ψ
0
ψ
¼
ϕ
0
þ 2ϕ=η
η À ϕ
:
ð5:10Þ
5.4.3 Energy Equation
Let us finally consider the equation of energy conservation in the following form (see
Eq. (1.67), Chap. 1);
∂
∂t
pρ
Àγ
ð
Þþu
∂
∂r
pρ
Àγ
ð
Þ¼0:
By carrying out the differentiation with respect to r and t it is straightforward to
show that this latter equation becomes
À
γp
ρ
∂ρ
∂t
þ
∂p
∂t
þ u À
γp
ρ
∂ρ
∂r
þ
∂p
∂r
¼ 0:
ð5:11Þ
We already have expressions for the density variations; ∂ρ/∂t and ∂ρ/∂r, however, the pressure variations are
∂p
∂t
¼ À3
ρ 0 A
2
γ
R
À4 f
dR
dt
þ
ρ 0 A
2
γ
R
À3 ∂f
∂η
∂η
∂R
dR
dt
¼ À3
ρ 0 A
2
γ
R
À4 fAR
À
3
2 À
ρ 0 A
2
γ
R
À4 f
0
ηAR
À
3
2
¼ À
ρ 0 A
2
γ
3R
À4 f þ R
À4 f
0
η
Â
Ã
AR
À
3
2
and
∂p
∂r
¼
∂
∂r
ρ 0 A
2
γ
R
À3 f
hence,
∂p
∂r
¼
ρ 0 A
2
γ
R
À3 ∂f
∂η
∂η
∂r
¼
ρ 0 A
2
γ
R
À4 f
0
:
When these expressions are substituted in Eq. (5.11) one finds that many common
multiplying factors cancel out and the resulting equation can be written in the form;
5.4 Taylor’s Analysis of Very Intense Shocks
225