Weierstrass Institute forApplied Analysis and Stochastics
PDE-constrained control, lecture 7
– Parabolic control problems and applications
Dietmar Hömberg
Mohrenstrasse 39 · 10117 Berlin · Germany · Tel. +49 30 20372 0 · www.wias-berlin.de
November 19, 2012
Outline
• A simplified phase transition model
• Optimal control of hot-rolling / SQP–methods
· November 19, 2012 · Page 2 (33)
A process route for dual phase steels
What happens
• cooling of steel
• leads to phase transition
• phase growth described
by ODE for volume
fraction of ferrite f
• coupling with heat
equation for
temperature θ
• coupling term: latent
heat
· November 19, 2012 · Page 3 (33)
cost functional
J(θ, f, u) =γ1
2
∫Ω
(f(x, T )−fd(x))2dx+γ2
2
T∫0
∫Ω
(θ−θd)2dxdt+γ3
2
T∫0
∫Ω
u2dxdt .
state equationsθt −∆θ = αu+ ft , in Q (1)
∂θ
∂ν= 0 , in ∂Ω× (0, T ) (2)
θ(0) = θ0 , in Ω (3)
ft = G(θ, f) , in Q (4)
f(0) = 0 , in Ω . (5)
(CP) Minimize J(θ, f, u)
subject to (1)–(5)
and control constraint u ∈ Uad· November 19, 2012 · Page 4 (33)
Linearization
• First, one has to show that
S : U = L2(Q)→ H1,1(Q)×W 1,∞(0, T ;L∞(Ω)), u 7→ (θ, z)
is well-defined• Existence of an optimal control: . . . as in elliptic case• F- differentiability of the solution operator S:
Lemma:
The solution operator S : L2(Q)→ H1,1(Q)×H1(0, T ; L2(Ω)) is F -diffb.
and S′h =: (ψ,ϕ) satisfies
ψt −∆ψ = ϕt + αh in Q (6)
∂ψ
∂ν= 0 in ∂Ω× (0, T ) (7)
ψ(0) = 0 in Ω (8)
ϕt = Gθ(θ, f)ψ +Gf (θ, f)ϕ in Q (9)
ϕ(0) = 0 in Ω . (10)
· November 19, 2012 · Page 5 (33)
Idea of proof
• Defineqh := θh − θ − ψrh := fh − f − ϕ
with (θh, fh) = S(u+ h) solution corresp. to u+ h
• (qh, rh) satisfy qht −∆qh = rht in Q (11)
∂qh
∂ν= 0 in ∂Ω× (0, T ) (12)
qh(0) = 0 in Ω (13)
rht = G(h) in Q (14)
rh(0) = 0 (15)
with G(h) = G(θh, fh)−G(θ, f)−Gθψ −Gfϕ . (16)
• show that qh = o(‖h‖) and qh = o(‖h‖)
· November 19, 2012 · Page 6 (33)
Derivation of adjoint equations – I
• Lagrangian:
L(θ, f ; η, ζ;u) =γ1
2
∫Ω
(f(x, T )− fd(x))2dx+γ2
2
∫ T
0
∫Ω
[θ − θn]2 dxdt
+γ3
2
∫ T
0
u2dt−∫ T
0
∫Ω
θtη dxdt−∫ T
0
∫Ω
∇θ∇η dxdt
+
∫ T
0
∫Ω
αuη dxdt+
∫ T
0
∫ Ω
G(θ, f)ηdx−∫ T
0
∫Ω
ftζdxt
+
∫ T
0
∫Ω
G(θ, f)ζ dxdt .
• we have to compute
Lθ(θ, f ; η, ζ; u)θ = 0 (17)
Lf (θ, f ; η, ζ; u)f = 0 . (18)
• Assume that θ and f are suff. smooth with θ(0) = f(0) = 0.
· November 19, 2012 · Page 7 (33)
Derivation of adjoint equations – II
• Evaluation of (17) yields:
0 = γ2
∫ T
0
∫Ω
(θ − θd)θ dxdt+
∫ T
0
∫Ω
ηtθ dxdt−∫
Ω
θη∣∣∣T0dxdt
−∫ T
0
∫Ω
∇θ∇η dxdt+
∫ T
0
∫Ω
fθ(θ, f)θζ dxdt
+
∫ T
0
∫Ω
fθ(θ, f)θη dxdt
( η(T ) = 0 and∂η
∂ν= 0 yield)
=
∫ T
0
∫Ω
(γ2(θ − θd) + ηt + ∆η + fθ(θ, f)ζ + fθ(θ, f)η
)θ dxdt
i.e., we have − ηt −∆η = γ2(θ − θd) + fθ(θ, f)(ζ + η) in Q (19)
∂η
∂ν= 0 in ∂Ω× (0, T ) (20)
η(T ) = 0 . (21)
· November 19, 2012 · Page 8 (33)
Derivation of adjoint equations – III
• Evaluating (18) gives
0 = γ1
∫Ω
(f(x, T )− fd(x))f(x, T )dx+
∫ T
0
∫Ω
Gf (θ, f)zη dxdt
+
∫ T
0
∫Ω
ζtfdxdt−∫
Ω
fζ∣∣∣T0
+
∫ T
0
∫Ω
Gf (θ, f)ζ dxdt
f(0)=0=
∫Ω
(f(x, T )− fd(x)− ζ(T )) f(T )dx
+
∫ T
0
∫Ω
(ζt +Gf (θ, f)(η + ζ)
)f dxdt
resulting in
− ζt = Gf (θ, f)(η + ζ) in Q (22)
ζ(T ) = f(x, T )− fd(x) in Ω . (23)
· November 19, 2012 · Page 9 (33)
The result
• Variational inequality
L(θ, f ; η, ζ; u)(u− u) ≥ 0 i.e.
γ3
∫ T
0
∫Ω
u(u− u)dxdt+
∫ T
0
∫Ω
αη(u− u)dxdt
=
∫ T
0
∫Ω
(γ3u+ αη)(u− u)dxdt ≥ 0 für alle Uad .(24)
• TheoremLet u an optimal control and (θ, f) corresponding state. Then there exist adjoint
variables (η, ζ), such that the adjoint system (19)–(21) and (22)–(23) and the
variational inequality (24) are satisfied.
· November 19, 2012 · Page 10 (33)
Box Constraints – I
We assume
Uad = u∣∣ ξ0(x, t) ≤ u(x, t) ≤ ξ1(x, t) a.e. in Q (25)
with ξ0, ξ1 ∈ L∞(Q), i.e. so-called box-constraints.
Then Uad ⊂ L2(Q) closed and convex and one can show that (24)is satisfied also
pointwise a.e., i.e.,
(αη(x, t) + γ3u(x, t))(ξ − u(x, t)) ≥ 0 für alle ξ ∈ [ξ0(x, t), ξ1(x)] for a.a. (x, t) ∈ Q . (26)
⇒ (αη(x, t) + γ3u(x, t))u(x, t) ≤ (αη(x, t) + γ3u(x, t))ξ for all ξ ∈[ξ0(x, t), ξ1(x, t)].
Corollary:
u(x, t) =
ξ0(x, t) if αη(x, t) + γ3u(x, t) > 0
∈ [ξ0(x, t), ξ1(x, t)] , if αη + γ3u(x, t) = 0
ξ1(x, t) , if αη(x, t) + γ3u(x, t) < 0 .
· November 19, 2012 · Page 11 (33)
Bang-Bang control
Remark: If γ3 = 0 there holds
u(x, t) =
ξ0(x, t) , if αη > 0
ξ1(x, t) , if αη < 0
for αη = 0 no conclusion possible.
If |αη(x, t)| 6= 0 a.e. in Ω, then u(x, t) only takes values at upper or lower bound of
Uad.
This behaviour is called bang-bang control.
· November 19, 2012 · Page 12 (33)
Box Constraints – II
TheoremLet u the solution to (CP) subject to box-constraints, i.e.
Uad = u ∈ U | ξ0(x, t) ≤ u(x, t) ≤ ξ1(x, t) a.e. in x ∈ Q
and γ3 > 0, then
u(x, t) = P[ξ0(x,t),ξ1(x,t)]
− αγ3η(x, t)
for a.a. x ∈ Ω.
Here,
P[a,b] : R→ R , x 7−→ minb,maxa, x
is the projection of R onto the interval [a, b].
· November 19, 2012 · Page 13 (33)
Pilot hot-rolling mill at IMF Freiberg
Identification of heat transfer coefficient
· November 19, 2012 · Page 14 (33)
Optimal control problem (P)
min J(θ, f, u) =α1
2
∫Ω
(f(x, T )− fd(x))2 dx+α2
2
∫∫Q
(θ − θd)2dx dt+α3
2
T∫0
u2dt
s.t. (θ, f, u) satisfies (27) and ua ≤ u ≤ ub, t ∈ (0, T ) a.e.
ft = G(θ, f), in Q = Ω× (0, T ) (27a)
f(0) = 0, in Ω (27b)
ρcθt −∇ · (k(x)∇θ) = ρLft, in Q (27c)
−k ∂θ∂n
= u(t)(θ − θw), on ΣR = Γ× (0, T ) (27d)
− k ∂θ∂n
= 0, on ΣD = (∂Ω \ Γ)× (0, T ) (27e)
θ(0) = θ0, in Ω (27f)
· November 19, 2012 · Page 15 (33)
Assumptions
(A1) Ω ⊂ Rd, d = 2, 3 denotes a bounded domain with Lipschitz boundary ∂Ω
(A2) 0 < k1 ≤ k(x) ≤ k2 a.e. in Ω
(A3) G = G(t, θ, f) : (0, T )× R2 satisfies Caratheodory condition, is C2 with
respect to θ and f for almost all (x, t) ∈ Q and derivatives of G w.r.t. (θ, f) up
to order two are uniformly Lipschitz on bounded sets
(A4) θw ∈ L∞(ΣR), θ0 ∈ Cγ(Ω) for some γ > 0 and θd ∈ L∞(Q)
(A5) fd ∈ L∞(Ω), 0 ≤ fd ≤ 1 a.e. in Ω
· November 19, 2012 · Page 16 (33)
A useful regularity result
Theorem 1 Let h ∈ Lp(Q), g ∈ Lp(ΣR) and θ0 ∈ Cδ(Ω) for some δ > 0, then for
every u ∈ Uad the solution θ to
ρcpθt −∇ · (k∇θ) = h in Q
k∂θ
∂n+ u(t)θ = g, on ΣR
∂θ
∂n= 0, on ΣD
θ(0) = θ0, in Ω.
belongs to W 1,1p (Q) = W 1,p(0, T ;Lp(Ω)) ∩ Lp(0, T,W 1,p(Ω)). If p > d, there
exists γ > 0 such that θ ∈ Cγ(Q) and there holds
‖θ‖W 1,1p (Q) + ‖θ‖Cγ(Q) ≤ c(‖h‖Lp(Q) + ‖g‖Lp(ΣR))
(see, e.g., S. Monniaux: Maximal regularity and applications to partial differential equations , in: E. Emmrich
and P. Wittbold (Eds.) Analytical and Numerical Aspects of Partial Differential Equations: Notes of a Lecture
Series, de Gruyter, Berlin, 2009)
· November 19, 2012 · Page 17 (33)
Solution to the state system
• Theorem 2 Let (A1)-(A5) be satisfied. Then, the state system (27) for every
control u ∈ Uad admits a unique solution
(θ, f) ∈W 1,1p (Q) ∩ C(Q)×W 1,∞(0, T ;L∞(Ω)).
• Theorem 3
Let Assumption (A1)-(A4) be satisfied. Then there exists at least one solution of
the optimal control problem (P).
• Control to state mapping
we define
Y = W 1,1p (Q) ∩ C(Q)
and introduce
S = (Sθ, Sf ) : L∞(0, T )→ Y ×W 1,p(0, T ;Lp(Ω)), 1 ≤ p <∞,
· November 19, 2012 · Page 18 (33)
Theorem 4 (Differentiability of solution operator)
Let Assumptions (A1)-(A4) be satisfied. Then, the solution operator S is twice
Frechét-differentiable from L∞(0, T ) to Y ×W 1,p(0, T ;Lp(Ω)), 1 ≤ p <∞.
The first derivative (θh, fh) = S′(u)h = (S′θ(u)h, S′f (u)h) at point
u ∈ L∞(0, T ) in direction h ∈ L∞(0, T ) is given by the solution of
fh,t = Gθ(θ, f)θh +Gf (θ, f)fh, in Q (28a)
fh(0) = 0, in Ω (28b)
ρcθh,t −∇ · (k∇θh) = ρL(fh)t, in Q (28c)
−k∂θh∂n
= u(t)θh + h(t)(θ − θw), on ΣR (28d)
−k∂θh∂n
= 0, on ΣD (28e)
θh(0) = 0, in Ω, (28f)
with (θ, f) = S(u).
· November 19, 2012 · Page 19 (33)
Theorem 4 (continued)
Furthermore, (θh1h2, fh1h2
) = S′′(u)[h1, h2] is the solution of
fh1h2,t = Gθ(θ, f)θh1h2 +Gf (θ, f)fh1h2 , in Q (29a)
+G′′(θ, f)[(θh1, fh1
), (θh2, fh2
)]
fh1h2(0) = 0, in Ω (29b)
ρcθh1h2,t −∇ · (k∇θh1h2) = ρLfh1h2,t in Q (29c)
−k∂θh1h2
∂n= u(t)θh1h2
+ h1(t)θh2+ h2(t)θh1
, on ΣR
(29d)
−k∂θh1h2
∂n= 0, on ΣD
(29e)
θh1h2(0) = 0, in Ω, (29f)
with (θhi , fhi) = S′(u)hi, i = 1, 2.
· November 19, 2012 · Page 20 (33)
Theorem 5 (Necessary optimality conditions)
Let u ∈ Uad be an optimal control of problem (P) and (θ, f) = S(u) the associated
solution of the state system (27). Then there exists a unique solution
(p, q) ∈ Y ×W 1,∞(0, T ;L∞(Ω)) such that
−qt = Gf (θ, f)(q + ρLp), in Q (30a)
q(T ) = α1(f(T )− fd), in Ω (30b)
− ρcpt −∇ · (k∇p) = Gθ(θ, f)(ρLp+ q) + α2(θ − θd), in Q (30c)
−k ∂p∂n
= u(t)p, on ΣR (30d)
−k ∂p∂n
= 0, on ΣD (30e)
p(T ) = 0, in Ω. (30f)
Moreover, the following variational inequality is valid∫∫ΣR
(−p(θ − θw) +α3
|Γ|u)(u− u)dσdt ≥ 0 ∀u ∈ Uad. (31)
· November 19, 2012 · Page 21 (33)
Sufficient optimality conditions – I
• Introduce reduced cost functional
j(u) = J(S(u), u)• compute second derivative
j′′(u)[h1, h2] = α1
∫Ω
fh1(T )fh2
(T )dx+ α2
∫∫Q
θh1θh2
dxdt
+ α3
T∫0
h1h2dt−∫∫ΣR
(θh1h2 + θh2
h1)pdσdt
+
∫∫Q
G′′(θ(u), f(u))[(θh1, fh1
), (θh2, fh2
)](ρLp+ q)dxdt,
with (θhi , fhi) = S′(u)hi, i = 1, 2 and (p, q) is the solution of the adjoint
system
· November 19, 2012 · Page 22 (33)
Sufficient optimality conditions – II
• strongly active set associated to u
For fixed τ > 0 define
Aτ (u) =
t ∈ (0, T ) :
∣∣∣∣∣∣∫Γ
−p(x, t)(θ(x, t)− θw(x, t))dσ + α3u(t)
∣∣∣∣∣∣ > τ
.
• Sufficient Second Order Conditions
There exist τ > 0 and δ > 0 such that
j′′(u)h2 ≥ δ‖h‖2L2(0,T )
holds for all h = u− u, u ∈ Uad with h = 0 on Aτ (u)
(SSC)
· November 19, 2012 · Page 23 (33)
Sufficient optimality conditions – III
• Theorem 6
Let u be an admissible control of problem (P) with associated state
(θ, f) = S(u) satisfying the first order necessary optimality conditions given in
Theorem 5 with associated adjoint states (p, q). Further, it is assumed that
(SSC) holds at u. Then there exist a δ > 0 and ρ > 0 such that
J(θ, f, u) ≥ J(θ, f , u) + δ‖u− u‖2L2(ΣR) (32)
holds for all U ∈ Uad with ‖u− u‖L∞(0,T ) ≤ ρ with associated states
(θ, f) = S(u).
• Ref.: F. Tröltzsch: Optimal Control of Partial Differential Equations: Theory,
Methods and Applications. American Mathematical Society (2010)
· November 19, 2012 · Page 24 (33)
Sequential Quadratic Programming (SQP) method – I
• Consider finite-dimensional problem
min f(u) u ∈ C ⊂ Rn convex and closed and f smooth enough
• Assume (for a moment) C = Rn, then necess. cond. reads
∇f(u) = 0 = f ′(u).
• Approximate u with Newton’s method: un+1 solves
f ′(un) + f ′′(un)(u− un) = 0 (33)
• interpret (33) as necessary opt. cond. of the quadratic problem
min(f ′(un)(u− un) +
1
2(u− un)f ′′(un)(u− un)
)(34)
• (34) has a unique solution if f ′′(un) is positive definite→ SQP method.
· November 19, 2012 · Page 25 (33)
Sequential Quadratic Programming (SQP) method – II
• Control constrained case∇f(u) = 0 is replaced with
∇f(u)(u− u) ≥ 0 for all u ∈ C
• no direct appl. of Newton’s method possible, but easy inclusion in (34):
minu∈C
f ′(un)(u− un) +1
2(u− un)f ′′(un)(u− un) .
• Remarks
• (SQP) converges quadratically (local) if f ∈ C2,1 and f ′′(u) is positive
definite.• Idea can be applied directly to (CP) by considering J(y(u), u) = f(u).
However, instead of computing yn+1 = S(un+1) we will linearize S and
define yn+1 = yn + S′(un)(un+1 − un).
· November 19, 2012 · Page 26 (33)
Sequential Quadratic Programming (SQP) method – III
• for Lagrangean L(x, p, q), with x = (θ, f, u)
• introduce quadratic subproblem (QP)k:
min1
2L′′(xk, pk, qk)[δx, δx] + J ′(xk)δx
such that δft = Gf (θk, fk)δf +Gθ(θk, fk)δθ − fkt +G(θk, fk)
δf(0) = −fk(0)
ρcδθt −∇ · (k∇δθ) = ρLδft − (ρcθkt −∇ · (k∇θk)− ρLfkt )
−k∂δθ∂n− χΣRu
k(t)δθ = χΣRδu(t)(θk − θw)
+k∂θk
∂n+ χΣRu
k(t)(θk − θw)
δθ(0) = θ0 − θk(0)and
ua ≤ δu+ uk ≤ ub
• use active set strategy to solve (QP)k
· November 19, 2012 · Page 27 (33)
A test case – I
min J(θ, u) =1
2
∫ T
0
∫Ω
(θ−θd,Ω)2dxdt+1
2
∫ T
0
∫Γ
(θ−θd,Γ)2dxdt+1
2
∫ T
0
(u−ud)2dt
subject to
θt −∆θ = θ5 + f(x, y), in Ω× (0, T )
∂θ
∂ν+ θ = (u(t)− u(t))g(x), in ∂Ω× (0, T )
θ(x, 0) = θ0(x), in Ω,
optimal solution:
u = Π[ua,ub](−e−t)
θ = e−t cosπx1 · cosπx2
p = (t− T ) cosπx1 · cosπx2
· November 19, 2012 · Page 28 (33)
A test case – II
Numerical results:
Iteration Ji ‖ui − u‖ ‖(ui−ui−1)‖‖ui−1‖ # PDAS-iterations
0 23.1458 0.228097
1 23.14518 0.001556 0.280573 2
2 23.14501 0.0014091 3 · 10−4 2
3 23.1450 0.0014090 8 · 10−8 2
· November 19, 2012 · Page 29 (33)
Transfer to real hot-rolling mill I – semi reality
• desired end temperature θd = 680C , ferrite phase fraction fd = 0.7
• with Newton cooling law BC, i.e., −k ∂θ∂n = u(t)(θ − θw)
• globalization with 5 iterations of Projected Gradient Method prior to SQP• objective function Ji and relative error in i-iteration of SQP Method
Iteration Ji ‖(ui, θi, f i)− (ui−1, θi−1, f i−1)‖ # PDAS-Loops
1 0.025359 0.24649 7
2 0.021225 0.016189 3
3 0.021237 0.000171 2
4 0.021237 2.16 · 10−6 2
• optimal control
· November 19, 2012 · Page 30 (33)
Transfer to real hot-rolling mill II – reality
• Data• steel H3• holding time on ROT 10 s• thickness of sample 2mm• desired temperature: 680C
• desired ferrite fraction: 90%
• Boundary condition
−k ∂θ∂n
= e−(1.48−0.28v2)(vt−0.8)2)( v
0.05
)−0.63(u(t)
100
)0.45
(θ(x, t)− θw)
with strip velocity v and amount of water u(t)• Solution of optimal control problem
· November 19, 2012 · Page 31 (33)
• Simulation with constant amount of water u = 70l/min
Experiment with u = 70l/min
· November 19, 2012 · Page 32 (33)
and finally . . .
• Thank you all for listening!
• For course assessment please send me solutions to problem sheets until 15
December.
• Contact: hoembergwias-berlin.de
· November 19, 2012 · Page 33 (33)
