Doubt about ordinary and partial derivative

I was reading this Wikipedia page Beltrami identity. There is this passage that I don't understand. According to the page, in the section "derivation":

$$ L[x, u(x), u'(x)] $$

In the second passage, the page writes this:

$$ \frac{dL}{dx} = \frac{\partial{L}}{\partial{u}}u' + \frac{\partial{L}}{\partial{u'}}u'' + \frac{\partial{L}}{\partial{x}} $$

I totally understand the chain rule applied here, and my question is: is there a difference between $\frac{d}{dx}$ operator and $\frac{\partial{}}{\partial{x}}$ operator? Shouldn't they mean the same thing?

Thanks in advance,

Dave

Solution 1:

Recall the basics of functions: roughly speaking a function is a triple of information $f:A \to B$, where $A$ is the domain, $B$ is the target space, and $f$ is the "rule".

In your case, you're given a function $L: \Bbb{R}^3 \to \Bbb{R}$ and a function $u: \Bbb{R} \to \Bbb{R}$ (i.e the lagrangian is a real valued function of three real variables, and you're given a real valued curve $u$). Now, using these two, you're constructing a new function, $\Lambda: \Bbb{R} \to \Bbb{R}$ defined by \begin{align} \Lambda(x) &= L(x,u(x), u'(x)) \end{align} What this means is that for a given real number $x$, $\Lambda(x)$ is that real number obtained by plugging the $3$-tuple of real numbers $(x,u(x),u'(x))$ into the function $L$. From here, you can clearly see that $L$ and $\Lambda$ are distinct functions. The domain of $L$ is $\Bbb{R}^3$, whereas the domain of $\Lambda$ is $\Bbb{R}$. However, it is very common to abuse notation in this context, and people often out of laziness simply denote $\Lambda$ as $L$.

What the chain rule actually says, with proper notation, and being explicit with where everything is being evaluated is that: for all $x \in \Bbb{R}$, \begin{align} \Lambda'(x) &= (\partial_1L)_{(x, u(x), u'(x))} + (\partial_2L)_{(x, u(x), u'(x))} \cdot u'(x) + (\partial_3L)_{(x, u(x), u'(x))} \cdot u''(x) \tag{$*$} \end{align} Here $\partial_iL$ means the partial derivative of the function $L$ with respect to the $i^{th}$ variable. This same equation written slightly more sloppily says: \begin{align} \dfrac{d \Lambda}{dx} &= \dfrac{\partial L}{\partial x} + \dfrac{\partial L}{\partial u} u' + \dfrac{\partial L}{\partial u'} u'' \end{align}

And lastly, if you want to be even more sloppy, you'll find the following equation: \begin{align} \dfrac{d L}{dx} &= \dfrac{\partial L}{\partial x} + \dfrac{\partial L}{\partial u} u' + \dfrac{\partial L}{\partial u'} u'' \end{align}

Again, strictly speaking, this is a nonsensical equation, because $L$ is a function of $3$ variables, so it doesn't make sense to take the ordinary derivative $\frac{d}{dx}$ of $L$; it only makes sense to take partial derivatives $\partial_iL$. But... this equation (and others like it which you'll encounter frequently in physics texts) is really just a very lazy way of writing $(*)$.

By the way, this abuse of notation is not only prevalent in multivariable calculus; it is also there in single variable calculus. For example, you may have learnt the single variable chain rule as:

If $f= f(y)$ and $y = y(x)$ are differentiable functions, then $f(y(x))$ is differentiable and \begin{align} \dfrac{df}{dx} = \dfrac{df}{dy} \cdot \dfrac{dy}{dx} \end{align}

well, again, the $f$ on the LHS and the $f$ on the RHS mean completely different things! You may not have paid much attention in this case, because it is a simple enough formula to remember, and it is very intuitive with the "$dy$'s cancelling". But strictly speaking it is a complete overload of notation. (you should try to understand properly what the proper meaning of the $f$'s on the two sides is)

Solution 2:

It is a short way to write something like this.

Consider a function of three variables, say $L[r,s,t]$, then plug in three functions $r = x, s = u(x), t=u'(x)$ to get $$ M(x) := L[x,u(x),u'(x)] $$ Then $$ \frac{dM}{dx} = \frac{\partial L}{\partial r}\;\frac{\partial r}{\partial x} +\frac{\partial L}{\partial s}\;\frac{\partial s}{\partial x} +\frac{\partial L}{\partial t}\;\frac{\partial t}{\partial x} $$ You question is whether there is any difference between $d M/d x$ and $\partial L /\partial r$. Yes, they are different. Even if we think $M = L$ and $r = x$. Try a simple examaple to see.