# Optimal Control Method of Parabolic Partial Differential Equations and Its Application to Heat Transfer Model in Continuous Cast Secondary Cooling Zone

Authors: Yuan Wang; Xiaochuan Luo; Sai Li
Journal: Advances in Mathematical Physics
 (2015)

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## Abstract

Our work is devoted to a class of optimal control problems 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 Lipschitz 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.


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## Body

## 1. Introduction

Continuous caster is a machine by which the molten steel is solidified to slabs by spraying on it the cooling water. Because the production situation of secondary cooling zone is very bad, the set point of water volume in secondary cooling plays an important role in the continuous casting [1]. Undercooling of the strand in continuous cast secondary cooling can result in a liquid pool that is too long. However, overcooling can lead to the formation of cracks. The quality of steel billet depends on the behaviour of the surface temperature [2]. Optimal control problem of partial differential equations is encountered [3–5] in many applications ranging from engineering to science. The mathematical model arises in many engineering and physical processes such as heat conduction [2], fluid mechanics [6], and material sciences [7]. Many researchers considered a similar optimal control problem of partial differential equations [8, 9]:
(1)min⁡u∈UJ=∫0T‍y0,t-yaimt2dtminu∈UJ=+α∫0T‍ut-u*t2dts.t.yt-k(x)yxx=0,(x,t)∈(0,l)×(0,T],s.t.yx,0=ϕx,x∈(0,l),s.t.k(0)yx(0,t)=u(t),t∈(0,T],s.t.k(l)yx(l,t)=0,t∈(0,T],
where y is temperature (K). y(0,t) is the surface temperature (K). yaim is the objective temperature. k(x) is a coefficient. u(t) is a controlled variable. u*(t) is a guess for the controlled variable by a 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. Farag et al. [12] used a modified partial quadratic interpolation method to solve parabolic optimal control problem. Based on the adjoint problem approach, Kaya and Erdem [13] studied an inverse parabolic problem. Based on the gradient of cost function that is Lipschitz continuous, Hasanov [14] proved existence of a quasisolution of the inverse problem and proposed a monotone iteration scheme. Hasanov [15] used the gradient method and proved the convergence of this method. However, the step size in their gradient algorithm is a constant. Step size largely infects the convergent speed of this gradient algorithm. Hasanov and Pektaş [16] considered an inverse source problem. They used a conjugate gradient algorithm to solve this problem, and the good results were obtained.In this paper, a class of optimal control problems of partial differential equations is abstracted from optimization problem for set-point values in secondary cooling zone process. In our optimal control of partial differential equations problem, the gradient of cost function is proved Lipschitz continuous. We present an improved conjugate gradient algorithm to solve this problem and prove convergence of this algorithm. The simulation experiment shows that this algorithm can effectively solve the optimization problem for set-point values in secondary cooling zone, and the quality of billet is ensured.


## 2. Continuous Casting Operation Model of Secondary Cooling Zone

Continuous casting makes a liquid metal by a special set of cooling devices into a certain section in the shape of a casting process. As shown in Figure1, x, y, and z are the thickness direction of steel billet, the width direction of steel billet, and the length direction of steel billet.Figure 1
Schematic representation of steel slab.


### 2.1. Continuous Casting Heat Transfer Model

#### 2.1.1. Assumptions

Based on the following key assumptions [17], continuous casting heat transfer model can be obtained.(1)
One-dimensional heat transfer is considered, so heat transfer can be ignored in the length and the width direction of billet.(2)
Steel billet at the same cooling zone cools uniformity.(3)
The meniscus surface is assumed to be flat.(4)
In mushy zone, we use the equivalent specific heat instead of specific heat.(5)
The convection heat transfer process in the liquid phase and two-phase zone is equal to conduction heat transfer process.


#### 2.1.2. The Differential Equation

Based on the above assumptions, we can get continuous casting heat transfer model for one-dimensional model.(2)∂y∂t=kx∂2y∂x2,
where y is temperature (K). t is times (s). x is slab thickness direction (m) and k(x) is a coefficient depending on physical parameters of the steel, which is calculated by
(3)kx=λρCe,
where ρ is steel density (kg/m3), Ce is specific heat (J/(kg·k)), and λ is thermal conductivity (W/m·K).


#### 2.1.3. Boundary Conditions

Steel billet goes through the crystallizer, secondary cooling zone, and air cooling zone. The different cooling sections have different cooling characteristics, so the boundary conditions are shown as(4)k0yx0,t=ut,
where u(t) is heat flux in continuous casting process which is calculated by
(5)ut=hy0,t-Tw,
where h is the equivalent heat transfer coefficient in secondary cooling zone W/(m2·K). y(0,t) is the steel billet surface temperature (K) and Tw is cooling water temperature (K):
(6)h=hxQwaterSWSLrwQwaterSWSLra+hr,
where hx and hr are constants. SW is the width of steel billet. SL is the length of steel billet. Qwater is the spray cooling water volume. Qair is the spray cooling air volume. The boundary condition at x=l can be written as
(7)k(l)yx(l,t)=0.


### 2.2. Optimal Control Model of Continuous Cast Secondary Cooling Zone

Based on continuous casting heat transfer model, we have optimal control model of continuous cast secondary cooling zone:(8)min⁡u∈UJ=∫0T‍y0,t-yaimt2dtmin⁡u∈UJ=+α∫0T‍ut-u*t2dts.t.yt-λρCeyxx=0,(x,t)∈(0,l)×(0,T],s.t.y(x,0)=y0,x∈(0,l),s.t.k(0)yx(0,t)=h[y(0,t)-Tw],t∈(0,T],s.t.k(l)yx(l,t)=0,t∈(0,T],
where y is steel slab temperature (K). y(0,t) is the surface temperature of steel slabs (K). yaim is the objective temperature of steel slabs (K), which depends on continuous casting technology. t is times (s). x is slab thickness direction (m). α is a positive number. y0 is initial casting temperature, which is usually a constant; U=L2(∂Ω), Ω∈(0,l)×(0,t). Other parameters have already been defined in continuous casting transfer model.Many researchers [18, 19] use y(0,t), the surface temperature of steel billet to evaluate the quality of billet. We hope that the surface temperature y(0,t) is close to the objective temperature yaim by controlling the boundary condition u(t). In the actual production of continuous casting, the controlled variable cannot be too large, so we also give a penalty for the controlled variable.According to (3) and (5), k(x) and u(t) in (1) are assigned, so optimal control problem in continuous casting secondary cooling zone transfers to a general optimal control problem in (1). Because of this, we only consider a general optimal control problem in later sections.


## 2.1. Continuous Casting Heat Transfer Model

### 2.1.1. Assumptions

Based on the following key assumptions [17], continuous casting heat transfer model can be obtained.(1)
One-dimensional heat transfer is considered, so heat transfer can be ignored in the length and the width direction of billet.(2)
Steel billet at the same cooling zone cools uniformity.(3)
The meniscus surface is assumed to be flat.(4)
In mushy zone, we use the equivalent specific heat instead of specific heat.(5)
The convection heat transfer process in the liquid phase and two-phase zone is equal to conduction heat transfer process.


### 2.1.2. The Differential Equation

Based on the above assumptions, we can get continuous casting heat transfer model for one-dimensional model.(2)∂y∂t=kx∂2y∂x2,
where y is temperature (K). t is times (s). x is slab thickness direction (m) and k(x) is a coefficient depending on physical parameters of the steel, which is calculated by
(3)kx=λρCe,
where ρ is steel density (kg/m3), Ce is specific heat (J/(kg·k)), and λ is thermal conductivity (W/m·K).


### 2.1.3. Boundary Conditions

Steel billet goes through the crystallizer, secondary cooling zone, and air cooling zone. The different cooling sections have different cooling characteristics, so the boundary conditions are shown as(4)k0yx0,t=ut,
where u(t) is heat flux in continuous casting process which is calculated by
(5)ut=hy0,t-Tw,
where h is the equivalent heat transfer coefficient in secondary cooling zone W/(m2·K). y(0,t) is the steel billet surface temperature (K) and Tw is cooling water temperature (K):
(6)h=hxQwaterSWSLrwQwaterSWSLra+hr,
where hx and hr are constants. SW is the width of steel billet. SL is the length of steel billet. Qwater is the spray cooling water volume. Qair is the spray cooling air volume. The boundary condition at x=l can be written as
(7)k(l)yx(l,t)=0.


## 2.1.1. Assumptions

Based on the following key assumptions [17], continuous casting heat transfer model can be obtained.(1)
One-dimensional heat transfer is considered, so heat transfer can be ignored in the length and the width direction of billet.(2)
Steel billet at the same cooling zone cools uniformity.(3)
The meniscus surface is assumed to be flat.(4)
In mushy zone, we use the equivalent specific heat instead of specific heat.(5)
The convection heat transfer process in the liquid phase and two-phase zone is equal to conduction heat transfer process.


## 2.1.2. The Differential Equation

Based on the above assumptions, we can get continuous casting heat transfer model for one-dimensional model.(2)∂y∂t=kx∂2y∂x2,
where y is temperature (K). t is times (s). x is slab thickness direction (m) and k(x) is a coefficient depending on physical parameters of the steel, which is calculated by
(3)kx=λρCe,
where ρ is steel density (kg/m3), Ce is specific heat (J/(kg·k)), and λ is thermal conductivity (W/m·K).


## 2.1.3. Boundary Conditions

Steel billet goes through the crystallizer, secondary cooling zone, and air cooling zone. The different cooling sections have different cooling characteristics, so the boundary conditions are shown as(4)k0yx0,t=ut,
where u(t) is heat flux in continuous casting process which is calculated by
(5)ut=hy0,t-Tw,
where h is the equivalent heat transfer coefficient in secondary cooling zone W/(m2·K). y(0,t) is the steel billet surface temperature (K) and Tw is cooling water temperature (K):
(6)h=hxQwaterSWSLrwQwaterSWSLra+hr,
where hx and hr are constants. SW is the width of steel billet. SL is the length of steel billet. Qwater is the spray cooling water volume. Qair is the spray cooling air volume. The boundary condition at x=l can be written as
(7)k(l)yx(l,t)=0.


## 2.2. Optimal Control Model of Continuous Cast Secondary Cooling Zone

Based on continuous casting heat transfer model, we have optimal control model of continuous cast secondary cooling zone:(8)min⁡u∈UJ=∫0T‍y0,t-yaimt2dtmin⁡u∈UJ=+α∫0T‍ut-u*t2dts.t.yt-λρCeyxx=0,(x,t)∈(0,l)×(0,T],s.t.y(x,0)=y0,x∈(0,l),s.t.k(0)yx(0,t)=h[y(0,t)-Tw],t∈(0,T],s.t.k(l)yx(l,t)=0,t∈(0,T],
where y is steel slab temperature (K). y(0,t) is the surface temperature of steel slabs (K). yaim is the objective temperature of steel slabs (K), which depends on continuous casting technology. t is times (s). x is slab thickness direction (m). α is a positive number. y0 is initial casting temperature, which is usually a constant; U=L2(∂Ω), Ω∈(0,l)×(0,t). Other parameters have already been defined in continuous casting transfer model.Many researchers [18, 19] use y(0,t), the surface temperature of steel billet to evaluate the quality of billet. We hope that the surface temperature y(0,t) is close to the objective temperature yaim by controlling the boundary condition u(t). In the actual production of continuous casting, the controlled variable cannot be too large, so we also give a penalty for the controlled variable.According to (3) and (5), k(x) and u(t) in (1) are assigned, so optimal control problem in continuous casting secondary cooling zone transfers to a general optimal control problem in (1). Because of this, we only consider a general optimal control problem in later sections.


## 3. Adjoint Problem and Lipschitz Continuity of the Gradient of the Cost Function

### 3.1. Adjoint Problem

Based on the first optimization and then discrete method, we can get the adjoint equation firstly and then derive the optimality conditions, using numerical methods to solve this problem. The gradient of the cost function with respect to the control variable can be efficiently calculated by the adjoint problem. Now we give the strong version of the Lagrange function, and then we transfer all constraints into cost function by separate multipliersp(x,t), p(0,t), p(l,t), and p(x,0) and perform partial integrations with respect to space and time, so the Lagrange function can be obtained:
(9)L(y,p,x,t)=α∫0T‍ut-u*t2dt-∫0l‍∫0T‍ptydtdx-∫0T‍∫0l‍k(x)pxx(x,t)y(x,t)dxdt+∫0T‍[k(l)yx(l,t)p(l,t)]dt+∫0T‍[k(0)yx(0,t)p(0,t)+y0,t-yaim2]dt+∫0l‍yx,Tpx,Tdt-∫0l‍ϕxpx,0dx-∫0T‍utp0,tdt.The vibration of the functionL(y,p,x,t) can be expressed as
(10)Ly+δy,p,x,t=α∫0T‍ut-u*t2dt+∫0l‍[y(x,T)+δy(x,T)]p(x,T)dt-∫0l‍∫0T‍ptyx,t+δyx,tdtdx-∫0T‍∫0l‍k(x)pxx(x,t)[y(x,t)+δy(x,t)]dxdt+∫0T‍k(0)px(0,t)[y(0,t)+δy(0,t)]dt+∫0T‍y0,t+δy0,t-yaim2dt+∫0T‍k(l)px(l,t)[y(l,t)+δy(l,t)]dt-∫0l‍ϕxpx,0dx-∫0T‍utp0,tdt.According to (9) and (10), the following equation can be obtained:
(11)ΔL=L(y+δy,p,x,t)-L(y,p,x,t)=∫0l‍px,Tδyx,Tdx-∫0l‍∫0T‍ptδyx,tdtdx-∫0T‍∫lT‍k(x)pxxδy(x,t)dxdt+∫0T‍k(0)px(0,t)δy(0,t)dt+∫0T‍[y(0,t)-yaim(t)]δy(0,t)dt+∫0T‍klpxl,tδyl,tdt.According to (11), the following dual-equation can be obtained:
(12)pt=-kxpxx,x,t∈0,l×0,T,p(x,T)=0,x∈(0,l),-k0px0,t=y0,t-yaimt,t∈0,T,klpxl,t=0,t∈0,T.Now we obtained adjoint problem in our optimal control problem.


### 3.2. Lipschitz Continuity of the Gradient of the Cost Function

We can consider the variation of the cost function:(13)ΔJu=J(u+Δu)-J(u)=∫0T‍y0,t-yaimtΔy0,tdt+∫0T‍Δy0,t2dt+∫0T[u(t)-u*(t)]Δu(t)dt+∫0TΔut2dt.In order to prove that the gradient of the cost function is Lipschitz continuous, we give two other partial differential equations. According to (8), if we let Δy=y(x,t;u+Δu)-y(x,t;u), we obtain the following partial differential equation:
(14)Δyt=k(x)Δyxx,(x,t)∈(0,l)×(0,T],Δy(x,0)=0,x∈(0,l),-k(0)Δyx(0,t)=Δu(t),t∈(0,T],klΔyxl,t=0,t∈0,T.According to (12), the following partial differential equation can be obtained in a similar way:
(15)Δpt=-k(x)Δpxx,(x,t)∈(0,l)×(0,T],Δp(x,0)=0,x∈(0,l),-k0Δpx0,t=Δy0,t,t∈(0,T],klΔpxl,t=0,t∈0,T.Lemma 1.
For allΔu, the following identity holds:
(16)∫0T‍[y(0,t;u)-yaim(t)]Δy(0,t)dt=∫0T‍p0,tΔutdt.Proof.
It is easy to know that the following identity holds:(17)∫0T‍∫0l‍kxpxx,tΔyx,txdxdt=∫0T‍klpxl,tΔyl,tdt-∫0T‍k0px0,tΔy0,tdt.
Thus boundary conditions of (12) show
(18)∫0T‍[y(0,t;u)-yaim(t)]Δy(0,t)dt=-∫0T‍k0px0,tΔy0,tdt.
The following equations can be obtained:(19)∫0T‍[y(0,t;u)-yaim(t)]Δy(0,t)dt=∫0T‍∫0l‍kxpxx,tΔyx,txdxdt-∫0T‍k(l)px(l,t)Δy(l,t)dt=∫0T‍∫0l‍kxpxxx,tΔyx,tdxdt+∫0T‍∫0l‍k(x)px(x,t)Δyx(x,t)dxdt-∫0T‍∫0l‍px,tΔyx,tdxdt-∫0T‍k0p0,tΔyx0,tdt=∫0T‍p0,tΔudt.
Lemma1 can be proved.Lemma 2.
There exists a constantc>0, so that the following equation can be obtained:
(20)∫0T‍Δy0,t2dt≤c∫0T‍Δut2dt.Proof.
If we multiply both sides of (14) by Δy(x,t) and integrate on time and space, the following equation can be obtained:
(21)∫0T‍∫0l‍Δyt-kxΔyxxΔydxdt=0.
It is easy to know that the following identity holds:(22)∫0T‍∫0l‍ΔytΔydxdt=12∫0T‍∫0l‍Δy2dxdt=12∫0l‍Δyx,T2dx.
According to the boundary conditions of (14), the following equation can be derived:
(23)∫0l‍∫0T‍kxΔyxxΔydtdx=∫0l‍∫0T‍kxΔyxΔyxdtdx-∫0l‍∫0T‍k(x)Δyx2dtdx=-∫0l‍k0Δyx0,tΔy0,tdt-∫0l‍∫0T‍k(x)Δyx2dtdx=∫0T‍Δy(0,t)Δu(t)dt-∫0l‍∫0T‍k(x)Δyx2dtdx.
Taking (22) and (23) into (21), we have the following equations:
(24)∫0TΔyx,T2+∫0l‍∫0T‍kxΔyx2dtdx=∫0T‍Δy0,tΔutdt.
We use theε-inequality
(25)αβ≤εα22+β22ε,∀α,β∈R,∀ε>0.
On the right-hand side of (24), then we have
(26)∫0T‍ΔuΔy0,tdt≤12ε∫0TΔu2dt+ε2∫0T‍Δy0,t2dt.
According to (24) and (26), the following equation can be derived:
(27)∫0l‍∫0T‍kxΔyx2dtdx+∫0l‍Δyx,T2dt≤12ε∫0T‍Δu2dt+ε2∫0TΔy0,t2dt.
From literature [14], the following equation can be obtained:
(28)2l2∫0l‍∫0T‍Δy2dtdx+2l∫0T‍Δy0,t2dt≤12ε∫0T‍Δu2dt+ε2∫0T‍Δy0,t2dt.
So the following equation can be obtained:(29)∫0T‍Δy0,t2dt≤c∫0T‍Δut2dt.
We can know that there is a constantc which makes (20) established. Lemma 2 can be proved.Lemma 3.
Consider(30)∫0T‍Δy0,t2dt=∫0T‍Δp0,tΔutdt.Proof.
According to (15), the weak solution of this partial differential equation is the following equation:
(31)∫0T‍∫0l‍ΔpΔytdxdt-∫0T‍∫0l‍ΔpΔytdxdt+∫0T‍kxΔyΔpx=0x=ldt-∫0T‍∫0l‍kxΔpxΔyxdxdt=0.
According to the boundary conditions of  (14) and (15), the following equation can be obtained:
(32)∫0T‍k(l)Δy(l,t)Δpx(l,x)dt-∫0T‍k(0)Δy(0,t)Δpx(0,x)dt-∫0T‍k(l)Δyx(l,t)Δp(l,x)dt+∫0T‍k(0)Δyx(0,t)Δp(0,x)dt=∫0T‍Δy0,t2=∫0T‍Δp0,tΔutdt.
Lemma3 can be proved.Theorem 4.
If the gradient of cost function is Lipschitz continuous, there is a constantL to make the following equation hold:
(33)J′u+Δu-J′(u)≤LΔu.Proof.
Taking into account Lemmas1, 2, and 3, we can obtain the following equation:
(34)J′u+Δu-J′u,Δu=∫0T‍Δp(0,t)Δudt=∫0T‍Δy0,t2≤L∫0T‍Δut2dt.


## 3.1. Adjoint Problem

Based on the first optimization and then discrete method, we can get the adjoint equation firstly and then derive the optimality conditions, using numerical methods to solve this problem. The gradient of the cost function with respect to the control variable can be efficiently calculated by the adjoint problem. Now we give the strong version of the Lagrange function, and then we transfer all constraints into cost function by separate multipliersp(x,t), p(0,t), p(l,t), and p(x,0) and perform partial integrations with respect to space and time, so the Lagrange function can be obtained:
(9)L(y,p,x,t)=α∫0T‍ut-u*t2dt-∫0l‍∫0T‍ptydtdx-∫0T‍∫0l‍k(x)pxx(x,t)y(x,t)dxdt+∫0T‍[k(l)yx(l,t)p(l,t)]dt+∫0T‍[k(0)yx(0,t)p(0,t)+y0,t-yaim2]dt+∫0l‍yx,Tpx,Tdt-∫0l‍ϕxpx,0dx-∫0T‍utp0,tdt.The vibration of the functionL(y,p,x,t) can be expressed as
(10)Ly+δy,p,x,t=α∫0T‍ut-u*t2dt+∫0l‍[y(x,T)+δy(x,T)]p(x,T)dt-∫0l‍∫0T‍ptyx,t+δyx,tdtdx-∫0T‍∫0l‍k(x)pxx(x,t)[y(x,t)+δy(x,t)]dxdt+∫0T‍k(0)px(0,t)[y(0,t)+δy(0,t)]dt+∫0T‍y0,t+δy0,t-yaim2dt+∫0T‍k(l)px(l,t)[y(l,t)+δy(l,t)]dt-∫0l‍ϕxpx,0dx-∫0T‍utp0,tdt.According to (9) and (10), the following equation can be obtained:
(11)ΔL=L(y+δy,p,x,t)-L(y,p,x,t)=∫0l‍px,Tδyx,Tdx-∫0l‍∫0T‍ptδyx,tdtdx-∫0T‍∫lT‍k(x)pxxδy(x,t)dxdt+∫0T‍k(0)px(0,t)δy(0,t)dt+∫0T‍[y(0,t)-yaim(t)]δy(0,t)dt+∫0T‍klpxl,tδyl,tdt.According to (11), the following dual-equation can be obtained:
(12)pt=-kxpxx,x,t∈0,l×0,T,p(x,T)=0,x∈(0,l),-k0px0,t=y0,t-yaimt,t∈0,T,klpxl,t=0,t∈0,T.Now we obtained adjoint problem in our optimal control problem.


## 3.2. Lipschitz Continuity of the Gradient of the Cost Function

We can consider the variation of the cost function:(13)ΔJu=J(u+Δu)-J(u)=∫0T‍y0,t-yaimtΔy0,tdt+∫0T‍Δy0,t2dt+∫0T[u(t)-u*(t)]Δu(t)dt+∫0TΔut2dt.In order to prove that the gradient of the cost function is Lipschitz continuous, we give two other partial differential equations. According to (8), if we let Δy=y(x,t;u+Δu)-y(x,t;u), we obtain the following partial differential equation:
(14)Δyt=k(x)Δyxx,(x,t)∈(0,l)×(0,T],Δy(x,0)=0,x∈(0,l),-k(0)Δyx(0,t)=Δu(t),t∈(0,T],klΔyxl,t=0,t∈0,T.According to (12), the following partial differential equation can be obtained in a similar way:
(15)Δpt=-k(x)Δpxx,(x,t)∈(0,l)×(0,T],Δp(x,0)=0,x∈(0,l),-k0Δpx0,t=Δy0,t,t∈(0,T],klΔpxl,t=0,t∈0,T.Lemma 1.
For allΔu, the following identity holds:
(16)∫0T‍[y(0,t;u)-yaim(t)]Δy(0,t)dt=∫0T‍p0,tΔutdt.Proof.
It is easy to know that the following identity holds:(17)∫0T‍∫0l‍kxpxx,tΔyx,txdxdt=∫0T‍klpxl,tΔyl,tdt-∫0T‍k0px0,tΔy0,tdt.
Thus boundary conditions of (12) show
(18)∫0T‍[y(0,t;u)-yaim(t)]Δy(0,t)dt=-∫0T‍k0px0,tΔy0,tdt.
The following equations can be obtained:(19)∫0T‍[y(0,t;u)-yaim(t)]Δy(0,t)dt=∫0T‍∫0l‍kxpxx,tΔyx,txdxdt-∫0T‍k(l)px(l,t)Δy(l,t)dt=∫0T‍∫0l‍kxpxxx,tΔyx,tdxdt+∫0T‍∫0l‍k(x)px(x,t)Δyx(x,t)dxdt-∫0T‍∫0l‍px,tΔyx,tdxdt-∫0T‍k0p0,tΔyx0,tdt=∫0T‍p0,tΔudt.
Lemma1 can be proved.Lemma 2.
There exists a constantc>0, so that the following equation can be obtained:
(20)∫0T‍Δy0,t2dt≤c∫0T‍Δut2dt.Proof.
If we multiply both sides of (14) by Δy(x,t) and integrate on time and space, the following equation can be obtained:
(21)∫0T‍∫0l‍Δyt-kxΔyxxΔydxdt=0.
It is easy to know that the following identity holds:(22)∫0T‍∫0l‍ΔytΔydxdt=12∫0T‍∫0l‍Δy2dxdt=12∫0l‍Δyx,T2dx.
According to the boundary conditions of (14), the following equation can be derived:
(23)∫0l‍∫0T‍kxΔyxxΔydtdx=∫0l‍∫0T‍kxΔyxΔyxdtdx-∫0l‍∫0T‍k(x)Δyx2dtdx=-∫0l‍k0Δyx0,tΔy0,tdt-∫0l‍∫0T‍k(x)Δyx2dtdx=∫0T‍Δy(0,t)Δu(t)dt-∫0l‍∫0T‍k(x)Δyx2dtdx.
Taking (22) and (23) into (21), we have the following equations:
(24)∫0TΔyx,T2+∫0l‍∫0T‍kxΔyx2dtdx=∫0T‍Δy0,tΔutdt.
We use theε-inequality
(25)αβ≤εα22+β22ε,∀α,β∈R,∀ε>0.
On the right-hand side of (24), then we have
(26)∫0T‍ΔuΔy0,tdt≤12ε∫0TΔu2dt+ε2∫0T‍Δy0,t2dt.
According to (24) and (26), the following equation can be derived:
(27)∫0l‍∫0T‍kxΔyx2dtdx+∫0l‍Δyx,T2dt≤12ε∫0T‍Δu2dt+ε2∫0TΔy0,t2dt.
From literature [14], the following equation can be obtained:
(28)2l2∫0l‍∫0T‍Δy2dtdx+2l∫0T‍Δy0,t2dt≤12ε∫0T‍Δu2dt+ε2∫0T‍Δy0,t2dt.
So the following equation can be obtained:(29)∫0T‍Δy0,t2dt≤c∫0T‍Δut2dt.
We can know that there is a constantc which makes (20) established. Lemma 2 can be proved.Lemma 3.
Consider(30)∫0T‍Δy0,t2dt=∫0T‍Δp0,tΔutdt.Proof.
According to (15), the weak solution of this partial differential equation is the following equation:
(31)∫0T‍∫0l‍ΔpΔytdxdt-∫0T‍∫0l‍ΔpΔytdxdt+∫0T‍kxΔyΔpx=0x=ldt-∫0T‍∫0l‍kxΔpxΔyxdxdt=0.
According to the boundary conditions of  (14) and (15), the following equation can be obtained:
(32)∫0T‍k(l)Δy(l,t)Δpx(l,x)dt-∫0T‍k(0)Δy(0,t)Δpx(0,x)dt-∫0T‍k(l)Δyx(l,t)Δp(l,x)dt+∫0T‍k(0)Δyx(0,t)Δp(0,x)dt=∫0T‍Δy0,t2=∫0T‍Δp0,tΔutdt.
Lemma3 can be proved.Theorem 4.
If the gradient of cost function is Lipschitz continuous, there is a constantL to make the following equation hold:
(33)J′u+Δu-J′(u)≤LΔu.Proof.
Taking into account Lemmas1, 2, and 3, we can obtain the following equation:
(34)J′u+Δu-J′u,Δu=∫0T‍Δp(0,t)Δudt=∫0T‍Δy0,t2≤L∫0T‍Δut2dt.


## 4. Conjugate Gradient Algorithm Solves Optimal Control of Parabolic Partial Differential Equations and Its Convergent Analysis

### 4.1. Conjugate Gradient Algorithm

After we prove that the gradient of cost function is Lipschitz continuous, we can use the conjugate gradient algorithm to solve this PDE optimization problem. If we choose the reasonable parameters in conjugate gradient algorithm, we can prove the convergence of conjugate gradient algorithm.As shown in Figure2, 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 parameter in conjugate gradient algorithm, and eps is a small constant.Figure 2
Conjugate gradient algorithm process.Because we use FR conjugate method to solve this optimal problem,dk and βk can be calculated by the following equations:
(35)dk=-gk+βkdk-1,βk=gkTgkgk-1Tgk-1.Step size largely infects the convergent speed of conjugate gradient algorithm. If we choose Wolfe line search rules, we have the following formula [20]:
(36)fxk+αkdk≤fxk+σ1αkgkTdk,∇fxk+αkdk≥σ2gkTdk,0<σ1<σ2<12.


### 4.2. Convergence Analysis of Conjugate Gradient Algorithm

Theorem 5.
Ifdk, βk, and αk are calculated by (35) and (36), the following equation holds:
(37)liminf⁡k→∞gk=0.Proof.
According to (36), the following equations can be obtained:
(38)-∑i=0k‍σ2i≤gkTdkgk2≤-2+∑i=0k‍σ2i,∀k=0,1,…,∞,(39)-∑i=0k‍σ2i<-∑i=0∞‍σ2i=11-σ2<2.
According to (38) and (39), the following equation can be obtained:
(40)gkTdk-1≤-σ2gk-1Tdk-1<σ21-σ2gk-1.
According to the recursion formula ofdk, the following equation can be represented:
(41)dk2=gk2-2βk-1gkTdk-1+βk-12dk-12≤gk2-2σ21-σ2gk2+βk-12dk-1=1+σ21-σ2gk2+βk-12dk-12=dk2≤1+σ21-σ2gk4∑i=1k‍1gi2+gk2g02.
If this algorithm is not convergent, (37) does not hold and there exists a constant ε0>0, so that the following identity holds:
(42)gk≥ε0,dk2≤τ1k+1.
According to (39), we can obtain the following equations:
(43)cos⁡θk=gkTdkgkdk≥(2-∑i=0k‍σ2j)gkdk≥1-2σ21-σ2gkdk,∑k=0∞‍cos⁡2θk≥1-2σ21-σ22∑k=0∞‍gk2dk2≥τ2∑k=0∞‍1k+1.
Sinceg(x) is the gradient of cost function, we have proven that g(x) is Lipschitz continuous. There exists a constant, so that the following inequality holds:
(44)σ2gkTdk≤gk+1Tdk=gkT+gk+1-gkTdk≤gkTdk+αkLdk2.
Then we can obtain the following equations:(45)αk≥-1-σ2Ldk2gkTdk,fk+1≤fk-σ11-σ2Lgk2cos⁡2θk.
The sequencefk is monotonic decline and has lower bound, so ∑k=0∞gk2cos⁡2θk is convergent. Since gk≥ε0, ∑k=0∞cos⁡2θk is convergent and we can conclude that cos⁡2θk is convergent and not convergent at the same time. Thus the assumption gk is invalid and we can prove that (37) holds, so this conjugate gradient algorithm is convergent.


### 4.3. An Optimal Choice of the Stopping in the Conjugate Gradient Algorithm

It is important to understand the behavior of the cost function value depending on the iteration number. For given smooth objective temperatureyaim, an optimal control problem is solved by conjugate gradient algorithm:
(46)yaim=1173.16+100sinπ10tK,k=7250×54050,eps=1×10-1,ux,0=1173.16K,L=0.25m,T=20s,u*=0,α=1×10-7.As shown in Figure3, the behaviour of cost function value as a function of the iteration number consists of three phases: the initial phase of rapid decrease but very short duration, the second phase of slow decrease, and the third phase of almost constant behaviour.Figure 3
Surface temperature error of billet.When dealing with optimal control problems, the stopping parameter plays an important role in conjugate gradient algorithm. However, it is difficult to choose a good stopping parametereps. If stopping parameter eps is too large, the result cannot converge to optimal solution. The iteration number is too large because stopping parameter is too small. As shown in Table 1, the best value of stopping parameter is achieved at eps=0.1.Table 1
Determination of an optimal value stopping parameter.
eps
Iteration number
Cost function value
1
42
394.80
0.1
115
258.15
0.01
170
258.15


## 4.1. Conjugate Gradient Algorithm

After we prove that the gradient of cost function is Lipschitz continuous, we can use the conjugate gradient algorithm to solve this PDE optimization problem. If we choose the reasonable parameters in conjugate gradient algorithm, we can prove the convergence of conjugate gradient algorithm.As shown in Figure2, 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 parameter in conjugate gradient algorithm, and eps is a small constant.Figure 2
Conjugate gradient algorithm process.Because we use FR conjugate method to solve this optimal problem,dk and βk can be calculated by the following equations:
(35)dk=-gk+βkdk-1,βk=gkTgkgk-1Tgk-1.Step size largely infects the convergent speed of conjugate gradient algorithm. If we choose Wolfe line search rules, we have the following formula [20]:
(36)fxk+αkdk≤fxk+σ1αkgkTdk,∇fxk+αkdk≥σ2gkTdk,0<σ1<σ2<12.


## 4.2. Convergence Analysis of Conjugate Gradient Algorithm

Theorem 5.
Ifdk, βk, and αk are calculated by (35) and (36), the following equation holds:
(37)liminf⁡k→∞gk=0.Proof.
According to (36), the following equations can be obtained:
(38)-∑i=0k‍σ2i≤gkTdkgk2≤-2+∑i=0k‍σ2i,∀k=0,1,…,∞,(39)-∑i=0k‍σ2i<-∑i=0∞‍σ2i=11-σ2<2.
According to (38) and (39), the following equation can be obtained:
(40)gkTdk-1≤-σ2gk-1Tdk-1<σ21-σ2gk-1.
According to the recursion formula ofdk, the following equation can be represented:
(41)dk2=gk2-2βk-1gkTdk-1+βk-12dk-12≤gk2-2σ21-σ2gk2+βk-12dk-1=1+σ21-σ2gk2+βk-12dk-12=dk2≤1+σ21-σ2gk4∑i=1k‍1gi2+gk2g02.
If this algorithm is not convergent, (37) does not hold and there exists a constant ε0>0, so that the following identity holds:
(42)gk≥ε0,dk2≤τ1k+1.
According to (39), we can obtain the following equations:
(43)cos⁡θk=gkTdkgkdk≥(2-∑i=0k‍σ2j)gkdk≥1-2σ21-σ2gkdk,∑k=0∞‍cos⁡2θk≥1-2σ21-σ22∑k=0∞‍gk2dk2≥τ2∑k=0∞‍1k+1.
Sinceg(x) is the gradient of cost function, we have proven that g(x) is Lipschitz continuous. There exists a constant, so that the following inequality holds:
(44)σ2gkTdk≤gk+1Tdk=gkT+gk+1-gkTdk≤gkTdk+αkLdk2.
Then we can obtain the following equations:(45)αk≥-1-σ2Ldk2gkTdk,fk+1≤fk-σ11-σ2Lgk2cos⁡2θk.
The sequencefk is monotonic decline and has lower bound, so ∑k=0∞gk2cos⁡2θk is convergent. Since gk≥ε0, ∑k=0∞cos⁡2θk is convergent and we can conclude that cos⁡2θk is convergent and not convergent at the same time. Thus the assumption gk is invalid and we can prove that (37) holds, so this conjugate gradient algorithm is convergent.


## 4.3. An Optimal Choice of the Stopping in the Conjugate Gradient Algorithm

It is important to understand the behavior of the cost function value depending on the iteration number. For given smooth objective temperatureyaim, an optimal control problem is solved by conjugate gradient algorithm:
(46)yaim=1173.16+100sinπ10tK,k=7250×54050,eps=1×10-1,ux,0=1173.16K,L=0.25m,T=20s,u*=0,α=1×10-7.As shown in Figure3, the behaviour of cost function value as a function of the iteration number consists of three phases: the initial phase of rapid decrease but very short duration, the second phase of slow decrease, and the third phase of almost constant behaviour.Figure 3
Surface temperature error of billet.When dealing with optimal control problems, the stopping parameter plays an important role in conjugate gradient algorithm. However, it is difficult to choose a good stopping parametereps. If stopping parameter eps is too large, the result cannot converge to optimal solution. The iteration number is too large because stopping parameter is too small. As shown in Table 1, the best value of stopping parameter is achieved at eps=0.1.Table 1
Determination of an optimal value stopping parameter.
eps
Iteration number
Cost function value
1
42
394.80
0.1
115
258.15
0.01
170
258.15


## 5. Simulation Experiment

To test the effectiveness of the method in this paper, the simulation experiments are done in actual plant data. The experiment result proves the effectiveness of the method in this paper. Physical parameters of steel grade are given in Table2.Table 2
Physical parameters of steel grade.
Items
Parameters
Specific heat
540 (J/(kg⋅k))
Density
7250 (kg/m3)
Thermal conductivity
30 (W/m⋅K)Example 1.
Experimental parameters are shown as(47)yaim=1173.16K,t<10s,1273.16K,t≥10s,k=7250×54050,eps=1×10-1,ux,0=1173.16K,L=0.25m,T=20s,u*=0,α=1×10-7.In this example, the objective temperature changes from 1173.16 (K) to 1273.16 (K), when time is equal to 10 seconds. Figure4 shows that the actual temperature can follow the objective temperature very well, so the quality of billet can be ensured. From Figure 5, we can know that the maximum error is about 50 (K). The temperature error is very small in most of the time. Based on technological requirements in continuous casting process, the behaviour of the surface temperature ensures that the billet has a good quality. Figure 6 shows the controlled variable. Because of penalty of controlled variable, the controlled variable cannot be very large. When the controlled variable is less than zero, this process is a heating process. When the controlled variable is larger than zero, this process is a cooling process.Figure 4
Surface temperature of billet.Figure 5
Surface temperature error of billet.Figure 6
Controlled variable.Example 2.
Experimental parameters are shown as(48)yaim=1173.16+100sin(π10t)(K)k=7250×54050,eps=1×10-1,ux,0=1173.16K,L=0.25m,T=20s,u*=0,α=1×10-7.In this example, the objective temperature is a sinusoidal curve which is continuous function. Figure7 shows that the actual temperature can follow the objective temperature very well, so the quality of billet can be ensured. From Figure 8, we can know that the maximum error is about 8 (K). Because the objective temperature is a sinusoidal curve, the maximum error becomes smaller than Example 1. The temperature error is very small in most of the time. Based on technological requirements in continuous casting process, the behaviour of the surface temperature ensures that the billet has a good quality. Figure 9 shows the controlled variable. Because of a penalty of controlled variable, the controlled variable cannot be very large.Figure 7
Surface temperature of billet.Figure 8
Surface temperature error of billet.Figure 9
Controlled variable.Example 3.
Experimental parameters are shown as(49)yaim=1673.16K,k=7250×54050,eps=1×10-1,ux,0=1173.16K,L=0.25m,T=20s,u*=0,α=1×10-7.In this example, the objective temperature is 1673.16 (K), but the initial temperature of billet is 1844.16 (K). Figure10 shows that the actual temperature can follow the objective temperature very well. While the billet quality can be ensured, the surface temperature of billet can approach the objective temperature rapidly. From Figure 11, we can know that the maximum error is about 160 (K). Due to the penalties of control variable, the actual temperature cannot step to the objective temperature. Figure 12 shows the controlled variable. Because of penalty of controlled variable, the controlled variable cannot be very large.Figure 10
Surface temperature of billet.Figure 11
Surface temperature error of billet.Figure 12
Controlled variable.


## 6. Conclusions

In order to solve optimization problem for set-point values in secondary cooling process, this paper applies first optimization and then discrete method to this optimal problem. We prove that the gradient of cost function is Lipschitz continuous. Base on this, we propose an improved conjugate algorithm to solve this problem and prove that this algorithm is convergent.



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