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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.)

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    $\begingroup$ 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. $\endgroup$
    – Math604
    Dec 25, 2021 at 22:52
  • $\begingroup$ @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. $\endgroup$
    – atzol
    Dec 27, 2021 at 4:57
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    $\begingroup$ 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. $\endgroup$
    – Math604
    Dec 27, 2021 at 5:24

1 Answer 1

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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.

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