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\}.

Consider the steady Stokes problem

\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.

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