(Edited)

If $u in H^2(B_1^+) cap {rm Lip}(B_1^+)$ satisfies

begin{cases}

{rm div}(F(x,u,nabla u)) = F_0(x,u,nabla u) quad & {rm in} B_1^+ \

u = 0 & {rm on} B_1′

end{cases}

where

$$F in C^{1,beta}(B_1^+timesmathbb{R}timesmathbb{R}^{n+1};mathbb{R}^{n+1}), quad F_0 in C^{0,beta}(B_1^+timesmathbb{R}timesmathbb{R}^{n+1};mathbb{R})$$

for some $0<beta<1$, and

$$langle D_p F(x,u,p) xi,xi rangle ge lambda(M) |xi|^2$$

for some $0 < lambda(M) < + infty$, for every $x in overline{B_1^+}$, $u in mathbb{R}$, and $|p| le M$,

then $u in C^{2,alpha}(overline{B_{1/2}^+})$ for some $0<alpha<1$.

$$B_1^+ = {x = (x’,x_{n+1}) in mathbb{R}^{n+1} : |x| < 1, , , x_{n+1} > 0}$$

is the half-ball and

$$B_1′ = {x = (x’,0) in mathbb{R}^{n+1} : |x’| < 1}$$

is the flat part of its boundary.

Also, we have $n ge 1$.

$H^2$ denotes the Sobolev Space of functions with second order weak derivatives in $L^2$ and ${rm Lip}$ is the space of Lipschitz-continuous funcions, whilst $C^{k,alpha}$ is the space of functions whose $k$-th order classical derivatives are Hölder-continuous of exponent $alpha$.

MathOverflow Asked by artful_dodger on January 3, 2022

1 AnswersThe discussion from Section 13.1 in the book of Gilbarg and Trudinger shows that $u in C^{1,,alpha}left(B_{3/4}^+right)$. From here one can apply Schauder estimates for linear equations. For example, one can pass the divergence on the left hand side and view $u$ as a solution to a non-divergence form linear equation with Hölder continuous coefficients (namely $F^i_j(nabla u)$, in the case that $F$ depends only on $nabla u$). For the relevant linear theory, see e.g. Section 5.5 from the book of Giaquinta and Martinazzi here.

Answered by Connor Mooney on January 3, 2022

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