Spectral Mapping Theorems for Neutron Transport, L1 -Theory Mustapha Mokhtar-Kharroubi and Mohammed Sbihi Laboratoire de Math´ematiques de Besan¸con. Universit´e de Franche Comt´e. 16, Route de Gray, 25030 Besan¸con. France. E-mail : [email protected]; [email protected] Abstract This work deals with spectral mapping theorems for neutron transport semigroups in unbounded geometries and L1 setting. The mathematical analysis relies on harmonic analysis of certain measure valued mappings related to Dyson-Phillips expansions and on some functional analytic results on the critical spectrum [2, 8].

1

Introduction

The investigation of spectral mapping theorems for neutron transport semigroups ¡ t(T +K) ¢ e t≥0 in unbounded geometries was initiated recently by the authors in the context of Lp spaces with 1 < p < ∞ [7], where T and K represent respectively the streaming and the collision operators. The mathematical analysis is based upon two ingredients: (a) Some functional analytic results on perturbation theory of the critical spectrum of C0 semigroups [1, 2, 8]. (b) The norm continuity of 0 ≤ t 7→ et(T +K) − etT , i.e. in the operator norm topology. The proof of (b) is based on Fourier integral analysis of et(T +K) − etT in the case p = 2 and on interpolation arguments. The present paper is devoted to the limiting case p = 1 which is not covered by [7] and which turns out to be the physical case for neutron transport. Besides the use of the properties of the critical spectrum of perturbed semigroups [2], our mathematical analysis relies on different tools. Moreover, we obtain very precise results which are different from those given in [7]. Before explaining the content of this paper, it is useful to recall some facts on the critical spectrum of C0 -semigroups [1, 2, 8]. Let X be a Banach space and τ = (U (t))t≥0 be a strongly continuous semigroup on e := `∞ (X) of all bounded sequences in X endowed with X. We consider the Banach space X the norm ke xk = sup kxn k n∈N

1

e and obtain a new semigroup where e =´(xn )n∈N . We extend the semigroup (U (t))t≥0 to X ³ x e (t) τe = U defined by t≥0

e (t)e U x := (U (t)xn )n∈N for x e = (xn )n∈N . eτ be the subspace of strong continuity of τe Let X ¾ ½ ° ° e e e ° ° x−x e =0 . Xτ := x e ∈ X; lim U (h)e h↓0

³ ´ e (t) b := X/ e X eτ , the This subspace is closed and U -invariant. On the quotient space X t≥0 ³ ´ ³ ´ e (t) b (t) semigroup U induces a quotient semigroup τb = U given by t≥0

t≥0

b (t)b e (t)e eτ for x eτ . U x=U x+X b=x e+X The critical spectrum of U (t) is then defined as ˆ (t)) σcrit (U (t)) = σ(U while its critical spectral radius is defined as b (t)). rcrit (U (t)) := r(U Moreover, the critical growth bound is defined as b (·)) ωcrit (U (·)) := ω0 (U where ω0 is the usual growth bound (type). We have: Theorem 1.1. [8] Let (U (t))t≥0 be a strongly continuous semigroup on a Banach space X with generator T . Then: (a) σcrit (U (t)) ⊂ σ(U (t)), (b) rcrit (U (t)) = eωcrit (U (·))t , (c) σ(U (t))\ {0} = etσ(T ) ∪σcrit (U (t))\ {0} , (d) ω0 (U (·)) = max {s(T ), ωcrit (U (·))} . Consider now the perturbed semigroup (V (t))t≥0 generated by T + K where K is a bounded operator: ∞ X V (t) = Uj (t), 0

where

Zt U0 (t) = U (t), Uj+1 (t) =

U0 (t − s)KUj (s)ds (j ≥ 0).

(1.1)

0

The following theorem provides a sufficient condition for the stability of critical growth bound. 2

Theorem 1.2. [2] Let (U (t))t≥0 be a C0 -semigroup with generator T and let (V (t))t≥0 be the C0 -semigroup generated by T + K. If for some k ∈ N 0 < t 7→ Rk (t) :=

∞ X

Ui (t)

i=k

is norm (right) continuous, then ωcrit (V (·)) = ωcrit (U (·)). The stability of critical spectrum is the subject of the next theorem. Theorem 1.3. [2] Let (U (t))t≥0 be a C0 -semigroup with generator T and let (V (t))t≥0 be the C0 -semigroup generated by T + K. If for some t0 ≥ 0 t0 ≤ t 7→ R1 (t) := V (t) − U (t) is norm (right) continuous, then σcrit (V (t)) = σcrit (U (t))

(t ≥ t0 ).

We give a sufficient condition for recognizing the critical spectrum. We first recall that the approximate spectrum of T is defined by σap (T ) := {λ ∈ C; ∃(xn )n ⊂ D(T ), kxn k = 1, kT xn − λxn k → 0 as n → ∞} . Theorem 1.4. [1] Let (U (t))t≥0 be a C0 -semigroup with generator T. Let (λn )n ⊂ σap (T ) be such that lim |Imλn | = ∞ and lim etλn = µ. Then µ ∈ σcrit (U (t)). n→∞

n→∞

We now present the neutron transport semigroup. Let Ω ⊂ RN be an open set and let µ be a positive Radon measure on RN with support V. We refer to V as the velocity space. The streaming semigroup is given by Z t − σ(x − sv, v)ds U (t) : L1 (Ω × V ) 3 ϕ 7→ e 0 ϕ(x − tv, v)χ{t 0; x − sv ∈ / Ω} and σ(·, ·) ∈ L∞ (Ω × V ) is the collision frequency. Here Ω × V is endowed with the product measure dx ⊗ dµ(v). We denote by T the generator of (U (t))t≥0 . The collision operator is the (partial) integral operator Z 1 K : L (Ω × V ) 3 ϕ 7→ k(x, v, v 0 )ϕ(x, v 0 )dµ(v 0 ), V

where the scattering kernel k(·, ·, ·) satisfies the estimate Z |k(·, v, ·)|dµ(v) ∈ L∞ (Ω × V ) V

ensuring the boundedness of K in L1 (Ω × V ). The neutron transport semigroup is the C0 semigroup generated by T + K. For all the sequel, the collision operator is assumed to be 3

compact ”with respect to velocities”, where the compactness is ”collective” with respect to the space variable. More precisely: (H1) The family nZ o k(x, v, v 0 )ϕ(v 0 )dµ(v 0 ); x ∈ Ω, ϕ ∈ L1 (V ), kϕkL1 (V ) ≤ 1 V

is relatively compact in L1 (V ). (H2) For each ψ ∈ L∞ (V ), the family nZ o k(x, v 0 , v)ψ(v 0 )dµ(v 0 ); x ∈ Ω V

is relatively compact in L∞ (V ). We note that under (H1) and (H2), K can be approximated in the norm operator topology of L(L1 (Ω × V )) by collision operators with separable kernels: X αi (x)fi (v)gi (v 0 ), (1.2) i∈I

where αi (·) ∈ L∞ (Ω), fi (·) ∈ L1 (V ), gi (·) ∈ L∞ (V ) and I finite (see [7]). We point out that when the scattering kernel is space homogeneous, (H1) and (H2) reduce simply to the compactness of the integral operator Z 1 L (V ) 3 ϕ 7→ k(v, v 0 )ϕ(v 0 )dµ(v 0 ) ∈ L1 (V ). V

In this paper, the collision frequency is assumed to be space homogeneous, i.e. σ(x, v) = σ(v). Our paper is organized as follows: Section 2 is devoted to the neutron transport semigroup in the whole space (Ω = RN ) with space homogeneous scattering kernels, i.e. k(x, v, v 0 ) = k(v, v 0 ). We show that if there exists α > 0 such that for all c > 0 there exists c0 > 0 such that sup µ ⊗ µ{(v, v 0 ); |v| ≤ c, |v 0 | ≤ c, |(v − v 0 ) · e| ≤ ε} ≤ c0 εα e∈S N −1

then 0 ≤ t 7→ Rj (t) ∈ L(L1 (Ω × V ))

(1.3)

is norm continuous where j depends on α and N . The proof is quite technical and is given in several steps: By a density argument, we can restrict ourselves to the separable case (1.2). In this case, the terms of the Dyson-Phillips expansion (1.1) are shown to be essentially iterated convolution of Radon measures depending on time t. In particular, the norm continuity of (1.3) amounts to the fact that such measures depend continuously on t with respect to the total variation norm. We show that for N ≥ 2, regardless of the choice of the velocity measure µ, 0 ≤ t 7→ R1 (t) = et(T +K) − etT 4

is never norm continuous. On the other hand, for N = 1, we show that 0 ≤ t 7→ et(T +K) − etT is norm continuous if and only if µ satisfies ©£ ¤ª sup µ v 0 − ε, v 0 + ε → 0 as ε → 0. v 0 ∈R

In section 3, we deal with general spatial domains and not (necessarily) space homogeneous scattering kernels under the assumption that the velocity measure µ is ”absolutely continuous v in speed |v| but arbitrary in directions |v| ” and prove that 0 ≤ t 7→ R2 (t) is norm continuous. We note however that such an assumption on µ covers the classical continuous model (dµ(v) = dv) but not the multigroup model (Lebesgue measure on spheres). Section 4 is devoted to spectral mapping theorems. We determine first the critical spectrum of the streaming semigroup; we essentially complement some results given in [6, 7]. We derive from the results of Section 2 spectral mapping theorems in the whole space for general velocity measures and space homogeneous scattering kernels. Similarly, we derive from the results of Section 3 a spectral mapping theorem for a restricted class of velocity measures but for general spatial domains and scattering kernels. The authors thank the referee for helpful remarks and suggestions.

2

On Dyson-Phillips expansions on the whole space

In this section devoted to the case Ω = RN , we assume that the scattering kernel is space homogeneous and that (H3) L1 (V ) 3 ϕ 7→

2.1

R

k(v, v 0 )ϕ(v 0 )dµ(v 0 ) ∈ L1 (V ) is compact.

V

Arbitrary dimension

Theorem 2.1. Let Ω = RN . Let µ be a positive (not necessarily finite) Radon measure on RN and let (H3) be satisfied. We assume that there exists α > 0 such that for all c > 0 there exists c0 > 0 such that sup µ ⊗ µ{(v, v 0 ); |v| ≤ c, |v 0 | ≤ c, |(v − v 0 ) · e| ≤ ε} ≤ c0 εα .

(2.1)

e∈S N −1

Let p1 be the smallest integer such that p1 >

(α+1) α

¡N 2

¢ + 1 . Then

0 < t 7→ Rj (t) ∈ L(L1 (RN × V )) is norm continuous for all j ≥ 2p1 . Remark 2.2. on spheres.

Condition (2.1) is obviously satisfied by Lebesgue measures on open sets or

5

The proof of Theorem 2.1 is quite technical and is given in several steps. We observe first that Uj = [U K]j ∗ U (j ≥ 1) where ∗ is the convolution operator which associates to strongly continuous (operator valued) mappings f, g : [0, ∞[ → L(L1 (RN × V )) the strongly continuous mapping Zt f (t − s)g(s)ds ∈ L(L1 (RN × V ))

f ∗ g : [0, ∞[ 3 t 7→ 0

and [U K]j := U K ∗ · · · ∗ U K (j times). Here U denotes the mapping 0 ≤ t 7→ U (t) and U K : 0 ≤ t 7→ U (t)K. We note that: f, g 7→ f ∗ g is associative. We recall that 0 ≤ t 7→ Rm (t) is norm continuous if and only if 0 ≤ t 7→ Um (t) is so [4, Theorem 2.7, p. 18]. Moreover, if 0 ≤ t 7→ [U K]m (t) is norm continuous then 0 < t 7→ Um (t) is also norm continuous. By the same arguments it suffices to show that 0 ≤ t 7→ K [U K]m−1 (t) is norm continuous. By density and linearity we may restrict ourselves to K1 U ∗ K2 U ∗ · · · ∗ Km−1 U Km where Ki (i = 1, · · · , m) has the form Z 1 N L (R × V ) 3 ϕ 7→ fi (v)gi (v 0 )ϕ(x, v 0 )dµ(v 0 ) ∈ L1 (RN × V ) V

where fi (·) ∈ L1 (V ) and gi (·) ∈ L∞ (V ). By density again and decomposition we can suppose that fi and gi are nonnegative, and fi are continuous with compact supports. We can then assume without loss of generality that µ has a compact support. Let Z Z ϕ(x, v 0 )gi (v 0 )dµ(v 0 ) = ϕ(x, v 0 )dµi (v 0 ) ∈ L1 (RN ) Mi : L1 (RN × RN ) 3 ϕ 7→ RN

RN

where µi = gi µ. We have Ki Mi = kµi k Ki and K1 U ∗ K2 U ∗ · · · ∗ Km−1 U Km =

m−1 Y

kµi k−1 K1 (M1 U K2 ∗ M2 U K3 ∗ · · · ∗ Mm−1 U Km ).

i=1

The latter operator is described in: Lemma 2.3.

Let m ≥ 2. There exists a finite Radon measure β m (t) on RN such that M1 U K2 ∗ · · · ∗ Mm U Km+1 ϕ = β m (t) ∗ Mm+1 ϕ.

Proof.

We first prove that Mi U Ki+1 ϕ = ηti ∗ Mi+1 ϕ 6

where ηti is a finite Radon measure on RN . Indeed, Z fi+1 (v)gi (v)e−tσ(v) Mi+1 ϕ(x − tv)dµ(v) Mi U Ki+1 ϕ = RZN

hi (v)e−tσ(v) Mi+1 ϕ(x − tv)dµ(v)

= RZN

Mi+1 ϕ(x − y)dηti (y) = ηti ∗ Mi+1 ϕ

= RN

where ηti is the image of e−tσ(v) hi (v)dµ under the dilation v 7→ tv and hi (v) = fi+1 (v)gi (v). Observe that the mapping 0 < t 7→ ηti ∈ M(RN ) (the space of finite Radon measures) is weak star continuous, i.e., for any ϕ ∈ L1 (RN ), Z ­ i ® 0 < t 7→ ηt , ϕ = ϕ(x − tv)e−tσ(v) hi (v)dµ RN

is continuous. We have Zt M1 U K2 ∗ M2 U K3 ϕ

=

M1 U (t − s)K2 M2 U (s)K3 ϕds 0

Zt 1 ηt−s ∗ M2 (M2 U (s)K3 ϕ)ds

= 0

Zt 1 ηt−s ∗ M2 (ηs2 ∗ M3 ϕ)ds

= 0

=

Zt 1 ηt−s ∗ (ηs2 ∗ M3 ϕ)ds

kµ2 k 0

Zt =

1 (ηt−s ∗ ηs2 ) ∗ M3 ϕds

kµ2 k 0

h Zt =

kµ2 k

i 1 (ηt−s ∗ ηs2 )ds ∗ M3 ϕ

0

=: β 2 (t) ∗ M3 ϕ where the integral

Zt 2

1 (ηt−s ∗ ηs2 )ds

β (t) = kµ2 k 0

is taken in the weak star sense, i.e. ­ 2 ® β (t), ϕ := kµ2 k

Zt 0

7

­ 1 ® ηt−s ∗ ηs2 , ϕ ds.

(2.2)

Now, suppose that M1 U K2 ∗ · · · ∗ Mm−1 U Km ϕ = β m−1 (t) ∗ Mm ϕ. Then Zt [M1 U K2 ∗ · · · ∗ Mm U Km+1 ] (t)ϕ

=

[M1 U K2 ∗ · · · ∗ Mm−1 U Km ] (t − s)(Mm U (s)Km+1 ϕ)ds 0

Zt β m−1 (t − s) ∗ Mm (Mm U (s)Km+1 ϕ)ds

= 0

Zt β m−1 (t − s) ∗ Mm (ηsm ∗ Mm+1 ϕ)ds

= 0

=

Zt β m−1 (t − s) ∗ (ηsm ∗ Mm+1 ϕ)ds

kµm k 0

h Zt =

kµm k

i β m−1 (t − s) ∗ ηsm ds ∗ Mm+1 ϕ

0

=: β m (t) ∗ Mm+1 ϕ where the integral

Zt m

β m−1 (t − s) ∗ ηsm ds

β (t) = kµm k

(2.3)

0

is taken in the weak star sense, i.e. Zt hβ m (t), ϕi = kµm k

­ m−1 ® β (t − s) ∗ ηsm , ϕ ds.

0

Then β m (t) is defined inductively by (2.3) which ends the proof.

¥

Thus, to prove Theorem 2.1, it suffices to prove: Lemma 2.4.

Let p1 be the smallest integer such that p1 >

α+1 N α (2

+ 1). Then for p ≥ p1

0 ≤ t 7→ β 2p (t) ∈ M(RN ) is continuous, where M(RN ) (the space of finite Radon measures) is endowed with the total variation norm. The proof of Lemma 2.4 is given in several steps. Before doing this, as in the proof of Lemma 2.3, we can show, for p > 1, that Zt β 2(p−1) (t − s) ∗ β 2p−1,2p (s)ds

2p

β (t) = kµ2p−1 k 0

8

(2.4)

where

Zt β

i,i+1

i ηt−s ∗ ηsi+1 ds.

(t) = kµi+1 k 0

Let us show first that for p large enough with respect to Lebesgue measure.

β 2p (t)

is a function, i.e. β 2p (t) is absolutely continuous

Lemma 2.5. Let p0 be the smallest integer such that p0 > have β 2p (t) ∈ L2 (RN ) ∩ L1 (RN ) for all t ∈ [0, T ] .

N (α+1) 2α .

Then for all p ≥ p0 we

We start with β 2 (t) (see (2.2)). We have

Proof.

2 (t)(ξ) kµ2 k−1 β[ Zt 1 b2 = (2π)N/2 ηd t−s (ξ)ηs (ξ)ds 0 −N/2

Zt h Z

= (2π)

e 0

Zt = (2π)−N/2 0

= (2π)−N/2

RN

hZ

−iv·ξ

ih Z

1 dηt−s (v)

i 0 e−iv ·ξ dηs2 (v 0 ) ds

RN

ih Z i 0 0 e−i(t−s)v·ξ e−(t−s)σ(v) h1 (v)dµ(v) e−isv ·ξ e−sσ(v ) h2 (v 0 )dµ(v 0 ) ds

RN

Z Z h Zt

RN

i 0 0 e−i(t−s)v·ξ e−(t−s)σ(v) e−isv ·ξ e−sσ(v ) ds h1 (v)h2 (v 0 )dµ(v)dµ(v 0 ).

0

RN RN

Introducing polar coordinates ξ = |ξ| e, e ∈ S N −1 , we decompose the last integral as ZZ

h Zt

|(v 0 −v)·e|≤ε

0

ZZ

h Zt

|(v 0 −v)·e|>ε

0

+

i 0 0 e−i(t−s)v·ξ e−(t−s)σ(v) e−isv ·ξ e−sσ(v ) ds h1 (v)h2 (v 0 )dµ(v)dµ(v 0 ) i 0 0 e−i(t−s)v·ξ e−(t−s)σ(v) e−isv ·ξ e−sσ(v ) ds h1 (v)h2 (v 0 )dµ(v)dµ(v 0 )

where ε > 0 is arbitrary. We have ZZ

h Zt

|(v 0 −v)·e|≤ε

0

¯ ¯ ¯ ≤ te

2tkσk∞

¯ i 0 0 ¯ e−i(t−s)v·ξ e−(t−s)σ(v) e−isv ·ξ e−sσ(v ) ds h1 (v)h2 (v 0 )dµ(v)dµ(v 0 )¯ ZZ dµ(v)dµ(v 0 ).

kh1 h2 k∞ |(v 0 −v)·e|≤ε

We estimate the second integral ¯ ¯ ¯

ZZ

|(v 0 −v)·e|>ε

h Zt e

−i(t−s)v·ξ −(t−s)σ(v) −isv 0 ·ξ −sσ(v 0 )

e

e

0

9

e

¯ i 0 0 ¯ ds h1 (v)h2 (v )dµ(v)dµ(v )¯

¯ ¯ ¯ −itv0 ·ξ −tσ(v0 ) ¯ e − e−itv·ξ e−tσ(v) ¯ ¯ ¯e ¯ ¯h1 (v)h2 (v 0 )¯ dµ(v)dµ(v 0 ) |i(v − v 0 ) · ξ + σ(v) − σ(v 0 )|

ZZ ≤ |(v 0 −v)·e|>ε

ZZ

2 |h1 (v)h2 (v 0 )| dµ(v)dµ(v 0 ) |(v − v 0 ) · e| |ξ| |(v 0 −v)·e|>ε ZZ ¯ ¯ 2 ¯h1 (v)h2 (v 0 )¯ dµ(v)dµ(v 0 ) ≤ 2kh1 k1 kh2 k1 . ≤ ε |ξ| ε |ξ| ≤

|(v 0 −v)·e|>ε

Thus ZZ

³ ¯ ¯ 2 (t)(ξ)¯ ≤ (2π)−N/2 kµ k T e2T kσk∞ kh h k ¯β[ 2 1 2 ∞

dµ(v)dµ(v 0 ) +

2 kh1 k1 kh2 k1 ´ . ε |ξ|

|(v 0 −v)·e|≤ε

Let 0 < τ < 1 and ε = |ξ|−τ then ¯ ¯ 2 (t)(ξ)¯ ¯β[

≤ (2π)−N/2 kµ2 k (T e2T kσk∞ kh1 h2 k∞ + 2 kh1 k1 kh2 k1 )(a(ξ) + b(ξ)) =: C1 (a(ξ) + b(ξ)),

where n o ¯ 0 ¯ a(ξ) := sup µ ⊗ µ (v, v 0 ); ¯(v − v) · e¯ ≤ |ξ|−τ and b(ξ) := e∈S N −1

Consider

1 . |ξ|1−τ

Zt β

2p−1,2p

2p−1 ηt−s ∗ ηs2p ds;

(t) = kµ2p k 0

as previously we can prove that ¯ ¯ \(t)(ξ)¯ ≤ (2π)−N/2 kµ2p k(T e2T kσk∞ kh2p−1 h2p k + 2 kh2p−1 k1 kh2p k )(a(ξ) + b(ξ)).(2.5) ¯β 2p−1,2p ∞ 1 Now let us prove by induction that there exists Cp > 0 (depending only on p) such that ¯ ¯ 2p (t)(ξ)¯ ≤ C (a(ξ) + b(ξ))p for all t ∈ [0, T ] . ¯β\ p Suppose that there exists Cp−1 > 0 such that ¯ \ ¯ ¯β 2(p−1) (t)(ξ)¯ ≤ Cp−1 (a(ξ) + b(ξ))p−1 for all t ∈ [0, T ] . Then by (2.4) and (2.5) we have ¯ ¯ 2p (t)(ξ)¯ kµ2p−1 k−1 ¯β\ Zt ¯ ¯ N/2 \ \(s)(ξ)¯ds ¯β 2(p−1) ≤ (2π) (t − s)(ξ)β 2p−1,2p 0

≤ tCp−1 (a(ξ) + b(ξ))p−1 kµ2p k(T e2T kσk∞ kh2p−1 h2p k∞ + 2kh2p−1 k1 kh2p k1 )(a(ξ) + b(ξ)). 10

Then

¯ ¯ 2p (t)(ξ)¯ ≤ C (a(ξ) + b(ξ))p for all t ∈ [0, T ] ¯β\ (2.6) p ³ ´ 1 where Cp := kµ2p−1 kkµ2p kT Cp−1 T e2T kσk∞ kh2p−1 h2p k∞ + 2kh2p−1 k1 kh2p k1 ) . Put τ = 1+α (clearly 0 < τ < 1) using (2.1) we obtain ¯ ¯ 2p Cp (max{c0 , 1})p 1 c0 2p (t)(ξ)¯ ≤ C [ ¯β\ + 1−τ ]p ≤ . pα p ατ |ξ| |ξ| |ξ| 1+α 2p (t) ∈ L2 (RN ) for all p ≥ p > N (1+α) and for all t ∈ [0, T ] . By Parseval’s identity Hence β\ 0 2α 2p 2 N we have β (t) ∈ L (R ) for all p ≥ p0 > N (1+α) and for all t ∈ [0, T ] . Since β 2p (t) is also a 2α N bounded Radon measure on R we conclude that β 2p (t) ∈ L1 (RN ). ¥

Now, the proof of Lemma 2.4 amounts to 0 ≤ t 7→ β 2p (t) ∈ L1 (RN ) is continuous.

(2.7)

We deal first with the continuity in L2 norm. Lemma 2.6. Let p1 be the smallest integer such that p1 > ]0, T ] 3 t 7→ β 2p (t) ∈ L2 (RN ) is continuous.

α+1 N α (2

+ 1). Then for all p ≥ p1 ,

Proof. We note the following elementary estimates which will be used repeatedly in the sequel. There exists C > 1 such that ¯ −itv·ξ−tσ(v) ¯ ¯ ¯ ¯e − e−itv·ξ−tσ(v) ¯ ≤ C max{1, |ξ|}¯t − t¯ (2.8) and

¯ −it(v−v0 )·ξ−t(σ(v)−σ(v0 )) ¯ ¯ 0 0 ¯ ¯e − e−it(v−v )·ξ−t(σ(v)−σ(v )) ¯ ≤ C max{1, |ξ|} ¯t − t¯

(2.9)

for all t, t ∈ [0, T ] , for almost all v, v 0 ∈ V and for all ξ ∈ RN . In a first step we prove inductively that ¯ ¯ ¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯ ≤ C ∗ ¯t − t¯ |ξ| (a(ξ) + b(ξ))p ¯β\ p for all ξ such that |ξ| ≥ 1, where Cp∗ is that © a(ξ) := sup µ ⊗ µ (v, v 0 );

some constant which depends only on p. We recall ¯ 0 ¯ ª ¯(v − v) · e¯ ≤ |ξ|−τ and b(ξ) :=

e∈S N −1

1 . |ξ|1−τ

2 (t)(ξ) given in the proof of Lemma 2.5 we have Using the expression of β[ ´ ³ [ 2 (t)(ξ) − β 2 (t)(ξ) kµ2 k−1 β[ Z Z −N/2 = (2π) h1 (v)h2 (v 0 )dµ(v)dµ(v 0 ) RN RN

h ×

Zt

Zt e

−i(t−s)v·ξ−(t−s)σ(v) −isv 0 ·ξ−sσ(v 0 )

e

ds −

0

0

11

i 0 0 e−i(t−s)v·ξ−(t−s)σ(v) e−isv ·ξ−sσ(v ) ds

Z Z = (2π)−N/2

h1 (v)h2 (v 0 )dµ(v)dµ(v 0 )

RN RN

Zt h 0 0 −itv·ξ−tσ(v) × e e−is(v −v)·ξ−s(σ(v )−σ(v)) ds t

Zt +(e

−itv·ξ−tσ(v)

−e

−itv·ξ−tσ(v)

)

i 0 0 e−is(v −v)·ξ−s(σ(v )−σ(v)) ds .

(2.10)

0

Introducing polar coordinates ξ = |ξ| e, e ∈ S N −1 , we decompose the last integral as ZZ h1 (v)h2 (v 0 )dµ(v)dµ(v 0 ) |(v 0 −v)·e|≤|ξ|−τ

Zt h 0 −itv·ξ−tσ(v) e−is(v−v)·ξ−s(σ(v )−σ(v)) ds × e t

Zt +(e−itv·ξ−tσ(v) − e−itv·ξ−tσ(v) )

i 0 0 e−is(v −v)·ξ−s(σ(v )−σ(v)) ds

0

ZZ h1 (v)h2 (v 0 )dµ(v)dµ(v 0 )

+ |(v 0 −v)·e|>|ξ|−τ

Zt h 0 −itv·ξ−tσ(v) × e e−is(v−v)·ξ−s(σ(v )−σ(v)) ds t

Zt +(e

−itv·ξ−tσ(v)

−e

−itv·ξ−tσ(v)

)

i 0 0 e−is(v −v)·ξ−s(σ(v )−σ(v)) ds

0

=: I1 + I2 . Clearly ¯ ¯ ¯

ZZ

Zt 0

0

h1 (v)h2 (v )dµ(v)dµ(v )e

|(v 0 −v)·e|≤|ξ|−τ

≤ e

2T kσk∞

≤ e

2T kσk∞

ZZ

|t − t|

−itv·ξ−tσ(v)

¯ 0 0 ¯ e−is(v −v)·ξ−s(σ(v )−σ(v)) ds¯

t

¯ ¯ ¯h1 (v)h2 (v 0 )¯ dµ(v)dµ(v 0 )

|(v 0 −v)·e|≤|ξ|−τ

¯ ¯ kh1 h2 k∞ ¯t − t¯a(ξ).

(2.11)

Using (2.8) we obtain ¯ ¯ ¯

ZZ h1 (v)h2 (v 0 )dµ(v)dµ(v 0 )(e−itv·ξ−tσ(v) − e−itv·ξ−tσ(v) )

|(v 0 −v).e|≤|ξ|−τ

12

Zt ×

¯ 0 0 ¯ e−is(v −v)·ξ−s(σ(v )−σ(v)) ds¯ ZZ

0

¯ ¯ ≤ C |ξ| ¯t − t¯T e2T kσk∞

¯ ¯ ¯h1 (v)h2 (v 0 )¯ dµ(v)dµ(v 0 )

|(v 0 −v)·e|≤|ξ|−τ

¯ ¯ kh1 h2 k∞ a(ξ) |ξ| ¯t − t¯.

2T kσk∞

≤ CT e

(2.12)

Then adding (2.11) and (2.12) we get ¯ ¯ |I1 | ≤ (1 + CT )e2T kσk∞ kh1 h2 k∞ a(ξ) |ξ| ¯t − t¯

∀ |ξ| ≥ 1.

(2.13)

Consider I2 . First we have by (2.9) ¯ ¯ ¯

Zt

ZZ 0

0

h1 (v)h2 (v )dµ(v)dµ(v )e

−itv·ξ−tσ(v)

|(v 0 −v)·e|>|ξ|−τ

t

ZZ

≤

≤

≤ ≤

¯ 0 0 ¯ e−is(v −v)·ξ−s(σ(v )−σ(v)) ds¯

¯ ¯ ¯ −it(v0 −v)·ξ−t(σ(v0 )−σ(v)) − e−it(v0 −v)·ξ−t(σ(v0 )−σ(v)) ¯¯ 0 ¯h1 (v)h2 (v 0 )¯ ¯¯ e ¯dµ(v)dµ(v ) i(v − v 0 ) · ξ + σ(v) − σ(v 0 ) |(v 0 −v)·e|>|ξ|−τ ¯ ¯ ZZ ¯ ¯ C |ξ| ¯t − t¯ 0 ¯h1 (v)h2 (v )¯ dµ(v)dµ(v 0 ) |(v − v 0 ) · e| |ξ| |(v 0 −v)·e|>|ξ|−τ ¯ ¯ ZZ ¯ ¯ C |ξ| ¯t − t¯ 0 ¯h1 (v)h2 (v )¯ dµ(v)dµ(v 0 ) −τ |ξ| |ξ| |(v 0 −v)·e|>|ξ|−τ ¯ ¯ Ckh1 k1 kh2 k1 b(ξ) |ξ| ¯t − t¯. (2.14)

Similarly, applying (2.8) we obtain ZZ ¯ ¯ h1 (v)h2 (v 0 )dµ(v)dµ(v 0 )(e−itv·ξ−tσ(v) − e−itv·ξ−tσ(v) ) ¯ |(v 0 −v)·e|>|ξ|−τ

Zt × ZZ

¯ 0 0 ¯ e−is(v −v)·ξ−s(σ(v )−σ(v)) ds¯

0 ¯ ¯ ¯ ¯¯ −itv·ξ−tσ(v) ¯ 0 ¯h1 (v)h2 (v )¯ ¯e − e−itv·ξ−tσ(v) ¯

≤ |(v 0 −v)·e|>|ξ|−τ

≤ (e2T kσk∞

¯ e−it(v0 −v)·ξ−t(σ(v0 )−σ(v)) − 1 ¯ ¯ ¯ ׯ ¯dµ(v)dµ(v 0 ) i(v − v 0 ) · ξ +¯ σ(v)¯− σ(v 0 ) + 1)Ckh1 k1 kh2 k1 b(ξ) |ξ| ¯t − t¯ .

(2.15)

Thus adding (2.14) and (2.15) ¯ ¯ |I2 | ≤ (e2T kσk∞ + 2)Ckh1 k1 kh2 k1 b(ξ) |ξ| ¯t − t¯ By (2.13) and (2.16) we have ¯ ¯ ¯ ¯ [ 2 (t)(ξ) − β 2 (t)(ξ)¯ ≤ C ∗ (a(ξ) + b(ξ)) |ξ| ¯t − t¯ ¯β[ 1 13

∀ |ξ| ≥ 1.

∀ |ξ| ≥ 1

(2.16)

h i where C1∗ := (2π)−N/2 kµ2 k (1 + CT )e2T kσk∞ kh1 h2 k∞ + (e2T kσk∞ + 2)Ckh1 k1 kh2 k1 . Let us show that there exists Cp∗ which depends only on p such that ¯ ¯ ¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯ ≤ C ∗ |ξ| ¯t − t¯(a(ξ) + b(ξ))p for |ξ| ≥ 1 and t, t ∈ [0, T ] . ¯β\ p Suppose that ¯ \ \(t)(ξ)¯¯ ≤ C ∗ |ξ| ¯¯t − t¯¯ (a(ξ) + b(ξ))p−1 ¯β 2(p−1) (t)(ξ) − β 2(p−1) p−1 for |ξ| ≥ 1 and t, t ∈ [0, T ] . Thanks to (2.4), (2.5) and (2.6) we have ¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯ kµ2p−1 k−1 ¯β\ Zt ¯ ¯¯ ¯ N/2 \ \ \(s)(ξ)¯ds ¯β 2(p−1) ≤ (2π) (t − s)(ξ) − β 2(p−1) (t − s)(ξ)¯¯β 2p−1,2p 0

Zt N/2

+(2π)

¯ ¯¯ ¯ \ \(s)(ξ)¯ds ¯β 2(p−1) (t − s)(ξ)¯¯β 2p−1,2p

t

≤

³ ´ ¯ ¯¯ ¯ T e2T kσk∞ kh2p−1 h2p k∞ + 2kh2p−1 k1 kh2p k1 ¯ξ ¯¯t − t¯(a(ξ) + b(ξ))p ³ ´¯ ¯ +Cp−1 kµ2p k T e2T kσk∞ kh2p−1 h2p k∞ + 2kh2p−1 k1 kh2p k1 ¯t − t¯(a(ξ) + b(ξ))p . ∗ tCp−1 kµ2p k

Then

¯ ¯ ¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯ ≤ C ∗ |ξ| ¯t − t¯(a(ξ) + b(ξ))p ¯β\ p

(2.17)

for |ξ| ≥ 1 and t, t ∈ [0, T ] where ³ ´¡ ¢ ∗ + Cp−1 . Cp∗ := kµ2p−1 kkµ2p k T e2T kσk∞ kh2p−1 h2p k∞ + 2kh2p−1 k1 kh2p k1 T Cp−1 We are going to estimate ¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯ for |ξ| < 1. ¯β\ Using (2.8) and decomposition (2.10) we obtain ³ ´¯ ¯ ¯ ¯ [ 2 (t)(ξ) − β 2 (t)(ξ)¯ ≤ (2π)−N/2 kµ k e2T kσk∞ kh k kh k + T Ce2T kσk∞ kh k kh k ¯β[ ¯ t − t¯ 2 1 1 2 1 1 1 2 1 ¯ ¯ =: C∗1 ¯t − t¯ for |ξ| < 1. Inductively we can obtain ¯ ¯ ¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯ ≤ C ¯t − t¯ for |ξ| < 1 ¯β\ ∗p where C∗p > 0 depends only on p. Considering (2.17), choosing τ = (2.18) we obtain Z ¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯2 dξ ¯β\ RN

14

1 1+α

(2.18) and using (2.1) and

Z

¯ ¯ \ 2p (t)(ξ) − β 2p (t)(ξ)¯2 dξ + ¯β\

≤ |ξ|

1+α N α (2

+ 1) where vol(B(0, 1)) is the volume of unit ball of RN .

¥

Before proving (2.7) we need the following two lemmas. Lemma 2.7. Let ν be a positive Radon measure on RN with compact support and (β(t))t≥0 be a family of positive measures uniformly bounded in t ∈ [0, T ] (for all T > 0) and such that β(t){RN \B(0, n)} → 0 as n → ∞

(2.19)

uniformly in t ∈ [δ, 1δ ] for all δ > 0 where B(0, n) is the ball centred at 0 with radius n. Then Zt β(t − s) ∗ νs ds{RN \B(0, n)} → 0 as n → ∞ 0

uniformly in t ∈ [0, T ] for all T > 0 where νs is the image of ν under the dilation v 7→ sv. Proof.

Let T > 0, ε > 0 and t ∈ [0, T ]. Set M = sup kβ(t)k . By definition we have t∈[0,T ]

ZZ β(t − s) ∗ νs {RN \B(0, n)} =

χRN \B(0,n) (x + y)dνs (x)dβ(t − s)(y) RN Z×RN

β(t − s){−x + RN \B(0, n)}dνs (x)

= RZN

β(t − s){−sx + RN \B(0, n)}dν(x).

= RN

We note that Zt β(t − s) ∗ νs ds{RN \B(0, n)} 0

Zε

Zt−ε β(t − s) ∗ νs ds{RN \B(0, n)} β(t − s) ∗ νs ds{R \B(0, n)} + N

= 0

ε

Zt +

=: I1 + I2 + I3 .

β(t − s) ∗ νs ds{RN \B(0, n)}

t−ε

15

(2.20)

From (2.20) I1 ≤ εM kνk

and

I3 ≤ εM kνk .

(2.21)

On the other hand, there exists a compact set Cε such that Supp νs ⊂ Cε

for all s ∈ [ε, T ],

where Supp ν denotes the support of ν. It follows that there exists an integer n0 such that n −x + RN \B(0, n) ⊂ RN \B(0, ) 2 for all n ≥ n0 and for all x ∈ Cε and then Zt−εZ I2 ≤ ε RN

n β(t − s){RN \B(0, )}dνs (x)ds 2

n ≤ sup β(r){R \B(0, )} 2 r∈[ε,T ]

Zt−εZ

N

dνs (x)ds ε RN

n ≤ T kνk sup β(r){R \B(0, )}. 2 r∈[ε,T ] N

Thanks to (2.19) there exists an integer n1 ≥ n0 such that I2 ≤ T kνkε

(2.22)

for all n ≥ n1 and for all t ∈ [ε, T ]. Using (2.21) and (2.22) we obtain I1 + I2 + I3 ≤ εkνk(2M + T ) for all n ≥ n1 and t ∈ [0, T ]. Lemma 2.8.

¥

For all m ≥ 2 we have β m (t){RN \B(0, n)} → 0 as n → ∞

(2.23)

uniformly in t ∈ [0, T ] for all T > 0. Proof.

Taking advantage of the expression (2.3), i.e. Zt m

β m−1 (t − s) ∗ ηsm ds

β (t) = kµm k 0

we apply inductively Lemma 2.7 to show (2.23). Indeed, we first observe that for T > 0 sup kβ m (t)k < ∞ t∈[0,T ]

which is true for m = 2 since (2.2) shows kβ 2 (t)k ≤ T kµ2 kkh1 k1 kh2 k1 16

(2.24)

for t ∈ [0, T ]. The proof of (2.24) follows by induction. Let us show that ηs1 satisfies (2.19). Since η 1 has compact support, there exists a compact set Cδ , which depends only on δ, such that Supp ηs1 ⊂ Cδ for all s ∈ [δ, 1δ ]. Thus Supp ηs1 ⊂ Cδ ⊂ B(0, n) for n large enough and then 1 ηs1 {RN \B(0, n)} = 0 for all s ∈ [δ, ]. δ 1 Combining this with the uniform boundedness of (ηs )s≥0 and Lemma 2.7 we get β 2 (t){RN \B(0, n)} → 0 as n → ∞

(2.25)

uniformly on compact intervals of [0, +∞[ . Now, using (2.3), (2.24) and (2.25), it is easy to end the proof by induction. ¥ We are now ready to prove Lemma 2.4. Proof of Lemma 2.4 there exists n0 such that

It suffices to show (2.7). Let t > 0 and ε > 0. From Lemma 2.8

β 2p1 (t){RN \B(0, n0 )} ≤

£ ¤ ε uniformly in t ∈ t − δ, t + δ 3

(with a suitable choice of δ). Then Z ¯ 2p ¯ ¯β 1 (t)(x) − β 2p1 (t)(x)¯dx RN

Z

≤

¯ ¯ 2p ¯β 1 (t)(x) − β 2p1 (t)(x)¯dx +

B(0,n0 ) 1

h

Z

≤ [vol(B(0, n0 ))] 2

i1 ¯ 2p ¯ ¯β 1 (t)(x) − β 2p1 (t)(x)¯2 dx 2 Z

β 2p1 (t)(x)dx +

+ RN \B(0,n0 )

1

hZ

≤ [vol(B(0, n0 ))] 2

¯ ¯ 2p ¯β 1 (t)(x) − β 2p1 (t)(x)¯dx

RN \B(0,n0 )

B(0,n0 )

Z

Z

β 2p1 (t)(x)dx

RN \B(0,n0 )

i1 ¯ 2p ¯ ¯β 1 (t)(x) − β 2p1 (t)(x)¯2 dx 2 + ε + ε 3 3

RN

£ ¤ for t ∈ t − δ, t + δ . By Lemma 2.6 there exists δ1 > 0 such that hZ ¯ i1 ¯ ε ¯β 2p1 (t)(x) − β 2p1 (t)(x)¯2 dx 2 ≤ 3vol(B(0, n0 )) RN

¤ £ for t ∈ t − δ1 , t + δ1 . Thus Z

¯ ¯ 2p ¯β 1 (t)(x) − β 2p1 (t)(x)¯dx ≤ ε

RN

£ ¤ for t ∈ t − δ1 , t + δ1 which ends the proof.

¥

17

Remark 2.9. It is an open problem to prove Theorem 2.1 when Ω 6= RN or when the scattering kernels are not space homogeneous. In the Lp theory (1 < p < ∞), 0 ≤ t 7→ R1 (t) is norm continuous [7]. We now show that this is never true in L1 . Theorem 2.10. Let µ be a positive Radon measure with compact support V 6= {0} and let Ω = RN with N ≥ 2. Let σ = 0 and Z 1 N K : L (R × V ) 3 ϕ 7→ ϕ(x, v 0 )dµ(v 0 ). V

Then there exists a sequence tn → 0 such that t 7→ R1 (t) is not norm continuous at tn for all n. Proof. If (for some T > 0) ]0, T ] 3 t 7→ R1 (t) is norm continuous, then ]0, T ] 3 t 7→ U1 (t) would be also norm continuous [4, Lemma 2.3, p. 16]. Thus it suffices to prove that 0 < t 7→ U1 (t) is not norm continuous. Let t > 0. We recall that Zt U1 (t) =

U (t − s)KU (s)ds. 0

To show that 0 < t 7→ U1 (t) is not norm continuous at t it suffices to show that Zt f1 (t) := 0 < t 7→ U

U (t − s)KU (s)ds 0

is not norm continuous at t. Let Lv := {αv; α ∈ R} be the line with direction v ∈ S N −1 . Without loss of generality, we may assume that µ{Lv } < kµk.

(2.26)

Indeed, if for some v ∈ S N −1 µ{Lv } = kµk, then for all v ∈ S N −1 with v 6= v we have Lv ∩ Lv = {0} and kµk ≥ µ{Lv ∪ Lv } = µ{Lv } + µ{Lv } − µ{0} = kµk + µ{Lv } − µ{0} so µ{Lv } ≤ µ{0} < kµk since µ is not supported by {0}. Let (fj )j ⊂ L1 (R × V ) be a normalized sequence converging in the weak star topology of measures to the Dirac mass δ(0,v) = δx=0 ⊗ δv=v . It is clear that f1 (t) − U f1 (t)k ≥ sup kU f1 (t)fj − U f1 (t)fj k kU j∈N

= sup

sup

j∈N {ϕ∈Cc (RN ×V ); kϕk∞ =1}

18

f1 (t)fj − U f1 (t)fj , ϕi hU

≥

sup {ϕ∈Cc (RN ×V ); kϕk∞ =1}

f1 (t)fj − U f1 (t)fj , ϕi, limj→∞ hU

where Cc (RN × V ) stands for the space of continuous functions with compact supports. On f1 (t)fj , ϕi is equal to the other hand hU Z

Z

Zt dxdµ(v)ϕ(x, v) 0

RN ×V

Z =

Z

V

Z fj (y, v 0 )ϕ(y + (t − s)v + sv 0 , v)dy

ds 0

V

RN

Z

Z 0

=

V

Zt dµ(v 0 )

dµ(v)

0

dydµ(v )fj (y, v )

Zt ϕ(y + (t − s)v + sv 0 , v)ds

dµ(v) 0

V

RN ×V

Zt

Z →

fj (x − (t − s)v − sv 0 , v 0 )dµ(v 0 )

ds

ϕ((t − s)v + sv, v)ds

dµ(v) V

0

as j → ∞ so Zt

Z f1 (t) − U f1 (t)k ≥ kU

sup

ϕ((t − s)v + sv, v) − ϕ((t − s)v + sv, v)ds.

dµ(v)

{ϕ∈Cc (RN ×V ); kϕk∞ =1}

0

V

Our goal, now, is to prove that for every t 6= t the last supremum is bounded away by a positive constant independent of t. Let Γt,v := {tv + s(v − v); s ∈ [0, t]} be a segment starting at tv with direction v − v. For t 6= t and v ∈ / Lv the segments Γt,v and Γt,v are disjoint. Let Lt,v := {tv + s(v − v); s ∈ R} be the line passing through tv and with direction v − v (it contains Γt,v ). Note that Lt,v and Lt,v are disjoint if v ∈ / Lv and are identical otherwise. Define the function ϕt on RN × (V \Lv ) by: d(x, Lt,v ) ϕt (x, v) = d(x, Lt,v ) + d(x, Lt,v ) where d(x, Lt,v ) = inf |x − y| = inf |x − tv − s(v − v)| is given by y∈Lt,v

s∈R

d(x, Lt,v ) = |x − tv − hx − tv, v − vi|v − v|−2 (v − v)| where h·, ·i denotes the scalar product in RN . Then ϕt is continuous on RN × (V \Lv ) and ½ 1 for x ∈ Γt,v t ϕ (x, v) = (2.27) 0 for x ∈ Γt,v . 19

Let θε : V → R be a continuous function satisfying 0 ≤ θε (v) ≤ 1, θε (v) = 0 if v ∈ Cε1 and θε (v) = 1 if v ∈ / Cε2 , where ε Cε1 = V ∩ {v ∈ RN ; d(v, Lv ) ≤ } 2 and Cε2 = V ∩ {v ∈ RN ; d(v, Lv ) ≤ ε}. N N The function θε ϕt : RN × V 3 (x, v) 7→ ϕt (x, v)θε (v) is continuous in R S × V . Let ψ ∈ Cc (R ) be a function verifying 0 ≤ ψ ≤ 1 and ψ = 1 on the compact set (t,v)∈Ξ Γt,v , where Ξ = [t − δ, t + δ] × V (for some fixed δ > 0). Using (2.27), the function

φtε : RN × V 3 (x, v) 7→ ϕt (x, v)θε (v)ψ(x), which is in Cc (RN × V ) because ψ has compact support in RN and V is compact, satisfies Z Zt φtε ((t − s)v + sv, v) − φtε ((t − s)v + sv, v)dsdµ(v) 0

V

Z Zt φtε ((t − s)v + sv, v)dsdµ(v)

= 0

V

Z

Zt φtε ((t − s)v + sv, v)dsdµ(v)

≥ V \Cε2 0

= tµ{V \Cε2 }. Finally, since µ{V \Cε2 } → µ{V \Lv } as ε → 0, it follows from (2.26) that f1 (t) − U f1 (t)k ≥ tµ{V \Lv } > 0 kU for all t 6= t.

2.2

¥

The dimension one The one dimensional theory (N = 1) is very different. We have:

Theorem 2.11.

Let µ be a positive Radon measure on R satisfying ©£ ¤ª sup µ v 0 − ε, v 0 + ε → 0 as ε → 0. v 0 ∈R

We assume that (H3) is satisfied. Then 0 < t 7→ R1 (t) ∈ L(L1 (R × R)) is norm continuous.

20

(2.28)

Proof.

We recall that 0 ≤ t 7→ R1 (t) is norm continuous if and only if Zt 0 ≤ t 7→ U1 (t) =

U (t − s)KU (s)ds 0

is so [4, Theorem 2.7, p. 18]. By density arguments we may suppose that K is of the form Z 1 K : L (R × R) 3 ϕ 7→ ϕ(x, v 0 )f (v)g(v 0 )dµ(v 0 ) ∈ L1 (R × R), R

where g(·) ∈ L∞ (R) and f (·) is continuous with compact support. We have Zt U1 (t)ϕ

=

U (t − s)KU (s)ϕds 0

Zt Z 0

f (v)e−(t−s)σ(v)−sσ(v ) ϕ(x − (t − s)v − sv 0 , v 0 )g(v 0 )dsdµ(v 0 )

= 0 R

Zv x−tv Z 0 ϕ(y, v 0 )Θ(t, v, v 0 , x, y)(v − v 0 )−1 dydµ(v 0 )

=

−∞ x−tv +∞ x−tv Z Z 0 + ϕ(y, v 0 )Θ(t, v, v 0 , x, y)(v − v 0 )−1 dydµ(v 0 ) v

x−tv

=: O1 (t)ϕ + O2 (t)ϕ, where

0

0 −1 σ(v)

Θ(t, v, v 0 , x, y) = f (v)g(v 0 )e−(x−y−tv )(v−v )

0 −1 σ(v 0 )

e−(y−x+tv)(v−v )

.

Let us show that both 0 < t 7→ O1 (t) and 0 < t 7→ O2 (t) are norm continuous. We restrict ourselves for instance to 0 < t 7→ O1 (t) since the same argument holds for 0 < t 7→ O2 (t). Note that Z Z O1 (t)ϕ = ϕ(y, v 0 )Θ(t, v, v 0 , x, y)E(t, v, v 0 , x, y)dydµ(v 0 ) R R

where E(t, v, v 0 , x, y) = χ{v0 ε so I2 → 0 as |z| → ∞.

¥

Theorem 4.7. Let Ω ⊂ RN be an open set and let (H1)-(H2) be satisfied. Let dµ(v) = dγ(ρ) ⊗ dβ(ω) where γ is a Radon measure on (0, ∞) absolutely continuous with respect to Lebesgue measure and β is a Radon measure on S N −1 . We assume that (3.1) is satisfied. Then ∗ ∗ σ(V (t)) ∩ {µ; |µ| > e−tλ } = et(σ(T +K)∩{λ; Reλ>−λ }) . (4.4) Moreover, if β is such that β{ω ∈ S N −1 ; ω · ω0 = 0} = 0 for all ω0 ∈ S N −1 and if Ω satisfies one of Assumptions (A1) or (A2), then σ(V (t)) = etσ(T +K) ∪ {0}. Proof. As in the proof of item (a) of Theorem 4.4, the use of Theorems 3.1, 1.2 and 1.1, and Lemma 4.1(i) gives (4.4). On the other hand, if Ω satisfies one of Assumptions (A1) or (A2) then by Lemmas 4.1(i), 4.2 and 4.6 and the inclusion etσap (T +K) ⊂ σap (V (t)) we obtain ∗

{µ; |µ| ≤ e−tλ } ⊂ etσ(T +K) ⊂ σ(V (t)) which ends the proof.

¥

28

Remark 4.8. In bounded geometries, the spectral mapping theorems are consequences of compactness results obtained under the assumption that the collision operator is ”weakly compact with respect to velocities”, where the weak compactness is ”collective” with respect to the space variable [3, 5]. This suggests that (H1)(H2) could probably be replaced by a collective weak compactness assumption.

References [1] Blake, M., S. Brendle, and R. Nagel, On the structure of the critical spectrum of strongly continuous semigroups, in Evolution equations and their applications in physical and life sciences (Bad Herrenalb, 1998), pp. 55 65, Lecture Notes in Pure and Appl. Math., vol. 215, Dekker, New York, 2001. [2] Brendle, S., R. Nagel, and J. Poland, On the spectral mapping theorem for perturbed strongly continuous semigroups, Archiv Math. 74 (2000), 365 378. [3] Lods, B., Th´eorie spectrale des ´equations cin´etiques, Th`ese de doctorat, Universit´e de Franche-Comt´e, 2002. [4] Mokhtar-Kharroubi, M., Mathematical Topics in Neutron Transport Theory. New Aspects, World Scientic, Vol. 46, 1997. [5] Mokhtar-Kharroubi, M., On L1 -spectral theory of neutron transport, J. Diff. Int. Eq., to appear. [6] Mokhtar-Kharroubi, M., Spectral properties of a class of positive semigroups on Banach lattices and streaming operators, Positivity, to appear. [7] Mokhtar-Kharroubi, M. and M. Sbihi, Critical spectrum and spectral mapping theorems in transport theory, Semigroup Forum 70 (2005) 406-435. [8] Nagel, R. and J. Poland, The critical spectrum of a strongly continuous semigroup, Advances in Mathematics 152 (2000), 120-133.

29