Intermittency on catalysts: three-dimensional simple symmetric exclusion

We continue our study of intermittency for the parabolic Anderson model $\partial u/\partial t = \kappa\Delta u + \xi u$ in a space-time random medium $\xi$, where $\kappa$ is a positive diffusion constant, $\Delta$ is the lattice Laplacian on $\Z^d$, $d \geq 1$, and $\xi$ is a simple symmetric exclusion process on $\Z^d$ in Bernoulli equilibrium. This model describes the evolution of a \emph{reactant} $u$ under the influence of a \emph{catalyst} $\xi$. In G\"artner, den Hollander and Maillard (2007) we investigated the behavior of the annealed Lyapunov exponents, i.e., the exponential growth rates as $t\to\infty$ of the successive moments of the solution $u$. This led to an almost complete picture of intermittency as a function of $d$ and $\kappa$. In the present paper we finish our study by focussing on the asymptotics of the Lyaponov exponents as $\kappa\to\infty$ in the \emph{critical} dimension $d=3$, which was left open in G\"artner, den Hollander and Maillard (2007) and which is the most challenging. We show that, interestingly, this asymptotics is characterized not only by a \emph{Green} term, as in $d\geq 4$, but also by a \emph{polaron} term. The presence of the latter implies intermittency of \emph{all} orders above a finite threshold for $\kappa$.


Introduction and main result 1.Model
In this paper we consider the parabolic Anderson model where κ is a positive diffusion constant, ∆ is the lattice Laplacian acting on u as ( · is the Euclidian norm), and is a space-time random field that drives the evolution. If ξ is given by an infinite particle system dynamics, then the solution u of the PAM may be interpreted as the concentration of a diffusing reactant under the influence of a catalyst performing such a dynamics.

Lyapunov exponents
For p ∈ , define the p-th annealed Lyapunov exponent of the PAM by We are interested in the asymptotic behavior of λ p (κ, ρ) as κ → ∞ for fixed ρ and p. To this end, let G denote the value at 0 of the Green function of simple random walk on 3 with jump rate 1 (i.e., the Markov process with generator 1 6 ∆), and let 3 be the value of the polaron variational problem where ∇ 3 and ∆ 3 are the continuous gradient and Laplacian, · 2 is the L 2 ( 3 )-norm, (1.8) (See Donsker and Varadhan [1] for background on how 3 arises in the context of a self-attracting Brownian motion referred to as the polaron model. See also Gärtner and den Hollander [2], Section 1.5.) We are now ready to formulate our main result (which was already announced in Gärtner, den Hollander and Maillard [4]).
Theorem 1.1. Let d = 3, ρ ∈ (0, 1) and p ∈ . Then 2 3 . (1.9) Note that the expression in the r.h.s. of (1.9) is the sum of a Green term and a polaron term. The existence, continuity, monotonicity and convexity of κ → λ p (κ, ρ) were proved in [3] for all d ≥ 1 for all exclusion processes with an irreducible and symmetric random walk transition kernel. It was further proved that λ p (κ, ρ) = 1 when the random walk is recurrent and ρ < λ p (κ, ρ) < 1 when the random walk is transient. Moreover, it was shown that for simple random walk in d ≥ 4 the asymptotics as κ → ∞ of λ p (κ, ρ) is similar to (1.9), but without the polaron term. In fact, the subtlety in d = 3 is caused by the appearance of this extra term which, as we will see in Section 5, is related to the large deviation behavior of the occupation time measure of a rescaled random walk that lies deeply hidden in the problem. For the heuristics behind Theorem 1.1 we refer the reader to [3], Section 1.5.
In [3] it was shown that for all d ≥ 3 the PAM is intermittent for small κ. We conjecture that in d = 3 it is in fact intermittent for all κ. Unfortunately, our analysis does not allow us to treat intermediate values of κ (see the figure).
The formulation of Theorem 1.1 coincides with the corresponding result in Gärtner and den Hollander [2], where the random potential ξ is given by independent simple random walks in a Poisson equilibrium in the so-called weakly catalytic regime. However, as we already pointed out in [3], the approach in [2] cannot be adapted to the exclusion process, since it relies on an explicit Feynman-Kac representation for the moments that is available only in the case of independent particle motion. We must therefore proceed in a totally different way. Only at the end of Section 5 will we be able to use some of the ideas in [2].

Outline
Each of Sections 2-5 is devoted to a major step in the proof of Theorem 1.1 for p = 1. The extension to p ≥ 2 will be indicated in Section 6.
In Section 2 we start with the Feynman-Kac representation for the first moment of the solution u, which involves a random walk sampling the exclusion process. After rescaling time, we transform the representation w.r.t. the old measure to a representation w.r.t. a new measure via an appropriate absolutely continuous transformation. This allows us to separate the parts responsible for, respectively, the Green term and the polaron term in the r.h.s. of (1.9). Since the Green term has already been handled in [3], we need only concentrate on the polaron term. In Section 3 we show that, in the limit as κ → ∞, the new measure may be replaced by the old measure. The resulting representation is used in Section 4 to prove the lower bound for the polaron term. This is done analytically with the help of a Rayleigh-Ritz formula. In Section 5, which is technical and takes up almost half of the paper, we prove the corresponding upper bound. This is done by freezing and defreezing the exclusion process over long time intervals, allowing us to approximate the representation in terms of the occupation time measures of the random walk over these time intervals. After applying spectral estimates and using a large deviation principle for these occupation time measures, we arrive at the polaron variational formula.

Separation of the Green term and the polaron term
In Section 2.1 we formulate the Feynman-Kac representation for the first moment of u and show how to split this into two parts after an appropriate change of measure. In Section 2.2 we formulate two propositions for the asymptotics of these two parts, which lead to, respectively, the Green term and the polaron term in (1.9). These two propositions will be proved in Sections 3-5. In Section 2.3 we state and prove three elementary lemmas that will be needed along the way.

Key objects
The solution u of the PAM in (1.1) admits the Feynman-Kac representation where X is simple random walk on 3 with step rate 6 (i.e., with generator ∆) and P X x and E X x denote probability and expectation with respect to X given X 0 = x. Since ξ is reversible w.r.t. ν ρ , we may reverse time in (2.1) to obtain where ν ρ ,0 is expectation w.r.t. ν ρ ,0 = ν ρ ⊗ P X 0 . As in [2] and [3], we rescale time and write and Here and in the rest of the paper we suppress the dependence on ρ ∈ (0, 1) from the notation. Under η,x = η ⊗ P X x , (Z t ) t≥0 is a Markov process with state space Ω × 3 and generator (acting on the Banach space of bounded continuous functions on Ω × 3 , equipped with the supremum norm). Let ( t ) t≥0 denote the semigroup generated by .
Our aim is to make an absolutely continuous transformation of the measure η,x with the help of an exponential martingale, in such a way that, under the new measure new η,x , (Z t ) t≥0 is a Markov process with generator new of the form This transformation leads to an interaction between the exclusion process part and the random walk part of (Z t ) t≥0 , controlled by ψ: Ω × 3 → . As explained in [3], Section 4.2, it will be expedient to choose ψ as with T a large constant (suppressed from the notation), implying that It was shown in [3], Lemma 4.3.1, that for all events A in the σ-algebra generated by (Z s ) s∈ [0,t] , then under new η,x indeed (Z s ) s≥0 is a Markov process with generator new . Using (2.11-2.13) and new ν ρ ,0 = Ω ν ρ (dη) new η,0 , it then follows that the expectation in (2.7) can be written in the form (2.14) The first term in the exponent in the r.h.s. of (2.14) stays bounded as t → ∞ and can therefore be discarded when computing λ * (κ) via (2.7). We will see later that the second term and the third term lead to the Green term and the polaron term in (2.6), respectively. These terms may be separated from each other with the help of Hölder's inequality, as stated in Proposition 2.1 below.

Key propositions
Proposition 2.1. For any κ > 0, where 1/q + 1/r = 1, with q > 0, r > 1 in the first inequality and q < 0, 0 < r < 1 in the second inequality. In the next proposition we write ψ T instead of ψ to indicate the dependence on the parameter T .

Proposition 2.2. For any
These propositions will be proved in Sections 3-5.

Preparatory lemmas
This section contains three elementary lemmas that will be used frequently in Sections 3-5.
Let p (1) t (x, y) and p t (x, y) = p (3) t (x, y) be the transition kernels of simple random walk in d = 1 and d = 3, respectively, with step rate 1.
(In the sequel we will frequently write p t (x − y) instead of p t (x, y).) From the graphical representation for SSE (Liggett [7], Chapter VIII, Theorem 1.1) it is immediate that Recalling (2.4-2.5) and (2.10), we therefore have where we abbreviate where C > 0 is the same constant as in Lemma 2.4, and G is the value at 0 of the Green function of simple random walk on 3 .

Reduction to the original measure
In this section we show that the expectations in Propositions 2.2-2.3 w.r.t. the new measure new ν ρ ,0 are asymptotically the same as the expectations w.r.t. the old measure ν ρ ,0 . In Section 3.1 we state a Rayleigh-Ritz formula from which we draw the desired comparison. In Section 3.2 we state the analogues of Propositions 2.2-2.3 whose proof will be the subject of Sections 4-5.

Rayleigh-Ritz formula
Recall the definition of ψ in (2.10). Let m denote the counting measure on 3 . It is easily checked that both µ ρ = ν ρ ⊗ m and µ new ρ given by are reversible invariant measures of the Markov processes with generators defined in (2.8), respectively, new defined in (2.9). In particular, and new are self-adjoint operators in L 2 (µ ρ ) and L 2 (µ new ρ ). Let ( ) and ( new ) denote their domains.
Proof. The limit in the l.h.s. of (3.2) coincides with the upper boundary of the spectrum of the operator new + V on L 2 (µ new ρ ), which may be represented by the Rayleigh-Ritz formula. The latter coincides with the expression in the r.h.s. of (3.2). The details are similar to [3], Section 2.2. Lemma 3.1 can be used to express the limits as t → ∞ in Propositions 2.2-2.3 as variational expressions involving the new measure. Lemma 3.2 below says that, for large κ, these variational expressions are close to the corresponding variational expressions for the old measure. Using Lemma 3.1 for the original measure, we may therefore arrive at the corresponding limit for the old measure.
For later use, in the statement of Lemma 3.2 we do not assume that ψ is given by (2.10). Instead, we only suppose that η → ψ(η) is bounded and measurable and that there is a constant K > 0 such that for all η ∈ Ω, a, b ∈ 3 with a − b = 1 and x ∈ 3 , but retain that new and µ new ρ are given by (2.9) and (3.1), respectively.

4)
where ± means + in the first inequality and − in the second inequality, and ∓ means the reverse.

Reduced key propositions
At this point we may combine the assertions in Lemmas 3.1-3.2 for the potentials with ψ given by (2.10). Because of (2.25-2.26), the constant K in (3.3) may be chosen to be the maximum of 2G and 2C T , resulting in K/κ → 0 as κ → ∞. Moreover, from (2.27) and a Taylor expansion of the r.h.s. of (3.9) we see that the potential in (3.9) is bounded for each κ and T , and the same is obviously true for the potential in (3.10) because of (2.4). In this way, using a moment inequality to replace the factor e ±K/κ α by a slightly larger, respectively, smaller factor α ′ independent of T and κ, we see that the limits in Propositions 2.2-2.3 do not change when we replace new ν ρ ,0 by ν ρ ,0 . Hence it will be enough to prove the following two propositions.

Proof of Proposition 3.4: lower bound
In this section we derive the lower bound in Proposition 3.4. We fix α, κ, T > 0 and use Lemma 3.1, to obtain (4.1) In Section 4.1 we choose a test function. In Section 4.2 we compute and estimate the resulting expression. In Section 4.3 we take the limit κ, T → ∞ and show that this gives the desired lower bound.

Choice of test function
To get the desired lower bound, we use test functions F of the form Before specifying F 1 and F 2 , we introduce some further notation. In addition to the counting measure m on 3 , consider the discrete Lebesgue measure m κ on 3 is the set of infinitely differentiable functions on 3 with compact support. Define and note that To choose F 1 , introduce the function where ( t ) t≥0 is the semigroup generated by the operator L in (1.4). Note that the construction of ψ from φ in (4.9) is similar to the construction of ψ from φ in (2.10). In particular, Combining the probabilistic representations of the semigroups ( t ) t≥0 (generated by in (2.8)) and ( t ) t≥0 (generated by L in (1.4)) with the graphical representation formulas (2.21-2.22), and using (4.4-4.5), we find that and Using the second inequality in (2.20), we have (4.14) Now choose F 1 as For the above choice of F 1 and F 2 , we have F 1 L 2 (ν ρ ) = F 2 l 2 ( 3 ) = 1 and, consequently, With F 1 , F 2 and φ as above, and as in (2.8), after scaling space by κ we arrive at the following lemma.

Computation of the r.h.s. of (4.16)
Clearly, as κ → ∞ the first summand in the r.h.s. of (4.16) converges to The computation of the second summand in the r.h.s. of (4.16) is more delicate: denotes the Gaussian transition kernel associated with ∆ 3 , the continuous Laplacian on 3 .
Proof. Using the probability measure in combination with (4.10), we may write the term under the lim inf in (4.18) in the form Recalling the expressions for ψ in (4.12-4.13) and using (4.14), we get for a, b ∈ 3 with a − b = 1, Hence, a Taylor expansion of the exponent in the r.h.s. of (4.22) gives (4.24) Using (4.12), we obtain Using (4.20), we have (after cancellation of factors not depending on a or b) Using (4.14), we obtain that On the other hand, by (4.13), where ∆ acts on the first spatial variable of p s (· , ·) and ∆p s = 6(∂ p s /∂ s). Therefore, (4.31) Combining (4.24-4.25) and (4.28-4.29) and (4.31), we arrive at (4.32) After replacing 2S in the first integral by 6εκ 2 1[κ], using a Gaussian approximation of the transition kernel p t (x, y) and recalling the definitions of S and U in (4.8), we get that, for any ε > 0, t (x, y) . (4.33) At this point it only remains to check that the U−S -term in (4.21) is nonnegative. By (4.11) and the probabilistic representation of the semigroup ( t ) t≥0 , we have and, by (4.20), which proves the claim.

Proof of the lower bound in Proposition 3.4
We finish by using Lemma 4.2 to prove the lower bound in Proposition 3.4.

Proof of Proposition 3.4: upper bound
In this section we prove the upper bound in Proposition 3.4. The proof is long and technical. In Sections 5.1 we "freeze" and "defreeze" the exclusion dynamics on long time intervals. This allows us to approximate the relevant functionals of the random walk in terms of its occupation time measures on those intervals. In Section 5.2 we use a spectral bound to reduce the study of the long-time asymptotics for the resulting time-dependent potentials to the investigation of timeindependent potentials. In Section 5.3 we make a cut-off for small times, showing that these times are negligible in the limit as κ → ∞, perform a space-time scaling and compactification of the underlying random walk, and apply a large deviation principle for the occupation time measures, culminating in the appearance of the variational expression for the polaron term 3 .

Freezing
We begin by deriving a preliminary upper bound for the expectation in Proposition 3.4 given by where, as before, T is a large constant. To this end, we divide the time interval [0, t] into ⌊t/R κ ⌋ intervals of length with R a large constant, and "freeze" the exclusion dynamics (ξ t/κ ) t≥0 on each of these intervals.
As will become clear later on, this procedure allows us to express the dependence of (5.1) on the random walk X in terms of objects that are close to integrals over occupation time measures of X on time intervals of length R κ . We will see that the resulting expression can be estimated from above by "defreezing" the exclusion dynamics. We will subsequently see that, after we have taken the limits t → ∞, κ → ∞ and T → ∞, the resulting estimate can be handled by applying a large deviation principle for the space-time rescaled occupation time measures in the limit as R → ∞. The latter will lead us to the polaron term.
Ignoring the negligible final time interval [⌊t/R κ ⌋R κ , t], using Hölder's inequality with p, q > 1 and 1/p + 1/q = 1, and inserting (5.2), we see that (5.1) may be estimated from above as and (2) R,α (t) = R,α (κ, T ; t) Therefore, by choosing p close to 1, the proof of the upper bound in Proposition 3.4 reduces to the proof of the following two lemmas.

Proof of Lemma 5.1
Proof. Fix R, α > 0 arbitrarily. Given a path X , an initial configuration η ∈ Ω and k ∈ , we first derive an upper bound for To this end, we use the independent random walk approximation ξ of ξ (cf. [3], Proposition 1.2.1), to obtain where Y is simple random walk on 3 with jump rate 1 (i.e., with generator 1 6 ∆), E Y 0 is expectation w.r.t. Y starting from 0, and Observe that the expectation w.r.t. Y of the expression in the exponent is zero. Therefore, a Taylor expansion of the exponential function yields the bound where s 0 = 0, y 0 = 0, and the product has to be understood in a noncommutative way. Using the Chapman-Kolmogorov equation and the inequality p t (z) ≤ p t (0), z ∈ 3 , we find that the cut-off Green function of simple random walk at 0 at time T . Substituting this into the above bound for l = n, n − 1, · · · , 3, computing the resulting geometric series, and using the inequality 1 + x ≤ e x , we obtain provided that 2αG T (0) < 1, which is true for T large enough. Note that C T → 1 as T → ∞. Substituting (5.16) into (5.11), we find that Using once more the Chapman-Kolmogorov equation and p t (x, y) = p t (x − y), we may compute the sums in the exponent, to arrive at Note that this bound does not depend on the initial configuration η and depends on the process X only via its increments on the time interval [(k − 1)R κ , kR κ ]. By (5.10), the increments over the time intervals labelled k = 1, 2, · · · , ⌊t/R κ ⌋ are independent and identically distributed. Using ν ρ ,0 = ν ρ (dη)E X 0 η , we can therefore apply the Markov property of the exclusion dynamics (ξ t/κ ) t≥0 at times R κ , 2R κ , · · · , (⌊t/R κ ⌋ − 1)R κ to the expectation in the r.h.s. of (5.5), insert the bound (5.19) and afterwards use that (X t ) t≥0 has independent increments, to arrive at Hence, recalling the definition of R κ in (5.3), we obtain lim sup and let E X 0 = E X 0 E Y 0 be the expectation w.r.t. X starting at 0. Observe that We next apply Jensen's inequality w.r.t. the first integral in the r.h.s. of (5.21), substitute s 2 = s 1 + s, take into account that X has independent increments, and afterwards apply Jensen's inequality w.r.t. E Y 0 , to arrive at the following upper bound for the expectation in (5.21): Applying Lemma 2.6, we can bound the last expression from above by where G 2T (0) is the cut-off at time 2T of the Green function G at 0 for X (which has generator 1[κ]∆). Since G 2T (0) → 1 6 G 12T (0) as κ → ∞, and since the latter converges to zero as T → ∞, a combination of the above estimates with (5.21) gives the claim.

Defreezing
To prove Lemma 5.2, we next "defreeze" the exclusion dynamics in (2) R,α (t). This can be done in a similar way as the "freezing" we did in Section 5.1.1, by taking into account the following remarks. In (5.6), each single summand is asymptotically negligible as t → ∞. Hence, we can safely remove a summand at the beginning and add a summand at the end. After that we can bound the resulting expression from above with the help of Hölder's inequality with weights p, q > 1, 1/p + 1/q = 1, namely, In this way, choosing p close to 1, we see that the proof of Lemma 5.2 reduces to the proof of the following two lemmas.

Proof of Lemma 5.3
Proof. The proof goes along the same lines as the proof of Lemma 5.1. Instead of (5.9), we consider Applying Jensen's inequality w.r.t. the first integral and the Markov property of the exclusion dynamics (ξ t/κ ) t≥0 at time u/κ, we see that it is enough to derive an appropriate upper bound for The main steps are the same as in the proof of Lemma 5.1. Instead of (5.19), we obtain and this expression may be bounded from above by (5.25).

Spectral bound
The advantage of Lemma 5.4 compared to the original upper bound in Proposition 3.4 is that, modulo a small time correction of the form (s − u)/κ, the expression under the expectation in (5.28) depends on X only via its occupation time measures on the time intervals [kR κ , (k + 1)R κ ], k = 1, 2, · · · , ⌊t/R κ ⌋. This will allow us in Section 5.3 to use a large deviation principle for these occupation time measures. The present section consists of five steps, organized in Sections 5.2.1-5.2.5, leading up to a final lemma that will be proved in Section 5.3.
We abbreviate and rewrite the expression for (4) R,α (t) in (5.28) in the form In (5.34) and subsequent expressions we suppress the dependence on T and R.

Reduction to a spectral bound
Let B(Ω) denote the Banach space of bounded measurable functions on Ω equipped with the supremum norm · ∞ . Given V ∈ B(Ω), let Proof. In the proof we will assume that s → V s is continuous. The extension to piecewise continuous s → V s will be straightforward. Let 0 = t 0 < t 1 < · · · < t r = t be a partition of the interval [0, t]. Then Let ( V t ) t≥0 denote the semigroup generated by L + V on L 2 (ν ρ ) with inner product (· , ·) and norm · . Then V t = e tλ(V ) . (5.40) Using the Markov property, we find that Combining (5.39) and (5.41), we arrive at Since the map V → λ(V ) from B(Ω) to is continuous (which can be seen e.g. from (5.40) and the Feynman-Kac representation of V t ), the claim follows by letting the mesh of the partition tend to zero. Lemma 5.6. For all α, T, R, t, κ > 0, Proof. Apply Lemma 5.5 to the potential replaced by (ξ u/κ ) u≥0 , and take the expectation w.r.t. E X 0 .

Proof of Lemma 5.8
Proof. Using the same arguments as in (5.62-5.63), we can replace new ν ρ by ν ρ in formula (5.54), to obtain Because of (5.49), this yields Now, using the independent random walk approximation ξ of ξ (see [3], Proposition 1.2.1), we find that where A η is given by (5.12) and Y is simple random walk with step rate 1. Define Kκ e Cα/T , where we again use the second inequality of Lemma 2.4. Substituting (5.84) into (5.79), we arrive at the claim with D α,T,K = 8α 2 C e 2Cα/T / K.

Further reduction of Lemma 5.4
To further estimate the expectation in Lemma 5.7 from above, we use the following two lemmas.
Proof. Using the bound in (5.58) for k,u 1 , we find that which tends to zero as T → ∞. Hence, we may apply Lemma 5.9 to (5.57) to get the claim.
At this point we may combine Lemmas 5.10 and 5.8 with (5.52), to get Note that the upper bound in (5.58) for k,u 1 depends on X only via its increments on the times interval [(k − 1)R κ , kR κ ] and that these increments are i.i.d. for k = 1, 2, · · · , ⌊t/R κ ⌋. Hence, combining (5.35) and Lemma 5.6 with (5.89) and splitting the resulting expectation w.r.t. E X 0 into ⌊t/R κ ⌋ equal factors with the help of the Markov property at times kR κ , k = 1, 2, · · · , ⌊t/R κ ⌋, we obtain, after also substituting (5.58), R,α (κ) = R,α (T, K; κ) Because of (5.60), we therefore conclude that the proof of Lemma 5.4 reduces to the following lemma. The proof of Lemma 5.11 will be reduced to two further lemmas in which we cut out small times. These lemmas will be proved in By Hölder's inequality with weights p, q > 1, 1/p + 1/q = 1, we have Hence, by choosing p close to 1, we see that the proof of Lemma 5.11 reduces to the following lemmas.  R,α (κ) we integrate the transition kernel over "small" times r ∈ [0, m]. What Lemma 5.12 shows is that the integral is asymptotically negligible.

Proof of Lemma 5.12
Proof. We need only prove the upper bound in (5.98). An application of Jensen's inequality yields As in (5.22), let X t = X t + Y t/κ and let E X 0 denote expectation w.r.t. X starting at 0. Then, using Jensen's inequality w.r.t. E Y 0 , we find that For further comments on Lemma 5.12, see the remark at the end of Section 5.3.3.

Scaling, compactification and large deviations
In this section we prove Lemma 5.13 with the help of scaling, compactification and large deviations.