Expressing the right-hand side of this integral in terms of partial fractions
according to the equation,
1
σ σ À 4
ð
Þ
¼
A
σ
þ
B
σ À 4
,
we find that A ¼ À (1/4) and B ¼ (1/4), hence,
Z t
0
dt
t þ t 1 À t R
ð
Þ
¼ 2
Z σ
σ 1
dσ
σ À 4
ð
Þ
À
Z σ
σ 1
dσ
σ
2
4
3
5
ð4:75Þ
and by carrying out the integration on both side of the latter equation we find that
ln t þ t 1 À t R
ð
Þ
½
ŠÀln t 1 À t R
ð
Þ¼2 ln σ À 4
ð
ÞÀ ln σ 1 À 4
ð
ÞÀ ln σ þ ln σ 1
½
Š ,
hence,
ln
t þ t 1 À t R
ð
Þ
t 1 À t R
ð
Þ
!
¼ ln
σ À 4
σ
σ 1
σ 1 À 4
! 2
and, therefore,
t þ t 1 À t R
ð
Þ¼ t 1 À t R
ð
Þ
σ À 4
σ
σ 1
σ 1 À 4
! 2
:
However, we have already noted that t ¼ t À t 1 , hence, we finally obtain the
equation,
t ¼ t R þ t 1 À t R
ð
Þ
σ À 4
σ
2
σ 1
σ 1 À 4
2
; σ 1 ! σ > 0:
ð4:76Þ
Substituting our previous relationships for (u + c) and ξ in terms of σ in the
equation, x ¼ u ξ
ð Þ þ c ξ
ð Þ
½
Š t þ ξ, we have
x ¼ c 0 1 þ σ
ð
Þ t À t 1
ð
Þþc 0 1 þ σ
ð
Þ t 1 À t R
ð
Þ:
Noting that x ¼ x À x R , we arrive at the following equation for x,
x ¼ x R þ c 0 1 þ σ
ð
Þ t À t R
ð
Þ:
ð4:77Þ
198
4 Numerical Treatment of Plane Shocks