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Let $k(x)$ be a function with lower bound $k_0\leq k(x)$ over the domain $\Omega\subset \mathbb{R}^2$.

Let us define the weighted norm $$ \|u\|_{1,k,\Omega}= \int_{\Omega} k(x) u^2dx+\int_{\Omega}k(x)\nabla u\cdot\nabla u dx. $$

In the Hilbert space $H^1(\Omega)$, if $u\in H ^1(\Omega)$, we already know that $\text{tr } u \in H^{\frac{1}{2}}(\partial\Omega)$.

Is there something similar in our weighted space? If $u\in H^1_k(\Omega)$, do we have $\text{tr } u\in H^\frac{1}{2}_k(\partial\Omega) $?

Moreover, if I want to define the dual space of $H^{\frac{1}{2}}_k(\partial\Omega)$, could the norm be

$$\|w\|_{H^{-\frac 1 2}_k(\partial\Omega)}=\sup_{u\in H^1_k(\Omega)} \frac{\langle u, w\rangle}{\|u\|_{H^1_k(\Omega)}}$$?

Ariel So
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  • Have you found any answers to this? I am interested in the dual question. Specifically, what duality product did you use? – Tarek Acila Sep 23 '23 at 14:22
  • @TarekAcila At the end I use the $$sup_{u\in H_k^1(\Omega)} \frac{\langle u, w\rangle}{|u|_{H^{\frac{1}{2}_k(\partial\Omega)}$$ where $$\langle \cdot,\cdot\rangle$$ is duality pair on the boundary. – Ariel So Oct 12 '23 at 20:57

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