GLOBAL DYNAMICS OF BOSE–EINSTEIN CONDENSATION4)-2016_2454_2494.pdfGLOBAL DYNAMICS OF BOSE–EINSTEIN CONDENSATION 2455 (1.1) @tf = 1 x2 @x ⇥ x4 @xf +f +f2 ⇤. This Fokker–Planck

SIAM J. MATH. ANAL. c� 2016 Society for Industrial and Applied Mathematics

Vol. 48, No. 4, pp. 2454–2494

GLOBAL DYNAMICS OF BOSE–EINSTEIN CONDENSATIONFOR A MODEL OF THE KOMPANEETS EQUATION⇤

C. DAVID LEVERMORE† , HAILIANG LIU‡ , AND ROBERT L. PEGO§

Abstract. The Kompaneets equation describes a field of photons exchanging energy by Comptonscattering with the free electrons of a homogeneous, isotropic, nonrelativistic, thermal plasma. Thispaper strives to advance our understanding of how this equation captures the phenomenon of Bose–Einstein condensation through the study of a model equation. For this model we prove existenceand uniqueness theorems for global weak solutions. In some cases a Bose–Einstein condensate willform in finite time, and we show that it will continue to gain photons forever afterward. Moreoverwe show that every solution approaches a stationary solution for large time. Key tools include auniversal super solution, a one-sided Oleinik type inequality, and an L1 contraction.

Key words. Kompaneets equation, Bose–Einstein condensate, quantum entropy, LaSalleinvariance principle

AMS subject classifications. 35K55, 35B40, 35Q85

DOI. 10.1137/15M1054377

1. Introduction. Photons can play a major role in the transport of energy in afully ionized plasma through the processes of emission, absorption, and scattering. Athigh temperatures or low densities, the dominant process can be Compton scatteringo↵ free electrons. We make the simplification that the plasma is spatially uniform,isotropic, nonrelativistic, and thermal at temperature T . We also neglect the heatcapacity of the photons and assume that T is fixed. If the photon field is also spatiallyuniform and isotropic, then it can be described by a nonnegative number densityf(x, t) over the unitless photon energy variable x 2 (0,1) given by

x =~|k|ckB

T,

where ~ is Planck’s constant, c is the speed of light, kB

is Boltzmann’s constant, andk is the photon wave vector. Because x is a unitless radial variable, the total photonnumber and (unitless) total photon energy associated with f(x, t) are then given by

N [f ] =

Z 1

0

f x2 dx , E[f ] =

Z 1

0

f x3 dx .

When the only energy exchange mechanism is Compton scattering of the photons bythe free electrons in the plasma, then the evolution of f is governed bythe Kompaneets equation [20]

⇤Received by the editors December 28, 2015; accepted for publication (in revised form) May 23,2016; published electronically July 19, 2016. This research was supported by the National ScienceFoundation under NSF Research Network grants RNMS11-07444 and RNMS11-07291(KI-Net).

http://www.siam.org/journals/sima/48-4/M105437.html†Department of Mathematics, University of Maryland, College Park, MD 20742-4015

([email protected]).‡Department of Mathematics, Iowa State University, Ames, IA 50011 ([email protected]). The

research of this author was supported by grants DMS 0907963 and DMS 1312636.§Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213-3890

([email protected]). The research of this author was partially supported by the Center for NonlinearAnalysis (CNA) under National Science Foundation PIRE grant OISE-0967140 and by grants DMS0905723, DMS 121161 and DMS 1515400.

2454

http://www.siam.org/journals/sima/48-4/M105437.html

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GLOBAL DYNAMICS OF BOSE–EINSTEIN CONDENSATION 2455

(1.1) @t

f =1

x2

@x

⇥x4

�@x

f + f + f2

�⇤.

This Fokker–Planck approximation to a quantum Boltzmann equation is justifiedphysically by arguing that little energy is exchanged by each photon-electron collision.

Because x is a radial variable, the associated divergence operator has the formx�2@

x

x2. Thereby we see from (1.1) that the di↵usion coe�cient in the Kompaneetsequation is x2, which vanishes at x = 0. This singular behavior allows the f2 convec-tion term to drive the creation of a photon concentration at x = 0. This hyperbolicmechanism models the phenomenon of Bose–Einstein condensation. Our goal is tobetter understand how the Kompaneets equation generally describes the process ofrelaxation to equilibrium over large time and how it captures the phenomenon ofBose–Einstein condensation in particular.

Rather than addressing these questions for the Kompaneets equation (1.1) we willconsider the model Fokker–Planck equation

(1.2) @t

f =1

x2

@x

⇥x4

�@x

f + f2

�⇤,

posed over x 2 (0, 1) and subject to a zero flux boundary condition at x = 1. Thismodel is obtained by simply dropping the f term that appears in the flux of theKompaneets equation (1.1) and reducing the x-domain to (0, 1). As we will see, thismodel shares many structural features with the Kompaneets equation. In particular,it shares the x2 di↵usion coe�cient and the f2 convection term that allow the onset ofBose–Einstein condensation. The neglect of the f term in the flux of the Kompaneetsequation is a reasonable approximation during the onset of Bose–Einstein condensa-tion when we expect f to be large. The advantage of model (1.2) is that we knowsome estimates for it that have no known analogues for (1.1) and which facilitatethe study of condensate dynamics and equilibration. A disadvantage of (1.2) is thatits equilibrium solutions di↵er from those of (1.1), so we may expect the long-timebehavior of its solutions to be similar to that of solutions of (1.1) only in a qualitativesense.

1.1. Structure of the Kompaneets equation. Here we describe some struc-tural features of the Kompaneets equation that will be shared by our model. First,solutions of (1.1) formally conserve total photon number N [f ]. Indeed, we formallycompute that

d

dtN [f ] = x4

�@x

f + f + f2

��1

0

= 0

under the expectation that the flux vanishes as x approaches 0 and 1. Second,solutions of (1.1) formally dissipate quantum entropy H[f ] given by

H[f ] =

Z 1

0

h(f, x)x2dx , h(f, x) = f log(f)� (1 + f) log(1 + f) + xf .

Indeed, because

hf

(f, x) = log(f)� log(1 + f) + x = log

✓exf

1 + f

◆,

@x

hf

= hff

@x

f + 1 =1

f(1 + f)(@

x

f + f + f2) ,

2456 C. D. LEVERMORE, H. LIU, AND R. L. PEGO

we formally compute that

d

dtH[f ] =

Z 1

0

hf

(f, x) (@t

f)x2dx = �Z 1

0

x4f(1 + f)�@x

hf

(f, x)�2

dx 0 .

By this “H theorem,” we expect solutions to approach an equilibrium for which@x

hf

(f, x) = 0. These equilibria have the Bose–Einstein form

f = fµ

(x) =1

ex+µ � 1for some µ � 0 .

At this point a paradox arises. The total photon number for the equilibriumfµ

is

N [fµ

] =

Z 1

0

x2

ex+µ � 1dx .

This is a decreasing function of µ over [0,1) and is thereby bounded above by N [f0

],which is finite. Because total number is supposed to be conserved, we expect anysolution to relax to an equilibrium f

µ

with the same total number as the initial data,satisfying N [f

µ

] = N [f in]. But if the initial number N [f in] > N [f0

], then no suchequilibrium exists!

1.2. Bose–Einstein condensation. The foregoing paradox indicates that theremust be a breakdown in the expectations given above. Previous studies ([6] in par-ticular) have shown that a breakdown in the no-flux condition at x = 0 can occur. Aphysical interpretation of a nonzero photon flux at x = 0 is that the photon distri-bution forms a concentration of photons at zero energy (i.e., energy that is negligibleon the scales described by the model). This Bose–Einstein condensate accounts forsome of the total photon number. See especially the works [31, 3, 6, 12] and the dis-cussion of related literature in subsection 1.4 below. As massless, chargeless particlesof integer spin, photons are the simplest bosons. Indeed, S. N. Bose had photons inmind in 1924 when he proposed his new way of counting indistinguishable particles,work soon followed by Einstein’s prediction of the existence of the condensate. Yet itwas not until 2010 that the first observation of a photon condensate was reported byMartin Weitz and colleagues [19].

In the present context, we can gain insight into this phenomenon by dropping thedi↵usion term in (1.1), as discussed by Zel’dovich and Levich [31]. In this case theKompaneets equation simplifies to the first-order hyperbolic equation

@t

f =1

x2

@x

⇥x4

�f + f2

�⇤.

Letting n = x2f , this becomes

(1.3) @t

n = @x

⇥x2n+ n2

⇤,

whose characteristic equations are

x = �x2 � 2n , n = 2xn .

Because n � 0, the origin x = 0 is an outflow boundary, and no boundary conditioncan be specified there. Clearly any nonzero entropy solution will develop a nonzeroflux of photons into the origin in finite time, leading to the formation of a condensate.


The fact the f2 convection term plays an essential role in the formation of Bose–Einstein condensates is illustrated by considering what happens when that term isdropped from the Kompaneets equation (1.1). This leads to the linear degenerateparabolic equation

(1.4) @t

f =1

x2

@x

⇥x4

�@x

f + f�⇤

.

This equation is the analogue of the Kompaneets equation for classical statistics. Itssolutions formally conserve N [f ] and dissipate the associated entropy

H[f ] =

Z 1

0

h(f, x)x2dx , where h(f, x) = f log(f)� f + xf .

Its family of equilibria is

fµ

(x) = e�x�µ for some µ 2 R .

The initial-value problem for (1.4) is well-posed in cones of nonnegative densities fsuch that Z 1

0

(exf)pe�xx2dx < 1 for some p 2 (1,1) .

Kavian and Levermore showed [18] (see [17]) that these solutions• are smooth over R+ ⇥ R+,• are positive over R+ ⇥ R+ provided that f in is nonzero,• satisfy all the expected boundary conditions,• conserve N [f ] and dissipate H[f ] as expected,• approach f

µ

as t ! 1, where N [f in] = N [fµ

].In particular, the no-flux boundary condition is satisfied at x = 0 without beingimposed! Therefore, no Bose–Einstein concentration happens!

1.3. Present investigation. Of course, solutions of the hyperbolic model (1.3)may develop shocks at any location. However, the di↵usion term in the Kompaneetsequation (1.1) prevents shock formation for x > 0. The results of Escobedo, Herrero,and Velazquez [6] prove that the degeneracy of its di↵usion does not prevent shockformation at x = 0. These authors proved that there exist solutions of (1.1) thatare regular and satisfy no-flux conditions for t on a bounded interval 0 < t < T

c

(which is solution-dependent), but at time t = Tc

the flux at x = 0 becomes nonzero.Such solutions exist for an arbitrarily small initial photon number. Moreover, globalexistence and uniqueness of solutions of (1.1) was proved subject to a boundednesscondition for x2f for x 2 [0, 1].

A number of interesting questions about solutions to the Kompaneets equationremain unanswered by previous studies: What happens to a condensate once it forms?Can it lose photons as well as gain them? Are there any boundary conditions at allthat we can impose near x = 0 that yield di↵erent condensate dynamics, allowing thecondensate to interact with other photons? Can we identify the long-time limit ofany initial density of photons?

In order to focus clearly on these questions, we have found it convenient to dropthe linear term x2f from the Kompaneets flux and consider the model equation (1.2),which retains the essential features of nonlinearity and degenerate di↵usion. Theequilibria of (1.2) are

(1.5) fµ

(x) =1

x+ µfor some µ � 0 .


We include these solutions in the class of functions considered by restricting ourattention to the interval 0 < x < 1 and imposing a no-flux boundary condition atx = 1. For these equilibria the maximal total photon number is N [f

0

] = 1

2

.For this model problem, we shall assemble a fairly detailed description of well-

posedness and long-time dynamics. We establish existence and uniqueness in a naturalclass of nonnegative weak solutions for initial data that simply has some finite moment

Z 1

0

xpf in dx , p � 2 .

These results are proved with essential use of estimates for hyperbolic (first-order)equations and establish that while the model Kompaneets equation (1.4) is parabolicfor x > 0, the point x = 0 remains an outflow boundary at which no boundarycondition can be specified.

The solution map is nonexpansive in L1 norm with weight x2. Therefore the totalphoton number N [f(t)] is nonincreasing in time. A condensate can gain photons butnever lose them and must form in finite time whenever N [f in] > N [f

0

]. Moreover,once it starts growing it never stops. Every solution relaxes to some equilibrium statefµ

in the long-time limit t ! 1. We cannot identify the limiting state in general, butthe solution must approach the maximal steady state f

0

if the initial data fin

� f0

everywhere.The proofs of the results on long-time behavior are greatly facilitated by two

features of the model problem (1.2). First, the problem admits a universal super-solution f

super

determined by

(1.6) x2fsuper

(x, t) = x+1� x

t+

2pt.

By consequence, for every solution, x2f is in fact bounded in x for each t > 0, andmoreover one has lim sup

t!1 x2f(x, t) x = x2f0

(x) for every solution, for example.Also, every solution satisfies

@x

(x2f) � �4

t,

which is Oleinik’s inequality for admissible solutions of the conservation law (1.3)after dropping the linear flux term x2n.

1.4. Literature on related problems. As indicated above, the Kompaneetsequation is derived from a Boltzmann–Compton kinetic equation for photons inter-acting with a gas of electrons in thermal equilibrium—see [20], and especially [8] for aderivation and links to some of the physical literature. Regarding the analysis of theBoltzmann–Compton equation itself, when a simplified regular and bounded kernel isadopted, Escobedo and Mischler [7] studied the asymptotic behavior of the solutionsand showed that the photon distribution function may form a condensate at zeroenergy asymptotically in infinite time. Further, Escobedo, Mischler, and Velazquez[9] showed that the asymptotic behavior of solutions is sensitive not only to the totalmass of the initial data but also to its precise behavior near the origin. In some cases,solutions develop a Dirac mass at the origin for long times (in the limit t ! 1) ina self-similar manner. For the Boltzmann–Compton equation with a physical kernel,some results concerning both global existence and nonexistence, depending on the sizeof initial data, were obtained by Ferrari and Nouri [11].

A natural question is whether results analogous to those obtained in the presentpaper concerning the development of condensates may hold for other kinetic equa-tions that govern boson gases, such as Boltzmann–Nordheim (aka Uehling–Uhlenbeck)


quantum kinetic equations. Concerning these issues we refer to the studies of Semikozand Tkachev [28, 27] and Lacaze et al. [22], the analyses of Spohn [29] and Lu [23],the recent analysis of blowup and condensation formation by Escobedo and Velazquez[10], and references cited therein. Higher-order Fokker–Planck-type approximationsto the Boltzmann–Nordheim equation were derived formally by Josserand, Pomeau,and Rica [12], and an analysis of the behavior of solutions has been performed recentlyby Jungel and Winkler [13, 14].

Bose–Einstein equilibria and condensation phenomena also appear in classicalFokker–Planck models that incorporate a quantum-type exclusion principle [15, 16].Concerning mathematical results on blowup and condensates for these models, werefer to work of Toscani [30] and Carrillo, Di Francesco, and Toscani [4] and referencestherein.

1.5. Plan of the paper. In section 2 we introduce our notion of weak solutionsfor (1.2) together with relevant notations, followed by precise statements of the mainresults and a discussion of related literature. In section 3 we prove the uniquenessof weak solutions for initial data with some finite moment. Existence is proved insection 4 by passing to the limit in a problem regularized by truncating the domainaway from x = 0.

In section 5 we establish that condensation must occur if the initial photon numberN [f

in

] > N [f0

], and we show that once a shock forms at x = 0 in finite time, it willpersist and continue growing for all later time. Large time convergence to equilibriumis proved for every solution in section 6, using arguments related to LaSalle’s invarianceprinciple.

The paper concludes with three appendices that deal with several technical butless central issues. A simple, self-contained treatment of some anisotropic Sobolevembedding estimates used in our analysis is contained in Appendix A. The truncatedproblem used in section 4 requires a special treatment due to the fact that the zero-fluxboundary condition at x = 1 is nonlinear—this treatment is carried out in AppendixB. A proof of interior regularity of the solution, su�cient to provide a classical solutionaway from x = 0 but up to the boundary x = 1, is established in Appendix C.

2. Main results.

2.1. Model initial-value problem. In light of the foregoing discussion, it isconvenient to work with the densities

(2.1) n = x2f , nin = x2f in .

The flux in our model equation (1.2) can be expressed as

(2.2) J = x2@x

n+ n2 � 2xn .

The initial-value problem for our model equation (1.2) that we will consider is

@t

n� @x

J = 0 , 0 < x < 1 , t > 0 ,(2.3a)

J(1, t) = 0 , t > 0 ,(2.3b)

n(x, 0) = nin(x), 0 < x < 1 .(2.3c)

Here we have imposed the no-flux boundary condition at x = 1 but do not imposeany boundary condition at x = 0, where the di↵usion coe�cient x2 vanishes.


We work with a weak formulation of the initial-value problem (2.3). We requirethe initial data nin to satisfy

(2.4) nin � 0 , xpnin 2 L1((0, 1]) for some p � 0 .

Let Q = (0, 1] ⇥ (0,1). We say n is a weak solution of the initial-value problem(2.3) if

n � 0 , n , @x

n 2 L2

loc

(Q) ,(2.5a)

xpn 2 L1((0, 1]⇥ (0, T )) for every T > 0 ,(2.5b)

n(·, t) ! nin in L1

loc

((0, 1]) as t ! 0+ ,(2.5c)Z

Q

�n @

t

� J @x

�dX = 0 (dX = dx dt)(2.5d)

for every C1 test function with compact support in Q. Condition (2.5a) is neededto make sense of the weak formulation (2.5d). Condition (2.5b) is an admissibilitycondition we need to establish uniqueness. Condition (2.5c) gives the sense in whichthe initial data is recovered.

2.2. Uniqueness, existence, and regularity. The following results establishthe basic uniqueness, existence, and regularity properties of weak solutions to (2.3).Henceforth we will use N [n] to denote the total photon number,

N [n] =

Z1

0

ndx ,

replacing the earlier notation N [f ]. We will also denote the positive part of a numbera by a

+

= max{a, 0}.Theorem 1 (stability and comparison). Let nin and nin satisfy (2.4) for some

p � 0. Let n and n be weak solutions of (2.3) associated with the initial data nin andnin, respectively, as defined by (2.5). Set c

p

= p(p+ 3). Then

(2.6)

Z1

0

xp(n� n)+

(x, t) dx ecptZ

1

0

xp(nin � nin)+

dx a.e. t > 0 .

Furthermore, if nin � nin a.e. on (0,1), then n � n a.e. on Q. In particular, ifnin = nin a.e. on (0,1), then n = n a.e. on Q.

From (2.6) we draw immediately the following conclusion on uniqueness.

Corollary 2 (uniqueness). Let n and n be two weak solutions to ( 2.3), subjectto initial data nin, nin, respectively, with xpnin, xpnin 2 L1((0, 1]). Then

(2.7)

Z1

0

xp|n(x, t)� n(x, t)| dx ecptZ

1

0

xp|nin � nin| dx a.e. t > 0.

For each initial data nin satisfying xpnin 2 L1(0, 1) for some p � 0, there exists atmost one weak solution of ( 2.3).

Remark 3. Because c0 = 0, if (2.4) holds with p = 0, then (2.7) is the L1 con-traction property

(2.8) k(n� n)(t)kL

1(0,1)

knin � ninkL

1(0,1)

.

In particular, the total photon number N [n] is nonincreasing in time.


Theorem 4 (existence and global bounds). Let nin satisfy (2.4) for some p � 0.Then there exists a unique global weak solution n of (2.3) as defined by (2.5). Moreoverxpn 2 C([0,1);L1(0, 1)) and we have the following bounds:

(i) (A universal upper bound) For every t > 0,

n x+1� x

t+

2pt

for a.e. x 2 (0, 1) ,

(ii) (Oleinik-type inequality) For a.e. (x, t) 2 Q,

@x

n � �4

t.

(iii) (Energy estimate) n 2 C((0,1), L2(0, 1)), and whenever 0 < s < t,(2.9)Z

1

0

n2(x, t) dx+

Zt

s

Z1

0

[n2 + x2(@x

n)2] dx d⌧ Z

1

0

n2(x, s) dx+8

3(t� s) .

Note that the Oleinik-type inequality allows for the formation of “shock waves”in n at x = 0 but rules out oscillations.

Theorem 5 (regularity away from x = 0). For the global weak solution n fromTheorem 4, the quantities n, @

x

n, @t

n, and @2x

n are locally Holder-continuous on Q.Furthermore, n is smooth in the interior of Q.

2.3. Dynamics of solutions. Next we state our main results concerning theformation of condensates and the large time behavior of solutions. Observe that thebounds in (i) and (ii) of Theorem 4 imply the existence of the right limit n(0+, t) foreach t > 0.

Theorem 6 (formation and growth of condensates). Let nin satisfy (2.4) forsome p � 0. Let n be the unique global weak solution to (2.3) associated with nin.Then

(i) (Conservation of photons) For every t > s > 0 we have

Z1

0

n(x, t) dx =

Z1

0

n(x, s) dx�Z

t

s

n(0+, ⌧)2 d⌧ .

(ii) (Persistence) There exists t⇤ 2 [0,1] such that n(0+, t) > 0 whenever t > t⇤

and n(0+, t) = 0 whenever 0 t < t⇤.(iii) (Formation) If N [nin] > 1

2

, then n(0+, t) > 0 whenever

1

2pt<

p1 + � � 1 , where 2� = N [nin]� 1

2.

(iv) (Absence) If nin x, then t⇤ = 1. That is, for every t > 0 we haven(0+, t) = 0 and N [n(·, t)] = N [nin].

The formula in part (i) justifies a physical description of the photon energy dis-tribution that contains a Dirac delta mass at x = 0, corresponding to a condensateof photons at zero energy that keeps total photon number conserved. By the formulain part (i), the quantity Z

t

s

n(0+, ⌧)2 d⌧


is the number of photons that have entered the condensate between times s and t.This quantity is nonnegative, meaning the condensate behaves like a “black hole”—photons go in but do not come out. Part (ii) shows that a condensate never stopsgrowing once it starts. Part (iii) states that a condensate must develop in finite timefor any initial data nin with more photons than the maximal equilibrium n

0

= x. Part(iv) means that for initial data bounded above by n

0

, a condensate does not form.According to Theorem 1 and Corollary 2, however, we see that the notion of a

condensate is not strictly required for mathematically discussing existence and unique-ness. The solution is determined by the conditions imposed for x 2 (0, 1], and it sohappens that the total photon number can decrease due to an outward flux at x = 0+.

Theorem 7 (large time convergence). Let nin satisfy (2.4) for some p � 0. Letn be the unique global weak solution to (2.3) associated with nin. Then there existsµ � 0 such that

limt!1

kn(·, t)� nµ

k1

= 0 , where nµ

(x) =x2

x+ µ.

The equilibrium nµ

to which a solution converges depends not only on N [nin] butalso on details of nin. In some special cases, µ can be explicitly determined.

Corollary 8. Let n be the global solution to (2.3), subject to initial data satis-fying nin(x) � x for x 2 (0, 1]. Then

limt!1

n(x, t) = n0

(x) = x.

Moreover,

(2.10) |n(x, t)� x| 1

t+

2pt

for every t > 0 .

If nin(x) x for x 2 (0, 1], then

limt!1

n(x, t) = nµ

(x) =x2

x+ µ

with µ uniquely determined by the relation

(2.11) N [nin] = N [nµ

] =1

2� µ+ µ2 log

✓1 +

1

µ

◆.

Remark 9. For the model equation (1.2), these results provide a definite answerto the main issues of concern. The main assertions are expected to hold true for thefull Kompaneets equation (1.1) and may be partially true for some extensions of theKompaneets equation [26, 5]. Theorem 1 and Corollary 2 improve upon Theorems1 and 2 of [6, p. 3839] for the Kompaneets equation, in the sense that we imposeno fixed growth condition near x = 0. However for a model equation, Theorems 6and 7 provide a theoretical justification of observations made previously, including thedetailed singularity analysis given in [6], the self-similar blowup of the Kompaneetsequation’s solution in finite time [12], as well as the classical result of Zel’dovich andLevich [31] on shock waves in photon spectra.

Remark 10. We remark that the quantum entropy defined by

H[n] =

Z1

0

[xn� x2 log(n)] dx


satisfies

H[n(t)] +

Zt

0

Z1

0

n2

✓1� @

x

✓x2

n

◆◆2

dx H[nin] 8t > 0,

provided H[nin] < 1. As we have no need for this entropy dissipation inequality inthis paper, we omit the proof. We mention, however, that the entropy H[n] is notsensitive to the presence of the Bose–Einstein condensate.

3. Uniqueness of weak solutions. This section is primarily devoted to theproof of Theorem 1. At the end of this section we include an additional result, astrict L1 contraction property, which will be used in section 6.

3.1. Proof of Theorem 1. Let w = n � n and n = n + n � 2x. Then from(2.5d) for both n and n it follows that

(3.1)

Z

Q

(w @t

� (x2@x

w + nw) @x

) = 0.

The estimate (2.6) can be derived formally by using a test function of form =xp

[0,t]

H(w), where H(w) is the usual Heaviside function andE

is the characteristicfunction of a set E. This is not an admissible test function, however, and instead weneed several approximation steps.

For use below, we fix a smooth, nondecreasing cut-o↵ function � : R ! [0, 1] withthe property �(x) = 0 for x 1, �(x) = 1 for x � 2 and set �

✏

(x) = �(x/✏) for ✏ > 0.For any interval I ⇢ [0,1) we define the space-time domains

(3.2) QI

= (0, 1]⇥ I, so Q = Q(0,1)

= (0, 1]⇥ (0,1).

1. (Steklov average in t.) For h 6= 0, the Steklov average uh

of a continuousfunction u on Q is defined by extending u(x, t) to be zero for t < 0 and setting

uh

(x, t) =1

h

Zt+h

t

u(x, s) ds, (x, t) 2 Q.

By density arguments, the Steklov average extends to an operator with the followingproperties. First, for 1 p < 1, if u 2 Lp

loc

(Q), then uh

2 Lp

loc

(Q) with weakderivative

@t

uh

=u(·, ·+ h)� u(·, ·)

h2 Lp

loc

(Q).

Moreover, one has uh

! u in Lp

loc

(Q) as h ! 0.Since

n, n 2 B+

:= {n | n, @x

n 2 L2

loc

(Q) with n � 0},it follows @j

x

@kt

wh

2 L2

loc

(Q) for j, k = 0, 1, whence wh

is continuous on Q. If is a C1

test function with compact support in Q, the same is true for �h

if |h| is su�cientlysmall, and a simple calculation with integration by parts and justified by density ofsmooth functions shows that

Z

Q

w @t

( �h

) =

Z

Q

w(@t

)�h

=

Z

Q

wh

@t

= �Z

Q

(@t

wh

) .

Substitution of this into (3.1) and treating the other term similarly, one finds

(3.3)

Z

Q

⇣(@

t

wh

) + (x2@x

w + nw)h

@x

⌘= 0.


Recall n, n 2 B+

, hence nw and @x

(nw) are in L1

loc

(Q), whence (nw)h

is contin-uous in Q. By replacing (x, t) by (x, t)�

✏

(t � �)�✏

(⌧ � t) and taking ✏ ! 0 usingdominated convergence, we find that for any C1 function with compact support inQ

[0,1)

,

(3.4)

Z

Q[�,⌧]

⇣(@

t

wh

) + (x2@x

w + nw)h

@x

⌘= 0 whenever [�, ⌧ ] ⇢ (0,1).

By approximation, (3.4) holds for any 2 W 1,2(Q) supported in Q[0,1)

.2. (Integrate in t.) Define ⇣(a) =

Ra

0

�✏

(u) du as a smooth, convex approximationto the function a 7! a

+

. Since ⇣ is Lipshitz with ⇣ 0(a) = 1 for a > 2✏, the composition⇣(w

h

) 2 W 1,2

loc

(Q), with the weak derivatives

@t

⇣(wh

) = ⇣ 0(wh

)@t

wh

, @x

⇣(wh

) = ⇣ 0(wh

)@x

wh

.

We may now set (x, t) = ⌘(x)(⇣ 0 �wh

)(x, t) in (3.4), where ⌘ is any C2 function withcompact support in (0, 1]. In what follows, we assume also that ⌘ � 0 and ⌘0 � 0 on

(0, 1]. The function t 7!R1

0

⌘(x)⇣(wh

(x, t)) dx is absolutely continuous for t > 0, with

(3.5)Z1

0

⌘(x)⇣(wh

(x, t)) dx��t=⌧

t=�

=

Z

Q[�,⌧]

⌘⇣ 0(wh

)@t

wh

= �Z

Q[�,⌧]

(x2@x

w + nw)h

@x

(⌘⇣ 0(wh

))

whenever [�, ⌧ ] ⇢ (0,1).3. (Take h ! 0.) As h ! 0, the hypotheses for weak solutions imply that n,

n 2 L1

loc

(Q[0,1)

). By consequence, in L1

loc

([0,1)) we have

(3.6)

Z1

0

⌘(x)⇣(wh

(x, ·)) dx !Z

1

0

⌘(x)⇣(w(x, ·)) dx.

In fact, we will show that the right-hand side here is absolutely continuous for t > 0,from studying the terms on the right-hand side of (3.5). First, as h ! 0, in L2

loc

(Q)we have

(3.7) x2@x

wh

! x2@x

w.

And along a subsequence hj

! 0, ⇣ 0(wh

) ! ⇣ 0(w) and ⇣ 00(wh

) ! ⇣ 00(w) boundedlya.e. on compact subsets of Q. Hence in L2

loc

(Q) we have

@x

(⌘ ⇣ 0(wh

)) = ⌘0⇣ 0(wh

) + ⌘ ⇣ 00(wh

)@x

wh

! ⌘0⇣ 0(w) + ⌘ ⇣ 00(w)@x

w.

Since ⇣ 0(w)@x

w = @x

⇣(w), we find

(3.8)

Z

Q[�,⌧]

(x2@x

wh

)@x

(⌘ ⇣ 0(wh

)) !Z

Q[�,⌧]

⇣x2⌘0@

x

⇣(w) + x2⌘ ⇣ 00(w)(@x

w)2⌘.

Next we deal with the nonlinear term in (3.5). ObserveZ

Q[�,⌧]

(nw)h

@x

(⌘ ⇣ 0(wh

))

=

Z⌧

0

⌘ ⇣ 0(wh

)(nw)h

(1, t) dt�Z

Q[�,⌧]

⌘ ⇣ 0(wh

)(n@x

w + w@x

n)h

.(3.9)


Because n, n 2 B+

we have nw, @x

(nw) 2 L1

loc

(Q), and it follows that (nw)h

(1, ·) !nw(1, ·) in L1

loc

((0,1)). Moreover, w(1, ·) 2 L2

loc

((0,1)) and ⇣ 0(wh

(1, ·)) ! ⇣ 0(w(1, ·))boundedly a.e. on compact subsets of (0,1) along a sub-subsequence of h ! 0. Thuswe may pass to the limit on the right-hand side of (3.9), integrate back by parts, andinfer that the limit is

Z⌧

�

⌘ ⇣ 0(w)(nw)(1, t) dt�Z

Q[�,⌧]

⌘ ⇣ 0(w)(n@x

w + w@x

n)

=

Z

Q[�,⌧]

(nw)(⌘0⇣ 0(w) + ⌘ ⇣ 00(w)@x

w).(3.10)

To justify this equality requires an additional argument that involves approximating nand n by smooth functions in B

+

: The equality is true for the smooth approximations,and one may pass to the limit along subsequences by essentially the same argumentsas were used to take h ! 0. Note that direct integration by parts does not work—theintermediate term

RQ[�,⌧]

(nw)@x

(⌘ ⇣ 0(w)) does not make sense with only the regularity

assumed for w, due to insu�cient integrability in time (L1 for one factor, L2 for theother. On the right-hand side of (3.10), however, ⇣ 0(w) and w⇣ 00(w) are bounded onthe support of the integrand, and one can pass to the limit a.e. along subsequencesfrom smooth approximations to deduce (3.10).

In sum, we find that in L1

loc

((0,1)) and for a.e. ⌧ > � > 0,Z

1

0

⌘(x)⇣(w(x, t)) dx��t=⌧

t=�

= �Z

Q[�,⌧]

⇣x2⌘0@

x

⇣(w) + x2⌘ ⇣ 00(w)(@x

w)2⌘

(3.11)

�Z

Q[�,⌧]

(nw)(⌘0⇣ 0(w) + ⌘ ⇣ 00(w)@x

w).

4. (Take ✏ ! 0.) Note that since ⇣(w), @x

⇣(w) 2 L2

loc

(Q), we have ⇣(w(1, ·)) 2L2

loc

((0,1)) and

�Z

Q[�,⌧]

x2⌘0@x

⇣(w) = �Z

⌧

�

⌘0(1)⇣(w(1, t)) dt+

Z

Q[�,⌧]

⇣(w)@x

(x2⌘0)

Z

Q[�,⌧]

⇣(w)@x

(x2⌘0).

Since ⌘, ⌘0 � 0, n � �2x, and w⇣ 0(w) � 0, therefore

(3.12)Z1

0

⌘(x)⇣(w(x, t)) dx��t=⌧

t=�

Z

Q[�,⌧]

⇣⇣(w)@

x

(x2⌘0) + 2x⌘0w⇣ 0(w)� nw⌘ ⇣ 00(w)(@x

w)⌘.

Now we take the limit ✏ # 0, for which we have ⇣ � w " w+

and w(⇣ 0 � w) " w+

pointwise. Moreover, w⇣ 00 � w = (w/✏)�0(w/✏) is bounded and converges to zero a.e.Since ⌘n@

x

w 2 L1(Q), by dominated convergence the last term in (3.12) tends tozero, and we derive

(3.13)

Z1

0

⌘w+

(x, ⌧) dx Z

1

0

⌘w+

(x,�) dx+

Z

Q[�,⌧]

(x2⌘00 + 4x⌘0)w+

for a.e. ⌧ > � > 0. Due to assumption (2.5c) on weak solutions, now we can take� ! 0 and conclude that this inequality holds also with � = 0.


5. (Make Gronwall estimate.) Finally, we take ⌘ of the form ⌘(x) = xp�✏

(x),where p � 0 is the exponent for which we assume xpnin 2 L1(0, 1). Observe that

(3.14) x⌘0 = xp(p�+ (x/✏)�0), x2⌘00 = xp(p(p� 1)�+ 2p(x/✏)�0 + (x/✏)2�00),

where the arguments of �, �0, and �00 are x/✏. As ✏! 0, since we assume xpn and xpnare in L1(Q

[0,T ]

) for any T > 0, we infer by monotone and dominated convergencethat

(3.15)

Z1

0

xpw+

(x, ⌧) dx Z

1

0

xpwin

+

(x) dx+ cp

Z⌧

0

Z1

0

xpw+

(x, t) dx dt

for a.e. ⌧ > 0, with cp

= p(p�1)+4p = p(p+3). Denoting the (absolutely continuous)right-hand side of (3.15) by U(⌧), we have

U(⌧) = U(0) +

Z⌧

0

U 0(s) ds U(0) + cp

Z⌧

0

U(s)ds,

and Gronwall’s inequality implies that

U(⌧) ecp⌧Z

1

0

xp win

+

(x) dx

for all ⌧ > 0. This proves (2.6). Clearly, nin nin implies n n, by virtue of (2.6).

3.2. Strict L1 contraction. From Corollary 2 it easily follows that weak so-lutions of (2.3) enjoy the L1 contraction property mentioned in (2.8). For use insection 6 below, we strengthen this to the following strict L1 contraction property forC1 solutions that cross transversely.

Lemma 11. Let n, n be nonnegative solutions to (2.3) with respect to initial datanin, nin that are in L1(0, 1) \ L1(0, 1); then for a.e. t > 0,

(3.16) kn(·, t)� nkL

1(0,1)

knin � ninkL

1(0,1)

.

Moreover, assuming the solutions n and n are C1 in (0, 1)⇥ [0,1) and that for somet0

� 0, n(·, t0

) and n(·, t0

) cross transversely at least once on (0, 1), then for all t > t0

we have

(3.17) kn(·, t)� n(·, t)kL

1(0,1)

< kn(·, t0

)� n(·, t0

)kL

1(0,1)

.

Proof. The L1 contraction estimate (3.16) follows directly from Corollary 2 withp = 0. In order to prove (3.17), it su�ces to treat the case t

0

= 0 for t > 0 su�cientlysmall and assume the right-hand side is finite. Let w = n� n and n = n+ n�2x. If ncrosses n transversely at (x

0

, 0), then the regularity of the solution implies that thereexists a nondegenerate rectangle ⌃

0

= [x0

� �, x0

+ �] ⇥ [0, �] such that w(x0

, 0) = 0and @

x

w 6= 0 in ⌃0

. We suppose @x

w(x0

, 0) > 0 (relabeling n and n if necessary),whence @

x

w � c1

> 0 in ⌃0

, so w(x0

+ �, t) > c1

� > 0 and w(x0

� �, t) < �c1

� < 0.We follow the proof of Theorem 1 up to (3.12), finding that for 0 < � < ⌧ < �,

Z1

0

⌘(x)⇣(w(x, t)) dx��t=⌧

t=�

(3.18)

Z

Q[�,⌧]

⇣�x2⌘⇣ 00(w)(@

x

w)2 + ⇣(w)@x

(x2⌘0) + 2x⌘0w⇣ 0(w)� nw⌘ ⇣ 00(w)(@x

w)⌘.


Here we include a term �x2⌘⇣ 00(w)(@x

w)2 from (3.11) that was dropped in (3.12).This identity is valid for any C2 function ⌘ � 0 with compact support in (0, 1] andwith ⌘0 � 0. We may require x2⌘ � c

2

> 0 in ⌃0

. Therefore, taking ✏ # 0, we findZ

Q[�,⌧]

x2⌘⇣ 00(w)(@x

w)2 �Z

⌃0\Q[�,⌧]

c1

c2

⇣ 00(w)@x

w

=

Z⌧

�

c1

c2

⇣ 0(w)��x0+

ˆ

�

x0�ˆ

�

dt ! c1

c2

(⌧ � �) > 0 .

When taking the limit ✏ # 0,we also have ⇣ � w " w+

and w(⇣ 0 � w) " w+

pointwise.Moreover, w⇣ 00 � w = (w/✏)�0(w/✏) is bounded and converges to zero a.e. Since⌘n@

x

w 2 L1(Q), by dominated convergence the last term in (3.18) tends to zero, andwe derive

(3.19)

Z1

0

⌘w+

(x, ⌧) dx Z

1

0

⌘w+

(x,�) dx+

Z

Q[�,⌧]

(x2⌘00+4x⌘0)w+

� c1

c2

(⌧ ��) .

Finally, we take ⌘ of the form ⌘(x) = �✓

(x) with ✓ < x0

� �. Observe that

(3.20) x⌘0 = (x/✓)�0, x2⌘00 = (x/✓)2�00,

where the arguments of �, �0, and �00 are x/✓. As ✓ ! 0, since we assume n and nare in L1(Q), we infer by monotone and dominated convergence that

(3.21)

Z1

0

w+

(x, ⌧) dx Z

1

0

w+

(x,�) dx� c1

c2

(⌧ � �) <

Z1

0

win

+

(x) dx ,

where the last inequality follows by applying Theorem 1 with t = � and p = 0. Addingthis result together with (2.6) with p = 0 and n interchanged with n, we obtain(3.17).

4. Existence of weak solutions. The existence result in Theorem 4 is provedthrough three main approximation steps:

(i) Approximate the rough initial data nin 2 L1(xpdx) by smooth data nin

thatis strictly positive and bounded.

(ii) Truncate the problem (2.3) to x 2 [✏, 1] with ✏ > 0, resulting in a strictlyparabolic problem at the cost of needing to impose an additional boundarycondition at x = ✏.

(iii) Further approximate by cutting o↵ the nonlinearity in the flux near theboundary x = 1, resulting in a problem with linear boundary conditions.

Passing to the limit in the various approximations involves compactness argumentsand uniform estimates that are based on energy estimates and Gronwall inequalities.Step (iii) is comparatively straightforward and its analysis is relegated to AppendixB. We deal with steps (i) and (ii) in the remainder of this section.

4.1. Smoothing the initial data. Consider fixed initial data nin in L1(xpdx).We regularize the given initial data to obtain a family of functions nin

for small > 0,which are smooth on [0, 1] and positive on (0, 1], with the following properties:

Z1

0

xp|nin

� nin| dx ! 0 as ! 0 ,(4.1)

nin

(x) = x2 , 0 < x < ,(4.2)

nin

(x) =x2

x+ 1, 1� < x < 1 .(4.3)


(The properties in (4.2) and (4.3) are conveniences so that we get compatible initialdata in the approximation steps to follow below.) The desired regularization can beachieved through mollification: Let ⇢ be a smooth, nonnegative function on R withsupport contained in (�1, 1) and total mass one. Define

(4.4) ⇢

(x) = �1⇢(x/), �(x) =

Zx

�1⇢(z) dz

(note �(x) = 0 for x < �1 and �(x) = 1 for x > 1), and require that

xpnin

(x) =

Z1�2

2

⇢

(x� y) ypnin(y) dy +x2+p

1 + x�(4x� 2).

The integral term vanishes when x < or x > 1� , and there is no singularity nearx = 0.

For this regularized initial data, our goal is to prove the following result.

Proposition 12. For every small enough > 0, there exists a weak solution n

of (2.3) with initial data nin = nin

, having n

2 C([0,1), L1((0, 1])).

4.2. Truncation. To obtain n

, we regularize by truncating the domain awayfrom the origin, thus removing the degenerate parabolic nature of the problem. Inother words, we will study classical solutions of the following problem for small ✏ > 0:In terms of the (left-oriented) flux

(4.5) J✏

= x2@x

n✏

� 2xn✏

+ n2

✏

,

we seek a solution to the problem

@t

n✏

= @x

J✏

, x 2 (✏, 1), t 2 (0,1) ,(4.6a)

n✏

= nin

, x 2 (✏, 1) , t = 0 ,(4.6b)

0 = J✏

, x = 1 , t 2 [0,1) ,(4.6c)

0 = ✏2@x

n✏

� 2✏n✏

, x = ✏ , t 2 [0,1) .(4.6d)

The boundary condition (4.6d) says J✏

= n2

✏

at x = ✏. As will be seen in section 5below, this boundary condition is well adapted to proving the conservation identityfor photon number in Theorem 6. An important point to note, however, is that theuniqueness result of Theorem 1 shows that the solution of (2.3) does not depend onthe choice of this boundary condition in (4.6d).

For fixed small ✏ > 0, the following global existence result for classical solutionsof (4.6) is proved in Appendix B. Note that due to (4.2) and (4.3), the boundaryconditions (4.6c)–(4.6d) hold at t = 0 whenever 0 < ✏ < .

Proposition 13. Let nin

be smooth and positive on (0, 1] and satisfy (4.2)–(4.3).Then for any su�ciently small ✏ > 0, there is a global classical solution n

✏

of (4.6),smooth in the domain

(4.7) Q✏ := (✏, 1)⇥ (0,1) ,

with n✏

, J✏

and @x

n✏

globally bounded and continuous on Q✏ = [✏, 1]⇥ [0,1).

From this result, we will derive Proposition 12 by taking ✏ # 0 after establishinga number of uniform bounds on the solution n

✏

of (4.6). The global bounds stated inTheorem 4 will follow directly from corresponding uniform bounds on n

✏

, which areproved in Lemmas 15 and 16 and are inherited by n

.


4.3. Uniform estimates for the truncation. The first few uniform estimatesthat we establish on the solution n

✏

of (4.6) are pointwise estimates that arise fromcomparison principles.

Lemma 14. We have n✏

(x, t) > 0 for every (x, t) 2 [✏, 1]⇥ [0,1) .

Proof. Recall min[✏,1]

nin

> 0. If we suppose the claim fails, then 0 < t⇤ < 1,where

t⇤ = sup{t | n✏

(x, t) > 0 8x 2 [✏, 1]} .

By continuity, there exists X⇤ = (x⇤, t⇤) with x⇤ 2 [✏, 1] such that n✏

(X⇤) = 0. Weclaim first that x⇤ 6= ✏ or 1. If x⇤ = ✏ or 1, by the strong maximum principle [25], wemust have 0 6= @

x

n✏

(X⇤), but this violates the boundary conditions (4.6c)–(4.6d).Thus ✏ < x⇤ < 1, but this is also not possible due to rather standard comparison

arguments: There exists � > 0 such that � is less than the minimum of n✏

(✏, t), n✏

(1, t),and n

✏

(x, 0) whenever 0 t t⇤ and x 2 [✏, 1]. Setting w = e3tn✏

, we find that w > �at t = 0, and there is some first time t 2 (0, t⇤) when w(X) = � for some X = (x, t)with x 2 (✏, 1). Then @

t

w 0, @x

w = 0, and @2x

w � 0 at X, but computation thenshows @

t

w � w = � > 0. This finishes the proof.

Next we establish a universal upper bound on our solution of (4.6). We do thisby establishing that the function defined by

(4.8) S(x, t) = x+1� x

t+

2pt

is a universal supersolution. This fact depends essentially on the hyperbolic natureof our problem at large amplitude—note that the middle term (1� x)/t is a centeredrarefaction wave solution of the equation @

t

n� 2n@x

n = 0.

Lemma 15. We have

n✏

(x, t) < S(x, t) 8(x, t) 2 Q✏ .

Furthermore, there exists ⌧1

> 0, depending only on supnin

, such that

n✏

(x, t) < S(x, t+ ⌧1

) 8(x, t) 2 Q✏ .

Proof. Let us write L[n] := @t

n�x2@2x

n�2n(@x

n�1). Then a simple calculationgives

L[S] =1� x

t2+

2x

t+ 3t�3/2 > 0 .

Hence with v = S � n✏

, we have

(4.9) L[S]� L[n✏

] = @t

v � x2@2x

v � @x

((n✏

+ S)v) + 2v > 0 .

By continuity we have minx

v(x, t) > 0 for small t > 0, and we claim that thiscontinues to hold for all t > 0. If not, there is a first time t when it fails, and someX = (x, t) with x 2 [✏, 1] where v(X) = 0. By (4.9) it is impossible that x 2 (✏, 1). Ifx = ✏, then v = 0 and @

x

v � 0 at X. But due to the boundary condition (4.6d) wefind that at (x, t) = (✏, t),

0 ✏@x

v = ✏@x

S � 2S = ✏

✓1� 1

t

◆� 2

✓✏+

1� ✏

t+

2pt

◆< 0 .


So x 6= ✏. On the other hand, if x = 1, we would have v = 0 and @x

v 0 at X. Butthen, at (x, t) = (1, t) we find by (4.6c) that

0 � @x

v = @x

S + S2 � 2S = 1� 1

t+

✓1 +

2pt

◆2

� 2

✓1 +

2pt

◆=

3

t> 0 .

Thus x 6= 1, and the result S � n✏

> 0 follows. Furthermore, if supnin

2/p⌧1

, then

minx

(S(x, ⌧1

)� nin

(x)) >2

p⌧1

� knin

kL

1 � 0 .

Then the above procedure shows that S(x, t+ ⌧1

) is also a supersolution.

The next result bounds @x

n✏

from below. This is again a typical kind of estimatefor the hyperbolic equation @

t

n� 2n@x

n = 0.

Lemma 16 (Oleinik-type inequality). We have

@x

n✏

(x, t) � �4

t8(x, t) 2 Q✏ .

Furthermore, there exists ⌧2

> 0, depending only on inf @x

nin

, such that

@x

n✏

(x, t) � � 4

t+ ⌧2

8(x, t) 2 Q✏ .

Proof. Let w = @x

n✏

with n✏

being the solution of (4.6). Di↵erentiation of (4.6a)shows that w satisfies

@t

w = x2@2x

w + 2(n✏

+ x)@x

w + 2w(w � 1) , (x, t) 2 Q✏ ,(4.10a)

✏2w(✏, t) = 2✏n✏

(✏, t) , t > 0 ,(4.10b)

w(1, t) = 2n✏

(1, t)� n✏

(1, t)2 , t > 0 .(4.10c)

We claim that z = �4/t is a subsolution of this problem. Set U = w � z = w + 4/t.A direct calculation gives

@t

U � x2@2x

U � 2(n✏

+ x)@x

U � 2(w � 4/t)U + 2U =4

t2(7 + 2t) > 0 .

Note that U(x, t) > 0 for t > 0 small, because @x

n✏

is continuous on Q✏. ThenU(x, t) > 0 for all x and t as long as it is so at x = ✏ and x = 1. At x = ✏, we have

U(✏, t) = w(✏, t) +4

t=

2

✏n✏

(✏, t) +4

t> 0 .

On the other hand, at x = 1, we have

U(1, t) = w(1, t) +4

t= 2n

✏

(1, t)� n✏

(1, t)2 +4

t.

If 0 n✏

(1, t) 2, then U(1, t) � 4/t, and otherwise, 2 < n✏

(1, t) S(1, t) =1 + 2t�1/2, hence

U(1, t) � 2S(1, t)� S(1, t)2 +4

t= 1 .

Therefore U(1, t) > 0.


Provided ⌧2

> 0 is su�ciently small so that @x

nin

> �4/⌧2

, we have

minx

⇢w(x, 0) +

4

⌧2

�> 0 ,

hence the above procedure shows that w(x, t) � �4/(t+ ⌧2

) for all (x, t) under con-sideration. This concludes the proof.

We now turn to obtain some compactness estimates that will be needed to estab-lish convergence as ✏ # 0. First we establish equicontinuity in the mean for solutionsof (4.6).

Lemma 17. For each t > 0, we have

(4.11)

Z1

✏

|@x

n✏

| dx K1

(t), K1

(t) = 1 +2p

t+ ⌧1

+8

t+ ⌧2

,

and for all t > 0 and all small h > 0,

(4.12)

Z1

✏

|n✏

(x, t+ h)� n✏

(x, t)| dx K2

(t)h1/2 ,

where K2

(t) is a decreasing function of t with K2

(0) bounded by a constant dependingonly on supnin

and inf @x

nin

.

Proof. With ⌧1

and ⌧2

determined by the previous two lemmas, set

u(x, t) = n✏

(x, t) +4x

t+ ⌧2

.

Then by Lemma 16, u is a nondecreasing function of x, satisfying @x

u > 0. We have

Z1

✏

|@x

n✏

| dx =

Z1

✏

��@xu� 4

t+ ⌧2

�� dx 4

t+ ⌧2

+

Z1

✏

@x

u dx(4.13)

8

t+ ⌧2

+ n✏

(1, t) 8

t+ ⌧2

+ S(1, t+ ⌧1

) = K1

(t) .

This proves the first estimate of the lemma.We next prove the bound (4.12). Fix any t > 0 and consider h > 0 small.

Suppressing the dependence on t and h, we set

v(x) = n✏

(x, t+ h)� n✏

(x, t) , x 2 [✏, 1] ,

and observe that

(4.14) kvk1 2kS(·, t+ ⌧1

)k1 ,

Z1

✏

|@x

v| dx 2K1

(t) .

We proceed by approximating |v(x)| by �(x)v(x), where � is obtained by mollifyingsgn v(x) as follows. Let ⇢ be a smooth, nonnegative function on R with supportcontained in (�1, 1) and total mass one, and let ↵ > 0 be a parameter. (We will take↵ = 1

2

below.) We define ⇢h

(x) = h�↵⇢(x/h↵) and set

(4.15) �(x) =

Z1

✏

⇢h

(x� z) sgn v(z) dz .


To bound the integral of |v(x)| over [✏, 1], we bound integrals over the sets

Ih

= [✏+ h↵, 1� h↵] , Ih

= [✏, ✏+ h↵] [ [1� h↵, 1] ,

writingZ

1

✏

|v(x)| dx =

Z1

✏

�(x)v(x) dx+

Z

Ih

⇣|v(x)|� �(x)v(x)

⌘dx

+

Z

ˆ

Ih

⇣|v(x)|� �(x)v(x)

⌘dx .

(4.16)

Since |�| 1, the third term is bounded using the first estimate in (4.14) as

(4.17)

Z

ˆ

Ih

��|v(x)|� �(x)v(x)�� dx 8h↵kS(·, t)k1 .

We next estimate the middle term in (4.16). For x 2 Ih

, we compute

|v(x)|� �(x)v(x) =

Z

R⇢h

(x� z)⇣|v(x)|� v(x) sgn v(z)

⌘dz .

Noting that ||a|� a sgn b| 2|a� b| for any real a, b, we have��|v(x)|� v(x) sgn v(z)

�� 2|v(x)� v(z)| .

Integrating over Ih

, we findZ

Ih

��|v(x)|� �(x)v(x)�� dx 2

Z

Ih

Z

|y|<h

↵

⇢h

(y)|v(x)� v(x� y)| dy dx

2

Z

Ih

Z

|y|<h

↵

⇢h

(y)|y|Z

1

0

|@x

v(x� ys)| ds dy dx .

Now we integrate first over x, note x�ys 2 [✏, 1] and use (4.14), and note that |y| h↵

and ⇢h

has unit integral. We infer that

(4.18)

Z

Ih

��|v(x)|� �(x)v(x)�� dx 2h↵

Z1

✏

|@x

v| dx 4h↵K1

(t) .

Finally, we bound the first term in (4.16). Multiply (4.6a) by � and integrateover (✏, 1)⇥ (t, t+ h). Integration by parts yields

Z1

✏

�v(x) dx =

Zt+h

t

Z1

✏

(@x

�)��x2@

x

n✏

� n2

✏

+ 2xn✏

�dx d⌧ �

Zt+h

t

�(✏)n2

✏

(✏, ⌧) d⌧ .

(4.19)

Note that |�| 1 and |@x

�| h�↵k⇢0k1

. By virtue of 0 n✏

S, we haveZ

1

✏

�v dx h�↵k⇢0k1

Zt+h

t

Z1

✏

(|@x

n✏

|+ S2 + 2S) dx d⌧ +

Zt+h

t

S(✏, ⌧)2 d⌧

h1�↵k⇢0k1

(K1

(t) + 3kS(·, t+ ⌧1

)k21) + hkS(·, t+ ⌧1

)k21 .

Assembling all the bounds on the terms in (4.16) above, we obtainZ

1

✏

|v(x)| dx kS(·, t+ ⌧1

)k21(8h↵ + h+ 3h1�↵k⇢0k1

) +K1

(t)(4h↵ + h1�↵k⇢0k1

) .

Choosing ↵ = 1

2

and determining K2

(t) to correspond, the result in the lemma fol-lows.


Finally, we have the following energy estimate.

Lemma 18. For any t > s > 0,

(4.20)

Z1

✏

n2

✏

(x, t) dx+

Zt

s

Z1

✏

[n2

✏

+ x2(@x

n✏

)2] dx d⌧ Z

1

✏

n2

✏

(x, s) dx+8

3(t� s) .

Proof. From (4.6a) and the boundary conditions (4.6c)–(4.6d) it follows that

d

dt

Z1

✏

n2

✏

dx = �2

Z1

✏

(@x

n✏

)J✏

dx� 2n3

✏

(✏, t) = �2

Z1

✏

[n2

✏

+ x2(@x

n✏

)2]dx+ �(t)

with

�(t) = �2

3n3

✏

(1, t)� 4

3n2

✏

(✏, t) + 2n2

✏

(1, t)� 2✏n2

✏

(✏, t) maxu>0

✓�2

3u3 + 2u2

◆=

8

3.

Hence, the claimed estimate follows by integration in time.

4.4. Proof of Proposition 12. We now show a solution of (2.3) does exist forinitial data nin

as prepared in subsection 4.1. Let n✏

be our solution of (4.6) for small✏ > 0.

Recalling the uniform estimates 0 < n✏

S(x, t + ⌧1

) from Lemmas 14 and 15,and using Lemma 17, we see the family{n

✏

} is uniformly bounded and equicontin-uous in the mean on any compact subset of (0, 1] ⇥ [0,1). Consequently, we mayextract a sequence ✏

k

# 0 as k ! 1, such that for each a 2 (0, 1) and T > 0, n✏k

converges to some function n, boundedly almost everywhere in [a, 1] ⇥ [0,1) and inC([0, T ];L1([a, 1])), with

(4.21)

Z1

a

|n(x, t+ h)� n(x, t)| dx Ch1/2 .

Actually, this C is independent of a, so (4.21) holds also with a = 0. Moreover, dueto (4.20) we can ensure that

x@x

n✏k ! x@

x

n weakly in L2

loc

(Q) .

We claim that n is a weak solution of (2.3). Multiply (4.6a) by a smooth testfunction with compact support in (0, 1]⇥ (0,1), and integrate over (✏, 1)⇥ (0,1)with integration by parts to obtain, for small enough ✏,

(4.22)

Z 1

0

Z1

✏

⇣n✏

@t

� (x2@x

n✏

+ n2

✏

� 2xn✏

)@x

⌘= 0 .

Setting ✏ = ✏k

and letting k ! 1, we conclude that

(4.23)

Z 1

0

Z1

0

⇣n@

t

� (x2@x

n+ n2 � 2xn)@x

⌘= 0

for all smooth test functions . By completion, we infer that (4.23) holds for all merely in H1 with compact support in Q. Hence n is a weak solution as claimed.

Since there may exist at most one such solution of (2.3), we conclude that thewhole family {n

✏

} converges to n, as ✏! 0. This ends the proof of Proposition 12.


4.5. Proof of Theorem 4. Our next task is to complete the proof of Theorem4 by studying the solutions n

from Proposition 12 in the limit ! 0.By the Gronwall inequality from (2.6), for any fixed T > 0, and small

1

,2

> 0,

(4.24) supt2[0,T ]

Z1

0

xp|(n1

� n2)(x, t)| dx ecpt

Z1

0

xp|(nin

1� nin

2)(x)| dx .

This with (4.1) implies that n

is a Cauchy sequence in C([0, T ];L1(xpdx)), andtherefore there is a function n 2 C([0, T ];L1(xpdx)) such that lim

!0

n

= n. Fromthe local-in-time energy estimate (4.20) we have

Z1

0

n2

(x, t) dx+

Zt

s

Z1

0

[n2

+ x2(@x

n

)2] dx d⌧ Z

1

0

n2

(x, s) dx+8

3(t� s) .

The right-hand side, by virtue of 0 n

(x, t) S(x, t), is bounded for any s > 0 by

Z1

0

S(x, s)2dx+8

3(t� s) < 1 .

Taking the limit ! 0, we deduce that n and @x

n lie in L2

loc

(Q(0,T ]

), and the limitn is nonnegative and satisfies (2.5d). This proves that n is indeed a weak solutionof (2.3), as claimed. Moreover, from Lemma 15 it follows that the limit n has theuniversal upper bound that for every t > 0,

n S(x, t) = x+1� x

t+

2pt

for a.e. x 2 (0, 1) ,

and Lemma 16 implies that for almost every (x, t) 2 Q(0,T ]

, its slope has the one-sidedbound

@x

n � �4

t.

Finally, the limit function n 2 C((0, T ], L2), due to the following estimate. With

!(h) = supsts+h

Z1

0

xp|n(x, t)� n(x, s)| dx , Cs

= supx2[0,1]

S(x, s) ,

whenever 0 < t� s < h is so small that ↵ = !(h)1/(p+1) < 1 we have

Z1

0

|n(x, t)� n(x, s)|2 dx C2

s

↵+ Cs

↵�p

Z1

↵

xp|n(x, t)� n(x, s)| dx (C2

s

+ Cs

)↵ .

By consequence, the energy estimate (2.9) follows from the one for n

. This finishesthe proof of Theorem 4.

5. Finite time condensation. The results of this section establish Theorem 6,demonstrating that loss of photons is due to the generation of a nonzero flux atx = 0+, that such a flux persists if ever formed, and that photon loss does occur ifthe initial photon number exceeds the maximum attained in steady state.

Throughout this section, we let n be any global weak solution of (2.3).


5.1. Formula for loss of photon number. First, we show how any possible de-crease of photon number in time is related to the nonvanishing of n(0, t)2 = n(0+, t)2,which is the formal limit of the flux J at the origin x = 0. The following result impliespart (i) of Theorem 6 in particular.

Lemma 19. For any fixed t > s > 0,

(5.1)

Z1

0

+

n(x, t) dx =

Z1

0

+

n(x, s) dx�Z

t

s

n2(0, ⌧) d⌧ .

Moreover, for any t > 0

(5.2)

Z1

0

+

n(x, t) dx 1

2+

1

2t+

2pt.

Proof. Integration of (4.6a) over (x, 1)⇥ (s, t), using J✏

(1, t) = 0, gives

Z1

x

n✏

(y, ⌧) dy��⌧=t

⌧=s

= �Z

t

s

(x2@x

n✏

(x, ⌧) + n2

✏

(x, ⌧)� 2xn✏

(x, ⌧)) d⌧ .

Taking an average in x over (✏, a) with 0 < ✏ < a < 1, we find

(5.3) �Z

a

✏

Z1

x

n✏

(y, ⌧) dy dx��t

s

+

Zt

s

�Z

a

✏

n2

✏

dx d⌧ =

Zt

s

�Z

a

✏

(2xn✏

� x2@x

n✏

) dx d⌧ .

For the first term on the left-hand side, integrating on [x, 1] = [x, a] [ [a, 1] we note

�Z

a

✏

Z1

x

n✏

(y, ⌧) dy dx =

Z1

a

n✏

(y, ⌧) dy +R,

where, because ✏ x,

R = �Z

a

✏

Za

x

n✏

(y, ⌧) dy ✓�Z

a

✏

Za

✏

n✏

(y, ⌧)2 dy dx

◆1/2

✓�Z

a

✏

Za

✏

1 dy dx

◆1/2

✓Z

a

✏

n2

✏

dy

◆1/2

(a� ✏)1/2

Cs

a1/2 ,

due to the energy estimate in Theorem 5. The right-hand side in (5.3) is boundedabove by

✓Zt

s

�Z

a

✏

(2n✏

� x@x

n✏

)2 dx d⌧

◆1/2

✓Zt

s

�Z

a

✏

x2 dx d⌧

◆1/2

✓Z

t

s

�Z

a

✏

8n2

✏

+ 2x2(@x

n✏

)2 dx d⌧

◆1/2 �

(t� s)a2�1/2

✓Z

t

s

Z1

✏

n2

✏

+ x2(@x

n✏

)2 dx d⌧

◆1/2

✓8(t� s)a2

a� ✏

◆1/2

Ct,s

✓a2

a� ✏

◆1/2

.


Passing to the limit ✏ # 0 first, we have

Z1

a

n(x, ⌧) dx��t

s

+

Zt

s

�Z

a

0

n2(x, ⌧) dx d⌧ = O(1)a1/2 .

The desired equality follows from further taking a # 0. Moreover, by virtue of n S,we have Z

1

0

n(x, t) dx Z

1

0

S(x, t) dx =1

2+

1

2t+ 2t�1/2

for any t > 0. The proof is complete.

Because n is a classical solution of @t

n = @x

J for x, t > 0, by integration overx 2 [a, 1], ⌧ 2 [s, t] we can infer that the loss term in (5.1) arises from the exit fluxfrom the interval [a, 1] in the limit a ! 0. Thus the following result provides a precisesense in which the flux J(a, t) converges to n2(0, t) as a ! 0.

Corollary 20. Whenever t > s > 0 we have

(5.4) lima!0

+

Zt

s

J(a, ⌧) d⌧ =

Zt

s

n2(0, ⌧) d⌧.

5.2. Persistence of condensate growth. Next we prove part (ii) of Theo-rem 6, showing that n(0, t) once positive will remain positive for all time. Moreprecisely, we have the following.

Lemma 21. If n(0, t⇤) > 0 for some t⇤ > 0, then n(0, t) � b(t)x for all t > t⇤,where

b(t) =⇣(1 + t⇤/4)e2(t�t

⇤) � 1

⌘�1

, x = min

⇢t⇤n(0, t⇤)

4, 1

�.

Proof. From @x

n(x, t⇤) � �4/t⇤ we have

n(x, t⇤) �✓n(0, t⇤)� 4x

t⇤

◆

+

� 4

t⇤(x� x)

+

= b(t⇤)(x� x)+

.

Hence n(x, t⇤) � n(x, t⇤) with

n(x, t) = b(t)(x� x)+

.

Note that n(1, t) = 0, and for 0 < x < x we have

L[n] := @t

n� x2@2x

n+ 2n� 2n@x

n = (x� x)(b0(t) + 2b(t) + 2b(t)2) = 0 .

We claim n � n on (0, x] for t > t⇤. Let ✏ > 0. Substitution of n = n + v + ✏ intothe equation L[n] = 0 gives

L[v] := @t

v � x2@2x

v + 2v � 2v@x

n� 2n@x

v = �✏L[ ] .

Choosing = �t+ log x, we have < 0 and

L[ ] = 2 + 2 b� 2n

x< 0 ,

hence L[v] > 0 for 0 < x < x, t � t⇤. For 0 < x � for � su�ciently small (dependingon ✏),

v = n� n� ✏ � �n+ ✏(t� log �) > 0 8t > t⇤ .


Moreover, at x = x we have

v(x, t) = n(x, t) + ✏(t� log x) > 0 .

These facts, together with the fact v(x, t⇤) � ✏(t� log x) > 0, ensure that

v(x, t) > 0 8t > t⇤, x 2 (0, x] .

Since ✏ > 0 is arbitrary, we infer

n(x, t) � n(x, t) 8t � t⇤, x 2 (0, 1] .

This gives the desired estimate upon taking x ! 0.

5.3. Formation of condensates. The next result shows that photon loss willoccur—meaning a condensate will form—in finite time if the initial photon numberN [nin] > 1

2

. This proves part (iii) of Theorem 6. Note that 1

2

= N [x] is the maximumphoton number for any steady state.

Proposition 22. If N [nin] > 1

2

, then photon loss begins in finite time. Moreprecisely, we have n(0, t) > 0 whenever

(5.5)1

2pt<

p1 + � � 1 , where 2� = N [nin]� 1

2.

Proof. From the supersolution obtained in Lemma 15 it follows that

n(x, t) x+1� x

t+ 2t�1/2 .

Integration in x over (0, 1) leads to

N [n(·, t)] 1

2+

1

2t+

2pt.

Using Lemma 19 we have

Zt

0

n(0, ⌧)2 d⌧ � N [nin]� 1

2� 2p

t� 1

2t= 2� + 2� 2

✓1 +

1

2pt

◆2

.

The right-hand side becomes positive when (5.5) holds. The conclusion then followsfrom Lemma 21.

5.4. Absence of condensates. Part (iv) of Theorem 6 follows from a simplecomparison: If nin(x) x, then since x = n

0

(x) is a steady weak solution, thecomparison property from Theorem 1 implies n(x, t) x for all t � 0. Then n(0+, t) =0, so by part (a), no condensate is formed and we have

N [n(·, t)] = N [nin] 8t > 0 .

6. Large time convergence. We now investigate the large time behavior ofsolutions with nontrivial initial data. In the system (2.3), the flux vanishes for anyequilibrium:

(6.1) 0 = J = n2@x

✓x� x2

n

◆.


Consequently n = nµ

for some constant µ � 0, where

nµ

(x) =x2

x+ µ.

Our main goal in this section is to prove Theorem 7, which means that for everysolution of (2.3) provided by Theorem 4 with nonzero initial data nin, there existsµ � 0 such that

(6.2) kn(·, t)� nµ

k1

=

Z1

0

|n(x, t)� nµ

(x)| dx ! 0 as t ! 1.

It will be convenient to denote by

n(·, t) = U(t)a

the solution of (2.3) with initial data nin(x) = a(x), x 2 (0, 1). Due to the boundn(x, t) S(x, t) that holds by Theorem 4(i), for t � 1 any solution U(t)nin will lie inthe set

A := {a 2 L1(0, 1) : 0 a(x) 3 for a.e. x 2 (0, 1)}

since S(x, 1) ⌘ 3. The set A is positively invariant under the semiflow induced by thesolution operator:

U(t)A ⇢ A , t � 0 .

With the metric induced by the L1 norm,

⇢(n1

, n2

) = kn1

� n2

k1

,

the set A is a complete metric space, and by Lemma 11, U(t) is a contraction: Wehave

kU(t)a� U(t)bk1

ka� bk1

whenever t � 0 and a, b 2 A.For present purposes it is important that a stronger contractivity property also

holds, as shown in Lemma 11: Namely, if the functions a and b are C1 and cross trans-versely, then for t > 0, U(t) strictly contracts the L1 distance between a and b. Basedon these contraction properties and the one-sided Oleinik bound in Theorem 4(ii), weproceed to establish the large time convergence (6.2).

We introduce the usual !-limit set of any element a 2 A as

!(a) = \s>0

{U(t)a | t � s} .

We have b 2 !(a) if and only if there is a sequence {tj

} ! 1 such that kU(tj

)a�bk1

!0.

Lemma 23 (the !-limit set). Let a 2 A. Then !(a) is not empty, and is invari-ant under U(t), with

(6.3) U(t)!(a) = !(a) , t > 0 .

Moreover, for any b 2 !(a), b is smooth (at least C2 on (0, 1]) and satisfies

(6.4) @x

b(x) � 0 , 0 b(x) x , 0 < x < 1 .


Proof. To show !(A) is not empty, note that for any sequence tj

! 1, theestimates from Lemma 17 show that {U(t

j

)a} is bounded in BV . By virtue of theHelley compactness theorem, some subsequence converges in L1, and this limit belongsto !(a).

Next we prove (6.3). Given b 2 !(a), there exists tj

such that

kU(tj

)a� bk1

! 0 , j ! 1 .

From L1 contractivity and the semigroup property it follows that k(U(t + tj

)a �U(t)bk

1

! 0, hence U(t)b 2 !(a). On the other hand, if b 2 U(t)!(a), we haveb = U(t)b⇤ with b⇤ 2 !(a). Then for some sequence t

j

! 1,

kU(t+ tj

)a� bk1

= kU(t)U(tj

)a� U(t)b⇤k1

kU(tj

)a� b⇤k1

! 0

as j ! 1, hence b 2 !(a).By relation (6.3), for each b 2 !(a) and t > 0, b = U(t)b⇤ for some b⇤ 2 !(a).

From this it follows b is smooth and that @x

b � �4/t and 0 b(x) S(x, t) byTheorem 4. Taking t ! 1, since S(x, t) ! x we infer (6.4).

Lemma 24 (equilibria and !(a)). (i) If nµ

2 !(a) for some µ � 0, then

(6.5) limt!1

kU(t)a� nµ

k1

= 0 .

(ii) Let b 2 !(A). Then for any µ � 0,

(6.6) kb� nµ

k1

= kU(t)b� nµ

k1

.

(iii) If a 6⌘ 0, then 0 /2 !(a).

Proof. (i) By definition, for any ✏ > 0, kU(tj

)a � nµ

k1

< ✏ for large tj

. Thisensures that for any t > t

j

,

kU(t)a� nµ

k1

= kU(t� tj

)U(tj

)a� U(t� tj

)nµ

k kU(tj

)a� nµ

k1

< ✏ ,

hence (6.5).(ii) Since b 2 !(A), there exists a 2 A and a sequence {t

j

} such that tj

! 1 asj ! 1 and

limt!1

kU(tj

)a� bk1

= 0 .

Given any µ � 0, by contraction of U(t) we know that

kU(t)a� nµ

k1

= kU(t)a� U(t)nµ

k1

is decreasing in time and thus admits a limit cµ

� 0 as t ! 1, i.e.,

limt!1

kU(t)a� nµ

k1

= cµ

, t ! 1 .

Letting t = tj

in the above equation and passing to the limit, we have

kb� nµ

k1

= cµ

.

Note that if b 2 !(a), then U(t)b 2 !(a); thereby

kU(t)b� nµ

k1

= cµ

.


Therefore (6.6) holds for for all t > 0, µ � 0.(iii) Suppose a 6⌘ 0, so that N [a] > 0. We claim 0 /2 !(a). Supposing 0 2 !(a)

instead, we write n(·, t) = U(t)a. Then N [n(·, t)] = kU(t)a � 0k1

is nonincreasingand approaches zero as t ! 1. By Lemmas 19 and 21, then, a condensate forms andn(0+, t) > 0 for all large t.

From the Oleinik-type lower bound of Theorem 4(ii), x < z < 1 entails n(x, t)�4

t

n(z, t). After integration from 1� x to 1 we find

(1� x)

✓n(x, t)� 4

t

◆ N [n(·, t)] .

For t large enough we have N [n(·, t)] < 1

16

and t > 32, and this ensures that

8x 21

4,1

2

�, n(x, t) 2N [n(·, t)] + 4

t<

1

4 x .

Then, because n(0+, t) > 0, the last crossing point defined by

(6.7) x1

= max

⇢x 2

✓0,

1

4

�: n(x, t) = x

�

is well defined. Using again Theorem 4(ii), it now follows

0 n(x, t) x1

+4

tx1

for 0 < x < x1

,

x � n(x, t) � x1

� 4

tx1

for x1

< x < 2x1

.

From these inequalities, we deduce respectively that

Zx1

0

|x� n(x, t)| dx x2

1

✓1 +

4

t

◆,

Z2x1

x1

|x� n(x, t)| dx Z

2x1

x1

x dx� x2

1

✓1� 4

t

◆.

We may also assume t is so large that S(x, t) < 2x for 1

2

x 1. Then since0 n(x, t) S(x, t), it follows that

Z1

2x1

|x� n(x, t)| dx Z

1

2x1

x dx .

Because x2

1

(8/t) <Rx1

0

x dx, after adding the last three inequalities we find kx �U(t)ak

1

< kx�0k1

. But then since kx�U(t)ak1

is nonincreasing in t, it is impossiblethat kU(t)a� 0k

1

! 0 as t ! 1. This proves 0 /2 !(a).

The following restatement of the result in Lemma 11 plays a critical role in proving(6.2).

Lemma 25. If a, b 2 A \ C1((0, 1)) and a and b cross transversely at least onceon (0, 1), then

kU(t)a� U(t)bk1

< ka� bk1

, t > 0 .


We are now ready to prove (6.2). Let a 2 A with a 6⌘ 0. By Lemma 23 we knowthat !(a) is not empty. Let b 2 !(a). We need to show there exists a µ � 0 such that

(6.8) b = nµ

.

Since b 6⌘ 0 and b is nondecreasing, the quantity

g(x) = x� x2

b(x),

which is the first variation �H/�n of entropy, is well defined in some nonempty interval(x

0

, 1). If g is not a constant, there exists some x⇤ 2 (x0

, 1) such that g0(x⇤) 6= 0.Then it follows that at x = x⇤, with µ⇤ = �g(x⇤) we have

b =x2

x� g(x)= n

µ

⇤ , @x

b = @x

nµ⇤ +

x2g0

(x� g(x))26= @

x

nµ

⇤ .

In other words, b and nµ

⇤ cross transversely at x⇤. Therefore by Lemma 25 we have

kU(t)b� U(t)nµ

⇤k1

< kb� nµ

⇤k1

.

This contradicts (6.6). We conclude that g must be a constant, i.e., g(x) = �µ, whichgives (6.8). From b 6= 0 and b x we see that µ � 0.

Remark 26. Due to loss of mass, determining µ quantitively for each given initialdata is not straightforward, except for some special cases as treated in Corollary 8.

Proof of Corollary 8. If nin � x, by the comparison result in Theorem 1, we have

x n(x, t) , t > 0 .

On the other hand, the supersolution bound from Theorem 4(i) ensures that

n(x, t) x+1� x

t+ 2t�1/2.

These together lead to (2.10), hence limt!1 n(x, t) = x.

In the case of nin x, we have n(x, t) x for all t. Then there is no mass loss,hence the limiting equilibrium state n

µ

satisfies

Z1

0

nµ

dx =

Z1

0

nin dx = N [nin] .

Integration of the left-hand side yields (2.11).

Appendix A. Anisotropic Sobolev estimates. For use in section 4 andAppendices B and C, we need some basic anisotropic Sobolev estimates that are noteasy to find in the extensive literature on the subject. The results that we need appearto be related to embedding results for anisotropic Besov spaces contained in [2]. Forthe reader’s convenience, however, we provide a self-contained treatment based onsimple estimates for Fourier transforms.

If ⌦ ⇢ R2, the typical anisotropic Sobolev space is

u 2 W 2k,k

2

(⌦) = {u | Ds

x

Dr

t

u 2 L2(⌦), 0 2r + s 2k}.


As usual, if a function u 2 W 2k,k

2

(⌦), it will automatically belong to certain otherspaces, which depend on k and the dimension. One such space is C�,�/2(⌦). We sayu 2 C�,�/2(⌦) if there is a constant K such that

|u(x, t)� u(y, ⌧)| K(|x� y|2 + |t� ⌧ |)�/2 8(x, t), (y, ⌧) 2 ⌦.

The space C�,�/2(⌦) is a Banach space with norm given by

kukC

�,�/2(⌦)

= max(x,t)2⌦

|u(x, t)|+ sup(x,t),(y,⌧)2⌦

|u(x, t)� u(y, ⌧)|(|x� y|2 + |t� ⌧ |)�/2

.

The results we need are contained in the following result.

Theorem 27. Let D = (a, b) ⇥ (c, d) be a rectangular domain in R2. Supposethat u and its distributional derivatives @

t

u and @2x

u are in L2(D), i.e., u 2 W 2,1

2

(D).Then u 2 C1/2,1/4(D), and there is a constant C depending on D and s such that

k@x

ukL

s(D)

CkukW

2,12 (D)

, 2 s < 6.

This result is proved in the remainder of this section. It seems interesting to pointout, however, that by using the same techniques and with very little more work, onecan discuss higher-order embeddings and arbitrary space dimensions.

Theorem 28. Let D be a bounded parabolic cylinder in Rn+1 with C1 spatialboundary. Suppose that u 2 W 2k,k

2

(D); then

(i) W 2k,k

2

!

8<

:

C�,�/2, � = 2k � n+2

2

, k > n+2

4

,Ls, 2 s < 1, k = n+2

4

,

Ls, 2 s < 2(n+2)

(n+2)�4k

, k < n+2

4

,

(ii) CkukW

2k,k2

�

8<

:

k@x

uk1, k > n

4

+ 1,k@

x

ukL

s(D)

, 2 s < 1, k = n

4

+ 1,

k@x

ukL

s(D)

, 2 s < 2(n+2)

(n+4)�4k

, k < n

4

+ 1.

The proof of Theorem 28 is a rather straightforward modification of the proof ofTheorem 27 that is given below. We omit the complete proof, however, since wemake no use of this theorem in this paper except for those cases contained already inTheorem 27. Those cases correspond to n = 1, k = 1, and the cases � = 1/2 in part(i) and s 2 [2, 6) in part (ii). The results of Theorem 28 are again related to results inthe comprehensive work [2], but it is not easy to cite precise statements with completeproofs.

A.1. Fourier estimates in R2. The Fourier transform for u 2 L1(R2) is

(A.1) u(⇠, l) =

Z

R2

u(x, t)e�2⇡i(x⇠+tl)dx dt,

which extends to a bounded linear map u ! u from Lp to Lp

0, for 1 p 2 and

1/p+ 1/p0 = 1. Moreover, the Hausdor↵–Young inequality holds:

(A.2) kukp

0 kukp

for u 2 Lp. This simply interpolates kuk1 kuk1

and the Plancherel theorem,kuk

2

= kuk2

. The continuity of u follows from the dominated convergence theorem.In case u is integrable, one may recover u from u by

(A.3) u(x, t) =

Z

R2

u(⇠, l)e2⇡i(x⇠+tl)d⇠dl.


We will deduce Theorem 27 from the corresponding result on all of R2.

Theorem 29. Suppose u 2 W 2,1

2

(R2). Then u 2 C1/2,1/4(R2). Moreover, @x

u 2Ls(R2) for 2 s < 6, with

k@x

ukL

s(R2

)

CkukW

2,12 (R2

)

.

To proceed, we first recall a characterization of W 2k,k

2

.

Lemma 30 (characterization of W 2k,k

2

(R2) by Fourier transform). Let k be a non-negative integer, and set

m(⇠, l) = (1 + l2 + |⇠|4)1/2.

Then u 2 W 2k,k

2

(R2) if and only if mku 2 L2(R2). In addition, there exists a constantC such that

C�1kukW

2k,k2

kmkukL

2 CkukW

2k,k2

.

The following two technical lemmas will be used as well.

Lemma 31. For 0 ↵ < 2� and � � 1, we have

A↵,�

:= |⇠|↵/m� 2 Ls(R2) if any only if s > max

⇢3

2� � ↵,1

�

�.

Proof. A direct calculation using the substitution l = y(1 + |⇠|4)1/2 gives

kA↵,�

kss

=

Z

R2

|⇠|↵s

(1 + l2 + |⇠|4)�s/2d⇠ dl

=

Z

R2

|⇠|↵s(1 + |⇠|4)1/2

(1 + y2)�s/2(1 + |⇠|4)�s/2d⇠ dy

=

Z

R

dy

(1 + y2)�s/2

Z

R

|⇠|↵s(1 + |⇠|4)1/2

(1 + |⇠|4)�s/2d⇠.

This is bounded if and only if �s > 1 and 2�s � 2 � ↵s > 1. That is, s� > 1 ands(2� � ↵) > 3.

Lemma 32. Let V (x) = |x| ^ 1 := min{|x|, 1}. Then for some constant C > 0,

km�1V (r⇠)k2

+ km�1V (r2l)k2

Cr1/2 8r > 0.(A.4)

Proof. For the first term, substituting l = y(1 + |⇠|4)1/2 again, we find

km�1V (r⇠)k22

=

Z

R2

(|r⇠| ^ 1)2

(1 + l2 + |⇠|4)d⇠ dl

Z

R2

(|r⇠| ^ 1)2(1 + |⇠|4)1/2

(1 + y2)(1 + |⇠|4) d⇠ dy

=

Z

R

dy

(1 + y2)

Z

R

(|r⇠| ^ 1)2

(1 + |⇠|4)1/2d⇠.

The first factor is finite. We proceed to decompose the last integral into two parts,one over {⇠ : |⇠| < r�1} and the other over {⇠ : |⇠| > r�1}: The integrand is even,and


Z 1

0

(|r⇠| ^ 1)2

(1 + |⇠|4)1/2d⇠ =

Zr

�1

0

(r|⇠|)2

(1 + |⇠|4)1/2d⇠ +

Z 1

r

�1

1

(1 + |⇠|4)1/2d⇠

r2Z

r

�1

0

d⇠ +

Z 1

r

�1

|⇠|�2d⇠ = 2r.

In a similar fashion, we estimate, using ⇠ = (1 + l2)1/4⌘,

km�1V (r2l)k22

=

Z

R2

||r2l| ^ 1|2

1 + l2 + |⇠|4 d⇠ dl

Z

R2

(|r2l| ^ 1)2

(1 + l2)3/4(1 + |⌘|4)d⌘ dl

=

Z

R

d⌘

(1 + |⌘|4)

Z

R

(|r2l| ^ 1)2

(1 + |l|2)3/4dl.

The first integral is bounded; the second integral is further estimated by

Z 1

0

(|r2l| ^ 1)2

(1 + |l|2)3/4dl

Zr

�2

0

(r2|l|)2

(1 + |l|2)3/4dl +

Z 1

r

�2

1

(1 + |l|2)3/4dl

r4Z

r

�2

0

|l|1/2dl +Z 1

r

�2

|l|�3/2dl

=

✓2

3+ 2

◆r =

8

3r.

These estimates together yield the bound (A.4) as claimed.

Proof of Theorem 29. From the inversion formula (A.3) it follows that

kuk1 kuk1

kmuk2

km�1k2

CkukW

2,12

,

where the bound on km�1k2

= kA0,1

k2

is ensured by Lemma 31.(i) Fix (x, t) 6= (y, ⌧) so that r =

p|y � x|2 + |⌧ � t| > 0. Using the inequalities

|e2ia � e2ib| 2|a� b| ^ 2 = 2V (a� b),

|(y � x) · ⇠ + (⌧ � t)l| r|⇠|+ r2|l|,we obtain from the inversion formula and Lemma 32 that

|u(x, t)� u(y, ⌧)| 2⇡

Z(V (r⇠) + V (r2l))|u(⇠, l)|d⇠ dl

2⇡(km�1V (r⇠)k2

+ km�1V (r2l)k2

)kmuk2

Cr1/2kukW

2,12

.

This proves the embedding W 2,1

2

(R2) ! C1/2,1/4(R2).(ii) For 2 s < 6 we have s0 = s

s�1

2 and r > 3, where

1

r=

1

2� 1

s.

We may then use the Hausdor↵–Young inequality (A.2) and Lemma 31 to obtain

k@x

uks

Cs

k⇠uks

0 kmuk2

kA1,1

kr

, CkukW

2,12

.


Proof of Theorem 27. Let D be the given closed rectangle in R2. For functionsu defined a.e. on D, we extend u from D to a larger rectangle D containing D inits interior, in two steps, using linear combinations of dilated reflections as shown inAdams [1, Theorem 4.26]. The extension is to be made so that the weak derivativesare preserved across @D.

For instance, we reflect across the faces of D sequentially. First, from x faces{a, b} with c t d, writing a = a� (b� a), b = b+ (b� a), let

Ex

u(x, t) =

8<

:

�3u(2a� x, t) + 4u(�x/2 + 3a/2), a x a,u(x, t), a x b,�3u(2b� x, t) + 4u(�x/2 + 3b/2), b x b.

and then from the t faces {c, d} in an entirely similar manner, such that

u = Et

Ex

u(x, t)

is an C1 extension when crossing @D and well defined in D = [a, b] ⇥ [c, d]. Thenmultiply by a fixed smooth cut-o↵ function �(x, t) that is 1 on D and 0 near theboundary of D to obtain

Eu = �(x, t)Et

Ex

u(x, t).

In this way, given u such that @t

u and @jx

u are in L2(D) for j = 0, 1, 2, we obtain Eusuch that @

t

Eu and @jx

Eu are in L2(R2) for j = 0, 1, 2. The extension E is thus abounded linear operator from 2 W 2,1

2

(D) to W 2,1

2

(R2). Moreover,

Eu = u a.e. in D,

Eu has support in D, and

kEukW

2,12 (R2

)

CkukW

2,12 (D)

.

This combined with Theorem 29 when applied to Eu proves Theorem 27.

Appendix B. Existence for the truncated problem. In this appendix, weestablish the existence of a classical solution to the truncated problem (4.6). We aimto prove Proposition 13. This global existence result does not appear to follow easilyfrom stated results in standard parabolic theories, due to the fact that the boundarycondition J = 0 at x = 1 is nonlinear. For the convenience of the reader, we indicatehow to establish Theorem 13 by use of an approximation method that involves cuttingo↵ the nonlinear term in the flux J together with interior regularity theory. This willresult in a problem with standard linear Robin-type boundary conditions that stillrespects a maximum principle which keeps the solution uniformly bounded.

B.1. Approximation by flux cut-o↵. We consider, then, the following prob-lem. Let �(x) be a smooth, nondecreasing function with �(x) = 0 for x < �1 and�(x) = 1 for x > 1 as in (4.4). For small h > 0 define �

h

(x) = �(1 + (x � 1)/h), sothat

(B.1) �h

(x) =

(0 , x < 1� 2h ,

1 , x = 1 .

Writing

(B.2) Jh

= x2@x

nh

� 2xnh

+ n2

h

+ (3nh

� n2

h

)�h

,


we consider the problem

@t

nh

= @x

Jh

, x 2 (✏, 1), t 2 (0,1),(B.3a)

nh

= nin

h

, x 2 (✏, 1), t = 0,(B.3b)

0 = Jh

, x = 1, t 2 [0,1),(B.3c)

0 = ✏2@x

nh

� 2✏nh

, x = ✏, t 2 [0,1) .(B.3d)

We construct the initial data nin

h

from the given nin

so that at t = 0, the cut-o↵ fluxJh

is the original J✏

. Namely, we require that at t = 0,

(B.4) Jh

= x2@x

nin

� 2xnin

+ (nin

)2 .

We make nin

h

(x) = nin

(x) for x < 1 � 2h and use (B.4) to determine nin

h

(x) forx 2 [1 � 2h, 1]. These initial data are compatible with the boundary conditions(B.3c)–(B.3d). Clearly in the limit h # 0, we have nin

h

! nin

uniformly on [✏, 1].

B.2. Uniform bounds on the cut-o↵ problem. Because �h

(1) = 1, theboundary condition in (B.3c) is linear in n

h

, taking the form

(B.5) @x

nh

= �nh

, x = 1, t 2 [0,1) .

Moreover, note that (B.3a) takes the explicit form

(B.6) @t

nh

= x2@2x

nh

� 2nh

+ (@x

nh

)(2nh

+ (3� 2nh

)�h

) + (3nh

� n2

h

)�0h

.

For this problem, comparison principles hold, whence we obtain positivity and uniformsup-norm bounds on solutions.

Lemma 33. Suppose min[✏,1]

nin

h

> 0 and max[✏,1]

nin

h

< M1

, where M1

� 3.Suppose n

h

is a classical solution of (B.3) in [✏, 1] ⇥ (0, T ] with nh

continuous on[✏, 1]⇥ [0, T ]. Then 0 < n

h

(x, t) < M1

for all (x, t) 2 [✏, 1]⇥ [0, T ].

Proof. The proof of strict positivity is similar to Lemma 14. To prove the upperbound, suppose n

h

(X⇤) = M with X⇤ = (x⇤, t⇤), where t⇤ > 0 is minimal. Because@x

nh

= �nh

< 0 holds at x = 1, and (B.3d) holds at x = ✏, x⇤ must lie strictlybetween ✏ and 1. But because (B.6) holds and �0

h

� 0, this is impossible.

We may obtain global existence of a classical solution to problem (B.3) with cut-o↵ flux from the proof of Proposition 7.3.6 of [24], due to the time-uniform boundson n

h

in this lemma and the fact that the nonlinear terms in (B.3a) appear in thedivergence form N

h

(nh

) := @x

(n2

h

(1 � �h

)), which enjoys a local Lipschitz bound inthe L1 norm of the form

(B.7) kNh

(u)�Nh

(v)k1 K⇣ku� vk1kuk

C

1 + kvk1ku� vkC

1

⌘

with K = 1 + k�0h

k1.From the proof of [24, Proposition 7.3.6], this solution n

h

is continuous on [✏, 1]⇥[0,1) = Q✏, and the quantities @

x

nh

, @t

nh

, and @2x

nh

are continuous on [✏, 1]⇥(0,1).However, these quantities are actually all continuous on Q✏ by the local-time existencetheorem 8.5.4 of [24], due to the fact that the initial data are C3 and satisfy thecompatibility conditions. (A simple energy estimate for the di↵erence, along the linesof step 1 in subsection B.3 below, shows that the local solution given by this theoremagrees with that given by Proposition 7.3.6.)


Additionally, these quantities are also locally Holder-continuous on [✏, 1]⇥ (0,1),due to the regularity results stated in [24, Proposition 7.3.3(iii)]. From standardinterior regularity theory for parabolic problems (e.g., based on Theorem 8.12.1 of[21] and bootstrapping), we infer that n

h

is smooth in Q✏. In particular, the flux Jh

is a classical solution in Q✏ of the equation

(B.8) @t

Jh

= x2@2x

Jh

+ (@x

Jh

)(�2x+ 2nh

+ (3� 2nh

)�h

) .

Since Jh

is continuous on Q✏, by the maximum principle it is bounded in terms of itsinitial and boundary values—recall J

h

= n2

h

at x = ✏. From this and the sup-normbound in the previous lemma, we obtain (✏-dependent) uniform bounds on @

x

nh

.

Lemma 34. There is a constant M2

depending on nin

and independent of h and✏ such that |J

h

|+ ✏2|@x

nh

| M2

for all (x, t) 2 Q✏,

B.3. Energy estimates. These are simpler than the corresponding ones insection 5, because here ✏ > 0 is fixed, and the initial data is smooth.

1. The basic energy estimate is (using that nh

is positive and bounded)

d

dt

Z1

✏

1

2n2

h

dx =

Z1

✏

nh

@x

Jh

dx = nh

Jh

��1

✏

�Z

1

✏

(@x

nh

)Jh

dx

= �nh

(✏, t)3 �Z

1

✏

(@x

nh

)(x2@x

nh

� 2xnh

+ n2

h

+ (3nh

� n2

h

)�h

) dx

�✏2

2

Z1

✏

(@x

nh

)2 dx+ C

Z1

✏

n2

h

dx.

Here C is independent of h and t, and after integration we conclude that @x

nh

(andalso J

h

) is uniformly bounded independent of h in L2 on [✏, 1]⇥ [0, T ] for any T .2. For (x, t) 2 Q✏, the flux J

h

satisfies (B.8), and we find

d

dt

Z1

✏

1

2J2

h

dx = Jh

(x2@x

Jh

)��1

✏

�Z

1

✏

(x@x

Jh

)2 dx

+

Z1

✏

Jh

(@x

Jh

)(�4x+ 2nh

+ (3� 2nh

)�h

)

�✏2

3@t

(nh

(✏, t)3)� ✏2

2

Z1

✏

(@x

Jh

)2 dx+ C

Z1

✏

J2

h

dx .

Upon integration in time, we conclude @x

Jh

= @t

nh

is uniformly bounded independentof h in L2 on [✏, 1]⇥ [0, T ] for any given T . And further, using (B.2) for x < 1� 2h,we deduce that @2

x

nh

is uniformly bounded independent of h in L2 on any compactset

(B.9) [✏, 1� ✏]⇥ [0, T ] ⇢ [✏, 1)⇥ [0,1)

fixed independent of h. (This does not work for ✏ = 0 because �0h

is not uniformlybounded.)

3. Next, we have

d

dt

Z1

✏

1

2(@

x

Jh

)2 dx = (@x

Jh

)(@t

Jh

)��1

✏

�Z

1

✏

(@2x

Jh

)(@t

Jh

) dx

�2nh

(✏, t)(@t

nh

(✏, t))2 � ✏2

2

Z1

✏

(@2x

Jh

)2 dx+ C

Z1

✏

(@x

Jh

)2 dx .


Because @x

Jh

is continuous on Q✏, we may integrate this inequality over t 2 [0, T ],and use the bound on @

x

Jh

from the previous step, to conclude that @2x

Jh

and @t

Jh

areuniformly bounded independent of h in L2 on [✏, 1]⇥[0, T ]. By the anisotropic Sobolevestimates in Appendix A, we deduce that @

x

Jh

is uniformly bounded independent ofh in L4 on [✏, 1]⇥ [0, T ], as well.

4. Last we derive an interior estimate on @3x

Jh

. We define

�(x, t) = �2x+ 2nh

+ (3� 2nh

)�h

, so @t

� = 2(1� �h

)@x

Jh

.

Then

(B.10) @2t

Jh

= x2@2x

@t

Jh

+ �@x

@t

Jh

+ 2(1� �h

)(@x

Jh

)2,

and we let ⌘(x) = x� ✏ so that ⌘(✏) = 0 and ⌘0 = 1,

d

dt

Z1

✏

1

2⌘2(@

t

Jh

)2 dx =

Z1

✏

⌘2(@t

Jh

)(@2t

Jh

) dx

=

Z1

✏

⌘2(@t

Jh

)(�@x

@t

Jh

+ 2(1� �h

)(@x

Jh

)2) dx

�Z

1

✏

⌘2x2(@x

@t

Jh

)2 dx�Z

1

✏

(@t

Jh

)(@x

@t

Jh

)@x

(⌘2x2) dx

�✏2

2

Z1

✏

⌘2(@x

@t

Jh

)2 dx+ C

Z1

✏

(@t

Jh

)2 + (@x

Jh

)4 dx.

Because we only know @t

Jh

is continuous for t > 0, we integrate this over t 2 [s, T ],then over s 2 [0, ⌧ ], and use the bounds from the previous step. We infer that⌘@

x

@t

Jh

is uniformly bounded independent of h in L2 on [✏, 1]⇥ [⌧, T ]. Due to (B.8)and (B.1), we infer that @

t

(@x

Jh

) and @2x

(@x

Jh

) are uniformly bounded independentof h in [✏+ ✏, 1� ✏]⇥ [⌧, T ] for any small fixed ✏ > 0 and compact [⌧, T ] ⇢ (0,1).

B.4. Compactness argument. By the anisotropic Sobolev estimates inAppendix A, from the bounds on @

t

Jh

and @2x

Jh

in step 3 above, we have that Jh

isuniformly Holder-continuous (independent of h) on any compact set

(B.11) [✏, 1]⇥ [0, T ] ⇢ [✏, 1]⇥ [0,1) = Q✏ .

Also, by the bounds on @t

nh

and @2x

nh

in step 2, nh

is uniformly Holder-continuouson any compact set of the form in (B.9). From this we infer by (B.2) for x < 1� 2hthe same for @

x

nh

. By step 4, @t

nh

= @x

Jh

and @2x

nh

are uniformly Holder-continuouson any compact set

(B.12) [✏+ ✏, 1� ✏]⇥ [⌧, T ] ⇢ (✏, 1)⇥ (0,1) .

From the Arzela–Ascoli theorem and a diagonalization argument, along a subse-quence of h ! 0 we get uniform convergence of J

h

to J✏

in sets of form (B.11), nh

and @x

nh

to respective limits n✏

and @x

n✏

in sets of form (B.9), and @t

nh

to @t

n✏

and@2x

nh

to @2x

n✏

in sets of form (B.12), with all limits Holder-continuous on the indicatedsets.

In the limit, the PDE @t

n✏

= @x

J✏

holds for (x, t) 2 Q✏, and

(B.13) J✏

= x2@x

n✏

� 2xn✏

+ n2

✏

, (x, t) 2 [✏, 1)⇥ [0,1) .


Because of the continuity of J✏

on the sets in (B.11) and n✏

on the sets in (B.9), byregarding (B.13) as an ODE for n

✏

we deduce that n✏

and @x

n✏

are continuous on thesets in (B.11) also (i.e., up to the boundary x = 1), and both boundary conditions in(4.6c)–(4.6d) hold.

From standard parabolic theory as before, we find that n✏

is smooth in Q✏. Thisconcludes the proof of Proposition 13.

Appendix C. Regularity away from the origin. What we seek to do inthis section is to prove Theorem 5, providing su�cient local regularity in the domainQ = (0, 1] ⇥ (0,1) to infer that the solutions n in Theorem 4 are classical, withat least the regularity needed for the strict contraction estimate in Lemma 11. Forhigher regularity in the interior of Q we will rely on standard parabolic theory viabootstrap arguments.

The idea is to obtain uniform local bounds on L2 norms of the solutions nh

ofthe flux-cut-o↵ problem, the associated fluxes J

h

in (B.2), and certain space-timederivatives @↵n

h

, @�Jh

. These bounds will be independent of h, ✏ and the smoothingparameter . The local L2 bounds on these derivatives are inherited by @↵n

✏

, @�J✏

inthe limit h ! 0, then by @↵n

, @�J

after taking ✏! 0, and then by @↵n, @�J aftertaking ! 0. Local Holder-norm bounds for each quantity v 2 {n, @

x

n, @t

n, @2x

n} inQ will follow from the local L2 bounds on @

t

v and @2x

v, due to Theorem 27.In order to achieve this, we proceed to first obtain the needed estimates for n

h

and Jh

, independent of h, ✏, and , and then pass to the limits. Select a smoothfunction ⌘ : R ! [0,1), convex and nondecreasing with 0 = ⌘(0) < ⌘(x) x forx > 0. Weighted energy estimates with weight ⌘(x) = ⌘(x �ma) will yield uniformestimates in L2(W

m

), where the sets Wm

⇢ [✏, 1]⇥ [s, T ] have the form

(C.1) Wm

= [(m+ 1)a, 1]⇥ [ms, T ] , m = 1, 2, . . . ,

for a, s > 0 arbitrary but fixed independent of h, ✏, and .0. As a preliminary step, we seek a uniform pointwise bound on n

h

independentof h, ✏, and , in domains of the form

(C.2) [✏, 1]⇥ [⌧,1), ⌧ > 0.

From the form of Jh

and J✏

it follows that

nh

(x, t)� n✏

(x, t) = nh

(1/2, t)� n✏

(1/2, t) +

Zx

1/2

1

y2(J

h

� J✏

) dy

+

Zx

1/2

2

y(n

h

� n✏

)� 1

y2(n2

h

� n2

✏

)

�dy +

Zx

1�2h

1

y2(n2

h

� 3nh

)�h

dy.

Using the uniform convergence of nh

to n✏

in [✏, 1 � ✏] ⇥ [0, T ] (proven previously)and of J

h

to J✏

in [✏, 1] ⇥ [0, T ], as well as the bounds on nh

in Lemma 33 and onn✏

S(x, t) in Lemma 15, we obtain the uniform convergence of nh

toward n✏

ash ! 0. Therefore, we get the following uniform pointwise bound independent of h, ✏,and : For any ⌧ > 0, for su�ciently small h > 0 we have

(C.3) 0 < nh

(x, t) M⌧

= maxx2[0,1]

S(x, ⌧) + 1, (x, t) 2 [✏, 1]⇥ [⌧,1).

(Here and below, the required smallness of h depends on , because the bound inLemma 15 depends on . But we will not mention this further.)


1. Next we proceed to obtain bounds using weighted energy estimates. Theweighted energy estimate with ⌘(x) = ⌘(x� a) is

d

dt

Z1

✏

1

2⌘2n2

h

dx =

Z1

✏

⌘2nh

@x

Jh

dx = �Z

1

✏

(⌘2@x

nh

+ 2⌘⌘0nh

)Jh

dx

= �Z

1

a

(⌘2@x

nh

+ 2⌘⌘0nh

)(x2@x

nh

� 2xnh

+ n2

h

+ (3nh

� n2

h

)�h

) dx

�1

2

Z1

a

(x⌘ @x

nh

)2 dx+ C

Z1

a

(nh

+ n2

h

)2 dx.

By integration over t 2 [s, T ] and using (C.3), we infer that

ZT

s

Z1

2a

(x⌘ @x

nh

)2dx dt

Z

1

✏

⌘2n2

h

(x, s) dx+ C

ZT

s

Z1

a

(nh

+ n2

h

)2 dx dt

Cs

,(C.4)

where Cs

may depend on s (and T , but we suppress this dependence) but is inde-pendent of h, ✏, . Because ⌘(2a) > 0, we conclude that @

x

nh

, hence also Jh

, is uni-formly bounded independent of h, ✏ and in L2(W

1

) (with a bound that depends ona and s).

2. The cut-o↵ flux Jh

satisfies

(C.5) @t

Jh

= x2@2x

Jh

+ �@x

Jh

,

with boundary condition Jh

(1, t) = 0 for t > 0, where

�(x, t) = �2x+ 2nh

+ (3� 2nh

)�h

.

Multiply by ⌘2Jh

with ⌘(x) = ⌘(x � 2a), integrate by parts, and use the inequalityuv 1

4

u2 + v2 to obtain

d

dt

Z1

✏

1

2⌘2J2

h

dx

= �Z

1

✏

(x⌘ @x

Jh

)2 dx+

Z1

✏

Jh

(@x

Jh

)(⌘2� � @x

(x2⌘2)) dx

�Z

1

✏

(x⌘ @x

Jh

)2 dx+

Z1

✏

|Jh

@x

Jh

| · 2x⌘Cs

dx

�1

2

Z1

✏

(x⌘ @x

Jh

)2 dx+ Cs

Z1

2a

J2

h

dx .

Integrating over t 2 [⌧, T ] first, then averaging over ⌧ 2 [s, 2s], we obtain

ZT

2s

Z1

3a

(x⌘ @x

Jh

)2dx dt

1

s

Z2s

s

Z1

2a

⌘2J2

h

(x, ⌧)dx d⌧ + Cs

ZT

s

Z1

2a

J2

h

dx dt

C(a, s).(C.6)


Here we have used |Jh

|2 C(x2|@x

nh

|2 + n2

h

+ n4

h

), (C.4), and the upper bound onnh

in (C.3). We conclude that @x

Jh

is uniformly bounded in L2(W2

), independent ofh, ✏, . Thus @

t

nh

(but not @2x

nh

) is uniformly bounded in the same L2 sense.3. Let us write n

1

= @t

nh

= @x

Jh

. Then for t > 0,

(C.7) @t

n1

= @x

J1

, J1

(1, t) = 0,

where

(C.8) J1

= @t

Jh

= x2@x

n1

+ �n1

.

Note that the validity of the zero-flux condition J1

(1, t) = 0 is implied by the Holdercontinuity of J

1

. To see this is valid, set v = Jh

� n2

h

(1� �h

). From (B.8) it followsthat v solves

@t

v � x2@2x

v = F

subject to homogeneous boundary conditions, where the source term

F = @x

Jh

(�2x+ 3�h

) + x2@2x

(n2

h

(1� �h

)).

From the results in Appendix B, F is locally Holder-continuous on [✏, 1] ⇥ (0,1).Hence, J

1

= @t

v + 2nh

@t

nh

(1� �h

) is continuous up to x = 1.Multiply (C.7) by ⌘2n

1

with ⌘(x) = ⌘(x � 3a), and integrate in x over [✏, 1] toobtain

d

dt

Z1

✏

1

2⌘2n2

1

dx = �Z

1

✏

(⌘2@x

n1

+ 2⌘⌘0n1

)(x2@x

n1

+ �n1

) dx

�Z

1

✏

(x⌘ @x

n1

)2 dx+

Z1

✏

⇣(2x2⌘⌘0 + |�|⌘2)|n

1

@x

n1

|+ 2⌘⌘0|�|n2

1

⌘dx

�1

2

Z1

✏

(x⌘ @x

n1

)2 dx+ Cs

Z1

3a

n2

1

dx .

Integrating over t 2 [⌧, T ] first, then over ⌧ 2 [2s, 3s], we obtain

ZT

3s

Z1

4a

(x⌘ @x

n1

)2dx d⌧

1

s

Z3s

2s

Z1

3a

⌘2n2

1

(x, ⌧) dx d⌧ + Cs

ZT

2s

Z1

3a

n2

1

dx d⌧

C(a, s),(C.9)

where the bound on n1

= @x

Jh

in (C.6) from step 2 has been used. We conclude that@2x

Jh

and @t

Jh

(by (C.5)) are uniformly bounded independent of h, ✏, and in L2 onW

3

, hence Jh

is uniformly Holder continuous on W3

.4. Next we compute @

t

J1

to complete the estimates for classical solutions. Dif-ferentiating (C.8) with respect to t we find that for t > 0,

(C.10) @t

J1

= x2@2x

J1

+ �@x

J1

+ @t

� n1

, J1

(1, t) = 0.

Recall that |�| Cs

and note @t

� = 2(1� �h

)@x

Jh

, hence |@t

�| 2|n1

|. Multiply by⌘2J

1

with ⌘(x) = ⌘(x� 4a), and integrate by parts to find


d

dt

Z1

✏

1

2⌘2J2

1

dx+

Z1

✏

(x⌘ @x

J1

)2 dx

=

Z1

✏

(�⌘2 � @x

(x2⌘2))J1

(@x

J1

) + ⌘2J1

@t

� n1

dx

1

2

Z1

✏

(x⌘ @x

J1

)2 dx+ Cs

✓Z1

4a

J2

1

dx+

Z1

4a

n4

1

dx

◆.

Integrating over t 2 [⌧, T ] first, then averaging over ⌧ 2 [3s, 4s], we obtain

ZT

4s

Z1

5a

(x⌘ @x

J1

)2dx dt 1

s

Z4s

3s

Z1

4a

⌘2J2

1

(x, ⌧) dx d⌧

+ Cs

ZT

3s

✓Z1

4a

J2

1

dx+

Z1

4a

n4

1

dx

◆dt

Cs

ZT

3s

Z1

4a

(|@x

n1

|2 + |@x

Jh

|2 + |n1

|4) dx dt.

The first two terms are bounded using the bounds from the previous steps, (C.6)and (C.9). Note also that n

1

= @x

Jh

is in L4(W3

) due to an anisotropic embeddingtheorem. Hence

ZT

4s

Z1

5a

(x⌘ @x

J1

)2dx d⌧ C(a, s).(C.11)

We can conclude that @x

J1

(= @x

@t

Jh

= @t

@x

Jh

= @2t

nh

) is bounded in L2(W4

)independent of h, ✏, and .

5. After taking the limit h ! 0 along a suitable subsequence, we conclude fromsteps 1 and 2 that @

t

n✏

= @x

J✏


), hence the same istrue of @2

x

n✏

due to the form of J✏

in (4.5). By Theorem 27, n✏

is uniformly Holder-continuous on W

2

, independent of ✏ and .Next we conclude from step 3 that J

✏

is uniformly Holder-continuous on W3

, andthe same is true of @

x

n✏

by (4.5).From step 4 we then conclude @

t

@x

J✏


) and bydi↵erentiating (4.5) we conclude the same for @3

x

J✏

. Therefore @x

J✏

= @t

n✏

is uniformlyHolder-continuous on W

4

, and the same holds for @2x

n✏

.After taking the limits ✏ ! 0, and finally ! 0, these estimates ensure that the

weak solution n of Theorem 4 is a classical solution in Q = (0, 1] ⇥ (0,1), with thelocal Holder regularity indicated in Theorem 5.

Acknowledgments. We want to thank IPAM for hospitality and support duringour visit in May and June 2009, when this work was initiated.

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