Hodge decomposition theorem that uses a symmetric positive-definite matrix

Question

Let $\Omega\subset \mathbb{R}^3$ be a bounded lipschitz domain, $n$ be the outward unit normal vector to the boundary $\partial \Omega$ and $A$ be a $3\times 3$ real symmetric positive-definite matrix.
Define then the scalar product $<u,v>_A:=<u,Av>$, $\forall u, v\in\mathbb{R}^3$.
I want to see if the following statement is true or not:

With respect to the scalar product $<.,.>_A$, the space of square integrable functions, denoted by $(L^2(\Omega))^3$ admits the following orthogonal decomposition $$ (L^2(\Omega))^3=\nabla H_0^1(\Omega) \oplus H_0(\operatorname{div}0,\Omega,A)\oplus W $$ where

• $H_0^1(\Omega)$ is the usual Sobolev space
• $H_0(\operatorname{div}0,\Omega,A)=\{u\in (L^2(\Omega))^3; \operatorname{div}(Au)=0,n\cdot Au=0\}$
• $W=\nabla H^1(\Omega)\cap \{u\in (L^2(\Omega))^3; \operatorname{div}(Au)=0\}$
  I note that it is easy to show that the three subspaces in that decomposition are two by two orthogonal with respect the scalar product $<.,.>_A$ thanks to the standard Helmholtz decomposition of $(L^2(\Omega))^3$ which is given by $$ (L^2(\Omega))^3=\nabla H_0^1(\Omega) \oplus H_0(\operatorname{div}0,\Omega)\oplus H $$ where
• $H_0(\operatorname{div}0,\Omega)=\{u\in (L^2(\Omega))^3; \operatorname{div}(u)=0,n\cdot u=0\}$
• $H=\nabla H^1(\Omega)\cap \{u\in (L^2(\Omega))^3; \operatorname{div}(u)=0\}$ (the subspace of harmonic functions).

Thank you

Answer 1

The idea is the same as the standard proof in the case $A=I$: Consider $f\in L^2$, multiply by $A$ and take the divergence (effectively killing any contribution from the last two summands), and solve the Dirichlet problem $$ \begin{cases} \text{div}(A\nabla u)= \text{div}(A f) & \text{ in } \Omega\\ \quad \quad \quad u=0 & \text{ on } \partial\Omega. \end{cases} $$ $v=\nabla u$ will be the first term. Call $g=f-\nabla u$. Notice that $ Ag\in L^2$ and $\text{div}(Ag)=0$ so we can define $n\cdot Ag\in H^{-1/2}(\partial\Omega)$. Now solve the Neumann problem $$ \begin{cases} \text{div}(A\nabla U)=0 & \text{ in } \Omega\\ \phantom{a} n\cdot A\nabla U = n\cdot Ag & \text{ on } \partial\Omega. \end{cases} $$ $w=\nabla U$ will be the element in your last space, and then $z:= f-v-w$ lies in the middle space as required. Since you've already checked the orthogonality, this gives the decomposition.