Ordinary Differential Equations whose solution has infinite jump discontinuities on its interval of existence I am just beginning to learn ordinary differential equation. My question:
Let : $t \in \mathbb{R}$, $x_0 \in \mathbb{R}$, 
Let $f:\mathbb{R} \times \mathbb{R}  \rightarrow \mathbb{R}: (t,x) \mapsto f(t,x)$.
Let $x:I \rightarrow \mathbb{R}:t \mapsto x(t)$, where  $I \subseteq \mathbb{R}$ is the maximal interval of existence such that the solution of the following ordinary differential equation initial value problem:
$\frac{d}{dt}x = f(t,x), x(0)=x_0$
exists and is unique on $I$ (well-posed in the sense of Hadamard).
Is it possible that the solution $x$ has infinitely jump discontinuities? 
Any comments, feedbacks, or inputs are very welcome. Thanks in advance. 
Note: I have removed $d$ as to make things clearer.
 A: Any solution $t\mapsto x(t)$ of the ODE $\dot x=f(t,x)$ carries with it a solution interval $I$ (you say it yourself) such that for all $t\in I$ we have $\dot x(t)=f(t,x(t))$. In particular, the function $x(\cdot)$ is continuous on $I$, so there is no room for a jump discontinuity.
Now what about the differential equation $\dot x= 1+x^2\ $? When an initial point $(t_0,x_0)$ is given there is a unique $\alpha_0\in\bigl]-{\pi\over2},{\pi\over2}\bigr[\ $ such that $\tan\alpha_0=x_0$, and it is easy to check that
$$x(t)\ :=\ \tan(t-t_0+\alpha_0)$$
satisfies the differential equation in some $t$-interval $I$ containing $t_0$ as well as the given initial condition. In order to determine $I$ we have to make sure that
$$-{\pi\over 2}<t- t_0+\alpha_0 <{\pi\over2}\ ,$$
from which we deduce $I=\ \bigl]t_0-(\alpha_0+{\pi\over2}),\ t_0+({\pi\over2}-\alpha_0)\bigr[\ $.
That the full graph of the $\tan$-function consists of infinitely many (disjoint) such curves is another matter and should not bother us here.
