Elliptic.tex

\section{Elliptic PDEs}
\subsection{Maximum Principle for Elliptic Operators}
Theorem: If $L$ is an elliptic differential operator on a connected and bounded domain $\Omega$,
and $u$ is a solution $Lu = 0$, then $u$ takes its maximum and minimum on the boundary of $\Omega$.

\subsection{Uniqueness of Solutions}
Theorem: If $L$ is an elliptic differential operator on a connected and bounded domain $\Omega$,
then there is at most one solution of $Lu = f$ with boundary conditions $u_{|\partial\Omega} = g$.

\symbolicsubsection{Green's Function}

\begin{addmargin}[1em]{0em}
    {\color{teal}
        Heaviside function: $\vartheta(x - \xi) = \left\{ \begin{matrix}
            1\quad \xi\leq x \\
            0\quad \xi > x
        \end{matrix} \right.$
        translates the integral $\int_0^xf(\xi)d\xi$ to $\int_0^1\vartheta(x-\xi)\cdot f(\xi)d\xi$ for $\xi$
        in $[0,1]$.
    }
\end{addmargin}

Given the Laplace equation $\nabla^2u=f$, we want to find an ``inverse'' Laplace such that
\begin{align*}
    u(x) = \int G(x,\xi)f(\xi)\ d\xi
\end{align*}

There is a function $G(x,\xi)$ on $\bar{\Omega}\times\bar{\Omega}$ called Green's function such that
\begin{align*}
    \Delta G(x,\xi) = \delta(x-\xi)\quad\text{ in }\Omega\text{ and }G(x,\xi) = 0\text{ for }\xi\in\partial\Omega
\end{align*}

the general Poisson problem $\Delta u = f$ in $\Omega$ with boundary conditions $u = g$ on $\partial\Omega$ has the solution
\begin{align*}
    u(x) = \int_\Omega G(x,\xi)f(\xi)\ d\xi + \int_{\partial\Omega} g(\xi)\cdot\mathrm{grad}_\xi G(x,\xi)\ dn
\end{align*}
with $n$ is an outside pointing normal.
To construct a particular solution $u_p$ of $\Delta u=f$ on $\Omega \subset \mathbb{R}^n$, there are the following Green's
functions (using Dirac-$\delta$ function):
\begin{align*}
    G(x,\xi) = \left\{
    \begin{matrix}
        \frac{1}{2}|x-\xi| & \text{for }n = 1 \\
        \frac{1}{2\pi}\log|x-\xi| & \text{for }n = 2 \\
        \frac{1}{4\pi}\frac{1}{|x-\xi|} & \text{for }n = 3
    \end{matrix}
    \right\}\Rightarrow\Delta G(x,\xi) = \delta(x-\xi)
\end{align*}

then $u_p(x) = \int_\Omega G(x,\xi)f(\xi)\ d\xi$.