Lyapunov Exponents of the Half-Line SHE

Abstract

We consider the half-line stochastic heat equation (SHE) with Robin boundary parameter $A = -\frac{1}{2}$. Under narrow wedge initial condition, we compute every positive (including non-integer) Lyapunov exponents of the half-line SHE. As a consequence, we prove a large deviation principle for the upper tail of the half-line KPZ equation under Neumann boundary parameter $A = -\frac{1}{2}$ with rate function $\Phi _+^{\text {hf}} (s) = \frac{2}{3} s^{\frac{3}{2}}$. This confirms the prediction of [44, 52] for the upper tail exponent of the half-line KPZ equation.

Appendices

Appendix A. Basic facts of Airy function

In this section, we review some basic properties of the Airy function. As a notational convention, we say $f(x) \sim g(x)$ as $x \rightarrow a$ (where a can be $\pm \infty$) if $\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = 1$.

Lemma 5.4

We have the following asymptotics for Airy function

$$\begin{aligned} Ai (x) \sim {\left\{ \begin{array}{ll} \frac{e^{-\frac{2}{3} x^{\frac{3}{2}}}}{2 \sqrt{\pi } x^{\frac{1}{4}}} &{} x \rightarrow +\infty ,\\ \frac{1}{\sqrt{\pi } |x|^{\frac{1}{4}}}\cos \Big (\frac{\pi }{4} - \frac{2 |x|^{\frac{3}{2}}}{3}\Big ) &{} x \rightarrow -\infty . \end{array}\right. } \quad Ai '(x) \sim {\left\{ \begin{array}{ll} -\frac{x^{\frac{1}{4}} e^{-\frac{2}{3} x^{\frac{3}{2}}}}{2 \sqrt{\pi }} &{} x \rightarrow +\infty ,\\ -\frac{|x|^{\frac{1}{4}}}{\sqrt{\pi } }\cos \Big (\frac{\pi }{4} + \frac{2 |x|^{\frac{3}{2}}}{3}\Big ) &{} x \rightarrow -\infty . \end{array}\right. } \end{aligned}$$

Proof

See Eq 10.4.59-10.4.62 of [3]. $\square$

Lemma 5.5

We have $\int _{-\infty }^{\infty } Ai (x) dx = 1$ and $\int _{-\infty }^0 Ai (x) dx = 1/3$.

Proof

See page 431 of [53]. $\square$

Lemma 5.6

There exists constant C such that

$$\begin{aligned}&\frac{1}{C (x+1)} e^{-\frac{4}{3} x^{\frac{3}{2}}}\le \int _0^{\infty } Ai (x+\lambda )^2 d\lambda \le \frac{C}{x+1} e^{-\frac{4}{3} x^{\frac{3}{2}}} \quad \, \forall \, x \ge 0 \\&\frac{1}{C} (\sqrt{|x|} + 1) \le \int _0^{\infty } Ai (x+\lambda )^2 d\lambda \le C (\sqrt{|x|} + 1) ,\qquad \forall \,x \le 0 \end{aligned}$$

Proof

This is Eq 2.8 and Eq 2.9 of [27]. $\square$

Appendix B. Estimate of the Pfaffian Kernel entries

In this section, we provide various bounds for the entries in the GOE Pfaffian kernel.

Lemma 5.7

There exists a constant $C > 0$ such that

• (i) $\frac{\exp (-\frac{2}{3} x^{\frac{3}{2}})}{C (1+x)^{\frac{1}{4}}} \le K_{12}(x,x) \le \frac{C \exp (-\frac{2}{3} x^{\frac{3}{2}})}{(1+x)^{\frac{1}{4}}}\qquad \forall \, x \ge 0,$

• (ii) $0 \le K_{12}(x, x) \le C\sqrt{1-x} \quad \forall \,\ x \le 0.$

Proof

We first prove (i). By setting $x = y$ in (2.2), we get

$$\begin{aligned} K_{12} (x, x) = \int _0^{\infty } \text {Ai}(x+\lambda )^2 d\lambda + \frac{1}{2} \text {Ai}(x) \int _{-\infty }^x \text {Ai}(\lambda ) d\lambda . \end{aligned}$$

(B.1)

For the second term in the above display, by Lemmas 5.4 and 5.5, we have as $x \rightarrow +\infty$

$$\begin{aligned} \text {Ai}(x) \int _{-\infty }^{x} \text {Ai}(\lambda ) d\lambda \sim \frac{e^{-\frac{2}{3} x^{\frac{3}{2}}}}{2 \sqrt{\pi } x^{\frac{1}{4}}} \end{aligned}$$

Combining this with the first inequality of Lemma 5.6, which controls the first term on the right hand side, the upper bound in (i) naturally follows. To prove the lower bound of (i), due to the above displayed asymptotic and the non-negativity of $\int _0^\infty \text {Ai}(x+\lambda )^2 d\lambda$, there exists constant M and C such that for $x > M$,

$$\begin{aligned} K_{12}(x, x) \ge C^{-1} x^{-\frac{1}{4}}\exp \left( -\frac{2}{3} x^{\frac{3}{2}}\right) . \end{aligned}$$

To conclude the lower bound in (i), it suffices to show that the minimum of $K_{12} (x, x)$ is positive over [0, M] ($K_{12}(x, x)$ is continuous, so admits a minimum). Due to Eq. (B.1) and Lemma 5.5, we can rewrite $K_{12}(x, x) = \int _{0}^\infty \text {Ai}(x+\lambda )^2 d\lambda + \frac{1}{3} \text {Ai}(x) + \frac{1}{2}\text {Ai} (x) \int _{0}^x \text {Ai} (\lambda ) d\lambda$. Since $\text {Ai}(x)$ is positive for $x \ge 0$, this implies $K_{12}(x, x) > 0$ for all $x > 0$, which completes the proof of the lower bound.

We move on proving (ii). The lower bound follows directly since $K_{12} (x, x)$ is the first order correlation function of a Pfaffian point process, thus is negative. For the upper bound, by the asymptotic of $\text {Ai}(x)$ at $-\infty$, there exists constant C such that for all $x \le 0$,

$$\begin{aligned} \bigg |\text {Ai}(x) \int _{-\infty }^{x} \text {Ai}(\lambda ) d\lambda \, \bigg | \le C (1+|x|)^{-\frac{1}{4}}. \end{aligned}$$

The result then follows from the second inequality of Lemma 5.6 and (B.1). $\square$

Recall that we defined $F_{\alpha , \beta }(x) = C\big ( e^{-\alpha x^{\frac{3}{2}}} {\mathbf {1}}_{\{x \ge 0\}} + (1-x)^\beta {\mathbf {1}}_{\{x < 0\}}\big )$.

Lemma 5.8

There exists a constant C, such that for all $x, y \in {\mathbb {R}}$, we have the following upper bounds for the Pfaffian kernel entries:

• (a) $|K_{11}(x, y)| \le C \big (F_{\frac{2}{3}, \frac{5}{4}} (x) \wedge F_{\frac{2}{3}, \frac{3}{4}}(x) F_{\frac{2}{3}, \frac{3}{4}}(y)\big )$

• (b) $|K_{12}(x, y)| \le C \big (F_{\frac{2}{3}, \frac{3}{4}}(x) \wedge F_{0, \frac{3}{4}} (y)\big )$

• (c) $|K_{22} (x, y)| \le C F_{0, \frac{3}{4}} (x)$

Proof

For (a), it suffices to show that $|K_{11}(x, y)| \le C F_{\frac{2}{3}, \frac{5}{4}} (x)$ and $|K_{11}(x, y)| \le CF_{\frac{2}{3}, \frac{3}{4}}(x) F_{\frac{2}{3}, \frac{3}{4}}(y)$. Recall the expression of $K_{11}(x, y)$ from (2.1). Using integration by parts for the right hand side of (2.1), we get $K_{11}(x, y) = \text {Ai}(x) \text {Ai}(y) - 2\int _0^{\infty } \text {Ai}(y + \lambda ) \text {Ai}'(x+\lambda ) d\lambda .$ This implies that $|K_{11}(x, y)| \le |\text {Ai}(x) \text {Ai}(y)| + 2 \int _0^\infty |\text {Ai}(y + \lambda ) \text {Ai}'(x+ \lambda )| d\lambda$. Since $|\text {Ai}(x)|$ is a bounded function, there exists constant C such that

$$\begin{aligned} |K_{11}(x, y)| \le C|\text {Ai}(x)| + C \int _0^{\infty } |\text {Ai}'(x+\lambda )| d\lambda = C + C \int _x^{\infty } |\text {Ai}'(\lambda )| d\lambda . \end{aligned}$$

To obtain the upper bound for $|\text {Ai}(x)|$ and $\int _x^{\infty } |\text {Ai}(\lambda )| d\lambda$, it suffices to look at their behavior as $x \rightarrow \pm \infty$. The asymptotic $\text {Ai}'(x)$ at $\pm \infty$ is specified in Lemma 5.4. Therefore,

$$\begin{aligned} \int _x^\infty |\text {Ai}'(\lambda )| d\lambda \le C e^{-\frac{2}{3} x^{\frac{3}{2}}}, \text { if } x \ge 0; \qquad \int _x^\infty |\text {Ai}'(\lambda )| d\lambda \le C(1 - x)^{\frac{5}{4}} \text { if } x \le 0. \end{aligned}$$

This implies that $|K_{11}(x, y)| \le C F_{\frac{2}{3}, \frac{5}{4}} (x)$. In addition, since

$$\begin{aligned} K_{11}(x, y) = \int _0^{\infty } \text {Ai}(x+\lambda ) \text {Ai}'(y+\lambda ) d\lambda - \int _0^{\infty } \text {Ai}'(x+\lambda ) \text {Ai}(y+\lambda ) d\lambda = A_1 - A_2 \end{aligned}$$

By Cauchy Schwartz inequality,

$$\begin{aligned} A_1^2 \le \int _0^{\infty } \text {Ai}(x + \lambda )^2 d\lambda \int _0^{\infty } \text {Ai}'(y + \lambda )^2 d\lambda = \int _x^{\infty } \text {Ai}(\lambda )^2 d\lambda \int _{y}^{\infty } \text {Ai}'(\lambda )^2 d\lambda . \end{aligned}$$

By Lemma 5.4, $\text {Ai}(x)^2$ decays asymptotically as $\exp (-\frac{4}{3} x^{\frac{3}{2}})$ as $x \rightarrow +\infty$ and is asymptotically upper bounded by $|x|^{-\frac{1}{2}}$ as $x \rightarrow -\infty$. This implies that $\int _x^{\infty } \text {Ai}(\lambda )^2 d\lambda \le C F_{\frac{4}{3}, \frac{1}{2}} (x)$. Similarly, $\text {Ai}'(y)^2$ decays asymptotically as $\exp (-\frac{4}{3} y^{\frac{3}{2}})$ and is asymptotically upper bounded by $|y|^{\frac{1}{2}}$, we get $\int _y^{\infty } \text {Ai}'(\lambda )^2 d\lambda \le C F_{\frac{4}{3}, \frac{3}{2}}(y)$. As a result,

$$\begin{aligned} |A_1|&\le \Big (\int _x^{\infty } \text {Ai}(\lambda )^2 d\lambda \Big )^{\frac{1}{2}} \Big (\int _y^{\infty } \text {Ai}'(\lambda )^2 d\lambda \Big )^{\frac{1}{2}} \le C F_{\frac{2}{3}, \frac{1}{4} }(x) F_{\frac{2}{3}, \frac{3}{4} }(y) \le C F_{\frac{2}{3}, \frac{3}{4} }(x) F_{\frac{2}{3}, \frac{3}{4} }(y). \end{aligned}$$

For the second inequality above, we use the property that $\sqrt{F_{\alpha , \beta }} = F_{\alpha /2, \beta /2}$ and for the third inequality, $F_{\alpha , \beta }(x)$ is increasing in $\beta$. Interchanging the role of x and y, we also have $|A_2| \le C F_{\frac{2}{3}, \frac{3}{4} }(x) F_{\frac{2}{3}, \frac{3}{4} }(y)$. Therefore, the same upper bound holds for $|K_{11} (x, y)|$ and we conclude the proof of (a).

We move on showing (b). We will prove $|K_{12}(x, y)| \le C F_{\frac{2}{3}, \frac{3}{4}}(x)$ and $|K_{12}(x, y)| \le C F_{0, \frac{3}{4}} (y)$ respectively. Recall $K_{12} (x, y)$ from (2.2). Note that both $|\text {Ai}(y + \lambda )|$ and $|\int _{-\infty }^y \text {Ai}(\lambda ) d\lambda |$ are bounded function of y (see Lemma 5.5), by using triangle inequality,

$$\begin{aligned} |K_{12} (x, y)|&\le \frac{1}{2} \int _0^{\infty } |\text {Ai}(x+\lambda ) \text {Ai} (y+\lambda )| d\lambda + \frac{1}{2} |\text {Ai}(x)| \\&\quad \cdot \Big |\int _{-\infty }^y \text {Ai}(\lambda ) d\lambda \Big | \le C \int _x^{\infty } |\text {Ai}(\lambda )| d\lambda + C |\text {Ai}(x)|. \end{aligned}$$

By the asymptotic of $\text {Ai}(x)$ at $\pm \infty$, (use the similar approach as in part (a)), we see that $|K_{12} (x, y)| \le C F_{\frac{2}{3}, \frac{3}{4}} (x).$

We proceed to obtain a different upper bound for $K_{12}$. Referring to the right hand side of the first inequality in the above display and upper bounding $|\text {Ai}(x +\lambda )|$ and $|\frac{1}{2} \text {Ai}(x) \int _{-\infty }^y \text {Ai}(\lambda ) d\lambda |$ by a constant, we find that

$$\begin{aligned} |K_{12}(x, y)| \le C \int _0^{\infty } |\text {Ai}(y+\lambda )| d\lambda + C \le C F_{0, \frac{3}{4}}(y). \end{aligned}$$

This concludes our proof of (b).

Finally, let us demonstrate (c). Recall from (2.3) that

$$\begin{aligned} K_{22} (x, y)&= \frac{1}{4} \int _0^\infty \text {Ai}(x + \lambda ) \Big (\int _\lambda ^\infty \text {Ai}(y+\mu )d\mu \Big ) d\lambda \nonumber \\&\quad - \frac{1}{4} \int _0^{\infty } \text {Ai}(y+ \lambda ) \Big (\int _{\lambda }^{\infty } \text {Ai}(x+\mu ) d \mu \Big ) d\lambda \nonumber \\&\quad - \frac{1}{4} \int _0^\infty \text {Ai}(x+\lambda ) d\lambda + \frac{1}{4} \int _0^{\infty } \text {Ai}(y+\lambda ) d\lambda - \frac{\text {sgn}(x-y)}{4} \end{aligned}$$

(B.2)

and recall that $\text {sgn}$ is the sign function. By Fubini’s theorem,

$$\begin{aligned} \int _0^\infty \text {Ai}(y+\lambda ) \Big (\int _{\lambda }^\infty \text {Ai}(x+\mu )d\mu \Big ) d\lambda&= \Big (\int _0^{\infty } \text {Ai}(x+\lambda ) d\lambda \Big ) \Big (\int _0^{\infty } \text {Ai}(y+\lambda ) d\lambda \Big ) \\&\quad - \int _0^\infty \text {Ai}(x + \lambda )\Big (\int _\lambda ^\infty \text {Ai}(y + \mu ) d\mu \Big ) d\lambda \end{aligned}$$

Replacing the term $\int _{0}^\infty \text {Ai}(y + \lambda ) \big (\int _{\lambda }^{\infty } \text {Ai}(x+ \mu ) d\mu \big ) d\lambda$ in (B.2) with the right hand side in the above display,

$$\begin{aligned} K_{22} (x, y)&= \frac{1}{2} \int _0^\infty \text {Ai}(x + \lambda ) \Big (\int _{\lambda }^\infty \text {Ai}(y + \mu )d\mu \Big )d\lambda \\&\quad - \frac{1}{4} \Big (\int _0^{\infty } \text {Ai}(x+\lambda ) d\lambda \Big ) \Big (\int _0^{\infty } \text {Ai}(y+\lambda ) d\lambda \Big )\\&\quad -\frac{1}{4} \int _0^{\infty } \text {Ai}(x+\lambda ) d\lambda + \frac{1}{4} \int _0^\infty \text {Ai}(y + \lambda ) d\lambda - \frac{\text {sgn}(x-y)}{4}. \end{aligned}$$

We know that $\big |\int _0^{\infty } \text {Ai}(x+\lambda ) d\lambda \big |, \big |\int _0^{\infty } \text {Ai}(y+\lambda ) d\lambda \big |$ can upper bounded by a constant. Applying triangle inequality to the above display,

$$\begin{aligned} |K_{22} (x, y)| \le C \int _0^{\infty } |\text {Ai}(x + \lambda )| d\lambda + C. \end{aligned}$$

Using the asymptotic of $\text {Ai}(x)$ at $\pm \infty$ in Lemma 5.4, we find that $|K_{22}(x, y) | \le C F_{0, \frac{3}{4}} (x)$, thus conclude (c). $\square$