How might we express a second order PDE as a system of first order PDE's? I have come across a question which asks the following of me:

By defining
  $$
\alpha = \frac{\partial \phi}{\partial x} \hspace{10mm} \beta = \frac{\partial \phi}{\partial y}
$$
  express the partial differential equation
  $$
\frac{\partial^2 \phi}{\partial x^2} - \frac{\partial^2 \phi}{\partial y^2} = 0
$$
  as a system of 2 first order partial differential equations in $\alpha$ and $\beta$.

I am not sure how I am to go about this one. Clearly, I can see that the PDE may be expressed in terms of $\alpha$ and $\beta$ as
$$
\frac{\partial \alpha}{\partial x} - \frac{\partial \beta}{\partial y} = 0
$$
and (I believe) we may write this as the system
$$
\gamma(x,y) = \frac{\partial \alpha}{\partial x} \hspace{10mm} \gamma (x,y) = \frac{\partial \beta}{\partial y}
$$
but I'm unsure of where to go from here.
Note: I have a check sheet for this question which states the answer as
$$
U_x + U_y = 0 \hspace{10mm} V_x - V_y = 0
$$
 A: Lauds to John Doe for an accurate and accepted answer.  In the following, I have tried to explain why both the relations
$\dfrac{\partial \alpha}{\partial x} = \dfrac{\partial \beta}{\partial y}, \; \dfrac{\partial \alpha}{\partial y} = \dfrac{\partial \beta}{\partial x} \tag 0$
are necessary and sufficient.
Given that
$\dfrac{\partial^2 \phi}{\partial x^2} - \dfrac{\partial^2 \phi}{\partial y^2} = 0, \tag 1$
or
$\dfrac{\partial^2 \phi}{\partial x^2} = \dfrac{\partial^2 \phi}{\partial y^2}, \tag 2$
then clearly by setting
$\alpha = \dfrac{\partial \phi}{\partial x} \tag 3$
and
$\beta = \dfrac{\partial \phi}{\partial y}, \tag 4$
we obtain the first-order equation
$\dfrac{\partial \alpha}{\partial x} = \dfrac{\partial^2 \phi}{\partial x^2} =  \dfrac{\partial^2 \phi}{\partial y^2} =  \dfrac{\partial \beta}{\partial y}, \tag 5$
i.e.
$\dfrac{\partial \alpha}{\partial x} - \dfrac{\partial \beta}{\partial y} = 0. \tag 6$
However, (6) by itself is not yet a system of first-order equations equivalent to (1), (2); indeed, so far all (6) tells us is that we have a vector field $(\alpha, \beta)^T$ on some region of $\Bbb R^2$ the components of which satisfy this equation.  What is not evident from (6), which we need to affirm the equivalence of (6) and (1), (2), is the existence of a scalar function $\phi$ such that (3) and (4) apply; that is, $(\alpha, \beta)^T$ is the gradient some $\phi$; we thus need $\phi$ such that
$(\alpha, \beta)^T = \nabla \phi. \tag 7$
It is well-known that such a $\phi$ exists if and only if
$\nabla \times (\alpha, \beta)^T = 0,  \tag 8$
or
$\dfrac{\partial \alpha}{\partial y} - \dfrac{\partial \beta}{\partial x} = 0; \tag 9$
thus the equation (1), (2) is equivalent to the system (6), (9); that is, to
$\dfrac{\partial \alpha}{\partial x} = \dfrac{\partial \beta}{\partial y}, \tag{10}$
$\dfrac{\partial \alpha}{\partial y} = \dfrac{\partial \beta}{\partial x}. \tag{11}$
To re-iterate, taking $\alpha$ and $\beta$ as in (3), (4), we see via (5) that (1)-(2) imply (6), (10); (11) follows from the fact that, for sufficiently differentiable $\phi$, the mixed partials satisfy 
$\dfrac{\partial^2 \phi}{\partial x \partial y} = \dfrac{\partial^2 \phi}{\partial y \partial x},  \tag{12}$
whence
$\dfrac{\partial \alpha}{\partial y} = \dfrac{\partial}{\partial y} \dfrac{\partial \phi}{\partial x} = \dfrac{\partial^2 \phi}{\partial y \partial x} = \dfrac{\partial^2 \phi}{\partial x \partial y} =  \dfrac{\partial}{\partial x} \dfrac{\partial \phi}{\partial y} = \dfrac{\partial \beta}{\partial x}; \tag{13}$
having arrived at (10), (11) starting from (1), (2), we may walk in the other direction:  for sufficiently differentiable functions $\alpha$, $\beta$, (11) implies that the vector field $(\alpha, \beta)^T$ is curl free on its region of definition in $\Bbb R^2$, hence must be the gradient of some function $\phi$, that is, (7) binds, and then we have
$\alpha = \dfrac{\partial \phi}{\partial x}, \tag{14}$
and
$\beta = \dfrac{\partial \phi}{\partial y}; \tag{15}$
thus, using (10),
$\dfrac{\partial^2 \phi}{\partial x^2} = \dfrac{\partial}{\partial x}\dfrac{\partial \phi}{\partial x} = \dfrac{\partial \alpha}{\partial x} = \dfrac{\partial \beta}{\partial y} = \dfrac{\partial}{\partial y}\dfrac{\partial \phi}{\partial y} = \dfrac{\partial^2 \phi}{\partial y^2}, \tag{16}$
whence follows (1).
I can't speak to our OP J Chapman's introduction of $U$ and $V$ with
$U_x + U_y = 0 = V_x - V_y \tag{17}$
since these functions are left undefined in the text of the question.
If we are give a sufficiently differentiable function $\phi$ of $x$ and $y$, then the symmetry relation of the mixed partials (12) immediately follows; thus (11) comes "for free", so to speak, when we define $(\alpha, \beta)^T = \nabla \phi$; it is in the attempt to define $\phi$ from $\alpha$ and $\beta$ satisfying (10), however, we discover that the assumption (11) is essential to ensure that the requisite $\phi$ exists.
Finally, I couldn't help but observe that the system (10)-(11) bears a strong resemblance to the Cauchy-Riemann equations of complex analysis; indeed, if instead of (11) we take
$\dfrac{\partial \alpha}{\partial y} = -\dfrac{\partial \beta}{\partial x}. \tag{18}$
then (10) and (18) are in fact a Cauchy-Riemann system for $\alpha + i \beta$, which is then holomorphic.  Indeed, it follows from (10) and (18) that $\alpha$ and $\beta$ are harmonic:
$\dfrac{\partial}{\partial x} \dfrac{\partial \alpha}{\partial x} = \dfrac{\partial}{\partial x} \dfrac{\partial \beta}{\partial y} = \dfrac{\partial}{\partial y} \dfrac{\partial \beta}{\partial x} = -\dfrac{\partial}{\partial y} \dfrac{\partial \alpha}{\partial y}, \tag{19}$
that is,
$\nabla^2 \alpha = 0, \tag{20}$
with a similar derivation holding to show that $\beta$ obeys 
$\nabla^2 \beta = 0. \tag{21}$
However, it the absence of the "$-$" sign occurring in (18), i.e. when (10) holds instead, there seems to be no analog of these facts for the resulting (hyperbolic) PDEs.
Finally, we note that $\alpha$ and $\beta$ also obey the equation (2); to see this, simply take $\partial / \partial x$ or $\partial / \partial y$ of that equation.
A: I believe the second equation required is $$\frac{\partial \alpha}{\partial y}=\frac{\partial \beta}{\partial x}$$by symmetry of mixed partials.
