| Author | Message | |||
     yafeng He (baoma) New member Username: baoma Post Number: 1 Registered: 12-2004 |
I have some trouble in the solution of Swift-Hohenberg equation: dt(u) = -dxxxx(u)-2*dxx(u)-(1+a)u + u^3 for 0 <x> 0 and u = 0 and dxx(u)=0 at x = 0,L, t > 0, u(x,0) = u0(x) for 0 < x < L, we will assume that u0 is symmetric with respect to the centre of the domain, i.e. u0(L-x) = u0(x) for 0 < x < L. so a periodic boundaries is needed. Would you like to giving me some advice in detail? | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 278 Registered: 06-2003 |
FlexPDE can treat equations of second order or less in space. Since yours is a fourth order equation, you will need to split it into a system of two second order equations: dt(u)=-dxx(v)-dxx(u)-(1+a)u-u^3 v=dxx(u) The boundary conditions are then simply value(u)=0 value(v)=0 at the ends of the interval. You don't need a periodic boundary condition, because the boundary conditions are already periodic. If you want to model only half the domain, you need a reflective boundary at the center point, not a periodic one. But this is only a 1D system, for heaven's sake, you don't need to economize unless you are running on a PDA. FlexPDE version 4 does not have a 1D option, so you will have to create a phoney second dimension, y, with dyy(u) and dyy(v) terms in the two equations to stabilize the solution in this fake dimension. See the example "Samples | Misc | OneDim.pde" for an example. The rest of the implementation is straightforward if you follow the discussion in the User Guide. | |||
| yafeng He (baoma) New member Username: baoma Post Number: 2 Registered: 12-2004 |
Thank you! I have tried again and again, but I can not obtain the result I wanted. The result is hexagon pattern. This is my file:
Help me please! | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 280 Registered: 06-2003 |
The value of U rises uniformly in time across the entire domain. This is inconsistent with the boundary condition, so there is some ringing at the discontinuity at the boundary. I don't see any hexagons. The implication of this is that you have your coefficients dimensionally inconsistent with your domain dimensions, so that all the spatial derivatives are too tiny to affect the time behavior. Check your dimensions. Plot each term of the equations independently, so you can see how they compare and which is dominant. | |||
| yafeng He (baoma) Junior Member Username: baoma Post Number: 3 Registered: 12-2004 |
I try again. I think there are some trouble with the boundary. This is the boundary, is it right? I don't know well it, can you help me? initial value u=0 v=0 equations u:dt(u)=u-u^3-v-b*(1+cos(2*pi*f*t))+da*dxx(u)+da*dyy(u) v:dt(v)=e*(u-aa*(v+b*(1+cos(2*pi*f*t)))-aaa)+db*dxx(v)+db*dyy(v) boundaries Region 1 start(-10,-10) value(v)=0 value(u)=0 line to (10,-10) value(v)=0 value(u)=0 line to (10,10) value(v)=0 value(u)=0 line to(-10,10) value(v)=0 value(u)=0 line to finish | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 283 Registered: 06-2003 |
Your boundary condition statements are syntactically correct. But see my previous posting for my comments on this problem. |
