# On the Regularized Solutions of Optimal Control Problem in a Hyperbolic System

Authors: Yeşim Saraç; Murat Subaşı
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

We use the initial condition on the state variable of a hyperbolic problem as control function and formulate a control problem whose solution implies the minimization at the final time of the distance measured in a suitable norm between the solution of the problem and given targets. We prove the existence and the uniqueness of the optimal solution and establish the optimality condition. An iterative algorithm is constructed to compute the required optimal control as limit of a suitable subsequence of controls. An iterative procedure is implemented and used to numerically solve some test problems.


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## Body

## 1. Introduction and Statement of the Problem

Optimal control problems for hyperbolic equations have been investigated by Lions in his famous book [1].  Lions examined the problems in detail when the control function is at the right hand side and in the boundary condition of the hyperbolic problem. Furthermore, when the control is in the boundaries [2–4], in the coefficient [5, 6], and at the right hand side  of the equation [7, 8], there have been some control problem studies for different types of cost functionals. As for the control of initial conditions, Lions mentioned the control of the initial velocity of the system in detail but stated briefly the control of initial status of the system solving the system in  L2.In this study, we consider the following problem of minimizing the cost functional:(1.1)Jα(φ)=∫0l[ut(x,T;φ)-y1(x)]2dx+∫0l[ux(x,T;φ)-y2(x)]2dx+α∫0lφx2dx,
under the following condition:
(1.2)utt-a2uxx=F(x,t),(x,t)∈Ω:=(0,l)×(0,T]u(x,0)=φ(x),ut(x,0)=ψ(x),x∈(0,l)u(0,t)=0,u(l,t)=0,t∈(0,T].
Since the problem is usually ill posed for α=0, we use the parameter α>0 as the regularization parameter which is the strong convexity constant, and this guaranties the uniqueness and stability of the regularized solution. The functional  Jα(φ) is called cost functional and the term ∥φ∥H012 is called penalization term; its role is, on one hand, to avoid using “too large” controls in the minimization of  Jα(φ) and, on the other hand, to assure coercivity for Jα(φ).Lions in [1] mentioned the observation of u(x,T;φ) in L2(0,l) and ut(x,T;φ) in H-1(0,l) for the control φ∈L2(0,l). Except this study, there is no investigation in the literature about the control of initial status of the hyperbolic system up to now. In this study, we investigate different targets. With the choice of the functional in (1.1), we use ut(x,T;φ) and ux(x,T;φ), which correspond to final velocity and force, respectively,  for the control φ∈H01(0,l). Since the Fréchet differential of the cost functional cannot be obtained with the usage of usual norm in H01, we get the differentiability with the only use of  H01-Poincare norm.The spaceH01(0,l) is a Hilbert subspace of H1(0,l) and the H01-Poincare inner product and the H01-Poincare norm are defined, respectively, as
(1.3)(u,v)H01=(∇u,∇v)L2,‖u‖H01=‖∇u‖L2.
Let
(1.4)Φad=closed,convexsubsetofH01(0,l).
We search for
(1.5)Infφ∈Φad⁡Jα(φ).
We organize this paper as follows. In Section 2, we establish the existence and the uniqueness of the optimal solution. In Section 3, we derive the necessary optimality condition. In Section 4, we construct an algorithm for the numerical approximation of the optimal solution according to steepest descent algorithm. In Section 5, we give symbolic representation for optimal solution by using this algorithm on some examples.


## 2. Existence and Uniqueness of the Optimal Solution

First we state the generalized solution of the hyperbolic problem (1.2).Definition 2.1.
The generalized solution of (1.2) will be defined as the function u∈H01(Ω),u(x,0)=φ(x)  which satisfies the following integral identity:
(2.1)∫0T∫0l(-utvt+a2uxvx)dxdt=∫0T∫0lFvdxdt+∫0lψ(x)v(x,0)dx,
for all v∈H01(Ω) with v(x,T)=0. To have this solution the following is needed:
(2.2)F∈L2(Ω),φ∈H1(0,l),ψ∈L2(0,l).Theorem 2.2.
Suppose that (2.2) holds, then (1.2) has a unique generalized solution and the following estimate is valid for the solution:
(2.3)‖u‖H01(Ω)2≤c1(‖φ‖H01(0,l)2+‖ψ‖L2(0,l)2+‖F‖L2(Ω)2).
Proof of this theorem can easily be obtained by Galerkin method used in [9].The strategy to prove existence and uniqueness of this optimal control is to use the relationship between minimization of quadratic functionals and variational problems corresponding to symmetric bilinear forms. The key point is to writeJα(φ) in the following way:
(2.4)Jα(φ)=π(φ,φ)-2Lφ+b.
Here
(2.5)π(φ,φ)=∫0l[ut(x,T;φ)-ut(x,T;0)]2dx+∫0l[ux(x,T;φ)-ux(x,T;0)]2dx+α∫0lφx2dx,
is bilinear (since the mapping φ→u[φ]-u[0] is linear) and symmetric.Also, the difference functionδu=u(x,t;φ)-u(x,t;0) is the solution of the following problem:
(2.6)δutt-a2δuxx=0,δu(x,0)=φ,δut(x,0)=0,δu(0,t)=0,δu(l,t)=0,
and for the solution of this problem the following estimates are valid:
(2.7)‖δut(x,T)‖L2(0,l)2≤c2‖φ‖H01(0,l)2,‖δux(x,T)‖L2(0,l)2≤c3‖φ‖H01(0,l)2.
Hence, we write the following:
(2.8)|π(φ,φ)|=∫0l(δut)2dx+∫0l(δux)2dx+α∫0lφx2dx=‖δut‖L2(0,l)2+‖δux‖L2(0,l)2+α‖φ‖H01(0,l)2≥α‖φ‖H01(0,l)2,
and this implies the coercivity of π(φ,φ). Since
(2.9)π(φ,η)=∫0l[ut(x,T;φ)-ut(x,T;0)][ut(x,T;η)-ut(x,T;0)]dx+∫0l[ux(x,T;φ)-ux(x,T;0)][ux(x,T;η)-ux(x,T;0)]dx+α∫0lφxηxdx
applying Cauchy-Schwartz inequality, we get
(2.10)|π(φ,η)|≤‖ut(x,T;φ)-ut(x,T;0)‖L2(0,l)‖ut(x,T;η)-ut(x,T;0)‖L2(0,l)+‖ux(x,T;φ)-ux(x,T;0)‖L2(0,l)‖ux(x,T;η)-ux(x,T;0)‖L2(0,l)+α‖φ‖H01(0,l)‖η‖H01(0,l),|π(φ,η)|≤‖δut(x,T;φ)‖L2(0,l)‖δut(x,T;η)‖L2(0,l)+‖δux(x,T;φ)‖L2(0,l)‖δux(x,T;η)‖L2(0,l)+α‖φ‖H01(0,l)‖η‖H01(0,l)
for  δu(x,T;φ)=u(x,T;φ)-u(x,T;0) and δu(x,T;η)=u(x,T;η)-u(x,T;0).So, we obtain(2.11)|π(φ,η)|≤c4‖φ‖H01(0,l)‖η‖H01(0,l)+c5‖φ‖H01(0,l)‖η‖H01(0,l)+α‖φ‖H01(0,l)‖η‖H01(0,l)
using (2.7) and write
(2.12)|π(φ,η)|≤c6‖φ‖H01(0,l)‖η‖H01(0,l),
for c6=max{c4,c5,α}. Then π(φ,η) is continuous.The functionalLφ in (2.4) is defined as
(2.13)Lφ=∫0l[ut(x,T;φ)-ut(x,T;0)][y1(x)-ut(x,T;0)]dx+∫0l[ux(x,T;φ)-ux(x,T;0)][y2(x)-ux(x,T;0)]dx.
We can easily write that
(2.14)Lφ≤‖δut(x,T)‖L2(0,l)‖y1(x)-ut(x,T;0)‖L2(0,l)+‖δux(x,T)‖L2(0,l)‖y2(x)-ux(x,T;0)‖L2(0,l),Lφ≤c7‖φ‖H01(0,l)
using (2.7). Hence we see that the functional Lφ is continuous.The numberb∈ℝ in (2.4) is defined as
(2.15)b=∫0l[y1(x)-ut(x,T;0)]2dx+∫0l[y2(x)-ux(x,T;0)]2dx.
Therefore we have established the conditions of the following existence and uniqueness theorem for the problem.Theorem 2.3.
Letπ(φ,φ) be a continuous symmetric bilinear coercive form and Lφ a continuous linear form on H01. Then there exists a unique element φ*∈Φad such that
(2.16)Jα(φ*)=Infφ∈ΦadJα(φ).
Proof of this theorem can easily be obtained by showing the weak lower semicontinuity of Jα(φ) as in [1].


## 3. Lagrange Multipliers and Optimality Condition

To derive the optimality condition, let us introduce the LagrangianL(u,φ,zt), given by
(3.1)L(u,φ,zt)=∫0l[ut(x,T;φ)-y1(x)]2dx+∫0l[ux(x,T;φ)-y2(x)]2dx+α∫0l(φx)2dx+∫0T∫0l[utt-a2uxx-F(x,t)]ztdxdt.
Notice that L is linear in zt, therefore
(3.2)Lzt′(u,φ,zt)=0
corresponds to the state equation (1.2). Moreover,
(3.3)Lu′(u,φ,zt)=0
generates the following adjoint problem:
(3.4)ztt-a2zxx=0,zt(x,T)=-2[ut(x,T;φ)-y1(x)],zx(x,T)=-2a2[ux(x,T;φ)-y2(x)],z(0,t)=0,z(l,t)=0,
while
(3.5)Lφ′(u,φ,zt)=0
constitutes the following Euler equation:
(3.6)〈Jα′(φ),δφ〉H01(0,l)=∫0l(-a2zx(x,0)+2αφx)(δφ)xdx=0,∀δφ∈Φad.
So, we can state the following theorem in view of [10].Theorem 3.1.
The controlφ* and the state u*=u(φ*) are optimal if there exists a multiplier zt*∈Φad such that z* and φ* satisfy the following optimality conditions:
(3.7)〈-a2z*(x,0)+2αφ*,φ-φ*〉H01(0,l)≥0,
for ∀φ∈Φad.


## 4. An Iterative Algorithm and Its Convergence

Now, we can apply standard steepest descent iteration.  Gradient ofJα at any φ is given by
(4.1)∇Jα(φ)=-a2z(x,0)+2αφ.
It turns out that -∇Jα(φ) plays the role of the steepest descent direction for Jα. This suggests an iterative procedure to compute a sequence of controls {φk} convergent to the optimal one.Select an initial controlφ0.  If  φk is known (k≥0) then φk+1 is computed according to the following scheme.(1)
Solve the state problem (1.2) in the sense (2.1) and get corresponding uk.(2)
Knowinguk solve the adjoint problem (3.4).(3)
Usingzk get the gradient (∇Jα)k.(4)
Set(4.2)φk+1=φk-βk∇Jα(φk),and select the relaxation parameterβk in order to assure that
(4.3)Jα(φk+1)-Jα(φk)=βk[-‖Jα′(φk)‖2+o(βk)βk]<0,for sufficiently smallβk>0.Concerning the choice of the relaxation parameter, there are several possibilities and these can be found in any optimization books.One of the following can be taken as a stopping criterion to the iteration process:(4.4)‖φk+1-φk‖<ε1,|Jα(φk+1)-Jα(φk)|<ε2,‖Jα′(φk)‖<ε3.Lemma 4.1.
The cost functional (1.1) is strongly convex with the strong convexity constant α.
From the following strongly convex functional definition:(4.5)Jα(βφ1+(1-β)φ2)≤βJα(φ1)+(1-β)Jα(φ2)-χβ(1-β)‖φ1-φ2‖H01(0,l)2,
we can see that the cost functional (1.1) is strongly convex with the constant χ=α.So, we can give the following theorem which states the convergence of the minimizer to optimal solution.Theorem 4.2.
Letφ* be optimum solution of the problem (1.1)–(1.5). Then the minimizer given in (4.2) satisfies the following inequality:
(4.6)‖φk-φ*‖2≤2α(Jα(φk)-Jα(φ*)),k=0,1,2,….Proof.
If we takeβ=1/2 in the definition of the strongly convex functional, we write
(4.7)Jα(12φk+12φ*)≤12Jα(φk)+12Jα(φ*)-α14‖φk-φ*‖L2(0,l)2.
Since
(4.8)Jα(φ*)≤Jα(12φk+12φ*),
we find
(4.9)Jα(φ*)≤12Jα(φk)+12Jα(φ*)-α14‖φk-φ*‖L2(0,l)2,‖φk-φ*‖2≤2α(Jα(φk)-Jα(φ*)).


## 5. Numerical Examples

Example 5.1.
Let us consider the following problem of minimizing the cost functional:(5.1)Jα(φ)=∫01[ut(x,1;φ)-(-sin⁡(1){14x20≤x≤12-x3+54x2-14x12≤x≤1)]2dx+∫01[ux(x,1;φ)-(cos⁡(1){12x0≤x≤12-3x2+52x-1412≤x≤1)]2dx+α∫01φx2dx
under the following condition:
(5.2)utt-uxx=cos⁡(t){-14(x2+2)0≤x<12,t∈(0,1]14(4x3-5x2+25x-10)12<x≤1,t∈(0,1],u(x,0)=φ(x),ut(x,0)=0,x∈(0,1),u(0,t)=0,u(1,t)=0,t∈(0,1].
Rewrite the functional as
(5.3)Jα(φ)=Jα1(φ)+αJα2(φ),
where
(5.4)Jα1(φ)=∫01[ut(x,1;φ)-(-sin⁡(1){14x20≤x≤12-x3+54x2-14x12≤x≤1)]2dx+∫01[ux(x,1;φ)-(cos⁡(1){12x0≤x≤12-3x2+52x-1412≤x≤1)]2dx,Jα2(φ)=∫01φx2dx
Choosing α=0.1 and starting the initial element φ0=sinπx, then we get the minimizing sequence. Here the relaxation parameter βk=0.01 assures the inequality J0.1(φk+1)<J0.1(φk). In this example if we use the stopping criteria J0.1(φk+1)-J0.1(φk)>-0.166×10-9, we get the following minimizing element after 250 iterations:
(5.5)φ250=0.00134242sin⁡(15.70796327x)+0.00859668sin⁡(9.424777962x)-0.00037644sin⁡(25.13274123x)-0.00148524sin⁡(18.84955592x)+0.06814824sin⁡(3.141592654x)-0.03968284sin⁡(6.283185308x)-0.00032149sin⁡(31.41592654x)+0.00024578sin⁡(28.27433389x)-0.00299971sin⁡(12.56637062x)+0.00061268sin⁡(21.99114858x)
and for this optimal control the values of the J0.11(φ250) and J0.12(φ250) are such as
(5.6)J0.11(φ250)=0.00005173,J0.12(φ250)=0.05881965.
For different α the values of Jα1(φ), Jα2(φ) and optimal controls φ* are given in Table 1.Table 1
α
J
α
1
(
φ
*
)
J
α
2
(
φ
*
)
φ
*
0.5
0.00066463
0.05016802
0.06066266
sin
(
3.14159265
x
)
-
0.03734277
sin
(
6.28318530
x
)
+
0.00825020
sin
(
9.42477796
x
)
-
0.00290788
sin
(
12.5663706
x
)
+
0.00130929
sin
(
15.7079632
x
)
-
0.00145454
sin
(
18.8495559
x
)
+
0.00060179
sin
(
21.9911485
x
)
-
0.00037057
sin
(
25.1327412
x
)
+
0.00024238
sin
(
28.2743338
x
)
-
0.00031746
sin
(
31.4159265
x
)
0.3
0.00027601
0.05420087
0.06418794
sin
(
3.14159265
x
)
-
0.03847726
sin
(
6.28318530
x
)
+
0.00841988
sin
(
9.42477796
x
)
-
0.00295308
sin
(
12.5663706
x
)
+
0.00132564
sin
(
15.7079632
x
)
-
0.00146973
sin
(
18.8495559
x
)
+
0.00060719
sin
(
21.9911485
x
)
-
0.00037348
sin
(
25.1327412
x
)
+
0.00024407
sin
(
28.2743338
x
)
-
0.00031946
sin
(
31.4159265
x
)
0.1
0.00005173
0.05881966
0.06814824
sin
(
3.14159265
x
)
-
0.03968284
sin
(
6.28318530
x
)
+
0.00859668
sin
(
9.42477796
x
)
-
0.00299971
sin
(
12.5663706
x
)
+
0.00134242
sin
(
15.7079632
x
)
-
0.00148524
sin
(
18.8495559
x
)
+
0.00061268
sin
(
21.9911485
x
)
-
0.00037644
sin
(
25.1327412
x
)
+
0.00024579
sin
(
28.2743338
x
)
-
0.00032149
sin
(
31.4159265
x
)
0.03
0.00002314
0.06059838
0.06965236
sin
(
3.14159265
x
)
-
0.04012284
sin
(
6.28318530
x
)
+
0.00866032
sin
(
9.42477796
x
)
-
0.00301638
sin
(
12.5663706
x
)
+
0.00134839
sin
(
15.7079632
x
)
-
0.00149075
sin
(
18.8495559
x
)
+
0.00061463
sin
(
21.9911485
x
)
-
0.00037749
sin
(
25.1327412
x
)
+
0.00024640
sin
(
28.2743338
x
)
-
0.00032220
sin
(
31.4159265
x
)
0.01
0.00002053
0.06112368
0.06814824
sin
(
3.14159265
x
)
-
0.03968284
sin
(
6.28318530
x
)
+
0.00859668
sin
(
9.42477796
x
)
-
0.00299971
sin
(
12.5663706
x
)
+
0.00134242
sin
(
15.7079632
x
)
-
0.00148524
sin
(
18.8495559
x
)
+
0.00061268
sin
(
21.9911485
x
)
-
0.00037644
sin
(
25.1327412
x
)
+
0.00024579
sin
(
28.2743338
x
)
-
0.00032149
sin
(
31.4159265
x
)
0.001
0.00002020
0.06133626
0.07029513
sin
(
3.14159265
x
)
-
0.04030800
sin
(
6.28318530
x
)
+
0.00868697
sin
(
9.42477796
x
)
-
0.00302334
sin
(
12.5663706
x
)
+
0.00135088
sin
(
15.7079632
x
)
-
0.00149304
sin
(
18.8495559
x
)
+
0.00061544
sin
(
21.9911485
x
)
-
0.00037793
sin
(
25.1327412
x
)
+
0.00024665
sin
(
28.2743338
x
)
-
0.00032250
sin
(
31.4159265
x
)Example 5.2.
We consider the following problem of minimizing the cost functional:(5.7)Jα(φ)=∫03[ut(x,2;φ)-({12(5x3-13x2+9x)0≤x≤112x2-2x+21≤x≤202≤x≤3)]2dx+∫03[ux(x,2;φ)-(3{12(15x2-26x+9)0≤x≤1x-21≤x≤202≤x≤3)]2dx+α∫03φx2dx
subject to
(5.8)utt-4uxx=-4(t+1){15x-130≤x<1,t∈(0,2]11≤x<2,t∈(0,2]02≤x<3,t∈(0,2]u(x,0)=φ(x),ut(x,0)={12(5x3-13x2+9x)0≤x≤112x2-2x+21≤x≤202≤x≤3u(0,t)=0,u(3,t)=0,t∈(0,2].
We can rewrite the cost functional as
(5.9)Jα(φ)=Jα1(φ)+αJα2(φ)
For
(5.10)Jα1(φ)=∫03[ut(x,2;φ)-({12(5x3-13x2+9x)0≤x≤112x2-2x+21≤x≤202≤x≤3)]2dx+∫03[ux(x,2;φ)-(3{12(15x2-26x+9)0≤x≤1x-21≤x≤202≤x≤3)]2dx,Jα2(φ)=∫03φx2dx.
Taking α=0.2 and the initial element φ0=0, we obtain a minimizing sequence. In this example β=0.015 and stopping criteria 0.5×10-6 are chosen.Optimal control function after 37 iterations is(5.11)φ37=0.02815886sin⁡(9.42477796x)+0.71530518sin⁡(3.14159265x)+0.60922355sin⁡(1.04719755x)+0.99053011sin⁡(2.09439510x)+0.36120182sin⁡(4.18879020x)+0.19210117sin⁡(5.23598775x)+0.02095115sin⁡(10.4719755x)+0.10849356sin⁡(6.28318530x)+0.05138069sin⁡(7.33038285x)+0.03214310sin⁡(8.37758041x).J0.21(φ37) and J0.22(φ37) are 0.7368431097 and 20.78884275, respectively, for this optimal control.  For different α, the values of Jα1(φ), Jα2(φ), and optimal controls φ* are given in Table 2.Table 2
α
J
α
1
(
φ
*
)
J
α
2
(
φ
*
)
φ
*
0.9
5.78860382
8.69671794
0.02889657
sin
(
8.37758041
x
)
+
0.04493685
sin
(
7.33038285
x
)
+
0.24991184
sin
(
1.04719755
x
)
+
0.52713734
sin
(
2.09439510
x
)
+
0.45790546
sin
(
3.14159265
x
)
+
0.26256973
sin
(
4.18879020
x
)
+
0.15223865
sin
(
5.23598775
x
)
+
0.09116834
sin
(
6.28318530
x
)
+
0.02580495
sin
(
9.42477796
x
)
+
0.01946190
sin
(
10.4719755
x
)
0.6
3.70770920
11.9063673
0.32482219
sin
(
1.04719755
x
)
+
0.03043404
sin
(
8.37758041
x
)
+
0.04796407
sin
(
7.33038285
x
)
+
0.65182517
sin
(
2.09439510
x
)
+
0.54263121
sin
(
3.14159265
x
)
+
0.30032478
sin
(
4.18879020
x
)
+
0.16919357
sin
(
5.23598775
x
)
+
0.02014690
sin
(
10.4719755
x
)
+
0.02690906
sin
(
9.42477796
x
)
+
0.09904027
sin
(
6.28318530
x
)
0.2
0.73684310
20.7888427
0.02815886
sin
(
9.42477796
x
)
+
0.71530518
sin
(
3.14159265
x
)
+
0.60922355
sin
(
1.04719755
x
)
+
0.99053011
sin
(
2.09439510
x
)
+
0.36120182
sin
(
4.18879020
x
)
+
0.19210117
sin
(
5.23598775
x
)
+
0.02095115
sin
(
10.4719755
x
)
+
0.10849356
sin
(
6.28318530
x
)
+
0.05138069
sin
(
7.33038285
x
)
+
0.03214310
sin
(
8.37758041
x
)
0.04
0.26519950
25.3541218
0.03275462
sin
(
8.37758041
x
)
+
0.92069918
sin
(
1.04719755
x
)
+
0.05250012
sin
(
7.33038285
x
)
+
1.14670525
sin
(
2.09439510
x
)
+
0.02863498
sin
(
9.42477796
x
)
+
0.19820000
sin
(
5.23598775
x
)
+
0.11127608
sin
(
6.28318530
x
)
+
0.02127006
sin
(
10.4719755
x
)
+
0.76666431
sin
(
3.14159265
x
)
+
0.37689906
sin
(
4.18879020
x
)
0.02
0.25964159
25.7401337
0.05264297
sin
(
7.33038285
x
)
+
0.02869561
sin
(
9.42477796
x
)
+
0.93896719
sin
(
1.04719755
x
)
+
1.15756155
sin
(
2.09439510
x
)
+
0.77151420
sin
(
3.14159265
x
)
+
0.02131060
sin
(
10.4719755
x
)
+
0.37869007
sin
(
4.18879020
x
)
+
0.03283263
sin
(
8.37758041
x
)
+
0.19895419
sin
(
5.23598775
x
)
+
0.11162916
sin
(
6.28318530
x
)
0.002
0.25769444
26.0957452
0.95568491
sin
(
1.04719755
x
)
+
0.02875041
sin
(
9.42477796
x
)
+
0.02134722
sin
(
10.4719755
x
)
+
0.19963788
sin
(
5.23598775
x
)
+
0.11194885
sin
(
6.28318530
x
)
+
0.05277220
sin
(
7.33038285
x
)
+
0.03290315
sin
(
8.37758041
x
)
+
0.77593184
sin
(
3.14159265
x
)
+
1.16750459
sin
(
2.09439510
x
)
+
0.38031659
sin
(
4.18879020
x
)


## 6. Conclusions

In a hyperbolic problem, the initial conditionu(x,0)=φ(x) can be controlled from the targets ut(x,T;φ) and ux(x,T;φ) using H01-Poincare norm. The Lagrange multiplier is zt while the function z(x,t) is the solution of adjoint problem. The symbolic optimal control function is easily obtained in numerical examples.



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*Source: 156541-2012-07-29.xml*