y) is nondecreasing in x for every fixed y, then we have u(x, Y ) ≤ a(x, Y )em(.,Y ) (x, x0 ). Proof : Since a(x, y) is nondecreasing in x for every fixed y, then from Lemma 18 we have ∫ x u(x, Y ) ≤ a(x, Y ) + em(.,Y ) (x, σ(s))a(s, Y )

p(x,y) (⊖p)(x, y) = − 1+µ(x)p(x,y) .

Theorem 15 If p(x, y) ∈ R with respect to x, x0 ∈ T is a fixed number, then the exponential f unction ep (x, x0 ) is the unique solution of the following initial value problem { (z(x, y))∆ x = p(x, y)z(x, y), z(x0 , y) = 1.

x0

∫ ≤ a(x, Y )[1 +

em(.,Y ) (x, σ(s))m(s, Y )∆s]. x0

On the other hand, from [22, Theorem 2.39 and 2.36 (i)], ∫ x 1+ em(.,Y ) (x, σ(s))m(s, Y )∆s = em(.,Y ) (x, x0 ).

Remark 16 Theorem 9 14 and 15 are extensions of [16, Theorem 2.2] and [21, Theorem 5.2, 5.1] respectively.

x0

Remark 17 For more details about time scales, we advise the reader to refer to [22].

2

m(s, Y )∆s x

Then Collecting the above information we can obtain the desired inequality. Lemma 20 [23] Assume that a ≥ 0, p ≥ q ≥ 0, and p ̸= 0, then for any K > 0

Main Results

For convenience of notation, in the rest of this paper we always assume that ∩ ∩ e0 T0 = [x0 , ∞) T, T = [y0 , ∞) T, where x0 , y0 ∈ T, and furthermore assume e 0 ⊆ Tκ . T0 ⊆ Tκ , T

q

a p ≤ pq K

q−p p

q

a+

p−q p p K .

Theorem 21 Suppose sup y = ∞, u, f, g, h, e0 y∈T

e 0 , R+ ), p is a constant with a, b ∈ Crd (T0 × T e 0 , u(x, y) satisfies the p ≥ 1. If for (x, y) ∈ T0 × T following inequality ∫ x∫ ∞ p u (x, y) ≤ a(x, y) + b(x, y) [f (s, t)u(s, t)

We will give some lemmas for further use. e 0 is an arbitrarily Lemma 18 Suppose Y ∈ T fixed number, and u(x, Y ) ∈ Crd , m(x, Y ) ∈ R+ with respect to x, m(x, Y ) ≥ 0, then ∫ x u(x, Y ) ≤ a(x, Y ) + m(s, Y )u(s, Y )∆s, x ∈ T0

∫

x0

s∫

∞

+g(s, t)+

x0

h(ξ, η)u(ξ, η)∆η∆ξ]∆t∆s, (1) x0

implies

y

t

then ∫

u(x, Y ) ≤ a(x, Y ) +

x

em(.,Y ) (x, σ(s))a(s, Y )

∫

x0

x

+

m(s, Y )∆s, x ∈ T0 ,

x0

1

eB2 (.,y) (x, σ(s))B2 (s, y)B1 (s, y)∆s]} p , e0, X(x, y) ∈ T0 × T

where em(.,Y ) (x, x0 ) is the unique solution of the following problem { (z(x, Y ))∆ x = m(x, Y )z(x, Y ), z(x0 , Y ) = 1.

ISSN: 1109-2769

u(x, y) ≤ {a(x, y) + b(x, y)[B1 (x, y)

where

∫

x∫ ∞

B1 (x, y) = x0

241

y

(2)

1 1−p p − 1 p1 [f (s, t)( K p a(s, t)+ K ) p p Issue 7, Volume 10, July 2011


∫

s

∫

∞

+g(s, t) + x0

t


1 1−p h(ξ, η)( K p a(ξ, η) p

+b(ξ, η)v(ξ, η)) + ∫

p − 1 p1 + K )∆η∆ξ]∆t∆s, (3) p ∫ ∞ ∫ x∫ ∞ [f (x, t) + h(ξ, η)∆η∆ξ] B2 (x, y) = y

x0

x

(4)

∫

s

x0 ∞

[f (s, t)u(s, t) + g(s, t) x0

∫

s

∫

h(ξ, η)u(ξ, η)∆η∆ξ]∆t∆s. x0

∫

x0

1 1−p h(ξ, η) K p b(ξ, η)v(ξ, η)∆η∆ξ]∆t∆s p t ∫ x ≤ B1 (x, Y ) + B2 (s, Y )v(s, Y )∆s, (9) x0

where B1 (x, y), B2 (x, y) are defined in (3) and (4) respectively. From Lemma 18, it follows

(5)

t

v(x, Y ) ≤ B1 (x, Y )

Then e0 u(x, y) ≤ (a(x, y)+b(x, y)v(x, y)) p , (x, y) ∈ T0 ×T (6) e 0 , and let y ∈ [Y, ∞) ∩ T, x ∈ T0 . Then Fix Y ∈ T ∫ x∫ ∞ [f (s, t)u(s, t) + g(s, t) v(x, Y ) =

∫

1

x0

∫

s∫

≤

Y

∞

h(ξ, η)u(ξ, η)∆η∆ξ]∆t∆s

+ ∫

x0

x∫ ∞

x0

Y

s

∞

∫

+

x

+ x0

eB2 (.,Y ) (x, σ(s))B2 (s, Y )B1 (s, Y )∆s, x ∈ T0 .

(10) e Since Y ∈ T0 is arbitrary, then in fact (10) holds e 0 ), that is, for ∀(x, y) ∈ (T0 × T ∫ x v(x, y) ≤ B1 (x, y) + eB2 (.,y) (x, σ(s))B2 (s, y) x0

t

e 0 ). B1 (s, y)∆s, (x, y) ∈ (T0 × T

1 p

[f (s, t)(a(s, t) + b(s, t)v(s, t)) + g(s, t) ∫

Y

1 1−p [f (s, t) K p b(s, t)v(s, t) p

∞

y

∞

+

x∫ ∞

+

Proof : Let v(x, y) =

1 1−p p − 1 p1 K ) + g(s, t) [f (s, t)( K p a(s, t) + p p Y ∞ 1 1−p p − 1 p1 h(ξ, η)( K p a(ξ, η)+ K )∆η∆ξ]∆t∆s p p t

∫

x0

+ x∫

∞

= ∫ xs0∫ +

t

1 1−p K p b(x, t)∆t. p ∫

∫

p − 1 p1 K )∆η∆ξ]∆t∆s p

(11)

Then combining (6) and (11) we obtain the desired inequality.

1

h(ξ, η)(a(ξ, η)+b(ξ, η)v(ξ, η)) p ∆η∆ξ]∆t∆s. x0

t

Theorem 22 Under the conditions of Theorem e 0 , u(x, y) satisfies (1), 2.1, if for (x, y) ∈ T0 × T then then

(7)

From Lemma 20, for ∀K > 0 we have 1

(a(x, y) + b(x, y)v(x, y)) p

1

u(x, y) ≤ {a(x, y)+b(x, y)B1 (x, y)eB2 (.,y) (x, x0 )} p ,

1 1−p p − 1 p1 ≤ K p (a(x, y) + b(x, y)v(x, y)) + K , p p e0. (x, y) ∈ T0 × T

e0, (x, y) ∈ T0 × T

(8)

where B1 (x, y), B2 (x, y) are defined the same as in Theorem 21.

So combining (7) and (8), considering v(x, y) is decreasing in y, it follows ∫ x∫ ∞ 1 1−p v(x, Y ) ≤ [f (s, t)( K p (a(s, t) p x0 Y

Proof : Considering B1 (x, y) is nondecreasing in x, from (2) we have u(x, y) ≤ {a(x, y) + b(x, y)[B1 (x, y)

p − 1 p1 K ) + g(s, t) +b(s, t)v(s, t)) + p ∫ s∫ ∞ 1 1−p + h(ξ, η)( K p (a(ξ, η) p x0 t ISSN: 1109-2769

(12)

∫

x

+ x0

1

eB2 (.,y) (x, σ(s))B2 (s, y)B1 (s, y)∆s]} p ≤ {a(x, y) + b(x, y)B1 (x, y)[1+

242



∫

x


e0, ef (.,y) (x, x0 )} p , (x, y) ∈ T0 × T 1

1 e0. eB2 (.,y) (x, σ(s))B2 (s, y)∆s]} p , (x, y) ∈ T0 × T

x0

where

(13)

On the other hand, according to [12, Theorem 2.39 and 2.36 (i)] we have ∫ x eB2 (x,y) (x, σ(s))B2 (s, y)∆s = eB2 (x,y) (x, x0 )−1. x0

(14) Combining (13) and (14) we obtain the desired inequality. Corollary 23 Under the conditions of Theorem e 0 , u(x, y) satisfies the 22, if for (x, y) ∈ T0 × T following inequality ∫ x∫ ∞ u(x, y) ≤ a(x, y) + [f (s, t)u(s, t) + g(s, t) ∫

x0

∫

s

1

∫

x∫ ∞

+ x0

y

y

p − 1 p1 K ] + h1 (x, y) p 1−p 1 1 h2 (ξ, η)(ef (.,η) (ξ, x0 )) p [ K p a(ξ, η) p

p − 1 p1 K ∆η∆ξ, p ∫ x∫ ∞ e1 (x, y) = ge(s, t)∆t∆s, B

(15)

t

then

x0

u(x, y) ≤ a(x, y) + B1 (x, y)eB2 (.,y) (x, x0 ), e0, (x, y) ∈ T0 × T

e2 (x, y) = B

∫

(16)

x∫ ∞

B1 (x, y) = s

∫

∫

h(ξ, η)a(ξ, η)∆η∆ξ]∆t∆s, t

x0

∫

∫

∞

B2 (x, y) = y

∞

fe(x, t)∆t.

(24)

x∫ ∞

[g(s, t)u(s, t) + h1 (s, t)

∫

∞

∫

u (x, y) ≤ v(x, y) +

x

p

(18)

(25)

t

Then

t

y

h2 (ξ, η)u(ξ, η)∆η∆ξ]∆t∆s. x0

h(ξ, η)∆η∆ξ]∆t. x0

s

+

(17)

x∫ ∞

[f (x, t)+

∫

x0

y

∞

+

(23)

y

v(x, y) = a(x, y) +

[f (s, t)a(s, t) + g(s, t) x0

∫

(22)

y

Proof : Let

where

∫

1−p

+

h(ξ, η)u(ξ, η)∆η∆ξ]∆t∆s, x0

1 fe(x, y) = g(x, y)(ef (.,y) (x, x0 )) p K p p ∫ x∫ ∞ 1−p 1 1 + h2 (ξ, η)(ef (.,η) (ξ, x0 )) p K p ∆η∆ξ, p x0 y (21) 1−p 1 1 ge(x, y) = g(x, y)(ef (.,y) (x, x0 )) p [ K p a(x, y)+ p

∞

+

(20)

[f (s, y)up (s, y)∆s, x0

Theorem 24 Suppose sup y = ∞, u, f, g, p

e0. (x, y) ∈ T0 × T (26) ∩ e 0 , and let y ∈ [Y, ∞) T. e Then Fix Y ∈ T ∫ x up (x, Y ) ≤ v(x, Y )+ [f (s, Y )up (s, Y )∆s, x ∈ T0 .

e0 y∈T

are the same as in Theorem 21, a, h1 , h2 ∈ e 0 , R+ ), and a(x, y) is nondecreasing in Crd (T0 × T e 0 , u(x, y) x for every fixed y. If for (x, y) ∈ T0 × T satisfies the following inequality ∫ x p u (x, y) ≤ a(x, y) + [f (s, y)up (s, y)∆s ∫

x∫

s

∞

up (x, Y ) ≤ v(x, Y )ef (.,Y ) (x, x0 ), x ∈ T0 .

[g(s, t)u(s, t) + h1 (s, t)

∫

+

x0 ∞

y

h2 (ξ, η)u(ξ, η)∆η∆ξ]∆t∆s, x0

(27) Considering v(x, Y ) is nondecreasing in x, by Lemma 19 we obtain

x0

+ ∫

x0

e 0 arbitrarily, then in Since Y is selected from T e fact (28) holds for ∀y ∈ T0 , that is

(19)

t

then

e 0 ). up (x, y) ≤ v(x, y)ef (.,y) (x, x0 ), (x, y) ∈ (T0 × T (29)

e1 (x, y)e e u(x, y) ≤ {[a(x, y) + B (x, x0 )] B2 (.,y) ISSN: 1109-2769

(28)

243



Let

∫

According to Corollary 23 we obtain

x∫

∞

c(x, y) = s

∫

y

h2 (ξ, η)u(ξ, η)∆η∆ξ]∆t∆s.

(30)

t

Then v(x, y) = a(x, y) + c(x, y),

(31)

Corollary 25 Under the conditions of Theorem e 0 , u(x, y) satisfies the 24, if for (x, y) ∈ T0 × T following inequality ∫ x u(x, y) ≤ a(x, y) + [f (s, y)u(s, y)∆s

and up (x, y) ≤ (a(x, y) + c(x, y))ef (.,y) (x, x0 ), e 0 ). (x, y) ∈ (T0 × T

(32)

∫

Combining (30) and (32) we have ∫ x∫ ∞ c(x, y) = [g(s, t)u(s, t) + h(s, t)]∆t∆s x0

∫

x∫

≤

x0

∞

∫

s

∫

s

∫

e1 (x, y)e e u(x, y) ≤ [a(x, y)+B B2 (x,y) (x, x0 )]ef (x,y) (x, x0 ),

∞

h2 (ξ, η)(a(ξ, η) x0

t 1

+c(ξ, η))ef (.,η) (ξ, x0 )] p ∆η∆ξ}∆t∆s.

e0. (x, y) ∈ T0 × T

(33)

where

On the other hand, From Lemma 2.3 the following inequality holds 1

ge(x, y) = g(x, y)ef (.,y) (x, x0 )a(x, y) + h1 (x, y) ∫ x∫ ∞ + h2 (ξ, η)ef (.,η) (ξ, x0 )a(ξ, η)∆η∆ξ, x0

(34)

x∫ ∞

= x0

∫

y ∞

ge(s, t)∆t∆s,

fe(x, t)∆t.

y

Theorem 26 Suppose sup y = ∞, u, a, f, p e0 y∈T

e0 are the same as in Theorem 24, L ∈ C(T0 × T ×R+ , R+ ), and 0 ≤ L(s, t, x) − L(s, t, y) ≤ M (s, t, y)(x − y) for x ≥ y ≥ 0, where M ∈ e 0 × R+ , R+ ). If for (x, y) ∈ T0 × T e0, C(T0 × T u(x, y) satisfies the following inequality ∫ x p u (x, y) ≤ a(x, y) + f (s, y)up (s, y)∆s

t

p − 1 p1 K ]∆η∆ξ}∆t∆s p

[fe(s, t)c(s, t) + ge(s, t)]∆t∆s,

(35)

y

∫

where fe(x, y), ge(x, y) are defined in (21) and (22) respectively.

ISSN: 1109-2769

x∫ ∞

e2 (x, y) = B

y

(a(ξ, η)+c(ξ, η))+

∫

x0

1 1−p p − 1 p1 [ K p (a(s, t) + c(s, t)) + K ] + h1 (s, t) p p ∫ s∫ ∞ 1 + h2 (ξ, η)(ef (.,η) (ξ, x0 )) p

∫

y

e1 (x, y) = B

So from (33) and (34) it follows ∫ x∫ ∞ 1 c(x, y) ≤ {g(s, t)(ef (.,t) (s, x0 )) p

1 [ K p

y

x0

p − 1 p1 1 1−p K , ≤ K p (a(x, y) + c(x, y)) + p p e0. (x, y) ∈ T0 × T

(38)

fe(x, y) = g(x, y)ef (.,y) (x, x0 ) ∫ x∫ ∞ h2 (ξ, η)ef (.,η) (ξ, x0 )∆η∆ξ, +

(a(x, y) + c(x, y)) p

x0

(37)

t

then

{g(s, t)[(a(s, t) + c(s, t))ef (.,t) (s, x0 )] p

+h1 (s, t) +

1−p p

∫

y

h2 (ξ, η)u(ξ, η)∆η∆ξ]∆t∆s, x0

x0

[g(s, t)u(s, t) + h1 (s, t) x0 ∞

+ 1

x0

x∫ ∞

+

y

y

(36)

e1 (x, y), B e2 (x, y) are defined in (23) and where B (24) respectively. Combining (32) and (36) we get the desired inequality (20).

∞

+ x0

e1 (x, y)e e c(x, y) ≤ B (x, x0 ), B2 (.,y)

[g(s, t)u(s, t) + h1 (s, t) x0

∫


x∫

+

∞

L(s, t, u(s, t))∆t∆s, x0

244

x0

y


(39)



then

Then we have

b1 (x, y)e b u(x, y) ≤ {[a(x, y) + B (x, x0 )] B2 (x,y) e0. ef (x,y) (x, x0 )} p , (x, y) ∈ T0 × T 1

v(x, y) = a(x, y) + c(x, y),

(51)

and

(40)

up (x, y) ≤ (a(x, y) + c(x, y))ef (x,y) (x, x0 ),

where 1 fb(x, y) = M (x, y, (ef (x,y) (x, x0 )) p ( K p 1

1−p p

e0. (x, y) ∈ T0 × T

a(x, y)

1−p 1 1 p − 1 p1 K ))(ef (x,y) (x, x0 )) p K p , + p p

Combining (34), (50) and (52) we have ∫ x∫ ∞ c(x, y) ≤ L(s, t, ((a(s, t)

(41)

1−p 1 1 gb(x, y) = L(x, y, (ef (x,y) (x, x0 )) p ( K p a(x, y) p

p − 1 p1 K )), + p ∫ x∫ ∞ b gb(s, t)∆t∆s, B1 (x, y) = x0

b2 (x, y) = B

∫

x0

fb(x, t)∆t.

1

(42)

∫ ≤

x∫ ∞

x0

(43)

y

(44)

∫

x∫ ∞

= Proof : Let

x0 x∫

y

+c(s, t)) +

L(s, t, u(s, t))∆t∆s. y

(45) ∫ u (x, y) ≤ v(x, y) +

x

1−p 1 1 p − 1 p1 +L(s, t, (ef (s,t) (s, x0 )) p ( K p a(s, t)+ K ))]∆t∆s p p

f (s, y)up (s, y)∆s, x0

e0. (x, y) ∈ T0 × T

∫

≤

(46)

e 0 , and let y ∈ [Y, ∞) ∩ T, e then Fix a Y ∈ T ∫ x p u (x, Y ) ≤ v(x, Y )+ f (s, Y )up (s, Y )∆s, x ∈ T0 . x0

(47) Considering v(x, Y ) is nondecreasing in x, then by Lemma 19 we obtain

x∫ ∞ y

x0

+

1−p 1 1 p − 1 p1 +L(s, t, (ef (s,t) (s, x0 )) p ( K p a(s, t)+ K ))]∆t∆s p p

∫

x∫ ∞

=

(48)

(53)

e0, b1 (x, y)e b c(x, y) ≤ B (x, x0 ), (x, y) ∈ T0 × T B2 (x,y) (54) b b where B1 (x, y), B2 (x, y) are defined in (43) and (44) respectively. Combining (52) and (54) we obtain the desired inequality (40).

e0. up (x, y) ≤ v(x, y)ef (x,y) (x, x0 ), (x, y) ∈ T0 × T (49) Let ∫ x∫ ∞ c(x, y) = L(s, t, u(s, t))∆t∆s. (50)

ISSN: 1109-2769

[fb(s, t)c(s, t) + gb(s, t)]∆t∆s,

y

where fb(x, y), gb(x, y) are defined in (41) and (42) respectively. By use of Corollary 2.3 we obtain

e 0 arbitrarily, then in Since Y is selected from T e fact (48) holds for ∀y ∈ T0 , that is,

x0

1−p 1 1 [M (s, t, (ef (s,t) (s, x0 )) p ( K p a(s, t) p

1−p 1 1 p − 1 p1 K ))(ef (s,t) (s, x0 )) p K p c(s, t) p p

x0

up (x, Y ) ≤ v(x, Y )ef (x,Y ) (x, x0 ), x ∈ T0 .

p − 1 p1 K )) p

1−p 1 1 p − 1 p1 −L(s, t, (ef (s,t) (s, x0 )) p ( K p a(s, t)+ K )) p p

Then p

p − 1 p1 K ))∆t∆s p

1−p 1 1 [L(s, t, (ef (s,t) (s, x0 )) p ( K p (a(s, t) p

∞

v(x, y) = a(x, y) + x0

1−p 1 1 L(s, t, (ef (s,t) (s, x0 )) p ( K p (a(s, t) p

+c(s, t)) +

y

∫

y

+c(s, t))ef (s,t) (s, x0 )) p )∆t∆s

y ∞

(52)

y

245



3


Some Applications

Theorem 30 If u(x, y) is a solution of (57), and |F1 (x, y)| ≤ f (x, y)u2 (x, y), |F2 (x, y, u)| ≤ g(x, y)|u| + h(x, y), where f, g, h ∈ Crd (T0 × e 0 , R+ ), then the following estimate holds T

In this section, we will present some applications for the established results above, and new explicit bounds of solutions of certain dynamic equations will be derived.

|u(x, y)| ≤ √

Example 27 Consider the following dynamic equation ∫ x∫ ∞ p u (x, y) = a(x, y) + F (s, t, u(s, t))∆t∆s, x0

e1 (x, y)e e [|φ(y)| + (B (x, x0 ))]ef (x,y) (x, x0 ), B2 (x,y) e0, (x, y) ∈ T0 × T

y

where

e0, (x, y) ∈ T0 × T

√

(55)

Theorem 28 If u(x, y) is a solution of (55), and |F (s, t, u)| ≤ f (s, t)|u| + g(s, t), where f, g ∈ e 0 , R+ ), then we have Crd (T0 × T ∫ x eB2 (x,y) (x, σ(s)) |u(x, y)| ≤ {|a(x, y)|+[B1 (x, y)+

√ |φ(y)| K ]+h(x, y), ge(x, y) = g(x, y)(ef (x,y) (x, x0 )) [ √ + 2 2 K 1 p

e1 (x, y) = B

∫

x∫ ∞

x0

x0

e 0 , (56) B2 (s, y)B1 (s, y)∆s]} , (x, y) ∈ T0 × T x∫ ∞

B1 (x, y) = x0

y

∫

1 1−p [f (s, t)( K p |a(s, t)| p

|u2 (x, y)| ≤ |φ(y)| +

x

|F1 (s, y)|∆s

x0

∫

x0

x∫ ∞

+ x0

y

≤ |φ(y)| +

Proof : From (55) we have ∫ x∫ ∞ |u(x, y)|p ≤ |a(x, y)|+ |F (s, t, u(s, t))|∆t∆s ≤ |a(x, y)|+

fe(x, t)∆t.

Proof : from (57) we have

p − 1 p1 K ) + g(s, t)]∆t∆s, p ∫ ∞ 1 1−p B2 (x, y) = f (x, t) K p ∆t. p y

x∫

∞

y

+

∫

∫

ge(s, t)∆t∆s,

y

B2 (x, y) =

1 p

∫

ef (x,y) (x, x0 ) √ , 2 K

fe(x, y) = g(x, y)

e 0 , R), and p is a constant where u, a ∈ Crd (T0 × T with p ≥ 1.

where

(58)

y

∫

x∫ ∞

+

|F2 (s, t, u(s, t))|∆t∆s ∫

x

f (s, y)|u(s, y)|2 ∆s x0

[g(s, t)|u(s, t)| + h(s, t)]∆t∆s, x0

y

and then a suitable application of Theorem 24 yields (58).

∞

[f (s, t)|u(s, t)|+g(s, t)]∆t∆s. x0

y

Remark 31 From these applications, one can see some explicit bounds for unknown functions to certain dynamic equations on time scales are established by the present theorems.

Then a suitable application of Theorem 2.1 yields the desired inequality (56). Example 29 Consider the following dynamic equation ∫ x 2 u (x, y) = φ(y) + F1 (s, y)∆s ∫

x∫

+

x0 ∞

F2 (s, t, u(s, t))∆t∆s, x0

Remark 32 If we take T = R, x0 = 0, y0 = 0, then our Theorem 22,24 and 26 reduce to [1, Theorem 1, 3, 5(α)] respectively. Remark 33 If we take T = Z, x0 = 0, y0 = 0, then our Theorem 22,24 and 26 reduce to [2, Theorem 1, 3, 5] respectively with slight difference.

(57)

y

e 0 , R), (x, y) ∈ T0 × T e0. where u ∈ Crd (T0 × T ISSN: 1109-2769

246



4


Conclusions

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In this paper, we established some new GronwallBellman Type dynamic inequalities in two independent variables on time scales containing integration on infinite intervals. As one can see from the presented examples, the established results provide a handy tool in the study of boundedness of solutions of certain dynamic equations on time scales. Furthermore, the established results generalize some known inequalities for continuous functions and their corresponding discrete analysis in the literature. References: [1] F. W. Meng, W. N. Li, On some new integral inequalities and their applications, Appl. Math. Comput. 148 (2004), pp. 381-392 [2] F. W. Meng, W. N. Li, On some newnonlinear discrete inequalities and their applications, J. Comput. Appl. Math. 158 (2003). pp.407-417. [3] O. Lipovan, Integral inequalities for retarded Volterra equations, J. Math. Anal. Appl. 322 (2006), pp.349-358. [4] 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), pp.736-751. [5] Rui A.C. Ferreira, Delfim F.M. Torres, Generalized retarded integral inequalities, Applied Mathematics Letters 22 (2009), pp.876881. [6] H. X. Zhang, F. W. Meng, Integral inequalities in two independent variables for retarded Volterra equations, Appl. Math. Comput. 199 (2008), pp.90-98. [7] W. S. Cheung, J. L. Ren, Discrete nonlinear inequalities and applications to boundary value problems, J. Math. Anal. Appl. 319 (2006), pp.708-724. [8] Y. H. Kim, Gronwall, Bellman and Pachpatte type integral inequalities with applications, Nonlinear Analysis 71 (2009), e2641e2656. [9] W. N. Li, M. A. Han, F.W. Meng, Some new delay integral inequalities and their applications, J. Comput. Appl. Math. 180 (2005), pp.191-200. [10] Q. H. Ma, E.H.Yang, Some new GronwallBellman-Bihari type integral inequalities with delay, Period. Math. Hungar. 44 (2002), No.2, pp.225-238. ISSN: 1109-2769

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