# Existence and uniqueness of Stokes flow

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.

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@ColinMcFaul, is it ok to reask it on math.SE or I should flag and hope for a migration? –  Juris May 18 '13 at 20:12
The etiquette is to flag for a migration. But I think the mods will migrate only if several people think it's off-topic here. –  Colin McFaul May 18 '13 at 22:10
The second condition should be $\nabla \cdot u= 0$, i.e., incompressible? –  Shuhao Cao Jun 6 '13 at 4:16
@ShuhaoCao, adding the dot changes anything there? –  Juris Jun 11 '13 at 14:43
The dot means dot product $\nabla \cdot$ is the divergence operator: en.wikipedia.org/wiki/Divergence What you wrote was gradient operator without that dot. The Stokesian flow we are learning is most of the time *incompressible*, using math terms, should be divergence free, i.e., $\nabla \cdot \vec{u} = 0$. –  Shuhao Cao Jun 11 '13 at 15:13

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.

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.

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).

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Thanks, this is quite the answer that I wanted, but I don't understand everything. The main problem is that I don't understand how could I check if particular problem is solvable or not - where do I plug in the boundary conditions? Maybe you could show an example? Could you please use this and show that the problem in the Stokes paradox is unsolvable? –  Juris Jun 11 '13 at 14:47
@Juris Hi, Juris. If you could show your particular problem, then please. For Stokes' paradox, the idea is that for an infinite cylinder, a sensible global solution can't be established for small Reynolds number. However, the good thing is, if we are dealing with bounded piecewise smooth domain, all those problems are gone. –  Shuhao Cao Jun 11 '13 at 15:21
OK, how about this one: a box of shear flow with given velocity $\vec{u}=u_0z\vec{e_x}$ for $x=0$,$x=L_x$,$y=0$,$y=L_y$,$z=0$,$z=L_z$ and stationary ($\vec{u}=\vec{0}$) sphere of radius $r<L_z<L_y<L_x$ at $\frac{1}{2}(L_x,L_y,L_z)$. –  Juris Jun 11 '13 at 19:38
@Juris In this problem, I am not sure what to do though, for the solution $\vec{u}$ is already given. Do you wanna check if the flow satisfying the Stokes equation? I suggest you post a new question about this. Based on your problem, guess my answer doesn't really suit your instant need, but the point is, in a bounded smooth domain, the Stokes equation is well posed with Dirichlet boundary condition. Also the boundary condition is incorporated into the function spaces. –  Shuhao Cao Jun 11 '13 at 20:06

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"

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Thanks, I know of Stokes' paradox, but has anyone specified conditions to distinguish solvable and unsolvable situations? –  Juris May 18 '13 at 12:37
Don't pin me down on it, but I believe the situation is only unsolvable if the far-field is unbounded –  Michiel May 18 '13 at 12:43
Dear @michielm: Please mention properly the author, title, etc, of the lecture notes (as opposed to just saying 'this document'). This serves at least 2 purposes: i) I'm sure the author appreciates the proper recognition, and ii) in case of future link rot, we can reconstruct your answer. –  Qmechanic May 18 '13 at 13:50
@Qmechanic Ok! I have corrected the referencing. –  Michiel May 18 '13 at 15:23