• $\Omega\subseteq\mathbb R^d$ be open ($d\in\mathbb N$)
  • $\mathcal D:=C_c^\infty(\Omega)^d$ and $$\mathfrak D:=\left\{\phi\in\mathcal D:\nabla\cdot\phi=0\right\}$$
  • $\mathcal H:=\overline{\mathfrak D}^{\langle\;\cdot\;,\;\cdot\;\rangle_{L^2(\Omega,\:\mathbb R^d)}}$ and $\mathcal V:=H_0^1(\Omega,\mathbb R^d)\cap\mathcal H$

If $$\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for }u,v\in H_0^1(\Omega,\mathbb R^d)\;,$$ then the Dirichlet-Laplace operator is defined to be $$\Delta_D:H_0^1(\Omega,\mathbb R^d)\to H^{-1}(\Omega,\mathbb R^d)\;,\;\;\;u\mapsto-\mathfrak a(\;\cdot\;,u)\;.$$ If $\tilde\iota:\mathcal V\to H_0^1(\Omega,\mathbb R^d)$ is the inclusion and $\tilde{\operatorname P}:H^{-1}(\Omega,\mathbb R^d)\to\mathcal V'$ is the adjoint of $\tilde\iota$, then $$A_0:\mathcal V\to\mathcal V'\;,\;\;\;u\mapsto-\tilde{\operatorname P}\Delta_D\tilde\iota u$$ is a well-defined bounded, linear and symmetric operator with $$A_0u=\left.(-\Delta_Du)\right|_{\mathcal V}\;\;\;\text{for all }u\in\mathcal V\tag 1$$ and the Stokes operator is defined to be the restriction $A$ of $A_0$ to $$D(A):=\left\{u\in\mathcal V:A_0u\in\mathcal H\right\}\;.\tag 2$$

Now, I've observed the following: If $u\in H_0^1(\Omega)$ is twice weak differentiable (e.g. $u\in H_0^2(\Omega)$), then $u$ has a weak Laplacian $\Delta u\in L_{\text{loc}}^1(\Omega)$ and $$(\Delta_Du)v=\langle v,\Delta u\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }v\in H_0^1(\Omega,\mathbb R^d)\;.\tag 3$$ If $\iota:\mathcal H\to L^2(\Omega,\mathbb R^d)$ is the inclusion and $\operatorname P:L^2(\Omega,\mathbb R^d)\to\mathcal H$ is the orthogonal projection, then $\iota^\ast=\operatorname P$ and hence (using $(3)$) $$(\Delta_Du)v=\langle v,\operatorname P\Delta u\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }v\in\mathcal V\;.\tag 4$$ Thus, if $u\in\mathcal V$, we obtain $$(A_0u)v=\langle v,-\operatorname P\Delta u\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }v\in\mathcal V\;.\tag 5$$ by $(1)$, i.e. $\Delta_Du\in\mathcal H$.

So, my question is: Shouldn't we be able to define the Stokes operator to be $$\tilde A:D(\tilde A)\to\mathcal H\;,\;\;\;u\mapsto-\operatorname P\Delta u$$ with $D(\tilde A)=H^2(\Omega,\mathbb R^d)\cap\mathcal V$? I've read that this isn't possible, unless $\Omega$ is regular enough, but I don't see why.


1 Answer 1


If $\Omega$ has reentrant corners, then solutions of the Stokes problem will in general have corner singularities which prevent $H^2$ regularity. Therefore, if you define the Stokes operator in the manner you suggest (which you can do, it is a free country), you will not get the "desirable" existence theory.

  • $\begingroup$ But there is no other manner in which we could define it?! Or is there a different definition of the Stokes operator $\tilde A$ for "regular" domains? If not, the operator stays the same. Of course, we might need these regularity assumptions in order to proof existence and regularity results for the Stokes problem, but from my point of view, that is another story. It confused me, that in almost every textbook, regularity assumptions on $\Omega$ are imposed in the section where $\tilde A$ is introduced. I first thought, we couldn't define it without the assumptions. $\endgroup$
    – 0xbadf00d
    Jul 21, 2016 at 16:58
  • $\begingroup$ sorry, what means $\Omega$ has reentrant corners ? do you have a basic example ? $\endgroup$
    – reuns
    Jul 22, 2016 at 15:56
  • $\begingroup$ @0xbadf00d I don't have access to this. what is "$\Omega$ has reentrant corners" in a few words ? $\endgroup$
    – reuns
    Jul 22, 2016 at 18:25
  • $\begingroup$ @user1952009 "If the direction of the boundary of a regionon changes suddenly by an angle exceeding $\pi$, then the resulting corner defined by the angle is called a reentrant corner." (from here). $\endgroup$
    – 0xbadf00d
    Jul 22, 2016 at 18:29
  • $\begingroup$ @0xbadf00d how is it a problem for the Laplacian and Stokes operator ? $\endgroup$
    – reuns
    Jul 22, 2016 at 18:31

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