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I have an ODE to which I want to introduce the new variable $\xi=ax$, where $a$ is a constant. How do I calculate the first and second derivatives of some function $f$? $$\frac{d}{d(ax)}f(x),~ \frac{d^2}{d(ax)^2}f(x)$$ In particular, I am dealing with Schrodinger eq for the harmonic oscillator $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{1}{2}m\omega^2x^2 \psi=E\psi$$ to which things look a little cleaner if the dimensionless variable is used:$$\xi=\sqrt{\frac{m\omega}{\hbar}}x.$$ In terms of $\xi$ the equation reads $$\frac{d^2\psi}{d\xi^2}=(\xi^2-\frac{2E}{\hbar\omega})\psi$$ Is $\psi$ still a function of $x$ or is it a function of $\xi$? I can't follow the steps to the "nicer" equation.

Evaluating the chain rule I get $$\frac{df}{dx}=\frac{df}{d\xi}\cdot \frac{d\xi}{dx}=\frac{df}{dx}\cdot a$$ but, again, is $f$ a function of $x$ or $\xi$? And what about the second derivative?

E Be
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  • Just a comment, not an answer: The original isn't that much worse. If you push things around, you have $$\frac{d^2\psi}{dx^2} = -\left(\frac{m\omega}{\hbar}\right)^2\left(x^2-\frac{2E}{m\omega^2}\right)\psi$$Also, it looks like either you or I dropped a minus sign somewhere. – MPW Jun 09 '16 at 17:46
  • I did. Edited original message. – E Be Jun 09 '16 at 18:01

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Ok so $\frac{d}{d ax} f(x)= \frac{d}{dx} f(x) \frac{dx}{dax}$ by the chain rule right? So then make the substitution $y=ax$, so then we have: $\frac{dx}{d ax}= \frac{ d}{dy} (\frac{y}{a})= \frac{1}{a}$, so then: $\frac{d}{d ax} f(x)= \frac{1}{a} \frac{df}{dx}$, assuming of course $a \neq 0$. $f$ is still a function of $x$. Reapply a second time to get the second derivative.

M10687
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