2Kinematics
IB Fluid Dynamics
2.4 Streamfunction for incompressible flow
We suppose our fluid is incompressible, i.e.
∇ · u = 0.
By IA Vector Calculus, this implies there is a vector potential A such that
Definition (Vector potential). A vector potential is an A such that
u = ∇ × A.
In the special case where the flow is two dimensional, say
u = (u(x, y, t), v(x, y, t), 0),
we can immediately know A is of the form
A = (0, 0, ψ(x, y, t)),
Taking the curl of this, we get
u =
∂ψ
∂y
, −
∂ψ
∂x
, 0
.
Definition
(Streamfunction)
.
The
ψ
such that
A
= (0
,
0
, ψ
) is the streamfunc-
tion.
This streamfunction is both physically significant, and mathematically con-
venient, as we will soon see.
We look at some properties of the streamfunction. The first thing we can do
is to look at the contours ψ = c. These have normal
n = ∇ψ = (ψ
x
, ψ
y
, 0) .
We immediately see that
u · n =
∂ψ
∂x
∂ψ
∂y
−
∂ψ
∂y
∂ψ
∂x
= 0.
So the flow is perpendicular to the normal, i.e. tangent to the contours of ψ.
Definition
(Streamlines)
.
The streamlines are the contours of the streamfunc-
tion ψ.
This gives an instantaneous picture of flow.
Note that if the flow is unsteady, then the streamlines are not particle paths.
Example.
Consider
u
= (
t,
1
,
0). When
t
= 0, the velocity is purely in the
y
direction, and the streamlines are also vertical; at
t
= 1, the velocity makes an
45
◦
angle with the horizontal, and the streamlines are slanted:
t = 0 t = 1
However, no particles will actually follow any of these streamlines. For a particle
released at x
0
= (x
0
, y
0
). Then we get
˙x(t) = u = t, ˙y = v = 1.
Hence we get
x =
1
2
t
2
+ x
0
, y = t + y
0
.
Eliminating t, we get that the path is given by
(x − x
0
) =
1
2
(y −y
0
)
2
.
So the particle paths are parabolas.
Typically, we draw streamlines that are “evenly spaced”, i.e. we pick the
streamlines ψ = c
0
, ψ = c
1
, ψ = c
2
etc. such that c
3
− c
2
= c
2
− c
1
.
Then we know the flow is faster where streamlines are closer together:
slow fast
This is since the fluid between any two stream lines must be between the stream
lines. So if the flow is incompressible, to conserve mass, they mast move faster
when the streamlines are closer.
We can also consider the volume flux (per unit length in the
z
-direction),
crossing any curve from x
0
to x
1
.
slow fast
x
0
x
1
u
Then the volume flux is
q =
Z
x
1
x
0
u · n d`.
We see that
n d` = (−dy, dx).
So we can write this as
q =
Z
x
1
x
0
−
∂ψ
∂y
dy −
∂ψ
∂x
dx = ψ(x
0
) − ψ(x
1
).
So the flux depends only the difference in the value of
ψ
. Hence, for closer
streamlines, to maintain the same volume flux, we need a higher speed.
Also, note that
ψ
is constant on a stationary rigid boundary, i.e. the boundary
is a streamline, since the flow is tangential at the boundary. This is a consequence
of u · n = 0. We often choose ψ = 0 as our boundary.
Sometimes it is convenient to consider the case when we have plane polars.
We embed these in cylindrical polars (r, θ, z). Then we have
u = ∇ × (0, 0, ψ) =
1
r
e
r
re
θ
e
z
∂
r
∂
θ
∂
z
0 0 ψ
=
r
∂ψ
∂θ
, −
∂ψ
∂r
, 0
.
As an exercise, we can verify that
∇ · u
= 0. It is convenient to note that in
plane polars,
∇ · u =
1
r
∂
∂r
(ru
r
) +
1
r
∂u
θ
∂θ
.