Finite-dimensional attractors for the Bertozzi-Esedoglu-Gillette-Cahn-Hilliard equation

In this article, we are interested in the study of the asymptotic behavior, in terms of ﬁnite-dimensional attractors, of a generalization of the Cahn-Hilliard equation with a ﬁdelity term (integrated over Ω \ D instead of the entire domain Ω , D ⊂⊂ Ω ). Such a model has, in particular, applications in image inpainting. The diﬃculty here is that we no longer have the conservation of mass, i


Introduction
The Cahn-Hilliard equation, is very important in materials science.This equation is a simple model for phase separation processes of a binary alloy at a fixed temperature.We refer the reader to [5,6] for more details.The function f : R → R is of "bistable type" with three simple zeros and is the derivative of a double-well potential F whose wells correspond to the phases of the material.A typical model nonlinearity is given by i.e. f (s) = s 3 − s.
The function u(x, t) represents the concentration of one of the metallic components of the alloy.
We are interested in this article in the following generalization of the Cahn-Hilliard equation introduced in [1] in view of applications in image inpainting: where h(x) is a given binary image, D ⊂⊂ Ω is the inpainting domain, and the last term on the left-hand side is added to keep the solution constructed close to the given image h(x) in the complement of the inpainting domain (Ω\D), where there is image information available.The idea here is to solve the equation up to equilibrium to have an inpainted version u(x) of h(x).
Image inpainting involves filling in parts of an image or video using information from the surrounding area.Its applications include restoration of old paintings by museum artists [16], removing scratches from old photographs [3], altering scenes in photographs [19], and restoration of motion pictures [21].
Well-posedness results for (1.2) have been obtained in [1] (see also [4] for the study of the stationary problem).
On the contrary, equation (1.2), which is also endowed with Neumann boundary conditions, does not satisfy this conservation property.We prove instead that < u > is dissipative, which then allows us to prove the existence of finite-dimensional attractors.
We also give numerical simulations which show that a dynamic one step scheme involving the diffuse interface ε (we note that, in [1] and [2], the authors first consider a larger value of ε and then a smaller one in order to obtain their numerical simulations) allows us to connect regions across large inpainting domains.While the simulations in [1] and [2] are programmed in MATLAB, we use FreeFem++.These simulations confirm the ones performed in [1] and [2] on the efficiency of the model.

Setting of the problem
Let Ω be an open bounded domain of R n , n = 1, 2, or 3, with a smooth boundary Γ, and D be an open bounded subset of Ω with a smooth boundary ∂D such that D ⊂⊂ Ω.The unknown function is a scalar u = u(x, t), x ∈ Ω, t ∈ R, and the equation reads (for simplicity, we set ε equal to 1) where f is the cubic function and h ∈ L 2 (Ω).We denote by F the antiderivative of f vanishing at s = 0, The equation is associated with the Neumann boundary conditions We finally supplement the equation with the initial condition We denote by .the L 2 −norm (with associated scalar product ((., .)))and set The space V in (2.8) makes sense by the trace theorem.Furthermore, V is a closed subspace of H 2 (Ω) and is equipped with the norm induced by H 2 (Ω) denoted by . 2 .We denote by < φ > the average over Ω of a function φ in L 1 (Ω), and we write φ = φ− < φ >.
We note that are norms in H −1 (Ω), L 2 (Ω), H 1 (Ω) and H 2 (Ω), respectively, which are equivalent to the usual ones.We finally set Throughout this article, the same letter c (and, sometimes, c ′ and c ′′ ) denotes constants which may vary from line to line, or even in a same line.Similarly, the same letter Q denotes monotone increasing functions which may vary from line to line, or even in a same line.

A priori estimates
The weak formulation of the problem is obtained by multiplying (2.3) by a test function v ∈ V , integrating over Ω, and using the Green formula and the boundary conditions.We find Now, we replace v by 1 in (3.10) to have Owing to (3.12), we can rewrite (2.3) in the form Recalling that u = ū+ < u >, we have (3.15)We take the scalar product of this equation by ū in L 2 (Ω) and obtain 1 2 (3.16) We have It follows from (3.16), (3.17) and (3.18) that 1 2 It thus follows from (3.19) that By Gronwall's lemma, we find Let B be a bounded subset of Ḣ−1 (Ω) and t 0 be such that ū0 ∈ B and t ≥ t 0 implies ū(t) ∈ B 0 , where for r > 0 fixed.We then multiply (3.14) by ū and find, noting that the inequality (3.25) Therefore, we obtain We finally deduce from (3.22), (3.26) and the uniform Gronwall's lemma that where the constant c is independent of ū0 and t, hence for some monotone increasing function Q.Now, setting u =< u > +ū in (3.11), we have Thus, . Here, Finally, we obtain where c and c ′′ are two constants which are nonnegative and independent of t and u 0 .
Finally, using the uniform Gronwall's lemma in (4.37), we deduce that where c is independent of u 0 and t, for 0 < r < 1 fixed.

Existence of the global attractor
We first have the Proposition 5.1.For every u 0 ∈ L 2 (Ω) and every T > 0, the initialboundary value problem (3.10) has a unique solution u which belongs to Proof.See [1], [27], and [24].
Theorem 5.3.The semigroup S(t) possesses the connected global attractor A such that A in compact in L 2 (Ω) and bounded in H 4 (Ω).
Remark 5.4.It is easy to see that we can assume, without loss of generality, that B ′ 0 is positively invariant by S(t), i.e.S(t)B ′ 0 ⊂ B ′ 0 , ∀t ≥ 0.

Existence of exponential attractors
Let u 1 and u 2 be two solutions of (2.3)-(2.6)with initial data u 0,1 and u 0,2 , respectively.We again set u = u 1 − u 2 and u 0 = u 0,1 − u 0,2 and have Furthermore, it is sufficient here to take initial data belonging to the bounded absorbing set B ′ 0 defined in the previous section.We rewrite (6.66) as Here, where the constant c depends only on B ′ 0 .Furthermore, where the constant c depends only on B ′ 0 , which yields and, owing to (5.63), we find We note that, integrating (5.64) over (0, t), we have where the constant α depends only on B ′ 0 , hence By (6.72), (6.74), and Gronwall's lemma, we obtain Now multiplying (6.69) by ∂ ū ∂t , we obtain, proceeding as above, where the constant c depends only on B ′ 0 .Therefore, integrating (6.76) over (1, t) and owing to (6.75) (for t = 1), we have where the constant c depends only on B ′ 0 .We note that, integrating (6.66) over (0, t), we easily obtain where the constant c depends only on B ′ 0 .Differentiating (6.69) with respect to time, we find where )ds and θ = ∂ ū ∂t .We multiply (6.79) by (t − 1)θ and have owing to the above estimates and a proper interpolation inequality.Furthermore, ) u H 1 (Ω) ≤ (thanks to (4.47) and (4.56)) (6.81)We thus deduce from (6.74), (6.75), (6.77), (6.78), and Gronwall's lemma that θ(t) where where the constant c depends only on B ′ 0 .It then follows from (6.75), (6.82), (6.85), and standard elliptic regularity results that (6.86) where the constant c depends only on B ′ 0 .we derive a Hölder (both with respect to space and time) estimate.Actually, owing to (5.65), it suffices to prove the Hölder continuity with respect to time.We have , (6.87) where u is solution of (2.3)-(2.6)-(2.7).
(i) M is compact in H −1 (Ω); (ii) M is positively invariant, S(t)M ⊂ M, ∀t ≥ 0; (iii) M has finite fractal dimension in H −1 (Ω); (iv) M attracts exponentially fast the bounded subsets of Φ, where the constant c is independent of B and dist H −1 (Ω) denotes the Hausdorff semidistance between sets defined by Remark 6.2.Setting M = S(1)M, we can prove that M is an exponential attractor for S(t), but now in the topology of H 2 (Ω) (see, e.g., [14]).
Since M (or M) is a compact attracting set, we deduce from Theorem 6.1 the following corollary.Remark 6.4.We can more generally consider a nonlinear term f of the form (see [8]).

Numerical simulations
As far as the numerical simulations are concerned, we rewrite the problem in the form ∂u ∂t which has the advantage of splitting the fourth-order (in space) equation into a system of two second-order ones (see [15] and [9]).Consequently, we use a P1-finite element for the space discretization, together with a semi-implicit Euler time disretization (i.e.implicit for the linear terms and explicit for the nonlinear ones).The numerical simulations are performed with the software Freefem++ [17].
In the numerical results presented below, Ω is a (0, 0.5)× (0, 0.5)-square.The triangulation is obtained by dividing Ω into 120 × 120 rectangles and by dividing every rectangle along the same diagonal.(c) Replacing the values larger than 1 2 by 1 and those smaller than 1 2 by 0.

Inpainting of a triangle
The gray region in Figure 1(a) denotes the inpainting region.We run the modified Cahn-Hilliard equation with f (s) = s 3 − s, ε = 0.03 and, at t = 1, we come close to a steady state, shown in Figure 1(b).We finally replace all the values larger than 1 2 by 1 and all those smaller than 1 2 by 0 to obtain the final inpainting result in Figure 1(c).The parameters are ∆t = 0.05, λ 0 = 900000.

Inpainting of four 3/4 circles
In Figure 2(a), the gray region denotes the region to be inpainted.The modified Cahn-Hilliard equation is run close to a steady state with ε = 0.05 and f (s) = s 3 − s, resulting in Figure 2(b) at t = 1.25.We replace all the values larger than 1  2 by 1 and all those smaller than 1 2 by 0 to obtain the final inpainting result in Figure 2(c).
Furthermore, we run again the modified Cahn-Hilliard equation with the same initial datum as in Figure 2(a) and the same ε = 0.05, but we now take f (s) = 4s 3 −6s 2 +2s as in [1].We are close to a steady state at t = 1.   2 by 1 and all those smaller than 1 2 by 0 to obtain the final inpainting in Figure 2(e).We finally deduce that, in the inpainting of a circle, the result obtained is better when considering the function f (s) = 4s 3 − 6s 2 + 2s than f (s) = s 3 − s.In this test, ∆t = 0.05, λ 0 = 900000.Remark 7.1.In the examples of the four circles, the choice f (s) = s 3 − s gives a bad inpainting result.We note that, if we take f (s) = 1 4 (s 3 − s) in ) owing to Young's inequality (e.g., 3ab ≤ 7 8 a 2 + 18 7 b 2 , a, b ≥ 0; here, a = ū2 and b = |ū < u > |), and