Learn how easy it is to sync an existing GitHub or Google Code repo to a SourceForge project!

## [Libmesh-users] What does the Newton term mean in ex13?

 [Libmesh-users] What does the Newton term mean in ex13? From: Bao Kai - 2012-01-29 19:47:55 ```Hi, I am just starting to lean Libmesh. I have a question on ex13. I want to know what the Newton term means in ex13. The code is listed as follows, for (unsigned int i=0; i

 [Libmesh-users] What does the Newton term mean in ex13? From: Bao Kai - 2012-01-29 19:47:55 ```Hi, I am just starting to lean Libmesh. I have a question on ex13. I want to know what the Newton term means in ex13. The code is listed as follows, for (unsigned int i=0; i
 Re: [Libmesh-users] What does the Newton term mean in ex13? From: John Peterson - 2012-01-29 21:27:57 ```On Sun, Jan 29, 2012 at 12:47 PM, Bao Kai wrote: > Hi, > > I am just starting to lean Libmesh. > > I have a question on ex13.  I want to know what the Newton term means in ex13. > theta*dt*(U*grad_u)*phi[i][qp]);              // Newton term The Newton method of Example 13 is atypical: rather than solving for the "update" du_{k+1} = u_{k+1} - u_{k}, it solves for u_{k+1} directly. So the "standard" Newton iteration: J_{k} du_{k+1} = -F_{k} is instead written as: J_{k} u_{k+1} = -F_{k} + J_{k} u_{k} This modified RHS can be computed by hand; it involves the term you asked about above. -- John ```
 Re: [Libmesh-users] What does the Newton term mean in ex13? From: Bao Kai - 2012-01-30 17:59:18 ```Hi, John, Thank you for your helpful response. Since the lid driven cavity problem is actually a steady problem, it seems that always the steady result will be obtained after the non-linear iteration, which also means we get the final result in the first time step. Is it right? Regards, Kai On Mon, Jan 30, 2012 at 12:27 AM, John Peterson wrote: > On Sun, Jan 29, 2012 at 12:47 PM, Bao Kai wrote: >> Hi, >> >> I am just starting to lean Libmesh. >> >> I have a question on ex13.  I want to know what the Newton term means in ex13. > >> theta*dt*(U*grad_u)*phi[i][qp]);              // Newton term > > The Newton method of Example 13 is atypical: rather than solving for > the "update" du_{k+1} = u_{k+1} - u_{k}, it solves for u_{k+1} > directly. > > So the "standard" Newton iteration: > > J_{k} du_{k+1} = -F_{k} > > is instead written as: > > J_{k} u_{k+1} = -F_{k} + J_{k} u_{k} > > This modified RHS can be computed by hand; it involves the term you > asked about above. > > -- > John ```
 Re: [Libmesh-users] What does the Newton term mean in ex13? From: John Peterson - 2012-01-30 18:04:29 ```On Mon, Jan 30, 2012 at 10:59 AM, Bao Kai wrote: > Hi, John, > > Thank you for your helpful response. > > Since the lid driven cavity problem is actually a steady problem, it > seems that always the steady result will be obtained after the > non-linear iteration, which also means we get the final result in the > first time step. Is it right? Not in general, no. The lid driven cavity has a steady state solution (for some Re) but we solve it in ex13 using a transient algorithm and time derivative terms are present in the PDE. The parameters (delta_t and Re) used in ex13 cause it to converge to steady state quite quickly, but certainly not in 1 timestep. -- John ```