dx
dt
¼
3 À γ
1 þ γ
x
t
À
4c 0
1 þ γ
ð2:76Þ
which determines the negative characteristics. In order to integrate this latter equation we let A ¼ (3 À γ)/(1 + γ) and B ¼ 4c 0 /(1 + γ), hence,
dx
dt
¼ A
x
t
À B
ð2:77Þ
and making the following change in variable; z ¼ x/t, it is easy to show that
dx
dt
¼ t
dz
dt
þ z:
Substituting this result in Eq. (2.77) gives
t
dz
dt
¼ A À 1
ð
Þz À B,
hence,
dz
A À 1
ð
Þz À B
¼
dt
t
ð2:78Þ
and by carrying out the integration we have
ln A À 1
ð
Þz À B
½
Š¼ A À 1
ð
Þln t þ constant:
Substituting back for z, yields,
ln A À 1
ð
Þx À Bt
½
Š¼A ln t þ constant:
ð2:79Þ
The constant of integration is obtained by noting that the negative characteristic
cuts the positive characteristic, x ¼ c 0 t, which is the boundary between the centred
expansion region and the air at rest. Suppose the negative characteristic crosses
x ¼ c 0 t at t ¼ t
0 as shown in Fig. 2.18, then the following equation is obtained;
ln A À 1
ð
Þc 0 t
0
À Bt
0
½
Š ¼ A ln t
0
þ constant,
which determines the constant of integration and substituting back, we obtain
ln
A À 1
ð
Þx À Bt
A À 1
ð
Þc 0 t 0 À Bt 0
!
¼ ln
t
t 0
3Àγ
1þγ :
ð2:80Þ
2.7 Application of Riemann Invariants to Simple Flow Problems
79