# Tag Info

## Hot answers tagged incompressible

5

When we say that the pressure is only defined "up to a constant", what we mean is this: If $u,p$ is a solution to the Navier-Stokes equations, then $u,p+c$ for any constant $c$ is also a solution. In other words, the solution is not unique, but the non-uniqueness has a very particular structure: two solutions $u,p$ and $u,p'$ can only differ in a way where $... 5 The problem was solved recently: Lederer/Schöberl: Polynomial robust stability analysis for H(div)-conforming finite elements for the Stokes equations 4 The projection scheme is indeed analytic, provided that the initial conditions are incompressible. You can derive it by taking the divergence of the evolution equation for$u$. Finding the projection P can be tricky; it's easier with some methods like spectral / pseudospectral methods. But one has to be careful that numerical error doesn't move the ... 4 It's important to realize that the original velocity field$\mathbf v_0$is also not divergence free in a pointwise sense. Rather, it is only divergence free when tested with the pressure test functions, i.e.,$(\nabla\cdot\mathbf v_0,q)=0$for all$q\in P_1(T_0)$where$T_0$is the original mesh. What you want to achieve is to get a velocity field$\mathbf ...

3

Water (and other fluids that are considered incompressible) are of course in reality compressible -- just not very much, at the pressures and speeds involved in the flow we are modeling. Within the ranges of pressures we consider, a good approximation is that $$\rho = \rho_0 (1+\alpha p)$$ where $\rho_0$ is the reference density at reference pressure, $p$...

3

All of the schemes I have seen are consistent. To be concrete, the first reference essentially rewrites their original equation $$\frac{\partial U}{{\partial t}} + \nabla H(U) = 0$$ into $$\frac{\partial U}{{\partial t}} + \nabla H(U) + h^3 \left( \frac{\partial}{\partial x} (\epsilon \frac{\partial ^3}{\partial x^3} U) + \frac{\partial}{\partial y} (\... 3 Note: I originally posted an answer that I was not 100% pleased with, so I have revised it heavily. 1/20/2017 The projection method is not in an exact approximation to the full system in general. There are a few reasons for this, but the most immediate has to do with the boundary conditions. Let's look at the continuous setting first. Consider the unsteady ... 2 It is a consequence of the continuity condition (\nabla \cdot u = 0), condition (3) at page 2. At chapter 4 at the beginning there is the assumption about the condition (3) is valid. You are at the \textit{n+1}, step and so$$\nabla \cdot U^{n+1} = 0$$This method is in the family of projection method. In this class of methods the pressure and ... 2 Perhaps your query is regarding "artificial compressibility" vs "artificial dissipation/viscosity". Artificial compressibility method is a way to solve the incompressible Navier-Stokes equations where the divergence free condition is satisfied by doing "inner iterations". This is an alternative to projection methods. It is unrelated to the artificial ... 2 That's what I usually do. Define streamline function:$$ \Psi = \left[ \begin{array}{c} \psi_x\\ \psi_y\\ \psi_z\\ \end{array} \right] $$the velocity equals:$$ \mathbf{u} = \nabla\times\Psi = \left[ \begin{array}{c} u_x =\partial_y \psi_z - \partial_z \psi_y\\ u_y =\partial_z \psi_x - \partial_x \psi_z\\ u_z =\partial_x \psi_y - \partial_y \psi_x\\ \...

2

I cannot comment, so as an answer may I recommend looking at something like: http://lorenabarba.com/blog/cfd-python-12-steps-to-navier-stokes/ form start to finish? This takes the reader from basic discretisation of convective and diffusive terms in 1d and 2d, over a way of dealing with incompressibility via an equation for pressure derived from the ...

2

It's of course hard to tell without knowing the code, the testcase, and your results. So I can't really help you out here. But it is worth pointing out that you are making yourself too much work. Any of the big, open source finite element libraries have these elements already built in and exceedingly well tested. They also have adaptive meshes, excellent ...

2

I would not bother too much. Chorin's method is known to be a consistent scheme, which means, that for smaller time-steps it will approximate the actual solution. Checkout Gresho's/Sani's book Incompressible Flow and the Finite Element Method. Vol. 2: Isothermal Laminar Flow there is the convergence theory and also Chorin's method with a forcing term. So, ...

1

We have recently proposed a mixed u-p formulation that results in a global symmetric matrix irrespective of the volumetric energy function (for hyperelasticity). The paper contains sufficient details on the linearisation (2nd derivative).

1

I think that both can be considered to be equally valid. In fact, asymptotically (with an infinite number of lattice) you will tend towards the incompressible athermal Navier-Stokes equations. In this limit, both measures of the flow should be equal. If I were you, I would initially report the two. Then I would measure the difference between them as a ...

1

I resolved my problem. I added a small pressure coefficient in the continuity equation, let's say $10^{-10}$. In the weak form, this quantity is multiplied by the integral {$N^T$ $N$} (or mass matrix), where $N$ are the shape functions of the 8 noded hexahedral. The code now works fine.

1

What sort of field is it? You've said "divergence free", but do you mean harmonic (zero divergence and curl), solenoidal (non-curl), or a mix of both? The distinction is important because it will affect the complexity of the algorithm. The goal is to do a Helmholtz decomposition numerically, interpolate the potentials, and then recalculate the velocities ...

1

Ironically, there is an iterative method called "SIMPLE" (semi-implicit method for pressure-linked equations) designed to resolve the steady state navier-stokes equations based on a predictor-corrector scheme. It works by solving a linearized form of the momentum equation (predictor step) which produces a velocity field which generally does not satisfy ...

1

With the continuity eqn only, you are missing all the mechanical balance: viscous and/or inertial effects will decide of the streamlines of such a flow. If your major aim is to keep it as simple as possible, I would go for a lubrication approximation. If you consider that your fluid is seeping between two flat plates with a very narrow gap, in the end the ...

1

@Wolfgang Bangerth has a very important point about what $f$ is. The constant external force will not affect your solution unless your BCs allow for motion. In your case, they do not. An easy conceptual proof of this is to look at the vorticity equation (by taking the curl of your equations). The source term disappears, which means no motion will ensue. ...

1

For the lid driven cavity problem, we apply the Dirichlet boundary condition (or, no-slip boundary condition) to the velocity field $\tt U$, that is, $\tt U = U_{\tt Lid}$ on the moving Lid and $\tt U=0$ on the rest of the boundary. Based on this, we should have ${\tt U}^{n+1} = {\tt U}^{n}$ on the boundaries. Hence, if we have \$\tt U^{n+1} = U^{n} - \Delta ...

1

In the meantime we did some more research into this issue and documented our findings in: M. Huber, U. Rüde, C. Waluga, B. Wohlmuth. Highly sparse surface couplings for subdomain-wise isoviscous Stokes finite element discretizations, Journal of Scientific Computing, 2017 (article, preprint). In summary, there are a couple of points that can be made: Yes, ...

Only top voted, non community-wiki answers of a minimum length are eligible