Controlling number of particles in fragmentation equations

Physica D 239 (2010) 1422–1435

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Controlling number of particles in fragmentation equations J. Banasiak ∗,1 , S.C. Oukouomi Noutchie School of Mathematical Sciences, University of KwaZulu-Natal, Durban, South Africa

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Article history: Available online 14 May 2009

One of the most interesting features of fragmentation models is the possibility of breaching the mass conservation principle through ‘shattering’; that is, the formation of a dust of zero-size particles. A similar phenomenon may occur in the evolution of the number of particles in the system which, to some extent, is intertwined with the evolution of its total mass. To investigate these phenomena, we consider the fragmentation equation in the space of densities yielding both a finite number of particles and a finite mass of the ensemble, and show, in particular, that in a non-shattering fragmentation one can typically control the total number of particles in the system. On the other hand, both mass and particles in the shattering fragmentation can disappear from the system. It is conjectured that in such a case the fragmentation equation alone does not offer the full description of the dynamics of the problem. © 2009 Elsevier B.V. All rights reserved.

Dedicated to Professor Aldo Belleni-Morante Keywords: Fragmentation processes Shattering Mass conservation Particle number conservation Substochastic semigroups

1. Introduction Fragmentation processes can be observed in natural sciences and engineering. To provide just a few examples, we mention the study of stellar fragments in astrophysics, rock fracture, degradation of large polymer chains, DNA fragmentation, evolution of phytoplankton aggregates, liquid droplet breakup or breakup of solid drugs in organisms. Though mathematical study of fragmentation processes can be traced back to papers by Melzak [1] (from the analytical point of view) and Filippov [2] (from the probabilistic one), it was not until the 80s that a systematic investigation of them was undertaken, mainly by Ziff and his students, e.g. [3,4], who provided explicit solutions to a large class of fragmentation equations of the form

∂t u(x, t ) = −a(x)u(x, t ) +

∞

Z

a(y)b(x|y)u(y, t )dy,

x ≥ 0, t > 0,

(1)

x

with power law fragmentation rates a(x) = xα , α ∈ R, and where b(x|y), the distribution of particle masses x spawned by the fragmentation of a particle of mass y > x, was also given by a power law xν b(x|y) = (ν + 2) ν+1 , y

(2)

with ν ∈ (−2, 0] (see also (50) for a more detailed discussion of this case). Here u(x, t ) is the density of particles having mass x at time t. Later a comprehensive probabilistic theory of fragmentation processes was developed by Bertoin and Haas, see e.g. [5–9], while a development of functional-analytic methods and, in particular, of the semigroup theory, helped to put many earlier phenomenological results on a firm mathematical ground, see e.g. [10,12–15]. Fragmentation processes are difficult to analyze as they involve the evolution of two intertwined quantities: the mass of the ensemble and the number of particles in it. That is why, though linear, they display nonlinear features such as phase transition which, in this case, is called shattering and consists of a formation of ‘dust’ particles of zero size, but, nevertheless, carrying a non-zero mass. We can identify this process quantitatively by the disappearance of mass from the system even though it is conserved in each fragmentation event. Probabilistically, shattering is an example of an explosive, or dishonest, Markov process, see e.g. [16,17] and from this point of view it

∗ Corresponding address: School of Mathematical Sciences, University of KwaZulu-Natal, Private Bag X54001, 4000 Durban, KwaZulu-Natal, South Africa. Tel.: +27 312601363; fax: +27 312608348. E-mail addresses: [email protected] (J. Banasiak), [email protected] (S.C.O. Noutchie). 1 Also: Institute of Mathematics, Technical University of Łódź. 0167-2789/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.physd.2009.05.002

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has been exhaustively analyzed in [2,5–9,18]. Shattering was rediscovered in natural sciences in [19] where the loss of mass was noticed by analyzing explicit solutions for fragmentation equations with power-law fragmentation rates. Shattering was explained analytically in a general case in [11–14] by linking it to characterization of the generator of the dynamical system associated with the fragmentation process; these results were compared with the probabilistic approach in [15]. R∞ If u is a solution to (1), the total mass of the ensemble at a time t is given by the first moment of u; that is, M (t ) = 0 xu(x, t )dx. From the physical point of view, the total mass of fragmenting particles cannot increase, thus fragmentation equations are usually investigated in the space X1 := L1 (R+ , xdx) =



∞

Z u;

 |u(x)|xdx < +∞ ,

(3)

0

where the norm gives the mass of the ensemble for nonnegative density u. The explanation for this is that the process in this space should be dissipative, which typically reduces the complexity of the analysis. However, as we mentioned earlier, fragmentation events result in an increase in the number of particles in the system, which is not tracked by the norm in X1 . Apart from an inherent interest in knowing how the number of particle evolves, there is also a practical angle to this question: fragmentation events are often coupled with, in some sense reverse, coagulation processes, which are most easily analyzed in the finite number of particles space: X0 := L1 (R+ , dx) =



∞

Z



|u(x)|dx < +∞ .

u;

(4)

0

Hence, analysis of the combined fragmentation–coagulation equation requires well-posedness of the fragmentation equation in



X0,1 := L1 (R+ , (1 + x)dx) =

∞

Z u;

 |u(x)|(1 + x)dx < +∞ .

(5)

0

The main aim of this paper is to analyze (1) in the space X0,1 for arbitrary a and a class of separable b(x|y) which is more general than (2). Of particular interest is the honesty of the process in X0,1 ; that is, whether the evolution of the mass and number of particles, given by the solution u to (1), coincides with the one predicted by the local laws (11) and (12), used to construct (1). One of the main results of the paper is that (1) is well-posed and honest in X0,1 for fragmentation rates a bounded at 0 and with b yielding a finite number of daughter particles at 0. On the other hand, shattering fragmentation (corresponding, roughly speaking, to a unbounded at 0) is associated with an accelerating infinite cascade of fragmentation events of smaller and smaller particles leading to the creation of a dust. Hence, intuitively, in shattering fragmentation we should observe an appearance of an infinite number of particles. We shall demonstrate that this intuition is not necessarily correct — for sufficiently fast fragmentation of small particles (e.g., for α < −1 and ν ∈ (−1, 0) in the power law case) the total number of them remains finite (though we do not know whether the process is honest in X0,1 ). We note that this phenomenon was noticed in [20] for power law rates by analyzing explicit solutions and in [6] for the so-called homogeneous fragmentation using probabilistic methods. To explain this phenomenon, one could conjecture that full description of the dynamics in the shattering regime requires two compartments: one for the ‘physical’ particles, which are visible within the model governed by (1), and one for the ‘dust’. In this interpretation, shattering would be a flow of particles from the former to the latter with the speed related to a close to zero. If a close to zero is just large enough for shattering to occur, we may observe an accumulation of small particles in the ‘physical’ compartment. If a becomes larger, then the flow between compartments becomes fast enough to keep the number of physical particles finite for all times. This approach is a subject of current research. Another counterintuitive result observed in this paper is related to the case when the number of daughter particles produced in each fragmentation event is infinite and, at the same time, the fragmentation is strongly shattering (e.g. if ν ∈ (−2, −1] and α < −1 in the power law case). Despite this, we observe that we still have evolution in X0,1 (at least for a class of initial densities). This phenomenon also could be explained by the conjecture discussed in the previous paragraph. We note that, while the theory for the non-shattering case has been developed up to a reasonably complete level and, as such, is presented here, the shattering case and the case with infinite production of daughter particles still contain gaps and open problems. Therefore we decided to present the results pertaining to the latter rather in the form of examples and comments; the research to fill the gaps and answer the open questions is ongoing. 2. Preliminaries As we mentioned in the introduction, we are concerned with the initial value problem for the kinetic type rate equation

∂t u(x, t ) = −a(x)u(x, t ) +

Z

∞

a(y)b(x|y)u(y, t )dy,

x, t > 0

x

◦

u(x, 0) =u,

(6)

which describes the evolution of the density u of particles having mass x at time t; the particles undergo fragmentation at a rate a. We assume that a is a positive and continuous function on (0, ∞). Further, b describes the distribution of daughter particle masses x spawned by the fragmentation of a parent particle of mass y > x. In the absence of any other mechanism, the mass of all daughter particles must be equal to the mass of the parent. This ‘local’ mass conservation principle mathematically is expressed by y

Z

xb(x|y)dx = y. 0

(7)

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Similarly, the expected number of particles produced by a particle of mass y is given by n(y) =

y

Z

b(x|y)dx.

(8)

0

We note that n(y) may be infinite. For a comprehensive discussion of the physical background of the model as well as the properties of the function b we refer the reader to [19,13] and [14, Section 8.2]. Solvability of (6) in X1 has been established by means of the substochastic semigroup theory, [14, Chapters 5 and 8]. Let us recall that a semigroup is called substochastic if it is a semigroup of positive contractions. To formulate the result, we introduce some additional notation which is also used in the remaining part of the paper. By A we denote the pointwise multiplication φ(x) → −a(x)φ(x) defined on the set of, say, measurable functions. Similarly, by B we denote the expression

[B φ](x) =

∞

Z

a(y)b(x|y)φ(y)dy,

(9)

x

defined first on all positive measurable functions for which the above integral is finite almost everywhere and then extended by linearity to a suitable linear subspace of measurable functions. The formal expressions A and B may define various operators. We start with A defined by Au = Au on D(A) = {u ∈ X1 ; Au ∈ X1 }; B restricted to D(A) is a well-defined positive operator, denoted by B. Then (6) can be written as an abstract Cauchy problem in X1 :

∂t u = Au + Bu,

t >0

◦

u(0) =u .

(10)

Unfortunately, (A + B, D(A)) does not generate a semigroup on X1 unless A is bounded. However, the substochastic semigroup theory yields the existence of the smallest substochastic semigroup (GK (t ))t ≥0 generated by an extension K of A + B. The semigroup (GK (t ))t ≥0 can be obtained as the strong limit in X1 of semigroups (Gr (t ))t ≥0 generated by (A + rB, D(A)) as r % 1− ; the limit is monotonic on non-negative data. The fact that, in general, K is a proper extension of A + B has far reaching consequences which we explain below. Local conservation principles (7) and (8) render, by formal integration of (6), global conservation principles: d dt d dt

M (t ) =

∞

Z

∂t u(x, t )xdx = 0,

(11)

0

N (t ) =

∞

Z

∂t u(x, t )dx = 0

∞

Z

a(x)(n(x) − 1)u(x, t )dx.

(12)

0

If the above conditions are satisfied by solutions to (6), then the process is called honest (in the respective space X1 , X0 or X0,1 ). However, validity of either condition depends on properties of u, namely whether each term on the right hand side of the integro-differential equation in (6) can be separately integrated. Since the solution semigroup is generated by an extension of A + B, the right hand side of (10) should be treated as a single operator K and thus such separation of integrals is not always possible. Hence the global conservation principles are not always satisfied, in which instance the process is called dishonest. The honesty of fragmentation processes in X1 has been extensively studied. Here we recall the main results specified to the particular form of the function b which is also our choice for analysis in this paper. Namely, we assume that b can be written as b(x|y) = β(x)γ (y)

(13)

where, to satisfy the local principle of mass conservation,

γ (y ) = R y 0

y sβ(s)ds

.

(14)

We assume that β is a non-negative continuous function on (0, ∞). Eq. (13) is a natural generalization of the power law b described in (2) and has the advantage of allowing the number of daughter particles

Ry

y β(s)ds n(y) = R y0 , sβ(s)ds 0

(15)

to vary with the parent size y, [13]. An important role in the analysis is played by the function xβ(x) d b(x|x) = β(x)γ (x) = R x = ln dx sβ(s)ds 0

x

Z

sβ(s)ds.

(16)

0

Honesty in X1 (that is, the validity of the global mass conservation principle (11)) turns out to be equivalent to K being equal to the closure of A + B, K = A + B. When b is given by (13), the following theorem settles this question. Theorem 1 ([13]). Assume that limx→0+ a(x) exists (finite or infinite). Then K = A + B if and only if there exists δ > 0 such that b(x|x)/a(x) 6∈ L1 ([0, δ]). Let us turn our attention to the spaces X0 and X0,1 . First we observe that, by direct integration of explicit solutions derived in [3,4], in general we cannot expect solvability in X0 (see also Remark 4 and Eq. (54)). However, we shall be able to prove solvability if the initial mass and number of particles is finite; that is, in the space X0,1 . We cannot, however, expect the process in X0,1 to be dissipative as the number of particles grows rapidly, see (12), and thus we shall not be able to employ the substochastic semigroup theory. Instead, we use the theory of resolvent positive operators. The starting point is to find the resolvent of the potential generator.

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

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3. Resolvent in X1 It is difficult to directly find the formula for the resolvent of the generator K since its domain is not known explicitly. To circumvent this difficulty, first we consider the Miyadera perturbation of Kr := A + rB of A, 0 < r < 1, with D(Kr ) = D(A), which is known (see [14, Theorem 5.2]) to generate a positive semigroup of contractions. Thus the resolvent R(λ, Kr ) = (λI − A − rB)−1 exists for all λ > 0. Let us define Qr λ := λI − A − rB. To find the formula for R(λ, Kr ), we start by solving ∞

Z

f (x) = λur (x) − a(x)ur (x) − r

a(y)b(x|y)ur (y)dy,

0 < r < 1.

x

This equation can be converted into a first order linear ordinary differential equation, similarly to [13]. Choosing the constant in the general solution of this equation so as to have solutions converging to zero for x → +∞ (at least for regular f ), we obtain f (x)

ur (x) = [Rr (λ)f ](x) :=

+

λ + a( x )

r β(x)

λ + a(x)

e−ξr (x)

∞

Z x

a(s)γ (s) ξr (s) e f (s)ds, λ + a( s )

(17)

where

ξr (x) = r

a(s)γ (s)β(s)

x

Z

ds.

λ + a(s)

1

To prove that (17) indeed represents the resolvent R(λ, Kr ), we re-write (17) in a more convenient form. Define b(x|x) , λ + a(x)

Bλ (x) =

λ > 0.

(18)

By (16)

ξr (x) = r

x

Z

b(s|s)ds − λ

1

and eξr (x) = e−C

Rx 0

x

Z

Bλ (s)ds




r

1 Bλ (z )dz

1

Z

0

Rx

e−λr

sβ(s)ds − ln

= r ln

1

sβ(s)ds

x

 Z

sβ(s)ds − λ 0

, where C = r ln

R1 0

x

Z



Bλ (s)ds ,

(19)

1

sβ(s)ds. Thus

ur (x) = [Rr (λ)f ](x) f (x)

=

λ + a( x )

Rx

where Γ (x) =

+

0

rBλ (x) x

Γ (x)1−r er λ

Rx

1 Bλ (s)ds

∞

Z

sa(s)f (s)

λ + a( s )

x

Γ (s)r −1 e−r λ

Rs

1 Bλ (z )dz

ds,

(20)

sβ(s)ds is a positive and increasing function. We start with

Lemma 1. For any f ∈ D(A), x > 0, 0 ≤ r ≤ 1, ∞

Z

a(s)γ (s)eξr (s) |f (s)|ds < ∞.

x

Proof. Since for f ∈ D(A) and x > 0 ∞

Z

∞

Z

a(s)γ (s)eξr (s) |f (s)|ds =

x

(sa(s)|f (s)|)

γ (s)

x

it is enough to show that

γ (s) ξr (s)

e

s

eξr (s) = e−c Γ r (s)e−λr

s

eξr (s) ds,

is bounded at ∞. For r > 0, s ≥ 1 we have

Rs

1 Bλ (z )dz

≤ e−c Γ r (s),

and for r = 0 it is bounded by 1. Using (14), the properties of Γ (s) and r ≤ 1, we obtain

γ (s) s

eξr (s) ≤ e−c Γ r −1 (s) ≤ e−c Γ r −1 (1). 

Proposition 1. Let 0 < r < 1, λ > 0. Then R(λ, A + rB) = Rr (λ). Proof. Consider (20). The first term on the right hand side, f → f /(λ + a), clearly defines a bounded operator provided λ > 0. Let f ≥ 0 and consider the norm of the second term: ∞

Z



E2 =

x

0

sf (s)

≤r 0

=

λ

λ + a(s)

sf (s) 0

sa(s)f (s)

∞

Rx

Z

Rs

Z s

1 Bλ (r )dr

λ + a( s )

x

∞

Z

Γ (x)1−r er λ a(s)

∞

Z 1

rBλ (x)

e− r λ

a(s)

λ + a(s)

1 Bλ (z )dz

Bλ (x)er λ

Γ (s)r −1 e−r λ

Rx

1 Bλ (z )dz

Rs

1 Bλ (z )dz

 ds xdx



dx ds

0

e

−r λ

Rs

1 Bλ (z )dz



Rs

er λ 1 Bλ (z )dz

− lim

→0+

R1 e−r λ  Bλ (z )dz



ds,

(21)

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J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

we changed the order of integration and used Γ (x)1−r ≤ Γ (s)1−r for 0 ≤ x ≤ s and r < 1. Now, Bλ is a positive function, so Rwhere s B ( z )dz always exists and can be either finite or +∞. In either case 0 λ  R  R R s −r λ 1s Bλ (z )dz r λ 1 Bλ (z )dz −r λ 1 Bλ (z )dz e e − lim e ≤1 →0+

and E2 ≤ λ−1 kf k1 , thus Rr (λ) is a bounded operator on X1 . Since we know that for 0 < r < 1 the resolvent R(λ, Kr ) exists, to show that R(λ, Kr ) = Rr (λ) it is enough to prove that Rr (λ) is the left inverse of Qr λ = λI − A − rB. For 0 ≤ f ∈ D(A) we have

[Qr λ f ](x) r β(x) −ξr (x) + e λ + a(x) λ + a(x) =: I1 (x) + I2 (x),

([Rr (λ)Qr λ ]f )(x) =

∞

Z x

a(s)γ (s) ξr (s) e [Qr λ f ](s)ds λ + a(s)

where we used the fact that each term in Rr (λ) is a bounded operator. Then

R∞ Z ∞ (λ + a(x))f (x) − r x a(y)b(x|y)f (y)dy r a(y)b(x|y)f (y)dy, I1 (x) = = f ( x) − λ + a(x) λ + a( x ) x and a(s)γ (s) ξr (s) e [Qr λ f ](s)ds λ + a(x) λ + a(s) Zx ∞ r β(x) −ξr (x) = e a(s)γ (s)eξr (s) f (s)ds λ + a(x) x Z ∞  Z r β(x) −ξr (x) ∞ a(s)γ (s) ξr (s) − r e e a(y)b(s|y)f (y)dy ds =: J1 (x) − J2 (x), λ + a(x) λ + a( s ) x s

I2 (x) =

r β(x)

e

−ξr (x)

Z

∞

where we could split the integral thanks to the integrability of the first term ensured by Lemma 1. Changing the order of integration by the Fubini theorem, we get that J2 (x) =

r β(x)

e−ξr (x)

Z

∞

a(y)γ (y)f (y)

Z

y

a(s)γ (s)β(s) ξr (s) e ds dy λ + a(s)



r

λ + a(x) x x Z
r β(x) −ξr (x) ∞ = e a(y)γ (y)f (y) eξr (y) − eξr (x) dy λ + a(x) Zx Z ∞ r β(x) −ξr (x) ∞ r β(x) = e a(y)γ (y)eξr (y) f (y)dy − a(y)γ (y)f (y)dy, λ + a(x) λ + a(x) x x

where again we used Lemma 1 to split the integral in the penultimate line. It follows that r β(x)

I2 (x) = J1 (x) − J2 (x) =

λ + a(x)

∞

Z

a(y)γ (y)f (y)dy. x

Thus, for 0 ≤ f ∈ D(A), ([Rr (λ)Qr λ ]f )(x) = I1 (x) + I2 (x) = f (x). Since D(A) is a weighted L1 space, an arbitrary f ∈ D(A) can be written as f = f+ − f− where f+ , f− ∈ D(A) are non-negative and the above equality extends to D(A), thus proving the proposition.  Let us introduce the formal expression

[R(λ)f ](x) =

f ( x)

λ + a(x)

+

Bλ (x) λ R x Bλ (s)ds e 1 x

∞

Z x

sa(s)f (s) −λ R s Bλ (r )dr 1 e ds. λ + a(s)

(22)

Theorem 2. Under the assumptions of this section, the resolvent R(λ, K ) of the generator K in X1 is given by

[R(λ, K )f ](x) = [R(λ)f ](x).

(23)

Proof. We use the fact that R(λ, K ) is the monotonic almost everywhere, as well as the strong in X1 , limit of R(λ, Kr ) as r % 1, see [14, Theorem 5.2]. Denote the right hand side of (22) by R(λ). First, using the same argument as in (21) with Γ (x) = 1, we see that R(λ) defines a bounded operator on X1 . Next, from Proposition 1 we know that Rr (λ) = R(λ, A + rB) is the resolvent of A + rB. Consider now limr %1 ur (x). It is clear that lim

rBλ (x)

r %1

x

Γ (x)1−r er λ

Rx

1 Bλ (s)ds

=

Bλ (x) λ R x Bλ (s)ds e 1 . x

Further, taking f with suppf ⊂ [x0 , M ] with 0 < x0 < M < ∞ we have

Z

∞ x0

sa(s)f (s)

λ + a(s)

Rs

r −1 −r λ 1 Bλ (z )dz e ds

Γ ( s)

Z

M

= x0

sa(s)f (s)

Γ (s)(λ + a(s))

Γ (s)

 eλ

Rs

1 Bλ (z )dz

r ds

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

1427

and the integrand is bounded by a constant which is integrable on this interval. Thus, we can pass to the limit for any x > 0 getting lim[R(λ, A + rB)f ](x) = [R(λ)f ](x) =

r %1

f ( x)

λ + a(x)

Bλ (x) λ R x Bλ (s)ds e 1 x

+

∞

Z x

sa(s)f (s) −λ R s Bλ (r )dr 1 e ds, λ + a(s)

for f ≥ 0 with bounded and separated from 0 support. Thus R(λ)f = R(λ, K )f on a dense subset of X1 and, since the expression for R(λ) defines a bounded positive operator on X1 , the representation (22) extends to the whole space.  4. Resolvent in X0,1 Our aim is to prove the existence of solutions to the fragmentation Eq. (6) in X0,1 = L1 (R+ , (1 + x)dx) = X0 ∩ X1 = L1 (R+ , dx) ∩ L1 (R+ , xdx). Since X0,1 ⊂ X1 , the resolvent, if it exists, must be given by (22). Define

∆λ (x) = e−λ

R1

x Bλ (s)ds

.

(24)

Theorem 3. The expression R(λ), λ > 0, defines an operator on X0,1 if and only if xa(x)

Ξλ (x) :=

λ + a(x)

1 ∆− λ (x)

x

Z

Bλ (s)(1 + s) s

0

∆λ (s)ds is bounded at 0.

(25)

Proof. First we observe that a necessary condition for (25) to hold is that the inner integral be finite; that is, Bλ (x)

Z

x

0+

∆λ (x)dx =

1

λ

Z

1 d 0+

x dx

∆λ (x)dx < +∞.

(26)

Since X0,1 inherits the lattice structure from X1 , R(λ) : X0,1 → X0,1 if and only if kR(λ)f k0,1 < +∞ for any 0 ≤ f ∈ X0,1 . In this case R(λ)f is a sum of two positive terms, hence for 0 ≤ f ∈ X0,1 we have ∞

Z

[R(λ)f ](x)(1 + x)dx = 0

∞

Z

f (x)(1 + x) 0



1

λ + a(x)

+

a(x)

x

λ + a(x) 1 + x

e−λ

Rx

1 Bλ (s)ds

x

Z 0

Bλ (s)(1 + s) λ R s Bλ (r )dr ds dx. e 1 s



(27)

Since 1/(λ + a(x)) is bounded, the first term is finite and kR(λ)f k0,1 < +∞ if and only if the second term is bounded on R. Hence, if kR(λ)f k0,1 < +∞ then, in particular, (25) is satisfied (as the behavior of x/(1 + x) is the same as that of x as x → 0). This, moreover, yields (26). To prove the opposite implication, assume (25). Then (26)Ris also satisfied. Consider the behavior of the second term in (27), which ∞ equals (1 + x)−1 Ξλ (x), as x → ∞. First observe that, by (26), α Bλ (x)(1 + x)x−1 ∆λ (x)dx either exists or does not exist irrespectively of α ≥ 0 and x

Z

Bλ (s)(1 + s)

1

s

∆λ (s)ds ≤

2

λ

(∆λ (x) − 1).

If the improper integral exists, then (1 + x)−1 Ξλ (x) is bounded at infinity on account of the boundedness of ∆λ (x)−1 . If the integral is infinite then, by the above, ∆λ (x) tends to infinity as x → ∞ and hence ∆λ (x)−1 tends to 0. Then the last two multipliers in the second summand of (27) are of 0 · ∞ type and we can use the l’Hospital rule getting lim e−λ

x→∞

Rx

1 Bλ (s)ds

x

Z 0

Bλ (s)(1 + s) λ R s Bλ (r )dr Bλ (x)(1 + x)∆λ (x) 1 e 1 ds = lim = . x →∞ s λxBλ (x)∆λ (x) λ

Since the other two multipliers are bounded, (1 + x)−1 Ξ (x) is bounded as x → ∞ and hence (25) alone ensures finiteness of kR(λ)f kX0,1 .  Corollary 1. If there is λ0 > 0 such that (25) is satisfied for all λ > λ0 , then R(λ) defines a positive resolvent of an operator K0,1 which is the part of K in X0,1 . Proof. Since R˜ (λ) := R(λ)|X0,1 is a resolvent, we can define the operator K˜ f = λf − R˜ (λ)−1 f for f ∈ D(K˜ ) := Range R˜ (λ). Recall that the part K0,1 is defined as the restriction of K to D(K0,1 ) = {f ∈ D(K ) ∩ X0,1 ; Kf ∈ X0,1 }. If f ∈ D(K˜ ) ⊂ X0,1 , then clearly f is in the range of R(λ, K ) (equal to D(K )). Hence f ∈ D(K0,1 ) and, since R(λ, K )−1 f = R˜ (λ)−1 f and K˜ f = K0,1 f , we have K˜ ⊂ K0,1 . On the other hand, if f ∈ D(K0,1 ), then f = R(λ, K ) = R(λ)g for some g ∈ X1 and f ∈ X0,1 . Then X0,1 3 Kf = λR(λ, K )g − g = λf − g , thus g ∈ X0,1 and f is in the range of R˜ (λ), which is D(K˜ ), and K0,1 f = λf − R(λ, K )−1 f = λf − R˜ (λ)−1 f = K˜ f . 

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J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

Due to the delicate interplay of possible singularities in Bλ and ∆λ , it seems to be difficult to give more explicit necessary and sufficient conditions ensuring that R(λ) defines an operator in X0,1 . We can, however, provide a set of easy-to -check sufficient conditions which cover most standard cases. Corollary 2. Let one of the conditions be satisfied: (1) lim xBλ (x) = Lλ

(28)

x→0+

with 1 < λLλ < +∞ and xB(x), extended by continuity to x = 0, be Hölder continuous, or (2) lim Bλ (x) = 0

(29)

x→0+

and Bλ (x), extended by continuity to 0, is Hölder continuous, or (3) lim xBλ (x) = ∞.

(30)

x→0+

with M1 /xβ ≤ xBλ (x) ≤ M2 /xα close to x = 0, where α, β > 0 and may depend on λ and M1 , M2 > 0 may depend on β and α , respectively. Then condition (25) is satisfied. Proof. First we consider (28) so that |xBλ (x) − Lλ | ≤ Mxα for some α > 0. We have e−λ

R1

x Bλ (s)ds

= e−λLλ

R1

1 x s ds

R 1

−λ

x

e



L Bλ (s)− sλ ds

(31)

with 1

Z

1

Z Bλ (s) − Lλ ds ≤ M s

sα−1 ds

bounded as x → 0+ . On the other hand, e−λLλ Bλ (s)

x

Z

(32)

x

x

s

0



∆λ (s)ds ≤ M 0 Lλ

x

Z

R1

1 x s ds

sλLλ −2 ds +

0

Z

= xλLλ and, by the above,  x |sB(s) − Lλ |sλLλ −2 ds < +∞

0

provided λLλ > 1. This gives (26). 1 1 −λLλ since the second factor in (31) is also bounded from below by (32), and thus x∆− Further, in this case, ∆− λ (x) tends to infinity as x λ also tends to infinity since λLλ > 1. Hence, by the l’Hospital rule 1 lim x∆− λ (x) +

x

Z

x→ 0

Bλ (s)(1 + s) s

0

x(1 + x)Bλ (x)

∆λ (s)ds = lim

x→0+

λxBλ (x) − 1

=

Lλ

λLλ − 1

< ∞.

(33)

Next assume that (29) is satisfied. Then, by Hölder continuity of Bλ , Bλ (x)/x (and also Bλ (x)) are integrable at 0 and both conditions (25) and (26) are satisfied. Finally, let (30) be satisfied. In this case we have C0

Cα e

− xαα

0

≤ ∆λ (x) ≤ Cβ e

for some constants Bλ (x) x

C − ββ

(34)

x

Cα , Cα0 , Cβ , Cβ0

and 0

∆λ (x) ≤ M2 Cβ

1 x1+α

e

C − ββ x

→0

1 for x → 0+ . Eq. (34) also ensures that ∆− λ (x) ≥ Cβ e the first part of the proof gives 1 lim x∆− λ (x)

x→0+

x

Z

Bλ (s)(1 + s)

0

on account of xBλ (x) → ∞.

s

∆λ (s)ds = lim

Cβ0 /xβ

x→0+

1 + and thus x∆− λ (x) → ∞ for x → 0 . Hence, applying the l’Hospital rule as in

x(1 + x)Bλ (x)

λxBλ (x) − 1

=

1

λ

,

(35)



In what follows we shall work with the additional assumption lim a(x) = a0 ∈ [0, ∞].

x→0+

(36)

We note that this assumption is mostly technical and several results below can be proved without it. This, however, would require a more detailed analysis of particular cases and obscure the main results of the paper. Lemma 2. Assume (36). If any the conditions (28), (29) or (30) holds for some λ0 > 0, then it holds for any λ ≥ λ0 .

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

1429

Proof. First consider (28). Then from xb(x|x)

−

λ + a(x)

xb(x|x)

µ + a(x)

(µ − λ)xb(x|x) (λ + a(x))(µ + a(x))

=

we see that if the limit exists for some λ > 0 then it exists for any λ > 0. Let a0 < ∞. Then

λ Lλ =

λ lim xb(x|x) x→0+

λ + a0

is an increasing function of λ so that if λLλ > 1 for some λ0 then it is true for any λ > λ0 . If a0 = ∞, then

λLλ = λ lim

xb(x|x)

x→0

a(x)

,

which is also monotonic in λ. Moving to (29) we see that 0 ≤ Bµ (x) ≤ Bλ (x) for µ > λ > 0 and the lemma holds true. If (30) holds, then either xb(x|x) → ∞ if a0 < +∞ or xb(x|x)/a(x) → +∞ if a0 = ∞ and in either case the condition is independent of λ.  Corollary 3. Under assumption (36), if any of the conditions (28)–(30) holds for some λ0 > 0, then R(λ) for λ ≥ λ0 is the resolvent of K0,1 . Proposition 2. Assume that (36) holds and let either (28), or (30) with the following additional condition: a(x) ≤ x−κ , κ > 0, x → 0+ , be satisfied. Then β(x) is integrable on [0, M ], M < ∞, and thus the number of particles produced in each fragmentation event is finite. Proof. Let us fix some λ for which (28) holds. First assume that (28) is satisfied and a(x) → a0 < ∞ as x → 0+ . Then lim xb(x|x) = Lλ (λ + a0 ) ≥ λLλ > 1

x→0+

xβ(x) and xb(x|x) ≥ L > 1 for x sufficiently close to zero. This yields R x

0 sβ(s)ds

α , gives Rα ln R0x 0

sβ(s)ds sβ(s)ds

≥ ln

which can be written as

β(x) = xb(x|x)

1

 α L x

Rx 0

L x

which, upon integration from x > 0 to some sufficiently small

,

(37)

sβ(s)ds ≤ Cα xL , for some constant Cα and small x > 0. Since

x

Z

x2

≥

sβ(s)ds,

(38)

0

for small x the function β(x) behaves as xL−2 which is integrable at 0. Let now a0 = ∞. We can write xBλ (x) = λx2 b(x|x) limx→0+ a(x)

Rα ln R0x 0

= λLλ > 1. Let us take α > 0 such a(x) > λ and

sβ(s)ds sβ(s)ds

≥

L

λ

α

Z

a( s ) s

x

α

Z ds ≥ L x

which is the same as (37). xβ(x) In the last case we have R x

≥

0 sβ(s)ds

ds s

M1 (λ+a(x)) xβ+1

= ln

R xxβ(x) 0 sβ(s)ds

≥

La(x) λx

b(x|x) x2 a(x) 1+λ/a(x)

so that

for some L > 1 for x ∈ (0, α). As before, this gives

 α L x

and, following the steps of the previous cases, we obtain

Rx 0

−β

sβ(s)ds ≤ C1 e−C2 x

for some

constants C1 and C2 , which, by (38), yields

β(x) ≤ C1 xb(x|x)x−2 e−C2 x

−β

−β

≤ Mx−κ−α−2 e−C2 x ,

for some M. It is now clear that β(x) → 0 as x → 0+ .



Remark 1. In general, (29) does not yield the result of the above proposition. Indeed, taking a(x) = x−2 and b(x|y) = 2−1 y1/2 x−3/2 , we see that Bλ (x) = 1/(2x(λ + x−2 )) = x/(λx2 + 1) → 0. However, β(x) = x−3/2 is not integrable and hence the expected number of particles in each fragmentation event is infinite. 5. Dynamics in X0,1 Theorem 4. Let assumptions of Proposition 2 be satisfied. Then (K0,1 , D(K0,1 )) generates a positive semigroup in X0,1 . Proof. First we prove that K0,1 is densely defined. We note that C0∞ (R+ ) ⊂ D(A) ∩ X0,1 ⊂ D(K ) ∩ X0,1 and thus, for φ ∈ C0∞ (R+ ) with support in [m, M ] we have K φ = Aφ + Bφ. It is clear that Aφ ∈ X0,1 and ∞

Z

[Bφ](x)(1 + x)dx = 0

where n(y) is given by (15).

Z

M

a(y)φ(y)(y + n(y))dy < ∞, m

1430

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

Let us return to the expression (27) for the norm of the resolvent in X0,1 for f ≥ 0: ∞

Z

∞

Z

[R(λ, K0,1 )f ](x)(1 + x)dx =

f (x)(1 + x)



λ + a(x)

0

0

1

+

a(x)

x

λ + a(x) 1 + x

e−λ

Rx

1 Bλ (s)ds

x

Z 0

Bλ (s)(1 + s) λ R s Bλ (r )dr e 1 ds dx. (39) s



The term within the brackets is a sum of two terms which are positive on (0, ∞). Note that a may not converge as x → ∞. However for s ≥ 1 we have 1 ≤ 1 + s−1 ≤ 2 and hence 1

λ

x

Z

(∆λ (x) − 1) ≤

Bλ (s)(1 + s) s

1

2

∆λ (s)ds ≤

λ

(∆λ (x) − 1).

Thus, ifR∆λ (x) is finite as x → ∞, then the improper integral exists and is non zero. If ∆λ (x) is infinite, then ∆λ (x)−1 → 0 and, by the x above, 1 Bλ (s)∆λ (s)(1 + s)s−1 ds tends to infinity as x → ∞. Hence we can use the l’Hospital rule obtaining lim e

−λ

1 Bλ (s)ds

x→∞

Bλ (s)(1 + s) λ R s Bλ (r )dr 1 Bλ (x)(1 + x)∆λ (x) e 1 ds = lim = . x→∞ s λxBλ (x)∆λ (x) λ

x

Z

Rx

0

Rx

B (s)(1+s)

Rs

Therefore at infinity 1+x e−λ 1 Bλ (s)ds 0 λ s eλ 1 Bλ (r )dr ds is bounded away from zero. As a result, for large x the expression within the a(x) 1 brackets in (39) is greater than λ+a(x) + α λ+a(x) for some α > 0. Furthermore x

1

λ + a(x)

+α

a(x)

Rx



λ + a( x )

≥ min

1

λ

,α



   λ a(x) 1 + = min ,α . λ + a(x) λ + a(x) λ

Consequently the expression within the brackets in (39) is bounded away from zero for large x. 1 −λLλ Next, assume a(x) to become infinite as x → 0. If (28) holds then, as (33), ∆− and λ (x) tends to infinity at the same rate as x 1 lim x∆− λ (x) +

x

Z

x→ 0

Bλ (s)(1 + s) s

0

∆λ (s)ds =

Lλ

λ Lλ − 1

,

which is finite and non-zero by (28). If a(x) is bounded as x → 0, then the first summand in (39) clearly is bounded away from zero for small x. Finally, if (30) is satisfied then, again following the proof of Corollary 2 and (35), we obtain 1 lim x∆− λ (x) +

x

Z

x→ 0

Bλ (s)(1 + s) s

0

∆λ (s)ds =

1

λ

> 0.

Therefore, there is c > 0 such that for any x ∈ [0, ∞) 1

λ + a(x)

+

a(x)

x

λ + a(x) 1 + x

e−λ

Rx

1 Bλ (s)ds

x

Z 0

Bλ (s)(1 + s) λ R s Bλ (r )dr e 1 ds ≥ c s

(40)

yielding

kR(λ, K0,1 )f kX0,1 ≥ c kf kX0,1 for λ > λ0 and f ≥ 0 and, by the Arendt–Robinson–Batty theorem, see e.g. [21, Theorem 2.5], K0,1 generates a positive semigroup.



Remark 2. An important role in the application of the Arendt-Batty-Robinson theorem in the previous proof is played by the density of the domain of K0,1 in X0,1 . We observe that, in general, it is far from obvious. Let us take a model with infinite n(y); that is, with a nonintegrable β . If a function 0 6= φ ≥ 0 belongs to D(K0,1 ), then X0,1 3 K0,1 φ = K φ . If φ has support in [m, M ], then K0,1 φ = K φ = Aφ + Bφ. However, [Aφ](x) = a(x)φ(x) has compact support hence it belongs to X0,1 . Thus, for φ ∈ D(K0,1 ), Bφ also must be in X0,1 . However,

[Bφ](x) = β(x)

M

Z

a(y)γ (y)φ(y)dy x

RM

RM

and close to zero x a(y)γ (y)φ(y)dy = m a(y)γ (y)φ(y)dy 6= 0, hence Bφ is integrable close to 0 if and only if β has the same property. Hence, in this case, positive compactly supported functions are not in D(K0,1 ) and, at present, we do not know whether K0,1 with nonintegrable β is densely defined in X0,1 . Thus, in general, we do not know whether K0,1 is densely defined if (29) holds. Also (29) does not allow the estimate (40) which was instrumental in getting the semigroup generation result. However, it follows that we still have some nontrivial dynamics in this case. To explain the result below, let us recall the concept of an n-times integrated semigroup, see e.g. [21]. Sometimes the operator A of the Cauchy problem in a Banach space X ◦

u0 = Au,

u(0) = u

(41) ◦

does not generate a semigroup in X and thus the problem does not have a classical solution for every u ∈ D(A). Then one can try to regularize (41) by integrating it with respect to time sufficiently many times, say n + 1 times. If such an integrated version of (41) has ◦

an exponentially bounded classical solution for any u ∈ X , then we say that A generates an exponentially bounded n-times integrated semigroup. The importance of this concept stems from the fact that in such a case (41) has a unique exponentially bounded classical ◦

solution for any u ∈ D(An+1 ) (which then can be obtained as the nth derivative with respect to time of the classical solution of the integrated version of (41)).

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

1431

Corollary 4. Let assumption (29) be satisfied. Then (K0,1 , D(K0,1 )) generates a once integrated positive semigroup X0,1 if D(K0,1 ) is dense in ◦

X0,1 and twice integrated semigroup if D(K0,1 ) is not dense in X0,1 . Consequently, the problem (6) has classical solutions for u∈ D(K12,0 ) in the ◦

X0,1

first case and if K12,0 u∈ D(K0,1 )

◦

(hence, in particular, u∈ D(K03,1 )), in the second.

Proof. The proof is a direct consequence of [22, Corollary 4.5, Proposition 5.5]. In particular, the second part follows from the fact that if X0,1

K0,1 is a resolvent positive operator with a non-dense domain, then its part in D(K0,1 ) argument from the first part can be applied for this restriction. 

generates a once-integrated semigroup and the

Theorem 5. Assume that either (28) or (30) are satisfied and n(y) is bounded as y → 0. Then K0,1 = (A + B)|X0,1

X0,1

.

(42) ◦

Moreover, if a(x) → a0 < +∞ as x → ∞, then for any 0 ≤ u ∈ D(K0,1 ) d dt

◦

kGK0,1 (t ) u kX0,1 =

∞

Z

◦

a(x)(n(x) − 1)[GK0,1 (t ) u](x)dx;

0

that is the semigroup is honest in X0,1 . Proof. To prove the first part, we use [14, Theorem 4.3]. Let us define A0,1 to be the part of A in X0,1 which, since X0,1 ⊂ X1 , is the restriction of the multiplication by a to D(A0,1 ) = {u ∈ X0,1 , au ∈ X0,1 }. Next we observe that the boundedness of n(y) at zero implies that n(y)/(1 + y) is bounded on [0, ∞). Indeed, let n(y) ≤ N1 for y ≤ 1. Then for y > 1 and some constant C we get n(y) y

Ry 0 y 0

= R

β(s)ds sβ(s)ds

C1 +

Ry

C2 +

R

=

1 y 1

β(s)ds sβ(s)ds

≤

C1 + C2 +

Ry R1y 1

β(s)ds

≤ C.

β(s)ds

If B denotes the integral expression (9), then for 0 ≤ u ∈ D(A0,1 ) we have ∞

Z

a(y)(y + n(y))u(y)dy ≤ sup

kB uk0,1 =

y + n(y) 1+y

y∈R+

0

∞

Z

a(y)u(y)(1 + y)dy < ∞, 0

hence we can define B0,1 by restricting B to D(A0,1 ). Let u ∈ D(A0,1 ) ⊂ D(A) ∩ X0,1 ⊂ D(K ) ∩ X0,1 . But then X0,1 3 (A0,1 + B0,1 )u = Au + Bu = Ku, hence u ∈ D(K0,1 ) and K0,1 ⊃ A0,1 + B0,1 . Since both K0,1 and A0,1 generate semigroups on X0,1 , we can use Theorem 4.3 X0,1

of [14], which states that K0,1 = (A + B)|X0,1 if and only if 1 is not an eigenvalue of (B0,1 R(λ, A0,1 ))∗ , in the same way as in the proof of Theorem 8.13 in [14] (see also [13]). We identify the dual to X0,1 with X∞ := L∞,(1+x)−1 (R+ ) so that the duality pairing is the same as between L1 and L∞ . Thanks to the considerations in the previous paragraph, the operator B0,1 R(λ, A0,1 ) is bounded and, as in op. cit., we find that the adjoint is given by the expression



B0,1 R(λ, A0,1 )


∗ 

g (y) =

a(y)γ (y)

y

Z

λ + a(y)

β(x)g (x)dx = 0

a(y)b(y|y)

β(y)(λ + a(y))

Z

y

β(x)g (x)dx.

0

Thus, assume 1 is an eigenvalue of (B0,1 R(λ, A0,1 ))∗ ; that is, there is 0 6= h ∈ X∞ satisfying

β(y)h(y) −

a(y)b(y|y)

λ + a(y)

y

Z

β(x)h(x)dx = 0.

(43)

0

Denoting H (y) =

y

Z

β(x)h(x)dx

(44)

0 a(y)b(y|y)

we find from (43) that H is differentiable on (0, ∞) and the equation can be converted to H 0 (y) = λ+a(y) H (y). It has a solution R y a(s)b(s|s) ds

H (y) = e

1

λ+a(s)

.

(45)

Differentiating and using b(s|s) = sβ(s)/ h(y) =

H 0 (y)

Rs 0

ya(y)

1

r β(r )dr and (19), we obtain Ry

R y b(s|s) b(s|s)ds −λ 1 λ+a(s) ds e

ya(y)

λ

R1

sBλ (s) ds s

, β(y) β(y) (λ + a(y)) 0 sβ(s)ds λ + a(y) R1 where C = 1/ ln 0 β(s)ds. Assume that (28) is satisfied. Then λsBλ (s) → λLλ > 1 as s → 0 and thus λsBλ (s) ≥ L > 1 on (0, α) for sufficiently small α . Hence e

λ

R1 y

sBλ (s) ds s

=

= eλ

Ry

R 1α y

sBλ (s) ds s

eλ

R1 α

sBλ (s) ds s

e

1

≥ Cα0 α L y−L = Cα y−L .

=C

e

y

1432

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435 a(y)y1−L

Thus h(y) ≥ λ+a(y) close to zero and clearly h is not bounded at 0 if a(y) → a0 > 0 (including a0 = ∞) as y → 0, contradicting the assumption that h ∈ X∞ . If a0 = 0, then we have two cases to consider: either a(y)y1−L is unbounded, which leads to the previous conclusion, or a(y)y1−L is bounded. Then the exponent in (45) can be written as y

Z 1

a(s)b(s|s)

λ + a(s)

a(s)s1−L sb(s|s)

y

Z ds =

which means that H (0) = e

λ + a(s)

s2−L

1

−

R 1 a(s)b(s|s)

6= 0, contrary to the construction (44).

λ+a(s)

0

ds

a(y)y

C2

If we assume (30), then in a similar way we obtain that h(y) ≥ C1 λ+a(y) e yβ close to zero, where C1 and C2 are constants. As before, as β long as a0 > 0, then h(y) is unbounded at zero since yeC2 /y is unbounded. If a0 = 0, then the numerator may be unbounded in which β

a(s)b(s|s)

−C2 /sβ

case the previous argument applies. Otherwise, we have a(y) ≤ Cy−1 e−C2 /y and λ+a(s) ≤ CM2 e sα+2 with the right hand side tending to zero as s → 0. This yields integrability on (0, 1) of the exponent in (45) and H (0) 6= 0, contradicting with (44). To prove the second part, first we note that (G0,1 (t ))t ≥0 is the restriction of the semigroup (G(t ))t ≥0 , generated by K on X1 , to X0,1 . This follows from the resolvent formula for semigroups and the analogous statement for the generators. Therefore (G0,1 (t ))t ≥0 leaves the spaces L1 ([0, M ], (1 + x)dx), M > 0, invariant (these are isometric to L1 ([0, M ], dx)). Clearly, if we take u ≥ 0 with supp u ⊂ [0, M ], then G0,1 (t )u ∈ D(A0,1 ) and, by direct integration, d dt

k[G0,1 (t )]ukX0,1 =

∞

Z

(n(x) − 1)a(x)[G0,1 (t )u](x)dx.

(46)

0

It will be more convenient to work with the integrated version of (46):

k[G0,1 (t )]ukX0,1 = kukX0,1 +

∞

Z

(n(x) − 1)a(x)

t

Z

0

 [G0,1 (s)u](x)ds dx,

(47)

0

where the change of the order of integration is justified by positivity of the integrands. Now, given 0 ≤ u0 ∈ X0,1 , we approximate it by un := χ[0,n] u0 % u0 in X0,1 . Then, for any t ≥ 0, we have G0,1 (t )un % G0,1 (t )u0 .

Rt

Consider S (t )un = 0 G0,1 (s)un ds. By the dominated convergence (or monotonic as well), we have S (t )un % S (t )u0 in X0,1 for any t ≥ 0. Rewriting (47) for un , we see that

k[G0,1 (t )]un kX0,1 = kun kX0,1 +

∞

Z

(n(x) − 1)a(x) 0

t

Z



[G0,1 (s)un ](x)ds dx,

(48)

0

hence the convergence of the norm terms imply the convergence of the integral and, since the multiplication by a(x) does not change the monotonicity of the sequence, we obtain

k[G0,1 (t )]u0 kX0,1 = ku0 kX0,1 +

∞

Z

(n(x) − 1)a(x) 0

t

Z

 [G0,1 (s)u0 ](x)ds dx.

(49)

0

Using the fact that X0,1 is an L-space, see [14, Theorem 2.39], we can represent [G(α0,1) (s)u0 ](x) as a measurable function of two variables φ(x, s) and the strong integral with respect to s as the Lebesgue integral with respect to one variable s. Multiplication by (n(x) − 1)a(x) does not change the measurability hence, by the Fubini theorem, we get

k[G0,1 (t )]u0 kX0,1 = ku0 kX0,1 + = ku0 kX0,1

∞

Z

Z

t



(n(x) − 1)a(x) [G0,1 (s)u0 ](x)ds dx 0 0  Z t Z ∞ + (n(x) − 1)a(x)[G0,1 (s)u0 ](x)dx ds. 0

0

If u0 ∈ D(K0,1 ), then the left hand side is differentiable and, since the inner integral in the last line is clearly integrable with respect to s, the derivative of the right hand side is this integral (at least almost everywhere); that is d dt

k[G0,1 (t )]u0 kX0,1 =

Z

∞

(n(x) − 1)a(x)[G0,1 (t )u0 ](x)dx,

a.e.

0

This, however, shows that t →

R∞ 0

(n(x) − 1)a(x)[G0,1 (t )u0 ](x)dx is continuous, and thus the above extends to all t.



6. Examples Conditions (28)–(30) seem to be quite technical but they prove to be sharp for a large, and best understood, class of fragmentation processes governed by power laws; that is, for a( x ) = x α ,

xν b(x|y) = (ν + 2) ν+1 . y

(50)

Here, α ∈ R; we exclude, however, the case α = 0 which yields boundedness of all involved operators. On the other hand, the range of the parameter ν is restricted to ν ∈ (−2, 0]. The reason for this is that for ν > 0 the expectedRnumber of daughter particles after each y fragmentation event is smaller than 2, which is nonphysical. On the other hand, if ν ≤ −2, then 0 xb(x|y)dx = ∞, yielding infinite mass of daughter particles after each split.

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

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We note that in the power law case the expected number of daughter particles in each fragmentation event does not depend on the size y of the parent and equals n(y) =

ν+2 . ν+1

(51)

In this framework we have Bλ (x) =

ν+2 b(x|x) = λ + a(x) x(λ + xα )

and (28) corresponds to α > 0 and −1 < ν ≤ 0, whereas (30) is yielded by α < −1 with arbitrary ν ∈ (−2, 0]. The case (30) cannot be realized in the present framework. We can state the following. Corollary 5. Let α > 0 and −1 < ν ≤ 0. Then K0,1 generates a positive semigroup (G0,1 (t ))t ≥0 on X0,1 which, moreover, is honest; that is, for ◦

any 0 ≤ , u∈ D(K0,1 ) d dt

ν+1

Furthermore, K0,1 = (A + B)|X0,1

∞

Z

1

◦

kGK0,1 (t ) ukX0,1 =

0

X0,1

◦

xα [GK0,1 (t ) u](x)dx.

.

A more interesting result is contained in the next corollary. Usually, shattering is associated with an infinite cascade of fragmentation events creating a dust of dimensionless particles which, however, carry some mass. Implicit in this interpretation is that we should have an infinite number of particles. The following result shows that such an interpretation is, in general, erroneous. Corollary 6. Let α < −1. ◦

◦

(1) If −1 < ν ≤ 0, then K0,1 generates a once integrated semigroup in X0,1 and therefore kG0,1 (t ) u k0,1 < +∞ for all t ≥ 0 and u ∈ D(K02,1 ) (e.g. with compact support). ◦

◦

(2) If −2 < ν ≤ −1, then K0,1 generates a twice integrated semigroup in X0,1 and thus kG0,1 (t ) u k0,1 < +∞ for all t ≥ 0 and u ∈ D(K03,1 ). Thus, in both cases there are (many) trajectories along which the number of particles in the system remains finite for all times. The fact that for α < −1 and ν < −1 we may have a finite number of particles in the system despite the expected number of particles in each split being infinite was noticed in [20] where the authors commented that

. . . a finite fraction of the total mass would be transferred to a finite number of particles with zero or infinitesimal mass! We conclude that a physically acceptable situation corresponds to α > −1. As we noted in the introduction, our interpretation of this case is different: in our opinion the fragmentation Eq. (6) ‘sees’ only a part of the system where only a finite number of ‘physical’ particles remain, while the mass is carried away by the dust, which is beyond the resolution of (6). Remark 3. The statement in Corollary 6 is rather due to our failure to prove the existence of the semigroup. Indeed, for the equation

∂t u(x, t ) = −x u(x, t ) + 2 −2

∞

Z

y−3 u(y, t )dy,

(52)

x

◦

with u(x, 0) = u (x), we have Bλ (x) = x(λ+2x−2 ) → 0. On the other hand, the formula for the solution of this equation is, [4], u(x, t ) = e

−

t x2

◦

u (x) + 2t

∞

Z

y−3 e

−

t y2

◦

u (y)dy

(53)

x

and, proceeding as in [15, Example 6.5], we can ascertain that the above formula is a representation of the fragmentation semigroup in X1 associated with (52). Moreover, let t ≥ 0 and 0 ≤ f ∈ X0,1 . Then we have

kG1 (t )f k1,0 ≤ kG1 (t )f k1 +

∞

Z

 e

−

t x2

f (x) + 2t

0

≤ 2kG1 (t )f k1 +

∞

Z

t −3 − y2

y

e



f (y)dy dx

x 1

Z

f (x)dx + 2tI , 0

where 1

Z

∞

Z

I =

t −3 − y2

y 0

e



|f (y)|dy dx ≤ (t

x

−1 −1

e

+ 1)

Z

∞

|f (y)|dy,

0 −2

◦

and where we used ty−2 e−ty ≤ e−1 . Hence kG1 (t )f k0,1 ≤ (2 + 2e−1 + 2t )kf k0,1 , so that the evolution takes place in X0,1 for any u ∈ X0,1 . In fact, we can prove that the restriction of (G1 (t ))t ≥0 to X0,1 is a C0 -semigroup. It is sufficient to prove strong continuity at t = 0. Moreover,

1434

J. Banasiak, S.C.O. Noutchie / Physica D 239 (2010) 1422–1435

due to the estimate above, and since the the semigroup is positive, it is sufficient to prove convergence for nonnegative f with compact 2 support, say [m, M ]. Let us look at (53). We have 0 ≤ e−t /x f (x) ≤ f (x) ∈ X0,1 and the pointwise convergence is monotonic hence, due 2 to the KB-structure of X0,1 , we have limt →0+ e−t /x f (x) = f (x) in norm. Thus, we have to prove that the second term converges to zero in norm. Indeed,

∞

Z

M

Z

y −3 e

2t

−

t y2



f (y)dy (1 + x)dx = 2t

x

0

M

Z

y−3 e

−

t y2



f (y) y +

m

1 2

y2

 dy → 0

as the integrand is bounded on [m, M ]. Remark 4. So far we have not discussed the case when α ∈ [−1, 0) and ν ∈ (−2, 0] or α > 0 and ν ∈ (−2, −1]. It is easy to see ν+2 that in both cases limx→0 λxBλ (x) = limx→0 λ+ ≤ 1 (0 in the first case and ν + 2 < 1 in the second one) and, at the same time, xα limx→0 Bλ (x) = ∞, so this case is not covered by Corollary 2. It turns out that there is a reason for this since the integral condition (26) reads now 

Z

Bλ (x) x

0

∆λ (x)dx = (ν + 2)



Z

e

−λ(ν+2)

dx

R1

ds x s(λ+sα )

0

x2 (λ + xα )

= (λ + 1)

ν+2 α

Z



0

xν

(λ + xα )1+

2+ν

α

dx,

with the integrand behaving as xν for α > 0 and as x−α−2 for α < 0. Hence (26) is not satisfied in either case discussed in this remark and therefore R(λ)X0,1 6⊆ X0,1 . Therefore, the restriction of K to X0,1 cannot generate any reasonable dynamics there. This is supported by the following example. The case with α = −1 and ν = 0 gives the equation

∂t u(x, t ) = −x−1 u(x, t ) + 2

∞

Z

y−2 u(y, t )dy, x

for which we have Bλ (x) = x(λ+2x−1 ) → 2 6= 0 and xBλ (x) = λ+2x−2 → 0 < 1. By [15, Example 6.5], the X1 -semigroup is represented by t ◦ u(x, t ) = e− x u (x) + 2t

∞

Z

e

− yt

y2

x

− yt ◦

e

u (y)dy + t 2

∞

Z x

e

− yt

y2



1 x

−

1 y



◦

u (y)dy.

◦

For u ≥ 0, each term is nonnegative. Taking the X0 norm of the last term we obtain 2t

2

∞

Z

e

0

− yt

y2

◦

u (y)

◦

Z y 

1

0

x

−

1 y



 dx dy = ∞

(54) ◦

for any nonzero u ≥ 0 and any t > 0, irrespectively whether k u k0 < ∞ or not. Hence, the number of particles in this case immediately becomes infinite and stays infinite for all times. 7. Conclusions We have shown that solutions to a large class of fragmentation equations with the fragmentation rate bounded at 0 satisfy the formal Eqs. (11) and (12) which govern the evolution of the number of particles in an ensemble and its mass, thus proving the honesty of these models. We have also proved that there exist solutions with a finite number of particles in the shattering regime, showing that shattering is not necessarily related to the creation of infinite numbers of particles. An interesting observation is that there are shattering models in which the expected number of particles in each fragmentation event is infinite, but the solutions of which still exhibit only finitely many particles. We conjecture that in such regimes the model (6) ‘sees’ only a fraction of the particles and thus the process should be described by a two-compartmental model. Acknowledgements This paper was presented at the symposium Evolution Equations in Pure and Applied Sciences dedicated to Professor Aldo BelleniMorante. The financial support from GNFM and from Professor Giovanni Frosali which enabled the first author to stay in Florence and complete the research leading to the results presented here is gratefully acknowledged. Both authors have been also supported by the National Research Foundation of South Africa under grant FA2007030300001. The first author is also grateful to Professor Jean Bertoin and Dr Benedicte Haas for stimulating discussions concerning probabilistic approach to fragmentation. References [1] [2] [3] [4] [5] [6] [7] [8] [9]

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