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\title[MATH 234 Spring 14]{Vanderbilt University, MATH 234 Spring 14: The Poisson equation 
in $\RR^n$.}
%\date{\today}

%\author[Disconzi]{Marcelo M. Disconzi}
%\address{Department of Mathematics\\
%Vanderbilt University, Nashville, TN 37240, USA}
%\email{marcelo.disconzi@vanderbilt.edu}
\urladdr{http://www.disconzi.net/Teaching/MAT234-Spring-14/MAT234-Spring-14.html}

%\date{\today}
%\pagestyle{empty}

\begin{document}


%\begin{abstract}
%Abstract. 
%\end{abstract}


\maketitle

This is a mix of class notes and homework assignment, whose goal is to 
solve
\begin{gather}
-\Delta u = f
\label{Poisson}
\end{gather}
in $\RR^n$. 

From now on, it is assumed that $n \geq 3$. 
Let $f \in C^2(\RR^n)$, and assume that $f$ has compact support. Recall that
in class we defined $\Phi: \RR^n \backslash \{ 0 \} \rar \RR$ by
\begin{gather}
\Phi(x) = \frac{1}{n(n-2) \alpha_n} \frac{1}{|x|^{n-2}},
\label{fs}
\end{gather}
where $\al_n$ is the volume of the unit ball in $\RR^n$.

\noindent \textbf{Problem 1.} Compute
\begin{gather}
\partial_i |x|,
\nonumber
\end{gather}
and use this to show that there exists a constant $C> 0$ such that
\begin{gather}
|\partial_i \Phi | \leq \frac{C}{|x|^{n-1}}, \, 
|\partial_{ij} \Phi | \leq \frac{C}{|x|^{n}}, \, i, j =1, \dots, n, \, x \neq 0.
\nonumber
\end{gather}

\noindent \textbf{Solution.} Since
\begin{gather}
|x| = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2},
\nonumber
\end{gather}
It follows at once that 
\begin{gather}
\partial_i |x| = \frac{x_i}{ \sqrt{x_1^2 + x_2^2 + \dots + x_n^2} } 
= \frac{x_i}{|x|}.
\nonumber
\end{gather}
Now use the chain rule to compute
\begin{gather}
\partial_i |x|^{2-n} = (2-n)|x|^{1-n} \frac{x_i}{|x|}.
\nonumber
\end{gather}
Noticing that 
\begin{gather}
\left| \frac{x_i}{|x|} \right| \leq 1,
\nonumber
\end{gather}
one obtains
\begin{gather}
\left| \partial_i |x|^{2-n|} \right| \leq \frac{n-2}{|x|^{1-n}},
\nonumber
\end{gather}
from which the first desired inequality follows.
Next, use the  chain rule again to compute
\begin{align}
\begin{split}
\partial_{ji} |x|^{2-n} &=
 \partial_j \left(  (2-n)|x|^{1-n} \frac{x_i}{|x|} \right) \\
& =  (2-n) \partial_j \left(  \frac{x_i}{|x|^{n}} \right) \\
& = (2-n) \frac{|x|^n \delta_{ij} - n |x|^{n-1}\frac{x_i x_j}{|x|} }{|x|^{2n} }\\
 & = \frac{2-n}{|x|^n} \left( \de_{ij} - \frac{n x_i x_j}{|x|^2} \right),
\end{split}
\nonumber
\end{align}
where we used that 
\begin{gather}
\partial_j x_i = 
\de_{ij} = 
\begin{cases}
0, & i \neq j,\\
1, & i=j.
\end{cases}
\nonumber
\end{gather}
The second desired inequality now follows from noticing that 
\begin{gather}
\left| \de_{ij} - \frac{n x_i x_j}{|x|^2}  \right| 
\leq 1 + n,
\nonumber
\end{gather}
since 
\begin{gather}
\left| \frac{ x_i x_j}{|x|^2} \right | \leq 1.
\nonumber
\end{gather}

Define $u: \RR^n \rar \RR$ by
\begin{gather}
u(x) = \int_{\RR^n} \Phi(x-y) f(y) \, dy.
\nonumber
\end{gather}
Recall that this can also be written as 
\begin{gather}
u(x) = \int_{\RR^n} \Phi(y) f(x-y) \, dy.
\label{conv}
\end{gather}
In class, we showed that $u$ is well-defined, and that the second derivatives 
of $u$ exist and satisfy
\begin{gather}
u_{x_i x_j}(x) = \int_{\RR^n} \Phi(y)f_{x_i x_j}(x-y) \, dy.
\nonumber
\end{gather}

\noindent \textbf{Problem 2.} Show that $u_{x_i x_j}$ is continuous. Recalling the
definition of continuity, you have to show that, given $x_0 \in \RR^n$ and
 $\varepsilon>0$, there exists
a $\de > 0$, such that if $|x - x_0| < \de$, then $|u_{x_i x_j}(x) - u_{x_i x_j}(x_0)|
< \varepsilon$. Do this as follows. Fix $\varepsilon > 0$. Write
\begin{align}
\begin{split}
|u_{x_i x_j}(x) - u_{x_i x_j}(x_0)| & = 
\Big | \int_{\RR^n} \Phi(y)f_{x_i x_j}(x-y) \, dy - 
\int_{\RR^n} \Phi(y)f_{x_i x_j}(x_0-y) \, dy \Big | \\
& = 
\Big | \int_{\RR^n} \Phi(y) ( f_{x_i x_j}(x-y) -  
f_{x_i x_j}(x_0 - y)  )\, dy \Big |
\end{split}
\nonumber
\end{align}
Use the continuity of $f_{x_i x_j}$, and the fact that $f$ has 
compact support (i.e., that $\operatorname{supp}(f) \subset B_R(0)$ for some
$R>0$), 
to show that given $\varepsilon^\prime > 0$, 
we can choose $\de > 0$ so that 
\begin{align}
\begin{split}
|u_{x_i x_j}(x) - u_{x_i x_j}(x_0)| & \leq  
\int_{\RR^n} \Phi(y) | f_{x_i x_j}(x-y) -  
f_{x_i x_j}(x_0- y)  | \, dy \\
& \leq \varepsilon^\prime \int_{B_R(0)} \Phi(y) \, dy,
\end{split}
\nonumber
\end{align}
provided that $|x - x_0| < \de$.
Next, use the expression (\ref{fs}), and 
integration in polar coordinates (in $n$ dimensions), to show that $\varepsilon^\prime$
can be  chosen so that 
\begin{gather}
\varepsilon^\prime \int_{B_R(0)} \Phi(y) \, dy, < \varepsilon,
\nonumber
\end{gather}
as desired.

\noindent \textbf{Solution.} By continuity, given $\varepsilon^\prime$, 
there exists a $\de > 0$ such that 
\begin{gather}
| f_{x_i x_j}(x-y) -  f_{x_i x_j}(x_0- y)  | < \varepsilon^\prime,
\nonumber
\end{gather}
provided that 
\begin{gather}
|x - y - (x_0 - y ) | = |x - x_0 | < \de.
\nonumber
\end{gather}
$\de$ may in principle depend on $y$, but since the support of $f$ (and hence
of $f_{x_i x_j}$) is compact, $\de$ can be chosen uniformly.
Since $\varepsilon^\prime$ can be chosen as small as we want, we set
\begin{gather}
\varepsilon^\prime = \left( \int_{B_R(0)} \Phi(y) \, dy \right)^{-1} \varepsilon.
\nonumber
\end{gather}

We will  now show that (\ref{Poisson}) holds. The argument here will be slightly simpler
than what we did in class, although conceptually it is the same.

From (\ref{conv}), compute
\begin{gather}
\Delta u(x) = \int_{\RR^n} \Phi(y) \Delta_x f(x-y) \, dy.
\nonumber
\end{gather}
Fix $\varepsilon > 0$, and write
\begin{gather}
\Delta u(x) = \int_{B_\varepsilon(0) } \Phi(y) \Delta_x f(x-y) \, dy +
\int_{\RR^n \backslash B_\varepsilon(0) } \Phi(y) \Delta_x f(x-y) \, dy.
\label{lap_u}
\end{gather}
Since $\Delta_x f$ is a continuous function and $f$ has compact support, it follows
that there exists a constant $M> 0$ such that 
\begin{gather}
|\Delta f (x) | \leq M \text{ for all } x \in \RR^n.
\nonumber
\end{gather}
Thus 
\begin{gather}
 \Big | \int_{B_\varepsilon(0) } \Phi(y) \Delta_x f(x-y) \, dy  \Big|
 \leq  M \int_{B_\varepsilon(0) } \Phi(y)  \, dy.
 \nonumber
 \end{gather}
 
\noindent \textbf{Problem 3.} Using polar coordinates, as done in class, estimate
the integral on the right-hand side of the previous expression and show that 
\begin{gather}
\lim_{\varepsilon \rar 0} 
\int_{B_\varepsilon(0) } \Phi(y) \Delta_x f(x-y) \, dy   = 0.
\nonumber
\end{gather}

\noindent \textbf{Solution.} It follows from 
\begin{gather}
0 \leq \int_{B_\varepsilon(0) } \Phi(y)  \, dy =
\frac{1}{n(n-2)\alpha_n} \int_0^\varepsilon \int_{S^{n-1}}
\frac{1}{r^{n-2}} r^{n-1} \, dr d\omega \leq C \varepsilon^2.
\nonumber
\end{gather}

\noindent \textbf{Problem 4.} As done in class, use the chain rule to show that 
\begin{gather}
\int_{\RR^n \backslash B_\varepsilon(0) } \Phi(y) \Delta_x f(x-y) \, dy
= 
\int_{\RR^n \backslash B_\varepsilon(0) } \Phi(y) \Delta_y f(x-y) \, dy.
\nonumber
\end{gather}
\noindent \textbf{Solution.} By the chain rule,
\begin{gather}
\frac{\partial}{\partial y_i} \left( f(x-y) \right)
= \sum_{j=1}^n 
\partial_j f(x-y) \frac{ \partial(x_j - y_j)}{\partial y_i }
= - \partial_i f(x-y),
\nonumber
\end{gather}
so that 
\begin{gather}
\frac{\partial^2}{\partial y_i^2} \left( f(x-y) \right)
=  \partial_{ii} f(x-y).
\nonumber
\end{gather}

Thus (\ref{lap_u}) becomes
\begin{gather}
\Delta u(x) = I + II,
\nonumber
\end{gather}
where
\begin{gather}
I = \int_{B_\varepsilon(0) } \Phi(y) \Delta_x f(x-y) \, dy.
\nonumber
\end{gather}
and 
\begin{gather}
II = \int_{\RR^n \backslash B_\varepsilon(0) } \Phi(y) \Delta_y f(x-y) \, dy.
\nonumber
\end{gather}
Integrating by parts, $II$ becomes
\begin{align}
\begin{split}
 \int_{\RR^n \backslash B_\varepsilon(0) } \Phi(y) \Delta_y f(x-y) \, dy 
 & = 
 - \int_{\RR^n \backslash B_\varepsilon(0) } \langle \nabla \Phi(y), \nabla_yf(x-y) 
 \rangle \, dy 
 \\
 &
 + \int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) } \Phi(y) \partial_\nu f(x-y) ds(y).
 \end{split} 
\nonumber
\end{align}

\noindent \textbf{Problem 5.} Arguing similarly to problem 3, show that 
\begin{gather}
\lim_{\varepsilon \rar 0} 
\int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) } \Phi(y) \partial_\nu f(x-y) ds(y)
= 0.
\nonumber
\end{gather}
\noindent \textbf{Solution.} Using continuity of the first derivatives of $f$
and the fact that it has compact support, we can find $M > 0$ such that
\begin{gather}
|\nabla f | \leq M.
\nonumber
\end{gather}
In particular,
\begin{gather}
|\partial_\nu f | \leq M.
\nonumber
\end{gather}
Thus
\begin{gather}
0 \leq 
\left |
\int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) } \Phi(y) \partial_\nu f(x-y) ds(y)
\right|
\leq M 
\int_{\partial B_\varepsilon(0) } \Phi(y)  \, ds(y)
\nonumber \\
= \frac{M}{n(n-2)\alpha_n} \int_{S^{n-1}} \frac{1}{\varepsilon^{n-2}} 
\varepsilon^{n-1} \ d\omega \leq C \varepsilon,
\nonumber
\end{gather}
which gives the result.


Integrating by parts again, $II$ can still be written as 
\begin{align}
\begin{split}
 \int_{\RR^n \backslash B_\varepsilon(0) } \Phi(y) \Delta_y f(x-y) \, dy 
 & = 
  \int_{\RR^n \backslash B_\varepsilon(0) }\Delta \Phi(y) f(x-y) \, dy 
 \\
 &
 - \int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) } \partial_\nu \Phi(y) f(x-y) ds(y),
 + III
 \end{split} 
\label{int}
\end{align}
where 
\begin{gather}
III = \int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) } \Phi(y) \partial_\nu f(x-y) ds(y).
\nonumber
\end{gather}
Since $\Phi$ satisfies $\Delta \Phi = 0$ in $\RR^n\backslash \{0\}$, the first integral
on the right-hand side of (\ref{int}) vanishes. Combining the above calculations then
gives
\begin{gather}
\Delta u(x) = I - \int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) } \partial_\nu \Phi(y) f(x-y) ds(y)
+ III.
\label{delta_u_I_III}
\end{gather}
Using (\ref{fs}) and problem 1, we can compute
\begin{gather}
\partial_i \Phi(y) = -\frac{1}{n\al_n} \frac{y_i}{|y|^n} = 
-\frac{1}{n\al_n |y|^{n-1}} \frac{y_i}{|y|}.
\nonumber
\end{gather}
On $\partial ( \RR^n \backslash B_\varepsilon(0) )$, we have
$|y|= \varepsilon$ and 
$ \frac{y_i}{|y|} = - \nu_i$ (recall that the negative sign appears
because $\nu$ is the outer normal to $\partial ( \RR^n \backslash B_\varepsilon(0) )$,
which is opposite to the normal to $B_\varepsilon(0)$). Thus, the above becomes
\begin{gather}
\partial_i \Phi(y) = 
\frac{1}{n\al_n \varepsilon^{n-1}} \nu_i,
\nonumber
\end{gather}
and then
\begin{align}
\begin{split}
\partial_\nu \Phi(y) & = \sum_{i=1}^n \partial_i \Phi(y) \nu_i 
\\
& =
\sum_{i=1}^n \frac{1}{n\al_n \varepsilon^{n-1}} \nu_i^2 \\
& = 
\frac{1}{n\al_n \varepsilon^{n-1}}  \sum_{i=1}^n \nu_i^2 \\ 
& = \frac{1}{n\al_n \varepsilon^{n-1}} |\nu|^2 \\ 
& = \frac{1}{n\al_n \varepsilon^{n-1}},
\end{split}
\nonumber
\end{align}
since $|\nu| = 1$.
Using this into (\ref{delta_u_I_III}) gives
\begin{gather}
\Delta u(x) = I - \frac{1}{n\al_n \varepsilon^{n-1}} \int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) }  f(x-y) ds(y)
+ III.
\nonumber
\end{gather}
Taking the limit $\varepsilon \rar 0$, using problems 3 and 5, and noticing that
$\Delta u (x)$ does not depend on $\varepsilon$, we obtain
\begin{gather}
\Delta u(x) =  - \lim_{\varepsilon \rar 0} 
\frac{1}{n\al_n \varepsilon^{n-1}} \int_{\partial ( \RR^n \backslash B_\varepsilon(0) ) }  f(x-y) ds(y).
\nonumber
\end{gather}
Notice that, as sets,
\begin{gather} 
\partial ( \RR^n \backslash B_\varepsilon(0) ) = \partial  B_\varepsilon(0),
\nonumber
\end{gather}
thus
\begin{gather}
\Delta u(x) =  - \lim_{\varepsilon \rar 0} 
\frac{1}{n\al_n \varepsilon^{n-1}} \int_{\partial  B_\varepsilon(0)  }  f(x-y) ds(y).
\nonumber
\end{gather}
\noindent \textbf{Problem 6.} Show that, upon changing variables, this last expression
becomes
\begin{gather}
\Delta u(x) =  - \lim_{\varepsilon \rar 0} 
\frac{1}{n\al_n \varepsilon^{n-1}} \int_{\partial  B_\varepsilon(x)  }  f(y).
\nonumber
\end{gather}

\noindent \textbf{Solution.} Set $z = x-y$ and notice that the Jacobian
of this change of variables is equal to one.

Notice that  $n\al_n \varepsilon^{n-1}$ is the volume of $\partial B_\varepsilon(x)$,
\begin{gather}
\Delta u(x) =  - \lim_{\varepsilon \rar 0} 
\frac{1}{\vol (\partial B_\varepsilon(x))} \int_{\partial  B_\varepsilon(x)  }  f(y).
\nonumber
\end{gather}
The right-hand side  is the average of $f$ over $\partial B_\varepsilon(x)$. But if we 
average $f$ over ever smaller concentric spheres, the value of the average approaches
the value of $f$ at the center. Hence,
\begin{gather}
   \lim_{\varepsilon \rar 0} 
\frac{1}{\vol (\partial B_\varepsilon(x))} \int_{\partial  B_\varepsilon(x)  }  f(y)
= f(x),
\nonumber
\end{gather}
finishing the proof.

\noindent \textbf{Problem 7.} Solve Poisson's equation, as above, in the case $n=2$.
\emph{Hint:} chapter 8 of the textbook.

\noindent \textbf{Solution.} This is done in chapter 8 of the textbook, see corollary 8.2.

\vskip 1cm

\end{document}
