Regularity of restricted function

Question

Let $\Omega =\Omega_1 \cup \Omega_2$ such that $\Gamma $ is the interface ( actually $\Omega_1$ is the fluid domain and $\Omega_2$ is the solid domain)

Let $v\in H^1(\Omega_1)$ and $u\in H^1(\Omega_2)$ such that $v=u_t$ on $\Gamma$. And let $\xi$ be such that $$\xi =v \quad\text{on} \: \Omega_1$$ and $$\xi =u_t \quad\text{on} \: \Omega_2$$ Why Can we say that $\xi\in H^1(\Omega)$ ??

Answer 1

I think this way it could work. Some arguments are a bit shortened, just ask if anything is unclear or wrong.

Let me assume for the beginning that $\Omega \subset \mathbb R^d$, $d\in \mathbb N$ and $\Omega_1 = \Omega \cap \mathbb R^{+} \times \mathbb R^{d-1} $, $\Omega_2 = \Omega \setminus \Omega_1$.

Then it is not hard to see, that for any $\varphi \in C^\infty_c(\Omega)$ \begin{align*} \int_{\Omega} \xi(x) \frac{\partial}{\partial x_1}\varphi(x) dx = \int_{\Omega_1} v(x) \frac{\partial}{\partial x_1} \varphi(x) dx + \int_{\Omega_2} u_t(x) \frac{\partial}{\partial x_1}\varphi(x) dx \\= -\int_{\Omega_1} \frac{\partial}{\partial x_1}v(x) \varphi(x) dx - \int_{\Omega_2} \frac{\partial}{\partial x_1}u_t(x) \varphi(x) dx, \end{align*} where I used partial integration in the last step and the boundary terms vanish due to the assumption that $u_t = v$ on $\Gamma$. In other words, the weak partial derivative of $\xi$ in $x_1$ is the piecewise derivative of $v$ and $u_t$. The same argument holds for the derivatives in other directions, but there no boundary terms appear due to the compact support of $\varphi$. Hence, we conclude that $\xi \in H^1(\Omega)$.

Now let us relax the assumption on $\Omega_1$ and $\Omega_2$. Observe that a function $f$ is in $H^1(\Omega)$ if there is a finite open cover $(U_i)_{i= 1}^N$ of $\Omega$ such that $f_{|U_i} \in H^1(U_i)$ for all $i$. (You can show this with a partition of unity.)

I need to assume some regularity of $\Gamma$, specifically, I require $\Gamma\in C^1$. This means that there is a finite cover $U_i$ of $\Omega$ (actually finite is not necessary, but I added this to make the argument easier) such that for every $U_i$ there is a $C^1$ diffeomorphism $\gamma_i: U_i \to A_i \subset \mathbb R^d$ such that $\gamma_i(\Gamma \cap U_i) \subset \{0\} \times \mathbb R^{d-1}$, $\gamma_i(\Omega_1 \cap U_i) \subset \mathbb R^+ \times \mathbb R^{d-1}$ and $\gamma_i(\Omega_2 \cap U_i)\subset \mathbb R^- \times \mathbb R^{d-1}$.

Using, that by the chain rule $f_{U_i} \in H^1(U_i)$ if $f_{U_i}\circ \gamma_i^{-1}\in H^1(A_i)$, we conclude by the first part of the argument, that $\xi \in H^1(\Omega)$.

Answer 2

If the interface $\Gamma$ has some regularity, and since $v=u_t$ on $\Gamma$ then $\xi \in L^2(\Omega)$.

Now to get the desired result, we need to show that the partial derivatives of $\xi$ are in $L^2(\Omega)$, it is sufficient to prove that the restriction of the partial derivatives of $\xi$ on $\Gamma$ are in $L^2(\Gamma)$. We have $$\frac{\partial\xi}{\partial x} |_{\Gamma} = \frac{\partial\xi|_{\Gamma}}{\partial x}$$

Consequently the result holds.