Characterizations of the hydrodynamic limit of the Dyson model

# Characterizations of the hydrodynamic limit of the Dyson model

Sergio Andraus Department of Physics, Graduate School of Science, The University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan.    Makoto Katori Department of Physics, Graduate School of Science and Engineering, Chuo University, 1-13-27 Kasuga, Bunkyo-ku, Tokyo 112-8551, Japan.
###### Abstract

Under appropriate conditions for the initial configuration, the empirical measure of the -particle Dyson model with parameter converges to a unique measure-valued process as goes to infinity, which is independent of . The limit process is characterized by its Stieltjes transform called the Green’s function. Since the Green’s function satisfies the complex Burgers equation in the inviscid limit, this is called the hydrodynamic limit of the Dyson model. We review the relations among the hydrodynamic equation of the Green’s function, the continuity equation of the probability density function, and the functional equation of the Green’s function. The basic tools to prove the relations are the Hilbert transform, a special case of the Sokhotski-Plemelj theorem, and the method of characteristics for solving partial differential equations. For two special initial configurations, we demonstrate how to characterize the limit processes using these relations.

Interacting Brownian motions, Dyson model, Hydrodynamic limit, Hilbert transform, Green’s function, Method of characteristics

## I Introduction

For , let be an -dimensional standard Brownian motion in a probability space with a filtration . Let denote the Weyl chamber of type A,

 WN={x=(x1,x2,…,xN)∈RN:x1

with closure . Dyson’s Brownian motion model of particles with parameter , , is defined as the solution of the following system of stochastic differential equations (SDEs) dyson62 (),

 dXNi(t)=dBi(t)+β2∑1≤j≤N,j≠idtXNi(t)−XNj(t),t≥0,1≤i≤N,

started at . Here we perform the time change

 βN2t→t

and consider the situation such that the SDEs are given as

 (1) dXNi(t)=√2βNdBi(t)+1N∑1≤j≤N,j≠idtXNi(t)−XNj(t),t≥0,1≤i≤N,

with . In this paper we simply call this system of interacting Brownian motions on the Dyson model AGZ10 (); katori16 ().

Let be the space of probability measures on equipped with its weak topology. For , denotes the space of continuous processes defined in the time period realized in . We regard the empirical measure of the solution of (1),

 (2) ΞN(t,⋅)=1NN∑i=1δXNi(t)(⋅),t∈[0,T],

as an element of . We express its initial configuration by

 (3) ξN(⋅)≡ΞN(0,⋅)=1NN∑i=1δxNi(⋅)∈M.

The following is proved.

###### Theorem I.1 (Anderson, Guionnet, Zeitouni (Agz10, , Proposition 4.3.10)).

Let be a sequence of initial configurations such that ,

 supN≥01NN∑i=1log{(xNi)2+1}<∞,

and converges weakly to a measure as . Then for any fixed ,

 (4) (ΞN(t,⋅))t∈[0,T]⟹∃!(μ(t,⋅))t∈[0,T]a.s. in C([0,T]→M),

where and the function

 (5) G(t,z)=∫Rdμ(t,x)z−x

satisfies the equation

 (6) ∂G(t,z)∂t+G(t,z)∂G(t,z)∂z=0,t∈[0,T],z∈C∖R.

The function defined by the Stieltjes transform (5) is called the Green’s function (or the resolvent) for the measure-valued process . The equation (6) can be regarded as the complex Burgers equation in the inviscid limit (i.e., the (complex) one-dimensional Euler equation). Thus the limit given by this theorem is called the hydrodynamic limit of the Dyson model AGZ10 (); blaizotnowak10 (); forrestergrela15 ().

For , assume that a function is bounded and integrable over and for any . We introduce the Cauchy principal value at for the integral of over as

if the limit exists and is finite. For , the Hilbert transform is defined by

 (7)

Note that is real-valued for all , by definition.

In this paper we consider the case in which has a probability density function for any . That is, we can write

 μ(t,A)=∫Aρ(t,x)dx,t∈[0,∞),

for any Borel set . We assume the following conditions for .

[C1]

is piecewise differentiable with respect to and ; the piecewise-defined derivatives are written as , .

[C2]

, , , for each . Thus, their Hilbert transforms are well-defined.

From now on, for piecewise differentiable functions, the derivatives are assumed to be piecewise-defined.

One of the purposes of the present paper is to give a precise proof to the following statement.

###### Theorem I.2.

Assume that the Green’s function for the process is given by

 (8) G(t,z)=∫Rρ(t,x)z−xdx,t∈[0,∞),z∈C∖R,

where satisfies the conditions [C1] and [C2]. Then the partial differential equation (6) for leads to the following two characterizations for the process.

(i)

The probability density function satisfies the following equation

 (9) ∂ρ(t,x)∂t+π∂∂x{ρ(t,x)H[ρ(t,⋅)](x)}=0.
(ii)

The following functional equation is solved by ,

 (10) G(t,z)=G(0,z−tG(t,z)),t∈[0,∞).

Equation (9) is the continuity equation of the density function of the system,

 ∂ρ(t,x)∂t=−∂J(t,x)∂x,t∈[0,∞),x∈R,

with the current density function

 J(t,x) = πρ(t,x)H[ρ(t,⋅)](x) = −ρ(t,x)∂∂x∫Rρ(t,y)V(x−y)dy,

associated with the logarithmic potential . In other words, (9) will provide the reaction-diffusion equation which governs the macroscopic behavior of the one-dimensional log-gas forrester10 (). An interesting and important fact is that the formula

 (11) ρ(t,x)=−I[limε↓01πG(t,x+√−1ε)],t∈[0,∞),x∈R,

is established (see the comment given above Proposition II.1 in the next section) and it is easier to obtain by solving the functional equation (10) for and using the formula (11) rather than by solving (9). Another purpose of the present paper is to demonstrate the usefulness of the functional equation (10).

The paper is organized as follows. In Section II we prove the formula (11) and Propositions concerning the basic properties of the Hilbert transform. Section III is devoted to proving Theorem I.2. There we will use the properties of the Hilbert transform given in Section II in order to prove (i) of Theorem I.2. Then the method of characteristics AGZ10 (); blaizotnowak08 (); blaizotnowak10 (); forrestergrela15 () is applied to prove (ii) of Theorem I.2. In Section IV we demonstrate how to solve the functional equation (10) for and determine through the formula (11) for two special cases of the initial data . In both cases, the support of is bounded on and the conditions [C1] and [C2] are clearly satisfied.

## Ii Basic properties of the Hilbert transform

The relation (11) will be obtained as the imaginary part of the upper equation in (12) in the following proposition. This proposition is a special case of the Sokhotski-Plemelj theorem.

###### Proposition II.1.

For any , the Hilbert transform of and the Green’s function are related by

 (12) limε↓01πG(t,x±√−1ε)=H[ρ(t,⋅)](x)∓√−1ρ(t,x),x∈R.
###### Proof.

Consider a closed, simple, and positively-oriented contour on and a complex function which is analytic on . Denote by the open region enclosed by . We define the function

 ϕ(z):=12π√−1∫Cf(w)w−zdw.

Clearly, is well-defined when , but it is discontinuous at . Choose and consider the limits when tends to from the inside and from the outside of . Assume, furthermore, that is smooth at . In the first case, there exists an angle such that

 ϕin(ζ):=limz→ζ,z∈Dϕ(z) =12π√−1\vbox−∫C∖{ζ}f(w)w−ζdw+limε↓012π√−1∫π0f(ζ+εe√−1(α+θ))εe√−1(α+θ)√−1εe√−1(α+θ)dθ =12π√−1\vbox−∫C∖{ζ}f(w)w−ζdw+12f(ζ).

The principal-value integral is taken along while excluding the point . The case where approaches from the outside of can be calculated similarly:

We assume that, for each fixed , the domain of the probability density function can be extended from into the complex plane. Then we specialize the above result for the case where with a fixed . We choose to be the contour given by the parameterizations with going from to , and with going from to , while letting tend to infinity (see Figure 1).

Taking , we have

 (13) ϕin(x) = 12π√−1\vbox−∫R∖{x}ρ(t,y)y−xdy+12ρ(t,x), (14) ϕout(x) = 12π√−1\vbox−∫R∖{x}ρ(t,y)y−xdy−12ρ(t,x).

The reason for this is the following. By the condition [C2], as , and we will be able to extend to a function on the complex upper half-plane so that . Hence, the part of the integral with a semi-circular contour vanishes:

 limR→∞∫π0ρ(t,Re√−1θ)Re√−1θ√−1Re√−1θdθ=limR→∞√−1∫π0ρ(t,Re√−1θ)dθ=0.

This also implies that, for ,

 (15) 2π√−1ϕ(z)=∫Rρ(t,y)y−zdy=−G(t,z),z∈C∖R.

The contour we have chosen covers the complex upper half-plane, so if approaches the real axis from above (resp., below), we must use (resp., ). Then, we obtain

 limε↓0G(t,x+√−1ε)=−2π√−1ϕin(x),limε↓0G(t,x−√−1ε)=−2π√−1ϕout(x),x∈R.

The result (12) follows by the definition (7) of the Hilbert transform applied to (13) and (14). ∎

Now we give the basic properties of the Hilbert transform, which will be used to prove Theorem I.2 in the next section.

###### Proposition II.2.

For , then the inverse Hilbert transform of is given by , that is,

 (16) H[H[f]](x)=−f(x).
###### Proof.

In the proof of Proposition II.1, we gave an expression (13) for for given by (15) for . This result implies the following expression for the Hilbert transform which is different from (7),

 (17) H[f](x)=√−1f(x)−limε↓01π∫Rf(y)y−(x+√−1ε)dy.

The Hilbert transform is doubly performed using this expression as

 (18) H[H[f]](x)=−f(x)−limε↓02√−1π∫Rdyf(y)y−(x+√−1ε) −limε1↓0limε2↓01π2∫Rdy∫Rdzf(z)[y−(x+√−1ε1)][y−(z−√−1ε2)].

In the double integral in the third term, we evaluate the integral over as the integral over the contour depicted in Figure 1, which encloses including a simple pole at . By Cauchy’s integral formula, the third term is calculated as

 −limε1↓0limε2↓01π2∫Rdz2π√−1f(z)(x+√−1ε1)−(z−√−1ε2) = limε1↓0limε2↓02√−1π∫Rdzf(z)z−{x+√−1(ε1+ε2)}.

This cancels the second term in (18) and (16) is obtained. ∎

###### Proposition II.3.

Assume that has a piecewise-defined derivative , and , . Then

 dH[f](x)dx=H[dfdx](x),x∈R.
###### Proof.

We use the expression (17) for the Hilbert transform,

 H[dfdx](x) = √−1df(x)dx−limε↓01π∫Rdf(y)dy1y−(x+√−1ε)dy = √−1df(x)dx+limε↓01π∫Rf(y)ddy[1y−(x+√−1ε)]dy.

The second equality is obtained from an integration by parts, where we use the assumption that , so vanishes at infinity. We change the variable of differentiation from to inside the integral. Then the above is equal to

where (17) is again used. Then the proof is completed. ∎

For the following proposition, it will be useful to know how the Hilbert transform behaves when a Fourier transform is present. Here we rewrite the Hilbert transform as

 H[f](x)=[f⋆g](x),

where denotes the convolution product

 [f⋆g](x):=∫Rf(y)g(x−y)dy,

and . The integral is interpreted as a principal value where necessary. It is clear that the Fourier transform defined by

 F[f](ν):=∫Rf(x)e−2π√−1xνdx

transforms the Hilbert transform of into , where and are the Fourier transforms of and , respectively. It is easy to verify that is given by

 (19) Γ(ν)=−√−1 sign(ν):=⎧⎪⎨⎪⎩√−1,if ν<0,0,if ν=0,−√−1,if % ν>0.

In summary, with (19), we obtain the formula

 (20) F[H[f]](ν)=Γ(ν)F[f](ν).

We will proceed with the proof of the following proposition.

###### Proposition II.4 (Carton-Lebrun cartonlebrun77 ()).

Assume that , with and . Then

 (21) H[f](x)H[h](x)−f(x)h(x)=H[fH[h]+H[f]h](x)a.e.

Before the proof of this statement, we establish a helpful equation cartonlebrun77 ().

###### Lemma II.5.

Assume that . Then with given by (19),

 [ΓΦ⋆ΓΘ](x)−[Φ⋆Θ](x)=Γ(x){[Φ⋆ΓΘ](x)+[ΓΦ⋆Θ](x)}.
###### Proof.

We perform a straightforward calculation. The first part of the LHS gives

 [ΓΦ⋆ΓΘ](x)=−∫Rsign[y(x−y)]Φ(y)Θ(x−y)dy,

so the LHS gives

 [ΓΦ⋆ΓΘ](x)−[Φ⋆Θ](x)=−2sign(x)∫x0Φ(y)Θ(x−y)dy.

For the RHS, we obtain

 Γ(x){[Φ⋆ΓΘ](x)+[ΓΦ⋆Θ](x)} =−sign(x)∫R[sign(y)+sign(x−y)]Φ(y)Θ(x−y)dy =−2sign(x)∫x0Φ(y)Θ(x−y)dy,

which is identical to the LHS, as desired. ∎

###### Proof of Proposition ii.4.

Here we consider the case . We take the Fourier transform of (21). Set and . By (20) the LHS gives

 F[H[f]H[h]−fh](ν)=[ΓΦ⋆ΓΘ](ν)−[Φ⋆Θ](ν).

By virtue of Lemma II.5, this is equal to . Again using (20), this is rewritten as

and thus we arrive at the equality,

Taking the inverse Fourier transform of this equation yields the result. For the general case , see the proof given in cartonlebrun77 (). ∎

Setting gives the following.

###### Corollary II.6.

For with ,

 (22) H[fH[f]](x)=12{H[f](x)2−f(x)2}.

## Iii Proof of Theorem i.2

In this section, we prove Theorem I.2.

###### Proof of (i).

For any , the Green’s function is analytic in . Then the following equation is guaranteed by (6),

 ∂G(t,x+√−1y)∂t+G(t,x+√−1y)∂G(t,x+√−1y)∂x=0,x∈R,y∈R∖{0}.

Put and take the limit . By Proposition II.1, we will obtain

 ∂∂t{H[ρ(t,⋅)](x)−√−1ρ(t,x)}

This is written as

 (23) {H[∂ρ(t,⋅)∂t](x)+π∂∂x12{H[ρ(t,⋅)](x)2−ρ(t,x)2}} −√−1{∂ρ(t,x)∂t+π∂∂x{ρ(t,x)H[ρ(t,⋅)](x)}}=0.

Applying Corollary II.6 and Proposition II.3, we have

 ∂∂x12{H[ρ(t,⋅)](x)2−ρ(t,x)2}=∂∂xH[ρ(t,⋅)H[ρ(t,⋅)]](x) =H[∂∂x{ρ(t,⋅)H[ρ(t,⋅)]}](x)

Therefore, (23) is equivalent with

 H[A(t,⋅)](x)−√−1A(t,x)=0⟺%$H[A(t,⋅)](x)=0$and$A(t,x)=0$,

with

for . Since the Hilbert transform is invertible by Proposition II.2, the equation (9) is obtained. ∎

###### Proof of (ii).

We apply the method of characteristics AGZ10 (); blaizotnowak08 (); blaizotnowak10 (); forrestergrela15 () to solve the partial differential equation (6). We consider a parametrization such that is constant for . That is, we construct a differentiable curve in along which . This construction leads to

 dtdr∂G(t,z)∂t+dzdr∂G(t,z)∂z=0.

Comparing this equation to (6) gives

 (24) dtdr=1,dzdr=G(t,z).

We can derive the explicit form of the parametrization from these equations. Clearly, and differ only by a constant, so we set , and use as the parametrization variable. For we have

 dzdt=G(t,z)=G(t0,z0).

The second equality is derived from the requirement that be constant along the curve , where is the value of at . Integrating this equation yields

 (25) z=tG(t0,z0)+z0.

Thus we obtain the equalities

 G(t,z)=G(t0,z−tG(t0,z0))=G(t0,z−tG(t,z)).

Without loss of generality, we can choose to obtain (10). ∎

## Iv Solutions of hydrodynamic equations for special initial configurations

### iv.1 Case with one source

We consider the case where all the particles are located at a single point when . Without loss of generality, we can choose the origin as the starting point, i.e.,

 μ(⋅)=δ0(⋅)⟺dμ(x)=ρ(0,x)dx=δ(x)dx,

where denotes Dirac’s delta function. Consequently,

 (26) G(0,z)=∫Rρ(x,0)dxz−x=∫Rδ(x)dxz−x=1z.

Then, (10) becomes

 tG2(t,z)−zG(t,z)+1=0.

This algebraic equation for is solved by

 (27) G(t,z)=12t(z±√z2−4t)=12t{z±√(z−2√t)(z+2√t)},t>0.

By the formula (11),

 ρ(t,x)=−12πtI{x±limε↓0√(x+√−1ε−2√t)(x+√−1ε+2√t)},

but it should be that , so we choose the lower (minus) sign. By taking the limit we obtain

 ρ(t,x)=⎧⎨⎩12πt√4t−x2,if |x|<2√t,0,if |x|≥2√t,

for . This is the time-dependent version of Wigner’s semicircle law AGZ10 (); katori16 ().

Note that the two edges of the spectrum, , coincide with the conditions under which the method of characteristics breaks down for the real characteristics. We fix and consider a map from to

 Mt(x0):=limε↓0tG(0,x0+√−1ε)+x0∈R.

If this map is injective for a domain , the real characteristic curves in the set do not cross and to each of them corresponds a distinct value of ; . That is, the method of characteristics works. In the present case, (26) gives

 Mt(x0)=tx0+x0.

The value of at which the method of characteristics breaks down, denoted by , is found by

 dMtdx0(x0,c)=−tx20,c+1=0⟺x0,c=±√t.

The domain on for which the method of characteristics works is given by . Actually, if , (27) with the lower (minus) sign gives

 G(s,Ms(x0)) = 12s⎧⎨⎩sx0+x0−√(sx0−x0)2⎫⎬⎭ = 1x0=G(0,x0),0≤s≤t.

The image of is given by

 It={Mt(x0):x0∈Dt}={x∈R:|x|≥|Mt(x0,c)|=2√t},

and the equality is established, . See Figure 2.

We put these observations in a more formal context in the following proposition.

###### Proposition IV.1.

At each define

 Dt:={x0∈R:x=Mt(x0) gives an % injection to R},It:={Mt(x0):x0∈Dt}.

Then for ,

 (28) supp[ρ(t,⋅)]=R∖It.
###### Proof.

Assume that . Then the method of characteristics works and

 ∃x0∈Dts.t.x=limε↓0tG(0,x0+√−1ε)+x0.

This implies that . By the equality (12) in Proposition II.1, is concluded, that is, .

Conversely, assume that . Then, and by Proposition II.1. Consider now the mapping

 (29) M−s(x):=x−sG(t,x), 0

By equation (6), it follows that

 ddsG(t−s,M−s(x))∣∣s=0=0.

Therefore, remains constant along the line . Note that is an injective mapping, because

 dM−s(x)dx=1−sddxG(t,x)=1+s∫supp[ρ(t,⋅)]ρ(t,u)(x−u)2du>0.

This means that the method of characteristics works for , so we set and find that

 G(t,x)=G(0,x0), and x=x0+tG(0,x0).

Consequently, and the statement (28) is proved. ∎

### iv.2 Case with two sources

Now we consider the case where, for ,

 μ(⋅)=12{δ−a(⋅)+δa(⋅)}⟺dμ(x)=ρ(0,x)dx=12{δ(x−a)+δ(x+a)}dx.

Then, the initial condition for is given by

 G(0,z)=12(1z−a+1z+a),

and (10) becomes

 (30) t2G3(t,z)−2ztG2(t,z)+(z2−a2+t)G(t,z)−z=0.

Note that, if we set and with , (30) is transformed into

 (31) τ2¯G3(τ,w)−2wτ¯G2(τ,w)+(w2−1+τ)¯G(τ,w)−w=0.

Therefore, without loss of generality we solve (30) for and assume that time and space are given in units of and , respectively.

Before we solve (30), we use Proposition IV.1 to determine the support of the particle density. The parametrization equation for the real characteristics is given by

 Mt(x0)=t2(1x0−1+1x0+1)+x0=x0(1−t1−x20),t∈[0,∞).

The breakdown condition for the injective map from to with fixed is

 (32) dMtdx(x0,c)=0⟺t=(1−x20,c)21+x20,c.

We show the plot of several characteristics and the breakdown curve in Figure 3.