What is the meaning of a differential in terms of an exact differential?

Question

As I understand it a differential is an outdated concept from the time of Liebniz which was used to define derivatives and integrals before limits came along. As such $dy$ or $dx$ don't really have any meaning on their own. I have seen in multiple places that the idea of thinking of a derivative as a ratio of two infinitesimal change while intuitive is wrong. I understand this, and besides I am not even really sure if there is a rigorous way of saying when a quantity is infinitesimal.

Now on the other hand, it have read that you can define these differentials as actual quantities that are approximations in the change of a function. For example for a function of one real variable the differential is the function $df$ of two independent real variables $x$ and $Δx$ given by:

$df(x,Δx)=f'(x)Δx$

How this then reduces to

$df = f'(x)dx$

and again what $dx$ means I dont understand. It seems to me that it is simply a linear approximation for the function at a point $x$. However there's no mention of how large or small $dx$ must be, it seems to be just as ill defined as before and I have still found other places referring to it as an infinitesimal even when it has been redefined as here.

Anyway ignoring this, I can see how this could then be extended to functions of more than one independent variable

$y = f(x_1,....,x_n)$

$dy = $$\frac{df}{dx_1}dx_1\ +\ .... \ +\frac{df}{dx_n}dx_n\ $

However then the notion of exact and inexact differentials are brought up. This seems like its unrelated but that raise the question of what a differential means in this case.

All this comes from a course I am taking in Thermal Physics.]2

If anyone can enlighten me as to what the concept of differentials means or perhaps direct me towards a book or website where I can study it myself I would be very grateful.

An explanation of Schwarz' Theorem in this context would be great too.

Answer 1

• Motivation

If we write $y = f(x)$, and $f'(x)$ exists for some particular $x$, then it is standard to write $\boxed{dy = f'(x) \ dx}$. If $x$ is a fixed value, then $dy$ is a dependent variable that depends on $dx$ as the independent variable.

• Geometric Intuition

Essentially, we're introducing $dy$ and $dx$ as new coordinate axes that meet (and form their origin) at a point that lies on a curve $f(x)$. Consequently, $dy$ and $dx$ act as scales for measurement of the variables $dy$ and $dx$, just as the $x$ and $y$ axes act as scales for measurement of $x$ and $y$. In this coordinate system, you can draw the line $dy = f'(x) \ dx$ -- it is tangent to $y = f(x)$ (at some fixed $x$) and has slope $f'(x)$. The thing is that we're not defining it within the $x$-$y$ coordinate system explicitly, but rather the $dx$-$dy$ coordinate system that we constructed.

By looking at it this way, it simplifies this notion down to nothing more than recognizing the slope of a line, because from this, you may derive the popular quotient relation, $\displaystyle{\frac{dy}{dx} = f'(x)}$ for $dx \neq 0$.

• Brief Historical Context

The $d$-notation we use for the differentials $dx$ and $dy$ goes back to Leibniz's work in the seventeenth century, but Leibniz never defined the derivative by the limit of a quotient.

• More In-Depth Discussion

We define the differential of a function $f(x)$ as a function of $x$ as well as another independent variable $dx$ whose value is given by $df = f'(x) \ dx$. This differential is, indeed, defined at each point where $f'(x)$ exists. It's worth noting that $dx$ can take any value. For a fixed $x$, the value of the differential is just a multiple of $dx$ (since, for a fixed $x$, $f'(x)$ is a constant).

Let's re-state differentiability in the following way: a function $f$ is called differentiable at $x$ if it is defined in a neighborhood of $x$ (as well as at $x$ itself) and if there exists some $\alpha$ so that

$$\lim_{\Delta x \to 0} \frac{|f(x + \Delta x) - f(x) - \alpha \Delta x|}{|\Delta x|} = 0$$

This can be reduced to say that

$$\lim_{\Delta x \to 0} \left|\frac{f(x + \Delta x) - f(x)}{\Delta x} - \alpha \right| = 0$$

Which is to say

$$\lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \alpha$$

Which just says that $f'(x)$ exists and is finite with value $\alpha$. Let the differential of $f$ be denoted by $df(x, dx)$. It's important to note that the dependence of $df$ on $x$ is not the same as its dependence on $dx$, so perhaps the notation I used is not quite clear. We're noting that $df$ depends linearly on $dx$. We can do some re-writing:

$$\lim_{\Delta x \to 0} \frac{|f(x + \Delta x) - f(x) - \alpha \Delta x|}{|\Delta x|} = \lim_{dx \to 0} \frac{|f(x + dx) - f(x) - df(x, dx)|}{|dx|}$$

because $\alpha \ dx = f'(x) \ dx = df(x, dx)$ (note that, for notational purposes, I've replaced $\Delta x$ with $dx$, since $dx$ can take any value). So, we've learned that, for a fixed $x$, $df$ is a multiple of $dx$ and that the limit relation we established above is valid. What the relation above tells us is that $df(x, dx)$ is a decent approximation to $f(x + dx) - f(x)$ in the sense that the difference $f(x + dx) - f(x)-df(x,dx)$ is very small compared to $|dx|$ as $|dx| \to 0$.

Answer 2

You can think of differential (1-forms) as new, independent variables. If $y=f(x)$, then it's convenient to place the origin of the $dx-dy$ axes on the curve given by $y=f(x)$. The new variables are related to $y$ and $x$ by the equation $dy = f'(x)\; dx$. (This is the equation of the tangent line at that point, but given in $dx-dy$ coordinates.) By leaving $dx$ and $dy$ as variable, it's possible to let $dx$ take on any value you like, e.g., $\Delta x$. Then the linear approximation theorem says that if $dx = \Delta x$, then $\Delta y \approx dy.$ And we can see that from the picture described above. This can be extended to functions of more than one variable.

Answer 3

One dimensional calculus is very special. For example

• If $f$ is any continuous function, then $f(x) \, \mathrm{d}x$ has an antiderivative
• If $y$ and $u$ depend differentiably on $x$, then $\mathrm{d}y$ and $\mathrm{d}u$ are proportional to each other. This means it makes sense to speak of the ratio $\frac{\mathrm{d}y}{\mathrm{d}u}$ (at least wherever $\mathrm{d}u$ is 'nonzero')
• To talk about definite integrals $\int_a^b f(x) \, \mathrm{d}x$ requires that we only need to specify the endpoints

In higher dimensions, things don't work out so nicely; when we do have these nice properties, it is a special thing deserving a name.

Exactness is the property of being antidifferentiable:

A differential $\omega$ is exact if and only if there exists a function $f$ such that $\omega = \mathrm{d}f$

It turns out that exact forms satisfy the fundamental theorem of calculus: if $\gamma$ is a path from $P$ to $Q$, then

$$ \int_\gamma \mathrm{d}f = f(Q) - f(P) $$

So hopefully you can see exactness being very important.

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We can define a product on differentials called the wedge product; it sort of behaves like you'd expect a product to behave, except that it has the weird feature that it's antisymmetric on ordinary differentials: $\mathrm{d}x \wedge \mathrm{d}y = -\mathrm{d}y \wedge \mathrm{d}x$. (and also $\mathrm{d}x \wedge \mathrm{d}x = 0$).

W can use this to differentiate a differential form: to get a 2-form:

$$ \mathrm{d}\left( u \, \mathrm{d}v \right) = \mathrm{d}u \wedge \mathrm{d} v $$ and furthermore to even higher degrees.

For example, in $\mathbb{R}^2$, we could compute $$\begin{align} \mathrm{d}\left( f(x,y) \, \mathrm{d}x + g(x,y) \, \mathrm{d} y \right) &= \mathrm{d}f(x,y) \wedge \mathrm{d}x + \mathrm{d}g(x,y) \wedge \mathrm{d} y \\&= (f_1(x,y) \mathrm{d}x + f_2(x,y) \mathrm{d}y) \wedge \mathrm{d}x + (g_1(x,y) \mathrm{d}x + g_2(x,y) \mathrm{d}y) \wedge \mathrm{d}y \\&= (g_1(x,y) - f_2(x,y)) \mathrm{d}x \wedge \mathrm{d}y \end{align}$$

(I use $h_n$ to refer to the partial derivative of a function with respect to its $n$-th argument while holding the other arguments constant)

You might notice that the test using Schwarz' rule that your image is alluding to boils down to checking if a differential $\omega$ satisfies $\mathrm{d} \omega = 0$. These are also special:

A differential form $\omega$ is closed if and only if $\mathrm{d} \omega = 0$

It turns out that every exact differential form is closed, and conversely, in extremely nice spaces like $\mathbb{R}^n$, every closed differential form is exact.

Since testing if a form is closed is relatively easy and computational, it provides a very nice first step in checking whether a differential is exact — and in very nice spaces like $\mathbb{R}^n$ it's the only step you need!