Content-Type: multipart/mixed; boundary="-------------0003311252890" This is a multi-part message in MIME format. ---------------0003311252890 Content-Type: text/plain; name="00-140.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-140.keywords" second order systems, blow-up ---------------0003311252890 Content-Type: application/x-tex; name="blowup.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="blowup.tex" \input vanilla.sty \def\rect{\hskip 0.1truecm\vbox{\hrule\hbox{\vrule \hskip .2truecm\vbox{\vskip .2truecm}\vrule} \hrule}\hskip 0.1truecm} \pagewidth{13cm} \pageheight{19cm} \normalbaselineskip=15pt \normalbaselines \parindent=15pt \parskip=3pt \mathsurround=1.7pt \overfullrule=0pt \scaledocument{\magstep1} \def\tit#1{\bigskip \bf\noindent #1 \medskip\rm} \def\lema#1{\medskip\smc\noindent #1\quad\sl} \def\demost#1{\smallskip\noindent\underbar{\it #1}\quad\rm} \def\note#1{\medskip \smc\noindent #1\quad\rm} \def\lra{\longrightarrow} \def\noi{\noindent} \def\li#1#2{\smash{\mathop{#1}\limits\sb{#2}}} \def\lie#1#2#3{\smash{\mathop{#1}\limits\sb{{\scriptstyle #2}\atop{\scriptstyle#3}}}} \def\cor{\allowmathbreak} \def\inb{\int\sb B } \def\dde{\sober{$d$}/{$d\va$}} \def\R{\text{{\rm I}\!{\rm R}}} \def\N{\text{{\rm I}\!{\rm N}}} \def \ee{\varepsilon} \def \a{\alpha} \def \sin {\text {\rm sin }} \def \cos {\text {\rm cos }} \title {Nonlinear problems for a second order ODE} \endtitle \centerline {P.Amster and M.C. Mariani} \medskip \medskip \lema{Abstract: } We study the general class of semilinear second order ordinary differential equations $u''(t)+r(t) u'(t) + g(t,u(t)) = f(t)$ with a fixed constraint $u(0) = u_0$. Under a growth condition on $g$ we prove the existence of solutions satisfying the nonlinear condition $u(T)=h(u'(T))$. Moreover, we give conditions in order to assure that any solution satisfying a Cauchy condition $u(0) = u_0, \quad u'(0)=v_0$ is defined over $[0,T]$. \tit {Introduction} We'll study the second order ODE $$u''(t)+r(t) u'(t) + g(t,u(t)) = f(t)\tag{*}$$ with a fixed initial data $u(0) = u_0$. In the first section we'll state the basic assumptions and results concerning the Dirichlet problem associated to (*), which will be applied in the following sections. In the second section we'll define a fixed point operator in order to solve a problem for the final position $u(T)$ depending on the speed at time $T$. Furthermore, we'll prove that if $g$ satisfies a growth condition (which holds, for example, if $g$ is {\it sublinear}) then there exist a class of $h's$ such that equation (*) admits at least one solution $u$ with $u(0)=u_0$, $u(T)=h(u'(T))$. Finally, in the third section we'll prove that the non-existence of a blow-up in the interval $[0,T]$ for a solution of (*) with initial value $u_0$ is equivalent to the solvability of the equation $\psi(s)=u'(0) \in I$, where $\psi: \R\to\R$ is a continuous function which depends on $u_0$. Furthermore, if $g$ is locally Lipschitz on $0\times \R^2$ then the disjoint union over $u_0$ of the sets $\{ u_0\} \times I(u_0)$ is an open subset of $\R^2$. \tit {1. Basic assumptions and unique solvability of the Dirichlet problem} Let $S:H^2(0,T)\to L^2(0,T)$ be the semilinear operator given by $Su= u''+ru' + g(t,u)$. We'll assume throughout the paper that $g$ is continuous and satisfies the condition $$\frac {g(t,u) - g(t,v)}{u-v}\le c < \left({\pi\over T}\right)^2 \qquad \text { for any } t\in [0,T], u,v \in \R, u\ne v \tag{G1}$$ Moreover, we'll assume that the friction coefficient $r\in H^1(0,T)$ is nondecreasing. We recall the following results, which establish the existence of an apriori bound for $S$ and the solvability of the problem $Su=f$ for arbitrary $f\in L^2(0,T)$ and Dirichlet conditions. Furthermore, the set $S^{-1}(f)\subset H^2(0,T)$ of all the solutions of $Su=f$ is homeomorphic to $\R^2$, thoght as the set of all possible Dirichlet boundary data. The proofs may be found in [AM]: \lema {Lemma 1} Let $u, v\in H^2(I)$ with $u=v$ on $\partial I$. Then $$\| Su-Sv\|_2 \ge \left(({\pi\over T})^2-c\right)\| u-v\|_2$$ and $$\| Su-Sv\|_2 \ge \frac {({\pi\over T})^2-c}{{\pi\over T}} \| u'-v'\|_2$$ \lema {Theorem 2} The Dirichlet problem $$ \cases Su=f(t) \qquad\text { in } (0,T) &\\ u(0)=u_0, \quad u(T)=u_T & \endcases $$ is uniquely solvable in $H^2(I)$ for any $f\in L^2(I)$. \lema {Theorem 3} Let $f\in L^2(0,T)$ and $Tr:S^{-1}(f)\to \R^2$ the restriction of the usual trace function, i.e. $Tr(u)= (u(0),u(T))$. Then $Tr$ is an homeomorphism. \rm \tit {2. Nonlinearities at the endpoint} In this section we'll study the problem $$\text{(1)}\cases u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\ u(0)=u_0,\quad u(T)=h(u'(T)) & \endcases $$ for fixed $f\in L^2(0,T)$ and continuous $h$. First we'll transform the problem in a one-dimensional fixed point problem: indeed, for $s\in \R$ theorem 2 allows us to define $u_s$ as the unique solution of the problem $$\cases u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\ u(0)=u_0,\quad u(T)=h(s) & \endcases $$ Hence, if $\varphi_s(t)= \frac {h(s)-u_0}T t + u_0$, we have that $$u_s(t)- \varphi_s(t) = \int_0^T (f -ru_s'-g(\theta), u_s'))G(t,\theta)d\theta$$ where $G$ is the Green function asociated to the second-derivative operator $\partial^2$, namely $$G(t,\theta) = \cases {t(\theta-T)\over T} \qquad \text { if } \theta \ge t &\\ {\theta(t-T)\over T} \qquad \text { if } \theta \le t & \endcases $$ By a simple computation we obtain that $$u_s'(T) = \frac {h(s)-u_0}T + \int_0^T (f -ru_s'-g(\theta), u_s')){\theta\over T}d\theta$$ and by theorem 3 we have: \lema {Lemma 4} Let $\xi:\R \to \R$ be given by $$\xi (s) = \frac {h(s)-u_0}T + \int_0^T (f -ru_s'-g(\theta, u_s')){\theta\over T}d\theta$$ Then $\xi$ is continuous. Moreover, $u$ is a solution of (1) if and only if there exists $s\in \R$ such that $s$ is a fixed point of $\xi$. \demost{Proof} Continuity of $\xi$ follows immediately from the continuity of $Tr^{-1}: \R^2 \to S^{-1}(f) $. Moreover, if $\xi(s)=s$, then $u_s(T)=h(u_s'(T))$, proving that $u_s$ is a solution of (1). Conversely, if $u$ is a solution of (1), then $u=u_s$ for $s=u'(T)$. \hfill\rect \medskip In order to state an existence result for (1), we'll give a different formulation for $\xi$. Indeed, as $$\int_0^Tr(\theta)u_s'(\theta) \theta d\theta= r(\theta) \theta u_s(\theta) \Big|_0^T - \int_0^T[r(\theta) \theta]' u_s(\theta) d\theta$$ $$=r(T)Th(s) - \int_0^T[r(\theta) + \theta r'(\theta)] u_s(\theta) d\theta $$ we obtain: $$\xi (s) = (\frac 1T - r(T)) h(s) + \frac 1T \left[\int_0^T \theta f(\theta)d\theta - u_0\right] +\frac 1T \int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta$$ Hence, $s$ is a fixed point of $\xi$ if and only if $$sT= (1-r(T)T)h(s) + \int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta + \int_0^T \theta f(\theta)d\theta - u_0 \tag2$$ \lema{Lemma 5} Let $g$ satisfy the growth condition $$|g(t,x)| \le \a |x| + \beta \tag{G2}$$ for some positive constants $\a$ and $\beta$. Then $$\| u_s\|_2 \le \sqrt{\frac T3} c(s) + \frac {T^2}{\pi^2 - cT^2} \left(\a \sqrt{\frac T3} c(s) + |h(s)-u_0| \frac {\| r\|_2}T +\| f\|_2 + \beta\right):=A(s)$$ where $c(s)= \sqrt{h(s)^2+h(s)u_0+u_0^2}$. In particular, $$\| u_s\|_2 \le c_T|h(s)| + \gamma |h(s)|^{1/2}+ \delta$$ for some constants $\gamma$, $\delta$ and $$c_T= \sqrt{\frac T3} + \frac {T^2}{\pi^2 - cT^2} \left(\a \sqrt{\frac T3} + \frac {\| r\|_2}T\right)$$ \demost {Proof} By lemma 1, we have that $$\| u_s-\varphi_s\|_2\le \frac {T^2}{\pi^2 - cT^2} \| Su_s - S\varphi_s\|_2 = \| f - r\varphi_s' - g(\cdot ,\varphi_s)\|_2 $$ A simple computation shows that $\|\varphi_s\|_2 = \sqrt{\frac T3} c(s)$, and as $\varphi_s' = \frac {h(s)-u_0}T$, and $\|g(\cdot ,\varphi_s)\|_2 \le \a \| \varphi_s\|_2 + \beta$, the result follows. \hfill\rect \lema {Theorem 6} Let us assume that (G2) holds, and define the functions $\xi^\pm:\R\to \R$ as $$\xi^\pm(s)= \frac 1T \left( (1-r(T)T)h(s)\pm \| r+\theta r'\|_2 A(s) + \sqrt{\frac {T^3}3} (\a A(s) + \beta) + \int_0^T\theta f(\theta)d\theta - u_0\right)$$ Moreover, assume that there exist $s_\pm \in \R$ such that $s_- \le \xi^-(s_-)$ and $\xi^+(s_+) \le s_+$. Then (1) admits at least one solution $u\in H^2(0,T)$. \demost{Proof} >From the previous computations it follows that $$|\int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta| \le \| r+\theta r'\|_2\| u_s\|_2 + \sqrt{\frac {T^3}3} (\a \| u_s\|_2 + \beta)\le$$ $$ \| r+\theta r'\|_2 A(s) + \sqrt{\frac {T^3}3} (\a A(s) + \beta)$$ and then $\xi^- \le \xi\le \xi^+$. Thus, the result follows from lemma 4. \hfill\rect \lema {Remarks and Examples } We also have that $$|\int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta| \le (\| r+\theta r'\|_2 + \sqrt{\frac {T^3}3} \a) c_T|h(s)| + B(s)$$ where $B(s)$ is a small-order term. \lema {Corollary 7} Let us assume that (G1) holds and that $h$ changes sign on $\R$. Then (1) admits a solution for any $T$ small enough. \demost {Proof} As $c_T\to 0$ for $T\to 0$, then for small $T$ we have from the previous remark that $$T\xi(s) \simeq \kappa(T) h(s) + B(s),$$ where $B(s)$ is a small-order term and $\kappa \to 1$. Choosing $s_\pm \in \R$ such that $h(s_+) > 0 > h(s_-)$, it suffices to take $T$ such that $h(s_+) > s_+T$, $h(s_-) < s_-T$. \tit {3. Blow-up results} In this section we'll study the behavior of the solutions of the Cauchy problem $$\text{(3)} \cases u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\ u(0)=u_0, \quad u'(0)= v_0 & \endcases $$ As in the previous section, for every $s\in \R$ we'll define $u_s$ as a solution of a two-point boundary value problem. In this case, we'll consider $\varphi_s(t)= k(s) t + u_0$ for some $k:\R\to \R$ and $u_s$ the unique solution of $$\cases u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\ u(0)=u_0, \quad u(T)=\varphi_s(T) & \endcases $$ Then $$u_s(t)- \varphi_s(t) = \int_0^T (f -ru_s'-g(\theta), u_s'))G(t,\theta)d\theta$$ and it's easy to conclude that $$u_s'(0) = k(s) + \int_0^T (f -ru_s'-g(\theta), u_s')){\theta-T\over T}d\theta := \psi(s) $$ If we define $I(u_0)= Range (\psi)$, then \lema{Theorem 8} Let us assume that $v_0\in I(u_0)$. Then at least one solution of (3) is defined over $[0,T]$. Conversely, if (3) admits a solution defined over $[0,T]$ then $v_0\in I(u_0)$. \demost{Proof} It's clear that if $v_0=\psi(s)$, then $u_s'(0)=v_0$. The converse is also immediate. \hfill\rect \medskip \lema {Remarks} a) In particular, if $g$ is Lipschitz with respect to $u$ in a neighborhood of $(0,u_0)$ the unique solution of (3) lasts up to $T$. In this case, we'll see in next theorem that $I(u_0)$ is open. b) The existence of an interval $I(u_0)$ such that theorem 8 holds may be deduced in a direct way, by proving that the set of initial data $v_0$ such that the solution of (3) does not blow-up in $[0,T]$ is $connected$. Indeed, it suffices to see that two solutions of $Su=f, u(0)=u_0$ do not intersect before the endpoint $T$, but this is a immediate consequence of the uniqueness of theorem 2, applied to any interval $[0,T_0] \subset [0,T]$. \lema {Theorem 9} Let us assume that $g$ is locally Lipschitz on $\{0\} \times A$, where $A$ is an open subset of $\R$. Then $$\bigcup_{u_0\in A} \{ u_0\}\times I(u_0)$$ is open in $\R^2$. \demost{Proof} Let $S_A =\{ u \in H^2(0,T) : Su = f, u(0)\in A\}$, and consider the continuous mapping $\rho:S_A \to \R^2$ given by $\ rho(u) = (u(0),u'(0))$. As $g$ is locally Lipschitz, $\kappa$ is injective, and hence $Tr^{-1} o \quad\ rho: S_A \to S^{-1}(f)$ is open. As a final remark, we'll give a result which avoids our basic assumption (G1) if $g$ grows at most linearly at infinity: \lema {Theorem 10} Let us assume that $|g(t,u)|\le \a |u| + \beta$, with $\a < \left( \frac \pi T\right)^2$. Then any solution of (3) is defined over $[0,T]$. \demost{Proof} Let $u$ be a solution of (3) such that $u(t) \to +\infty$ for $t\to t_0 \le T$ and consider the linear operator $Lw:= w'' + rw'+\a w$. Then $Lu\ge f - \beta$. Choosing $v:\R\to \R$ such that $$Lv = f - \beta,\qquad v(0)=u_0, \quad v'(0) \ge v_0$$ then $v > u$ in a maximal interval $(0,A)$ with $AFrom the previous computations it follows that $$|\int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta| \le \| r+\theta r'\|_2\| u_s\|_2 + \sqrt{\frac {T^3}3} (\a \| u_s\|_2 + \beta)\le$$ $$ \| r+\theta r'\|_2 A(s) + \sqrt{\frac {T^3}3} (\a A(s) + \beta)$$ and then $\xi^- \le \xi\le \xi^+$. Thus, $s_- \le \xi(s_-)$, $\xi(s_+) \le s_+$ and the result follows from lemma 4. \hfill\rect \lema {Remarks and Examples } We also have that $$|\int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta| \le (\| r+\theta r'\|_2 + \sqrt{\frac {T^3}3} \a) c_T|h(s)| + R(s)$$ where $R(s)$ is a small-order term. Calling $M = (\| r+\theta r'\|_2 + \sqrt{\frac {T^3}3} \a) c_T$ we obtain: $$[(1-r(T)T) -M sg(h(s))] h(s) + R(s) \le T\xi(s) \le [(1-r(T)T) + M sg(h(s))] h(s) + R(s)$$ For example, the assumptions of theorem 6 are fulfilled in the following cases: i) $h$ subquadratic (i.e. $\frac {h(s)}{s^2}\to 0$ for $s\to \pm \infty$) and $$\liminf_{s\to \pm \infty} \left(sg \left(\frac {h(s)}s\right)(1-r(T)T)+M\right) |\frac {h(s)}s| < T$$ We remark that for $M > |1-r(T)T|$ this condition implies that $h$ grows at most linearly at infinity, more precisely: $$|h(s)|\le {T\over M-|1-r(T)T|} |s| + N$$ for some constant $N$. ii) $$\limsup_{s\to \pm \infty} \left(sg \left(\frac {h(s)}s\right)(1-r(T)T)-M\right) |\frac {h(s)}s| > T$$ \medskip \rm From i) and ii) we obtain the following corollary: \lema {Corollary 7} With the previous notations, let us assume that one of the following conditions holds: \noi A) $|h(s)|\le {T\over M+|1-r(T)T|} |s| + N$ for some constant $N$. \medskip \noi B) $M < |1-r(T)T|$, and B1) $sg \left(\frac {h(s)}s\right) = sg(1-r(T)T)$ for $|s|$ large B2) $|h(s)|\ge {T\over |1-r(T)T|-M} |s| - N$ for some constant $N$. \noi Then (1) admits at least one solution $u\in H^2(0,T)$. \lema {Remark:} In particular, A (resp. B2) holds if $h$ is sublinear (superlinear). \medskip \rm Moreover, if $h$ crosses the constant $u_0$ then (1) is solvable for $T$ small: \lema {Corollary 8} Let us assume that (G1) holds and that $h - u_0$ changes sign on $\R$. Then (1) admits a solution for any $T$ small enough. \demost {Proof} Writing $$T\xi^\pm (s) = (1-r(T)T) h(s) \pm B(s) + \int_0^T\theta f(\theta)d\theta - u_0$$ for fixed $s\in \R$ it's clear that $B(s)\to 0$ when $T\to 0$. Choosing $s_\pm \in \R$ such that $h(s_+) > u_0 > h(s_-)$, it follows, for small $T$, that $$T\xi (s_-) \le (1- r(T)T) h(s_-) + \left(\int_0^T\theta f(\theta)d\theta - u_0\right) + B(s_-) \le A_- + B(s_-) $$ and $$T\xi (s_+) \ge (1- r(T)T) h(s_+) + \left(\int_0^T\theta f(\theta)d\theta - u_0\right) + B(s_+) \ge A_+ - B(s_+) $$ for some $A_- < 0 < A_+$. Thus, it suffices to take $T$ such that $$A_- + B(s_-) \le \delta A_- \le Ts_-, \qquad\qquad A_+ - B(s_+) \le \delta A_+ \ge Ts_+$$ for some $\delta < 1$. \tit {3. Blow-up results} In this section we'll study the behavior of the solutions of the Cauchy problem $$\text{(3)} \cases u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\ u(0)=u_0, \quad u'(0)= v_0 & \endcases $$ As in the previous section, for every $s\in \R$ we'll define $u_s$ as a solution of a two-point boundary value problem. In this case, we'll consider $\varphi_s(t)= k(s) t + u_0$ for some $k:\R\to \R$ and $u_s$ the unique solution of $$\cases u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\ u(0)=u_0, \quad u(T)=\varphi_s(T) & \endcases $$ Then $$u_s(t)- \varphi_s(t) = \int_0^T (f -ru_s'-g(\theta, u_s'))G(t,\theta)d\theta$$ and it's easy to conclude that $$u_s'(0) = k(s) + \int_0^T (f -ru_s'-g(\theta, u_s')){\theta-T\over T}d\theta := \psi(s) $$ If we define $I(u_0)= Range (\psi)$, then \lema{Theorem 9} Let us assume that $v_0\in I(u_0)$. Then at least one solution of (3) is defined over $[0,T]$. Conversely, if (3) admits a solution defined over $[0,T]$ then $v_0\in I(u_0)$. \demost{Proof} It's clear that if $v_0=\psi(s)$, then $u_s'(0)=v_0$. The converse is also immediate. \hfill\rect \medskip \lema {Remarks} a) In particular, if $g$ is Lipschitz with respect to $u$ in a neighborhood of $(0,u_0)$ the unique solution of (3) lasts up to $T$. In this case, we'll see in next theorem that $I(u_0)$ is open. b) The existence of an interval $I(u_0)$ such that theorem 9 holds may be deduced in a direct way, by proving that the set of initial data $v_0$ such that the solution of (3) does not blow-up in $[0,T]$ is {\rm connected}. Indeed, it suffices to see that two solutions of $Su=f, u(0)=u_0$ do not intersect before the endpoint $T$, but this is an immediate consequence of the uniqueness in theorem 2, applied to any interval $[0,T_0] \subset [0,T]$. \lema {Theorem 10} Let us assume that $g$ is locally Lipschitz on $\{0\} \times A$, where $A$ is an open subset of $\R$. Then $$\bigcup_{u_0\in A} \{ u_0\}\times I(u_0)$$ is open in $\R^2$. \demost{Proof} Let $S_A =\{ u \in H^2(0,T) : Su = f, u(0)\in A\}$, and consider the continuous mapping $\rho:S_A \to \R^2$ given by $\rho(u) = (u(0),u'(0))$. As $g$ is locally Lipschitz, $\rho$ is injective, and hence $Tr^{-1} \text{o }\rho: S_A \to S^{-1}(f)$ is open. \hfill\rect As a final remark, we'll give a result which avoids our basic assumption (G1) if $g$ grows at most linearly at infinity: \lema {Theorem 11} Let us assume that $|g(t,u)|\le \a |u| + \beta$, with $\a < \left( \frac \pi T\right)^2$. Then any solution of (3) is defined over $[0,T]$. \demost{Proof} Let $u$ be a solution of (3) such that $u(t) \to +\infty$ for $t\to t_0 \le T$ and consider the linear operator $Lw:= w'' + rw'+\a w$. Taking $A>0$ such that $Lu\ge f - \beta$ for $t\ge A$ and $v:\R\to \R$ such that $$Lv = f - \beta,\qquad v(A)=u(A), \quad v'(A) \ge u'(A)$$ then $v > u$ in a maximal interval $(A,B)$ with $B