What does $\frac{\partial}{\partial x}(\frac{\partial f}{\partial u})$ mean when $f(x,t), u=x+ct, v=x-ct $? I'm trying to transform the wave equation $\frac{\partial^2f}{\partial t^2}-c^2\frac{\partial^2f}{\partial x^2}$ using the substitution:
$
\\u=x+ct
\\v=x-ct
$
and using the chain rule for the partial derivatives $\frac{\partial f}{\partial x}$ and$\frac{\partial f}{\partial t}$ I get:
$
\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f}{\partial u}*1+\frac{\partial f}{\partial v}*1 $ and similarly:
$:
\frac{\partial f}{\partial t}=\frac{\partial f}{\partial u}*c+\frac{\partial f}{\partial v}*(-c)
$
To express the second partial derivative $\frac{\partial^2f}{\partial x^2}$:
$
\frac{\partial^2f}{\partial x^2}=\frac{\partial}{\partial x}(\frac{\partial f}{\partial x})=\frac{\partial}{\partial x}(\frac{\partial f}{\partial u})+\frac{\partial}{\partial x}(\frac{\partial f}{\partial v})
$
Now the notation says I should take the derivative with respect to $x$ of $\frac{\partial f}{\partial u}$, but what then is $\frac{\partial f}{\partial u}$? If $f$ is a function of $x$ and $t$, which both are in expressed in both $u$ and $v$, how does the chain rule apply here (as compared to the first step above, which I can handle)?
Thanks!
Alexander
 A: We have
$$f(x,t)=g(u,v)=g(x+ct,x-ct)$$
so
$$\frac{\partial f}{\partial x}=\frac{\partial g}{\partial u}+\frac{\partial g}{\partial v}\quad\text{and}\quad \frac{\partial^2 f}{\partial x^2}=\frac{\partial^2 g}{\partial^2 u}+2\frac{\partial^2 g}{\partial u\partial v}+\frac{\partial^2 g}{\partial v^2}$$
and
$$\frac{\partial f}{\partial t}=c\frac{\partial g}{\partial u}-c\frac{\partial g}{\partial v}\quad\text{and}\quad \frac{\partial^2 f}{\partial t^2}=c^2\frac{\partial^2 g}{\partial u}-2c^2\frac{\partial^2 g}{\partial u\partial v}+c^2\frac{\partial^2 g}{\partial v^2}$$
hence the wave equation
$$\frac{\partial^2f}{\partial t^2}-c^2\frac{\partial^2f}{\partial x^2}=0$$
becomes
$$\frac{\partial^2 g}{\partial u\partial v}=0$$
so
$$g(u,v)=\phi(u)+\psi(v)$$
and finally we have
$$f(x,t)=\phi(x+ct)+\psi(x-ct)$$
A: I think that the notation is the following.
Let $\varphi: (x,t)\mapsto \varphi(x,t)=(\varphi_1,\varphi_2):=(u(x,t),v(x,t))$ be the coordinate transformation with 
$$u(x,t)=x+ct $$
$$v(x,t)=x-ct;$$ 
What you have computed is the partial derivative $\frac{\partial g}{\partial x}$ with
$$g=f\circ \varphi.$$
Then
$$\frac{\partial g}{\partial x}=\frac{\partial g}{\partial \varphi_1}\frac{\partial\varphi_1}{\partial x}   + 
\frac{\partial g}{\partial \varphi_2}\frac{\partial\varphi_2}{\partial x} =  
\frac{\partial g}{\partial u}\frac{\partial u}{\partial x}   + 
\frac{\partial g}{\partial v}\frac{\partial v}{\partial x}.$$
So $\frac{\partial f}{\partial u}$ is nothing but $\frac{\partial g}{\partial u}$, and similarly for $\frac{\partial f}{\partial v}$.
A: This is a common (albeit slight) abuse of notation.  Let $s(u,v) = (x,t)$.  Then there is a function $\bar f(u,v) = (f \circ s)(u,v)$.  What is meant by $\partial f/\partial u$ is really, in a strict sense, $\partial \bar f/\partial u$.
From here it's easier to write things in terms of vectors.  Let $\mathbf r = (x,t)$ and $\mathbf w = (u,v)$.  That is, we have a vector function $\mathbf s(\mathbf w) = \mathbf r$, and a scalar field $f(\mathbf r) = \bar f (\mathbf w)   = (f \circ \mathbf s)(\mathbf w)$.  Crucially see that $f = \bar f \circ \mathbf s^{-1}$.  I will use this immediately.
Let $\partial_\mathbf{r} = (\partial_x, \partial_t)$ and $\partial_{\mathbf w} = (\partial_u, \partial_v)$.  The chain rule takes the form
$$a \cdot \partial_{\mathbf r} f = a \cdot \partial_{\mathbf r} (\bar f \circ \mathbf s^{-1})= (a \cdot \partial_{\mathbf r} \mathbf s^{-1}) \cdot \partial_{\mathbf w} \bar f$$
When $a = (1,0)$, this gets us
$$\partial_x f = (\partial_x \mathbf s^{-1}) \cdot \partial_\mathbf{w} \bar f = \frac{\partial \mathbf s^{-1}}{\partial x} \cdot (\partial_u \bar f, \partial_v \bar f)$$
The function $\mathbf s^{-1}$ is just the vector field $(u(x,t), v(x,t))$, and this produces the chain rule as you know it.
$$\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} \frac{\partial \bar f}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial \bar f}{\partial v}$$
It's clearer, however, to expand $\bar f$ back into $f$ and $\mathbf s$.  In doing so, let me just write things in terms of a dummy function $\alpha$:
$$\frac{\partial}{\partial x} \alpha  =  \left(\frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} \right) (\alpha \circ \mathbf s^{-1})$$
This is true for any possible $\alpha$.  That includes  $\alpha = f$ or, crucially, $\alpha = \partial f/\partial x$!  So when you take the second derivative, you just do the same as you did before.  Instead of plugging $f$ into the expression, you plug in $\partial f/\partial x$, and you're done.
