Finding when Cauchy data is characteristic I have been stuck on the following problem:

Consider the Cauchy problem 
  \begin{equation} \frac{\partial^2 u}{\partial x_1 \partial x_2} - 4\frac{\partial^2 u}{\partial x_3^2} + \frac{\partial u}{\partial x_1} = 0
\end{equation}
  in the half-space $x_1 \cos(\theta) + x_2 \sin(\theta) \geq 0,$ i.e. the problem of finding the equation in the half space with prescribed values and normal derivative on the plane $x_1 \cos(\theta) + x_2 \sin(\theta) = 0.$
  For what values $\theta$ is this problem characteristic? For what values is this problem strictly hyperbolic with respect to the normal plane?

It seems like for this problem, the Cauchy–Kowalevski theorem should be applied.  Since this theorem is usually applied on flat surfaces (like $x \in \mathbb{R}^n$ with $x_n = 0$), I am not sure how I'd explicitly calculate the conditions on $\theta.$
I've tried reading about how an "analytic surface $S$" is the zero-level set of an analytic function $\phi$, and if we have a change of coordinates (somehow involving $\phi : \mathbb{R^3} \to \mathbb{R^3}$) from $(x_1,x_2,x_3) \to (y_1,y_2,y_3)$ such that $S = \{y_3 = 0\},$ then all we need is for the coefficient attached to $\partial^2_{y_3} u$ to be nonzero for a local solution to exist assuming everything else is analytic which is likely based off of the PDE.
I feel like we should have $y_3 := x_1 \cos(\theta) + x_2 \sin(\theta),$ so then the plane on which the initial data is defined is $\{y_3 = 0\},$ but I am not sure what to do about the other coordinates (do I even need to do anything about them?).
If anyone sees how to do this using Cauchy–Kowalevski (CK), or even has a good reference for CK, anything is appriciated! Thank you!
 A: Consider the boundary value problem
\begin{equation}
\left\{
\begin{array}
[c]{ll}%
\boldsymbol{F}\left(  \boldsymbol{x},\boldsymbol{u}\left(  \boldsymbol{x}%
\right)  ,\nabla\boldsymbol{u}\left(  \boldsymbol{x}\right)  \right)
=\boldsymbol{0} & \text{in }\Omega,\\
\boldsymbol{u}=\boldsymbol{g} & \text{on }\Gamma,
\end{array}
\right.  \label{c-k1}%
\end{equation}
where $\Omega\subseteq\mathbb{R}^{N}$ is an open set of class $C^{1}$,
$\Gamma\subseteq\partial\Omega$, $\boldsymbol{u}:\Omega\rightarrow
\mathbb{R}^{d}$, $\boldsymbol{F}:\Omega\times\mathbb{R}^{d}\times
\mathbb{R}^{d\times N}\rightarrow\mathbb{R}^{d}$ is $C^{1}$ and
$\boldsymbol{g}:\mathbb{R}^{N}\rightarrow\mathbb{R}^{d}$ is of class $C^{1}$.
Here $\nabla\boldsymbol{u}\left(  \boldsymbol{x}\right)  $ is the $d\times N$
matrix given by
$$
\nabla\boldsymbol{u}\left(  \boldsymbol{x}\right)  =\left(
\begin{array}
[c]{c}%
\nabla u_{1}\left(  \boldsymbol{x}\right)  \\
\vdots\\
\nabla u_{d}\left(  \boldsymbol{x}\right)
\end{array}
\right)  .
$$
Given a $d\times N$ matrix $\boldsymbol{P}$, we write
$$
\boldsymbol{P}=\left(
\begin{array}
[c]{ccc}%
p_{1,1} & \cdots & p_{1,N}\\
\vdots & \vdots & \vdots\\
p_{d,1} & \cdots & p_{d,N}%
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\boldsymbol{p}^{\left(  1\right)  }\\
\vdots\\
\boldsymbol{p}^{\left(  d\right)  }%
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
\boldsymbol{p}_{\left(  1\right)  } & \cdots & \boldsymbol{p}_{\left(
N\right)  }%
\end{array}
\right)  .
$$
A triple $\left(  \boldsymbol{x},\boldsymbol{z},\boldsymbol{P}\right)
\in\Gamma\times\mathbb{R}^{d}\times\mathbb{R}^{d\times N}$ is called
admissible if
$$
\boldsymbol{z}=\boldsymbol{g}\left(  \boldsymbol{x}\right)  ,\quad
\boldsymbol{F}\left(  \boldsymbol{x},\boldsymbol{z},\boldsymbol{P}\right)
=\boldsymbol{0},\quad\boldsymbol{p}^{\left(  k\right)  }\cdot\boldsymbol{t}%
=\nabla g_{k}\left(  \boldsymbol{x}\right)  \cdot\boldsymbol{t}%
$$
for all $\boldsymbol{t}\in T_{\boldsymbol{x}}\partial\Omega$ and all
$i=k,\ldots,d$, while it is non-characteristic if
\begin{equation}
\det\left(  \left(  \nabla_{\boldsymbol{p}^{\left(  k\right)  }}F_{j}\left(
\boldsymbol{x},\boldsymbol{z},\boldsymbol{P}\right)  \cdot\boldsymbol{n}%
\right)  _{k,j}\right)  \neq0,\label{non characteristic systems}%
\end{equation}
where $\boldsymbol{n}$ is a normal unit vector to $\partial\Omega$ at
$\boldsymbol{x}$.
In the case of a quasilinear first order partial differential system
\begin{equation}
\left\{
\begin{array}
[c]{ll}%
\sum_{i=1}^{N}\boldsymbol{B}_{i}\left(  \boldsymbol{x},\boldsymbol{u}\left(
\boldsymbol{x}\right)  \right)  \frac{\partial\boldsymbol{u}}{\partial x_{i}%
}\left(  \boldsymbol{x}\right)  +\boldsymbol{C}\left(  \boldsymbol{x}%
,\boldsymbol{u}\left(  \boldsymbol{x}\right)  \right)  =\boldsymbol{0} &
\text{in }\Omega,\\
\boldsymbol{u}=\boldsymbol{g} & \text{on }\Gamma,
\end{array}
\right.  \label{SBVP}%
\end{equation}
where $\Omega\subseteq\mathbb{R}^{N}$ is an open set, $\Gamma\subseteq
\partial\Omega$, $\boldsymbol{u}:\Omega\rightarrow\mathbb{R}^{d}$,
$\boldsymbol{B}_{i}:\Omega\times\mathbb{R}^{d}\rightarrow\mathbb{R}^{d\times
d}$, $\boldsymbol{C}:\Omega\times\mathbb{R}^{d}\rightarrow\mathbb{R}^{d}$, we
have that
$$
\boldsymbol{F}\left(  \boldsymbol{x},\boldsymbol{z},\boldsymbol{P}\right)
=\sum_{i=1}^{N}\boldsymbol{B}_{i}\left(  \boldsymbol{x},\boldsymbol{z}\right)
\boldsymbol{p}_{\left(  i\right)  }+\boldsymbol{C}\left(  \boldsymbol{x}%
,\boldsymbol{z}\right)  ,
$$
and so the condition for non-characteristic reduces to
\begin{equation}
\det\left(  \sum_{i=1}^{N}\boldsymbol{B}_{i}\left(  \boldsymbol{x}%
,\boldsymbol{g}\left(  \boldsymbol{x}\right)  \right)  n_{i}\left(
\boldsymbol{x}\right)  \right)  \neq
0.\label{non characteristic quasilinear system}%
\end{equation}
Theorem [Cauchy-Kowalevski] Consider the boundary value problem above, where $\Omega\subseteq\mathbb{R}^{N}$ is an open set of
class $C^{\infty}$, $\Gamma\subseteq\partial\Omega$, $\boldsymbol{F}%
:\mathbb{R}^{N}\times\mathbb{R}^{d}\times\mathbb{R}^{d\times N}\rightarrow
\mathbb{R}^{d}$ is $C^{\infty}$ and $\boldsymbol{g}:\mathbb{R}^{N}%
\rightarrow\mathbb{R}^{d}$ is of class $C^{\infty}$. Let $\boldsymbol{x}%
_{0}\in\Gamma$ and assume that the triple $\left(  \boldsymbol{x}%
_{0},\boldsymbol{g}\left(  \boldsymbol{x}_{0}\right)  ,\boldsymbol{P}%
_{0}\right)  $ is non-characteristic, with $\partial\Omega$ and
$\boldsymbol{g}$ analytic at $\boldsymbol{x}_{0}$, and $\boldsymbol{F}$ is
analytic at $\left(  \boldsymbol{x}_{0},\boldsymbol{g}\left(  \boldsymbol{x}%
_{0}\right)  ,\boldsymbol{P}_{0}\right)  $. Then there exists a neighborhood
$V$ of $\boldsymbol{x}_{0}$ and a function $\boldsymbol{u}\in C^{\infty
}\left(  V;\mathbb{R}^{d}\right)  $, analytic at $\boldsymbol{x}_{0}$, such
that
\begin{equation}
\left\{
\begin{array}
[c]{ll}%
\boldsymbol{F}\left(  \boldsymbol{x},\boldsymbol{u}\left(  \boldsymbol{x}%
\right)  ,\nabla\boldsymbol{u}\left(  \boldsymbol{x}\right)  \right)
=\boldsymbol{0} & \text{in }\Omega\cap V,\\
\boldsymbol{u}=\boldsymbol{g} & \text{on }\Gamma\cap V.
\end{array}
\right.  \label{local system}%
\end{equation}
Moreover, $\boldsymbol{u}$ is the unique solution of the system which is analytic at $\boldsymbol{x}_{0}$.
In your case
$$
\Omega=\{\boldsymbol{x}\in\mathbb{R}^{3}:\,\boldsymbol{x}\cdot(\cos\theta
,\sin\theta,0)\geq0\}
$$
and taking $\boldsymbol{u}=(u,v_{1},v_{2},v_{3})$, where $v_{1}:=\frac
{\partial u}{\partial x_{1}}$, $v_{2}:=\frac{\partial u}{\partial x_{2}}$,
$v_{3}:=\frac{\partial u}{\partial x_{3}}$ 
and \begin{equation} \frac{\partial v_2}{\partial x_1} - 4\frac{\partial v_3}{\partial x_3} + \frac{\partial u}{\partial x_1} = 0
\end{equation}
you have%
\begin{gather*}
\left(
\begin{array}
[c]{cccc}%
1 & 0 & 1 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\right)  \frac{\partial\boldsymbol{u}}{\partial x_{1}}+\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\right)  \frac{\partial\boldsymbol{u}}{\partial x_{2}}\\
+\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & -4\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}
\right)  \frac{\partial\boldsymbol{u}}{\partial x_{3}}+\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{array}
\right)  \boldsymbol{u}=\boldsymbol{0}%
\end{gather*}
and the normal $\boldsymbol{n}=(\cos\theta,\sin\theta,0)$. Can you continue from here?
