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III-TRAJECTORIES

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:

ue1= -1/3-(1/3)^1/2

ue2= -1/3

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 :
• K=0.01:
• K=0.06:

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: