Taylor expansions for the HJB equation associated with a bilinear control problem

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1 Taylor expansions for the HJB equation associated with a bilinear control problem Tobias Breiten, Karl Kunisch and Laurent Pfeiffer University of Graz, Austria Rome, June 217

2 Motivation dragged Brownian particle following Langevin equation dx s = Y (s)ds, dy s = βy s ds+f (X, s)ds+ 2βkT /mdb(s) particle confined by potential V : force F (X, s) = V (X, s) β >>, t = s β, ν = kt m, Smoluchowski equation dx t = V (X, t) + 2νdB t atomic force microscopy, single molecule pulling, optical trapping collinate light from laser into aperature of microscope objetive control by optical tweezer V (X t, t) = W (X t ) + u(x t, t).

3 The Fokker-Planck equation Consider probability distribution function Fokker-Planck equation ρ(x, t)dx = P[X t [x, x + dx)] ρ = ν ρ + (ρ V ) t in Ω (, ), = (ν ρ + ρ V ) n on Γ (, ), ρ(x, ) = ρ (x) in Ω, Ω R n bounded open set with boundary Γ = Ω, ρ initial probability distribution with Ω ρ (x)dx = 1. [ e.g. Borzi et al., Hartmann et al.

4 Control of the Fokker-Planck equation ρ = ν ρ + (ρ V ) t in Ω (, ), = (ν ρ + ρ V ) n on Γ (, ), ρ(x, ) = ρ (x) in Ω, system converges to stationary distribution ρ (x), (Boltzmann distribution) particles have to cross energy barrier between potential wells, may be inadequately slow, exp( W /ν) can we control the potential V (x, t) = W (x) + α(x)u(t), if we can, what are good choices for α? Assumptions: W W 2,max(2,n+ε) (Ω), ε > α W 2,max(2,n+ε) (Ω), ε > with α n = on Γ

5 Control of the Fokker-Planck equation ρ = ν ρ + (ρ V ) t in Ω (, ), = (ν ρ + ρ V ) n on Γ (, ), ρ(x, ) = ρ (x) in Ω, system converges to stationary distribution ρ (x), (Boltzmann distribution) particles have to cross energy barrier between potential wells, may be inadequately slow, exp( W /ν) can we control the potential V (x, t) = W (x) + α(x)u(t), if we can, what are good choices for α? Assumptions: W W 2,max(2,n+ε) (Ω), ε > α W 2,max(2,n+ε) (Ω), ε > with α n = on Γ

6 Solutions to the Fokker-Planck equation For T > we call ρ a (variational) solution on (, T ) if and ρ W (, T ) = L 2 (, T ; H 1 (Ω)) H 1 (, T ; (H 1 (Ω)) ) ρ t (t), v + ν ρ(t) + ρ(t) W, v + u(t) ρ(t) α, v =, v H 1 (Ω) ρ() = ρ. Proposition (i) u L 2 (, T ), ρ L 2 (Ω)! solution ρ W (, T ) (ii) If moreover α L (Ω), ρ H 1 (Ω) ρ t L 2 (, T ; L 2 (Ω)), ρ C([, T ]; H 1 (Ω)) Proposition Let u L 2 (, T ) and ρ L 2 (Ω). (i) For every t [, T ] we have Ω ρ(t) dx = Ω ρ dx. (ii) If ρ a.e. on Ω, then ρ(x, t) t > and a.e. on Ω.

7 Solutions to the Fokker-Planck equation For T > we call ρ a (variational) solution on (, T ) if and ρ W (, T ) = L 2 (, T ; H 1 (Ω)) H 1 (, T ; (H 1 (Ω)) ) ρ t (t), v + ν ρ(t) + ρ(t) W, v + u(t) ρ(t) α, v =, v H 1 (Ω) ρ() = ρ. Proposition (i) u L 2 (, T ), ρ L 2 (Ω)! solution ρ W (, T ) (ii) If moreover α L (Ω), ρ H 1 (Ω) ρ t L 2 (, T ; L 2 (Ω)), ρ C([, T ]; H 1 (Ω)) Proposition Let u L 2 (, T ) and ρ L 2 (Ω). (i) For every t [, T ] we have Ω ρ(t) dx = Ω ρ dx. (ii) If ρ a.e. on Ω, then ρ(x, t) t > and a.e. on Ω.

8 A bilinear control system Consider the bilinear control system ρ(t) = Aρ(t) + u(t)n ρ(t), ρ() = ρ, where the operators A and N are defined as follows A: D(A) L 2 (Ω) L 2 (Ω), D(A) = { ρ H 2 (Ω) (ν ρ + ρ W ) n = on Γ }, Aρ = ν ρ + (ρ W ), N : H 1 (Ω) L 2 (Ω), N ρ = (ρ α). σ(a) R, Φ(x) = log (ν) + W (x) ν ρ = e Φ is an eigenfunction of A with Aρ =

9 A shifted problem Instead of ρ, consider the shifted state y := ρ ρ ẏ(t) = Ay(t) + u(t)n y(t) + Bu(t), y() = ρ ρ, with B = N ρ, thus Bu(t) = u(t)n ρ = u(t) (ρ α). We decompose our state space L 2 (Ω) =: Y P Y Q Y P = {v L 2 (Ω): y = y P + y Q = Py + Qy. Ω v dx = }, Y Q = span {ρ }

10 Decoupling the problem applying P and Q yields ) ( ) ( ) ( ) ( ) (ẏp PA PA yp PN PN yp = + u + u QA QA QN QN ẏ Q y Q y Q ( ) PB. QB simplifies to ) ( (ẏp PA = ẏ Q simplifies to ) ( yp y Q ) + u ( ) ( ) PN PN yp + u y Q ẏ P = Ây P + u N y P + Bu, y P () = Pρ, y Q (t) = Qρ ρ =, t. ( ) PB.

11 An LQR problem for the linearized system For δ > let us focus on the linearized system ẏ P = (Â + δi )y P(t) + Bu, y P () = Pρ, together with the quadratic cost functional J(y P, u) = 1 2 y P (t), My P (t) L 2 (Ω) dt u(t) 2 dt, where M L(Y P ) is a self-adjoint nonnegative operator on Y P. Riccati-based feedback law: u = B ΠyP (Â + δi ) Π + Π(Â + δi ) Π B B Π + M =, Π L(YP ). The -dimensional Hautus test: The pair (Â, B) is δ-stabilizable if ker(λi Â ) ker( B ) = {} for λ C δ σ(â ). into nonlinear system: compare e.g.: M. Badra, I. Lasiecka, J.-P. Raymond, R. Triggiani,...

12 Fokker-Planck equation Confining potential W(x) Figure: 1D Fokker-Planck equation, n = 124.

13 Fokker-Planck equation Stationary distribution ρ (x) Figure: 1D Fokker-Planck equation, n = 124.

14 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

15 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

16 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

17 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

18 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

19 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

20 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

21 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

22 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

23 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

24 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

25 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

26 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

27 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

28 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

29 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

30 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

31 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

32 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

33 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

34 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

35 Fokker-Planck equation Uncontrolled solution at t = Figure: 1D Fokker-Planck equation, n = 124.

36 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

37 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

38 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

39 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

40 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

41 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

42 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

43 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

44 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

45 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

46 Back to the 1D Fokker-Planck equation (Un)controlled solution at t = Figure: Fokker-Planck, n = 124, r = 1, β = 1 5.

47 2-D and movies removed

48 Comparison of control laws u(t) 15 1 Optimal (open loop) 2nd order Taylor exp. (Riccati) 3rd order Taylor exp. 7th order Taylor exp Figure: 1D Fokker-Planck equation, n = 124.

49 Bilinear quadratic optimization Consider a bilinear control system ẋ(t) = Ax(t) + Nx(t)u(t) + Bu(t), x() = x, y(t) = Cx(t), A, N R n n, B R, control u : [, ) R and output y : [, ) R p of the system, (A, B) stabilizable. For this system, we introduce the minimal value functional V(x ) = 1 inf u L 2 (, ) 2 y(t) 2 dt + α 2 u(t) 2 dt.

50 Dynamic programming By the dynamic programming principle, for any x and τ > : V(x ) = τ inf u L 2 (,τ) henceforth l(y, u) = 1 2 y 2 + α 2 u2. l(y(u, x ; t), u(t))dt + V(x(u, x ; τ)), Under smoothness assumptions on V, we obtain [(Ax + (Nx + B)u) T V(x) + 12 Cx 2 + α2 ] u2 =, V() =. min u R

51 Dynamic programming By the dynamic programming principle, for any x and τ > : V(x ) = τ inf u L 2 (,τ) henceforth l(y, u) = 1 2 y 2 + α 2 u2. l(y(u, x ; t), u(t))dt + V(x(u, x ; τ)), Under smoothness assumptions on V, we obtain [(Ax + (Nx + B)u) T V(x) + 12 Cx 2 + α2 ] u2 =, V() =. min u R

52 The Hamilton-Jacobi-Bellman equation Consider again [(Ax + (Nx + B)u) T V(x) + 12 y 2 + α2 ] u2 =, V() =. min u R Minimization yields Hamilton-Jacobi-Bellman (HJB) equation x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =, V() =. Optimal feedback law via solving HJB equation u opt (x) = 1 α (Nx + B)T V(x). Problem: The HJB equation is a nonlinear n-dimensional PDE...

53 The Hamilton-Jacobi-Bellman equation Consider again [(Ax + (Nx + B)u) T V(x) + 12 y 2 + α2 ] u2 =, V() =. min u R Minimization yields Hamilton-Jacobi-Bellman (HJB) equation x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =, V() =. Optimal feedback law via solving HJB equation u opt (x) = 1 α (Nx + B)T V(x). Problem: The HJB equation is a nonlinear n-dimensional PDE...

54 HJB equation: the linear case For the linear case, we obtain x T A T V(x) Cx 2 1 2α (BT V(x)) 2 =, V() =. The ansatz V(x) = 1 2 x T Πx, with Π = Π T R n n yields The more familiar expression is 1 2 x T algebraic Riccati equation x T A T Πx Cx 2 1 2α (BT Πx) 2 =. ( A T Π + ΠA + C T C 1 α ΠBBT Π ) x =.

55 Taylor expansions basic idea Assume that V can be expanded around as follows 1 1 V(x) = V() + DV() (x) + }{{}}{{} 2! D2 V() (x, x) + }{{} 3! D3 V() (x, x, x) +... }{{} R R n R n n R n n n Feedback law can be determined via u = 1 α k=2 1 (k 1)! Dk V()(Nx + B, x,..., x) Finite-dimensional case: [Lukes, Cebuhar/Costanza, Krener] Infinite-dimensional case: [Thevenet/Buchot/Raymond] Question: Precise structure of D k V()?

56 Taylor expansions basic idea Assume that V can be expanded around as follows 1 1 V(x) = V() + DV() (x) + }{{}}{{} 2! D2 V() (x, x) + }{{} 3! D3 V() (x, x, x) +... }{{} R R n R n n R n n n Feedback law can be determined via u = 1 α k=2 1 (k 1)! Dk V()(Nx + B, x,..., x) Finite-dimensional case: [Lukes, Cebuhar/Costanza, Krener] Infinite-dimensional case: [Thevenet/Buchot/Raymond] Question: Precise structure of D k V()?

57 Taylor expansions basic idea Assume that V can be expanded around as follows 1 1 V(x) = V() + DV() (x) + }{{}}{{} 2! D2 V() (x, x) + }{{} 3! D3 V() (x, x, x) +... }{{} R R n R n n R n n n Feedback law can be determined via u = 1 α k=2 1 (k 1)! Dk V()(Nx + B, x,..., x) Finite-dimensional case: [Lukes, Cebuhar/Costanza, Krener] Infinite-dimensional case: [Thevenet/Buchot/Raymond] Question: Precise structure of D k V()?

58 Taylor expansions Differentiating HJB Let us come back to x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =,

59 Taylor expansions Differentiating HJB Let us come back to x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =, one differentiation in direction z 1 R n yields D 2 V(x)(Ax, z 1 ) + DV(x)Az 1 + Cx, Cz 1 1 α( D 2 V(x)(Nx + B, z 1 ) + DV(x)Nz 1 )( DV(x)(Nx + B) ) ] =.

60 Taylor expansions Differentiating HJB Let us come back to x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =, two differentiations in directions z 1, z 2 R n yield D 3 V(x)(Ax, z 1, z 2 ) + D 2 V(x)(Az 2, z 1 ) + D 2 V(x)(Az 1, z 2 ) + Cz 1, Cz 2 1 ) (D 2 V(x)(Nx + B, z 1 ) + DV(x)Nz 1 α ) (D 2 V(x)(Nx + B, z 2 ) + DV(x)Nz 2 1 ( ) D 3 V(x)(Nx + B, z 1, z 2 ) + D 2 V(x)(Nz 2, z 1 ) + D 2 V(x)(Nz 1, z 2 ) α ( ) DV(x)(Nx + B) =.

61 Taylor expansions Differentiating HJB Let us come back to x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =, two differentiations in directions z 1, z 2 R n yield D 3 V()(A, z 1, z 2 ) + D 2 V()(Az 2, z 1 ) + D 2 V()(Az 1, z 2 ) + Cz 1, Cz 2 1 ( D 2 V()( N + B, z 1 ) + α DV()Nz 1 ) ( D 2 V()( N + B, z 2 ) + ) DV()Nz 2 1 ( ) D 3 V()( N + B, z 1, z 2 ) + D 2 V()(Nz 2, z 1 ) + D 2 V()(Nz 1, z 2 ) α ( ) DV()(N + B) =. This is the Riccati equation...

62 Taylor expansions Differentiating HJB Let us come back to x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =, three differentiations in directions z 1, z 2, z 3 R n yield D 3 V()(Az 3, z 1, z 2) + D 3 V()(Az 2, z 1, z 3) + D 3 V()(Az 1, z 2, z 3) 1 ( )( ) D 3 V ()(B, z 1, z 3) + D 2 V()(Nz 3, z 1) + D 2 V()(Nz 1, z 3) D 2 V()(B, z 2) α 1 ( )( ) D 3 V()(B, z 2, z 3) + D 2 V()(Nz 3, z 2) + D 2 V()(Nz 2, z 3) D 2 V()(B, z 1) α 1 ( )( ) D 3 V()(B, z 1, z 2) + D 2 V()(Nz 2, z 1) + D 2 V()(Nz 1, z 2) D 2 V()(B, z 3) α =.

63 Taylor expansions Differentiating HJB Let us come back to x T A T V(x) Cx 2 1 2α ((Nx + B)T V(x)) 2 =, three differentiations in directions z 1, z 2, z 3 R n yield D 3 V()(Az 3, z 1, z 2) + D 3 V()(Az 2, z 1, z 3) + D 3 V()(Az 1, z 2, z 3) 1 ( )( ) D 3 V ()(B, z 1, z 3) + D 2 V()(Nz 3, z 1) + D 2 V()(Nz 1, z 3) D 2 V()(B, z 2) α 1 ( )( ) D 3 V()(B, z 2, z 3) + D 2 V()(Nz 3, z 2) + D 2 V()(Nz 2, z 3) D 2 V()(B, z 1) α 1 ( )( ) D 3 V()(B, z 1, z 2) + D 2 V()(Nz 2, z 1) + D 2 V()(Nz 1, z 2) D 2 V()(B, z 3) α =. Looks complicated Linear in D 3 V()

64 The general structure For i, j N, consider the following set of permutations: S i,j = { σ S i+j σ(1) <... < σ(i) and σ(i + 1) <... < σ(i + j) }, where S i+j is the set of permutations of {1,..., i + j}. Example S 2,2 = { σ S 4 σ(1) < σ(2) and σ(3) < σ(4) } = { (1, 2, 3, 4), (1, 3, 2, 4), (1, 4, 2, 3), (2, 3, 1, 4), (2, 4, 1, 3), (3, 4, 1, 2) } For given multilinear form T (of order i + j), we define Sym i,j (T )(z 1,..., z i+j ) := ( ) 1 [ i + j ] T (z i σ(1),..., z σ(i+j) ). σ S i,j

65 The general structure Define A Π = A 1 α BB Π. For k 3 and z 1,..., z k R n consider k i=1 D k V()(z 1,..., z i 1, A Π z i, z i+1,..., z k ) = 1 2α R k(z 1,..., z k ) where R k (z 1,..., z k ) is given by: ( ), R k (z 1,..., z k ) = 2k(k 1)Sym 1,k 1 (C 1 (z 1 )G k 1 (z 2,..., z k )) k 2 ( ) ( k + Sym i i,k i (C i (z 1,..., z i ) + ig i (z 1,..., z i )) i=2 where: C i (z 1,..., z i ) = D i+1 V()(B, z 1,..., z i ) G i (z 1,..., z i ) = 1 i ) (C k i (z i+1,..., z k ) + (k i)g k i (z i+1,..., z k )), [ i j=1 ] D i V()(z 1,..., z j 1, Nz j, z j+1,..., z i ).

66 Tensor calculus Main numerical task: ( k ) Solve I k i A T Π I i 1 T k = R k (T 2,..., T k 1 ). }{{} i=1 }{{}? low rank? A Since A is stable: A 1 = e ta dt = Approximate by quadrature formula k i=1 e tat Π dt. [Grasedyck,Hackbusch,Stenger] r A 1 j= r w j k e t j A T Π, i=1 with suitable quadrature weights w j and points t j.

67 The infinite dimensional setup Let us focus on the abstract bilinear control system ẋ(t) = Ax(t) + N x(t)u(t) + Bu(t), x() = x X, V X V Gelfand triple of Hilbert spaces a: V V R a bounded V -X bilinear form, i.e., ν > and λ R a(v, v) ν v 2 V λ v 2 X v V N L(V, X ) L(D(A), V ), N L(V, X ), B X For β > large enough define A := A + βi (A, B) stabilizable [D(A ), X ] 1 2 = [D(A ), X ] 1 2 = V

68 A multilinear operator equation Well-posedness of T k D k V() For k 3, and z 1,..., z k X define the multilinear form T k (z 1,..., z k ) = 1 2α T k : X X R, R k (e A Πt z 1,..., e A Πt z k ) dt. Then T k is the unique solution of ( ). Moreover, it holds that k T k (z 1,..., z k ) C z i X. Definition: V p : Y R, V p (y) = p k=2 i=1 1 k! T k(y,..., y).

69 A multilinear operator equation Well-posedness of T k D k V() For k 3, and z 1,..., z k X define the multilinear form T k (z 1,..., z k ) = 1 2α T k : X X R, R k (e A Πt z 1,..., e A Πt z k ) dt. Then T k is the unique solution of ( ). Moreover, it holds that k T k (z 1,..., z k ) C z i X. Definition: V p : Y R, V p (y) = p k=2 i=1 1 k! T k(y,..., y).

70 A suboptimal feedback law Consider now the polynomial feedback law u p (x) = p k=2 1 (k 1)! T k(n x + B, x,..., x) and the corresponding (nonlinear) closed-loop system (CL) ẋ = Ax (N x + B)u p (x), x() = x. Local well-posedness There exist constants C 1, C 2 > such that: if x X C 1, then (CL) admits a unique solution x W (, ) = {ϕ L 2 (, ; X ) ϕ t L 2 (, ; V )} this solution satisfies x W (, ) C 2 and lim t x(t) X =

71 A suboptimal feedback law Consider now the polynomial feedback law u p (x) = p k=2 1 (k 1)! T k(n x + B, x,..., x) and the corresponding (nonlinear) closed-loop system (CL) ẋ = Ax (N x + B)u p (x), x() = x. Local suboptimality There exists a constant C 3 >, C 4 such that: if x X C 3, l(x(u p, x ; t), u p (t)) dt V(x ) + C 4 ( x p+1 X ) V(x ) V p (x ) C y p+1 X x(ū, x ) x(u p, x ) W (, ) C 4 x p+1 2 X ū u p W (, ) C 4 x p+1 2 X

72 Comparison of control laws u(t) 15 1 Optimal (open loop) 2nd order Taylor exp. (Riccati) 3rd order Taylor exp. 7th order Taylor exp Figure: 1D Fokker-Planck equation, n = 124.

73 Comparison of control laws u(t) nd order Taylor exp. (Riccati) 3rd order Taylor exp. 4th order Taylor exp. 5th order Taylor exp. 6th order Taylor exp. 7th order Taylor exp Figure: Fokker-Planck, n = 124, r = 1, β = 1 4.

74 Miscellanea Applicable to Fokker Planck Feasible for general infinite dimensional control systems Combined with balanced truncation When is higher order useful? Efficient tensor numerics

75 Finite-dimensional properties Assume spatial discretization yields A, N, ρ d, e = 1 ( ) T d 1,..., 1. ρ(t) dx = ρ dx e T ρ d (t) = e T ρ d = 1 Ω Ω ρ ρ(x, t) t A is a Metzler matrix A s = DAD 1 with D = diag(e Φd 2 ) is a symmetric matrix A T e = N T e = = Aρ d µ = max(ε, ε λ max(â + ÂT ))

76 A two dimensional double well potential As a numerical example, let us consider ρ = ν ρ + (ρ W ) + u (ρ α) t in Ω (, ), = (ν ρ + ρ W ) n on Γ (, ), ρ(x, ) = ρ (x) in Ω. Ω = ( 1.5, 1.5) ( 1.1) W (x) = 3(x1 2 1)2 + 6x2 2 finite differences with n x n y = = 6144 grid points upwind scheme for convective terms

77 Conclusion Fokker-Planck equation yields a bilinear control system. Decoupled the system by spectral projections. Riccati-based feedback law local stabilization. Lyapunov-based feedback law global stabilization. Numerically efficient approaches? Optimal feedback law for bilinear system?

78 Conclusion Fokker-Planck equation yields a bilinear control system. Decoupled the system by spectral projections. Riccati-based feedback law local stabilization. Lyapunov-based feedback law global stabilization. Numerically efficient approaches? Optimal feedback law for bilinear system? Thank you for your attention!

A reduction method for Riccati-based control of the Fokker-Planck equation

A reduction method for Riccati-based control of the Fokker-Planck equation Tobias Breiten Karl Kunisch, Laurent Pfeiffer Institute for Mathematics and Scientific Computing, Karl-Franzens-Universität, Heinrichstrasse