| Author | Message | ||
     luyang.han@gmail.com (hans_da) New member Username: hans_da Post Number: 1 Registered: 05-2007 |
Rather than NATURAL(phi) = 0, I would like to realize: DOT(A**grad(phi), ds) = 0 , where ds is the unit vector normal to the surface and A a matrix. Is this possible with FlexPDE? | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 858 Registered: 06-2003 |
As described in the FlexPDE documentation, the meaning of NATURAL(phi)=<value> depends on the PDE. Since you have not stated your equation, I cannot explain what the NATURAL means in your system. FlexPDE will not do the matrix multiply, so you will have to write out the components of the product. DOT(<something>*grad(phi),ds) = <value> is what NATURAL(phi)=<value> means. It just remains to determine what <something> is. Once you determine the meaning of NATURAL in your system, you can do the algebra to find what <value> needs to be. Refer to the entries under NATURAL in the Help Index. | ||
| luyang.han@gmail.com (hans_da) New member Username: hans_da Post Number: 2 Registered: 05-2007 |
Thank you for your reply. The problem I have is just a normal Poisson equation div(-sigma*grad(phi)) = 0 , representing the potential within a conductor. However here the conductivity is rather anisotropic, so that sigma here should be expressed as a tensor. As the boundary condition, the current should be zero along the surface, which means DOT(-sigma*grad(phi), ds) = 0, and here sigma is a 3x3 matrix. If I understand it right, the natural boundary condition in FlexPDE means DOT(grad(phi), ds), or saying, the component of grad(phi) in direction of ds is zero. But in my problem is rather that the component of sigma*grad(phi) in direction of ds is zero, since the matrix acts as rotating the grad(phi). | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 859 Registered: 06-2003 |
In your system, NATURAL(phi) provides a value for the outward normal component of -sigma*grad(phi) at the boundary. See NATURAL in the Help Index. |