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second order systems, blowup
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\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 nonexistence
of a blowup 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
$$\frac {g(t,u)  g(t,v)}{uv}\le c < \left({\pi\over T}\right)^2
\qquad \text { for any } t\in [0,T], u,v \in \R, u\ne v \tag{G1}$$
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
$$\ SuSv\_2 \ge \left(({\pi\over T})^2c\right)\ uv\_2$$
and
$$\ SuSv\_2 \ge \frac {({\pi\over T})^2c}{{\pi\over T}}
\ u'v'\_2$$
\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 onedimensional 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 secondderivative operator $\partial^2$, namely
$$G(t,\theta) =
\cases
{t(\thetaT)\over T} \qquad \text { if } \theta \ge t &\\
{\theta(tT)\over T} \qquad \text { if } \theta \le t &
\endcases
$$
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= (1r(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_Th(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( (1r(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_Th(s)
+ B(s)$$
where
$B(s)$ is a smallorder term.
\lema {Corollary 7}
Let us assume that (G1) holds and that $h$
changes sign on $\R$. Then (1) admits a solution for any
$T$ small enough.
\demost {Proof}
As
$c_T\to 0$ for
$T\to 0$, then
for small $T$ we have
from the previous
remark that
$$T\xi(s) \simeq \kappa(T) h(s) + B(s),$$
where $B(s)$ is a smallorder term and $\kappa \to 1$.
Choosing $s_\pm \in \R$ such that $h(s_+) > 0 > h(s_)$, it suffices to
take $T$ such that $h(s_+) > s_+T$, $h(s_) < s_T$.
\tit {3. Blowup results}
In this section we'll
study the behavior of
the solutions of the Cauchy problem
$$\text{(3)}
\cases
u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\
u(0)=u_0, \quad u'(0)= v_0 &
\endcases
$$
As in the previous section,
for every
$s\in \R$ we'll define $u_s$ as a solution
of a twopoint boundary
value problem. In this case, we'll consider
$\varphi_s(t)= k(s) t + u_0$ for some $k:\R\to \R$
and $u_s$ the unique solution of
$$\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')){\thetaT\over T}d\theta
:= \psi(s) $$
If we define $I(u_0)= Range (\psi)$, then
\lema{Theorem 8}
Let us assume that
$v_0\in I(u_0)$.
Then at least one
solution of (3)
is defined over
$[0,T]$. Conversely, if
(3) admits a solution defined over $[0,T]$ then
$v_0\in I(u_0)$.
\demost{Proof}
It's clear that if
$v_0=\psi(s)$, then $u_s'(0)=v_0$. The converse is also immediate.
\hfill\rect
\medskip
\lema {Remarks}
a) In particular,
if $g$ is
Lipschitz with
respect to $u$ in a neighborhood
of
$(0,u_0)$ the unique solution of
(3) lasts up to $T$.
In this
case, we'll see in next theorem that $I(u_0)$
is open.
b) The existence of an interval
$I(u_0)$ such that theorem 8 holds
may be deduced in a direct way, by proving that
the set of initial data $v_0$
such that the solution of (3)
does not blowup in $[0,T]$
is $connected$.
Indeed, it suffices to see
that two solutions of $Su=f, u(0)=u_0$
do not intersect
before the endpoint $T$, but this is a immediate consequence of
the uniqueness of theorem 2, applied to any interval $[0,T_0] \subset [0,T]$.
\lema {Theorem 9}
Let us assume that $g$ is locally Lipschitz on $\{0\} \times A$, where $A$ is
an open subset of $\R$.
Then
$$\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_Th(s)
+ R(s)$$
where
$R(s)$
is a smallorder term.
Calling $M = (\ r+\theta r'\_2 + \sqrt{\frac {T^3}3} \a) c_T$
we obtain:
$$[(1r(T)T) M sg(h(s))] h(s) + R(s) \le T\xi(s) \le
[(1r(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)(1r(T)T)+M\right) \frac {h(s)}s < T$$
We remark that for
$M > 1r(T)T$ this condition implies that $h$ grows at most
linearly at infinity, more precisely:
$$h(s)\le {T\over M1r(T)T} s + N$$
for some constant $N$.
ii) $$\limsup_{s\to \pm \infty}
\left(sg \left(\frac {h(s)}s\right)(1r(T)T)M\right) \frac {h(s)}s > T$$
\medskip
\rm From i) and ii) we obtain the following corollary:
\lema {Corollary 7}
With the previous notations, let us assume that one of the
following conditions holds:
\noi A) $h(s)\le {T\over M+1r(T)T} s + N$
for some constant $N$.
\medskip
\noi B) $M < 1r(T)T$, and
B1) $sg \left(\frac {h(s)}s\right) = sg(1r(T)T)$ for $s$ large
B2) $h(s)\ge {T\over 1r(T)TM} s  N$
for some constant $N$.
\noi Then (1) admits at least one solution $u\in H^2(0,T)$.
\lema {Remark:}
In particular, A (resp. B2) holds
if $h$ is sublinear (superlinear).
\medskip
\rm
Moreover, if $h$
crosses the constant $u_0$
then (1) is solvable for $T$ small:
\lema {Corollary 8}
Let us assume that (G1)
holds and that
$h  u_0$
changes sign on $\R$. Then (1) admits a solution for any
$T$ small enough.
\demost {Proof}
Writing
$$T\xi^\pm (s) = (1r(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_+$$
for some $\delta < 1$.
\tit {3. Blowup results}
In this section we'll
study the behavior of
the solutions of the Cauchy problem
$$\text{(3)}
\cases
u''+ru'+g(t,u) = f \qquad \text { in } (0,T) &\\
u(0)=u_0, \quad u'(0)= v_0 &
\endcases
$$
As in the previous section,
for every
$s\in \R$ we'll define $u_s$ as a solution
of a twopoint boundary
value problem. In this case, we'll consider
$\varphi_s(t)= k(s) t + u_0$ for some $k:\R\to \R$
and $u_s$ the unique solution of
$$\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')){\thetaT\over T}d\theta
:= \psi(s) $$
If we define $I(u_0)= Range (\psi)$, then
\lema{Theorem 9}
Let us assume that
$v_0\in I(u_0)$.
Then at least one
solution of (3)
is defined over
$[0,T]$. Conversely, if
(3) admits a solution defined over $[0,T]$ then
$v_0\in I(u_0)$.
\demost{Proof}
It's clear that if
$v_0=\psi(s)$, then $u_s'(0)=v_0$. The converse is also immediate.
\hfill\rect
\medskip
\lema {Remarks}
a) In particular,
if $g$ is
Lipschitz with
respect to $u$ in a neighborhood
of
$(0,u_0)$ the unique solution of
(3) lasts up to $T$.
In this
case, we'll see in next theorem that $I(u_0)$
is open.
b) The existence of an interval
$I(u_0)$ such that theorem 9 holds
may be deduced in a direct way, by proving that
the set of initial data $v_0$
such that the solution of (3)
does not blowup in $[0,T]$
is {\rm connected}.
Indeed, it suffices to see
that two solutions of $Su=f, u(0)=u_0$
do not intersect
before the endpoint $T$, but this is an immediate consequence of
the uniqueness in theorem 2, applied to any interval $[0,T_0] \subset [0,T]$.
\lema {Theorem 10}
Let us assume that $g$ is locally Lipschitz on $\{0\} \times A$, where $A$ is
an open subset of $\R$.
Then
$$\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