κΔx
ð
Þ
2 dυ
dω
2
¼
γ þ 1
2
υ À υ f
À
Á υ i À υ
ð
Þ
ð4:34Þ
In order to solve Eq. (4.34) Von Neumann and Richtmyer used the following
substitutions;
ψ ¼ υ À
υ i þ υ f
2
, ψ 0 ¼
υ i À υ f
2
and φ ¼
ψ
ψ 0
,
hence,
κΔx
dφ
dω
¼ Æ
γ þ 1
2
1=2
1 À φ
2
À
Á 1=2 :
Consequently,
ω ¼ Æ
2
γ þ 1
1=2
κΔx
Z
dφ
1 À φ 2
ð
Þ
1=2
and by making the substitution, φ ¼ Sin(x), one can integrate the latter equation to
obtain,
ω ¼ Æ
2
γ þ 1
1=2
κΔx
ð
ÞarcSinφ
¼ Æω 0 arcSinφ,
ð4:35Þ
where
ω 0 ¼
2
γ þ 1
1=2
κΔx
ð4:36Þ
and the constant of integration has been set to zero as it only shifts the ω-axis by that
constant amount. Eq. (4.35) gives
φ ¼ ÆSin
ω
ω 0
or ψ ¼ Æ
υ i À υ f
2
Sin
ω
ω 0
,
hence,
υ ¼
υ i þ υ f
2
Æ
υ i À υ f
2
Sin
ω
ω 0
:
ð4:37Þ
Differentiating this latter equation, we obtain,
4.4 Artificial Viscosity
143