Control of ferromagnetic nanowires

Transcription

1 Control of ferromagnetic nanowires Y. Privat E. Trélat 1 1 Univ. Paris 6 (Labo. J.-L. Lions) et Institut Universitaire de France Workshop on Control and Observation of Nonlinear Control Systems with Application to Medicine September 5 7, 2013 East-West Center, University of Hawai i

2 Longstanding project on the simulation, modelling and control of ferromagnetic materials, with F. Alouges, G. Carbou, S. Labbé, C. Prieur, Y. Privat. Aims Model and control ferromagnetic materials, optimize the shape and composition of ferromagnetic particles for technological applications. Possible applications : waves guides magnetic micro particles for antiradar paints behavior of magnetic components for nano-electronic sciences rapid magnetic storage for hard disks, magnetic memories MRAMs or mobile phones (spin injection speed access) Idea : use the steady-states of the magnetic moment to encode information. Models available : micromagnetism Description of the magnetization (vector field) : static configurations : micromagnetism theory of Brown (60 s) dynamics : Landau-Lifshitz equation (30 s)

3 Micromagnetism Thermodynamical description of ferromagnetic materials : theory of micromagnetism, W.F. BROWN, in the 60 s. Magnetic domain Open subset Ω IR 3 occupied by the ferromagnetic material. spontaneous magnetization, only depending on the temperature : Magnetization Vector field m on Ω, such that m(x) = 1 a.e. in Ω. Remanent magnetization under external field Critical temperature dividing linear and nonlinear behavior Emergence of microstructures : walls and domains

4 Micromagnetism Static micromagnetism model where min E(m) m H 1 (Ω,S 2 ) Z Z Z E(m) =A m 2 + H d (m) 2 m H ext Ω IR 3 Ω A R Ω m 2 : exchange term H d (m) : demagnetizing field solution of j curl(hd ) = 0 div(h d ) = div(m) in D (IR 3, IR 3 ) (m extended by 0 outside Ω) H ext : Zeeman, action of an external (magnetic) field

5 Dynamical model Magnetization vector : m(t, x) Landau-Lifshitz equation m t = m H(m) m (m H(m)) where H(m) = de dm = 2A m + H d (m) + H ext (effective field), and : H d (m) (demagnetization field) H ext (external magnetic field) Remark Z d dt (E(m(t)) = H(m) (H(m) m)m 2 Ω Every steady-state satisfies m H(m) = 0

6 Dynamical model Magnetization vector : m(t, x) Landau-Lifshitz equation m t = m H(m) m (m H(m)) where H(m) = de dm = 2A m + H d (m) + H ext (effective field), and : H d (m) (demagnetization field) H ext (external magnetic field) Existence of global weak solutions : Visintin (1985), Alouges-Soyeur (1992), Melcher (2013) Strong local solutions : Carbou-Fabrie (2001), Melcher (2013) Modelization, homogenization : Carbou, Fabrie, De Simone, Haddar, Santugini, Sanchez,... Numerical analysis : Alouges, Labbé, Merlet, Santugini,... Controllability / stabilization issues (control = external magnetic field H ext (t)) : Carbou, Labbé, Prieur, Privat, Trélat (2009, 2011), Alouges, Beauchard, Sigalotti (2009)

7 Nanowires 1D finite nanowire submitted to an external magnetic field : First possibility : u 1 (t)χ (a,b) (x)e 1 Second possibility : u 1 (t)χ (a,b) (x)e 1 u 2 (t)χ (c,d) (x) d 0 a b L 0 a b c d L Landau-Lifshitz equation on an interval? Same questions with a 1D network of nanowires (connected with semiconductor devices) :

8 From a 3-D model to a 1-D model The magnetic moment m : IR Ω ε S 2 of the nanowire (represented by Ω ε IR 2, cylinder of thickness ε) verifies the Landau-Lifshitz equation : m t = m H(m) m (m H(m)). where H(m) = m + H d (m) + H ext (m) (effective field) ε e 3 e 2 e 1 nanowire : cylinder with internal radius ε As ε 0 : (e 1, e 2, e 3 ) : orthonormal basis m ε 0 2 m x 2 Γ-convergence result : H d (m) h d (m) = (m e 1 )e 1 ε 0 (Sanchez 2006, Carbou Labbé 2012)

9 1-D model with periodic boundary conditions Landau-Lifshitz PDE for a 1D finite-length nanowire 8 >< >: m t = m h u(m) m (m h u(m)) x (0, L), t > 0 m m (t, 0) = (t, L) = 0 t > 0 x x m(t, x) = 1 m : IR + [0, L] S 2 : magnetic moment m(t, x) = m 1 (t, x)e 1 + m 2 (t, x)e 2 + m 3 (t, x)e 3 L : length of the nanowire h u(m) = xx 2 m (m e 2)e 2 (m e 3 )e 3 + u d u : external field applied to the nanowire control

10 Steady-states A steady-state (with u 0) must satisfy m h 0 (m) = 0. Hence every steady-state is of the form where 8 < : m 1 (x) = cos θ(x) m 2 (x) = cos ω sin θ(x) m 3 (x) = sin ω sin θ(x) ω [0, 2π) (rotation constant parameter around the nanowire) θ is solution of the pendulum equation θ (x) = sin θ(x) cos θ(x) x (0, L) θ (0) = θ (L) = 0

11 Steady-states Pendulum equation θ (x) = sin θ(x) cos θ(x), x (0, L) elliptic functions, period : 4K (k) (Jacobi function) with θ 2 (x) + cos 2 θ(x) = Cst = k 2 0 k < 1 Z π/2 dθ K (k) = p. 0 1 k 2 sin 2 θ (Jacobi elliptic integral of the first kind)

12 Steady-states Pendulum equation θ (x) = sin θ(x) cos θ(x), x (0, L) elliptic functions with period 4K (k) (Jacobi function) with θ 2 (x) + cos 2 θ(x) = Cst = k 2 Boundary conditions θ (0) = θ (L) = 0 This imposes that L = 2nK (k) with n IN (and hence L π)

13 Steady-states Therefore : The energies of steady-states are quantized. More precisely : h i Setting N 0 = L, there exist real numbers π in (0, 1) such that E 1, E 2,..., E N0 θ 2 + cos 2 θ = Cst = E n {E 1, E 2,..., E N0 } Note that there is no steady-state if L < 2π.

14 Steady-states Conclusion : Quantified set of steady-states 8 < m 1 (x) = cos θ(x) m : 2 (x) = cos ω sin θ(x) m 3 (x) = sin ω sin θ(x) with θ such that θ 2 + cos 2 θ = Cst = E n {E 1, E 2,..., E N0 } and ω IR. θ (0) = θ (L) = 0 N 0 one-parameter families h of i steady-states, with N 0 = L π

15 Local stability properties of the steady states

16 Linearization around a steady state m t = m h u(m) m (m h u(m)), x (0, L), t > cos θ(x) M 0 (x) sin θ(x) A steady state (ω = 0) with energy E n. 0 Mobile frame (M 0 (x), M 1 (x), M 2 ) : m(t, x) S 2 new coordinates r = (r 1, r 2 ) such that sin θ(x) M 1 (x) cos θ(x) A, M A 0 1 q m(t, x) = 1 r1 2(t, x) r 2 2(t, x) M 0(x) + r 1 (t, x) M 1 (x) + r 2 (t, x) M 2 Projection of the PDE on (M 1, M 2 ). Note that x m(t, 0) = x m(t, L) = 0 leads to x r 1 (t, 0) = x r 2 (t, L) = 0.

17 Linearization around a steady state In these new coordinates, for r + u small : Landau-Lifshitz equation «r1 r r = r2 t = Ar + ub(x, r) + R(x, r, r x, r xx, u) «A + Id A + EnId A = (A + Id) A + E nid with (not diagonalizable) A = xx 2 cos 2 θ Id on D(A) = {f H 2 (0, L) f (0) = f (L) = 0} C > 0 s.t. R(x, r, p, q, u) C( r 2 q + r p + r p 2 + r 2 + r u ).

18 Spectral study of A = 2 xx 2 cos 2 θ Id (A + E n Id)f, f L 2 = f θ cotan θ f 2 L 2 0 A sin θ = E n sin θ A cos θ = (1 + E n) cos θ (recall that 0 < E n < 1) (hence in particular E n is the largest eigenvalue of A) There exists a Hilbert basis (e k ) k IN of L 2 (0, L), consisting of eigenvectors of A, associated with simple eigenvalues λ k satisfying < < λ k < < λ 1 < λ 0 = E n. Moreover : - x e k (x) vanishes exactly k times on (0, L). - λ 0 = E n is associated with the eigenfunction e 0 = sin θ. - Since cos(θ) vanishes n times over (0, L) (with L = 2nK (E n)), it follows that λ n = (1 + E n) and e n = cos θ. - If n > 1 then λ n 1 = 1.

19 Spectral study of A = 2 xx 2 cos 2 θ Id (A + E n Id)f, f L 2 = f θ cotan θ f 2 L 2 0 A sin θ = E n sin θ A cos θ = (1 + E n) cos θ (recall that 0 < E n < 1) (hence in particular E n is the largest eigenvalue of A) There exists a Hilbert basis (e k ) k IN of L 2 (0, L), consisting of eigenvectors of A, associated with simple eigenvalues λ k satisfying < < λ k < < λ 1 < λ 0 = E n. Moreover : - x e k (x) vanishes exactly k times on (0, L). - λ 0 = E n is associated with the eigenfunction e 0 = sin θ. - Since cos(θ) vanishes n times over (0, L) (with L = 2nK (E n)), it follows that λ n = (1 + E n) and e n = cos θ. - If n > 1 then λ n 1 = 1.

20 Spectral study of A = 2 xx 2 cos 2 θ Id (A + E n Id)f, f L 2 = f θ cotan θ f 2 L 2 0 A sin θ = E n sin θ A cos θ = (1 + E n) cos θ (recall that 0 < E n < 1) (hence in particular E n is the largest eigenvalue of A) There exists a Hilbert basis (e k ) k IN of L 2 (0, L), consisting of eigenvectors of A, associated with simple eigenvalues λ k satisfying < < λ k < < λ 1 < λ 0 = E n. Moreover : - x e k (x) vanishes exactly k times on (0, L). - λ 0 = E n is associated with the eigenfunction e 0 = sin θ. - Since cos(θ) vanishes n times over (0, L) (with L = 2nK (E n)), it follows that λ n = (1 + E n) and e n = cos θ. - If n > 1 then λ n 1 = 1.

21 Spectral study of A = 2 xx 2 cos 2 θ Id Ae 0 = λ 0 e 0, e 0 = sin θ, λ 0 = E n... Ae n 1 = λ n 1 e n 1, λ n 1 = 1 Ae n = λ ne n, e n = cos θ, λ n = 1 E n... Ae j = λ j e j, λ j.. (e n 1 is a linear combination of θ and of θ R x 1 θ 2 )

22 Linearization around a steady state «A + Id A + EnId Consequences for A = (A + Id) A + E nid Diagonalization into blocks of 2 2 linear systems, k IN : «ṙ 1k = (λ k + 1)r 1k + (λ k + E n)r 2k λk + 1 λ A k = k + E n ṙ 2k = (λ k + 1)r 1k + (λ k + E n)r 2k (λ k + 1) λ k + E n A k is Hurwitz λ k < 1 Therefore A sin θ = E n sin θ. Ae n 1 = e n 1 A cos θ = (1 + E n) cos θ. 9 >= n unstable modes >; 9 = ; stable modes

23 Control Linearized Landau-Lifshitz equation (around a steady-state) with r t = Ar + u 1 B 1 + u 2 B 2 + R(x, r, r x, r xx, u) «A + Id A + En Id A = (A + Id) A + E n Id ««sin θ(x) d3 d B 1 (x) = χ (a,b) (x) B sin θ(x) 2 (x) = χ (c,d) (x) 1 sin θ(x) d 3 + d 1 sin θ(x) u 1 (t)χ (a,b) (x)e 1 u 2 (t)χ (c,d) (x) d 0 a b c d L

24 Control Using the spectral decomposition linear systems : with ««d r1k r1k = A dt r k + u 2k r 1 B 1k + u 2 B 2k k IN 2k «λk + 1 λ A k = k + E n (λ k + 1) λ k + E n 0 Z b 1 0Z d 1 sin θ(x)e k (x) dx (d 3 d 1 sin θ(x))e k (x) dx B 1k = B Z C b A B 2k = B Z C d A sin θ(x)e k (x) dx (d 3 + d 1 sin θ(x))e k (x) dx a c

25 Control First possibility : external magnetic field generated by a single solenoid rolling around the nanowire : u 1 (t)χ (a,b) (x)e 1 0 a b L Kalman condition is not satisfied for the n first systems.

26 Control Second possibility : external magnetic field generated by two solenoids rolling around the nanowire, one of which having an angle with the axis of the nanowire : u 1 (t)χ (a,b) (x)e 1 u 2 (t)χ (c,d) (x) d 0 a b c d L Let (a, d) (0, L) 2 be such that a < d. Assume that d 3 0. For every N IN there exists a finite discrete set Z (0, L) such that, for all b and c in (0, L) \ Z such that a < b < c < d, the Kalman condition is satisfied for the N first systems. Consequence It is possible to design explicitly a feedback control stabilizing any steady state. This feedback is designed from the n first systems. A Lyapunov function is built to prove that this feedback stabilizes actually the whole system.

27 Control Consequence It is possible to design explicitly a feedback control passing approximately from any steady state to any other one having the same level of energy. Given two steady-states m 1 and m 2 having the same level of energy θ 2 + cos 2 θ = E n, for some n {1,..., N 0 }, for every ε > 0 there exist a time T > 0 and feedback controls u 1 L 2 (0, T ) and u 2 L 2 (0, T ) such that every solution such that m(0, ) m 1 ε satisfies at time T m(t, ) m 2 ε. Strategy : asymptotically track a path of steady-states.

28 Control Given a control system ẋ(t) = f (x(t), u(t)) and a path of steady states f ( x(τ), ū(τ)) = 0 we set so that y(t) = x(t) x(εt), v(t) = u(t) ū(εt), ẏ(t) = f f ( x(εt), ū(εt)) y(t) + ( x(εt), ū(εt)) v(t) x u + O(ε) + o(y(t), v(t)) slowly varying in time linear control system : ẏ(t) = A(εt)y(t) + B(εt)v(t) + remainder terms

29 Control Under Kalman condition ẏ(t) = A(εt)y(t) + B(εt)v(t) rank(b(τ), A(τ)B(τ),..., A(τ) n 1 B(τ)) = n τ (0, 1) this instationary system is stabilizable by usual pole placement A(τ) + B(τ)K (τ) Hurwitz ẏ(t) = (A(εt) + B(εt)K (εt))y(t) asymptotically stable : The feedback K (εt))y(t) stabilizes the control system along the path of steady-states. End of the proof : design a Lyapunov function in order to prove that this feedback acts appropriately on the whole system (in infinite dimension)

30 Control Open questions 1 Is it possible to jump to other levels of energy? 2 2d or 3d case : analysis, control and simulation of the Landau-Lifshitz equation on a thin layer (with Bloch and Néel walls)