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I was playing a little bit with the Runge-Kutta procedure for the Incompressible Navier-Stokes equation and came up with something strange, so I would like to know where I'm wrong or doing something I shouldn't.

Let's consider $u_i = u^n + \Delta t \sum_{j=0}^{i-1}a_{ij}F(u_j)u_j - c_i \Delta t G \phi_i $

where $F$ is the right hand side operator of the Navier-Stokes equation and $G$ is the discrete gradient operator.

So if I want to find a pressure equation, I should apply the divergence $M$ operator both sides

$Mu_i = Mu^n + \Delta t \sum_{j=0}^{i-1}a_{ij}MF(u_j)u_j - c_i \Delta t MG \phi_i $

and by employing the incompressibility constraint in the inner stages and of course for the $u^n$ (i.e. $M u^n = 0$ and $Mu_i = 0$), I get

$ MG \phi_i = \frac{1}{c_i} \sum_{j=0}^{i-1}a_{ij}MF(u_j)u_j$

Let's define $L = MG$ as the discrete laplacian operator, so that

$ L \phi_i = \frac{1}{c_i} \sum_{j=0}^{i-1}a_{ij}MF(u_j)u_j$

I know that if I solve this pressure poisson equation, I would get the $\phi_i$ lagrange multiplier that allows me to project the velocity into a divergence free field.

But since I need the gradient of the pseudo-pressure, what if I write something like

$ \phi_i = \frac{1}{c_i} \sum_{j=0}^{i-1}a_{ij}L^{-1}MF(u_j)u_j$

$ G\phi_i = \frac{1}{c_i} \sum_{j=0}^{i-1}a_{ij}GL^{-1}MF(u_j)u_j$

but

$GL^{-1}M = G(MG)^{-1}M = GG^{-1}M^{-1}M = I$

so I get that

$G\phi_i = \frac{1}{c_i} \sum_{j=0}^{i-1}a_{ij}F(u_j)u_j$

that implies that $u_i = u_n$, hence I'm not advancing in time ?? I don't know where am I wrong. Thank you guys in advance.

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1 Answer 1

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Applying the inverse of the Laplacian $L$ and $G$

$$ G\phi_i = \frac{1}{c_i} \sum_{j=0}^{i-1}a_{ij}G L^{-1}MF(u_j)u_j, $$

would violate some previous assumpions

$$ M u^n = 0 ,\quad Mu_i = 0, $$

since

$$ \begin{align} L^{-1}M u^n &\ne 0 ,\quad L^{-1}Mu_i \ne 0,\\ (MG)^{-1}M u^n &\ne 0 ,\quad (MG)^{-1}Mu_i \ne 0,\\ G^{-1}M^{-1}M u^n &\ne 0 ,\quad G^{-1}M^{-1}Mu_i \ne 0,\\ G^{-1} u^n &\ne 0 ,\quad G^{-1}u_i \ne 0.\\ \end{align} $$

and

$$ \begin{align} GG^{-1} u^n &\ne 0 ,\quad GG^{-1}u_i \ne 0,\\ u^n &\ne 0 ,\quad u_i \ne 0. \quad \square \\ \end{align} $$

The correct equation should be

$$ G\phi_i = \frac{1}{c_i} \sum_{j=0}^{i-1}a_{ij}F(u_j)u_j - 1/\Delta t (u_i - u^n). $$

See your first equation.

Edit (see the comment):

Suppose you have a divergence/gradient operator, we call it $D$, with a divergence free $u = c$ (constant).

Then it holds

$$ D u = 0. $$

Now we apply the inverse of the divergence/gradient operator and naively assume that a matrix-vector multiplication with a zero vector results also in a zero vector

$$ \begin{align} D^{-1} D u &= D^{-1} 0,\\ u &= 0. \end{align} $$

However, this contradicts the assumption $u=c$.

The only logical conclusion from this is that the discrete divergence/gradient operator is not invertible on its own without further assumptions, e.g. boundary conditions.

Or in other words: The indefinite integral of zero is not zero but constant c.

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  • $\begingroup$ Thanks for the answer, I have a few questions though. If $M u^n = 0$ and the matrix multiplication is associative, shouldn't be $L^{-1}Mu^n = 0$ too ? Also, in the last equation you wrote, is it missing a $\Delta t$ at the denominator for $u_i$ and $u^n$ ? $\endgroup$
    – Marco
    May 1, 2023 at 15:04
  • $\begingroup$ That is an interesting question, see my edit. $\endgroup$
    – ConvexHull
    May 1, 2023 at 17:11
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    $\begingroup$ I got it thanks, indeed $D$ is singular, otherwise there will be no chance other than $u = 0$ for the equation $Du = 0$ to have a solution, as well as $L$ his $L^{-1}$ does not exist, and writing it as $L^{-1}$ is just only "a way to say" solve the following system, but I'm not allowed to do any particular mathematical manipulation with it, right? big thanks anyway $\endgroup$
    – Marco
    May 1, 2023 at 17:28
  • $\begingroup$ You are welcome. By the way, I added the $\Delta t$ $\endgroup$
    – ConvexHull
    May 1, 2023 at 18:44

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