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Article

On Opial’s Type Integral Inequalities

Department of Mathematics, China Jiliang University, Hangzhou 310018, China
Mathematics 2019, 7(4), 375; https://doi.org/10.3390/math7040375
Submission received: 17 March 2019 / Revised: 17 April 2019 / Accepted: 18 April 2019 / Published: 25 April 2019

Abstract

:
In the article we establish some new Opial’s type inequalities involving higher order partial derivatives. These new inequalities, in special cases, yield Agarwal-Pang’s, Pachpatte’s and Das’s type inequalities.

1. Introduction and Statement of Results

In 1960, Opial [1] established the following integral inequality:
Theorem 1.
Suppose x C 1 [ 0 , a ] satisfies x ( 0 ) = x ( a ) = 0 and x ( t ) > 0 for all t ( 0 , a ) . Then the integral inequality holds
0 a x ( t ) x ( t ) d t a 4 0 a ( x ( t ) ) 2 d t ,
where this constant a 4 is best possible.
The first natural extension of Opial’s inequality (1) involving the higher order derivatives x ( n ) ( s ) ( n 1 ) instead of x ( s ) is embodied in the following:
Theorem 2 ([2]).
Let x ( t ) C ( n ) [ 0 , a ] be such that x ( i ) ( 0 ) = 0 , 0 i n 1 ( n 1 ) . Then the following inequality holds
0 a x ( t ) x ( n ) ( t ) d t 1 2 a n 0 a x ( n ) ( t ) 2 d t .
A sharp version of (2) is the following:
Theorem 3 ([3]).
Let x ( t ) C ( n 1 ) [ 0 , a ] be such that x ( i ) ( 0 ) = 0 , 0 i n 1 ( n 1 ) . Further, let x ( n 1 ) ( t ) be absolutely continuous, and 0 a ( x ( n ) ( t ) 2 d t < . Then the following inequality holds
0 a x ( t ) x ( n ) ( t ) d t 1 2 n ! n 2 n 1 1 / 2 a n 0 a x ( n ) ( t ) 2 d t .
A more general version of (3) was established in [4] as follows:
Theorem 4.
Let r k , 0 k n 1 and ℓ be non-negative real numbers such that σ 1 , where σ = k = 0 n 1 r k , and let p ( t ) be a non-negative continuous function on [ 0 , h ] . Further, let x ( t ) C ( n 1 ) [ 0 , h ] be such that x ( i ) ( 0 ) = x ( i ) ( h ) = 0 , 0 i n 1 , and let x ( n 1 ) ( t ) be absolutely continuous. Then the following inequality holds
0 h p ( t ) κ = 0 n x ( κ ) ( t ) r k d t 1 2 σ 0 h [ t ( h t ) ] ( σ 1 ) / 2 p ( t ) d t k = 0 n 1 r k x ( k + 1 ) ( t ) σ d t .
Opial’s inequality and its generalizations, extensions and discretizations play a fundamental role in establishing the existence and uniqueness of initial and boundary value problems for ordinary and partial differential equations as well as difference equations [2,5,6,7,8]. The inequality (1) has received considerable attention and a large number of papers dealing with new proofs, extensions, generalizations, variants and discrete analogues of Opial’s inequality have appeared in the literature [1,3,4,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31]. For an extensive survey on these inequalities, see [2,8].
The first aim of the present paper is to establish a new Opial’s type inequality involving higher order partial derivatives which is a generalization of inequality (4).
Theorem 5.
Let r k 1 , , k n , 0 k i n i 1 , 1 i n and ℓ be non-negative real numbers such that σ 1 , where σ = k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n , and let p ( s 1 , , s n ) be a non-negative continuous function on [ 0 , a 1 ] × × [ 0 , a n ] . Let x ( s 1 , , s n ) be a continuous function on [ 0 , a 1 ] × × [ 0 , a n ] such that k i s i k i x ( s 1 , , s i , , s n ) | s i = 0 and s i = a i = 0 , 1 i n and 0 k i n i 1 . Further, let n i s i n i n i 1 s i n i 1 x ( s 1 , , s i , , s n ) and n i 1 s i n i 1 n i s i n i x ( s 1 , , s i , , s n ) be absolutely continuous on [ 0 , a 1 ] × × [ 0 , a n ] . Then
0 a 1 0 a n p ( s 1 , , s n ) k 1 = 0 n 1 1 k n = 0 n n 1 k 1 + + k n s 1 k 1 s n k n x ( s 1 , , s n ) r k 1 , , k n d s 1 d s n 1 2 σ 0 a 1 0 a n i = 1 n s i ( a i s i ) ( σ 1 ) / 2 p ( s 1 , , s n ) d s 1 d s n × × k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n 0 a 1 0 a n ( k 1 + 1 ) + + ( k n + 1 ) s 1 k 1 + 1 s n k n + 1 x ( s 1 , , s n ) σ d s 1 d s n .
Remark 1.
Let x ( s 1 , , s n ) and p ( s 1 , , s n ) reduce to x ( t ) and p ( t ) , respectively, and with suitable modifications, then (5) changes to (4). A special case of Theorem 5 was proved in [31].
A result involving two functions and their higher order derivatives is embodied in [18]:
Theorem 6.
For j = 1 , 2 let x j ( t ) C n 1 [ 0 , a ] be such that x j ( i ) ( 0 ) = 0 , 0 i n 1 . Further, let x j ( n 1 ) be absolutely continuous, and 0 a | x j ( n ) ( t ) | 2 d t < . Then
0 a x 2 ( t ) x 1 ( n ) ( t ) + x 1 ( t ) x 2 ( n ) ( t ) d t 1 2 n ! n 2 n 1 1 / 2 a n 0 a x 1 ( n ) ( t ) 2 + x 2 ( n ) ( t ) 2 d t ,
The second aim of the present paper is to establish a new Opial’s type inequality involving higher order partial derivatives which is a generalization of inequality (6).
Theorem 7.
For j = 1 , 2 , let x j ( s 1 , , s n ) be a continuous function on [ 0 , a 1 ] × × [ 0 , a n ] such that κ 1 + + κ i 1 + κ i + 1 + + κ n s 1 κ 1 s i 1 κ i 1 s i + 1 κ i + 1 s 1 κ n x j ( s 1 , , s i 1 , s i , s i + 1 , , s n ) | s i = 0 and s i = a i = 0 , 0 κ i n i 1 , and i = 1 , , n . Further, let n i s i n i n i 1 s i n i 1 x j ( s 1 , , s i , , s n ) and n i 1 s i n i 1 n i s i n i x j ( s 1 , , s i , , s n ) be absolutely continuous on [ 0 , a 1 ] × × [ 0 , a n ] , and 0 a 1 0 a n n 1 + + n n s 1 n 1 s n n n x j ( s 1 , , s n ) 2 d s 1 d s n , exist. Then
0 a 1 0 a n { κ 1 + + κ n s 1 κ 1 s n κ n x 1 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) + κ 1 + + κ n s 1 κ 1 s n κ n x 2 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) } d s 1 d s n 2 D ( n 1 , , n n , κ 1 , , κ n ) i = 1 n a i n i κ i 0 a 1 0 a n { n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) 2 + n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 2 } d s 1 d s n ,
where
D ( n 1 , , n n , κ 1 , , κ n ) = 1 4 i = 1 n ( n i κ i ) ! i = 1 n ( n i κ i ) i = 1 n ( 2 n i 2 κ i 1 ) 1 / 2 .
Remark 2.
Taking for κ 1 = = κ n = 0 and n 1 = = n n = 1 in (7) and for j = 1 , 2 , let x j ( s 1 , , s n ) reduce to x j ( t ) , respectively and with suitable modifications, then (7) changes to (6). A special case of Theorem 7 was proved in [31].

2. Proofs of Main Results

In order to prove Theorem 5, we need the following lemma.
Lemma 1.
Let λ 1 be a real number, and let p ( s 1 , , s n ) be a nonnegative and continuous functions on [ 0 , a 1 ] × × [ 0 , a n ] . Further, let x ( s 1 , , s n ) be an absolutely continuous function on [ 0 , a 1 ] × × [ 0 , a n ] , with x ( s 1 , , s i 1 , 0 , s i + 1 , , s n ) = 0 , x ( s 1 , , s i 1 , a i , s i + 1 , , s n ) = 0 and i = 1 , , n . Then
0 a 1 0 a n p ( s 1 , , s n ) | x ( s 1 , , s n ) | λ d s 1 d s n 1 2 0 a 1 0 a n i = 1 n s i ( a i s i ) ( λ 1 ) / 2 p ( s 1 , , s n ) d s 1 d s n × 0 a 1 0 a n n s 1 s n x ( s 1 , , s n ) λ d s d t .
Proof. 
From the hypotheses, we have
x ( s 1 , , s n ) = 0 s 1 0 s n n s 1 s n x ( s 1 , , s n ) d s 1 d s n .
By Hölder’s inequality with indices λ and λ / ( λ 1 ) , it follows that
| x ( s 1 , , s n ) | λ / 2 0 s 1 0 s n n s 1 s n x ( s 1 , , s n ) d s 1 d s n λ 1 / 2 i = 1 n s i ( λ 1 ) / 2 0 s 1 0 s n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n 1 / 2 .
Similarly
| x ( s 1 , , s n ) | λ / 2 i = 1 n ( a i s i ) ( λ 1 ) / 2 s 1 a 1 s n a n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n 1 / 2 .
Now a multiplication of (9) and (10), and by the elementary inequality 2 α β α + β , α 0 , β 0 gives
| x ( s 1 , , s n ) | λ i = 1 n s i ( a i s i ) ( λ 1 ) / 2 0 s 1 0 s n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n 1 / 2 × s 1 a 1 s n a n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n 1 / 2 1 2 i = 1 n s i ( a i s i ) ( λ 1 ) / 2 { 0 s 1 0 s n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n + s 1 a 1 s n a n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n } = 1 2 i = 1 n s i ( a i s i ) ( λ 1 ) / 2 0 a 1 0 a n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n .
Multiplying the both sides of (11) by p ( s 1 , , s n ) and integrating both sides over s i from 0 to a i , i = 1 , , n respectively, we obtain
0 a 1 0 a n p ( s 1 , , s n ) | x ( s 1 , , s n ) | λ d s 1 d s n 1 2 0 a 1 0 a n ( i = 1 n s i ( a i s i ) ( λ 1 ) / 2 p ( s 1 , , s n ) × × 0 a 1 0 a n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n ) d s 1 d s n = 1 2 0 a 1 0 a n i = 1 n s i ( a i s i ) ( λ 1 ) / 2 p ( s 1 , , s n ) d s 1 d s n × 0 a 1 0 a n n s 1 s n x ( s 1 , , s n ) λ d s 1 d s n .
 □
Proof of Theorem 5.
We recall that for the real numbers α k 1 , , k n , r k 1 , , k n 0 , 0 k i n i 1 , 1 i n , and any 1 , the following elementary inequality holds
k 1 = 0 n 1 1 k n = 0 n n 1 α k 1 , , k n r k 1 , , k n k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n α k 1 , , k n k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n α k 1 , , k n 1 / .
From the inequality (12), we have
k 1 = 0 n 1 1 k n = 0 n n 1 k 1 + + k n s 1 k 1 s n k n x ( s 1 , , s n ) r k 1 , , k n = k 1 = 0 n 1 1 k n = 0 n n 1 k 1 + + k n s 1 k 1 s n k n x ( s 1 , , s n ) r k 1 , , k n / σ σ k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n σ k 1 + + k n s 1 k 1 s n k n x ( s 1 , , s n ) σ k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n σ k 1 + + k n s 1 k 1 s n k n x ( s 1 , , s n ) σ .
Multiplying (13) by p ( s 1 , , s n ) , integrating the two sides of (13) over s i from 0 to a i , i , , n , respectively, and then applying the Lemma 1 to the right side again, we observe
0 a 1 0 a n p ( s 1 , , s n ) k 1 = 0 n 1 1 k n = 0 n n 1 k 1 + + k n s 1 k 1 s n k n x ( s 1 , , s n ) r k 1 , , k n d s 1 d s n 1 σ k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n 0 a 1 0 a n p ( s 1 , , s n ) k 1 + + k n s 1 k 1 s n k n x ( s 1 , , s n ) σ d s 1 d s n 1 σ k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n 2 0 a 1 0 a n i = 1 n s i ( a i s i ) ( σ 1 ) / 2 p ( s 1 , , s n ) d s 1 d s n × × 0 a 1 0 a n ( k 1 + 1 ) + + ( k n + 1 ) s 1 k 1 + 1 s n k n + 1 x ( s 1 , , s n ) σ d s 1 d s n = 1 2 σ 0 a 1 0 a n i = 1 n s i ( a i s i ) ( σ 1 ) / 2 p ( s 1 , , s n ) d s 1 d s n × × k 1 = 0 n 1 1 k n = 0 n n 1 r k 1 , , k n 0 a 1 0 a n ( k 1 + 1 ) + + ( k n + 1 ) s 1 k 1 + 1 s n k n + 1 x ( s 1 , , s n ) σ d s 1 d s n .
This completes the proof of Theorem 5.
Proof of Theorem 7.
From the hypotheses of the Theorem 7, we have for 0 κ i n i 1 , i = 1 , , n
κ 1 + + κ n s 1 κ 1 s n κ n x 1 ( s 1 , , s n ) 1 i = 1 n ( n i κ i 1 ) ! × 0 s 1 0 s n i = 1 n ( s i σ i ) n i κ i 1 n 1 + + n n σ 1 n 1 σ n n n x 1 ( σ 1 , , σ n ) d σ 1 d σ n .
Multiplying (14) by n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) and using Cauchy-Schwarz inequality, we obtain
κ 1 + + κ n s 1 κ 1 s n κ n x 1 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 1 i = 1 n ( n i κ i 1 ) ! n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 0 s 1 0 s n i = 1 n ( s i σ i ) 2 ( n i κ i 1 ) d σ 1 d σ n 1 / 2 × 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 1 ( σ 1 , , σ n ) 2 d σ 1 d σ n 1 / 2 = i = 1 n s i n i κ i 1 / 2 i = 1 n ( n i κ i 1 ) ! ( 2 n i 2 κ i 1 ) 1 / 2 n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) × 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 1 ( σ 1 , , σ n ) 2 d σ 1 d σ n 1 / 2 .
Integrating the two sides of (15) over s i from 0 to a i , i = 1 , , n , respectively and then applying the Cauchy-Schwarz inequality to the right side again, we observe
0 a 1 0 a n κ 1 + + κ n s 1 κ 1 s n κ n x 1 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) d s 1 d s n 1 i = 1 n ( n i κ i 1 ) ! ( 2 n i 2 κ i 1 ) 1 / 2 0 a 1 0 a n i = 1 n s i 2 n i 2 κ i 1 d s 1 d s n 1 / 2 × { 0 a 1 0 a n n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 2 × 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 1 ( σ 1 , , σ n ) 2 d σ 1 d σ n d s 1 d s n } 1 / 2 .
Similarly
0 a 1 0 a n κ 1 + + κ n s 1 κ 1 s n κ n x 2 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) d s 1 d s n 1 i = 1 n ( n i κ i 1 ) ! ( 2 n i 2 κ i 1 ) 1 / 2 0 a 1 0 a n i = 1 n s i 2 n i 2 κ i 1 d s 1 d s n 1 / 2 × { 0 a 1 0 a n n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) 2 × 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 2 ( σ 1 , , σ n ) 2 d σ 1 d σ n d s 1 d s n } 1 / 2 .
Taking the sum of the two sides of (16) and (17), and in view of the elementary inequality α 1 / 2 + β 1 / 2 [ 2 ( α + β ) ] 1 / 2 , α 0 , β 0 , we have
0 a 1 0 a n { κ 1 + + κ n s 1 κ 1 s n κ n x 1 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) + κ 1 + + κ n s 1 κ 1 s n κ n x 2 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) } d s 1 d s n 1 i = 1 n ( n i κ i 1 ) ! ( 2 n i 2 κ i 1 ) 1 / 2 0 a 1 0 a n i = 1 n s i 2 n i 2 κ i 1 d s 1 d s n 1 / 2 × { 2 0 a 1 0 a n n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 2 × 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 1 ( σ 1 , , σ n ) 2 d σ 1 d σ n d s 1 d s n + 2 0 a 1 0 a n n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) 2 × 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 2 ( σ 1 , , σ n ) 2 d σ 1 d σ n d s 1 d s n } 1 / 2 .
On the other hand, note the derivation rule of integral upper bound function and the derivation rule of product function, we have
n s 1 s n [ 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 1 ( σ 1 , , σ n ) 2 d σ 1 d σ n × 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 2 ( σ 1 , , σ n ) 2 d σ 1 d σ n ] = n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) 2 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 2 ( σ 1 , , σ n ) 2 d σ 1 d σ n + n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 2 0 s 1 0 s n n 1 + + n n σ 1 n 1 σ n n n x 1 ( σ 1 , , σ n ) 2 d σ 1 d σ n
and
0 a 1 0 a n i = 1 n s i 2 n i 2 κ i 1 d s 1 d s n 1 / 2 = 0 a 1 s 1 2 n 1 2 k 1 1 d s 1 · 0 a 2 s 2 2 n 2 2 k 2 1 d s 2 · · 0 a n s n 2 n n 2 k n 1 d s n 1 / 2 = i = 1 n a i n i κ i 2 i = 1 n ( n i κ i ) 1 / 2 .
From (18)–(20) and in view the elementary inequality ( α β ) 1 / 2 1 2 ( α + β ) , α 0 , β 0 , we have
0 a 1 0 a n { κ 1 + + κ n s 1 κ 1 s n κ n x 1 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) + κ 1 + + κ n s 1 κ 1 s n κ n x 2 ( s 1 , , s n ) n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) } d s 1 d s n 2 D ( n 1 , , n n , κ 1 , , κ n ) i = 1 n a i n i κ i [ 2 0 a 1 0 a n n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) 2 d s 1 d s n × 0 a 1 0 a n n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 2 d s 1 d s n ] 1 / 2 2 D ( n 1 , , n n , κ 1 , , κ n ) i = 1 n a i n i κ i 0 a 1 0 a n { n 1 + + n n s 1 n 1 s n n n x 1 ( s 1 , , s n ) 2 + n 1 + + n n s 1 n 1 s n n n x 2 ( s 1 , , s n ) 2 } d s 1 d s n ,
where
D ( n 1 , , n n , κ 1 , , κ n ) = 1 4 i = 1 n ( n i κ i ) ! i = 1 n ( n i κ i ) i = 1 n ( 2 n i 2 κ i 1 ) 1 / 2 , i 1 , , n .

Funding

Research is supported by National Natural Science Foundation of China (11371334, 10971205).

Conflicts of Interest

The author declares no conflict of interest.

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