\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 98, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/98\hfil Wirtinger-Beesack integral inequalities] {Wirtinger-Beesack integral inequalities} \author[Gulomjon M. Muminov\hfil EJDE-2005/98\hfilneg] {Gulomjon M. Muminov} \address{Gulomjon M. Muminov \hfill\break Andijon State University, Faculty of Mathematics, 129, Universitetskaya str., Andijon, Uzbekistan} \email{muminov\_g@rambler.ru} \date{} \thanks{Submitted April 22, 2005. Published September 19, 2005.} \subjclass[2000]{26D10} \keywords{Integral inequality; absolutely continuous function} \begin{abstract} A uniform method of obtaining various types of integral inequalities involving a function and its first or second derivative is extended to integral inequalities involving a function and its third derivative \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \section{Introduction} Integral inequalities of the form \begin{equation} \label{e1} \int_I sh^{2}dt\leq\int_I rh''^{2}dt, \quad h\in H, \end{equation} have appeared in publications such as \cite{f1,f2}. In the above equation $I$ is the interval $(\alpha,\beta)$, with $-\infty\leq\alpha<\beta\leq\infty$, $r>0$, $r\in AC(I)$, \begin{equation} \label{e2} s=(r\varphi'')''\varphi^{-1} \end{equation} with a given function $\varphi\in AC^{1}(I)$ such that $\varphi>0$ on the interval $I$, $r\varphi''\in AC^{1}(I)$, $\omega=(r\varphi')'\varphi+2r\varphi\varphi''-2r\varphi'^{2}\leq0$ and $H$ is the class of functions $h\in AC^{1}(I)$ satisfying some integral and limit conditions. In this article, we assume that $r\in AC^{1}(I)$, $\varphi\in AC^{2}(I)$ and $r\varphi'''\in AC^{2}(I)$ are such that $r>0$, $\varphi>0$ on the interval $I$. Putting \begin{equation} \label{e3} s=-(r\varphi''')'''\varphi^{-1}, \end{equation} we obtain the integral inequality \begin{equation} \label{e4} \int_Ish^2dt\le \int_Ir{h'''}^2dt,\quad h\in H. \end{equation} The method used here consists in determining auxiliary functions depending on the given function $r$ and the auxiliary function $\varphi$ so that these functions determine the class $H$ for which the inequality \eqref{e4} holds. \section{Main result} Let $I=(\alpha,\beta)$ be an arbitrary open interval with $-\infty\le\alpha<\beta\le\infty$. We denote by $AC^k(I)$ the set of functions whose $k$ derivative is absolutely continuous on the interval $I$. Let $r\in AC^1(I)$ and $\varphi\in AC^2(I)$ be given functions such that $r>0$, $\varphi>0$ on the interval $I$ and $r\varphi'''\in AC^2(I)$. Let us put \[ s=-( r\varphi''')'''\varphi^{-1}, \] Let us denote by $H$ the set of functions $h\in AC^2(I)$ for which \begin{equation} \label{e5} \int_Ir{h'''}^2dt<\infty, \quad \int_Ish^2dt>-\infty \end{equation} and satisfy the limit conditions \begin{gather} \label{e6} \lim_{t\to\alpha}\inf S(t,h,h',h'')<\infty ,\quad \lim_{t\to\beta}\sup S(t,h,h',h'')>-\infty, \\ \label{e7} \lim_{t\to\alpha}\inf S(t,h,h',h'')\leq \lim_{t\to\beta}\sup S(t,h,h',h''), \end{gather} where \begin{equation} \begin{aligned} &S(t,h,h',h'')\\ &=\nu_0(t)h^2+\nu_1(t){h'}^2+\nu_2(t){h''}^2 +2\varepsilon_{01}(t)hh'+2\varepsilon_{02}(t)hh''+2\varepsilon_{12}h'h'', \end{aligned}\label{e8} \end{equation} \begin{gather} \nu_0(t)=[(r\varphi''')'\varphi]'\varphi^{-2}-\frac{ 1}{ 2}r\varphi'''\varphi^{-3}(\varphi^2)''- 3(\varphi'\varphi^{-1})^3\big(\frac{r\varphi''}{ \varphi'}\big)' -2r{\varphi'}^3\varphi^{-2}(\varphi'\varphi^{-2})', \label{e9}\\ \nu_1(t)=-6(r\varphi''\varphi^{-1})'-2r(\varphi''\varphi^{-1})' +4r(\varphi'\varphi^{-1})^3, \label{e10} \\ \nu_2(t)=r\varphi'\varphi^{-1}, \label{e11} \\ \varepsilon_{01}(t)=-(r\varphi'''\varphi)'\varphi^{-2}+3(r\varphi''\varphi^{-2})'\varphi' +r[(\varphi''\varphi^{-1})^2-4(\varphi'\varphi^{-1})^4], \label{e12}\\ \varepsilon_{02}(t)=r[(\varphi'\varphi^{-1})''+\varphi'\varphi''\varphi^{-2}], \label{e13}\\ \varepsilon_{12}(t)=r\varphi(\varphi'\varphi^{-2})'. \label{e14} \end{gather} These assumptions apply that $\nu_0\in AC(I)$, $\nu_1,\varepsilon_{01}\in AC^1(I)$ and $\nu_2,\varepsilon_{02}, \varepsilon_{12}\in AC^2(I)$. The following theorem is the main result of this paper. \begin{theorem} \label{thm1} Let \begin{gather} \omega_0(t)=[(r\varphi'''+(r\varphi'')'\varphi^{-1}]\varphi^2+r{\varphi''}^{2}\ge 0, \label{e15}\\ \omega_1(t)=2r{\varphi'}^2-2r\varphi''\varphi-(r\varphi')'\varphi\ge 0 \label{e16} \end{gather} almost everywhere on the interval $I$. Then for every function $h\in H$ the inequality \eqref{e4} holds. If $\omega_0\ne 0$, $\omega_1\ne 0$ and $h\ne 0$ then \eqref{e4} becomes an equality if and only if $h=c\varphi$ with $c$ a non-zero constant, $\varphi\in H$, and and \begin{equation} \lim_{t\to\alpha}S(t,h,h',h'')= \lim_{t\to\beta}S(t,h,h',h'')\,. \end{equation} \end{theorem} \begin{proof} For this proof, we use a standard method for obtaining various types of integral inequalities involving a function and its third derivative. See, for example, \cite{f1,f2} and the references cited there in. Let $h\in AC^2(I)$. From \eqref{e8}--\eqref{e14} and the assumptions, we have $\varphi^{-1}h\in AC^2(I)$ and $S(t,h,h',h'') \in AC(I)$. If we substitute $h=\varphi f$, where $f\in AC^2(I)$, in the expression $r{h'''}^2$, then, after simple calculations, we obtain \begin{align*} r{h'''}^2&=r\left(\varphi'''f+3\varphi''f'+3\varphi'f''+\varphi f'''\right)^2\\ &=rh'''[\varphi'''f^2+3\varphi''(f^2)'+3\varphi'(f^2)'' +\varphi(f^2)''']+r(3\varphi''f'+3\varphi'f''+\varphi f''')^2\\ &\quad -6r\varphi'''(\varphi'{f'}^2+\varphi f'f'')\\ &=r\varphi'''(\varphi f^2)'''-3(r\varphi'''\varphi{f'}^2)'+ 3[(r\varphi''')'\varphi-r\varphi'''\varphi']{f'}^2\\ &\quad + r\big(3\varphi''f'+3\varphi'f''+\varphi f'''\big)^2. \end{align*} Then, using the obvious identity \[ r\varphi'''(\varphi f^2)'''+(r\varphi''')'''\varphi f^2=[r\varphi'''(\varphi f^2)'' -(r \varphi''')'(\varphi f^2)'+( r\varphi''')''\varphi f^2]', \] and \begin{align*} &r\big(3\varphi''f'+3\varphi'f''+\varphi f'''\big)^2\\ &=3[r{\varphi''}^2+(r\varphi'')''\varphi-(r\varphi'')'\varphi']{f'}^2 +3[2r{\varphi'}^2- 2r\varphi''\varphi -(r\varphi')'\varphi]{f''}^2+r\varphi^2{f'''}^2\\ &\quad +3[2r\varphi''\varphi'{f'}^2+r\varphi'\varphi{f''}^2 +2r\varphi''\varphi f'f''-(r\varphi'')'\varphi{f'}^2]', \end{align*} we obtain \begin{align*} r{h'''}^2&=sh^2+3\omega_0{f'}^2+3\omega_1{f''}^2+r\varphi^2{f'''}^2\\ &\quad +\Big\{[ r\varphi'''(\varphi f^2)''- (r\varphi''')'\cdot (\varphi f^2)'+(r\varphi''')''\varphi f^2] \\ &\quad +3[2r\varphi''\varphi'- (r\varphi'')'\varphi-r\varphi'''\varphi]{f'}^2+6r\varphi''\varphi f'f''+3r\varphi'\varphi {f''}^2\Big\}'. \end{align*} Now substituting $f=\varphi^{-1}h$ on the right hand side of the above identity, and using \begin{gather*} \varphi f^2 =\varphi^{-1} h^2, \\ (\varphi f^2)'=(\varphi^{-1})'h^2+2\varphi^{-1}hh',\\ (\varphi f^2)''=(\varphi^{-1})''h^2+4(\varphi^{-1})'h h'+2\varphi^{-1}{h'}^2+2\varphi^{-1}h h'',\\ f'=(\varphi^{-1})'h+\varphi^{-1}h', \\ f''=(\varphi^{-1})''h+2(\varphi^{-1})'h'+\varphi^{-1}h'', \end{gather*} we obtain the identity \begin{equation} \label{e18} r{h'''}^2-sh^2=\left[ S(t,h,h',h'')\right]' +3\omega_0{(\varphi^{-1} h)'}^2+3\omega_1{( \varphi^{-1}h)''}^2+r\varphi^2{(\varphi^{-1}h)'''}^2. \end{equation} Now let $h\in H$. Condition \eqref{e3} implies that the function $r{h'''}^2$ is summable on $I$ since $r{h'''}^2\ge 0$ on $I$. It follows from assumptions that the function $sh^2$ and $[S(t,h,h',h'')]'$ are summable on each compact interval $[a,b]\subset I$. Thus by \eqref{e18} we get the summability of the function \begin{equation} \label{e19} 3\omega_0{(\varphi^{-1}h)'}^2+3\omega_1{(\varphi^{-1} h)''}^2+r\varphi^2{(\varphi^{-1}h)'''}^2 \end{equation} on each compact interval $[a,b]\subset I$ and we obtain the equality \begin{equation} \label{e20} \int_a^br{h'''}^2dt=\int_a^bsh^2dt+S(t,h,h',h'')\Big|_a^b +\int_a^bg(t)dt. \end{equation} for arbitrary $\alpha-\infty. \] Thus, there is a constant $C$ such that \[ -S(t,h,h',h'')\Big|_{a_n}^{b_n}\le C<\infty. \] By condition \eqref{e19}, $g\ge0$ a.e. on $I$. From \eqref{e20}, we infer that \[ \int_{a_n}^{b_n} sh^2dt\le \int_{a_n}^{b_n} r{h'''}^2t +C\le \int_{I_n}r{h'''}^2dt+C, \] and from this, letting $n\to\infty$, we obtain \[ \int_Ish^2dt\le\int_Ir{h'''}^2dt+C<\infty. \] From this estimate and by the second condition of \eqref{e5}, we conclude that $sh^2$ is summable on $I$. Next, in a similar way, using \eqref{e20} and the sum ability of the function $sh^2$ on $I$, we prove that the function $g$ is sum able on $I$. Thus all the integrals in \eqref{e20} have finite limits as $a\to\alpha$ or $b\to\beta$, and hence both of the limits in \eqref{e6} are proper and finite. Therefore the conditions \eqref{e6} and \eqref{e7} may be written in the equivalent form \[ -\infty<\lim_{t\to\alpha}S(t,h,h',h'')\le \lim_{t\to\beta}S(t,h,h',h'')<\infty. \] Now by \eqref{e20} as $a\to\alpha$ and $b\to\beta$, we obtain the equality \begin{equation} \label{e21} \int_I r{h'''}^2dt-\int_I sh^2dt= \lim_{t\to\beta}S(t,h,h',h'') -\lim_{t\to\alpha}S(t,h,h',h'')+\int_Igdt, \end{equation} hence, in view of \eqref{e19}, the inequality \eqref{e4} follows, since $g\ge0$ a.e. on $I$. If \eqref{e4} becomes an equality for a non-vanishing function $h\in H$, then by \eqref{e19} and \eqref{e21}, we have \begin{equation} \label{e22} \int_I gdt=0,\quad \lim_{t\to\alpha}S(t,h,h',h'')= \lim_{t\to\beta}S(t,h,h',h''). \end{equation} Since $g\ge 0$ a.e. on $I$, we obtain $g=0$ a.e. on $I$. In view of $g$ it follows from assumptions that it $g=0$ a.e. on $I$, then $(\varphi^{-1}h)'(t_0)=0$ for some $t_0\in I$, and we get that $(\varphi^{-1}h)'=0$ on $I$, since $(\varphi^{-1}h)'\in AC^2(I)$. This implies that $h=C\varphi$, where $C=const\ne 0$, since $\varphi^{-1}h\in AC^2(I)$. Thus $\varphi\in H$, so that we obtain from the condition \eqref{e22} we get the condition (17). Now let \eqref{e21} be satisfied and let $h=C\varphi$, where $C=const\ne 0$. This implies $g=0$ a.e. on $I$, so that $\int_Igdt=0$. In view of \eqref{e19}-\eqref{e22}, \eqref{e4} becomes equality. The theorem is proved. \end{proof} \section{Example} Let $I=(-1,1)$, $r=(1-t^2)^a$ and $\varphi=(1-t^2)^{3-a}$ on $I$, where $a$ is an arbitrary constant such that the case $a\in(-\infty;1]$ is considered. Then by \eqref{e3}, \eqref{e15} and \eqref{e16}, we have \begin{gather*} s=-(r\varphi''')'''\varphi^{-1}=24(3-a)(2-a)(5-2a)(1-t^2)^{a-3}>0,\\ \omega_0=4-(3-a)(1-t^2)^{2-a}[(15-6a)+(12a-30)t^2+(15-29a)t^4]>0, \\ \omega_1=2(3-a)(1-t^2)^{5-a} >0 \mbox{ on } I. \end{gather*} From Theorem \ref{thm1}. we obtain that the inequality \eqref{e4} holds for every function $h\in H$, where $H$ is the class of function $h\in AC^2((-1,1))$ satisfying the integral condition \begin{equation} \label{e23} \int_{-1}^1(1-t^2)^a{h'''}^2dt<\infty \end{equation} and the limit condition \begin{equation} \label{e24} -\infty<\lim_{t\to -1} S(t,h,h',h'')\le \lim_{t\to 1} S(t,h,h',h'')<\infty, \end{equation} where by \eqref{e7}-\eqref{e14} \begin{equation} \label{e25} \begin{aligned} S(t,h,h',h'') &=\nu_0(t)(1-t^2)^{a-5}h^2+\nu_1(t)(1-t^2)^{a-3}{h'}^2\\ &\quad+ \nu_2(t)(1-t^2)^{a-1}{h''}^2+2\varepsilon_{01}(t)(1-t^2)^{a-4}hh'\\ &\quad +2\varepsilon_{02}(t)(1-t^2)^{a-3} hh''+2\varepsilon_{12}(t)(1-t^2)^{a-2}hh'', \end{aligned} \end{equation} \begin{align*} \nu_0(t)=&8(3-a)t[-3(a^2-3a+1)+(12a^3-90a^2+238a-222)]t^2\\ &\quad +(-4a^4+60a^3-319a^2+811a-528)t^4], \end{align*} \begin{gather*} \nu_1(t)=-8(3-a)t[6-a+2(7-2a)t^2], \\ \nu_2(t)=-2(3-a)t,\\ \varepsilon_{01}(t)=4(3-a)[2a-3+(-10a^2+52a-66)t^2+(28a^3-238a^2+728a-803)t^4], \\ \varepsilon_{02}(t)=-4(3-a)[a+(2a^2-11a+16)t^2], \\ \varepsilon_{12}(t)=-4(3-a)[1+(7-2a)t^2]. \end{gather*} Since the second condition of \eqref{e5} is satisfied trivially. Now we show that a function $h\in AC^2((-1,1))$ that satisfies the integral condition \eqref{e23} and limit conditions $h(\pm 1)=h'(\pm 1)=h''(\pm 1)=0$ belongs to the class $H$. At first we show that, if $h(1)= h'(1)= h"(1)=0$ and \eqref{e23} hold, then \[ \lim_{t\to 1} S(t,h,h',h'')=0\,. \] Let us consider the right-hand neighborhood $U$ of the point $1$. In \cite{f1}, it has been shown that \begin{equation} \label{e26} |h'(t)|\le k(t) (1-t)^{\frac{1-a}{2}} \end{equation} for $t\in U$, where \[ k(t)=\big\{\frac{A}{1-a}\int_{t}^1(1-\tau^2)^a{h''}^2(\tau)d\tau\big\} ^{1/2} >0, \quad t\in U\,. \] This function is a continuous function on $I$, $\lim_{t\to 1} k(t)\equiv k(1)=0$, and \begin{equation} \label{e27} |h(t)|\le\frac{k(\theta)}{\sqrt{2-a}}(1-t)^{\frac{3-a}{2}}, \end{equation} for $t\in U$, where $t<\theta<1$ and $\lim_{t\to1}k(t)\equiv k(1)=0$. It is easy to see that if we write $h'''$ instead of $h''$, $h''$ instead of $h'$, and $h'$ instead of $h$ in \eqref{e26} and \eqref{e27} then we obtain \begin{equation} \label{e28} |h''(t)|\le k(t)(1-t)^{\frac{1-a}{2}} \end{equation} for $t\in U$, with $k$ as above and \begin{equation} \label{e29} |h'(t)|\le\frac{k(\theta)}{\sqrt{2-a}}(1-t)^{\frac{3-a}{2}}, \end{equation} for $t\in U$, where $t<\theta<1$ and $\lim_{t\to1}k(t)\equiv k(1)=0$. From \eqref{e27} we have \begin{equation} \label{e30} |h(t)|\le\frac{2k(\theta)}{(5-a)\sqrt{2-a}}(1-t)^{\frac{5-a}{2}}\,. \end{equation} Based on the estimates \eqref{e28}, \eqref{e29} and \eqref{e30}, from \eqref{e25}, we obtain \begin{align*} |S(t,h,h',h'')| &\le\frac{4k^2(\theta)}{(2-a)(5-a)^2}|\nu_0(t)| +\frac{k^2(\theta)}{2-a}|\nu_1(t)|\\ &\quad +k^2(\theta)|\nu_2(t)|+\frac{2k^2(\theta)}{(2-a)(5-a)} |\varepsilon_{01}(t)|\\ &\quad +\frac{2k^2(\theta)}{(5-a)\sqrt{2-a}}|\varepsilon_{02}(t)| +\frac{k^2(\theta)}{\sqrt{2-a}}|\varepsilon_{12}(t)|=m(t) \end{align*} Whence it follows that $\lim_{t\to1}S(t,h,h',h'')=0$. In an analogous way we show that if $h(-1)=h'(-1)=h''(-1)=0$ and \eqref{e23} hold then $\lim_{t\to-1}S(t,h,h',h'')=0$. Therefore we get the following result. \begin{theorem} \label{thm2} If $a<1$ and the function $h\in AC^2((-1,1))$ satisfies the integral condition \[ \int_{-1}^1(1-t^2)^a{h'''}^2dt<\infty \] and the limit condition $h(\pm 1)=h'(\pm 1)=h''(\pm 1)=0$, then \[ \int_{-1}^1(1-t^2)^a{h'''}^2dt\ge 24(3-a)(2-a)(5-2a)\int_{-1}^1\frac{h^2dt}{(1-t^2)^{3-a}}\,. \] holds. The inequality \eqref{e26} becomes on equality if and only if $h=C(1-t^2)^{3-a}$, where $C$ is a constant. \end{theorem} In the particular case for $a=0$ we obtain \[ \int_{-1}^1{h'''}^2dt\ge 720 \int_{-1}^1\frac{h^2dt}{(1-t^2)^3} \] as deduced in \cite{k1}. \subsection*{Acknowledgments} The author wishes to express his gratitude to the anonymous referee for his/her valuable remarks and comments on the original version of this paper. \begin{thebibliography}{0} \bibitem{f1} B. Florkiewicz and K. Wojteczek; \emph{On some further Wirtinger-Beesack integral inequalities}, Demonstratio. Math. 32(1999), 495-502. \bibitem{f2} B. Florkiewicz and K. Wojteczek; \emph{Some second order integral inequalities of generalized Hardy-type}, Proc. Royal Soc. Edinburgh. 129 A. Part 5(1999), 947-958. \bibitem{k1} W. J. Kim; \emph{Disconjugacy and Nonoscillation Criteria for Linear Differential Equations}. J. of Diff. Equation 8 (1970), 163-172. \end{thebibliography} \end{document}