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Advances in Difference Equations

, 2018:206 | Cite as

Wirtinger inequality using Bessel functions

  • Tatjana Z. Mirković
Open Access
Research
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Part of the following topical collections:
  1. Advances in Fractional Differential Equations and Their Real World Applications

Abstract

This paper presents of some new Wirtinger-type integral inequalities by using Bessel functions. We establish one weighted Wirtinger inequality.

Keywords

Wirtinger’s inequality Bessel functions 

MSC

26D20 33C10 

1 Introduction

The Wirtinger inequality plays a very important role in the theory of approximation, the theory of Sobolev’s spaces, the theory of function of several variables and functional analysis. In 1916. Wirtinger established an integral inequality.

Theorem 1.1

(Wirtinger inequality)

Let\(f:\mathbb{R} \to \mathbb{R}\)be a continuous periodic function with period 2πand let\(f' \in L^{2}\). Then, if\(\int_{0}^{2\pi} {f ( x )}\,dx = 0\)the following inequality holds:
$$\int_{0}^{2\pi} {f^{2} ( x )\,dx} \le \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx}, $$
with equality if and only if\(f ( x ) = a\cos x + b\sin x\), whereaandbare constants.

Theorem 1.2

Let\(f ( x )\)be a smooth function with period 2π. Then, for all realt,
$$ \int_{0}^{2\pi} { \bigl[ {f ( x ) - f ( {x + t} )} \bigr]^{2} \,dx} \le4\sin^{2} \frac{t}{2} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx}. $$
(1)
Equality is attained if and only if\(f ( x ) = a\cos x + b\sin x + c\), where\(a, b, c\)are real constants (for\(t = 0\)equality holds always).
In [1], Beesack obtained the following generalization of the Wirtinger inequality: If \(k > 1\), \(f(x) \in C^{1} ( { [ {0,\pi} ]} )\), \(f ( 0 ) = 0\), then
$$ \int_{0}^{\pi}{ \bigl( {f' ( x )} \bigr)^{2k} }\,dx \ge\frac{{2k - 1}}{{ ( {k\sin\frac{\pi}{{2k}}} )^{2k} }} \int_{0}^{\pi}{f^{2k} ( x )\,dx} ,\quad k \ge1. $$
(2)

In [2], Hall proved the following theorem:

Theorem 1.3

Suppose that\(k \in N\), \(f ( x ) \in C^{2} [ {0,\pi} ]\)and\(f ( 0 ) = f ( \pi ) = 0\). Let\(H ( u )\)be an even function, increasing and strictly convex on\(R^{+}\), and such that\(H ( 0 ) = H' ( 0 ) = 0\); moreover, \(uH^{\prime\prime} ( u ) \to0\)as\(u \to 0\). Then we have
$$ \int_{0}^{\pi}{H \bigl( {{{f' ( x )} / {f ( x )}}} \bigr)f^{2k} ( x ) \ge ( {2k - 1} )\lambda } \int_{0}^{\pi}{f^{2k} ( x )\,dx}, \quad \lambda = \lambda ( {k,H} ), $$
(3)
where\(\lambda = \lambda ( {k,H} )\)is determined by the equation
$$ \int_{0}^{\infty}{\frac{{G' ( u )}}{{G ( u ) + ( {2k - 1} )\lambda}}} \frac{{du}}{u} = k\pi ,\quad G ( u ): = uH' ( u ) - H ( u ). $$
(4)
For each non-negative constant p, the associated Bessel equation is
$$ x^{2} \frac{{d^{2} y}}{{\,dx^{2} }} + x\frac{{dy}}{{\,dx}} + \bigl( {x^{2} - p^{2} } \bigr)y = 0. $$
(5)
Since Bessel’s differential equation is a second-order equation, there must be two linearly independent solutions, which are called Bessel functions. These functions play important roles in many areas of applied mathematics (see [3, 4]). Typically the general solution is given as
$$y = a_{1} J_{p} ( x ) + a_{2} Y_{p} ( x ), $$
where \(a_{1}\) and \(a_{2}\) are arbitrary constants.

Special functions \(J_{p} ( x )\) are Bessel functions of the first kind, which are finite at \(x = 0\) for all real values of p, and \(Y_{p} ( x )\) are Bessel functions of the second kind, which are singular at \(x = 0\).

The Bessel function of the first kind of order p can be determined using an infinite power series expansion as follows: \(J_{p} ( x ) = \sum_{k = 0}^{ + \infty} {\frac{{ ( { - 1} )^{k} }}{{k!\Gamma ( {k + p + 1} )}}} ( {\frac{x}{2}} )^{2k + p}\). Since \(\Gamma ( {k + 1} ) = k!\), it follows that
$$ J_{p} ( x ) = \sum_{k = 0}^{ + \infty} {\frac{{ ( { - 1} )^{k} }}{{k! ( {k + p} )!}}} \biggl( {\frac{x}{2}} \biggr)^{2k + p}. $$
(6)

For integer order p, functions \(J_{p}\) and \(J_{ - p}\) are not linearly independent, \(J_{ - p} = ( { - 1} )^{p} J_{p}\). In contrast, for non-integer orders, \(J_{p}\) and \(J_{ - p}\) are linearly independent.

The most important Bessel functions are \(J_{0} ( x )\) and \(J_{1} ( x )\). For \(p = - \frac{1}{2}\) and \(p = \frac{1}{2}\), this functions expansion as follows:
$$\begin{aligned} &J_{{{ - 1} / 2}} ( x ) = \sqrt{\frac{2}{{\pi x}}} \cos x, \end{aligned}$$
(7)
$$\begin{aligned} &J_{{1 / 2}} ( x ) = \sqrt{\frac{2}{{\pi x}}} \sin x. \end{aligned}$$
(8)

2 Main results

Theorem 2.1

Let\(f' \in L^{2k}\)on\([ {0,\pi} ]\), with\(f ( 0 ) = f ( \pi ) = 0\). Then the following inequality holds:
$$ \int_{0}^{\pi}{f^{2k} ( x )\,dx} \le \frac{1}{{2k - 1}}\biggl(\frac{\pi}{2}\biggr)^{2k} \biggl( { \int_{0}^{\frac{\pi}{2}} {J_{0} \biggl( { \frac{\pi }{{2k}}\cos t} \biggr)\cos t\,dt} } \biggr)^{2k} \int_{0}^{\pi}{f^{\prime2k} ( x )\,dx}. $$
(9)

Proof

Since \(J_{n} ( z ) = ( {\frac{z}{2}} )^{n} \sum_{r = 0}^{ + \infty} { ( { - 1} )^{r} \frac{{ ( {\frac{z}{2}} )^{2r} }}{{r! ( {n + r} )!}}}\), it follows that \(J_{0} ( {\frac{\pi }{{2k}}\cos t} ) = \sum_{r = 0}^{ + \infty} { ( { - 1} )^{r} \frac{{ ( {\frac{\pi}{{2k}}\frac{{\cos t}}{2}} )^{2r} }}{{ ( {r!} )^{2} }}}\).
$$\begin{aligned} \int_{0}^{\frac{\pi}{2}} {J_{0} \biggl( { \frac {\pi}{{2k}}\cos t} \biggr)\cos t\,dt} =& \int_{0}^{\frac{\pi}{2}} {\sum_{r = 0}^{ + \infty} { ( { - 1} )^{r} \frac{{ ( {\frac{\pi}{{2k}}\frac{{\cos t}}{2}} )^{2r} }}{{ ( {r!} )^{2} }}} \cos t\,dt} \\ =& \sum_{r = 0}^{ + \infty} { ( { - 1} )^{r} \frac{{\pi^{2r} }}{{ ( {2k} )^{2r} 2^{2r} ( {r!} )^{2} }}} \int_{0}^{\frac{\pi}{2}} {\cos^{2r + 1} t\,dt}. \end{aligned}$$

Using the integration by parts formula on the integral \(I_{2r + 1} = \int_{0}^{\frac{\pi}{2}} {\cos^{2r + 1} t\,dt}\) and the fact that \(\int_{0}^{{\pi / 2}} {\cos t\,dt} = 1\), we obtained the recurrence relation \(I_{2r + 1} = \frac{{2r}}{{2r + 1}}I_{2r - 1}\), which implies \(I_{2r + 1} = \frac{{ ( {2r} )!!}}{{ ( {2r + 1} )!!}} = \frac{{ ( { ( {2r} )!!} )^{2} }}{{ ( {2r + 1} )!}} = \frac{{ ( {2^{r} r!} )^{2} }}{{ ( {2r + 1} )!}}\).

The above equality becomes
$$\begin{aligned} \int_{0}^{\frac{\pi}{2}} {J_{0} \biggl( { \frac{\pi}{{2k}}\cos t} \biggr)\cos t} \,dt =& \sum _{r = 0}^{ + \infty} { ( { - 1} )^{r} \frac{{\pi^{2r} }}{{ ( {2k} )^{2r} 2^{2r} ( {r!} )^{2} }}\frac{{ ( {2^{r} r!} )^{2} }}{{ ( {2r + 1} )!}}} \\ = &\frac{{2k}}{\pi}\sum_{r = 0}^{ + \infty} { ( { - 1} )^{r} \frac{{ ( {\frac{\pi}{{2k}}} )^{2r + 1} }}{{ ( {2r + 1} )!}}} = \frac{{2k}}{\pi}\sin \frac{\pi }{{2k}}, \end{aligned}$$
which implies
$$\sin^{2k} \frac{\pi}{{2k}} = \biggl( {\frac{\pi}{{2k}}} \biggr)^{2k} \biggl( { \int_{0}^{\frac{\pi}{2}} {J_{0} \biggl( { \frac{\pi}{{2k}}\cos t} \biggr)\cos t\,dt} } \biggr)^{2k}. $$
By (2) it follows that
$$\begin{aligned} & \int_{0}^{\pi}{f^{2k} ( x )\,dx} \\ &\quad \le \frac{1}{{2k - 1}}k^{2k} \frac{{\pi^{2k} }}{{4^{k} k^{2k} }} \biggl( { \int_{0}^{\frac{\pi}{2}} {J_{0} \biggl( { \frac{\pi}{{2k}}\cos t} \biggr)\cos t\,dt} } \biggr)^{2k} \int_{0}^{\pi}{f^{\prime2k} ( x )\,dx}, \\ & \int_{0}^{\pi}{f^{2k} ( x )\,dx} \\ &\quad \le \frac{1}{{2k - 1}} \biggl( {\frac{\pi}{2}} \biggr)^{2k} \biggl( { \int_{0}^{\frac{\pi}{2}} {J_{0} \biggl( { \frac{\pi}{{2k}}\cos t} \biggr)\cos t\,dt} } \biggr)^{2k} \int_{0}^{\pi}{f^{\prime2k} ( x )\,dx}. \end{aligned}$$
 □

Theorem 2.2

If\(f' \in L^{2k}\)is absolutely continuous on\([ {0,\pi} ]\), with\(f ( 0 ) = f ( \pi ) = 0\)then
$$\begin{aligned} \int_{0}^{\pi}{f^{2k} ( x )\,dx} \le \frac{{\pi ^{2k} }}{{2k + 1}}C ( k ) \int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx}, \end{aligned}$$
(10)
where\(C ( k ): = \frac{{\int_{0}^{\frac{\pi}{2}} {x^{k} J_{{1 / 2}}^{2k} ( x )\,dx} }}{{\int _{0}^{\frac{\pi}{2}} {x^{k + 1} J_{{1/ 2}}^{2k + 1} ( x )\,dx} }}\).

Proof

Starting with the right side of (10), we obtain
$$\begin{aligned} &\frac{{\pi^{2k} }}{{2k + 1}}C ( k ) \int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx} \\ &\quad = \frac{{\pi^{2k} }}{{2k + 1}}\frac{{\int_{0}^{\frac{\pi}{2}} {x^{k} J_{{1 / 2}}^{2k} ( x )\,dx} }}{{\int _{0}^{\frac{\pi}{2}} {x^{k + 1} J_{{1 / 2}}^{2k + 1} ( x )\,dx} }} \int _{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx} \\ &\quad = \frac{{\pi^{2k} \int_{0}^{\frac{\pi}{2}} {x^{k} ( {\sqrt {\frac{2}{{\pi x}}} } )^{2k} \sin^{2k} x\,dx} }}{{ ( {2k + 1} )\int_{0}^{\frac{\pi}{2}} {x^{k + 1} ( {\sqrt {\frac{2}{{\pi x}}} } )^{2k + 1} \sin^{2k + 1} x\,dx} }} \int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx} \\ &\quad = \frac{{\pi^{2k} }}{{2k + 1}}\sqrt{\frac{\pi}{2}} \frac{{\int_{0}^{\frac{\pi}{2}} {\sin^{2k} x\,dx} }}{{\int_{0}^{\frac{\pi}{2}} {\sin^{2k + 1} x\,dx} }} \int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx}. \end{aligned}$$

Since \(\int_{0}^{\frac{\pi}{2}} {\sin^{p} x\cos^{q} x\,dx} = \frac{{\Gamma ( {\frac{{p + 1}}{2}} )\Gamma ( {\frac{{q + 1}}{2}} )}}{{2\Gamma ( {\frac{{p + q}}{2} + 1} )}}\), for \(p = 2k\) and \(q = 0\), we get \(\int_{0}^{\frac{\pi}{2}} {\sin^{2k} x\,dx} = \frac{{\Gamma ( {\frac{1}{2}} )\Gamma ( {\frac{{2k + 1}}{2}} )}}{{2\Gamma ( {k + 1} )}}\); For \(p = 2k + 1\) and \(q = 0\), we get \(\int_{0}^{\frac{\pi}{2}} {\sin^{2k + 1} x\,dx} = \frac{{\Gamma ( {\frac{1}{2}} )\Gamma ( {k + 1} )}}{{2\Gamma ( {\frac{{2k + 3}}{2}} )}}\).

By integrating by parts, we obtain \(\Gamma ( {\frac{{2k + 1}}{2}} ) = \frac{{ ( {2k} )!}}{{2^{k} k!}}\sqrt {\pi}\), \(\Gamma ( {\frac{{2k + 3}}{2}} ) = \frac{{ ( {2k + 1} )!}}{{2^{2k + 1} k!}}\sqrt{\pi}\), and since \(\Gamma ( {\frac{1}{2}} ) = \sqrt{\pi}\), \(\Gamma ( {k + 1} ) = k!\), it follows
$$\begin{aligned} &\frac{{\pi^{2k} }}{{2k + 1}}\sqrt{\frac{\pi}{2}} \frac{{\frac{\pi}{2}\frac{{ ( {2k} )!}}{{2^{2k} ( {k!} )^{2} }}}}{{\frac{{ ( {k!} )^{2} }}{{ ( {2k + 1} )!}}}} \int_{0}^{\pi}{f^{\prime2} ( x ) f^{2 ( {k - 1} )} ( x )\,dx} \\ &\quad = \frac{1}{{2k + 1}} \biggl( {\sqrt{\frac{\pi}{2}} } \biggr)^{2k + \frac{3}{2}} \frac{{ ( { ( {2k} )!} )^{2} ( {2k + 1} )}}{{ ( {2^{k} k!} )^{2} ( {k!} )^{2} }} \int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx} \\ &\quad = \frac{1}{{2k + 1}} \biggl( {\sqrt{\frac{\pi}{2}} } \biggr)^{2k + \frac{3}{2}} \biggl( {\frac{{ ( {2k - 1} )!}}{{k ( {k!} )}}} \biggr)^{2} k^{2} \int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx}. \end{aligned}$$
If in (4) we put \(H ( u ) = u^{2}\), \(G ( u ) = u^{2}\), then (6) gives \(\lambda = \frac{1}{{k^{2} ( {2k - 1} )}}\), so (3) becomes
$$\int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx} \ge\frac{1}{{k^{2} }} \int_{0}^{\pi}{f^{2k} ( x )\,dx}, $$
which implies
$$\frac{{\pi^{2k} }}{{2k + 1}}C ( k ) \int_{0}^{\pi}{f^{\prime2} ( x )f^{2 ( {k - 1} )} ( x )\,dx} \ge \biggl( {\sqrt{\frac{\pi}{2}} } \biggr)^{2k + \frac{3}{2}} \biggl( {\frac{{ ( {2k - 1} )!}}{{k ( {k!} )}}} \biggr)^{2} \int_{0}^{\pi}{f^{2k} ( x )\,dx}. $$

Since \(( {\sqrt{\frac{\pi}{2}} } )^{2k + \frac{3}{2}} > 1\) and \(( {\frac{{ ( {2k - 1} )!}}{{k ( {k!} )}}} )^{2} > 1\), inequality (10) is established. □

Theorem 2.3

Let\(f ( x )\)be a smooth function with period 2π. Then, for all realt,
$$ \int_{0}^{2\pi} { \bigl[ {f ( x ) - f ( {x + t} )} \bigr]^{2} \,dx} \le t\pi J_{{1 / 2}}^{2} \biggl( { \frac{t}{2}} \biggr) \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx}. $$
(11)
Equality is attained if and only if\(f ( x ) = A\cos x + B\sin x + C\), where\(A,B,C\)are real constants (for\(t = 0\)equality holds always).

Proof

From the equation \(J_{n}^{2} ( t ) = \frac{2}{\pi }\int_{0}^{\frac{\pi}{2}} {J_{2n} ( {2t\cos x} )\,dx}\), for \(n = \frac{1}{2}\), the right side of (11) becomes
$$\begin{aligned} &2t \int_{0}^{\frac{\pi}{2}} {J_{1} ( {t\cos x} )\,dx} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = 2t \int_{0}^{\frac{\pi}{2}} {\sum_{n = 0}^{ + \infty} {\frac{{ ( { - 1} )^{n} }}{{n! ( {n + 1} )!}} \biggl( {\frac{{t\cos x}}{2}} \biggr)^{2n + 1} } } \,dx \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = 2t\sum_{n = 0}^{ + \infty} { \frac{{ ( { - 1} )^{n} }}{{n! ( {n + 1} )!}} \biggl( {\frac{t}{2}} \biggr)^{2n + 1} } \int_{0}^{\frac{\pi}{2}} {\cos^{2n + 1} x\,dx} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = 2t\sum_{n = 0}^{ + \infty} { \frac{{ ( { - 1} )^{n} }}{{n! ( {n + 1} )!}}\frac{{t^{2n + 1} }}{{2 \cdot2^{2n} }}\frac{{2^{2n} n!n!}}{{ ( {2n + 1} )!}}} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = t\sum_{n = 0}^{ + \infty} { \frac{{ ( { - 1} )^{n} n!t^{2n + 1} }}{{ ( {n + 1} )n! ( {2n + 1} )!}}} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = t\sum_{n = 0}^{ + \infty} { \frac{{ ( { - 1} )^{n} t^{2n + 1} }}{{ ( {n + 1} ) ( {2n + 1} )!}}} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = t \biggl[ {\frac{t}{{1!}} - \frac{{t^{3} }}{{2 \cdot3!}} + \frac{{t^{5} }}{{3 \cdot5!}} - \frac{{t^{7} }}{{4 \cdot7!}} + \cdots} \biggr] \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = t \biggl[ {\frac{{2t^{2} }}{{2!t}} - \frac{{4t^{4} }}{{2 \cdot 4!t}} + \frac{{6t^{6} }}{{3 \cdot6!t}} - \frac{{8t^{8} }}{{4 \cdot 8!t}} + \frac{{10t^{10} }}{{5 \cdot10!t}} - \cdots} \biggr] \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = 2t \biggl[ {\frac{{t^{2} }}{{2!t}} - \frac{{t^{4} }}{{4!t}} + \frac{{t^{6} }}{{6!t}} - \frac{{t^{8} }}{{8!t}} + \frac{{t^{10} }}{{10!t}} - \cdots} \biggr] \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = 2t\sum_{n = 1}^{ + \infty} { ( { - 1} )^{n + 1} \frac{{t^{2n} }}{{t ( {2n} )!}}} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = t \Biggl( {1 - \sum_{n = 0}^{ + \infty} { ( { - 1} )^{n} \frac{{t^{2n} }}{{ ( {2n} )!}}} } \Biggr) \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} \\ &\quad = 2 ( {1 - \cos t} ) \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx} = 4 \sin^{2} \frac{t}{2} \int_{0}^{2\pi} {f^{\prime2} ( x )\,dx}. \end{aligned}$$

Equation (1) implies the desired inequality (11). □

Notes

Acknowledgements

The author would like to thank the anonymous referees for their constructive comments. The author states that no funding source or sponsor has participated in the realization of this work.

Authors’ contributions

The work as a whole is a contribution of the author. All authors read and approved the final manuscript.

Competing interests

The author declares that he has no competing interests.

References

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    Beesack, P.: Hardy’s inequality and its extension. Pac. J. Math. 11, 39–61 (1961) MathSciNetCrossRefzbMATHGoogle Scholar
  2. 2.
    Hall, R.: Generalized Wirtinger inequalities, random matrix theory, and the zeros of the Riemann zeta-function. J. Number Theory 97, 397–409 (2002) MathSciNetCrossRefzbMATHGoogle Scholar
  3. 3.
    Lavoie, J., Osler, T., Tremblay, R.: Fractional derivatives and special functions. SIAM Rev. 18(2), 240–268 (1976) MathSciNetCrossRefzbMATHGoogle Scholar
  4. 4.
    Watson, G.: A Treatise on the Theory of Bessel Functions. Cambridge University Press, Cambridge (1922) zbMATHGoogle Scholar

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Authors and Affiliations

  1. 1.College of Applied Professional StudiesVranjeSerbia

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