A comparative study of the control of two beam models

We study the well posedness and the controllability of a realistic beam model. It turns out that for some values of the papameters it is controllable, while for other values it is not even well posed. In order to solve this problem we also give a general abstract necessary and sufficient condition of well posedness of linear distributed systems.


On the well posedness of linear systems
Consider the linear evolutionary problem y = Ay + f, y(0) = y 0 (1) in a complex Hilbert space H. Assume that H has a Riesz basis e 1 , e 2 ,. . .formed by eigenvectors of A, corresponding to complex eigenvalues λ 1 , λ 2 , . . .: We are going to study the well posedness of this problem.It is rather well-known that the condition C := sup λ k < ∞ is sufficient for the well-posedness.It seems to be less known that this condition is also necessary.For the reader's convenience we prove both results.Fix T > 0 arbitrarily.
Theorem 1 (a) If C < ∞, then the problem (1) has a unique solution for every y 0 ∈ H and f ∈ L 2 (0, T ; H).It is given by the series converging in C([0, T ], H), where the coefficients c k and f k (s) are given by the expansions of y 0 and f (s) according to the Riesz basis (e k ): (The last expansion holds true for almost every s ∈ (0, T ).)(b) If C = ∞, then there exists y 0 ∈ H such that the problem (1) has no solution (defined on the whole interval [0, T ]) for any f ∈ L 2 (0, T ; H).

Remark 1 The proof will show
then there still exists a unique solution y ∈ C([0, T ], H).
Proof:[Proof of part (a)] Assume that (1) has a solution y and expand it into a series according to the Riesz basis (e k ): we obtain for each t ∈ [0, T ] an expansion Using the definition of the solution, we obtain for each k that This proves the uniqueness of the solution and also the validity of the series representation (2).
It remains to verify that under the assumption on y 0 and f the series (2) converges in C([0, T ], H).One can then easily show that its sum is a solution of (1).Since the space C([0, T ], H) is complete, it suffices to prove that the partial sums of the series Since (e k ) is a Riesz basis in H, there exists a constant c > 0 such that for every x ∈ H, its expansion x = x k e k satisfies the inequalities We shall frequently use these inequalities in the sequel.
For n > m we have We conclude by noting that the last expression does not depend on t ∈ [0, T ] and that it converges to zero as m, n → ∞ because Similarly, for n > m we also have The last expression does not depend on the particular choice of t ∈ [0, T ].Moreover, thanks to our assumption f ∈ L 1 (0, T ; H), it tends to zero as m, n → ∞.
Assume on the contrary that for some f ∈ L 2 (0, T ; H) there exists a solution y ∈ C([0, T ]; H).Then, repeating the beginning of the proof of part (a) above, y is given by the series (2) with c k = 1/ for all and with c k = 0 otherwise.
Observe that since Hence all but finitely many terms of the sum are smaller than 1.Since we conclude that for all sufficiently large we have But this is impossible because the right hand side tends to infinity by the choice of λ k . 2

Control of Kirchhoff beams
Given three numbers a, b, c > 0, consider for a given (time) T > 0 the following problem: We may rewrite it in the form (1) by setting for brevity.Equivalently, D j is the completion of the vector space generated by the functions sin kx, k = 1, 2, . . ., with respect to the scalar product where (Note that this definition enables us to define a Hilbert space D j for each real j. Consider the Hilbert space H := D 4 × D 3 .One can readily verify that f ∈ L 1 (0, T ; H) if and only if g ∈ L 1 (0, T ; D 1 ).Furthermore, setting for k = 1, 2, . . ., we have a Riesz basis (e ± k ) of H, formed by eigenvectors of A with the eigenvalues λ ± k .Since the eigenvalues λ ± k are purely imaginary for all sufficiently large k, applying Theorem 1 we conclude that our problem is well posed.Hence for every u 0 ∈ D 4 , u 1 ∈ D 3 and g ∈ L 1 (0, T ; D 1 ), the problem (4) has a unique solution satisfying We are going to prove the following exact internal controllability result: Theorem 2 Given u 0 ∈ D 4 and u 1 ∈ D 3 arbitrarily, there exists a function g ∈ L ∞ (0, T ; D 1 ) such that the solution of (4) satisfies u(x, T ) = u t (x, T ) = 0 for all 0 < x < π. (5) Proof: Let us write the solution of (4) in an explicit form.Developing u, g, u 0 and u 1 into Fourier series according to the orthogonal basis (sin kx) of L 2 (0, π) (which is also an orthogonal basis in H 1 0 (0, π) and in (H 2 ∩ H 1 0 )(0, π)), we have where k runs from 1 to ∞ in all sums.Substituting these expressions into (4), we obtain for each k the initial-value problem Let us solve (6) for u k .Since the characteristic equation applying the variation of constants formula we find that Let us try to find controls of the form with suitable coefficients α k and β k .Substituting this expression into (8), the conditions (5) are equivalent to u k (T ) = u k (T ) = 0 for each k, which leads to the system of linear equations This system can be solved uniquely for α k and β k because its determinant is strictly negative.Indeed, this follows from the Cauchy-Schwarz inequality because the functions sin γ k s and cos γ k s are linearly independent in L 2 (0, T ).
Let us investigate the asymptotic behavior of α k and β k for large k.From (7) we deduce that Here and in the sequel the notation a k b k means the existence of two positive constants c 1 and c 2 such that Hence α k αk and β k βk where αk and βk denote the solutions of the system We conclude that for all k and for all 0 ≤ t ≤ T , and then for all 0 ≤ t ≤ T .This proves that g satisfies (6). 2 One can show that the choice (9) leads to the control of minimal norm.Indeed, putting R k (s) = sin γ k (T − s) for brevity, for any control g k (t) we deduce from the equalities u k (T ) = u k (T ) = 0 that ), with equality if and only if g k is a multiple of R k by the condition of equality in the Cauchy-Schwarz inequality.

A beam model of bending waves. The well-posed case
Given three positive numbers a, b, c, consider the problem for some given time T > 0. Let us rewrite it in the form (1) by setting Introducing the same Hilbert spaces D j as in the preceding section, now consider the Hilbert space ).Furthermore, denoting for each positive integer k by λ k1 ,. . ., λ k4 the four numbers given by the formula and setting one can readily verify that every e kj is an eigenvector of A with the eigenvalue λ kj and that these vectors form a Riesz basis in H.
For large k we have If c < b, then these formulas show that the real parts of the eigenvalues are not bounded from above, so that the problem is ill-posed.On the other hand, if c ≥ b, then they are bounded from above (indeed, they are purely imaginary apart from maybe a finite number of terms), so that the problem is well posed.More precisely, applying Theorem 1 we obtain that for every given the problem (10) has a unique solution u in with D 0 := L 2 (0, π).In the rest of this section we restrict ourselves to the case c > b.Then the problem is well posed and the four roots λ kj are pairwise distinct for each k.We study the exact internal controllability of this system.We are going to prove the Theorem 3 Assume that c > b.Given u 0 ∈ D 4 , u 1 ∈ D 3 , u 2 ∈ D 2 and u 3 ∈ D 1 arbitrarily, there exists g ∈ L ∞ (0, T ; D 1 ) such that the solution of (10) satisfies u(x, T ) = u t (x, T ) = u tt (x, T ) = u ttt (x, T ) = 0 for all 0 < x < π.
Proof: Developing u, g, u 0 , u 1 , u 2 , u 3 into Fourier series according to the functions sin kx as in the preceding section, now we find for each k = 0, 1, . . . the following system: Since the four roots λ kj are distinct , the solution of this system has the form where R k is the resolvent of the system given by the formula and the complex coefficients v kj depend on the initial data via the linear system It is natural to seek controls of the form with suitable coefficients g k .Substituting into (12), the "final" conditions u (p) k (T ) = 0, p = 0, . . ., 3 are equivalent to the following linear system: The determinant of this system is different from zero by the general theory of linear ordinary differential equations.Furthermore, since the eigenvalues are purely imaginary for all sufficiently large k, using also the asymptotic relations (11) we obtain that where the r 1 ,. . ., r 4 are strictly positive numbers depending on a, b, c.Using these relations we deduce from (14) that for all k, j and 0 < t < T with some uniform constant c.Using the last two estimates, we deduce from (13) that with another constant c , independent of k.Finally, since 4. The beam model of bending waves.The ill-posed case If c < b, then the problem (10) is generally ill-posed.However, according to the remark following the formulation of Theorem 1, for certain initial data there still exist global solutions.It is then natural to study the existence of controllable states.
Using the notations of the preceding section, we may still assume that the four roots λ kj are pairwise distinct for each k.Let us order them such that λ k1 ≥ λ k2 ≥ λ k3 ≥ λ k4 for each k.We may repeat the computations of the preceding section, by only changing the asymptotic relations for the integrals in ( 14).Now we have k 6 e λ k − λ kj

T 0 R 4 j=1|g kj | ≤ c k 3 4 j=1k 2 k 8 4 j=1
kj (t)R k (t) dt → r j if j = , 0 if j = ,where the r 1 ,. . ., r 4 are nonzero numbers depending on a, b, c.Using these relations we deduce from (14) thatg kj ≈ − k 6 r j e λ kj T v kjif k is sufficiently large.It follows from the definition of the functions R kj , from (11) and from the fact that the real parts of the eigenvalues are bounded from above, that|R kj (t)| ≤ ck −3for all k, j and 0 < t < T with some uniform constant c.Using the last two estimates, we deduce from (13) that|g k (t)| ≤ ck −3 |v kj |with another constant c , independent of k.Finally, sinceg(t) 2 D 1 ≤ c ∞ k=1 |g k (t)| 2 ≤ c ∞ k=1|v kj | 2 < ∞ so that g ∈ L ∞ (0, T ; D 1 ) indeed.
is sufficiently large.It follows from the definition of the functions R kj , from (11) and from the fact that the real parts of the eigenvalues are bounded from above, that |R kj (t)| ≤ ck −3