Vol. 18 No. 3

Journal of Systems Science and Complexity

Jul., 2005

MAXIMUM PRINCIPLE FOR THE OPTIMAL CONTROL OF AN ABLATION-TRANSPIRATION COOLING SYSTEM∗ SUN Bing (Institute of Systems Science, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100080, China; Graduate School of the Chinese Academy of Sciences, Beijing 100049, China)

GUO Baozhu (Institute of Systems Science, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100080, China. Email: [email protected]) Abstract. This paper is concerned with an optimal control problem of an ablationtranspiration cooling control system with Stefan-Signorini boundary condition. The existence of weak solution of the system is considered. The Dubovitskii and Milyutin approach is adopted in the investigation of the Pontryagin’s maximum principle of the system. The optimality necessary condition is presented for the problem with fixed final horizon and phase constraints. Key words. Ablation-transpiration cooling process, Stefan-Signorini problem, weak solution, optimal control, maximum principle.

1 Introduction When a high speed space vehicle flies through the air, the high temperature due to the friction between the vehicle and the air may cause ablation of the material in its front surface which could result in the damage of the vehicle. The problem is even more serious in the sidewall when launching the electromagnetic gun. In engineering, a thermal shield must be designed to prevent this event from happening [1–3]. Ablation problems are not new. They have attracted considerable attention both in engineering and in mathematics since the beginning of the last century (see e.g. [1–15]). Basically, this process can be described by a heat conduction equation with moving boundary condition, which is the so-called Stefan problem dating back to 1890. The earliest research on ablation cooling process can be traced back to L.Prandtl in 1904[4]. Traditionally, the study of these kinds of problems is limited to the ablation process itself by either using numerical methods[5−7] and analytic approach[8,9,12], or imposing some indirectly realized control design as the ablation shield. In [16] the ablation was prerented by controlling the velocity and attitude of the vehicle so that the ablation rate does not exceed certain maximal allowable value at any time during the re-entry flight for the ablation shield of re-entry vehicle with ablated surface. Other control methods such as the heat exchange coefficient can be found in [17] and the references therein. On the other hand, because of its easy realization through porous materials, there has been a rapid growth of interest in the Received February 10, 2004. Revised November 11, 2004. *This research is supported by the National Natural Science Foundation of China.

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application of transpiration-cooling control to the ablation process in the last two decades. A first partial differential equation model of ablation-transpiration cooling control system was established in [3] (see also [11]) and later the same partial differential equation model with Stefan-Signorini boundary condition (from [5] and [12]) appeared in [14]. This model can be demonstrated by a one-dimensional version of a solid thermal shield as shown in Figure 1. 6 m(t) —gaseous coolant flux

x

????? ? b b b b b b b b b b b b b b b b b b b b b b b b b b b b

? b b b b

66666 6 6

? ? ????? x=l b b b b b b b b b b b b b porous medium slab b b b b b b b b x = s(t) b b b b 6 6 66666 x=0

Q(m) —– heated air flow

Figure 1 Schematic representation of the ablation-transpiration cooling control system of a porous medium slab

The thermal shield consists of a porous solid slab of thickness of `. The gaseous coolant is poured into the structure at x = `, flows through the air hole of the slab, and enters the heated air flow Q(m), e a specified bounded function of the coolant flux m(t) e (mass of coolant per unit time flowing through per unit area) on the outer layer of the structure[3,11] . The heated air flows enter the structure on the opposite of the coolant flows from x = 0. When the temperature of the front face exceeds the melting temperature um of the material, the outer layer melts and recedes to the new position x = s(t) after time t. Denote by u(x, t) the temperature of the medium at position x and time t, and let uc be the temperature of the coolant at the inside of the tank. Then the ablation-transpiration cooling control process can be described by the following heat equation with moving boundary of Stefan-Signorini condition[15] :  k   ut (x, t) = uxx (x, t) + β(t)ux (x, t), s(t) < x < `,   ρc        k ux (`, t) + β(t)u(`, t) = β(t)uc , uc > 0, ρc (1.1)   0 0  u(s(t), t) ≤ u , s (t) ≥ 0, [u − u(s(t), t)]s (t) = 0, m m      k    s0 (t) = ux (s(t), t) + Q(β(t)), s(0) = 0, u(x, 0) = u0 (x), ρL

where L is the latent heat of melting. The thermal conductivity k, density ρ, and the specific heat c are all constants. The nonnegative bounded function β(t) is the function of m(t) e which is considered as the control variable for the system (1.1). u0 is the initial condition. Since the following discussion also holds for the case k/ρc 6= 1 and k 6= ρL, for simplicity in notation we set, without loss of generality, that k/ρc = 1 and k = ρL. Thus, a simpler form of model is as follows:  ut (x, t) = uxx (x, t) + β(t)ux (x, t), s(t) < x < `, t ∈ [0, T ],       ux (`, t) + β(t)u(`, t) = β(t)uc , uc > 0, (1.2) 0 0    u(s(t), t) ≤ um , s (t) ≥ 0, [um − u(s(t), t)]s (t) = 0,    s0 (t) = u (s(t), t) + Q(β(t)), s(0) = 0, u(x, 0) = u (x). x 0

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The following well-posedness of the system (1.2) was proven in [15]. Theorem 1 For a fixed control β(t) ≥ 0 satisfying β(·) ∈ C 1 [0, T ] and the initial condition u(x, 0) = u0 (x) satisfying compatible conditions  u0 (·) ∈ C 2 [0, `], uc ≤ u0 (·) ≤ um , Q(β(·)) ∈ C 1 [0, ∞), Q(β(·)) ≥ 0,     0 u0 (`) + β(0)u0 (`) = uc β(0), (1.3)     u00 (0) + Q(β(0)) ≥ 0, [um − u0 (0)][u00 (0) + Q(β(0))] = 0, equation (1.2) admits a unique classical solution u(x, t) for (x, t) ∈ [s(t), `]×[0, T ] which depends continuously on β(t) with C 1 norm, and either T = ∞ and s(t) < ` for all t > 0, or T < ∞ and s(T ) = `. Moreover, the following properties are satisfied: (i) uc ≤ u(x, t) ≤ um and ux (x, t) ≤ 0 provided that u00 (x) ≤ 0. (ii) if uc < u0 < um , then u(x, t) > uc for all x < ` and u(x, t) < um for all x > s(t). In this article, we consider an optimal control problem for the system (1.1) with the general cost functional for some finite T > 0: Z TZ ` ˜ min J(u, β, s) = min L(u(x, t), β(t), s(t), x, t)dxdt (1.4) β(·)∈Uad

β(·)∈Uad

0

s(t)

and the control constraint: n o Uad = β ∈ L∞ (0, T ) | 0 ≤ β0 ≤ β(t) ≤ β1 , t ∈ [0, T ] a.e. .

(1.5)

˜ is quite general in the sense that it contains most practically concerned The cost function L cost functional like quadratic cost functional of the following form: J(u, β, s) = α1

Z TZ 0

` s(t)

[u(x, t) − u∗ (x, t)]2 dxdt + α2

Z

T 0

|β(t) − β ∗ |2 dt + α3

Z

T 0

|s(t) − s∗ |2 dt,

where αi > 0 for i = 1, 2, 3 are constants, and u∗ , β ∗ , s∗ are ideal temperature profile, coolant flux and boundary position, respectively. Essentially speaking, the investigated problem in this paper is an optimal control problem of infinite dimensional system and one can refer to [18] for the general theory. The paper is organized as follows. In Section 2, the existence of weak solution of the system (1.1) is derived so that the system makes sense for the control β in the L∞ space. In Section 3, the system (1.1) is transformed into a fixed interval problem with respect to spatial variables, and a general optimal control problem of the form (1.4) is formulated. Subsections of Section 3 are devoted to the Pontryagin’s maximum principle of the optimal control problem via the Dubovitskii and Milyutin’s approach ([19]).

2 Weak Solution In order to discuss the optimal control, we need to relax the constraint set of β(·) from C 1 [0, T ] into L∞ (0, T ). To this purpose, we have to consider the weak solution of the system. Actually, a wide variety of methods for weak formulation of Stefan problem have been developed in literature, see for instance [20–30], name just a few. However, the system studied in this article is of speciality for its Stefan-Signorini boundary condition which makes the problem even more complex than the problem with nonlinear boundary conditions[31,32] . Here, we give a definition of weak solution and show its existence.

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By the standard transformation: θ(x, t) = um − u(x, t), θ0 (x) = um − u0 (x);

y=

x − s(t) `, w(y, t) = θ(x, t), ` − s(t)

the system (1.2) is transformed into the following controlled nonlinear heat equation defined on the fixed domain D = [0, `] × [0, T ]:   w (y, t) = α2 (s(t))wyy (y, t) + f (β(t), s(t), y)wy (y, t),   t     α(s(t))wy (`, t) + β(t)w(`, t) = qβ(t), q = um − uc > 0,   (2.1) w(0, t) ≥ 0, −α(s(t))wy (0, t) + Q(β(t)) ≥ 0,     w(0, t)[−α(s(t))wy (0, t) + Q(β(t))] = 0,      s0 (t) = −α(s(t))w (0, t) + Q(β(t)), s(0) = 0, w(y, 0) = w (y), y 0 where

α(s(t)) =

` , ` − s(t)

f (β(t), s(t), y) = β(t)

` `−y + s0 (t) . ` − s(t) ` − s(t)

Definition 1 For any T > 0 and β(t) ∈ L∞ (0, T ), select a sequence βn ∈ C 1 [0, T ] so that kβn − βkL∞ (0,T ) → 0. Let {wn (y, t), sn (t)} be the classical solution of (2.1) corresponding to βn . Suppose that supn sn (T ) < `. If there exists a subsequence {wnk (y, t), snk (t)} of {wn (y, t), sn (t)} such that ( wnk → w in C[0, T ; L2 (0, `)] as k → ∞, (2.2) snk → s in C[0, T ] as k → ∞ and s(·) ∈ C 1 [0, T ], then the pair {w(y, t), s(t)} is said to be a weak solution of (2.1). Obviously, the classical solution must be a weak solution. In order to prove the existence of weak solution, we need the following lemma which was proved in [14]. Lemma 1 (i) 0 ≤ wn (y, t) ≤ q for all (t, y) ∈ [0, ∞) × [0, `]; (ii) |wny (y, t)| ≤ max{kw0 kC 0 , qkβkL∞ , kQkC 0 } for all (t, y) ∈ [0, ∞) × [0, `]; (iii) 0 ≤ s0n (t) ≤ kQkC 0 for all t ∈ [0, ∞). In addition, the following result is needed (Proposition 1 of [33]). Proposition 1 Let η ∈ C 1 [0, 1] have the following properties: (i) ∃ M > 0, |η 0 | ≤ M on [0, 1]; Rb (ii) ∀ ε > 0, | a η(y)dy| ≤ ε for any a, b ∈ [0, 1]. Then n √ o |η(y)| ≤ max 2ε, 2M ε for y ∈ [0, 1]. Lemma 2 Let wn (y, t) be the sequence of Definition 1. Then there exists a constant C > 0 independent of n, y, t1 , t2 such that |wn (y, t1 ) − wn (y, t2 )| ≤ C|t1 − t2 |1/2 , ∀ n and t1 , t2 ∈ [0, T ], y ∈ [0, `]. Proof

Integrate the governing equation wnt (y, t) = α2 (sn (t))wnyy (y, t) + f (βn (t), sn (t), y)wny (y, t)

(2.3)

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over [a, b] × [t1 , t2 ] ⊂ D with respect to (y, t), to obtain Z

= =

b

[wn (y, t2 ) − wn (y, t1 )] dy

a

Z bZ Z

t2

a t1 t2

t1





 α2 (sn (t))wnyy (y, t) + f (βn (t), sn (t), y)wny (y, t) dydt

 2  b α (sn (t))wny (y, t) + f (βn (t), sn (t), y)wn (y, t) a +

s0n (t) ` − sn (t)

Z

b



wn (y, t)dy dt. a

Let η(y) = wn (y, t2 ) − wn (y, t1 ). By Lemma 1, one can find constants M1 , M2 independent of y, t1 , t2 , n such that Z b 0 |η (y)| ≤ 2M1 , η(y)dy ≤ M2 |t2 − t1 |. a

Applying Proposition 1 to η, we can obtain that n o p |η(y)| ≤ max 2M2 (t2 − t1 ), 4M1 M2 (t2 − t1 ) .

(2.4)

(2.3) then follows from (2.4). Theorem 2(Existence of weak solution) There exists a weak solution to the system (2.1). Proof By virtue of (iii) of Lemma 1 and the Ascoli-Arzela theorem, there exists a subsequence of {sn } which is still denoted by itself without confusion so that sn → s in C[0, T ], s0n → p in C[0, T ] weak∗ as n → ∞.

Rt Rt Rt Whence, for any t ∈ (0, T ], 0 s0n (τ )dτ → 0 p(τ )dτ as n → ∞. However, 0 s0n (τ )dτ = sn (t) → Rt s(t), we thereby obtain that s(t) = 0 p(τ )dτ or s0 (t) = p(t), that is to say, s(·) ∈ C 1 [0, T ]. Next by Lemma 2 and the Ascoli-Arzela theorem, it follows that {wn (y, t)} is a compact sequence in the space C[0, T ; L2(0, `)]. Accordingly, there exists a subsequence of {wn (y, t)}, which is still denoted by itself without confusion, so that wn → w in C[0, T ; L2 (0, `)] as n → ∞. The proof is complete.

3 Optimal Control Formulation Now we are in a position to consider the optimal control of the system. Unless otherwise stated, in what follows when we speak of a solution of (2.1) we shall always mean the weak solution. Instead of control problem (1.4), we put forward the following optimal control problem for the system (2.1): min J(w, β, s) =

β(·)∈Uad

min

β(·)∈Uad

Z TZ 0

`

L(w(y, t), β(t), s(t), y, t)dydt,

(3.1)

0

where Uad is the admissible control set given in (1.5). The following assumptions are assumed throughout the paper:

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(a) L is a functional defined on L2 (0, `) × [β0 , β1 ] × [0, `]2 × [0, T ] and ∂L(w(y, t), β(t), s(t), y, t) ∂L(w(y, t), β(t), s(t), y, t) ∂L(w(y, t), β(t), s(t), y, t) , , ∂w ∂β ∂s exist for every (w, β, s) ∈ L2 (0, `) × [β0 , β1 ] × [0, `] and L is continuous in its variables. (b) Z ` Z ` ∂L(w(y, t), β(t), s(t), y, t) ∂L(w(y, t), β(t), s(t), y, t) dy, dy, ∂w ∂β 0 0 Z ` ∂L(w(y, t), β(t), s(t), y, t) dy ∂s 0

are bounded for t ∈ [0, T ]. By Theorem 2, we define the state space X = C[0, T ; L2 (0, `)] × L∞ (0, T ) × C 1 [0, T ]. Let ∗ (w , β ∗ , s∗ ) be the optimal solution to the control problem (3.1) subject to the equation (2.1). Set  Ω1 = (w, β, s) ∈ X | β0 ≤ β(t) ≤ β1 , t ∈ [0, T ] a. e. , Ω2 = (w, β, s) ∈ X | w(0, t) ≥ 0, t ∈ [0, T ] a. e. , Ω3 = {(w,  β, s) ∈ X | −α(s(t))wy (0, t) + Q(β(t)) ≥ 0, t ∈ [0, T ] a. e.}, Ω4 = (w, β, s) ∈ X | wt (y, t) = α2 (s(t))wyy (y, t) + f (β(t), s(t), y)wy (y, t), (3.2) α(s(t))wy (`, t) + β(t)w(`, t) = qβ(t), w(0, t)[−α(s(t))wy (0, t) + Q(β(t))] = 0, s0 (t) = −α(s(t))wy (0, t) + Q(β(t)), s(0) = 0, w(y, 0) = w0 (y), w(y, T ) = w ∗ (y, T ) .

Then the problem (3.1) is equivalent to finding (w ∗ , β ∗ , s∗ ) ∈ Ω = ∩4i=1 Ωi such that J(w∗ , β ∗ , s∗ ) =

min

(w,β,s)∈Ω

J(w, β, s).

(3.3)

It is seen that the problem (3.3) is an extremum problem on inequality constraints ∩3i=1 Ωi and the equality constraint Ω4 . In this situation, the Dubovitskii and Milyutin’s functional approach has been turned out to be very powerful to solve such kinds of extremum problems (see e.g. [34, 35]). The general Dubovitskii and Milyutin’s theorem for the problem (3.3) can be stated as the following Theorem 3. Theorem 3 Suppose the functional J(w, β, s) assumes a minimum at the point (w ∗ , β ∗ , s∗ ) in Ω . Assume that J(w, β, s) is regularly decreasing at (w ∗ , β ∗ , s∗ ) with the cone of directions of decrease K0 and inequality constraints are regular at (w ∗ , β ∗ , s∗ ) with the cones of feasible directions K1 , K2 , K3 ; and that the equality constraint is also regular at (w ∗ , β ∗ , s∗ ) with the cone of tangent directions K4 . Then there exist continuous linear functionals f0 , f1 , f2 , f3 , f4 , not all identically zero, such that fi ∈ Ki∗ , the dual cone of Ki , i = 0, 1, 2, 3, 4, which satisfy the condition f0 + f1 + f2 + f3 + f4 = 0. 3.1 Determination of the cone of directions of decrease K0 In order to apply Theorem 3, we have to determine all cones Ki , i = 0, 1, 2, 3, 4. First, let us find K0 . By assumption, J(w, β, s) is differentiable at any point (w0 , β0 , s0 ) in any direction (w, β, s) and its directional derivative is J 0 (w0 , β0 , s0 ; w, β, s) = lim

ε→0+

1 [J(w0 + εw, β0 + εβ, s0 + εs) − J(w0 , β0 , s0 )] ε

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291

) (Z Z T ` 1 = lim [L(w0 + εw, β0 + εβ, s0 + εs, y, t) − L(w0 , β0 , s0 , y, t)] dydt ε→0+ ε 0 0  Z TZ `  ∂L(w0 , β0 , s0 , y, t) ∂L(w0 , β0 , s0 , y, t) ∂L(w0 , β0 , s0 , y, t) = w+ β+ s dydt. ∂w ∂β ∂s 0 0 Hence the cone of directions of decrease of the functional J(w, β, s) at the point (w ∗ , β ∗ , s∗ ) is determined by ([Theorem 7.5, [19], p. 48]) o n K0 = (w, β, s) ∈ X J 0 (w∗ , β ∗ , s∗ ; w, β, s) < 0  Z TZ `  ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) = (w, β, s) ∈ X w+ β ∂w ∂β 0 0   ∂L(w∗ , β ∗ , s∗ , y, t) + s dydt < 0 . (3.4) ∂s If K0 6= φ, then for any f0 ∈ K0∗ , there exists a λ0 ≥ 0 such that ([Theorem 10.2, [19], p. 69]) f0 (w, β, s) = −λ0

Z TZ `  0

0

∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) w+ β ∂w ∂β  ∂L(w∗ , β ∗ , s∗ , y, t) + s dydt. ∂s

(3.5)

3.2 Determination of the cone of feasible directions K1 e 1 × C 1 [0, T ], here Ω e 1 = {β ∈ L∞ (0, T ) | β0 ≤ β(t) ≤ β1 , t ∈ Since Ω1 = C[0, T ; L2 (0, `)] × Ω o

[0, T ] a. e.} is a closed convex subset of L∞ (0, T ), so the interior Ω 1 of Ω1 is not empty and at the point(w∗ , β ∗ , s∗ ), the cone of feasible directions K1 of Ω1 is determined by ([Theorem 8.2, [19], p. 59]) n o o K1 = λ(Ω 1 −(w∗ , β ∗ , s∗ )) | λ > 0 n o o = h | h = λ(w − w∗ , β − β ∗ , s − s∗ ), (w, β, s) ∈Ω 1 , λ > 0 . (3.6) ¯(t) ∈ L(0, T ), such that the linear functional Therefore, for an arbitrary f1 ∈ K1∗ , if there is an a defined by Z T f1 (w, β, s) = a ¯(t)β(t)dt (3.7) 0

e 1 at the point β ∗ , then ( [19], p. 76–77) is a support to Ω

a ¯(t)[β(t) − β ∗ (t)] ≥ 0, ∀ β(t) ∈ [β0 , β1 ], t ∈ [0, T ] a. e.

(3.8)

3.3 Determination of the cone of feasible directions K2 Note that the second inequality constraint is Ω2 = {(w, β, s) ∈ X |w(0, t) ≥ 0, t ∈ [0, T ]a. e.} . Let F1 (w) = max {−w(0, t)}. (3.9) 0≤t≤T

Then Ω2 = {(w, β, s) ∈ X | F1 (w) ≤ 0},

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here we only consider F1 (w∗ ) = max {−w∗ (0, t)} = 0, since otherwise F1 (w∗ ) < 0 and 0≤t≤T

(w∗ , β ∗ , s∗ ) is an inner point of Ω2 . In this case any direction is feasible and hence the cone of feasible directions K2 of Ω2 at (w∗ , β ∗ , s∗ ) is the whole space, i.e., K2 = X. So Ω2 = {(w, β, s) ∈ X | F1 (w) ≤ F1 (w∗ )}. Similar to [19] on p. 52, we have the following Lemma 3. Lemma 3 Let F1 (w) be defined by (3.9). Then F1 (w) is differentiable at any point in any b w) = max{−w(0, t)}, here S = {t ∈ [0, T ] | − direction and its directional derivative F10 (w; t∈S

w(0, b t) = F1 (w)}, b and F1 (w) satisfies the Lipschitz condition in any ball. Here we assume that F10 (w; h) 6= 0 provided that F1 (w) = 0. Note that for F1 (w) given by (3.9), the directional derivative of F1 at w∗ in the direction w + 1 is F10 (w∗ ; w + 1) = max{−w(0, t) − 1} < 0. t∈S

Hence ([Theorem 7.3, [19], p. 45]) K2 = {(w, β, s) ∈ X | F10 (w∗ ; w) < 0}.

(3.10)

Define the linear operator A : X → C[0, T ] by Aw = w(0, t) and K = {ξ ∈ C[0, T ] | ξ(t) ≥ 0, ∀ t ∈ S}. Then K2 = {(w, β, s) ∈ X | Aw(y, t) ∈ K}. o

In view of A(w + 1) = w(0, t) + 1 ∈K , the interior of K, one has ([19], p. 72–73) K2∗ = A∗ K ∗ , i.e., for any f2 ∈ K2∗ , there exists a nonnegative measure dm(t) with support on S such that f2 (w, β, s) = =

Z

Z

T

Aw(y, t)dm(t) = 0

w(0, t)dm(t) = S

Z

Z

Aw(y, t)dm(t) S

T

w(0, t)dm(t).

(3.11)

0

3.4 Determination of the cone of feasible directions K3 The arguments for the determination of the cone of feasible directions of Ω3 is similar to that of Ω2 . Let F2 (w, β, s) = max {α(s(t))wy (0, t) − Q(β(t))}. (3.12) 0≤t≤T

Then

Ω3 = {(w, β, s) ∈ X | α(s(t))wy (0, t) − Q(β(t)) ≤ 0, t ∈ [0, T ] a. e.} = {(w, β, s) ∈ X | F2 (w, β, s) ≤ 0}. ∗

∗

∗

Again we only consider F2 (w , β , s ) = max {α(s 0≤t≤T

∗

∗

∗

∗

∗

∗

∗

(t))wy∗ (0, t)

(3.13)

∗

− Q(β (t))} = 0, since

otherwise F2 (w , β , s ) < 0 and (w , β , s ) is an interior point of Ω3 . Hence any direction is feasible and the cone of feasible directions K3 of Ω3 at (w∗ , β ∗ , s∗ ) is the whole space, i.e.,

No. 3

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293

K3 = X. So Ω3 = {(w, β, s) ∈ X | F2 (w, β, s) ≤ F2 (w∗ , β ∗ , s∗ )}. Similar to Section 5, we have the following Lemma 4. Lemma 4 Let F2 (w, β, s) be given by (3.12). Then F2 is differentiable at any point in any direction and its directional derivative is given by ˆ sˆ; w, β, s) = max{α(ˆ ˆ F20 (w, b β, s(t))wy (0, t) + α0 (ˆ s)s(t)w by (0, t) − Q0 (β(t))β(t)} t∈Ξ

ˆ ˆ sˆ)} and F2 (w, β, s) satisfies the where Ξ = {t ∈ [0, T ] | α(ˆ s(t))w by (0, t) − Q(β(t)) = F2 (w, b β, Lipschitz condition in any ball, here again we assume that F20 (w, β, s; h1 , h2 , h3 ) 6= 0 provided that F2 (w, β, s) = 0. Next since F20 (w∗ , β ∗ , s∗ ; −α(s∗ ), Q0 (β ∗ ), −α0 (s∗ )wy∗ (0, t)) < 0, we have K3 = {(w, β, s) ∈ X | F20 (w∗ , β ∗ , s∗ ; w, β, s) < 0} .

(3.14)

Define the linear operator B : X → C[0, T ] by B(w, β, s) = −α(s∗ (t))wy (0, t) − α0 (s∗ (t))s(t)wy∗ (0, t) + Q0 (β ∗ (t))β(t) and Y = {ξ ∈ C[0, T ] | ξ(t) ≥ 0, ∀ t ∈ Ξ}. Then K3 = {(w, β, s) ∈ X | B(w(y, t), β(t), s(t)) ∈ Y } . o

By virtue of the fact that B(−α(s∗ ), Q0 (β ∗ ), −α0 (s∗ )wy∗ (0, t)) ∈Y , we have K3∗ = B ∗ Y ∗ . Namely, for any f3 ∈ K3∗ , there exists a nonnegative measure dn(t) with support on Ξ such that f3 (w, β, s) = = =

Z

Z

T

B(w(y, t), β(t), s(t))dn(t) = 0

Z

B(w(y, t), β(t), s(t))dn(t) Ξ

[−α(s∗ (t))wy (0, t) − α0 (s∗ (t))s(t)wy∗ (0, t) + Q0 (β ∗ (t))β(t)]dn(t)

Ξ Z T 0

[−α(s∗ (t))wy (0, t) − α0 (s∗ (t))s(t)wy∗ (0, t) + Q0 (β ∗ (t))β(t)]dn(t).

(3.15)

3.5 Determination of the cone of tangent directions K4 Define the operator G : X → C[0, T ; L2 (0, `)] × L2 (0, `) × (L∞ (0, T ))2 × C 1 [0, T ] by  Z th    w(y, t) − w (y) − α2 (s(τ ))wyy (y, τ ). 0    0 i    + f (β(τ ), s(τ ), y)wy (y, τ ) dτ,    w(y, T ) − w∗ (y, T ), G(w, β, s) = (3.16)  α(s(t))wy (`, t) + β(t)w(`, t) − qβ(t),     w(0, t)[−α(s(t))wy (0, t) + Q(β(t))],   Z t      s(t) + [α(s(τ ))wy (0, τ ) − Q(β(τ ))] dτ. 0

Then

Ω4 = {(w, β, s) ∈ X | G(w(y, t), β(t), s(t)) = 0}.

(3.17)

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Since

=

ˆ s + sˆ) − G(w, β, s) G(w + w, b β + β, Z t   2   α (s(τ ) + sˆ(τ ))(wyy (y, τ ) + w byy (y, τ )) w(y, t) + w(y, b t) − w (y) −  0   0      ˆ ), s(τ ) + sˆ(τ ), y)(wy (y, τ ) + w  + f (β(τ ) + β(τ by (y, τ )) dτ     Z t     − w(y, t) + w (y) + [α2 (s(τ ))wyy (y, τ ) + f (β(τ ), s(τ ), y)wy (y, τ )]dτ,  0   0     ∗  w0 (y, T ) + w(y, b T ) − w (y, T ) − w0 (y, T ) + w∗ (y, T ),       ˆ by (`, t)] + (β(t) + β(t))[w(`, t) + w(`, b t)]  α(s(t) + sˆ(t))[wy (`, t) + w ˆ  − q(β(t) + β(t)) − α(s(t))wy (`, t) − β(t)w(`, t) + qβ(t),      ˆ  (w(0, t) + w(0, b t))[−α(s(t) + sˆ(t))(wy (0, t) + w by (0, t)) + Q(β(t) + β(t))]       − w(0, t)[−α(s(t))wy (0, t) + Q(β(t))],     Z t    ˆ ))]dτ  [α(s(τ ) + sˆ(τ ))(wy (0, τ ) + w by (0, τ )) − Q(β(τ ) + β(τ s(t) + sˆ(t) +    0    Z t     − s(t) − [α(s(τ ))wy (0, τ ) − Q(β(τ ))]dτ,   0

the Fr´echet-derivative of the operator G(w, β, s) is

=

ˆ sˆ) G0 (w, β, s)(w, b β,  Z t   w(y, b t) − [α2 (s(τ ))w byy (y, τ ) + f (β(τ ), s(τ ), y)w by (y, τ )]dτ    0     Z t   ∂f (β, s, y)   ∂f (β, s, y)  0  ˆ 2α(s(τ ))α (s(τ ))ˆ s(τ )wyy (y, τ ) + sˆ(τ ) wy (y, τ ) dτ, β(τ ) + −   ∂β ∂s 0      b T ),  w(y,  ˆ + α0 (s(t))ˆ ˆ α(s(t))w by (`, t) + β(t)w(`, b t) − q β(t) s(t)wy (`, t) + β(t)w(`, t),      ˆ  w(0, t)[−α0 (s(t))ˆ s(t)wy (0, t) − α(s(t))w by (0, t) + Q0 (β(t))β(t)]       +w(0, b t)[−α(s(t))wy (0, t) + Q(β(t))],    Z t     ˆ )]dτ.  s ˆ (t) + [α0 (s(τ ))ˆ s(τ )wy (0, τ ) +α(s(τ ))w by (0, τ ) −Q0 (β(τ ))β(τ   0

Since (w∗ , β ∗ , s∗ ) is the solution to the problem (3.1), it has G(w ∗ , β ∗ , s∗ ) = 0. Choosing arbitrary (g, g0 , g1 , g2 , g3 ) ∈ C[0, T ; L2 (0, `)] × L2 (0, `) × (L∞ (0, T ))2 × C 1 [0, T ] and solving the equation ˆ sˆ) = (g(y, t), g0 (y), g1 (t), g2 (t), g3 (t)), G0 (w∗ , β ∗ , s∗ )(w, b β,

No. 3

OPTIMAL CONTROL OF AN ABLATION PROCESS

we obtain Z t    α2 (s∗ (τ ))w byy (y, τ ) + f (β ∗ (τ ), s∗ (τ ), y)w by (y, τ ) w(y, b t) −    0   ∗ ∗  − 2α(s (τ ))α0 (s∗ (τ )) · sˆ(τ )wyy (y, τ )       ∗ ∗  ∂f (β , s , y) ˆ ∂f (β ∗ , s∗ , y)  ∗  − (y, τ ) dτ = g(y, t), s ˆ (τ ) w β(τ ) +  y  ∂β ∂s       b T ) = g0 (y),   w(y, ˆ + α0 (s∗ (t))ˆ α(s∗ (t))w by (`, t) + β ∗ (t)w(`, b t) − q β(t) s(t)wy∗ (`, t)  ∗ ˆ  + β(t)w (`, t) = g1 (t),      ∗ 0 ∗ ˆ  s(t)wy∗ (0, t) − α(s∗ (t))w by (0, t) + Q0 (β ∗ (t))β(t)]   w (0, t)[−α (s (t))ˆ     + w(0, b t)[−α(s∗ (t))wy∗ (0, t) + Q(β ∗ (t))] = g2 (t),     Z t    ˆ )]dτ = g3 (t).  s ˆ (t) + [α0 (s∗ (τ ))ˆ s(τ )wy∗ (0, τ ) + α(s∗ (τ ))w by (0, τ ) − Q0 (β ∗ (τ ))β(τ  

295

(3.18)

0

Next, assume that the linearized system  wt (y, t) = α2 (s∗ )wyy (y, t) + f (β ∗ , s∗ , y)wy (y, t)         ∂f (β ∗ , s∗ , y) ∂f (β ∗ , s∗ , y)  ∗ 0 ∗ ∗  + 2α(s )α (s )s(t)wyy (y, t) + β(t) + s(t) wy∗ (y, t),    ∂β ∂s    α(s∗ )wy (`, t) + β ∗ (t)w(`, t) = qβ(t) − α0 (s∗ )s(t)wy∗ (`, t) − β(t)w∗ (`, t),    w(0, t)[−α(s∗ )wy∗ (0, t) + Q(β ∗ (t))] + w∗ (0, t)s0 (t) = 0,       s0 (t) = −α0 (s∗ )wy∗ (0, t)s(t) − α(s∗ )wy (0, t) + Q0 (β ∗ )β(t),      w(y, 0) = 0, s(0) = 0

(3.19)

ˆ ∈ L∞ (0, T ) such that w(y, T ) = g0 (y) − γ(y, T ) and let is controllable. Then choose β(t) = β(t) (w, s) be the solution to the linearized system (3.19). Choose w(y, b t) = w(y, t) + γ(y, t), sˆ(t) = s(t) + (t), where (γ, ) satisfies the following equation:  Z t    γ(y, t) = g(y, t) + α2 (s∗ )γyy (y, τ ) + f (β ∗ , s∗ , y)γy (y, τ )    0      ∂f (β ∗ , s∗ , y)  ∗ 0 ∗ ∗ ∗  + 2α(s )α (s )(τ )wyy (y, τ ) + (τ )wy (y, τ ) dτ,    ∂s  α(s∗ )γy (`, t) + β ∗ (t)γ(`, t) = −α0 (s∗ )(t)wy∗ (`, t) + g1 (t),      w∗ (0, t)[−α0 (s∗ )(t)wy∗ (0, t) − α(s∗ )γy (0, t)] + γ(0, t)[−α(s∗ )wy∗ (0, t) + Q(β ∗ )] = g2 (t),      Z t    0 ∗ ∗ ∗   α (s )(τ )w (t) = − (0, τ ) + α(s )γ (0, τ ) dτ + g3 (t). y y   0

ˆ sˆ) satisfying (3.18). Therefore G0 (w∗ , β ∗ , s∗ ) maps X onto In this way, it suffices for (w, b β, 2 2 ∞ C[0, T ; L (0, `)] × L (0, `) × (L (0, T ))2 × C 1 [0, T ]. Moreover the cone of the tangent directions K4 to the constraint Ω4 at the point (w ∗ , β ∗ , s∗ ) consists of the kernel of G0 (w∗ , β ∗ , s∗ ), i.e.,

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(w, β, s) satisfies the following equation in X ([Theorem 9.1, [19], p. 61]):   wt (y, t) = α2 (s∗ )wyy (y, t) + f (β ∗ , s∗ , y)wy (y, t)        ∂f (β ∗ , s∗ , y) ∂f (β ∗ , s∗ , y)  ∗ 0 ∗ ∗  +2α(s )α (s )s(t)wyy (y, t) + β(t) + s(t) wy∗ (y, t),    ∂β ∂s    α(s∗ )wy (`, t) + β ∗ (t)w(`, t) = qβ(t) − α0 (s∗ )s(t)wy∗ (`, t) − β(t)w∗ (`, t),    w(0, t)[−α(s∗ )wy∗ (0, t) + Q(β ∗ (t))] + w∗ (0, t)s0 (t) = 0,     0  0 ∗ ∗ ∗ 0 ∗    s (t) = −α (s )wy (0, t)s(t) − α(s )wy (0, t) + Q (β )β(t),    w(y, 0) = 0, s(0) = 0,

Vol. 18

(3.20)

and

w(y, T ) = 0.

(3.21)

Define K41 = {(w, β, s) ∈ X | (w(y, t), β(t), s(t)) satisfies (3.20)}, K42 = {(w, β, s) ∈ X | (w(y, t), β(t), s(t)) satisfies (3.21)}. T Then the cone of the tangent directions K4 = K41 K42 . Hence ∗ ∗ K4∗ = K41 + K42 .

For any f4 ∈ K4∗ , decompose ∗ f4 = f41 + f42 , f4i ∈ K4i , i = 1, 2.

Then f41 (w, β, s) = 0 and for all w(y, t) ∈ C[0, T ; L2(0, `)] satisfying w(y, T ) = 0, there exists a %(y) ∈ L2 (0, `) such that Z ` f42 (w(y, t), β(t), s(t)) = w(y, T )%(y)dy. 0

It then follows from Theorem 3 that there exist continuous linear functionals, not all identically zero, such that f0 + f1 + f2 + f3 + f41 + f42 = 0. Therefore, when selecting (w, β, s) satisfying (3.19), f41 (w, β, s) = 0. Moreover, f1 (w(y, t), β(t), s(t)) = −f0 (w(y, t), β(t), s(t)) − f2 (w(y, t), β(t), s(t)) −f3 (w(y, t), β(t), s(t)) − f42 (w(y, t), β(t), s(t)) Z T Z ` ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) = λ0 w(y, t) + β(t) ∂w ∂β 0 0  ∂L(w∗ , β ∗ , s∗ , y, t) + s(t) dydt ∂s Z T  ∗  α(s )wy (0, t) + α0 (s∗ )s(t)wy∗ (0, t) − Q0 (β ∗ )β(t) dn(t) + −

Z

0

T

0

w(0, t)dm(t) −

Z

`

w(y, T )%(y)dy. 0

(3.22)

No. 3

OPTIMAL CONTROL OF AN ABLATION PROCESS

3.6 Maximum principle for problem (3.1) Define the adjoint system of (3.19) as  ∂f (β ∗ , s∗ , y)   vt (y, t) = −α2 (s∗ )vyy (y, t) + f (β ∗ , s∗ , y)vy (y, t) + v(y, t)    ∂y      ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) s(t)   + λ0 + λ0   ∂w ∂s w(y, t)   α(s∗ )wy (0, t) − Q0 (β ∗ )β(t) dn(t) w(0, t) dm(t)  + − ,   `w(y, t) dt `w(y, t) dt      α(s∗ )vy (0, t) = [β ∗ (t) + s∗0 (t)]v(0, t), vy (`, t) = 0, v(y, T ) = %(y),         r0 (t) = α0 (s∗ )wy∗ (0, t)r(t) + α0 (s∗ )wy∗ (0, t) dn(t) , r(T ) = 0 dt

with

s(t)

Z

`

0

297

(3.23)

  ∂f (β ∗ , s∗ , y) ∗ ∗ −2α(s∗ )α0 (s∗ )wyy (y, t) − wy (y, t) v(y, t)dy ∂s

+ r(t)α(s∗ )wy (0, t)

= α(s∗ )α0 (s∗ )s(t)[wy∗ (0, t)v(0, t) + wy∗ (`, t)v(`, t)].

(3.24)

As with (2.1), when the solution of (3.24), (3.25) are mentioned, we mean the weak solution. Lemma 5 The solution of the system (3.19) and the solution of its adjoint system (3.23), (3, 24) have the following relation:  Z TZ `  ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) λ0 w(y, t) + s(t) dydt ∂w ∂s 0 0 Z T  ∗  + α(s )wy (0, t) + α0 (s∗ )s(t)wy∗ (0, t) − Q0 (β ∗ )β(t) dn(t) − =

Z

Z T

0

0

T

0

(

w(0, t)dm(t) − −

Z

` 0

Z

`

w(y, T )%(y)dy 0

∂f (β ∗ , s∗ , y) ∗ wy (y, t)v(y, t)dy − r(t)Q0 (β ∗ ) ∂β )

+ α(s∗ )Q0 (β ∗ )v(0, t) − α(s∗ )[q − w∗ (`, t)]v(`, t) β(t)dt.

(3.25)

Proof Multiplying the first equation in (3.23) by w(y, t) and integrating from 0 to T and 0 to ` with respect to t and y respectively yields Z TZ ` vt (y, t)w(y, t)dydt 0

=

0

Z TZ ` n

∂f (β ∗ , s∗ , y) v(y, t)w(y, t) ∂y 0 0 o ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) + λ0 w(y, t) + λ0 s(t) dydt ∂w ∂s Z T Z T + [α(s∗ )wy (0, t) − Q0 (β ∗ )β(t)]dn(t) − w(0, t)dm(t). 0

− α2 (s∗ )vyy (y, t)w(y, t) + f (β ∗ , s∗ , y)vy (y, t)w(y, t) +

0

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Similarly, we have Z T Z r0 (t)s(t)dt = 0

T 0

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α0 (s∗ )wy∗ (0, t)r(t)s(t)dt +

Z

T 0

α0 (s∗ )wy∗ (0, t)s(t)dn(t).

For these two integrals, we integrate them by parts and transfer the derivatives from v(y, t) and r(t) to w(y, t) and s(t) respectively. The proof then follows. Now, by virtue of Lemma 5, we can rewrite f1 (w, β, s) as  Z T (Z `  ∂L(w∗ , β ∗ , s∗ , y, t) ∂f (β ∗ , s∗ , y) ∗ λ0 f1 (w, β, s) = − wy (y, t)v(y, t) dy ∂β ∂β 0 0 ) −r(t)Q0 (β ∗ ) + α(s∗ )Q0 (β ∗ )v(0, t) − α(s∗ )[q − w∗ (`, t)]v(`, t) β(t)dt.

(3.26)

Therefore a ¯(t) =

 Z ` ∂L(w∗ , β ∗ , s∗ , y, t) ∂f (β ∗ , s∗ , y) ∗ λ0 − wy (y, t)v(y, t) dy − r(t)Q0 (β ∗ ) ∂β ∂β 0

+α(s∗ )Q0 (β ∗ )v(0, t) − α(s∗ )[q − w∗ (`, t)]v(`, t),

and (3.8) then reads (Z   ` ∂L(w∗ , β ∗ , s∗ , y, t) ∂f (β ∗ , s∗ , y) ∗ λ0 − wy (y, t)v(y, t) dy − r(t)Q0 (β ∗ ) ∂β ∂β 0 ) +α(s∗ )Q0 (β ∗ )v(0, t) − α(s∗ )[q − w∗ (`, t)]v(`, t) ∀ β(t) ∈ [β0 , β1 ],

t ∈ [0, T ],

· (β(t) − β ∗ (t)) ≥ 0,

a. e.,

(3.27)

where λ0 , v(y, t), and r(t) are not identical to zero simultaneously, since otherwise, we must have dn(t) ≡ 0, contradicting the choice of dn(t). On the other hand, if K0 is a null set, then we have  Z TZ ` ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) w(y, t) + β(t) + s(t) dydt = 0, ∂w ∂β ∂s 0 0 ∀ (w, β, s) ∈ X. In particular, if we choose λ0 = 1, %(y) = 0, dm(t), dn(t) satisfying Z T Z [α(s∗ )wy (0, t) + α0 (s∗ )s(t)wy∗ (0, t) − Q0 (β ∗ )β(t)]dn(t) = 0

T

w(0, t)dm(t), 0

it then follows from Lemma 5 that  Z TZ `  ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) w(y, t) + s(t) dydt ∂w ∂s 0 0 Z T( Z ` ∂f (β ∗ , s∗ , y) ∗ = − wy (y, t)v(y, t)dy − r(t)Q0 (β ∗ ) ∂β 0 0 ) + α(s∗ )Q0 (β ∗ )v(0, t) − α(s∗ )[q − w∗ (`, t)]v(`, t) β(t)dt.

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Therefore, Z T (Z ` 0

0

 ∂L(w∗ , β ∗ , s∗ , y, t) ∂f (β ∗ , s∗ , y) ∗ − wy (y, t)v(y, t) dy − r(t)Q0 (β ∗ ) ∂β ∂β )

+ α(s∗ )Q0 (β ∗ )v(0, t) − α(s∗ )[q − w∗ (`, t)]v(`, t) β(t)dt = 0, ∀ β(t) ∈ L∞ (0, T ), from which we obtain  Z ` ∂L(w∗ , β ∗ , s∗ , y, t) ∂f (β ∗ , s∗ , y) ∗ − wy (y, t)v(y, t) dy − r(t)Q0 (β ∗ ) ∂β ∂β 0 +α(s∗ )Q0 (β ∗ )v(0, t) − α(s∗ )[q − w∗ (`, t)]v(`, t) = 0,

∀ t ∈ [0, T ] a. e.

Therefore (3.27) still holds. Finally, if there is a nonzero solution to the adjoint system  ∂f (β ∗ , s∗ , y)  2 ∗ ∗ ∗  v ˆ (y, t) = −α (s )ˆ v (y, t) + f (β , s , y)ˆ v (y, t) + vˆ(y, t)  t yy y   ∂y      ∂L(w∗ , β ∗ , s∗ , y, t) ∂L(w∗ , β ∗ , s∗ , y, t) s(t)   + λ0 + λ0   ∂w ∂s w(y, t)   ∗ 0 ∗ α(s )wy (0, t) − Q (β )β(t) dn(t) w(0, t) dm(t)  + − ,   `w(y, t) dt `w(y, t) dt      α(s∗ )ˆ vy (0, t) = [β ∗ (t) + s∗0 (t)]ˆ v (0, t), vˆy (`, t) = 0, vˆ(y, T ) = %(y),       dn(t)   rˆ0 (t) = α0 (s∗ )wy∗ (0, t)ˆ r (t) + α0 (s∗ )wy∗ (0, t) , rˆ(T ) = 0, dt with

s(t)

Z ` 0

∗ −2α(s∗ )α0 (s∗ )wyy (y, t) −

(3.28)

 ∂f (β ∗ , s∗ , y) ∗ wy (y, t) vˆ(y, t)dy ∂s

+ˆ r(t)α(s∗ )wy (0, t) = α(s∗ )α0 (s∗ )s(t)[wy∗ (0, t)ˆ v (0, t) + wy∗ (`, t)ˆ v (`, t)],

(3.29)

such that the following equality holds true Z ` ∂f (β ∗ , s∗ , y) ∗ − wy (y, t)ˆ v (y, t)dy − rˆ(t)Q0 (β ∗ ) + α(s∗ )Q0 (β ∗ )ˆ v (0, t) ∂β 0 −α(s∗ )[q − w∗ (`, t)]ˆ v (`, t) = 0,

∀ t ∈ [0, T ] a. e.,

then when we choose λ0 = 0, %(y) = vˆ(y, T ), (3.27) is still valid. Otherwise, for an any nonzero solution (ˆ v , rˆ) of (3.28) and (3.29), it has Z ` ∂f (β ∗ , s∗ , y) ∗ − wy (y, t)ˆ v (y, t)dy − rˆ(t)Q0 (β ∗ ) + α(s∗ )Q0 (β ∗ )ˆ v (0, t) ∂β 0 −α(s∗ )[q − w∗ (`, t)]ˆ v (`, t) 6≡ 0,

(3.30)

in this case we say that the situation is non-degenerate, and the linearized system (3.19) is controllable. In fact, if (3.19) is not controllable, then there exists a %(y) ∈ L2 (0, `) such that Z ` %(y)w(y, T )dy = 0, %(y) 6≡ 0. 0

300

SUN BING

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GUO BAOZHU

Choose λ0 = 0, (ˆ v , rˆ) to be the solution of (3.28) and (3.29) and dm(t), dn(t) satisfying Z

T 0

[α(s∗ )wy (0, t) + α0 (s∗ )s(t)wy∗ (0, t) − Q0 (β ∗ )β(t)]dn(t) =

Z

T

w(0, t)dm(t), 0

then it follows from Lemma 5 that Z T Z ` ∂f (β ∗ , s∗ , y) ∗ − wy (y, t)ˆ v (y, t)dy − rˆ(t)Q0 (β ∗ ) + α(s∗ )Q0 (β ∗ )ˆ v (0, t) ∂β 0 0  ∗ ∗ −α(s )[q − w (`, t)]ˆ v (`, t) β(t)dt = 0, ∀ β(t) ∈ L∞ (0, T ). Hence −

Z

` 0

∂f (β ∗ , s∗ , y) ∗ wy (y, t)ˆ v (y, t)dy − rˆ(t)Q0 (β ∗ ) + α(s∗ )Q0 (β ∗ )ˆ v (0, t) ∂β

−α(s∗ )[q − w∗ (`, t)]ˆ v (`, t) = 0,

∀ t ∈ [0, T ] a.e.

This is a contradiction. Therefore, under the assumption of (3.30), the system (3.19) is controllable. Combining the results above, we have obtained the Pontryagin’s maximum principle for the problem (3.1) subject to the system (2.1). Theorem 4 Suppose (w ∗ , β ∗ , s∗ ) is a solution to the optimal control problem (3.1). Then there exist λ0 ≥ 0 and v(y, t), r(t), not identically zero, such that the following maximum principle holds: Z `  ∂L(w∗ , β ∗ , s∗ , y, t) ∂f (β ∗ , s∗ , y) ∗ λ0 − wy (y, t)v(y, t) dy − r(t)Q0 (β ∗ ) ∂β ∂β 0  ∗ 0 ∗ ∗ ∗ +α(s )Q (β )v(0, t) − α(s )[q − w (`, t)]v(`, t) · [β(t) − β ∗ (t)] ≥ 0, ∀ β(t) ∈ [β0 , β1 ] t ∈ [0, T ] a. e.,

(3.31)

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