Vanishing corrections for the position in a linear model of FKPP fronts

Take the linearised FKPP equation \[\partial_t h =\partial^2_x h +h\] with boundary condition $h(m(t),t)=0$. Depending on the behaviour of the initial condition $h_0(x)=h(x,0)$ we obtain the asymptotics - up to a $o(1)$ term $r(t)$ - of the absorbing boundary $m(t)$ such that $\omega(x):=\lim_t h(x+m(t) ,t)$ exists and is non-trivial. In particular, as in Bramson's results for the non-linear FKPP equation, we recover the celebrated $-(3/2)\log t$ correction for initial conditions decaying faster than $x^\nu e^{-x}$ for some $\nu<-2$. Furthermore, when we are in this regime, the main result of the present work is the identification (to first order) of the $r(t)$ term which ensures the fastest convergence to $\omega(x)$. When $h_0(x)$ decays faster than $x^\nu e^{-x}$ for some $\nu<-3$, we show that $r(t)$ must be chosen to be $-3\sqrt{\pi/t}$ which is precisely the term predicted heuristically by Ebert-van Saarloos in the non-linear case. When the initial condition decays as $x^\nu e^{-x}$ for some $\nu \in [-3,-2)$, we show that even though we are still in the regime where Bramson's correction is $-(3/2)\log t$, the Ebert-van Saarloos correction has to be modified. Similar results were recently obtained by Henderson using an analytical approach and only for compactly supported initial conditions.


Introduction
The celebrated Fisher-Kolmogorov-Petrovsky-Piscounof equation (FKPP) in one dimension for h : R × R + → R is: This equation is a natural description of a reaction-diffusion model [Fis37,KPP37,AW78]. It is also related to branching Brownian motion: for the Heaviside initial condition h 0 (x) = ½ {x<0} , h(x, t) is the probability that the rightmost particle at time t in a branching Brownian motion (BBM) is to the right of x. For suitable initial conditions where h 0 (x) ∈ [0, 1], h 0 (x) goes to 1 fast enough as x → −∞ and h 0 (x) goes to 0 fast enough as x → ∞, it is known that h(x, t) develops into a travelling wave: there exists a centring term m(t) and an asymptotic shape ω v (x) such that where m(t)/t → v and ω v (x) is a travelling wave solution to (1) with velocity v: that is, the unique (up to translation) non-trivial solution to with ω v (−∞) = 1 and ω v (+∞) = 0.
In his seminal works [Bra83], Bramson showed how the initial condition h 0 (and in particular its large x asymptotic behaviour) determines m(t) in (2). For the important example h 0 (x) = ½ {x<0} corresponding to the rightmost particle in BBM, he finds for some constant a, and a limiting travelling wave with (critical) speed v = 2. (Here and throughout, we use the notation f (t) = o(1) to mean that f (t) → 0 as t → ∞.) What makes Bramson's results extremely interesting is their universality; for instance Bramson proves [Bra83] that the previous result still holds if the reaction term h−h 2 in (1) is replaced by f (h) with f (0) = f (1) = 0, f ′ (0) = 1 and f (x) ≤ x. The universality goes further than that, and for many other front equations, it is believed and sometimes known that the centring term m(t) follows the same kind of behaviour as for (1): one needs to compute a function v(γ) which has a minimum v c at a point γ c (in the FKPP case (1), v(γ) = γ + 1/γ, γ c = 1, v c = 2); then for an initial condition decreasing like e −γx , the front converges to a travelling wave with velocity v(γ) if γ ≤ γ c and critical velocity v c if γ ≥ γ c .
When the centring term m(t) is defined as in (2), it is not uniquely determined: if m(t) is any suitable centring term, then m(t) + o(1) is also a suitable centring term. Instead one can try to give a more precise definition for m(t). For example, one could reasonably ask for h m(t), t = α for some α ∈ (0, 1) or ∂ 2 x h m(t), t = 0 or m(t) = − dx x∂ x h(x, t) (5) in addition to (2). In the case h 0 (x) = ½ {x<0} , so that h(x, t) = P(R t > x) where R t is the position of the rightmost particle in a BBM at time t, the first definition in (5) would be the α-quantile of R t , the second definition would be the mode of the distribution of R t , and the third definition would be the expectation of R t . It has been heuristically argued [EvS00,MM14,Hen14,BD15] that any quantity m(t) defined as in (5) behaves for large t as for any front equation of the FKPP type and for any initial condition that decays fast enough. In the FKPP case (1), one has γ c = 1 and v ′′ (γ c ) = 2 so that m(t) = 2t − (3/2) log t + a − 3 π/t + o(1/ √ t). Heuristically, the coefficient of the 1/ √ t term does not depend on the precise definition of m(t) because the front h(x, t) converges very quickly to its limiting shape in the region where h is neither very close to 0 nor very close to 1, so that the difference between any two reasonable definitions of m(t) converges quickly (faster than 1/ √ t) to some constant. Note that the constant term "a" is expected to be non-universal and to depend on the model, the initial condition and the precise definition of m(t).
As argued in [EvS00], the reason why the "log t" and the "1/ √ t" terms in (6) are so universal is that they are driven by the way the front develops very far on the right, in a region where it is exponentially small and where understanding the position m(t) of the front is largely a matter of solving the linearised front equation.
However there is a catch: solving directly the linearised equation ing the position m(t) by h linear m(t), t = 1 gives m(t) = 2t − 1 2 log t + a + O (log 2 t)/t rather than (4); the linearised equation has the same velocity 2 as for the FKPP equation, a logarithmic correction but with a different prefactor and no 1/ √ t correction. The problem is that with the linearised equation, the h linear (x, t) increases exponentially on the left of m(t) and this "mass" pushes the front forward, leading to a − 1 2 log t rather than a − 3 2 log t correction. This means that in order to recover the behaviour of m(t) for the FKPP equation, one must have a front equation with some saturation mechanism on the left. The behaviour of m(t) is not expected to depend on which saturation mechanism is chosen, but one must be present. For these reasons, we consider in this paper a linearised FKPP with a boundary on the left, as in [Hen14].
We emphasize that, in the present work, the FKPP equation is only a motivation: we do not attempt to establish the equivalence between the FKPP equation and the linear model with a boundary. Our results are proved only for the linear model with boundary, and we can only conjecture that they do apply to the FKPP equation.

Statement of the problem and main results
We study the following linear partial differential equation with initial condition h 0 (x) and a given boundary m : [0, ∞) → R: Observe that without loss of generality we can (and will) insist that m(0) = 0 since otherwise we can simply shift the reference frame by m(0) by the change of coordinate x → x − m(0).
The same system was studied in [Hen14] by PDE methods for compactly supported initial conditions. In this paper, we use probabilistic methods, writing the solution of the heat equation as an expectation involving Brownian motion with a killing boundary. We give more general results, in particular lifting the compactly supported hypothesis.
If the boundary is linear, m(t) = vt, the problem is easily solved explicitly. However, as soon as m(t) is no longer linear, gaining any explicit information about the solution is known to be hard (see for instance [HE15]) and there are few available results.
Motivated by the earlier FKPP discussion about convergence to a travelling wave as in (2), we are looking for functions m : with ω non-trivial, ω(0) = 0 and ω(x) > 0 for all x > 0. Note that such a function ω necessarily satisfies In this case, the boundary condition anchors the front. Requiring the convergence of h(m(t) + x, t) to a limiting shape means that m(t) must increase fast enough to prevent the mass near the front from growing exponentially, but not so fast that it tends to zero. This provides a saturation mechanism, and even though it might seem very unlike FKPP fronts to have h m(t), t = 0, as discussed earlier we do expect the two systems to behave similarly.
Throughout the article we use the following notation: • A random variable G is said to have "Gaussian tails" if there exist two positive constants c 1 , c 2 such that P(|G| > z) ≤ c 1 exp(−c 2 z 2 ) for all z ≥ 0.
Our first theorem recovers the analogue of Bramson's results for the system (7), (8 for large x and such that the value of α below is non-zero, where ∆ and ψ ∞ are quantities depending on the whole function m (and not only the asymptotics) which are introduced (in (61) and (68)) in the proofs.

Remarks.
• From the probabilistic representation of h(x, t) written later in the paper (21), it is clear that the solution h(x, t) to (7) must be an increasing function of h 0 and a decreasing function of m (in the sense that if for all x and t). This implies that the α given in Theorem 1 must be increasing functions of h 0 and decreasing functions of m. This was obvious from the explicit expression of α in cases (a), (b) and (c). In case (d), given the complicated expressions for ∆ and ψ ∞ , it is not obvious at all from its expression that α decreases with m.
• Consider now a twice differentiable function m without the assumption that m ′′ (t) = O(1/t 2 ). The monotonicity of h(x, t) with respect to m still holds, and by sandwiching such a m between two sequences of increasingly close functions that satisfy the O(1/t 2 ) condition, one can show easily in cases (a), (b) and (c) that if m has the correct asymptotics, then h m(t) + x, t converges as in Theorem 1. Case (d) is more difficult as both ∆ and ψ ∞ might be ill defined when one does not assume m ′′ (t) = O(1/t 2 ).
We now turn to the analogue of the Ebert-van Saarloos correction (6) for our model (7). As explained in the introduction and shown in Theorem 1, with a characterization as in (8), m(t) is only determined up to o(1). If we wish to improve upon Theorem 1, then we need a more precise definition for m(t), analogous to (5). Natural possible definitions could be However, it is not obvious that such a function m(t) even exists, would be unique or differentiable.
We are furthermore interested only in the long time asymptotics of m(t). Therefore, instead of requiring something like (11) we rather look, as in [Hen14], for the function m(t) such that the convergence (8) is as fast as possible.
Our main result, Theorem 2, tells us how fast h m(t) + x, t converges for suitable choices of m in case (d) of Theorem 1. This case is the most classical as it contains, for example, initial conditions with bounded support. It is the case studied by Ebert-Van Saarloos and Henderson, and is the case for which universal behaviour is expected. Theorem 2 is followed by two corollaries that highlight important consequences.
Theorem 2. Suppose that h 0 is a bounded function such that h 0 (x) = O x ν e −x for large x for some ν < −2, and such that α defined in (10d) is non-zero. Suppose also that m is twice continuously differentiable with where r(0) = −a, r(t) → 0 as t → ∞ and r ′′ (t) = O(t −2−η ) for large t for some η > 0. Then for any x ≥ 0, (13) with α as in (10d). If we further assume that h 0 (x) ∼ Ax ν e −x for large x for some A > 0 and −4 < ν < −2, then This result allows us to bound the rate of convergence h m(t)+x, t to αxe −x : it is generically of order max 1/ √ t, |r(t)|, t 1+ν/2 . This also suggests that for m(t) defined as in either choice of (11), one should have r(t) ∼ −3 √ π/ √ t for ν < −3 and r(t) ≍ t 1+ν/2 for −3 ≤ ν < −2. Note however that we are not sure that such a m(t) exists and, if it exists, we do not know whether it satisfies the hypothesis on m ′′ (t) that we used in the Theorem.
In the following two corollaries we highlight the best rates of convergence of h m(t) + x, t → xe −x that we can obtain from Theorem 2. For simplicity, we dropped the technical requirement that m(0) = 0 in the corollaries; the expression for α must therefore be adapted.
Note in particular that we have recovered the result of [Hen14], but with more general initial conditions ( [Hen14] only considered compactly supported initial conditions).
Notice that for h 0 (x) ∼ Ax −3 e −x the position m(t) still features a first order correction in 1/ √ t but with a coefficient − 3 √ π + 1 4α Ae −a which is different from the ν < −3 case.

Writing the solution as an expectation of a Bessel
In this section, we write the solution to (7) as an expectation of a Bessel process. We only consider functions m(t) that are twice continuously differentiable. For each given m(t), (7) is a linear problem. We first study the fundamental solutions q(t, x, y) defined as where δ is the Dirac distribution. Then It is clear that e −t q(t, x, y) is the solution to the heat equation with boundary, and therefore Then by Girsanov's theorem, Plugging into (21) at position m(t) + x instead of x, we get We recall that, by the reflection principle, the probability that a Brownian path started from y stays positive and ends in dx is: Using (25), we write (24) as a conditional expectation: where ξ is a Brownian motion (normalized as in (22)) started from y and conditioned not to hit zero for any s ∈ (0, t) and to be at x at time t. Such a process is called a Bessel-3 bridge, and we recall some properties of Bessel processes and bridges in Section 4.
It is convenient to think of the path s → ξ (t:y→x) s as the straight line s → y + (x − y)s/t plus some fluctuations. This leads us to define where we have used integration by parts. With this quantity, (26) now reads and the main part of the present work is to estimate ψ t (y, x).

The Bessel toolbox
Before we begin our main task, we need some fairly standard estimates on Bessel-3 processes and Bessel-3 bridges. From here on, we refer to these simply as Bessel processes and Bessel bridges; the "3" will be implicit. We include proofs for completeness. We build most of our processes on the same probability space. We fix a driving Brownian motion (B s , s ≥ 0) started from 0 under a probability measure P, with the normalization E[B 2 t ] = 2t.
For each y ≥ 0 we introduce a Bessel process ξ (y) started from y as the strong solution to the SDE It is well-known that ξ (y) s has the law of a Brownian motion conditioned to never hit zero. We also introduce, for each t ≥ 0 and y ≥ 0 This process is a Bessel bridge from y to 0 in time t, which is a Brownian motion started from y and conditioned to hit 0 for the first time at time t. One can check by direct substitution that where for each t, and is thus itself a Brownian motion. One can compute directly the law of the Brownian motion conditioned to hit zero for the first time at time t using (25) and check that this law solves the forward Kolmogorov equation (or Fokker Planck equation) associated with the SDE (or Langevin equation) (31).
Similarly, we construct the Bessel bridge from y to x in time t, the Brownian motion conditioned not to hit zero for any s ∈ (0, t) and to be at x at time t, through The advantages of constructing all the processes from a single Brownian path s → B s is that they can be compared directly, realization by realization. In particular we use the following comparisons: Lemma 5. For any y ≥ z ≥ 0 and s ≥ 0, Proof. To prove (34) we make three observations.
-The processes ξ (y) s and y + B s both start from y and 0 , so the two processes must remain ordered at all times (see for instance [Kun97]).
and since ξ (y) The inequalities in the left part of (35) are a direct consequence of (34) through the change of time (30). We now focus on the inequalities in the right part of (35). First we assume that z > 0.
The fact that for x ≥ z we have ξ follows from the fact that x coth(ax) ≥ z coth(az) for any a > 0 and x ≥ z.
For the other inequality, the fact that u(coth u − 1) is decreasing yields that so that, writing ζ s := ξ But the solution to dφs which concludes the proof for z > 0. For the case z = 0 the proof is the same but uses the inequalities 1 ≤ u coth u ≤ 1 + u for u ≥ 0.
We note that, intuitively, as the length of a Bessel bridge tends to infinity, on any compact time interval the bridge looks more and more like a Bessel process. Similarly, as the start point of a Bessel process tends to infinity, on any compact interval it looks more and more like a Brownian motion relative to its start position. We make this precise in the lemma below.
Lemma 6. For all s ≥ 0 and y ≥ 0, For all s ≥ 0 and any y t → ∞ as t → ∞, Proof. For (40), we simply recall (30) which defined Finally, for (42), we write Using our coupling between the Bessel processes and Brownian motion we have dB u ≤ dξ so by continuity of paths, which concludes the proof of (42).
We need the fact that the increments of a Bessel process over time s are roughly of order s 1/2 . By paying a small price on the exponent, we obtain the following uniform bounds: Furthermore, uniformly in x ≥ 0, y ≥ 0, t ≥ 0 and 0 ≤ s ≤ t, s . Also by symmetry P(|B s | > x) = 2P(B s > x). Thus to prove (49), it is sufficient to show that for some positive c 1 and c 2 . The proof is is elementary and we defer it to an appendix. To prove (50), notice that from (35) we have But from the change of time (30) and (49), where the last step is obtained by pushing the (t − s)/t inside the max. This provides the upper bound of (50). For the lower bound, we introduce Brownian bridges s → B (t:y→x) s started from y and conditioned to be at x at time t. We couple the Brownian bridge to the Bessel bridges by building them over the familyB t,s of Brownian motions defined in (32): One can check directly that Furthermore, by comparing (54) to (33), it is immediate from the fact that coth u ≥ 1 for all . Therefore Also, as in (30), we can relate B s and B (t:0→0) s through a time change: and, as in (53), We define which is finite because of (59).

Simple properties of ψ t (y, x)
We recall from (27) that the main quantity we are interested in is with where we recall that ξ (t:y→x) s , s ∈ [0, t] is a Bessel bridge from y to x over time t. We mainly need to consider x = 0 so we use the shorthand ψ t (y) := ψ t (y, 0).
We also define where ξ (y) s , s ≥ 0 is a Bessel process started from y.
Proposition 8. The function ψ t (y, x) has the following properties: • It is bounded away from zero and infinity: there exist two positive constants 0 < K 1 < K 2 depending on the function m ′′ (s) such that for any x, y, t, • It hardly depends on x for large times: recalling that ψ t (y) := ψ t (y, 0), • For fixed y, it has a finite and positive limit as t → ∞: • The large time limit ψ ∞ (y) has a well-behaved large y limit: for any function t → y t that goes to infinity as t → ∞, Proof. For the first result, Lemma 7 tells us that where G > 0 is a random variable with Gaussian tail independent of t, y and x. Then, since m ′′ (s) = O(1/s 2 ), For the second result, we compare paths going to x with paths going to 0: we know from Lemma 5 that 0 ≤ ξ We now turn to the third result. For any fixed s and y, Lemma 6 tells us that ξ Then, by dominated convergence using again a uniform Gaussian bound from Lemma 7, It now remains to compute the right-hand-side. Let By integration by parts,

Proof of Theorem 1
Since ds where we recall that ∆ = 1 4 ∞ 0 ds δ ′ (s) 2 . We now fix x > 0, so that any terms written as O(f (t)) might depend on x; since x is fixed this will not matter. For instance, instead of (67) we simply write that ψ t (y, x) = ψ t (y)e O( log t t ) . We recall (28): Substituting in the estimate above we get Then since h(x, t) = ∞ 0 dy q(t, x, y)h 0 (y)-see (20)-we have with We now must choose v and δ(t), depending on the initial condition, such that (83) has a finite and non-zero limit as t → ∞. We use the following simple calculus lemma to evaluate H(x, t). We defer the proof to the end of this section.

Lemma 9. Let φ(y) a bounded function such that
for some A > 0 and some α. We now continue with the proof of Theorem 1. We distinguish two cases.
Case 1: h 0 (y) = O y ν e − v 2 y for some ν We introduce H 1 (t) such that xH 1 (t) is the same as H(x, t) with the sinh expanded to first order: For any z ≥ 0, by Taylor's theorem (with the Lagrange remainder), there exists w ∈ [0, z] such that 0 ≤ sinh(z) − z = z 3 6 cosh(w) ≤ z 3 6 e z . It follows that By applying Lemma 9 to φ(y) = h 0 (y) e v 2 y y 3 with α = ν + 3 we obtain We now apply Lemma 9 to H 1 (t) with α = ν + 1 and obtain if h 0 (y) ∼ Ay ν e − v 2 y with A > 0 and ν > −2, where we assumed that in the third case the right hand side is non-zero. As the difference (89) between H(x, t) and xH 1 (t) is always asymptotically small compared to the values in the right hand side of (90), it follows that (90) also gives the asymptotic behaviour of H(x, t).
We now plug this estimate of H(x, t) into (83). To prevent h m(t) + x, t from growing exponentially fast we need to take v = 2. Then δ(t) must be adjusted (up to a constant a) to kill the remaining time dependence. We find Case 2: h 0 (y) ∼ Ay ν e −γy with γ < v/2 We write h 0 (y) = g 0 (y)e −γy with g 0 (y) ∼ Ay ν so that (84) becomes The terms in the second exponential reach a maximum at y = λt with λ = v − 2γ. We make the change of variable y = λt + u √ t; after rearranging we have We bound each term in the integral with the goal of applying dominated convergence.
• As g 0 is bounded for small y and g 0 ∼ Ay ν for large y, we can takeÃ such that g 0 (y) ≤ A(y + 1) ν . Then (94) • We have the simple bound • ψ t (·) is bounded by Proposition 8.
We have bounded the integrand in (93) by a constant times exp(3u − u 2 /4) for t large enough, so we can apply dominated convergence. As t → ∞, the g 0 (·)/t ν term converges to Aλ ν , the sinh(·) term to sinh(λx/2), the ψ t (·) term to e ∆ and the exponential to e −u 2 /4 . We are left with some constants and the integral of e −u 2 /4 , which is √ 4π, and finally: In (83), this gives Recall that λ = v − 2γ. To avoid exponential growth, we need 1 − v 2 /4 + λ 2 /4 = 0, which implies v = γ + 1/γ with γ < 1 because we started with the assumption γ < v/2. As v−λ 2 = γ, to have convergence of h m(t) + x, t we need δ(t) to be of the form Writing the sinh(·) as the difference of two exponentials leads to 2 sinh(λx/2)e −vx/2 = e −γx − e −(1/γ)x ; we then recover case (a) of Theorem 1 with the claimed value of ω(x) and α. This completes the proof of Theorem 1, subject to proving Lemma 9.
For α > −1, cut the integral at y = 1. The integral from 0 to 1 is bounded, and in the integral from 1 to ∞ we make the substitution y = u √ t: A simple application of dominated convergence then leads to and the substitution t = u 2 /4 gives (86a). For α = −1, we cut the integral at y = √ t and again make the change of variable y = u √ t in the second part: Again by dominated convergence, the second integral has a limit; we simply write it as O(1). For the first, the integrand is bounded so the integral from 0 to 1 is certainly O(1), and we may concentrate on the integral from 1 to √ t. Making the substitution y = t x , we have √ t 1 dy φ(y)e − y 2 4t +ǫty ψ t (y) = (log t) The integrand on the right converges for each x ∈ (0, 1/2) to Ae ∆ so by dominated convergence, as required.

Estimating ψ t : finer bounds, and Proof of Theorem 2
We want to refine Proposition 8 and estimate the speed of convergence of ψ t (y, x) to its limit as t → ∞. As we are only interested up to errors of order log t t , it suffices to consider the case x = 0 since by (67), ψ t (y, x) = ψ t (y)e xO( log t t ) . Recall that where, introducing I t (y) := I t (y, 0), We have used the change of time (30) to give the second expression of I t (y). As in the hypothesis (12) of Theorem 2, we suppose that m is twice continuously differentiable and Our estimate of ψ t (y) is based on the following two propositions. By writing I t (y) = I(y) − (I(y) − I t (y)) in the definition of ψ t (y), and expanding the exponential in the small correction term I(y) − I t (y), we show that: Proposition 10. Assuming (106), the following holds uniformly in y: Further, some straightforward computations give that: Proposition 11. Assuming (106), the following holds uniformly in y: We prove Propositions 10 and 11 in Sections 6.2 and 6.3, after some preparatory work in Section 6.1. We now show how to prove Theorem 2 from these two propositions.
Proof of Theorem 2. We assume that m(t) satisfies the hypothesis (12) of Theorem 2: As in the proof of Theorem 1, we recall that h m(t) + x, t is related to H(x, t) through (83) and that H(x, t) is given by (84). With v = 2 and δ(t) = −(3/2) log(t + 1) + a + r(t), these two equations read: H(x, t) = ∞ 0 dy h 0 (y)e y 2t sinh xy 2t e −(3/2) log(t+1)+a+r(t) 2t We compute H(x, t) for an initial condition h 0 (x) = O x ν e −x for some ν < −2. In (87) in the proof of Theorem 1, we introduced H 1 (t) which is H(x, t)/x with the sinh replaced by its first order expansion: and we showed in (89) that the difference between H(x, t) and xH 1 (t) is very small. We continue to simplify the integral by introducing successive simplifications and by writing We now bound the successive differences in the above expression, as we did in (89), for the first one.
An application of Lemma 9 with φ(y) = h 0 (y)e y y 2 and hence α = ν + 2 then gives For the difference involving H 2 and H 3 , we use Propositions 10 and 11 which give that uniformly in y, We get = O 1 Indeed, the yO log t t gives the same correction as in (116) by another application of Lemma 9 with α = ν + 2. As dy h 0 (y) e y y < ∞ because ν < −2, the contribution of the yO log t t term subsumes the other O in (117) except in the case η < 1 2 . Finally, notice that H 4 < ∞ because we supposed ν < −2. Recalling ψ ∞ (y) ≤ K 2 , one has where we used h 0 (y)e y = O(y ν ). The end result comes from the integral from 0 to √ t; the other integral is always O(t 1+ν/2 ).
Finally, collecting the differences (89), (116), (118) and (119) leads with (114) to Substituting into (110) and expanding e −r(t) leads to the main expression (13) of Theorem 2, with the value α given in Theorem 1. We now turn to the second part of Theorem 2 and assume that h 0 (y) ∼ Ay ν e −y with −4 < ν < −2. We look for an estimate of H 4 − H 3 (t) which is more precise than (119).
Writing H 4 − H 3 (t) as a single integral and doing the change of variable y = u √ t one gets A simple application of dominated convergence then gives and (120) becomes This leads with (110) to (13).

Decorrelation between I(y) and ξ (y)
s A large part of our argument relies on a statement that roughly says "I(y) and ξ (y) s are almost independent for large s". The following proposition makes this precise.
The proof of this result is quite involved. The first step is to prove two fairly accurate estimates on the difference between two bridges with different end points, the first of which is best when the starting point y is large and the second of which is more accurate when y is small.
It is well-known that a Bessel process started from y and conditioned to be at position x at time t is equal in law to a Bessel bridge from y to x in time t followed by an independent Bessel process started from x at time t. We defined ξ (t:y→x) s for s ∈ [0, t] as a Bessel bridge from y to x in a time t. In this section, we extend the definition of ξ (t:y→x) s for s > t by interpreting it as an independent Bessel started from x at time t, so that ξ (t:y→x) s , s ≥ 0 is a Bessel process conditioned to be at x at time t. We assume that the Bessel processes attached to ξ (t:y→x) s for s ≥ t are built for all x and t with the same noise, so that we can compare them to each other. In particular, we apply (34) and (49) to these Bessel processes.
Recall that I(y) = 1 u − y and definẽ where we used that E e I(y) ξ (y) By the mean value theorem, |e a − e b | ≤ |a − b|e max(a,b) ≤ |a − b|e b+|a−b| . Thus where we applied (127) of Lemma 13 in the exponential. Now, by Cauchy-Schwarz, (151) DecomposeĨ s (y, x) in the following way: The first integral is 2I s (y, x). Using (50) it can be bounded uniformly in y, x and s by a variable with Gaussian tails. The second integral, which does not depend on y, can also be bounded uniformly in x and s using (49) by an independent variable with Gaussian tails. The third integral is (x − y)O log s s and the fourth is (x − y)O 1 s ; they can be bounded together by 2c|x − y| log(s+1) s for some constant c. Finally, there exists a C 1 and a c such that, uniformly in s, y and x: Substituting back into (151), we get |w(y, s)| ≤ C 1 e c|x−y| log(1+s) First we concentrate on showing the first line of (125), i.e. that |w(y, s)| ≤ C log(s + 1). Using (127) again, It remains to bound the expectations above. Note from (34) that for all z ≥ 0 we have 1 and therefore so, with Γ the positive random variable with Gaussian tail defined by we have, uniformly in z, so using Lemma 14 again we obtain |w(y, s)| ≤ C 1 + y log(s+1) √ s for some constant C, which is the second line of (125).
Finally we turn to the last line of (125) and bound the increments of w(y, s). Our approach is very similar to the above, conditioning on the value of ξ = E e I(y) − E(µ) (X − µ) , where Γ was defined in (158) and is a non-negative random variable with Gaussian tail. Therefore for some constant C provided δ ≤ s 2 , and one may check similarly to (153) that E(µ) is also bounded uniformly in y, s and δ. This establishes the last line of (125) and completes the proof.

Proof of Proposition 10
To prove Proposition 10 we proceed via three lemmas. We first write I t (y) = I(y)− I(y)−I t (y) , and show that the correction I(y) − I t (y) is small in the following sense: Lemma 15. Suppose that m is twice continuously differentiable and satisfies (106). Then there exist positive random variables G and G t with Gaussian tails, where all the G t have the same distribution, such that uniformly in y, I(y) = GO(1) and Unsurprisingly, for random variables with Gaussian tails we can make series expansions rather easily: Lemma 16. Let G and G t be positive random variables with Gaussian tails such that all the G t have the same distribution. Suppose that A t and B t are random variables such that where ǫ t ≥ 0 is a deterministic function with ǫ t → 0 as t → ∞. Then for any integer n ≥ 0, Taking n = 1, ǫ t = t −1/2 , A t = I(y) and B t = − I(y) − I t (y) , we find