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Dear all, <br>
I hope you had a great Thanksgiving. <br>
Currently, I tested the order of accuracy for
/src/ksp/ksp/tutorial/example/ex45.c. Since the 2nd-order
discretization is used in this program and ksp solver is converged
to 10^-7, I expected that the solution should provides a 2nd-order
in L2 norm. However, as I tested (even with a Laplace equation), the
L2 norm slope is much less than 2. Sometime, if the grid size is
reduced, the L2 norm increases. Could anyone help me about this
issue, please?<br>
<br>
Here is the L2 norm outputted:<br>
<br>
<table style="border-collapse: collapse;" border="0" cellpadding="0"
cellspacing="0" height="126" width="278">
<colgroup><col style="width:48pt" span="2" width="64"> </colgroup><tbody>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt;width:48pt" height="20" width="64">Grid</td>
<td style="width:48pt" width="64">L2 norm (10^-8)</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20" align="right">0.05</td>
<td align="right">4.36242</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20" align="right">0.025</td>
<td align="right">2.20794</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20" align="right">0.0125</td>
<td align="right">7.02749</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20" align="right">0.00625</td>
<td align="right">12.64</td>
</tr>
</tbody>
</table>
<br>
Once the grid size is reduced to half, the number of the grid
will be multiplied by 8 in order to keep the same size of the
computational domain.<br>
The code is also attached. It is from ex45.c with very little
modifications. <br>
<br>
thanks in advance,<br>
Alan<br>
<br>
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