| Author | Message | ||
     Harvey (jones) New member Username: jones Post Number: 1 Registered: 06-2006 |
How does one set up the equation in flexpde to handle the following problem: du/dt=D*(d^2 u/dx^2) where D=k*u, with k=constant So basically the "traditional diffusion constant" D is no longer a constant but now a function of u. The corresponding boundary conditions are: at x=0, du/dx=0 (so in flexpde natural(u)=0)) at x=L, h*u=D*du/dx where D is defined as above (function of u), and h is a constant. (what is the boundary condition here in flexpde?) | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 626 Registered: 06-2003 |
The equation should be written much as you have written it in your post, using the DEFINITIONS section to define D: VARIABLES u DEFINITIONS k=something D = k*u EQUATIONS U: dt(u)=DIV(D*GRAD(u)) The NATURAL BC for this equation defines the outward normal component of -D*GRAD(u) [negative because it is on the right side of the equation]. (See NATURAL in the Help Index). At x=0 this is NATURAL(u)=0, at x=L, NATURAL(u) = -h*u |