Problem: Let $p = \frac{\partial z}{\partial x} $ and $q = \frac{\partial z}{\partial y} $. Show that the equations
\begin{align*}
xp - yq = x \text{ and } x^2p + q = xz,
\end{align*}
are compatible.
Solution: We recall that two PDEs
\begin{align*}
f(x,y,z, p,q) = 0 \text{ and } g(x,y,z,p,q) = 0
\end{align*}
are said to be
compatibleif they have a common solution. We have the following necessary and sufficient condition for compatibility.
Two PDEs
\begin{align*}
f(x,y,z, p,q) = 0 \text{ and } g(x,y,z,p,q) = 0
\end{align*}
will be compatible if and only if
\[
J = \frac{\partial (f,g)}{\partial (p,q)} \neq 0 \text{ and } [f,g] = 0,
\]
where
\[
[f,g] = \frac{\partial (f,g)}{\partial (x,p)} + p \frac{\partial (f,g)}{\partial (z,p)} + \frac{\partial (f,g)}{\partial (y,q)} + q \frac{\partial (f,g)}{\partial (z,q)} .
\]
Let
\begin{align}
f(x,y,z,p,q) & = xp - yq - x = 0 \label{eq:28June2024-1} \\
g(x,y,z,p,q) & = x^2 p + q - xz = 0 \label{eq:28June2024-2}.
\end{align}
So,
\begin{align*}
\frac{\partial (f,g)}{\partial (x,p)} & =
\begin{vmatrix}
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial p} \\[2ex]
\frac{\partial g}{\partial x} & \frac{\partial g}{\partial p}
\end{vmatrix} \\
& = \begin{vmatrix}
p - 1 & x \\
2xp - z & x^2 \\
\end{vmatrix} \\
& = (p-1)x^2 - x(2xp - z).
\end{align*}
Similarly, we can compute
\begin{align*}
\frac{\partial (f,g)}{\partial (z,p)} = x^2, \quad \frac{\partial (f,g)}{\partial (y,q)} = -q \quad \text{ and } \frac{\partial (f,g)}{\partial (z,q)} = -xy.
\end{align*}
Therefore,
\begin{align*}
[f,g] & = \frac{\partial (f,g)}{\partial (x,p)} + p \frac{\partial (f,g)}{\partial (z,p)} + \frac{\partial (f,g)}{\partial (y,q)} + q \frac{\partial (f,g)}{\partial (z,q)} \\
& = (p-1)x^2 - x(2xp - z) + px^2 - q - qxy \\
& = x^2 p - x^2 - 2x^2 p + xz + x^2p - q - qxy \\
& = x(xp - yq - x) - (x^2p + q - xz) \\
& = 0 \quad (\text{ from \eqref{eq:28June2024-1} and \eqref{eq:28June2024-2}}).
\end{align*}
Hence, \eqref{eq:28June2024-1} and \eqref{eq:28June2024-2} are compatible.