\input amstex \documentstyle{amsppt} \loadmsbm \magnification=\magstephalf \hcorrection{1cm} \vcorrection{-6mm} \nologo \TagsOnRight \NoBlackBoxes \headline={\ifnum\pageno=1 \hfill\else% {\tenrm\ifodd\pageno\rightheadline \else \leftheadline\fi}\fi} \def\rightheadline{\hfil On oscillatory solutions \hfil\folio} \def\leftheadline{\folio\hfil Miroslav Bartu\v sek \hfil} \def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt % Fourth Mississippi State Conference on Differential Equations and \hfill\break Computational Simulations, Electronic Journal of Differential Equations, \hfill\break Conference 03, 1999, pp 1--11.\hfill\break URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \hfill\break ftp ejde.math.swt.edu (login: ftp) \bigskip} } \topmatter \title On oscillatory solutions of third order differential equation with quasiderivatives \endtitle \thanks {\it Mathematics Subject Classifications:} 34C10.\hfil\break\indent {\it Key words:} Oscillatory solutions, third order differential equations. \hfil\break\indent \copyright 2000 Southwest Texas State University and University of North Texas. \hfil\break\indent Published July 10, 2000. \endthanks \author Miroslav Bartu\v sek \endauthor \address Miroslav Bartu\v sek \hfill\break\indent Department of Mathematics, Masaryk University \hfill\break\indent Jan\'a\v ckovo n\'am. 2a, 662 95 Brno, Czech Republic \endaddress \email bartusek\@math.muni.cz \endemail \abstract This paper gives sufficient conditions under which all oscillatory solutions of a third order nonlinear differential equation with quasiderivatives vanish at infinity. Applications to third order differentials equation with a middle term are also given. \endabstract \endtopmatter \def\e{\operatorname{e}} \def\sgn{\operatorname{sgn}} \head I. Introduction \endhead Consider the differential equation $$y^{[3]}= \left(\frac{1}{a_2} \left( \frac{1}{a_1} y'\right)'\right)' = r(t)\, f(y) \tag1$$ where $J =[0, T)$, $T\leqslant \infty$, $r\in C^\circ (J)$, $f\in C^\circ (R)$, $R=(-\infty, \infty)$, $a_i \in C^1 (J)$, $i=1,2$, \ $a_i$ are positive on $J$, $$r(t)> 0 \text{\ on\ } J , \quad \quad f(x) x>0\quad \text{for\ }\ x\ne 0\,, \tag{H1}$$ and $y^{[i]}$, $i=0,1,2,3$, is the $i$-th quasiderivative of $y$ defined by $$y^{[0]}= y\,, \quad y^{[i]}=\frac{1}{a_i(t)} \left( y^{[i-1]}\right)'\,, \quad i=1,2, \quad y^{[3]}= \left(y^{[2]}\right)'\,. \tag2$$ Let a function $y: I \to R$ have the continuous quasiderivatives up to the order $3$ on $I$ and let (1) hold on $I$. Then $y$ is called a solution of (1). A solution $y$ is called oscillatory if it is defined on $J$, $\sup\limits_{\tau\leqslant t 0$ for an arbitrary $\tau \in J$ and if there exists a sequence of its zeros tending to $T$. Denote by $\Cal O$ the set of all oscillatory solutions of (1). Due to the methods used, we will study two cases: $$\left(\frac{a_2(t)}{a_1(t)}\right)'\leqslant 0\,, \quad t\in J\,, \tag{H2}$$ and $$\left(\frac{a_2(t)}{a_1(t)}\right)'\geqslant 0\,, \quad t\in J \,. \tag{H3}$$ A great effort has been exerted to the study of the asymptotic behaviour of oscillatory solutions of (1) and its special cases, see e.g. \cite{1--6, 8, 10, 12}. If $a_2 \equiv a_1 \equiv 1$ and $T=\infty$, sufficient conditions are given in \cite{1,10} for every oscillatory solution $y$ of (1) to vanish at infinity, i.e. $$\lim_{t\to \infty} y(t) =0\,. \tag3$$ \proclaim{Theorem A {\rm(\cite{10})}} Let $T=\infty$, {\rm(H1)} hold, $a_1 \equiv a_2 \equiv 1$, and $00$ ($<0$) on $I$, then $y^{[i-1]}$ is increasing (decreasing) on $I$. \endproclaim \remark{Remark 1} Note that $<$ and increasing ($>$ and decreasing) can be replaced by $\leqslant$ and nondecreasing ($\geqslant$ and non-increasing). \endremark The following lemma describes the structure of oscillatory solutions of (1). \proclaim{Lemma 2 {\rm(\cite{2})}} Let $y\in {\Cal O}$. Then sequences $\{t_k^i\}$, $i=0,1,2$, $k=1,2,\dots$ exist such that $\lim\limits_{k\to \infty} t_k^0 =T$, $$t_k^0 < t_k^1 < t_k^2 < t_{k+1}^0\,, \quad y^{[i]}(t_k^i)=0\,, \quad i=0,1,2\,,$$ \alignedat3 (-1)^{j+1} y^{[j]}(t)\, y(t)&>0 \quad &\text{on}\quad &(t_k^0, t_k^j)\,,\\ &<0 \quad &\text{on}\quad &(t_k^j, t_{k+1}^0)\,, \quad j=1,2;\ k=1,2,\dots \endalignedat \endproclaim \remark{Remark 2} Note that according to Lemmas 1 and 2, the sequences $\{|y(t_k^1)|\}_1^\infty$, $\{|y^{[1]}(t_k^2)|\}_1^\infty$ and $\{|y^{[2]}(t_k^0)|\}_1^\infty$ are the sequences of the absolute values of all local extrema of $y$, $y^{[1]}$ and $y^{[3]}$ on $[t_0^0, T)$, respectively. Sometimes it is useful to express (1) in an equivalent form. \endremark \proclaim{Lemma 3} Let $a_0 \in C^\circ (J)$ be positive. Then the transformation $$x(t)= \int_0^t a_0 (s)\, ds\,, \ Y(x)=y(t)\,, \ t\in J, \ x\in [0, x^\ast)\,, \ x^\ast = x(T)$$ transforms (1) into $$\left(\frac{1}{A_2}\left(\frac{1}{A_1}\overset{\bullet}\to{Y} \right)^\bullet\right)^\bullet= R(x)\, f(Y) \tag5$$ where $A_i (x)=\frac{a_i(t(x))}{a_0(t(x))}\,, \ i=1,2\,, \quad R(x)=\frac{r(t(x))}{a_0(t(x))}\,, \frac{d}{dx} = {}^\bullet$ and $t(x)$ is the inverse function to $x(t)$. At the same time, $$Y^{\{i\}} (x)= y^{[i]} (t)\,, \quad i=0,1,2,3\,, \tag6$$ where $$Y^{\{0\}} = Y\,, \quad Y^{\{j\}}=\frac{1}{A_{j}(x)} \left( Y^{\{j-1\}}\right)^\bullet\,, \ j=1,2\,, \quad Y^{\{3\}}= \left(Y^{\{2\}}\right)^\bullet\,.$$ \endproclaim \demo{Proof} Use a direct computation or see \cite{4}.\hfill$\square$ \enddemo \head 2. Case (H2) \endhead Some results will be used that are obtained for (1) under a different assumptions than (H1). Consider $$\left(\frac{1}{b(\sigma)} Z^{\prime\prime}\right)' + \bar r (\sigma)\, f(Z)=0 \tag7$$ where $I\subset {\Bbb R}_+$, $b\in C^1 (I)$, $\bar r\in C^1(I)$, $f\in C^\circ (I)$, $f(x) x>0$ for $x\ne 0$, $$b(\sigma)>0\,, \quad \bar r(\sigma)\geqslant 0 \text{\ on\ } I\,, \quad f'(x)\geqslant 0 \ \text{\ on\ } R\,.$$ The quasiderivatives are given by $$Z ^{[0]}=Z\,, \ Z^{[1]}=Z'\,, \ Z^{[2]}=\frac{Z^{\prime\prime}}{b(\sigma)}\,.$$ Note that the sign of $\bar r$ is opposite to the one of $r$. \proclaim{Lemma 4} Let $b'\geqslant 0$ and $\bar r'\geqslant 0$ on $I$. Let $Z$ be a solution of {\rm(7)} the second quasiderivatives $Z^{[2]}$ of which has three consecutive zeros $\sigma_0$, $\sigma_1$ and $\sigma_2 \in I$, $\sigma_0<\sigma_1<\sigma_2$. Then $$\sqrt{2}|Z' (\sigma_2)|< |Z'(\sigma_1)|\,.$$ \endproclaim \demo{Proof} The assertion is proved for $T=\infty$ and for an oscillatory solution in \cite{4} (see Lemma 2.4 and Definition 2.1). But it follows from the proof that only information on $[\sigma_1, \sigma_2]$ and the existence of the zero $\sigma_0$ were used. Thus, the statement is valid under our assumptions as well.\hfill$\square$ \enddemo The following theorem investigates the asymptotic behaviour of the first and the second quasiderivatives of an oscillatory solution of (1). \proclaim{Theorem 1} Let {\rm (H1)} and {\rm(H2)} hold. Let $y\in {\Cal O}$ and $\{t_k^i\}$, $i=0,1,2$, $k=1,2,\dots$, be given by Lemma 2. \vskip 1mm \noindent {\rm(i)} Then the sequence $\{|y^{[1]} (t_k^2)|\}_1^\infty$ of the absolute values of all local extrema of $y^{[1]}$ on $[t_1^0, T)$ is decreasing. \vskip 1mm \noindent {\rm(ii)} Let $r\in C^1 (J)$, $f\in C^1(R)$, $f'\geqslant 0$ on $R$ and $\left(\frac{r(t)}{a_1(t)}\right)'\leqslant 0$ on $J$. \newline Then \ $\lim\limits_{t\to T} y^{[1]}(t)=0$ \ and $$\left| y^{[1]}(t_k^2)\right|\leqslant 2^{\frac{1-k}{2}}\left| y^{[1]}(t_1^2)\right|\,, \quad k=1,2,\dots \tag8$$ {\rm(iii)} Let $r\in C^1(J)$, $f\in C^1(R)$, $f'\geqslant 0$ on $R$ and $\left(\frac{r(t)}{a_2(t)}\right)'\leqslant 0$ on $J$. \newline Then the sequence $\left\{|y^{[2]}(t_k^0)|\right\}_1^\infty$ of the absolute values of all local extrema of $y^{[2]}$ on $[t_1^0, T)$ is decreasing. \endproclaim \demo{Proof} Note that according to Remark 2, the sequences $\left\{|y^{[1]}(t_k^2)|\right\}_1^\infty$ and \hfil\break $\left\{|y^{[2]}(t_k^0)|\right\}_1^\infty$ are the sequences of the absolute values of all local extrema of $y^{[1]}$ and $y^{[2]}$, respectively. \vskip 2mm (i) Let $k\in \{ 2,3, \dots\}$ and suppose, without loss of generality, that $$y(t)>0 \quad \text{on}\quad (t_k^0, t_{k+1}^0)\,.$$ Thus, according to Lemmas 1 and 2 there exists $t_k^\ast$ such that $$\gathered t_k^\ast \in (t_k^0, t_k^1)\,, \ y(t_k^\ast)= y(t_k^2)\,,\\ y\ \text{is increasing (decreasing) on} \ [t_k^\ast, t_k^1] \text{\ (on [t_k^1, t_k^2])}\,,\\ y^{[1]}(t)>0 \ \text{(<0) on [t_k^\ast, t_k^1) (on t_k^1, t_k^2)}\,,\\ y^{[1]}(t_k^1)=0\,, \quad y^{[1]} (t_{k-1}^2)>y^{[1]}(t_k^\ast)>0\,,\\ y^{[2]}(t)<0 \ \text{and } \ |y^{[2]}| \text{\ is decreasing on \ } [t_k^\ast, t_k^2)\,, y^{[2]}(t_k^2)=0\,. \endgathered \tag9$$ Let $\varphi$ and $\psi$ be the inverse functions to $y$: $$\gathered t_k^\ast \leqslant \varphi (v)\leqslant t_k^1\,, \quad y(\varphi(v))=v\,, \\ t_k^1 \leqslant \psi (v)\leqslant t_k^2\,, \quad y(\psi(v))=v\,, \\ v\in I=[y(t_k^\ast), y(t_k^1)]\,. \endgathered$$ We prove by an indirect proof that $$y^{[1]}(\varphi(v))\geqslant \left| y^{[1]}(\psi(v))\right|\,, \quad v\in I\,. \tag10$$ \noindent Observe that (9) yields $y^{[1]}(\varphi(v))>0$ and $y^{[1]}(\psi(v))<0$ for $v\in I_1=\left[ y(t_k^\ast), y(t_k^1)\right)$. Define $$S(v)= y^{[1]}(\varphi (v))-|y^{[1]}(\psi(v))|\,, \quad v\in I\,.$$ Suppose, contrarily, that there exists $\bar v\in I_1$ such that $$S(\bar v)<0\,. \tag11$$ Then using (2), (9) and (H2), we have \align \frac{d}{dv} S(v)&= \frac{y^{[2]} (\varphi(v))a_2 (\varphi(v))}{y'(\varphi(v))} + \frac{y^{[2]}(\psi(v))a_2(\psi(v))}{y'(\psi(v))}\\ &=\frac{y^{[2]}(\varphi(v))}{y^{[1]}(\varphi(v))} \ \frac{a_2(\varphi(v))}{a_1(\varphi(v))}+\frac{y^{[2]}(\psi(v))}{y^{[1]}(\psi(v))} \ \frac{a_2(\psi(v))}{a_1(\psi(v))}\\ &\leqslant y^{[2]} (\psi(v)) \frac{a_2(\psi(v))}{a_1(\psi(v))} \left[ \frac{1}{y^{[1]}(\varphi(v))}+\frac{1}{y^{[1]}(\psi(v))}\right]\,, \quad v\in I_1\,. \endalign Thus $$v\in I_1\,, \ S(v)<0 \ {\Bbb R}_+ightarrow \frac{d}{dv} S(v)<0\,.$$ >From this and from (11), it is clear that $$S(v)<0 \ \text{on\ } [\bar v, y(t_k^1)]\,,$$ and this contradicts $S(y(t_k^1))=0$. Thus, (10) holds and using $v=y(t_k^\ast)$ in (10) and (9), $y^{[1]}(t_{k-1}^2)> |y^{[1]}(t_k^2)|$. \vskip 2mm (ii) Let $t_0 < t_1From this the inequality (8) holds and$\lim\limits_{t\to T_-} y^{[1]}(t)=0$. \vskip 2mm (iii) We prove the third statement for (5) with$a_0\equiv a_2$$$\gather \left(\left( \frac{1}{A_1}\overset{\bullet}\to{Y}\right)^\bullet\right)^\bullet = R(x)\, f(Y)\,,\\ A_1(x)=\frac{a_1(t(x))}{a_2(t(x))} \,, \ R(x)=\frac{r(t(x))}{a_2(t(x))}\,,\ Y^{\{1\}}=\frac{1}{A_1(x)} Y^\bullet\,, \ Y^{\{2\}} = (Y^{\{1\}})^\bullet\,; \endgather$$ then according to (6), it will hold for (1) too. \vskip 2mm Applying Lemma 2 to (5), sequences$\{x_k^i\}$,$k=1,2,\dots$,$i=0,1,2exist such that \alignedat2 &x_k^0 < x_k^1 < x_k^2 < x_{k+1}^0\,, \ k=1,2,\dots , \ &&\lim_{k\to \infty} x_k^0 = x(T)\,,\\ &Y^{\{i\}}(x_k^i)=0\,, \ (-1)^{j+1} Y^{\{j\}}(x) Y(x)&&>0 \ \text{on\ } (x_k^0, x_k^j)\,,\\ &&&<0 \ \text{on\ } (x_k^j, x_{k+1}^0)\,,\\ & k=1,2,\dots\,; \quad j=1,2\,. && \endalignedat \tag14 Letk\in \{1,2,\dots\}$. Put$\tau_0 = x_k^1$,$\tau_1 = x_k^2$,$\tau_2= x_{k+1}^0$,$\Delta_1 = [\tau_0, \tau_1]$,$\Delta_2=[\tau_1, \tau_2]$,$\delta_1=\tau_1-\tau_0$,$\delta_2=\tau_2-\tau_1$and suppose, for simplicity, that$Y^{\{1\}}(x)\leqslant 0$on$\Delta_1. Then (14) and Lemma 1 yield \aligned &Y(x)>0\,, \ Y^{\{1\}}(x)<0\,, \ Y^{\{2\}}(x)<0\,, \ Y\ \text{and\ } |Y^{\{2\}}| \ \text{are decreasing}\\ & \text{and}\ |Y^{\{1\}}|\ \text{is increasing on\ } (\tau_0, \tau_1)\,;\\ & Y(x)>0\,, \ Y^{\{1\}}(x)<0\,, \ Y^{\{2\}}(x)>0\,, \ Y \ \text{and\ } |Y^{\{1\}}| \ \text{are decreasing}\\ &\text{and\ } Y^{\{2\}} \ \text{is increasing on \ } (\tau_1, \tau_2)\,. \endaligned \tag15 The statement will be valid if we prove that $$|Y^{\{2\}}(x_k^0)| > |Y^{\{2\}}(\tau_0)| > Y^{\{2\}}(\tau_2)\,.$$ As the first inequality follows from (14) and Lemma 1, the second one only must be proved. Thus, suppose that $$|Y^{\{2\}}(\tau_0)| \leqslant Y^{\{2\}}(\tau_2)\,. \tag16$$ According to (15) and the assumptions of the theorem, the functionY^{\{2\}}$is concave on$[\tau_0, \tau_2]: \align \left(Y^{\{2\}}(x)\right)^{\bullet\bullet} &=\left( Y^{\{3\}}(x)\right)^{\bullet}=\left[ \frac{r(t(x))}{a_2(t(x))}\, f(Y(x))\right]^\bullet =\\ &=\left(\frac{r(t(x))}{a_2(t(x))}\right)^\bullet\, f(Y(x))+\frac{r(t(x))}{a_2(t(x))}\, f' (Y(x)) Y^{\{1\}}(x) \frac{a_1(t(x))}{a_2(t(x))}\leqslant 0\,, \tag17\\ &\quad \quad\quad \quad x\in \Delta_1\cup \Delta_2\,. \endalign Thus,Y^{\{2\}}$is above the secant line on$\Delta_1 \cup \Delta_2, and using (14) and (15), we have \align |Y^{\{1\}}(\tau_1)| =&\int_{\Delta_1}|(Y^{\{ 1\}}(x))^\bullet |\, dx=\int_{\Delta_1}|Y^{\{ 2\}}(x)|\, dx \leqslant |Y^{\{ 2\}}(\tau_0)|\, \frac{\delta_1}{2}\,,\\ |Y^{\{ 1\}}(\tau_1)| &\geqslant Y^{\{ 1\}} (\tau_2)- Y^{\{ 1\}}(\tau_1)= \int_{\Delta_2}Y^{\{ 2\}}(x)\, dx \geqslant Y^{\{ 2\}}(\tau_2)\, \frac{\delta_2}{2}\,. \endalign >From this and (16), $$\delta_1\geqslant \delta_2\,. \tag18$$ Furthermore, according to (1), (15) and (17), \Y^{\{ 3\}}\geqslant 0$is decreasing on$\Delta_1\cup \Delta_2. From this it follows that \align |Y^{\{ 2\}}(\tau_0)| &= \int_{\Delta_1} Y^{\{ 3\}}(x)\, dx > Y^{\{ 3\}}(\tau_1)\, \delta_1\,,\\ Y^{\{ 2\}}(\tau_2)&= \int_{\Delta_2} Y^{\{ 3\}}(x)\, dx < Y^{\{ 3\}} (\tau_1)\, \delta_2\,. \endalign Thus, with respect to (16), \\delta_1 < \delta_2$and this contradicts (18). \hfill$\square$\enddemo \vskip 2mm The following theorem states a sufficient condition under which oscillatory solutions tend to zero as$t\to T$. \proclaim{Theorem 2} Let {\rm (H1)} and {\rm (H2)} hold,$r\in C^1(J)$,$f\in C^1(R)$,$f' \geqslant 0$on$R$, $$\left(\frac{r(t)}{a_1(t)}\right)' \leqslant 0\,, \tag19$$ and let one of the following assumptions hold: \vskip 2mm {\rm(i)}$\left(\frac{r(t)}{a_2(t)}\right)'\leqslant 0$,\quad$00$\quad for$t\in J$; \vskip 1mm {\rm(iii)}$\int\limits_{0}^T a_1 (s)\, ds <\infty$. \vskip 2mm \noindent If$y\in {\Cal O}$, then$\lim\limits_{t\to \infty} y^{(j)}(t)=0$\ for$j=0,1$. \endproclaim \demo{Proof} Let$y\in {\Cal O}$. According to Lemma 3 with$a_0 \equiv a_1$, it is sufficient to prove the results for(5) only: $$\gather \left( \frac{1}{A_2(x)} Y^{\bullet\bullet}\right)^\bullet = R (x)\, f(Y)\,, \ \frac{d}{dx}= {}^\bullet\,, \tag20\\ A_1\equiv 1\,, \ A_2(x)=\frac{a_2(t(x))}{a_1(t(x))}\,, \ R(x)=\frac{r(t(x))}{a_1(t(x))}\,,\ x\in I=[0, x^\ast)\,, \ x^\ast = x(T)\,,\\ Y^{\{ 1\}} = Y^\bullet\,, \ Y^{\{ 2\}}= \frac{1}{A_2(x)} Y^{\bullet\bullet}\,. \tag21 \endgather$$ Denote by$\{x_k^i\}$,$i=0,1,2$,$k=1,2,\dots$, the sequences given by Lemma 2 for (20) (i.e.$x_k^i=t_k^i) and put $$\Delta_k = [x_k^0, x_k^1]\,.$$ Then, according to Lemmas 1 and 2, \align &Y^{\{ 1\}} (x)\,Y(x)\geqslant 0\,, \ Y^{\{ 2\}} (x)\, Y(x)\leqslant 0\ \text{for\ } x\in \Delta_k\,,\tag22\\ &|Y^{\{ 1\}}| \ \text{and\ } |Y^{\{ 2\}}| \ \text{are decreasing on \ } \Delta_k\,. \endalign Furthermore, using (19), $$\left(\frac{R(x)}{A_1(x)}\right)^\bullet = R^\bullet (x)=\left(\frac{r(t)}{a_1(t)}\right)' t^\bullet (x)\leqslant 0 \ \ \text{on\ } I\,,$$ the assumptions of Th. 1 (ii), applied to (20), are fulfilled. Thus,\lim\limits_{x\to x^\ast} Y^{\{ 1\}}(x)=0$and $$|Y^{\{ 1\}}(x_k^0)| \leqslant |Y^{\{ 1\}}(x_{k-1}^2)| \leqslant 2^{\frac{2-k}{2}} |Y^{\{ 1\}}(x_1^2)|\,, \quad k\geqslant 2\,; \tag23$$ note that the first inequality follows from Lemmas 1 and 2. We prove indirectly that $$\lim_{t\to T} Y(t)=0\,. \tag24$$ Thus suppose, without loss of generality, that $$|Y(x_k^1)| \geqslant M_1 >0\,, \quad k=1,2,\dots$$ Then, according to Lemmas 1 and 2, there exists a sequence$\bar x_k \in (x_k^0, x_k^1)$such that $$|Y(\bar x_k)| =\frac{M_1}{2}\,, \frac{M_1}{2}\leqslant |Y(x)| \leqslant M_1\ \text{on\ } \bar \Delta_k= [\bar x_k, x_k^1]\,. \tag25$$ Let$\delta_k = x_k^1 - \bar x_k. Using (22) and (23), we have \align \frac{M_1}{2}&\leqslant|Y(x_k^1)-Y(\bar x_k)| = \int_{\bar\Delta_k} |Y^{\{ 1\}}(x)|\, dx \\ & \leqslant | Y^{\{ 1\}}(x_k^0)|\, \delta_k\leqslant \leqslant 2^{\frac{2-k}{2}}\delta_k |Y^{\{ 1\}}(x_1^2)| \endalign and thus $$\lim_{k\to \infty} \delta_k =\infty\,.\tag26$$ (i) According to (19) and (22), \align |Y^{\{ 2\}}(x_k^0)|&\geqslant \left[ Y^{\{ 2\}}(x_k^1) - Y^{\{ 2\}}(\bar x_k)\right] \sgn \,Y(x_k^1)=\int_{\bar \Delta_k}Y^{\{ 3\}}(x)\sgn Y(x)\, dx\\ &=\int_{\bar\Delta_k} R(x)\, f(Y(x))\sgn \, Y(x)\, dx \geqslant M \delta_k \min_{\frac{M_1}{2}\leqslant s \leqslant M_1}|f(s)|>0 \endalign and thus (26) yields\lim_{k\to \infty} Y^{\{ 2\}} (x_k^0)=\infty$which contradicts Theorem 1 (iii). \vskip 1mm (ii) Using (22), (H2) and the assumptions, we have for$x\in \bar \Delta_k$: $$\gather A_2 (x)|Y^{\{ 2\}}(x)|\geqslant A_2 (x)\left[ Y^{\{ 2\}}(x_k^1)- Y^{\{ 2\}}(x)\right] \sgn \, Y (x_k^1) =\\ =A_2 (x)\int_x^{x_k^1}|Y^{\{ 3\}}(s)|\, ds \geqslant \int_x^{x_k^1} R(s)\, A_2(s)|f(Y(s))|\, ds \geqslant\\ \geqslant M M_2(x_k^1 -x)\,, \quad \quad M_2=\min_{\frac{M_1}{2}\leqslant s \leqslant M_1} |f(s)|>0\,. \endgather$$ >From this and from (21), $$Y^{\{ 1\}} (\bar x_k)= \int_{\bar\Delta_k} A_2 (x) |Y^{\{ 2\}}(x)|\, dx \geqslant MM_2 \int_{\bar\Delta_k}(x_k^1 -x)\, dx = \frac{MM_2}{2}\, \delta_k^2\,.$$ Since$\lim\limits_{x\to x^\ast}Y^{\{ 1\}}(x)=0$, \$Y^{\{ 1\}}(\bar x_k)$is bounded, say $$|Y^{\{ 1\}}(\bar x_k)| \leqslant M_3\,, \quad k=1,2,\dots\,,$$ and we can conclude that$\delta_k$is bounded as well. This contradiction to (26) proves the statement. \vskip 1mm (iii) In this case,$x^\ast<\infty$and$I$is bounded which contradicts (26).\hfill$\square$\enddemo \remark{Remark 3} (i) Note that$\left(\frac{r(t)}{a_1(t)}\right)'\leqslant 0$follows from (H2) and the fact that \linebreak$\left(\frac{r(t)}{a_2(t)}\right)'\leqslant 0$\,: $$\left( \frac{r}{a_1}\right)'=\left(\frac{r}{a_2}\, \frac{a_2}{a_1} \right)' = \left(\frac{r} {a_2} \right)'\, \frac{a_2} {a_1}+\frac{r}{a_2}\left(\frac{a_2}{a_1}\right)'\leqslant 0\,.$$ (ii) The differential equation in Ex. 1 fulfills all assumptions of Th. 2 (i) with the exception of$00 \text{\ on\ } {\Bbb R}_+\,, \ f(x)x>0\ \text{for\ } x\ne 0\,. \tag{H4} \endgather $$A solution y of (27) is called oscillatory if it is defined on {\Bbb R}_+, \sup\limits_{\tau\leqslant t<\infty} |y(t)|>0 for every \tau\in {\Bbb R}_+ and there exists a sequence of zeros of y tending to \infty. \vskip 2mm Let h be a positive solution on [\tau, \infty), \tau\in {\Bbb R}_+, of the equation$$ h^{\prime\prime} + q(t)h=0\,. \tag28 $$Then (27) is equivalent to (1) (see \cite{5} or make a direct computation) on J=[\tau, \infty), where T=\infty,$$ \gathered a_1(t)=h(t)\,, \quad a_2(t)=\frac{1}{h^2(t)}\,, \quad r(t)=s(t)\,h(t)\,, \\ y^{[1]}=\frac{y'}{h}\,, \quad y^{[2]} = h^2 (y^{[1]})'\,. \endgathered \tag29 $$Thus (H1) is satisfied, (H2) holds if h is increasing, and (H3) holds if h is decreasing. \medskip \proclaim{Theorem 4} Let {\rm (H4)} hold,$$ q(t)\leqslant 0\,, s(t)\geqslant M>0\ \quad \text{for \quad } t\in [M_1, \infty) $$and \int\limits_0^\infty t|q(t)|\, dt <\infty where M and M_1 are positive constants. Then every oscillatory solution of {\rm (27)} tends to zero as t\to \infty. \endproclaim \demo{Proof} If follows from \cite{11} and from \int\limits_0^\infty t|q(t)|\, dt <\infty that (28) is non-oscillatory and there exists a positive solution h of (27) that is decreasing for large t and \lim\limits_{t\to \infty} h(t)=h_0 \in (0, \infty). Thus, the conclusion follows from Theorem 3.\hfill\square \enddemo \proclaim{Theorem 5} Let {\rm(H4)} hold, s\in C^1({\Bbb R}_+), f\in C^1(R), f'\geqslant 0 on R,$$ q(t)\geqslant 0\,, \quad 0