## Brunovsky decomposition and motion planning for diﬀusion equation with boundary control

01/10/2017**OAI :**oai:www.see.asso.fr:545:2004-1:20064

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Brunovsky decomposition and motion planning for diffusion equation with boundary control Beatrice Laroche Philippe Martin Laboratoire des Signaux et Systèmes Centre Automatique et Systèmes, École des Mines de Paris Supélec, 3 rue Joliot Curie 91190 Gif sur Yvette, France 35 rue Saint-Honoré, 77305 Fontainebleau, France laroche@lss.supelec.fr martin@cas.ensmp.fr Introduction Motion planning, i.e., the construction of an open-loop control connecting an initial state to a final state, is a fundamental problem of control theory both from a practical and theoretical point of view. For systems governed by ordinary differential equations the notion of flatness [7, 15] provides a constructive solution to this problem: flat systems are such that, at least locally, all the systems variables can be expressed as finite linear combinations of a (possibly vectorial) “flat output” and its derivatives. This expression provides an explicit open-loop control for motion planning. For finite-dimensional linear systems, flatness is equivalent to controllability, which is itself equivalent to the existence of a so-called canonical Brunovsky form. As noticed in [15], the idea underlying equivalence and flatness –the existence of a one-to-one correspondence between trajectories of systems– can be adapted to trajectories parametrization and motion planning for systems described by 1 − D partial differential equations [8, 2, 9] with boundary control. A first approach consists in looking for solutions taking the form of series of powers of the space variables, whose coefficients are functions of time depending on a “flat output” and its derivatives (see e.g. [13]). Another approach is to extend Brunovsky decomposition to infinite dimensional systems. For general 1 − D linear time-invariant diffusion equations, we show that the systems variables can be expressed as infinite linear combinations of a “flat output and its derivatives. These series can be seen as a decomposition on dense families of functions playing the role of a “Brunovsky basis”, and flat outputs are restricted to be Gevrey of order ≤ 2 to ensure convergence. This provides an explicit open-loop control achieving approximate motion planning. 1 Basics tools 1.1 Gevrey functions In all that follows, we will call “smooth functions” on [0, T], where T > 0 functions that have a continuous derivative on [0, T] (left and right derivatives at T and zero) at any order. The Taylor expansion of a smooth function is not convergent, unless the function is analytic. The notion of Gevrey order is a way of estimating this divergence. Definition 1. A smooth function t ∈ [0, T] 7→ y(t) is Gevrey of order α if ∃M, R > 0, ∀m ∈ N, sup t∈[0,T ] y(m) (t) ≤ M Γ(αm + 1) Rm . 1 Γ is the Euler Gamma function, which interpolates the factorial function: if n ∈ (N), Γ(n + 1) = n!. We denote by Gα(0, T) the set of all Gevrey functions of order α on [0, T]. By definition, a Gevrey function of order α is also of order β for any β ≥ α. A classical result (the Cauchy estimates) asserts that Gevrey functions of order 1 are analytic (entire functions if α < 1). Gevrey functions of order α > 1 have a divergent Taylor expansion; the larger α, the “more divergent” the Taylor expansion. Important properties of analytic functions generalize to Gevrey functions of order α > 1: the scaling, integration, addition, multiplication and composition of Gevrey functions of order α > 1 is of order α [11]. But contrary to analytic functions, functions of order α > 1 may be constant on an open set without being constant everywhere. For example the “bump function” φγ(t) = 0 if t 6∈]0, 1[, exp −1 (1 − t)t γ ! if t ∈]0, 1[. is of order 1 + 1/γ whatever γ > 0 [17]. Similarly, Φγ(t) = 0 if t ≤ 0 1 if t ≥ 1 R t 0 φγ(τ)dτ R 1 0 φγ(τ)dτ if t ∈]0, 1[, that will be used for motion planning, has order 1 + 1/γ. 1.2 Brunovsky decomposition for finite dimensional systems Consider the linear, single-input system on Rn defined by its coordinates in a reference basis as: (1) Ẋ = AX + Bu, where A is an (n, n) matrix, B is a n-dimensional vector. For all t, u(t) is a scalar and X(t) an the n-dimensional coordinate vector of the system’s state expressed in the reference basis. We suppose that this system is controllable, so that: span{B, AB, A2 B, . . . , An−1 B} = Rn . We would like to parametrize the solutions of (1) by linear combinations of the derivatives of an arbitrary function y(t). This parametrization takes the form X(t) = n−1 X i=0 eiy(i) (t) (2) u(t) = n X i=0 aiy(i) (t), (3) where the ei’s form a basis of R and the ai’s are real coefficients. To determine the ai’s and the ei’s we substitute relations (2) and (3) in the system equation (1): n X i=1 ei−1y(i) (t) = n−1 X i=1 Aeiy(i) (t) + n X i=0 aiBy(i) (t) and sort along the derivatives of y(t), which leads to 0 = Ae0 + a0B ei−1 = Aei + aiB, i = 1, . . . , n − 1 en−1 = anB. 2 We deduce that ei−1 = n X j=i ajAj−i B, i = 1, . . . , n − 1 0 = n X i=0 aiAi B As we want the ei’s to be a basis, we must have an 6= 0; we assume without loss of generality that an = 1. The above expression of the ei’s implies that span{e0, . . . , en−1} = span{B, AB, A2 B, . . . , An−1 B}, which proves that the ei’s form a basis of the state space Rn if and only if system (1) controllable. In this case the ai’s are necessarily the coefficients of the normalized characteristic polynomial of A, P(λ) = n−1 X i=0 aiλi + λn Expressed in basis (e1, e2, · · · , en), system (1) takes the well-known Brunovsky form ˙ X̃ = 0 1 · · · 0 . . . 0 0 · · · 1 −a0 −a1 · · · −an−1 X̃ + 0 . . . 0 1 u. The trajectories are obviously parametrized by y(t) = X̃1(t), the first component of coordinate vector X̃; y(t) = X̃1(t) is thus the Brunovsky or flat output of the system. Motion planning is now quite easy to perform. Suppose we want a control law U(t) that steers the system from initial vector state X(0) = X0 at time 0 to final state X(T) = XT (both expressed in the reference basis) at time T. Let us call M the matrix of the Brunovsky basis. The flat output y is the first coordinate of the vector state expressed in this basis, so that at each time t we have y(t) = 1 · · · 0 M−1 X(t) = m̃11 · · · m̃1n X(t) Values of y(0), ẏ(0), . . . , y(n−1) (0) are therefore linear combinations of the components of vector X0 , so that they are known at time 0. We denote them by d0, d1, . . . , dn−1. In the same way, values of y(T), ẏ(T), . . . , y(n−1) (T) are known at time T. We denote them by f0, f1, . . . , fn−1. We then choose y to be any polynomial function Y with degree ≥ 2n − 1 such that: p(0) = d0 p(T) = f0 p(i) (0) = di p(i) (T) = fi, i = 1, . . . , n − 1. Notice that function Y (t) = n−1 X i=0 fi (t − T)i i! ! ψ( t T ) + n−1 X i=0 di ti i! ! ψ(1 − t T ), where ψ(z) = zn ((z − 1)n + 1) is such that ψ(0) = 0 ψ(T) = 1 ψ(i) (0) = 0 ψ(i) (T) = 0, i = 1, . . . , n − 1, 3 is a possible solution to this problem. The open-loop control law U(t) = n X i=0 aiY (i) (t) obtained from (3) steers the system from X(0) = n−1 X i=0 Y (i) (0)ei = X0 to X(T) = n−1 X i=0 Y (i) (T)ei = XT . We end this section with a remark that will happen to be very useful when trying to generalize the Brunovsky decomposition (2) - (3) to partial differential systems. Remark 1. If A is invertible (which means that a0 6= 0), basis (ei)i=0,n−1 can be obtained by : e0 = −a0A−1 B ei = A−1 ei−1 + ai a0 e0, i = 1, . . . , n − 1, (4) and the controllability condition reads span{A−1 B, A−2 B, . . . , A−n B} = Rn 1.3 General properties of f00 + q(x)f = λf equation Consider, for any λ ∈ C, equation (5) f00 + q(x)f = λf, where q is a smooth function on an open neighborhood of [0, 1]. Let (e1(x, λ), e2(x, λ)) and (f1(x, λ), f2(x, λ)) be the systems of fundamental solutions of (5) at x = 0 and x = 1 respectively. This means that e1(0, λ) = 1e0 1(0, λ) = 0, e2(0, λ) = 0e0 2(0, λ) = 1, f1(1, λ) = 1f0 1(1, λ) = 0, f2(0, λ) = 0f0 2(1, λ) = 1, We recall that when λ is fixed in C, e1, e2, f1 and f2 are smooth functions un on an open neighborhood of [0, 1] in C, and that when x is fixed in [0, 1], e1, e2, f1 and f2 are entire functions of λ (see for instance [4]). For i = 1, 2 we define ei(x) = ei(x, 0) and fi(x) = fi(x, 0). The ei’s and fi’s are (obviously) related by e1 = f0 2(0)f1 − f0 1(0)f2 e2 = −f2(0)f1 + f1(0)f2 f1 = e0 2(1)e1 − e0 1(1)e2 (6) f2 = −e2(1)e1 + e1(1)e2 4 Let V0 be the operator defined on L2 (0, 1) by f = V0g ⇔ f00 + qf = g f(0 = f0 (0) = 0, It is easy to show that V0 is an integral operator whose expression is V0g(x) = Z 1 0 Z(x, ξ)g(ξ)dξ where Z(x, ξ) = χ(x − ξ)(e1(ξ, 0)e2(x, 0) − e1(x, 0)e2(ξ, 0)), χ being the characteristic function of R+ . Hence it follows, using relations (6), that (7) V0g(1) = Z 1 0 (e2(1)e1(ξ) − e1(1)e2(ξ))g(ξ)dξ = − < f2, g > . Similarly let V1 be the operator defined on L2 (0, 1) by f = V1g ⇔ f00 + qf = g f(1) = f0 (1) = 0, It is again easy to show that V1 is an integral operator whose expression is V1g(x) = Z 1 0 U(x, ξ)g(ξ)dξ, where U(x, ξ) = χ(ξ − x)(−f1(ξ, 0)f2(x, 0) + f1(x, 0)f2(ξ, 0)), so that (8) V1g(0) = Z 1 0 (−f2(0)e1(ξ) + f1(0)f2(ξ))g(ξ)dξ =< e2, g > In the same way we get (V0g)0 (1) = Z 1 0 (e0 2(1)e1(ξ) − e1(1)0 e2(ξ))g(ξ)dξ =< f1, g > (9) (V1g)0 (0) = Z 1 0 (f0 1(0)f2(ξ) − f2(0)0 f1(ξ))g(ξ)dξ = − < e1, g > . (10) Moreover, an easy calculation shows that (11) V ? 0 = V1 which leads to the useful relations Lemma 1. ∀i ∈ N, V i 0 e1(1) = (V i 1 f2)0 (0) (12) V i 0 e2(1) = −V i 1 f2(0) (13) (V i 0 e1)0 (1) = −(V i 1 f1)0 (0) (14) (V i 0 e2)0 (1) = V i 1 f1(0) (15) 5 Proof. The property is obvious for i = 0, after relations (6). Suppose i ≥ 1, we have V i 0 e1(1) = V0(V i−1 0 e1)(1) = − < f2, V i−1 0 e1 > after relation (7), = − < V i−1 1 f2, e1 > after relation (11), = (V i 1 f2)0 (0) after relation (10). The three other relations are demonstrated in the same way. We will finally make in the sequel a great use of the following technical lemmas Lemma 2. ∃M > 0/ ∀j ∈ {1, 2}, ∀k ∈ {0, 1}, ∀i ∈ N, ∀x ∈ [0, 1], dk V i 0 (ej) dxk (x) ≤ M x2i−k (2i − k)! . and Lemma 3. For any fixed x in [0, 1], ej(x, λ) and e0 j(x, λ) are entire functions of λ with Taylor expansions (k=0,1) dk ej dxk (x, λ) = ∞ X i=0 dk V i 0 (ej) dxk (x)λi We refer the reader to [12] for a detailed proof of these lemmas. 1.4 1-D second order linear equations with boundary conditions We briefly recall in this paragraph some important properties of 1-D second order linear equations with boundary conditions. Many books have been written on this subject, we refer the reader for instance to [4] and [6]. We consider on interval [0, 1] the homogeneous boundary conditions problem: y00 + c1(x)y0 + c0(x)y = v l11y(0) + l12y0 (0) + l13y(1) + l14y0 (1) = 0 (16) l21y(0) + l22y0 (0) + l23y(1) + l24y0 (1) = 0 Functions c0 and c1 are assumed to be C∞ on an open neighborhood of [0, 1] in R, to take real values, and coefficients lij are real constants, function v is given in L2 (0, 1). Problem (16) reduces to a problem of the following form (the so-called self-adjoint form for the differential equation): y00 + q(x)y = v a11y(0) + a12y0 (0) + a13y(1) + a14y0 (1) = 0 (17) a21y(0) + a22y0 (0) + a23y(1) + a24y0 (1) = 0 thanks to the change of variable: y(x) ← e R x 0 c1(ξ) 2 dξ y(x). All the results obtained for this system are still true for the original system (16) since the change of coordinates is only a multiplication by a function of x which is smooth and non zero on a neighborhood of [0, 1] . We now define the functional setting in which we try to solve problem (17). We set D(A) = {y ∈ H2 (0, 1)/ a11y(0) + a12y0 (0) + a13y(1) + a14y0 (1) = 0, a21y(0) + a22y0 (0) + a23y(1) + a24y0 (1) = 0} 6 and define operator A : D(A) → L2 (0, 1) by A(y) = y00 + q(x)y. Solving (17) is therefore equivalent to, given v ∈ L2 (0, 1), finding y ∈ D(A) such that A(y) = v. Properties of operator A depend on the values of coefficients aij. We set in all that follows: Aij = a1i a1j a2i a2j 2 Self-adjoint problems Operator A is self-adjoint if and only if A12 = A34 (the general case of complex coefficients in boundary conditions is left as an exercise in [4], chap.7). 2.1 Split boundary conditions We suppose that a13 = a14 = a21 = a22 = 0, and that a2 11 +a2 12 as well as a2 23 +a2 24 are different from zero. Operator A is called a mono-dimensional symmetric Schrödinger operator. The Sturm-Liouville problem for A, which means finding its eigenvalues and eigenvectors, is completely solved and we have (see [6], chap.2,§8 or [4]): Proposition 1. The eigenvalues of A are the zeros of function (18) D(λ) = −a12a23e1(1, λ) − a12a24e0 1(1, λ) + a11a23e2(1, λ) + a11a24e0 2(1, λ). Proposition 2. If λ is such that D(λ) 6= 0, functions (with the same notations as in paragraph 1.3) y1(x, λ) = −a12e1(x, λ) + a11e2(x, λ), and y2(x, λ) = (a23e2(1, λ) + a24e0 2(1, λ))e1(x, λ), are two independent solutions of y00 + q(x)y = λy from which we define Green’s function of A : (19) G(x, ξ, λ) = 1 D(λ) (χ(x − ξ)y1(x)y2(ξ) + χ(ξ − x)y1(ξ)y2(x)) Operator A − λI is invertible, and it’s inverse is the integral operator defined on L2 [0, 1] by: (A − λI)−1 (f)(x) = Z 1 0 G(x, ξ, λ)f(ξ)dξ. Finally if λ0 is real and D(λ0) 6= 0 ( then after the above property, it is in the resolvent set of A )we have Theorem 1. a) Operator A − λ0I is the inverse of an integral, self-adjoint compact operator. b) There exists a Hilbert basis of L2 (0, 1) formed with eigenfunctions of A − λ0I. c) A − λ0I has a countable infinity of simple real eigenvalues , whose only accumulation point is −∞. d) If (µi)i∈N is the sequence of A − λ0I’s eigenvalues and ψi is the associated normalized eigenfunction, then: ∀y ∈ D(A), (A − λ0I)(y) = ∞ X i=0 µi(y, ψi)ψi ¿From theorem 1 we deduce theorem 2: Theorem 2. 7 a) There exists a Hilbert basis of L2 (0, 1) formed with eigenfunctions A. b) A has a countable infinity of simple real eigenvalues , whose only accumulation point is −∞. c) If (λi)i∈N is the sequence of A’s eigenvalues and φi is the associated normalized eigenfunction, then: ∀y ∈ D(A), A(y) = ∞ X i=0 λi(y, φi)φi Remark 2. In theorem 2 A may have a (simple) zero eigenvalue, and hence may not be invertible. We also have in this case an asymptotic estimation of the eigenvalues (see [6], chap.2, §8): Theorem 3. if a12 6= 0 and a24 6= 0 then λn = −σ2 n with σn = nπ + 1 nπ − a11 a12 + a23 a24 − 1 2 Z 1 0 q(ξ)dξ + O( 1 n2 ) if a12 = 0 and a24 6= 0 then λn = −σ2 n with σn = (n + 1 2 )π + 1 (n + 1 2 )π a23 a24 − 1 2 Z 1 0 q(ξ)dξ + O( 1 n2 ) if a12 6= 0 and a24 = 0 then λn = −σ2 n with σn = (n + 1 2 )π + 1 (n + 1 2 )π −a11 a12 − 1 2 Z 1 0 q(ξ)dξ + O( 1 n2 ) if a12 = 0 and a24 = 0 then λn = −σ2 n with σn = nπ − 1 2nπ Z 1 0 q(ξ)dξ + O( 1 n2 ) 2.2 Periodic boundary conditions Symmetric operators with periodic boundary conditions are much more delicate to handle. Theorems 1 and 2 are still true, except from the fact that operator A’s eigenvalues are not necessarily simple any more. The asymptotic estimation in |λi| ∼i→∞ Ki2 is still true, although much more difficult to obtain. 2.3 Non self-adjoint problems We now consider problem (17), and the associated operator A , without any particular assumption on the boundary conditions. Like in paragraph 2.1, we can compute function DA(λ) whose zeros are the eigenvalues of A, and Green’s function of A. We obtain (20) DA(λ) = A12 + A34 + A32e1(1, λ) + A13e2(1, λ) + A42e0 1(1, λ) + A14e0 2(1, λ). and G(x, ξ, λ) = χ(x − ξ)(e1(ξ, λ)e2(x, λ) − e1(x, λ)e2(ξ, λ)) − (a13(e1(ξ, λ)e2(1, λ) − e1(1, λ)e2(ξ, λ)) + a14(e1(ξ, λ)e0 2(1, λ) − e0 1(1, λ)e2(ξ, λ))) × y1(x, λ) − (a23(e1(ξ, λ)e2(1, λ) − e1(1, λ)e2(ξ, λ)) + a24(e1(ξ, λ)e0 2(1, λ) − e0 1(1, λ)e2(ξ, λ))) × y2(x, λ) 8 where y1(x, λ) = (a22 + a23e2(1, λ) + a24e0 2(1, λ))e1(x, λ) − (a21 + a23e1(1, λ) + a24e0 1(1, λ))e2(x, λ) DA(λ) y2(x, λ) = (a14e0 2(1, λ) − a12 + a13e2(1, λ))e1(x, λ) + (a11 + a13e1(1, λ) + a14e0 1(1, λ))e2(x, λ) DA(λ) A is thus a closed operator whose spectrum reduces to ρ(A) = {λ ∈ C/ DA(λ) = 0} The resolvent of A is defined for λ ∈ ρ(A) by (21) (R(λ : A)v) (.) = Z 1 0 G(., ξ, λ)v(ξ)dξ For all complex numbers such that DA(λ) 6= 0, A − λI is invertible and its inverse Kλ is the integral operator of L2 (0, 1) with kernel G(x, ξ, λ). This kernel is, for a fixed λ , a continuously differentiable function, bounded on [0, 1] × [0, 1], hence DA is, up to multiplication by a constant, the Fredholm determinant (see [14]) of K0 if 0 is not an eigenvalue of A. As a direct consequence of lemmas 3 and 2 we have Proposition 3. DA(λ) =A12 + A34 + ∞ X i=0 (A23V i (e1(., 0))(1) + A13V i (e2(., 0))(1) + A42V i (e1(., 0))0 (1) + A14V i (e2(., 0))0 (1))λi , and DA is an entire function of Weierstrass order less than 1 2 . Moreover, coefficients ai of the expansion of DA on powers of λ are such that there exists strictly positive constants C and P with |ai| ≤ C Pi (2i)! Moreover (see [4], chap.6, sec.5) we have an asymptotic estimation of the eigenvalues: Proposition 4. If A24 6= 0 or (A24 = 0 and A14 − A23 6= 0), the eigenvalues of A are asymptotically equivalent to − K n2 Finally we have (after theorem 5.2, sec.5, chap.12 of [4]) : Theorem 4. If A is such that • C = A24 6= 0, • or: C = 0, E = A23 − A14 6= 0, and A12 = ±A34), • or all the Aij are zero, except A13, and if all the poles of G are simple, then for all function f in L2 (0, 1), we have (22) lim n→+∞

f − n X k=0 < f, φ? k > φk

2 = 0 9 where φk is an eigenfunction of A, associated to λk, φ? k is an eigenfunction of A? , associated to ¯ λk and for all i,j we have < φi, φ? j >= δij. Moreover, operator A is Riesz-spectral. Riesz-spectral operators are described in [5], chap.2. We briefly recall the definition of a Riesz basis of a Hilbert space X (see [5], chap.2, section 2.3) Definition 2. A family (φi)i∈N in X is a Riesz basis of X if 1. V ect(φi)i∈N = X 2. ∃ M > 0 and m > 0 such that: ∀N ∈ N, ∀(α1, . . . , αN ) ∈ RN m N X i=1 |αi| 2 ≤

N X i=1 αiφi

2 ≤ M N X i=1 |αi| 2 (23) Riesz-spectral operators are defined by Definition 3. A closed operator A is Riesz-spectral if and only if 1. it has a countable infinity of eigenvalues denoted by (λi)i∈N,and they are simple, 2. the associated eigenvectors form a Riesz basis of X, 3. for all λ < µ in (λi)i∈N, we have [λ, µ] * (λi)i∈N. 3 Linear diffusion 1-D partial differential equations In all that follows, we study the linear evolution problem with boundary control, where (x, t) ∈ [0, 1] × [0, T]: ∂ty(x, t) = ∂xxy(x, t) + c1(x)∂xy(x, t) + c0(x)y(x, t) (24) l11y(0, t) + l12∂xy(0, t) + l13y(1, t) + l14∂xy(1, t) = u(t) (25) l21y(0, t) + l22∂xy(0, t) + l23y(1, t) + l24∂xy(1, t) = u(t) (26) Functions c0 and c1 are supposed to be C∞ on an open neighborhood of [0, 1] in R, to take real values, and coefficients lij are real constants. Control law t 7→ u(t) is assumed to be either in C2 (0, T), or in C∞ (0, T). 3.1 Standard form of the problem Combining boundary conditions, problem (24)-(26) can always be brought to the following form ∂ty(x, t) = ∂xxy(x, t) + c1(x)∂xy(x, t) + c0(x)y(x, t) (27) l11y(0, t) + l12∂xy(0, t) + l13y(1, t) + l14∂xy(1, t) = 0 (28) l21y(0, t) + l22∂xy(0, t) + l23y(1, t) + l24∂xy(1, t) = u(t) (29) where the control appears only in one of the two boundary equations. Equation (28) will be called “constraint equation ”, and equation (29) will be called “control equation”. As before, transformation θ(x, t) := e R x 0 c1(ξ) 2 dξ y(x, t), 10 (see [16]) brings problem (27)-(29) to the standard form ∂tθ(x, t) = ∂xxθ(x, t) + q(x)θ(x, t) (30) a11θ(0, t) + a12∂xθ(0, t) + a13θ(1, t) + a14∂xθ(1, t) = 0 (31) a21θ(0, t) + a22∂xθ(0, t) + a23θ(1, t) + a24∂xθ(1, t) = u(t) (32) where a11 = l11 − c1(0) 2 l12 a12 = l12 a13 = l13 − c1(0) 2 l14 e− R 1 0 c1(ξ) 2 dξ a14 = c1(0) 2 l14e− R 1 0 c1(ξ) 2 dξ a21 = l21 − c1(0) 2 l22 a22 = l22 a23 = l23 − c1(0) 2 l24 e− R 1 0 c1(ξ) 2 dξ a24 = c1(0) 2 l24e− R 1 0 c1(ξ) 2 dξ and q(x) = c0(x) − c2 1(x) 4 − c0 1(x) 2 . We now study problems under the form (30)-(32). 3.2 Well-posedness and related transformations We briefly give sufficient conditions for problem (30)-(32) to be well posed. We use concepts developed in [5] (chap.3, sec.3) in a much more general framework. Definition 4. Boundary control problem (30)-(32) is well posed if spatial differential operator E : D(E) → L2 (0, 1) E(f) = f00 + q(x)f where D(E) = {f ∈ H2 (0, 1)/ a11f(0) + a12f0 (0) + a13f(1) + a14f0 (1) = 0, a21f(0) + a22f0 (0) + a23f(1) + a24f0 (1) = 0}, is the infinitesimal generator of a strongly continuous semi-group, and if there exists a function F in C∞ (0, 1) such that a11F(0) + a12F0 (0) + a13F(1) + a14F0 (1) = 0, a21F(0) + a22F0 (0) + a23F(1) + a24F0 (1) = 1 Remark 3. With this definition, problem ∂tθ(x, t) = −∂xxθ(x, t) (33) ∂xθ(0, t) = 0 (34) θ(1, t) = u(t) (35) is ill-posed, since the associated spatial operator has real positive eigenvalues going to infinity: according to Hille- Yosida theorem (see [5]) it cannot generate a strongly continuous semi-group. For instance, setting Aij = a1i a1j a2i a2j , and following the results of section 1.4, the problem will be well posed when A12 = A34 ( operator E is up to a shift of a constant times the identity the inverse of self adjoint compact operator and has a positively bounded spectrum ), and when A24 6= 0 ( operator E is Riesz-spectral with positively bounded spectrum ). In the case of a well-posed problem, if u is in C2 (0, T), problem (36) ż(t) = E(z)(t) − Fu̇(t) + Gu(t) where z is viewed as a function of t with values in D(E), and G = F00 + qF, is well posed in the sense defined in [5]. It is obtained setting formally z(t) = f(t) − Fu(t), and we have indeed: 11 Proposition 5 ([5] th. 3.3.3). Suppose that u is in C2 (0, T), and that for t = 0 we have z0 = f0 −Fu(0) ∈ D(E). Then the classical (smooth enough) solution of boundary control problem (30)-(32) and the solution of (36) are related by (37) z(t) = f(t) − Fu(t). Remark 4. When dealing with 1-D problems, function F always exists (it can be taken polynomial with high enough degree). Moreover we have Proposition 6. If E is invertible, there exits a unique function B solution of B00 + q(x)B = 0 a11B(0) + a12B0 (0) + a13B(1) + a14B0 (1) = 0 (38) a21B(0) + a22B0 (0) + a23B(1) + a24B0 (1) = 1. Proof. If B does not exist, it follows that all the solutions of f00 + q(x)f = 0 (39) a11f(0) + a12f0 (0) + a13f(1) + a14f0 (1) = 0 (40) are such that f00 + q(x)f = 0 a11f(0) + a12f0 (0) + a13f(1) + a14f0 (1) = 0 a21f(0) + a22f0 (0) + a23f(1) + a24f0 (1) = 0. But problem (39)-(40) has at least one solution. Hence operator E has a zero eigenvalue, with an associated eigenspace of dimension ≥ 1, which proves the E is not invertible. So, in the particular case where E is invertible, we can choose F = B and transform once more our problem into a new problem, which splits into an integrator and a problem u̇(t) = e(t) ż = E(z) − Fe. This last transformation is only introduced to bring our problem to the familiar finite dimensional form (1). 3.3 Transformation by boundary terms feedback It is fundamental to notice that since our goal is to obtain a parametrization of θ(x, t) and u(t), we can transform system (30)-(32) by boundary feedback to obtain more easily a parametrization for θ(x, t), and then deduce from this a parametrization of u(t), since u(t) = a21θ(0, t)+a22∂xθ(0, t)+a23θ(1, t)+a24∂xθ(1, t). We can then transform as we like by boundary feedback (provided the resulting problem remains well-posed) the coefficients of control equation (32), keeping constraint equation (31) unchanged. This means that we study the following family of problems, where coefficients defining the constraint equation (a11, a12, a13, a14) are fixed, and coefficients defining the control equation (ã21, ã22, ã23, ã24) take any possible value in R4 : (41) θt = θxx + q(x)θ a11θ(0, t) + a12θx(0, t) + a13θ(1, t) + a14θx(1, t) = 0 ã21θ(0, t) + ã22θx(0, t) + ã23θ(1, t) + ã24θx(1, t) = u(t) 12 We drop the “ã” notation and use “a” instead for simplicity. Consider the associated orbit of operator under the action of boundary terms feed-backs, whose generic element is operator A defined as: D(A) → L2 (0, 1) f 7→ f00 + qf on its domain D(A) = {f ∈ H2 (0, 1) a11f(0) + a12f0 (0) + a13f(1) + a14f0 (1) = 0 a21f(0) + a22f0 (0) + a23f(1) + a24f0 (1) = 0}. We admit that generically, we can pick in this orbit an operator A which is invertible and is the infinitesimal generator of a strongly continuous semi-group, so that boundary control problem (41) is well-posed. We call B the solution of B00 + q(x)B = 0 a11B(0) + a12B0 (0) + a13B(1) + a14B0 (1) = 0 (42) ã21B(0) + ã22B0 (0) + ã23B(1) + ã24B0 (1) = 1. 4 Parametrizing trajectories 4.1 Brunovsky form We suppose that we have found in the orbit of the original standard problem an operator A Riesz-spectral, invertible as defined in paragraph 3.3, and the associated function B. The new control equation is denoted by ã21θ(0, t) + ã22∂xθ(0, t) + ã23θ(1, t) + ã24∂xθ(1, t) = ũ(t). We set z(x, t) = θ(x, t) − B(x)ũ(t), and supposing ũ is smooth enough, we have ∂tz(x, t) = ∂xxz(x, t) + q(x)z(x, t) − B(x) ˙ ũ(t) a11z(0, t) + a12∂xz(0, t) + a13z(1, t) + a14∂xz(1, t) = 0 (43) ã21z(0, t) + ã22∂xz(0, t) + ã23z(1, t) + ã24∂xz(1, t) = 0, which we write, setting ˙ ũ = e, as: (44) ż = A(z) − Be. Like we did in the finite dimensional case, we look for a formal parametrization of trajectories of (30) under the form: θ(x, t) = ∞ X i=0 αi(x)y(i) (t) (45) ũ(t) = ∞ X i=0 aiy(i) (t) (46) u(t) = ∞ X i=0 biy(i) (t), (47) with α0 6= 0. We have: 13 Proposition 7. Functions (αi)i∈N and coefficients (ai)i∈N and (bi)i∈N must be related by: α0 = a0B, where a0 6= 0, A(αi+1 − ai+1B) = αi, bi = a21αi(0) + a22α0 i(0) + a23αi(1) + a24α0 i(1). Proof. Let us express z and e as functions of y, B and αi: z(x, t) = ∞ X i=0 (αi(x) − aiB(x))y(i) (t) e(t) = ∞ X i=0 aiy(i+1) (t), hence, applying formal differentiation: ∂tz − ∂xxz − qz + Be = (−α00 0 + a0B00 − q(α0 − a0B))y + ∞ X i=0 (αi − aiB − α00 i+1 + ai+1B00 − q(αi+1 − ai+1B) + aiB)y(i+1) , a11z(0, .) + a12z0 (0, .) + a13z(1, .) + a14z0 (1, .) = ∞ X i=0 (a11(αi(0) − aiB(0)) + a12(α0 i(0) − aiB0 (0)) + a13(αi(1) − aiB(1)) + a14(α0 i(1) − aiB0 (1)))y(i) ã21z(0, t) + ã22z0 (0, t) + ã23z(1, t) + ã24z0 (1, t) = ∞ X i=0 ã21(αi(0) − aiB(0)) + ã22(α0 i(0) − aiB0 (0)) + ã23(αi(1) − aiB(1)) + ã24(α0 i(1) − aiB0 (1)))y(i) , We want, after (44), those three quantities to be zero at least when y is any polynomial function, therefore α0 must be such that α0 − a0B ∈ D(A), A(α0 − a0B) = 0, and for all i ≥ 0, αi+1 must be such that αi+1 − ai+1B ∈ D(A), A(αi+1 − ai+1B) = αi. As 0 is not an eigenvalue of A, it follows that α0 = a0B, and the proposition is proved. One possible degree of freedom is the choice of coefficients (ai)i∈N. One particular choice consists in taking for the (ai)i∈N the coefficients of the series that defines entire function DA(λ) (the Fredholm determinant of K. This choice amounts to the (almost) exact transposition to infinite dimension of Brunovsky algorithm as given in remark (1) : α0 = a0B αi+1 = K(αi) + ai+1B = i+1 X j=0 ajKi+1−j B Parametrization corresponding to this choice will be referred to as Brunovsky parametrization, and family (αi)i∈N will be referred to as Brunovsky family. 14 4.2 Convergence of formal trajectories 4.2.1 “Weak” and classical convergence of formal trajectories: first proof We first prove a “weak” convergence result in the following sense: formal trajectories are in C∞ ([0, T], L2 (0, 1)). As A is Riesz-spectral and invertible we call (φi)i∈N a Riesz basis of L2 (0, 1) formed with eigenvectors of A; λi is the corresponding eigenvalue (non zero, by assumption). Moreover, we suppose that the eigenvalues are indexed so that |λi| ≤ |λi+1|, and we set: µi = 1 λi µi is therefore the eigenvalue of K associated to φi, and we have |µi| ≥ |µi+1|. General expression of the Fredholm determinant of the inverse of A is given in paragraph 2.3 by formula (20). The notations are the same as in this paragraph. The following is true: Proposition 8. If y is in Gσ (0, T), where 0 < σ ≤ 2, then (45) defines a function t 7→ θ(x, t) in C∞ ([0, T], L2 (0, 1)). Proof. In Riesz basis (φi)i∈N we have : α0 = ∞ X k=0 xkφk, and by definition αn = ∞ X k=0 n X i=0 aiµn−i k ! xkφk = ∞ X k=0 n X i=0 aiλi k ! µn k xkφk. After the definition of Riesz basis, there exits two real constants C > 0 and c > 0 such that: kαnk ≤ C v u u t ∞ X k=0 n X i=0 aiλi k 2 |µk| 2n |xk| 2 , ≤ C c sup k∈N n X i=0 aiλi k |µk| n ! kα0k . We notice that expression Pn i=0 aiλi k is the Taylor expansion at zero and at order n of DA(λk). After proposition 3, and applying Weierstrass factorization theorem (voir [18]) we know that : (48) DA(λ) = ∞ Y i=0 (1 − λ λk ), and this product is convergent in norm, since 3 implies that: (49) ∀r > 1 2 , ∞ X k=0 1 |λk| r ) < +∞. 15 Termwise identification after developing the above expression leads to: aa1 = − ∞ X i=0 1 λi , a2 = ∞ X ij 1 λiλj , al = (−1)l ∞ X i1i2...il−1il 1 λi1 λi2 . . . λil−1 λil . Hence : ∀k ∈ N?, |1 + a1λk| = 1 − ∞ X i=0 λk λi = |λk| ∞ X i6=k 1 λi ≤ |λk| M ∞ X i=1 1 i2 . in the same way : ∀k ∈ N?, 1 + a1λk + a2λ2 k = − ∞ X i6=k λk λi + ∞ X i1i2 λ2 k λi1 λi2 = |λk| 2 ∞ X i1,i26=k i1i2 1 λi1 λi2 ≤ |λk| 2 M2 ∞ X i1,i26=0 i1i2 1 i2 1i2 2 , and by induction : ∀(k, l) ∈ N ? ×N?, 1 + a1λk + . . . + alλl k = |λk| l ∞ X i1,...,il6=k i1l 1 λi1 . . . λil ≤ |λk| l M2l ∞ X i1,...,il6=0 i1l 1 i2 1 . . . i2 l . But from the classic result (see for instance [19]): sin(z) z = ∞ Y i=1 (1 − z2 i2π2 ) = ∞ X i=0 z2i (2i + 1)! , we deduce that: ∞ X i1,...,il6=0 i1l 1 i2 1 . . . i2 l ≤ π2l (2l + 1)! , and finally: ∀(k, l) ∈N ? ×N?, 1 + a1λk + . . . + alλl k ≤ |λk| l (Mπ)2l (2l + 1)! , 16 and kαnk ≤ C c sup k∈N n X i=0 aiλi k |µk| n ! kα0k ≤ C c kα0k (Mπ)2n (2n + 1)! . (50) It is now clear that series (45) defines a C∞ ([0, T], L2 (0, 1)) function whenever y is a Gevrey function of order σ2, and also when y is a Gevrey function of order 2, if we impose restrictions on the radius of convergence of y, which ends the proof of the proposition. This result leads to a “solution” of the problem in a too weak sense because formal expressions (46) and (47) do not make sense. This is why we go on and show the following regularity result Proposition 9. If operator A is Riesz-spectral invertible, and if y is in Gσ (0, T), then (45) converges normally in C∞ ([0, T], C2 (0, 1)). Proof. Thanks to majoration (50) and 3, we know that we can choose C, M > 0 big enough so that for all αn we have: kαnk2 ≤ C M2n (2n)! , |an| ≤ C M2n (2n)! . Remember (see paragraph 2.3) that the inverse K of A is an integral operator with C1 ([0, 1]2 ) kernel G(x, ξ, 0), and that αn = K(αn−1) + anB. Hence, applying Schwarz inequality ∀x ∈ [0, 1], |αn(x)| ≤ |K(αn−1)(x)| + |an| |B(x)| ≤ kG(x, ., 0)k2 kαn−1k2 + |an| |B(x)| ≤ kG(x, ., 0)k2 C M2n−2 (2n − 2)! + C M2n (2n)! |B(x)| ≤ C̃ M2(n−1) (2(n − 1))! ≤˜ ˜ C M̃2n) (2n)! where we set C̃ = C sup s,x∈[0,1] (|G(x, s, 0)| + |B(x)|), M̃ = kM, k > 1 ˜ ˜ C = sup n∈N (C̃ n(n − 1) k(2n) ). Hence the series of continuous functions ∞ X n=0 y(i) (t)αn(x) is normally convergent on [0, 1] × [0, T], whenever y is a Gevrey function of order σ < 2, and also when y is a Gevrey function of order 2, if we impose restrictions on the radius of convergence of y. In the same way, we have for all x ∈ [0, 1] |α0 n(x)| ≤ |K(αn−1)0 (x)| + |anB0 (x)| ≤

∂G ∂x (x, ., 0)

2 kαn−1k2 + |an| |B0 (x)| , 17 so that ∞ X n=0 y(i) (t)α0 n(x) is normally convergent on [0, 1] × [0, T]. Normal convergence of the series of second derivatives is now obvious, which ends the proof of the proposition. 4.2.2 Classical convergence of formal trajectories using a Brunovsky decomposition: second proof Let’s go back the associated orbit of operator under the action of boundary terms feed-backs, whose generic element is operator E defined as: D(E) → L2 (0, 1) f 7→ f00 + qf on its domain D(E) = {f ∈ H2 (0, 1) a11f(0) + a12f0 (0) + a13f(1) + a14f0 (1) = 0 ã21f(0) + ã22f0 (0) + ã23f(1) + ã24f0 (1) = 0}. We have picked (we suppose that we can) in this orbit an operator A which is invertible and is the infinitesimal generator of a C0-semigroup, so that boundary control problem is well-posed. Using the notations defined in section 1.3 and relations (6), it is easy to check that function G(x) := a11e2(x) − a12e1(x) + a13f2(x) − a14f1(x) satisfies to G00 + qG = 0 a11G(0) + a12G0 (0) + a13G(1) + a14G0 (1) = 0. As A is invertible, we must have (dropping the “ã” notation for simplicity): c0 = a21G(0) + a22G0 (0) + a23G(1) + a24G0 (1) 6= 0, and we define B = G c0 . Moreover the inverse of A, which we denote by K, is a compact integral operator with continuous kernel, defined on L2 (0, 1) (see 2.3). The following is true: Proposition 10. The Fredholm determinant of A is given by: D(λ) = a11a22 − a12a21 + a13a24 − a14a23 (51) + (a11a23 − a13a21)e2(1, λ) + (a11a24 − a14a21)e0 2(1, λ) + (a13a22 − a12a23)e1(1, λ) + (a14a22 − a12a24)e0 1(1, λ) It’s Taylor expansion is D(λ) = ∞ X i=0 ciλi , where the ci’s have expression c0 =a11a22 − a12a21 + a13a24 − a14a23 (52) +a21(a13f2(0) − a14f1(0)) + a22(a13f0 2(0) − a14f0 1(0)) +a23(a11e2(1) − a12e1(1)) + a24(a11e0 2(1) − a12e0 1(1)) =a21G(0) + a22G0 (0) + a23G(1) + a24G0 (1). ci =a21(a13V i 1 (f2)(0) − a14V i 1 (f1)(0)) + a22(a13V i 1 (f2)0 (0) − a14V i 1 (f1)0 (0)) (53) +a23(a11V i 0 (e2)(1) − a12V i 0 (e1)(1)) + a24(a11V i 0 (e2)0 (1) − a12V i 0 (e1)0 (1)). 18 Proof. Expression (51) has been given in 2.3. Substituting the Taylor expansions in lemma 3 into (51), and using relations (12)- (15) in lemma 1 provides expression (53) for i ≥ 1: ci =(a11a23 − a13a21)V i 0 (e2)(1) + (a11a24 − a14a21)V i 0 (e2)0 (1) +(a13a22 − a12a23)V i 0 (e1)(1) + (a14a22 − a12a24)V i 0 (e1)0 (1) =a11a23V i 0 (e2)(1) − a12a23V i 0 (e1)(1) + a11a24V i 0 (e2)0 (1) − a12a24V i 0 (e1)0 (1) +a13a21V i 1 (f2)(0) − a14a21V i 1 (f1)(0) + a13a22V i 1 (f2)0 (0) − a14a22V i 1 (f1)0 (0) =a21(a13V i 1 (f2)(0) − a14V i 1 (f1)(0)) + a22(a13V i 1 (f2)0 (0) − a14V i 1 (f1)0 (0)) +a23(a11V i 0 (e2)(1) − a12V i 0 (e1)(1)) + a24(a11V i 0 (e2)0 (1) − a12V i 0 (e1)0 (1)). In the same way, we obtain the expression of c0. We now build the (αi)’s associated to A applying: α0 = c0B αi+1 = K(αi) + ci+1B The following result is true: Proposition 11. The (αi)’s are independent of the choice of operator A in the associated orbit of operators under the action of boundary terms feed-backs. Moreover, we have majorations ∃(C̃, M)/∀x ∈ [0, 1], |αi(x)| ≤ C̃ Mi (2i)! (54) |α0 i(x)| ≤ C̃ Mi−1 (2i − 1)! (55) Proof. We prove by induction that (56) αi(x) = V i 0 (a11e2 − a12e1)(x) + V i 1 (a13f2 − a14f1)(x) As this expression does not depend on coefficients a2i any more, this will prove the first part of the proposition. (56) is true for i = 0, since α0 = a0B = G = a11e2 − a12e1 + a13f2 − a14f1. Let us suppose it is true for rank i, and define βi+1 = V i+1 0 (a11e2 − a12e1) + V i+1 1 (a13f2 − a14f1). By definition of operators V0 and V1 and from the induction hypothesis we have: β00 i+1 + q(x)βi+1 = V i 0 (a11e2 − a12e1) + V i 1 (a13f2 − a14f1) = αi and: a11βi+1(0) + a12β0 i+1(0) + a13βi+1(1) + a14β0 i+1(1) = 0. Moreover a21βi+1(0) + a22β0 i+1(0) + a23βi+1(1) + a24β0 i+1(1) = a21V i+1 1 (a13f2 − a14f1)(0) + a22V i+1 1 (a13f2 − a14f1)0 (0) + a23V i+1 0 (a11e2 − a12e1)(1) + a24V i+1 0 (a11e2 − a12e1)0 (1) = ci+1 This means, by definition of B and K, that: βi+1 − ci+1B = K(αi). From this, we deduce that: αi+1 = K(αi) + ci+1B = βi+1 = V i+1 0 (a11e2 − a12e1) + V i+1 1 (a13f2 − a14f1), 19 which ends the induction. Majorations (54) and (55) result from lemma 2: ∀x ∈ [0, 1], |αi(x)| ≤ |a11| V i 0 (e2)(x) + |a12| V i 0 (e1)(x) + |a13| V i 1 (f2)(x) + |a14| V i 1 (f1)(x) ≤ max(|a11| , |a12| , |a13| , |a14|)M 1 (2i)! which proves that (54) holds with C̃ = max(|a11| , |a12| , |a13| , |a14|)M and M = 1. (55) is proved in the same way. Proposition 12. ∀y ∈ C∞ (0, T) with Gevrey order (at most) 2, the series X(x, t) = ∞ X i=0 αi(x)y(i) (t) u(t) = ∞ X i=0 (a21αi(0) + a22α0 i(0) + a23αi(1) + a24α0 i(1))y(i) (t) are normally convergent and define a classical solution of (41) on [0, 1] × [0, T]. Proof. This is a straightforward consequence of proposition 11. 4.3 L2 (0, 1) density of iterates of K on B We start with an elementary proof of a general result. Let K be a bounded operator defined on a Hilbert space H, Riesz spectral ( as defined in [5], def. 2.3.4 p. 41 ). We denote by (λn)n∈N the eigenvalues of K, which are simple, and we suppose that lim n→∞ λn = 0. The (λn)n∈N are ordered so that |λn| ≥ |λn+1|. Denote by (φn)n∈N a Riesz basis of H formed with eigenfunctions of K, and (ψn)n∈N a Riesz basis of H formed with eigenfunctions of K∗ such that: ∀k ∈ (N), K(φk) = λkφk, K∗ (ψk) = λkψk, and ∀(i, j) ∈ N2 , < φk, ψk >= δij We know that (see [5], lemma 2.3.2) any f in H decomposes in f = ∞ X k=0 < f, ψk > φk, that there exists m > 0 and M > 0 such that ∀f ∈ H, m ∞ X k=0 < f, ψk >2 ≤ kfk 2 ≤ M ∞ X k=0 < f, ψk >2 and that K(f) is expressed as K(f) = ∞ X k=0 λk < f, ψk > φk. Then we have Proposition 13. If B in H is such that ∀k ∈ N, < B, ψk >6= 0, then (Kn (B))n∈N is dense in H. 20 Proof. As (φk)k∈N is dense is H, we only have to prove that ∀k ∈ N, ∀ > 0, ∃m ∈ N, ∃(α0, · · · , αm) ∈ Cm+1 /

φk − m X i=0 αiKi (B)

≤ . Let us fix k ∈ N, and for all integer m > k, let us define: Pm,k(λ) = m Y i=0,i6=k λ − λi λk − λi . P is the Legendre polynomial which takes value 1 for λk and 0 for λi with i 6= k, i ∈ {1, . . . , m}. For m > k, if n is an integer strictly greater than m, we have: |Pm,k(λn)| ≤ m Y i=0,i6=k |λn − λi| |λk − λi| ≤ m Y i=0,i6=k 2 |λi| |λk − λi| . Let us set Cm,k = m Y i=0,i6=k 2 |λi| |λk − λi| . Hence ∀n > m, |Pm,k(λn)| ≤ Cm,k. But, for a fixed k we have lim m→∞ Cm+1,k Cm,k = lim m→∞ 2 |λm+1| |λk − λm+1| = 0, from which we deduce that there exists Nk > 0, independent of m, such that: ∀m ∈ N, Cm,k ≤ Nk. It follows that ∀m ∈ N, ∀n > m, |Pm,k(λn)| ≤ Nk. Denote by α0, · · · , αm the coefficients of the powers of λ in Pm,k(λ), we have

φk − m X i=0 αiKi (B)

2 = kφk − Pm,k(K)(B)k 2 =

φk − ∞ X n=0 Pm,k(λn) < B, ψn > φn

2 =

φk − φk − ∞ X n=m+1 Pm,k(λn) < B, ψn > φn

2 ≤ M ∞ X n=m+1 |Pm,k(λn)| 2 |< B, ψn >| 2 ≤ MN2 k ∞ X n=m+1 |< B, ψn >| 2 . The last expression is the remainder of order m of a convergent series: we can then, for a fixed k, choose m such that this remainder is less than MN2 k , which leads to the desired result. Whenever K is self-adjoint this result is classical (see [10] or [1]). Now with A, B, (ai)(i∈N), (αi)(i∈N) and K = A−1 as above we have: Proposition 14. If ∀n ∈ N, < ψn, B >6= 0, where the φn are the normalized eigenfunctions of K and the ψn are the normalized eigenfunctions of K? , the (αi)(i∈N) form a dense family of functions in L2 (0, 1). 21 Proof. By assumption, A is Riesz spectral invertible, hence K is also Riesz-spectral with zero as only accumulation point for its eigenvalues. Moreover, we know that K is compact, hence bounded. Applying proposition 13 we get [ n∈N vect(B, KB, · · · , KnB) = L2 (0, 1) But β0 = a0B ⇒ vect(B) = vect(β0), β1 = a0KB + a1B ⇒ vect(B, KB) = vect(β0, β1), βn = n X i=0 aiKn−i B ⇒ vect((Ki B)i=0→n) = vect((βi)i=0→n), for all order n, hence [ n∈N vect(β0, β1, · · · , βn) = [ n∈N vect(B, KB, · · · , KnB) = L2 (0, 1) which is the desired result. 4.4 Motion planning We suppose that we have built a dense Brunovsky parametrization of the system. Given fixed initial and final states of system (30)-(32), we wish to find a control law u(t) that steers the system from the initial to the final state in a given time T. We have Proposition 15. Suppose that the initial and final states take the form: θini = ∞ X i=0 diαi(x) (57) θfinal = ∞ X i=0 fiαi(x), (58) where coefficients di and fi are real, and there exist three positive constants P,Q and 0 < σ ≤ 2 such that: ∀i ∈ N, |di| ≤ PQi Γ(σi + 1) |fi| ≤ PQi Γ(σi + 1) Then there exists a control law [0, T] 3 t 7→ U(t) in Gσ (0, T) that steers system (30)-(32) from initial state θini at time 0 to final state θfinal at time T. Proof. Control law: (59) u(t) = ∞ X i=0 (a21αi(0) + a22α0 i(0) + a23αi(1) + a24α0 i(1))y(i) (t) where y is a Gevrey function of order σ ≤ 2, taken to be compatible with initial and final states: since we have a well posed control problem, semi-group properties (see [5]) guarantee the existence and uniqueness of the classical trajectory generated by the control and the initial state which is in D(A). The system follows trajectory θ(x, t) = ∞ X i=0 y(i) (t)αi(x). y must therefore be such that (60) ∀i ∈ N, y(i) (0) = di, and y(i) (T) = fi. Ritt’s theorem and its corollaries (see [3]) tell us that such a y exists. 22 The second result guarantees approximate motion planning: Proposition 16. If θini at time 0 and θfinal at time T are in L2 (0, 1) then there exists a control law [0, T] 3 t 7→ ū(t) in Gσ (0, T), where σ ≤ 2, that steers system (30) from initial state θini at time 0 to a state as close as desired to the final state θfinal at time T. Proof. The (αi)i∈N family is dense in L2 (0, 1), so that we can choose two states of the form (57)-(58) as close as we wish to the initial and final states (we take polynomials for (57)-(58)). we take a control law 59 as in 15 that would exactly drive the system from (57) to (58). Thanks to the semi-group property of the system, this control law answers to our problem. Remark 5. The control law obtained by truncation of 59 at a suitable order also achieves approximate motion planning. References [1] N. Akhiezer and Glazman I. Theory of Linear Operators in Hilbert Space. 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