Regularity of BCs VS regularity of solutions

Question

Given a PDE, is there a relation between the regularity of the BCs and the regularity of the solutions ? I mean, can we have for example discontinuous BCs and at the same time solutions with some regularity (like for example, being $C^1$ with respect to time or something like that) ? I am very new to this field. From what I read, it seems like having the BCs in $L^2$ works for many problems, and being in $L^2$ does not mean that we are continuous.

Answer 1

This question is a bit too broad in the sense that you're asking about all PDEs, and the behavior of solutions relative to data changes greatly based on the type of PDE in question. For instance, elliptic and parabolic equations will generally have their solutions gain regularity relative to data, but this is generally not true (or at least is more subtle) for hyperbolic equations or equations without a definite type. Since you say you're new to PDE, then perhaps this is a reasonable answer: it depends, and part of our classification scheme for PDEs is based on what happens in this regard.

To give you a more concrete example, here is a classic. Given a bounded and integrable function $f : \mathbb{R}^ \to \mathbb{R}$ we can define $u : \mathbb{R}^d \times (0,\infty) \to \mathbb{R}$ via $$ u(x,t) = \int_{\mathbb{R}^d}f(y) \frac{e^{-|x-y|^2 / (4t)}}{(4\pi t)^{d/2}} dy. $$ This function can be shown (see any book on PDE) to be smooth on $\mathbb{R}^d \times (0,\infty)$ and solves the heat equation: $$ \partial_t u(x,t) = \Delta u(x,t), $$ and obeys the initial condition $$ u(x,t) \to f(x) \text{ as } t \to 0 $$ in some sense (see the books). We can certainly take $f$ to be discontinuous here, say $f = \mathbb{1}_{B(0,R)}$, but the solution will still be smooth, which is referred to as instantaneous smoothing since, in particular, $u(\cdot,t)$ is smooth (as a function of space) for all $t >0$.

On the other hand, given $a \in \mathbb{R}^d$, the function $v : \mathbb{R}^d \times [0,\infty) \to \mathbb{R}$ given by $$ v(x,y) = f(x-t a) $$ is a (weak) solution to the transport equation $$ \partial_t v(x,t) + a \cdot \nabla v(x,t) =0 $$ with initial data $v(x,0) = f(x)$. This $v$ clearly does not gain regularity on the data.

The transport equation is classified as hyperbolic, but the heat equation is parabolic.