Inverse problems for general second order hyperbolic equations with time-dependent coefficients

We study the inverse problems for the second order hyperbolic equations of general form with time-dependent coefficients assuming that the boundary data are given on a part of the boundary. The main result of this paper is the determination of the time-dependent Lorentzian metric by the boundary measurements. This is achieved by the adaptation of a variant of the Boundary Control method developed by the author in [E1], [E2].

We consider the following class of domains D ⊂ R n+1 . Let D t = D∩{x 0 = t} be the intersection of D with the plane {t = x 0 }, t ∈ R. We assume that ∂D t is a smooth closed bounded domain in R n smoothly dependent and uniformly bounded in t. Let S(x 0 , x 1 , ..., x n ) = 0 be the equation of ∂D = t∈R ∂D t . we assume that S is a time-like smooth surface in R n+1 , i.e.
Consider the initial-boundary value problem where f is a smooth function on S = 0 with compact support, Lu = 0 is the same as in (1.1).
Let S 0 ⊂ S be a part of S such that (see Fig. 1.1) ∂D t ∩S has a nonempty interior for all t ∈ R. We assume also that for any x (0) ∈ ∂S 0 a vector τ (1) , tangent to S and normal to S 0 , is not parallel to (1, 0, ..., 0). The Dirichlet-to-Neumann operator Λ that maps the Dirichlet data to the Neumann data on ∂D is defined as where u(x) is the solution of (1.8), (1.9),(1.10) and (ν 0 (x), ..., ν n (x)) is the unit outward normal to S. Denote by (1.12) y = y(x) a proper diffeomorphism of D ontoD such that (1.13) y = x on S 0 .
We call a diffeomorphism of D ontoD proper if for any [t 1 , t 2 ] ⊂ R the image of D∩{t 1 ≤ x 0 ≤ t 2 } is a domainD∩{S where y and x are related by (1.12). Metric tensors [g jk (x)] n j,k=0 and [ĝ jk (x)] n j,k=0 , related by (1.18), are called isometric.
Note that y =ŷ(x) maps D t ontoD t for any t ∈ R. One can find y =ŷ(x) such thatD t =D 0 andŜ 0t = Γ 0 for all t ∈ R, i.e.D =D 0 × R,Ŝ 0 = Γ 0 × R, i.e.D andŜ 0 are cylindrical domains (see Fig. 1.2). HereŜ 0 is the image of S 0 under the map y =ŷ(x). Note that we can also arrange that Γ 0 is contained in the planeŷ n = 0 ifΓ 0 is small.
Let c(x) ∈ C ∞ (D) be such that The group G 0 (D) of such c(x) is called the gauge group. If Lu = 0 then u ′ = c −1 (x)u(x) satisfies the equation of the form (1.1) with A j (x) replaced by We shall call potentials (A ′ 0 , .., A ′ n (x)) and (A 0 (x), .., A n (x))) related by (1.23) gauge equivalent. Note that when D is simply connected then c(x) = exp iϕ where ϕ(x) ∈ C ∞ (D), ϕ(x) is real-valued and ϕ(x) = 0 on S 0 .
We shall formulate now some conditions which will be required to solve the inverse problem.

1) Real analyticity in the time variable
One of the crucial steps in solving the inverse problem will be the use of the following unique continuation theorem of Tataru and Robbiano and Zuily (cf [T], [RZ]) that requires the analyticity in x 0 : Theorem 1.1. Let the coefficients of (1.1) be analytic in x 0 . Consider the equation Lu = 0 in a neighborhood U 0 of a point P 0 . Let Σ = 0 be a noncharacteristic surface with respect to L passing through P 0 . If u = 0 in U 0 ∩ {Σ < 0} then u = 0 in U 0 ∩ {Σ > 0}.
We shall assume not only that the coefficients of L are analytic in x 0 but also the boundary S(x) = 0, the maps (1.12) and (1.20) and the gauges c(x) are all analytic in x 0 . Thus the coefficients of operatorsL, L ′ are also analytic in y 0 .
Let y = ϕ(x) be a diffeomorphism of neighborhood U 0 onto the neighborhood V 0 = ϕ(U 0 ). Here ϕ(x) is smooth but not analytic in any variable. It is clear that if the unique continuation property for the operator L holds in U 0 then it holds in V 0 for the operatorL = ϕ • L, though the coefficients of L are not analytic. Therefore the following more general class of operators L with non-analytic coefficients has the unique continuation property: For each point x (0) on D there is a neighborhood U 0 and the diffeomorphism Ψ(x) of U 0 onto V 0 = ψ(U 0 ) such that the coefficients of the operators ψ • L in V 0 are analytic in x 0 . Thus, the unique continuation property holds for L in U 0 .
Let t 0 be arbitrary. It is shown in [BLR] that the BLR condition holds on 2) Domains of influence −1 . We say that is called the backward time-like ray.
One can show (cf [CH]) that the forward domain of influence D + (F ) of a closed set F ⊂ D 0 × R is the closure of the union of all piece-wise smooth forward time-like rays in D 0 × R starting on F .
Analogously, the backward domain of influence D − (F ) of the closed set F ⊂ D 0 × R is the closure of the union of all backward time-like piece-wise smooth rays in [KKL1]). Here ∂ ∂ν is the normal derivative to Γ.
In particular, we shall use the following fact: is arbitrary large. Then the domain of dependence of Γ × [t 1 , t 2 ] contains D 0 × [t 1 + δ, t 2 − δ] for some δ > 0 dependent of the metric and the domain.
In this paper we will not attempt to estimate δ > 0 since [t 0 + δ, t 2 − δ] is also arbitrary large if [t 1 , t 2 ] is arbitrary large. Now we shall state the main result of this paper. Making the change of variables of the form (1.20) we can, without loss of generality, assume that the domain D is cylindrical (cf Fig. 1.2 We consider the initial-boundary value problem where L has the form (1.1) and f has a compact support in Γ 0 × R. Consider another initial-boundary problem (1.39) where L ′ u ′ has the form (1.30), f ′ has a compact support in Γ 0 × R.
nonempty and open. Let Λ and Λ ′ be the corresponding DN operators for L and L ′ . Assume that Λf Γ 0 ×R = Λ ′ f Γ 0 ×R for all smooth f with compact support in Γ 0 × R. Suppose the conditions (1.2), (1.3) hold for L and L ′ . Assume that the coefficients of L and L ′ are analytic in x 0 and y 0 , respectively. Suppose also that BLR condition holds for andÂ k (y), 0 ≤ k ≤ n as in (1.15), (1.16), c ′ • y * • L is the operator with potentials A ′ j (y), 0 ≤ j ≤ n, gauge equivalent toÂ k (y), 0 ≤ k ≤ n. We end the introduction with the outline of the previous work and a short description of the content of the paper.
The first result on inverse hyperbolic problems with the data on the part of the boundary was obtained by Isakov in [I1]. The powerful Boundary Control (BC) method was discovered by Belishev [B1] and was further developed by Belishev [B2], [B3], [B4], Belishev and Kurylev [BK], Kurylev and Lassas [KL1], [KL2] and others (see [KKL1], [KKL2]). In [E1], [E2] the author proposed a new approach to hyperbolic inverse problems that uses substantially the idea of BC method. This approach was extended in [E3] to a class of time-dependent hyperbolic problems and in [E4] to the case of hyperbolic equations of general form with time-independent coefficients. The generalization to the case of Yang-Mills potentials was considered in [E7]. The inverse problems for the D'Alambert equation with the time-dependent scalar potentials were considered earlier by Stefanov [S] and Ramm and Sjostrand [RS] (see also Isakov [I2]). The case of the D'Alambert equation with time-dependent vector potentials was studied by Salazar [S1], [S2]: The following observation of the importance of studying hyperbolic equations with time-dependent coefficients was made by Bardos-Lebeau-Rauch in [BLR]: The linearizations of basic nonlinear evolution equations of mathematical physics are linear hyperbolic equations with time-dependent coefficients.
In this paper we study the inverse problem for general second order hyperbolic equations with time-dependent coefficients. The proofs are an extension and simplification of corresponding proofs in [E1], [E2], [E3], [E4] for a more general case.
The main step in the proof is the local step of solving the inverse problem in a small neighborhood near the boundary. This is done in § §2-6. In the last §7 we consider the global step leading to the proof of Theorem 1.2.

The Goursat coordinates
We shall prove first the Theorem 1.2 in the small neighborhood of the boundary ∂D.
Let x (0) ∈ S 0 and let U 0 ⊂ R n+1 be a small neighborhood of x (0) . Suppose that we already did the change of variables to make ∂D and S 0 cylindrical, i.e. ∂D = ∂D 0 × R and S 0 = Γ 0 × R. We assume that we have chosen the coordinates (x 0 , x ′ , x n ), x ′ = (x 1 , ..., x n−1 ) in U 0 such that x n = 0 is the equation of U 0 ∩ ∂D and U 0 ∩ D is contained in the half-space n−1 , 0) be the coordinates of the point x (0) . Let Consider the initial-boundary value problem in U 0 ∩ D: We assume that L has the form (1.1). For the simplicity, we shall not change the notations when choosing the local coordinates such that the equation of We introduce new coordinates to simplify the operator L (cf. [E4], pages 327-329) that we called the Goursat coordinates.
Denote by ψ ± (x), x = (x 0 , x ′ , x n ) the solutions of the eikonal equations xn one has to specify the sign of the square root. We have We will need ψ + xn + ψ − xn < 0 for x n > 0 (cf. (2.16) below). So we choose the plus sign of the square root: . The solutions ψ ± (x) exists for 0 < x n < δ, δ is small. We assume that δ is such that surfaces ψ + = 0 and ψ − = 0 intersect when x n < δ and are inside U 0 when x n < δ.
and the derivatives of (2.31) in y n at y n = 0.
There is a parametrix of the Dirichlet problem in the elliptic region and DN operator microlocally in Σ is a pseudodifferential operator on y n = 0. We shall find the principal symbol of this operator in Σ. Let λ ± be the roots in η n of p(y, η 0 , η ′ , η n ) = 0: (2.34) where ℑλ + > 0 in Σ. Therefore the symbol of DN in Σ is (cf. [E5], §57): Knowing Λ ′ we know the symbol (2.35) for all η 0 , η ′ . In particular, we can find (2.31). Computing the next term of the parametrix (cf. [E], §57) we can find the normal derivatives of (2.31).
We have It follows from the Lemma 2.1 that g 1 , ∂g 1 ∂yn ,ĝ +,− , g +,j 0 are known on y n = 0 if Λ ′ is known. Therefore knowing Λ ′ we can determine Λ 1 f for any f .

The Green's formula
First, we introduce some notations.
We assume also that D 3, Consider the following initial-boundary value problem: T 2 ]. Also let v g be such that Integrating by parts in τ we get We used in (3.4), (3.5) that u f , v g are equal to zero on Z 3T 1 . Note that s = y 0 −T 1 , τ = T 2 −y 0 on y n = 0, and ∂ ∂s = 1 Therefore, making changes of variable τ = T 2 − y 0 in the first integral and s = y 0 − T 1 in the second, we get Analogously, integrating by parts other terms of where Λ 1 has the form (2.30) It follows from (3.7) that is determined by the boundary data, i.e. by the DN operator on Γ (3) ×(T 1 , T 2 ). We shall denote the L 2 inner product in Proof: The proof of Lemma 3.1 is a simplification of the proof of Lemmas 2.2 and 3.2 in [E3]. We shall prove Lemma 3.1 for the case s 0 = T 1 . The proof for the case T 1 < s 0 < T 2 is identical.
Denote by ∆ ′ 2 the domain bounded by the half-plaines Γ ′ where F −1 is the inverse Fourier transform, z 0 is the dual variable to s. The distribution θ(−τ )u satisfies the equation in the half-space y n > 0 with the boundary condition (3.14) θ(−τ )u yn=0 = 0.
We look for θ(−τ )u in the form where w satisfies We impose the zero initial conditions on w requiring that Therefore w is the solution of the hyperbolic equation L 1 w = ϕ in the halfspace y n > 0 with the boundary condition (3.17) and the zero initial conditions (3.19). It follows from ( [H] and [E6]) that initial-boundary value problem has a unique solution in appropriate Sobolev space of negative order.
Since ϕ belongs to L 2 in τ and to Sobolev spaces of negative order in s and y ′ , we get that w belongs to H 1 in τ . Therefore w τ =τ 0 is continuous function of τ 0 with the values in Sobolev's spaces of negative order in (s, y ′ ). Since ϕ = 0 for τ < 0 we have that w = 0 for τ < 0 by the domain of influence argument. Therefore by the continuity w τ =0 = 0 and ∂w ∂s τ =0 = 0. Therefore u = v(s) + w(s, τ, y ′ ) is the distribution solution of (3.10), (3.11), (3.12) in ∆ ′ 2 . Note that the restrictions of any distribution solution of L 1 u = 0 to y n = 0 exists since y n = 0 is not a characteristic surface for L 1 . This property is called the partial hypoellipticity (cf., for example, [E5]). Now using Lemma 3.2 we can prove Lemma 3.1.
where u 1 is an arbitrary extension of u f for s > T 2 . Analogously for the right hand side or (3.20). (Note that v = 0 for y 0 > T 2 ).
To justify (3.20) we take a sequence Applying the Green's formula (3.7) to u f and v j we get since v j yn=0 = 0. Integrating by parts we get Since v j = 0 for s > T 2 we have that v j yn=0,y 0 =T 2 = 0. Therefore, taking the limit when j → ∞ we get (3.20).
Since f is arbitrary and u f , ∂v 0 ∂s = 0 we get that ∂v ∂yn = 0 on Γ jT 1 .
Thus v 0 = 0 on R jT 1 and this contradicts the assumption that h = 0.
We shall prove two more theorem in this section that will be used in §4.
We shall need some known results on the initial-boundary hyperbolic problem. The following theorem holds: . Then for any s ≥ 0 and any f ∈ H s+1 Moreover, .
We assume that F (y) and f (y 0 , y ′ ) have compact supports in y ′ . Note that f = 0, F = 0, u = 0 for y 0 < T 1 .
Then u(y 0 , y ′ , y n ) has also a compact support in y ′ .
The proof of Lemma 3.3 in the case of time-dependent coefficients is given in [H] and [E6].
Note that Lemma 3.3 holds also in the case when R n + is replaced by an arbitrary smooth domain Ω ⊂ R n .
The following lemma follows from Lemma 3.3.
Lemma 3.4. Let, for the simplicity, Integrating by parts the identity |u y k | 2 dy 0 dy ′ dy n .
Note that I = 0 when the coefficients of L 1 do not depend on y 0 .
Combining Lemmas 3.4 and 3.5 we can prove the following lemma: Proof: It follows from Lemma 3.4 that Then (3.34) gives Combining (3.42) and (3.43) and taking into account that supp u f Γ 2 = Y js 0 , we get (3.42).
2T 1 is the intersection of y n = 0 with the closure of the union supp u f i where the union is taken over all ) and also f = 0 in a neighborhood of x ′ 0 . Then by the uniqueness of the Cauchy problem (see [T], [RZ]) we have that all j . It follows from (3.7) that (3.9) is determined by the boundary data. Integrating by part we have is also determined by the boundary data. Fig. 4.1).
. We shall show that still (u 0 , v g ) is determined by the DN operator. The following theorem holds.
. Here u f i , v g i are the same as in (3.1), (3.2) for i = 1, 2, respectively. Operators L (i) and, consequently, L To prove Theorem 4.3 we will need the Density Lemma 3.1 and the following lemma that uses the BLR condition: Proof of Lemma 4.4 (cf. Lemma 2.3 in [E3]): Suppose that BLR condition (see [BLR]) is satisfied for [H] (cf. also Lemma 3.6) that By the closed graph theorem we have It follows from (4.7) that there exists Note that (4.10) Let Λ (i) be the DN operator corresponding to L (i) , i = 1, 2,. We can introduce the Goursat coordinates for 1 be the operator L (i) in these coordinates and let Λ where v g is the same as in (3.2). Since supp v g ⊂ D(Γ js 0 ), where D(Γ js 0 ) is the domain of influence of Γ js 0 for y 0 ≤ T 2 , we have that v g i = 0 on Z js 0 . Therefore integrating by parts in (4.11) we get as in ( 3.7): We have that Λ (1) Note that v g i s=0 = 0 and u f 0 i v g 1 s=T 2 −T 1 = f 0 (T 2 , y ′ )g(T 2 , y ′ ). Therefore (4.12) (u f 0 1s , v g 1 ) = (u f 0 2s , v g 2 ) for all g ∈ H 1 0 (Γ 3s 0 ). Let Γ 2 , Γ 3 , Γ 4 be the same as in Lemma 3.5. It was proven there that by the domain of dependence argument. Comparing (4.12) with (u f 1s , v g 1 ) = (u f 2s , v g 2 ) and taking into account (4.15) we get (4.16) 2s 0 , ∀g ∈ H 1 0 (Γ 3 × (s 0 , T 2 )).
The following formula will be the main tool in solving the inverse problem.
Theorem 4.5. For any T 1 ≤ s 0 ≤ T 2 the integral (4.29) ∂s v g dsdy ′ is determined by the DN operator on Γ jT 1 . The integral (4.29) is the difference (u f s , v g ) − (u 0s , v g ) thus (4.29) is determined by DN operator.

Equation (5.34) is a linear homogeneous equation for
It follows from (5.35) and (5.22) that
Theorem 6.2 (Local step). Consider two initial boundary value problems and suppose the BLR condition holds for Suppose coefficients of L (1) and L (2) are analytic in x 0 .

The global step
f has a compact support in Γ 0 × R. First we extend the Theorem 6.2 for a larger time interval. Let [t 1 , t 2 ] be an arbitrary time interval. Let [t 0 , T t 0 ] be such that T t 0 ≤ t 1 and the BLR condition holds on [t 0 , T t 0 ]. Thus the BLR condition is satisfied on [t 0 , t] for any t ∈ [t 1 , t 2 ]. Let Γ 1 be arbitrary connected part of Γ 0 , Γ 1 ⊂ Γ 0 . Note that we do not require Γ 1 to be small.
We impose the following initial conditions on ψ ± 0i , i = 1, 2, 0i such that (2.6) is satisfied and we choose ε 1 > 0 such that ε 1 ≤ ε 0 and {0 < x n < ε 1 , with initial conditions Similarly to (2.14) consider the map (y As in (2.15) we have that the map (x 0 , x ′ , x n ) → (y 0 , y ′ , y n ) is the identity when x n = 0: Let u s = 1 2 (u y 0 − u yn ), u τ = − 1 2 (u y 0 + u yn ). Making the change of variables (7.4) in L i u i = 0, the gauge transformation (2.18), (2.21) and the change of unknown function (2.26), we get in t 0 ≤ y 0 ≤ t 2 , 0 ≤ y n ≤ T 0 , y ′ ∈ Γ ′ , T 0 is small, the equation of the form 1 has the form (2.28). Here We assume that u We also assume that DN operators for L i and subsequently for L Note that the change of variables where α (i) are the same as in (5.9), (5.11), are also defined globally onΩ 0 .
Therefore we proved Assume that the coefficients of L (i) are analytic in x 0 , i = 1, 2. Then where c 3 is the gauge transformation. Therefore Fig. 7.1). It follows from [Hi], Chapter 8, that there exists an extensionΦ 2 of Φ 2 from S 2 ⊂ D 1 ). There exists also an extensionc 3 of the gauge c 3 from S 2 to D (2) 1  The following lemma is the key lemma of this section. It allows to reduce the solution of the inverse problem to an inverse problem in a smaller domain.
Lemma 7.2. Consider two initial-boundary value problem We assume that (∂D Fig. 7.2).
are DN operators for the initial-boundary value problem To prove Lemma 7.2 we will need the following version of the Runge theorem about the approximation of solutions of the equation in a smaller domain by solutions of the same equation in a larger domain.
Then the closure of the restrictions of W to L 2 ((D t 2 )) norm by the functions in W . t 2 )). Take any g ∈ K ⊥ and denote by g 0 the extension of g by zero outside (D (1) 0 \ B) × (t 1 , t 2 ). Let w be the solution of the initial-boundary value problem Fig.7.2).

=L
(1) 1 inΩ 2 . Then L (2) has also Goursat coordinates in 1 =L (2) 1 in Ω 3 . Proof: Let y (i) = ψ (i) (x) be the transformation to the Goursat coordinates, and let Dψ (i) (x) Dx be the Jacobi matrix of this transformation. We have where [ĝ jk i ] −1 is the metric tensor in the Goursat coordinates. The Goursat coordinates degenerate at point y (2.11)). We call such point a focal point. We shall prove that there is no focal points for L (2) in Ω 3 .
We shall proceed with the enlargement of the domain Ω 2 using Lemmas 7.2 and 7.6. It follows from Lemmas 7.4, 7.5 that this enlargement depends only on L (1) and does not depend on L (2) . Therefore after finite number of steps (cf. [E2]) we get a domain D 0 . Note that [t 1 , t 2 ] is arbitrary large and therefore [t ′ 1 , t ′ 2 ] = [t 1 + δ, t 2 − δ] is also arbitrary large. Therefore we obtained the following theorem: Theorem 7.7. Let L (1) and L (2) be two operators in D ]. Now we shall use Theorem 7.7 to prove Theorem 1.2. Proof of Theorem 1.2 Let L (i) be two operators in D Let (t j1 , t j2 ) be an interval as in Theorem 7.7 and ∞ j=−∞ (t j1 , t j2 ) = R. We have D (1) . It follows from Theorem 7.7 that for each j ∈ Z there exists a diffeomorphism Φ j on D (1) j × [t j1 , t j2 ] and a gauge transformation c j such that Φ j = I and c j = 1 on Γ 0 × [t j1 , t j2 ], and In (7.26) Φ j is a diffeomorphism of D (1) 0 × [t j+1,1 , t j2 ] where [t j+1,1 , t j2 ] is the intersection of [t j1 , t j2 ] and [t j+1,1 , t j+1,2 ].