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Let $Q_1$ and $Q_2$ be two open squares in $\mathbb{R}^2$ whose closures have an edge - say L - in common. Let be $u_i\in W^{1,p}(Q_i)$ for $i=1,2$ and for such $p\in[1,+\infty]$.

Suppose that $Tr(u_1)=Tr(u_2)$ on $L$

Given $\Omega:=Q_1\cup Q_2\cup L$, show that the function $u:\Omega\to\mathbb{R}$ which coincides with $u_i$ on $Q_i$ belongs to $W^{1,p}(\Omega)$.

My book gives the following theorem about Tr (without proof):

Let $\Omega\subset\mathbb{R}^N$ be a bounded open set with $C^1$ boundary and $p\in[1,+\infty]$.

There exists a linear and continuous operator $Tr:W^{1,p}(\Omega)\to L^p(\delta\Omega,\lambda^{N-1})$ such that:

  1. $Tr(u)=u_{|\delta\Omega}$ if $u\in C^0(\overline\Omega)$

  2. Exists $C=C(\Omega,p)$ such that $||Tr(u)||_{L^p(\delta\Omega)}\le C||u||_{W^{1,p}(\Omega)} \quad \forall u\in W^{1,p}(\Omega)$

So, I have two questions:

The subsets of $\mathbb{R}^2$ in the exercise don't have $C^1$ boundary. So I assume the theorem above holds even if $\delta\Omega$ is piecewise $C^1$.

Anyway I don't see the point of the exercise since the definition of $u$ on $L$ does not affect the $W^{1,p}$ norm value of $u$ due to the fact that $\lambda^2(L)=0$. What is wrong with this argument?

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  • $\begingroup$ The problem is to show that $u$ has weak derivatives on the combined domain. $\endgroup$ – daw Feb 5 '19 at 15:34
  • $\begingroup$ Thanks daw, your hint helped me :) $\endgroup$ – framago Feb 5 '19 at 17:51
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My last argument was wrong, it is true that the definition of $u$ on $L$ does not affect its $L^p(\Omega)$ norm value though. The point, as suggested in the comment by daw, is to show that $u$ has weak derivative in $L^p(\Omega)$.

Pick $\phi\in C^\infty_c\left(\Omega;\mathbb{R}^2\right)$

$$\int_\Omega\nabla u \cdot\phi\,dx=-\sum_i\int_{Q_i} u_i \,div(\phi)\,dx=\sum_i\left[-\int_{\delta Q_i} Tr(u_i\phi)\cdot\nu_i\,dS+\int_{Q_i}\nabla u_i\cdot\phi\,dx\right]=\sum_i\left[-\int_{\delta Q_i} Tr(u_i)\,\phi\cdot\nu_i\,dS+\int_{Q_i}\nabla u_i\cdot\phi\,dx\right]=\sum_i \left[-\int_{\delta Q_i\setminus L} Tr(u_i)\,\phi\cdot\nu_i\,dS+\int_{Q_i}\nabla u_i\cdot\phi\,dx\right]=\\ 0+\sum_i\int_{Q_i}\nabla u_i\cdot\phi\,dx=\int_\Omega \left(\chi_{Q_1}\nabla u_1+\chi_{Q_2}\nabla u_2\right)\cdot\phi\,dx$$

The fact that $Tr(u_i\phi)=Tr(u_i)\,\phi$ follows by the extension theorem, the density of $C^\infty_c\left(\mathbb{R}^2\right)$ in $W^{1,p}\left(\mathbb{R}^2\right)$, the regularity of $\phi$ and the continuity of $Tr$.

The integral on the boundary is $0$ because $\nu_1+\nu_2=0$ on $L$ and because $\phi=0$ on $\delta Q_i\setminus L$.

So we find out that $\nabla u=\chi_{Q_1}\nabla u_1+\chi_{Q_2}\nabla u_2$ which is in $L^p\left(\Omega;\mathbb{R}^2\right)$.

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