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Idea

Given a $W^{1,2}$ solution to a linear divergence form uniformly elliptic pde with bounded coefficients, standard De Giorgi-Nash-Moser theory tells us that the solution is infact (Holder) continuous. If you have better regularity away from one isolated point, say you are $C^1$ on the puncutered ball, can the solution still fail to be differentiable at that isolated point?

The situations I am most familiar with tend to be ones in which one can go from boundedness and continuity to smoothness via Schauder theory and bootstrapping. Here, we already have continuity: But can differentiability fail at a single point? TIt seems out of reach of all the theorems I've seen, which makes me suspect it is false, but I cannot be sure without a counterexample. Does anyone have any ideas?

Details

In the specific situation I am interested in, I know a bit more. I am considering the following:

$u \in W^{1,\infty}(B_1(0)) \cap C^{1,1}_{loc}(B_1(0)\setminus ${0}$)$ satisfies weakly the equation

$D_i(A_{ij}(x)D_ju) = D_ig^i$

in $B_1(0)$, where $A_{ij},g^i \in L^{\infty}(B_1(0))\cap W^{1,2}_{loc}(B_1(0)\setminus ${0}$)$.

Questions

1) Must $u$ in fact be a $C^1$ solution on $B_1(0)$?

2) What about just being differentiable at 0?

3) How about even just $u \in W^{2,p}(B_1(0))$ for some $p > 1$?

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3 Answers 3

up vote 3 down vote accepted

I believe the following is a counterexample:

Let $N=1$, $B_1(0)=(-1,1)$, $u(x)=|x|$, then $|u^\prime(x)| = 1$, $u^{\prime \prime}(x) = 2\delta_0$,

and

$u^\prime(x) = 2H(x)-1$,

where $H$ is the Heaviside function, $H \in L^\infty \cap W^{1,2}_{loc}((-1,1)\setminus \{0\})$.

Thus $u$ solves $(\frac{1}{2} u^\prime)^\prime = H^\prime$,

which is the PDE in one dimension, with $H$ in the right space, $u$ is smooth away from the origin and $u \in W^{1,\infty}(-1,1)$, but $u^{\prime\prime}$ is only a measure and so not in $L^p(-1,1)$ for any $p>1$.

In particular, $u$ is not $C^1$, $u$ is not smooth at the origin, and $u^{\prime\prime} \notin L^p(-1,1)$.

More generally, the computation extends into more dimensions, as far as I can tell.

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How do you extend it to N=2? –  Jung Wen Chen Nov 1 '13 at 18:52

The theory of removable singularities began with Laurent Véron. It was continued by H. Brézis and others.

When the elliptic equation is linear, the important ingredient is the codimension of the subset where you are lacking information. Codimension one is too big (see the counter-example in Dan Spector' answer). Usually, codimension 2 is OK, because then the subset has zero capacity.

When the equation is non-linear, for instance $\Delta u+|u|^{p-1}u=0$, the situation depends upon the exponent $p>1$ and the integrability that you a priori know about $\nabla u$ or $u$. Say that the equation is posed the punctured ball. It admits a radial solution $$u=\omega r^{-\alpha},\qquad \alpha=\frac2{p-1},\qquad\omega=(\alpha(\alpha+2-d))^{\frac1{p-1}}.$$ If you know (or assume) for instance that $u\in W^{1,2}$, then $\alpha<\frac d2-1$, that is $\frac2{p-1}<\frac d2-1$. In other words, if $p\le\frac{d+2}{d-2}$, then the singularity at origin is removable.

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With a linear elliptic PDE, there's no way to bootstrap. What you see is what you get. The regularity of $A^{ij}\partial_j u$ cannot be made any better than the regularity of $g^i$. So if all you assume about $g^i$ is that it is $L^\infty$, then that's all you get for $A^{ij}\partial_j u$. Once this observation is made, it's easy to find the simplest possible example (where $u$ is Lipschitz, $\partial u$ is $L^\infty$, and you just set $A^{ij} = \delta^{ij}$ and $g^i = \partial_iu$). And that's exactly the one found by Daniel, which works in all dimensions.

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That's a contradiction to Denis Serre's answer.. –  Jung Wen Chen Nov 1 '13 at 18:51

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