Computer Algebra Recipes

4.2. SECOND-ORDER MODELS 171
In vel, the total energy is identi¯ed to be E =11=9. To get a qualitative feeling
for the eardrum motion, Mike plots the potential energy U as a thick red line
over the range x = ¡1:6 to0:6,
> U:=plot(9*x^2+6*x^3,x=-1.6..0.6,color=red,thickness=2):
and the total energy E as a thick green horizontal line between the points
(x = ¡2; E=11=9) and (x =1;E=11=9).
> E:=plot([[-2,11/9],[1,11/9]],style=line,color=green,
thickness=2):
The two energy curves are superimposed in Figure 4.6.
> display(fU,Eg,tickmarks=[3,3],labels=["x","Energy"]);
-1 x
4
Energy
Figure 4.6: Asymmetric potential well for the eardrum model equation.
The horizontal total energy line in Figure 4.6 intersects the potential energy
curve at three x values. The two intersection points inside the potential well
centered at x =0aretheturning points, the eardrum system oscillating between
these two points if it is placed at one of these two points with zero velocity.
The turning point on the right is the input value x(0) = A = 1
3 . Because the
potential well is slightly asymmetric, the other turning point on the left is not
at x = ¡ 1.
The other turning point is easily determined by ¯rst ¯nding all the
3
intersection points using the solve command.
> points:=solve(vel=0,y);
points := 1 11
; ¡
3 12 +
p
33 11
; ¡
12 12 ¡
p
33
12
Three solutions are produced, corresponding to the three intersection points
of the total energy line with the potential energy curve. The ¯rst answer corresponds
to the turning point on the far right, and the third answer to the
intersection on the far left. So the second answer is selected as corresponding
to the second turning point for motion inside the potential well. This turning
point is labeled B.
2
0