Let $L$ be a second order linear elliptic differential operator on an open bounded subset $U\subset \mathbb R^n$, with smooth uniformly bounded coefficients. Suppose the boundary of $U$ is $C^\infty$. Suppose $f\in C_c^\infty(U)$ ($f$ is smooth and has compact support in $U$). Must there exist a solution $u$ to the PDE $Lu = f$, $u|_{\partial U} = 0$ such that $u$ extends to be $C^\infty$ on the closure $\bar U$?

  • $\begingroup$ I guess the ellipticity assumption is uniform too? $\endgroup$
    – user53153
    Feb 2, 2013 at 7:41
  • $\begingroup$ Yeah, the coefficients are uniformly elliptic on $U$. $\endgroup$
    – user15464
    Feb 2, 2013 at 13:35
  • 1
    $\begingroup$ How about $L=-\Delta+\lambda I$ (meaning $Lu=-\Delta u +\lambda u$, for $\lambda \in \mathbb{R}$) and $f=0$. The only $\lambda$ for which there is a solution are the Laplacian's eigenvalues, which form a countable set. On the other hand if a solution exists then it's a weak solution so regularity gives the $C^\infty$ extension up to the boundary. $\endgroup$
    – Jose27
    Feb 2, 2013 at 18:10

1 Answer 1


Yes, this is the Schauder existence theory (see, for example, Gilbarg and Trudinger, Section 6.3 or so, or if that is not readily available, the Wikipedia article http://en.wikipedia.org/wiki/Schauder_estimates has a good summary).

Applying it once will give you $C^{2,\alpha}$ estimates. Then you subsequently apply the theory to the first derivatives, recognizing that differentiating the equation gives you 2nd order elliptic operator in the first derivative, and so on for higher order terms.

  • $\begingroup$ Can we necessarily find a solution in $C_c^\infty(U)$? $\endgroup$
    – user15464
    Feb 10, 2013 at 23:08
  • $\begingroup$ The solution will probably not be compactly supported in $U$, but will take on 0 along $\partial U$. In fact, it's probably not possible for $u$ to be compactly supported inside $U$, due to the unique continuation properties of solutions to elliptic PDEs. $\endgroup$
    – Ray Yang
    Feb 10, 2013 at 23:18
  • $\begingroup$ I take that last sentence back. It's possible for $u$ to be compactly supported, but it would not be the general case - the unique continuation theorems I was thinking of do not apply when there is a source term. In general, think of what happens if $f$ is your favorite bump function, and if the domain $U$ is really large - then as you get far away from the support of $f$, you would expect $u$ to resemble an appropriate multiple of the fundamental solution .... $\endgroup$
    – Ray Yang
    Feb 11, 2013 at 19:27

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