\section{Question 3}

\subsection{Part A}

Linear 1st-order PDE is the equation
$$ \sum_{i=1}^n a_i(x) \pdv{u}{x_i}(x) = f(x) $$
which we consider for $x \in \Omega \subset \R^n$.
By its solution we mean a continuously differentiable function $u(x)$ such that the previous equation holds pointwise for all $x \in \Omega$.

Functions $u_1(x_1, \dots, x_n), u_2(x_1, \dots, x_N), u_n(x_1, \dots, x_N)$ are dependent on the set $\bar O$ ($O$ is open and bounded), if there exists a continuously differentiable function $F(u_1, \dots, u_n)$ such that
\begin{enumerate}
    \item $F$ is not identically zero on any region $G \subset \R^n$.
    \item for all $x = (x_1, \dots, x_N) \in \bar O$ we have that $F(u_1(x), \dots, u_n(x)) = 0$.
\end{enumerate}

Let $\psi_1, \dots, \psi_{N-1}$ are solutions of the 1st-order PDE. For every subregion $\Omega' \subset \Omega$ let there be a point $x_0 \in \Omega'$ such that the matrix
$$ \begin{pmatrix}\pdv{\psi_1}{x_1} & \cdots & \pdv{\psi_1}{x_N}\\
\vdots & \ddots & \vdots \\
\pdv{\psi_{N-1}}{x_1} & \cdots & \pdv{\psi_{N-1}}{x_N} \end{pmatrix}$$
has rank $N - 1$. Then $\psi_1, \psi_{N-1}$ make the fundamental system of the linear PDE.

\begin{theorem}[Maximum Number of Independent Solutions]
	Let there exist a point $x \in \Omega'$ for every subregion $\Omega' \subset \Omega$ such that $\sum |a_i(x)| > 0$. Let $\psi_1, \dots, \psi_{N-1}$ be the fundamental system of said PDE. Then $\theta \in C^1(\Omega)$ solves the PDE iff $\psi_1, \dots, \psi_{N - 1}, \theta$ are dependent in $\Omega$.
\end{theorem}

\begin{proof}
	$\implies$ was already proven before
	$\impliedby$: We assume that $J_{\vec\psi, \theta} = 0$. Therefore the system
	$y_1\pdv{\psi_1}{x_i} + \dots + y_{N-1}\pdv{\psi_{N-1}}{x_i} + y_N\pdv{\theta}{x_i} = 0$ for $i=1, \dots, N$ has a non-trivial solution for every $x \in \Omega$. Multiplying this equation by $a_i(x)$ and summing all of the together gives us
	$$ \sum_j y_j(x)\sum_i a_i(x)\pdv{\psi_j}{x_i} + y_N(x) \sum_i a_i(x) \pdv{\theta}{x_i} = 0. $$
	Since $\psi_i$ solve the equation, we get that
	$$ y_N(x) \sum_i a_i(x) \pdv{\theta}{x_i} = 0. $$
	Assume that $\theta$ does not solve are equation. Therefore we have an $x_0$ such that $\sum_{i=1}^N a_i(x_0)\pdv{\theta(x_0)}{x_i} \neq 0$.
	By the usual notion of continuity we get that there exists a neighborhood of $x_0$ where this formula is non-zero. Therefore $y_N = 0$ on some $U(x_0)$. Therefore $y_1\pdv{\psi_1}{x_i} + \dots + y_{N-1}\pdv{\psi_{N-1}}{x_i} = 0$. Since all the original functions are independent, there is a point $x_1$ where the system only has a trivial solution and we get that $J_{\vec\phi, \theta}(x_1) \neq 0$, which contradicts our assumption of independence.
\end{proof}

\subsection{Part B}

Consider the following system: $-\Delta u = f$ on some $\Omega$, $u = g$ on $\delta\Omega$. If $\Omega$ is bounded, we talk about the inner Dirichlet problem, if $\R^n \setminus \bar \Omega$ is bounded, we talk about the outer Dirichlet problem (in this case we only care about functions $u(x) = O\left(\frac{1}{|x|^{N - 2}}\right)$ for $|x| \to \infty$.

For the uniqueness of the solution for the inner problem, we only need to prove that the zero problem for $f = g = 0$ only has the trivial solution $u = 0$, which in fact follows from the weak maximum principle ($0 = \min_{\delta\Omega} u = \min_\Omega u \leq \max_\Omega u = \max_{\delta\Omega} u = 0$).

Now we shall prove the uniqueness for the outer problem. Assume that $u_1$ and $u_2$ are two solutions. Then $u_1 - u_2 =: w$ solves
$\Delta w = 0$ on $\Omega$, $w = 0$ on the boundary and, $w(x) = O\left(\frac{1}{|x|^{N-2}}\right)$.

First assume that $N \geq 3$, therefore $w(x) \to 0$ as $|x|$ goes to infinity. Let $x_0 \in \Omega$ and $\varepsilon > 0$ be given. Then there exists $R > 0$ such that $x_0, G \subset B_R$, where $G := \R^N \setminus \bar\Omega$ and $|w| \leq \varepsilon$ on $\delta B_R$. Now apply the maximum principle for the set $\Omega_R := \Omega \cup B_R$. We have that $w \leq \varepsilon$ on $\Omega_R$. The same can be said for $-w$, which means that we are done.

Let now $N = 2$. WLOG let the origin be in $G := \R^N \setminus \bar\Omega$. Fix an open ball $B_R \subset G$. Define a mapping $F: \R^N \setminus \{0\} \to \R^N \setminus \{0\}$ defined as $F(x) = x' := \frac{xR^2}{|x|^2}$. This means that $x$ and $F(x)$ lie on the same half-line originating in the origin. Then $F$ maps $\Omega$ to some region $\Omega^* \subset B_R$, in fact, $F(\Omega) = \Omega^* \setminus \{0\}$. Define $w'(x') = w(F^{-1}(x')) = w(x)$. Obviously $w' = 0$ on $\delta\Omega^*$. We will show that $w'$ is harmonic on $\Omega^* \setminus \{0\}$. Set $r = |x|$ ad $\rho = |x'|$. Then set
$\tilde w(r, \phi) = w(x_1, x_2)$ and $\tilde w'(\rho, \phi') = w'(x_1', x_2')$ and moving to the polar coordinates we get the desired equality. Now the function is obviously bounded, therefore we can extend it harmonically. By what was already proven for the inner problem, this means that $w' = 0$, which in turn implies that $w = 0$ on $\bar \Omega$.
