A Riemann Hilbert Approach to the Study of the Generating Function Associated to the Pearcey Process

Abstract

Using Riemann–Hilbert methods, we establish a Tracy–Widom like formula for the generating function of the occupancy numbers of the Pearcey process. This formula is linked to a coupled vector differential equation of order three. We also obtain a non linear coupled heat equation. Combining these two equations we obtain a PDE for the logarithm of the the generating function of the Pearcey process.

Appendix

1.1 Asymptotics of P and Q

As $s\rightarrow \infty$, P and Q have the following asymptotic.

Proposition 5.1

Let P and Q defined respectively as in Eqs. (4.7) and (4.6). Then, as $s\rightarrow \infty$:

$$\begin{aligned}{} & {} Q(s)\displaystyle \sim _{s\rightarrow \infty }\sqrt{\dfrac{2}{3\pi }}s^{-1/3}e^{-\frac{3}{8}s^{4/3}-\frac{\tau }{4}s^{2/3}+\frac{\tau ^2}{6}}\cos \left( \frac{3}{4}\sin \left( \frac{2\pi }{3}\right) s^{4/3}-\frac{\tau }{2}\sin \left( \frac{2\pi }{3}\right) s^{2/3}-\frac{\pi }{6}\right) \nonumber \\ \end{aligned}$$

(5.1)

$$\begin{aligned}{} & {} P(s)\displaystyle \sim _{s\rightarrow \infty }\sqrt{\dfrac{2}{3\pi }}s^{-1/3}e^{\frac{3}{8}s^{4/3}+\frac{\tau }{4}s^{2/3}-\frac{\tau ^2}{6}}\cos \left( \frac{3}{4}\sin \left( \frac{2\pi }{3}\right) s^{4/3}-\frac{\tau }{2}\sin \left( \frac{2\pi }{3}\right) s^{2/3}-\frac{\pi }{6}\right) \nonumber \\ \end{aligned}$$

(5.2)

Proof

By definition,

$$\begin{aligned}Q(s)=\dfrac{1}{2i\pi }\displaystyle \int _{i\mathbb {R}}e^{-\frac{\mu ^4}{4}+\tau \frac{\mu ^2}{2}+s\mu }d\mu \end{aligned}$$

The derivative with respect to $\mu$ of $\theta _s(\mu )$ is $\frac{d}{ds}\theta _s(\mu )=\mu ^3-\tau \mu -s$ whose roots are

$$\begin{aligned} \mu _k:=\mu _k(s,\tau )=j^k\root 3 \of {\frac{1}{2}\left( s+\sqrt{s^2-\frac{2^2\tau ^3}{3^3}}\right) }+j^{-k}\root 3 \of {\frac{1}{2}\left( s-\sqrt{s^2-\frac{2^2\tau ^3}{3^3}}\right) } \nonumber \\ \end{aligned}$$

(5.3)

where $j:=e^{\frac{2\pi }{3}}$ and $k\in \{0,1,2\}$. We only consider saddle points for $k=1$ and 2.

Denote by $\mu _*$ either $\mu _1$ or $\mu _2$. One can deform the contour $i\mathbb {R}$ into a contour $\mathcal {C}_Q$ so that it passes through $\mu _1$ and $\mu _2$ and the following holds $\Re \left( \theta _s(\mu )-\theta _s(\mu _*)\right) <0$.

See Fig. 3 for the study of the sign of $\Re \left( \theta _s(\mu )-\theta _s(\mu _*)\right)$ for $s=100$ and $\tau =1$. As $s\rightarrow \infty$ and $\tau$ is fixed, the algebraic curve $\Re \left( \theta _s(\mu )-\theta _s(\mu _*)\right) =0$ keeps a similar shape and it is always possible to deform $i\mathbb {R}$ into $\mathcal {C}_Q$.

Fig. 3

The sign of $\Re (\theta _s(\mu )-\theta _s(\mu _*))$ for $\tau =1$ and $s=100$

Then

$$\begin{aligned} \begin{array}{rl} Q(s)&{}=\dfrac{e^{-\theta _s(\mu _*)}}{2i\pi }\displaystyle \int _{\mathcal {C}_Q}e^{-\left( \theta _s(\mu )-\theta _s(\mu _*)\right) }d\mu \\ &{}=\dfrac{e^{-\theta _s(\mu _*)}}{2i\pi }\displaystyle \int _{\mathcal {C}_Q}e^{-\left( \frac{3}{2}\mu _*^2(\mu -\mu _*)^2\left( 1-\frac{\tau }{3\mu _*^2}+\frac{2}{3\mu _*}(\mu -\mu _*)+\frac{1}{6\mu _*^2}(\mu -\mu _*)^2\right) \right) }d\mu \end{array} \end{aligned}$$

(5.4)

and one can approximate Q as $s\rightarrow \infty$ by expanding $\mu _*$ and by approximating the integral. It follows

$$\begin{aligned} Q(s)\sim \sqrt{\dfrac{2}{3\pi }}s^{-1/3}e^{-\frac{3}{8}s^{4/3}-\frac{\tau }{4}s^{2/3}+\frac{\tau ^2}{6}}\cos \left( \frac{3}{4}\sin \left( \frac{2\pi }{3}\right) s^{4/3}-\frac{\tau }{2}\sin \left( \frac{2\pi }{3}\right) s^{2/3}-\frac{\pi }{6}\right) \nonumber \\ \end{aligned}$$

(5.5)

The same method leads to the asymptotic for P as $s\rightarrow \infty$. $\square$

1.2 Computations for Equation (1.6)

We now prove equation (1.6).

Define $u(s,\tau ):=\log \left( F(\overrightarrow{\textbf{a}}+s,\tau ,\overrightarrow{\textbf{k}})\right)$ and $v(s,\tau ):=\dfrac{\partial ^2}{\partial s^2}u(s,\tau )$. According to Proposition 4.2, $v(s,\tau )=p^T(s)q(s)$. Differentiating v with respect to $\tau$ and using Eq. (3.11) to express $\partial _\tau p^T$ and $\partial _\tau q$, we obtain:

$$\begin{aligned} \partial _\tau v=\dfrac{1}{2}\partial _s\left( p^T\partial _sq-\partial _sp^Tq\right) \end{aligned}$$

(5.6)

Differentiating a second time v with respect to $\tau$ (again using Eq. (3.11)) yields to

$$\begin{aligned} \partial _{\tau \tau }v=\dfrac{1}{2}\left( 2v\partial _sv+\dfrac{1}{2}\left( p^T\partial _{sss}q+\partial _{sss}p^Tq\right) -\dfrac{1}{2}\partial _s\left( \partial _sp^T\partial _sq\right) \right) \end{aligned}$$

(5.7)

Recall $v(s,\tau )=p^T(s)q(s)$, then differentiating three times with respect to s the following equation holds:

$$\begin{aligned} \partial _s\left( \partial _sp^T\partial _sq\right) =\dfrac{1}{3}\left( \partial _{sss}v-\left( p^T\partial _{sss}q+\partial _{sss}p^Tq\right) \right) \end{aligned}$$

(5.8)

Replacing $\partial _s\left( \partial _sp^T\partial _sq\right)$ in Eq. (5.7) and using Eq. (3.10) we obtain:

$$\begin{aligned} \partial _{\tau \tau }v=\partial _s\left( -v\partial _sv-\dfrac{1}{12}\partial _{sss}v+\dfrac{1}{3}\tau \partial _sv\right) . \end{aligned}$$

(5.9)

Replacing v by $\partial _{ss}u$ and integrating twice with respect to s we prove u satisfies Eq. (1.6).