Research article Special Issues

Nonlocal diffusion of smooth sets

  • We consider normal velocity of smooth sets evolving by the sfractional diffusion. We prove that for small time, the normal velocity of such sets is nearly proportional to the mean curvature of the boundary of the initial set for s[12,1) while, for s(0,12), it is nearly proportional to the fractional mean curvature of the initial set. Our results show that the motion by (fractional) mean curvature flow can be approximated by fractional heat diffusion and by a diffusion by means of harmonic extension of smooth sets.

    Citation: Anoumou Attiogbe, Mouhamed Moustapha Fall, El Hadji Abdoulaye Thiam. Nonlocal diffusion of smooth sets[J]. Mathematics in Engineering, 2022, 4(2): 1-22. doi: 10.3934/mine.2022009

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  • We consider normal velocity of smooth sets evolving by the sfractional diffusion. We prove that for small time, the normal velocity of such sets is nearly proportional to the mean curvature of the boundary of the initial set for s[12,1) while, for s(0,12), it is nearly proportional to the fractional mean curvature of the initial set. Our results show that the motion by (fractional) mean curvature flow can be approximated by fractional heat diffusion and by a diffusion by means of harmonic extension of smooth sets.



    For N2, we let Ω0 be a bounded open set of RN with boundary Γ0. Consider the heat equation with initial data the indicator function of the set Ω0:

    {utΔu=0in   RN×(0,t1]u(x,0)=1Ω0(x) on   RN. (1.1)

    for some time t1>0. In 1992, Bence-Merriman-Osher [5] provided a computational algorithm for tracking the evolution in time of the set Ω0 whose boundary Γ0 moves with normal velocity proportional to its classical mean curvature. At time t1>0, they considered

    Ω1={xRN:u(x,t1)1/2}.

    Bence-Merriman-Osher [5] applied iteratively this procedure to generate a sequence of sets (Ωj)j0 and conjectured in [5] that their boundaries Γj evolved by mean curvature flow.

    Later Evans [16] provided a rigorous proof for the Bence-Merriman-Osher algorithm by means of the level-set approach to mean curvature flow developed by Osher-Sethian [32], Evans-Spruck [17,18,19,20] and Chen-Giga-Goto [10]. For related works in this direction, we refer the reader to [4,25,28,29,31,34,36] and references therein.

    Recently Caffarelli and Souganadis considered in [9] nonlocal diffusion of open sets ERN given by

    {ut+(Δ)su=0in   RN×(0,)u(x,0)=τE(x)in   RN×{t=0}, (1.2)

    where

    τE(x)=1E(x)1RN¯E(x).

    We consider the fractional heat kernel Ks with Fourier transform given by ˆK(ξ,t)=et|ξ|2s. It satisfies

    {Kst+(Δ)sKs=0 in RN×(0,)Ks=δ0 on   RN×{t=0}.

    It follows that the unique bounded solution to (1.2) is given by

    u(x,t)=Ks(,t)τE(x)=RN Ks(xy,t)τE(y)dy. (1.3)

    By solving a finite number of times (1.2) for a small fixed time step σs(h), the authors in [9] find a discrete family of sets

    Eh0=E,Ehnh={xRN:Ks(,σs(h))τEh(n1)h(x)>0},

    for a suitable scaling function σs to be defined below. It is proved in [9] that as nht, Ehnh converges, in a suitable sense, to Γt. Here, the family of hypersurface {Γt}t>0, with Γ0=E, evolves under generalized mean curvature flow for s[12,1) and under generalized fractional mean curvature flow for s(0,12). We refer the reader to [9,16,27] for the notion generalized (nonlocal) mean curvature flow which considers the level sets of viscosity solutions to quasilinear parabolic integro-differential equations.

    In the present paper, we are interested in the normal velocity of the sets

    Et:={xRN:κs(,σs(t))τE(x)>0} (1.4)

    as they depart from a sufficiently smooth initial set E0:=E. We consider here and in the following

    Ks(x,t)=tN2sPs(t12sx), for some radially symmetric function  PsC1(RN). (1.5)

    We make the following assumptions:

    C1N,s1+|y|N+2sPs(y)CN,s1+|y|N+2s,|Ps(y)|CN,s1+|y|N+2s+1. (1.6)

    and

    limt0t1κs(y,t)=CN,s|y|N+2s locally uniformly in  RN{0}, (1.7)

    for some constants CN,s,CN,s>0. In the Section 1.1 below, we provide examples of valuable kernels Ks satisfying the above properties.

    Now, as we shall see below (Lemma 2.1), for t>0 small, xu(x,σs(t))0 for all xB(y,σs(t)12s) and yE. Hence Et is a C1 hypersurface, for small t>0. For t>0 and yE, we let v=v(t,y) be such that

    y+vtν(y)EtB(y,σs(t)12s), (1.8)

    where ν(y) is the unit exterior normal of E at y. In the spirit of the work of Evans [16] on diffusion of smooth sets, we provide in this paper an expansion of v(t,y) as t0. It turns out that v(0,y) is proportional to the fractional mean curvature of E at y for s(0,1/2) and v(0,y) is proportional to the classical mean curvature of E at y for s[1/2,1).

    We notice that it is not a priori clear from (1.8), that v remains finite as t0. This is where the (unique) appropriate choice of σs(t) enters during our estimates. Here and in the following, we define

    σs(t)={t2s1+2sfor s(0,1/2),tsfor s(1/2,1) (1.9)

    and for s=1/2, σs(t) is the unique positive solution to

    t=σ1/2  2(t)|log(σ1/2  (t))|. (1.10)

    Before stating our main result, we recall that for s(0,12) and E is of class C1,β for some β>2s, the fractional mean curvature of E is defined for xE as

    Hs(x):=P.V.RN τE(y)|xy|N+2sdy=limε0RNBε(x)τE(y)|xy|N+2sdy.

    On the other hand, if E is of class C2 then the normalized mean curvature of E is given, for xE, by

    H(x):=N1N+1limε01ε|Bε(x)|Bε(x)τE(y)dy,

    see also (2.4) and [21]. Having fixed the above definitions, we now state our main result.

    Theorem 1.1. We let s(0,1) and ERN, N2. We assume, for s(0,1/2), that E is of class C1,β for some β>2s and that E is of class C3, for s[1/2,1). Then, as t0, the expansion of v(t,y), defined in (1.8), is given, locally uniformly in yE, by

    v(t,y)={aN,sHs(y)+ot(1)for  s(0,1/2)bN(t)H(y)+O(1log(σ1/2  (t)))for  s=1/2cN,sH(y)+O(t2s12)for  s(1/2,1),

    where Hs and H are respectively the fractional and the classical mean curvatures of E and the positive constants aN,s, bN,1/2 and cN,s are given by

    aN,s=CN,s2RN1  Ps(y,0)dy,bN(t)=BN1σ12(t)1  |y|2P1/2  (y,0)dy2|log(σ12(t))|RN1  P1/2  (y,0)dy,cN,s=RN1  |y|2Ps(y,0)dy2RN1  Ps(y,0)dy

    and Ps(y):=κs(y,1).

    Some remarks are in order. The assumption of E being of class C3 in Theorem 1.1 is motivated by the result of Evans in [16], where in the case s=1 and K1, the heat kernel, he obtained v=(N1)H(0)+O(t12). We notice that from our argument below, we cannot improve the error term ot(1) in the case s(0,1/2) even if E is of class C. This is due to the definition of the fractional mean curvature Hs as a principal value integral. Using polar coordinates and the estimates in (1.6), it easy follows that

    |SN2|C1N,1/22RN1  P1/2  (y,0)dylimt0bN(t)|SN2|CN,1/22RN1  P1/2  (y,0)dy.

    This then justify the choice of the scaling when s=12. In the particular case, that κ1/2  (y,t)=CNt(t2+|y|2)N+12, we have that

    bN(t)=|SN2|CN2RN1  P1/2  (y,0)dy+O(1log(σ1/2  (t))).

    For the general case, we get the following

    We next put emphasis on two valuable examples where Theorem 1.1 applies.

    1) Fractional heat diffusion of smooth sets. We recall, see e.g., [6,35], that the fractional heat kernel Ks satisfies (1.5), (1.6) and (1.7) with

    CN,s=s22ssin(sπ)Γ(N2+s)Γ(s)π1+N2. (1.11)

    We recall that Ks is known explicitly only in the case s=1/2, where K1/2  (y,t)=CN,1/2t(t2+|y|2)N+12. In this case Theorem 1.1 provides an approximation of the (fractional) mean curvature motion by fractional heat diffusion of smooths sets, thereby extending, in the fractional setting, Evan's result in [16] on heat diffusion of smooth sets.

    2) Diffusion of smooth sets by Harmonic extension. We consider the Poisson kernel on the half space RN+1+:=RN×(0,), given by

    ¯Ks(x,t):=tNPs(x/t),Ps(x)=pN,s(1+|x|2)N+2s2, (1.12)

    where pN,s:=1RN (1+|y|2)N+2s2dy. Thanks to the result of Caffarelli and Silvestre in [8], the function

    w(x,t)=¯Ks(,t)τE(x)=pN,st2sRN τE(y)(t2+|yx|2)N+2s2dy

    solves

    {div(t12sw)=0in   RN×(0,)w=τEon RN×{t=0}.

    It is clear that Ks(x,t):=¯Ks(x,t12s) satisfies (1.5), (1.6) and (1.7) with CN,s=pN,s. Hence, Theorem 1.1 provides an expansion of the normal velocities of the boundary of the sets

    Et:={xRN:¯Ks(,σs(t)12s)τE(x)>0},

    where σs(t) is given by (1.9) and (1.10). Therefore this Harmonic extension yields an approximation of (fractional) mean curvature motion of smooth sets.

    We conclude Section 1 by noting that the notion of nonlocal curvature appeared for the first time in [9]. Later on, the study of geometric problems involving fractional mean curvature has attracted a lot of interest, see [1,7], the survey paper [21] and the references therein. While the mean curvature flow is well studied, see e.g., [2,3,15,24,26], its fractional counterpart appeared only recently in the literature, see e.g., [11,12,13,14,27,30,33].

    We finally remark that the changes of normal velocity of the nonlocal diffused sets as s varies in (0, 1/2) and [1/2, 1), appeared analogously in phases transition problems, see e.g., [22,23,33].

    Unless otherwise stated, we assume for the following that E is an open set of class C1,β, with 0E and the unit normal of E at 0 coincides with eN. We denote by Qr=BN1r×(r,r) the cylinder of RN centred at the origin with BN1r the ball of RN1 centred at the origin with radius r>0. Decreasing r, if necessary, we may assume that

    EQr={(y,yN)BN1r×R:yN>γ(y)}, (2.1)

    with γC1,β(BN1r) satisfying

    γ(y)=O(|y|1+β). (2.2)

    In the following, for f,g:RR, we write g(t):=O(f(t)) if

    |g(t)|C|f(t)|.

    We also write g(t)=o(f(t)) if g(t)=O(f(t)) and moreover when f(t)0, we have

    limt0|g(t)||f(t)|=0.

    We denote by ot(1) any function that tends to zero when t0.

    If in addition, E is of class C3, then for yBN1r, we have

    γ(y)=12D2γ(0)[y,y]+O(|y|3) (2.3)

    and the normalized mean curvature of E at 0 is given by

    H(0)=Δγ(0)N1=N1N+1limε01ε|Qε(0)|Qε(0)τE(y)dy. (2.4)

    Recall that the unit exterior normal ν(y):=ν(y,γ(y)) of E and the volume element dσ(y) on EQr are given by

    ν(y)=(γ(y),1)1+|γ(y)|2anddσ(y)=1+|γ(y)|2dy. (2.5)

    We finally note, in view of (1.5) and (1.6), that we have

    0<κs(y,t)Ct|y|N+2s for all  yRN{0},t>0, (2.6)

    for some positive constant C=C(N,s). We start with the following result.

    Lemma 2.1. Let s(0,1) and E be a set of class C1,β satisfying (2.1). Define

    w(z,t)=RN κs(zy,t)τE(y)dy.

    Then there exist t0,C>0, only depending on N,s,β and E, such that for all t(0,t0) and zBt12s,

    wzN(z,t)Ct1/2s. (2.7)

    As a consequence, for all t(0,t0), the set

    {zRN:w(z,t)=0}Bt12sis  of  class  C1. (2.8)

    Proof. We fix t>0 small so that t12s<r8 and let zBt12s. We write

    wzN(z,t)=RN κszN(zy,t)τE(y)dy=Br2(z)κsyN(zy,t)τE(y)dy+RNBr2(z)κsyN(zy,t)τE(y)dy. (2.9)

    By a change of variable, (1.5) and (1.6), we have

    RNBr2(z)κsyN(zy,t)dy=t12sRNt12sBr2(z)PsyN(t12szy)dy=O(t12sRNt12sBr2(z)|t12szy|N12sdy)=O(t). (2.10)

    Integrating by parts, we have

    Br2(z)κsyN(zy,t)τE(y)dy=Br2(z)EκsyN(zy,t)τE(y)dyBr2(z)EcκsyN(zy,t)dy=2Br2(z)Eκs(zy,t)eNνE(y)dσ(y)+Br(z)κs(zy,t)eNνBr2(y)τE(y)dσ(y).

    By a change of variable, (1.5), (1.6) and the fact that Qr/8Br/4Br2(z)Qr, we have

    Br2(z)Eκs(zy,t)eNνE(y)dσ(y)CBN1r/8  κs(zy,zNγ(y),t)dy=Ct12sBN1r8t12sPs(t12szy,t12szNt12sγ(t12sy))dyCt12sBN1r8t12sB2dy1+|y|N+2s, (2.11)

    provided r8>t12s. Next, using (2.6) and recalling that zBt12s, we then have

    |Br2(z)κs(zy,t)eNνBr2(z)(y)dσ(y)|Br2(z)κs(zy,t)dσ(y)tCBr2(z)1|y|N+2sdσ(y)=O(t).

    From this and (2.11), we deduce that

    Br(z)κsyN(zy,t)τE(y)dyCt1/2s.

    Combining this with (2.9) and (2.10), we get

    wzN(z,t)Ct1/2s.

    Therefore (2.7) follows. Finally (2.8) follows from the inverse function theorem and the fact that w is of class C1 on RN×(0,).

    In the sequel, we will need the following lemmas to estimate some error terms.

    Lemma 2.2. For s(0,1), we let ERN be a set of class C1,β, for some β>2s, as in Section 2. For r>0, we set

    Jr(t):=Qrκs(y,t)τE(y)dyandIr(t)=RNQr(t1κs(y,t)CN,s|y|N+2s)τE(y)dy.

    Then we have

    |Jr(t)|Ctrβ2sandlimt0Ir(t)=0,

    where C is a positive constant depending only on N, β, s and E.

    Proof. Since τE=1E1RN¯E, we get

    Jr(t)=QrEκs(y,t)dyQrEcκs(y,t)dy=BN1r  rγ(y)κs((y,yN),t)dyNdyBN1r  γ(y)rκs((y,yN),t)dyNdy=BN1r  (rγ(y)κs((y,yN),t)dyNγ(y)rκs((y,yN),t)dyN)dy=BN1r  (rγ(y)κs((y,yN),t)dyNγ(y)rκs((y,yN),t)dyNγ(y)γ(y)κs((y,yN),t)dyN)dy=BN1r  (rγ(y)κs((y,yN),t)dyN+γ(y)rκs((y,yN),t)dyNγ(y)γ(y)κs((y,yN),t)dyN)dy.

    Since the map yKs(y,t) is radial, we have κs(y,yN,t)=κs(y,yN,t) so that

    rγ(y)κs((y,yN),t)dyN+γ(y)rκs((y,yN),t)dyN=0.

    Therefore

    Jr(t)=BN1r  γ(y)γ(y)κs((y,yN),t)dyNdy=2BN1r  γ(y)0κs((y,yN),t)dyNdy.

    Then, by (2.6),

    |Jr(t)|2CN,st|BN1r  γ(y)01|(y,yN)|N+2sdyNdy|Ctrβ2s.

    where C is a positive constant depending on N, β and s and which may change from a line to another. Next, using (2.6), (1.7) and the dominate convergence theorem, we obtain

    limt0Ir(t)=0.

    This then ends the proof.

    Lemma 2.3. Let s(0,1) and let x=vteNEtBσs(t)12s, with Et given by (1.4). Then

    vt=O((σs(t))1+2s2s)as  t0.

    Proof. Since x=vteNEt, we have that u(x,σs(t))=0. By the fundamental theorem of calculus, we have

    u(x,σs(t))=u(0,σs(t))+vt10uxN(θvteN,σs(t))dθ=0

    so that

    vt10uxN(θvteN,σs(t))dθ=u(0,σs(t)). (2.12)

    We write

    u(0,σs(t))=RN κs(y,σs(t))τE(y)dy=Qrκs(y,σs(t))τE(y)dy+Qcrκs(y,σs(t))τE(y)dy.

    Then by Lemma 2.2 and (2.6), we have

    |Qrκs(y,σs(t))τE(y)dy|Cσs(t) and Qcrκs(y,σs(t))τE(y)dy=O(σs(t)) (2.13)

    for some constant C depending on r. Furthermore by (2.7), we have

    uxN(θvteN,σs(t))C(σs(t))1/2s. (2.14)

    Therefore, by (2.12), (2.13) and (2.14), we obtain

    |vt|=u(0,σs(t))|10uxN(θvten,σs(t))dθ|Cσs(t)(σs(t))12s=C(σs(t))1+2s2s.

    This then ends the proof.

    Lemma 2.4. Under the assumptions of Lemma 2.3, we have

    10BN1rγ(y)vt0yNKsyN(y,θyN,σs(t))dydθ=O(σs(t))as  t0.

    Proof. Let θ[0,1]. By (1.5), (1.6) and a change of variable, we have

    BN1rγ(y)vt0yNKsyN(y,θyN,σs(t))dy=(σs(t))N12sBN1rγ(y)vt0yNPsyN(y(σs(t))12s,θyN(σs(t))12)dyCBN1r(σs(t))1/2s(σs(t))1/2s(γ((σs(t))1/2sy)vt)0yN1+|y|N+2s+1dyC(σs(t))1/sBN1r(σs(t))1/2s(γ((σs(t))1/2sy)vt)21+|y|N+2s+1dy=O(σs(t))+O(v2t2(σs(t))1/s)+O(vt).

    Applying Lemma 2.3, we get

    γ(y)vt0yNKsyN(y,θyN,σs(t))dydθ=O(σs(t)),

    as t0. This then ends the proof.

    In this section, we start by the following preliminary result.

    Lemma 3.1. Let s(0,1/2). We assume that E is of class C1,β for some β>2s satisfying (2.1). Then, for all θ[0,1], we have

    RN κsyN(y,yNvtθ,σs(t))τE(y)dy=2(σs(t))1/2sRN1  Ps(y,0)dy+O((σs(t))2s12s). (3.1)

    Proof. We have

    RN κsyN(y,yNvtθ,σs(t))τE(y)dy=BrκsyN(y,yNvtθ,σs(t))τE(y)dy+BcrκsyN(y,yNvtθ,σs(t))τE(y)dy,

    where Br is the ball of RN centered at the origin and of radius r>0. By integration by parts, we have

    BrκsyN(y,yNvtθ,σs(t))τE(y)dy=2EBrκs(y,yNvtθ,σs(t))νN(y)dσ(y)+Brκs(y,yNvtθ,σs(t))eNνBr(y)τE(y)dσ(y).

    Therefore

    RN κsyN(y,yNvtθ,σs(t))τE(y)dy=2EBrκs(y,yNvtθ,σs(t))νN(y)dσ(y) +BcrκsyN(y,yNvtθ,σs(t))τE(y)dy+Brκs(y,yNvtθ,σs(t))yNrτE(y)dσ(y). (3.2)

    Then by a change of variable and (2.5), we have

    EBrκs(y,yNvtθ,σs(t))νN(y)dσ(y)=BN1r  κs(y,γ(y)vtθ,σs(t))dy.

    By the Fundamental Theorem of calculus, we can write

    κs(y,γ(y)vtθ,σs(t))=κs(y,0,σs(t))+(γ(y)vtθ)10κsyN(y,θ(γ(y)vtθ)),σs(t))dθ. (3.3)

    In the following, we let

    β(y):=γ(y)vtθ. (3.4)

    Then we have

    EBrκs(y,yNvtθ,σs(t))νN(y)dσ(y)=BN1r  κs(y,β(y),σs(t))dy=BN1r  κs(y,0,σs(t))dy+10BN1r  β(y)κsyN(y,θβ(y),σs(t))dydθ. (3.5)

    Therefore By a change of variable, (1.5) and (1.6), we have

    BN1r  κs(y,0,σs(t))dy=BN1r  (σs(t))N2sPs(y(σs(t))1/2s,0)dy=(σs(t))1/2sRN1  Ps(y,0)dy+(σs(t))1/2sRN1BN1r(σs(t))1/2s  Ps(y,0)dy=(σs(t))1/2sRN1  Ps(y,0)dy+O((σs(t))1/2s(σs(t))1+2s2s). (3.6)

    By a change of variable, (1.6) and (3.4), we have

    BN1r  β(y)κsyN(y,θβ(y),σs(t))dy=(σs(t))1/sBN1r(σs(t))1/2s  β(y(σs(t))1/2s)PsyN(y,θ(σs(t))1/2sβ(y(σs(t))1/2s))dy.

    We use (2.2), (3.4) and Lemma 2.3 to get

    (σs(t))1/sβ(y(σs(t))1/2s)=O(|y|1+β(σs(t))β12s)vtθ(σs(t))1/s=O(|y|1+β(σs(t))2s12s)+O((σs(t))2s12s)in   BN1r(σs(t))1/2s.

    Then by (1.6), we have

    BN1r  β(y)κsyN(y,θβ(y),σs(t))dy=O((σs(t))2s12sBN1r(σs(t))1/2s  |y|1+βPsyN(y,θ(σs(t))1/2sβ(y(σs(t))1/2s))dy)+O((σs(t))2s12sBN1r(σs(t))1/2s  PsyN(y,θ(σs(t))1/2sβ(y(σs(t))1/2s))dy)=O((σs(t))2s12sRN1  1+|y|1+β1+|y|N+2s+1dy)=O((σs(t))2s12s).

    Hence

    10BN1r  β(y)κsyN(y,θβ(y),σs(t))dydθ=O((σs(t))2s12s)as t0. (3.7)

    It follows from (3.5), (3.6) and (3.7) that

    EBrκs(y,yNvtθ,σs(t))νN(y)dσ(y)=(σs(t))1/2sRN1  Ps(y,0)dy+O((σs(t))2s12s)as t0. (3.8)

    By a change of variable and the fact that |τE(y)|1, we have

    BcrκsyN(y,yNvtθ,σs(t))τE(y)dy=O((σs(t))1/2sBcr(σs(t))1/2sPsyN(y,yNvt(σs(t))1/2sθ)dy) =O((σs(t))1/2sBcr(σs(t))1/2s11+|y|N+2s+1dy)=O(σs(t)). (3.9)

    We use (1.7) to get, as t0,

    |Brκs(y,yNvtθ,σs(t))yNrτE(y)dσ(y)|Brκs(y,yNvtθ,σs(t))dσ(y) σs(t)BrCN,s|(y,yNvtθ)|N+2sdσ(y)=O(σs(t)).

    Therefore, the expansion (3.1) follows immediately from (3.2), (3.8), (3.9) and the above estimate. This ends the proof.

    The following result completes the proof of Theorem 1.1 in the case s(0,1/2).

    Proposition 3.2. Under the assumptions of Lemma 3.1, we have

    v=aN,sHs(0)+ot(1)as  t0, (3.10)

    where Hs(0) is the fractional mean curvature of E at the point 0 and the positive constant aN,s is given by

    aN,s=CN,s2RN1  Ps(y,0)dy.

    Proof. We put with x=vteN and we recall that

    u(x,σs(t))=RN κs(yx,σs(t))τE(y)dy=0.

    By the fundamental theorem of calculus, we have

    κs(yx,σs(t))=κs(y,σs(t))vt10κsyN(y,yNvtθ,σs(t))dθ.

    Then

    u(x,σs(t))=˜Jr(t)+σs(t)˜Ir(t)+σs(t)CN,sRNQrτE(y)|y|N+2sdyvtRN 10κsyN(y,yNvtθ,σs(t))τE(y)dθdy, (3.11)

    where ˜Jr(t)=Jr(σs(t)), ˜Ir(t)=Ir(σs(t)), while Ir(t) and Jr(t) are given by Lemma 2.2. Moreover, by Lemma 3.1, we have

    RN κsyN(y,yNvtθ,σs(t))τE(y)dy=2(σs(t))1/2sRN1  Ps(y,0)dy+O((σs(t))2s12s).

    Therefore

    10RN κsyN(y,yNvtθ,σs(t))τE(y)dydθ=2(σs(t))1/2sRN1  Ps(y,0)dy+O((σs(t))2s12s) as t0. (3.12)

    Putting (3.12) in (3.11), we obtain that

    u(x,σs(t))=˜Jr(t)+σs(t)˜Ir(t)+σs(t)CN,sRNQrτE(y)|y|N+2sdyvt(2(σs(t))1/2sRN1  Ps(y,0)dy+O((σs(t))2s12s))=σs(t)[(σs(t))1˜Jr(t)+˜Ir(t)+CN,sRNQrτE(y)|y|N+2sdy 2vt(σs(t))2s12sRN1  Ps(y,0)dy+O(σs(t))].

    Recalling that σs(t)=t2s1+2s and using the fact that u(x,σs(t))=0, we have

    0=t2s1+2s˜Jr(t)+˜Ir(t)+CN,sRNQrτE(y)|y|N+2sdy2vRN1  Ps(y,0)dy+O(t2s1+2s) as  t0.

    As a consequence,

    |CN,sHs(0)2vRN1  Ps(y,0)dy||CN,sHs(0)CN,sRNQrτE(y)|y|N+2sdy|+|t2s1+2s˜Jr(t)|+˜Ir(t)+O(t2s1+2s).

    Therefore by Lemma 2.2, taking the limsups as t0 and as r0 respectively, we obtain

    lim supt0|CN,sHs(0)2vRN1  Ps(y,0)dy|=0.

    Hence

    v=CN,sHs(0)2RN1  Ps(y,0)dy+ot(1) as   t0.

    This then ends the proof.

    We have the following result.

    Proposition 4.1. We consider E a set of class C3 satisfying the condition in Section 2. For s(1/2,1), we have

    v=cN,sH(0)+O(t2s12),as  t0. (4.1)

    where H(0) is the normalized mean curvature of E at 0 and the positive constant cN,s is given by

    cN,s=RN1  |y|2Ps(y,0)dy2RN1  Ps(y,0)dy.

    Proof. We let x=vteNEt and we expand

    u(x,σs(t))=RN κs(yx,σs(t))τE(y)dy=Qrκs(yx,σs(t))τE(y)dy+Qcrκs(yx,σs(t))τE(y)dy, (4.2)

    By (2.6) and Lemma 2.3, we have

    Qcrκs(yx,σs(t))τE(y)dy=O(σs(t)) as   t0.

    Therefore

    u(x,σs(t))=Qrκs(yx,σs(t))τE(y)dy+O(σs(t)). (4.3)

    By a change of variable, the fact that τE=1E(x)1RN¯E(x) and x=vteN, we have

    Qrκs(yx,σs(t))τE(y)dy=EQrκs(yx,σs(t))dyEcQrκs(yx,σs(t))dy=BN1r  γ(y)rκs(yx,σs(t))dyBN1r  rγ(y)κs(yx,σs(t))dy=BN1r  γ(y)vtrvtκs(y,σs(t))dyBN1r  rvtγ(y)vtκs(y,σs(t))dy=2BN1r  γ(y)vt0κs(y,σs(t))dy+BN1r  0rvtκs(y,σs(t))dyBN1r  rvt0κs(y,σs(t))dy=2BN1r  γ(y)vt0κs(y,σs(t))dy+BN1r  r+vtrvtκs(y,σs(t))dy.

    The last line is due to the fact that the map yNκs(y,σs(t)) is even so that

    rvt0κs(y,σs(t))dyN=r+vt0κs(y,σs(t))dyN.

    Therefore we have

    Qrκs(yx,σs(t))τE(y)dy=2BN1r  γ(y)vt0κs(y,σs(t))dy+BN1r  r+vtrvtκs(y,σs(t))dy. (4.4)

    By (2.6) and the fact that vt=ot(1), we have

    BN1r  r+vtrvtκs(y,σs(t))dy=O(σs(t)). (4.5)

    By a change of variable, the Fundamental Theorem of Calculus, (1.5) and (2.3), we have

    BN1r  γ(y)vt0κs(y,σs(t))dy=BN1r  γ(y)vt0κs(y,0,σs(t))dy+BN1r  γ(y)vt010yNκsyN(y,θyN,σs(t))dydθ=BN1r  κs(y,0,σs(t))(12γyiyj(0)yiyj+O(|y|3)vt)dy+BN1r  γ(y)vt010yNκsyN(y,θyN,σs(t))dydθ=Δγ(0)2(N1)BN1r  |y|2κs(y,0,σs(t))dyvtBN1r  κs(y,0,σs(t))dy+BN1r  γ(y)vt010yNκsyN(y,θyN,σs(t))dydθ+O(BN1r  |y|3κs(y,0,σs(t))dy).

    Therefore, recalling (2.4),

    BN1r  γ(y)vt0κs(y,σs(t))dy=H(0)2BN1r  |y|2κs(y,0,σs(t))dyvtBN1r  κs(y,0,σs(t))dy+BN1r  γ(y)vt010yNκsyN(y,θyN,σs(t))dydθ+O(BN1r  |y|3κs(y,0,σs(t))dy). (4.6)

    By a change of variable and (1.5), we have

    BN1r  |y|2κs(y,0,σs(t))dy=(σs(t))12sBN1r(σs(t))12s  |y|2Ps(y,0)dy (4.7)

    and

    BN1r  κs(y,0,σs(t))dy=(σs(t))12sBN1r(σs(t))12s  Ps(y,0)dy. (4.8)

    Moreover by (1.6), we get

    (σs(t))12sRN1BN1r(σs(t))12s  |y|2Ps(y,0)dy+(σs(t))12sRN1BN1r(σs(t))12s  Ps(y,0)dy=O(σs(t))as  t0 (4.9)

    and

    BN1r  |y|3κs(y,0,σs(t))dy=O(σs(t)). (4.10)

    By Lemma 2.4, we get

    BN1rγ(y)vt010yNκsyN(y,θyN,σs(t))dydθ=O(σs(t)). (4.11)

    Combining (4.4), (4.6), (4.7), (4.8), (4.9) and (4.11), we obtain

    Qrκs(yx,σs(t))τE(y)dy=(σs(t))12sH(0)2RN1  |y|2Ps(y,0)dyvt(σs(t))12sRN1  Ps(y,0)dy+O(σs(t)). (4.12)

    By (4.3) and (4.12), we obtain

    u(x,σs(t))=(σs(t))1/2sH(0)RN1  |y|2Ps(y,0)dy2vt(σs(t))1/2sRN1  Ps(y,0)dy+O(σs(t))=(σs(t))1/2s[(σs(t))1/sH(0)RN1  |y|2Ps(y,0)dy2vtRN1  Ps(y,0)dy+O((σs(t))1+2s2s)].

    Since x=vtνEt, we have u(x,σs(t))=0. Now, from the definition of σs(t)=ts, we deduce that

    H(0)RN1  |y|2Ps(y,0)dy2vRN1  Ps(y,0)dy+O(t2s12)=0.

    Thus

    v=cN,sH(0)+O(t2s12),

    where

    cN,s=RN1  |y|2Ps(y,0)dy2RN1  Ps(y,0)dy.

    This then ends the proof.

    As usual, we consider the function

    u(x,t)=κ1/2  (,t)τE(x)

    and recall that

    Et:={xRN:u(x,σ1/2  (t))0}.

    To alleviate the notations, for the following of this section, we write σ1/2  (t):=σ(t).

    Proposition 5.1. For s=1/2, we have

    v=bN(t)H(0)+O(1log(σ1/2  (t)))as  t0, (5.1)

    where H(0) is the mean curvature of E at 0.

    Proof. Recall that x=vtν0 as t0, thanks to Lemma 2.3. We write

    u(x,σ(t))=RN κ1/2  (yx,σ(t))τE(y)dy=Qrκ1/2  (yx,σ(t))τE(y)dy+Qcrκ1/2  (yx,σ(t))τE(y)dy, (5.2)

    where Qr=BN1r×(r,r). By (2.6), we have

    Qcrκ1/2  (yx,σ(t))τE(y)dy=O(σ(t)) as   t0.

    Then, we have

    u(x,σ(t))=Qrκ1/2  (yx,σ(t))τE(y)dy+O(σ(t)). (5.3)

    By a change of variable and (2.3), we have

    Qrκ1/2  (yx,σ(t))τE(y)dy=EQrκ1/2  (yx,σ(t))dyEcQrκ1/2  (yx,σ(t))dy=BN1r  γ(y)rκ1/2  (yx,σ(t))dyBN1r  rγ(y)κ1/2  (yx,σ(t))dy=BN1r  γ(y)vtrvtκ1/2  (y,σ(t))dyBN1r  rvtγ(y)vtκ1/2  (y,σ(t))dy=2BN1r  γ(y)vt0κ1/2  (y,σ(t))dy+BN1r  0rvtκ1/2  (y,σ(t))dyBN1r  rvt0κ1/2  (y,σ(t))dy=2BN1r  γ(y)vt0κ1/2  (y,σ(t))dy+BN1r  r+vtrvtκ1/2  (y,σ(t))dy.

    The last line is due to the fact that the map yNκ1/2  (y,σ(t)) is even so that

    rvt0κ1/2  (y,σ(t))dyN=r+vt0κ1/2  (y,σ(t))dyN.

    Therefore we have

    Qrκ1/2  (yx,σ(t))τE(y)dy=2BN1r  γ(y)vt0κ1/2  (y,σ(t))dy+BN1r  r+vtrvtκ1/2  (y,σ(t))dy. (5.4)

    Using (2.6), we find that

    BN1r  r+vtrvtκ1/2  (y,σ(t))dy=O(σ(t)). (5.5)

    By a change of variable, the fundamental theorem of calculus, (1.5) and (2.3), we have

    BN1r  γ(y)vt0κ1/2  (y,σ(t))dy=BN1r  γ(y)vt0κ1/2  (y,0,σ(t))dy+BN1r  γ(y)vt010yNκ1/2  yN(y,θyN,σ(t))dydθ=BN1r  κ1/2  (y,0,σ(t))(12γyiyj(0)yiyj+O(|y|3)vt)dy+BN1r  γ(y)vt010yNκ1/2  yN(y,θyN,σ(t))dydθ=Δγ(0)2(N1)BN1r  |y|2κ1/2  (y,0,σ(t))dyvtBN1r  κ1/2  (y,0,σ(t))dy+BN1r  γ(y)vt010yNκ1/2  yN(y,θyN,σ(t))dydθ+O(BN1r  |y|3κ1/2  (y,0,σ(t))dy).

    Therefore

    BN1r  γ(y)vt0κ1/2  (y,σ(t))dy=H(0)2BN1r  |y|2κ1/2  (y,0,σ(t))dy vtBN1r  κ1/2  (y,0,σ(t))dy+BN1r  γ(y)vt010yNκ1/2  yN(y,θyN,σ(t))dydθ+O(σ(t)). (5.6)

    By (1.5), (1.6) and a change of variable, we have

    BN1rγ(y)vt0yNκ1/2  yN(y,θyN,σ(t))dy=O(BN1rσ(t)1  σ(t)1(γ(yσ(t))vt)0yN(1+|(y,θyN)|2)N+22dy)=O(BN1rσ(t)1  σ(t)2(γ(yσ(t))vt)2(1+|y|2)N+22dy)=O(σ2(t)BN1rσ(t)1  (σ(t)2|y|2vt)2(1+|y|2)N+22dy)=O(σ(t))+O(vt)+O(v2t2(σ(t))2).

    Now Lemma 2.3 yields vt=O(σ(t)2) and thus

    BN1rγ(y)vt010yNκ1/2  yN(y,θyN,σ(t))dydθ=O(σ(t)). (5.7)

    We get from (5.3), (5.4), (5.5), (5.6) and (5.7) that

    u(x,σ(t))=σ(t)H(0)BN1rσ(t)1  |y|2P1/2  (y,0)dy2vtBN1rσ(t)1  P1/2  (y,0)dy+O(σ(t)) as   t0.

    Thanks to (1.6), we have

    RN1BN1rσ(t)1  P1/2  (y,0)dy=O(σ(t)2)andBN1σ(t)1BN1rσ(t)1  |y|2P1/2  (y,0)dy=O(1)as t0.

    This implies that

    u(x,σ(t))=σ(t)H(0)BN1σ(t)1  |y|2P1/2  (y,0)dy2vt(σ(t))1RN1  P1/2  (y,0)dy+O(σ(t)). (5.8)

    Using polar coordinates and (1.6), we then have

    BN1σ(t)1  |y|2P1/2  (y,0)dyCN,1/2ωN21/σ(t)0mN(1+m2)N+12dm, (5.9)

    where ωN2:=|SN2|. By the change of variable ρ=1m, we have

    1/σ(t)0mN(1+m2)N+12dm=+σ(t)1ρ(1+ρ2)N+12dρ =1σ(t)1ρ(1+ρ2)N+12dρ++11ρ(1+ρ2)N+12dρ =1σ(t)1ρ(1+ρ2)N+12dρ+O(1)=log(σ(t))+O(1).

    Letting bN(t):=BN1σ(t)1  |y|2P1/2  (y,0)dy2log(σ(t))RN1  P1/2  (y,0)dy, by (5.8), (5.9) and the above estimate, we obtain, as t0,

    u(x,σ(t))=σ(t)H(0)BN1σ(t)1  |y|2P1/2  (y,0)dy2vt(σ(t))1RN1  P1/2  (y,0)dy+O(σ(t))=(σ(t))1[σ2(t)H(0)BN1σ(t)1  |y|2P1/2  (y,0)dy2vtRN1  P1/2  (y,0)dy+O(σ2(t))].

    Since x=vtνEt, we have u(x,σ(t))=0. Recalling that t=σ2(t)|log(σ(t))|, we finally get

    0=σ(t)log(σ(t))[H(0)BN1σ(t)1  |y|2P1/2  (y,0)dy|log(σ(t))|2vRN1  P1/2  (y,0)dy+O(1log(σ(t)))].

    Hence

    v=bN(t)H(0)+O(1log(σ(t))) as  t0.

    The proof is then ended.

    This work is supported by the Alexander von Humboldt foundation and the German Academic Exchange Service (DAAD). Part of this work was done while the authors were visiting the International Center for Theoretical Physics (ICTP) in December 2019 within the Simons associateship program.

    The authors declare no conflict of interest.



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