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(1)

El e c t ro nic

Journ a l of

Pr

ob a b il i t y

Vol. 16 (2011), Paper no. 36, pages 1048–1064.

Journal URL

http://www.math.washington.edu/~ejpecp/

From the Pearcey to the Airy process

M. Adler M. Cafasso P. van Moerbeke§¶

Abstract

Putting dynamics into random matrix models leads to finitely many nonintersecting Brownian motions on the real line for the eigenvalues, as was discovered by Dyson. Applying scaling limits to the random matrix models, combined with Dyson’s dynamics, then leads to interesting, infinite-dimensional diffusions for the eigenvalues. This paper studies the relationship between two of the models, namely the Airy and Pearcey processes and more precisely shows how to approximate the multi-time statistics for the Pearcey process by the one of the Airy process with the help of a PDE governing the gap probabilities for the Pearcey process.

Key words:Airy process, Pearcey process, Dyson’s Brownian motions.

AMS 2010 Subject Classification:Primary 60K35; Secondary: 60B20.

Submitted to EJP on December 19, 2010, final version accepted May 1, 2011.

Department of Mathematics, Brandeis University, Waltham, Mass 02454, USA.adler@brandeis.edu

The support of a National Science Foundation grant # DMS-07- 04271 is gratefully acknowledged

Centre de Recherches Mathématiques, Université de Montréal C. P. 6128, succ. centre ville, Montréal, Québec, Canada H3C 3J7 and Department of Mathematics and Statistics, Concordia University 1455 de Maisonneuve W., Montréal, Québec, Canada H3G 1M8.cafasso@crm.umontreal.ca

§Département de Mathématiques, Université Catholique de Louvain, 1348 Louvain-la-Neuve, Belgium and Brandeis Uni- versity, Waltham, Mass 02454, USA.vanmoerbeke@math.ucl.ac.be

The support of a National Science Foundation grant # DMS-07-04271, a European Science Foundation grant (MIS- GAM), a Marie Curie Grant (ENIGMA), Nato, FNRS and Francqui Foundation grants is gratefully acknowledged.

(2)

1 Introduction

Putting dynamics into random matrix models leads to finitely many nonintersecting Brownian mo- tions onRfor the eigenvalues, as was discovered by Dyson[13]. Applying scaling limits to the ran- dom matrix models, combined with Dyson’s dynamics, then leads to interesting, infinite-dimensional diffusions for the eigenvalues. This paper studies the relationship between two of the models, namely the Airy and Pearcey processes and more precisely shows how to approximate the Pearcey process by the Airy process with the help of a PDE[2]governing the gap probabilities for the Pearcey process. The Airy process was introduced by Prähofer-Spohn[18]and further developed by K. Jo- hansson[14, 15]. A simple non-linear 3rd order PDE for the transition probabilities for this process was found in[3]; see also[19, 20]. The Pearcey process was introduced in[21, 17]in the context of non-intersecting Brownian motions and plane partitions, also based on prior work on matrix models with external source[16, 10, 22, 23, 11, 12, 5, 7, 8, 9].

Considernnonintersecting Brownian particles on the real lineR,

−∞< x1(t)<...<xn(t)<∞ with (local) Brownian transition probability given by

p(t;x,y):= 1

pπte(yx)

2

t ,

all starting from the originx =0 at timet=0 and such that n

2 particles are forced to±pn

2 att=1.

For very large n, the average mean density of particles has its support, for t12, on one interval centered about x =0, and for 12 < t ≤1 on two intervals symmetrically located about x =0. The end points of the interval(s) of support describe a heart-shaped region in(x,t)-space, with a cusp at (x0,t0) = (0, 1/2). The Pearcey process P(τ) is defined (see Figure 1) as the motion of these nonintersecting Brownian motions for largen, about(x0,t0) = (0, 1/2) (i.e., near the cusp), with space microscopically rescaled by a factor ofn1/4and time rescaled by a factorn1/2, in tune with the Brownian motion rescaling. A partial differential equation for the Pearcey process was found by Adler-van Moerbeke[4]and in[2]a much better version was obtained, namely a simple third order non-linear PDE for the transition probabilities; it was obtained by a scaling limit on a PDE for non-intersecting Brownian motions with target points. This PDE is related to the Boussinesq equation and its hierarchy; this is part of a general result on integrable kernels, as explained in the paper[1].

Near the boundary of the heart-shaped region of Figure 1, but away from the cusp, the local fluctu- ations behave as the so-called Airy process, which describe the non-intersecting Brownian motions with space stretched by the customary GUE edge rescalingn1/6and time rescaled by the factorn1/3, again in tune with the Brownian motion space-time rescaling.

This paper shows how the Pearcey process statistics tends to the Airy process statistics when one is moving out of the cuspx = 272 (3(t−t0))3/2 very near the boundary, that is at a distance of(3τ)1/6 forτvery large, withτ being the Pearcey time. To be precise, in the two-time case, the timesτi

must be sufficiently near -in a very precise way- for the limit to hold. The main result of the paper can be summarized as follows:

(3)

Theorem 1.1. Given finite parameters t1<t2, let bothτ1, τ2→ ∞, such thatτ2τ1→ ∞behaves in the following precise way:

τ2τ1

2(t2t1) = (3τ1)1/3+ t2t1

(3τ1)1/3 +2t1t2 3τ1

+O τ15/3

. (1)

Let also E1,E2 be two arbitrary finite intervals. The parameters t1 and t2 provide the Airy times in the following approximation of the Airy process by the Pearcey process:

P

2

\

i=1

(P(τi)−272(3τi)3/2

(3τi)1/6 ∩ −Ei

=; )!

=P

2

\

i=1

A(ti)∩(−Ei) =;

!

1+O 1 τ4/31

! .

(2)

The same estimate holds as well for the one-time case. A similar (but different) result was then obtained in[6]for the one–time case in the situation when one is moving out of the cusp following both the two branches of the cusp simultaneously. It is also worth recalling that, in this paper as well as in[6], the statistics of the processes are studied on intervals rather than on general Borel subsets.

The proof of this Theorem proceeds in two steps. At first, Proposition 2.1, stated later, shows that, using the scaling above, the Pearcey kernel tends to the Airy kernel. In a second step, we show, using the PDE for the Pearcey process[2]and Proposition 2.1, the result of Theorem 1.1.

It is a fact that both the Airy and Pearcey processes are determinantal processes, for which the multi- time gap probabilities is given by the matrix Fredholm determinant of the matrix kernel, which will be described here.

TheAiry processis a determinantal process, for which the multi-time gap probabilities1 are given by the matrix Fredholm determinant fort1<. . .<t`,

P

`

\

i=1

(A(ti)∩Ei=;)

=det

I−[χEiKtA

itjχEj]

1i,j≤` (3)

of the matrix kernel in~x = (x1, . . . ,x`)and~y= (y1, . . . ,y`), denoted as follows:

KAt1,...,t`(~x,~y)p d~x d~y

:=

KtA

1,t1(x1,y1)p

d x1d y1 . . . KtA

1,t`(x1,y`)p

d x1d y`

... ...

KtA

`,t1(x`,y1)p

d x`d y1 . . . KtA

`,t`(x`,y`)p

d x`d y`

 ,

(4)

where, for arbitrary ti andtj, the extended Airy kernelKAt

i,tj(x,y)is given by (see Johansson[14, 15])

KtA

i,tj(x,y):=K˜At

i,tj(x,y)−1I(ti <tj)pA(tjti,x,y), (5)

1χEis the indicator function for the setE

(4)

# of particles

t=1/2 t=1

x t

n/2 n/2

-p

n/2 p

n/2

Figure 1: The Pearcey process.

where

K˜tA

i,tj(x,y):= Z

0

e−λ(ti−tj)A(x+λ)A(y+λ)dλ

= 1

(2πi)2 Z

Γ>

d v Z

Γ<

du e−v3/3+y v eu3/3+xu

1

(v+tj)−(u+ti)

pA(t,x,y):= 1 p4πtet

3 12(xy)2

4t 2t(y+x), for t>0.

The contouru∈Γ< consists of two rays emanating from the origin with anglesθ1 andθ10 with the positive real axis, and the contourv ∈Γ> also consists of two rays with anglesθ2 andθ20 with the negative real axis, as indicated in Figure 2. As is well known, one may chooseθ1,θ2, θ10,θ20=π/3.

In particular, forti=tj, this is the customary Airy kernel KAt

i,ti :=KA(x,y):=

Z

0

dλA(x+λ)A(y+λ) = A(x)A0(y)−A(y)A0(x)

xy . (6)

ThePearcey processis determinantal as well, for which the multi-time gap probability is given by the Fredholm determinant forτ1<. . .< τ`,

P

`

\

i=1

(P(τi)∩Ei =;)

=det

1I−[χEiKτP

iτjχEj]

1≤i,j≤` (7)

(5)

for thePearcey matrix kernelin = (ξ1, . . . ,ξ`)and= (η1, . . . ,η`), denoted as follows:

KPτ1,...,τ`(~ξ,~η)Æ d~ξd~η

:=

KτP

1,τ11,η1)p

11 . . . KτP

1,τ`1,η`)p

1`

... ...

KτP

`,τ1`,η1)p

`1 . . . KτP

`,τ``,η`)p

``

 ,

(8)

where for arbitraryτi andτj, (see Tracy-Widom[21]) KτP

i,τj(ξ,η) =K˜τP

i,τj(ξ,η)−1I(τi< τj)pPjτi;ξ,η), (9) with

K˜τP

i,τj(ξ,η):=− 1

2 Z

X

d U Z

Y

d V VU

eV

4

4+τj V22Vη

eU

4

4+τi U22−Uξ

pP(τ;ξ,η):= 1

p2πτe(ξ−η)

2

, forτ >0,

where X and Y are the following contours: X =X1X2 consists of four rays emanating from the origin with anglesσ1,σ01with the positive real axis andσ2,σ02with the negative real axis, as given in Figure 2. The contourY consists of two rays emanating from the origin with anglesτ,τ0with the negative real axis; it is customary to pickσ1=σ2=σ10 =σ02=π/4 andτ=τ0=π/2.

p I

R

O

@

@

@

@

@

E E E E E E E EE

UX2

UX2

σ02 σ2

τ0 τ

UX1 UX1

VY VY

σ01 σ1

@

@

@

@

@

π/8< σ12,σ0102<3π/8 3π/8< τ,τ0<5π/8

Pearcey contour

p M

N A A A A A A

v∈Γ>

v∈Γ>

θ20 θ2

u∈Γ<

u∈Γ<

θ10 θ1

A A

A A

A A

π/6< θ1210,θ20< π/2

Airy contour Figure 2: Integration paths for the kernels.

(6)

In[4, 2], it was shown that, given intervals

Ei = (ξ(i)1 ,ξ(i)2 ), the log of the probability

Q=Q1, . . . ,τ`;E1, . . . ,E`):=logP

`

\

i=1

{P(τi)∩Ei=;}

satisfies a non-linear PDE, which we describe here. Given the times and the intervals above, one defines the following operators:

τ:=X

i

∂ τi

, Ei :=X

k

∂ ξ(i)k , E:=X

i

Ei

"τ:=X

i

τi

∂ τi

, "E:=X

i

X

k

ξ(i)k

∂ ξ(i)k . (10)

ThenQ satisfies the Pearcey partial differential equation in its arguments and the boundary of the intervals2:

2τ3Q+1

4(2"τ+"E−2)∂E2Q−(X

i

τiEi)∂τEQ

τEQ,∂E2Q©

∂E =0. (11)

2 From the Pearcey to the Airy kernel

Define the rational functions

Φ(x,t;u):=− 1

4(3u)3t

(3u)2 + xt2 3u +4

3t x+u 6t2x h(x,t;u):= ux

4 (x+6t2),

(12)

and the diagonal matrix S:=

‚ eΦ(x,t1;z4)−h(x,t1;z4) 0

0 eΦ(x,t2;z4)−h(x,t2;z4)

Π. We now state:

Proposition 2.1. Given a parameter z → 0, define two times τ1 and τ2 blowing up like z6 and depending on two parameters t1and t2,

τi= 1

3z6(1+6tiz4) +O(z10). (13)

2Here and below, given two arbitrary functionsf,gand a vector fieldX, we denote{f,g}X:=g X(f)f X(g)

(7)

Given largeξi andηj, define new space variables xi and yj, usingτi above, ξi= 2

27(3τi)3/2−(3τi)1/6xi ηj= 2

27(3τj)3/2−(3τj)1/6yj.

(14)

With this space-time rescaling, the following asymptotic expansion holds for the Pearcey kernel in powers of z4, with polynomial coefficients in t1+t2,

SKPτ12(~ξ,~η)Æ

d~ξd~ηS−1

1+ (t1+t2)O1(z4

KAt1,t2(~x,~y

~d x~d y+O(z8), (15) where O1 refers to a differential operator in∂ /∂x,∂ /∂y, with polynomial coefficients in x,y,t1±t2 and(t1t2)1 and varying with the four matrix entries of the matrix kernel.

On general grounds, due to the Fredholm determinant formula, this would give Theorem 1.1, but with a much poorer estimate in (2), namely instead ofO−4/31 ), the estimate would beO−2/31 ). The existence of the PDE for the Fredholm determinant of the Pearcey process enables us to boot- strap theO(τ12/3)estimate toO(τ14/3), without tears!

Also note that conjugating a kernel does not change its Fredholm determinant. Before proving Proposition 2.1, we shall need the following identities:

Lemma 2.2. Introducing the polynomial Ψ(x,s;ω):= 1

4ω4+3

2ω2s2−4s(ω3), (16) the Airy kernel (6) and the (double) integral part of the extended Airy kernel (5), at t1 =s, t2 =−s, satisfy the following differential equations,

Ψ(x,s,−∂x)−Ψ(y,s,∂y) +4s

KA(x,y)

= 1

4(xy)(x+y+6s2)KA(x,y),

(17)

and

Ψ(x,s,−∂x)−Ψ(y,−s,∂y)−3 2s∂

∂s(∂xy)

K˜A

s,−s(x,y)

= 1

4(x−y)(x+ y+6s2)K˜A

s,−s(x,y).

(18)

Proof. The operator on the left hand side of (17) reads Ψ((x,s,−∂x)−Ψ((y,s,∂y) +4s

= 4s(1+y∂yy3+x∂xx3) +3

2s2(∂x2y2) +1

4(∂x4y4). (19) Using the first representation (6) of the Airy kernel, one checks using the differential equation for the Airy kernel, A00(x) = xA(x) and thusA000(x) = xA0(x) +A(x) andA(i v)(x) =2A0(x) +x2A(x),

(8)

and using differentiation by parts to establish the last equality, (y∂yy3) + (x∂xx3)

KA(x,y)

= −

Z

0

dz

z d dz +2

A(x+z)A(y+z) =−KA(x,y).

In order to take care of the other pieces in (19), one uses the second representation (6) of the kernel KA(x,y), yielding

x2y2

KA(x,y) = (x−y)KA(x,y)

x4y4

KA(x,y) = (x2y2)KA(x,y).

This establishes the first identity (17) of Lemma 2.2. The operator on the left hand side of the second identity (18) reads:

Ψ(x,s,−∂x)−Ψ(y,s,∂y)−3 2s∂

∂s(∂xy)

= 1

4(∂x4−∂y4) +3

2s2(∂x2−∂y2) +4s(xxx3y∂y+y3)−3 2s∂

∂s(∂x−∂y),

of which we will evaluate all the different terms acting on the integral. Notice at first that, since the expression under the differentiation vanishes at 0 and∞, one has

0= Z

0

dz

∂z

€ze2szKA(x+z,y+z)Š

= Z

0

dz e2szKA(x+z,y+z)− Z

0

zdz e2szA(x+z)A(y+z)

−2s Z

0

zdz e−2szKA(x+z,y+z),

(20)

Again, using the second representation (6) of the kernelKA(x,y), one checks

xy

K˜A

s,−s(x,y) =−(xy) Z

0

dz e2szKA(x+z,y+z)

x2y2K˜A

s,−s(x,y) = (x−y)K˜A

s,−s(x,y), and also

s∂

∂s x−∂y

K˜A

s,−s(x,y) = 2s(xy)

Z

0

zdz e2szKA(x+z,y+z) and, using the differential equation x∂xx3

A(x) =−A(x)and (20),

−2s

x∂xx3

y∂yy3 K˜A

s,s(x,y)

= −2s(x−y) Z

0

zdz e2szKA(x+z,y+z)

(9)

UsingA(i v)(x) =2A0(x) +x2A(x)and the Darboux-Christoffel representation (6) of the Airy kernel, one checks

x4y4K˜A

s,−s(x,y)

= Z

0

e2sz(A(i v)(x+z)A(y+z)A(x+z)A(i v)(y+z))dz

= −2(xy) Z

0

dz e2szKA(x+z,y+z) + (x2y2)K˜A

s,s(x,y) +2(x−y)

Z

0

zdz e−2szA(x+z)A(y+z). Adding all these different pieces and using the expression for

−2s(xy) Z

0

zdz e2szKA(x+z,y+z), given by (20), leads to the statement of Lemma 2.2.

For the sake of notational convenience in the proof below, set

KPτ12(~ξ,~η)Æ

d~ξd~η=

K11P1,η1)p

11 K12P1,η2)p

12

K21P2,η1)p

21 K22P2,η2)p

22

(21)

and similarly for the Airy kernel KAt1t2; the ˜Ki jP and ˜Ki jA refer, as before, to the (double) integral part.

Proof of Proposition 2.1: Notice that acting withx andy on the kernelsKeA

11 22

(x,y)andKeA

12 21

(x,y), amounts to multiplication of the integrand of the kernels with−uandvrespectively; i.e.,u↔ −∂x

andvy.

Given the (small) parameter z ∈R, consider the following (i,j)-dependent change of integration variables(U,V)7→(u,v)in the four Pearcey kernelsKi jP in (21), namely

U= 1

3z3(1+3uz4)(1+3tiz4), V = 1

3z3(1+3vz4)(1+3tjz4), (22) together with the changes of variables (ξ,η,τi,τj) 7→ (x,y,ti,tj), in accordance with (13) and (14),

τi= 1

3z6(1+6tiz4) +O(z10), τj= 1

3z6(1+6tjz4) +O(z10) ξ= 2

27(3τi)3/2−(3τi)1/6x, η= 2

27(3τj)3/2−(3τj)1/6y.

(23)

To be precise, forKkkP, one sets in the transformations above i= j = k, forK12P andK21P, one sets i=1, j=2 and i=2, j=1 respectively. Then, remembering the expressionsΦandΨ, defined in

(10)

(12) and (16), the following estimate holds for smallz:

e−Φ(y,tj;z4) e−Φ(x,ti;z4)

eV

4

4+τj V22Vη

eU

4

4 +τi U22−Uξ = ez4Ψ(y,tj;v) e−z4Ψ(x,ti;u)

ev

3 3+y v

eu

3

3+xu(1+tiO(z8)+tjO(z8))

= e−z4Ψ(y,tj;∂y) ez4Ψ(x,ti;−∂x)

ev

3 3+y v

eu

3

3+xu(1+tiO(z8)+tjO(z8));

(24)

replacing in the latter expression u and v by differentiations x and y has the advantage that upon doubly integrating inu and v, the fraction of exponentials in front can be taken out of the integration. Moreover, settingt1=t+sandt2=ts, one also checks3:

e−Φ(y,tj;z4)

e−Φ(x,ti;z4)pPjτi;ξ,η)p dξdη

=pA(−2s,x,y)p d x d y

‚ 1+z4

4

(x−y)(x+y+6s2) +t

sr(x,y,s)

+O(z8)

Π.

(25)

The multiplication by the quotient of the exponentials on the left hand side of the expressions above will amount to a conjugation of the kernel, which will not change the Fredholm determinant. Setting t1=t+sandt2=ts, witht=0, one finds for the non-exponential part in the kernel

pdξdηd U d V VU

=

pd x d y dud v(1+72(ti+tj)z4)

v+tjuti+3z4(v tjuti) +O(z8)

=









pd x d y dud v v−u

€1±4sz4+O(z8)Š for

¨ i=1, j=1 i=2, j=2 pd x d y dud v

vu2s

3zv4s(u+v)u2s +O(z8) for

¨ i=1, j=2 i=2, j=1.

(26) Along the same vein as the remark in the beginning of the proof of this Theorem, one notices that multiplication of the integrand of the kernel ˜KA

12 21

(x,y)with the fraction, appearing in the last formula of (26), amounts to an appropriate differentiation, to wit:

± 3s(u+v)

vu∓2s ←→3 2s∂

∂s(∂yx). (27)

3The precise expression forr(x,y,s):= (xy)28s2(x+y2s2)will be irrelevant in the sequel.

(11)

So we have, using (24), (25), (26), and the differential equation forKA of Lemma 2.2, e−Φ(y,t12;z

4)

e−Φ(x,t12;z

4) KP11

22

(ξ,η)p dξdη

t=0

KA(x,y)p d x d y

=z4€

±4s+ Ψ((x,±s,−∂x)−Ψ((y,±s,∂y

KA(x,y)p

d x d y+O(z8)

= z4

4 (x−y)(x+y+6s2)KA(x,y)p

d x d y+O(z8);

(28)

the upper(lower)-indices correspond to the upper(lower)-signs. Using (27), and the differential equation for ˜KA

12 21

of Lemma 2.2,

e−Φ(y,t21;z

4)

e−Φ(x,t12;z

4) K˜P

12 21

(ξ,η)p

dξdηK˜A

12 21

(x,y)p d x d y

t=0

=z43 2s∂

∂s(∂y−∂x) + Ψ(xs,−∂x)−Ψ(y,s,∂y) K˜A

12 21

(x,y)p d x d y

t=0

+O(z8)

= z4

4 (x−y)(x+y+6s2)K˜A12

21

(x,y)p d x d y

t=0

+O(z8).

(29)

Also, using (25),

e−Φ(y,t2;z4)

e−Φ(x,t1;z4)pP2τ1;ξ,η)p

dξdηpA(−2s,x,y)p d x d y

t=0

= z4

4(x−y)(x+y+6s2)pA(−2s,x,y)p

d x d y+O(z8).

(30)

Sinceh(y,ti,z4)−h(x,tj;z4) =−z44(xy)(x+y+6s2)for arbitraryti,j=t±satt=0, and thus eh(y,ti;z4)

e−h(x,tj;z4)=1+z4

4(xy)(x+y+6s2) +O(z8) (31) Then the following approximations follow upon combining the three estimates above (28), (29), (30) and using (31):

e−Φ(y,t12;z

4)+h(y,t1 2

;z4)

e−Φ(x,t12;z

4)+h(x,t1 2

;z4) KP

11 22

(ξ,η)p dξdη

t=0

KA(x,y)p

d x d y=O(z8)

e−Φ(y,t21;z

4)+h(y,t2 1

;z4)

e−Φ(x,t12;z

4)+h(x,t1 2

;z4) K˜P

12 21

(ξ,η)p dξdη

t=0

K˜A

12 21

(x,y)p

d x d y=O(z8) e−Φ(y,t2;z4)+h(y,t2;z4)

e−Φ(x,t1;z4)+h(x,t1;z4)pP2−τ1;ξ,η)p dξdη

t=0pA(−2s,x,y)p

d x d y=O(z8).

(12)

The reader is reminded that this estimate sofar is done att =0. It then follows for t 6=0 from the estimates (22), (23), (24), (25), (26), (27) that the right hand side of (15) is an asymptotic series inz4 with polynomial coefficients int1+t2=2t, proving the claim aboutO1.

So far an important point was omitted, namely to analyze how the Pearcey contour turns in the limit into an Airy contour; a detailed description of the possible contours was given in Figure 2.

The Pearcey-rays X1, with angles σ1,σ01 ∈(π/8, 3π/8)can be deformed into two acceptable rays θ1,θ10 ∈(π/6,π/2)for theΓ<-Airy contour, since the two intervals have a non-empty intersection.

Also, since(3π/8, 5π/8)∩(π/6,π/2)6=;, the PearceyY-contour can be deformed into an acceptable v ∈Γ>-Airy contour. Again, since the admissible interval (π/8, 3π/8) for σ2 and σ02 in the X2- Pearcey contour containsπ/6, one ends up in the limit integrating the functioneu3/3along a contour of the formΓ>, which we may choose to have an angle of π/6−δ with the negative real axis, for small arbitrary δ > 0. In the sector Γ>, with 0< θ2 = θ20π/6δ, the function eu3/3 decays exponentially fast. Therefore, the contribution, due to the u-integration of eu3/3×(lower order terms)overΓ>, having an angle ofπ/6−δwith the negative real axis, vanishes by applying Cauchy’s Theorem in that sector. This ends the proof of Proposition 2.1.

3 From the Pearcey to the Airy statistics

This section concerns itself with proving Theorem 1.1.

Proof of Theorem 1.1: For any intervalsE1 andE2, the function Q(τ1,τ2;E1,E2) =logP

2

\

i=1

(P(τi)∩Ei=;)

!

(32) satisfies the Pearcey PDE (11). Reparametrizing, without loss of generality,

τ1=τ+σ, τ2=τσ, E1= (ξ+η+µ,ξ+ηµ), E2= (ξ−η+ν,ξην), (33) leads to a manageable PDE for the function

F(τ,σ;ξ,η,µ,ν):=Q(τ1,τ2;E1,E2), (34) namely the PDE:

23F

∂ τ3 +1 4

2(σ

∂ σ−τ

∂ τ) +ξ

∂ ξ+η

∂ η +µ

∂ µ+ν

∂ ν−2 2F

∂ ξ2

σ 3F

∂ τ∂ ξ∂ η+

¨ 2F

∂ τ∂ ξ,2F

∂ ξ2

«

ξ

=0.

(35)

To go from the Pearcey PDE (11) to the PDE (35), one notices that the operators (10), appearing in the Pearcey PDE (11), have simple expressions in terms of the variablesτ,σ,ξ,η,µ,ν,

τQ= ∂F

∂ τ, EQ= ∂F

∂ ξ, "EQ= ξ

∂ ξ+η

∂ η+µ

∂ µ+ν

∂ ν F,

(13)

"τQ:= σ

∂ σ+τ

∂ τ

F, X

i

τiEiQ= τ

∂ ξ+σ

∂ η F.

The change of variables, considered in (13) and (14), combined with a linear change of variables, in parallel with (33),

τi= 1

3z6(1+6tiz4) with

¨ t1=t+s t2=ts Ei= 2

27(3τi)3/2−(3τi)1/6E˜i with

¨ E˜1= (x+y+u,x+ yu) E˜2= (x−y+v,xyv)

(36)

yields az-dependent invertible map,

T : (t,s,x,y,u,v)7→(τ,σ,ξ,η,µ,ν), (37) and thus the functionF in (34) leads to a newz-dependent functionG:

F(τ,σ;ξ,η,µ,ν) =F(T(t,s,x,y,u,v)) =:G(t,s,x,y,u,v),

of which we compute, in principle, the series inz. From Proposition 2.1, it follows that thez4-term in the asymptotic expansion in (15) vanishes whent=0; so, omittingp

d x d y, one has SKPτ1,τ2S1=KAt1,t2+z4K1+

X

2

z4iKi, withK1=tH, for some kernelH, with theKi polynomial in t. Then defining

Li:= (I−KAt1,t2)1Ki, (38) one finds:

G(t,s,x,y,u,v)

=F(τ,σ;ξ,η,µ,ν)

=logP \2

i=1

{P(τi)∩Ei =;}

=log det(I−KPτ1,τ2)EE

2

=log det(I−KAt1,t2)E˜E˜2z4TrL1z8Tr(L2+1

2L21)−. . .

=:G0(s; ˜E1, ˜E2)−G1(t,s; ˜E1, ˜E2)z4G2(t,s; ˜E1, ˜E2)z8+O(z12),

(39)

where

G0(s; ˜E1, ˜E2) =log det(I−KAt1,t2)˜EE˜2,

G1(t,s; ˜E1, ˜E2) =tTr((I−KAt1,t2)−1H)E˜E˜

2.

(40)

NoteG0(s; ˜E1, ˜E2)ist-independent, since the Airy process is stationary. Then setting F(τ,σ;ξ,η,µ,ν) =G(T1(τ,σ;ξ,η,µ,ν)) =G(t,s,x,y,u,v)

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