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A rigorous derivation of mean-field models describing 2D micro phase separation

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Abstract

We study the free boundary problem describing the micro phase separation of diblock copolymer melts in the regime that one component has small volume fraction \(\rho \) such that the micro phase separation results in an ensemble of small disks of one component. We consider the two dimensional case in this paper, whereas the three dimensional case was already considered in Niethammer and Oshita (Calc Var PDE 39:273–305, 2010). Starting from the free boundary problem restricted to disks we rigorously derive the heterogeneous mean-field equations on a time scale of the order of \({\mathcal {R}}^{3}\ln (1/\rho )\), where \({\mathcal {R}}\) is the mean radius of disks. On this time scale, the evolution is dominated by coarsening and stabilization of the radii of the disks, whereas migration of disks becomes only relevant on a larger time scale.

Introduction

Diblock copolymer molecules consist of subchains of two different type of monomers, say A- and B-monomers. The different type of subchains tend to segregate, and hence the phase separation take place. However since the subchains are chemically bonded, the two subchains mix on a macroscopic scale, while on a molecular scale, A- and B-subchains still segregate and the micro-domains are formed. This is called micro phase separation. For more physical background on this phenomenon we refer to [2, 8].

In the strong segregation regime, energetically favorable configurations have been characterized in the Ohta–Kawasaki theory [18] by minimizers of an energy functional, which is in the two dimensional case of the form

$$\begin{aligned} E(\Omega )= \mathcal{H}^{1}(\partial \Omega )\,+\, \frac{\sigma }{2}\int _{(0,L)^2} \big |(-\Delta )^{-1/2} (\chi - \rho )\big |^2\,dx. \end{aligned}$$
(1)

Here \((0,L)^2 \subset {\mathbb {R}}^2\) is the domain covered by the copolymers, \(\Omega \subset [0,L)^2\) denotes the region covered by, say, A-monomers, \(\rho =\frac{|\Omega |}{L^2} \in (0,1)\) the average density, \(\sigma \in {\mathbb {R}}_+=(0,\infty )\) is a parameter related to the polymerization index, \(\chi \) is the characteristic function of \(\Omega \), and \(\mathcal{H}^1\) denotes one dimensional Hausdorff measure.

The first term in the energy prefers large blocks of monomers, the second favors a very fine mixture. Competition between these terms leads to minimizers of E which represent micro phase separation.

Starting with the pioneering work [15], where the Ohta–Kawasaki theory is formulated on a bounded domain as a singularly perturbed problem and the limiting sharp interface problem is identified, there has been a large body of analytical work. Minimizers of the energy functionals have been characterized in [1, 3, 4, 20], the existence/stability of stationary solutions has been investigated in [16, 17, 19, 21] and a time dependent model has been considered in [7, 9]. The mean field models in the three dimensional case have been derived in [6, 10, 12].

We consider the gradient flow of the energy, which is a standard way to set up a model for the evolution of the copolymer configuration that decreases energy and preserves the volume fraction. Then the evolution equation becomes the following extension of the Mullins–Sekerka evolution for phase separation in binary alloys [11]. The normal velocity v of the interface \(\partial \Omega =\partial \Omega (t)\) satisfies

$$\begin{aligned} v= [ \nabla w \cdot \vec n] \qquad \hbox { on } \partial \Omega , \end{aligned}$$
(2)

where \([ \nabla w \cdot \vec n]\) denotes the jump of the normal component of the gradient of the potential across the interface. Here \({\vec {n}}\) denotes the outer normal to \(\Omega \) and

$$\begin{aligned} {[}f]= \lim _{\begin{array}{c} x \notin \Omega \\ x \rightarrow \partial \Omega \end{array}} f(x) -\lim _{\begin{array}{c} x \in \Omega \\ x \rightarrow \partial \Omega \end{array}} f(x) . \end{aligned}$$

The potential w is for each time determined via

$$\begin{aligned} -\Delta w&= 0&\quad&\hbox { in } (0,L)^2 \backslash \partial {\Omega }, \end{aligned}$$
(3)
$$\begin{aligned} w&= \kappa + \sigma (-\Delta )^{-1}(\chi - \rho )&\hbox { on } \partial \Omega , \end{aligned}$$
(4)

where \(\kappa \) is the curvature of \(\partial \Omega \). We are interested in the case that the volume of \(\Omega (t)\) is preserved in time and can thus impose Neumann or periodic boundary conditions for w on \(\partial (0,L)^2\). In what follows we will consider a periodic setting and hence always require that the potential w is \((0,L)^2\)-periodic. For local well-posedness of this evolution see [5].

The evolution defined by (2)–(4) has a formal interpretation as a gradient flow of the energy (1) on a Riemannian manifold. Indeed, consider the manifold of subsets of the 2-dimensional flat torus \({\mathbb {T}}\) of length L with fixed volume, that is \(\mathcal{M}:=\{ \Omega \subset {\mathbb {T}}\,;\, |\Omega |= L^2 \rho \}\), whose tangent space \(T_\Omega \mathcal{M}\) at an element \(\Omega \in \mathcal{M}\) is described by all kinematically admissible normal velocities of \(\partial \Omega \), that is,

$$\begin{aligned} T_\Omega \mathcal{M}\;=\;\left\{ v :\partial \Omega \rightarrow {\mathbb {R}}\, ; \, \int _{\partial \Omega }v\, dS\;=\;0\right\} . \end{aligned}$$

The Riemannian structure is given by the following metric tensor on the tangent space:

$$\begin{aligned} g_\Omega (v^{1},v^{2})\;=\;\int _{{\mathbb {T}}}\nabla w^{1}\cdot \nabla w^{2}\,d x, \end{aligned}$$
(5)

where \(w^{\alpha } :{\mathbb {T}}\rightarrow {\mathbb {R}}\) \((\alpha =1,2)\) solves

$$\begin{aligned} \begin{aligned} -\Delta w^{\alpha }&= 0 \\ \left[ \nabla w^{\alpha } \cdot \vec n\right]&= v^{\alpha } \end{aligned}\qquad \begin{aligned}&\hbox { in }\; {\mathbb {T}}\,\backslash \partial \Omega ,\\&\hbox { on }\;\partial \Omega \end{aligned} \end{aligned}$$
(6)

for \(v^\alpha \in T_{\Omega }{\mathcal {M}}\) \((\alpha =1,2)\). The gradient flow of the energy (1) is now the dynamical system where at each time the velocity is the element of the tangent space in the direction of steepest descent of the energy. In other words, v is such that

$$\begin{aligned} g_{\Omega (t)}(v,{\tilde{v}}) = - \langle DE(\Omega (t)),{\tilde{v}}\rangle \end{aligned}$$
(7)

for all \({\tilde{v}} \in T_{\Omega (t)}\mathcal{M}\). Choosing \({\tilde{v}} =v\) we immediately obtain the energy estimate associated with each gradient flow, which is

$$\begin{aligned} \int _0^T g_{\Omega (t)}(v, v)\,d t + E(\Omega (T))= E(\Omega (0)) \quad \text {for all } T>0 . \end{aligned}$$
(8)

In what follows we consider the micro phase separation in the two dimensional case in the regime where the fraction of A-monomers is much smaller than the one of B-monomers. In this case A-phase consists of an ensemble of many small approximately circular particles. We reduce the evolution to the gradient flow on circular particles.

For that purpose we define the submanifold \(\mathcal{N}\subset \mathcal{M}\) of all sets \(\Omega \) which are the union of disjoint balls \(\Omega \;=\;\bigcup _i B_{R_i}(X_i)\), where the centers \(\{X_i\}_i\) and the radii \(\{R_i\}_i\) are variables. Hence \(\mathcal{N}\) can be identified with an open subspace of the hypersurface \( \{ {{\mathbf {Y}}}=\{R_i,X_i\}_i \,;\, (R_i,X_i) \in {\mathbb {R}}_+ \times {\mathbb {T}}, \pi \sum _i R_i^2=L^2 \rho \}\) in \({\mathbb {R}}^{3N}\), where N is the number and \(i=1,\ldots ,N\) an enumeration of the particles with centers in the torus \({\mathbb {T}}\). Since the normal velocity v satisfies \(v=\frac{d R_i}{d t}+\frac{d X_i}{d t}\cdot {\vec {n}}\) on \(\partial B_{R_i}(X_i)\), the tangent space can be identified with the hyperplane

$$\begin{aligned} T_{{{\mathbf {Y}}}}\mathcal{N}\;=\; \Big \{{{\mathbf {Z}}}= \sum _i \left( V_i \frac{\partial }{\partial R_i} + \xi _i \cdot \frac{\partial }{\partial X_i}\right) \,;\, (V_i,\xi _i)\in {\mathbb {R}}\times {\mathbb {R}}^2 ,\sum _i R_iV_i\;=\;0\Big \}\;\subset \;{\mathbb {R}}^{3N}, \end{aligned}$$

such that \(V_i\) describes the rate of change of the radius of particle i and \(\xi _i\) the rate of change of its center. We use the abbreviation \({{\mathbf {Z}}}=\{V_i,\xi _i\}_i\) for \({{\mathbf {Z}}}=\sum _i (V_i \frac{\partial }{\partial R_i} + \xi _i \cdot \frac{\partial }{\partial X_i})\).

The metric tensor is then given by

$$\begin{aligned} g_{\mathbf{Y}}({{\mathbf {Z}}}^{1},{{\mathbf {Z}}}^{2})\;=\; \int _{{\mathbb {T}}}\nabla w^{1}\cdot \nabla w^{2}\,d x, \end{aligned}$$

where the function \( w^{\alpha }:{\mathbb {T}}\rightarrow {\mathbb {R}}\) solves

$$\begin{aligned} \begin{aligned} -\Delta w^{\alpha }&=0 \\ \left[ \nabla w^{\alpha }\cdot \vec n\right]&=V_i^{\alpha }+ \xi _i^{\alpha }\cdot \vec n\end{aligned}\quad \begin{aligned}&\hbox { in }\; {\mathbb {T}}\,\backslash \cup _i\partial B_{R_i}(X_i), \\&\hbox { on }\; \partial B_{R_i}(X_i). \end{aligned} \end{aligned}$$
(9)

for \({\mathbf {Z}}^\alpha =\{V_i^\alpha ,\xi _i^\alpha \}_i \in T_{{\mathbf {Y}}} {\mathcal {N}}\), \(\alpha =1,2\). For the following it will be convenient to split the metric tensor into the radial and shift part respectively. For any \(\{V_i,\xi _i\}_i\), we write

$$\begin{aligned} w=u+\phi \end{aligned}$$

where u and \(\phi \) are harmonic in- and outside the particles and where

$$\begin{aligned} \begin{aligned} \left[ \nabla u\cdot \vec n\right]&=V_i\\ \left[ \nabla \phi \cdot \vec n\right]&=\xi _i\cdot \vec n\end{aligned}\qquad \begin{aligned}&\hbox { on } \partial B_{R_i}(X_i), \\&\hbox { on }\partial B_{R_i}(X_i). \end{aligned} \end{aligned}$$
(10)

We consider the energy \(E({{\mathbf {Y}}})=E_{\text {surf}}({{\mathbf {Y}}})+\sigma E_{\text {nl}}({{\mathbf {Y}}})\), where

$$\begin{aligned} E_{\text {surf}}({{\mathbf {Y}}})= 2 \pi \sum _iR_i\qquad \hbox { and } \qquad E_{\text {nl}} ({{\mathbf {Y}}}) = \int _{{\mathbb {T}}} |\nabla \mu |^2\,dx \end{aligned}$$

with \(\mu :{\mathbb {T}}\rightarrow {\mathbb {R}}\) solving \( - \Delta \mu = \chi _{\cup B_{R_i}} - \rho \). We obtain the differentials of the energies in the direction of a tangent vector \({\tilde{{{\mathbf {Z}}}}}=\{{\tilde{V_i}},{\tilde{\xi _i}}\}_i\) as

$$\begin{aligned} \langle DE_{\text {surf}}({{\mathbf {Y}}}),{\tilde{{{\mathbf {Z}}}}}\rangle \;=\; 2 \pi \sum _i {\tilde{V_i}}\end{aligned}$$

and

$$\begin{aligned} \langle DE_{\text {nl}}({{\mathbf {Y}}}),{\tilde{{{\mathbf {Z}}}}}\rangle \;=\; - 2\int _{{\mathbb {T}}} \nabla \mu \cdot \nabla {\tilde{w}} \, d x= 2 \sum _i\int _{\partial B_{R_i}(X_i)} \mu \big ( {\tilde{V_i}}+ {\tilde{\xi _i}}\cdot \vec n\big )\,dS. \end{aligned}$$

Here \({\tilde{w}}:{\mathbb {T}}\rightarrow {\mathbb {R}}\) is a function \(w^\alpha \) satisfying (9) for \({{\mathbf {Z}}}^\alpha ={\tilde{{{\mathbf {Z}}}}}\). The integration by parts yields

$$\begin{aligned} g_{{{\mathbf {Y}}}}({{\mathbf {Z}}},{\tilde{{{\mathbf {Z}}}}}) = -\sum _i \int _{\partial B_{R_i}(X_i)} w (\tilde{V}_i+{\tilde{\xi }}_i \cdot {\vec {n}})\, d S. \end{aligned}$$

From now on we consider an arrangement of particles as described above which evolves according to the gradient flow equation. This means that for any \(t \ge 0\), it holds for \({{\mathbf {Z}}}(t)=\frac{d}{d t}{\mathbf {Y}}(t)\) that

$$\begin{aligned} g_{{{\mathbf {Y}}}}({{\mathbf {Z}}},{\tilde{{{\mathbf {Z}}}}})=-\langle DE({{\mathbf {Y}}}) , {\tilde{{{\mathbf {Z}}}}}\rangle , \end{aligned}$$

that is,

$$\begin{aligned} \sum _i \int _{\partial B_{R_i}(X_i)} w ({\tilde{V}}_i+{\tilde{\xi }}_i \cdot {\vec {n}})\, d S = 2 \pi \sum _i{\tilde{V_i}}\,+\, 2\sigma \sum _i\int _{\partial B_{R_i}(X_i)} \mu \big ( {\tilde{V_i}}+ {\tilde{\xi _i}}\cdot \vec n\big )\,dS\, \end{aligned}$$
(11)

for all \({\tilde{{{\mathbf {Z}}}}}\in T_{{{\mathbf {Y}}}}\mathcal{N}\). Since \({\tilde{{{\mathbf {Z}}}}}\) is an arbitrary element of the tangent space we conclude from (11) that w satisfies

$$\begin{aligned} \frac{1}{2 \pi R_i} \int _{\partial B_{R_i}(X_i)} \big ( w - \tfrac{1}{R_i} - 2\sigma \mu \big ) \,dS = \lambda (t) \end{aligned}$$
(12)

and

$$\begin{aligned} \int _{\partial B_{R_i}(X_i)} \big ( w - 2 \sigma \mu \big ) {\vec {n}} \,dS=0 \end{aligned}$$
(13)

for all i such that \(R_i>0\), with a Lagrange parameter \(\lambda (t)\) that ensures volume conservation. Equations (12) and (13) are the analogue of (4) in the restricted setting.

Our aim is to identify the evolution in the limit of vanishing volume fraction of particles. More precisely, we consider a sequence of systems characterized by the parameter

$$\begin{aligned} \varepsilon := \left( \ln \left( \frac{d}{\mathcal{R}}\right) \right) ^{-1/2} \end{aligned}$$
(14)

in the limit \(\varepsilon \rightarrow 0\). Here d is defined by

$$\begin{aligned} d^2 \sum _i1 = L^2 \end{aligned}$$
(15)

and \(\mathcal{R}\) by

$$\begin{aligned} \mathcal{R}^2\sum _i1 = \sum _iR_i(0)^2 . \end{aligned}$$
(16)

Then \(\frac{1}{d^2}\) denotes the number density of particles, and \(\pi \,\mathcal{R}^2\) the average volume of particles. Here and throughout this paper we use the abbreviation \(\sum _i = \sum _{i:R_i>0}\).

Our main result informally says that when \(L \sim L_{sc}\), with

$$\begin{aligned} L_{sc}^2 := d^2 \ln (d/\mathcal{R}) \sim d^2 \ln (1/\rho ), \end{aligned}$$
(17)

on the time scale of order \({\mathcal{R}}^3 \ln (1/\rho )\), the number density of particles with radius r and center x, denoted by \(\nu =\nu (t,r,x)\) (suitably normalized), satisfies

$$\begin{aligned} \partial _t \nu + \partial _r \Big (\frac{1}{r^2}\big \{r \psi -1 - \sigma r^3 \big \} \nu \Big ) =0 \, , \end{aligned}$$
(18)

where \(\psi =\psi (t,x)\) satisfies for each t that

$$\begin{aligned} -\Delta \psi +2\pi \psi \int _0^\infty \nu \, d r= 2 \pi \bigg (\int _0^\infty \frac{1}{r} \, \nu \, d r + \frac{\sigma }{L^2} \int _{{\mathbb {T}}}\int _0^\infty r^2 \, \nu \, d r \, d y\bigg ) \quad \text {in }{\mathbb {T}} \end{aligned}$$
(19)

in the limit \(\varepsilon \rightarrow 0\). Here \(\sigma \) is also suitably normalized.

We remark that on the other hand, in the case that \(L \ll L_{sc}\), that is, in the very dilute case, one obtains a homogeneous version where \( \psi \) is constant in space, and is replaced by \(\lambda (t)\). More precisely that the number density of particles with radius r, denoted by \(\nu (t,r)\) (suitably normalized), satisfies

$$\begin{aligned} \partial _t\nu + \partial _r \Big ( \frac{1}{r^2}\big (\lambda r -1 -\sigma r^3\big ) \nu \Big )=0 \end{aligned}$$
(20)

with

$$\begin{aligned} \lambda (t)= \frac{\int _0^\infty \frac{1}{r} \nu \,dr + \sigma \int _0^\infty r^2 \,\nu \, d r}{ \int _0^\infty \nu \,dr}. \end{aligned}$$
(21)

The result

In this section, we will introduce suitably rescaled variables, state the precise assumption on our initial particle arrangement, and present the statement of our main result.

We assume from now on that \(L=L_{sc}\) for the ease of presentation, and we will rescale the spatial variables by \(L_{sc}\) such that

$$\begin{aligned} L_{sc}\;=\;L\;=\;1\quad \hbox {and hence}\quad d\;=\;\varepsilon ,\;\mathcal{R}\;=\;\varepsilon \exp (-1/\varepsilon ^2) =:\alpha _{\varepsilon }. \end{aligned}$$

Notice that \(\rho =\pi \alpha _{\varepsilon }^2\varepsilon ^{-2}\) and \(\ln (1/\rho ) \sim \varepsilon ^{-2}\). We introduce \({\hat{R_i}}\), \({\hat{t}}\), \({\hat{V_i}}\), \(\hat{\xi }\), \({\hat{w}}\), \({\hat{\sigma }}\) and \({\hat{\mu }}\) via

$$\begin{aligned} \begin{aligned} R_i(t)&=\alpha _{\varepsilon }{\hat{R_i}}({\hat{t}}), \qquad t = \alpha _{\varepsilon }^3 \ln (1/\rho ) {\hat{t}}, \qquad w(t,x) = \alpha _{\varepsilon }^{-1}{\hat{w}}({\hat{t}} , x),\\ V_i(t)&= \frac{1}{\alpha _{\varepsilon }^2 \ln (1/\rho )} {\hat{V_i}} ({\hat{t}}) \sim \frac{\varepsilon ^2}{\alpha _{\varepsilon }^2} {\hat{V_i}}({\hat{t}}), \qquad \xi _i (t)= \frac{\varepsilon }{\alpha _{\varepsilon }^2} {\hat{\xi }}_i ({\hat{t}}), \\ \sigma&= \frac{1}{\alpha _{\varepsilon }^3 \ln (1/\rho ) }{\hat{\sigma }} \sim \frac{\varepsilon ^2}{\alpha _{\varepsilon }^3} {\hat{\sigma }}, \qquad \mu (t,x)=\frac{\alpha _{\varepsilon }^2}{\varepsilon ^2} {\hat{\mu }}({\hat{t}}, x). \end{aligned} \end{aligned}$$

From now on we only deal with the rescaled quantities and drop the hats in the notation.

We denote the joint distribution of particle centers and radii at a given time t by \(\nu _t^\varepsilon \in (C^0_p)^*\), which is given by

$$\begin{aligned} \int \zeta \,d\nu _t^\varepsilon \;=\; \sum _i\varepsilon ^2\zeta \left( R_i(t),X_i(t) \right) \quad \hbox {for}\;\zeta \in C^0_p\,, \end{aligned}$$
(22)

where \(C^0_p\) stands for the space of continuous functions on \({\mathbb {R}}_+ \times {\mathbb {T}}\) with compact support contained in \({\mathbb {R}}_+ \times {\mathbb {T}}\). Here \({\mathbb {T}}\) denotes the unit flat torus, and \({\mathbb {R}}_+=(0,\infty )\). Note that since \(\zeta (r,x)=0\) for \(r=0\), particles which have vanished do not enter the distribution. The natural space for \(\nu _t^\varepsilon \) and its limit \(\nu _t\) is the space \((C^0_p)^*\) of Borel measures on \({\mathbb {R}}_+ \times {\mathbb {T}}\).

We are now going to make the assumptions on our initial particle arrangement precise. Notice first, that in view of (15) and (16) we have

$$\begin{aligned} \int \,d\nu _0^{\varepsilon } = \sum _i \varepsilon ^2 = 1 \qquad \hbox { and } \qquad \int r^2 \,d\nu _0^{\varepsilon } = \sum _i\varepsilon ^2 R_i^2(0) = 1. \end{aligned}$$
(23)

It follows immediately, that

$$\begin{aligned} \int r \,d\nu _0^{\varepsilon } = \sum _i\varepsilon ^2 R_i(0) \le 1\,, \end{aligned}$$
(24)

that is the surface energy of the initial particle arrangement is finite.

Furthermore it is natural to assume that initially the nonlocal energy is uniformly bounded in \(\varepsilon \), that is

$$\begin{aligned} \int _{{\mathbb {T}}} | \nabla \mu ^{\varepsilon }(0,x) |^2 \,dx \le C\,, \end{aligned}$$
(25)

where C is independent of \(\varepsilon \) and where \(\mu ^{\varepsilon }(0,x)\) satisfies \(-\Delta \mu ^{\varepsilon }(0,\cdot ) = \frac{\varepsilon ^2}{\alpha _{\varepsilon }^2} \chi _{\cup _i B_i(0)} - \pi \) and \(\int _{{\mathbb {T}}} \mu ^{\varepsilon }\,dx =0\).

We will see later [cf. (84)], that the nonlocal energy controls \( \sum _i \varepsilon ^2 R_i^{\,4}\). Hence, finiteness of the nonlocal energy initially also implies \(\sum _i \varepsilon ^2 R_i^{\, 4}(0) \le C\). For our analysis we need a little more than this. We need a certain tightness assumption which ensures, that not too much mass is contained in very large particles as \(\varepsilon \rightarrow 0\). More precisely, we assume that

$$\begin{aligned} \sup _{\varepsilon } \sum _{R_i\ge M} \varepsilon ^2 R_i^{\, 4}(0) \rightarrow 0 \qquad \hbox { as } M \rightarrow \infty . \end{aligned}$$
(26)

Finally, we assume that initially particles are well separated in the sense that we assume that there is \(\gamma >0\), such that

$$\begin{aligned} \Big \{ B_{2 \gamma \varepsilon }(X_i(0)) \Big \}_i \qquad \hbox { are disjoint}. \end{aligned}$$
(27)

In accordance with the notation in (22) we will use in the following the abbreviation \(\int \zeta \,d\nu _t:= \int _0^{\infty } \int _{{\mathbb {T}}} \zeta (r,x)\,d\nu _t(r,x)\) for \(\nu _t \in (C^0_p)^*\). Otherwise the domain of integration is specified.

The natural space for potentials of diffusion fields is \(H^{1}({\mathbb {T}})\). Furthermore we will denote by \(\mathring{H}^{1}({\mathbb {T}})\) the subspace of \(H^{1}({\mathbb {T}})\) of functions with mean value zero.

We can now state our main result which informally says that \(\nu _t^{\varepsilon }\) converges as \(\varepsilon \rightarrow 0\) to a weak solution of (18)–(19).

Theorem 2.1

Let \(T>0\) be given and assume that the assumptions in Sect. 2 are satisfied. Then there exists a subsequence, again denoted by \(\varepsilon \rightarrow 0\), and a weakly continuous map \([0,T]\ni t\mapsto \nu _t\in (C^0_p)^*\) with

$$\begin{aligned} \int \zeta \,d\nu _t^\varepsilon \;\rightarrow \int \zeta \,d\nu _t\quad \hbox { uniformly in} \;t\in [0,T]\quad \hbox {for all}\;\zeta \in C^0_p , \end{aligned}$$

\(\int r^2\,d\nu _t\;=\;1\) for all \(t\in [0,T]\). Furthermore, there exists a measurable map \((0,T)\ni t\mapsto \psi (t)\in H^{1}({\mathbb {T}})\) such that (18) and (19) hold in the following weak sense

$$\begin{aligned} \frac{d}{dt}\int \zeta \,d\nu _t\;=\;\int \partial _r\zeta \, \frac{1}{r^2}\,\bigg (r\,\psi (t,x)-1- \sigma r^3 \bigg )\, d\nu _t \end{aligned}$$
(28)

distributionally on (0, T) for all \(\zeta \in C^0_p\) with \(\partial _r\zeta \in C^0_p\). Here

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla \psi (t,x)\cdot \nabla \zeta -2\pi \sigma \zeta \,dx\,+\,2\pi \int \zeta \,\bigg (\psi (t,x) -\tfrac{1}{r} \bigg ) \,d\nu _t=0 \end{aligned}$$
(29)

for all \(\zeta \in H^{1}({\mathbb {T}})\) and almost all \(t \in (0,T)\).

The proof of Theorem 2.1 goes similarly to the approach for the three dimensional case in [12]. However in contrast to the three dimensional case we need to estimate \(1/R_i\) term in the proof that the tightness property is preserved in time (see Lemmas 3.9 and 3.10) since the Lagrange multiplier diverges when particles disappear.

Proof of Theorem 2.1

We can deduce Theorem 2.1 by the homogenization of Rayleigh Principle (see Theorem 3.11). This will be obtained from the homogenization of metric tensor (Lemmas 3.5, 3.6) and the limit of the differential of the energy (Lemma 3.8). Also we need some a-priori estimates, which are given by a series of lemmas. The proof of Lemmas in this section will be given in Sect. 4. For readers convenience we will not abbreviate the arguments.

Gradient flow structure

In rescaled variables the submanifold \(\mathcal{N}\) turns into

$$\begin{aligned} \mathcal{N}^{\varepsilon }=\bigg \{ {{\mathbf {Y}}}^{\varepsilon }= \{R_i,X_i\}_i \, ; \, \sum _i\varepsilon ^2R_i^2 =1 \bigg \} \end{aligned}$$

and the tangent space

$$\begin{aligned} T_{{{\mathbf {Y}}}^{\varepsilon }}\mathcal{N}^{\varepsilon }=\bigg \{ {{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }= \sum _i \left( {\tilde{V_i}}\frac{\partial }{\partial R_i}+\varepsilon \alpha _{\varepsilon }\ln (1/\rho ) {\tilde{\xi _i}}\cdot \frac{\partial }{\partial X_i}\right) \,;\, \sum _iR_i{\tilde{V_i}}=0 \bigg \}. \end{aligned}$$

We use the abbreviation \({{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }=\{{\tilde{V_i}},{\tilde{\xi _i}}\}_i\) for \({{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }= \sum _i ({\tilde{V_i}}\frac{\partial }{\partial R_i}+\varepsilon \alpha _{\varepsilon }\ln (1/\rho ) {\tilde{\xi _i}}\cdot \frac{\partial }{\partial X_i})\), and regard \(\{{\tilde{V_i}},{\tilde{\xi _i}}\}_i\) as the component of a tangent vector \({{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }\) with respect to a basis

$$\begin{aligned} \bigg \{ \frac{\partial }{\partial R_i}, \, \varepsilon \alpha _{\varepsilon }\ln (1/\rho )\frac{\partial }{\partial X_i} \bigg \}_i . \end{aligned}$$

We will always denote by \({{\mathbf {Z}}}^{\varepsilon }=\{V_i,\xi _i\}_i\) the direction of steepest descent. Recall that \(V_i=\frac{d R_i}{d t }\), but \(\xi _i=(\varepsilon \alpha _{\varepsilon }\ln (1/\rho ))^{-1} \frac{d X_i}{d t}\). The notation \({{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }\) will be used for an arbitrary element of the tangent space. Furthermore we use the abbreviation \(B_i:=B_{\alpha _{\varepsilon }R_i}(X_i)\).

We define the energy in rescaled variables as

$$\begin{aligned}&E_\varepsilon ({\mathbf {Y}}^\varepsilon )=E_{\text {surf}, \varepsilon }({\mathbf {Y}}^\varepsilon )+\sigma E_{\text {nl},\varepsilon } ({\mathbf {Y}}^\varepsilon ) \\&E_{\text {surf},\varepsilon }({\mathbf {Y}}^\varepsilon ) = 2 \pi \sum _{i} \varepsilon ^2 R_i,\quad E_{\text {nl},\varepsilon }({\mathbf {Y}}^\varepsilon ) = \int _{{\mathbb {T}}} | \nabla \mu ^\varepsilon |^2 \, d x , \end{aligned}$$

where \(\mu ^{\varepsilon }=\mu ^{\varepsilon }(t,x)\) solves \(-\Delta \mu ^{\varepsilon } = \frac{\varepsilon ^2}{\alpha _{\varepsilon }^2}\chi _{\cup B_i} - \pi \) and \(\int _{{\mathbb {T}}}\mu ^{\varepsilon }\,dx=0\), and the metric tensor for \({{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }\in T_{{{\mathbf {Y}}}^{\varepsilon }} \mathcal{N}^{\varepsilon }\) is computed via

$$\begin{aligned} g_{{{\mathbf {Y}}}^{\varepsilon }}\big ( {{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon },{{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }\big ) =\int _{{\mathbb {T}}} \vert \nabla {{\tilde{w}}}^{\varepsilon }\vert ^2 \,dx = \int _{{\mathbb {T}}} \vert \nabla {{\tilde{u}}}^{\varepsilon }+ \nabla {{\tilde{\phi }}}^{\varepsilon }\vert ^2\,dx , \end{aligned}$$
(30)

where \({\tilde{w}}^\varepsilon ={\tilde{u}}^\varepsilon +{\tilde{\phi }}^\varepsilon \),

$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {T}}} \nabla {{\tilde{u}}}^{\varepsilon }\cdot \nabla \zeta \,dx \, + \, \sum _i\int _{\partial B_i} \frac{\varepsilon ^2}{\alpha _{\varepsilon }} {{\tilde{V_i}}} \, \zeta \,dS =0\,,\\&\int _{{\mathbb {T}}} \nabla {{\tilde{\phi }}}^{\varepsilon }\cdot \nabla \zeta \,dx \, + \, \sum _i\int _{\partial B_i} \frac{\varepsilon }{\alpha _{\varepsilon }} {\tilde{\xi _i}}\cdot \vec n\, \zeta \,dS =0 \end{aligned} \end{aligned}$$
(31)

for all \(\zeta \in \mathring{H}^{1}({\mathbb {T}})\). Notice that the potentials are only determined up to additive constants. In what follows we fix this constant by requiring that \(\int _{{\mathbb {T}}} {\tilde{u}}^{\varepsilon }\,dx = \int _{{\mathbb {T}}} {\tilde{\phi }}^{\varepsilon }\,dx=0\). For the steepest descent directions \({\mathbf {Z}}^\varepsilon =\{V_i,\xi _i\}_i\), we define the potentials \(w^\varepsilon \), \(u^\varepsilon \), \(\phi ^\varepsilon \) analogously.

Equations (12) and (13) for the direction of steepest descent, turn into

$$\begin{aligned} \frac{1}{|\partial B_i|} \int _{\partial B_i} (u^{\varepsilon }+ \phi ^{\varepsilon }- 2\sigma \mu ^{\varepsilon } ) \,dS = \frac{1}{R_i} + \lambda ^\varepsilon (t) \end{aligned}$$
(32)

for some \(\lambda ^\varepsilon (t) \in {\mathbb {R}}\) and

$$\begin{aligned} \int _{\partial B_i} \big ( u^\varepsilon + \phi ^{\varepsilon } - 2 \sigma \mu ^{\varepsilon } \big ) {\vec {n}} \,dS = 0\, \end{aligned}$$
(33)

for all i such that \(R_i>0\). Here and in what follows we abbreviate, with some abuse of notations, for a disk \(B_R(X)\) the perimeter by \(|\partial B_R|\) and its area by \(|B_R|\). Now the energy estimate (8) reads

$$\begin{aligned} \begin{aligned}&\int _0^{t_1} \!\! \int _{{\mathbb {T}}} |\nabla w^{\varepsilon }|^2\,dx\, dt \,+\, 4 \pi \sum _i\varepsilon ^2 \frac{R_i^2}{2}(t_1) \,+\, \sigma \int _{{\mathbb {T}}} |\nabla \mu ^{\varepsilon }(t_1)|^2\,dx \\&\quad = 4 \pi \sum _i\varepsilon ^2 \frac{R_i^2}{2}(0) \,+\, \sigma \int _{{\mathbb {T}}} |\nabla \mu ^{\varepsilon }(0)|^2\,dx \end{aligned} \end{aligned}$$
(34)

for all \(t_1 >0\). Finally, the Rayleigh principle says that \({\mathbf {Z}}^\varepsilon \) satisfies

$$\begin{aligned}&\int _0^T \beta (t) \Big ( \tfrac{1}{2} g_{{\mathbf {Y}}^\varepsilon (t)}({\mathbf {Z}}^\varepsilon , {\mathbf {Z}}^\varepsilon ) + \langle DE_\varepsilon ({\mathbf {Y}}^{\varepsilon }(t)),{\mathbf {Z}}^\varepsilon \rangle \Big )\,d t \nonumber \\&\quad \le \int _0^T \beta (t) \Big ( \tfrac{1}{2} g_{{\mathbf {Y}}^\varepsilon (t)}(\tilde{{\mathbf {Z}}}^\varepsilon , \tilde{{\mathbf {Z}}}^\varepsilon ) + \langle DE_\varepsilon ({\mathbf {Y}}^\varepsilon (t)),\tilde{{\mathbf {Z}}}^\varepsilon \rangle \Big )\,d t \end{aligned}$$
(35)

for all \(\tilde{{\mathbf {Z}}}^\varepsilon \in T_{{\mathbf {Y}}} {\mathcal {N}}\) and nonnegative \(\beta \in C^\infty ([0,T])\).

A priori estimates and weak limits

It follows from definitions (15), (16) and the facts that volume of particles is conserved and the number can only decrease, that

$$\begin{aligned} \int \,d\nu _t^\varepsilon\le & {} 1\quad \hbox {for all}\;t\in [0,T], \end{aligned}$$
(36)
$$\begin{aligned} \int r^2\,d\nu _t^\varepsilon= & {} 1\quad \hbox {for all}\;t\in [0,T] . \end{aligned}$$
(37)

On the other hand the uniform bound on the energy in (34) implies the following.

Lemma 3.1

$$\begin{aligned} \int r^4 \,d\nu _t^{\varepsilon }\le C \quad \hbox { for all }\;t\in [0,T]. \end{aligned}$$
(38)

Next to \(\nu _t^\varepsilon \), we introduce signed Borel measures \(\rho _t^\varepsilon \in (C^0_p)^*\) and \(\psi _t^{\varepsilon } \in ((C^0_p)^2)^*\) on \({\mathbb {R}}_+\times {\mathbb {T}}\) via

$$\begin{aligned} \begin{aligned} \int \zeta \,d\rho _t^\varepsilon&=\varepsilon ^2\sum _i\zeta (R_i(t),X_i(t))\,V_i(t)\quad \hbox {for}\;\zeta \in C^0_p,\\ \int \eta \cdot \,d\psi _t^\varepsilon&=\varepsilon ^2\sum _i\eta (R_i(t),X_i(t))\cdot \xi _i(t)\quad \hbox {for}\;\eta \in ( C^0_p)^2 . \end{aligned} \end{aligned}$$

The measures satisfies \(\partial _t \nu ^{\varepsilon } + \partial _r \rho ^{\varepsilon } + \varepsilon \alpha _{\varepsilon }\ln (1/\rho ) \hbox {div}\, \psi ^{\varepsilon }=0\) in the sense of distributions, that is

$$\begin{aligned} \int _0^T \left( \partial _t\beta (t)\int \zeta \,d\nu _t^\varepsilon + \beta (t)\Big \{\int \partial _r\zeta \,d\rho _t^\varepsilon \,+\, \varepsilon \alpha _{\varepsilon }\ln (1/\rho ) \int \nabla _x \zeta \cdot d\psi _t^{\varepsilon }\Big \} \right) \,dt\;=\;0 \end{aligned}$$
(39)

for all \(\zeta \in C^0_p\cap C^\infty \) and \(\beta \in C_0^\infty ([0,T])\). As will be shown in Sect. 4, we have

$$\begin{aligned} \mathcal{D}^\varepsilon \; := \; \int _0^T 2 \pi \sum _i\varepsilon ^2R_i^2\,\big (V_i^2+ \tfrac{1}{4} |\xi _i|^2\big )\,dt \le (1+o(1)) \int _0^T \!\!\! \int _{{\mathbb {T}}} |\nabla w^{\varepsilon }|^2\,dx\,dt \le C, \end{aligned}$$
(40)

which yields in particular

$$\begin{aligned} \mathcal{D} \; := \; \liminf _{\varepsilon \rightarrow 0}\mathcal{D}^\varepsilon \le \limsup _{\varepsilon \rightarrow 0}\mathcal{D}^\varepsilon <\infty . \end{aligned}$$
(41)

Bounds (39) and (40) yield a weak Hölder regularity in t of \(\{\nu _t^\varepsilon \}_t\):

$$\begin{aligned}&\bigg | \int \zeta \,d \nu _{t_1}^\varepsilon - \int \zeta \,d \nu _{t_1}^\varepsilon \bigg | \le |t_1-t_2|^{1/2} \nonumber \\&\quad \bigg ( \frac{{\mathcal {D}}^\varepsilon }{2\pi } \sup _t \int \Big ( |\partial _r \zeta |^2 + 4\varepsilon ^2 \alpha _\varepsilon ^2 (\ln \tfrac{1}{\rho })^2 |\nabla _x \zeta |^2 \Big )\frac{1}{r^2}\, d \nu _t^\varepsilon \bigg )^{1/2} . \end{aligned}$$
(42)

Using Arzela–Ascoli’s Theorem, (36) and (42) imply that there exists a weakly continuous family \(\{\nu _t\}_t\) of nonnegative Borel measures on \({\mathbb {R}}_+ \times {\mathbb {T}}\) such that for a subsequence

$$\begin{aligned} \int \zeta \,d\nu _t^\varepsilon \;\rightarrow \;\int \zeta \,d\nu _t\quad \hbox {uniformly in} \;t\in [0,T] \end{aligned}$$
(43)

for \(\zeta \) in a countable subset of \(C^0_p\cap C^\infty \). Again by (36), we see that we can extend the locally uniform convergence in (43) to all \(\zeta \in C^0_p\). Obviously, the bound (36) is conserved

$$\begin{aligned} \int \,d\nu _t\le 1\quad \hbox {for all}\;t\in [0,T] , \end{aligned}$$
(44)

and due to (37) and (38) we have

$$\begin{aligned}&\int r^2\,d\nu _t = 1 \quad \hbox {for all}\;t\in [0,T] , \end{aligned}$$
(45)
$$\begin{aligned}&\int r^4\,d \nu _t \le C \quad \text {for all } t \in [0,T]. \end{aligned}$$
(46)

The uniform control of the signed Borel measures \(\{ r^2 \, d\rho _t^\varepsilon \, d t \}_\varepsilon \) and \(\{ r^2 \, d\psi _t^{\varepsilon }\, d t\}_\varepsilon \) on \({\mathbb {R}}_+ \times {\mathbb {T}}\times [0,T]\) implied by (40) ensures the weak convergence, where the limits can be regarded as bounded linear functionals on \(L^2(r^2\, d \nu _t \, d t)\) and \(L^2(r^2\, d \nu _t \, d t)^2\) respectively, and hence by Riesz Representation Theorem, there exist \(v\in L^2(r^2\,d\nu _t\,dt)\) and \(\xi \in L^2(r^2\,d\nu _t\,dt)^2\) with

$$\begin{aligned} \int _0^T \int \big (|v(t)|^2+ |\xi (t)|^2\big )\,r^2\,d\nu _t\,dt\;\le \; \frac{{\mathcal {D}}}{2 \pi } , \end{aligned}$$
(47)

such that for a subsequence

$$\begin{aligned} \begin{aligned} \int _0^T\beta (t)\int \zeta \,d\rho _t^\varepsilon \,dt&\rightarrow \; \int _0^T\beta (t)\int \zeta \,v(t)\,d\nu _t\,dt,\\ \int _0^T\beta (t)\int \eta \cdot d\psi _t^\varepsilon \,dt&\rightarrow \; \int _0^T\beta (t)\int \eta \cdot \xi (t)\,d\nu _t\,dt, \end{aligned} \end{aligned}$$
(48)

for all \(\beta \in C^0([0,T])\), \(\zeta \in C^0_p\) and \(\eta \in (C^0_p)^2\), and

$$\begin{aligned} \int _0^T \beta \int |\xi (t)|^2\,r^2\,d\nu _t\,dt\;\le \; \liminf _{\varepsilon \rightarrow 0}\int _0^T \beta \sum _i \varepsilon ^2 R_i^2 |\xi _i|^2\, d t \end{aligned}$$
(49)

for all nonnegative \(\beta \in C^0([0,T])\). Thus the limit of (39) is

$$\begin{aligned} \int _0^T \left( \partial _t\beta (t)\int \zeta \,d\nu _t+ \beta (t)\int \partial _r\zeta \,v(t)\,d\nu _t\,\right) \,dt\;=\;0 \end{aligned}$$
(50)

for all \(\zeta \in C^0_p\cap C^\infty \) and \(\beta \in C_0^\infty ([0,T])\).

Definition 3.2

For \(\tilde{{\mathbf {Z}}}^\varepsilon \) satisfying \(\int _0^T g_{{\mathbf {Y}}^\varepsilon } (\tilde{{\mathbf {Z}}}^\varepsilon ,\tilde{{\mathbf {Z}}}^\varepsilon )\, d t \le C\), we can define \({\tilde{v}} \in L^2(r^3 \,d\nu _t\,d t)\) and \({\tilde{\xi }} \in L^2(r^3 \,d\nu _t\,d t)^2\) analogously. We say that \(\tilde{{\mathbf {Z}}}^\varepsilon \) converges weakly to \(({\tilde{v}},\tilde{\xi })\).

We are going to show below that for any \(T>0\), particles do not collide on a time interval [0, T] for sufficiently small \(\varepsilon \). More precisely

Lemma 3.3

For any \(T>0\) we can find \(\varepsilon _0>0\) such that

$$\begin{aligned} \{B_{\gamma \varepsilon }(X_i(t)) \}_i \qquad \hbox {are disjoint for all } t \in [0,T], \end{aligned}$$
(51)

for all \(\varepsilon \in (0,\varepsilon _0]\).

Thus it follows that the marginal of \(\nu _t\) with respect to x has a bounded Lebesgue density. Hence it follows from (46) that the functional

$$\begin{aligned} \langle L(t), \zeta \rangle := \int \zeta r^2 \, d \nu _t -\int _{{\mathbb {T}}}\zeta \, d x ,\quad \zeta \in H^{1}({\mathbb {T}})\end{aligned}$$

is an element of \(H^{-1}({\mathbb {T}})=(H^{1}({\mathbb {T}}))^*\) for all \(t\in [0,T]\). Hence \(K(t,\cdot ) \in H^{1}({\mathbb {T}})\) is uniquely determined via (29) up to additive constants.

In order to prove Lemma 3.3 we show the following.

Lemma 3.4

(slow motion of the particle centers) As long as (51) is satisfied, we have

$$\begin{aligned} |\xi _i(t)| \le C \frac{\alpha _{\varepsilon }}{\varepsilon ^2} \left( \int _{{\mathbb {T}}} |\nabla \phi ^{\varepsilon }|^2 + |\nabla u^{\varepsilon }|^2 + |\nabla \mu ^{\varepsilon }|^2 \,dx \right) ^{1/2}. \end{aligned}$$

Homogenization of the metric tensor

We identify the \(\Gamma \)-limit for the metric tensor and provide the necessary results to pass to the limit in the metric tensor. The following is a lower semicontinuity result.

Lemma 3.5

(lower semicontinuity) For all nonnegative \(\beta =\beta (t)\in C^{\infty }([0,T])\) we have

$$\begin{aligned} \begin{aligned}&\liminf _{\varepsilon \rightarrow 0} \int _0^{T} \beta \,g_{{{\mathbf {Y}}}^{\varepsilon }}\big ( {{\mathbf {Z}}}^{\varepsilon },{{\mathbf {Z}}}^{\varepsilon }\big ) \, d t = \liminf _{\varepsilon \rightarrow 0} \int _0^{T} \beta \int _{{\mathbb {T}}} \vert \nabla w^{\varepsilon }\vert ^2 \, dx \,dt \\&\quad = \liminf _{\varepsilon \rightarrow 0} \int _0^{T} \beta \int _{{\mathbb {T}}} \vert \nabla u^{\varepsilon }\vert ^2 + 2 \nabla u^{\varepsilon }\cdot \nabla \phi ^{\varepsilon }+ \vert \nabla \phi ^{\varepsilon }\vert ^2 \, dx \,dt\\&\quad \ge \int _0^{T} \beta \int _{{\mathbb {T}}} \vert \nabla u \vert ^2 \, dx \,dt \, + \, 2 \pi \int _0^T \beta \int \big (\vert v \vert ^2\, + \tfrac{1}{4} |\xi |^2\big ) r^2 \, d \nu _t \, d t , \end{aligned} \end{aligned}$$
(52)

where for almost all t the function \(u(t,\cdot )\in \mathring{H}^{1}({\mathbb {T}})\) is determined via

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla u(t) \cdot \nabla \zeta \,dx \;+\; 2 \pi \int \zeta \, r \, v(t) \,d\nu _t \,=0 \end{aligned}$$
(53)

for all \(\zeta \in \mathring{H}^{1}({\mathbb {T}})\).

Furthermore we show that for any tangent vector of the limit manifold \(({\tilde{v}}, {\tilde{\xi }})\), there exists an approximating sequence along which the metric tensor is continuous.

Lemma 3.6

(construction) For any \({\tilde{v}}\in L^2(r^2 \,d\nu _t \, d t)\) with \( \int {\tilde{v}}\,r^2\,d\nu _t=0\) for almost all t and any \({\tilde{\xi }} \in L^2(r^2 \,d\nu _t \, d t)^3\) there exists a sequence \({{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }\) with \({\tilde{{{\mathbf {Z}}}}}^{\varepsilon } \in T_{{{\mathbf {Y}}}^{\varepsilon }}\mathcal{N}^{\varepsilon }\) such that \(\tilde{{\mathbf {Z}}}^\varepsilon \) weakly converges to \(({\tilde{v}},{\tilde{\xi }})\), and

$$\begin{aligned} \begin{aligned}&\limsup _{\varepsilon \rightarrow 0} \int _0^{T} \beta \, g_{{{\mathbf {Y}}}^{\varepsilon }}\big ( {{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon }, {{\tilde{{{\mathbf {Z}}}}}}^{\varepsilon } \big ) \,dt \\&\quad \le \;\int _0^{T} \beta \int _{{\mathbb {T}}} \vert \nabla {\tilde{u}} \vert ^2 \, dx \, dt \, + \,2 \pi \int _0^T \beta \int \big (\vert {\tilde{v}}\vert ^2 + \tfrac{1}{4} |{\tilde{\xi }}|^2\big ) \,r^2 \,d\nu _t \, d t \end{aligned} \end{aligned}$$
(54)

for all nonnegative \(\beta =\beta (t) \in C^{\infty }([0,T])\) and with \({\tilde{u}}(t,\cdot ) \in \mathring{H}^{1}({\mathbb {T}})\) determined via

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla {\tilde{u}}(t) \cdot \nabla \zeta \,dx \;+\; 2 \pi \int \zeta \, r \,{\tilde{v}}(t) \,d\nu _t \,=0 \end{aligned}$$
(55)

for all \(\zeta \in \mathring{H}^{1}({\mathbb {T}})\).

Note that the contribution from the drift term and the radial part do not interact in the limit \(\varepsilon \rightarrow 0\).

The limit of the differential of the energy

We identify the limit of the differential of the energy. To that aim we identify the limit of the potentials \(\mu ^{\varepsilon }\) so as to prove the convergence for the nonlocal part of the energy.

Note that

$$\begin{aligned} \langle DE_{\text {nl},\varepsilon }( {\mathbf {Y}}^\varepsilon ), \tilde{ {\mathbf {Z}}}^\varepsilon \rangle = - 2 \int _{{\mathbb {T}}} \nabla \mu ^{\varepsilon } \cdot \nabla \tilde{w}^{\varepsilon } dx . \end{aligned}$$

Here, as also in the homogenization of the metric tensor, the key idea is that the potentials can be represented as a sum of monopoles, which represent the self-interaction of particles, plus a slowly varying field, which represents the interaction between different particles.

We set \(l= \gamma \varepsilon \), where \(\gamma >0\) is as in (51). We write \(\mu ^\varepsilon =\sum _i {\tilde{\mu }}_i + {\tilde{\mu }}^\varepsilon \) with

(56)

Here \({\tilde{\mu }}_i\) is chosen such that \(-\Delta {\tilde{\mu }}_i = \frac{\varepsilon ^2}{\alpha _{\varepsilon }^2}\) in \(B_i\) and \(\Delta {\tilde{\mu }}_i=0\) in \(B_l (X_i){\setminus } B_i\). Since

(57)

it holds that

$$\begin{aligned} {[}\nabla {\tilde{\mu }}_i \cdot {\vec {n}}]=0 \quad \text {on }\partial B_i \quad \text {and }\quad [\nabla {\tilde{\mu }}_i \cdot {\vec {n}}]=\frac{\varepsilon ^2 R_i^2}{2 l} \quad \text {on }\partial B_l (X_i). \end{aligned}$$
(58)

The slowly varying field \({\tilde{\mu }}^\varepsilon \) converges strongly to K and this enables us to pass to the limit in the differential of the energy.

Lemma 3.7

$$\begin{aligned} \sup _{t\in [0,T]}\Vert \nabla {\tilde{\mu }}^\varepsilon - \tfrac{1}{2} \nabla K \Vert _{L^2({\mathbb {T}})} \rightarrow 0 \end{aligned}$$

as \(\varepsilon \rightarrow 0\). Here \(K(t,\cdot ) \in \mathring{H}^{1}({\mathbb {T}})\) satisfies for each t

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla K\cdot \nabla \zeta \, d x = 2 \pi \bigg (\int r^2 \zeta \, d \nu _t - \int _{{\mathbb {T}}} \zeta \, d x \bigg ) \end{aligned}$$
(59)

for all \(\zeta \in H^{1}({\mathbb {T}})\).

Lemma 3.8

Assume that \({\tilde{{{\mathbf {Z}}}}}^\varepsilon \) satisfies

$$\begin{aligned} \int _0^T g_{{\mathbf {Y}} ^\varepsilon }(\tilde{{\mathbf {Z}}}^\varepsilon ,\tilde{{\mathbf {Z}}}^\varepsilon )\,d t\le C \end{aligned}$$
(60)

and converges weakly to \(({\tilde{v}},{\tilde{\xi }})\). Then for all \(\beta \in C^{\infty }([0,T])\),

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\int _0^T \beta \langle DE_{{\mathrm{nl}},\varepsilon } ({\mathbf {Y}}^\varepsilon ), \tilde{ {\mathbf {Z}}}^\varepsilon \rangle \, d t = \int _0^T \beta \langle DE_{\mathrm{nl}}(\nu _t),({\tilde{v}},\tilde{\xi }) \rangle \, d t . \end{aligned}$$

where

$$\begin{aligned} \langle DE_{\mathrm{nl}}(\nu _t),({\tilde{v}},{\tilde{\xi }})\rangle = 2 \pi \int (r^3 + r K(t,x)) {\tilde{v}} \,d\nu _t \, . \end{aligned}$$

Tightness

In order to prove Lemma 3.8 we first need to show that the tightness property (26) is preserved in time so that no mass is lost at infinity in the limit \(\varepsilon \rightarrow 0\).

Lemma 3.9

(tightness) For any \(t>0\) we have

$$\begin{aligned} \sum _{R_i\ge M} \varepsilon ^2 R_i^{\, 4}(t) \rightarrow 0 \qquad \hbox { as } M \rightarrow \infty \hbox { uniformly in } \varepsilon . \end{aligned}$$

This lemma is crucial to our proof. The proof is much more difficult than three dimensional case. In fact, the main idea of the proof is to show by asymptotics that, at least in some average sense, \(V_i\) satisfies approximately \(R_i^2 V_i \sim u R_i - 1 - \sigma R_i^3\) where u is the Lagrange multiplier that ensures the volume conservation. In contrast to the three dimensional case, the Lagrange multiplier may diverge. However we can still show the following a-priori estimate, and thus, at least on average, \(V_i \le 0\), if \(R_i\) is sufficiently large, and no mass can escape to infinity as \(\varepsilon \rightarrow 0\), from which one deduces Lemma 3.9.

Lemma 3.10

For any \(T>0\), there exist constants \(C_T>0\) and \(\varepsilon _1>0\) such that

$$\begin{aligned}&\int _0^T \bigg ( \int \frac{1}{1-\varepsilon ^2 \ln r} \frac{1}{r^2} \, d\nu _t^\varepsilon \bigg )^{1/2}\, d t \nonumber \\&\quad \le C\bigg ( \int _0^T \bigg (1+\int _{\mathbb {T}}|\nabla (w^\varepsilon -2\sigma \mu ^\varepsilon )|^2 \, d x \bigg ) \, d t\bigg ) ^{1/2} \le C_T \end{aligned}$$
(61)

for all \(\varepsilon \in (0,\varepsilon _1]\).

Homogenization of Rayleigh principle

The main task that remains to be done now is to determine the equation for the velocity function v. It will be characterized as the minimizer in the Rayleigh principle. Thus our task is to characterize the limits of \({\mathbf {Z}}^{\varepsilon }\) that satisfy (35).

Notice that we can use Lemma 3.8 with \({\mathbf {Z}}^{\varepsilon } - \tilde{{\mathbf {Z}}}^{\varepsilon }\) where \({\mathbf {Z}}^{\varepsilon }\) is the direction of steepest descent and \(\tilde{{\mathbf {Z}}}^{\varepsilon }\) is as in Lemma 3.6. The main result from which one deduces Theorem 2.1 is the following.

Theorem 3.11

For all nonnegative \(\beta \in C^{\infty }([0,T])\) we have

$$\begin{aligned} \begin{aligned}&\int _0^{T} \beta \,\Big (\tfrac{1}{2} \int _{{\mathbb {T}}} \vert \nabla u \vert ^2\,dx \, + \, 2 \pi \int \, \tfrac{1}{2} \big (\vert v \vert ^2 + \tfrac{1}{4} |\xi |^2\big )\,r^2 \,d\nu _t\\&\qquad + \langle DE(\nu _t), (v,\xi )-({\tilde{v}},{\tilde{\xi }})\rangle \Big )\,dt\\&\quad \le \int _0^T \beta \,\Big (\tfrac{1}{2} \int _{{\mathbb {T}}} \vert \nabla {\tilde{u}} \vert ^2 \, dx \, + \, 2 \pi \int \, \tfrac{1}{2} \big (\vert {\tilde{v}}\vert ^2 + \tfrac{1}{4} |{\tilde{\xi }}|^2\big )\,r^2 \,d\nu _t\,\Big ) \,dt \end{aligned} \end{aligned}$$
(62)

for all \({\tilde{v}}\in L^2(r^2 \, d\nu _t \, d t)\) and \({\tilde{\xi }} \in L^2(r^2 \, d\nu _t \, d t)^2\) such that \(\int r \, {\tilde{v}}\,d\nu _t=0\) for almost all t and such that \((v - {\tilde{v}})(t,x,\cdot ) =0\) in a neighborhood of \(r=0\). Here

$$\begin{aligned} \langle DE(\nu _t),({\tilde{v}},{\tilde{\xi }})\rangle = 2 \pi \int \bigg ( 1 +\sigma \Big ( r^3 + r K(t,x) \Big ) \bigg ) {\tilde{v}} \,d\nu _t \end{aligned}$$

for \({\tilde{v}}\) which vanish in a neighborhood of \(r=0\), and \(u(t,\cdot ) \), \({\tilde{u}}(t,\cdot )\), \(K(t,\cdot )\in \mathring{H}^{1}({\mathbb {T}})\) are determined for a.a. t via (53), (55), and (59) respectively.

The Euler–Lagrange equation for (62) becomes \(r^2 v= r(u-\sigma K-\lambda )-1-\sigma r^3\) with \(\lambda \) being a Lagrange multiplier that ensures the constraint \(\int r v \,d \nu _t=0\). Setting \(\psi (t,x)\equiv u-\sigma K -\lambda \), we can then derive (28)–(29). The proof is basically straightforward and goes similarly to the one in Chapter 6 of [13]. We omit the details here.

Notice that in the formulation (62) we need that \(v-{\tilde{v}}\) has compact support in the r-variable. This is due to the fact that we cannot guarantee that the term \(\int r v \,d\nu _t\) which appears in the differential of the surface energy is well-defined.

Proof of Lemmas

Proof of Lemma 3.4

We set \(l=\gamma \varepsilon \), where \(\gamma \) is as in (51). Notice that due to (51) and \(\sum _i\varepsilon ^2 R_i^2=1\), the balls \(\{B_{l}(X_i)\}_i\) are disjoint. For given \(\{\xi _i\}_i\) define

(63)

For further use we collect some properties of \(\phi _i\). It is easily checked that \(\phi _i\) is continuous in \({\mathbb {T}}\), harmonic in \( B_{l}(X_i) {\setminus }\partial B_i \) and satisfies

$$\begin{aligned} \begin{aligned} {[}\nabla \phi _i \cdot \vec n]&= \frac{\varepsilon }{\alpha _{\varepsilon }} \xi _i\cdot \vec n\quad \hbox { on } \partial B_i,\\ {[}\nabla \phi _i \cdot \vec n]&= -\frac{\varepsilon \alpha _{\varepsilon }R_i^2}{ l^2} \xi _i\cdot \vec n\quad \hbox { on } \partial B_{l} (X_i). \end{aligned} \end{aligned}$$
(64)

Furthermore, \(\phi _i=0\) on \(\partial B_{l}(X_i)\) and

$$\begin{aligned} {\phi _i} = -\frac{\varepsilon R_i}{2}\big (1+o(1)) \xi _i\cdot \vec n\quad \text {on } \partial B_i . \end{aligned}$$
(65)

Then

$$\begin{aligned} \begin{aligned} \tfrac{1}{2} |\xi _i|^2&= \frac{1}{|\partial B_i|} \int _{\partial B_i} |\xi _i\cdot {\vec {n}}|^2 \,dS\\&= \frac{1}{|\partial B_i|} \int _{\partial B_i} \frac{\alpha _{\varepsilon }}{\varepsilon } [ \nabla \phi ^{\varepsilon }\cdot {\vec {n}}] \xi _i \cdot {\vec {n}}\,dS\\&= -\frac{2 \alpha _{\varepsilon }}{\varepsilon ^2 R_i|\partial B_i|} \int _{\partial B_i} [ \nabla \phi ^{\varepsilon } \cdot {\vec {n}}] \phi _i \,d S \big ( 1+o(1) \big ). \end{aligned} \end{aligned}$$
(66)

Integration by parts yields

$$\begin{aligned} \begin{aligned} \int _{\partial B_i} [ \nabla \phi ^{\varepsilon }\cdot {\vec {n}}] \phi _i \,dS&=- \int _{B_l (X_i) \backslash B_i} \nabla \phi ^{\varepsilon } \cdot \nabla \phi _i\,dS\\&= \int _{\partial B_i} [ \nabla \phi _i\cdot {\vec {n}}] \phi ^{\varepsilon } \,dS + \int _{\partial B_l (X_i)} [ \nabla \phi _i\cdot {\vec {n}}] \phi ^{\varepsilon } \,dS\\&=\frac{\varepsilon \xi _i }{\alpha _{\varepsilon }} \cdot \int _{\partial B_i} \phi ^{\varepsilon } {\vec {n}} \,dS - \frac{ \varepsilon \alpha _{\varepsilon }R_i^2}{l^2} \int _{\partial B_l (X_i)} \phi ^{\varepsilon } \xi _i \cdot {\vec {n}}\,dS. \end{aligned} \end{aligned}$$
(67)

Together with (66), (67) and (33) we find

$$\begin{aligned} \begin{aligned} |\xi _i|^2&\le C\Big ( \frac{1}{\varepsilon R_i} \xi _i \cdot \frac{1}{|\partial B_i|} \int _{\partial B_i} \phi ^{\varepsilon } {\vec {n}}\,dS + \frac{\alpha _{\varepsilon }}{\varepsilon l^2}\int _{\partial B_l (X_i)} \phi ^{\varepsilon } \xi _i \cdot {\vec {n}} \,dS \Big )\\&\le C \Big ( \frac{1}{\varepsilon R_i |\partial B_i|} \big | \xi _i \cdot \int _{\partial B_i} \mu ^\varepsilon {\vec {n}}\,dS \big |+ \frac{1}{\varepsilon R_i |\partial B_i|}\big | \xi _i \\&\quad \cdot \int _{\partial B_i} u^\varepsilon {\vec {n}} \, d S \big |+\frac{\alpha _{\varepsilon }}{\varepsilon l^2}\big | \int _{\partial B_l (X_i)} \phi ^{\varepsilon } \xi _i \cdot {\vec {n}} \,dS \big | \Big ) . \end{aligned} \end{aligned}$$
(68)

We estimate the last term on the right hand side of (68) via

$$\begin{aligned} \begin{aligned} \frac{\alpha _{\varepsilon }}{\varepsilon l^2}\big |\int _{\partial B_l (X_i)} \phi ^{\varepsilon } \xi _i \cdot {\vec {n}} \,dS\big |&= \frac{\alpha _{\varepsilon }}{\varepsilon l^2}\big |\int _{B_l (X_i)} \nabla \phi ^{\varepsilon } \cdot \xi _i \,dx\big |\\&\le |\xi _i| \frac{\alpha _{\varepsilon }}{\varepsilon l^2}\int _{B_l (X_i)} | \nabla \phi ^{\varepsilon } |\,dx\\&\le C |\xi _i| \frac{\alpha _{\varepsilon }}{\varepsilon l} \Big ( \int _{B_l (X_i)} |\nabla \phi ^{\varepsilon }|^2\,dx \Big )^{1/2}. \end{aligned} \end{aligned}$$
(69)

We can write \(\mu ^\varepsilon = \sum _i \mu _i +{\bar{\mu }}^\varepsilon \) with

(70)

Notice that \(\mu _i\) is continuous in \({\mathbb {T}}\), and satisfies \(-\Delta \mu _i =\frac{\varepsilon ^2}{\alpha _{\varepsilon }^2}-\pi \) in \(B_i\), \(-\Delta \mu _i= -\pi \) in \(B_l (X_i) {\setminus } \overline{B_i}\). These imply that \({\bar{\mu }}^\varepsilon \), and hence \(\nabla {\bar{\mu }}^ \varepsilon \) also are harmonic in \( \cup _i B_l (X_i)\). Then, due to the properties of \(\mu _i\) and the mean value theorem, we have

$$\begin{aligned} \begin{aligned} \int _{\partial B_i} \mu ^\varepsilon {\vec {n}} \,d S&= \int _{B_i} \nabla \mu ^\varepsilon \,d x = \int _{B_i} \nabla \mu _i \,d x + \int _{B_i} \nabla {\bar{\mu }}^\varepsilon \, d x\\&= \nabla \mu _i(X_i) |B_i| + \int _{B_i} \nabla {\bar{\mu }}^\varepsilon \, d x = \int _{B_i} \nabla {\bar{\mu }}^\varepsilon \, d x\\&= \frac{ |B_i|}{|B_l (X_i)|} \int _{B_l (X_i)} \nabla {\bar{\mu }}^\varepsilon \, d x . \end{aligned} \end{aligned}$$
(71)

But

$$\begin{aligned} \int _{B_l (X_i)} \nabla {\bar{\mu }}^\varepsilon \, d x = \int _{\partial B_l (X_i)} {\bar{\mu }}^\varepsilon {\vec {n}} \, d S = \int _{\partial B_l (X_i)} \mu ^\varepsilon {\vec {n}} \, d S = \int _{B_l (X_i)} \nabla \mu ^\varepsilon \, d x , \end{aligned}$$

so

$$\begin{aligned} \left| \int _{\partial B_i} \mu ^\varepsilon {\vec {n}} \,d S \right|&\le C |B_i| l^{-1 } \left( \int _{B_l (X_i)} |\nabla \mu ^\varepsilon |^2 \, d x \right) ^{1/2}. \end{aligned}$$

Next we write \(u^\varepsilon =\sum _i u_i + {\bar{u}}^\varepsilon \) where

(72)

It is easily checked that \(u_i\) is continuous in \({\mathbb {T}}\), harmonic in \(B_l (X_i) {\setminus } \partial B_i \) and satisfies

$$\begin{aligned} {[}\nabla u_i \cdot {\vec {n}}]=\frac{\varepsilon ^2}{\alpha _{\varepsilon }}V_i \quad \text {on } \partial B_i . \end{aligned}$$

Then from a similar argument as above, we see that

$$\begin{aligned} \left| \int _{\partial B_i} u^\varepsilon {\vec {n}} \,d S \right|&\le C |B_i| l^{-1 } \left( \int _{B_l (X_i)} |\nabla u^\varepsilon |^2 \, d x \right) ^{1/2}. \end{aligned}$$

Thus, in summary we find

$$\begin{aligned} |\xi _i| \le C \frac{\alpha _{\varepsilon }}{\varepsilon l} \left( \int _{B_l (X_i)} |\nabla \mu ^\varepsilon |^2 + |\nabla u^\varepsilon |^2+ |\nabla \phi ^{\varepsilon }|^2\,dx \right) ^{1/2}. \end{aligned}$$
(73)

This completes the proof of Lemma. \(\square \)

Proof of Lemma 3.3

In Sect. 3.3 we will see that \(\int _0^{T} \int _{{\mathbb {T}}} | \nabla \phi ^{\varepsilon }|^2 + |\nabla u^{\varepsilon }|^2\,dx\,dt \le C\) (cf. (81)). Furthermore, due to (34), we have \(\int _{{\mathbb {T}}} |\nabla \mu ^{\varepsilon }|^2 \,dx \le C\). Then the statement follows from Lemma 3.4 and

$$\begin{aligned} \begin{aligned} |X_i(t)-X_j(t)|&= |X_i(0) - X_j(0) + \frac{\alpha _{\varepsilon }}{\varepsilon ^2} \int _0^t \xi _i(s) - \xi _j(s) \,ds| \\&\ge |X_i(0) - X_j(0) | - \frac{\alpha _{\varepsilon }}{\varepsilon ^2} \int _0^t |\xi _i(s)|+|\xi _j(s)|\,ds. \end{aligned} \end{aligned}$$

\(\square \)

Proof of Lemma 3.5

We can prove

$$\begin{aligned} \liminf _{\varepsilon \rightarrow 0} \int _0^{T} \beta \int _{{\mathbb {T}}} \vert \nabla u^{\varepsilon }\vert ^2 \, dx \,dt\ge \int _0^{T} \beta \int _{{\mathbb {T}}} \vert \nabla u \vert ^2 \, dx \,dt \; + \; 2 \pi \int _0^T \beta \int \vert v \vert ^2\, r^2 \, d \nu _t \, d t \end{aligned}$$

in the same way as in [13] since (51) is satisfied.

Hence, it remains to show that

$$\begin{aligned} \liminf _{\varepsilon \rightarrow 0} \int _0^{T} \beta \int _{{\mathbb {T}}} \vert \nabla \phi ^{\varepsilon }\vert ^2 \, dx \,dt\ge 2 \pi \int \beta \int \tfrac{1}{4} |\xi |^2 r^2 \, d \nu _t \, d t \end{aligned}$$
(74)

and

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \int _0^{T} \beta \int _{{\mathbb {T}}} \nabla u^{\varepsilon }\cdot \nabla \phi ^{\varepsilon }\,dx\,dt =0. \end{aligned}$$
(75)

Step 1: Monopoles. Our goal is to construct a good approximation of \(\phi ^{\varepsilon }\) which is based on cutting off the single monopole solutions. To that aim we define \(\phi _i\) as in (63) with \(\gamma \) is as in (51) and thus the balls \(\{B_l (X_i)\}_i\) are disjoint. Using (64) and (65), we find

$$\begin{aligned} \begin{aligned} \int _{B_{l} (X_i)} |\nabla \phi _i|^2\,dx&= - \int _{\partial B_i} \phi _i \frac{\varepsilon }{\alpha _{\varepsilon }} \xi _i\cdot \vec n\,dS\\&= \frac{1}{2}\big (1+o(1)\big ) \frac{\varepsilon ^2R_i}{\alpha _{\varepsilon }} \int _{\partial B_i} |\xi _i\cdot \vec n|^2\,dS\\&= \frac{1}{2}\big (1+o(1)\big ) 2 \pi \varepsilon ^2 R_i^2 \fint _{\partial B_i} |\xi _i\cdot \vec n|^2\,dS\\&= \frac{\pi }{2} \varepsilon ^2 R_i^2 |\xi _i|^2 \big (1+o(1)\big ). \end{aligned} \end{aligned}$$
(76)

Step 2: A lower bound. We will show that for any given \(\delta >0\) we have

$$\begin{aligned} \int _{{\mathbb {T}}} |\nabla \phi ^{\varepsilon }|^2\,dx \ge (1-\delta ) \frac{\pi }{2} \sum _i\varepsilon ^2 R_i^2 |\xi _i|^2 \end{aligned}$$
(77)

if \(\varepsilon \) is sufficiently small. Indeed, due to the fact that \(\{B_l(X_i)\}_i\) are disjoint we have

$$\begin{aligned} \int _{{\mathbb {T}}} |\nabla \phi ^{\varepsilon }|^2\,dx \ge \sum _i\int _{B_l(X_i)} |\nabla \phi ^{\varepsilon }| ^2\,dx. \end{aligned}$$

On the other hand

$$\begin{aligned} \int _{B_l(X_i)} |\nabla \phi ^{\varepsilon }|^2\,d x \ge \inf _{\psi } \int _{B_l(X_i)}|\psi |^2\,d x, \end{aligned}$$

where

$$\begin{aligned} \int _{B_l(X_i)} \psi \cdot \nabla \zeta + \int _{\partial B_i} \frac{\varepsilon }{\alpha _{\varepsilon }} \xi _i\cdot \vec n\,\zeta \,dS=0 \end{aligned}$$

for all \(\zeta \in C_0^{\infty }(B_l (X_i))\). From the corresponding Euler–Lagrange equation we see that the minimizer \({\hat{\psi }}\) is orthogonal to all divergence-free function and hence a gradient. We find that \({\hat{\psi }} = \nabla \phi _i\) and thus (77) follows from (76).

Step 3: Approximation of \(\phi ^{\varepsilon }\). With

$$\begin{aligned} {\hat{\phi }}^\varepsilon (x):=\sum _i\phi _i(x)\, , \end{aligned}$$
(78)

we have

$$\begin{aligned} \int _{{\mathbb {T}}} |\nabla \phi ^{\varepsilon }- \nabla {\hat{\phi }}^\varepsilon |^2\,dx \le C \varepsilon ^k \sum _i\varepsilon ^2 R_i^2 |\xi _i|^2 \end{aligned}$$
(79)

for any \(k>0\). In fact, for \(\bar{\phi }^{\varepsilon }:= \phi ^{\varepsilon }- \hat{\phi }^\varepsilon \) it follows from the definitions and (64) that

$$\begin{aligned} \int _{{\mathbb {T}}}\nabla \bar{\phi }^{\varepsilon } \cdot \nabla \zeta \,dx = \pm \sum _i\frac{\alpha _{\varepsilon }R_i^2}{ l^2} \int _{\partial B_l(X_i)} \xi _i\cdot \vec n\,\zeta \, dS \end{aligned}$$
(80)

for all \(\zeta \in \mathring{H}^{1}({\mathbb {T}})\). We now define \(L_{\varepsilon } \in ({H^{1}({\mathbb {T}})})^*\) via

$$\begin{aligned} \langle L_{\varepsilon },\zeta \rangle := \sum _i\frac{\alpha _{\varepsilon }R_i^2}{ l^2} \int _{\partial B_l(X_i)} \zeta \, \xi _i\cdot \vec n\, dS. \end{aligned}$$

We observe that due to \( -\! \! \! \! \!\int _{\partial B_l(X_i)} \xi _i\cdot \vec n\,dS =0\) we can write

$$\begin{aligned} \langle L_{\varepsilon },\zeta \rangle = \sum _i\frac{\alpha _{\varepsilon }R_i^2}{ l^2} \int _{\partial B_l(X_i)} \left( \zeta - \fint _{\partial B_l(X_i)} \zeta \right) \xi _i\cdot \vec ndS. \end{aligned}$$

We can then estimate

$$\begin{aligned} \begin{aligned} \big | \langle L_{\varepsilon },\zeta \rangle \big |&\le C \sum _i\frac{\alpha _{\varepsilon }R_i^2}{ l^2} \Bigg ( \int _{\partial B_l(X_i)} |\xi _i\cdot \vec n| ^2\,dS \Bigg )^{1/2} \Bigg ( \int _{\partial B_l(X_i)}\big | \zeta - \fint _{\partial B_l(X_i)}\zeta |^2\,dS\Bigg )^{1/2}\\&\le C \sum _i\frac{\alpha _{\varepsilon }R_i^2}{ l^2} \Bigg ( \pi l |\xi _i|^2 \Bigg )^{1/2} \Bigg ( C l^2 \int _{B_l(X_i)}|\nabla \zeta |^2\,dx\Bigg )^{1/2}, \end{aligned} \end{aligned}$$

where the last estimate follows from Trace Theorem and Poincaré’s inequality. This gives

$$\begin{aligned} \begin{aligned} \big | \langle L_{\varepsilon },\zeta \rangle \big |&\le C \sum _i\frac{\alpha _{\varepsilon }R_i^2}{ l^2} l^{3/2} |\xi _i| \Bigg (\int _{B_l (X_i) }|\nabla \zeta |^2\,dx\Bigg )^{1/2}\\&\le C \Bigg ( \sum _i\frac{\alpha _{\varepsilon }^2 R_i^2}{\varepsilon ^3} \varepsilon ^2 R_i^2 |\xi _i| ^2\Bigg )^{1/2} \Bigg ( \sum _i\int _{B_l (X_i) } |\nabla \zeta |^2\,dx\Bigg )^{1/2}. \end{aligned} \end{aligned}$$

Due to the simple estimate \(R_i \le C \varepsilon ^{-1/2}\) we find

$$\begin{aligned} \sup _{\zeta \ne 0}\frac{\big | \langle L_{\varepsilon },\zeta \rangle \big |}{\Vert \nabla \zeta \Vert _{L^2({\mathbb {T}})}} \le C \alpha _{\varepsilon }\varepsilon ^{-2} \Bigg ( \sum _i\varepsilon ^2 R_i^2 |\xi _i|^2\Bigg )^{1/2} \le C \varepsilon ^k \Bigg ( \sum _i\varepsilon ^2 R_i^2 |\xi _i|^2\Bigg )^{1/2} \quad \text {for any } k>0, \end{aligned}$$

which proves (79).

Step 4: Bounds on the individual terms.

$$\begin{aligned} \int _0^T\int _{{\mathbb {T}}} |\nabla u^{\varepsilon }|^2 + |\nabla \phi ^{\varepsilon }|^2\,dx\,dt \le 2 \int _0^T \int _{{\mathbb {T}}} |\nabla w^{\varepsilon }|^2\,dx\,dt \le C. \end{aligned}$$
(81)

We write

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla u^{\varepsilon }\cdot \nabla \phi ^{\varepsilon }\,d x = \int _{{\mathbb {T}}} \nabla u^{\varepsilon }\cdot \nabla \bar{\phi }^{\varepsilon }\,dx + \int _{{\mathbb {T}}} \nabla u^{\varepsilon }\cdot \nabla \hat{\phi }^{\varepsilon } \,dx. \end{aligned}$$

Now \(\nabla u^{\varepsilon }\) is orthogonal to \(\nabla {\hat{\phi }}^\varepsilon \) since

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla u^{\varepsilon }\cdot \nabla \hat{\phi }^{\varepsilon }\,dx = -\sum _i\frac{\varepsilon ^3V_i}{\alpha _{\varepsilon }} \int _{\partial B_i} \phi _i\,dS, \end{aligned}$$

but \(\int _{\partial B_i} \phi _i\,dS=0\). Thus

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla u^\varepsilon \cdot \nabla \phi ^\varepsilon \, d x =\int _{{\mathbb {T}}} \nabla u^\varepsilon \cdot \nabla {\bar{\phi }}^\varepsilon \, d x . \end{aligned}$$
(82)

Furthermore, for some small \(\delta >0\), using (77),

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {T}}} \nabla u^{\varepsilon }\cdot \nabla \bar{\phi }^{\varepsilon }\,dx&\ge - \delta \Vert \nabla u^{\varepsilon }\Vert ^2_{L^2({\mathbb {T}})} - C_{\delta } \Vert \nabla \bar{\phi }^{\varepsilon }\Vert _{ L^2({\mathbb {T}})}\\&\ge - \delta \Vert \nabla u^{\varepsilon }\Vert ^2_{L^2({\mathbb {T}})} - C_{\delta } \varepsilon ^4 \sum _i\varepsilon ^3R_i^3 |\xi _i|^2\\&\ge - \delta \Vert \nabla u^{\varepsilon }\Vert ^2_{L^2({\mathbb {T}})} - C_{\delta }\varepsilon ^4 \Vert \nabla \phi ^{\varepsilon }\Vert _{L^2({\mathbb {T}})}. \end{aligned} \end{aligned}$$

We choose \(\delta = \frac{1}{4}\) and then \(\varepsilon \) so small such that \(C_{\delta }\varepsilon ^4\le \frac{1}{4}\). Then (81) follows from (34) and (82).

Step 5: Lower semicontinuity. Since we have now established (81), we obtain the existence of the weak limit \(\xi \) as explained before in Sect. 3.2, and hence the assertion (74) follows from (49) and (77).

Step 6: The mixed term vanishes in the limit.

As an immediate consequence of (77), (79), (81) and (82) we find

$$\begin{aligned} \int _0^T \Vert \nabla \bar{\phi }^{\varepsilon }\Vert ^2_{L^2({\mathbb {T}})} \,dt \le C \varepsilon ^{4} \end{aligned}$$

and thus that (75) is valid. \(\square \)

Proof of Lemma 3.6

We need only to show that for given \({\tilde{\xi }}\) as above we can find \( {{\tilde{\phi }}}^{\varepsilon }\) such that

$$\begin{aligned} \limsup _{\varepsilon \rightarrow 0} \int _0^{T} \beta \int _{{\mathbb {T}}} |\nabla {{\tilde{\phi }}}^{\varepsilon }|^2\,dx\,dt \le 2 \pi \int _0^T \beta \int \tfrac{1}{4} |\tilde{\xi }|^2\,r^2 \,d\nu _t \, d t \end{aligned}$$

and

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \int _0^{T} \beta \int _{{\mathbb {T}}} \nabla {{\tilde{u}}}^{\varepsilon }\cdot \nabla {{\tilde{\phi }}}^{\varepsilon }\,dx\,dt =0 \end{aligned}$$

since the corresponding result for \({\tilde{u}}^\varepsilon \) can be proved in the same way as in [13]. The proof is in fact quite similar to the proof of Lemma 3.5, since there the minimization property of \(\{\xi _i\}_i\) is not used in the construction, which is henceforth quite general.

Step 1: Construction for smooth \({\tilde{\xi }}\). We first assume that \({\tilde{\xi }}\) is smooth. Then we define \({\tilde{\xi _i}}:= \tilde{\xi }(X_i)\) and construct \(\phi _i\) as in (63) for \(\xi _i={\tilde{\xi }}_i\), and a corresponding \({\hat{\phi }}^\varepsilon \) as in (78). The property (76) implies

$$\begin{aligned} \int _{{\mathbb {T}}} |\nabla {\hat{\phi }}^\varepsilon |^2\,dx = \frac{\pi }{2} \sum _i\varepsilon ^2 |{\tilde{\xi _i}}| ^2R_i^2 \big (1+o(1)\big ) \end{aligned}$$

and hence

$$\begin{aligned} \begin{aligned} \int _0^T \beta \int _{{\mathbb {T}}} |\nabla {\hat{\phi ^{\varepsilon }}}|^2\,dx\,dt&= \pi \int _0^T \beta \int \tfrac{1}{2} |{\tilde{\xi }}|^2 r^2 \,d\nu ^{\varepsilon }_t\,dt \big (1+o(1)\big )\\&\quad \rightarrow \pi \int _0^T \beta \int \tfrac{1}{2} |{\tilde{\xi }}|^2 r^2 \,d\nu _t\,dt \end{aligned} \end{aligned}$$

as \(\varepsilon \rightarrow 0\). Here we used (43) and that \({\tilde{\xi }}\) has compact support.

Furthermore, we obtain also exactly as in the proof of Lemma 3.5 that \({\bar{\phi }}^\varepsilon := {\tilde{\phi }}^\varepsilon - \hat{\phi }^\varepsilon \) converges to zero strongly in \(L^2((0,T);H^{1}({\mathbb {T}}))\), and as a consequence the mixed term vanishes in the limit.

Step 2: Construction for general \({\tilde{\xi }}\).

In order to finish the proof of the lemma we have to show that we can approximate \({\tilde{\xi }} \in L^2(r^2\,d\nu _t\,d t)^3\) by smooth functions \({\tilde{\xi }}_n\) with compact support such that

$$\begin{aligned} \int _0^T \beta \int |{\tilde{\xi }}_n|^2 r^2 \,d\nu _t\,dt \rightarrow \int _0^T \beta \int |{\tilde{\xi }}|^2 r^2 \,d\nu _t\,dt \end{aligned}$$

as \(n \rightarrow \infty \). But this follows from a density argument. \(\square \)

Proof of Lemma 3.1

We first prove (38). Set \(l=\gamma \varepsilon \), where \(\gamma \) is as in (51). Due to the fact that \(\{B_l (X_i)\}_i\) are disjoint we have

$$\begin{aligned} \int _{{\mathbb {T}}} | \nabla \mu ^\varepsilon |^2 \, d x \ge \sum _i \int _{B_l (X_i)} | \nabla \mu ^\varepsilon |^2 \, d x . \end{aligned}$$
(83)

On the other hand, for each i,

$$\begin{aligned} \int _{B_l (X_i)} | \nabla \mu ^\varepsilon |^2 \, d x \ge \inf _{\psi } \int _{B_l (X_i)} | \psi |^2 \, d x , \end{aligned}$$

where the infimum is taken over all \(\psi \in (L^2(B_l (X_i)))^2\) which satisfy

$$\begin{aligned} \int _{B_l (X_i)} \psi \cdot \nabla \zeta \, d x = \int _{B_l (X_i)} \left( \tfrac{\varepsilon ^2}{\alpha _{\varepsilon }^2}\chi _{B_i}-\pi \right) \zeta \, d x \end{aligned}$$

for all \(\zeta \in C_0^\infty (B_l (X_i))\). We see that the minimizer \({\hat{\psi }}\) is orthogonal to all divergence-free vector-valued functions. Hence \({\hat{\psi }}=\nabla \phi \), \(\phi =\text {const.}\) on \(\partial B_l (X_i)\), where \(-\Delta \phi =\tfrac{\varepsilon ^2}{\alpha _{\varepsilon }^2}\chi _{B_i}-\pi \). We find that \({\hat{\psi }}=\nabla \mu _i\), where \(\mu _i\) is defined as in (70), and we have

$$\begin{aligned} \int _{B_l (X_i)} | {\hat{\psi }} |^2 \, d x&= \int _{B_l (X_i)} | \nabla \mu _i |^2 \, d x \\&\ge \int _{B_l (X_i) {\setminus } B_i} \frac{(\varepsilon R_i)^4}{4|x-X_i|^2}-\frac{\pi \varepsilon ^2 R_i^2}{2} \, d x\\&= 2\pi \frac{(\varepsilon R_i)^4}{4} \ln \frac{l}{\alpha _{\varepsilon }R_i} -\frac{\pi ^2}{2} (l^2 - (\alpha _{\varepsilon }R_i)^2 )(\varepsilon R_i)^2 . \end{aligned}$$

Then it follows from (34), \(\sum _i \varepsilon ^2 R_i^2 =1\) and (83) that

$$\begin{aligned} \sum _i \varepsilon ^2 R_i^{\,4} (t)\le C \quad \text {for all } t \in [0,T], \end{aligned}$$
(84)

which proves (38). \(\square \)

Proof of Lemma 3.10

We can prove this in a similar way as in [14, section 3.4] by making use of (32) and Lemma A.1 although we are using Neumann boundary condition on \(\partial B_i\) instead of Dirichlet boundary condition.

Note \(0<\frac{1}{1-\varepsilon ^2 \ln R_i} \le \frac{1}{1+\varepsilon ^2 \ln \varepsilon } \le \frac{2e}{2e-1}\) by \(0<R_i \le 1/\varepsilon \) and \(|B_l(X_i)|/|A_i|\le 2\) for small \(\varepsilon \) with \(l=\gamma \varepsilon \) and

$$\begin{aligned} A_i=\{x\in {\mathbb {T}} \,;\, \alpha _\varepsilon R_i<|x-R_i| <\gamma \varepsilon \}. \end{aligned}$$

Setting \(v^\varepsilon =u^\varepsilon +\phi ^\varepsilon -2\sigma \mu ^\varepsilon -\lambda ^\varepsilon \), we have

$$\begin{aligned}&\int \frac{1}{1-\varepsilon ^2 \ln r}\frac{1}{r^2} \, d \nu _t^\varepsilon = \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln R_i}\frac{1}{R_i^2}\\&\quad \le C \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln R_i} \bigg | \frac{1}{R_i} - \frac{1}{|A_i|}\int _{A_i} v^\varepsilon \, d x \bigg |^2 + C \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln R_i} \bigg | \frac{1}{|A_i|}\int _{A_i} v^\varepsilon \, d x\bigg |^2\\&\quad \le C \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln R_i} \bigg | \frac{1}{R_i} - \frac{1}{|A_i|}\int _{A_i} v^\varepsilon \, d x \bigg |^2 + C \sum _i \frac{1}{|A_i|} \bigg |\int _{A_i} v^\varepsilon \, d x\bigg |^2 \end{aligned}$$

since \(|B_l(X_i)|/|A_i|\le 2\) for small \(\varepsilon \). Using

$$\begin{aligned} \bigg |\int _{A_i} v^\varepsilon \, d x\bigg |^2 \le |A_i| \int _{A_i} (v^\varepsilon )^2 \, d x , \end{aligned}$$

we see that

$$\begin{aligned} \int \frac{1}{1-\varepsilon ^2 \ln r}\frac{1}{r^2} \, d \nu _t^\varepsilon&\le C \int _{\mathbb {T}}|v^\varepsilon |^2 \, d x + C \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln R_i} \bigg | \frac{1}{R_i} - \frac{1}{|A_i|}\int _{A_i} v^\varepsilon \, d x \bigg |^2 . \end{aligned}$$
(85)

Now we estimate the second term of the right hand side of (85). Note that \(v^\varepsilon =u^\varepsilon +\phi ^\varepsilon -2\sigma \mu ^\varepsilon -\lambda ^\varepsilon \) satisfies

$$\begin{aligned} \int _{\partial B_i} v^\varepsilon - \frac{1}{R_i} \, d S = 0 \end{aligned}$$

for each i. Using Lemma A.1 for \(f=v^\varepsilon -\frac{1}{R_i}\), we have

$$\begin{aligned} \bigg |\frac{1}{|A_i|}\int _{A_i} \bigg ( \frac{1}{R_i} - v^\varepsilon \bigg )\, d x \bigg |^2 \le C \frac{|B_l(X_i)|^2}{|A_i|^2} \ln \bigg ( \frac{\gamma \varepsilon }{\alpha _ \varepsilon R_i}\bigg ) \int _{A_i} |\nabla v^\varepsilon |^2 \, d x, \end{aligned}$$

and since \(|B_l(X_i)|/|A_i|\le 2\) for small \(\varepsilon \),

$$\begin{aligned} \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln R_i} \bigg | \frac{1}{R_i} - \frac{1}{|A_i|}\int _{A_i} v^\varepsilon \, d x \bigg |^2 \le C \int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x. \end{aligned}$$
(86)

It follows from (85) and (86) that

$$\begin{aligned} \int \frac{1}{1-\varepsilon ^2 \ln r}\frac{1}{r^2} \, d \nu _t^\varepsilon&\le C \bigg ( \int _{\mathbb {T}}|v^\varepsilon |^2 \, d x + \int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x \bigg ) . \end{aligned}$$
(87)

We estimate the \(L^2\) norm of \(v^\varepsilon \).

$$\begin{aligned} \int _{\mathbb {T}}|v^\varepsilon |^2 \, d x&\le C \int _{\mathbb {T}}|v^\varepsilon - \int _{\mathbb {T}}v^\varepsilon |^2 \, d x + C \bigg | \int _{\mathbb {T}}v^\varepsilon \, d x\bigg |^2 \nonumber \\&\le C \int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x + C \bigg | \int _{\mathbb {T}}v^\varepsilon \, d x\bigg |^2. \end{aligned}$$
(88)

Using

$$\begin{aligned} \bigg | |A_i| \int _{\mathbb {T}}v^\varepsilon \, d x\bigg |&\le \bigg | \int _{A_i} \bigg (\int _{\mathbb {T}}v^\varepsilon - v^\varepsilon \bigg )\bigg | + \bigg | \int _{A_i} v^\varepsilon \bigg | \nonumber \\&\le \int _{A_i} \bigg | \int _{\mathbb {T}}v^\varepsilon - v^\varepsilon \bigg | + \bigg | \int _{A_i} v^\varepsilon \bigg | \nonumber \\&\le C \varepsilon \bigg [ \int _{A_i} \bigg | \int _{\mathbb {T}}v^\varepsilon - v^\varepsilon \bigg | ^2 \bigg ]^{1/2} + \bigg | \int _{A_i} v^\varepsilon \bigg | , \end{aligned}$$
(89)

we see that

$$\begin{aligned}&\bigg ( \int \frac{r}{1-\varepsilon ^2 \ln r} \, d \nu _t^\varepsilon \bigg )\bigg | \int _{\mathbb {T}}v^\varepsilon \bigg | \nonumber \\&\quad \le C\sum _i \frac{R_i}{1-\varepsilon ^2 \ln R_i} \bigg | |A_i| \int _{\mathbb {T}}v^\varepsilon \, d x\bigg | \nonumber \\&\quad \le C \Bigg ( \sum _i \varepsilon ^2R_i^2 \Bigg )^{1/2} \Bigg ( \sum _i \int _{A_i} \bigg | \int _{\mathbb {T}}v^\varepsilon - v^\varepsilon \bigg | ^2 \Bigg )^{1/2} + C \sum _i \frac{R_i}{1-\varepsilon ^2 \ln R_i} \bigg | \int _{A_i} v^\varepsilon \bigg | \nonumber \\&\quad \le C \Bigg (\int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x \Bigg )^{1/2} + C \sum _i \frac{R_i}{1-\varepsilon ^2 \ln R_i} \bigg | \int _{A_i} v^\varepsilon \bigg |. \end{aligned}$$
(90)

We estimate the second term of the right hand side of (90).

$$\begin{aligned}&\sum _i \frac{R_i}{1-\varepsilon ^2 \ln R_i} \bigg | \int _{A_i} v^\varepsilon \bigg |\nonumber \\&\quad \le \sum _i \frac{R_i |A_i|}{1-\varepsilon ^2 \ln R_i} \bigg | \frac{1}{|A_i|}\int _{A_i} v^\varepsilon - \frac{1}{R_i}\bigg | + \sum _i \frac{|A_i|}{1-\varepsilon ^2 \ln R_i}\nonumber \\&\quad \le C \bigg ( \sum _i \varepsilon ^2 R_i^2 \bigg )^{1/2} \bigg ( \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln R_i} \bigg | \frac{1}{|A_i|}\int _{A_i} v^\varepsilon - \frac{1}{R_i}\bigg |^2 \bigg )^{1/2} +C\sum _i \varepsilon ^2 \nonumber \\&\quad \le C \bigg (1+\bigg (\int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x \bigg )^{1/2} \bigg ). \end{aligned}$$
(91)

Note that due to

$$\begin{aligned} 1=\Bigg ( \sum _i \varepsilon ^2R_i^2\Bigg )^2 \le \Bigg ( \sum _i\varepsilon ^2 R_i^{3} (1-\varepsilon ^2 \ln R_i) \Bigg )\Bigg (\sum _i\frac{\varepsilon ^2 R_i}{1-\varepsilon ^2 \ln R_i} \Bigg ), \end{aligned}$$
(92)

\(-\ln R_i \le \frac{1}{e\alpha R_i^\alpha }\), \(\sum _i \varepsilon ^4 R_i^2 \le C\) and \(\sum _i \varepsilon ^2 R_i^3 \le C\), there is some c, independent of \(\varepsilon \), such that

$$\begin{aligned} \sum _i \frac{\varepsilon ^2 R_i}{1-\varepsilon ^2 \ln R_i} \ge c > 0 . \end{aligned}$$

Then it follows from (90) and (91) that

$$\begin{aligned} \bigg | \int _{{\mathbb {T}}} v^\varepsilon \bigg | \le C \bigg (1+\bigg (\int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x \bigg )^{1/2} \bigg ). \end{aligned}$$
(93)

Then by (88),

$$\begin{aligned} \int _{\mathbb {T}}|v^\varepsilon |^2 \, d x \le C \bigg ( \int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x +1 \bigg ). \end{aligned}$$

Thus we obtain from (87),

$$\begin{aligned} \int \frac{1}{1-\varepsilon ^2 \ln r}\frac{1}{r^2} \, d \nu _t^\varepsilon \le C \bigg ( \int _{\mathbb {T}}|\nabla v^\varepsilon |^2 \, d x +1 \bigg ). \end{aligned}$$

\(\square \)

Proof of Lemma 3.9

Step 1: An expression for \(V_i\). Similarly to \(\phi _i\) in the proof of Lemma 3.4 we introduce a suitable test function \(v_i\). Here it is the capacity potential of \(B_i\) with respect to \(B_l(X_i)\), with \(l = \gamma \varepsilon \). This gives that

with \(a^{\varepsilon } = (\ln \frac{l}{\alpha _{\varepsilon }R_i} )^{-1} \). With this definition we also have

$$\begin{aligned} {[}\nabla v_i \cdot {\vec {n}} ]= - \frac{a^{\varepsilon }}{\alpha _{\varepsilon }R_i} \quad \hbox { on } \partial B_i \qquad \hbox { and } \qquad {[}\nabla v_i \cdot {\vec {n}} ]= \frac{a^{\varepsilon }}{l} \quad \hbox { on } \partial B_l (X_i) . \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} 2 \pi R_i V_i&= 2 \pi R_i \frac{\alpha _{\varepsilon }}{\varepsilon ^2} [\nabla u^{\varepsilon } \cdot {\vec {n}}]\\&= \frac{1}{\varepsilon ^2} \int _{\partial B_i} [\nabla u^{\varepsilon } \cdot {\vec {n}}] \,dS\\&= \frac{1}{\varepsilon ^2} \int _{\partial B_i} [\nabla u^{\varepsilon } \cdot {\vec {n}}] v_i \,dS\\&= \frac{1}{\varepsilon ^2} \int _{\partial B_i} [\nabla v_i \cdot {\vec {n}}] u^{\varepsilon } \,dS +\frac{1}{\varepsilon ^2} \int _{\partial B_l (X_i)} [\nabla v_i\cdot {\vec {n}}] u^{\varepsilon } \,dS\\&= - \frac{a^{\varepsilon }}{\varepsilon ^2 \alpha _{\varepsilon }R_i} \int _{\partial B_i} u^{\varepsilon }\,dS + \frac{a^{\varepsilon }}{\varepsilon ^2 l} \int _{\partial B_l (X_i)} u^{\varepsilon } \,dS. \end{aligned} \end{aligned}$$
(94)

Due to (32) it follows that

$$\begin{aligned} R_i V_i = - \frac{a^{\varepsilon }}{\varepsilon ^2 } \Big ( \frac{1}{R_i} + \frac{1}{|\partial B_i|}\int _{\partial B_i} (2 \sigma \mu ^\varepsilon -\phi ^ \varepsilon ) \, d x + \lambda ^\varepsilon (t) \Big ) + \frac{a^{\varepsilon }}{2\pi \varepsilon ^2 l} \int _{\partial B_l (X_i)} u^{\varepsilon } \,dS. \end{aligned}$$
(95)

Using that \(\mu ^\varepsilon = \sum _i \mu _i + {\bar{\mu }}^\varepsilon \) with \( \mu _i\) as in (70) we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{|\partial B_i|} \int _{\partial B_i} \mu ^\varepsilon \,dx&= \frac{1}{|\partial B_i|}\int _{\partial B_i} \mu _i \,dx + \frac{1}{|\partial B_i|}\int _{\partial B_i} {\bar{\mu }}^\varepsilon \,dx \\&= \frac{\varepsilon ^2 R_i^2}{2}\ln \frac{l}{\alpha _{\varepsilon }R_i} +O(\varepsilon ^2) + \frac{1}{|\partial B_l (X_i)|}\int _{\partial B_l (X_i)} {\bar{\mu }}^\varepsilon \,dx \\&= \frac{\varepsilon ^2 R_i^2}{2}\ln \frac{l}{\alpha _{\varepsilon }R_i} +O(\varepsilon ^2) + \frac{1}{|\partial B_l (X_i)|}\int _{\partial B_l (X_i)} \mu ^\varepsilon \,dx . \end{aligned} \end{aligned}$$

Similarly using that \(\phi ^\varepsilon =\sum _i \phi _i+{\bar{\phi }}^\varepsilon \) with \(\phi _i\) as in (63) we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{|\partial B_i|} \int _{\partial B_i} \phi ^\varepsilon \,dx&= \frac{1}{|\partial B_i|}\int _{\partial B_i} \phi _i \,dx + \frac{1}{|\partial B_i|}\int _{\partial B_i} {\bar{\phi }}^\varepsilon \,dx \\&= \frac{1}{|\partial B_l (X_i)|}\int _{\partial B_l (X_i)} {\bar{\phi }}^\varepsilon \,dx = \frac{1}{|\partial B_l (X_i)|}\int _{\partial B_l (X_i)} \phi ^\varepsilon \,dx . \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} R_i V_i = - \frac{a^{\varepsilon }}{\varepsilon ^2} \Big ( \frac{1}{R_i} +O(\varepsilon ^2) + \lambda ^\varepsilon (t) \Big ) -\sigma R_i^2 + \frac{a^{\varepsilon }}{2\pi \varepsilon ^2 l} \int _{\partial B_l (X_i)} u^{\varepsilon } +\phi ^\varepsilon - 2 \sigma \mu ^\varepsilon \,dS. \end{aligned}$$
(96)

Step 2: A bound on \(\int _{\partial B_l (X_i)} u^{\varepsilon } +\phi ^\varepsilon - 2 \sigma \mu ^\varepsilon \,d S\).

We define \(\psi (x)= \frac{1}{2} (l^2 - |x-X_i|^2)\) in \(B_l (X_i)\) such that

$$\begin{aligned} -\Delta \psi = 2 \; \hbox { in } B_l (X_i)\,, \qquad -\nabla \psi = x-X_i \hbox { in } B_l (X_i)\, \qquad \hbox { and } \; \nabla \psi \cdot {\vec {n}} = -l \; \hbox { on } \partial B_l (X_i). \end{aligned}$$

Thus, abbreviating \(v^{\varepsilon }= u^{\varepsilon } + \phi ^{\varepsilon } - 2 \sigma \mu ^{\varepsilon } \), we have

$$\begin{aligned} \begin{aligned} \frac{1}{l} \int _{\partial B_l (X_i)} v^{\varepsilon } \,dS&= - \frac{1}{l^2} \int _{\partial B_l (X_i)} v^{\varepsilon } \nabla \psi \cdot {\vec {n}}\,dS\\&= - \frac{1}{l^2} \int _{B_l (X_i)} \Delta \psi v^{\varepsilon } + \nabla \psi \cdot \nabla v^{\varepsilon } \,dx\\&=\frac{3}{l^2} \int _{B_l (X_i)} v^{\varepsilon } - \frac{1}{l^2} \int _{B_l (X_i)} \nabla \psi \cdot \nabla v^{\varepsilon } \,dx. \end{aligned} \end{aligned}$$
(97)

We multiply (97) by \(R_i^m a^\varepsilon \) with \(m\ge 0\), and sum over \(i \in I\) for some \(I \subset \{i:R_i>0\}\). Then due to \(a^{\varepsilon } \le 2 \varepsilon ^2\), we find, using \(l=\gamma \varepsilon \) and Hölder’s inequality, that

$$\begin{aligned} \begin{aligned}&\Big | \sum _{i \in I} R_i^m \frac{a^{\varepsilon }}{l} \int _{\partial B_l (X_i)} v^{\varepsilon } \,dS\Big | \\&\quad \le C \Bigg ( \sum _{i \in I} R_i^{m} \int _{B_l (X_i)} |v^{\varepsilon }| \,+\, \sum _{i \in I} R_i^{m} \int _{B_l (X_i)} |\nabla \psi \cdot \nabla v^{\varepsilon } |\,dx \Bigg )\\&\quad \le C \left( \sum _{i \in I} R_i^{m} l^{5/3} \left( \int _{B_l (X_i)} |v^{\varepsilon }|^6 \right) ^{1/6} + \sum _{i \in I} R_i^{m} l^{2} \left( \int _{B_l (X_i)} |\nabla v^{\varepsilon }|^2 \right) ^{1/2}\right) \\&\quad \le C \bigg ( \sum _{i \in I} \varepsilon ^2 R_i^{6m/5} \bigg )^{5/6} \bigg ( \int _{{\mathbb {T}}} | v^\varepsilon |^6\bigg )^{1/6} + C \bigg ( \sum _{i \in I} \varepsilon ^4 R_i^{2m} \bigg )^{1/2} \left( \int _{{\mathbb {T}}} |\nabla v^{\varepsilon }|^2\right) ^{1/2}. \end{aligned} \end{aligned}$$

Due to Sobolev embedding and Poincare’s inequality (recall that \(v^{\varepsilon }\) has mean value zero) we finally find

$$\begin{aligned} \begin{aligned}&\Bigg | \sum _{i \in I} R_i^m \frac{a^{\varepsilon }}{l} \int _{\partial B_l (X_i)} u^\varepsilon +\phi ^\varepsilon - 2 \sigma \mu ^\varepsilon \,dS\Bigg | \\&\quad \le C \bigg ( \int _{{\mathbb {T}}} |\nabla \mu ^\varepsilon |^2 +|\nabla \phi ^\varepsilon |^2 + |\nabla u^\varepsilon |^2 \bigg )^{1/2} \Bigg \{ \bigg ( \sum _{i \in I} \varepsilon ^2 R_i^{6m/5} \bigg )^{5/6} + \bigg ( \sum _{i \in I} \varepsilon ^4 R_i^{2m} \bigg )^{1/2} \Bigg \}. \end{aligned} \end{aligned}$$
(98)

Step 3: A bound for \(\lambda ^\varepsilon (t)\). We go back to (96) to estimate \(\lambda ^\varepsilon (t)\). For that purpose we multiply (96) by \(\varepsilon ^2 \) and sum over i. We find

$$\begin{aligned} \begin{aligned}&0= - \sum _i a^{\varepsilon } \Bigg ( \frac{1}{R_i} + O(\varepsilon ^2) \Bigg ) - \sigma \sum _i \varepsilon ^2 R_i^2 \\&\quad - \lambda ^\varepsilon (t) \sum _i a^{\varepsilon } + \sum _i \frac{ a^{\varepsilon }}{2\pi l} \int _{\partial B_l (X_i)} u^{\varepsilon } + \phi ^\varepsilon - 2\sigma \mu ^\varepsilon \,dS. \end{aligned} \end{aligned}$$
(99)

First notice that due to

$$\begin{aligned} 1= & {} \Bigg ( \sum _i \varepsilon ^2 R_i^2(t) \Bigg )^5 \le \Bigg (\sum _i \varepsilon ^2 \Bigg ) \Bigg ( \sum _i \varepsilon ^4 R_i^4 \ln \frac{l}{\alpha _{\varepsilon }R_i} \Bigg )^2 \Bigg ( \sum _i \frac{R_i}{\ln \frac{l}{\alpha _{\varepsilon }R_i}} \Bigg )^2\\ 1= & {} \Bigg ( \sum _i \varepsilon ^2 R_i^2(t) \Bigg )^2 \le \Bigg ( \sum _i \varepsilon ^4 R_i^4 \ln \frac{l}{\alpha _{\varepsilon }R_i} \Bigg ) \Bigg ( \sum _i \frac{1}{\ln \frac{l}{\alpha _{\varepsilon }R_i}} \Bigg ) \end{aligned}$$

we have \(\left( \sum _i \frac{1}{ \ln \frac{l}{\alpha _{\varepsilon }R_i}} \right) ^{-1} \le \sum _i \varepsilon ^4 R_i^4 \ln \frac{l}{\alpha _{\varepsilon }R_i} \le C \) due to (84). As a consequence, we also have \(\frac{1}{\sum _i a^{\varepsilon }} \le C\).

Taking \(m=0\) in (98), we find

$$\begin{aligned} \begin{aligned} |\lambda ^\varepsilon (t)|&\le C \Bigg ( \sum _i \frac{a^\varepsilon }{R_i}+ \sum _i \varepsilon ^2 (\varepsilon ^2+R_i^2) + \bigg ( \int _{{\mathbb {T}}} |\nabla u^{\varepsilon }|^2 + |\nabla \mu ^\varepsilon |^2 + | \nabla \phi ^\varepsilon |^2 \bigg )^{1/2} \Bigg ) \\&\le C \Bigg ( 1+ \sum _i \frac{a^\varepsilon }{R_i}+ \big ( \int _{{\mathbb {T}}} |\nabla u^{\varepsilon }|^2 + |\nabla \mu ^ \varepsilon |^2 + |\nabla \phi ^\varepsilon |^2 \big )^{1/2}\Bigg ). \end{aligned} \end{aligned}$$
(100)

Using

$$\begin{aligned} \sum _i \frac{a^\varepsilon }{R_i} \le \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln \frac{R_i}{\gamma }}\frac{1}{R_i}\le & {} \Bigg ( \sum _i \varepsilon ^2 \Bigg )^{1/2} \bigg ( \sum _i \frac{\varepsilon ^2}{1-\varepsilon ^2 \ln \frac{R_i}{\gamma }}\frac{1}{R_i^2} \bigg )^{1/2} \\\le & {} C \bigg ( \int \frac{1}{1-\varepsilon ^2 \ln r} \frac{1}{r^2} \, d \nu _t^\varepsilon \bigg )^{1/2}, \end{aligned}$$

and Lemma 3.10, it follows [cf. (81) in Sect. 3.3] that

$$\begin{aligned} \int _0^T |\lambda ^\varepsilon (t)|\,dt \le C_T. \end{aligned}$$
(101)

Step 4: Completion of Proof. Now we go back to (96), multiply with \(\varepsilon ^2 R_i^2\) and sum, but only over i such that \(R_i \ge M\). We find

$$\begin{aligned} \begin{aligned} \sum _{R_i \ge M} \varepsilon ^2 R_i^3 V_i&= \Bigg ( - \sum _{R_i \ge M} \varepsilon ^2 R_i - \lambda ^\varepsilon (t) \sum _{R_i \ge M} \varepsilon ^2 R_i^2 \\&\quad + \sum _{R_i \ge M} \frac{\varepsilon ^2 R_i^2}{2\pi l} \int _{\partial B_l (X_i)} u^{\varepsilon }+\phi ^\varepsilon - 2\sigma \mu ^\varepsilon \,dS \Bigg ) (1+O(\varepsilon )) - \sigma \sum _{R_i \ge M} \varepsilon ^2 R_i^4 . \end{aligned} \end{aligned}$$
(102)

Here we used \(\frac{a^\varepsilon }{ \varepsilon ^{2}} =\frac{1}{1+ \varepsilon ^2 \ln \frac{\gamma }{ R_i}}=1+O(\varepsilon )\) as \(\varepsilon \rightarrow 0\) for i such that \(R_i\ge M\), \(\varepsilon R_i\le 1\) and \(M>\gamma \).

The key feature of (102) is that we have a term of the form \(- \sum _{R_i \ge M} \varepsilon ^2 R_i^4\) on the right hand side and otherwise terms which converge to zero as \(M\rightarrow \infty \) uniformly in \(\varepsilon \). Now define

$$\begin{aligned} y(t):= \frac{ 1}{4} \sum _{R_i \ge M} \varepsilon ^2 R_i^4(t) \end{aligned}$$

such that it follows from (102) together with (98) with \(m=2\) that

$$\begin{aligned} \begin{aligned}&y'(t) +4 \sigma y(t) \le C \Bigg \{ (|\lambda ^\varepsilon (t)| +1)\sum _{R_i \ge M} \varepsilon ^2 R_i^2 \\&\quad + \big ( \int _{{\mathbb {T}}} |\nabla u^{\varepsilon }|^2+|\nabla \mu ^{\varepsilon }|^2 +|\nabla \phi ^{\varepsilon }|^2 \big )^{1/2} \Bigg ( \bigg ( \sum _{R_i \ge M} \varepsilon ^2 R_i^{12/5} \bigg )^{5/6} + \varepsilon \big ( \sum _{R_i \ge M} \varepsilon ^2 R_i^4 \big )^{1/2} \Bigg ) \Bigg \} . \end{aligned} \end{aligned}$$
(103)

Now

$$\begin{aligned}&\sum _{R_i \ge M} \varepsilon ^2 R_i^2 \le \frac{1}{M^2} \sum _{R_i \ge M} \varepsilon ^2 R_i^4 \le \frac{C}{M^2}\,, \end{aligned}$$
(104)
$$\begin{aligned}&\sum _{R_i \ge M} \varepsilon ^2 R_i^{12/5} \le \frac{1}{M^{8/5}} \sum _{R_i \ge M} \varepsilon ^2 R_i^4 \le \frac{C}{M^{8/5}}. \end{aligned}$$
(105)

To estimate the last term on the right-hand side of (103) we recall that we have the simple estimate \(R_i \le C\varepsilon ^{-1/2}\) such that

$$\begin{aligned}&\varepsilon \bigg ( \sum _{R_i \ge M} \varepsilon ^2 R_i^4\bigg )^{1/2} \le \varepsilon ^{1/2} \bigg ( \sum _{R_i \ge M} \varepsilon ^2 R_i^2\bigg )^{1/2} \le C \frac{\varepsilon ^{1/2}}{M}\,, \end{aligned}$$
(106)
$$\begin{aligned}&\sum _{R_i \ge M} \varepsilon ^4 R_i^4 = \varepsilon ^2 \sum _{R_i \ge M} \varepsilon ^2 R_i^4 \le C\frac{\varepsilon }{M^2}. \end{aligned}$$
(107)

Thus, (103)–(107) imply that we have an estimate of the form

$$\begin{aligned} y'(t) + 4 \sigma y(t) \le C \frac{ f(t)}{M^{\delta }} \qquad \hbox { with } \int _0^T |f(t)|\,dt \le C_T\, \quad \hbox { and some } \delta >0. \end{aligned}$$
(108)

A simple comparison argument now implies that indeed \(y(t) \le C \big (y(0)+ M^{-\delta } \big )\) and thus the statement of the lemma follows with (26). \(\square \)

Proof of Lemma 3.7

We use an argument similar to Step 2 of the proof of [13, Lemma 5.4], and give only an outline of the proof. It follows from the definition of K (59) that

$$\begin{aligned} \int _{{\mathbb {T}}} \nabla K\cdot \nabla \zeta \, d x = 2 \pi \bigg (\int r^2 \zeta \, d \nu _t - \int _{{\mathbb {T}}} \zeta \, d x \bigg ) = 2 \pi \int r^2 \zeta \, d \nu _t \end{aligned}$$
(109)

for all \(\zeta \in \mathring{H}^{1}({\mathbb {T}})\). Furthermore since \(\Delta {\tilde{\mu }}^\varepsilon = \pi \) in \({\mathbb {T}}{\setminus } \cup _i ( \partial B_l (X_i) \cup \partial B_i)\), \([\nabla {\tilde{\mu }}^\varepsilon \cdot {\vec {n}}]= 0\) on \(\partial B_i\), and \([\nabla {\tilde{\mu }}^\varepsilon \cdot {\vec {n}}]=- \frac{\varepsilon ^2 R_i^2}{2 l}\) on \(\partial B_l (X_i)\), we see that \({\tilde{\mu }}^\varepsilon \) solves

$$\begin{aligned} \int _{{\mathbb {T}}}\nabla {\tilde{\mu }}^\varepsilon \cdot \nabla \zeta \, d x&= \sum _i \int _{\partial B_l (X_i)} [\nabla {\tilde{\mu }}^\varepsilon \cdot {\vec {n}}] \zeta \, d S - \pi \int _{{\mathbb {T}}} \zeta \, d x \\&=\sum _i \int _{\partial B_l (X_i)} \frac{l R_i^2 }{ 2 \gamma ^2} \zeta \, d S \end{aligned}$$

for all \(\zeta \in \mathring{H}^{1}({\mathbb {T}})\). We define

$$\begin{aligned} \langle L_\varepsilon (t),\zeta \rangle := \sum _{i} \int _{\partial B_{l} (X_i)} \frac{l R_i^2 }{ \gamma ^2} \zeta \, d S , \qquad \langle L(t),\zeta \rangle := 2 \pi \int r^2 \zeta \, d \nu _t \end{aligned}$$

for \(\zeta \in \mathring{H}^{1}({\mathbb {T}})\). Then \(L_\varepsilon (t)\) and L(t) are elements of \(\mathring{H}^{-1}({\mathbb {T}})=(\mathring{H}^{1}({\mathbb {T}}))^*\) for all \(t \in [0,T]\). Since \(\langle L_\varepsilon (t),\zeta \rangle = 2 \int _{{\mathbb {T}}}\nabla {\tilde{\mu }}^\varepsilon \cdot \nabla \zeta \, d x\) and \( \langle L(t),\zeta \rangle =\int _{{\mathbb {T}}} \nabla K\cdot \nabla \zeta \, d x\), we only need to prove that \(L_\varepsilon \rightarrow L\) strongly in \(C^0([0,T];\mathring{H}^{-1}({\mathbb {T}}))\).

In order to prove this, define

$$\begin{aligned} \langle {\tilde{L}}_\varepsilon (t),\zeta \rangle := \sum _{i} \int _{B_{l} (X_i)} \frac{2 R_i^2 }{ \gamma ^2} \zeta \, d x . \end{aligned}$$

Then we see that \({\tilde{L}}_\varepsilon (t)\in (L^{4}({\mathbb {T}}))^* \cong L^{4/3}({\mathbb {T}})\) and satisfies \(\Vert {\tilde{L}}_\varepsilon (t)\Vert _{L^{4/3}} \le C\) for all \(t \in [0,T]\) and small \(\varepsilon >0\). Here \(L^2\) estimate does not work. Furthermore for all \(t_1,t_2 \in [0,T]\) we have

$$\begin{aligned} \Vert {\tilde{L}}_\varepsilon (t_1)-{\tilde{L}}_\varepsilon (t_2) \Vert _{L^{4/3}} \le C |t_1-t_2 |^{1/2} \big ( {\mathcal {D}}^\varepsilon \big )^{1/2}. \end{aligned}$$

These mean that \({\tilde{L}}_\varepsilon \in C^0([0,T];L^{4/3}({\mathbb {T}}))\) is uniformly bounded and equicontinuous. It follows from the compact embedding of \(L^{4/3}({\mathbb {T}})\) into \(\mathring{H}^{-1}({\mathbb {T}})\) and the generalized Arzela–Ascoli Theorem that \({\tilde{L}}_\varepsilon \) is relatively compact in \(C^0([0,T];\mathring{H}^{-1}({\mathbb {T}}))\).

Since we can deduce that \({\tilde{L}}_\varepsilon (t) \rightharpoonup L(t)\) weakly in \((C_p^0)^*\) for each \(t \in [0,T]\), and that \(L_\varepsilon -{\tilde{L}}_\varepsilon \rightarrow 0\) in \(C^0([0,T];\mathring{H}^{-1}({\mathbb {T}}))\), the proof of Lemma is completed. \(\square \)

Proof of Lemma 3.8

The differential of the nonlocal energy is given by

$$\begin{aligned} \begin{aligned} \langle DE_{\text {nl},\varepsilon }({\mathbf {Y}}^\varepsilon ), \tilde{{\mathbf {Z}}}^\varepsilon \rangle&= - 2 \int _{{\mathbb {T}}} \nabla \mu ^\varepsilon \cdot \nabla {\tilde{w}}^\varepsilon \, d x \\&= - 2 \int _{{\mathbb {T}}} \nabla \mu ^\varepsilon \cdot (\nabla \tilde{u}^\varepsilon +\nabla {\tilde{\phi }}^\varepsilon ) \, d x \end{aligned} \end{aligned}$$
(110)

with \({\tilde{u}}^\varepsilon \) and \({\tilde{\phi }}^\varepsilon \) determined via (31).

Define \(u_i\) as in (72), however we take \(V_i={\tilde{V}}_i\), and set \({\tilde{u}}^\varepsilon =\sum _i u_i + {\bar{u}}^\varepsilon \). Note that

(111)

Also we use the same notations \({\tilde{\mu }}_i,{\tilde{\mu }}^\varepsilon \) as in Sect. 3.4.

We write

$$\begin{aligned}&\int _0^T \beta \int _{{\mathbb {T}}} \nabla {\tilde{u}}^\varepsilon \cdot \nabla \mu ^\varepsilon \, d x \, d t\nonumber \\&\quad =\int _0^T \beta \int _{{\mathbb {T}}} \nabla {\tilde{u}}^\varepsilon \cdot \left( \nabla \sum _j {\tilde{\mu }}_j + \nabla \tfrac{K}{2} \right) \,d x\,d t + \int _0^T \beta \int _{{\mathbb {T}}} \nabla {\tilde{u}}^\varepsilon \cdot \nabla \left( {\tilde{\mu }}^\varepsilon -\tfrac{K}{2}\right) \,d x\,d t\nonumber \\&\quad = - \int _0^T \beta \sum _{i} \int _{\partial B_i } \frac{\varepsilon ^2 {\tilde{V}}_i}{\alpha _{\varepsilon }} \sum _{j} {\tilde{\mu }}_j \,d S\,d t - \int _0^T \tfrac{\beta }{2}\sum _{i} \int _{B_l (X_i)} \nabla u_i \cdot \nabla K \,d x\,d t \nonumber \\&\quad \quad +\int _0^T \tfrac{\beta }{2} \int _{{\mathbb {T}}} \nabla \bar{u}^\varepsilon \cdot \nabla K \,d x\,d t +\int _0^T \beta \int _{{\mathbb {T}}} \nabla {\tilde{u}}^\varepsilon \cdot \nabla \left( {\tilde{\mu }}^\varepsilon -\tfrac{K}{2}\right) \,d x\,d t . \end{aligned}$$
(112)

It follows from the condition (60) that

$$\begin{aligned} (1-o(1)) 4 \pi \int _0^T \sum _i \varepsilon ^2 R_i^2 {\tilde{V}}_i^2 \, d t \le \int _0^T \int _{{\mathbb {T}}} | \nabla {\tilde{u}}^\varepsilon |^2 \, d x \,d t \le C \end{aligned}$$
(113)

and that \({\bar{u}}^\varepsilon \) converges to \({\tilde{u}}\) strongly in \(L^2 ((0,T)\times {\mathbb {T}})\), where \({\tilde{u}}\) is determined via (53). (See the proof of [13, Lemma 5.4] for details.) Using

$$\begin{aligned} \bigg |\sum _{R_i \ge M} \varepsilon ^2 R_i^3 {\tilde{V}}_i \bigg |\le \left( \sum _{R_i \ge M} \varepsilon ^2 R_i^4\right) ^{1/2} \left( \sum _i \varepsilon ^2 R_i^2 \tilde{V}_i^2 \right) ^{1/2} , \end{aligned}$$

(113), Lemma 3.9, \(|R_i\ln 1/R_i|\le C \) for \(R_i\ll 1\) and \(\varepsilon R_i \le 1\), we see that

$$\begin{aligned} -\int _0^T \beta \sum _{i} \int _{\partial B_i } \frac{\varepsilon ^2 \tilde{V}_i}{\alpha _{\varepsilon }}\sum _{j} {\tilde{\mu }}_j \,d S\,d t&=-\int _0^T \beta \sum _{i} \int _{\partial B_i } \frac{\varepsilon ^2 {\tilde{V}}_i}{\alpha _{\varepsilon }} {\tilde{\mu }}_i \,d S\,d t \nonumber \\&= -\int _0^T \beta \sum _{i} \pi \varepsilon ^4 R_i^3 \tilde{V}_i \ln \frac{l}{\alpha _{\varepsilon }R_i} \,d t\nonumber \\&\quad \rightarrow -\pi \int _0^T \beta \int r^3 {\tilde{v}} \, d \nu _t \,d t \end{aligned}$$
(114)

as \(\varepsilon \rightarrow 0\). In addition, we obtain

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \int _0^T \beta \int _{{\mathbb {T}}} \nabla \bar{u}^\varepsilon \cdot \nabla K \,d x\,d t&= \int _0^T \beta \int _{{\mathbb {T}}} \nabla {\tilde{u}} \cdot \nabla K \,d x\,d t \nonumber \\&=-2\pi \int _0^T \beta \int r K(t,x) {\tilde{v}} \, d \nu _t \,d t , \end{aligned}$$
(115)

and it follows from Lemma 3.7 that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \int _0^T \beta \int _{{\mathbb {T}}} \nabla \tilde{u}^\varepsilon \cdot \nabla \left( {\tilde{\mu }}^\varepsilon -\tfrac{K}{2}\right) \,d x\,d t = 0 . \end{aligned}$$
(116)

Finally from \(\sum _i |B_l (X_i)| \le C \gamma ^2 \sum _i \varepsilon ^2 \le C \gamma ^2\) and (111),

$$\begin{aligned} \bigg | \sum _i \int _{B_l (X_i)} \nabla u_i \cdot \nabla K \, d x \bigg |&\le \sum _i \left( \int _{B_l (X_i)} |\nabla u_i |^2 \, d x \right) ^{1/2} \left( \int _{B_l (X_i)} |\nabla K |^2 \, d x \right) ^{1/2} \nonumber \\&\le \left( \sum _i \int _{B_l (X_i)} |\nabla u_i |^2 \, d x \right) ^{1/2} \big ( \sum _i \int _{B_l (X_i)} |\nabla K |^2 \, d x \big )^{1/2}\nonumber \\&\le C \omega (C \gamma ^2) , \end{aligned}$$
(117)

where

$$\begin{aligned} \omega (z):=\sup _{|E| \le z} \bigg (\int _{E} | \nabla K|^2 \, d x \bigg )^{1/2}. \end{aligned}$$
(118)

Since we can choose \(\gamma \) arbitrarily small, we obtain

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \int _0^T \beta \int _{{\mathbb {T}}} \nabla {\tilde{u}}^ \varepsilon \cdot \nabla \mu ^\varepsilon \,d x\,d t = -\pi \int _0^T \beta \int (r^3+r K(x)) {\tilde{v}} \, d \nu _t \,d t \end{aligned}$$

from (112), (114), (115), (116) and (117).

Similarly we can prove that \(\int _0^T \beta \int _{{\mathbb {T}}} \nabla \tilde{\phi }^ \varepsilon \cdot \nabla \mu ^\varepsilon \,d x\,d t \rightarrow 0\) as \(\varepsilon \rightarrow 0\). This completes the proof. \(\square \)

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Communicated by J. Ball.

A Appendix

A Appendix

Lemma A.1

\(A=\{x\in {\mathbb {R}}^2\,;\, a<|x|<b\}\). There holds

$$\begin{aligned} \left| \frac{1}{\pi b^2} \int _A f \, d x \right| ^2 \le \frac{1}{2\pi } \left( \ln \frac{b}{a} \right) \int _A|\nabla f|^2 \, d x \end{aligned}$$
(119)

for all \(f \in H^1(A)\) such that \(\int _{\partial B_a} f\, d S=0\).

Proof

Using the polar coordinate \((r,\theta )\), we write \(f=f(r,\theta )\). Let

$$\begin{aligned} g(r)= \int _0^{2\pi } f(r,\theta )\, d \theta \end{aligned}$$

for \(r \in [a,b]\). Then by

$$\begin{aligned} g'(s)=\int _0^{2\pi } f_r (s,\theta )\, d\theta , \end{aligned}$$

and

$$\begin{aligned} g'(s)^2 \le 2\pi \int _0^{2\pi } f_r (s,\theta )^2 \, d\theta , \end{aligned}$$

for \(s \in [a,b]\), we get

$$\begin{aligned} \int _a^b g'(s)^2 s \, d s \le 2\pi \int _a^b s \int _0^{2\pi } f_r (s,\theta )^2 \, d\theta \, d s \le 2 \pi \Vert \nabla f\Vert _2^2 . \end{aligned}$$

Note that \(g(r)=\int _a^r g'(s)\, d s\) by \(g(a)=0\). Hence

$$\begin{aligned} |g(r)|&=\bigg |\int _a^r g'(s)\, d s \bigg | \le \left( \int _a^b g'(s)^2 s \, d s\right) ^{1/2} \left( \int _a^r \frac{1}{s} \, d s \right) ^{1/2} \\&\le \sqrt{2\pi } \Vert \nabla f\Vert _2 \left( \ln \frac{r}{a}\right) ^{1/2} . \end{aligned}$$

We find

$$\begin{aligned} \left| \int _A f \, d x \right| = \left| \int _a^b g(r) r \, d r \right|&\le \sqrt{2\pi } \Vert \nabla f\Vert _2 \int _a^b r \left( \ln \frac{r}{a}\right) ^{1/2}\, d r \\&\le \sqrt{2\pi } \Vert \nabla f\Vert _2 \frac{b^2}{2} \left( \ln \frac{b}{a}\right) ^{1/2} . \end{aligned}$$

\(\square \)

Remark: Similarly we can also show that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} E({\mathbf {Y}}^\varepsilon ) =E(\nu _t), \quad \text {uniformly in } t\in [0,T], \end{aligned}$$

where

$$\begin{aligned} E(\nu _t)=2 \pi \int \bigg ( r+\sigma \frac{r^4}{4} \bigg )\, d \nu _t + \frac{\sigma }{4} \int _{{\mathbb {T}}} | \nabla K |^2 \, d x . \end{aligned}$$

This statement is of interest in itself, but since we do not need it here we omit the proof for the sake of brevity.

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Niethammer, B., Oshita, Y. A rigorous derivation of mean-field models describing 2D micro phase separation. Calc. Var. 59, 54 (2020). https://doi.org/10.1007/s00526-020-1706-x

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Mathematics Subject Classification

  • 35B27
  • 35R35
  • 82D60