Computational Optimization and Applications, 18, 115–140, 2001 c 2001 Kluwer Academic Publishers. Manufactured in The Netherlands. °

Distributed Control Problems for the Burgers Equation S. VOLKWEIN [email protected] Karl-Franzens-Universi¨at Graz, Institut f¨ur Mathematik, Heinrichstraße 36, A-8010 Graz, Austria Received July 27, 1999; Accepted October 20, 1999

Abstract. In this work a distributed optimal control problem for time-dependent Burgers equation is analyzed. To solve the nonlinear control problems the augmented Lagrangian-SQP technique is used depending upon a second-order sufficient optimality condition. Numerical test examples are presented. Keywords: nonlinear programming, multiplier methods, Burgers equation

1.

Introduction

In the recent past there has been given considerable attention to the problem of active control of fluids or combustion, where nonlinear effects actually can improve mixing. There are a lot of papers concerned with the study of asymptotic or steady state properties of solutions of nonlinear distributed parameter systems, such as Navier-Stokes equations, which contains both diffusion terms and nonlinear convection terms. A one dimensional simple model for convection-diffusion phenomena is the Burgers equation, such as shock waves, supersonic flow about airfoils, traffic flows, acoustic transmission etc. We mention that for small viscosity parameter the solution of Burgers’ equation produces steep gradients due to the nonlinear nature of the convection. Burgers introduced the equation yt + yyx = νyx x

(1)

as a simple model for turbulence [2–4], where ν > 0 denotes a viscosity parameter. In [11] general properties of solutions to (1) were studied and related to some applications. Since Eq. (1) was introduced, the conservation law yt + yyx = 0 and the so-called viscosity solution y = lim y ν , ν→0

where y ν satisfies Eq. (1), considered, see for example in [16]. A complete discussion of results concerned with the conservation law and the viscosity solution may be found

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in [30]. Further, in [29] existence and uniqueness of a solution to (1) with homogeneous Dirichlet boundary conditions was proved by using the theory of semigroups. We refer to the book [23], where the Cauchy problem for the viscous and inviscid (ν = 0 in (1)) Burgers’ equation was investigated. We analyze control problems governed by the Burgers equation. Let T > 0, Q = (0, T ) × (0, 1), σ > 0 and let C be a continuous and linear operator mapping the state space into the real Hilbert space H. We consider optimal distributed control minimizing J (y, u) =

σ 1 kC y − zk2H + kuk2L 2 (Q) 2 2

subject to    yt − νyx x + yyx = f + u y(t, 0) = y(t, 1) = 0   y(0, x) = φ(x)

in Q, in (0, T ),

(2a)

(2b)

in (0, 1),

where f ∈ L 2 (Q) and φ ∈ L 2 (0, 1) are given and u ∈ L 2 (Q) denotes the control variable. To solve (2) we use the augmented Lagrangian-SQP method introduced in [19]. In this method the Burgers equation is treated as an equality constraint which is realized by a Lagrangian term together with a penalty functional. The presented algorithm has secondorder convergence rate and depends upon a second-order sufficient optimality condition. In [33, 34] the augmented Lagrangian-SQP method and its discretization was analyzed. It was proved that the algorithm satisfies a mesh-independence principle. Let us mention some papers concerning control problems for Burgers’ equation. Chen, Wang and Weerakoon handled an initial value control problem for the Burgers equation and showed numerical results in [8, 35]. Burns, Ito and Kang investigated several control problems for Burgers’ equation [5, 6, 20, 21]. Using a linearized model feedback control laws were computed, which were applied to the full nonlinear model. The problems, which were under consideration, contain both unbounded inputs and observations. Numerical examples illustrated the theoretical results. Dean and Gubernatis considered the Burgers equation with homogeneous Dirichlet-Neumann boundary conditions on the time boundary [13]. To obtain a given output function they established a time control v(t) at a fixed point x◦ ∈ (0, 1). In the article [9] the authors described the role of incremental unknowns for approximating attractors and inertial manifolds arising in fluid dynamics. Especially, they were concerned with the Burgers equation with stochastic forces. In [10] a feedback control method of minimizing a cost function was developed and applied to the Burgers equation as a first step towards application to fluid mechanics problems. Ito and Kang constructed a dissipative feedback control to regulate the systems arising in fluid dynamics [18]. For the numerical experiments they used the Burgers equation, but also the Navier-Stokes equations. We mention the paper of Ikeda [17], where a method for active fluid flow control based on control theory was analyzed and applied to stochastic Burgers’ equation. Numerical results were also presented. Diaz was concerned with approximate controllability for a final observation [14]. He considered the Burgers equation over the space interval (0, L), where the control took place at x = L. In [7] a boundary control problem for a forced Burgers

DISTRIBUTED CONTROL PROBLEMS

117

equation in a Hilbert space was considered. For the unforced problem, the stability of equilibria was investigated, and for the forced problem the authors concerned with steady state behavior of solutions. In the paper [27] the results of Burns et al. and Byrnes at al. were generalized. Further, the Burgers equation was treated with other nonlinear boundary conditions. The control problem (2) was also studied utilizing suboptimal control strategies. In [25] a reduced basis was computed by the proper orthogonal decomposition (POD). Then (2) was solved numerically by projecting the dynamics on the POD basis. It turns out that the POD-solutions have a surprisingly good agreement with the finite element solutions, but the CPU-time were decreased rapidly. In [15] an instantaneous control approach was analyzed. It was proved that the instantaneous solution was asymptotically stable. The paper is organized in the following manner. In §2 the augmented Lagrangian-SQP algorithm is proposed and a convergence result is reviewed. The distributed control problem for the Burgers equation and the associated Lagrangian framework is developed in §3. Finally, some numerical experiments are presented in §4. It is appropriate to introduce some notations that will be used throughout the paper. Let (V, k·kV ) and (W, k·kW ) be normed linear spaces. The set B(v; r ) denotes an open ball of radius r > 0 centered at the point v ∈ V . The open set U ⊂ V is called a neighborhood of v ∈ U if B(v; r ) ⊂ U for some r > 0. By L(V, W ) we denote the normed linear space of all bounded linear operators from V into W and set L(V ) = L(V, V ). For A ∈ L(V, W ) the set {v ∈ V : Av = 0} is called the kernel of A and is denoted by ker A. Let V and H be two real, separable Hilbert spaces with duals V ∗ and H ∗ . It is supposed that V is dense in H so that, by identifying H and H ∗ , we have V ,→ H = H ∗ ,→ V ∗ , each embedding being dense. For T > 0 the spaces L 2 (0, T ; V ) and C(0, T ; V ) denote the space of square integrable and continuous functions, respectively, in the sense of Bochner from [0, T ] to V . The space W (0, T ; V ) is defined by W (0, T ; V ) = {ϕ : ϕ ∈ L 2 (V ), ϕt ∈ L 2 (V ∗ )}, which is a Hilbert space endowed with the common inner product, see [12]. For brevity we write L 2 (V ), C(H ), and W (V ) in place of L 2 (0, T ; V ), C(0, T ; H ) and W (0, T ; V ), respectively. Since W (V ) is continuously embedded into C(H ), [12], there exists an embedding constant c E > 0 such that kϕkC(H ) ≤ c E kϕkW (V )

for all ϕ ∈ W (V ).

(3)

Because the following inequalities are used frequently in this work, we shall give complete formulations of them here and only refer to their names whenever necessary. Young’s inequality [1]: Let a, b ∈ [0, ∞) and ε > 0. Then we have ab ≤

q 1 1 1 ap q b + ε , where 1 < p < ∞ and + = 1. p ε p q p q

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Agmon’s inequality [32]: For every ϕ ∈ H 1 (0, 1), there exists a constant c A > 0 such that 1/2

1/2

kϕk L ∞ (0,1) ≤ c A kϕk L 2 (0,1) kϕk H 1 (0,1) . 2.

Review of the augmented lagrangian-SQP method

In this section we propose the augmented Lagrangian-SQP method which is used in the following sections to solve optimal control problems for the Burgers equation. We consider the constrained minimizing problem minimize

J (x)

subject to e(x) = 0,

(P)

where J : X → R, e : X → Y and X , Y denote real Hilbert spaces. The Hilbert space X × Y is endowed with the Hilbert space product topology. Concerning the notation we shall not distinguish between a functional in the dual space and its Riesz representation in the Hilbert space. For c ≥ 0 let us introduce the augmented Lagrangian functional L c : X × Y → R associated with (P): L c (x, ξ ) = J (x) + he(x), ξ iY +

c ke(x)k2Y , 2

where h· , ·iY denotes the inner product in Y . The following assumption is rather standard for SQP methods in Hilbert spaces. Assumption 1. Let x ∗ ∈ X be a reference point such that a) J and e are twice continuously Fr´echet-differentiable, and the mappings J 00 and e00 are Lipschitz-continuous in a neighborhood of x ∗ , b) the linearization e0 (x ∗ ) of the operator e at x ∗ is surjective, c) there exists a Lagrange multiplier ξ ∗ ∈ Y satisfying the first-order necessary optimality condition L 0c (x ∗ , ξ ∗ ) = L 00 (x ∗ , ξ ∗ + ce(x ∗ )) = 0, e(x ∗ ) = 0

for all c ≥ 0,

(OS)

where the Fr´echet-derivative with respect to the variable x is denoted by a prime, and d) the operator L 000 (x ∗ , ξ ∗ ) is coercive on the kernel of e0 (x ∗ ), i.e. there exists a constant κ > 0 such that L 000 (x ∗ , ξ ∗ )χ 2 ≥ κ kχk2X

for all χ ∈ ker e0 (x ∗ ).

The next proposition follows directly from Assumption 1. For a proof we refer to [26] and [28].

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DISTRIBUTED CONTROL PROBLEMS

Proposition 1. With Assumption 1 holding x ∗ is a local solution to (P). Furthermore, there exists a neighborhood of (x ∗ , ξ ∗ ) such that (x ∗ , ξ ∗ ) is the unique solution of (OS) in this neighborhood. Due to Assumption 1 there are neighborhoods U (x ∗ ) ⊂ X of x ∗ and U (ξ ∗ ) ⊂ Y of ξ ∗ such that a) J (x) and e(x) are twice Fr´echet-differentiable and their second Fr´echet-derivatives are Lipschitz-continuous in U (x ∗ ), b) e0 (x) is surjective, c) L 000 (x, ξ ) is coercive on the kernel of e0 (x) for all (x, ξ ) ∈ U = U (x ∗ ) × U (ξ ∗ ), and d) the point (x ∗ , λ∗ ) is the unique solution to (OS) in U . To find x ∗ numerically we solve (OS) by the Newton method, where we avoid the explicit calculation of the augmented Lagrangian functional and its derivatives. To formulate the algorithm it will be convenient to introduce the matrix of operators µ M(x, ξ ) =

L 000 (x, ξ ) e0 (x)

e0 (x)? 0

¶ for all (x, ξ ) ∈ U,

(4)

where e0 (x)? : Y → X denotes the adjoint of e0 (x). Remark 1.

With Assumptions 1 holding there exists a constant C > 0 satisfying

kM(x, ξ )−1 kL(X ×Y ) ≤ C

for all (x, ξ ) ∈ U.

Algorithm 1 (Augmented Lagrangian-SQP method). a) Choose (x 0 , ξ 0 ) ∈ U, c ≥ 0 and put n = 0. b) Set ξ˜ n = ξ n + ce(x n ). c) Solve for (x, ¯ ξ¯ ) the linear system Ã

x¯ − x n M(x , ξ ) ξ¯ − ξ˜ n n

˜n

!

à =−

L 00 (x n , ξ˜ n ) e(x n )

! .

(5)

¯ ξ¯ ), n = n + 1 and go back to b). d) Set (x n+1 , ξ n+1 ) = (x, Remark 2. a) Since X and Y are Hilbert spaces, it can be shown that the update (x, ¯ λ¯ ) of Algorithm 1 can equivalently be obtained from Ã

L 00c (x n , λn ) e0 (x n )

e0 (x n )? 0

!Ã

x¯ − x n λ¯ − λn

!

! L 0c (x n , λn ) , =− e(x n ) Ã

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VOLKWEIN

which corresponds to a Newton step applied to (OS). This form of the iteration requires the implementation of e0 (x n )? e0 (x n ), whereas steps b) and c) of Algorithm 1 do not. b) If we set c = 0 Algorithm 1 becomes the SQP-method. It turns out that the augmented Lagrangian-SQP method has second order rate of convergence if Assumption 1 holds true. The following result was proved in [19]. Theorem 2. Let Assumption 1 hold. If c k(x 0 , ξ 0 ) − (x ∗ , ξ ∗ )k X ×Y is sufficiently small, then Algorithm 1 is well-defined and k(x n+1 , ξ n+1 ) − (x ∗ , ξ ∗ )k X ×Y ≤ C k(x n , ξ n ) − (x ∗ , ξ ∗ )k2X ×Y , where the constant C > 0 does not depend on n. 3.

The optimal control problem

In this section we study a distributed control problem for the stationary Burgers equation with Dirichlet boundary conditions. We shall prove that the first-order necessary optimality conditions possesses a locally unique solution, where the second-order sufficient optimality condition holds. 3.1.

Formulation of the optimal control problem

For fixed T > 0 we set Q = (0, T ) × (0, 1). Let Q ◦ ⊆ Q be an open set with positive measure, let V = H01 (0, 1), H = L 2 (0, 1), f ∈ L 2 (V ∗ ) and φ ∈ H . We supply V with the inner product hϕ, ψiV = hϕ 0 , ψ 0 i H

for ϕ, ψ ∈ V.

In the dual space V ∗ = H −1 (0, 1) we use the inner product hg, hiV ∗ = h(−1)−1 g, (−1)−1 hi H

for g, h ∈ V ∗ ,

where 1 : V → V ∗ denotes the Laplace operator with homogeneous Dirichlet boundary conditions. Further, the extension operator B ∈ L(L 2 (Q ◦ ), L 2 (H )) is given by ( q Bq = 0

in Q ◦ , in Q\Q ◦ .

We define the continuous trilinear form b(ϕ, ψ, φ) =

1 3

Z 0

1

(ϕψ)0 φ + ϕψ 0 φ d x

for ϕ, ψ, φ ∈ H 1 (0, 1)

DISTRIBUTED CONTROL PROBLEMS

121

and denote by y(t) and f (t) the functions y(t, ·) and f (t, ·), respectively, considered as functions of x only when t is fixed. For a control u ∈ L 2 (Q ◦ ) the state y ∈ W (V ) is given by the weak solution of the Burgers equation (

yt − νyx x + yyx = f + Bu y(0) = φ

Definition 1.

in L 2 (V ∗ ), in H.

(6)

A function y ∈ W (V ) is called a weak solution to (6) if

hyt (t), ϕiV ∗ ,V + νhy(t), ϕiV + b(y(t), y(t), ϕ) = h f (t) + Bu(t), ϕiV ∗ ,V for all ϕ ∈ V and a.e. t ∈ [0, T ] and y(0) = φ in H are valid. The next theorem ensures the existence of a unique weak solution to the Burgers equation. Theorem 3. With f ∈ L 2 (V ∗ ) and φ ∈ H holding there exists a unique weak solution y ∈ W (V ) to (6). Moreover, there exists a constant C > 0, which depends on ν, φ and f , such that ¡ ¢ kykW (V ) ≤ C 1 + kuk L 2 (Q ◦ ) + kuk2L 2 (Q ◦ ) , Proof: As in the proof for the unsteady Navier-Stokes equations in [31] we derive that there exists a unique y ∈ W (V ) satisfying (6) in the weak sense. Thus, we have to prove the estimate. Let us first prove that ¡ ¢2 kyk2L 2 (V ) ≤ c1 kφk H + k f k L 2 (V ∗ )

(7)

holds for a constant c1 > 0. For this purpose we multiply the differential equation in (6) by y(t) and integrate over the interval (0, 1). Then we get 1 d ky(t)k2H + ν ky(t)k2V = h f (t) + Bu(t), y(t)iV ∗ ,V . 2 dt

(8)

Applying the H¨older inequality we obtain Z

T 0

h f (t) + Bu(t), y(t)iV ∗ ,V dt ≤ k f + Buk L 2 (V ∗ ) kyk L 2 (V ) .

Integrating (8) over the interval (0, T ) we thus derive 1 1 ky(T )k2H + ν kyk2L 2 (V ) ≤ kφk2H + k f + Buk L 2 (V ∗ ) kyk L 2 (V ) . 2 2

(9)

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VOLKWEIN

Hence, Young’s inequality gives 2ν kyk2L 2 (V ) ≤ kφk2H +

1 k f + Buk2L 2 (V ∗ ) + ν kyk2L 2 (V ) , ν

and therefore 1 1 kφk2H + 2 k f + Buk2L 2 (V ∗ ) ν ν µ ¶ ¢2 1 1 ¡ ≤ max , 2 kφk H + k f + Buk L 2 (V ∗ ) ν ν {z } |

kyk2L 2 (V ) ≤

(10)

=c1

such that estimate (7) holds. To prove that kyt k2L 2 (V ∗ ) is bounded in terms of kφk H and k f + Buk L 2 (V ∗ ) we show that kykC(H ) ≤ max(1,

¢2 √ ¡ c1 ) kφk H + k f + Buk L 2 (V ∗ ) .

(11)

Due to (8) ky(t)k2H ≤ kφk2H + 2 k f + Buk L 2 (V ∗ ) kyk L 2 (V ) ¢ (7) √ ¡ ≤ kφk2H + 2 k f + Buk L 2 (V ∗ ) c1 kφk H + k f + Buk L 2 (V ∗ ) ¢2 √ ¡ ≤ max(1, 2 c1 ) kφk H + k f + Buk L 2 (V ∗ ) , which gives (11). By (6) we get kyt (t)kV ∗ ≤ k f (t) + Bu(t)kV ∗ + ν ky(t)kV + ky(t)k H ky(t)kV ¡ ¢ ≤ ν + kykC(H ) ky(t)kV + k f (t) + Bu(t)kV ∗ . Hence, we have Z kyt k2L 2 (V ∗ ) ≤ 2

0

T

¡ ¢ 2 2 2 2 ν 2 + kykC(H ) ky(t)kV + k f (t) + Bu(t)kV ∗ dt

¢¡ ¢2 ¡ ≤ 2 + 4ν 2 c1 kφk H + k f + Buk L 2 (V ∗ ) ¢4 √ ¡ + 32c1 max(1, c1 ) kφk H + k f + Buk L 2 (V ∗ ) . Finally, we conclude from (10) and (12) that kyk2W (V ) = kyk2L 2 (V ) + kyt k2L 2 (V ∗ ) ¡ ¢¡ ¢2 ≤ c1 + 2 + 4ν 2 c1 kφk H + k f + Buk L 2 (V ∗ ) ¢4 √ ¡ + 4c1 max(1, c1 ) kφk H + k f + Buk L 2 (V ∗ )

(12)

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DISTRIBUTED CONTROL PROBLEMS

¡ ¢¡¡ ¢2 ¢ ≤ 2 c1 + 2 + 4ν 2 c1 kφk H + k f k L 2 (V ∗ ) + kBuk2L 2 (V ∗ ) ¢4 ¢ √ ¡¡ + 4c1 max(1, c1 ) kφk H + k f k L 2 (V ∗ ) + kBuk4L 2 (V ∗ ) , 2

which gives the claim.

Now we proceed by introducing the cost functional. Let C ∈ L(W (V ), H) be a given observation operator, where H is a real Hilbert space. With every control we associate the cost of tracking type J (y, u) =

σ 1 kC y − zk2H + kuk2L 2 (Q ◦ ) , 2 2

where z ∈ H is a desired state and σ > 0 is fixed. The control problem then is J (y, u)

minimize

such that

(y, u) is a weak solution to (6).

(13)

In the context of §2 we set X = W (V ) × L 2 (Q ◦ ) and Y = L 2 (V ) × H . We introduce the operator e = (e1 , e2 ) : X → Y by ! (−1)−1 (yt − νyx x + yyx − f − Bu) . e(y, u) = y(0) − φ Ã

Then we write (13) in the following form minimize

J (y, u)

subject to e(y, u) = 0.

(P)

Note that both J and e are twice continuously Fr´echet-differentiable and their second Fr´echet-derivatives are Lipschitz-continuous on X . In particular, Assumption 1-a) holds. In the following the Fr´echet-derivatives with respect to the variable x = (y, u) are denoted by a prime.

3.2.

Existence of a solution to (OS)

The following proposition guarantees that the optimal control problem (P) has a solution. Proposition 4. Proof:

There exists an optimal solution (y ∗ , u ∗ ) of problem (P).

Let (y, u) ∈ X satisfy the equation e(y, u) = 0. We have

J (y, u) ≥

σ kuk2L 2 (Q ◦ ) . 2

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VOLKWEIN

From Proposition 3 we conclude that kykW (V ) → ∞ yields kuk L 2 (Q ◦ ) → ∞. Hence, J (y, u) → +∞

for

k(y, u)k X → ∞.

(14)

As the norm is weakly lower semi-continuous [36], we achieve that J is weakly lower semi-continuous. Since J (y, u) ≥ 0 for all (y, u) ∈ X there exists ζ ≥ 0 with ζ = inf{J (y, u) : (y, u) ∈ X

with e(y, u) = 0}.

This implies the existence of a minimizing sequence {(y n , u n )}n∈N in X such that ζ = limn→∞ J (y n , u n ) and e(y n , u n ) = 0 for all n ∈ N. From property (14) we infer that there exists an element (y ∗ , u ∗ ) ∈ X with (y n , u n ) * (y ∗ , u ∗ ) as n → ∞ in X.

(15)

We infer from (15) that Z lim

T

­

n→∞ 0

® ytn (t) − yt∗ (t), ϕ(t) V ∗ ,V dt = 0

for all ϕ ∈ L 2 (V ).

Since W (V ) is compactly embedded into L 2 (L ∞ ), [31], we derive that y n → y ∗ strongly in L 2 (L ∞ ). As the sequence {y n }n∈N converges weakly, ky n kW (V ) is bounded [24]. Consequently, ky n kC(H ) is also bounded [12]. Thus, it follows by H¨older’s inequality that ¯Z Z ¯ ¯ T 1¡ ¯ ¢ ¯ ¯ n n ∗ ∗ y yx − y yx ϕ dx dt¯ ¯ ¯ 0 0 ¯ ¯Z Z ¯ ¯ 1 ¯¯ T 1 ∗ ∗ ¯ n n (y y − y y )ϕx dx dt¯ = ¯ ¯ 2¯ 0 0 Z Z Z Z 1 T 1 ∗ 1 T 1 ∗ |(y − y n )y n ϕx | dx dt + |(y − y n )y ∗ ϕx | dx dt ≤ 2 0 0 2 0 0 Z 1 T ∗ ≤ ky (t) − y n (t)k L ∞ ky n (t)k H kϕ(t)k H01 dt 2 0 Z 1 T ∗ + ky (t) − y n (t)k L ∞ ky ∗ (t)k H kϕ(t)k H01 dt 2 0 ≤

1 ∗ ky − y n k L 2 (L ∞ ) ky n kC(H ) kϕk L 2 (V ) 2 1 + ky ∗ − y n k L 2 (L ∞ ) ky ∗ kC(H ) kϕk L 2 (V ) 2 n→∞

−→ 0

for all ϕ ∈ L 2 (V ).

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DISTRIBUTED CONTROL PROBLEMS

As (15) implies Z

T 0

Z

1

n→∞

(Bu n − Bu ∗ )ϕ dx dt −→ 0

for all ϕ ∈ L 2 (V )

0

we conclude that e1 (y ∗ , u ∗ ) = 0 in L 2 (V ). From y ∗ ∈ X we derive that y ∗ (0) ∈ H . Since n→∞ n→∞ y n * y ∗ in W (V ) we infer that y n (0) * y ∗ (0) in H and thus n→∞

hy n (0) − y ∗ (0), ψi H −→ 0

for all ψ ∈ H.

Thus, e(y ∗ , u ∗ ) = 0 in Y .

2

Let (y ∗ , u ∗ ) ∈ X be an optimal solution to (P). If the operator e0 (y ∗ , u ∗ ) is surjective then there exist unique Lagrange multipliers λ∗ ∈ L 2 (V ) and µ∗ ∈ H satisfying the first-order necessary optimality condition L 0c (y ∗ , u ∗ , λ∗ , µ∗ ) = 0, Proposition 5. is valid.

e(y ∗ , u ∗ ) = 0

for c ≥ 0.

The operator e0 (y, u) is surjective for all (y, u) ∈ X , i.e. Assumption 1-b)

Proof: The operator e0 (y, u) is surjective if and only if for all (g, h) ∈ Y there exists (v, q) ∈ X such that e10 (y, u)(v, q) = g in L 2 (V ), e20 (y, u)(v, q) = h in H.

(16a) (16b)

Due to the definition of the operator e Eq. (16a) is equivalent with vt − νvx x + (yv)x = (−1)g + Bq in L 2 (V ∗ ). We introduce the bilinear form a(t; · , ·) : V × V → R for t ∈ [0, T ] a.e. by Z a(t; η, ξ ) =

1

νη0 ξ 0 − y(t)ηξ 0 d x.

0

Thus, setting q = 0 Eq. (16) are equivalent with d hv(t), ξ i H + a(t; v(t), ξ ) = h−1g(t), ξ iV ∗ ,V dt

(17a)

for all ξ ∈ V and t ∈ [0, T ] a.e. and the initial condition v(0) = h.

(17b)

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VOLKWEIN

Here v(t) and g(t) denote the functions g(t, ·) and v(t, ·) considered as a function of x only when t is fixed. For the existence of a solution v to problem (17) we have to prove that the bilinear form a(t; · , ·) is continuous for t ∈ [0, T ] a.e. and satisfies a(t, η, η) ≥ ρ kηk2V − τ kηk2H

(18)

for all η ∈ V , for two constants ρ > 0, τ ≥ 0 and for t ∈ [0, T ] a.e. As |a(t; η, ξ )| ≤ (ν + kykC(H ) )kηkV kξ kV

for t ∈ [0, T ] a.e.

holds, the bilinear form a(t; · , ·) is continuous for t ∈ [0, T ] a.e. If y = 0 holds we conclude that a(t; η, η) = ν kηk2V , i.e. setting ρ = ν and τ = 0 condition (18) is valid. On the other hand y 6= 0 leads to kykC(H ) > 0. Thus, Young’s and Agmon’s inequality result in Z 1 2 y(t)ηη0 dx a(t; η, η) = ν kηkV − 0

≥ ν kηk2V − ky(t)k H kηk L ∞ kηkV 1/2

3/2

≥ ν kηk2V − c A kykC(H ) kηk H kηkV √ µ ¶ 3 3 ε4 1 2 2 2 kηk ≥ ν kηkV − c A kykC(H ) kηk + H V 4ε 4 4 for all η ∈ V and t ∈ [0, T ] a.e. Choosing ! 34 Ã 2ν 6= 0 ε= 3c A kykC(H ) we infer that a(t; η, η) ≥

cA ν kηk2V − 4 kykC(H ) kηk2H 2 4ε

cA for all η ∈ V and t ∈ [0, T ] a.e. Setting ρ = ν2 and τ = 4ε 4 kykC(H ) condition (18) is satisfied. Applying Theorems 1 and 2 in [12, Chapter XVIII] yields the existence of an element v ∈ W (V ) satisfying (17), which is even unique. 2

Remark 3.

The proof of Proposition 5 shows that e0 (y, u) | X ×{0} is already surjective.

Propositions 4 and 5 imply directly the next theorem, which guarantees that Assumption 1-c) is satisfied. Theorem 6. There exists an element (y ∗ , u ∗ , λ∗ , µ∗ ) ∈ X × Y satisfying the first-order necessary optimality conditions (OS), in particular hold. ∂L ∗ ∗ ∗ ∗ (y , u , λ , µ ) = 0 in W (V ), ∂y ∂L ∗ ∗ ∗ ∗ (y , u , λ , µ ) = 0 in L 2 (Q ◦ ). ∂u

(19) (20)

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DISTRIBUTED CONTROL PROBLEMS

Note that Eq. (20) leads to σ u ∗ = λ∗ in Q ◦ . Equation (19) is equivalent to Z T hvt (t), λ∗ (t)iV ∗ ,V dt 0 = hC y ∗ − z, CviH + 0 Z TZ 1 Z 1 + νvx λ∗x + (vy ∗ )x λ∗ dx dt + v(0)µ∗ dx 0

0

(21)

0

for all v ∈ W (V ), specifically for v(t) = χ(t)w, where w ∈ V and χ ∈ C0∞ (0, T ). We find Z

T

*Z

∗

hvt (t), λ (t)iV ∗ ,V dt =

0

T

+

*Z

∗

λ (t)χ˙ dt, w

0

=− 0

H

T

+ λ∗t (t)χ

dt, w

, V ∗ ,V

where λ∗t denotes the distributional derivatives of λ∗ with respect to t. The remaining terms in (21) lead to Z

∗

T

Z

1

hC y − z, CviH + νλ∗x vx − y ∗ vλ∗x dx dt 0 0 + *Z T ? ∗ ∗ ∗ ∗ (C (C y − z) − νλx x − y λx )χ dt, w = 0

Inserting these expressions into (21) yields *Z + *Z T

0

. V ∗ ,V

λ∗t (t)χ, w

T

= V ∗ ,V

?

+ ∗

(C (C y − z) −

0

νλ∗x x

−

y ∗ λ∗x )χ

dt, w V ∗ ,V

for all w ∈ V and χ ∈ C0∞ (0, T ). Since C ? (C y ∗ − z) − νλ∗x x − y ∗ λ∗x ∈ L 2 (V ∗ ) holds and the set {v : v(t) = χ(t)w

for χ ∈ C0∞ (0, 1) and w ∈ V }

is dense in L 2 (V ), we conclude that λ∗t ∈ L 2 (V ∗ ) and consequently λ∗ ∈ W (V ). We notice that for all v ∈ W (V ) Z 0

T

d ∗ hλ (t), v(t)i H dt = hλ∗t , vi L 2 (V ∗ ),L 2 (V ) + hvt , λ∗ i L 2 (V ∗ ),L 2 (V ) . dt

Hence, we may apply (21) and (22) and have 0 = h−λ∗t − νλ∗x x − y ∗ vλ∗x + C ? (C y ∗ − z), vi L 2 (V ∗ ),L 2 (V ) Z T d ∗ hλ (t), v(t)i H dt + hv(0), µ∗ i H + 0 dt

(22)

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VOLKWEIN

= hλ∗ (T ), v(T )i H − hλ∗ (0), v(0)i H + hv(0), µ∗ i H = hλ∗ (T ), v(T )i H + hµ∗ − λ∗ (0), v(0)i H . Choosing v appropriately we conclude that λ∗ (T ) = 0

and µ∗ = λ∗ (0).

Thus, λ∗ is the weak solution of the backward differential equation λ∗t = −νλ∗x x − y ∗ λ∗x + C ? (C y ∗ − z) in L 2 (V ∗ )

(23a)

with the initial condition at t = T λ∗ (T ) = 0.

(23b)

We infer from (23a) that λ∗ ∈ W (V ) holds. Let τ = T − t and ρ ∗ (τ ) = λ∗ (t) for t ∈ [0, T ]. Then problem (23) can be transformed into the forward differential equation ρτ∗ − νρx∗x − y ∗ ρx∗ = −C ? (C y ∗ − z) in L 2 (V ∗ )

(24a)

with the initial condition at τ = 0 ρ ∗ (0) = 0.

(24b)

Now we are going to derive an estimate for the Lagrange-multiplier λ∗ . If y ∗ = 0 holds we directly obtain kλ∗ k L 2 (V ) = kρ ∗ k L 2 (V ) ≤

1 ? kC (C y ∗ − z)k L 2 (V ∗ ) ν

(25)

from ρ ∗ (τ ) = λ∗ (t) and (24a). On the other hand, if y ∗ 6= 0 holds we set c◦ =

27 4 ∗ 4 c ky kC(H ) , 32ν 3 A

(26)

where the constant c A is given by Agmon’s inequality. Let w∗ be given by ρ ∗ (τ ) = w ∗ (τ ) exp(c◦ τ ). Then w∗ satisfies ν

kw∗ k2L 2 (V )

Z

T

− 0

Z

1 0

|y ∗ wx∗ w∗ | dx dτ

+ c◦ kw∗ k2L 2 (H ) ≤ kC ? (C y ∗ − z)k L 2 (V ∗ ) kw∗ k L 2 (V ) . Analogously as in the proof of Proposition 5 we derive that Z

T 0

Z

1 0

|y ∗ wx∗ w ∗ | dx dτ ≤

ν kw∗ k2L 2 (V ) + c◦ kw∗ k2L 2 (H ) 2

(27)

129

DISTRIBUTED CONTROL PROBLEMS

such that kw ∗ k L 2 (V ) ≤

2 ? kC (C y ∗ − z)k L 2 (V ∗ ) ν

follows. From ρx∗ (τ ) = λ∗x (t) and ρ ∗ (τ ) = w∗ (τ ) exp(c◦ τ ) we conclude that kλ∗ k L 2 (V ) ≤

2 exp(c◦ τ ) kC ? (C y ∗ − z)k L 2 (V ∗ ) . ν

(28)

Hence, combining (25) and (28) we have proved the next lemma. Proposition 7.

The Lagrange multiplier λ∗ satisfies λ∗ ∈ W (V ) and

kλ∗ k L 2 (V ) ≤

2 exp(c◦ T ) kC ? (C y ∗ − z)k L 2 (V ∗ ) , ν

where c◦ > 0 is given by (26). 3.3.

Second-order sufficient optimality condition

In this section we prove the second-order sufficient optimality condition for problem (P) for two choices of the operator C and for the Hilbert space H. Before we discuss the second-order sufficient optimality condition we prove an auxiliary lemma. Lemma 8. such that

Let (y, u) ∈ X and (v, q) ∈ ker e0 (y, u). Then there exists a constant cker > 0

kvk2W (V ) ≤ cker kqk2L 2 (Q ◦ ) holds. Proof:

If y = 0 then (v, q) ∈ ker e0 (y, u) leads to v(0) = 0 and Z

hvt , ϕi L 2 (V ∗ ),L 2 (V ) +

T 0

Z

1

νvx ϕx − (Bq)ϕ dx dt = 0

for all ϕ ∈ L 2 (V ).

0

In particular, the choice ϕ = v implies that kvk L 2 (V ) ≤

1 kqk L 2 (Q ◦ ) . ν

(29)

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VOLKWEIN

Thus, we conclude from (29) that Z kvt k L 2 (V ∗ ) =

T

Z

1

sup

kϕk L 2 (V ) =1 0

−νvx ϕx + (Bq)ϕ dx dt

0

≤ ν kvk L 2 (V ) + kqk L 2 (Q ◦ ) ≤ 2 kqk L 2 (Q ◦ ) .

(30)

Combining (29) and (30) the claim follows with cker = 4 + 1/ν 2 . On the other hand, if y 6= 0, then (v, q) ∈ ker e0 (y, u) implies v(0) = 0 and Z hvt , ϕi L 2 (V ∗ ),L 2 (V ) +

T 0

Z

1

νvx ϕx − yvϕx − (Bq)ϕ dx dt = 0

(31)

0

for all ϕ ∈ L 2 (V ). Let c◦ > 0 be given c◦ =

27 4 4 c kykC(H ), 32ν 3 A

where c A is determined by Agmon’s inequality. Then we derive as in the proof of Lemma 7 that kvk L 2 (V ) ≤

2 exp(c◦ T ) kqk L 2 (Q ◦ ) . ν

(32)

Using (31) and (32) we derive that kvt k L 2 (V ∗ ) ≤ (ν + kykC(H ) )kvk L 2 (V ) + kqk L 2 (Q ◦ ) , ¶ ¶ µµ 2 ≤ 2 + kykC(H ) exp(c◦ T ) + 1 kqk L 2 (Q ◦ ) . ν

(33)

Combining (32) and (33) the claim follows with µµ cker =

¶2 ¶ 2 4 2 + kykC(H ) exp(c◦ T ) + 1 + 2 exp(2c◦ T ). ν ν

2

Now we consider the second-order sufficient optimality condition, i.e. we give a sufficient criterion which implies that Assumption 1-d) holds. Proposition 9. Let the constants c E and c◦ be given by (3) and (26), respectively. a) If H = W (V ), C = I and ν exp(−c◦ T ) ky ∗ − zk2L 2 (H ) < 2c E hold, then Assumption 1-d) is satisfied. Moreover, L 00 (y ∗ , u ∗ , λ∗ , µ∗ ) is coercive on the whole space X .

131

DISTRIBUTED CONTROL PROBLEMS

b) Let H = L 2 (H ) and C be the injection of W (V ) into L 2 (H ). If σν exp(−c◦ T ) ky ∗ − zk2L 2 (H ) < 4cker c E is valid, then Assumption 1-d) holds. Proof: a) Using (3) we conclude that hL 00 (y ∗ , u ∗ , λ∗ , µ∗ )(v, q), (v, q)i X Z TZ 1 = kvk2W (V ) + σ kqk2L 2 (Q ◦ ) − λ∗x v 2 dx dt 0 0 Z T kλ∗ (t)kV kv(t)k H kv(t)k L ∞ dt ≥ kvk2W (V ) + σ kqk2L 2 (Q ◦ ) − 0

≥ kvk2W (V ) + σ kqk2L 2 (Q ◦ ) − kvkC(H ) kλ∗ k L 2 (V ) kvk L 2 (V ) ≥ (1 − c E kλ∗ k L 2 (V ) )kvk2W (V ) + σ kqk2L 2 (Q ◦ ) for all (v, q) ∈ X . Applying Proposition 7 leads to kλ∗ k L 2 (V ) ≤

2 exp(c◦ T ) ky ∗ − zk L 2 (H ) . ν

Setting ¶ µ 2 κ = min 1 − c E exp(c◦ T )ky ∗ − zk L 2 (H ) , σ > 0 ν we derive that hL 00 (y ∗ , u ∗ , λ∗ , µ∗ )(v, q), (v, q)i X ≥ κ k(v, q)k2X

for all (v, q) ∈ X.

b) Using Lemma 8 we obtain for all (v, q) ∈ ker e0 (y ∗ , u ∗ ) hL 00 (y ∗ , u ∗ , λ∗ , µ∗ )(v, q), (v, q)i X 2 ≥ kvk2L 2 (H ) + σ kqk2L 2 (Q ◦ ) − c E exp(c◦ T ) ky ∗ − zk L 2 (H ) kvk2W (V ) ν µ ¶ σ σ 2 ≥ kqk2L 2 (Q ◦ ) + − c E exp(c◦ T ) ky ∗ − zk L 2 (H ) kvk2W (V ) . 2 2cker ν Setting µ κ = min

2 σ σ − c E exp(c◦ T ) ky ∗ − zk L 2 (H ) , 2cker ν 2

¶ >0

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VOLKWEIN

we conclude that hL 00 (y ∗ , u ∗ , λ∗ , µ∗ )(v, q), (v, q)i X ≥ κ k(v, q)k2X 2

for all (v, q) ∈ ker e0 (y ∗ , u ∗ ).

Remark 4. For optimal control problems with cost functional of tracking type, smallness of the residuum ky ∗ − zk L 2 (H ) helps for the second-order sufficient optimality condition to hold. 4.

Numerical examples

For the numerical realization of (P) we have to discretize the optimal control problem. We utilize piecewise linear finite elements for the spatial discretization and the implicit Euler method for the time integration. The grid is chosen to be xi =

i for i = 0, . . . , n + 1 n+1

tj =

and

jT m

for j = 0, . . . , m − 1.

The state and control variable in the control problem is approximated by finite sums of time n satisfying ψi (x j ) = dependent modal coefficients multiplied by the finite elements {ψ}i=1 δi j for 1 ≤ i ≤ n and 0 ≤ j ≤ n + 1. To simplify the notation we consider here the case that B is just the identity. yn (t, x) =

n X

yi (t)ψ i (x),

i=1

u n (t, x) =

n X

u i (t)ψ i (x).

i=1

We introduce the tensor T = (((Ti, j,l ))) ∈ Rn×n×n

with Ti, j,l = (ψ j (ψ l )0 , ψ i ) L 2 ,

the mass matrix M = ((Mi, j )) ∈ Rn×n

with Mi, j = (ψ j , ψ i ) L 2 ,

the stiffness matrix A = ((Ai, j )) ∈ Rn×n

with Ai, j = (ψ j , ψ i ) H01 ,

the vectors of the time dependent modal coefficients yE(t) = (yi (t)) ∈ Rn ,

uE(t) = (u i (t)) ∈ Rn

and the vectors of the data φE = ((φ, ψ i ) L 2 ) ∈ Rn ,

¡ ¢ fE(t) = h f (t), ψ i i H −1 ,H01 ∈ Rn .

133

DISTRIBUTED CONTROL PROBLEMS

Then the Galerkin approximation of the optimal control problem (P) is given by 1 min Jn (Ey , uE) = 2

Z

T

(Ey (t) − 2Ez (t))T yE(t) + ζ (t) + α uE(t)T uE(t) dt

(34a)

0

subject to the n-dimensional system of ordinary differential equations M yE0 (t) + ν A yE(t) + (T : yE(t))Ey (t) = fE(t) + M uE(t) in (0, T )

(34b)

with the initial condition at t = 0 E M yE(0) = φ,

(34c)

where Ez (t) = ((ψ i , z(t)) L 2 ) ∈ Rn , ζ (t) = kz(t)k2L 2 , and the tensor product is defined by ÃÃ ((T : yE(t)))1≤i, j≤n =

n X

!! Ti, j,l yl (t)

l=1

. 1≤i, j≤n

To solve (34b) we applied the implicit Euler method. The programs were written in MATLAB version 5.3 executed on a DIGITAL Alpha 21264 computer. Run 1.

For the Burgers equation we chose the parameters T = 1, ν = 0.01, f = 0 and

φ=

 1  0

¸ 1 in 0, , 2 otherwise. µ

For n, m ∈ N the grid was given by xi =

i n

for i = 0, . . . , n

and

tj =

jT m

for j = 0, . . . , m − 1.

To solve (6) with u = 0 we apply Newton’s method at each time-step. The algorithm needed 0.25 seconds in case of n = m = 30. The numerical solution to (6) is shown in figure 1. Now we turn to the optimal control problem. Let H = L 2 (H ) and C be the injection of W (V ) into L 2 (H ). The control acted on the located support Q ◦ = (0, T ) × ( 14 , 34 ). As the desired state we took z(t) = φ in (0, T ). In view of figure 1 we can interpret this problem as determining u in such a way that it counteracts the uncontrolled dynamics which smoothes the discontinuity at x = 12 and transports it to the left as t increases. The regularization parameter was σ = 0.001. We solved the optimal control problem by applying

134

VOLKWEIN

Figure 1.

Solution to (6) for n = m = 30.

the augmented Lagrangian-SQP method with c = 4000. As start-up values we took y 0 = z, u 0 = 0 and λ0 = 0. We stopped the iterative method if °Ã !° ° L 0 (y n , u n , λn , µn ) ° ° c ° ° ° ° ° e(y n , u n )

≤ 10−10 . X ×Y

To solve the linear indefinite system (5) in each level of the iteration we use the MATLAB function gmres, which is an implementation of the Generalized Minimum Residual Method. As a preconditioner we took the LU-factorization of the matrix 

I  B = 0 P

0 σI 0

 P?  0 . 0

Here, the matrix P is the discretization of the heat operator yt − νyx x with homogeneous Dirichlet boundary conditions at x = 0 and x = 1, and the matrix P ? is the discretization of the adjoint heat operator −yt − νyx x with homogeneous Neumann boundary conditions at x = 0 and x = 1. The operator B is the constant part of the operator M(x n , ξ˜ n ) in (5) that does not change during the iterations. For numerical reasons it is wise to compute the LU-decomposition of the matrix B. We did computations using the complete and the incomplete factorization. Let us also refer to [22] where among other techniques the preconditioned conjugate gradient method was used to solve the SQP step for an optimal control problem for the solid fuel ignition model. For n = m = 30 Algorithm 1 needed 6 iterations and 15 seconds CPU-time if we decomposed B with the MATLAB function lu to compute the LU-decompositon, whereas the method stopped after

DISTRIBUTED CONTROL PROBLEMS

Figure 2.

135

Optimal state and optimal control for n = m = 30.

34 iterations and 172 seconds CPU-time, if we did not use any preconditioning. Furthermore, we utilized the MATLAB function luinc(B,1e-05) to compute an incomplete LU-factorization of the matrix B. Again 6 iteration were necessary to reach the stopping criteria, but the CPU-time reduced to 11 seconds. Instead of B we also tested the preconditioner 

I  C = 0 −1n

0 σI 0

 −1n  0 , 0

(35)

where 1n denotes the discretization of the Laplace operator 1. It turns out that the algorithm using the incomplete LU factorization needed also 6 iterations but the CPU time was 28 seconds. In figure 2 the discrete solution is shown. The value of the cost functional is 0.056, whereas the costs are 0.143 in the uncontrolled case. The reduction of the residuum ky(t) − z(t)k H is plotted in figure 3. In Table 1 the number of iteration and the used CPU-time for different values of c is presented. It was shown in [19] that if one starts sufficiently close to the optimal solution, then the augmented Lagrangian-SQP-method converges most fastly for c = 0. Furthermore, the second-order rate of convergence depends on the second-order sufficient optimality condition. Since the start-up values for the Lagrange-multiplier were zero, the operator L 00 (y 0 , u 0 , λ0 , µ0 ) was coercive on the kernel of the linearized constraints. This gives us a hint that we were close to the solution. Mesh-independence is an important feature of iterative approximation schemes for infinite dimensional problems. It asserts that the number of iterations to reach a specified approximation quality ε is independent of the mesh-size. We require the following

136

VOLKWEIN

Table 1.

Behavior of the algorithm for different values of c. c

Figure 3.

Iterations

CPU-time

0

5

9.5 seconds

1000

8

14.9 seconds

2000

7

13.2 seconds

3000

7

13.4 seconds

4000

6

11.0 seconds

5000

6

11.3 seconds

6000

6

11.4 seconds

7000

6

11.1 seconds

8000

6

11.3 seconds

9000

6

11.3 seconds

10000

6

11.5 seconds

L 2 -norms of the residuum for n = m = 30.

notation: °Ã !° ° L ¡ y ` , u ` , λ` , µ` ¢ ° ° c,h h¡ h h¢ h ° `h (ε) = min `0 | for ` ≥ `0 : ° ° ° ° eh yh` , u `h (

) <ε . X ×Y

We point out, that `h (ε) depend on the start-up values (yh0 , u 0h , λ0h , µ0h ) of the finite dimensional methods. In [33] it was proved that the augmented Lagrangian-SQP method is mesh-independent if one chooses appropriate approximation schemes. That means that for sufficiently small mesh-size there is at most a difference of one iteration step between

137

DISTRIBUTED CONTROL PROBLEMS Table 2. `h (ε) for different mesh-sizes and ε. m=n

10

20

30

40

50

60

70

ε = 10−0

1

1

1

1

1

1

1

10−2

3

2

2

2

2

2

2

ε = 10−4

4

4

4

4

4

4

4

ε = 10−6

4

4

4

4

4

5

5

ε = 10−8

5

5

5

5

5

5

5

6

6

6

6

6

6

6

ε=

ε=

10−10

the number of steps required by the finite dimensional methods for different mesh-sizes to converge within a given tolerance ε > 0. But our discretization scheme does not generally satisfy the assumptions stated in [33]. However, Table 2 illustrates that mesh-independence can be observed numerically. Run 2. Let us investigate a second example which shows that for c > 0 the augmented Lagrangian-SQP method is less sensitive with respect to the start-up values. We took T = 1, ν = 0.01, φ = 0. Obviously, the solution of (6) is zero. Let H = L 2 (H ) and C be the injection of W (V ) into L 2 (H ). By applying a control on the set Q ◦ = (0, T2 ) × ( 14 , 34 ) we wanted to track the dynamics to the desired state presented in figure 4. Let σ = 0.0175. We chose the same stopping criteria as in the previous example. x in (0, T ) and µ0 = 0. Then The start-up values were y 0 = z, u 0 = 0, λ0 (·, x) = 11 10

Figure 4.

Desired state for n = m = 30.

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VOLKWEIN

Figure 5.

Optimal state and optimal control for n = m = 30.

Figure 6.

L 2 -norms of the residuum for n = m = 30.

we have hL 00 (y 0 , u 0 , λ0 , µ0 )(v, q), (v, q)i X = −

1 kvk2L 2 (V ) + σ kqk2L 2 (Q ◦ ) 10

for all (v, q) ∈ ker e0 (y 0 , u 0 ). We can not expect that L 00 is coercive on the kernel of the linearized constraints. But the discretization of the second derivative of the augmented

DISTRIBUTED CONTROL PROBLEMS

139

Lagrangian with c = 30 is positive definite in the whole space. The smallest eigenvalue is 0.0962. As a preconditioner for the linear system (5) we used the same matrix B introduced in Run 1, where we again computed the incomplete LU-factorization of B. For c = 30 and n = m = 30 Algorithm 1 reached the stopping criteria after 17 iterations. The CPU-time was 22 seconds. In figure 5 the numerical solution for the state as well as for the control variable are shown. The reduction of the residuum ky(t) − z(t)kH is presented in figure 6. The value of the costs were reduced from 0.111 to 0.085. In case of c = 0 no convergence was observed. Taking the preconditioner C given by (35) the algorithm needed 22 iterations and 185 seconds.

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