Level Set Regularization Using Geometric Flows

Level Set Regularization Using Geometric Flows

1

Luis Alvarez† , Carmelo Cuenca† , Jes´ us Ildefonso D´ıaz‡ ,

2

and

Esther González†

3 4 5 6 7 8 9 10 11 12

Abstract. In this paper we study a geometric partial differential equation including a forcing term. This equation defines an hypersurface evolution that we use for level set regularization. We study the shape of the radial solutions of the equation and some qualitative properties about the level set propagations. We show that under a suitable choice of the forcing term, the geometric equation has non trivial asymptotic states and it represents a model for level set regularization. We show that by using a forcing term which is merely a bounded H¨ older continuous function, we can obtain finite time stabilization of the solutions. We introduce an explicit finite difference scheme to compute numerically the solution of the equation and we present some applications of the model to nonlinear 2D image filtering and 3D segmentation in the context of medical imaging.

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Key words. Geometric flows, level sets evolution, partial differential equations.

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AMS subject classifications. 35K55, 35D40, 35R37.

This is a pre-print of an article published in SIAM Journal on Imaging Sciences. The final 16 authenticated version is available at: https://epubs.siam.org/doi/abs/10.1137/17M1139722

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1. Introduction. In this paper we consider the Cauchy problem for the parabolic perturbed 18 mean curvature type equation ∂u 2 n ∂t = F (∇u, ∇ u) + k(x)|∇u| in (0, T ) × R 19 (1) n u(0, x) = u0 (x) on R , 17

where ∇u and ∇2 u denotes respectively the gradient and the Hessian of u in spatial variables, 21 k : Rn → R is a bounded H¨ older continuous function which introduces a forcing term in 22 equation (1) and F : Rn − {0} × Sn → R is a given continuous function satisfying 20

− F (p, X) ≤ −F (p, Y ) ∀X, Y ∈ Sn with X ≥ Y, ∀p ∈ Rn ,

23 24

(2)

25

(3)

26

where ⊗ denotes the tensor product in Rn . Here Sn denotes the space of the n × n real and symmetric matrices with the usual Loewner order. We say equation (1) generates a geometric flow if F satisfies (3) (we point out that as k|λp| = λk|p|, the forcing term does not change the geometric character of (3)). In this paper, we pay special attention to the mean curvature operator given by n n X X uxi uxj ∇u 2 (4) F (∇u, ∇ u) = div |∇u| = uxi xi − ux x , |∇u| |∇u|2 i j

27 28 29 30 31

F (λp, λX + σp ⊗ p) = λF (p, X)

∀λ > 0, σ ∈ R, ∀p ∈ Rn , ∀X ∈ Sn ,

i=1

i,j=1

† CTIM (Centro de Tecnolog´ıas de la Imagen), Departamento de Inform´ atica y Sistemas. Universidad de Las Palmas de Gran Canaria, 35017. Las Palmas de G.C., Spain. email:{lalvarez,carmelo.cuenca,esther.gonzalez}.ulpgc.es ‡ Instituto de Matem´ atica Interdisciplinar, Departamento de Matem´ atica Aplicada. Universidad Complutense de Madrid, Parque de Ciencias 3, 28040-Madrid, Spain. email:[email protected]

1

This manuscript is for review purposes only.

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which satisfies (2)-(3). One of the main interests of the geometric flows in the context of Computer Vision is that they are closely related to the following level set formulation of front 34 propagations introduced in [27], [30]: given an hypersurface Γ0 ⊂ Rn , we embed Γ0 in a 35 function u0 : Rn → R in such a way that Γ0 = ∂{x : u0 (x) < 0}, and we make u0 evolve using 36 the partial differential equation 32 33

37

(5)

∂u + Fext |∇u| = 0. ∂t

The hypersurface evolution is then defined by the set Γt = ∂{x : u(t, x) < 0}. Fext is a function representing the front propagation velocity in the hypersurface normal direction. This elegant 40 formulation has been extensively applied in the literature to address a variety of problems. 41 For instance, in [10], authors propose an active contour model where Fext is given by ∇u 42 (6) (x) + ν , Fext (x) = −g(x) div |∇u| 38 39

where ν ∈ R and g(x) ≥ 0. In [13], [14] authors introduce the ”geodesic snake model” where Fext is given by: ∇u ∇u Fext (x) = −g(x) div 45 (7) (x) + ν − ∇g(x) (x). |∇u| |∇u| 43

44

In the context of 3D medical image segmentation, in [22], authors propose a model for 3D 47 vessel segmentation based on the front propagation velocity

46

48

(8)

Fext (x) = g(x)(1 − κmin (x)),

where ∈ R, g(x) ∈ [−1, 1] and κmin (x) represents the minimum value of the curvatures of the level set surface of u passing by x. 51 We point out that if we choose the mean curvature operator (4) in equation (1), then 52 equation (1) is equivalent to (5) with ∇u 53 (9) Fext (x) = − div (x) + k(x) . |∇u| 49

50

54 55 56 57 58 59 60 61 62 63

From the above examples, we can observe that equation (5) is very flexible and that it can be adapted to many different problems accordingly to the choice of the velocity term Fext . From a mathematical point of view there are two main important issues related to the hypersurface evolution defined by equation (5). The first one is that the evolution of Γt should be independent of the choice of the function u0 where Γ0 is embedded. The main interest of the geometric assumption (3) is that it ensures, as we will see below, that under the action of (1), Γt is independent of the choice of u0 . The second one is whether equation (5) defines a wellposed problem, that is, we would like to show that equation (5) has a unique solution in some sense and that we have some other nice properties like comparison principle, L∞ stability of the solutions, etc. The usual framework to study these issues is the viscosity solutions theory


3

84

introduced at the end of the last century in some seminal papers as, for instance, [17], [16] and [18] (see also the presentation made in the more recent monograph [20]). In the case of second order differential operators, the well-posedness of the problem is closely related to the ellipticity condition (2) and to the regularity of functions like g(x) and k(x) in the examples above. For instance, the existence and uniqueness of the viscosity solutions for equations (6), 1 (7) with g(x) ≥ 0 and g, g 2 ∈ W 1,∞ (Rn ) (for n = 2, 3), have been shown in [10, 14, 8]. In most of the papers concerning the case in which a forcing term, k(x), is involved, it is assumed that k ∈ W 1,∞ (Rn ). In the Section 2 of this paper we study the case in which k(x) is merely a bounded H¨ older continuous function. Equation (8) is an heuristic model that has not been well studied mathematically yet (notice that g(x) ∈ [−1, 1] and thus the associated differential operator loses its elliptic character). In fact, given the generality of equation (5), the possible absence of a correct mathematical treatment in terms of PDEs may justify many strange properties of the front propagation. Comparison principle is a very important topic in PDE theory. In particular, in terms of ˆt = front propagation, it means that if the evolution of two level sets Ut = {x : u(t, x) < 0}, U {x : u ˆ(t, x) < 0} (where u(t, x), u ˆ(t, x) are solutions of the associated PDE) satisfying that ˆ ˆ U0 ⊆ U0 , then Ut ⊆ Ut ∀t ≥ 0. This is a nice and useful behavior of the front propagation. For instance, in this paper, we make use of this comparison principle to compare the behavior of radial solutions with the one of non-radial solutions of the equation. As we shall comment with more details in Section 2, the comparison principle for the viscosity solutions of the more general equation

85

(10)

86

proved in [21] under very general assumptions on G, can be applied to our mentioned framework. A characterization of the profile of the function G using a ”principle” approach including the geometric character of the equation was introduced in [3, 4, 5]. As pointed out in [11] and [12], the topographic map, based on the image level set collection is a robust and invariant image description suitable for image analysis. In terms of curve evolution, the geometric flows were studied in [6, 7, 28, 29] (see also the presentation made in the monograph [20]). In this paper we focus our attention on the study of the evolution of the level sets under the action of the geometric flow (1). Our main motivation is that, as we will show in this paper, with a suitable choice of the forcing term k(x), equation (1) can be used as a level set regularization model. Moreover (see e.g. Lemma 8), if k(x) is a non-Lipschitz function (k being merely H¨ older continuous) then we can obtain finite time stabilization of the solutions. The main contributions of the paper are: (i) the qualitative study of the solutions of equation (1) with a forcing term k(x) and specially the proof of some qualitative properties which are exclusive to the case in which k(x) is merely a Hölder continuous function (based on the study of the level set propagation by using a sharp analysis of radial solutions), (ii) the proposal of equation (1) as a level set regularization model using an appropriate choice of the forcing term k(x), and (iii) the illustration of the results of the proposed model in a variety of situations including nonlinear filtering of 2D images or 3D image segmentation in the context of medical imaging.

64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83

87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105

∂u + G(t, x, u, ∇u, ∇2 u) = 0, ∂t


4 106 107 108 109 110 111 112 113 114

The rest of the paper is organized as follows: in Section 2, we apply some previous results in the literature to get the existence and comparison of solutions of (1) in the context of the viscosity solution framework. In Section 3, we study the hypersurface evolution obtained from the solutions of the equation (1). Section 4 is devoted to the study of equation (1) in radial coordinates. In Section 5, we present some qualitative properties of the hypersurface evolution. An explicit finite difference discretization scheme for equation (1)-(4) is presented in Section 6 and some applications of equation (1)-(4) to image level set regularization including some experimental results are presented in Section 7. Finally, in Section 8 we summarize the main conclusions of the paper.

124

2. On the existence and comparison of solutions. As we shall see in the next sections, one of the main qualitative properties of solutions of equation (1), i.e. the finite time stabilization of radial solutions, holds only when k(x) is merely a Hölder continuous function (i.e. the property is not satisfied when k(x) is assumed to be Lipschitz continuous). In most of the presentations on the existence and comparison of solutions of general formulations of the equation given in terms of equation (10) it is assumed that G(t, x, u, ∇u, ∇2 u) is independent of x or, at least G(t, x, u, p, O) is Lipschitz continuous on x. The motivation of this section is simply to recall the general results obtained in [21] and to check that they apply without any special difficulty to our framework: the equation (1) with (9) and when k(x) is a bounded Hölder continuous function. Notice that this corresponds to the choice

125

(11)

126

(see (4)). So we can write

115 116 117 118 119 120 121 122 123

G(t, x, u, p, X) = −F (p, X) − k(x) |p| with F (p, X) := trace((I − p ⊗ p)X).

127

128

G(t, x, u, p, X) = −trace((I − p ⊗ p)X) − k(x) |p| , p := p/ |p| ,

We start by recalling the notion of viscosity solution:

Definition 1. i) A function u : [0, T ] × Rn → R is a (viscosity) subsolution of (1) if 130 u is upper semicontinuous (usc), u(0, x) ≤ u0 (x) for all x ∈ Rn , and for all ϕ ∈ C2 , if 131 (b t, x b) ∈ [0, T ] × Rn is a maximum of u − ϕ then 129

132

(12)

ϕt (b t, x b) + F (Dϕ(b t, x b), D2 ϕ(b t, x b)) − k(b x)|Dϕ(b t, x b)| ≤ 0,

where F is the lower semicontinuous envelope of −F. ii) A function u : [0, T ] × Rn → R is a (viscosity) supersolution of (1) if u is lower 135 semicontinuous (lsc), u(0, x) ≥ u0 (x) for all x ∈ Rn , and for all ϕ ∈ C2 , if (b t, x b) ∈ [0, T ] × Rn 136 is a minimum of u − ϕ then

133

134

137

(13)

ϕt (b t, x b) + F (Dϕ(b t, x b), D2 ϕ(b t, x b)) − k(b x)|Dϕ(b t, x b)| ≥ 0,

where F is the upper semicontinuous envelope of −F. iii) A function u : [0, T ] × Rn → R is a (viscosity) solution of (1) if u is both a (viscosity) 140 subsolution and a (viscosity) supersolution of (1). 138

139


5 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172

We recall that there are some other equivalent definitions expressed in terms of the so called parabolic super 2-jets (see, e.g. [17] and [20]) but they will not be used in this paper. The following existence and uniqueness of solutions result is a special case of Theorem 4.1 and Theorem 4.9 of [21]. We recall some previous notations: given a constant β ∈ R and a metric space K, Cβ (K) denotes the space of continuous functions v such that v(.)− β is compactly supported in K. Proposition 2. Let k(x) be a bounded H¨ older continuous function and G given by (11). Assume that u0 ∈ Cβ (Rn ) for some constant β ∈ R. Then problem (1) has a unique viscosity solution u ∈ Cβ ([0, T ] × Rn ). Moreover the comparison principle holds under the above conditions, i.e. if u(t, x) and u ¯(t, x) are, respectively, a subsolution and a supersolution of (1) satisfying that u(0, x) ≤ u ¯(0, x) ∀x ∈ Rn , then ¯ u(t, x) ≤ u ¯(t, x) ∀(t, x) ∈ [0, T ] × Rn . ¯ Proof. It reduces to check the assumptions made in Theorem 4.1 and Theorem 4.9 of [21]. Keeping the same notation as in Theorem 4.1 of [21] applied to the operator G(t, x, u, p, X) given by (11), it is obvious that: i) G(t, x, u, p, X) is a continuous function over J0 := [0, T ] × Rn × R × (Rn \ {0}) × Sn . ii) G(t, x, u, p, X) is degenerate elliptic (thanks to (2)). iii) −∞ 0 cR := sup{|G(t, x, u, p, X)| :|p| , |X| ≤ R, (t, x, u, p, X) ∈ J0 } < ∞ (since k is bounded). v) G(t, x, u, p, X) is independent on u. vi) For every R > ρ there is a modulus σ (i.e. σ : [0, ∞) → [0, ∞), σ(0) = 0 and α is nondecreasing), σ = σRρ , such that |G(t, x, u, p, X) − G(t, x, u, q, Y )| ≤ σRρ (|p − q| + |X − Y |), for all (t, x, u) ∈ [0, T ] × Rn × R, ρ ≤ |p| , |q| ≤ R, |X| , |Y | ≤ R (again due to the boundedness of k). vii) There are ρ0 > 0 and a modulus σ1 such that F (p, X) − F (0, O) − k(x) |p| ≤ σ1 (|p| + |X|) F (p, X) − F (0, O) − k(x) |p| ≥ −σ1 (|p| + |X|), provided x ∈ Rn and |p| , |X| ≤ ρ0 (once again by the boundedness of k) and, finally viii) There is a modulus σ2 such that |G(t, x, u, p, X) − G(t, y, u, q, Y )| ≤ σ2 (|x − y|)(|p| + 1),

for (t, x, u, p, X), (t, y, u, p, X) ∈ J0 (take σ2 (s) = M sα , with α ∈ (0, 1) the Hölder 174 exponent of k(x)). 175 The assumptions of Theorem 4.9 of [21] requires additionally the condition G geometric, but 176 this holds obviously since 173

177

G(x, λp, λX + σp ⊗ p) =λG(x, p, X) ∀λ > 0, σ ∈ R, ∀p∈ Rn and ∀x∈ Rn .


6 178 179 180 181 182 183 184

Remark 1. As a matter of fact, since the conclusion in the above result shows that u ∈ Cβ ([0, T ] × Rn ), we could consider also data k(x) which are H¨ older continuous but unbounded n in R . Indeed, in that case we can truncate function k(x) outside a ball BR (0), with radius R > 0 large enough, and to apply Theorem 1 to this truncation kR (x) which now is bounded and H¨ older continuous in Rn . Obviously, the speed of propagation would increase with the “size” of the (compact) support of the initial data, but in any case the speed would be finite for any given u0 ∈ Cβ (Rn ).

An important property (direct consequence of the results of [18] and [16]), unique to the 186 geometric flows, is that the composition of any solution with a nondecreasing function, remains 187 a solution of the equation. So we have the following useful corollary 185

Corollary 3. If u(t, x), is a bounded uniformly continuous viscosity solution of (1), then for any nondecreasing function φ : R → R the function v(t, x) = φ(u(t, x)) is a viscosity solution 190 of (1). 188 189

191 192 193 194 195 196

197 198 199 200 201 202

Remark 2. In [25],[23], [24] the authors study the problem (1) with obstacles for a forcing term k(x) assumed to be bounded and Lipschitz continuous. We conjecture that it seems possible to extend the existence and uniqueness of solutions also to this context and under the mere assumption of k(x) being bounded H¨ older continuous. We also mention that weaker assumptions on k(x) (as k ∈ L∞ (Rn )) were assumed in [15] for the study of (1)-(4) in terms of the curve evolution context. Remark 3. The assumption u0 (.)− β is compactly supported in Rn is not restrictive at all when we consider bounded hypersurface evolution. Indeed, if Γ0 is a bounded hypersurface, using the signed distance function we can embed Γ0 in a function u0 ∈ Cβ (Rn ). Moreover, as we shall see in the next section, the hypersurface evolution is independent of the particular choice of function u0 where Γ0 is embedded and therefore the size of the support of u0 (.) − β is not relevant in terms of the hypersurface evolution.

3. Hypersurface evolution using the parabolic perturbed mean curvature equation. 204 We will consider an initial hypersurface Γ0 of Rn as the boundary of a bounded open set 205 U0 . We choose any u0 : Rn → R such that u0 ∈ Cβ (Rn ) for some constant β ∈ R satisfying 203

206

(14)

 x ∈ U0  u0 (x) < 0 if u0 (x) = u (x) > 0 if x ∈ Rn − U0  0 u(x) = 0 if x ∈ Γ0 = ∂U0 .

It is not difficult to see ([20]) that, in fact, we can choose as u0 any bounded uniformly 208 continuous function if the datum k(x) is Lipschitz continuous. Then, we define {Ut }t≥0 and 209 {Γt }t>0 as 207

210

(15)

Ut = {x ∈ Rn : u(t, x) < 0},


7

211

(16)

Γt = ∂Ut ,

where u(t, x) is the (unique) viscosity solution of (1) for the initial datum u0 (x). One important remark is that, in contrast to the case where no forcing term is included (k ≡ 0), the sign 214 of u0 (x) in U0 matters in the evolution of U0 . Indeed if u(t, x) is a solution of (1)-(4), then 215 v(t, x) = −u(t, x) is the solution of (1)-(4) but changing k(x) by −k(x) and therefore Ut , Γt 216 will be different using u0 (x) or −u0 (x) as initial surface to embed Γ0 .

212 213

217 218

Let us prove first that Ut and Γt are well defined in the sense that they are independent of the choice of u0 .

Theorem 4. Let U0 be an open bounded set, Γ0 = ∂U0 , F satisfying (2)-(3) and u0 ∈ 220 Cβ (Rn ), u ˆ0 ∈ Cβb(Rn ), for some constants β, βb ∈ R, satisfying (14). Then, for any t > 0 219

221

{x ∈ Rn : u(t, x) < 0} = {x ∈ Rn : u ˆ(t, x) < 0},

222

where u(t, x), u ˆ(t, x) are the viscosity solutions of (1) associated to the initial data u0 (x), u ˆ0 (x).

Proof : (This proof is based on the techniques presented in [18],[23]). First, we are going to 224 show that if u0 (x), u ˆ0 (x) satisfy (14) then we can build a nondecreasing function φ : R → R 225 satisfying that {φ ≡ 0} = {0} and u ˆ0 (x) ≤ φ(u0 (x)). We define E0 = Rn and Em = {x ∈ Rn : 1 226 u0 (x) ≤ m }, for m ∈ Z − {0} then we have for m > 0 223

227

(17)

E−1 ⊆ · · · ⊆ E−m ⊆ · · · ⊂ U0 ⊂ · · · ⊆ Em ⊆ · · · ⊆ E0 ,

next we define a0 = supx∈Rn u ˆ0 and supx∈Em−1 u ˆ0 if Em−1 6= ∅ 229 am = n inf x∈R u ˆ0 if Em−1 = ∅,

228

230 231 232

233

for m ∈ Z − {0}. Using (14) -(17) we obtain for m > 0 a−1 ≤ · · · ≤ a−m ≤ · · · < 0 < · · · ≤ am ≤ · · · ≤ a1 . Finally, we define φ(s) be a continuous nondecreasing function satisfying φ(0) = 0 1 φ( m ) = am ∀m ∈ Z − {0} 1 1 φ linear on [ m+1 ,m ] m = 1, 2, . . . 1 1 φ linear on [ m , m−1 ] m = −1, −2, . . . φ constant outside [−1, 1],

by construction, we can easily show that u ˆ0 (x) ≤ φ(u0 (x)) and by the comparison principle 235 showed in Proposition 2 we obtain 234

236

u ˆ(t, x) ≤ φ(u(t, x)),


8 237

and then {x ∈ Rn : u ˆ(t, x) < 0} ⊆ {x ∈ Rn : φ(u(t, x)) < 0} = {x ∈ Rn : u(t, x) < 0},

238 239

the opposite inclusion is proven in a similar way switching u(t, x) and u ˆ(t, x).

Let U0 a bounded open set, Γ0 = ∂U0 . A typical choice of function u0 (x) satisfying (14) is the following truncated signed distance function  x∈ U0  −dist(x, Γ0 ) if n 242 (18) dU0 ,β (x) = dist(x, Γ0 ) if x ∈ R − U0 ∧ dist(x, Γ0 ) ≤ β  β otherwise, 240

241

243 244 245 246 247

for some β > 0. We point out that if U0 is bounded, then dU0 ,β ∈ Cβ (Rn ). Moreover, following the results of the above theorem, the evolution of Γ0 is independent of the choice of β > 0 used to define the truncated signed distance. Next we show that the comparison principle holds for Ut . ˆ0 be bounded open sets satisfying U0 ⊆ U ˆ0 , then for t ≥ 0 Ut ⊆ U ˆt . Corollary 5. Let U0 , U

Proof : Using Theorem 4, without loss of generality, we can assume that u0 (x) and u ˆ0 (x) are ˆ 249 given by the distance function (18). If U0 ⊆ U0 then

248

u ˆ0 (x) ≤ u0 (x),

250 251

and by the comparison of the viscosity solutions given in Proposition 2, we obtain that ˆt Ut = {x ∈ Rn : u(t, x) < 0} ⊆ {x ∈ Rn : u ˆ(t, x) < 0} = U

252 253

which completes the proof.

254

4. Radial solutions of the parabolic perturbed mean curvature equation. In the case U0 is a ball, Br0 (x0 ), of radius r0 , then the truncated signed distance function (18) becomes

255

256

(19)

dBr0 (x0 ),β (x) =

|x − x0 | − r0 if |x − x0 | < r0 + β β otherwise.

Let u(t, r) be a radial solution of equation (1) with F given by (4). In radial coordinates this equation becomes n−1 ut (t, r) = 259 (20) sgn(ur (t, r)) + k(r) |ur (t, r)|, r 257 258

261

where sgn(.) is the sign function. In this case Ut is given by the ball Br(t) (x0 ), where r(t) satisfies

262

(21)

260

u(t, r(t)) = 0,


9 263

by computing the derivatives of this expression we obtain

264

(22)

ut (t, r(t)) +

dr (t)ur (t, r(t)) = 0. dt

Using equations (20)-(22) and assuming that ur (t, r(t)) > 0 (the case ur (t, r(t)) < 0 is similar) 266 we obtain that r(t) satisfies the equation 265

dr n−1 (t) = − − k(r(t)). dt r(t)

267

(23)

268 269

Notice, once again, that the mere assumption of k Hölder continuous is enough for the correct treatment of this ODE. Therefore by a straightforward integration we obtain that if

270

(24)

271

then there exists a strictly monotone function ψ : (a, b) → R such that

−

n−1 − k(r(t)) 6= 0 for r(t) ∈ (a, b), r(t)

ψ 0 (r) =

272

− n−1 r

1 − k(r)

∀r ∈ (a, b),

273

and for any r0 ∈ (a, b),

274

(25)

275

where η ≡ ψ −1 and t∗ = sup{t : η(t + ψ(r0 )) ∈ (a, b)}.

276

280

Remark 4. We point out that if we know the evolution of the radius of the ball, r(t), for any r0 > 0 then we can construct an explicit formula for the solution of (1) with F given by (4). Indeed, for −r0 < c < r0 + 1 we define U0c = {x : dBr0 (x0 ),β (x) < c} = Br0 +c (x0 ) and Utc = Brc (t) (x0 ) where rc (t) is the solution of (23) for the radius r0 + c. Then we can recover the solution of (1)-(4) using the expression:

281

u(t, x) = inf{c : x ∈ Brc (t) (x0 )}.

277 278 279

r(t) = η(t + ψ(r0 ))

for t ∈ [0, t∗ ],

Let us study the shape of r(t), the solution of (23), for some particular choices of the forcing 283 term k(r). The proofs of the following lemmas are straighforward using basic integration 284 techniques. 282

285 286

Lemma 6. (Case k(x) constant) Let k ≡ k0 ∈ R − {0} and let r(t) be the solution of (23). (i) If k0 > 0 then (26)

288

where η ≡ ψ −1 with ψ : [0, ∞) → (−∞, ψ(0)] the decreasing function defined by

289

r(t) =

η(t + ψ(r0 )) if t ∈ [0, ψ(0) − ψ(r0 )) 0 if t ≥ ψ(0) − ψ(r0 ),

287

ψ(r) =

n − 1 (n − 1) + rk0 r ln − . 2 |k0 | k0 k0


4

10

2 1

x10[[2]]

3

η2(s)

0

η1(s)

−4

−2

0

2

4

s

Figure 1. plot of η1 (s) and η2 (s) for k(r) ≡ −1 and n = 2.

290

291

(ii) If k0 < 0 then

(27)

r(t) = η1 (t + ψ1 (r0 )) r(t) = 0 r(t) ≡ r0 r(t) = η2 (t + ψ2 (r0 ))

if r0 < 1−n k0 and t ∈ [0, ψ1 (0) − ψ1 (r0 )) if r0 < 1−n k0 and t ≥ ψ1 (0) − ψ1 (r0 ) if r0 = 1−n k0 and t ≥ 0 1−n if r0 > k0 and t ≥ 0,

1−n where ψ1 : [0, 1−n k0 ) → (−∞, ψ1 (0)] is a decreasing function, ψ2 : ( k0 , ∞) → (−∞, ∞) is an −1 −1 293 increasing function, η1 = ψ1 , η2 = ψ2 and

292

ψ1 (r) =

n−1 k02 n−1 k02

0 ln 1−n−rk − k0

r k0 r k0

if r0
1−n k0 then r(t) is an increasing function such that limt→∞ r(t) = ∞. That is, the ball Br(t) (x0 ) tends to expand across the time.

296 297 298 299 300 301 302

r∗

ψ2 (r) =

0 ln n−1+rk − k0

if r0 >

Lemma 7. (Case k(r) linear) Assume k(r) ≡ mr + c for r ∈ [0, r∗ ] with m > 0, c < 0 and > max{r0 , −m/c}. (i) If −4m(n − 1) + c2 > 0 then the function v(r) = − n−1 r − mr − c has


11 303

the roots R1 > R0 > 0 given by −c − m

r

4(n − 1) c 2 + −m m

!

304

1 R0 = R0 (n, m, c) = 2

−c + m

r

4(n − 1) c 2 + −m m

!

305

1 R1 = R1 (n, m, c) = 2

306 307 308

309

310

In consequence, the solution r(t) of equation (23) is given by r(t) = η(t + ψ(r0 )), where η ≡ ψ −1 with ! √ √ 2 2 1 −c− √ −4m(n−1)+c −4m(n−1)+c2

ψ(r) = ln |r − R0 | 2m

1 − 2m

−c+

√

−4m(n−1)+c −4m(n−1)+c2

.

−c (−c − 2mr) ln c+2mr 2m ψ(r) = . m (c + 2mr) (iii) Assume −4m(n − 1) + c2 < 0. Then r(t) = η(t + ψ(r0 )) where η ≡ ψ −1 with ψ(r) = −

313

|r − R1 |

(ii) Assume −4m(n − 1) + c2 = 0. Then r(t) = η(t + ψ(r0 )) where η ≡ ψ −1 with

311 312

.

b

1 3

2(m) 2

r

−4m(n−1)+c2 −4m

arctan

q

√ −4m mr −4m(n−1)+c2

+

q

−4m √c −4m(n−1)+c2 2 m

1 − 2m ln n − 1 + mr2 + cr . 314 315 316 317 318 319 320

321

322

323

Remark. We notice that as in Lemma 6, functions ψ(r) and η(r) should be distinguished on intervals where ψ(r) changes its monotonicity. Therefore, in practice, we have to deal with ψi (r) and ηi (r) = ψi−1 (r) in different intervals. In Fig. 2, 3, 4, we show the profile of functions ηi (r) for the different cases of the previous lemma. Remark. We point out that the solution r(t) of the equation (23) is given by r(t) = ηi (t + ψi (r0 )) and satisfies the following qualitative properties: In the case (i) of Lemma 7:  limt→∞ r(t) = R1 if r0 ∈ (R1 , r∗ )     r(t) ≡ R1 if r0 = R1    limt→∞ r(t) = R1 if r0 ∈ (R0 , R1 ) r(t) ≡ R if r0 = R0  0     limt→t∞ r(t) = 0 if r0 < R0 with t∞ = ψ(0) − ψ(r0 )   r(t) = 0 if r0 < R0 and t ≥ t∞ . In the case (ii) of Lemma 7:  limt→∞ r(t) = R1    r(t) ≡ R0  limt→t∞ r(t) = 0   r(t) = 0

if if if if

r0 ∈ (R0 , r∗ ) r0 = R0 r0 < R0 with t∞ = ψ(0) − ψ(r0 ) r0 < R0 and t ≥ t∞ .


3

η3(s) η2(s)

1

2

x20[[2]]

4

5

12

0

η1(s) −2

−1

0

1

2

3

4

5

s

2

η2(s)

1

x30[[2]]

3

4

Figure 2. (Case (i) of lemma 7). Plot of η1 (s), η2 (s) and η3 (s) for k(r) = r − 3 and n = 2.

0

η1(s) −4

−2

0

2

4

6

8

s

Figure 3. (Case (ii) of lemma 7). Plot of η1 (s) and η2 (s) for k(r) = r − 2 and n = 2. 324 325

In the case (iii) of Lemma 7: limt→t∞ r(t) = 0 if r(t) = 0 if

r0 ∈ (0, r∗ ) with t∞ = ψ(0) − ψ(r0 ) t ≥ t∞ .

As mentioned before, in the case (ii) of Lemma 7 we observe that if r0 < R0 then r(t) converges in finite time to 0 and limt→∞ r(t) = R0 when r0 > R0 . One interesting point is to 328 know how to choose the forcing term k(r) in order to have r(t) converging to R0 in finite time 329 (assumed r0 > R0 ). In the next lemma we study this issue.

326

327


2

η(s)

0

1

x40[[2]]

3

4

13

−2.0

−1.5

−1.0

−0.5

0.0

0.5

1.0

s

Figure 4. (Case (iii) of lemma 7). Plot of η(s) for k(r) = r − 1 and n = 2.

Lemma 8 (stabilization in finite time of radial solutions). Let r∗ , R0 be such that r∗ > R0 > 0 331 and k(r) satisfying 330

332

(29)

−

n−1 − k(r) < 0 r

for r ∈ (R0 , r∗ ),

333 334

−

(30)

n−1 − k(R0 ) = 0, R0

335

Z 336

(31)

r∗

dr < ∞. − k(r)

0 ψ(R0 ) − ψ(r0 ),

337

339

where η ≡ ψ −1 with ψ(r) given by decreasing function Z

340

341 342

ψ(r) =

r

ds

(n−1) R0 − s

− k(s)

.

In particular, the above conditions are fulfilled for the following Hölder continuous function: k(r) = −

n−1 + C(r − R0 )γ R0

∀r ∈ [R0 , r∗ ],


1.5

η2(t)

0.5

1.0

x50[[2]]

2.0

2.5

3.0

14

0.0

η1(t) −4

−2

0

2

4

s

Figure 5. Plot of η1 (s) and η2 (s) for k(r) given by (32) with R0 = 1, m = 1, c = −2, γ = 1/2, C = 1 and n = 2.

with γ ∈ (0, 1) and C > 0 big enough. In fact, for any R0 > 0 and γ ∈ (0, 1) there exists 344 C ∗ > 0 such that, if C ≥ C ∗ , function ( n−1 (r − R0 ) − n−1 if r < R0 R0 R02 kR0 ,γ,C (r) = 345 (32) n−1 γ − R0 + C(r − R0 ) if r ∈ [R0 , r∗ ], 343

satisfies conditions (29)-(30)-(31) for r ∈ [R0 , r∗ ], k(r) = mr + c, with c2 = 4m(n − 1) and −c 347 R0 = 2m , for r < R0 . Therefore the corresponding solution of equation (23) is given by 348 r(t) = η(t + ψ(r0 )) as before and verifies  limt→t+ r(t) = R0 if r0 ∈ (R0 , r∗ ) with t+  ∞ = ψ(R0 ) − ψ(r0 ) ∞   ∗ r(t) ≡ R0 if r0 ∈ [R0 , r ) and t ≥ t+ ∞ 349 − = ψ(0) − ψ(r ) r(t) = 0 if r < R with t  limt→t− 0 0 0 ∞  ∞  r(t) = 0 if r0 < R0 and t ≥ t− . ∞ 346

Remark. We point out that the function k(r) given by (32) is Hölder but not Lipschitz 351 continuous since limr↓R0 k 0 (r) = ∞. In fact, it is easy to show (as in [2]) that if, for R0 > 0, 352 conditions (30)-(31) are satisfied then k(r) can not be Lipschitz in [R0 , r ∗ ). 350

353

5. Some qualitative properties of the evolution of Ut , Γt .

Theorem 9. Let {Ut }t≥0 be the set evolution given by (15) under the action of the geometric 355 flow (1)-(4) and k0 = minx∈Rn k. Then if for some x0 ∈ Rn and r0 > 0, U0 ⊆ Br0 (x0 ) then 356 Ut ⊆ Br(t) (x0 ) ∀ t ≥ 0 where r(t) is given by (26) or (27) according to the sign of k0 . 354

Proof : Let v(t, x) be the corresponding solution of the equation ∇v ∂v 358 = div |∇v| + k0 |∇v|, ∂t |∇v|

357


15 359 360

with the same initial datum as indicated before. Then v is a subsolution of the original problem and, by the comparison principle given in Theorem 1, we conclude that Ut ⊆ Br(t) (x0 ).

Remark 6. From the above theorem we obtain that Ut is bounded ∀t ≥ 0 and that Br(t) (x0 ) provides an upper estimate of the set evolution. In particular, according to the expressions 363 (26) - (27), Ut vanishes in finite time if k0 is positive or if k0 is negative and r0 < 1−n k0 . 364 Therefore, in any case, if U0 is small enough, Ut vanishes in a finite time. 361 362

Theorem 10. Let {Ut1 }t≥0 , {Ut2 }t≥0 be two set evolutions given by (15) under the action of 366 the geometric flow (1)-(4). If Ut1 ∩ Ut2 = ∅ for t ∈ [0, T ], then the evolution level set Ut for 367 U0 = U01 ∪ U02 is given by Ut = Ut1 ∪ Ut2 for t ∈ [0, T ]. 365

368 369 370 371 372

Proof : Let u1 (t, x) and u2 (t, x) be the solutions of (1)-(4) taking as initial data the distance function (18) for U0 and U1 respectively. By the definition of the distance function (18) and Theorem 9 we obtain that for < 1, Ut1, = {x ∈ Rn : u1 (t, x) ≤ } and U21, = {x ∈ Rn : u2 (t, x) ≤ } are compact sets in Rn . We are going to show that if is small enough then Ut1, ∩ Ut2, = ∅ for t ∈ [0, T ]. Indeed, if we assume the opposite, that is, that for any < 1

there exists (t , x ) ∈ Ut1, ∩ Ut2, , then as {(t , x )} 0 such that Ut1, ∩ Ut2, = ∅. We define the nondecreasing 377 function s if s < 378 φ (s) = if s ≥ . 376

Then, for any > 0, functions φ (u1 (t, x)) and φ (u2 (t, x)) are solutions of (1)-(4). We point 1, 2, 380 out that as Ut , Ut are compact sets then the distance between them is positive and then 381 the function defined by   φ (u1 (t, x)) if φ (u1 (t, x)) < φ (u2 (t, x)) if φ (u2 (t, x)) < u(t, x) = 382  otherwise, 379

383 384 385

is well defined for t ∈ [0, T ], it is a solution of (1)-(4) and satisfies that {(t, x) ∈ [0, T ] × Rn : u(t, x) < 0} =Ut1 ∪ Ut2 . This concludes the proof of the theorem.

Remark 7. From the above theorem we obtain that, for small times, the evolution of the set Ut is local in the sense that if U0 = ∪k U0k , where {U0k } are connected sets such that 0 0 388 distance(U0k , U0k ) > δ > 0 for k 6= k 0 , then Utk ∩ Utk = ∅ for t > 0 small enough. In 389 consequence, each connected set U0k evolves in an independent way during the interval [0, t∗ ] 0 390 with t∗ such that distance( Utk , Utk ) > 0 for all t ∈ [0, t∗ ]. 386 387


16 391 392 393 394 395 396 397 398 399 400

401

6. Numerical discretization of equation (1)-(4). The discretization scheme we propose is based on the general approach introduced in [26]. More sophisticated implicit numerical schemes for mean curvature motion have been proposed in [31]. We use an explicit finite difference method where the mean curvature term is discretized using central differencing and the forcing term is discretized using the upwind scheme proposed in [26] (Sections 6.2 and 6.3). A more detailed study about this kind of upwind schemes for first order geometric flows has been presented in [9]. Let tn = n · ∆t and {xk } be the time and spatial discretization lattice. The ways the forcing term and the mean curvature term are discretized is different. For the forcing term we use the following upwind scheme:  q 2 Pn + −  k if km,k ≥ 0 m,k i=1 max max{uxi , 0}, min{uxi , 0} k q km,k |∇u|(tm , x ) ≈ P 2 n  k max min{u+ , 0}, max{u− , 0} if k < 0, m,k

402

i=1

xi

m,k

where km,k = k(tm , xk ) and u(tm , xk ) − u(tm , xk − ∆xi ei ) ∆xi k u(tm , x + ∆xi ei ) − u(tm , xk ) , = ∆xi

403

u− xi =

404 405

u+ xi

406

xi

∆xi is the discretization step in the spatial variable xi and {ei = (0, . . . , |{z} 1 , . . . , 0)}i=1,...,n i

are the vectors of the standard basis. For the mean curvature operator we use the following 408 discretization ( Pn Pn Pn uxi uxj 2 u − u if P x x x x 2 ∇u i i i j i,j=1 i=1 i=1 (uxi ) > 0 n k u ( ) x 409 (tm , x ) ≈ div i=1 i |∇u| 0 otherwise, 407

410

where

411

uxi =

412

uxi xj =

413

u(tm , xk + ∆xi ei ) − u(tm , xk − ∆xi ei ) 2∆xi  k k k  u(tm ,x +∆xi ei )−2u(tm ,x2 )+u(tm ,x −∆xi ei ) P



(∆xi ) k h1 ,h2 ∈{−1,1} h1 h2 u(tm ,x +h1 ∆xi ei +h2 ∆xj ej ) 4∆xi ∆xj

if

i=j otherwise.

By combining the discretization of the forcing term and the mean curvature term we obtain the following explicit discretization scheme for the full PDE : ∇u u(tm+1 , xk ) − u(tm , xk ) 416 (33) = div (tm , xk ) + km,k |∇u|(tm , xk ). ∆t |∇u| 414 415

In the image boundary we assume the usual homogeneous Neumann boundary condition. 418 From the above numerical scheme, we observe that u(tm+1 , xk ) is computed as a non linear

417


17 0

0

combination of {u(tm , xk )}k0 ∈N 0 where {xk }k0 ∈N 0 represents a neighborhood of xk . We are 420 going to fix a CFL condition in order to the diffusion coefficient associated to the term u(tm , xk ) 421 is positive. First, in the case of the forcing term, we take into account that if we denote by

419

422 423

424

00

0

00

umax = {u(tm , xk ) : k 00 ∈ N 0 : |u(tm , xk ) − u(tm , xk )| ≤ |u(tm , xk ) − u(tm , xk )| then v v u n u n uX uX 2 + − k t max max{uxi , 0}, min{uxi , 0} ≤ |u(tm , x ) − umax |t i=1

425 426

427

∀k 0 ∈ N 0 }

i=1

1 , (∆xi )2

on the other hand, in the case of the mean curvature operator the absolute value of the diffusion coefficient associated to u(tm , xk ) is given by n X i=1

n

n

X X −2 2 2 2 (uxi )2 − ≥ > 0, P 2 2 2 2 + n (∆xi ) (∆xi ) mini (∆xi )2 i=1 (uxi ) (∆xi ) i=1 i=1

therefore, based on these expressions we propose the following CFL type condition for the choice of the time step ∆t v   u n n X X u 2 2 1  t ∆t  430 < 1, 2 − 2 + kkk∞ (∆xi ) mini (∆xi ) (∆xi )2 i=1 i=1 428 429

431 432 433 434 435 436

we point out that this CFL type condition does not guarantee the L∞ stability of the scheme. It is well-known that, in contrast with the solution of the continuous problem, in the discrete case, the explicit finite difference schemes for the mean curvature operator does not preserve the L∞ norm. So, as proposed in [1], we force a local L∞ norm preservation condition by introducing the following modification in the scheme ∇u M (u,n,k) u(tn+1 , xk ) = Tm(u,n,k) u(tn , xk ) + ∆t div |∇u| + k|∇u| (tn , xk ) , |∇u|

where m(u, n, k) = minB1 (xk ) u(tn , x), M (u, n, k) = maxB1 (xk ) u(tn , x), B1 (xk ) is the ball 438 centered in xk with radius 1, and Tba (s) is the truncation function  s b. 437

440 441 442 443 444 445

For simplicity in the exposition we choose the radius equal to 1 but, of course, we can use a radius bigger than 1. We choose as initial datum u0 (x) an approximation of the signed distance function dU0 ,β (.). To compute such approximation we use the following fast scheme based on the morphological operators erosion, Et (U0 ), and dilation, Dt (U0 ), using a ball of radius t as structuring element: − sup{t : x ∈ Et (U0 )} if x ∈ U0 dU0 ,β (x) ≈ inf{β, inf{t : x ∈ Dt (U0 )}} if x ∈ / U0 .


18 446 447

We use t ∈ N and Et (U0 ), Dt (U0 ) are estimated by iterations of erosion and dilations computed with t = 1, that is we use that Et (U0 ) = E1 ◦ · · · ◦ E1 (U0 ) | {z }

448

t

Dt (U0 ) = D1 ◦ · · · ◦ D1 (U0 ). | {z } t

In general, as explained in [26], the distance function is usually computed only in a neighbor450 hood of ∂U0 and recomputed after a number of iterations. However, in this paper we do not 451 recompute the distance function because we are interested in the long time behavior of the 452 solution u(t, x) of equation (1). 449

7. Application of equation (1) to level set regularization. Let U0 ⊂ Rn be an open 454 bounded set and let h(x) be a Lipschitz bounded function satisfying 453

U0 = {x ∈ Rn : h(x) < 0}.

455 456

We are going to choose k(x) in the following way: kα,γ (x) = φα,γ (h(x)),

457

where φα,γ : R → R is the increasing function given by α·s if s ∈ [−1, 1] 459 (34) φα,γ (s) = α · sgn(s)(|s − 1|γ + 1) otherwise.

458

468

The parameter α > 0 is introduced to balance the mean curvature term and the forcing term. Here sgn(.) denotes the sign function. Parameter γ > 0 represents the Hölder continuity coefficient of the forcing term when |h(x)| > 1. Taking into account Remark 5 we observe that in the points where k(x) is nonnegative (that is in Rn − U0 ), Ut tends to shrink and in the case k(x) is negative (that is in U0 ), Ut tends to expand if the mean curvature in such point is small enough, otherwise it tends to shrink. In this way Ut tends to regularize the boundary of U0 and it also removes the connected components of U0 with a small size. Following the type of problem we deal with, there are different ways to choose function h(x). A basic choice is to take h(x) = dU0 ,β (x) and then kα,γ (x) given by

469

(35)

470

We point out that, in radial coordinates, if U0 = Br0 (x0 ) and if we choose as forcing term 1 (x), with γ = 1, then U = B kα,γ t r(t) (x0 ), where r(t) satisfies

460 461 462 463 464 465 466 467

471

1 kα,γ (x) = φα,γ (dU0 ,β (x)).

( 472

(36)

dr dt (t)

= − n−1 r(t) − α(r(t) − r0 ) f or r ∈ [0, r0 + S) r(0) = r0 ,

for some S > 0. We observe that this is a particular case of the forcing term considered in 474 Lemma 7 with m = α and c = −αr0 . Then, using the result quoted in Lemma 7 we obtain: 473


19

481

Theorem 11. Let Ut be the evolution of the set U0 = Br0 (x0 ) under the action of the geometric flow (1)-(4) with k(r) = φα,1 (dU0 ,β (x)). Then Ut = Br(t) (x0 ) where r(t) is a nonincreasing function satisfying that r 1 n−1 1 lim r(t) = r0 + r02 − 4 , t→∞ 2 2 α q if r0 ≥ 2 n−1 α , or Ut vanishes in a finite time, q if r0 < 2 n−1 α .

482

Proof : It is a straightforward application of Lemma 7.

483

Remark 8. We point out that from the above theorem and Theorems 5 and 10 we obtain that k k if we use as forcing term k(r) = φα,1 (dU0 ,β (x)) and if U0 = ∪k U connected 0 , where {U0 } are

475 476 477 478 479 480

484 485

0

sets such that there exist Brk (xk0 ) ⊆ U0k ⊆ Brk (xk1 ) with distance Brk (xk1 ), Brk0 (xk1 ) > δ > 0 0

1

1

1

for k 6= k 0 , then each U0k evolves in an independent way for all t > 0 and satisfies that 487 Brk (t) (xk0 ) ⊆ Utk ⊆ Brk (t) (xk1 ), where r0k (t) and r1k (t) are the solutions of (36) for r0 = r0k and 1 0 q k 488 r0 = r1k , respectively. In particular, if r1k < 2 n−1 α , then Ut vanishes in a finite time and if q k k k q 489 r0k ≥ 2 n−1 α then Ut 6= ∅ and B 1 rk + 1 (rk )2 −4 n−1 (x0 ) ⊆ Ut for all t > 0. 0 2 0 2 α 486

In the case in which U0 is an image level set, that is, if U0 = U0c = {x ∈ Rn : I(x) < c}, 491 where I : Rn → R is an image, then we can choose h(x) depending of the image intensity 492 value. That is, we can choose h(x) = I(x) − c and take as forcing term 490

493

(37)

2 kα,γ (x) = φα,γ (I(x) − c).

We notice that in this case, the forcing term depends on the image contrast: the higher is the level set contrast in the image, the closer will be Ut to U0 . As we will see later, in some c ,c 496 applications we are interested in the regularization of the image level set U0 = U0 1 2 = {x ∈ 497 Rn : c1 < I(x) < c2 }. In such case we can choose 494 495

3 kα,γ (x) = φα,γ ((I(x) − c1 )(I(x) − c2 )) .

498

(38)

499

In Fig. 6 we show a 2D example of set U0 , and the associated distance function dU0 ,β (x), which we use to illustrate the shape of the set evolution Ut under the action of the geometric flow (1)-(4). In Fig. (7)-(14) we show the shape of Ut for different choices of the forcing term k(x). In each figure, we include a link to a video to show the evolution of Ut . We point out 1 (x), the solution, u(t, x), of (1)-(4) tends to a that in the case we choose as forcing term kα,γ non-trivial asymptotic state. We have included in U0 circles of different sizes and we observe, as predicted by the theoretical results, that according to the values of α and γ and the circles size, the circles vanish in finite time or they remain in the set Ut . Next we present an experiment using the real 2D image, I : Ω ⊂ R2 → {0, 1, 2, .., 255}, showed in Fig. 16.a. In this figure, we also show some level sets U0c = {x : I(x) ≤ c}

500 501 502 503 504 505 506 507 508


20

Figure 6. From left to right we show a silhouette (in black) used as initial set U0 and the signed distance function dU0 ,β (x) scaled between 0 and 255 for visualization purposes.

Figure 7. Evolution of Ut for equation (1)-(4) without forcing term (k ≡ 0). From left to right we show: (i) the original set U0 where we illustrate with blue arrows the initial motion direction of Γ0 , (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).

513

for different values of levels c. Using the approach presented in [12], we consider the image topographic map, that is, the collection of all its level sets {U0c }c=0,1,..,255 . Then, for each level c, we compute the level set evolution Utc = {x : u(t, x) ≤ c} where u(t, x) is the solution of the equation (1)-(4). Then, for each time t we can recover a graylevel image It using the expression

514

(39)

509 510 511 512

It (x) = inf{c : x ∈ Utc }.

In Fig. 17, we show, for the level set U0144 = {x : I(x) ≤ 144}, the distance function 516 dU 144 ,β (x) and the solution u(t, x) of equation (1)-(4) for t = 10 and different choices of the 0 517 forcing term k(x). In Fig. 18 we present the evolution of Ut144 for t = 10 and in Fig. 19 the 518 reconstruction of the image from the level sets using (39). 515


21

= 0.1|Du|. From left to write we show: (i) the original set U0 Figure 8. Evolution of Ut for equation ∂u ∂t where we illustrate with red arrows the initial motion direction of Γ0 , (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of the equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).

= −0.1|Du|. From left to write we show: (i) the original set U0 Figure 9. Evolution of Ut for equation ∂u ∂t where we illustrate with red arrows the initial motion direction of Γ0 , (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of the equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).

Figure 10. Evolution of Ut for equation (1)-(4) using as forcing term k(x) ≡ 0.1. From left to right we show: (i) the original set U0 where we illustrate with blue arrows the initial motion direction with respect to the mean curvature term and with red arrows with respect to the forcing term, (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).


22

Figure 11. Evolution of Ut for equation (1)-(4) using as forcing term k(x) ≡ −0.1. From left to right we show: (i) the original set U0 where we illustrate with blue arrows the initial motion direction with respect to the mean curvature term and with red arrows with respect to the forcing term, (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).

1 Figure 12. Evolution of Ut for equation (1)-(4) using as forcing term kα,γ (x) with α = 0.1 and γ = 0.5. From left to right we show: (i) the original set U0 where we illustrate with blue arrows the initial motion direction with respect to the mean curvature term and with red arrows with respect to the forcing term, (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).

1 Figure 13. Evolution of Ut for equation (1)-(4) using as forcing term kα,γ (x) with α = 0.1 and γ = 1. From left to right we show: (i) the original set U0 where we illustrate with blue arrows the initial motion direction with respect to the mean curvature term and with red arrows with respect to the forcing term, (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).


23

1 Figure 14. Evolution of Ut for equation (1)-(4) using as forcing term kα,γ (x) with α = 0.5 and γ = 1.. From left to right we show: (i) the original set U0 where we illustrate with blue arrows the initial motion direction with respect to the mean curvature term and with red arrows with respect to the forcing term, (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).

Figure 15. Evolution of Ut for equation (1)-(4) using as forcing term k(x) = kR0 ,γ,C (dU0 ,β (x) + R0 ) where kR0 ,γ,C (.) is given by (32) and R0 = 3, γ = 0.5 and C = 1. From left to right we show: (i) the original set U0 where we illustrate with blue arrows the initial motion direction with respect to the mean curvature term and with red arrows with respect to the forcing term, (ii) the set Ut for t = 50 and (iii) the solution u(t, x) of equation equation (1)-(4) for t = 50. (video with the evolution of Ut and u(t, x)).


24

(a) Original image

(d) level set U094 = {x : I(x) ≤ 94}

(e) level set U0144 = {x : I(x) ≤ 144}

(b) level set U058 = {x : I(x) ≤ 58}

(e) level set U0121 = {x : I(x) ≤ 121}

(f) level set U0175 = {x : I(x) ≤ 175}

Figure 16. We show a real image used in the experiments and some of its level sets U0c = {x : I(x) ≤ c} for different values of levels.


25

(a) distance function for the level set U0144

(b) regularization with

∂u ∂t

= div

∇u |∇u|

|∇u|

(c)

∂u ∂t

= div

∇u |∇u|

1 |∇u| + k1,0.2 (x)|∇u|

(d)

∂u ∂t

= div

∇u |∇u|

2 |∇u| + k1,0.2 (x)|∇u|

(e)

∂u ∂t

= div

∇u |∇u|

1 |∇u| + k1,0.5 (x)|∇u|

(f)

∂u ∂t

= div

∇u |∇u|

2 |∇u| + k1,0.2 (x)|∇u|

Figure 17. (a) distance function of the real image showed in fig. 16 for the level set U0144 . (b)-(f ) evolution of the distance function dU 144 ,β (x) under the action of the equation (1)-(4) for t = 10 and different choices of 0 the forcing term k(x).


26

(a) level set U0144


∂u ∂t

= div

∇u |∇u|

|∇u|

(c)

∂u ∂t

= div

∇u |∇u|

1 |∇u| + k1,0.2 (x)|∇u|

(d)

∂u ∂t

= div

∇u |∇u|

2 |∇u| + k1,0.2 (x)|∇u|

(e)

∂u ∂t

= div

∇u |∇u|

1 |∇u| + k1,0.5 (x)|∇u|

(f)

∂u ∂t

= div

∇u |∇u|

2 |∇u| + k1,0.2 (x)|∇u|

Figure 18. (a) level set U0144 of the real image showed in fig. 16. (b)-(f ) regularization of the level set U0144 under the action of the equation (1)-(4) for t = 10 and different choices of the forcing term k(x).


27

(a) Original image


∂u ∂t

= div

∇u |∇u|

|∇u|

(c)

∂u ∂t

= div

∇u |∇u|

1 (x)|∇u| |∇u| + k1,0.2

(d)

∂u ∂t

= div

∇u |∇u|

2 (x)|∇u| |∇u| + k1,0.2

(e)

∂u ∂t

= div

∇u |∇u|

1 |∇u| + k1,0.5 (x)|∇u|

(f)

∂u ∂t

= div

∇u |∇u|

2 |∇u| + k1,0.2 (x)|∇u|

Figure 19. (a) real image showed in fig. 16. (b)-(f ) reconstruction of the image from the level sets (using (39)) after regularization of the level sets using equation (1)-(4) for t = 10 and different choices of the forcing term k(x).


28

Figure 20. From left to right we show the level set U0 and Ut for t = 10 using equation (1)-(4) with the 3 3 forcing term k1.,0.2 (x), k0.1,0.2 (x) and k ≡ 0 respectively.

519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546

Next, we present an experiment on a real 3D image in the context of medical image segmentation. We use a CT (computed tomography) 3D scan, I(x) (showed in Fig. 21), of a patient suffering from severe aortic elongations. For medical diagnosis purpose, a segmentation of the aortic lumen is required. Usually, an initial segmentation is obtained by thresholding the intensity value of the CT image between 2 values c1 and c2 , that is, the set U0 = {x : c1 < I(x) < c2 } is an initial segmentation of the aortic lumen. However, as showed in Fig. 21), this initial segmentation is inaccurate because other human organs in the CT image have similar intensity values and we can not fix thresholds c1 and c2 such that U0 contains properly just the aorta lumen. Therefore some kind of processing is required to improve the quality of the initial estimation of U0 . We are going to use equation (1)-(4) to regularize the set U0 and provide a better segmentation of the aortic lumen. In Fig. 20 we show a 3D representation of the sets Ut for different forcing terms. We use as forcing term k(x), the one given in (38). In Fig. 21 we show Γ0 = ∂U0 the initial contours as well as Γt = ∂Ut for different choices of the forcing terms. We observe that by adjusting the parameters α and γ in the forcing term (38) we obtain different degrees of regularization, we point out that in the case of no forcing term (k ≡ 0) we regularize too much and the contour Γt tends to move away from aorta lumen contour actual location. In Fig. 22 and 23 we show the level set and distances associated to the different forcing terms. 8. Conclusions. The main limitation of the usual geometric flows without forcing term is that they tend to regularize too much the level sets and the asymptotic states are trivial. By including an appropriate forcing term in the equation we obtain more suitable models for level set regularization which can be useful for nonlinear filtering and image segmentation. In this paper we show some mathematical results on the solutions of these equations for a forcing term k(x) being merely a bounded H¨ older continuous function. We also studied in details the shape of some radial solutions with non-trivial asymptotic states and we also present some results on the evolution of the level sets. We perform some experiments, using a basic explicit finite difference numerical scheme, to illustrate the level set regularization obtained with the proposed equations and the results are very promising. The level set is regularized by


29

(a) original image and Γ0 (in black) (video)

3 (c) Γ10 for k(x) = k0.1,0.2 (x) (video)

3 (b) Γ10 for k(x) = k1.,0.2 (x) (video)

(d) Γ10 for k(x) ≡ 0 (video)

Figure 21. We show one slice of the original image and Γt = ∂Ut (in black) using equation (1)-(4) with different forcing terms. For each image we also include a link to a video with the full 3D image.

removing high curvature areas but at the same time it remains close to the original level set 548 shape. As showed in the experiments, this behavior is quite useful in the context of nonlinear 549 filtering or image segmentation. In particular, we show some interesting results on medical 550 image segmentation. 547

Acknowledgments. This research has partially been supported by the MINECO projects references MTM2016-75339-P and MTM2014-57113-P (AEI/FEDER, UE) of the DGISPI 553 (Spain) and the Research Group MOMAT (Ref. 910480) supported by the Universidad Com551 552


30

(a) U0 = {x : c1 < I(x) < c2 } (video)

3 (c) U10 using (1)-(4) for k0.1,0.2 (x) (video)

3 (b) U10 using (1)-(4) for k1.,0.2 (x) (video)

(d) U10 using (1)-(4) for k(x) ≡ 0 (video)

Figure 22. We show one slice of the level set U0 = {x : c1 < I(x) < c2 } as well as Ut = {x : u(10, x) ≤ 0} where u(t, x) is the solution of equation (1)-(4) for t = 10 with different forcing terms. For each image we also include a link to a video with the full 3D sequence.

554

plutense de Madrid.


31

(a) signed distance dU0 ,β (x) (video)

3 (b) u(10, x) sol. of (1)-(4) for k1.,0.2 (x) (video)

3 (c) u(10, x) sol. (1)-(4) for k0.1,0.2 (x) (video) (d) u(10, x) sol. (1)-(4) for k(x) ≡ 0 (video)

Figure 23. We show one slice of the signed distance function dU0 ,β (x) as well as the solution of equation (1)-(4) for t = 0 with different forcing terms. For each image we also include a link to a video with the full 3D sequence.


32 555

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