Advantages for controls imposed in a proper subset

In this paper, we study time optimal control problems for heat equations on $\Omega\times \mathbb{R}^+$. Two properties under consideration are the existence and the bang-bang properties of time optimal controls. It is proved that those two properties hold when controls are imposed on some proper subsets of $\Omega$; while they do not stand when controls are active on the whole $\Omega$. Besides, a new property for eigenfunctions of $-\Delta$ with Dirichlet boundary condition is revealed.


Introduction.
Let Ω be a bounded domain in R n . Let ω be a non empty and open subset of Ω. Write χ ω for its characteristic function. Consider the following controlled heat equation: y(x, t) = 0 on ∂Ω × R + , y(x, 0) = y 0 (x) in Ω, (1.1) where u is a control function taken from a control constraint set and y 0 is an initial state taken from L 2 (Ω). The solution of (1.1) corresponding to u and y 0 will be treated as a function from R + to L 2 (Ω) and denoted by y(·; u, y 0 ). The purpose of this study is to reveal the following fact: Some properties hold for some time optimal control problems of (1.1) when ω is a proper subset of Ω, but do not stand when ω = Ω. Consequently, the local control may be more effective than the global control for heat equations in some cases.
We begin with introducing time optimal control problems. Let ξ i ∞ i=1 be a complete set of eigenfunctions for −∆ with Dirichlet boundary condition such that it serves as a normalized orthonormal basis of L 2 (Ω). Write λ i ∞ i=1 , with 0 < λ 1 < λ 2 ≤ · · · < +∞, for the corresponding set of eigenvalues. Then, we take the following target set: S m = span ξ m+1 , ξ m+2 , · · · , where m ≥ 2 is arbitrarily fixed.
Next, we define, for each natural number k and each finite sequence of positive numbers {ā i } k i=1 , the following control constraint set: Consider the following time optimal control problem: (P) inf t ≥ 0 y(t; u, y 0 ) ∈ S m , where the infimum is taken over all .
Two properties of Problem (P) under consideration are as follows: (i) The existence of time optimal controls; (ii) The bang-bang property: any optimal control u * = k i=1 α * i ξ i satisfies that for each i, |α * i (t)| =ā i for almost every t ∈ (0, t * ), where t * is the optimal time. In the case that Ω, ω, k and y 0 / ∈ S are fixed, we say Problem (P) has optimal controls if for any finite sequence of positive numbers {ā i } k i=1 , it has optimal controls. When ω = Ω, y 0 / ∈ S m and k are given, Problem (P) has optimal controls if and only if there is a finite sequence of positive numbers , has optimal controls (see Remark 2.3). The main results of this paper are broadly stated as follows: (a) Suppose that ω = Ω and y 0 / ∈ S m . Then, k and y 0 are such that Problem (P) has no optimal control if and only if k < m and y 0 satisfies ( y 0 , ξ k+1 , y 0 , ξ k+2 , · · · , y 0 , ξ m ) T = 0; (1.2) (b) Suppose that ω = Ω and y 0 / ∈ S m . Assume that either k ≥ m or k < m and y 0 does not satisfy (1.2). Then, in general, Problem (P) does not hold the bang-bang property; (c) Suppose that Ω and ω satisfy accordingly the following conditions: • (D1) The eigenvalues λ 1 · · · λ m are simple, i.e., λ 1 < λ 2 < · · · < λ m , and • (D2) χ ω ξ i , ξ j = 0 for all i ∈ {1, 2, · · · , m} and j ∈ {1, 2, · · · , k}.
Then, for each k ≥ 1, each y 0 / ∈ S m and each finite sequence of positive numbers {ā i } k i=1 , Problem (P) has optimal controls and holds the bang-bang property.
It is worth mentioning that for any fixed bounded domain Ω, there are a lot of open subsets ω in Ω such that χ ω ξ i , ξ j = 0 for all i, j = 1, 2, · · · (see Theorem 4.2 for a new property of the eigenfunctions {ξ i } ∞ i=1 ); while there are a lot of bounded domains Ω such that the property (D1) holds (see Remark 4.1)).

Studies of Problem (P) where Ω = ω
The following result is another version of Theorem 2.5 in [1]. It will be used later.
Lemma 2.1. LetÂ ∈ R d×d andB ∈ R d×l , where d and l are natural numbers. Suppose that rank B ,ÂB,Â 2B , · · · ,Â d−1B = d, (2.1) and the spectrum ofÂ belongs to the left half plane of C. Then, for each finite sequence of positive numbers {b i } l i=1 and each w 0 in R d , there are at ≥ 0 and a controlβ in the set:
Theorem 2.2. Suppose ω = Ω and y 0 / ∈ S m . Then, k and y 0 are such that Problem (P) has no optimal control if and only if k < m and y 0 satisfies (1.2).
Proof. The proof will be organized in three steps as follows: Step 1. Suppose that k < m and y 0 satisfies (1.2). Then, for any finite sequence of positive numbers {ā i } k i=1 , Problem (P) has no optimal control.
y i (t)ξ i . Clearly, the controlled equation (1.1) is equivalent to the following system: Write Consider the following time optimal control problem: where the infimum is taken over all α from the control constraint set: and z(·; α, z 0 ) is the solution to the following equation: (2.5) Clearly, Problems (P) and ( P) are equivalent, i.e., t * and u * = k i=1 α * i ξ i are accordingly the optimal time and an optimal control to Problem (P) if and only if t * and (α * 1 , · · · , α * k ) T are the optimal time and an optimal control to Problem ( P) respectively.
Since ω = Ω and k < m, it follows from (2.4) that B = I k×k 0 in this case. Let Then, Equation (2.5) can be written as d dt together with the initial condition: This, along with the condition (1.2), indicates that z 2 (t) = 0, for each t > 0 and each control α . Consequently, Problem (P) has no time optimal control.
Step 2. Suppose that k < m and y 0 does not satisfy (1.2). Then, Problem (P) has optimal controls.
Let {ā i } k i=1 be a finite sequence of positive numbers. Since y 0 does not satisfy (1.2), it holds that z 2 (0) = 0. Thus, it follows from (2.6) that z 2 (t) = 0 for all t ≥ 0. Hence, Problem ( P) shares the same optimal time and optimal controls with the following time optimal control problem: , and z 1 (·; α) is the solution to the equation: According to Lemma 2.1, Problem ( P 1 ) has admissible controls. Then, by the standard argument (see either Theorem 13 and the note after it in Chapter III on Page 130 in [13] or Theorem 3.1 on Page 31 in [1]), one can easily verify that Problem ( P 1 ) has optimal controls. Consequently, Problem (P) has optimal controls.
Step 3. Suppose that k ≥ m. Then, Problem (P) admits optimal controls.
be a finite sequence of positive numbers. Since B = I m×m , 0 m×(k−m) in the case that k ≥ m, control variables α m+1 (·), · · · , α k (·) play no role in Equation (2.5) when k > m. Hence, in the case that k ≥ m, the effective controls in Problem ( P ) have the form: α = (α 1 (·), · · · , α m (·)) T . Therefore, Problem ( P) shares the same optimal time and optimal controls with the following time optimal control problem: where the infimum is taken over allα (α 1 , · · · , α m ) T from the control constraint set: and z(·;α) is the solution of the following equation: Then, by Lemma 2.1, using the same argument as that in Step 2, one can prove that Problem ( P) has optimal controls.
In summary, we complete the proof.
, has optimal controls. (b) In the case that ω = Ω and y 0 / ∈ S m , Problem (P) has optimal controls, provided either k ≥ m or k < m and y 0 does not satisfy (1.2).
be a finite sequence of positive numbers. For each i ∈ {1, · · · , k}, write , does not have the bang-bang property, if either of the following conditions stands: (i) k ≥ m and the numbers T 1 , · · · , T m are not the same; (ii) k < m, y 0 does not satisfy (1.2) and the numbers T 1 , · · · , T k are not the same.
Proof. Simply write (P) for the problem (P), with We first prove the following property (H 1 ): When k ≥ m, T and u are the optimal time and an optimal control to Problem (P) respectively, where By the equivalence of Problems (P) and ( P 2 ) (see Step 3 in the proof of Theorem 2.2), we need only to verify that T and α are the optimal time and an optimal control to Problem ( P 2 ) respectively, where α ( α 1 , · · · , α m ) T . For this purpose, we observe from direct computation that for each i ∈ {1, · · · , m}, T i and α i (·) are the optimal time and the optimal control to the following time optimal control problem: where the infimum is taken over all α i (·) from the set of all measurable functions from R + to [−ā i ,ā i ], and z i (·; α i ) solves the following equation: and ((z 1 (·; α 1 ) , · · · , z m (·; α m )) T is the solution z (·; α) to Equation (2.7) withα = α. Since z i (T i ; α i ) = 0, it holds that z i T ; α i = 0 for all i ∈ {1, 2, · · · , m}, i.e., z T ; u = 0. (2.10) Hence, the optimal time to Problem ( P 2 ) is not bigger than T . On the other hand, ifα andT > 0 are such that z T ;α = 0, then it stands that z i T ;α i = 0 for all i ∈ {1, · · · , m}.
By the optimality of T i to Problem (P i ), we see thatT ≥ T i for all i ∈ {1, · · · , m}, from which, it follows thatT ≥ T . Therefore, T is the optimal time to Problem ( P 2 ). Along with (2.10), this yields that α is an optimal control to this problem. Hence, the property (H 1 ) stands.
Since y 0 / ∈ S m , it holds that T > 0. Because T 1 , · · · , T m are not the same, there is an i 0 ∈ {1, 2, · · · , m} such that T i 0 < T . Then, it follows from (2.9) that α i 0 (t) = 0 for all t ∈ (T i 0 , T ]. Thus, the optimal control u does not satisfy the bang-bang property. Using the very similar argument to that in the proof of the property (H 1 ), one can easily show the following property (H 2 ): When k < m, y 0 does not satisfy (1.2),T andû are the optimal time and an optimal control to Problem (P), wherê Then, by the property (H 2 ), (2.9) and the assumptions that y 0 / ∈ S m and the numbers T 1 , · · · , T k are not the same, one can easily show that the optimal controlû does not satisfy the bang-bang property. This completes the proof.

Studies of Problem (P)
where ω is a proper subset of Ω Theorem 3.1. Let Ω satisfy the condition (D1). Suppose that ω holds the condition (D2). Then, for each k ≥ 1, each y 0 / ∈ S m and each finite sequence of positive numbers {ā i } k i=1 , Problem (P) has optimal controls.
Proof. By the same way as that in Step 1 of the proof of Theorem 2.2, we define the matrices A and B, and the problem ( P ). Write B ij for the element in i-th row and j-th column of B, namely, B ij = χ ω ξ i , ξ j . Let B 1 = (B 11 , · · · , B m1 ) T . We first claim that rank(B 1 , AB 1 , A 2 B 1 , · · · , A m−1 B 1 ) = m. (3.1) In fact, since which is a determinant of Vandermonde' type and equals to m i=1 B i1 k>l (λ k − λ l ). Because of conditions (D1) and (D2), this determinant is not zero, which implies (3.1). Now, according to Lemma 2.1, Problem ( P) has admissible controls. Then, by the standard argument (see either Theorem 13 and the note after it in Chapter III on Page 130 in [13] or Theorem 3.1 on Page 31 in [1]), one can easily show that Problem ( P) admits optimal controls. This, along with the equivalence of Problems (P) and ( P), completes the proof.
Remark 3.2. From the proof of the above theorem,it follows that Theorem 3.1 still stands when the condition (D2) is replaced by the following condition: Before studying the bang-bang property for Problem (P) where ω is a proper open subset of Ω, we recall the general position condition which plays an important role in the studies of the bang-bang property for linear controlled ordinary differential equations. LetÂ andB be m × m and m × k matrices respectively. LetV be a closed polyhedron in R k . We say thatV satisfies the general position condition with respect to (Â,B), if for each nonzero vector v, which is parallel to one of the edges ofV , the vectorŝ Bv,ÂBv, · · ·Â m−1B v are linearly independent. Consider the following time optimal control problem: where the infimum is taken over all measurable functions v from R + to the polyhedronV , and z(·; v, z 0 ) is the solution to the following equation: with z 0 a non-zero vector in R m . Lemma 3.3. (see [13], [4]) Suppose that the closed polyhedronV satisfies the general position condition with respect to (Â,B). Then any optimal controlū(t) to Problem (P ), if exists, takes values on the vertices ofV and has a finite number of switchings.

Theorem 3.4.
Let Ω satisfy the condition (D1). Suppose that ω satisfies the condition (D2). Then, for each k ≥ 1, each y 0 / ∈ S m and each finite sequence of positive numbers {ā i } k i=1 , Problem (P) holds the bang-bang property.
Proof. By the same way as that in Step 1 of the proof of Theorem 2.2, we define the matrices A and B, and the problem ( P). According to Lemma 3.3, Theorem 3.1 and the equivalence of Problems (P) and ( P), it suffices to prove the general condition of with respect to (A, B). Clearly, the later is equivalent to the statement that for each j ∈ {1, · · · , k}, the vectors Be j , ABe j , · · · , A m−1 Be j are linearly independent, where {e 1 , · · · , e k } is the standard basis of R k .
satisfies the general position condition with respect to (A, B). This completes the proof.

Further studies on the conditions (D1) and (D2)
In this section, we first give a remark and a theorem, which reveal accordingly some properties for eigenvalues and eigenfunctions of −∆ with Dirichlet boundary condition. From the remark, it follows that there are a lot of Ω satisfying the property (D1). From the theorem, it follows that for any bounded domain Ω in R n , there are a lot of ω ⊂ Ω where the property (D2) holds. We end this section with another remark which provides an open problem.
Remark 4.1. It is presented in [9] (see also [15], [11]) that there are a lot of Ω of class C 3 satisfies the condition (D1) in the following sense: Let Ω be a bounded open set of class C 3 in R n . For each ε ∈ (0, 1), an ε−neighborhood of Ω is defined to be the image (I +ψ)(Ω), where I is the identity map over R n and ψ ∈ C 3 (R n ; R n ), with the C 3 −norm less than ε. For each bounded open set Ω of class C 3 in R n , Write ∆ Ω for the self-adjoint operator in L 2 ( Ω) generated by the Laplacian on Ω with the homogeneous Dirichlet boundary condition. Then, for each ε ∈ (0, 1), there is an ε−neighborhood of Ω ε such that −∆ Ω ε has only simple eigenvalues.
Before presenting the theorem, we introduce the following notations: for each x ∈ R n and each ρ > 0, B ρ (x) stands for the open ball in R n , centered at x and of radius ρ; B ρ (x) denotes the closure of the ball B ρ (x); for each ρ > 0, x 1 − x 2 R n for any subsets E 1 and E 2 in R n .

Theorem 4.2.
Let Ω be a bounded domain in R n and ω be an open subset of Ω such that Ω \ ω = ∅. Then, for any ε > 0, there exists an ε 0 ∈ (0, ε) such that Ω ε 0 = ∅ and for almost every x ∈ Ω ε 0 , Proof. We recall that each eigenfunction ξ i belongs to C ∞ (Ω) (see Page 335 in [2]). Let It is obvious that By the property of harmonic functions (see Page 6 in [5]), the function ϕ(·, ·) is real analytic over Ω × R. Thus, each eigenfunction ξ i is real analytic over Ω. Write Then, for each pair (i, j), we define a function F i,j (·.·) from D to R by setting: Clearly, it is well defined. The rest of the proof will be carried out by the following three steps: Step 1. Suppose that f is a real analytic function over Ω. Define the function F : D → R by Then F is real analytic over D.
We need only to explain that F is real analytic in a small neighborhood of (x 0 , ρ 0 ) for any point (x 0 , ρ 0 ) ∈ D. First, there is a neighborhood U of (x 0 , ρ 0 ) in R n × R + such that B ρ (x) ⊂ Ω for any (x, ρ) ∈ U . Hence, the function f (x + ρη) is real analytic in (x, ρ, η) over U × B 1 (0). Extend f to a complex-valued function in (z, w, η) over a small neighborhood U c × B 1 (0) of U × B 1 (0) in C n × C × B 1 (0) by making use of the power series expansion. We then get f c (z + wη), which is real analytic over (z, w, η) ∈ U c × B 1 (0) and holomorphic in (z, w) ∈ U c for each fixed η ∈ B 1 (0). Clearly, it holds that for all (x, ρ, η) ∈ U × B 1 (0). Now we define a function F c : U c → C by setting: and define the operator∂ in the standard way: are the standard Cauchy-Riemann operators (see [6]). It follows from the holomorphic property of f c in (z, w) that∂ Hence, F c is holomorphic in U c . In particular, the function F c (·, ·) is real analytic, i.e. F (·, ·) is analytic in U . Thus, F (·, ·) is real analytic over D. Consequently, for each pair (i, j), the function F i,j (·, ·) is real analytic over D.
Step 2. For each pair (i, j), F i,j (·, ·) is not identically a constant over D.
Since (x, δ 1 ) and (x, δ 2 ) belong to D, F i,j is not identically zero over D for each pair (i, j).
Since each F i,j is real analytic and is not identically a constant over D, the set Thus, there is a subset E ⊂ (0, ∞) of zero measure such that m(W ρ ) = 0 for each ρ ∈ (0, ∞)\E.
for all i . Consider the following time optimal control problem (P ): inf {t : y(t; u, y 0 ) = 0}, where the infimum is taken over all u from the set: u i (t)ξ i each u i (·) is measurable from R + to [−a i , a i ] , and y(·; u, y 0 ) is the solution to Equation (1.1). The set U ad is called a control constraint set of the rectangular type. We say Problem (P ) has the bang-bang property if any optimal control u * = ∞ i=1 u * i (t)ξ i satisfies that for each i, u * i (t) = a i for a.e. t ∈ (0, t * ), where t * is the optimal time.
It is not clear to us what conditions are needed to obtain the bang-bang property for Problem (P ). With regard to this question, we would like to mention the following: (i) It is necessary to impose certain conditions on {a i } ∞ i=1 ∈ l 2 + to ensure the existence of optimal controls for Problem (P ) (see [8]); (ii) When U ad is replaced by the following control constraint sets of the ball type: U ad u(·) ∈ L ∞ (R + ; L 2 (Ω)) u(t) ∈ B(0, r) , where B(0, r) is the ball in L 2 (Ω), centered at the origin and of radius r > 0, the bang-bang property for the corresponding time optimal control problem (P ) has been studied (see [3], [10], [16] and [12]).