why $u \in H^2$ imply zero value of jump $[u] = u^{-}-u^{+}$

97 Views Asked by At

in page 13 of simple DG tutorial,

it gives following conclusion:

  • Define average $w=\frac{1}{2}\left(w^{-}+w^{+}\right)$ and recall jump $[w]=w^{-}*w^{+}$ then $$ [a b]=[a]\{b\}+\{a\}[b] $$

  • confusing conclusion : since $u \in H^2(\Omega)$ and $v \in H^1(\Omega)$, $$ \begin{aligned} & -\sum_{i=1}^m \int_{e_i}\left[\nabla u \cdot n_i\right]\{v\}=0, \\ & -\sum_{i=1}^m \int_{e_i}\left\{\nabla v \cdot n_i\right\}[u]=0 \end{aligned} $$

and in book Discontinuous Galerkin Method Analysis and Applications to Compressible Flow it gives the same conclusion

  • let triangulation $\mathscr{T}_h$ ( $h>0$ is a parameter) be a partition of the closure $\bar{\Omega}$ of the domain $\Omega$ into a finite number of closed $d$-dimensional simplexes $K$ with mutually disjoint interiors such that $$ \bar{\Omega}=\bigcup_{K \in \mathscr{T}_h} K $$

  • The discontinuous Galerkin method is based on the use of discontinuous approximations. This is the reason that over a triangulation $\mathscr{T}_h$, for any $k \in \mathbb{N}$, we define the so-called broken Sobolev space $$ H^k\left(\Omega, \mathscr{T}_h\right)=\left\{v \in L^2(\Omega) ;\left.v\right|_K \in H^k(K) \forall K \in \mathscr{T}_h\right\}, $$ which consists of functions, whose restrictions on $K \in \mathscr{T}_h$ belong to the Sobolev space $H^k(K)$. On the other hand, functions from $H^k\left(\Omega, \mathscr{T}_h\right)$ are, in general, discontinuous on inner faces of elements $K \in \mathscr{T}_h$.

  • Let $\Gamma \in \mathscr{F}_h^I$ and let $K_{\Gamma}^{(L)}, K_{\Gamma}^{(R)} \in \mathscr{T}_h$ be elements adjacent to $\Gamma$. For $v \in$ $H^1\left(\Omega, \mathscr{T}_h\right)$ we introduce the following notation: $$ \begin{aligned} & v_{\Gamma}^{(L)}=\text { the trace of }\left.v\right|_{K_{\Gamma}^{(L)}} \text { on } \Gamma, \\ & v_{\Gamma}^{(R)}=\text { the trace of }\left.v\right|_{K_{\Gamma}^{(R)}} \text { on } \Gamma, \\ & \langle v\rangle_{\Gamma}=\frac{1}{2}\left(v_{\Gamma}^{(L)}+v_{\Gamma}^{(R)}\right) \quad(\text { mean value of the traces of } v \text { on } \Gamma), \\ & {[v]_{\Gamma}=v_{\Gamma}^{(L)}-v_{\Gamma}^{(R)} \quad(\text { jump of } v \text { on } \Gamma) .} \end{aligned} $$

  • Due to the assumption that $u \in H^2(\Omega)$, we have $$ [u]_{\Gamma}=[\nabla u]_{\Gamma}=0, \quad \nabla u_{\Gamma}^{(L)}=\nabla u_{\Gamma}^{(R)}=\langle\nabla u\rangle_{\Gamma}, \quad \Gamma \in \mathscr{F}_h^I $$

My question is how to derive the above conclusion for jump and average from the properties of the $H^2$ and $H^1$ space when functions are known to be discontinuous on all inner faces of elements in triangulation

1

There are 1 best solutions below

0
On

First notice that if $u\in H^2$, then $\nabla u\in H^1$. Therefore if we can show that $[u]=0$ a.e. for internal faces when $u\in H^1$, we can deduce the analogous property for $[\nabla u\cdot n]$ if $u\in H^2$.

Now, as $C^\infty$ is dense in $H^1$, there is a sequence $u_k\in C^\infty$ such that $\|u-u_k\|_{H^1}\to0$ as $k\to\infty$. Since these functions are smooth it is clear that they have zero jump over internal faces. Therefore, by applying the trace theorem we find that for an internal face $F$, $$\|[u]\|_{L^2(F)}=\|[u-u_k]\|_{L^2(F)}\le C\sum_{K\in\mathscr{T}_h:K\cap F\ne\emptyset}\|u-u_k\|_{H^1(K)}\to 0$$ as $k\to\infty$.

Notice that this result holds more generally than relying on Sobolev embeddings.