What does smooth curve mean?

Question

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?

Answer 1

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.

Answer 2

I stumbled upon this old question and I'd like to add something: there is a difference of perspective on smoothness depending whether you look at the geometric object or its parametrization.

Look at the standard example: the real cusp. It is a curve in the real plane parametrized $f:t\to (t^2,t^3)$. Of course, the mapping $f$ is smooth (of any order), and the graph of $f$ is a smooth manifold in $\mathbb{R}^3$, but its image is singular: it is the zero set $x^3=y^2$. It is "worse than a corner"!

So you need to be always clear what you want: do you need only differentiability of the parametrization or do you want the image to be a differentiable manifold (typically in such a case you would assume that the derivative of $f$ does not vanish).

Answer 3

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.

Answer 4

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

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