Let $\Omega\subset\mathbb{R}^d$ be open,and $P(D)=\operatorname{\sum_{|\alpha|\le N}}f_{\alpha}D^{\alpha}$ be an elliptic differential operator. Rudin proves in Functional Analysis Part II the Regularity Theorem for Elliptic Operators that if $f_{\alpha}$ are smooth and $f_{\alpha}$ are constants for $|\alpha|=N$, then $P(D)u$ is locally in $H_s(\Omega)$ if and only if $u$ is locally in $H_{s+N}(\Omega)$, where \begin{equation} H_s=\{u\in D'(\mathbb{R}^d):(1+|y|^2)^{s/2}\hat{u}\in\mathcal{L}^2\}. \end{equation}

I know almost nothing about partial differential equations, and am ignorant of special examples especially. I need some examples showing the conclusion is false for other types of differential operators. That is, something like $Lu=v$ where $v$ is good but $u$ is bad when $L$ is not elliptic.



Some trivial examples are always useful. We work on $\mathbb{R}^2$, with coordinates $(x,y)$. One can apply partial differential operators to distributions, and in particular to functions in $L^1_{\text{loc}}(\mathbb{R}^2)$.

  1. Take $D = \frac{\partial}{\partial x}$. Take any function as bad as you want, that depends only on $y$, and still belongs to $L^1_{\text{loc}}(\mathbb{R}^2)$. Then, $Du = 0$, the nicest, smoothest function you could possibly imagine, but still, $u$ is ugly.
  2. Take $D = \frac{\partial^2}{\partial x \partial y}$. Then any distribution $u$ has $Du = 0$, but this says nothing about $u$.
  3. Take $D = \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2}$. The equation $Du = 0$ is the one dimensional wave equation. Any function $u$ of the form $u(x,y) = F(x - y) + G(x + y)$, where $F$ and $G$ are arbitrary, ugly as you want, functions solves $Du = 0$ formally. Not all such functions can be understood as distributions, but those that can, and for which you can perform integration by parts, yield examples of non-smooth distribution solutions. $D$ is an example of a hyperbolic operator.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.