PDE without finite time blow up for small initial data? Is there a pde (or a class of pde), for which, having small initial data, is a necessary and sufficient condition for its solution to not have a finite time blow up?
 A: We can reverse engineer such a PDE by using ODE and the method of characteristics.
Let's start with the ODE part.  Consider the function $f: \mathbb{R} \to \mathbb{R}$ given by 
$$
f(x) = 
\begin{cases}
-(x+\alpha)^2 & \text{if }x \le -\alpha \\
0 & \text{if }-\alpha < x < \alpha \\
(x-\alpha)^2 & \text{if } x \ge \alpha
\end{cases}
$$
for some fixed $\alpha >0$.  It's easy to see that a solution to 
$$
\begin{cases}
\dot{z}(t) = f(z(t)) \\
z(0) = z_0
\end{cases}
$$
will exist for all time (and be stationary) if $z_0 \in [-\alpha,\alpha]$ and will blow up in finite time if $\vert z_0 \vert > \alpha$.
For the PDE part we build a first order equation whose characteristic ODE is as above.  For the sake of simplicity we do this by picking $a \in \mathbb{R}^n \backslash \{0\}$.  Then the PDE 
$$
\begin{cases}
\partial_t u(x,t) + a \cdot \nabla u(x,t) = f(u(x,t)) \\
u(x,0) = g(x)
\end{cases}
$$
will not blow-up if and only if $\Vert g \Vert_{L^\infty} \le \alpha$.  This can be verified using the method of characteristics.  One can build other examples using this template and playing with the operator on the LHS.  For example if $a = a(x)$ is smooth with bounded derivative, then a similar argument will work.
