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get
Ω
∇·
u
=0
.
Now, here's the magical part: this equation should be true for any choice
of Ω (any region of fluid). The only continuous function that integrates
to zero independent of the volume of integration is zero itself. Thus the
integrand has to be zero everywhere:
∇·
u
=0
.
This is the
incompressibility condition
, the other part of the incompressible
Navier-Stokes equations.
A vector field that satisfies the incompressibility condition is called
divergence-free
for obvious reasons. One of the tricky parts of simulating
incompressible fluids is making sure that the velocity field stays divergence-
free. This is where the pressure comes in.
One way to think about pressure is that it's whatever it takes to keep
the velocity divergence-free. If you're familiar with constrained dynamics,
you can think of the incompressibility condition as a constraint and the
pressure field as the Lagrange multiplier needed to satisfy that constraint
subject to the principle of zero virtual work. If you're not, don't worry.
Let's derive exactly what the pressure has to be.
The pressure only shows up in the momentum equation, and we want
to somehow relate it to the divergence of the velocity. Therefore, let's take
the divergence of both sides of the momentum equation:
∂u
∂t
+
1
ρ
∇
∇·
∇·
(
u
·∇
u
)+
∇·
p
=
∇·
(
g
+
ν
∇·∇
u
)
.
(1.3)
We can change the order of differentiation in the first term to the time
derivative of divergence:
∂
∂t
∇·
u.
If the incompressibility condition always holds, this had better be zero.
Subsequently, rearranging Equation (1.3) gives us an equation for pressure:
1
ρ
∇
∇·
p
=
∇·
(
−
u
·∇
u
+
g
+
ν
∇·∇
u
)
This isn't exactly relevant for our numerical simulation, but it's worth
seeing because we'll go through almost exactly the same steps, from looking
at how fast a volume is changing to an equation for pressure, when we
discretize.
