1. science
2. mathematics
3. algebra

Optimal Control Method of Parabolic Partial Differential
Equations and Its Application to Heat Transfer Model in
Continuous Cast Secondary Cooling Zone
Yuan Wang, Xiao chuan Luo ∗, Sai Li
State Key Laboratory of Synthetically Automation for Process industries, Northeastern University, Shenyang 110004, PR China
——————————————————————————————————————————————–
Abstract
Our work is devoted to a class of optimal control problem of parabolic partial differential equations. Because of the partial
differential equations constraints, it is rather difficult to solve the optimization problem. The gradient of the cost function can
be found by the adjoint problem approach. Based on the adjoint problem approach, the gradient of cost function is proved to be
Lipchitz continuous. An improved conjugate method is applied to solve this optimization problem and this algorithm is proved to
be convergent. This method is applied to set-point values in continuous cast secondary cooling zone. Based on the real data in
a plant, the simulation experiments show that the method can ensure the steel billet quality. From these experiment results, it is
concluded that the improved conjugate gradient algorithm is convergent and the method is effective in optimal control problem of
partial differential equations.
Keywords: Partial differential equations; Continuous cast; Heat transfer; Lipchitz continuous; An improved conjugate algorithm.
——————————————————————————————————————————–
1. Introduction
conduction [2], fluid mechanics [6] and material sciences [7].
Many researchers considered a similar optimal control problem
Continuous caster is a machine by which the molten steel
of partial differential equations [8–9].
Z T
min J =
[y(0, t) − yaim (t)]2 dt+
is solidified to slabs by spraying on it the cooling water. Because
the production situation of secondary cooling zone is very bad,
u∈U
the set point of water volume in secondary cooling plays an im-
Z
0
T
[u(t) − u∗ (t)]2 dt
portant role in the continuous casting [1]. Undercooling of the
α
strand in continuous cast secondary cooling can result in a liquid
s.t. yt − k(x)yxx = 0,
pool is too long. However, overcooling can lead to the formation
y(x, 0) = φ(x),
of cracks. The quality of steel billet depends on the behaviour of
0
(x, t) ∈ (0, l) × (0, T ],
x ∈ (0, l),
k(0)yx (0, t) = u(t),
t ∈ (0, T ],
the surface temperature [2]. Optimal control problem of partial
k(l)yx (l, t) = 0,
differential equations is encountered [3–5] in many applications
t ∈ (0, T ],
(1)
ranging from engineering to science. The mathematical model
Where y is temperature(K). y(0, t) is the surface temper-
arises in many engineering and physical processes such as heat
ature (K). yaim is the objective temperature. k(x) is a coeffici-
∗ Corresponding
author. E-mail address: [email protected] (Xiao chuan. Luo)
1
ent. u(t) is a controlled variable. u∗ (t) is a guess for the controlled variable by priori knowledge. U = L2 (∂Ω), Ω ∈ (0, l)×
(0, t).
Li [10] applied conjugate gradient algorithm to estimate
boundary condition. Lee et al.[11] proposed a repulsive particle
swarm optimization algorithm to solve this problem. M.H.Fara
[12] used a modified partial quadratic interpolation method to
solve parabolic optimal control problem. Based on the adjoint problem approach, Mujdat and Arzu [13] studied an inverse
parabolic problem. Based on the gradient of cost function is
Lipchitz continuous, Alemdar Hasanov [14] proved existence of
a quasi-solution of the inverse problem and proposed a mono-
Figure 1: schematic representation of steel slab
tone iteration scheme. Alemdar Hasanov [15] used the gradient
method and proved the convergence of this method. However,
2.1. Continuous casting heat transfer model
the step size in their gradient algorithm is a constant. Step size
largely infects the convergent speed of this gradient algorithm.
2.1.1. Assumptions
Alemdar Hasanov and Burhan Pektas [16] considered an inverse
Based on the following key assumptions [17]:
source problem. They used a conjugate gradient algorithm to
(1)One-dimensional heat transfer is considered, so heat trans-
solve this problem, and the good results were obtained.
fer can be ignored in the length and the width direction of billet.
In this paper, a class of optimal control problem of partial d-
(2)Steel billet at the same cooling zone cools uniformity.
ifferential equations is abstracted from optimization problem for
set-point values in secondary cooling zone process. In our opti-
(3)The meniscus surface is assumed to be flat.
mal control of partial differential equations problem, the gradi-
(4)In two-phase, we use the equivalent specific heat instead
of specific heat.
ent of cost function is proved Lipchitz continuous. We present an
improved conjugate gradient algorithm to solve this problem and
(5)The convection heat transfer process in the liquid phase
prove convergence of this algorithm. The simulation experiment
and two-phase zone are equal to conduction heat transfer pro-
shows that this algorithm can effectively solve the optimization
cess.
problem for set-point values in secondary cooling zone, and the
2.1.2. The differential equation
quality of billet is ensured.
Based on the above assumptions, we can get continuous
casting heat transfer model for one dimensional.
2. Continuous casting operation model of secondary cooling zone
∂y
∂2y
= k(x) 2
∂t
∂x
Continuous casting makes a liquid metal by a special set
(2)
Where y is temperature(K). t is times(s). x is slab thickness
of cooling device into a certain section in the shape of a casting
direction(m) and k(x) is a coefficient depending on physical
process. As shown in Figure 1, x, y and z is the thickness di-
parameters of the steel, which is calculated by
rection of steel billet, the width direction of steel billet and the
k(x) =
length direction of steel billet.
2
λ
ρCe
(3)
Where ρ is steel density (kg/m3 ), Ce is specific heat (J/(kg·
Where y is steel slab temperature(K). y(0, t) is the surface
temperature of steel slabs(K). yaim is the objective temperature
k)) and λ is thermal conductivity(W/m · K).
of steel slabs(K), which depends on continuous casting technol-
2.1.3. Boundary conditions
ogy. t is times(s). x is slab thickness direction (m). α is a pos-
Steel billet goes through the crystallizer, secondary cooling
itive number. y0 is initial casting temperature, which is usually
zone and air cooling zone. The different cooling sections have
a constant, U = L2 (∂Ω), Ω ∈ (0, l) × (0, t). Other parameters
have already defined in continuous casting transfer model.
different cooling characteristics, so the boundary conditions are
Many researchers [18–19] use y(0, t) the surface tempera-
shown below.
ture of steel billet to evaluate the quality of billet. We hope that
k(0)yx (0, t) = u(t)
(4)
the surface temperature y(0, t) is close to the objective temperature yaim by controlling the boundary condition u(t). In the
Where u(t) is heat flux in continuous casting process which
actual production of continuous casting, the controlled variable
is calculated by
cannot be too large, so we also give a penalty for the controlled
u(t) = h[y(0, t) − Tw ]
(5)
variable.
According to equation(3) and equation(5), k(x) and u(t)
Where h is the equivalent heat transfer coefficient in sec-
in equation(1) are assigned, so optimal control problem in con-
ondary cooling zone W/(m2 · K). y(0, t) is the steel billet sur-
tinuous casting secondary cooling zone transfers to a general op-
face temperature(K) and Tw is cooling water temperature(K).
Qwater rw Qwater ra
h = hx(
) (
) + hr
SW SL
SW SL
timal control problem in equation(1). Because of this, we only
(6)
consider a general optimal control problem in later sections.
Where hx and hr are constants. SW is the width of steel
3. Adjoint problem and Lipchitz con-
billet. SL is the length of steel billet. Qwater is the spray cooling water volume. Qair is the spray cooling air volume. The
tinuity of the gradient of the cost func-
boundary condition at x = l can be written as
k(l)yx (l, t) = 0
tion
(7)
3.1. Adjoint problem
2.2. Optimal control model of continuous cast
Based on first optimization then discrete method, we can
secondary cooling zone
get the adjoint equation firstly and then derives the optimality
Based on continuous casting heat transfer model, we have
conditions, using numerical methods to solve this problem. The
optimal control model of continuous cast secondary cooling zone.
gradient of the cost function with respect to the control variable
Z T
can be efficiently calculated by the adjoint problem. Now we
min J =
[y(0, t) − yaim (t)]2 dt+
u∈U
0
give the strong version of the Lagrange function, and then we
Z T
∗
2
transfer all constraints into cost function by separate multipliers
α
[u(t) − u (t)] dt
0
s.t. yt −
λ
yxx = 0,
ρCe
y(x, 0) = y0 ,
p(x, t), p(0, t), p(l, t), p(x, 0) and perform partial integrations
(x, t) ∈ (0, l) × (0, T ],
x ∈ (0, l),
k(0)yx (0, t) = h[y(0, t) − Tw ],
k(l)yx (l, t) = 0,
with respect to space and time, so the Lagrange function can be
t ∈ (0, T ],
obtained:
t ∈ (0, T ],
L(y, p, x, t) =
Z T
Z lZ
α
[u(t) − u∗ (t)]2 dt −
(8)
0
3
0
T
pt ydtdx
0
−
Z
+
Z
+
Z
+
−
Z
Z
T
0
Z
l
p(x, T ) = 0,
k(x)pxx (x, t)y(x, t)dxdt
x ∈ (0, l),
0
−k(0)px (0, t) = [y(0, t) − yaim (t)],
T
t ∈ (0, T ],
[k(l)yx (l, t)p(l, t)]dt
k(l)px (l, t) = 0,
0
T
[k(0)yx (0, t)p(0, t) + (y(0, t) − yaim )2 ]dt
t ∈ (0, T ],
(12)
Now we obtained adjoint problem in our optimal control
0
l
problem.
y(x, T )p(x, T )dt
0
l
φ(x)p(x, 0)dx −
0
Z
T
u(t)p(0, t)dt
3.2. Lipchitz continuity of the gradient of the
(9)
0
The vibrational of the function L(y, p, x, t) can be expressed
as
cost function
We can consider the variation of the cost function.
L(y + δy, p, x, t) = α
+
−
Z
+
Z
+
Z
+
Z
−
Z
[u(t) − u∗ (t)]2 dt
∆J(u) = J(u + ∆u) − J(u) =
Z T
Z
[y(0, t) − yaim (t)]∆y(0, t)dt +
0
[y(x, T ) + δy(x, T )]p(x, T )dt
0
0
Z lZ
Z
T
l
0
−
Z
T
Z
pt [y(x, t) + δy(x, t)]dtdx
0
T
0
Z
l
+
Z
+
Z
[∆u(t)]2 dt
(13)
0
tion. According to equation (8), if we let ∆y = y(x, t; u +
k(0)px (0, t)[y(0, t) + δy(0, t)]dt
0
T
∆u) − y(x, t; u), we obtain the following partial differential e[y(0, t) + δy(0, t) − yaim ]2 dt
quation.
0
T
k(l)px (l, t)[y(l, t) + δy(l, t)]dt
∆yt = k(x)∆yxx ,
0
l
φ(x)p(x, 0)dx −
0
Z
u(t)p(0, t)dt
(10)
∆y(x, 0) = 0,
0
Z
x ∈ (0, l),
0
0
T
(x, t) ∈ (0, l) × (0, T ],
T
−k(0)∆yx (0, t) = ∆u(t),
k(l)∆yx (l, t) = 0,
0
+
T
chitz continuous, we give two others partial differential equa-
∆L = L(y + δy, p, x, t) − L(y, p, x, t) =
Z l
Z lZ T
p(x, T )δy(x, T )dx −
pt δy(x, t)dtdx
Z
0
Z
T
can be obtained:
−
[u(t) − u∗ (t)]∆u(t)dt +
[∆y(0, t)]2 dt
0
In order to prove the gradient of the cost function is Lip-
k(x)pxx (x, t)[y(x, t) + δy(x, t)]dxdt
0
According to equation (9) and (10), the following equation
Z
T
T
t ∈ (0, T ],
t ∈ (0, T ],
(14)
According to equation (12), the following partial differential equation can be obtained by similar way.
0
∆pt = −k(x)∆pxx ,
T
(x, t) ∈ (0, l) × (0, T ],
k(x)pxx δy(x, t)dxdt
l
∆p(x, 0) = 0,
T
k(0)px (0, t)δy(0, t)dt
x ∈ (0, l),
−k(0)∆px (0, t) = ∆y(0, t),
0
t ∈ (0, T ],
T
[y(0, t) − yaim (t)]δy(0, t)dt
k(l)∆px (l, t) = 0,
0
T
k(l)px (l, t)δy(l, t)dt
t ∈ (0, T ],
(15)
Lemma 3.1: For all ∆u, the following identity holds:
Z T
[y(0, t; u) − yaim (t)]∆y(0, t)dt
(11)
0
According to equation (11), the following dual-equation
0
can be obtained:
=
pt = −k(x)pxx ,
(x, t) ∈ (0, l) × (0, T ]
4
Z
T
p(0, t)∆u(t)dt
0
(16)
Proof: It is easy to know that the following identity holds:
Z TZ l
[k(x)px (x, t)∆y(x, t)]x dxdt =
0
Z
According to the boundary conditions of equation (14), the
following equation can be derived:
Z lZ T
Z lZ
(k(x)∆yx )x ∆ydtdx =
0
T
0
k(l)px (l, t)∆y(l, t)dt
0
0
0
−
Z
−
T
k(0)px (0, t)∆y(0, t)dt
0
(17)
0
−
Thus boundary conditions of equation (12) show
Z T
[y(0, t; u) − yaim (t)]∆y(0, t)dt =
−
−
=
(18)
0
T
0
Z
Z
0
T
k(l)px (l, t)∆y(l, t)dt
T
Z
T
l
Z
l
0
+
−
Z
−
Z
k(x)pxx (x, t)∆y(x, t)dxdt
T
0
Z
k(x)px (x, t)∆yx (x, t)dxdt
l
p(x, t)∆y(x, t)dxdt
0
k(0)p(0, t)∆yx (0, t)dt =
0
Z
T
p(0, t)∆udt (19)
0
Lemma 3.1 can be proved.
following equation can be obtained:
Z T
Z T
|∆y(0, t)|2 dt ≤ c
|∆u(t)|2 dt
k(x)(∆yx )2 dtdx
0
T
∆y(0, t)∆u(t)dt
(24)
0
0
(20)
0
≤
0
∀ε > 0
(25)
1
2ε
0
Z
T
|∆u)|2 dt +
0
0
(21)
0
It is easy to know that the following identity holds:
Z Z
Z TZ l
1 T l
[(∆y)2 ]dxdt
∆yt ∆ydxdt =
2 0 0
0
0
Z
1 l
=
∆y(x, T )2 dx
2 0
∀α, β ∈ R,
ε
2
Z
T
|∆y(0, t)|2 dt
(27)
0
From literate[14], the following equation can be obtained:
Z lZ T
Z T
2l2
(∆y)2 dtdx + 2l
[∆y(0, t)]2 dt
and also simultaneously in time and space integration, the following equation can be obtained:
Z TZ l
[∆yt − (k(x)∆yx )x ]∆ydxdt = 0
β2
α2
+
,
2
2ε
tion can be derived:
Z lZ T
Z l
k(x)(∆yx )2 dtdx +
[∆y(x, T )]2 dt
Proof: If we multiply both sides of equation (14) by ∆y(x, t)
0
Z
T
According to equation (24) and (26), the following equa-
Lemma 3.2: There exists a constant c > 0, so that the
0
k(x)(∆yx )2 dtdx (23)
0
On the right-hand side of equation(24), then we have
Z T
Z T
1
∆u∆y(0, t)dt ≤
|∆u|2 dt
2ε 0
0
Z
ε T
|∆y(0, t)|2 dt
(26)
+
2 0
0
T
0
αβ ≤ ε
0
0
0
T
We use the ε-inequality.
0
=
Z lZ
∆y(0, t)∆u(t)dt −
=
Z
k(x)(∆yx )2 dtdx
0
T
[k(x)px (x, t)∆y(x, t)]x dxdt−
0
Z
T
0
l
Z
Z lZ
have the following equations:
Z T
Z lZ
[∆y(x, T )]2 +
0
Z
k(0)∆yx (0, t)∆y(0, t)dt
Taking the equations (22) and (23) into equation (21), we
The following equations can be obtained:
Z T
[y(0, t; u) − yaim (t)]∆y(0, t)dt
=
k(x)(∆yx )2 dtdx =
0
l
Z
0
T
k(0)px (0, t)∆y(0, t)dt
T
(k(x)∆yx ∆y)x dtdx
0
0
0
Z
Z lZ
T
≤
1
2ε
0
Z
0
T
0
(∆u)2 dt +
ε
2
Z
T
[∆y(0, t)]2 dt
So the following equation can be obtained
Z T
Z T
|∆y(0, t)|2 dt ≤ c
|∆u(t)|2 dt
0
(28)
0
(29)
0
We can know there is a constant c which makes equation
(22)
(20) established. Lemma 3.2 can be proved.
5
Lemma 3.3
Z T
Z
|∆y(0, t)|2 dt =
0
4. Conjugate gradient algorithm solve
T
∆p(0, t)∆u(t)dt
optimal control of parabolic partial d-
(30)
0
Proof: According to equation (15), the weak solution of
ifferential equations and its convergen-
this partial differential equation is the following equation.
Z TZ l
Z TZ l
(∆p∆y)t dxdt −
∆p∆yt dxdt
0
0
+
Z
−
Z
0
t analysis
0
4.1. Conjugate gradient algorithm
T
k(x)∆y∆p|x=l
x=0 dt
0
After we prove the gradient of cost function is Lipchitz conT
Z
l
k(x)∆px ∆yx dxdt = 0
(31)
tinuous, we can use the conjugate gradient algorithm to solve
According to the boundary conditions of equation(14) and
this PDE optimization problem. If we choose the reasonable pa-
0
0
rameters in conjugate gradient algorithm, we can prove that the
equation(15), the following equation can be obtain:
Z T
k(l)∆y(l, t)∆px (l, x)dt
convergence of conjugate gradient algorithm.
0
−
−
+
Z
Z
Z
Z
T
Set k=1
k(0)∆y(0, t)∆px (0, x)dt
0
T
YES
k(l)∆yx (l, t)∆p(l, x)dt
k=1?
0
T
k(0)∆yx (0, t)∆p(0, x)dt =
NO
0
T
[∆y(0, t)]2 =
0
Z
T
∆p(0, t)∆u(t)dt
(32)
dk
0
gk
Calculate the gradient
of cost function g k
dk
g k E k d k 1
Lemma3.3 can be proved.
k=k+1
Theorem1: If the gradient of cost function is Lipchitz con-
Determine step size by
wolf line search rules
tinuous, there is a constant L to make the following equation
holds:
uk D k d k
uk 1
kJ ′ (u + ∆u) − J ′ (u)k ≤ Lk∆uk
(33)
NO
Proof: Taking into account lemma3.1, lemma 3.2 and lemJ (uk 1 ) J (uk ) d eps ?
ma 3.3, we can obtain the following equation:
Z T
(J ′ (u + ∆u) − J ′ (u), ∆u) =
∆p(0, t)∆udt
YES
END
0
=
Z
0
T
[∆y(0, t)]2 ≤ L
Z
T
[∆u(t)]2 dt
Figure 2: conjugate gradient algorithm process
(34)
0
As shown in Figure 2, k is iteration number. dk is iteration
direction. gk is gradient of cost function. α is step size. uk is
controlled quantity. J(u) is cost function. βk is a parameters in
conjugate gradient algorithm, and eps is a small constant.
Because of that we use FR conjugate method to solve this
optimal problem, dk and βk can be calculated by the following
6
equations:
If this algorithm is not convergent, the equation (38) does
dk = −gk + βk dk−1
(35)
g T gk
βk = T k
gk−1 gk−1
(36)
not hold, there exists a constant ε0 > 0, so that the following
identity holds:
kgk k ≥ ε0 , kdk k2 ≤ τ1 (k + 1)
(44)
Step size largely infects the convergent speed of conjugate
According to equation (40), we can obtain the following
gradient algorithm. If we choose wolf line search rules, we have
equations:
the following formula [20]:
cos θk =
f (xk + αk dk ) ≤ f (xk ) + σ1 α( k)gkT dk
|∇f (xk + αk dk )| ≥ σ2 |gkT dk |
0 < σ1 < σ2 <
(2 −
1
2
k
X
gkT dk
≥
kgk kkdk k
σ2j )
i=0
(37)
∞
X
cos2 θk ≥ (
k=0
4.2. Convergence analysis of conjugate gradi-
1 − 2σ2 kgk k
kgk k
≥
kdk k
1 − σ2 kdk k
(45)
∞
∞
k=0
k=0
X 1
1 − 2σ2 2 X kgk k2
)
≥ τ2
(46)
2
1 − σ2
kdk k
k+1
Since g(x) is the gradient of cost function, we have proven
ent algorithm
g(x) is Lipchitz continuous. There exists a constant, so that the
following inequality holds:
Theorem2: If dk , βk and αk are calculated by equation
(35), equation (36) and equation (37), the following equation
T
σ2 gkT dk ≤ gk+1
dk = gkT + (gk+1 − gk )T dk
holds:
≤ gkT dk + αk Lkdk k2
lim inf kgk k = 0
k→∞
Then we can obtain the following equations:
Proof: According to equation (37), the following equations
αk ≥ −
can be obtained:
−
k
X
k
σ2i ≤
i=0
X i
gkT dk
≤ −2 +
σ2
2
kgk k
i=0
∀k = 0, 1, . . . , ∞
−
k
X
σ2i < −
∞
X
σ2i =
1
<2
1 − σ2
1 − σ2
kgk k2 cos2 θk
L
(49)
k=0
k=0
is convergent, we can conclude that cos2 θk is convergent and
(40)
not convergent at the same time. Thus the assumption kgk k is
invalid, we can prove that equation (38) holds, so this conjugate
tion can be obtained:
T
|gkT dk−1 | ≤ −σ2 gk−1
dk−1 <
σ2
kgk−1 k
1 − σ2
gradient algorithm is convergent.
(41)
4.3. An optimal choice of the stopping in the
According to the recursion formula of dk , the following
conjugate gradient algorithm
equation can be represented as
2
kdk k2 = kgk k2 − 2βk−1 gkT dk−1 + βk−1
kdk−1 k2
It is important to understand the behavior of the cost function value depending on the iteration number. For given smooth
2σ2
2
kgk k2 + βk−1
kdk−1 k
1 − σ2
1 + σ2
2
=
kgk k2 + βk−1
kdk−1 k2
1 − σ2
objective temperature yaim, an optimal control problem is solved
by conjugate gradient algorithm.
(42)
k
= kdk k2 ≤
(48)
The sequence fk is monotonic decline and has lower bound,
∞
∞
P
P
cos2 θk
kgk k2 cos2 θk is convergent. Since kgk k ≥ ε0 ,
so
According to equation (39) and (40), the following equa-
≤ kgk k2 −
1 − σ2 T
g k dk
Lkdk k2
fk+1 ≤ fk − σ1
(39)
i=0
i=0
(47)
(38)
X 1 2 kgk k2
1 + σ2
kgk k4
+
1 − σ2
kgi k
kg0 k2
i=1
yaim = 1173.16 + 100 sin(
(43)
7
π
t)(K)
10
5. Simulation experiment
k = 7250 × 540/50, eps = 1 × 10−1
u(x, 0) = 1173.16K, L = 0.25m
To test the effectiveness of the method in this paper, the
T = 20s, u∗ = 0, α = 1 × 10−7
(50)
simulation experiments are done in actual plant data. The experiment result proves the effectiveness of the method in this paper.
6
10
Physical parameters of steel grade is given in Table 2.
Table 2: Physical parameters of steel grade
5
cost function
10
Items
Parameters
4
10
540(J/(kg · k))
Specific heat
Density
7250(kg/m3)
Thermal conductivity
30(W/m · K)
3
10
2
10
0
20
40
60
80
iteration times
100
Example 1: Experimental parameters are shown below:



1173.16(K) t < 10(s)
yaim =


1273.16(K) t ≥ 10(s)
120
Figure 3: surface temperature error of billet
As shown in Figure 3, the behaviour of cost function value
k = 7250 × 540/50, eps = 1 × 10−1
as a function of the iteration number, consists of three phases:
the initial phase of rapid decrease but very short duration, the
u(x, 0) = 1173.16K, L = 0.25m
second phase of slow decrease, and the third phase of almost
T = 20s, u∗ = 0, α = 1 × 10−7
(51)
constant behaviour.
1280
Table 1: Determination of an optimal value stopping pa-
objective temperature
rameter
iteration number
surface temperature(K)
eps
1260
cost function value
1
42
394.80
0.1
115
258.15
0.01
170
258.15
actual temperature
1240
1220
1200
1180
When dealing with optimal control problems, the stopping
1160
parameter plays an important role in conjugate gradient algorithm. However, it is difficult to choose a good stopping parameter
0
5
10
time(s)
15
20
Figure 4: surface temperature of billet
eps. If stopping parameter eps is too large, the result can’t converge to optimal solution. The iteration number is too large be-
In this example, the objective temperature changes from
cause stopping parameter is too small. As shown in Table 1, the
1173.16(K) to 1273.16(K), when time is equal to 10 seconds.
best value of stopping parameter is achieved at eps = 0.1
8
ing process.
40
Example 2: Experimental parameters are shown below:
error(K)
30
π
t)(K)
10
20
yaim = 1173.16 + 100 sin(
10
k = 7250 × 540/50, eps = 1 × 10−1
0
u(x, 0) = 1173.16K, L = 0.25m
−10
T = 20s, u∗ = 0, α = 1 × 10−7
(52)
−20
−30
1300
objective temperature
actual temperature
−40
0
5
10
time(s)
15
1250
20
surface temperature(K)
−50
Figure 5: surface temperature error of billet
4
2
x 10
conotroled variable(W/m2)
0
1200
1150
1100
−2
−4
1050
−6
0
5
10
time(s)
15
20
−8
Figure 7: surface temperature of billet
−10
−12
−14
−16
2
0
5
10
time(s)
15
1
20
0
Figure 6: controlled variable
error(K)
−1
Figure 4 shows that the actual temperature can follow the objective temperature very well, so the quality of billet can be en-
−2
−3
−4
sured. From Figure 5, we can know that the maximum error is
−5
about 50(K). The temperature error is very small in most of the
−6
time. Based on technological requirements in continuous cast-
−7
ing process, the behaviour of the surface temperature ensures
that the billet has a good quality. Figure 6 shows that the con-
0
5
10
time(s)
15
20
Figure 8: surface temperature error of billet
trolled variable. Because of penalty of controlled variable, the
controlled variable cannot be very large. When the controlled
In this example, the objective temperature is a sinusoidal
variable is less than zero, this process is a heating process. When
curve which is continuous function. Figure 7 shows that the ac-
the controlled variable is larger than zero, this process is a cool-
tual temperature can follow the objective temperature very well,
9
4
x 10
1840
3
1820
2
1800
surface temperature(K)
conotroled variable(W/m2)
4
1
0
−1
−2
1780
1760
1740
1720
−3
1700
−4
1680
−5
0
5
10
time(s)
15
1660
20
objective temperature
actual temperature
0
Figure 9: controlled variable
5
10
time(s)
15
20
Figure 10: surface temperature of billet
so the quality of billet can be ensured. From Figure 8, we can
160
know that the maximum error is about 8(K). Because the objec-
140
tive temperature is a sinusoidal curve, the maximum error be-
120
comes smaller than example 1. The temperature error is very s-
100
error(K)
mall in most of the time. Based on technological requirements in
continuous casting process, the behaviour of the surface temper-
80
60
ature ensures that the billet has a good quality. Figure 9 shows
40
that the controlled variable. Because of a penalty of controlled
20
variable, the controlled variable cannot be very large.
0
Example 3: Experimental parameters are shown below:
−20
yaim = 1673.16(K)
0
5
10
time(s)
15
20
Figure 11: surface temperature error of billet
k = 7250 × 540/50, eps = 1 × 10−1
u(x, 0) = 1173.16K, L = 0.25m
T = 20s, u∗ = 0, α = 1 × 10−7
6. Conclusions
(53)
In this example, the objective temperature is 1673.16(K),
In order to solve optimization problem for set-point values
but the initial temperature of billet is 1844.16(K). Figure 10
in secondary cooling process, this paper apply first optimize then
shows that the actual temperature can follow the objective tem-
discrete method to this optimal problem. We prove the gradient
perature very well. While the billet quality can be ensured, the
of cost function is Lipchitz continuous. Base on this, we pro-
surface temperature of billet can approach the objective temper-
pose an improved conjugate algorithm to solve this problem and
ature rapidly. From Figure 11, we can know that the maximum
prove that this algorithm is convergent.
error is about 160(K). Since the penalties of control variable, the
actual temperature cannot step to the objective temperature. Figure 12 shows that the controlled variable. Because of penalty of
controlled variable, the controlled variable cannot be very large.
10
5
5
based multi-criteria approach, Applied Mathematical Mod-
x 10
elling , (2005), 29, pp:653–672.
4.5
[3] Karl kunisch, Primal-Dual Strategy for State-Constrained
conotroled variable(W/m2)
4
Optimal Control Problems, Computational Optimization
3.5
and Applications , (2002), 22, pp:193–224.
3
2.5
[4] Wenbin Liu, Ningning Yan, Adaptive Finite Element Meth-
2
ods for Optimal Control Governed by PDEs, science,
2008.
1.5
1
[5] M.Hinze, R.Pinnau, M.Ulbrich, Optimization with PDE
0.5
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Figure 12: controlled variable
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Conflict of interests
[7] M.Renardy, W.J.Hursa, J.A.Nohel, Mathematical in Viscoelasticity, Wiley, 1987.
The authors declare that there is no conflict of interests regarding the publication of this paper.
[8] D.H¨omberg, K. Krumbiegel, J. Rehberg, Optimal Control
of a Parabolic Equation with Dynamic Boundary Condition, Appl Math Optim, (2013), 1, pp:3–31.
Acknowledgements
[9] M.M. Kostreva, A.L. Ward, Optimal control of a system
The authors wish to thank the reviewer for their careful
governed by an elliptic partial differential equation, Jour-
reading and very helpful feedback, which helped improve the
nal of Computational and Applied Mathematics , (2000),
paper as well as the important guiding significance to our re-
114, pp:173–187.
searches. This work was supported by National Natural Science
[10] H.Y. Li, Estimation of thermal properties in combined con-
Foundation of China (61333006, 60974091) and the Fundamen-
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