φ ¼ ϕ=η,
ð5:121Þ
consequently, we can write Eq. (5.120) as
f ¼
γ γ À 1
ð
Þ
2
ψη
2
φ
2 1 À φ
γφ À 1
!
ð5:122Þ
¼ η
2
Ψψ
ð5:123Þ
where
Ψ ¼
γ γ À 1
ð
Þ
2
φ
2 1 À φ
γφ À 1
:
ð5:124Þ
and, as a result, Ψ is expressed in terms of the velocity variable φ. We will now
dispense with the momentum equation, namely Eq. (5.13), and, instead, use
Eq. (5.122) in addition the continuity and energy equations. Firstly, the continuity
equation, namely Eq. (5.10), is
ψ
0
ψ
¼
ϕ
0
þ 2ϕ=η
η À ϕ
,
which can be written as,
ϕ À η
ð
Þd ln ψ
dη
¼ À
1
η 2
d η
2
ϕ
ð Þ
dη
,
where, for example, lnψ implies log e ψ, and in view of the above definitions we can
write the latter equation as,
η φ À 1
ð
Þd ln ψ
dη
¼ À
1
η 2
d η
3
φ
ð Þ
dη
,
so that
d ln ψ
d ln η
¼ À
1
φ À 1
dφ
d ln η
À
3φ
φ À 1
,
ð5:125Þ
which is an alternative form of the continuity equation. The energy equation,
namely, Eq. (5.12), is
3f þ η f
0
þ γ
ψ
0
ψ
f ϕ À η
ð
ÞÀϕ f
0
¼ 0,
with ϕ ¼ ηφ, this latter equation becomes
5.17 Route to an Analytical Solution
267