Consider the spring equation from before.
\begin{equation*}
my^{\prime \prime} + by^\prime + ky = 0
\end{equation*}
Now, instead of all constants, lets assume this is an aging spring. That is, at time passes, the spring constant decreases. One model is exponential decay, so let \(k(t) = ke^{-\alpha
t}\) for \(\alpha > 0\text{.}\) Let’s see what happens without friction.
\begin{equation*}
m y^{\prime \prime} + ke^{-\alpha t} = 0
\end{equation*}
Here’s a strange change of variables.
\begin{align*}
s \amp = \frac{2}{\alpha} \sqrt{\frac{k}{m}}
e^{\frac{-\alpha t}{2}}
\end{align*}
I want to know how to change the differential equation. This is pretty difficult to do; I have to go through some derivative acrobatics to make it work. First I differentiate the change of variables function \(s(t)\text{.}\)
\begin{align*}
\frac{ds}{dt} \amp = \frac{2}{\alpha} \frac{-\alpha}{2}
\sqrt{ \frac{4k}{m\alpha^2}} e^{\frac{-\alpha t}{2}} = -
\sqrt{\frac{k}{m}} e^{-\frac{\alpha t}{2}}\\
\frac{d^2s}{dt^2} \amp = \frac{\alpha}{2} \sqrt{\frac{k}{m}}
e^{-\frac{\alpha t}{2}}
\end{align*}
Now I want to find an expression for \(\frac{d^2 y}{dt^2}\) in terms of some new function \(x(s)\) instead of \(y(t)\text{.}\) The function I need is implicity defined as follows.
\begin{equation*}
x(s(t)) = y(t)
\end{equation*}
This is implicit; \(x\) is the function that, when compsed with the change of variables \(s(t)\text{,}\) give the original function \(y\text{.}\) Since the composition is invertible, this function will always exist (assuming reasonable differentiability condition of \(y\text{,}\) which are already required since it is the solution of a differential equation.) There is a technical theorem in analysis called the Implicit Function Theorem that guarantees existence of functions like this. In any case, I want to differentiate this new function. I need the chain rule to do so.
\begin{align*}
\frac{dx}{dt} \amp = \frac{dx}{ds} \frac{ds}{dt}
\end{align*}
Then I want to differentiate again. Here I need the produce rule and the chain rule. After the derivatives, I bracket the result in a useful way.
\begin{align*}
\frac{d^2 x}{dt^2} \amp = \frac{d}{dt} \left( \frac{dx}{ds}
\frac{ds}{dt} \right)\\
\amp = \frac{d^2 s}{dt^2} \frac{dx}{ds} + \frac{ds}{dt}
\frac{d}{dt} \frac{dx}{ds}\\
\amp = \frac{d^2 s}{dt^2} \frac{dy}{ds} + \frac{ds}{dt}
\frac{d^2y}{ds^2} \frac{ds}{dt}\\
\amp = \frac{d^2 s}{dt^2} \frac{dx}{ds} + \left(
\frac{ds}{dt} \right)^2 \frac{d^2 x}{ds^2}
\end{align*}
I can calculate the derivatives of the change of variables \(s(t)\) and replace them in this expression.
\begin{align*}
\amp = \frac{\alpha}{2} \sqrt{\frac{k}{m}} e^{-\frac{\alpha
t}{2}} \frac{dx}{ds} + \frac{k}{m} e^{-\alpha t} \frac{d^2
x}{ds^2}
\end{align*}
Now, the function \(x(s(t)) = y(t)\) is the same as the function \(y(t)\text{,}\) as a function of \(t\text{;}\) that was exactly how it was defined. Therefore, I can replace \(y\) and \(y^{\prime \prime}\) in the original equation with the \(t\) derivatives of \(x\text{.}\) Let me do so, and proceed to do some algebra with the result.
\begin{align*}
m y^{\prime \prime} + ke^{-\alpha t} \amp = 0\\
m \left(\frac{d^2 s}{dt^2} \frac{dx}{ds} + \left(
\frac{ds}{dt} \right)^2 \frac{d^2 x}{ds^2} \right) +
ke^{-\alpha t} \amp = 0\\
\frac{m\alpha}{2} \sqrt{\frac{k}{m}} e^{-\frac{\alpha t}{2}}
\frac{dx}{ds} + m \frac{k}{m} e^{-\alpha t} + k e^{-\alpha
t} x \amp = 0
\end{align*}
To write this in nicer way, I divide by \(\alpha^2\) and multiply by \(4\text{.}\)
\begin{align*}
\frac{2}{\alpha} \sqrt{\frac{k}{m}} e^{-\frac{\alpha t}{2}}
\frac{dx}{ds} + \frac{4}{\alpha^2} \frac{k}{m} e^{-\alpha t}
\frac{d^2 x}{ds^2} + \frac{4k}{m\alpha^2} e^{-\alpha t} y \amp
= 0
\end{align*}
Then the coefifcient of the second derivative of precisely \(s(t)^2\) and the coefficient of the first derivative is precisely \(s(t)\text{.}\)
\begin{align*}
s^2 \frac{d^2x}{ds^2} + s \frac{dx}{ds} + s^2 y \amp = 0
\end{align*}
I have now produced a new differential equation for \(x(t)\) in the new variable \(s\text{.}\) If I can solve this equation, then by simply replace \(s\) with the change of variables expression \(s(t)\text{,}\) I can recover \(y(t) = x(s(t))\text{.}\)
