σ ξ
ð Þ ¼
γ þ 1
2
u ξ
ð Þ
c 0
¼
ξ
c 0 t 1 À t R
ð
Þ
À 1
ð4:72Þ
and with ξ in the range, c 0 t 1 À t R
ð
Þ< ξ
c 0 þ
γþ1
2
À Á
u p
Â
Ã
t 1 À t R
ð
Þ , we have σ in the
range,
0 < σ σ 1 ,
where σ 1 ¼
γþ1
2
À Á u p
c 0
. Using Eq. (4.69) for the shock velocity, we can write it in terms
of σ, hence, it becomes,
U ¼ c 0 1 þ
1
2
σ þ
1
8
σ
2
h
i
,
similarly, we have
u þ c ¼ c 0 þ
γ þ 1
2
u ¼ c 0 1 þ σ
ð
Þ
and, therefore,
U À u þ c
ð
Þ¼
c 0 σ
8
σ À 4
ð
Þ:
Substituting these latter relationships for U and (u + c) in Eq. (4.68) and using
Eq. (4.70) we find that
c 0 σ
8
σ À 4
ð
Þ
dt
dξ
À
t
t 1 À t R
À 1 ¼ 0,
hence,
c 0 σ
8
σ À 4
ð
Þ
dt
dσ
dσ
dξ
À
t
t 1 À t R
À 1 ¼ 0
ð4:73Þ
and after using Eq. (4.72) this latter equation reduces to
σ σ À 4
ð
Þ
dt
dσ
¼ 8t þ 8 t 1 À t R
ð
Þ:
ð4:74Þ
Integrating, we obtain,
1
8
Z t
0
dt
t þ t 1 À t R
ð
Þ
¼
Z σ
σ 1
dσ
σ σ À 4
ð
Þ
:
4.8 Numerical Examples of Plane Shocks
197