# What does smooth curve mean?

In this problem, I know that the hypothesis of Green's theorem must ensure that the simple closed curve is smooth, but what is smooth? Could you give a definition and an intuitive explanation?

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Intuitive: "no corners". Rigorous: "a sufficient number of its derivatives are continuous." – J. M. Nov 27 '10 at 6:05
@J.M.: What is "sufficient"? – Jichao Nov 27 '10 at 6:08
Sivaram elaborated a bit; the working definition I have is "whatever the application needs." – J. M. Nov 27 '10 at 6:40
@J.M. The "no corners" characterization is good for a regular curve (where the tangent vector doesn't vanish). However, for a smooth curve, as defined in the accepted definition, we can have cusps, as in the deltoid, for example en.wikipedia.org/wiki/Deltoid_curve – yasmar Nov 27 '10 at 16:11
@yas: So, "piecewise smooth" then. :) – J. M. Nov 28 '10 at 1:22

There are many ways you can characterize the smoothness of a curve.

Typically, people use the notation $C^{(n)}(\Omega)$ where $n \in \mathbb{N}$.

So when we say $f(x) \in C^{(n)}(\Omega)$, we mean that $f(x)$ has $n$ derivatives in the entire domain ($\Omega$ denotes the domain of the function) and the $n^{th}$ derivative of $f(x)$ is continuous i.e. $f^{n}(x)$ is continuous.

Also by convention, if $f(x)$ is just continuous, then we say $f(x) \in C^{(0)}(\Omega)$.

Also, $f(x) \in C^{(\infty)}$ if the function is differentiable any number of times. For instance, $e^{x} \in C^{(\infty)}$

An example to illustrate is to consider the following function $f: \mathbb{R} \rightarrow \mathbb{R}$. $$f(x) = \begin{cases}0, &\mbox{if }x \leq 0 \\ x^2, &\mbox{if }x>0\end{cases}$$

This function is in $C^{(1)}(\mathbb{R})$ but not in $C^{(2)}(\mathbb{R})$.

When the domain of the function is the largest set over which the function definition makes sense, we omit $\Omega$ and write that $f \in C^{(n)}$, the domain being understood as the largest set over which the function definition makes sense.

Note that $C^{(n)} \subseteq C^{(m)}$ whenever $n>m$.

EDIT:

In case of Green's theorem, when we apply the formula $$\oint_c (L\,dx + M\,dy) = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\,dx\,dy$$ we need $L,M \in C^1{(\Omega)}$, where $\Omega$ is a domain containing the curve and the interior of the curve viz $D$.

The simple closed curve $C$ should be piecewise smooth or more generally the curve $C$ should be in $C^{(0)}$.

We say that the curve $C$ is piecewise smooth curve when the two conditions below are satisfied:

(i) $C \in C^{(0)}$

(ii) The domain over which the curve is defined can be partitioned into disjoint subsets such that the curve is in $C^{(\infty)}$ (or sufficiently smooth i.e. the curve is in $C^{(n)}$ for some $n$ till which we are interested) over each of these subsets.

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@Sivaram: Do you mean if curve $L$ is smooth, then $L \in C^{(0)}$? – Jichao Nov 27 '10 at 7:10
@Jichao: In the Green's theorem, you want the curve over which you are integrating i.e. the curve $C$ to be piecewise smooth. It depends on to what extent of smoothness you want. In the case of Green's theorem, you want the functions $L$ and $M$ to be smooth up to their first derivative i.e. $L,M \in C^{(1)}$ and the curve over which you are integrating i.e. $C$ to be smooth up to being continuous i.e. $C \in C^{(0)}$. – user17762 Nov 27 '10 at 7:14
@Sivaram: So smoothness is a relative concept and the smoothest curve should belong to $C^{(\infty)}$? – Jichao Nov 27 '10 at 7:17
@Jichao: Exactly. Smoothness is a relative concept and is problem specific. $C^{(\infty)}$ is as smooth as smooth can be. In applications, when you say the curve is smooth it means till the derivatives you are interested in the curve has to be continuous. So for instance in Green's theorem, smoothness would mean the functions $L,M \in C^{(1)}$ and the curve $C \in C^{(0)}$. – user17762 Nov 27 '10 at 7:24
@Sivaram: So considering parametric representation of 2-dimensional curve $C$, $C \in C^{(n)}$ means $x(t) \in C^{(n)}$ and $y(t) \in C^{(n)}$ ? – Jichao Nov 27 '10 at 7:44

Consider the following curve in the plane, $(x(t),y(t))$, this curve is called smooth if the functions $x(t)$ and $y(t)$ are smooth, which simply means that for all $N$, the derivatives $\frac{d^Nx}{dt^N}$ and $\frac{d^Ny}{dt^N}$ exist.

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Smooth means differentiable (at least once). In other words, smooth means continuous and without "corners".

Edit: changed "edges" to "corners" after J.M.

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So smooth is different with piecewise smooth? – Jichao Nov 27 '10 at 6:07
A piecewise smooth curve could have corners. Think of a square. The individual sides are smooth, but the square itself is a nonsmooth curve at the corners. – TKN Nov 27 '10 at 6:14
This is wrong. A smooth curve must be defined by a function which has infinite derivatives. – Matt Calhoun Nov 27 '10 at 6:18
@Matt: It depends on the author. Some authors define "smooth" to mean $C^\infty$, others define it to mean $C^1$, and still others define it to mean $C^1$ and has non-vanishing derivative. – Jesse Madnick Jan 31 '11 at 0:52

a. A curve is ‘smooth’ at a given point p on the curve if a unique line exists in the curve's plane at p that contains a segment that is centered at p, intersects the curve at the single point p and lies entirely on one side of the curve [such a line is called a ‘tangent line’ at p]

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That's not true. Consider the curve which is the graph of $f(x) = x^3$. The curve does not lie on one side of the tangent line at $0$. – Michael Albanese Jan 20 at 1:15

I have heard the term "smooth" being used in the context of the Hölder condition.

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For any derivitive f'(x) on the smooth curve f(x) = y there is one value of x that indicates the slope of a line tanget to y. And, if the curve is jagged it cannot have a tangent at the sharp edges of the line f(x) = y, and therefore is not smooth.

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I don't think this is a good survey of what meaning "smooth" can have for curves or for functions. In posting a new answer so long after the Question (five years or so, in this case), it would help you as well as your Readers to carefully review the answers already given (and accepted, in this case) so that new material can be highlighted. – hardmath Feb 24 at 16:19