The Total Variation Flow with Measure Initial Data

F. Andreu^{∗}, V. Caselles^{∗∗}, J. M. Mazón^{∗} and S. Moll^{∗}

(*) Dept. de Análisis Matemático, Universitat de Valencia. (**) Dept. de Tecnologia, Universitat Pompeu-Fabra,

We consider the minimizing total variation flow in ℝ^{N}

\frac{∂u}{∂t} = div (\frac{Du}{|Du|})  in Q_{T} =]0, T [×ℝ^{N},  (1)

coupled with the initial condition

u(0) = μ,  μ being a Radon measure in ℝ^{N}.  (2)

This PDE appears (in a bounded domain Ω) in the steepest descent method for minimizing the total variation, a method introduced by L. Rudin, S. Osher and E. Fatemi in the context of image denoising and reconstruction.

In previous papers thes gradient descent flow (1) has been studied in a bounded domain under Neumann and Dirichlet boundary conditions, obtaining existence and uniqueness results for initial data in L^{1}(Ω). Our aim is to continue the study of the flow (1) when the initial conditions are Radon measures in ℝ^{N}. We shall not consider general measures, instead we shall restrict ourselves to the case of measures

μ = h + αH^{k} S,  (3)

where h ∈ L^{1}(ℝ^{N}) ∩ L^{∞}(ℝ^{N}), α ≥ 0, H^{k} is the k-dimensional Hausdorff measure in ℝ^{N} and S is a k-manifold in ℝ^{N} of class W^{3,∞}.

We study limit solutions obtained by weakly approximating the initial measure μ by functions in L^{1}(ℝ^{N}), characterizing these limit solutions when the initial condition μ = h + μ_{s} where h ∈ L^{1}(ℝ^{N}) ∩ L^{∞}(ℝ^{N}), and μ_{s} = αH^{k} S, α ≥ 0, k is an integer and S is a k-dimensional manifold with bounded principal curvatures. In case k < N − 1 we prove that the singular part of the solution does not move, it remains equal to μ_{s} for all t ≥ 0 and its absolutely continuous part is the unique strong solution of the Cauchy problem associated with (1) with initial datum h. In particular, u(t) = δ_{0} when u(0) = δ_{0}. In case k = N − 1 we prove that the singular part of the limit solution is (1 − \frac{2}{α}t)^{+}μ_{s} and we also characterize its absolutely continuous part.