## Uniqueness and nonuniqueness of steady Stokes equations

Simply take $u$ as $\overset{\rightharpoonup}{u}$, $u$ is understood as a vector function.

$\displaystyle D(\Omega)=\{\overset{\rightharpoonup}{u}\in C^\infty_c(\Omega)| \nabla\cdot \overset{\rightharpoonup}{u}=0 \}$

$\displaystyle H(\Omega)$ is  the closure of $D(\Omega)$ under the norm of $H^1_0(\Omega)$

$\langle\displaystyle \overset{\rightharpoonup}{u}, \overset{\rightharpoonup}{v}\rangle_{H(\Omega)}=\int_{\Omega}\nabla\overset{\rightharpoonup}{u}\cdot \nabla\overset{\rightharpoonup}{v}dx$

$H(\Omega)$ is defined for arbitrary domain $\Omega\subset \mathbb{R}^3$ or $\mathbb{R}^2$, unless $\Omega=\mathbb{R}^2$

$\displaystyle \widehat{H}(\Omega)=\{\overset{\rightharpoonup}{u}|\overset{\rightharpoonup}{u} \in H^1_0(\Omega), \nabla\cdot \overset{\rightharpoonup}{u}=0\}$.

$\displaystyle (1)\begin{cases}\nu\Delta \overset{\rightharpoonup}{v}=\nabla p-\overset{\rightharpoonup}{f}\\ \nabla\cdot \overset{\rightharpoonup}{v}=0\\ \overset{\rightharpoonup}{v}|_{\partial \Omega}=0\end{cases}$

Reasonable solution of this problem is $\overset{\rightharpoonup}{v}_x\in L^2(\Omega)$ from physical reason. So it is also “natural” to require $\overset{\rightharpoonup}{v}\in W^{1,2}(\Omega)$. However, if the $\partial\, \Omega$ is too bad(say not lipschitz), we can not talk about the $\overset{\rightharpoonup}{v}|_{\partial \Omega}$ in a right sense. So we have to require $\overset{\rightharpoonup}{v}$ is in a reasonable space and $\partial \Omega$ is has some regularities. In order to overcome this difficulty, we force $\overset{\rightharpoonup}{v}\in H(\Omega)$ for any domain, bounded or unbounded, except $\Omega=\mathbb{R}^2$.

Formulation I: A generalized solution is $\overset{\rightharpoonup}{v}\in H(\Omega)$ and $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$, for all $\overset{\rightharpoonup}{u}\in H(\Omega)$.

In this formulation, we force $\overset{\rightharpoonup}{v}\in H(\Omega)$. If $\overset{\rightharpoonup}{f}\in H'(\Omega)$, which means

$\displaystyle \int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u}dx\leq C||u||_{H(\Omega)}$

then $(1)$ is uniquely solved. Additionally, we do not need to make any assumption on $p(x)$ near the boundary of $\Omega$.

Formulation II: A generalized solution is $\overset{\rightharpoonup}{v}\in \widehat{H}(\Omega)$ and $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$, for all $\overset{\rightharpoonup}{u}\in \widehat{H}(\Omega)$.

In this formulation, in order to get $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$ from $(1)$, we need $\int_{\Omega}\nabla p\cdot \overset{\rightharpoonup}{u}dx=0$, $\forall \overset{\rightharpoonup}{u}\in \widehat{H}(\Omega)$. But this require $p(x)\in L^2(\Omega)$. We can not isolate finding of $\overset{\rightharpoonup}{v}$ from the of $p$. Moreover, it is not natural to require $p\in L^2(\Omega)$ for unbounded domain from the physical point of view.

Formulation III: A generalized solution is $\overset{\rightharpoonup}{v}\in \widehat{H}(\Omega)$ and $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$, for all $\overset{\rightharpoonup}{u}\in {H}(\Omega)$.(see[2])

In this formulation, we are able to isolate $\overset{\rightharpoonup}{v}$ and $p$. But this may results in $(1)$ has more than one solutions. According to the paper of Heywood, the number of solution is the number of cosets $\widehat{H}(\Omega)/H(\Omega)$. He actually gives an example such that $\widehat{H}(\Omega)/H(\Omega)$ is one-dimensional.

So the real question is whether $H(\Omega)$ coincides with $\widehat{H}(\Omega)$. [1] says that if $\Omega$ is “star shaped” bounded domains, or $\Omega$ has a compact lipschitz boundary.

[2] proves that if $\Omega$ is the exterior domain with $C^2$ boundaries, then they are the same space.

Application:

$\displaystyle (2)\begin{cases}\nu\Delta \overset{\rightharpoonup}{v}=\nabla p-\overset{\rightharpoonup}{f}\\ \nabla\cdot \overset{\rightharpoonup}{v}=0\\ \overset{\rightharpoonup}{v}|_{\partial \Omega}=\alpha(x)\end{cases}$

When one try to solve $(2)$, first one takes some solenoidal extension $\overset{\rightharpoonup}{a}$  of $\alpha(x)$ inside of $\Omega$. If one forces $\overset{\rightharpoonup}{a}\in W^{1,2}(\Omega)$, one can find a solution of $(2)$ in the form of $\overset{\rightharpoonup}{a}+\overset{\rightharpoonup}{v}$, where $\overset{\rightharpoonup}{v}\in H(\Omega)$. However, there is question here, if we find another extension $\overset{\rightharpoonup}{a}$ of $\alpha(x)$, are we necessarily find the same solution?

In another way, we can find solution of $(2)$ by hydrodynamic-potentials method, is this solution the same with what we have obtained before?

All of these question is related to $\widehat{H}(\Omega)/H(\Omega)$.

$\textbf{Remark:}$

[1]O.A. Ladyzhenskaya, V.A. Solonnikov. Some problems for vector analysis and generalized formulations of boundary-value problems for the navier-stokes equations.

[2]J.G. Heywood. On uniqueness questions in the theory of viscous flow.