0

Wikipedia defines the line integral of a scalar field $\int_C f({\bf s})\, ds$ for a "piecewise smooth curve $C$". Unfortunately, there does not seem to be a consistent definition across the mathematical community for the term "piecewise smooth curve":

  1. Wikipedia says that "The meaning of a function being piecewise $P$, for a property $P$ is roughly that the domain of the function can be partitioned into pieces on which the property $P$ holds, but is used slightly differently by different authors."
  2. This professor said that "a piecewise smooth function [is] a vector-valued function ${\bf r}$ on an interval $I$ that is composed of a finite number of subintervals [$I$] where ${\bf r}$ is smooth, [i.e.] has a continuous and nonzero ${\bf r}'$ on $I$".
  3. Another natural definition would be that a "piecewise smooth function" is a piecewise function that is smooth (in the standard sense of "infinitely differentiable") on each piece.
  4. This textbook says that "A path $C$ is called piecewise smooth if it has a bounded derivative $C'$ which is continuous everywhere on $[a, b]$ except (possibly) a finite number of points for which it is required that both right- and left-hand derivatives exist."
  5. This question repeats the previous definition - but then also separately states a stronger definition that also requires that the function have a continuous derivative on each compact subinterval. I.e. at the interval endpoints, the function's one-sided derivatives must not only exist but must also equal the limit of the function's derivative at it approaches the endpoint.
  6. This paper gives some convoluted definition of "piecewise smooth" that I've never seen anywhere else.

What is the precise requirement on the curve $C$ for the line integral of a scalar field $\int_C f({\bf s})\, ds$ to exist, or at least to "make sense" conceptually and potentially exist?

(For simplicity, assume that the integrand $f({\bf s})$ itself is smooth; I'm just wondering about the requirements for the curve $C$ over which the integral is taken. I assume that the answer is probably the same for smooth scalar field and smooth vector field integrands, but please correct me if I'm wrong.)

Lee Mosher
  • 135,265
tparker
  • 6,756
  • 1
    Well, integration can be performed on a much wider class of curves than these, and as you can see, different authors have different opinions on what is the right level of generality to introduce integration. So, I'm not certain of what would be a "correct answer" to your question. If you are looking for the "most general class of curve-like functions" over which you can integrate, the answer would probably involve measure theory, absolutely continuous measures and Radon-Nikodym derivatives, but I'm not sure this is what you want. – Jackozee Hakkiuz Mar 28 '25 at 04:13
  • @JackozeeHakkiuz For concreteness, what are the minimal requirements for the boundary curve used in Stokes's theorem? (I realize that that's a much more specific question, which should maybe be a separate question.) – tparker Mar 28 '25 at 04:29
  • @tparker Regarding your last comment, I suggest reading through a clearly written, valid proof of Stoke's Theorem (which rules out the proofs in most undergraduate textbooks) to see exactly what definition of piecewise smoothness is needed for that proof. – Lee Mosher Mar 31 '25 at 14:28

1 Answers1

-1

The difficulty arises from what a "curve" means. The most efficient way to define a "curve", for your purposes, is how it is done in topology. A "curve", in this case, is also known as a "1-chain".

So what is a 1-chain? We define a "simple" $1$-chain as a smooth map, $$ f:[0,1]\to \mathbb{R}^2$$ Here smooth means that $f$ can be extended into an infinitely differentiable function from $(-\varepsilon, 1 + \varepsilon)\to \mathbb{R}^2$, so basically we can thicken the closed interval into an open interval on which all component functions of $f$ are infinitely differentiable.

Intuitively, we can visualize $f$ as a curve in $\mathbb{R}^2$ which has no sharp corners.

But sometimes we integrate over triangles, squares, ect.

Thus, it is convenient to enlarge the class of objects we can integrate over. So we define a "1-chain" to be a collection simple 1-chains with multiplicities. For instance, a $1$-chain would look like this: $$ \begin{bmatrix} f_1 & f_2 & f_3 \\ 2 & -3 & 5 \end{bmatrix} $$ The intuition here is that we have a curve consisting of three segments, $f_1$, $f_2$, and $f_3$. The numbers below the segments indicate how many times we travel over each segment. Thus, $2$ below the $f_1$ simply says that repeat the segment $f_1$ twice. The negative sign, $-3$, is useful for saying that we are moving along $f_2$ but backwards.

The above matrix notation is fairly tedious to write. So we prefer to express that matrix as a formal sum, $$ 2f_1 - 3f_2 + 5f_3 $$ By a "formal" sum we simply mean that the $+$ and $-$ signs are just placeholders. We are not actually adding the functions up, instead we just use the notation of $+$ and $-$ to indicate the segments which make up the paths and the required multiplicities along each path.

Suppose then that $c$ is a formal $1$-chain, so, $$ c = \sum_i a_i f_i $$ where each $a_i\in \mathbb{Z}$ and $f_i:[0,1]\to \mathbb{R}^2$ is a smooth map.

Given another smooth function $\phi$ we can then define $$ \int_c \phi = \sum_i a_i \int_0^1 (\phi\circ f_i)||f_i'|| $$

Nicolas Bourbaki
  • 5,553