Existence and uniqueness of Stokes flow

Question

What are the solution existence and uniqueness conditions for Stokes' flow?

$$\begin{gathered} \nabla p = \mu \Delta \vec{u} + \vec{f} \\ \nabla \cdot \vec{u} = 0 \end{gathered}$$

Maybe you could also provide some articles or books about the topic? Most physics books seem not to care about these details.

Answer 1

Michiel's answer is more from the aspects of physics. Here is the pde style answer.

Short answer: The Stokes flow's variational problem is well-posed (uniqueness and existence) in certain Hilbert spaces pair which relies on inf-sup condition.

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Functional equations:$\newcommand{\b}{\boldsymbol}$ Suppose $\mu=1$, the Stokesian flow can be written in the following way (I believe this is called the Pressure-Velocity formulation): $$\left\{ \begin{aligned}&-\Delta \b{u} + \nabla p =\b{f}, \\ &\nabla\cdot \b{u} =0.\end{aligned}\right.\tag{1} $$ In operator form this can be written as the following abstract problem: $$ \begin{pmatrix}A & B' \\ B& 0\end{pmatrix}\begin{pmatrix}\b{u}\\p \end{pmatrix} = \begin{pmatrix}\b{f}\\0 \end{pmatrix}, $$ where $A = -\Delta$ is the vector Laplacian, and $B = -\nabla\cdot$ with $B' = \nabla$, $\b{u}\in X$, and $p\in Y$, the operators: $$ A: X\to X',\quad B:X\to Y', \quad \text{and}\quad B': Y\to X'. $$

Stokes problem is well-posed when the follpwing operator is an isomorphism: $$ \mathscr{S}:(\b{v},q)\in X\times Y \mapsto (A\b{v}+B'q,B\b{v})\in X'\times Y'. $$

Normally, the isomorphism is either proved using Lax-Milgram through coercivity to pin down a fixed point, or using Fredholm alternative.

The sufficient condition for this is a weak version of coercivity (you can view it as invertibility of an operator):

$B: \mathrm{ker}(B)^{\perp}\subset X \to Y'$ is an isomorphism and $\|\b{v}\|_X \leq \beta \|B\b{v}\|_{Y'}$.

$B': Y \to (\mathrm{ker}(B)^{\perp})'$ is an isomorphism and $\|q\|_Y \leq \beta \|B'q\|_{X'}$.

Then by closed range theorem, Babuska proved an equivalence of these conditions with the inf-sup condition(in that pdf link 1.1). Whenever that condition holds for certain Hilbert spaces pair $X\times Y$, (1)'s variational problem has a unique solution.

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Weak formulation:

The weak formulation for the abstract version of (1) is then:

$$\left\{ \begin{aligned}\langle A \b{u},\b{v}\rangle + \langle p,B\b{v}\rangle =\langle\b{f},\b{v}\rangle,\quad \forall \b{v} \in X&, \\ \langle q,B\b{u}\rangle =0,\;\;\;\quad\quad \forall q \in Y.&\end{aligned}\right.\tag{2} $$ Possible pairs of Hilbert spaces $X\times Y$ mentioned above are: $$\begin{gathered} H^1_0(\Omega)\times \{q\in L^2(\Omega):\int_{\Omega}q=0\}, \\ H(\mathrm{div})= \{\b{v}\in L^2(\Omega):\nabla\cdot \b{v}\in L^2(\Omega)\}\times L^2(\Omega). \end{gathered}$$ Using integration by parts for (2) leads to: $$\left\{ \begin{aligned}\int_{\Omega} \mathrm{tr}\big((\nabla \b{u})^T \nabla \b{v}\big) +\int_{\Omega}p(\nabla \cdot \b{v}) =\int_{\Omega}\b{f}\cdot\b{v},\quad \forall \b{v} \in X&, \\ \int_{\Omega}q(\nabla \cdot \b{u}) =0,\quad\quad\quad \forall q \in Y.&\end{aligned}\right.\tag{3} $$

Problem (3) can be viewed as a constraint minimization problem for the following conjugate functionals also (viewing the pressure $p$ as a Lagrange multiplier): denote $E(\b{v}) = (\nabla \b{v}^T +\nabla \b{v})/2$ (symmetric part of the Jacobian), the stationary strain tensor, then $\mathrm{tr}\big((\nabla \b{v})^T \nabla \b{v}\big)= |E(\b{v})|^2 $ (a notation usually used in elasticity PDEs). Let $$ \mathcal{L}(\b{v},q) = \int_{\Omega}|E(\b{v})|^2 - \int_{\Omega} \b{f}\cdot \b{v} - \int_{\Omega} q\nabla\cdot \b{v}, $$ and $$\mathcal{J}(\b{v}) = \sup_{q\in Y}\mathcal{L}(\b{v},q) ,$$ then our goal is to minimize $\mathcal{J}$ in $X$ (like looking for a saddle point).

Answer 2

There is one famous example in which there is no solution for the Stokes' flow case: Stokes flow around a cylinder, which is approriately named Stokes' paradox. In this case it is impossible to match the boundary conditions both at infinity (uniform flow) and at the cylinder surface (no-slip) with Stokes flow dynamics. See e.g. paragraph 6.4 of the Fluid Dynamics Lecture Notes of Jacques Lewalle. The breakdown of a solution means that there will always be some inertial effect, regardless how small the Reynolds number is.

An approximate solution to this issue was first proposed by Oseen who introduced a linearized inertial term to account for inertial contributions in the far field.

Later, Proudman and Pearson calculated a more precise solution through asymptotic expansions and matching of the far field solution and the near-cylinder solution.

So to answer your question: existence for Stokes' flow is not guaranteed even though the criterion $Re << 1$ is satisfied. A pretty good explanation for the precise reasons of non-existence is given in chapter 7 of the Fluid Dynamics I lecture notes by Prof. Childress. In the same document they also show (in paragraph 7.2) that Stokes' flow does exhibit uniqueness for non-trivial cases (i.e. $\textbf{u}\neq0$).

For more details on conditions for solvability: there is quite a lot of mathematical fluid dynamics literature on the solvability of Stokes' flow. Nazarov and Pileckas reference a number of them in their paper with the telling title "On the Solvability of the Stokes and Navier-Stokes Problems in the Domains That Are Layer-Like at Infinity"