Help needed showing that $f(x,y)=U(x+y)+V(x-y)$ Let $h(u,v)=f(u+v,u-v)$ and $f_{xx}=f_{yy}$ for every $(x,y)\in\mathbb{R}^2$. In addition, $f\in{C^2}.$ Show that $f(x,y)=U(x+y)+V(x-y)$.
I think applying the Taylor theorem could be useful.
$$f(x,y)=f(x+h_1,y+h_2)-\left(\frac{\partial{f(x,y)}}{\partial{x}}h_1+\frac{\partial{f(x,y)}}{\partial{y}}h_2\right)-\frac 1 2\left(\frac{\partial^2{f(x,y)}}{\partial^2{x}}h_1^2+\frac{\partial^2{f(x,y)}}{\partial{x}\partial{y}}h_1h_2+\frac{\partial^2{f(x,y)}}{\partial^2{y}}h_2^2\right)-R(h_1,h_2)$$
 A: This is a d'Alembert form solution for the hyperbolic PDE
$$
f_{xx} - f_{yy} = 0
$$ 
One changes to variables
$$
\xi = x - y \\
\eta = x + y
$$
and uses the chain rule to get
$$
\frac{\partial f}{\partial x} =
\left(
\frac{\partial f}{\partial \xi}
\right)
\frac{\partial \xi}{\partial x}
+ 
\left(
\frac{\partial f}{\partial \eta}
\right)
\frac{\partial \eta}{\partial x}
=
\frac{\partial f}{\partial \xi}
+ 
\frac{\partial f}{\partial \eta}
\\
\frac{\partial f}{\partial y} =
\left(
\frac{\partial f}{\partial \xi}
\right)
\frac{\partial \xi}{\partial y}
+ 
\left(
\frac{\partial f}{\partial \eta}
\right)
\frac{\partial \eta}{\partial y}
=
-\frac{\partial f}{\partial \xi}
+ 
\frac{\partial f}{\partial \eta}
$$
and
$$
\left(
\frac{\partial}{\partial x}\frac{\partial}{\partial x}
\right) f
=
\left(
\frac{\partial}{\partial \xi}
+ 
\frac{\partial}{\partial \eta}
\right)
\left(
\frac{\partial}{\partial \xi} 
+ 
\frac{\partial}{\partial \eta}
\right) f
=
\left(
\frac{\partial}{\partial \xi}
\right)^2 f
+
2
\left(
\frac{\partial}{\partial \xi}
\frac{\partial}{\partial \eta}
\right) f
+
\left(
\frac{\partial}{\partial \eta}
\right)^2 f \iff \\
f_{xx} = f_{\xi\xi} + 2 f_{\xi\eta} + f_{\eta\eta}
\\
\left(
\frac{\partial}{\partial y}\frac{\partial}{\partial y}
\right) f
=
\left(
-\frac{\partial}{\partial \xi}
+ 
\frac{\partial}{\partial \eta}
\right)
\left(
-\frac{\partial}{\partial \xi}
+ 
\frac{\partial}{\partial \eta}
\right) f
=
\left(
\frac{\partial}{\partial \xi}
\right)^2 f
-
2
\left(
\frac{\partial}{\partial \xi}
\frac{\partial}{\partial \eta}
\right) f
+
\left(
\frac{\partial}{\partial \eta}
\right)^2 f \iff \\
f_{yy} = f_{\xi\xi} - 2 f_{\xi\eta} + f_{\eta\eta}
$$
This gives the transformed PDE:
$$
f_{\xi\eta} = 0
$$
Integration regarding $\eta$ gives
$$
f_\xi = C(\xi)
$$
where $C = C(\xi)$, as this is only constant regarding $\eta$, so it can still be dependent on $\xi$, thus $C(\xi)$ instead of just $C$.
Another integration, now regarding $\xi$, gives
$$
f = \underbrace{\int C(\xi) d\xi}_{E(\xi)} + D(\eta) 
= E(\xi) + D(\eta)
= E(x - y) + D(x + y)
$$
where $E(\xi)$ is an antiderivative of $C(\xi)$ and $D$ is constant regarding $\xi$, so it can still be $D(\eta)$.
If we rename the introduced functions, we get
$$
f = V(x-y) + U(x+y)
$$
A: You define
$$
\xi=x+y\\
\eta=x-y
$$
then the partial differential equation becomes
$$
\frac{\partial^2f}{\partial \xi\partial \eta}=0
$$
The only way you can satisfy it is that it is a sum of 2 functions 
$$
U(\xi)+V(\eta)
$$
Such combination is always identically zero by the application of a second mixed derivative.
