2Kinematics
IB Fluid Dynamics
2.3 Kinematic boundary conditions
Suppose our system has a boundary. There are two possible cases — either the
boundary is rigid, like a wall, or it moves with the fluid. In either case, fluids
are not allowed to pass through the boundary.
To deal with boundaries, we first suppose it has a velocity
U
(which may
vary with position if the boundary extends through space). We define a local
reference frame moving with velocity
U
, so that in this frame of reference, the
boundary is stationary.
In this frame of reference, the fluid has relative velocity
u
0
= u − U.
As we mentioned, fluids are not allowed to cross the boundary. Let the normal
to the boundary be n, then we must have
u
0
· n = 0.
In terms of the outside frame, we get
u · n = U · n.
This is the condition we have for the boundary.
If we have a stationary rigid boundary, e.g. the wall, then
U
=
0
. So our
boundary condition is
u · n = 0.
Free boundaries are more complicated. A good example of a free material
boundary is the surface of a water wave, or interface between two immiscible
fluids — say oil and water.
oil
water
n
ζ(x, y, t)
We can define the surface by a function
z = ζ(x, y, t).
We can alternatively define the surface as a contour of
F (x, y, z, t) = z − ζ(x, y, t).
Then the surface is defined by
F
= 0. By IA Vector Calculus, the normal is
parallel to
n k ∇F = (−ζ
x
, −ζ
y
, 1).
Also, we have
U = (0, 0, ζ
t
).
We now let the velocity of the fluid be
u = (u, v, w).
Then the boundary condition requires
u · n = U · n.
In other words, we have
−uζ
x
− vζ
y
+ w = ζ
t
.
So we get
w = uζ
x
+ vζ
y
+ ζ
t
= u · ∇ζ +
∂ζ
∂t
=
Dζ
Dt
.
So all the boundary condition says is
Dζ
Dt
= w.
Alternatively, since
F
is a material surface, its material derivative must vanish.
So
DF
Dt
= 0,
and this just gives the same result as above. We will discuss surface waves
towards the end of the course, and then come back and use this condition.