Ann. I. H. Poincaré – AN 31 (2014) 477–499
www.elsevier.com/locate/anihpc
Bang-bang property for time optimal control of semilinear heat equation
Kim Dang Phung
a,∗, Lijuan Wang
b,1, Can Zhang
baUniversité d’Orléans, Laboratoire MAPMO, CNRS UMR 7349, Fédération Denis Poisson, FR CNRS 2964, Bâtiment de Mathématiques, B.P. 6759, 45067 Orléans Cedex 2, France
bSchool of Mathematics and Statistics, Wuhan University, Wuhan 430072, China Received 5 December 2012; received in revised form 3 April 2013; accepted 25 April 2013
Available online 31 May 2013
Abstract
This paper studies the bang-bang property for time optimal controls governed by semilinear heat equation in a bounded domain with control acting locally in a subset. Also, we present the null controllability cost for semilinear heat equation and an observability estimate from a positive measurable set in time for the linear heat equation with potential.
©2013 Elsevier Masson SAS. All rights reserved.
Keywords:Semilinear heat equation; Time optimal control; Bang-bang property; Observability estimate from measurable sets
1. Introduction and main result
This paper continues the investigations carried out in[14]. Our main result deals with the bang-bang property for time optimal controls governed by semilinear heat equations with control acting locally. We complete the result in[14]
in two directions: the nonlinearity of the equation; the geometry on which the equation takes place.
LetΩ be a bounded connected open set ofRn,n1, with boundary ∂Ω of classC2. Let ωbe an open and non-empty subset ofΩand denote 1|·for the characteristic function of a set in the place where·stays. Lety0∈L2(Ω) andv∈L∞(0,+∞;L2(Ω)). Consider the following semilinear heat equation with initial datay0and external forcev:
⎧⎨
⎩
∂ty−y+f (y)=1|ωv inΩ×(0,+∞),
y=0 on∂Ω×(0,+∞),
y(·,0)=y0 inΩ.
Existence and uniqueness of the solutionyis ensured with the following assumptions:f :R→Ris globally Lipschitz and satisfies the “good-sign” conditionf (s)s0 for alls∈R(and consequently,f (0)=0). In such a case, for any T >0, the solutionyis inC([0, T];L2(Ω))and the above equation holds in the sense of distributions inΩ×(0, T ).
* Corresponding author.
E-mail addresses:kim_dang_phung@yahoo.fr(K.D. Phung),ljwang.math@whu.edu.cn(L. Wang),canzhang@whu.edu.cn(C. Zhang).
1 This work was partially supported by the National Science Foundation of China under grants 10971158 and 11161130003.
0294-1449/$ – see front matter ©2013 Elsevier Masson SAS. All rights reserved.
http://dx.doi.org/10.1016/j.anihpc.2013.04.005
Our motivation is a null control problem for semilinear heat equations which means that our goal consists in finding v∈L∞(0,+∞;L2(Ω))such thaty(·, T )=0 inΩ.
The first natural null control problem solved in the literature is the following.
Question 1.What are the assumptions onf in order that the property ∀y0∈L2(Ω), ∀T >0, ∃M >0, ∃v∈L∞
0,+∞;L2(Ω) , such thaty(·, T )=0 inΩandvL∞(0,+∞;L2(Ω))M
holds? Notice that the existence of a null controlvgives the one of the boundM. This property is intensively studied in the literature (see e.g.[2,8,7]) and is called null controllability for semilinear heat equation. It holds for any nonlinear terms which are locally Lipschitz and slightly superlinear. Precisely, it is enough forf to satisfyf (0)=0 and
|slim|→∞
|f (s)|
|s|ln3/2(1+ |s|)=0.
In particular, if we assume thatf is globally Lipschitz withf (0)=0, then null controllability for the corresponding semilinear heat equation holds.
However, we can formulate another type of null control problem as follows.
Question 2.What are the assumptions onf in order that the property ∀y0∈L2(Ω), ∀M >0, ∃T >0, ∃v∈L∞
0,+∞;L2(Ω) , such thaty(·, T )=0 inΩandvL∞(0,+∞;L2(Ω))M
holds? In this article, we will prove the existence ofT andvunder the assumption thatf is globally Lipschitz and satisfies the “good-sign” condition. Once existence of a couple(y, v)is established fory0∈L2(Ω)\{0}andM >0 given, via suitable assumption onf, we introduce the following admissible set of controls
VM=
v∈L∞
0,+∞;L2(Ω)
; vL∞(0,+∞;L2(Ω))Mand the solutiony corresponding tovsatisfiesy(·, T )=0 inΩfor someT >0 . Among all the control functionsv∈VM, we select the infimum of all such time:
T∗=inf{T;v∈VM},
i.e., the minimal time needed to drive the system to rest with control functions inVM.
A controlv∗ such that the corresponding solutiony satisfiesy(·, T∗)=0 inΩ is called time optimal control. In this article, we shall prove the existence of a time optimal controlv∗under the assumption thatf is globally Lipschitz and satisfies the “good-sign” condition.
Now, we are able to state our main result.
Theorem 1.Letf :R→Rbe a globally Lipschitz function satisfyingf (s)s0 for alls∈R. Then for anyy0∈ L2(Ω)\{0}and anyM >0, any time optimal controlv∗ satisfies the bang-bang property:v∗(·, t )L2(Ω)=Mfor a.e.t∈(0, T∗).
Clearly, bang-bang property is of high importance in optimal control theory as mentioned in [6] and [11]. In particular, the bang-bang property for certain time optimal controls governed by parabolic equations can be provided by making use of Pontryagin’s maximum principle (see[9,10,17]). Another approach to get bang-bang property for linear heat equation consists in following a strategy based on null controllability with control functions acting on measurable set in time variable as in[12]and[16]. Recently, the authors in[1]established an observability inequality for the linear heat equation, where the observation is a subset of positive measure in space and time, and from which they obtained another kind of bang-bang property of time optimal problem for the linear heat equation with bounded
controls in space and time. Naturally, the extension of this strategy for nonlinear parabolic equations requires a fixed point argument and an observability inequality for heat equations with space- and time-dependent potentials.
This paper is organized as follows. Section2is devoted to the null controllability for semilinear heat equation with control functions acting onω×Ewhere|E|>0. We present (seeTheorem 2) and prove an estimate of the cost of the control functions whenf is globally Lipschitz. Before giving the proof ofTheorem 2, we recall the linear case and the observability estimate needed (seeTheorem 4). In Section3, applyingTheorem 2in a very special case, we prove the existence for admissible control (seeTheorem 5) whenf is globally Lipschitz and satisfies the “good-sign”
condition. Next we deduce the existence of time optimal (seeTheorem 6). The proof of our main result,Theorem 1, concerning the bang-bang property for time optimal controls governed by semilinear heat equation with local control is given in Section4. Finally, in Section5, we prove the observability estimate ofTheorem 4.
2. Null controllability for semilinear heat equation
The goal of this section is to present the null controllability for semilinear heat equation with control functions acting onω×Ewhere|E|>0. A particular attention is given to the cost estimate.
Theorem 2. Let f :R→ R be a globally Lipschitz function. Let 0T0< T1 < T2 and E⊂(T1, T2) with
|E|>0. Then for any φ∈C([T0, T2], L2(Ω)) and any w0∈L2(Ω), there are a constant κ >0 and a function v1∈L∞(0,+∞;L2(Ω))such that
v1L∞(0,+∞;L2(Ω))κw0L2(Ω)
and the solutionw=w(x, t)of
⎧⎨
⎩
∂tw−w+f (φ+w)−f (φ)=1|ω×Ev1 inΩ×(T0, T2),
w=0 on∂Ω×(T0, T2),
w(·, T0)=w0 inΩ,
satisfiesw(·, T2)=0inL2(Ω). Further, κ=eKec(T1−T0)eK(1+c+c(T2−T1)).
Here,c=c(f ),K=K(Ω, ω) >1andK=K(Ω, ω, E) are positive constants which do not depend onT0. Remark 1.WhenE=(T1, T2), thenκ=ec(T1−T0)eK(1+
1
T2−T1+c+c(T2−T1))
. 2.1. Linear case
In this section, we treat the casef (φ+w)=aw+f (φ)that is the linear heat equation with potential.
Theorem 3.Let 0T0< T1< T2and E⊂(T1, T2)with|E|>0. Let a∈L∞(Ω×(T0, T2)). Then for anyz0∈ L2(Ω), there is a functionv0∈L∞(Ω×(0,+∞))such that the solutionz=z(x, t )of
⎧⎨
⎩
∂tz−z+az=1|ω×Ev0 inΩ×(T0, T2),
z=0 on∂Ω×(T0, T2),
z(·, T0)=z0 inΩ, satisfiesz(·, T2)=0inL2(Ω). Further,
v0L∞(Ω×(0,+∞))eKe(T1−T0)aL∞(Ω×(T0,T1))
×eK(1+(T2−T1)aL∞(Ω×(T1,T2))+a
2/3L∞(Ω×(T1,T2)))
z0L2(Ω).
Here,K=K(Ω, ω) >1andK=K(Ω, ω, E) are positive constants which do not depend onT0. Remark 2.WhenE=(T1, T2), thenK=KT 1
2−T1.
Proof of Theorem 3. We divide its proof into three steps. In the first step, we start to solve
⎧⎨
⎩
∂tz−z+az=0 inΩ×(T0, T1), z=0 on∂Ω×(T0, T1), z(·, T0)=z0 inΩ.
Therefore,z(·, T1)∈L2(Ω)and it is well-known that z(·, T1)
L2(Ω)e(T1−T0)aL∞(Ω×(T0,T1))z0L2(Ω).
The second step consists in establishing the existence of a function v∈L∞(Ω×(0,+∞))such that the solution
˜
z= ˜z(x, t)of
⎧⎨
⎩
∂tz˜−z˜+az˜=1|ω×Ev inΩ×(T1, T2),
˜
z=0 on∂Ω×(T1, T2),
˜
z(·, T1)=z(·, T1) inΩ, satisfiesz(˜ ·, T2)=0 inL2(Ω). Further,
vL∞(Ω×(T1,T2))eKeK(1+(T2−T1)aL∞(Ω×(T1,T2))+a
2/3L∞(Ω×(T1,T2)))z(·, T1)
L2(Ω).
Here,K=K(Ω, ω) >1 andK=K(Ω, ω, E) are positive constants which do not depend onT0. Finally, in the last step, we choose
v0(·, t )=
0 ift∈(0, T1)∪ [T2,+∞), v(·, t ) ift∈ [T1, T2).
Since
v0L∞(Ω×(0,+∞))= vL∞(Ω×(T1,T2)),
the desired result holds. It is standard to get the existence of the above functionvfrom an observability estimate. More precisely, we apply the following result. Its proof is provided in Section5. 2
Theorem 4(Observability estimate). Letωbe an open and non-empty subset ofΩ. LetT >0andEbe a subset of positive measure in(0, T ). Then there are two constantsK=K(Ω, ω)andK=K(Ω, ω, E) > 0such that for any a=a(x, t )∈L∞(Ω×(0, T ))and anyϕ0∈L2(Ω), the solutionϕ=ϕ(x, t )of
⎧⎨
⎩
∂tϕ−ϕ+aϕ=0 inΩ×(0, T ),
ϕ=0 on∂Ω×(0, T ),
ϕ(·,0)=ϕ0 inΩ, satisfies
ϕ(·, T )
L2(Ω)eKeK(1+TaL∞(Ω×(0,T ))+a
2/3L∞(Ω×(0,T )))
ω×E
ϕ(x, t )dx dt.
Remark 3.This is a refined observability estimate. WhenE=(0, T ), then the observability constant becomes eK(1+T1+TaL∞(Ω×(0,T ))+a
2/3L∞(Ω×(0,T )))
.
This is in accordance with the work of[4]. WhenEis a positive measurable set with 0 its Lebesgue point, then the observability constant becomes, for some 1∈E∩(0, T ),
eK(1+
1
1+1aL∞(Ω×(0,T ))+a2/3L∞(Ω×(0,T )))
.
2.2. Nonlinear case with Kakutani’s fixed point
In this section, we prove Theorem 2. Let 0 T0 < T1 < T2 and E ⊂(T1, T2) with |E| >0. Let φ ∈ C([T0, T2], L2(Ω))andw0∈L2(Ω).
By a classical density argument, we may assume thatf ∈C1. We shall use Kakutani’s fixed point theorem to prove the result. First, define for any(x, t)∈Ω×(T0, T2),
a(x, t, r)=
f (φ (x,t)+r)−f (φ (x,t ))
r ifr=0,
f(φ(x, t)) ifr=0.
And consider K=
ξ ∈L2
Ω×(T0, T2)
; ξL2(T0,T2;H1
0(Ω))∩H1(T0,T2;H−1(Ω))κˆ
whereκ >ˆ 0 will be determined later. Sincef:R→Ris a globally Lipschitz function, we have that for a.e.(x, t)∈ Ω×(T0, T2)and anyr∈R
a(x, t, r)L(f )
whereL(f ) >0 is the Lipschitz constant of the functionf.
Next, using the factL∞(Ω×(0,+∞))⊂L∞(0,+∞;L2(Ω)), we know byTheorem 3that for anyξ∈L2(Ω× (T0, T2)), there are a functionv0∈L∞(0,+∞;L2(Ω))and a corresponding solutionz=z(x, t )of
⎧⎪
⎨
⎪⎩
∂tz−z+a
·,·, ξ(·,·)
z=1|ω×Ev0 inΩ×(T0, T2),
z=0 on∂Ω×(T0, T2),
z(·, T0)=w0 inΩ,
(2.1)
such that
z(·, T2)=0 inL2(Ω) (2.2)
and
v0L∞(0,+∞;L2(Ω))Kw0L2(Ω). (2.3)
Here and throughout the proof ofTheorem 2, K=eKe(T1−T0)L(f )eK(1+(T2−T1)L(f )+L(f )2/3)
whereK=K(Ω, ω) >1 andK=K(Ω, ω, E) are positive constants which do not depend onT0. Therefore, we can define the map
Λ:K→L2
Ω×(T0, T2) , ξ→z
where(2.1), (2.2), (2.3)hold.
Now, we check that Kakutani’s fixed point theorem is applicable. For convenience, let us state this result (see e.g.[3]).
Theorem (Kakutani’s fixed point). LetZbe a Banach space andΠbe a non-empty convex compact subset ofZ. Let Λ:Π→Zbe a set-valued mapping satisfying the following assumptions:
i) Λ(ξ )is a non-empty convex set ofZfor everyξ∈Π. ii) Λ(Π )⊂Π.
iii) Λ:Π→Zis upper semicontinuous inZ.
ThenΛpossesses a fixed point in the setΠ.
Here Z=L2(Ω×(T0, T2)) andΠ=K with an adequate choice ofκˆ given below. Clearly, Kis a non-empty convex compact set in L2(Ω×(T0, T2)). Further, from the above arguments,Λ(ξ ) is a non-empty convex set in L2(Ω×(T0, T2)). Thus i) holds.
Let us prove that ii) holds with an adequate choice ofκ. By a standard energy method, using the fact thatˆ |a|L(f ) and(2.1), (2.2), (2.3), there existsC >0 such that
z2C([T
0,T2];L2(Ω))+
T2
T0
z(·, t )2
H01(Ω)dtCw02L2(Ω).
Combining the latter with the fact that|a|L(f ), we deduce that the solutionzsatisfies zL2(T0,T2;H01(Ω))∩H1(T0,T2;H−1(Ω))Cw0L2(Ω),
for someC=C(Ω, ω, E, T2, L(f ))which is a positive constant which does not depend onT0. Hence, if we takeκˆ as follows
ˆ
κ=Cw0L2(Ω) thenΛ(K)⊂K.
Let us finally prove the upper semicontinuity ofΛ:K→L2(Ω×(T0, T2)). We need to prove that ifξm∈K→ξ strongly in L2(Ω×(T0, T2)) and ifpm∈Λ(ξm)→p strongly inL2(Ω×(T0, T2)), thenp∈Λ(ξ ). To this end, firstly, we claim that there exists a subsequence of(m)m1, still denoted in the same manner, such that
a
·,·, ξm(·,·)
pm→a
·,·, ξ(·,·)
p strongly inL2
Ω×(T0, T2)
. (2.4)
Indeed, sinceξm→ξ strongly inL2(Ω×(T0, T2)), we have that there exists a subsequence of(m)m1, still denoted by itself, such that
ξm(x, t)→ξ(x, t ) for a.e.(x, t)∈Ω×(T0, T2).
On one hand, for(x, t)withξ(x, t )=0, by the above, there exists a positive integerm0depending on(x, t)such that ξm(x, t)=0 ∀mm0,
which implies by the definition ofa, a
x, t, ξm(x, t)
→a
x, t, ξ(x, t)
asm→ +∞. (2.5)
On the other hand, for any (x, t) such that ξ(x, t )=0, by the definition of a, we have that a(x, t, ξ(x, t))= f(φ(x, t)). Since
a
x, t, ξm(x, t)
=
f (φ (x,t)+ξm(x,t ))−f (φ (x,t ))
ξm(x,t ) ifξm(x, t)=0, f(φ(x, t)) ifξm(x, t)=0, it gives
a
x, t, ξm(x, t)
→a
x, t, ξ(x, t)
asm→ +∞. This, combined with(2.5), implies
a
x, t, ξm(x, t)
→a
x, t, ξ(x, t)
for a.e.(x, t)∈Ω×(T0, T2).
From the latter, the fact that|a|L(f )and the Lebesgue dominated convergence theorem it follows that a
·,·, ξm(·,·)
pm−a
·,·, ξ(·,·)p2
L2(Ω×(T0,T2))
2a
·,·, ξm(·,·)
(pm−p)2
L2(Ω×(T0,T2))+2a
·,·, ξm(·,·)
−a
·,·, ξ(·,·)p2
L2(Ω×(T0,T2))
2L(f )2pm−p2L2(Ω×(T0,T2))+2a
·,·, ξm(·,·)
−a
·,·, ξ(·,·)p2
L2(Ω×(T0,T2))
→0.
This completes the proof of(2.4). Secondly, sincepm∈Λ(ξm), there exists(vm)m1satisfying
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
(pm)t−pm+a
·,·, ξm(·,·)
pm=1|ω×Evm inΩ×(T0, T2),
pm=0 on∂Ω×(T0, T2),
pm(·, T0)=w0 inΩ,
pm(·, T2)=0 inΩ,
(2.6)
pmL2(T0,T2;H1
0(Ω))∩H1(T0,T2;H−1(Ω))+ pmL2(T1,T2;H2(Ω)∩H1
0(Ω))∩H1(T1,T2;L2(Ω))C, whereC >0 is a constant independent ofm, and
vmL∞(0,+∞;L2(Ω))Kw0L2(Ω). (2.7)
Thus, we deduce the existence ofvand subsequences(vm)m1and(pm)m1such that vm→v weakly star inL∞
0,+∞;L2(Ω)
, (2.8)
pm→p weakly inL2
T0, T2;H01(Ω)
∩H1
T0, T2;H−1(Ω)
, (2.9)
pm(·, T2)→p(·, T2) strongly inL2(Ω). (2.10)
Finally, passing to the limit form→ +∞in(2.6) and (2.7), by(2.4) and (2.8), (2.9), (2.10), we obtain thatp∈Λ(ξ ).
By Kakutani’s fixed point theorem, we conclude that there existsw∈K with an adequate choice ofκˆ such that w∈Λ(w), i.e., there is a controlv1∈L∞(0,+∞;L2(Ω))satisfying
v1L∞(0,+∞;L2(Ω))Kw0L2(Ω),
with the sameKgiven in(2.3), and the corresponding solutionw=w(x, t)solves
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
∂tw−w+a
·,·, w(·,·)
w=1|ω×Ev1 inΩ×(T0, T2),
w=0 on∂Ω×(T0, T2),
w(·, T0)=w0 inΩ,
w(·, T2)=0 inΩ.
Since for any(x, t)∈Ω×(T0, T2), a
x, t, w(x, t)
w(x, t)=f
φ(x, t)+w(x, t)
−f φ(x, t)
, we finally get
⎧⎪
⎪⎨
⎪⎪
⎩
∂tw−w+f (φ+w)−f (φ)=1|ω×Ev1 inΩ×(T0, T2),
w=0 on∂Ω×(T0, T2),
w(·, T0)=w0 inΩ,
w(·, T2)=0 inΩ.
This completes the proof ofTheorem 2.
3. Existence of time optimal control
In this section, we start to prove the existence of admissible controls (see e.g.[15]). In other words we prove that Theorem 5.Letf :R→Rbe a globally Lipschitz function satisfyingf (s)s0for all s∈R. Then for anyy0∈ L2(Ω)\{0}and anyM >0, there are a timeT >0 and an admissible controlv∈L∞(0,+∞;L2(Ω)) such that vL∞(0,+∞;L2(Ω))Mand the solutionycorresponding tovsatisfiesy(·, T )=0inΩ.
Proof. We divide its proof into three steps.
Step 1. We consider the following equation
⎧⎨
⎩
∂ty−y+f (y)=0 inΩ×(0, T0),
y=0 on∂Ω×(0, T0),
y(·,0)=y0 inΩ,
whereT0>0 will be determined later. By a standard energy method, using the fact thatf (s)s0, we have y(·, T0)
L2(Ω)e−λ1T0y0L2(Ω),
whereλ1>0 is the first eigenvalue of−with Dirichlet boundary condition.
Step 2. We applyTheorem 2withT1=T0+1,T2=T0+2,E=(T1, T2),φ=0 andw0=y(·, T0)in order that there are a constantκ >0 and a functionv˜∈L∞(0,+∞;L2(Ω))such that the solutionw=w(x, t)of
⎧⎨
⎩
∂tw−w+f (w)=1|ω×Ev˜ inΩ×(T0, T0+2),
w=0 on∂Ω×(T0, T0+2),
w(·, T0)=y(·, T0) inΩ, satisfiesw(·, T0+2)=0 inL2(Ω). Further,
˜vL∞(0,+∞;L2(Ω))κy(·, T0)
L2(Ω), andκdoes not depend onT0.
Step 3. We can easily check that the function v(·, t)=
0 ift∈(0, T0+1] ∪ [T0+2,+∞),
˜
v(·, t ) ift∈(T0+1, T0+2),
is an admissible control withT =T0+2 whenT0>0 is taken such that T0= 1
λ1ln
1+κy0L2(Ω)
M
in order that
vL∞(0,+∞;L2(Ω))= ˜vL∞(T0+1,T0+2;L2(Ω))κe−λ1T0y0L2(Ω)M.
This completes the proof ofTheorem 5. 2
Now, we establish the existence of time optimal controls (see e.g.[15]). In other words, we shall prove that Theorem 6. Let f : R→ R be a globally Lipschitz function satisfying f (s)s 0 for all s ∈ R. Then for any y0 ∈ L2(Ω)\{0} and any M > 0, there is a time optimal control v∗ ∈ L∞(0,+∞;L2(Ω)) such that v∗L∞(0,+∞;L2(Ω)) M and the solution y corresponding to v∗ satisfies y(·, T∗)=0 in Ω where T∗ = inf{T;v∈VM}.
Proof. By Theorem 5 and the definition of T∗, 0T∗< T for some T >0. Therefore, there exist sequences (Tm)m1 of positive real number and (vm)m1 of function in L∞(0,+∞;L2(Ω)) such that T∗ =limm→∞Tm, vmL∞(0,+∞;L2(Ω))Mand the solutionym=ym(x, t)corresponding tovmsatisfies
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
∂tym−ym+f (ym)=1|ωvm inΩ×(0, T ),
ym=0 on∂Ω×(0, T ),
ym(·,0)=y0 inΩ,
ym(·, Tm)=0 inΩ.
We have by a standard energy method, using the boundMonvmand the “good-sign” condition onf, ym2C([0,T];L2(Ω))+
T 0
ym(·, t )2
H01(Ω)dtC.
Here and throughout the proof,C denotes a generic constant independent ofm. Sincef is globally Lipschitz and f (0)=0, the above inequality implies
f (ym)
L2(Ω×(0,T ))=f (ym)−f (0)
L2(Ω×(0,T ))C.
Therefore, from the boundedness of−f (ym)+1|ωvm, the sequence(ym)m1is bounded inH1(0, T;H−1(Ω)).
Now, we deduce the existence ofv∗∈L∞(0,+∞;L2(Ω))and subsequences(vm)m1and(ym)m1such that vm→v∗ weakly star inL∞
0,+∞;L2(Ω)
withv∗
L∞(0,+∞;L2(Ω))M, ym→y∗ weakly inL2
0, T;H01(Ω)
∩H1
0, T;H−1(Ω)
, strongly inC
[0, T];L2(Ω) . Further,
⎧⎪
⎨
⎪⎩
∂ty∗−y∗+f y∗
=1|ωv∗ inΩ×(0, T ),
y∗=0 on∂Ω×(0, T ),
y∗(·,0)=y0 inΩ, and y∗
·, T∗
L2(Ω) y∗
·, T∗
−y∗(·, Tm)
L2(Ω)+y∗(·, Tm)−ym(·, Tm)
L2(Ω)
→0 whenm→ ∞.
This givesy∗(·, T∗)=0 inΩand consequently,v∗is a time optimal control. This completes the proof. 2 4. Bang-bang property for time optimal control (proof ofTheorem 1)
We want to prove that ifv∗is a time optimal control corresponding to the optimal timeT∗=inf{T;v∈VM}, then v∗(·, t )L2(Ω)=Mfor a.e.t∈(0, T∗). To prove this, we work by contradiction. Suppose that there areε∈(0, M) and a positive measurable subsetE∗⊂(0, T∗)such that
v∗(·, t )
L2(Ω)M−ε ∀t∈E∗
and the solutiony∗=y∗(x, t)corresponding tov∗satisfies
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
∂ty∗−y∗+f y∗
=1|ωv∗ inΩ× 0, T∗
,
y∗=0 on∂Ω×
0, T∗ , y∗(·,0)=y0 inΩ,
y∗(·, T∗)=0 inΩ.
We claim that there exist a real numberδ∈(0, T∗)and a couple(y, v)such that
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
∂ty−y+f (y)=1|ωv inΩ×
0, T∗−δ ,
y=0 on∂Ω×
0, T∗−δ , y(·,0)=y0 inΩ,
y(·, T∗−δ)=0 inΩ,
andv∈L∞(0,+∞;L2(Ω)) withvL∞(0,+∞;L2(Ω))M. This is clearly a contradiction with the time optimal assumptionT∗=inf{T;v∈VM}.
Now, we prove our claim. We divide its proof into four steps.
Step 1.T∗>0 and 0<|E∗|T∗being given, letδ0= |E∗|/2 and denote E=E∗∩
δ0, T∗ . Then
|E|>0.
Indeed,|E∗∩(δ0, T∗)||E∗| −δ0|E∗|/2.
Step 2. We applyTheorem 2with 0< T0< T1< T2,E⊂(T1, T2)with|E|>0 andφ=y∗, in order that there are a constantκ >0 and a functionv1∈L∞(0,+∞;L2(Ω))such that
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩
∂tw−w+f y∗+w
−f y∗
=1|ω×Ev1 inΩ×(T0, T2),
w=0 on∂Ω×(T0, T2),
w(·, T0)=w0 inΩ,
w(·, T2)=0 inΩ,
v1L∞(0,+∞;L2(Ω))κw0L2(Ω), and furtherκ does not depend onT0.
Step 3. We apply step 2 withT0=δ,T1=δ0,T2=T∗,w0=y0−y∗(·, δ), in order thatz=y∗+wsolves
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
∂tz−z+f (z)=1|ω
v∗+1|Ev1
inΩ× δ, T∗
,
z=0 on∂Ω×
δ, T∗ ,
z(·, δ)=y0 inΩ,
z(·, T∗)=0 inΩ.
Denotev2=v∗+1|Ev1. On one hand, ift∈(0,+∞)\E, thenv2(·, t )L2(Ω)= v∗(·, t )L2(Ω)M. On the other hand, ift∈E, then
v2(·, t )
L2(Ω)v∗(·, t )
L2(Ω)+v1(·, t )
L2(Ω)
M−ε+κy∗(·,0)−y∗(·, δ)
L2(Ω). Now, we chooseδsufficiently closed to 0 in order that
y∗(·,0)−y∗(·, δ)
L2(Ω)ε/κ.
This is possible becausey∗∈C([0, T∗], L2(Ω)). Consequently,v2(·, t )L2(Ω)Mfor a.e.t∈(0,+∞).
Step 4. Letv(·, t )=v2(·, t+δ). Thenv∈L∞(0,+∞;L2(Ω))and further we can check thatv(·, t )L2(Ω)M for a.e.t∈(0,+∞). Lety(x, t)=z(x, t+δ). Then it solves
⎧⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎩
∂ty−y+f (y)=1|ωv inΩ×
0, T∗−δ ,
y=0 on∂Ω×
0, T∗−δ , y(·,0)=y0 inΩ,
y
·, T∗−δ
=0 inΩ.
This is the desired claim.
5. The heat equation with potential (proof ofTheorem 4)
The proof ofTheorem 4is based on five lemmas. From now,ϕdenotes the solution of
⎧⎨
⎩
∂tϕ−ϕ+aϕ=0 inΩ×(0, T ),
ϕ=0 on∂Ω×(0, T ),
ϕ(·,0)=ϕ0 inΩ,
wherea=a(x, t )∈L∞(Ω×(0, T )). We also denotea∞= aL∞(Ω×(0,T )).
Lemma 1.For anyϕ0∈L2(Ω), the solutionϕsatisfies the two following estimates for anyt∈(0, T],
Ω
ϕ(x, t )2dxe2ta∞
Ω
ϕ0(x)2dx and
Ω
∇ϕ(x, t )2dxe3ta∞ t
Ω
ϕ0(x)2dx.
This result is deduced by energy estimate and is standard. Its proof is omitted here.
Letx0∈Ω. Denote byBR=B(x0, R)the ball of centerx0and radiusR.
Lemma 2.LetR0>0andλ >0. Introduce fort∈ [0, T]andx0∈Ω, Gλ(x, t)= 1
(T−t+λ)n/2e− |
x−x0|2 4(T−t+λ).
Define foru∈H1(0, T;L2(Ω∩BR0))∩L2(0, T;H2∩H01(Ω∩BR0))andt∈(0, T],
Nλ(t)=
Ω∩BR0|∇u(x, t )|2Gλ(x, t) dx
Ω∩BR
0|u(x, t )|2Gλ(x, t) dx , whenever
Ω∩BR0
u(x, t )2dx=0.
The following two properties hold.
i)
1 2
d dt
Ω∩BR0
u(x, t )2Gλ(x, t) dx+
Ω∩BR0
∇u(x, t )2Gλ(x, t) dx
=
Ω∩BR0
u(x, t)(∂t−)u(x, t )Gλ(x, t) dx. (5.1)
ii) WhenΩ∩BR0 is star-shaped with respect tox0, d
dtNλ(t) 1
T −t+λNλ(t)+
Ω∩BR0|(∂t−)u(x, t )|2Gλ(x, t) dx
Ω∩BR0|u(x, t )|2Gλ(x, t) dx . (5.2) Proof. The identity follows from some direct computations. The proof of the second one is the same as that in [13, pp. 1240–1245]or[5, Lemma 2]. 2
Lemma 3.LetR >0andδ∈(0,1]. Then there are two constantsC1, C2>0, only dependent on(R, δ)such that for anyϕ0∈L2(Ω)withϕ0=0, the quantity
h0= C1
ln
(1+C2) e1+
2C1
T +3Ta∞+a2/3∞ Ω|ϕ0(x)|2dx
Ω∩BR|ϕ(x,T )|2dx
(5.3)
has the following two properties.
i)
0<
1+2C1
T +Ta∞+ a2/3∞
h0< C1. (5.4)
ii) For anyt∈ [T−h0, T], it holds e3Ta∞
Ω
ϕ0(x)2dxe1+C3
1 h0
Ω∩B(1+δ)R
ϕ(x, t )2dx (5.5)
for someC3> C1only dependent on(R, δ).
Remark 4.By the strong unique continuation property for parabolic equations with zero Dirichlet boundary condition, it is impossible to have
Ω∩BR|ϕ(x, T )|2dx=0 ifϕ0∈L2(Ω)withϕ0=0.
Remark 5.From(5.4), we haveh0< T /2 and thereforeT /2< T−h0< T. Here,(5.5)says that for anytsufficiently closed toT, the following Hölder interpolation estimate holds.
