Prove that $y'$ is a function of $y$

Question

First, some context:

This occurred to me while I was learning how to do order reduction for $2^{nd}$ order ODEs of the form $F(y, y', y'') = 0$.

Apparently, I'm supposed to set $p = y'$, say that $y'' = p_x = p_y * y' = p_y * p$ (according to the chain rule), and then solve $F(y, p, p_y * p) = 0$, which is now a $1^{st}$ order ODE.

However, I have to ask: What is $p$? My first instinct would be to say that it is a function of $x$. But here it is being treated as a function of $y$ (unless there is some variation to the chain rule I'm unaware of).

So this leads me to believe that $p$, and by extension $y'$, are functions of $y$. But why? We can't necessarily solve the original equation to prove that it is the case - so how can we prove it?

Answer 1

Unless $y$ is a constant, $y'=p$ is a function of $x$. Consequently, there is a functional relation between $p$ and $y$ (via $y=f(x)\leftrightarrow x=f^{-1}(y)$).

E.g. $y=e^x\to p=e^x=y$.

Answer 2

I'm just trying to explain it from a physical perspective. Maybe that helps you.

In short: Is $p$ in general a function of $y$?

No!

$p$ is computed as the derivative of $y(x)$. Hence, if we only know a value $y_1 \in \mathbb{R}$, then we cannot compute $p_1$ as a function of $y_1$. There are ways to recover $p$ in special cases. But I guess that these cases are not important for you at the moment.

Long answer:

Case 1: We known $y(x)$ explicitly.

Let us assume first, that we know $y(x)$ already, we interpret $x$ as time and $y(x)$ as the position at time $x$.

To make it explicit, we take: $y(x) := x^3$.

Now, we can compute the velocity of the trajectory as $y'(x) = 3x^2$. When we define $p = y'$ is means that $p$ is the velocity of the trajectory.

Question: If you know the position of an object, do you automatically know it's velocity?

Since we know $y(x) = x^3$ we can compute for a given position $y_1$ the corresponding time $x_1$. Namely, $x_1 = \sqrt[3]{y_1}$.

In this case there is a global one-to-one relation between the current velocity $p_1 = p(x_1)$ and the position$y_1 = y(x_1)$: $$ p_1 = 3 x_1^2 = 3 ( \sqrt[3]{y_1})^2. $$

But this is not always the case, for example if $y(x) := x^2$, then no such relation exists.

For a given position $y_1 := y(x_1)$ the corresponding time could be $\pm \sqrt{y_1}$. Hence, knowing the position $y_1$ at one time-point is not sufficient anymore to find the velocity $p_1$!

Remark: If we know the positions for a whole time-interval, i.e. $y(x)$ for $x \in [x_1,x_2]$, then we have enough data to determine the velocity uniquely. But this is not very useful for us.

Case 2: $y(x)$ is a solution of $F(y,y',y'',x)$

A trajectory $y(x)$ which solves the equation $F(y(x),y'(x),y''(x),x) = 0$ (for all $x$) is often a very special function. However, it depends on $F$ if we can determine $y(x)$ uniquely from that equation!

For example if we define $F(y,y',y'',x) := 0$ than any trajectory $y(x)$ solves $F = 0$. Hence, in this case there are no condition on the trajectory at all.

Therefore, an underlying condition for implicit differential equations is that we can solve $F(y,y',y'',x) = 0$ with respect to $y''$.

In physical terms: If we know the time $x$, the position $y$ and the velocity $y'$, then we can determine the acceleration $y''$ (typically via Newton's law).

However, if we can determine the velocity already from the positions, i.e. if for a fixed values $y_1, x_1 \in \mathbb R$ there is some acceleration $a_1$ and a unique value $p_1 \in \mathbb R$ such that $F(y_1,p_1,a_1,x_1) = 0$, then the ODE is in fact a first-order ODE. So, if a functional relation $p(y)$ follows directly by algebraic manipulation of $F = 0$, then the system is a first-order ODE, namely $y' = p(y)$.

Let me shortly come back to the example 1 form the link you sent in the comments: $$F(y,y',y'',x) = x y'' - y'.$$

We see, that we can solve $F = 0$ for $y''$, i.e. $$y'' = \frac{y'}{x}.$$ (i.e. we satisfy the invertibility assumption on $F$.)

If we introduce $p(x) = y'(x)$ this equation becomes $p'(x) = \frac{p(x)}{x}$. The reason why the example ask you to consider $x p'(x) = p$ is only to make the computations easier. But this is not the chain rule!