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Computational aspects of the crop field segmentation problem based on anisotropic active contour model
            the form                                          problem:

                      Z
                                                                                           f   0   !
                                        f

                     ∂     V 2 φ 0 ε,τ  φ |M Dφ 0 ε,τ  | +                  0          f  M ε ∇φ ε,τ
                                        ε
                                                                                           f
                         Ω                                            −L(φ  ε,τ  ) div M ε  |M ε ∇φ 0  |
                     ∂J (w, φ)|   0 + V 1 φ 0    ∋ 0.  (32)                                   ε,τ
                              w=φ ε,τ       ε,τ                                  0             ′
                                                                           +V 1 φ    = 0   in D (Ω),     (39)
                                                                                 ε,τ
                                                                        f          !
                                                                      M ε ∇φ 0 ε,τ
                                                                  M ε f  f      , ν Ω  = 0  a.e. on ∂Ω. (40)
                Following Anzellotti, 32  for an open bounded         |M ε ∇φ 0 ε,τ |
                                              2
            set with Lipschitz boundary Ω ⊂ R , we denote
                             ∞      2             2
               X 2 (Ω) = z ∈ L (Ω; R ) : div(z) ∈ L (Ω)
                                                       (33)
            to give a correct interpretation of the Anzelotti  4. Numerical approximation of the
            pairing (z, Du) with u ∈ BV (Ω). In this case, if    optimality conditions
            u ∈ BV (Ω) and z ∈ X 2 (Ω), then for every test
                           ∞
            function ψ ∈ C (Ω), we have                       In this section, we briefly describe the numeri-
                           0
                                                              cal scheme and a procedure to solve the Euler–
                                  Z
                 ⟨(z, Du) , ψ⟩ := −  u div(ψz) dx =           Lagrange system (39)–(40) numerically. Since for
                                   Ω                          implementations, it is reasonable to define the so-
                   Z                Z
                                                              lution of the problem (39)–(40) using a “gradi-
                 −    uψ div(z) dx −   u(z, ∇ψ) dx.    (34)
                    Ω                Ω                        ent descent”strategy, denoted by S a given circle
                                                              and by d(x, S) the Euclidean distance of the point
                                                              x ∈ Ω from the circle S, we suggest starting with
                                                              the initial level function φ 0 ∈ C(Ω) that can be
                                                                                                          19
                Arguing as in, ?Proposition 4.1.]Vese  it can be  defined as follows (see also the other variants in )
            shown that ξ ∈ ∂J (w, φ)|      0  if and only if              
                                       w=φ ε,τ                             +d(x, S), x ∈ inside S,
            there exists a vector field z ∈ X 2 (Ω) (see (33))   φ 0 (x) =  0,         x ∈ S,            (41)
            such that the following conditions hold:                        −d(x, S), x ∈ outside S.
                            ∥z∥ ∞    2 ≤ 1,                   Next, for numerical purposes, we calculate u ε,τ ,
                               L (Ω;R )

                                                            the continuous approximation of the BV solution
                                   f
                     f
                z, M Dφ  0   = M Dφ   0    as measures,

                     ε   ε,τ       ε   ε,τ                    0
                                                              φ ε,τ , with ε > 0 and τ > 0 small enough, as a

                                    f
                    − div L(φ  0 ε,τ )M z = ξ  in Ω,          solution of the problem
                                    ε
                                                                                             !
                                                                                        f
                  h               i                              ∂φ                f  M ε ∇φ
                             f
                   L(φ 0 ε,τ )M z, ν Ω = 0  a.e. on ∂Ω.          ∂t  = L(φ) div M  ε  |M ε ∇φ|  − V 1 (φ)
                            ε
                                                                                        f
            Then taking into account the Anzelotti represen-
                                                                                         in (0, ∞) × Ω, (42)
            tation (34) and the fact that
                                                                                 !
                                                                        f
                                                                      M ε ∇φ
                                                                M f              = 0 on (0, ∞) × ∂Ω, (43)
                               f
               − div L(φ 0  )M z                                    ε   f    , ν Ω
                         ε,τ   ε                                     |M ε ∇φ|

                                       f

                       = −V 2 φ 0 ε,τ  M Dφ 0 ε,τ  , z                             φ(0, ·) = φ 0 (·) in Ω. (44)
                                       ε
                                                                The second point that we realize in this algo-
                                          f
                          − L(φ 0 ε,τ  ) div M z              rithm is to consider the adaptive version of the
                                          ε
                                                                                     19     34
                                                     f
                             f

             = −V 2 φ 0 ε,τ  M Dφ 0   − L(φ 0 ε,τ ) div M z  model (42)–(44). Following  and,   we substi-

                                                     ε
                             ε
                                 ε,τ
                                                              tute the multipliers δ ε (−φ 0 ε,τ  + l j+1 + τ) in the
            we can finally rewrite (32) as follows            external force V 1 (φ) with 1. It has been recently
                                                           shown in 34  that, in this way, the external force
                                f
                −L(φ  0 ε,τ ) div M z + V 1 φ 0 ε,τ  ∋ 0,  (35)  without δ ε (−φ 0  + l j+1 + τ) can take action on
                                ε
                h               i                                           ε,τ
                           f
                 L(φ 0  )M z, ν Ω = 0   a.e. on ∂Ω,    (36)   φ everywhere in Ω (not simply on the prescribed
                     ε,τ  ε
                                                              level sets of φ), and so can keep φ moving more


                            f
                                         f
                       z, M Dφ  0 ε,τ  = M Dφ 0   ,  (37)  quickly towards the desired segment edges. As a

                                         ε
                                              ε,τ
                           ε
                                                              result, the adaptive version of the external force
                            ∥z∥ ∞    2 ≤ 1.            (38)   V 1 (φ) takes the form
                               L (Ω;R )
            If in addition φ 0 ε,τ  is differentiable in the weak                               f  !
            sense, then the Euler–Lagrange system (35)–(38)          V 1 (φ) = − f − c 0 ε,τ,0  log  0
                                                                     b
            assumes the form of the following boundary value                                   c ε,τ,0
                                                           455
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