Computer Algebra Recipes

172 CHAPTER 4. NONLINEAR ODE MODELS
> B:=points[2];
B := ¡ 11
12 +
p
33
12
If ¯ = 0, the potential well would have been symmetric about x =0,andthe
second turning point would have been at x = ¡ 1
3 ¼¡0:333. For ¯ =1,the
ppotential well is asymmetric and the second turning point is at B =(¡11 +
33)=12 ¼¡0:438. Clearly, the period of oscillation will be somewhat longer
than the period 2 ¼=!0 ¼ 6:28forthelinearcase.
Mike calculates the period T for the nonlinear situation by multiplying the
time it takes the system to go from B to A by 2, i.e, T =2 R A
(1=vel) dy.
B
> T:=2*int(1/vel,y=B..A);
Ãs
15 ¡
24 EllipticK
T :=
p 33
15 + p !
33
p p
30 + 2 33
The period is expressed in terms of the complete elliptic integral K(k) ofthe
¯rst kind, which is de¯ned as
Z ¼=2
d®
K(k) ´ p
0 1 ¡ k2 sin 2 : (4.10)
®
Maple expresses
q
the complete elliptic integral in the form EllipticK(k), so in
this case k = (15 ¡ p 33)=(15 + p 33): The elliptic integral can be evaluated
numerically,
> T:=evalf(T);
T := 6:747679332
so the period is T ¼ 6:75, which is about 7 1%
longer than for the linear case.
2
To perform the necessary integration to obtain the analytic solution, it is
necessary to assume that the displacement satis¯es x>Band x assume(x>B,x sol:=t=int(1/vel,y=B..x); #implicit solution
à s
45 x +33¡
12 EllipticF 2
sol := t =
p 33 + 3 p 33 x
(15 ¡ p 33) (11 + p 33 + 12 x) ;
s
15 ¡ p 33
15 + p !
33
p p
30 + 2 33
The implicit solution is expressed in terms of the incomplete elliptic integral of
the ¯rst kind, u ´ F (Á; k), which is de¯ned as follows:
Z Á
Z sin Á
d®
dy
u ´ F (Á; k) = p = p
0 1 ¡ k2 2
sin ® 0 1 ¡ y2 p 1 ¡ k2 : (4.11)
y2 Maple expresses the incomplete elliptic integral in the form EllipticF(z; k) with
z ´ sin Á. The complete elliptic integral K(k) corresponds to taking Á = ¼=2.