discrete-to-continuumlimitsoflong ... - prashant k. jha

DISCRETE-TO-CONTINUUM LIMITS OF LONG-RANGE ELECTRICAL1INTERACTIONS IN NANOSTRUCTURES∗2

PRASHANT K. JHA†, TIMOTHY BREITZMAN‡, AND KAUSHIK DAYAL§3

Abstract. We consider electrostatic interactions in two classes of nanostructures: (1) helical nanotubes, and (2) thin films4with uniform bending (i.e., constant mean curvature). Starting from the atomic scale with a discrete distribution of dipoles,5we obtain the continuum limit of the electrostatic energy; the continuum energy depends on the geometric parameters that6define the nanostructure, such as the pitch and twist of the helical nanotubes and the curvature of the thin film. We find that7the limiting energy is local in nature. This can be rationalized by noticing that the decay of the dipole kernel is sufficiently8fast when the lattice sums run over one and two dimensions, and is also consistent with prior work on dimension reduction of9continuum micromagnetic bodies to the thin film limit.10

1. Introduction. Electrical and magnetic interactions are long-range; that is, a charge or dipole11interacts with all the other charges and dipoles in the system, and the interactions cannot be truncated12because the decay with distance is slow [Tou56, Bro63, JM94,MD14]. We consider such electrostatic13interactions in nanostructures, specifically helical geometries and thin films with uniform bending.14These geometries are ubiquitous in nanotechnology; while not periodic, their structure has significant15symmetry that we exploit in this paper, using the framework of Objective Structures [Jam06]. We16exploit this symmetry to adapt periodic calculations of the continuum energy to the setting of these17nanostructures. Specifically, starting from a discrete atomic-scale description of the electrostatic18energy, we find the limit energy when the discrete lengthscale of the nanostructures goes to zero.19

For simplicity, we assume in this paper that the charge density can be approximated as composed20of discrete dipoles. The electrostatic energy of such a system is the sum of all pairwise dipole-21dipole interactions. Unlike short-range bonded atomic interactions that typically scale as r−6 with22distance r, the dipole-dipole interactions decay slowly with distance as r−3. Consequently, we23cannot simply truncate after a few neighbors, and naive truncation can lead to qualitatively incorrect24results in numerical calculations [MD14, GD20a, GD20b]. While we use the setting of discrete25electrical dipoles, the setting of magnetic dipoles has an identical mathematical structure and physical26interpretation [Bro63, JM94, MS02, SS09], and we borrow key ideas from that literature. Further,27while discrete dipoles provide the simplest setting to illustrate the physics, it can be extended to28the more realistic and general setting of a charge density field following ideas from [Xia05]. A29key physical distinction between the electrical and magnetic situations is the possibility of electrical30monopoles that does not exist for magnetic case, but we examine this elsewhere [SWB+21] and31assume here that there are no free charges.32

We turn to the question of dealing with the non-periodic geometry of the nanostructures. While33neither helices nor thin films with curvature are periodic, the framework of Objective Structures34(OS) introduced in [Jam06] provides a powerful approach to deal with such geometries. In brief,35OS provides a group-theoretic description of these nanostructures that enables a parallel to be made36with periodic lattices. This parallel to periodic lattices has enabled the adaptation of various methods37developed for lattices to the setting of helices and thin-films, e.g. [DJ07, HTJ12, ADE13a, ADE13b].38Our strategy in this work is to use the OS framework to adapt continuum limit calculations from the39

∗Submitted to the editors December 16, 2020.Funding: We acknowledge financial support from NSF (1635407), ARO (W911NF-17-1-0084, MURI W911NF-19-

1-0245), ONR (N00014-18-1-2528), BSF (2018183), AFOSR (MURI FA9550-18-1-0095), and the AFRL Visiting Facultyprogram.†Oden Institute for Computational Engineering and Sciences, The University of Texas at Austin ([email protected]).‡Air Force Research Laboratory ([email protected]).§Center for Nonlinear Analysis, Department of Mathematical Sciences, Carnegie Mellon University; Department of

Materials Science and Engineering, Carnegie Mellon University; Department of Civil and Environmental Engineering,Carnegie Mellon University ([email protected]).

1

This manuscript is for review purposes only.

mailto:[email protected]



setting of periodic lattices to the setting of nanostructures.40Our work is focused on obtaining discrete-to-continuum limits of the energy. This multiscale41

approach has proven very powerful in enabling the systematic reduction of the very large number of42degrees of freedom associated to the discrete problem to a much more tractable continuum problem.43This overall idea has played an important role in developing models, often in conjunction with44variational tools such as Γ -convergence, both for bulk crystals [BLBL02, BLBL07, Sch09] as well45as for thin films [Sch06]. Further, these ideas have played a role in the development of numerical46multiscale atomistic methods such as the quasicontinuum method [TOP96, MT02, TM11, KO01,47LLO12, DLO10].48

However, all the work in the previous paragraph is to restricted to the setting of short-range49bonded atomic interactions. In the context of electrical and magnetic interactions, the calculation of50continuum limit energies in the context of discrete-to-continuum has been examined both formally51and rigorously using a discrete dipoles on a periodic 3-d lattice [Tou56, Bro63, JM94, MS02, SS09].52Further, this has been examined formally for periodic charge distributions, also in 3-d, [MD14].53All of these works show that the continuum limit energy consists of a local part and a nonlocal54part. In contrast, in this work, we consider topologically low-dimensional structures: a 1-d helical55nanotube and a 2-d thin filmwith constant bending curvature. In the limit that the discrete lengthscale56characterizing the nanotube and thin film goes to 0, we find that the limit continuum energy is entirely57local.58

The absence of nonlocality in the limit can be rationalized by observing that the decay of the59interactions as r−3 is sufficiently fast that we obtain a local limit if summed over a (topologically)601-d or 2-d object. We highlight a complementary body of work that applies dimension reduction61techniques to go from a 3-d continuum to a 2-d or 1-d continuum. In the context of electrical and62magnetic interactions, [GJ97] and subsequent works [Car01, KS05, KSZ15] (for thin films) and63[GH15, CH15] (for thin wires) find, as we do, that the limit energy is not nonlocal.64

The techniques employed in this work are broadly based on the rigorous results provided in65[JM94] on the continuum limit of magnetic dipole interactions on a 3-d lattice, with appropriate66generalizations and modifications for our setting. The overall strategy of [JM94] is as follows. First,67the operator that associates the discrete dipole lattice to the generated electric field is shown to be68bounded for smooth test functions; next, the pointwise limit of the action of the operator on smooth69test functions is obtained; and, finally, using the boundedness of the operator and the density of the70test functions, the limit of the energy density is obtained. For the helical and thin film nanostructures71considered in this work, we adapt this strategy to account for the fact that the lattice sites and dipoles72are not related by a translation transformation, but by a more general isometric transformation.73

The key results of this work are as follows. First, the limit energy is rigorously derived and found74to be local. Second, the limiting energy density depends on the macroscopic geometric parameters,75such as the pitch, radius and so on for the helical nanotube, and on the stretch and curvature for76the thin film. These parameters can be related to macroscopic measures of deformation, and link77the macroscopic deformation to the small-scale structure. Third, while the limiting energy is local,78there are energetic contributions from both the normal and the tangential components of the dipole.79This is in contrast to the result obtained by dimension reduction, and is due to the fact that we start80with a single unit cell thickness in the normal direction and take the limit along the length (for the81helix) or in the plane (for the thin film). In contrast, in dimension reduction approaches, there is no82discreteness at all, and the limit is related to the ratio of the dimensions in and out of plane.83

Organization. In section 2, we discuss prior work, primarily on dimension reduction from a843-d continuum to a 2-d continuum, and highlight the local nature of the limiting energy. We then85discuss heuristically the scaling of electrostatic interactions that lead to this locality generically for86topologically low-dimensional nanostructures. In section 3, we present the main results for helical87nanotubes and thin films with constant bending curvature. We prove various claims in section 4. In88section 5, we summarize the results.89

2


Notation. We denote the real line and set of integers by R and Z respectively; Rd,Zd denote90these in dimension d = 1, 2, 3. For any c, c1, c2 ∈ R, cZd denotes the set cz; z ∈ Zd and c1Z×c2Z91denotes the set (c1z1, c2z2); z1, z2 ∈ Z. L and U denote the set of lattice sites and the lattice unit92cell respectively; Lλ and Uλ denote these in the lattice scaled by λ, with L1, U1 denoting Lλ, Uλ93for λ = 1. We use x = (x1, x2, x3) ∈ R3 to denote the point in space with components xi in the94orthonormal basis e1, e2, e3 for R3. We follow the standard notation wherein scalars are denoted95by lowercase letters, vectors by bold lowercase letters, and second order tensors by bold uppercase96

letters. |x| =

√√√√ n∑i=1

x2i denotes the Euclidean norm of vector x ∈ Rn; |A| =

√A : A denotes the97

norm of tensorA; andA : B = AijBij denotes the inner product of the two tensors. For any vector98a ∈ Rn and tensor A, we have |Aa| ≤ |A||a|. We use |Ω| to denote the Lebesgue measure of the99setΩ ∈ Rn. We use χA = χA(x) to denote the indicator function of set A ⊂ Rd. We use L2(A,B)100to denote the space of Lebesgue square-integrable functions u : A ⊂ Rn → B ⊂ Rm; (u, v)L2(A,B)101

for the inner product of functions u, v ∈ L2(A,B); and ||u||L2(A,B) the L2 norm of u ∈ L2(A,B).102

When there is no ambiguity, we will suppress L2(A,B) and write (u, v) and ||u||. C∞0 (Rn,Rm)103denotes the space of infinitely differentiable test functions u : Rn → Rm with compact support in104Rn. L(V,W ) is the space of bounded linear maps T : V →W where V,W are Hilbert spaces. The105norm of the map T ∈ L(V,W ) is denoted by ||T ||L(V,W ), and is given by the expression:106

(1.1) ||T ||L(V,W ) = sup||f ||V 6=0

||Tf ||W||f ||V

.107

We use uλ −−−→λ→0

u to denote the strong convergence of uλ ∈ V to u ∈ V as λ → 0, i.e.,108

||uλ − u||V → 0 as λ→ 0.109

2. Prior Results on Dimension Reduction, and Dipole Interaction Scalings. We briefly110revisit the results of [GJ97] and [CH15]. Respectively, they performed dimension reduction from the1113-d continuum to the 2-d thin film and 1-d thin wire limits for the magnetostatic energy.112

Consider a material domainΩh = S× [0, h], where S ⊂ R2 is a 2-d domain in the plane spanned113by (e1, e2), and h > 0 is the material thickness in the normal direction e3. Suppose d : Ωh → R3,114with d = 0 on R3\Ωh, is the dipole field in the material. The electrostatic energy density is given by115

eh(d) =1

|Ωh|

∫Ωh

1

2∇φ(x) · d dx,116

where |Ωh| is the volume ofΩh, and φ is the electric potential that satisfies the electrostatic equation:117

div(−∇φ+ d) = 0 on R3, |∇φ(x)| → 0 as |x| → ∞118

with |d| = d. Let Ω1 = S × [0, 1], and y(x) = (x1, x2, x3/h) ∈ Ω1 for x ∈ Ωh is the map from119Ωh to Ω1. For fixed h > 0, consider the dipole field dh : Ωh → R3 and dh : Ω1 → R3 such that120

dh(y(x)) = dh(x), ∀x ∈ Ωh.121

Let dh be the sequence of dipole field for h > 0, and dh is defined as above. Assume that dipole122field dh is such that, first, dh = 0 onR3\Ω1, and second, it converges to d0 in L2(R3); then the limit123of the energy density eh = eh(dh) is [GJ97]:124

eh(dh)→ e0(d0) =1

2|Ω1|

∫Ω1

|d03|2 dx125

3


That is, the limiting energy e0 is local, and only the normal component of the dipole moment appears126in the expression.127

Next, consider a thin straight wire with axis along e1, denoted by Ωh = (−1, 1) × B2(0, h),128whereB2(0, h) is a ball of size h centered at 0 in the plane spanned by (e2, e3). The limiting energy129density is [CH15]:130

1

2|Ω1|

∫Ω1

(|d02 |2 + |d03 |2

)dx,131

whereΩ1 = (−1, 1)×B2(0, 1) is the rescaled domain ofΩh, and d0 is the limiting field. We notice132that the limiting energy is again local, and only the components of the dipole moment perpendicular133to the wire appear.134

(a) (b)

(c) (d)

Fig. 2.1: A schematic of the unit cell with a dipole (a), and generic 1-d, 2-d, 3-d periodic lattices (b,c, d).

The absence of nonlocality in the limiting energy in the results above, as well as in our results135in section 3 below, can be physically understood through the fact that these structures are 1-d or 2-d136topologically. To see this, we consider a system of discrete dipoles associated to the uniform 1-d, 2-d,137and 3-d periodic lattices with the unit cell of size 1 (Figure 2.1). The energy of a lattice of dipoles is138given by [Bro63, JM94]:139

(2.1) E = −1

2

∑i

∑j,j 6=i

di ·K(xj − xi)dj =∑i

|Ui|

− 1

|Ui|1

2

∑j,j 6=i

di ·K(xj − xi)dj

,140

where the sum is over the cells in the lattice and the term inside square bracket denotes the energy141density of a cell i. di denotes the dipole in cell i, xi denotes the coordinate of lattice site i, andK is142the dipole field kernel defined as:143

(2.2) K(x) = − 1

4π|x|3

(I − 3

x

|x|⊗ x

|x|

), x 6= 0.144

We use these expressions to heuristically understand the scaling of the energy for systems with145different topological dimensions. For simplicity, we assume below that the volume of the unit cell146and the magnitude of the dipole are both 1, i.e., |Ui| = 1 and |di| = 1 for each i, and some constant147factors are neglected.148

4


Remark 2.1 (1-d lattice). We can estimate an upper bound on the energy density e of a typical149unit cell as follows:150

e ≤∞∑r=1

1

r3× |d| × (number of dipoles at r) ≤

∞∑r=1

1

r3× 1× 1 =

∞∑r=1

1

r3.151

We use that the total dipole moment at a distance r from a given unit cell is, at most, that of another152dipole in the unit cell at a distance r. This sum is well-behaved and bounded.153

Remark 2.2 (2-d lattice). As in the 1-d setting, we first bound the net dipole at a distance r from154a given unit cell. Since the structure is a 2-d lattice, the number of unit cells at a distance r is of155order 2πr. Therefore, an upper bound on the energy density is:156

e ≤∞∑r=1

1


∞∑r=1

1

r3× 1× 2πr = 2π

∞∑r=1

1

r2.157

This sum is also well-behaved and bounded.158

Remark 2.3 (3-d lattice). Following the argument of the 2-d lattice, we now have that the net159dipole at a distance r from a given unit cell is, at most, of the order 4πr2. Therefore, an upper bound160on the energy density is:161

e ≤∞∑r=1

1


∞∑r=1

1

r3× 1× 4πr2 = 4π

∞∑r=1

1

r.162

This sum is divergent. However, through a more careful analysis that accounts for the signs – not163just the magnitudes – of the dipole interactions, the energy density can be shown to be conditionally164convergent [Tou56, JM94].165

When the lattice sum is bounded and converges unconditionally, it is possible to truncate after166a finite distance and obtain sufficient numerical accuracy. When the lattice sum is conditionally167convergent, that can be physically related to nonlocality; specifically, the slow convergence does not168allow for truncation, and the far-field values play an important role. [MD14] discusses this from a169physical perspective.170

3. Results on Continuum Limits of the Electrostatic Energy. We consider two classes of171nanostructures: helical nanotubes and thin films, the latter allowing for a constant bending curvature172(i.e., nonzero constant mean curvature and zero Gauss curvature), and obtain the corresponding con-173tinuum limit electrostatic energy. In both cases, we start with discrete dipoles, where the discreteness174is parametrized by the scale λ > 0, and examine the limit λ → 0. We show that the dipole-dipole175interaction energy density – per unit cross-sectional area in the case of nanotubes, and per unit176thickness in the case of films – converges to a local energy density in the limit.177

3.1. Helical Nanotube. We consider a discrete helix with axis e3 characterized by the angle θ178and length δ. Suppose x0 ∈ R3 is a point on the helix. Then, the other points on the helix are related179by an isometric transformation of x0. Let s ∈ R be the parametric coordinate of a point on the helix.180Then, the map x : R → R3 that takes a point in the parametric space to a unique point on the helix181can be expressed as182

(3.1) x(s) = Q(sθ)x0 + sδe3.183

5


HereQ(θ) is the rotational tensor represented by the matrix in the orthonormal basis e1, e2, e3 as:184

Q(θ) :=

cos(θ) − sin(θ) 0sin(θ) cos(θ) 0

0 0 1

.(3.2)185

186

We notice that these definitions imply that the pitch of the helix is 2πδ/θ.187Without loss of generality, we assume x0 = e1. The tangent vector to the helix at s is given by188

(3.3) t(s) =dx(s)

ds= θQ′(sθ)e1 + δe3.189

Let t(s) = t(s)/√θ2 + δ2 denote the unit tangent vector. We define the second order projection190

tensors P || = P ||(s) and P⊥ = P⊥(s), for s ∈ R, as follows191

(3.4) P ||(s) = t(s)⊗t(s), P⊥(s) = I − P ||(s).192

For any vector a and any s ∈ R, we have193

(3.5) a = P ||(s)a + P⊥(s)a, with P ||(s)a · P⊥(s)a = 0.194

3.1.1. Lattice Geometry and Dipole Moment. Let L = Z denote the set of parametric coordi-195nates of the points on helix. We consider a discrete system of dipole moments d : L → R3 associated196to the points on the helix given by L, see Figure 3.1. The magnitudes of the dipoles at the lattice sites197are equal, but they are oriented differently; in particular, the orientations of dipoles at lattice sites198follow the relation:199

(3.6) d(s+ 1) = Q(θ)d(s), s ∈ L200

We associate a unit cell to each lattice site. LetU(s) = [s, s+1) denote the unit cell in the parametricspace at the site s, for s ∈ L. Let S(r), r ∈ R, be given by

S(r) =x; (x− x(r)) · t(r) = 0, |x− x(r)|2 < R2

.

Note that |S(r)| = |S(0)| = πR2. The unit cell in real space is defined by U(s) = x ∈ S(r); r ∈ U(s).201

We take, without loss of generality,R2 = 1/(π√θ2 + δ2) so that |U(s)| = area(S)×length(x(r); r ∈202

U(s)) = πR2√θ2 + δ2 = 1.203

We now consider the setting in which the cells are of size λ > 0 so that as λ → 0 the density204of cells in the helix increases. For λ > 0, suppose Lλ = λZ denotes the parametric coordinates205of the sites in a scaled lattice, and dλ : Lλ → R3 denotes the corresponding system of dipole206moments. Associated to s ∈ Lλ, let Uλ(s) = [s, s + λ) denote the cell in the parametric space.207The 3-d cell is given by Uλ(s) = x ∈ Sλ(r); r ∈ Uλ(s), where Sλ(r) = x; (x − x(r)) ·208t(r) = 0, |x − x(r)|2 < λ2R2 is the scaled cross-section. Note that area(Sλ(r)) = πλ2R2 and209

|Uλ(s)| = πλ2R2 × λ√θ2 + δ2 = λ3. Let dλ : R→ R3 be the piecewise constant extension of dλ210

given by211

(3.7) dλ(s) =dλ(i)

|Uλ(s)|=

dλ(i)

λ3, ∀s ∈ Uλ(i), ∀i ∈ Lλ.212

To compute the limit of the dipole-dipole interaction energy as λ→ 0, we assume that dipole moment213density field dλ converges to some field f ∈ L2(R,R3) in the L2 norm. As in [JM94], instead of214

6


working with dλ, as defined above, we could assume that the dipole moment dλ(i), for i ∈ Lλ, is215due to the background dipole moment density field fλ ∈ L2(R,R3) such that216

(3.8) dλ(i) =√θ2 + δ2

∫Uλ(i)

∫Sλ(r)

fλ(r) dSλ(r) dr = λ2

∫Uλ(i)

fλ(r) dr,217

where dSλ(r) is the area measure for surface Sλ. We assumed that the background field is uniform218

in Sλ(r) for all r ∈ R and used R2 = 1/(π√θ2 + δ2). The existence of one such background field219

fλ is evident: we can define fλ = dλ. The physical dimension of fλ is dipole moment per unit220volume. We have the following lemma that relates the convergence of the background dipole moment221field and the piecewise constant extension.222

Lemma 3.1. Let fλ, λ > 0, be the sequence of L2(R,R3) functions and let f ∈ L2(R,R3) such223that fλ → f in L2(R,R3). Let dλ : Lλ → R3 be given by (3.8) and let dλ be a piecewise constant224L2 extension of dλ given by (3.7). Then, dλ → f in L2(R,R3).225

On the other hand, if dλ : Lλ → R3 is such that dλ → f in L2(R,R3), then there exists a226background field fλ ∈ L2(R,R3) such that dλ is given by (3.8).227

The proof is similar to the proof of Theorem 4.1 from [JM94].228

Remark 3.1. Since the discrete dipole field dλ has helical symmetry, from (3.8) we can see that229f will also have helical symmetry.230

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x(s)

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e2

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e3

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o


(a) (b)

Fig. 3.1: Discrete dipole moments (red arrows) lying on the helix. (a) and (b) show the view in(e1, e3) and (e1, e2) planes respectively. The dipole moments corresponding to different sites arerelated by (3.6). For the parametric coordinate s, x(s) gives the coordinate of the point on the helix.

3.1.2. Electrostatic Energy. For λ > 0, the energy associated to the system of dipole moments231dλ can be expressed as [Bro63, JM94]:232

Eλ = −1

2

∑s,s′∈Lλ,s 6=s′

dλ(s) ·K(x(s′)− x(s))dλ(s′) = |Sλ|eλ,233

7


where eλ is the energy per unit area given by234

(3.9) eλ = − 1

2|Sλ|∑

s,s′∈Lλ,s6=s′

dλ(s) ·K(x(s′)− x(s))dλ(s′).235

Substituting (3.8) above and proceeding similar to Section 6 of [JM94], eλ can be written as236

(3.10) eλ = (fλ, Tλfλ)L2(R,R3) ,237

where Tλ : L2(R,R3)→ L2(R,R3) is the map given by238

(3.11) (Tλf)(s) = λ2

∫RKλ(s′, s)f(s′) ds′239

and Kλ(s′, s), for s, s′ ∈ R, is the discrete dipole field kernel given by240

(3.12) Kλ(s′, s) =∑

u,v∈Lλ,u6=v

χUλ(v)(s′)K(x(v)− x(u))χUλ(u)(s).241

Scaling Property of Kλ. For any a, b ∈ R, we have242

x(λa)− x(λb) = Q(λaθ)e1 + δλae3 −Q(λbθ)e1 − δλbe3

= λ

x(a)− x(b) +

[Q(λaθ)−Q(λbθ)− (λQ(aθ)− λQ(bθ))

λ

]︸︷︷︸

=:Aλ(a,b)

e1

.(3.13)243

Using the relation above, it is easy to show244

Kλ(s′, s) =1

λ3

∑u,v∈L1,u 6=v

χU1(v)(s′/λ)K(x(v)− x(u) + Aλ(v, u)e1)χU1(u)(s/λ),245

where we recall that U1(u) = [u, u+ 1), u ∈ L1, is the lattice cell in the parametric space for λ = 1.246We define a discrete kernel K1,λ(s, s′), for s, s′ ∈ R, as follows247

(3.14) K1,λ(s′, s) =∑

u,v∈L1,u6=v

χU1(v)(s′)K(x(v)− x(u) + Aλ(v, u)e1)χU1(u)(s).248

We then have249

Kλ(s′, s) =1

λ3K1,λ(s′/λ, s/λ).250

3.1.3. Limit of Electrostatics Energy. In this section, we obtain the limit of the energy per251unit surface area eλ as λ→ 0 assuming that the background dipole field density (or equivalently the252dipole moment density dλ) fλ converges to some density field f in L2. The idea is to first show that253the map Tλ in (3.11) is bounded and obtain its limit. With that, the limit of eλ follows.254

8


Limit of Discrete Electric Field. Let T1,λ be the map with kernel K1,λ. For any function255f ∈ L2(R,R3), we have256

(3.15) (T1,λf)(s) =

∫RK1,λ(s′, s)f(s′) ds′.257

We have the following main result on the map Tλ.258

Proposition 3.2. The map T1,λ and Tλ are bounded in L2(R,R3) for all λ > 0 and satisfy259

(3.16) ‖Tλ‖L(L2,L2) = ‖T1,λ‖L(L2,L2) .260

Further, for f ∈ C∞0 (R,R3),261

(Tλf)(s) −−−→λ→0

−h0(I − 3P ||(s))f(s) = −h0(P⊥(s)− 2P ||(s))f(s)262

pointwise, where P⊥(s) and P ||(s) are projection tensors that project onto the normal plane and263the tangent line to the helix respectively (see (3.4)). h0 is a constant given by264

(3.17) h0 =∑v∈Z,v 6=0

1

4π|v|3(θ2 + δ2)3/2.265

We provide the proof of Proposition 3.2 in subsection 4.1.266Limit of Energy.267

Theorem 3.3. Let fλ ∈ L2(R,R3) be a sequence of functions for λ > 0 with f ∈ L2(R,R3)268such that fλ −−−→

λ→0f in L2. Let the system of dipole moments dλ : Lλ → R3 be given by (3.8). Then269

eλ −−−→λ→0

1

2h0

[||P⊥f ||2L2(R,R3) − 2||P ||f ||2L2(R,R3)

],270

271

where h0 is a constant defined in (3.17).272

Proof. Since Tλ is bounded and fλ → f , we have273

limλ→0

(Tλfλ) = limλ→0

(Tλf) + limλ→0

(Tλ(fλ − f)) = limλ→0

(Tλf).274

We only need to analyze Tλf in the limit for f ∈ L2(R,R3). Let fk ∈ C∞0 (R,R3) be a sequence275of functions such that fk → f . Using Proposition 3.2, we have276

limλ→0

(Tλf) = limk→∞

limλ→0

(Tλfk) + lim

k→∞limλ→0

(Tλ(f − fk)) = limk→∞

limλ→0

(Tλfk)277

= limk→∞

(H0f

k)

= H0f .(3.18)278279

Using the expression in (3.10) for eλ, we have280

eλ = −1

2(fλ, Tλfλ)L2(R,R3) = −1

2

[(fλ − f , Tλfλ)L2(R,R3) + (f , Tλfλ)L2(R,R3)

]281

= −1

2

[(fλ − f , Tλfλ)L2(R,R3) + (f , Tλf)L2(R,R3) + (f , Tλ(fλ − f))L2(R,R3)

].282

283

The first and third terms are zero in the limit. Taking the limit of the remaining term and using (3.18),284we have285

limλ→0

eλ = limλ→0−1

2(fλ, Tλfλ)L2(R,R3) =

1

2h0

[||P⊥f ||2L2(R,R3) − 2||P ||f ||2L2(R,R3)

].286

This completes the proof.287

9


Remark 3.2. The limiting energy only comprises of a local self-field energy. In the limit, any288point on the helix sees the uniform 1-d system of dipole moments along the tangent line. Further, we289see that both the normal components and the tangential component of the dipole moment contribute290to the energy and electric field. This is in contrast to [CH15] where the thin wire limit of the291magnetostatic energy has contribution only from the normal components.292

3.2. Nanofilm with Constant Bending Curvature. Let S = (−θ, θ) × R be the parametric293space for a surface with a constant bending curvature κ. The map that takes a point in the parametric294space to a unique point on the film is given by295

(3.19) x(s1, s2) = RQ(s1)e1 + s2δe3,296

whereR = 1/κ is the inverse of curvature, θ > 0 is the angular size of the film, and δ is the spacing297in the flat direction. κ, δ, θ are fixed parameters for a given film. Here, Q(θ) is the rotational tensor298with the axis e3, see definition (3.2). The tangent vectors at s := s1, s2 ∈ S are299

(3.20) t1(s) =dx

ds1= RQ′(s1)e1, t2(s) =

dx

ds2= δe3300

and the normal vector is301

(3.21) n(s) = Q(s1)e1.302

3.2.1. Lattice Geometry and Dipole Moment. We consider a lattice embedded on the film x.303We assume that the lattice is one lattice cell thick in the direction n normal to the film. Suppose304L ⊂ S is the set of parametric coordinates of the discrete lattice sites. Let L and the lattice cell U305(in the parametric space S) be given by306

L = s = (s1, s2) ∈ S; s1 = iθl, s2 = j, i, j ∈ Z = (−θ, θ) ∩ θlZ× Z307

U(s) = [s1, s1 + θl)× [s2, s2 + 1), ∀s ∈ L.(3.22)308309

Here, θl is the angular width of the lattice cell. We assume that θl is such that the set of sites310in the angular direction, (−θ, θ) ∩ θlZ, is not empty, and in fact is sufficiently large so that the311continuum limit approximation of the energy density is justified. We assume that the lattice has unit312thickness in the normal direction, and suppose that the film given by x passes through the center of313the lattice in the normal direction. Then, the unit cell for a given site s ∈ L is U(s) = x; x =314x(s) + tn(s), s ∈ U(s), t ∈ (−1/2, 1/2). On the lattice L, we define a discrete system of dipole315moments d : L → R3. Similar to the case of the helical nanotube, the lattice cells in real space are316related by an isometric transformation, so the magnitudes of the dipoles at the lattice sites are equal,317but they are oriented differently. In particular, we have318

(3.23) d(s + r) = Q(r1)d(s), s ∈ L,319

where r = (r1, r2) ∈ L such that r + s ∈ L (i.e. all the translations within L). We see that the320dipole orientation depends only on the angular (first) parameter and is invariant with respect to the321second parameter.322

We next consider the setting when the lattice is scaled by λ > 0. The scaled lattice Lλ and the323associated lattice cell Uλ are defined by the natural scaling of L and U as follows324

Lλ = s ∈ S; s1 = iθlλ, s2 = jλ, i, j ∈ Z = (−θ, θ) ∩ λθlZ× λZ325

Uλ(s) = [s1, s1 + λθl)× [s2, s2 + λ), ∀s ∈ Lλ.(3.24)326327

After scaling, the thickness of the lattice cell in the normal direction is λ and the unit cell for s ∈ Lλ328is x ∈ R3; x = x(s) + tn(s), s ∈ Uλ(s), t ∈ (−λ/2, λ/2). We can show that the unit cell in329

10


e2


e3

<latexit sha1_base64="6BvVCAoUce8DBBsbm1OV7mvJ9bg=">AAAcuXicvZlbc9y2FYA3SS/p9uY0j9EDXVkzacf2rNR40mlfIusSRZZiWfdYlHcAErukxZtBcK0Nhz+ir23/WP9NzwG4SxKHcuuOp6tZLoDvHBxcDw4onkVhrkajf3308Sc/+enPfv7pL4a//NWvf/Pbe5/97jxPC+mJMy+NUnnJWS6iMBFnKlSRuMykYDGPxAW/2UJ+MRMyD9PkVM0zcR2zaRJOQo8pKLp44HIx/tOD8b3V0eOR/jg0sV4nVgf152j82Rf/dP3UK2KRKC9ieX61PsrUdcmkCr1IVEO3yEXGvBs2FVeQTFgs8utSt7dy1qDEdyaphG+iHF3a1ihZnOfzmINkzFSQ2wwL+9hVoSZ/vi7DJCuUSDxjaFJEjkod7Lzjh1J4KppDgnkyhLY6XsAk8xQM0XDoSpGIt14axyzx/+hOWBxGc19MWBGp0s0ndbJyhsO1ttnT9esSe4I2u8RqUFevDEQ0E6qrUrIZS1RnAEsv5jKcBlDaKcb+qzSN8ocOj7saStwq6EfWnQh/FmZ5PRW3Zi6GUGXT59KNYOn4LKsWqVele9/F2oqszGRaVVWfvN8r74spyncVRJaD8IxJSIRRmlj1zbIgNBwTXbazdViV5Y4b5NAhUT56PHoi4mqrya9j/rDCAW3rXZycgN6FJXfSymPW7pnR0n0pMW3zw+0d4IdWrdvdWndsrdPNXdA6tbQ2u1q7tpbwYhgUGJGxGQULKzZZYjRg4dgXS4yttrFgSVUuJk3nLAlINwKYsUc4jfxGADNWBV6KFoI0vdHrmEmZviUi0MW1oQMfFxe2FFFpEhHLYB5iM0BfxUXlRlhTRWu0zfqtVpE2+QrgwwYrm+ddntv8pMtPiHE0MAHnUjfE9x2FjWg7GWgeNjGMfGGp4xbM0J2yyCaoko3t9qazCh9C4lFAWNaCZWaPRVpvu6UErhYyy0o2wwlp0hU/nLXGGzL2MiX6bbpZmfn2WFRu2nCrBbdsuN2C2zbcb8F9srAZj1jEqjo1Rud1oP0YsSEiZSR1CiQbue4oHLQMHlC818J7tpXjGnJeHttsp2FkC+82jOz+7xv2PbG3d9JMCWYsLsViB+nTosS8JcKjQnRkdIG9viRLpl2xusj2AvOlLzJiusASmkohulK65P7Xo/swid4NWTx76Mib5dMef3tAttuix1V7XdnbMAg5NILHi81i09nC6ZKNAKXJRM1r7pqMtWxBex9WWR5OY+aQGvYzjTMCDsKsmVHMGK4CkUoRwwTGTN5U5bH57TCI4oo4w7CwKjebdEcmEmCnKg/0T4fUv1V5Wic6FCKnMAlN3dtNuiMD4UWW5rXQUSvTkfJSmUYRkzB6W8ukFZOkyVQ7wfIBTs+jCnY1/pH0AwzIHn2oz9pwzYFmOHU7cufLI8nyAKK5P3xgK2kmJFOpzB2w4+SZ8OCUcDBihmBwCB4KQnIpDmHpPq8lIS4A5wtfcOm9GH0zfO9QhnB5WpX6eYcEhA1wXurn3XXEuo74LgkogLWln3dIzNKoKvFxpw3cdXj03CEA2wkWl37e2QjYcLfYDP1rOTt0N4nZ9X8hnpC/y31z/i5HzPnLhr6k9FlDn1H6rqPBn+voXOMJeGIVQvhPqvhRQGy/lBrZnPG8uoJLjhuJiXLh0iKVs7ruuDr2ciXm7ZM1lXFb5dxSOe9XuayuNq7LdyqNy9UN0kkR1a5Yn+N0DHxwLn4t0j8EaNHDAEWbLu2DEBugsWkKGcDbEA8OrB5ShHrBEmOS8KzhWR+HoLfGkCL0KF9qY5JwoZiJ4FECMj2jd9sKgdCJaFfil7c0pBbRvF92TmTh3ORPa7N4hj7tldhuSdBTFkqPsHNIj/ro7oLu9tG9BaVHPZTuLOhOH91c0M2eCdXK2O6mAvumC9G7L6Re0PU1oI7nYUlXrQxdzW1N/71UVaq6Nn2t4fdZacv6/0E4F15/X15t2E161dcoou/Xin6vht+W9/8bg3CQOzPhgQt33hRw5ELkIHJr0piZMbL+eR3HcTrNBngE1L6ERsPCABLY8snEkAkhUwOmBAQGBASEBpCgk7824DUBNwaQ0JjXfjOiLic2JCakPv9IWM5TA1Lq3gwg107+xoA3BEgDyCWR5wbQE0wZoAgoDCgImBlAQmheu8q31MkbcEvA3IA5PVcN+JE2N0BHbDyfSZNXcXzTcHIZXjjVpwRsGUBuyAsnS27HvPZiJGzguwaQ6yT/1oBvCdgzgFxq+XcGfEfAvgHkWq7DHQA01Dkw4ICAQwPIqzEddQGgEddzA57T49SAIwJeGPCCgGMDaMx3YgC5V/NTA04JODPgjIBzA84JuDDggoBLAy4J+MGAHwh4aQCNP4X2OGJsv5HlRcBw48FT7zSCOdM7mcl+bF581S/AekVmbQN0q87aBnqwbQBF4KSYRMWto4SMu/JYDJGhCUHhAry8f8MV3VmeOM8qzKsFrJxHGJGakMgp4XByXM9PlQOeclw+e+gsxHzitNBebe3DVMh8PANNrP4+PUAjtT0Yo/c0V1v7Hyv8sDdjAffhdPJ/uYYP9VuWt6GvAudJmEA+FyoSyVQFOjoJE18ksHZH+C6kO25Fws3dqAAZySW+2IYBs1+azWopfCvbEhrecXMVEkIMfAyH43ur6/Z/72jifOPx+lePn7zYWP3maf2fvU8HXwx+P/hysD74evDNYG9wNDgbeIObwd8Gfx/8Y+WvK2wlWHltRD/+qNb5fND5rOT/BhDc6Sc=</latexit>

o


e2


o


(a) (b)

Fig. 3.2: Discrete dipole moments on a nanofilm with uniform bending curvature. (a) and (b) showthe view from different perspectives.

the scaled lattice has volume λ3Rθl. Suppose dλ : S → R denotes the piecewise constant extension330of dλ and is given by331

(3.25) dλ(s) =dλ(a)

λ3Rθl, ∀s ∈ Uλ(a), ∀a ∈ Lλ.332

We are interested in the limit of the energy when dλ converges to f in L2(S,R3). As in the case of333the helix and following [JM94], we suppose that there exists a background dipole moment density334field fλ ∈ L2(S,R3) such that the dipole moment at site s ∈ Lλ is given by335

(3.26) dλ(s) =

∫ λ/2

−λ/2

[∫Uλ(s)

fλ(t)R dt1 dt2

]dt3 = Rλ

∫Uλ(s)

fλ(t) dt,336

where dt = dt1 dt2 is the area measure (note that dt does not include R). The existence of one337such background field fλ is evident: We can define fλ = dλ. Similar to the case of the helix, we338have the following lemma that relates the convergence of the background dipole moment field and339the piecewise constant extension.340

Lemma 3.4. Let fλ, λ > 0, be a sequence of L2(S,R3) functions and let f ∈ L2(S,R3) be341such that fλ → f in L2(S,R3). Let dλ : Lλ → R3 be given by (3.26) and let dλ be a piecewise342constant L2 extension of dλ given by (3.25). Then, dλ → f in L2(S,R3).343

On the other hand, if dλ : Lλ → R3 is such that dλ → f in L2(S,R3) then there exists a344background field fλ ∈ L2(S,R3) such that dλ is given by (3.26).345

The proof follows directly from the proof of Theorem 4.1 of [JM94].346

Remark 3.3. As different unit cells are related by isometric transformations, the dipole moments347in different unit cells are related by the rotational part of the isometric transformation.348

3.2.2. Electrostatic Energy. As before, the energy associated to the system of dipole moments349dλ, for λ > 0, is given by350

Eλ = −1

2

∑s,s′∈Lλ,

s 6=s′

dλ(s) ·K(x(s′)− x(s))dλ(s′) = |(−λ/2, λ/2)|eλ,351

11


where |(−λ/2, λ/2)| = λ is the thickness of the lattice in normal direction, and eλ is the energy per352unit length given by353

(3.27) eλ = − 1

2λ

∑s,s′∈Lλ,

s 6=s′

dλ(s) ·K(x(s′)− x(s))dλ(s′).354

For convenience, we normalize eλ by Rθl, where Rθl is independent of λ and gives the size of355original lattice in the angular direction. We let356

(3.28) eλ =eλRθl

⇒ Eλ = λ(Rθl)eλ.357

Substituting (3.26) and proceeding similar to the case of the helix, we can express eλ as358

(3.29) eλ = (fλ, Tλfλ)L2(S,R3) ,359

where Tλ : L2(S,R3)→ L2(S,R3) is the map defined as360

(3.30) (Tλf)(s) =Rθlλ

∫SKλ(s′, s)f(s′) ds′361

and Kλ(s′, s), for s, s′ ∈ S, is the discrete dipole field kernel given by362

(3.31) Kλ(s′, s) =∑

u,v∈Lλ,u6=v

χUλ(v)(s′)K(x(v)− x(u))χUλ(u)(s).363

Scaling Property of Kλ. As in the case of the helix, it is convenient to first rescale the lattice364Lλ such that the lattice cell size is independent of λ after rescaling, and define a new map on the365rescaled lattice. This is considered next.366

Let S1,λ = (−θ/λ, θ/λ) × R so that s ∈ S implies s/λ ∈ S1,λ. We define a rescaled lattice367L1,λ such that s ∈ Lλ implies s/λ ∈ L1,λ. It is given by368

(3.32) L1,λ = s ∈ S1,λ; s1 = iθl, s2 = j, i, j ∈ Z = (−θ/λ, θ/λ) ∩ θlZ× Z.369

The lattice cell for s ∈ L1,λ is given by U1(s), where U1(s) is defined in (3.22) (using λ = 1 in Uλ).370For a, b ∈ S1,λ, we have371

(3.33) x(λa)− x(λb) = λ (x(a)− x(b) + Aλ(a, b)e1) ,372

where373

(3.34) Aλ(a, b) =Rλ

[Q(λa1)−Q(λb1)− λQ(a1) + λQ(b1)] .374

Keeping in mind these definitions, for u ∈ L1,λ, we also note375

χUλ(λu)(s) =

1 if s ∈ Uλ(λu),

0 otherwise=

1 if s/λ ∈ U1(u),

0 otherwise= χU1(u)(s/λ).

(3.35)

376377

Using the above relation and (3.33), we can show, for any s, s′ ∈ S,378

Kλ(s′, s) =1

λ3

∑u,v∈L1,λ,

u6=v

χU1(v)(s′/λ)K(x(v)− x(u) + Aλ(v,u)e1)χU1(u)(s/λ).379

12


If we introduce the discrete dipole field kernel K1,λ(s′, s), for s, s′ ∈ S1,λ, defined on L1,λ as:380

(3.36) K1,λ(s′, s) =∑

u,v∈L1,λ,u6=v

χU1(v)(s′)K(x(v)− x(u) + Aλ(v,u)e1)χU1(u)(s),381

we have shown that:382

Kλ(s′, s) =1

λ3K1,λ(s′/λ, s/λ), ∀s, s′ ∈ S.383

3.2.3. Limit of Electrostatics Energy. In this section, we obtain the limit of the energy per384unit length eλ. The broad strategy is similar to the helical nanotube. We first show that the map Tλ385is bounded and obtain its limit. The continuum limit of the energy density eλ then follows easily.386

Limit of Discrete Electric Field. Let T1,λ : L2(S1,λ,R3)→ L2(S1,λ,R3) be themapwith kernel387K1,λ. For any function f ∈ L2(S1,λ,R3), we have388

(3.37) (T1,λf)(s) =Rθl

∫S1,λ

K1,λ(s′, s)f(s′) ds′, ∀s ∈ S1,λ.389

LetHλ = Hλ(s) be the zeroth order moment (with respect to the first argument) of kernelKλ390given by391

(3.38) Hλ(s) =Rλθl

∫s′∈S

Kλ(s′, s) ds′, ∀s ∈ S.392

We now state the limit result of Tλ.393

Proposition 3.5. Suppose 0 < θ < π/4. The maps T1,λ and Tλ are bounded in L2 for all394λ > 0 and satisfy395

‖Tλ‖L(L2(S,R3),L2(S,R3)) = ‖T1,λ‖L(L2(S1,λ,R3),L2(S1,λ,R3)) .396

Further, for f ∈ C∞0 (R,R3),397

(Tλf)(s) −−−→λ→0

H0(s)f(s),398

pointwise, where H0(s), for s ∈ S, is given by399

H0(s) = limλ→0

Hλ(s) = R∑

u=(u1,u2)∈θlZ×Z,u6=0

K (u1t1(s) + u2t2(s)) .400

ti(s) =dx(s)

dsi, i = 1, 2, are tangent vectors on the film.401

We provide the proof of Proposition 3.5 in subsection 4.2. Based on the proposition above, we402state the main result for the thin film.403

Limit of Energy.404

Theorem 3.6. Let fλ ∈ L2(S,R3) be a sequence of functions for λ > 0 with f ∈ L2(S,R3)405such that fλ → f in L2(S,R3). Let the system of dipole moments dλ : Lλ → R3 be given by (3.26).406Let eλ, given by (3.29), be the energy per unit length normalized byRθl. Then407

eλ −−−→λ→0

−1

2(f ,H0f)L2(S,R3) .408

H0 = H0(s) is defined in Proposition 3.5.409

13


The proof of Theorem 3.6 follows from the proof of Theorem 3.3 and using Proposition 3.5.410

Remark 3.4. Note that, for s ∈ S,411

(3.39) H0(s) = Q(s1)H0(0)Q(−s1).412

Thus, if the limiting dipole moment field f is uniform in the e3 direction, the electric fieldH0(s)f(s)413will be independent of the e3-coordinate. It is easy to see from the expression of H0 that both the414normal component and the tangential components of the dipole field contribute to the electric field415and energy. This is in contrast to [GJ97] where the thin film limit of magnetostatic energy has416contribution only from the normal component.417

4. Proof of Claims.418

4.1. Helical Nanotube. In this section, we prove Proposition 3.2. First, we collect some419important results, and then show that Tλ is bounded and extends from f ∈ C∞0 (R,R3) toL2(R,R3).420We then obtain the limit of the map Tλ.421

Lemma 4.1. 1. For any a, b ∈ R,422

(4.1) x(b)− x(a) = Q(aθ)[(Q((b− a)θ)− I)e1 + δ(b− a)e3],423

where x is the map (3.1),Q is the rotational tensor (3.2), θ and δ define the helix.4242. For any θ ∈ (0, π),425

(4.2) δ ≤ mina,b∈L1,a 6=b

|x(b)− x(a)|,426

where L1 is Lλ = λZ for λ = 1.4273. For any a, b ∈ L1 and λ > 0,428

(4.3) δ|a− b| ≤ |x(a)− x(b) + Aλ(a, b)e1|,429

whereAλ(a, b) is given by430

Aλ(a, b) =Q(λaθ)−Q(λbθ)− (λQ(aθ)− λQ(bθ))

λ.431

4. For any s, s′ ∈ R such that |s− s′| ≥ 1, suppose a, b ∈ L1 are such that s ∈ [a, a+ 1), s′ ∈432[b, b+ 1), then433

(4.4)|s− s′||a− b|

< 3.434

Proof. 1. For any α, β ∈ R, we have the identities435

(4.5) QT (α) = Q(−α), Q(α)Q(β) = Q(α+ β), Q(α)e3 = e3,436

where the last relation shows that e3 is the axis of Q. By noting the definition of x in (3.1)437and using the identities above, (4.1) follows.438

2. To show (4.2), we use (4.1) to get439

|x(b)− x(a)|2 = |(Q((b− a)θ)− I)e1|2 + δ2|b− a|2 ≥ δ2|b− a|2 ≥ δ2,440

where we used the fact that |b− a| ≥ 1 for a, b ∈ L1, a 6= b.441

14


3. To show (4.3), we substitute the definition of Aλ to get442

(4.6) x(a)− x(b) + Aλ(a, b)e1 =Q(λaθ)−Q(λbθ)

λe1 + (a− b)δe3.443

Since Q(α)e1 is orthogonal to e3 for any α, we have444

|x(a)− x(b) + Aλ(a, b)e1| ≥ δ|a− b|.445

4. To show (4.4), we note that for s, s′ ∈ R such that |s − s′| ≥ 1 with a, b ∈ L1 and s ∈446[a, a+1), s′ ∈ [b, b+1), we can write s = a+∆s and s′ = b+∆s′ with 0 ≤ ∆s,∆s′ < 1.447Thus448

|s− s′||a− b|

=|a− b+ (∆s−∆s′)|

|a− b|≤ |a− b|+ |∆s−∆s

′||a− b|

< 1 +2

|a− b|≤ 3,449

where in the last step we used the fact that |a− b| ≥ 1 for a, b ∈ L1, a 6= b (which is ensured450when |s− s′| ≥ 1).451

This completes the proof.452

4.1.1. Boundedness. We next show that Tλ is a bounded map. Let Sλ : L2(R,R3) →453L2(R,R3) be an isometry defined as454

(Sλf)(s) := λ1/2f(λs).455

It is easy to see that ||Sλf ||L2 = ||f ||L2 . The inverse of Sλ is given by456

(S−1λ f)(s) = λ−1/2f(s/λ).457

Using Sλ, we can show – noting the definition of Tλ in (3.11) – for f ∈ L2(R,R3),458

(Tλf)(s) = λ2

∫RKλ(s′, s)f(s′) ds′ = λ2

∫R

1

λ3K1,λ(s′/λ, s/λ)

(λ−1/2(Sλf)(s′/λ)

)ds′459

= λ−3/2

∫RK1,λ(s′, s/λ)(Sλf)(s′)λ ds′ = λ−1/2(T1,λ(Sλf))(s/λ)460

= (S−1λ T1,λSλf)(s),461462

where we used a change of variable and the fact that (S−1λ f)(s) = λ(−1/2)f(s/λ). It follows from463

the above equation that464

‖Tλ‖L(L2,L2) = sup‖f‖6=0

‖Tλf‖L2(R,R3)

‖f‖L2(R,R3)

= sup‖f‖6=0

∥∥S−1λ (T1,λSλf)

∥∥L2(R,R3)

‖f‖L2(R,R3)

465

= sup‖f‖6=0

‖T1,λ(Sλf)‖L2(R,R3)

‖f‖L2(R,R3)

= sup‖Sλf‖6=0

‖T1,λ(Sλf)‖L2(R,R3)

‖Sλf‖L2(R,R3)

466

= ‖T1,λ‖L(L2,L2) .467468

This completes the proof of (3.16) in Proposition 3.2. Next, we show that T1,λ is a bounded map to469prove the boundedness of Tλ. We first analyze the discrete dipole field kernelK1,λ, which is defined470as471

(4.7) K1,λ(s′, s) =∑

u,v∈L1,u6=v

χU1(v)(s′)K(x(v)− x(u) + Aλ(v, u)e1)χU1(u)(s),472

15


where Uλ(s) = [s, s+ λ) for s ∈ Lλ, and Aλ(a, b) is given by (3.13).473Consider some typical s, s′ ∈ R and the corresponding a, b ∈ L1 such that s ∈ [a, a+ 1), s′ ∈474

[b, b+ 1). From (4.7), we have, for all s, s′ ∈ R such that |s− s′| < 1,475• If a = b, then K1,λ(s, s′) = 0.476• If a 6= b, then from (4.2), we have

|K1,λ(s, s′)| ≤√

6/(4πδ3)

using |Aa| ≤ |A| |a| and |I − 3(x/|x|)⊗(x/|x|)| ≤√

6, ∀x 6= 0 .477

Combining the two cases above, |K1,λ(s, s′)| ≤√

6/(4πδ3).478We now consider a case when |s − s′| ≥ 1. Noting that for this case, a 6= b. We proceed as479

follows480

|K1,λ(s, s′)| ≤√

6

4π|s− s′|3|s− s′|3

|a− b|3|a− b|3

|x(a)− x(b) + Aλ(a, b)e1|3481

≤√

6

4π|s− s′|333 |a− b|3

δ3|a− b|3=

33√

6

4πδ3

1

|s− s′|3482483

where we used the bounds (4.3) and (4.4). Combining the above bound for |s − s′| ≥ 1 with the484bound for |s− s′| < 1, and renaming the constants, we can write485

(4.8) |K1,λ(s, s′)| ≤ C1

C2 + |s− s′|3.486

Since the kernel K1,λ satisfies (4.8), we have487

(4.9)∫R|K1,λ(s′, s)| ds′ ≤ C3,

∫R|K1,λ(s′, s)| ds ≤ C3,488

for some fixed C3 <∞ independent of λ. Using Young’s inequality (e.g., Theorem 0.3.1 [Sog17]),489we have490

(4.10) ||T1,λf ||L2(R,R3) ≤ C3||f ||L2(R,R3)491

showing that T1,λ is a bounded linear map for all f ∈ C∞0 (R,R3). Since C∞0 (R,R3) is dense in492L2(R,R3), it follows that Tλ is also bounded in L2(R,R3), and extends as a bounded linear map493from C∞0 (R,R3) to L2(R,R3).494

4.1.2. Limit of the Map Tλ. Let f ∈ C∞0 (R,R3). Write Tλf as:495

Tλ(f)(s) = λ2

∫RKλ(s′, s)f(s′) ds′496

=

[λ2

∫RKλ(s′, s) ds′

]︸︷︷︸

=:Hλ(s)

f(s) + λ2

∫RKλ(s′, s)(f(s′)− f(s)) ds′,(4.11)497

498

The second term above is zero in the limit λ → 0. To see this, consider any R > 0, and then using499the bound on K1,λ in (4.8), we have500

1

λ

∫|s−s′|≥Rλ

K1,λ(s′/λ, s/λ) ds′ ≤ 1

λ

∫|s−s′|≥Rλ

C1

C2 + |s/λ− s′/λ|3ds′501

= λ2

∫|s−s′|≥Rλ

C1

C2λ3 + |s− s′|3ds′ = λ2

∫|t|≥R

C1

C2λ3 + |λt|3λ dt502

=

∫|t|≥R

C1

C2 + |t|3dt,(4.12)503

504

16


In the intermediate step, we changed variables t = (s′ − s)/λ. Thus, the upper bound on

1

λ

∫|s−s′|≥Rλ

K1,λ(s′/λ, s/λ) ds′

is independent of λ, and goes to zero as R → ∞. For the second term in (4.11), using that505

Kλ(s′, s) =1

λ3K1,λ(s′/λ, s/λ), we get506 ∣∣∣∣λ2

∫RKλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣507

=

∣∣∣∣ 1λ∫RK1,λ(s′/λ, s/λ)(f(s′)− f(s)) ds′

∣∣∣∣508

=

∣∣∣∣∣ 1λ∫|s−s′|≥Rλ

K1,λ(s′/λ, s/λ)f(s′) ds′ − 1

λ

∫|s−s′|≥Rλ

K1,λ(s′/λ, s/λ) ds′f(s)509

+1

λ

∫|s−s′|≤Rλ

K1,λ(s′/λ, s/λ)(f(s′)− f(s)) ds′

∣∣∣∣∣510

=

∣∣∣∣∣ 1λ∫|s−s′|≥Rλ

K1,λ(s′/λ, s/λ)f(s′) ds′

∣∣∣∣∣+

∣∣∣∣∣ 1λ∫|s−s′|≥Rλ

K1,λ(s′/λ, s/λ) ds′f(s)

∣∣∣∣∣511

+

∣∣∣∣∣ 1λ∫|s−s′|≤Rλ

K1,λ(s′/λ, s/λ)(f(s′)− f(s)) ds′

∣∣∣∣∣ .512513

We let λ → 0 and then R → ∞. In the limit λ → 0, the third term above will go to zero, as514f ∈ C∞0 (R,R3) andK1,λ is bounded. The first and second terms will go to zero using (4.12). Thus,515we have from (4.11) that516

(4.13) limλ→0

Tλ(f)(s) =

[limλ→0

Hλ(s)

]f(s),517

We next compute the limit of Hλ(s). Fix s ∈ R and suppose a ∈ Lλ such that s ∈ Uλ(a).518Using the definition of Kλ(s′, s), we have519

Hλ(s) = λ2

∫RK(s′, s) ds′ = λ2

∑u∈L1,u 6=a

K(x(u)− x(a))

∫Uλ(u)

dt520

= λ3∑u∈λZ,u 6=a

K(x(u)− x(a)).521

522

From (4.5), we have x(u)− x(a) = Q(aθ)((Q((u− a)θ)− I)e1 + (u− a)δe3). Using the identity523K(Qx) = QK(x)QT and K(λx) = K(x)/λ3, we get524

Hλ(s) = Q(aθ)

∑u∈λZ,u 6=a

K((Q((u− a)θ)− I)/λe1 + (u− a)δ/λe3)

Q(−aθ)525

= Q(aθ)

∑i∈Z,i6=0

K((Q(iλθ)− I)/λe1 + iδe3)

Q(−aθ),526

527

17


where we changed variables i = (u − a)/λ. Note that a ∈ λZ, and, therefore, (u − a) ∈ λZ for528u ∈ λZ, which implies i ∈ Z. Since s is related to a by s ∈ Uλ(a), we have a → s in the limit529λ→ 0. Therefore, we get530

H0(s) := limλ→0

Hλ(s) = Q(sθ)

limλ→0

∑i∈Z−0

K((Q(iλθ)− I)/λe1 + iδe3)

Q(−sθ).531

To take the limit inside the summation, we show that the sum is absolutely convergent for all λ > 0532as follows:533

aλ :=∑

i∈Z−0

|K ((Q(iλθ)− I)/λe1 + iδe3)| ≤∑

i∈Z−0

c

|(Q(iλθ)− I)/λe1 + iδe3)|3534

=∑

i∈Z−0

c

(4 sin2(iλθ/2)/λ2 + i2δ2)3/2≤

∑i∈Z−0

c

|i|3<∞, ∀λ > 0.535

536

Now, we can write537

H0(s) = Q(sθ)

∑i∈Z−0

limλ→0

K

(Q(iλθ)− I

iλθ(iθe1) + iδe3

)Q(−sθ).538

Note that for a fixed i ∈ Z539

limλ→0

(Q(iλθ)− I)

iλθiθe1 + iδe3 = lim

h=iλθ→0

(Q(h)− I)

hiθe1 + iδe3 = iθQ′(0)e1 + iδe3,540

where Q′(0) = d/ dxQ(x)|x=0. Now, using the equation above, and the fact that K(x) is smooth541away from x = 0 (which is ensured in the summation), we get542

H0(s) = Q(sθ)

∑i∈Z−0

K(iθQ′(0)e1 + iδe3

)Q(−sθ) = Q(sθ)H0(0)Q(−sθ).543

Combining this with (4.13), we get544

limλ→0

Tλ(f)(s) = H0(s)f(s) = Q(sθ)H0(0)Q(−sθ)f(s).545

Next, we simplify H0(s). Using QK(x)QT = K(Qx) and Q(sθ)Q′(0) = Q′(sθ), we can show546

H0(s) =∑i∈Z−0

K(iθQ′(sθ)e1 + iδe3) =∑i∈Z−0

K(i|t(s)| t(s)

),547

where t(s) = θQ′(sθ)e1 +δe3 is the tangent vector, and t(s) = t(s)/|t(s)|with |t(s)| =√θ2 + δ2.548

Using above and substituting the form of dipole field kernel K, it is easy to show that549

H0(s) = −h0

[I − 3t(s)⊗t(s)

]= −h0

[P⊥f(s)− 2P ||(s)

],550

with h0 defined as551

(4.14) h0 =∑i∈Z−0

1

4π|i|3(θ2 + δ2)3/2552

and projection tensors P ||(s) = t(s)⊗t(s) and P⊥(s) = I − P ||(s).553

18


4.2. Nanofilm with Uniform Bending. In this section, we prove Proposition 3.5. The outline554of the proof is similar to the case of the helix in subsection 4.1.555

Lemma 4.2. 1. Suppose s, s′ ∈ S1,λ = (−θ/λ, θ/λ) × R such that a, b ∈ L1,λ =556(−θ/λ, θ/λ) ∩ θlZ× Z with s ∈ U1(a) = [a1, a1 + θl)× [a2, a2 + 1), s′ ∈ U1(b). When557|s− s′| ≥ minθl, 1, we have a 6= b and558

(4.15)|s− s′||a− b|

< 1 +θl + 1

minθl, 1=: cL.559

2. For any a, b ∈ L1,λ, we have560

(4.16) cA|a− b| ≤ |x(a)− x(b) + Aλ(a, b)e1|,561

where x is given by (3.19) and Aλ(a, b) is defined as562

Aλ(a, b) =Rλ

[Q(λa1)−Q(λb1)− λQ(a1) + λQ(b1)] .563

Here cA = minδ,R√

1− θ2/3 is the constant independent of λ. Note that cA > 0 for564

0 < θ < π/2.565

Proof. To show (4.15), we proceed as follows. For s, s′ ∈ S1,λ and corresponding a, b ∈ L1,λ,566there exists∆s, ∆s′ such thats = a+∆s, s′ = b+∆bwith 0 ≤ ∆s1, ∆s

′1 < θl, 0 ≤ ∆s2, ∆s

′2 < 1.567

We have the bound568

(4.17)|s− s′||a− b|

≤ 1 +|∆s1 −∆s′1|+ |∆s2 −∆s′2|

|a− b|< 1 +

θl + 1

|a− b|≤ 1 +

θl + 1

minθl, 1,569

where in the last step we used the fact that any a, b ∈ L1,λ, satisfying a 6= b, are at least minθl, 1570distance apart.571

We next show (4.16). Using572

x(a)− x(b) + Aλ(a, b)e1 =Rλ

(Q(λa1)−Q(λb1))e1 + δ(a2 − b2)e3573

and574

|(Q(θ1)−Q(θ2))e1|2 = (cos θ1 − cos θ2)2 + (sin θ1 − sin θ2)2 = 2(1− cos(θ1 − θ2)),575

we have that576

(4.18) |x(a)− x(b) + Aλ(a, b)e1|2 = δ2|a2 − b2|2 +2R2

λ2(1− cos(λa1 − λb1)).577

Let r = a1− b1. Then, using a Taylor expansion and the mean value theorem, there exists ξ such that578

1− cos(λr) =1

2λ2r2 − 1

24λ4r4 cos(ξ).579

Since −1 ≤ cos(ξ) ≤ 1, it follows580

1− cos(λrξ) ≥ 1

2λ2r2 − 1

24λ4r4.581

Substituting the relation above in (4.18), we get582

|x(a)− x(b) + Aλ(a, b)e1|2 ≥ δ2|a2 − b2|2 +R2r2

(1− 1

12λ2r2

).583

19


Since a, b ∈ L1,λ, we have −2θ < λr < 2θ, and584

1− 1

12λ2r2 ≥ 1− 1

12θ24 = 1− θ2

3.585

Using the two equations above and defining the constant cA as in Lemma 4.2(2), (4.16) can be easily586shown.587

4.2.1. Boundedness. Let Sλ : L2(S,R3) → L2(S1,λ,R3) be a map such that, for any f ∈588L2(S,R3),589

(Sλf)(s) = λf(λs), ∀s ∈ S1,λ.590

It is easy to see that Sλ is an isometry. The inverse of Sλ is given by591

(S−1λ f)(s) = λ−1f(s/λ), ∀s ∈ S.592

Following the similar steps in subsubsection 4.1.1, we can show that593

||Tλ||L(L2,L2) = ||T1,λ||L(L2,L2).594

Thus, to show that Tλ is a bounded map, it is sufficient to show that T1,λ is bounded. Towards that595goal, we first establish that596

(4.19) |K1,λ(s, s′)| ≤ C1

C2 + |s− s′|3, ∀s′, s ∈ S1,λ,597

where C1, C2 are constants that may depend on the parameters R, θ, δ defining the surface S and are598independent of λ. Similar to the case of the helix (subsubsection 4.1.1), we apply Theorem 0.3.1599of [Sog17] to show that, for any λ > 0, T1,λ is a bounded linear map on C∞0 (S1,λ,R3). Using the600density of C∞0 (S1,λ,R3) in L2(S1,λ,R3), T1,λ extends as a bounded linear map to L2(S1,λ,R3).601

It remains to show (4.19). We recall that θ is the fixed angular extent of the film and satisfies the602bound 0 < θ < π/2 (in fact we restrict it such that 0 < θ < π/4). Let s, s′ ∈ S1,λ be any two generic603points, and let a, b ∈ L1,λ be such that s ∈ U1(a) and s′ ∈ U1(b). We refer to subsubsection 3.2.1604and subsubsection 3.2.2 for the notation appearing in this section.605

First, consider s, s′ such that |s− s′| ≥ minθl, 1. For this case, we have a 6= b. Noting that606

|I − 3(x/|x|)⊗(x/|x|)| =√

6, ∀x 6= 0, we have607

|K1,λ(s, s′)| ≤√

6

4π|x(a)− x(b) + Aλ(a, b)e1|3608

=

√6

4π|s− s′|3|s− s′|3

|a− b|3|a− b|3

|x(a)− x(b) + Aλ(a, b)e1|3609

≤√

6

4π|s− s′|3c3L

1

c3A,(4.20)610

611

where we used the bounds (4.15) and (4.16).612Next, we consider the case when |s−s′| < minθl, 1. This can be further divided in two cases:613• Case 1: a = b which implies |K1,λ(s′, s)| = 0.614• Case 2: a 6= b. For this case, we have615

(4.21) |K1,λ(s, s′)| ≤√

6

4π|x(a)− x(b) + Aλ(a, b)e1|3≤

√6

4πc3A|a− b|3.616

20


Note that when a 6= b, we can have (A) a1 = b1, a2 = b2± 1, or (B) a1 = b1± θl, a2 = b2,617or (C) a1 = b1 ± θl, a2 = b2 ± 1. For all the cases mentioned above, the denominator in618(4.21) is bounded from below as follows |a− b| ≥ minθl, 1. Thus, we have619

(4.22) |K1,λ(s, s′)| ≤√

6

4πc3A(minθl, 1)3.620

In summary (4.22) holds for any s, s′ such that |s− s′| < minθl, 1.621Combining the bound for the case |s− s′| < minθl, 1 with the bound for the case |s− s′| ≥622

minθl, 1, we can write623

(4.23) |K1,λ(s, s′)| ≤ C1

C2 + |s− s′|3,624

where we have renamed the constants for convenience. This completes the proof of the boundedness625of T1,λ. We next obtain the limit of the map Tλ.626

4.2.2. Limit of Map Tλ. Let f ∈ C∞0 (S,R3). We write Tλf as follows627

(Tλf)(s) =Rλθl

∫SKλ(s′, s)f(s′) ds′628

=

[Rλθl

∫SKλ(s′, s) ds′

]︸︷︷︸

=:Hλ(s)

f(s) +Rλθl

∫SKλ(s′, s)(f(s′)− f(s)) ds′,(4.24)629

630

We next show that the second term in (4.24) is zero in the limit λ → 0. Fix R > 0, then using the631relation Kλ(s′, s) = K1,λ(s′/λ, s/λ)/λ3 and the bound on K1,λ in (4.23), we have632 ∣∣∣∣λ ∫

Sχ|s−s′|≥Rλ(s′)Kλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣ ≤ λ ∫R2

χ|s−s′|≥Rλ(s′)|Kλ(s′, s)(f(s′)− f(s))| ds′633

≤ 1

λ2

∫R2

χ|s−s′|≥Rλ(s′)C1

C2 + |s/λ− s′/λ|3ds′.

(4.25)

634635

Using the change of variable t′ = s′/λ (so ds′ = λ2 dt′) and t = s/λ to have636

∣∣∣∣λ ∫Sχ|s−s′|≥Rλ(s′)Kλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣ ≤∫R2

χ|t′−t|≥R(t′)C1

C2 + |t′ − t|3dt′.

(4.26)

637638

From above, we have that639

limR→∞

[limλ→0

∣∣∣∣λ ∫Sχ|s−s′|≥Rλ(s′)Kλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣] = 0.(4.27)640641

We bound the second term in (4.24) by splitting it into two parts as follows:642 ∣∣∣∣λ ∫SKλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣ ≤ ∣∣∣∣λ ∫Sχ|s−s′|≥Rλ(s′)Kλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣+

∣∣∣∣λ ∫Sχ|s−s′|≤Rλ(s′)Kλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣ .643

21


Since |f(s′)− f(s)| ≤ 2||f ||L∞ < ∞ and from (4.27), the first term in the equation above is zero644in the limit R→∞ following the limit λ→ 0. For the second term, we proceed as follows645 ∣∣∣∣λ ∫

Sχ|s−s′|≤Rλ(s′)Kλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣646

≤ 1

λ2

∫Sχ|s−s′|≤Rλ(s′)|K1,λ(s′/λ, s/λ)| |f(s′)− f(s)| ds′647

≤ 1

λ2

∫R2

χ|s−s′|≤Rλ(s′)|K1,λ(s′/λ, s/λ)| |f(s′)− f(s)| ds′.(4.28)648649

Since K1,λ is bounded as s′ → s (see (4.23)), |s′ ∈ R2; |s − s′| ≤ Rλ| = πR2λ2, and650sup

|s′−s|≤Rλ|f(s′)− f(s)| → 0 as λ→ 0, we have651

(4.29) limR→∞

[limλ→0

∣∣∣∣λ ∫Sχ|s−s′|≤RλKλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣] = 0.652

We have shown653

(4.30) limλ→0

∣∣∣∣λ ∫SKλ(s′, s)(f(s′)− f(s)) ds′

∣∣∣∣ = 0.654

Thus, from (4.24), we have655

(4.31) limλ→0

(Tλf)(s) =

[limλ→0

Hλ(s)

]f(s).656

Limit of Hλ. Consider a typical s ∈ S such that s ∈ Uλ(a) where a ∈ Lλ. Recall that Lλ is657the lattice for λ > 0 and Uλ(a) = [a1, a1 + θlλ) × [a2, a2 + λ) is the lattice cell. In the definition658of Hλ, we substitute Kλ, to get659

(4.32) Hλ(s) =Rλθl

∑u∈Lλ,u6=a

K(x(u)−x(a))

∫Uλ(u)

ds′ = Rλ3∑

u∈Lλ,u6=a

K(x(u)−x(a)).660

Substituting the definition of transformation x in (3.19), we can show for a,u ∈ Lλ that661

x(u)− x(a) = Q(a1θ) [R(Q(u1 − a1)− I)e1 + (u2 − a2)δe3] .662

Using the identitiesK(Q(t)x) = Q(t)K(x)QT (t) andK(λx) = K(x)/λ3, from (4.32), we have663

Hλ(s) = RQ(a1)

∑u∈Lλ,u6=a

K (R(Q(u1 − a1)− I)/λe1 + (u2 − a2)δ/λe3)

︸︷︷︸

=:Hλ(s)

Q(−a1),

(4.33)

664

665

We analyze Hλ as follows. First, we expand the sum u ∈ Lλ666

Hλ(s) =∑u2∈λZ

∑u1∈λθlZ∩(−θ,θ),

(u1,u2)6=a

K

(RQ(u1 − a1)− I

λe1 + δ

u2 − a2

λe3

)667

=∑t′2∈Z


(u1,t′2) 6=(a1,0)

K

(RQ(u1 − a1)− I

λe1 + δt′2e3

) ,(4.34)668

669

22


where we introduced the new variable t′2 = (u2− a2)/λ. Since u2, a2 ∈ λZ, we have t′2 ∈ Z. Using670a Taylor expansion and the mean value theorem, we have an identity671

Q(u1 − a1)− I = Q′(ξ)(u1 − a1),(4.35)672673

where ξ = ξ(u1−a1) ∈ (−θ, θ) depends on u1−a1. Based on the above observation, we decompose674(4.34) as follows675

Hλ(s)676

=∑t′2∈Z


(u1,t′2)6=(a1,0)

K

(RQ′(0)

u1 − a1

λe1 + δt′2e3

)︸︷︷︸

=:H(1)λ (s)

677

+∑t′2∈Z


(u1,t′2)6=(a1,0)

K

(RQ(u1 − a1)− I

λe1 + δt′2e3

)−K

(RQ′(0)

u1 − a1

λe1 + δt′2e3

)︸︷︷︸

=:H(2)λ (s)

(4.36)

678

679

where two new terms are defined for convenience.680Step 1: We show H

(2)λ goes to zero in the limit λ→ 0. Let681

(4.37) x1 = RQ(u1 − a1)− I

λe1, x2 = RQ′(0)

u1 − a1

λe1, z = δt′2e3.682

Consider a function y : [0, 1]→ R3 defined as683

(4.38) y(r) = x1 + r(x2 − x1).684

Note that, since (u1, t′2) 6= (a1, 0) and t′2 ∈ Z,685

(4.39) |y(r) + z| ≥ minδ, minr∈[0,1],u1∈λθlZ−a1∩(−θ,θ)

|y(r)|.686

We show that minr∈[0,1],u1∈λθlZ−a1∩(−θ,θ)

|y(r)| > 0 and the lower bound is independent of λ.687

For convenience, let t = u1 − a1. Since u1, a1 ∈ λθlZ ∩ (−θ, θ), and u1 6= a1, we have t ∈688λθlZ− 0 ∩ (−2θ, 2θ). The hypothesis of Proposition 3.5 restricts θ such that689

(4.40) 0 < θ < π/4 ⇒ 1 > cos(2θ) > 0.690

With t = u1 − a1, writing out the action of Q(t) and Q′(0) on e1, we get691

x1 = RQ(t)− I

λe1 =

Rλ

[(cos(t)− 1)e1 + sin(t)e2]692

x2 = RQ′(0)t

λe1 =

Rtλ

[− sin(t)e1 + cos(t)e2].(4.41)693694

Through elementary calculations, we can show695

(4.42) |y(r)|2 = |x1 + r(x2 −x1)|2 =R2

λ2

[2(1− r)2(1− cos(t)) + r2t2 + 2r(1− r)t sin(t)

].696

23


Using a Taylor expansion and noting that t ∈ λθlZ−0∩ (−2θ, 2θ), there exists ξ1, ξ2 ∈ (−2θ, 2θ)697with ξ1 = ξ1(t), ξ2 = ξ2(t) such that698

(4.43) 1− cos(t) = cos(ξ1)t2/2, sin(t) = t cos(ξ2).699

Thus700

|y(r)|2 =R2

λ2

[2(1− r)2 cos(ξ1)t2/2 + r2t2 + 2r(1− r)t cos(ξ2)t

]701

=R2t2

λ2

[(1− r)2 cos(ξ1) + r2 + 2r(1− r) cos(ξ2)

]702

≥ R2t2

λ2

[(1− r)2 min

ξ∈(−2θ,2θ)cos(ξ) + r2 + 2r(1− r) min

ξ∈(−2θ,2θ)cos(ξ)

]703

=R2t2

λ2

[(1− r)2 cos(2θ) + r2 + 2r(1− r) cos(2θ)

]704

≥ R2t2

λ2minr∈[0,1]

[(1− r)2 cos(2θ) + r2 + 2r(1− r) cos(2θ)

]705

=R2t2

λ2cos(2θ),(4.44)706

707

where we used the fact that minξ∈(−2θ,2θ)

cos(ξ) = cos(2θ) in the fourth equation, and cos(2θ) is the708

minimum with respect to r ∈ [0, 1] of the function in the square bracket in the fifth equation. Further,709since t ∈ λθlZ− 0 ∩ (−2θ, 2θ), we have710

(4.45) 0 < Cy :=R2λ2θ2

l

λ2cos(2θ) = (Rθl)2 cos(2θ) ≤ |y(r)|2,711

for any t ∈ λθlZ − 0 ∩ (−2θ, 2θ) and r ∈ [0, 1]. The lower bound on |y(r)| is independent of λ712and r. Finally, combining (4.45) with (4.39) to get713

(4.46) 0 < Cyz := minδ,Rθl√

cos(2θ) ≤ |y(r) + z|.714

Proceeding further, we have, from the fundamental theorem of calculus,715

K(x1 + z)−K(x2 + z) =

∫ 1

0

d

drK(y(r) + z) dr =

∫ 1

0

∇K(y(r) + z)d

dry(r) dr716

=

∫ 1

0

∇K(y(r) + z)(x2 − x1) dr.(4.47)717718

Note that because of (4.46), ∇K(y(r) + z) exists and is bounded. From the definition of H(2)λ in719

(4.36), a change of variable t = u1 − a1, the definition of x1,x2, z in (4.37) and (4.41), and noting720

24


the identity (4.47), we have721

|H(2)λ (s)|722

≤∑t′2∈Z


(u1,t′2)6=(a1,0)

∣∣∣∣K (RQ(u1 − a1)− I

λe1 + δt′2e3

)−K

(RQ′(0)

u1 − a1

λe1 + δt′2e3

)∣∣∣∣723

≤∑t′2∈Z

∑t∈λθlZ−0∩(−2θ,2θ)

|K (x1 + z)−K (x2 + z)|

724

≤∑t′2∈Z

∑t∈λθlZ−0∩(−2θ,2θ)

∫ 1

0

|∇K(y(r) + z)| |x2 − x1| dr

725

≤∑t′2∈Z

∑t∈λθlZ−0∩(−2θ,2θ)

∫ 1

0

C

|y(r) + z|4|x2 − x1| dr

726

=∑t′2∈Z

∑t∈λθlZ−0∩(−2θ,2θ)

∫ 1

0

C

(|y(r)|2 + |z|2)2|x2 − x1| dr

,(4.48)

727728

where we utilized the bound on the gradient of K with constant C > 0 fixed.729Next, we get an upper bound on |x1 − x2| in terms of t. From (4.41), we have

x2 − x1 =Rλ

[tQ′(0)−Q(t) + I

]e1.

By a Taylor expansion and the mean value theorem, we have Q(t) = I + tQ′(0) + (t2/2)Q′′(ξ)730where ξ = ξ(t) ∈ (−2θ, 2θ) depends on t. Substituting this and using the bound |Q′′ij(ξ)| ≤ 1 gives731

(4.49) |x2 − x1| =Rλ

|t|2

2|Q′′(ξ)| ≤ R

λ

|t|2

2.732

Combining the equation above with (4.48), we get733

|H(2)λ (s)|734

≤∑t′2∈Z

∑t∈λθlZ−0∩(−2θ,2θ)

∫ 1

0

C

(|y(r)|2 + |z|2)2

Rλ

|t|2

2dr

735

=∑t′2∈Z

∑t′∈θlZ−0∩(−2θ/λ,2θ/λ)

∫ 1

0

C

(|y(r)|2 + |z|2)2

Rλ

λ2|t′|2

2dr

736

≤ λ

∑t′2∈Z

∑t′∈θlZ−0∩(−2θ/λ,2θ/λ)

∫ 1

0

C

(|y(r)|2 + |z|2)2

R|t′|2

2dr

,737

738

where in the third line we introduced the variable t′ = t/λ. We only have to show that the term inside739

the braces is bounded as λ → 0 to conclude that |H(2)λ (s)| → 0 as λ → 0. First, note from (4.44),740

25


we have741

(4.50) |y(r)|2 ≥ R2

λ2|t|2 cos(2θ) = R2|t′|2 cos(2θ).742

Therefore,743

(4.51)C

(|y(r)|2 + |z|2)2≤ C

(R2|t′|2 cos(2θ) + |z|2)2.744

Thus745

|H(2)λ (s)| ≤ λ

∑t′2∈Z

∑t′∈θlZ−0∩(−2θ/λ,2θ/λ)

∫ 1

0

C

(R2|t′|2 cos(2θ) + |z|2)2

R|t′|2

2dr

.

(4.52)

746

747

Note that the integrand is independent of r. Further, the numerator has |t′|2 whereas the denominator748has (|t′|2c + |z|2)2, therefore, the sum inside the braces is absolutely convergent and finite. Hence,749

due to the factor λ, we have shown limλ→0|H(2)

λ (s)| = 0.750

This completes the step 1. We next study H(1)λ .751

Step 2: We have from (4.36)752

H(1)λ (s) =

∑t′2∈Z


(u1,t′2)6=(a1,0)

K

(RQ′(0)

u1 − a1

λe1 + δt′2e3

)753

=∑t′2∈Z

∑u1∈λθlZ,

(u1,t′2)6=(a1,0)

K

(RQ′(0)

u1 − a1

λe1 + δt′2e3

)︸︷︷︸

=:I1

754

−∑t′2∈Z

∑u1∈[λθlZ]−[λθlZ∩(−θ,θ)],

(u1,t′2)6=(a1,0)

K

(RQ′(0)

u1 − a1

λe1 + δt′2e3

)︸︷︷︸

=:I2

,(4.53)755

756

where we have used the notation [λθlZ] − [λθlZ ∩ (−θ, θ)] to denote the set t ∈ λθlZ; t /∈757λθlZ ∩ (−θ, θ). Using the decay property of the dipole field kernel K, we can show that |I2| → 0758in the limit λ→ 0. Therefore, we have759

limλ→0

H(1)λ (s) = lim

λ→0

∑t′2∈Z

∑u1∈λθlZ,

(u1,t′2)6=(a1,0)

K

(RQ′(0)

u1 − a1

λe1 + δt′2e3

)760

=∑t′2∈Z

∑t′1∈θlZ,

(t′1,t′2)6=(0,0)

K(RQ′(0)t′1e1 + δt′2e3

)761

(4.54)762763

26


where we introduced the new variable t′1 = (u1−a1)/λ. Since u1 ∈ λθ1Z and a1 ∈ λθlZ∩ (−θ, θ),764we have t′1 ∈ θlZ. This completes step 2. Note that lim

λ→0Hλ(s) is independent of s ∈ S.765

Upon substituting the limit of H(1)λ and H

(2)λ in (4.36), we have shown766

(4.55) limλ→0

Hλ(s) = limλ→0

Hλ(0) =∑


K(RQ′(0)u1e1 + δu2e3

).767

Recall that s ∈ S was fixed such that s ∈ Uλ(a), which implies that a → s as λ → 0. With this768observation and (4.55), we have from (4.33),769(4.56)

H0(s) = limλ→0

Hλ(s) = RQ(s1)

∑u=(u1,u2)∈θlZ×Z,

u6=0

K(RQ′(0)u1e1 + δu2e3

)Q(−s1).770

Next we simplify H0(s). Given the parametric map x = x(s), the two tangent vectors at771s = (s1, s2) are772

(4.57) t1(s) =dx

ds1= RQ′(s1)e1, t2(s) =

dx

ds2= δe3.773

UsingQK(x)QT = K(Qx) and Q(r)Q′(0) = Q′(r), we write,774

H0(s) =∑


K (u1t1(s) + u2t2(s)) .775

This completes the proof of Proposition 3.5.776

5. Summary ofResults. Wehave shown rigorously that certain low-dimensional nanostructures777do not have long-range dipole-dipole interaction in the continuum limit. The energy density in the778limit is entirely due to Maxwell-self field. In 1-d and 2-d lattices (in a 3-d ambient space), the dipole779field kernel decay is sufficiently fast that long-range interactions do not contribute to the limit energy.780

While our calculations show that the energy is local in the continuum limit for 1-d and 2-ddiscrete systems, in agreement with dimension reduction approaches that reduce a 3-d continuum toa 1-d or 2-d continuum (e.g., [GJ97, CH15] and others), we note an interesting difference. As shownin [GJ97] and other work following it, the thin film limit of the continuum electrostatic energy is dueto the component of dipole moment field along the normal direction to the film. Similarly, the thinwire limit of the energy is due to components of dipole moment field in the plane normal to the wire,see [CH15]. This is different from the limit energy in the discrete-to-continuum limit: for the case ofa helical nanotube, the limiting energy density is given by

h0

∫R

[|P⊥f |2 − 2|P ||f |2

]ds,

where h0 is a constant, and P ||f and P⊥f are the projections of the dipole moment field f along781the axis of the helix and in the plane normal to the axis of the helix respectively. Therefore, unlike the782thin wire limit using dimension reduction, the discrete-to-continuum energy has contributions from783both the normal and tangential components of the dipole moment field. For the case of a thin film784with curvature, the limiting energy density is given by785

−1

2

∫Sf(s) ·H0(s)f(s) ds,786

27


where787

H0(s) = R∑


K (u1t1(s) + u2t2(s)) .788

Here, S is the parametric domain of the film,R is the inverse of the curvature, θl is the angular width789of the unit cell, and ti(s), i = 1, 2, are the tangent vectors at coordinate s ∈ S. For simplicity, we fix790s ∈ S and assume t1 = e1 and t2 = e2; then the lattice sum above is over a 2-d lattice in (e1, e2)791plane. By substituting the form ofK and computingH0(s)f(s), we can show that both the normal792and the tangential components of f are present in the final expression for the energy above. The793difference between the dimension reduction approach on the one hand, and the discrete-to-continuum794limit on the other hand, is due to the fact that we assume that the discrete structures consist of a finite795unit cell in the thickness direction and do not take the limit in the thickness direction.796

Acknowledgments. This paper draws from the doctoral dissertation of Prashant K. Jha at797Carnegie Mellon University [Jha16]. We thank Richard D. James for useful discussions, and AFRL798for hosting a visit by Kaushik Dayal.799

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