Stability Estimates for an Inverse Problem for the Linear Boltzmann Equation


 Alexandrina Carter
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1 Stability Estimates for an Inverse Problem for the Linear Boltzmann Equation Rolci CIPOLATTI, CarlosM.MOTTA, and Nilson C. ROBERTY Departamento de Métodos Matemáticos Instituto de Matemática Universidade Federal do Rio de Janeiro C.P. 6853, CEP Rio de Janeiro Brazil ICE/P1/DEMAT Universidade Federal Rural do Rio de Janeiro Rio de Janeiro Brazil Programa de Engenharia Nuclear COPPE, Universidade Federal do Rio de Janeiro C.P. 6859, CEP Rio de Janeiro Brazil Received: March 18, 25 Accepted: June 14, 25 ABSTRACT In this paper we consider the inverse problem of recovering the total extinction coefficient and the collision kernel for the timedependent Boltzmann equation via boundary measurements. We obtain stability estimates for the extinction coefficient in terms of the albedo operator and also an identification result for the collision kernel. Key words: inverse problems, albedo operator, stability estimates. 2 Mathematics Subject Classification: 35R3, 83D Introduction In this paper we consider an inverse problem for the linear Boltzmann equation t u + ω x u + qu = K f [u] in (,T) S Ω, (1) Rev. Mat. Complut. 19 (26), no. 1, ISSN:
2 where T>, Ω is a bounded and convex domain of R N, N 2, S = S N 1 denotes the unit sphere of R N, q L (Ω) and K f is the integral operator with kernel f(x, ω,ω) defined by K f [u](t, ω, x) = f(x, ω,ω)u(t, ω,x) dω. S In applications (N = 2 or 3), (1) describes the dynamics of a monokinetic flow of particles in a body Ω under the assumption that the interaction between them is negligible (which leads us to discard nonlinear terms). For instance, in the case of a lowdensity flux of neutrons (see [7, 13]), q is the total extinction coefficient and the collision kernel f is given by f(x, ω,ω)=q(x)c(x)h(x, ω ω), where the coefficient c corresponds to the withingroup scattering probability and the kernel h describes the anisotropy of the scattering process. In this case, q(x)u(t, ω, x) describes the loss of particles at x in the direction ω at time t due to absorption or scattering, while the integral on the right hand side of (1) represents the production of particles at x in the direction ω from those coming from directions ω. The inverse problem associated with (1) that we are interested here is recovering q and f by boundary measurements. That will lead us to consider the albedo operator A, that maps the incoming flux on the boundary Ω into the outgoing one. There is a lot of papers devoted to this problem and we specially mention the general results for the identification one, obtained by Choulli and Stefanov [5] (see also [11]), that state that q and f are uniquely determined by A. There is also a wide bibliography concerning the stationary case (See for instance those by V. G. Romanov [14,15], P. Stefanov and G. Uhlmann [16], Tamasan [17], J. N. Wang [18], and also the references therein.) In this paper, we are mainly concerned with stability estimates for the timedependent equation (1). We consider q j, f j, j =1, 2, and A j : L 1 (,T; L 1 (Σ ; dξ)) L 1 (,T; L 1 (Σ + ; dξ)) the corresponding albedo operator (that will be precisely described in section 2 below, as well as the spaces concerned). Since A j is linear and bounded, we consider the usual norm A j 1. Our main result is the following: Theorem 1.1. We assume that q j M for some M > and f j L (Ω; L 2 (S S)), j =1, 2. IfT>diam(Ω), there exists C = C(M) > such that q 1 q 2 H 1 C A 1 A (2) (Ω) Moreover, if q 1,q 2 H N 2 +s (Ω) for some s> and q j N H +s 2 M, j =1, 2, then, (Ω) for each <r<s, there exists C r > such that q 1 q 2 N C H +r 2 r A 1 A 2 θ(r) (Ω) 1, 26: vol. 19, num. 1, pags
3 where θ(r) =2(s r)/(n +2s +1). In particular, for each <r<sthere exists C r such that q 1 q 2 C r A 1 A 2 θ(r) 1. Using the same ideas considered in the proof of Theorem 1.1, we obtain the following partial result on the identification of both q and f in (1): Theorem 1.2. Let q 1,q 2 L (Ω) and f 1,f 2 L ( Ω; L 2 (S S) ).IfT>diam(Ω) and A 1 = A 2, then q 1 = q 2. Moreover, if f j (x, ω,ω)=g j (x)h(ω,ω), with h L (S S) such that h(ω, ω),a.e.ons, then g 1 = g 2. We organize the paper as follows: in section 2 we introduce the functional framework in which the initialboundary value problem for (1) is well posed in the sense of the semigroup theory and where the albedo operator is defined over suitable L p spaces; in section 3, following an argument used in [18] and an analogous strategy used for identifying coefficients in the wave equation (see [6, 12]), we construct special oscillatory solutions of (1) that allow us to prove Theorem 1.1. Although the result established in the Theorem 1.2 is not new, we also present here its proof to highlight the method. 2. Functional framework and wellposedness results For the reader s convenience we gather below a few more or less well known results concerning the functional framework for the operator ω x and the semigroup it generates. Let Ω R N (N 2) be a convex and bounded domain of class C 1 and S = S N 1 the unit sphere of R N. We denote by = S Ω and Σ its boundary, i.e., Σ = S Ω. For p [1, + ) we consider the space L p () with the usual norm ( 1/p u Lp () = u(ω, x) p dx dω), where dω denotes the surface measure on S associated to the Lebesgue measure in R N 1. For each u L p () we define A u by (A u)(ω, x) =ω x u(ω, x) = N k=1 ω k u x k (ω, x), ω =(ω 1,...,ω N ) where the derivatives are taken in the sense of distributions in Ω. One checks easily that setting W p = { u L p (); A u L p () }, the operator (A, W p ) is a closed densely defined operator and W p with the graph norm is a Banach space : vol. 19, num. 1, pags
4 For every σ Ω, we denote ν(σ) the unit outward normal at σ Ω andwe consider the sets Σ ± = { (ω, σ) S Ω; ± ω ν(σ) > }. We also denote Σ ± ω = { σ Ω; (ω, σ) Σ ± } (respectively, the incoming and outcoming boundaries in the direction ω). Remark. It is well known that functions u W p may present singularities on Σ at points (ω, σ) such that ω ν(σ) =, in such a way that u Σ / L p (Σ). For instance, consider α>, Ω = { (x 1,x 2 ) R 2 ; x x 2 2 < 1 } and u : S Ω R defined by u(ω, x) = 1 (ω 2 x 1 ω 1 x 2 +1) α. It is easy to check that if α<3/4 one has u W 2 but u / Σ + L2 (Σ + ). Yet worse, if α [1/2, 3/4) one has (ω ν(σ)) u(ω, σ) 2 dωdσ =+. Σ + In order to well define the albedo operator as a trace operator on the outcoming boundary, we consider L p (Σ ± ; dξ), where dξ = ω ν(σ) dσdω, and we introduce the spaces W p ± = {u W p ; u Σ ± Lp (Σ ± ; dξ)}, which are Banach spaces if equipped with the norms ( ) 1/p u W± p = u p W p + ω ν(σ) u(ω, σ) p dσdω. Σ ± The next two lemmas concern the continuity and surjectivity of the trace operators (see [3, 4]): γ ± : W p ± L p (Σ ; dξ), γ ± (u) =u. Σ Lemma 2.1. Let 1 p<+. Then there exists C> (depending only on p) such that ω ν(σ) u(ω, σ) p dσdω C u p W±, u W p ±. (3) Σ p Moreover, if p>1 and 1/p +1/p =1, we have the Gauss identity div x (uvω) dxdω = ω ν(σ)u(ω, σ)v(ω, σ) dσdω, (4) for all u W ± p and v W ± p. Σ 26: vol. 19, num. 1, pags
5 Proof. Let ϕ C 1 (R) andu C(S; C 1 (Ω)). Then, for each ω S, wehavefrom Gauss Theorem, (ω ν)uϕ(u) dσ = (ω u)[ϕ(u)+uϕ (u)] dx + ω ν uϕ(u) dσ. (5) Σ + ω Ω If p 2, we consider ϕ(s) =s s p 2. Then ϕ C 1 (R) and (5) takes the form (ω ν) u p dσ = p (ω u)u u p 2 dx + ω ν u p dσ. (6) Σ + ω Ω From the Young inequality, we get u u Ω ω p 1 dx 1 ω u p dx + 1 p Ω p u p. (7) Ω By substituting (7) in (6) and integrating over S, we obtain (3). The general case follows by density. For 1 p<2, we consider ϕ ε (s) =s(ε + s 2 ) (p 2)/2. Then ϕ C 1 (R) and since ϕ ε (s) ε s s p 2, Σ ω ϕ ε(s) ε (p 1) s p 2, ϕ ε (s) s p 1, ϕ ε(s) (p 1) s p 2 + ε p/2 1 we obtain (6) from (5) by application of Lebesgue Theorem. The conclusion follows, as before, by density. As an immediate consequence of Lemma 2.1, we can introduce the space { } W p = f W p ; ω ν(σ) f(ω, σ) p dωdσ < + andwehavethat Corollary 2.2. W + p = W p = W p with equivalent norms. Σ Lemma 2.3. The trace operators γ ± : W± p L p (Σ ; dξ) are surjective. More precisely, for each f L p (Σ ; dξ), there exists h W p ± such that γ ± (h) =f and Σ ω h W± p C f L p (Σ,dξ), where C> is independent of f. Proof. See [3, 4]. We consider the operator A : D(A) L p (), defined by (Au)(ω, x) =ω u(ω, x), with D(A) ={ u W p ; γ + (u) =} : vol. 19, num. 1, pags
6 Theorem 2.4. The operator A is maccretive in L p (), forp [1, + ). Proof. It follows from the following two lemmas. Lemma 2.5. The operator A is accretive in L p (), i.e., u + λau L p () u L p (), u D(A), λ >. Proof. It is sufficient to prove that u + Au L p () 1, u D(A), such that u L p () =1. We prove first the case 1 <p<. Letu C(S; C 1 (Ω)) with u =onσ and such that u Lp () = 1. Then { u + Au Lp () =sup (u + ω u)vdxdω; v L p (), v L p () }, =1 where 1/p +1/p =1. Ifv = u p 2 u, then v L p () and v L p () = 1. Hence u + Au L p () ( u(ω, x) p + u(ω, x) p 2 u(ω, x)ω u(ω, x)) dxdω. Since u(ω, x) p 2 u(ω, x)ω u(ω, x) = 1 p div x(ω u(ω, x) p ), we have, from Gauss Theorem, x) Ω u(ω, p 2 u(ω, x)ω u(ω, x) dx = 1 ω ν(σ) u(ω, σ) p dσ. p Σ + ω Therefore, u + Au L p () u(ω, x) p dxdω =1. The conclusion follows by density. The case p = 1 follows easily by a density argument and the fact that u p u 1 as p 1, for all u L 2 () L 1 (). (See also the Corollary 2.7 below.) In order to prove the maximality of A, weconsider,foru L p (), the extension of u, that is the function ũ : S R N R, { u(ω, x) if (ω, x) S Ω, ũ(ω, x) = otherwise. It defines the extension operator u ũ which is a continuous operator from L p () into L p (S R N ). 26: vol. 19, num. 1, pags
7 Let L λ : L p () L p () be the operator defined by (L λ u)(ω, x) = e s ũ(ω, x λsω) ds, (ω, x) S Ω. Lemma 2.6. L λ satisfies the following properties, for each p [1 + ): (i) L λ L(L p (),L p ()), L λ L(Lp (),L p ()) 1; (ii) u L p (), L λ u u in L p (); λ (iii) (I + λa)(l λ u)=u, u L p (); (iv) L λ L(L p (), W p ), L λ L(L p (),W p) 2/λ; (v) L λ (u) D(A), u L p (). Proof. Let u L p (). Then, (L λ u)(ω, x) e s ũ(ω, x λsω) ds ( ) 1/p ( ) 1/p e s ds e s ũ(ω, x λsω) p ds ( ) 1/p = e s ũ(ω, x λsω) p ds which yields Ω (L λ u)(ω, x) p dx e s ( Ω ) ũ(ω, x λsω) p dx ds ) ds e s (R N ũ(ω, y) p dy = u(ω, ) p L p (Ω). By integrating the expression above over S we obtain L λ u Lp () u Lp () and the item (i) is proved. In order to prove (ii), we first note that L λ u(ω, x) u(ω, x) = So, thanks to Hölder inequality, we have L λ u(ω, x) u(ω, x) p e s [ũ(ω, x λsω) ũ(ω, x)] ds. e s ũ(ω, x λsω) ũ(ω, x) p ds : vol. 19, num. 1, pags
8 By integrating this last inequality over Ω and applying Fubini s theorem, we obtain ) L λ u(ω, x) u(ω, x) p dx e (R s ũ(ω, x λsω) ũ(ω, x) p dx ds N Ω and the conclusion follows from the Lebesgue Theorem, since the translation operator (τ λ g)(x) =g(x λsω) defines a continuous group of isometries in L p (R N ), that is, for 1 p<, lim τ λg g p =. λ By a direct application of Fubini s Theorem, one checks easily that the adjoint of L λ is given by (L λu)(ω, x) = e s ũ(ω, x + λsω) ds, since (L λ u)(ω, x)v(ω, x) dxdω = (L λv)(ω, x)u(ω, x) dxdω, for each u L p () andv L p (). Moreover, for u L p (), L λ u(ω, ),ϕ = u(ω, x)l λ(ϕ)(x) dx, ϕ C (Ω). On the other hand, Ω L λ(λaϕ)(x) =λ e s ω ϕ(x + λsω) ds = e s d ϕ(x + λsω) ds ds = ϕ(x)+(l λϕ)(x), ϕ C (Ω) andwehave L λ u; ϕ = ul λϕ = ũl λϕ = Ω R N = u; ϕ + ũl λ(λω ϕ) R N = u; ϕ L λ u; λω ϕ = u λω (L λ u); ϕ R N ũ[ϕ + L λ(λω ϕ)] which means that (I + λa)(l λ u)=u in the sense of distributions in Ω. Since λa(l λ u)=u L λ u L p (), we obtain L λ u W p.moreover, Au Lp () 1 λ ( u L p () + L λ u Lp ()) 2 λ u L p (). In order to prove that L λ u D(A), we note that if u C(S Ω), then it is obvious that u L p () andl λ u =onσ. Again, the general case follows by density. 26: vol. 19, num. 1, pags
9 Corollary 2.7. Let f L p (), p [1, + ) and assume that u D(A) is a solution of u + Au = f. Iff a.e. in, then u a.e. in. In particular, it follows that u L1 () f L1 (). Proof. Since (L 1 f)(ω, x) = e s f(ω, x sω) ds iff, we obtain u from items (iii) and (v) in Lemma 2.6. Now assume that f L 1 () and let v D(A) the solution of v + Av = f. Then v and f f f implies that u v. Therefore, by integrating over, we obtain u L1 () v(ω, x) dxdω [v(ω, x)+av(ω, x)] dxdω = f L1 (). It follows from Theorem 2.4 and Corollary 2.7 that the operator A generates a positive semigroup {U (t)} t of contractions acting on L p (). Let q L (Ω) and f :Ω S S R be a real measurable function satisfying { S f(x, ω,ω) dω M 1 a.e. Ω S, S f(x, (8) ω,ω) dω M 2 a.e. Ω S. Associated to these functions, we define the following continuous operators: (a) B L(L p (),L p ()) defined by B[u](ω, x) =q(x)u(ω, x), (b) K f [u](ω, x) = S f(x, ω,ω)u(ω,x) dω. It follows from (8) that K f L(L p (),L p ()) p [1, + ) and (see [7]) K f [u] Lp () M 1/p 1 M 1/p 2 u Lp (). The operator A + B K f : D(A) L p () generates a c semigroup {U(t)} t on L p () satisfying U(t) L e Ct, C = q + M Stability and identification of parameters We consider the initialboundary value problem for the linear Boltzmann equation t u(t, ω, x)+ω u(t, ω, x)+q(x)u(t, ω, x) =K f [u](t, ω, x) u(,ω,x)=, (ω, x) S Ω (9) u(t, ω, σ) =ϕ(t, ω, σ), (ω, σ) Σ, t (,T), where q L (Ω) and K f [u](t, ω, x) = S f(x, ω,ω)u(t, ω,x) dω, : vol. 19, num. 1, pags
10 with f satisfying (8). By the results stated in section 2, it follows that, for any p [1, + ), if ϕ L p (,T; L p (Σ,dξ)), there exists a unique solution u C([,T]; W p ) C 1 ([,T]; L p ()) of (9). This solution u allows us to define the albedo operator A q,f : L p(,t; L p (Σ,dξ) ) L p(,t; L p (Σ +,dξ) ) A q,f [ϕ](t, ω, σ) =u(t, ω, σ), (ω, σ) Σ +. As a consequence of Lemmas 2.1 and 2.3, A q,f is a linear and bounded operator. So, we denote by A q,f p its norm. In order to simplify our notation, we will consider from now on L ± p = Lp (,T ;L p (Σ ±,dξ)). We also consider the following backwardboundary value problem, called the adjoint problem of (9): t v(t, ω, x)+ω v(t, ω, x) q(x)v(t, ω, x) = Kf [v](t, ω, x) v(t,ω,x)=, (ω, x) S Ω (1) v(t, ω, σ) =ψ(t, ω, σ), (ω, σ) Σ +,t (,T), ( where ψ L p,t; L p (Σ +,dξ) ), p [1, + ), Kf [v](t, ω,x)= f(x, ω,ω)v(t, ω, x) dω with the corresponding albedo operator A q,f S A q,f : L p (,T; L p (Σ +,dξ)) L p (,T; L p (Σ,dξ)) A q,f[ψ](t, ω, σ) =v(t, ω, σ), (ω, σ) Σ. The operators A q,f and A q,f satisfy the following property: Lemma 3.1. Let ϕ L p (,T ; L p (Σ ; dξ)) and ψ L p (,T ; L p (Σ + ; dξ)), where p, p (1, + ) are such that 1/p +1/p =1. Then, we have T (ω ν(σ))ϕ(t, ω, σ)a q,f[ψ](t, ω, σ) dσdωdt = Σ T = (ω ν(σ))ψ(t, ω, σ)a q,f [ϕ](t, ω, σ) dσdωdt. Σ + Proof. It is a direct consequence of Lemma 2.1. Let u(t, ω, x) the solution of (9) with boundary condition ϕ and v(t, ω, x) the solution of (1) with boundary condition ψ. We obtain the result by using (4), once the equation in (9) is multiplied by v and integrated over. 26: vol. 19, num. 1, pags
11 Lemma 3.2. Let T >, q 1,q 2 L (Ω) and f 1, f 2 satisfying (8). Assume that u 1 is the solution of (9) with (q, f) =(q 1,f 1 ) and boundary condition ϕ L p (,T; L p (Σ,dξ)), p (1, + ) and that ( v 2 is the solution of (1), with(q, f) =(q 2,f 2 ) and boundary condition ψ L p,t; L p (Σ +,dξ) ), 1/p +1/p =1. Then we have T ( q2 (x) q 1 (x) ) u 1 (t, ω, x)v 2 (t, ω, x) dxdωdt T + K f1 f 2 [u 1 ](t, ω, x)v 2 (t, ω, x) dxdωdt T = (ω ν(σ)) [ A 1 [ϕ] A 2 [ϕ] ] (t, ω, σ)ψ(t, ω, σ) dσdωdt, Σ + where A j = A qj,f j, j =1, 2. Proof. It is a direct consequence of Lemma 3.1. Lemma 3.3. Let T >, q 1,q 2 L (Ω), andf 1,f 2 satisfying (8). ψ 1,ψ 2 C ( S, C (R N ) ) such that We consider supp ψ 1 (ω, ) Ω = (supp ψ 2 (ω, )+Tω) Ω=, ω S. (11) Then, there exists C > such that, for each λ>, there exist R 1,λ C([,T]; W p ) and R 2,λ C([,T]; W p ) satisfying R 1,λ C([,T ];L p ()) C, R 2,λ C([,T ];L p ()) C, p, p (1, + ), (12) with 1/p +1/p =1, for which the functions u 1,v 2 defined by ( t ) u 1 (t, ω, x) = ψ 1 (ω, x tω)exp q 1 (x sω) ds +iλ(t ω x) +R 1,λ (t, ω, x) ( t ) v 2 (t, ω, x) = ψ 2 (ω, x tω)exp q 2 (x sω) ds iλ(t ω x) +R2,λ (t, ω, x) (13) are solutions of (9) with (q, f) =(q 1,f 1 ) and (1) with (q, f) =(q 2,f 2 ) respectively. Moreover, if f 1,f 2 L (Ω; L 2 (S S)), then we have lim R 1,λ L λ + 2 ([,T ] ) = lim λ + R 2,λ L 2 ([,T ] ) =. (14) Proof. Let u be the function ( t ) u(t, ω, x) =ψ 1 (ω, x tω)exp q 1 (x sω) ds +iλ(t ω x) + R(t, ω, x). (15) : vol. 19, num. 1, pags
12 By direct calculations, we easily verify that t u + ω u + q 1 u K f1 [u] = t R + ω R + q 1 R K f1 [R] exp(iλt)z 1,λ, where z 1,λ (t, ω, x) = S f 1 (x, ω,ω)ψ 1 (ω,x tω )e t q1(x sω ) ds iλω x dω. By choosing R 1,λ C 1 ([,T]; L p ()) C([,T]; D(A)) the solution of t R + ω R + q 1 R = K f1 [R] + exp(iλt)z 1,λ, R(,ω,x) =, (ω, x) S Ω, R(t, ω, σ) =, (ω, σ) Σ, (16) we see that (11) implies that the function u defined by (15) satisfies (9) with boundary condition ϕ(t, ω, σ) =ψ 1 (ω, σ tω)e t q1(σ sω) ds+iλ(t ω σ), (ω, σ) Σ. Multiplying both sides of the equation in (16) by the complex conjugate of R p 2 R, integrating it over and taking its real part, we get, from Lemma 2.1, 1 d p dt R(t) p dωdx + 1 ω ν(σ) R(t) p dωdσ + q 1 R(t) p dωdx p Σ + ] R K f1 [R(t)] R(t) p 2 R(t)dωdx = R [e iλt z 1,λ (t) R(t) p 2 R(t) dωdx. It follows from the Hölder inequality and (8) that K f1 [R(t)] R(t) p 1 dxdω C p R(t) p L p (), where C p = M 1/p 1 M 1/p 2 max{m 1,M 2 }. Therefore, considering the decomposition q 1 = q 1 + q 1, we obtain, d dt R(t) p L p () pc 1 R(t) p L p () + z 1,λ(t) p L p (), where C 1 = q1 + max{m 1,M 2 } + 1 and the Gronwall inequality implies that R(t) p L p () z 1,λ(t) p L p () ept C1, t [,T]. The first inequality in (12) follows easily because z 1,λ (t, ω, x) ψ 1 e q 1 T M 1. 26: vol. 19, num. 1, pags
13 Since the same arguments hold for v 2 and R 2,λ with p in place of p, we obtain the second inequality in (12). We assume now f L (Ω; L 2 (S S)). Following the same steps as before, we obtain R(t) 2 L 2 () z 1,λ(t) 2 L 2 () e2c1t, t [,T], (17) where C 1 = q 1 + f L (Ω;L 2 (S S)) + 1. Since for each x R N, the map ω exp(iλω x) converges weakly to zero in L 2 (S) when λ + and the integral operator with kernel f 1 (x,, ) is compact in L 2 (S), we obtain lim z 1,λ(t,,x) L2 (S) = a.e. in [,T] Ω. λ + Moreover, z 1,λ (t,,x) L2 (S) C, where C> is a constant that does not depend on λ. The Lebesgue s Dominated Convergence Theorem (Lebesgue Theorem in short) implies that lim z 1,λ L2 ([,T ] ) =. (18) λ + From (18) and (17) we obtain (14), and our proof is complete. We are now in position to prove our main result. Proof of Theorem 1.1. Let ε =(T diam(ω))/2 and consider Ω ε = { x R N \ Ω ; dist(x, Ω) <ε}. Let χ C(S) and Φ, Ψ C (Ω ε ). We define ψ 1 (ω, x) =χ(ω)φ(x) andψ 2 (ω, x) = Ψ(x). Since T > diam(ω) implies that ψ 1,ψ 2 satisfy (11), we may consider the solutions u 1 and v 2 defined by (13). Denoting by f = f 1 f 2 and ρ = q 1 q 2 ( q j being the zero extension of q j in the exterior of Ω), we have by Lemma 3.2, T ρ(x)e t ρ(x sω) ds χ(ω)φ(x tω)ψ(x tω) dxdωdt T + z λ (t, ω, x)ψ(x tω)e t q2(x sω) ds+iλω x dxdωdt + I λ A 1 A 2 p ϕ L p ψ L +, p where z λ (t, ω, x) = f(x, ω,ω)χ(ω )Φ(x tω )e iλω x t q1(x sω )ds dω, S ϕ(t, ω, σ) =χ(ω)φ(σ tω)e iλ(t ω σ) t q1(σ sω) ds, ψ(t, ω, σ) =Ψ(σ tω)e t q2(σ sω) ds iλ(t ω σ) and I λ represents the sum of the integrals that contain terms with R 1,λ and R 2,λ : vol. 19, num. 1, pags
14 Because R 1,λ, R2,λ and z λ converge to zero in L 2 ([,T] ), we get, after taking the limit as λ +, T ρ(x)e t ρ(x sω) ds χ(ω)φ(x tω)ψ(x tω) dxdωdt S R N A 1 A 2 p ϕ L p ψ L + p Since T ρ(x)e t ρ(x sω) ds χ(ω)φ(x tω)ψ(x tω) dxdωdt S R N T = ρ(y + tω)e t ρ(y+sω) ds χ(ω)φ(y)ψ(y) dydωdt S R N [ = 1 e T ρ(y+sω) ds] χ(ω)φ(y)ψ(y) dydω, S R N it follows that [ 1 e T ρ(y+sω) ds] χ(ω)φ(y)ψ(y) dydω A 1 A 2 p ϕ L p ψ L +. S R N p Note that ψ L + C 1 Ψ L p (R N ), C 1 = C(T,M, S, Ω ), lim ϕ p 1 L = ϕ p L, 1 lim sup A 1 A 2 p A 1 A 2 1 p 1 which yields [ 1 e T ρ(y+sω) ds] χ(ω)φ(y)ψ(y) dydω C 1 A 1 A 2 1 ϕ L Ψ L. S R N 1 Note also that ϕ L C 2 Φ L1 (R N ) χ 1 L 1 (S), for some C 2 that depends as C 1 on T, M, S, and Ω. Hence, [ 1 e T ρ(y+sω) ds] χ(ω)φ(y)ψ(y) dydω S R N C 3 A 1 A 2 1 Φ L1 (R N ) Ψ L (R N ) χ L 1 (S). Let O Ω ε be an open set such that supp Φ, supp Ψ O. Assume that Ψ 1. Taking a sequence {Ψ k } k such that Ψ k converges a.e. in R N to 1 O, the characteristic function of O, we obtain from Lebesgue Theorem [ 1 e T ρ(y+sω) ds] χ(ω)φ(y) dydω C 3 A 1 A 2 1 Φ L 1 (R N ) χ L1 (S). S O 26: vol. 19, num. 1, pags
15 Now, taking the supremum of the above inequality among all Φ with Φ L1 (R N ) =1, we get [ 1 e T ρ(y+sω) ds] χ(ω) dω C 3 A 1 A 2 1 χ L1 (S). L (O) S In particular, [ 1 e T ρ(y+sω) ds] χ(ω) dω C 3 A 1 A 2 1 χ L1 (S), (19) S for a.e. y O. Since O can be chosen arbitrarily in Ω ε, we obtain (19) for a.e. y Ω ε. Again, taking the supremum among all χ C(S), with χ L1 (S) =1,weget e T ρ(y+sω) ds 1 C 3 A 1 A 2 1 for a.e. (ω, y) S Ω ε. Since q j M, j =1, 2, the Mean Value Theorem gives e T ρ(y+sω) ds T 1 e 2TM ρ(y + sω) ds. and consequently T ρ(y + sω) ds e2tm C 3 A 1 A 2 1 for a.e. (ω, y) S Ω ε. Since T>diam(Ω) and supp ρ Ω, we obtain ρ(y + sω) ds e2tm C 3 A 1 A 2 1 (2) for a.e. (ω, y) S R N. Since Ω is bounded, there exists a R> such that Ω B R,sothatwecan rewrite (2) as P [ρ](ω, y) C 4 A 1 A 2 1 (21) for a.e. ω S and a.e. y ω B R, where P [ρ] denotes the Xray transform of ρ. Taking the square on both sides of (21), we obtain P [ρ] 2 L 2 (T ) = P [ρ](ω, y) 2 dydω C 5 A 1 A 2 2 1, (22) S ω B R where T = { (ω, y); ω S, y ω } denotes the tangent bundle : vol. 19, num. 1, pags
16 For the Xray transform, we have the following well known estimate (see [1]) ρ H 1/2 C P [ρ] L2 (T ), (23) where C > depends only on N. Combining (22) and (23) we obtain (2). The conclusion follows from interpolation formulæ and classical Sobolev imbedding theorems. Proof of Theorem 1.2. Since we are assuming that A q1,f 1 = A q2,f 2, q 1 = q 2 follows trivially from (2). As a consequence, the identity in Lemma 3.2 is reduced to T K f1 f 2 [u 1 ](t, ω, x)v 2 (t, ω, x) dxdωdt =. (24) Assuming that f j (x, ω,ω)=g j (x)h(ω,ω), where h L (S S) andh(ω, ω) a.e. on S, wehavethatf j satisfies (8) with M 1 = M 2 = α N g j h, where α N =2π N/2 /Γ(N/2) is the measure of the unit sphere of R N. For <r<1 we define χ r : S S R as χ r (ω, ω )=P (rω, ω ), where P is the Poisson kernel for B 1 (), i.e., P (x, y) = 1 x 2 α N x y N. From the well known properties of P (see Theorem 2.46 of [8]), we have χ r (ω, ω 2 ) α N (1 r) N 1, ω, ω S, χ r (ω, ω ) dω =1, <r<1, ω S S χ r (ω, ω )ψ(ω ) dω = ψ(ω), lim r 1 S where the limit is taken in the topology of L p (S), p [1, + ) and uniformly on S if ψ C(S). As before, we take Ω ε = { x R N \Ω ; dist(x, Ω) <ε}, where ε =(T diam(ω))/2. For ω S, < r,r < 1, we define the functions ψ 1 (ω, x) = χ r ( ω, ω)φ(x) and ψ 2 (ω, x) =χ r ( ω, ω)ψ(x), where Φ, Ψ C (Ω ε ), It follows from Lemma 3.3 that there exist functions u 1,r and v 2,r in such way that (24) takes the form T [ ] η(x) h(ω, ω )χ r ( ω, ω )Φ(x tω )e t q(x sω )ds iλx ω dω S χ r ( ω, ω)ψ(x tω)e t q(x sω)ds+iλx ω dxdωdt = I λ,r,r( ω), (25) 26: vol. 19, num. 1, pags
17 where q = q 1 = q 2, η = g 1 g 2 and I λ,r,r denotes the sum of the integrals that contain the terms with R 1,λ,r and R2,λ,r. First of all, we write the left hand side of (25) as T [ ] J λ,r,r( ω) = ηk h χr ( ω, )u,λ χr ( ω, )v,λ dxdωdt, where [ t ] u,λ (t, ω, x) =Φ(x tω)exp q(x sω)ds iλω x, [ t ] v,λ (t, ω, x) =Ψ(x tω)exp q(x sω)ds +iλω x. Taking into account the properties of χ r,wehave,asr 1, K h [χ r ( ω, )u,λ ](t, ω, x) h(ω, ω)u,λ (t, ω, x) a.e. in [,T] S. Moreover, since S χ r( ω, ω )dω =1for<r<1, it follows that ηk h [χ r ( ω, )u,λ ]χ r ( ω, )v,λ C N (1 r ) 1 N, a.e. in [,T], where C N > is independent of r. Therefore, as an application of Lebesgue Theorem, we have where J λ,r ( ω) = = T T Ω J λ,r,r( ω) J λ,r ( ω), a.e. ω S, η(x)h( ω, ω)u,λ (t, ω, x)χ r ( ω, ω)v,λ (t, ω, x) dxdωdt η(x)k h[ χr ( ω, )v,λ ] (t, ω, x)u,λ (t, ω, x) dxdt. We repeat the same procedure with r 1 to obtain J λ,r ( ω) J λ ( ω) a.e. ω S, where T J λ ( ω) = η(x)h( ω, ω)v,λ (t, ω, x)u,λ (t, ω, x) dxdt Ω T (26) = h( ω, ω) η(x)φ(x t ω)ψ(x t ω) dxdt. Ω Now we rewrite the right hand side of (25) as I λ,r,r( ω) = T η ( K h [R 1,λ,r ]χ r ( ω, )v,λ + K h [χ r ( ω, )u,λ ]R 2,λ,r + K h[r 1,λ,r ]R 2,λ,r ) dxdωdt, : vol. 19, num. 1, pags
18 where R 1,λ,r is the solution of the initialboundary value problem t R + ω R + q 1 R = g 1 K h [R] + exp(iλt)g 1 z 1,λ,r, R(,ω,x) =, (ω, x) S Ω, R(t, ω, σ) =, (ω, σ) Σ, with z 1,λ,r (t, ω, ω, x) = S h(ω, ω )χ r ( ω, ω )u,λ (t, ω,x) dω and R 2,λ,r is the solution of the corresponding adjoint problem with z 2,λ,r. Again, the properties of χ r and Lebesgue Theorem give, as r 1, z 1,λ,r z 1,λ in L 2 ([,T] S ), where z 1,λ (t, ω, ω, x) =h(ω, ω)u,λ (t, ω, x). From the continuity of the operators U(t) and K h we obtain, as r 1, K h [R 1,λ,r ]( ω) K h [R 1,λ ]( ω) in C([,T]; L 2 ()) for almost every ω S, where R 1,λ is the solution of (27) with z 1,λ taking the place of z 1,λ,r. Therefore, taking the limit as r 1, we have I λ,r,r( ω) I λ,r ( ω) a.e. ω S, where T I λ,r ( ω) = η ( K h [R 1,λ ]χ r ( ω, )v,λ + h(, ω)u,λ R2,λ,r + K h [R 1,λ ]R2,λ,r ) dxdωdt T = η(kh[χ r ( ω, )v,λ ]R 1,λ + Kh[ R 2,λ,r ] R1,λ )dxdω dt+ T + ηkh[r 2,λ,r ](t, ω, x)u,λ(t, ω, x) dxdt Ω (27) We repeat the same procedure with r 1 to obtain I λ,r ( ω) Iλ ( ω) a.e. ω, where T Iλ( ω) = η ( [ ] K h R1,λ v,λ + K ] ) h[ R 2,λ u,λ dxdt Ω T [ ] + ηk h R1,λ R 2,λ dxdωdt. (28) In order to take the limit as λ + in (28), we point out that x e iλ ω x converges weakly to zero in L 2 (Ω) for all ω S. Therefore, using the Lebesgue Theorem and the continuity of the operators U(t) andk h we may conclude that e iλt g 1 z 1,λ (t, ω,, ) R 1,λ (t, ω,, ) K h [R 1,λ ](t, ω,, ), (29) 26: vol. 19, num. 1, pags
19 weakly in L 2 (), t [,T], a.e. ω S. On another hand, it follows from (12) that ω R 1,λ (t, ω, ω, x) is bounded in L 2 (S) a.e. in [,T]. Since K h is compact, we have (by taking a sequence λ n if necessary) that K h [R 1,λ ](t, ω,,x) converges strongly in L 2 (S) asλ +. As a consequence of the Lebesgue Theorem, K h [R 1,λ ](t, ω,, ) converges strongly in L 2 (), t [,T], a.e. ω S. In particular, we have from (29) that K h [R 1,λ ](t, ω,, ) strongly in L 2 (), t [,T], a.e. ω S. (3) With the same arguments, we also obtain K h [R 2,λ](t, ω,, ) strongly in L 2 (), t [,T], a.e. ω S. (31) Since u,λ and v,λ are bounded functions, we see from (3), (31), and (28) that lim λ + I λ( ω) =, a.e. ω S. (32) Since we have by (25), J λ,r,r( ω) =I λ,r,r( ω) a.e. ω S, we take the limit in r, r and λ and we get, from (26), (32), and the assumption that h( ω, ω), that T T η(x + t ω)φ(x)ψ(x) dxdt = η(x)φ(x t ω)ψ(x t ω) dxdt =. R N Ω Since Φ and Ψ are arbitrary and T>diam(Ω), we obtain + η(x + t ω) dt =, a.e. x R N, ω S, which means, by standard arguments for the Xray transform, that η(x) = a.e. in R N. Hence the result. References [1] A.P. Calderón, On an inverse boundary value problem, Seminar on Numerical Analysis and its Applications to Continuum Physics (Rio de Janeiro, 198), 198, pp [2] T. Cazenave and A. Haraux, Equations d évolution linéaires: Semigroupes, applications e approximations, DEA d Analyse Numérique [3] M. Cessenat, Théorèmes de trace L p pour des espaces de fonctions de la neutronique, C.R. Acad. Sci. ParisSér. I Math. 299 (1984), no. 16, (French, with English summary). [4], Théorèmes de trace pour des espaces de fonctions de la neutronique, C. R. Acad. Sci. Paris Sér. I Math. 3 (1985), no. 3, (French, with English summary). [5] M. Choulli and P. Stefanov, Inverse scattering and inverse boundary value problems for the linear Boltzmann equation, Comm. Partial Differential Equations 21 (1996), no. 56, [6] R. Cipolatti and I. F. Lopez, Determination of coefficients for a dissipative wave equation via boundary measurements, J. Math. Anal. Appl. 36 (25), no. 1, : vol. 19, num. 1, pags
20 [7] R. Dautray and J.L. Lions, Mathematical analysis and numerical methods for science and technology. Vol. 6, SpringerVerlag, Berlin, [8] G. B. Folland, Introduction to partial differential equations, Mathematical Notes, Princeton University Press, Princeton, N.J., [9] V. Isakov, Inverse problems for partial differential equations, Applied Mathematical Sciences, vol. 127, SpringerVerlag, New York, [1] A. Louis and F. Natterer, Mathematical problems of computarized tomography, Procc. IEEE 71 (1983), no. 1, [11] M. MokhtarKharroubi, Mathematical topics in neutron transport theory, Series on Advances in Mathematics for Applied Sciences, vol. 46, World Scientific Publishing Co. Inc., River Edge, NJ, [12] Rakesh and W. W. Symes, Uniqueness for an inverse problem for the wave equation, Comm. Partial Differential Equations 13 (1988), no. 1, [13] M. Reed and B. Simon, Methods of modern mathematical physics. III, Academic Press, New York, [14] V. G. Romanov, Estimation of stability in the problem of determining the attenuation coefficient and the scattering indicatrix for the transport equation, Sibirsk. Mat. Zh. 37 (1996), no. 2, (Russian); English transl., Siberian Math. J. 37 (1996), no. 2, [15], Stability estimates in the threedimensional inverse problem for the transport equation, J. Inverse IllPosed Probl. 5 (1997), no. 5, [16] P. Stefanov and G. Uhlmann, Optical tomography in two dimensions, Methods Appl. Anal. 1 (23), no. 1, 1 9. [17] A. Tamasan, An inverse boundary value problem in twodimensional transport, Inverse Problems 18 (22), no. 1, [18] J.N. Wang, Stability estimates of an inverse problem for the stationary transport equation, Ann. Inst. H. Poincaré Phys. Théor. 7 (1999), no. 5, : vol. 19, num. 1, pags
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