4.6
The method of images
4.6.1
A vortex next to a wall
1
2
b
Γ
4
πb
ψ
= 0
Consider the motion of a vortex located at some position
z
0
=
a
+
ib
in the upper half of
the complex plane. The
x
axis is the location of a solid wall. Thus we have to satisfy the
boundary condition of no flow through the wall:
v
= 0 for
y
= 0. Simply taking the flow
field of the vortex,
u

iv
=

i
Γ
2
π
x

a

i
(
y

b
)
(
x

a
)
2
+ (
y

b
)
2
,
(39)
does not satisfy this condition: the flow field must be modified by the presence of the
wall.
The method of images is a technique that permits to find solutions to Laplace’s equa
tion which satisfy the right boundary condition on the solid wall.
Near the vortex at
z
0
, the solution should look like (39). The method of solution is indicated in the figure:
imagine replacing the wall by another
image
vortex that is placed at an equal distance
on the other side of the wall. Its position is where an image in a mirror would appear.
And equally like a mirror image, the sense of rotation of the image vortex is reversed.
By symmetry it is clear that the
v
component of the image vortex has opposite sign, so
on the line of symmetry (the locus of the wall), the two contributions cancel and
v
= 0.
The boundary condition on the wall is satisfied and we have solved our problem! In other
words, the flow problem we want to solve (
one
vortex in the presence of a wall) is exactly
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the same as the flow problem of two vortices (the original vortex and its image)
without
the wall.
Let’s check if our reasoning was correct.
It is easiest to do the calculation using
complex notation. If the vortex is at
z
0
, it’s image is at
z
0
. Since the image also has the
opposite sense of rotation, there is a minus sign in front of it. The total potential (vortex
+ image) becomes
w
(
z
) =

i
Γ
2
π
log(
z

z
0
) +
i
Γ
2
π
log(
z

z
0
)
.
We have to verify that
Im
{
w
}
=
ψ
=
const
for
z
=
x
real!
Now since
ln(
z
) = ln(
z
)
(why?) we have
w
(
z
) =
i
Γ
2
π
log(
z

z
0
)

i
Γ
2
π
log(
z

z
0
)
.
But if
z
is real,
z
=
z
, and thus
w
(
x
)

w
(
x
) = 0, and so
ψ
= 0 along the wall.
In fact, we have shown that the problem of a vortex near a wall is mathematically
equivalent to the problem of two counterrotating vortices of equal strength, which in the
previous section we showed to move at constant speed in the direction perpendicular to
the line connecting them, in other words, parallel to the wall. Let us check this fact as
well, using complex notation. The vortex in question moves in the flow field of its image,
i.e.
u

iv
=
i
Γ
2
π
x

a

i
(
y
+
b
)
(
x

a
)
2
+ (
y
+
b
)
2
,
evaluated at
z
=
a
+
ib
. Thus
u

iv
=
i
Γ
2
π
parenleftbigg

2
ib
4
b
2
parenrightbigg
=
Γ
4
πb
,
and the vortex moves with a horizontal speed of
Γ
4
πb
, which corresponds to the earlier
result with
h
= 2
b
.
4.6.2
A recipe
This leads us to the following result, which permits to find the flow generated by
any
combination of singularities next to a wall. Let the wall be the real axis of the complex
plain, and let
f
(
z
) describe all singularities (vortices, sources, etc.) present in the flow
domain
y >
0. For example,
f
(
z
) =
i
Γ
2
π
ln(
z

z
0
) for a vortex and
f
(
z
) =

μ
2
π
(
z

z
0
)
for a dipole.
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 Fall '11
 Eggers
 Complex number, Herr Hiemenz

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