dψ
ψ
¼
3dη
γ À 1
ð
ÞÀαγη nÀ1
½
Š η
þ
n þ 2
ð
Þαγη
nÀ2 dη
γ À 1
ð
ÞÀαγη nÀ1 :
ð5:42Þ
Writing the first term on the right-hand side in Eq. (5.42) as
C 1
η
þ
C 2 αγη
nÀ2
γ À 1
ð
ÞÀαγη nÀ1 ,
we find that C 1 ¼ C 2 ¼ 3/(γ À 1). Substituting these values back in Eq. (5.42) we
obtain
dψ
ψ
¼
3
γ À 1
dη
η
þ
αγ
3
γÀ1 þ n þ 2
ð
Þ
h
i
η
nÀ2 dη
γ À 1
ð
ÞÀαγη nÀ1
:
Substituting for α and n yields,
dψ
ψ
¼
3
γ À 1
dη
η
þ
2γ γ þ 5
ð
Þ
γ þ 1
ð
Þ
2
η
nÀ2 dη
γ À 1
ð
ÞÀαγη nÀ1 ,
ð5:43Þ
hence,
ln ψ ¼
3
γ À 1
ln η þ
2γ γ þ 5
ð
Þ
γ þ 1
ð
Þ
2
Z
η
nÀ2 dη
γ À 1
ð
ÞÀαγη nÀ1 þ constant:
ð5:44Þ
In order to evaluate the integral we let x ¼ (γ À 1) À αγη
n À 1 , hence, we find that
η
nÀ2 dη ¼ À
γ þ 1
ð
Þ
2
γ 7 À γ
ð
Þ
dx:
ð5:45Þ
Substituting Eq. (5.45) in Eq. (5.44) and integrating yields,
ln ψ ¼
3
γ À 1
ln η À
2 γ þ 5
ð
Þ
7 À γ
ð
Þ
ln γ À 1
ð
ÞÀαγη
nÀ1
Â
à þ constant:
In order to determine the constant of integration we note that ψ ¼ (γ + 1)/(γ À 1)
when η ¼ 1, and with these substitutions we find the constant of integration to be
ln
γ þ 1
γ À 1
þ
2 γ þ 5
ð
Þ
7 À γ
ln
γ γ À 1
ð
Þ
γ þ 1
ð
Þ
:
Upon substituting back and simplifying the resulting equation we obtain,
240
5 Spherical Shock Waves: The Self-similar Solution