| Author | Message | ||
     Patricio A. Greco (pagreco) Member Username: pagreco Post Number: 8 Registered: 08-2003 |
If Dtangential(A) is different to 0 then , as is equivalent to n x grad(A). I´ve a couple of questions: 1)If n=(nx,ny,nz)then nx^2+ny^2+nz^2=1 ? 2) The result of n x grad(A) is a vector then how can I define each component at the bondary surface. Suppose I´ve a plane surface in the x,y plane then nx=ny=0 ,nz=1, n x grad(A)=(-dy(A),dx(A),0), How can I define the value for each component. When I write , Dtangential(A)=vector(1,0,1) the program reports a type error. If I define Dtangential(A)=1 it works , then with component I'm defining? Thank you very much. | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 26 Registered: 06-2003 |
FlexPDE cannot deal with vector variables. You must present a scalar PDE for each component of a vector. Each scalar equation is able to impose a single scalar condition at each node on a boundary. In 2D, a tangential condition is possible, because the cross product of a normal and a vector in the plane has only a single component out of the plane. In 3D, the cross product is a 3-component vector, and therefore cannot be fully specified. If you use Dtangential() in 3D, FlexPDE will impose the condition on the magnitude of the tangential vector, and the condition will be ambiguous in azimuthal orientation. Try some algebraic manipulations, and see if you can use knowledge of the tangential derivative to replace terms and generate a definition of the normal. |