Existence of an extremum for the solution of the ODE $\ddot{x}+\frac32x^2=0$ Consider the 2nd order ODE
$$
\ddot{x}+\frac32x^2=0.
$$
Denote by $u$ the maximal solution of the associated Cauchy problem with initial condition $(x(0),\dot{x}(0))=(0,1)$.
The problem is to prove the existence of some $T>0$ such that $\dot{u}(T)=0$.
Any help will be appreciated.

Remark: So far I know that if $I$ denotes the domain of existence of $u$, then
$$
\dot{u}^2(t)+u^3(t)=1\quad \forall t \in I.
$$
In particular, if such a $T$ exists then $u(T)=1$.
 A: Let $(T_u^-,T_u^+) \ni 0$ be the maximal domain of existence for the solution $u$, and suppose $\dot{u}(t)\ne 0$ for every $t \in [0,T_u^+)$. Since $\dot{u}(0)>0$ and $\dot{u}$ is continuous, it follows that $\dot{u}>0$ on $[0,T_u^+)$.
Using the Hamiltonian nature of the original DE we have
$$\tag{1}
\dot{u}^2(t)+u^3(t)=1 \quad \forall t \in (T_u^-,T_u^+).
$$
It follows that the restriction $u_0$ of $u$ to $[0,T_u^+)$ solves the problem
$$\tag{2}
\dot{x}=\sqrt{1-x^3},\ x(0)=0.
$$
If $u_1:(t_1,t_2) \to \mathbb{R}$ denotes the maximal solution for (2), then its restriction to $[0,T_u^+)$ is precisely $u_0$. Hence
$$
T_u^+\le t_2=\int_0^1\frac{dx}{\sqrt{1-x^3}} \in (0,\infty).
$$
If $T_u^+<t_2$, then 
$$
\tilde{u}:(T_u^-,t_2) \to \mathbb{R},\ \tilde{u}(t)=\begin{cases}
u(t)&\text{ for }\ t \in (T_u^-,0]\\
u_1(t)&\text{ for }\ t \in (0,t_2)
\end{cases}
$$
is a solution of the original problem which is defined on a larger interval than $(T_u^-,T_u^+)$, contradicting the maximality of $(T_u^-,T_u^+)$. Hence we necessarily have $t_2=T_u^+$, and on
$[0,T_u^+)$ the solution is given by $u(t)=G^{-1}(t)$, where 
$$
G(x)=\int_0^x\frac{ds}{\sqrt{1-s^3}}
$$
NB. $G$ is continuous on $(-\infty,1]$ and differentiable on $(-\infty,1)$.
Using the definition of $u$ on $[0,T_u^+)$, we have
$$
\lim_{t\uparrow T_u^+}u(t)=1,
$$
and thanks to (1) we get
$$ 
\lim_{t\uparrow T_u^+}\dot{u}(t)=0.
$$
Let us now consider the IVP
$$\tag{3}
\ddot{x}+\frac32x^2=0,\ x(T_u^+)=1,\ \dot{x}(T_u^+)=0,
$$ 
or equivalenty
$$\tag{4}
(\dot{x},\dot{y})=f(x,y),\ (x(T_u^+),y(T_u^+))=(1,0),
$$
where 
$$
f: \mathbb{R}^2 \to \mathbb{R}^2,\ f(x,y)=(y,-\frac32x^2)
$$
is $C^1$ ($C^\infty$ actually), and therefore locally Lipschitz. Thus we can solve (4) (at least locally), i.e. we have a solution 
$$
u_2:(T_u^+-\delta, T_u^++\delta) \to \mathbb{R}
$$
for (4), where $\delta>0$. Gluing $u$ with $u_2$ at $t=T_u^+$ we obtain a solution of the original problem which is defined on $(T_u^-,T_u^++\delta)$. Again this contradicts the maximality of $(T_u^-,T_u^+)$. Thus, there must exists some $T>0$ such that $\dot{u}(T)=0$.
A: As $\dot x(0)=1$ there is a neighborhood $U$ of $t=0$ such that $\dot x(t)\ne0$ for $t\in U$. Within $U$ the given DE is equivalent with
$$0=2\dot x\ddot x + 3x^2\dot x=(\dot x^2+ x^3)^\cdot \ ,$$
therefore $\dot x^2+x^3$ is constant in $U$, and as $x(0)=0$, $\ \dot x(0)=1$ this constant has to be $1$. It follows that the solution of the given initial problem is the solution of the first order initial value problem
$$\dot x=\sqrt{1-x^3}\ ,\qquad x(0)=0\ ,\tag{1}$$
at least as long as $\dot x(t)>0$.
From looking at the DE  it is obvious that the  solution $t\mapsto x(t)$ of $(1)$ will in a primary phase $[0,T[\ $ increase from $0$ to $1$. During this phase we have $\dot x(t)>0$. If $T=\infty$ we are done; otherwise we have to investigate what happens as $t\to T\!-$. In order to determine $T$ we separate the variables in $(1)$, i.e., we write the DE  in the form
$$dt={dx\over\sqrt{1-x^3}}\ .$$
From this we derive
$$T=\lim_{\xi\to 1-}\int_0^\xi{dx\over\sqrt{1-x^3}}<\int_0^1{dx\over\sqrt{1-x^2}}={\pi\over2}<\infty\ .$$
(In his answer Did has given the exact value of $T$.)
So we are left with the question what happens at time $T$. Note that $\lim_{t\to T-} x(t)=1$ and $\lim_{t\to T-}\dot x(t)=0$; furthermore it is easy to verify that $\lim_{t\to T-}\ddot x(t)=-{3\over2}$. Our solution is a solution of the original DE $\ddot x +{3\over2}x^2=0$; it therefore should be possible (see below) to identify it with the solution to the initial value problem
$$\left\{\eqalign{\dot x&=y\cr \dot y&=-{3\over2} x^2\cr}\right.\quad ,\quad x(T)=1, \quad y(T)=0\ .\tag{2}$$
 The solution of $(2)$ is by general principles an analytical function of $\tau:=t-T$ in a neighborhood of $\tau=0$. Comparing coefficients one gets the expansion
$$x(T+\tau)=1-{3\over4}\tau^2 +{3\over16}\tau^4 -{3\over64}\tau^6 +{9\over896}\tau^8 + \ ? \tau^{10}\ .$$
If the indicated identification can be done this would mean that our solution $x(\cdot)$ can be continued past $t=T$ (albeit with another constant when we reduce to a first order problem).
As a final note I'd like to remark that even for the linear DE $\ddot x+x=0$ we run into the same difficulties if we don't know about trigonometric functions.
A: Multiply by $2\dot{x}$
$$
2\dot{x}\ddot{x}+3x^2\dot{x}=0\tag{1}
$$
and integrate (using the initial conditions) to get
$$
\dot{x}^2+x^3=1\tag{2}
$$
Since $\ddot{x}\le0$, $\dot{x}$ starts at $1$ and decreases. While $\dot{x}>0$, $x$ increases from $0$. In fact, $\dot{x}=\sqrt{1-x^3}$. Should $x$ increase to $1$ at finite time $T$, we then have $\dot{x}=0$ and $\ddot{x}=-\frac32$. This precludes the solution to $(2)$ where $x=1$ for $t\ge T$. Thus, for $t\gt T$, $\dot{x}<0$ and $x$ decreases from $1$. In fact, $\dot{x}=-\sqrt{1-x^3}$ and we get that $x(2T)=0$.
$x$ does indeed increase to $1$ in finite time. In fact, the time from $x=0$ to $x=1$ is
$$
\begin{align}
T
&=\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^3}}\\
&=\frac13\int_0^1u^{-2/3}(1-u)^{-1/2}\,\mathrm{d}u\\
&=\frac13\mathrm{B}\left(\frac13,\frac12\right)\\
&=\frac13\frac{\Gamma\left(\frac13\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac56\right)}\\
&\stackrel.=1.4021821053254542612\tag{3}
\end{align}
$$
As mentioned above, $\dot{x}(T)=0$ and $x(2T)=0$.
