Odd-Even Decoupling on 1D 7 pt. Stencil Finite Differences?
 

Dear All,

I am writing my code from scratch for incompressible Navier-Stokes. I will use 7 pt. stencil Finite Differences on a given direction, so a total 13 pt stencil on 2D. What are your ideas on odd-even decoupling?

My guess is that since odd and even nodes are able to communicate with each other because the only zero entry will be the center node for which I have to calculate the first derivative. However, I cannot be sure. Any insights or advices?

Best Regards,

Ali

But is your stencil used for the pressure equation? how do you apply the div grad product in discrete form?

Yes, the stencil is also used for the pressure equation, no staggered grid. The paper I read suggests applying upwind discretization for div and downwind for grad or vice versa. Thus, odd-even decoupling would not occur.

My thinking is that div dot grad product is mathematically Laplace operator. So, I thought of using 7 pt. stencil on the Laplace operator.

This brings another question for which I have not looked up the answer, why bother with div dot grad when it is simply Laplacian?

Matemathically the product corresponds to the Laplacian. However, numericall there are several well known issues that drive to an Approximate or an Exact projection method.
Consider that you have to compute the gradient of the pressure to ensure the continuity constraint. I suggest to see by hand in 1D the differences if you discretize the div grad and the Lap operators.
You can find many references in the CFD literature

Thank you for the explanation. Any ideas on odd-even decoupling?

It happens for some type of discretizations on non-staggered grids. A classical example is the second order discretization. Other examples can be found in the literature. I never used a 7-point stencil for the gradient operator so you have to develop the scheme. But I immagine that the implementation will be very difficult close to the boundaries..