\section{Question 4}

\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}

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}
	Let $m < N - 1$, $\psi_i$ are $C^1$ for $i = 1,\dots,m$ and solve the ODE such that
	$$ \frac{D(\psi_1, \dots, \psi_m)}{D(x_1, \dots, x_m)}(x_1, \dots, x_m) \neq 0. $$

	Let $\tilde x_i = \psi_i(x), i \leq m$ and $\tilde x_i = x_i$ otherwise. Let $\tilde \Omega = \tilde x(\Omega)$ and\\$\tilde a_i(\tilde x_1, \dots, \tilde x_N) = a_i(x_1(\tilde x), \dots, x_n(\tilde x))$.

	Now let $\tilde\psi_{m+1},\dots,\tilde\psi_{N-1}$ be solutions of the equation
	$$ \sum_{i=m+1}^N \tilde a_i(\tilde x) \pdv{\tilde\psi}{\tilde x_i} = 0. $$

	Then functions $\psi_1, \dots, \psi_{N-1}$ ($\psi_k(x) = \tilde\psi_k(\tilde x(x))$) make up the fundamental system of the PDE.
\end{theorem}

\begin{proof}
	First let's show that the functions $\psi_k$ solve the equation.
	$$ \pdv{\psi_k}{x_i} =
	\sum_j^m \pdv{\tilde\psi_k}{\tilde x_j}\pdv{\tilde x_j}{x_i} +
	\sum_j^N \pdv{\tilde\psi_k}{\tilde x_j}\pdv{\tilde x_j}{x_i} $$
	Multiplying this equation by $a_i(x)$ and summing them up across all $i$s, we get
	$$ \sum_i a_i(x) \pdv{\psi_k}{x_i} = \sum_j^m \pdv{\tilde\psi_k(\tilde x(x))}{\tilde x_j} \sum_i \pdv{\psi_j(x)}{x_i}a_i(x) + \sum_j^N \tilde a_j(\tilde x) \pdv{\tilde\psi}{\tilde x_j} = 0. $$

	It remains to prove that the set is indeed the fundamental system. This however comes from rewriting the determinant in question using the chain rule as two determinants of the non-trivial solutions given in the statement.
\end{proof}

\subsection{Part B}

We are given the equation $\pdv[order={2}]{u}{t} - c^2\Delta u = f$ (on $(0, T]\times\Omega$), $u(0, x) = g(x)$, $\pdv{u(0, x)}{t} = h(x)$ and
\begin{itemize}
	\item Dirichlet: $u(t, x) = w(t, x)$ on the boundary $(0, T]\times\delta\Omega$);
	\item Neumann: $\pdv{u}{\vec n} = (\nabla u \cdot \vec n)(t, x) = w(t, x)$ on the same boundary.
\end{itemize}

For $\Omega$ ``behaved" enough ($C^{0,1}$ or $C^1$ boundary for D. and N. respectively), the solution $u \in C^2$ is determined uniquely (if exists).

\begin{proof}
	Let $u_1$ and $u_2$ be two such solutions. Then $u := u_1 = u_2$ solves the following problem:
	$$ \pdv[order={2}]{u}{t} - c^2\Delta u = 0, u(0, x) = \pdv{u}{t}(0, x) = 0$$
	and $u = 0$ ($\pdv{u}{\vec{n}} = 0$) on $(0, T]\times\delta\Omega$.

	Now fix $T_0 \in (0, T)$. For every $t \in (0, T_0)$ multiply the PDE by $\pdv{u}{t}$ and integrate:
	$$ 0 = \int_\Omega (\pdv[order={2}]{u}{t} - c^2\Delta u) \pdv{u}{t} dx = \int_\Omega \pdv[order={2}]{u}{t} \pdv{u}{t} dx - c^2 \int_\Omega \Delta u \pdv{u}{t}dx. $$
	Using differentiation under the integral sign and Green's identities, we get
	$$ \frac{1}{2} \odv*{\int_\Omega \pdv{u}{t}^2 dx}{t} + \frac{1}{2} c^2 \odv*{\int_\Omega |\nabla u|^2 dx}{t} = 0. $$
	Integrating over $(0, T_0)$, we get
	$$ \frac{1}{2} \int_\Omega \pdv{u(T_0, x)}{t}^2 dx + \frac{1}{2} c^2 \int_\Omega |\nabla u(T_0, x)|^2 dx = 0. $$
	From this (both integrals must be zero) we get that $\pdv{u}{t} = 0$ and also $u = 0$ on the entire region $\Omega$.
\end{proof}
