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A Fractional Notion of Length and an Associated Nonlocal Curvature

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Abstract

In this paper, a new notion of fractional length of a smooth curve, which depends on a parameter \(\sigma ,\) is introduced that is analogous to the fractional perimeter functional of sets. It is shown that in an appropriate limit, the fractional length converges to the traditional notion of length up to a multiplicative constant. Since a curve that connects two points of minimal length must have zero curvature, the Euler–Lagrange equation associated with the fractional length is used to motivate a nonlocal notion of curvature for a curve. This is analogous to how the fractional perimeter has been used to define a nonlocal mean-curvature.

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Notes

  1. For the definition of sets of finite perimeter and essential boundary, see, for example, Ambrosio et al. [2].

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Correspondence to Brian Seguin.

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Dedicated to Eliot Fried, whose guidance following my time as a graduate student will always be appreciated.

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Appendix: Some Change of Variables

Appendix: Some Change of Variables

Let \(\mathcal{C}\) be a \(C^1,\) compact curve in \(\mathbb {R}^n.\)

Lemma 5.1

Consider the function

$$\begin{aligned} \Phi :\mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+\rightarrow \mathbb {R}^n\times \mathbb {R}^n \end{aligned}$$

defined by

$$\begin{aligned} \Phi (z,\mathbf{a},\mathbf{b},\xi ):=(z+\xi \mathbf{a},\mathbf{b})\quad \text {for all}\ (z,\mathbf{a},\mathbf{b},\xi )\in \mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+. \end{aligned}$$
(64)

If \(\mathcal{A}\) is a subset of \(\mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+\) and \(f:\Phi (\mathcal{A})\rightarrow \mathbb {R}\) is an integrable function, then

$$\begin{aligned}&\int _{\Phi (\mathcal{A})}\left[ \sum _{(z,\mathbf{a},\mathbf{b},\xi )\in \Phi ^{-1}(p,\mathbf{u})} f(p,\mathbf{u})\right] \mathrm{d}\mathcal{H}^{2n-1}(p,\mathbf{u}) \nonumber \\&\quad = \int _\mathcal{A}f(z+\xi \mathbf{a},\mathbf{b})\xi ^{n-2} |\mathbf{b}\cdot \mathbf{t}(z)| \mathrm{d}\mathcal{H}^{2n-1}(z,\mathbf{a},\mathbf{b},\xi ), \end{aligned}$$
(65)

where \(\mathbf{t}(z)\) is a unit-vector tangent to the curve \(\mathcal{C}\) at the point z.

Proof

It suffices to prove the result for a set of the form \(\mathcal{A}=\mathcal{C}_\mathcal{A}\times \mathcal{U}_\mathcal{A}\times \mathcal{I}_\mathcal{A},\) where \(\mathcal{C}_\mathcal{A}\subseteq \mathcal{C},\)\(\mathcal{U}_\mathcal{A}\subseteq \mathcal{U}_\perp ^2,\) and \(\mathcal{I}_\mathcal{A}\subseteq \mathbb {R}^+.\) Moreover, by employing a partition of unity, we can reduce the problem to the case where \(\mathcal{U}_\mathcal{A}\) is covered by one chart. Thus, there is a set \(U_\mathcal{A}\subseteq \mathbb {R}^{2n-3}\) and a diffeomorphism \(\varvec{\chi }:U_{\mathcal{A}}\rightarrow \mathcal{U}_\mathcal{A}.\) Since \(\mathcal{U}_\mathcal{A}\subseteq \mathbb {R}^n\times \mathbb {R}^n,\) we can view this function as \(\varvec{\chi }=(\varvec{\chi }_1,\varvec{\chi }_2),\) where \(\varvec{\chi }_1,\varvec{\chi }_2:\mathbb {R}^{2n-3}\rightarrow \mathbb {R}^n.\) Recall that if g is an integrable function defined on \(\mathcal{U}_\mathcal{A},\) then

$$\begin{aligned} \int _{\mathcal{U}_\mathcal{A}} g(\mathbf{a},\mathbf{b})\, \mathrm{d}\mathcal{H}^{2n-3}(\mathbf{a},\mathbf{b})=\int _{U_\mathcal{A}} g(\varvec{\chi }_1(w),\varvec{\chi }_2(w)) J_{\varvec{\chi }}(w)\, \mathrm{d}w, \end{aligned}$$
(66)

where \(J_{\varvec{\chi }}=\sqrt{\text {det} (\nabla \varvec{\chi }^\top \nabla \varvec{\chi })}\) is the Jacobin of \(\varvec{\chi }.\) Also, if h is an integrable function defined on \(\mathcal{C}_\mathcal{A}\) and \(\phi \) is a parameterization of \(\mathcal{C},\) then

$$\begin{aligned} \int _{\mathcal{C}_\mathcal{A}} h(z)\, \mathrm{d}z=\int _{C_\mathcal{A}} h(\phi (s)) |\phi '(s)|\, \mathrm{d}s, \end{aligned}$$
(67)

where \(C_\mathcal{A}\) is the subset of \(\mathbb {R}\) such that \(\phi (C_\mathcal{A})=\mathcal{C}_{\mathcal{A}}.\)

Set \(A:=C_\mathcal{A}\times U_\mathcal{A}\times \mathcal{I}_\mathcal{A}\) and define \(\tilde{\Phi }:A\rightarrow \Phi (\mathcal{A})\) by

$$\begin{aligned} \tilde{\Phi }(s,w,\xi ):=\Phi (\phi (s) ,\varvec{\chi }_1(w),\varvec{\chi }_2(w),\xi )=(\phi (s)+\xi \varvec{\chi }_1(w), \varvec{\chi }_1(w)) \end{aligned}$$
(68)

and \({\tilde{f}}:A\rightarrow \mathbb {R}\) by

$$\begin{aligned} {\tilde{f}}(s,w,\xi )=f(\phi (s)+\xi \varvec{\chi }_1(w),\varvec{\chi }_1(w)). \end{aligned}$$
(69)

By the Area Formula (see Ambrosio et al. [2, Theorem 2.71]), it follows that

$$\begin{aligned}&\int _{\Phi (\mathcal{A})}\left[ \sum _{(s,w,\xi )\in \Phi ^{-1}(p,\mathbf{u})} \tilde{f}(s,w,\xi )\right] \mathrm{d}\mathcal{H}^{2n-1}(p,\mathbf{u})\nonumber \\&\quad =\int {\tilde{f}}(s,w,\xi )J_{\tilde{\Phi }}(s,w,\xi ) \mathrm{d}\mathcal{H}^{2n-1}(s,w,\xi ). \end{aligned}$$
(70)

Thus, by (66) and (67), the result will follow once it is shown that

$$\begin{aligned} J_{\tilde{\Phi }}=\xi ^{n-2}|\varvec{\chi }_2\cdot \mathbf{t}| |\phi '| J_{\varvec{\chi }}. \end{aligned}$$
(71)

To establish this, first notice that

(72)

and hence, noting that \(|\varvec{\chi }_1|^2=1\) and \(\nabla \varvec{\chi }_1^\top \varvec{\chi }_1=\mathbf{0},\) we have

(73)

Since switching rows or columns of a matrix does not change the absolute value of its determinant, we have

(74)

Recall that the determinant of a block matrix can be computed using

This identity will be used with \(\mathbf{D}=\xi ^2\nabla \varvec{\chi }_1^\top \nabla \varvec{\chi }_1+\nabla \varvec{\chi }_2^\top \nabla \varvec{\chi }_2.\) Let \(P_{\varvec{\chi }_1}\) denote the projection onto the plane orthogonal to the vector \(\varvec{\chi }_1.\) After some computation, using the identity \(|P_{\varvec{\chi }_1} \phi '|^2=|\phi '|^2-(\varvec{\chi }_1\cdot \phi ')^2,\) one finds that

$$\begin{aligned}&|\text {det}(\nabla \tilde{\Phi }^\top \nabla \tilde{\Phi }) |\nonumber \\&\quad =\left| P_{\varvec{\chi }_1}\phi '\right| ^2 \left| \text {det}\left[ \nabla \varvec{\chi }_1^\top \left( \xi ^2\mathbf 1 _n - \frac{\xi ^2}{|P_{\varvec{\chi }_1}\phi '|^2} P_{\varvec{\chi }_1}\phi '\otimes P_{\varvec{\chi }_1}\phi '\right) \nabla \varvec{\chi }_1 + \nabla \varvec{\chi }_2^\top \nabla \varvec{\chi }_2\right] \right| ,\nonumber \\ \end{aligned}$$
(75)

where \(\mathbf 1 _n\) is the identity function on \(\mathbb {R}^n.\) It follows that

(76)

where \(\tilde{\mathbf{t}}=P_{\varvec{\chi }_1}\phi '/|P_{\varvec{\chi }_1}\phi '|.\) To simplify the right-hand side of the previous equation, set \(\mathbf{L}\) equal to the square \(2n\times 2n\) matrix in the previous equation between \(\nabla \varvec{\chi }^\top \) and \(\nabla \varvec{\chi }\) and recall the fact that

$$\begin{aligned} \text {det}(\nabla \varvec{\chi }^\top \mathbf{L}\nabla \varvec{\chi })=J_{\varvec{\chi }}^2\text {det}(\mathbf{I}^\top \mathbf{L}\mathbf{I}), \end{aligned}$$
(77)

where \(\mathbf{I}\) is the natural injection of the range of \(\nabla \varvec{\chi }\) into \(\mathbb {R}^{2n}.\) One can find an orthonormal basis for \(\mathbb {R}^n\) of the form

$$\begin{aligned} (\mathbf{e}_1,\mathbf{e}_2,\ldots ,\mathbf{e}_{n-2},\varvec{\chi }_1,\varvec{\chi }_2) \end{aligned}$$

such that \(\tilde{\mathbf{t}}\) can be written as a linear combination of \(\mathbf{e}_{n-2}\) and \(\varvec{\chi }_2.\) Notice that

$$\begin{aligned} \left\{ (\mathbf{e}_1,\mathbf{0}),\ldots , (\mathbf{e}_{n-2},\mathbf{0}),(\mathbf{0},\mathbf{e}_1),\ldots ,(\mathbf{0},\mathbf{e}_{n-2}),\frac{1}{\sqrt{2}}(\varvec{\chi }_2,-\varvec{\chi }_1)\right\} \end{aligned}$$
(78)

is an orthonormal basis for the tangent space of \(\mathcal{U}_\perp ^2\) at \((\varvec{\chi }_1,\varvec{\chi }_2).\) By writing the matrix

relative to the basis (78), one can compute its determinant, and hence, putting together (76) and (77), we have that

$$\begin{aligned} \left| \text {det}\left( \nabla \varvec{\chi }^\top \mathbf{L}\nabla \varvec{\chi }\right) \right| =\xi ^{2n-4}\left| P_{\varvec{\chi }_1}\phi '\right| ^2 |\varvec{\chi }_2\cdot \tilde{\mathbf{t}}|^2J^2_{\varvec{\chi }}. \end{aligned}$$
(79)

Since \(|P_{\varvec{\chi }_1}\phi '|^2 |\varvec{\chi }_2\cdot \tilde{\mathbf{t}}|^2=|\varvec{\chi }_2 \cdot P_{\varvec{\chi }_1}\mathbf{t}|^2|\phi '|^2=|\varvec{\chi }_2\cdot \mathbf{t}|^2|\phi '|^2,\) this proves (71). \(\square \)

A slight variation on the change of variables formula (65) will be needed. Namely, we will require a change of variables associated with the function

$$\begin{aligned} \varXi :\mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+\times \mathbb {R}^+\rightarrow \mathbb {R}^n\times \mathbb {R}^n\times \mathbb {R}\end{aligned}$$

defined by

$$\begin{aligned} \Xi (z,\mathbf{a},\mathbf{b},\xi ,r):=(z+\xi \mathbf{a},\mathbf{b},r)\quad \text {for all}\ (z,\mathbf{a},\mathbf{b},\xi ,r)\in \mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+\times \mathbb {R}^+.\nonumber \\ \end{aligned}$$
(80)

The function \(\varXi \) allows us to describe disks using points on the curve \(\mathcal{C};\) see Fig. 2. Since \(\varXi \) acts like the identity on the last variable r,  the previous lemma immediately implies that if \(\mathcal{A}\) is a subset of \(\mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+\times \mathbb {R}^+\) and \(f:\varXi (\mathcal{A})\rightarrow \mathbb {R}\) is an integrable function, then

$$\begin{aligned}&\int _{\varXi (\mathcal{A})}\left[ \sum _{(z,\mathbf{a},\mathbf{b},\xi ,r)\in \varXi ^{-1}(p,\mathbf{u},r)} f(p,\mathbf{u},r)\right] \mathrm{d}\mathcal{H}^{2n}(p,\mathbf{u}, r) \nonumber \\&\quad = \int _\mathcal{A}f(z+\xi \mathbf{a},\mathbf{b},r)\xi ^{n-2} |\mathbf{b}\cdot \mathbf{t}(z)| \mathrm{d}\mathcal{H}^{2n}(z, \mathbf{a},\mathbf{b},\xi ,r). \end{aligned}$$
(81)
Fig. 2
figure 2

A depiction of the disk associated with \(\varXi (z,\mathbf{a},\mathbf{b},\xi ,r)\)

Fig. 3
figure 3

A depiction of the disk associated with \(\varPsi (z,\mathbf{a},\mathbf{b},r)\)

Another useful change of variables will be needed. This time, the function, which is closely related to \(\varXi ,\) will allow us to describe all disks whose boundary touches the curve \(\mathcal{C};\) see Fig. 3. Define the function

$$\begin{aligned} \varPsi :\mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+\rightarrow \mathbb {R}^n\times \mathbb {R}^n\times \mathbb {R}\end{aligned}$$

by

$$\begin{aligned} \varPsi (z,\mathbf{a},\mathbf{b},r):=(z+r\mathbf{a},\mathbf{b},r)\quad \text {for all}\ (z,\mathbf{a},\mathbf{b},r)\in \mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+. \end{aligned}$$
(82)

It is possible to show that if \(\mathcal{A}\) is a subset of \(\mathcal{C}\times \mathcal{U}_\perp ^2\times \mathbb {R}^+\) and \(f:\varPsi (\mathcal{A})\rightarrow \mathbb {R}\) is an integrable function, then

$$\begin{aligned}&\int _{\varPsi (\mathcal{A})}\left[ \sum _{(z,\mathbf{a},\mathbf{b},r)\in \varPsi ^{-1}(p,\mathbf{u},r)} f(p,\mathbf{u},r)\right] \mathrm{d}\mathcal{H}^{2n}(p,\mathbf{u}, r)\nonumber \\&\quad = \int _\mathcal{A}f(z+r\mathbf{a},\mathbf{b},r)(2r)^{n-2} \sqrt{(\mathbf{a}\cdot \mathbf{t}(z))^2+(\mathbf{b}\cdot \mathbf{t}(z))^2} \mathrm{d}\mathcal{H}^{2n}(z,\mathbf{a},\mathbf{b},r). \end{aligned}$$
(83)

The proof of this result is similar to that of Lemma 5.1 and thus will not be presented.

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Seguin, B. A Fractional Notion of Length and an Associated Nonlocal Curvature. J Geom Anal 30, 161–181 (2020). https://doi.org/10.1007/s12220-018-00140-9

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