Correlation-length bounds, and estimates for intermittent islands in parabolic SPDEs

We consider the nonlinear stochastic heat equation in one dimension. Under some conditions on the nonlinearity, we show that the"peaks"of the solution are rare, almost fractal like. We also provide an upper bound on the length of the"islands,"the regions of large values. These results are obtained by analyzing the correlation length of the solution.


Introduction
LetẆ := {Ẇ t (x)} t>0,x∈R denote space-time white noise, and consider the nonlinear stochastic heat equation, for (t , x) ∈ (0 , ∞) × R, subject to u 0 (x) := 1 for all x ∈ R. Throughout we consider only the case that σ : R → R is Lipschitz continuous. In that case, the theory of Walsh [11] explains the meaning of (1.1) and shows that (1.1) has a unique [strong] solution that is continuous for all (t , x) ∈ [0 , ∞) × R. The goal of this article is to make some observations about the geometric structure of the random function x → u t (x) for t > 0 fixed. Notice that a consequence of the fact that u 0 (x) is constant is that the law of u t (x) doesn't depend on x ( [3]).
Although we will have some results that are valid for (1.1) in general, we are mainly motivated by the following two special cases of Eq. (1.1): Case 1. There exists q > 0 such that σ(z) = qz for all z ∈ R. In this case, (1.1) is known as the parabolic Anderson model ; Case 2. 0 < inf z∈R σ(z) sup z∈R σ(z) < ∞. An important special case of this case occurs when σ is a constant; then (1.1) is the linear SPDE whose solution is a stationary Gaussian process.
If R −1 E α (R) did look like a random fractal subset of [0 , 1] with Hausdorff dimension d(α) ∈ (0 , 1), then we would expect its Lebesgue measure to behave as R −d(α)+o(1) as R → ∞. Or stated in more precise terms, we would expect that if α is sufficiently small, then The first theorem of this paper comes close to proving this last assertion.
The results of [2] imply that E α (R) is eventually empty a.s. when α > α * . Therefore, α 0 cannot be made to be arbitrarily large.
Choose and fix a time t > 0. Given two numbers 0 < a < b, we say that a closed interval I ⊂ R + is an (a , b)-island [at time t] if: 2. u t (x) > a for all x ∈ int(I); and (1.6) The following result shows that the relative length of the largest "tall island" in [0 , R]-also known as "intermittency islands"-is vanishingly small as R → ∞. This phenomenon has been predicted [particularly for Case 1] and is a part of a property that is referred to somewhat loosely as "physical intermittency" [2,7,8].
If Case 2 occurs, then the preceding can be improved to the following: Let us make a few remarks before we continue our introduction.
During the course of the proof of this theorem, we will establish that exists b > 1 such that P{u t (0) > b} > 0; therefore, the result always has content.
2. The condition σ(1) = 0 is necessary. Indeed, if σ(1) were zero, then u t (x) = 1 for all t > 0 and x ∈ R [this is because u 0 ≡ 1]. Theorems 1.1 and 1.2 both rely on a fairly good estimation of "correlation length" for the random field x → u t (x). There are many ways one can understand the loose term, "correlation length." Let us describe one next.
Let {X x } x∈R be a random field on (Ω , F , P), and let L(ℓ) denote the collection of all weakly stationary random fields {Y x } x∈R on (Ω , F , P) such that Y has "lag" ℓ; that is, Y z is independent of (Y xi ) N i=1 for all z, x 1 , . . . , x N ∈ R that satisfy min 1 j N |z − x j | ℓ. Then, the correlation length of X is the function where ǫ, δ > 0 can be thought of as fidelity parameters. Informally speaking, when we find L X (ǫ ; δ), we seek to find the smallest lag-length ℓ for which there exists a coupling of X with a lag-ℓ process Y , such that the coupling is good to within δ units with probability at least 1 − ǫ.
The following is the main technical result of this paper. It states that the correlation length of the solution to (1.1) is logarithmic in the fidelity parameter ǫ; and the fidelity parameter δ can be as small as exp(−K |log ǫ| 2 /3 ) for a universal K = K(t). (1.10) If σ is a bounded function, then in fact there exists θ ∈ (0 , 1) such that Our notion of correlation length is stronger than other, somewhat simpler, notions of this general type. For instance, consider the following: Let {X x } x∈R be a random field, and define L * X (ǫ ; δ) to be the smallest ℓ > 0 for which we can find-on some probability space-a coupling (X * , Y * ), where X * has the same law as X and Y * has lag ℓ, and sup x∈R P{|X * Open Problem. Is it true that L * ut (ǫ ; 0) = O(|log ǫ|)? This is equivalent to asking whether or not x → u t (x) is exponentially mixing.
Although we do not know how to prove that x → u t (x) is exponentially mixing, we are able to prove that the coupling in Theorem 1.4 is "good on all scales." In order to interpret this, note that if ℓ := L ut (ǫ; δ) then we can basically approximate u t well enough by a random field Y in L(ℓ) such that Y replicates u t to within δ units. According to (1.10) this can be done with -ℓ = O(|log ǫ|)-with δ having the form exp{−K |log ǫ| 2 /3 } for some K := K(t). Thus, for example, if we wanted to know how small x → u t (x) can possibly get, then we could study instead Y provided that "how small" means "exp{−K |log ǫ| 2 /3 } or more." Our next result shows that this notion of "how small" is generic [and not at all a restriction]. Our proof borrows several important ideas from a paper by Mueller and Nualart [10].
(1.13) Throughout this paper, "log" denotes the natural logarithm, p t (x) denotes the standard heat kernel for ( 1 /2)∆, and Let us conclude the Introduction with a brief outline of the paper. In Section 2 we prove Theorem 1.4, whose corollaries, Theorems 1.1 and 1.2, are proved respectively in §3 and §4. In a final Section 5 we state and prove an improved version of Theorem 1.5, which might turn out to be a first step in answering the mentioned Open Problem.

Proof of Theorem 1.4
First of all, recall that the solution to the stochastic PDE (1.1) is the unique continuous solution to the following random evolution equation [11]: For all β > 0, let U (β) solve the following closely-related stochastic evolution equation: It has been observed in [2] that the same methods as in [11] can be used to show that there exists a unique continuous random field U (β) that solves the preceding. The following result of [2] shows that U (β) ≈ u if β is large.
It is easy to adapt the arguments of [2] to improve the preceding in the case that σ is bounded. Because all of the key steps are already in Ref. [2], we state the end result without proof.
The process U (β) is useful only as a first step in a better coupling, which we describe next. Define U (β, 0) t (x) := 1. Then, once U (β,l) is defined [for some l 0] we define U (β,l+1) as follows: In other words, U (β,l) is the l th step in the Picard-iteration approximation to U (β) . The following result of [2] tells us that if l is large then U (β,l) ≈ U (β) .
Once again, we state-without proof-an improvement in the case that σ is bounded.
Now we are ready to establish Theorem 1.4.
Proof of Theorem 1.4. Choose and fix t > 0. The final assertion of Lemma 2.3 implies that the process for every β > 0 and n 0. Therefore, we may apply Lemmas 2.1 and 2.3 in conjunction with Chebyshev's inequality to see that for all k ∈ [1 , ∞) and δ > 0, where n , k , δ). Now we choose β = n := 1 + ⌊(2/c 2 )k 2 ⌋ in order to find that there existsc ∈ (1 , ∞) such that for all k sufficiently large, Becausec does not depend on δ, we can set δ := exp(−k 2 /c) to deduce from the preceding that for every ν ∈ (0 , 1) fixed, In the case that ǫ is a general positive number, (1.10) follows from the preceding and a simple monotonicity argument.
In the case that σ is bounded, we proceed similarly as in the general case, but apply Lemmas 2.2 and 2.4 in place of Lemmas 2.1 and 2.3, and then select the various parameters accordingly. In this way, we find the following improvement to (2.8) in the case that σ is bounded: where c ′ 1 , c ′ 2 do not depend on (β , n , k , δ). Now we choose β = n := 1 + ⌊(2/c ′ 2 ) log k⌋ in order to deduce the existence of a constant c ′′ ∈ (1 , ∞) such that for all suffiently large k, (2.13) This is our improvement to (2.9) in the case that σ is bounded. In particular, for all k large, (2.14) If ǫ := exp{−(1/c ′′ )k log k} is small, then k ≈ c ′′ |log ǫ| / log |log ǫ| and (1.11) follows from (2.14) for every θ ∈ (0 , 1/c ′′ ). We apply monotonicity in order to deduce (1.11) for general [small] ǫ.
3 Proof of Theorem 1.1 Before we proceed with the proof we need a few technical results. Suppose Y ∈ L(ℓ) for some ℓ > 0, and define, for all integers n 1 and real numbers α > 0, Lemma 3.1. Assume that Y ∈ L(ℓ) for some ℓ 1. Then for every integer k 3 there exists a universal constant C k ∈ (0 , ∞) such that for all α ∈ (0 , 1 /2) and n 2, Proof. We can write The processes S (o) and S (e) are mean-zero random walks, and hence martingales [in their respective filtrations]. Define to be the increments of S (x) for x ∈ {o , e}, and G (x) k the sigma-algebra generated by {X j | < ℓ for every j, and since ℓ 1, an application of Burkholder's inequality [1] (specifically, see Hall and Heyde [6, Theorem 2.10, p. 23]) implies that for every k 1 there exists a universal constant c k ∈ (0 , 1) such that for every k 2 and n 2 and x ∈ {o , e}, c k k E S (x) n k n k/2 ℓ k + ℓ k 2n k/2 ℓ k . (3.8) The lemma follows from the above, (3.6), and Minkowski's inequality together with the observation that EY α (n) (nℓ) 1−α .
Proof of Theorem 1.1. Throughout the demonstration, we choose and fix a time t > 0. Consider first Case 1. According to [2], there exist constants A 1 , . . . , A 4 ∈ (0 , ∞) such that for all λ 1 and x ∈ R, (3.9) [One could prove that the preceding probability does not depend on x ∈ R.] According to Theorem 1.4, for every m > 1 there exists c ∈ (0 , ∞) such that for all R large enough, we can find a process Y ∈ L(c log R) such that (3.10) [One could prove that the preceding probability does not depend on x ∈ R.] Note, in particular, that For all α ∈ (0 , 1) and R large enough, provided that m > A 4 . Similarly, provided that m > (2A 2 ) 3 /2 . We combine the preceding two bounds, and then relabel α to see that there exist B 1 , B 2 , B 3 , B 4 ∈ (0 , ∞) and α 0 ∈ (0 , 1 /4) such that for all α ∈ (0 , α 0 ), whereḠ was defined in (3.2). According to (3.11), We emphasize that"const" does not depend on R in the previous two displays. We may apply Lemma 3.1 with ℓ := c log R, n := R/ℓ and Y α (n) = R 0 1 {Yx Ḡ (R −α )} dx to see that for all k 2, R sufficiently large, Let us pause and recall that α < α 0 < 1 /4, so that the right-hand side is at most R −2 provided that we have chosen k sufficiently large. We also note that there exists γ ∈ (0 , 1) such that for all sufficiently large R; see (3.9) and (3.12). Therefore, we can combine (3.14), (3.15), and (3.16), together with the Borel-Cantelli lemma to see that as long as α 0 were selected sufficiently small, A monotonicity argument finishes the proof for Case 1. Case 2 is proved similarly, but we apply the following estimate [2] in place of (3.9): C 1 exp(−C 2 λ 2 ) P{u t (x) > λ} C 3 exp(−C 4 λ 2 ) (for λ 1). We omit the details.

Proof of Theorem 1.2
First of all, let us note that the conditions of the theorem are non vacuous. In other words, we need to prove that there exists b > 1 such that P{u t (0) b} > 0. Because Eu t (0) = 1, it follows that there exists b 1 such that P{u t (0) b} > 0. Suppose to the contrary that P{u t (0) > 1} = 0. Then, u t (0) is a.s. equal to 1. It follows that that the stochastic integral in (2.1) vanishes a.s. for x = 0. The corresponding quadratic variation must too; that is, Since the heat kernel never vanishes, we find that σ (u s (y)) = 0 for almost all (s , y) ∈ (0 , t) × R, whence for all (s , y) ∈ (0 , t) × R by continuity. This is a contradiction since u 0 ≡ 1. Therefore, there exists b > 1 such that P{u t (0) b} > 0. Now we proceed with our proof of the bulk of Theorem 1.2. Theorem 1.2 is a simple consequence of Theorem 1.4 together with ideas that are borrowed from a classical paper by Erdős and Rényi [4] on the length of the longest run of heads in an infinitely-long sequence of independent coin tosses.
Choose and fix two integer R, m ≫ 1 and a real δ ∈ (0 , 1) small enough that a − 2δ > 1 and P{u t (0) > b + 2δ} > 0. According to Theorem 1.4 we can find a constant c ∈ (0 , ∞)-independent of R-and a random field Y ∈ L(c log R) such that [One can prove that the probability does not depend on x.] Define x j := cj log R for all non negative integers j, and observe that Let us call the index j "good" if Y xj , Y xj+2 < a − δ and Y xj+1 > b + δ. Otherwise j is deemed "bad." Clearly, We may observe that p does not depend on j. Moreover, P{u t (0) < a − 2δ} ∧ P{u t (0) > b + 2δ} > 0 because of the choice of (b , δ) and the fact that Eu t (0) = 1 < a − 2δ. Therefore, we may choose m large enough to ensure that p > 0; note that we may also choose m independently of R ≫ 1. Because P {j , j + 3 , . . . , j + 3n are all bad} = (1 − p) n , (4.5) it follows that provided that γ is a sufficiently-large universal constant. This and the Borel-Cantelli lemma together imply that a.s. for all sufficiently-large integers R, the maximum distance between two good points is at most 6γ log R · c log R. Combined with (4.3), we can conclude that the size of the largest island is at most 6cγ(log R) 2 . This proves the theorem for Case 1.
If Case 2 holds, then we proceed exactly as we did above, but can find our random field Y ∈ L(c[log log R] 3 /2 ) instead of L(c log R). The remaining details are omitted.

Proof of Theorem 1.5
We conclude by proving Theorem 1.5. Throughout we assume that σ(0) = 0. (5.1) [This of course includes Case 1.] In that case Mueller's comparison principle [2,9] guarantees that u t (x) 0 for all t > 0 and x ∈ R a.s. We offer the following quantitative improvement, which clearly implies Theorem 1.5: Theorem 5.1. For every t > 0 there exist A, B ∈ (0 , ∞) such that uniformly for all ǫ ∈ (0 , 1) and x ∈ R, Before we prove this result, let us state and prove two corollories to Theorem 5.1. The corollaries are of independent interest, but also showcase the usefulness of quantitative estimates in this area. The first corollary identifies an upper bound for the exponential growth of the high negative moments of u t (x) when σ(0) = 0. We believe that the rate provided below is sharp. Proof. Theorem 5.1 implies that u t (x) > 0 a.s., whence X := 1/u t (x) is well defined. Because E(X k ) = k ∞ 0 λ k−1 P{X > λ} dλ, we can divide the integral into two pieces where: (i) λ < e; and (ii) λ e. In this way we find that Laplace's method [and/or the method of stationary phase] tells us that The corollary follows from this and a series of elementary estimates which we omit.
We mention [and verify] the second corollary to Theorem 5.1 next. This corollary describes a bound for how close u t (x) can come to zero, as x → ∞. According to [5,Lemma A.3], there exists an c ∈ (0 , ∞) such that for all t > 0, k ∈ [2 , ∞), and x, y ∈ R, Therefore, a careful appeal to the Kolmogorov continuity theorem implies that for all k ∈ [2 , ∞), η ∈ (0 , 1), and t > 0 where "sup I " denotes the supremum over all closed intervals I ⊂ R of length one. We omit the details, as they are standard.
Now we choose and fix k > 4/(ηγ) so that the left-hand side of (5.12) sums [in n]. It follows from (5.9), (5.12), and the triangle inequality that, for every ζ > 0 large enough, The Borel-Cantelli lemma completes the proof.
Proof of Theorem 5.1. We are going to prove that for all n 1, Since the distribution of u t (x) does not depend on x, this is a stronger result than the one advertised by the statement of the theorem.
Let v t (x) denote the unique continuous solution to (1.1) subject to v 0 (x) = 1 (−1,1) (x). Because v 0 (x) 1 = u 0 (x) for all x, Mueller's comparison principle [9] tells us that there exists a null set off which u t (x) v t (x). Therefore, it suffices to prove that for all n 1, Set T 0 := 0, and then define iteratively where inf ∅ := ∞. Evidently, the T k 's are {F t } t>0 -stopping times, where F t denotes the filtration generated by time t by all the values of the white noise. Without loss of any generality we may assume that {F t } t>0 is augmented in the usual way, so that t → v t is a C(R)-valued strong Markov process. Next we observe that for every k 1, for all x ∈ R, a.s. on {T k < ∞}. (5.17) Therefore, we apply first the strong Markov property, and then Mueller's comparison principle, in order to see that the following holds a.s. on {T k < t}:  We apply this inequality with τ := t/(2n) and optimize over m in order to deduce from (5.22) that there exists a constant L := L(η, t) ∈ (0 , ∞) such that for all integers n > t/2 the following holds a.s. on {T k < t}: Finally, we notice that if T n < t, then certainly there are at least ⌊n/2⌋-many distinct values of k ∈ {0 , . . . , n − 1} such that T k+1 − T k t/(2n). Therefore, (5.26) implies that for all n > t/2, P {T n < t} n ⌊n/2⌋ L ⌊n/2⌋ exp −L ⌊n/2⌋ n 1 /2 (log n) 3 /2 . This, Stirling's formula, and monotonicity together imply (5.15).