\documentclass[twoside]{article} \usepackage{amsfonts, amsmath} \pagestyle{myheadings} \markboth{\hfil A note on the singular Sturm-Liouville problem \hfil EJDE--2002/80} {EJDE--2002/80\hfil Nickolai Kosmatov \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 80, pp. 1--10. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % A note on the singular Sturm-Liouville problem with infinitely many solutions % \thanks{ {\em Mathematics Subject Classifications:} 34B16, 34B18. \hfil\break\indent {\em Key words:} Sturm-Liouville problem, Green's function, fixed point theorem, \hfil\break\indent H\"{o}lder's inequality, multiple solutions. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Submitted November 13, 2001. Published September 27, 2002.} } \date{} % \author{Nickolai Kosmatov} \maketitle \begin{abstract} We consider the Sturm-Liouville nonlinear boundary-value problem $$\displaylines{ -u''(t) = a(t)f(u(t)), \quad 0 < t < 1, \cr \alpha u(0) - \beta u'(0) =0, \quad \gamma u(1) + \delta u'(1) = 0, }$$ where $\alpha$, $\beta$, $\gamma$, $\delta \geq 0$, $\alpha \gamma + \alpha \delta + \beta \gamma > 0$ and $a(t)$ is in a class of singular functions. Using a fixed point theorem we show that under certain growth conditions imposed on $f(u)$ the problem admits infinitely many solutions. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \section{Introduction} In this paper we are interested in the Sturm-Liouville nonlinear boundary-value problem \begin{gather} -u''(t) = a(t)f(u(t)), \quad 0 < t < 1, \label{1}\\ \alpha u(0) - \beta u'(0) = 0, \quad \gamma u(1) + \delta u'(1) = 0. \label{2} \end{gather} The paper is organized in the following manner. In the introduction we briefly discuss the background of the problem, make standing assumptions on the right side of \eqref{1} and state the theorems that will be used to obtain our main results presented in Section 3. The approach is based on the properties of the Green's function of the homogeneous \eqref{1}-\eqref{2}. They will be presented in Section 2. Fixed point theorems have been applied to various boundary value problems to establish the existence of multiple positive solutions. Just recently there have been obtained several results concerning the existence of countably infinitely many positive solutions, e.g., Ehme \cite{e1}, Eloe, Henderson and Kosmatov \cite{e2}, Kaufmann and Kosmatov \cite{k1} and Kosmatov \cite{k2}. They cover the cases of $(k, \ n-k)$ and second order conjugate type boundary value problems. In addition, [5, 6] deal with a singular BVP. Singular boundary value problems have been considered by many authors, e.g., Agarwal, O'Regan and Wong \cite{a1} and Baxley and Thompson \cite{b1}. This paper complements the results of \cite{k2} in which we considered the conjugate type boundary value problem \begin{gather*} -u''(t) = a(t)f(u(t)), \quad 0 < t < 1,\\ u(0) = u(1) =0. \end{gather*} It needs to be mentioned that \cite{k2} only treats the class of symmetric about $t = \frac{1}{2}$ solutions. The assumption of symmetry imposes a constraint on $a(t)$: it must also be chosen to be symmetric about $t = \frac{1}{2}$. In our situation we can more generally locate the point of singularity anywhere in $[0,1]$. We will analyze a family of singular functions $$a(t) = \vert t - t' \vert^{-\epsilon},\label{3}$$ where $t' \in (0,1)$ and $0 < \epsilon < 1$. We will show that it is possible to construct $f(u)$ in such a fashion that it can be uniformly used for a range of value of the parameter $\epsilon$. To achieve a result that does not involve $\epsilon$ in growth conditions H\"{o}lder's inequality is to our advantage. It is utilized to yield norm-estimates from above" of Theorem \ref{thm1.2} presented below. The Green's function of $$-u'' = 0$$ satisfying \eqref{2} is $$G(t,s) = \begin{cases} \sigma(\alpha t + \beta)(\gamma + \delta - \gamma s), & t \leq s \leq 1, \\ \sigma(\alpha s + \beta)(\gamma + \delta - \gamma t), & 0 \leq s \leq t, \end{cases} \label{4}$$ where $\sigma = 1/(\alpha \gamma + \alpha \delta + \beta \gamma)$ (note that $\alpha \gamma + \alpha \delta + \beta \gamma >0$). At this point we assume that $f(u)$ is a continuous nonnegative function and introduce an integral operator, $T$, associated with the BVP \eqref{1}, \eqref{2} as follows $$Tu(t) = \int_{0}^{1} G(t,s) a(s)f(u(s)) ds,\quad 0 \leq t \leq 1. \label{5}$$ Fixed points of \eqref{5} are in fact (positive) solutions of \eqref{1}, \eqref{2}. The main tools used in this paper are the Krasnosel'ski\u i's fixed point theorem \cite{k3} and H\"{o}lder's inequality stated below. Now we define a cone in a Banach space. \begin{definition} \rm Let $\cal B$ be a real Banach space. A nonempty, closed set $\cal C\subset \cal B$ is said to be a cone provided: \begin{itemize} \item[(i)] $\alpha u + \beta v \in \cal C$ for all $u,v \in \cal C$ and $\alpha,\beta \geq 0$, \item[(ii)] $u,-u \in \cal C$ implies $u = 0$. \end{itemize} \end{definition} \begin{theorem} \label{thm1.2} Let $\cal B$ be a Banach space and let $\cal C\subset \cal B$ be a cone in $\cal B$. Assume $\Omega_{1}$, $\Omega_{2}$ are open bounded subsets of $\cal B$ with $0 \in \Omega_{1}$, $\bar \Omega_{1} \subset \Omega_{2}$, and let $$T\colon {\cal C}\cap(\bar \Omega_{2} \setminus \Omega_{1})\rightarrow \cal C$$ be a completely continuous operator such that either \begin{itemize} \item[(i)] $\| Tu \| \leq \| u\|$, $u \in {\cal C} \cap \partial \Omega_1$, and $\| Tu \| \geq \| u\|$, $u \in {\cal C} \cap \partial \Omega_2$, or \item[(ii)] $\| Tu \| \geq \| u\|$, $u \in {\cal C} \cap \partial \Omega_1$, and $\| Tu \| \leq \| u\|$, $u \in {\cal C} \cap \partial \Omega_2$. \end{itemize} Then $T$ has a fixed point in ${\cal C}\cap(\bar \Omega_{2} \setminus \Omega_{1})$. \end{theorem} Under appropriate growth assumptions on $f$, Theorem \ref{thm1.2} guarantees the existence of fixed points of \eqref{5}. With $a(t)$ specified by \eqref{3}, \eqref{5} becomes $$Tu(t) = \int_{0}^{1} G(t,s) \vert t' - s \vert^{-\epsilon}f(u(s)) ds,\quad 0 \leq t \leq 1.\label{6}$$ Evidently, $T$ is a completely continuous operator. We say that $f$ is in the Lebesgue space of (real valued) functions, $L^p[a,b]$, if $$\int_{a}^{b} {\vert f \vert}^p dx < \infty.$$ The norm on $L^p[a,b]$ is $${\Vert f \Vert}_p = \big(\int_{a}^{b} {\vert f \vert}^p dx\big)^{1/p}.$$ Now we state H\"{o}lder's inequality. \begin{theorem} \label{thm1.3} Let $f \in L^p$ and $g \in L^q$, where $p > 1$ and $q = \frac{p}{p-1}$. Then $fg \in L^1$ and we have $${\Vert fg \Vert}_1 \leq {\Vert f \Vert}_p {\Vert g \Vert}_q. \label{7}$$ \end{theorem} \section{Technical Results} For $\tau \in [0,\frac{1}{2})$, denote the interval $[\tau, 1 - \tau]$ by $I_{\tau}$. Note that, for each $\tau \in (0, \frac{1}{2})$, \eqref{4} satisfies $$\min_{t \in I_{\tau}}G(t,s) \geq L_{\tau}G(t',s),$$ where $L_{\tau} = \min\{ \frac{\delta + \gamma \tau}{\delta + \gamma}, \frac{\beta + \alpha \tau}{\beta + \alpha} \}$ for all $t',s \in [0,1]$. Let ${\cal B} = C[0,1]$ endowed with the norm $\|u\| = \max_{t\in[0,1]}\vert u(t) \vert$ and define $\cal {C}_{\tau} \subset \cal B$ by $${\cal C}_{\tau} = \{u(t) \in {\cal B} \, \vert u(t) \geq 0 \quad \mathrm{on } [0,1], \; \min_{t \in I_{\tau}} u(t) \geq L_{\tau} \| u \| \}.$$ Clearly, $\cal {C}_{\tau}$ is a cone and it can be shown that \eqref{6} preserves $\cal C_{\tau}$. At this point we would like to establish the $L^p$-norm estimates on \eqref{3} and \eqref{4} which will be used in Section 3. It is easy to see that $a \in L^p$ for all $p < \frac{1}{\epsilon}$ with $${\Vert a \Vert}_p = \frac{1}{(1 - \epsilon p)^{1/p}} ({t'}^{1-\epsilon p} + (1 - t')^{1-\epsilon p})^{1/p} \leq \frac{2^{\epsilon}}{(1 - \epsilon p)^{1/p}}. \label{8}$$ To obtain the required estimates on \eqref{4} we consider the following three cases: (1) $\alpha = 0$, $\gamma > 0$, (2) $\gamma = 0$, $\alpha > 0$, and (3) $\alpha\gamma > 0$ . If $\alpha = 0$ and $\gamma > 0$, then \eqref{4} becomes $$G(t,s) = \begin{cases} 1 + \frac{\delta}{\gamma} - s, & t \leq s \leq 1, \\ 1 + \frac{\delta}{\gamma} - t, & 0 \leq s \leq t. \end{cases} \label{9}$$ Now, $$\int_{\tau}^{1-\tau} G(t,s) ds = (1 + \frac{\delta}{\gamma})(1 - 2\tau) + \frac{1}{2}{\tau}^2 - t(1-\tau) - \frac{1}{2}t^2$$ attains its maximum at $t=\tau$ and $$\max_{t \in [0,1]} {\int_{\tau}^{1-\tau} G(t,s) ds} = 2(1 -\tau + \frac{\delta}{\gamma})(\frac{1}{2}-\tau) \\ \geq \frac{1}{2}-\tau. \label{10}$$ If $\gamma = 0$ and $\alpha > 0$, then \eqref{4} becomes $$G(t,s) = \begin{cases} \frac{\beta}{\alpha} + t, & t \leq s \leq 1, \\ \frac{\beta}{\alpha} + s, & 0 \leq s \leq t. \end{cases} \label{11}$$ Then $$\int_{\tau}^{1-\tau} G(t,s) ds = \frac{\beta}{\alpha}(1 - 2\tau) - \frac{1}{2}{\tau}^2 +t(1 - \tau) - \frac{1}{2}t^2$$ has its maximum at $t=1-\tau$ and $$\max_{t \in [0,1]} {\int_{\tau}^{1-\tau} G(t,s) ds} = (1+2\frac{\beta}{\alpha})(\frac{1}{2}-\tau) \geq \frac{1}{2}-\tau. \label{12}$$ If $\alpha\gamma > 0$, then \eqref{4} takes shape of $$G(t,s) = \begin{cases} \sigma'(t + \beta')(1 + \delta' - s), & t \leq s \leq 1, \\ \sigma'(s + \beta')(1 + \delta' - t), & 0 \leq s \leq t, \end{cases} \label{13}$$ where $\beta' = \beta/\alpha$, $\delta' = \delta/\gamma$, $\sigma' = 1/(\beta' + \delta' +1)$. A direct computation gives that for all $t \in [0,1]$, $$\int_{\tau}^{1-\tau} G(t,s) ds = \sigma' (-2 \beta' \delta' \tau - \frac{1}{2 \sigma'} +\beta'(\frac{1}{2} - \tau) + \beta' \delta' + (\beta' \tau + \delta' - \delta' \tau + \frac{1}{2})t - \frac{1}{2 \sigma'} t^2).$$ The right side of the equation above attains its max at $t_m = {\sigma'}^2 (\beta' \tau + \delta' - \delta' \tau + \frac{1}{2}) \in (\tau,1-\tau)$ and we obtain that \begin{eqnarray} \max_{t \in [0,1]} {\int_{\tau}^{1-\tau} G(t,s) ds} &=& \sigma'[\beta'(1+2\delta')(\frac{1}{2}-\tau)+\frac{1}{2\sigma'}({t_m}^2-{\tau}^2)] \nonumber \\ &=& \sigma'[\beta'(1+2\delta')(\frac{1}{2}-\tau)+\frac{1}{2} (2\tau+\sigma'(1+2\delta')(\frac{1}{2}-\tau))] \nonumber \\ &\geq& \sigma'[\beta'(1+2\delta')(\frac{1}{2}-\tau)+ \frac{\sigma'}{4}{(1+2\alpha')}^{2}(\frac{1}{2}-\tau)] \nonumber \\ &=& \frac{{\sigma'}^2}{4}(1+2\beta')(2\beta'+1+2\delta')(2\delta'+1) (\frac{1}{2}-\tau) \label{14} \\ &\geq& \frac{1}{4}(\frac{1}{2}-\tau) \nonumber \end{eqnarray} Combining \eqref{10}, \eqref{12}, and \eqref{14} we get that \eqref{4} satisfies $$\max_{t \in [0,1]} {\int_{\tau}^{1-\tau} G(t,s) ds} \geq \frac{1}{4}(\frac{1}{2}-\tau). \label{15}$$ Now we establish an estimate from above for \eqref{9}. Let $q \geq 1$. It is easy to see that $G(t,s) \leq G(s,s)$ for all $t,s \in [0,1]$. If $\alpha = 0$, $\gamma > 0$, then \begin{eqnarray*} {\Vert G(t,\cdot) \Vert}_{q}^q &=& \int_{0}^{1} G^q(t,s) ds \nonumber \\ &\leq& \int_{0}^{1} G^q(s,s) ds \nonumber \\ &=& \int_{0}^{1} (1+\frac{\delta}{\gamma}-s)^q ds \nonumber \\ &\leq& \frac{1}{q+1} (1+\frac{\delta}{\gamma})^{q+1} \nonumber \\ &<& {(1+\frac{\delta}{\gamma})}^{2q} \nonumber \end{eqnarray*} so that $$\max_{t \in [0,1]}{\Vert G(t,\cdot) \Vert}_q < {(1+\frac{\delta}{\gamma})}^2. \label{16}$$ By the same argument applied to the case of $\gamma = 0$, $\alpha > 0$ we get for \eqref{11} $$\max_{t \in [0,1]}{\Vert G(t,\cdot) \Vert}_q < {(1+\frac{\beta}{\alpha})}^2. \label{17}$$ If $\alpha\gamma > 0$, then we have for \eqref{13} $$G(t,s) \leq G(s,s) = \sigma'(\beta'\delta'+\beta'+(\delta'+1-\beta')s-s^2).$$ Consider the function $g(s) = \beta'\delta'+\beta'+(\delta'+1-\beta')s-s^2$ on $[0,1]$. There are three cases: (i) $\delta'-\beta'\leq -1$, (ii) $\delta' - \beta' \geq 1$ and (iii) $\vert \beta' - \delta' \vert < 1$. In cases (i) and (ii), the maximum of $g(s)$ occurs at $s=0$ and $s=1$, respectively. So that $\max_{s \in [0,1]}{g(s)} = g(0) = \beta' (1+\delta') < \frac{1}{{\sigma'}^2}$ and $\max_{s \in [0,1]}{g(s)} = g(1) = \delta'(1+\beta') < \frac{1}{{\sigma'}^2}$ as corresponding to the above two cases. Combining cases (i) and (ii), we get that if $\vert \delta' - \beta' \vert \geq 1$, then $$\max_{s \in [0,1]}{g(s)} < \frac{1}{{\sigma'}^2}.$$ In case (iii) the maximum is attained at $s = \frac{\delta'+1-\beta'}{2} \in (0,1)$ and $\max_{s \in [0,1]}{g(s)} = g(\frac{\delta'+1-\beta'}{2}) = \frac{1}{4}{(\beta'+\delta'+1)}^2 = \frac{1}{{\sigma'}^2}$. Pasting all the cases together, $$\max_{s \in [0,1]}{g(s)} < \frac{1}{{\sigma'}^2}$$ and so since $G(s,s) = \sigma' g(s)$, $$G(t,s) < \frac{1}{\sigma'}, \ t,s \in [0,1].$$ In particular, \begin{eqnarray*} {\Vert G(t,\cdot) \Vert}_{q}^q &=& \int_{0}^{1} G^q(t,s) ds \\ &\leq& \int_{0}^{1} G^q(s,s) ds \\ &=& \frac{1}{{\sigma'}^{2q}} \\ &=& {(1+\frac{\beta}{\alpha}+\frac{\delta}{\gamma})}^{2q} \end{eqnarray*} and hence $$\max_{t \in [0,1]}{\Vert G(t,\cdot) \Vert}_q < {(1+\frac{\beta}{\alpha}+\frac{\delta}{\gamma})}^2. \label{18}$$ Finally, combining \eqref{16}, \eqref{17}, and \eqref{18} we get $$\max_{t \in [0,1]}{\Vert G(t,\cdot) \Vert}_q < A, \label{19}$$ where $$A = \begin{cases} (1+\frac{\delta}{\gamma})^2, & \alpha = 0, \gamma > 0 \\ (1+\frac{\beta}{\alpha})^2, & \gamma = 0, \alpha > 0 \\ (1+\frac{\beta}{\alpha}+\frac{\delta}{\gamma})^2, & \alpha\gamma > 0. \end{cases} \label{20}$$ \section{Main Results} Without loss of generality, let $0 < t' < 1/2$. First, we are going to consider the case of $0 < \epsilon < 1/2$. \begin{theorem} \label{thm3.1} Suppose the sequence $\{t_i\}_{i=1}^{\infty}$ satisfies $0 < t_i < t_{i+1}$, $i \in N$, and $$\lim_{i \rightarrow \infty} t_i = t' < \frac{1}{2}.$$ Suppose $\{a_i\}_{i=1}^{\infty}$ and $\{b_i\}_{i=1}^{\infty}$ are the sequences satisfying $a_{i+1} < L_{t_i}b_i < b_i < \frac{16A}{\frac{1}{2}-t_i} b_i < a_i \quad i \in N,$ where $A$ is given by $\eqref{20}$. Assume also that $f$ satisfies the following conditions: \begin{itemize} \item[(H1)] $f(z) \leq \frac{1}{4A} a_i$ for all $z \in [0,a_i]$, $i \in N$, \item[(H2)] $f(z) \geq \frac{4}{{\frac{1}{2}} - t'} b_i$ for all $z \in [L_{t_i} b_i, b_i]$, $i \in N$. \end{itemize} Then \eqref{1}, \eqref{2} has infinitely many fixed points $\{u_i\}_{i=1}^{\infty}$ such that $b_i < \Vert u_i \Vert < a_i$, $i \in N$. \end{theorem} \paragraph{Proof.} Consider the sequences $\{\Omega_{1,i}\}_{i=1}^{\infty}$ and $\{\Omega_{2,i}\}_{i=1}^{\infty}$ of open sets in $\cal B$ defined, for each $i \in N$, by \begin{gather*} \Omega_{1,i} = \{u\in {\cal B}\colon \Vert u \Vert < a_i\},\\ \Omega_{2,i} = \{u\in {\cal B}\colon \Vert u \Vert < b_i\}, \end{gather*} Consider also the sequence of cones $\{{\cal C}_i\}_{i=1}^{\infty}$ defined by $${\cal C}_i = \{u(t) \in {\cal B} \vert u(t) \geq 0 \text{ on [0,1] with } \min_{t \in I_{t_i}} u(t) \geq L(t_i) \| u \|\}.$$ Let $i \in N$ and $u \in {\cal C}_i \cap \partial \Omega_{1,i}$, then $$u(s) \leq \Vert u \Vert = a_i$$ for all $s \in [0,1]$. So, by (H1) \label{21} \begin{aligned} \Vert Tu \Vert =& \max_{t \in [0,1]}\int_{0}^{1} a(s)G(t,s)f(u(s)) ds \\ \leq& \max_{t \in [0,1]}\int_{0}^{1} a(s)G(t,s)) ds \frac{1}{4A} a_i \end{aligned} Let $1 \frac{1}{(\frac{1}{2} - t_i)^{\epsilon}}$, $s \in I_{t_i}$, the inequality above becomes $$\Vert Tu \Vert \geq \max_{t \in [0,1]}\int_{t_i}^{1-t_i} G(t,s) ds \frac{1}{(\frac{1}{2} - t_i)^{\epsilon}} \ \frac{4}{{\frac{1}{2}} - t'} \ b_i$$ Now \eqref{15} applies and we get \begin{eqnarray*} \Vert Tu \Vert &=& {\frac{1}{4}}(\frac{1}{2} -t_i) \ \frac{1}{(\frac{1}{2} - t_i)^{\epsilon}} \ \frac{4}{{\frac{1}{2}} - t'} \ b_i \\ &>& (\frac{1}{2} - t_i)^{1-\epsilon} \frac{1}{{\frac{1}{2}} - t'} b_i > b_i; \end{eqnarray*} that is, $$\Vert Tu \Vert > \Vert u \Vert \label{24}$$ for all $u \in {\cal C}_i \cap \partial \Omega_{2,i}$, $i \in N$. Note that since $0 \in \Omega_{2,i} \subset \bar \Omega_{2,i} \subset \Omega_{1,i}$ and \eqref{23} and \eqref{24} hold, we can apply Theorem \ref{thm1.2} to conclude that the operator $T$ has a fixed point $u_i \in {\cal C}_i \cap (\bar \Omega_{1,i} \setminus \Omega_{2,i})$ such that $b_i < \Vert u_i \Vert < a_i, \ i \in N$. The proof is complete. \hfill$\square$ \smallskip Now, let us deal with the case $\frac{1}{2} \leq \epsilon < 1$. \begin{theorem} \label{thm3.2} Suppose the sequence $\{t_i\}_{i=1}^{\infty}$ satisfies $0 < t_i \Vert u \Vert \label{28} for all$u \in {\cal C}_i \cap \partial \Omega_{2,i}$,$i \in N.$Note that since$\Omega_{2,i} \subset \bar \Omega_{2,i} \subset \Omega_{1,i}$and \eqref{27} and \eqref{28} hold, we can apply Theorem \ref{thm1.2} to conclude that the operator$T$has a fixed point$u_i \in {\cal C}_i \cap (\bar \Omega_{1,i} \setminus \Omega_{2,i})$such that$b_i < \Vert u_i \Vert \leq a_i, \ i \in N$, which completes the proof. \hfill$\square\$ \begin{thebibliography}{00} \frenchspacing \bibitem{a1} R. P. Agarwal, D. O'Regan and P. J. Y. Wong, Positive Solutions of Differential, Difference and Integral Equations'', Kluwer, Dordrecht, 1999. \bibitem{b1} J. V. Baxley and H. B. Thompson, {\em Boundary\ Behavior\ and\ Computation\ of\ Solutions\ of\ Singular\ Nonlinear\ Boundary\ Value\ Problems}, Commun. Appl. Anal. {\bf 4} (2000), 207-226. \bibitem{e1} J. Ehme, {\em Denumerable\ Symmetric\ Positive\ Solutions\ for\ a\ Two\ Point\ Conjugate\ Problem}, preprint. \bibitem{e2} P. W. Eloe, J. L. Henderson and N. Kosmatov, {\em Countable\ Positive\ Solutions\ of\ a\ Conjugate\ Type\ Boundary\ Value\ Problem}, Commun. Appl. Nonliner Anal. {\bf 7} (2000), 47-55. \bibitem{k1} E. R. Kaufmann and N. Kosmatov, {\em A\ Multiplicity\ Result\ for\ a\ Boundary\ Value\ Problem\ with\ Infinitely\ Many\ Singularities}, J. Math. Anal. Appl. {\bf 269} (2002), 444-453. \bibitem{k2} N. Kosmatov, {\em On\ a\ Singular\ Conjugate\ Boundary\ Value\ Problem\ with\ Infinitely\ Many\ Solutions}, Math. Sci. Res. Hot-Line {\bf 4} (2000), 9-17. \bibitem{k3} M. A. Krasnosel'ski\u i, Positive Solutions of Operator Equations'', (English) Translated by R. E. Flaherty, Noordhoff Ltd., Groningen, 1964. \end{thebibliography} \noindent\textsc{Nickolai Kosmatov} \\ Department of Mathematics and Statistics\\ University of Arkansas at Little Rock\\ Little Rock, Arkansas 72204-1099, USA\\ e-mail: nxkosmatov@ualr.edu \end{document}