Substituting this back into the equation for x(u, t) gives
x u, t
ð Þ ¼ c 0 t þ
1
2
γ þ 1
ð
Þut þ
1
n þ 1
1
a 1=n u
nþ1
ð
Þ=n
À
c 0
a 1=n u
1=n
À
1
2
γ þ 1
ð
Þ
a 1=n u
nþ1
ð
Þ=n
:
Differentiating and setting the result equal to zero we have
∂x
∂u
t
¼
1
2
γ þ 1
ð
Þt þ
n þ 1
n
1
n þ 1
1
a 1=n u
1=n
À
1
n
c 0
a 1=n u
1
n À1
À
1
2
γ þ 1
ð
Þ
a 1=n
Â
n þ 1
n
u
1=n
¼ 0:
Similarly, the second derivative equated to zero gives
∂
2 x
∂u 2
t
¼
n þ 1
n
1
n þ 1
1
a 1=n
1
n
u
1
n À1 À
1
n
c 0
a 1=n
1
n
À 1
u
1
n À2 À
1
2
γ þ 1
ð
Þ
a 1=n
n þ 1
n
1
n
u
1
n À1 ¼ 0:
By simplifying the latter equation we obtain
1
n 2 À
c 0
n
1
n
À 1
1
u
À
1
2
γ þ 1
ð
Þ
n þ 1
n 2
¼ 0
and by multiplying across by n
2 we obtain
c 0
u
n À 1
ð
Þ¼
1
2
γ þ 1
ð
Þ n þ 1
ð
ÞÀ1,
ð2:29Þ
so that
1
u
¼
γ
nþ1
nÀ1
À Á þ 1
Â
Ã
2c 0
:
ð2:30Þ
The equation for (∂x/∂u) t ¼ 0 gives
1
2
γ þ 1
ð
Þt ¼
1
n
u
1=n
a 1=n
c 0
u
þ
1
2
γ þ 1
ð
Þ n þ 1
ð
ÞÀ1
h
i
and by using Eq. (2.29) this latter equation becomes
1
2
γ þ 1
ð
Þt ¼
u
1=n
a 1=n
c 0
u
¼
c 0
a 1=n
1
u
nÀ1
n
:
58
2 Waves of Finite Amplitude