Graduate Nonlinear first order Differential equation

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SUMMARY

The discussion centers on solving the momentum equation in 3D cylindrical coordinates, specifically the stationary solution of the equation ρ(∂v/∂t +(v.∇)v)=A, where A and velocity v are local vector variables. The user, Jay, is facing convergence issues while employing successive overrelaxation methods and is using IDL for computations. Suggestions include utilizing a stream function to reduce equation complexity and visualizing the 3D field to identify non-convergence causes. Additionally, the importance of considering coordinate singularities and refining mesh resolution is emphasized.

PREREQUISITES
  • Understanding of 3D cylindrical coordinates and their properties
  • Familiarity with the momentum equation in fluid dynamics
  • Experience with iterative numerical methods, specifically successive overrelaxation
  • Proficiency in IDL for computational modeling
NEXT STEPS
  • Explore the use of stream functions in fluid dynamics to simplify equations
  • Research techniques for visualizing 3D vector fields to aid in debugging
  • Investigate the impact of mesh resolution on convergence in numerical simulations
  • Learn about potential coordinate transformations to Cartesian coordinates for easier computations
USEFUL FOR

Fluid dynamics researchers, computational physicists, and engineers working on numerical simulations of nonlinear differential equations in cylindrical coordinates.

say_cheese
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I need to solve the well known momentum equation in 3D cylindrical coordinates:
ρ(∂v/∂t +(v.∇)v)=A

where A and the velocity v are both local vector variables.

I am actually looking for the stationary solution to the equation, i.e. no ∂/∂t term)

I have tried evolving the velocity and tried successive overrelaxation method but separately calculating derivatives. I can't get convergence.

Any one with a suggestion? I actually use IDL. So a suggestion in that would be even better.

Thanks
Jay
 
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What are your boundary conditions?
 
The theta and phi coordinates are cyclic (v(0)=v(2*pi)), v=0, dv/dr=0 at v=0 and a.
 
you can write u*(du/dx) as 1/2 (d(u^2)/dx). Try solving it that way.
 
I will try this, but that is only one term in (v.∇)v. The parameters have 3 components and the components of the equation have other component terms such as vr∂vθ/∂r.
 
say_cheese said:
I will try this, but that is only one term in (v.∇)v. The parameters have 3 components and the components of the equation have other component terms such as vr∂vθ/∂r.

Is there any chance your iterative methods are getting thrown off by coordinate and/or basis vector singularities? Did you include the derivative of the radial and angular basis vectors in the nonlinear term? Also, what's the resolution of your mesh?
 
Good questions. My present mesh sizes are about 5% of the variation length of the driving term A. The driving term is sinusoidal in theta and phi with a mild radial dependence. (180,50,50) mesh. There is a sharper variation in one area and I have remeshed there 5 times. It is quite possible that the divergence of iteration is seeded in one location. I need to interrupt the iteration and see. As you can see 3D problems are expensive in finer mesh. But yes, I will try doubling the mesh and see.

I realized that the proper way to solve it is to use a stream function - v=∇xψ (divergence free flow). This would reduce the number of equations and couplings, but would make it a higher order equation.
 
Good plan. I would also recommend developing some way of visualizing the 3D field so that you can look at it when you interrupt. You're going to need some way of seeing what's making the algorithm cough in your data and a good bit of luck to find what's causing non-convergence. It's still possible that it's just not converging fast enough.

The calculations might be less expensive in Cartesian. I'd try that if you get stumped.
 
say_cheese said:
The theta and phi coordinates are cyclic (v(0)=v(2*pi)), v=0, dv/dr=0 at v=0 and a.
In your first post you said "in cylindrical coordinates". Cylindrical coordinates has only a "theta" coordinate, not phi.
 

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