Optimal Control Algorithm for Complex Heat Transfer Model

Abstract

An optimal control problem for a nonlinear steady-state heat transfer model accounting for heat radiation e ects is considered. The problem consists in minimization of a given cost functional by controlling the sources in the heat equation. The solvability of this control problem is proved, optimality conditions are derived, and an iterative algorithm for solving the optimal control problem is constructed.

Keywords

• optimal control
• radiative heat transfer
• conductive heat transfer
• gradient descent method

Introduction

The interest in studying problems of complex heat transfer (where the radiative, convective, and conductive contributions are simultaneously taken into account) is motivated by their importance for many engineering applications. Here, the following examples can be mentioned: modeling and predicting the heat transfer in molten glass [1{3], nano uids [4, 5], etc.

A considerable number of works devoted to optimal control problems of complex heat transfer models consider the evolutionary systems (see, e.g., [1{3, 6{10]). In the mentioned works, the radiation transfer is described by steady-state radiative transfer equation. The temperature eld is simulated by the conventional evolutionary heat transfer equation with additional source terms describing the contribution of the radiative heat transfer.

Theoretical analysis of optimal control problems for steady-state systems of complex heat transfer with source terms in the heat equation is an open question. It is worth to mention the work [11], where the problem of optimal boundary multiplicative control for a steady-state complex heat transfer model was considered. The problem was formulated as the maximization of the energy out ow from the model domain by ? The research was supported by the Ministry of education and science of Russian Federation (project 14.Y26.31.0003) Copyright c by the paper's authors. Copying permitted for private and academic purposes. In: A. Kononov et al. (eds.): DOOR 2016, Vladivostok, Russia, published at http://ceur-ws.org controlling re ection properties of the boundary. On the basis of new a priori estimates of solutions of the control system, the solvability of the optimal control problem was proved. The main result there was the proof of an analogue of the bang-bang principle arising in control theory for ordinary di erential equations.

In this paper, an optimal control problem of obtaining a desired temperature and(or) radiative intensity distributions in a part of the model domain by controlling the sources in the heat equation is considered. Analogous problems appear in many engineering applications and draw attention of many researchers. For example, similar optimal control problems for non-stationary complex heat transfer models were studied in [1{3, 8] in context of glass manufacturing. In the current work, the optimal control problem for a steady-state model is studied. The solvability of this problem is proved, and an optimality system is derived. Moreover, an iterative algorithm based on the gradient descent method is constructed, and results of numerical experiments are presented. 2

Problem formulation

The following steady-state normalized di usion (P1) model (see [12{15]) describing radiative and conductive heat transfer in a bounded domain R3 is under consideration: a

+ b a(j j 3 ') = u; ' + a('

j j 3) = 0; b)j = 0;

3 j bj b )j = 0: Here, is the normalized temperature, ' the normalized radiation intensity averaged over all directions, and a the absorption coe cient. The physical sense of the parameters a, b, , , can be found in [13{15]. The control function u describes the internal sources of heat. The symbol @n denotes the derivative in the outward normal direction n on the boundary := @ .

The problem of optimal control consists in the determination of functions u, , and ' which satisfy the conditions ( 1 ), ( 2 ) and minimize a cost functional J ( ; '; u), i.e.

J ( ; '; u) = J ( ; ') + 2 kuk2L2( ) ! inf; u 2 Uad: Here, Uad L2( ) is the set of admissible controls, 0 is a given cost parameter. In particular, the functional J can describe the L2-deviation of the temperature and radiation elds from prescribed distributions, say d and 'd. Thus, e.g.

J ( ; ') = a k

2 dkL2( ) + a' k'

2 'dkL2( ); where a and a' are nonnegative weights. ( 1 ) ( 2 ) ( 3 ) 3

Formalization of the optimal control problem

Suppose that the model data satisfy the following conditions:

(i) ; 2 L1( ), 0 > 0, 0 > 0, 0; 0 = const, b 2 L1( ); (ii) Uad is a closed convex set; Uad is a bounded set, if = 0; (iii) The cost functional J : H1( ) H1( ) ! R is weakly lower semicontinuous and bounded from below.

Here and further, the Sobolev space W2s( ) is denoted by Hs( ), s 0, and (f; g) and kf k denote respectively the inner product and the norm of the space L2( ).

Denote H = L2( ), V = H1( ), Y = V V . Identifying H with the dual space H0 yields the Gelfand triple V H = H0 V 0. Let the value of a functional f 2 V 0 on an element v 2 V be denoted by (f; v). Notice that (f; v) is the inner product in H if f and v are elements of H.

Assuming that , ', v are arbitrary elements of V , de ne operators and functionals

A1; A2 : V ! V 0, f; g 2 V 0 by the following relations:

Z

Z (A1 ; v) = a(r ; rv) + vd ; (A2'; v) = (r'; rv) + Z 'vd ; (f; v) = bvd ; (g; v) =

Z j bj b3vd :

A pair f ; 'g 2 V is called weak solution of the problem (1), (2) if

A1 + b a(j j 3 ') = f + u; A2' + a(' j j 3) = g:

The optimal control problem consists in the minimization of a functional J de ned on solutions of system ( 4 ) provided that u 2 Uad. That is,

J ( ; '; u) ! inf; f ; 'g are solutions of (4) yielded by u 2 Uad:

A pair fb; 'bg minimizing J and corresponding to a function ub is called optimal state, and ub is called optimal control. 4

Solvability of the optimal control problem

To prove the solvability of the problem ( 5 ), establish some properties of the boundary value problem ( 1 ), ( 2 ).

Lemma 1. If the conditions (i) hold and u 2 Uad, then for a weak solution, f ; 'g, of the problem ( 1 ), ( 2 ) the following estimate is true: Here, a positive constant C depends only on a, b, , a, , , kuk, and .

Proof. Let hp(s) := jsjpsigns, p > 0, s 2 R. Denote '1 = h1=4(') and, for " > 0, de ne k k2V + k'k2V

C: w" = >8'1 < 0;

j'1j >:'1 + "; '1 < "; '1 > "; "; ": ( 4 ) ( 5 ) ( 6 ) Notice that if ' 2 V , then '1 2 L24( ), '1j 2 L16( ), w" 2 V , and rw" = 1 (j'j 3=4r'; j'1j > "; 4 0; otherwise:

It is important that

Z 'w"d a(h4( ) '; w") (g; w") = =

Z (' h4( b))'1d a(h4( ) '; '1) + c": ( 7 )

In this expression jc"j C", where C > 0 does not depend on ":

Multiply, in the sense of the inner product of H, the rst equation of ( 4 ) by , the second equation by bw", and add the equalities. Then, taking into account monotonicity of (h4( ) ')( h1=4(')) 0, we obtain the inequality akr k2 +

Z 2d + Here, K1 depends only on a, , b, 0, 0, k kL1( ), and the domain .

The estimate of k kV allows to obtain the estimate of k'kV . Multiplying the second equation of ( 4 ) by ' in the sense of the inner product of H, and denoting k3 = minf ; 0g, we obtain the inequality k3k'k2V + ak'k2

Z aj(h4( ); ')j + jh4( b)'jd :

Using Holder and Young inequalities with parameter > 0, we estimate: (9) (10)

j(h4( ); ')j 2 k'k2L3( ) + 21 k k8L6( ); Taking into account the continuity of the embedding of V into L6( ), the continuity of the trace operator from V into L4( ), and a su ciently small , we obtain the estimate of k'kV :

8 8

K2 k bkL16=3( ) + k kV :

Here, K2 depends only on , 0, a k kL1( ), and the domain

and ( 11 ) prove the lemma. . The estimates ( 10 ) ( 11 ) tu

On the base of the estimate ( 6 ), similarly as in [11], the solvability of the problem ( 5 ) is proved.

Theorem 1. If the conditions (i){(iii) hold, then there exists at least one solution of the problem ( 5 ). 5

The necessary conditions of optimality

To derive optimality relations, add to conditions (i)-(iii) the following assumption: (iv) J : Y ! R is Frechet di erentiable.

Introduce a constraint operator F : Y

H ! Y 0 as follows:

F (y; u) = fA1 + b a(j j 3 ') f u; A2' + a(' j j 3) gg; where y = f ; 'g 2 Y; u 2 H: Lemma 2. For any y 2 Y the map Fy0 : Y ! Y 0 is epimorphic, Im Fy0 = Y 0: Proof. Equation F 0 q = z = fz1; z2g 2 Y 0 is equivalent to the following boundary value y problem:

A1q1 + b a(4j j3q1 q2) = z1; A2q2 + a(q2 4j j3q1) = z2; q = fq1; q2g 2 Y: ( 12 ) To prove the solvability of a Fredholm problem ( 12 ), it su ces to prove the uniqueness of its solutions. Let sign s = s=jsj, if s 6= 0, sign 0 = [ 1; 1]. Let us consider the function which is regularization of multivalued function sign, (s) = s=jsj, if jsj , (s) = s= , if jsj < .

Let z = 0, h = 4j j3: Multiplying, in the sense of the inner product of H, the rst equation of ( 12 ) by (q1), the second equation by b (q2), and adding these equalities, we obtain (A1q1; (q1)) + b(A2q2; (q2)) + b a(hq1 q2; (q1) (q2)) = 0: Notice that (A1q1; (q1)) = a(rq1; 0 (q1)rq1) + and

Therefore,

Z

Z

Z

Z (A2q2; (q2)) = (rq2; 0 (q2)rq2) +

q2 (q2)d Z

Z q2 (q2)d : q1 (q1)d + b q2 (q2)d + b a(hq1 q2; (q1) (q2)) 0: ( 13 ) Passing to the limit as

! 0, from inequality ( 13 ), we obtain jq1jd + b jq2jd + b a(hq1 q2; sign q1 sign q2) 0: ( 14 )

tu ( 15 ) ( 16 ) ( 17 ) From monotonicity of the function sign, it follows the conditions q1j = q2j = 0:

Further, notice that A1q1 + bA2q2 = 0. Scalarly multiplying this equation by aq1 + bq2, and taking into account zero boundary values of q1; q2, we obtain kr(aq1 + bq2)k2 = 0: Hence, aq1 + bq2 = 0: Therefore,

a(rq1; rv) + b a((h + a= b)q1; v) = 0 8v 2 V: Assuming v = q1 in ( 14 ), we obtain q1 = 0, and therefore q2 = 0:

Applying the principle of Lagrange for smooth convex extremal problems [16], we can prove the following result.

Theorem 2. Let yb = fb; 'bg 2 Y , ub 2 Uad be a solution of the control problem ( 5 ). Then there exists an adjoint state p = fp1; p2g 2 Y such that the triple fyb; ub; pg satis es the conditions

A1p1 + 4jbj3 a(bp1 p2) =

J 0 (y); A2p2 + a(p2 b

be subdomains of . Consider the following cost functional: where d 2 L2(G1) and 'd 2 L2(G2) are given functions. It is easy to see that (J 0 ( ; '); ) = ( d) dx; (J '0( ; '); ) = (' 'd) dx; In this case, if Uad = L2( ) and

> 0, the optimality system assumes the form a b + b a(jbjb3 ') = ub; b ' + a('b b b)j = 0; jbjb3) = 0; bp1) =

G2 ('b = 0;

'd); = 0; (19) and u = p1= : b

Here, G1;2 are the characteristic functions of the subdomains G1;2, respectively. 7

Iterative algorithm

For the numerical solution of the optimality system ( 18 ), (19), we can apply the method of gradient descent: uk+1 = uk k uk p(k) ; k = 0; 1; 2; : : : ; 1 where u0 2 H is given.

Here k > 0 is a step size, p(k) = fp(1k); p(2k)g is a pair satisfying the system ( 18 ), (19), where u := uk:

b If Uad 6= H we can apply the gradient projection method (see, e.g, [17]): uk+1 = PUad uk k uk where PUad : H ! Uad is the projection operator.

The method of choosing the step size k is adjusted as required for decreasing the cost functional. Unlike methods based on the Armijo rule, this method does not need an inner loop for adjustment of k that requires computing the value of J and, therefore, solving the problem ( 18 ).

De ne Jb(u) = J ( (u); '(u); u), where f (u); '(u)g is a solution of system ( 4 ). The method is as follows. If Jb(uk+1) Jb(uk), then return back to the control uk and reduce k by a factor of 2. Additionally, if Jb(uk+1) < Jb(uk); k = s; s + 1; : : : ; s + m0 1, then k is increased by a factor of 2. Here, m0 1 is a prescribed integer parameter of a quantity of decreases of the cost functional, which is enough for increasing the step size k.

The pseudocode of the algorithm is presented below.

Algorithm 1: Gradient descent method with a variable step size

Consider an example for the two-dimensional domain = f(x; y) : 0 x; y Lg which can be interpreted as a long rectangular channel in the three-dimensional space. The parameters values are taken as follow: L = 10 [cm], = 3:3 : : : [cm], a = 0:01 [cm 1], = 1:5 [cm/s], and = "=2(2 "), where " = 0:7 is the emissivity coe cient of the boundary. The thermodynamical characteristics of the medium correspond to air at the normal atmospheric pressure and the temperature of 400 C. The maximum temperature is chosen as Tmax = 773 K. This yields a = 0:92 [cm2/s] and b = 18:7 [cm/s]. Notice that the absolute temperature is T = Tmax . The boundary temperature b = 0:5.

The cost functional is de ned by ( 3 ) and ( 17 ), where G1 = n S, G2 = ;, d = 0:7, and = 0:01. Let the set of admissible controls be Uad = fu 2 L2( ) : u1 u u2g, where u1 = 0, u2 = 1 in S = [x1; x2] [y1; y2], and u1 = u2 = 0 in n S. Assume x1 = 0:65L, x2 = 0:85L, y1 = L=12, y2 = 5L=12.

For the numerical solution we use the software FreeFem++ [18]. The boundaryvalue problem ( 18 ) is solved by Newton's method. The initial guess for the optimal control is chosen as u0 = 0, and parameters of the optimization algorithm are 0 = 5, m0 = 3.

The computed optimal control is presented in Fig. 1. The graph of the optimal temperature is depicted in Fig. 2. The values of Jb(uk) and k for di erent k are indicated in Figs. 3, 4. As it is seen in Fig. 4, the most frequent value of k is 10, and the step size is adjusted as needed.

The optimal controls for = 0:1 and = 0:001 are presented in Figs. 5, 6 for comparison. It can be easily proved from ( 16 ) that in the case of = 0 the optimal control satis es an analog of the bang-bang principle, that is ub(x) = u1(x) or u2(x) for a.e. x 2 where p1(x) 6= 0. Notice that the optimal control comes near to a bang-bang control as ! 0. The optimal control for = 0 is depicted in Fig. 7. This bang-bang control was computed by an optimization algorithm of the gradient descent type, see [9, 10].

4.0 3.5 3.0 2.5 2.0 1.5 1.0 1 0 1 7

8 3

4 Iterations

5 8 0.9

5 10 Itera1ti5ons 20 Fig. 4. Step size k at di erent iterations ( 25

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