• $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be nonnempty and open
  • $\langle\;\cdot\;,\;\cdot\;\rangle:=\langle\;\cdot\;,\;\cdot\;\rangle_{L^2(\Lambda,\:\mathbb R^d)}$

I've found a thesis where the author is suggesting (see Algorithm 3.1 on page 26) the scheme $$\langle u^n-u^{n-1},v\rangle+k\nu\sum_{i=1}^d\langle\nabla u^n,\nabla v\rangle+k\langle(u^n\cdot\nabla)u^n,v\rangle\color{red}{-k\langle p^n,\nabla\cdot v\rangle+\frac k2\langle(\nabla\cdot u^n)u^n,v\rangle}=k\langle f^n,v\rangle\tag1$$ for the time discretization of the (stochastic) Navier-Stokes equation (the stochastic forcing is omitted in $(1)$) $$u'-\nu\Delta u+(u\cdot\nabla)u+\nabla p=f+g(u)\dot W\tag2$$. Above $u^n=u(t_n)$, $p^n=p(t_n)$, $f^n=f(t_n)$ and $(t_n)_n$ is an equidistant partition of the time interval of size $k>0$.

What is the idea behind the red terms in $(1)$?

  • 1
    $\begingroup$ The first term in red is obtained by integration by parts of the pressure term (multiplied by test function $v$ ) in 2. $\endgroup$ – Piyush Grover Feb 24 '17 at 22:11
  • 1
    $\begingroup$ The second terms is intended to enforce incompressibility of $u$ $\endgroup$ – Piyush Grover Feb 24 '17 at 22:17
  • $\begingroup$ @PiyushGrover The first term is clear, but how does the second term enforce incompressibility? Instead of the second term, I had expected $$k\langle(u^n\cdot\nabla)u^n,v\rangle\;.$$ Why don't we use that? (And let me note that one obviously uses this relation.) $\endgroup$ – 0xbadf00d Feb 24 '17 at 22:29
  • 1
    $\begingroup$ That term is already present (third term in black) in 3. $\endgroup$ – Piyush Grover Feb 24 '17 at 22:52

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