Hamilton-Jacobi Equation and Optimality conditions for control systems governed by semilinear parabolic equations with boundary control

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1 Hamilton-Jacobi Equation and Optimality conditions for control systems governed by semilinear parabolic equations with boundary control Marc Quincampoix and Catalin Trenchea Laboratoire de Mathématiques, Unité CNRS, FRE-228 Université de Bretagne Occidentale 6 avenue Victor Le Gorgeu BP 809, Brest cedex, FRANCE Marc-Quincampoix@univ-brest.fr, Catalin.Trenchea@univ-brest.fr January 7, 2004 Abstract : We characterize the value function by an appropriate Hamilton Jacobi Bellman equation in viscosity sense and derive optimality conditions from the knowledge of the value function. Keywords : value function, viscosity solutions. AMS Classification. 49K20, 49L25, 49L20, 35B37, 93C20, 35K20. Introduction Let be an open bounded subset in R N N 2 of class C 2,β for some β > 0 and Q = 0, T, Σ = Γ 0, T, Γ is the boundary of, T > 0. We consider a control system described by the following parabolic equation y + Ay + fx, t, y = 0 in Q, t y + gs, t, y, v = 0 on Σ, y, t 0 = y 0 in,. where A is a second-order differential operator defined by N Ayx = D i a ij xd j yx i,j= with coefficients a ij belonging to C,β and satisfying the conditions a ij x = a ji x for every i, j {,..., N}, Work supported by the TMR No. HPRN-CT Evolution Equations and Deterministic and Stochastic Systems

2 m 0 ξ 2 N a ij xξ j ξ i M ξ 2 i,j= for all x and all ξ R N, with 0 < m 0 < M D i denotes the partial derivative with respect to x i. Let y/ s, t = i,j a ijsd j ys, tn i s denote the normal derivative of y associated with A, and n = n,..., n N is the outward unit normal to Γ. In the present paper we will investigate the following optimal control problem { } inf Lx, yt dx y, v L Q L σ Σ, y, v satisfies.. P With the above control problem we associate a value function { } V t 0, y 0 = inf Lx, yt ; t 0, y 0, vdx y is a solution to., v L σ Σ.2 letting t 0, y 0 range over [0, T ] Y, with Y = L H. Our main aim in this article is twofold: we wish first to characterize the value function by a appropriate Hamilton Jacobi Bellman equation in viscosity sense, second to derive optimality conditions from the knowledge of the value function. This problem has been considered in the early literature in the case of smooth value function [] and for distributed control problems, namely the control does not act only on the boundary but on the set cf [2], [3], [3], [4]. Since the control acts on the system through a boundary condition, one cannot apply the existing results of characterization of the value function by viscosity solutions of Hamilton Jacobi equations [4], [5], [7], [8], [9], [2] but one has to provide a new proof. We derive necessary and sufficient optimality conditions using the -Lipschitz- value function. Note again that because this Mayer control problem has a boundary control, our technique slightly differ from those employed in [2] and [4]. Let us explain how our paper is organized. In the first section we state our assumptions and recall some basic estimations on the control dynamics. The second section is devoted to study the value function: we prove a dynamic programming principle and we obtain the Lipschitz regularity and the semiconcavity of the value-function. Section 3 establishes a characterization of the value function in terms of the unique solution in viscosity sense of an Hamilton Jacobi Bellman equation. The last section is intended to prove necessary conditions of optimality; sufficient conditions are also obtained. 2 Preliminaries We denote by q, σ positive numbers satisfying q > N 2 + and σ > N +. We make the following assumptions: 2

3 A For every y R, f, y is continuous on Q. For almost every x, t Q, fx, t, is of class C on R. The following estimates hold: fx, t, 0 M x, t, f yx, t, y f yx, t, z M x, tη y η z ω y z, C 0 f y x, t, y M x, tη y, f t x, t, y M xηtη y, where M belongs to L q Q, η is a nondecreasing function from R to R, and C 0 R. A2 For every y, v R 2, g, y, v is continuous on Σ. For almost every s, t Σ and every v R, gs, t,, v is of class C on R. For almost every s, t Σ, gs, t, and g s, t, are continuous on R R. The following estimates hold: gs, t, 0, v M 2 s, t C 0 g y s, t, y, v M 2s, t + m v η y, g y s, t, y, v g y s, t, z, v M 2s, tη y η z ω y z, where M 2 belongs to L σ Σ, m > 0, and C 0 and η are as in A. A3 For every y R, L, y is measurable on. For almost every x, Lx, is of class C on R. The following estimates are verified: L y x, y M 3η y, L y x, y L y x, z M 3η y η z ω y z, where M 3 R +, η is as in A and ω is an increasing continuous function from R + to R + such that ω0 = 0. We recall the following results see [3] on the existence, uniqueness and regularity of the state variable. Proposition 2. Under assumptions A and A2, if v L σ Σ and y 0 L, then. admits a unique weak solution y v in W 0, T L Q. This solution belongs to CQ ε,t and satisfies the estimates y v L Q + y v L Σ C g, 0, v L σ Σ + y 0 L +, y v CQε,T C 2ε g, 0, v Lσ Σ + y 0 L where C = C T,, N, q, σ, C 0 and C 2 ε = C 2 ε, T,, N, q, σ, C 0. Moreover, the mapping v y v is continuous from L σ Σ into L Q. Here W 0, T = {y L 2 0, T ; H y t L2 0, T ; H } and CQ ε,t = {y L 2 Q y is continuous on [ε, T ]}. Proposition 2.2 For every k > 0 and every ε > 0, there exist C 3 = C 3 T,, N, q, σ, C 0, k, C 4 = C 4 T,, N, q, σ, C 0, k, ε, and α > 0 such that, for every v, y 0 L σ Σ C satisfying v L σ Σ + y 0 L k, the weak solution y v of. corresponding to v, y 0 is Hölder continuous on [ε, T ] and obeys y v CQ C 3, y v C α, α 2 [ε,t ] C 4. Moreover, if y 0 is Hölder continuous on, then y v is Hölder continuous on Q. 3

4 3 Value function In this section we establish some basic property of the value function. Proposition 3. V is continuous in [0, T ] L and bounded on bounded subsets of [0, T ] L. Moreover, y 0 V t, y 0 is Lipschitz on all bounded subsets of L, uniformly in t [0, T ]. Furthermore, for every v L σ Σ the function V, y ; t 0, y 0, v is nondecreasing in [t 0, T ] and the dynamic programming principle hold { } V t 0, y 0 = inf V t, yt; t 0, y 0, v v L σ Σ for all t [t 0, T ]. 3. Finally, t V t, yt; t 0, y 0, v is constant if and only if v is optimal for problem P. Proof. These basic properties are well known in the Hilbert space case see [] and can be proved by similar arguments in the present context. We will sketch the proof for reader s convenience. Let denote by rb, r > 0, the closed ball in L with radius r and center at 0. We note first that from 2. there exists K r > 0 such that yt; t 0, ξ, v K r, ξ rb, t [t 0, T ], and for all controls v L σ Σ. Furthermore, there exists a constant C r > 0 such that yt; t 0, ξ, v yt; t 0, ξ 2, v L C r ξ ξ 2 L, 3.2 ξ, ξ 2 rb, t [t 0, T ] and for all controls v L σ Σ. Indeed yt; t 0, ξ, v yt; t 0, ξ 2, v is also a solution to the following equation z + Az + az = 0 in Q, t z + b v z = 0 on Σ, z, t 0 = ξ ξ 2 in, where ax, t = 0 f y x, t, y 2 + θy y 2 dθ C 0, b v s, t = 0 g y s, t, y 2 + θy y 2, vdθ C 0 and the estimate yields from Proposition 2.. Let 0 t 0 t T and let ṽ L σ [t 0, T ] be defined by ṽθ = v 0 for t 0 θ < t ; ṽ = v on [t, T ] where v 0 is an element of L σ [t 0, t ] and v is optimal in P [t,t ] with the initial datum yt = y 0. Then we have V t 0, y 0 V t, y 0 Lx, yt ; t 0, y 0, ṽdx Lx, yt ; t, y 0, v dx M 3 ηk r yt ; t, yt ; t 0, y 0, ṽ, v yt ; t, y 0, v dx M 3 ηk r mc r yt ; t 0, y 0, ṽ y 0 L 4

5 and the t-continuity of V yields from Proposition 2.. Now consider y 0, y two initial data for P and, for a fixed ε > 0, a control v ε such that Lx, yt ; t 0, y, v ε dx < V t 0, y + ε. We have V t 0, y 0 V t 0, y M 3 ηk r Lx, yt ; t 0, y 0, v ε dx Lx, yt ; t 0, y, v ε dx + ε yt ; t 0, y 0, v ε yt ; t 0, y, v ε dx + ε M 3 ηk r C r m y 0 y L + ε so V t, is Lipschitz continuous, uniformly in t [t 0, T ]. Now we fix v L σ Σ and prove that V, y, t 0, y 0, v is nondecreasing in [t 0, T ]. These yields from the fact that {u L σ Σ; u [t0,t 2 = v} {u ] L σ Σ; u = v} for t [t0,t ] 0 t t 2 T, and the definition.2. V t, yt ; t 0, y 0, v = inf u Lx, yt ; t, yt ; t 0, y 0, v, udx inf Lx, yt ; t 2, yt 2 ; t 0, y 0, v, udx = V t 2, yt 2 ; t 0, y 0, v. u Let prove first that V t 0, y 0 inf V t, yt; t 0, y 0, v v L σ Σ and consider an arbitrary control v L σ Σ. Then for V t, yt; t 0, y 0, v there exists an control ε-optimal, v ε L σ Σ, such that V t, yt; t 0, y 0, v + ε Lx, y ε T ; t, yt; t 0, y 0, v, v ε dx where y ε is the trajectory starting at yt; t 0, y 0, v, associated to the control v ε. We define now for V t 0, y 0 a control ṽ in the following manner { vs, θ if t0 θ < t, ṽs, θ = v ε s, θ if t θ T for all s. If ỹ denote the trajectory associated to ṽ, we have V t 0, y 0 Lx, ỹt ; t 0, y 0, ṽdx = Lx, y ε T ; t, yt; t 0, y 0, v, v ε dx. Hence V t 0, y 0 ε + V t, yt; t 0, y 0, v and we conclude by letting ε tend to 0 and taking the infimum with respect to v, which is arbitrary in L σ Σ. 5

6 For the opposed inequality we consider an control v ε, ε-optimal for V t 0, y 0, and denote by y ε the associated trajectory. We have then V t 0, y 0 + ε Lx, y ε T ; t 0, y 0, v ε dx = Lx, y ε T ; t, y ε t; t 0, y 0, v ε, v ε dx V t, y ε t; t 0, y 0, v ε inf v L σ Σ V t, yt; t 0, y 0, v. We get the second inequality by letting ε 0 and therefore the desired 3.. To prove the final assertion let denote by y, v the optimal pair for problem P. Then by 3. and.2 we have V t 0, y 0 = inf {V t, yt; t 0, y 0, v} V t, y t; t 0, y 0, v v L σ Σ = inf Lx, yt ; t, y t; t 0, y 0, v, v v L σ Σ [t,t ] dx Lx, y T ; t, y t; t 0, y 0, v, v dx = Lx, y T ; t [t,t ] 0, y 0, v dx = V t 0, y 0 for all t [0, T ] and therefore V, y ; t 0, y 0, v is constant. Conversely, let v be a control such that t V t, ȳt; t 0, y 0, v is constant, i.e., V t, yt; t 0, y 0, v = V t 0, y 0, t [0, T ]. Using the definition of the value function and the boundary condition for V we get { } inf Lx, yt ; t 0, y 0, vdx = V t 0, y 0 = V T, yt ; t 0, y 0, v v = Lx, ȳt dx and therefore v = v. We recall next some generalizations of the notion of gradient for nonsmooth functions see [2], [0] for more details. Let K be an open subset of Y and ϕ : K R. For any y 0 K, the superdifferential D + ϕy 0 is defined as follows D + ϕy 0 = { p Y ϕy ϕy 0 py y 0 y y 0 y y 0 } 0. If ϕ is Fréchet differentiable at y 0, then D + ϕy 0 is a singleton and coincides with the gradient ϕy 0. We denote by D ϕy 0 the set of all p Y for which there exists a sequence {y n } n N of points in Y such that i y n converges to y 0 as n, ii ϕ is Fréchet differentiable at y n for all n N, iii ϕy n weakly-* converges to p as n. 6

7 Let K Y be convex and ϕ : K R. We say that ϕ is semiconcave if there exists a function { r R, s S, ωr, s ωr, S, ω : R + R + R + satisfying r 0, lim ωr, s = 0, s 0 such that, for every r > 0, λ [0, ], and ζ, ζ 2 K rb, λϕζ + λϕζ 2 ϕλζ + λζ 2 λ λ ζ ζ 2 ωr, ζ ζ 2. For instance, by A-A3, we have that fx, t,, gs, t,, and Lx, are semiconcave. Superdifferentials of semiconcave functions enjoy the following regularity property. If ϕ is semiconcave and Lipschitz in y 0 + rb for some r > 0 then D + ϕy 0 = co D ϕy 0, where co denotes the closed convex hull. Next we prove a semiconcavity result for the value function V. Theorem 3. Assume A-A3. [0, T ]. Then V t, is semiconcave for all t Proof. Fix r > 0, t 0 [0, T ], and y0, y0 2 rb. Now let λ 0, and define y λ = λy0 2 + λy 0. Fix ε > 0 and consider a control v ε such that Lx, yt ; t 0, y λ, v ε dx < V t 0, y λ + ε. Set y i = y ; t 0, y i 0, v ε, i =, 2. Then using A3 we get λv t 0, y0 2 + λv t 0, y0 V t 0, y λ λ Lx, y 2 T dx + λ Lx, ȳ T dx Lx, yt ; t 0, y λ, v ε dx + ε λ λ y 2 T ȳ T ω ȳ 2 T ȳ T dx + [Lx, λȳ 2 T + λy T Lx, yt ; t 0, y λ, v ε ] dx + ε λ λ y 2 T ȳ T ω ȳ 2 T ȳ T dx + ε +M 3 ηk r λȳ2 T + λy T yt ; t 0, y λ, v ε. It remains to estimate the last term in the above inequality. Set y λ t = λȳ 2 t + λȳ t yt; t 0, y λ, v ε. As previously we have that y λ satisfies the equation z t + A z + ã z = fx, t, λȳ 2 + λy λfx, t, ȳ 2 λfx, t, ȳ in Q, z + bz = gs, t, λȳ 2 + λȳ, v ε λgs, t, y 2, v ε λgs, t, ȳ, v ε on Σ, z, t 0 = 0 in, 7

8 where ãx, t = 0 f y x, t, yt; t0, y λ, v ε +θλy 2 t+ λȳ t yt; t 0, y λ, v ε dθ C 0, and bs, t = 0 y g s, t, yt; t0, y λ, v ε +θλy 2 t+ λȳ t yt; t 0, y λ, v ε dθ C 0. Therefore y λ C Q C f, λy 2 + λȳ λf, ȳ 2 λf, ȳ L q Q + g, λy 2 + λȳ, v ε λg, ȳ 2, v ε λg, ȳ, v ε Lσ Σ λ λc r y 0 y 2 0 L mq /q + mσ /σ η 2 K r ωc r ȳ 0 ȳ 2 0 L which completes the proof. 4 Hamilton-Jacobi-Bellman Equation We define the Hamiltonian function for the Mayer problem P by Ht, y, p = min v L σ Γ fx, t, yp dx a ij xd j yd i p dx i,j } 4. gs, t, ys, vsps ds for all t, y, p [t 0, T ] Y Y. Theorem 4. The value function V is the viscosity solution to the Hamilton- Jacobi-Bellman equation V V + Ht, y, = 0 on [0, T Y t y V T, y = Lx, yt dx. 4.2 We recall see [9] that V C0, T Y is a viscosity subsolution of 4.2 on [0, T Y if and only if for every φ C0, T Y φ φ + Ht, y, t y 0 at each local maximum point t, y 0, T Y of V φ at which φ is differentiable. Similarly, V is a viscosity supersolution of 4.2 when φ φ + Ht, y, t y 0 at each local minimum point t, y 0, T Y of V φ at which φ is differentiable. Finally, V is a viscosity solution if it is both a viscosity subsolution and a viscosity supersolution. Proof. Let t 0, y 0 be a local maximum point of V φ and assume without loss of generality that V t 0, y 0 = φt 0, y 0, i.e., V t 0 + h, yt 0 + h φt 0 + h, yt 0 + h. 8

9 By the Dynamic Programming Principle 3. with a constant control v, t v L σ we have and therefore Hence V t 0, y 0 V t 0 + h, yt 0 + h 0 φt 0 + h, yt 0 + h φt 0, y 0. 0 φ t t φ 0, y 0 h + y t 0, y 0 yt 0 + h yt 0 dx + oh Dividing by h and letting h 0 we get 0 φ t t 0, y 0 + lim h 0 h t0+h t 0 Ht 0, yt, φ y t 0, y 0 dt so V is a viscosity supersolution of 4.2. Assume now that t 0, y 0 is a local minimum point of V φ. If y, v is the optimal pair for problem P, then again by Proposition 3. we have which implies V t 0, y 0 = V t 0 + h, yt 0 + h φt 0 + h, yt 0 + h φt 0, y 0 0. Since φ is differentiable we conclude that 0 φ φ t t 0, y 0 + lim h 0 y t 0, y 0, y t 0 + h y 0 h and therefore V is a viscosity supersolution of 4.2. Let assume through the last part of Section 4 that gt, x, y, v = Cxy v, Cx > 0. We define the unbounded operator A in L 2 by { } DA = y H 2 y ; + Cxy = 0 Ay = i,j j a ij i y + y, and the lifting map M : L 2 L 2 by j a ij i y = y in i,j Mv = y y + Cxy = v on. Then we may rewrite the equation. as { y t + Ay + F t, y = AMvt yt 0 = y 0 9

10 where F t, y = ft, y y. We note that the right-hand side of the above equation is not well defined, since the range of M is not contained in DA. Therefore we shall proceed as in [5]-[7] and use the fact that M has a regularizing effect. Indeed, M : L 2 H 3/2, which may be expressed in abstract form using the fractional powers of A. Actually { H 2θ DA θ { } if 0 θ < 3 4 = y H 2 y ; + Cxy = 0 if 3 4 < θ. By classical results M : L 2 DA α for all α 0, 3/4. Consequently equation. can be written as { y + Ay + F y = A β M β v yt 0 = y 0, where β 3/4, ] and M β = A β M. Now the value function satisfies the Hamilton Jacobi Bellman equation V t + Ht, y, D yv = 0 V T, y = Lx, yt dx where Ht, y, p = inf v { } ft, yp Cσypdσ a ij i y j p + M β va β p. We give now a comparison result. Theorem 4.2 Assume that A-A3 hold true. Let U,U be respectively a viscosity subsolution, and a viscosity supersolution of 4.2. Let and moreover Ut, y, Ut, y C + y L 2 Q lim t T lim t T Ut, y Ut, y Lx, yt dx + = 0 Lx, yt dx = 0 uniformly on bounded subsets of Y. Then U U. Proof. Given σ > 0, define U σ t, y = Ut, y σ t, U σt, y = Ut, y + σ t. It is easy to see that U σ and U σ satisfy respectively U σ t, y + Ht, y, D y U σ t, y σ t T 2 U σ T, y = Lx, yt dx σ T 4.3 0

11 and For ε, δ, γ > 0 we define the function U σ t, y + Ht, y, D y U σ t, y σ t T 2 U σ T, y = Lx, yt dx + σ 4.4 T. Φt, θ, y, z, ε, δ, γ = U σ t, y U σ θ, z 2ε δ 2 y 2 L 2 + z 2 L 2 t θ 2γ A y z, y z t, y, θ, z t 0, T L 2. Because of the continuity assumptions on U, U and the fact that A is compact, the function above is weakly sequentially upper-semicontinuous and so it attains a global maximum at some points t, θ, ȳ, z, where 0 < t, θ and ȳ, z are bounded independently of ε for a fixed δ. Moreover we have as for instance in [4],[],[2] δ 0 ε 0 δ ȳ 2 L 2 + z 2 L 2 = 0, 4.5 γ 0 and To see this we set A /2 ȳ z 2 L 2 = 0 for fixed δ, 4.6 ε 0 γ 0 2ε γ 0 t θ 2 = 0 for fixed δ, ε γ m ε, δ, γ = sup Φt, θ, y, z, ε, δ, γ y,z { m 2 ε, δ = sup U σ t, y U σ θ, z y,z 2ε A /2 y z 2 L 2 δ } 2 y 2 L 2 + z 2 L 2 m 3 δ = sup {U σ t, y U σ θ, z δ2 } y 2L2 + z 2L2. y,z We note that we have Now lim m ε, δ, γ = m 2 ε, δ, lim m 2 ε, δ = m 3 δ, lim m 3 δ = m. γ 0 ε 0 δ 0 m ε, δ, γ = Φt, θ, y, z, ε, δ, γ = U σ t, y U σ θ, z 2ε A /2 y z 2 L 2 δ 2 y 2 L 2 + t z 2 θ2 L 2 2γ

12 and for δ, γ fixed m ε, δ, γ + 4ε A y z 2 L 2 = U σt, y U σ θ, z 4ε A /2 y z 2 L 2 δ 2 y 2 L 2 + t z 2 θ2 L 2 2γ m 2ε, δ, γ. Thus 4ε A y z 2 L 2 m 2ε, δ, γ m ε, δ, γ. This gives 4.5. Similarly we have m ε, δ, γ+ δ 2 y 2 L 2 + z 2 L 2 = U σt, y U σ θ, z 2ε A /2 y z 2 L 2 which gives δ 2 y 2 L 2 + t z 2 θ2 L 2 m ε, δ 2γ 2, γ δ y 2 L z 2 L 2 m ε, δ 2, γ m ε, δ, γ from which we obtain 4.6. Finally we have then m ε, δ, γ + t θ 4γ δ 2 = U σt, y U σ θ, z 2ε A /2 y z 2 L 2 y 2 L 2 + t z 2 θ2 L 2 4γ t θ 2 m ε, δ, γ m ε, δ, γ 4γ m ε, δ, 2γ, and 4.7. If U is not less than or equal to U it then follows from the above that for small σ and δ, and for T δ sufficiently close to T we have t, θ < T δ if γ and ε are sufficiently small. Now define the following functions ϕt, y = U σ θ, z+ 2ε and ψθ, z = U σ t, y 2ε A y z, y z + δ y 2 L t z 2 θ2 L 2 + 2γ A y z, y z δ y 2 L t z 2 θ2 L 2 2γ Since t, y argmax U σ ϕ and θ, z argmin U σ ψ, we can conclude that D y ϕt, y, D y ψθ, z DA θ, θ [0,. Therefore recalling that U σ is a viscosity subsolution of 4.3 we obtain σ T 2 t θ + H t, ȳ, D y U σ t, ȳ. γ 2

13 Similarly, recalling that U σ is a viscosity supersolution of 4.4 we have t θ γ Combining these two inequalities we find + H θ, z, D y U σ θ, z σ T 2. 2σ T 2 H θ, z, D y U σ θ, z H t, ȳ, D y U σ t, ȳ = F θ, z ε A ȳ z δ z A z ε A ȳ z δ z +inf M β va β v ε A ȳ z δ z + F t, ȳ ε A ȳ z + δȳ + Aȳ ε A ȳ z + δȳ inf M β va β v ε A ȳ z + δȳ = ε A ȳ z F t, ȳ F θ, z ȳf + δ t, ȳ + zf θ, z + ε Aȳ za ȳ z + δ Aȳȳ + A z z inf M β va β v ε A ȳ z + δȳ + inf M β va β v ε A ȳ z δ z. The first term writes A ȳ z F t, ȳ F θ, z = A ȳ z f t, ȳ ȳ f θ, z + z ε ε = A ȳ zȳ z + A ȳ z f t, ȳ f t, z + f t, z f θ, z ε ε = A /2 ȳ z 2 ε + A ȳ z L 2 ε f y t, λȳ + λ zȳ z + f tλ t + λ θ, z t θ t θ A /2 ȳ z 2 ε L 2 ε A /2 A /2 ȳ z L 2 [ /2 ] /2 f y t, λȳ + λ z 2 ȳ z 2 + f t λ t + λ θ, z 2 t θ 2. The second term gives δ ȳf t, ȳ = δ ȳ f t, ȳ ȳ while the third one yields δ M β v A β ȳ + z = δ So finally we get = δ ȳ 2 + δ ȳ f t, 0 + f y t, λȳȳ δ ȳ 2 L 2 δ z L 2 A β M β v ȳ + z δc ȳ L2 + z L2. 2σ T 2 ε ȳ z 2 L 2 C ε A /2 ȳ z 2 δ ȳ 2 + z 2. This inequality yields a contradiction if we let ε 0 and δ 0. 3

14 5 Optimality Conditions 5. Necessary Conditions for Optimality In order to derive the optimality conditions for problem P, we introduce the function Hs, t, y, v, p = pgs, t, y, v for all s, t, y, v, p [0, T ] R 3. Theorem 5. If A-A3 holds true and y, v is an optimal trajectorycontrol pair for problem P, then there exists p W 0, T C b Q \ {T } satisfying the equation p t + Ap + f y x, t, y p = 0 in Q, p + g y s, t, y, v p = 0 on Σ, p T = L yx, y T in, and such that 5. Hs, t, y s, t, v s, t, p s, t = for a.e. s, t Σ. Moreover, we have min Hs, t, v L σ Σ y s, t, v, p s, t 5.2 p t D + y V t, y t, t [t 0, T ] 5.3 and there exists a subset L [t 0, T ], of full measure, such that, for all t L, Ht, y t, p t, p t D + V t, y t. 5.4 Proof. The equality 5.2 is a well-known result see for instance [4]-[5]. Let y, v be a solution of P and let p be its associated adjoint state. To show 5.3, fix y Y and t [t 0, T ]. If yτ denote yτ; t, y, v for all t τ T, then by. we have t y y + Ay y + fx, τ, y fx, τ, y = 0 in [t, T ], y y + gs, τ, y, v gs, τ, y, v = 0 on [t, T ], y y t = y y t in. Multiplying this equation by p, the solution of 5., integrating over Q t = [t, T ] and applying the Green formula, we get subsequently that L y y T yt y T dx = y y tp tdx + p y y g y gy, v gy, v dsdτ+ p y y f y fy fy dxdτ. Σ t t 4

15 We have y y t V t, y V t, y < p, y y t > y y t y y t L y x, yθ T yt ; t, y, v yt ; t, y t, v dx < p, y y t > y y t where y θ denote y + θy y, for all θ 0, p T yt ; t, y, v yt ; t, y t, v dx y y t y y t + y y t A3 y y t p ty y tdx L y x, yθ T L yt y x, y T ; t, y, v yt ; t, y t, v dx y y t { yτ; y y p τ t, y, v y τ; t, y t, v f y t Q t fyτ; t, y, v fy τ; t, y t, v dxdτ yτ; + p τ t, y, v y τ; t, y t, v g y Σ t gyτ; t, y, v, v gy τ; t, y t, v, v } dsdτ y θ T y T yt ; t, y, v yt ; t, y t, v dx + y y t y y t { = y y t y y p τ yτ y τ t Q t f yx, t, y θ τdθ 0 + p Σt τ yτ y τ g y x, t, y θ τ, v dθ f yx, t, y ζ τdζ dxdτ { y y t y y p τ yτ y τ t Q t 0 0 M x, tη y θ η y ζ ω θ ζyτ y τ dθdζ + p τ yτ y τ Σ t M 2 s, tη y θ η y ζ ω θ ζyτ y τ dθdζ 0 g y x, t, y ζ τ, v } dζ dsdτ dxdτ } dsdτ 5

16 { CM, M 2 y y t y y p yτ y τ ω yτ y τ dxdτ t Q t } + p yτ y τ ω yτ y τ dsdτ Σ t { C 5 CM, M 2 y y t y y p L t Q yτ y τ ω yτ y τ dxdτ Q t } + yτ y τ ω yτ y τ dsdτ Σ t 3.2 y y t C 5 CM, M 2 y y y y t L ωc y y t L t and 5.3 follows. Finally, we prove 5.4. We denote by L the set of Lebesgue points of y and recall that L has a full measure in [t 0, T ]. Let t t 0, T L. Fix r t 0, T and y Y. As previously we have t y y + Ay y + fx, τ, y fx, τ, y = 0 in r, T, y y + gs, τ, y, v gs, τ, y, v = 0 on r, T, y y r = y y r in. Thus multiplying by p and integrating over r, T we get p T yt ; r y T ; tdx = p ry y rdx + f y p y y dxdτ Q r fy fy p dxdτ + Ap y y dxdτ Ay y p dxdτ. 5.5 Q r Q r Q r Therefore { V r, y V t, y t y y t,r t r t + y y t Ht, y t, p tr t+ < p t, y y } t > r t + y y t { = y y t, r t r t + y y t L yx, y T ; t, y t, v, yt ; r, y, v yt ; r, y, v y T ; t, y t, v dx L y x, y θ T L y x, y T yt ; r, y, v y T ; t, y t, v dx + r t + y y t Ht, y t, p tr t+ < p t, y y t > r t + y y t } 6

17 where y θ T = y T ; t, y t, v + θyt ; r, y, v y T ; t, y t, v ; we also use that y T ; t, y t, v = y T ; r, y r, v A3,5. p T yt ; r, y, v y T ; t, y t, v dx y y t, r t r t + y y t and using 5.5 we get + Ht, y t, p tr t+ < p t, y y } t > r t + y y t = y y t, r t + Σ r Q r f y p y y fy fy p dxdτ r t + y y t g y p y y gy, v gy, v p dsdτ r t + y y t p r y y r dx p t y y t dx r t + y y t = y y t, r t y y t, r t y y t, r t p r p t y y t dx + r t r t + y y t Ht, y t, p tr t r t + y y t r t + y y t p r y t y r dx + Ht, y t, p } tr t r t + y y t p r y t y r dx p r p t dx + r t + y y t + Ht, y t, p } tr t r t + y y t 5.2 Sufficient Conditions for Optimality p r y t y r + Ht, y t, p t r t Theorem 5.2 Assume A A3 and consider a trajectory-control pair y, v for the system.. If there is pt such that Ht, yt, pt, pt D + V t, yt 5.6 for a.e. t [t 0, T ], then u is optimal for problem P. In particular, v is optimal if for a.e. t [t 0, T ] there is pt satisfying 5.2 and 5.4. dx = 0. 7

18 Proof. Consider the function ψt = V t, yt, which is continuous and nondecreasing in [t 0, T ] by Proposition 3.. We will show that ψ is also nonincreasing and the conclusion will follow. We obtain { V t + h, yt + h V t, yt V t + h, yt + h V t, yt = h 0 h h 0 h } Consequently, Ht, yt, pt + h 0 p, yt + h yt h ψt + h ψt h 0 for a.e. t [0, T ]. Next we claim that ψ is absolutely continuous on [t 0, T ]. Recalling 3., there exists v t such that 0. 0 V θ, yθ V t, yt V θ, yθ V θ, yθ; t, yt, v t + θ t for all t 0 t θ T. Now set y t = y ; t, yt, v t and note that, in view of Proposition 2.2, y t θ L C 2 for some constant C 2 > 0 and all t 0 < t θ T. Moreover, by Proposition 2.2 there exists C 3 > 0 such that V θ, yθ V θ, y t θ C 3 yθ y t θ L2 for all t 0 < t θ T. Therefore, again by Proposition 2. we get 0 ψθ ψt C 3 yθ y t θ L 2 + θ t Indeed, we have t y y t + Ay y t + fx, t, y fx, t, y t = 0 in t, T, y y t + gs, t, y, v gs, t, y t, v t = 0 on Γ t, T, y y t t = 0 in θ t C +. and as above, by multiplying with y y t and integrating over t, θ we get θ t 2 yθ y tθ 2 L 2 + θ t θ t y y t 2 f yx, τ, y ζ dxdτ y y t 2 g y s, τ, yξ, v dsdτ gs, τ, y t, v gs, τ, y t, v t y y t dsdτ C, m, M, M 2 θ t and the proof is complete. 8

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