# Second-order elliptic regularity with rough coefficients

Let $$\Omega \subseteq \mathbb R^n$$ be a bounded open set with smooth boundary. Let $$k\geq 1$$, $$\alpha\in(0,1)$$, $$a_{ij},b_i,f \in C^{k,\alpha}(\Omega)$$ for $$i,j=1,...,n$$, and define the operator

$$L = \sum_{i,j=1}^n a_{ij} \partial_{ij} + \sum_{i=1}^n b_i \partial_i.$$

Assume further that $$L$$ is uniformly elliptic, i.e. $$\sum_{i,j} a_{ij}(x) \xi_i\xi_j\geq \lambda \|\xi\|^2$$ for all $$x \in \Omega$$, $$\xi\in\mathbb R^n$$, and some $$\lambda > 0$$. Standard elliptic regularity theory ensures that the Dirichlet problem $$L u =f$$ in $$\Omega$$, with $$u=0$$ on $$\partial\Omega$$, admits a solution $$u \in C^{k+2,\alpha}$$, whose Holder norm depends only on that of the above data.

I was wondering whether there are any known conditions under which the regularity of the coefficients $$b_i$$ can be weakened from $$C^{k,\alpha}$$ to $$C^{k-1,\alpha}$$, while retaining the existence of a solution $$u \in C^{k+2,\alpha}$$. It is unclear to me whether the proof of Schauder's estimates based on reduction to constant-coefficient equations (e.g. Theorem 13.2.1. of Jost's text) can be adapted to this setting under any sensible conditions. (While my question is general, I will note that $$f$$ happens to be of class $$C^{k+2,\alpha}$$ in my particular use case.)

• I had exactly something like this come up once. I had solutions on a cube that were reflections and the $b$ term didn't have enough regularity to get what I wanted. But when you viewed the product of $b$ and the first derivatives you could apply the Holder theory you wanted. But of course this was very special case. Dec 25, 2021 at 22:52
• @Math604 Thank you. Even though your example is quite specific, I would be interested to read it in more detail if you have time to write it as an answer. Dec 27, 2021 at 4:57
• I think i had something like $-\Delta u(x) + b(x) u_{x_1} = f(u)$ and $b$ was not smooth enough across say a hyperplane like $x_1=0$ to apply the regularity theory to get $u$ was $C^{2,\alpha}$. But $u$ was an even reflection across $x_1=0$ and hence $u_{x_1}=0$ on the hyperplane. WHen you examined exactly the term $b(x) u_{x_1}$ it was holder continuous and hence you could get what you wanted. Dec 27, 2021 at 5:24

In dimension $$d=1$$, let's try $$u^{\prime\prime}+b u^\prime =0,$$ a solution is $$u^\prime = \exp\left({-\int_0^x b(t) \textrm{d} t}\right)$$ So the regularity of $$u^\prime$$ is that of $$b$$,+1, and that of $$u$$ is that of $$b$$,+2. So you cannot get regularity of $$b$$+3 in general.
Based on this example, you would need truly miraculous cancelations between $$b$$ and $$a$$ for things to work out exactly the right way. In fact, $$a_{ij}u_{,ij}=-b_{i}u_{,i}+f$$ means that if $$u_{,ij}$$ and $$a_{ij}$$ are $$C^{k,\alpha}$$ as well as $$f$$, then so is $$b_{i}u_{,i}$$. So $$u_{,i}$$ should cancel at every not $$C^{k,\alpha}$$ point for $$b_i$$, that's asking a lot. Limiting to the case of finitely many such points, your problem locally very much look like the one dimensional problem I wrote above, and the solutions will behave accordingly..so I venture that the answer is no.