[arjun]

Originally Posted by FMDenaro

The residual is the residual. Is a unique definition. I would not consider any other definition



https://www.sciencedirect.com/scienc...67610521003366

Nope. It depends on the solver. Here in the OP we are talking about residual remaining after the solution is obtained and descretisation is applied.

Accuracy is with respect to exact or benchmark solution. So it is possible for solver to produce a solution that has higher residual but is more accurate. This is the whole point.


The residual in this context is different than what you have in mind.

[arjun]

Yes very much.

Originally Posted by LuckyTran

Within the realm of linear algebra the difference between x_k or x_k+1 and the x_ex imparts a bias onto the solution and hence the residual. Whether you know the exact solution or not, the norm of the residual remains a measure assuming the difference x_k and x_k+1 is small (i.e. the solution is converging to something). Here is where higher order schemes can impart a bias that can give more accurate solution despite higher absolute level of residuals. Lower residual does not guarantee a better/worse solution but lowering of residual is quite necessary. The exceptions are the null cases: 1) you aren't converging, 2) the initial condition/guess used the converged solution, or 3) it is a case that requires non-linear estimators

So I tend to agree with the philosophy that in essence you are choosing to accept a less-converged solution. This is also apparent since you can manipulate the spectral radius to get arbitrarily small x_k+1 - x_k (i.e. set the urf's to 0). That being said, of course there have been many examples of when I have chosen to do exactly such, accept the solution as-is because there is indeed some type of error in system that prevents it from converging better–one bad cell in the mesh, etc.


Of course I also understand the point that solvers obfuscate the residual a bit by normalizing it about the solution. But the impact of this is mostly making it hard to compare convergence quality between two different cases. Although you can manipulate solver-reported residuals by the initial guess or normalization options, most people have habits that make it consistent, i.e. initial velocity is always U=0 every time they run a case, etc.

[FMDenaro]

Originally Posted by arjun

Nope. It depends on the solver. Here in the OP we are talking about residual remaining after the solution is obtained and descretisation is applied.

Accuracy is with respect to exact or benchmark solution. So it is possible for solver to produce a solution that has higher residual but is more accurate. This is the whole point.


The residual in this context is different than what you have in mind.

There is no sense in this discussion if we do not formalize the object of the issue.


- The numerical solution, by definition it satisfies the discrete set of equations A_d(fnum)=q



- The exact solution, by definition it satisfies the original PDE A_pde(fex)=q.



- Order of accuracy is the main term in the local truncation error of the whole equations.




It seems you are talking about point #1, that is you select a numerical solution f_num_app (that is not fnum) and does not satisfy the discrete set of equations


A_d(f_num_app)-q=res.

And you are stating that f_num_app is a better approximation of fex than fnum. And that a solution f_num_app2, producing a res2>res would be still better. Therefore, we are talking about the discretization error, not about the local truncation error.

Am I right?


But that has nothing to do with the accuracy order.