Inverse scattering at fixed energy for the multidimensional Newton equation in short range radial potentials
aa r X i v : . [ m a t h - ph ] O c t Inverse scattering at fixed energy for themultidimensional Newton equation in shortrange radial potentials
Alexandre JollivetJuly 8, 2018
Abstract
We consider the inverse scattering problem at fixed and sufficientlylarge energy for the nonrelativistic and relativistic Newton equationin R n , n ≥
2, with a smooth and short range electromagnetic field(
V, B ). Using results of [Firsov, 1953] or [Keller-Kay-Shmoys, 1956]we obtain a uniqueness result when B is assumed to be zero in aneighborhood of infinity and V is assumed to be spherically symmetricin a neighborhood of infinity. Consider the following second order differential equation that is the multidi-mensional nonrelativistic Newton equation with electromagnetic field¨ x ( t ) = F ( x ( t ) , ˙ x ( t )) := −∇ V ( x ( t )) + B ( x ( t )) ˙ x ( t ) , (1.1)where x ( t ) ∈ R n , ˙ x ( t ) = d x d t ( t ). In this equation we assume that V ∈ C ( R n , R ) and for any x ∈ R n , B ( x ) is a n × n antisymmetric matrix withelements B i,k ( x ) , B i,k ∈ C ( R n , R ), which satisfy ∂B i,k ∂x l ( x ) + ∂B l,i ∂x k ( x ) + ∂B k,l ∂x i ( x ) = 0 , (1.2)for x = ( x , . . . , x n ) ∈ R n and for l, i, k = 1 . . . n .For n = 3, the equation (1.1) is the equation of motion in R n of a nonrel-ativistic particle of mass m = 1 and charge e = 1 in an external and staticelectromagnetic field described by ( V, B ) (see, for example, [14, Section 17]).1or the electromagnetic field the function V is an electric potential and B isthe magnetic field. Then x denotes the position of the particle, ˙ x denotes itsvelocity, ¨ x denotes its acceleration and t denotes the time.For the equation (1.1) the energy E = 12 | ˙ x ( t ) | + V ( x ( t )) (1.3)is an integral of motion.We assume that the electromagnetic coefficients V and B are short range.More precisely we assume that ( V, B ) satisfies the following conditions | ∂ j x V ( x ) | ≤ β | j | (1 + | x | ) − α −| j | , x ∈ R n , (1.4) | ∂ j x B i,k ( x ) | ≤ β | j | +1 (1 + | x | ) − α − −| j | , x ∈ R n , (1.5)for | j | ≤ , | j | ≤ , i, k = 1 . . . n and some α > j l is the multiindex j l = ( j l, , . . . , j l,n ) ∈ ( N ∪ { } ) n , | j l | = P nk =1 j l,k and β | j l | are positive realconstants). We denote by k . k the norm on the short range electromagneticfields defined by k ( V, B ) k = sup x ∈ R n, j ∈ N n | j |≤ (cid:16) (1 + | x | ) α + | j | | ∂ j x V ( x ) | (cid:17) (1.6)+ sup x ∈ R n, j ∈ N n | j |≤ , i,k =1 ...n (cid:16) (1 + | x | ) α +1+ | j | | ∂ j x B i,k ( x ) | (cid:17) . Under conditions (1.4)–(1.5), we have the following properties (see, forexample, [17] and [15] where classical scattering of particles in a short-rangeelectric field and in a long-range magnetic field are studied respectively): forany ( v − , x − ) ∈ R n × R n , v − = 0 , the equation (1.1) has a unique solution x ∈ C ( R , R n ) such that x ( t ) = tv − + x − + y − ( t ) , (1.7)where | ˙ y − ( t ) | + | y − ( t ) | → , as t → −∞ ; in addition for almost any ( v − , x − ) ∈ R n × R n , v − = 0 , the unique solution x ( t ) of equation (1.1) that satisfies (1.7)also satisfies the following asymptotics x ( t ) = tv + + x + + y + ( t ) , (1.8)where v + = 0 , | ˙ y + ( t ) | + | y + ( t ) | → , as t → + ∞ . At fixed energy E > S E the set { v − ∈ R n | | v − | = 2 E } and we denote by D ( S E )the set of ( v − , x − ) ∈ S E × R n for which the unique solution x ( t ) of equation21.1) that satisfies (1.7) also satisfies (1.8). We have that D ( S E ) is an openset of S E × R n and Mes(( S E × R n ) \D ( S E )) = 0 for the Lebesgue measure on S E × R n . The map S E : D ( S E ) → S E × R n given by the formula S E ( v − , x − ) = ( v + , x + ) , (1.9)is called the scattering map at fixed energy E > V ( x ) ≡ B ( x ) ≡
0, then v + = v − , x + = x − , ( v − , x − ) ∈ R n × R n , v − = 0.In this paper we consider the following inverse scattering problem at fixedenergy Given S E at fixed energy E > , find ( V, B ) . (1.10)Note that using the conservation of energy we obtain that if E < sup R n V then S E does not determine uniquely V .We mention results on Problem (1.10). When B ≡ V is assumedto be spherically symmetric and monotonuous decreasing in | x | ( V is notassumed to be short range), uniqueness results for Problem (1.10) were ob-tained in [5, 12]. The scattering map S E also uniquely determines ( V, B ) atfixed and sufficiently large energy when (
V, B ) is assumed to be compactlysupported inside a fixed domain of R n (see [16] for B ≡ B ≡
0, and see [3, 9]) andconnection between this boundary value problem and the inverse scatteringproblem on R n (see [16] for B ≡
0, and see [9]).To our knowledge it is still unknown whether the scattering map at fixedand sufficiently large energy uniquely determine the electromagnetic fieldunder the regularity and short range conditions (1.4) and (1.5) (see [16,Conjecture B] for B ≡ Theorem 1.1.
Let ( λ, R ) ∈ (0 , + ∞ ) and let ( V, B ) be an electromagneticfield that satisfies the assumptions (1.2) , (1.4) and (1.5) and k ( V, B ) k ≤ λ .Assume that B ≡ outside B (0 , R ) and that V is spherical symmetric outside B (0 , R ) . Then there exists a positive constant E ( λ, R ) (which does not dependon ( V, B ) ) so that the scattering map at fixed energy E > E ( λ, R ) uniquelydetermines ( V, B ) on R n . The proof of Theorem 1.1 is obtained by recovering first the electric po-tential in a neighborhood of infinity using Firsov or Keller-Kay-Shmoys’ re-sult [5, 12] and then by recovering the electromagnetic field on R n using thefollowing proposition which generalizes [9, Theorem 7.2].3 roposition 1.2. Let ( λ, R ) ∈ (0 , + ∞ ) and let ( V, B ) be an electromagneticfield that satisfies the assumptions (1.2) , (1.4) and (1.5) and k ( V, B ) k ≤ λ .Assume that ( V, B ) is known outside B (0 , R ) . Then there exists a positiveconstant E ( λ, R ) so that the scattering map at fixed energy E > E ( λ, R ) uniquely determines ( V, B ) on R n . Concerning the inverse scattering problem for the classical multidimen-sional nonrelativistic Newton equation at high energies and the inverse scat-tering problem for a particle in electromagnetic field (with B B ≡ (1.1) We will use the standard Lemma 2.1 on nontrapped solutions of equation(1.1). For sake of consistency its proof is given in Appendix.
Lemma 2.1.
Let
E > and let R E and C E be defined by C E := 2 E ( nβ + 2 β )(1 + p E + β )) , (2.1)sup | x |≥ R E (1 + | x | ) − α ≤ C E . (2.2) If x ( t ) is a solution of equation (1.1) of energy E such that | x (0) | < R E andif there exists a time T > such that x ( T ) = R E then | x ( t ) | ≥ R E + E | t − T | for t ∈ ( T, + ∞ ) , (2.3) and there exists a unique ( x + , v + ) ∈ R n × S E so that x ( t ) = x + + tv + + y + ( t ) , t ∈ R , where | y + ( t ) | + | ˙ y + ( t ) | → as t → + ∞ . Note that C E → + ∞ as E → + ∞ while sup | x |≥ R (1+ | x | ) − α is a decreasingfunction of R that goes to 0 as R → + ∞ . Note that Lemma 2.1 is stated forpositive times t but a similar result hold for negative times t .4 .2 The inverse kinematic problem for equation (1.1) We first formulate the inverse kinematic problem for equation (1.1) inside aball of center 0 and radius
R > B (0 , R ). For ( m, l ) ∈ ( N \{ } ) and for a function f from B (0 , R ) to R m of class C l we define the C l normof f by k f k C l ,R = sup x ∈ B (0 ,R ) , α ∈ N n | α |≤ l | ∂ αx f ( x ) | . We denote by ∂B (0 , R ) the boundary of the ball B (0 , R ).Then we recall that there exists a constant E ( R, k V k C ,R , k B k C ,R ) sothat at fixed energy E > E ( R, k V k C ,R , k B k C ,R ) the solutions x of equation(1.1) in B (0 , R ) at energy E have the following properties (see for example[9]):for each solution x ( t ) there are t , t ∈ R , t < t , such that x ∈ C ([ t , t ] , R n ) , ( x ( t ) , x ( t )) ∈ ∂B (0 , R ) , x ( t ) ∈ B (0 , R ) for t ∈ ] t , t [ ,x ( s ) = x ( s ) for s , s ∈ [ t , t ] , s = s ; (2.4)andfor any two distinct points q , q ∈ ∂B (0 , R ) , there is one and only one solution x ( t ) = x ( t, E, q , q ) such that x (0) = q , x ( s ) = q for some s > . (2.5)This is closely related to the property that at fixed and sufficiently largeenergy E , the compact set B (0 , R ) endowed with the riemannian metric p E − V ( x ) | dx | and the magnetic field defined by B is simple (see [3]).For ( q , q ) two distinct points of ∂B (0 , R ) we denote by s ( E, q , q ) the timeat which x ( t, E, q , q ) reaches q from q and we denote by k ( E, q , q ) the ve-locity vector ˙ x (0 , E, q , q ) and by k ( E, q , q ) the velocity vector ˙ x ( s ( E, q , q ) ,E, q , q ) . The inverse kinematic problem is thenGiven k ( E, q , q ) , k ( E, q , q ) for all q , q ∈ ∂B (0 , R ) ,q = q, at fixed sufficiently large energy E, find ( V, B ) in B (0 , R ) . The data k ( E, q , q ) , k ( E, q , q ) , q , q ∈ ∂B (0 , R ) , q = q, are the bound-ary value data of the inverse kinematic problem, and we recall the followingresult. Lemma 2.2 (see, for example, Theorem 7.1 in [9]) . At fixed
E > E ( k V k C ,R , k B k C ,R , R ) , the boundary data k ( E, q , q ) , ( q , q ) ∈ ∂B (0 , R ) × ∂B (0 , R ) ,q = q , uniquely determine ( V, B ) in B (0 , R ) . .3 Relation between boundary data of the inversekinematic problem and the scattering map S E We will prove that at fixed and sufficiently large energy E the scatter-ing map S E determines the boundary data k ( E, q , q ) , k ( E, q , q ) , q , q ∈ ∂B (0 , R ) , q = q . This will prove that S E uniquely determines ( V, B ) in B (0 , R ), which will prove Proposition 1.2.Let R > λ >
V, B ) are known outside B (0 , R ) and k ( V, B ) k < λ . Note that max( k V k C ,R , k B k C ,R ) ≤ k ( V, B ) k < λ . Thus thereexists a constant E ( λ, R ) such that at fixed energy E > E ( λ, R ) solutions x ( t ) of equation (1.1) in B (0 , R ) at energy E have properties (2.4) and (2.5)and such that at fixed E > E ( λ, R ) the boundary data k ( E, q , q ), ( q , q ) ∈ ∂B (0 , R ) × ∂B (0 , R ) , q = q , uniquely determine ( V, B ) in B (0 , R ). Thenusing that the constant C E → + ∞ as E → + ∞ in Lemma 2.1 we obtain thatthere exists E ( λ, R ) such that for E > E ( λ, R ) we have sup | x |≥ R (1+ | x | ) − α ≤ C E so that R E can be replaced by R in Lemma 2.1.Set E ( λ, R ) = max( E ( λ, R ) , E ( λ, R )) and fix E > E ( λ, R ). Let ( x − , v − ) ∈D ( S E ) and ( v + , x + ) = S E ( v − , x − ). We denote by x ( ., v − , x − ) the solution ofequation (1.1) that satisfies (1.7) (and (1.8)). Set t − ( x − , v − ) = sup { t ∈ R | | x ( s, x − , v − ) | ≥ R, s ∈ ( −∞ , t ) } ,t + ( x − , v − ) = inf { t ∈ R | | x ( s, x − , v − ) | ≥ R, s ∈ ( t, + ∞ ) } . Since (
V, B ) is known outside B (0 , R ) we can solve equation (1.1) with ini-tial conditions (1.7) and (1.8) and we obtain that x ( ., x − , v − ) is known on( −∞ , t − ( x − , v − )] ∪ [ t + ( x − , v − ) , ∞ ).If x ( s, x − , v − ) B (0 , R ) for any s ∈ R , then t ± ( x − , v − ) = ∓∞ . If thereexists s ∈ R such that x ( s, x − , v − ) ∈ B (0 , R ) then set q = x ( t − ( x − , v − )) , q = x ( t + ( x − , v − )) . (2.6)Using Lemma 2.1 and E > E ( λ, R ) we obtain that | x ( s, x − , v − ) | < R for s ∈ ( t − ( x − , v − ) , t + ( x − , v − )) and q = q . (Note that if x ( t ) satisfies equation(1.1) then x ( t + t ) also satisfies (1.1) for any t ∈ R .) Therefore we have x ( s, x − , v − ) = x ( s − t − ( x − , v − ) , E, q , q ) for s ∈ ( t − ( x − , v − ) , t + ( x − , v − )) where x ( t, E, q , q ) is the solution of (1.1) given by (2.5), and we have k ( E, q , q ) = ˙ x ( t − ( x − , v − )) , k ( E, q , q ) = ˙ x ( t + ( x − , v − )) . (2.7)We proved that the scattering map S E uniquely determines the data k ( E, q , q ) ,k ( E, q , q ) , ( q , q ) ∈ ∂B (0 , R ) , q = q , when ( q , q ) = ( x ( t − ( x − , v − )) , x ( t + ( x − , v − )))for ( x − , v − ) ∈ D ( S E ). And using again Lemma 2.1 we know that for any6 q , q ) ∈ ∂B (0 , R ) , q = q , the solution x ( t, E, q , q ) given by (2.5) satisfies(1.7) and (1.8) for some ( x ± , v ± ) ∈ R n × S E and that | x ( t, E, q , q ) | > R for t < t > s ( E, q , q ). Thus S E uniquely determines the data k ( E, q , q ) ,k ( E, q , q ) , ( q , q ) ∈ ∂B (0 , R ) , q = q . (cid:3) In this section we assume that the electromagnetic field (
V, B ) in equation(1.1) satisfies (1.4) and (1.5) and is so that B ≡ V is sphericallysymmetric outside B (0 , R ) for some R >
0. Let W ∈ C ([ R, + ∞ ) , R ) bedefined by V ( x ) = W ( | x | ) for x B (0 , R ). From (1.4) it follows thatsup r>R (1 + r ) α | W ( r ) | ≤ β and sup r>R (1 + r ) α +1 | W ′ ( r ) | ≤ β , (3.1)where W ′ denotes the derivative of W . r min ,. Set β := (2 E + 2 β ) max (cid:0) R, (cid:0) β + 2 β E (cid:1) α (cid:1) . (3.2)Then for q ≥ β consider the real number r min ,q defined by r min ,q = sup { r ∈ ( R, + ∞ ) | W ( r ) + q r = E } . (3.3)The function r min ,. has the following properties. Lemma 3.1.
The function r min ,. is a C strictly increasing function from [ β, + ∞ ) to ( R, + ∞ ) and we have r ,q W ′ ( r min ,q ) < q for q ≥ β and W ( r min ,q ) + q r ,q = E, dr min ,q dq = qr min ,q q − r ,q W ′ ( r min ,q ) > . (3.4) In addition the following estimates and asymptotics at + ∞ hold q √ E + 2 β ≤ r min ,q ≤ q (cid:16) E − β q − α (2 β + 2 E ) α (cid:17) , for q ∈ [ β, + ∞ ) , (3.5) r min ,q = q √ E + O ( q − α ) , as q → + ∞ . (3.6)Lemma 3.1 and works [5, 12] allow the reconstruction of the force F in aneighborhood of infinity. In sections 3.2 and 3.3 we develop the reconstructionprocedure given in [5, 12]. 7 .2 The scattering angle Let P be a plane of R n containing 0 and let ( e , e ) be an orthonormalbasis of P . For ( v , v ) ∈ R and for v = v e + v e we define v ⊥ ∈ P by v ⊥ = − v e + v e .Let q ≥ β . Then set x − := (2 E ) − qe and v − = √ Ee . We have x − · v − =0 and x − · v ⊥− = − q , and for such couple ( x − , v − ) we consider x q ( t ) thesolution of (1.1) with energy E and with initial conditions (1.4) at t → −∞ .Let t − := sup { t ∈ R | | x q ( s ) | ≥ R for s ∈ ( −∞ , t ) } . We will prove that t − = + ∞ . Since the force F in (1.1) is radial outside B (0 , R ) we obtain that x q ( t ) ∈ P for t ∈ ( −∞ , t − ).We introduce polar coordinates in P . We write x q ( t ) = r q ( t )(cos( θ q ( t )) e ( t )+sin( θ q ( t )) e ) for t ∈ ( −∞ , t − ) where the functions r q , θ q , satisfy the followingordinary differential equations¨ r q ( t ) = − W ′ ( r q ( t )) + q ′ r q ( t ) , (3.7) r q ( t ) ˙ θ q ( t ) = q ′ , for some q ′ ∈ R . (3.8)Asymptotic analysis of x q ( t ) at t = −∞ using the initial conditions (1.4) and˙ y − ( t ) = o ( t − ) as t → −∞ (see for example [16, Theorem 3.1] for this latterproperty) shows that q ′ = q . We refer the reader to the Appendix for details.The energy E defined by (1.3) is then written as follows2 E = ˙ r q ( t ) + q r q ( t ) + 2 W ( r q ( t )) . (3.9)Let t q = inf { t ∈ ( −∞ , t − ) | r q ( t ) = r min ,q } . Then using (3.3) and (3.9) wehave ˙ r q ( t q ) = 0. Since r q satisfies the second order differential equation (3.7)we obtain that t − = + ∞ , r q ( t q + t ) = r q ( t q − t ) for t ∈ R , and ± ˙ r q ( t ) > ± t > t q . We thus define g ( q ) = Z + ∞−∞ dtr q ( t ) = 2 Z + ∞ t q dtr q ( t ) . (3.10)The integral (3.10) is absolutely convergent and from (3.8) it follows that qg ( q ) = R R ˙ θ q ( s ) ds is the scattering angle of x q ( t ), t ∈ R .Note also that from (3.8) we have S ,E ( √ Ee , (2 E ) − qe ) = √ E (cid:16) cos (cid:0) qg ( q ) + π (cid:1) , sin (cid:0) qg ( q ) + π (cid:1)(cid:17) , (3.11)for q ∈ [ β, + ∞ ) and where S ,E is the first component of the scattering map S E . Note that qg ( q ) → q → + ∞ and that g is continuous on [ β, + ∞ )8these properties can be proven by using [16, Theorem 3.1] and continuity ofthe flow of (1.1)). Hence using (3.11) we obtain that S ,E uniquely determinesthe function g . Let χ be the strictly increasing function from [0 , β − ) to [0 , r − ,β ), continuouson [0 , β − ) and C on (0 , β − ), defined by χ (0) = 0 , and χ ( σ ) = r − ,σ − , for σ ∈ (0 , β − ) . (3.12)Let φ : (0 , χ ( β − )) → (0 , β − ) denote the inverse function of χ . From (3.4)and (3.12) it follows that2( E − V ( χ ( u ) − )) = χ ( u ) u , for u ∈ (0 , β − ) , (3.13)2( E − V ( s − )) φ ( s ) = s , for s ∈ (0 , χ ( β − )) . (3.14)Define the function H from (0 , β − ) to R by H ( σ ) := Z σ g ( u − ) du √ u √ σ − u for σ ∈ (0 , β − ) , (3.15)Hence H is known from the first component of the scattering map S E .The following formulas are valid (see Appendix for more details) H ( σ ) = π Z χ ( σ )0 ds p E − V ( s − )) , π √ σ dHdσ ( σ ) = ddσ ln( χ ( σ )) , (3.16)for σ ∈ (0 , β − ).Then note that from (3.6) it follows that χ ( σ ) = (cid:16) (2 E ) σ − + O ( σ − α ) (cid:17) − =(2 E ) − σ + O ( σ α +12 ) as σ → + , and ln (cid:16) (2 E ) χ ( σ ) σ − (cid:17) = ln(1+ O ( σ α )) → σ → + (note that we just need the assumption α > χ ( σ ) = (2 E ) − σ e R σ (cid:0) π √ s dHds ( s ) − s (cid:1) ds for σ ∈ (0 , β − ) , (3.17) W ( s ) = E − s φ ( s − ) for s ∈ ( r min ,β , + ∞ ) . (3.18)Set β ′ = β (cid:16) E − β β − α (2 β + 2 E ) α (cid:17) . (3.19)9hen note that from (3.5) and (3.2), it follows that r min ,β ≤ β ′ . (3.20)Therefore from (3.17) and (3.18) we obtain that W is determined by the firstcomponent of the scattering map S E on ( β ′ , + ∞ ).The proof of Theorem 1.1 then relies on this latter statement and onProposition 1.2. (cid:3) Let c >
0. Consider the relativistic multidimensional Newton equation in anelectromagnetic field˙ p = F ( x, ˙ x ) := −∇ V ( x ) + 1 c B ( x ) ˙ x, (4.1) p = ˙ x q − | ˙ x | c , ˙ p = dpdt , ˙ x = dxdt , x ∈ C ( R , R n ) , where ( V, B ) satisfy (1.2), (1.4) and (1.5). The equation (4.1) is an equationfor x = x ( t ) and is the equation of motion in R n of a relativistic particleof mass m = 1 and charge e = 1 in an external static electromagnetic fielddescribed by the scalar potential V and the magnetic field B (see [4] and, forexample, [14, Section 17]). In this equation x is the position of the particle, p is its impulse, F is the force acting on the particle, t is the time and c isthe speed of light.For the equation (4.1) the energy E = c r | p ( t ) | c + V ( x ( t )) = c q − | ˙ x ( t ) | c + V ( x ( t )) , (4.2)is an integral of motion. We denote by B c the euclidean open ball whoseradius is c and whose centre is 0.Under the conditions (1.3), we have the following properties (see [18]): forany ( v − , x − ) ∈ B c × R n , v − = 0 , the equation (4.1) has a unique solution x ∈ C ( R , R n ) that satisfies (1.7) where y − in (1.7) satisfies | ˙ y − ( t ) | + | y − ( t ) | → , as t → −∞ ; in addition for almost any ( v − , x − ) ∈ B c × R n , v − = 0 , the10nique solution x ( t ) of equation (1.1) that satisfies (1.7) also satisfies theasymptotics x ( t ) = tv + + x + + y + ( t ) , (4.3)where v + = 0, | ˙ y + ( t ) | + | y + ( t ) | →
0, as t → + ∞ . At fixed energy E > c ,we denote by S E,c the set { v − ∈ R n | | v − | = c q − c E } and we denoteby D ( S rel E ) the set of ( v − , x − ) ∈ S E,c × R n for which the unique solution x ( t ) of equation (4.1) that satisfies (1.7) also satisfies (1.8). We have that D ( S rel E ) is an open set of S E × R n and Mes(( S E,c × R n ) \D ( S rel E )) = 0 forthe Lebesgue measure on S E,c × R n . The map S rel E : D ( S rel E ) → S E,c × R n given by S rel E ( v − , x − ) = ( v + , x + ) , is called the scattering map at fixed energy E > c for the equation (4.1). Note that if V ( x ) ≡ B ( x ) ≡
0, then v + = v − , x + = x − , ( v − , x − ) ∈ B c × R n , v − = 0.We consider the inverse scattering problem at fixed energy for equation(4.1) that is similar to the inverse problem (1.10)Given S rel E at fixed energy E > c , find ( V, B ) . (4.4)Note that using the conservation of energy we obtain that if E < c +sup R n V then S E does not determine uniquely V .For problem (4.4) Theorem 1.1 and Proposition 1.2 still hold. In Sections4.2 and 4.3 we sketch the proof of Theorem 1.1 and Proposition 1.2 forequation (4.1).For inverse scattering at high energies for the relativistic multidimensionalNewton equation and inverse scattering in relativistic quantum mechanics see[8] and references therein.Concerning the inverse problem for (4.1) in the one-dimensional case, wecan mention the work [6]. (4.1) We first consider the analog of Lemma 2.1.
Lemma 4.1.
Let
E > c and let R E and C rel E be defined by C rel E := min (cid:16) E − c β , c (cid:0)(cid:0) E − c c + 1 (cid:1) − (cid:1) β n (cid:0) E − c )2 c + 1 (cid:1) (cid:17) , sup | x |≥ R E (1 + | x | ) − α ≤ C rel E . (4.5) If x ( t ) is a solution of equation (1.1) of energy E such that | x (0) | < R E andif there exists a time T > such that x ( T ) = R E then | x ( t ) | ≥ R E + 12 c (cid:0)(cid:0) E − c c + 1 (cid:1) − (cid:1)(cid:0) E − c )2 c + 1 (cid:1) | t − T | for t ∈ ( T, + ∞ ) , (4.6)11 nd there exists a unique ( x + , v + ) ∈ R n × S E,c so that x ( t ) = x + + tv + + y + ( t ) , t ∈ R , where | y + ( t ) | + | ˙ y + ( t ) | → as t → + ∞ . The proof of Lemma 4.1 is similar to the proof of Lemma 2.1.The solutions x ( t ) of equation (4.1) in B (0 , R ) for some R > B (0 , R ) for equation (4.1) similar to the inverse kinematic problemgiven in Section 2.2. Then Lemma 2.2 still holds (see [9, Theorem 1.2]) andthe connection between boundary data of the inverse kinematic problem andthe scattering map S rel E is similar to the one given for the nonrelativisticcase in Section 2.3 (note that the radius R has also to be chosen so thatsup | x |≥ R (1 + | x | ) − α < c / (144 β n ) = lim E → + ∞ C rel E ). This proves Proposition1.2 for equation (4.1). (cid:3) (4.1) We assume that the electromagnetic field (
V, B ) in equation (4.1) satisfies(1.4) and (1.5) and is so that B ≡ V is spherically symmetric out-side B (0 , R ) for some R >
0. Let W ∈ C ([ R, + ∞ ) , R ) be defined by V ( x ) = W ( | x | ) for x B (0 , R ). We give the analog of Lemma 3.1. Let˜ β = (2 β ) α (cid:16) E (2 β + β )+ β β − (cid:16) ( E (2 β + β )+ β β ) − β ( E − c ) (cid:17) (cid:17) − α and set β := max ˜ β, (cid:18) β E − c (cid:19) α E − c p ( E + β ) − c , cR p ( E + β ) − c E ! . (4.7)Then for q ≥ β consider the real number r min ,q defined by r min ,q = sup { r ∈ ( R, + ∞ ) | ( E − W ( r )) − c − q E c r = 0 } . (4.8)The function r min ,. has the following properties. Lemma 4.2.
The function r min ,. is a C strictly increasing function from [ β, + ∞ ) to ( R, + ∞ ) and we have r ,q W ′ ( r min ,q ) c ( E − W ( r min ,q )) E < q for q ≥ β and ( E − W ( r min ,q )) − c − q E c r ,q = 0 , (4.9) dr min ,q dq = E qr min ,q − c ( E − W ( r min ,q )) r ,q W ′ ( r min ,q ) + q E > . (4.10)12 n addition the following estimates and asymptotics at + ∞ hold qEc p ( E + β ) − c ≤ r min ,q ≤ Eqc r(cid:16) E − β (cid:0) qEc √ ( E + β ) − c (cid:1) − α (cid:17) − c , (4.11) for q ≥ β , and r min ,q = qEc √ E − c + O ( q − α ) , as q → + ∞ . (4.12)Then take any plane P containing 0 and keep notations of Section 3.2.Let q ≥ β . Then set x − := ( c p − c /E ) − qe and v − = c p − c /E e ,and consider x q ( t ) the solution of (4.1) with energy E and with initialconditions (1.4) at t → −∞ . We write x q in polar coordinates: x q ( t ) = r q ( t )(cos( θ q ( t )) e ( t ) + sin( θ q ( t )) e ) for t ∈ ( −∞ , t − ) where t − := sup { t ∈ R | | x q ( s ) | ≥ R for s ∈ ( −∞ , t ) } and the functions r q , θ q , satisfy¨ r q ( t ) = − W ′ ( r q ( t )) (cid:16) E − W ( r q ( t )) c (cid:17) + q E E − W ( r q ( t )) − rW ′ ( r q ( t )) r ( E − W ( r q ( t ))) , (4.13) r q ( t ) ˙ θ q ( t ) q − ˙ r q ( t ) + r q ( t ) ˙ θ q ( t ) c = qEc . (4.14)The energy E defined by (4.2) is then written as follows1 − ˙ r q ( t ) c − c ( E − W ( r q ( t ))) − q E c r q ( t ) ( E − W ( r q ( t ))) = 0 . (4.15)We also have ˙ θ q ( t ) = qEr q ( t ) ( E − W ( r q ( t ))) . (4.16)Similarly to Section 3.2 we have t − = + ∞ , r q ( t q + t ) = r q ( t q − t ) for t ∈ R ,and ± ˙ r q ( t ) > ± t > t q where t q = inf { t ∈ ( −∞ , t − ) | r q ( t ) = r min ,q } .We thus define g ( q ) = Z + ∞−∞ dtr q ( t ) ( E − W ( r q ( t ))) = 2 Z + ∞ t q dtr q ( t ) ( E − W ( r q ( t ))) . (4.17)From (4.16) Eqg ( q ) = R R ˙ θ q ( s ) ds is the scattering angle of x q ( t ), t ∈ R , and S rel1 ,E ( v − , x − ) = c p − c /E (cid:16) cos (cid:0) Eqg ( q ) + π (cid:1) , sin (cid:0) Eqg ( q ) + π (cid:1)(cid:17) , (4.18)13or q ∈ [ β, + ∞ ) and where S rel1 ,E is the first component of the scattering map S rel E . Since qg ( q ) → q → + ∞ and that g is continuous on [ β, + ∞ ) S rel1 ,E uniquely determines the function g .We now provide the reconstruction formulas for W from g in a neighbor-hood of infinity . Let χ be the strictly increasing function from [0 , β − ) to[0 , r − ,β ), continuous on [0 , β − ) and C on (0 , β − ), defined by χ (0) = 0 and χ ( σ ) = r − ,σ − for σ ∈ (0 , β − ). Let φ : (0 , χ ( β − )) → (0 , β − ) denote the in-verse function of χ . From (4.9) it follows that ( E − V ( χ ( u ) − )) − c − E χ ( u ) c u =0 for u ∈ (0 , β − ), and (cid:0) ( E − V ( s − )) − c (cid:1) φ ( s ) − E s c = 0 for s ∈ (0 , χ ( β − )).Define the function H from (0 , β − ) to R by (3.15). The following formulasare valid H ( σ ) = π Z χ ( σ )0 ds p ( E − V ( s − ) − c , Ecπ √ σ dHdσ ( σ ) = ddσ ln( χ ( σ )) , (4.19)for σ ∈ (0 , β − ). The proof of formulas (4.19) is similar to the proof offormulas (3.16).Then note that from (4.12) it follows that χ ( σ ) = (cid:16) Ec √ E − c σ − + O ( σ − α ) (cid:17) − = c √ E − c E σ + O ( σ α +12 ) , σ → + , andln (cid:16) Eχ ( σ ) c √ E − c σ (cid:17) = ln(1 + O ( σ α )) → , as σ → + . (4.20)Therefore we obtain the following reconstruction formulas χ ( σ ) = c √ E − c E σ e R σ (cid:0) Ecπ √ s dHds ( s ) − s (cid:1) ds , for σ ∈ (0 , β − ) , (4.21) W ( s ) = E − (cid:18) c + E c s φ ( s − ) (cid:19) , for s ∈ ( r min ,β , + ∞ ) . (4.22)Set β ′ = β (cid:16) E − β β − α (2 β +2 E ) α (cid:17) . Then note that from (4.7) and (4.11) itfollows that r min ,β ≤ β ′ . Therefore using (4.21) and (4.22) we obtain that W is determined by the first component of the scattering map S rel E on ( β ′ , + ∞ ).The proof of Theorem 1.1 for equation (4.1) then relies on this latterstatement and on Proposition 1.2 for equation (4.1). (cid:3) A Proof of Lemmas 2.1 and 3.1
In this Section we give a proof of Lemmas 2.1 and 3.1, and we give detailson the derivation of formulas (3.16) and the equality ” q = q ′ ” in Section 3.2.14 roof of Lemma 2.1. We will use the following estimate. Under conditions(1.4) and (1.5) we have | F ( x, v ) | ≤ β n (1 + | x | ) − ( α +1) (1 + | v | ) , for ( x, v ) ∈ R n × R n . (A.1)Let I ( t ) = 12 | x ( t ) | . (A.2)Then using the conservation of energy and equation (1.1) we have˙ I ( t ) = x ( t ) · ˙ x ( t ) , (A.3)¨ I ( t ) = 2 E − V ( x ( t )) − x ( t ) · F ( x ( t ) , ˙ x ( t )) , (A.4)for t ∈ R . Using (A.1) and estimate on V and | ˙ x | ( t ) = p E − V ( x ( t ))) ≤ p E + β ) for t ∈ R , we obtain that¨ I ( t ) ≥ E − β (1 + | x ( t ) | ) − α − nβ (1 + p E + β )) | x ( t ) | (1 + | x ( t ) | ) − α − ≥ (2 β + nβ )(1 + p E + β ))( C E − (1 + | x ( t ) | ) − α ) , (A.5)for t ∈ R . Hence we have¨ I ( t ) > (2 β + nβ )(1 + p E + β )) C E E whenever | x ( t ) | ≥ R E . (A.6)Let t = inf { t ∈ [0 , T ] | | x ( t ) | > R E } . Then using (A.2) we have˙ I ( t ) = lim h → + I ( t ) − I ( t − h ) h ≥ . (A.7)Combining (A.6) and (A.7) we obtain that | x ( t ) | ≥ R E , ˙ I ( t ) ≥
0, ¨ I ( t ) ≥ E for t ∈ [ t , + ∞ ). Moreover we have I ( t ) = I ( t ) + ˙ I ( t )( t − t ) + Z tt ( s − t ) ¨ I ( s ) ds ≥ R E + E t − t ) , (A.8)for t ∈ [ t , + ∞ ). This proves (2.3). Then using that | ˙ x ( t ) | ≤ p E + β ) for t ∈ R and using (2.3) we have | F ( x ( τ ) , ˙ x ( τ )) | ≤ nβ (1 + √ E | τ − T | ) − ( α +1) (1 + p E + β )) , (A.9)for τ ∈ [ T, + ∞ [. Equation (1.1) then gives x ( t ) = x + + tv + + y + ( t ) , (A.10)15or t ∈ (0 , + ∞ ), where v + = ˙ x (0) + Z + ∞ F ( x ( τ ) , ˙ x ( τ )) dτ, (A.11) x + = x (0) − Z + ∞ Z + ∞ σ F ( x ( τ ) , ˙ x ( τ )) dτ dσ (A.12) y + ( t ) = Z + ∞ t Z + ∞ σ F ( x ( τ ) , ˙ x ( τ ))) dτ dσ, (A.13)for t ∈ (0 , + ∞ ), where by (A.9) the integrals in (A.11), (A.12) and (A.13)are absolutely convergent ( α >
1) and | y + ( t ) | + | ˙ y + ( t ) | → t → + ∞ . Proof Lemma 3.1.
Note that for q ≥ β we have q / (2 R ) > E + β (1+ R ) − α ≥ E − W ( R ) and lim r → + ∞ q / (2 r ) = 0 < E = lim r → + ∞ ( E − W ( r )). Henceusing (3.3) we obtain r min ,q ∈ ( R, + ∞ ) and W ( r min ,q ) + q r ,q = E, (A.14)for q ∈ [ β, + ∞ ).Let q ∈ [ β, + ∞ ). From (A.14) it follows that q / (2 r ,q ) ≤ E + β andthen r min ,q ≥ q/ p E + β ). Combining this latter estimate and estimate(3.1) we obtain that2 (cid:16) E − sup r ∈ (cid:0) q √ E + β , + ∞ (cid:1) W ( r ) (cid:17) ≤ q r ,q ≤ (cid:16) E − inf r ∈ (cid:0) q √ E + β , + ∞ (cid:1) W ( r ) (cid:17) . (A.15)Then using again (3.1) we obtainsup r ∈ (cid:0) q √ E + β , + ∞ (cid:1) | W ( r ) | ≤ β q p E + β ) ! − α . (A.16)Combining (A.15) and (A.16) we obtain q q E + 2 β E + β )) α q α ≤ r min ,q ≤ q q E − β E + β )) α q α . (A.17)Estimates (3.5) and the asymptotics (3.6) follows from (A.17).Note that using (3.1) we have rW ′ ( r ) < β r − α ≤ β q − α (2( E + β )) α , (A.18)2 E − W ( r ) > E − β r − α ≥ E − β q − α (2( E + β )) α , (A.19)16or r ≥ q (2( E + β )) − and q ≥ β . From (A.18) and (A.19) we obtain rW ′ ( r ) < E − W ( r ) , (A.20)for r ≥ q (2( E + β )) − and q ≥ β . Then consider the function f ∈ C ([ β, + ∞ ) × ( R, + ∞ ) , R ) defined by f ( q, r ) = 2 E − W ( r ) − q r for ( q, r ) ∈ [ β, + ∞ ) × ( R, + ∞ ) ( ∂f∂r ( q, r ) = − W ′ ( r ) + 2 q /r ). We have f ( q, r min ,q ) = 0 for q ≥ β and from the implicit function theorem and (A.17) it follows that r min ,. is a C strictly increasing function from [ β, + ∞ ) to ( R, + ∞ ) so that r ,q W ′ ( r min ,q ) β. (A.21)Performing the change of variables “ r − = s ” in (A.21) we obtain g ( u − ) = 2 Z χ ( u )0 ds q E − s u − V ( s − ) , for u ∈ (0 , β − ) . (A.22)Let σ ∈ (0 , β − ). From (A.22) and (3.15) it follows that H ( σ ) = Z χ ( σ )0 Z σφ ( s ) du √ σ − u p E − V ( s − )) u − s ! ds. (A.23)And performing the change of variables u = φ ( s ) + ε ( σ − φ ( s )) in (A.23)( du = ( σ − φ ( s )) dε ) and using the equality (3.14) we obtain Z σφ ( s ) du √ σ − u p E − V ( s − )) u − s = φ ( s ) s − Z σφ ( s ) du √ σ − u p u − φ ( s )= φ ( s ) s − Z dε √ ε √ − ε = φ ( s ) s − π, (A.24)for s ∈ (0 , χ ( β − )) (we used the integral value π = R dε √ ε √ − ε ). Using (A.23),(A.24) and (3.14) we obtain the first equality in (3.16) H ( σ ) = π Z χ ( σ )0 φ ( s ) s − ds = π Z χ ( σ )0 (2( E − V ( s − ))) − ds. (A.25)17rom (A.25) it follows that dHdσ ( σ ) = π p E − V ( χ ( σ ) − )) dχdσ ( σ ) . (A.26)Then combining (3.13) and (A.26) we obtain the second equality in (3.16).We end this appendix by giving details on the equality q ′ = q in section3.2. We keep the notations of section 3.2. We set u θ = (cid:0) xr (cid:1) ⊥ and we have r ˙ θu θ = r ˙ x − ˙ rx = r ( v − + ˙ y − ) − ˙ r ( x − + tv − + y − ) . (A.27)Using x − · v − = 0 and y − = o (1), ˙ y − = o ( t − ), as t → −∞ , we have r ( t ) = | x − + tv − + y − | = (cid:0) t | v − | + 2 ty − · v − + | x − + y − | (cid:1) = − t | v − | + o (1) , t → −∞ , (A.28)˙ r ( t ) = ( v − + ˙ y − ) · ( x − + tv − + y − ) r ( t ) = t | v − | + o (1) − t | v − | + o (1)= −| v − | + o ( t − ) , t → −∞ , (A.29)and we obtain r ˙ θu θ = ( −| v − | t + o (1))( v − + ˙ y − ) − ( −| v − | + o ( t − ))( x − + tv − + y − )= | v − | x − + o (1) , t → −∞ , (A.30) u θ = (cid:16) x ( t ) r ( t ) (cid:17) ⊥ = (cid:16) x − + tv − + o (1) − t | v − | + o (1) (cid:17) ⊥ = − c v −⊥ + o (1) , t → −∞ (A.31)where ˆ w = w | w | for w = 0. Using (A.30) and (A.31) we obtain r ˙ θu θ · c v −⊥ = x − · v ⊥− + o (1) , u θ · c v −⊥ = − o (1) , t → −∞ , (A.32)which proves that q ′ = r ˙ θ = − x − · v ⊥− . (cid:3) References [1]
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