Existence of solution for 1-D two-body problem Consider the 1-D two-body problem 
$$
\ddot x(t)=-\frac{m}{x(t)^{2}},\, t>0,\quad x(0)>0,\quad \dot x(0)=0,\quad m>0,
$$
If a solution $x$ exists, then 
\begin{equation}
x(t) = x(0)-m\int_0^t \frac{(t-s)}{x(s)^{2}}\,ds,
\end{equation}
for $0<t<\inf\{s:x(s)=0\}$.
Is either of the two formulations wellposed in some sense?
I have seen related questions, like this one, but I did not find a clear answer to the wellposedness aspect.
 A: First, we will rewrite your ODE into a first-order ODE.
Let be $X = \begin{bmatrix} x \\ \dot{x} \end{bmatrix}$ such that $\dot{X} = \begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix}$, notice that $X : I \to \Bbb R^2$.
Now, let be $f : \begin{bmatrix} x \\ y \end{bmatrix} \mapsto \begin{bmatrix} 0 \\ -\dfrac{m}{x^2} \end{bmatrix}$.
Therefore: $\dot X(t) = f\left(X(t)\right)\quad (1)$ and $X(0) = \begin{bmatrix} a \\ 0 \end{bmatrix}\quad (2)$, for some $a \in \Bbb R$.
Now, we have a Cauchy non-linear problem defined by $(1)$ and $(2)$.
In order for it, to have: a unique solution, we must verify if $f$ is locally lipschitz.
But, as $f$ is $C^1$ over $\Bbb R \times \Bbb R \setminus \{ 0 \}$, it is locally lipschitz over all compacts of $I \times K$ where $K$ is a compact of $\Bbb R \setminus \{ 0 \}$ and $I$ a subset of $\Bbb R$.
Thus, by Cauchy-Lipschitz theorem, we have a unique solution to the problem $X$ defined over some $[-\varepsilon, \varepsilon]$, now using your closed formula, you may be able to extend this solution by continuity over its maximal interval.
We checked: existence, uniqueness, all that remains is to prove the continuity of the following map, by denoting $X_{u}$ the unique solution of $(1)$ and $X(0) = u$,
\begin{equation*}
Y : u \mapsto X_{u}
\end{equation*}
We should fix some norm but we're going to show that $Y$ is Lipschitz which will be enough for any norm.
We will make us of Gronwall lemma:

Let be $f : \Bbb R \to \Bbb R$ differentiable over an interval $I$.
If: $\forall t \in I$, $\dot u(t) \leq a u(t)$, then, $t \mapsto e^{-at} u(t)$ is a nonincreasing function of $t$.

Let be $u, v$ two initial data, let be $I \times K$ some subset of $\Bbb R$ on which $f$ is locally $L$-lipschitz (those exist, we have seen it earlier) :
\begin{equation*}
\lVert \dot X_u(t) - \dot X_v(t) \rVert = \lVert f(X_u(t)) - f(X_v(t)) \rVert \leq L \lVert X_u(t) - X_v(t) \rVert
\end{equation*}
Thus, apply Gronwall's Lemma to $t \mapsto \lVert X_u(t) - X_v(t) \rVert$.
We have: $e^{-Lt} \lVert X_u(t) - X_v(t) \rVert$ non increasing function of $t$.
Thus, for all $t \in I$ on which $X_u$ and $X_v$ exists:
\begin{equation*}
\lVert X_u(t) - X_v(t) \rVert \leq e^{Lt} \lVert X_u(0) - X_v(0) \rVert = \lVert u - v \rVert
\end{equation*}
Let be $K$ a compact on which $X_u$ and $X_v$ exists, we equip our space of the sup-norm to prove the continuity of $Y$ over all compacts of $I$, let us denote $M = \sup K$.
\begin{equation*}
\lVert X_u - X_v \rVert_{\infty} \leq e^{LM} \lVert u - v \rVert
\end{equation*}
Thus:
\begin{equation*}
\lVert Y(u) - Y(v) \rVert_{\infty} \leq e^{LM} \lVert u - v \rVert
\end{equation*}
That is, $Y$ is $e^{LM}$-Lipschitz over $K$.
But, this is true for all compact $K$ of $I$.
Thus, $Y$ is locally Lipschitz over $I$, thus: $Y$ is continuous over $I$.
We have well-posedness of the two-body-problem, this could have been solved faster by just saying this is an autonomous ODE with a $f$ locally lipschitz, all those results are connected to the Cauchy-Lipschitz existence theorem, but for the sake of clarity, here they are.
