| Author | Message | ||
     Wilson Chen (emptyempty) Junior Member Username: emptyempty Post Number: 3 Registered: 09-2003 |
Hello, Mr. Nelson, I tried to solve a second order ODE, which has both boundary conditions in the same point (both function value and the first derivative of the function are specified at the left handside boundary). But it turns out that flexPDE does not respond to the change of natural condition. I met the same problem several times whenever the natural condition and value condition are specified in the same boundary. Could you please tell me what is the problem? Thank you very much! - Wilson !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! TITLE ' Test BVP' COORDINATES XCYLINDER SELECT ERRLIM=1e-3 VARIABLES y DEFINITIONS D Rd length=100 r1 =5 INITIAL VALUES EQUATIONS div(D*grad(y)) - Rd = 0 BOUNDARIES { The domain definition } Region 1 D=1 Rd=0.1 START(1.0,1.0) NATURAL(y)= 10 { Change of Natural condition will not change the solution !!! } value(y)=30 LINE TO(length,1.0) LINE TO(length,r1) LINE TO(1.0,r1) LINE TO FINISH PLOTS ELEVATION(y ) FROM (length/2,0) TO (length/2,r1) END | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 56 Registered: 06-2003 |
FlexPDE uses a Galerkin finite element method to convert your continuous PDE's to discrete equations at the mesh nodes. When you state a boundary condition, the equation at the affected boundary nodes is replaced by the boundary condition equation. The result is that you can only have a single boundary condition for any variable on a given boundary. (There is a gap in the checking logic in FlexPDE -- it should have issued an error diagnostic.) |