A partial data inverse problem for the Convection-diffusion equation

In this article, we study the unique determination of convection term and the time-dependent density coefficient appearing in a convection-diffusion equation from partial Dirichlet to Neumann map measured on boundary.

As shown in [14] (see also [45]) that for f ∈ K 0 , Equation (1.1) admits a unique solution u ∈ H 1 (0, T ; H −1 (Ω)) ∩ L 2 (0, T ; H 1 (Ω)) and the operator N A,q u given by N A,q u, w| Σ := Q −u∂ t w + ∇ x u · ∇ x w + 2uA · ∇ x w + (∇ x · A)uw − |A| 2 uw + quw dxdt 1 is well-defined for w ∈ H 1 (Q) such that w(T, x) = 0, for x ∈ Ω. Note that if A, q and f are smooth enough then N A,q u is given by where ν stands for the outward unit normal vector to ∂Ω and u solution to (1.1). Motivated by this and [14], we define the Dirichlet to Neumann (DN) map Λ A,q : where H * T denotes the dual of space H T and u is solution to (1.1) with Dirichlet boundary data equal to f . Then from ( [14], see Section 2), we have that DN map Λ A,q defined by (1.2) is continuous from K 0 to H * T . In the present article we first consider the problem of unique recovery of coefficients A(x) and q(t, x) appearing in (1.1) from the information of DN map Λ A,q measured on a subset of Σ. It is wellknown [see [53]] that one cannot determine coefficient A(x) uniquely from DN map Λ A,q measured on Σ and this is because of the gauge invariance associated with A(x). So one can only hope to recover A(x) uniquely upto a potential term however the coefficient q(t, x) can be determined uniquely (see Theorem 2.1 in §2 for more details). Later as a corollary of Theorem 2.1, we consider the problem of determining time-dependent coefficients A(t, x) and q(t, x) appearing in (1.1) from the partial information of DN map Λ A,q . Using some extra assumption on A(t, x) and Theorem 2.1, we show that time-dependent coefficients A(t, x) and q(t, x) can be determined uniquely from the knowledge of DN map Λ A,q measured on a part of Σ (see Corollary 2.2 below in §2 for more details).
The initial boundary value problem (1.1) is known as a convection-diffusion equation with constant diffusion. The coefficients A and q are called convection term and density coefficient respectively. The convection-diffusion equations appear in chemical engineering, heat transfer and probabilistic study of diffusion process etc.
Determination of the coefficients from boundary measurements appearing in parabolic partial differential equations have been studied by several authors. Isakov in [33] considered the problem of determining time-independent coefficient q for the case when A = 0 in (1.1) from the DN map and he proved the uniqueness result by showing the density of product of solutions (inspired by the work of [54]) in some Lebesgue space. Avdonin and Seidman in [2] studied the problem of determining time-independent density coefficient q(x) appearing in (1.1) by using the boundary control method pioneered by Belishev, Kurylev, Lassas and others see [1,4,37] and references therein. In [23] Choulli proved the stability estimate analogous to the uniqueness problem considered in [33]. In [25] problem of determining the first order coefficients appearing in a parabolic equations in one dimension from the data measured at final time is studied. Cheng and Yamamoto in [16] proved the unique determination of convection term A(x) (when q = 0 in (1.1)) from a single boundary measurement in two dimension. Gaitan and Kian [30] using the global Carleman estimate used for hyperbolic equations [see [12]] proved the stable determination of time-dependent coefficient q(t, x) in a bounded waveguide. Choulli and Kian in [21] proved the stability estimate for determining time-dependent coefficient q from the partial DN map. For more works related to parabolic inverse problems, we refer to [8,18,19,20,22,23,25,30,34,35,46] and the references therein. We also mention the work of [3,5,9,11,27,41,42,43] related to dynamical Schrödinger equation and the work of [10,28,29,49,50,51] for hyperbolic inverse problems. We refer to [15,17,47] for steady state convection-diffusion equation. Recently Caro and Kian in [14] established the unique determination of convection coefficient together with non-linearity term appearing in the equation from the knowledge of DN map measured on Σ.
Inspired by the work of [21], we consider the problem of determining the full first order space derivative perturbation of heat operator from the partial DN map. We have proved our uniqueness result by using the geometric optics solutions constructed using a Carleman estimate in a Sobolev space of negative order and inverting the ray transform of a vector field which is known only in a very small neighbourhood of fixed direction ω 0 ∈ S n−1 := {x ∈ R n : x = 1}. For elliptic and hyperbolic inverse problems these kind of techniques have been used by several authors. Related to our work, we refer to [13,26] for the elliptic case and to [6,7,32,36,38,39,40,44] for the hyperbolic case.
The article is organized as follows. In §2 we give the statement of the main result. §3 contains the boundary Carleman estimate. In §4 we construct the geometric optics solutions using a Carleman estimate in a Sobolev space of negative order. In §5 we derive an integral identity and §6 contains the proof of main Theorem 2.1 and Corollary 2.2.
Since Ω is bounded and T < ∞, so we can choose a smallest R > 0 such that Q ⊂ B(0, R) where B(0, R) ⊂ R 1+n is a ball of radius R with center at origin. Now we define admissible set A of vector fields A(t, x) appearing in (1.1) by We first prove the uniqueness result for time-independent convection coefficient A ∈ A and timedependent density coefficient q. More precisely we prove the following theorem: Theorem 2.1. Let A (1) , q 1 and A (2) , q 2 be two sets of coefficients such that A (i) ∈ A are timeindependent and q i ∈ L ∞ (Q) for i = 1, 2. Let u i be the solutions to (1.1) when (A, q) = A (i) , q i and Λ A (i) ,q i for i = 1, 2 be the DN maps defined by (1.2) corresponding to u i . Now if then there exists a function Φ ∈ W 2,∞ 0 (Ω) such that In Theorem 2.1 if we take some extra assumption on convection term A (i) then we can prove the uniqueness result for full recovery of A (i) even for the case when A (i) ∈ A for i = 1, 2 are time-dependent. The precise statement of this is given in the following Corollary.
Corollary 2.2. Let A (1) , q 1 and A (2) , q 2 be two sets of time-dependent coefficients such that A (i) ∈ A and q i ∈ L ∞ (Q) for i = 1, 2. Let u i be the solutions to (1.1) when (A, q) = A (i) , q i and Λ A (i) ,q i for i = 1, 2 be the DN maps defined by (1.2) corresponding to u i . Now if and Remark 2.3. The additional assumption (2.3) on convection term A (i) in Corollary 2.2 have been considered in prior works as well. See for example [9,24] for the determination of vector field term appearing in the dynamical Schrödinger equation and also in [14] for non-linear parabolic equation.

Boundary Carleman estimate
In this section we prove a Carleman estimate involving the boundary terms for the operator L A,q . We will use this estimate to control the boundary terms appearing in integral identity given by (5.9) where no information is given.
If A ∈ A and q ∈ L ∞ (Q) then there exists C > 0 depending only on Ω, T, q and A such that holds for λ large.
Proof. Let and denote by Then where Now let where Next we estimate each of I j for j = 1, 2. Now I 1 is We consider each term separately on right hand side of the above equation. Using integration by parts and the fact that v| Σ = 0, we have holds for any s > 0. Thus we have Following the proof of [21, Lemma 3.1] we have where R > 0 is the radius of smallest ball Combining Equations (3.6),(3.7) and (3.8) we get (3.9) Next we estimate I 2 . (3.10) Using (3.9) and (3.10) in (3.5) we get , therefore taking λ large enough and using the Poincaré inequality, we have holds for large λ and C > 0 depending only on Q, A and q. Now after substituting v(t, This completes the proof of Carleman estimate given by (3.1).

Construction of geometric optics solutions
In this section, we construct the exponentially growing solution to and exponentially decaying solution to is a formal L 2 adjoint of the operator L A,q . We construct these solutions by using a Carleman estimate in a Sobolev space of negative order as used in [26] for elliptic case and in [39,44] for hyperbolic case. Before going further following [39] we will give some definition and notation, which will be used later.
with the norm where S ′ (R n ) denote the space of all tempered distribution on R n and u(t, ξ) is the Fourier transform with respect to space variable x ∈ R n . We define by here F x and F −1 x denote the Fourier transform and inverse Fourier transform respectively with respect to space variable x ∈ R n . With this we define the symbol class S m λ (R n ) of order m by With these notations and definitions, we state the main theorem of this section. (1) (Exponentially growing solutions) Let L A,q be as defined above. Then for λ where for χ ∈ C ∞ c ((0, T )) arbitrary, we have and R g (t, x, λ) satisfies the following (2) (Exponentially decaying solutions) Let L * A,q be as before. Then for λ large there exists v ∈ H 1 (0, T ; H −1 (Ω)) ∩ L 2 (0, T ; H 1 (Ω)) a solution to and R d (t, x, λ) satisfies the following Proof of the above theorem is based on a Carleman estimate in a Sobolev space of negative order. To prove the Carleman estimate stated in Proposition 4.2, we follow the arguments similar to one used in [26,39,44] for elliptic and hyperbolic inverse problems.
(2) (Interior Carleman estimate for L ϕ ) Let L ϕ be as before then there exists a constant C > 0 independent of λ and v such that (1) (Proof for (4.7)) Since Writing L * ϕ as Now from (3.4), we have P * 1 = P 1 and P * 2 = −P 2 . Hence using the arguments similar to Theorem 3.1, we have for some constant C > 0 independent of λ and v. The above estimate can be written in compact form as for some constant C independent of λ and v.
Proof. We will give the proof for existence of solution to L A,q and the proof for L * A,q follows by using similar arguments. The proof is based on the standard functional analysis arguments. Consider the space S := {L * ϕ u : u ∈ C 1 ([0, T ]; C ∞ c (Ω)) and u(T, x) = 0} as a subspace of L 2 0, T ; H −1 λ (R n ) . Define the linear operator T on S by Now using the Carleman estimates (4.7), we have holds for z ∈ C 1 ([0, T ]; C ∞ c (Ω)) with z(T, x) = 0. Hence using the Hahn-Banach theorem, we can extend the linear operator T to L 2 0, T ; H −1 λ (R n ) . We denote the extended map as T and it satisfies T ≤ C v L 2 (Q) . Since T is bounded linear functional on L 2 0, T ; H −1 λ (R n ) therefore using the Riesz representation theorem there exists a unique u ∈ L 2 (0, T ; H 1 λ (R n )) such that ) satisfying z(T, x) = 0. Choosing f = L * ϕ z in the above equation, we get L ϕ u = v. Using the expression for L ϕ from (3.3) and the fact that u ∈ L 2 (0, T ; H 1 (Ω)) and v ∈ L 2 (Q), we get that ∂ t u ∈ L 2 (0, T ; H −1 (Ω)). Hence we have u ∈ H 1 (0, T ; H −1 (Ω)) ∩ L 2 (0, T ; H 1 (Ω)).
Next we will show that u(0, x) = 0 for x ∈ Ω. To prove this we choose f = L * ϕ z where z ∈ C 1 ([0, T ]; C ∞ c (Ω)) and z(T, x) = 0. Using this choice of f in (4.15), we have

Now using integration by parts and the fact that
The above identity holds for any z ∈ C 1 ([0, T ]; C ∞ c (Ω)) satisfying z(T, x) = 0. Therefore, we conclude that u(0, x) = 0 for x ∈ Ω. This completes the proof of first part of Proposition 4.3 .

4.1.
Proof of the Theorem 4.1. Using expressions v g and B g from (4.1) and (4.2) respectively and where L A,q B g (t, x) ∈ L 2 (Q). Next using Proposition 4.3, there exists R g ∈ L 2 (0, T ; H 1 (Ω)) ∩ H 1 (0, T ; H −1 (Ω)) solution to L ϕ R g (t, x, λ) = −L A,q B g (t, x), (t, x) ∈ Q, R g (0, x, λ) = 0, x ∈ Ω and it satisfies the following estimate R g L 2 (0,T ;H 1 (Ω)) ≤ C where C is a constant independent of λ. This completes the construction of solution for L A,q u = 0 and existence of the solution for L * A,q v = 0, follows in a similar way.

Integral identity
This section is devoted to proving an integral identity which will be used to prove the main result of this article. We derive this identity by using the geometric optics solutions constructed in §4. Let u i be the solutions to the following initial boundary value problems with vector field coefficient A (i) and scalar potential q i for i = 1, 2.
Let us denote Then u is solution to the following initial boundary value problem: Let v(t, x) of the form given by (4.4) be the solution to following equation Also let u 2 of the form given by (4.1) be solution to the following equation Since the right hand side of (5.3) lies in L 2 (Q) therefore using ( [23], Theorem 1.43) we have u ∈ L 2 (0, T ; H 2 (Ω)) ∩ H 1 (0, T ; L 2 (Ω)) and ∂ ν u ∈ L 2 0, T ; H 1/2 (Σ) . Next consider the following After following the arguments used in [ [14], see Proposition 2.3], we get that Also multiplying (5.3) by v(t, x) and integrating over Q, we have where in deriving the above identity we have used the following: u| Σ = 0, u| t=0 = 0 and A (1) = A (2) on Σ. Now using Equation (5.6) and the fact that L * This gives us Using (2.2), we have ∂ ν u| G = 0. Finally using Equations (5.7), (5.8) and ∂ ν u| G = 0, in (5.6), we get Next we need to estimate the right hand side of above equation. This we will do in the following lemma: for all ω ∈ S n−1 such that |ω − ω 0 | ≤ ǫ.
Hence using this in (5.11), we get This completes the proof of lemma.

Proof of theorem 2.1 and Corollary 2.2
In this section, we prove the uniqueness results. Since from (5.9), we have Now using Equation (5.10), we have After dividing the above equation by λ and taking λ → ∞, we have Next using the expression for u 2 and v from (4.1) and (4.4) respectively, we have This after using the expressions for B g (t, x) and B d (t, x) from Equations (4.2) and (4.5) respectively, we get for ω ∈ S n−1 with |ω − ω 0 | ≤ ǫ. Since the above identity holds for all χ ∈ C ∞ c (0, T ), therefore we get where ξ · ω = 0 for all ω with |ω − ω 0 | ≤ ǫ. Now decompose R n = Rω ⊕ ω ⊥ and using this in the above equation, we have here dk denotes the Lebesgue measure on ω ⊥ . After substituting τ + s =s, we get Combining this with (6.4), we get R ω · A(t, k + sω)ds = 0, for k ∈ ω ⊥ with |ω − ω 0 | ≤ ǫ. Now using the decomposition R n = Rω ⊕ ω ⊥ in the above equation, we get R ω · A(t, x + sω)ds = 0, for x ∈ R n with |ω − ω 0 | ≤ ǫ. (6.5) Thus we have the ray transform of vector field A is vanishing in a very small enough neighbourhood of fixed direction ω 0 . In order to get the uniqueness for vector field term A, we need to invert this ray transform which we will do in the following lemma: Lemma 6.1. Let n ≥ 2 and F = (F 1 , F 2 , · · · , F n ) be a real-valued time-dependent vector field with for all ω ∈ S n−1 with |ω − ω 0 | ≤ ǫ, for some ǫ > 0 and for all x ∈ R n . Then x). Proof. The proof uses the arguments similar to the one used in [44,48,52] for the case of light ray transforms. We assume that t ∈ (0, T ) is arbitrary but fixed. We have the ray transform of F at x ∈ R n in the direction of ω ∈ S n−1 is given by Now let η := (η 1 , η 2 , · · · , η n ) ∈ R n be arbitrary and denote ω := (ω 1 , ω 2 , · · · , ω n ) ∈ S n−1 . Then we have Since F has compact support therefore using the Fundamental theorem of calculus, we have Subtracting (6.7) from (6.6), we get n i,j=1 R ω i η j h ij (t, x + sω)ds = 0, for x ∈ R n and ω ∈ S n−1 near a fixed ω 0 ∈ S n−1 (6.8) where h ij is an n × n matrix with entries Define the Fourier transform of n i,j=1 ω i η j h ij (t, x) with respect to space variable x by n i,j=1 Now decomposing R n = Rω ⊕ ω ⊥ and using (6.8), we get n i,j=1 ω i η j h ij (t, ξ) = 0, for all η ∈ R n , ξ ∈ ω ⊥ and ω near ω 0 . (6.9) The goal is to prove that h ij (t, ξ) = 0, for ξ ∈ ω ⊥ with ω near ω 0 and for each t ∈ (0, T ). From the definition of h ij (t, ξ), it is clear that For n = 2, equation 6.9 gives us (ω 1 η 2 − ω 2 η 1 ) h 12 (t, ξ) = 0, for η ∈ R 2 , ξ ∈ ω ⊥ and ω near ω 0 . (6.10) Now choosing η = (ω 2 , −ω 1 ) ∈ ω ⊥ in (6.10), we get h 12 (ξ) = 0. Next we show that h ij (t, ξ) = 0 when n ≥ 3. Let {e j : 1 ≤ j ≤ n} be the standard basis for R n where e j is is given by e j := (0, 0, · · · , 0, 1 jth , 0, · · · , 0) and for simplicity we fix ω 0 = e 1 . Now let ξ 0 = e 2 be a fixed vector in R n . Our first aim is to show that h ij (ξ 0 ) = 0, for all 1 ≤ i, j ≤ n, then later we will prove that h ij (t, ξ) = 0 for 1 ≤ i, j ≤ n and ξ near ξ 0 . Following [44], consider a small perturbation ω 0 (a) of vector ω 0 = e 1 by ω 0 (a) := cos ae 1 + sin ae k where 3 ≤ k ≤ n.
After using the fact that h ij = − h ji for 1 ≤ i, j ≤ n, we get h ij (t, ξ 0 ) = 0, for 1 ≤ i, j ≤ n.
Let A be an orthogonal matrix such that Aξ = e 2 , where A is given by a k1 a k2 a k3 · · · a kn a k+11 a k+11 a k+13 · · · a k+1n . . .
then ω is near ω 0 = e 1 when φ i 's and θ are close to 0. Next choose ω 0 (a) = cos ae 1 + sin ae l with l = 2, then ω 0 (a) is close to e 1 when a is close to zero. Now define ω(a) by cos a cos φ 1 cos φ 2 + a l1 sin a − cos a sin φ 1 + a l2 sin a . . . cos a cos φ 1 sin φ 2 · · · sin φ n−2 cos θ + a ln sin a     .
Then we have ω(a) is close ω for a near 0 andω is close to ω 0 when φ i and θ are close to zero. Also we can see that ω(a) · ξ = A T ω 0 (a) · ξ = ω 0 (a) · Aξ = ω 0 (a) · e 2 = 0, hold because of the choice of ω 0 (a). Hence using these choices of ω(a) and choosing η = e j in (6.9), we have cos a n i=1 ∂ξ i ∂φ 1 h ij (t, ξ) + sin a n i=1 a li h ij (t, ξ) = 0, for1 ≤ j ≤ n and a near 0. Now since sin a and cos a are linearly independent, therefore we get a li h ij (t, ξ) = 0, for all 1 ≤ j ≤ n and l = 2. (6.12) Above equations can be written as Now let us define matrix B and a n-vector h j as follows: a k1 a k2 a k3 · · · a kn a k+1,1 a k+1,2 a k+1,3 · · · a k+1,n . . .
Using these, we have equation (6.13), can be written as Bh j (t, ξ) = 0, for 1 ≤ j ≤ n. (6.14) Note that the matrix B is obtained from A by removing the second row and it is (n − 1) × n matrix. From the definition of A it is clear that rank of A is n, so the rank of B is n − 1. i.e. there exists at-least one non-zero minor of order n − 1 of the matrix B. Without loss of generality assume B ′ is non-zero minor of order n − 1, where B ′ is given by . Now using the fact h 11 (t, ξ) = 0 in (6.14), we have B ′ h ′ 1 (t, ξ) = 0 (6.15) where h ′ 1 = h 21 (t, ξ), h 31 (t, ξ), · · · , h n1 (t, ξ) T . Since B ′ has full rank therefore h ′ 1 (t, ξ) = 0. Also using the fact h ij (t, ξ) = − h ji (t, ξ) and h 1 (t, ξ) = 0, we have B ′ h ′ j (t, ξ) = 0, for 2 ≤ j ≤ n (6. 16) where h ′ j is an (n − 1) vector obtained after deleting jth entry from h j . Now using (6.15) and (6.16) in (6.14), we get h j (t, ξ) = 0, for 1 ≤ j ≤ n which gives us h ij (t, ξ) = 0, for 1 ≤ i, j ≤ n and ξ near e 2 .
Since h ij are compactly supported therefore using the Paley-Wiener theorem, we have h ij (t, ξ) = 0, for 1 ≤ i, j ≤ n ξ ∈ R n and t ∈ (0, T ).
Fourier inversion formula gives us h ij (t, x) = 0 for x ∈ R n and for each t ∈ (0, T ). Finally after using the definition of h ij (t, x) and the Poincaré lemma, there exists a Φ(t, ·) ∈ C ∞ c (R n ) such that F (t, x) = ∇ x Φ(t, x) for x ∈ Ω and for each t ∈ (0, T ). This completes the proof of Lemma 6.1.
Since q ∈ L ∞ (Q) is zero outside Q therefore by using the Paley-Wiener theorem we have q 1 (t, x) = q 2 (t, x) for (t, x) ∈ Q. This completes the proof of Theorem 2.1.