Computer Algebra Recipes

7.2. ANALYTIC SOLITON SOLUTIONS 305
The positive square root is chosen (the negative square root yields an antikink)
and the trig identity cos U =1¡ 2sin 2 (U=2) substituted,
> sol1:=subs(cos(U)=1-2*sin(U/2)^2,sol[1]);
sol1 := V (U) = 4
s
μ 2
p U
60 sin
15
2
and the result simpli¯ed using the square root and symbolic options.
> sol1:=simplify(sol1,sqrt,symbolic);
sol1 := V (U) = 8
μ
p U
15 sin
15 2
Since V = dU=dz, thenz ´ x ¡ ct is equal to the integral with respect to U of
the reciprocal of the right-hand side of sol1 .
> z:=x-c*t=int(1/rhs(sol1),U);
z := x ¡ t
μ μ μ
1 p U U
= 15 ln csc ¡ cot
4 4
2 2
Then, solving for U and recalling that U(z) ´ Ã(x ¡ ct),
> solwave:=psi(x,t)=solve(z,U);
solwave := Ã(x; t) =
0
B
2arctanB
Ã
B
@
1+
2 e ( (4 x¡t) p 15
15
e ( (4 x¡t) p 15
15
Ã
)
! 2 ; ¡ Ã
)
1+
e ( (4 x¡t) p 15
15
! 2
)
e ( (4 x¡t) p 15
15
1
¡ 1 C
! 2 C
) A
yields the analytic solitary-wave solution shown in solwave. The arctan function
is expressed in terms of two arguments separated by a comma. The term to the
left (right) of the comma is the numerator (denominator) of arctan.
The common denominators in the arguments can be removed and the numerator
of the second argument simpli¯ed with the following assumption:
> solwave:=simplify(solwave,assume=real);
solwave := Ã(x; t) =2arctan(2e ( (4 x¡t) p 15
15
)
; ¡e
( 2(4x¡t) p 15
) 15 +1)
If you can't instantly see that this still complicated-appearing result is a kink
solitary-wave solution of the SGE, you can be excused. However, that it is a
solution can be con¯rmed by applying the following pdetest command.
> pdetest(solwave,SGE);
0
To see that it is a kink solitary wave, let's animate the rhs of solwave.
> animate(plot,[rhs(solwave),x=-10..50],t=0..200,frames=50,
thickness=2,axes=framed);
On running the animation, we observe that a kink solitary wave travels to the