Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 31, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXACT CONTROLLABILITY PROBLEM OF A WAVE
EQUATION IN NON-CYLINDRICAL DOMAINS
HUA WANG, YIJUN HE, SHENGJIA LI
Abstract. Let α : [0, ∞) → (0, ∞) be a twice continuous differentiable function which satisfies that α(0) = 1, α0 is monotone and 0 < c1 ≤ α0 (t) ≤ c2 < 1
for some constants c1 , c2 . The exact controllability of a one-dimensional wave
equation in a non-cylindrical domain is proved. This equation characterizes
small vibrations of a string with one of its endpoint fixed and the other moving with speed α0 (t). By using the Hilbert Uniqueness Method, we obtain the
exact controllability results of this equation with Dirichlet boundary control
on one endpoint. We also give an estimate on the controllability time that
depends only on c1 and c2 .
1. Introduction and main results
Suppose α : [0, ∞) → (0, ∞) is a twice continuous differentiable function satisfying the following assumptions:
(A1) 0 < c1 ≤ α0 (t) ≤ c2 < 1 for all 0 ≤ t < ∞;
(A2) α0 is monotone;
(A3) α(0) = 1.
b α by
Let T > 0. We define the non-cylindrical domain Q
T
2
bα
Q
T = {(y, t) ∈ R : 0 < y < α(t), t ∈ (0, T )}.
This article concerns the exact controllability of the one-dimensional wave equation
bα
(y, t) ∈ Q
T,
utt (y, t) − uyy (y, t) = 0,
u(0, t) = 0,
0
u(y, 0) = u (y),
u(α(t), t) = v(t),
1
ut (y, 0) = u (y),
t ∈ (0, T ),
(1.1)
y ∈ (0, 1),
where v ∈ L2 (0, T ) and (u0 , u1 ) ∈ L2 (0, 1) × H −1 (0, 1). Since supt∈(0,T ) |α0 (t)| < 1,
by [9], the system of (1.1) admits a unique solution in the sense of transposition.
Here, as in [10], u ∈ L∞ (0, T ; L2 (0, α(t)) is called a solution by transposition of
2000 Mathematics Subject Classification. 35L05, 93B05.
Key words and phrases. Exact controllability; non-cylindrical domain;
Hilbert uniqueness method.
c
2015
Texas State University - San Marcos.
Submitted December 6, 2014. Published January 30, 2015.
1
2
H. WANG, Y. HE, S. LI
EJDE-2015/31
problem (1.1) if u verifies
Z T Z α(t)
u(y, t)ĥ(y, t) dy dt
0
Z
0
1
1
Z
0
[u (y)θ(y, 0) − u (y)θt (y, 0)]dy −
=
0
(1.2)
T
v(t)θy (α(t), t)dt,
0
for all ĥ ∈ L1 (0, T ; L2 (0, α(t)), where θ is the weak solution of the problem
θtt (y, t) − θyy (y, t) = ĥ,
bα ,
(y, t) ∈ Q
T
t ∈ (0, T ),
θ(0, t) = θ(α(t), 0) = 0,
0
θ(T ) = θ (T ) = 0,
(1.3)
x ∈ (0, 1).
The exact controllability problem of system (1.1) is stated as follows.
Definition 1.1. We say system (1.1) is exactly controllable at time T , if for any
(u0 , u1 ) ∈ L2 (0, 1) × H −1 (0, 1), (u0d , u1d ) ∈ L2 (0, α(T )) × H −1 (0, α(T )), there exists
v ∈ L2 (0, T ) such that the solution by transposition u of (1.1) satisfies u(T ) = u0d
and ut (T ) = u1d .
For a function α satisfying conditions (A1)–(A3), we define
1
2c22 (1 − c1 )(1 + c2 ) exp
−1 ,
2
c2
c1 (1 − c2 )
1
2c22 (1 − c1 ) T1∗ =
exp
−1 .
c2
c1 (1 − c2 )3
T∗ =
(1.4)
(1.5)
One of the main results of this article as follows.
Theorem 1.2. For any given T > T ∗ , (1.1) is exactly controllable at time T .
Similarly, for the exact controllability problem, when the control is acting on the
fixed endpoint,
bα
(y, t) ∈ Q
T,
utt (y, t) − uyy (y, t) = 0,
u(0, t) = v(t), u(α(t), t) = 0,
0
u(y, 0) = u (y),
1
t ∈ (0, T ),
ut (y, 0) = u (y),
(1.6)
y ∈ (0, 1),
we have the following result.
Theorem 1.3. For any given T > T1∗ , (1.6) is exactly controllable at time T .
Remark 1.4. When α(t) = 1 + kt for some constant k ∈ (0, 1), T ∗ is reduced to
Tk∗ defined in [4], and Theorem 1.2 is reduced to [4, Theorem 1.1].
Remark 1.5. Theorem 1.3 extends the results in [5] and [6]. In fact, when α(t) =
1 + kt, an exact controllability result of system (1.6) has been proved for 0 < k <
2
1 − √1e in [5] and for 0 < k < 1 − 1+e
2 in [6]. We also note that the controllability
∗
time T1 given here is better than the constants Tk∗ in [5] and [6] in this case.
Remark 1.6. We note that there are many functions α(t) satisfying conditions
(A1)–(A3) but are not the form 1 + kt, for example α(t) = 1 + (t + arctan t)/c
where c is any constant that is greater than 2.
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EXACT CONTROLLABILITY OF A WAVE EQUATION
3
Recently, several works on the controllability problems of wave equations in noncylindrical domains have been published. The existence of solutions of the initial
boundary value problem for the nonlinear wave equation in non-cylindrical domains
has been studied in [3, 8]. The controllability problem for a multi-dimensional wave
equation in a non-cylindrical domain has been investigated in [2, 9, 10]. About the
one-dimension cases, there have been extensive study of the controllability problem
in a non-cylindrical domain. We refer the reader to [1, 4, 5, 6].
When α(t) = 1 + kt for some constant 0 < k < 1, in [4], the exact controllability
of the system (1.1) has been acquired. When α(t) = 1 + kt, Cui and Song obtained
that the system (1.6) is exactly controllable for 0 < k < 1 − √1e in [5] and is exactly
2
controllable for 0 < k < 1 − 1+e
2 in [6].
There are also other results on the exact controllability problem for wave equations of variable coefficients in cylindrical domains, see [7, 10, 11, 12] and the references therein. So, our first aim is to transform (1.1) and (1.6) into wave equations
with variable coefficients in a cylindrical domain.
y
b α . Then, it
Let x = α(t)
and w(x, t) = u(y, t) = u(α(t)x, t) for (y, t) ∈ Q
T
is straightforward to show that (x, t) varies in QT := (0, 1) × (0, T ) and (1.1) is
transformed into the wave equation with variable coefficients,
wtt −
β(x, t) γ(x, t)
τ (x, t)
wx x +
wtx +
wx = 0,
α(t)
α(t)
α(t)
w(0, t) = 0, w(1, t) = v(t) t ∈ (0, T ),
w(x, 0) = w0 (x),
02
wt (x, 0) = w1 (x),
in QT ,
(1.7)
x ∈ (0, 1),
2
(t)x
where β(x, t) = 1−αα(t)
, γ(x, t) = −2α0 (t)x, τ (x, t) = −α00 (t)x, w0 = u0 , w1 =
u1 + α0 (0)xu0x .
From [10], we know that for (u0 , u1 ) ∈ L2 (0, 1) × H −1 (0, 1) and v ∈ L2 (0, T ),
(1.7) admits a unique solution w ∈ C([0, T ]; L2 (0, 1)) ∩ C 1 ([0, T ]; H −1 (0, 1)) in the
sense of transposition, where w is called a solution by transposition of problem (1.7)
if
Z TZ 1
wh dx dt
0
0
1
Z
[−w0 (x)zt (x, 0) + α0 (0)w0 (x)z(x, 0) + w0 (x)z(x, 0)]dx
=
0
Z
−
T
Z
β(1, t)zx (1, t)v(t)dt +
0
1
[γx (x, 0)w0 (x)z(x, 0) + γ(x, 0)wx0 (x)z(x, 0)]dx,
0
1
2
for every h ∈ L (0, T ; L (0, 1)) and z is the weak solution of the problem
L∗ z = h,
in QT ,
z(0, t) = z(1, t) = 0,
z(x, T ) = zt (x, T ) = 0,
t ∈ (0, T ),
(1.8)
x ∈ (0, 1),
∗
where the formal adjoint L of L is defined by
L∗ z = α(t)ztt − [β(x, t)zx ]x + γ(x, t)zxt + τ (x, t)zx .
(1.9)
Thus, Theorem 1.2 can be restated as the following exact controllability result
for equation (1.7).
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H. WANG, Y. HE, S. LI
EJDE-2015/31
Theorem 1.7. For any T > T ∗ where T ∗ is given by (1.4), any (w0 , w1 ) ∈
L2 (0, 1) × H −1 (0, 1) and (wd0 , wd1 ) ∈ L2 (0, 1) × H −1 (0, 1), we can always find a
control v ∈ L2 (0, T ) such that the corresponding solution by transposition w of
(1.7) satisfies w(T ) = wd0 , wt (T ) = wd1 .
Similarly, (1.6) can be transformed into the wave equation with variable coefficients,
β(x, t) γ(x, t)
τ (x, t)
wx x +
wtx +
wx = 0, in QT ,
wtt −
α(t)
α(t)
α(t)
(1.10)
w(0, t) = v(t), w(1, t) = 0, t ∈ (0, T ),
w(x, 0) = w0 (x),
wt (x, 0) = w1 (x),
x ∈ (0, 1),
and Theorem 1.3 can be restated as the following exact controllability result for
equation(1.10).
Theorem 1.8. For any T > T1∗ where T1∗ is given by (1.5), any (w0 , w1 ) ∈
L2 (0, 1) × H −1 (0, 1) and (wd0 , wd1 ) ∈ L2 (0, 1) × H −1 (0, 1), we can always find a
control v ∈ L2 (0, T ) such that the corresponding solution by transposition w of
(1.10) satisfies w(T ) = wd0 , wt (T ) = wd1 .
2. Description of the Hilbert uniqueness method
In this section, we describe the Hilbert uniqueness method which is used in the
proof of Theorems 1.7 and 1.8. Next, we consider Theorem 1.7 in detail.
Firstly, for any (wd0 , wd1 ) ∈ L2 (0, 1) × H −1 (0, 1), the system
α(t)ξtt − β(x, t)ξx x + γ(x, t)ξtx + τ (x, t)ξx = 0, in QT ,
ξ(0, t) = 0,
ξ(x, T ) =
wd0 (x),
t ∈ (0, T ),
ξ(1, t) = 0,
ξt (x, T ) =
wd1 (x),
(2.1)
x ∈ (0, 1)
has a unique solution ξ ∈ C([0, T ]; L2 (0, 1)) ∩ C 1 ([0, T ]; H −1 (0, 1)) in the sense of
transportation.
Secondly, for any (z 0 , z 1 ) ∈ H01 (0, 1) × L2 (0, 1), we solve
α(t)ztt − [β(x, t)zx ]x + γ(x, t)zxt + τ (x, t)zx = 0,
z(0, t) = z(1, t) = 0,
0
z(x, 0) = z (x),
in QT ,
t ∈ (0, T ),
1
zt (x, 0) = z (x),
(2.2)
x ∈ (0, 1),
and
α(t)ηtt − β(x, t)ηx x + γ(x, t)ηtx + τ (x, t)ηx = 0,
η(0, t) = 0,
η(x, T ) = 0,
η(1, t) = zx (1, t),
ηt (x, T ) = 0,
in QT ,
t ∈ (0, T ),
(2.3)
x ∈ (0, 1).
Then we define a linear operator Λ : H01 (0, 1) × L2 (0, 1) → H −1 (0, 1) × L2 (0, 1), by
(z 0 , z 1 ) 7→ (ηt (·, 0) + γ(·, 0)ηx (·, 0) − α0 (0)η(·, 0), −η(·, 0)),
Lastly, the problem is reduced to prove the existence of some (z 0 , z 1 ) ∈ H01 (0, 1) ×
L2 (0, 1) such that
Λ(z 0 , z 1 ) = ([w1 − ξt (0)] − α0 (0)[w0 − ξ(0)] + γ(0)[wx0 − ξx (0)] − [w0 − ξ(0)]). (2.4)
EJDE-2015/31
EXACT CONTROLLABILITY OF A WAVE EQUATION
5
To solve (2.4), we observe that
Z 1
β(1, t)|zx (1, t)|2 dt = hΛ(z 0 , z 1 ), (z 0 , z 1 )iH −1 (0,1)×L2 (0,1),H01 (0,1)×L2 (0,1) . (2.5)
0
In section 3, we prove the following observability inequality for system (2.2):
there exists a constant C > 0 such that
Z T
(2.6)
β(1, t)|zx (1, t)|2 dt ≥ C kz 0 k2H 1 (0,1) + kz 1 k2L2 (0,1) .
0
0
Also, we prove that Λ is a bounded linear operator; i.e., there exists a constant
C > 0 such that
Z T
β(1, t)|zx (1, t)|2 dt ≤ C(||z 0 ||2H 1 (0,1) + ||z 1 ||2L2 (0,1) ).
(2.7)
0
0
Combining (2.6), (2.7) and the Lax-Milgram Theorem, we can show that Λ is an
isomorphism.
Then, the equation(2.4) has a unique solution (z 0 , z 1 ) ∈ H01 (0, 1) × L2 (0, 1), and
the function zx (1, t) is the desired control such that the solution w of (1.7) satisfies
w(T ) = wd0 , wt (T ) = wd1 .
For the proof of Theorem 1.8, the steps are similar to those of Theorem 1.7. In
this case, instead of (2.3), we consider the following homogeneous wave equation
α(t)ηtt − β(x, t)ηx x + γ(x, t)ηtx + τ (x, t)ηx = 0, in QT ,
η(0, t) = zx (0, t),
η(x, T ) = 0,
η(1, t) = 0,
ηt (x, T ) = 0,
t ∈ (0, T ),
(2.8)
x ∈ (0, 1),
and define a linear operator Λ just same as (2.4), then we observe that
Z 1
β(0, t)|zx (0, t)|2 dt = −hΛ(z 0 , z 1 ), (z 0 , z 1 )iH −1 (0,1)×L2 (0,1),H01 (0,1)×L2 (0,1) . (2.9)
0
We omit the details of the proof here.
3. Observability estimates
The main purpose of this section is to prove the observability inequalities for
system (2.2). To prove those estimates, we need some technical lemmas.
From [10], we know that: for any (z 0 , z 1 ) ∈ H01 (0, 1) × L2 (0, 1), the equation
(2.2) has a unique weak solution z ∈ C([0, T ]; H01 (0, 1)) ∩ C 1 ([0, T ]; L2 (0, 1)) in the
sense of transportation.
The energy for (2.2) is defined as
Z
1 1
[α(t)|zt (x, t)|2 + β(x, t)|zx (x, t)|2 ]dx, for t ≥ 0,
(3.1)
E(t) =
2 0
where z is the solution of (2.2). Since α(0) = 1, we have
Z
1 1 1
E0 := E(0) =
|z (x)|2 + β(x, 0)|zx0 (x)|2 dx.
2 0
(3.2)
First, we prove a lemma which is related to the decay rate of the energy E(t).
6
H. WANG, Y. HE, S. LI
EJDE-2015/31
Lemma 3.1. If α(0) = 1, 0 < c1 ≤ α0 (t) ≤ c2 < 1 and α0 is monotone, then
c3 E 0
c4 E0
≤ E(t) ≤
,
α(t)
α(t)
(3.3)
where
1−c2 c2
1−c1 , c1
(
(c3 , c4 ) =
,
if α0 is increasing,
1−c1
,
if α0 is decreasing.
c1
c2 , 1−c2
(3.4)
Proof. For any 0 < t ≤ T , through multiplying the first equation of (2.2) by zt and
integrating the result on (0, 1) × (0, t), we conclude that
Z tZ 1
α(s)ztt (x, s)zt (x, s) − [β(x, s)zx (x, s)]x zt (x, s)
0=
0
0
+ γ(x, s)zxt (x, s)zt (x, s) + τ (x, s)zx (x, s)zt (x, s) dx ds
:= I1 + I2 + I3 + I4 ,
where
Z
Z Z
t 1 t 1 0
1 1
2
I1 =
α(s)|zt (x, s)| dx 0 −
α (s)|zt (x, s)|2 dx ds,
2 0
2 0 0
Z
Z Z
t 1 t 1
1 1
I2 =
β(x, s)|zx (x, s)|2 dx0 −
βt (x, s)|zx (x, s)|2 dx ds
2 0
2 0 0
Z
Z Z
t 1 t 1 α0 (s)
1 1
β(x, s)|zx (x, s)|2 dx0 +
β(x, s)|zx (x, s)|2 dx ds
=
2 0
2 0 0 α(s)
Z tZ 1 0
α (s)α00 (s) 2
+
x |zx (x, s)|2 dx ds,
α(s)
0
0
Z tZ 1
I3 =
α0 (s)|zt (x, s)|2 dx ds,
0
0
Z tZ
I4 = −
0
1
xα00 (s)zx (x, s)zt (x, s) dx ds.
0
We thereby obtain:
Z
t
E(t) = E0 −
0
Z tZ
+
0
α0 (s)
E(s)ds −
α(s)
Z tZ
0
0
1
α0 (s)α00 (s) 2
x |zx (x, s)|2 dx ds
α(s)
1
α00 (s)xzx (x, s)zt (x, s) dx ds.
0
α0 (t)
E 0 (t) = −
E(t) −
α(t)
Z
0
1
α0 (t)α00 (t) 2
x |zx (x, t)|2 dx +
α(t)
Z
1
α00 (t)xzx (x, t)zt (x, t)dx.
0
(3.5)
We subdivide the proof into two cases:
(1) α0 is increasing; that is, α00 (t) ≥ 0. By using the inequalities
α0 (t)α00 (t) 2
(t)α(t)α00 (t)
x |zx (x, t)|2 −
|zt (x, t)|2
2(t)α(t)
2α0 (t)
≤ α00 (t)xzx (x, t)zt (x, t)
−
≤
α(t)α00 (t)
α0 (t)α00 (t) 2
x |zx (x, t)|2 +
|zt (x, t)|2 ,
2α(t)
2α0 (t)
EJDE-2015/31
EXACT CONTROLLABILITY OF A WAVE EQUATION
where (t) =
α0 (t)
1−α0 (t) ,
−(
7
we easily obtain
α0 (t)
α00 (t)
α0 (t) α00 (t)
+
)E(t) ≤ E 0 (t) ≤ −(
− 0 )E(t),
0
α(t)
1 − α (t)
α(t)
α (t)
(3.6)
so
(1 − α0 (t))E0
α0 (t)E0
≤
E(t)
≤
.
(1 − α0 (0))α(t)
α0 (0)α(t)
Using 0 < c1 ≤ α0 (t) ≤ c2 < 1, we conclude that
c4 E0
c3 E 0
≤ E(t) ≤
,
α(t)
α(t)
where c3 =
1−c2
1−c1 ,
c4 =
(3.7)
(3.8)
c2
c1 .
(2) α0 is decreasing; that is, α00 (t) ≤ 0. By using the inequalities
α0 (t)α00 (t) 2
α(t)α00 (t)
x |zx (x, t)|2 +
|zt (x, t)|2
2α(t)
2α0 (t)
≤ α00 (t)xzx (x, t)zt (x, t)
≤−
where (t) =
α0 (t)α00 (t) 2
(t)α(t)α00 (t)
x |zx (x, t)|2 −
|zt (x, t)|2 ,
2(t)α(t)
2α0 (t)
α0 (t)
1−α0 (t) ,
we easily get
α0 (t)
α00 (t)
α (t) α00 (t)
− 0 )E(t) ≤ E 0 (t) ≤ −(
+
)E(t),
α(t)
α (t)
α(t)
1 − α0 (t)
0
−(
(3.9)
so
α0 (t)E0
(1 − α0 (t))E0
≤
E(t)
≤
.
α0 (0)α(t)
(1 − α0 (0))α(t)
Using 0 < c1 ≤ α0 (t) ≤ c2 < 1, we conclude that
c3 E 0
c4 E0
≤ E(t) ≤
,
α(t)
α(t)
where c3 =
c1
c2 ,
c4 =
1−c1
1−c2 .
(3.10)
(3.11)
Remark 3.2. When α00 (t) ≡ 0, that is, α(t) = 1 + kt for some constant k ∈ (0, 1),
then c3 = c4 = 1, Lemma 3.1 is reduced to Lemma 3.1 in [4].
Next, similar to the proof of [4, Lemma 3.2], we can get the following estimate
for each weak solution z of (2.2) by the multiplier method.
Lemma 3.3. For any function q ∈ C 1 ([0, 1]), the solution z of (2.2) satisfies the
estimate
Z
1
1 T
β(x, t)q(x)|zx (x, t)|2 dt
2 0
0
Z TZ 1
1
=
q 0 (x)[α(t)|zt (x, t)|2 + β(x, t)|zx (x, t)|2 ] dx dt
2 0 0
Z TZ 1
Z Z
1 T 1
−
α0 (t)q(x)zx (x, t)zt (x, t) dx dt −
βx (x, t)q(x)|zx (x, t)|2 dx dt
2
0
0
0
0
Z 1
T
+
[α(t)q(x)zx (x, t)zt (x, t) − xα0 (t)q(x)|zx (x, t)|2 ]dx .
0
0
(3.12)
8
H. WANG, Y. HE, S. LI
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Finally, we derive the continuity estimate.
Theorem 3.4. Assume T > 0, for any (z 0 , z 1 ) ∈ H01 (0, 1) × L2 (0, 1), there exists a
constant C > 0 such that the solution of (2.2) satisfies the following two estimates:
Z T
(3.13)
β(1, t)|zx (1, t)|2 dt ≤ C(||z 0 ||2H 1 (0,1) + ||z 1 ||2L2 (0,1) ),
0
0
Z
0
T
β(0, t)|zx (0, t)|2 dt ≤ C(||z 0 ||2H 1 (0,1) + ||z 1 ||2L2 (0,1) );
0
(3.14)
so zx (0, ·) ∈ L2 (0, T ) and zx (1, ·) ∈ L2 (0, T ).
Proof. First, we prove inequality (3.13). Let q(x) = x for x ∈ [0, 1] in (3.12) and
02
(t)x
noticing that βx (x, t) = − 2αα(t)
, γ(x, t) = −2α0 (t)x, it follows that
Z
1 T
β(1, t)|zx (1, t)|2 dt
2 0
Z T
Z TZ 1
=
E(t)dt −
α0 (t)xzt (x, t)zx (x, t) dx dt
0
0
0
(3.15)
Z T Z 1 02
α (t) 2
2
x |zx (x, t)| dx dt
+
α(t)
0
0
Z 1
T
+
[α(t)xzt (x, t)zx (x, t) − α0 (t)x2 |zx (x, t)|2 ]dx .
0
0
We estimate every terms on the right side of (3.15). By the assumption for α,
1−c2
we have 1 ≤ α(t) ≤ 1 + c2 T and 0 < 1+c22T ≤ β(x, t) ≤ 1 for any (x, t) ∈ QT , these
inequalities together with (3.3) and the boundedness of α0 (t) imply
Z T
Z TZ 1
E(t)dt −
α0 (t)xzt (x, t)zx (x, t) dx dt
0
0
T
Z
1
Z
+
0
0
02
α (t) 2
x |zx (x, t)|2 dx dt
α(t)
Z TZ 1
E(t)dt + C
[|zt (x, t)|2 + |zx (x, t)|2 ] dx dt
0
T
Z
≤
0
0
T
Z
≤
Z
0
T
Z
E(t)dt + C
0
0
(3.16)
1
[α(t)|zt (x, t)|2 + β(x, t)|zx (x, t)|2 ] dx dt
0
≤ CE0 .
For each t ∈ [0, T ] and (t) > 0, it holds that
Z 1
[α(t)xzt (x, t)zx (x, t) − α0 (t)x2 |zx (x, t)|2 ]dx
0
Z
≤
1
[α(t)|zt (x, t)||zx (x, t)| + α0 (t)|zx (x, t)|2 ]dx
0
1
2(t)
Z
α(t)
≤
2(t)
Z
≤
1
α2 (t)|zt (x, t)|2 dx +
0
0
1
(t)
2
Z
1
|zx (x, t)|2 dx +
Z
0
(t)
α(t)
α(t)|zt (x, t)|2 dx + [
+ α0 (t)]
2
1 − α02 (t)
Z
1
α0 (t)|zx (x, t)|2 dx
0
1
0
β(x, t)|zx (x, t)|2 dx.
EJDE-2015/31
EXACT CONTROLLABILITY OF A WAVE EQUATION
9
Choosing (t) = 1 − α0 (t), then it is easy to see
(t) > 0 and
α(t)
2α(t)
α(t)
= [ + α0 (t)]
=
.
02
2
1 − α (t)
1 − α0 (t)
This implies that
Z 1
[α(t)xzt (x, t)zx (x, t) − α0 (t)x2 |zx (x, t)|2 ]dx ≤
0
α(t)
α(t)
E(t) ≤
E(t).
1 − α0 (t)
1 − c2
Then, using (3.3), it follows that
T Z 1
[α(t)xzt (x, t)zx (x, t) − α0 (t)x2 |zx (x, t)|2 ]dx ≤ c5 E0 ,
(3.17)
0
0
2c4
where c5 = 1−c
. Therefore, combining (3.15), (3.16) and (3.17), it follows that
2
Z T
β(1, t)|zx (1, t)|2 dt ≤ CE0 ≤ C kz 0 k2H 1 (0,1) + kz 1 k2L2 (0,1) .
0
0
Next, we prove the inequality (3.14). Let q(x) = x − 1 for x ∈ [0, 1] in (3.12)
02
(t)x
and noticing that βx (x, t) = − 2αα(t)
, γ(x, t) = −2α0 (t)x, it follows that
1
2
T
Z
β(0, t)|zx (0, t)|2 dt
0
T
Z
Z
T
Z
E(t)dt −
=
0
0
Z
T
Z
+
0
Z
+
0
0
1
1
1
α0 (t)(x − 1)zt (x, t)zx (x, t) dx dt
0
(3.18)
α02 (t)
x(x − 1)|zx (x, t)|2 dx dt
α(t)
T
[α(t)(x − 1)zt (x, t)zx (x, t) − α0 (t)x(x − 1)|zx (x, t)|2 ]dx .
0
Through estimating every terms on the right side of (3.18), similar to the derive
of (3.16), it follows that
Z T
Z TZ 1
E(t)dt −
α0 (t)(x − 1)zt (x, t)zx (x, t) dx dt
0
0
0
(3.19)
Z T Z 1 02
α (t)
2
+
x(x − 1)|zx (x, t)| dx dt ≤ CE0 .
α(t)
0
0
Since
Z 1
[α(t)(x − 1)zt (x, t)zx (x, t) − α0 (t)x(x − 1)|zx (x, t)|2 ]dx
0
Z 1
≤
[α(t)|zt (x, t)||zx (x, t)| + α0 (t)|zx (x, t)|2 ]dx,
0
similar to the derive of (3.17), it follows that
Z 1
T [α(t)(x − 1)zt (x, t)zx (x, t) − α0 (t)x(x − 1)|zx (x, t)|2 ]dx0 ≤ c5 E(0), (3.20)
0
where c5 =
2c4
1−c2 .
10
H. WANG, Y. HE, S. LI
EJDE-2015/31
From (3.18), (3.19) and (3.20), it follows that
Z T
β(0, t)|zx (0, t)|2 dt ≤ CE0 ≤ (||z 0 ||2H 1 (0,1) + ||z 1 ||2L2 (0,1) ).
0
0
Now, we give the proof of the observability inequalities.
Theorem 3.5. For T > T ∗ where T ∗ satisfies (1.4) and any (z 0 , z 1 ) ∈ H01 (0, 1) ×
L2 (0, 1), there exists a constant C > 0 such that the corresponding solution of (2.2)
satisfies
Z T
β(1, t)|zx (1, t)|2 dt ≥ C kz 0 k2H 1 (0,1) + kz 1 k2L2 (0,1) .
(3.21)
0
0
Proof. If we choose (t) =
1
,
2
0 < (t) <
α0 (t)
1+α0 (t) ,
then it is obvious that
1 α02 (t)
1
1 − (t) = 1 + 2 −
=
.
02
(t) 1 − α (t)
1 + α0 (t)
and
Thus, using
x2 =
α(t)x2
α(t)
β(x, t) ≤
β(x, t)
02
2
1 − α (t)x
1 − α02 (t)
and (3.3), it follows that
Z T
Z TZ
E(t)dt −
0
0
Z
T
Z
T
Z
1
+
0
0
1
α0 (t)xzt (x, t)zx (x, t) dx dt
0
α02 (t) 2
x |zx (x, t)|2 dx dt
α(t)
1
1 − (t)
α(t)|zt (x, t)|2 dx dt
2
0
0
Z TZ 1
1
1 α02 (t) 2 +
β(x, t) + (1 −
)
x |zx (x, t)|2 dx dt
2
2(t) α(t)
0
0
Z TZ 1
1 − (t)
≥
α(t)|zt (x, t)|2 dx dt
2
0
0
Z TZ 1
1
)α02 (t) 1
(2 − (t)
+
1+
β(x, t)|zx (x, t)|2 dx dt
1 − α02 (t)
2
0
0
Z T
1
=
E(t)dt
0
0 1 + α (t)
Z T
1
≥ c6 E0
dt,
α(t)
0
Z
≥
where c6 = c3 /(1 + c2 ). By (3.15), (3.17) and (3.22), we obtain
Z
Z T
1
1 T
2
β(1, t)|zx (1, t)| dt ≥ c6 E0
dt − c5 E0
2 0
0 1 + c2 t
c6
=
log(1 + c2 T ) − c5 E0 .
c2
(3.22)
EJDE-2015/31
EXACT CONTROLLABILITY OF A WAVE EQUATION
If we choose T ∗ as in (1.4), then it is easy to see that
1
c2 c5 T∗ =
exp
−1 ,
c2
c6
11
(3.23)
so for T > T ∗ ,
T
Z
β(1, t)|zx (1, t)|2 dt ≥ C kz 0 k2H 1 (0,1) + kz 1 k2L2 (0,1) ,
0
0
holds for C =
2c6
c2
log(1 + c2 T ) − 2c5 > 0.
Theorem 3.6. For T > T1∗ where T1∗ satisfies (1.5) and any (z 0 , z 1 ) ∈ H01 (0, 1) ×
L2 (0, 1), there exists a constant C > 0 such that the corresponding solution of (2.2)
satisfies
Z T
β(0, t)|zx (0, t)|2 dt ≥ C(||z 0 ||2H 1 (0,1) + ||z 1 ||2L2 (0,1) ).
(3.24)
0
0
Proof. Choosing (x, t) =
0
0
α (t)(1−α (t)x)
,
α(t)
it is easy to see that
|α0 (t)zx (x, t)zt (x, t)|
α02 (t)
(x, t)
|zt (x, t)|2 +
|zx (x, t)|2
2(x, t)
2
α0 (t) α(t)
α0 (t) β(x, t)
2
=
|z
(x,
t)|
+
|zx (x, t)|2 .
t
1 − α0 (t)x 2
1 + α0 (t)x 2
≤
Since x − 1 ≤ 0 for x ∈ [0, 1], we have
Z T
Z TZ 1
E(t)dt −
α0 (t)(x − 1)zt (x, t)zx (x, t) dx dt
0
Z
T
Z
1
0
02
0
α (t)
x(x − 1)|zx (x, t)|2 dx dt
α(t)
0
0
Z T
Z TZ 1 0
α (t)(x − 1) α(t)
|zt (x, t)|2 dx dt
≥
E(t)dt +
1 − α0 (t)x 2
0
0
0
α0 (t)(x − 1) β(x, t)
+
|zx (x, t)|2 dx dt
1 + α0 (t)x
2
Z T Z 1 02
2α (t)x(x − 1) β(x, t)
+
|zx (x, t)|2 dx dt
1 − α02 (t)x2
2
0
0
Z Z
1 T 1 1 − α0 (t)
=
[α(t)|zt (x, t)|2 + β(x, t)|zx (x, t)|2 ] dx dt
2 0 0 1 − α0 (t)x
Z T
≥
(1 − α0 (t))E(t)dt
+
0
≥
c∗6 E0
Z
0
T
1
dt,
α(t)
where c∗6 = (1 − c2 )c3 .
From (3.18), (3.20) and (3.25), we arrive at
Z
Z T
1 T
1
2
∗
β(0, t)|zx (0, t)| dt ≥ c6 E0
dt − c5 E0
2 0
α(t)
0
(3.25)
12
H. WANG, Y. HE, S. LI
≥ c∗6 E0
Z
0
T
EJDE-2015/31
1
c∗
dt − c5 E0 = ( 6 log(1 + c2 T ) − c5 )E0 .
1 + c2 t
c2
If we choose T1∗ as in (1.5), then it is easy to see that
c2 c5
1
T1∗ =
exp( ∗ ) − 1 ;
c2
c6
thus, when T > T1∗ ,
Z T
β(0, t)|zx (0, t)|2 dt ≥ C kz 0 k2H 1 (0,1) + kz 1 k2L2 (0,1) .
0
0
holds for C =
2c∗
6
c2
log(1 + c2 T ) − 2c5 > 0.
Acknowledgments. The first author was partially supported by the grant No.
61403239 of the NSFC. The second author was partially supported by the grant
No. 11401351 of the NSFC and by the grant No. 2014011005-2 of the Science
Foundation of Shanxi Province, China. The third author was partially supported
by the grant No. 61174082 of the NSFC, and by the grant No. 2011-006 of Shanxi
Scholarship Council of China.
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Hua Wang
School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China
E-mail address: 197wang@163.com
Yijun He
School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China
E-mail address: heyijun@sxu.edu.cn
EJDE-2015/31
EXACT CONTROLLABILITY OF A WAVE EQUATION
Shengjia Li (corresponding author)
School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China
E-mail address: shjiali@sxu.edu.cn
13