Research Article Optimal Control Problems for Nonlinear Variational Evolution Inequalities

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Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 724190, 10 pages
http://dx.doi.org/10.1155/2013/724190
Research Article
Optimal Control Problems for Nonlinear Variational
Evolution Inequalities
Eun-Young Ju and Jin-Mun Jeong
Department of Applied Mathematics, Pukyong National University, Busan 608-737, Republic of Korea
Correspondence should be addressed to Jin-Mun Jeong; jmjeong@pknu.ac.kr
Received 12 November 2012; Revised 4 January 2013; Accepted 6 January 2013
Academic Editor: Ryan Loxton
Copyright © 2013 E.-Y. Ju and J.-M. Jeong. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
We deal with optimal control problems governed by semilinear parabolic type equations and in particular described by variational
inequalities. We will also characterize the optimal controls by giving necessary conditions for optimality by proving the GaΜ‚teaux
differentiability of solution mapping on control variables.
1. Introduction
In this paper, we deal with optimal control problems governed by the following variational inequality in a Hilbert
space 𝐻:
(π‘₯σΈ€  (𝑑) + 𝐴π‘₯ (𝑑) , π‘₯ (𝑑) − 𝑧) + πœ™ (π‘₯ (𝑑)) − πœ™ (𝑧)
≤ (𝑓 (𝑑, π‘₯ (𝑑)) + 𝐡𝑒 (𝑑) , π‘₯ (𝑑) − 𝑧) ,
a.e., 0 < 𝑑 ≤ 𝑇, 𝑧 ∈ 𝑉,
(1)
π‘₯ (0) = π‘₯0 .
Here, 𝐴 is a continuous linear operator from 𝑉 into 𝑉∗ which
is assumed to satisfy Gårding’s inequality, where 𝑉 is dense
subspace in 𝐻. Let πœ™ : 𝑉 → (−∞, +∞] be a lower semicontinuous, proper convex function. Let U be a Hilbert space of
control variables, and let 𝐡 be a bounded linear operator from
U into 𝐿2 (0, 𝑇; 𝐻). Let Uad be a closed convex subset of U,
which is called the admissible set. Let 𝐽 = 𝐽(𝑣) be a given
quadratic cost function (see (61) or (103)). Then we will find
an element 𝑒 ∈ Uad which attains minimum of 𝐽(𝑣) over Uad
subject to (1).
Recently, initial and boundary value problems for permanent magnet technologies have been introduced via variational inequalities in [1, 2] and nonlinear variational inequalities of semilinear parabolic type in [3, 4]. The papers treating
the variational inequalities with nonlinear perturbations are
not many. First of all, we deal with the existence and a variation of constant formula for solutions of the nonlinear functional differential equation (1) governed by the variational
inequality in Hilbert spaces in Section 2.
Based on the regularity results for solution of (1), we
intend to establish the optimal control problem for the cost
problems in Section 3. For the optimal control problem of
systems governed by variational inequalities, see [1, 5]. We
refer to [6, 7] to see the applications of nonlinear variational
inequalities. Necessary conditions for state constraint optimal
control problems governed by semilinear elliptic problems
have been obtained by Bonnans and Tiba [8] using methods
of convex analysis (see also [9]).
Let π‘₯𝑒 stand for solution of (1) associated with the control
𝑒 ∈ U. When the nonlinear mapping 𝑓 is Lipschitz continuous from R × π‘‰ into 𝐻, we will obtain the regularity for
solutions of (1) and the norm estimate of a solution of the
above nonlinear equation on desired solution space. Consequently, in view of the monotonicity of πœ•πœ™, we show that
the mapping 𝑒 󳨃→ π‘₯𝑒 is continuous in order to establish
the necessary conditions of optimality of optimal controls for
various observation cases.
In Section 4, we will characterize the optimal controls
by giving necessary conditions for optimality. For this, it is
necessary to write down the necessary optimal condition due
to the theory of Lions [9]. The most important objective of
such a treatment is to derive necessary optimality conditions
2
Abstract and Applied Analysis
that are able to give complete information on the optimal
control.
Since the optimal control problems governed by nonlinear equations are nonsmooth and nonconvex, the standard
methods of deriving necessary conditions of optimality are
inapplicable here. So we approximate the given problem by a
family of smooth optimization problems and afterwards tend
to consider the limit in the corresponding optimal control
problems. An attractive feature of this approach is that it
allows the treatment of optimal control problems governed
by a large class of nonlinear systems with general cost criteria.
If 𝐻 is identified with its dual space we may write 𝑉 ⊂ 𝐻 ⊂ 𝑉∗
densely and the corresponding injections are continuous. The
norm on 𝑉, 𝐻, and 𝑉∗ will be denoted by || ⋅ ||, | ⋅ |, and || ⋅ ||∗ ,
respectively. The duality pairing between the element 𝑣1 of
𝑉∗ and the element 𝑣2 of 𝑉 is denoted by (𝑣1 , 𝑣2 ), which is
the ordinary inner product in 𝐻 if 𝑣1 , 𝑣2 ∈ 𝐻.
For 𝑙 ∈ 𝑉∗ we denote (𝑙, 𝑣) by the value 𝑙(𝑣) of 𝑙 at 𝑣 ∈ 𝑉.
The norm of 𝑙 as element of 𝑉∗ is given by
𝑣∈𝑉
|(𝑙, 𝑣)|
.
‖𝑣‖
(2)
Therefore, we assume that 𝑉 has a stronger topology than 𝐻
and, for brevity, we may regard that
‖𝑒‖∗ ≤ |𝑒| ≤ ‖𝑒‖ ,
∀𝑒 ∈ 𝑉.
(3)
Let π‘Ž(⋅, ⋅) be a bounded sesquilinear form defined in 𝑉×𝑉
and satisfying Gårding’s inequality
Re π‘Ž (𝑒, 𝑒) ≥ πœ”1 ‖𝑒‖2 − πœ”2 |𝑒|2 ,
(4)
where πœ”1 > 0 and πœ”2 is a real number. Let 𝐴 be the operator
associated with this sesquilinear form:
(𝐴𝑒, 𝑣) = π‘Ž (𝑒, 𝑣) ,
𝑒, 𝑣 ∈ 𝑉.
(5)
Then −𝐴 is a bounded linear operator from 𝑉 to 𝑉∗ by the
Lax-Milgram Theorem. The realization of 𝐴 in 𝐻 which is
the restriction of 𝐴 to
𝐷 (𝐴) = {𝑒 ∈ 𝑉 : 𝐴𝑒 ∈ 𝐻}
(6)
is also denoted by 𝐴. From the following inequalities
πœ”1 ‖𝑒‖2 ≤ Re π‘Ž (𝑒, 𝑒) + πœ”2 |𝑒|2 ≤ 𝐢 |𝐴𝑒| |𝑒| + πœ”2 |𝑒|2
≤ (𝐢 |𝐴𝑒| + πœ”2 |𝑒|) |𝑒| ≤ max {𝐢, πœ”2 } ‖𝑒‖𝐷(𝐴) |𝑒| ,
(7)
(10)
where each space is dense in the next one with continuous
injection.
Lemma 1. With the notations (9) and (10), we have
(𝑉, 𝑉∗ )1/2,2 = 𝐻,
(𝐷(𝐴), 𝐻)1/2,2 = 𝑉,
(11)
2 1/2
‖𝑒‖𝐷(𝐴) = (|𝐴𝑒| + |𝑒| )
(8)
is the graph norm of 𝐷(𝐴), it follows that there exists a
constant 𝐢0 > 0 such that
1/2
‖𝑒‖ ≤ 𝐢0 ‖𝑒‖1/2
𝐷(𝐴) |𝑒| .
It is also well known that 𝐴 generates an analytic semigroup 𝑆(𝑑) in both 𝐻 and 𝑉∗ . For the sake of simplicity we
assume that πœ”2 = 0 and hence the closed half plane {πœ† :
Re πœ† ≥ 0} is contained in the resolvent set of 𝐴.
If 𝑋 is a Banach space, 𝐿2 (0, 𝑇; 𝑋) is the collection of all
strongly measurable square integrable functions from (0, 𝑇)
into 𝑋 and π‘Š1,2 (0, 𝑇; 𝑋) is the set of all absolutely continuous
functions on [0, 𝑇] such that their derivative belongs to
𝐿2 (0, 𝑇; 𝑋). 𝐢([0, 𝑇]; 𝑋) will denote the set of all continuously
functions from [0, 𝑇] into 𝑋 with the supremum norm. If 𝑋
and π‘Œ are two Banach spaces, L(𝑋, π‘Œ) is the collection of all
bounded linear operators from 𝑋 into π‘Œ, and L(𝑋, 𝑋) is simply written as L(𝑋). Here, we note that by using interpolation
theory we have
𝐿2 (0, 𝑇; 𝑉) ∩ π‘Š1,2 (0, 𝑇; 𝑉∗ ) ⊂ 𝐢 ([0, 𝑇] ; 𝐻) .
(9)
(12)
First of all, consider the following linear system:
π‘₯σΈ€  (𝑑) + 𝐴π‘₯ (𝑑) = π‘˜ (𝑑) ,
π‘₯ (0) = π‘₯0 .
(13)
By virtue of Theorem 3.3 of [11] (or Theorem 3.1 of [12,
13]), we have the following result on the corresponding linear
equation of (13).
Lemma 2. Suppose that the assumptions for the principal
operator 𝐴 stated above are satisfied. Then the following properties hold.
(1) For π‘₯0 ∈ 𝑉 = (𝐷(𝐴), 𝐻)1/2,2 (see Lemma 1) and π‘˜ ∈
𝐿2 (0, 𝑇; 𝐻), 𝑇 > 0, there exists a unique solution π‘₯ of
(13) belonging to
𝐿2 (0, 𝑇; 𝐷 (𝐴)) ∩ π‘Š1,2 (0, 𝑇; 𝐻) ⊂ 𝐢 ([0, 𝑇] ; 𝑉)
(14)
and satisfying
σ΅„© σ΅„©
β€–π‘₯‖𝐿2 (0,𝑇;𝐷(𝐴))∩π‘Š1,2 (0,𝑇;𝐻) ≤ 𝐢1 (σ΅„©σ΅„©σ΅„©π‘₯0 σ΅„©σ΅„©σ΅„© + β€–π‘˜β€–πΏ2 (0,𝑇;𝐻) ) ,
where
2
𝐷 (𝐴) ⊂ 𝑉 ⊂ 𝐻 ⊂ 𝑉∗ ⊂ 𝐷(𝐴)∗ ,
where (𝑉, 𝑉∗ )1/2,2 denotes the real interpolation space between
𝑉 and 𝑉∗ (Section 1.3.3 of [10]).
2. Regularity for Solutions
‖𝑙‖∗ = sup
Thus we have the following sequence
(15)
where 𝐢1 is a constant depending on 𝑇.
(2) Let π‘₯0 ∈ 𝐻 and π‘˜ ∈ 𝐿2 (0, 𝑇; 𝑉∗ ), 𝑇 > 0. Then there
exists a unique solution π‘₯ of (13) belonging to
𝐿2 (0, 𝑇; 𝑉) ∩ π‘Š1,2 (0, 𝑇; 𝑉∗ ) ⊂ 𝐢 ([0, 𝑇] ; 𝐻)
(16)
Abstract and Applied Analysis
3
and satisfying
β€–π‘₯‖𝐿2 (0,𝑇;𝑉)∩π‘Š1,2 (0,𝑇;𝑉∗ )
(2) We assume the following.
󡄨 󡄨
≤ 𝐢1 (󡄨󡄨󡄨π‘₯0 󡄨󡄨󡄨 + β€–π‘˜β€–πΏ2 (0,𝑇;𝑉∗ ) ) ,
(17)
where 𝐢1 is a constant depending on 𝑇.
Let 𝑓 be a nonlinear single valued mapping from [0, ∞)×
𝑉 into 𝐻.
(F) We assume that
σ΅„©
󡄨󡄨
󡄨
σ΅„©
󡄨󡄨𝑓 (𝑑, π‘₯1 ) − 𝑓 (𝑑, π‘₯2 )󡄨󡄨󡄨 ≤ 𝐿 σ΅„©σ΅„©σ΅„©π‘₯1 − π‘₯2 σ΅„©σ΅„©σ΅„© ,
(18)
(A) 𝐴 is symmetric and there exists β„Ž ∈ 𝐻 such that for
every πœ– > 0 and any 𝑦 ∈ 𝐷(πœ™)
π½πœ– (𝑦 + πœ–β„Ž) ∈ 𝐷 (πœ™) ,
πœ™ (π½πœ– (𝑦 + πœ–β„Ž)) ≤ πœ™ (𝑦) ,
(26)
where π½πœ– = (𝐼 + πœ–π΄)−1 .
Then for 𝑒 ∈ 𝐿2 (0, 𝑇; π‘Œ), 𝐡 ∈ L(π‘Œ, 𝐻), and π‘₯0 ∈ 𝐷(πœ™) ∩ 𝑉
(1) has a unique solution
π‘₯ ∈ 𝐿2 (0, 𝑇; 𝐷 (𝐴)) ∩ π‘Š1,2 (0, 𝑇; 𝐻) ∩ 𝐢 ([0, 𝑇] ; 𝐻) , (27)
for every π‘₯1 , π‘₯2 ∈ 𝑉.
Let π‘Œ be another Hilbert space of control variables and
take U = 𝐿2 (0, 𝑇; π‘Œ) as stated in the Introduction. Choose a
bounded subset π‘ˆ of π‘Œ and call it a control set. Let us define
an admissible control Uad as
Uad = {𝑒 ∈ 𝐿2 (0, 𝑇; π‘Œ) : 𝑒 is strongly measurable
function satisfying 𝑒 (t) ∈ π‘ˆ for almost all 𝑑} .
(19)
Noting that the subdifferential operator πœ•πœ™ is defined by
πœ•πœ™ (π‘₯) = {π‘₯∗ ∈ 𝑉∗ ; πœ™ (π‘₯) ≤ πœ™ (𝑦) + (π‘₯∗ , π‘₯ − 𝑦) , 𝑦 ∈ 𝑉} ,
(20)
the problem (1) is represented by the following nonlinear
functional differential problem on 𝐻:
which satisfies
σ΅„© σ΅„©
β€–π‘₯‖𝐿2 ∩π‘Š1,2 ∩𝐢 ≤ 𝐢2 (1 + σ΅„©σ΅„©σ΅„©π‘₯0 σ΅„©σ΅„©σ΅„© + ‖𝐡𝑒‖𝐿2 (0,𝑇;𝐻) ) .
Remark 4. In terms of Lemma 1, the following inclusion
𝐿2 (0, 𝑇; 𝑉) ∩ π‘Š1,2 (0, 𝑇; 𝑉∗ ) ⊂ 𝐢 ([0, 𝑇] ; 𝐻)
The following Lemma is from Brézis [14, Lemma A.5].
Lemma 5. Let π‘š ∈ 𝐿1 (0, 𝑇; R) satisfying π‘š(𝑑) ≥ 0 for all 𝑑 ∈
(0, 𝑇) and π‘Ž ≥ 0 be a constant. Let 𝑏 be a continuous function
on [0, 𝑇] ⊂ R satisfying the following inequality:
𝑑
1
1 2
𝑏 (𝑑) ≤ π‘Ž2 + ∫ π‘š (𝑠) 𝑏 (𝑠) 𝑑𝑠,
2
2
0
(21)
π‘₯ (0) = π‘₯0 .
(29)
is well known as seen in (9) and is an easy consequence of
the definition of real interpolation spaces by the trace method
(see [4, 13]).
π‘₯σΈ€  (𝑑) + 𝐴π‘₯ (𝑑) + πœ•πœ™ (π‘₯ (𝑑)) ∋ 𝑓 (𝑑, π‘₯ (𝑑)) + 𝐡𝑒 (𝑑) ,
0 < 𝑑 ≤ 𝑇,
(28)
𝑑 ∈ [0, 𝑇] .
(30)
Then,
Referring to Theorem 3.1 of [3], we establish the following
results on the solvability of (1).
Proposition 3. (1) Let the assumption (F) be satisfied. Assume
that 𝑒 ∈ 𝐿2 (0, 𝑇; π‘Œ), 𝐡 ∈ L(π‘Œ, 𝑉∗ ), and π‘₯0 ∈ 𝐷(πœ™) where
𝐷(πœ™) is the closure in 𝐻 of the set 𝐷(πœ™) = {𝑒 ∈ 𝑉 : πœ™(𝑒) < ∞}.
Then, (1) has a unique solution
π‘₯ ∈ 𝐿2 (0, 𝑇; 𝑉) ∩ 𝐢 ([0, 𝑇] ; 𝐻)
(22)
which satisfies
0
π‘₯σΈ€  (𝑑) = 𝐡𝑒 (𝑑) − 𝐴π‘₯ (𝑑) − (πœ•πœ™) (π‘₯ (𝑑)) + 𝑓 (𝑑, π‘₯ (𝑑)) ,
(23)
0
where (πœ•πœ™) : 𝐻 → 𝐻 is the minimum element of πœ•πœ™ and
there exists a constant 𝐢2 depending on 𝑇 such that
󡄨 󡄨
β€–π‘₯‖𝐿2 ∩𝐢 ≤ 𝐢2 (1 + 󡄨󡄨󡄨π‘₯0 󡄨󡄨󡄨 + ‖𝐡𝑒‖𝐿2 (0,𝑇;𝑉∗ ) ) ,
(24)
where 𝐢2 is some positive constant and 𝐿2 ∩ 𝐢 = 𝐿2 (0, 𝑇; 𝑉) ∩
𝐢([0, 𝑇]; 𝐻).
Furthermore, if 𝐡 ∈ L(π‘Œ, 𝐻) then the solution π‘₯ belongs to
π‘Š1,2 (0, 𝑇; 𝐻) and satisfies
󡄨 󡄨
β€–π‘₯β€–π‘Š1,2 (0,𝑇;𝐻) ≤ 𝐢2 (1 + 󡄨󡄨󡄨π‘₯0 󡄨󡄨󡄨 + ‖𝐡𝑒‖𝐿2 (0,𝑇;𝐻) ) .
(25)
𝑑
|𝑏 (𝑑)| ≤ π‘Ž + ∫ π‘š (𝑠) 𝑑𝑠,
0
𝑑 ∈ [0, 𝑇] .
(31)
For each (π‘₯0 , 𝑒) ∈ 𝐻 × πΏ2 (0, 𝑇; π‘Œ), we can define the
continuous solution mapping (π‘₯0 , 𝑒) 󳨃→ π‘₯. Now, we can state
the following theorem.
Theorem 6. (1) Let the assumption (F) be satisfied, π‘₯0 ∈ 𝐻,
and 𝐡 ∈ L(π‘Œ, 𝑉∗ ). Then the solution π‘₯ of (1) belongs to π‘₯ ∈
𝐿2 (0, 𝑇; 𝑉) ∩ 𝐢([0, 𝑇]; 𝐻) and the mapping
𝐻 × πΏ2 (0, 𝑇; π‘Œ) ∋ (π‘₯0 , 𝑒)
󳨃󳨀→ π‘₯ ∈ 𝐿2 (0, 𝑇; 𝑉) ∩ 𝐢 ([0, 𝑇] ; 𝐻)
(32)
is Lipschtz continuous; that is, suppose that (π‘₯0𝑖 , 𝑒𝑖 ) ∈ 𝐻 ×
𝐿2 (0, 𝑇; π‘Œ) and π‘₯𝑖 be the solution of (1) with (π‘₯0𝑖 , 𝑒𝑖 ) in place
of (π‘₯0 , 𝑒) for 𝑖 = 1, 2,
σ΅„©
σ΅„©σ΅„©
σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝑉)∩𝐢([0,𝑇];𝐻)
󡄨 σ΅„©
σ΅„©
󡄨
≤ 𝐢 {󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨 + 󡄩󡄩󡄩𝑒1 − 𝑒2 󡄩󡄩󡄩𝐿2 (0,𝑇;π‘Œ) } ,
where 𝐢 is a constant.
(33)
4
Abstract and Applied Analysis
(2) Let the assumptions (A) and (F) be satisfied and let
𝐡 ∈ L(π‘Œ, 𝐻) and π‘₯0 ∈ 𝐷(πœ™) ∩ 𝑉. Then π‘₯ ∈ 𝐿2 (0, 𝑇; 𝐷(𝐴)) ∩
π‘Š1,2 (0, 𝑇; 𝐻), and the mapping
integrating the above inequality over (0, 𝑑), we have
𝑑
𝑒−2πœ”2 𝑑 ∫ |𝑋(𝑠)|2 𝑑𝑠
0
𝑉 × πΏ2 (0, 𝑇; π‘Œ) ∋ (π‘₯0 , 𝑒)
2
1,2
󳨃󳨀→ π‘₯ ∈ 𝐿 (0, 𝑇; 𝐷 (𝐴)) ∩ π‘Š
(0, 𝑇; 𝐻)
𝑑
𝜏
1󡄨
󡄨2
≤ 2 ∫ 𝑒−2πœ”2 𝜏 { 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨 + ∫ 𝐻 (𝑠) 𝑑𝑠} π‘‘πœ
2
0
0
(34)
is continuous.
Proof. (1) Due to Proposition 3, we can infer that (1) possesses
a unique solution π‘₯ ∈ 𝐿2 (0, 𝑇; 𝑉) ∩ 𝐢([0, 𝑇]; 𝐻) with the data
condition (π‘₯0 , 𝑒) ∈ 𝐻 × πΏ2 (0, 𝑇; π‘Œ). Now, we will prove the
inequality (33). For that purpose, we denote π‘₯1 − π‘₯2 by 𝑋.
Then
+ 𝐡 (𝑒1 (𝑑) − 𝑒2 (𝑑)) ,
0 < 𝑑 ≤ 𝑇,
𝑑 𝑑
1 − 𝑒−2πœ”2 𝑑 󡄨󡄨
󡄨2
−2πœ” 𝜏
󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨 + 2 ∫ ∫ 𝑒 2 π‘‘πœπ» (𝑠) 𝑑𝑠
2πœ”2
0 𝑠
=
1 − 𝑒−2πœ”2 𝑑 󡄨󡄨
󡄨2
󡄨π‘₯ − π‘₯02 󡄨󡄨󡄨
2πœ”2 󡄨 01
1 𝑑 −2πœ”2 𝑠 −2πœ”2 𝑑
−𝑒
) 𝐻 (𝑠) 𝑑𝑠.
∫ (𝑒
πœ”2 0
+
𝑋󸀠 (𝑑) + 𝐴𝑋 (𝑑) + πœ•πœ™ (π‘₯1 (𝑑)) − πœ•πœ™ (π‘₯2 (𝑑))
∋ 𝑓 (𝑑, π‘₯1 (𝑑)) − 𝑓 (𝑑, π‘₯2 (𝑑))
=
(35)
Thus, we get
𝑋 (0) = π‘₯01 − π‘₯02 .
𝑑
πœ”2 ∫ |𝑋(𝑠)|2 𝑑𝑠 ≤
Multiplying on the above equation by 𝑋(𝑑), we have
0
𝑑
1 𝑑
|𝑋 (𝑑)|2 + πœ”1 ‖𝑋 (𝑑)β€–2
2 𝑑𝑑
2πœ”2 (𝑑−𝑠)
+ ∫ (𝑒
0
≤ πœ”2 |𝑋 (𝑑)|2
󡄨 󡄨
󡄨
󡄨
+ {󡄨󡄨󡄨𝑓 (𝑑, π‘₯1 (𝑑)) − 𝑓 (𝑑, π‘₯2 (𝑑))󡄨󡄨󡄨 + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑑) − 𝑒2 (𝑑))󡄨󡄨󡄨}
× |𝑋 (𝑑)| .
(36)
1 2πœ”2 𝑑
󡄨
󡄨2
− 1) 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨
(𝑒
2
(41)
− 1) 𝐻 (𝑠) 𝑑𝑠.
Combining this with (38) it holds that
𝑑
1
1
󡄨
󡄨2
|𝑋 (𝑑)|2 + πœ”1 ∫ ‖𝑋 (𝑠)β€–2 𝑑𝑠 ≤ 𝑒2πœ”2 𝑑 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨
2
2
0
𝑑
2πœ”2 (𝑑−𝑠)
+∫ 𝑒
Put
󡄨
󡄨
𝐻 (𝑑) = (𝐿 ‖𝑋 (𝑑)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑑) − 𝑒2 (𝑑))󡄨󡄨󡄨) |𝑋 (𝑑)| .
(40)
0
(42)
𝐻 (𝑠) 𝑑𝑠.
(37)
By Lemma 5, the following inequality
By integrating the above inequality over [0, 𝑑], we have
𝑑
2
1 −πœ”2 𝑑
(𝑒
|𝑋 (𝑑)|) + πœ”1 𝑒−2πœ”2 𝑑 ∫ ‖𝑋(𝑠)β€–2 𝑑𝑠
2
0
𝑑
1
|𝑋(𝑑)|2 + πœ”1 ∫ ‖𝑋 (𝑠)β€–2 𝑑𝑠
2
0
𝑑
𝑑
1󡄨
󡄨2
≤ 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨 + πœ”2 ∫ |𝑋 (𝑠)|2 𝑑𝑠 + ∫ 𝐻 (𝑠) 𝑑𝑠.
2
0
0
1󡄨
󡄨2
≤ 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨
2
(38)
𝑑
󡄨
󡄨
+ ∫ 𝑒−πœ”2 𝑠 (𝐿 ‖𝑋 (𝑠)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨) 𝑒−πœ”2 𝑠 |𝑋 (𝑠)| 𝑑𝑠
0
Note that
𝑑 −2πœ”2 𝑑 𝑑
{𝑒
∫ |𝑋 (𝑠)|2 𝑑𝑠}
𝑑𝑑
0
(43)
implies that
𝑑
1
≤ 2𝑒−2πœ”2 𝑑 { |𝑋 (𝑑)|2 − πœ”2 ∫ |𝑋 (𝑠)|2 𝑑𝑠}
2
0
𝑑
1󡄨
󡄨2
≤ 2𝑒−2πœ”2 𝑑 { 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨 + ∫ 𝐻 (𝑠) 𝑑𝑠} ,
2
0
(39)
󡄨
󡄨
𝑒−πœ”2 𝑑 |𝑋 (𝑑)| ≤ 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨
𝑑
−πœ”2 𝑠
+∫ 𝑒
0
󡄨
󡄨
(𝐿 ‖𝑋 (𝑠)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨) 𝑑𝑠.
(44)
Abstract and Applied Analysis
5
From (42) and (44) it follows that
𝑑
1
1
󡄨
󡄨2
|𝑋(𝑑)|2 + πœ”1 ∫ ‖𝑋(𝑠)β€–2 𝑑𝑠 ≤ 𝑒2πœ”2 𝑑 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨
2
2
0
𝑑
󡄨
󡄨
+ ∫ 𝑒2πœ”2 (𝑑−𝑠) (𝐿 ‖𝑋 (𝑠)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨) π‘’πœ”2 𝑠
0
󡄨
󡄨
× σ΅„¨σ΅„¨σ΅„¨π‘₯01 − π‘₯02 󡄨󡄨󡄨 𝑑𝑠
𝑑
2πœ”2 (𝑑−𝑠)
+∫ 𝑒
0
󡄨
󡄨
(𝐿 ‖𝑋 (𝑠)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨)
(45)
𝑠
󡄨
󡄨
× ∫ π‘’πœ”2 (𝑠−𝜏) (𝐿 ‖𝑋 (𝜏)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝜏) − 𝑒2 (𝜏))󡄨󡄨󡄨) π‘‘πœπ‘‘π‘ 
0
= 𝐼 + 𝐼𝐼 + 𝐼𝐼𝐼.
Putting
󡄨
󡄨
𝐺 (𝑠) = ‖𝑋 (𝑠)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨 .
(46)
The third term of the right hand side of (45) is estimated as
𝑑
𝑠
= 𝐿2 𝑒2πœ”2 𝑑 ∫
𝑑
0
π‘₯1 (𝑑) − π‘₯2 (𝑑) = π‘₯01 − π‘₯02 + ∫ (π‘₯1Μ‡ (𝑠) − π‘₯2Μ‡ (𝑠)) 𝑑𝑠,
0
0
2
𝑠
1 𝑑
{∫ 𝑒−πœ”2 𝜏 ‖𝐺(𝜏)β€– π‘‘πœ} 𝑑𝑠
2 𝑑𝑠 0
1 2 2πœ”2 𝑑 𝑑 −πœ”2 𝜏
𝐿 𝑒 {∫ 𝑒
‖𝐺(𝜏)β€– π‘‘πœ}
2
0
≤
1 2 2πœ”2 𝑑 1 − 𝑒−2πœ”2 𝑑 𝑑
𝐿𝑒
∫ ‖𝐺 (𝜏)β€–2 π‘‘πœ
2
2πœ”2
0
=
≤
+
σ΅„©
σ΅„©1/2
≤ 𝐢0 σ΅„©σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐷(𝐴))
𝑑
󡄨2
󡄨
∫ (‖𝑋 (𝑠)β€–2 + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨 ) 𝑑𝑠.
0
(47)
The second term of the right hand side of (45) is estimated as
𝑑
󡄨
󡄨
𝐼𝐼 = 𝑒2πœ”2 𝑑 ∫ 𝑒−πœ”2 𝑠 (𝐿 ‖𝑋 (𝑠)β€– + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨) 𝑑𝑠
0
1 2πœ”2 𝑑 2 𝑑
󡄨2
󡄨
𝑒 𝐿 ∫ (‖𝑋 (𝑠)β€–2 + 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨 ) 𝑑𝑠
2
0
(48)
1
󡄨
󡄨2
+ 𝑒2πœ”2 𝑑 󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨 .
2
Thus, from (47) and (48), we apply Gronwall’s inequality to
(15), and we arrive at
𝑑
1
|𝑋 (𝑑)|2 + πœ”1 ∫ ‖𝑋 (𝑠)β€–2 𝑑𝑠
2
0
𝑇1
𝑇 1/2 σ΅„©
σ΅„©
σ΅„©1/2
σ΅„©1/2
× {𝑇1/4 σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„© + ( ) σ΅„©σ΅„©σ΅„©π‘₯1 − π‘₯2 σ΅„©σ΅„©σ΅„©π‘Š1,2 (0,𝑇;𝐻) }
√2
σ΅„©
σ΅„©1/2 σ΅„©
σ΅„©1/2
≤ 𝐢0 𝑇1/4 σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„© σ΅„©σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐷(𝐴))
𝑇 1/2 σ΅„©σ΅„©
σ΅„©
) σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐷(𝐴))∩π‘Š1,2 (0,𝑇;𝐻)
√2
σ΅„©
σ΅„©
≤ 2−7/4 𝐢0 σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„©
+ 𝐢0 (
󡄨
󡄨
× σ΅„¨σ΅„¨σ΅„¨π‘₯01 − π‘₯02 󡄨󡄨󡄨
≤
(52)
σ΅„©σ΅„©
σ΅„©
σ΅„©
σ΅„©1/2
σ΅„©
σ΅„©1/2
σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝑉) ≤ 𝐢0 σ΅„©σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐷(𝐴)) σ΅„©σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐻)
𝐿
(𝑒2πœ”2 𝑑 − 1) ∫ ‖𝐺 (𝑠)β€–2 𝑑𝑠
4πœ”2
0
2πœ”2
𝑇 σ΅„©σ΅„©
σ΅„©
σ΅„©π‘₯ − π‘₯2 σ΅„©σ΅„©σ΅„©π‘Š1,2 (0,𝑇;𝐻) .
√2 σ΅„© 1
Hence arguing as in (9) we get
𝑑
𝐿2 (𝑒2πœ”2 𝑑 − 1)
(51)
we get, noting that | ⋅ | ≤ || ⋅ ||,
σ΅„©
σ΅„©
σ΅„©
σ΅„©σ΅„©
σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐻) ≤ √𝑇 σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„©
2
=
2
Since
𝑑
𝐼𝐼𝐼 = 𝐿2 𝑒2πœ”2 𝑑 ∫ 𝑒−πœ”2 𝑠 ‖𝐺 (𝑠)β€– ∫ 𝑒−πœ”2 𝜏 ‖𝐺 (𝜏)β€– π‘‘πœπ‘‘π‘ 
0
where 𝐢 > 0 is a constant. Suppose (π‘₯0𝑛 , 𝑒𝑛 ) → (π‘₯0 , 𝑒) in 𝐻×
𝐿2 (0, 𝑇; π‘Œ), and let π‘₯𝑛 and π‘₯ be the solutions (1) with (π‘₯0𝑛 , 𝑒𝑛 )
and (π‘₯0 , 𝑒), respectively. Then, by virtue of (49), we see that
π‘₯𝑛 → π‘₯ in 𝐿2 (0, 𝑇, 𝑉) ∩ 𝐢([0, 𝑇]; 𝐻).
(2) It is easy to show that if π‘₯0 ∈ 𝑉 and 𝐡 ∈ L(π‘Œ, 𝐻),
then π‘₯ belongs to 𝐿2 (0, 𝑇; 𝐷(𝐴))∩π‘Š1,2 (0, 𝑇; 𝐻). Let (π‘₯0𝑖 , 𝑒𝑖 ) ∈
𝑉 × πΏ2 (0, 𝑇; 𝐻), and π‘₯𝑖 be the solution of (1) with (π‘₯0𝑖 , 𝑒𝑖 ) in
place of (π‘₯0 , 𝑒) for 𝑖 = 1, 2. Then in view of Lemma 2 and
assumption (F), we have
σ΅„©
σ΅„©σ΅„©
σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐷(𝐴))∩π‘Š1,2 (0,𝑇;𝐻)
σ΅„© σ΅„©
σ΅„©
σ΅„©
≤ 𝐢1 {σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„© + 󡄩󡄩󡄩𝑓 (⋅, π‘₯1 ) − 𝑓(⋅, π‘₯2 )󡄩󡄩󡄩𝐿2 (0,𝑇;𝐻)
σ΅„©
σ΅„©
+󡄩󡄩󡄩𝐡 (𝑒1 − 𝑒2 )󡄩󡄩󡄩𝐿2 (0,𝑇;𝐻) }
(50)
σ΅„© σ΅„©
σ΅„©
σ΅„©
≤ 𝐢1 {σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„© + 󡄩󡄩󡄩𝐡 (𝑒1 − 𝑒2 )󡄩󡄩󡄩𝐿2 (0,𝑇;𝐻)
σ΅„©
σ΅„©
+𝐿󡄩󡄩󡄩π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇:𝑉) } .
󡄨2
󡄨
󡄨
󡄨2
≤ 𝐢 (󡄨󡄨󡄨π‘₯01 − π‘₯02 󡄨󡄨󡄨 + ∫ 󡄨󡄨󡄨𝐡 (𝑒1 (𝑠) − 𝑒2 (𝑠))󡄨󡄨󡄨 𝑑𝑠) ,
0
(49)
+ 2𝐢0 (
𝑇 1/2 σ΅„©σ΅„©
σ΅„©
) σ΅„©σ΅„©π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐷(𝐴))∩π‘Š1,2 (0,𝑇;𝐻) .
√2
Combining (50) and (53) we obtain
σ΅„©σ΅„©σ΅„©π‘₯1 − π‘₯2 σ΅„©σ΅„©σ΅„© 2
σ΅„©
󡄩𝐿 (0,𝑇;𝐷(𝐴))∩π‘Š1,2 (0,𝑇;𝐻)
σ΅„© σ΅„©
σ΅„©
σ΅„©
≤ 𝐢1 {σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„©} + 󡄩󡄩󡄩𝐡𝑒1 − 𝐡𝑒2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐻)
σ΅„©
σ΅„©
+ 2−7/4 𝐢0 𝐢1 𝐿 σ΅„©σ΅„©σ΅„©π‘₯01 − π‘₯02 σ΅„©σ΅„©σ΅„©
+ 2𝐢0 𝐢1 (
𝑇 1/2 σ΅„©σ΅„©
σ΅„©
) 𝐿󡄩󡄩π‘₯1 − π‘₯2 󡄩󡄩󡄩𝐿2 (0,𝑇;𝐷(𝐴))∩π‘Š1,2 (0,𝑇;𝐻) .
√2
(53)
(54)
6
Abstract and Applied Analysis
Suppose that
(π‘₯0𝑛 , 𝑒𝑛 ) 󳨃󳨀→ (π‘₯0 , 𝑒) ∈ 𝑉 × πΏ2 (0, 𝑇; π‘Œ) ,
(55)
and let π‘₯𝑛 and π‘₯ be the solutions (1) with (π‘₯0𝑛 , 𝑒𝑛 ) and (π‘₯0 , 𝑒),
respectively. Let 0 < 𝑇1 ≤ 𝑇 be such that
2𝐢0 𝐢1 (
𝑇1 1/2
) 𝐿 < 1.
√2
(56)
Then by virtue of (54) with 𝑇 replaced by 𝑇1 we see that
π‘₯𝑛 󳨀→ π‘₯ ∈ 𝐿2 (0, 𝑇1 ; 𝐷 (𝐴)) ∩ π‘Š1,2 (0, 𝑇1 ; 𝐻) .
(57)
This implies that (π‘₯𝑛 (𝑇1 ), (π‘₯𝑛 )𝑇1 ) 󳨃→ (π‘₯(𝑇1 ), π‘₯𝑇1 ) in 𝑉 ×
𝐿2 (0, 𝑇; 𝐷(𝐴)). Hence the same argument shows that π‘₯𝑛 󳨃→ π‘₯
in
𝐿2 (𝑇1 , min {2𝑇1 , 𝑇} ; 𝐷 (𝐴)) ∩ π‘Š1,2 (𝑇1 , min {2𝑇1 , 𝑇} ; 𝐻) .
(58)
Repeating this process we conclude that π‘₯𝑛
𝐿2 (0, 𝑇; 𝐷(𝐴)) ∩ π‘Š1,2 (0, 𝑇; 𝐻).
󳨃→
π‘₯ in
In this section we study the optimal control problems for the
quadratic cost function in the framework of Lions [9]. In what
follows we assume that the embedding 𝐷(𝐴) ⊂ 𝑉 ⊂ 𝐻 is
compact.
Let π‘Œ be another Hilbert space of control variables, and 𝐡
be a bounded linear operator from π‘Œ into 𝐻; that is,
(59)
which is called a controller. By virtue of Theorem 6, we can
define uniquely the solution map 𝑒 󳨃→ π‘₯(𝑒) of 𝐿2 (0, 𝑇; π‘Œ) into
𝐿2 (0, 𝑇; 𝑉) ∩ 𝐢([0, 𝑇]; 𝐻). We will call the solution π‘₯(𝑒) the
state of the control system (1).
Let 𝑀 be a Hilbert space of observation variables. The
observation of state is assumed to be given by
𝑧 (𝑒) = 𝐺π‘₯ (𝑒) ,
∗
𝐺 ∈ L (𝐢 (0, 𝑇; 𝑉 ) , 𝑀) ,
(60)
where 𝐺 is an operator called the observer. The quadratic cost
function associated with the control system (1) is given by
σ΅„©2
σ΅„©
𝐽 (𝑣) = 󡄩󡄩󡄩𝐺π‘₯ (𝑣) − 𝑧𝑑 󡄩󡄩󡄩𝑀
+ (𝑅𝑣, 𝑣)𝐿2 (0,𝑇;π‘Œ)
for 𝑣 ∈ 𝐿2 (0, 𝑇; π‘Œ) ,
(61)
where 𝑧𝑑 ∈ 𝑀 is a desire value of π‘₯(𝑣) and 𝑅 ∈ L(𝐿2 (0, 𝑇; π‘Œ))
is symmetric and positive; that is,
(𝑅𝑣, 𝑣)𝐿2 (0,𝑇;π‘Œ) = (𝑣, 𝑅𝑣)𝐿2 (0,𝑇;π‘Œ) ≥
𝑑‖𝑣‖2𝐿2 (0,𝑇;π‘Œ)
W = 𝐿2 (0, 𝑇; 𝑉) ∩ π‘Š1,2 (0, 𝑇; 𝑉∗ ) ⊂ 𝐢 ([0, 𝑇] ; 𝐻)
(63)
endowed with the norm
β€–⋅β€–W = max {β€–⋅‖𝐿2 (0,𝑇;𝑉) , β€–⋅β€–π‘Š1,2 (0,𝑇;𝑉∗ ) } .
(64)
Let Ω be an open bounded and connected set of R𝑛
with smooth boundary. We consider the observation 𝐺 of
distributive and terminal values (see [15, 16]).
(1) We take 𝑀 = 𝐿2 ((0, 𝑇)×Ω)×𝐿2 (Ω) and 𝐺 ∈ L(W, 𝑀)
and observe
𝑧 (𝑣) = 𝐺π‘₯ (𝑣)
= (π‘₯ (𝑣; ⋅) , π‘₯ (𝑣, 𝑇)) ∈ 𝐿2 ((0, 𝑇) × Ω) × πΏ2 (Ω) .
(65)
(2) We take 𝑀 = 𝐿2 ((0, 𝑇) × Ω) and 𝐺 ∈ L(W, 𝑀) and
observe
𝑧 (𝑣) = 𝐺π‘₯ (𝑣) = 𝑦󸀠 (𝑣; ⋅) ∈ 𝐿2 ((0, 𝑇) × Ω) .
(66)
The above observations are meaningful in view of the regularity of (1) by Proposition 3.
3. Optimal Control Problems
𝐡 ∈ L (π‘Œ, 𝐻) ,
Remark 7. The solution space W of strong solutions of (1) is
defined by
(62)
for some 𝑑 > 0. Let Uad be a closed convex subset of
𝐿2 (0, 𝑇; π‘Œ), which is called the admissible set. An element
𝑒 ∈ Uad which attains minimum of 𝐽(𝑣) over Uad is called
an optimal control for the cost function (61).
Theorem 8. (1) Let the assumption (F) be satisfied. Assume
that 𝐡 ∈ L(π‘Œ, 𝑉∗ ) and π‘₯0 ∈ 𝐷(πœ™). Let π‘₯(𝑒) be the solution of
(1) corresponding to 𝑒. Then the mapping 𝑒 󳨃→ π‘₯(𝑒) is compact
from 𝐿2 (0, 𝑇; π‘Œ) to 𝐿2 (0, 𝑇; 𝐻).
(2) Let the assumptions (A) and (F) be satisfied. If 𝐡 ∈
L(π‘Œ, 𝐻) and π‘₯0 ∈ 𝐷(πœ™) ∩ 𝑉, then the mapping 𝑒 󳨃→ π‘₯(𝑒)
is compact from 𝐿2 (0, 𝑇; π‘Œ) to 𝐿2 (0, 𝑇; 𝑉).
Proof. (1) We define the solution mapping 𝑆 from 𝐿2 (0, 𝑇; π‘Œ)
to 𝐿2 (0, 𝑇; 𝐻) by
𝑆𝑒 = π‘₯ (𝑒) ,
𝑒 ∈ 𝐿2 (0, 𝑇; π‘Œ) .
(67)
In virtue of Lemma 2, we have
‖𝑆𝑒‖𝐿2 (0,𝑇;𝑉)∩π‘Š1,2 (0,𝑇;𝑉∗ )
󡄨 󡄨
= β€–π‘₯ (𝑒)β€– ≤ 𝐢1 {󡄨󡄨󡄨π‘₯0 󡄨󡄨󡄨 + ‖𝐡𝑒‖𝐿2 (0,𝑇;𝑉∗ ) } .
(68)
Hence if 𝑒 is bounded in 𝐿2 (0, 𝑇; π‘Œ), then so is π‘₯(𝑒) in
𝐿2 (0, 𝑇; 𝑉) ∩ π‘Š1,2 (0, 𝑇; 𝑉∗ ). Since 𝑉 is compactly embedded in 𝐻 by assumption, the embedding 𝐿2 (0, 𝑇; 𝑉) ∩
π‘Š1,2 (0, 𝑇; 𝑉∗ ) ⊂ 𝐿2 (0, 𝑇; 𝐻) is also compact in view of Theorem 2 of Aubin [17]. Hence, the mapping 𝑒 󳨃→ 𝑆𝑒 = π‘₯(𝑒) is
compact from 𝐿2 (0, 𝑇; π‘Œ) to 𝐿2 (0, 𝑇; 𝐻).
(2) If 𝐷(𝐴) is compactly embedded in 𝑉 by assumption,
the embedding
𝐿2 (0, 𝑇; 𝐷 (𝐴)) ∩ π‘Š1,2 (0, 𝑇; 𝐻) ⊂ 𝐿2 (0, 𝑇; 𝑉)
(69)
is compact. Hence, the proof of (2) is complete.
As indicated in the Introduction we need to show the existence of an optimal control and to give the characterizations
of them. The existence of an optimal control 𝑒 for the cost
function (61) can be stated by the following theorem.
Abstract and Applied Analysis
7
Theorem 9. Let the assumptions (A) and (F) be satisfied and
π‘₯0 ∈ 𝐷(πœ™) ∩ 𝑉. Then there exists at least one optimal control
𝑒 for the control problem (1) associated with the cost function
(61); that is, there exists 𝑒 ∈ Uad such that
𝐽 (𝑒) = inf 𝐽 (𝑣) := 𝐽.
π‘₯σΈ€  (𝑑) + 𝐴π‘₯ (𝑑) + πœ•πœ™ (π‘₯ (𝑑)) ∋ 𝑓 (𝑑, π‘₯ (𝑑))
+ 𝐡𝑒 (𝑑) ,
(70)
𝑣∈Uad
Proof. Since Uad is nonempty, there is a sequence {𝑒𝑛 } ⊂ Uad
such that minimizing sequence for the problem (70) satisfies
inf 𝐽 (𝑣) = lim 𝐽 (𝑒𝑛 ) = π‘š.
𝑛→∞
𝑣∈Uad
Thus we have proved that π‘₯(𝑑) satisfies a.e. on (0, 𝑇) the following equation:
(71)
Obviously, {𝐽(𝑒𝑛 )} is bounded. Hence by (62) there is a positive constant 𝐾0 such that
σ΅„© σ΅„©2
𝑑󡄩󡄩󡄩𝑒𝑛 σ΅„©σ΅„©σ΅„© ≤ (𝑅𝑒𝑛 , 𝑒𝑛 ) ≤ 𝐽 (𝑒𝑛 ) ≤ 𝐾0 .
π‘₯𝑛󸀠 (𝑑) + 𝐴π‘₯𝑛 (𝑑) + πœ•πœ™ (π‘₯𝑛 (𝑑)) ∋ 𝑓 (𝑑, π‘₯𝑛 (𝑑)) + 𝐡𝑒𝑛 (𝑑) ,
0 < 𝑑 ≤ 𝑇, (73)
π‘₯𝑛 (0) = π‘₯0 .
(80)
π‘₯ (0) = π‘₯0 .
Since 𝐺 is continuous and || ⋅ ||𝑀 is lower semicontinuous, it
holds that
󡄩󡄩󡄩𝐺π‘₯(𝑒) − 𝑧𝑑 σ΅„©σ΅„©σ΅„© ≤ lim inf 󡄩󡄩󡄩𝐺π‘₯ (𝑒𝑛 ) − 𝑧𝑑 σ΅„©σ΅„©σ΅„© .
(81)
󡄩𝑀
󡄩𝑀
σ΅„©
𝑛→∞ σ΅„©
It is also clear from lim inf 𝑛 → ∞ ‖𝑅1/2 𝑒𝑛 ‖𝐿2 (0,𝑇;π‘Œ)
‖𝑅1/2 𝑒‖𝐿2 (0,𝑇;π‘Œ) that
(72)
This shows that {𝑒𝑛 } is bounded in Uad . So we can extract a
subsequence (denoted again by {𝑒𝑛 }) of {𝑒𝑛 } and find a 𝑒 ∈
Uad such that 𝑀 − lim 𝑒𝑛 = 𝑒 in π‘ˆ. Let π‘₯𝑛 = π‘₯(𝑒𝑛 ) be the
solution of the following equation corresponding to 𝑒𝑛 :
a.e., 0 < 𝑑 ≤ 𝑇,
≥
lim inf (𝑅𝑒𝑛 , 𝑒𝑛 )𝐿2 (0,𝑇;π‘Œ) ≥ (𝑅𝑒, 𝑒)𝐿2 (0,𝑇;π‘Œ) .
(82)
π‘š = lim 𝐽 (𝑒𝑛 ) ≥ 𝐽 (𝑒) .
(83)
𝑛→∞
Thus,
𝑛→∞
But since 𝐽(𝑒) ≥ π‘š by definition, we conclude 𝑒 ∈ Uad is a
desired optimal control.
4. Necessary Conditions for Optimality
By (15) and (17) we know that {π‘₯𝑛 } and {π‘₯𝑛󸀠 } are bounded
in 𝐿2 (0, 𝑇; 𝑉) and 𝐿2 (0, 𝑇; 𝑉∗ ), respectively. Therefore, by the
extraction theorem of Rellich’s, we can find a subsequence of
{π‘₯𝑛 }, say again {π‘₯𝑛 }, and find π‘₯ such that
2
π‘₯𝑛 (⋅) 󳨀→ π‘₯ (⋅) weakly in 𝐿 (0, 𝑇; 𝑉) ∩ 𝐢 ([0, 𝑇] ; 𝐻) ,
π‘₯𝑛󸀠 󳨀→ π‘₯σΈ€  , weakly in 𝐿2 (0, 𝑇; 𝑉∗ ) .
(74)
However, by Theorem 8, we know that
π‘₯𝑛 (⋅) 󳨀→ π‘₯ (⋅) , strongly in 𝐿2 (0, 𝑇; 𝑉) .
(75)
(76)
By the boundedness of 𝐴 we have
𝐴π‘₯𝑛 󳨀→ 𝐴π‘₯, strongly in 𝐿2 (0, 𝑇; 𝑉∗ ) .
(77)
Since πœ•πœ™(π‘₯𝑛 ) are uniformly bounded from (73)–(77) it follows
that
πœ•πœ™ (π‘₯𝑛 ) 󳨀→ 𝑓 (⋅, π‘₯) + 𝐡𝑒 − π‘₯σΈ€  − 𝐴π‘₯,
weakly in 𝐿2 (0, 𝑇; 𝑉∗ ) ,
(78)
and noting that πœ•πœ™ is demiclosed, we have that
𝑓 (⋅, π‘₯) + 𝐡𝑒 − π‘₯σΈ€  − 𝐴π‘₯ ∈ πœ•πœ™ (π‘₯) in 𝐿2 (0, 𝑇; 𝑉∗ ) .
𝐷𝐽 (𝑒) (𝑣 − 𝑒) ≥ 0,
𝑣 ∈ Uad
(84)
and to analyze (84) in view of the proper adjoint state system,
where 𝐷𝐽(𝑒) denote the GaΜ‚teaux derivative of 𝐽(𝑣) at 𝑣 = 𝑒.
Therefore, we have to prove that the solution mapping 𝑣 󳨃→
π‘₯(𝑣) is GaΜ‚teaux differentiable at 𝑣 = 𝑒. Here we note that from
Theorem 6 it follows immediately that
lim π‘₯ (𝑒 + πœ†π‘€)
πœ†→0
= π‘₯ (𝑒) , strongly in 𝐿2 (0, 𝑇; 𝑉) ∩ 𝐢 ([0, 𝑇] ; 𝐻) .
From (F) it follows that
𝑓 (⋅, π‘₯𝑛 ) 󳨀→ 𝑓 (⋅, π‘₯) , strongly in 𝐿2 (0, 𝑇; 𝐻) .
In this section we will characterize the optimal controls by
giving necessary conditions for optimality. For this it is necessary to write down the necessary optimal condition
(79)
(85)
The solution map 𝑣 󳨃→ π‘₯(𝑣) of 𝐿2 (0, 𝑇; π‘Œ) into 𝐿2 (0, 𝑇; 𝑉) ∩
𝐢([0, 𝑇]; 𝐻) is said to be GaΜ‚teaux differentiable at 𝑣 =
𝑒 if for any 𝑀 ∈ 𝐿2 (0, 𝑇; π‘Œ) there exists a 𝐷π‘₯(𝑒) ∈
L(𝐿2 (0, 𝑇; π‘Œ), 𝐿2 (0, 𝑇; 𝑉) ∩ 𝐢([0, 𝑇]; 𝐻) such that
σ΅„©σ΅„© 1
σ΅„©
σ΅„©σ΅„© (π‘₯ (𝑒 + πœ†π‘€) − π‘₯ (𝑒)) − 𝐷π‘₯ (𝑒) 𝑀󡄩󡄩󡄩 󳨀→ 0
σ΅„©σ΅„©
σ΅„©σ΅„©
σ΅„©πœ†
σ΅„©
as πœ† 󳨀→ 0.
(86)
The operator 𝐷π‘₯(𝑒) denotes the GaΜ‚teaux derivative of π‘₯(𝑒) at
𝑣 = 𝑒 and the function 𝐷π‘₯(𝑒)𝑀 ∈ 𝐿2 (0, 𝑇; 𝑉)∩𝐢([0, 𝑇]; 𝐻)) is
called the GaΜ‚teaux derivative in the direction 𝑀 ∈ 𝐿2 (0, 𝑇; π‘Œ),
which plays an important part in the nonlinear optimal
control problems.
First, as is seen in Corollary 2.2 of Chapter II of [18], let
us introduce the regularization of πœ™ as follows.
8
Abstract and Applied Analysis
Lemma 10. For every πœ– > 0, define
and, by the relation (12),
σ΅„©σ΅„©
σ΅„©2
σ΅„©π‘₯ − π½πœ– π‘₯σ΅„©σ΅„©σ΅„©∗
+ πœ™ (π½πœ– π‘₯) : ∀πœ– > 0, π‘₯ ∈ 𝐻} ,
πœ™πœ– (π‘₯) = { σ΅„©
2πœ–
(87)
where π½πœ– = (𝐼 + πœ–πœ™)−1 . Then the function πœ™πœ– is Fréchet differentiable on 𝐻 and its Frećhet differential πœ•πœ™πœ– is Lipschitz continuous on 𝐻 with Lipschitz constant πœ–−1 . In addition,
lim πœ™πœ– (π‘₯) = πœ™ (π‘₯) ,
∀π‘₯ ∈ 𝐻,
πœ–→0
πœ™ (π½πœ– π‘₯) ≤ πœ™πœ– (π‘₯) ≤ πœ™ (π‘₯) ,
0
lim πœ• πœ™πœ– (π‘₯) = (πœ•πœ™) (π‘₯) ,
(88)
∀π‘₯ ∈ 𝐻,
where (πœ•πœ™)0 (π‘₯) is the element of minimum norm in the set
πœ•πœ™(π‘₯).
Now, we introduce the smoothing system corresponding
to (1) as follows.
π‘₯σΈ€  (𝑑) + 𝐴π‘₯ (𝑑) + πœ•πœ™πœ– (π‘₯ (𝑑))
= 𝑓 (𝑑, π‘₯ (𝑑)) + 𝐡𝑒 (𝑑) ,
0 < 𝑑 ≤ 𝑇,
(89)
π‘₯ (0) = π‘₯0 .
Lemma 11. Let the assumption (F) be satisfied. Then the solution map 𝑣 󳨃→ π‘₯(𝑣) of 𝐿2 (0, 𝑇; π‘Œ) into 𝐿2 (0, 𝑇; 𝑉)∩𝐢([0, 𝑇]; 𝐻)
is Lipschtz continuous.
Moreover, let us assume the condition (A) in Proposition 3.
Then the map 𝑣 󳨃→ πœ•πœ™πœ– (π‘₯(𝑣)) of 𝐿2 (0, 𝑇; π‘Œ) into 𝐿2 (0, 𝑇; 𝐻) ∩
𝐢([0, 𝑇]; 𝑉∗ ) is also Lipschtz continuous.
Proof. We set 𝑀 = 𝑣 − 𝑒. From Theorem 6, it follows immediately that
β€–π‘₯ (𝑒 + πœ†π‘€) − π‘₯ (𝑒)‖𝐢([0,𝑇];𝐻) ≤ const. |πœ†| ‖𝑀‖𝐿2 (0,𝑇;π‘Œ) ,
2
(90)
2
so the solution map 𝑣 󳨃→ π‘₯(𝑣) of 𝐿 (0, 𝑇; π‘Œ) into 𝐿 (0, 𝑇; 𝑉) ∩
𝐢([0, 𝑇]; 𝐻) is Lipschtz continuous. Moreover, since
πœ•πœ™πœ– (π‘₯ (𝑒; 𝑑)) − πœ•πœ™πœ– (π‘₯ (𝑒 + πœ†π‘€; 𝑑))
= π‘₯σΈ€  (𝑒 + πœ†π‘€; 𝑑) − π‘₯σΈ€  (𝑒; 𝑑) + 𝐴 (π‘₯ (𝑒 + πœ†π‘€; 𝑑) − π‘₯ (𝑒; 𝑑))
− {𝑓 (𝑑, π‘₯ (𝑒 + πœ†π‘€; 𝑑)) − 𝑓 (𝑑, π‘₯ (𝑒; 𝑑))} − πœ†π΅π‘€ (𝑑) ,
(91)
by the assumption (A) and (2) of Theorem 6, it holds
σ΅„©σ΅„©σ΅„©πœ•πœ™πœ– (π‘₯ (𝑒 + πœ†π‘€)) − πœ•πœ™πœ– (π‘₯ (𝑒))σ΅„©σ΅„©σ΅„© 2
σ΅„©
󡄩𝐿 (0,𝑇;𝐻)
σ΅„©
σ΅„©σ΅„© σΈ€ 
≤ σ΅„©σ΅„©σ΅„©π‘₯ (𝑒 + πœ†π‘€) − π‘₯σΈ€  (𝑒)󡄩󡄩󡄩󡄩𝐿2 (0,𝑇;𝐻)
+ β€–π‘₯ (𝑒 + πœ†π‘€) − π‘₯ (𝑒)‖𝐿2 (0,𝑇;𝐷(𝐴))
+ 𝐿‖π‘₯ (𝑒 + πœ†π‘€) − π‘₯ (𝑒)‖𝐿2 (0,𝑇;𝑉)
+ |πœ†|‖𝐡‖‖𝑀‖𝐿2 (0,𝑇;π‘ˆ)
≤ const. |πœ†|‖𝑀‖𝐿2 (0,𝑇;π‘Œ)
+ ‖𝐴‖L(𝑉,𝑉∗ ) β€–(π‘₯ (𝑒 + πœ†π‘€; 𝑑) − π‘₯ (𝑒; 𝑑))β€–
(93)
+ 𝐿 β€–π‘₯ (𝑒 + πœ†π‘€; 𝑑) − π‘₯ (𝑒; 𝑑)β€– + |πœ†| ‖𝐡‖|𝑀 (𝑑)|
≤ const. |πœ†|‖𝑀‖𝐿2 (0,𝑇;π‘Œ) .
∀πœ– > 0, π‘₯ ∈ 𝐻,
πœ–→0
σ΅„©
σ΅„©σ΅„©
σ΅„©σ΅„©πœ•πœ™πœ– (π‘₯ (𝑒 + πœ†π‘€; 𝑑)) − πœ•πœ™πœ– (π‘₯ (𝑒; 𝑑))σ΅„©σ΅„©σ΅„©∗
σ΅„©
σ΅„©
≤ σ΅„©σ΅„©σ΅„©σ΅„©π‘₯σΈ€  (𝑒 + πœ†π‘€; 𝑑) − π‘₯σΈ€  (𝑒; 𝑑)σ΅„©σ΅„©σ΅„©σ΅„©∗
So we know that the map 𝑣 󳨃→ πœ•πœ™πœ– (π‘₯(𝑣)) of 𝐿2 (0, 𝑇; π‘Œ) into
𝐿2 (0, 𝑇; 𝐻) ∩ 𝐢([0, 𝑇]; 𝑉∗ ) is also Lipschtz continuous.
Let the solution space W1 of (1) of strong solutions is
defined by
W1 = 𝐿2 (0, 𝑇; 𝐷 (𝐴)) ∩ π‘Š1,2 (0, 𝑇; 𝐻)
(94)
as stated in Remark 7.
In order to obtain the optimality conditions, we require
the following assumptions.
(F1) The GaΜ‚teaux derivative πœ•2 𝑓(𝑑, π‘₯) in the second argument for (𝑑, π‘₯) ∈ (0, 𝑇) × π‘‰ is measurable in 𝑑 ∈ (0, 𝑇)
for π‘₯ ∈ 𝑉 and continuous in π‘₯ ∈ 𝑉 for a.e. 𝑑 ∈ (0, 𝑇),
and there exist functions πœƒ1 , πœƒ2 ∈ 𝐿2 (R+ ; R) such that
σ΅„©σ΅„©
σ΅„©
σ΅„©σ΅„©πœ•2 𝑓 (𝑑, π‘₯)σ΅„©σ΅„©σ΅„©∗ ≤ πœƒ1 (𝑑) + πœƒ2 (β€–π‘₯β€–) ,
∀ (𝑑, π‘₯) ∈ (0, 𝑇) × π‘‰.
(95)
(F2) The map π‘₯ → πœ•πœ™πœ– (π‘₯) is GaΜ‚teaux differentiable, and
the value π·πœ•πœ™πœ– (π‘₯)𝐷π‘₯(𝑒) is the GaΜ‚teaux derivative of
πœ•πœ™πœ– (π‘₯)π‘₯(𝑒) at 𝑒 ∈ 𝐿2 (0, 𝑇; π‘ˆ) such that there exist
functions πœƒ3 , πœƒ4 ∈ 𝐿2 (R+ ; R) such that
σ΅„©
σ΅„©σ΅„©
σ΅„©σ΅„©π·πœ•πœ™πœ– (π‘₯) 𝐷π‘₯ (𝑒)σ΅„©σ΅„©σ΅„©∗
≤ πœƒ3 (𝑑) + πœƒ4 (‖𝑒‖𝐿2 (0,𝑇;π‘Œ) ) ,
∀𝑒 ∈ 𝐿2 (0, 𝑇; π‘Œ) .
(96)
Theorem 12. Let the assumptions (A), (F1), and (F2) be satisfied. Let 𝑒 ∈ Uad be an optimal control for the cost function 𝐽
in (61). Then the following inequality holds:
(𝐢∗ Λ π‘€ (𝐢π‘₯ (𝑒) − 𝑧𝑑 ) , 𝑦)W1
+ (𝑅𝑒, 𝑣 − 𝑒)𝐿2 (0,𝑇;π‘Œ) ≥ 0,
∀𝑣 ∈ Uad ,
(97)
where 𝑦 = 𝐷π‘₯(𝑒)(𝑣 − 𝑒) ∈ 𝐢([0, 𝑇]; 𝑉∗ ) is a unique solution
of the following equation:
0
𝑦󸀠 (𝑑) + 𝐴𝑦 (𝑑) + 𝐷(πœ•πœ™) (π‘₯) (𝑦 (𝑑))
= πœ•2 𝑓 (𝑑, π‘₯) 𝑦 (𝑑) + 𝐡𝑀 (𝑑) ,
(92)
0 < 𝑑 ≤ 𝑇,
(98)
𝑦 (0) = 0.
Proof. We set 𝑀 = 𝑣 − 𝑒. Let πœ† ∈ (−1, 1), πœ† =ΜΈ 0. We set
𝑦 = lim πœ†−1 (π‘₯ (𝑒 + πœ†π‘€) − π‘₯ (𝑒)) = 𝐷π‘₯ (𝑒) 𝑀.
πœ†→0
(99)
Abstract and Applied Analysis
9
Proof. Let π‘₯(𝑑) = π‘₯0 (𝑑) be a solution of (1) associated with the
control 0. Then it holds that
From (89), we have
σΈ€ 
σΈ€ 
π‘₯ (𝑒 + πœ†π‘€) − π‘₯ (𝑒) + 𝐴 (π‘₯ (𝑒 + πœ†π‘€) − π‘₯ (𝑒))
+ πœ•πœ™πœ– (π‘₯ (𝑒 + πœ†π‘€)) − πœ•πœ™πœ– (π‘₯ (𝑒))
(100)
= 𝑓 (⋅, π‘₯ (𝑒 + πœ†π‘€)) − 𝑓 (⋅, π‘₯ (𝑒)) + πœ†π΅π‘€.
𝑇
𝑇
σ΅„©2
σ΅„©
= ∫ 󡄩󡄩󡄩𝐢 (π‘₯𝑣 (𝑑) − π‘₯ (𝑑)) + 𝐢π‘₯ (𝑑) − 𝑧𝑑 (𝑑)󡄩󡄩󡄩𝑋 𝑑𝑑
0
Then as an immediate consequence of Lemma 11 one obtains
lim
1
πœ†→0πœ†
{πœ•πœ™πœ– (π‘₯ (𝑒 + πœ†π‘€; 𝑑)) − πœ•πœ™πœ– (π‘₯ (𝑒; 𝑑))} = π·πœ•πœ™πœ– (π‘₯) 𝑦 (𝑑) ,
𝑇
+ ∫ (𝑅𝑣 (𝑑) , 𝑣 (𝑑)) 𝑑𝑑
0
1
{𝑓 (𝑑, π‘₯ (𝑒 + πœ†π‘€; 𝑑)) − 𝑓 (𝑑, π‘₯ (𝑒; 𝑑))} = πœ•2 𝑓 (𝑑, π‘₯)𝑦 (𝑑) ,
πœ†→0πœ†
(101)
lim
thus, in the sense of (F2), we have that 𝑦 = 𝐷π‘₯(𝑒)(𝑣 − 𝑒)
satisfies (98) and the cost 𝐽(𝑣) is GaΜ‚teaux differentiable at 𝑒
in the direction 𝑀 = 𝑣 − 𝑒. The optimal condition (84) is
rewritten as
(𝐢π‘₯ (𝑒) − 𝑧𝑑 , 𝑦)𝑀 + (𝑅𝑒, 𝑣 − 𝑒)𝐿2 (0,𝑇;π‘Œ)
= (𝐢∗ Λ π‘€ (𝐢π‘₯ (𝑒) − 𝑧𝑑 ) , 𝑦)W1
+ (𝑅𝑒, 𝑣 − 𝑒)𝐿2 (0,𝑇;π‘Œ) ≥ 0,
𝑇
σ΅„©
σ΅„©2
𝐽1 (𝑣) = ∫ 󡄩󡄩󡄩𝐢π‘₯𝑣 (𝑑) − 𝑧𝑑 (𝑑)󡄩󡄩󡄩𝑋 𝑑𝑑 + ∫ (𝑅𝑣 (𝑑) , 𝑣 (𝑑)) 𝑑𝑑
0
0
𝑇
σ΅„©2
σ΅„©
= πœ‹ (𝑣, 𝑣) − 2𝐿 (𝑣) + ∫ 󡄩󡄩󡄩𝑧𝑑 (𝑑) − 𝐢π‘₯ (𝑑)󡄩󡄩󡄩𝑋 𝑑𝑑,
0
(108)
where
𝑇
πœ‹ (𝑒, 𝑣) = ∫ (𝐢 (π‘₯𝑒 (𝑑) − π‘₯ (𝑑)) , 𝐢 (π‘₯𝑣 (𝑑) − π‘₯ (𝑑)))𝑋 𝑑𝑑
0
(102)
𝑇
+ ∫ (𝑅𝑒 (𝑑) , 𝑣 (𝑑)) 𝑑𝑑,
∀𝑣 ∈ Uad .
0
𝑇
With every control 𝑒 ∈ 𝐿2 (0, 𝑇; π‘Œ), we consider the
following distributional cost function expressed by
𝑇
0
(109)
𝑇
σ΅„©2
σ΅„©
𝐽1 (𝑒) = ∫ 󡄩󡄩󡄩𝐢π‘₯𝑒 (𝑑) − 𝑧𝑑 (𝑑)󡄩󡄩󡄩𝑋 𝑑𝑑 + ∫ (𝑅𝑒 (𝑑) , 𝑒 (𝑑)) 𝑑𝑑,
0
0
(103)
where the operator 𝐢 is bounded from 𝐻 to another Hilbert
space 𝑋 and 𝑧𝑑 ∈ 𝐿2 (0, 𝑇; 𝑋). Finally we are given that 𝑅 is a
self adjoint and positive definite:
𝑅 ∈ L (𝑋) ,
𝐿 (𝑣) = ∫ (𝑧𝑑 (𝑑) − 𝐢π‘₯ (𝑑) , 𝐢 (π‘₯𝑣 (𝑑) − π‘₯ (𝑑)))𝑋 𝑑𝑑.
(𝑅𝑒, 𝑒) ≥ 𝑐 ‖𝑒‖ ,
𝑐 > 0.
(104)
Let π‘₯𝑒 (𝑑) stand for solution of (1) associated with the control
𝑒 ∈ 𝐿2 (0, 𝑇; π‘Œ). Let Uad be a closed convex subset of
𝐿2 (0, 𝑇; π‘Œ).
Theorem 13. Let the assumptions in Theorem 12 be satisfied
and let the operators 𝐢 and 𝑁 satisfy the conditions mentioned
above. Then there exists an element 𝑒 ∈ Uad such that
𝐽1 (𝑒) = inf 𝐽1 (𝑣) .
0
𝑣 ∈ 𝐿2 (0, 𝑇; π‘Œ) .
(110)
If 𝑒 is an optimal control, similarly for (97), (84) is equivalent
to
𝑇
∫ (𝐢∗ Λ π‘‹ (𝐢π‘₯𝑒 (𝑑) − 𝑧𝑑 (𝑑)) , 𝑦 (𝑑)) 𝑑𝑑
0
(111)
𝑇
+ ∫ (𝑅𝑒 (𝑑) , (𝑣 − 𝑒) (𝑑)) 𝑑𝑑 ≥ 0.
0
Now we formulate the adjoint system to describe the optimal
condition:
Furthermore, the following inequality holds:
𝑇
πœ‹ (𝑣, 𝑣) ≥ 𝑐‖𝑣‖2 ,
(105)
𝑣∈Uad
∗
∫ (Λ−1
π‘Œ 𝐡 𝑝𝑒 (𝑑) + 𝑅𝑒 (𝑑) , (𝑣 − 𝑒) (𝑑)) 𝑑𝑑 ≥ 0,
The form πœ‹(𝑒, 𝑣) is a continuous form in 𝐿2 (0, 𝑇; π‘Œ) ×
𝐿2 (0, 𝑇; π‘Œ) and from assumption of the positive definite of the
operator 𝑅, we have
∀𝑣 ∈ Uad ,
(106)
∗
holds, where Λ π‘Œ is the canonical isomorphism π‘Œ onto π‘Œ and
𝑝𝑒 satisfies the following equation:
𝑝𝑒󸀠 (𝑑) − 𝐴∗ 𝑝𝑒 (𝑑) − π·πœ•πœ™πœ– (π‘₯)∗ 𝑝𝑒 (𝑑) + πœ•2 𝑓(𝑑, π‘₯)∗ 𝑝𝑒 (𝑑)
= − (𝐢∗ Λ π‘‹ 𝐢π‘₯𝑒 (𝑑) − 𝑧𝑑 (𝑑)) ,
for 0 < 𝑑 ≤ 𝑇,
(112)
𝑃𝑒 (𝑇) = 0.
0
𝑝𝑒󸀠 (𝑑) − 𝐴∗ 𝑝𝑒 (𝑑) − 𝐷(πœ•πœ™) (π‘₯)∗ 𝑝𝑒 (𝑑) + πœ•2 𝑓(𝑑, π‘₯)∗ 𝑝𝑒 (𝑑)
∗
= −𝐢 Λ π‘‹ (𝐢π‘₯𝑒 (𝑑) − 𝑧𝑑 (𝑑)) ,
for 0 < 𝑑 ≤ 𝑇,
𝑃𝑒 (𝑇) = 0.
(107)
Taking into account the regularity result of Proposition 3
and the observation conditions, we can assert that (112)
admits a unique weak solution 𝑝𝑒 reversing the direction of
time 𝑑 → 𝑇 − 𝑑 by referring to the well-posedness result of
Dautray and Lions [19, pages 558–570].
10
Abstract and Applied Analysis
We multiply both sides of (112) by 𝑦(𝑑) of (98) and integrate it over [0, 𝑇]. Then we have
𝑇
∫ (𝐢∗ Λ π‘‹ (𝐢π‘₯𝑒 (𝑑) − 𝑧𝑑 (𝑑)) , 𝑦 (𝑑)) 𝑑𝑑
0
𝑇
𝑇
0
0
= − ∫ (𝑝𝑒󸀠 (𝑑) , 𝑦 (𝑑)) 𝑑𝑑 + ∫ (𝐴∗ 𝑝𝑒 (𝑑) , 𝑦 (𝑑)) 𝑑𝑑
𝑇
(113)
∗
+ ∫ (π·πœ•πœ™πœ– (π‘₯) 𝑝𝑒 (𝑑) , 𝑦 (𝑑)) 𝑑𝑑
0
𝑇
− ∫ (πœ•2 𝑓(𝑑, π‘₯)∗ 𝑝𝑒 (𝑑) , 𝑦 (𝑑)) 𝑑𝑑.
0
By the initial value condition of 𝑦 and the terminal value
condition of 𝑝𝑒 , the left hand side of (113) yields
− (𝑝𝑒 (𝑇) , 𝑦 (𝑇)) + (𝑝𝑒 (0) , 𝑦 (0))
𝑇
𝑇
0
0
+ ∫ (𝑝𝑒 (𝑑) , 𝑦󸀠 (𝑑)) 𝑑𝑑 + ∫ (𝑝𝑒 (𝑑) , 𝐴𝑦 (𝑑)) 𝑑𝑑
𝑇
+ ∫ (𝑝𝑒 (𝑑) , π·πœ•πœ™πœ– (π‘₯) 𝑦 (𝑑)) 𝑑𝑑
0
(114)
𝑇
− ∫ (𝑝𝑒 (𝑑) , πœ•2 𝑓 (𝑑, π‘₯) 𝑦 (𝑑)) 𝑑𝑑
0
𝑇
= ∫ (𝑝𝑒 (𝑑) , 𝐡 (𝑣 − 𝑒) (𝑑)) 𝑑𝑑.
0
Let 𝑒 be the optimal control subject to (103). Then (111) is
represented by
𝑇
∫ (𝑝𝑒 (𝑑) , 𝐡 (𝑣 − 𝑒) (𝑑)) 𝑑𝑑 + ∫ (𝑅𝑒 (𝑑) , (𝑣 − 𝑒) (𝑑)) 𝑑𝑑 ≥ 0,
Ω
0
(115)
which is rewritten by (106). Note that 𝐢∗ ∈ 𝐡(𝑋∗ , 𝐻) and for
πœ™ and πœ“ in 𝐻 we have (𝐢∗ Λ π‘‹ πΆπœ“, πœ™) = βŸ¨πΆπœ“, πΆπœ™βŸ©π‘‹ , where
duality pairing is also denoted by (⋅, ⋅).
Remark 14. Identifying the antidual 𝑋 with 𝑋 we need not
use the canonical isomorphism Λ π‘‹ . However, in case where
𝑋 ⊂ 𝑉∗ this leads to difficulties since 𝐻 has already been
identified with its dual.
Acknowledgment
This research was supported by Basic Science Research
Program through the National research Foundation of Korea
(NRF) funded by the Ministry of Education, Science and
Technology (2012-0007560).
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