Use trace theorem to define $H^2_0$ space and the requirement of the boundary

Question

For homogeneous biharmonic problems, the solution space is in general defined as

$$H^2_0(\Omega):=\{u\in H^2(\Omega): u=\frac{\partial u}{\partial \mathbf{n}}=0\text{ on }\partial \Omega\}.$$

With the help of the trace operator $T:H^1(\Omega)\rightarrow H^{\frac{1}{2}}(\partial \Omega) $, one may have the restriction of $u$ on $\partial \Omega$. And it requires the boundary of $\Omega$ to be Lipschitz continuous.

My question is how could I define $\frac{\partial u}{\partial \mathbf{n}}$ on the boundary. And if I want $\frac{\partial u}{\partial \mathbf{n}} \in H^{\frac{1}{2}}(\partial \Omega)$, what boundary condition do I need?

Answer 1

In order to satisfactorily define normal and tangential components of functions in $\mathbf{H}^{1/2}(\Gamma)=(H^{1/2})^d(\Gamma)$, where $\Gamma = \partial \Omega$, you need two things:

1. Let $v \in \mathbf{W}^{s, p} (\Omega), s \in \mathbb{N}$ and let $h \in C^{\kappa,\lambda} (\overline{\Omega})$ with $\kappa \in \mathbb{N}_0, \lambda \in (0,1]$ and $\kappa + \lambda \geqslant s$. The product $hv$ is in $\mathbf{W}^{s, p} (\Omega)$ and there exists a constant $C = C (h, p, \Omega)$ such that $$ \| hv \|_{\mathbf{W}^{s, p} (\Omega)} \leqslant c \| v \|_{\mathbf{W}^{s, p} (\Omega)} . $$
2. Let $\Omega$ be bounded and with $C^{\kappa, 1}$ boundary $\Gamma$, for $\kappa = 0, 1, \ldots$ Then the unit normal vector $\nu$ to $\Gamma$ is well defined and in $C^{\kappa - 1, 1}$ for $\kappa > 0$, or $L^{\infty}$ for $\kappa = 0$.

With 2. we have for $\Omega$ of class $C^{1, 1}$ that $\nu \in C^{0, 1}$ and we can properly define for any $v \in \mathbf{H}^{1 / 2} (\Gamma)$ the decomposition $$ v = v_{\nu} \nu + v_{\tau} \text{ where } v_{\nu} = v \cdot \nu \text{ and } v_{\tau} = v - v_{\nu} \nu, $$ and thanks to 1. (applied to the function $\nu$ and some lifting $\tilde{v} \in \mathbf{H}^1 (\Omega)$ of $v$) we know that $v_{\nu} \in H^{1 / 2} (\Gamma)$ and $v_{\nu} \nu \in \mathbf{H}^{1 / 2} (\Gamma)$

In order to prove 2. you need only Rademacher's theorem on Lipschitz functions. For 1. the case $\kappa = s$ follows easily from the boundedness of all derivatives of the function $h$. The case $\kappa = s -1, \lambda = 1$ is also easy.

The theorem also holds for $\kappa = 0$, $1 < s < \lambda \leqslant 1$, but it's trickier to prove.