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.)

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.