A Liouville theorem for a heat equation and applications for quenching

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1 A Liouville theorem for a heat equation and applications for quenching Nejla Nouaili To cite this version: Nejla Nouaili. A Liouville theorem for a heat equation and applications for quenching. Nonlinearity, IOP Publishing, 0, 4 3, pp <hal-00538> HAL Id: hal Submitted on Nov 00 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

2 A Liouville theorem for a heat equation and applications for quenching Nejla Nouaili Université Paris Dauphine July 6, 00 Abstract: We prove a Liouville Theorem for a semilinear heat equation with absorption term in one dimension. We then derive from this theorem uniform estimates for quenching solutions of that equation. Mathematical Subject classification: 35K05, 35K55, 74H35. Keywords: Quenching, Liouville theorem, heat equation. Introduction This paper is concerned with quenching solutions of the following nonlinear heat equation where β 3 and { t u = xu in [0, T, uβ ux, 0 = u 0 x > 0 for x, u 0, u 0 L. We say that ut quenches in finite time T if u exists for t [0, T and lim inf ux, t = 0. 3 t T x Hereafter we consider a solution u of which quenches at finite time T. A point a is said to be a quenching point if there is a sequence {a n, t n } such that a n a, t n T and ua n, t n 0 as n. Quenching phenomena play an important role and have many applications in physics, chemistry and biology see Salin [Sal04]. It may arise for instance in the theory of combustion and population genetics see Galaktionov and Vázquez [GV0], in population dynamics see Gaucel and Langlais [GL07], in connection with the diffusion equation generated by a polarization phenomena in ionic conductors see Kawarada [Kaw75], in connection with phase transition, when we study the motion of the borderline between liquids and solids see Fila and Kawohl [FK9a] and also in vortex reconnection with the

3 boundary in type II superconductors see Chapman and all [CHO98] and Merle and Zaag [MZ97]. The study of the quenching problem was initiated by Kawarada [Kaw75]. There is a considerable litterature on parabolic problems with strong absorption that deal with the precise description of the qualitative behavior of solutions near quenching. Typical questions in this respect concern the existence of quenching solutions as well as the quenching rates and profiles see Deng and Levine [DL89], Fila and Kawohl [FK9a], Fila, Kawohl and Levine [FKBL9], Merle and Zaag [MZ97], Levine [Lev93] and Dávila and Montenegro [DM05]. ecently, equation was numerically studied by Liang, Lin and Tan [LLT07]. The quenching rate as t T of the solution u of near a quenching point is among the important issues. It was obtained by Guo on bounded domains [Guo90], [Guo9a], Fila and Kawohl [FK9a]. For the higher-dimensional radial problem in [Guo9b] and Fila, Kawohl and Quittner [FHQ9], under some restrictions on initial data. In [Guo90], Guo showed that if and a is an arbitrary quenching point, then u 0 u β 0 0, lim t T ux, t = U T t, uniformly for x a CT t, where C > 0 and U T t = β+ β+ T t β+ is the solution of U T = and U U β T T = 0. T In [FG93], Filippas and Guo were able to obtain a precise description of the profile of the solution in a neighborhood of the quenching point. They proved the following: Let u be a solution of defined for x,, with the boundary conditions u±, t =, which quenches at the point a at time T. Moreover, we assume that u 0 0 and that u u β 0 has a single minimum. Then, we have that 0 ua, t 0, as x a 0. ux, t u x as t T if x a, [ β + u ] β+ x a x = 8β log x a β+ + o, However, the result of Filippas and Guo is not uniform with respect to the quenching point or initial data. We aim in this paper at obtaining uniform estimates on ut defined on at or near the singularity, that is as t T. In order to do so, we introduce for each a the following similarity variables: y = x a T t, s = logt t, w a y, s = T t β+ ux, t. 5 The function w a = w satisfies for all s log T, and y : s w = yw y yw + 4 w β + w β, 6

4 The study of ut near a, T, where a is a quenching point, is equivalent to the study of the long time behavior of w a. If we introduce v defined by v = u, 7 then, we can see easily from and 3, that v satisfies the following equation: t v = xv xv + v +β in [0, T, 8 v We note that u quenches at time T if and only if v blows up at time T, and that a is a quenching point for u if and only if a is a blow-up point for v. As we did for u, we introduce for v the following function: z a y, s = T t β+ vx, t, where y and s are defined as in 5. 9 Note from 5 and 7 that z a = w a. We get by 8 and 9, that z a = z satisfies for all s log T, and y, s z = yz y yz yz z z β + + z+β. 0 We note by L ρ the Hilbert space defined by; L ρ = {g L loc N, C, N g e e y 4 4 dy < + } where ρy = 4π N/. y If g depends only on the variable y N, we use the notation g L = gy e y 4 ρ dy. N If g depends only on y, s N, we use the notation g., s L = gy, s e y 4 ρ dy. N. A Liouville Theorem Our aim in this paper is to prove estimates for quenching solution of 6, uniform with respect to the space variable x. The following Liouville theorem is crucial for that purpose. Theorem A Liouville Theorem for equation 6 Assume that β 3. We consider w, a global nonnegative continuous solution of 6, satisfying y wy, s + M, for all y, s, where M > 0. wy, s 3

5 Then either i w κ, or ii there exists s 0 such that for all y, s, wy, s = ϕs s 0, where is a solution of 6 independent of y and satisfies ϕ = ϕs = κ + e s β+ ϕ + β ϕ β, ϕ = κ, ϕ+ = +. 3 emark: If 0 < β < 3, we don t have an explicit result for the global stationary solution of 6 as for the case β > 3, the only result that we know is that there exist slow orbit, for more details see Guo [Guo9a], [Guo9b] and Fila, Kawohl and Quittner [FHQ9]. Theorem has an equivalent formulation for solutions of equation : Corollary A Liouville Theorem for equation Assume that u is a nonnegative continuous solution of defined for x, t, T. Assume in addition that ux, t BT t β+ for some B > 0 and x u., t L M T t β+, for M > 0. Then, there exists T 0 T such that for all x, t, T, ux, T = κt 0 t. Application to quenching β+. We note that in the blow-up recent litterature [MZ00], [MM00] and [MZ08], Liouville Theorems have important applications to blow-up. In the quenching problem we can get similar results. In the folllowing, we will say that a solution u is of type I, if it satisfy u., t L where M is a positive constant. T t β+ + x u., t L T t β+ + M, type I 4 Note that several authors proved the existence of type I solution see for example Guo [Guo90] and [Guo9a]. We wander whether the bound on x u L in 4 follows from the bound on u L. In other words, we wander whether we can charecterize the set of type I solution only by the fact that T t /β+ M for some M > 0. u L In the following, we derive from the Liouville Theorem some uniform estimates of type I quenching solutions of equation 6. Theorem 3 i Uniform bounds on ut on quenching time Assume that ut is a nonnegative solution of equation that quenches in finite time T > 0 with initial data u 0 satisfying. Then, we have inf ux, t U T t = β + β+ T t β+ as t T, x 4

6 and T t β+ + x v., t L + T t β+ + xv., t L 0 as t T. Equivalently, for any a, inf w ay, s κ as s + and y z a., s L + yz a., s L 0 as s +, y where κ = + β β+. ii ODE-type behavior For all ε > 0, there exists C ε > 0, such that, for all t [ T, T and x, u t u β ε u β + C ε. 5.3 Strategy of the proof of the Liouville theorem Our method is inspired by the one of Merle and Zaag [MZ98a] and [MZ00] developed for the semilinear heat equation where t u = u + u p u 6 u : x, t N [0, T and p >, p < N + N if N 3. However, our contribution is not a simple adaptation of the proof of [MZ98a] and [MZ00]. In fact, in these papers the authors strongly rely on two blow-up criteria for the selfsimilar version of 6 w s = w y yw w p + w p w. 7 Criterion : For nonnegative solution of 7 such that N wy, s 0 ρydy > p p, for some s 0, then w blows up in some finite time s > s 0. Criterion : with no sign condition Assume that w is a solution of 7 such that Ews 0 p wy, s 0 ρydy p + N p+ for some s 0, where E is the following Lyapunov functional defined by Ew = yw + w p w p+ ρydy with ρy = p + Then, w blows up in finite time s > s 0. 5, y e 4 4π N/.

7 Note that in [Zaa0], Zaag obtained a Liouville theorem for the non-gradient structure system { t u = u + v p, t v = v + u q 8 u., 0 = u 0, v., 0 = v 0, where p q is small. There, he adapts the proof of [MZ98a] and uses an infinite time blow-up criterion in similarity variables. In our quenching problem, there is no way to find any equivalent blow-up criterion. We need new ideas which make the originality of our paper. From this point of view the difficulty level is similar to the case of the complex valued equation with no gradient structure t u = u + + iδ u p u 9 where u : N C. treated in [NZ0]. We proceed in five parts. Part, we show that z = /w has a limit z ± as s ±, where z ± is a critical point of the stationary version of 0 or 0. That is, z ± κ or z ± 0, thanks to Guo. Then, we rule out the case where z 0 with sophisticated energy argument. The following parts are dedicated to the non trivial case where z = κ. in Part, we investigate the linear problem in w around κ as s and show that w behaves at most in three ways: i wy, s = κ + C 0 e s + oe s as s, for some constant C 0. ii wy, s = κ + C e s/ + oe s/ as s, for some constant C \ {0}. iii wy, s = κ + κ βs y + o as s. s In these cases, convergence take place in L [, ] for any > 0 and in L ρ. In Part 3, we show that i corresponds to wy, s = ϕs s 0 or wy, s = κ, for some s 0, where ϕ is defined by 3. In Parts 4 and 5, we rule out cases ii and iii. In [MZ98a] and [MZ00], the authors shows that for some a 0 and s 0, w a0 y, s = wy + a 0 e s 0, s 0 satisfies one of the blow-up criteria stated in page 5, which contradicts the fact that w exists for all s. In our case we don t have any blow-up criterion. It turns out that this is the major difficulty in our paper, as in [NZ0] for equation 9. Following [NZ0], we will use a geometrical method where the key idea is to extend the convergence stated in ii and iii from compact sets to larger zones, so that we find the singular profile for w. It appears that in both cases, for larger y, this profile becomes strictly inferior to M, where M is defined in, which is a contradiction. The originality of our paper is based on Velázquez s work in [Vel9], where he extends the convergence from compact sets to larger sets to find the profile for solutions of 7. Note that the fact that w is not in L makes it delicate to use the estimate of [Vel9]. More precisely, we obtain the following singular profiles 6

8 If case ii holds, then lim sup s y e s/ with 0 < < β+ C κ β and C 0. If case iii holds, then with 0 < < lim sup s y s 4β β+. wy, s β + C κ β ye s/ wy, s β + + β + 4β β+ y s = 0 Then using condition, case ii and iii are ruled out, which ends the proof of our Liouville Theorem. Our paper is organised as follows: In section, we prove the Liouville theorem. In section 3, we prove the uniform estimates for quenching solutions of 6 in Theorem 3. Proof of the Liouville Theorem for equation 6 We assume that β 3 and consider wy, s a nonnegative, global solution of 6 satisfying, defined for all y, s. Introducing z = w, we know that z satisfies 0 on. Our goal is to show that w depends only on the variable s.. Part : Behavior of w as s ± The main results of this part are consequences of the parabolic estimates and the gradient structure of equation 6. Let us recall them. Lemma. Parabolic estimates There exists M 0 > 0 such that for all y, s : i y wy, s + wy, s M 0 and wy, s w0, s + M 0 y, ii zy, s + y zy, s + yzy, s M 0 and s z M 0 + y. iii For all > 0, z, y zy, s, yz and s z are bounded in Cα[, 0 ] for some α 0,, where { } Cα[, 0 ] = ψ L [, ψξ, τ ψξ, τ ] sup ξ,τ,ξ,τ [,] ξ ξ + τ τ / α < iv s wy, s M 0 + y w0, s + y. β+ 7

9 Proof: i See. iiand iii See the proof of Lemma 3. from [Guo90]. iv If we write s w = s z/z = s zw, then we get the result by i and ii. Lemma. Gradient Structure i Gradient Structure for equation 6. If we introduce Ews = y w ρdy then, for all s < s, such that we have w ρdy β + β y w β ρdy with ρy = e 4 0 < wy, s c s + y for all s s s with c s > 0, s s, 4π 0 s w ρdy = Ews Ews ii Gradient Structure for equation 0.If we define for each z solution of 0 y z Ezs = z 4 ρdy β + z ρdy z β ρdy, β under the same conditions in i, we obtain: b s z z ρdy = Eza Ezb, for any real a < b. 3 a Proof: i One may multiply equation 0 by s wρ and integrate over the ball B = BO, with > 0. Then using Lemma. and the Lebesgue s theorem yields the result. see Proposition 3 from Giga and Kohn [GK85] for more details. ii Since Ez = E z, this immediately follows from i. We now give the limits of z as s ± in the following: Proposition.3 Limit of z as s ± The limit z + y = lim s + zy, s exists and equals 0 or κ. The convergence is uniform on every compact subset of. The corresponding statements also hold for the limit z y = lim s zy, s. Proof: It follows from the following: Lemma.4 Consider any increasing respectively decreasing sequence such that s j ± as j. Then: i there is a subsequence still denoted by s j such that zy, s + s j l in C α[, ] for all > 0, where l = 0 or κ. ii If l = 0, then Ezs j, if l = κ, then Ezs j Eκ. 8

10 Indeed, from Lemma.4, in order to get the conclusion of Proposition.3, it is enough to show that the limit in Lemma.4 is independent of the choice of the sequence. We consider the case s, the other case being similar. Suppose that s j and s j both tend to infinity. Up to extracting subsequence, we assume that for all j N, s j > s j and proceeding by contradiction, we assume that z j y, s = zy, s + s j κ and z j y, s = zy, s + s j 0 as j. By ii of Lemma.4, we have hence, for j large enough, we have Ez j s Eκ and Ez j s, s j > s j, Ezs j > Eκ > Ezs j, which contradicts the monotonicity of E. Thus the limit in Lemma.4 is independent of the choice of the sequence and the whole function zy, s converges as s. It remains to prove Lemma.4 to finish the proof. Proof of Lemma.4: i We only present the case s j +, the analysis for s j is the same. Let us recall that for all y, s, zy, s satisfies 0, which is: s z = yz y yz yz z z β + + z+β. 4 From condition, we get 0 yz obtain: z = yw w z Cz and 0 z +β Cz, hence we L 0 z Cz s z L 0 z + Cz where L 0 z = yz y yz. 5 Using the semigroup S 0 τ associated to L 0 : S 0 τφy = we see that for all s s, 4π e τ / exp ye τ λ 4 e τ φλdλ, 6 e Cs s S 0 s s z., s z., s L e Cs s z., s L. 7 Let s j be a sequence tending to +, and let z j y, s = zy, s + s j. From Lemma., up to extracting a sub-sequence still denoted by s j, z j converges to z + in C,, for any > 0 and we obtain that z + satisfy 5. In the following, we will prove that the limit z + is either 0 or κ. We have to consider two cases.. Case : There exists y 0, s 0 such that z + y 0, s 0 = 0 We claim first that y, z + y, s 0 =

11 Indeed, if for some y, we have z + y, s 0 > 0, then since we have from z j y 0, s 0 z j y, s 0 + M y y 0, letting j, we get a contradiction. Thus 8 holds. Using 7, we conclude that z 0 on indeed, by 7 for s s 0, z., s 0, and for s s 0, S 0 s 0 sz., s 0, hence z., s 0. Case : For all y, s, z + y, s > 0 Let us introduce w j y, s = wy, s + s j. In this case, w j y, s = /z j y, s w + = / z + y, s in C,,, for any > 0. 9 Since z j and w j are solutions of 4 and 6 respectively, the same holds for z + and w + respectively. Our goal in this step is to prove that z + y, s = κ. First, we prove that z + is independent of s, then we conclude using the result of Guo concerning the stationary global solution of 6. We will proceed as in Proposition 4, page 308 in [GK85]. We claim the following: Claim.5 i For all y, s y w j y, s + /w j., s C, w j y, s C y + w + 0, s and s w j y, s C + y y + w + 0, s, where C > 0 independent of j and w + is a solution of 6. ii For all y, s, y w + y, s + / w + y, s C, w + y, s C y + w + 0, s and s w y, s C + y y + w + 0, s. 30 emark: To make the notation legible, we will note the partial derivative in time and space of the sequence w j or z j and the limit w + or z + respectively s w j, y w j and s w +, y w +. Proof of Claim.5: i Integrating the inequality y w j y, s M by i of Lemma. in space between 0 and y, we obtain w j y, s M y + w j 0, s. Then using the fact that w j 0, s w + 0, s= / z + 0, s < +, as j, we get the second inequality of i in Claim.5. Using Lemma. and the identity s w j y, s = s z j y, s/z j y, s = s z j y, sw j y, s, we obtain the third inequality in i. ii Using i and the convergence 9, the result comes immediately. Up to extracting a subsequence, we can assume that s j+ s j. Therefore, using with w = w j, a = m, b = m + s j+ s j, where m Z, we get: 0

12 si+ s j +m m s w j ρdy = Ew j m Ew j m + s j+ s j = Ew j m Ew j+ m. Using i, ii and iii of Claim.5 together with Lesbegue s Theorem we have y wj ρdy y w + ρdy, wj ρdy w + ρdy, and w β j ρdy w β + ρdy as j. 3 Therefore, Ew j m E w + m as j well defined by iii of Claim.5. In particular, the right-hand side of 3 tends to zero. Since s j+ s j +, it follows that for every integers m < M on the one hand M lim s w j ρdyds = 0, 3 j m On the other hand, by the third inequality in i of Claim.5 and the continuity of w + 0, s, we obtain y, s m, M s w j y, s Cm, M + y 3. Since s w j converges simply to s w + by iii of Claim.5, we conclude that: M m s w j ρdyds M m s w + ρdyds. 33 Using 3 and 33, we get M m s w + ρdyds = 0. Using the fact that m and M are arbitrary, we conclude that s w + = 0 on and w + y, s = w + y is independent of s. Since w + is a solution of 6, it follows that w + solves the following stationary equation y, 0 = w yw + β + w w β. 34 We recall that by ii of Claim.5 we have w + s > /C and w + s C + y. Now we recall the result of Guo gived by Theorem. in [Guo90]: The only global solution of w yw + w β + εw β = 0, y, which is greater than or equal to some positive constant c and which grows at most polynomially as y, is w εβ + β+.

13 Using this result, it follows that w + = κ and we conclude that z + κ. This concludes the proof of i of Lemma.4. ii We recall from Lemma. that Ezs = y z z 4 ρdy β + z ρdy z β ρdy. 35 β If z j converges to κ, using Lesbegue s Theorem and Claim.5, we obtain Ez j s = Ew j s Eκ = Eκ. In the case, where z j converges to 0, we write from Lemma. and Claim.5 Ez j s = Ew j s C w β ρdy 36 β Since the integral in 36 tends to infinity as j + we have: Ez j s. This concludes the proof of Lemma.4 and Proposition.3. To end this part, we recall the result obtained by Proposition.3, which says that z + y = lim s + zy, s and z y = lim s zy, s exist and equal 0 or κ. Now, we will prove that the case z = 0 can t occur. Indeed, suppose that zy, s 0 as s, by ii of Lemma.4, we have: Ezs as s. Therefore, there exists s < 0, such that for all s s, we have s < 0 and Ezs < Ez0. which contradicts the monotonicity of E. Thus, the case z = 0 is ruled out. In the following, we will study the case where w = κ.. Part : Linear behavior of w near κ as s In this part, we assume that w κ as s 37 in L ρ and uniformly on every compact sets, and we classify the L ρ behavior of w κ as s. Let us introduce v = w κ. From 6, v satisfies the following equation: where y, s, s v = Lv fv, 38 Lv = yv y yv + v and fv = v + κ β κ β + βκ β+ v. 39 Concerning the non-linear term of equation 39, we have:

14 Lemma.6 There exists s such that for all s s : 0 fv M v and fv M v. 40 Proof: Using Taylor s formula, we obtain v + κ β = κ β for some θ between 0 and v. Therefore β β + v + ββ + θ + κ β v, 4 fv = cθ, βv, cθ, β = ββ + θ + κ β. If v > 0 then 0 cθ, β β/β + κ β = β/κ. If v < 0, recalling that B v + κ θ + κ we obtain 0 cθ, β β/β + B β. Thus 40 was established. To show the second inequality in 40 we observe that if v < κ, then it follows from 40. If v > κ then from 39 and the lower bound of v, we have for some constant C. fv β β + v β + κ + B β M v, In the following, we will discuss general properties of the operator L. At first we note that it is self-adjoint on L ρ. Its spectrum is specl = { m m N}; it consists of eigenvalues. The eigenfunctions of L are simple and derived from Hermite polynomials. For m corresponds the eigenfunction h m y = [ m ] n=0 m! n!m n! n y m n. The polynomial h m satisfies h n h m ρdy = n n!δ nm. Let us introduce k m = the eigenfunctions of L span all the space L ρ, we expand v as follows: hm h m L ρ. Since vy, s = v m s h m y + v y, s 4 m=0 where v m is the projection of v on h m and v y, s = P v, with P is the orthogonal projector on the negative subspace of L. Now we show that as s, either v 0 s, v s or v s is predominant with respect to the expansion 4 of v in L ρ. We have the following: Proposition.7 Classification of the behavior of vy, s as s As s, one of the following situations occurs: i v s + v., s + v., s L ρ = ov 0 s, v., s C 0 e s L ρ = Oe 3 s for some C 0. ii v 0 s + v., s + v., s L ρ = ov s, v., s C ye s/ L ρ = Oe εs for some C \ 0 and any ε > 0. iii v 0 s + v s + v., s L ρ = o v., s L ρ, v., s κ 4β y log s L = O. ρ s 3

15 .3 Part 3: Case i of Proposition.7: s 0 such that wy, s = ϕs s 0 In this part, we prove the following: Proposition.8 The relevant case i of Proposition.7 Assume that case i of Proposition.7 holds. Then there exists s 0 such that for all y, s, wy, s = ϕs s 0, where ϕs is introduced in. Proof: First we recall from i of Proposition.7 and the definition of v that w., s {κ + C 0 e s } L ρ Ce 3/s as s. 43 Let us remark that we already have a solution ˆϕ of 6 defined in, s ] for some s and which satisfies the same expansion as wy, s when s : a if C 0 = 0, just take ˆϕ κ, β + b if C 0 > 0, take ˆϕ ϕs s 0 where s 0 = log κ c if C 0 < 0, take ˆϕ ϕ 0 s s 0 where s 0 = log and ϕ 0 s = κ e s β+ C 0, β + C 0 κ 44 is a solution of 6 that quenches at s = 0 but there exist C > 0 such that ϕ 0 C for all s. If we introduce V = w ˆϕ, then we see from 6 that V is defined for all y, s, s ] and satisfies where L is given in 39 V s = L + lsv F V, 45 and F V = ˆϕ + V β ˆϕ β + β ˆϕ β+ V ls = 0 if ˆϕ = κ, ls = βe s β + + e s if ˆϕ = ϕs s 0, ls = βe s β + e s if ˆϕ = ϕ 0 s s 0. We note that ls Ce s for all s s and some C > 0 and there exist M i=, such that 46 0 F V M V and 0 F V M V. 47 We omit the proof of 47 since it is quite similar to the proof of 40. Let us introduce Is = V., s L ρ, multiply 45 by V ρ and integrate over. Using the fact that is the greatest eigenvalue of L and 47, we obtain I s + lsis + C V 4 y, sρdy. Now, we recall the following from [Vel93]: 4

16 Lemma.9 egularizing effect of the operator L Assume that ψy, s satisfies for some a b and σ, where s [a, b], y, s ψ L + σψ, 0 ψy, s, Lψ = yψ y yψ + ψ = ρ div ρ yψ + ψ. 48 Then for any r >, there exists C = C r, σ > 0 and s r, such that s [a + s, b], /r ψy, s ρydy r C ψ., s s L. 49 ρ Proof of Lemma: See Lemma.3 in [HV93] Using the lemma above, we obtain the existence of C > 0 and s > 0, such that V., s L ρ C V., s s L ρ. Then we obtain for some s s : s s I s 5 4 Is + CIs s. 50 Since Is Ce 3/s from 43, the following lemma from [NZ0] allows us to conclude. Lemma.0 Lemma 3.6 of [NZ0] Consider Is a positive C function such that 50 is satisfied and 0 Is Ce 3/s for all s s, for some s. Then, for some s 3 s, we have Is = 0 for all s s 3. Using Lemma.0, we obtain V 0 on, s 3 ]. Consequently, we have y, s, s 3 ], wy, s = ˆϕs. 5 From the uniqueness of the Cauchy problem for equation 6 and since w is defined for all y, s, ˆϕ is defined for all y, s and 5 holds for all y, s. Therefore, case c in 46 cannot hold and for all y, s, wy, s = κ or wy, s = ϕs s 0. This concludes the proof of Proposition.8 and finishes Part 3..4 Part 4: Irrelevance of the case iii of Proposition.7 We consider case iii of Proposition.7. In this part, we will proceed like in Step 4 in [NZ0]. The following proposition allows us to reach a contradiction. Proposition. Assume that case iii of Proposition.7 holds. ε 0 > 0 such that lim sup y wy, s G = 0, s s y ε 0 s where Gξ = β + β+ 4β ξ β+. Then, there exists 5 5

17 Indeed, let us first use Proposition. to find a contradiction ruling out case iii of Proposition.7, and then prove Proposition.. We fundamentally rely on the following Lemma: Lemma. Lemma. from [MZ98b] Assume that ψξ, τ satisfies for all ξ 4B and τ [0, τ ]: { τ ψ yyψ + λψ + µ, ψξ, 0 ψ 0, ψξ, τ B, where τ. Then, for all ξ B and τ [0, τ ], ψξ, τ e λτ ψ 0 + µ + CB e B /4. Let us define u s0 by u s0 ξ, τ = τ β+ wy, s where y = ξ + ε 0 s0 τ and s = s 0 log τ. 53 We note that u s0 is defined for all τ [0, and ξ, and that u s0 satisfies equation. From Lemma., we have ξ <4 s 0 /4 τ [0,, u s0., τ M τ β+. 54 The initial condition at time τ = 0 is u s0 ξ, 0 = wξ + ε 0 s0, s 0. Using Proposition., we get: ε0 sup u s0 ξ, 0 G gs0 0 as s If we define v, the solution of: then v = v β and v0 = G ε0, vτ = κ β + ε β+ 0 τ, 56 6β which quenches at time β+ε 0 6β <. Therefore, there exists τ 0 = τ 0 ε 0 <, such that Now, if we consider the function then the following claim allows us to conclude: vτ 0 = M 3 τ 0 β+. 57 ψ = u s0 v, 58 Claim.3 For s 0 large enough, τ [0, τ 0 ] and ξ 4 s 0 /4, we have: i τ ψ ξ ψ + Cε 0ψ, ii ψξ, 0 gs 0, iii ψξ, τ Cε 0 s 0 /. 6

18 Indeed, using Lemma. with B = s 0 /4, B = Cε 0 s 0 /, τ = τ 0, ψ 0 = gs 0, λ = Cε 0 and µ = 0, we get for all τ [0, τ 0 ], sup ψξ, τ Cε 0 gs 0 + s 0 / e s 0 / 4 0 as s 0. ξ s 0 /4 For s 0 large enough and ξ = 0, we get: ψ0, τ 0 M 3 τ 0 /β+ and by 57 u s0 0, τ 0 vτ 0 + ψ0, τ 0 3 M τ 0 /β+, which is in contradiction with 54. To conclude, it remains to prove Claim.3. Proof of Claim.3: s Ψ i If we note by Ψ = u s0 v, then we get = yψ u β s 0 v β, = yψ + βψθ β+, for some θ between u s0 and v. Using 54, 56 and 57, we have for all τ [0, τ 0 ] and ξ θ max u 0 ξ, τ, Cε 0, for some positive θ. vτ Since ψ = Ψ, using Kato s inequality, we conclude the proof of i. ii It is directly obtained from 55. iii Using the definition 58 of ψ, 53 and 56, we write for all τ [0, τ 0 ] and ξ 4 s 0 /4, ψξ, τ u s0 ξ, τ + vτ wy, s + κ, where y, s are defined in Since we have from Lemma., 37 and 53 for s 0 large enough wy, s w0, s + M 0 y C + y and y ξ + ε 0 s0 τ 4 s 0 /4 + ε 0 s0 τ0 Cε 0 s 0 /, the bound on ψ follows from 59. This concludes the proof of claim.3. It remains to prove Propossition. to conclude Part 4. Proof of Proposition.: Consider some arbitrary ε 0 0,, where 4β = β The parameter ε 0 will be fixed later in the proof small enough. If we note y fy, s = G, 6 s 7

19 and then f satisfies F y, s = fy, s y yf + β + f f β = 0. κ βs, 6 and we see from iii of Proposition.7 that log s F., s w., s χ ε0 L ρ = O as s, 63 s where χ ε0 y, s = if y s 3ε 0 and zero otherwise. 64 The formal idea of this proof is that F solves in an approximate way the same equation as w for s. By 63, w and F are very close in the region y. Our task is to prove that they remain close in the larger region y ε 0 s, for some ε0 chosen later. Let us consider a cut-off function y γy, s = γ 0, 65 s where γ 0 C is such that γ 0 ξ = if ξ 3ε 0 and γ 0 ξ = 0 if ξ 4ε 0. We introduce ν = w F and Z = γ ν. 66 Our proof is the same as Velázquez [Vel9] and Nouaili and Zaag [NZ0]. As in [NZ0], we need to multiply by the cut-off, since our profile F y, s defined by 6 is singular on the parabola y = s. The cut-off function will generate an extra term, difficult to handle. Let us present the major steps of the proof in the following. The proof of the presented Lemmas will be given at the end of this step. Lemma.4 Estimates in modified L ρ spaces There exists ε 0 > 0 such that the function Z satisfies for all s s and y : s Z yz + y yz + σz C Z + y + s + + sχ ε0 div ν y γ, 67 where s, σ = /00 and χ ε0 is defined in 64. Moreover, N ε 0 s Zs = o as s, 68 where the norm N q r ψ is defined, for all r > 0 and q <, by Nr q ψ = sup ξ r ψy q exp /q y ξ dy

20 Using the regularizing effect of the operator L, we derive the following pointwise estimate, which allows us to conclude the proof of Proposition.: Lemma.5 An upper bound for Zy, s in { y ε 0 s} We have: sup y ε 0 s Zy, s = o as s. Thus, Proposition. follows from Lemma.5 by 66, 6 and 6. It remains to prove Lemmas.4 and.5. Proof of Lemma.4: The proof of 67 is straightforward and a bit technical. We leave it to Appendix B. Let us then prove 68. We take s 0 < s and s 0 s < s such that e s s 0 s. We use the variation of constant formula in 67 to write Zy, s S σ s s 0 Z., s 0 + s s 0 S σ s τ C { Z + y + τ + + } τχ ε0 div ν y γ dτ, where S σ is the semigroup associated to the operator L σ φ = yφ y yφ + + σφ, defined on L ρ. The kernel of the semigroup S σ τ is Setting S σ τ, y, z = and taking the N r norm we obtain +C s [ ] e +στ 4π e τ exp ye τ/ z / 4 e τ. 70 r rs, s 0 = ε 0 e s s 0 = e s s 0 7 Nr Z., s Nr S σ s s 0 Z., s 0 + C Nr S σ s τz., τ dτ s 0 s y +C Nr + S σ s τ s 0 τ dτ s 0 N r S σ s τ + τχ ε0 y, τdτ + C s s s 0 N r S σ s τdiv ν y γdτ J + J + J 3 + J 4 + J 5. In comparison with [Vel9], we have a new term J 5 coming from the cut-off terms. Therefore, we just recall in the following claim the estimates on J...J 4 from [Vel9], and treat J and J 5, which are new ingredients in our proof: 9

21 Claim.6 We obtain as s J Ce +σs s 0 log s 0 s 0, J C s0 +s 0 s 0 + s 0 e +σs τ 0 e s τ 0 /0 with 0 = 4ε 0, J 3 C es s 0+σ + s s 0 s 0, s 0 J 4 Ce s s 0+σ e αs, where α > 0, J 5 Ce s s 0+σ e βs, where β > 0. L r Z., τ dτ + C es s 0+σ, s 0 Proof: See page 578 in [Vel9] for J, J 3 and J 4. To obtain the bound on J, we need the following inequality, which will be proved in Appendix B s Z yz + y Z C Z + y + s + + sχ ε0 div ν y γ, 7 then, we proceed exactly as in page in [Vel9] and we obtain the wanted estimation on J. Now, we treat J 5. We have from 70: S σ s τ div ν y γ, = Ces τ+σ e s τ / exp ye s τ/ λ 4 e s τ div ν y γdλ, = Ces τ+σ e s τ / ye s τ/ λ e s τ exp ye s τ/ λ 4 e s τ ν y γdλ. 73 We have from Lemma. and 37 Since F is bounded for wy, s w0, s + M 0 y C + y. y τ 4ε 0 τ, 3ε0 τ 3ε0 τ, 4ε0 τ, we have /, where is defined by 60 and supp y γ ν y γ C ν I {3ε0 y τ 4ε 0 }, C + τi 3ε0 y τ 4ε 0 C + τχ ε0. Using Cauchy-Schwartz inequality, we obtain: S σ s τ div ν y γ Ces τ+σ + τ e s τ 3/ I I, 0

22 where, I = I = / ye s τ/ λ exp ye s τ/ λ dλ 4 e s τ, / exp ye s τ/ λ 4 e s τ χ ε0 dλ. Doing a change of variables, we obtain I = C e s τ 3/4. Furthermore, we have: where, I 3 = exp I I 3 χ ε0 e λ 4 dλ /, ye s τ/ λ e s τ / + dλ λ. 4 We introduce θ = ye s τ/, by completing squares, we readily check that: λ 4 θ λ e s τ = + e s τ 4 e s τ λ θ θ + e s τ + + e s τ, then we obtain: / I3 e s τ = C + e s τ exp θ e s τ. Therefore, N r S σ s τdiv ν y γ e s τ+σ + τ C + e s τ /8 e s τ 5/8 χ ε 0 / I L 4, ρ where I 4 = Nr y e s τ exp 8 e s τ. Let us compute I 4. Using the fact that y µ y e s τ 4 e s τ = y + e s τ / µ + e s τ / + µ e s τ, and doing a change of variables, we obtain: y µ exp + y e s τ 4 4 e s τ dy µ e s τ Cexp exp y + e s τ / µ + e s τ / dy. 4 4

23 Hence I 4 C + e s τ /8 and This gives N r S σ s τdiv ν y γ C es τ+σ + τ e s τ 5/8 J 5 = s e λ 4 dλ. λ τ s 0 N r S σ s τdiv ν y γ dτ Cηe s s 0+σ e αs 0, where α > 0. This concludes the proof of the claim.6. This concludes the proof of Claim.6. Summing up J i=..5, from claim.6 we obtain N r Z., s e s s 0+σ C log s 0 s 0 + C s0 +s 0 s 0 + Now, we recall the following from [Vel9]: s 0 e s τ 0+σ e s τ 0 /0 N r Z., τ dτ. Lemma.7 Let ε, C,, σ and α be positives constants, 0 < α < and assume that Hs is a family of continuous functions satisfying: Hs εe s+σ + C s + 0 e s τ+σ Hτ e s τ dτ for s > 0. α Then there exists ξ = ξ, C, α such that for any ε 0, ε and any s for which εe s+σ ξ, we have Hs εe s+σ. Proof: See the proof of Lemma. from [Vel9]. Note that the proof of [Vel9] is done in the case σ = 0, but it can be adapted to some σ > 0 with no difficulty. We conclude that N rτ,s 0 Z., s Ces s 0+σ log s 0 s 0 as s. If we fix s = e s s 0, then we obtain s s 0, log s log s 0 and N s Z., s Cs +σ log s 0 s 0 as s. Since σ = 00, we get N s Z., s = o, as s. This concludes the proof of Lemma.4. log s C 0 s σ Now we give the proof of Lemma.5. Proof of Lemma.5. We aim at bounding Zy, s for y s in terms of N s Zs, where = ε 0 and = ε 0, for some s < s. Starting from equation 7, we do as in [Vel9]: Z., s { e C 0 S 0 Z., s 0 } { s y + C e Cs τ + Ss τ s 0 τ + + τχ ε0 { s } e Cs τ Ss τ div ν y γ dτ s 0 = M + M + M 3, where 0 = 4ε 0, } dτ

24 where S is the semigroup associated to the operator L defined in 48. The terms M and M are estimated in the following: Claim.8 Velázquez There exists s 0, such that for all s s 0 sup M = sup y s e C 0 S 0 Z., s 0 = o as s, y s sup M = sup s y + y s s τχ s ε0 C s. 74 y s Proof: See page 58 from [Vel9] and Lemma 6.5 in [HV93] in a similar case. It remains to estimate M 3. Proceeding as in page 73 and using the fact that νy, s C + s for all y C s obtained from i of Lemma. we write Ss τ div ν y γ Ce s τ = e s τ / exp ye s τ/ λ 4 e s τ div ν y γdλ, Ce s τ = e s τ / ye s τ/ λ e s τ exp ye s τ/ λ 4 e s τ ν y γdλ, Ce s τ e s τ 3/ ye s τ/ λ exp ye s τ/ λ 4 e s τ χ ε0 dλ, Ces τ τ + τ e s τ 3/ exp ye s τ/ λ 4 e s τ χ ε0 dλ. We make the change of variables z = e s τ / λ e τ s/ y and we obtain exp ye s τ/ λ 4 e s τ χ ε0 dλ e s τ / e z /4 dz, Σ where, { Σ = z : z + e τ s/ e s τ / y 3ε 0 e s τ / } τ. Since ye τ s/ ε 0 s, we readily see that Σ { z : z ε0 s }. Then we conclude that and we obtain Ss τ div ν y γ Ces τ e s τ eβs, where β > 0, sup M 3 = o as s. y s s Putting together M i=..3, the proof of lemma.5 is complete. This concludes also the proof of Proposition. and rules out case iii of Proposition.7. 3

25 .5 Part 5: Irrelevance of the case ii of Proposition.7 To conclude the proof of Theorem, we consider case ii of Proposition.7. We claim that the following proposition allows us to reach a contradiction in this case. Proposition.9 lim sup s y ε 0 e s/ wy, s Gye s/ = 0, where Gξ = κ C κ β β+ ξ. 75 As in the previous part, first, we will find a contradiction ruling out case ii of Proposition.7 and then prove Proposition.9. Let us define u s0 by u s0 ξ, τ = τ β+ wy, s where y = ξ + ε 0 e s 0/ τ and s = s 0 log τ. 76 We note that u s0 is defined for all τ [0, and ξ, and that u s0 satisfies equation. From Lemma., we have ξ <4e s 0 /4 τ [0,, u s0., τ M τ β+. 77 The initial condition at time τ = 0 is u s0 ξ, 0 = wξ + ε 0 e s0/, s 0. Using Proposition., we get: ε0 sup u s0 ξ, 0 G gs0 0 as s If we define v, the solution of: then which quenches at time C κ β ε 0 Now, if we consider the function v = v β and v0 = G ε0, vτ = κ C κ β ε 0 τ β+, 79 <. Therefore, there exists τ 0 = τ 0 ε 0 <, such that vτ 0 = M 3 τ 0 β+. 80 ψ = u s0 v, 8 then the following claim allows us to conclude we omit the proof since it is the same as the proof of Lemma.3: Claim.0 For s 0 large enough and for all τ [0, τ 0 ] and ξ 4e s 0/4, we have: i τ ψ ξ ψ + Cε 0ψ, ii ψξ, 0 gs 0, iii ψξ, τ Cε 0 e s 0/. 4

26 Indeed, using Lemma. with B = e s0/4, B = Cε 0 e s0/, τ = τ 0, ψ 0 = gs 0, λ = Cε 0 and µ = 0, we get for all τ [0, τ 0 ], sup ψξ, τ Cε 0 gs 0 + e s0/ e e s 0 / 4 0 as s 0. ξ e s 0 /4 For s 0 large enough and ξ = 0, we get: ψ0, τ 0 M 3 τ 0 /β+ and by 57 which is in contradiction with 54. u s0 0, τ 0 vτ 0 + ψ0, τ 0 3 M τ 0 /β+, Proof of Proposition.9: The proof is very similar to that of Proposition.. If we note fy, s = Gye s/, then f satisfies s f y yf + f β + f β = 0. 8 Consider an arbitrary ε 0 0, 0, where = κp C. ε 0 will be fixed small enough later. Let us consider a cut-off function γy, s = γ 0 ye s/, where γ 0 C such that γ 0 ξ = if ξ 3ε 0 and γ 0 ξ = 0 if ξ 4ε 0. We note ν = w f and Z = γ ν. From ii of Proposition.7, we have Z L ρ Ce s ε as s, for some ε > As in the previous part, we divide our proof in two parts given in the following lemmas. Lemma. Estimates in the modified L ρ spaces There exists ε 0 > 0 such that the function Z satisfies for all s s and y, s Z yz + y yz + σz CZ + e s + + e s/ χ ε0 div ν y γ, 84 where s, σ = 00 and χ ε0 y, s = if y e s/ 3ε 0 and zero otherwise. 85 Moreover, we have N ε 0 e s/ Zs = o as s. 86 As in Part 4, the following lemma allows us to conclude the proof of Proposition.9: Lemma. An upper bound for Zy, s in y ε 0 e s/ We have: sup Zy, s = o as s. 87 y ε 0 e s/ 5

27 emains to prove Lemmas. and. to conclude the proof of Proposition.9. Here, we only sketch the proof of Lemma., since it is completely similar to Part 4. We don t give the proof of Lemma.. We refer the reader to Part 4 and Proposition.4 from Velázquez [Vel9] for similar situations. Proof of Lemma.: As in the previous step, we leave the proof of 84 to Appendix B. Let us now apply the variation of constants formula and take the norm N rs,s 0, where rs, s 0 is as in 7. Assume that s 0 < s, then for all s 0 s s 0, we have Nr Z., s Nr S σ s s 0 Z., s 0 + C s s 0 Nr S σ s τz., τ dτ +C +C s s s s 0 N r Ss τe τ dτ s 0 N r S σ s τ + e τ/ χ ε0., τdτ Nr S σ s τdiv ν y γdτ s 0 = J + J + J 3 + J 4 + J 5. Arguing as in Part 4 and using 83, we prove: Claim.3 J Ce s s0+σ e s0 ε, s0 +s 0 s 0 + e s τ 0+σ J C e s τ 0 /0 L r Z., s dτ s 0 +Ce s s 0+σ e s with 0 = 4ε 0, J 3 Ce s s 0+σ e s + e s/, J 4 Ce s s 0+σ e αe s where α > 0, J 5 Ce s s 0+σ e βe s where β > 0. Proof: To estimate J i=,3,4, see page 584 in [Vel9]. To treat J and J 5, we proceed as in the proof of Lemma.4 of the previous part. Summing up J i=..5, we obtain: N r Z., s Ce s s 0+σ e εs + C s 0 +s 0 s 0 + s 0 e s τ 0 +σ e s τ 0 /0 L r Z., s dτ, then using Proposition.7, we get N rs,s 0 Z., s Ces s 0+σ e εs as s for s 0 s s 0. If we fix s = s 0 /, then we obtain N rs,s 0 Z., s Ces ε +σ Ce s ε+σ 0 as s, since ɛ is small enough and σ = 00. This concludes the proof of Lemma.. As announced earlier, we don t give the proof of Lemma. and refer the reader to Part 4 and Section from [Vel9]. This concludes the proof of Proposition.9 and rules out case ii of Proposition.7. Conclusion of Part 3 and the proof of the Liouville theorem We conclude from Part 4 and 5 that cases ii and iii of Proposition.7 are ruled out. By Part 3, we obtain that w κ or w ϕs s 0 for some real s 0, where ϕ is defined in, which is the desired conclusion of Theorem. 6

28 3 Uniform bounds on ut on quenching time and ODE type behavior We prove here Theorem 3. Let us first state the following convergence result which derives from [Guo90]. Proposition 4 Guo Consider ux, t a type I solution of, which quenches in finite time T. Then, for any quenching point a lim t T ua, tt t β+ = κ, Proof : See Theorem 3.0 page 77 in [Guo90]. Although Guo studied equation on a finite interval with Dirichlet boundary conditions. One can easily check that this method holds on the whole space, whenever the solution is of type I. Consider ut a nonnegative solution of type I of equation that quenches at time T and y wy, s + M, for all y, s, where M > wy, s In this section we proceed as in the proof of Theorem. in [MZ98a]. In the following, Part is devoted to the proof of i of Theorem 3. In Part, we will prove ii of Theorem 3. Part : Uniform quenching estimates From the definition 6 and 0 of w a and z a, it is enough to prove the second statement of i for some a. Consider a a quenching point of u. For simplicity, we assume that a = 0 and write w respectively z instead of w a respectively z a. We want to prove that inf y wy, s κ and y z., s L + y z., s L 0 as s + where z /w. Using Proposition 4, we know that w0, s κ as s +. Since inf y wy, s w0, s, we obtain lim sup inf wy, s κ. 89 s + y and hence, lim inf z., s L s + κ, 90 lim inf z., s L s + + yz., s L + yz., s L κ. The conclusion follows if we prove that lim sup z., s L + y z., s L + yz., s L κ. s + We argue by contradiction and assume that there exists a sequence s n n such that s n + as n + and lim n + z., s n L + y z., s n L + yz., s n L = κ + 3ε 0, 7

29 where ε 0 > 0. Up to extracting a subsequence, we have either lim z., s n L κ +ε 0, or n + We deduce the existence of y i n i=..3 such that lim yz., s n L ε 0, or n + z., s n L ε 0 zy n, s n, or y z., s n L ε 0 y zy n, s n, or y z., s n L ε 0 yzy 3 n, s n. Introducing y n = y i n, where i =, or 3, we define for each n N W n y, s = wy + y n e s/, s + s n and Z n y, s = /W n y, s. lim yyz., s n L ε 0. n + We can see that W n satisfies 6 and Z n satisfies 0. following: More precisely, we claim the Lemma 3. Z n n is a sequence of solutions of 0 with the following properties: i lim n + Z n 0, 0 κ +ε 0 /, or lim n + y Z n 0, 0 ε 0 /, or lim n + yz n ε 0 /. ii For all > 0, there exists n 0 N, such that for all n n 0, - Z n is defined for all y, s [, ] and W n = /Z n satisfies y W n y, s + W n y, s M. - Z n 0 and Z n., s L [,] + y Z n., s L [,] M. - There exists m > 0 such that Z n., s C, [,] m, where C, [, ] stands for the functions which are continuously differentiable twice in space and once in time. Proof of Lemma 3.: We give the following version of Lemma., Claim 3. Parabolic estimates For all y, s [, ] : i Z n y, s + y Z n y, s + yz n y, s M and s Z n M + y. ii For all > 0, Z n, y Z n y, s, yz n and s Z n are bounded in Cα[, 0 ] for some α 0,, where { } Cα[, 0 ] = ψ L [, ψξ, τ ψξ, τ ] sup ξ,τ,ξ,τ [,] ξ ξ + τ τ / α < Using the fact that u is of type I and claim 3., we proceed as in the proof of Lemma. in [MZ98a]. Now, using the compactness property of Z n, shown in Lemma 3., we find Z C, [, ] such that up to extracting a subsequence, Z n Z C, [, ]. Using a diagonal process, we find Z C, such that Z n Z in C, loc. From Lemma 3., we have 8

30 a Z satisfies 0 for y, s. b Z 0 and Z L + y Z L C. c Z0, 0 κ + ε 0 /, or y Z0, 0 ε 0 /, or yz0, 0 ε 0 /. Now, we have to consider two cases: Case : If there exits y 0, s 0 such that Zy 0, s 0 = 0. Then, doing as in the case of proof of Lemma.4, we obtain Z 0, which is impossible because of c. Case : For all y, s, Zy, s > 0. In this case, we have W n = /Z n W = /V in C, loc and y Z n 0, 0 /Z n 0, 0 y Z0, 0 /Z0, 0. Since W n satisfies, the same holds for W. Using ii of claim 3., we see that y, s y W y, s + W y, s M. Using our Liouville theorem, we get W κ or W y, s = ϕs s 0, where ϕs s 0 = κ+e s /β+. In all cases, this contradicts c. This concludes the proof of i of Theorem 3. Part : ODE type behavior Consider a type I solution ut of, that quenches in finite time T > 0, with initial condition satisfying. We know from 88 that: t [0, T, x, u x x, t T t β+ + T t β+ M, where M > 0. 9 ux, t Let us prove now the uniform pointwise control of the diffusion term by the nonlinear term, which asserts that the solution ut behaves everywhere like the ODE u = u β up to a constant. Since u 0 L, hence ut,. L u 0 L, from equation, it is enough to prove that ε, c ε > 0 such that for all x, t [T/, T, t v v +β εv +β + c ε, where v = /u. The plan of the proof is the same as in [MZ98a] and [MZ00]. However, the Giga-Kohn property small local energy implies no blow-up locally breaks down. We proceed as in [NZ0], where the authors proved that small L ρ norm implies no blow-up locally. In our case we have the following Lemma: Lemma 3.3 For all M > 0, there exist positive ɛ 0, C 0 and M 0 such that if U is a type I solution of and for some 0 < ɛ η 0, where x 0, Z x0., 0 L ρ ɛ, 9 y = x x 0, s = log t, Z x0 y, s = t β+ t Ux, t, 9

31 then: i For all x 0 and s [0, +, ii For all x and t [0,, we have Ux, t M 0. s Z x0., s L ρ C 0 ɛe β+. 93 Proof: The idea of the proof is the same as in the proof of Proposition 3.3 in [NZ0]. i For simplicity, we write Z instead of Z x0. Since Z is a solution of 0, we multiply 0 by Zρ and integrate to get I s β + Is + Zy, s β+3 ρydy, where Is = Zy, s ρydy. 94 We note by y Z /Z = y w Z 3, 0 and 9 that Z satisfies s Z yz y yz + Z + σz, where σ is a positive constant. Using Lemma.9 for Z, we see that there exist C M > 0 and s = s p + > 0 such that for all s s Zy, s β+3 ρydy C Z., s s β+3 95 L ρ Now, we divide the proof in two steps: Step : 0 s s. Using 94 and the fact that Z is bounded by M > 0 see 9, we get I s λis for some λ = λm > 0, hence Is e λs I0 e λs ɛ C 0 ɛ e s β+, where we define C0 = e λ+ β+ s. This gives 93 for 0 s s. Step : s s. In this step, we argue by contradiction to prove 93 for all s s. We suppose that there exists s > s, such that Is < C 0 ɛ e s β+, for all s s < s 96 Is = C 0 ɛ e s β+. 97 Let F s = IsC 0 ɛ e s β+. From 94, 95, 96 and Step, we have for all s s s, F s C C 0 ɛ e s β+ Is s β+3 C C 0 ɛ β+ e s β+ e s s β+3 β+ C C 0 ɛ β e β+3 β+ s e s. Since F s from the step above, we integrate the last inequality to obtain F s C Cɛ β e s e s e s + F s, C Cɛ β e s β , 30

32 for ɛ ɛ 0 M small enough. This contradicts 97. Therefore, 93 holds. ii Applying parabolic regularity to equation 6 and using estimate 9, we get for all x 0, > 0 and y <, Z x0 y, s M 0 e s β+, hence for all t [0,, Ux 0, t M 0, for some M 0 = M 0 M. This ends the proof of Lemma 3.3. We argue by contradiction and assume that for some ε 0 > 0, there exists x n, t n n N, a sequence of elements of [ T, T, such that n N, xvx n, t n ε 0 vx n, t n +β + n. 98 From the uniform estimates of i of Theorem 3 and the parabolic regularity, since xx v., t L is bounded on compact sets of [ T, T, we have T t n 0, as n. We note that i of Theorem 3 implies that vx n, t n T t n β+ is uniformly bounded, therefore, we can assume that it converges as n +. Let us consider two cases: iestimates in the very singular region: vx n, t n T t n β+ κ 0 > 0. From 98, it follows that xv., t n L xvx κ 0 n, t n ε 0 with t n T, which contradicts i Theorem 3. +βt tn β+ β+, iiestimates in the singular region: vx n, t n T t n β+ 0. We consider n large enough, such that vx n, t n T t n β+ η 0 /3, where η 0 is defined in Lemma 3.3. We take t 0 n T such that Using 98 and i of Theorem 3, we obtain: T t 0 n β+ β+ = n. 99 n xvx n, t n C 0 T t n β+ β+, hence t 0 n < t n. Now we distinguish two cases: Case. We assume that up to extracting a subsequence there exists t n t 0 n, t n, such that vx n, t nt t n β+ = 3 η 0. If we consider v n ξ, τ = T t n β+ vx n + ξ T t n, t n + τt t n, 00 then, we have from i of Theorem 3, v n 0, 0 = 3 η 0, ξ v n., 0 L + ξ v n., 0 L 0. 0 If we note by u n = /v n, then we have from and 9: τ <, ξ, ξ u n ξ, τ τ β+ + u n ξ, τ τ β+ M and τ u n = xu n u β 3 n. 0

33 Then, we can see that v n is bounded and satisfies 8. Using parabolic regularity, we can extract a subsequence still denoted by t n such that, v n ξ, τ ˆvξ, τ in C, of every compact set of,. We note that if there exists ξ 0, τ 0 such that ˆvξ 0, τ 0 = 0, then ˆv 0 see case in the proof of Lemma.4. This case is imposible because ˆv0, 0 = 3/η 0. Therefore, it holds that for all ξ, τ, ˆvξ, τ > 0. ecalling that u n ξ, τ = /v n ξ, τ, we have u n ξ, τ = /v n ξ, τ ûξ, τ = /ˆvξ, τ in C, of every compact set of, and by 0, we get τ û = ξ û û β, û0, 0 = 3 and ξ ûξ, τ τ β+ + τ β+ M. η 0 ûξ, τ Using the Liouville Theorem see Theorem, we get ûξ, τ = κ 3κ η 0 β+ τ β+. We claim that it is enough to prove that the convergence of u n û takes place in C,, [0, to conclude. Indeed, if we have this extended convergence, then we write from 98 and the definition 00 of v n, with τ n = tn t n. Letting n, we obtain T t n Hence ξ v n0, τ n = T t n β+ β+ ξξ v0, t n ε 0 v0, t n β+ T t n β+ β+ = ε 0 v n 0, τ n +β ε 0 min τ [0,] v n 0, τ +β, ξ v n0, τ n 0 and min v n0, τ +β min τ [0,] τ [0,] ˆvτ +β. 0 ε 0 min τ [0,] ˆvτ +β = ε 0 ˆ0τ +β > 0 03 which is a contradiction. Let us then extend the convergence. If we consider the following similarity variables, y = ξ ξ 0 τ, s = log τ, Z n,ξ0 y, s = τ β+ v n ξ, τ, 04 then, we see from 0 that for all ξ 0, Z n,ξ0., 0 L ρ η 0, for n large enough. Using Lemma 3.3, we get for all ξ and τ [0,, v n ξ, τ M 0. Using the parabolic regularity, we can extend the convergence, and then reach the contradiction 03. This concludes Case. Case. We assume that for some n 0 N, for all n n 0 and t [t 0 n, t n ], we have: vx n, tt t β+ < 3 η 0. 3

34 Then, we take t n = t 0 n and introduce v n by 00. As in Case, we obtain by Lemma 3.3 and the parabolic regularity: ξ and τ [0,, v n ξ, τ /M 0, ξ v n0, τ n C 0 η 0 where τ n = t n t 0 n T t 0. n Therefore, we get from 98, 00 and 99: n ξ v nx n, t n = T t 0 n β+ β+ ξ v n0, τ n 0 η 0 T t 0 n β+ β+ = C 0 η 0 n, which is a contradiction, as n. This ends Case and concludes the proof of Theorem 3. A Proof of Proposition.7 We proceed as in Appendix A [MZ98a], the most important difference, is that in our case we have not w bounded. Let us introduce some notations, v + y, s = v 0 sh 0 y + v sh y, v null y, s = v sh y. We divide the proof in two parts: in part, we show that either v null or v + is predominant in L ρ as s. In part, we show that in the case where v + is predominant, then either v 0 s or v s predominates the other. Step of the Proof: Competition between v +, v null and v First, we recall from 38 Let us introduce some notations y, s, s v = Lv fv. 05 zs = v +., s L ρ, xs = v null., s L ρ and ys = v., s L ρ. Projection 05 onto the unstable subspace of L forming the L ρ-inner product with v +, and using standard inequalities, Lemma.6 and 40. We get ż z N, where N = v L ρ. Working similarly with v null and v we arrive at the system If we knew for s large enough ż z N ẋ N, ẏ y + N. 06 N εx + y + z, 07 we could use ODE techniques to conclude the step. However, we do not have this information at this stage. We thus estimate N proceeding as in Section 4 in [FK9b]. We 33

35 note that in [FK9b] 07 was proved under the additional assumption that vy, s is uniformly bounded. One can check in [FK9b] that this assumption is only used for the derivation of the estimate fv c v, which in our case is true by 40. Then, we get by 07 and 06, ż ε z εx + y ẋ εx + y + z, ẏ 08 + ε y + εx + z. Now, let us recall Lemma A. from [NZ0]. Lemma A. Merle-Zaag Let xs, ys, and zs be absolutely continuous, real-valued functions that are nonnegative and satisfy ix, y, zs 0 as s, and s s, xs + ys + zs 0, and ii ε > 0, s 0 such that s s 0 ż c 0 z εx + y ẋ εx + y + z ẏ c 0 y + εx + y Then, either x + y = oz or y + z = ox as s. Applying Lemma above to 08, we get either v., s L ρ + v +., s L ρ = o v null., s L ρ, 09 or v s L ρ + v null., s L ρ = o v +., s L ρ. Step of the Proof: Competition between v 0 and v In this step we focus on the case where v., s L ρ + v null., s L ρ = o v +., s L ρ. We will show that it leads to either case ii or iii of Proposition.7. We want to derive from 05 the equations satisfied by v 0 and v. For this we will estimate in the following fvkm yρydy, for m = 0, where k m y = h m y/ h m y L ρ. Lemma A. There is 0 > 0 and an integer k > 4 such that for all δ 0, δ 0, s 0 such that s s 0, v y k ρdy c 0 k δ 4 k zs Poof: This lemma is analogous to Lemma A.3 p 75 from [MZ98a], it have been proved under the additional assumption that vy, s is uniformly bounded. One can check that this assumption is only used for the derivation of the estimate fv < Cv, which is in our case true. Proceeding as in Appendix A from [MZ98a] and doing the projection of equation 05 respectively on k 0 y and k y, we obtain v 0s = v 0 s β κ zs + αs. 0 34