Solution 2.
See text and/or instructor's solution manual.
Answer. The
transformation
, of
Exercise 6 in Section
11.9,
is known to map a horizontal flow in the upper half
plane
onto
flow the upper half plane
that moves around the exterior of the infinitely long rectangular
barrier.
Solution. Along
the x-axis use the
points
and
in the w-plane
use
respectively.
The exterior angles are
,
and the formula for the derivative
is given by the Schwarz-Christoffel
formula
Integrate and get
The images of
, are
, respectively.
Use
and
, and
obtain thel system of equations
Which simplifies to yield
![[Graphics:../Images/FluidFlowImageMod_gr_58.gif]](../Images/FluidFlowImageMod_gr_58.gif)
The solution to this system of equations is easily found to
be
.
Therefore,
.
Or if you desire, you can use the equivalent formula
.
We are done.
Aside. The
image of horizontal streamlines in the z
plane are curves in the w plane given by the parametric equation
![]()
for
. Or
if you desire,
![]()
for
.
We are really done.
Aside. We can let Mathematica double check our work.
We are really really done.
We can let Mathematica graph some of the streamlines.
![[Graphics:../Images/FluidFlowImageMod_gr_75.gif]](../Images/FluidFlowImageMod_gr_75.gif)
The
streamlines in the w-plane given by
the parametric equation
, for
.
We are really really really done.
Aside. We can let
Mathematica graph
.
![[Graphics:../Images/FluidFlowImageMod_gr_80.gif]](../Images/FluidFlowImageMod_gr_80.gif)
The
image of the upper half plane
under
the mapping
.
We are really really really really done.
Aside. We can
replace
with
in
the formula for
.
![[Graphics:../Images/FluidFlowImageMod_gr_87.gif]](../Images/FluidFlowImageMod_gr_87.gif)
The
image of the upper half plane
under
the mapping
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell