Dependence of the blow-up time with respect - Universidad de ...

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10 P. GROISMAN AND J.D. ROSSI The inequalities are understood coordinate by coordinate. Lemma 2.1. (Comparison Lemma) Let U and U be a super and a subsolution of (2.2) respectively, such that the initial data verify Then U(0) ≥ U(0). U(t) ≥ U(t). Proof: Let W = U − U. We can assume, by an approximation argument, that we have strict inequalities in Definition 2.2 and that W (0) > δ > 0. We observe that W verifies MW ′ > −AW + M U p − U p U = −AW + M p − U p W. U − U Now, suppose that the conclusion of the Lemma is false. Thus, let t0 . At that time, there must be be the first time such that min W (t) = δ 2 a node j such that wj(t0) = δ 2 . As wj has a minimum there we must have w ′ j(t0) ≤ 0, but, on the other hand, by our hypotheses on A, mjw ′ j > − ≥ − N u aijwi + mj i=1 p j − upj wj uj − uj N i=1 δ aij 2 p−1 + mjpuj wj ≥ 0, a contradiction that completes the proof. Next we prove a lemma that ensures that the blow-up rate of the numerical solutions has the same upper bound than the continuous ones. And that this bound does not depend on h, the mesh size. We need to assume a technical hypothesis on the initial data. Lemma 2.2. Assume ∆u0 + u p 0 > 0 in Ω, then there exists a constant C, independent of h such that uh(x, t) ≤ C (Th − t) 1 p−1
DEPENDENCE OF THE BLOW-UP TIME 11 Proof: First we observe that the hypothesis ∆u0 +u p 0 > 0 implies that there exist δ > 0, independent of h, such that u ′ k (0) ≥ δup k (0), 1 ≤ k ≤ N. Next we show that this inequality holds for every 0 ≤ t < Th. Let wk(t) = u ′ k (t) − δup k (t). We want to use the minimum principle to show that wk(t) is positive. To this end, we observe that wk verifies mkw ′ N k + akjwj = mk(u ′′ k − δpu p−1 k u′ N k) + akj(u ′ j − δu p j ) j=1 = −δmkpu p−1 k u′ k + mkpu p−1 k u′ k − δ = −δpu p−1 k N j=1 = mkpu p−1 k wk − δ = mkpu p−1 k wk − δ + N j=1 akj(1 − p)u p k akjuj + mku p k j=k j=k . N j=1 akju p j j=1 + mkpu p−1 k u′ k − δ N j=1 akju p j akj(u p j − pup−1 k uj) + akk(1 − p)u p k akj(u p j − pup−1 k (uj − uk) − u p k ) As f(u) = u p is convex (p > 1) and from our hypotheses on the matrix A it follows that W = (w1, . . . , wN) verifies MW ′ ≥ −AW + MpU p−1 W. Since W (0) > 0 and the minimum principle holds for this equation, we obtain that every node verify u ′ k(t) ≤ δuk(t) p . If xk is a node that blows up we can integrate the above inequality between t and Th to get so that Th u t ′ k u p k (s) ds ≥ δ(Th − t), 1 − uk(t) ≤ C(Th − t) p−1 , where C depends only on p and δ (but not on h). This proves the result.
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