\section{Question 2}

\subsection{Part A}

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}

\begin{theorem}[Jacobi's criterion for function dependence]
	Let $\Omega \subset \R^N$ be open, $u_i \in C^1(\Omega)$ for $i = 1,\dots,N$. Then $u_1(x), \dots, u_n(x)$ are dependent in $\Omega$ if and only if
	$$ J_u(x) = \det
	\begin{pmatrix}
		\pdv{u_1(x)}{x_1} & \cdots & \pdv{u_1(x)}{x_N} \\
		\vdots & \ddots & \vdots \\
		\pdv{u_N(x)}{x_1} & \cdots & \pdv{u_N(x)}{x_N}
	\end{pmatrix} = 0.
	$$
\end{theorem}

\begin{theorem}[Dependence of $N$ solutions]
	Let $\psi_1, \dots, \psi_N$ be solutions of a non-trivial linear homogeneous 1st-order PDE in $\Omega \subset \R^N$. Then they are dependent in $\Omega$.
\end{theorem}

\begin{proof}
	Assume that $\psi_1, \dots, \psi_N$ are not dependent in $\Omega$. Then there exists a point $x_0 \in \Omega$ such that $J_\psi(x_0) \neq 0$. From continuity of solutions we have that the determinant in question is non-zero on some neighborhood $U(x_0)$. This implies that the system 
	$ \begin{pmatrix}
		\pdv{u_1(x)}{x_1} & \cdots & \pdv{u_1(x)}{x_N} \\
		\vdots & \ddots & \vdots \\
		\pdv{u_N(x)}{x_1} & \cdots & \pdv{u_N(x)}{x_N}
	\end{pmatrix} y = 0 $
	only has a trivial solution, which is in contradiction with the assumption that our original PDE was non-trivial.
\end{proof}

\subsection{Part B}

Consider the following problem: $\pdv[order={2}]{v(t, x; s)}{t} - c^2\pdv[order={2}]{v(t, x; s)}{x} = 0$, $v(s, x; s) = 0$ and $\pdv{f(t, x; s)}{t}_{t=s} = f(t, x)$.

Then $u(t, x) := \int_0^t v(t, x; s)ds$ solves general wave equation with RHS $f$.

For $N = 1$ we have $v(t, x; s) = \frac{1}{2c}\int_{B_{c(t-s)}(x)} f(s, y)dy$ and thus 
$$ u(t, x) = \frac{1}{2c} \int_0^t \int_{x-c(t-s)}^{x+c(t-s)}f(s, y) dy ds = \frac{1}{2c} \int_0^t \int_{x-c\tau}^{x+c\tau} f(t - \tau, y)dy d\tau. $$

For $N = 2$ we have $v(t, x; s) = \frac{1}{2\pi c} \int_{B_{c(t-s)}(x)} \frac{f(s, y)}{\sqrt{c^2(t-s)^2}-|y-x|^2} dy$.
Thus $$u(t, x) = \frac{1}{2\pi c} \int_{B_{c(t-s)}(x)} \frac{f(s, y)}{\sqrt{c^2(t-s)^2}-|y-x|^2} dy ds.$$