DSpace at VNU: Lower semicontinuity of the solution set to a parametric optimal control problem

c 2012 Society for Industrial and Applied Mathematics

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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. WONG§ , AND J. C. YAO¶
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

1. 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 1 < p < ∞ and a trajectory
x ∈ W 1,1 ([0, 1], Rn ) which minimizes the cost
1

(1)
0

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˙ 1 .

The couple (μ, λ) ∈ L∞ ([0, 1], Rk ) × Lr ([0, 1], Rl ) with 1 ≤ 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.
/>† Department of Information and Technology, Hanoi National University of Civil Engineering,
Hanoi, Vietnam (). This author’s research was partially supported by grant
NAFOSTED 101.01-2011.23.

‡ Department of Mathematics, Vinh University, Vinh City, Vietnam ().
§ Department of Applied Mathematics, Chung Yuan Christian University, Chung Li, Taiwan
().
¶ Corresponding author. Center for General Education, Kaohsiung Medical University, Kaohsiung
807, Taiwan (). This author’s research was partially supported by grants NSC 992221-E-037-007-MY3 and NSC 99-2115-M-037-002-MY3.

2889

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2890

B. T. KIEN, N. T. TOAN, M. M. WONG, AND J. C. YAO

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(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
1

(5)
(6)

J(x, u, μ) =

0

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) = ∅}

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STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM

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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, 1 , a constant 0 ≤ γ ≤ p, and a nonnegative continuous function ϕ : [0, 1] × R ×
R × R → R such that for all (t, μ) ∈ Ω 1 (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 ≥ 0 with i = 1, 2, . . . , 6 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)| ≤ 1 .
(A3) There exist continuous functions kj : [0, 1] × R3 → R, positive numbers sj
with j = 1, 2, and 0 ≤ 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)| ≤ 1 .
(A4) The function (x, u) → f (t, x, u, μ(t)) is strongly convex, that is, there exists
a constant ρ > 0 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, 2 and for a.e. t ∈ [0, 1].

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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

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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 2. In section 3, we will
consider some examples of optimal control problems satisfying conditions (A1)–(A5)
of Theorem 1.1.
2. 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 > 0 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 > 0 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].

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STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM

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By subtracting (9) and (10) and putting z = x − y, we get z(0) = 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

0

0

t
0

z(s)ds,
˙
we obtain

(T1 |z(s)| + T2 |λ1 (s) − λ2 (s)|)ds

t

t

t

0

T1 |z(s)|ds +

1
0

T2 |λ1 (t) − λ2 (t)|dt

T1 |z(s)|ds + T2 λ1 − λ2

1

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

0

≤ T2 λ1 − λ2

r

exp(T1 ).

Combining this with (12), we have
z˙
(14)

1

≤ T1 z 1 + 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˙

1

≤ k0 λ1 − λ2 r .

The proof of the lemma is now complete.
Let us choose 0 > 0 and β0 > 0 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 ).

> 0 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, 0 ] and 0 < β < min{β0 , 4k0 }, one has
(16)

K (λ1 ) ⊆ K (λ2 ) + 5k0 λ1 − λ2 B Z ∀λ1 , λ2 ∈ BΛ (λ, β).

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2894

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Fix = 0 > 0 and β > 0 as in Lemma 2.2 such that (16) is satisfied. For each
(μ, λ) ∈ B M (μ, 1 ) × 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, 1 ) × 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
0

x˙ i (s)ds, we have from above

|xi (t)| ≤ |x0 | +
≤ |x0 | +
≤
(19)

≤

t
0

0

t

t
0

0

t

T1 |xi (s)|ds +
T1 |xi (s)|ds +

t
0

(|B(s)||ui (s)| + T2 |λ(s)|)ds

1

0

T1 |xi (s)|ds + (|x0 | + φ

φ(s)|ui (s)| + T2 |λ(s)| ds
q

ui

p

+ T2 λ 1 )

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)|.

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2895

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Since {ui } is bounded in Lp , we have
x˙ i

1

(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

1
q

K1 + T2

E

|λ(t)|r dt

1
r

and the second term of the right-hand side of the above inequality approaches zeros
as |E| → 0 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 + 0 x˙ i (s)ds we get
t
x(t) = x0 + 0 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) +
1

(22)

+
0

1
0

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)

1
0

0

1

|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
1

0

|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

+

1 ]}.

By (A2) and using similar arguments as in Claim 1, we can show that for each
μ ∈ B M (μ, 1 ), Fz (·, μ) is uniformly bounded on B Z (z, ). Hence there exists a
constant C > 0 such that
Fz (z, μ) ≤ C

∀z ∈ B Z (z, ).

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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, 1 ) × 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, 1 ) 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
0 ∈ Jz (zi , μi ) + N zi ; K(λi ) ∩ B Z (z, )
which is equivalent to the variational inequality
(24)

Jz (zi , μi ), z − zi ≥ 0

∀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
1

=
0

(25)

fx (t, x0 (t), u0 (t), μ(t)), h1 (t) dt
1

+
0

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].

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STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM

2897

Since λi → λ strongly in Λ, there exists K4 > 0 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
0

x˙ i (s)ds we obtain

|xi (t)| ≤ |x0 | +
≤
≤

(27)

t
0

0

t

t
0

t

T1 |xi (s)|ds +

0

T1 |xi (s)|ds + |x0 | +

0

(φ(s)|ui (s)| + T2 h(s))ds
1

(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

1

≤ 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

1
q

K3 + T2

r

E

|h| dt

1
r

and the second part of (28) approaches to 0 as |E| → 0. 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˙ 1 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 | → 0.
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.

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2898

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Claim 1. {Jz (zi , μ)} is bounded.
Indeed, for each h ∈ Lp ([0, 1], Rm ) we have
1

|Ju (xi , ui , μ)h| =

fu (t, xi (t), ui (t), μ(t))h(t)dt .

0

Hence
1

Ju (xi , ui , μ) ≤

0

1
q

q

|fu (t, xi (t), ui (t), μ(t))| dt

.

By (A2) we have
1

Ju (xi , ui , μ) ≤ α4

0

(|ω2 (t)| + |xi (t)|

α5

+ |ui (t)|

p−1

α6 q

+ |μ(t)| ) dt

1
q

.

Using the inequality
(a + b)q ≤ 2q−1 (aq + bq )

(30)

∀a, b ≥ 0, q ≥ 1

(see [1, Lemma 2.24]), we get
Ju (xi , ui , μ) ≤ α4

1

4

q−1

0

q

(|ω2 (t)| + |xi (t)|

Since xi → x uniformly and ui
xi

α5 q

+ |ui (t)|

(p−1)q

+ |μ(t)|

α6 q

1
q

)dt

.

u, there exist constants γ1 , γ2 > 0 such that
∞

≤ γ1 , u i

p

≤ γ2

∀i ≥ 1.

It follows that
Ju (xi , ui , μ) ≤ α4 4q−1 ( ω2

q
q

+ γ1α5 q + γ2p + μ

α6 q
∞ )

1
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 2. Jz (zi , μ) − Jz (zi , μi ) → 0 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
∞

≤

1

∞]

kj (t1 , t2 , t3 , t4 ),

for i > 0 large enough, assumption (A3)

| Jx (xi , ui , μ) − Jx (xi , ui , μi ), x
1

=
0

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
∞

1

x

0

ui

η
p

0

x

|ui (t)|η dt
1,1

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2899

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for some constant c > 0. 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

0

≤

as i → ∞.

It remains to be shown that Ju (xi , ui , μ) − Ju (xi , ui , μi ) → 0 as i → ∞. It is easy
to see that
Ju (xi , ui , μ) − Ju (xi , ui , μi )
1

≤

0

|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 )
≤
≤

1
0

0

1

|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
∞

→ 0 as i → ∞.

Combining this with (31) we obtain the proof of Claim 2.
We now notice that by Claim 1 and Claim 2, {Jz (zi , μi )} is bounded. Hence from
(29) and z2i → z we obtain

lim Jz (zi , μi ), zi − z ≤ 0.

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 ≤ 0.
i→∞

It follows that
lim Jz (zi , μ) − Jz (z, μ), zi − z ≤ 0.

(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)

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2900

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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

0

0

t

t

t
(x˙ i (s)
0

− x(s))ds,
˙
we have

T1 |xi (s) − x(s)|ds +
T1 |xi (s) − x(s)|ds +

t
0

(φ(s)|ui (s) − u(s)| + T2 |λi (s) − λ(s)|)ds

1

0

(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

1

= x˙ i − x˙

1

+ T2 λi − λ r ) exp(T1 )

+ T2 λi − λ r .

Since ui − u p → 0 and λi − λ r → 0, we see that xi − x 1,1 → 0. 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 ≥ 0.

(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 ≥ 0.

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STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM

2901

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On the other hand, since z ∈ S(μ, λ) and z ∈ K(λ) ∩ B Z (z, ), it follows that
Jz (z, μ), z − z ≥ 0.

(38)

Claim 3. Jz (·, μ) is strictly monotone on K(λ) ∩ B Z (z, ).
In fact, suppose that
Jz (z1 , μ) − Jz (z2 , μ), z1 − z2 = 0

(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˙ 1 − x˙ 2

1

≤ 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 > 0.
This and (38) imply that
Jz (z, μ), z − z > Jz (z, μ), z − z ≥ 0
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 3 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 ≥ 0 ∀z ∈ K(λ) ∩ B Z (z, )

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2902

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and

Jz (z0 , μ), z − z0 ≥ 0 ∀z ∈ K(λ) ∩ B Z (z, ).
Consequently,
Jz (z, μ), z0 − z ≥ 0
and
Jz (z0 , μ), z − z0 ≥ 0.
Hence
Jz (z0 , μ) − Jz (z, μ), z0 − z ≤ 0
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 > 0 such that BZ (z, ) ⊂ G1 and K(λ)∩B Z (z, ) =
∅ for all λ ∈ BΛ (λ, δ). Taking β such that 0 < 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

3. 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)
⎧
1
μ2
⎪
J(x, u, μ) = 0 μ(t)(x1 (t) + x2 (t)) + u2 (t) + μ+u
2 (t) dt → inf,
⎪
⎪
⎪
⎪
⎪x˙ 1 = x1 + tu + λ1 ,
⎨
P (μ, λ)
x˙ 2 = 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)), 0 ;
(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.

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STABILITY OF A PARAMETRIC OPTIMAL CONTROL PROBLEM

2903

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Solution. From the above we have
f (t, x, u, μ) = μ(x1 + x2 ) + u2 +

1
.
1 + μ2 u 2

(a) For (μ, λ) = (0, 0), problem (42) becomes
1

J(x, u, 0) = 0 u2 (t)dt → inf,
(x, u) ∈ K(0),

P (μ, λ)
where

K(0) = {(x, u)|x˙ 1 = x1 + tu, x˙ 2 = x1 + x2 + u, x1 (0) = x2 (0) = 1, u ∈ [−1, 1] a.e.}.

We have J(x, u, μ) ≥ 0 for all (x, u) ∈ K(0). Hence
min

(x,u)∈K(0)

J(x, u, μ) = J(x, 0, μ) = 0.

Thus u = 0. Since (x, u) satisfies the equation
⎧
⎪
⎨x˙ 1 = x1 + tu,
x˙ 2 = 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 +

1
.
1 + μ2 u 2

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

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 )
≥
≥ 0 ∀μ ∈ B(μ, 1).
(1 + μ2 u2 )3

ϕu (u, μ) = 2 +

(43)

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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)

⎛

0
f(x,u) (t, x, u, μ) = ⎝ 0
0

⎞
0
0
⎠.
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 2
(μ + |x|2 ).
f (t, x, u, μ) ≥ − 2|μ||x| ≥ −
2
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| +
1 + 2μ2 u2 + μ4 u4
2μ2 |u|
2μ2 |u|
≤ 2|u| +

≤ 2|u| + √
2
2
1 + 2μ u
2 2|μ||u|
1
≤ 2|u| + √ |μ|.
2

|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| ≤ 1 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| ≤ 1. Therefore, assumption (A3) is valid.
Finally, we have
(45)

A=

1
0

0
1

,B =

t
1

,T =

1 0
0 1

.

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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
1

J(x, u, μ) =

0

[(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 ≤ 1.
Here function sig(a) is defined by
⎧
⎪
⎨1
sig(a) = 0
⎪
⎩
−1

if
if
if

a > 0,
a = 0,
a < 0.

It can be verified that all conditions of Theorem 1.1 are fulfilled. When (μ, λ) = (0, 0)
the problem becomes
1

J(x, u, μ) =

0

2u(t)2 dt → inf

subject to
⎧
⎪
⎨x˙ = x + u,
x(0) = 1,
⎪
⎩
−1 ≤ u ≤ 1.
In this case we have S(0, 0) = {(et , 0)}. However, when μ = 0, we get inf J(x, u, μ) = 0
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.

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2906

B. T. KIEN, N. T. TOAN, M. M. WONG, AND J. C. YAO

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