Processing math: 84%
Research article Special Issues

The fourth-order total variation flow in Rn

  • We define rigorously a solution to the fourth-order total variation flow equation in Rn. If n3, it can be understood as a gradient flow of the total variation energy in D1, the dual space of D10, which is the completion of the space of compactly supported smooth functions in the Dirichlet norm. However, in the low dimensional case n2, the space D1 does not contain characteristic functions of sets of positive measure, so we extend the notion of solution to a larger space. We characterize the solution in terms of what is called the Cahn-Hoffman vector field, based on a duality argument. This argument relies on an approximation lemma which itself is interesting. We introduce a notion of calibrability of a set in our fourth-order setting. This notion is related to whether a characteristic function preserves its form throughout the evolution. It turns out that all balls are calibrable. However, unlike in the second-order total variation flow, the outside of a ball is calibrable if and only if n2. If n2, all annuli are calibrable, while in the case n=2, if an annulus is too thick, it is not calibrable. We compute explicitly the solution emanating from the characteristic function of a ball. We also provide a description of the solution emanating from any piecewise constant, radially symmetric datum in terms of a system of ODEs.

    Citation: Yoshikazu Giga, Hirotoshi Kuroda, Michał Łasica. The fourth-order total variation flow in Rn[J]. Mathematics in Engineering, 2023, 5(6): 1-45. doi: 10.3934/mine.2023091

    Related Papers:

    [1] José M. Mazón . The total variation flow in metric graphs. Mathematics in Engineering, 2023, 5(1): 1-38. doi: 10.3934/mine.2023009
    [2] Filippo Gazzola, Gianmarco Sperone . Remarks on radial symmetry and monotonicity for solutions of semilinear higher order elliptic equations. Mathematics in Engineering, 2022, 4(5): 1-24. doi: 10.3934/mine.2022040
    [3] Rupert L. Frank, Tobias König, Hynek Kovařík . Energy asymptotics in the Brezis–Nirenberg problem: The higher-dimensional case. Mathematics in Engineering, 2020, 2(1): 119-140. doi: 10.3934/mine.2020007
    [4] Qiang Du, Tadele Mengesha, Xiaochuan Tian . $ L^{p} $ compactness criteria with an application to variational convergence of some nonlocal energy functionals. Mathematics in Engineering, 2023, 5(6): 1-31. doi: 10.3934/mine.2023097
    [5] Giuseppe Procopio, Massimiliano Giona . Bitensorial formulation of the singularity method for Stokes flows. Mathematics in Engineering, 2023, 5(2): 1-34. doi: 10.3934/mine.2023046
    [6] Márcio Batista, Giovanni Molica Bisci, Henrique de Lima . Spacelike translating solitons of the mean curvature flow in Lorentzian product spaces with density. Mathematics in Engineering, 2023, 5(3): 1-18. doi: 10.3934/mine.2023054
    [7] Juan-Carlos Felipe-Navarro, Tomás Sanz-Perela . Semilinear integro-differential equations, Ⅱ: one-dimensional and saddle-shaped solutions to the Allen-Cahn equation. Mathematics in Engineering, 2021, 3(5): 1-36. doi: 10.3934/mine.2021037
    [8] Konstantinos T. Gkikas . Nonlinear nonlocal equations involving subcritical or power nonlinearities and measure data. Mathematics in Engineering, 2024, 6(1): 45-80. doi: 10.3934/mine.2024003
    [9] Yoshihisa Kaga, Shinya Okabe . A remark on the first p-buckling eigenvalue with an adhesive constraint. Mathematics in Engineering, 2021, 3(4): 1-15. doi: 10.3934/mine.2021035
    [10] Nguyen Cong Phuc, Igor E. Verbitsky . Uniqueness of entire solutions to quasilinear equations of $ p $-Laplace type. Mathematics in Engineering, 2023, 5(3): 1-33. doi: 10.3934/mine.2023068
  • We define rigorously a solution to the fourth-order total variation flow equation in Rn. If n3, it can be understood as a gradient flow of the total variation energy in D1, the dual space of D10, which is the completion of the space of compactly supported smooth functions in the Dirichlet norm. However, in the low dimensional case n2, the space D1 does not contain characteristic functions of sets of positive measure, so we extend the notion of solution to a larger space. We characterize the solution in terms of what is called the Cahn-Hoffman vector field, based on a duality argument. This argument relies on an approximation lemma which itself is interesting. We introduce a notion of calibrability of a set in our fourth-order setting. This notion is related to whether a characteristic function preserves its form throughout the evolution. It turns out that all balls are calibrable. However, unlike in the second-order total variation flow, the outside of a ball is calibrable if and only if n2. If n2, all annuli are calibrable, while in the case n=2, if an annulus is too thick, it is not calibrable. We compute explicitly the solution emanating from the characteristic function of a ball. We also provide a description of the solution emanating from any piecewise constant, radially symmetric datum in terms of a system of ODEs.



    Dedicated to Professor Neil Trudinger on the occasion of his 80th birthday.

    We consider the fourth-order total variation flow equation in Rn of the form

    ut=Δdivu|u|. (1.1)

    We aim to give explicit description of its solutions emanating from piecewise constant radial data. However, it turns out that the definition of a solution is itself non-trivial since Δ does not have a bounded inverse on L2(Rn). Our first goal is thus to provide a rigorous definition of a solution. Our second goal is to find explicit formula for the solution to (1.1) when the initial datum u(0,x)=u0(x) is the characteristic function of a ball or an annulus. In other words,

    u0=a01BR0oru0=a01AR10R00a0R,

    where 1K is the characteristic function of a set KRn, i.e.,

    1K(x)={1,xK0,xRnK.

    Here BR denotes the open ball of radius R centered at 0Rn and AR1R0 denotes the annulus defined by AR1R0=BR1¯BR0. Our major concern is whether or not the solution remains a characteristic function throughout the evolution. For example, in the case u0=a01BR0, whether or not the solution u of (1.1) is of the form

    u(t,x)=a(t)1BR(t)

    with a function a=a(t). In other words, we are asking whether the speed ut on the ball BR(t) and on its complement are constant in the spatial variable. As in the second-order problem [3] (see also [17]), this leads to the notion of calibrability of a set. In the case of the second-order problem ut=div(u/|u|), a ball and its complement are always calibrable and R(t)R0, i.e., the ball does not expand nor shrink [3]. In our problem, R(t) may be non-constant.

    We first note that the definition of a solution itself is non-trivial. The fourth-order total variation flow has been mainly studied in the periodic setting [12,14] or in a bounded domain with some boundary conditions [15]. Formally, it is a gradient flow of the total variation functional

    TV(u):=Ω|u|

    with respect to the inner product

    (u,v)1=Ωu(Δ)1v

    when Ω is a domain in Rn or a flat torus Tn. In the periodic setting, i.e., Ω=Tn as in [12,14], it is interpreted as a gradient flow in H1av which is the dual space of H1av, the space of average-free H1 functions equipped with the inner product

    (u,v)1=Ωuv.

    For the homogeneous Dirichlet boundary condition with bounded Ω, H1av is replaced by D1, the dual space of D10=D10(Ω), which is the completion of Cc(Ω) in the norm associated with the inner product (u,v)1; here Cc(Ω) denotes the space of all smooth functions compactly supported in Ω. By the Poincaré inequality, both H1av and D10(Ω) can be regarded as subspaces of L2(Ω). However, if Ω equals Rn, the situation is more involved. If n3, D10(Rn) is continuously and densely embedded in L2(Rn), where p=np/(np) so that 2=2n/(n2), by the Sobolev inequality. In fact,

    D10(Rn)=D1(Rn)L2(Rn),D1(Rn)={uL1loc(Rn)|uL2(Rn)}

    see e.g., [11]. On the other hand, if n2, D10 is isometrically identified with the quotient space ˙D1(Rn):=D1(Rn)/R, when D1(Rn) is equipped with inner product (u,v)1 [11]. Thus, we need to be careful when n2 because an element of D10(Rn) is determined only up to a constant. In any case, D10(Rn) is a Hilbert space with the scalar product

    (u,v)D10(Rn)=Rnuv.

    Therefore, we can identify D10(Rn) with its dual space by means of the isometry

    Δ:u(u,)D10(Rn).

    On the other hand, let us define a subspace ˜D1(Rn)D10(Rn) by

    ˜D1(Rn)={wRnuw:uCc(Rn)}if n3,
    ˜D1(Rn)={wRnuw:uCc,av(Rn)}if n=1 or n=2,

    where

    Cc,av(Rn)={uCc(Rn):Rnu=0}.

    Then the closure D1(Rn) of ˜D1(Rn) coincides with D10(Rn) [11]. Note that the restriction to Cc,av(Rn) in the definition of ˜D1(Rn) in n=1,2 is necessary for the functionals to be well-posed on D1(Rn)/R. In any case, since (by definition) the space of test functions D(Rn) is continuously and densely embedded in D10(Rn), we also have a continuous embedding D1(Rn)=D10(Rn)D(Rn). Throughout the paper, we will often drop (Rn) in the notation for spaces of functions on Rn, e.g., D1=D1(Rn).

    In the first step, we give a rigorous definition of the total variation functional TV on D1. Then we calculate the subdifferential of TV in D1 space. Since it is a homogeneous functional, we are able to apply a duality method [3] to characterize the subdifferential, provided that TV is well approximated by nice functions in D1. We know that Cc,av(Rn) is dense in D1 for n2; see e.g., [11]. However, it is not immediately clear whether TV is simultaneously approximable. Fortunately, it turns out that for any wD1, there is a sequence wkCc,av(Rn) which converges to w in D1 and TV(wk)TV(w) as k. This approximation part is relatively involved since we have to use an efficient cut-off function. Using the approximation, we are able to characterize the subdifferential D1TV of TV in D1 by adapting the argument in [3]. Namely, we have

    D1TV(u)={v=ΔdivZ|ZL(Rn), |Z|1, u,divZ=TV(u)},

    where ,  denotes the canonical paring of D1 and D10. A vector field Z corresponding to an element of the subdifferential is often called a Cahn-Hoffman vector field. The equation (1.1) should be interpreted as the gradient flow of TV in D1, i.e.,

    utD1TV(u), (1.2)

    and its unique solvability for any initial datum u0D1 is guaranteed by the classical theory of maximal monotone operators ([6,24]). By our characterization of the subdifferential, we are able to give a more explicit definition of a solution which is consistent with that proposed in [15]. Namely, for u0D1 with TV(u0)<, a function uC([0,T[,D1) is a solution to (1.2) with u(0)=u0 if and only if there exists ZL(]0,T[×Rn) satisfying divZL2(0,T;D10(Rn)) such that

    ut=ΔdivZinD1(Rn) (1.3)
    |Z(t,x)|1for a.e. xRn (1.4)
    u,divZ=TV(u) (1.5)

    for a.e. t]0,T[. This is convenient for calculating explicit solutions.

    Unfortunately, in n2, for a compactly supported square integrable function u0, we know that u0D1 if and only if u0 is average-free, i.e., Rnu0=0 (see Lemma 17). Thus, the characteristic function of any bounded, measurable set of positive measure does not belong to D1. We have to extend a class of initial data u0 such that u0=ψ+w0 with w0D1 while ψ is a fixed compactly supported L2 function. We consider a gradient flow utD1TV(u) in the affine space ψ+D1. Since D1 is a directional partial derivative in the direction of D1, it is more convenient to consider solutions to evolutionary variational inequality

    12ddtu(t)g2D1TV(g)TV(u(t))for a.e. t>0 (1.6)

    for any gψ+D1 [2]. In the case ψ=0, it is easy to show that the evolutionary variational inequality is equivalent to (1.2). Indeed, by definition of the subdifferential, (1.2) is equivalent to

    (ut,gu(t))D1TV(g)TV(u(t))

    for any gD1. The left-hand side equals (d/dt)(ug2/2). Thus, the equivalence follows if ψ=0. From now on we assume that Rnψ0.

    It is easy to check that there is at most one solution to the evolutionary variational inequality (1.6). The solution u is constructed by solving

    wtD1TV(w+ψ)withw(0)=w0=u0ψ

    and setting u=w+ψ. Characterization of the (directional) subdifferential is more involved since wTV(w+ψ) is no more positively one-homogeneous. We identify the one dimensional space {cψ|cR} with R and consider the Hilbert space E1 defined as the orthogonal sum D1R. We calculate the subdifferential by the duality method since TV is now positively one-homogeneous on E1. We then project this subdifferential onto D1 to get a characterization of a (directional) subdifferential D1TV. We end up with a characterization of solution to (1.2) similar to (1.3)–(1.5), with (1.5) adjusted in a suitable way. If we also denote E10=D1 in n2, E10=D10, E1=D1 in n3 and

    u,vE={u,v if n3,w,[v]+cψv, where u=w+cψ, wD1 if n2, (1.7)

    for uE1, vE10, we end up with the following definition of solution.

    Definition 1. Assume that u0E1. We say that uC([0,[,E1) with utL2loc(]0,[,D1) is a solution to (1.1) with initial datum u0 if there exists ZL(]0,[×Rn) with divZ(t,)E10 for a.e. t>0 such that

    ut=ΔdivZinD1(Rn) (1.8)
    |Z(t,x)|1fora.e.xRn (1.9)
    u,divZE=TV(u) (1.10)

    for a.e. t>0.

    and associated well-posedness result.

    Theorem 2. Let u0E1. There exists a unique solution to (1.1) with initial datum u0.

    Our next problem is whether or not the speed of a characteristic function of a set is spatially constant inside and outside of the set. By the general theory ([6,24]), the speed is determined by the minimal section (canonical restriction) 0D1TV of D1TV. In other words, 0D1TV(u)=v0 minimizes vD1 for vD1TV(u), i.e.,

    0D1TV(u):=argmin{vD1|vD1TV(u)}.

    To motivate the notion of calibrability, we consider a smooth function u such that

    ¯U={xRn|u(x)=0},

    where U is a smooth open set. Outside ¯U, we assume that u0. To fix the idea, we assume that U has negative signature (orientation) in the sense that u<0 outside ¯U. By our specification of u, we see that

    0D1TV(u)=argmin{divZD10||Z|1 in U, Z=u/|u| in ¯Uc, divZD10}.

    Since divZ is locally integrable, Zν does not jump across U, where ν is the exterior unit normal of U. In this case,

    Zν=Zu/|u|=1on U. (1.11)

    Since divZ does not have a singular part, divZ does not jump across U. In this case,

    divZ=divνon U. (1.12)

    However, v=ΔdivZ may have a non-zero singular part concentrated on U even if v=v0, i.e., v is the minimizer. This phenomenon is observed in [12,20,21] in a one-dimensional periodic setting. Different from the second-order problem, this causes expansion or shrinking of the ball when the solution u is of the form u(t,x)=a(t)1BR(t)(x). If u>0 outside U, the minus in (1.11) and (1.12) should be replaced by the plus.

    If 0D1TV(u) is constant on BR(t) and (¯BR(t))c, this property is preserved under the evolution, which leads us to definition of calibrability. Note that the value of 0D1TV(u) on U is determined by U and its signature does not depend on particular value of u. We say that U (with negative signature) is calibrable if there exists Z0L(U,Rn) such that divZ0L2(U,Rn), Z0 satisfies (1.11), (1.12), |Z0|1 a.e. in U and ΔdivZ0 is constant on U. We call any such Z0 a calibration for U.

    Recall that in the case of the second-order problem, we say that U (with negative signature) is calibrable if there exists ˜Z0L(U,Rn) satisfying (1.11), |˜Z0|1 a.e. in U and div˜Z0 is a constant function on U. This is formally equivalent to calibrability in [3,4]. It can be shown that ˜Z0 is a calibration for U if and only if

    ˜Z0argmin{divZL2(U)|z satisfies (1.11) and |Z|1 a.e. in U}

    and div˜Z0 is a constant function on U, which is the definition of calibrability in [17].

    Going back to our fourth-order problem, if Z0 is a calibration for U, then w=divZ0 must satisfy

    Δw=λin U (1.13)
    w=divνon U (1.14)

    with some constant λ. If U is bounded, λ is determined by (1.11) since

    UwdLn=UdivZ1dLn=UZ1νdHn1=Hn1(U), (1.15)

    where Hn1 denotes the n1 dimensional Hausdorff and Ln denotes the Lebesgue measure in Rn. Using this fact, in Section 5 we prove that if Z0 is a calibration for a bounded U, then

    Z0argmin{divZL2(U)|Z satisfies (1.11), (1.12) and |Z|1 a.e. in U}. (1.16)

    Moreover, we obtain an "explicit" formula for the constant λ in terms of the Saint-Venant problem in U.

    In the radially symmetric setting, it is not difficult to show that Z0 in (1.16) can be chosen in the form z(|x|)x|x|. Indeed, if Z0 is belongs to the set of minimizers (1.16), then its rotational average ¯Z0 belongs to (1.16) as well, because averaging preserves (1.11), (1.12) and the inequality |Z|1. Since the angular part of ¯Z0 does not contribute to the divergence, it is possible to delete this part (Lemma 31). We thus conclude that there is an element of (1.16) of form Z(x)=z(|x|)x|x|. Thus, the Eq (1.13) can be written as the third-order ODE of the form

    r1n(rn1(r1n(rn1z)))=λ (1.17)

    since divZ=r1n(rn1z). If U is BR with negative signature, condition (1.11) implies

    z(R)=1. (1.18)

    Since divZ=z+(n1)z/r, condition (1.12) implies that

    z(R)=0. (1.19)

    Solving (1.17) under the assumption that z is smooth near zero under conditions (1.18), (1.19), we eventually get a unique solution (1.17)–(1.19) of the form

    z(r)=12(rR)332rR,λ=n(n+2)R3

    for all n1. It is easy to see that Z(x)=z(|x|)x|x| satisfies the constraint |Z|1 in BR. We conclude that all balls are calibrable. More careful argument is necessary, but we are able to discuss calibrability of an annulus as well as a complement of a ball.

    Theorem 3. (i) All balls are calibrable for all n1.

    (ii) All complement of balls are calibrable except n=2.

    (iii) If n=2, all complement of balls are not calibrable.

    (iv) All annuli (with definite signature) are calibrable except in n=2.

    (v) For n=2, there is Q>1 such that an annulus (with definite signature) is calibrable if and only if the ratio of the exterior radius over the interior radius is smaller than or equal to Q. In other words, AR1R0 is calibrable if and only if R1/R0Q.

    Theorem 3(v) is consistent with (iii) since R1 implies AR1R0 converges to ¯BR0c, a complement of the closure of the ball BR0. Note that in the case of an annulus, there is a possibility we take a signature which is different on the exterior boundary BR1 and the interior boundary BR0. We also study such indefinite cases.

    We now calculate an explicit solution of (1.1) starting from u0=a01BR0. We first discuss the case n2. Since a ball and its complement is calibrable, the solution is of the form

    u(t,x)=a(t)1BR(t). (1.20)

    We take the (radial) calibration Zin in BR(t) and Zout in Rn¯BR(t) and set

    Z(x,t)={Zin(x),xBR(t)Zout(x),xRn¯BR(t).

    Here Zout(x)=zout(|x|)x|x| can be calculated as

    zout(r)=n12(rR)3+n32(rR)1n

    while, as we already discussed, zin for Zin(x)=zin(|x|)x|x| is of the form

    zin(r)=12(rR)332rR.

    This Z satisfies (1.9) and (1.10), and moreover divZD10 for any t>0. Moreover, divZ is continuous across BR(t). However, divZ may jump across BR(t). Actually,

    ΔdivZ=λ1BR(t)+ν(divZindivZout)δBR(t),

    where δΓ(φ)=ΓφdHn1 or δΓ=Hn1Γ for a hypersurface Γ and ν is the exterior unit normal of BR(t), i.e., ν=x/R(t). Here λ=n(n+2)R3. By a direct calculation, the quantity ν(divZindivZout)=n(n4)R2. Since ut=ΔdivZ, by

    t(a1BR)=dadt1BR+adRdtδBR,

    we conclude that

    dadt=n(n+2)R3,dRdt=n(n4)aR2.

    Since

    ddt(aR3)=n(n+2)3n(n4)=n(4n10),

    an explicit form of a solution is given as

    a(t)=a0(1n(4n10)a0R30t)n+24n10,R(t)=R0(1n(4n10)a0R30t)n44n10.

    As noticed earlier, in the case n=2, the complement of the disk is not calibrable. If u is a radially strictly decreasing function outside BR, we expect Zout(x)=x/|x| for |x|>R(t). In [15], it is proposed that a solution u to (1.1) must satisfy

    ut=ΔdivZout.

    Since divZout=(n1)/|x|2 and divZout=(n1)x|x|3, this implies

    ut(t,x)=(n1)(n3)|x|3,x(¯BR(t))c=Rn¯BR(t). (1.21)

    In the case n=2, divZoutL2((¯BR(t))c) so Zout is a Cahn-Hoffman vector field.

    If we start with u0=a01BR0 with a0>0 for n=2, the expected form of a solution is

    u(t,x)=a(t)1BR(t)(x)+t|x|31¯BR(t)c(x), (1.22)

    where

    dadt=24R3,(a(t)tR(t)3)dRdt=22R2. (1.23)

    Analyzing this ODE system, we can deduce qualitative properties of the solution. Summing up our results yields.

    Theorem 4. Let u0=a01BR0 with a0>0.

    If n3, then the solution u to (1.1) with initial datum u0 is of the form

    u(t,x)=a(t)1BR(t)fort<t=a0R30/(n(4n10))

    and u(t,x)0 for tt. (The time t is called the extinction time.) Moreover, a(t) is decreasing and a(t)0 as tt.

    (i) R(t) is increasing and R(t) as tt for n=3.

    (ii) R(t)=R0 for n=4.

    (iii) R(t) is decreasing and R(t)0 as tt for n5.

    If n=2, then the solution is not a characteristic function for t>0. It is of the form (1.21) and moves by (1.23). In particular, there is no extinction time, R(t) is increasing and a(t) is decreasing. Moreover, R(t) and a(t)0 as t. The gap a(t)tR(t)3 is always positive.

    If n=1, then the solution is of the form u(t,x)=a(t)1BR(t) for t>0. Moreover, R(t) is increasing and a(t) is decreasing with R(t) and a(t)0 as t.

    We note that the infinite extinction time observed in n2 is related to the fact that 0 is not an element of the affine space u0+D1 where the flow lives if u00. In [14], finite time extinction for solution to (1.1) is proved in a periodic setting for average zero initial data when the space dimension n4 Our result is unrelated to their result because we consider (1.1) in Rn.

    The formula (1.21) does not give a solution to (1.1) when n4 since divZout does not belong to L2((¯BR(t))c). In the case n=3, this formula is consistent with our definition. If we consider u0 strictly radially decreasing for |x|>R0 and u0(x)=u for |x|R0, then u0 does not belong to the domain of D1TV for n4. In other words, there is no Cahn-Hoffman vector field.

    These results contrast with the second-order total variation flow

    ut=div(u/|u|).

    In the second-order problem, a ball and an annulus are always calibrable with their complements, see e.g., [3] or [17, Section 5]. Furthermore, ut(t,) is a locally integrable function without singular part for t>0. Thus, for example, the solution starting from u0=a01BR0 (a0>0) must be u(t,x)=a(t)1BR0 with a(t)=λt+a0, where λ is the Cheeger ratio, i.e., λ=Hn1(BR0)/Ln(BR0). In particular, the extinction time t equals t=a0/λ.

    We conclude this paper by deriving a system of ODEs prescribing the solution in the case when the initial datum is a piecewise constant, radially symmetric function, which we call a stack. To be precise, we say that wE1 is a stack if it is of the form

    w=a01BR0+a11AR1R0++aN11ARN1RN2+aN1RnBRN1,

    0<R0<R1<<RN1, akR. In particular, we obtain

    Theorem 5. Let n2 and let u0 be a stack. If u is the solution to (1.1), then u(t,) is a stack for t>0.

    In the case n=2, this result is no longer true, as evidenced by Theorem 4. However the solution can still be prescribed by a finite system of ODEs.

    A total variation flow type equation

    wt=Δ(div(w/|w|)+βdiv(w|w|)) (1.24)

    was introduced by [31] to describe the height of crystal surface moved by relaxation dynamics below the roughening temperature, where β>0. For this equation, characterization of the subdifferential of the corresponding energy was given by Y. Kashima in periodic setting [20,21] and under Dirichlet condition on a bounded domain [21]. The speed of a facet (a flat part of the graph) is calculated for n=1 in [20] and for a ball with the Dirichlet condition under radial symmetry [21]. Different from the second-order problem, the speed of a facet is determined not only by the shape of facet. Also it has been already observed in [20], that the minimal section may have a delta part although the behavior of the corresponding solution was not studied there. A numerical computation was given in [23]. The Eq (1.24) was derived as a continuum limit of models describing motion of steps on crystal surface as discussed in [27], where numerical simulation was given; see also [22].

    In [7], a crystalline diffusion flow was proposed and calculated numerically. In a special case, it is of the form wt=2x(W(wx)), where W is a piecewise linear convex function, when the curve is given as the graph of a function. This equation was analyzed in [13] in a class of piecewise linear (in space) solutions.

    Fourth-order equations of type (1.1) were proposed for image denoising as an improvement over the second-order total variation flow. For example, the equation

    wt=Δdiv(w/|w|)+λ(fw),

    where f is an original image which is given and λ>0, corresponds to the Osher-Solé-Vese model [28]. The well-posedness of this equation was proved by using the Galerkin method by [8].

    For (1.1), an extinction time estimate was given in [14] for n=1,2,3,4 in the periodic setting. It was extended to the Dirichlet problem in a bounded domain by [15]. In the review paper [12], it was proved that the solution u of (1.1) in n=1 may become discontinuous instantaneously even if the initial datum is Lipschitz continuous, because the speed may have a delta part.

    There are a few numerical studies for (1.1) in the periodic setting. A duality-based numerical scheme which applies the forward-backward splitting has been proposed in [16]. A split Bregman method was adjusted to (1.1) and also (1.24) in [18]. In these methods, the singularity of the equation at u=0 is not regularized. However, all above studies deal with either periodic, Dirichlet or Neumann boundary condition for a bounded domain. It has never been rigorously studied in Rn, although in [15] there are some preliminary calculations for radial solution in Rn.

    This paper is organized as follows. In Section 2, we discuss basic properties of the total variation on D1, notably we show strict density of Cc,av. In Section 3, we give a rigorous definition of a solution to (1.1) and obtain a verifiable characterization of solutions. In Section 4, we extend the results of the previous section to include initial data with non-zero average in n=1,2. In Section 5, we introduce the notion of calibrability. In Section 6, we discuss calibrability of rotationally symmetric sets in Rn. In Section 7, we study solutions emanating from piecewise constant, radially symmetric data.

    In this section, we give a rigorous definition of the total variation TV on D1 and relate it to the usual total variation defined on L1loc. The main tool that we use here as well as in the following section is an approximation lemma, which for a given wD1 produces a sequence of nice functions wkD1 that converges to w in D1 and TV(wk)TV(w).

    Let us denote

    X1={ψCc(Rn,Rn), ψL(Rn,Rn)1}.

    We define TV:D1(Rn)[0,] by

    TV(u)=supψX1u,divψ.

    Let us compare this definition with the usual total variation, which we denote here by ¯TV:L1loc(Rn)[0,], defined by

    ¯TV(u)=supψX1Rnudivψ.

    First of all, as in the case ¯TV, we easily check that TV is lower semicontinuous with respect to the weak-* (and, a fortiori, strong) convergence in D1(Rn). Indeed, if vkv in D1(Rn),

    TV(v)=supψX1{v,divψ}=supψX1lim infk{vk,divψ}lim infksupψX1{vk,divψ}=lim infkTV(vk).

    In fact, we have

    Lemma 6. We have D(TV)L1loc, and so D(TV)D(¯TV) with TV and ¯TV coinciding on D(TV). In particular, if n2, D(TV)L1(Rn). If n=1,

    D(TV)L0(R)={wL(R):esslimx±w(x)=0}.

    The proof of this fact is a consequence of the lemma below and we postpone it.

    Lemma 7. For any wD1(Rn) there exists a sequence wkCc,av(Rn) such that

    wkwinD1(Rn)

    and

    TV(wk)TV(w).

    To prove it, we will use a special choice of cut-off function and associated variant of the Sobolev-Poincaré inequality. For R>0, let us denote by ϑR the element of minimal norm in D10(Rn) among those wD10(Rn) that satisfy w(x)=1 if |x|R2, w(x)=0 if |x|R. It is an easy exercise to show that for R2|x|R

    ϑR(x)=(2n21)1((|x|R)2n1) if n2,ϑR(x)=logR|x|log2 if n=2.

    In either case,

    ϑR(x)=Cn|x|nxR2n if R2|x|R. (2.1)

    Lemma 8. If p[1,n[ and q[1,p], then for all wC1(Rn), R>0 there holds

    wϑRwϑRLq(BR)CR1+nqnpwLp(BR) (2.2)

    with C=C(n,p) and

    ϑRLp(Rn)=CR1p(n1)(2p) (2.3)

    with a different C=C(n,p).

    Proof. Let vC1(Rn). Following the proof of the standard Poincaré inequality by contradiction using Rellich-Kondrachov theorem, we obtain

    vϑ1vϑ1Lp(B1)CvLp(B1).

    Applying the Sobolev inequality in B1 to the function vϑ1vϑ1, we upgrade this to

    vϑ1vϑ1Lq(B1)CvLp(B1) (2.4)

    Next, let v(x)=w(Rx) for a given wC1(Rn). We observe that

    ϑ1(x)=ϑR(Rx)for xRn

    and so, by a change of variables x=y/R,

    ϑ1=1RnϑR,ϑ1v=1RnϑRw.

    Applying the same change of variables to both sides of (2.4) we conclude the proof of (2.2).

    The proof of (2.3) is a matter of direct calculation.

    Let us now return to the proof of the approximation lemma.

    Proof of Lemma 7. Given wD1, let

    wε,R=(ϱεwϑRϱεwϑR)ϑR.

    Equivalently, for φD10(Rn),

    wε,R,φ=w,ϱε((φϑRφϑR)ϑR).

    Denoting ˜w=(Δ)1w,

    wε,Rw,φ=ϱε˜w((φϑRφϑR)ϑRφ)+(ϱε˜w˜w)φ=ϱε˜w(ϑR1)φ+ϱε˜w(φϑRφϑR)ϑR+(ϱε˜w˜w)φ.

    We estimate the second term on the r. h. s. using the Poincaré inequality from Lemma 8, taking into account that the support of the integrand is contained in ¯AR, where AR=BR¯BR/2,

    |ϱε˜w(φϑRφϑR)ϑR|Cϱε˜w1ARL2(Rn)φϑRφϑRL2(Rn)ϑRL(Rn)Cϱε˜w1ARL2(Rn)φL2(Rn).

    Thus,

    wε,RwD1(Rn)=supφD10(Rn)1wε,Rw,φ(1ϑR)ϱε˜wL2(Rn)+Cϱε˜w1ARL2(Rn)+ϱε˜w˜wL2(Rn)

    and so

    limε0+limRwε,RwD1(Rn)=0. (2.5)

    Next we estimate TV(wε,R). Due to lower semicontinuity of TV, we can assume without loss of generality that TV(w)<. First, we note that ϱεwD1(Rn)C(Rn) ([29,30]) and

    ϱεψL(Rn,Rn)ψL(Rn,Rn).

    Thus, for any ψL(Rn,Rn),

    TV(ϱεw)=supψX1{w,divϱεψ}TV(w).

    In particular, this implies that ϱεwL1(Rn) for ε>0 and |ϱεw|=TV(ϱεw)TV(w). By Lemma 8,

    TV(wε,R)=|wε,R|ϑR|ϱεw|+|(ϱεwϑRϱεwϑR)ϑR||ϱεw|+CϱεwϑRϱεwϑRL1(Rn)ϑRLn(Rn)ϱεwL1(Rn)+CϱεwL1(Rn)R(n1)(n2)n(1+CR(n1)(n2)n)TV(w). (2.6)

    If n3, together with lower semicontinuity of TV, this yields

    lim(ε,R)(0,)TV(wε,R)=TV(w).

    Taking into account (2.5), by a diagonal procedure we can select sequences (εk), (Rk) such that wk:=wεk,Rk satisfies both requirements in the assertion. On the other hand, if n=1 or n=2, (2.6) only implies uniform boundedness of TV(wε,R).

    Let now n=2. Since wε,R have compact support, uniform boundedness of TV(wε,R) implies uniform bound on wε,R in L2(Rn) by the Sobolev inequality. As wε,R converges to ϱεw in D1(Rn), we have ϱεwL2(Rn). This allows us to improve (2.6):

    TV(wε,R)=|wε,R|ϑR|ϱεw|+|ϱεwϑR|+ϑRϱεwϑR|ϑR|ϱεwL1(R2)+Cϱεw1ARL2(R2)ϑRL2(R2)+CϑRD10(R2)ϱεwD1(R2)ϑRL1(R2)ϑRL1(R2)TV(w)+Cϱεw1ARL2(R2)+CRϱεwD1(R2). (2.7)

    The r. h. s. of (2.7) converges to TV(w) as R and we conclude as before.

    Next, consider n=1. In this case, finiteness of TV(w) implies that ϱεwL(Rn) and there exist g±εR such that

    limx±ϱεw(x)=g±ε.

    Now, let η±R be the element of minimal norm in D10(R) under constraints

    η±R(±x)=1 if x[2R,3R],η±R(±x)=0 if x]R,4R[.

    (Clearly, η±R is a continuous, piecewise affine function.) We have

    |ϱεw,η±R|ϱεwD1(R)η±RD10(R)0 as R.

    On the other hand, since η±R are compactly supported and ϱεw coincides as distribution with a locally integrable function, we can calculate

    ϱεw,η±R=ϱεwη±Rg±ε as R,

    so g±ε=0. Therefore, we can estimate

    TV(wε,R)=|wε,R|ϑR|ϱεw|+|ϱεwϑR|+ϑRϱεwϑR|ϑR|TV(w)+2RAR|ϱεw|+2ϑRϱεwϑR. (2.8)

    Since we have shown that ϱεw(x)0 as x±, the averages on the r. h. s. converge to 0 and we conclude as before.

    As a first application of the approximation lemma, we demonstrate Lemma 6 announced before.

    Proof of Lemma 6. Let wD(TV) and let (wk)Cc,av(Rn) be the sequence provided by Lemma 7. Let first n>1. Since wk is uniformly bounded in L1(Rn,Rn), by the Sobolev embedding wk is uniformly bounded in L1(Rn). Therefore, wL1(Rn).

    In case n=1, since wk are compactly supported and uniformly bounded in L1(R), wk is uniformly bounded in L(R). From these two bounds it follows that wL(R) and that w is a finite signed measure on R. In particular, w has essential limits at ±. Reasoning as in the final part of the proof of Lemma 7, we show that these limits vanish.

    For a gradient flow of a convex functional, there is a general theory initiated by Y. K¯omura [24] and developed by H. Brézis [6] and others. It is summarized as follows.

    Proposition 9 ([6]). Let H be a real Hilbert space. Let E be a lower semicontinuous, convex functional on H with values in ],]. Assume that D(E) is dense in H. Then, for any u0H, there exists a unique solution uC([0,[,H) which is absolutely continuous in (δ,T) (for any δ<T<) satisfying

    {utE(u)for a.e.t>0u(0)=u0. (3.1)

    Moreover,

    tsut2HdτE(u(s))E(u(t))for allts>0.

    If E(u0)<, then s=0 is allowed. In particular, utL2(0,;H).

    As in [2], this solution satisfies the evolutionary variational inequality

    12ddtuf2HE(f)E(u)for a.e. t>0

    for any fH. Indeed, by definition, utE(u) is equivalent to saying

    12ddtuf2H=(ut,fu)HE(f)E(u)

    for any fH. The evolutionary variational inequality is not only an equivalent formulation of the gradient flow utE(u), but also apply to a gradient flow of a metric space by replacing ufH by distance between u and f; see [2] for the theory.

    To be able to actually find solutions to (3.1), we need to characterize the subdifferential of the total variation in the space D1=D1(Rn). The basic idea of the proof is a duality argument, which has been carried out in the case of L2 subdifferentials. In the case of L2 setting, the idea goes back to the unpublished note of F. Alter and a detailed proof is given in [3]. Let E be a functional on a real Hilbert space H equipped with an inner product (,)H. The main idea is to characterize the subdifferential E by the polar E0 of E:H[,] which is defined by

    E0(v):=sup{(u,v)H|uH, E(u)1}=sup{(u,v)/E(u)|uD(E), E(u)0},

    where D(E)={uH||E(u)|<}. We first recall a lemma [3, Lemma 1.7].

    Lemma 10. Let E be convex. Assume that E is positively one-homogeneous, i.e.,

    E(λu)=λE(u)

    for all λ>0, uH. Then, vE(u) if and only if E0(v)1 and (u,v)H=E(u).

    Remark 11. By general theory of convex functionals, we know that

    (E0)0=E

    if E is a non-negative, lower semicontinuous, convex, positively one-homogeneous functional [3, Proposition 1.6].

    This property is essential for the proof of

    Theorem 12. Let Ψ:D1[0,] by defined by

    Ψ(v)=inf{Z|v=ΔdivZ, ZL(Rn), divZD10}.

    Then (TV)0=Ψ.

    Remark 13. (i) By definition, Ψ is a convex, lower semi-continuous, positively one-homogeneous function, so (Ψ0)0=Ψ.

    (ii) if Ψ(v)<, the infimum is attained. Theorem 12 together with Lemma 10 implies the following characterization of the subdifferential of TV.

    Theorem 14. An element vD1 belongs to TV(u) if and only if there is ZL(Rn) with divZD10 such that

    (i) |Z|1

    (ii) v=ΔdivZ

    (iii) u,divZ=TV(u).

    Proof. By Lemma 10 and Theorem 12,

    vTV(u)  Ψ(v)1 and (v,u)D1=TV(u).

    The property Ψ(v)1 together with Remark 13(ii) implies (i), (ii) and divZD10.

    (v,u)D1=u,(Δ)1v=u,divZ.

    It is not difficult to check the converse.

    Proof of Theorem 12. The inequality TV0Ψ:

    We take vD1 with Ψ(v)<. By Remark 13(ii), there is ZL(Rn) with v=ΔdivZ with divZD10 such that Ψ(v)=Z. By Lemma 7, there is ukCc,av such that TV(uk)TV(u), uku in D1. We observe that

    (uk,v)D1=uk,(Δ)1v=uk,divZ=RnZukZTV(uk).

    Sending k, we conclude that

    (u,v)D1Zfor alluD1withTV(u)1.

    By definition of Ψ, this implies TV0Ψ.

    The inequality ΨTV0:

    By definition,

    TV(u)=sup{u,divZ|ZCc(Rn), |Z|1}=sup{u,divZZ|ZCc(Rn), Z0}.

    Since

    u,divZ=u,(Δ)1ΔdivZ=(u,ΔdivZ)D1,

    we observe that

    TV(u)=sup{(u,ΔdivZ)D1Z|ZCc(Rn), Z0}sup{(u,ΔdivZ)D1Ψ(ΔdivZ)|ZCc(Rn), Ψ(ΔdivZ)0}Ψ0(u).

    This implies that TV0(Ψ0)0=Ψ.

    Now that the subdifferential of TV in D1 is calculated, we are able to justify an explicit definition of a solution proposed in [15].

    Theorem 15. Assume that uC([0,[,D1). Then u is a solution of utD1TV(u) with u0=u(0) in the sense of Proposition 9 if and only if there exists ZL(]0,[×Rn) satisfying

    divZL2(δ,;D10(Rn))for anyδ>0

    such that for a.e. t]0,[ there holds

    ut=ΔdivZinD1(Rn),
    |Z|1Ln-a.e.

    and

    u,divZ=TV(u).

    (If TV(u0)<, δ=0 is allowed.)

    The Theorem essentially follows from Theorem 14. We only need to justify that a Cahn-Hoffman vector field Z defined separately for every time instance by Theorem 14 can be chosen to be jointly measurable, i.e., ZL(]0,[×Rn). As in the second-order case [3, section 2.4], this can be done by recalling that Bochner functions can be well approximated by piecewise constant functions. Since our situation is slightly different, let us include the argument for completeness. We begin with a lemma which is a version of [3, Lemma A.8] suited to our needs.

    Lemma 16. Let vL1loc(]0,[,X), where X is a Banach space and let N]0,[ have Lebesgue measure 0. Then for each ε>0 there exists a countable family G of disjoint closed intervals Ik=¯B(tk,rk), kN, such that tk is a Lebesgue point for v, tkN,

    L1(]0,[kNIk)=0

    and

    0vvεXε,wherevε(t)=v(tk)fortIk, kN.

    Proof. Referring e.g., to [6, p. 140], L1-a.e. point t]0,[ is a Lebesgue point for v, i.e.,

    limh0+12ht+hthvv(t)X=0.

    Let A be the set of Lebesgue points of v contained in ]0,[N and let us take

    F={¯B(t,r) | tA, r<min(ε,t), 12rt+rtrvv(t)Xεet}.

    Using Besicovitch covering theorem [9, Corollary 1 on p. 35] with U=]0,[ and (for example) dμ(t)=etdt, we obtain a candidate for the family G. We check that indeed

    0vvεX=kNIkvv(tk)XεkN2rketkε0etdt=ε.

    Proof of Theorem 15. Applying Lemma 16 to ut, for each ε>0 we obtain a partition of ]0,[ (up to a set of Lebesgue measure 0) into disjoint closed intervals Ik=¯B(tk,rk), kN such that

    ut(tk)D1TV(u(tk))for kN (3.2)

    and

    0utvεD1ε,where vε(t)=ut(tk)for tIk, kN.

    By (3.2) and Theorem 14, for kZ there exist ZkL(Rn) such that

    |Zk|1,ut(tk)=ΔdivZk,u(tk),divZk=TV(u(tk)).

    Further denoting

    uε(t)=u(tk),Zε(t)=Zkfor tIk, kN,

    we have

    |Zε|1,vε=ΔdivZε,vε,divZε=TV(uε)a.e. in [0,[.

    We immediately deduce that there exists ZL([0,[×Rn) with |Z|1 a.e. and a subsequence Zεj such that Zεj converges to Z weakly-* in ZL([0,[×Rn). Moreover, for any φCc([0,[×Rn),

    0vεj,φ=0RndivZεjφ=0RnZεjΔφ0RnZΔφ,

    at least along a subsequence. Since on the other hand

    0vε,φ0ut,φ,

    we infer that ut=ΔdivZ and in particular divZεjdivZ in L1loc(]0,[,D10). Finally, we observe that uεu in Lloc(]0,[,D1). Moreover, since tTV(u(t)) is non-increasing, we have TV(uε)TV(u) in L1loc(]0,[). Therefore, for any φCc(]0,[),

    0uεj,divZεjφ0u,divZφ,0TV(uε)φ0TV(u)φ.

    Unfortunately, in the case n2, the characteristic function 1A of a set A of positive measure is not in D1 since 1A0. We shall define a new space containing 1A as follows. We take a function ψL2(Rn) with compact support such that Rnψ=1. We introduce a vector space

    E1ψ={w+cψ|wD1(Rn), cR}.

    This space is independent of the choice of ψ. Indeed, let ψiL2(Rn) be compactly supported and ψi=1 (i=1,2). An element w+cψ1E1ψ1 can be rewritten as

    w+cψ1=w+c(ψ1ψ2)+cψ2.

    The next lemma implies q=c(ψ1ψ2)D1(Rn) since q=0. We then conclude that w+cψ1E1ψ2.

    Lemma 17. Assume that n2. A compactly supported function qL2(Rn) belongs to D1(Rn) if and only if Rnq=0.

    Proof. If qD1L1, then

    q=q,[1]=0,

    where [1] stands for the element of D10 whose representatives are 1 as well as 0.

    Now suppose that a compactly supported function qL2(Rn) satisfies Rnq=0. Given a [φ]D10, by the Poincaré inequality, we have for any R>0, independently of the representative φD1,

    φ1|BR|BRφL2(BR)CRφL2(BR) CR[φ]D10.

    In particular, φL2loc. Taking into account this and the assumption q=0, we see that the linear functional

    q,[φ]=qφ (4.1)

    on D10 is well defined. Moreover, if R>0 is large enough that suppqBR,

    |qφ|=|BRq(φ1|BR|BRφ)|qL2(BR)φ1|BR|BRφL2(BR)CRqL2[φ]D10.

    Thus, the functional defined by (4.1) is bounded, i.e., qD1.

    Since E1ψ is independent of the choice of ψ, we suppress ψ and denote this space by E1. In case n3, we will use notation E1=D1. We also denote E10=D1 if n2, E10=D10 if n3. For uE1, vE10, we denote

    u,vE={w,v if n3,w,[v]+cψv if n2, (4.2)

    where u=w+cψ, wD1, ψL2c, ψ=1. As before, we check that the value of u,vE does not depend on the choice of this decomposition.

    We recall that if n3, E10=D10 and E1=D1 come with a Hilbert space structure. We also define inner products on E10, E1 in case n2 by

    (v1,v2)E10:=([v1],[v2])D10+ψv1ψv2,
    (u1,u2)E1:=(w1,w2)D1+c1c2

    for ui=wi+ciψ, wiD1, ciR (i=1,2). This gives an orthogonal decomposition

    E1=D1R.

    We note that although the values of those products may depend on the choice of ψ, the topologies they induce on E10, E1 do not. Formula (4.2) associates to any uE1 a continuous linear functional on E10. The resulting mapping is an isometric isomorphism between E1 and the continuous dual to E10.

    We extend TV onto E1 by defining

    TV(u):=supψX1u,divψE.

    As usual, we check that TV is a convex, weakly-* (and strongly) lower semicontinuous functional. In particular, for a fixed gE1, the functional wTV(w+g) is convex and lower semicontinuous on D1. We next give a definition of a solution of utTV(u) in the space E1. It turns out the idea of evolutionary variational inequality is very convenient since it is a flow in an affine space g+D1 for some gE1.

    Definition 18. Assume that u0E1. We say that u:[0,[E1 is a solution to

    utD1TV(u) (4.3)

    in the sense of EVI (evolutionary variational inequality) with initial datum u0 if

    (i) ug is absolutely continuous on [δ,T] (for any 0<δ<T<) with values in D1 and continuous up to zero with u(0)=u0 and

    (ii) ug satisfies the evolutionary variational inequality, i.e.,

    12ddtu(t)g2D1TV(g)TV(u(t))

    holds for a.e. t>0, provided that gE1 is such that u0gD1.

    Theorem 19. For any u0E1, there exists a unique solution u of (4.3) in the sense of EVI. Moreover, if ui is the solution to (4.3) in the sense of EVI with ui(0)=ui0E1 for i=1,2, then

    u1(t)u2(t)D1u10u20D1for allt0, (4.4)

    provided that u10u20D1.

    Proof. Uniqueness follows from contractivity (4.4), which is established by a standard reasoning [2]. We give a short proof for the reader's convenience and for completeness. Let ui (i=1,2) be a solution to (4.3) in the sense of EVI with initial datum ui0 such that u10u20D1. Since ui are solutions, we also have ui(t)ui0D1 for t0 and so u10u2(t)D1, u20u1(t)D1. Thus, EVI yields

    12ddtu1u22D1=(u1t,u1u2)D1+(u2t,u2u1)D1=12ddtu1(t)g2D1|g=u2(t)+12ddtu2(t)g2D1|g=u1(t)TV(u2(t))TV(u1(t))+TV(u1(t))TV(u2(t))=0

    for a.e. t>0. We conclude that u1(t)u2(t)2 is non-increasing, in particular (4.4) holds.

    The existence is a bit more involved. For u0=w0+g0E1 with w0D1, we consider the gradient flow of the form

    wtD1TV(w+g0),w(0)=w0. (4.5)

    Applying Proposition 9, there is a unique solution w to (4.5) for w0D1. This solution satisfies the evolutionary variational inequality

    12ddtwf2D1TV(f+g0)TV(w+g0)for a.e. t>0

    for any fD1. Setting u=w+g0, g=f+g0, we end up with

    12ddtwg2D1TV(g)TV(u(t)).

    Since g can be taken arbitrary such that u0gD1, this shows that u is the solution of (4.3) in the sense of EVI; condition (i) follows easily from Proposition 9.

    It is non-trivial to characterize the subdifferential D1TV. For this purpose, we introduce a mapping I which plays a role analogous to Δ in n3.

    Lemma 20. Let n2. The mapping I:E10E1 defined by

    I(f)=(Δ)[f]+(Rnfψ)ψ

    is an isometric isomorphism.

    Proof. It is clear that I(f)E1 and I is linear. For given u=w+cψE1 with wD1, cR, there is ¯fD10 such that (Δ)¯f=w. Since a representative f of ¯f is determined up to an additive constant, there is a unique representative f such that

    Rnfψ=c.

    Thus, the mapping I is surjective. If I(f)=0, then (Δ)[f]=0 so [f]=0. Thus f is a constant. Since fψ=0, this constant must be zero, so f=0. Thus, I is injective. Recalling our definitions of inner products on E10, E1, it is easy to check that I is an isometry.

    We have a characterization of the polar of TV in E1 as in Theorem 12.

    Theorem 21. Let n2. Let Ψ be given by

    Ψ(v)=inf{Z|v=I(divZ), ZL(Rn), divZE10}

    for vE1. Then (TV)0=Ψ.

    Admitting this fact, we are able to give a characterization of the subdifferential.

    Theorem 22. Let n2. An element vE1 belongs to E1TV(u) if and only if there is ZL(Rn) with divZE10 such that

    (i) |Z|1

    (ii) v=I(divZ)

    (iii) u,divZE=TV(u).

    Proof of Theorem 22. The proof parallels that of Theorem 14. By Lemma 10 and Theorem 21

    vTV(u)  Ψ(v)1 and (u,v)E1=TV(u).

    The properties (i), (ii) together with divZE10 are equivalent to Ψ(v)1. Since

    (u,v)E1=(w,(Δ)[v])D1+cRnvψ=w,[v]+cRnvψ, (4.6)

    the Euler equation (u,v)E1=TV(u) is equivalent to (iii).

    Proof of Theorem 21. The proof parallels that of Theorem 12. We first prove that

    (u,v)E1Zfor alluE1withTV(u)1

    for v=I(divZ). This implies TV0Ψ. The estimate (u,v)E1Z formally follows from the identity (4.6). Indeed, by (4.6), we see

    (u,v)E1=w,[divZ]cRnψdivZ.

    If u is in Cc(Rn), then, by this formula, we obtain

    (u,v)E1=RnudivZ=RnuZZTV(u).

    By approximation, as in the proof of Theorem 12, we conclude the desired estimate.

    The other inequality ΨTV0 follows from TVΨ0. The proof of TVΨ0 is parallel to that of Theorem 12 by replacing ΔdivZ by I(divZ) and the D1 inner product by the E1 inner product, respectively, if one notes the identity (4.6). Since I is an isometry, Ψ is lower semicontinuous, and we conclude that Ψ=TV0 by Remark 11.

    We have to be careful, since the E1 gradient flow

    utE1TV(u)

    does not correspond to the total variation flow ut=(Δ)div(u/|u|). By Theorem 22(iii), Z=u/|u| if u0. Thus the E1 gradient flow is formally of the form

    ut=(Δ)div(u/|u|)+ψRnψdiv(u/|u|).

    To recover the original total variation flow, we consider "partial" subdifferential in the direction of D1. Let P be the orthogonal projection from E1 to D1. Then, by definition,

    D1TV(w+cψ)=PE1TV(u).

    The equation

    wtD1TV(w+cψ)

    is now formally of the form

    ut=(Δ)div(u/|u|)

    since cψ is time-independent. Here is a precise statement.

    Theorem 23. Let n2. Consider the functional F:wTV(w+cψ) on D1 for a fixed cR and ψ. Then, D1F(w)=PE1TV(u) for u=w+cψ. In particular, an element vD1 belongs to D1F(w) if and only if there is ZL(Rn) with divZE10 such that

    (i) |Z|1,

    (ii) v=ΔdivZ,

    (iii) u,divZE=TV(u).

    (In case n2, by ΔdivZ we understand Δ[divZ].) This characterization is important to calculate the solution of ut=(Δ)div(u/|u|) for n2 explicitly. In fact, we can recover the characterization of a solution in the sense of EVI analogous to the one in Theorem 15, amounting to Theorem 2.

    Proof of Theorem 2. This is almost immediate. However, like in the case of Theorem 15, we need to justify that the vector field Z can be chosen to be jointly measurable with respect to (t,x). We proceed as in the proof of Theorem 15. The difference is, now we need to pass to the limit with

    uε,divZεE=TV(uε),

    but convergence of vε in L1loc(]0,[,D1) gives only [divZε][divZ]L1loc(]0,[,D10). To deal with this problem, let us choose ψC1c(Rn) and let cR be such that u(t)cψD1 for t>0. Then we also have uε(t)cψD1 for t>0, ε>0 and

    uε,divZεE=uεcψ,[divZε]+cRnψdivZε=uεcψ,[divZε]cRnψZε.

    Testing with φCc(]0,[) and using weak-* convergence of Zε in L(]0,[×Rn) we get

    0uε,divZεEφ=0uεcψ,[divZε]φc0RnψZεφ0ucψ,[divZ]φc0RnψZφ=0u,divZEφ,

    at which point we conclude as in the proof of Theorem 15.

    We are interested in sets where the speed of solution ut is spatially constant. The speed is given as minus the minimal section of the subdifferential, i.e.,

    0D1TV(u):=argmin{vD1|vD1TV(u)}.

    Since D1TV(u) is closed and convex, 0D1TV(u) is uniquely determined if D1TV(u). Since we have characterized the subdifferential, we end up with

    0D1TV(u)=argmin{vD1|v=ΔdivZ, ZL(Rn,Rn), |Z|1,divZE10(Rn), u,divZE=TV(u)}.

    Although the minimizer v is unique, the corresponding Z may not be unique. Let U be a smooth open set in Rn. We consider a smooth function u such that

    ¯U={xRn|u(x)=0}

    and D1TV(u). Such a closed set is often called a facet. Assume further that u0 outside ¯U. Let Z be a vector field satisfying v=ΔdivZ for vTV(u). It is easy to see that outside the facet ¯U,

    Z(x)=u(x)/|u(x)|

    by u,divZE=TV(u). Since vD1=divZD10, we see that

    0D1TV(u)=argmin{divZD10||Z|1 in U, Z=u/|u| in ¯Uc, divZE10}.

    Since divZ is locally integrable, the normal trace is well-defined from inside as an element of L(U) [3] and it must agree with that from outside, i.e.,

    νZ=νu/|u|=νχν=χ, (5.1)

    where ν(x) is the exterior unit normal of U and

    χ(x)={1if u>0 outside ¯U near xU,1otherwise.

    Since divZ is in E10, its trace from inside and outside must agree, i.e.,

    divZ=div(u/|u|)=χdivν=:χκon U.

    Note that κ(x) is the sum of all principal curvatures, equal to n1 times the (inward) mean curvature, of U at x.

    Let Z0 be a minimizer corresponding to v=0D1TV(u). Since the value Z0 outside ¯U is always the same, we consider its restriction on U and still denote by Z0. Then, Z0 is a minimizer of

    {U|divZ|2||Z|1 in U, νZ=χ on U, divZ=χκ on U}. (5.2)

    Although divZ0D10(Rn) so that divZ0L2(Rn,Rn), the quantity divZ0 may jump across U. Thus ΔdivZ0 may contain singular part which is a driving force to move the facet boundary "horizontally" during its evolution under the fourth-order total variation equation as observed in the previous section and earlier in [12]. In the second-order problem, the speed does not contain any singular part so the jump discontinuity does not move.

    We are interested in a situation where ΔdivZ0 is constant over U. In the spirit of [25], we call any continuous function χ:U{1,1} a signature for U.

    Definition 24. Let U be a smooth open set in Rn with signature χ. We say that U is (D1-)calibrable (with signature χ) if there exists Z0 satisfying the constraint

    |Z0|1onU (5.3)

    with boundary conditions

    νZ0=χ,divZ0=χκ onU, (5.4)

    with the property that

    ΔdivZ0isconstantoverU. (5.5)

    We call any such Z0 a (D1-)calibration for U (with signature χ).

    From the definition of calibration, we easily deduce

    Proposition 25. Let U be a smooth bounded domain in Rn. Assume that Z0 is a calibration for U with signature χ. Then, w0=divZ0 is a solution to the Saint-Venant problem

    {Δw=λinU(5.6)w=χKonU(5.7)

    with the constraint

    UwdLn=UχdHn1, (5.8)

    where λ is some constant.

    Appealing to this relation between calibrability and the Saint-Venant problem, we can prove the following

    Theorem 26. Let U be a smooth bounded domain in Rn. Suppose that Z is a calibration for U with signature χ. Then Z is a minimizer of (5.2).

    Proof. We first note that w=divZ must satisfy (5.8). We consider the minimization problem for

    e(w)=U|w|2

    under the Dirichlet condition (5.7) and the constraint (5.8). Since the problem is strictly convex, there is a unique minimizer ¯w in D10(U). By Lagrange's multiplier method, ¯w must satisfy (5.6) because of the constraint (5.8). (Actually, a weak solution ¯w of (5.6) is a smooth solution of (5.6), (5.7) by the standard regularity theory of linear elliptic partial differential equations [19, Chapter 6].) As we see in Lemma 27 below, the constant λ is uniquely determined by (5.6) and (5.8). For ¯w, there always exists ZC(¯U) such that

    divZ=¯winU,νZ=χonU. (5.9)

    Indeed, let p be a solution of the Neumann problem

    Δp=¯winU,νp=χonU.

    Such a solution p always exists since ¯w satisfies the compatibility condition (5.8) and it is smooth up to ¯U; see e.g., [11,19]. If we set Z=p, then Z satisfies the desired property (5.9). Thus, the minimum e(¯w) of the Dirichlet energy under the constraint (5.8) agrees with

    min{U|divZ|2|νZ=χ, divZ=χκ on U}.

    Since w=¯w and |Z|1, this shows that Z is a minimizer of (5.2).

    Lemma 27. Let U be a smooth bounded domain in Rn. Let w solve

    Δw=λinUw=fonU

    for λR, fC(U). This solution can be written as

    w=λwsv+hf

    where wsv solves the Saint-Venant problem

    Δwsv=1inU (5.10)
    wsv=0onU (5.11)

    and hf is the harmonic extension of f to U. In particular, if Uw=c is given then λ is uniquely determined by

    λUwsv+Uhf=c, (5.12)

    since wsv>0 in U.

    Proof. The decomposition w=λwsv+hf is rather clear. The property wsv>0 in U follows from the maximum principle [19].

    From Lemma 27 we also deduce the following formula for vertical speed of calibrable facets in terms of the solution to the Saint-Venant problem.

    Proposition 28. Let U be a smooth bounded domain in Rn and λR. If Z is a calibration for U with signature χ satisfying

    ΔdivZ=λ,

    then

    λUwsv=Uχκνwsv+Uχ,

    where wsv is the solution to the Saint-Venant problem (5.10).

    Proof. We recall (5.12) and calculate

    Uhf=UΔwsvhf=Uνwsvχdivν+Uwsvhf,
    Uwsvhf=UwsvνhfUwsvΔhf=0,
    c=UdivZ=Uχ.

    We now compare the definition of calibrability for the second-order problem.

    Definition 29. Let U be a smooth open set in Rn with signature χ. We say that ¯U is (L2-)calibrable if there is Z0 satisfying the constraint |Z0|1 in U and the boundary condition νZ0=χ with the property that divZ0 is a constant over U.

    This definition is slightly weaker than the calibrability used in [26], where a(t)1U is a solution of the total variation flow in Rn with some function a(t) of t; see also [3]. This requires that 0L2TV(u) is constant not only on U but also Uc. Our definition follows from that of [5].

    Definition 30. We say that a Lebesgue measurable subset U (defined up to a set of measure zero) of Rn is a generalized annulus if U is non-empty, open, connected and rotationally symmetric, i.e., invariant under the linear action of SO(n) on Rn.

    It is easy to see that any generalized annulus is a ball, an annulus, the complement of a ball or the whole space Rn. In other words, any generalized annulus is of form

    AR1R0={xRn:R0<|x|<R1} with 0<R0<R1orAR0=BR with R>0.

    In this section we will settle the question which generalized annuli are calibrable.

    Lemma 31. Let U be a generalized annulus. Suppose that U is calibrable with signature χ. Then there exists a calibration ¯Z for (U,χ) of form ¯Z(x)=z(|x|)x|x|.

    Proof. Let Z be any calibration for (U,χ). Let μn be the Haar measure on SO(n). We define ¯Z as the average

    ¯Z(x)=LZ(L1x)dμn(L).

    It is an exercise in vector calculus to check that ¯Z satisfies boundary conditions (5.4) and that Δdiv¯Z is a constant (equal to ΔdivZ) on U. By convexity, |¯Z|1. Thus ¯Z is a calibration for (U,χ). By definition, it is invariant under rotations, i.e.,

    L¯Z(L1x)=¯Z(x)

    for LSO(n), xRn. In the case n=1 this already shows that ¯Z is in the desired form. In higher dimensions, we consider the orthogonal decomposition

    ¯Z(x)=¯ZZ(x)+¯ZZT(x):=x|x|x|x|¯Z(x)+(Ix|x|x|x|)¯Z(x).

    Both ¯ZZ and ¯ZZT are invariant under rotations. In particular, for any given R>0, the restriction of ¯ZZT to Sn1R is an invariant tangent vector field on Sn1R. Note that any such vector field is smooth. If n=3, it follows by the hedgehog uncombability theorem [10, Proposition 7.15] that ¯ZZT0. If n>3, any vector field invariant on Sn1 is in particular invariant on a sphere S2 containing any given point in Sn1, so the same conclusion follows. Thus, we have

    ¯Z(x)=¯ZZ(x)=x|x|x|x|¯Z(x)=x|x|¯Z(x)x|x|=:¯z(x)x|x|.

    By rotational invariance, we have ¯z(x)=z(|x|), which concludes the proof.

    We are left with the case n=2 in which there exists a one-dimensional space of invariant tangent fields on Sn1=S1 spanned by eT(x):=(x2,x1). Thus, we have

    ¯ZZT(x)=zT(|x|)eT(x).

    We calculate

    div¯ZZT(x)=zT(|x|)diveT(x)+(zT)(|x|)x|x|eT(x)=0.

    Thus, we can disregard ¯ZZT and choose ¯ZZ as our calibration, since it satisfies conditions (5.3)–(5.5) (recall that ¯ZZT and ¯ZZ are orthogonal). As before, we see that

    ¯ZZ(x)=z(|x|)x|x|.

    Let U be a generalized annulus. By Lemma 31, if U is calibrable, then there exists a calibration Z for U of form Z=z(|x|)x|x|. It follows from (5.10) that z needs to satisfy the ODE

    r1n(rn1(r1n(rn1z)))=λ. (6.1)

    The general solution to this ODE is

    z(r)=c0r3+c1r3n+c2r+c3r1n (6.2)

    where c0=λ2n(n+2) if n2 and

    z(r)=c0r3+c1rlogr+c2r+c3r1. (6.3)

    where c0=λ16 if n=2. We will now try to find a calibration for U by solving a suitable boundary value problem for (6.1).

    Let U=BR(0). To focus attention, we choose χ=1 on U. In this case, boundary conditions (5.4) lead to

    z(R)=1,z(R)=0. (6.4)

    If n2, in order to satisfy the requirements |Z|1 and divZ, we need to restrict to c1=c3=0 in (6.2). We make the same choice also in case n=1, as it leads to the right result. Then, applying (6.4) in (6.2) or (6.3), we obtain a system of two affine equations for two unknowns λ,c2. We solve it obtaining

    z(r)=12(rR)332rR, (6.5)
    λ=n(n+2)R3. (6.6)

    We check that z satisfies |z|1 on [0,R], so Z is a calibration for BR. Thus, all balls are calibrable in any dimension.

    Let U=RnBR. For consistency with the previous case, we choose χ=1 on U. In this case, boundary conditions (5.4) also lead to

    z(R)=1,z(R)=0. (6.7)

    Let us first assume that n3. In order to satisfy the requirement |Z|1, we need to restrict to λ=c2=0 in (6.2). Again, applying (6.4) in (6.2) leads to a system of two affine equations for two unknowns c1,c3. We solve it obtaining

    z(r)=n12(rR)3n+n32(rR)1n. (6.8)

    Again, we easily check that z satisfies |z|1 on [0,R], so Z is a calibration for BR.

    In the omitted cases n=1,2, requirement |Z|1 implies λ=c1=c2=0 in (6.2). If n=1, there exists z of such form satisfying (6.7): z(r)1, consistently with (6.8). On the other hand, if n=2, applying (6.7) to (6.2) with λ=c1=c2=0 leads to a contradiction.

    Summing up, all complements of balls are calibrable if n2. On the other hand, if n=2 all complements of balls turn out not to be calibrable.

    Let now U=AR1R0=BR1BR0, 0<R0<R1. In this case U has two connected components, so there exist two distinct choices of signature: constant and non-constant. Let us first consider the former. To focus attention, we choose χ1. Then, boundary conditions (5.4) take form

    z(R0)=1,z(R1)=1,z(R0)=z(R1)=0. (6.9)

    Applying (6.9) to (6.2) or (6.3) leads to a system of four affine equations with four unknowns. In the case n2, the solution is

    c0=(R1R0)Rn1Rn0((n1)(n2)(R1+R0)22R0R1)+2R31R2n02R30R2n1R1R0(Rn1Rn0(n2(R21R20)2+8R21R20)4R21R2n+204R20R2n+21),c1=(R1+R0)(Rn1(2(n1)R21+(n+2)R1R0(n+2)R20)+Rn0((n+2)R21(n+2)R1R02(n1)R20))2(n24)R31R30+4R3n1Rn+30+4Rn+31R3n0n2R51R0n2R1R50,c2=(R1R0)Rn1Rn0(6R21R20+(n1)nR31R0+(n1)nR41+(n1)nR1R30+(n1)nR40)6R31R2n+20+6R30R2n+21R1R0(Rn1Rn0(n2(R21R20)2+8R21R20)4R21R2n+204R20R2n+21),c3=(R1+R0)(R20Rn1(2(n3)R21nR1R0+nR20)R21Rn0(nR21R0(nR1+2(n3)R0)))2(n24)R31R30+4R3n1Rn+30+4Rn+31R3n0n2R51R0n2R1R50. (6.10)

    This can be rewritten in a form emphasizing homogeneity:

    c0=(Q1)Qn((n1)(n2)(Q+1)22Q)+2Q32Q2nQn2(n2(Q21)2+8Q2)44Q2nR31,c1=(Q+1)(Qn(2(n1)Q2+(n+2)Q(n+2))+((n+2)Q2(n+2)Q2(n1)))2(n24)Qn+4+4Q2nn2Qn+2n2Qn2Rn31,c2=(Q1)Qn(6Q2+(n1)nQ3+(n1)nQ4+(n1)nQ+(n1)n)6Q3+6Q2n+2Qn(n2(Q21)2+8Q2)4Q24Q2n+2R11,c3=(Q+1)(Qn(2(n3)Q2nQ+n)Q2(nQ2(nQ+2(n3))))2(n24)Qn+2+4Q2+4Q2n+2n2Qn+4n2QnRn11, (6.11)

    where we denoted Q=R1/R0. We can further simplify it to

    c0=2Q3(Q2n31)+(Q1)Qn((n1)(n2)(Q+1)22Q)4(Qn1)2n2(Q21)2Qn2R31,c1=(Q+1)(2(n1)(Qn+21)+(n+2)Q(Q1)(Qn1+1))4(Qn1)2n2(Q21)2Qn2Rn31,c2=6Q(Q2n11)+(Q1)Qn2(6Q2+n(n1)(1+Q)(1+Q3))4(Qn1)2n2(Q21)2Qn2R11,c3=(Q+1)(2(n3)(Qn1)+nQ(Q1)(Qn3+1))4(Qn1)2n2(Q21)2Qn2Rn11. (6.12)

    We need to check whether condition |Z|1 is satisfied. We calculate

    z(r)=6c0r+(n3)(n2)c1r1n+n(n1)c3rn1=rn1(6c0rn+2+(n3)(n2)c1r2+n(n1)c3)=:rn1w(r). (6.13)

    Using the form (6.12), we can check that c0>0, c1>0 for all Q>1. Therefore, w has at most one zero on the half-line r>0. Consequently, z has at most one zero, so z has at most one inflection point. Taking into account (6.9), z cannot have a local extremum on ]R0,R1[. Thus, |z|1 on ]R0,R1[ and Z is a valid calibration.

    In the case n=2, the solution is

    c0=R21R20+2R1R0log(R1/R0)4R1(R1R0)R0(R21+R20+(R21+R20)log(R1/R0))c1=R313R21R03R1R20R302R1R0(R21+R20+(R21+R20)log(R1/R0))c2=3R41+3R40+2(R41R31R0+R21R203R1R30)log(R1)+2(3R31R0R21R20+R1R30R40)log(R0)4R1(R1R0)R0(R21+R20+(R21+R20)log(R1/R0))c3=R1(3R21R0+3R30+2(R21R0R1R20+R30)log(R1/R0))4(R1R0)(R21+R20+(R21+R20)log(R1/R0))

    which can be rewritten (again, denoting Q=R1/R0) as

    c0=Q2(Q21+2QlogQ)4(Q1)(Q2+1+(Q2+1)logQ)R31,c1=(Q+1)32(Q2+1+(Q2+1)logQ)R11,c2=3(Q41)2(3Q3Q2+Q1)logQ4(Q1)(Q2+1+(Q2+1)logQ)R11+(Q+1)32(Q2+1+(Q2+1)logQ)R11logR1,c3=3Q2+3+2(Q2Q+1)logQ4(Q1)(Q2+1+(Q2+1)logQ)R1. (6.14)

    As before, we calculate the second derivative of z:

    z(r)=6c0r+c1r1+2c3r3=r3(6c0r4+c1r2+2c3)=:r3w(r).

    The polynomial w has at most 2 positive roots, and so does z. By (6.9), at least one of them belongs to ]R0,R1[. Furthermore, since c0>0 for Q>1, z(r) is positive for large values of r. Taking into account these observations, we deduce that |z|1 on [R0,R1] if and only if z(R0)0 (compare Figure 1). This inequality is equivalent to

    m(Q):=logQ(Q21)(2Q1)Q(Q22Q+3)0.
    Figure 1.  Plots of z for an annulus with constant signature for two different values of Q in case n=2.

    We compute

    m(1)=0,limQ+m(Q)=+,m(Q)=(Q3)(Q1)(Q+1)3Q2(Q22Q+3)2. (6.15)

    We observe that m has exactly one zero Q on ]1,+[, and m(Q)0 if and only if QQ. Therefore, Z is a valid calibration for AR1R0 with signature 1 if and only if R1/R0Q. By (6.15) it is evident that Q>3. Numerical computation using Wolfram Mathematica shows that Q9.7. Thus, AR1R0 with constant signature is calibrable if and only if R1/R0Q. This concludes the proof of Theorem 3.

    Now, let us consider non-constant signature. We assume that χ=1 on BR0 and χ=1 on BR1. This choice leads to

    z(R0)=1,z(R1)=1,z(R0)=z(R1)=0. (6.16)

    If n2, the solution to the resulting affine system is

    c0=2R31R2n02R30R2n1+Rn0Rn1(R0+R1)((n2)(n1)R202((n3)n+1)R0R1+(n2)(n1)R21)R0R1(Rn0Rn1(n2(R20R21)2+8R20R21)4R21R2n+204R20R2n+21)c1=3nR1Rn+20+2(n1)Rn+30+(n+2)R31Rn0+(n+2)R30Rn13nR0Rn+21+2(n1)Rn+314R3n0R3n1(Rn0Rn1)2n2R0R1(R20R21)2c2=6R31R2n+20+6R30R2n+21Rn0Rn1(R0+R1)((n1)nR40(n1)nR30R1(n1)nR0R31+(n1)nR41+6R20R21)R0R1(Rn0Rn1(n2(R20R21)2+8R20R21)4R21R2n+204R20R2n+21)c3=(R0R1)(R20Rn1(n(R0R1)(R0+2R1)+6R21)R21Rn0(2(n3)R20+nR0R1+nR21))4R3n0R3n1(Rn0Rn1)2n2R0R1(R20R21)2 (6.17)

    which we rewrite as

    c0=2Q32Q2n+Qn(1+Q)((n2)(n1)2((n3)n+1)Q+(n2)(n1)Q2)Qn2(n2(1Q2)2+8Q2)44Q2nR31c1=3nQ+2(n1)+(n+2)Q3+(n+2)Qn3nQn+2+2(n1)Qn+34(1Qn)2n2Qn2(1Q2)2Rn31c2=6Q3+6Q2n+2Qn(1+Q)((n1)n(n1)nQ(n1)nQ3+(n1)nQ4+6Q2)Qn(n2(1Q2)2+8Q2)4Q24Q2n+2R11c3=(1Q)(Qn(n(1Q)(1+2Q)+6Q2)Q2(2(n3)+nQ+nQ2))4Q2(1Qn)2n2Qn(1Q2)2Rn11 (6.18)

    We note that in case n=1 the solution reduces to

    c0=0,c1=0,c2=0,c3=1,

    while in case n=3 it reduces to

    c0=0,c1=1,c2=0,c3=0.

    In both of these cases z is constant and we have λ=c0=0. On the other hand, if n4, we can check that c0>0 for Q>1. Recalling (6.13), we observe that z has at most two zeros on the positive half-line and z(r)>0 for large values of r. On the other hand, by (6.16), if z has N local extrema on ]R0,R1[, it needs to have at least N+1 inflection points. We deduce from these conditions that z has exactly one local maximum and no local minima, and therefore z1 on ]R0,R1[ (compare Figure 2). It remains to check whether z1 on ]R0,R1[. Let now

    f(r)=r1n(rn1z(r))=f(r)+(n1)f(r)r.
    Figure 2.  Plots of z for an annulus with non-constant signature for two values of n, with R0=1, R1=10.

    Then, by (6.1), (6.16), f is a solution to the second-order elliptic problem

    Af=λ,f(R0)=n1R0,f(R1)=n1R1,

    where

    Af=r1n(rn1f(r))=f(r)(n1)f(r)r.

    Since c00, we have λ<0 for Q>1. By the classical weak maximum principle [19, Theorem 3.1.],

    max[R0,R1]f=max{R0,R1}f=n1R1.

    Now, if z has a local maximum at r0, then z(r0)=0, so f(r0)=z(r0)r0. Consequently,

    z(r0)r0n1R1<0,

    so z<0 on [R0,R1]. Thus, if n2, all annuli with non-constant signature are calibrable.

    We move to the case n=2. Now, the solution to the affine system for coefficients of z is

    c0=R202R0R1log(R1/R0)+R214R0R1(R0+R1)(R20+R21(R20+R21)log(R1/R0)),c1=(R0R1)32R0R1(R20+R21(R20+R21)log(R1/R0)),c2=(R0R1)(R0+R1)(R20+4R0R1+R21)2R0(R30+R20R1+R0R21+3R31)log(R0)+2R1(3R30+R20R1+R0R21+R31)log(R1)4R0R1(R0+R1)(R20+R21(R20+R21)log(R1/R0)),c3=R0R1(2(R20+R0R1+R21)log(R1/R0))+3(R0R1)(R0+R1))4(R0+R1)(R20+R21(R20+R21)log(R1/R0))

    or equivalently

    c0=Q2(12QlogQ+Q2)4(1+Q)(1+Q2(1+Q2)logQ)R31,c1=(1Q)32(1+Q2(1+Q2)logQ)R11,c2=(1Q)(1+Q)(1+4Q+Q2)+2(1+Q+Q2+3Q3)logQ4(1+Q)(1+Q2(1+Q2)logQ)R11+(Q1)32(1+Q2(1+Q2)logQ)R11logR1,c3=2(1+Q+Q2)logQ+3(1Q)(1+Q)4(1+Q)(1+Q2(1+Q2)logQ)R1.

    We can check that in this case c_0 < 0 for Q > 1 . By the same argument as in the previous case, we show that z < -1 in ]R_0, R_1[ (compare Figure 2), so it does not define a valid calibration. Thus, in the case n = 2 all annuli with non-constant signature are not calibrable.

    In this section, our goal is to provide explicit description of solutions to (1.1) emanating from the characteristic function of a ball

    \begin{equation} u_0 = a_0 \mathbf 1_{B_{R_0}}. \end{equation} (7.1)

    In the case of second-order total variation flow, the solutions with initial datum (7.1) are known to be of form

    u(t) = a(t)\mathbf 1_{B_{R_0}}

    with finite extinction time, i.e., there exists t_* > 0 such that a(t) = 0 for t \geq t_* . In the fourth order case, based on the treatment of case n = 1 in [12], we would expect the solutions to have the form

    \begin{equation} u(t) = a(t)\mathbf 1_{B_{R(t)}}, \end{equation} (7.2)

    at least until an extinction time beyond which u(t, \cdot) \equiv 0 . This intuition turns out to be correct in every dimension except n = 2 .

    Let first n \geq 3 . As we have checked in Section 6, in this case both balls and complements of balls are calibrable. Thus, as long as the solution is of form (7.2) in time instance t \geq 0 , we expect a valid Cahn-Hoffman vector field Z to be given by

    \begin{equation} Z(x) = \left\{\begin{array}{l} Z_{in}(x) \text{ if } |x| \in [0, R[ \\ Z_{out}(x) \text{ if } |x| > R, \end{array} \right. \end{equation} (7.3)

    where Z_{in} is the calibration w constructed for a ball B_R and Z_{out} is the calibration we constructed for the complement of that ball, recall:

    \begin{equation} \lambda = - \frac{n(n+2)}{R^3}, \end{equation} (7.4)
    \begin{equation} Z_{in}(x) = \frac{1}{2} \left(\frac{|x|}{R}\right)^3 \frac{x}{|x|} -\frac{3}{2} \frac{x}{R}, \qquad Z_{out}(x) = - \frac{n-1}{2} \left(\frac{|x|}{R}\right)^{3-n}\frac{x}{|x|} + \frac{n-3}{2} \left(\frac{|x|}{R}\right)^{1-n}\frac{x}{|x|}. \end{equation} (7.5)

    We further calculate:

    \begin{equation*} \operatorname{div} Z_{in}(x) = \frac{n+2}{2} \frac{|x|^2}{R^3} - \frac{3n}{2} \frac{1}{R}, \qquad \operatorname{div} Z_{out}(x) = - (n-1) \frac{|x|^{2-n}}{R^{3-n}}, \end{equation*}
    \begin{equation*} \nabla \operatorname{div} Z_{in}(x) = (n+2) \frac{x}{R^3}, \qquad \nabla \operatorname{div} Z_{out}(x) = (n-1)(n-2) \frac{|x|^{-n} x}{R^{3-n}}. \end{equation*}

    It is straightforward to check that \operatorname{div} Z \in D^1(\mathbb{R}^n) \cap L^{2^*}(\mathbb{R}^n) = D_0^1(\mathbb{R}^n) . Next, we deduce

    \begin{eqnarray} u_t = -\Delta \operatorname{div} Z \\ = - \Delta \operatorname{div} Z_{in}\, \mathcal{L}^n \, ∟\, _{B_R} - \Delta \operatorname{div} Z_{out}\, \mathcal{L}^n \, ∟\, _{\mathbb{R}^n\setminus B_R} + \frac{x}{|x|} \cdot(\nabla \operatorname{div} Z_{in} - \nabla \operatorname{div} Z_{out})\, \mathcal{H}^{n-1} \, ∟\, _{\partial B_R} \\ = - \frac{n(n+2)}{R^3}\, \mathcal{L}^n \, ∟\, _{B_R} - \frac{n(n-4)}{R^2}\, \mathcal{H}^{n-1} \, ∟\, _{\partial B_R}. \end{eqnarray} (7.6)

    Then, using the identity \frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbb{R}^n} u = \int_{\mathbb{R}^n} u_t , we obtain (recall notation (7.2))

    \begin{equation*} a(t) \mathcal{H}^{n-1}\left(\partial B_{R(t)}\right) \frac{\mathrm{d} R}{\mathrm{d} t} = a(t) \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{L}^n\left(B_{R(t)}\right) = - \frac{n(n-4)}{R^2} \mathcal{H}^{n-1}\left(\partial B_{R(t)}\right). \end{equation*}

    Summing up, evolution of initial datum (7.1) is given by (7.2) with a , R satisfying

    \begin{equation} \frac{\mathrm{d} a}{\mathrm{d} t} = - \frac{n(n+2)}{R^3}, \qquad \frac{\mathrm{d} R}{\mathrm{d} t} = - \frac{n(n-4)}{R^2 a}. \end{equation} (7.7)

    This system can be explicitly solved by noticing that

    \begin{equation*} \frac{\mathrm{d}}{\mathrm{d} t} (a R^3) = -n(n+2) - 3n(n-4) = -n (4n - 10) \end{equation*}

    and therefore

    \begin{equation*} a R^3 = a_0 R_0^3 - n (4n - 10) t \end{equation*}

    along trajectories. The solution is

    \begin{equation} a(t) = a_0 \left(1 - \frac{n(4n-10)}{a_0 R_0^3} t\right)^{\frac{n+2}{4n-10}}, \qquad R(t) = R_0 \left(1 - \frac{n(4n-10)}{a_0 R_0^3} t\right)^{\frac{n-4}{4n-10}}. \end{equation} (7.8)

    We note that the solution satisfies

    \left(\frac{a}{a_0}\right)^{n-4} = \left(\frac{R}{R_0}\right)^{n+2}

    along trajectories. (This "first integral" could also have been used to solve the system (7.7).) Let us point out a few observations concerning the solutions (compare Figure 3):

    Figure 3.  Plots of the solution u(t, x) emanating from the characteristic function of the unit ball as a function of |x| for chosen values of t .

    ● the extinction time is equal to t_* = \frac{a_0 R_0^3}{n(4n-10)} ,

    ● if n = 3 , R(t) is increasing and R(t) \to + \infty as t \to t_*^- ,

    ● if n = 4 , R(t) = R_0 is constant,

    ● in higher dimensions, R(t) is decreasing and R(t) \to 0 as t \to t_*^- .

    In the case n = 2 we were able to exhibit a calibration for the ball B_R , but not for its complement. Another possible ansatz on the Cahn-Hoffman vector field of form (7.3) is one where Z_{in} is the calibration we constructed for B_R and Z_{out} is the choice considered in [15]:

    \begin{equation} Z_{in}(x) = \frac{1}{2} \left(\frac{|x|}{R}\right)^3 \frac{x}{|x|} -\frac{3}{2} \frac{x}{R}, \qquad Z_{out}(x) = - \frac{x}{|x|}. \end{equation} (7.9)

    We calculate

    \begin{equation} \operatorname{div} Z_{out} = - \frac{(n-1)}{|x|}, \qquad \nabla \operatorname{div} Z_{out} = \frac{(n-1)x}{|x|^3}, \end{equation} (7.10)

    hence

    \begin{equation} u_t(t, x) = - \frac{(n-1)(n-3)}{|x|^3} \quad \text{in } \mathcal{D}'\left(\mathbb{R}^n \setminus \overline{B}_{R(t)}\right). \end{equation} (7.11)

    If n \geq 4 , this would lead to u(t) being radially strictly increasing for positive t and large values of |x| , which would be at odds with our choice of Z_{out} . In fact, if n \geq 4 , \operatorname{div} Z \not \in D_0^1(\mathbb{R}^n) for any Z of this form. However, in smaller dimensions this ansatz remains a viable option. If n = 3 , it leads to the same solution as before. On the other hand, if n = 2 , we obtain a solution which is not of form (7.2). Instead, we are led to assume

    \begin{equation} u(t, x) = a(t) \mathbf 1_{B_{R(t)}} + \frac{t}{|x|^3}\mathbf 1_{\mathbb{R}^2\setminus B_{R(t)}}. \end{equation} (7.12)

    We have:

    \begin{equation*} \operatorname{div} Z_{in}(x) = 2 \frac{|x|^2}{R^3} - 3 \frac{1}{R}, \qquad \operatorname{div} Z_{out}(x) = - \frac{1}{|x|}, \end{equation*}
    \begin{equation*} \nabla \operatorname{div} Z_{in}(x) = 4 \frac{x}{R^3}, \qquad \nabla \operatorname{div} Z_{out}(x) = \frac{x}{|x|^3}, \end{equation*}
    \begin{equation*} u_t = -\Delta \operatorname{div} Z = - \frac{8}{R^3}\, \mathcal{L}^2 \, ∟\, _{B_R} + \frac{1}{|x|^3}\, \mathcal{L}^2 \, ∟\, _{\mathbb{R}^2 \setminus B_R} + \frac{3}{R^2}\, \mathcal{H}^1 \, ∟\, _{\partial B_R} \end{equation*}

    and, recalling (7.12),

    \begin{equation*} \left(a(t) - \frac{t}{R(t)^3}\right) \mathcal{H}^1(\partial B_{R(t)}) \frac{\mathrm{d} R}{\mathrm{d} t} = \left(a(t) - \frac{t}{R(t)^3}\right) \frac{\mathrm{d} }{\mathrm{d} t} \mathcal{L}^2(B_{R(t)}) = \frac{3}{R(t)^2} \mathcal{H}^1(\partial B_{R(t)}). \end{equation*}

    Thus, we arrive at ODE system

    \begin{equation} \frac{\mathrm{d} a}{\mathrm{d} t} = -\frac{8}{R^3}, \qquad \frac{\mathrm{d} R}{\mathrm{d} t} = \frac{3R}{aR^3 - t}. \end{equation} (7.13)

    This system is not autonomous, but it can be integrated by noticing that along trajectories

    \begin{equation*} \frac{\mathrm{d}}{\mathrm{d} t} \left(aR^3 - t\right) = \frac{9aR^3}{aR^3 - t} - 8 - 1 = \frac{9t}{aR^3 - t} \end{equation*}

    and so

    \begin{equation*} aR^3 = \sqrt{a_0^2 R_0^6 + 9 t^2} + t. \end{equation*}

    This implies, first of all, that

    \begin{equation*} a(t) > \frac{t}{R(t)^3} \end{equation*}

    for all t > 0 and the form of solution (7.12) is preserved as long as the solution does not vanish. Furthermore, we can rewrite the system (7.13) in decoupled form

    \begin{equation} \frac{\mathrm{d}}{\mathrm{d} t} \log a = \frac{- 8}{\sqrt{a_0^2 R_0^6 + 9 t^2} + t}, \qquad \frac{\mathrm{d}}{\mathrm{d} t} \log R^3 = \frac{9}{\sqrt{a_0^2 R_0^6 + 9 t^2}}. \end{equation} (7.14)

    These equations can be explicitly integrated:

    \begin{equation*} a(t) = a_0 \frac{a_0^4 R_0^{12} + 8 a_0^2 R_0^{12}t^2}{\left(\sqrt{a_0^2 R_0^6 + 9 t^2} + 3 t \right)^2 \left(a_0^2 R_0^6 + 6 t^2 + 2 t \sqrt{a_0^2 R_0^6 + 9 t^2}\right)}, \end{equation*}
    \begin{equation*} R(t) = R_0 \sqrt{1 + 6t \frac{3 t + \sqrt{a_0^2 R_0^6 + 9 t^2}}{a_0^2 R_0^6}}. \end{equation*}

    We observe that the solutions exist globally and

    \begin{equation*} \lim\limits_{t \to \infty} a(t) = 0, \qquad \lim\limits_{t \to \infty} R(t) = \infty. \end{equation*}

    In particular u stays in the form (7.12) for all t > 0 .

    Finally we consider n = 1 . In this case, both ansätze considered before lead to the same solution:

    \begin{equation} Z_{in}(x) = \frac{1}{2} \left(\frac{x}{R}\right)^3 -\frac{3}{2} \frac{x}{R}, \qquad Z_{out}(x) = - \mathrm{sgn}\, x \end{equation} (7.15)

    which coincides with (7.5). Repeating the calculations following (7.5), we obtain a solution of form (7.2) satisfying (7.8), i.e.,

    \begin{equation} u(t) = a(t)\mathbf 1_{B_{R(t)}}, \qquad a(t) = a_0 \left(1 + \frac{6}{a_0 R_0^3} t\right)^{-\frac{1}{2}}, \qquad R(t) = R_0 \left(1 + \frac{6}{a_0 R_0^3} t\right)^{\frac{1}{2}}. \end{equation} (7.16)

    Note that now, as opposed to the case n \geq 3 , the coefficient multiplying t is positive. Like in n = 2 , the extinction time is infinite and we have

    \begin{equation*} \lim\limits_{t \to \infty} a(t) = 0, \qquad \lim\limits_{t \to \infty} R(t) = \infty. \end{equation*}

    This concludes the proof of Theorem 4.

    Using the calibrations we constructed for generalized annuli, we will now derive a system of ODEs locally prescribing the solution emanating from any piecewise constant, radially symmetric datum (a stack).

    Definition 32. Let w \in D(TV) . We say that w is a stack if there exists a number N \in \mathbb{N} and sequences 0 < R^0 < R^1 < \ldots < R^{N-1} , a^0, a^1, \ldots, a^N with a^k \in \mathbb{R} such that

    w = a^0 \mathbf{1}_{B_{R^0}} + a^1 \mathbf{1}_{A_{R^0}^{R^1}} +\ldots + a^{N-1} \mathbf{1}_{A_{R^{N-2}}^{R^{N-1}}} + a^N \mathbf{1}_{\mathbb{R}^n \setminus B_{R^{N-1}}}.

    Suppose first that n\neq 2 , in which case all connected components of level sets of any stack w are calibrable. Let u_0 be a stack

    \begin{equation} u_0 = a^0_0 \mathbf{1}_{B_{R^0_0}} + a^1_0 \mathbf{1}_{A_{R^0_0}^{R^1_0}} +\ldots + a^{N-1}_0 \mathbf{1}_{A_{R^{N-2}_0}^{R^{N-1}_0}} + a^N_0 \mathbf{1}_{\mathbb{R}^n \setminus B_{R^{N-1}_0}}, \end{equation} (7.17)

    where a^{k-1}\neq a^k for k = 1, \ldots, N , a^N_0 = 0 . We expect that if u is the solution emanating from u_0 , then u(t, \cdot) is a stack of form

    \begin{equation} u(t, \cdot) = a^0(t) \mathbf{1}_{B_{R^0(t)}} + a^1(t) \mathbf{1}_{A_{R^0(t)}^{R^1(t)}} +\ldots + a^{N-1}(t) \mathbf{1}_{A_{R^{N-2}(t)}^{R^{N-1}(t)}} + a^N(t) \mathbf{1}_{\mathbb{R}^n \setminus B_{R^{N-1}(t)}}, \end{equation} (7.18)

    with a^N(t) = 0 for all t > 0 , and that a^{k-1}\neq a^k , k = 1, \ldots, N for small t . We construct a Cahn-Hoffman vector field Z(t, \cdot) for u(t, \cdot) by pasting together calibrations Z^k for B_{R^0(t)} , A_{R^k(t)}^{R^{k+1}(t)} , \mathbb{R}^n \setminus B_{R^{N-1}(t)} with suitable choice of signatures. We have

    \begin{eqnarray} u_t = -\Delta \operatorname{div} Z = - \Delta \operatorname{div} Z^0\, \mathcal{L}^n \, ∟\, _{B_{R^0}} - \sum\limits_{k = 1}^{n} \Delta \operatorname{div} Z^k\, \mathcal{L}^n \, ∟\, _{A_{R^{k-1}}^{R^k}} - \Delta \operatorname{div} Z^N\, \mathcal{L}^n \, ∟\, _{\mathbb{R}^n\setminus B_{R^N}} \\ + \sum\limits_{k = 0}^{n} \frac{x}{|x|} \cdot(\nabla \operatorname{div} Z^k - \nabla \operatorname{div} Z^{k+1})\, \mathcal{H}^{n-1} \, ∟\, _{S_{R^k}}. \end{eqnarray} (7.19)

    We denote

    \begin{eqnarray} \left. \frac{x}{|x|} \cdot(\nabla \operatorname{div} Z^k - \nabla \operatorname{div} Z^{k+1}) \right|_{S_{R^k}} \\ = z^k_{rr}(R^k) - z^{k+1}_{rr}(R^k) + \frac{n-1}{R^k}(z^k_{r}(R^k)- z^{k+1}_{r}(R^k))-\frac{n-1}{(R^k)^2}(z^k(R^k)- z^{k+1}(R^k))\\ = z^k_{rr}(R^k) - z^{k+1}_{rr}(R^k) = : d^k. \end{eqnarray}

    The values of d^k are functions of R^0, \ldots, R^{N-1} . Assuming that R^k are regular enough and \varepsilon , |t-s| are small enough, we have

    \frac{\mathrm{d}}{\mathrm{d} t}\int_{A_{R^k(s)-\varepsilon}^{R^k(s)+\varepsilon}} u = \int_{A_{R^k(s)-\varepsilon}^{R^k(s)+\varepsilon}} u_t,

    whence

    (a^k(t) - a^{k-1}(t)) \mathcal{H}^{n-1}(S_{R^k(t)}) \frac{\mathrm{d} R^k}{\mathrm{d} t} = (a^k(t) - a^{k-1}(t)) \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{L}^n(A_{R^k(s)-\varepsilon}^{R^k(t)}) = d^k(t) \mathcal{H}^{n-1}(S_{R^k(t)}).

    Further, for k = 0, \ldots, N , we denote by \lambda^k the value of -\Delta \operatorname{div} Z^k(t, \cdot) which is constant since Z^k is a calibration. Then, we can write down the system of ODEs for a^k and R^k :

    \begin{equation} \frac{\mathrm{d} a^k}{\mathrm{d} t} = \lambda^k \text{ for } k = 0, \ldots N, \qquad \frac{\mathrm{d} R^k}{\mathrm{d} t} = \frac{d^k}{a^k - a^{k+1}} \text{ for } k = 0, \ldots N-1. \end{equation} (7.20)

    Let c_0^k denote c_0 given by (6.17) if \mathrm{sgn}\, (a^{k+1} - a^k) = \mathrm{sgn}\, (a^k - a^{k-1}) or by (6.10) if \mathrm{sgn}\, (a^{k+1} - a^k) \neq \mathrm{sgn}\, (a^k - a^{k-1}) , with R^{k+1} and R^k in place of R_1 and R_0 . Then, we have

    \begin{equation} \lambda^0 = \mathrm{sgn}\, (a^1-a^0)\frac{n(n+2)}{(R^0)^3}, \quad \lambda^k = 2n(n+2)\mathrm{sgn}\, (a^{k+1} - a^k) c_0^k \text{ for } k = 1, \ldots, N-1, \quad \lambda^N = 0. \end{equation} (7.21)

    We observe that in a neighborhood of any initial datum R^0_0, \ldots, R^{N-1}_0 , a^0_0, \ldots, a^N_0 , R^k_0 < R^{k+1}_0 , a^k_0 \neq a^{k+1}_0 , the r. h. s. of (7.20) is regular in R^0, \ldots, R^{N-1} , a^0, \ldots, a^N , so locally the system has a unique solution. Unique solvability fails when a time instance t > 0 is reached such that a^k(t) = a^{k+1}(t) , R^k(t) = R^{k+1}(t) or R_0 = 0 . In such case u(t, \cdot) is again a stack with a smaller N , and we can restart our procedure. This concludes the proof of Theorem 5.

    Next we deal with the remaining case of dimension n = 2 . In this case, our attempt to obtain a radial calibration failed for complements of balls and for some annuli. Again, let u_0 be a stack of form (7.17). For k = 1, \ldots, N , let \sigma^k = \mathrm{sgn}\, (a^k_0 - a^{k-1}_0) . We assume the following ansatz on the solution u and the associated field Z for small t > 0 :

    \begin{equation} u(t, \cdot) = a^0(t) \text{ on } B_{R^0(t)}, u(t, \cdot) = a^k(t) \text{ on } A_{\max(R^{k-1}(t), R^k(t)/Q_*)}^{R^k(t)} \text{ if } \sigma^{k+1} \neq \sigma^k, \ k = 1, \ldots, {N-1}, \end{equation} (7.22)
    \begin{eqnarray} Z(t, x) = \sigma^k \tfrac{x}{|x|} \text{ on } A_{R^{k-1}(t)}^{R^k(t)} \text{ if } \sigma^{k+1} \neq \sigma^k \text{ or on } A_{R^{k-1}(t)}^{R^k(t)/Q_*} \text{ if }\sigma^{k+1} = \sigma^k, \ k = 1, \ldots, {N-1}, \\ Z(t, x) = \sigma^{k+1} \tfrac{x}{|x|} \text{ on } \mathbb{R}^n \setminus B_{R^{N-1}(t)}. \end{eqnarray} (7.23)

    We complete the definition of a Cahn-Hoffman field Z consistent with (7.22), (7.23) by pasting the calibrations Z^k with suitable choice of signatures into the gaps left in (7.23). This leads to

    u_t(t, \cdot) = \lambda^0(t) \text{ in } \mathcal{D}'(B_{R^0(t)}),
    \begin{align*} u_t(t, x) & = \lambda^k(t) \text{ in } \mathcal{D}'\left(A_{\max(R^{k-1}(t), R^k(t)/Q_*)}^{R_k(t)}\right), \\ u_t(t, x) & = \frac{\sigma^k}{|x|^3} \text{ in } \mathcal{D}'\left(A_{R^{k-1}(t)}^{R^k(t)/Q_*}\right) \text{ if } \sigma^k \neq \sigma^{k+1} \text{ or in } \mathcal{D}'\left(A_{R^{k-1}(t)}^{R^k(t)}\right) \text{ if } \sigma^k = \sigma^{k+1}, \ k = 1, \ldots, N-1, \\ u_t(t, x) & = \frac{\sigma^N}{|x|^3} \text{ in } \mathcal{D}'(\mathbb{R}^2 \setminus B_{R^N(t)}). \end{align*}

    Moreover, u_t(t, \cdot) \in M(\mathbb{R}^2) and

    u_t \, ∟\, _{S_{R^k}} = \frac{x}{|x|}\cdot ((\nabla \operatorname{div} Z)^- - (\nabla \operatorname{div} Z)^+) \mathcal{H}^{1} \, ∟\, _{S_{R^k}} = : d^k

    for k = 0, \ldots, N-1 , where (\nabla \operatorname{div} Z)^\pm are the one sided limits as |x| \to (R^k)^\pm . The values of d^k are functions of R^0, \ldots, R^{N-1} . Reasoning as in the case n \neq 2 , the evolution of R^k is governed by equations

    \begin{equation} \frac{\mathrm{d} R^k}{\mathrm{d} t} = \frac{d^k}{u(t, x)\big|_{|x| = (R^k)^-} - u(t, x)\big|_{|x| = (R^k)^+}}. \end{equation} (7.24)

    The values u(t, x)\big|_{|x| = (R^k)^+} are either prescribed by ODEs

    \begin{equation} \frac{\mathrm{d} a^k}{\mathrm{d} t} = \lambda^k \end{equation} (7.25)

    with \lambda^k functions of R^0, \ldots, R^{N-1} in calibrable regions where u(t, x) = a^k(t) , or explicitly determined by u_t(t, x) = \sigma^k /|x|^3 in bending regions. It is important to note that in the case \sigma^k \neq \sigma^{k+1} , R^{k-1} \leq R^k/Q_* the functions d^k, \lambda^k do not depend on R^{k-1} . Thus, one can first solve a part of the system (7.24), (7.25) for the outer annuli, then calculate u in the bending region (without knowing a priori its inner boundary) and move on to solving innermore parts of (7.24), (7.25). This way, finding the solution is indeed again reduced to solving a system of ODEs. We include Figure 4 illustrating the evolution of stacks on the example of the characteristic function of an annulus.

    Figure 4.  Plots of the solution u(t, x) emanating from the characteristic function of annulus A_{R^1}^{R^2} as a function of |x| for chosen values of t with R^1 = 1 , R^2 = 2 .

    In the case of thick annuli in n = 2 , the qualitative behavior resulting from this procedure is rather intricate and may be surprising. To showcase this, we include Figures 5 and 6 depicting the evolution emanating from the characteristic function of a thick annulus.

    Figure 5.  Plots of the solution u(t, x) emanating from the characteristic function of annulus A_{R^1}^{R^2} with R_0 = 1 , R_1 = 20 in n = 2 as a function of |x| for t = 0 , t = 2 , t = 4 , t = 6 .
    Figure 6.  Plot of the Cahn-Hoffman vector field Z(t, x) for the characteristic function of annulus A_{R^1}^{R^2} with R_0 = 1 , R_1 = 20 in n = 2 as a function of |x| .

    Let us explain the evolution in a few words. The inner part of the initial facet corresponding to the annulus instantaneously bends downwards. Meanwhile, the outer boundary of the facet expands outwards, at relatively low speed (practically invisible in the picture). Since the ratio of outer to inner radius of the facet is constant, this means that the whole facet slowly moves outwards. The combined effect of this and the bending results in the very steep (but continuous!) part of the graph between the facet and the bending part. At the same time, the facet corresponding to the inside ball also expands outwards, gradually consuming the bending part. In the final pictured time instance, the whole bending part has disappeared and the solution is a stack again.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors would like to thank the anonymous reviewers for their helpful comments.

    The first author of the work was partly supported by the Japan Society for the Promotion of Science (JSPS) through the grants Kakenhi: No. 19H00639, No. 18H05323, No. 17H01091, and by Arithmer Inc. and Daikin Industries, Ltd. through collaborative grants.

    The second author of the work was partly supported by JSPS through the grant Kakenhi No. 18H05323.

    This work was created during the last author's JSPS Postdoctoral Research Fellowship at the University of Tokyo. The last author was partly supported by the Kakenhi Grant-in-Aid No. 21F20811.

    The authors declare no conflict of interest.



    [1] F. Alter, V. Caselles, A. Chambolle, A characterization of convex calibrable sets in \mathbb{R}^N, Math. Ann., 332 (2005), 329–366. http://doi.org/10.1007/s00208-004-0628-9 doi: 10.1007/s00208-004-0628-9
    [2] L. Ambrosio, N. Gigli, G. Savaré, Gradient flows: in metric spaces and in the space of probability measures, 2 Eds., Basel: Birkhäuser, 2008. https://doi.org/10.1007/978-3-7643-8722-8
    [3] F. Andreu-Vaillo, V. Caselles, J. M. Mazón, Parabolic quasilinear equations minimizing linear growth functionals, Basel: Birkhäuser, 2004. http://doi.org/10.1007/978-3-0348-7928-6
    [4] G. Bellettini, V. Caselles, M. Novaga, The total variation flow in \mathbb{R}^N, J. Differ. Equations, 184 (2002), 475–525. http://doi.org/10.1006/jdeq.2001.4150 doi: 10.1006/jdeq.2001.4150
    [5] G. Bellettini, M. Novaga, M. Paolini, Characterization of facet breaking for nonsmooth mean curvature flow in the convex case, Interfaces Free Bound., 3 (2001), 415–446. http://doi.org/10.4171/IFB/47 doi: 10.4171/IFB/47
    [6] H. Brézis, Opérateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert, New York: American Elsevier Publishing Co., Inc., 1973.
    [7] W. C. Carter, A. R. Roosen, J. W. Cahn, J. E. Taylor, Shape evolution by surface diffusion and surface attachment limited kinetics on completely faceted surfaces, Acta Metallurgica et Materialia, 43 (1995), 4309–4323. http://doi.org/10.1016/0956-7151(95)00134-H doi: 10.1016/0956-7151(95)00134-H
    [8] C. M. Elliott, S. A. Smitheman, Analysis of the TV regularization and H^{-1} fidelity model for decomposing an image into cartoon plus texture, Commun. Pure Appl. Anal., 6 (2007), 917–936. http://doi.org/10.3934/cpaa.2007.6.917 doi: 10.3934/cpaa.2007.6.917
    [9] L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions, New York: CRC Press, 2015. https://doi.org/10.1201/b18333
    [10] W. Fulton, Algebraic topology. A first course, New York, NY: Springer, 1995. https://doi.org/10.1007/978-1-4612-4180-5
    [11] G. P. Galdi, An introduction to the mathematical theory of the Navier-Stokes equations. Steady-state problems, 2 Eds., New York, NY: Springer, 2011. https://doi.org/10.1007/978-0-387-09620-9
    [12] M.-H. Giga, Y. Giga, Very singular diffusion equations: second and fourth order problems, Japan J. Indust. Appl. Math., 27 (2010), 323–345. http://doi.org/10.1007/s13160-010-0020-y doi: 10.1007/s13160-010-0020-y
    [13] M.-H. Giga, Y. Giga, Crystalline surface diffusion flow for graph-like curves, Discrete Cont. Dyn. Syst., 43 (2023), 1436–1468. http://doi.org/10.3934/dcds.2022160 doi: 10.3934/dcds.2022160
    [14] Y. Giga, R. V. Kohn, Scale-invariant extinction time estimates for some singular diffusion equations, Discrete Cont. Dyn. Syst., 30 (2011), 509–535. http://doi.org/10.3934/dcds.2011.30.509 doi: 10.3934/dcds.2011.30.509
    [15] Y. Giga, H. Kuroda, H. Matsuoka, Fourth-order total variation flow with Dirichlet condition: Characterization of evolution and extinction time estimates, Adv. Math. Sci. Appl., 24 (2014), 499–534.
    [16] Y. Giga, M. Muszkieta, P. Rybka, A duality based approach to the minimizing total variation flow in the space H^{-s}, Japan J. Indust. Appl. Math., 36 (2019), 261–286. http://doi.org/10.1007/s13160-018-00340-4 doi: 10.1007/s13160-018-00340-4
    [17] Y. Giga, N. Požár, Motion by crystalline-like mean curvature: a survey, Bull. Math. Sci., 12 (2022), 2230004. http://doi.org/10.1142/S1664360722300043 doi: 10.1142/S1664360722300043
    [18] Y. Giga, Y. Ueda, Numerical computations of split Bregman method for fourth order total variation flow, J. Comput. Phys., 405 (2020), 109114. http://doi.org/10.1016/j.jcp.2019.109114 doi: 10.1016/j.jcp.2019.109114
    [19] D. Gilbarg, N. S. Trudinger, Elliptic partial differential equations of second order, 2 Eds., Berlin, Heidelberg: Springer, 2001. https://doi.org/10.1007/978-3-642-61798-0
    [20] Y. Kashima, A subdifferential formulation of fourth order singular diffusion equations, Adv. Math. Sci. Appl., 14 (2004), 49–74.
    [21] Y. Kashima, Characterization of subdifferentials of a singular convex functional in Sobolev spaces of order minus one, J. Funct. Anal., 262 (2012), 2833–2860. http://doi.org/10.1016/j.jfa.2012.01.005 doi: 10.1016/j.jfa.2012.01.005
    [22] R. V. Kohn, Surface relaxation below the roughening temperature: some recent progress and open questions, In: Nonlinear partial differential equations, Berlin, Heidelberg: Springer, 2012,207–221. http://doi.org/10.1007/978-3-642-25361-4_11
    [23] R. V. Kohn, H. M. Versieux, Numerical analysis of a steepest-descent PDE model for surface relaxation below the roughening temperature, SIAM J. Numer. Anal., 48 (2010), 1781–1800. http://doi.org/10.1137/090750378 doi: 10.1137/090750378
    [24] Y. K\overline{\rm o}mura, Nonlinear semi-groups in Hilbert space, J. Math. Soc. Japan, 19 (1967), 493–507. http://doi.org/10.2969/jmsj/01940493 doi: 10.2969/jmsj/01940493
    [25] M. Łasica, S. Moll, P. B. Mucha, Total variation denoising in l^1 anisotropy, SIAM J. Imaging Sci., 10 (2017), 1691–1723. http://doi.org/10.1137/16M1103610 doi: 10.1137/16M1103610
    [26] G. P. Leonardi, An overview on the Cheeger problem, In: New trends in shape optimization, Cham: Birkhäuser, 2015,117–139. http://doi.org/10.1007/978-3-319-17563-8_6
    [27] I. V. Odisharia, Simulation and analysis of the relaxation of a crystalline surface, Ph.D. thesis, New York University, New York, 2006.
    [28] S. Osher, A. Solé, L. Vese, Image decomposition and restoration using total variation minimization and the H^{-1} norm, Multiscale Model. Sim., 1 (2003), 349–370. http://doi.org/10.1137/S1540345902416247 doi: 10.1137/S1540345902416247
    [29] W. Rudin, Functional analysis, 2 Eds., New York: McGraw-Hill, Inc., 1991.
    [30] L. Schwartz, Théorie des distributions, Publications de l'Institut de Mathématique de l'Université de Strasbourg, Paris, 1966.
    [31] H. Spohn, Surface dynamics below the roughening transition, J. Phys. I France, 3 (1993), 69–81. http://doi.org/10.1051/jp1:1993117 doi: 10.1051/jp1:1993117
  • This article has been cited by:

    1. Julie Clutterbuck, Jiakun Liu, Preface to the Special Issue: Nonlinear PDEs and geometric analysis – Dedicated to Neil Trudinger on the occasion of his 80th birthday, 2023, 5, 2640-3501, 1, 10.3934/mine.2023095
  • Reader Comments
  • © 2023 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)
通讯作者: 陈斌, bchen63@163.com
  • 1. 

    沈阳化工大学材料科学与工程学院 沈阳 110142

  1. 本站搜索
  2. 百度学术搜索
  3. 万方数据库搜索
  4. CNKI搜索

Metrics

Article views(1864) PDF downloads(380) Cited by(1)

Figures and Tables

Figures(6)

Other Articles By Authors

/

DownLoad:  Full-Size Img  PowerPoint
Return
Return

Catalog