| Author | Message | ||
     Olexiy Orel (scorpion) Member Username: scorpion Post Number: 12 Registered: 03-2007 |
Dear sirs, I have following issue. I need to set natural BC as a function of the charge stored in my region: natural(C)=1-dt(integral(C,'region')). Rather essential BC: flux is a function of stored charge. If I use plot to see integral(C,'region'), it shows me sinusoid - not a constant. But if I use plot to see dt(integral(C,'region'), it shows me eternal 0. Why? Can I somehow calculate derivative of a integral of a function of a region? Also another question. If I have an equation: dt(C)=D*del2(C), what will natural(C) on a vertical boundary (left) be? dx(C) or D*dx(C)? Thank you very much. Sorry for disturbing. | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1244 Registered: 06-2003 |
1. FlexPDE does not store a history of integrals, so it cannot compute a time derivative. You have two alternatives: You can declare a GLOBAL variable to hold the integral and use the Global variable in the BC, or you can move the time derivative inside the integral operator. 2. First, FlexPDE moves all variable-dependent terms to the left side of the equation, resulting in dt(C) - D*div(grad(C)) = 0. It then moves the D* inside the divergence, giving dt(C) - div(D*grad(C)) + grad(D)<dot>grad(C) = 0 It then integrates the equation over the cell volume, reducing the second-order spatial terms by integrating by parts. This results in Vol_integral(dt(C)) - Surf_integral(normal(D*grad(C))) + Vol_Integral(grad(D)<dot>grad(C)) = 0. The Natural BC specifies the integrand of the surface integral. In your case, this is normal(-D*grad(C)). The Natural BC on your left side boundary must specify the value of D*dx(C), because the outward normal vector points toward negative x. The proper form of the equation is, of course, dt(C) - div(D*grad(C)) =0 {rate of change equals divergence of flux}, in which case the spurious grad(D).grad(C) disappears. |