Lipschitz stability for a coefficient inverse problem for the non-stationary transport equation via Carleman estimate

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1 Lipschiz sabiliy for a coefficien inverse problem for he non-saionary ranspor equaion via Carleman esimae Michael V. Klibanov Sergey E. Pamyanykh Deparmen of Mahemaics Saisics, The Universiy of Norh Carolina a Charloe, Charloe, NC 83, U.S.A. s: mklibanv@ .uncc.edu, spamyan@uncc.edu Absrac The Lipschiz sabiliy esimae for a coefficien inverse problem for he non-saionary single-speed ranspor equaion wih he laeral boundary daa is obained. The mehod of Carleman esimaes is used. Uniqueness of he soluion follows.. Inroducion The ranspor equaion is used o model a variey of diffusion processes, such as diffusion of neurons in medium, scaering of ligh in he urbulen amosphere, propagaion of rays in a scaering medium, ec. (see, e.g., he book of Case Zweifel [6]). Coefficien inverse problems (CIPs) for he ranspor equaion are he problems of deermining of he absorpion coefficien, angular densiy of sources or scaering indicarix from an exra boundary daa. They find a variey of applicaions in opical omography, heory of nuclear reacors, ec. (see, e.g., he book of Anikonov, Kovanyuk Prokhorov [], [6]). This paper addresses he quesion of he Lipschiz sabiliy for a CIP for he non-saionary single-speed ranspor equaion wih he laeral boundary daa. In general, sabiliy esimaes for CIPs provide guidelines for he sabiliy of corresponding numerical mehods. Sabiliy, uniqueness exisence resuls references o such resuls for CIPs for he saionary ranspor equaion can be found, e. g., in [] in he book of Romanov [3]. Uniqueness exisence resuls for CIPs for he non-saionary ranspor equaion were obained in he works of Prilepko Ivankov [], [] []. Uniqueness exisence resuls in [] [] were obained for special forms of he unknown coefficien using he overdeerminaion a a poin. Also, uniqueness exisence resuls were obained for an inverse problem wih he final overdeerminaion, i.e. where complee laeral boundary daa is no presen bu boh iniial end condiions (a T) are given; see []. For some recen publicaions on inverse problems for he ranspor equaion see Tamasan [5] Sefanov [4]. A derivaion of he ranspor equaion for he non-saionary case can be found, for example, in [6]. The proof of he main resul of his paper is based on a Carleman esimae, obained by Klibanov Pamyanykh [6]. Tradiionally, Carleman esimaes have been used for proofs of sabiliy uniqueness resuls for non-sard Cauchy problems for PDEs. They were firs inroduced by Carleman in 939 [5], also see, e.g., books of Hörmer [7], Klibanov Timonov [7] Lavrenev, Romanov Shishaskii [9]. Bukhgeim Klibanov [4] have inroduced he ool of Carleman esimaes in he field of CIPs for proofs of global uniqueness sabiliy resuls; also, see Klibanov [], [3] [4], Klibanov Timonov [7], [8]. This mehod works for

2 non-overdeermined CIPs, as long as he iniial condiion is no vanishing he Carleman esimae holds for he corresponding differenial operaor (see Chaper in [7] for he definiion of non-overdeermined CIPs). Recenly, Klibanov Timonov have exended he original idea of [4] [] - [4] for he consrucion of numerical mehods for CIPs, including he case when he iniial condiion is he -funcion; see [7] for deails more references. Klibanov Malinsky [5] Kazemi Klibanov [] have proposed o use he Carleman esimaes for proofs of he Lipschiz sabiliy esimaes for hyperbolic equaions wih he laeral Cauchy daa; also see [7]. The mehod of [4], []-[4] [7] has generaed many publicaions, see, for example, Bellassoued [], [3], Imanuvilov Yamamoo [8], [9] [] he references cied herein. The Lipschiz sabiliy of he soluion of he non-saionary ranspor equaion wih he laeral daa was proved in [6]. In his paper he ideas of [] [5] are combined wih he ideas of [8], [9], [6]-[8]. In Secion he saemens of he resuls are given; in Secion 3, 4 5 he proofs of hese resuls are provided.. Saemens of resuls.. Saemens of resuls Le T R be posiive numbers. Denoe x R n : x R, R n :, H T,T, T,T, Z. Also, denoe C k H s C k H : D x, u x,, v C H, k The ranspor equaion in H has he form [6] u, u a x, u g x,,, u x,, d F x,,, (.) where is he uni vecor of paricle velociy, u x,,v C 3 H is he densiy of paricle flow, a x,v is he absorpion coefficien, F x,, v is he angular densiy of sources, g x,, v, is he scaering indicarix, u denoes he scalar produc of wo vecors u. Consider he following boundary condiion u p x,,, where! x,, v T, T" n, v#. (.) Here n,v is he scalar produc of he ouer uni normal vecor n o he surface he direcion v of he velociy. So, only incoming radiaion is given a he boundary in his case. Equaion (.) wih he boundary condiion (.) he iniial condiion a u x,, v f x,, $ x,v Z, form he classical forward problem for he ranspor equaion in any direcion of (posiive or negaive). Uniqueness, exisence sabiliy resuls for his problem are well known, see, e. g., Prilepko Ivankov []. Suppose now ha he absorpion coefficien a x,v is unknown, bu he following addiional (.3)

3 boundary condiion is given: u q x,,, where x,, v! T, T" n,v. The funcion q x,,v describes he ougoing radiaion a he boundary. Inroduce he funcion x,, x,, p x,,, q x,,, if if n, v#, n, v. (.4) Hence u x,,, $! x,, v T, T". (.5) Thus, we obain he following coefficien inverse problem for he non-saionary ranspor equaion: Inverse Problem: Given he iniial condiion (.3) he laeral daa (.5), deermine he coefficien a x,v of he equaion (.). For a posiive consan M, denoe D M s x C Z : s C Z M. Theorem. [Lipschiz sabiliy uniqueness] Le T R. Suppose ha derivaives k g exis in H k g C H r for k,,, where r is a posiive consan. Le f x,v r f x,v C Z r 3, where r 3 r. Suppose ha he coefficiens a,a D M correspond o he boundary daa x,, v x,,v, respecively, funcions k i,. Then he following Lipschiz sabiliy esimae holds a a L Z K L i L L for k,,, L ", where K K,T,r,r,r 3,M is he posiive consan depending on, T, r, r, r 3, M independen on he funcions a, a,,. In paricular, when, hen a x,v a x,v which implies ha he Inverse Problem has a mos one soluion. (.6) Below K K,T,r, r,r 3,M denoes differen posiive consans, depending on, T, r, r, r 3, M independen on funcions a, a,,, condiions of Theorem are assumed o be saisfied. The proof of Theorem is based on he Carleman esimae formulaed in Lemma. Le L u u, u u where u i u/ x i. Le x R n. Inroduce he funcion n iu i, i 3

4 x, x x, cons,. Le c cons. Denoe G c x x, : x x c x R. (.7) Obviously, G c G c if c c. (.7_) Inroduce he Carleman Weigh Funcion (CWF) as C x, exp x, ". Lemma. Choose he number such ha, T R/. Also, choose he consan c such ha G c x T,T. Then here exis posiive consans G c x B B G c x, depending only on he domain G c x, such ha he following poinwise Carleman esimae holds in G c x for all funcions u x,, C G c x C for all G c x : where he vecor funcion U,V saisfies he esimae L u C u C U V, (.8) U,V B u C. (.9) The proof of his lemma can be found in [6]. Also, we will use he following Lipschiz sabiliy resul, proved in [6] Theorem. Suppose ha he funcion u C! T, T" C saisfies he condiions (.) (.4). Le funcions a x,, g x,,, be bounded, i.e. a x,, r 5 $ x,,v H g x,,, r 6 $ x,, v, H, where r 5 r 6 are posiive consans. Le funcions x,,v L, F x,,v L H le T R. Then he following Lipschiz sabiliy esimae holds: u L H K L F L H ", where K K,T,r 5,r 6 is he posiive consan independen on funcions u, F... Preliminaries Before proceeding wih he proof of he Theorem, we inroduce some new funcions formulae necessary lemmaa. Le funcions u u be soluions of equaion (.) wih he iniial condiion (.3) he laeral daa (.5) for a x,v a x,v, x,,v x,, v a x,v a x,v, x,,v x,, v, respecively. Denoe u u u, 4

5 a a a, (.). From relaions (.), (.3), (.5) (.), noicing ha a u a u a u u, u a x, u g x,,, u x,, d au, au, we obain (.) u x,,v, $ x,v Z, (.) u x,,, $ x,, v! T, T". (.3) Applying he Theorem o he equaion (.) wih laeral daa (.3), we obain he following esimae for he funcion u Denoe u L H K L a L Z. u. Differeniaing (.) (.3) wih respec o, we obain, a x, g u g d au (.4) (.5) x,,, $ x,,v! T, T" Seing in (.), we obain x,,v au x,,v a x, v f x,v, where x, v Z. Differeniaing (.5) (.6) wih respec o denoing w, we obain w, w a x, w. g u g gw d au, (.6) (.7) (.8) w x,,, $ x,,v! T,T" We will need he following lemma Lemma. Le funcions a x,, a x, D M. The following Lipschiz sabiliy esimaes. (.9) hold: L H K a L Z L L ", (.) w L H K a L Z L L L ". (.) These esimaes are similar o he Lipschiz sabiliy esimae ha was obained in [6], bu do no follow direcly from he resul of [6] due o he presence of he funcion u in (.5) (.8). 5

6 The following lemma provides an esimae from he above for an inegral conaining he CWF. Lemma 3. For all funcions s C G c x for all, he following esimae holds G c x s x, d C x, dxd G c x s C x, dxd. See Secion 3. in [7] for he proof. Lemma 4. Le T R. Then for any c, R here exiss a G c T,T for all R,T,c,. Proof. By he definiion of he domain G c G c T,T max x, T c., R,T,c, such ha i.e. when which leads o he following inequaliy R T c, Since c,r R T hen 3. Proof of Lemma R c T., we can choose. Denoe G c G c R, T, such ha Choose so ha Choose a small number for arbirary c cons. Since T R, we can choose a small number T R 3 x 3 R,T, / ( Lemma 4 ) le, for he sake of definieness, G / 3 G /, /, such ha R,T, / T,T. T, T ". x 3 Consider he domains G / 3 G / G / G /. (See (.7_) Fig. for a schemaic represenaion in he - D case),.. (3.) (3.) (3.3) 6

7 Fig.. Ses G / 3 G / G / G /. Also, consider he cu-off funcion x, C T, T, such ha x, beween in in G /, T,T \G /, in G / \G /. The equaions (.5) (.8) imply ha, K u d d a, (3.4) w, w K w u d d w d a (3.5) Le x,, x,, x,. Then 7

8 n n i i i i i i. i i i Derivaives, i, i,..., n equal o zero in G / in T,T \G / are bounded in G / \G /. So, using he inequaliy (3.4), we obain n i i i n K u d d a. (3.6) Similarly, for w x,, w x,, x,, we obain from (3.5) w n iw i i K w u d d w d a w (3.7) Denoe u u x,, x,. Then (3.6) (3.7) become n i i i K u d d a (3.8) w n iw i i K w u d d w d a w (3.9) Muliplying (3.8) (3.9) by he CWF squaring boh sides, we obain 8

9 n i i C K u d d a C i K " C, w n iw i i C K w u d d w d a w C. The Carleman esimae (.8) leads o C U V (3.) K u d d w C U a V C (3.) K w u d d w d a w C where x,, H /, H / G / funcions U, V U, V are he funcions U, V from he Carleman esimae (.8)-(.9) for he case, when he funcion u is replaced by he funcions w, respecively. Inegraing over H / applying he Gauss formula, we obain C dh K u d d a C dh H / H / K C dh U,V ds (3.) H / M / Similarly, we obain for w w C dh (3.3) H / 9

10 K w u d d w d a C dh H / w C dh U,V ds. H / where dh dxd vd, M / G / Noicing ha for any funcion s x,,v M / G / denoes he boundary of he domain G /. C H s d C dh A s C dh, H / H / where A is he area of he uni sphere, we remove he inner inegrals over in (3.) (3.3). So, (3.) (3.3) become C dh K u a C dh H / H / K C dh U,V ds H / M / w C dh H / K w u a C dh H / w C dh U,V ds. H / Choose such ha K/ /. Then for all M / C dh we have H / K a C dh u C dh C dh U,V ds H / H / H / M /

11 w C dh H / K a C dh u C dh C dh w C dh H / H / H / H / U,V ds. M / Using (.9), we obain C dh (3.4) H / K a C dh u C dh C ds H / H / C dh K M H / / w C dh (3.5) H / K a C dh u C dh C dh w C dh H / H / H / H / K M w C ds. The boundary M / of he domain G / consiss of wo pars M / M / M /, where M / / x,,v : x R G / M / x,,v : x / G /. Since x,,v x,,v w x,,v x,,v, for x,,v M /,

12 x,,v w x,, v, for x,, v M /, hen C ds C ds w C ds C ds. M / M / M / M / Esimae boh sides of he inequaliy (3.4). Noe ha since in H / H / 3 H /, hen C dh C dh e / 3 dh. (3.6) Also, since Hence, H / H / 3 $ x, in G /, hen C e /, H / 3 x, H /. C dh e / H / dh. H / Therefore (3.4) (3.6) lead o e / 3 dh (3.7) H / 3 K a C dh u C dh e / dh C ds. H / H / H / M / Similarly, from (3.5) we obain e / 3 w dh (3.8) H / 3 K a C dh u C dh C dh e / w dh C ds. H / H / H / H / M / Le m sup x. Then (3.7) (3.8) yield G / e / 3 L H / 3 (3.9)

13 K e / L H / e m a L H / u L H / L M / e / 3 w L H / 3 (3.) K e / w L H / e m a L H / u L H / L H / /. L M Since u x,,v u x,,v x,,v x,, v $ x,, v H, hen (3.9) (3.) become e / 3 L H / 3 K e / L H / e m a L H / u L H / L M / e / 3 w L H / 3 K e / w L H / e m a L H / u L H / L H / /. L M Dividing hese inequaliies by exp / 3 ", we obain L H / 3 (3.) K e 5 L H / e m a L H / u L H / /, L M w L H / 3 K e 5 w L H / (3.) m K e a L H / u L H / L H / /. L M An inconvenience of he domain H / 3 for our goal is ha alhough he domain H / 3, bu H / 3. Thus, we now shif his domain. Choose an x such ha x 3 / consider he domain G / x, which is obained by a shif of he domain G /. Clearly one can choose R,T, / so small ha in addiion o (3.)-(3.3) Then G / x T, T G / 3 x T, T ". 3

14 G / 3 G / 3 x x x x 3 3 Consider now he ball B, / 3 : x : x / 3. By (3.) B, / 3, /. We prove now ha B G / 3 x. Le x of he ball B. Then Since Hence, by (3.4) B x x x x 3 x 3, /, hen 3 / 3. Hence, G / 3 Hence, here exiss a number E x x (3.3) (3.4), since B be an arbirary poin x. Therefore, using (3.3) (3.4), we obain ha G / 3 G / 3,T such ha he layer x, : x, The schemaic represenaion of he domains G /, G /. x. G / G / x. (3.5) x E in -D case is provided on Fig. 4

15 Fig.. G / Solid line, G / x Dashed line, E Shaded area. Since he Carleman esimae (.8)-(.9) is valid for he domain G / x, we can obain esimaes similar o (3.) (3.) L H / 3 x (3.6) K e 5 L H / x e m a L H / x u L H / x L M x / w L H / 3 x K e 5 w L H / x (3.7) m K e a L H / x u L H / x L H / x / x. L M where H / x G / x M / x G / x x, : x R. Consider now he layer E defined by (3.5) (see Fig.). Esimaes (3.), (3.6) (3.), (3.7) lead o he following esimaes in E : L E K e 5 L H e m w L E a L Z u L H L (3.8) (3.9) K e 5 w L H e m a L Z u L H L H L. Since for any funcion s x,,v C H here exiss, such ha hen (3.8) (3.9) lead o s x,,v dxd v s L E, x,, v dxd v N, (3.3) w x,,v dxd v N, 5

16 where N K e 5 L H e m a L Z u L H L (3.3) N K e 5 w L H e m a L Z u L H L H L. Le S, T, H, T, S T,, H T,. Denoe Y x,, n i i, (3.3) i x,, x,, S x,,. Esimae he L H norm of he funcion. Muliplying (3.3) by Z,, where, T, we obain inegraing over Consider he vecor funcion B so (3.33) becomes dxd vd n i v i idxd vd,,..., n. Then n i i i B, Ydxd vd. (3.33) x,,v dxd v x,,v dxd v B,n dsd vd K dxd vd Y dxd vd. Here B,n denoes he scalar produc of vecors B n, where n is he ouward normal vecor on. 6

17 Noicing ha B, where using he Cauchy-Schwarz inequaliy, we obain x,,v dxd v x,, v dxd v dsd vd (3.34) K dxd vd Y dxd vd, Esimae Y using (3.4) (3.3) Y K u d d a. (3.35) Esimaes (3.34) (3.35) lead o x,,v dxd v x,,v dxd v dsd vd K dxd vd u dxd vd a dxd vd. Using he Gronwall s inequaliy, we obain x,, v dxd v (3.36) K x,,v dxd v dsd vd u dxd vd a dxd vd. Subsiuing (3.3) (3.3) in he righ-h side of (3.36), we ge x,,v dxd v K N dsd vd u dxd vd a dxd vd K e 5 L H e m a L Z u L H L 7

18 K dsd vd u dxd vd a dxd vd K e 5 L H e m a L Z u L H L. Thus, L H K e 5 L H e m a L Z One can obain similar esimae for L H. Summing up ha esimae wih (3.37), we obain L H K e 5 e L H m a L Z u L H u L H L. L. (3.37) To remove he erm wih u from he laer formula we apply he esimae (.4). Hence L H K e 5 L H Consider, such ha e m a L Z Ke 5. L L. Then Choosing max,, we obain m L H K e a L Z 5 ln K. e m L L, (3.38) which implies he desired esimae (.). Applying he procedure, similar o (3.3)-(3.38), o he equaions depending on w, using he esimae (3.38), one can similarly obain he esimae (.). 4. Proof of he Theorem. This secion consiss of hree subsecions. In he subsecion 4. geomery is defined he proof of he Theorem is sared. In he subsecion 4. he supplemenary fac is proved. In he subsecion 4.3 he proof of he Theorem is finished. 4.. Beginning of he Proof of Theorem. The proof of he heorem is based on he Carleman esimae (.8)-(.9). The values of he 8

19 parameers, ha are used in he proof of his heorem are independen on he values of hese parameers used in he proof of he Lemma. Consider he problem (.)-(.3) in H. Also, consider he relaions, (.5)-(.7) (.8)-(.9). A equaion (.) becomes u x,,v au x,, v, (4.) Since u x,, v f x,v hen (4.) leads o Since f x,v r, a x,v K u x,, v. (4.) u x,,v u x,, v u x,, v d, we have u x,,v u x,,v u x,, v d. (4.3) $ $ Choose a poin x R n, R x R. Choose he number, such ha T R/ Denoe he domains P c G c x G c x, c, where he domains G c x are defined by (.7). Choose he consan c such ha x x c G c T, x R n : x R. Hence, x T. Define he domain b,b choose consans b such ha b P c 3 P c. (See fig. 3 for a schemaic represenaion in he - D case). 9

20 Fig. 3. The shaded area schemaically represens he domain P c. Consider he domains P c 3 P c P c P c. Also, consider he funcion x, C T,T, such ha x, beween in in P c, T, T \P c, in P c \P c, le x, be a non-increasing funcion of in he domain P c \P c non-decreasing funcion of in he domain P c \P c holds for any funcion s x,,v C H any x,,v H, a, so ha he following inequaliy x, s x,, v d x, s x,,v d. An example of such funcion is consruced in Appendix A. Denoe x,, x,, x, w x,, w x,, x,. Following he proof of Lemma from (3.4) o (3.5), we obain he analogs o esimaes (3.4) (3.5) for he domains C dh (4.4)

21 K a C dh u C dh C dh K Bc C ds, w C dh (4.5) K a C dh u C dh C dh w C dh K Bc w C ds, where B c is he boundary of he domain. Represen he inegrals u C dh C dh as a sums of inegrals u C dh u C dh u C dh \ C dh C dh C dh, consider he inegrals over he domain firs. Since u x,, v u x,, v \ u x,,v d x,,v x,,v x,,v d, we obain, using (.) (.7), u x,,v u x,, v u x,, v d u x,, v d (4.6)

22 x,, v x,,v x,, v d a f x,,v d, (4.7) Since u x,,v u x,, v, x,, v x,, v, $ x,,v, hen, applying (4.6) (4.7) o he inegrals u C dh C dh, we obain u C dh (4.8) K u x,,v d C dh K x,,v d C dh C dh K a C dh x,,v d C dh K a C dh w x,, v d C dh. (4.9) Applying Lemma 3 o (4.8) (4.9), we obain u C dh K Qc C dh (4.)

23 C dh K a C dh w C dh. (4.) Also, applying he esimae (4.7) o he righ-h side of (4.) using Lemma 3, we obain K u C dh Qc a C dh w C dh. (4.) Applying he esimaes (4.), (4.) (4.) o (4.4) (4.5), choosing sufficienly large, we obain C dh o be (4.3) K a C dh u C dh C dh K Bc C ds, \ w C dh (4.4) K a C dh u C dh C dh w C dh \ \ K Bc w C ds, Since u x,,v u x,,v x,,v x,, v $ x,, v H, (4.3) (4.4) become C dh (4.5) K a C dh u C dh C dh K Bc C ds \ \ 3

24 w C dh (4.6) K a C dh u C dh C dh w C dh \ \ K Bc w C ds. 4.. Proof of he Inegral Inequaliy. Here we esimae he inegral from he above hrough he inegral Consider he funcion c x Qc a C dh b a C dh. x x c. Then for any funcion s x,,v Hence, s x,,v dh C, which is even wih respec o he variable, we have Z c x s x,,v dd vdx c x a C dh a C dh Z c x s x,,v dd vdx. a C dh (4.7) (4.8) \ a C dh b a C dd vdx c x a C dd vdx. \ Z Z b Noe ha, since a x,v is independen of, we have b a C dd vdx c x a C d d vdx Z Z b 4

25 $ a b C x, dd vdx a c x C x, d d vdx. (4.9) Z Since he funcion is decreasing when, we have c x C x, d e x x b c x e b Z b e d c x b e x x e b c x b e x x b b e b d. Since hen b e b d b e d, c x b e x x b b e b d c So, by (4.8) (4.9), we obain x b e x x b a C dh b e a C dh d K b b K C x, d. a C dh. (4.) \ Noe ha C x, e c x, P c \P c. (4.) From (4.7), (4.9) (4.), we obain a C dh e c a dh e c a dh \ \ 5

26 e c a d vdx Z c x c x d Thus, we have Ke c Since b P c 3 P c, hen So, (4.) implies ha \ e c a d vdx Z a C dh b d Ke c a C dh Ke c e c 3 C x, $ K e c a dh b b b a dh. x, K b b. a dh. a C dh. (4.) (4.3) \ Finally, by (4.) (4.3), we have a C dh 4.3. The End of he Proof of he Theorem. K b a C dh. (4.4) Consider now he esimaes (4.), (4.3) (4.5). By (4.) we have a x,v K x,, v, (4.5) (4.3) leads o x,,v x,, v w x,, v d. (4.6) Combining (4.5) (4.6), we obain a x,v x,, v w x,,v d. Muliplying he las inequaliy by he C x, inegraing over 3, we obain a x, v C dh (4.7) 3 6

27 x,,v C dh w x,,v d C dh. 3 3 Since 3, he esimaes (4.5) (4.6) lead o C dh 3 (4.8) K a C dh u C dh C dh K Bc C ds \ w C dh 3 (4.9) K a C dh u C dh C dh w C dh \ \ K Bc w C ds. Since x,,v x,,v, $ x,, v 3, hen, combining he esimaes (4.7) (4.8), we obain a C dh w x,,v d C dh (4.3) 3 3 K a C dh u C dh C dh K Bc C ds. \ By Lemma 3 w x,, v d C x, dh w x,,v C x, dh. 3 3 Hence, (4.3) leads o 7

28 3 a C dh Qc 3 w C dh (4.3) K a C dh u C dh C dh K Bc C ds. \ Summing up he esimaes (4.3) (4.9), noicing ha w x,,v w x,, v, $ x,,v 3, aking, we obain a C dh w C dh (4.3) 3 3 K a C dh u C dh C dh \ \ K C dh w C dh K Bc C ds K Bc w C ds. The boundary B c consiss of wo pars. Denoe B c x, : x R P c, B c x, : x x c P c. Then B c B c B c. Since x,,v x,,v w x,,v x,, v, if x,, v B c, x,,v w x,, v, if x,,v B c, hen C ds C ds w C ds C ds. B c B c B c B c Thus, (4.3) leads o 8

29 3 a x, v C dh K a C dh u C dh C dh w C dh \ \ \ Noicing ha b K Bc C ds K B c C ds. 3 applying (4.4) o he las inequaliy, we obain a x, v C dh b (4.33) K b a C dh \ u C dh C dh \ w C dh \ K Bc C ds K B c C ds. Taking K in (4.33), we obain a x, v C dh b (4.34) K u C dh C dh w C dh \ \ \ Le m sup x x K Bc. Then, since C ds K B c C ds. max C x, : x, \ e c, inequaliy (4.34) yields b a x, v C dh 9

30 K e c u L H L H w L H Le d inf x x b K e m. Then (4.35) becomes L e d a L Z L. (4.35) K e c u L H L H w L H K e m L L. Using he esimaes for L H w L H, given by Lemma he esimae (.4) for u L H, we obain e d a L Z K e c a L Z L L L Since d K e m L c, hen dividing (4.36) by e d aking we obain he desired esimae (.6). K exp d L. c ", o be so large ha (4.36) Appendix A. Here we consruc supplemenary funcion. Consider consans C i, i,...,6, ha will be chosen laer, denoe he surfaces in R n, corresponding o hese consans, S i x, : x Le C C. Consider he funcion C C i, i,..., 6. C, C C e exp C C C C C C, C C C, C C. This is a non-increasing funcion of he parameer C. Consider he funcion 3

31 x, x, x, R n T,T. Consider any x R n, such ha he line x x in R n T,T crosses boh surfaces S S. Le firs. Choose arbirary,, T",, such ha he poins x, x, are locaed beween he surfaces S S. Clearly, he poins x, x, correspond o differen level surfaces of he funcion x,, S 3 S 4, respecively, ha have corresponding consans C 3 C 4, such ha C C 4 C 3 C (see. Fig.4). Since Fig.4. Schemaic represenaion of level surfaces for -D case. C is a non-increasing funcion, we have x, x,. Thus, he funcion x, is non-increasing wih respec o, when. Le. Choose arbirary 3, 4 T,", 3 4, such ha he poins x, 3 x, 4 are locaed beween he surfaces S S. Clearly, he poins x, 3 x, 4 correspond o differen level surfaces of funcion x,, S 5 S 6, respecively, ha have corresponding consans C 5 C 6, such ha C C 6 C 5 C (see Fig. 4). Since he funcion C is a non-increasing funcion, we have x, 3 x, 4. Thus, he funcion x, is non-decreasing wih respec o, when. So, since he funcion x, is coninuously differeniable in R n T,T, we can ake i as he funcion. Acknowledgmen The work of M.V. Klibanov was suppored by, or in par by, he U.S. Army Research Laboraory U.S. Army Research Office under conrac/gran number W9NF His work was also 3

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