Remark on the Helmholtz decomposition in domains with noncompact boundary

Abstract

Let $\varOmega$ be a domain in $\mathbb {R}^{d+1}$ whose boundary is given as a uniform Lipschitz graph $x_{d+1}=\eta (x)$ for $x \in \mathbb {R}^d$. For such a domain, it is known that the Helmholtz decomposition is not always valid in $L^p(\varOmega )$ except for the energy space $L^2 (\varOmega )$. In this paper we show that the Helmholtz decomposition still holds in certain anisotropic spaces which include vector fields decaying slowly in the $x_{d+1}$ variable. In particular, these classes include some infinite energy vector fields. For the purpose, we develop a new approach based on a factorization of divergence form elliptic operators whose coefficients are independent of one variable.

Appendix

1.1 Semigroup $\{e^{-t\varLambda }\}_{t\ge 0}$ in $L^r(\mathbb {R}^d)$ for $r\in (1,\infty )$

Proposition 3

Let $r\in (1,\infty )$. Then the restrictions of $\{e^{-t\varLambda }\}_{t\ge 0}$ on $L^2(\mathbb {R}^d)\cap L^r(\mathbb {R}^d)$ is extended as a strongly continuous and bounded semigroup in $L^r (\mathbb {R}^d)$.

Proof

Here we give only a sketch of the proof. We first consider the case $r\in [2,\infty )$. Set $u(t) = e^{-t\varLambda } f, f\in H^1 (\mathbb {R}^d) \cap L^r (\mathbb {R}^d)$, and set $v(s;t) = e^{-s\mathcal {P}} u(t), s\ge 0$. Then we have

$$\begin{aligned} \frac{\,\mathrm{d}}{\,\mathrm{d}t} \Vert u(t) \Vert _{L^r(\mathbb {R}^d)}^r&= - r \langle \varLambda u(t), |u(t)|^{r-2} u(t) \rangle _{L^2(\mathbb {R}^d)} \\&= - r \int _0^\infty \langle A\nabla v(s;t), \nabla \big ( | v(s;t)|^{r-2} v(s;t) \big ) \rangle _{L^2 (\mathbb {R}^d)} \,\mathrm{d}s \\&\le -c r \Vert \nabla ( |v(\cdot ; t) |^{\frac{r}{2}} ) \Vert _{L^2 (\mathbb {R}^{d+1}_+)}^2\le -c r \Vert | u(t) |^{\frac{r}{2}} \Vert _{\dot{H}^\frac{1}{2}(\mathbb {R}^d)}^2. \end{aligned}$$

In particular, we have $\Vert e^{-t\varLambda } f \Vert _{L^r(\mathbb {R}^d)}\le \Vert f \Vert _{L^r(\mathbb {R}^d)}$ when $r\in [2,\infty )$, and thus, when $r\in [2,\infty ]$ (by taking the limit $r\rightarrow \infty$). By the dual relation $\langle e^{-t\varLambda } f, g\rangle _{L^2 (\mathbb {R}^d)} = \langle f, e^{-t\varLambda }g \rangle _{L^2 (\mathbb {R}^d)}$ we have this uniform bound also for $r\in [1,2]$. Hence, by the density argument $\{e^{-t\varLambda }\}_{t\ge 0}$ is extended as a bounded semigroup acting on $L^r(\mathbb {R}^d)$ for all $r\in [1,\infty ]$ (note that this uniform bound holds also for $r=1,\infty$). As for the strong continuity, let $r\in (2,\infty )$, and for any $f\in L^r(\mathbb {R}^d)$ we take $\{f_n\}\subset C_0^\infty (\mathbb {R}^d)$ such that $f_n\rightarrow f$ as $n\rightarrow \infty$ in $L^r(\mathbb {R}^d)$. Then we see

$$\begin{aligned}&\Vert e^{-t\varLambda } f - f\Vert _{L^r (\mathbb {R}^d)}\\&\quad \le \Vert e^{-t\varLambda } (f-f_n)\Vert _{L^r(\mathbb {R}^d)} + \Vert f-f_n \Vert _{L^r(\mathbb {R}^d)} + \Vert e^{-t\varLambda } f_n-f_n\Vert _{L^r(\mathbb {R}^d)}\\&\quad \le 2 \Vert f -f_n \Vert _{L^r(\mathbb {R}^d)} + \Vert e^{-t\varLambda } f_n-f_n\Vert _{L^2(\mathbb {R}^d)}^{\frac{2}{r}} \Vert e^{-t\varLambda } f_n-f_n\Vert _{L^\infty (\mathbb {R}^d)}^{1-\frac{2}{r}}\\&\quad \le 2 \Vert f -f_n \Vert _{L^r(\mathbb {R}^d)} + 2^{1-\frac{2}{r}} \Vert e^{-t\varLambda } f_n-f_n\Vert _{L^2(\mathbb {R}^d)}^{\frac{2}{r}} \Vert f_n \Vert _{L^\infty (\mathbb {R}^d)}^{1-\frac{2}{r}}. \end{aligned}$$

Since we have already known that $\{e^{-t\varLambda }\}_{t\ge 0}$ is strongly continuous in $L^2 (\mathbb {R}^d)$, the last estimate implies the strong continuity in $L^r(\mathbb {R}^d)$ for $r\in (2,\infty )$. The case $r\in (1,2)$ is proved in the same manner. The proof is complete. $\square$

1.2 Semigroup $\{e^{-t\mathcal {P}}\}_{t\ge 0}$ in $L^r(\mathbb {R}^d)$ for $r\in [2,\infty )$

Proposition 4

Let $r\in [2,\infty )$. Then the restriction of $\{e^{-t\mathcal {P} }\}_{t\ge 0}$ on $L^2(\mathbb {R}^d)\cap L^r(\mathbb {R}^d)$ is extended as a strongly continuous and bounded semigroup in $L^r (\mathbb {R}^d)$.

Proof

Again we give only a sketch of the proof. Let $f\in L^2 (\mathbb {R}^d)\cap L^\infty (\mathbb {R}^d)$ and set $u(t) = e^{-t\mathcal {P}} f$. Then, since $u$ satisfies $\mathcal {A}u=0$ in $\mathbb {R}^{d+1}_+$, the maximum principle implies that

$$\begin{aligned} \Vert u(t) \Vert _{L^\infty (\mathbb {R}^d)} \le \Vert u \Vert _{L^\infty (\mathbb {R}^{d+1}_+)} \le \Vert f \Vert _{L^\infty (\mathbb {R}^d)}. \end{aligned}$$

This estimate gives the boundedness of $e^{-t\mathcal {P}}$ in $L^\infty (\mathbb {R}^d)$. Since $e^{-t\mathcal {P}}$ is bounded in $L^2 (\mathbb {R}^d)$, the interpolation inequality yields the boundedness of $e^{-t\mathcal {P}}$ in $L^r(\mathbb {R}^d)$ for each $r\in (2,\infty )$. The strong continuity in $L^r(\mathbb {R}^d)$ is shown as in the proof of Proposition 3. The proof is complete.$\square$