If I have a function $ u(x,t) = p(x+ct) + q(x-ct) $ (which is the d'Alembert solution to the $1D$-wave equation), I can make the substitutions

$$ \xi(x,t) = x + ct\\ \eta(x,t) = x - ct $$

So I am now left with

$$ u(x,t) = p(\xi) + q(\eta) $$

I could then take that a step further and write

$$ u(x,t) = p(\xi) + q(\eta) = u(\xi, \eta) $$

but I suspect the latter equality simply is wrong... Is that the case? Should I instead be defining a new function, say $v(\xi, \eta)$, so that

$$ u(x,t) = v(\xi, \eta) $$

Even in that last equality, I suspect you can not really 'equate' two functions of different variables. What is the correct approach/terminology here, to make it clear that two functions are equal, save for an appropriate change of variables?

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    $\begingroup$ And is why the concept of 'function of a variable' is a pedagogical cancer. $\endgroup$ – Git Gud Jan 5 '15 at 14:45
  • $\begingroup$ @GitGud I'm not sure I follow. What would an alternative view of a function be, that doesn't focus on variables? $\endgroup$ – funklute Jan 5 '15 at 15:57
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    $\begingroup$ The standard definitions of function do not depend on variables. The function $\sin$ is the function $\sin$, it doesn't matter if you write $x\mapsto \sin(x)$ or $\theta\mapsto \sin(\theta)$. $\endgroup$ – Git Gud Jan 5 '15 at 16:18

Once you've defined $u$ by the equation $u(x,t)=p(x+ct)+q(x-ct)$, you have this same equation for any other quantities that you might substitute for $x$ and $t$. In particular, $u(\xi,\eta)=p(\xi+x\eta)+q(\xi-c\eta)$. So you're quite right that you cannot claim $p(\xi)+q(\eta)=u(\xi,\eta)$. (If you made that claim, what would $u(3,7)$ mean? Would you substitute 3 and 7 for $x$ and $t$ in the original equation or substitute them for $\xi$ and $\eta$ in the new claim?) So you definitely need a new name, like $v$, for the new function.

You correctly observed that the resulting equation $u(x,t)=v(\xi,\eta)$ looks suspicious, because the two sides are functions of different variables. That problem, however, resolves itself when you remember that $\xi$ and $\eta$ were introduced as certain functions of $x$ and $t$. So, although the equation $u(x,t)=v(\xi,\eta)$ is wrong (or nonsense) if you think of $x,t,\xi,\eta$ as just variables, it becomes correct if you take the functional dependence into account and write $u(x,t)=v(\xi(x,t),\eta(x,t))$.

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  • $p(\xi) + q(\eta) = u(\xi, \eta)$ is incorrect, because you had already defined $u$ as "$\text{$p$([first_variable] + $c$*[second_variable]) + $q$([first_variable] - $c$*[second_variable])}$", while the function here written is "$\text{$p$([first_variable]) + $q$([second_variable])}"$

  • $u(x,t) = p(\xi) + q(\eta)$ is almost correct (or totally correct if we agree to this writing convention) becuase it is shorthand for $u(x,t) = p(\xi(x,t)) + q(\eta(x,t))$

  • $u(x,t)=v(ξ,η)$ is, again, almost correct because (supposing $v$ is defined adequately) it is short for $u(x,t) = v(\xi(x,t),\eta(x,t))$
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