Chain rule when taking non-dimensionalising an ODE

Question

I have a silly question. So let's say we have:

$$\frac{d^{2}x}{dt^{2}} = kx$$

Now let's say we pick $X = \frac{x}{x_{c}}$ and $T = \frac{t}{t_{c}}$. What I don't understand is, if we plug in $Xx_{c}$ for x, and $Tt_{c}$ for t, how come we get:

$$\frac{x_{c}d^{2}X}{t_{c}^{2}dT^{2}}$$

How come the $t_{c}$ is squared? Can someone do the math and explain it to me? Our professo r just said. it was. chain rule but I am not sure how

Answer 1

$$\frac{d^{2}x}{dt^{2}} =\frac{d\left(dx/dt\right)}{dt}$$

$$ =\frac{d\left(d[x_c X]/d[t_cT]\right)}{d[tcT]}$$ Since $x_c,t_c$are constant multipliers pull them out of parentheses. And so it becomes

$$ = \dfrac{x_c}{t_c^2}\cdot\frac{d^2X}{dT^2}$$

Stimmt?

Answer 2

You could write the chain rule down as

$$ \frac{d X}{dT} = \frac{d X}{dx} \frac{d x}{dt} \frac{d t}{dT} $$

and from your relationships $X = \frac{x}{x_{c}}$ and $T = \frac{t}{t_{c}}$ obtain

$$ \frac{d x}{dt} = \frac{d X}{dT} \frac{x_c}{t_c} $$

The calculation for the second derivative is analogous, just take the first derivative of the above, $$ \frac{\bullet}{dt} = \frac{\bullet}{dT} \frac{dT}{dt} = \frac{\bullet}{dT} \frac{1}{t_c}$$ and the $t_c$ term gets squared.