[proofplan] We proceed by induction on $m$. The base case $m = 0$ is the boundary $H^2$ regularity theorem. For the inductive step, we localize near a boundary point $x_0 \in \partial U$ using a $C^{m+3}$-diffeomorphism $\Phi$ that flattens $\partial U$ to $\{y_n = 0\}$, transforming the equation to a half-ball $B_1^+$. Tangential regularity is obtained by differentiating the flattened PDE $m+1$ times in tangential directions and applying the base case to the resulting equation. Normal regularity is recovered by solving the PDE for the highest pure normal derivative $\partial_{y_n}^{m+3} v$, expressing it in terms of mixed derivatives already controlled. A partition of unity and interior regularity complete the global result. [/proofplan] [step:Establish the base case $m = 0$] For $m = 0$, the hypotheses require $\partial U \in C^2$, $a_{ij}, b_i, c \in C^1(\bar{U})$, and $f \in L^2(U)$. The conclusion $u \in H^2(U)$ with the estimate $\|u\|_{H^2(U)} \le C(\|f\|_{L^2(U)} + \|u\|_{L^2(U)})$ is the boundary $H^2$ regularity theorem (Theorem 99). [/step] [step:State the inductive hypothesis and localize near a boundary point] Assume the theorem holds for a non-negative integer $m$. For the inductive step, assume $\partial U \in C^{m+3}$, $a_{ij}, b_i, c \in C^{m+2}(\bar{U})$, $f \in H^{m+1}(U)$, and $u \in H^1_0(U)$ solves $Lu = f$. By the inductive hypothesis (since $C^{m+2} \subset C^{m+1}$ and $H^{m+1} \subset H^m$), we already have $u \in H^{m+2}(U)$. The goal is $u \in H^{m+3}(U)$. Since regularity is a local property, we use a [partition of unity](/page/Partition%20of%20Unity) subordinate to a covering of $\bar{U}$ by finitely many balls. Interior points are handled by the [Higher Interior Elliptic Regularity](/theorems/96) theorem. It suffices to prove $u \in H^{m+3}$ near each boundary point. Fix $x_0 \in \partial U$. Since $\partial U \in C^{m+3}$, there exist a radius $r > 0$ and a $C^{m+3}$-diffeomorphism $\Phi: U \cap B(x_0, r) \to B_1^+ := \{y \in B_1(0) : y_n > 0\}$ with $\Phi(\partial U \cap B(x_0, r)) \subset \{y_n = 0\}$. [/step] [step:Transform the PDE to the half-ball and verify ellipticity is preserved] Define $v(y) := u(\Phi^{-1}(y))$ for $y \in B_1^+$. The function $v \in H^{m+2}(B_1^+)$ satisfies the transformed equation: \begin{align*} L'v = -\sum_{k,l=1}^n \partial_{y_k}(A_{kl}\, \partial_{y_l} v) + \sum_{k=1}^n B_k\, \partial_{y_k} v + C\, v = \tilde{f}, \end{align*} where $\tilde{f} \in H^{m+1}(B_1^+)$ and the new coefficients satisfy $A_{kl} \in C^{m+2}(\overline{B_1^+})$. The boundary condition becomes $v(y', 0) = 0$. Uniform ellipticity is preserved: for any $\xi \in \mathbb{R}^n$, setting $\eta = (D\Phi)^\top \xi$ and using the non-degeneracy of $D\Phi$ (since $\Phi$ is a diffeomorphism), there exists $c > 0$ such that: \begin{align*} \sum_{k,l=1}^n A_{kl}(y)\, \xi_k\, \xi_l \ge \theta\, c\, |\xi|^2 \qquad \text{for all } y \in B_1^+. \end{align*} [/step] [step:Obtain tangential regularity by differentiating the flattened PDE] Let $\alpha = (\alpha_1, \dots, \alpha_{n-1}, 0)$ be a multi-index of order $|\alpha| = m + 1$ with $\alpha_n = 0$ (purely tangential). Define $w := \partial_y^\alpha v$. Since tangential derivatives commute with the flat boundary condition, $w = 0$ on $\{y_n = 0\}$. Applying $\partial_y^\alpha$ to $L'v = \tilde{f}$ gives: \begin{align*} L'w = \tilde{f}_\alpha := \partial_y^\alpha \tilde{f} + [L', \partial_y^\alpha]\, v. \end{align*} The commutator $[L', \partial_y^\alpha]\, v$ involves derivatives of $v$ of order at most $|\alpha| + 1 = m + 2$ multiplied by derivatives of the coefficients. Since $v \in H^{m+2}(B_1^+)$ and the coefficients are in $C^{m+2}$, the commutator terms lie in $L^2(B_1^+)$. Since $\tilde{f} \in H^{m+1}(B_1^+)$, we have $\partial_y^\alpha \tilde{f} \in L^2(B_1^+)$. Therefore $\tilde{f}_\alpha \in L^2(B_1^+)$. The function $w$ is a weak solution of $L'w = \tilde{f}_\alpha$ with $w = 0$ on $\{y_n = 0\}$. By the base case (boundary $H^2$ regularity): \begin{align*} w = \partial_y^\alpha v \in H^2(B_1^+). \end{align*} Since $|\alpha| = m + 1$, this shows that any derivative $D^\beta v$ with $|\beta| \le m + 3$ and at most $2$ normal components belongs to $L^2(B_1^+)$. [guided] Why do tangential derivatives behave well at the boundary? Because the boundary has been flattened to $\{y_n = 0\}$, tangential translations $y' \mapsto y' + h e_k$ (for $k < n$) preserve the half-space $\{y_n > 0\}$. Concretely, if $y = (y', y_n) \in B_1^+$, then for small $h$ the shifted point $(y' + h e_k, y_n)$ also lies in $B_1^+$ when $k < n$. The zero boundary condition $v(y', 0) = 0$ is therefore inherited by tangential derivatives: \begin{align*} \partial_{y_k} v(y', 0) = \lim_{h \to 0} \frac{v(y' + h e_k, 0) - v(y', 0)}{h} = 0 \quad \text{for } k < n. \end{align*} By induction on $|\alpha|$, every purely tangential derivative $\partial_y^\alpha v$ (with $\alpha_n = 0$) vanishes on $\{y_n = 0\}$. This means $w = \partial_y^\alpha v$ satisfies homogeneous Dirichlet conditions on the flat boundary, making it a valid candidate for the base-case boundary $H^2$ regularity theorem. The commutator $[L', \partial_y^\alpha]$ arises because the coefficients of $L'$ are not constant. When we move $\partial_y^\alpha$ past a term like $\partial_{y_k}(A_{kl} \partial_{y_l} v)$, the Leibniz rule generates contributions of the form: \begin{align*} \sum_{\beta < \alpha} \binom{\alpha}{\beta} \partial_{y_k}\bigl((\partial_y^{\alpha - \beta} A_{kl}) \, \partial_y^\beta \partial_{y_l} v\bigr), \end{align*} where each term involves a derivative of $A_{kl}$ of order $|\alpha - \beta| \ge 1$ multiplied by a derivative of $v$ of order $|\beta| + 1 \le |\alpha| + 1 = m + 2$. The smoothness assumption $A_{kl} \in C^{m+2}$ ensures that $\partial_y^{\alpha - \beta} A_{kl}$ is bounded (and in $C^1$), while the known regularity $v \in H^{m+2}$ ensures that $\partial_y^\beta \partial_{y_l} v \in L^2$. After the outer divergence $\partial_{y_k}$, the entire commutator lies in $H^{-1}$ at worst — but a more careful count using $|\beta| \le m$ shows the commutator terms actually belong to $L^2(B_1^+)$, which is what we need. [/guided] [/step] [step:Recover normal regularity by solving for $\partial_{y_n}^{m+3} v$ from the PDE] It remains to control derivatives $D^\beta v$ with $|\beta| = m + 3$ and $3$ or more normal components. Rearrange the PDE to isolate the pure normal second derivative: \begin{align*} A_{nn}\, \partial_{y_n y_n} v = \tilde{f} + \sum_{(k,l) \ne (n,n)} \partial_{y_k}(A_{kl}\, \partial_{y_l} v) - \sum_{k=1}^n B_k\, \partial_{y_k} v - C\, v. \end{align*} Since $A_{nn}(y) \ge \theta\, c > 0$ (by ellipticity), we can divide by $A_{nn}$. To show $\partial_{y_n}^{m+3} v \in L^2$, apply $\partial_{y_n}^{m+1}$ to both sides: \begin{align*} \partial_{y_n}^{m+3} v = \partial_{y_n}^{m+1}\left[\frac{1}{A_{nn}}\left(\tilde{f} + \sum_{(k,l) \ne (n,n)} \partial_{y_k}(A_{kl}\, \partial_{y_l} v) - \text{lower-order terms}\right)\right]. \end{align*} The right-hand side involves: - $\partial_{y_n}^{m+1} \tilde{f} \in L^2$, since $\tilde{f} \in H^{m+1}$. - Mixed derivatives $\partial_{y_n}^{m+1} \partial_{y_k} \partial_{y_l} v$ with $(k, l) \ne (n, n)$: at least one of $k, l$ is tangential, so the total derivative of order $m + 3$ has at most $m + 2$ normal components. By the tangential regularity established above, these terms lie in $L^2$. - Products of derivatives of $A_{nn}$, $A_{kl}$ (bounded, since coefficients are $C^{m+2}$) with lower-order derivatives of $v$ (in $L^2$). Therefore $\partial_{y_n}^{m+3} v \in L^2(B_1^+)$. [guided] This is the bootstrap argument specific to boundary regularity. The tangential step gave us all derivatives of order $m + 3$ with at most $2$ normal components. For derivatives with $3$ or more normal components, we use the PDE as an algebraic relation. The equation $L'v = \tilde{f}$ lets us express $\partial_{y_n y_n} v$ in terms of tangential and mixed derivatives (plus lower-order terms and $\tilde{f}$): \begin{align*} \partial_{y_n y_n} v = \frac{1}{A_{nn}}\Bigl(\tilde{f} + \sum_{(k,l) \ne (n,n)} \partial_{y_k}(A_{kl}\, \partial_{y_l} v) - \sum_{k=1}^n B_k\, \partial_{y_k} v - C\, v\Bigr). \end{align*} Differentiating this relation $m + 1$ times in the normal direction yields: \begin{align*} \partial_{y_n}^{m+3} v = \partial_{y_n}^{m+1}\Bigl[\frac{1}{A_{nn}}\Bigl(\tilde{f} + \sum_{(k,l) \ne (n,n)} \partial_{y_k}(A_{kl}\, \partial_{y_l} v) - \text{l.o.t.}\Bigr)\Bigr], \end{align*} which expresses $\partial_{y_n}^{m+3} v$ in terms of derivatives that have at least one tangential component — which are already controlled. Why does $(k, l) \ne (n, n)$ guarantee a tangential component? If both $k$ and $l$ were $n$, the term would be $\partial_{y_n}(A_{nn}\, \partial_{y_n} v)$ — but that is exactly the term we isolated on the left. So every remaining term $\partial_{y_k}(A_{kl}\, \partial_{y_l} v)$ has at least one index $k$ or $l$ different from $n$, contributing at least one tangential derivative. When $\partial_{y_n}^{m+1}$ hits such a term, the resulting derivative of order $m + 3$ still carries at least one tangential index, so it falls within the scope of the tangential regularity already established. The Leibniz rule applied to $\partial_{y_n}^{m+1}\bigl[\frac{1}{A_{nn}} \cdot (\cdots)\bigr]$ produces terms of the form: \begin{align*} \sum_{j=0}^{m+1} \binom{m+1}{j} \bigl(\partial_{y_n}^j \tfrac{1}{A_{nn}}\bigr) \cdot \partial_{y_n}^{m+1-j}(\cdots). \end{align*} Each factor $\partial_{y_n}^j \frac{1}{A_{nn}}$ is bounded because $A_{nn} \in C^{m+2}(\overline{B_1^+})$ and $A_{nn} \ge \theta' > 0$. The remaining factors involve derivatives of $v$ of total order at most $m + 3$ with at least one tangential component, plus derivatives of $\tilde{f}$ of order at most $m + 1$, all of which lie in $L^2(B_1^+)$. [/guided] [/step] [step:Transform back and assemble the global estimate] We have shown $v \in H^{m+3}(B_1^+)$. Since $\Phi$ is a $C^{m+3}$-diffeomorphism, the chain rule for $(m+3)$-th order derivatives involves products of derivatives of $\Phi$ (bounded) and derivatives of $v$ (in $L^2$). Transforming back gives $u \in H^{m+3}(U \cap B(x_0, r))$. Covering $\partial U$ with finitely many such balls and using [Higher Interior Elliptic Regularity](/theorems/96) for points away from $\partial U$, we obtain $u \in H^{m+3}(U)$ with the estimate: \begin{align*} \|u\|_{H^{m+3}(U)} \le C\bigl(\|f\|_{H^{m+1}(U)} + \|u\|_{L^2(U)}\bigr), \end{align*} completing the induction. [/step]