# New Characterizations of Weights in Hardy and Opial Type Inequalities via Solvability of Dynamic Equations

Authors: S. H. Saker; A. G. Sayed; A. Sikorska-Nowak; I. Abohela
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

In this paper, we prove that the solvability of dynamic equations of second order is sufficient for the validity of some Hardy and Opial type inequalities with two different weights on time scales. In particular, the results give new characterizations of two different weights in inequalities containing Hardy and Opial operators. The main contribution in this paper is the characterizations of weights in discrete inequalities that will be formulated from our results as special cases.


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## Body

## 1. Introduction

In 1925, Hardy [1] proved that(1)
∫
0
∞
1
t
∫
0
t
f
s
d
s
p
d
t
≤
p
p
-
1
p
∫
0
∞
f
p
t
d
t
,where f is a positive integrable function over any finite interval (
0
,
t
), f
p is an integrable function over (
0
,
∞
), and p
>
1
. The constant (
p
/
(
p
-
1
)
)
p in (1) is the best possible. Hardy proved his inequality by employing the calculus of variations which reduces the validation of the inequality to the existence of a positive solution of an Euler-Lagrange differential equation. In 1961 Beesack [2] followed the same technique and proved a generalized inequality of the form(2)
∫
a
b
s
t
∫
a
t
f
τ
d
τ
p
d
t
≤
∫
a
b
r
t
f
p
t
d
t
,where the weighted functions r and s satisfy the Euler-Lagrange differential equation(3)
r
t
ϕ
′
t
p
-
1
′
+
s
t
ϕ
p
-
1
t
=
0
.For more related results for Hardy’s type inequalities that improved or extended the results of Beesack, we refer the reader to the papers by Bennett [3], Beesack [4], Beesack and Heing [5], Bliss [6], Bloom and Kerman [7], Bradly [8], Florkiewicz [9], Leindler [10, 11], Muckenhoupt [12], Shum [13], Sinnamon [14], Tomaselli [15], and Talenti [16, 17] and the references they cited.In 1960, Opial [18] proved a new type of inequalities involving the relation between the function and its derivatives. In particular, he proved that if h is an absolutely continuous function on [
a
,
b
] with h
(
a
)
=
h
(
b
)
=
0, then(4)
∫
a
b
h
t
h
′
t
d
t
≤
b
-
a
4
∫
a
b
h
′
t
2
d
t
.The constant b
-
a
/
4 in (4) is the best possible. In further the proof of the Opial inequality had already been simplified by Olech [19], Beesack [20], Levinson [21], Mallows [22], and Pederson [23]. In particular, it is proved that if h is real absolutely continuous on (
0
,
b
) and with h
(
0
)
=
0, then(5)
∫
0
b
h
t
h
′
t
d
t
≤
b
2
∫
0
b
h
′
t
2
d
t
.Since then much work has been done, and many papers which deal with new proofs, various generalizations, extensions, and their discrete analogues have appeared in the literature. The discrete analogy of (4) has been proved in [24] and the discrete analogy of (5) has been proved in [25, Theorem 5.2.2].In 1976, Florkiewicz and Rybarski [26] used a new method to derive inequalities involving a function h and its first derivative of Opial’s type with weights of the form(6)
∫
0
b
s
t
h
t
h
′
t
d
t
≤
∫
0
b
r
t
h
′
t
2
d
t
,where r and s satisfy a second order differential equation of the form (7)
r
t
ϕ
′
t
′
+
s
t
ϕ
t
=
0
.This method makes it possible to obtain, in a natural way, the equality conditions important in differential equations and also makes it is possible, given a function r and an auxiliary function ϕ, to define a function s and an additional function v and next using r, s, and v to define a new class H of functions h for which (2) or (6) holds. Moreover it allows avoiding some assumptions on weights that have to be given in other methods.In the last decades, some authors interested to find certain discrete inequalities analogues to integral inequalities in the analysis of different fields and as a result this subject becomes a topic of ongoing research. One reason for this upsurge of interest in discrete case is also due to the fact that discrete operators may even behave differently from their continuous counterparts. But the main challenge is that there are no general methods to study these questions on discrete spaces. Instead, these methods have to be developed starting from the basic definitions. In some cases it is possible to translate or adapt almost straightforward the objects and results from the continuous setting to the discrete setting or vice versa, however, in some other cases that is far from be trivial. In recent years, some authors have been interested in finding some characterizations of discrete Hardy’s type inequalities analogues for characterizations of Hardy’s type inequalities via differential equations; see, for example, the results by Chen [27, 28] and Liao [29] and the references they cited. To the best of the authors’ knowledge there are no characterizations obtained for discrete Opial type inequalities via difference equations.In recent years, the study of dynamic inequalities on time scales has received a lot of attention and becomes a major field in pure and applied mathematics. The general idea is to prove a result for an inequality where the domain of the unknown function is a so-called time scaleT, which is an arbitrary nonempty closed subset of the real numbers R. The study of dynamic inequalities on time scales helps avoid proving results twice, for differential inequalities as well as for difference inequalities. For more details of Hardy’s type inequalities on time scales, we refer the reader to the papers [30–39] and to the book [40] and to more results about Opial’s type inequalities on time scales, we refer to papers [41–44] and to the book [45].The natural question which is expected now isIs it possible to prove new characterizations of the weighted functions in dynamic Hardy’s and Opial’s type inequalities by reducing the problem to the solvability of dynamic equations on time scales?Our aim is to give an affirmative answer to this question in our present paper, which is organized as follows: In Section2, we present some basic definition concerning the delta calculus on time scales. In Section 3, we prove a new characterization of weights on dynamic Hardy’s type inequality and prove that the characterization reduces to the solvability of dynamic equation (8)
r
t
ϕ
Δ
t
Δ
+
s
t
ϕ
σ
t
=
0
,
for
t
∈
α
,
β
T
,of second order, where σ
:
T
→
T is the forward jump operator which is defined by σ
(
t
)
≔
inf
⁡
s
∈
T
:
s
>
t, for t
∈
T, and f
σ
≔
f
∘
σ
. In Section 4, we prove a new characterization of weighs on dynamic Opial’s type inequality and prove that the characterization reduces to the solvability of dynamic equation (9)
r
t
ϕ
Δ
t
Δ
-
1
2
s
Δ
t
ϕ
σ
t
=
0
,
for
t
∈
α
,
β
T
.In Section 5, we prove that the characterization of weights in an inequality involving Hardy and Opial operator will be obtained from the solvability of a generalized dynamic equation of the form (10)
r
t
ϕ
Δ
t
Δ
+
s
Δ
t
ϕ
σ
t
-
u
t
ϕ
σ
t
=
0
,
for
t
∈
α
,
β
T
.We will formulate the classical integral and discrete inequalities in this paper as special cases. Throughout this paper, we will assume that t, s
t, and f
t are nonnegative rd-continuous functions and the integrals considered are assumed to exist.


## 2. Preliminaries

In this section, we present some basic definition concerning the delta calculus on time scales. For more details on time scale analysis, we refer the reader to the two books by Bohner and Peterson [46, 47] which summarize and organize much of time scales calculus. A time scale T is an arbitrary nonempty closed subset of the real numbers R. The forward jump operator and the backward jump operator are defined by σ
(
t
)
≔
inf
⁡
s
∈
T
:
s
>
t, and ρ
(
t
)
≔
sup
⁡
s
∈
T
:
s
<
t, where sup
⁡
∅
=
inf
⁡
T. A point t
∈
T, which is said to be left-dense if ρ
(
t
)
=
t and t
>
inf
⁡
T, is right-dense if σ
(
t
)
=
t, is left-scattered if ρ
(
t
)
<
t, and is right-scattered if σ
(
t
)
>
t
. A function f
:
T
→
R is said to be right-dense continuous (rd-continuous) provided f is continuous at right-dense points and at left-dense points in T, and left hand limits exist and are finite. The set of all such rd-continuous functions is denoted by C
r
d
(
T
)
. Also, the set of functions that are differentiable and whose derivative is rd-continuous is denoted by C
r
d
1
(
T
)
=
C
r
d
1
(
T
,
R
)
. The graininess function μ for a time scale T is defined by μ
(
t
)
≔
σ
(
t
)
-
t, and for any function f
:
T
→
R the notation f
σ
(
t
) denotes f
(
σ
(
t
)
)
. Without loss of generality, we assume that sup
⁡
T
=
∞ and define the time scale interval [
a
,
b
]
T by [
a
,
b
]
T
≔
[
a
,
b
]
∩
T
. Recall of the following product and quotient rules for the derivative of the product f
g and the quotient f
/
g (where g
g
σ
≠
0, here g
σ
=
g
∘
σ) of two differentiable functions f and g
(11)
f
g
Δ
=
f
Δ
g
+
f
σ
g
Δ
=
f
g
Δ
+
f
Δ
g
σ
,
and
f
g
Δ
=
f
Δ
g
-
f
g
Δ
g
g
σ
.The first chain rule that we will use in this paper is(12)
f
γ
t
Δ
=
γ
∫
0
1
h
f
σ
+
1
-
h
f
γ
-
1
d
h
f
Δ
t
,
γ
∈
R
,which is a simple consequence of Keller’s chain rule [46, Theorem 1.90]. The second chain rule that we will use in this paper is given in the following. Let f
:
R
→
R be continuously differentiable and suppose g
:
T
→
R is delta differentiable, then f
∘
g
:
T
→
R is delta differentiable and(13)
f
Δ
g
t
=
f
′
g
d
g
Δ
t
,
for
d
∈
t
,
σ
t
.In this paper we will refer to the (delta) integral which we can define as follows. If F
Δ
(
t
)
=
f
(
t
), then the Cauchy (delta) integral of f is defined by ∫
t
0
t
f
(
s
)
Δ
s
≔
F
(
t
)
-
F
(
t
0
)
. It can be shown (see [46]) that if f
∈
C
r
d
(
T
), then the Cauchy integral F
(
t
)
≔
∫
t
0
t
f
(
s
)
Δ
s exists, t
0
∈
T, and it satisfies F
Δ
(
t
)
=
f
(
t
), t
∈
T
. An infinite integral is defined as ∫
a
∞
f
(
t
)
Δ
t
=
l
i
m
b
→
∞
∫
a
b
f
(
t
)
Δ
t
. Integration on discrete time scales is defined by (14)
∫
a
b
f
t
Δ
t
=
∑
t
∈
a
,
b
μ
t
f
t
.The integration by parts formula on time scales reads(15)
∫
a
b
u
t
υ
Δ
t
Δ
t
=
u
t
υ
t
a
b
-
∫
a
b
u
Δ
t
υ
σ
t
Δ
t
.Hölder’s inequality [46, Theorem 6.13] states that, for f
,
g
∈
C
r
d
(
[
a
,
b
]
T
,
R
), we have(16)
∫
a
b
f
t
g
t
Δ
t
≤
∫
a
b
f
t
p
Δ
t
1
/
p
∫
a
b
g
t
q
Δ
t
1
/
q
,where >
1, 1
/
p
+
1
/
q
=
1, and a, b
∈
T
.


## 3. Characterizations of Weights in Hardy’s Type Inequalities

In this section, we will establish some characterizations of two different weights in Hardy’s type inequalities on time scales and via the solvability of a dynamic equation of the form(17)
r
t
ϕ
Δ
t
Δ
+
s
t
ϕ
σ
t
=
0
,
for
t
∈
α
,
β
T
,where r and s are r
d
-continuous functions defined on [
α
,
β
]
T
. In this section, we will define(18)
v
t
≔
r
t
ϕ
Δ
t
ϕ
t
,
for
t
∈
α
,
β
T
,where r
>
0 and ϕ is a positive solution of (17).Definition 1.
We say thatg belongs to the class H
^
1 if(19)
∫
α
β
s
t
g
σ
t
2
Δ
t
>
-
∞
,
and
∫
α
β
r
t
g
Δ
t
2
Δ
t
<
∞
,and(20)
l
i
m
t
→
α
inf
⁡
v
g
2
t
<
∞
,
l
i
m
t
→
β
sup
⁡
v
g
2
t
>
-
∞
.Lemma 2.
Assume thatT is a time scale and g
∈
H
^
1. If ϕ is a positive solution of the dynamic equation (17), then(21)
v
g
2
Δ
=
-
s
g
σ
2
+
r
ϕ
Δ
ω
2
+
r
ϕ
Δ
ϕ
σ
ω
2
Δ
,where ω
=
g
/
ϕ
.Proof.
For convenience, we skip the argumentt sometimes in the computations. From the definition of v and by using the product and the quotient rules of derivative on time scales(22)
u
v
Δ
=
u
v
Δ
+
u
Δ
v
σ
,
and
u
v
Δ
=
u
Δ
v
-
u
v
Δ
v
v
σ
,we have for t
≥
α that(23)
v
Δ
=
r
ϕ
Δ
Δ
1
ϕ
σ
+
r
ϕ
Δ
1
ϕ
Δ
=
r
ϕ
Δ
Δ
ϕ
σ
+
r
ϕ
Δ
-
ϕ
Δ
ϕ
ϕ
σ
=
r
ϕ
Δ
Δ
ϕ
σ
-
r
ϕ
Δ
2
ϕ
ϕ
σ
,and(24)
v
g
2
Δ
=
v
Δ
g
σ
2
+
v
g
2
Δ
.From (17) and (23), we get that(25)
v
Δ
=
-
s
-
r
ϕ
Δ
2
ϕ
ϕ
σ
.Since g
=
ϕ
ω, we get that(26)
g
Δ
=
ϕ
ω
Δ
=
ϕ
Δ
ω
+
ϕ
σ
ω
Δ
,and then(27)
g
2
Δ
=
g
+
g
σ
g
Δ
=
ϕ
ω
+
ϕ
σ
ω
σ
ϕ
Δ
ω
+
ϕ
σ
ω
Δ
=
ϕ
ω
ϕ
Δ
ω
+
ϕ
σ
ω
Δ
+
ϕ
σ
ω
σ
ϕ
Δ
ω
+
ϕ
σ
ω
Δ
=
ϕ
ϕ
Δ
ω
2
+
ϕ
ϕ
σ
ω
ω
Δ
+
ϕ
σ
ϕ
Δ
ω
ω
σ
+
ϕ
σ
2
ω
σ
ω
Δ
.Substituting (25) and (27) into (24), we obtain that(28)
v
g
2
Δ
=
v
Δ
g
σ
2
+
v
g
2
Δ
=
-
s
-
r
ϕ
Δ
2
ϕ
ϕ
σ
g
σ
2
+
r
ϕ
Δ
ϕ
ϕ
ϕ
Δ
ω
2
+
ϕ
ϕ
σ
ω
ω
Δ
+
ϕ
σ
ϕ
Δ
ω
ω
σ
+
ϕ
σ
2
ω
σ
ω
Δ
=
-
s
g
σ
2
-
r
ϕ
Δ
2
ϕ
ϕ
σ
ϕ
σ
ω
σ
2
+
r
ϕ
Δ
ω
2
+
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
+
r
ϕ
Δ
2
ϕ
σ
ω
ω
σ
ϕ
+
r
ϕ
Δ
ϕ
σ
2
ω
σ
ω
Δ
ϕ
=
-
s
g
σ
2
+
r
ϕ
Δ
ω
2
+
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
-
r
ϕ
σ
ϕ
Δ
ω
σ
2
ϕ
+
r
ϕ
Δ
2
ϕ
σ
ω
ω
σ
ϕ
+
r
ϕ
Δ
ϕ
σ
2
ω
σ
ω
Δ
ϕ
.Since ω
σ
=
ω
+
μ
ω
Δ, then(29)
-
r
ϕ
σ
ϕ
Δ
ω
σ
2
ϕ
+
r
ϕ
σ
ϕ
Δ
2
ω
ω
σ
ϕ
+
r
ϕ
Δ
ϕ
σ
2
ω
σ
ω
Δ
ϕ
=
r
ϕ
σ
ϕ
Δ
ω
σ
ϕ
-
ϕ
Δ
ω
σ
+
ϕ
Δ
ω
+
ϕ
σ
ω
Δ
=
r
ϕ
σ
ϕ
Δ
ω
σ
ϕ
-
μ
ϕ
Δ
ω
Δ
+
ϕ
σ
ω
Δ
=
r
ϕ
σ
ϕ
Δ
ω
σ
ω
Δ
ϕ
-
μ
ϕ
Δ
+
ϕ
σ
=
r
ϕ
σ
ϕ
Δ
ω
σ
ω
Δ
.From (28) and (29), we have (30)
v
g
2
Δ
=
-
s
g
σ
2
+
r
ϕ
Δ
ω
2
+
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
+
r
ϕ
σ
ϕ
Δ
ω
σ
ω
Δ
=
-
s
g
σ
2
+
r
ϕ
Δ
ω
2
+
r
ϕ
Δ
ϕ
σ
ω
Δ
ω
+
ω
σ
=
-
s
g
σ
2
+
r
ϕ
Δ
ω
2
+
r
ϕ
Δ
ϕ
σ
ω
2
Δ
,which is the desired equation (21). The proof is complete.Theorem 3.
LetT be a time scale and g
∈
H
^
1. If ϕ is a positive solution of (17), then(31)
r
g
Δ
2
=
s
g
σ
2
+
v
g
2
Δ
+
r
ϕ
ϕ
σ
ω
Δ
2
,where ω
=
g
/
ϕ.Proof.
For convenience, we skip the argumentt sometimes in the computations. Since ω
=
g
/
ϕ and by using (26) we get(32)
r
g
Δ
2
=
r
ϕ
Δ
ω
+
ϕ
σ
ω
Δ
2
=
r
ϕ
Δ
ω
2
+
2
ϕ
Δ
ϕ
σ
ω
ω
Δ
+
ϕ
σ
ω
Δ
2
=
r
ϕ
Δ
ω
2
+
2
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
+
r
ϕ
σ
ω
Δ
2
.By applying Lemma 2, we get that(33)
r
ϕ
Δ
ω
2
=
v
g
2
Δ
+
s
g
σ
2
-
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
-
r
ϕ
σ
ϕ
Δ
ω
σ
ω
Δ
.Substituting (33) into (32), we obtain(34)
r
g
Δ
2
=
v
g
2
Δ
+
s
g
σ
2
-
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
-
r
ϕ
σ
ϕ
Δ
ω
σ
ω
Δ
+
2
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
+
r
ϕ
σ
ω
Δ
2
=
v
g
2
Δ
+
s
g
σ
2
-
r
ϕ
σ
ϕ
Δ
ω
σ
ω
Δ
+
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
+
r
ϕ
σ
ω
Δ
2
.Since ω
σ
=
ω
+
μ
ω
Δ, then(35)
-
r
ϕ
σ
ϕ
Δ
ω
σ
ω
Δ
+
r
ϕ
Δ
ϕ
σ
ω
ω
Δ
+
r
ϕ
σ
ω
Δ
2
=
r
ϕ
σ
ω
Δ
-
ϕ
Δ
ω
σ
+
ϕ
Δ
ω
+
ϕ
σ
ω
Δ
=
r
ϕ
σ
ω
Δ
-
μ
ϕ
Δ
ω
Δ
+
ϕ
σ
ω
Δ
=
r
ϕ
σ
ω
Δ
2
ϕ
σ
-
μ
ϕ
Δ
=
r
ϕ
ϕ
σ
ω
Δ
2
.Then, from (34) and (35), we have that (36)
r
g
Δ
2
=
s
g
σ
2
+
v
g
2
Δ
+
r
ϕ
ϕ
σ
ω
Δ
2
,which is the desired identity (31). The proof is complete.Theorem 4.
LetT be a time scale and r and s satisfy the dynamic equation (17). Then for every g
∈
H
^
1, we have that(37)
∫
α
β
s
g
σ
2
Δ
t
+
l
i
m
t
→
β
v
g
2
-
l
i
m
t
→
α
v
g
2
≤
∫
α
β
r
g
Δ
2
Δ
t
.Furthermore (37) becomes an equality if and only if g
=
c
ϕ where c
≠
0 is a constant.Proof.
From Theorem3, we have that (38)
s
g
σ
2
+
v
g
2
Δ
+
r
ϕ
ϕ
σ
ω
Δ
2
=
r
g
Δ
2
.Then (39)
∫
α
β
s
g
σ
2
Δ
t
+
∫
α
β
v
g
2
Δ
Δ
t
+
∫
α
β
r
ϕ
ϕ
σ
ω
Δ
2
Δ
t
=
∫
α
β
r
g
Δ
2
Δ
t
.Since r, ϕ are nonnegative, we see that r
ϕ
ϕ
σ
ω
Δ
2
>
0, which implies that(40)
∫
α
β
s
g
σ
2
Δ
t
+
l
i
m
t
→
β
v
g
2
-
l
i
m
t
→
α
v
g
2
≤
∫
α
β
r
g
Δ
2
Δ
t
,which is the desired inequality (37). Furthermore inequality (37) becomes equality if and only if r
ϕ
ϕ
σ
ω
Δ
2
=
0, where ω
=
g
/
ϕ
. This in fact gives us that ω
Δ
=
g
/
ϕ
Δ
=
0, and then g
=
c
ϕ with c
≠
0
. The proof is complete.Definition 5.
We say thatg belongs to the class H
1 if(41)
∫
α
β
s
t
g
σ
t
2
Δ
t
>
-
∞
,
∫
α
β
r
t
g
Δ
t
2
Δ
t
<
∞
,and(42)
l
i
m
i
n
f
t
→
α
v
t
g
2
t
≤
0
,
lim
⁡
s
u
p
t
→
β
v
t
g
2
t
≥
0
.Remark 6.
ObviouslyH
1
⊂
H
^
1
.Theorem 7.
LetT be a time scale and r and s satisfy the dynamic equation (17). Then for every function g
∈
H
1, we have(43)
∫
α
β
s
t
g
σ
t
2
Δ
t
≤
∫
α
β
r
t
g
Δ
t
2
Δ
t
.If g
≠
0, then (43) becomes an equality if and only if g
/
ϕ
=
c
=
c
o
n
s
t
≠
0, and(44)
ϕ
∈
H
^
1
,
l
i
m
t
→
α
r
ϕ
Δ
ϕ
=
l
i
m
t
→
β
r
ϕ
Δ
ϕ
=
0
.Proof.
From (42), we have l
i
m
t
→
β
v
g
2
-
l
i
m
t
→
α
v
g
2
≥
0
. By applying Theorem 4, then we obtain (45)
∫
α
β
s
g
σ
2
Δ
t
≤
∫
α
β
r
g
Δ
2
Δ
t
,which is the desired inequality (43). Furthermore inequality (43) becomes equality if and only if (46)
r
ϕ
ϕ
σ
ω
Δ
2
=
0
,where ω
=
g
/
ϕ
. This implies that ω
Δ
=
g
/
ϕ
Δ
=
0, and g
=
c
ϕ with c
≠
0 and(47)
l
i
m
t
→
β
v
g
2
=
l
i
m
t
→
α
v
g
2
=
0
.Since v
=
r
ϕ
Δ
/
ϕ, and g
=
c
ϕ, this implies that (48)
l
i
m
t
→
β
v
g
2
=
l
i
m
t
→
β
r
ϕ
Δ
ϕ
-
1
c
ϕ
2
=
c
2
l
i
m
t
→
β
r
ϕ
Δ
ϕ
=
0
,and (49)
l
i
m
t
→
α
v
g
2
=
l
i
m
t
→
α
r
ϕ
Δ
ϕ
-
1
c
ϕ
2
=
c
2
l
i
m
t
→
α
r
ϕ
Δ
ϕ
=
0
,then ϕ
∈
H
^
1 and l
i
m
t
→
α
r
ϕ
Δ
ϕ
=
l
i
m
t
→
β
r
ϕ
Δ
ϕ
=
0
. The proof is complete.Remark 8.
As a special case of Theorem7 when T
=
R, we obtain the inequality (50)
∫
α
β
s
t
∫
0
t
u
τ
d
τ
2
d
t
≤
∫
α
β
r
t
u
2
t
d
t
,where r and s satisfy the differential equation(51)
r
t
ϕ
′
t
′
+
s
t
ϕ
t
=
0
,
for
t
∈
α
,
β
.Remark 9.
As a special case of Theorem7 when T
=
N, we obtain the inequality (52)
∑
n
=
0
N
s
n
∑
k
=
0
n
u
k
2
≤
∑
n
=
0
N
r
n
u
2
n
,where r and s satisfy the difference equation(53)
Δ
r
n
Δ
ϕ
n
+
s
n
ϕ
n
+
1
=
0
,
for
n
∈
0
,
N
.Remark 10.
For the differential form we get the Wirtinger type inequality(54)
∫
α
β
s
t
u
2
t
d
t
≤
∫
α
β
r
t
u
′
t
2
d
t
,where r and s satisfy the differential equation (51) with u
(
0
)
=
0
.Example 11.
LetT
=
R with α
,
β
∈
R, and 0
≤
α
<
β
≤
∞. Put r
(
t
)
=
1, and ϕ
=
t, then r, ϕ
>
0, and (55)
s
=
-
r
ϕ
′
′
ϕ
=
-
ϕ
′
′
ϕ
=
1
4
t
2
.By applying Theorem 7, we see that (56)
∫
0
∞
1
t
∫
0
t
u
s
d
s
2
d
t
≤
4
∫
0
∞
u
t
2
d
t
,which is Hardy’s inequality with a sharp constant 4.Example 12.
LetT
=
R with α
,
β
∈
R, and 0
≤
α
<
β
≤
∞. Put r
(
t
)
=
t
2, and ϕ
=
t
-
1, we see that s
=
-
r
ϕ
′
′
/
ϕ
=
1
/
4. By applying Theorem 7, we get the inequality (57)
∫
0
∞
∫
0
t
u
s
d
s
2
d
t
≤
4
∫
0
∞
t
u
t
2
d
t
.which is the Hardy-type inequality.


## 4. Characterizations of Weights in Opial’s Type Inequalities

In this section, we will establish some characterizations of two different weights in Opial’s type inequalities on time scales via the solvability of a dynamic equation of the form(58)
r
t
ϕ
Δ
t
Δ
-
1
2
s
Δ
t
ϕ
σ
t
=
0
,
for
t
∈
α
,
β
T
.In this section, we will define(59)
v
t
≔
-
s
t
2
+
r
t
ϕ
Δ
t
ϕ
t
,where ϕ is a positive solution of (58).Definition 13.
We say thath belongs to the class H
^
2, if(60)
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
>
-
∞
,
and
∫
α
β
r
h
Δ
2
Δ
t
<
∞
,and(61)
l
i
m
t
→
α
inf
⁡
v
h
2
<
∞
,
l
i
m
t
→
β
sup
⁡
v
h
2
>
-
∞
.Lemma 14.
Assume thatT is a time scale and h
∈
H
^
2
. If ϕ is a solution of the dynamic equation (58), then(62)
v
h
2
Δ
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
-
s
2
h
+
h
σ
h
Δ
+
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
.Proof.
For convenience we skip the argumentt sometimes in the computations. From the definition of v and by using the product and the quotient rules (22) for t
≥
α, we have(63)
v
Δ
=
-
s
2
+
r
ϕ
Δ
ϕ
Δ
=
-
s
Δ
2
+
r
ϕ
Δ
ϕ
Δ
=
-
s
Δ
2
+
r
ϕ
Δ
Δ
ϕ
-
r
ϕ
Δ
ϕ
Δ
ϕ
ϕ
σ
=
-
s
Δ
2
+
r
ϕ
Δ
Δ
ϕ
σ
-
r
ϕ
Δ
2
ϕ
ϕ
σ
.Since h
σ
=
h
+
μ
h
Δ, then(64)
h
2
Δ
=
h
+
h
σ
h
Δ
=
2
h
σ
-
μ
h
Δ
h
Δ
.Substituting (63) and (64) into(65)
v
h
2
Δ
=
v
Δ
h
σ
2
+
v
h
2
Δ
,we obtain that (66)
v
h
2
Δ
=
v
Δ
h
2
σ
+
v
h
2
Δ
=
-
s
Δ
2
+
r
ϕ
Δ
Δ
ϕ
σ
-
r
ϕ
Δ
2
ϕ
ϕ
σ
h
2
σ
+
-
s
2
+
r
ϕ
Δ
ϕ
h
2
Δ
=
-
s
Δ
2
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
-
s
2
h
2
Δ
+
r
ϕ
Δ
ϕ
h
2
Δ
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
-
s
2
h
+
h
σ
h
Δ
+
r
ϕ
Δ
ϕ
2
h
σ
h
Δ
-
μ
h
Δ
2
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
-
s
2
h
+
h
σ
h
Δ
+
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
.which is (62). The proof is complete.Lemma 15.
LetT be a time scale and h
∈
H
^
2
. If ϕ is a solution of the dynamic equation (58), then(67)
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
r
h
Δ
2
ϕ
σ
ϕ
-
2
r
h
Δ
h
σ
ϕ
Δ
ϕ
+
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
.Proof.
For convenience we skip the argumentt sometimes in the computations. By using the product and the quotient rules (22) for t
≥
α, we have that (68)
h
ϕ
Δ
=
h
×
1
ϕ
Δ
=
h
Δ
×
1
ϕ
+
h
σ
×
1
ϕ
Δ
=
h
Δ
ϕ
+
h
σ
×
-
ϕ
Δ
ϕ
ϕ
σ
=
h
Δ
ϕ
σ
-
h
σ
ϕ
Δ
ϕ
ϕ
σ
,and this implies that(69)
h
ϕ
Δ
2
=
h
Δ
ϕ
σ
-
h
σ
ϕ
Δ
ϕ
ϕ
σ
2
=
1
ϕ
ϕ
σ
2
h
Δ
ϕ
σ
-
h
σ
ϕ
Δ
2
=
1
ϕ
ϕ
σ
2
h
Δ
ϕ
σ
2
-
2
h
Δ
h
σ
ϕ
σ
ϕ
Δ
+
h
σ
ϕ
Δ
2
.Multiplying (69) by r
ϕ
ϕ
σ, we have that (70)
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
r
ϕ
ϕ
σ
h
Δ
ϕ
σ
2
-
2
h
σ
h
Δ
ϕ
σ
ϕ
Δ
+
h
σ
ϕ
Δ
2
=
r
h
Δ
2
ϕ
σ
ϕ
-
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
+
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
,which is (67). The proof is complete.Theorem 16.
Assume thatT is a time scale and r and s satisfy the dynamic equation (58). Then for every and h
∈
H
^
2, we have that(71)
∫
α
β
s
2
h
+
h
σ
h
Δ
Δ
t
+
l
i
m
t
→
β
v
h
2
-
l
i
m
t
→
α
v
h
2
≤
∫
α
β
r
h
Δ
2
Δ
t
.Furthermore inequality (71) becomes equality if and only if h
=
c
ϕ with a constant c
≠
0
.Proof.
From Lemmas14 and 15, we get that (72)
v
h
2
Δ
+
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
-
s
2
h
+
h
σ
h
Δ
+
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
+
r
h
Δ
2
ϕ
σ
ϕ
-
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
+
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
s
2
h
+
h
σ
h
Δ
+
r
h
Δ
2
ϕ
σ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
.Since ϕ
σ
=
ϕ
+
μ
ϕ
Δ, then we obtain (73)
v
h
2
Δ
+
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
s
2
h
+
h
σ
h
Δ
+
r
h
Δ
2
ϕ
σ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
s
2
h
+
h
σ
h
Δ
+
r
h
Δ
2
ϕ
σ
ϕ
-
μ
ϕ
Δ
ϕ
=
-
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
s
2
h
+
h
σ
h
Δ
+
r
h
Δ
2
.This implies that(74)
v
h
2
Δ
+
g
t
+
s
Δ
2
-
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
+
s
2
h
+
h
σ
h
Δ
=
r
h
Δ
2
,where g
=
r
ϕ
ϕ
σ
h
ϕ
-
1
Δ
2
≥
0
. By using the dynamic equations (58) and (74) and g
≥
0, we see that (75)
v
h
2
Δ
+
s
2
h
+
h
σ
h
Δ
≤
r
h
Δ
2
.This implies that(76)
∫
α
β
s
2
h
+
h
σ
h
Δ
Δ
t
+
l
i
m
t
→
β
v
h
2
-
l
i
m
t
→
α
v
h
2
≤
∫
α
β
r
h
Δ
2
Δ
t
,which is the desired inequality (71). Furthermore inequality (71) becomes equality if and only if g
=
r
ϕ
ϕ
σ
f
Δ
2
=
0, where f
=
h
ϕ
-
1
. This in fact gives us that (77)
f
Δ
=
h
ϕ
Δ
=
0
,and then h
=
c
ϕ with a constant c
≠
0
. The proof is complete.Definition 17.
We say thath belongs to the class H
2 if(78)
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
>
-
∞
,
and
∫
α
β
r
h
Δ
2
Δ
t
<
∞
,and(79)
l
i
m
t
→
α
inf
⁡
v
h
2
≤
0
,
l
i
m
t
→
β
sup
⁡
v
h
2
≥
0
.Remark 18.
ObviouslyH
2
⊂
H
^
2
.Theorem 19.
Assume thatT is a time scale and r and s satisfy the dynamic equation (58). Then for every function h
∈
H
2 the inequality(80)
∫
α
β
s
2
h
+
h
σ
h
Δ
Δ
t
≤
∫
α
β
r
h
Δ
2
Δ
tholds. If h
≠
0, inequality (80) becomes equality if and only h
/
ϕ
=
c
≠
0, ϕ
∈
H
^
2, and (81)
l
i
m
t
→
α
r
ϕ
Δ
ϕ
-
1
2
s
ϕ
2
=
l
i
m
t
→
β
r
ϕ
Δ
ϕ
-
1
2
s
ϕ
2
=
0
.Proof.
From (79), we have l
i
m
t
→
β
v
h
2
-
l
i
m
t
→
α
v
h
2
≥
0
. By applying Theorem 16, then we obtain (82)
∫
α
β
s
2
h
+
h
σ
h
Δ
Δ
t
≤
∫
α
β
r
h
Δ
2
Δ
t
,which is the desired inequality (80). Furthermore inequality (80) becomes equality if and only if g
=
r
ϕ
ϕ
σ
h
ϕ
-
1
Δ
2
=
0, where f
=
h
/
ϕ
. This in fact gives us that (83)
f
Δ
=
h
ϕ
Δ
=
0
,and then h
=
c
ϕ with a constant c
≠
0, and l
i
m
t
→
β
v
h
2
=
l
i
m
t
→
α
v
h
2
=
0
. Since v
=
-
s
/
2
+
r
ϕ
Δ
ϕ
-
1, and h
=
c
ϕ with a constant c
≠
0, this implies that (84)
l
i
m
t
→
β
v
h
2
=
lim
t
→
β
⁡
-
s
2
+
r
ϕ
Δ
ϕ
c
ϕ
2
=
c
2
lim
t
→
β
⁡
r
ϕ
Δ
ϕ
-
1
2
s
ϕ
2
=
0
,and (85)
l
i
m
t
→
α
v
h
2
=
l
i
m
t
→
α
-
s
2
+
r
ϕ
Δ
ϕ
c
ϕ
2
=
c
2
l
i
m
t
→
α
r
ϕ
Δ
ϕ
-
1
2
s
ϕ
2
=
0
,then ϕ
∈
H
^
2, and (86)
lim
t
→
α
⁡
r
ϕ
Δ
ϕ
-
1
2
s
ϕ
2
=
lim
t
→
β
⁡
r
ϕ
Δ
ϕ
-
1
2
s
ϕ
2
=
0
.The proof is complete.Remark 20.
As a special case of Theorem19, we see that(87)
∫
α
β
s
t
h
t
+
h
σ
t
h
Δ
t
Δ
t
≤
2
∫
α
β
r
t
h
Δ
t
2
Δ
t
.Remark 21.
As a special case of Theorem19 when T
=
R, we get the Opial’s type inequality (88)
∫
α
β
s
t
h
t
h
′
t
d
t
≤
∫
α
β
r
t
h
′
t
2
d
t
,where the weighted functions r and s satisfy the differential equation(89)
r
t
ϕ
′
t
′
-
1
2
s
′
t
ϕ
t
=
0
,
for
t
∈
α
,
β
.


## 5. A Combination between Hardy’s and Opial’s Inequalities

In this section, we will prove some characterizations of weights on a general inequality which contains Hardy’s type inequality and Opial’s type inequalities as special cases on time scales and reduces the characterizations of these weights to the solvability of a generalized Euler dynamic equation of the form(90)
r
t
ϕ
Δ
t
Δ
+
s
Δ
t
ϕ
σ
t
-
u
t
ϕ
σ
t
=
0
,
for
t
∈
α
,
β
T
.In this section, we define(91)
v
t
≔
s
t
+
r
t
ϕ
Δ
t
ϕ
t
,
for
t
∈
α
,
β
T
,where ϕ is a positive solution of (90).Definition 22.
We say that the functionh belongs to the class H
^
3 if(92)
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
<
∞
,
∫
α
β
u
h
σ
2
Δ
t
<
∞
,
and
∫
α
β
r
h
Δ
2
Δ
t
<
∞
,and(93)
l
i
m
t
→
α
inf
⁡
v
h
2
<
∞
,
l
i
m
t
→
β
sup
⁡
v
h
2
>
-
∞
.Lemma 23.
Assume thatT is a time scale and h
∈
H
^
3
. If ϕ is a positive solution of (90), then(94)
v
h
2
Δ
=
s
Δ
-
u
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
+
u
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
+
s
h
+
h
σ
h
Δ
+
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
.Proof.
For convenience, we skip the argumentt sometimes in the computations. By using the product rule in(22) for t
≥
α, we see that(95)
h
2
Δ
=
h
+
h
σ
h
Δ
=
2
h
σ
-
μ
h
Δ
h
Δ
,and(96)
v
h
2
Δ
=
v
Δ
h
σ
2
+
v
h
2
Δ
.From the definition of v and by using the quotient rule in (22) for t
≥
α, we have that(97)
v
Δ
=
s
+
r
ϕ
Δ
ϕ
Δ
=
s
Δ
+
r
ϕ
Δ
ϕ
Δ
=
s
Δ
+
r
ϕ
Δ
Δ
ϕ
-
r
ϕ
Δ
ϕ
Δ
ϕ
ϕ
σ
=
s
Δ
+
r
ϕ
Δ
Δ
ϕ
σ
-
r
ϕ
Δ
2
ϕ
ϕ
σ
.Substituting (97) and (95), we obtain that (98)
v
h
2
Δ
=
v
Δ
h
2
σ
+
v
h
2
Δ
=
s
Δ
+
r
ϕ
Δ
Δ
ϕ
σ
-
r
ϕ
Δ
2
ϕ
ϕ
σ
h
2
σ
+
s
+
r
ϕ
Δ
ϕ
h
2
Δ
=
s
Δ
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
+
s
h
2
Δ
+
r
ϕ
Δ
ϕ
h
2
Δ
=
s
Δ
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
+
s
h
+
h
σ
h
Δ
+
r
ϕ
Δ
ϕ
2
h
σ
h
Δ
-
μ
h
Δ
2
=
s
Δ
-
u
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
+
u
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
+
s
h
+
h
σ
h
Δ
+
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
,which is (94). The proof is complete.Lemma 24.
Assume thatT be a time scale and let h
∈
H
^
3
. If ϕ is a positive solution of (90), then(99)
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
r
h
Δ
2
ϕ
σ
ϕ
-
2
r
h
Δ
h
σ
ϕ
Δ
ϕ
+
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
.Proof.
For convenience, we skip the argumentt sometimes in the computations. Since h
=
ϕ
f, and by using the product and the quotient rules (22) for t
≥
α, we have that (100)
h
ϕ
Δ
=
h
×
1
ϕ
Δ
=
h
Δ
×
1
ϕ
+
h
σ
×
1
ϕ
Δ
=
h
Δ
ϕ
+
h
σ
×
-
ϕ
Δ
ϕ
ϕ
σ
=
h
Δ
ϕ
σ
-
h
σ
ϕ
Δ
ϕ
ϕ
σ
.This implies that(101)
h
ϕ
Δ
2
=
h
Δ
ϕ
σ
-
h
σ
ϕ
Δ
ϕ
ϕ
σ
2
=
1
ϕ
ϕ
σ
2
h
Δ
ϕ
σ
-
h
σ
ϕ
Δ
2
=
1
ϕ
ϕ
σ
2
h
Δ
ϕ
σ
2
-
2
h
Δ
h
σ
ϕ
σ
ϕ
Δ
+
h
σ
ϕ
Δ
2
.Multiplying (101) by r
ϕ
ϕ
σ, we have that (102)
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
r
ϕ
ϕ
σ
h
Δ
ϕ
σ
2
-
2
h
σ
h
Δ
ϕ
σ
ϕ
Δ
+
h
σ
ϕ
Δ
2
=
r
h
Δ
2
ϕ
σ
ϕ
-
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
+
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
,which is (99). The proof is complete.Theorem 25.
LetT be a time scale and let h
∈
H
^
3
. If , s, and u satisfy (90), then(103)
lim
t
→
β
⁡
v
h
2
-
lim
t
→
α
⁡
v
h
2
≤
∫
α
β
r
h
Δ
2
Δ
t
+
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
+
∫
α
β
u
h
σ
2
Δ
t
.Proof.
From Lemmas23 and 24 we get that (104)
v
h
2
Δ
+
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
s
Δ
-
u
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
+
u
h
2
σ
-
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
+
s
h
+
h
σ
h
Δ
+
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
+
r
h
Δ
2
ϕ
σ
ϕ
-
2
r
h
σ
h
Δ
ϕ
Δ
ϕ
+
r
h
σ
ϕ
Δ
2
ϕ
ϕ
σ
.Since ϕ
σ
=
ϕ
+
μ
ϕ
Δ, then we obtain (105)
v
h
2
Δ
+
r
ϕ
ϕ
σ
h
ϕ
Δ
2
=
s
Δ
-
u
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
+
u
h
2
σ
+
s
h
+
h
σ
h
Δ
+
r
h
Δ
2
ϕ
σ
ϕ
-
r
h
Δ
2
μ
ϕ
Δ
ϕ
=
s
Δ
-
u
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
+
u
h
2
σ
+
s
h
+
h
σ
h
Δ
+
r
h
Δ
2
ϕ
σ
ϕ
-
μ
ϕ
Δ
ϕ
=
s
Δ
-
u
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
+
u
h
2
σ
+
s
h
+
h
σ
h
Δ
+
r
h
Δ
2
.This implies that(106)
v
h
2
Δ
+
g
t
+
f
t
=
r
h
Δ
2
+
s
h
+
h
σ
h
Δ
+
u
h
σ
2
,where g
=
r
ϕ
ϕ
σ
h
ϕ
-
1
Δ
2
≥
0, and by using dynamic equation (90), we obtain (107)
f
t
=
-
s
Δ
-
u
+
r
ϕ
Δ
Δ
ϕ
σ
h
2
σ
=
0
.From (106), we have that(108)
lim
t
→
β
⁡
v
h
2
-
lim
t
→
α
⁡
v
h
2
≤
∫
α
β
r
h
Δ
2
Δ
t
+
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
+
∫
α
β
u
h
σ
2
Δ
t
,which is the desired inequality (103). The proof is complete.Definition 26.
We say thath belongs to the class H
3 if(109)
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
<
∞
,
∫
α
β
u
h
σ
2
Δ
t
<
∞
,
and
∫
α
β
r
h
Δ
2
Δ
t
<
∞
,and(110)
lim
t
→
α
⁡
v
h
2
≤
0
,
lim
t
→
β
⁡
v
h
2
≥
0
.Remark 27.
ObviouslyH
3
⊂
H
^
3
.Theorem 28.
Assume thatT is a time scale and , s, and u satisfy the dynamic equation (90). Then for every function h
∈
H
3 the inequality(111)
0
≤
∫
α
β
r
h
Δ
2
Δ
t
+
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
+
∫
α
β
u
h
σ
2
Δ
t
,holds. Furthermore inequality (111) becomes equality if and only if h
=
c
ϕ with c
=
c
o
n
s
t
≠
0, and(112)
ϕ
∈
H
^
3
,
l
i
m
t
→
α
r
ϕ
Δ
ϕ
+
s
ϕ
2
=
l
i
m
t
→
β
r
ϕ
Δ
ϕ
+
s
ϕ
2
=
0
.Proof.
From (110), we have l
i
m
t
→
β
v
h
2
-
l
i
m
t
→
α
v
h
2
≥
0
. By applying Theorem 25, we obtain (113)
0
≤
∫
α
β
r
h
Δ
2
Δ
t
+
∫
α
β
s
h
+
h
σ
h
Δ
Δ
t
+
∫
α
β
u
h
σ
2
Δ
t
,which is the desired inequality (111). Furthermore inequality (111) becomes equality if and only if (114)
g
=
r
ϕ
ϕ
σ
f
Δ
2
=
0
,where f
=
h
/
ϕ
. This in fact gives us that f
Δ
=
h
/
ϕ
Δ
=
0, and then h
=
c
ϕ with c
=
c
o
n
s
t
≥
0
. And (115)
l
i
m
t
→
β
v
h
2
=
l
i
m
t
→
α
v
h
2
=
0
.Since v
=
s
+
r
ϕ
Δ
ϕ
-
1, and h
=
c
ϕ with c
=
c
o
n
s
t
≥
0, this implies that (116)
lim
t
→
β
⁡
v
h
2
=
lim
t
→
β
⁡
s
+
r
ϕ
Δ
ϕ
-
1
c
ϕ
2
=
c
2
lim
t
→
β
⁡
r
ϕ
Δ
ϕ
+
s
ϕ
2
=
0
,and (117)
lim
t
→
α
⁡
v
h
2
=
lim
t
→
α
⁡
s
+
r
ϕ
Δ
ϕ
-
1
c
ϕ
2
=
c
2
lim
t
→
α
⁡
r
ϕ
Δ
ϕ
+
s
ϕ
2
=
0
,then ϕ
∈
H
^
3 and (118)
lim
t
→
α
⁡
r
ϕ
Δ
ϕ
+
s
ϕ
2
=
lim
t
→
β
⁡
r
ϕ
Δ
ϕ
+
s
ϕ
2
=
0
.The proof is complete.Remark 29.
As a special case of Theorem28 when T
=
R, we obtain inequality (119)
0
≤
∫
α
β
r
h
′
2
d
t
+
∫
α
β
2
s
h
h
′
d
t
+
∫
α
β
u
h
2
d
t
,where r
,
s, and u satisfy the differential equation (120)
r
t
ϕ
′
t
′
+
s
′
t
ϕ
t
-
u
t
ϕ
t
=
0
,
for
t
∈
α
,
βRemark 30.
As a special case of Theorem28 when T
=
R, let s
=
0 and u
<
0 on [
α
,
β
]
T, then we get the Sturm-Liouville type inequality (121)
∫
α
β
u
t
h
2
t
d
t
≤
∫
α
β
r
t
h
′
t
2
d
t
,where r and u satisfy the differential equation (122)
r
t
ϕ
′
t
′
-
u
t
ϕ
t
=
0
,
for
t
∈
α
,
β
.Remark 31.
As a special case of Theorem28 when T
=
R, let s
(
t
)
≤
0, u
(
t
)
=
0 on [
α
,
β
]
T, and (123)
h
t
=
∫
α
t
h
′
u
d
u
,or let s
(
t
)
≥
0, u
(
t
)
=
0 on [
α
,
β
]
T, and (124)
h
t
=
∫
t
β
h
′
u
d
u
.Then we get the Opial type inequality (125)
2
∫
α
β
s
h
h
′
d
t
≤
∫
α
β
r
h
′
2
d
t
,where r and s satisfy the differential equation (126)
r
t
ϕ
′
t
′
+
s
′
t
ϕ
t
=
0
,
for
t
∈
α
,
β
.Example 32.
LetT
=
R with α
,
β
∈
R, and 0
≤
α
<
β
≤
∞. Put r
(
t
)
=
t
a, and ϕ
=
t
1
-
a
/
2, where a
≠
1 arbitrary constant, and let s
(
t
)
=
A
t
a
-
1, and u
(
t
)
=
1
/
4
a
-
1
4
A
-
a
+
1
t
a
-
2, where A is an arbitrary constant. This in fact satisfies the equation (127)
r
t
ϕ
′
t
′
+
s
′
t
ϕ
t
-
u
t
ϕ
t
=
0
,
for
t
∈
α
,
β
.By applying Theorem 28, if h
≠
0, we get(128)
0
<
∫
α
β
t
a
h
′
2
+
2
A
t
a
-
1
h
h
′
+
1
4
a
-
1
4
A
-
a
+
1
t
a
-
2
h
2
d
t
.If A
=
0, then inequality (128) becomes (129)
0
<
∫
α
β
t
a
h
′
2
-
1
4
a
-
1
2
t
a
-
2
h
2
d
t
.If α
=
0, β
=
∞ and a
=
0, we get that (130)
∫
0
∞
1
t
2
h
2
t
Δ
t
≤
4
∫
0
∞
h
′
t
2
Δ
t
,which is Hardy’s inequality.



---

*Source: 6757080-2019-08-01.xml*