Solution 5.
See text and/or instructor's solution manual.
Answer.
.
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
Now integrate
![]()
The integral can be found using the suggested change of
variable
![]()
![[Graphics:../Images/FluidFlowImageMod_gr_199.gif]](../Images/FluidFlowImageMod_gr_199.gif)
Make substitutions in the integral
Now use the substitution
,
![]()
![]()
![]()
Solve for
, and
, and
get the following solution
.
The first logarithm term can be rewritten in the following
forms
,
.
We are done.
Aside. We can let Mathematica double check our work.
We are really done.
We can let Mathematica graph some of the streamlines.
The flow up an inclined ramp, as shown in Figure 11.92(a).
![[Graphics:../Images/FluidFlowImageMod_gr_215.gif]](../Images/FluidFlowImageMod_gr_215.gif)
The
streamlines in the w-plane given by
the parametric equation
, for
.
We are really really done.
Aside. We can extend the flow into the third and fourth quadrants using symmetry.
The flow around a pointed object, as shown in Figure 11.92(b).
![[Graphics:../Images/FluidFlowImageMod_gr_219.gif]](../Images/FluidFlowImageMod_gr_219.gif)
The
streamlines in the w-plane given by
the parametric equation
, for
.
We are really really really done.
Aside. We can use
Mathematica to graph
.
![[Graphics:../Images/FluidFlowImageMod_gr_224.gif]](../Images/FluidFlowImageMod_gr_224.gif)
The
image of the upper half plane
under
the mapping
.
We are really really really really done.
Aside. We can extend the mapping into the third and fourth quadrants using symmetry.
![[Graphics:../Images/FluidFlowImageMod_gr_228.gif]](../Images/FluidFlowImageMod_gr_228.gif)
The
image of the complex plane under the mapping
.
Aside. We can graph
the other version involving logarithms
.
![[Graphics:../Images/FluidFlowImageMod_gr_232.gif]](../Images/FluidFlowImageMod_gr_232.gif)
The
image of the upper half plane
under
the mapping
.
We are really really really really really done.
Remark 1. If the
computer algebra Mathematica is used to perform the
integration then the answer is
,
which gives the correct result but uses the a specialized
hypergeometric function.
Aside. Here are two
extra graphs using two different mappings:
and
.
![[Graphics:../Images/FluidFlowImageMod_gr_241.gif]](../Images/FluidFlowImageMod_gr_241.gif)
The
image of the upper half plane
under
the mapping
.
![[Graphics:../Images/FluidFlowImageMod_gr_245.gif]](../Images/FluidFlowImageMod_gr_245.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