What is intuitive meaning of $\frac{d}{dx}(\frac{dx}{dt})$?

Question

In a Nutshell...

What is the intuitive meaning of $\frac{d}{dx}(\frac{dx}{dt})$ - of differentiating a velocity function with respect to the position?

Longer Version:

I've been trying to understand the Picard–Lindelöf theorem.

It states that given a differential equation $\frac{dx}{dt}=f(t,x)$, a solution-curve $x(t)$ passing through some initial point $(t_0,x_0)$ exists and is unique if $f(t,x)$ & $\frac{d}{dx}(f(t,x))$ are both continuous around that point.

I've looked at some great Stack Exchange posts, including these two:

Logic/Intuition behind the Uniqueness Theorem

What is the intuition behind uniqueness of differential equation condition that $f$ and $\frac{\partial f}{\partial y}$ are continuous?

And, I've read the section in Arnold where the theorem is introduced.

Although I can see from examples such as $f(x,t)=\frac{x}{t}$ and $f(x,t)=|x|^{\frac{1}{2}}$ that a non-differentiable velocity function implies multiple solutions...(for the former, through the point $(0,0)$ and for the latter, through any $x=0$)...I still can't quite wrap my head around what it means, intuitively, to differentiate the velocity $\frac{dx}{dt}$ with respect to the position $x$.

I'm afraid that if I can't even do that...then, it doesn't matter how many examples I look at...I'll never be able to understand this theorem.

Can someone please help? Thanks!

Answer 1

A flaw in the premise

$x=\tan t$ satisfies the differential equation $\dfrac{\mathrm{d}x}{\mathrm{d}t}=x^{2}+1$. So it would seem that "$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)$ for $x=\tan t$" is $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{2}+1\right)=\boxed{2x}$ and/or $\boxed{2\tan t}$.

But $x=\tan t$ also satisfies the differential equation $\dfrac{\mathrm{d}x}{\mathrm{d}t}=\sec^{2}t$. So it would seem that "$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)$ for $x=\tan t$" is $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sec^{2}t\right)=\boxed{0}$.

Those calculations can't both be right. And there's not really a reason to choose one over the other. So what would make the most sense, and what is the case, is that they're both wrong. "$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)$ for $x=\tan t$" is meaningless, so it's actually good that the OP didn't have intuition for it.

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How can we think about things?

As with many things in calculus, this is a situation where the notation can cause confusion. I'll rephrase things without any Leibniz notation and then come back to it later.

$\varphi'(t)=\varphi(t)-t$ is a differential equation, and the functions $\varphi(t)$ that satisfy it on an interval are of the form $\varphi(t)=1+t+ce^{t}$ for a constant $c$. There is a function of two variables that can help us talk about this differential equation: $f\left(x,t\right)=x-t$. It lets us write the differential equation as $\varphi'(t)=f\left(\varphi(t),t\right)$.

By changing the function $f$, we can get different differential equations $\varphi'(t)=f\left(\varphi(t),t\right)$. And it turns out there are theorems like Picard-Lindelöf that tell us things about solving the differential equation if we know things about the function $f$.

Since $f$ is a function of two variables, it doesn't have a derivative we can call $f'$. Instead, it has two "partial" derivatives: $f_{1}$ where we only care about changes in the first coordinate (and treat the second as a constant) and $f_{2}$ for the second coordinate. We have, for every pair of numbers $(a,b)$, $f_{1}\left(a,b\right)={\displaystyle \lim_{h\to0}}\dfrac{f\left(a+h,b\right)-f\left(a,b\right)}{h}$. By letting $a$ and $b$ vary, we can think of $f_{1}$ as its own function of two variables, that you might write $f_{1}\left(x,t\right)$.

Picard-Lindelöf says that if $f$ and $f_{1}$ are continuous (which isn't as simple as "continuous" for a function of one variable) in a region around a point $\left(x_{0},t_{0}\right)$ then we are guaranteed a unique solution to $\varphi'(t)=f\left(\varphi(t),t\right)$ at least on a tiny interval around $t_{0}$.

Note that $f\left(\varphi(t),t\right)$ depends only on $t$. It only has one input, so we cannot talk about different partial derivatives of $g_{f,\varphi}(t)=f\left(\varphi(t),t\right)$. We could look at $g'(t)$, but that would be $\varphi''(t)$.

Translating to Leibniz notation

If we write $X=\varphi(t)$, then $\varphi'(t)=\dfrac{\mathrm{d}}{\mathrm{d}t}\varphi(t)=\dfrac{\mathrm{d}}{\mathrm{d}t}X=\dfrac{\mathrm{d}X}{\mathrm{d}t}$. I'm using a capital $X$ because I don't want to confuse this "dependent variable" with the independent variable $x$ in $f\left(x,t\right)$. If $X=\varphi(t)$ is a solution to the differential equation, we have that $X$ satisfies $\dfrac{\mathrm{d}X}{\mathrm{d}t}=f(X,t)$.

The partial derivative $f_{1}(x,t)$ is often written $\dfrac{\partial}{\partial x}f(x,t)$. The "curly $d$" is used to suggest that there is at least one independent variable other than $x$. (Also note that nothing in a discussion of the Picard-Lindelöf theorem talks about substituting $X=\varphi(t)$ in for $x$ in the expression $\dfrac{\partial}{\partial x}f(x,t)$.)

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What went wrong?

We need to be very clear on what a phrase like "$x=\varphi\left(t\right)$ satisfies $\dfrac{\mathrm{d}x}{\mathrm{d}t}=f\left(t,x\right)$" means. Something like $\dfrac{\mathrm{d}}{\mathrm{d}t}\varphi\left(t\right)$ (e.g. $\dfrac{\mathrm{d}}{\mathrm{d}t}\tan t=\sec^{2}t$) will just have $t$s in it and never have $x$s, so the phrase can't generally mean "if you substitute in $\varphi(t)$ for $x$ on the left side, you get the right side". It has to mean "if you substitute in $\varphi(t)$ for $x$ on both sides, it's true". Since a solution for $x$ like $\varphi(t)$ is a function of $t$, and $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ has no $x$ in it, we shouldn't write $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)$. Nor should we use the partial derivative notation $\dfrac{\partial}{\partial x}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)$. While we may write the differential equation as "$\dfrac{\mathrm{d}x}{\mathrm{d}t}=f(x,t)$", the only independent variable in $\dfrac{\mathrm{d}X}{\mathrm{d}t}=f(X,t)$ is $t$, since $X=\varphi(t)$ depends on $t$. So $\dfrac{\partial}{\partial x}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)$ is either completely undefined, or always $0$ (if you interpret something like $\sec^{2}t$ as depending on both an independent variable $x$ and $t$).

The point of confusion

The key point is that $x$ is used as an independent variable in "$\dfrac{\partial}{\partial x}f(x,t)$", but used as a dependent variable (hiding an expression $\varphi(t)$) in "$\dfrac{\mathrm{d}x}{\mathrm{d}t}=f(x,t)$". That difference is subtle and can lead to confusion. But Leibniz notation is common and nicer for doing calculations, so we just have to watch out for issues like this.

Answer 2

You are mixing up the context. The condition that started your question is not a condition on the solution of the equation, but on the function $f$ as function, before it is a part of the differential equation. So your conclusion does not make much sense.

Now if the equation has a parameter $p$ like for instance the initial point, then you can consider $u=\frac{∂x}{∂p}$, and that then satisfies a differential equation $\dot u=\frac{∂f}{∂x}u$, so that in this situation there is indeed a connection of the partial derivative of the solution and the derivative of $f$.

Answer 3

If you interpret the differential equation so that $x$ is position and $t$ is time (and this is a big "if", conceptually), the key thing is that the equation does a lot more than just track the motion of some particular individual particle.

The equation is capable of tracking the straight-line motion of an entire swarm of single-point particles arranged densely along a single line. At any given instant $t$, the particle at position $x$ on the line has velocity $f(x,t),$ but a different particle at position $x+h$ at that same instant has velocity $f(x+h,t),$ which is not necessarily the same as $f(x,t).$

On the other hand, if you sit and watch the particles passing through point $x,$ at time $t+h$ then whatever particle happens to be at that point at time $t+h$ will have velocity $f(x,t+h),$ which again is not necessarily the same as $f(x,t).$

Note that $f(x,t)$ does not tell you directly what happens to an individual particle. In particular, $\frac{d}{dt} f(x,t)$ is not the acceleration of the particle at position $x$ at time $t.$ Rather, $\frac{d}{dt} f(x,t)$ is an observation across the velocities of all the particles passing through position $x$ in an interval of time around the instant $t.$

Now suppose you have two detectors that you can place on the line, each of which will read off the velocity of whatever particle is at the exact position of the detector. You put one detector at $x$ and the other at $x+h.$ And you will take measurements with both detectors at time $t$ and compare them.

If you set $h = 0$ you obviously get the same velocity from both detectors, but if you set $h$ to a small positive or negative value you may get two different velocities. The existence of $\frac{d}{dx}\left(\frac{dx}{dt}\right) = \frac{d}{dx} f(x,t)$ tells you that the difference in measured velocities is approximately a linear function of $h$ if $h$ is small enough. If you plotted the velocities that might be measured all along the line at a particular instant $t$ as a function of $x,$ $\frac{d}{dx}\left(\frac{dx}{dt}\right)$ tells you the slope of the plot.

But if the change in velocity is not necessarily perfectly linear over signficant distances, $\frac{d}{dx}\left(\frac{dx}{dt}\right)$ may vary with $x$ at any given time $t.$ Also, as time passes the velocities at all points along the line can change (since $f(x,t)$ is a function of $t$ as well as $x$), not necessarily perfectly linearly, so $\frac{d}{dx}\left(\frac{dx}{dt}\right)$ at a given position $x$ may vary with the time $t.$ The condition that $\frac{d}{dx}\left(\frac{dx}{dt}\right)$ is continuous tells you that these variations are not discontinuous.

The theorem tells you that the particles never collide and no particle ever has a choice of two trajectories it can follow. Personally, I find these conclusions intuitively acceptable, but by no means can I intuitively see why they must be true based only on the given conditions. That's what makes this an interesting theorem.

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Personally, I find this visualization difficult, so I would like to point out that there is nothing in the theorem that says $x$ is position and $t$ is time. The theorem would apply equally well to something like the magnetic field near one end of a large bar magnet. If we take $t$ as the horizontal distance and $x$ as the vertical distance of a point in the field, then $f(x,t)$ is the slope of the "field line" at that point.

Now let's suppose you have a magnetic compass whose size is negligible. If you place this compass in this field, the needle of the compass will point in the direction of the field line through that point. The function $f(x,t)$ tells you the slope of the needle, that is, $f(x,t) = \tan \theta$, where $\theta$ is the angle between the needle and the $t$ axis.

Now if you move the compass left or right (in the $t$ direction), $\frac{d}{dt}\left(\frac{dx}{dt}\right)$ tells you how quickly the slope of the needle changes. If you move the compass up or down (in the $x$ direction), $\frac{d}{dx}\left(\frac{dx}{dt}\right)$ tells you how quickly the slope of the needle changes.

Now the theorem tells you that the field lines do not fork or cross each other anywhere.