Now that we know the equilibriums of the equation (1) :
We are going to plot the curves in the plane (x,t,u) with Matlab in order to know its behaviour.
During all this study, we fixe the parameters: = -2/27, =1,=1.So, with the same notations that in the previous section, we have:
ue3= -1/3 +(1/3)^1/2
We consider the following initials conditions:
And the limits conditions : u(0)=ue3 and u(L)=ue1.
1-Curves of the equation (1) in the plane (x,t,u):
- K=0 :
The more we increase K, the more we tend to an equilibrium curve ue(x) we have seen in previous section.
2-Curve u(x) for the last value of t:
In order to obain the best results, we have taken a time t=20 and K=0.24.
- Curve for L=1:
This curve corespond to an equilibrium curve ue(x) of equation :
but details are not enough important to see that is really an equilibrium curve.
- Curve for L=3:
We have taken L=3 instead of L=1 inorder to see that this limit curve corespond very well to an equilibrium curve ue(x) we have seen in previous section. Indeed, the relative gap of P is enough important to see the variations of P on this curve and so understand that it is an equilibrium curve if we compare with the graph of previous section. But to have a better proof, we are going to compare the curve (P(x),u(x)) obtained for the last value of t with the curve (Pe(x),ue(x)) which corespond to the equilibrium curve of previous section.
3-Curve (P(x),u(x)) for the last value of t:
In order to obain the best results, we have taken a time t=20, K=0.24 and L=1.
4-Curve (Pe(x),ue(x)) of the corresponding equilibrium:
We have plot the curve corresponding to:
for L=1, u(0)=ue3 and P(0)= -1.21
We can see that the 2 curves are the same. So, we have shown that the initials conditions tend to an equilibrium curve (Pe(x),ue(x)) given by the initials conditions and the equation: