\documentclass[twoside]{article} \usepackage{amsfonts} % used for R in Real numbers \pagestyle{myheadings} \markboth{ Heteroclinic connections } { L. Sanchez } \begin{document} \setcounter{page}{257} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent USA-Chile Workshop on Nonlinear Analysis, \newline Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 257--266.\newline http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Heteroclinic connections for a class of non-autonomous systems % \thanks{ {\em Mathematics Subject Classifications:} 34B15, 34C37. \hfil\break\indent {\em Key words:} Heteroclinics, Fisher equation. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Published January 8, 2001. \hfil\break\indent Supported by Funda\c c\~ao para a Ci\^encia e a Tecnologia and PRAXIS XXI.} } \date{} \author{ L. Sanchez } \maketitle \begin{abstract} We prove the existence of heteroclinic connections for a system of ordinary differential equations, with time-dependent coefficients, which is reminiscent of the ODE arising in connection with traveling waves for the Fisher equation. The approach is elementary and it allows in particular the study of the existence of positive solutions for the same system that vanish on the boundary of an interval $(t_0,+\infty)$. \end{abstract} \section{Introduction} When one looks for one-dimensional traveling waves $u(x-ct)$ for the Fisher equation $$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+f(u)$$ (that models a diffusion phenomenon in biomathematics), one finds the ordinary differential equation $$u''+cu'+f(u)=0\,.\eqno(1)$$ Here $c>0$ represents the admissible wave speed; the function $f$ takes positive values between two zeros, say 0 and $a$ ($a>0$): see for example \cite{kpp,aw}. In this paper we consider the following system, which is a non-autonomous multi-dimensional analogue of (1): $$u_i''+p_i(t)u_i'+f_i(u)=0,\quad i=1,\dots, n\,,\eqno(2)$$ where $u=(u_1,\cdots u_n)$. The vector field $f=(f_1,\cdots,f_n)$ is assumed to be defined in some $n$-dimensional box $[0,a_1]\times\cdots\times[0,a_n]$ ($a_i>0,\forall i=1,\dots, n$), the vertices $(0,\cdots,0)$ and $(a_1,\cdots,a_n)$ being its only zeros. More precisely, we state the following basic assumptions: \begin{enumerate} \item[(H1)] For each $i\in{1,\cdots, n}$, $f_i:[0,a_1]\times\cdots\times[0,a_n]\rightarrow{{\mathbb R}_+}$ is a Lipschitz continuous function such that $f_i(0,\cdots,0)=0=f_i(a_1,\cdots a_n)$ and $f_i(u)>0$ if $u_i>0$ and $u\neq a:=(a_1,\cdots a_n)$. \item[(H2)] The functions $p_i:{\mathbb R}\to{{\mathbb R}_+}$ will be assumed throughout to be continuous and $$c_i:={\displaystyle\inf_{t\in{\mathbb R}}p_i(t)}>0.$$ \end{enumerate} We look for {\it positive solutions} $u(t)=(u_1(t);\cdots, u_n(t))$, i.e., solutions that have positive components. Accordingly, the word {\it solution} will be used to mean {\it positive solution} throughout. The relevant problem is to find monotonic'' heteroclinics (in the sense that their components are decreasing functions) that connect the equilibria $(a_1,\cdots a_n)$ and 0. This problem has been very much studied for the autonomous scalar equation, various approaches being available in a vast literature: we refer the reader to \cite{aw,as,k} and the bibliography in those papers. The autonomous system has been dealt with in \cite{al}; we owe a lot to the ideas there, and we would like to stress that our approach, which is also elementary, works in a slightly more general setting in the sense that it allows not only time dependence but also consideration of models where the vector field $f$ may vanish to a higher order at $u=0$. In addition we could equally consider a more general form of (2) where nonlinear terms $b_i(t)f_i(u)$ replace $f_i(u)$ and the functions $b_i$ are bounded above and below by positive numbers (see \cite{s}). An important role is played by the functions $g_i(u)$, defined (for those $u$ such that $00,$ $\mu>0$, $c^2\geq4M$ and $0\leq \epsilon\leq \mu(\frac{c}{2}+\frac{\sqrt{c^2-4M}}{2})$. Then the solution $u(t)$ of the initial value problem $$\displaylines{ \hfill u''+cu'+Mu=0\hfill\llap{(3)}\cr \hfill u(0)=\mu,\quad u'(0)=-\epsilon\hfill\llap{(4)} }$$ is positive in $[0,+\infty)$ and tends to zero as $t\to+\infty$. (See \cite{s}.) \paragraph{\bf Lemma 2.1} {\sl Let continuous functions $p,\,q,\, l,\, m$ be given such that $p(t)\geq q(t)>0$, $0\leq l(t)\leq m(t)$ in the interval $[t_0,t_1]$. Let $u$ and $v$ be the respective solutions of $$\displaylines{ \hfill u''+p(t)u'+l(t)u=0,\hfill\llap{(5)}\cr \hfill v''+q(t)v'+m(t)v=0\hfill\llap{(6)} }$$ such that $u(t_0)=v(t_0)\geq0$ and $u'(t_0)=v'(t_0)$. Assume in addition that $p(t)\equiv q(t)$ in case $u'(t_0)=v'(t_0)>0$. Then if $v(t)\geq0$ in $[t_0,t_1]$ we have $u(t)\geq v(t)$ in $[t_0, t_1]$. } \paragraph{Proof.} If $u(t_0)=v(t_0)=u'(t_0)=v'(t_0)=0$ or $p\equiv q$ and $l\equiv m$ there is nothing to prove. Otherwise, starting with $u(t_0)=v(t_0)\geq0$ and $u'(t_0)=v'(t_0)+\epsilon$ ($\epsilon>0$) it follows that $u>v$ in some interval $(t_0,t_0+\delta)$. Suppose that there exists $\bar t\leq t_1$ such that $u(t)>v(t)$ $\forall t\in (t_0,\bar t)$ and $u(\bar t)=v(\bar t)$. Set $P(t):=\int_{t_0}^tp(s)\,ds$. Multiplying (5) and (6) by $e^{P(t)}$, then (5) by $v$, (6) by $u$, integrating in $[t_0,\bar t]$ and subtracting we obtain, since $v'(t)<0$ in case $v'(t_0)\leq 0$ (according to Remark 2) \begin{eqnarray*} 0&>&[e^{P(t)}(u'(t)v(t)-u(t)v'(t)]_{t_0}^{\bar t}+\int_{t_0}^{\bar t}e^{P(t)}(p(t)-q(t))v'(t)u(t)\,dt\\ &&+\int_{t_0}^{\bar t}e^{P(t)}(l(t)-m(t))u(t)v(t)\,dt=0, \end{eqnarray*} a contradiction. Hence $u\geq v$ in $[t_0,t_1]$. In the limit as $\epsilon\to0^+$ the statement follows. \smallskip Let us write, in accordance with previous notation for the $n$-dimensional case, $$f(u)=ug(u),\quad 0\leq u\leq a.$$ \paragraph{Lemma 2.2} {\sl Assume $p$ is continuous in ${\mathbb R}$, $f$ is a Lipschitz continuous function in $[0,\mu]$ such that $f(0)=0$ and $f(u)>0$ if $0t_0$. Then the solution of the Cauchy problem $$u''+p(t)u'+f(u)=0,\quad u(t_0)=\mu,\;\;u'(t_0)=-\epsilon$$ where we assume that $\mu$ and $\epsilon>0$ are as in Remark 3, is positive and strictly decreasing in $[t_0,\infty)$ and vanishes at $+\infty$.} \paragraph{Proof.} Apply Lemma 2.1 with $q(t)=c$, $l(t)=g(u(t))$, $m(t)=M$. Take Remarks 2 and 3 into account. The fact that $u(+\infty)=0$ is an easy consequence of the boundedness of $p$, $u$ and $u'$ since for $t>t_0$ and some $t^*\in(t_0,t)$ $$u'(t)+\epsilon+p(t^*)(u(t)-u_0)+\int_{t_0}^tf(u(s))\,ds=0,\quad t>t_0$$ and we infer that $\int_{t_0}^{+\infty} f(u(s))\,ds$ converges. \paragraph{Remark.} It is immediately recognized that the above result still holds if $\epsilon=0$ provided $f(u)>0$ $\forall u\in(0,\mu]$. \section{Heteroclinics} In this section we give a simple analytic argument to prove the existence of heteroclinics under hypotheses (H1)-(H2). For the sake of clarity we start with the case of the scalar equation $$u''+p(t)u'+f(u)=0\,,\eqno(2_1)$$ where $f:[0,a]\to{\mathbb R}_+$ has the property (H1) for $n=1$ and we write accordingly $$c:={\displaystyle\inf_{t\in{\mathbb R}}p(t)}>0;\quad f(u)=ug(u).$$ A basic assumption, which cannot be improved when $p\equiv c$ is a constant and $g(u)$ is decreasing, is $c\geq2\sqrt{{\displaystyle\sup_{00$ $0t_0$. Then for each sufficiently small $\epsilon>0$ the solution $u(t,t_0,\epsilon)$ of (2) such that $u(t_0,t_0,\epsilon)=\mu$ and $u'(t_0,t_0,\epsilon)=-\epsilon$ is positive in $[t_0,+\infty)$ and $$\lim_{t\to+\infty}u(t,t_0,\epsilon)=0.$$} \paragraph{Proof.} In case (i) holds, this is only Lemma 2.2. Otherwise note that the solution $u(t,t_0,\epsilon)$ has no critical points and therefore is strictly decreasing. It cannot remain above a positive constant by the argument used at the end of the proof of Lemma 2.2. Let $t_1$ be such that $u(t_1, t_0,\epsilon)=\nu$. The equation itself shows that $u'( t_1, t_0,\epsilon)\geq -N/c$ (consider separately the cases where $t_1$ lies in an interval of convexity or of concavity of the solution) and therefore Lemma 2.2 can be applied. \paragraph{Remark.} According to the remark after Lemma 2.1 it is obvious, via the same arguments, that the Proposition holds even if $\epsilon=0$ except in case $f(\mu)=0$. \paragraph{Theorem 3.2} {\sl Assume (H1)-(H2) with $n=1$ and, in addition to the hypotheses of proposition 3.1 with $\mu=a$, that $p(t)$ is bounded. Then ($2_1$) has a strictly decreasing heteroclinic solution connecting $a$ and 0.} \paragraph{Proof.} With respect to $\mu=a$ in Proposition 3.1 take a sequence $t_m$ decreasing to $-\infty$ and consider the solution $u(.,t_1,\epsilon_1)$ where $\epsilon_1$ is a small positive number. According to proposition 3.1, $01$ such that $$u(\bar t, t_{m_2},\epsilon_1)1 and, by a t_m translation, this can be written$$u_m(\bar t-t_m)\geq a/2\eqno(8) $$in terms of the solution of$$u''_m+p_m(t)u'_m+f(u_m)=0,\quad u_m(0)=a,\;\;u'_m(0)=-\epsilon_1\,, $$where p_m(t)=p(t+t_m). The boundedness of p_m, u_m and u'_m and Ascoli's theorem enable us, by extracting subsequences and a diagonal procedure, to suppose that (where we set d:={\displaystyle\sup_{t\in{\mathbb R}}p(t)})$$p_m\to p_{\infty}\;\;\mbox{in}\;\;L^{\infty}\mbox{weak-*},\;\;c\leq p_{\infty}(t)\leq d u_m\to u\;\;\mbox{in}\;\; C^1(K),\;\; K\;\; \mbox{any compact interval in}\;[0,+\infty). $$Since$$u''+p_\infty(t) u'+f(u)=0,\quad u(0)=1,\;\;u'(0)=-\epsilon_1 $$(and it is easy to see that proposition 3.1 still applies to solutions in the Carath\'eodory sense) there exists \tilde t such that u(\tilde t)=a/4. Since u_m\to u uniformly in [0,\tilde t] and \bar t-t_m\to+\infty this contradicts (8) and so the Claim holds. To go on with the proof we observe that if \delta>0 is sufficiently small we have u(\bar t, t_{m_2},\delta)>a/2, since u(.,t_{m_2},\delta)\to a as \delta\to0^+ in [t_{m_2},\bar t]. By the intermediate value theorem we can pick up 0<\epsilon_2<\epsilon_1  such that u(\bar t, t_{m_2},\epsilon_2)=a/2. This argument can be iterated so as to construct decreasing sequences \tau_k=t_{m_k} and \epsilon_k with the property that u(\bar t, \tau_k,\epsilon_k)=a/2. Using again the boundedness of u(.,\tau_k,\epsilon_k) and u'(.,\tau_k,\epsilon_k) and the diagonal procedure we can pass to a subsequence (which for convenience is denoted by the same symbol) so that for any compact interval K\subset{{\mathbb R}},$$u(.,\tau_k,\epsilon_k)\to u\;\;\mbox{in}\;\; C^1(K).$$The limit function u thus obtained is, of course, a decreasing solution to (1), such that u(\bar t)=a/2 and 00 sufficiently small such solutions are defined in [t_0,\infty), their components being strictly decreasing and vanishing at +\infty. Now take a sequence t_m\to-\infty and u(.,t_1,\epsilon_1) where \epsilon_1 is small. Selecting the first component, u_1, we easily establish, as in the proof of theorem 3.2, that there exists \bar t and subsequences \tau_k=t_{m_k}\to-\infty,\;\epsilon_k\to0^+) so that$$u_1(\bar t, \tau_k,\epsilon_k)=a_1/2\eqno(9) $$(it is sufficient to argue as in the proof of theorem 3.2 with respect to the equation for the first component). Next consider the sequence u_2(.,\tau_k,\epsilon_k) and let s_k be numbers such that$$u_2(s_k, \tau_k,\epsilon_k)=a_2/2.$$We claim that the sequence s_k-\bar t is bounded: for suppose for instance that along a subsequence s_k-\bar t\to+\infty (the case s_k-\bar t\to-\infty is analogous); integrating the second equation of the system (2) in [\bar t, s_k] we obtain \begin{eqnarray*} u'_2(s_k,\tau_k,\epsilon_k)-u'_2(\bar t,\tau_k,\epsilon_k)+p_2(t^*_k)(u_2(s_k,\tau_k,\epsilon_k)-u_2(\bar t, \tau_k,\epsilon_k))&&\\ +\int_{\bar t}^{s_k}u_2(t,\tau_k,\epsilon_k)g_2(u(t,\tau_k,\epsilon_k)) \,dt&=&0\,, \end{eqnarray*} where t^*_k\in[\bar t,s_k]. Now the first factor in the integrand is greater than a_2/2; using (H1) we see that the second is bounded away from zero (because u_1 takes values a_2/2); therefore we have reached a contradiction. Since this argument can be repeated with respect to the remaining components, along with (9) we construct sequences s_k^{(j)}, j=2,\cdots,n such that$$u_j(s_k^{(j)}, \tau_k,\epsilon_k)=a_j/2\eqno(10)$$and s_k^{(j)}-\bar t is bounded. Now, as in theorem 3.2 we go to the limit through a diagonal subsequence: u(.,\tau_k,\epsilon_k)\to v uniformly in compact intervals, and v is a solution of (2) with decreasing components. Moreover we may assume that s_k^{(j)}\to t_j, j=2,\cdots,n and therefore on account of (9)-(10) we obtain$$v_1(\bar t)=a_1/2,\quad v_j(t_j)=a_j/2,\;\;j=2,\cdots n.\eqno(11) $$We assert that 00 with v'_1(t_k)\to0 and \vert v_1'(s_k)\vert\geq\delta. Multiplying the first equation in (2) by v_1' and integrating yields$$\frac{1}{2}[v_1'^2(s_k)-v_1'^2(t_k)]+\int_{t_k}^{s_k}p_1v_1'^2 +\int_{t_k}^{s_k}f_1(v)v_1'=0, $$where, by the mean value theorem and what has been already proved, the last summand tends to 0 as k\to\infty; this is a contradiction and the proof is complete. \paragraph{Remark.} As in theorem 3.2, we could use a set of conditions like (7) to improve the lower bounds on c_i,\quad i=1,\dots, n in case the functions g_i approach 0 as u_i\to0. \section{Positive solutions vanishing at the endpoints of an unbounded interval} In this section we consider f satisfying \begin{enumerate} \item[(H3)] For each i\in{1,\cdots n}, f_i:{R_+}^n\rightarrow{{\mathbb R}_+} is a locally Lipschitz continuous function such that f_i(0)=0 and f_i(u)>0 if u_i>0. \end{enumerate} We consider the problem of finding {\it nontrivial} positive solutions to$$\displaylines{ \hfill u_i''+p_i(t)u_i'+f_i(u)=0,\quad i=1,\dots, n\hfill\llap{(12)}\cr \hfill u(0)=0=u(+\infty)\hfill\llap{(13)} }$$where by {\it nontrivial} we mean that each component u_i of such solution is positive in (0,+\infty). For definiteness the initial endpoint is taken to be t_0=0, but our results can obviously be restated with an arbitrary left endpoint. \smallskip Before stating the result we note the following fact: denote by u(\cdot ,A) the solution of the Cauchy problem$$u_i''+p_i(t)u_i'+f_i(u)=0,\quad u_i(0)=0,\;\;u_i'(0)=A;$$then u_i(\cdot , A) has, for every A>0, a maximum \mu_i(A) depending continuously on A and \mu_i(0^+)=0, \mu_i(+\infty)=+\infty. To see this, take t^*>0 in a neighborhood of 0, A_i^*=u_i'(t^*,A)>0, u_i^*=u_i(t^*,A)>0. Let K_i be the least upper bound of the (scalar) solution of$$z''+p_i(t)z'=0, \quad z(t^*)=u_i^*,\;\;z'(t^*)=A_i^*, $$and define \delta_i:=\inf\{g_i(x):\;\;u_i^*\leq x_i \leq K_i;\;x_j\leq K_j \;\;\mbox{if}\;j\neq i\}>0. Comparing u_i(\cdot,A) with the solution to$$v''+p_i(t)v'+\delta_i v=0, \quad v(t^*)=u_i^*,\;\;v'(t^*)=A_i^*,  it is easy to see, using Lemma 2.1, that (since $v\leq z$) $u_i(t,A)\leq v(t)$ as long as $u_i(t,A)>u_i^*$. But the behavior of $v(t)$ implies that $v(t)$ returns to the value $u_i^*$ and therefore $u_i(t,A)$ attains a maximum. The assertion about $\mu_i(0^+)$ comes from the fact that $K_i\to0$ as $A\to0^+$. The other assertion is straightforward. \paragraph{Proposition 4.1} {\sl Assume (H2)-(H3). Assume $p$ is bounded in $[0,\infty)$ and let $c_i:={\displaystyle\inf_{t\geq0}p_i(t)}$. Given positive numbers $\alpha_i$, $i=1,\cdots,n$ such that $M_i:=\sup\{g_i(u): \;\;00$ such that whenever $00:\;\;(\mu_1(A),\cdots,\mu_n(A))\in(0,\alpha_1] \times\cdots\times(0,\alpha_n]\}$. Then if \$0