3
$\begingroup$

My finite difference scheme for the 2D Euler equations is second order accurate in theory, since all the terms are second order accurate, with the advective terms being third order. So I expect a rate of convergence of about 2 when using successive refinements. However, I know that the discretization at the boundary reduces to first order (I am working on that). These are the values I compute (Dx=4 m, and coarser grid is 40X40 cells):

|------|-----------------------------------------------------------------
| grid | Norm L1  |  Rate L1 |  Norm L2  |  Rate L2 |  Norm Linf | Rate Linf  
|------|----------------------------------------------------------------
|Dx    | 0.0264   |    /     |0.0017799  |   /      |   0.149    |   /
|Dx/2  | 0.0088   |   1.58   |0.0003712  |  2.26    |   0.075    |  1.00 
|Dx/4  | 0.0032   |   1.40   |0.0000858  |  2.09    |   0.043    |  0.78 
|Dx/8  | 0.0009   |   1.70   |0.0000159  |  2.38    |   0.024    |  0.90 
|Dx/16 | 0.0003   |   1.45   |0.0000038  |  2.10    |   0.013    |  0.74 
|Dx/32 | 0.0001   |   1.51   |0.0000007  |  2.19    |   0.009    |  0.82 

Therefore, I do understand that my Linf norm is first order, since the max errors are at the boundary where the scheme is only first order accurate. Also, from my results it looks like the L2 norm averages the boundary error through the domain and blurs the information, therefore still giving a global second order of accuracy (or slightly above 2). What I do not understand is: why is the L1 norm converging with a rate of about 1.5? Or better why does the norm L2 blur the error at the boundary why the L1 does not?

Note: I compute the norms with:

\begin{align} L_1 &= \frac{1}{N}\sum_{j=1}^N|u^{numerical}_j-u^{exact}_j| \\ L_2 &= \frac{1}{N}\sqrt{\sum_{j=1}^N(u^{numerical}_j-u^{exact}_j)^2} \;\;\;\;\;\;\;\;\;\;\;\; (1)\\ L_{inf} &= \max|u^{numerical}_j-u^{exact}_j| \end{align}

where $u^{numerical}$ is the computed along-x velocity at velocity point $j$, with $j=1..N$, and $u^{exact}$ is the analytical solution.

EDIT: I am thinking that the way I am computiong $L_2$ norm might be wrong. See my related question Correct way of computing norm $L_2$ for a finite difference scheme . If I compute $L_2$ as \begin{align} L_2 &= \sqrt{\frac{1}{N}\sum_{j=1}^N(u^{numerical}_j-u^{exact}_j)^2} \;\;\;\;\;\;\;\;\;\;\;\; (2)\\ \end{align}

I obtain this convergence rates:

|------|----------------------| 
| grid |   Norm L2 |  Rate L2 |  
|------|----------------------| 
|Dx    | 0.0308232 |   /      |  
|Dx/2  | 0.0133451 |  1.17    |    
|Dx/4  | 0.0065001 |  1.01    |   
|Dx/8  | 0.0024768 |  1.36    |    
|Dx/16 | 0.0011984 |  1.08    |     
|Dx/32 | 0.0005142 |  1.20    |   

Question is: either if I use equation (1) or (2) the rate of convergence in $L_2$ is pretty different (respectively higher or lower) than the one $L_1$, which clashes with Leveque's statement "Except in fairly rare cases, the 1-norm and 2-norm will give similar results" (https://books.google.com/books/about/Finite_Volume_Methods_for_Hyperbolic_Pro.html?id=QazcnD7GUoUC&hl=en). Is this a rare case, in which the accuracy is locally smaller at boundary?

$\endgroup$
3
$\begingroup$

Your initial method for computing the norms was incorrect
(see my answer to your related question)
The norm was incorrect by a factor of $1/\sqrt{N}$, leading to underestimating the error as $N$ increases and giving the appearance of a higher-than-expected convergence rate. I would expect that both L1 and L2 norms give somewhere between orders 1 and 2 for convergence which appears to be the case in your update.

$\endgroup$
  • $\begingroup$ Thanks. Do you think that the convergence rate of L2 is markedly smaller because the norm L2 gives more weight to the larger errors at the boundaries (since the errors are squared)? Or is there some better explanation I am missing? $\endgroup$ – Millemila May 26 '15 at 14:52
  • $\begingroup$ @Albert, Yes, that would be my suspicion. $\endgroup$ – Doug Lipinski May 26 '15 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.