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Consider $\Omega\subset\mathbb{R}^n$, open bounded, $$ Lu=f\text{ in }\Omega,\quad u=0\text{ on }{\partial\Omega}, $$ with $Lu=a^{ij}D_{ij}u+b^{i}(x)D_iu+c(x)u=f(x)$, $a^{ij}=a^{ji}$, $L$: strictly elliptic.

Q: How smooth can $u$ get if $f$, $a$, $b$, $c$ is just continuous?

Boundary can be smooth.

I am asking this because: I was studying Gilbarg--Trudinger (and Evans), and thought if there is $C$ equivalent of Sobolev estimates, and bumped into these questions, answers of which says we cannot get $u\in C^2$ if $f$ is just $C$.

Elliptic Regularity Theorem

Counterexample for the solvability of $-\Delta u = f$ for $f\in C^{0}$

Then I found Theorem 11.1.2 (a) in Partial Differential Equations, 2nd by Jost, GTM 214, which says pretty much

$\Omega\subset{\mathbb{R}^n}$: open, bounded. $\Omega_0\subset\subset\Omega$. Let $u$ be a weak solution of $\Delta u=f$ in $\Omega$.

If $f\in C(\Omega)$, then $u\in C^{1,\alpha}(\Omega)$ ($0<\alpha<1$) and $$ \|u\|_{C^{1,\alpha}}(\Omega_0)\le c(\|f\|_{C(\Omega)}+\|u\|_{L^2(\Omega)}) $$

Then Jost proceed to the discussion on variable coefficients, where $f\in C^{\alpha}$ ($0<\alpha<1$) is assumed. Can we have a similar result for non-Poisson case as well? Not only interior but also global estimate?

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shall.i.am
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  • If $f$ is continuous, you cannot expect solutions to be regular "enough". I do not remember a precise reference, but I remember that a collegue of mine began a talk with a counterexample a few years ago... – Siminore Sep 03 '15 at 11:06
  • @Siminore not regular enough as in not even $C^1$ (I said $C^1$ particularly bc in that case the weak formulation makes sense without introducing weak derivatives)? Sad. I would really love some references! – shall.i.am Sep 04 '15 at 03:10

3 Answers3

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An easy way to see this is to consider $f \in L^{\infty}_{loc}$, then the standard $W^{2,p}$ local estimates implies that $u \in W^{2,p}_{loc}$ for all $p<\infty$ and the standard Morrey embedding implies $u \in C^{1,\alpha}$ for any $\alpha<1$.

Note that the coefficients should be sufficiently regular.

Adi
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  • Just to make sure you are saying $u\in W^{2,p}(\Omega_0)$ for any $\Omega_0\subset\subset \Omega$ implies $u\in C^{1,\alpha}(\Omega)$? – shall.i.am Feb 01 '17 at 00:54
  • Sorry, I meant everything locally. Of course, you can also get estimates upto the boundary provided $\partial \Omega$ is $C^{1,1}$. I only write my answer for the local version. – Adi Feb 12 '17 at 04:32
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It's possible to construct an $f$ that is continuous such that that the equation $-\Delta u = f$ does not admit a $C^2$ solution. Rather than write out the proof, let me give you the reference you asked for above. Take a look at Exercise 4.9 in Elliptic Partial Differential Equations by Gilbarg and Trudinger.

Glitch
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    Thank you for your answer. That we cannot hope for a classical solution is understood, from the question quoted in my question. My question is more, given that we cannot get $C^2$, for example can we get $C^{1,\alpha}$, etc. – shall.i.am Sep 06 '15 at 02:03
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If $a$, $b$, $c$, and $f$ are continuous, then $u$ is $C^{1,\alpha}$ for every $\alpha<1$. This follows from the Cordes-Nirenberg estimate; see Proof of an elliptic equation.

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Xavi
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