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\centerline{Mathematics 373 -- Applied Mathematics (PDE)}
\centerline{Problem Set 3}
\centerline{March 15, 2001}
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\noindent
I.  The first step in our proof of the Dirichlet pointwise convergence
theorem for Fourier series was the rather ``magical'' formula 
for the partial sum of the Fourier series of a function $f$:
$$s_N(x) = {1\over L}\int_{-L}^L f(x+u)D_N(u)\ du,\leqno(1)$$
where $D_N$ is the {\it Dirichlet kernel function}:
$$D_N(u) = {1\over 2} + \sum_{n=1}^N \cos\left({n\pi u\over L}\right).$$
In this problem, we want to see that this really works, at least in 
a particular example(!)  You'll probably want to use Maple for all
parts here.
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\item{A)}  Plot $D_N(u)$ for $[-L,L] = [-1,1]$, for $N = 5,10,20$.
What seems to be happening as $N$ increases?
\item{B)}  Using Maple, construct and plot a function giving the 
part of the periodic extension of 
$$f(x) = \cases{0& if $x < 0$\cr
                1& if $0 \le x \le 1$\cr}$$
for $-3\le x \le 3$.  Call your function $f$.  Now
use the following commands to construct a function as in (1) above
with $N = 5$ and plot it:
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\centerline{\tt D5:=u->1/2+add(cos(n*Pi*u),n=1..5);}
\centerline{\tt PS:=x->evalf(Int(f(x+u)*D5(u),u=-1..1));}    
\centerline{\tt plot(PS,-2..2);}
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\item{} Warning: This is a rather large computation because integrals
of rapidly oscillating functions are difficult by numerical methods.
Expect the plot to take about 10 minutes to display.
\item{C)} Now compute the 5th partial sum of the Fourier series for this
function on $[-1,1]$ directly from the integral formulas for 
the Fourier coefficients (as in Lab/Demo 1), and plot that.  
Are your results the same?   
\item{D)}  What happens if you only use $f$ on $[-1,1]$ and not
the periodic extension in B above?  Explain.
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\noindent
II.  Evaluate the line integrals of each of the following
vector fields along the indicated curves, and answer the 
accompanying questions.
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\item{A)} ${\bf F} = (x,x+y)$, $\alpha(t) = (t-t^2,t^2+t^3)$, $t \in [0,1]$.
{\it Answer: 139/60}.
\item{B)} ${\bf F} = (y^2,2xy)$, $\alpha$ the path from 
$(-1,2)$ to $(1,5)$ made up of two straight line segments, the
first parallel to the $x$-axis, the second parallel to the $y$-axis.
\item{C)} ${\bf F} =$ same vector field as in part B, but now
$\alpha$ the line segment directly from $(-1,2)$ to $(1,5)$.
\item{D)}  Your answers for B and C should be equal.  Explain why this 
is true.  {\it Hint:}  Is ${\bf F}$ a gradient vector field?
\item{E)}  The vector field
$${\bf F} = \left( {-y \over x^2 + y^2}, 
{x \over x^2+y^2}\right)$$
$\alpha$ the circle of radius 1 centered at $(0,0)$ in ${\bf R}^2$,
traversed counterclockwise.  
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\noindent
III.  Use Green's Theorem to evaluate each of the following
line integrals around the specified simple closed curve, and 
answer the other questions.
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\item{A)}  ${\bf F} = (\sin(\pi xy),\cos(\pi xy))$, $\alpha$ the 
positively-oriented boundary of the square
$[0,1]\times [0,1]$.  {\it Answer: $-{2+ \pi \over \pi}$}.
\item{B)}  ${\bf F} = (y^2,xy)$, $\alpha$ the ellipse with equation
$x^2/9 + y^2/4 = 1$.
\item{C)}  Compute 
$$\int\!\!\!\int_D {\partial Q\over \partial x} - {\partial P\over \partial y} dA$$
for the vector field ${\bf F}$ and the region $D$ inside $\alpha$ 
from part E of question II.  Your answer should be different from what
you saw in that part.  Is this a contradiction to Green's Theorem?? Explain.
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\noindent
IV.  The statement of Green's Theorem is also valid in more general
situations.
For example, consider the {\it circular ring\/} $D$ in the plane between the 
two circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.  The positively
oriented boundary of $D$ consists of the the circle of radius 2 traversed 
counterclockwise, and the circle of radius 1 traversed clockwise. 
Let ${\bf F} = (P,Q)$ be a vector field whose component
functions have continous first order partial derivatives 
on some open set in ${\bf R}^2$ containing the whole ring. 
Show that the line integral of ${\bf F}$ over the positively oriented
boundary is equal to the double integral of 
${\partial Q\over \partial x} - {\partial P\over \partial y}$ over the 
ring.  {\it Hint: Consider the curve $\alpha_1$ which consists of 
the line segment from $(1,0)$ to $(2,0)$, followed by the upper semicircle 
of radius 2 traversed counterclockwise, followed the line segment from 
$(-2,0)$ to $(-1,0)$, followed by the upper semicircle of radius 1 traversed 
clockwise. Similarly let $\alpha_2$ be the curve consisting
of the line segment from 
$(-1,0)$ to $(-2,0)$, followed by the lower semicircle of radius 2 traversed 
counterclockwise, followed by the line segment from $(2,0)$ to $(1,0)$, followed
by the lower semicircle of radius 1 traversed clockwise.}
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\noindent
{\it Assignment}
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Individual solutions, due Thursday, March 22.
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