# Lecture notes on free-discontinuity problems by Braides A. PDF

By Braides A.

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Example text

11 If in addition to the hypotheses of the previous theorem, we have lims→±∞ θ(s) = +∞ and θ(s) = 0 only if s = 0, then the functional F , extended to +∞ on L1 (I) \ BV (I), is lower semicontinuous with respect to the L1 (I)-convergence. Proof If uj → u in L1 (I) and supj F (uj ) < +∞, then we have sup{|uj (t+) − uj (t−)| : t ∈ S(uj )} ≤ c, independent of j. This implies that supj uj BV (I) < +∞, so that the previous theorem can be applied. 1 Give a direct proof that if ϑ : R2 → [1, +∞) is lower semicontinuous and subadditive then the functional b |u |2 dt + F (u) = a ϑ(u(t−), u(t+)) t∈S(u) is lower semicontinuous on SBV (I) with respect to the weak BV convergence.

1); (d) in condition (I) we can replace the requirement that (aij ), (bij ) be equibounded in R by the stronger condition that aij → ai and bij → bi as j → +∞, with ai , bi ﬁnite and ai = bi . Proof Suppose that F be lower semicontinuous. Let u = a + (b − a)χ[t,+∞) , and let uj be deﬁned by General lower semicontinuity conditions in one dimension 31  0 if s < t0j aj      if s ≥ tN uj (s) = bN j j      v i (t) if ti−1 ≤ s < ti , i = 1, . . , N , j j j i where vji ∈ W 1,1 (ti−1 j , tj ) is a minimum point for the problem deﬁning the quani−1 i−1 i i tity d(tj , bj ; tj , aj ).

2) u) (Da u + Dc u) (Ω) ≤ D(φ(u)) − φ (˜ φ∈X(f ) S(u) This inequality in a sense characterizes the measure Da u + Dc u, as precised below. 5 Let ψ and X(ψ) be as above. Let u ∈ BV (Ω), and let λ be an Rn -valued measure on Ω such that |λ|(S(u)) = 0 and sup u) λ (Ω) < +∞ . 3) φ∈X(ψ) Then λ = Da u + Dc u. Proof Set µ = Da u + Dc u − λ. We have to prove that µ = 0. 3) we get sup φ∈X(ψ) φ (˜ u) d|µ| = sup Ω φ∈X(ψ) φ (˜ u) µ (Ω) General lower semicontinuity conditions in one dimension ≤ sup u) µ + φ(u+ ) − φ(u− ) Hn−1 φ (˜ 29 S(u) (Ω) < +∞ .