Chain Rule/Second Partial Derivative Problem If you're given the arbitrary function:
$f(u(x,y),v(x,y))$
Solve for functions $u$ and $v$ that satisfy the following equation (for any function $f$).
$ \frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u \ \partial v} $
I attempted to solve this by first solving for $f_{xx}$ and $f_{yy}$ then subtracting them to see if something cancels.
$ f_{xx} = f_u u_{xx} + f_v v_{xx} + f_{uu} u_x^2 + f_{vv} v_x^2 + 2f_{uv} u_x v_x $
$ f_{yy} = f_u u_{yy} + f_v v_{yy} + f_{uu} u_y^2 + f_{vv} v_y^2 + 2f_{uv} u_y v_y $
Subtracting these results in a long nonsense equation that doesn't have much meaning.
$ f_{xx} - f_{yy} = f_u (u_{xx} - u_{yy}) + f_v(v_{xx} - v_{yy}) + f_{uu} (u_x^2 - u_y^2) + f_{vv} (v_x^2 - v_y^2) + 2f_{uv} (u_x v_x - u_y v_y) $
How would I go about solving this? Is this a partial differential equation problem?
 A: Choosing $f(u,v)=u,v,u^2,v^2,uv$ successively implies that all coefficients of $f$ derivatives in your equation vanish.  That is, we obtain an overdetermined system
\begin{align}
u_{xx}-u_{yy}&=0,\\
v_{xx}-v_{yy}&=0,\\
u_x^2-u_y^2&=0,\\
v_x^2-v_y^2&=0,\\
2(u_xv_x-u_yv_y)&=1.
\end{align}
Let us focus on the first and third equations:
\begin{align}
u_{xx}=u_{yy},\\
u_x=\pm u_y,
\end{align}
for some sign $\pm$.  Differentiating the second equation implies $u_{xx}=\pm u_{yy}$.  If $+$ is chosen, then this is consistent with the first equation, i.e. we are left with $u_x=u_y$, and thus 
$$
u(x,y)=U(x+y)
$$
for an arbitrary function $U$.  But if $-$ is chosen, then the first equation implies $u_{xx}=u_{yy}=0$, and we get
$$
u(x,y)=a+bx+cy+dxy
$$
for some constants $a,b,c,d$.  To summarize, the general solution of the "$u$ equations" is
$$
u(x,y)=U(x+y),\qquad a+bx+cy+dxy.
$$
Similarly,
$$
v(x,y)=V(x+y),\qquad e+fx+gy+hxy.
$$
We are thus left with solving the fifth equation
$$
u_xv_x-u_yv_y=1/2.
$$
Up to the natural $u\leftrightarrow v$ exchange symmetry, we have three possible combinations of solutions:
Case 1: $u=U,v=V$:
Here, the remaining equation reduces to $0=1/2$, so this combination yields no solutions.
Case 2: $u=a+bx+cy+dxy,v=V$.
In this case, the remaining equation becomes
$$
V'(x+y)(b-c+dy+dx)=1/2.
$$
This gives $V(z)=C+\frac{1}{2d}\ln|b-c+dz|$ if $d\neq 0$, or $V(z)=C+\frac{z}{2(b-c)}$ otherwise.
Case 3: $u=a+bx+cy+dxy,v=e+fx+gy+hxy$.
The remaining equation becomes
$$
(b+dy)(f+hy)-(c+dx)(g+hx)=1/2.
$$
Differentiating this equation in $x$ and $y$ implies $d=h=0$, so we are left with
$$
bf-cg=1/2.
$$
This determines $u,v$.
