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I've asked this question on Math.SE too.

Let

  • $d\in\left\{1,\ldots,4\right\}$
  • $\Lambda\subseteq\mathbb R^d$ be bounded, nonempty and open and $\partial\Lambda$ be Lipschitz
  • $V:=\left\{u\in H_0^1(\Lambda,\mathbb R^d):\nabla\cdot u=0\right\}$
  • $W:=\left\{p\in L^2(\Lambda):\int_\Lambda p=0\right\}$
  • $\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2}$ for $u,v\in H^1(\Lambda,\mathbb R^d)$
  • $\mathfrak b(p,v):=\langle p,\nabla\cdot v\rangle_{L^2}$ for $p\in L^2(\Lambda)$ and $v\in H^1(\Lambda,\mathbb R^d)$
  • $\mathfrak c(u,v,w):=\langle((u\cdot\nabla)v,w\rangle_{L^2}$ for $u,v,w\in H^1(\Lambda,\mathbb R^d)$

The usually studied variational formulation of the steady Navier-Stokes equation is

\begin{equation}\left\{ \begin{split} \mathfrak a(u,v)+\mathfrak b(p,v)+\mathfrak c(u,u,v)&=0\;\;\;\text{for all }v\in H_0^1(\Lambda,\mathbb R^d)\\ \mathfrak b(u,q)&=0\;\;\;\text{for all }q\in W \end{split}\tag1\right. \end{equation}

where $(u,p)\in H_0^1(\Lambda,\mathbb R^d)\times W$ is the searched solution.

I want to solve $(1)$ numerically. Usually, $(1)$ is linearized in some way and then a mixed finite element method is used to approximate a solution.

However, I'm not interested in $p$. Now, $$\mathfrak a(u,v)+\mathfrak c(u,u,v)=0\;\;\;\text{for all }v\in V\tag2$$ (where $u\in V$ is the searched solution) is a variational formulation of the steady Navier-Stokes equation which is equivalent to $(1)$ and which doesn't contain $p$.

I wonder if it might be better for me to find a numeric scheme which solves $(2)$. A linearization, e.g. an Oseen iteration, is possible for $(2)$ too. I think the crucial point is the choice of a (conforming) finite element.

So, the question if I can benefit from the fact that I'm not interested in $p$ and hence don't need to care about its approximation. I've read many papers, but I couldn't find any which tries to solve the steady Navier-Stokes equation for the velocity only.

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0xbadf00d
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    It very much depends on what you are modelling -- (2) is not equivalent to (1) unless the pressure gradient is zero (otherwise you'd get a non-zero internal thermodynamic source driving the fluid flow). So even if you are not interested in $p$, the fluid might be, and if you don't solve for it accurately enough, your approximation of $u$ will suffer (and, in fact, be limited by your approximation of $p$, see http://scicomp.stackexchange.com/a/14157). – Christian Clason Feb 27 '17 at 20:08
  • @ChristianClason If $(u,p)$ solves the first equation in $(1)$, then $u$ solves $(2)$, since $\mathfrak b(p,v)=0$ for all $v\in V$. On the other hand, if $u\in V$ is a solution of $(2)$, then there is a unique $p\in W$ with $(1)$. So, the problems are equivalent. – 0xbadf00d Feb 27 '17 at 20:20
  • This might just be because this isn't quite the notation I'm used to, but where are your boundary conditions? – origimbo Feb 27 '17 at 20:42
  • Wouldnt that require test functions that are a-priori divergence free? This way you would already be projecting the velocity on a divergence free subspace? How could you construct such test functions? – BlaB Feb 27 '17 at 20:58
  • @origimbo The Dirichlet boundary condition is implicit by the choice of $H_0^1(\Lambda,\mathbb R^d)$ as the solution and test function space of $(1)$. – 0xbadf00d Feb 27 '17 at 21:12
  • @BlaisB The test functions in $(2)$ are chosen from $V$ and hence are divergence-free. I don't know what you mean by "constructing" such test functions. $V$ is a separable Hilbert space and hence has a countable orthonormal basis. So, we can construct finite-dimensional subspaces of $V$ by choosing a finite subset of this basis. – 0xbadf00d Feb 27 '17 at 21:13
  • You mean you want to do something like this: http://www.math.udel.edu/~szhang/research/p/2dpowell.pdf? – Christian Clason Feb 27 '17 at 22:52
  • The pressure $p$ is nothing more than a Lagrange multiplier imposing $\nabla \cdot u = 0$. You must impose $\nabla \cdot u = 0$ somehow. I think Lagrange multiplier is a better approach than any others I've seen. – knl Feb 28 '17 at 11:23
  • @knl $\nabla\cdot u=0$ is enforced by the choice of $V$ as the solution and test function space of $(2)$. – 0xbadf00d Feb 28 '17 at 12:26
  • @ChristianClason I will take a look at that. Meanwhile, please take note of my comments below the answer of Wolfgang Bangerth. Maybe they clarify. – 0xbadf00d Feb 28 '17 at 12:53
  • I'm not talking about anything abstract, like the definition of $V$, but about practical constraints posed on the floating point numbers existing on the memory of a computer. – knl Feb 28 '17 at 12:56
  • @knl So, what you mean is that it's hard to code test functions $v$ with $\nabla\cdot v=0$ due to floating point accuracy, right? We could do so, but that would lead to a poorer approximation than the Lagrange multiplier approach. Is that what you mean? If so, this is rather vague. Is there some exploration of this topic (in the context of finite element methods)? – 0xbadf00d Feb 28 '17 at 13:09
  • Sorry for not being clear. I was not trying to say that the cause is insufficient floating point accuracy. I was merely trying to stress the fact that the definition of $V$, as you wrote it, is somewhat abstract and it is not immediately clear how to implement it practice. On the other hand, Lagrange multipliers are very well established in finite dimensional optimization, etc., so it is quite clear how to do the practical implementation. AFAIK, Lagrange multipliers were originally introduced as a trick to impose constraints in finite dimensional optimization problems. – knl Feb 28 '17 at 13:14

1 Answers1

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It is difficult, in practice, to construct finite dimensional subspaces of $V$ that are conforming and that still satisfy appropriate approximation properties. For example, finite element spaces lack appropriate continuity properties across faces between cells. But, it is not difficult to use Fourier approximations that are divergence free and that allow you what you want to do. Of course, Fourier approximations require you to work in a box geometry -- and that is their main practical drawback, because the Navier-Stokes equations are just not very interesting from an applied perspective in a box geometry.

Wolfgang Bangerth
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  • Just to make sure I've understood what you mean: If we deal with $(1)$, we search a solution $u\in H_0^1(\Lambda,\mathbb R^d)$ of the first equation of $(1)$ (for some $p\in W$) and enforce $\nabla\cdot u=0$ by the second equation of $(1)$. This yields $u\in V$. I've asked the question: If we're not interested in $p$, why don't we search the solution of the first equation of $(1)$ in $V$ in the first place? And in order to state this problem completely in $V$, we should test against functions from $V$. This immediately leads $(2)$, cause $(\mathfrak b(p,v)=0$ for all $v\in V$). – 0xbadf00d Feb 28 '17 at 12:50
  • Doing so, we wouldn't need to deal with mixed finite elements. We would only need to deal with the space $V$. Now, you say the problem is that it's "difficult" to construct finite-dimensional subspaces of $V$. Why? And it's not clear to me, why they have poorer approximation properties that the subspace of $H_0^1$. – 0xbadf00d Feb 28 '17 at 12:51
  • What I wanted to say in my answer is that it is possible to do as you suggest, i.e., search for solutions in $V$ (with test functions in $V$). The problem is constructing finite dimensional subspaces. For example, you could use $Q_1$ elements for the $x$ and $y$ velocities, and then choose the $z$ velocity so that the divergence is zero in the interior of the element. But the $z$ velocity may not be continuous across cell interfaces, and so the velocity is not in a subspace of $H^1$, and consequently not in a subspace of $V$. – Wolfgang Bangerth Feb 28 '17 at 13:11
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    In other words, the challenge is to construct finite element spaces whose functions are both divergence free, and sufficiently regular to be a subspace of $H^1$. I don't think I've ever seen that worked out on arbitrary meshes. (Which doesn't mean that it can't be done, but I suspect that you need fairly high polynomial degrees.) – Wolfgang Bangerth Feb 28 '17 at 13:12
  • I'm only interested in triangulations of polyhedrons. So, the mesh I've got in mind is not arbitrary. – 0xbadf00d Feb 28 '17 at 13:24
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    I'm quite sure that by the term 'arbitrary mesh' Wolfgang does not refer to something more general than 'triangulations of polyhedrons'. Even arbitrary polyhedron is something far more general than we typically consider in computational science. When a computational scientist uses the term 'arbitary mesh' he usually describes tetrahedral/triangular or hexahedral/quadrilateral mesh that does not have any constraints regarding the shape of the individual elements. There are examples of numerical methods that work only, e.g., with square elements, and hence the notion. – knl Feb 28 '17 at 19:41
  • @knl is right. When I say "arbitrary mesh" I mean "unstructured mesh", as opposed to "a mesh that only contains axi-parallel rectangular cells" or something equally simple with triangles or tetrahedra. – Wolfgang Bangerth Feb 28 '17 at 23:17