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What is the necessary and sufficient condition for a curve to have infinite length in a compact interval? Say the curve is restricted to $[0, 1]$. I vaguely remember that it is related to the boundedness of the total variation. I checked already the answers here but they are related to specific examples.

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    $\begingroup$ The total variation is exactly the same thing as the length of the curve. $\endgroup$ Mar 13, 2016 at 12:44
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    $\begingroup$ You can have a look here: whitman.edu/Documents/Academics/Mathematics/grady.pdf Theorem 5.1 says that a curve given by $f\colon [a,b]\to\mathbb R$ has finite length if and only if $f$ has bounded variation. $\endgroup$ Mar 17, 2016 at 8:06
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    $\begingroup$ @GiuseppeNegro: If the curve is given parametrically, this is true. However, if the curve is given as the graph of a function, which is where total variation is usually mentioned, there is a slight difference. The total variation is $$\int_a^b\sqrt{f'(x)^2}\,\mathrm{d}x$$ and the length is $$\int_a^b\sqrt{\color{#C00000}{1+}f'(x)^2}\,\mathrm{d}x$$ $\endgroup$
    – robjohn
    Mar 17, 2016 at 11:17
  • $\begingroup$ @robjohn: Of course. Thank you for the clarification. $\endgroup$ Mar 17, 2016 at 11:46

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Given a function $$f:\quad [a,b]\to{\mathbb R}^n, \qquad t\mapsto f(t)\ ,$$ the total variation of $f$ over $[a,b]$ is defined by $$V(f):=\sup_{\cal P}\sum_{k=1}^N|f(t_k)-f(t_{k-1})|\leq\infty\ ,\tag{1}$$ whereby the $\sup$ ranges over all partitions $${\cal P}:\qquad a=t_0<t_1<\ldots<t_N=b\ ,\qquad N=N_{\cal P}\ .$$ If $V(f)<\infty$ the function $f$ is called of bounded variation.

Consider now a curve $$\gamma:\quad [a,b]\to {\bf z}(t):=\bigl(x(t), y(t)\bigr)\tag{2}$$ in the plane. Then the length $L(\gamma)$ is by definition the total variation of the vector-valued function ${\bf z}(\cdot)$ used for the parametrization of $\gamma$: $$L(\gamma):=V({\bf z}(\cdot))\leq\infty\ .$$ Note that for any two points ${\bf z}_0$, ${\bf z}_1$ one has $$\max\{|x_1-x_0|,\>|y_1-y_0|\}\leq|{\bf z}_1-{\bf z}_0|\leq|x_1-x_0|+|y_1-y_0|\ .$$ From $(1)$ it then easily follows that the function ${\bf z}(\cdot)$ in $(2)$ is of bounded variation iff both $x(\cdot)$ and $y(\cdot)$ are of bounded variation.This allows to conclude that a graph $$\gamma:\quad [a,b]\to{\mathbb R}^2,\qquad x\mapsto\bigl(x,f(x)\bigr)$$ has finite length iff $V(f)<\infty$, since the total variation of the first coordinate is $=b-a<\infty$ in any case.

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The total variation of a differentiable function on $[0,1]$ is given by $$ \int_0^1\left|f'(x)\right|\mathrm{d}x =\int_0^1\sqrt{f'(x)^2}\,\mathrm{d}x $$ The length of the graph of a differentiable function on $[0,1]$ is given by $$ \int_0^1\sqrt{1+f'(x)^2}\,\mathrm{d}x $$ Furthermore, the triangle inequality tells us that $$ \int_0^1\sqrt{f'(x)^2}\,\mathrm{d}x \le\int_0^1\sqrt{1+f'(x)^2}\,\mathrm{d}x \le1+\int_0^1\sqrt{f'(x)^2}\,\mathrm{d}x $$ That is, the length is finite if and only if the total variation is finite.

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A curve will have infinite length in an interval $[a, b]$ if the integral: $$ \int_a^b \sqrt{\left(1 + (f'(x))^2\right)}\mathrm{d}x$$ evaluates to $\infty$.

Here $f'(x)$ represents the derivative of $f(x)$. This can be derived as follows.

Consider a length of the curve $\text ds$. It's position in the $XY$ plane is $(x,y)$. Then: $$\mathrm ds = \sqrt{(\mathrm dx)^2 + (\mathrm dy)^2}$$

where $y=f(x)$. As $\displaystyle \frac {\text dy}{\text dx} = f'(x)$, so: $\text dy = f'(x)\,\text dx$. Thus,

$$\large \int_0^s\text ds = s = \int_a^b\sqrt{\left(1 + (f'(x))^2\right)}\text dx$$

where $s$ is the total length of the curve $f(x)$ in the interval $[a,b]$. If $s\to\infty$, then the curve will have infinite length in $[a,b]$.

Note: I had help from here. Also, from @Martin Sleziak's comment, this technique works only for differentiable functions.

Please correct me if I'm wrong. Hope this helps!

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    $\begingroup$ Actually for $sin(1/x)$ this is not true. It has infinite length in (0 1]. $\endgroup$
    – user48672
    Mar 13, 2016 at 16:11
  • $\begingroup$ I would like to know what is wrong with my answer...i thought the edit covered it? $\endgroup$ Mar 16, 2016 at 4:44
  • $\begingroup$ You probably mean $s=+\infty$ rather than $s\to+\infty$. (You denote by $s$ the total length, so it should be a number or $+\infty$.) And you should probably make clear in your answer that you only work with differentiable functions. Length can also be defined in a greater generality. (And since the OP explicitly mentioned bounded variation, this is probably what they are mostly interested in.) $\endgroup$ Mar 17, 2016 at 8:11
  • $\begingroup$ The statement $dy=f′(x)dx$ is true only for a linear function. You should write $dy=f′(x)dx + \mathcal{O} (dx)$ in the general case. Where the last term is understood as $ \lim_{dx \rightarrow 0} \frac{\mathcal{O} (dx) }{dx} =0$ $\endgroup$
    – user48672
    Mar 17, 2016 at 16:56

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