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\begin{document}


\begin{center}
Laplace Equation
\end{center}

\bigskip

\begin{equation*}
\Delta u=\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{%
\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}=0
\end{equation*}%
Also Cauchy-Riemann equations (2D)%
\begin{align*}
\frac{\partial u}{\partial x}& =\frac{\partial v}{\partial y} \\
\frac{\partial u}{\partial y}& =-\frac{\partial v}{\partial x}
\end{align*}

\bigskip

\begin{center}
Maximum Principle
\end{center}

\begin{theorem}
Let $\Delta u=0$ in D. Then the maximum and minimum occur on the boundary
and not inside unless $u$=constant.
\end{theorem}

\begin{proof}
proof \#1 (weak version)

same as for heat equation consider $v(x,y)=u(x,y)+\varepsilon (x^{2}+y^{2})$.

Then $\Delta v>0$.

However, at a maximum $\Delta v\leq 0$ and so we can't have an interior
maximum for $v$

and then one can estimate the same for $u$

\proof\#2

By the first Green formula%
\begin{equation*}
\tiint\limits_{D}v\Delta udV+\tiint\limits_{D}\nabla u\cdot \nabla
vdV=\toint\limits_{\partial D}v\frac{\partial u}{\partial n}dS
\end{equation*}

First choose $v=1$. Then%
\begin{equation*}
(\ast )\qquad 0=\tiint\limits_{D}\Delta udV=\toint\limits_{\partial D}\frac{%
\partial u}{\partial n}dS
\end{equation*}

Next choose $v=u$. Then%
\begin{equation*}
\tiint\limits_{D}u\Delta udV+\tiint\limits_{D}|\nabla
u|^{2}dV=\toint\limits_{\partial D}u\frac{\partial u}{\partial n}dS
\end{equation*}

Assume that there is a maximum at an interior point ($x_{0},y_{0})$. Then we
can draw a circle about this point so that $u$ is constant on this circle.
Then using we have $\Delta u=0$ we have%
\begin{equation*}
\tiint\limits_{D}|\nabla u|^{2}dV=u\toint\limits_{\partial D}\frac{\partial u%
}{\partial n}dS\overset{by\;(\ast )}{=}0
\end{equation*}

Since $|\nabla u|^{2}\geq 0$ it follows that $|\nabla u|^{2}=0$ and so $u$
is constant
\end{proof}

\begin{center}
\bigskip

Neumann problem
\end{center}

\begin{align*}
\Delta u &= 0\qquad \text{in \ D} \\
\frac{\partial u}{\partial n} &= h\qquad \text{on }\partial D
\end{align*}

\bigskip

\noindent By (*)%
\begin{equation*}
0=\tiint\limits_{D}\Delta udV=\toint\limits_{\partial D}\frac{\partial u}{%
\partial n}dS=\toint\limits_{\partial D}hdS
\end{equation*}

Fredholm alternative:

\noindent Hence, a necessary condition for a solution to exist is that%
\begin{equation*}
\toint\limits_{\partial D}hdS=0
\end{equation*}

\noindent If this is true then there are an infinite number of solutions
since we can add an arbitrary constant to any solution.

\bigskip

\begin{center}
Uniqueness
\end{center}

\bigskip

\begin{theorem}
\begin{align*}
\Delta u& =f\qquad \text{in \ D} \\
u& =h\qquad \text{on }\partial D
\end{align*}

Then the solution is unique
\end{theorem}

\begin{proof}
Follows from maximum principle
\end{proof}

Existence is hard to prove.

\bigskip

\begin{center}
Dirichlet Principle
\end{center}

\bigskip 
\begin{equation*}
E(w)=\frac{1}{2}\tiint\limits_{D}|\nabla w|^{2}dV
\end{equation*}

\noindent Consider all functions $w$ that satisfy $w=h$ on $\partial D$ .
Then the minimum of $w$ is reached for $w=u$ and

\begin{align*}
\Delta u& =0\qquad \text{in \ D} \\
u& =h\qquad \text{on }\partial D
\end{align*}

\noindent Problem with the proof is that it is not clear the minimum is
achieved !

\bigskip

\begin{center}
Energy
\end{center}

\bigskip 
\begin{equation*}
\tiint\limits_{D}u\Delta udV+\tiint\limits_{D}|\nabla
u|^{2}dV=\toint\limits_{\partial D}u\frac{\partial u}{\partial n}dS
\end{equation*}

\noindent So if $\Delta u=0$ in D and either $u=0$ or $\frac{\partial u}{%
\partial n}=0$ on $\partial D$ then $\tiint\limits_{D}|\nabla u|^{2}dV=0$ .
For the Dirichlet boundary condition this implies that $u=0$ everywhere and
for the Neumann boundary condition that $u$ is a constant in D.

\end{document}
