On Hermite-Hadamard-type inequalities for systems of partial differential inequalities in the plane

Mohamed Jleli; Bessem Samet

doi:10.1515/math-2023-0115

Article Open Access

On Hermite-Hadamard-type inequalities for systems of partial differential inequalities in the plane

Mohamed Jleli and Bessem Samet

Published/Copyright: September 22, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

We establish necessary conditions for the existence of solutions to various systems of partial differential inequalities in the plane. The obtained conditions provide new Hermite-Hadamard-type inequalities for differentiable functions in the plane. In particular, we obtain a refinement of an inequality due to Dragomir [On the Hadamard’s inequality for convex functions on the co-ordinates in a rectangle from the plane, Taiwan. J. Math. 5 (2001), no. 4, 775–788] for convex functions on the coordinates.

Keywords: systems of partial differential inequalities; Hermite-Hadamard-type inequalities; convex functions; convex functions on the coordinates

MSC 2010: 26B25; 26D10; 35A23

1 Introduction

Let I be an interval of R and f : I → R be a convex function, i.e.,

f ( η x + ( 1 − η ) y ) ≤ η f ( x ) + ( 1 − η ) f ( y ) ,

for all η ∈ ] 0 , 1 [ and x , y ∈ I . Then, for all a , b ∈ I with a < b , it holds that:

f a + b 2 ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( a ) + f ( b ) 2 .

The aforementioned result is known in the literature as Hermite-Hadamard inequality. This inequality dates back to an 1883 observation of Hermite [1] with an independent use by Hadamard [2] in 1893. Hermite-Hadamard inequality has been refined and generalized in various directions. For instance, in 1906, Fejér [3] established a weighted version of this inequality. More recent generalizations can be found in Abramovich and Persson [4], Cerone and Dragomir [5], Dragomir [6], Dragomir and Pearce [7], Niculescu and Persson [8,9], and Samet [10] (see also references therein).

Hermite-Hadamard inequality has also been studied in higher dimensions, i.e., for convex functions f : Ω → R , where Ω ⊂ R N is a convex domain. For instance, we refer to Dragomir [11] ( Ω is a disk of the plane), Dragomir [12] ( Ω is the three-dimensional ball), de la Cal and Carcamo [13] ( Ω is the N -dimensional ball), Chen [14] ( Ω is a triangle) and Bessenyei [15] ( Ω is a simplex), Steinerberger [16] ( Ω is a convex domain), Merino González [17] ( Ω is a 0-symmetric convex compact set), and Kravitz and Lee [18] ( Ω is a nearly spherical domain) (see also references therein). In most cases, it was shown that Hermite-Hadamard inequality takes the form:

1 ∣ Ω ∣ ∫ Ω f ( x ) d x ≤ 1 ∣ ∂ Ω ∣ ∫ ∂ Ω f ( x ) d S x .

It is interesting to point out that the aforementioned inequality cannot hold with constant 1 in higher dimensions uniformly over all convex bodies (see, e.g., [19]).

This present work is motivated by the study [20], where Dragomir established a Hermite-Hadamard-type inequality for convex functions on the coordinates defined in a rectangle from the plane. Before presenting the main result in [20], we need to recall some notions related to convexity. Let a , b , c , d ∈ R with a < b and c < d . A function f : [ a , b ] × [ c , d ] → R is said to be convex on the coordinates, if

for all a ≤ x ≤ b , the partial mapping f x : [ c , d ] ∋ y ↦ f ( x , y ) is convex;
for all c ≤ y ≤ d , the partial mapping f y : [ a , b ] ∋ x ↦ f ( x , y ) is convex.

We point out that every convex function f : [ a , b ] × [ c , d ] → R is convex on the coordinates, but the converse is not generally true (see [20, Lemma 1]). The main result obtained in [20] can be stated as follows.

Theorem 1.1

Let a , b , c , d ∈ R with a < b and c < d . If f : [ a , b ] × [ c , d ] → R is convex on the coordinates, then

(1.1) f a + b 2 , c + d 2 ≤ 1 2 1 b − a ∫ a b f x , c + d 2 d x + 1 d − c ∫ c d f a + b 2 , y d y , 1 2 1 b − a ∫ a b f x , c + d 2 d x + 1 d − c ∫ c d f a + b 2 , y d y ≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x , 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 4 1 b − a ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 d − c ∫ c d ( f ( a , y ) + f ( b , y ) ) d y , 1 4 1 b − a ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 d − c ∫ c d ( f ( a , y ) + f ( b , y ) ) d y ≤ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 .

On the other hand, observe that if f : R 2 → R is a twice continuously differentiable function that satisfies the system of partial differential inequalities:

(1.2) ∂ 2 f ∂ x 2 ( x , y ) ≥ 0 , ( x , y ) ∈ R 2 , ∂ 2 f ∂ y 2 ( x , y ) ≥ 0 , ( x , y ) ∈ R 2 ,

then f is convex on the coordinates, and consequently, f verifies (1.1) for all a , b , c , d ∈ R with a < b and c < d . From this observation, it is natural to ask whether it is possible to obtain new Hermite-Hadamard-type inequalities for functions f satisfying other systems of partial differential inequalities. In this study, we provide a positive answer to this question. Namely, we obtain new Hermite-Hadamard-type inequalities for various systems of second- and fourth-order partial differential inequalities in the plane. Moreover, an improvement of (1.1) is obtained under an additional condition on f .

The rest of this article is organized as follows. Section 2 is devoted to the study of some systems of second-order partial differential inequalities. In particular, a system of Legendre differential inequalities is investigated. In Section 3, some systems of fourth-order partial differential inequalities are considered.

2 Systems of second-order partial differential inequalities

In this section, we establish new Hermite-Hadamard-type inequalities for some systems of second-order partial differential inequalities in the plane.

We first consider the set of functions f = f ( x , y ) ∈ C 2 ( R 2 ) satisfying systems of the form:

(2.1) ∂ 2 f ∂ x 2 ( x , y ) − α f ( x , y ) ≥ 0 , ( x , y ) ∈ R 2 , ∂ 2 f ∂ y 2 ( x , y ) − β f ( x , y ) ≥ 0 , ( x , y ) ∈ R 2 ,

where α and β > 0 are the constants.

We have the following result.

Theorem 2.1

Let f ∈ C 2 ( R 2 ) verify (2.1), where α and β > 0 are the constants. Then, for all a , b , c , d ∈ R with a < b and c < d , it holds that:

(2.2) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 2 β tanh β 2 ( d − c ) ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 2 α tanh α 2 ( b − a ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y .

Proof

Let a , b , c , d ∈ R with a < b and c < d . Let us introduce the nonnegative function:

k ( x ) = 1 α 1 − cosh α 2 ( a + b − 2 x ) cosh α 2 ( b − a ) , a ≤ x ≤ b .

Elementary calculations show that:

(2.3) k ″ ( x ) − α k ( x ) = − 1 , a < x < b

and

(2.4) k ( a ) = k ( b ) = 0 , k ′ ( a ) = − k ′ ( b ) = 1 α tanh α 2 ( b − a ) .

On the other hand, for all c ≤ y ≤ d , integrating by parts, we obtain

− ∫ a b f ( x , y ) k ″ ( x ) d x = − [ k ′ ( x ) f ( x , y ) ] x = a b + ∫ a b ∂ f ∂ x ( x , y ) k ′ ( x ) d x = − [ k ′ ( x ) f ( x , y ) ] x = a b + k ( x ) ∂ f ∂ x ( x , y ) x = a b − ∫ a b ∂ 2 f ∂ x 2 ( x , y ) k ( x ) d x ,

that is,

(2.5) − ∫ a b f ( x , y ) k ″ ( x ) d x = − [ k ′ ( x ) f ( x , y ) ] x = a b + k ( x ) ∂ f ∂ x ( x , y ) x = a b − ∫ a b ∂ 2 f ∂ x 2 ( x , y ) k ( x ) d x .

Moreover, by (2.3) and (2.4), we have

(2.6) − ∫ a b f ( x , y ) k ″ ( x ) d x = ∫ a b f ( x , y ) d x − α ∫ a b f ( x , y ) k ( x ) d x

and

(2.7) − [ k ′ ( x ) f ( x , y ) ] x = a b + k ( x ) ∂ f ∂ x ( x , y ) x = a b = 1 α tanh α 2 ( b − a ) ( f ( a , y ) + f ( b , y ) ) .

Hence, it follows from (2.5), (2.6), and (2.7) that

∫ a b f ( x , y ) d x = 1 α tanh α 2 ( b − a ) ( f ( a , y ) + f ( b , y ) ) + ∫ a b α f ( x , y ) − ∂ 2 f ∂ x 2 ( x , y ) k ( x ) d x .

Since

k ≥ 0 , α f − ∂ 2 f ∂ x 2 ≤ 0 ,

we obtain

∫ a b f ( x , y ) d x ≤ 1 α tanh α 2 ( b − a ) ( f ( a , y ) + f ( b , y ) ) .

Integrating the aforementioned inequality over y ∈ [ c , d ] , we obtain

(2.8) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 α tanh α 2 ( b − a ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y .

Similarly, we introduce the function

l ( y ) = 1 β 1 − cosh β 2 ( c + d − 2 y ) cosh β 2 ( d − c ) , c ≤ y ≤ d .

For all a ≤ x ≤ b , multiplying − f ( x , y ) by l ″ ( y ) and integrating by parts over y ∈ [ c , d ] , we obtain

∫ c d f ( x , y ) d y ≤ 1 β tanh β 2 ( d − c ) ( f ( x , c ) + f ( x , d ) ) .

Integrating the aforementioned inequality over x ∈ [ a , b ] , we obtain

(2.9) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 β tanh β 2 ( d − c ) ∫ a b ( f ( x , c ) + f ( x , d ) ) d x .

Finally, summing (2.8) and (2.9), we obtain (2.2).□

Remark 2.2

Observe that in the limit case α → 0 + and β → 0 + , System (2.1) reduces to System (1.2). It is interesting to remark that in this case, Inequality (2.2) reduces to Dragomir Inequality (1.1) that holds for convex functions on the coordinates.

We next consider the set of functions f = f ( x , y ) ∈ C 2 ( [ 0 , h ] × [ 0 , ℓ ] ) satisfying the system of Legendre differential inequalities:

(2.10) ( 1 − x 2 ) ∂ 2 f ∂ x 2 ( x , y ) − 2 x ∂ f ∂ x ( x , y ) + 2 f ( x , y ) ≥ 0 , ( x , y ) ∈ [ 0 , h ] × [ 0 , ℓ ] , ( 1 − y 2 ) ∂ 2 f ∂ y 2 ( x , y ) − 2 y ∂ f ∂ y ( x , y ) + 2 f ( x , y ) ≥ 0 , ( x , y ) ∈ [ 0 , h ] × [ 0 , ℓ ] ,

where 0 < h , ℓ < 1 .

We have the following result.

Theorem 2.3

Let f ∈ C 2 ( [ 0 , h ] × [ 0 , ℓ ] ) verify (2.10), where 0 < h , ℓ < 1 . Then, it holds that:

(2.11) ∫ 0 h ∫ 0 ℓ f ( x , y ) d y d x ≤ 1 4 ∫ 0 h ( tanh − 1 ( ℓ ) f ( x , 0 ) + ℓ f ( x , ℓ ) ) d x + 1 4 ∫ 0 ℓ ( tanh − 1 ( h ) f ( 0 , y ) + h f ( h , y ) ) d y .

Proof

Let us introduce the nonnegative function:

μ ( x ) = x ( tanh − 1 ( h ) − tanh − 1 ( x ) ) 2 , 0 ≤ x ≤ h .

Elementary calculations show that:

(2.12) ( 1 − x 2 ) μ ″ ( x ) − 2 x μ ′ ( x ) + 2 μ ( x ) = − 1 , 0 < x < h

and

(2.13) μ ( 0 ) = μ ( h ) = 0 , μ ′ ( 0 ) = tanh − 1 ( h ) 2 , and μ ′ ( h ) = − h 2 ( 1 − h 2 ) .

On the other hand, for all 0 ≤ y ≤ ℓ , integrating by parts, we obtain

∫ 0 h f ( x , y ) [ ( 1 − x 2 ) μ ″ ( x ) − 2 x μ ′ ( x ) ] d x = ∫ 0 h f ( x , y ) ( ( 1 − x 2 ) μ ′ ( x ) ) ′ d x = [ ( 1 − x 2 ) μ ′ ( x ) f ( x , y ) ] x = 0 h − ∫ 0 h ∂ f ∂ x ( x , y ) ( 1 − x 2 ) μ ′ ( x ) d x = [ ( 1 − x 2 ) μ ′ ( x ) f ( x , y ) ] x = 0 h − ( 1 − x ) 2 μ ( x ) ∂ f ∂ x ( x , y ) x = 0 h + ∫ 0 h μ ( x ) ∂ ∂ x ∂ f ∂ x ( x , y ) ( 1 − x 2 ) d x = [ ( 1 − x 2 ) μ ′ ( x ) f ( x , y ) ] x = 0 h − ( 1 − x ) 2 μ ( x ) ∂ f ∂ x ( x , y ) x = 0 h + ∫ 0 h − 2 x ∂ f ∂ x ( x , y ) + ( 1 − x 2 ) ∂ 2 f ∂ x 2 ( x , y ) μ ( x ) d x ,

that is,

(2.14) ∫ 0 h f ( x , y ) [ ( 1 − x 2 ) μ ″ ( x ) − 2 x μ ′ ( x ) ] d x = [ ( 1 − x 2 ) μ ′ ( x ) f ( x , y ) ] x = 0 h − ( 1 − x ) 2 μ ( x ) ∂ f ∂ x ( x , y ) x = 0 h + ∫ 0 h − 2 x ∂ f ∂ x ( x , y ) + ( 1 − x 2 ) ∂ 2 f ∂ x 2 ( x , y ) μ ( x ) d x .

Moreover, by (2.12) and (2.13), we have

(2.15) ∫ 0 h f ( x , y ) [ ( 1 − x 2 ) μ ″ ( x ) − 2 x μ ′ ( x ) ] d x = − ∫ 0 h f ( x , y ) d x − 2 ∫ 0 h μ ( x ) f ( x , y ) d x

and

(2.16) [ ( 1 − x 2 ) μ ′ ( x ) f ( x , y ) ] x = 0 h − ( 1 − x ) 2 μ ( x ) ∂ f ∂ x ( x , y ) x = 0 h = − h 2 f ( h , y ) − tanh − 1 ( h ) 2 f ( 0 , y ) .

Hence, it follows from (2.14), (2.15), and (2.16) that:

− ∫ 0 h f ( x , y ) d x − 2 ∫ 0 h μ ( x ) f ( x , y ) d x = − h 2 f ( h , y ) − tanh − 1 ( h ) 2 f ( 0 , y ) + ∫ 0 h − 2 x ∂ f ∂ x ( x , y ) + ( 1 − x 2 ) ∂ 2 f ∂ x 2 ( x , y ) μ ( x ) d x ,

that is,

∫ 0 h f ( x , y ) d x = h 2 f ( h , y ) + tanh − 1 ( h ) 2 f ( 0 , y ) − ∫ 0 h ( 1 − x 2 ) ∂ 2 f ∂ x 2 ( x , y ) − 2 x ∂ f ∂ x ( x , y ) + 2 f ( x , y ) μ ( x ) d x .

Since

μ ≥ 0 , ( 1 − x 2 ) ∂ 2 f ∂ x 2 − 2 x ∂ f ∂ x + 2 f ≥ 0 ,

it holds that:

∫ 0 h f ( x , y ) d x ≤ h f ( h , y ) + tanh − 1 ( h ) f ( 0 , y ) 2 .

Integrating the aforementioned inequality over y ∈ [ 0 , ℓ ] , we obtain

(2.17) ∫ 0 h ∫ 0 ℓ f ( x , y ) d y d x ≤ 1 2 ∫ 0 ℓ ( h f ( h , y ) + tanh − 1 ( h ) f ( 0 , y ) ) d y .

Similarly, we introduce the function:

ν ( y ) = y ( tanh − 1 ( ℓ ) − tanh − 1 ( y ) ) 2 , 0 ≤ y ≤ ℓ .

For all 0 ≤ x ≤ h , multiplying f ( x , y ) by ( 1 − y 2 ) ν ″ ( y ) − 2 y ν ′ ( y ) and integrating by parts over y ∈ [ 0 , ℓ ] , we obtain

∫ 0 ℓ f ( x , y ) d y ≤ ℓ f ( x , ℓ ) + tanh − 1 ( ℓ ) f ( x , 0 ) 2 .

Integrating the aforementioned inequality over x ∈ [ 0 , h ] , we obtain

(2.18) ∫ 0 h ∫ 0 ℓ f ( x , y ) d y d x ≤ 1 2 ∫ 0 h ( ℓ f ( x , ℓ ) + tanh − 1 ( ℓ ) f ( x , 0 ) ) d x .

Then, summing (2.17) and (2.18), we obtain (2.11).□

3 Systems of fourth-order partial differential inequalities

In this section, we study some systems of fourth-order partial differential inequalities in the plane.

We first consider the set of functions f = f ( x , y ) ∈ C 4 ( R 2 ) satisfying the system:

(3.1) ∂ 4 f ∂ x 4 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 , ∂ 4 f ∂ y 4 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 .

We have the following result.

Theorem 3.1

Let f ∈ C 4 ( R 2 ) verify (3.1). Then, for all a , b , c , d ∈ R with a < b and c < d , it holds that:

(3.2) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 4 1 b − a ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 d − c ∫ c d ( f ( a , y ) + f ( b , y ) ) d y + 1 24 d − c b − a ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x + b − a d − c ∫ c d ∂ f ∂ x ( a , y ) d y − ∂ f ∂ x ( b , y ) d y .

Proof

Let a , b , c , d ∈ R with a < b and c < d . We introduce the function:

L ( y ) = − 1 24 ( y − c ) 2 ( y − d ) 2 , c ≤ y ≤ d .

For all a ≤ x ≤ b , integrating by parts, we obtain

∫ c d f ( x , y ) L ″ ″ ( y ) d y = [ L ′ ′ ′ ( y ) f ( x , y ) ] y = c d − ∫ c d ∂ f ∂ y ( x , y ) L ′ ′ ′ ( y ) d y = [ L ′ ′ ′ ( y ) f ( x , y ) ] y = c d − L ′ ′ ( y ) ∂ f ∂ y ( x , y ) y = c d + ∫ c d ∂ 2 f ∂ y 2 ( x , y ) L ′ ′ ( y ) d y = [ L ′ ′ ′ ( y ) f ( x , y ) ] y = c d − L ′ ′ ( y ) ∂ f ∂ y ( x , y ) y = c d + L ′ ( y ) ∂ 2 f ∂ y 2 ( x , y ) y = c d − ∫ c d ∂ 3 f ∂ y 3 ( x , y ) L ′ ( y ) d y = L ′ ′ ′ ( y ) f ( x , y ) − L ′ ′ ( y ) ∂ f ∂ y ( x , y ) + L ′ ( y ) ∂ 2 f ∂ y 2 ( x , y ) − L ( y ) ∂ 3 f ∂ y 3 ( x , y ) y = c d + ∫ c d ∂ 4 f ∂ y 4 ( x , y ) L ( y ) d y ,

that is,

(3.3) ∫ c d f ( x , y ) L ′ ′ ′ ′ ( y ) d y = L ′ ′ ′ ( y ) f ( x , y ) − L ′ ′ ( y ) ∂ f ∂ y ( x , y ) + L ′ ( y ) ∂ 2 f ∂ y 2 ( x , y ) − L ( y ) ∂ 3 f ∂ y 3 ( x , y ) y = c d + ∫ c d ∂ 4 f ∂ y 4 ( x , y ) L ( y ) d y .

On the other hand, elementary calculations show that:

(3.4) L ′ ′ ′ ′ ( y ) = − 1 , c < y < d ,

(3.5) L ( c ) = L ( d ) = 0 , L ′ ( c ) = L ′ ( d ) = 0

and

(3.6) L ′ ′ ( c ) = L ′ ′ ( d ) = − 1 12 ( d − c ) 2 and L ′ ′ ′ ( c ) = − L ′ ′ ′ ( d ) = 1 2 ( d − c ) .

Hence, in view of (3.4), (3.5), and (3.6), we obtain

L ′ ′ ′ ( y ) f ( x , y ) − L ′ ′ ( y ) ∂ f ∂ y ( x , y ) + L ′ ( y ) ∂ 2 f ∂ y 2 ( x , y ) − L ( y ) ∂ 3 f ∂ y 3 ( x , y ) y = c d = L ′ ′ ′ ( y ) f ( x , y ) − L ′ ′ ( y ) ∂ f ∂ y ( x , y ) y = c d = L ′ ′ ′ ( d ) f ( x , d ) − L ′ ′ ( d ) ∂ f ∂ y ( x , d ) − L ′ ′ ′ ( c ) f ( x , c ) + L ′ ′ ( c ) ∂ f ∂ y ( x , c ) = − 1 2 ( d − c ) ( f ( x , d ) + f ( x , c ) ) + 1 12 ( d − c ) 2 ∂ f ∂ y ( x , d ) − ∂ f ∂ y ( x , c ) ,

that is,

(3.7) L ′ ′ ′ ( y ) f ( x , y ) − L ′ ′ ( y ) ∂ f ∂ y ( x , y ) + L ′ ( y ) ∂ 2 f ∂ y 2 ( x , y ) − L ( y ) ∂ 3 f ∂ y 3 ( x , y ) y = c d = − 1 2 ( d − c ) ( f ( x , d ) + f ( x , c ) ) + 1 12 ( d − c ) 2 ∂ f ∂ y ( x , d ) − ∂ f ∂ y ( x , c ) .

Moreover, we have

(3.8) ∫ c d f ( x , y ) L ′ ′ ′ ′ ( y ) d y = − ∫ c d f ( x , y ) d y .

Then, it follows from (3.3), (3.7), and (3.8) that:

∫ c d f ( x , y ) d y = 1 2 ( d − c ) ( f ( x , d ) + f ( x , c ) ) − 1 12 ( d − c ) 2 ∂ f ∂ y ( x , d ) − ∂ f ∂ y ( x , c ) − ∫ c d ∂ 4 f ∂ y 4 ( x , y ) L ( y ) d y .

Since

L ≤ 0 , ∂ 4 f ∂ y 4 ≤ 0 ,

it holds that:

1 d − c ∫ c d f ( x , y ) d y ≤ f ( x , c ) + f ( x , d ) 2 − 1 12 ( d − c ) ∂ f ∂ y ( x , d ) − ∂ f ∂ y ( x , c ) .

Integrating the aforementioned inequality over x ∈ [ a , b ] , we obtain

(3.9) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 2 ( b − a ) ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + d − c 12 ( b − a ) ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x .

Similarly, we introduce the function:

M ( x ) = − 1 24 ( x − a ) 2 ( x − b ) 2 , a ≤ x ≤ b .

For all c ≤ y ≤ d , multiplying f ( x , y ) by M ′ ′ ′ ′ ( x ) and integrating by parts over x ∈ [ a , b ] , we obtain

1 b − a ∫ a b f ( x , y ) d x ≤ f ( a , y ) + f ( b , y ) 2 − 1 12 ( b − a ) ∂ f ∂ x ( b , y ) − ∂ f ∂ x ( a , y ) .

Then, integrating the aforementioned inequality over y ∈ [ c , d ] , we obtain:

(3.10) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 2 ( d − c ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y + b − a 12 ( d − c ) ∫ c d ∂ f ∂ x ( a , y ) − ∂ f ∂ x ( b , y ) d y .

Finally, summing (3.9) and (3.10), we obtain (3.2).□

We next consider the set of functions f = f ( x , y ) ∈ C 4 ( R 2 ) satisfying systems of the form:

(3.11) ∂ 4 f ∂ x 4 ( x , y ) ≤ p , ( x , y ) ∈ R 2 , ∂ 4 f ∂ y 4 ( x , y ) ≤ q , ( x , y ) ∈ R 2 ,

where p and q ∈ R are the constants.

From Theorem 3.1, we deduce the following result.

Corollary 3.2

Let f ∈ C 4 ( R 2 ) verify (3.11), where p and q ∈ R are the constants. Then, for all a , b , c , d ∈ R with a < b and c < d , it holds that:

(3.12) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 4 1 b − a ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 d − c ∫ c d ( f ( a , y ) + f ( b , y ) ) d y + 1 24 d − c b − a ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x + b − a d − c ∫ c d ∂ f ∂ x ( a , y ) d y − ∂ f ∂ x ( b , y ) d y + ( b − a ) 4 p + ( d − c ) 4 q 1440 .

Proof

We introduce the function:

g ( x , y ) = f ( x , y ) − p 24 x 4 − q 24 y 4 , ( x , y ) ∈ R 2 .

Observe that g verifies (3.1). Namely, one has

∂ 4 g ∂ x 4 ( x , y ) = ∂ 4 f ∂ x 4 ( x , y ) − p ≤ 0

and

∂ 4 g ∂ y 4 ( x , y ) = ∂ 4 f ∂ y 4 ( x , y ) − q ≤ 0 .

Hence, from Theorem 3.1, we deduce that for all a , b , c , d ∈ R with a < b and c < d , it holds that:

(3.13) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d g ( x , y ) d y d x ≤ 1 4 1 b − a ∫ a b ( g ( x , c ) + g ( x , d ) ) d x + 1 d − c ∫ c d ( g ( a , y ) + g ( b , y ) ) d y + 1 24 d − c b − a ∫ a b ∂ g ∂ y ( x , c ) − ∂ g ∂ y ( x , d ) d x + b − a d − c ∫ c d ∂ g ∂ x ( a , y ) d y − ∂ g ∂ x ( b , y ) d y .

On the other hand, we have

∫ a b ∫ c d g ( x , y ) d y d x = ∫ a b ∫ c d f ( x , y ) d y d x − 1 24 ∫ a b ∫ c d ( p x 4 + q y 4 ) d y d x = ∫ a b ∫ c d f ( x , y ) d y d x − p ( b 5 − a 5 ) ( d − c ) + q ( b − a ) ( d 5 − c 5 ) 120 .

We also have

∫ a b ( g ( x , c ) + g ( x , d ) ) d x = ∫ a b ( f ( x , c ) + f ( x , d ) ) d x − p ( b 5 − a 5 ) 60 − q ( c 4 + d 4 ) ( b − a ) 24 , ∫ c d ( g ( a , y ) + g ( b , y ) ) d y = ∫ c d ( f ( a , y ) + f ( b , y ) ) d y − q ( d 5 − c 5 ) 60 − p ( a 4 + b 4 ) ( d − c ) 24 , ∫ a b ∂ g ∂ y ( x , c ) − ∂ g ∂ y ( x , d ) d x = ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x + q 6 ( d 3 − c 3 ) , ∫ c d ∂ g ∂ x ( a , y ) d y − ∂ g ∂ x ( b , y ) d y = ∫ c d ∂ f ∂ x ( a , y ) d y − ∂ ∂ x ( b , y ) d y + p 6 ( b 3 − a 3 ) .

Substituting the aforementioned quantities in (3.13), we obtain (3.12).□

We now consider the set of functions f = f ( x , y ) ∈ C 4 ( R 2 ) satisfying the system:

(3.14) ∂ 2 f ∂ x 2 ( x , y ) ≥ 0 , ( x , y ) ∈ R 2 , ∂ 4 f ∂ x 4 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 , ∂ 2 f ∂ y 2 ( x , y ) ≥ 0 , ( x , y ) ∈ R 2 , ∂ 4 f ∂ y 4 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 .

Observe that if f verifies (3.14), then f is convex on the coordinates, and consequently, f satisfies (1.1).

From Theorem 3.1, we deduce the following interesting refinement of Dragomir Inequality (1.1).

Corollary 3.3

Let f ∈ C 4 ( R 2 ) verify (3.14). Then, for all a , b , c , d ∈ R with a < b and c < d , it holds that:

(3.15) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 4 1 b − a ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 d − c ∫ c d ( f ( a , y ) + f ( b , y ) ) d y + 1 24 d − c b − a ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x + b − a d − c ∫ c d ∂ f ∂ x ( a , y ) d y − ∂ f ∂ x ( b , y ) d y ≤ 1 4 1 b − a ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 d − c ∫ c d ( f ( a , y ) + f ( b , y ) ) d y .

Proof

By (3.14), we observe that for all c ≤ y ≤ d , the function

∂ f ∂ x ( ⋅ , y ) : [ a , b ] ∋ x ↦ ∂ f ∂ x ( x , y )

is nondecreasing. Then, for all c ≤ y ≤ d , it holds that:

∂ f ∂ x ( a , y ) − ∂ f ∂ x ( b , y ) ≤ 0 ,

which implies that:

(3.16) ∫ c d ∂ f ∂ x ( a , y ) − ∂ f ∂ x ( b , y ) d y ≤ 0 .

Similarly, for all a ≤ x ≤ b , the function

∂ f ∂ y ( x , ⋅ ) : [ c , d ] ∋ y ↦ ∂ f ∂ y ( x , y )

is nondecreasing, which yields

(3.17) ∫ c d ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d y ≤ 0 .

Hence, (3.15) follows from (3.2), (3.16), and (3.17).□

We now study the set of functions f = f ( x , y ) ∈ C 4 ( R 2 ) satisfying the system:

(3.18) ∂ 2 f ∂ x 2 ( x , y ) ≥ 0 , ( x , y ) ∈ R 2 , ∂ 4 f ∂ y 4 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 .

We have the following result.

Theorem 3.4

Let f ∈ C 4 ( R 2 ) verify (3.18). Then, for all a , b , c , d ∈ R with a < b and c < d , it holds that:

(3.19) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 4 1 b − a ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 d − c ∫ c d ( f ( a , y ) + f ( b , y ) ) d y + d − c 24 ( b − a ) ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x .

Proof

Let a , b , c , d ∈ R with a < b and c < d . By (3.18), for all c ≤ y ≤ d , the function

f y : R ∋ x ↦ f ( x , y )

is convex. Then, by Hermite-Hadamard inequality, it holds that:

1 b − a ∫ a b f y ( x ) d x ≤ f y ( a ) + f y ( b ) 2 ,

that is,

1 b − a ∫ a b f ( x , y ) d x ≤ f ( a , y ) + f ( b , y ) 2 .

Integrating the aforementioned inequality over y ∈ [ c , d ] , we obtain

(3.20) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 2 ( d − c ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y .

On the other hand, since ∂ 4 f ∂ y 4 ≤ 0 , by (3.9), we have

(3.21) 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 2 ( b − a ) ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + d − c 12 ( b − a ) ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x .

Then, summing (3.20) and (3.21), we obtain (3.19).□

Let us now consider the set of functions f = f ( x , y ) ∈ C 4 ( R 2 ) satisfying the system:

(3.22) ∂ 2 f ∂ x 2 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 , ∂ 4 f ∂ y 4 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 .

We have the following result.

Theorem 3.5

Let f ∈ C 4 ( R 2 ) verify (3.22). Then, for all a , b , c , d ∈ R with a < b and c < d , it holds that:

(3.23) 1 2 ( d − c ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y ≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 2 ( b − a ) ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + d − c 12 ( b − a ) ∫ a b ∂ f ∂ y ( x , c ) − ∂ f ∂ y ( x , d ) d x .

Proof

Let a , b , c , d ∈ R with a < b and c < d . By (3.22), for all c ≤ y ≤ d , the function

f y : R ∋ x ↦ f ( x , y )

is concave. Then, by Hermite-Hadamard inequality, it holds that:

f y ( a ) + f y ( b ) 2 ≤ 1 b − a ∫ a b f y ( x ) d x ,

that is,

f ( a , y ) + f ( b , y ) 2 ≤ 1 b − a ∫ a b f ( x , y ) d x .

Integrating the aforementioned inequality over y ∈ [ c , d ] , we obtain

(3.24) 1 2 ( d − c ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y ≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x .

On the other hand, since ∂ 4 f ∂ y 4 ≤ 0 , (3.21) holds. Hence, (3.23) follows from (3.21) and (3.24).□

Similarly, if we consider the set of functions f = f ( x , y ) ∈ C 4 ( R 2 ) satisfying the system:

(3.25) ∂ 4 f ∂ x 4 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 , ∂ 2 f ∂ y 2 ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 ,

we obtain the following result.

Theorem 3.6

Let f ∈ C 4 ( R 2 ) verify (3.25). Then, for all a , b , c , d ∈ R with a < b and c < d , it holds that:

1 2 ( b − a ) ∫ a b ( f ( x , c ) + f ( x , d ) ) d x ≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 2 ( d − c ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y + b − a 12 ( d − c ) ∫ c d ∂ f ∂ x ( a , y ) − ∂ f ∂ x ( b , y ) d y .

4 Conclusion

Various kinds of systems of partial differential inequalities posed in the plane are investigated. For each type of system, a necessary condition for the existence of solutions is obtained. Namely, we proved that, if a solution exists, then it satisfies an inequality of Hermite-Hadamard type. Our approach makes use of some tools from ordinary differential equations. The considered approach is general, and it can be adapted for studying other types of systems of partial differential inequalities. For instance, it will be interesting to study the system of higher-order differential inequalities:

( − 1 ) n ∂ 2 n f ∂ x 2 n ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 , ( − 1 ) n ∂ 2 n f ∂ y 2 n ( x , y ) ≤ 0 , ( x , y ) ∈ R 2 ,

where n is a positive integer. Note that in the case n = 1 , the aforementioned system reduces to System (1.2).

Acknowledgement

Bessem Samet was supported by Researchers Supporting Project number (RSP2023R4), King Saud University, Riyadh, Saudi Arabia.

Conflict of interest: The authors state no conflicts of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2023-05-05

Revised: 2023-08-16

Accepted: 2023-08-16

Published Online: 2023-09-22

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2023-0115

Keywords for this article

systems of partial differential inequalities; Hermite-Hadamard-type inequalities; convex functions; convex functions on the coordinates

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