Solution 3.
See text and/or instructor's solution manual.
Answer. Both
and
are
harmonic in the upper half plane and satisfy the boundary
conditions
.
Also,
.
The Poisson integral formula defines a bounded function in the upper
half plane, therefore the desired solution
is
.
Solution. The
Poisson's integral formula
defines
a harmonic in the upper half plane
and has the boundary values
,
wherever
is continuous. Hence,
,
is harmonic in the upper half plane and has the boundary conditions
.
There are two possible choices for
, both
are harmonic in the upper half plane and satisfy the boundary
conditions
.
Also,
all
values of x, but
does
not go to zero for all values of x.
The Poisson integral formula defines a bounded function in the upper
half plane, therefore the desired solution is the
choice
.
We are done.
Aside. We can graph
the solution
.
![[Graphics:../Images/PoissonIntegralModHome_gr_149.gif]](../Images/PoissonIntegralModHome_gr_149.gif)
A
contour graph of the function
where
for
.
![[Graphics:../Images/PoissonIntegralModHome_gr_153.gif]](../Images/PoissonIntegralModHome_gr_153.gif)
A
graph of the function
.
We are really done.
Aside. We can let Mathematica 4.0 investigate this situation.
We are really really done.
Aside. We can construct an integral.
The special functions CosIntegral and SinIntegral are not part of our course.
Aside. We can construct another integral.
Aside. If you use
Mathematica 7.0 you will get a slightly different answer.
![[Graphics:../Images/PoissonIntegralModHome_gr_166.gif]](../Images/PoissonIntegralModHome_gr_166.gif)
We can expand the integral and get
The imaginary part is identically zero in the upper
half-plane.
The functions
and
take
on the values
![[Graphics:../Images/PoissonIntegralModHome_gr_171.gif]](../Images/PoissonIntegralModHome_gr_171.gif)
Thus, for values of z in the upper half-plane the real part
becomes
![[Graphics:../Images/PoissonIntegralModHome_gr_172.gif]](../Images/PoissonIntegralModHome_gr_172.gif)
Unfortunately, it's value is zero on the real axis.
The software Maple will construct an equally complicated
expression for the integral
.
Caveat. Be careful
when using computer algebra systems, often times they might use
"special functions."
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell