No giant component in the limit

3.3 No giant component in the limit

The aim of this subsection is to prove the following proposition: Proposition 2. If n −1 ≪ λn ≪ 1 and m 2 0 +∞ holds for v0 on the right-hand side of 32 then any weak limit point P of the sequence of probability measures P n is concentrated on the set of conservative forest fire evolutions: P ∞ X k=1 v k t ≡ 1 = 1 75 We are going to prove Proposition 2 by contradiction: in Lemma 6 we show that if θ · 6≡ 0 in the limit, then there is a positive time interval such that θ t has a positive lower bound, and that this implies that even in the convergent sequence of finite-volume models, a lot of mass is contained in arbitrarily big components on this interval. Than in subsequent Lemmas we prove that these big components indeed burn, which produces such a big increase in the value of the burnt mass r · that is in contradiction with E rT ≤ 2 + E qT ≤ 2 + T . By Proposition 1 the random FFE obtained as a weak limit point is almost deterministic: 37 holds with a possibly random control function r ·. Also, by 33 we P-almost surely have qt ≤ t from which 40 follows. Thus 71 and 73 hold P-almost surely for the random flow obtained as a weak limit point with a deterministic constant C ∗ . Lemma 6. If P n ⇒ P where P does not satisfy 75 on [0, T ], then there exist ǫ 1 , ǫ 2 , ǫ 3 0 and a deterministic t ∗ ∈ [ǫ 1 , T ] such that for every K +∞, every m +∞ and every sequence t ∗ − ǫ 1 α 1 β 1 α 2 β 2 · · · α m β m t ∗ there exists an n +∞ such that for every n ≥ n and 1 ≤ i ≤ m we have P n max α i ≤t≤β i 1 − K −1 X k=1 v n,k t ǫ 2 ǫ 3 . 76 Proof. First we prove that if P does not satisfy 75 then there exist ǫ 1 , ǫ 2 , ǫ 3 0 and ǫ 1 ≤ t ∗ ≤ T such that P inf t ∗ −ǫ 1 ≤t≤t ∗ θ t ǫ 2 ǫ 3 . 77 Since 75 is violated, we have P sup ≤t≤T θ t ǫ ǫ for some ǫ 0. Let L := ⌊ 2C ∗ T ǫ ⌋ and t i := ǫi 2C ∗ for 1 ≤ i ≤ L where C ∗ is the constant in 73. Since θ 0 = 0 we have sup ≤t≤T θ t ǫ ⊆ L [ i=1 θ t i ǫ 2 almost surely with respect to P. Thus P θ t ∗ ǫ 2 ǫ L for some t ∗ ∈ {t 1 , . . . t L }. Using 73 again 77 follows with ǫ 1 := ǫ 4C ∗ , ǫ 2 := ǫ 4 , ǫ 3 = ǫ L . 1311 Now given K and the intervals [ α i , β i ], 1 ≤ i ≤ m we define the continuous functionals f i : D[0, T ] → R by f i v0, q ·, r· := 1 β i − α i Z β i α i 1 − K X k=1 v k t d t where v k t is defined by 19. Thus for all i H i := { v0, q·, r· ∈ D[0, T ] : f i v0, q ·, r· ǫ 2 } is an open subset of D[0, T ] with respect to the topology of Definition 1. Thus by the definition of weak convergence of probability measures we have lim n →∞ P n H i ≥ PH i ≥ P inf t ∗ −ǫ 1 ≤t≤t ∗ θ t ǫ 2 ǫ 3 from which the claim of the lemma easily follows. Lemma 7. If n −1 ≪ λn then for every ǫ 2 0 there is a ǫ 4 0 such that for every ˜t 0 there is a K and an n 1 such that for all n ≥ n 1 1 − P K −1 k=1 v n,k 0 ≥ ǫ 2 implies E n r n ˜t ≥ ǫ 4 78 The proof of Lemma 7 will follow as a consequence of the Lemmas 8 and 9. Proof of Proposition 2. We are going to show that if there is a sequence P n such that the weak limit point P violates 75 then for some n we have E n r n T T + 2 79 which is in contradiction with 35 and 26. In fact, T +2 could be replaced with any finite constant in 79, but T + 2 is big enough to have a contradiction. We define ǫ 1 , ǫ 2 , ǫ 3 0 and t ∗ using Lemma 6. Next, we define ǫ 4 using this ǫ 2 and Lemma 7. Given these, we choose ˜t be so small that › ǫ 1 2˜t ž ǫ 3 ǫ 4 T + 2. We choose K and n 1 big enough so that 78 holds. Further on, we fix the intervals [ α i , β i ], 1 ≤ i ≤ m = ⌊ ǫ 1 2˜t ⌋ so that α i+1 − β i ˜t holds for all i and also T − β m ˜t holds. We choose n such that 76 holds and let n := max {n , n 1 }. Finally, we define the stopping times τ 1 , τ 2 , . . . , τ m by τ i := β i ∧ min{t : t ≥ α i and 1 − K −1 X k=1 v n,k t ≥ ǫ 2 }. We have τ i + t ∗ ≤ β i + t ∗ α i+1 ≤ τ i+1 . Using the strong Markov property, 78 and 76, the inequality 79 follows: E r n T ≥ m X i=1 E r n τ i + ˜t − r n τ i τ i β i P τ i β i ≥ mǫ 4 ǫ 3 . 1312 Lemma 7 stated that if initially a lot of mass is contained in big components, then in a short time a lot of mass burns. We prove this statement in two steps: in Lemma 8 we prove that if we start with a lot of mass contained in big components, then in a short time either a lot of this mass is burnt or the big components coagulate, so a lot of mass is contained in components of size n 1 3 the same proof works if we replace the exponent α = 13 by any 0 α 12. Then in Lemma 9 we prove that if we start with a lot of components of size n 1 3 then in a short time a lot of mass burns. We will make use of the following generating function estimates in the proof of Lemma 8. If V x is defined as in 41 and if v ∈ V 1 then for ǫ ≤ 1 2 1 − K −1 X k=1 v k ≥ ǫ =⇒ V 1 K ≤ e −1 − 1ǫ 80 V 1 K ≤ −ǫ =⇒ 1 − ǫK2 X k=1 v k ≥ ǫ4. 81 Lemma 8. There are constants C 1 +∞, C 2 0, C 3 0 such that if 1 − K −1 X k=1 v n,k 0 ≥ ǫ 2 82 for all n then lim n →∞ P n X k=C 3 ǫ 2 n 1 3 v n,k ¯t + r n ¯t ≥ C 2 ǫ 2 = 1 83 Where ¯t = C 1 K ǫ 2 . Sketch proof. If we let n → ∞ immediately, we get that the limiting functions v 1 t, v 2 t, . . . solve 37, 38, 39 with a possibly random control function rt ≡ r ∞ t. The n → ∞ limit of 83 is θ ¯t + r ¯t ≥ C 2 ǫ 2 84 Now we prove that if v · is a solution of 37, 38, 39 then 1 − P K −1 k=1 v k 0 ≥ ǫ 2 implies 84 with C 1 = 4 and C 2 = 1 4 . This proof will also serve as an outline of the proof of Lemma 8. In order to prove 84 define V t, x by 41. Thus V t, x solves the integrated Burgers control problem 43, 44, 45. Define Ut, x := V t, x − rte −x . Thus U ′ t, x = V ′ t, x + rte −x and by 43 we have ˙ Ut, x = −V t, xV ′ t, x. Define the characteristic curve ξ· by ˙ ξt = V t, ξt ξ0 = 1 K 85 Let ut := Ut, ξt − V 0, 1 K . Thus u0 = 0, and ˙ ut = ˙ Ut, ξt + U ′ t, ξt ˙ ξt = −V t, ξtV ′ t, ξt+ € V ′ t, ξt + rte −ξt Š V t, ξt = rte −ξt V t, ξt ≤ 0. 86 1313 Thus ut ≤ 0, moreover V t, ξt = V 0, 1 K + rte −ξt + ut ≤ V 0, 1 K + rt, 87 ξt = 1 K + Z t us ds + Z t rse −ξs ds + t V 0, 1 K ≤ 1 K + t · rt + tV 0, 1 K . 88 By 80 we have V 0, 1 K ≤ − 1 2 ǫ 2 . In order to prove that θ ¯t+ r ¯t ≥ 1 4 ǫ 2 with ¯t = 4 K ǫ 2 we consider two cases: If r ¯t ≥ 1 4 ǫ 2 then we are done. If r ¯t 1 4 ǫ 2 define τ := min{t : ξt = 0}. By 88 we have ξ¯t ≤ 1 K + ¯t · r¯t + ¯t · − 1 2 ǫ 2 1 K + 1 K − 2 K = 0 Thus τ ≤ ¯t. By 87 we get −θ τ = V τ, 0 = V τ, ξτ ≤ − 1 2 ǫ 2 + 1 4 ǫ 2 = − 1 4 ǫ 2 Thus 1 4 ǫ 2 ≤ θ τ ≤ θ τ+ rτ ≤ θ ¯t+ r ¯t because by 21 the function θ t+ rt is increasing. To make this proof work for Lemma 8 we have to deal with the fluctuations caused by randomness, combinatorial error terms and the fact that λn only disappears in the limit. Proof of Lemma 8. Given a FFF obtained from a forest fire Markov process by 29,30 and 31, define U n t, x := n X k=1  v n,k 0 + k 2 k −1 X l=1 q n,l,k −l t − kq n,k t − r n,k t   e −kx − 1 − λn By 19 we have U n t, x + r n te −x = n X k=1 v n,k te −kx − 1 − λn =: V n t, x − λn =: W n t, x. W ′ t, x = − X k ≥1 k · v n,k te −kx − 1 2 ∂ x W t, x + 1 + λn 2 = X k ≥1 k 2 k −1 X l=1 v n,l tv n,k −l te −kx W ′′ t, x = X k ≥1 k 2 · v n,k te −kx W ′′ t, 2x = X k ≥1 k 2 2 · 11[2 | k] · v n, k 2 te −kx 1314 If X t is a process adapted to the filtration F t, let L X t := lim d t →0 1 d t E X t + d t − X t F t Using the martingales of Proposition 1 we get L U n t, x = X k ≥1   k 2 k −1 X l=1 L q n,l,k −l t − k · L q n,k t − L r n,k t   e −kx = X k ≥1   k 2 k −1 X l=1 v n,l tv n,k −l t − l · 11[2l = k] n v n,l t − k · v n,k t − k n v n,k t − € λn · k · v n,k t Š e −kx = − 1 2 ∂ x W t, x + 1 + λn 2 − 1 n W ′′ t, 2x+ W ′ t, x + 1 n W ′′ t, x + λnW ′ t, x = − W ′ n t, xW n t, x + 1 n € W ′′ n t, x − W ′′ n t, 2x Š 89 Given the random function W n t, x we define the random characteristic curve ξ n t similarly to 85: ˙ ξ n t = W n t, ξ n t, ξ n 0 := 1 K 90 This ODE is well-defined although W n t, x is not continuous in t, but almost surely it is a step function with finitely many steps which is a sufficient condition to have well-posedness for the solution of 90. Define u n t := U n t, ξ n t − W n 0, 1 K . Thus u n 0 = 0 and u n t = W n t, ξ n t − W n 0, 1 K − r n te −ξ n t = V n t, ξ n t − V n 0, 1 K − r n te −ξ n t 91 The solution of 90 is ξ n t = 1 K + Z t u n s ds + Z t r n se −ξ n s ds + tW n 0, 1 K 92 Putting together 89 and 90 similarly to 86 and using 56 we get L u n t ≤ 1 n € W ′′ n t, ξ n t − W ′′ n t, 2ξ n t Š ≤ n −1 · ξ n t −2 93 Now e u n t = u n t − R t L u n sds is a martingale and L e u n t 2 = lim h →0 + 1 h E U n t + h, ξ n t − U n t, ξ n t 2 F t ≤ 1 2 n X k,l=1 k + l n e −k+lξ n t − k n e −kξ n t − l n e −lξ n t 2 v n,k tv n,l tn + n X l=1 l n e −lξ n t 2 λnv n,l tn = O 1 n W ′′ n t, ξ n t = O € n −1 · ξ n t −2 Š 94 1315 Define the stopping time τ n := min {t : ξ n t = n −α } α = 13. In fact any 0 α 12 would be just as good to make the right-hand side of 93 and 94 disappear when t ≤ τ n and n → ∞. It follows from 94 and Doob’s maximal inequality that sup t eu n t ∧ τ n ∧ T ⇒ 0 as n → ∞ By 93 we have e u n t + R t n −1 · ξ n s −2 ds ≥ u n t thus sup t u n t ∧ τ n ∧ T ⇒ 0 as n → ∞ 95 By 80 and 82 we have V n 0, 1 K ≤ e −1 − 1ǫ 2 =: −ǫ 5 96 Define the events A n , B n and the time ¯t n by A n := sup t ≤τ n ∧T Z t u n sds ≤ 1 K ∩ u n τ n ∧ T ≤ ǫ 5 3 , B n := r n τ n ≤ ǫ 5 3 , ¯t n := 3 K W n 0, ξ n ≤ 3 K ǫ 5 , We are going to show that that there are constants C 2 , C 3 +∞ such that A n ⊆ n X k=C 3 ǫ 2 n 1 3 v n,k ¯t + r n ¯t ≥ C 2 ǫ 2 97 which, since 95 implies that lim n →∞ P A n = 1, gives 83. First we show that A n ∩ B n ⊆ {τ n ≤ ¯t n }. 98 If we assume indirectly that A n , B n and τ n ¯t n hold then R ¯t n u n sds ≤ 1 K , so by 92 we get ξ n ¯t n ≤ 1 K + 1 K + Z ¯t n r n se −ξ n s ds + ¯t n W n 0, ξ n 0 ≤ − 1 K + ¯t n · r n τ n ≤ 0. But ξ n ¯t n ≤ 0 is in contradiction with τ n ¯t n , thus 98 holds. Now, by 91 we have V n τ n , n −13 = u n τ n + V n 0, 1 K + r n τ n e −n −13 . Thus by 96, the definition of A n and B n and 81 we get A n ∩ B n ⊆ u n τ n ≤ ǫ 5 3 ∩ V n 0, 1 K ≤ −ǫ 5 ∩ r n τ n e −n −13 ≤ ǫ 5 3 ⊆ V n τ n , n −13 ≤ −ǫ 5 3 ⊆ n X k=n 1 3 ǫ 5 6 v n,k τ n ≥ ǫ 5 12 1316 Thus we have A n ⊆ A n ∩ B n ∪ B c n ⊆ n X k=n 1 3 ǫ 5 6 v n,k τ n ≥ ǫ 5 12 ∪ r n τ n ǫ 5 3 ⊆ n X k=C 3 ǫ 2 n 1 3 v n,k τ n + r n τ n ≥ C 2 ǫ 2 with C 3 = 1 − e −1 6 and C 2 = 1 − e −1 12. But P n k=C 3 ǫ 2 n 1 3 v n,k t + r n t increases with time, from which 97 follows. Lemma 9. There are constants C 4 +∞, C 5 0 such that if n X k=C 3 ǫ 2 n 1 3 v n,k 0 ≥ C 2 ǫ 2 2 for all n then with ¯t n := C 4 ǫ −2 2 n −13 logn + n λn −1 99 we have lim n →∞ E r n ¯t n ≥ C 5 ǫ 2 . 100 Remark. The upper bound 99 is technical: on one hand it is not optimal, on the other hand, for the proof of Lemma 7 we only need ¯t n ≪ 1 as n → ∞. Proof. If v is a vertex of the graph Gn, t let C n v, t denote the connected component of v at time t. Denote by τ b v the first burning time of v: τ b v := inf{t : C n v, t + C n v, t − } Of course C n v, τ b v + = 1. Define ¯n := C 3 ǫ 2 n 1 3 and H n t := {v : C n v, 0 ≥ ¯n and τ b v t} Fix a vertex v ∈ H n 0. c n t := 1 n C n v, t ∧ τ b v − w n t := 1 n H n t z n t := 1 n X w ∈H n 11 {τ b w≤t} = w n 0 − w n t Thus c n t is an increasing process we freeze c n t when it burns. We consider the right- continuous versions of the processes c n t, w n t, z n t. w n 0 ≥ C 2 ǫ 2 2 =: ǫ 6 . 1317 We are going to prove that there are constants C 4 +∞, C 5 0 such that lim n →∞ E z n ¯t n ≥ C 5 ǫ 2 101 which implies 100. Define the stopping times τ w := inf {t : w n t ǫ 6 2} τ g := inf {t : c n t ǫ 6 4} τ := τ b v ∧ τ w ∧ τ g Since v ∈ H n 0 we have c n t ≥ c n 0 = C n v, 0 n ≥ ¯ n n If C n v, t is connected to a vertex in H n t by a new edge at time t then c n t + − c n t − ≥ ¯ n n , logc n t + − logc n t − ≥ log 1 + ¯ n nc n t − ≥ log2¯ n nc n t − L logc n t ≥ log2¯ n nc n t lim d t →0 1 d t P c n t + d t − c n t ≥ ¯ n n F t ≥ log2¯ n nc n t · 1 n C n v, t € H n t − C n v, t Š 11 {t ≤τ b v} ≥ log2¯n · w n t − c n t 11 {t ≤τ b v} ≥ log2¯ n ǫ 6 4 11 {t≤τ} = n 1 3 log2 8 · C 2 · C 3 · ǫ 2 2 · 11 {t≤τ} =: n 1 3 ǫ 7 11 {t≤τ} Thus logc n t − ǫ 7 · n 1 3 t ∧ τ is a submartingale. Using the optional sampling theorem we get −ǫ 7 · n 1 3 E τ ≥ E logc n τ − ǫ 7 · n 1 3 E τ ≥ logc n 0 ≥ − logn By Markov’s inequality we obtain that for some constant C +∞ P τ ≤ C n −13 ǫ −2 2 logn ≥ 1 2 If τ g ≤ τ b v ∧ τ w , then C n v, τ g ǫ 6 4 n, so E τ b v − τ g ≤ nλn −1 4 ǫ 6 , which implies P τ w ∧ τ b ≤ C n −13 ǫ −2 2 logn + C ′ nλn −1 ǫ −1 2 ≥ 1 4 . for some constant C ′ . We define ¯t of 99 with C 4 := max {C, C ′ }. Using the linearity of expectation we get E z n ¯t = E 1 n X w ∈H n 11 {τ b w≤¯t} ≥ ǫ 6 P τ b v ≤ ¯t . The inequality 11 {τ w ≤¯t} ǫ 6 2 ≤ z n ¯t follows from the definition of τ w . 1 4 ≤ P τ w ∧ τ b ≤ ¯t ≤ P τ w ≤ ¯t + P τ b ≤ ¯t ≤ E z n ¯t 2 ǫ 6 + E z n ¯t 1 ǫ 6 From this 101 follows. 1318 4 The critical equation

4.1 Elementary properties