\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 90, pp. 1--5.\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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/90\hfil A Neumann boundary-value problem] {A Neumann boundary-value problem on an unbounded interval} \author[P. Amster, A. Deboli\hfil EJDE-2008/90\hfilneg] {Pablo Amster, Alberto Deboli} % in alphabetical order \address{Pablo Amster\newline Departamento de Matem\'atica, Universidad de Buenos Aires, Ciudad Universitaria, Pabell\'on I (1428) Buenos Aires, Argentina. \newline Consejo Nacional de Investigaciones Cient\'\i ficas y T\'ecnicas (CONICET)} \email{pamster@dm.uba.ar} \address{Alberto Deboli \newline Departamento de Matem\'atica, Universidad de Buenos Aires, Ciudad Universitaria, Pabell\'on I (1428) Buenos Aires, Argentina } \email{adeboli@yahoo.com.ar} \thanks{Submitted April 11, 2008. Published June 21, 2008.} \subjclass[2000]{34B40, 34B15} \keywords{Boundary-value problem on the half line; Neumann conditions; \hfill\break\indent upper and lower solutions; diagonal argument} \begin{abstract} We study a Neumann boundary-value problem on the half line for a second order equation, in which the nonlinearity depends on the (unknown) Dirichlet boundary data of the solution. An existence result is obtained by an adapted version of the method of upper and lower solutions, together with a diagonal argument. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{example}[theorem]{Example} \section{Introduction} We study the Neumann boundary-value problem on the half line: $$\label{problem} \begin{gathered} y''(x) = f(x,y(x),y(0),y(\infty)) \quad x\in (0,+\infty),\\ y'(0)=v_0,\quad y'(\infty)=0, \end{gathered}$$ where $$y(\infty):=\lim_{x\to +\infty} y(x), \quad y'(\infty):=\lim_{x\to +\infty} y'(x)$$ and $f:[0,+\infty)\times \mathbb{R}^3\to \mathbb{R}$ is continuous. Note that the nonlinearity $f$ depends on the (unknown) Dirichlet boundary data of the solution. For a bounded interval, this kind of problem has been considered for example in \cite{T}, where a problem on a two-ion electro-diffusion model is discussed. It is worth to observe that problem \eqref{problem} is resonant, since the kernel of the associated linear operator $\mathcal L y:=y''$ is non-trivial. For the case of a bounded interval, the Neumann problem has been widely studied in the literature; in this situation, the operator $\mathcal L$ is Fredholm of index $0$, and the problem can be solved by the use of coincidence degree \cite{Ma}. If the interval is unbounded then $\mathcal L$ is not a Fredholm operator anymore, and an alternative method of proof is needed (see e.g. \cite{Andres,Const, Furi,Przera,Rabier,Regan, S}). In contrast with the above mentioned works, we deal with an extra difficulty, which arises on the fact that the nonlinear term in problem \eqref{problem} depends also on the (unknown) Dirichlet boundary values of the solution. In particular, it is implicitly assumed that the limit value $y(\infty)$ exists, and it is finite; in this sense, the problem is over-determined. We shall prove the existence of solutions in presence of an ordered couple of a lower and an upper solution that converge at infinity to the same limit $L$. More precisely, we shall assume the existence of smooth functions $\alpha,\beta:[0,\infty)\to \mathbb{R}$ such that $\alpha\le \beta$, and $\alpha'(0)\ge v_0\ge \beta'(0)$, $\alpha(\infty)=\beta(\infty)=L$ and $$\alpha''(x)\ge f(x,\alpha(x),u,L),\quad \beta''(x)\le f(x,\beta(x),u,L)$$ for each $x\in [0,\infty)$ and $u\in [\alpha(0),\beta(0)]$. Then, we shall proceed in the following way: firstly, we prove the existence of solutions of a mixed boundary-value problem for any bounded interval $[0,N]$; then we apply a diagonal argument in order to obtain a subsequence of $\{ u_N\}$ that converges to a solution of the equation. Finally, under an extra assumption on $f$ we shall prove that the Neumann boundary condition at infinity is satisfied. Our main theorem reads as follows: \begin{theorem}\label{main} Let $(\alpha,\beta)$ be an ordered couple of a lower and an upper solution as before. Furthermore, assume that $$|f(x,y,u,L)|\le \varphi(x)$$ for $x\ge x_0$, $\alpha(0)\le u\le \beta(0)$, $\alpha(x)\le y\le \beta(x)$ and some $\varphi\in L^1(x_0,+\infty)$. Then \eqref{problem} admits at least one solution $y$, with $\alpha\le y\le \beta$. \end{theorem} \section{Mixed conditions on a bounded interval. Upper and lower solutions} In this section, we consider the problem with mixed conditions $$\label{problem-bounded} \begin{gathered} y''(x) = f(x,y(x),y(0),L) \quad x\in (0,T),\\ y'(0)=v_0,\quad y(T)=y_T, \end{gathered}$$ for some arbitrary constants $y_T,L\in \mathbb{R}$ and $T>0$. We shall prove the existence of solutions of \eqref{problem-bounded} in presence of an ordered couple of a lower and an upper solution. More precisely, we shall assume the existence of some smooth functions $\alpha\le \beta$ satisfying $$\label{ul1} \alpha''(x)\ge f(x,\alpha(x),u,L),\quad \beta''(x)\le f(x,\beta(x),u,L)$$ for $0\le x\le T$ and $\alpha(0)\le u\le \beta(0)$, and $$\label{ul2} \alpha'(0)\ge v_0\ge\beta'(0), \quad \alpha(T)\le y_T\le\beta(T).$$ The following theorem is a slight variation of the standard results in the theory of upper and lower solutions; for the sake of completeness we give a simple proof. \begin{theorem}\label{bounded} Let $\alpha\le \beta$ satisfy \eqref{ul1} and \eqref{ul2}. Then \eqref{problem-bounded} admits at least one solution $y$, with $\alpha\le y\le \beta$. \end{theorem} \begin{proof} Let $\lambda >0$ be a fixed constant and consider, for each $w\in C([0,T])$, the modified linear problem $$\label{modif.linear problem} \begin{gathered} y''(x)-\lambda y(x)=f(x,P_x(w(x)),P_0(w(0)),L)-\lambda P_x(w(x)) \\ y'(0)=v_0,\quad y(T)=y_T \end{gathered}$$ where $P:[0,T]\times \mathbb{R}\to \mathbb{R}$ is the truncation function $$\label{truncation} P_x(w):=P(x,w)= \begin{cases} \alpha &\text{if } w < \alpha(x) \\ w &\text{if } \alpha(x)\leq w\leq \beta(x) \\ \beta &\text{if } w > \beta(x) \end{cases}$$ Problem (\ref{modif.linear problem}) has a unique solution $y\in C^2([0,T])$, given by the integral representation $$y(x)=y_0(x)+\int_0^T G(x,s)\phi(s)ds,$$ where \begin{gather*} y_0(x)=\frac{v_0 \sinh(\sqrt{\lambda}(x-T)) +\sqrt{\lambda} y_T \cosh(\sqrt{\lambda}x)}{\sqrt{\lambda} \cosh(\sqrt{\lambda}T)},\\ \phi(s)=\phi(w,s):= f(s,P_s(w(s)),P_0(w(0)),L)-\lambda P_s(w(s)) \end{gather*} and $G$ is the Green function $$G(x,s)=\begin{cases} \frac{\cosh(\sqrt{\lambda}s)\sinh(\sqrt{\lambda}(x-T))}{\sqrt{\lambda} \cosh(\sqrt{\lambda}T)} & \text{if } 0\leq s0 such that \|Kw\|_\infty\le R for every w\in C([0,T]). From Schauder's Theorem it follows that K has a fixed point y\in \overline {B_R(0)}. Next, we shall prove that \alpha(x)\leq y(x)\leq \beta(x) for all x\in [0,T], and hence y is a solution of the original problem. It suffices to prove that \alpha\le y, since the other inequality is analogous. By contradiction, suppose there exists x_0\in [0,T] such that y(x_0)<\alpha(x_0). We may assume that x_0 is an absolute maximum value of the function \alpha-y. We consider the two possible cases: \noindent\textbf{Case 1.} x_0\in (0,T). From the definition of P, it follows that P_{x_0}(y(x_0))=\alpha(x_0). Moreover, \alpha(0)\leq P_0(y(0))\leq \beta(0) and hence $\alpha''(x_0)\geq f(x_0,\alpha(x_0),P_0(y(0)),L).$ On the other hand,$$ 0\geq (\alpha- y)''(x_0) =\alpha''(x_0)- f(x_0,\alpha(x_0),P_0(y(0)),L) +\lambda (\alpha(x_0)-y(x_0))>0, $$which is a contradiction. \noindent\textbf{Case 2.} x_0=0 or x_0=T. As y(T)=y_T \ge \alpha(T), it follows that x_0=0. Moreover, \alpha'(0)\geq y'(0)=v_0, and by continuity$$ (\alpha-y)''=\alpha''- f(\cdot,\alpha,P_0(y(0)),L) +\lambda (\alpha-y(\cdot))>0 $$over some interval (0,\delta) with \delta small enough. This implies (\alpha-y)' > 0 on (0,\delta), and then \alpha-y > (\alpha - y)(0) on (0,\delta). This contradicts the fact that 0 is an absolute maximum of the function \alpha-y. \end{proof} \section{A diagonal argument for problem \eqref{problem}} In this section we give a proof of Theorem \ref{main}. From Theorem \ref{bounded}, for every N\in \mathbb{N} we consider a solution y_N of $$\label{problem-N} \begin{gathered} y_N''(x) = f(x,y_N(x),y_N(0),L) \quad x\in (0,N),\\ y_N'(0)=v_0,\quad y_N(N)=\frac{\alpha(N)+\beta(N)}2, \end{gathered}$$ such that \alpha|_{[0,N]}\leq y_N\leq\beta|_{[0,N]}. For fixed M, let us observe firstly that if N\ge M then$$ \begin{array}{l} \|y''_N|_{[0,M]}\|_\infty=\sup_{x\in [0,M]}\{|f(x,y_N(x),y_N(0),L)|\}\leq C_2 \end{array} for some positive constant C_2 independent of N. Moreover, writing $y'_N(x)=v_0+\int_0^{x} y''_N(s)\ ds$ it is also seen that \|y'_N|_{[0,M]}\|_\infty\leq |v_0|+M C_2:=C_1. Finally, from the equality $y_N(x)=y_N(M) - \int_x^{M} y'_N(s)\ ds$ we conclude that \|y_N|_{[0,M]}\|_\infty\leq C_0 for some constant C_0 depending only on M. Hence, from the Arzel\'a-Ascoli theorem we deduce the existence of a subsequence of \{y_N\}_{N\ge M} that converges on [0,M] for the C^1-norm. Thus, we may proceed as follows: for M=1, let us choose a subsequence (still denoted \{y_N\}) that converges in C^1([0,1]) to some function y^1. Repeating the procedure for M=2, 3, \ldots, we may assume that y_N|_{[0,M]} converges to some function y^M in the C^1-sense. It follows from the construction that y^{M+1}|_{[0,M]}=y^M, and this implies that the function y:[0,+\infty)\to \mathbb{R} given by y(x)=y^M(x) if 0\le x\leq M is well defined. Moreover, y'(0)=v_0, and y''_N converges uniformly in [0,M] to f(\cdot,y(\cdot),y(0),L). Thus, for any test function \xi\in C_0^\infty(0,M) we obtain: \begin{align*} \int_0^M f(x,y,y(0),L)\xi(x)\,dx &= \lim_{N\to\infty} \int_0^M y_N''(x)\xi(x)\,dx\\ &= \lim_{N\to\infty}\int_0^My_N(x)\xi''(x)\,dx \\ &=\int_0^M y(x)\xi''(x)\,dx, \end{align*} whence y''(x)=f(x,y(x),y(0),L) $$in [0,M], in the weak sense. Moreover, it is clear that y(\infty)=L, and as y is twice continuously differentiable we deduce that y is a classical solution of the equation. Finally, for each N\ge x_0 we may establish a point x_N\in (N,N+1) such that y'(x_N)=y(N+1)-y(N)\to 0; thus, for x > x_N we have$$ |y'(x)-y'(x_N)| =\big|\int_{x_N}^x f(t,y(t),y(0),L)\,dt \big| \le \int_{x_N}^x \varphi(t)\,dt\le \int_{x_N}^{+\infty} \varphi(t)\,dt. $$As \varphi is integrable, we conclude that y'(x)\to 0 as x\to \infty, and so completes the proof. \begin{example} \label{exa3.1} \rm Consider the problem \begin{gather*} y''(x) = 2(x+1)y(x)^4 + g(x,y(x),y(0),y(\infty)) \quad x\in (0,+\infty),\\ y'(0)=v_0,\quad y'(\infty)=0, \end{gather*} where g:[0,+\infty)\times \mathbb{R}^3\to \mathbb{R} is continuous and satisfies$$ g(x,0, u, 0) = 0,\quad 0\le g(x,y,u,0)\le \varphi(x) \quad \text{for $x\ge 0$, $0\le y\le\frac 1{x+1}$ and $u \in [0,1]$}  for some $\varphi\in L^1([0,+\infty))$. Then, if $v_0\in [-1,0]$ the assumptions of Theorem \ref{main} are fulfilled, taking $\alpha\equiv 0$ and $\beta(x)=\frac 1{x+1}$. \end{example} \begin{thebibliography}{0} \bibitem{Andres} J. Andres, G. Gabor, L. G\'orniewicz; Boundary value problems on infinite intervals, Trans. Amer. Math. Soc. 351 (1999) 4861-4903. \bibitem{Const} A. Constantin; On an infinite interval boundary-value problem, Annali di Matematica pura ed applicata (IV) Vol. CLXXVI (1999) 379-394. \bibitem{Furi} M. Furi, P. Pera; A continuation method on locally convex spaces and applications to ordinary differential equations on noncompact intervals, Ann. Polon. Math. XLVII (1987) 331-346. \bibitem{Ma} J. Mawhin; Topological degree methods in nonlinear boundary-value problems, Regional Conf. Series in Math. 40, Amer. Math. Soc., Providence R.I., (1979). \bibitem{Przera} B. Przeradzki; A new continuation method for the study of nonlinear equations at resonance, J. Math. Anal. Appl. 180 No 2 (1993) 553-565. \bibitem{Rabier} P. J. Rabier, C.A. Stuart; A Sobolev space approach to boundary-value problems on the half-line. Comm. in Contemp. Math. 7 No.1 (2005) 1-36. \bibitem{Regan} D. O'Regan; Solvability of some singular boundary-value problems on the semi-infinite intevral, Can. J. Math. 48 (1) (1996) 143-158. \bibitem{S} Szyma\'nska, K.: Resonant problem for some second-order differential equation on the half-line. Electronic Journal of Differential Equations Vol. 2007(2007), No. 160, pp. 1--9. \bibitem{T} Thompson H. B.: Existence for Two-Point Boundary Value Problems in Two Ion Electrodiffusion. Journal of Mathematical Analysis and Applications 184, No. 1 (1994) 82-94. \end{thebibliography} \end{document}