| Author | Message | |||
     rplcs (rpl) Member Username: rpl Post Number: 4 Registered: 04-2009 |
Can I solve a PDE with periodic boundary conditions such as U(x,0)=U(x,L)?? Thank you | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1271 Registered: 06-2003 |
See "Periodic" in the Help Index. | |||
| rplcs (rpl) Member Username: rpl Post Number: 5 Registered: 04-2009 |
I went through the exaples of periodic BCs. Should the shape of the boundaries be same in a periodic BC? Can I solve a PDE on triangle where the unknown u takes same value on the base as well as on the diagonal taking it as a Periodic Boundary Condition?? (u(x,0)=u(x,1-x))? I tried with the script START(0,0)periodic(x,1-x) LINE TO (0,1) LINE TO (1,0)value(u)=1 LINE TO CLOSE getting :Periodic boundary Transformation Singualr. Any suugestions?? | |||
| Marek Nelson (mgnelson) Moderator Username: mgnelson Post Number: 126 Registered: 07-2007 |
Yes, the shape and size of the periodic segments should be the same. They can be in translated or rotated positions, but the shape and size need to be the same. Otherwise it would not make physical sense anyway. There is an additional difficulty in the case you are proposing. The point at the intersection of the triangle's sides cannot map into itself. So you would need a small amount of the tip of the triangle cut off. PS - The way you have posed your domain, the periodic boundary can only be on the segment from (0,0) to (1,0) but you have applied it from (0,0) to (0,1). | |||
| rplcs (rpl) Member Username: rpl Post Number: 6 Registered: 04-2009 |
Thanks Dr Nelson for your suggestions. I tried out as you suggested but got same message Periodic boundary Transformation Singular. start (0,0) value(u)=1 line to (0.1,0) periodic(x,1-x)line to (0.9,0) value(u)=1 line to (0.9,0.1) line to (0.1,0.9) value(u)=1 line to (0,0.9) to close start(0,0.9) value(u)=1 line to (0,0) periodic (x,1-x) line to (0.9,0) value(u)=1 line to (0.1,0.9) line to close start(0,0.9) value(u)=1 line to (0,0) to (0,0.1)periodic (x,1-x) line to (0.9,0) value(u)=1 line to (0.1,0.9) line to close start (0,0) periodic(x,1-x)line to (0.9,0) value(u)=1 line to (0.9,0.1) line to (0.1,0.9) value(u)=1 line to (0,0.9) to close | |||
| Marek Nelson (mgnelson) Moderator Username: mgnelson Post Number: 132 Registered: 07-2007 |
The error reads: "Error: Periodic Boundary Transformation is Singular Transformations must be invertable" The key part of the error diagnostic is the second part: "Transformations must be invertable" Your periodic statement "periodic(x,1-x)" states the transformation for point (x1,y1) to point (x2,y2) using x2=x1 and y2=1-x1. The point (x2,y2) can be computed from a given point (x1,y1), but (x1,y1) cannot be computed from the point (x2,y2), because y1 appears nowhere in the transformation. Hence the transformation is not "invertable". You should look again at the periodic examples. In all these cases, there is a transformation that can be calculated in both directions. I have also attached a modification of our "azimuthal_periodic.pde" example (it is named "periodicaz.pde" in version 5 releases.) It is periodic on two sides of a triangle.
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