Change of variables in differential equation and the order of the arguments This question comes from Boas, Mathematical Methods in the Physical Sciences, chapter 4, section 11: "Change of variables". Exercise question 3.

Suppose that $w=f(x,y)$ satisfies
  $$
\frac{\partial^2 w}{\partial x^2}-\frac{\partial^2 w}{\partial y^2}=1. \tag{1}
$$
Put $x=u+v$, $y=u-v$ and show that $w$ satisfies $\dfrac{\partial^2 w}{\partial u\partial v}=1$. Hence solve the equation.

I can obtain an answer by a combination of the two new variables,
\begin{align}
u&=\dfrac{x+y}{2},\\
v&=\dfrac{x-y}{2}.
\end{align}
Because then I can write $$\dfrac{\partial w}{\partial x}=\dfrac{\partial w}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial w}{\partial v}\dfrac{\partial v}{\partial x}=\frac{1}{2}\left(\dfrac{\partial w}{\partial u}+\dfrac{\partial w}{\partial v}\right)~,$$ and thus it also follows that $$\frac{\partial^2 w}{\partial x^2}=\frac{1}{4}\frac{\partial^2 w}{\partial u^2}+\frac{1}{2}\frac{\partial^2 w}{\partial u\partial v}+\frac{1}{4}\frac{\partial^2 w}{\partial v^2}.\tag{2a}$$ If you do the same for the other equation, $$\frac{\partial^2 w}{\partial y^2}=\frac{1}{4}\frac{\partial^2 w}{\partial u^2}-\frac{1}{2}\frac{\partial^2 w}{\partial u\partial v}+\frac{1}{4}\frac{\partial^2 w}{\partial v^2},\tag{2b}$$ you can combine them with the given equation to find the desired relationship.
But I'm not very satisfied with the answer: we were given $w(x,y)$ and we use that that is equal to $w(x(u,v),y(u,v))$. However, the chain rule as I applied it above is applied to a function of the form $w(u(x,y),v(x,y))$. It thus seems to me that equation (1) is:
\begin{align}
\frac{\partial^2 w(x(u,v),y(u,v))}{\partial x^2} - \frac{\partial^2 w(x(u,v),y(u,v))}{\partial y^2}=1,
\end{align}
while a subtraction of eq (2b) from (2a) gives:
\begin{align}
\frac{\partial^2 w(u(x,y),v(x,y))}{\partial x^2} - \frac{\partial^2 w(u(x,y),v(x,y))}{\partial y^2} = \frac{\partial^2 w(u(x,y),v(x,y))}{\partial u\partial v}.\tag{3}
\end{align}
It is not clear to me that the left-hand side in equation (3) equally satisfies equation 1, because the argument of $w$ is in reverse order.
Does my thinking go wrong somewhere? Or is my proposed answer wrong?
(My question also applies to changes of variables in general -- it is still quite unclear to me what the exact rules are.)
 A: Your difficulty stems from an abuse of notation which is very common in these circles: The same letter $w$ is used for two different functions, one in the $(x,y)$-world, and one in the $(u,v)$-world. 
At the start you are given a function
$$(x,y)\mapsto f(x,y)\ ,$$
satisfying a certain pde. Now you introduce a new function of other variables $u$, $v$ by putting
$$(u,v)\mapsto F(u,v):=f(u+v,u-v)\ .$$
Of course you have some physical interpretation, e.g., $w:=f(x,y)$ could be the temperature at the point $(x,y)$, and you introduce new coordinates $(u,v)$ in the $(x,y)$-plane. You still call the temperature at the points of the plane $w$, and say "$w$ is now a function of $u$ and $v$". But the chain rule does not know about such sentiments. It says that
$$F_u(u,v)=f_x(u+v,u-v)\cdot1+f_y(u+v,u-v)\cdot1\ ,$$
and that
$$\eqalign{F_{uv}(u,v)&=f_{xx}(u+v,u-v)\cdot1+f_{xy}(u+v,u-v)\cdot(-1)\cr &\qquad+f_{yx}(u+v,u-v)\cdot1+f_{yy}(u+v,u-v)\cdot(-1)\ .\cr
&=f_{xx}(u+v,u-v)-f_{yy}(u+v,u-v)\cr
&=1\ .\cr}$$
A: Another way to think about it which surely is the same as Christian Blatter wrote but from other perspective would be to notice that you are using the same name for two different function.
You're using $w$ for the function that send $(x,y)$ to $w(x,y)$ as in 
$\frac{\partial^2 w(x(u,v),y(u,v))}{\partial x^2} - \frac{\partial^2 w(x(u,v),y(u,v))}{\partial y^2}=1$
But also $w$ for function that sends (u,v) to $w(u,v)$ as in 
$\frac{\partial^2 w(u(x,y),v(x,y))}{\partial x^2} - \frac{\partial^2 w(u(x,y),v(x,y))}{\partial y^2} = \frac{\partial^2 w(u(x,y),v(x,y))}{\partial u\partial v}$
We can't write $w(x,y)=w(u,v)$ but we can rename ours functions to write $w(x,y)=\tilde{w}(u,v)\ (1)$
Now lets call $h:(x,y) \to h(x,y)=(u,v)$ then we have that $\tilde{w}(u,v)=(\tilde{w}\circ h)(x,y)\ (2) $
From (1) and (2) we have that $w(x,y)=_{*}(\tilde{w}\circ h)(x,y)=_{**}\tilde{w}(u,v)\ $ so,
$\frac{\partial^2 w(x,y)}{\partial x^2} - \frac{\partial^2 w(x,y)}{\partial y^2}
\\=_{*}\frac{\partial^2 (\tilde{w}\circ h)(x,y)}{\partial x^2} - \frac{\partial^2 (\tilde{w}\circ h)(x,y)}{\partial y^2}
\\=_{**}(\frac{1}{4}\frac{\partial^2 \tilde{w}(u,v)}{\partial u^2}+\frac{1}{2}\frac{\partial^2 \tilde{w}(u,v)}{\partial u\partial v}+\frac{1}{4}\frac{\partial^2 \tilde{w}(u,v)}{\partial v^2})-(\frac{1}{4}\frac{\partial^2 \tilde{w}(u,v)}{\partial u^2}-\frac{1}{2}\frac{\partial^2 \tilde{w}(u,v)}{\partial u\partial v}+\frac{1}{4}\frac{\partial^2 \tilde{w}(u,v)}{\partial v^2})
\\=\frac{\partial^2 \tilde{w}(u,v)}{\partial u\partial v}$
