c 2012 Society for Industrial and Applied Mathematics Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php SIAM J CONTROL OPTIM Vol 50, No 5, pp 2889–2906 LOWER SEMICONTINUITY OF THE SOLUTION SET TO A PARAMETRIC OPTIMAL CONTROL PROBLEM∗ B T KIEN† , N T TOAN‡ , M M WONGĐ , AND J C YAOả Abstract This paper studies the solution stability of a parametric optimal control problem with linear state equation, control constraints, and convex cost functions By reducing the problem to a parametric programming problem and a parametric variational inequality, we obtain the lower semicontinuity of the solution map to a parametric optimal control problem Key words parametric optimal control, solution stability, lower semicontinuity, variational inequality AMS subject classifications 49K40, 49K30, 49K15 DOI 10.1137/110842491 Introduction A wide variety of problems in optimal control can be posed in the following form Determine a control vector u ∈ Lp ([0, 1], Rm ) with < p < ∞ and a trajectory x ∈ W 1,1 ([0, 1], Rn ) which minimizes the cost (1) f (t, x(t), u(t), μ(t))dt with the state equation x(t) ˙ = A(t)x(t) + B(t)u(t) + T (t)λ(t) a.e t ∈ [0, 1], (2) the initial value x(0) = x0 , (3) and the control constraint (4) u(t) ∈ U a.e t ∈ [0, 1] Here W 1,1 ([0, 1], Rn ) is a Sobolev space which consists of absolutely continuous functions x : [0, 1] → Rn such that x˙ ∈ L1 ([0, 1], Rn ) Its norm is given by x 1,1 = |x(0)| + x˙ The couple (μ, λ) ∈ L∞ ([0, 1], Rk ) × Lr ([0, 1], Rl ) with ≤ r ≤ ∞ are parameters, f : [0, 1] × Rn × Rm × Rk → R ∪ {+∞} is a function, A(t) = (aij (t))n×n , B(t) = ∗ Received by the editors July 27, 2011; accepted for publication (in revised form) July 26, 2012; published electronically September 20, 2012 http://www.siam.org/journals/sicon/50-5/84249.html † Department of Information and Technology, Hanoi National University of Civil Engineering, Hanoi, Vietnam (btkien@gmail.com) This author’s research was partially supported by grant NAFOSTED 101.01-2011.23 ‡ Department of Mathematics, Vinh University, Vinh City, Vietnam (toandhv@gmail.com) § Department of Applied Mathematics, Chung Yuan Christian University, Chung Li, Taiwan (mmwong@cycu.edu.tw) ¶ Corresponding author Center for General Education, Kaohsiung Medical University, Kaohsiung 807, Taiwan (yaojc@kmu.edu.tw) This author’s research was partially supported by grants NSC 992221-E-037-007-MY3 and NSC 99-2115-M-037-002-MY3 2889 Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2890 B T KIEN, N T TOAN, M M WONG, AND J C YAO Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php (bij (t))n×m and T (t) = (cij (t))n×l are matrix-valued functions, and U is a closed convex set in Rm Let us put X = W 1,1 ([0, 1], Rn ), U = Lp ([0, 1], Rm ), Z = X × U and M = L∞ ([0, 1], Rk ), Λ = Lr ([0, 1], Rl ) In what follows, we denote by BX and B X the open unit ball and the closed unit ball in a norm space X, respectively Also, given x ∈ X and δ > 0, BX (x, δ) and B X (x, δ) stand for an open ball and a closed ball, respectively, with center x and radius δ We now return to the problem (1)–(4) For each (μ, λ) ∈ M × Λ, we define (5) (6) J(x, u, μ) = f (t, x(t), u(t), μ(t))dt, K(λ) = {z = (x, u) ∈ X × U |(2), (3), and (4) are satisfied} Then (1)–(4) can be reformulated in the form (7) P (μ, λ) J(z, μ) → inf, z ∈ K(λ) Throughout this paper we denote by S(μ, λ) the solution set of (1)–(4) or P (μ, λ) corresponding to parameter (μ, λ) and assume that z = (x, u) is a solution of P (μ, λ), that is, (x, u) ∈ S(μ, λ) Our main concern is to investigate the behavior of S(μ, λ) when (μ, λ) varies around (μ, λ) This problem has been interesting to several authors in the last decade For papers which have a close connection to the present work, we refer the readers to [7], [8], [11], [12], [13], [14], [15], and the references given therein It is known that when J(·, μ) is strongly convex for all μ then the solution map of (7) is single-valued In this case, under certain conditions, Dontchev [8] showed that the solution map is continuous with respect to parameters In other ways, by using techniques of the implicit function theorem, Malanowski (see [11], [12], [13], [14], [15]) showed that if weak second-order optimality conditions and standard constraints qualifications are satisfied at the reference point, then the solution map is a Lipschitz continuous function of the parameter Note that the obtained results in [11], [12], [13], [14], [15] are of problems subject to state constraints without control constraints When conditions mentioned above are invalid, the solution map is not single valued in general In this situation, we have to use tools of set-valued analysis to treat the problem The aim of this paper is to study the lower semicontinuous property of the solution map to problem (1)–(4) in this situation We show that if the unperturbed problem is good enough, then the perturbed problems are stable Namely, by reducing the problem to a parametric programming problem and a parametric variational inequality and using techniques of set-valued analysis, we show that under certain conditions, the solution map S(μ, λ) of (1)–(4) is lower semicontinuous at the reference point (μ, λ) Let us assume that F : E1 ⇒ E2 is a multifunction between Banach spaces We denote by domF and gphF the effective domain and the graph of F , respectively, where domF := {z ∈ E1 |F (z) = ∅} Copyright © by SIAM Unauthorized reproduction of this article is prohibited STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM 2891 Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php and the graph gphF := {(z, v) ∈ E1 × E2 |v ∈ F (z)} A multifunction F is said to be lower semicontinuous at z0 ∈ E1 if for any open set V0 in E2 satisfying F (z0 ) ∩ V0 = ∅, there exists a neighborhood G0 of z0 such that F (z) ∩ V0 = ∅ for all z ∈ G0 (see [4, Definition 5.1.15, p 173]) Let E be a Banach space and Ω be a nonempty closed convex set in E The normal cone to Ω at a point z0 ∈ Ω is defined by N (z0 ; Ω) = {z ∗ ∈ E ∗ | z ∗ , z − z0 ≤ 0, ∀z ∈ Ω} For a definition of normal cones and their properties, we refer the readers to [9, Chapter 4] Given μ ∈ M and > 0, we define the set Ω (μ) = {(t, μ) ∈ [0, 1] × Rk : |μ − μ(t)| ≤ } To deal with our problem, we impose the following assumptions (A1) f (·, x, u, μ) is measurable for all (x, u, μ) ∈ Rn × Rm × Rk and f (t, ·, ·, ·) is continuous for almost everywhere t ∈ [0, 1] Besides, there exist positive constants α, , a constant ≤ γ ≤ p, and a nonnegative continuous function ϕ : [0, 1] × R × R × R → R such that for all (t, μ) ∈ Ω (μ), the function Rn × Rm (x, u) → f (t, x, u, μ) is continuously differentiable and convex on Rn × U and |f (t, x(t), u(t), μ) − f (t, x(t), u(t), μ(x))| ≤ ϕ(t, x(t), μ(t), μ)|u(t)|γ (A2) There exist constants αi ≥ with i = 1, 2, , and nonnegative functions ω1 ∈ L∞ ([0, 1], R), ω2 ∈ Lq ([0, 1], R), where q is the conjugate number of p, such that |fx (t, x, u, μ)| ≤ α1 (ω1 (t) + |x|α2 + |μ|α3 ) and |fu (t, x, u, μ)| ≤ α4 (ω2 (t) + |x|α5 + |u|p−1 + |μ|α6 ) for all (t, x, u, μ) ∈ [0, 1] × Rn × Rm × Rk satisfying |μ − μ(t)| ≤ (A3) There exist continuous functions kj : [0, 1] × R3 → R, positive numbers sj with j = 1, 2, and ≤ η ≤ p such that |fx (t, x, u, μ) − fx (t, x, u, μ(t))| ≤ k1 (t, |x|, |μ|, |μ(t)|)|u|η |μ − μ(t)|s1 and |fu (t, x, u, μ) − fu (t, x, u, μ(t))| ≤ k2 (t, |x|, |μ|, |μ(t)|)|u|p−1 |μ − μ(t)|s2 for all (t, x, u, μ) ∈ [0, 1] × Rn × U × Rk satisfying |μ − μ(t)| ≤ (A4) The function (x, u) → f (t, x, u, μ(t)) is strongly convex, that is, there exists a constant ρ > such that fz (t, z1 , μ(t)) − fz (t, z2 , μ(t)), z1 − z2 ≥ ρ|u1 − u2 |p for all zi = (xi , ui ) ∈ Rn × Rm with i = 1, and for a.e t ∈ [0, 1] Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2892 B T KIEN, N T TOAN, M M WONG, AND J C YAO (A5) There exist positive constants T1 , T2 and a nonnegative function φ ∈ Lq ([0, 1], R ) such that Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php m |A(t)| ≤ T1 , |T (t)| ≤ T2 , and |B(t)| ≤ φ(t), a.e t ∈ [0, 1] The following theorem is the main result of our paper Theorem 1.1 Suppose that assumptions (A1)–(A5) are fulfilled Then there exist a bounded neighborhood Z0 = X0 × U0 of (x, u) and a neighborhood M0 × Λ0 of ˆ λ) = (μ, λ) such that the solution map Sˆ : M0 × Λ0 → 2Z which is defined by S(μ, S(μ, λ) ∩ Z0 satisfies the following properties: ˆ λ) = ∅ for all (μ, λ) ∈ M0 × Λ0 ; (a) S(μ, (b) Sˆ is lower semicontinuous at (μ, λ) The proof of Theorem 1.1 will be presented in section In section 3, we will consider some examples of optimal control problems satisfying conditions (A1)–(A5) of Theorem 1.1 Proof of the main result The proof of Theorem 1.1 will be proceeded by proving some lemmas The first lemma gives the Lipschitz property of the set-valued map K(·) Lemma 2.1 Suppose that assumption (A5) is fulfilled Then the set-valued map K(·) which is defined by (6) has closed convex values and there exists a constant k0 > such that (8) K(λ1 ) ⊂ K(λ2 ) + k0 λ1 − λ2 r B Z ∀λ1 , λ2 ∈ Λ Proof It is obvious that for each λ ∈ Λ, K(λ) is a convex set Let zi = (xi , ui ) ∈ K(λ) such that zi → z = (x, u) Then xi → x uniformly, x˙ i → x, ˙ and un → u strongly in L1 By passing to subsequence if necessary, we may assume that x˙ i → x˙ and ui → u almost everywhere in t ∈ [0, 1] Since x˙ i (t) = A(t)xi (t) + B(t)ui (t) + T (t)λ(t), xi (0) = x0 , by letting i → ∞ we obtain x(t) ˙ = A(t)x(t) + B(t)u(t) + T (t)λ(t), x(0) = x0 Besides, we have u(t) ∈ U for a.e t ∈ [0, 1] Consequently, z ∈ K(λ) and so K(λ) is a closed set Fixing any λ1 , λ2 ∈ Λ, we show that there exists a constant k0 > such that (8) is satisfied Notice that by the solution existence theorem for linear differential equations (see [2, Lemma 2.5.4, p 121]), K(λ) = ∅ for all λ ∈ Λ Let (x, u) ∈ K(λ1 ) Then one has (9) x(t) ˙ = A(t)x(t) + B(t)u(t) + T (t)λ1 (t), a.e t ∈ [0, 1] We have to prove that there exists (y, v) ∈ K(λ2 ) such that (x, u) − (y, v) ≤ k0 λ1 − λ2 r Choose v = u By the solution existence theorem for the Cauchy problem of linear differential equations (see [2, Lemma 2.5.4, p 121]), there exists y ∈ X such that (10) y(t) ˙ = A(t)y(t) + B(t)v(t) + T (t)λ2 (t), a.e t ∈ [0, 1] Copyright © by SIAM Unauthorized reproduction of this article is prohibited STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM 2893 Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php By subtracting (9) and (10) and putting z = x − y, we get z(0) = and z˙ = A(t)z + T (t)(λ1 − λ2 ) (11) From this we get |z(t)| ˙ ≤ T1 |z(t)| + T2 |λ1 (t) − λ2 (t)|, a.e t ∈ [0, 1] (12) for suitable constants T1 and T2 Since z(t) = |z(t)| ≤ ≤ ≤ ≤ (13) t 0 t z(s)ds, ˙ we obtain (T1 |z(s)| + T2 |λ1 (s) − λ2 (s)|)ds t t t T1 |z(s)|ds + T2 |λ1 (t) − λ2 (t)|dt T1 |z(s)|ds + T2 λ1 − λ2 T1 |z(s)|ds + T2 λ1 − λ2 r By the Gronwall inequality (see [5, Lemma 18.1.i]), we obtain |z(t)| ≤ T2 λ1 − λ2 t r exp T1 ds ≤ T2 λ1 − λ2 r exp(T1 ) Combining this with (12), we have z˙ (14) ≤ T1 z + T2 λ1 − λ2 r ≤ T1 (T2 λ1 − λ2 r exp(T1 )) + T2 λ1 − λ2 r = k0 λ1 − λ2 r , where k0 := T1 T2 exp(T1 ) + T2 Thus we have shown that (x, u) − (y, v) = x − y 1,1 = z 1,1 = |z(0)| + z˙ ≤ k0 λ1 − λ2 r The proof of the lemma is now complete Let us choose > and β0 > such that K(λ) ∩ (z + 0BZ ) =∅ ∀λ ∈ B Λ (λ, β0 ) By Lemma 2.1, we have K(λ) ∩ (z + (15) For each 0BZ ) ⊂ K(λ ) + k0 λ − λ B Z ∀λ, λ ∈ B Λ (λ, β0 ) > we put K (λ) = K(λ) ∩ (z + B Z ) The following lemma plays an important role in the proof of the main result Its proof can be found in [3, Lemma 2.3] and [10, Lemma 2.1] Lemma 2.2 For any ∈ (0, ] and < β < min{β0 , 4k0 }, one has (16) K (λ1 ) ⊆ K (λ2 ) + 5k0 λ1 − λ2 B Z ∀λ1 , λ2 ∈ BΛ (λ, β) Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2894 B T KIEN, N T TOAN, M M WONG, AND J C YAO Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Fix = > and β > as in Lemma 2.2 such that (16) is satisfied For each (μ, λ) ∈ B M (μ, ) × B Λ (λ, β), we consider the problem J(z, μ) → inf, z ∈ K(λ) ∩ B Z (z, ) (17) Lemma 2.3 Suppose that assumptions (A1)–(A5) are satisfied Then for each (μ, λ) ∈ B M (μ, ) × B Λ (λ, β), problem (17) has a solution Proof Put ξ= inf z∈K(λ)∩B Z (z, ) J(z, μ) Then there exists a sequence zi = (xi , ui ) ∈ K(λ) ∩ B Z (z, ) such that ξ = lim J(xi , ui , μ) i→∞ Since xi − x 1,1 + ui − u p ≤ and U is a reflexive Banach space, we can assume that ui other hand, we have u ∈ B U (u, ) On the x˙ i (t) = A(t)xi (t) + B(t)ui (t) + T (t)λ(t) It follows that |x˙ i (t)| ≤ T1 |xi (t)| + |B(t)||ui (t)| + T2 |λ(t)| (18) Since xi (t) = x0 + t x˙ i (s)ds, we have from above |xi (t)| ≤ |x0 | + ≤ |x0 | + ≤ (19) ≤ t 0 t t 0 t T1 |xi (s)|ds + T1 |xi (s)|ds + t (|B(s)||ui (s)| + T2 |λ(s)|)ds T1 |xi (s)|ds + (|x0 | + φ φ(s)|ui (s)| + T2 |λ(s)| ds q ui p + T2 λ ) T1 |xi (s)|ds + (|x0 | + φ q K1 + T2 λ r ) Here we used the fact that ui p ≤ K1 for some positive constant K1 By the Gronwall inequality we have from (19) that |xi (t)| ≤ (|x0 | + φ q K1 + T2 λ r ) exp(T1 ) = K2 Combining this with (18), we obtain (20) |x˙ i (t)| ≤ T1 |xi (t)| + |B(t)||ui (t)| + T2 |λ(t)| ≤ T1 K2 + φ(t)|ui (t)| + T2 |λ(t)| Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2895 STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Since {ui } is bounded in Lp , we have x˙ i (21) ≤ T1 K2 + φ q ui p + T2 λ ≤ T1 K2 + φ q K1 + T2 λ r r Moreover, x˙ i is equiabsolutely integrable Indeed, if E is any measurable subset of [0, 1], then E |x˙ i (t)|dt ≤ T1 K2 |E| + E |φ|q dt q K1 + T2 E |λ(t)|r dt r and the second term of the right-hand side of the above inequality approaches zeros as |E| → uniformly with respect to i By the Dunford–Pettis theorem (see [5, Theorem 10.3.i]) there exists a function v ∈ L1 ([0, 1], Rn ) such that x˙ i → v weakly By Lemma 10.2.i in [5], we have {xi } is equiabsolutely continuous Since {xi } is equibounded, by Ascoli’s theorem (see [5, Theorem 9.1.i]) there is an absolutely continuous t function x on [0,1] such that xi → x uniformly Since xi (t) = x0 + x˙ i (s)ds we get t x(t) = x0 + v(s)ds Hence x(t) ˙ = v(t) a.e t ∈ [0, 1] Since K(λ) is also weakly closed and xi x in X, we see that z = (x, u) ∈ K(λ) ∩ B Z (z, ) In particular, we z in Z showed that zi By a property of convex functions, we have f (t, xi (t), ui (t), μ(t)) ≥ f (t, x(t), u(t), μ(t)) + fx (t, x(t), u(t), μ(t)), xi (t) − x(t) + fu (t, x(t), u(t), μ(t)), ui (t) − u(t) , a.e t ∈ [0, 1] Hence J(xi , ui , μ) ≥ J(x, u, μ) + (22) + fx (t, x(t), u(t), μ(t)), xi (t) − x(t) dt fu (t, x(t), u(t), μ(t)), ui (t) − u(t) dt Letting i → ∞ and using weak convergence of zi , we obtain from (22) that ξ = J(x, u, μ) It remains to show that ξ is finite By (A1) we have |J(z, μ) − J(z, μ)| ≤ ≤ ≤ (23) 0 |f (t, x(t), u(t), μ(t) − f (t, x(t), u(t), μ(t))|dt ϕ(t, x(t), μ(t), μ(t))|u(t)|γ |u(t)|γ dt ≤ c u p for some constant c and = max{ϕ(t, t1 , t2 , t3 )|(t, t1 , t2 , t3 ) ∈ [0, 1] × [0, x ∞] × [0, μ M] × [0, μ M + ]} By (A2) and using similar arguments as in Claim 1, we can show that for each μ ∈ B M (μ, ), Fz (·, μ) is uniformly bounded on B Z (z, ) Hence there exists a constant C > such that Fz (z, μ) ≤ C ∀z ∈ B Z (z, ) Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2896 B T KIEN, N T TOAN, M M WONG, AND J C YAO Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Combining this with (23) and using the mean value theorem, we get |J(z, μ) − J(z, μ)| ≤ |J(z, μ) − J(z, μ)| + |J(z, μ) − J(z, μ) Jz (sz + (1 − s)z, μ) ≤ c μ−μ α M u p + sup 0≤s≤1 ≤c u p + C( x − x 1,1 z−z + u − u p ) < +∞ This implies that |ξ| = |J(z, μ)| < +∞ The proof of the lemma is complete Lemma 2.4 There exists a neighborhood M0 × Λ0 ⊂ BM (μ, ) × BΛ (λ, β) of (μ, λ) such that for each (μ, λ) ∈ M0 × Λ0 , all solutions z(μ, λ) = (x(μ, λ), u(μ, λ)) of problem (17) satisfy z(μ, λ) ∈ / ∂BZ (z, ) Proof Assume that the conclusion of the lemma is false Then we can find a sequence (μi , λi ) → (μ, λ) and a sequence (xi , ui ) = (x(μi , λi ), u(μi , λi )) which is a solution of (17) such that (xi , ui ) ∈ ∂BZ (z, ) By assumptions (A1) and (A2), for each μ ∈ B M (μ, ) the functional J(·, μ) has the Gˆateaux derivative at z ∈ Z (see [5, Lemma 2.3.i and Theorem 2.3.ii]) Besides, problem (17) is a convex programming problem Hence zi = (xi , ui ) satisfies the generalized equation ∈ Jz (zi , μi ) + N zi ; K(λi ) ∩ B Z (z, ) which is equivalent to the variational inequality (24) Jz (zi , μi ), z − zi ≥ ∀z ∈ K(λi ) ∩ B Z (z, ) Here Jz (z0 , μ) denotes the Gˆ ateaux derivative of J in z at a point z0 = (x0 , u0 ) ∈ Z, which is defined by Jz (z0 , μ), h = Jx (x0 , u0 , μ), h1 + Ju (x0 , u0 , μ), h2 = (25) fx (t, x0 (t), u0 (t), μ(t)), h1 (t) dt + fu (t, x0 (t), u0 (t), μ(t)), h2 (t) dt for all h = (h1 , h2 ) ∈ Z Moreover, since J is a convex function, Jz is a monotone operator Putting λ1 = λi and λ2 = λ in (16), we see that there exists yi = (x1i , u1i ) ∈ K(λ) ∩ B Z (z, ) such that (26) yi − zi ≤ 5k λi − λ r We now claim that there exists z = (x, u) ∈ B Z (z, ) such that ui uniformly and x˙ i x˙ in L1 and so we have zi z Indeed, since xi − x 1,1 + ui − u p u in U , xi → x = , we see that ui p ≤ K3 for some positive constant K3 Noting that U is a reflexive u for some u ∈ U Since zi ∈ K(λi ), we have Banach space, we have ui x˙ i (t) = A(t)xi (t) + B(t)ui (t) + T (t)λi (t) a.e t ∈ [0, 1] Copyright © by SIAM Unauthorized reproduction of this article is prohibited Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM 2897 Since λi → λ strongly in Λ, there exists K4 > and a function h ∈ Lr ([0, 1], R) such that λi r ≤ K4 and |λi (t)| ≤ h(t) a.e t ∈ [0, 1] (see [6, Theorem 1.20]) Hence, we have |x˙ i (t)| ≤ T1 |xi (t)| + φ(t)|ui (t)| + T2 h(t) Since xi (t) = x0 + t x˙ i (s)ds we obtain |xi (t)| ≤ |x0 | + ≤ ≤ (27) t 0 t t t T1 |xi (s)|ds + T1 |xi (s)|ds + |x0 | + (φ(s)|ui (s)| + T2 h(s))ds (φ(s)|ui (s)| + T2 h(s))ds T1 |xi (s)|ds + |x0 | + φ q K3 + T2 h r By the Grownwall inequality, we get from (27) that |xi (t)| ≤ (|x0 | + φ q K3 + T2 h r ) exp(T1 ) It follows that x˙ i ≤ T1 (|x0 | + φ q K3 + T2 h r ) exp(T1 ) + φ q K3 + T2 h r Besides, if E is a measurable set of [0, 1], then E (28) |x˙ i (t)|dt ≤ T1 (|x0 | + φ q K3 + T2 h r ) exp(T1 )|E| q + E |φ| dt q K3 + T2 r E |h| dt r and the second part of (28) approaches to as |E| → Hence x˙ i is equiabsolutely integrable By the Dunford–Pettis theorem (see [5, Theorem 10.3.i]) there exists a x˙ in L1 Note that z = (x, u) ∈ function x ∈ X such that xi → x uniformly and x˙ i B Z (z, ) The claim is proved Using similar arguments as in the above, we can show that there exists y = x˙ in L1 , and u1i u1 for (x1 , u1 ) ∈ X × U such that x1i → x1 uniformly, x˙ 1i ∗ ∗ some y = (x1 , u1 ) ∈ K(λ) ∩ B Z (z, ) Taking any z ∈ Z and using (26), we have | z ∗ , zi − z ∗ , y | = | z ∗ , zi − z ∗ , yi − z ∗ , y + z ∗ , yi | ≤ z ∗ zi − yi + | z ∗ , yi − y | → Consequently, zi y and so z = y or (x, u) = (x1 , u1 ) ∈ K(λ) ∩ B Z (z, ) Putting λ1 = λ and λ2 = λi in (16), we see that for each i there exists z2i = (x2i , u2i ) ∈ K (λi ) = K(λi ) ∩ B Z (z, ) such that z − z2i ≤ 5k0 λ − λi r Hence z2i → z strongly in Z Substituting z = z2i into (24) we obtain (29) Jz (zi , μi ), z2i − z ≥ Jz (zi , μi ), zi − z We now prove some claims below Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2898 B T KIEN, N T TOAN, M M WONG, AND J C YAO Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Claim {Jz (zi , μ)} is bounded Indeed, for each h ∈ Lp ([0, 1], Rm ) we have |Ju (xi , ui , μ)h| = fu (t, xi (t), ui (t), μ(t))h(t)dt Hence Ju (xi , ui , μ) ≤ q q |fu (t, xi (t), ui (t), μ(t))| dt By (A2) we have Ju (xi , ui , μ) ≤ α4 (|ω2 (t)| + |xi (t)| α5 + |ui (t)| p−1 α6 q + |μ(t)| ) dt q Using the inequality (a + b)q ≤ 2q−1 (aq + bq ) (30) ∀a, b ≥ 0, q ≥ (see [1, Lemma 2.24]), we get Ju (xi , ui , μ) ≤ α4 q−1 q (|ω2 (t)| + |xi (t)| Since xi → x uniformly and ui xi α5 q + |ui (t)| (p−1)q + |μ(t)| α6 q q )dt u, there exist constants γ1 , γ2 > such that ∞ ≤ γ1 , u i p ≤ γ2 ∀i ≥ It follows that Ju (xi , ui , μ) ≤ α4 4q−1 ( ω2 q q + γ1α5 q + γ2p + μ α6 q ∞ ) q , where we used the fact that if 1/p + 1/q = 1, then (p − 1)q = p Hence Ju (xi , ui , μ) is bounded Using similar arguments, we also show that Jx (xi , ui , μ) is bounded Consequently, Jz (zi , μ) is bounded, which justifies our claim Claim Jz (zi , μ) − Jz (zi , μi ) → as i → ∞ In fact, we have Jz (zi , μ)−Jz (zi , μi ) ≤ Jx (xi , ui , μ)−Jx (xi , ui , μi ) + Ju (xi , ui , μ)−Ju (xi , ui , μi ) Since xi → x and μi → μ uniformly, there exist γ3 , γ4 > 0, and i0 ∈ N such that xi ∞ ≤ γ3 , μi ∞ ≤ γ4 ∀i ≥ i0 Since kj is continuous, we obtain kj (t, |xi (t)|, |μi (t)|, |μ(t)|) ≤ ξj := (j = 1, 2, i ≥ i0 ) Since μi − μ implies that max (t1 ,t2 ,t3 ,t4 )∈[0,1]×[0,γ3 ]×[0,γ4 ]×[0, μ ∞ ≤ ∞] kj (t1 , t2 , t3 , t4 ), for i > large enough, assumption (A3) | Jx (xi , ui , μ) − Jx (xi , ui , μi ), x = fx (t, xi (t), ui (t), μ(t)) − fx (t, xi (t), ui (t), μi (t)), x(t) dt ≤ sup k1 (t, |xi (t)|, |μi (t)|, |μ(t)|) μi − μ t∈[0,1] (31) ≤ ξ1 μi − μ s1 ∞ ui η η x 1,1 ≤ cξ1 μi − μ s1 ∞ s1 ∞ x ui η p x |ui (t)|η dt 1,1 Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2899 STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php for some constant c > Here we used the facts that ui x 1,1 Hence (31) implies that Jx (xi , ui , μ) − Jx (xi , ui , μi ) ≤ cξ1 μi − μ s1 ∞ ui η p η ≤ c ui →0 p and x ≤ as i → ∞ It remains to be shown that Ju (xi , ui , μ) − Ju (xi , ui , μi ) → as i → ∞ It is easy to see that Ju (xi , ui , μ) − Ju (xi , ui , μi ) ≤ |fu (t, xi (t), ui (t), μ(t)) − fu (t, xi (t), ui (t), μi (t))|q dt 1/q By (A3), we get Ju (xi , ui , μ) − Ju (xi , ui , μi ) ≤ ≤ 0 |fu (t, xi (t), ui (t), μ(t)) − fu (t, xi (t), ui (t), μi (t))|q dt q k2 (t, xi (t), μi (t), μ(t)) |ui (t)|(p−1)q |μi (t) − μ(t)|qs2 dt ≤ ξ2q μi − μ (32) q qs2 ∞ ui p p ≤ ξ2q γ2p μi − μ qs2 ∞ → as i → ∞ Combining this with (31) we obtain the proof of Claim We now notice that by Claim and Claim 2, {Jz (zi , μi )} is bounded Hence from (29) and z2i → z we obtain lim Jz (zi , μi ), zi − z ≤ i→∞ From this and Claim 2, one has lim Jz (zi , μ), zi − z = lim Jz (zi , μ) − Jz (zi , μi ), zi − z i→∞ i→∞ + lim Jz (zi , μi ), zi − z ≤ i→∞ It follows that lim Jz (zi , μ) − Jz (z, μ), zi − z ≤ (33) i→∞ By (A4) we have fx (t, xi (t), ui (t), μ(t)) − fx (t, x(t), u(t), μ(t)), xi (t) − x(t) + fu (t, xi (t), ui (t), μ(t)) − fu (t, x(t), u(t), μ(t)), ui (t) − u(t) ≥ ρ|ui (t) − u(t)|p Hence Jz (zi , μ) − Jz (z, μ), zi − z ≥ ρ ui − u pp Combining this with (33), we obtain that ui → u strongly in Lp Since zi ∈ K(λi ) and z ∈ K(λ), we have x˙ i (t) = A(t)xi (t) + B(t)ui (t) + T (t)λi (t) Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2900 B T KIEN, N T TOAN, M M WONG, AND J C YAO Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php and x(t) ˙ = A(t)x(t) + B(t)u(t) + T (t)λ(t) It follows that x˙ i (t) − x(t) ˙ = A(t)(xi (t) − x(t)) + B(t)(ui (t) − u(t)) + T (t)(λi (t) − λ(t)) This implies that |x˙ i (t) − x(t)| ˙ ≤ T1 |xi (t) − x(t)| + φ(t)|ui (t) − u(t)| + T2 |λi (t) − λ(t)| (34) Since xi (t) − x(t) = |xi (t) − x(t)| ≤ ≤ ≤ t 0 t t t (x˙ i (s) − x(s))ds, ˙ we have T1 |xi (s) − x(s)|ds + T1 |xi (s) − x(s)|ds + t (φ(s)|ui (s) − u(s)| + T2 |λi (s) − λ(s)|)ds (φ(s)|ui (s) − u(s)| + T2 |λi (s) − λ(s)|)ds T1 |xi (s) − x(s)|ds + φ q ui − u p + T2 λi − λ r ) exp(T1 ) p + T2 λi − λ r The Gronwall inequality implies that |xi (t) − x(t)| ≤ ( φ q ui − u Combining this with (34) yields |x˙ i (t) − x(t)| ˙ ≤ T1 ( φ q ui − u p + T2 λi − λ r ) exp(T1 ) + φ(t)|ui (t) − u(t)| + T2 |λi (t) − λ(t)| Consequently, xi − x 1,1 = |xi (0) − x(0)| + x˙ i − x˙ ≤ T1 ( φ + φ (35) q q ui − u ui − u p p = x˙ i − x˙ + T2 λi − λ r ) exp(T1 ) + T2 λi − λ r Since ui − u p → and λi − λ r → 0, we see that xi − x 1,1 → Thus we have shown that zi = (xi , ui ) → z = (x, u) strongly in Z Since zi ∈ ∂BZ (z, ) and ∂BZ (z, ) is closed in the strong topology of Z, we obtain that z ∈ ∂BZ (z, ) In particular, z = z Notice that z ∈ K(λ) ∩ B Z (z, ) By (16), there exists z˜i ∈ K(λi ) ∩ B Z (z, ) such that z˜i − z ≤ 5k λi − λ r and so z˜i → z strongly in Z Substituting z = z˜i into (24), we have Jz (zi , μi ), z˜i − zi ≥ (36) Rewrite Jz (zi , μi ), z˜i − zi = Jz (zi , μi ) − Jz (zi , μ), z˜i − zi + Jz (zi , μ), z˜i − zi Letting i → ∞ and using (36) and Claim 1, we obtain (37) Jz (z, μ), z − z ≥ Copyright © by SIAM Unauthorized reproduction of this article is prohibited STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM 2901 Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php On the other hand, since z ∈ S(μ, λ) and z ∈ K(λ) ∩ B Z (z, ), it follows that Jz (z, μ), z − z ≥ (38) Claim Jz (·, μ) is strictly monotone on K(λ) ∩ B Z (z, ) In fact, suppose that Jz (z1 , μ) − Jz (z2 , μ), z1 − z2 = (39) for z1 = (x1 , u1 ), z2 = (x2 , u2 ) ∈ K(λ) ∩ B Z (z, ) Since Jz (z1 , μ) − Jz (z2 , μ), z1 − z2 ≥ ρ u1 − u2 (40) p p, we have u1 = u2 Using similar arguments as in the proof of inequality (35), we obtain x1 − x2 (41) 1,1 = x˙ − x˙ ≤ T1 φ q u1 − u2 p exp(T1 ) + φ q u1 − u2 p Hence x1 = x2 and so z1 = z2 Our claim is justified Since Jz (·, μ) is strictly monotone and z = z, we have Jz (z, μ) − Jz (z, μ), z − z > This and (38) imply that Jz (z, μ), z − z > Jz (z, μ), z − z ≥ which contradicts to (37) The proof of this lemma is complete We now return to the proof of Theorem 1.1 (a) Choose M0 × Λ0 as in the Lemma 2.4 and put Z0 = BZ (z, ) By Lemma 2.3, for each (μ, λ) ∈ M0 × Λ0 problem (17) has a solution z = z(μ, λ) ∈ B Z (z, ) Moreover, by Lemma 2.4, z(μ, λ) ∈ intBZ (z, ) Fixing any z ∈ K(λ) we see that for γ ∈ (0, 1) small enough, one has J(z, μ) ≤ J(z + γ(z − z), μ) ≤ γJ(z , μ) + (1 − γ)J(z, μ) This implies that J(z, μ) ≤ J(z , μ) Consequently, z = z(μ, λ) is also a solution of the problem P (μ, λ) J(z, μ) → inf, z ∈ K(λ) ˆ λ) = S(μ, λ) ∩ BZ (z, ) = ∅ for all (μ, λ) ∈ M0 × Λ0 Assertion (a) of Hence S(μ, Theorem 1.1 is proved ˆ λ) ∩ G0 = ∅ By Claim in the (b) Suppose G0 is an open set in Z such that S(μ, proof of Lemma 2.4, Jz (·, μ) is strictly monotone on K(λ) ∩ B Z (z, ) Assume that ˆ λ) such that z0 = z Since z0 and z are solutions of the problem there exists z0 ∈ S(μ, J(z, μ) → inf, z ∈ K(λ) ∩ B Z (z, ), we have Jz (z, μ), z − z ≥ ∀z ∈ K(λ) ∩ B Z (z, ) Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2902 B T KIEN, N T TOAN, M M WONG, AND J C YAO Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php and Jz (z0 , μ), z − z0 ≥ ∀z ∈ K(λ) ∩ B Z (z, ) Consequently, Jz (z, μ), z0 − z ≥ and Jz (z0 , μ), z − z0 ≥ Hence Jz (z0 , μ) − Jz (z, μ), z0 − z ≤ ˆ λ) = which contradicts the strict monotonicity of Jz (·, μ) Thus we must have S(μ, ˆ {z} Since S(μ, λ) ∩ G0 = ∅, we get z ∈ G0 and so z ∈ G1 := Z0 ∩ G0 We now choose ∈ (0, ) and δ > such that BZ (z, ) ⊂ G1 and K(λ)∩B Z (z, ) = ∅ for all λ ∈ BΛ (λ, δ) Taking β such that < β < min{δ, 4k0 } and using Lemma 2.2 again, we see that K(λ1 ) ∩ (z + B Z ) ⊆ K(λ2 ) ∩ (z + B Z ) + 5k0 λ1 − λ2 r B Z ∀λ1 , λ2 ∈ BΛ (λ, β) By using similar arguments as above, we can show that there exists a neighborhood M1 × Λ1 ⊂ M0 × Λ0 of (μ, λ) such that for all (μ, λ) ∈ M1 × Λ1 , the problem J(z, μ) → inf, z ∈ K(λ) ∩ B Z (z, ) has a solution and its solutions z(μ, λ) satisfy z(μ, λ) ∈ / ∂BZ (z, ) From this we can show that z(μ, λ) is also a solution of P (μ, λ) Consequently, ˆ λ) ∩ G0 = S(μ, λ) ∩ G1 ⊃ S(μ, λ) ∩ BZ (z, ) = ∅ S(μ, Hence Sˆ is lower semicontinuous at (μ, λ) completed ∀(μ, λ) ∈ M1 × Λ1 The proof of the theorem is now Some examples Let us give an example which illustrates Theorem 1.1 Let us give an example which illustrates Theorem 1.1 Example 3.1 Suppose that n = 2, m = 1, k = l = 1, p = 2, and (μ, λ) = (0, 0) For each (μ, λ) ∈ B M (μ, 1) × Λ, we consider the problem (42) ⎧ μ2 ⎪ J(x, u, μ) = μ(t)(x1 (t) + x2 (t)) + u2 (t) + μ+u (t) dt → inf, ⎪ ⎪ ⎪ ⎪ ⎪x˙ = x1 + tu + λ1 , ⎨ P (μ, λ) x˙ = x2 + u + λ2 , ⎪ ⎪ ⎪ ⎪ x1 (0) = x2 (0) = 1, ⎪ ⎪ ⎩ u ∈ [−1, 1], a.e t ∈ [0, 1] Then we have the following assertions: (a) P (μ, λ) has the unique solution (x, u) = (exp(t), exp(t)), ; (b) J(x, u, μ) satisfies conditions (A1)–(A5) of Theorem 1.1 Moreover, for all μ = 0, the function (x, u) → J(x, u, μ) is not strongly convex Copyright © by SIAM Unauthorized reproduction of this article is prohibited STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM 2903 Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Solution From the above we have f (t, x, u, μ) = μ(x1 + x2 ) + u2 + + μ2 u (a) For (μ, λ) = (0, 0), problem (42) becomes J(x, u, 0) = u2 (t)dt → inf, (x, u) ∈ K(0), P (μ, λ) where K(0) = {(x, u)|x˙ = x1 + tu, x˙ = x1 + x2 + u, x1 (0) = x2 (0) = 1, u ∈ [−1, 1] a.e.} We have J(x, u, μ) ≥ for all (x, u) ∈ K(0) Hence (x,u)∈K(0) J(x, u, μ) = J(x, 0, μ) = Thus u = Since (x, u) satisfies the equation ⎧ ⎪ ⎨x˙ = x1 + tu, x˙ = x2 + u, ⎪ ⎩ x1 (0) = x2 (0) = 1, we obtain x1 (t) = x2 (t) = exp(t) Thus (x(t), u(t)) = ((exp(t), exp(t)), 0) is the unique solution of P (μ, λ) (b) Let us verify assumptions (A1)–(A5) of Theorem 1.1 From the above, we have f (t, x, u, μ(t)) = u2 It follows that for all zi = (xi , ui ) ∈ Rn × Rm with i = 1, 2, we get fz (t, x1 , u1 , μ(t)) − fz (t, x2 , u2 , μ(t)), z1 − z2 = 2|u1 − u2 |2 Hence (A4) is valid Let us consider functions φ(x, μ) = μ(x1 + x2 ), x = (x1 , x2 ) and ϕ(u, μ) = u2 + + μ2 u Then f (t, x, u, μ) = φ(x, μ) + ϕ(u, μ) We have ∇x φ(x, μ) = μ(1, 1), ϕu (u, μ) = 2u + −2μ2 u (1 + μ2 u2 )2 Also, we have φx (x, μ) = 0 0 and 6μ4 x2 − 2μ2 (1 + μ2 x2 )3 2(μ2 u2 )3 + 6(μ2 u2 )2 + (6 + 6μ2 )(μ2 u2 ) + 2(1 − μ2 ) = (1 + μ2 u2 )3 2(1 − μ2 ) ≥ ≥ ∀μ ∈ B(μ, 1) (1 + μ2 u2 )3 ϕu (u, μ) = + (43) Copyright © by SIAM Unauthorized reproduction of this article is prohibited Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php 2904 B T KIEN, N T TOAN, M M WONG, AND J C YAO Therefore, we can conclude that for each μ, ϕ(u, μ) and φ(x, μ) are convex but they are not strongly convex Moreover, ∇x φ(·) and ϕu (·) are monotone It is easy to see that ∇f (t, x, u, μ) = (∇x φ(x, μ), ϕu (u, μ)) and (44) ⎛ f(x,u) (t, x, u, μ) = ⎝ 0 ⎞ 0 ⎠ 0 ϕ (u) Since Det(f(x,u) (t, x, u, μ)) = 0, f(x,u) (t, x, u, μ) is nonnegative definite This implies that (x, u) → f (t, x, u, μ) is convex but not strongly convex for all (t, μ) ∈ Ω1 (μ) with Ω1 (μ) = {(t, μ) ∈ [0, 1] × R : |μ| ≤ 1} It follows that for each μ ∈ B(μ, 1), J(x, u, μ) is convex in (x, u) but it is not strongly convex Moreover, we have √ √ 2 (μ + |x|2 ) f (t, x, u, μ) ≥ − 2|μ||x| ≥ − Hence assumption (A1) is fulfilled Let us verify (A2) For this, we have |fx (t, x, u, μ)| = |∇φ(x)| = √ 2|μ| and 2μ2 |u| (1 + μ2 u2 )2 2μ2 |u| = 2|u| + + 2μ2 u2 + μ4 u4 2μ2 |u| 2μ2 |u| ≤ 2|u| + ≤ 2|u| + √ 2 + 2μ u 2|μ||u| ≤ 2|u| + √ |μ| |fu (t, x, u, μ)| = |ϕ (u)| ≤ 2|u| + Hence (A2) is valid For assumption (A3), we have |fx (t, x, u, μ) − fx (t, x, u, μ(t))| = √ 2|μ − μ(t)| for all (t, μ) with |μ| ≤ and t ∈ [0, 1] Also, we have |fu (t, x, u, μ) − fu (t, x, u, μ(t))| = 2μ2 |u| ≤ 2μ2 |u| (1 + μ2 u2 )2 for all (t, x, u, μ) with |μ| ≤ Therefore, assumption (A3) is valid Finally, we have (45) A= 0 ,B = t ,T = 0 Copyright © by SIAM Unauthorized reproduction of this article is prohibited Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM 2905 Hence (A5) is automatically fulfilled Thus we have shown that assumptions (A1)– (A5) are satisfied Notice that although assumption (A4) of Theorem 1.1 requires that unperturbed problem P (μ, λ) has a unique solution, perturbed problems may have several solutions The following example illustrates this situation Example 3.2 Suppose that m = n = l = k = 1, p = 2, and (μ, λ) = (0, 0) For each (μ, λ) ∈ B M (μ, 1) × Λ with μ(t) = μ ∈ R and λ(t) = λ ∈ R for all t ∈ [0, 1], we consider the problem J(x, u, μ) = [(u(t)+μ2 )2 (1−sig(u(t) + μ2 ))+(u(t)−μ2 )2 (1+sig(u(t) − μ2 ))]dt → inf subject to ⎧ ⎪ ⎨x˙ = x + u + λ, x(0) = 1, ⎪ ⎩ −1 ≤ u ≤ Here function sig(a) is defined by ⎧ ⎪ ⎨1 sig(a) = ⎪ ⎩ −1 if if if a > 0, a = 0, a < It can be verified that all conditions of Theorem 1.1 are fulfilled When (μ, λ) = (0, 0) the problem becomes J(x, u, μ) = 2u(t)2 dt → inf subject to ⎧ ⎪ ⎨x˙ = x + u, x(0) = 1, ⎪ ⎩ −1 ≤ u ≤ In this case we have S(0, 0) = {(et , 0)} However, when μ = 0, we get inf J(x, u, μ) = which holds at all u(·) = ξ with ξ ∈ [−μ2 , μ2 ] and corresponding state x = −(λ + ξ) + (λ + ξ + 1)et Hence S(μ, λ) = {(x, u) = ((λ + ξ + 1)et − ξ − λ, ξ)| − μ2 ≤ ξ ≤ μ2 } It easy to see that S(μ, λ) is lower semicontinuous at (0, 0) Remark 3.3 Theorem 1.1 shows that under certain conditions, the solution map S(μ, λ) of P (μ, λ) is lower semicontinuous at a reference point (μ, λ) Since it is easy to see that S(μ, λ) is upper semicontinuous at (μ, λ), we can conclude from Theorem 1.1 that the solution map S(·, ·) is continuous at (μ, λ) Acknowledgments The authors thank three anonymous reviewers for their helpful comments and suggestions which improved the original manuscript greatly Copyright © by SIAM Unauthorized reproduction of this article is prohibited 2906 B T KIEN, N T TOAN, M M WONG, AND J C YAO Downloaded 01/02/13 to 132.206.27.25 Redistribution subject to SIAM license or copyright; 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Problems, North-Holland, Amsterdam, 1979 [10] B T Kien, Lower semicontinuity of the solution set to a parametric generalized variational inequality in reflexive Banach spaces, Set- Valued Anal., 16 (2008),