(1)We have the equation: (1)Back II-

EQUILIBRIUMS

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.

: 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.Rem

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

and 2 stable equilibriumsue2= -1/3andue1= -1/3-(1/3)^1/2ue3= -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.