First- and Second-Order Derivatives

Z

Ω

y(T)−y_d

2dx, J₂:U →R, J₂(u) := 1

2

k

X

k=1

γ_k Z T

0

|u_k(t)|²dt.

Since yn → y¯ but un * u¯ as n → ∞ we treat the functions J1 and J2 separately.

Note that nonlinear continuous functions are not neccessarily weakly continuous.

The function J₂ is convex and consequently weakly lower semicontinuous, see [45, Theorem 2.12]. I.e. we have

lim inf_n→∞ J₂(u_n)≥J₂(¯u) as u_n*u.¯ The following estimate finishes the proof:

J^∗= lim_n→∞J(y_n, u_n) = lim_n→∞J₁(y_n) + lim inf_n→∞ J₂(u_n)

=J₁(¯y) + lim inf_n→∞ J₂(u_n)≥J₁(¯y) +J₂(¯u) =J(¯y,u¯).

Obviously, to show optimality of (¯y,u¯) we did not need the specific structure of the cost functional. Lower semicontinuity of J would have been sufficient.

Remark 2.8. The cost functional is convex. If (SE) is linear, problem (P) is strictly convex with respect to u. But in case of a semilinear equation it might be non-convex.

Hence, there might exist several (local) optimal controls and further assumptions would be necessary to prove uniqueness; compare, e.g., [45].

2.4 First- and Second-Order Derivatives

For the numerical solution of the original problem (P) we will work with the equivalent nonlinear reduced problem (Pˆ). We turn to the first- and second-order derivatives of the reduced cost functional. For an introduction to the generalization of the notion of differentiability to Banach spaces we refer to [21, 45].

First, we name the derivatives of the cost functional J : W(0, T)×U → R. Further below, the derivatives of the reduced cost functional ˆJ are computed following a Lagrangian function based approach. The chain rule, see [45, Theorem 2.20], is applied. This requires the derivatives of the control-to-state operator. We restrict the dimension of the spatial domain tomΩ := 2 but we will point out when this restriction is actually needed.

Recall the Hilbert spaceU =L²(0, T;R^k). We identifyU⁰ withU viah·,·i_U⁰_,U =h·,·i_U. LetY₁ :=W(0, T) and U₁ :=L^∞(0, T;R^k).

Cost functional:

Lety, v, v₁, v₂∈W(0, T) and u, w, w₁, w₂∈U. The Fréchet derivatives ofJ are given by hJ_y(y, u), vi_Y⁰

1,Y1 =hy(T)−y_d, v(T)i_H, hJ_yy(y, u)v₂, v₁i_Y⁰

1,Y1 =hv₁(T), v₂(T)i_H, hJ_u(y, u), wi_U =^{Z T}

0 k

X

k=1

γ_ku_k(t)w_k(t) dt, hJ_uu(y, u)w₂, w₁i_U =^{Z T}

0 k

X

k=1

γ_kw_2k(t)w_1k(t) dt,

while the second-order mixed derivatives vanish. The linear mapping y ∈ W(0, T) 7→

y(T)∈H is continuous due to the embeddingW(0, T),→C([0, T];H).

The Riesz representation for J_u(y, u) is directly visible within the third of the above equations: Fort∈[0, T] a.e. it holds

J_u(y, u)(t) =D_γu(t) withD_γ := diag(γ₁, . . . , γ_k), (2.8) and thus

J_uu(y, u)w(t) =D_γw(t). (2.9) Remark 2.9. J is twice continuously Fréchet differentiable with Lipschitz continuous second-order derivative becauseJuu(y, u) and Jyy(y, u) are both independent of (y, u).

We continue by considering the control-to-state operator. To obtain the desired differen-tiability we require

(A3) N is twice differentiable with locally Lipschitz continuous second-order derivative.

Assumption (A3) implies local Lipschitz continuity of N⁰ and N by using the mean value theorem. Therefore assumption (A1) holds.

Remark 2.10. The second-order derivative N⁰⁰(y) = 6y of the function N(y) = y³ is obviously globally Lipschitz continuous.

The given control-to-state operator is differentiable as a mapping from L^s(0, T;R^k) to Y with s > m_Ω+ 1; compare [45, Chapter 5]. In advance of the derivative computation we give the following theorem, where we restrict ourselves toL^∞(0, T;R^k)⊃U_ad.

Theorem 2.11. Suppose that (A2)-(A3) hold. Then, the control-to-state operator is twice continuously Fréchet differentiable as a function fromL^∞(0, T;R^k) toY.

Proof. We have the continuous linear operatorB∞ from the proof of Theorem 2.4 so that Theorem 5.15 in [45] yields the claim.

By the chain rule it follows:

Corollary 2.12. With (A2)-(A3) holding the reduced cost functional Jˆ is twice con-tinuously Fréchet differentiable on L^∞(0, T;R^k).

Unfortunately, the control-to-state operator and hence the reduced cost functional are not differentiable from the Hilbert spaceU toY and toRrespectively. Here, we encounter

the two-norm discrepancy being well-known for occuring in optimal control problems go-verned by semilinear parabolic PDEs; see [24, 45].

In [24] the problem is overcome by using continuous extensions. Let u∈ L^∞(0, T;R^k) arbitrary but fixed. Motivated by the argumentation given in [24] the following will be shown:

• We can view y⁰(u) ∈ L(L^∞(0, T;R^k), Y) as continuous linear operator from U to W(0, T) so that its dual operator maps continuouslyW(0, T)⁰ toU⁰∼U.

• ˆJ⁰(u)∈L^∞(0, T;R^k)⁰ belongs to U⁰ ∼U.

• ˆJ⁰⁰(u) maps continuously U toU⁰∼U.

For the derivative computation we write (SE) elegantly as a nonlinear operator equa-tion ‘e(y, u) = 0’. We use the two abbreviations L²(V) := L²(0, T;V) and L²(V⁰) :=

L²(0, T;V⁰) so thatL²(V⁰)⁰ =L²(V) holds. We introduce the required mappings:

• DefineF ∈L²(V⁰) by

hF, ϕi_L2(V⁰),L²(V)=^{Z T}

0

hF(t), ϕ(t)i_V⁰_,V dt:=^{Z T}

0

Z

Ω

f(t,x)ϕ(t,x) dxdt for ϕ∈L²(V).

• The continuous linear operator A:L²(V)→L²(V⁰), hAy, ϕi_L2(V⁰),L²(V)=^{Z T}

0

h Ay(t), ϕ(t)i_V⁰_,V dt:=^{Z T}

0

a(y(t), ϕ(t)) dt for y, ϕ∈L²(V), with the symmetric and bounded bilinear form a:V ×V →R,

a(ϕ1, ϕ2) :=^Z

Ω

∇ϕ₁(x)· ∇ϕ₂(x) dx+q Z

Γ

ϕ1(x)ϕ2(x) dx forϕ1, ϕ2∈V.

The boundedness of a is transferred to A. Therefore, the operator A is indeed continuous; compare [21, p. 90] or see also [45].

• The continuous linear operator B:U →L²(0, T;V⁰), hBu, ϕi_L2(V⁰),L²(V)=^{Z T}

0

h Bu(t), ϕ(t)i_V⁰_,V dt :=^{Z T}

0 k

X

k=1

u_k(t)^Z

Γ

χ_k(x)ϕ(t,x) dxdt foru∈U, ϕ∈L²(V).

• The nonlinear operatorN :L^∞(Q)→L²(V⁰), hN(y), ϕi_L2(V⁰),L²(V)=^{Z T}

0

h N(y)(t), ϕ(t)i_V⁰_,V dt :=^{Z T}

0

Z

Ω

N y(t,x)ϕ(t,x) dxdt fory∈L^∞(Q), ϕ∈L²(V).

In addition,y ∈W(0, T) implies yt ∈L²(V⁰). Hence, the weak formulation (2.3) defines

Recall that a linear and bounded operator is Fréchet differentiable and that the deriva-tive is given by the operator itself; see [45]. Thus, concerning differentiability the only delicate term in the above definition is the nonlinear operatorN.

Lemma 2.13. With(A3)holding the operatorN :L^∞(Q)→L²(V⁰)is twice continuously Fréchet differentiable. The action of the derivatives reads

hN⁰(y)v, ϕi_L2(V⁰),L²(V) =^{Z T}

Proof. First, one has to verify that the expressions above represent the Fréchet derivatives.

This can be shown by using the estimation techniques from [45, Sections 4.3, 4.9] where Tröltzsch considers Nemytskii operators and their first- and second-order derivatives as mappings fromL^∞(Q) to L^∞(Q).

We briefly show that the derivatives can be continuously extended. Let y ∈ L^∞(Q).

Local Lipschitz continuity of N⁰ and N⁰⁰ implies N⁰(y(·,·)), N⁰⁰(y(·,·))∈L^∞(Q). This is of Hölders inequality; see [21, Lemma 1.13]. Boundedness is given due to W(0, T) ,→ C([0, T];H) andL²(V),→L²(0, T;L^q(Ω)) for 2≤q≤6. The latter embedding is true for m_Ω≤3 by the Sobolev embedding theorem; see [21, Theorem 1.14].

Now, we can name the derivatives of theoperator e:

e⁰(y, u)(v, w) = c_pv_t+Av+N⁰(y)v v(0)

!

| {z }

=ey(y,u)v

+ −Bw

0

!

| {z }

=eu(y,u)w

,

e⁰⁰(y, u)((v1, w1),(v2, w2)) = N⁰⁰(y)(v₁, v₂) 0

! , fory, v, v₁, v₂∈Y and u, w, w₁, w₂ ∈U.

Remark 2.14. Let (y, u) ∈ Y ×U. The above formula for ey(y, u)v and Lemma 2.13 show that ey(y, u) can be viewed as a continuous linear operator from W(0, T) to Z = L²(V⁰)×H. So, its dual operator maps continuouslyZ⁰ =L²(V)×H toW(0, T)⁰. Control-to-state operator:

From the chain rule and Theorem 2.11, it follows that the equation e(y(u), u) = 0

can be differentiated in a directionw∈L^∞(0, T;R^k). This yields

e_y(y(u), u)y⁰(u)w+e_u(y(u), u)w= 0. (2.13) Thus, the sensitivityv:=y⁰(u)w is given by the solution to the

linearized state equation

ey(y(u), u)v=−e_u(y(u), u)w.

Letyu :=y(u). Written in expanded form the linearized state equation reads c_pv_t+Av+N⁰(y_u)v

v(0)

!

= Bw

0

! . This is the weak formulation of

(LSE)















c_pv_t−∆v+N⁰(y_u(·,·))v= 0 inQ,

∂v

∂ν +qv=

k

X

k=1

w_kχ_k on Σ, v(0) = 0 in Ω. We investigate the solvability of (LSE):

1. w∈U ⇒v∈W(0, T):

The operator B is linear and continuous. From (A2)-(A3) we can deduce that the function (t,x) 7→ N⁰(yu(t,x)) ≥ 0 belongs to L^∞(Q). Hence, the linearized state equation has a continuous linear solution operator w 7→ v from U to W(0, T); see [11, Chapter XVIII].

2. w∈L^∞(0, T;R^k)⇒v∈Y:

If the controlwbelongs toL^∞(0, T;R^k), we even obtainv∈C( ¯Q) and hencey∈Y; see [45, Chapter 5].

Remark 2.15. (1) The above point 2 would allow to apply the implicit function theo-rem, see [21, Theorem 1.41], to prove Theorem 2.11, i.e. differentiability of the control-to-state operator.

(2) Let u∈L^∞(0, T;R^k). Point 1 above justifies to write

y⁰(u) =−e_y(y(u), u)⁻¹e_u(y(u), u)∈ L(U, W(0, T)). (2.14) Consequently, the dual operator y⁰(u)^∗ maps continuously W(0, T)⁰ to U. Point 2 above yields y⁰(u)w∈Y, ifw∈L^∞(0, T;R^k).

(3) The operatore⁰(y, u) = (e_y(y, u), e_u(y, u)) is surjective for all (y, u)∈Y ×U because the operator e_y(y, u) is bijective. In order to see this, we consider the linearized state equation with an arbitrary right-hand side. Surjectivity of ey(y, u) follows if and only if for all (g, v₀) ∈Z there exists v ∈Y such that e_y(y, u)v = (g, v₀). The reference from point 1 above yields the existence of a weak solution v ∈ W(0, T) which is even unique. By a bootstrap argument the regularity of v can be improved such that v∈C( ¯Q) is satisfied.

Hence, a so-called regular point condition is fulfilled and provides the existence of a Lagrange muliplier p= (p1, p2)∈Z⁰ associated with (SE) in the context of Karush-Kuhn-Tucker theory; see [32, Theorem 4.1]. By variational arguments it follows that p_1,t belongs to L²(V⁰). Thus, we even havep₁∈W(0, T).

(4) Differentiating equation (2.13) with w₁ := w once again in another direction w₂ ∈ L^∞(0, T;R^k) yields an equation for the second-order derivative y⁰⁰(u)(w₁, w₂). We will not have to compute this derivative. But we need thaty⁰⁰(u) can also be applied to elements w₁, w₂ ∈ U. Therefore, let us name a representation formula which is also stated in [45, Theorem 5.16]. The application v:=y⁰⁰(u)(w₁, w₂) is the solution

to 









c_pv_t−∆v+N⁰(y_u(·,·))v=−N⁰⁰(y_u(·,·))v₁v₂ inQ,

∂v

∂ν +qv= 0 on Σ,

v(0) = 0 in Ω,

(2.15)

withv_i =y⁰(u)w_i,i= 1,2. Forv₁, v₂ ∈W(0, T) we obtainN⁰⁰(y_u(·,·))v₁v₂∈L²(Q):

The embedding W(0, T),→L⁴(0, T;L⁴(Ω))∼L⁴(Q) formΩ= 2 gives Z T

0

Z

Ω

v1(t,x)v2(t,x)²dxdt≤ Z T

0

v1(t)²_L4(Ω)

v2(t)²_L4(Ω)dt

≤v₁

2 L⁴(Q)

v₂

2 L⁴(Q).

Thus, the reference from point 1 above ensures that equation (2.15) has a unique weak solutionv∈W(0, T). Let us mention that the use of bootstrapping even yields v ∈L^∞(0, T;H²(Ω))∩H¹(0, T;H).

Reduced cost functional:

Now, we can compute ˆJ⁰(u) for any u∈L^∞(0, T;R^k) using a Lagrangian function based approach.

We introduce the Lagrange functionL:Y ×L^∞(0, T;R^k)×Z⁰ →Rby L(y, u, p) :=J(y, u) +hp, e(y, u)i_Z⁰_,Z

=J(y, u) +he¹(y, u), p1i_L2(V⁰),L²(V)+hp₂, e²(y, u)i_H, wherep= (p1, p2)∈Z⁰ =L²(V)×H.

For anyu∈L^∞(0, T;R^k) andp∈Z⁰ we have

Jˆ(u) =J(y(u), u) =J(y(u), u) +hp, e(y(u), u)i_Z⁰_,Z =L(y(u), u, p), becausee(y(u), u) = 0 holds. The first-order derivative of ˆJ thus reads

hJˆ⁰(u), w₁i_U⁰

1,U1 =hL_u(y(u), u, p), w₁i_U⁰

1,U1+hL_y(y(u), u, p), y⁰(u)w₁i_Y⁰_,Y (2.16) forw1 ∈L^∞(0, T;R^k).

The left termLu(y(u), u, p) is given by hL_u(y(u), u, p), w₁i_U⁰

1,U1 =hJ_u(y(u), u), w₁i_U+he¹_u(y(u), u)w₁, p₁i_L2(V⁰),L²(V)

=hJ_u(y(u), u), w1i_U+h−Bw₁, p1i_L2(V⁰),L²(V)

=hJ_u(y(u), u)− B^∗p1, w1i_U,

with the dual operatorB^∗ of B. Thus, Lu(y(u), u, p) can be identified with the element Lu(y(u), u, p) =Ju(y(u), u)− B^∗p1 ∈U. (2.17) We determine the dual operatorB^∗ :L²(V)→U ofB, satisfying

hBu, ϕi_L2(V⁰),L²(V)=hu,B^∗ϕi_U for all (u, ϕ)∈U ×L²(V). Actually, it can be directly read off from the definition ofB. We obtain

(B^∗ϕ) (t) =





 R

Γχ₁(x)ϕ(t,x) dx ...

R

Γχ_k(x)ϕ(t,x) dx





 for all ϕ∈L²(V), a.e. in [0, T]. (2.18) For the second term in (2.16) we introduce the adjoint statep(u) ∈Z⁰ associated with the controlu: It is given by the solution to

L_y(y(u), u, p(u)) = 0. Letv∈Y. We have

hL_y(y(u), u, p(u)), vi_Y⁰_,Y =hJ_y(y(u), u), vi_Y⁰

1,Y1+hp(u), ey(y(u), u)vi_Z⁰_,Z

=hJ_y(y(u), u) +ey(y(u), u)^∗p(u), viY₁⁰,Y1. (2.19) The second equality holds since e_y(y(u), u) maps from W(0, T) to Z, see Remark 2.14.

This shows thatLy(y(u), u, p(u)) belongs toW(0, T)⁰. We can deduce ˆJ⁰(u)∈U⁰ ∼U via equation (2.16) and equation (2.17) withy⁰(u)∈ L(U, W(0, T)).

Let us determine the adjoint statep(u) = p(u)1, p(u)2

. For clarity we writep1 :=p(u)1

and p₂ := p(u)2. The adjoint state can be viewed as a Lagrange multiplier associated with (SE). Hence, its existence withp1 ∈W(0, T) follows from Remark 2.15(3). Equation (2.19) says that the adjoint state is given by the solution to the

adjoint equation

e_y(y(u), u)^∗p(u) =−J_y(y(u), u).

Letyu :=y(u). Written in variational form, the adjoint equation reads he¹_y(yu, u)v, p1i_L2(V⁰),L²(V)+he²_y(yu, u)v, p2i_H =−hJ_y(yu, u), vi_Y⁰

1,Y1 for all v∈Y1. Expandinge¹_y(yu, u), e²_y(yu, u) and insertingJy(yu, u) gives

hc_pv_t+Av+N⁰(y_u)v, p₁i_L2(V⁰),L²(V)+hp₂, v(0)i_H =−hy_u(T)−y_d, v(T)i_H. The formula of integration by parts applied on the termhc_pv_t, p₁i_L2(V⁰),L²(V) yields

hc_pp₁(T), v(T)i_H − hc_pp₁(0), v(0)i_H

+h−c_pp1,t+Ap₁+N⁰(yu)p1, vi_L2(V⁰),L²(V)

+hp₂, v(0)i_H =−hy_u(T)−yd, v(T)i_H.

Notice that the equalityA =A^∗ was used, which follows from symmetry of the bilinear forma.

Recall the space C_c^∞((0, T);V) which consists of all functions in C^∞((0, T);V) with compact support in (0, T). The space C_c^∞((0, T);V) ⊂ W(0, T) is dense in L²(V); see [21, p. 80]. Therefore, the above equation is equivalent to

h−c_pp_1,t+Ap₁+N⁰(y_u)p₁, vi_L2(V⁰),L²(V)= 0 for all v∈L²(0, T;V), c_pp₁(T) =−(y_u(T)−y_d) ∈H, p₂=c_pp₁(0).

This is the weak formulation of

(AE1)











−c_pp1,t−∆p1+N⁰(yu(·,·))p1 = 0 inQ,

∂p₁

∂ν +qp₁ = 0 on Σ,

cpp1(T) =−(yu(T)−yd) in Ω.

We already know thatp1 ∈W(0, T) exists and Lemma 3.17 in [45] provides unique exis-tence of a weak solution to (AE1). From y_u(T)−y_d∈ C(¯Ω), it even follows p₁ ∈C( ¯Q).

This impliesp1 ∈Y; see [45, p. 279]. Hence, the unique adjoint state associated with uis given byp(u) = (p1, cpp1(0)).

We insert the adjoint statep(u) into equation (2.16). Thus, the derivative ˆJ⁰(u) is given by formula (2.17) and only the first component of p(u) is needed. This is why we set p_u:=p(u)1. In the following we will refer to only this first component as the adjoint state

associated with u. Since we identified ˆJ⁰(u) ∈ U⁰ with an element in U we refer to it as the gradient of the reduced cost functional ˆJ atu∈L^∞(0, T;R^k).

We derived the following computation scheme:

Computation of the representation of ˆJ⁰(u) in U:

Require: u∈L^∞(0, T;R^k);

1. Solve (SE) to get yu=y(u);

2. Solve (AE1) to getp_u =p(u)1; 3. Insert u and p_u into

Jˆ⁰(u)(t) = Dγu(t) − (B^∗pu) (t)

=





 γ₁u₁(t)

...

γ_ku_k(t)





 −





 R

Γχ₁(x)pu(t,x) dx ...

R

Γχ_k(x)p_u(t,x) dx





; (2.20)

Let us turn to the second-order derivative. We differentiate equation (2.16) in the directionw₂∈L^∞(0, T;R^k), omit the mixed derivatives which vanish for the given problem and use point (4) from Remark 2.15:

hJˆ⁰⁰(u)w₂, w₁i_U⁰

1,U1 =hL_uu(y(u), u, p)w₂, w₁i_U⁰

1,U1

+hL_yy(y(u), u, p)y⁰(u)w2, y⁰(u)w1i_Y⁰_,Y +hL_y(y(u), u, p), y⁰⁰(u)(w1, w2)i_Y⁰

1,Y1.

By inserting the adjoint statep=p(u), satisfying L_y(y(u), u, p(u)) = 0, we obtain hJˆ⁰⁰(u)w2, w1i_U⁰

1,U1 =hL_uu(y(u), u, p(u))w2, w1i_U⁰

1,U1

+hL_yy(y(u), u, p(u))y⁰(u)w₂, y⁰(u)w₁i_Y⁰_,Y. (2.21) We verify that the controlsw1, w2 in equation (2.21) can also belong toU:

The first termL_uu(y(u), u, p(u)) equalsJ_uu(y(u), u) ∈U⁰ ∼U since the state equation is linear inu; compare equation (2.17). This gives

L_uu(y(u), u, p(u))w(t) =D_γw(t) forw∈U,a.e. in [0, T]. (2.22) The second term in (2.21) requires some effort. For anyv, ϕ∈Y we have

hL_yy(y(u), u, p(u))v, ϕiY⁰,Y =hJ_yy(y(u), u)v, ϕiY₁⁰,Y1+he¹_yy(y(u), u)(v, ϕ), p(u)1i_L2(V⁰),L²(V)

=hv(T), ϕ(T)i_H +hN⁰⁰(y(u))(v, ϕ), p(u)1i_L2(V⁰),L²(V).

Hence,v andϕcan also belong toW(0, T) and we can viewL_yy(y(u), u, p(u)) as mapping from W(0, T) to W(0, T)⁰. Consequently, the controls w₁, w₂ in (2.21) can belong to U and we have

hL_yy(y(u), u, p(u))y⁰(u)w₂, y⁰(u)w₁i_Y⁰

1,Y1 =hy⁰(u)^∗L_yy(y(u), u, p(u))y⁰(u)w₂, w₁i_U.

Letw∈U. The computation ofy⁰(u)^∗Lyy(y(u), u, p(u))y⁰(u)wneeds to be done in several steps. We define

v:=y⁰(u)w^(2.14)= −e_y(y(u), u)⁻¹eu(y(u), u)w, h:=Lyy(y(u), u, p)v.

Thus, we obtain

y⁰(u)^∗Lyy(y(u), u, p(u))y⁰(u)w=y⁰(u)^∗h.

Withyu :=y(u) andpu:=p(u)1 this means:

1. The sensitivityv=y⁰(u)w is the solution to (LSE).

2. The application ofh is given by hh, ϕi_Y⁰

1,Y1 =hv(T), ϕ(T)i_H +hN⁰⁰(yu)(v, ϕ), pui_L2(V⁰),L²(V). 3. The formula fory⁰(u)^∗h reads

y⁰(u)^∗h=−e_u(y(u), u)^∗ey(y(u), u)^−∗h.

Hence, this requires one adjoint equation solve. We define p := −e_y(y(u), u)^−∗h. This gives the adjoint equation with Jy(y(u), u) replaced byh. Letp1∈W(0, T) be the unique weak solution to

(AE2)











−c_pp_1,t−∆p₁+N⁰(y_u(·,·))p₁=−N⁰⁰(y_u(·,·))vp_u inQ,

∂p1

∂ν +qp1= 0 on Σ,

c_pp₁(T) =−v(T) in Ω.

Thus,pis given byp= (p1, cpp1(0)). Note thatw∈L^∞(0, T;R^k) first impliesv∈Y and hence p₁ ∈Y. Consequently,

y⁰(u)^∗h=eu(y(u), u)^∗h=−B^∗p1.

We summarize the computation of the application of the Hessian ˆJ⁰⁰(u),u∈L^∞(0, T;R^k), on a vectorw∈U:

Computation of Hessian-vector products ˆJ⁰⁰(u)w:

Require: w∈U,y_u =y(u) andp_u=p(u)1 (with u∈L^∞(0, T;R^k));

1. Solve (LSE) to get v; 2. Solve (AE2) to getp1; 3. Insert w andp1 into

( ˆJ⁰⁰(u)w)(t) = D_γw(t) − (B^∗p₁)(t)

=







γ₁w₁(t) ...

γ_kw_k(t)





 −





 R

Γχ₁(x)p₁(t,x) dx ...

R

Γχ_k(x)p₁(t,x) dx





; (2.23)

Remark 2.16. Consider the linear case, i.e. let N be a linear function so that N⁰ is constant and N⁰⁰ equals zero. Then, via (SE) and (AE1), ˆJ⁰(·) is linear and, via (LSE) and (AE2), ˆJ⁰⁰(·) is constant as expected.

Im Dokument Trust Region POD for Optimal Boundary Control of a Semilinear Heat Equation (Seite 14-24)