Numerical Solution of Riemann–Hilbert Problems: Random Matrix Theory and Orthogonal Polynomials

Abstract

Recently, a general approach to solving Riemann–Hilbert problems numerically has been developed. We review this numerical framework and apply it to the calculation of orthogonal polynomials on the real line. Combining this numerical algorithm with the approach of Bornemann to compute Fredholm determinants, we are able to calculate spectral densities and gap statistics for a broad class of finite-dimensional unitary invariant ensembles. We show that the accuracy of the numerical algorithm for approximating orthogonal polynomials is uniform as the degree grows, extending the existing theory to handle g-functions. As another example, we compute the Hastings–McLeod solution of the homogeneous Painlevé II equation.

Appendix A: Computing a Hastings–McLeod Solution of the Painlevé II Transcendent

Here we focus on the (homogeneous) Painlevé II ODE, which is as follows:

$$\begin{aligned} u''(x) = x u(x) + 2 u^3(x). \end{aligned}$$

(22)

(For brevity we refer to the homogeneous Painlevé II simply as Painlevé II.) There are many important applications of this equation. As mentioned above, the Tracy–Widom distribution [33] is written in terms of the Hastings–McLeod solution q [20]:

$$\begin{aligned} &\det(I - {\mathcal{A}}|_{L^2(s,\infty)}) = \exp\biggl({-\int _s^\infty (x-s) q^2(x) \,\mathrm {d}x}\biggr)\\ &\quad \mathrm{for}\ q''(x) = x q(x) + 2 q^3(x) \ \mathrm{and}\ q(x) \sim-\mathrm{Ai}(x) \ \mathrm{as}\ x \rightarrow \infty. \end{aligned}$$

Furthermore, it has also been shown that asymptotic solutions to the Korteweg–de Vries and modified Korteweg–de Vries equations can be written in terms of Ablowitz–Segur solutions [1]. The aim of this section is to demonstrate that the RH formulation can indeed be used effectively to compute solutions to Painlevé II, even in the asymptotic regime.

Solutions to differential equations such as (22) are typically defined by initial conditions: at a point x we are given u(x) and u′(x). In the RH formulation, however, we do not specify initial conditions. Rather, the solution is specified by the Stokes’ constants; constants s ₁,s ₂,s ₃ which satisfy the following condition:

$$\begin{aligned} s_1 - s_2 + s_3 +s_1s_2s_3 = 0. \end{aligned}$$

(23)

We treat the Stokes’ constants as given, as in many applications they arise naturally while initial conditions do not. Given such constants, we denote the associated solution to (22) by

$$\begin{aligned} P_{\mathrm{II}}(s_1,s_2,s_3;z). \end{aligned}$$

(24)

P _II and its derivative can be viewed as the special functions that map Stokes’ constants to initial conditions.

At first glance, computing solutions to (22) appears trivial: given initial conditions, simply use one’s favourite time-stepping algorithm, e.g., input it into an ODE toolbox such as Matlab’s ode45 or Mathematica’s NDSolve. Unfortunately, several difficulties immediately become apparent. In Fig. 15, we plot several solutions to (22) (computed using the approach we are advocating): the Hastings–McLeod solution and perturbations of the Hastings–McLeod solution. Note that the solution is inherently unstable, and small perturbations cause oscillations—which make standard ODE solvers inefficient—and poles—which completely breaks such ODE initial value problem solvers (the issue of poles can be resolved using the methodology of [18]).

Fig. 15

Solutions to Painlevé II, with the Hastings–McLeod solution in solid, showing that an initial value solver for Hastings–McLeod must be unstable. (a) Small perturbations at the right end integrated left. (b) Small perturbations at the left end integrated right

Remark

There are many other methods for computing the Tracy–Widom distribution itself as well as the Hastings–McLeod solution [4, 5], based on the Fredholm determinant formulation or solving a boundary value problem. Moreover, accurate data values have been tabulated using high precision arithmetic with a Taylor series method [29, 30]. However, we see that there is a whole family of solutions to Painlevé II which exhibit similar sensitivity to initial conditions, and thus a reliable, general numerical method is needed even for this case. We note that the approach [18], which combines boundary value problem solvers with initial value problem solvers, may also be successful for calculating such solutions. However, it only works with initial conditions, not with Stokes’ constants.

Let Φ(x;λ) solve the RH problem depicted in Fig. 16: let Γ=Γ ₁∪⋯∪Γ ₆ for $\varGamma_{\kappa}\{ s \mathrm{e}^{\mathrm{i}\pi(\kappa/3-1/6)}: s \in\mathbb{R}^{+}\}$, i.e., Γ consists of six rays emanating from the origin, as seen in Fig. 16. Then the jump matrix is defined by G(x;λ)=G _κ (x;λ) for z∈Γ _κ , where

$$\begin{aligned} G_\kappa(x;\lambda) = G_\kappa(\lambda) = \left\{\begin{array}{l@{\quad}l} \big(\begin{array}{c@{\ }c} \scriptstyle 1 & \scriptstyle s_\kappa \mathrm{e}^{-\mathrm{i}8/3 \lambda^3 -2\mathrm{i} x \lambda} \\ \scriptstyle 0 & \scriptstyle 1 \end{array}\big) & \mathrm{if}\ \kappa\ \mbox{even},\\ \big(\begin{array}{c@{\ }c} \scriptstyle 1 & \scriptstyle 0\\ \scriptstyle s_\kappa \mathrm{e}^{\mathrm{i}8/3 \lambda^3 + 2\mathrm{i}x\lambda} & \scriptstyle 1 \end{array}\big) & \mathrm{if}\ \kappa\mbox{ odd}, \end{array}\right. \end{aligned}$$

with s ₄=−s ₁, s ₅=−s ₂, and s ₆=−s ₃. This is the RH problem that was solved numerically in [26]. We recover the corresponding solution to Painlevé II from Φ by [16]:

$$\begin{aligned} P_{\mathrm{II}}(s_1,s_2,s_3;x) = 2 {\mathrm {i}}\lim_{\lambda \rightarrow \infty} \lambda\varPhi(x;\lambda)_{12}. \end{aligned}$$

Fig. 16

The contour and jump matrix for the Painlevé II RH problem

As |x| becomes large, the jump matrices G are increasingly oscillatory. We combat this issue by deforming the contour so that these oscillations become exponential decay. To simplify this procedure, we first rescale the RH problem. Note that if we let $z = \sqrt{|x|} \lambda$, then the jump contour Γ remains unchanged, and

$$\begin{aligned} \tilde{\varPhi}^+(z) = \varPhi^+\bigl(x;z/\sqrt{|x|} \,\bigr) = \varPhi^-\bigl(x;z/\sqrt {|x|}\,\bigr)G\bigl(z/ \sqrt{|x|}\,\bigr) = \tilde{\varPhi}^-(z) \tilde{G}(z), \end{aligned}$$

where $\tilde{G}(z) = G_{\kappa}(z/\sqrt{|x|})$ on Γ _κ for

$$\begin{aligned} G_\kappa\bigl(z/\sqrt{|x|}\,\bigr) = \left\{\begin{array}{l@{\quad}l} \big(\begin{array}{c@{\ }c} \scriptstyle 1 & \scriptstyle s_\kappa \mathrm{e}^{-\mathrm{i}|x|^{3/2} \theta(z)} \\ \scriptstyle 0& \scriptstyle 1 \end{array}\big)& \mathrm{if}\ \kappa\ \mathrm{even},\\ \big(\begin{array}{c@{\ }c} \scriptstyle 1 &\scriptstyle 0 \\ \scriptstyle s_\kappa \mathrm{e}^{\mathrm{i}|x|^{3/2}\theta(z)} & \scriptstyle 1 \end{array}\big)& \mathrm{if}\ \kappa\ \mathrm{odd}, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \theta(z) = \frac{2}{3} \bigl( 4z^3 + 2 \mathrm{e}^{\mathrm{i} \arg x} z \bigr). \end{aligned}$$

Then

$$\begin{aligned} P_{\mathrm{II}}(s_1,s_2,s_3;x) = 2\mathrm{i} \lim_{\lambda \rightarrow \infty} \lambda \varPhi(x;\lambda)_{12} = 2\mathrm{i} \sqrt{|x|} \lim_{z \rightarrow \infty} z \tilde{\varPhi}(z)_{12}. \end{aligned}$$

1.1 A.1 Positive x with s ₂=0

We deform the RH problem for Painlevé II so that numerics are asymptotically stable for positive x. The deformation is extremely simple under the following special case:

$$ s_2 = 0. $$

(25)

We remark that, unlike other deformations, the following deformation can be easily extended to achieve asymptotic stability for x in the complex plane such that $-\frac{\pi}{3} < \arg x < \frac{\pi}{6}$.

On the undeformed contour, the terms $\mathrm{e}^{\pm \mathrm{i} |x|^{3/2} \theta(z)}$ become oscillatory as |x| becomes large. However, with the right choice of curve h(t), e^±iθ(h(t)) has no oscillations; instead, it decays exponentially fast as t→∞. But h is precisely the path of steepest descent, which passes through the stationary points of θ, i.e., the points where the derivative of θ vanishes. We readily find that

$$\begin{aligned} \theta'(z) = 2\bigl(4z^2+1\bigr), \end{aligned}$$

and the stationary points are z=±i/2.

We note that, since G ₂=I, when we deform Γ ₁ and Γ ₃ through i/2, they become completely disjoint from Γ ₄ and Γ ₆, which we then deform through −i/2. We also point out that $G_{3}^{-1} = G_{1}$ and $G_{6}^{-1} = G_{4}$; thus we reverse the orientation of Γ ₃ and Γ ₄. Define Γ _↑ to be the curve G ₁ is defined on and Γ _↓ to be the contour G ₄ is defined on, as seen in Fig. 17.

Fig. 17

Deforming the RH problem for positive x, assuming (25)

Now recall that

$$\begin{aligned} \theta \biggl(\pm\frac{\mathrm{i}}{2} \biggr) = \pm\frac{\mathrm{i}}{3}. \end{aligned}$$

However, we only have Γ _↑ emanating from i/2, with jump matrix

$$\begin{aligned} G_1 = \left(\begin{array}{c@{\quad}c} 1 & 0\\ s_1 \mathrm{e}^{\mathrm{i}|x|^{3/2} \theta(z)} & 1 \end{array}\right) . \end{aligned}$$

This is exponentially decaying to the identity along Γ _↑, as is G ₄ along Γ _↓. We employ the approach of Sect. 3.2. We first use Lemma 3 to truncate the contours near the stationary point. What remains is to determine what near means. Because θ behaves like $\mathcal {O}(z \pm \mathrm{i}/2)^{2}$ near the stationary points, Assumption 2 implies that we should choose the shifting of β ₁=i/2 and β ₂=−i/2, the scalings α ₁=α ₂=r|x|^−3/4, and the canonical domains Ω ₁=Ω ₂=[−1,1]. Here r is chosen so that what is truncated is negligible in the sense of Lemma 3. G ₆ is similar. The complete proof of asymptotic stability follows from Theorem 2.

1.2 A.2 Negative x with s ₁=−s ₃=±i and s ₂=0

We now develop deformations for the Hastings–McLeod solution for negative x, which corresponds to s ₁=±i, s ₂=0, and s ₃=∓i [16]. We realise numerical asymptotic stability in the aforementioned sense for the following special case:

$$ s_1 = - s_3 = \pm \mathrm{i}\quad\mathrm{and}\quad s_2 = 0. $$

(26)

We begin by deforming the RH problem (Fig. 16) to the one shown in Fig. 18. The horizontal contour extends from −α to α for α>0. We determine α below. Define

$$\begin{aligned} G_0 = G_6G_1 = \left(\begin{array}{c@{\quad}c} 0& s_1\mathrm{e}^{-\mathrm{i} |x|^{3/2} \theta(z)} \\ s_1\mathrm{e}^{\mathrm{i} |x|^{3/2} \theta (z)} & 1 \end{array}\right) . \end{aligned}$$

Note that the assumption s ₂=0 simplifies the form of the RH problem substantially, see Fig. 18(b). We use an approach similar to that of the equilibrium measure to replace θ with a function possessing more desirable properties. Define

$$\begin{aligned} \varTheta(z) = \mathrm{e}^{\mathrm{i} |x|^{3/2} \frac{g(z)-\theta(z)}{2} \sigma_3},\qquad \sigma _3 = \left(\begin{array}{c@{\quad}c} 1 &0\\ 0& -1 \end{array}\right) , \qquad g(z) = \bigl(z^2-\alpha^2\bigr)^{3/2}. \end{aligned}$$

The branch cut for g(z) is chosen along [−α,α]. If we set $\alpha= 1/\sqrt{2}$, the branch of g is chosen so that $g(z) -\theta (z) \sim \mathcal {O}(z^{-1})$. Furthermore, g ₊(z)+g ₋(z)=0, and Im(g ₋(z)−g ₊(z))>0 on (−α,α). Define $\hat{G}_{i} = \varTheta^{-1}_{-}G_{i} \varTheta_{+}$, and note that

$$\begin{aligned} \hat{G}_0(z)& = \left(\begin{array}{c@{\quad}c} 0& s_1 \mathrm{e}^{-\mathrm{i}|x|^{3/2} \frac{g_+(z) + g_-(z)}{2}} \\ s_1 \mathrm{e}^{\mathrm{i}|x|^{3/2} \frac{g_+(z) + g_-(z)}{2}} & \mathrm{e}^{\mathrm{i}|x|^{3/2} \frac {g_-(z) - g_-(z)}{2}} \end{array}\right)\\ &= \left(\begin{array}{c@{\quad}c} 0 & s_1 \\ s_1 & \mathrm{e}^{\mathrm{i}|x|^{3/2} \frac{g_-(z) - g_-(z)}{2}} \end{array}\right) . \end{aligned}$$

As x→−∞, G ₀ tends to the matrix

$$\begin{aligned} J = \left(\begin{array}{c@{\quad}c} 0 & s_1 \\ s_1 & 0 \end{array}\right) . \end{aligned}$$

Fig. 18

Deforming the RH problem for negative x, assuming (26). The black dots represent ±α. (a) Initial deformation for s ₂≠0. (b) Simplification following from s ₂=0

The solution of the RH problem

$$\begin{aligned} \varPsi^+(z) = \varPsi^-(z)J,\quad z \in[-\alpha,\alpha],\qquad \varPsi(\infty) = I, \end{aligned}$$

is given by

$$\begin{aligned} \varPsi _{\mathrm {HM}}^{\text {out}}(z) &= \frac{1}{2} \left(\begin{array}{c@{\quad}c} \beta(z) + \beta(z)^{-1} & -\mathrm{i} s_1(\beta(z) - \beta (z)^{-1}) \\ -\mathrm{i} s_1(\beta(z) - \beta(z)^{-1}) & \beta(z) + \beta(z)^{-1} \end{array}\right) , \\ \beta(z) &= \biggl( \frac{z-\alpha}{z+\alpha} \biggr)^{1/4}. \end{aligned}$$

Here β has a branch cut on [−α,α] and satisfies β(z)→1 as z→∞. It is clear that $(\varPsi _{\mathrm {HM}}^{\text {out}})_{+} \hat{G}_{0} (\varPsi _{\mathrm {HM}}^{\text {out}})^{-1}_{-} \rightarrow I$ uniformly on every closed subinterval of (−α,α).

We define local parametrises near ±α:

$$\begin{aligned} \varPsi _{\mathrm {HM}}^{\alpha }&= \left\{\begin{array}{l@{\quad}l} I & \mathrm{if}\ -\frac {\pi }{3}< \arg(z-\alpha) < \frac {\pi }{3},\\ \hat{G}_1^{-1}& \mathrm{if}\ \frac {\pi }{3}< \arg(z-\alpha) < \pi,\\ \hat{G}_1 & \mathrm{if}\ - \pi< \arg(z-\alpha) < - \frac {\pi }{3}, \end{array}\right.\\ \varPsi _{\mathrm {HM}}^{-\alpha }&= \left\{\begin{array}{l@{\quad}l} I & \mathrm{if}\ \frac {2\pi }{3}< \arg(z+\alpha) < \pi\ \mathrm{or}\ - \pi< \arg(z+\alpha) < - \frac {2\pi }{3},\\ \hat{G}_1^{-1} & \mathrm{if}\ 0 < \arg(z+\alpha) < \frac {2\pi }{3},\\ \hat{G}_1 & \mathrm{if}\ -\frac {2\pi }{3}< \arg(z+\alpha) < \frac {2\pi }{3}. \end{array}\right. \end{aligned}$$

We are ready to define the global parametrix. Given r>0, define

$$\begin{aligned} \varPsi _{\mathrm {HM}}= \left\{\begin{array}{l@{\quad}l} \varPsi _{\mathrm {HM}}^{\alpha }& \mathrm{if}\ |z-\alpha| < r,\\ \varPsi _{\mathrm {HM}}^{-\alpha }& \mathrm{if}\ |z+\alpha| < r,\\ \varPsi _{\mathrm {HM}}^{\text {out}}& \mathrm{if}\ |z+\alpha| > r\ \mathrm{and}\ |z-\alpha| > r. \end{array}\right. \end{aligned}$$

It follows that Ψ _HM satisfies the RH problem shown in Fig. 19.

Fig. 19

The jump contours and jump matrices for the RH problem solved by Ψ _HM. The radius for the two circles is r

Let $\hat{\varPhi}$ be the solution of the RH problem shown in Fig. 20(a). It follows that $\varDelta = \hat{\varPhi} \varPsi _{\mathrm {HM}}^{-1}$ solves the RH problem shown in Fig. 20(b). The RH problem for Δ has jump matrices that decay to the identity away from ±α. We use Assumption 2 to determine that we should use r=|x|⁻¹. We solve the RH problem for Δ numerically. To compute the solution of Painlevé II, we use the formula

$$\begin{aligned} P_{\mathrm{II}}(\pm{ \mathrm {i}},0,\mp{ \mathrm {i}};x) = 2\mathrm{i} \sqrt{|x|}\lim_{z\rightarrow \infty} z \varDelta (z)_{12}. \end{aligned}$$

Fig. 20

The final deformation of the RH problem for negative x, assuming (26). The black dots represent ±α. (a) After conjugation by Θ. (b) Bounding the contours away from the singularities of g and β using Ψ _HM

See Fig. 15 for a plot of the Hastings–McLeod solution with s ₁=−i. To verify our computations, we may we use the asymptotics [16]:

$$\begin{aligned} P_{\mathrm{II}}(-\mathrm{i},0,\mathrm{i};x) \sim \sqrt{\frac{-x}{2}} + \mathcal {O}\bigl(x^{-5/2} \bigr). \end{aligned}$$

(27)

We use

$$\begin{aligned} D_{\mathrm{HM}}(x) = \bigg|\frac{P_{\mathrm{II}}(-{\mathrm {i}},0,{\mathrm {i}};x)}{\sqrt {\frac{-x}{2}}} - 1\bigg| \end{aligned}$$

as an indicator of the relative error, which should tend to zero for x large and negative. We demonstrate this in Fig. 21.

Fig. 21

Analysis of the numerical approximation of P _II(−i,0,i;x). (See Fig. 15 for a plot of the solution for positive and negative x. For small |x| we solve the undeformed RH problem.) A verification the numerical approximation using the asymptotics (27)

Remark

Since β has unbounded singularities, we expect that a similar issue as in Sect. 5.2.2 will arise. We do not go though the details of this, but this approach produces accurate numerics for all x on the real line.