The logarithmic anti-derivative of the Baik-Rains distribution satisﬁes the KP equation

It has been discovered that the Kadomtsev-Petviashvili (KP) equation governs the distribution of the ﬂuctuation of many random growth models. In particular, the Tracy-Widom distributions appear as special self-similar solutions of the KP equation. We prove that the anti-derivative of the Baik-Rains distribution, which governs the ﬂuctuation of the models in the KPZ universality class starting with stationary initial data, satisﬁes the KP equation. The result is ﬁrst derived formally by taking a limit of the generating function of the KPZ equation, which satisﬁes the KP equation. Then we prove it directly using the explicit Painlevé II formulation of the Baik-Rains distribution. the long-time ﬂuctuations of models which belong to the KPZ universality class. A 1 ( x ) is a stationary process, whose one-point distribution is the Tracy-Widom GOE distribution. The one point marginal of A 2 ( x ) is given by the Tracy-Widom GUE distribution. The one point marginal of A stat ( x ) is given by the Baik-Rains distribution.


Introduction
The fluctuations for the KPZ universality class depend on the initial data. Let h(x, t) be the solution of the Kardar-Parisi-Zhang equation, ∂ t h(x, t) = 1 2 (∂ x h(x, t)) 2 + 1 2 ∂ 2 x h(x, t) + ξ(x, t). (1.1) Here ξ(x, t) is space-time Gaussian white noise, E(ξ(x, t)ξ(y, s)) = δ(x − y)δ(t − s). (1. 2) The equation is ill-posed because the quadratic non-linear term does not make sense for a realization of a solution. A typical solution h(x, t) looks like a Brownian motion in x variable. One way to make sense of the equation is through the Hopf-Cole transformation.
In the paper [QR20], it was stated that the GUE and GOE Tracy-Widom distributions are seen to arise as special similarity solutions of the scalar Kadomtsev-Petviashvili (KP) equation, (1.8) In this paper, we explain how the Baik-Rains distribution can be seen as a similarity solution of the KP equation. We first explain how the GUE and GOE distributions arise as similarity solutions of the KP equation, then give the definition of the Baik-Rains distribution and then state our main results.
Example 1.1. Tracy-Widom GUE distribution [QR20]: We consider a self-similar solution of (1.8) in the following form, (1.9) This turns (1.8) to ψ + 12ψψ − 4rψ − 2ψ = 0. (1.10) If we look for solutions of the form ψ = −q 2 , then the above equation becomes the Painlevé II equation, q = rq + 2q 3 . (1.11) In this way, we recover the GUE distribution, since F GUE (r) = exp{− −∞ r du(u − r)q 2 (u)}. φ(t, r) = (t/4) −2/3 ψ((t/4) −1/3 r). (1.12) This turns (1.8) to ψ + 12ψ ψ − rψ − 2ψ = 0. (1.13) If we look for solutions of the form ψ = 1 2 (q − q 2 ), we get the Painlevé II equation again, thus we recover the GOE distribution, (1.14) Now let us look at the definition of the Baik-Rains distribution. Here we present two definitions which turn out to be equivalent. One definition appears in [BCFV15], as the large time fluctuation of the stationary KPZ equation; the other definition will be the main tool to prove our results. Definition 1.3. [FS06] For w, s ∈ R, we define the following functions, where P 0 (x) = I x≥0 is the projection operator, Ai(x) is the Airy function, and K Ai,s is the shifted Airy kernel, K Ai,s (x, y) = ∞ 0 dλAi(x + λ + s)Ai(y + λ + s). (1.16) The Tracy-Widom GUE distribution can be written as Then we define the function g(s, w) which appears as a component in the Baik-Rains distribution, Finally, the Baik-Rains distribution is defined to be F τ (r) = ∂ ∂r (g(r + τ 2 , τ )F GUE (r + τ 2 )). (1.19) The function g(s, w) can also be derived by solving the PNG model using the Riemann-Hilbert technique. We also present this equivalent definition here [BR00]. Let u(x) be the solution of the Painlevé II equation, with the boundary condition, Then the Tracy-Widom distributions can be defined in terms of u and v. Set (1.23) Then we can define the GUE and GOE distributions as (1.24) (1.26) Here functions a(x; w) and b(x; w) arise in the Painlevé II Riemann-Hilbert problem.
In this paper, we do not need the exact definition of a(x; w), b(x; w), thus we skip the formulation of the Riemann-Hilbert problem here. What we need are the following identities [BR00], (1.27) Remark 1.4. It is proven in the appendix A of [FS06] that y(s + w 2 , w /2) = g(s + w 2 , w) when w = 2w, which establishes the equivalence of the two definitions presented above.
We first explain where the result comes from in Section 2. The Baik-Rains distribution governs the fluctuation of the stationary KPZ equation in the large time limit. The generating function of the KPZ equation starting with two sided Brownian motion with drifts satisfies the KP equation. One way of seeing this is by checking the cumulants [Dou20], which is formally derived using Bethe ansatz method; another way is by checking its determinantal formula [QR20]. Then we conjecture that the same equation should still be satisfied when the drift goes to zero and time goes to infinity, which gives Theorem 1.5. This is only formal since we assume that if a sequence of functions satisfy an equation, its limit also satisfies the same equation. In section 3, we will prove Theorem 1.5 by directly differentiating y(t −1/3 r + t −4/3 x 2 , t −2/3 x/2). Using the identities in (1.27), we find that equation (1.30) holds.

Motivation for the result
A key fact we will use is the following theorem, (2.1) Suppose in addition that det(I − K) > 0 for all finite t, x, r, and K is real analytic in t, x and r, and that the trace norm K 1 < 1 for r in some open real interval. Then in that interval, φ(x, t, r) = ∂ 2 r log F (x, t, r) satisfies the scalar KP equation, Here we check it using a different method, which agrees with the results in [Dou20]. We begin with the following theorem.
where K ν (z) is the modified Bessel function of order ν and the kernel on the right-hand side is given by where σ = (2/t) 1/3 . (2.5) The integration contour for w is from − 1 4σ − i∞ to − 1 4σ + i∞ and crosses the real axis between b and β. The other contour for z goes from 1 4σ − i∞ to 1 4σ + i∞, it also crosses the real axis between b and β and it does not intersect the contour for w.
In order to get the formula for stationary initial data, we need to take the limit as β → b and set b = 0. To do so, we need to rewrite the kernel K b,β . Two contours in the integral kernel of K b,β intersect the real axis between the pole at b/σ and β/σ, so when β → b, two contours will collide. We move the contour of w cross the pole at b/σ and move the contour of z cross the pole at β/σ. Using the residue theorem, we have Notice that the only difference between K b,β andK b,β is that they have different contours. We can write with suitable f i , g i . Then for the Fredholm determinant, we have the following formula, (2.9) Both kernels K b,β (x, y) andK b,β (x, y) depend on S. For S = e τ 2 +σr , where τ is related to x, b, t as x = bt + 2τ σ 2 and when b = 0, we observe that both K b+ x clear from the later context; this is the scaling that gives the Baik-Rains distribution. We will do some transformations on the integral kernel so that equations (2.1) become obvious while the determinant of the operator remains unchanged. When S = e τ 2 +σr , . (2.10) ECP 27 (2022), paper 30.
(2.11) Let u → 1 σ u, v → 1 σ v and conjugate the kernel by e ux/t (which is equivalent to multiplying the kernel by e (u−v)x/t ). It becomes When b = 0, the only term contains x, t, r, u, v is e (tz 3 +3z 2 x)/6−z(v+r) e (tw 3 +3w 2 x)/6−w(u+r) , which clearly satisfies equations (2.1). When b = 0, it will have extra terms when differentiating t, x coming from e −(z−w)(b 2 t/2+bx) , which fail to satisfy equations (2.1). The reason we can directly take derivatives under the integral sign and the kernel being analytic in t, x, r come from the following lemma, Lemma 2.4. [BCFV15]Let f (z, ζ) be a complex function in two variables and suppose that 1. f is defined on (z, ζ) ∈ A × C where A is an open set and C is a contour, 2. For each z ∈ A, define the contour γ z = {z + re it : 0 ≤ t ≤ 2π} with a sufficiently small r such that also the disc around z with radius r lies in A. Suppose that for 3. For each ζ ∈ C, z → f (z, ζ) is analytic in A, 4. For each z ∈ A, ζ → f (z, ζ) is continuous on C. Then is analytic in A with F (z) = C ∂ ∂z f (z, ζ)dζ.
It can be easily seen from the form of the kernel that conditions (1), (3), (4) are satisfied. Since e z 3 /3 decays along C z as e −c|Im(z)| 2 , e w 3 /3 decays along C w as e −c|Im(w)| 2 , using the gamma ratio formula, as |z| → ∞, we have [BCFV15] Γ(β − σz) (2.15) Similarly we have the same bound for large w. Thus if we integrate β on some finite contour, we have the same polynomial bounds. Thus the whole integrand decays exponentially on the contour, so condition (2) is satisfied.
HereK b,β also satisfies equations (2.1) if S = e τ 2 +σr , because the only difference betweenK b,β and K b,β is that they have different contours, for which Lemma 2.4 still applies. We denote φ b,β = ∂ 2 r log det( (2.16) The Baik-Rains distribution and the KP equation Combining these two equations, we obtain an equation for α involvingφ b,β , This is a similar equation to (1.30), except thatφ b,β is not ∂ 2 r log F GUE . Now we want to see how equation (2.9) leads us to the Baik-Rains distribution.
In a purely formal way, we assume all the partial derivatives converge to the derivatives of the limit, thus we obtain that φ(x, t, r), ψ(x, t, r) also satisfy equations (2.16).
Here the relation of τ to x, t is Under this scaling, we have the following result.