Some estimate in a growing domain

Question

Consider a growing family of domains $\Omega_t$ in $\mathbb{R}^n$ with boundary $\Gamma_t$, for $t \in [0,1]$. Examples that I have in mind are $\Omega_t = \{ x = (x_1,x_2) \in \mathbb{R}^2: x_1 < t \}$ or $\Omega_t = B_t(0)$ a ball of radius $t$. An advantage of the latter case is that for $0 \leq s < t \leq 1$ one has $|\Omega_t \setminus \Omega_s| \leq C |t-s|$ (recall for later).

Also, write $-\Delta_s$ for the Dirichlet Laplacian over $L^2(\Omega_s)$ and $C^\infty_0(\Omega_t)$ for the test functions in $\Omega_t$.

For $0\leq s < t \leq 1$, $u\in D(-\Delta_s) =H^1_0(\Omega_s) \cap H^2(\Omega_s)$ and $v\in C^\infty_0(\Omega_t)$ I want to show the following estimate: $$\int_{\Gamma_s} \partial_\nu u \cdot v d \sigma \leq C \omega(t-s) \| \Delta_s u \|_{L^2(\Omega_s)} \| v \|_{H^1(\Omega_t)},$$ where $\omega(t-s)$ is a modulus of continuity that should be at least as good as $1/2 + \varepsilon$ Hölder, and $\partial_\nu$ is the co-normal derivative with respect to $\Gamma_s$.

The question comes from some application to evolution problems in a domain that changes over time.

I tried the following things so far:

1. In the situation with the moving half-planes I calculated with the fundamental theorem of calculus (using that $v$ vanishes on $\Gamma_t$) that $$|v(x,s)|^2 \leq (\int_s^t |\partial_1 v(x,r)| d r)^2 \leq (t-s) \int_s^t |\partial_1 v(x,r)|^2 d r.$$ Hence, $\| v \|_{L^2(\Gamma_s)} \leq (t-s)^{1/2} \| v \|_{H^1(\Omega_t)}$, so Hölder in $\Gamma_s$ followed by the trace operator $H^{1/2}(\Omega_s) \to L^2(\Gamma_s)$ give $$\int_{\Gamma_s} \partial_\nu u \cdot v d \sigma \leq C \| \partial_\nu u \|_{H^{1/2}(\Omega_s)} (t-s)^{1/2} \| v \|_{H^1(\Omega_t)} \leq C (t-s)^{1/2} \| u \|_{H^{3/2}(\Omega_s)} \| v \|_{H^1(\Omega_t)}.$$ That is, in this case I'm missing the $\varepsilon$ in the Hölder condition but on $u$ I still have half an order in the Sobolev scale to use for my estimate.

2. Formal calculation! Pretend taking traces would just work from $L^2(\Omega_t \setminus \Omega_s) \to H^{-1/2}(\Gamma_s)$. Then with this trace and Poincaré $$\| v \|_{H^{-1/2}(\Gamma_s)} \leq C \| v \|_{L^2(\Omega_t \setminus \Omega_s)} \leq C (t-s) \| \nabla v \|_{L^2(\Omega_t \setminus \Omega_s)}.$$ So if I estimate the boundary integral with an $H^{1/2} - H^{-1/2}$ pairing instead of $L^2 - L^2$ (as in approach 1) it could heuristically happen that the order of $(t-s)$ improves.

3. I tried to work in direction two, for instance using a fractional Sobolev embedding $L^r(\Gamma_s) \subseteq H^{-s}(\Gamma_s)$ with $r < 2$ and perform the argument from approach 1. However, to get back to $L^2$ I would have to use Hölder's inequality for another time and if I have the estimate $|\Omega_t \setminus \Omega_s| \leq C (t-s)$ then this just let's me recover the order $1/2$ from approach 1.

I'm happy about all suggestions!

Answer 1

I do not think this is possible without assuming more regularity of $v_t$ with respect to $t$.

Take $\Omega_t = (-t,+t)$.

Construct $v_t$ to be piecewise linear with $v_t(\pm t)=0$ and $v_t(\pm s)=1$. Then $\|v_t'\|_{L^2} =c (t-s)^{-1/2}$, and $v_t'$ is concentrated on $\Omega_t \setminus \Omega_s$.

Set $u_s(x) := 1-x^2/s^2$. Then $\partial_\nu u =c s^{-1}$, $\|\Delta u\|_{L^2} = c s^{-3/2}$. This implies $$ \frac{ \int_{\partial \Omega_s} \partial_\nu u v_t }{ \|v_t'\|_{L^2} \|\Delta u\|_{L^2} } = c \frac{s^{-1}}{ (t-s)^{-1/2} s^{-3/2}} = c s^{1/2}(t-s)^{1/2}. $$