| Author | Message | ||
     Mohammad Rahmani (mrahmani) Member Username: mrahmani Post Number: 5 Registered: 10-2004 |
Dear Robert, Would you please drop a note about the ODE solver in FlexPDE. Has it the capability to handle the stiff problems. Can it solve DAE set of equations. In the provided texbooks and helps there is a little about ODE solver. The other question is, I frequently encountered set of PDEs accompanying a coupled ODE. For example consider the following case: U: -div(grad(U))+F(Psi)*U*V=0 V: -div(grad(V))+G(Psi)*V*U=0 Psi: dt(Psi)=H(Psi,U,V) Where F, G, and H represent function symbols. FlexPDE simply solve the above set but I don't understand how it treats the last ODE. Normally I leave the BC for the last ODE to the FlexPDE defaults (Natural()=0). Thanks in advance Mohammad | ||
| abbasali Member Username: abbasali Post Number: 5 Registered: 04-2010 |
Dear Robert I want to solve following initial and boundary value problems y"=2y^3+xy+.25, y(0)=1, y'(0)=0 how can I set this initial conditions in flexpde? As the second problem y"=2y^3+xy+.25, y(0)=1, y(2)=1 when I change the errlim in this case the solution changes dramatically. Thanks in advance abbasali | ||
| rgnelson Moderator Username: rgnelson Post Number: 1389 Registered: 06-2003 |
The first equation can be posed as an initial value problem, using time as the independent variable. FlexPDE accepts only first-order time derivatives, so you will need to declare a second variable, say p=dt(y). This second variable allows you to impose p=0 as an initial condition as well as y=1 as required. The equations are then P: dt(p)=2*y^3+y*t+0.25 Y: dt(y)=p (The solution appears to grow without bound.) The second equation can be posed as a boundary value problem in X, with VALUE boundary conditions at both ends. |