I was given a question on partial derivatives to solve, which I was able to handle overall - but I realised with one of the intermediary steps, I knew it was correct, but I didn't know why?
Tl;dr - My Question
Jumping straight to where I was stuck - if $u = x + t$ and $v = x -t$, and you want to calculate $\frac{\partial{x}}{\partial{u}}, \frac{\partial{x}}{\partial{v}}, \frac{\partial{t}}{\partial{u}}, \frac{\partial{t}}{\partial{v}}$, why must you rearrange $x$ to be in terms of $u$ and $v$, i.e. $x = \frac{1}{2}(u + v)$, to then calculate the partials wrt $u$ and $v$, rather than using $x = u - t$ and $x=v+t$? Likewise of course for solving for t, for its partials?
I know that using $x = \frac{1}{2}(u + v)$ (as I went on to do in my answer) is correct, but I can't quote the proper/concise reason as to why.
- For context, I'll lay out the original question I was asked below, and the steps up to where I got confused -
Original Question:
Let $f(x,t) = F(u,v)$, and let $u= x+t $ and $v=x-t$. Show that $\frac{\partial^2{f}}{\partial{x^2}} - \frac{\partial^2{f}}{\partial{2^2}} = k\cdot\frac{\partial^2{F}}{\partial{u}\partial{v}}$, and find $k$.
Answer:
Overall then, since $f(x,t) = F(u,v)$, then $\frac{\partial^2{F}}{\partial{u}\partial{v}} = \frac{\partial^2{f}}{\partial{u}\partial{v}}$. I'll show that $\frac{\partial^2{f}}{\partial{u}\partial{v}} = (\frac{\partial^2{f}}{\partial{x^2}} - \frac{\partial^2{f}}{\partial{2^2}})\cdot C$, and thus that $K \cdot \frac{\partial^2{F}}{\partial{u}\partial{v}} = (\frac{\partial^2{f}}{\partial{x^2}} - \frac{\partial^2{f}}{\partial{2^2}})$
So, the first step I took was to calculate $\frac{\partial{f}}{\partial{v}}$:
$\frac{\partial{f}}{\partial{v}} = \frac{\partial{f}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{v}} + \frac{\partial{f}}{\partial{t}}\cdot\frac{\partial{t}}{\partial{v}}$
And thus $\frac{\partial^2{f}}{\partial{u}\partial{v}}$:
$\frac{\partial^2{f}}{\partial{u}\partial{v}} = \frac{\partial{}}{\partial{u}}(\frac{\partial{f}}{\partial{v}}) = \frac{\partial{}}{\partial{u}} (\frac{\partial{f}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{v}} + \frac{\partial{f}}{\partial{t}}\cdot\frac{\partial{t}}{\partial{v}}) = \frac{\partial{}}{\partial{u}}(\frac{\partial{f}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{v}}) + \frac{\partial{}}{\partial{u}}(\frac{\partial{f}}{\partial{t}}\cdot\frac{\partial{t}}{\partial{v}})$
$= \frac{\partial{}}{\partial{x}}(\frac{\partial{f}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{v}}) \frac{\partial{x}}{\partial{u}} + \frac{\partial{}}{\partial{t}}(\frac{\partial{f}}{\partial{t}}\cdot\frac{\partial{t}}{\partial{v}})\frac{\partial{t}}{\partial{u}} $
$= \frac{\partial{f}}{\partial{x^2}} \cdot (\frac{\partial{x}}{\partial{v}}\cdot\frac{\partial{x}}{\partial{u}}) + \frac{\partial{f}}{\partial{t^2}} \cdot (\frac{\partial{t}}{\partial{v}}\cdot\frac{\partial{t}}{\partial{u}}) $
So far, so good. All that remains is to workout $\frac{\partial{x}}{\partial{u}}, \frac{\partial{x}}{\partial{v}}, \frac{\partial{t}}{\partial{u}}, \frac{\partial{t}}{\partial{v}}$, then substitute them into the last line, and complete the remaining algebra.
But I refer you to my tl;dr above for where I was/am slightly confused. As said, I had an instinct to use and did use the correct definition of $x$ in terms of $u,v$, i.e. $x= \frac{1}{2}(u+v)$, to work out the partials, and likewise for $t$ with $u,v$, and went on to complete the question.
Just can't quite say definitively why the $u,v$ versions of $x$ and $t$ are the right ones.
Apologies for the very basic question, I'm just teaching myself university calculus, and am very new at this.
Many thanks, indeed!
The old-fashioned reason is this.
Symbols like $\frac{\partial x}{\partial u}$ have additional concealed information: in the context of your question it means "the derivative of $x$ w.r.t. $u$ keeping $v$ constant".
It doesn't mean "the derivative of $x$ w.r.t. $u$ keeping $t$ (or whatever) constant".
Sometimes people clarify this by writing $\frac{\partial x}{\partial u}\left|_{v\text{ constant}}\right.$.