Time-dependent angularly averaged inverse transport

Time-dependent angularly averaged inverse transport (extended version) Guillaume Bal and Alexandre Jollivet ∗ arXiv:0902.3432v2 [math-ph] 7 May 2009 May 7, 2009 Abstract This paper concerns the reconstruction of the absorption and scattering parameters in a time-dependent linear transport equation from knowledge of angularly averaged measurements performed at the boundary of a domain of interest. We show that the absorption coefficient and the spatial component of the scattering coefficient are uniquely determined by such measurements. We obtain stability results on the reconstruction of the absorption and scattering parameters with respect to the measured albedo operator. The stability results are obtained by a precise decomposition of the measurements into components with different singular behavior in the time domain. 1 Introduction Inverse transport theory has many applications in e.g. medical and geophysical imaging. It consists of reconstructing optical parameters in a domain of interest from measurements of the transport solution at the boundary of that domain. The optical parameters are the total absorption (extinction) parameter σ(x) and the scattering parameter k(x, v ′ , v), which measures the probability of a particle at position x ∈ X ⊂ Rn to scatter from direction v ′ ∈ Sn−1 to direction v ∈ Sn−1 , where Sn−1 is the unit sphere in Rn . The domain of interest is probed as follows. A known flux of particles enters the domain and the flux of outgoing particles is measured at the domain’s boundary. Several inverse theories may then be envisioned based on available data. In this paper, we assume availability of time dependent measurements that are angularly averaged. Also the source term used to probe the domain is not resolved angularly in order to e.g. save time in the acquisition of data. More precisely, the incoming density of particles φ(t, x, v) as a function of time t, at position x ∈ ∂X at the boundary of the domain of interest, and for incoming directions v, is of the form φS (t, x, v) = φ(t, x)S(x, v), where φ(t, x) is arbitrary but S(x, v) is fixed. This paper is concerned with the reconstruction of the optical parameters from such measurements. We show that the attenuation coefficient is uniquely determined and that the spatial structure of the scattering coefficient can be reconstructed provided that scattering vanishes in the vicinity of the domain’s boundary (except in dimension n = 2 and when X is a disc, where our theory does not require k to vanish in the vicinity of ∂X). For instance, when k(x, v ′ , v) = k0 (x)g(v ′, v) with g(v ′, v) known a priori, then k0 (x) is uniquely determined by the measurements. Similar ∗ Department of Applied Physics and Applied Mathematics, Columbia University, New York NY, 10027; [email protected] and [email protected] 1 results were announced in [1] when measurements are available in the modulation frequency variable, which is the dual (Fourier) variable to the time variable. Several other regimes have been considered in the literature. The uniqueness of the reconstruction of the optical parameters from knowledge of angularly resolved measurements both in the time-dependent and time-independent settings was proved in [9, 10]; see also [17] for a review. Stability in the time-independent case has been analyzed in dimension n = 2, 3 under smallness assumptions for the optical parameters in [14, 15] and in dimension n = 2 in [18]. Stability results in the presence of full, angularly resolved, measurements have been obtained in [3, 4, 19]. The intermediate case of angularly averaged measurements with angularly resolved sources was considered in [13]. The lack of stability of the reconstruction in the time independent setting with angularly averaged measurements and isotropic sources is treated in [6]. See also [2] for a recent review of results in inverse transport theory. The rest of the paper is structured as follows. Section 2 recalls known results on the transport equation and the decomposition of the albedo operator. In section 3 we define and decompose the averaged albedo operator (Proposition 3.1) and we study its distributional kernel (Theorems 3.2–3.5). Our main results on uniqueness and stability are presented in section 4 (Theorems 4.1–4.2, Theorems 4.4–4.5 and Corollary 4.6). We show that the absorption coefficient and the spatial structure of the scattering coefficient (the phase function describing scattering from v to v ′ has to be known in advance) can be reconstructed stably from angularly averaged time dependent data. The reconstruction of the scattering coefficient requires invertion of a weighted Radon transform in the general case. In the specific case of a spherical geometry (measurements are performed at the boundary of a sphere), then the scattering coefficient may be obtained by inverting a classical Radon transform. In section 5 we prove Theorems 3.4–3.5. In section 6 we prove Theorems 4.1–4.2, Theorem 4.5 and Theorem 4.4 (4.13). In section 7 we prove Theorem 3.2. In section 8 we prove Theorem 3.3. In section 9 we prove Lemmas 8.1–8.4 that are used in section 8. In section 10 we prove Proposition 3.1. The derivation of the results is fairly technical and is based on a careful analysis of the temporal behavior of the decomposition of the albedo operator into components that are multilinear in the scattering coefficient. Our results are based on showing that the ballistic and single scattering components may be separated from the rest of the data. These two components are then used to obtain our uniqueness and stability results. It turns out that the structure of single scattering is different depending on whether k vanishes on ∂X or not. When k does not vanish on ∂X, the main singularities of the single scattering component do not allow us to “see inside” the domain as they only depend on values of k at the domain’s boundary in dimension n ≥ 3. The singular structure of single scattering and the resulting stability estimates are presented in detail when both k vanishes and does not vanish on ∂X. This is the extended version of a submitted paper [5]. 2 2.1 The forward problem and albedo operator The linear Boltzmann transport equation We now introduce notation and recall some known results on the linear transport equation. Let X be a bounded open subset of Rn , n ≥ 2, with a C 1 boundary ∂X. Let ν(x) denote the outward normal unit vector to ∂X at x ∈ ∂X. Let Γ± = {(x, v) ∈ ∂X × Sn−1 | ± ν(x) · v > 0} be the sets of incoming and outgoing conditions. For (x, v) ∈ X̄ × Sn−1 we define τ± (x, v) and τ (x, v) by τ± (x, v) := inf{s ∈ (0, +∞) | x ± sv 6∈ X} and τ (x, v) := τ− (x, v) + τ+ (x, v). For n−1 x ∈ ∂X we define Sx,± := {v ∈ Sn−1 | ± ν(x) · v > 0}. 2 Consider σ : X × Sn−1 → R and k : X × Sn−1 × Sn−1 → R two nonnegative measurable functions. We assume that (σ, k) is admissible when 0 ≤ σ ∈ L∞ (X × Sn−1 ), 0 ≤ k(x, v ′ ,Z.) ∈ L1 (Sn−1 ) for a.e. (x, v ′ ) ∈ X × Sn−1 , ′ ′ σp (x, v ) = Sn−1 ∞ n−1 k(x, v , v)dv belongs to L (X × S (2.1) ). Let T > η > 0. We consider the following linear Boltzmann transport equation ∂u (t, x, v) + v · ∇x u(t, x, v) + σ(x, v)u(t, x, v) ∂tZ = k(x, v ′ , v)u(t, x, v ′)dv ′, (t, x, v) ∈ (0, T ) × X × Sn−1 , Sn−1 (2.2) u|(0,T )×Γ− (t, x, v) = φ(t, x, v), u(0, x, v) = 0, (x, v) ∈ X × Sn−1 , where φ ∈ L1 ((0, T ), L1(Γ− , dξ)) and suppφ ⊆ [0, η]. Here, dξ(x, v) = |v · ν(x)|dvdµ(x), where dµ is the surface measure on ∂X and dv is the surface measure on Sn−1 . In other words, we assume that the initial condition is concentrated in the η-vicinity of t = 0 and measurements are performed for time T , which we will choose sufficiently large so that particles have the time to travel through X and be measured. 2.2 Semigroups and unbounded operators We introduce the following space Z := {f ∈ L1 (X × Sn−1 ) | v · ∇x f ∈ L1 (X × Sn−1 )}, kf kZ := kf kL1 (X×Sn−1 ) + kv · ∇x f kL1 (X×Sn−1 ) ; (2.3) (2.4) where v · ∇x is understood in the distributional sense. It is known (see [7, 8]) that the trace map γ− from C 1 (X̄ × Sn−1 ) to C(Γ− ) defined by γ− (f ) = f|Γ− (2.5) extends to a continuous operator from Z onto L1 (Γ− , τ+ (x, v)dξ(x, v)) and admits a continuous lifting. Note that L1 (Γ− , dξ) is a subset of the spaces L1 (Γ− , τ+ (x, v)dξ(x, v)). We introduce the following notation Z A1 f = −σf, A2 f = k(x, v ′ , v)f (x, v ′)dv ′ . (2.6) Sn−1 As (σ, k) is admissible, the operators A1 and A2 are bounded operators in L1 (X × Sn−1 ). Consider the following unbounded operators T1 f = −v · ∇x f + A1 f, D(T1 ) = {f ∈ Z | f|Γ− = 0}, T f = T1 f + A2 f, D(T ) = D(T1 ). (2.7) (2.8) It is known that the unbounded operators T1 and T are generators of strongly continuous semigroups in L1 (X × Sn−1 ) U1 (t), U(t) respectively (see for example [11, Proposition 2 pp 3 226]). In addition U1 (t) and U(t) preserve the cone of positive functions, and U1 (t) is given explicitly by the following formula U1 (t)f = e− Rt 0 σ(x−sv,v)ds f (x − tv, v)Θ(x, x − tv), for a.e. (x, v) ∈ X × Sn−1 , (2.9) for f ∈ L1 (X × Sn−1 ), where Θ(x, y) =  1 if x + p(y − x) ∈ X for all p ∈ (0, 1], 0 otherwise, (2.10) for (x, y) ∈ Rn × Rn . We recall the Dyson-Phillips formula U(t) = +∞ X Hm (t) (2.11) m=0 for t ≥ 0, where Z Hm (t) := U1 (t − s1 − . . . − sm )A2 U1 (s1 ) . . . A2 U1 (sm )ds1 . . . dsm , m ≥ 1, (2.12) s1 ≥0,...,sm ≥0 s1 +...+sm ≤t Hm (t) = Z t 0 Hm−1 (t − s)A2 U1 (s)ds, m ≥ 1, H0 (t) := U1 (t). 2.3 (2.13) (2.14) Trace results We introduce the following space  ∂ + v · ∇x u ∈ L1 ((0, T ) × X × Sn−1 )}, W := {u ∈ L ((0, T ) × X × S ) | ∂t   ∂ ; + v · ∇x u kukW := kukL1 ((0,T )×X×Sn−1 ) + ∂t L1 ((0,T )×X×Sn−1 ) 1 n−1  (2.15) (2.16) ∂ where ∂t and v · ∇x are understood in the distributional sense. It is known (see [7, 8]) that the trace map γ− (respectively γ+ ) from C 1 ([0, T ] × X̄ × Sn−1 ) to C(X × Sn−1 ) × C((0, T ) × Γ± ) defined by γ− (ψ) = (ψ(0, .), ψ|(0,T )×Γ− ) (respectively γ+ (ψ) = (ψ(T, .), ψ|(0,T )×Γ+ ) (2.17) extends to a continuous operator from W onto L1 (X × Sn−1 , τ+ (x, v)dxdv) × L1 ((0, T ) × Γ− , min(T − t, τ+ (x, v))dtdξ(x, v)) (respectively L1 (X × Sn−1 , τ− (x, v)dxdv) × L1 ((0, T ) × Γ+ , min(t, τ− (x, v))dtdξ(x, v))). In addition γ± admits a continuous lifting. Note that L1 (X × Sn−1 ) is a subset of L1 (X ×Sn−1 , τ+ (x, v)dxdv). Note also that L1 ((0, T ) ×Γ− , dtdξ) (resp. L1 ((0, T ) × Γ+ , dtdξ)) is a subset of L1 ((0, T )×Γ− , min(T −t, τ+ (x, v))dtdξ(x, v)) (resp. L1 ((0, T )×Γ+ , min(t, τ− (x, v))dtdξ(x, v))). We now introduce the space W̃ := {u ∈ W | γ− (u) ∈ L1 (X × Sn−1 ) × L1 ((0, T ) × Γ− , dtdξ)}. We recall the following trace results (owed to [7, 8] in a more general setting). 4 (2.18) Lemma 2.1. The following equality is valid W̃ = {u ∈ W | γ+ (u) ∈ L1 (X × Sn−1 ) × L1 ((0, T ) × Γ+ , dtdξ)}. (2.19) In addition the trace maps γ± : W̃ → L1 (X×Sn−1 )×L1 ((0, T )×Γ± , dtdξ) are continuous, onto, and admit continuous lifting. (2.20) 2.4 Solution to equation (2.2) We identify the space L1 ((0, r), L1(Γ± , dξ)) with the space L1 ((0, r) × Γ± , dtdξ) for any r > 0. We extend by 0 on R outside the interval (0, η) any function φ ∈ L1 ((0, η), L1(Γ− , dξ)). Let φ ∈ L1 ((0, η), L1 (Γ− , dξ)). Then we consider the lifting G− (t)φ ∈ W̃ of (0, φ) defined by G− (t)φ(x, v) := e− R τ− (x,v) φ− (t−τ− (x, v), x−τ− (x, v)v, v), for a.e. (t, x, v) ∈ (0, T )×X×Sn−1 . (2.21) ∂ + v · ∇x )u + σu = 0 Note that G− (.)φ is a solution in the distributional sense of the equation ( ∂t in (0, T ) × X × Sn−1 and 0 σ(x−sv,v)ds kG− (.)φkW ≤ (1 + kσk∞ )kG− (.)φkL1 ((0,T )×X×Sn−1 ) ≤ (1 + kσk∞ )T kφ− kL1 ((0,η)×Γ− ,dtdξ) . (2.22) To prove this two latter statements, one can use the following change of variables (see [10]). Lemma 2.2. We have Z f (x, v)dxdv = X×Sn−1 Z Γ∓ Z 0 τ± (x,v) f (x ± tv)dtdξ(x, v), (2.23) for f ∈ L1 (X × V ). From (2.22) we obtain that the map i : L1 ((0, η), L1 (Γ− , dξ)) → W̃ defined by i(φ) = G− (.)φ, φ ∈ L1 ((0, η), L1(Γ− , dξ)), (2.24) is continuous. The following result holds (see [11, Theorem 3 p. 229]). Lemma 2.3. The equation (2.2) admits a unique solution u in W̃ which is given by Z t u(t) = G− (t)φ + U(t − s)A2 G− (s)φds. (2.25) 0 where U(t) is the strongly continuous semigroup in L1 (X × Sn−1 ) introduced in subsection 2.2. Using (2.25) and the Dyson-Phillips expansion (2.11) we obtain that the solution u of (2.2) may be decomposed as ∞ Z t X u(t) = G− (t)φ + Hm (t − s)A2 G− (s)φds, (2.26) m=0 −∞ for t ≥ 0 and φ ∈ L1 ((0, η) × ∂X, dtdµ(x)). The first term in the above series G− (t)φ is the ballistic part of u(t) while the term corresponding to m ≥ 1 is m-linear in the scattering kernel k. The term corresponding to m = 1 is the single scattering term. From (2.24), Lemma 2.3 and (2.20), we also obtain the existence of the albedo operator. 5 Lemma 2.4. The albedo operator A given by the formula Aφ = u|(0,T )×Γ+ , for φ ∈ L1 ((0, η), L1(Γ− , dξ)) where u is given by (2.25), (2.27) is well-defined and is a bounded operator from L1 ((0, η), L1(Γ− , dξ)) to L1 ((0, T ), L1(Γ+ , dξ)). We refer the reader to [9] for the reconstruction of the optical parameters when the full albedo operator is known. We assume here that only partial knowledge of the albedo operator is available from measurements. The operator AS,W and its distributional kernel 3 3.1 Angularly averaged measurements We now define more precisely the type of measurements we consider in this paper. The directional behavior of the source term is determined by a fixed function S(x, v), which is bounded and continuous on Γ− . We assume that the incoming conditions have the following structure φS (t′ , x′ , v ′) = S(x′ , v ′ )φ(t′ , x′ ), t′ ∈ (0, η), (x′ , v ′) ∈ Γ− , (3.1) where φ(t, x) is an arbitrary function in L1 ((0, η) × ∂X). We model the detectors by the kernel W (x, v), which we assume is a continuous and bounded function on Γ+ . The available measurements are therefore modeled by the availability of the averaged albedo operator AS,W from L1 ((0, η) × ∂X, dtdµ(x)) to L1 ((0, T ) × ∂X, dtdµ(x)) and defined by Z AS,W φ(t, x) = A(φS )(t, x, v)W (x, v)(ν(x) · v)dv, for a.e. (t, x) ∈ (0, T ) × ∂X. (3.2) Sn−1 x,+ The functions S and W are fixed throughout the paper. The case W ≡ 1 corresponds to measurements of the current of exiting particles at the domain’s boundary. The decomposition of the transport solution (2.26) translates into a similar decomposition of the albedo operator of the form AS,W φ(t, x) = +∞ X Am,S,W φ(t, x), (3.3) m=0 for (t, x) ∈ (0, T ) × ∂X, where we have defined Z A0,S,W φ(t, x) = (ν(x) · v)W (x, v) (G− (.)φS )|(0,T )×Γ+ (t, x, v)dv, Sn−1 x,+ Am,S,W φ(t, x) = Z Sn−1 x,+ (ν(x) · v)W (x, v) Z t Hm−1 (t − s)A2 G− (s)φS ds −∞  (t, x, v)dv, (3.4) (3.5) |(0,T )×Γ+ for a.e. (t, x) ∈ (0, T ) × ∂X where φS is defined by (3.1). The kernels of the operators Am,S,W can be written explicitly. 6 3.2 Distributional kernel of the operators Am,S,W Consider the nonnegative measurable E from ∂X × ∂X → R defined by ( R |x −x | x −x x −x − 0 1 2 σ(x1 −s |x1 −x2 | , |x1 −x2 | )ds 1 2 1 2 if x1 + p(x2 − x1 ) ∈ X for all p ∈ (0, 1), e E(x1 , x2 ) = 0 otherwise, (3.6) for a.e. (x1 , x2 ) ∈ ∂X ×∂X. For m ≥ 3, we also define the nonnegative measurable real function E(x1 , . . . , xm ) by the formula − E(x1 , . . . , xm ) = e Pm−1 R |xi −xi+1 | i=1 0 x −x x −x σ(xi −s |xi −xi+1 | , |xi −xi+1 | )ds i i+1 i i+1 Θ(xm , xm−1 )Πm−2 i=1 Θ(xi , xi+1 ), (3.7) for a.e. (x1 , . . . , xm ) ∈ ∂X × (Rn )m−2 × ∂X, where Θ is defined by (2.10). The function E(x1 , . . . , xm ) measures the total attenuation along the broken path (x1 , . . . , xm ) ∈ ∂X × Rm−2 × ∂X provided (x2 , . . . , xm−1 ) ∈ X m−2 . For m ∈ N, m ≥ 1 and for any subset U of Rm we denote by χU the characteristic function from Rm to R defined by χU (y) = 1 when y ∈ U and χU (y) = 0 otherwise. Using (3.4)–(3.5), (2.21) and (2.13)–(2.14) we then obtain the following result on the structure of the kernels of the albedo operator. Proposition 3.1. We have Am,S,W (φ)(t, x) = Z (0,η)×∂X γm (t − t′ , x, x′ )φ(t′ , x′ )dt′ dµ(x′ ), (3.8) for m ≥ 0 and for a.e. (t, x) ∈ (0, T ) × ∂X, where E(x, x′ ) [W (x, v)S(x′ , v)(ν(x) · v)|ν(x′ ) · v|]v= x−x′ δ(τ − |x − x′ |), (3.9) |x−x′ | |x − x′ |n−1 Z γ1 (τ, x, x′ ) := χ(0,+∞) (τ − |x′ − x|) (ν(x) · v)W (x, v) [E(x, x − sv, x′ )k(x − sv, v ′, v) γ0 (τ, x, x′ ) := Sn−1 x,+ × χ(0,τ− (x,v)) (s)S(x′ , v ′ )|ν(x′ ) · v ′ |  x−x′ −sv v′ = |x−x ′ −sv| τ 2 −|x−x′ |2 ; s= 2(τ −v·(x−x′ )) 2n−2 (τ − (x − x′ ) · v)n−3 dv, (3.10) |x − x′ − τ v|2n−4 for (τ, x, x′ ) ∈ R×∂X ×∂X and where γm for m ≥ 2 admits a similar, more complex, expression given in Section 8 (see (8.12)–(8.13)). Because the above formulas are central in our uniqueness and stability results, we briefly present their derivation and refer the reader to Section 10 for the rest of the proof of Proposition 3.1. Derivation of (3.9) and (3.10). From (3.4) and the definition of G− , we obtain A0,S,W φ(t, x) = R (ν(x)·v)W (x, v)E(x, x−τ− (x, v)v)S(x−τ− (x, v)v, v)φ(t−τ− (x, v), x−τ− (x, v)v)dv, (t, x) ∈ Sn−1 x,+ (0, T ) × ∂X and for φ ∈ L1 ((0, η) × ∂X). Therefore, performing the change of variables “x′ ”= |ν(x′ )·v| ′ ′ x − τ (x, v)v (dv = |x−x ′ |n−1 dµ(x ) and τ (x, v) = |x − x |), we obtain (3.9). R From the definition of A2 and G− we note that A2 G− (s)φS (z, w) := Sn−1 k(z, v ′ , w)E(z, z − τ− (z, v ′ )v ′ )S(z − τ− (z, v ′ )v ′ , v ′)φ(s − τ− (z, v ′ ), z − τ− (z, v ′ )v ′ )dv ′, for a.e. (z, w) ∈ X × Sn−1 and for φ ∈ L1 ((0, η) × ∂X). Performing the change of variables “x′ = z − τ− (z, v ′ )v ′ ”, we obtain R E(z,x′ ) the equality (A2 G− (s)φS ) (z, w) = ∂X [k(z, v ′ , w)S(x′ , v ′)|ν(x′ ) · v ′ |]v′ = z−x′ |z−x ′ |n−1 φ(s − |z − |z−x′ | x′ |, x′ )dµ(x′ ), for a.e. (z, w) ∈ X × Sn−1 and φ ∈ L1 ((0, η) × ∂X). Using also the definition of 7 A1,S,W (see (3.5) for m = 1) we obtain the following equality for any φ ∈ L1 ((0, η) × ∂X) and for a.e. (t, x) ∈ (0, T ) × ∂X Z Z t Z A1,S,W (φ)(t, x) = [k(x − (t − s)v, v ′ , v)S(x′, v ′ )|ν(x′ ) · v ′ |]v′ = x−(t−s)v−x′ (ν(x) · v) Sn−1 x,+ × −∞ |x−(t−s)v−x′ | ∂X E(x, x − (t − s)v, x′ ) Θ(x, x − (t − s)v)φ(s − |x − (t − s)v − x′ |, x′ )W (x, v)dµ(x′ )dsdv. (3.11) |x − (t − s)v − x′ |n−1 Then performing the changes of variables “s”= t − s and “t′ ”= t − s − |x − sv − x′ | (s = ′ )−(x−x′ )·v)2 ′ (t−t′ )2 −|x−x′ |2 , dt = 2((t−t ), we obtain (3.10). 2(t−t′ −v·(x−x′ )) ds |x−x′ −(t−t′ )v|2 To simplify notation, we define the multiple scattering kernels Γk = ∞ X γm . (3.12) m=k 3.3 Regularity of the albedo kernels The reconstruction of the optical parameters is based on an analysis of the behavior in time of the kernels of the albedo operator. Our first result in this direction is the following. Theorem 3.2. Assume that k ∈ L∞ (X × Sn−1 × Sn−1 ). Then the following holds: p τ 2 − |x − x′ |2 γ1 (τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X) when n = 2; ′ τ |x − x |  γ1 (τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X)  when n = 3; τ +|x−x′ | ln τ −|x−x′| τ |x − x′ |n−2 γ1 (τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X) when n ≥ 4. (3.13) (3.14) (3.15) In addition, assume that k ∈ L∞ (X × Sn−1 × Sn−1 ) and that there exists δ > 0 such that suppk ⊆ {y ∈ X | inf x∈∂X |x − y| ≥ δ}, i.e., the scattering coefficient vanishes in the vicinity of ∂X. Then, the following holds (τ − |x − x′ |) 3−n 2 γ1 (τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X) when n ≥ 2. (3.16) Theorem 3.2 is proved in Section 7. The results (3.14) and (3.15) of Theorem 3.2 correspond to singularities of the single scattering contribution that depend on the values of k on ∂X. The above theorem shows that the structure of the single scattering coefficient is quite different depending on whether k vanishes on ∂X or not. The following result describes some regularity properties of the multiple scattering. It is because multiple scattering is more regular than single scattering, in an appropriate sense, that we can reconstruct the scattering coefficient in a stable manner. Theorem 3.3. Assume that k ∈ L∞ (X × Sn−1 × Sn−1 ). Then the following holds: Γ2 (τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X), τ |x − x′ |Γ2 (τ, x, x′ ) ∞   2 ∈ L ((0, T ) × ∂X × ∂X), ′| τ +|x−x (τ − |x − x′ |) 1 + ln τ −|x−x′| τ |x − x′ |n−2 Γ2 (τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X), τ − |x − x′ | 8 when n = 2; (3.17) when n = 3; (3.18) when n ≥ 4. (3.19) In addition, assume that k ∈ L∞ (X × Sn−1 × Sn−1 ) and that there exists δ > 0 such that suppk ⊆ {y ∈ X | inf x∈∂X |x − y| ≥ δ}. Then the following holds: −1   τ + |x − x′ | Γ2 (τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X), when n = 3; 1 + ln (τ − |x − x |) ′ τ − |x − x | (3.20) ′ 1−n ′ ∞ 2 (τ − |x − x |) Γ2 (τ, x, x ) ∈ L ((0, T ) × ∂X × ∂X), when n ≥ 4. (3.21) ′ −1 Theorem 3.3 is proved in Section 8. These results quantify how “smoother” multiple scattering is compared to the single scattering contribution considered in Theorem 3.2. 3.4 Asymptotics of the single scattering term In this subsection we assume that X is also convex. We give limits for the single scattering term in two configurations given by: the nonnegative function σ is bounded and continuous on X × Sn−1 , the nonnegative function k is continuous on X̄ × Sn−1 × Sn−1 ; (3.22) or there exists a convex open subset Y ⊆ X with C 1 boundary such that σ(x, v) = 0 for (x, v) ∈ (X\Y ) × Sn−1 and the nonnegative function σ is bounded and continuous on Y × Sn−1 ; and there exists a convex open subset Z ⊆ Y ⊆ X with C 1 boundary such that δ := inf (x,z)∈∂X×Z |x − z| > 0 and k(x, v ′ , v) = 0 for (x, v ′ , v) ∈ (X\Z) × Sn−1 × Sn−1 , and the nonnegative function k is bounded and continuous on Z × Sn−1 × Sn−1 , (3.23) When either (3.22) or (3.23) is satisfied, we want to analyze the behavior of the function γ1 (τ, x, x′ ) given by the right hand side of (3.10) for all (τ, x, x′ ) ∈ R × ∂X × ∂X. We need to introduce some notation. Let ϑ0 : Sn−1 × X → R be the function defined by ϑ0 (v, x) = (τ− (x, v)τ+ (x, v))− n−1 2 , (v, x) ∈ Sn−1 × X, (3.24) and consider the weighted X-ray transform Pϑ0 defined by Pϑ0 f (v, x) = Z τ+ (x,v) ϑ0 (v, tv + x)f (tv + x)dt, (3.25) τ− (x,v) for a.e. (v, x) ∈ Sn−1 × ∂X and f ∈ L2 (X, supv∈Sn−1 ϑ0 (v, x)dx). The first result analyzes the behavior of γ1 under hypothesis (3.22). Theorem 3.4. Assume that the open subset X of Rn with C 1 boundary is convex. Let (x, x′0 ) ∈ x−x′ ∂X 2 be such that x + s(x − x′0 ) ∈ X for some s ∈ (0, 1). Set v0 = |x−x0′ | and t0 = |x − x′0 |. Then 0 under condition (3.22), we have the following results. When n = 2, then √ 1 2W (x, v0 )S(x′0 , v0 )(ν(x) · v0 )|ν(x′0 ) · v0 |E(x, x′0 ) ′ √ γ1 (τ, x, x0 ) = √ τ − t0 t0 (3.26)   1 + ×Pϑ0 kv0 (v0 , x) + o √ , as τ → t0 , τ − t0 9 where Pϑ0 is defined by (3.25) and kv0 (y) := k(y, v0 , v0 ) for y ∈ X. When n = 3, then 1 π ) 2 W (x, v0 )S(x′0 , v0 )(ν(x) · v0 )|ν(x′0 ) · v0 |E(x, x′0 ) τ − t0 t0  1  ) , as τ → t+ × (k(x, v0 , v0 ) + k(x′0 , v0 , v0 )) + o ln( 0. τ − t0 γ1 (τ, x, x′0 ) = ln( When n ≥ 4, then γ1 (τ, x, x′0 ) = t01−n E(x, x′0 ) " W (x, v)(ν(x) · v)k(x, v0 , v) dv 1 − v · v 0 Sn−1 x,+  ′ ′ ′ ′ ′ ′ k(x0 , v , v0 )S(x0 , v )|ν(x0 ) · v | ′ dv 1 − v ′ · v0 S(x′0 , v0 )|ν(x′0 ) +W (x, v0 )(ν(x) · v0 ) +o(1), as τ → t+ 0. Z Sn−1 ′ x ,− 0 (3.27) · v0 | Z (3.28) Theorem 3.4 is proved in Section 5. Note that γ1 depends on the value of k on ∂X in dimension n ≥ 3. Under hypothesis (3.23), i.e., when the scattering coefficient vanishes in the vicinity of where measurements are collected, we have the quite different behavior: Theorem 3.5. Assume that the open subset X of Rn with C 1 boundary is also convex and assume that condition (3.23) is fulfilled. Let (x, x′0 ) ∈ ∂X 2 be such that x′0 + s(x − x′0 ) ∈ Z for x−x′ some s ∈ (0, 1). Set v0 = |x−x0′ | and t0 = |x − x′0 |. Then we have the following. 0 When n = 2, then (3.26) still holds. When n ≥ 3, then γ1 (τ, x, x′0 ) = (τ − t0 ) n−3 2 (2t0 ) 1−n 2 Voln−2 (Sn−2 )S(x′0 , v0 )W (x, v0 )|ν(x′0 ) · v0 |(ν(x) · v0 ) ×E(x, x′0 )Pϑ0 kv0 (v0 , x) + o((τ − t0 ) n−3 2 ), as τ → t+ 0, (3.29) where Pϑ0 is defined by (3.25) and kv0 (y) := k(y, v0 , v0 ) for y ∈ X. Theorem 3.5 is proved in Section 5. Theorem 3.5 may remain valid under different conditions from those stated in (3.23). For instance, when σ is bounded and continuous on X and k is continuous on X × Sn−1 × Sn−1 and k(x, ., .) decays sufficiently rapidly as x get closer and closer to the boundary ∂X for any x ∈ X, then the same asymptotics of γ1 holds. 4 Uniqueness and stability results P+∞ We denote by γ := Γ0 = m=0 γm the distributional kernel of AS,W . Then γ − γ0 = Γ1 denotes the distributional kernel of the multiple scattering of AS,W . For the rest of the paper, we assume that the duration of measurement T > diam(X) := sup(x,y)∈X 2 |x − y| so that the singularities of the ballistic and single scattering contributions are indeed captured by the available measurements. Let (σ̃, k̃) be a pair of absorption and scattering coefficients that also satisfy (2.1). We denote by a superscript˜any object (such as the albedo operator à or the distributional kernels γ̃ and γ̃0 ) associated to (σ̃, k̃). Moreover if (σ, k) satisfies (3.23) for some (Y, Z) and (σ̃, k̃) also satisfies (3.23) for some (Ỹ , Z̃), then we always make the additional assumption Y = Ỹ and Z = Z̃. Let k.kη,T := k.kL(L1 ((0,η)×∂X)),L1 ((0,T )×∂X)) . 10 4.1 Stability estimates under condition (3.22) or (3.23) Theorem 4.1. Assume that the open subset X of Rn with C 1 boundary is also convex. Let (σ, k) and (σ̃, k̃) satisfy condition (3.22). Let x′0 ∈ ∂X. Then we have: # Z " |E − Ẽ|(x, x′0 ) dµ(x) ≤ kAS,W − ÃS,W kη,T . W (x, v0 )S(x′ , v0 )(ν(x) · v0 )|ν(x′0 ) · v0 | ′ n−1 t0 =|x−x′ | |x − x | ∂X 0 0 x−x′0 v0 = |x−x′ | 0 px′0 +(1−p)x Let x ∈ ∂X be such that When n = 2, we have ∈ X for some p ∈ (0, 1). Set v0 = x−x′0 |x−x′0 | (4.1) and t0 = |x−x′0 |. W (x, v0 )S(x′0 , v0 )(ν(x) · v0 )|ν(x′0 ) · v0 | E(x, x′0 )Pϑ0 kv0 (v0 , x) − Ẽ(x, x′0 )Pϑ0 k̃v0 (v0 , x) 1 p 2 τ − |z − z ′ |2 (Γ1 − Γ̃1 )(τ, z, z ′ ) , (4.2) ≤ 2 L∞ where k · kL∞ := k · kL∞ ((0,T )×∂X×∂X) , Pϑ0 is defined by (3.25) and kv0 (y) := k(y, v0, v0 ) for y ∈ X (k̃v0 is defined similarly). When n = 3, then E(x, x′0 )(k(x, v0 , v0 ) + k(x′0 , v0 , v0 )) − Ẽ(x, x′0 )(k̃(x, v0 , v0 ) + k̃(x′0 , v0 , v0 )) ×W (x, v0 )S(x′0 , v0 )(ν(x) · v0 )|ν(x′0 ) When n ≥ 4, then S(x′0 , v0 )|ν(x′0 ) · v0 | Z +W (x, v0 )(ν(x) · v0 ) Sn−1 x,+ Z Sn−1 ′ 1 τ |z − z ′ |   (Γ1 − Γ̃1 )(τ, z, z ′ ) · v0 | ≤ ′| τ +|z−z π ln τ −|z−z ′ | .(4.3) L∞  W (x, v)(ν(x) · v)  E(x, x′0 )k(x, v0 , v) − Ẽ(x, x′0 )k̃(x, v0 , v) dv 1 − v · v0 x0 ,−  S(x′0 , v ′)|ν(x′0 ) · v ′ |  ′ ′ ′ ′ ′ ′ E(x, x0 )k(x0 , v , v0 ) − Ẽ(x, x0 )k̃(x0 , v , v0 ) dv ′ 1 − v ′ · v0 ≤ τ |z − z ′ |n−2 (Γ1 − Γ̃1 )(τ, z, z ′ ) L∞ . (4.4) Theorem 4.1 is proved in Section 7. It shows that the spatial structure of k may be stably reconstructed at the domain’s boundary. More interesting is the following theorem, which provides some stability of the reconstruction of the scattering coefficient when it vanishes in the vicinity of the boundary ∂X. Theorem 4.2. Assume that the open subset X of Rn with C 1 boundary is also convex. Assume also that inf (x′ ,v′ )∈Γ− S(x′ , v ′ ) > 0 and inf (x,v)∈Γ+ W (x, v) > 0. Let (σ, k) and (σ̃, k̃) satisfy condition (3.23). Let x′0 ∈ ∂X. Then there exist constants C1 = C1 (S, W, X, Y ) and C2 = C2 (S, W, X, Z) such that Z |E − Ẽ|(x′0 + τ+ (x′0 , v0 )v0 , x′0 )|ν(x′0 ) · v0 |dv0 ≤ C1 kAS,W − ÃS,W kη,T , (4.5) Sn−1 ′ x ,− 0 E(x, x′0 )Pϑ0 kv0′ (v0′ , x′0 ) − Ẽ(x, x′0 )Pϑ0 k̃v0′ (v0′ , x′0 ) ≤ C2 (τ − |z − z ′ |) 11 3−n 2 (Γ1 − Γ̃1 )(τ, z, z ′ ) L∞ , (4.6) for x ∈ ∂X such that px′0 + (1 − p)x ∈ Z for some p ∈ (0, 1) where v0′ = in (3.25), and kv0′ (y) := k(y, v0′ , v0′ ) for y ∈ X (k̃v0′ is defined similarly). x−x′0 , |x−x′0 | Pϑ0 is defined Theorem 4.2, which is one of the main results of this paper, is proved in Section 8. 4.2 The case when X is a ball of Rn When X is an open Euclidean ball of Rn , which is important from the practical point of view in medical imaging as it is relatively straightforward to place sources and detectors on a sphere, we are able to invert the weighted X-ray transform Pϑ0 f , f (x, v) := f (x) ∈ L2 (X, supv∈Sn−1 ϑ0 (v, x)dx) using the classical inverse X-ray transform (inverse Radon transform in dimension n = 2). In the next subsection we shall consider a larger class of domains X, which requires one to solve more complex weighted X-ray transforms. Up to rescaling, we assume X = Bn (0, 1), the ball in Rn centered at 0 of radius 1. Consider the X-ray transform P defined by P f (v, x) = Z τ+ (x,v) τ− (x,v) f (sv + x)ds for a.e. (v, x) ∈ Sn−1 × ∂X, (4.7) for f ∈ L2 (X) (we extend f by 0 outside X). We have the following Proposition 4.3. Proposition 4.3. When X = Bn (0, 1) we have Pϑ0 f (v, x) = P (̺f )(v, x), for a.e. (v, x) ∈ Sn−1 × ∂X, for f ∈ L2 (X, supv∈Sn−1 ϑ0 (v, x)dx) where ̺(y) := (1 − |y|2)− Proof of Proposition 4.3. It is easy to see that p 1 − q 2 ∓ t, τ± (tv + qv ⊥ , v) = ϑ0 (v, x) = (1 − q 2 − t2 ) − n−1 2 n−1 2 (4.8) , y ∈ X. (4.9) = (1 − |x|2 ) − n−1 2 , (4.10) for (t, q) ∈ R2 , t2 + q 2 ≤ 1 and for (v, v ⊥ ) ∈ Sn−1 × Sn−1 , v · v ⊥ = 0, where x = tv + qv ⊥ (we remind that ϑ0 is defined by (3.24)). Then Proposition 4.3 follows from the definition (3.25). Assume that (σ, k) satisfies condition (3.22) when n = 2 or (3.23) when n ≥ 2. Assume also that k(x, v, v ′) = k0 (x)g(v, v ′) for a.e. (x, v, v ′) ∈ X × Sn−1 × Sn−1 where g is a given continuous function on Sn−1 × Sn−1 , inf v∈Sn−1 g(v, v) > 0, and where k0 ∈ L∞ (X). Then from the decomposition of the angularly averaged albedo operator AS,W (Proposition 3.1) and from Theorems 3.2, 3.3, 3.4 and 3.5, and from Proposition 4.3 and methods of reconstruction of a function from its X-ray transform, it follows that (σ, k0 ) can be reconstructed from the asymptotic expansion in time of AS,W provided that σ = σ(x) and inf (x′ ,v′ )∈Γ− S(x′ , v ′) > 0 and inf (x,v)∈Γ+ W (x, v) > 0. In addition we have the following stability estimates. Theorem 4.4. Assume X = Bn (0, 1) and inf (x′ ,v′ )∈Γ− S(x′ , v ′ ) > 0 and inf (x,v)∈Γ+ W (x, v) > 0. Let (σ, k) and (σ̃, k̃) satisfy either condition (3.23) or (3.22). Assume that σ, σ̃ do not depend on the velocity variable (σ(x, v) = σ(x)) and suppσ∪suppσ̃ ⊆ Y , where Y ⊆ X is a convex open subset of Rn with C 1 boundary, and let M = max(kσkL∞ (Y ) , kσ̃kL∞ (Y ) ). Assume k(x, v, v ′ ) = k0 (x)g(v, v ′) and k̃(x, v, v ′) = k̃0 (x)g(v, v ′), g(v, v) > 0, for (x, v, v ′) ∈ X × Sn−1 × Sn−1 where g is an a priori known continuous function on Sn−1 × Sn−1 . 12 Then there exists C3 = C3 (S, W, X, Y, M) such that 1 1 2 . kσ − σ̃kH − 12 (Y ) ≤ C3 kσ − σ̃kL2 ∞ (Y ) kAS,W − ÃS,W kη,T (4.11) When n ≥ 2 and (σ, k) satisfies (3.23), there exists C4,1 = C4,1 (S, W, X, Y, Z, M, g) such that 1  2 k̺(k0 − k̃0 )kH − 21 (Z) ≤ C4,1 kk0 − k̃0 k∞ kk̃0 k∞ kAS,W − ÃS,W kη,T (4.12)  21 3−n + (τ − |z − z ′ |) 2 (Γ1 − Γ̃1 )(τ, z, z ′ ) . L∞ When n = 2 and (σ, k) satisfies (3.22), there exists C4,2 = C4,2 (S, W, X, M, g) such that 3  4 kk̃0 k∞ kAS,W − ÃS,W kη,T (4.13) k̺(k0 − k̃0 )kH − 21 (X) ≤ C4,2 kk0 − k̃0 k∞  41 p + τ 2 − |z − z ′ |2 (Γ1 − Γ̃1 )(τ, z, z ′ ) . L∞ Theorem 4.4 can be proved by mimicking the proof of Theorem 4.5 given below for a larger class of domains X. However we give a proof of estimate (4.13) in Section 6. Note that the left-hand side k̺(k0 − k̃0 )kH − 12 (Z) of (4.12) can be replaced by kk0 − k̃0 kH − 21 (Z) 1 since ̺−1 ∈ C ∞ (Z̄) and the operator f → ̺−1 f is bounded in H − 2 (Z) for any open convex subset Z (with C 1 boundary) of X which satisfies Z̄ ⊆ X. Under the assumptions of Theorem 4.4 and additional regularity assumptions on (σ, k) one obtains stability estimates similar to those given in Corollary 4.6 given below for a larger class of domains X. 4.3 Uniqueness and stability estimates for more general domains X Theorem 4.5. Assume that the open subset X of Rn is convex with a real analytic boundary and that inf (x′ ,v′ )∈Γ− S(x′ , v ′ ) > 0 and inf (x,v)∈Γ+ W (x, v) > 0. Let (σ, k) and (σ̃, k̃) satisfy condition (3.23). Assume also that σ, σ̃ do not depend on the velocity variable (σ(x, v) = σ(x)) and k(x, v, v ′ ) = k0 (x)g(x, v, v ′ ) and k̃(x, v, v ′) = k̃0 (x)g(x, v, v ′), g(x, v, v ′) > 0, for (x, v, v ′ ) ∈ X × Sn−1 × Sn−1 where g is an a priori known real analytic function on X × Sn−1 × Sn−1 and where suppk0 ∪ suppk̃0 ⊆ Z̄, (k0 , k̃0 ) ∈ L∞ (Z). Then estimate (4.11) still holds and there exists C4 = C4 (S, W, X, Y, Z, M, g) such that 1  2 kk0 − k̃0 kH − 21 (Z) ≤ C4 kk − k̃k∞ kk̃k∞ kAS,W − ÃS,W kη,T (4.14)  12 3−n , + (τ − |z − z ′ |) 2 (Γ1 − Γ̃1 )(τ, z, z ′ ) L∞ where M = max(kσkL∞ (Y ) , kσ̃kL∞ (Y ) ). Theorem 4.5 is proved in Section 6. Assume that X is convex with a real analytic boundary and that inf (x′ ,v′ )∈Γ− S(x′ , v ′) > 0 and inf (x,v)∈Γ+ W (x, v) > 0. Let Y and Z be open convex subsets of X, Z̄ ⊂ X, Z ⊆ Y ⊆ X, with a C 1 boundary. Let g be an a priori known real analytic function on X × Sn−1 × Sn−1 , g(x, v, v ′) > 0 for (x, v, v ′ ) ∈ X × Sn−1 × Sn−1 . Let r1 > 0, r2 > 0. Consider the class  n N := (σ, k) ∈ H 2 +r1 (Y ) × L∞ (Z × Sn−1 × Sn−1 ) | kσkH n2 +r1 (Y ) ≤ M1 , k = k0 g, suppk0 ⊆ Z̄, kk0 kH n2 +r2 (Z) ≤ M2 . 13 (4.15) Note that there exist a function D1 : N × (0, +∞) → (0, +∞) such that kσkL∞ (Y ) ≤ D1 (n, r1 )kσkH n2 +r1 (Y ) ≤ D1 (n, r1 )M1 , kk0 kL∞ (Z) ≤ D1 (n, r2 )kk0 kH n2 +r2 (Z) ≤ D1 (n, r2 )M2 , kkkL∞ (Z) ≤ kgkL∞ (Z) kk0 kL∞ (Z) ≤ D1 (n, r2 )M2 kgkL∞ (Z) , (4.16) for (σ, k) ∈ N. We also use the interpolation formula kf kH s (O) ≤ kf k s2 −s s2 −s1 H s1 (O) kf k s−s1 s2 −s1 H s2 (O) , (4.17) for s1 < s < s2 and for (O, s1 , s2 ) ∈ {(Y, − 12 , n2 + r1 ), (Z, − 12 , n2 + r2 )}. Using Theorem 4.5 and (4.16), and applying (4.17) on f = σ − σ̃ and f = k0 − k̃0 we obtain the following result. Corollary 4.6. Let (σ, k), (σ̃, k̃) ∈ N. Then, for − 12 ≤ s ≤ exists C5 = C5 (S, W, X, Y, M1 , r1 , s) such that n 2 + r1 and for 0 < r < r1 , there κ κ 2 kσ − σ̃kH s (Y ) ≤ C5 kσ − σ̃kL2 ∞ (Y ) kAS,W − ÃS,W kη,T , (4.18) κ′ 2−κ′ kσ − σ̃kH n2 +r (Y ) ≤ C6 kAS,W − ÃS,W kη,T ,  ′ n+2(r1 −s) 2(r1 −r) , n+1+2r1 n+1+2r1 2  ′ (4.19) κ′ where (κ, κ ) = and C6 = C52−κ D1 (n, r) 2−κ′ (D1 (n, r) is defined by (4.16)). In addition, there exists C7 = C7 (S, W, X, Y, Z, g, M1, r1 , M2 , r2 , s) such that  κ (4.20) kk0 − k̃0 kH s (Z) ≤ C7 kk0 − k̃0 kL2 ∞ (Z) kAS,W − ÃS,W kη,T  κ2 3−n + (τ − |z − z ′ |) 2 (Γ1 − Γ̃1 )(τ, z, z ′ ) , L∞ kk0 − k̃0 kH for 2 − 12 n +r 2 (Z) ≤ s ≤ n 2  3−n ≤ C8 kAS,W − ÃS,W kη,T + (τ − |z − z ′ |) 2 (Γ1 − Γ̃1 )(τ, z, z ′ ) + r2 and 0 < r < r2 , where (κ, κ′ ) = κ′ ′ C72−κ D1 (n, r) 2−κ′ (D1 (n, r) is defined by (4.16)).  n+2(r2 −s) 2(r2 −r) , n+1+2r2 n+1+2r2  L∞  κ′ 2−κ′ , (4.21) and C8 = Remark 4.7. (i.) Theorem 4.5 and Corollary 4.6 remain valid when: X is only assumed to be convex with C 2 boundary; the weight ϑo defined by (3.24) (resp. the function g which appears in the assumptions of Theorem 4.5 and Corollary 4.6) is sufficiently close (in the C 2 norm) to an analytic weight θ0,a on the vicinity of Z̄ × Sn−1 (resp. an analytic function ga on the vicinity of Z̄ × Sn−1 × Sn−1 ); see proof of Theorem 4.5 and [12, Theorem 2.3]. (ii.) When n = 3 then under hypothesis (3.23), we have ′ (τ − |z − z |) 3−n 2 ′ (Γ1 − Γ̃1 )(τ, z, z ) L∞ = ∞ X (Am,S,W − Ãm,S,W ) m=1 . L(L1 ((0,η)×∂X),L∞ ((0,T )×∂X)) P 1 where the distributional kernel of the bounded operator +∞ m=1 (Am,S,W −Ãm,S,W ) from L ((0, η)× ∂X) to L1 ((0, T ) × ∂X) is given by Γ1 − Γ̃1 . Therefore when n = 3 and under condition (3.23), the right-hand side of the stability estimates (4.14) and (4.20) can be expressed with operator norms only (instead of using a norm on the distributional kernel of the multiple scattering). 14 5 Proof of Theorems 3.4, 3.5 Proof of Theorem 3.5. For the sake of simplicity and without loss of generality we assume v0 = (1, 0, . . . , 0). Assume that condition (3.23) is satisfied. For n ≥ 2 consider the following open subset of (0, +∞) × Sn−1 × Sn−1 n−1 D := {(s, v, v ′) ∈ (0, +∞) × Sx,+ × Sxn−1 ′ ,− | s ∈ (0, τ− (x, v))}. (5.1) 0 Then we introduce the bounded function Ψn on D defined by Ψn (s, v, v ′) = 2n−2 W (x, v)(ν(x) · v)E(x, x − sv, x′0 )k(x − sv, v ′ , v)S(x′0 , v ′)|ν(x′0 ) · v ′ |, (5.2) for (s, v, v ′) ∈ D. Note that from convexity of X it follows that τ± is continuous on Γ∓ and sv, x′0 ) − Rs σ(x−pv,v)dp− R |x−x′0 −sv| 0 E(x, x − =e τ− (x, v). Under (3.23) we obtain that 0 x−x′ −sv x−x′ −sv 0 0 σ(x−sv−p |x−x′0 −sv| , |x−x′0 −sv| )dp n−1 for v ∈ Sx,+ and 0 < s < n−1 Ψn (s, v, v ′) = 0 for (s, v, v ′) ∈ (0, +∞) × Sx,+ × Sxn−1 ′ ,− such that x − sv 6∈ Z̄, 0 and the function Ψn is continuous at any point (s, v, v ′) ∈ D such that x − sv ∈ Z. (5.3) We first prove (3.26) for n = 2. Let τ > t0 . From (5.2), (3.10), it follows that Z π−α0   1 ′ χ(0,τ− (x,v)) (s)Ψ2 (s, v, v ′) v=(cos Ω,sin Ω) dΩ γ1 (τ, x, x0 ) = t (1,0)−sv τ − t0 cos(Ω) −α0 v ′ = 0 τ −s s= τ 2 −t2 0 2(τ −t0 cos(Ω)) = γ1,1 (τ, x, x′0 ) + γ1,2 (τ, x, x′0 ), (5.4) n−1 where Sx,+ = {(cos Ω, sin Ω) | − α0 < Ω < π − α0 } (0 < α0 < π) and γ1,1 (τ, x, x′0 ) γ1,2 (τ, x, x′0 ) = = Z Z π  χ(0,π−α0 ) (Ω)  χ(0,τ− (x,v)) (s)Ψ2 (s, v, v ′) τ − t0 cos(Ω) 0 0 −π v=(cos Ω,sin Ω) t (1,0)−sv v ′ = 0 τ −s τ 2 −t2 0 s= 2(τ −t0 cos(Ω))  χ(−α0 ,0) (Ω)  χ(0,τ− (x,v)) (s)Ψ2 (s, v, v ′) τ − t0 cos(Ω) v=(cos Ω,sin Ω) t (1,0)−sv v ′ = 0 τ −s τ 2 −t2 0 s= 2(τ −t0 cos(Ω)) dΩ, (5.5) dΩ. (5.6) We shall prove that √ τ− t0 γ1,i(τ, x, x′0 ) as τ → W (x, v0 )S(x′0 , v0 )(ν(x) · v0 )|ν(x′0 ) · v0 |E(x′0 , x) √ → 2t0 t+ 0, Z 0 t0 k(x − sv0 , v0 , v0 ) p ds, s(t0 − s) (5.7) for i = 1, 2. Then adding (5.7) for i = 1 and i = 2, we obtain (3.26). We only prove (5.7) for i = 1 since the proof for i = 2 is similar. Let τ > t0 . Using the change of variables τ 2 −t20 0 − τ −t , we obtain s = 2(τ −t0 cos(Ω)) 2 γ1,1 (τ, x, x′0 ) =p 1 τ 2 − t20 Z 0 t0 0 )Ψ2 (s, v(s, τ ), v ′(s, τ )) χ(0,τ− (x,v(s,τ ))) (s + τ −t 2 p χ(0,π−α0 ) (Ω(s, τ )) dτ, s(t0 − s) (5.8) 15 where τ− v(s, τ ) = (cos Ω(s, τ ), sin Ω(s, τ )), Ω(s, τ ) = arccos
0 t0 (1, 0) − s + τ −t v(s, τ ) ′ 2 . v (s, τ ) = τ +t0 − s 2 τ 2 −t20 2s+τ −t0 t0 ! , (5.9) Let s ∈ (0, t0 ). From (5.9), it follows that ′ + v(s, τ ) → (1, 0) as τ → t+ 0 , v (s, τ ) → (1, 0) as τ → t0 . (5.10) Note that using the definition of v0 and using the assumption x′0 + ε(x − x′0 ) ∈ X for some , s ∈ (0, t0 ), is ε ∈ (0, 1) we obtain t0 = τ− (x, v0 ). Note also that the function s 7→ √ 1 s(t0 −s) integrable in (0, t0 ). Therefore, using (5.3), the boundedness of Ψ2 on D and the Lebesgue dominated convergence theorem, we obtain (5.7). This proves (3.26) when n = 2. Let n ≥ 3 and prove (3.29). From (5.2) and (3.10), it follows that Z (τ − t0 v0 · v)n−3 ′ χ(0,+∞) (ν(x) · v)Ψn (s, v, v ′) v′ = t0 v0 −sv dv, (5.11) γ1 (τ, x, x0 ) = 2n−4 τ −s |t v − τ v| n−1 0 0 S τ 2 −t2 s= 0 2(τ −t0 v·v0 ) for τ > |x − x′0 |. Let Φ(Ω, ω) = (sin Ω, cos(Ω)ω1 , . . . , cos(Ω)ωn−1 ) for Ω ∈ (− π2 , π2 ) and ω = (ω1 , . . . , ωn−1 ) ∈ n−2 S . Using spherical coordinates we obtain Z π/2 (τ − t0 sin(Ω))n−3 ′ γ1 (τ, x, x0 ) = (5.12) cos(Ω)n−2 2 (t0 + τ 2 − 2t0 τ sin(Ω))n−2 −π/2 Z χ(0,+∞) (ν(x) · Φ(Ω, ω))Ψn (s, Φ(Ω, ω), v ′) v′ = t0 v0 −sΦ(Ω,ω) dωdΩ, Sn−2 s= for τ > t0 . Performing the change of variables “r = the right-hand side of (5.12), we obtain γ1 (τ, x, x′0 ) τ 2 −t20 2(τ −t0 sin(Ω)) p t0 − τ −t0 ” 2 τ −s τ 2 −t2 0 2(τ −t0 sin(Ω)) on the first integral on n−3 r(t0 − r) = − (5.13) τ −t0 0 − r)n−2 ( 2 + r)n−2 ( τ +t 0 2   Z τ − t0 ′ χ(Φ(Ω, ω))Ψn (r + dωdr. , Φ(Ω, ω), v ) (τ 2 −t2 0) Ω=arcsin(t−1 (τ − 2 τ −t0 )) 0 Sn−2 2(r+ ) 22−n t02−n (τ 2 t20 ) n−3 2 Z τ −t s=r+ 2 0 t v −sΦ(Ω,ω) v ′ = 0 0 τ −s 2 Therefore using (5.13), (5.3) and (5.2) and using Lebesgue dominated convergence theorem, we obtain (3.29). This concludes the proof of Theorem 3.5. Proof of Theorem 3.4. For the sake of simplicity and without loss of generality we assume v0 = (1, 0, . . . , 0). Assume that condition (3.22) is satisfied. We consider the measurable function Ψn defined by (5.2) for all (s, v, v ′) ∈ D where D is defined by (5.1). Under (3.22) we obtain that the function Ψn is continuous at any point (s, v, v ′) ∈ D (5.14) (i.e. for any (s, v, v ′) ∈ (0, +∞) × Sxn−1 × Sxn−1 , x − sv ∈ X), ′ 0 16 The proof of (3.26) under condition (3.22) is actually similar to the proof of (3.26) under condition (3.23). Note that (5.4)–(5.6) still hold so that we have to prove that (5.7) still holds for i = 1, 2. Again we only sketch the proof of (5.7) for i = 1. Note also that (5.8)–(5.10) still hold. Then using (5.8)–(5.10), (5.14) and (5.2) and using Lebesgue dominated convergence theorem, we obtain (5.7) for i = 1. This proves (3.26). Let n ≥ 3. Formula (5.12) still holds. Now assume that n = 3. We shall prove (3.27). Let ln(t20 +τ 2 −2t0 τ sin(Ω))−ln((τ −t0 )2 ) τ > t0 . Using the change of variables “ε = ”, the equality (5.12) ln((τ +t0 )2 )−ln((τ −t0 )2 ) gives   +t0 ln ττ −t 0 γ1 (τ, x, x′0 ) = (γ1,1 (τ, x, x′0 ) + γ1,2(τ, x, x′0 )) (5.15) 2t0 τ where Z 1Z 2 ′ γ1,1 (τ, x, x0 ) = χ(0,+∞) (ν(x) · Φ(Ω(τ, ε), ω))Ψ3 (s(τ, ε), Φ(Ω(τ, ε), ω), v ′(τ, ω, ε))dωdε 0 γ1,2 (τ, x, x′0 ) = Z 1Z 1 2 S1 (5.16) S1 χ(0,+∞) (ν(x) · Φ(Ω(τ, ε), ω))Ψ3(s(τ, ε), Φ(Ω(τ, ε), ω), v ′(τ, ω, ε))dωdε (5.17) and τ 2 + t20 − (τ − t0 )2(1−ε) (τ + t0 )2ε Ω(τ, ε) := arcsin 2t0 τ 2 2 τ − t0 , s(τ, ε) := 2(τ − t0 sin(Ω(τ, ε))) t0 v0 − s(τ, ε)Φ(Ω(τ, ε), ω) v ′ (τ, ω, ε) := , τ − s(τ, ε)   , (5.18) (5.19) (5.20) for 0 < ε < 1, ω ∈ S1 . We shall give some properties of Ω(τ, ε), s(τ, ε) and v ′ (τ, ω, ε) for 0 < ε < 1 and ω ∈ Sn−1 . From (5.18), it follows that Ω(τ, ε) → π , as τ → t+ 0 , for all ε ∈ (0, 1). 2 (5.21) From (5.18) and (5.19), it follows that τ (t0 + τ )(τ − t0 )2ε−1 τ (t0 + τ ) s(τ, ε) = , s(τ, ε) = 2ε−1 2ε (t0 + τ )(τ − t0 ) + (τ + t0 ) t0 + τ + (τ − t0 )1−2ε (τ + t0 )2ε (5.22) for ε ∈ (0, 1). From (5.22) it follows that − + s(τ, ε) → 0+ as τ → t+ 0 , when ε ∈ (1/2, 1), s(τ, ε) → t0 as τ → t0 , when ε ∈ (0, 1/2). (5.23) In addition, from (5.18)–(5.20), it follows that  −τ (τ + t0 )2 (τ − t0 )2ε + 2t20 τ (τ + t0 )2ε + τ (τ + t0 )1+2ε (τ − t0 ) ′ v (τ, ω, ε) = , 2t0 τ 2 (τ + t0 )2ε q −(τ − t0 )4ε (τ + t0 )2 − (τ − t0 )2 (τ + t0 )4ε + 2(τ 2 + t20 )(τ − t0 )2ε (τ + t0 )2ε  (τ + t0 + (τ − t0 )1−2ε (τ + t0 )2ε ) ω (5.24) × 2t0 τ (τ + t0 )2ε 17 for 0 < ε < 1. Therefore v ′ (τ, ω, ε) → v0 = (1, 0, 0) as τ → t+ 0, (5.25) for 0 < ε < 1. Using (5.16), (5.21), (5.23), (5.25) and Lebesgue dominated convergence theorem, we obtain Z 1Z 2 ′ lim− Ψ3 (s, v0 , v0 )dωdε γ1,1 (τ, x, x0 ) →+ S1 s→t0 2πE(x′0 , x)W (x, v0 )S(x′0 , v0 )(ν(x′0 )v0 )(ν(x)v0 )k(x′0 , v0 , v0 ) τ →t0 0 = (5.26) (we also used (5.2)). Similarly, using (5.16), (5.21), (5.23), (5.25) and Lebesgue dominated convergence theorem, we obtain γ1,2(τ, x, x′0 ) → 2πE(x′0 , x)W (x, v0 )S(x′0 , v0 )(ν(x′0 )v0 )(ν(x)v0 )k(x, v0 , v0 ), as τ → t+ 0. (5.27) Statement (3.27) follows from (5.15) and (5.26)–(5.27). Let n ≥ 4. We shall prove (3.28). From (5.12) it follows that γ1 (τ, x, x′0 ) = γ1,1 (τ, x, x′0 ) + γ1,2 (τ, x, x′0 ), (5.28) where γ1,1 (τ, x, x′0 ) × Z Sn−2 Z π 2 − π2 (τ − t0 sin(Ω))n−3 χ(− π ,arcsin( t0 )) (Ω) cos(Ω)n−2 2 2 τ (t + τ 2 − 2t0 τ sin(Ω))n−2 = Sn−2 (5.29) 0 χ(0,+∞) (Φ(Ω, ω) · ν(x))χ(0,τ− (x,Φ(Ω,ω)))) (s)Ψn (s, Φ(Ω, ω), v ′) v′ = t0 v0 −sΦ(Ω,ω) dωdΩ, s= γ1,2 (τ, x, x′0 ) × = Z Z π 2 arcsin( t0 τ ) cos(Ω)n−2 τ −s τ 2 −t2 0 2(τ −t0 sin(Ω)) (τ − t0 sin(Ω))n−3 (t20 + τ 2 − 2t0 τ sin(Ω))n−2 (5.30) χ(0,+∞) (ν(x) · Φ(Ω, ω))χ(0,τ− (x,Φ(Ω,ω)))) (s)Ψn (s, Φ(Ω, ω), v ′) v′ = t0 v0 −sΦ(Ω,ω) dωdΩ. s= τ −s τ 2 −t2 0 2(τ −t0 sin(Ω)) First we study γ1,1 . Note that t20 for − π2 ≤ Ω1 < Ω2 ≤ π 2 τ − t0 sin(Ω2 ) τ − t0 sin(Ω1 ) < 2 , 2 + τ − 2t0 τ sin(Ω1 ) t0 + τ 2 − 2t0 τ sin(Ω2 ) (5.31) and for τ > t0 . Therefore using also the estimate cos(Ω) ≤ 1 we obtain (τ − t0 sin(Ω))n−3 (t20 + τ 2 − 2t0 τ sin(Ω))n−2  n−3 cos(Ω)2 τ − t0 sin(Ω) C02 n−4 ≤ cos(Ω) , ≤ t20 + τ 2 − 2t0 τ sin(Ω) t20 + τ 2 − 2t0 τ sin(Ω) 2t0 τ n−2 cos(Ω)n−2 for Ω ∈ (− π2 , π2 ) and sin(Ω) ≤ tτ0 (we used (5.31) with “Ω1 ”= Ω and “Ω2 ”= estimate t20 + τ 2 − 2t0 τ sin(Ω) ≥ 2t0 τ (1 − sin(Ω))), where sin2 (ϕ) cos2 (Ω) = sup . ϕ∈(0,2π) 1 − cos(ϕ) Ω∈(− 3π , π ) 1 − sin(Ω) C0 := sup 2 18 2 t0 , τ (5.32) and we used the (5.33) Using (5.29), (5.32), (5.2) and Lebesgue dominated convergence theorem, we obtain γ1,1 (τ, x, x′0 ) −→ τ →t+ 0 22−n t1−n 0 Z π 2 cos(Ω)n−2 1 − sin(Ω) −π 2 Z Sn−2 χ(ν(x) · Φ(Ω, ω)) lim+ Ψn (s, Φ(Ω, ω), v0 )dωdΩ s→0 1 lim Ψn (s, v, v0 )dv s→0+ 1 − v · v 0 Sn−1 x,+ Z 1 1−n ′ ′ ′ W (x, v)(ν(x) · v)k(x, v0 , v)dv. = t0 E(x0 , x)S(x0 , v0 )(ν(x0 ) · v0 ) 1 − v · v0 Sn−1 x,+ = 22−n t1−n 0 Z (5.34) Now we shall study γ1,2 defined by (5.30). Note that using the convexity of X we obtain x−sv−x′ n−1 v = |x−sv−x0′ | ∈ Sxn−1 ′ ,− whenever v ∈ Sx,+ and s ∈ (0, τ− (x, v)). Therefore using the change ′ 0 0 (τ 2 −t2 )2 cos(Ω′ ) 2τ t −(τ 2 +t2 ) sin(Ω) dΩ 0 of variables “sin(Ω′ ) = τ 20+t2 −2t0 τ0 sin(Ω) ”, Ω ∈ (− π2 , π2 ) (“cos(Ω) dΩ ′ = (τ 2 +t2 −2t τ sin(Ω′ ))2 ”), we 0 0 0 obtain Z arcsin( t0 ) τ (τ − t0 sin(Ω′ ))n−3 ′ cos(Ω′ )n−2 2 γ1,2 (τ, x, x0 ) := (5.35) 2 − 2t τ sin(Ω′ ))n−2 (t + τ 0 −π 0 2 Z χ(0,τ− (x,Φ(Ω(τ,Ω′ ),ω))) (s(τ, Ω′ ))Ψn (s(τ, Ω′ ), Φ(Ω(τ, Ω′ ), ω), v ′)v′ =Φ(Ω′ ,−ω) dωdΩ′ , ω∈Sn−2 ν(x)·Φ(Ω(τ,Ω′ ),ω)>0 −ν(x′ )·Φ(Ω′ ,−ω)>0 0 where ′ Ω(τ, Ω ) = arcsin   τ 2 + t20 − 2t0 τ sin(Ω′ ) 2τ t0 − (τ 2 + t20 ) sin(Ω′ ) ′ , s(τ, Ω ) = , τ 2 + t20 − 2t0 τ sin(Ω′ ) 2(τ − t0 sin(Ω′ )) (5.36) for Ω′ ∈ (− π2 , π2 ). Note that Ω(τ, Ω′ ) s(τ, Ω′ ) →+ π , Φ(Ω(τ, Ω′ ), ω) →+ v0 , 2 τ →t0 (5.37) → t0 , (5.38) τ →t0 τ →t+ 0 for (Ω′ , ω) ∈ (− π2 , π2 )×Sn−2 . Note also that from (5.37), it follows that at fixed v ′ = Φ(Ω′ , −ω) ∈ ′ ′ ′ Sxn−1 ′ ,− the condition χ(0,τ− (x,Φ(Ω(τ,Ω′ ),ω))) (s(τ, Ω )) = 1 (which is equivalent to x−s(τ, Ω )Φ(Ω(τ, Ω ), ω) ∈ 0 n−1 X due to convexity of X) for Φ(Ω(τ, Ω′ ), ω) ∈ Sx,+ is satisfied when τ − t0 > 0 is sufficiently small. Therefore from (5.32) (with “Ω” replaced by “Ω′ ”) and from (5.35) and Lebesgue dominated convergence theorem we obtain Z π 2 cos(Ω′ )n−2 1 − sin(Ω′ ) Z χ(0,+∞) (−ν(x′0 ) · Φ(Ω′ , −ω)) τ →t0 Sn−2 Z 1 ′ ′ 2−n 1−n × lim− Ψn (s, v0 , Φ(Ω , −ω))dωdΩ = 2 t0 lim Ψn (s, v0 , v ′ )dv ′ n−1 1 − v ′ · v0 s→t− s→t0 S ′ 0 x0 ,− Z 1 S(x′0 , v ′ )|ν(x′0 ) · v ′ |k(x, v ′ , v0 )dv ′ . (5.39) = t01−n E(x′0 , x)W (x, v0 )(ν(x) · v0 ) n−1 1 − v ′ · v0 S ′ γ1,2 (τ, x, x′0 ) −→+ 22−n t1−n 0 − π2 x ,− 0 Statement (3.28) follows from (5.28), (5.34) and (5.39). Theorem 3.4 is proved. 19 6 Proof of Theorems 4.1, 4.2, 4.5 and Theorem 4.4 (4.13) Proof of Theorem 4.2. We now prove (4.5). Let x′0 ∈ ∂X. For ε = (ε1 , ε2 ) ∈ (0, +∞)2 and ε3 ∈ (0, +∞) let (fε1 , gε2 ) ∈ C 1 (∂X) × C 1 (R) satisfy gε2 ≥ 0, fε1 ≥ 0, suppgε2 ⊆ (0, min(ε2 , η)), suppf ⊆ {x′ ∈ ∂X | |x′ − x′0 | < ε1 }, Z η ε1 Z ′ ′ gε2 (t )dt = 1, fε1 (x′ )dµ(x′ ) = 1, 0 (6.1) (6.2) (6.3) ∂X for ε = (ε1 , ε2 ) ∈ (0, +∞)2. Therefore φε := gε2 fε1 is an approximation of the delta function at (0, x′0 ) ∈ R × ∂X for ε := (ε1 , ε2 ) ∈ (0, +∞)2. Let ψε3 ∈ L∞ ((0, T ) × ∂X) be defined by ψε3 (t, x) = χ(−ε3 ,ε3) (t − |x − x′0 |)(2χ(0,+∞) ((E − Ẽ)(x, x′0 )) − 1), (t, x) ∈ (0, T ) × ∂X, for ε3 > 0. From (3.8) and (3.12) it follows that Z ψε3 (t, x)(AS,W − ÃS,W )φε (t, x)dtdµ(x) = I0 (ψε3 , φε ) (0,T )×∂X Z + ψε3 (t, x)φε (t′ , x′ )(Γ1 − Γ̃1 )(t − t′ , x, x′ )dtdµ(x)dt′ dµ(x′ ), (6.4) (6.5) (0,T )×∂X×(0,η)×∂X for ε = (ε1 , ε2) ∈ (0, +∞) and ε3 ∈ (0, +∞), where Z ′ ′ ′ ′ E(x, x ) − Ẽ(x, x ) I0 (ψε3 , φε ) = ψε3 (t, x)φε (t − |x − x |, x ) (0,T )t ×∂Xx ×∂Xx′ |x − x′ |n−1 ′ |x−x |<t × [W (x, v)S(x′ , v)(ν(x) · v)|ν(x′ ) · v|]v= x−x′ dtdµ(x)dµ(x′). |x−x′ | (6.6) From (3.16), (3.17), (3.20) and (3.21) it follows that (τ − |x − x′ |) 3−n 2 (Γ1 − Γ̃1 )(τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X). (6.7) Combining (6.5) and the equality kφε kL1 ((0,η)×∂X) = 1 and the estimate kψε3 kL∞ ((0,T )×∂X) ≤ 1 and (6.7) we obtain I0 (ψε3 , φε ) ≤ kAS,W − ÃS,W kη,T + C∆1 (ψε3 , φε ), (6.8) 3−n for ε = (ε1 , ε2 ) ∈ (0, +∞) and ε3 ∈ (0, +∞), where C = k(τ − |x − x′ |) 2 (Γ1 − Γ̃1 )(τ, x, x′ ) kL∞ ((0,T )×∂Xx ×∂Xx′ ) and Z n−3 ∆1 (ψε3 , φε ) = ψε3 (t, x)φε (t′ , x′ )(t − t′ − |x − x′ |) 2 dtdµ(x)dt′ dµ(x′ ). (6.9) (0,T )t ×∂Xx ×(0,η) ′ ×∂X ′ t x |x−x′ |<t−t′ Note that the function Φ1,ε3 : [0, η) × ∂X → R defined by Z n−3 ′ ′ Φ1,ε3 (t , x ) := ψε3 (t, x)(t − t′ − |x − x′ |) 2 dtdµ(x), (t′ , x′ ) ∈ [0, η) × ∂X, (0,T )t ×∂Xx |x−x′ |<t−t′ (6.10) is continuous Ron [0, η) × ∂X for ε3 ∈ (0, +∞). Therefore from (6.1)–(6.3) and the equality ∆1 (ψε3 , φε ) = (0,η)×∂X φε (t′ , x′ )Φ1,ε3 (t′ , x′ )dt′ dµ(x′ ) it follows that lim lim lim ∆1 (ψε3 , φε ) = lim+ Φ1,ε3 (0, x′0 ) = 0 ε3 →0+ ε2 →0+ ε1 →0+ ε3 →0 20 (6.11) (we also used (6.10), (6.4) and the Lebesgue dominated convergence theorem to prove that limε3 →0+ Φ1,ε3 (0, x′0 ) = 0). Note that under condition (3.23) the function Φ0,ε2 ,ε3 : ∂X → R defined by Z E(x, x′ ) − Ẽ(x, x′ ) ′ ψε3 (t, x)gε2 (t − |x − x′ |) Φ0,ε2 ,ε3 (x ) = (0,T )t ×∂Xx |x − x′ |n−1 ′ |x−x |<t × [W (x, v)S(x′ , v)(ν(x) · v)|ν(x′ ) · v|]v= x−x′ |x−x′ | dtdµ(x), is continuous on ∂X, for (ε2 , ε3 ) ∈ (0, +∞)2 . Therefore from the equality I0 (ψε3 , φε ) = ×fε1 (x′ )dµ(x′ ) (see (6.6)) it follows that (6.12) Φ0,ε2 ,ε3 (x′ ) R ∂X lim I0 (ψε3 , φε ) = Φ0,ε2 ,ε3 (x′0 ), for (ε2 , ε3 ) ∈ (0, +∞)2. ε1 →0+ (6.13) Therefore using the Lebesgue dominated convergence theorem and (6.12) we obtain lim lim E(x, x′0 ) − Ẽ(x, x′0 ) Z lim I0 (ψε3 , φε ) = ε3 →0+ ε2 →0+ ε1 →0+ |x − x′0 |n−1 ∂Xx [W (x, v)S(x′0 , v)(ν(x) · v)|ν(x′0 ) · v|] x−x′ v= |x−x0′ | (6.14) Combining (6.14), (6.11) and (6.8) we obtain the formula (4.1). Using (4.1) and the estimates inf (x′ ,v′ )∈Γ− S(x′ , v ′ ) > 0 and inf (x,v)∈Γ+ W (x, v) > 0 and the change of variables x = x′0 + ν(x)·v0 τ+ (x′0 , v0 )v0 ( |x−x ′ |n−1 dµ(x) = dv0 ) we obtain (4.5) where the constant C1 which appears on the 0
−1 right-hand side of (4.5) is given by C1 = inf (x′ ,v′ )∈Γ− S(x′ , v ′) inf (x,v)∈Γ+ W (x, v) . We now prove (4.6). Let x ∈ ∂X be such that px′0 + (1 − p)x ∈ Z for some p ∈ (0, 1). We x−x′ set t0 = |x − x′0 | and v0 = |x−x0′ | . From (3.16), (3.17), (3.20) and (3.21) it follows that 0 (τ − |x − x′0 |) 3−n 2 ′ +k(s − |z − z |) |γ1 − γ̃1 |(τ, x, x′0 ) ≤ (τ − |x − x′0 |) 3−n 2 3−n 2 ′ |Γ2 − Γ̃2 |(τ, z, z ′ ) (6.15) (Γ1 − Γ̃1 )(s, z, z )kL∞ ((0,T )s ×∂Xz ×∂Xz′ ) , 3−n for τ > |x − x′0 |. From (3.17), (3.20)–(3.21) it turns out that limτ →|x−x′0|+ (τ − |x − x′0 |) 2 |Γ2 − Γ̃2 |(τ, z, z ′ ) = 0. Therefore applying (3.26) and (3.29) on the left-hand side of (6.15) we obtain 2 1−n 2 × Z |x − x′0 |− t0 e− R t0 0 n−1 2 Cn S(x′0 , v0 )W (x, v0 )|ν(x′0 ) · v0 |(ν(x) · v0 ) σ(x′0 +sv0 ,v0 )ds k(x − pv0 , v0 , v0 ) − e− p 0 ≤ k(s − |z − z ′ |) 3−n 2 n−1 2 R t0 (t0 − p) 0 σ̃(x′0 +sv0 ,v0 )ds n−1 2 k̃(x − pv0 , v0 , v0 ) (Γ1 − Γ̃1 )(s, z, z ′ ))kL∞ ((0,T )s ×∂Xz ×∂Xz′ ) , dp (6.16) where Cn = 2 if n = 2 and Cn = Voln−2 (Sn−2 ) if n ≥ 3. Then note that CX := inf x1 ∈∂X, z∈Z̄ ν(x1 )· x1 −z > 0 since X is a bounded convex subset of Rn with C 1 boundary and Z̄ ⊂ X. Therefore |x1 −z| (4.6) follows from (6.16) where the constant C2 which appears on the right-hand side of (4.6) is given by C2 = n−1 n−1 2 diam(X) 2 2 inf ′ ′ Cn CX (x′ ,v ′ )∈Γ− S(x ,v ) inf (x1 ,v1 )∈Γ+ 2 W (x1 ,v1 ) . Theorem 4.2 is proved. Proof of Theorem 4.1. We prove (4.1). Let x′0 ∈ ∂X. For ε = (ε1 , ε2 ) ∈ (0, +∞)2 and ε3 ∈ (0, +∞) let (fε1 , gε2 ) ∈ C 1 (∂X) × C 1 (R) satisfy (6.1)–(6.3) and ψε3 be defined by (6.4). First note that (6.5)–(6.6) still hold. From Theorems 3.2 and 3.3 it follows that 7 3 |x − x′ |n− 4 (τ − |x − x′ |) 4 (Γ1 − Γ̃1 )(τ, x, x′ ) ∈ L∞ ((0, T ) × ∂X × ∂X). 21 (6.17) 0 dµ(x). Combining (6.5), (6.17) and the equality kφε kL1 ((0,η)×∂X) and the estimate kψε3 kL∞ ((0,T )×∂X) ≤ 1 we obtain I0 (ψε3 , φε ) ≤ kAS,W − ÃS,W kη,T + C∆2 (ψε3 , φε ), (6.18) 3 7 for ε = (ε1 , ε2 ) ∈ (0, +∞), ε3 ∈ (0, +∞), where C = k|x − x′ |n− 4 (τ − |x − x′ |) 4 (Γ1 − Γ̃1 )(τ, x, x′ ) kL∞ ((0,T )×∂Xx ×∂Xx′ and Z ψε3 (t, x)φε (t′ , x′ ) ′ ′ ∆2 (ψε3 , φε ) = (6.19) 3 dtdµ(x)dt dµ(x ). 7 (0,T )t ×∂Xx ×(0,η)t′ ×∂Xx′ ′ |n− 4 (t − t′ − |x − x′ |) 4 |x − x ′ ′ |x−x |<t−t Note that the function Φ1,ε3 : [0, T ) × ∂X → R defined by Z ψε3 (t, x) ′ ′ ′ ′ Φ3,ε3 (t , x ) := (6.20) 3 dtdµ(x), (t , x ) ∈ [0, η) × ∂X, 7 n− (0,T )t ×∂Xx ′ ′ ′ 4 4 (t − t − |x − x |) |x − x | ′ ′ |x−x |<t−t is Therefore using (6.19), (6.1)–(6.3) and using the equality ∆2 (ψε3 , φε ) = R continuous on [0,′ η)×∂X. ′ ′ ′ Φ (t , x )φ (t , x )dt′ dµ(x′ ) we obtain ε (0,η) ′ ×∂X ′ 3,ε3 t x lim+ lim+ lim+ ∆2 (ψε3 , φε ) = lim+ Φ3,ε3 (0, x′0 ) = 0. ε3 →0 ε2 →0 (6.21) ε3 →0 ε1 →0 (we also used (6.20), (6.4) and Lebesgue dominated convergence theorem to prove limε3 →0+ Φ3,ε3 (0, x′0 ) = 0). Note that under condition (3.22) the function Φ4,ε3 ,ε2 : ∂X → R defined by Z E∂X×∂X (x, x′ ) − Ẽ∂X×∂X (x, x′ ) ′ ψε3 (t, x)gε2 (t − |x − x′ |) Φ4,ε3 ,ε2 (x ) = (0,T )t ×∂Xx |x − x′ |n−1 ′ |x−x |<t × [W (x, v)S(x′, v)(ν(x) · v)|ν(x′ ) · v|]v= x−x′ |x−x′ | dtdµ(x), (6.22) is continuous on ∂X. Therefore from (6.1)–(6.3) and the equality I0 (ψε3 , φε ) = (see (6.6)) it follows that R ∂X fε1 (x′ )Φ4,ε3 ,ε2 (x′ )dµ(x′ ) lim I0 (ψε3 , φε ) = Φ4,ε3 ,ε2 (x′0 ), for (ε2 , ε3 ) ∈ (0, +∞). (6.23) ε1 →0+ Then using Lebesgue dominated convergence theorem and (6.22) we obtain lim lim lim I0 (ψε3 , φε ) = ε3 →0+ ε2 →0+ ε1 →0+ Z E∂X×∂X (x, x′0 ) − Ẽ∂X×∂X (x, x′0 ) |x − x′0 |n−1 × [W (x, v)S(x′0 , v)(ν(x) · v)|ν(x′0 ) · v|] (6.24) ∂Xx x−x′ v= |x−x′0 | dµ(x). 0 Combining (6.24), (6.21) and (6.18) we obtain (4.1). We prove (4.2)–(4.4). Let x ∈ ∂X be such that px′0 + (1 − p)x ∈ X for some p ∈ (0, 1). Let βn : {(τ, z, z ′ ) ∈ (0, T ) × ∂X × ∂X | z 6= z ′ , τ > |z − z ′ |} → R be defined by  p 2 − |z − z ′ |2 , if n = 2,   ττ |z−z ′| ” “ (6.25) βn (τ, z, z ′ ) = τ +|z−z ′ | , if n = 3, ln τ −|z−z ′ |   τ |z − z ′ |n−2 , if n ≥ 4, for (τ, z, z ′ ) ∈ (0, T ) × ∂X × ∂X, z 6= z ′ , τ > |z − z ′ |. From (3.13)–(3.15) and (3.17)–(3.19) it follows that βn (τ, x, x′0 )|(γ1 − γ̃1 )(τ, x, x′0 )| ≤ βn (τ, x, x′0 )|(Γ2 − Γ̃2 )(τ, z, z ′ )| (6.26) ′ ′ +kβn (s, z, z )(Γ1 − Γ̃1 )(s, z, z )kL∞ ((0,T )s ×∂Xz ×∂Xz′ ) . 22 From (6.25) and (3.17)–(3.19) it turns out that limτ →|x−x′0|+ βn (τ, x, x′0 )|(Γ2 − Γ̃2 )(τ, z, z ′ )| = 0. Therefore applying (3.26) (resp. (3.27), (3.28)) on the left-hand side of (6.26) we obtain (4.2) (resp. (4.3), (4.4)). Theorem 4.1 is proved. Proof of Theorem 4.5. We first prove (4.11). We extend σ and σ̃ by 0 outside Y . For a bounded and continuous function f on Y consider the X-ray transform P f : Sn−1 × Rn → R defined by (4.7) (we extend f by 0 outside Y ). We recall the following estimate kf kH − 12 (Y ) ≤ Z Sn−1 Z 2 Πv |P f (v, x)| dxdv  21 , (6.27) where Πv := {x ∈ Rn | v · x = 0} for v ∈ Sn−1 . Note that using the estimate kσk∞ ≤ M, we obtain Z τ+ (x′0 ,v) σ(x′0 + sv, v)ds ≤ Mτ+ (x′0 , v) ≤ Mdiam(X), for (x′0 , v) ∈ Γ− . (6.28) 0 Replacing σ by σ̃ on the left-hand side of (6.28) we obtain an estimate similar to (6.28) for σ̃. Therefore using the estimate |et1 −et2 | ≥ e−M diam(X) |t1 −t2 | for (t1 , t2 ) ∈ [0, +∞)2, max(t1 , t2 ) ≤ Mdiam(X), we obtain e− R τ+ (x′0 ,v) 0 σ(x′0 +sv,v)ds − e− R τ+ (x′0 ,v) 0 σ̃(x′0 +sv,v)ds ≥ e−M diam(X) |P (σ − σ̃)(v, x′0 )| , (6.29) for (x′0 , v) ∈ Γ− . Integrating the left-hand side of (4.5) over ∂X and using (6.29), we obtain Z |P (σ − σ̃)(v, x′0 )| dξ(v, x′0) ≤ eM diam(X) Vol(∂X)C1 kAS,W − ÃS,W kη,T , (6.30) Γ− where C1 is the constant that appears on the right-hand side of R (4.5). Note that using that X is a convex open subset of Rn with C 1 boundary we obtain Γ− |P (σ − σ̃)(v, x′0 )| dξ(v, x′0) = R R |P (σ − σ̃)(v, x)|dxdv. Therefore using (6.30) and the estimate |P (σ − σ̃)(v, x)|2 ≤ n−1 S Πv kσ − σ̃kL∞ (Y ) diam(X)|P (σ − σ̃)(v, x)| for (v, x) ∈ T Sn−1 (see (6.28) and the estimates σ ≥ 0, σ̃ ≥ 0) we obtain Z Sn−1 Z Πv 2 |P (σ − σ̃)(v, x)| dxdv  12 1 1 2 2 ≤ C3 kσ − σ̃k∞ kAS,W − ÃS,W kη,T . (6.31)
1 where C3 = diam(X)eM diam(X) Vol(∂X)C1 2 . Combining (6.31) and (6.27) we obtain (4.11). We now prove (4.14). Let f ∈ L2 (X), suppf ⊆ Z̄. We consider the weighted X-ray transform of f , Pϑ f , defined by Z τ+ (v,x) Pϑ f (x, v) = f (pv + x)ϑ(pv + x, v)dp, for a.e. (x, v) ∈ Γ− , (6.32) 0 where ϑ : X × Sn−1 → (0, +∞) is the analytic function given by ϑ(x, v) = (τ− (x, v)τ+ (x, v))− n−1 2 g(x, v, v), for (x, v) ∈ X × Sn−1 . (6.33) From [12, theorem 2.2] and from [16, theorem 4] we obtain kf kH − 21 (Z) ≤ CkPϑ f kL2 (Γ− ,dξ) , 23 (6.34) where C = C(X, Z, g) is a constant that does not depend on f . Let x′0 ∈ ∂X and let x ∈ ∂X x−x′ such that px′0 + (1 − p)x ∈ Z for some p ∈ (0, 1) where v0 = |x−x′0 | and t0 = |x − x′0 |. Note that 0 using (3.23) (since k̃ ∈ L∞ (Z) and suppk̃ ⊆ Z̄ ⊆ {x ∈ X | inf x′ ∈∂X |x − x′ | ≥ δ}), we obtain Z τ+ (x0 ,v0′ )−δ Z τ+ (x0 ,v0′ ) 1 k̃(x′0 + pv0′ , v0′ , v0′ ) ≤ kk̃kL∞ (Z) n−1 dp n−1 dp n−1 n−1 ′ ′ ′ p 2 (τ+ (x0 , v0 ) − p) 2 p 2 (τ+ (x0 , v0′ ) − p) 2 δ 0 ≤ kk̃kL∞ (Z) δ −(n−1) τ+ (x′0 , v0′ ) ≤ kk̃kL∞ (Z) δ −(n−1) diam(X). (6.35) We use the estimate ′ ′ ′ ′ ′ ′ |Pϑ (k0 − k̃0 )(x′0 , v0′ )| ≤ eP σ(v0 ,x0 ) |Pϑ k̃0 (x′0 , v0′ )| e−P σ(v0 ,x0 ) − e−P σ̃(v0 ,x0 ) ′ ′ ′ ′ ′ (6.36) ′ +eP σ(v0 ,x0 ) e−P σ(v0 ,x0 ) Pϑ k0 (x′0 , v0′ ) − e−P σ̃(v0 ,x0 ) Pϑ k̃0 (x′0 , v0′ ) . ′ ′ P σ(v0 ,x0 ) Integrating both sides of inequality (6.36) over v0′ ∈ Sxn−1 ≤ ′ ,− and using the estimate e 0 M diam(X) e , and using (6.35), (4.5)–(4.6), we obtain Z |Pϑ (k0 − k̃0 )|(x′0 , v0′ )|ν(x′0 ) · v|dv ≤ δ −(n−1) diam(X)eM diam(X) C1 kk̃k∞ kAS,W − ÃS,W kη,T Sn−1 ′ x ,− 0 + n−3 Vol(Sn−1 )eM diam(X) C2 (τ − |z − z ′ |) 2 (Γ1 − Γ̃1 )(τ, z, z ′ ) 2 L∞ ((0,T )×∂X×∂X) , (6.37) where C1 and C2 are the constants that appear on the right-hand side of (4.5) and (4.6). From the estimate |Pϑ (k0 − k̃0 )(v0′ , x′0 )| ≤ kk − k̃kL∞ (Z) δ −(n−1) diam(X) for a.e. (x′0 , v0′ ) ∈ Γ− (see (6.35)), it follows that Z Z 2 −(n−1) |Pϑ (k0 −k̃0 )(x′0 , v0′ )||ν(x′0 )·v|dvdµ(x′0). kPϑ (k0 −k̃0 )kL2 (Γ− ,dξ) ≤ kk−k̃kL∞ (Z) δ diam(X) ∂X Sn−1 ′ x0 ,− (6.38) Combining (6.37)–(6.38) and (6.34) we obtain (4.14). Proof of Theorem 4.4 (4.13). We first prove (6.42) given below. Note that from (3.24)–(3.25), it follows that Z τ+ (x,v) 1 p |Pϑ0 f (v, x)| ≤ kf k∞ dt = Ckf k∞ , (6.39) t(τ+ (x, v) − t) 0 R1 for (v, x) ∈ Γ− and for f ∈ C(X) ∩ L∞ (X), where C = 0 √ 1 dt. t(1−t) Note that ν(x) = x and ν(x) · (x − x′0 ) = |ν(x′0 ) · (x − x′0 )| for (x, x′0 ) ∈ ∂X = S1 . Therefore from (4.2), it follows that p |ν(x′0 )·v0′ |2 E(x, x′0 )Pϑ0 k0 (v0′ , x) − Ẽ(x, x′0 )Pϑ0 k̃0′ (v0′ , x) ≤ C ′ τ 2 − |z − z ′ |2 (Γ1 − Γ̃1 )(τ, z, z ′ ) for (x, x′0 ) 2 ∈ ∂X , x 6= x′0 and v0′ = x−x′0 |x−x′0 | ′ and the similar identity for k̃), where C = (6.40) (we also used Pϑ0 kv0 (v0 , x) = g(v0 , v0 )Pϑ0 k0 (v0 , x) 1 . 2 inf Γ− S inf Γ+ W inf v∈Sn−1 g(v,v) In addition, from (6.40), (6.36) (with Pϑ0 in place of “Pϑ ”), and from (6.28) and (6.39) (with k̃0 in place of f ), it follows that ′ ′ ′ ′ |ν(x′0 ) · v0′ |2 |Pϑ0 (k0 − k̃0 )(x′0 , v0′ )| ≤ eM diam(X) Ckk̃0 k∞ e−P σ(v0 ,x0 ) − e−P σ̃(v0 ,x0 ) |ν(x′0 ) · v0 | ′ ′ ′ ′ +eM diam(X) e−P σ(v0 ,x0 ) Pϑ k0 (x′0 , v0′ ) − e−P σ̃(v0 ,x0 ) Pϑ k̃0 (x′0 , v0′ ) |ν(x′0 ) · v0′ |2 . 24 (6.41) L∞ , for (x′0 , v0′ ) ∈ Γ− (we also used the estimate |ν(x′0 ) · v0′ | ≤ 1). Performing the change of variables ν(x)·v0′ dµ(x) = dv0′ ) on the left-hand side of (4.1), we obtain that the “x = x′0 + τ+ (x′0 , v0′ )v0′ ” ( |x−x′ |n−1 0 estimate (4.5) still holds. Using (4.5), (6.41) and (6.40) (and (4.8)), we obtain that there exists a constant C ′′ such that Z  |ν(x′0 ) · v0′ |2 |P (ρ(k0 − k̃0 ))(x′0 , v0′ )|dv0′ ≤ C ′′ kk̃0 k∞ kAS,W − ÃS,W kη,T Sn−1 ′ x 0 + p τ 2 − |z − z ′ |2 (Γ1 − Γ̃1 )(τ, z, z ′ ) L∞  (6.42) for x′0 ∈ ∂X. Moreover, using (6.39) (with k0 − k̃0 in place of “f ”) and Cauchy-BunyakovskiSchwarz estimate, we obtain  12 Z 2 ′ ′ |P (ρ(k0 − k̃0 ))| (v0 , x0 )dξ(x0 , v0 ) Γ−  Z 3 4  ≤ Ckk0 − k̃0 k∞ ∂X Z S1 ′ x ,− 0  Z 3 √ 4 ≤ Ckk0 − k̃0 k∞ 2π  ∂X Z  12   12 |P (ρ(k0 − k̃0 ))|(v0 , x′0 )|v0 · ν(x′0 )|2 dv0′ dµ(x′0 ) S1 ′ x0 ,−  14 |P (ρ(k0 − k̃0 ))|(v0 , x′0 )|v0 · ν(x′0 )|2 dv0′ dµ(x′0 ) . (6.43) Finally combining (6.42)–(6.43), (6.27) (and the identity R R |P (k0 − k̃0 )(v, x)|2 dxdv), we obtain (4.13). n−1 S Πv 7 R 2 Γ− P (ρ(k0 − k̃0 )(v, x′0 ) dξ(v, x′0 ) = Proof of Theorem 3.2 For 0 < b < a we remind that Z 2π 0 1 2π dΩ = √ . 2 a − b sin(Ω) a − b2 (7.1) We will use the following Lemma 7.1 to prove Theorem 3.2 (3.13), (3.14) and (3.15). Lemma 7.1. Let n ≥ 2. Let N denote the nonnegative measurable function from (0, T ) × ∂X × Rn to [0, +∞[ defined by Z (τ − (x − x′ ) · v)n−3 ′ ′ N(τ, x, x ) = χ(0,+∞) (τ − |x − x |) dv, (7.2) |x − x′ − τ v|2n−4 Sn−1 for (τ, x, x′ ) ∈ (0, T ) × ∂X × Rn . When n = 2, then 2π , N(τ, x, x′ ) = χ(0,+∞) (τ − |x − x′ |) p τ 2 − |x − x′ |2 for (τ, x, x′ ) ∈ (0, T ) × ∂X × Rn . When n = 3, then χ(0,+∞) (τ − |x − x′ |) ln N(τ, x, x ) = 2π τ |x − x′ | ′ for (τ, x, x′ ) ∈ (0, T ) × ∂X × Rn . When n ≥ 4, then sup (τ,x,x′ )∈(0,T )×∂X×Rn   τ + |x − x′ | , τ − |x − x′ | τ |x − x′ |n−2 N(τ, x, x′ ) < ∞. 25 (7.3) (7.4) (7.5) Proof of Lemma 7.1. Let (τ, x, x′ ) ∈ (0, T ) × ∂X × Rn . We first prove (7.3). Let n = 2. Note that Z 2π 1 ′ ′ dΩ. N(τ, x, x ) = χ(0,+∞) (τ − |x − x |) τ − |x − x′ | sin(Ω) 0 Therefore using (7.1) we obtain (7.3). We prove (7.4). Let n = 3. Note that χ(0,+∞) (τ − |x − x′ |) N(τ, x, x ) = 2π 2τ |x − x′ | ′ Z π 2 − π2
d ln τ 2 + |x − x′ |2 − 2τ |x − x′ | sin(Ω) dΩ, dΩ which gives (7.4). We prove (7.5). Let n ≥ 4 and let (τ, x, x′ ) ∈ (0, T ) × ∂X × Rn be such that τ > |x − x′ | (we remind that N(τ, x, x′ ) = 0 if τ ≤ |x − x′ |). Using spherical coordinates, we obtain ′ n−2 N(τ, x, x ) = Voln−2 (S ) Z π 2 −π 2 (τ − |x − x′ | sin(Ω))n−3 n−2 dΩ. n−2 cos(Ω) ′ 2 2 ′ (|x − x | + τ − 2τ |x − x | sin(Ω)) Performing the change of variables “r = τ 2 −|x−x′ |2 2(τ −|x−x′ | sin(Ω)) Voln−2 (Sn−2 )(τ 2 − |x − x′ |2 ) N(τ, x, x′ ) = |x − x′ |n−2 n−3 2 |x−x′ | Z p ′ n−3 2 r(|x − x′ | − r) n−3 ′ ′ ′ 0 2Voln−2 (Sn−2 )|(τ 2 − |x − x′ |2 ) ≤ |x − x′ |n−2 n−3 2 2Voln−2 (Sn−2 )(τ 2 − |x − x′ |2 ) ≤ |x − x′ |n−2 n−3 2 Z |x−x′ | 2 Z dr 1 ′| ( τ −|x−x 2 τ 1−n 2 2 dr | | + r)n−2( τ +|x−x − r)n−2 ( τ −|x−x 2 2 0 + r) |x−x′ | 2 2 τ + |x − x | Voln−2 (Sn−2 )|x − x′ |2−n n−3 2τ n−1 2 ≤ Voln−2 (Sn−2 )|x − x′ |2−n τ −1 , n−3 n−1 2 ′ | ( τ +|x−x − r) 2 1 ( 0 ′  ≤ we obtain | | + r)n−2 ( τ +|x−x − r)n−2 ( τ −|x−x 2 2 ′ p n−3 Z |x−x | 2 r(|x − x′ | − r) 0 2Voln−2 (Sn−2 )(τ 2 − |x − x′ |2 ) = |x − x′ |n−2 n−1 τ −|x−x′ | ”, 2 − (7.6) τ −|x−x′ |  n−3 2 2 + r) n−1 2 n−1 2 dr dr τ −1 (7.7) which proves (7.5). We are ready to prove Theorem 3.2. First we give an estimate on the simple scattering term. From (3.10) it follows that |γ1(τ, x, x′ )| ≤ 2n−2 kW k∞ kSk∞ kkk∞ I1 (τ, x, x′ ) (7.8) for a.e. (τ, x, x′ ) ∈ R × ∂X × ∂X, where ′ ′ I1 (τ, x, x ) = χ(0,+∞) (τ − |x − x |) × Z Sn−1 (τ − (x − x′ ) · v)n−3 dv. |x − x′ − τ v|2n−4 26 χsuppk (x − sv)|s= τ 2 −|x−x′ |2 2(τ −v·(x−x′ )) (7.9) Let (τ, x, x′ ) ∈ (0, T ) × ∂X × ∂X be such that x 6= x′ and τ > |x − x′ |. Assume without loss of generality x′ − x = |x′ − x|(1, 0 . . . 0). First we prove (3.13)–(3.15). From (7.9) and (7.2), it follows that I1 (τ, x, x′ ) ≤ N(τ, x, x′ ). (7.10) Combining (7.3) (respectively (7.4), (7.5)) with (7.8) and (7.10), we obtain (3.13) (respectively (3.14), (3.15)). Now assume that k ∈ L∞ (X × Sn−1 × Sn−1 ) and suppk ⊆ {x ∈ X | inf |y − x| ≥ δ} for some 0 < δ < ∞. (7.11) y∈∂X 2 ′ 2 −|x−x | ′ Let v ∈ Sn−1 and s := 2(ττ −(x−x ′ )·v) . Straightforward computations give s + |x − x − sv| = τ . Using (7.11) we obtain that if τ < δ or s > τ − δ, then x − sv 6∈ suppk. (7.12) Using (7.9) and (7.12), we obtain if τ < δ then I1 (τ, x, x′ ) = 0. (7.13) We prove (3.16) for n = 2. Using (7.10), we obtain that I1 (τ, x, x′ ) ≤ √ δ √ 2π τ −|x−x′ | for τ ≥ δ. Combining (7.8) with this latter estimate and (7.13), we obtain (3.16) for n = 2. We now prove (3.16) for n ≥ 3. Let n ≥ 3 and τ ≥ δ (the case τ < δ is already considered in τ 2 −|x−x′ |2 τ −|x−x′ | (7.13)). Performing the change of variables “r = 2(τ −|x−x ” with “v = Φ(Ω, ω) := ′ | sin(Ω)) − 2 π π n−2 (sin(Ω), cos(Ω)ω), Ω ∈ (− 2 , 2 ), ω ∈ S ” on the right-hand side of (7.9), we obtain p n−3 Z |x−x′| 2 ′ 2 n−3 2 r(|x − x′ | − r) ′ 2−n (τ − |x − x | ) I1 (τ, x, x ) = 2 ′| ′| |x − x′ |n−2 + r)n−2 ( τ +|x−x − r)n−2 ( τ −|x−x 0 2 2 Z [χsuppk (x − sv)] Ω=arcsin(|x−x′ |−1 (τ − (τ 2 −|x−x′ |2 ) )) dωdr. (7.14) Sn−2 Now assume τ > ′ 2−n |x − x | Z 2(r+ τ −|x−x′ | s=r+ 2 v=Φ(Ω,ω) δ 2 τ −|x−x′ | ) 2 + |x − x′ |. Then p n−3 r(|x − x′ | − r) |x−x′ | dr ′| ′| + r)n−2 ( τ +|x−x − r)n−2 ( τ −|x−x 2 2  4−2n Z 1  4−2n  4−2n Z |x−x′| p p n−3 δ δ δ ′ 2−n ′ |x − x | . r(|x − x | − r) dr = r(1 − r)dr ≤ ≤ 4 4 4 0 0 0 Therefore using (7.14) we obtain ′ (τ − |x − x |) − n−3 2 I1 (τ, x, x ) ≤ 2 Finally assume δ ≤ τ ≤ ′ (τ −|x−x |) − n−3 2 ′ δ 2 n−2 n−2 Voln−2 (S )(T + diam(X))  4−2n δ . 2 n−3 2 (7.15) + |x − x′ | and |x − x′ | < τ ≤ T . From (7.12), it follows that Voln−2 (Sn−2 )(T + diam(X)) I1 (τ, x, x′ ) ≤ 2n−2 |x − x′ |n−2 n−3 2 ′ r+ (τ,x,x Z ) p ′ r(|x − x′ | − r) n−3 ′ | | ( τ′−|x−x + r)n−2 ( τ +|x−x − r)n−2 2 2 r− (τ,x,x ) (7.16) 27 dr, where r− (τ, x, x′ ) := τ − δ + |x − x′ | |x − x′ | + δ − τ , r+ (τ, x, x′ ) := . 2 2 (7.17) Note that Z p n−3 r(|x − x′ | − r) r+ (τ,x,x′ ) ′ r− (τ,x,x′ ) ≤2 ′ | | ( τ −|x−x + r)n−2 ( τ +|x−x − r)n−2 2 2  τ 2−n 2 Z ′ n−3 |x−x | |x−x′ | 2 dr = 2 + |x−x′ | 2 r)n−2 p ′ r− (τ,x,x′ ) 1 τ −|x−x′ | r− (τ,x,x′ ) ( 2 Z dr = 2 r(|x − x′ | − r) n−3 ′ | | ( τ −|x−x + r)n−2 ( τ +|x−x − r)n−2 2 2 n−1 2−n τ ′ n−3 |x−x | Z |x−x′ | 2 dr 1 τ −|x−x′ | r− (τ,x,x′ ) ( 2 + r)n−2 (7.18) Using (7.17) we obtain Z |x−x′ | 2 r− (τ,x,x′ ) 1 ( τ −|x−x′ | 2 + r)n−2 dr = C(n, τ ) := ( ln 1 n−3  From (7.16), (7.18), (7.19) and the estimates δ ≤ τ < (τ − |x − x′ |)− n−3 2 δ 2
δ 3−n 2 τ δ
, if n = 3,
 τ 3−n otherwise. − 2 (7.19) + |x − x′ |, it follows that I1 (τ, x, x′ ) ≤ 2n Voln−2 (Sn−2 )δ −(n−1) (T + diam(X)) n−3 2 C(n, T ), (7.20) where the constant C(n, T ) is defined in (7.19). Combining (7.8) with (7.13), (7.15) and (7.20), we obtain (3.16) for n ≥ 3.  8 Proof of Theorem 3.3 We shall use the following Lemmas 8.1, 8.2, 8.3 and 8.4. Lemmas 8.1, 8.2, 8.3 and 8.4 are proved in Section 9. We introduce some notation first. Let m ≥ 1 and z ′ , z ∈ Rn such that z 6= z ′ . Let µ ≥ 0. We denote by Em,n (µ, z, z ′ ) the subset of (Rn )m defined by Em,n (µ, z, z ′ ) = {(y1 , . . . , ym ) ∈ (Rn )m | |y1| + . . . + |ym| + |z − z ′ − y1 − . . . − ym | < µ}. (8.1) When µ ≤ |z − z ′ |, then Em,n (µ, z, z ′ ) = ∅. Lemma 8.1. Let J2 be the function from (0, T ) × ∂X × Rn defined by Z 1 ′ J2 (µ, z, z ) = N(µ − |y|, z, z ′ + y)dy, n−1 |y| ′ E1,n (µ,z,z ) (8.2) where N is defined by (7.2). Then the following statements are valid: sup (µ,z,z ′ )∈(0,T )×∂X×Rn µ>|z−z ′ | J2 (µ, z, z ′ ) < ∞, when n = 2; (8.3) −2   µ + |z − z ′ | J2 (µ, z, z ′ ) < ∞, when n = 3; sup (µ−|z−z |) µ|z−z | 1 + ln ′| ′ n µ − |z − z (µ,z,z )∈(0,T )×∂X×R ′ −1 ′ µ>|z−z ′ | sup (µ,z,z ′ )∈(0,T )×∂X×Rn µ>|z−z ′ | (µ − |z − z ′ |)−1 µ|z − z ′ |n−2J2 (µ, z, z ′ ) < ∞, when n ≥ 4. 28 (8.4) (8.5) dr. Lemma 8.2. Let m ≥ 3 and let J˜m be the function from {(τ, x, x′ ) ∈ (0, T ) × Rn × Rn | 0 < |x − x′ | < τ } to R defined by Z dy2 . . . dym−1 ′ ˜ , (8.6) Jm (τ, x, x ) = n−1 n−1 . . . |ym−1 | |x − x′ − y2 − . . . − ym−1 |n−2 Em−2,n (τ,x,x′ ) |y2 | for (τ, x, x′ ) ∈ (0, T ) × Rn × Rn , 0 < |x − x′ | < τ . When n = 3, then there exists a constant C̃ which does not depend on m such that    n−1 m−3 ′ ′ τ − |x − x | τ + |x − x | m (Voln−1 (Sn−1 )τ ) J˜m (τ, x, x ) ≤ C̃ 1 + ln , |x − x′ | τ − |x − x′ | (m − 3)! ′ (8.7) for (τ, x, x′ ) ∈ (0, T ) × Rn × Rn , 0 < |x − x′ | < τ . When n ≥ 4, then there exists a constant C̃ which does not depend on m such that n−1 m J˜m (τ, x, x′ ) ≤ C̃(τ − |x − x′ |)|x − x′ |2−n (Voln−1 (Sn−1 )τ ) (m − 3)! m−3 , (8.8) for (τ, x, x′ ) ∈ (0, T ) × Rn × Rn , 0 < |x − x′ | < τ . Lemma 8.3. Let n ≥ 2. Let (τ, x, x′ ) ∈ R × Rn × Rn be such that τ > |x − x′ | > 0, the following estimate is valid: !n−1 p n−2 ′ 2 − |x − x′ |2 τ Vol (S )π(τ + |x − x |) n−2 , (8.9) Voln (E1,n (τ, x, x′ )) ≤ 4 2 where E1,n is defined by (8.1). Lemma 8.4. Let B be the function from {(µ, z, z ′ ) ∈ (0, T ) × ∂X × R3 | µ > |z − z ′ | > 0} to R defined by     Z Z µ − |y| + |z − z ′ − y| µ − |y| + |(t0 , 0, 0) − y| ′ dy = ln dy, B(µ, z, z ) := ln µ − |y| − |z − z ′ − y| µ − |y| − |(t0 , 0, 0) − y| E1,3 (µ,t0 (1,0,0),0) E1,3 (µ,z,z ′ ) (8.10) ′ n ′ ′ ′ for (µ, z, z ) ∈ (0, T ) × ∂X × R where t0 = |z − z |, z 6= z , µ > |z − z |. Then we have: ′ sup (µ,z,z ′ )∈(0,T )×∂X×R3 µ>|z−z ′ |>0 (µ − |z − z |) −1  µ + |z − z ′ | 1 + ln µ − |z − z ′ |  −1 B(µ, z, z ′ ) < ∞. (8.11) We also need the explicit expression of γm , m ≥ 2, to prove Theorem 3.3 Z Z ′ γ2 (τ, x, x ) := (ν(x) · v)W (x, v) [E(x, x − (τ − |y| − s1 )v, x′ + y, x′) ′ y∈E1,n (τ,x,x ) x′ +y∈X Sn−1 x,+ χ(0,τ− (x,v)) (τ − |y| − s1 )k(x − (τ − s1 − |y|)v, v1, v)k(x′ + y, v ′, v1 )S(x′ , v ′) 2n−2(τ − |y| − (x − x′ − y) · v)n−3 |ν(x′ ) · v ′ |] s = |x−x′ −y−(τ −|y|)v|2 dydv, 1 2(τ −|y|−(x−x′ −y)·v) |x − x′ − y − (τ − |y|)v|2n−4 |y|n−1 x−x′ −y−(τ −s1 −|y|)v v1 = s1 y v′ = |y| 29 (8.12) and ′ γm (τ, x, x ) := Z (y2 ,...,ym )∈Em−1,n (τ,x,x′ ) (x′ +ym ,...,x′ +ym +...+y2 )∈X m−1 Z Sn−1 x,+ (ν(x) · v)W (x, v) 2n−2 (τ − |y2| − . . . − |ym | − (x − x′ − y2 − . . . − ym ) · v)n−3 |y2 |n−1 . . . |ym |n−1 |x − x′ − y2 − . . . − ym − (τ − |y2 | − . . . − |ym |)v|2n−4  × χ(0,τ− (x,v)) (τ − s1 − |y2| − . . . − |ym |)E(x, x − (τ − s1 − |y2 | − . . . − |ym|)v, x′ + ym . . . + y2 , . . . , x′ + ym , x′ )k(x − (τ − s1 − |y2 | − . . . − |ym |)v, v1 , v) ×k(x′ + ym + . . . + y2 , v2 , v1 ) . . . k(x′ + ym + . . . + yi+1 , vi+1 , vi ) . . . k(x′ + ym + ym−1 , vm−1 , vm−2 )k(x′ + ym , v ′ , vm−1 )S(x′ , v ′) |ν(x′ ) · v ′ |] v = x−x′ −y2 −...−ym −(τ −s1 −|y2 |−...−|ym |)v dy2 . . . dym dv, (8.13) × 1 s1 |x−x′ −y2 −...−ym −(τ −|y2 |−...−|ym |)v|2 s1 = 2(t−|y2 |−...−|ym−1 |−(x−x′ −y2 −...ym−1 )·v) y v′ = m |ym | y vi = i , i=2...m−1 |yi | for τ ∈ R and a.e. (x, x′ ) ∈ ∂X × ∂X and for m ≥ 3. We are ready to prove Theorem 3.3. We prove (3.17), (3.18) and (3.19). Let τ ∈ (0, T ) and let x ∈ ∂X, x′ ∈ ∂X and x 6= x′ . Set t0 = |x − x′ |. We first look for an upper bound on |γ2(τ, x, x′ )|. Using (8.12) and the fact that σ is a nonnegative function, we obtain |γ2(τ, x, x′ )| ≤ 2n−2 kW k∞ kSk∞ kkk2∞ J2 (τ, x, x′ ), (8.14) where J2 and E1,n (τ, x, x′ ) are defined by (8.2) and (8.1). From (8.14) and (8.3)–(8.5) it follows that there exists a real constant C such that |γ2(τ, x, x′ )| ≤ CkW k∞ kSk∞ kkk2∞ sup (s,z,z ′ )∈(0,T )×Rn ×Rn s>|z−z ′ | J2 (s, z, z ′ ), when n = 2, (8.15) and τ |x − x′ | ′ 2 2 |γ2(τ, x, x )| ≤ CkW k∞ kSk∞ kkk∞ , when n = 3,   ′| τ +|x−x (τ − |x − x′ |) 1 + ln τ −|x−x′| and (8.16) τ |x − x′ |n−2 |γ2 (τ, x, x′ )| ≤ CkW k∞ kSk∞ kkk2∞ , when n ≥ 4. ′ τ − |x − x | (8.17) ′ |γm (τ, x, x′ )| ≤ 2n−2 kW k∞ kSk∞ kkkm ∞ Jm (τ, x, x ), (8.18) Let m ≥ 3. Using (8.13) we obtain where ′ Jm (τ, x, x ) = Z Em−2,n (τ,x,x′ ) J2 (τ (ȳ), w(ȳ), 0) dy2 . . . dym−1 , |y2 |n−1 . . . |ym−1 |n−1 (8.19) and J2 (resp. Em−2,n (τ, x, x′ )) is defined by (8.2) (resp. (8.1)) and where ȳ = (y2 , . . . , ym−1 ), τ (ȳ) = τ −|y2 | −. . .−|ym−1 |, t0 (ȳ) = |x−x′ −y2 −. . .−ym−1 | and w(ȳ) = x−x′ −y2 −. . .−ym−1 for ȳ ∈ (Rn )m−2 . 30 Assume n = 2. Then using (8.19), (8.3) and spherical coordinates (and (8.1)), we obtain Z 1 ′ ′ Jm (τ, x, x ) ≤ sup J2 (s, z, z ) dȳ (s,z,z ′ )∈(0,T )×Rn ×Rn Em−2,n (τ,x,x′ ) |y2 | . . . |ym−1 | s>|z−z ′ | Z m−2 ′ ≤ (2π) sup J2 (s, z, z ) ds2 . . . dsm−1 s2 +...+sm−1 ≤τ si ≥0, i=2...m−1 (s,z,z ′ )∈(0,T )×Rn ×Rn s>|z−z ′ | = (2π)m−2 τ m−2 (m − 2)! sup (s,z,z ′ )∈(0,T )×Rn ×Rn s>|z−z ′ | J2 (s, z, z ′ ). (8.20) Finally combining (8.20) and (8.18), we obtain |γm(τ, x, x′ )| ≤ (2π)m−2 kW k∞ kSk∞ kkkm ∞ τ m−2 (m − 2)! sup (s,z,z ′ )∈(0,T )×Rn ×Rn s>|z−z ′ | J2 (s, z, z ′ ), (8.21) Statement (3.17) follows from (8.15), (8.21). Assume n ≥ 3. Note that µ − |z − z ′ | |z − z ′ | =1− ≤ 1, µ µ (8.22) and µ − |z − z ′ | µ  1 + ln  µ + |z − z ′ | µ − |z − z ′ | 2 2   1+s ≤ sup (1 − s) 1 + ln , 1−s s∈(0,1) (8.23) for (µ, z, z ′ ) ∈ (0, T ) × Rn × Rn such that |z − z ′ | < µ. From (8.19), (8.22) and (8.4), (8.23) and (8.5), it follows that there exists a real constant C such that Jm (τ, x, x′ ) ≤ C J˜m (τ, x, x′ ), (8.24) where J˜m is defined by (8.6). Assume n = 3. Combining (8.18), (8.24), (8.7), we obtain that there exists a real constant C ′ (which does not depend on τ , x, x′ and m) such that ′ |γm (τ, x, x )| ≤ C ′ τ kW k∞ kSk∞ kkkm ∞ − |x − x′ | |x − x′ |  1 + ln  τ + |x − x′ | τ − |x − x′ |  mn−1 (Voln−1 (Sn−1 )τ ) (m − 3)! (8.25) Statement (3.18) follows from (8.16) and (8.25). Now assume n ≥ 4. Combining (8.18), (8.24), (8.8), we obtain that there exists a real constant C ′ (which does not depend on τ , x, x′ and m) such that ′ ′ 2−n |γm (τ, x, x′ )| ≤ C ′ kW k∞ kSk∞ kkkm ∞ (τ − |x − x |)|x − x | mn−1 (Voln−1 (Sn−1 )τ ) (m − 3)! m−3 . (8.26) Statement (3.19) follows from (8.17) and (8.26). We now prove (3.20)–(3.21). Let n ≥ 3 and m ≥ 2. From the expression of γm (see (8.12)– (8.13)), it follows that ′ |γm (τ, x, x′ )| ≤ 2n−2 kW k∞ kSk∞ kkkm ∞ Im (τ, x, x ) 31 (8.27) m−3 . where N(τ − Z ′ Im (τ, x, x ) := (y2 ,...,ym )∈Em−1,n (τ,x,x′0 ) P (x′ +ym ,...,x′ + m yi )∈(suppk)m−1 i=2 Pm i=2 P |yi|, x, x′ + m i=2 yi )dym . . . dy2 , (8.28) |y2 |n−1 . . . |ym|n−1 where N and Em−1,n (τ, x, x′ ) are defined by (7.2) and (8.1). Note that |ym | ≥ δ and τ− m X i=2 |yi| ≥ |x − x′ − y2 − . . . − ym | ≥ δ, (8.29) for (y2 , .., ym ) ∈ Em−1,n (τ, x, x′ ) such that x′ + ym ∈ suppk and x′ + y2 + . . . + ym ∈ suppk since suppk ⊆ {z ∈ X | inf y∈∂X |y − z| ≥ δ}. We prove (3.20). Assume n = 3. Using (8.28)–(8.29), (7.4) and (8.10) we obtain I2 (τ, x, x′ ) ≤ 2πδ −4 B(τ, x, x′ ). Therefore using (8.11) and (8.30) we obtain −1   s + |z − z ′ | ′ −1 I2 (s, z, z ′ ) < ∞. sup 1 + ln (s − |z − z |) ′| ′ s − |z − z (s,z,z )∈(0,T )×∂X×∂X (8.30) (8.31) s>|z−z ′ |>0 Now assume m ≥ 3. Using (8.28), we obtain Pm P Z ′ J2 (τ − m ′ i=3 |yi |, x, x + i=3 yi )dym . . . dy3 Im (τ, x, x ) ≤ , n−1 |y3| . . . |ym |n−1 (8.32) (y3 ,...,ym )∈Em−2,n (τ,x,x′ ) 0 P (x′ + m yi ,x′ +ym )∈(suppk)2 i=3 where J2 is defined by (8.2). Using (8.32) and (8.4) and the estimate supr∈(0,1) r(1−ln(r))2 < ∞ we obtain P Z (|x − x′ − y3 − . . . − ym | + τ − m ′ i=3 |yi |)dym . . . dy3 Pm , Im (τ, x, x ) ≤ D n−1 n−1 |y3 | . . . |ym | (τ − i=3 |yi |)|x − x′ − y3 − . . . − ym | (y3 ,...,ym )∈Em−2,n (τ,x,x′0 ) P (x′ + m yi ,x′ +ym )∈(suppk)2 i=3 (8.33)   −2 ′| s+|z−z where D := supr∈(0,1) r(1−ln(r))2 sup (s,z,z′ )∈(0,T )×∂X×∂X (s−|z−z ′ |)−1 s|z−z ′ | 1 + ln s−|z−z ′| s>|z−z ′ |>0 J2 (s, z, z ′ ). If m = 3, then using (8.29) with “(y2 , . . . , ym )” replaced by “(y3 , . . . , ym )”, we obtain I3 (τ, x, x′ ) ≤ 2τ δ −4 DVol(E1,3 (τ, x, x′ )) (8.34) (we also used the estimate |x − x′ − y3 | + τ − |y3 | ≤ 2τ for y3 ∈ E1,3 (τ, x, x′ )). If m ≥ 4, then using (8.29) with “(y2 , . . . , ym )” replaced by “(y3 , . . . , ym )”, we obtain Z dy3 . . . dym−1 ′ −4 ′ Im (τ, x, x ) ≤ 2τ δ DVol(E1,3 (τ, x, x )) |y3|n−1 . . . |ym−1 |n−1 (y3 ,...,ym−1 )∈Em−3,3 (τ,x,x′ ) Z −4 ′ n−1 m−3 ds3 . . . dsm−1m−3 ≤ 2τ δ DVol(E1,3 (τ, x, x ))Vol(S ) (s3 ,...,sm−1 )∈(0,+∞) s3 +...+sm−1 <τ = 2τ δ −4 DVol(Sn−1 )m−3 τ m−3 Vol(E1,3 (τ, x, x′ ) (m − 3)! 32 (8.35) (we also used the estimate |x − x′ − y3 − . . . − ym | + τ − |y3| − . . . − |ym | ≤ 2τ for (y3 , . . . , ym ) ∈ Em−2,3 (τ, x, x′ ) and we performed the changes of variables yi = si ωi , (si , ωi) ∈ (0, +∞) × Sn−1 ). Statement (3.20) follows from (8.27), (8.31) and (8.34)–(8.35) (and (8.9)). We prove (3.21). Let n ≥ 4. Using (8.28) and (7.5), we obtain Im (τ, x, x′ ) ≤ ks|z − z ′ |n−2 N(s, z, z ′ )kL∞ (Rs ×∂Xz ×Rnz′ ) Z dym . . . dy2 Pm P × . (8.36) n−2 (τ − n−1 n−1 |y2 | . . . |ym | |x − x′ − m i=2 |yi |) i=2 yi | (y2 ,...,ym )∈Em−1,n (τ,x,x′ ) 0 ′ +ym ∈suppk xP m ′ y ∈suppk x + i=2 i Assume m = 2. Using (8.29) and (8.36), we obtain Im (τ, x, x′ ) ≤ δ −2n+2 ks|z − z ′ |n−2N(s, z, z ′ )kL∞ (Rs ×∂Xz ×Rnz′ ) Vol(E1,n (τ, x, x′ )). (8.37) Therefore using (8.9), we obtain ′ Im (τ, x, x ) ≤ δ −2n+2 ′ n−2 ks|z − z | n−2 ′ N (s, z, z )kL∞ (Rs ×∂Xz ×Rnz′ ) Voln−2 (S ′ )π(τ + |x− x |) !n−1 p τ 2 − |x − x′ |2 . 2 (8.38) Assume m ≥ 3. Using (8.29) and (8.36), we obtain ′ −2n+2 Im (τ, x, x ) ≤ δ ′ n−2 ks|z − z | dym . . . dy2 . n−1 |y2 | . . . |ym−1 |n−1 (y2 ,...,ym )∈Em−1,n (τ,x,x′ ) ′ N(s, z, z )kL∞ (Rs ×∂Xz ×Rnz′ ) Z (8.39) Note that |ym |+|x−x′ −ym | ≤ |y2 |+. . .+|ym |+|x−x′ −y2 −. . .−ym | for (y2 , . . . , ym ) ∈ (Rn )m−1 . Hence |y2 | + . . . + |ym−1 | < τ − |ym | and |ym| + |x − x′ − ym | < τ, (8.40) for (y2 , . . . , ym) ∈ Em−1,n (τ, x, x′ ) (see (8.1)). Therefore Im (τ, x, x′ ) ≤ δ −2n+2 ks|z − z ′ |n−2 N(s, z, z ′ )kL∞ (Rs ×∂Xz ×Rnz′ ) Z Z dym . . . dy2 × P n−1 m−1 |y2 | . . . |ym−1 |n−1 ym ∈E1,n (τ,x,x′ ) i=2 |yi |<τ −|ym | =δ −2n+2 n−1 m−2 Vol(S ) ′ n−2 ks|z − z | ′ N(s, z, z )kL∞ (Rs ×∂Xz ×Rnz′ ) Z ym ∈E1,n (τ,x,x′ ) (τ − |ym |)m−2 dym (m − 2)! τ m−2 (m − 2)! −2n+2 n−1 m−2 ′ n−2 ′ n−2 =δ Vol(S ) ks|z − z | N(s, z, z )kL∞ (Rs ×∂Xz ×Rnz′ ) Voln−2 (S )π(τ + |x − x′ |) !n−1 p τ 2 − |x − x′ |2 τ m−2 × . (8.41) 2 (m − 2)! ≤ δ −2n+2 Vol(Sn−1 )m−2 ks|z − z ′ |n−2 N(s, z, z ′ )kL∞ (Rs ×∂Xz ×Rnz′ ) Vol(E1,n (τ, x, x′ )) Statement (3.21) follows from (8.27), (8.38) and (8.41). 33  9 Proof of Lemmas 8.1, 8.2, 8.3 and 8.4 We remind the following change of variables for the proof of Lemmas 8.1, 8.2, 8.3 and 8.4. Z f (y)dy E1,n (τ,t0 v,0)) !  Z p 2 − t2 s t + s cos(ϕ)  0 0   f , sin ϕ   2 2  (0,2π)×(t ,τ ) 0     (s2 − t20 cos2 (ϕ))   p × dsdϕ, if n = 2,   4 s2 − t20 ! p Z = (9.1) 2 − t2 s t + s cos(ϕ)  0 0  , sin ϕω f    2 2 Sn−2 ×(0,π)×(t0 ,τ )   !  p n−2    s2 − t20 cos2 (ϕ) sin(ϕ) s2 − t20   p dωdsdϕ, if n ≥ 3,  × 2 4 s2 − t20 for f ∈ L1 (Rn ) and (τ, t0 , v) ∈ (0, +∞) × (0, +∞) × Sn−1 such that τ > t0 . Proof of Lemma 8.1. We prove (8.3). Let (µ, z, z ′ ) ∈ (0, T ) × ∂X × R2 be such that µ > |z − z ′ |. From (8.2) and (7.3) it follows that Z 2π ′ p dy J2 (µ, z, z ) = 2 ′ 2 E1,2 (µ,z,z ′ ) |y| (µ − |y|) − |z − z − y| Z 2π p = dy, (9.2) 2 2 E1,2 (µ,t0 (1,0),0) |y| (µ − |y|) − |t0 (1, 0) − y| where t0 = |z − z ′ |. Using the change of variables y = s ∈ (t0 , µ), we obtain t0 (1, 0)+(s cos(ϕ), 2 ′ J2 (µ, z, z ) = 4π Z µ t0 where J2,1 (µ, s, ϕ) = p Z √ s2 −t20 2 sin(ϕ)) (see (9.1)), ϕ ∈ (0, 2π), 2π J2,1 (µ, s, ϕ)dϕds, (9.3) 0 s − t0 cos(ϕ) p , √ s2 − µ − s µ − t0 cos(ϕ) t20 (9.4) for ϕ ∈ (0, 2π) and s ∈ (t0 , µ). We give an estimate on J2,1 . From (9.4) and the estimates µ − t0 cos(ϕ) ≥ s − t0 cos(ϕ), s + t0 ≥ s − t0 cos(ϕ), it follows that J2,1 (µ, s, ϕ) ≤ √ 1 √ s − t0 µ − s , (9.5) for ϕ ∈ (0, π2 ) and s ∈ (t0 , µ). Performing the change of variables s = t0 + ε(µ − t0 ) we have Z µ Z 1 1 1 √ p dε < +∞, (9.6) ds = √ s − t0 µ − s ε(1 − ε) t0 0 for s ∈ (t0 , µ). Combining (9.3), (9.5), (9.6), we obtain Z 1 1 ′ 2 p sup J2 (µ, z, z ) ≤ 8π dε < ∞. (µ,z,z ′ )∈(0,T )×∂X×R2 ε(1 − ε) 0 µ>|z−z ′ | 34 (9.7) Statement (8.3) follows from (9.7). We prove (8.4). Let (µ, z, z ′ ) ∈ (0, T ) × ∂X × R3 be such that µ > |z − z ′ |. Set t0 = |z − z ′ |. From (8.2), (7.4) and (9.1), it follows that   ′ −y| Z 2π ln µ−|y|+|z−z µ−|y|−|z−z ′ −y| J2 (µ, z, z ′ ) ≤ dy 2 ′ E1,3 (µ,z,z ′ ) |y| (µ − |y|)|z − z − y|   0 (1,0,0)−y| Z 2π ln µ−|y|+|t µ−|y|−|t0 (1,0,0)−y| = dy 2 E1,3 (µ,t0 (1,0,0),0) |y| (µ − |y|)|t0 (1, 0, 0) − y| Z µZ π 2 = 8π J2,1 (µ, s, ϕ)dϕds, (9.8) t0 0 where sin(ϕ) ln J2,1 (µ, s, ϕ) =  µ−t0 cos(ϕ) µ−s  (s + t0 cos(ϕ))(2µ − s − t0 cos(ϕ))    µ − t0 cos(ϕ) sin(ϕ) sin(ϕ) = ln , (9.9) + µ−s 2µ(s + t0 cos(ϕ)) 2µ(2µ − s − t0 cos(ϕ)) for ϕ ∈(0, π) and s ∈ (t0 , µ). From (9.9) and the estimates 2µ − s − t0 cos(ϕ) ≥ µ − t0 cos(ϕ), 0 cos(ϕ) 0 0 ≤ ln µ−tµ−s ≤ ln µ+t , it follows that µ−s J2,1 (µ, s, ϕ) ≤ ln  µ + t0 µ−s  sin(ϕ) sin(ϕ) + 2µ(s − t0 cos(ϕ)) 2µ(µ − t0 cos(ϕ))  , for ϕ ∈ (0, π) and s ∈ (t0 , µ). Therefore Z π 0 ln  µ+t0 µ−s       s + t0 µ + t0 ln + ln J2,1 (µ, s, ϕ)dϕ ≤ 2µt0 s − t0 µ − t0        0 ln µ+t µ−s µ + t0 µ + t0 ≤ ln + ln . 2µt0 s − t0 µ − t0 We remind the following integral value    Z µ  µ + t0 µ + t0 ln ds = (µ − t0 ) ln + µ − t0 . µ−s µ − t0 t0     2(µ+t0 ) 0 Using the estimate ln µ+t for s ∈ (t0 , t0 +µ ≤ ln ), we obtain µ−s µ−t0 2     Z t0 +µ  2 µ + t0 µ + t0 µ + t0 ln ln ds = 2 ln ds ln µ−s s − t0 µ−s t0 t0  Z t0 +µ    2 µ + t0 2(µ + t0 ) ln ds ≤ 2 ln µ − t0 s − t0 t0        (µ + t0 ) µ + t0 ≤ 2(µ − t0 ) ln + ln(2) ln +1 . µ − t0 µ − t0 Z µ  µ + t0 s − t0  (9.10) (9.11)  35 (9.12) Combining (9.8)–(9.12) and (9.11), we obtain        (µ + t0 ) µ + t0 ′ 2 µ − t0 3 ln + 2 ln(2) ln +1 . J2 (µ, z, z ) ≤ 4π µt0 µ − t0 µ − t0 (9.13) Statement (8.4) follows from (9.13). We prove (8.5). Let n ≥ 4. Let (µ, z, z ′ ) ∈ (0, T ) × ∂X × Rn be such that µ > |z − z ′ |. From (8.2) and (7.5) it follows that Z Z ′ 1−n −1 ′ 2−n J2 (µ, z, z ) ≤ C |y| (µ−|y|) |z−z −y| dy = C |y|1−n(µ−|y|)−1|t0 (1, 0, 0)−y|2−ndy, E1,n (µ,z,z ′ ) E1,n (µ,t0 (1,0...0),0) (9.14) where t0 = |z − z ′ | and C = sup(µ̃,z̃,z̃ ′)∈(0,T )×∂X×Rn µ̃|z̃ − z̃ ′ |n−2 N(µ̃, z̃, z̃ ′ ). √2 2 s −t t0 Using the change of variables y = 2 (1, 0)+(s cos(ϕ), 2 0 sin(ϕ)ω) (see (9.1)), ϕ ∈ (0, π), s ∈ (t0 , µ), ω ∈ Sn−2 , we obtain Z µZ π ′ ′ J2 (µ, z, z ) ≤ C J2,1 (µ, s, ϕ)dϕds, (9.15) t0 0 where C ′ = 2n−2 Voln−2 (Sn−2 )C and n−3 (s2 − t20 ) 2 sinn−2 (ϕ) , J2,1 (µ, s, ϕ) = (s + t0 cos(ϕ))n−2 (2µ − s − t0 cos(ϕ))(s − t0 cos(ϕ))n−3 (9.16) for ϕ ∈ (0, 2π) and s ∈ (t0 , µ). We give estimates on J2,1 . Let ϕ ∈ (0, π2 ) and s ∈ (t0 , µ). From (9.16) and the estimates p s + t0 cos(ϕ) ≥ s, s2 − t20 sin(ϕ) ≤ s − t0 cos(ϕ) and the estimate 2µ − s − t0 cos(ϕ) ≥ s − t0 cos(ϕ), it follows that p s2 − t20 sin2 (ϕ) J2,1 (µ, s, ϕ) ≤ n−2 s (s − t0 cos(ϕ))2 p s2 − t20 ≤ C0 n−1 , (9.17) s (s − t0 cos(ϕ)) where C0 is defined by (5.33) (we also used the estimate s − t0 cos(ϕ) ≥ s(1 p − cos(ϕ))). Let π ϕ ∈ ( 2 , π) and s ∈ (t0 , µ). From (9.16) and the estimates s − t0 cos(ϕ) ≥ s, s2 − t20 sin(ϕ) ≤ s + t0 cos(ϕ) and the estimate 2µ − s − t0 cos(ϕ) ≥ s, it follows that p p s2 − t20 sin2 (ϕ) s2 − t20 ≤ C0 n−1 , (9.18) J2,1 (µ, s, ϕ) ≤ n−2 s (s + t0 cos(ϕ))2 s (s + t0 cos(ϕ)) where C0 is defined by (5.33) (we also used the estimate s + t0 cos(ϕ) ≥ s(1 + cos(ϕ))). Combining (9.17) and (9.18) and (7.1), we obtain Z π 2πC0 J2,1 (µ, s, ϕ)dϕ ≤ n−1 , for s ∈ (t0 , µ). (9.19) s 0 Note that Z µ t0 1 ds = n−1 s n−2 1 ≤  µn−2 − t0n−2 t0n−2 µn−2 µ − t0 . µt0n−2  = n−3 µ − t0 X −1−i i+2−n µ t0 n − 2 i=0 (9.20) 36 (we used the estimate t0 < µ which gives µ−1−i t0i+2−n ≤ µ−1 t0n−2 for i = 0 . . . n − 3). Combining (9.15), (9.19)–(9.20), we obtain J2 (µ, z, z ′ ) ≤ C ′ µ − t0 . µt0n−2 (9.21) where C ′ does not depend on µ and z, z ′ . Statement (8.5) follows from (9.21). Proof of Lemma 8.2. Let (τ, x, x′ ) ∈ (0, T )×Rn ×Rn such that τ > |x−x′ | > 0. Set t0 = |x−x′ |. Let (y2 , . . . , ym−1 ) ∈ Em−2,n (τ, x, x′ ). Then t0 ≤ |y2 | + . . . + |ym−1 | + |x − x′ − y2 − . . . ym−1 |. t0 t0 Therefore either |x − x′ − y2 − . . . − ym−1 | ≥ m−1 or |x − x′ − y2 − . . . − ym−1 | < m−1 and there t0 exists j ∈ N, j = 1 . . . n such that |yj | ≥ m−1 . Therefore using (8.6) we obtain J˜m (τ, x, x′ ) ≤ m−1 XZ j=2 (y2 ,...,ym−1 )∈Em−2,n (τ,x,x′ ) P t0 |x−x′ − m−1 yi |< m−1 i=2 t0 |yj |≥ m−1 dy2 . . . dym−1 P n−2 |y2|n−1 . . . |ym−1 |n−1 |x − x′ − m−1 i=2 yi | +J˜m,0 (τ, x, x′ ) = (m − 2)J˜m,1 (τ, x, x′ ) + J˜m,0 (τ, x, x′ ), (9.22) where J˜m,0 (τ, x, x′ ) = Z J˜m,1 (τ, x, x′ ) = Z (y2 ,...,ym−1 )∈Em−2,n (τ,x,x′ ) P t0 |x−x′ − m−1 yi |> m−1 i=2 |y2 (y2 ,...,ym−1 )∈Em−2,n (τ,x,x′ ) P t0 |x−x′ − m−1 yi |< m−1 i=2 t0 |y2 |≥ m−1 |n−1 dy2 . . . dym−1 , P n−2 . . . |ym−1 |n−1 |x − x′ − m−1 i=2 yi | dy2 . . . dym−1 . (9.24) P n−2 |y2 |n−1 . . . |ym−1 |n−1|x − x′ − m−1 i=2 yi | We first reduce the estimate of J˜m,0 and J˜m,1 to an estimate on Z dy2 . . . dym−1 ′ . Pm (τ, x, x ) := n−1 . . . |ym−1 |n−1 (y2 ,...,ym−1 )∈Em−2,n (τ,x,x′ ) |y2 | From (9.23) and the estimate |x − x′ − y2 − . . . − ym−1 | > J˜m,0 (τ, x, x′ ) ≤  (9.23) m−1 t0 n−2 t0 m−1 (9.25) it follows that Pm (τ, x, x′ ). t0 From (9.24) and the estimates |y2| ≥ m−1 ≥ |x − x′ − y2 − . . . − ym−1 | it follows that Z dy2 . . . dym−1 ′ ˜ Jm,1 (τ, x, x ) ≤ (y2 ,...,ym−1 )∈Em−2,n (τ,x,x′) Pm−1 n−1 . n−2 . . . |y n−1 |x − x′ − P t0 |y | | 2 m−1 |x−x′ − m−1 yi |< m−1 i=2 yi | i=2 (9.26) (9.27) t0 |y2 |≥ m−1 Therefore performing the change of variables “y2 ”= x − x′ − y2 − . . . − ym−1 we obtain  n−2 m−1 ′ ˜ ˜ Jm,1 (τ, x, x ) ≤ Jm,0 ≤ Pm (τ, x, x′ ). (9.28) t0 Now we estimate P3 (τ, x, x′ ). From (9.1) and (9.25) it follows that P3 (τ, x, x′ ) = Voln−2 (Sn−2 ) Z 0 π Z τ t0 n−3 sinn−2 (ϕ) (s2 − t20 ) 2 (s − t0 cos(ϕ)) dsdϕ. 2(s + t0 cos(ϕ))n−2 37 (9.29) Let n = 3. Then using the estimate cos(ϕ) ≥ −1 (and the fact that t0 (1−cos(ϕ)) s+t0 cos(ϕ)) ≤1+ 2t0 ) s+t0 cos(ϕ) s−t0 cos(ϕ) s+t0 cos(ϕ) = + we obtain τ π   d ln(s + t0 cos(ϕ) dϕds P3 (τ, x, x ) ≤ π sin(ϕ) − 2 dϕ t0 0   Z τ Z τ ln(s − t0 )ds ln(s + t0 )ds − = 2π τ − t0 + t0 t0    τ + t0 = 2π(τ − t0 ) 2 + ln τ − t0 ′ s−t0 s+t0 cos(ϕ) Z Z (9.30) (we used the estimate ln(s + t0 ) ≤ ln(τ + t0 ) for s ∈ (t0 , τ )pand we used the integral value (9.11)). Let n ≥ 4. Using (9.29) and using the estimates s2 − t20 sin(ϕ) ≤ s + t0 cos(ϕ), s + t0 cos(ϕ) ≥ s(1 + cos(ϕ)) and s − t0 cos(ϕ) ≤ s + t0 ≤ 2s for (s, ϕ) ∈ (t0 , τ ) × (0, π), we obtain p Z πZ τ 2 s2 − t20 (s − t0 cos(ϕ)) sin (ϕ) dsdϕ P3 (τ, x, x′ ) ≤ Voln−2 (Sn−2 ) 2(s + t0 cos(ϕ))2 0 t0 Z τZ π p 2 s − t20 n−2 ≤ C0 Voln−2 (S ) dϕds ≤ πC0 Voln−2 (Sn−2 )(τ − t0 ), (9.31) t0 0 s + t0 cos(ϕ) where C0 is defined by (5.33) (we also used (7.1)). Finally let m ≥ 4 then (y2 , . . . , ym−1 ) ∈ Em−2,n (τ, x, x′ ) ⇒ y2 ∈ E1,n (τ, x, x′ ) and m−1 X i=3 |yi | < τ ! . (9.32) Therefore using (9.25), spherical coordinates (“yi”= si ωi, si ∈ (0, +∞), ωi ∈ Sn−1 for i = 3, . . . , m − 1) we obtain Z dy3 . . . dym−1 ′ ′ Pm (τ, x, x ) ≤ P3 (τ, x, x ) n−1 . . . |y n−1 m−1 | |y3 |+...+|ym−1 |<τ |y3 | Z m−3 m−1 ′ n−1 m−3 τ ′ n−1 m−3 Πi=3 dsi = P3 (τ, x, x )Voln−1 (S ) ≤ P3 (τ, x, x )Voln−1 (S ) . (9.33) s3 +...+sm−1 <τ (m − 3)! 0<si , i=3,...,m−1 Finally statement (8.7) follows from (9.22), (9.26), (9.28), (9.30) and (9.33), and statement (8.8) follows from (9.22), (9.26), (9.28), (9.31) and (9.33). Proof of Lemma 8.3. Let n ≥ 2. Using a rotation and (8.1), we have Voln (E1,n (τ, x, x′ )) = Voln (E1,n (τ, t0 e1 , 0)), (9.34) where t0 = |x − x′ | and e1 = (0, . . . , 0) ∈ Rn . From (9.1), it follows that Voln (E1,n (τ, t0 e1 , 0)) = Voln−2 (Sn−2 ) Z τ t0 Z π 0 38 !n−2 p s2 − t20 cos2 (ϕ) sin(ϕ) s2 − t20 p dsdϕ. 2 4 s2 − t20 (9.35) p p From (9.35) and the estimate sin(ϕ) s2 − t20 ≤ τ 2 − t20 for s ∈ (t0 , τ ), we obtain !n−2 Z Z τ π 2 τ 2 − t20 s − t20 cos2 (φ) p dsdϕ 2 4 s2 − t20 t0 0 !n−2 p τ 2 − t20 Vol(E1,2(τ, t0 e1 , 0)). (9.36) 2 p Voln (E1,n (τ, t0 e1 , 0)) ≤ Voln−2 (Sn−2 ) ≤ 1 Voln−2 (Sn−2 ) 2 We remind that Vol(E1,2 (τ, t0 e1 , 0)) = 8.3 is proved. √ π(t0 +τ ) τ 2 −t20 . 4 Therefore (8.9) follows from (9.36). Lemma Proof of Lemma 8.4. Let (µ, z, z ′ ) ∈ (0, T ) ×√∂X × R3 be such that µ > |z − z ′ | > 0. Using the change of variables y = ω ∈ S1 , we obtain t0 (1, 0) 2 s2 −t20 2 + (s cos(ϕ), π B(µ, z, z ) = 4 ′ where Z µ t0 Z sin(ϕ)ω) (see (9.1)), ϕ ∈ (0, π), s ∈ (t0 , µ), π B1 (µ, s, ϕ)dϕds, (9.37) 0  µ − t0 cos(ϕ) , (9.38) B1 (µ, s, ϕ) = (s − cos (ϕ)) sin(ϕ) ln µ−s     µ−t0 cos(ϕ) µ+t0 for ϕ ∈ (0, 2π) and s ∈ (t0 , µ). Using (9.38) and the estimates ln ≤ ln µ−s , µ−s 2 t20  2 s2 − t20 cos2 (ϕ) ≤ µ2 , we obtain Z π Z 2 B1 (µ, s, ϕ)dϕ ≤ µ 0 π 0  µ + t0 sin(ϕ)dϕ ln µ−s  , for s ∈ (t0 , µ). Combining (9.37), (9.39) and (9.11) we obtain     µ2 π µ + t0 ′ B(µ, z, z ) ≤ +1 , (µ − t0 ) ln 2 µ − t0 (9.39) (9.40) which proves (8.11). 10 The distributional kernel of the operators Hm and the proof of Proposition 3.1 Before we prove Proposition 3.1 we shall introduce and prove Proposition 10.1 given below, which gives the distributional kernel of the operators Hm defined by (2.12). Let Ē denotes the nonnegative mesurable function from Rn × Rn to R defined by − Ē(x1 , x2 ) = e R |x1 −x2 | 0 x −x x −x σ(x1 −s |x1 −x2 | , |x1 −x2 | )ds 1 2 1 2 Θ(x1 , x2 ), for a.e. (x1 , x2 ) ∈ Rn × Rn , (10.1) where Θ is defined by (2.10). For m ≥ 3, we define recursively the nonnegative measurable real function Ē(x1 , . . . , xm ) by the formula Ē(x1 , . . . , xm ) = Ē(x1 , . . . , xm−1 )Ē(xm−1 , xm ), (10.2) for (x1 , . . . , xm ) ∈ (Rn )m . Concerning the distributional kernel of the Hm , m ≥ 2, we have the following result. 39 Proposition 10.1. We have Hm (t)φ(x, v) = Z βm (t, x, v, x′ , v ′ )φ(x′ , v ′ )dx′ dv ′ , (10.3) X×Sn−1 for t ∈ (0, T ) and a.e. (x, v) ∈ X × Sn−1 and for m ≥ 2, where Z t 2n−2 (t − s2 − (x − s2 v ′ − x′ ) · v)n−3 ′ ′ ′ ′ β2 (t, x, v, x , v ) = χ(0,t−s2 ) (|x − (x − s2 v )|) |x − s2 v ′ − x′ − (t − s2 )v|2n−4 0  × Ē(x, x − (t − s1 − s2 )v, x′ + s2 v ′ , x′ )k(x − (t − s1 − s2 )v, v1 , v) (10.4) ×k(x′ + s2 v ′ , v ′, v1 )] v = x−s2 v′ −x′ −(t−s1 −s2 )v ds2 , 1 s1 |x−s2 v ′ −x′ −(t−s2 )v|2 s1 = 2(t−s2 −(x−x′ −s2 v ′ )·v) for t ∈ (0, T ) and a.e. (x, v, x′ , v ′ ) ∈ X × Sn−1 × X × Sn−1 , and where Z Z ′ ′ χ(0,t−sm −...−s2 ) (|x′ + sm v ′ + . . . + s2 v2 − x|) βm (t, x, v, x , v ) = s2 +...+sm ≤t si ≥0, i=2...m (Sn−1 )m−2 2n−2 (t − s2 − . . . − sm − (x − x′ − s2 v2 − . . . − sm−1 vm−1 − sm v ′ ) · v)n−3  Ē(x, x − (t − s1 − . . . − sm )v, |x − x′ − s2 v2 − . . . − sm−1 vm−1 − sm v ′ − (t − s2 − . . . − sm )v|2n−4 x′ + sm v ′ + sm−1 vm−1 + . . . + s2 v2 , x′ + sm v ′ + sm−1 vm−1 + . . . + s3 v3 , . . . , x′ + sm v ′ , x′ )k(x − (t − s1 − . . . − sm )v, v1 , v)k(x′ + sm v ′ + sm−1 vm−1 + . . . + s2 v2 , v2 , v1 ) . . . k(x′ + sm v ′ + sm−1 vm−1 + . . . + si+1 vi+1 , vi+1 , vi ) . . . (10.5) k(x′ + sm v ′ , v ′, vm−1 )] v = x−x′−s2 v2 −...−sm−1 vm−1 −sm v′ −(t−s1 −...−sm )v ds2 . . . dsm dv2 . . . dvm−1 , × 1 s1 |x−x′ −s2 v2 −...−sm−1 vm−1 −sm v ′ −(t−s2 −...−sm )v|2 s1 = ′ 2(t−s2 −...−sm −(x−x −s2 v2 −...sm−1 vm−1 −sm v ′ )·v) for t ∈ (0, T ) and a.e. (x, v, x′ , v ′ ) ∈ X × Sn−1 × X × Sn−1 , m ≥ 3. Proof of Proposition 10.1. Note that Z t Z t−s1  H2 (t)φ(x, v) = U1 (t − s1 − s2 )A2 U1 (s1 )A2 U1 (s2 )φds2 ds1 (x, v) 0 0 Z t Z t−s2   = U1 (t − s1 − s2 )A2 U1 (s1 )A2 ds1 U1 (s2 )φds2 (x, v) 0 0 Z t Z t−s2 Z = Ē(x, x − (t − s1 − s2 )v) k(x − (t − s1 − s2 )v, v1 , v) 0 Sn−1 0 ×Ē(x − (t − s1 − s2 )v, x − (t − s1 − s2 )v − s1 v1 ) Z × k(x − (t − s2 − s1 )v − s1 v1 , v2 , v1 ) Sn−1 ×Ē(x − (t − s1 − s2 )v − s1 v1 , x − (t − s1 − s2 )v − s1 v1 − s2 v2 ) ×φ(x − (t − s1 − s2 )v − s1 v1 − s2 v2 , v2 )dv2 dv1 ds1 ds2 , for t ∈ (0, T ) and (x, v) ∈ X × Sn−1 , where functions Ē are defined by (10.1)–(10.2). Using the change of variables “y(s1, v1 ) = (t − s2 − s1 )v + s1 v1 ” we obtain Z tZ  H2 (t)φ(x, v) = Ē(x, x − (t − s1 − s2 )v, x − y, x − y − s2 v2 )k(x − (t − s1 − s2 )v, v1 , v) 0 Sn−1 × k(x − y, v2, v1 )] v1 = y−(t−s1 −s2 )v s1 |y−(t−s2 )v|2 s1 = 2(t−s2 −y·v) 2n−2 ((t − s2 ) − y · v)n−3 × φ(x − y − s2 v2 , v2 )dydv2ds2 . |y − (t − s2 )v|2n−4 40 Hence we obtain (10.3). Note that Z t (H3 (t)φ)(x, v) = H2 (t − s3 )A2 U1 (s3 )φds3 0 Z tZ = β2 (t − s3 , x, v, x2 , v2 )(A2 U1 (s3 ))φ(x2 , v2 )dx2 dv2 ds3 0 X×Sn−1 Z Z t Z = β2 (t − s3 , x, v, x2 , v2 ) k(x2 , v ′ , v2 )Ē(x2 , x2 − s3 v ′ ) X×Sn−1 Sn−1 0 ′ ′ ′ ×φ(x2 − s3 v , v )dv ds3 dx2 dv2 . Hence Z (H3 (t)φ)(x, v) = where ′ ′ β3 (t, x, v, x , v ) = Z Sn−1 Z tZ 0 β3 (t, x, v, x′ , v ′ )φ(x′ , v ′ )dx′ dv ′ , (10.6) X×Sn−1 t−s3 χ(0,t−s3 −s2 ) (|x′ + s3 v ′ − x + s2 v2 |) 0 2n−2 (t − s2 − s3 − (x − s2 v2 − x′ − s3 v ′ ) · v)n−3 × |x − x′ − s2 v2 − s3 v ′ − (t − s2 − s3 )v|2n−4  × Ē(x, x − (t − s1 − s2 − s3 )v, x′ + s2 v2 + s3 v ′ , x′ + s3 v ′ , x′ ) ×k(x − (t − s3 − s2 − s1 )v, v1 , v)k(x′ + s2 v2 + s3 v ′ , v2 , v1 ) (10.7) × k(x′ + s3 v ′ , v ′, v2 )] v = x−x′ −s2 v2 −s3 v′ −(t−s1 −s2 −s3 )v ds2 ds3 dv2 . 1 s1 |x−x′ −s2 v2 −s3 v ′ −(t−s2 −s3 )v|2 s1 = 2(t−s2 −s3 −(x−x′ −s2 v2 −s3 v ′ )·v) The proof of (10.5) follows by induction from (10.6) and (2.13). Proof of (8.12)–(8.13). We recall that Z (A2 G− (s)φS ) (z, w) = [k(z, v ′ , w)S(x′, v ′ )|ν(x′ ) · v ′ |]v′ = z−x′ E(z, x′ ) φ(s−|z−x′ |, x′ )dµ(x′ ), ′ | |z − x′ |n−1 |z−x ∂X (10.8) for a.e. (z, w) ∈ X × Sn−1 and φ ∈ L1 ((0, η) × ∂X) (see the derivation of (3.9) and (3.10) given in Section 3). Let m = 2. Then from (2.12) and (2.13) it follows that Z Z t Z t−s Z Z A2,S,W (φ)(t, x) = (ν(x) · v)W (x, v) [k(x − (t − s − s1 )v, v1 , v) Sn−1 x,+ × k(x − (t − s − x − (t − s − s1 )v Sn−1 ∂X s1 )v − s1 v1 , v , v1 )S(x , v )|ν(x ) · v ′ |]v′ = x−(t−s−s1 )v−s1 v1 −x′ E(x, x − (t − |x−(t−s−s1 )v−s1 v1 −x′ | ′ ′ φ(s − |x − (t − s − s )v − s 1 1 v1 − x |, x ) − s1 v1 , x′ ) dµ(x′ )dv1 ds1 dsdv. |x − (t − s − s1 )v − s1 v1 − x′ |n−1 ′ ′ −∞ ′ 0 ′ s − s1 )v, Performing the change of variables y(s1, v1 ) = (t − s − s1 )v + s1 v1 , we obtain Z Z t A2,S,W (φ)(t, x) = (ν(x) · v)W (x, v) χ(0,t−s) (|y|) n Sn−1 x,+ ×∂X×R −∞ ′ × [E(x, x − (t − s − s1 )v, x − y, x )k(x − (t − s − s1 )v, v1 , v)k(x − y, v ′, v1 )S(x′ , v ′ ) 2n−2 (t − s − y · v)n−3 φ(s − |x − y − x′ |, x′ ) dsdydµ(x′)dv. × |ν(x′ ) · v ′ |] s = |(t−s)v−y|2 1 2(t−s−y·v) |(t − s)v − y|2n−4 |x − y − x′ |n−1 y−(t−s−s1 )v s1 x−y−x′ v′ = |x−y−x′ | v1 = 41 (10.9) Performing the change of variables “y”= x − x′ − y and t′ = s − |y| we obtain (8.12). Let m = 3. Then from (3.5), (10.3) and (10.8) it follows that Z Z t Z A3,S,W (φ)(t, x) = (ν(x) · v)W (x, v) β2 (t − s, x, v, x2 , v2 ) (10.10) Sn−1 x,+ Z ∂X X×Sn−1 −∞ [k(x2 , v ′, v2 )S(x′ , v ′ )|ν(x′ ) · v ′ |]v′ = x2 −x′ |x2 −x′ | ′ E(x2 , x ) φ(s − |x2 − x′ |, x′ )dµ(x′)dx2 dv2 dsdv, ′ n−1 |x2 − x | for t ∈ (0, T ) and x ∈ ∂X. From (10.10) and (10.4) we obtain Z Z Z t Z t−s A3,S,W (φ)(t, x) = (ν(x) · v)W (x, v) χ(0,t−s−s2 ) (|x2 − (x − s2 v2 )|) Sn−1 x,+ X×Sn−1 ×∂X −∞ n−3 n−2 0 2 (t − s − s2 − (x − s2 v2 − x2 ) · v) |x2 − x′ |n−1 |x − s2 v2 − x2 − (t − s − s2 )v|2n−4 × [E(x, x − (t − s1 − s2 )v, x2 + s2 v2 , x2 , x′ )k(x − (t − s1 − s2 )v, v1 , v)k(x2 + s2 v2 , v2 , v1 ) k(x2 , v ′ , v2 )S(x′ , v ′)|ν(x′ ) · v ′ |] v1 = x−s2 v2 −x2 −(t−s−s1 −s2 )v φ(s − |x2 − x′ |, x′ )ds2 dsdx2 dv2 dµ(x′ )dv. s1 |x−s2 v2 −x2 −(t−s−s2 )v|2 s1 = 2(t−s−s2 −(x−x2 −s2 v2 )·v) x −x′ v′ = 2 |x2 −x′ | Performing the change of variables y2 = s2 v2 and y3 = x2 − x′ we obtain (8.13) for “m = 3”. Let m ≥ 3. From (3.5), (10.3), (10.5) and (10.8) it follows that Z Z Z Z Am+1,S,W (φ)(t, x) = (ν(x) · v)W (x, v) βm (t − t′ − |xm − x′ |, x, v, xm , vm ) Sn−1 x,+ ∂X X (−∞,t−|xm −x′ |)×Sn−1 E(xm , x′ ) [k(xm , v ′ , vm )S(x′ , v ′ )|ν(x′ ) · v ′ |]v′ = xm −x′ φ(t′ , x′ )dµ(x′ )dt′ dxm dvm dv ′ n−1 |xm −x′ | |xm − x | Z = γm+1 (t − t′ , x, x′ )φ(t′ , x′ )dt′ dµ(x′ ), (10.11) (0,η)×∂X where ′ γm+1 (τ, x, x ) := Z (Sn−1 )m−2 Z Z Sn−1 x,+ (ν(x) · v)W (x, v) s2 +...+sm ≤τ −|xm −x′ | si ≥0, i=2...m Z X×Sn−1 χ(0,+∞) (τ − |xm − x′ |) χ(0,τ −|xm −x′ |−sm −...−s2 ) (|xm + sm vm + . . . + s2 v2 − x|) 2n−2 (τ − |xm − x′ | − s2 − . . . − sm − (x − xm − s2 v2 − . . . − sm vm ) · v)n−3 × |xm − x′ |n−1 |x − xm − s2 v2 − . . . − sm vm − (τ − |xm − x′ | − s2 − . . . − sm )v|2n−4 × [E(x, x − (τ − |xm − x′ | − s1 − . . . − sm )v, xm + sm vm . . . + s2 v2 , xm + sm vm + . . . + s3 v3 , . . . , xm + sm vm , xm , x′ )k(x − (τ − |xm − x′ | − s1 − . . . − sm )v, v1 , v)k(xm + sm vm + . . . + s2 v2 , v2 , v1 ) . . . k(xm + sm vm + . . . + si+1 vi+1 , vi+1 , vi ) . . . k(xm + sm vm , vm , vm−1 )k(xm , v ′ , vm )S(x′ , v ′)|ν(x′ ) · v ′ |] v = x−xm −s2 v2 −...−sm vm −(τ −|xm −x′ |−s1 −...−sm )v 1 s1 = ds2 . . . dsm dv2 . . . dvm−1 dxm dvm dv. s1 |x−xm −s2 v2 −...−sm vm −(τ −|xm −x′ |−s2 −...−sm )v|2 2(t−s2 −...−sm −(x−x′ −s2 v2 −...sm vm )·v) x −x′ v′ = m |xm −x′ | (10.12) Performing the change of variables yi = si vi , i = 2 . . . m, and ym+1 = xm+1 − x′ , we obtain (8.13) for “m ≥ 4”. 42 Acknowledgment This paper was funded in part by grant NSF DMS-0554097. 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