Graphics Reference
In-Depth Information
Only the advection term is left. Writing it out in component form
⎛
⎝
⎞
⎠
∂
2
φ
∂x
2
∂
2
φ
∂
2
φ
∂x∂z
∂φ
∂x
+
∂φ
∂y
∂x∂y
+
∂φ
∂z
∂φ
∂x
∂x∂y
+
∂φ
∂
2
φ
∂
2
φ
∂y
2
+
∂φ
∂z
∂
2
φ
∂y∂z
(
∇
φ
)
·
(
∇∇
φ
)=
,
∂y
∂
2
φ
∂
2
φ
∂
2
φ
∂z
2
∂φ
∂x
∂x∂z
+
∂φ
∂y∂z
+
∂φ
∂y
∂z
and it becomes clear that this is actually the same as
⎛
⎞
∂φ
∂x
2
∂φ
∂y
2
∂φ
∂z
2
∂
∂x
∂
∂x
∂
∂x
1
2
1
2
1
2
+
+
⎝
⎠
∂φ
∂x
2
∂φ
∂y
2
∂φ
∂z
2
∂
∂y
∂
∂y
∂
∂y
(
∇
φ
)
·
(
∇∇
φ
)=
1
2
1
2
1
2
+
+
∂φ
∂x
2
∂φ
∂y
2
∂φ
∂z
2
∂
∂z
∂
∂z
∂
∂z
1
2
1
2
1
2
+
+
∇
2
∇
2
.
=
φ
Using this now brings the momentum equation to
∂t
+
∇
2
∇
2
+
∂φ
p
ρ
+
∇
(
gy
)=0
∇
φ
∇
ρ
+
gy
=0
.
The only function whose gradient is everywhere zero is a constant, and
since constants added to
φ
(a theoretical abstraction) have no effect on the
velocity field (the real physical thing), we can assume that the constant is
just zero for simplicity. This gives
Bernoulli's equation
:
∂φ
∂t
+
2
∇
2
+
p
⇒∇
φ
∂t
+
2
∇
2
+
p
∂φ
φ
ρ
+
gy
=0
.
You may have already heard of this. For example, in the steady-state case,
where
∂φ/∂t
= 0, and after subtracting out the hydrostatic component of
pressure, we end up with the pressure variation Δ
p
=
2
.Many
simple experiments, such as blowing over the top ofa sheet of paper held
1
−
2
ρ
u
