Moving and oblique observations of beams and plates

We study the observability of the one-dimensional Schr{\"o}dinger equation and of the beam and plate equations by moving or oblique observations. Applying different versions and adaptations of Ingham's theorem on nonharmonic Fourier series, we obtain various observability and non-observability theorems. Several open problems are also formulated at the end of the paper.


Introduction
Fourier series methods have been applied for a long time in control theory [5,19,20,3]. Since Haraux [6] recognized the usefulness of a classical theorem of Ingham [7] in this context, many new results have been obtained by applying multiple variants of Ingham's theorem [1,2,4,11,17].
The purpose of this paper is to investigate the observability of beams and plates by moving or oblique observations.
Moving point observability theorems for parabolic and hyperbolic equations have been obtained earlier by Khapalov by different methods [12,13].
Another motivation for this paper was the following recent result of the first author with K. Kellay [10]: Theorem 1.1. Let µ be a bounded measure on R 2 and let u = µ be its Fourier transform. Assume that u is a solution of the Schrödinger equation ∂ t u(t, x)+i∂ 2 x u(t, x) = 0 on R + ×R and assume that, for some a = b ∈ R, u(t, at) = u(t, bt) = 0 for every t > 0 then u = 0.
In other words, a solution of the Schrödinger equation is uniquely determined by its value in two moving points x = at and x = bt, t > 0. The proof however does not provide any quantitative estimate on u from its values on these points.
We first consider the one-dimensional Schrödinger equation u t + iu xx = 0 in a bounded interval I with periodic boundary conditions and initial data u 0 ∈ L 2 (I). We prove among other things the observability relations fail for any choice of finitely many integers a i and for any T > 0.
Here and in the sequel the notation A ≍ B means that c 1 A ≤ B ≤ c 2 A with some positive constants c 1 , c 2 that do not depend on the particular choice of the initial data.
Next we carry over a similar study for the one-dimensional beam equation u tt + u xxxx = 0 in a bounded interval I with periodic boundary conditions and initial data u 0 ∈ L 2 (I), u 1 ∈ H −2 (I). For example, we have T 0 |u(t, at)| 2 dt ≍ u 0 2 L 2 (I) + u 1 2 H −2 (I) if and only if the circle centered in ( −a 2 , a 2 ) and passing through the origin contains no other points with integer coordinates. This is the case whenever a is irrational. Furthermore, we give a necessary and sufficient geometric condition for the validity of the estimates Finally we consider vibrating rectangular plates. Improving several earlier theorems given in [6,8,9,14], it was shown in [17] that these plates may be observed on an arbitrarily small segment which is parallel to one of the sides of the rectangle. Using a different tool we prove that the observability still holds for oblique segments.
The paper is organized as follows. In Section 2 we recall some Ingham type theorems that we need in the subsequent proofs. Section 3 is then devoted to the one-dimensional Schrödinger equation while Section 4 is devoted to the one-dimensional beam equation and Section 5 to vibrating rectangular plates. We end the paper with a list of open questions related to the problems studied here.

A review of Ingham type inequalities
Ingham type inequalities play a central role is this study. We therefore devote this section to summarize the results we use.
If I is an interval of length |I| = 2π, then Parseval's equality holds for all square summable sequences (c k ) of complex numbers. This equality remains valid if the length of I is a positive multiple of 2π. It follows by an elementary argument that if 2kπ < |I| < (2k + 2)π for some nonnegative integer k, then for all square summable sequences (c k ), and the constants 2kπ, (2k + 2)π are the best possible here. Hence for every bounded interval I of length ≥ 2π. Here and in the sequel we use the notation A ≪ B if there exists a positive constant α such that A ≤ αB for all sequences (c k ), and A ≍ B if A ≪ B and B ≪ A. Ingham [7] proved an important generalization of the last relation. A set Λ of real numbers is called uniformly separated if then γ(Λ) is called the uniform gap of Λ. For example, Z is uniformly separated with γ(Z) = 1. Note that the empty set and the one-point sets are uniformly separated with γ(Λ) = ∞.
Theorem A (Ingham). Let Λ ⊂ R be a uniformly separated set.
(i) λ∈Λ c λ e iλx is a well-defined locally square summable function on R for every square summable sequence (c λ ). Remark 2.1. If Λ is not uniformly separated, but it is the union of finitely many, say m uniformly separated sets, then a simple application of the inequality shows that the direct inequality still holds.
The condition |I| > 2π γ(Λ) is the best uniform condition for all uniformly separated sets, but it can be weakened for individual uniformly separated sets. We illustrate this by recalling from [6] the following Theorem B (Haraux). If Λ ⊂ R is a uniformly separated set and F ⊂ Λ is a finite subset, then the inverse inequality of Theorem A holds under the condition |I| > 2π γ(Λ\F ) .
Remark 2.3. The proof of Theorem B shows that the direct and inverse inequalities remain valid under the same assumptions for more general sums of the form where Λ ′ is some finite subset of Λ; see [16,Theorem 4.5].
For a particular uniformly separated set the optimal condition for the inverse inequality has been determined by Beurling [4]; see also [2] for a generalization to weakly separated sets.

One-dimensional Schrödinger equation
We consider the one-dimensional Schrödinger equation on a bounded interval with periodic boundary condition. Up to an affine change of variable, we may assume that the interval is (0, 2π). Thus we consider the following system for x ∈ (0, 2π).
Setting L 2 := L 2 (0, 2π) for brevity and introducing the Sobolev space . Furthermore, u has a Fourier series representation where the c k 's are the Fourier coefficients of u 0 : In particular, the c k satisfy Parseval's identity Using (3.2) we extend the solutions to R 2 by 2π-periodicity in x and t. First we ask whether the observability of the solutions on a fixed line segment of R 2 allows us to identify the unknown initial datum.
The case of vertical segments is easy: since the exponential functions e ikx form an orthogonal basis in L 2 (I) on every interval I of length 2π, we infer from the formula u(t 1 , x) = k∈Z c k e ik 2 t 1 e ikx that the knowledge of u on a segment {t 1 } × I determines u 0 if and only if |I| ≥ 2π. Moreover, in the latter case we also have the quantitative relation The case of horizontal segments (pointwise observability) is different: we infer from the equality that the knowledge of u even on the line on a segment R × {x 1 } does not determine u 0 . For example, if u 0 (x) = e −ix 1 e ix − e ix 1 e −ix , that is c 1 = e −ix 1 , c −1 = −e ix 1 and c k = 0 for all other k's, then u(t, x 1 ) = 0 for all t ∈ R, although u(t, x) is not the zero solution.
The situation is much better for most oblique segments: 4 Theorem 3.1. Fix (t 1 , x 1 ) ∈ R 2 , a ∈ R and T > 0 arbitrarily, and consider the solutions of (3.1).
where we use the notations In particular, then there exist non-trivial solutions satisfying and therefore the inverse inequality (3.3) fails.
Changing u(t, x) to v(t, x) := u(−t, x) we see that analogous results hold if we change the equation in (3.1) to u t − iu xx = 0.
Proof. (i) For any fixed a ∈ R a straightforward computation shows that Since Λ := {k 2 − ak : k ∈ Z} is the union of it suffices to show that latter two sets are uniformly discrete. (In view of Theorem A (i) this will also show that the restrictions of the solutions for segments are well defined.) This follows from the following inequalities: if k ≥ a/2 (ii) If a / ∈ Z, then the set {k 2 − ak : k ∈ Z} itself is uniformly discrete. Indeed, if k and m are different integers, then where [a] is the integer part of a. If, for some positive integer N, k = m and k, m / ∈ {−N, . . . , N}, then, using again the identity for all integers N > |a|. Letting N → ∞ and applying Theorem B we get the inverse inequality (3.3).
(iii) If a ∈ Z, then we may rewrite (3.6) in the form is uniformly separated, and It follows from (3.4) that all solutions satisfying d a/2 = 0 and d k + d a−k = 0 for all k ∈ Z satisfy the equality (3.5). If at least one of these coefficients is different from zero, then the right side of (3.3) vanishes, while the left side is positive.
A concrete nonzero function satisfying (3.5) may be given as follows. We choose an integer k = a/2 and then two nonzero numbers c k , c a−k satisfying the equality Then the function has the required properties.
Next we investigate what happens if we observe the solutions on two or more segments. In view of Theorem 3.1 (i) we only investigate the validity of the inverse inequalities.
For the second part, we construct a sequence (u 0,n ) ⊂ L 2 (0, 2π) of initial data such that the corresponding solutions u n satisfy the relation By the previous theorem the solutions satisfy the relation Assuming by symmetry that a 2 > a 1 , and setting p = a 2 −a 1 the relation may be rewritten in the form with suitable unimodular complex numbers ω k .
Fix an integer q > a 1 2 . For any fixed positive integer n we define consecutively the following numbers d k : Setting d k := for all other indices, we obtain a trigonometric polynomial This proves (3.7).
(ii) Note that in this part, (x 1 , t 1 ) is the same for each a 1 , . . . , a m . We will take advantage of this to construct the sequence d k .
Now we are looking for a sequence (u 0,n ) ⊂ L 2 (0, 2π) of initial data such that the corresponding solutions u n satisfy the relation For any fixed positive integer n we consider the numbers and we define On the other hand, using (3.4), Therefore we will reach a contradiction if we bound Fix i arbitrarily and write a := a i for brevity. The sequence (d k + d a−k ) takes only the values −2, −1, 0, 1, 2. It suffices to show that the number of k's for which d k + d a−k = 0 is ≪ 1 + |a|. By the symmetry of the sequence (d k ) it suffices to consider the values 1 and 2.
We have so that we have either no such k if a < 0 or at most 1 + a such indices k if a > 0. Next, we have d k + d a−k = 1 in the following three cases: and in three other symmetric cases by exchanging k and a − k.
Since the first two cases above may only occur for a > 0, while the third case only for a < 0, at most max {(a − 1 − n) + 1, −a} ≤ |a| indices k satisfy one of them. We have the same upper bound for the three symmetric cases, so that there are at most 2 |a| indices k for which d k + d a−k = 1.
We may also consider other boundary conditions. Let us consider for example the Dirichlet condition: The problem is well posed for every u 0 ∈ H 2 0 (0, π). Let us observe that extending an arbitrary solution of (3.8) to a 2π-periodic odd function in the x variable we obtain a solution of (3.1). Therefore Theorem 3.1 (i), (ii), (iii) and Theorem 3.2 (i) remain valid for the solutions of (3.8).
The remaining parts were based on the construction of special solutions, so we need some additional arguments. We have the following Proposition 3.3. Fix T > 0 arbitrarily, and consider the solutions of (3.8).
(i) For any given (t 1 , x 1 ) ∈ R 2 and a ∈ Z there exist non-trivial solutions of (3.8) satisfying with suitable square summable complex coefficients c k satisfying the relations c k +c −k = 0.
(i) We choose an integer k for which the four numbers k, a − k, −k, k − a are different that and then two nonzero numbers c k , c a−k satisfying the equality Then the function has the required properties by the same arguments as in the proof of Theorem 3.1 (iv).
(ii) Since (t 1 , x 1 ) = (0, 0) and therefore d k = c k for all k, the sequences constructed in the proof of Theorem 3.2 (ii) define solutions of not only (3.1), but also of (3.8).

Beam equation
We consider the one-dimensional linear beam equation with periodic boundary conditions: For any given initial data u 0 ∈ H 2 p and u 1 ∈ L 2 there is a unique weak solution u ∈ C(R, H 2 p ) ∩ C 1 (R, L 2 ). Furthermore, u has a Fourier series representation with suitable square summable complex coefficients c + k , c − k satisfying the relations Using (4.2) we extend the solutions to R 2 by 2π-periodicity in x. Theorem 4.2. Fix two distinct nonzero real numbers t 1 , t 2 , a number T > 0, and consider the solutions of (4.1).
(i) The direct inequality and the weakened inverse inequality always hold. (ii) If (t 2 − t 1 )/π is irrational, then the right hand side of (4.4) does not vanish for any non-trivial solution. (iii) If (t 2 − t 1 )/π is rational, there exist non-trivial solutions for which the right hand side of (4.4) vanishes. (iv) The inverse inequality

Proof. (i) Since
for j = 1, 2, applying Theorem B we get the relations They imply (4.3) by using the elementary inequality |a + b| 2 ≤ 2 |a| 2 + 2 |b| 2 . The relations (4.4) follows by adding (4.6) for j = 1, 2, and using for each k ∈ Z * the following estimates with a = k 2 t 1 and b = k 2 t 2 : (ii) If (t 2 − t 1 )/π is irrational and the right side of (4.4) vanishes for some solution u, then we infer from (4.4) that all coefficients c ± k are equal to zero because sin 2 k 2 (t 1 −t 2 ) = 0 for all k ∈ Z * , so that u is the trivial solution.
(iii) If (t 2 − t 1 )/π is rational, there exists a nonzero integer k such that k 2 (t 1 − t 2 ) is a multiple of 2π. Then the formula u(t, x) = e −ik 2 t 1 e ik 2 t − e ik 2 t 1 e −ik 2 t e ikx defines a non-trivial solution of (4.1) such that u(t 1 , x) = u(t 2 , x) = 0 for all x ∈ R.
(iv) It follows from (4.6) that the inverse inequality (4.5) holds if and only if the matrices A k := e ik 2 t 1 e −ik 2 t 1 e ik 2 t 2 e −ik 2 t 2 are invertible, and the norms of their inverses are bounded by some uniform constant.
If (t 2 − t 1 )/π is rational, then not all matrices A k are invertible by (iii). Otherwise, by the irrationality there exists a sequence (k j ) of positive integers such that k 2 j (t 1 − t 2 ) → 0 mod 2π, and then the above norms tend to ∞ as j → ∞. Now we turn to the case of oblique segments. Given a real number a, if u is a solution of (4.1), then a straightforward computation shows that where we use the notations Observe that In order to state our results we introduce the circle S a ⊂ R 2 centered in (a/2, −a/2) and passing through the origin. Its cartesian equation is or equivalently x 2 − ax + y 2 + ay = 0.

Remarks 4.3.
(i) Since the distance between distinct elements of A a is at least one, A a cannot have more elements than the perimeter of the circle S a : |A a | ≤ √ 2πa.
If a is not an integer, then no element of A a has any zero coordinate, and hence A a = (Z * ) 2 ∩ S a . Indeed, if (k, 0) ∈ A a , then a = k ∈ Z from the above equation of S a .
Theorem 4.4. Fix (t 1 , x 1 ) ∈ R 2 , a ∈ R and T > 0 arbitrarily, and consider the solutions of (4.1). (ii) Since (a, −a) ∈ A a for all nonzero integers, a / ∈ Z by our assumptions, and therefore both sets k 2 − ak : k ∈ Z and −k 2 − ak : k ∈ Z are uniformly discrete by the proof of Theorem 3.1 (ii). In view of (4.7) we have to show that their union is also uniformly discrete. This amounts to show that this means that the circle S a has a positive distance from the set Z 2 \ {(0, 0)}. Since the latter set is discrete, this is satisfied by our assumption A a = ∅.

Now let us investigate the inverse inequality
when we observe the solutions on two segments.
We start with some simple observations. We write A j instead of A a j for brevity, and we denote by A + j , A − j its projection on the first and second coordinate axis, respectively. Remarks 4.6.
(i) If a 1 = 0 and A 1 = ∅, then (4.8) holds by the preceding theorem. The same conclusion holds by symmetry if a 2 = 0 and A 2 = ∅. (ii) The proof of Theorem 4.4 (ii) shows that (4.9) and c ∈ R, then changing d + k and +d − m to d + k + c and +d − m − c the right side of (4.9) remains unchanged. Lemma 4.7. If a 1 , a 2 are different nonzero integers, then Proof. (i) Since all circles S a have the same tangent line in the origin, S 1 ∩ S 2 = {(0, 0)}, and hence A 1 ∩ A 2 = ∅.
(ii) If for example (k, m), (k ′ , m) ∈ A 1 with k = k ′ , then both k and k ′ solve the equation x 2 − a 1 x + m 2 + a 1 m = 0, and hence a 1 = k + k ′ ∈ Z. The other cases are similar.
At this stage it is convenient to associate a graph G(a 1 , a 2 ) to a pair of distinct integers a 1 , a 2 . The vertices of this graph form the set A 1 ∪ A 2 . Two vertices are adjacent if they have a common coordinate. The previous lemma then states that this graph is bipartite, namely a vertex in A 1 can only be adjacent to a vertex in A 2 and vice versa. A direct consequence of this is that a vertex has at most two neighbours.
A (simple) path is a sequence of distinct vertices v 1 , v 2 , . . . , v n where v j , v j+1 are adjacent for every j = 1, . . . , n − 1. In particular, if the first coordinate is common in v j , v j+1 then v j+1 , v j+2 have the second coordinate in common, and vice versa.
A simple path has at most |A 1 | + |A 2 | ≤ √ 2π(a 1 + a 2 ) elements. Furthermore, every v ∈ A 1 ∪ A 2 belongs to a unique maximal simple path v −ℓ 1 , . . . , v 0 , . . . , v ℓ 2 (see the figure). This maximal path is called a cycle if v −ℓ 1 and v ℓ 2 are adjacent, that is, they have a common component. Note that a cycle has necessarily an even number of points.
Proof. Adding the relations (4.9) for j = 1, 2 we see that the right hand side of (4.8) is Since A 1 ∪ A 2 is finite, the difference between (4.10) and the left hand side of (4.8) is a quadratic form in a finite number of variables. Therefore (4.8) is equivalent to the following uniqueness property: if the expression in (4.10) is zero, then all coefficients d ± k vanish.
Up to exchanging A 1 and A 2 we may assume that (k 1 , To prove the other direction, assume that G(a 1 , a 2 ) has no cycle, and consider an arbitrary maximal simple path. By symmetry between a 1 and a 2 , we may assume that this path starts in A 1 . Depending on whether the first move is horizontal or vertical, and whether the path ends in A 1 or A 2 , there are four possibilities: • The first move is horizontal and the path ends in A 2 , so that the path has the form (k 1 , m 1 ), (k 2 , m 1 ), (k 2 , m 2 ), . . . , (k n , m n−1 ) with (k 1 , m 1 ) ∈ A 1 and (k n , m n−1 ) ∈ A 2 . Since the path is maximal, the only element of A 2 adjacent to (k 1 , m 1 ) is (k 2 , m 1 ); hence k 1 / ∈ A + 2 . Similarly, k n / ∈ A + 1 . • The first move is horizontal and the path ends in A 1 : we have (k 1 , m 1 ), (k 2 , m 1 ), (k 2 , m 2 ), . . . , (k n , m n−1 ), (k n , m n ) with (k 1 , m 1 ) ∈ A 1 and (k n , m n ) ∈ A 1 . The maximality implies that k 1 / ∈ A + 2 and m n / ∈ A − 2 . • The first move is vertical and the path ends in A 2 : we have Let us consider the first case; the others are similar. If the expression in (4.10) vanishes, then d ± k = 0 whenever k / ∈ A ± 1 or k / ∈ A ± 2 , and In particular, d + k 1 = 0, d + kn = 0 and . . , n − 1. Finally, since every (k, j) ∈ A 1 ∪ A 2 belongs to a maximal simple path, we conclude that d + k = d − m = 0 for all k, m as claimed.    G(a 1 , a 2 ) has no cycle in the following cases: (i) a 1 and a 2 have opposite nonzero signs; (ii) a 1 and a 2 have equal signs, and a 1 /a 2 ≥ 3/2.
Proof. (i) Without loss of generality, we may assume that a ≥ b > 0 and a 1 = −a, a 2 = b. Let us start with the observation that, to belong to a cycle, a point (k, m) ∈ S −a must have two neighbours (k, m ′ ), (k ′ , m) ∈ S b . Setb and denote byã + andã − the positive and negative root of x 2 + ax +b 2 − ab = 0, that is Finally, let P = (−b,ã − ) (resp. Q = (−ã − ,b)) be the point on S −a such that the (vertical) (resp. horizontal) line through P and (−b, −b/2) (resp. Q and (b/2,b)) is tangent to S b . Now observe that a vertex of G(−a, b) on S −a \ Γ 0 has at most one neighbour, so that it cannot belong to a cycle. Furthermore, a vertex of G(−a, b) on Γ 0 + has no neighbour on Γ 0 − . Therefore, if there exists a cycle, then its points on S −a should all belong to Γ 0 + , or should all belong to Γ 0 − . Since the reflection of a cycle with respect to the anti-diagonal is also a cycle, it remains to prove that there is no cycle all of whose points on S −a belong to Γ 0 + . The geometric property of Γ 0 + that we use is the following: consider the arcΓ of S b joining (0, 0) to (b/2,b). Then if we start at (x, y) ∈ Γ 0 + , draw a vertical line till we reach Γ at some point (x, y ′ ) and then draw an horizontal line ℓ. Then ℓ will intersect S −a at a point (x ′ , y ′ ) ∈ Γ 0 + with 0 < x ′ < x (and at a second point (x ′′ , y ′ ) / ∈ Γ 0 ). Now it is easy to see that Γ 0 + contains no cycle. Indeed, if A 0 = (k, l) ∈ G(−a, b) ∩ Γ 0 + belonged to a cycle, then it would have a neighbour of the form A 1 = (k, l ′ ) ∈Γ. Then the second neighbour of A 1 would be of the form (k ′ , l ′ ) ∈ Γ 0 + with 0 < k ′ < k. Thus this path cannot return to A 0 , a contradiction.
(ii) The condition a ≥ 3 2 b > 0 was chosen so that the situation is exactly the same as previously. Again, S a splits into 2 arcs. On one of them the vertices have no neighbours, and on the other arc after two steps we always get strictly closer to the origin.
The results of this section may be extended to general rectangular domains by a simple linear change of variables.
Let us consider the orthonormal basis Using (4.2) we extend the solutions to R 3 (by 2π-periodicity in x and y).