% This is a sample file for the Undergraduate Faculty Program at
% PCMI, containing one lecture 
%
%   The Wave Equation and Separation of Variables
%
%  from the Park City Lectures.
%
% The sample file illustrates the use of epsf.tex to include postscript 
% graphics, as well as various "AmS-LaTeX" constructions from the amsmath
% package (which is automatically loaded by the pcms-l document class).
% To run this file you need the files 
%
%        pcms-l.cls and pcmslmod.tex
%


\documentclass[lecture,12pt,]{pcms-l}
\input pcmslmod.tex  % v.1.2
%\input epsf.tex

%\usepackage{amssymb} % this command would have loaded all the extra symbols,



% authors should not define these, they will be defined by the volume editors
%\def\currentvolume{3}
%\def\currentyear{1993}


% EQUATION NUMBERING AND THEOREM SETUP

\numberwithin{section}{chapter}
\numberwithin{equation}{chapter}

\theoremstyle{plain}
\newtheorem{theorem}[equation]{Theorem}
\newtheorem{lemma}[equation]{Lemma}

\theoremstyle{definition}
\newtheorem{definition}[equation]{Definition}

\newtheorem{exercise}{Exercise}
\newtheorem{problem}{Problem}
\newtheorem*{remark}{Remark}

% Set enumerate to use letters, not numbers for problem parts.

\renewcommand{\theenumi}{\alph{enumi}}
\renewcommand{\labelenumi}{(\theenumi)}

% AUTHOR-DEFINED MACROS:

\newcommand{\cA}{\mathcal{A}}
\newcommand{\cB}{\mathcal{B}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\cC}{\mathcal{C}}
\newcommand{\cD}{\mathcal{D}}

\begin{document}

\mainmatter
\setcounter{page}{1}

%\LogoOn

\lectureseries[An Introduction to PDE's]{An Introduction to Partial Differential Equations in the Undergraduate Curriculum}


\auth[J. Tolosa \& M. Vajiac]{J. Tolosa \& M. Vajiac} 

%\address{Chapman University}
%\email{mbvajiac@chapman.edu}

%The following items will become first page footnotes; they are optional.
%\date{\today}
%\setaddress


% the following hack starts the lecture numbering at 1
\setcounter{lecture}{10}
\setcounter{chapter}{10}

\lecture{Laplace's Equation in a Disk}

\section{Outline of Lecture}
\begin{itemize}
\item{The Laplacian in Polar Coordinates}
\item{Separation of Variables}
\item{The Poisson Kernel}
\item{Validity of the Solution}
\item{Interpretation of the Poisson Kernel}
\item{Examples}
\item{Challenge Problems for Lecture 11}
\end{itemize}


\vfill
\eject

\section{The Laplacian in Polar Coordinates}

When a problem has rotational symmetry, it is often convenient to change from Cartesian to polar coordinates. It is then useful to know the expression of the Laplacian 
$\Delta u = u_{xx} + u_{yy}$ in polar coordinates. Recall that  
$$ x = r \cos \theta, \qquad y = r \sin \theta . $$ 
Using the chain rule,  
$$ u_x = u_r r_x + u_\theta \theta_x . $$ 
Let us find $r_x$ and $\theta_x$. Implicit functions help simplify the computations a little bit. 
Differentiating both sides of 
$$ r^2 = x^2 + y^2 $$ 
with respect to $x$, we get  
$$ 2 r r_x = 2x , $$ 
whence 
$r_x = \frac{x}{r}$. 
You can find $r_y$ in a similar way. 
 
To compute $\theta_x$, we can start from the 
relation 
%  
$$ y = r \sin \theta . $$ 
Differentiating both sides with respect to $x$, we get 
$$ 0 = r_x \sin \theta + r \cos \theta \cdot \theta_x , $$ 
whence 
$$ \theta_x = -\frac{r_x \sin \theta}{r \cos \theta} = -\frac{r_x}{r} \tan \theta =  
     - \frac{r_x}{r} \cdot \frac{y}{x} . $$ 
Substituting here the value of $r_x$ we have just found, we get 
$$ \theta_x = - \frac{y}{r^2} . $$ 
You can find $\theta_y$ in a similar way, starting from the equation $x = r \cos \theta$.
{\bf Exercise 1.}  
Find $r_y$ and $\theta_y$ using the given hints.  
 
Differentiating once again, we can show that 
$$ r_{xx}= \frac{y^2}{r^3} , \qquad \theta_{xx} = \frac{2xy}{r^4} , $$ 
and find similar values for $r_{yy}$, $\theta_{xx}$,  and $\theta_{yy}$. 
 
This will yield the following expressions for $u_{xx}$ and $u_{yy}$: 
 
$$u_{xx} = \frac{x^2}{r^2} u_{rr}-\frac{2xy}{r^3} u_{r\theta} +  
                    \frac{y^2}{r^4} u_{\theta \theta} + \frac{y^2}{r^3} u_r + \frac{2xy}{r^4} u_{\theta} ; $$ 
$$u_{yy} = \frac{y^2}{r^2} u_{rr} + \frac{2xy}{r^3} u_{r\theta} +  
                    \frac{x^2}{r^4} u_{\theta \theta} + \frac{x^2}{r^3} u_r - \frac{2xy}{r^4} u_{\theta} . $$ 
 
{\bf Exercise 2.}  
Compute $r_{xx}$, $r_{yy}$, $\theta_{xx}$, $\theta_{yy}$, and obtain the above expressions for $u_{xx}$ and $u_{yy}$ using the chain rule. 
 
Adding up both expressions, doing a couple of cancellations and regrouping, we obtain 
% 
$$ u_{xx} + u_{yy} = u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta\theta} . $$ 
% 
This is the desired expression of the Laplacian in polar coordinates.  
Sometimes it is convenient to write it in a slightly different way:  

\begin{eqnarray} 
\Delta u = u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta\theta}  
                  =  \frac{1}{r}(r  u_r)_r + \frac{1}{r^2} u_{\theta\theta} ;  
\label{10.1} 
\end{eqnarray} 
for the second expression we combined the first two terms, using the product rule.  



%\vfill
%\eject

\section{Separation of Variables}
We will solve the Dirichlet problem for the Laplace equation on a circle, that is, the 
problem of finding a function that is harmonic inside a circle and has a prescribed value on the boundary: 
$$ \Delta u = 0 \qquad \hbox{inside the circle;} $$ 
$$ u = f            \qquad \hbox{on the boundary.} $$ 
Let us call $a$ the radius of the circle.  
If we put the center of the circle at the origin and use polar coordinates, we can be more specific: 
% 
\begin{eqnarray*}
\Delta u(r, \theta)&=&0 \quad \hbox{for every $\theta$ and for $r < a$;}  
                                                             \quad \text{\bf PDE} \\ 
\Delta u(a, \theta)&=&f(\theta) \quad \hbox{for every $\theta$,} \quad \text{\bf BC} 
\end{eqnarray*}
% 
where $f(\theta)$ is a specified periodic function with period $2 \pi$.  
(Periodicity is required because $\theta$ represents the polar angle, so $\theta + 2 \pi$ and $\theta$ are measures of the same angle.) 
 
Using the expression~(\ref{10.1}) of the Laplacian in polar coordinates, we can rewrite the problem as 
% 
\begin{eqnarray*}
u_{rr} +\frac{1}{r} u_r + \frac{1}{r^2} u_{\theta\theta} &=&0  
           \quad \hbox{for every $\theta$ and for $r < a$;}   \quad  \text{\bf PDE} \\ 
 \Delta u(a, \theta) &=& f(\theta) \quad \hbox{for every $\theta$,} \quad \text{\bf BC} 
\end{eqnarray*}
% 
Using the method of separation of variables, we will first forget for a moment all about the boundary condition and seek nontrivial solutions (eigenfunctions) of the Laplace equation on the circle as a product: 
% 
$$ u(r, \theta) = R(r) \Theta(\theta). $$
 
Substituting into the Laplace equation, we get 
% 
$$ R'' \Theta+\frac{1}{r}R' \Theta + \frac{1}{r^2} R \Theta''  = 0 . $$ 
% 
Multiplying both sides by $r^2$ we can rewrite this as 
$$  (r^2 R'' + r R')\Theta = - \Theta'' R .  $$ 
Dividing both sides by $R \Theta$, which is assumed to be nonzero, this becomes 
% 
$$ \frac{r^2 R'' + r R'}{R} = -\frac{\Theta''}{\Theta} .      $$ 
% 
Now we use an argument you have already heard many times: since the left-hand side depends only on $r$, and the right-hand side depends only on  $\theta$, both sides must be constant, call it $k$: 

\begin{eqnarray}
\frac{r^2 R'' + r R'}{R} = - \frac{\Theta''}{\Theta} = k .    
\label{10.2}  
\end{eqnarray} 
 
The equation for $\Theta$ reads 
% 
$$ \Theta'' + k \Theta = 0, $$ 
with the additional conditions that 

\begin{eqnarray}
\Theta(2 \pi) = \Theta(0),  \quad   
     \Theta'(2\pi) = \Theta'(0) ; \label{10.3}    
\end{eqnarray}
%
these conditions arise, again, from the fact that $\theta$ represents the polar angle. 
 
{\bf Exercise 3.} 
Prove that under these conditions $k$ cannot be negative. 
 
If $k=0$, we get  $\Theta(\theta) = A \theta + B$. Condition~(\ref{10.3}) implies $A = 0$, so the only solution is 
$\Theta = {\rm const}$. 
 
If $k>0$, we rewrite $k = \lambda^2$, just for convenience, and the equation for $\Theta$ becomes 

$$ \Theta'' + \lambda^2  \Theta = 0. $$ 
 
The general solution is 
$$ \Theta(\theta) = A \cos \lambda \theta + B \sin \lambda \theta . $$ 
These functions have period $\lambda$; conditions~(\ref{10.3}) now force $\lambda$ to be an integer: 
$ \lambda=n$. 
 
{\bf Exercise 4.} 
Prove this fact. 
 
So, if we want nonzero solutions, we must have  
$$
\lambda = \lambda_n = n^2; 
$$ 
these are the {\it eigenvalues\/} of the given problem. We can assume that $n$ is non-negative, since for negative $n$'s we get essentially the same solutions. 
 
{\bf Exercise 5.}  
Why? 
 
Thus, the $\Theta$ factor is 
% 
$$ 
\Theta_n(\theta) = A_n \cos n \theta + B_n \sin n \theta ; \quad n = 0,1,2,3 \ldots , 
$$ 
% 
where $A_n$ and $B_n$ are arbitrary constants. 
Notice that for $n = 0$ we recover the solution $\Theta = A_0 = {\rm const}$, obtained for the case when $k=0$.  
 
Substituting $k = \lambda^2 = n^2$ into~(\ref{10.2}) produces 

\begin{eqnarray}
\frac{r^2 R'' + r R'}{R} = - \frac{\Theta''}{\Theta} =  n^2; \quad n = 0, 1, 2, \ldots     
\label{10.4}  
\end{eqnarray}

 
The equation for $R$ is now $ r^2 R'' + r R' =  n^2 R$, or 
$$ r^2 R'' + rR'-n^2 R = 0 . $$ 
 
This is an ordinary differential equation which you probably have seen in your ODE course; it is called an 
{\it Euler equation}. The main feature of an Euler equation is that each term contains a power of $r$ that coincides with the order of the derivative of $R$.  
 
Euler equations always admit particular solutions of the form $R(r) = r^\alpha$, where $\alpha$ is a suitable (possibly fractional) power, which we will now find.  
Differentiating twice and substituting into the equation, we get 
 
$$ \alpha(\alpha-1) r^\alpha + \alpha r^\alpha-n^2 r^\alpha = 0, $$ 
or, cancelling the common (positive) factor $r^\alpha$, 
$$ \alpha(\alpha-1) + \alpha -n^2 = 0, $$ 
or just 
$$ \alpha^2 -n^2 = 0 . $$ 
% 
This is the {\it characteristic equation\/} for the Euler equation. In this case, the solutions are particularly easy to find: $\alpha = \pm n $.  
 
Accordingly, we have two particular solutions for our equation: 
$R_1(r) = r^n $ and $R_2(r) = r^{-n}$.  
The general solution of the factor $R(r)$ is a linear combination of these: 
% 
$$ R(r) = P_n r^n + Q_n r^{-n}, n = 0, 1, 2, \ldots , $$ 
where $P_n$ and $Q_n$ are arbitrary constants. 
 
For $n=0$ the two solutions coincide and are equal to $r^0 = 1 = {\rm const}$. In this case a second solution of the corresponding equation $ r^2 R'' + rR' = 0 $ is $R(r) = \ln r$.  
 
 
{\bf Exercise 6.} 
Find the general solution of the Euler equation $r^2R'' + r R' = 0$, corresponding to the case $n=0$. 
 
{\it Hint:\/} We already know one particular solution, $R(r) = 1 = {\rm const}$.  
To find the second solution, do the change of variables $R' = z$ and solve a first-order separable equation for $z$. You should get a second solution looking like $R(r) = \ln r$. 
 
We will ignore this second solution because it is not bounded at the center of the circle, when $r=0$. 
 
If  $Q_n$ is nonzero for some positive $n$, then $R(r)$ will contain the term $r^{-n}$, which blows up at the center of the circle. We don't want this kind of behavior, so we ask that $Q_n = 0$ for all nonzero $n$'s. 
 
{\bf Exercise 7.} 
Can you imagine a problem involving $\Delta u = 0$ somewhere and $u = f$ on the boundary of a circle, for which $Q_n$ nonzero could be useful? (Think outside the  box!) 
 
We have thus obtained the radial factor:
$$ R_n(r) = P_n r^n , \quad n = 0, 1, 2, \ldots $$ 
 
Combining the two factors we get the {\it eigenfunctions\/} of the given problem  
$u_n(r, \theta) = R_n(r) \Theta_n(\theta)$: 
% 
$$ u_n(r, \theta) = r^n(A_n \cos n \theta + B_n \sin n \theta) , \quad n = 0, 1, 2, \ldots $$ 
 
All these functions satisfy the Laplace equation, that is, all these functions are {\it harmonic}.  
In particular, we get the following interesting result: the functions 
% 
$$ r^n \cos n \theta \qquad \hbox{and}\qquad  r^n \sin n \theta $$ 
are harmonic for every $n = 0, 1, 2, \ldots$ 
 
{\bf Exercise 8.} 
Have you seen these functions before? Where and when?  
Do you know another way of proving that they are harmonic? 
 
Now we can get back to the original problem (if we still remember what it was!).  
Each $u_n$ satisfies the Laplace equation.  
Since this equation is linear and homogeneous, any linear combination of the $u_n$'s will also satisfy the same equation. This is even true for an infinite linear combination, provided the series converges nicely enough. So, 

\begin{eqnarray}
\qquad u(r, \theta)= \sum_{n=0}^{\infty} u_n(r, \theta)
             =A_0 + \sum_{n=1}^{\infty}   r^n(A_n \cos n \theta + B_n \sin n \theta)
\label{10.5} 
\end{eqnarray}
%
will be a solution of $\Delta u = 0$. Our hope is to be able to pick the coefficients $A_n$ and $B_n$ so as to also satisfy the boundary condition $u(a, \theta) = f(\theta)$.  
 
We have 

\begin{eqnarray}
u(a, \theta) = A_0 +  \sum_{n=1}^{\infty}   a^n (A_n \cos n \theta + B_n \sin n \theta). 
\label{10.6} 
\end{eqnarray}
 
To compare this with $f(\theta)$ it helps to expand this function in Fourier series: 

\begin{eqnarray}
 f(\theta) = \frac{a_0}{2} +  
    \sum_{n=0}^{\infty} (a_n \cos n \theta + b_n \sin n \theta), 
\label{10.7}
\end{eqnarray}
where 
\begin{eqnarray}
a_n &=& \frac{1}{\pi} \int_{-\pi}^{\pi} f(\phi) \cos n \phi \, d \phi, \quad n = 0, 1, 2, \ldots ;  
                                              \label{10.8} \\
b_n &=& \frac{1}{\pi} \int_{-\pi}^{\pi} f(\phi) \sin n \phi \, d \phi, \quad n =  1, 2, \ldots 
                                              \label{10.9} 
\end{eqnarray}

Expansion~(\ref{10.7}) is valid under mild assumptions on $f(\phi)$; for example, it holds if $f$ is piecewise smooth. Unfortunately, just the assumption that $f$ be continuous is not enough. We will ignore this question for now, and assume that everything converges. (As D'Alembert said, ``allez en avant et la fois fous viendra''---``just do it now, and prove it later''.) 
 
We want $u(a, \theta)$ to coincide with $f(\theta)$. Comparing~(\ref{10.6}) and~(\ref{10.7}), we get 
% 
$$ A_0 = \frac{a_0}{2}; \qquad a^n A_n = a_n;   
                                        \qquad a^n B_n = b_n, \quad n = 1, 2, \ldots $$ 
% 
In turn, using~(\ref{10.8}) and~(\ref{10.9}), this implies that 
% 
$$ A_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\phi) \, d \phi; $$ 
$$ A_n = \frac{1}{a^n \pi} \int_{-\pi}^{\pi} f(\phi) \cos n \phi \, d \phi, \quad  n = 1, 2, \ldots ; $$ 
$$ B_n = \frac{1}{a^n \pi} \int_{-\pi}^{\pi} f(\phi) \sin n \phi \, d \phi, \quad  n = 1, 2, \ldots . $$ 
% 
By substituting all these expressions into~(\ref{10.5}) we obtain a formula for the solution of our problem: 
% 
$$ u(r, \theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(\phi) \, d \phi +  
     \frac{1}{\pi}  \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n  
     \int_{-\pi}^{\pi} (\cos n\theta \cos n\phi + \sin n\theta \sin n\phi) f(\phi) \, d\phi . $$ 
% 
Using the addition formula for the cosine, we can simplify a little this expression: 
\begin{eqnarray*}
u(r, \theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(\phi) \, d \phi 
  +\frac{1}{\pi}  \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n  
     \int_{-\pi}^{\pi}  \cos n(\theta-  \phi) f(\phi) \, d\phi .  
\end{eqnarray*} 
\begin{eqnarray}
\label{10.10}
\end{eqnarray}
%\vfill
%\eject



\section{The Poisson Kernel}
 
Formula~(\ref{10.10}) is usually as far as one can go, using the method of separation of variables. 
Sometimes, though, one gets lucky 
(are you feeling lucky today?), 
and one can obtain a more compact expression.  
That's what we will do now.  
Let us work a little more with~(\ref{10.10}). To begin with, let us assume that it is legal to interchange summation and integration. Again, you need some assumptions on $f(\phi)$ for this to hold, but we will again ignore this fact for the time being. Then we can rewrite~(\ref{10.10}) as  
\begin{eqnarray}
\quad u(r, \theta) = \frac{1}{\pi} \int_{-\pi}^{\pi} \left[ \frac{1}{2} + 
       \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n  
        \cos n(\theta- \phi) \right] f(\phi) \, d\phi .  
\label{10.11} 
\end{eqnarray}

The surprising fact is that you can actually compute the sum of this series. We don't want to spoil you the fun of actually doing it yourself. 
 
{\bf Exercise 9.}  
Compute the sum: 
\begin{eqnarray}
\frac{1}{2} + \sum_{n=1}^{\infty} t^n \cos n \alpha, \qquad |t| < 1. 
\label{10.12} 
\end{eqnarray}
The first term $\frac{1}{2}$ is here for convenience; besides, this is the series we need. 
 
Here is a hint: Use complex variables to relate this to a geometric series. The idea is that this series is the real part of the series 

\begin{eqnarray}
\frac{1}{2} + \sum_{n=1}^{\infty} t^n e^{i n \alpha },  \label{10.13} 
\end{eqnarray}
%
which converges for every $t$ less than $1$ in absolute value.  
Find the sum of~(\ref{10.13}) using geometric series.  
Then the real part of your answer will be the sum of~(\ref{10.12}). 
 
As another hint, here is the answer you should get: 

\begin{eqnarray}
\qquad \frac{1}{2} + \sum_{n=1}^{\infty} t^n \cos n \alpha  
 = \frac{1-t^2}{2(1 - 2t\cos \alpha+t^2)}         , \qquad |t| < 1. 
\label{10.14}  
\end{eqnarray}
 
{\bf Exercise 10.}  
(This is an aside.) When $t = 1$, expression~(\ref{10.14}) no longer makes sense. However, you can still find the  
{\it finite\/} sum 
% 
$$  \sum_{n=1}^{N }  \cos n \alpha = \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha , $$ 
% 
which is interesting by itself, and has many applications. Find this sum, using similar ideas. 
 
Continuing with our computations, let us apply~(\ref{10.14}) to find the sum of the series in~(\ref{10.11}): 
% 
\begin{eqnarray*}
u(r, \theta) = \frac{1}{2 \pi} \int_{-\pi}^{\pi}  
      \frac{1 - \displaystyle \left(\frac{r}{a}\right)^2}{1-2 \displaystyle \left(\frac{r}{a}\right) \cos(\theta-\phi) + r^2}  f(\phi) \, d\phi ,
\end{eqnarray*}  
% 
or, multiplying top and bottom by $a^2$, 

\begin{eqnarray}
u(r, \theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi}  
      \frac{a^2 - r^2}{a^2-2 ar \cos(\theta-\phi) + r^2} f(\phi) \, d\phi . 
\label{10.15}  
\end{eqnarray}
 
This formula makes sense for every $\theta$ and every $r < a$. When $r=a$, expression~(\ref{10.15}) no longer makes sense. 
 
The function 
\begin{eqnarray}
K(r, \theta, a, \phi) = \frac{1}{2\pi} \frac{a^2 - r^2}{a^2-2 ar \cos(\theta-\phi) + r^2} 
                              \label{10.16} 
\end{eqnarray}
is called the {\it Poisson kernel}. Using it, we can write the solution to the problem of finding $u$ such that  
$\Delta u = 0$ inside the circle, and $u = f$ on the boundary, in a very compact form: 

\begin{eqnarray}
u(r, \theta) = \int_{-\pi}^{\pi} K(r, \theta, a, \phi) f(\phi) \, d \phi . 
\label{10.17}  
\end{eqnarray} 

%\vfill
%\eject



\section{Validity of the Solution}

 
Formula~(\ref{10.15}) or, equivalently,~(\ref{10.17}), looks very nice. However, we obtained it under unspecified assumptions for $f$, lighthearted assumptions that the infinite sum of solutions is still a solution, and a careless swapping of integration.  
 
What one can try to do now is look at~(\ref{10.15}), forget how we obtained this formula, and see if it satisfies the given problem.  
One immediate problem we have already noticed is that~(\ref{10.15}) no longer makes sense when $r=a$, so there is little hope that~(\ref{10.15}) will directly satisfy the boundary condition.  
However, one can prove the following surprising result. 
 
\begin{theorem} 
If $f(\theta)$ is continuous and periodic with period $2\pi$ then the function
$u(r, \theta)$ given by~(\ref{10.15}) or~(\ref{10.17}) satisfies 
$\Delta u = 0$ for $r <a$ and:
\begin{eqnarray}
\lim_{r \to a^{-}, \theta \to \theta_0} = f(\theta_0),  
\label{10.18}
\end{eqnarray}
for every $\theta_0$. 
\end{theorem}

\begin{remark} 
In other words, if we define a function $u(r, \theta)$ for $r \le a$ and every $\theta$ as: 
\begin{eqnarray}
u(r, \theta)=
\begin{cases}
\int_{-\pi}^{\pi} K(r,\theta,a,\phi) f(\phi) \, d\phi,& \text{if}\, r<a,\\ 
  f(\theta),& \text{if}\,  r = a,  \, \label{10.19}
\end{cases}
\end{eqnarray}
then we obtain a function that is continuous on the closed circle $r \le a$ and harmonic inside it. 
 
One of the reasons why this result is surprising is the fact that  
we know that the Fourier expansion for $f$, which was one of our key assumptions along the way, in general is not valid when $f$ is just continuous. 
 
And important addition to this result is that {\bf the solution is unique}. Indeed, this is a consequence of the Maximum Principle (Lecture 3, Challenge Problem 5). 
 
What is more, one can prove that, in fact, if $f$ is only {\it piecewise continuous\/} and has only jump discontinuities, then~(\ref{10.15}) {\it is still a harmonic function\/} inside the circle and satisfies the limit condition~(\ref{10.18}) at every point $\theta_0$ at which $f(\theta)$ is continuous.  
\end{remark} 
The proof of the theorem is very beautiful and uses several results of analysis and harmonic functions: 
%Here is an outline of the proof. 
\begin{proof}
\begin{enumerate} 
\item
If $f(\theta)$ is continuous or, even weaker, if $f(\theta)$ is bounded an integrable, the function  
$u(r, \theta)$ given by:

\begin{eqnarray}
 u(r, \theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n 
         (a_n \cos n \theta + b_n \sin n \theta), 
\label{10.20}    
\end{eqnarray}
where $a_n$ and $b_n$ are given by~(\ref{10.8}) and~(\ref{10.9}), is infinitely differentiable inside the circle. Indeed, the Fourier coefficients $a_n$ and $b_n$ are bounded, so any derivative of $u(r, \theta)$ is bounded above in absolute value by 
$$ 
\sum_{n=0}^{\infty} \left(\frac{r}{a}\right)^n (|a_n| + |b_n|) , 
$$ 
which converges inside the circle. A theorem from analysis (the ``Weierstrass $M$-series theorem'') shows that $u(r, \theta)$ has as many derivatives as we want, and in fact satisfies $\Delta u = 0$ inside the circle, that is, it is harmonic inside the circle. 
The same applies to all subsequent formulas we got for $u(r ,\theta)$; in particular, this proves that~(\ref{10.15}) is harmonic inside the circle, under these very weak assumptions on $f$. 
 
\item
If $f(\theta)$ has many continuous derivatives, let us say three, to be on the safe side, then  
equality~(\ref{10.20}) is true, with $a_n$ and $b_n$ given by~(\ref{10.8}) and~(\ref{10.9}). Moreover, the coefficients $a_n$ and $b_n$ can be bounded by terms of the form $\displaystyle \frac{M}{n^3}$. 
(Remember Lecture 6.) 
This, again by the ``Weierstrass $M$-series'' result will imply that the series~(\ref{10.20}) converges uniformly on the {\it closed\/} circle $r \le a$, and on the boundary is equal to $f(\theta)$.  
In other words, if $f$ is sufficiently smooth, then~(\ref{10.20}) indeed produces a solution $u(r, \theta)$ to our problem: a function $u$ harmonic inside the circle and equal to $f$ on the boundary. This in turn implies that the same can be said about the function $u(r, \theta)$ defined by~(\ref{10.15}). 
 
\item
If $f(\theta)$ is (only) continuous, then it can be uniformly approximated by a sequence of 
functions $f_k(\theta)$ that have as many derivatives as desired. This is a powerful result from analysis, the Weierstrass approximation theorem (this guy Weierstrass keeps popping up awfully often, don't you think?). In fact, one can choose $f_k$ to be even a finite sum of trigonometric functions $\sin(m \theta)$ and $\cos(m \theta)$ (in general, it will {\it not\/} be a truncation of the Fourier series, though). 
 
 
\item
For every $f_k(\theta)$, which is now as smooth as we want, the corresponding function $u_k(r, \theta)$ given by~(\ref{10.15}) or by~(\ref{10.17}) for $f_k$ instead of $f$, will be a function harmonic inside the circle, and equal to $f_k$ on the boundary of the circle. 
This means that $u_k-u_m$ will also be harmonic on the circle and equal to $f_k-f_m$ on the boundary of the circle, for every $k$ and $m$. 
 
\item
Remember now the Maximum Principle for harmonic functions (Lecture 3): the maximum and the minimum of $u_k-u_m$ on the closed circle can only be attained on the boundary. This will mean that if 
$|f_k-f_m|$ can be made less than $\epsilon$, then also $|u_k(r, \theta) -u_m(r, \theta)|$ will be 
less than $\epsilon$ for all $r \le a$ and all $\theta$.  
Since $f_k$ converges uniformly (to $f$) on the boundary of the circle, this will imply that $u_k$ converges uniformly to some function $u$ on the whole closed circle. 
 
\item
Finally, since $f_k$ converges uniformly to $f$, this means that  
when taking the limit as $k \to \infty$ in 
% 
$$ u_k(r, \theta) = \frac{1}{2 \pi} \int_{-\pi}^{\pi}  
      \frac{a^2 - r^2}{a^2-2 ar \cos(\theta-\phi) + r^2} f_k(\phi) \, d\phi  $$  
% 
we can switch ``$\lim_{k \to \infty}$'' with integration at the right-hand side.  
This proves that the limit function $u$ of the sequence $u_k$ is precisely given  
by~(\ref{10.15}), and concludes the proof. 
\end{enumerate}
TaDaaa!! 
\end{proof} 
 
The proof of the more general assertion when $f$ is piecewise continuous is omitted. 
We will only observe that  
one important ingredient in this proof if the fact that the Poisson kernel  
$K(r, \theta, a, \phi)$ is {\it positive\/} for $r < a$, a fact that we invite you to prove. 
 
{\bf Exercise 11.} 
Prove that the Poisson kernel $K(r, \theta, a, \phi)$ is always positive for $r < a$. 

%
%\vfill
%\eject

\section{Interpretation of the Poisson Kernel}
 
Every time you get a solution of a linear problem in the form~(\ref{10.17}), 
where $f(\theta)$ may be either a non-homogeneous boundary conditions (as in 
this case) or a non-homogeneous right-hand side (as in the case of a force 
acting on the system), the kernel $K$ has an important mathematical and 
physical interpretation. 
 
To save a little in the notation, since the radius of the circle is fixed and 
the function $K$ depends on the {\it difference\/} $\theta-\phi$ rather than 
on the variables $\theta$ and $\phi$ independently, let us call: 
% 
$$ k(r, s) =  \frac{1}{2\pi} \frac{a^2 - r^2}{a^2-2 ar \cos s + r^2} , $$ 
% 
so that $K(r, \phi, a, \theta) = k(r, \phi-\theta)$. 
Then~(\ref{10.17}) can be rewritten as 

\begin{eqnarray}
u(r, \theta) = \int_{-\pi}^{\pi} k(r, \theta -\phi) f(\phi) \, d \phi . \label{10.21}  
\end{eqnarray}
Imagine now a constant boundary condition $f$ of height $\frac{1}{\Delta}$, concentrated on a small interval of length $\Delta$ with center at a point $\phi_0$. 
 
\vglue 2 truein  % PICTURE OF CONCENTRATED f(\theta) 
 
In other words, $f$ is zero everywhere, except near $\phi_0$, at which it is constant and equal to 
$\frac{1}{\Delta}$ on an interval of length $\Delta$. 
The height of $f$ has been chosen so as to have $\int_{-\pi}^{\pi} f(\phi) \, d \phi = 1$. 
Call $\delta_{\phi_0}$ this function. (It also depends on $\Delta$, but let us not complicate the notation.) 
According to~(\ref{10.21}), the solution of the Dirichlet problem on the circle for this $f$ is given by 
% 
$$ u(r, \theta) = \int_{-\pi}^{\pi} k(r, \theta -\phi) \delta_{\phi_0}(\phi) \, d \phi . $$ 
% 
Applying the intermediate value theorem for integrals, we can rewrite this as 
% 
$$ u(r, \theta) = k(r,\theta - \phi^*) \int_{-\pi}^{\pi}  \delta_{\phi_0}(\phi) \, d \phi, $$ 
% 
where $\phi^*$ is a number on the interval with length $\Delta$ and center $\phi_0$. 
Since the integral of $\delta_{\phi_0}$ is equal to $1$, this becomes 
% 
$$ u(r, \theta) = k(r, \theta - \phi^*) . $$ 
% 
Taking the limit as $\Delta \to 0$, we get 
% 
$$ u(r,\theta) = k(r, \theta-\phi_0) . $$ 
% 
As $\Delta \to 0$, the function $\delta_{\phi_0}$ itself tends to infinity in such a way that its integral 
is kept equal to $1$ all the time. This is the so-called {\it the delta function} with center $\phi_0$, and is denoted as $\delta(\theta - \phi_0)$.  
It is not a function but rather a generalized function, what mathematicians call a {\it distribution}.  
It is one of those extremely useful beasts that was born in the deranged mind of physicists and mathematicians had to work very hard to make any sense out of them. 
\begin{lemma} 
We conclude that 
{\it the Poisson  kernel} 
$$K(r, \theta, a, \phi) = k(r, \theta-\phi) 
       = \frac{1}{2\pi} \frac{a^2 - r^2}{a^2-2 ar \cos(\theta-\phi) + r^2} $$ 
{\it is a harmonic function inside the circle $r<a$ that is equal to zero everywhere on the boundary, except at the point $\phi$, at which it is equal to $\infty$}. Not only that, but $k(r, \theta-\phi)$ is 
a harmonic function that on the boundary coincides with the delta function with center at $\phi$. 
\end{lemma} 
Knowing the meaning of the Poisson kernel, one can now use an heuristic argument to obtain again formula~(\ref{10.21}). 
 
Assume we are given a function $f(\theta)$ on the boundary of the circle. 
We divide the interval $[0, 2\pi]$ into $N$ intervals of length $\Delta$ and approximate $f(\theta)$ by a piecewise constant function, writing it down as the sum of functions 
$f(\phi_i) \delta_{\phi_i} \Delta$: 
% 
$$ f(\phi) \approx \sum_{i=1}^N f(\phi_i) \delta_{\phi_i} \Delta . $$ 
 
\vglue 2 truein  % PIECEWISE APPROXIMATION OF f(\theta) 
 
The factor $\Delta$ is needed to get the correct height $f(\theta_i)$; 
recall that $\delta_{\phi_i}$  had height $\frac{1}{\Delta}$. 
 
Since the Dirichlet problem is linear, the solution to the boundary condition  
$f(\phi_i) \delta_{\phi_i} \Delta$ will be $f(\phi_i) \Delta$  
times the solution to the boundary condition $\delta_{\theta_i}$.  
Therefore, the solution will be 
% 
$$ k(r, \theta - \phi_i) f(\phi_i) \Delta . $$ 
% 
Again, since the Dirichlet problem is linear, the solution of a sum of boundary conditions 
$f(\phi_i) \delta_{\phi_i} \Delta$ will be the sum of the solutions for each boundary condition separately, that is, 
$$ \sum_{i=1}^N  k(r, \theta - \phi_i) f(\phi_i) \Delta . $$ 
% 
We recognize this as a Riemann sum, so if we take the limit as $\Delta \to 0$, we get the solution to the problem with boundary condition $f(\theta)$ as 
% 
$$  \lim_{\Delta\to0} \sum_{i=1}^N  k(r, \theta - \phi_i) f(\phi_i) \Delta  
     = \int_{-\pi}^{\pi} k(r, \theta-\phi) f(\phi) \, d \phi . $$ 
% 
TaDaaaa!! 
 


%\vfill
%\eject

\section{Examples}
 
In practice, Laplace equations arise when seeking steady-state distributions. In the one-dimensional heat equation, for example, $u_t = c^2 u_{xx}$, for a finite bar, $0 \leq x \leq L$, a steady state distribution of temperature is a time-independent solution. This implies $u_t=0$, so for $u = u(x)$ we get the one-dimensional Laplace equation $u_{xx} = 0$, whose general solution is a bunch of lines, 
$u = Ax+ B$. 
 
If we seek the steady-state distribution of temperature for a circular membrane, this leads from the heat equation $u_t = \Delta u$ to the Laplace equation $\Delta u = 0$, since $u_t=0$ for the steady-state distribution. 
 
In Electrostatics, according to Maxwell's equations the electrostatic potential $\phi$ satisfies the equation 
$\Delta \phi= - 4 \pi \rho$, where $\rho$ is the density of the charges. Thus, if there are no electric charges inside the region, the potential will satisfy Laplace's equation $\Delta \phi = 0$. 
 
 
{\bf Example 1.} 
As a concrete example, let us consider the problem of finding the steady-state distribution of temperature of a circular membrane, if the temperature is kept fixed and equal to $1$ in half the boundary, and $-1$ in the other half. Namely, we will consider the problem of finding $u(r, \theta)$ on the circle $r \le a$ such that 
% 
$$ \Delta u (r, \theta) = 0 \qquad \hbox{ for $r<a$},  \quad \text{\bf PDE} $$ 
\begin{eqnarray*}
u(a,\theta) = \begin{cases}
-1, \quad \text{if} & \quad  -\pi<\theta<0,\\ 
1, \quad \text{if}  & \quad  0<\theta<\pi. \qquad \text{\bf BC}
\end{cases}
\end{eqnarray*}
 
Rephrasing a little our computations in Section 10.3, as in~(\ref{10.20}) 
we can say that the solution will look like:
\begin{eqnarray*}
 u(r, \theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n 
         (a_n \cos n \theta + b_n \sin n \theta), 
\end{eqnarray*}
where: 
\begin{eqnarray}
a_n &=& \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \cos n \theta \, d \theta   
                                    \quad n = 0, 1, 2, \ldots      \label{10.22} \\ 
b_n &=& \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \sin n \theta \, d \theta , 
                                      \quad n = 1,2, \ldots  \label{10.23} 
\end{eqnarray}
In our case $f(\theta)$ is odd; therefore, all the coefficients $a_n$ will vanish. 
 
As regards the $b_n$, using the symmetry of the integrand and the interval of integration, we can write 
\begin{eqnarray*}
b_n &=& \frac{2}{\pi} \int_{0}^{\pi} f(\theta) \sin n \theta \, d \theta  \\
         &=& - \frac{2}{n\pi}(\cos n \pi - \cos 0) = \frac{2}{n\pi} \left( 1-(-1)^n \right)\\ 
       &=&\begin{cases} 0, \quad \text{if} \, n \,\text{is even},\\ 
                        \frac{4}{n\pi}, \quad \text{if} \, n \,\text{is odd}.
          \end{cases}
\end{eqnarray*}
Therefore, the solution is 
% 
\begin{eqnarray*}
u(r, \theta) &=& \frac{4}{\pi}\left( \frac{r \sin \theta}{a} + \frac{r^3 \sin 3 \theta}{3a^3} 
           + \frac{r^5 \sin 5 \theta}{ 5 a^5} + \cdots \right)\\ 
      &=&\frac{4}{\pi} \sum_{n=0}^{\infty}  \left(\frac{r}{a}\right)^{2n+1} \frac{\sin (2n+1)\theta}{2n+1} .
\end{eqnarray*} 
 
{\bf Example 2.} 
We seek a function harmonic inside the unit circle and equal to (the restriction of) the 
polynomial $x^3-y^3$ on the boundary of the unit circle.  
Let us write the problem in polar coordinates. 
Since on the unit circle we have $x = \cos \theta$ and $y = \sin \theta$, 
we must solve the problem 
% 
$$  \Delta u(r, \theta) = 0  \quad\hbox{if $r<1$}, \quad \text{\bf PDE} $$ 
$$ u(1, \theta)  =  \cos^3 \theta - \sin^3 \theta . \quad \text{\bf BC} $$ 
To expand the boundary condition, $\cos^3 \theta - \sin^3 \theta$, in Fourier series, 
it is perhaps faster to use trigonometric identities. 
 
{\bf Exercise 12.} 
Use de Moivre's formula: 
$$(\cos \theta + i \sin \theta)^n = \cos n \theta + i \sin n \theta$$ 
with $n=3$ to obtain the formulas 
\begin{eqnarray}
\cos 3 \theta &=& \cos^3 \theta-3 \cos \theta \sin^2 \theta \label{10.24}\\ 
\sin 3 \theta &=& - \sin3 \theta + 3 \cos^2 \theta \sin \theta .  
\label{10.25}   
\end{eqnarray}
In turn, use these identities, plus a little ``massaging'', to prove that 
\begin{eqnarray}
\cos^3 \theta &=& \frac{3}{4} \cos \theta + \frac{1}{4} \cos(3\theta), \label{10.26}\\
     \sin^3 \theta &=& \frac{3}{4} \sin \theta  -  \frac{1}{4} \sin(3\theta) . 
\label{10.27}  
\end{eqnarray} Identities~(\ref{10.26},\ref{10.27}) yield the desired Fourier expansion of the boundary condition: 

\begin{eqnarray}
\qquad \, \cos^3 \theta -   \sin^3 \theta =\frac{3}{4} \cos \theta + \frac{1}{4} \cos(3\theta) 
          -  \frac{3}{4} \sin \theta  +  \frac{1}{4} \sin(3\theta) . 
\label{10.28} 
\end{eqnarray} 
By~(\ref{10.20}), the solution will look like 
% 
$$ u(r, \theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} r^n 
         (a_n \cos n \theta + b_n \sin n \theta) .  $$ 
At $r=1$, the solution will be 
$$ u(1, \theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty}  
         (a_n \cos n \theta + b_n \sin n \theta) .  $$ 
Comparing this with~(\ref{10.28}) we conclude that 
 
$$ a_1 = \frac{3}{4}, \qquad b_1 = - \frac{3}{4}, \qquad a_3 = \frac{1}{4},\qquad b_3 = \frac{1}{4} , $$ 
and all the other coefficients will be equal to zero. 
In conclusion, the solution is 
% 
$$ u(r,\theta) = \frac{3}{4}r(\cos \theta - \sin \theta) + \frac{1}{4}(\cos 3 \theta + \sin 3 \theta) . $$ 
 
This solution can be written back in Cartesian coordinates, recalling that 
$r \cos \theta = x$ and $r \sin \theta = y$. Thus, the first two terms are just 
$\displaystyle \frac{3}{4} (x-y)$. To get the other two terms in Cartesians, we use again the identities~(\ref{10.24}) and~(\ref{10.25}): 
% 
$$ r^3\cos 3\theta = r^3(\cos^3\theta-3 \cos \theta \sin^2 \theta) = x^3 -3xy^3; $$ 
$$ r^3\sin 3\theta = r^3(-\sin^3\theta+3 \cos^2 \theta \sin \theta) = -y^3+3x^2y. $$ 
Therefore, in Cartesian coordinates the solution is 

\begin{eqnarray}
u(x,y) = \frac{3}{4}(x-y) + \frac{1}{4}(x^3-3xy^2-y^3+3x^2y) . \label{10.29} 
\end{eqnarray} 
 
{\bf Exercise 13.} 
Check directly that~(\ref{10.29}) is harmonic inside the unit circle (in fact, it is harmonic 
{\it everywhere\/}), and that it is equal to $x^3 -y^3$ on the boundary of the circle. 
 
 


%\vfill
%\eject


\section{Challenge Problems for Lecture 10}





\begin{problem}
{\bf Integrating the Poisson Kernel. } 
Prove that 
% 
$$ \int_{-\pi}^{\pi} K(r, \theta, a, \phi) \, d \phi $$ 
% 
is a harmonic function $u(r, \theta)$ inside the circle $r < a$, and tends to $1$ for every $\theta$, 
as $r \to a^{-}$. 
 
{\it Hint:\/} solve the Dirichlet problem $\Delta u = 0$ inside the circle and $u = 1$ on the boundary, and use uniqueness of the solution.
\end{problem}




\begin{problem}
Solve the problem 
% 
$$ \Delta u(r, \theta) = 0  \quad\hbox{if $r<a$}, \quad \text{\bf PDE} $$ 
\begin{eqnarray*}
u(a, \theta)= 
\begin{cases} 
0, & \text{if} \quad  -\pi<\theta<0  ,\\ 
1, & \text{if} \quad  0<\theta<\pi . \quad \text{\bf BC}
\end{cases}
\end{eqnarray*}


\end{problem}



\begin{problem}

Using the result you found in Problem 2, plus uniqueness of the solution of 
the Dirichlet problem for the Laplace equation, write down the integral 
 
$$ 
\int_{0 }^{\pi} K(r, \theta, a, \phi) \, d \phi  
     = \frac{1}{2\pi} \int_{0}^{\pi}  
      \frac{a^2 - r^2}{a^2 -2 ar \cos(\theta-\phi) + r^2}  \, d\phi  ,
$$   
as an infinite series, assuming $r < a$.
\end{problem}

\begin{problem}
% Articolo, P. 274, Example 5.7.1 
One can use polar coordinates, and separation of variables, for ``pizza slices'', that is, for sectors of  a circle. As an example, find the steady-state temperature distribution of a thin plate over the sector 
% 
$$\Omega = \{ (r, \theta) \, | \, 0<r<1, \> 0< \theta < \frac{\pi}{3} \} $$ 
given that 
$$ u(r, 0) = 0, \qquad u \left( r, \frac{\pi}{3}\right) = 0; $$ 
$$ u(1, \theta) = \theta\left( \frac{\pi}{3} - \theta \right) . $$ 
We assume that the temperature at zero is bounded. 
 
Repeat the process of separation of variables. What will be the boundary conditions for $\Theta(\theta)$? 
Notice that the eigenvalues will {\bf not} be the same as for the case of the full circle.  

\end{problem}


\begin{problem}
% Articolo, P. 296, Exercise 5.23. 
Find the steady-state temperature distribution of a thin plate over the sector 
% 
$$\Omega = \{ (r, \theta) \, | \, 0<r<1, \> 0< \theta < \frac{\pi}{3} \} $$ 
given that 
$$ 
u(r, 0) = 0, \qquad u_\theta \left( r, \frac{\pi}{3}\right) = 0; 
$$ 

$$ 
u(1, \theta) = \theta\left(1- \frac{3\theta}{2 \pi}  \right) . 
$$ 
Assume that the temperature at zero is bounded. 
 
(See the comments in Problem 4.)
\end{problem}



\vfill


\end{document}