\section{Question 1}

\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 $x \in \Omega$.

Characteristic system for this equation is a system of ODEs $\odv{x_i}{t}(t) = a_i(x(t))$ for $i = 1, \dots, N$ and $x \in \Omega$.

\begin{theorem}[characterization of solutions of 1st order linear PDE]
	Function $\psi(x_1, \dots, x_N)$ is a solution for a linear homogeneous PDE if and only if it is constant along each of the solutions of the characteristic system. 
\end{theorem}

\begin{proof}
	$\Rightarrow$: Let $\psi$ be such that $\sum_{i=1}^N a_i(x) \odv{\psi(x)}{x_i} = 0$.
	Let $\{\varphi_1(t), \dots, \varphi_N(t)\}$ be one of its characteristics. Then
	$$ \odv*{\psi(\varphi_1(t),\dots,\varphi_N(t))}{t} = \sum_{i=1}^N \pdv{\psi}{x_i}(\varphi_1(t), \dots, \varphi_N(t)) \odv{\varphi_i(t)}{t} = $$
	$$ \sum_{i=1}^N \pdv{\psi}{x_i}(\varphi_1(t), \dots, \varphi_N(t)) a_i(\varphi_1(t), \dots, \varphi_N(t)) = 0. $$

	$\Leftarrow$: Let $\psi$ be constant along every characteristic. Choose $\xi \in \Omega$. Since $\{a_i\}_{i=1}^N$ are continuous, at least one characteristic goes through the point $\xi$ which corresponds to some interval $(t_1, t_2)$. Fix such characteristic $(\varphi_1(t), \dots, \varphi_N(t))$ and a point $\tau$ such that $(\varphi_1(\tau), \dots, \varphi_N(\tau)) = (\xi_1, \dots, \xi_N)$.

	Since $\psi(\varphi_1(t), \dots, \varphi_N(t))$ is constant on $(t_1, t_2)$, we have that
	$$ 0 = \odv{\psi}{t}(\varphi_1(t), \dots, \varphi_N(t)) = \sum_{i=1}^N \pdv{\psi}{x_i}(\varphi_1(t),\dots, \varphi_N(t))a_i(\varphi_1(t), \dots, \varphi_N(t)). $$
	Now it is sufficient to choose $t = \tau$ and we get that $\psi$ is a solution at point $\xi$. Since this point was chosen arbitrarily, we conclude that it is a solution on the entire $\Omega$.
\end{proof}

\subsection{Part B}

Fundamental solution of the Poisson equation is the function 
$$ \mathcal{E}(x) := \begin{cases} - \frac{1}{2\pi} \log |x|, N = 2;\\\frac{1}{(N - 2}\kappa_N|x|^{N - 2}, N \geq 3,\end{cases} $$
where $\kappa_N$ is the measure of the $n$-dimensional unit ball.

\begin{theorem}[Three Potentials Theorem]
	Let $\omega \subset \R^N$ be a bounded region with at least Lipschitz boundary, $u \in C^2(\bar \Omega)$ and $\mathcal{E}$ is the fundamental solution of the Poisson equation. Then for all $x \in \Omega$,
	$$ u(x) = -\int_\omega \mathcal{E}(x - y) \Delta u(y) dy + \int_{\delta\Omega} (\mathcal{E}(x - y) \pdv{u}{n}(y) - u(y) \pdv{\mathcal{E}}{n_y}(x - y)) dS(y), $$
	where
	$$ \pdv{u}{n}(y) := \nabla u(y) n(y), \pdv{\mathcal{E}}{n_y}(x - y) := \nabla_y \mathcal{E}(x - y) n(y) $$
	and $n$ is the outer normal to $\delta\Omega$.
\end{theorem}

\begin{proof}
	Let $x \in \Omega$ and let $B_\varepsilon(x) \subset \Omega$. Let $V_\varepsilon := \Omega \setminus \overline{B_\varepsilon(x)}$. Using Green's theorem, we obtain
	$$ \int_{V_\varepsilon} \mathcal{E}(x - y)\Delta u(y) dy = \int_{V_\varepsilon} \delta_y \mathcal{E}(x - y) u(y) dy + \int_{\delta V_\varepsilon} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y) - $$
	$$ \int_{\delta V_\varepsilon} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y). $$
	Denote the individual integrals $I_1 = I_2 + I_3 - I_4$ and send $\varepsilon$ to zero.

	First we can write (the functions in question are ``nice" enough)
	$$I_1 \to \int_{\Omega} \mathcal{E}(x - y)\Delta u(y) dy$$

	$I_2$ is obviously zero (fundamental solution).

	$I_3$ we can rewrite in the following manner:
	$$ I_3 = \int_{\delta \Omega} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y) - \int_{\delta B_\varepsilon(x)} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y) \to $$
	$$ \int_{\delta \Omega} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y). $$
	
	For the last integral, we can write
	$$ I_4 = \int_{\delta \Omega} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y) - \int_{\delta B_\varepsilon(x)} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y) \to $$
	$$ \int_{\delta \Omega} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y) + u(x). $$

	From this follows the equality in question.
\end{proof}