CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > General Forums > Main CFD Forum

Inner non linear iterations and Newton method

Register Blogs Community New Posts Updated Threads Search

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old   March 11, 2010, 01:31
Default Inner non linear iterations and Newton method
  #1
New Member
 
Isabel
Join Date: Mar 2009
Posts: 4
Rep Power: 17
misabel is on a distinguished road
Hello:

Not obtaining the desired results in two different problems, I would like
to clarify a concept that is the “Newton method”.

In both cases I want to simulate Navier Stokes equations: the first one is
a time dependent problem (which has an analytical solution) and the second
one is a steady state problem which uses time as a way to get the steady
state, so I will always have a temporal term in the equation.

In both cases, the NS equations are:

du/dt+(u*grad)u-viscosity(grad²)u+(grad)p=0;
(div)u=0;

To implement these equations I use different schemes: coupled method,
pressure projection method, consistent pressure method... and I am using
Finite Element Method (FEM).

The way I advance in time is the following one:

Assemble the unchanged matrices (i.e. mass matrix, diffusion matrix,
gradient matrix, divergence matrix)

for(unsigned int i=0; i<maximum_num_iterations;i++)
{
step value++;

Assemble the matrices (i.e. convection matrix and right hand side vector)

Solver the linear system (if working with coupled method) or the linear
systems (if working with projection method).

Steady state case: if the residual is less than a tolerance value I give I
stop the loop.
Time dependent problem: I continue until the total time I want to simulate
is reached.
}

I mean, inside this loop, I don't create any inner nonlinear loop.

I am not sure if I must also apply the Newton method in every case since I
always use the convection term in a previous time step or I linearize it
using (u^(n)*gradu)^(n+1). Anyway, it is supposed that convection term is
the only term that introduces a non linearity in the equation, isn't it?
And it is because of it, we should introduce the Newton method to reach
the right solution or on the contrary, I would like to know when and why
we should use Newton method to get the right solution. Is it always
compulsory?

I would be really grateful if you could clarify me this concept.

Thank-you very much in advance.

Isa
misabel is offline   Reply With Quote

Old   March 11, 2010, 11:14
Default
  #2
agd
Senior Member
 
Join Date: Jul 2009
Posts: 358
Rep Power: 19
agd is on a distinguished road
Don't know if this will help, but I'll try - it's difficult to express some concepts on a message board. I find it very useful to think of the Newton iteration (which occurs inside the time step loop) as a search over solution space. If you discretize the governing equations at the new time level, you can write the system as

R(Q(N+1)) = 0

This is prior to any application of a Newton scheme, so you have essentially a semi-discrete form of the equation. Apply a Newton iterative scheme to this system, i.e. we solve for DQ where

A*DQ = -R(Q)

Here A is a system Jacobian, and Q is a possible solution out of a solution space, call it Q(N+1,M). The next Newton iterate of Q, call it Q(N+1,M+1) = Q(N+1,M) + DQ. If we drive DQ to zero (which is the purpose of the Newton scheme) then at convergence we have the "correct" solution at the new time step N+1.

One of the first applications I remember of this type of scheme was in conjunction with a Beam-Warming approximate factorization scheme. In the original scheme the splitting error completely destroys time accuracy in 3D for reasonable timestep sizes, but by using a Newton iterative solver the factorization error is contained within the Newton iteration and does not affect the converged value of Q(N+1). So it is not necessary, but it is a very useful approach for creating a scheme that is stable and accurate at very large timesteps. Hope that helps.
agd is offline   Reply With Quote

Old   March 11, 2010, 11:42
Default
  #3
New Member
 
Isabel
Join Date: Mar 2009
Posts: 4
Rep Power: 17
misabel is on a distinguished road
Agd,tThank-you very much!

But even being usefual approach, calculating the system Jacobian (A) for a NS equations must be really computationally expensive, mustn't it?

Best
Isa
misabel is offline   Reply With Quote

Old   March 11, 2010, 12:19
Default
  #4
agd
Senior Member
 
Join Date: Jul 2009
Posts: 358
Rep Power: 19
agd is on a distinguished road
Depending on the way you construct the left-hand side, the system Jacobian is a combination of the inviscid flux Jacobians and/or viscous flux Jacobians. These have to be determined for an implicit scheme anyway. So the overall cost of using the Newton method versus a standard temporal marching approach is just roughly the product of the FLOPS for the temporal stepping and the number of Newton iterations. But the expense is offset by the improved convergence, the fact that you can impose boundary conditions within the Newton loop to get quasi-implicit BC implementation using simple BCs, and greatly enhanced time-accurate behavior.
agd is offline   Reply With Quote

Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Newton's method failed to converge Mohan CFX 7 September 23, 2020 07:50
how to increase "Newton Pressure Iteration Limit" kus CFX 9 April 21, 2013 02:54
About Newton iteration Lixian CFX 3 November 27, 2008 08:57
Newtons method failed(Error) Chetan Prashant CFX 4 May 17, 2007 01:58
newtons method failed/fatal overflow ioana CFX 0 March 2, 2005 03:57


All times are GMT -4. The time now is 04:30.