PDE without finite time blow up for small initial data?

Question

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?

Answer 1

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.