VARIATIONAL EVOLUTION PROBLEMS AND NONLOCAL GEOMETRIC MOTION

MIKHAIL FELDMAN

Abstract. We consider two variational evolution problems related to Monge-Kantorovich mass transfer. These problems provide models for collapsing sandpiles and for compression molding. We prove the following connection between these problems and nonlocal geometric curvature motion: The distance functions to surfaces moving according to certain nonlocal geometric laws are solutions of the variational evolution problems. Thus we do the first step of the proof of heuristics developed in earlier works. The main techniques we use are differential equations methods in the Monge-Kantorovich theory.

In this paper we study two models involving limits as p → ∞ of solutions of p-Laplacian evolution problems. One is a model of collapsing sandpiles proposed by L. C. Evans, R. F. Gariepy and the author [EFG97]. Another is a model of compression molding proposed by G. Aronsson and L. C. Evans [AE]. In both models the limits were characterized as solutions of variational evolution problems related to Monge-Kantorovich mass transfer. Solutions that have the form of the distance function to a moving boundary arise naturally in the both models. The equations of the motion of the boundary for both models were derived heuristically in [EFG97] and [AE]. According to the equations the outer normal velocity of the boundary at a point depends on both the local geometry (curvatures) and the nonlocal geometry of the boundary. Thus, the motion of the boundary is a nonlocal geometric motion.

In this paper we prove rigorously the connection between geometric and variational evolution problems. Namely, assuming that a moving surface satisfying the geometric equation is given, we prove that the distance function defines a solution of the corresponding variational evolution problem. The main assumption is that the surface remains convex (or, more generally, semiconvex) during the evolution. For the collapsing sandpiles model we also prove the corresponding result for the solutions that have the form of the maximum of several distance functions (such solutions represent several sand cones interacting in the process of collapse). In the proofs we utilize the connection between the models and the Monge-Kantorovich mass transfer. This allows us to use the differential equations methods in the Monge-Kantorovich theory, which have been developed recently by L. C. Evans and W. Gangbo [EGan].

The examples suggest that solutions of the geometric equations derived in [EFG97] can develop singularities, even if the initial data is smooth. Thus we do not assume below that the moving surface is smooth. This however makes the technique more involved.

Research is supported in part by NSF grants DMS-9701755 (MSRI) and DMS-9623276.

In the forthcoming paper [Fel] we construct convexity preserving viscosity solutions of the geometric equations given convex initial data, thus finishing the proof of the heuristics developed in [EFG97] and [AE].

This paper is organized as following. In Sections 2 - 7 we work on the collapsing sandpiles model. In Section 8 we work on the compression molding model. In Appendix A we estimate the local Lipschitz constant of the gradient of the distance function of a set.

1. Notation

Let Ω ∈ R^{n} be a bounded open set. Let Γ = ∂Ω. We will use the following two distance functions of the set Ω. The first, interior distance function, is

(1) d_{Ω}(x) = \{ \begin{matrix}dist(x, ∂Ω), & x ∈ Ω, \\ 0, & x ∈ R^{n} \ Ω. \end{matrix}

The second, signed distance function, is

(2) d_{Ω}^{s}(x) = \{ \begin{matrix}dist(x, ∂Ω), & x ∈ Ω, \\ −dist(x, ∂Ω), & x ∉ Ω. \end{matrix}

Let x ∈ R^{n}. We denote N_{∂Ω}(x) the set of all points of ∂Ω nearest to x, i.e.,

N_{∂Ω}(x) = {y ∈ ∂Ω, |x − y| = inf_{z∈∂Ω} |x − z|}.

Note that N_{∂Ω}(x) may have more then one point.

Let y ∈ ∂Ω. Then ∂Ω is differentiable at y if there exists δ > 0 such that in a suitable coordinate system (x_{1}, ..., x_{n}) in R^{n} we have

(3) y = 0, Ω ∩ B_{δ}(0) = {x_{n} > Ψ(x′)} ∩ B_{δ}(0) where Ψ(x′) = o(|x′|).

Here x′ = (x_{1}, ..., x_{n−1}), and B_{δ}(0) = B_{δ}^{n}(0) is the ball in R^{n} with radius δ and center at 0. ∂Ω is twice differentiable at y if in a suitable coordinate system in R^{n−1} the function Ψ has the expression

Ψ(x′) = ∑_{i=1}^{n−1} κ_{i}x_{i}^{2} + o(|x′|^{2}). (4)

The numbers κ_{1}, ..., κ_{n−1} are the principal curvatures of ∂Ω at y.

Let {Ω_{t}}, t ∈ R_{+}^{1}, be a family of open bounded sets in R^{n}. We call a family of sets {Ω_{t}} Lipschitz (with respect to t) if there exists a constant M > 0 such that for any t_{1}, t_{2} the set (Ω_{t_{1}} \ Ω_{t_{2}}) ∪ (Ω_{t_{2}} \ Ω_{t_{1}}) is subset of both M | t_{1} − t_{2} | neighborhood (in R^{n}) of ∂Ω_{t_{1}} and M | t_{1} − t_{2} | neighborhood of ∂Ω_{t_{2}}.

Let

E = ∪_{t} (Ω_{t} × {t}) ⊂ R^{n} × R_{+}^{1},

Γ = ∪_{t} (∂Ω_{t} × {t}) ⊂ R^{n} × R_{+}^{1}, (5)

Γ_{t} = ∂Ω_{t} ⊂ R^{n}.

Let (y, t) ∈ R^{n} × R_{+}^{1} be such that y ∈ Γ_{t} and the function (x, τ) → d_{Ω_{τ}}^{s} (x) is differentiable at (y, t). We define the outer normal velocity of Γ_{t} at y as

V (y) = \frac{∂}{∂t}d_{Ω_{t}}^{s}(y). (6)

Let (y, t_{0}) ∈ Γ. The surface Γ is (2,1) differentiable at (y, t_{0}) if in a suitable coordinate system (x_{1}, ..., x_{n}) in R^{n} we have y = 0, E ∩ B_{δ}^{n+1}(0) = {x_{n} > Ψ(x′, t)} ∩ B_{δ}^{n+1}(0) for small δ > 0 and

Ψ(x′, t) = v(t − t_{0}) + ∑_{i=1}^{n−1} κ_{i}x_{i}^{2} + o(| x′ − y′ |^{2} + | t − t_{0} |).

L^{n} is n-dimensional Lebesgue measure.

H^{n} is n-dimensional Hausdorff measure.

2. The collapsing sandpiles model

In the paper [EFG97] the p-Laplacian evolution problems

\{ \begin{matrix}∂_{t}u_{p} − div(| Du_{p} |^{p−2} Du_{p}) = 0, & in R^{n} × (0, ∞), \\ u_{p} = g, & on R^{n} × \{0\}, \end{matrix} (7)

are considered in the “infinitely fast/slow diffusion limit” p → ∞. Du_{p} denotes the gradient of u_{p} with respect to the spatial variables x_{1}, ..., x_{n}. The operator

Δ_{p}(u) = div(| Du_{p} |^{p−2} Du_{p})

is p-Laplacian. The initial data g is nonnegative, Lipschitz continuous function with compact support. We assume that

Lip[g] = L > 1 (8)

where Lip[g] is the Lipschitz constant of g.

The initial data g satisfying (8) is unstable in the following sense. It is shown in [EFG97] that u_{p} → u uniformly on R^{n} × (0, ∞), and that

| Du |≤ 1 a.e. in R^{n} × (0, ∞). (9)

This limit u does not depend on t for t > 0, i.e., u = u(x). By (8) and (9) we get u ≠ g. Thus the limit solution is discontinuous at t = 0. The transformation g → u takes place at t = 0.

The described problem can be interpreted as a crude model of a collapsing sandpile. The function u(x, t) is the height of the pile at the location x, time t. The main physical assumption is that a sandpile is stable if its slope does not exceed 1, i.e., if Lip[u(·, t)] ≤ 1. The condition (8) implies that the initial profile g is unstable. According to the model above, the initially unstable sandpile collapses instantaneously to a stable one, with the height function u.

The transformation g → u was studied in [EFG97] by relating it to the evolution problem governed by Monge-Kantorovich mass transfer. Introduce the convex functional I_{∞} : L^{2}(R^{n}) → R^{1} ∪ {∞} defined by

I_{∞}[v] = \{ \begin{matrix}0 & if v ∈ L^{2}(R^{n}), | Dv |≤ 1 a.e. \\ +∞ & otherwise. \end{matrix}

Then u(x) = w(x, 1), where w is the unique solution of the evolution problem

\{ \begin{matrix}\frac{w}{t} − ∂_{t}w ∈ ∂I_{∞}[w], & for a.e. t ∈ [\frac{1}{L}, 1], \\ w = \frac{g}{L} = ĝ, & at t = \frac{1}{L}, \end{matrix} (10)

with L = Lip[g] > 1. Note that Lip[ĝ] = 1, and thus the evolution problem (10) is well-posed. The connection between this evolution problem and Monge-Kantorovich mass transfer is explained in [EFG97].

In the static case the limits as p → ∞ of the boundary value problems for p-Laplacian type equations were considered in the papers [BDM89], [Jan93] and references therein. In many cases the limits have the form of the distance function to the boundary of the domain.

A limit evolution problem that represents a stable case of the problem (7) was considered in the paper [AEW96] as a model of a growing stable sandpile. Let u(x, t) be, as before, the height of the pile, and let f(x, t) ≥ 0 be the sand source function which describes the rate at which sand is coming at location x at time t. The dynamics is defined as following. Consider the nonhomogeneous evolution problem

\{ \begin{matrix}∂_{t}u_{p} − div(| Du_{p} |^{p−2} Du_{p}) = f, & in R^{n} × (0, ∞), \\ u_{p} = 0, & on R^{n} × \{0\}, \end{matrix} (11)

and let p → ∞. Then for any T > 0

u_{p} → u uniformly on R^{n} × [0, T]

and this limit u satisfies the equation

f − ∂_{t}u ∈ ∂I_{∞}[u], for a.e. t ≥ 0 (12)

with zero initial condition. In particular the case was considered when sand sources are concentrated in m points, i.e., if

f(x, t) = ∑_{k=1}^{m} f_{k}(t)δ_{d_{k}}(x),

where δ_{d_{k}}(·) is the Dirac mass at the point d_{k} ∈ R^{n}. The solution has the explicit form

u = max(0, z_{1}(t)− | x − d_{1} |, ..., z_{m}(t)− | x − d_{m} |),

where the height functions {z_{k}(t)}_{k=1}^{m} satisfy a system of ODE. Thus the solution has the form of growing and interacting cones centered at the points {d_{k}}_{k=1}^{m}.

By analogy with the static problems and the model of growing sandpiles [AEW96], one can expect to find solutions of (10) that have the form of distance function to the moving boundary or superposition of such, i.e., growing and interacting cones. However, unlike the situation considered in [AEW96], interacting spherical cones do not maintain their shape under the evolution defined by (10). A more suitable class of solutions is the following class of more general “cones”. A single cone solution has the form

w(x, t) = d_{Ω_{t}}(x), (13)

where {Ω_{t}} is a suitable expanding family of sets. A multiple interacting cones solution has the form (for simplicity we consider the case of two cones)

w(x, t) = max(d_{Ω_{t}^{1}}(x), d_{Ω_{t}^{2}}(x)), (14)

where {Ω_{t}^{1}}, {Ω_{t}^{2}} are suitable expanding families of sets. Thus in order to find such solutions one has to find the appropriate families of sets.

The following equations of motion of the boundaries of the sets in (13) and (14) are derived heuristically in [EFG97]. Assume that Γ defined by (5) is a smooth surface in R^{n} × R^{1}. Let y ∈ Γ_{t} and let κ_{1}, ..., κ_{n−1} be the principal curvatures of Γ_{t} at y. Denote κ = (κ_{1}, ..., κ_{n−1}). Let γ(y) = γ_{Ω_{t}}(y) be the radius of the largest ball touching Γ_{t} at y from within Ω_{t}. Let V (y) be the outer normal velocity of Γ_{t} at the point y defined by (6). Then in the case of a single cone solution (13) the equation for the surface Γ_{t} is:

V (y) = \frac{1}{t}F (κ(y), γ(y)) for y ∈ Γ_{t},

where (15)

F (κ, γ) = \frac{∫_{0}^{γ} s ∏_{i=1}^{n−1}(1 − κ_{i}s)ds}{∫_{0}^{γ} ∏_{i=1}^{n−1}(1 − κ_{i}s)ds}.

To write equations for the motion of the surfaces Γ_{t}^{k} = ∂Ω_{t}^{k}, k = 1, 2 in the case of the interacting cones solution (14), we extend the function γ(·) = γ_{Ω}(·) of a set Ω ∈ R^{n} to a function R^{n} → R^{1} by setting γ ≡ 0 on R^{n}\\overline{Ω}, and γ(x) = γ(y) for x ∈ Ω such that y ∈ N_{∂Ω}(x). Then the equations are:

V (y)  = \frac{1}{t}F (κ(y), γ_{Ω_{t}^{k}}(y), γ_{Ω_{t}^{1}∩Ω_{t}^{2}}(y)) for y ∈ Γ_{t}^{k}, k = 1, 2,

where (16)

F (κ, γ, δ) = \frac{∫_{δ}^{γ} s ∏_{i=1}^{n−1}(1 − κ_{i}s)ds}{∫_{δ}^{γ} ∏_{i=1}^{n−1}(1 − κ_{i}s)ds}.

The equations (15) and (16) have the form of nonlocal geometric curvature motion. Geometric nonlocality is caused by the functions γ(·).

In this paper we prove that if the moving surfaces satisfy geometric equations (15) or (16) and some regularity conditions, then the functions (13) or (14) respectively are solutions of the evolution problem (10). Namely, we prove the following.

Theorem 1. Let {Ω_{t}} be a locally Lipschitz continuous family of open convex bounded sets, where t ∈ [a, b], b > a > 0. Suppose that the equation (15) is satisfied at every point of (2, 1) differentiability of the surface Γ defined by (5). Then the function (13) is a solution of the evolution equation

\frac{w}{t} − ∂_{t}w ∈ ∂I_{∞}[w] for a.e. t ∈ [a, b]. (17)

Theorem 2. Let {Ω_{t}^{1}}, {Ω_{t}^{2}} be two Lipschitz continuous families of open bounded convex sets, where t ∈ [a, b], b > a > 0. Suppose that the equation (16) is satisfied at every point of (2, 1) differentiability of the surfaces Γ^{k} = ∪_{t}(Γ_{t}^{k} × {t}), k = 1, 2. Then the function (14) is a solution of the evolution equation (17).

The convexity condition in Theorems 1, 2 can be relaxed to the following ”semiconvexity” condition.

Condition 3. (lower curvature bound condition with radius r ). An open set Ω ⊂ R^{n} satisfies this condition if γ_{R^{n}\\overline{Ω}}(x) ≥ 2r for any x ∈ R^{n} \ \overline{Ω}.

It is easy to check that if a set is either convex or has C^{1,1} boundary, then the set satisfies the Condition 3.

Remark 4. Let the set Ω satisfy Condition 3. Let ∂Ω be twice differentiable at a point y, i.e., (3) and (4) are satisfied in an appropriate coordinate system on R^{n}. Then the principal curvatures of ∂Ω at y satisfy

κ_{i} ≥ −\frac{1}{2r} for i = 1, ..., n − 1.

This follows from the fact that the graph of the function Ψ defined by (3) and (4) lies above the ball B_{2r}^{n}(0, −2r) by Condition 3.

The following two theorems state that convexity can be replaced by the weaker Condition 3 in Theorems 1 and 2.

Theorem 5. Let {Ω_{t}} be a locally Lipschitz continuous family of open bounded sets, where t ∈ [a, b], b > a > 0. Let for every t ∈ [a, b] the set Ω_{t} satisfy Condition 3 with radius r_{0} > 0. Suppose that the equation (15) is satisfied at every point of (2, 1) differentiability of the surface Γ defined by (5). Then the function (13) is a solution of the evolution equation (17).

Theorem 6. Let {Ω_{t}^{1}}, {Ω_{t}^{2}} be two Lipschitz continuous families of open bounded sets, where t ∈ [a, b], b > a > 0. Let for every t ∈ [a, b] the sets Ω_{t}^{1}, Ω_{t}^{2} satisfy Condition 3 with radius r_{0} > 0. Suppose that the equation (16) is satisfied at every point of (2, 1) differentiability of the surfaces Γ^{k} = ∪_{t}(Γ_{t}^{k} × {t}), k = 1, 2. Then the function (14) is a solution of the evolution equation (17).

Theorems 1 and 2 follow from Theorems 5 and 6 since convex sets satisfy Condition 3 with any radius.

The proof of Theorems 5 and 6 utilizes the following relation of the equation (17) and the Monge-Kantorovich mass transfer problem. Fix t. Equation (17) implies that that the function w(·, t) is the Monge’s potential for the optimal transfer of the measure with density \frac{w}{t}(·, t) into the measure with density ∂_{t}w(·, t). Then, according to [EGan], there exists a measurable function a(x), the mass transport density, that satisfies the following properties:

a ≥ 0,  supp(a) ⊂ {x | | Dw |= 1},

(18)

−div(aDw) = \frac{w}{t} − ∂_{t}w,

where the last equation is understood in the weak sense and t is fixed. Conversely, if there exists a function a(x) satisfying (18), then the function w is a solution of (17) for the given t, see Section 6 below.

The idea of the proof of Theorem 5 is as follows. Let a family of sets {Ω_{t}} satisfy (13). An open bounded set can be represented as the union of distance rays, i.e., maximal intervals that start at the boundary, on which distance to the boundary is the linear function with slope 1. The Condition 3 implies that the collection of distance rays possesses ”nice” measure-theoretic properties. Fix t and consider a distance ray R_{y} of Ω_{t} starting at a smooth point y ∈ ∂Ω_{t} . Introduce the coordinate s on R_{y}, the distance to the boundary. The equation (18) can be formally rewritten as:

Da · Dw + aΔw − \frac{w}{t} − ∂_{t}w = 0. (19)

Since w(·, t) = dist(·, Γ_{t}), we have on R_{y}

Da · Dw = \frac{da}{ds}, w = s, −Δw = ∑_{i=1}^{n−1} \frac{κ_{i}}{1 − κ_{i}s}, ∂_{t}w = V (y),

where κ_{1}, ..., κ_{n−1} are the principal curvatures of Γ_{t} at y, and V (y) is the outer normal velocity of Γ_{t} at y. Thus the equation (19) can be formally rewritten on the ray R_{y} as the following ODE

a′(s) − a(s) ∑_{i=1}^{n−1} \frac{κ_{i}}{1 − κ_{i}s} − \frac{s}{t} + V (y) = 0, (20)

This ODE has a solution a(s) with zero boundary conditions at both ends of the ray R_{y} if V (y) satisfies (15). Define a(x, t) by ODE (20) with zero boundary conditions on each distance ray that starts at a smooth point of the boundary. We prove that a(·, t) is a measurable function that satisfies (18). In order to do this we examine the properties of distance function and distance rays, and use a nonsmooth (Lipschitz) change of coordinates on the set Ω_{t} with n − 1 coordinates along the boundary and the n-th coordinate along distance rays.

We prove Theorems 5 and 6 in Sections 3 - 7. In Section 3 we examine properties of sets satisfying Condition 3, and in particular we describe the change of coordinates mentioned above. In Section 4 we define the mass transport density function and examine some properties of this function. In Section 5 we prove that the mass balance equation (15) is satisfied. In Section 6 we conclude the proof of Theorem 5. Finally, in Section 7 we sketch the proof of Theorem 6.

3. Properties of distance function and ridge sets of the sets satisfying the lower curvature bound condition

Let Ω be an open set. When there is no possible confusion, we write d(x) for d_{Ω}(x) defined by (1).

Let x ∈ \overline{Ω}. Denote by R_{x} the longest line segment through x in \overline{Ω} along which d_{Ω}(·) is a linear function with slope 1. We call R_{x} the distance ray of x. Note that a point x can have more than one distance ray, in such case denote any of them by R_{x}.

If x ∈ Ω then all such rays R_{x} have equal length. If x ∈ ∂Ω then rays R_{x} in the case of general open set Ω can have different length. However if Ω satisfies Condition 3 then there exists at most one R_{x} for x ∈ ∂Ω, see Proposition 9 below.

For any x ∈ Ω one endpoint of R_{x} lies on ∂Ω and belongs to the set N_{∂Ω}(x). Call this endpoint the lower end of R_{x}. Call another endpoint of R_{x} the upper end of R_{x} (the names “lower” and “upper” correspond to the positions of the endpoints of R_{x} on the graph of y = d_{Ω}(x)). Let y, z be upper and lower ends of the ray R, and let x = λy + (1 − λ)z where λ ∈ (0, 1). Then we say that the point x lies in the relative interior of the ray R.

Remark 7. If x ∈ Ω lies in the relative interior of a distance ray R_{x}, then the ray R_{x} is the unique distance ray that contains x, and the set N_{∂Ω}(x) consists of one point, see [EH87].

Remark 8. It follows from Remark 7 that sets satisfying Condition 3 are sets with positive reach defined by H. Federer [Fed59]. More precisely, if a set Ω satisfies Condition 3 with radius r then reach(Ω)=2r.

The function γ_{Ω}(·) in the equations (15), (16) is called ridge function and can be expressed as following:

(21) γ_{Ω}(x) = \{ \begin{matrix}0 & x ∈ R^{n} \ \overline{Ω}, \\ | R_{x} | & x ∈ Ω, \\ sup | R_{x} | & x ∈ ∂Ω. \end{matrix}

The function γ_{Ω}(·) is uppersemicontinuous in R^{n}, cf. [EH87], Proposition 3.2.

Define the set

R = \{ x ∈ \overline{Ω} | d_{Ω}(x) = γ_{Ω}(x) \}.

The set R is called ridge set of Ω. Note that if x ∈ R then either x is the upper end of some distance ray of Ω, or otherwise x lies at ∂Ω and there are no points y ∈ Ω such that x ∈ N_{∂Ω}(y).

Proposition 9. Let Ω ⊂ R^{n} be a bounded open set satisfying Condition 3 with radius r > 0. Let Ω_{r} be the r-neighborhood of Ω. Then

(a) γ_{R^{n}\\overline{Ω}_{r}}(x) ≥ r for all x ∈ R^{n} \ Ω_{r}, i.e., the set Ω_{r} satisfies Condition 3 with radius \frac{r}{2}.

(b) For any y_{r} ∈ ∂Ω_{r} there exists a unique y ∈ ∂Ω such that |y_{r} − y| = r, and |y_{r} − x| > r for all x ∈ \overline{Ω}\\{y\}. For any y ∈ ∂Ω there exists y_{r} ∈ ∂Ω_{r} such that |y − y_{r}| = r. If ∂Ω is differentiable at y then y − y_{r} is orthogonal to ∂Ω at y. If there exists a ball inside Ω touching ∂Ω at y, then ∂Ω is differentiable at y. In particular, ∂Ω_{r} is everywhere differentiable.

(c) For any x ∈ Ω we have

(22) dist(x, ∂Ω_{r}) = dist(x, ∂Ω) + r.

If x ∈ Ω_{r} \ \overline{Ω} then dist(x, ∂Ω_{r}) < r and x ∉ R_{r}, where R_{r} is ridge set of Ω_{r}. (d) Sets Ω and Ω_{r} have the same ridge set, i.e., R = R_{r}.

Proof. (a) Let x ∈ R^{n} \ \overline{Ω}_{r}. Then dist(x, Ω) = R > r. Let y ∈ ∂Ω be a point such that | x − y |= R. Let y_{r} be such point of the interval connecting x and y that | y − y_{r} |= r. Then y_{r} ∈ \overline{Ω}_{r}.

Consider first the case R ≥ 2r. Then dist(x, Ω_{r}) ≥ r. Since if this is not true then there exists such point z_{1} ∈ \overline{Ω}_{r} that | x − z_{1} |< r, and there exists z_{2} ∈ \overline{Ω} such that | z_{1} − z_{2} |≤ r. Now we get R ≤| x − z_{2} |≤| x − z_{1} | + | z_{1} − z_{2} |< 2r, a contradiction. So γ_{R^{n}\\overline{Ω}_{r}}(x) ≥ dist(x, Ω_{r}) ≥ r.

Consider now the case r < R < 2r. Then by Condition 3 we have γ_{R^{n}\\overline{Ω}}(x) ≥ 2r and so R = dist(x, Ω) < γ_{R^{n}\\overline{Ω}}(x). Thus y is the unique point on ∂Ω such that | x − y |= dist(x, Ω). Let z = y + 2r\frac{x−y}{|x−y|}. Then from the condition 3 and [EH87], Lemma 3.4, it follows that for any x̃ of the form tx + (1 − t)z where 0 < t < 1 the point y is the unique point on ∂Ω such that dist(x̃, ∂Ω) =| x̃−y |. Then for each such point x̃, the point y_{r} is the unique point on ∂Ω_{r} such that dist(x̃, ∂Ω_{r}) =| x̃ − y_{r} |. For suppose on the contrary that there exists z_{1} ∈ ∂Ω_{r} such that z_{1} ≠ y_{r} and | x̃ − z_{1} |≤| x̃ − y_{r} |. Then there exists z_{2} ∈ ∂Ω such that | z_{1} − z_{2} |≤ r. Consider first the case z_{2} ≠ y. Then

| x̃ − y |=| x̃ − y_{r} | + | y_{r} − y |=| x̃ − y_{r} | +r ≥| x̃ − z_{1} | + | z_{1} − z_{2} |≥| x̃ − z_{2} |,

a contradiction with the fact that y is the unique nearest to x̃ point on ∂Ω. Consider now the case z_{2} = y. Since the ball of radius R and center x̃ lies outside Ω and |z_{1} − y| ≤ r, it follows that z_{1} does not lie on the interval connecting x̃ and y, and so

| x̃ − y |<| x̃ − z_{1} | + | z_{1} − y |≤| x̃ − y_{r} | +r =| x̃ − y |,

a contradiction. So, y_{r} is the unique nearest to x̃ point on ∂Ω_{r} for any x̃ on the interval connecting z and y_{r}, so γ_{R^{n}\\overline{Ω}_{r}}(x) ≥| z − y_{r} |= r.

(b) Let y_{r} ∈ ∂Ω_{r}. Then dist(y_{r}, Ω) = r. Thus there exists y ∈ ∂Ω such that | y − y_{r} |= r. By Condition 3 , γ_{R^{n}\\overline{Ω}}(y_{r}) ≥ 2r > dist(y_{r}, ∂Ω), and thus by [EH87], Lemma 3.4, y is the unique point of ∂Ω nearest to y_{r}.

Let y ∈ ∂Ω. Let y_{i} ∈ R^{n} \ \overline{Ω} for i = 1, 2, ..., and let y_{i} → y. By condition 3, γ_{R^{n}\\overline{Ω}}(y_{i}) ≥ 2r, and by uppersemicontinuity of γ_{R^{n}\\overline{Ω}}(·) we get γ_{R^{n}\\overline{Ω}}(y) ≥ 2r. Thus we have shown that every point of ∂Ω has an exterior tangent ball of the radius 2r. So if a point y of ∂Ω has an interior tangent ball, then ∂Ω is differentiable at y.

Let y ∈ ∂Ω. Then, as we have shown, γ_{R^{n}\\overline{Ω}}(y) ≥ 2r. Thus there exists a distance ray R_{y} of the set R^{n} \ \overline{Ω} starting at y such that the length of R_{y} is at least 2r. Let e be the unit vector in the direction of R_{y}, let y_{r} = y + re. Then N_{∂Ω}(y_{r}) = {y} by Remark 7. Since | y − y_{r} |= r we get y_{r} ∈ ∂Ω_{r}.

We have shown that the interval connecting y and y_{r} lies on the distance ray of the set R^{n} \ \overline{Ω} starting at y, and on distance ray of the set Ω_{r} starting at y_{r}. It follows that the vector y − y_{r} is orthogonal to ∂Ω_{r} at y_{r} and also to ∂Ω at y if ∂Ω is differentiable at y.

(c) Let x ∈ Ω^{0}, y ∈ ∂Ω, and |x − y| = dist(x, ∂Ω). Then, by (b),

y_{r} := y + \frac{r}{|y − x|}(y − x) ∈ ∂Ω_{r}.

Clearly,

|x − y_{r}| = |x − y| + r = d_{Ω}(x) + r.

Assume that there exists z_{r} ∈ ∂Ω_{r} such that

|x − z_{r}| < d_{Ω}(x) + r.

Let z be the point of intersection of ∂Ω with the interval connecting z_{r} and x (such point exists since x ∈ Ω, z_{r} ∉ \overline{Ω}). Then |z − z_{r}| ≥ r and so

|x − z| < d_{Ω}(x).

This contradicts the fact that z ∈ ∂Ω. Thus y_{r} ∈ N_{∂Ω_{r}}(x) and thus (22) is proved. It remains to prove the last assertion of the statement (c). Let x ∈ Ω_{r}\\overline{Ω} . Let y_{r} ∈ N_{∂Ω_{r}}(x). Let R_{y_{r}} be the ray orthogonal to ∂Ω_{r} at y_{r} and

let y = R_{y_{r}} ∩ ∂Ω. Then by (b) we have the following: x ∈ R_{y_{r}}, x lies on the interval connecting y and y_{r}, and |y − y_{r}| = r = dist(y, ∂Ω_{r}). Then it follows from [EH87], Lemma 3.4 that x ∉ R_{r}. Also, dist(x, ∂Ω_{r}) =| x − y_{r} |< |y − y_{r}| = r.

(d) We show first that R ⊂ R_{r}. If x ∈ Ω and B_{ρ}(x) ⊂ Ω, then B_{ρ+r}(x) ⊂ \overline{Ω}_{r}. Let x ∈ R\∂Ω. Let y ∈ N_{∂Ω}(x) and let y_{r} = y + r\frac{y−x}{|y−x|}. Then by (b),(c) we see that y_{r} ∈ ∂Ω_{r}, and |x − y_{r}| = dist(x, ∂Ω) + r = dist(x, ∂Ω_{r}). Let x̃ = y + (1 + ε)(x − y) where ε > 0 is small enough so that x̃ ∈ Ω. Since x ∈ R, it follows that dist(x̃, ∂Ω) < |x̃ − y| = (1 + ε)dist(x, ∂Ω). Let ỹ ∈ ∂Ω be such point that dist(x̃, ∂Ω) = |x̃ − ỹ|, let ỹ_{r} = ỹ + r\frac{ỹ−x̃}{|ỹ−x̃|}, then ỹ_{r} ∈ ∂Ω_{r} and |x̃ − ỹ_{r}| = dist(x̃, ∂Ω_{r}). Then

dist(x̃, ∂Ω_{r}) =| ỹ − x̃ | +r < (1 + ε)dist(x, ∂Ω) + r = |x̃ − y_{r}|,

so γ_{Ω_{r}}(x) = |x − y_{r}| = d_{Ω_{r}}(x). Thus x ∈ R_{r}.

Let x ∈ R ∩ ∂Ω, i.e., there does not exist a ball inside Ω that touches ∂Ω at x. Let B_{r}(y_{r}) be a ball of radius r and center at such point y_{r} that \overline{B}_{r}(y_{r}) ∩ \overline{Ω} = \{x\}, such ball exists by Condition 3. Then dist(y_{r}, ∂Ω) = r, and by (b) x is the unique point of ∂Ω such that |x − y_{r}| = r. Let x̃ = y_{r} + (1 + ε)(x − y_{r}) for ε ∈ (0, \frac{r}{2}). Given ε, there are two possibilities: either x̃ ∈ Ω or x̃ ∉ Ω.

Let x̃ ∈ Ω. Then, since x ∈ R ∩ ∂Ω, dist(x̃, ∂Ω) < |x − x̃| = εr. Let ỹ ∈ ∂Ω be such that |x̃ − ỹ| = dist(x̃, ∂Ω), let ỹ_{r} = ỹ + r\frac{ỹ−x̃}{|ỹ−x̃|}. Then ỹ_{r} ∈ ∂Ω_{r} and by (c) we get |x̃ − ỹ_{r}| = dist(x̃, ∂Ω) + r = dist(x̃, ∂Ω_{r}), and so dist(x̃, ∂Ω_{r}) < (ε + 1)r = |x̃ − y_{r}|, so y_{r} ∉ N_{∂Ω_{r}}(x̃).

Let now x̃ ∉ Ω, i.e., x̃ ∈ Ω_{r} \ \overline{Ω} since ε < \frac{r}{2}. Then, by (c), dist(x̃, ∂Ω_{r}) < r, but |x̃ − y_{r}| = (ε + 1)r, so y_{r} ∉ N_{∂Ω_{r}}(x̃).

Thus y_{r} ∉ N_{∂Ω_{r}}(x̃) for any small ε > 0. This implies x ∈ R_{r}.

Thus R ⊂ R_{r}.

Now we show that R_{r} ⊂ R. By (b), (c) we get R_{r} ∩ (\overline{Ω}_{r} \ \overline{Ω}) = ∅.

It remains to consider x ∈ \overline{Ω} ∩ R_{r}. Let first x ∈ Ω ∩ R_{r}. Let as above y ∈ N_{∂Ω}(x), y_{r} = y + r\frac{y−x}{|y−x|} ∈ N_{∂Ω_{r}}(x), x̃ = y + (1 + ε)(x − y) where ε > 0 is small. From x ∈ R_{r} we conclude that dist(x̃, ∂Ω_{r}) < |x̃ − y_{r}| = (1 + ε)dist(x, ∂Ω) + r. Let ỹ_{r} ∈ ∂Ω_{r} be such that |x̃ − ỹ_{r}| = dist(x̃, ∂Ω_{r}) and let ỹ be the unique point of ∂Ω nearest to ỹ_{r}, then |ỹ − ỹ_{r}| = r (uniqueness of ỹ follows from (b)). Then by (c), (b) the points x̃, ỹ, ỹ_{r} lie on one distance ray and dist(x̃, ∂Ω) = |x̃ − ỹ|, so

dist(x̃, ∂Ω) = |x̃ − ỹ| = |ỹ_{r} − x̃| − |ỹ_{r} − y| < |x̃ − y_{r}| − r =
(1 + ε)dist(x, ∂Ω) = (1 + ε)|x − y| = |x̃ − y|

and so x̃ ∈ R. The case of x ∈ ∂Ω ∩ R_{r} is similar to the case above.

Remark 10. Examples of nonconvex polygons on a plane show that without assuming some condition of the type of Condition 3 the statements (b)-(d) of Proposition 9 are not true in general.

Proposition 11. Let Ω ∈ R^{n} satisfy Condition 3 with radius r. Then ∂Ω_{ρ} ∈ C^{1,1} for any ρ ∈ (0, \frac{r}{M}).

Proof. The assertion follows from inequality 4.8(8) of [Fed59] and Proposition 9 (d). Also, inequality (96) proved in Appendix A1 can be applied instead of 4.8(8) of [Fed59].

Recall the following facts (see e.g. [GT83], Lemma 14.16, 14.17). Let Ω ⊂ R^{n} be an open bounded set with C^{2} boundary ∂Ω. Let Ω_{r} be, as above, r-neighborhood of Ω. Then ∂Ω_{r} ∈ C^{2} for small r. If y ∈ ∂Ω and if y_{r} ∈ ∂Ω_{r} is the unique point on ∂Ω_{r} such that |y − y_{r}| = r then the principal coordinate systems of ∂Ω at y and of ∂Ω_{r} at y_{r} are parallel. Denote by κ_{i} and κ_{i,r}, (i = 1, ...n − 1) the principal curvatures of ∂Ω at y and of ∂Ω_{r} at y_{r} respectively. We have

κ_{i,r} = \frac{κ_{i}}{1 + κ_{i}r} for i = 1, ..., n − 1. (23)

In the next two propositions we show similar facts for sets with nonsmooth boundaries satisfying Condition 3 and for Lipschitz families of such sets.

Proposition 12. Let Ω ⊂ R^{n} be an open set satisfying Condition 3 with radius r_{0}. Let 0 < r ≤ r_{0}. Fix y ∈ ∂Ω. Let y_{r} ∈ ∂Ω_{r} be such point that |y − y_{r}| = r. Then:

a) Let ∂Ω be twice differentiable at y, i.e., in a suitable coordinate system in R^{n} we have y = 0, Ω ∩ B_{δ}(0) = {x_{n} > Ψ(x′)} ∩ B_{δ}(0) for small δ > 0 and Ψ(x′) = ∑_{i=1}^{n−1} κ_{i}x_{i}^{2} + o(|x′|^{2}). Then the point y_{r} is unique given the point y, and in the coordinate system introduced above y_{r} = (0, −r), Ω_{r} ∩ B_{δ_{1}}(y_{r}) = {x_{n} > Ψ_{r}(x′)} ∩ B_{δ_{1}}(y_{r}) for sufficiently small δ_{1} > 0, where Ψ_{r}(x′) = −r + ∑_{i=1}^{n−1} \frac{κ_{i}}{1+κ_{i}r}x_{i}^{2} + o(|x′|^{2}), i.e., Ω_{r} is twice differentiable at y_{r} and (23) holds.

b) If ∂Ω_{r} is twice differentiable at y_{r} and γ_{Ω}(y) > 0 then ∂Ω is twice differentiable at y and (23) holds (i.e., the functions Ψ and Ψ_{r} have same expansions at 0 as in a)).

Proof. a) Uniqueness of y_{r} and the fact that y_{r} = (0, −r) follow from smoothness of ∂Ω at y. So the interval connecting y and y_{r} lies on x_{n}-axis. By Propositions 9 and 11 the surface ∂Ω_{r} is of class C^{1,1}, and the interval connecting y and y_{r} is orthogonal to ∂Ω_{r}. So Ω_{r} ∩ B_{δ_{1}}(y_{r}) = {x_{n} > Ψ_{r}(x′)} ∩ B_{δ_{1}}(y_{r}) for some δ_{1} > 0, where Ψ_{r} is a C^{1,1} function.

Let ε > 0. There exists σ > 0 such that

|Ψ(x′) − ∑_{i=1}^{n−1} κ_{i}x_{i}^{2}| < ε|x′|^{2} if |x′| < σ.

Denote A_{ε} = {(z′, z_{n}) ∈ R^{n} | z_{n} ≥ ∑_{i=1}^{n−1}(κ_{i} − ε)z_{i}^{2}}. Then we get:

∂Ω ∩ B_{σ}(0) ⊂ A_{ε}\A_{−ε}.

Since γ_{R^{n}\\overline{Ω}}(y) ≥ 2r, the point y_{r} lies in the relative interior of the distance ray R_{y} of the set R^{n} \ \overline{Ω}. Then there exists σ_{1} > 0 such that for any point x_{r} ∈ ∩B_{σ_{1}}(y_{r}) we have N_{∂Ω}(x_{r}) ⊂ ∂Ω∩B_{σ}(0). (Proof: suppose this is false, then there exist σ_{1} > 0 and a sequence x_{r}^{k} → y_{r} such that for x^{k} ∈ N_{∂Ω}(x_{r}^{k}) we get |x^{k} − y| > σ_{1}. Then, passing to a subsequence, we get x^{k} → x where x ∈ ∂Ω and x ≠ y. By continuity of distance function we get x ∈ N_{∂Ω}(y_{r}) and x ≠ y. By Condition 3 y_{r} lies in the relative interior of the distance ray R_{y} of the set R^{n} \ \overline{Ω}. Thus y_{r} has a unique nearest point on ∂Ω. A contradiction.). So

(24) x ∈ A_{ε}\A_{−ε} if x ∈ N_{∂Ω}(x_{r}) where x_{r} ∈ B_{σ_{1}}(y_{r}).

Let N_{ε} be r-neighborhood of the set A_{ε}. By Remark 4 we have κ_{i} ≥ −\frac{1}{2r}. Let ε < \frac{1}{4r}. Then (by [GT83] Lemma 14.16, 14.17), ∂N_{ε} is the graph of a C^{∞} function g_{ε}, and \frac{∂^{2}g_{ε}}{∂x_{i}∂x_{j}} (0) = \frac{κ_{i}−ε}{1+(κ_{i}−ε)r}δ_{ij}. Let

Ñ_{ε,δ} = \{(z′, z_{n}) | z_{n} ≥ ∑_{i=1}^{n−1}(\frac{κ_{i} − ε}{1 + (κ_{i} − ε)r} − δ)z_{1}^{2} − r\}.

If σ_{2} > 0 is small enough then

Ñ_{±ε,−ε} ∩ B_{σ_{2}}(y_{r}) ⊂ N_{±ε} ∩ B_{σ_{2}}(y_{r}) ⊂ Ñ_{±ε,ε} ∩ B_{σ_{2}}(y_{r}). (25)

Thus if we choose σ_{3} small, then we get by (24), (25)

∂Ω_{r} ∩ B_{σ_{3}}(y_{r}) ⊂ Ñ_{ε,ε}\Ñ_{−ε,−ε},

which means that for x_{r} = (x′_{r}, Ψ_{r}(x′_{r})) ∈ ∂Ω_{r} ∩ B_{σ_{3}}(y_{r}) we have

∑_{i=1}^{n−1}(\frac{κ_{i} − ε}{1 + (κ_{i} − ε)r} − ε)x_{r,i}^{2} − r ≤ Ψ_{r}(x′_{r}) ≤ ∑_{i=1}^{n−1}(\frac{κ_{i} + ε}{1 + (κ_{i} + ε)r} + ε)x_{r,i}^{2} − r.

Sending ε to 0 we conclude the proof.

b) By Proposition 9 the condition γ_{Ω}(y) > 0 implies that γ_{Ω_{r}}(y_{r}) > r. Now we can perform calculation similar to the proof of assertion a).

We need similar facts for families of sets. Let {Ω_{t}} be a family of open sets in R^{n}. Let (Ω_{t})_{r} be r-neighborhood of Ω_{t} in R^{n}. Define

E_{r} = ∪_{t}[(Ω_{t})_{r} × {t}], Γ_{r} = ∪_{t}[∂(Ω_{t})_{r} × {t}]. (26)

Proposition 13. Let {Ω_{t}} be a locally Lipschitz continuous family of open sets in R^{n}. Let for every t ∈ R_{+}^{1} the set Ω_{t} satisfy Condition 3 with radius r_{0} > 0. Let 0 < r ≤ r_{0}. Fix y ∈ ∂Ω_{t_{0}}. Let y_{r} ∈ ∂(Ω_{t_{0}})_{r} be such point that |y − y_{r}| = r. Then:

a) Let Γ, defined by (5), be (2,1) differentiable at (y, t_{0}). That is in a suitable coordinate system (x_{1}, ..., x_{n}) in R^{n} we have

y = 0, Ω ∩ B_{δ}^{n+1}(0) = {x_{n} > Ψ(x′, t)} ∩ B_{δ}^{n+1}(0)

for small δ > 0 and

Ψ(x′, t) = v(t − t_{0}) + ∑_{i=1}^{n−1} κ_{i}x_{1}^{2} + o(|t − t_{0}| + |x′|^{2}).

Then the point y_{r} is unique given the point y, and in the coordinate system introduced above y_{r} = (0, −r),

Ω_{r} ∩ B_{δ_{1}}^{n+1}(y_{r}) = {x_{n} > Ψ_{r}(x′, t)} ∩ B_{δ_{1}}^{n+1}(y_{r})

for sufficiently small δ_{1} > 0, where

Ψ_{r}(x′, t) = −r + v(t − t_{0}) + ∑_{i=1}^{n−1} \frac{κ_{i}}{1 + κ_{i}r}x_{1}^{2} + o(|t − t_{0}| + |x′|^{2}).

b) If Γ_{r} is (2,1) differentiable at (y_{r}, t_{0}) and γ_{Ω_{t_{0}}}(y) > 0 then Γ is (2,1) differentiable at y and the functions ψ and ψ_{r} have same expansions at (0, t_{0}) as in a).

Proof. The proof is similar to the one of Proposition 12. In particular, the sets A_{ε}, N_{ε} and Ñ_{ε,δ} are defined as following.

A_{ε} = { (z′, z_{n}, t) | z_{n} ≥ v(t − t_{0}) + ∑_{i=1}^{n−1}(κ_{i} − ε)z_{i}^{2} − ε|t − t_{0}| }.

N_{ε} is defined as following: for each t = t^{∗} the set N_{ε} ∩ {t = t^{∗}} is ε-neighborhood in R^{n} of the set {x | (x, t^{∗}) ∈ A_{ε}}. Finally,

Ñ_{ε,δ} = \{ (z′, z_{n}) | z_{n} ≥ v(t − t_{0}) + ∑_{i=1}^{n−1}(\frac{κ_{i} − ε}{1 + (κ_{i} − ε)r} − δ)z_{1}^{2} − r − (ε + δ)|t − t_{0}| \}.

The rest of the argument does not change.

To derive further properties of the distance function of a set Ω satisfying the Condition 3, we need to integrate over such sets. To do this it is convenient to decompose the set by suitable subsets, and define a local coordinate system in each subset. The construction is carried out below and can be roughly described as following. The subsets are unions of distance rays passing through subsets of the boundary. The coordinate systems consist of variables x_{1}, ..., x_{n−1} on the boundary, and the variable x_{n} along the distance rays. However, since ∂Ω is not smooth enough, we have to use ∂Ω_{r}. Now we turn to the construction.

By Proposition 11 the set Ω_{r} for small enough r > 0 has C^{1,1} boundary. Fix such r. We can choose sets U_{1}, ..., U_{N}, bounded and open in R^{n}, so that ∂Ω_{r} ⊂ ∪_{k=1}^{N}U_{k}, and for each k = 1, ..., N in a suitable coordinate system in R^{n} we have

Ω_{r} ∩ U_{k} = { (x′, x_{n}) | x′ ∈ Ũ_{k}, x_{n} > Φ_{k}(x′) } ∩ U_{k},

where Ũ_{k} = { x′ | (x′, x_{n}) ∈ U_{k} } ⊂ R^{n−1} is an open set, and Φ_{k} is a C^{1,1} function on R^{n−1}.

We can also choose open sets U_{0} and U_{N+1} ⊂ R^{n} such that U_{0} ⊂ Ω_{r}, U_{N+1} ⊂ R^{n}\Ω_{r}, both U_{0} and U_{N+1} do not intersect ∂Ω, and R^{n} = ∪_{k=0}^{N+1}U_{k}. Let Ψ̃_{k}, k = 1, ..., N, be a smooth partition of unity on R^{n} related to the sets U_{0}, U_{1}, ..., U_{N+1}, i.e., ∑_{k=0}^{N+1} Ψ̃_{k} ≡ 1 on R^{n} and Ψ̃_{k} ∈ C_{0}^{∞}(U_{k}). Then by our choice of U_{0} and U_{N+1} we get ∑_{k=1}^{N} Ψ̃_{k} ≡ 1 on ∂Ω_{r}.

Define maps G_{k} : Ũ_{k} × R^{1} → R^{n} (k = 1, ..., N) by

G_{k}(x′, x_{n}) = y + x_{n}Dd_{Ω_{r}}^{s}(y) (27)

where y = (x′, Φ_{k}(x′)) ∈ ∂Ω_{r}, and d_{Ω_{r}}^{s}(·) is the signed distance to ∂Ω_{r}.

For each z ∈ ∂Ω_{r} the vector Dd_{Ω_{r}}^{s}(z) is the inner unit normal to ∂Ω_{r} in z. Since ∂Ω_{r} is a C^{1,1} manifold, Dd_{Ω_{r}}^{s} is Lipschitz on ∂Ω_{r}. Then since Φ_{k} is C^{1,1} on R^{n−1}, the map G_{k} is Lipschitz on bounded subsets of Ũ_{k} × R^{1}. In particular, the map Ũ_{k} → ∂Ω_{r} defined by x′ → G_{k}(x′, 0) is Lipschitz.

From that by explicit computation we get the following

Lemma 14. Let the map G_{k} : Ũ_{k} × R^{1} → R^{n} be defined by (27). Then the Jacobian JG_{k} is a locally bounded measurable function. For L^{n−1} a.e. point x′ ∈ Ũ_{k} the surface ∂Ω_{r} is twice differentiable at the point G_{k}(x′, 0). For every such point x′ the map G_{k} is differentiable at (x′, x_{n}) for every x_{n} ∈ R^{1}. At every such point (x′, x_{n}) the Jacobian of G_{k} is given by

J_{n}G_{k}(x′, x_{n}) = \sqrt{1 + |DΦ_{k}(x′)|^{2}} ∏_{i=1}^{n−1} [1 − κ_{r,i}x_{n}] , (28)

where κ_{r,1}, ..., κ_{r,n−1} are the principal curvatures of ∂Ω_{r} at G_{k}(x′, 0).

Denote by U_{k} the sets

U_{k} = ∪_{x′∈Ũ_{k}}{ (x′, x_{n}) | x_{n} ∈ [0, γ_{Ω_{r}}(y)) } ⊂ R^{n}, (29)

where γ_{Ω_{r}} : R^{n} → R^{1} is the ridge function (21) of the set Ω_{r}. Note that γ_{Ω_{r}}(z) ≥ r for all z ∈ \overline{Ω}_{r} by Proposition 9 (c, d).

The map G_{k} is one-to-one on U_{k} (by definition of γ_{Ω_{r}} and Lemma 3.4 of [EH87]). In addition Ω\R ⊂ ∪_{k=1}^{N}G_{k}(U_{k}), since every point of Ω_{r}\R lies inside some distance ray. For each k = 1, ..., N define a function Ψ_{k} : R^{n} → R^{1} by extending the function Ψ̃_{k} from the boundary inside Ω_{r} as a constant along distance rays and defining Ψ_{k} to be the 0 on the ridge of Ω_{r} and outside Ω_{r}, i.e.,

(30) Ψ_{k}(z) = \{ \begin{matrix}Ψ̃_{k}(y), & where y ∈ N_{∂Ω_{r}}(z), & if z ∈ Ω_{r} \ R \\ 0, & & otherwise. \end{matrix} Note that y is uniquely defuned by z in the first case of (30). We have

suppΨ_{k} ⊂ G_{k}(U_{k}); ∑_{k=1}^{N} Ψ̃_{k}(y(z)) ≡ 1 for all z ∈ Ω_{r}\R.

We thus decomposed the set Ω_{r} \ R by the sets G_{k}(U_{k}) and defined a related partition of unity Ψ_{k}. Note that in general the sets U_{k} and G_{k}(U_{k}) are neither open nor closed, and the functions Ψ_{k} are discontinuous. In the next propositions we prove that these sets and functions are measurable.

Proposition 15. Define a function γ̃_{r,k} : R^{n−1} → R^{1} by

(31)  γ̃_{r,k}(x′) = \{ \begin{matrix}γ_{Ω_{r}}(G_{k}(x′, 0)), & if x′ ∈ cl(Ũ_{k}) \\ 0, & if x′ ∉ cl(Ũ_{k}) \end{matrix}

where cl(·) denotes closure of a set. Then γ̃_{r,k} is uppersemicontinuous (usc). The sets U_{k} are L^{n} measurable. The ridge set of Ω_{r} has L^{n} measure 0.

Proof. It is enough to show that for any sequence {x′_{m}}_{m=1}^{∞} such that x′_{m} ∈ cl(Ũ_{k}), x′_{m} → x′ and γ̃_{r,k}(x′_{m}) → L we have γ_{r,k}(x′) ≥ L.

The function γ_{Ω_{r}} is usc on R^{n}. Since x ∈ cl(Ũ_{k}) and G_{k} is Lipschitz on cl(Ũ_{k}) × 0),

γ̃_{r,k}(x′) = γ_{Ω_{r}}(G_{k}(x′, 0)) ≥ lim_{m→∞} γ_{Ω_{r}}(G_{k}((x′_{m})).

Thus, γ̃_{r,k} is usc.

We have U_{k} = { (x′, x_{n}) | x′ ∈ Ũ_{k}, 0 ≤ x_{n} < γ̃_{r,k}(x′) }. Thus U_{k} is L^{n}-measurable (note that Ũ_{k} is open in R^{n−1}).

To prove that the ridge has L^{n} measure 0, we note that G_{k}^{−1}(R)∩\overline{U}_{k} = \{(x′, x_{n})|x′ ∈ Ũ_{k}, x_{n} = γ̃_{r,k}(x′)\}. Then L^{n}(G_{k}^{−1}(R) ∩ \overline{U}_{k}) = 0 because γ̃_{r,k}(·) is usc. Since G_{k} is Lipschitz on cl(Ũ_{k}) × [0, diamΩ_{r}] and since ∪_{k=1}^{N}\overline{U}_{k} = \overline{Ω}_{r}, we have L^{n}(R) = 0.

We will use repeatedly the following fact:

Lemma 16. Let f : A ⊂ R^{m} → R^{n}(m ≤ n) be a Lipschitz function, A be a L^{m}- measurable set, and let g : R^{m} → R^{1} be a L^{m}-measurable function. Then the function g ∘ f^{−1} : R^{n} → R^{1} is L^{n}-measurable.

Proof. We can assume that g ≡ 0 outside A. Let V ⊂ R^{1} ∪ {+∞} be an open set. Then W = g^{−1}(V) ∩ A is L^{m}-measurable. Then (g ∘ f^{−1})^{−1}V = f(g^{−1}(U)) ∩ A = f(W), which is a measurable set since f is Lipschitz and m ≤ n (see [EGar92], Lemma 2 of 3.3.1).

Proposition 17. Functions Ψ_{k} and Ψ_{k} ∘ G_{k} are L^{n}-measurable.

Proof. The functions G_{k} are Lipschitz. Then by Lemma 16 it is enough to prove that functions Ψ_{k} ∘ G_{k} are L^{n}-measurable (since then Ψ_{k} = (Ψ_{k} ∘ G_{k}) ∘ G_{k}^{−1}). Let (x′, x_{n}) ∈ U_{k}. Then

Ψ_{k}(G_{k}(x′, x_{n})) = Ψ̃_{k}(G_{k}(x′, 0)) = Ψ̃_{k}(x′, Φ_{k}(x′)).

Consider the function Θ(x′, x_{n}) = Ψ̃_{k}(x′, Φ_{k}(x′)) . Since Ψ̃_{k} ∈ C^{∞}(R^{n}) and Φ_{k} ∈ C^{1,1}(R^{n−1}), we see that Θ is a C^{1,1} function R^{n} → R^{1}. Now since Ψ_{k} ∘ G_{k} = Θ on U_{k}, Ψ_{k} ∘ G_{k} ≡ 0 outside U_{k}, and by Proposition 15 the set U_{k} is measurable, the assertion follows.

Proposition 18. Let Ω be a bounded open set satisfying the Condition 3. Let X ⊂ ∂Ω and H^{n−1}(X) = 0. Let Y be the union of distance rays passing through X, i.e., Y = ∪_{x∈X}R_{x}. Then L^{n}(Y) = 0.

Proof. We can assume that X ⊂ G_{k}(U_{k}) ∩ ∂Ω for some k, and then Y ⊂ G_{k}(U_{k}). We also can assume that

γ_{Ω}(x) > 0 for all x ∈ X. (32)

If (32) is not true then we can replace X by X ∩ {x | γ_{Ω}(x) > 0} and Y does not change.

Since γ̃_{r,k}(x′) is a u.s.c. function of x′, we get L^{n}({x_{n} = γ_{r,k}(x′, 0)}) = 0. We also know that J_{n}G_{k} > 0 on the set U_{k}. The map G_{k} is one-to-one on U_{k}, and so N(G_{k}, y) = 1 for y ∈ Y ⊂ G_{k}(U_{k}), where N(f, z) denotes the multiplicity function, which is the number of elements of f^{−1}(z). Then by area formula we obtain

L^{n}(Y) = ∫_{R^{n}} χ_{Y} (y)dy = ∫_{U_{k}} χ_{Y} (G_{k}(x))J_{n}G_{k}(x)dx, (33)

where χ_{A}(·) is the characteristic function of the set A, i.e., χ_{A}(·) equals 1 on A and 0 outside A. By Lemma 14 the Jacobian J_{n}G_{k} is a locally bounded measurable function. Let x′ ∈ Ũ_{k} be such point that the surface ∂Ω_{r} is twice differentiable at the point G_{k}(x′, 0). By Lemma 14 the Jacobian J_{n}G_{k} is defined by the expression (28) at (x′, x_{n}) for every x_{n} ∈ R^{1}. The principal curvatures of ∂Ω_{r} at the point G_{k}(x′, 0) satisfy

κ_{r,i} > −\frac{1}{r}, κ_{r,i}γ̃_{r,k}(x′) ≤ 1 (34)

by definition of γ̃_{r,k}(·), and by Condition 3. We also have

γ̃_{r,k}(x′) ≤ diamΩ_{r}. (35)

Let m ≤ n be such that −\frac{1}{r} ≤ κ_{r,1}, ..., κ_{r,m} < 0 and κ_{r,m+1}, ..., κ_{r,n−1} ≥ 0. Then for x_{n} ∈ (r, γ̃_{r,k}(x′)) we calculate using (34), (35):

0 ≤ \frac{J_{n}G_{k}(x′, x_{n})}{J_{n}G_{k}(x′, r)} ≤ \begin{matrix}∏ & {m \atop i=1} & [1 − κ_{r,i}x_{n}] \\ ∏ & {m \atop i=1} & [1 − κ_{r,i}r] \end{matrix} ≤ [1 + \frac{1}{r}γ̃_{r,k}(x′)]^{m} ≤ C, (36)

where C depends on r, n and diam(Ω). By Lemma 14 the inequalities (36) hold for a.e. x ∈ U_{k} ∩ {(x′, x_{n}) | x_{n} > r}.

Define a map G̃_{k} : Ũ_{k} → ∂Ω ⊂ R^{n} by G̃_{k}(x′) = G_{k}(x′, r). Then by (28)

J_{n−1}G̃_{k}(x′) = J_{n}G_{k}(x′, r) for a.e. x′ ∈ Ũ_{k}. (37)

So using the fact that the relation Y ⊂ \overline{Ω} implies χ_{Y} (G_{k}(x′, x_{n})) = 0 for x_{n} < r, we get from (33), (35), (36) and (37) using area formula:

L^{n}(Y) ≤ C ∫_{U_{k}} χ_{Y} (G_{k}(x)) J_{n−1}G̃_{k}(x′)dx

= C ∫Ũ_{k} J_{n−1}G̃_{k}(x′)[∫_{r}^{γ̃_{r,k}(x′)} χ_{Y} (G_{k}(x′, x_{n}))dx_{n}]dx′
= C ∫Ũ_{k} χ_{X}(G̃_{k}(x′))[γ̃_{r,k}(x′) − r] · J_{n−1}G̃_{k}(x′)dx′
≤ C ∫Ũ_{k} χ_{X}(G̃_{k}(x′)) J_{n−1}G̃(x′)dx′
= C ∫_{∂Ω} χ_{X}(z)N(G̃_{k}, z)dH^{n−1}(z) = CH^{n−1}(X) = 0,

where we have used the fact that N(G̃_{k}, z) = 1 for z ∈ X (which follows from (32) and Proposition 9 (b,c) ). Remark 19. Without assuming a condition of the type of Condition 3 the assertion of Proposition 18 is not true. For example consider a nonconvex polygon in R^{2} and take X to be one point, a vertex of a re-enterant corner. Remark 20. Note that it follows from the proof above that

L^{n}(G_{k}^{−1}(Y)) = ∫Ũ_{k} J_{n−1}G̃_{k}(x′)[∫_{r}^{γ̃_{r,k}(x′)} χ_{Y} (G_{k}(x′, x_{n}))dx_{n}]dx′ = 0.

The following proposition describes the structure of the boundary of a set that satisfies Condition 3. Proposition 21. Let Ω be a bounded set. Let Ω satisfy the condition 3 with radius r_{0}. Then ∂Ω is (H^{n−1}, n − 1) rectifiable subset of R^{n}. In addition, ∂Ω is twice differentiable H^{n−1} a.e. on the set

B = {x ∈ ∂Ω | γ_{Ω}(x) > 0}. (38)

Proof. Let r > 0 be such number that ∂Ω_{r} is C^{1,1}. Existence of such r follows from Proposition 11. It follows from Proposition 9 b) that the nearest point projection mapping P : Ω_{r} → Ω is well-defined and onto. Moreover, by inequality 4.8(8) of [Fed59] (or by inequality (95) proved in Appendix A below) the map P is Lipschitz. Since ∂Ω_{r} is C^{1,1}, we get that ∂Ω is (H^{n−1}, n − 1) rectifiable.

By Proposition 9 every point of the set B has a unique nearest point on ∂Ω_{r}. Since ∂Ω_{r} is C^{1,1}, it follows that ∂Ω_{r} is twice differentiable H^{n−1} a.e. Since the map P is Lipschitz it follows that for H^{n−1} a.e. y ∈ B the surface ∂Ω_{r} is twice differentiable at the corresponding point y_{r}. Applying Proposition 12 b), we conclude the proof.

Propositions 15, 18 and 21 imply Corollary 22. Let Ω ⊂ R^{n} satisfy Condition 3. Then L^{n} a.e point of Ω lies in the relative interior of a distance ray that intersects ∂Ω at a point at which ∂Ω is twice differentiable.

We need an analogue of the Proposition 21 for a Lipschitz family {Ω_{t}} of sets that satisfy Condition 3 with a radius r_{0} independent of t. Let E, Γ be defined by (5). Let E_{r}, Γ_{r} be defined by (26). Denote

D(x, t) = d_{Ω_{t}}^{s}(x), (39)

where d_{Ω_{t}}^{s}(x) is the signed distance function (2).

Proposition 23. Let {Ω_{t}} be a locally Lipschitz family of open bounded sets that satisfy the Condition 3 with radius r_{0} independent of t. Then

a) L^{n+1} a.e. (x_{0}, t_{0}) ∈ E lies in the relative interior of a distance ray of the set Ω_{t_{0}} that intersects ∂Ω_{t_{0}} at such point y ∈ ∂Ω_{t_{0}} that Γ is (2,1) differentiable at (y, t_{0});

b) There exists a constant C depending only on r_{0} such that the function D(x, t) − C|x|^{2} on the set E_{r_{0}} is concave in x for every t.

Proof. First we prove the following fact.

Lemma 24. Let Ω satisfies Condition 3 with radius r_{0}. Then the distance function d_{Ω}(x) is semiconcave in Ω, i.e. there exists a constant C depending only on r_{0} such that the function d_{Ω}(x) − C|x|^{2} is concave in Ω.

Proof. Let x ∈ Ω. Let w ∈ R^{n} satisfy

|w| ≤ d_{Ω}(x).

Let y ∈ N_{∂Ω_{r_{0}}}(x), where Ω_{r_{0}} is r_{0}- neighborhood of Ω. Then by Proposition 9(c),

|x − y| = d_{Ω}(x) + r_{0},
|x + w − y| ≥ d_{Ω}(x + w) + r_{0},
|x − w − y| ≥ d_{Ω}(x − w) + r_{0}.

Then we have

d_{Ω}(x + w) − 2d_{Ω}(x) + d_{Ω}(x − w) ≤ |x + w − y| − 2|x − y| + |x − w − y| ≤ \frac{C}{r_{0}} |w|^{2},

and so d_{Ω} − 2\frac{C}{r_{0}} |x|^{2} is concave.

Now we prove Proposition 23. The function D(x, t) defined in (39) is Lipschitz since {Ω_{t}} is a Lipschitz family of sets. By Proposition 9 the set (Ω_{t})_{r_{0}} satisfies Condition 3 with radius \frac{r_{0}}{2} , and

d_{(Ω_{t})_{r_{0}}}^{s}(x) = d_{Ω_{t}}^{s}(x) + r_{0} for (x, t) ∈ E_{r_{0}}.

Then by Lemma 24 the function D(x, t) − C|x|^{2} on the set E_{r_{0}} is concave in x for every t. Now by Theorem 1 of Appendix 2 of [Kry87] the function D(x, t) is (2,1) differentiable at L^{n+1} a.e. point of E_{r_{0}}. It follows from Proposition 13 that if D(·) is (2,1) differentiable at a point (x_{0}, t_{0}) ∈ E, then Γ is (2,1) differentiable at (y, t_{0}), where y ∈ N_{∂Ω_{t_{0}}}(x_{0}). Note that x_{0} lies in the relative interior of its distance ray if d_{Ω_{t_{0}}} is twice x-differentiable at x_{0}. The Proposition is proved.

4. Properties of mass transport density.

Let Ω be a bounded set that satisfies the condition 3 with radius r_{0}. Let t > 0. Define a function a = a_{Ω,t} : R^{n} → R^{1} as following. Let x ∈ ∂Ω be such a point that the surface ∂Ω is twice differentiable at x and let κ_{1}, ..., κ_{n−1} be the principal curvatures of ∂Ω at x. Introduce the coordinate s on the distance ray R_{x} by s(z) = |z − x| for z ∈ R_{x}. Then s changes in the interval (0, γ(x)). Define the function a on R_{x} as the following:

a(s) = \frac{1}{t} ∏_{i=1}^{n−1} \frac{1}{1 − sκ_{i}} ∫_{0}^{s} ∏_{i=1}^{n−1}(1 − ξκ_{i}) [\frac{∫_{0}^{γ(x)} ν ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν}{∫_{0}^{γ(x)} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν} − ξ] dξ (40)

By Corollary 22 the function a is now defined a.e. in Ω.

Define a ≡ 0 on R^{n}\Ω. Now a is defined a.e. in R^{n}.

Definition 25. The function a(·) = a_{Ω,t}(·) is called mass transport density.

Define the function V : ∂Ω → R at the point of twice differentiability of ∂Ω as following.

V (x) = \frac{1}{t} \frac{∫_{0}^{γ(x)} s ∏_{i=1}^{n−1}(1 − κ_{i}s)ds}{∫_{0}^{γ(x)} ∏_{i=1}^{n−1}(1 − κ_{i}s)ds}. (41)

Proposition 26. Let Ω ∈ R^{n} be a bounded open set that satisfies Condition 3 with radius r_{0}. Then there exists a constant C depending only on r_{0}, n, diam Ω such that the following is true. Let x ∈ ∂Ω be such point that ∂Ω be twice differentiable at x. Then the mass transport density on the distance ray R_{x}, defined by (40) for s ∈ [0, γ(x)], satisfies:

(a) |a(s)| ≤ C. Derivative a′(s) exists and is continuous for s ∈ (0, γ(x)).

(b) a(s) satisfies the equation

a′(s) − a(s) ∑_{i=1}^{n−1} \frac{κ_{i}}{1 − κ_{i}s} − \frac{s}{t} + V (x) = 0

on (0, γ(x)), the boundary conditions a(0) = 0, a(γ(x)) = 0, and a(s) > 0 on (0, γ(x)).

(c) For s ∈ (0, γ(x)) the inequalities hold

|a(s)| ≤ C(γ(x) − s), |a(s)| ≤ Cs.

(d) |a′(s)| ≤ C on (0, γ(x)).

Proof. (a) Existence and continuity of a′(s) on (0, γ(x)) are checked explicitly. The bound |a(s)| ≤ C will follow from (c).

(b) The equation is checked explicitly. The boundary conditions follow from (c). The inequality a(s) > 0 on (0, γ(x)) holds since either κ_{i} < 0, or 0 ≤ γ(x) ≤ \frac{1}{κ_{i}} and then, defining

Ṽ (s) = \frac{∫_{0}^{s} ν ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν}{∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν}

20 MIKHAIL FELDMAN

we calculate:

\frac{dṼ}{ds} = \frac{s ∏_{i=1}^{n−1}(1 − κ_{i}s) ∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν − ∏_{i=1}^{n−1}(1 − κ_{i}s) ∫_{0}^{s} ν ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν}{[∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν)]^{2}}

= \frac{∏_{i=1}^{n−1}(1 − κ_{i}s)}{[∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν)]^{2}} ∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)(s − ν)dν > 0 for s ∈ (0, γ(x)),

and so

a(s) = \frac{1}{t}[Ṽ (γ(x)) − Ṽ (s)] ∏_{i=1}^{n−1} \frac{1}{1 − κ_{i}s} ∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν > 0 for s ∈ (0, γ(x)).

(c)We write γ for γ(x): Denote Φ(s) = ∫_{0}^{s} ν ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν, Ψ(s) = ∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν. For s ∈ (0, γ) we calculate:

ta(s) = \frac{[Φ(γ)Ψ(s) − Φ(γ)Ψ(γ)] + [Φ(γ)Ψ(γ) − Φ(s)Ψ(γ)]}{Ψ(γ)} ∏_{i=1}^{n−1} \frac{1}{1 − κ_{i}s}
= (γ − s)\frac{−Φ(γ)Ψ′(s_{1}) + Φ′(s_{2})Ψ(γ)}{Ψ(γ)} ∏_{i=1}^{n−1} \frac{1}{1 − κ_{i}s},

where s_{1}, s_{2} ∈ [s, γ]. Note, that Ṽ (ξ) = \frac{Φ(ξ)}{Ψ(ξ)}. So

ta(s) = ∏_{i=1}^{n−1} \frac{1}{1 − κ_{i}s} [−Ṽ (γ) ∏_{i=1}^{n−1}(1 − κ_{i}s_{1}) + ∏_{i=1}^{n−1}(a − κ_{i}s_{2}) · s_{2}] (γ − s)
= [−Ṽ (γ) ∏_{i=1}^{n−1} \frac{1 − κ_{i}s_{1}}{1 − κ_{i}s} + s_{2} ∏_{i=1}^{n−1} \frac{1 − κ_{i}s_{2}}{1 − κ_{i}s} ] (γ − s).

If κ_{i} ≥ 0 then for s^{∗} ∈ (s, γ) we have 0 ≤ sκ_{i} < s^{∗}κ_{i} < 1, and so

0 ≤ \frac{1 − κ_{i}s^{∗}}{1 − κ_{i}s} ≤ 1.

If κ_{i} < 0 then it follows from the Condition 3 that |κ_{i}| ≤ C, and so

0 ≤ \frac{1 − κ_{i}s^{∗}}{1 − κ_{i}s} = \frac{1 + |κ_{i}|s^{∗}}{1 + |κ_{i}|s} ≤ 1 + Cγ ≤ 1 + CdiamΩ.

It remains to estimate Ṽ (s) for s ∈ [0, γ].

Ṽ (s) ≤ \frac{∫_{0}^{s} γ ∏_{i=1}^{n−1}(1 − νκ_{i})dν}{∫_{0}^{s} ∏_{i=1}^{n−1}(1 − νκ_{i})dν} = γ ≤ diamΩ. (42)

So, ta(s) ≤ Cdiam Ω (γ − s), and, since t > 0, 0 < a(s) ≤ C(γ − s) for s ∈ (0, γ(x)). (43)

VARIATIONAL EVOLUTION PROBLEMS AND NONLOCAL GEOMETRIC MOTION 21

To estimate a(s) near s = 0, we compute using (42)

|ta(s)| = |[Ṽ (γ) − Ṽ (s)] ∏_{i=1}^{n−1} \frac{1}{1 − κ_{i}s} ∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν|
≤ γ |∏_{i=1}^{n−1} \frac{1}{1 − κ_{i}s} ∫_{0}^{s} ∏_{i=1}^{n−1}(1 − κ_{i}ν)dν| = γs ∏_{i=1}^{n−1} \frac{1 − κ_{i}s^{∗}}{1 − κ_{i}s} ,

where s^{∗} ∈ [0, s]. Since κ_{i}γ ≤ 1, then for s ≤ \frac{γ}{2} we get

0 ≤ \frac{1 − κ_{i}s^{∗}}{1 − κ_{i}s} ≤ 2 if κ_{i} ≥ 0,

and since κ_{i} > −\frac{1}{r} we get

0 ≤ \frac{1 − κ_{i}s^{∗}}{1 − κ_{i}s} ≤ C(diam(Ω), r) if κ_{i} < 0.

So |ta(s)| ≤ Cγs, and, since t > 0, we get

|a(s)| ≤ Cs.

If s ≥ \frac{γ}{2}, then s ≥ γ − s, and from (43)

|a(s)| ≤ C(γ − s) ≤ Cs.

(d) We prove that |a′(s)| ≤ C for s ∈ (0, γ).

a′(s) = a(s) ∑_{i=1}^{n−1} \frac{κ_{i}}{1 − κ_{i}s} + \frac{s}{t} − V (x).

By (42), |V (x)| ≤ \frac{γ}{t}. It remains to prove that

|a(s) ∑_{i=1}^{n−1} \frac{κ_{i}}{1 − κ_{i}s}| ≤ C (44)

If κ_{i} ≥ 0 then 0 ≤ κ_{i}γ ≤ 1, and since γ > s we get:

0 ≤ \frac{κ_{i}}{1 − κ_{i}s} ≤ \frac{κ_{i}}{κ_{i}γ − κ_{i}s} = \frac{1}{γ − s},

If κ_{i} < 0, then

|\frac{κ_{i}}{1 − κ_{i}s}| ≤ \frac{1}{s}.

Now (44) follows from (c).

5. Mass balance equation

Let {Ω_{t}} be a Lipschitz family of open sets satisfying Condition 3 with radius r_{0} > 0. Let {Ω_{t}} satisfy equation (15) on the time-interval [a, b] in the sense described in Theorem 5. Let the function w(x, t) be defined by (13). Let a(x, t) be the function a_{Ω_{t},t}(x) from Definition 25.

The purpose of this section is to show that for a.e. t ∈ [a, b]

∫_{R^{n}} aDwDϕdz = ∫_{R^{n}}(\frac{w}{t} − w_{t})ϕdz (45)

for any ϕ ∈ C^{∞}(R^{n}), where w_{t} denotes ∂_{t}w. Thus we show that the function a(·, t) satisfies (18) for a.e t.

It follows from Proposition 23 that for a. e. t the following is true: L^{n} a. e. point x ∈ Ω_{t} lies in the relative interior of the distance ray that intersects ∂Ω_{t} at such point y ∈ ∂Ω_{t} that the surface Γ is (2,1) differentiable at (y, t), where Γ is defined by (5). Fix such t. For the rest of this section we drop t in the notation, i.e., we write Ω, a(x), w(x), w_{t}(x) for Ω_{t}, a(x, t), w(x, t), w_{t}(x, t).

We use same notation a(·) for both mass transport density a(x), a function defined on R^{n}, and for mass transport density on a ray R_{x}, the function a(s) defined on [0, γ(x)].

Denote by Γ̃ the subset of ∂Ω that consists of all points at which the surface Γ is (2, 1)-differentiable. By choice of t and Proposition 21 we get

H^{n−1}(B \ Γ̃) = 0,

where the set B is defined by (38). From definition of B it follows that no distance rays of Ω have their lower ends in the subset ∂Ω \ B of the boundary. Denote by R(Γ̃) the union of all distance rays that have lower ends in Γ̃, i.e., R(Γ̃) = ∪_{x∈Γ̃}R_{x}. Then it follows from Proposition 18 and Remark 20 that

L^{n}(Ω \ R(Γ̃)) = 0, L^{n}[G_{k}^{−1}(Ω \ R(Γ̃))] = 0 for k = 1, ..., N. (46)

Let V (·) be the function (41). It follows from (46) that at a. e. point x of Ω the function w_{t}(x) is given by the expression \frac{1}{t}V (y) where y ∈ N_{∂Ω}(x). In the next proposition we prove that w_{t} is measurable.

We continue to use local coordinate systems on Ω_{r}, described in Section 3 and notation introduced there, in particular sets U_{k}, partition of unity Ψ_{k} and coordinate mappings G_{k} defined by (29), (30) and (27) respectively.

Proposition 27. a) For k = 1, ..., n there exists a bounded L^{n} measurable function v_{k} : U_{k} → R^{1} such that

v_{k}(x′, x_{n}) = V (G_{k}(x′, r)) if G_{k}(x′, r) ∈ Γ̃, x_{n} ∈ [r, γ̃_{r,k}(x′)]. (47)

b) The function w_{t} is L^{n} measurable and for any bounded measurable function φ : U_{k} → R^{1}

\frac{1}{t} ∫_{U_{k}∩\{x_{n}≥r\}} v_{k}φJ_{n}G_{k}dx = ∫_{G_{k}(U_{k})} w_{t}(φ ∘ G_{k}^{−1})dx (48)

Proof. a) ∂Ω_{r} is a C^{1,1} manifold, and so the second fundamental form of ∂Ω_{r} is defined as a differential form with L^{∞} coefficients. At all points where ∂Ω_{r} is twice differentiable the principal curvatures of ∂Ω_{r} are the roots of the characteristic polynomial of the second fundamental form. To write this in coordinates, we recall that ∂Ω_{r}∩G_{k}(U_{k}) is the graph of a C^{1,1} function x_{n} = Φ_{k}(x′) defined on the set Ũ_{k} ∈ R^{n−1}. Then there exist L^{∞} functions [u_{ij}(x′)]_{i,j=1}^{n} such that at the points where D^{2}Φ_{k} exists, \frac{∂^{2}Φ_{k}}{∂x_{i}∂x_{j}} = u_{ij}.

Denote by P_{x′}(s) the characteristic polynomial of the matrix [u_{ij}(x′)] i.e. P_{x′}(s) = det([u_{ij}(x′)] − sI_{n−1}). Note that the coefficients of the polynomial P_{x′}(·) are measurable functions of x′ since they are compositions of polynomials and measurable functions u_{ij}(x′).

The eigenvalues of the matrix [u_{ij}(x′)]_{i,j=1}^{n} are principal curvatures of ∂Ω_{r} at the points of twice differentiability of Φ_{k}. Let x′ be such a point, let κ_{r,1}, ..., κ_{r,n−1} be principal curvatures of ∂Ω_{r} in G_{k}(x′, 0), and let m be such number that κ_{r,i} ≠ 0 for i = 1, ..., m and κ_{r,i} = 0 for i = m + 1, ..., n − 1. Then P_{x′}(s) = ∏_{i=1}^{n−1}(κ_{r,i} − s), and so

(−1)^{n−1}s^{n−1}P_{x′}(\frac{1}{s}) = ∏_{i=1}^{n−1} (1 − κ_{r,i}s) = ∏_{i=1}^{m} κ_{r,i} ∏_{i=1}^{m}(\frac{1}{κ_{r,i}} − s).

For any x′ ∈ U_{k} the point G_{k}(x′, r) belongs to ∂Ω and |G_{k}(x′, 0)−G_{k}(x′, r)| = r. If the surface ∂Ω is twice differentiable at the point y = G_{k}(x′, r), then, by Proposition 12, the surface ∂Ω_{r} is twice differentiable at the point y_{r} = G_{k}(x′, 0), and, denoting by {κ_{i}}_{i=1}^{n} and {κ_{r,i}}_{i=1}^{n} the principal curvatures of ∂Ω at y and of ∂Ω_{r} at y_{r}, we get

κ_{r,i} = \frac{κ_{i}}{1 + κ_{i}r}, so \frac{1}{κ_{r,i}} = \frac{1}{κ_{i}} + r (if κ_{i} ≠ 0).

Then, assuming as before that κ_{1}, ..., κ_{m} ≠ 0 and κ_{m+1}, ..., κ_{n−1} = 0, we get if m ≠ 0

∏_{i=1}^{n−1}(1 − κ_{i}s) = ∏_{i=1}^{m} κ_{i} ∏_{i=1}^{m}(\frac{1}{κ_{i}} − s) = ∏_{i=1}^{m} κ_{i} ∏_{i=1}^{m} [\frac{1}{κ_{r,i}} − (s + r)]

= (−1)^{n−1} \frac{∏_{i=1}^{m} κ_{i}}{∏_{i=1}^{m} κ_{r,i}} (s + r)^{n−1}P_{x′}(\frac{1}{s + r}).

If m = 0 (i.e., all κ_{r,i} ≡ 0), then ∏_{i=1}^{n−1}(1 − κ_{i}s) = (−1)^{n−1}(s + r)^{n−1}P_{x′}(\frac{1}{s+r}) = 1.

Now we see that at all x′ as above, the function V defined in (41) can be expressed at the point G_{k}(x′, r) ∈ ∂Ω as

V (G_{k}(x′, r)) = \frac{∫_{0}^{γ̃_{r,k}(x′)−r} s(s + r)^{n−1}P_{x′}(\frac{1}{s+r})ds}{∫_{0}^{γ̃_{r,k}(x′)−r}(s + r)^{n−1}P_{x′}(\frac{1}{s+r})ds} (49)

since, by Proposition 9, γ_{Ω_{r}}(z) = γ(z)_{Ω} + r for z ∈ Ω.

Define a function η = η(x′, x_{n}) on Ũ_{k} × R^{1} by the right-hand side of (49). This function η is L^{n}-measurable: η does not depend on x_{n}, the coefficients of the polynomial P_{x′}(·) are measurable functions of x′ and by Proposition 15, the functions γ̃_{r,k} are measurable functions of x′.

Then it follows from Proposition 15 that the function

v_{k}(x′, x_{n}) = η(x′, x_{n})χ_{U_{k}}(x′, x_{n})

satisfies all properties asserted in (a) (note that the right-hand side of (47) is bounded by diam Ω).

By Lemma 16 the function v_{k} ∘ G_{k}^{−1} is L^{n}-measurable. By (46) we get

\frac{1}{t}v_{k} ∘ G_{k}^{−1} = w_{t} at L^{n} a.e. point of G_{k}(U_{k} ∩ \{ x_{n} ≥ r \}),
and w_{t} ≡ 0 on G_{k}(U_{k} ∩ { x_{n} < r })

(50)

and so w_{t} is measurable.

Since map G_{k} is one-to-one and Lipschitz on U_{k}, and since U_{k} is measurable, we get by Theorem 3.2.5 of [Fed69] and Lemma 16 that for every bounded measurable function φ

∫_{U_{k}} v_{k}φ J_{n}G_{k}dz = ∫_{G_{k}(U_{k})}(v_{k} ∘ G_{k}^{−1})(x)(φ ∘ G_{k}^{−1})(x)dx (51)

Now (48) follows from (51), (50).

Corollary 28. The function \frac{w}{t} − w_{t} is measurable, and for each φ ∈ C^{∞}(R^{n})

∫_{R^{n}}(\frac{w}{t} − w_{t})φdz = \frac{1}{t} ∑_{k=1}^{N} ∫_{G_{k}^{−1}(R(Γ̃))}(w ∘ G_{k} − v_{k})(φ ∘ G_{k})(Ψ_{k} ∘ G_{k})J_{n}G_{k}dx. (52)

Proof. We have

\frac{w}{t} − w_{t} = ∑_{k=1}^{N}(\frac{w}{t} − w_{t})Ψ_{k}.

Since supp Ψ_{k} ⊂ G_{k}(U_{k}), the integrand on the right-hand side of (52) is measurable by Propositions 27 and 18. It is also clear that the integrand on the right-hand side is bounded.

Then from Proposition 27 (b) and area formula, we get

∫_{R^{n}}(\frac{w}{t} − w_{t})φdz = \frac{1}{t} ∑_{k=1}^{N} ∫_{R^{n}}(w ∘ G_{k} − v_{k})(φ ∘ G_{k})(Ψ_{k} ∘ G_{k})J_{n}G_{k}dx.

This formula is true for each φ ∈ C^{∞}, and, by approximation, for every bounded measurable φ. Now (46) implies (52).

Now we prove that the mass transport density a(·) is measurable. The proof is similar to the proof of Proposition 27.

Proposition 29. The function a(·) is L^{n}-measurable function on R^{n}. The function a ∘ G_{k}^{−1} is L^{n}-measurable function on U_{k}.

Proof. For (x′, s) such that

G_{k}(x′, r) ∈ Γ̃, s ∈ [r, γ̃_{r,k}(x′, r)],

the function a ∘ G_{k}(x′, s) is given by expression (40), where the curvatures are computed at the point G_{k}(x′, r). Calculations similar to those in Proposition 27 show that this expression can be rewritten using the characteristic polynomial P_{x′}(·) of the matrix D^{2}Φ_{k}(x′) as following:

a ∘ G_{k}(x′, s) = \frac{1}{t} \frac{∫_{0}^{s−r}(ν + r)^{n−1}P_{x′}(\frac{1}{ν+r})dν}{(s + r)^{n−1}P_{x′}(\frac{1}{s+r})} ×

[\frac{∫_{0}^{γ̃_{k,r}(x′)−r} ν(ν + r)^{n−1}P_{x′}(\frac{1}{ν+r})dν}{∫_{0}^{γ̃_{k,r}(x′)−r}(ν + r)^{n−1}P_{x′}(\frac{1}{ν+r})dν} − \frac{∫_{0}^{s−r} ν(ν + r)^{n−1}P_{x′}(\frac{1}{ν+r})dν}{∫_{0}^{s−r}(ν + r)^{n−1}P_{x′}(\frac{1}{ν+r})dν} ] (53)

By (46), the function a ∘ G_{k}^{−1}(·) is defined by the above formula at a.e. point of U_{k}, and is L^{n}-measurable function of variables x′, s on U_{k} since it is a rational function of variables s and γ̃_{k,r}(x′) with measurable coefficients and γ̃_{k,r}(x′) is L^{n−1} measurable function of x′. By Lemma 16, the function a(·) is measurable.

Proof of (45). We will transform the right-hand side of (52). From (30) we get

Ψ_{k}(G_{k}(x′, s)) = Ψ_{k}(G_{k}(x′, r)), s ∈ (r, γ̃_{r,k}(x′)). (54)

Let G̃_{k} : Ũ_{k} → ∂Ω be the map defined by

G̃_{k}(x′) = G_{k}(x′, r).

Then for k = 1, ..., N, using definitions of w, v_{k}, Ψ_{k} we compute

\frac{1}{t} ∫_{G_{k}^{−1}(R(Γ̃))}(w ∘ G_{k} − v_{k})(φ ∘ G_{k})(Ψ_{k} ∘ G_{k})J_{n}G_{k}dx

= ∫_{G̃_{k}^{−1}(Γ̃)} Ψ_{k}(G_{k}(x′, r)) \{∫_{r}^{γ̃_{r,k}(x′)} \frac{1}{t}[s − r (55)

− v_{k}(x′, s)]φ(G_{k}(x′, s))J_{n}G_{k}(x′, s)ds} dx′

Let x′ ∈ R^{n−1} be such that G_{k}(x′, r) ∈ Γ̃. Then ∂Ω is twice differentiable at G_{k}(x′, r), and so

\frac{1}{t}v_{k}(x′, s) = V (G_{k}(x′, r)) for s ∈ [r, γ̃_{r,k}(x′)].

By Proposition 12, ∂Ω_{r} is twice differentiable at G_{k}(x′, 0). Let κ_{1}, ..., κ_{n−1} be principal curvatures of ∂Ω at G_{k}(x′, r), and κ_{r,1}, ..., κ_{r,n−1} be principal curvatures of ∂Ω_{r} at G_{k}(x′, 0). Then κ_{r,i} = \frac{κ_{i}}{1+rκ_{i}}. Let y = G_{k}(x′, r). Then the mass transport density on the ray R_{y} satisfies ODE from Proposition 26(b). Thus we get for s ∈ [r, γ̃_{r,k}(x′)] = [r, γ_{Ω}(y) + r]:

\frac{1}{t}(s − r − v_{k}) = −\frac{da(s − r)}{ds} + a(s − r) ∑_{i=1}^{n−1} \frac{κ_{i}}{1 − κ_{i}(s − r)}

= −\frac{da(s − r)}{ds} + a(s − r) ∑_{i=1}^{n−1} \frac{κ_{r,i}}{1 − κ_{r,i}s}

where a(s) is mass transport density on the ray R_{y}.

By Proposition 26, \frac{da}{ds} exists and is continuous and bounded on the interval (0, γ̃_{r,k}(x′)− r), and

a(0) = a(γ̃_{r,k}(x′) − r) = 0.

We also see by (28) that

\frac{∂J_{n}G_{k}}{∂s} (x′, s) = J_{n}G_{k}(x′, s) ∑_{i=1}^{n−1} \frac{−κ_{r,i}}{1 − κ_{r,i}s}. (56)

Then we calculate, integrating by parts:

∫_{r}^{γ̃_{r,k}(x′)} \frac{1}{t}[s − r − v_{k}(x′, s)]φ(G_{k}(x′, s))J_{n}G_{k}(x′, s)ds =

∫_{r}^{γ̃_{r,k}(x′)} [−\frac{da(s − r)}{ds} − a(s − r) \frac{\frac{∂}{∂s}J_{n}G_{k}(x′, s)}{J_{n}G_{k}(x′, s)}] φ(G_{k}(x′, s))J_{n}G_{k}(x′, s)ds

= ∫_{r}^{γ̃_{r,k}(x′)} a(s − r)\frac{∂}{∂s}φ(G_{k}(x′, s))J_{n}G_{k}(x′, s)ds

This allows to derive from (52), (55) and (54) the following equality:

∫_{R^{n}}(\frac{w}{t} − w_{t})φdz = (57)

∑_{k=1}^{N} ∫_{G̃_{k}^{−1}(Γ̃)} ∫_{r}^{γ̃_{r,k}(x′)} Ψ_{k}(G_{k}(x′, s))a(G_{k}(x′, s))\frac{∂}{∂s}φ(G_{k}(x′, s)J_{n}G_{k}(x′, s)dsdx′

Let z = G_{k}(x′, s). Using the fact that distance function is twice differentiable on R(Γ̃), we get:

\frac{∂}{∂s}(φ(G_{k}(x′, s)) = (Dφ)(G_{k}(x′, s)) · \frac{∂}{∂s}(G_{k}(x′, s))
(58)  = Dφ(z) · Dd_{Ω}(z).

We insert the right-hand side of (58) into the right-hand side of (57) and use area formula and (46) to change variables from (x′, s) to z in the integral on the righthand side of (57). Then using that ∑_{k} Ψ_{k} ≡ 1 on Ω^{0}\R and that a ≡ 0 on R and in R^{n} \ Ω, we get (45).

6. Proof of Theorem 5

To conclude the proof of Theorem 5 we need to demonstrate the following. Let, as above, w(x) = d_{Ω}(x), and let v ∈ W^{1,∞}(R^{n}) satisfy |Dv| ≤ 1 a.e. Then

∫_{R^{n}}(\frac{w}{t} − w_{t})w ≥ ∫_{R^{n}}(\frac{w}{t} − w_{t})v. (59)

We show that this follows from (45). By approximation, (45) is true for Lipschitz φ. Let φ = w − v. Then by (45) that it is enough to prove that

∫_{R^{n}} aDw(Dw − Dv) ≥ 0.

But, since a ≥ 0 in R^{n} and a ≡ 0 outside Ω^{0} and on R, and |Dw| ≡ 1 on Ω\R,

∫_{R^{n}} aDwDw = ∫_{R^{n}} a ≥ ∫_{R^{n}} aDwDv

since |DwDv| ≤ \sqrt{|Dw|^{2}|Dv|^{2}} ≤ 1.

Theorem 5 is proved.

7. Proof of Theorem 6

The proof of Theorem 6 is similar to the proof Theorem 5. We give a formal calculation. Each step can be justified the same way as in the proof of Theorem 5.

Fix t ∈ [a, b]. Define

D_{1} = D_{1}(t) = {x ∈ Ω_{t}^{1} | d_{Ω_{t}^{1}}(x) > d_{Ω_{t}^{2}}(x)},
D_{2} = D_{2}(t) = {x ∈ Ω_{t}^{2} | d_{Ω_{t}^{2}}(x) > d_{Ω_{t}^{1}}(x)}.

Let r > 0, let Ω_{r}^{1} = Ω_{t,r}^{1} and Ω_{r}^{2} = Ω_{t,r}^{2} be r-neighborhoods of Ω^{1} = Ω_{t}^{1} and Ω^{2} = Ω_{t}^{2}, and let r be so small that ∂Ω_{r}^{1} and ∂Ω_{r}^{2} are C^{1,1} manifolds.

Let l=1,2. Let {Ũ_{k}^{l}}_{k=1}^{N_{l}} and {U_{k}^{l}}_{k=1}^{N_{l}} be the coordinate neighborhoods associated the set Ω_{r}^{l} same way as {Ũ_{k}}, {U_{k}} with Ω_{r} in the proof of Theorem 5. Let G_{k}^{l} : U_{k}^{l} → R^{n} be the corresponding coordinate mappings (27) and let {Ψ_{k}^{l}}_{k=0}^{N_{l}+1} be the corresponding partitions of unity (30). Let γ̃_{r,k}^{l} be the functions defined as in (31) from the functions γ_{Ω_{r}^{l}} using the coordinate mappings G_{k}^{l} for k = 1, ..., N_{l}. Define the following functions δ̃_{r,k}^{l} for k = 1, ..., N_{l}. Let δ(·) ≡ γ_{Ω^{1}∩Ω^{2}}(·). Then

δ̃_{r,k}^{l}(x′) = \{ \begin{matrix}δ(G_{k}^{l}(x′, r)) + r, & if x′ ∈ cl(Ũ_{k}^{l}) and G_{k}^{l}(x′, r) ∈ ∂(Ω^{1} ∩ Ω^{2}); \\ r, & otherwise. \end{matrix}

Note that

(G_{k}^{l})^{−1}(D_{l}) ∩ U_{k}^{l} = { (x′, s) | x′ ∈ Ũ_{k}^{l}, s ∈ [ δ̃_{r,k}^{l}(x′), γ̃_{r,k}^{l}(x′) ] }.

The functions δ̃_{r,k}^{l} are uppersemicontinuous, the proof is similar to the proof of Proposition 15, using the properties of sets satisfying Condition 3.

Define the mass transport density a(·). Let a ≡ 0 on R^{n}\(Ω^{1} ∪ Ω^{2}). To define a in Ω^{1} ∪ Ω^{2} it is enough to define a on D_{1} ∪ D_{2}. Let x ∈ D_{l}\R_{l}, where l = 1, 2. Let y ∈ N_{∂Ω_{l}}(x) and s = |x − y|. Then we define

a(x) = \frac{1}{t} ∏_{i=1}^{n−1} \frac{1}{1 − sκ_{i}} ∫_{δ(y)}^{s} ∏_{j=1}^{n−1}(1 − κ_{j}ξ) [\frac{∫_{δ(y)}^{γ^{l}(y)} ν ∏_{k=1}^{n−1}(1 − κ_{k}ν)dν}{∫_{δ(y)}^{γ^{l}(y)} ∏_{k=1}^{n−1}(1 − κ_{k}ν)dν} − ξ] dξ,

where κ_{k} are curvatures of ∂Ω_{t}^{l} at y. Then a satisfies the equation (20) on subintervals of distance rays of Ω_{t}^{l} that lie in D_{l}, and a ≡ 0 on ∂D_{l} and on R_{l}, l = 1, 2. We calculate, denoting by κ_{r,i} the principal curvatures of ∂Ω_{r}^{l} at the point y_{r} ∈ Ω_{r}^{l} nearest to x

∫_{R^{n}}(\frac{w}{t} − w_{t})φdz

= \frac{1}{t} ∑_{l=1}^{2} ∑_{k=1}^{N_{l}} ∫_{U_{k}^{l}∩(G_{k}^{l})^{−1}(D_{l})}(w ∘ G_{k}^{l} − V ∘ G_{k}^{l})(φ ∘ G_{k}^{l})(Ψ_{k}^{l} ∘ G_{k}^{l})J_{n}G_{k}^{l} dx

= ∑_{l=1}^{2} ∑_{k=1}^{N_{l}} ∫Ũ_{k}^{l} Ψ_{k}^{l}(G_{k}^{l}(x′, 0))

× ∫_{δ̃_{r,k}^{l}(x′)}^{γ̃_{r,k}^{l}(x′)} \frac{1}{t} [(s − r) − V (G_{k}^{l}(x′, 0))] (φ(G_{k}^{l}(x′, s))J_{n}G_{k}^{l}(x′, s) dsdx′

= ∑_{l=1}^{2} ∑_{k=1}^{N_{l}} ∫Ũ_{k}^{l} Ψ_{k}^{l}(G_{k}^{l}(x′, 0)) ∫_{δ̃_{r,k}^{l}(x′)}^{γ̃_{r,k}^{l}(x′)} (−\frac{d a(G_{k}^{l}(x′, s))}{d s}

+a(G_{k}^{l}(x′, s)) ∑_{i=1}^{n−1} \frac{κ_{r,i}}{1 − κ_{r,i}s}) φ J_{n}G_{k}^{l}(x′, s) dsdx′

= ∑_{l=1}^{2} ∑_{k=1}^{N_{l}} ∫Ũ_{k}^{l} Ψ_{k}^{l}(G_{k}^{l}(x′, 0))

× ∫_{δ̃_{r,k}^{l}(x′)}^{γ̃_{r,k}^{l}(x′)} a(G_{k}^{l}(x′, s)) \frac{∂φ(G_{k}^{l}(x′, s))}{∂s}JG_{k}^{l}(x′, s) dsdx′

= ∑_{l=1}^{2} ∑_{k=1}^{N_{l}} ∫_{G_{k}^{l}(U_{k}^{l})∩D_{l}} a(z)Ψ_{k}^{l}(z) Dφ(z) Dw(z) dz

= ∫_{R^{n}} aDwDφ dz.

Thus we have proved that the mass balance equation is satisfied. This implies Theorem 6.

8. Compression molding model

Compression molding is the process of deformation of an incompressible plastic material between two horizontal plates. The following simplified mathematical model of the process was derived by G. Aronsson [Ar95] based on Hele-Show model for non-Newtonian fluid. Suppose that the distance between the horizontal plates is small. Then we can assume that the region occupied by plastic at each time t has the form U_{t} = Ω_{t} × [0, h_{t}] where Ω_{t} ⊂ R^{2}, and that the pressure in plastic does not depend on the vertical coordinate, i.e., pressure is the function u(x, t) where x ∈ Ω_{t}. Evolution of rescaled Ω_{t} and u(x, t) is described by the following free boundary problem. Given an open set Ω_{0} ∈ R^{2} find an expanding family of open sets Ω_{t} ∈ R^{2}, t ≥ 0, and a function u(x, t) defined on ∪_{t}(Ω_{t} × {t}) such that

(60)  −div(|Du|^{p−2}Du) = 1  in Ω_{t},
(61)  u = 0  on Γ_{t},
(62)  V = |Du|^{p−2}  on Γ_{t},

where Γ_{t} = ∂Ω_{t}, V denotes the outer normal velocity of Γ_{t}. Condition (62) means that the free boundary Γ_{t} moves with the velocity of the flow.

In the paper [AE] the asymptotic limit as p → ∞ in the problem (60) - (62) was considered. This limit corresponds to the case of highly non-Newtonian fluid.

It was shown in [AE] that formally sending to a limit in (60) - (62) one obtains the following problem. Find a family {Ω_{t}} of open subsets of R^{2}, a function u(x, t), and a mass transport density function a(x, t) satisfying:

|Du| ≤ 1  a.e. in Ω_{t},
a ≥ 0  a.e. in Ω_{t},
supp(a) ∈ {|Du| = 1},
−div(aDu) = 1  in Ω_{t},

(63)

where the last equation is understood in the weak sense, and

u = 0, V = a on Γ_{t}, (64)

where V is the outer normal velocity of Γ_{t}.

It was shown in [AE] that solutions of (63), (64) have the form

u(x, t) = d_{Ω_{t}}(x), (65)

where the right-hand side is defined by (1), and that formally the problem (63) - (64) can be rewritten as following. Find {Ω_{t}} such that

w − ∂_{t}w ∈ ∂I_{∞}[u], (66)

where

w(·, t) = χ_{Ω_{t}}(·), (67)

where χ_{Ω_{t}}(·) is the indicator function of Ω_{t} that equals 1 inside and 0 outside Ω_{t}. Existence of a weak solution of (65) - (66) was proved in [AE]. Namely, there exists a family {Ω_{t}} of sets of finite perimeter such that ∂_{t}w is a nonnegative Radon measure and

∫_{R^{2}} w(x, t)(v(x) − u(x, t))dx ≤ ∫_{R^{2}}(v − u(·, t))d(∂_{t}w(·, t))

(68)

for every v with |Dv| ≤ 1, for a.e. t.

The following law of motion of the free boundary Γ_{t} was derived in [AE] by a formal calculation:

V = γ(1 − \frac{κγ}{2} ), (69)

where V = outer normal velocity of Γ_{t}, γ = function γ_{Ω_{t}}(·) defined by (21), κ = curvature of Γ_{t}.

The equation (69) was derived as following. Starting from (63), we perform calculations similar to the ones that lead from (18) to (20). Thus we deduce that equation (63) can be formally rewritten on each distance ray R_{x}, where x ∈ Γ_{t}, as ODE

a′(s) − a(s)\frac{κ}{1 − κs} + 1 = 0, s ∈ (0, γ_{Ω_{t}}(x)). (70)

By the nature of mass transfer process in the compression molding model (i.e., mass transfer from within the set onto the boundary), we expect that the mass transport density equals to zero on the ridge of Ω_{t} and equals to the outer normal velocity at the boundary. This translates into the following boundary conditions for the ODE (70) on R_{x}:

a(0) = V (x), a(γ(x)) = 0. (71)

The function a(·) and the constant V can be found from (70), (71). V has the expression (69).

In this section we prove the connection between the variational equation (66) and the geometric equation (69).

Let E, Γ be defined by (5).

Theorem 30. Let {Ω_{t}}, t ∈ R_{+}^{1}, be an expanding, locally Lipschitz continuous family of open bounded sets. Let for every t ∈ R_{+}^{1} the set Ω_{t} satisfy Condition 3 with radius r_{0} > 0. Suppose that the equation (69) is satisfied at every point of (2, 1) differentiability of the surface Γ. Then the equation (66) is satisfied for a.e. t ∈ R^{1}, where the functions u(x, t), w(x, t) are defined by (65), (67).

Proof. The proof goes along the lines of the proof of Theorem 5. We will sketch the proof and present some details for the parts that are different from the proof of Theorem 5.

The main steps of the proof are following: Step 1. Definition and properties of mass transport density. Step 2. Show that the main mass balance equation is satisfied.

Step 3. Show that the assertion of the Theorem follows from the main mass balance equation.

We discuss each step.

Step 1.

Let (x, t) be a point of (2, 1) differentiability of the surface Γ in R^{2} × R^{1}. Define mass transport density a(y, t) = a(s) on the distance ray R_{x} of Ω_{t} as the solution of (70), (71). Do this for every such point (x, t). It follows from the hypothesis of the Theorem and from Proposition 23 that a(y, t) is now defined at L^{3} a.e. point of the set E = ∪_{t}(Ω_{t} × {t}) ⊂ R^{2} × R_{+}^{1}. Define a(y, t) by zero at all other points of R^{2} × R_{+}^{1}.

Solving (70), (71) (and taking into account (69)) we see that on the ray R_{x}

a(s) = \frac{1 − κs}{2κ} − \frac{(1 − κγ)^{2}}{2κ(1 − κs)}. (72)

This function satisfies assertions (a) and (d) of Proposition 26. Instead of assertion (c) of Proposition 26 we have

|a(s)| ≤ C(γ(x) − s), |a(s) − V (x)| ≤ Cs. (73)

To see this, we note that from the conditions of the Theorem and (69) it follows that

V (x) ≤ C(r_{0}, diamΩ_{t}).

Then the assertions (a) and (d) of Proposition 26 for the function a(s) defined by (72) follow from (73) like in the proof of Proposition 26.

Let us prove (73). We can rewrite (72) as

a(s) = \frac{γ − s}{2} (1 + \frac{1 − κγ}{1 − κs} ) .

Now, since 0 ≤ s ≤ γ ≤ diamΩ_{t} and κγ ≤ 1 and κ ≥ −\frac{1}{r_{0}} , we get the estimate

0 ≤ \frac{1 − κγ}{1 − κs} ≤ C(r_{0}, diamΩ_{t}), (74)

and the first inequality of (73) follows. To prove the second, we calculate using (71)

a(s) − V = a(s) − a(0) = \frac{s}{2} [−1 − \frac{(1 − κγ)^{2}}{1 − κs} ] ,

and use (74) and the inequality

0 ≤ 1 − κγ ≤ C(r_{0}, diamΩ_{t})

to finish the proof of (73).

Step 2.

The purpose of this step is to prove that for any smooth function φ ∈ C_{c}^{1}(R^{2} × (0, T)) we have

∫_{0}^{T} ∫_{R^{2}} w(φ + φ_{t}) − aDuDφ = 0. (75)

This is main mass balance equation for compression molding model.

The proof follows Section 5.

Fix t. We use local coordinate systems on (Ω_{t})_{r} where 0 < r ≤ r_{0}, defined in Section 3 and notation introduced there, in particular sets Ũ_{k}, U_{k}, partition of unity Ψ_{k}, coordinate mappings G_{k} and functions γ̃_{r,k} defined by (29), (30), (27) and (31) respectively (k=1,...,N).

We use notation E_{r}, Γ_{r} introduced in (26). From Proposition 9 it follows that for r ∈ [0, r_{0}]

Γ_{r} = {(x, t) ∈ R^{2} × (0, T) | d_{Ω_{t}}^{s}(x) = −r}.

The function

(x, t) → d_{Ω_{t}}^{s}(x)

is Lipschitz since {Ω_{t}} is a Lipschitz family of sets. Then by Proposition 3.2.15 of [Fed69], for a.e. r ∈ [0, r_{0}]

set Γ_{r} is (H^{2}, 2) rectifiable. (76)

In addition, by Lemma 24, the function (x, t) → d_{Ω_{t}}^{s}(x) − C|x|^{2}, where C is large enough depending on r_{0}, is concave in x for every t in the set E_{r_{0}}. Then it follows from Theorem 1 of Appendix 2 of [Kry87] and Propositions 23(b) and 11 that for a.e. r ∈ [0, r_{0}]

surface Γ_{r} is (2, 1) differentiable at H^{2} a.e. point; (77)

functions ∂_{t}d_{Ω_{t}}^{s}(x), Dd_{Ω_{t}}^{s}(x), D^{2}d_{Ω_{t}}^{s}(x) are bounded and

H^{2} measurable on Γ_{r}. (78)

Function (x, t) → γ_{Ω_{t}}(x) is uppersemicontinuous since the family of sets {Ω_{t}} is Lipschitz. Then similar to Propositions 27, 29 and Corollary 28 we prove that the mass transport density a(y, t) defined at Step 1 is Lebesgue measurable in R^{2} × R_{+}^{1}. In the proof we use (78) (i.e., we choose such r in the definition of local coordinate systems that (78) is satisfied).

It follows from Proposition 23 that for a.e. t we have the following: a.e. point x ∈ Ω_{t} lies in the relative interior of the distance ray of the set Ω_{t} that intersects ∂Ω_{t} at a point of (2,1) differentiability of the surface Γ. Fix such t. Let r ∈ (0, r_{0}]. Let y ∈ ∂Ω_{t} be a point of (2,1) differentiability of Γ. Let κ^{t} be the curvature of ∂Ω_{t} at y, let denote κ_{r}^{t} be the curvature of ∂(Ω_{t})_{r} at the unique point y ∈ ∂(Ω_{t})_{r} nearest to x, then

\frac{1}{κ_{r}^{t}} = \frac{1}{κ^{t}} + r.

We calculate using properties of the function a on the rays proved in the Step 1, equation (70), and Lemma 14 with n = 2:

∫_{R^{2}} wφ dz = ∫_{Ω_{t}} φ dz

= ∑_{k=1}^{N} ∫_{U_{k}}(φ ∘ G_{k})(Ψ_{k} ∘ G_{k})J_{2}G_{k}dx

= ∑_{k=1}^{N} ∫Ũ_{k} Ψ_{k}(G_{k}(x′, r)) ∫_{r}^{γ̃_{r,k}(x′)} [\frac{da(s − r)}{ds}

+a(s − r)\frac{κ_{r}^{t}}{1 − κ_{r}^{t}s}] (φ(G_{k}(x′, s))J_{2}G_{k}(x′, s)dsdx′.

In the last expression we integrate by parts and use (56), (71) and (58). Then we get

∫_{R^{2}} wφ dz = ∑_{k=1}^{N} ∫Ũ_{k} Ψ_{k}(G_{k}(x′, r)) [ a(0)φ(G_{k}(x′, r))J_{2}G_{k}(x′, r)

+ ∫_{r}^{γ̃_{r,k}(x′)} a(s − r)\frac{∂φ(G_{k}(x′, s))}{∂s}J_{2}G_{k}(x′, s)ds ] dx′.

= ∫_{ˆ∂(Ω_{t})_{r}} φ(P_{r}^{t}(y))∂_{t}d_{Ω_{t}}^{s}(P_{r}^{t}(y))(1 − rκ_{r}(y))dH^{1}(y) + ∫_{Ω_{t}} aDφDu dz,

where

ˆ∂(Ω_{t})_{r} = {y ∈ ∂(Ω_{t})_{r} |γ_{(Ω_{t})_{r}}(y) > r}, (79)

and

P_{r}^{t} : (Ω_{t})_{r} → Ω_{t} (80)

is the nearest point projection mapping. By Proposition 9 b) the mapping P_{r}^{t} is well-defined and onto.

By Proposition 13, if Γ_{r} is (2,1) differentiable at y ∈ ˆ∂(Ω_{t})_{r}, then Γ is (2,1) differentiable at y′ = P_{r}^{t}(y) and

∂_{t}d_{Ω_{t}}^{s}(P_{r}^{t}(y)) = ∂_{t}d_{Ω_{t}}^{s}(y)

By the equality

d_{(Ω_{t})_{r}}^{s}(x) = d_{Ω_{t}}^{s}(x) + r for (x, t) ∈ E_{r_{0}}, r ∈ (0, r_{0}]

we see that

∂_{t}d_{Ω_{t}}^{s}(P_{r}^{t}(y)) = ∂_{t}d_{(Ω_{t})_{r}}^{s}(y)

for y ∈ ˆ∂(Ω_{t})_{r}.

Thus we have showed that for a.e. t ∈ [0, T], every r ∈ (0, r_{0}] the equality holds

∫_{R^{2}} wφ dz = ∫_{ˆ∂(Ω_{t})_{r}}(φ ∘ P_{r}^{t}) ∂_{t}d_{(Ω_{t})_{r}}^{s}(1 − rκ_{r}^{t})dH^{1} + ∫_{Ω_{t}} aDφDu dz. (81)

In the next lemma we show that in the first integral at the right-hand side of (81) we can integrate over the whole boundary ∂(Ω_{t})_{r}.

Lemma 31. Let r ∈ (0, r_{0}] be such that properties (76), (77), (78) are satisfied. Then for a.e. t ∈ (0, T)

∫_{∂(Ω_{t})_{r} ∩{γ_{(Ω_{t})r} ≤r}} φ∂_{t}d_{(Ω_{t})_{r}}^{s}(1 − rκ_{r}^{t})dH^{1} = 0

Proof. We will show that for a.e. t

∂_{t}d_{(Ω_{t})_{r}}^{s}(y)(1 − rκ_{r}^{t}(y)) = 0 H^{1} a.e. on ∂(Ω_{t})_{r} ∩ {γ_{(Ω_{t})_{r}} ≤ r}. (82)

By (77) for a.e. t the surface Γ_{r} is (2,1) differentiable H^{1} a.e. on ∂(Ω_{t})_{r}. Fix such t. Then in (82) we can consider only y at which Γ_{r} is (2,1) differentiable. Fix such y.

By Proposition 9,

γ_{(Ω_{t})_{r}} ≥ r on ∂(Ω_{t})_{r}.

Thus

γ_{(Ω_{t})_{r}}(y) = r.

It follows that

κ_{r}^{t}(y) ≤ \frac{1}{r}.

Let y′ = P_{r}^{t}(y), i.e., y′ is the point of ∂Ω_{t} nearest to y. Consider 3 cases. Case 1. y is the unique point of ∂(Ω_{t})_{r} for which y′ the nearest point on ∂Ω_{t}. Then the calculations of [EGan], Proposition 7.1, Steps 2,3 imply that

κ_{r}^{t}(y) = \frac{1}{r}.

Thus we have (82) in this case.

Case 2.There exists z ∈ ∂(Ω_{t})_{r}, z ≠ y, such that y′ = P_{r}^{t}(z), and

B_{r}(y) ∩ B_{r}(z) ≠ ∅.

Note that

|y − y′| = |z − y′| = r.

Then, denoting w the point \frac{y+z}{2}, we get

w ∈ B_{r}(y) ∪ B_{r}(z).

We have

B_{r}(y), B_{r}(z) ⊂ R^{2} \ \overline{Ω}_{t}.

Thus w ∈ R^{2} \ \overline{Ω}_{t}, and y′ is the point of ∂Ω_{t} nearest to w. Denote

v = y′ + \frac{r}{|w − y′|}(w − y′).

Then, by Condition 3, we get the following:

v ∉ Ω_{t}, y′ is the unique point of ∂Ω_{t} nearest to v.

But

|v − y′| = r.

Thus we proved that

v ∈ ∂(Ω_{t})_{r}.

The points y and z divide the circle ∂B_{r}(y′) on two arcs, and the point v is the middle point of one of these arcs. Denote this arc C_{1}. Repeating the above argument inductively and using continuity of distance function, we prove that

C_{1} ⊂ ∂(Ω_{t})_{r}.

But then

κ_{r}^{t}(y) = \frac{1}{r}.

Thus we have (82) in the Case 2.

Case 3.There exists z ∈ ∂(Ω_{t})_{r}, z ≠ y, such that y′ = P_{r}^{t}(z), and

B_{r}(y) ∩ B_{r}(z) = ∅.

Then

∂B_{r}(y) ∩ ∂B_{r}(z) = y′.

Since {Ω_{t}} is an expanding family of sets, we have

B_{r}(y), B_{r}(z) ⊂ R^{2} \ \overline{Ω}_{τ} for τ < t. (83)

If

(84) y′ ∈ ∂Ω_{τ^{∗}} for some τ^{∗} < t,

then the same is true for all τ ∈ [τ^{∗}, t], and (83) implies that

y ∈ ∂(Ω_{τ})_{r} for all τ ∈ [τ^{∗}, t].

Then

∂_{t}d_{(Ω_{t})_{r}}^{s}(y) = 0.

Thus we have (82) if (84) is satisfied.

The remaining case is

y′ ∉ ∂Ω_{τ} for all τ < t. (85)

Introduce a coordinate system (x_{1}, x_{2}) on R^{2} in which

y′ = (0, 0), y = (0, r).

Then

z = (0, −r).

Since {Ω_{t}} is a continuous expanding family of sets, the function

ϕ(τ) = dist(y′, Ω_{τ})

is continuous and nonincreasing, and

ϕ(t) = 0.

Let w_{τ} ∈ N_{∂Ω_{τ}} (y′). Then by (83)

w_{τ} ∈ ∂B_{ϕ(τ)}(0, 0) \ [B_{r}(0, r) ∪ B_{r}(0, −r))] . (86)

In particular,

w_{τ} → y′ as τ → t. (87)

By Condition 3 we get

B_{r}(v_{τ}) ⊂ R^{2} \ \overline{Ω}_{τ} where v_{τ} = w_{τ} + \frac{r}{|y′ − w_{τ}|}(y′ − w_{τ}). (88)

By (86), (87) we see that there exists a sequence τ_{j} → t such that

v_{τ_{j}} → p, where p is either (−r, 0) or (r, 0).

Then by (88) and continuity of the family {Ω_{t}} we conclude

either B_{r}(−r, 0) ⊂ R^{2} \ \overline{Ω}_{t} or B_{r}(r, 0) ⊂ R^{2} \ \overline{Ω}_{t}.

Let in fact

B_{r}(r, 0) ⊂ R^{2} \ \overline{Ω}_{t}.

Then y′ = (0, 0) is the point of ∂Ω_{t} nearest to the point (r, 0). Thus

(r, 0) ∈ ∂(Ω_{t})_{r}.

But y = (0, r), and thus

B_{r}(y) ∩ B_{r}(r, 0) ≠ ∅.

Thus the points y and (r, 0) satisfy the conditions of Case 2. Thus we get

κ_{r}^{t}(y) = \frac{1}{r}.

Case 3 is proved. Then Lemma 31 is proved.

The properties (76), (77), (78) are satisfied for a.e. r ∈ (0, r_{0}]. Then there exists a decreasing sequence r_{i} → 0, where i = 1, 2, ... and r_{i} ∈ (0, r_{0}], such that (76), (77),

(78) are satisfied for each r_{i}. Note that by (78), the function

(y, t) → κ_{r_{i}}^{t}(y)

defined on Γ_{r_{i}} is H^{2} measurable.

Now we can integrate (81) by t and use Lemma 31 to get

∫_{0}^{T} ∫_{R^{2}} wφ dzdt = ∫_{0}^{T} ∫_{∂(Ω_{t})_{r_{i}}}(φ ∘ P_{r_{i}}^{t}) ∂_{t}d_{(Ω_{t})_{r_{i}}}^{s}(1 − r_{i}κ_{r_{i}}^{t})dH^{1} dt

+ ∫_{0}^{T} ∫_{Ω_{t}} aDφDu dzdt. (89)

It follows that the first integral at the right-hand side does not depend on r_{i}. Thus it is enough to compute the limit as i → ∞. We will prove the following:

lim_{i→∞} ∫_{0}^{T} ∫_{∂(Ω_{t})_{r_{i}}}(φ ∘ P_{r_{i}}^{t}) ∂_{t}d_{(Ω_{t})_{r_{i}}}^{s}(1 − r_{i}κ_{r_{i}}^{t})dH^{1} dt = − ∫_{0}^{T} ∫_{R^{2}} χ_{Ω_{t}}∂_{t}φ dxdt (90)

We first prove that such equality is true if the boundary satisfies additional regularity assumptions.

Lemma 32. Let {Ω_{t}} satisfy (76) - (78) with r = 0. Let V (x, t) be the outer normal velocity of Γ_{t}, defined by (6) at every point (x, t) of differentiability of Γ and defined as 0 at all points where Γ is not differentiable. Then for every φ ∈ C_{c}^{1}(R^{2} × (0, T))

∫_{0}^{T} ∫_{∂Ω_{t}} φV dH^{1}dt = − ∫_{0}^{T} ∫_{R^{2}} χ_{Ω_{t}}∂_{t}φ dxdt (91)

Proof. Let the set E be defined by (5). Then by (76) E has locally finite perimeter. By (77)

H^{2}(∂E \ ∂_{∗}E) = 0,

where ∂_{∗}E is the reduced boundary of E. Let Φ : R^{2} × R^{1} → R^{2} × R^{1} be defined by

Φ(x, t) = (0, ..., 0, φ(x, t))

Let ν_{E}(x, t) be the measure-theoretical outer normal to Γ at (x, t) ∈ Γ. Then by Green-Gauss theorem for sets with finite perimeter ([EGar92], section 5.8)

∫_{0}^{T} ∫_{R^{2}} χ_{Ω_{t}}∂_{t}φ dxdt = ∫_{E} divΦ dxdt = ∫_{Γ} Φ ν_{E} dH^{2},

where

divΦ = ∑_{i=1}^{3} \frac{∂Φ_{i}}{∂x_{i}} , where x_{3} = t.

At every point (x, t) of differentiability of Γ we have

ν_{E}(x, t) = −D_{(x,t)}d_{E}^{s}(x, t),

where d_{E}^{s} is the signed distance to the boundary of the set E in the (x, t)-space. Thus we get

∫_{E} ∂_{t}φ dxdt = − ∫_{Γ} φ ∂_{t}d_{E}^{s} dH^{2}.

Let f : Γ → R^{1} be the mapping defined by

f(x, t) = t

Then f is Lipschitz, and f and Γ are differentiable at H^{2} a.e. point (x, t) ∈ Γ. At such point (x, t) the gradient of f is a linear mapping Df(x, t) : T_{x,t}Γ → R^{1}, where T_{x,t}Γ is the tangent to Γ at (x, t) space. Let e_{1}, e_{2} be such orthonormal basis in R^{2} that e_{2} is the inner normal to ∂Ω_{t} at x. Let τ be the unit vector in the t-direction. Then the vectors

e_{1}, ẽ_{2} = −∂_{t}d_{E}^{s}(x, t)e_{2} + |D_{x}d_{E}^{s}(x, t)|τ

form an orthonormal basis in T_{x,t}Γ. We calculate:

Df(x, t)e_{1} = 0,
Df(x, t)ẽ_{2} = |D_{x}d_{E}^{s}(x, t)|.

Thus

|Df(x, t)| = |D_{x}d_{E}^{s}(x, t)|.

Also, the following relation holds:

D_{x,t}d_{Ω_{t}}^{s}(x, t) = \frac{1}{|D_{x}d_{E}^{s}(x, t)|}D_{x,t}d_{E}^{s}(x, t).

Now, applying formula 3.2.22 of [Fed69] (which is applicable by (76)), we get:

∫_{Γ} φ ∂_{t}d_{E}^{s} dH^{2} = ∫_{Γ} φ ∂_{t}d_{Ω_{t}}^{s}|Df(x, t)| dH^{2}
= ∫_{0}^{T} ∫_{f^{−1}(t)} φ ∂_{t}d_{Ω_{t}}^{s} dH^{1}dt

= ∫_{0}^{T} ∫_{∂Ω_{t}} φ V dH^{1}dt.

The lemma is proved.

Now we can prove (90). Each {(Ω_{t})_{r_{i}}} satisfies (76) - (78). Thus we have

∫_{0}^{T} ∫_{∂(Ω_{t})_{r_{i}}} φV^{(r_{i})}dH^{1}dt = − ∫_{0}^{T} ∫_{R^{2}} χ_{(Ω_{t})_{r_{i}}}∂_{t}φ dxdt, (92)

where V^{r_{i}} is velocity of (Ω_{t})_{r_{i}}. It follows from Proposition 21 that L^{3}(Γ) = 0, and thus

χ_{(Ω_{t})_{r_{i}}}(x) → χ_{Ω_{t}}(x) for a.e. (x, t) ∈ R^{n} × (0, T).

Then by Dominated Convergence Theorem the right-hand side of (92) converges to the right-hand side of (90) as i → ∞. Thus it remains to prove that the left-hand side of (92) and the left-hand side of (90) converge to the same limit.

Let P_{r,r_{i}}^{t} : ∂(Ω_{t})_{r} → ∂(Ω_{t})_{r_{i}} be the nearest point projection (well-defined by Proposition 9). P_{r,r_{i}}^{t} is a Lipschitz map by Propositions 9, 11. Then using Lemma 14 and identity (23), the difference between the left-hand sides of (92) and (90) can be transformed to

∫_{0}^{T} ∫_{∂(Ω_{t})_{r}}(φ ∘ P_{r,r_{i}}^{t} − φ ∘ P_{r}^{t}) V^{r_{i}} ∘ P_{r,r_{i}}^{t} (1 − rκ_{r}^{t})dH^{1} dt + (93)

r_{i} ∫_{0}^{T} ∫_{∂(Ω_{t})_{r}} φ ∘ P_{r,r_{i}}^{t} V^{r_{i}} ∘ P_{r,r_{i}}^{t} κ_{r}^{t} dH^{1} dt = I_{1,i} + r_{i}I_{2,i}.

We have |V^{r_{i}}| < C, where C does not depend on i, and |κ_{r}^{t}| ≤ \frac{1}{r}. Using the fact that P_{r,r_{i}}^{t} → P_{r}^{t} as r_{i} → 0 and Dominated Convergence Theorem we see that I_{1,i} → 0. We also have |I_{2,i}| < C. Thus the expression (93) converges to zero. Thus (90) is proved.

The equalities (89) and (90) imply (75).

Step 3.

The family {Ω_{t}} is expanding. Thus the left-hand side of (90) defines a nonnegative linear functional of φ ∈ C^{∞}(R^{n} × [0, T]). By [EGar92], Chapter 1.8, Corollary 1, it follows from (90) that ∂_{t}χ_{Ω_{t}} is a nonnegative Radon measure. Thus the mass balance equation (75) can be rewritten as

∫_{0}^{T} ∫_{R^{2}} wφ dxtdt + ∫_{R^{2}×(0,T)} φ dw_{t} − ∫_{0}^{T} ∫_{R^{2}} aDuDφ dxtdt = 0.

From this equation it follows that for a.e. t ∈ [0, T], every φ ∈ C_{c}^{1}(R^{2})

∫_{R^{2}} w(·, t)φ dx + ∫_{R^{2}} φ dw_{t}(·, t) − ∫_{R^{2}} a(·, t)Du(·, t)Dφ dx = 0.

Now, repeating argument of Section 6, we conclude the proof of Theorem 30.

Appendix A. Appendix

Let Ω be an open set. We write d(x) for d_{Ω}(x) below. The purpose of this section is to prove that the gradient of d_{Ω} is locally Lipschitz at x ∈ Ω\R and give an estimate of the Lipschitz constant in the terms of the distance between x and endpoints of the distance ray R_{x}.

Proposition 33 should be compared with Proposition 4.1 of [EGan] and with inequality 4.8(8) of [Fed59].

In the inequality (95) below the only assumption regarding the point x_{1} is that it is close enough to x. In particular it is possible that x_{1} ∈ R. In the inequality 4.8(8) of [Fed59] the conditions on x and x_{1} are symmetric and exclude the possibility that x_{1} ∈ R.

Proposition 33 improves the estimates of Proposition 4.1 of [EGan] in the following. Two quantities are estimated explicitly in Proposition 33: the local Lipschitz constant of Dd(·) at x and the size of the neighborhood of x in which the estimates (95) and (96) hold.

Proposition 33. There exist constants C and M depending only on n such that the following is true. Let Ω ∈ R^{n} be an open set. Let ε > 0, x ∈ Ω\R, and let

d_{Ω}(x) ≥ Mε, γ_{Ω}(x) − d_{Ω}(x) ≥ Mε. (94)

Then by Remark 7 there exists a unique ray R_{x}. Denote y and v the lower and upper ends of R_{x}, i.e., y = R_{x} ∩ ∂Ω and v = R_{x} ∩ R. Then for every x_{1} ∈ Ω satisfying | x − x_{1} |< ε, the inequality holds

| y − y_{1} |≤ C (1 + \frac{| x − y |}{| x − v |}) | x − x_{1} |, (95)

where y_{1} ∈ N_{∂Ω}(x_{1}). If in addition the function d(·) is differentiable at x_{1} then

|Dd(x) − Dd(x_{1})| ≤ \frac{C}{ε} |x − x_{1}|. (96)

Proof. We first prove the inequalities (95), (96) assuming that

| x − v |≥| x − y | . (97)

Denote d := d_{Ω}(x). By (97), we can find a point O on the interval of R_{x} between v and x such that | x − O |= d. Let y ∈ N_{∂Ω}(x), then we also have y ∈ N_{∂Ω}(O).

In the calculations below C will denote different constants depending only on n. We assume that

M > 10. (98)

Choose x_{1} ∈ B_{ε}(x). Let x̂_{1} be projection of x_{1} onto R_{x}, then

|x − x̂_{1}| ≤ ε ≤ \frac{1}{10}d

by (94), (98). Thus x̂_{1} lies between y and O on R_{x}. Let d_{1} = |x̂_{1} − O|, then

\frac{11}{10} ≥ \frac{d}{d_{1}} ≥ \frac{9}{10}, |d − d_{1}| < ε (99)

Introduce the following coordinate system in R^{n}: let the point O be the origin, let e_{n} = \frac{x−O}{|x−O|} (thus e_{n} is the unit vector along the ray R_{x}), let e_{1}, ..., e_{n−1}, e_{n} be an orthonormal basis in R^{n}. Then in these coordinates

x = (0, d)_{,} x̂_{1} = (0, d_{1}), y = (0, 2d),

where 0 ∈ R^{n−1}. We also have x_{1} = (x′_{1}, d_{1}), where x′_{1} ∈ R^{n−1}. Let y_{1} ∈ ∂Ω be such that |x_{1} − y_{1}| = d(x_{1}). Let the coordinates of y_{1} be (y′_{1}, y_{1,n}) where y′_{1} ∈ R^{n−1}, y_{1,n} ∈ R^{1}. Then

|x_{1} − y_{1}| ≤ |x_{1} − y| (100)

or

|x′_{1} − y′_{1}|^{2} + |d_{1} − y_{1,n}|^{2} ≤ |x′_{1}|^{2} + |d_{1} − 2d|^{2} (101)

Claim 34. There exists M_{1}, depending on n, such that for any δ ∈ (0, 2) the following is true:

if d > \frac{M_{1}}{δ^{2}} ε then |y′_{1}| < δd. (102)

Proof. Note that

|x′_{1}| < ε, |y′_{1}| < 2d. (103)

Indeed, the first inequality is true since x_{1} ∈ B_{ε}(x). To prove the second inequality of (103) we use (101) and (99) to get

|y′_{1} − x′_{1}|^{2} ≤ ε^{2} + |d + ε|^{2} < (d + \sqrt{2}ε)^{2},

so

|y′_{1}| ≤ |x′_{1}| + d + \sqrt{2}ε < d + 3ε < 2d (104)

since d ≥ 10ε. Thus (103) is proved.

Suppose that the assertion of Claim 34 is false, that is

|y′_{1}| ≥ δd. (105)

By (102)

δd > \frac{M_{1}}{δ} ε > 2ε (106)

if M_{1} > 4. Then we get from (101), (103), (99), (105)

(y_{1,n} − d_{1})^{2} ≤ ε^{2} + (2d − d_{1})^{2} − (|y′_{1}| − |x′_{1}|)^{2}
(107)  ≤ ε^{2} + (\frac{11}{10}d)^{2} − (δd − ε)^{2}
< 4d^{2}(1 − \frac{δ^{2}}{4} ) − 2d^{2} + 2δdε
≤ 4d^{2}(1 − \frac{δ^{2}}{4} )

The last inequality follows from −2d^{2} + 2δdε < 0 which holds by (102) if δ < 2 and M_{1} > 8.

Consider two cases: \underline{Case 1}: y_{1,n} ≥ d_{1}. Denote by α the angle between vectors y_{1} − x_{1} and (x′_{1}, y_{1,n}) − x_{1}. In the triangle x_{1}, y_{1}, (x′_{1}, y_{1,n}) the side (x′_{1}, y_{1,n}) − y_{1} is orthogonal to x_{1} − (x′_{1}, y_{1,n}), since (x′_{1}, y_{1,n}) − y_{1} = (x′_{1} − y′_{1}, 0), and x_{1} − (x′_{1}, y_{1,n}) = (0, d_{1} − y_{1,n}). Thus we get

tan α = \frac{|y′_{1} − x′_{1}|}{y_{1,n} − d_{1}}.

But

|y′_{1} − x′_{1}| ≥ ||y′_{1}| − |x′_{1}|| = |δd − ε| = δd − ε

by (106). By (107) we get

tan α ≥ \frac{δd − ε}{2d\sqrt{1 − \frac{δ^{2}}{4}}} = \frac{δ}{2\sqrt{1 − \frac{δ^{2}}{4}}} (1 − \frac{1}{δ} · \frac{ε}{d}) ≥ \frac{δ}{4\sqrt{1 − \frac{δ^{2}}{4}}} .

where the last inequality follows from (106). Thus, using the condition δ ∈ (0, 2) we get

| cos α| = \frac{1}{\sqrt{1 + tan^{2} α}} ≤ \sqrt{\frac{4(4 − δ^{2})}{16 − 3δ^{2}}} ≤ 1 − \frac{δ^{2}}{2(16 − 3δ^{2})}. (108)

Consider the triangle x_{1}, y_{1}, (x′_{1}, 0). The angle between the vectors (x′_{1}, 0) − x_{1} and y_{1} − x_{1} is π − α, and so we get

|y_{1} − (x′_{1}, 0)|^{2} = |x_{1} − (x′_{1}, 0)|^{2} + |y_{1} − x_{1}|^{2} − 2|x_{1} − (x′_{1}, 0)||y_{1} − x_{1}| cos(π − α). (109)

Since x_{1} = (x′_{1}, d_{1}), we have

|x_{1} − (x′_{1}, 0)| = d_{1}.

From (101) we get

|y_{1} − x_{1}| ≤ \sqrt{ε^{2} + (2d − d_{1})^{2}}.

Then by (109), (108) we get

|y_{1} − (x′_{1}, 0)|^{2} ≤ d_{1}^{2} + [ε^{2} + (2d − d_{1})^{2}] + 2d_{1}\sqrt{ε^{2} + (2d − d_{1})^{2}} (1 − \frac{δ^{2}}{32})
≤ d_{1}^{2} + (2d − d_{1})^{2} + 2d_{1}(2d − d_{1})(1 + \frac{ε^{2}}{2(2d − d_{1})^{2}} )(1 − \frac{δ^{2}}{32}) + ε^{2}

≤ 4d^{2} + 2d_{1} \frac{ε^{2}(1 − \frac{δ^{2}}{32})}{2(2d − d_{1})} + ε^{2} − 2d_{1}(2d − d_{1})\frac{δ^{2}}{32}
≤ 4d^{2} + d_{1} \frac{ε^{2}}{2d − d_{1}} + ε^{2} − \frac{1}{16}d_{1}(2d − d_{1})δ^{2}.

Thus we see that

|y_{1} − (x′_{1}, 0)|^{2} < (2d − ε)^{2}, (110)

provided

d_{1} \frac{ε^{2}}{2d − d_{1}} − \frac{1}{16}d_{1}(2d − d_{1})δ^{2} < −4dε. (111)

Using (99) we see that (111) is satisfied if

d^{2}δ^{2} > C(ε^{2} + dε) (112)

where C is a large enough constant. In its turn, (112) is true if d > \frac{M_{1}}{δ^{2}} ε, where M_{1} is large enough. Thus, for such d, the inequality (110) holds, and so

|y_{1} − (x′_{1}, 0)| < 2d − ε.

Then

|y_{1}| ≤ |x′_{1}| + |y_{1} − (x′_{1}, 0)| < 2d = |y|.

This contradicts the fact that y ∈ N_{∂Ω}(O). Thus (105) is false. Thus Case 1 of Claim 34 is proved.

\underline{Case 2:} y_{1,n} ≤ d_{1}.

Using this condition, we get

|(x′_{1}, 0) − y_{1}|^{2} = |(x′_{1}, 0) − x_{1} + x_{1} − y_{1}|^{2} = | − (0, d_{1}) + (x′_{1} − y′_{1}, d_{1} − y_{1,n})|^{2}
= d_{1}^{2} + |x_{1} − y_{1}|^{2} + 2(y_{1,n} − d_{1})d_{1}
≤ d_{1}^{2} + |x_{1} − y_{1}|^{2} ≤ d_{1}^{2} + ε^{2} + (2d − d_{1})^{2}
= 4d^{2} − 2d_{1}(2d − d_{1}) + ε^{2} < (2d − ε)^{2},
if d_{1}(2d − d_{1}) > 2dε.

The last inequality is in fact true by (99) and d > 10ε. Then, as in the Case 1, we arrive to a contradiction. Thus Case 2 cannot happen for d and ε defined in Claim 34. Thus, Claim 34 is proved.

Now the rest of the proof of part (a) of Proposition 11 follows the proof of Proposition 4.1 in [EGan].

The main step is to show that there exist constants C and M depending only on n such that the inequality d > Mε implies

|y − y_{1}| ≤ C|x̂_{1} − x_{1}|. (113)

Let δ ∈ (0, 1) be chosen later, and let d > \frac{M_{1}}{δ^{2}} ε where M_{1} is the constant from Claim 34. Then by Claim 34 we get

|y′_{1}| < δd. (114)

Since y ∈ N_{∂Ω}(O), we get |y_{1}| > |y| = 2d, and so

|y′_{1}|^{2} + |y_{1,n}|^{2} ≥ 4d^{2}.

Thus we estimate using (114)

y_{1,n} ≥ \sqrt{4d^{2} − |y′_{1}|^{2}} = 2d\sqrt{1 − \frac{|y′_{1}|^{2}}{4d^{2}}} ≥ 2d(1 − \frac{|y′_{1}|^{2}}{8d^{2}} − C \frac{|y′_{1}|^{4}}{d^{4}} )
(115)  ≥ 2d − \frac{|y′_{1}|^{2}}{4d} − C \frac{|y′_{1}|^{4}}{4d^{3}} ≥ 2d − \frac{|y′_{1}|^{2}}{4d} (1 + Cδ^{2}),

where all constants C do not depend on y′_{1} and d. So

0 < 2d − d_{1} = y_{1,n} − d_{1} + 2d − y_{1,n} ≤ (y_{1,n} − d_{1}) + \frac{|y′_{1}|^{2}}{4d} (1 + Cδ^{2}).

Now we can estimate

(2d − d_{1})^{2} ≤ (y_{1,n} − d_{1})^{2} + (y_{1,n} − d_{1})\frac{|y′_{1}|^{2}}{2d} (1 + Cδ^{2}) + \frac{|y′_{1}|^{4}}{16d^{2}} (1 + Cδ^{2})
(116)  ≤ (y_{1,n} − d_{1})^{2} + (y_{1,n} − d_{1})\frac{|y′_{1}|^{2}}{2d} (1 + Cδ^{2}) + Cδ^{2}|y′_{1}|^{2}

where we again used (114). Using (101), (99) and (103) we get the following estimate

|y_{1,n} − d_{1}| ≤ \sqrt{ε^{2} + (2d − d_{1})^{2}} = (2d − d_{1})\sqrt{1 + \frac{ε^{2}}{(2d − d_{1})^{2}}}

≤ (2d − d_{1})(1 + \frac{ε^{2}}{2(2d − d_{1})^{2}} ) ≤ (2d − d_{1})(1 + C \frac{ε^{2}}{d^{2}} )
≤ (2d − d_{1})(1 + Cδ^{4}) since d ≥ \frac{M_{1}}{δ^{2}} ε.

Thus, by (116):

(2d − d_{1})^{2} ≤ (y_{1,n} − d_{1})^{2} + (2d − d_{1})\frac{|y′_{1}|^{2}}{2d} (1 + Cδ^{4}) + C|y′_{1}|^{2}δ^{2}
(117)  ≤ (y_{1,n} − d_{1})^{2} + Θ|y′_{1}|^{2},

where

Θ = (1 − \frac{d_{1}}{2d})(1 + Cδ^{4}) + Cδ^{2}.

Using (99) we estimate

Θ ≤ \frac{11}{20}(1 + Cδ^{4}) + Cδ^{2} < 1,

if δ > 0 is small enough. Fix such δ. Let M_{1} be the constant from Claim 34 defined by this δ. Let M = \frac{M_{1}}{δ^{2}} be our choice of the constant M in the inequalities (94). Then the inequalities (94) imply (117) with Θ < 1. Thus (117) and (101) give:

|y′_{1} − x′_{1}|^{2} ≤ |x′_{1}|^{2} + Θ|y′_{1}|^{2}. (118)

Now we estimate:

| y′_{1} |≤| y′_{1} − x′_{1} | + | x′_{1} |

and from this, using (118), we get

| y′_{1} |^{2} ≤ | y′_{1} − x′_{1} |^{2} +2 | y′_{1} − x′_{1} || x′_{1} | + | x′_{1} |^{2}
≤ (1 + μ) | y′_{1} − x′_{1} |^{2} +(1 + \frac{C}{μ}) | x′_{1} |^{2}
≤ (1 + μ)(|x′_{1}|^{2} + Θ|y′_{1}|^{2}) + (1 + \frac{C}{μ}) | x′_{1} |^{2} .

Choose μ > 0 so small that (1 + μ)Θ < 1, then we get

| y′_{1} |≤ C | x′_{1} |= C | x_{1} − x̂_{1} | . (119)

It remains to estimate | y_{1,n} − y_{n} | where y_{n} = 2d.

From (115) we get

2d − y_{1,n} ≤ C \frac{| y′_{1} |^{2}}{d} ≤ C | y′_{1} |, (120)

since | y′_{1} |< δd.

Since x_{1} = (x′_{1}, d_{1}), d(x_{1}) =| x_{1} −y_{1} |, and d(x̂_{1})) =| x̂_{1} −y |= 2d−d_{1}, we calculate (using the fact that Lip[d(·)] = 1)

y_{1,n} − d_{1} ≤| y_{1} − x_{1} |= 2d − d_{1} + d(x_{1}) − d(x̂_{1}) ≤ 2d − d_{1}+ | x_{1} − x̂_{1} |

and so

y_{1,n} − 2d ≤| x_{1} − x̂_{1} | . (121)

The estimates (119), (120), and (121) imply (113).

Now since

Dd(x_{1}) = \frac{x_{1} − y_{1}}{| x_{1} − y_{1} |}, Dd(x̂_{1}) = \frac{x̂_{1} − y}{| x̂_{1} − y |},

we get the following estimate

|Dd(x_{1}) − Dd(x̂_{1})| ≤ \frac{|x_{1} − x̂_{1}| + |y_{1} − y|}{|x̂_{1} − y|},

then using the equality |x̂_{1} − y| = 2d − d_{1} and the inequalities (99) and (117) we deduce that

|Dd(x_{1}) − Dd(x̂_{1})| ≤ \frac{C}{d} |x_{1} − x̂_{1}|.

Note that Dd(x̂_{1}) = Dd(x) since the points x and x̂_{1} are on the same ray R_{x}^{1}. Now from |x̂_{1} − x_{1}| ≤ |x − x_{1}|, we get (96) in the case of (97). Note also that (113) and inequality |x̂_{1} − x_{1}| ≤ |x − x_{1}| imply (95).

In the case | x − v |<| x − y | we can make the following reduction to the case (97). Let

Q =| x − y | − | x − v |, (122)

and let Ω̃ be the set {z ∈ Ω | dist(z, ∂Ω) > Q}. Then B_{ε}(x) ⊂ Ω̃ since the function dist(·, ∂Ω) is Lipschitz with constant 1. For z ∈ Ω̃ we have

(123) dist(z, ∂Ω)) = dist(z, ∂Ω̃) + Q.

To see this consider z_{1} ∈ N_{∂Ω}(z), and let z_{2} be the point of intersection of the interval connecting z and z_{1} with ∂Ω̃. Then z_{1} ∈ N_{∂Ω}(z_{2}) since z_{2} lies in the relative interior of the distance ray R_{z_{1}} that intersects ∂Ω in z_{1}. Thus | z_{1} − z_{2} |= Q. Now suppose that there exists a point ẑ_{2} ∈ ∂Ω̃ such that | z − ẑ_{2} |<| z − z_{2} |. Let ẑ_{1} ∈ N_{∂Ω}(ẑ_{2}), then | ẑ_{1} − ẑ_{2} |= Q. Then

| z − ẑ_{1} |≤| z − ẑ_{2} | + | ẑ_{2} − ẑ_{1} |<| z − z_{2} | +Q =| z − z_{1} |,

a contradiction to the fact that z_{1} ∈ N_{∂Ω}(z). This proves (123). So the distance rays and the ridge set of the set Ω̃ are intersections of those of the set Ω with the set Ω̃.

Then in (96) we can consider d(·) as the distance to ∂Ω̃, so we can consider Ω̃ instead of Ω, and then the inequality (97) is satisfied.

To show (95), denote by w and w_{1} the points of intersection of ∂Ω̃ with the intervals (x, y) and (x_{1}, y_{1}) respectively. Then it follows from (123) w ∈ N_{∂Ω̃}(x) and w_{1} ∈ N_{∂Ω̃}(x_{1}), and

| w − y |=| w_{1} − y_{1} |= Q.

Then it follows from (122) that

| x − v |=| x − w | .

Using (122) and letting A = \frac{|x−y|}{|x−v|}, we get

y = x + A(w − x),
y_{1} = x_{1} + [1 + (A − 1)\frac{| x − w |}{| x_{1} − w_{1} |}] (w_{1} − x_{1}).

The inequality (95) can be applied to the set Ω̃, so

| w − w_{1} |≤ C | x − x_{1} | .

Now we estimate:

| y − y_{1} | = |x − x_{1} + A [(w − x) − \frac{| x − w |}{| x_{1} − w_{1} |}(w_{1} − x_{1})]

+ (\frac{| x − w |}{| x_{1} − w_{1} |} − 1) (w_{1} − x_{1})|

≤ C(1 + A)(| x − x_{1} | + | w − w_{1} |)
≤ C(1 + A) | x − x_{1} | .

This implies (95).

Acknowledgement

It is the author’s pleasure to thank L. C. Evans and R. F. Gariepy for their interest, insight and and encouragement.

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