• $T>0$
  • $I:=(0,T]$
  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be nonempty and open, $$\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda,\mathbb R^d):\nabla\cdot\phi=0\right\}$$ and $$V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda,\:\mathbb R^d)}}\;,\;\;\;H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}$$
  • $\operatorname P_H$ denote the orthogonal projection from $L^2(\Lambda,\mathbb R^d)$ onto $H$
  • $A_0u:=-\Delta u$ for $u\in\mathcal D(A_0):=H_0^1(\Lambda,\mathbb R^d)\cap H^2(\Lambda,\mathbb R^d)$, $$Au:=\operatorname P_HA_0u\;\;\;\text{for }u\in\mathcal D(A):=\mathcal D(A_0)\cap V$$ and $$B(u,v):=(u\cdot\nabla)v\;\;\;\text{for }u\in L^2(\Lambda,\mathbb R^d)\text{ and }v\in H^1(\Lambda,\mathbb R^d)$$
  • $f:I\to H$
  • $u\in L^2(I,\mathcal D(A))$ with $u'\in L^2(I,H)$ and $$u'(t)+A_0u(t)+B(u(t),u(t))+\nabla p(t)=f(t)\;\;\;\text{for all }t\in I\tag1$$ for some $p:I\to H^1(\Lambda)$

Assuming that $\Lambda$ is sufficiently regular such that $(1)$ is well-defined, it can be shown that $(1)$ is equivalent to $$u'(t)+Au(t)+\operatorname P_HB(u(t),u(t))=f(t)\;\;\;\text{for all }t\in I\;.\tag2$$ I want to solve $(2)$ numerically and I'm only interested in $u$ (and not in $p$).

I know that there are many references for the numerical study of $(1)$. However, it seems to me that all the considered schemes don't use $(2)$. They only use $(2)$ for theoretical results like existence and uniqueness of solutions. Maybe I'm wrong and I just don't see that these schemes use $(2)$.

In any case, my question is: Are we able to provide a numerical scheme which solves $(2)$ directly?

Or is there something which prevents us from doing that? My idea is to apply, for example, a semi-implicit Oseen discretization in time, i.e. consider $$\frac{u(t_n)-u(t_{n-1})}h+Au(t_n)+\operatorname P_HB(u(t_{n-1}),u(t_n))=f(t_n)\;\;\;\text{for all }n\in\left\{1,\ldots,N\right\}\tag3$$ with $$t_n:=nh\;\;\;\text{for }n\in\left\{0,\ldots,N\right\}$$ and $h:=T/N$ for some $N\in\mathbb N$. After that for each $n\in\left\{1,\ldots,N\right\}$ $(3)$ should be solvable by a finite element method (or is there some problem that I don't see?).

  • $\begingroup$ Where is the projection operator $P_H$ in eq. 3 ? $\endgroup$ Feb 24, 2017 at 14:26
  • $\begingroup$ @PiyushGrover Fixed that. $\endgroup$
    – 0xbadf00d
    Feb 24, 2017 at 22:25
  • $\begingroup$ So, your suggestion is to project the equation onto divergence free functions, thus throwing $p$ away, and solve it there, is it right? If yes, then I think, one can do that, but the main difficulty will be to construct a finite element space within divergence-free functional space. $\endgroup$
    – VorKir
    Feb 25, 2017 at 2:16
  • $\begingroup$ I recommend asking on scicomp.SE. $\endgroup$ Feb 25, 2017 at 4:35
  • $\begingroup$ @VorKir Yes, that's the idea. If we test $(3)$ against $v\in V$, we obtain $$\frac1h\langle u^n-u^{n-1},v\rangle_H+\mathfrak a(u^n,v)+\mathfrak b(u^{n-1},u^n,v)=\langle f^n,v\rangle_H\;,$$ where $u^n:=u(t_n)$, $f^n:=f(t_n)$, $$\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v\in H_0^1(\Lambda,\mathbb R^d)$$ and $$\mathfrak b(u,v,w):=\langle B(u,v),w\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v,w\in H_0^1(\Lambda,\mathbb R^d)\;.$$ You see a problem in choosing a (conforming) finite element space $V_h\subseteq V$, right? $\endgroup$
    – 0xbadf00d
    Feb 25, 2017 at 10:14


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