Imaginary component in example 22 #4807
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v-dobrev
LoudspeakerEngineering
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If one runs example 22 in the plot the exact solution is on the boundary of the real component but the imaginary component is zero (I would expect the exact imaginary component to be on the boundary). If "-r 1" is passed to refine the grid then both the real and imaginary components have the exact solution on the boundary (as I would expect). Is this expected behaviour? If it is can someone suggest how to handle the imaginary component of essential boundary conditions on unrefined grids. |
Answered by
v-dobrev
Apr 23, 2025
Replies: 2 comments 2 replies
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This behavior does not seem expected. Can you please provide more details: are you running |
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The issue is that the coarse mesh (with the default order
-o 1) does not capture the boundary conditions well -- they are very small but not quite zero, see the top row with the projections of the exact solution on the coarse and a finer mesh in the screenshot below.The discrete solutions (bottom row) use the boundary conditions from the projected exact solution, so for the coarse mesh, they are very small but not zero. The top and the respective bottom plots have the exact same boundary values on both meshes. The big error in the interior on the coarse mesh is due to the coupling of the imaginary part to the real part and that decreases with mesh refinement (or increasing the order of t…