1 Introduction

In the year 1960, Opial [1] established the following integral inequality:

Theorem 1.1. Suppose f ∈ C1[0, h] satisfies f(0) = f(h) = 0 and f(x) > 0 for all x ∈ (0, h). Then the integral inequality holds

(1.1)

where this constantis best possible.

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 [2–6]. The inequality (1.1) 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 [7–22]. For an extensive survey on these inequalities, see [2, 6]. For Opial-type integral inequalities involving high-order partial derivatives see [23–27]. The main purpose of the present paper is to establish some new Opial-type inequalities involving higher-order partial derivatives by an extension of Das's idea [28]. Our results in special cases yield some of the recent results on Opial's-type inequalities and provide some new estimates on such types of inequalities.

2 Main results

Let n ≥ 1, k ≥ 1. Our main results are given in the following theorems.

Theorem 2.1 Let x(s, t) ∈ C(n - 1)[0, a] × C(k - 1)[0, b] be such that, , σ ∈ [0, s], τ ∈ [0, t], 0 ≤ i ≤ n - 1, 0 ≤ j ≤ k - 1. Further, let, be absolutely continuous, and. Then

(2.1)

where

and

Proof. For σ integration by parts (n - 1)-times and in view of , , 0 ≤ i ≤ n - 1, 0 ≤ j ≤ k - 1 we have

(2.2)

Multiplying both sides of (2.2) by x(n,k)(s, t) and using the Cauchy-Schwarz inequality, we have

(2.3)

Thus, integrating both sides of (2.3) over t from 0 to b first and then integrating the resulting inequality over s from 0 to a and applying the Cauchy-Schwarz inequality again, we obtain

This completes the proof.

Remark 2.1. Let x(s, t) reduce to s(t) and with suitable modifications, Then (2.1) becomes the following inequality:

(2.4)

This is just an inequality established by Das [28]. Obviously, for n ≥ 2, (2.4) is sharper than the following inequality established by Willett [29].

(2.5)

Remark 2.2. Taking for n = k = 1 in (2.1), (2.1) reduces to

(2.6)

Let x(s, t) reduce to s(t) and with suitable modifications. Then (2.6) becomes the following inequality: If x(t) is absolutely continuous in [0, a] and x(0) = 0, then

This is just an inequality established by Beesack [30].

Remark 2.3. Let 0 ≤ α, β < n, but fixed, and let g(s, t) ∈ C(n-α- 1)[0, a] × C(k-β-1)[0, b] be such that , 0 ≤ i ≤ n - α - 1, 0 ≤ i ≤ k - β -1 and suppose that , are absolutely continuous, and .

Then from (2.1) it follows that

Thus, for g(s, t) = x(α, β)(s, t), where x(s, t) ∈ C(n- 1)[0, a] × C(k- 1)[0, b], , , α ≤ i ≤ n - 1, β ≤ j ≤ k - 1, and x(n- 1, k-1)(s, t) are absolutely continuous, and , then

(2.7)

Obviously, a special case of (2.7) is the following inequality:

(2.8)

Let x(s, t) reduce to s(t) and with suitable modifications. Then (2.8) becomes the following inequality:

This is just an inequality established by Agarwal and Thandapani [31].

Theorem 2.2. Let l and m be positive numbers satisfying l + m > 1. Further, let x(s, t) ∈ C(n- 1)[0, a] × C(k- 1)[0, b] be such that, , σ ∈ [0, s], τ ∈ [0, t], 0 ≤ i ≤ n - 1, 0 ≤ j ≤ k - 1 and assume that, are absolutely continuous, and . Then

(2.9)

where

Proof. From (2.2), we have

by Hölder's inequality with indices l + m and , it follows that

where

Multiplying the both sides of above inequality by |x(n,k)(s, t)|m and integrating both sides over t from 0 to b first and then integrating the resulting inequality over s from 0 to a, we obtain

Now, applying Hölder's inequality with indices and to the integral on the right-side, we obtain

This completes the proof.

Remark 2.4. Let x(s, t) reduce to s(t) and with suitable modifications. Then (2.9) becomes the following inequality:

(2.10)

This is an inequality given by Das [28]. Taking for n = 1 in (2.10), we have

(2.11)

For m, l ≥ 1 Yang [32] established the following inequality:

(2.12)

Obviously, for m, l ≥ 1, (2.11) is sharper than (2.12).

Remark 2.5. For n = k = 1; (2.9) reduces to

Let x(s, t) reduce to s(t) and with suitable modifications. Then above inequality becomes the following inequality:

This is just an inequality established by Yang [32].

Remark 2.6. Following Remark 2.3, for x(s, t) ∈ C(n - 1)[0, a] × C(k - 1)[0, b], , , α ≤ i ≤ n - 1, β ≤ j ≤ k - 1 and x(n - 1, k - 1)(s, t) are absolutely continuous, and , it is easy to obtain that

(2.13)

Obviously, a special case of (2.14) is the following inequality:

(2.14)

Let x(s, t) reduce to s(t) and with suitable modifications, then (2.14) becomes the following inequality:

This is just an inequality established by Agarwal and Thandapani [31].

Theorem 2.3. Let l and m be positive numbers satisfying l + m = 1. Further, let x(s, t) ∈ C(n - 1)[0, a] × C(k - 1)[0, b] be such that, , σ ∈ [0, s], τ ∈ [0, t], 0 ≤ i ≤ n - 1, 0 ≤ j ≤ k - 1 and assume that, are absolutely continuous, and. Then

(2.15)

Proof. It is clear that

and hence

Now applying Hölder inequality with indices and , we obtain

This completes the proof.

Remark 2.7. Let x(s, t) reduce to s(t) and with suitable modifications. Then (2.16) becomes the following inequality:

This is an inequality given by Das [28].

Remark 2.8. Following Remark 2.3, for x(s, t) ∈ C(n - 1)[0, a] × C(k - 1)[0, b], , , α ≤ i ≤ n - 1, β ≤ j ≤ k - 1, and x(n - 1, k - 1)(s, t) are absolutely continuous, and , from (2.16), it is easy to obtain that

(2.16)

Let x(s, t) reduce to s(t) and with suitable modifications, then (2.16) becomes the following inequality:

This is an inequality given by Das [28].