\section{Proof that B\"acklund implies Darboux}\label{wprovedarboux}

Now suppose that we have a \BT/ as in \S \ref{wsetup}, and that the Monge-Amp\`ere system $(\scM^5, \I)$ is contact equivalent to the standard wave equation $Z_{XY}=0$.  We can choose local coordinates $(X,Y,Z,P,Q)$ on $\scM$ such that $\I$ is generated by the forms
\[ \theta = dZ - P\, dX - Q\, dY, \ \ \ \ \ 
\Omega_1 = dX \& dP, \ \ \ \ \ 
\Omega_2 = dY \& dQ. \]
There is a unique local section $\sigma = (\theta, \thetabar, \om^1, \om^2, \om^3, \om^4):\B \to \calP$ satisfying
\begin{align}
\theta & = dZ - P\, dX - Q\, dY, \notag \\
\om^1 & = dX + C_1 \theta, \notag \\
\om^2 & = dP + C_2 \theta, \label{wavecoframe} \\
\om^3 & = dY + C_3 \theta, \notag \\
\om^4 & = dQ + C_4 \theta \notag
\end{align}
for some functions $C_i$ on $\B$.  (Note that, because specifying this portion of the coframing determines a unique local section of $\calP$, the 1-form $\thetabar$  is also uniquely determined.)  The functions $C_i$ are the pullbacks under $\sigma$ of the corresponding torsion functions, and this coframing has $A_1 \equiv 1$.  When the structure equations \eqref{G0struct} are pulled back to $\B$ via $\sigma$, the 1-forms $\alp_i, \bet_i$---which were linearly independent from the 1-forms $\theta, \thetabar, \om^i$ on $\calP$---must pull back to some linear combinations of these 1-forms.

Now we embark on the process of comparing the structure equations \eqref{G0struct} to those for the explicit coframing above.  First, note that
\[  0 = d(dX) = d(\om^1 - C_1 \theta) \equiv -(\alp_1 \& \om^1 + \alp_2 \& \om^2 + C_1 \om^1 \& \om^2)  \mod{\theta}. \]
Therefore, by choosing $r_2, r_3$ appropriately in \eqref{G0connfreedom}, we may assume that
\[ \alp_1 = a_{10}\theta + a_{11}\om^1 + \tfrac{1}{2} C_1 \om^2, \qquad
\alp_2 = a_{20}\theta - \tfrac{1}{2} C_1 \om^1 \]
for some functions $a_{10}, a_{11}, a_{20}$ on $\B$.
Similar considerations of $d(dP), d(dY), d(dQ)$ modulo $\theta$ yield similar expressions for the remaining connection forms:
\begin{alignat*}{2}
\alp_1 &= a_{10}\theta + a_{11}\om^1 + \tfrac{1}{2} C_1 \om^2,  \qquad  \qquad & \bet_1 &= b_{10}\theta + b_{13}\om^3 + \tfrac{1}{2} C_3 \om^4, \\
\alp_2 &= a_{20}\theta - \tfrac{1}{2} C_1 \om^1, \qquad \qquad & \bet_2 &= b_{20}\theta - \tfrac{1}{2} C_3 \om^3,\\
\alp_3 &= a_{30}\theta + \tfrac{1}{2} C_2 \om^2, \qquad \qquad & \bet_3 &= b_{30}\theta + \tfrac{1}{2} C_4 \om^4,\\
\alp_4 &= a_{40}\theta - \tfrac{1}{2} C_2 \om^1 + a_{42}\om^2,  \qquad \qquad & \bet_4 &= b_{40}\theta - \tfrac{1}{2} C_4 \om^3 + b_{44}\om^4 .
\end{alignat*}

Next, a straightforward computation shows that for the coframing \eqref{wavecoframe},
\[ d\theta = \theta \& (C_2 \om^1 - C_1 \om^2 + C_4 \om^3 - C_3 \om^4)
+ \om^1 \& \om^2 + \om^3 \& \om^4. \]
Comparing this with the first structure equation in \eqref{G0struct} yields
\[ b_{13} = \tfrac{3}{2} C_4 , \qquad
b_{44} = -\tfrac{3}{2} C_3.  \]
In order to continue this comparison, we need to introduce derivatives of the functions $A_2, C_i$.  So, set
\begin{align}
dA_2 & = A_{2,0}\theta + A_{2,\overline{0}}\thetabar + A_{2,1}\om^1 + A_{2,2}\om^2 + A_{2,3}\om^3 + A_{2,4}\om^4, \notag \\
dC_1 & = C_{1,0}\theta + C_{1,\overline{0}}\thetabar + C_{1,1}\om^1 + C_{1,2}\om^2 + C_{1,3}\om^3 + C_{1,4}\om^4, \notag \\
dC_2 & = C_{2,0}\theta + C_{2,\overline{0}}\thetabar + C_{2,1}\om^1 + C_{2,2}\om^2 + C_{2,3}\om^3 + C_{2,4}\om^4, \label{first-derivs} \\
dC_3 & = C_{3,0}\theta + C_{3,\overline{0}}\thetabar + C_{3,1}\om^1 + C_{3,2}\om^2 + C_{3,3}\om^3 + C_{3,4}\om^4, \notag \\
dC_4 & = C_{4,0}\theta + C_{4,\overline{0}}\thetabar + C_{4,1}\om^1 + C_{4,2}\om^2 + C_{4,3}\om^3 + C_{4,4}\om^4 . \notag
\end{align}
Comparing the structure equations \eqref{G0struct} for $d\om^i$ with the derivatives of the explicit forms $\om^i$ in \eqref{wavecoframe} yields
\begin{alignat}{2}
a_{10} & = C_{1,1} - C_1 C_2, \qquad \qquad & b_{10} & = C_{3,3} - C_3 C_4 \notag \\
a_{20} & = C_{1,2} + C_1^2, \qquad \qquad & b_{20} & = C_{3,4} + C_3^2, \notag \\
a_{30} & = C_{2,1} - C_2^2, \qquad \qquad & b_{30} & = C_{4,3} - C_4^2, \notag \\
a_{40} & = C_{2,2} + C_1 C_2, \qquad \qquad & b_{40} & = C_{4,4} + C_3 C_4,\notag
\end{alignat}
in addition to the following relations among the torsion and its derivatives:
\begin{alignat}{3}
B_1 & = -C_{1,\overline{0}}, \qquad \qquad & C_{1,3} & = C_1 C_4, \qquad \qquad & C_{3,1} & = C_2 C_3, \notag  \\
B_2 & = -C_{2,\overline{0}}, \qquad \qquad & C_{1,4} & = -C_1 C_3, \qquad \qquad & C_{3,2} & = -C_1 C_3, \label{B-relations} \\
B_3 & = -C_{3,\overline{0}}, \qquad \qquad & C_{2,3} & = C_2 C_4, \qquad \qquad & C_{4,1} & = C_2 C_4,  \notag \\
B_4 & = -C_{4,\overline{0}}, \qquad \qquad & C_{2,4} & = -C_2 C_3, \qquad \qquad & C_{4,2} & = -C_1 C_4. \notag
\end{alignat}

While we don't have an explicit coordinate representation for $\thetabar$, we can still explore the consequences of $d(d\thetabar) = 0$.  Computing $d(d\thetabar) \equiv 0$ modulo $\thetabar$ yields
\begin{gather*}
a_{11} = \frac{-2A_{2,1} + C_2 (A_2 + 2)}{2 A_2}, \qquad a_{42} = \frac{-2A_{2,2} - C_1 (A_2 + 2)}{2 A_2}, \\[0.2in]
A_{2,0} = A_2 (C_{3,3} + C_{4,4} - C_{1,1} - C_{2,2}).
\end{gather*}
Then computing $d(d\thetabar) \equiv 0$ modulo $\theta, \om^1, \om^2$ yields
\[ A_{2,\overline{0}} = C_{1,2} + C_{2,2} - A_2 (C_{3,3} + C_{4,4}) - \frac{C_1}{A_2} A_{2,1} - \frac{C_2}{A_2} A_{2,2} - C_3 A_{2,3} - C_4 A_{2,4}. \]
(Note that the fact that $\thetabar$ is a contact form implies that
$A_2$ cannot be zero.)




At this point, 
all coefficients in the structure equations \eqref{G0struct} have been expressed in terms of the functions $A_2, C_1, C_2, C_3, C_4$ and their first derivatives.  In addition, we have relations among the derivatives that amount to an overdetermined PDE system for these five functions on $\B$.  Necessary compatibility conditions for this system may be found by computing $d(dA_2) = d(dC_i) = 0$.  In particular, computing
\begin{align*}
d(dC_1) & \equiv d(dC_2) \equiv 0 \mod{\theta, \thetabar, \om^1, \om^2}, \\
d(dC_3) & \equiv d(dC_4) \equiv 0 \mod{\theta, \thetabar, \om^3, \om^4}
\end{align*}
yields
\begin{align}
C_{1,0} & = -A_2 C_{1,\overline{0}} - C_2 C_{1,2} + C_1 (-C_{1,1} + C_{3,3} + C_{4,4}), \notag \\
C_{2,0} & = -A_2 C_{2,\overline{0}} - C_1 C_{2,1} + C_2 (-C_{2,2} + C_{3,3} + C_{4,4}), \label{left-off-here} \\
C_{3,0} & = -C_{3,\overline{0}} - C_4 C_{3,4} + C_3 (C_{1,1} + C_{2,2} - C_{3,3}) , \notag \\
C_{4,0} & = -C_{4,\overline{0}} - C_3 C_{4,3} + C_4 (C_{1,1} + C_{2,2} - C_{4,4}). \notag
\end{align}

At this point, we have derived all the relations among the torsion functions on $\B$ and their derivatives that will be needed in order to prove that $(\Mbar, \Ibar)$ is Darboux-integrable after at most one prolongation.  The proof of Darboux-integrability is divided into two main cases.  In \S \ref{nondegenerate-case}, we prove Darboux-integrability under the assumption that the vectors $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ are both nonzero.  This case is further divided into three subcases, depending on the ranks of certain Frobenius systems that arise during the proof.  Precise statements of the results are contained in Propositions \ref{Case-1-prop}, \ref{Case-2-prop}, and \ref{Case-3-prop}.  In \S \ref{degenerate-case}, we prove Darboux-integrability under the assumption that exactly one of the vectors $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ vanishes; the precise result is contained in Proposition \ref{one-C-vector-zero-prop}.  As noted in \S \ref{wsetup}, it is not necessary to consider the case where $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ both vanish, since in that case both \MA/ systems are contact equivalent to the standard wave equation.

\subsection{Case 1: $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ are both nonzero}\label{nondegenerate-case}
Without loss of generality, we may assume that $C_2$ and $C_4$ are nonzero.  Consider the exterior derivatives of the ratios $\frac{C_1}{C_2}$ and $\frac{C_3}{C_4}$.  A straightforward computation shows that
\[
d\left(\frac{C_1}{C_2} \right) \equiv 0 \mod{\theta, \thetabar, \om^1, \om^2};
\]
therefore, $d\left(\frac{C_1}{C_2}\right)$ must lie in the last derived system  of $\calK_1 = \{\theta, \thetabar, \om^1, \om^2\}$---i.e., the largest integrable subsystem of $\calK_1$, denoted by $\calK_1^{(\infty)}$.  Similarly,
\[
d\left(\frac{C_3}{C_4} \right) \equiv 0 \mod{\theta, \thetabar, \om^3, \om^4},
\]
so $d\left(\frac{C_3}{C_4}\right)$ must lie in the last derived system of $\calK_2 = \{\theta, \thetabar, \om^3, \om^4\}$.

First, consider the system $\calK_1$.  In order to compute its first derived system $\calK_1^{(1)}$, we must find those 1-forms in $\calK_1$ whose exterior derivatives are zero modulo the linear span of the 1-forms in $\calK_1$.  To this end, we compute:
\[
\left. \begin{aligned}
d\theta & \equiv \om^3 \& \om^4 \\[0.1in]
d\thetabar & \equiv A_2\, \om^3 \& \om^4 \\[0.1in]
d\om^1 & \equiv C_1\, \om^3 \& \om^4 \\[0.1in]
d\om^2 & \equiv C_2\, \om^3 \& \om^4
\end{aligned} \right\} \mod{\calK_1}.
\]
Therefore,
%\[ 
$\calK_1^{(1)} = \{ \thetabar - A_2 \theta, \om^1 - C_1\theta, \om^2 - C_2\theta \}.$
% \]
Observe that
\[ \om^1 - C_1\theta = dX, \qquad  \om^2 - C_2\theta = dP. \]
Therefore, the rank 2 subsystem $\{\om^1 - C_1\theta, \om^2 - C_2\theta \} = \{dX, dP\}$ of $\calK_1^{(1)}$ is integrable, and the next derived system $\calK_1^{(2)}$ (i.e., the first derived system of $\calK_1^{(1)}$)  contains this rank 2 system.  The only question is whether, in fact, $\calK_1^{(2)} = \calK_1^{(1)}$---i.e., whether $\calK_1^{(1)}$ itself is integrable.  In either case, we will have $\calK_1^{(2)} = \calK_1^{(\infty)}$.  A computation shows that
\begin{equation}
 d(\thetabar - A_2 \theta) \equiv \theta \& [(A_{2,3} + A_2 C_4(A_2 - 1))\, \om^3 + (A_{2,4} - A_2 C_3(A_2 - 1))\, \om^4]  \mod \calK_1^{(1)}, \label{int-cond-1}
\end{equation}
so the rank of $\calK_1^{(\infty)}$ is either 3 or 2, depending on whether or not the 1-form in brackets vanishes.

Similarly, we can compute that
\[ \calK_2^{(1)} = \{ \thetabar - \theta, \om^3 - C_3\theta, \om^4 - C_4\theta \} = \{ \thetabar - \theta, dY, dQ \}. \]
So $\calK_2^{(\infty)}$ contains the rank 2 subsystem $\{\om^3 - C_3\theta, \om^4 - C_4\theta \} = \{dY, dQ\}$, and a computation shows that
\begin{equation}
 d(\thetabar - \theta) \equiv -\theta \& \left[\frac{(A_{2,1} + C_2(A_2 - 1))}{A_2}\, \om^1 + \frac{(A_{2,2} - C_1(A_2 - 1))}{A_2}\, \om^2 \right]  \mod \calK_2^{(1)}. \label{int-cond-2}
\end{equation}
So the rank of $\calK_2^{(\infty)}$ is either 3 or 2, 
depending on whether or not the 1-form in brackets vanishes.

Now we must divide into cases depending on the ranks of these derived systems.

\subsubsection{Case 1.1: $\calK_1^{(\infty)}$ and $\calK_2^{(\infty)}$ both have rank 3.}

In this case, we have the following result:

\begin{prop}\label{Case-1-prop}
If $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ are both nonzero and $\calK_1^{(\infty)}$ and $\calK_2^{(\infty)}$ both have rank 3, then the system $(\Mbar, \Ibar)$ is contact equivalent to the standard wave equation.
\end{prop}

\begin{proof}
By Theorem 2.1 of \cite{BGG}, it suffices to show that $(\Mbar, \Ibar)$ is Darboux-integrable, i.e., that each of the characteristic systems $\{\thetabar, \om^1, \om^2\}$ and $\{\thetabar, \om^3, \om^4\}$---which are well-defined on $\Mbar$ even though the 1-forms $\om^i$ are not---contains a rank 2 integrable subsystem.

The hypothesis that $\calK_1^{(\infty)}$ and $\calK_2^{(\infty)}$ both have rank 3 implies that the expressions \eqref{int-cond-1} and \eqref{int-cond-2} must both vanish identically; therefore,
%\begin{align*}
$$
A_{2,1}  = -C_2(A_2-1), 
\qquad
%\\
A_{2,2}  = C_1(A_2-1), 
\qquad
%\\
A_{2,3}  = -A_2 C_4(A_2-1), 
\qquad
%\\
A_{2,4}  = A_2 C_3(A_2-1) .
$$
%\end{align*}
Using these conditions, a straightforward computation shows that
\begin{align*}
\{\thetabar, \om^1, \om^2\}^{(1)} & = \{ A_2 \om^1 - C_1 \thetabar, \, A_2 \om^2 - C_2 \thetabar \}, \\
\{\thetabar, \om^3, \om^4\}^{(1)} & = \{ A_2 \om^3 - C_3 \thetabar, \, A_2 \om^4 - C_4 \thetabar \},
\end{align*}
and that each of these derived systems is integrable.
\end{proof}

\subsubsection{Case 1.2: Exactly one of $\calK_1^{(\infty)}$ and $\calK_2^{(\infty)}$ has rank 3.}\label{middle-case}

Without loss of generality, we may assume that $\calK_2^{(\infty)}$ has rank 2 and is equal to $\{dY, dQ\}$, and that $\calK_1^{(\infty)}$ has rank 3.  It follows that \eqref{int-cond-1} vanishes identically and that \eqref{int-cond-2} does not.  Since all our results are local, we will assume that we are working on an open set where \eqref{int-cond-2} is nonzero.  The vanishing of \eqref{int-cond-1} implies that
\[ A_{2,3} = -A_2 C_4(A_2-1), \qquad
A_{2,4} = A_2 C_3 (A_2-1)  . \]

Recall that the function $\frac{C_3}{C_4}$ satisfies
\[ d \left(\frac{C_3}{C_4} \right) \in \calK_2^{(\infty)} = \{dY, dQ\}. \]
It follows that $\frac{C_3}{C_4}$ is a function of $Y$ and $Q$ alone.  Now consider the 1-form
\[ \widetilde{\om}^3 = \om^3 - \frac{C_3}{C_4}\, \om^4 = dY - \frac{C_3}{C_4} dQ. \]
This 1-form is contained in the span of $\om^3, \om^4$, and we have
\[ d\widetilde{\om}^3 \equiv 0 \mod \widetilde{\om}^3; \]
so $\widetilde{\om}^3$ is a multiple of an exact 1-form, say $\widetilde{\om}^3 = \lambda\, d\widetilde{Y}.$  Moreover, because $\widetilde{\om}^3$ is expressed solely in terms of $Y$ and $Q$, $\lambda$ and $\widetilde{Y}$ may be chosen to be functions depending only on $Y$ and $Q$, and which are therefore well-defined on $\scM$.  The crucial point here is that {\em there exists an exact 1-form in the span of $\{ \om^3, \om^4 \}$ which is well-defined on $\scM$.}  Then we have
\[ dY \& dQ = \widetilde{\om}^3 \& dQ = \lambda\, d\widetilde{Y} \& dQ = d\widetilde{Y} \& d\widetilde{Q}, \]
where
\[ \widetilde{Q}(\widetilde{Y}, Q) = \int_0^Q \lambda(\widetilde{Y},t)\, dt. \]
Since
\[ d\theta = dX \& dP + d\widetilde{Y} \& d\widetilde{Q} , \]
Pfaff's Theorem (see Ch. 1 of \cite{CFB}) implies that there exists a function $\widetilde{Z}$ on $\scM$ such that
\[ \theta = d\widetilde{Z} - P\, dX - \widetilde{Q}\, d\widetilde{Y}. \]

We can now repeat all our constructions starting with the coordinate system $(X,\widetilde{Y},\widetilde{Z},P,\widetilde{Q})$, but now our adapted coframing $\sigma$ will have
the additional property that $\om^3 = d\widetilde{Y}$ and $C_3 = 0$.  Thus we will drop the tildes and assume that $C_3 = 0$ for the remainder of this subsection.

\begin{prop}\label{Case-2-prop}
If $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ are both nonzero, $\calK_1^{(\infty)}$ has rank 3, and $\calK_2^{(\infty)}$ has rank 2, then the system $(\Mbar, \Ibar)$ is Monge-integrable, and it becomes Darboux-integrable after one prolongation.
\end{prop}

\begin{proof}
The same argument as that given in Case 1.1 shows that the characteristic system $\{\thetabar, \om^1, \om^2\}$ on $\Mbar$ contains a rank 2 integrable subsystem; therefore, $(\Mbar, \Ibar)$ is Monge-integrable.

In order to prove the second half of the Proposition, we will need to make use of relations among the second derivatives of the functions $A_2, C_1, C_2, C_4$.  These will be denoted as, e.g.,
\[ dA_{2,0} = A_{2,0 0}\theta + A_{2,0 \overline{0}}\thetabar + A_{2,0 1}\om^1 + A_{2,0 2}\om^2 + A_{2,0 3}\om^3 + A_{2,0 4}\om^4. \]
Note that, although (for example) the $A_{2,ij}$ are second derivatives of $A_2$, because we are working in a coframing rather than in coordinates, we cannot assume that $A_{2,ij} = A_{2,ji}.$
%Note that mixed ``partials" with respect to the 1-forms in the coframing do not necessarily commute.

Computing $d(dA_2) \equiv 0 \mod{\{\thetabar, \om^3, \om^4\}}$ shows that
\begin{align*}
A_{2,10} & = A_2 (C_{4,41} - C_{1,11} - C_{2,21}  + C_2 (C_{1,1} + C_{2,2} - C_{4,4})) \\
& \qquad \qquad + A_{2,1} (C_{4,4} - C_{2,2} - C_1 C_2) + A_{2,2} (C_{2,1} - C_2^2),\\
A_{2,20} & = A_2 (C_{4,42} - C_{1,12} - C_{2,22} + C_1 (C_{4,4} - C_{1,1} - C_{2,2})) \\
& \qquad \qquad  + A_{2,1} (C_{1,2} + C_1^2) + A_{2,2} (C_{4,4} - C_{1,1} + C_1 C_2),   \\
A_{2,21} & =  A_{2,12} + (A_2-1)\left(C_{1,1} + C_{2,2}  - \frac{A_{2,1}}{A_2} C_1 - \frac{A_{2,2}}{A_2} C_2\right).
\end{align*}
Next, computing $d(dC_4) \equiv 0 \mod{\{\theta, \thetabar\}}$ shows that
\begin{align*}
C_{4,34} & = C_{4,43} + (A_2-1)C_{4,\overline{0}} + C_4(C_{1,1} + C_{2,2}), \\
C_{4,31} & = C_2 (C_{4,3} + C_4^2),\\
C_{4,32} & = -C_1 (C_{4,3} + C_4^2), \\
C_{4,41} & = C_2 C_{4,4}, \\
C_{4,42} & = -C_1 C_{4,4}.
\end{align*}
Now computing $d(dC_4) \equiv 0 \mod{\theta}$ shows that
\begin{align*}
C_{4, \overline{0} 1} & = C_4 C_{2,\overline{0}} + \frac{((A_2+1) C_1 - A_{2,1})}{A_2} C_{4,\overline{0}} , \\
C_{4, \overline{0} 2} & = -C_4 C_{1,\overline{0}} - \frac{((A_2+1) C_2 + A_{2,2})}{A_2} C_{4,\overline{0}} , \\
C_{4, 3\overline{0}} & = C_{4,\overline{0}3} - A_2 C_4 C_{4,\overline{0}}, \\
C_{4, 4\overline{0}} & = C_{4,\overline{0}4},
\end{align*}
and then computing $d(dC_4) \equiv 0 \mod{\thetabar}$ shows that
\begin{align*}
C_{4, 30} & = C_4 (C_{1,13} + C_{2,23} - C_{4,43} + C_{4,\overline{0}}) + (C_{4,3} - C_4^2)(C_{1,1} + C_{2,2}) - C_{4,\overline{0}3} , \\
C_{4, 40} & = C_4 (C_{1,14} + C_{2,24} - C_{4,44})  + C_{4,4}(C_{1,1} + C_{2,2}) - C_{4,\overline{0}4}, \\
C_{1,12} & = -C_{2,22} + (A_2-1) C_{1,\overline{0}} - C_1 C_{4,4} + \frac{((A_2 - 1) C_1 - A_{2,2})}{A_2 C_4} C_{4,\overline{0}},  \\
C_{2,21} & =   -C_{1,11} - (A_2-1) C_{2,\overline{0}} + C_2 C_{4,4} - \frac{((A_2 - 1) C_2 + A_{2,1})}{A_2 C_4} C_{4,\overline{0}} .
\end{align*}
Next, computing $d(dC_1) \equiv 0 \mod{\{\theta, \thetabar\}}$ yields
\begin{align*}
C_{1,13} & = C_4 (C_{1,1} + C_1 C_2), \\
C_{1,14} & = 0, \\
C_{1,23} & = C_4 (C_{1,2} - C_1^2), \\
C_{1,24} & = 0, \\
C_{1,21} & = -C_{2,22} - 2 C_1 C_{4,4} + 2 (A_2 - 1) C_{1,\overline{0}} + \frac{((A_2 - 1) C_1 - A_{2,2})}{A_2 C_4}   C_{4,\overline{0}}.
\end{align*}
Similarly, computing $d(dC_2) \equiv 0 \mod{\{\theta, \thetabar\}}$ yields
\begin{align*}
C_{2,13} & = C_4 (C_{42,1} + C_2^2), \\
C_{2,14} & = 0, \\
C_{2,23} & = C_4 (C_{2,2} - C_1 C_2), \\
C_{2,24} & = 0 ,\\
C_{2,12} & = -C_{1,11} + 2 C_2 C_{4,4} - 2 (A_2 - 1) C_{2,\overline{0}} - \frac{((A_2 - 1) C_2 + A_{2,1})}{A_2 C_4}   C_{4,\overline{0}}.
\end{align*}
Computing $d(dC_1) \equiv 0 \mod{\{\om^2, \om^3, \om^4\}}$ yields
\begin{align*}
C_{1,10} & = C_2 C_{2,22} - C_1 C_{1,11} - A_2 C_{1,\overline{0}1} + ((2 - A_2) C_2 - A_{2,1}) C_{1,\overline{0}} + C_{4,4} (C_{1,1} + 2 C_1 C_2 ) \\
& \qquad \qquad  - \frac{((A_2 - 1) C_1 - A_{2,2})}{A_2 C_{4}}  C_2 C_{4,\overline{0}},\\
C_{1,1\overline{0}} & = C_{1,\overline{0} 1} + \frac{(A_{2,1} - C_2)}{A_2} C_{1,\overline{0}},
\end{align*}
and computing $d(dC_2) \equiv 0 \mod{\{\om^1, \om^3, \om^4\}}$ yields
\begin{align*}
C_{2,20} & = C_1 C_{1,11} - C_2 C_{2,22} - A_2 C_{2,\overline{0}2} + ((A_2 - 2) C_1 - A_{2,2}) C_{2,\overline{0}} + C_{4,4} (C_{2,2} - 2 C_1 C_2 ) \\
& \qquad \qquad  + \frac{((A_2 - 1) C_2 + A_{2,1})}{A_2 C_{4}}  C_1 C_{4,\overline{0}},\\
C_{2,2\overline{0}} & = C_{2,\overline{0} 2} + \frac{(A_{2,2} + C_1)}{A_2} C_{2,\overline{0}},
\end{align*}
Now computing $d(dA_2) \equiv 0 \mod{\om^3}$ yields
\begin{align*}
A_{2,1\overline{0}} & = \frac{1}{A_2} \left[ -C_1 A_{2,11} - C_2 A_{2,12} + A_{2,1}(C_1 C_2 + C_{2,2}) + A_{2,2} (-C_{2,1} + C_2^2) \right]  \\
& \qquad \qquad  - (A_2 - 1) C_{2,\overline{0}} - C_2 (C_{1,1} + C_{2,2}) - ((A_2 - 2)C_2 + 2 A_{2,1}) C_{4,4}  \\
& \qquad \qquad - \frac{((A_2 - 1)C_2 + A_{2,1})}{A_2 C_4} C_{4,\overline{0}} ,\\
A_{2,2\overline{0}} & = \frac{1}{A_2} \left[ -C_1 A_{2,12} - C_2 A_{2,22} - A_{2,1}\left( C_{1,2} + \frac{C_1^2}{A_2} \right) + A_{2,2} \left( C_{1,1} - \frac{C_1 C_2}{A_2}\right) \right]  \\
& \qquad \qquad  + (A_2 - 1) C_{1,\overline{0}} + \frac{C_1}{A_2} (C_{1,1} + C_{2,2}) + ((A_2 - 2)C_1 - 2 A_{2,2}) C_{4,4}  \\
& \qquad \qquad + \frac{((A_2 - 1)C_1 - A_{2,2})}{A_2 C_4} C_{4,\overline{0}} ,
\end{align*}
and
\[ A_{2,14} = A_{2,24} = C_{4,44} = 0. \]

Finally, we need two additional relations which do not become apparent until we differentiate the equations for $dC_{4,3}$ and $dC_{4,4}$.  Computing $d(dC_{4,3}) \equiv 0 \mod{ \{\thetabar - \theta, \om^3, \om^4 - C_4 \theta\} }$ yields
\[ (C_4 C_{4,\overline{0}3} - ((A_2+1)C_4^2 + C_{4,3}) C_{4,\overline{0}}) \theta \& [(A_{2,1} + C_2(A_2 - 1))\, \om^1 + (A_{2,2} - C_1(A_2 - 1))\, \om^2]  . \]
Note that the right-hand factor is precisely \eqref{int-cond-2}, which we have assumed is nonzero.  Therefore,
\[ C_{4,\overline{0}3} = \frac{C_{4,\overline{0}}(C_{4,3} + (A_2+1) C_4^2)}{C_4} . \]
Precisely the same argument applied to $d(dC_{4,4})$ shows that
\[ C_{4,\overline{0}4} = \frac{C_{4,\overline{0}} C_{4,4}}{C_4}. \]

With these relations in hand, consider the characteristic system $\overline{\calK} = \{\thetabar, \om^3, \om^4\}$ of $\Ibar$---which is well-defined on $\Mbar$, even though $\om^4$ is not.  We need to show that after one prolongation, the corresponding characteristic system $\overline{\calK}'$ of the prolongation contains a rank 2 integrable subsystem.  In order to perform this computation, we need to construct a basis for $\overline{\calK}$ consisting of 1-forms which are well-defined on $\Mbar$.  Fortunately, $\thetabar$ and $\om^3$ are already well-defined on $\Mbar$.  For the remaining 1-form, it will be convenient to choose a 1-form which is contained in the first derived system $\overline{\calK}^{(1)} = \{\om^4 - C_4\, \thetabar, \om^3\}$.  To this end, introduce functions $\tau, g$ on $\B$ such that the 1-form
\[ \psi = e^{\tau}(\om^4 - C_4\, \thetabar - g\, \om^3) \]
is well-defined on $\Mbar$.  (The fact that $\overline{\calK}^{(1)}$ is well-defined on $\Mbar$ guarantees the existence of such functions.)
As before, we denote the derivatives of these functions by
\begin{align*}
d\tau & = \tau_0 \theta + \tau_{\overline{0}} \thetabar + \tau_1 \om^1 + \tau_2 \om^2 + \tau_3 \om^3 + \tau_4 \om^4 ,\\
dg & = g_0 \theta + g_{\overline{0}} \thetabar + g_1 \om^1 + g_2 \om^2 + g_3 \om^3 + g_4 \om^4,
\end{align*}
and similarly for second derivatives.

Because $\psi$ is well-defined on $\Mbar$,  $d\psi$ contains no terms involving $\theta$.  This, in turn, determines the partial derivatives $\tau_0, g_0$:
\begin{align*}
\tau_0 & = C_{4,4}, \\
g_0 & = C_4^2 - C_{4,3} - g C_{4,4}.
\end{align*}
We will also need to make use of relations among the second derivatives of $\tau, g$.  These are determined by computing $d(d\tau) = d(dg) = 0$; this is a straightforward computation, which we omit here for the sake of brevity.

We can define a {\em partial prolongation} $\Ibar'$ of $\Ibar$ on $\Mbar \times \R$ as follows.  (Note that Darboux-integrability of the partial prolongation implies Darboux-integrability of the full prolongation.)  Let $t$ be a new coordinate on the $\R$ factor; then the partial prolongation $\Ibar'$ is  generated by the 1-forms $\thetabar, \thetabar' = \psi - t \om^3,$ and the 2-form $\om^1 \& \om^2$.  Again, this system is well-defined on $\Mbar \times \R$, even though $\om^1, \om^2$ are not.

A straightforward computation shows that
 \[
d\thetabar' \equiv -\pi_1 \& \om^3
 \mod{\{\thetabar, \thetabar'\}},
\]
where
\[ \pi_1 = dt - \left(t \tau_1 - e^{\tau} g_1\right)\om^1 - \left(t \tau_2 - e^{\tau} g_2  \right)\om^2. \]
The corresponding characteristic system of $\Ibar'$ is
\[ \overline{\calK}' = \{\thetabar, \thetabar', \pi_1, \om^3 \}. \]

We will now compute the derived systems of $\overline{\calK}'$ and show that the second derived system $\overline{\calK}'^{(2)}$ is a Frobenius system of rank 2; this will complete the proof of the Proposition.  In order to compute the first derived system, we compute:
 \begin{equation*}
\left. \begin{aligned}
d\thetabar & \equiv \om^1 \& \om^2 \\[0.1in]
d\thetabar' & \equiv 0 \\[0.1in]
d\pi_1 & \equiv E \om^1 \& \om^2 \\[0.1in]
d\om^3 & \equiv 0
\end{aligned} \right\} \mod{\{\thetabar, \thetabar', \pi_1, \om^3\}}
\end{equation*}
(the last line following from $C_3=0$), where
\[
E = e^{\tau} C_4^2 + (e^{\tau}g_4 - t \tau_4) C_4 - e^{\tau} C_{4,3} - (e^{\tau} g + t) C_{4,4} + (e^{\tau} g_{\overline{0}} - t \tau_{\overline{0}}) .
\]
Let $\pi_2=\pi_1 - E \thetabar$, so that
\[ \overline{\calK}'^{(1)} = \{\thetabar', \pi_2, \om^3 \}. \]
Next we compute the derived system of $\overline{\calK}'^{(1)}$:
\[
\left. \begin{aligned}
d\thetabar' & \equiv \frac{C_4}{A_2} \thetabar \& \left[(A_{2,1} + C_2(A_2 - 1))\, \om^1 + (A_{2,2} - C_1(A_2 - 1))\, \om^2 \right]  \\[0.1in]
d\pi_2 & \equiv  \frac{F}{A_2} \thetabar \& \left[(A_{2,1} + C_2(A_2 - 1))\, \om^1 + (A_{2,2} - C_1(A_2 - 1))\, \om^2 \right] \\[0.1in]
d\om^3 & \equiv 0
\end{aligned} \right\} \mod{\{\thetabar', \pi_2, \om^3\}},
\]
where
\[  F = e^{\tau} C_4^2 + (e^{\tau}g_4 - t \tau_4) C_4 - e^{\tau} C_{4,3} - (e^{\tau} g + t) C_{4,4}. \]
Once again, we see the bracketed 1-form in \eqref{int-cond-2} appearing.  Since this 1-form is assumed to be nonzero, the derived system $\overline{\calK}'^{(2)}$ has rank 2 and is spanned by the forms $\om^3$ and
\[ \pi_3 =  C_4 \pi_2 - F\thetabar.  \]
Finally, another computation shows that
\[
\left. \begin{aligned}
d\pi_3 & \equiv  0 \\[0.1in]
d\om^3 & \equiv 0
\end{aligned} \right\} \mod{\{\pi_3, \om^3\}};
\]
therefore, $\overline{\calK}'^{(2)}$ is integrable.
\end{proof}


\subsubsection{Case 1.3: $\calK_1^{(\infty)}$ and $\calK_2^{(\infty)}$ both have rank 2.}
Now we assume that the bracketed 1-forms in both \eqref{int-cond-1} and \eqref{int-cond-2} are nonzero.  By the same argument as that given in the previous case, we may assume that $C_1 = C_3 = 0$, with $\om^1 = dX, \ \om^3 = dY$.

\begin{prop}\label{Case-3-prop}
If $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ are both nonzero and $\calK_1^{(\infty)}$ and $\calK_2^{(\infty)}$ both have rank 2, then the system $(\Mbar, \Ibar)$ becomes Darboux-integrable after one prolongation.
\end{prop}

\begin{proof}
The proof is very similar to that of Proposition \ref{Case-2-prop}.  We must now consider both characteristic systems
\[ \overline{\calK}_1 = \{\thetabar, \om^1, \om^2\}, \qquad \overline{\calK}_2 = \{\thetabar, \om^3, \om^4\} \]
of $\Ibar$.  As before, these systems are both well-defined on $\Mbar$, even though $\om^2$ and $\om^4$ are not.  We introduce functions $\rho, \tau, f, g$ on $\B$ such that the 1-forms
\[ \eta = e^{\rho}(A_2\om^2 - C_2\thetabar - f\om^1), \qquad \psi = e^{\tau}(\om^4 - C_4\thetabar - g\om^3) \]
are well-defined on $\Mbar$.  These forms have the property that
\[ \overline{\calK}_1^{(1)} = \{\eta, \om^1\}, \qquad \overline{\calK}_2^{(1)} = \{\psi, \om^3\}. \]

We construct the prolongation $\Ibar'$ of $\Ibar$ on $\Mbar \times \R^2$ as follows. Let $r, t$ be new coordinates on the $\R^2$ factor; then the prolongation $\Ibar'$ is  generated by the 1-forms $\thetabar, \thetabar_1 = \eta - r \om^1, \thetabar_2 = \psi - t \om^3$, and their exterior derivatives.


The remainder of the proof consists of applying the argument of Proposition \ref{Case-2-prop} to each of the characteristic systems $\overline{\calK}_1', \overline{\calK}_2'$ of the prolongation $\Ibar'$.  The argument varies only in the details of the calculations, and so we omit it for the sake of brevity.
\end{proof}

\subsection{Case 2: One of the $C$-vectors vanishes}\label{degenerate-case}

%Finally, we consider the case where exactly one of the vectors $[C_1 \ \ C_2], \, [C_3 \ \ C_4]$ vanishes.
Without loss of generality, assume that $[C_1 \ \ C_2] = [0 \ \ 0]$, and that $C_4 \neq 0$.  By Proposition \ref{one-C-vector-zero}, it follows that $[B_1 \ \ B_2] = [0 \ \ 0]$ as well.

\begin{prop}\label{one-C-vector-zero-prop}
If $[C_1 \ \ C_2] = [0 \ \ 0]$, then the system $(\Mbar, \Ibar)$ is Monge-integrable, and it becomes Darboux-integrable after at most one prolongation. Furthermore, the \BT/ $\B \subset \scM \times \Mbar$ is holonomic.
\end{prop}

\begin{proof}
It follows from Proposition \ref{one-C-vector-zero} that $(\Mbar, \Ibar)$ is Monge-integrable; in fact, the characteristic system $\{\thetabar, \om^1, \om^2\}$ contains $\{\om^1, \om^2\} = \{dX, dP\}$ as a rank 2 integrable subsystem.

Now consider the other characteristic system $\calK = \{\thetabar, \om^3, \om^4\}$ of $\Ibar$.  One easily computes that the first derived system of $\calK$ is
\[ \calK^{(1)} = \{\om^3 - C_3 \thetabar, \om^4 - C_4 \thetabar\}. \]
In order to find the second derived system, we compute:
\[
\left.  \begin{aligned}
d(\om^3 - C_3 \thetabar) & \equiv \frac{C_3}{A_2} \thetabar \& (A_{2,1} \om^1 + A_{2,2}\om^2) \\
d(\om^4 - C_4 \thetabar) & \equiv \frac{C_4}{A_2} \thetabar \& (A_{2,1} \om^1 + A_{2,2}\om^2)
\end{aligned} \right\} \mod{\calK^{(1)}}.
\]
If $A_{2,1} = A_{2,2} = 0$, then $\calK^{(1)}$ is integrable; in this case, $\Ibar$ is Darboux-integrable and hence contact equivalent to the standard wave equation.  Therefore, we assume that $A_{2,1}$ and $A_{2,2}$ are not both zero.

In order to prove the second statement, we will construct a partial prolongation of $\Ibar$ and proceed as in \S \ref{middle-case}.  But first we need to derive relations among the derivatives of the torsion functions.

Picking up where we left off at \eqref{left-off-here}, consider $d(dC_3), d(dC_4)$.  Computing $d(dC_3) \equiv d(dC_4) \equiv 0 \mod{\{\theta - \thetabar, \om^3 - C_3 \thetabar, \om^4 - C_4 \thetabar\} }$ yields
\[ C_{3,\overline{0}} = C_{4,\overline{0}} = 0. \]
From \eqref{B-relations}, it follows that $B_3 = B_4 = 0$; therefore, the \BT/ is holonomic, as claimed.

We now have
\begin{align}
dC_{3} & = C_{3,3}(\om^3 - C_3 \theta) + C_{3,4} (\om^4 - C_4 \theta) = C_{3,3} dY + C_{3,4} dQ, \notag \\
dC_{4} & = C_{4,3}(\om^3 - C_3 \theta) + C_{4,4} (\om^4 - C_4 \theta) = C_{4,3} dY + C_{4,4} dQ. \label{dC4-exprn}
\end{align}
It follows that $C_3, C_4$ are functions of $Y$ and $Q$ alone.  Now the same argument as that given in \S \ref{middle-case} shows that we may assume $C_3 = 0$; moreover, $C_4$ remains a function of $Y$ and $Q$ alone when we do so.
Computing $d(dC_4) = 0$ yields the following relations among the second derivatives of $C_4$:
\[ C_{4,30} = - C_4 C_{4,34}, \qquad C_{4,40} = - C_4 C_{4,44}, \qquad C_{4,43} = C_{4,34}, \]
\[ C_{4,3\overline{0}} = C_{4,4\overline{0}} = C_{4,31} = C_{4,32} = C_{4,41} = C_{4,42} = 0. \]

Now consider the characteristic system $\calK = \{\thetabar, \om^3, \om^4\}$.  As we computed above (recalling that $C_3=0$), its first derived system is
\[ \calK^{(1)} = \{\om^3, \om^4 - C_4 \thetabar\}. \]
As in \S \ref{middle-case}, choose functions $g, \tau$ so that the 1-form
\[ \psi =  e^{\tau}(\om^4 - C_4\, \thetabar - g\, \om^3) \]
is well-defined on $\Mbar$, and construct the partial prolongation $\Ibar'$ of $\Ibar$ and the 1-form $\thetabar'$ as we did there.  Similar calculations to those of \S \ref{middle-case} show that the corresponding characteristic system $\overline{\calK}'$ of $\Ibar'$ has a rank 2 integrable subsystem.  This completes the proof.
\end{proof}