(1)We have the equation: (1)
So, the equilibriums of this equation (1) result from the equation (2):(2)
1- Equilibriums and stability of (2)
We reconize a dynamic system with a potentiel. So, we know that the equilibriums of (2) are:
(P,V'(ue))=(0,0). Futhermore, we know that stability of this equilibres result from the sign of V''(ue).
So, we are going to study V'(u) for differents values of the parameters.
- Graph of V' for >0 and >0:
If (>0 or ) we have one stable equilibrium because V''(ue)>0.
If (<0 and ) we have 2 stable equilibriums (ue1,ue3) because V''(ue1)>0 and V''(ue3)>0 and one instable equilibrium ue2 because V''(ue2)<0.
- Graph of V' for >0 and <0:
If (>0 or ) we have one instable equilibrium because V''(ue)<0.
If (<0 and ) we have 2 instable equilibriums (ue1,ue3) because V''(ue1)<0 and V''(ue3)<0 and one stable equilibrium ue2 because V''(ue2)>0.
Rem : We don't plot V' for (<0 and <0) and (<0 and >0) because we find the same graphs with a symetry of the axe u.
2-Graph of equation (2) in the plane (u,P)
We considere the case = -2/27, =1, =1, K=0,06 .
So, we have one instable equilibrium ue2= -1/3 and 2 stable equilibriums ue1= -1/3-(1/3)^1/2 and ue3= -1/3+(1/3)^1/2.
With Matlab, we obtain the 3 differents curves for each equilibrium.
- Graph for ue3:
- Graph for ue1:
- Graph for ue2:
These curves corespond very well with the analytic theory of dynamic system with a potentiel.