Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 239, pp. 1–19.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
BOUNDARY CONTROLLABILITY FOR A NONLINEAR
BEAM EQUATION
XIAO-MIN CAO
Abstract. This article concerns a nonlinear system modeling the bending
vibrations of a nonlinear beam of length L > 0. First, we derive the existence
of long time solutions near an equilibrium. Then we prove that the nonlinear
beam is locally exact controllable around the equilibrium in H 4 (0, L) and
with control functions in H 2 (0, T ). The approach we used are open mapping
theorem, local controllability established by linearization, and the induction.
1. Introduction and statement of main results
We consider a controllability problem for a system modeling the bending vibrations of a nonlinear beam of length L > 0. Let φ denote the deflection of the beam,
the left end of the beam is fixed, and an appropriate shear force u exerted on the
right end of the beam, then the equations of motion describing beam bending are
φtt + (a(x, φ)φ00 )00 = b(x, φ, φ0 , φ00 ),
φ(x, 0) = φ0 (x),
0
φ(0, t) = 0,
φ(L, t) = 0,
x ∈ (0, L), t ∈ (0, T ),
φt (x, 0) = φ1 (x),
φ (0, t) = 0,
φ0 (L, t) = u(t),
x ∈ (0, L),
t ∈ (0, T ),
(1.1)
t ∈ (0, T ),
where a(x, y), b(x, y1 , y2 , y3 ) are smooth functions on [0, L] × R and [0, L] × R3 ,
respectively, such that
a(x, y) > 0,
∀(x, y) ∈ [0, L] × R
b(x, 0, 0, 0) = 0,
∀x ∈ [0, L]
(1.2)
(1.3)
Let φ0 , φ1 , φb0 , φb1 be given functions and T > 0 be given. If there is boundary
function u on (0, T ) such that the solution of (1.1) satisfies φ(T ) = φb0 , φt (T ) = φb1
on [0, L], we say the system (1.1) is exactly controllable from (φ0 , φ1 ) to (φb0 , φb1 ) at
time T by controlled moment on the right end.
Boundary exact controllability on linear beam problems has been studied for
many years, see [8, 9, 11, 12, 15, 17, 18], and many others.
To the best of our knowledge, there is little about the boundary control of nonlinear beam equation in the existing related papers. Recently, Yao and G. Weiss [27]
2010 Mathematics Subject Classification. 93B05, 35L75, 93C10, 93C20.
Key words and phrases. Nonlinear beam equation; locally exact controllability;
equilibrium; smooth control.
c
2015
Texas State University - San Marcos.
Submitted May 4, 2015. Published September 17, 2015.
1
2
X.-M. CAO
EJDE-2015/239
consider a dynamical system with boundary input and output describing the bending vibrations of a quasi-linear beam, where the nonlinearity comes from Hooke’s
law. they show that the structure of the boundary input and output forces the
system to admit global solutions at least when the initial data and the boundary
input are small in a certain sense. And they prove that the norm of the state of the
system decays exponentially if the input becomes zero after a finite time. Cindea
and Tucsnak [3] study the exact controllability of a nonlinear plate equation by
the means of a control which acts on an internal region of the plate. For rectangular domains, they obtain that the Berger system is locally exactly controllable in
arbitrarily small time and for every open and nonempty control region.
Recently, using moment theory and Nash-Moser theorem, Beauchard [2] prove
that the linear beam equation with clamped ends is locally controllable in a H 5+ε ×
H 3+ε ((0, 1), R) neighborhood of a particular trajectory of the free system, with
ε > 0 and with control functions in H01 ((0, T ), R).
In the present work we will study boundary controllability for the nonlinear
beam equation (1.1) by using some ideas of [21] for the nonlinear system of wave
equations. First, we will derive the existence of long time solutions near an equilibrium. Then, the locally exact controllability of the system around the equilibrium
will be established.
Let us choose some Sobolev spaces to formulate the problems. To get a smooth
control, we assume initial data φ0 ∈ H 4 (0, L), φ1 ∈ H 2 (0, L) to study the controllability of the system around the equilibrium in H 4 (0, L) at time T via a boundary
control u ∈ H 2 (0, T ). Obviously, control function we obtained is more smooth than
the previous results [2, 9].
We say ω ∈ H 4 (0, L) is an equilibrium of the system (1.1) if
(a(x, ω)ω 00 )00 = b(x, ω, ω 0 , ω 00 ).
(1.4)
Let φ0 ∈ H 4 (0, L), φ1 ∈ H 2 (0, L), u ∈ H 2 (0, T ). We say these functions satisfy
compatibility conditions if
φ0 (0) = 0, φ00 (0) = 0, φ1 (0) = 0, φ01 (0) = 0,
φ0 (L) = 0, φ00 (L) = u(0), φ1 (L) = 0, φ01 (L) = u̇(0).
Set
V (0, L) = {ϕ|ϕ ∈ H 2 (0, L), ϕ(0) = ϕ(L) = ϕ0 (0) = 0.}
(1.5)
The next result shows that near one equilibrium, the system has solutions of
long time.
Theorem 1.1. Let w ∈ H 4 (0, L) ∩ V (0, L) be an equilibrium of (1.1). Let T > 0
be arbitrary given. Then there is εT > 0, which depends on the time T , such that
if φ0 ∈ H 4 (0, L) ∩ V (0, L), φ1 ∈ V (0, L) satisfy
kφ0 − ωk4 < εT , kφ1 k2 < εT ,
and u ∈ H 2 (0, T ) satisfies the compatibility conditions with (φ0 , φ1 ) and kuk2 < εT ,
then system (1.1) has a solution
φ ∈ C([0, T ], H 4 (0, L)) ∩ C 1 ([0, T ], H 2 (0, L)) ∩ C 2 ([0, T ], L2 (0, L)).
Near an equilibrium, we have the following locally exact controllability results.
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BOUNDARY CONTROLLABILITY
3
Theorem 1.2. Let ω ∈ H 4 (0, L) ∩ V (0, L) be an equilibrium of (1.1). Then, for
T > 0 given, there is εT > 0 such that for any (φi0 , φi1 ) ∈ (H 4 (0, L) ∩ V (0, L)) ×
V (0, L)(i = 1, 2) with
kφi0 − ωk4 < εT , kφi1 k2 < εT ,
we can find u ∈ H 2 (0, T ) which is compatible with (φ10 , φ11 ) such that the solution
of the system (1.1) with the initial data (φ10 , φ11 ) satisfies
φ(T ) = φ20 ,
φt (T ) = φ21 .
2. Existence of long time solutions near equilibria
The existence of short time solutions for the nonlinear beam equation can be
proved using standard arguments such as the nonlinear semigroups theory or the
Galerkin method and fixed point arguments [1, 4, 16, 20]. We only study some
energy estimates of the short time solutions to have long time solutions when initial
data are close to an equilibrium here.
We suppose that the equilibrium is the zero in this section. If an equilibrium
ω ∈ H 4 (0, L) ∩ V (0, L) is not zero, we can make a transform by φ = ω + ψ, and
consider the problem
ψtt + (a(x, ω + ψ)(ψ 00 + ω 00 ))00 = b(x, ω + ψ, ω 0 + ψ 0 , ω 00 + ψ 00 ),
x ∈ (0, L), t ∈ (0, T ),
ψ(x, 0) = φ0 − ω,
ψt (x, 0) = φ1 (x),
0
ψ(0, t) = 0, ψ (0, t) = 0,
ψ(L, t) = 0,
0
x ∈ (0, L),
t ∈ (0, T ),
0
ψ (L, t) = u(t) − ω (L),
t ∈ (0, T ),
Let φ ∈ C([0, T ], H 4 (0, L)) ∩ C 1 ([0, T ], H 2 (0, L)) ∩ C 2 ([0, T ], L2 (0, L)) be a solution of (1.1) for some T > 0. Suppose that u ∈ H 2 (0, T ). We introduce
E(t) = kφk24 + kφt k22 + kφtt k2 ,
EΓ (t) = u2 (t) + u̇2 (t) + ü2 (t),
Q(t) = kφt k2 + kφ00 k2 + kφ0t k2 + kφ000 k2 + kφtt k2 + kφ00t k2 .
For solutions of (1.1) near the zero equilibrium, we have the following result.
Theorem 2.1. Let γ > 0 be given and φ be a solution of (1.1) on the interval
[0, T ] for some T > 0 such that the condition
sup kφ(t)k4 ≤ γ.
(2.1)
0≤t≤T
holds. Then there is cγ > 0, which depends on the γ but is independent of initial
data (φ0 , φ1 ) and boundary functions u, such that
Z th
i
3
(2.2)
Q(t) ≤ cγ Q(0) + cγ
(1 + E 1/2 (τ ) + E(τ ) + E 2 (τ ))E(τ ) + EΓ (τ ) dτ.
0
and
Q(t) ≤ E(t) ≤ cγ Q(t) + cγ EΓ (t),
for t ∈ [0, T ].
(2.3)
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X.-M. CAO
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Here we list a few basic properties of Sobolev spaces to be invoked in the sequel.
(i) Let s1 > s2 ≥ 0. For any ε > 0 there is cε > 0 such that
kϕk2s2 ≤ εkϕk2s1 + cε kϕk2 ,
∀ϕ ∈ H s1 (0, L).
(ii) If m > 12 , then for each k = 0, 1, · · · , we have H m+k (0, L) ⊂ C k ([0, L]) with
continuous inclusion.
(iii) If r := min{s1 , s2 , s1 + s2 − 1} ≥ 0, then there is a constant c > 0 such that
kf gkr ≤ ckf ks1 kgks2 , ∀f ∈ H s1 (0, L), g ∈ H s2 (0, L).
(2.4)
Lemma 2.2. (i) Let f (x, y) be a smooth function on [0, L]×R. Set F (x) = f (x, φ),
for 0 ≤ k ≤ 4, then there is c = c(supx∈[0,L] |φ|) > 0, such that
k
X
kF kk ≤ c
(1 + kφk4 )j .
(2.5)
j=0
(ii) Let f (x, y1 , y2 , y3 ) be a smooth function on [0, L] × R3 . Set
G(x) = f (x, φ, φ0 , φ00 ),
then there is c = c supx∈[0,L] |φ|, supx∈[0,L] |φ0 |, supx∈[0,L] |φ00 | > 0, such that
kGk ≤ c.
(2.6)
Proof. (i) Inequality (2.5) is clearly true for k = 0. By using (2.4) and the induction
of the order k, inequality (2.5) follows.
(ii) A standard method as to the linearly elliptic problem can give inequality
(2.6), for example see Taylor [22].
Observing the partial differential of the function a(x, φ) and b(x, φ, φ0 , φ00 ), using
the formula (2.4) and Lemma 2.2, we have the following lemma.
Lemma 2.3. Let γ > 0 be given and φ be a solution of (1.1) on the interval [0,T]
for some T > 0 such that the condition (2.1) holds true. Then there is cγ > 0,
which depends on the γ, such that
kat k2 ≤ cγ E 1/2 (t), ka0t k1 ≤ cγ E 1/2 (t) + E(t) ,
3
ka00t k ≤ cγ E 1/2 (t) + E(t) + E 2 (t) , kbt k ≤ cγ E 1/2 (t).
Lemma 2.4. Let γ > 0 be given and φ be a solution of (1.1) on the interval [0,T]
for some T > 0 such that the condition (2.1) holds true. Set
Υ1 (t) = kφt k2 + kφ00 k2 ,
Υ2 (t) = kφ0t k2 + kφ000 k2 ,
Υ3 (t) = kφtt k2 + kφ00t k2 .
Then there is cγ > 0, which depends on the γ, such that
Z t
1
Υ1 (t) ≤ cγ Υ1 (0) + cγ
(1 + E 2 (τ ))E(τ ) + EΓ (τ ) dτ,
(2.7)
0
Υ2 (t) + Υ3 (t)
≤ cγ Υ2 (0) + Υ3 (0) + cγ
Z t h
i
3
1 + E 1/2 (τ ) + E(τ ) + E 2 (τ ) E(τ ) + EΓ (τ ) dτ
0
(2.8)
for 0 ≤ t ≤ T .
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BOUNDARY CONTROLLABILITY
5
Proof. Let
P1 (t) = kφt k2 + (aφ00 , φ00 ).
We obtain
Ṗ1 (t) = 2(φt , φtt ) + 2(aφ00 , φ00t ) + (at φ00 , φ00 )
= 2(φt , b − (aφ00 )00 ) + 2(aφ00 , φ00t ) + (at φ00 , φ00 )
= 2(φt , b) + 2u̇(t)(aφ00 )(L, t) + (at φ00 , φ00 ).
It follows that
Υ1 (t) ≤ cγ P1 (t)
t
Z
(kφt kkbk + kat k2 kφ00 k2 )dt + cγ
≤ cγ P1 (0) + cγ
Z
0
≤ cγ Υ1 (0) + cγ
t
2
(u̇2 (t) + φ00 (L, t))dt
0
Z th
i
(1 + E (τ ))E(τ ) + EΓ (τ ) dτ.
1
2
0
To obtain inequality (2.8), first we differentiate equation (1.1) with respect to x,
then we have
φ0tt + (aφ00 )000 = b0 .
(2.9)
Let
P2 (t) = kφ0t k2 + (aφ000 , φ000 ).
(2.10)
We obtain
000
000
Ṗ2 (t) = 2(φ0t , φ0tt ) + 2(aφ000 , φ000
t ) + (at φ , φ )
000
000
= 2(φ0t , b0 − (aφ00 )000 ) + 2(aφ000 , φ000
t ) + (at φ , φ )
= 2(φ0t , b0 ) − 2u̇(t)b(L, t) + 2φ00t (L, t)(aφ000 )(L, t) − 2φ00t (0, t)(aφ000 )(0, t)
− 2(φ00t , a0 φ000 ) + 2(φ00t , 2a0 φ000 + a00 φ00 ) + (at φ000 , φ000 ).
So
Υ2 (t) ≤ cγ P2 (t)
Z th
i
≤ cγ P2 (0) + cγ
(1 + E 1/2 (τ ))E(τ ) + u̇2 (τ ) dτ
0
Z t
Z t
2
2
2
00
00 2
+ε
φt (0, t) + φt (L, t) dt + cγ,ε
φ000 (0, t) + φ000 (L, t) dt.
0
0
(2.11)
Furthermore, we differentiate (1.1) with respect to t, then we have
φttt + (aφ00t )00 + (at φ00 )00 = bt .
(2.12)
P3 (t) = kφtt k2 + (aφ00t , φ00t ).
(2.13)
Let
We deduce that
Ṗ3 (t) = 2(φtt , φttt ) + 2(aφ00t , φ00tt ) + (at φ00t , φ00t )
= 2(φtt , bt − (at φ00 )00 − (aφ00t )00 ) + 2(aφ00t , φ00tt ) + (at φ00t , φ00t )
= 2(φtt , bt ) − 2(φtt , at φ(4) ) − 4(φtt , a0t φ000 ) − 2(φtt , a00t φ00 )
+ 2ü(t)(aφ00t )(L, t) + (at φ00t , φ00t ).
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X.-M. CAO
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So
Υ3 (t) ≤ cγ P3 (t)
Z t
3
(1 + E 1/2 (τ ) + E(τ ) + E 2 (τ ))E(τ ) dτ
≤ cγ P3 (0) + cγ
0
Z t
Z t
00 2
ü2 (t)dt.
φt (L, t)dt + cγ,ε
+ε
(2.14)
0
0
From (2.11) and (2.14), we conclude that
Z t
2
2
Υ2 (t) + Υ3 (t) ≤ cγ (P2 (0) + P3 (0)) + ε
φ00t (0, t) + φ00t (L, t) dt
0
Z t
3
+ cγ
(1 + E 1/2 (τ ) + E(τ ) + E 2 (τ ))E(τ ) + u̇2 (t) dτ
0
Z t
2
2
+ cγ,ε
φ000 (0, t) + φ000 (L, t) + ü2 (t) dt.
(2.15)
0
Rt 2
Rt 2
Next, we estimate the terms 0 φ00t (0, t)dt and 0 φ00t (L, t)dt.
Differentiating twice (1.1) with respect to x, multiplying the two sides of the
equation by (L − x)φ000 and integrating from 0 to L by parts, we obtain
φ00tt + (aφ00 )(4) = b00 ,
Z
L
φ00tt (L − x)φ000 dx =
0
1
1
∂ 00
2
(φ , (L − x)φ000 ) + Lφ00t (0, t) −
∂t t
2
2
(2.16)
Z
L
2
φ00t dx,
0
L
Z
(b00 − (aφ00 )(4) )(L − x)φ000 dx
0
2
1
= (b00 , (L − x)φ000 ) − b0 (0, t)Lφ000 (0, t) + ü1 (t)Lφ000 (0, t) − L(aφ(4) )(0, t)
2
Z
Z L
2
1 L 0
(a (L − x) − a)φ(4) dx +
−
(3a0 φ(4) + 3a00 φ000 + a000 φ00 )(L − x)φ(4) dx
2 0
0
Z L
− b(L, t)φ000 (L, t) + b(0, t)φ000 (0, t) +
(aφ00 )00 φ(4) dx.
0
Whence
1 00 2
Lφ (0, t)
2 t
Z
∂
1 L 00 2
φ dx + (b00 , (L − x)φ000 )
= − (φ00t , (L − x)φ000 ) +
∂t
2 0 t
Z
2
2
1
1 L 0
− b0 (0, t)Lφ000 (0, t) − L(aφ(4) )(0, t) −
(a (L − x) − a)φ(4) dx
2
2 0
Z L
+
(3a0 φ(4) + 3a00 φ000 + a000 φ00 )(L − x)φ(4) dx
0
000
000
Z
− b(L, t)φ (L, t) + b(0, t)φ (0, t) +
0
L
(aφ00 )00 φ(4) dx.
(2.17)
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BOUNDARY CONTROLLABILITY
It follows that
Z t
2
φ00t (0, t)dt ≤ cγ Υ2 (0) + Υ3 (0) + Υ2 (t) + Υ3 (t)
0
Z t h
i
3
+ cγ
1 + E 1/2 (τ ) + E(τ ) + E 2 (τ ) E(τ ) dτ.
7
(2.18)
0
Rt 2
Similarly, with respect to the term 0 φ00t (L, t)dt, multiplying the two sides of
(2.16) by xφ000 and integrating from 0 to L by parts, we obtain
Z L
Z
∂
1 L 00 2
1
2
φ00tt xφ000 dx = (φ00t , xφ000 ) + Lφ00t (L, t) −
(2.19)
φ dx,
∂t
2
2 0 t
0
and
Z L
(b00 − (aφ00 )(4) )xφ000 dx
0
2
1
= (b00 , xφ000 ) − b0 (L, t)Lφ000 (L, t) + ü(t)Lφ000 (L, t) + L(aφ(4) )(L, t)
2
Z
Z L
2
1 L 0
−
(a x + a)φ(4) dx +
(3a0 φ(4) + 3a00 φ000 + a000 φ00 )xφ(4) dx
2 0
0
Z L
+ b(L, t)φ000 (L, t) − b(0, t)φ000 (0, t) −
(aφ00 )00 φ(4) dx.
(2.20)
0
Using (2.19) and (2.20) we have
1 00 2
Lφ (L, t)
2 t
Z
1 L 00 2
∂ 00
000
φ dx + (b00 , xφ000 ) − b0 (L, t)Lφ000 (L, t)
= − (φt , xφ ) +
∂t
2 0 t
Z
2
1 L 0
1
(4) 2
000
)(L, t) −
(a x + a)φ(4) dx
+ ü(t)Lφ (L, t) + L(aφ
2
2 0
Z L
+
(3a0 φ(4) + 3a00 φ000 + a000 φ00 )xφ(4) dx
(2.21)
0
+ b(L, t)φ000 (L, t) − b(0, t)φ000 (0, t) −
Z
L
(aφ00 )00 φ(4) dx.
0
We deduce the inequality
Z t
2
φ00t (L, t)dt ≤ cγ Υ2 (0) + Υ3 (0) + Υ2 (t) + Υ3 (t)
0
(2.22)
Z t h
i
3
+ cγ
1 + E 1/2 (τ ) + E(τ ) + E 2 (τ ) E(τ ) + ü2 (t) dτ.
0
Finally, Substituting inequality (2.18) and (2.22) in (2.15), choosing ε > 0 small
enough such that the term εcγ [Υ2 (t) + Υ3 (t)] can be moved to the left hand side
of the inequality, then (2.8) is obtained.
Proof of Theorem 2.1. Lemma 2.4 gives inequality (2.2). To prove inequality (2.3),
we notice that aφ(4) = b − φtt − 2a0 φ000 − a00 φ00 , then there is cγ > 0, such that
kφ(4) k2 ≤ cγ (aφ(4) , φ(4) )
= cγ (b − φtt − 2a0 φ000 − a00 φ00 , φ(4) )
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X.-M. CAO
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≤ cγ,ε (kbk2 + kφtt k2 + kφ000 k2 + kφ00 k2 ) + εkφ(4) k2 .
Choosing ε sufficiently small, then there is cγ > 0, such that
kφ(4) k2 ≤ cγ (kbk2 + kφtt k2 + kφ000 k2 + kφ00 k2 ) ≤ cγ Q(t).
From the definition of Q(t) and E(t) and using the Poincaré inequality, inequality
(2.3) holds.
Using the standard method as for the global solution of partial differential equation, from Theorem 2.1 we have the following proof.
Proof of Theorem 1.1. Clearly, it will suffice to prove Theorem 1.1 for the zero
equilibrium ω = 0.
Let T1 > 0 be arbitrary given. We take γ = 1. Let
c1 = cγ ≥ 1
(2.23)
be fixed such that the corresponding inequalities (2.2) and (2.3) of Theorem 2.1
hold for t in the existence interval of the solution φ.
We shall prove that, if the initial data (φ0 , φ1 ) and boundary value u are compatible and satisfy
Z T1
1 −4c21 T1
EΓ (t)dt ≤
E(0) + max EΓ (t) +
,
(2.24)
3e
0≤t≤T1
16c
0
1
then the solution of problem (1.1) exists at least on the interval [0, T1 ].
We set
1
1
< .
η=
4c1
2
Since E(0) ≤ η4 , the solution of short time must satisfy
(2.25)
1
(2.26)
2
for some interval [0, δ]. Let δ0 be the largest number such that (2.26) is true for
t ∈ [0, δ0 ). We shall prove δ0 ≥ T1 by contradiction.
Suppose that δ0 < T1 . In this interval t ∈ [0, δ0 ], the condition (2.1) is true. We
apply Theorem 2.1 and the inequalities (2.2) and (2.3), via (2.23)–(2.26). Then we
conclude that
Z T1
Z t
2
2
E(t) ≤ 2c1 [E(0) + max EΓ (t) +
EΓ (t)dt] + 4c1
E(t)dt,
(2.27)
E(t) ≤ η ≤
0≤t≤T1
0
0
for t ∈ [0, δ0 ]. By (2.24) and (2.27), the Gronwall inequality yields
E(δ0 ) ≤ η/2 < η.
This is a contradiction.
3. Locally exact controllability
Let T > 0 given. The first step of the proof for the local exact controllability
depends on the following fact: Let X and Y be Banach spaces and Φ : U → Y ,
where U is an open subset of X, be Frechét differentiable. If Φ0 (x0 ) : X → Y
is surjective, then there is an open neighborhood of y0 = Φ(x0 ) contained in the
image Φ(U ).
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BOUNDARY CONTROLLABILITY
9
Given an equilibrium ω ∈ H 4 (0, L) ∩ V (0, L), We invoke Theorem 1.1 to define
the map for u ∈ H 2 (0, T ) by setting
F (u) = (ψ(·, T ), ψt (·, T )),
where ψ is the solution to
ψtt + (a(x, ψ)ψ 00 )00 = b(x, ψ, ψ 0 , ψ 00 ),
ψ(x, 0) = ω(x),
x ∈ (0, L),
ψt (x, 0) = 0,
0
ψ(0, t) = 0,
ψ(L, t) = 0,
x ∈ (0, L), t ∈ (0, T ),
t ∈ (0, T ),
ψ (0, t) = 0,
ψ 0 (L, t) = u + ω 0 (L),
(3.1)
t ∈ (0, T ),
Let εT > 0 be given by Theorem 1.1. Then
F : B(0, εT ) → (H 4 (0, L) ∩ V (0, L)) × V (0, L),
(3.2)
2
where B(0, εT ) ⊂ H (0, T ) is the ball with the radius εT centered at 0. We note
that F (0) = (ω(x), 0).
We need to evaluate F 0 (0)(u) = Dλ F (λu)|λ=0 . It is easy to check that F 0 (0)(u) =
(ψ(·, T ), ψt (·, T )), where ψ(x, t) is the solution of the linear system
ψtt + (p(x)ψ 00 )00 = b1 (x)ψ + b2 (x)ψ 0 + b3 (x)ψ 00 ,
ψ(x, 0) = 0,
ψ(0, t) = 0,
ψ(L, t) = 0,
x ∈ (0, L),
ψt (x, 0) = 0,
0
ψ (0, t) = 0,
x ∈ (0, L), t ∈ (0, T ),
t ∈ (0, T ),
ψ 0 (L, t) = u(t),
(3.3)
t ∈ (0, T ),
and
p(x) = a(x, ω),
b1 (x) = by1 (x, ω, ω 0 , ω 00 ) − (ay (x, ω)ω 00 )00 ,
b2 (x) = by2 (x, ω, ω 0 , ω 00 ) − 2(ay (x, ω)ω 00 )0 ,
b3 (x) = by3 (x, ω, ω 0 , ω 00 ) − ay (x, ω)ω 00 .
We now verify that F 0 (0) is surjective. In the language of control theory surjectivity
is just exact controllability, which for a reversible system such as (3.3) is equivalent
to null controllability.
Explicitly one has to show that, for T > 0 given, given ψ0 ∈ H 4 (0, L) ∩
V (0, L), ψ1 ∈ V (0, L), one can find u ∈ H 2 (0, T ) such that the solution to
ψtt + (p(x)ψ 00 )00 = b1 (x)ψ + b2 (x)ψ 0 + b3 (x)ψ 00 ,
ψ(x, 0) = ψ0 (x),
ψ(0, t) = 0,
ψ(L, t) = 0,
ψt (x, 0) = ψ1 (x),
0
ψ (0, t) = 0,
ψ 0 (L, t) = u(t),
x ∈ (0, L), t ∈ (0, T ),
x ∈ (0, L),
t ∈ (0, T ),
(3.4)
t ∈ (0, T ),
satisfies ψ(·, T ) = 0 and ψt (·, T ) = 0.
Theorem 3.1. Given an equilibrium ω ∈ H 4 (0, L) ∩ V (0, L). Let T > 0 given.
Then for any
(ψ 0 , ψ 1 ) ∈ (H 4 (0, L) ∩ V (0, L)) × V (0, L),
there is a control u ∈ H 2 (0, T ) such that the solution
ψ ∈ C([0, T ], H 4 (0, L)) ∩ C 1 ([0, T ], H 2 (0, L)) ∩ C 2 ([0, T ], L2 (0, L))
of problem (3.4) satisfies ψ(·, T ) = 0 and ψt (·, T ) = 0.
10
X.-M. CAO
EJDE-2015/239
Unfortunately, our proof does not give any information on the time T needed
for the controlled motion.
3.1. Distributed control. As to the exact controllability of linear systems by distributed control, there is a long history and a lot of results where many approaches
are involved. Here the distributed control means that solutions (ψ(t), ψt (t)) of the
controlled system (3.4) are only in the space L2 (0, L) × H −2 (0, L) for t ∈ [0, T ].
One of the useful approaches is the multiplier method, which introduced by Ho
[6] where Lions [15] provided a key technique of multipliers for observability estimates, to control the linear system by its duality system. In the case of constant
coefficients, Lagnese [9, 11], study the controllability of linear beam by using the
multiplier method. In this subsection, we will consider the controllability of linear
beam with variable coefficients.
As noted in [9], everything is therefore going to rely on a prior inequality
Z T
2
k(φ0 , φ1 )k2H 2 ×L2 ≤ CT
φ00 (L, t)dt, ∀(φ0 , φ1 ) ∈ H02 (0, L) × L2 (0, L) (3.5)
0
0
where φ is the solution of the system
φtt + (p(x)φ00 )00 = b1 (x)φ − (b2 (x)φ)0 + (b3 (x)φ)00 ,
φ(x, 0) = φ0 (x),
x ∈ (0, L),
φt (x, 0) = φ1 (x),
0
x ∈ (0, L), t ∈ (0, T ),
φ(0, t) = 0, φ (0, t) = 0,
t ∈ (0, T ),
φ(L, t) = 0, φ0 (L, t) = 0,
t ∈ (0, T ),
(3.6)
Given φ0 ∈ H02 (0, L), φ1 ∈ L2 (0, L). We define φ as the solution to (3.6) and we
define η as the solution of
ηtt + (p(x)η 00 )00 = b1 (x)η + b2 (x)η 0 + b3 (x)η 00
η(x, T ) = 0,
η(0, t) = 0,
0
η (0, t) = 0,
x ∈ (0, L),
t ∈ (0, T ),
η 0 (L, t) = φ00 (L, t),
η(L, t) = 0,
We define the operator Λ :
ηt (x, T ) = 0,
H02 (0, L)
x ∈ (0, L), t ∈ (0, T ),
(3.7)
t ∈ (0, T ),
× L (0, L) → H −2 (0, L) × L2 (0, L) by
2
λ(φ0 , φ1 ) = (ηt (0), −η(0))
(3.8)
The solution η of (3.7) is a weak solution defined by transposition, in such a way
that Green’s formula makes sense. Therefore
Z T
2
hΛ(φ0 , φ1 ), (φ0 , φ1 )i =
p(L)φ00 (L, t)dt.
(3.9)
0
Lemma 3.2. Let φ be a solution of (3.6). Then there exist constants c, c0 > 0
such that for T ≥ t > 0:
e−ct E(0) − N (T ) ≤ E(t) ≤ [E(0) + N (T )]ect ,
where we introduced the energy
Z
E(t) =
0
and set N (T ) = c0
RT
0
kφk21 dt.
L
φ2t + p(x)φ00
2
dx
(3.10)
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BOUNDARY CONTROLLABILITY
11
Proof. Multiplying equation (3.6) by φt and integrating over (s, t) × (0, L) by parts
in t on the left-hand side we obtain that for all s, t:
Z tZ L φt b1 (x)φ − (b2 (x)φ)0 + (b3 (x)φ)00 dx dt.
(3.11)
E(t) = E(s) + 2
s
0
By Schwartz inequality, we obtain for t ≥ s ≥ 0:
Z t
Z tZ L E(τ )dτ.
φt b1 (x)φ − (b2 (x)φ)0 + (b3 (x)φ)00 dx dt ≤ N (T ) + c
2
s
s
0
So
Z
t
E(t) ≤ [E(s) + N (T )] + c
E(τ )dτ,
(3.12)
E(τ )dτ.
(3.13)
s
Z
t
E(s) ≤ [E(t) + N (T )] + c
s
We apply the classical argument of the Gronwall’s inequality to (3.12) and (3.13),
where we note that the terms into the square brackets are independent of s in
(3.13), we thus obtain for t ≥ s ≥ 0:
E(t) ≤ [E(s) + N (T )]ec(t−s) ;
E(s) ≤ [E(t) + N (T )]ec(t−s) .
Setting s = 0 and thus t > 0 in (3.14) yields (3.10).
(3.14)
The observability inequality for the system (3.6) is covered in the following
lemma.
Lemma 3.3. Let an equilibrium ω ∈ H 4 (0, L) ∩ V (0, L) be given. Then for T >
0, Λ is an isomorphism from H02 (0, L) × L2 (0, L) onto H −2 (0, L) × L2 (0, L). In
particular, there are constants C1 > 0 and C2 > 0 such that the inequality
Z T
2
C1 k(φ0 , φ1 )k2H 2 ×L2 ≤
p(L)φ00 (L, t)dt ≤ C2 k(φ0 , φ1 )k2H 2 ×L2
(3.15)
0
holds true for (φ0 , φ1 ) ∈
0
H02 (0, L)
0
2
× L (0, L).
Proof. Assume that φ0 ∈ H02 (0, L), φ1 ∈ L2 (0, L). Then (3.6) admits a unique
solution φ ∈ C([0, T ]; H02 (0, L)) ∩ C 1 ([0, T ]; L2 (0, L)) and the energy E(t) of the
system (3.6) satisfies
Z L
2
E(0) = E(φ0 , φ1 ) =
(φ21 + p(x)φ000 )dx.
0
That is, we are going to prove that there exist two constants C1 , C2 > 0 such that:
Z T
2
C1 E(0) ≤
p(L)φ00 (L, t)dt ≤ C2 E(0).
(3.16)
0
We always have the right side of (3.16), so we just need to prove that
Z T
2
p(L)φ00 (L, t)dt ≥ C1 E(0).
0
We use multiplier h(x)φ0 , where h(x) is a function on the interval [0, L], and we
obtain
Z LZ T
Z Z
1 T L 0
0
0
T
φtt h(x)φ dx dt = (h(x)φ , φt )|0 +
h (x)φ2t dx dt.
2 0 0
0
0
12
X.-M. CAO
T
Z
EJDE-2015/239
L
Z
(p(x)φ00 )00 h(x)φ0 dx dt
0
0
T
1
=−
2
Z
1
−
2
Z
2
p(x)h(x)φ00 |L
0 dt
0
T
L
Z
Z
0
T
0
Z
Z
T
L
Z
2
p(x)h0 (x)φ00 dx dt
0
0
1
2
p0 (x)h(x)φ00 dx dt −
2
0
T Z L
0
3
+
2
Z
T
L
Z
2
(p(x)h00 (x))0 φ0 dx dt.
0
0
(b1 (x)φ − (b2 (x)φ)0 + (b3 (x)φ)00 )hφ0 dx dt
L
Z
=
m1 (x)φ2 + m2 (x)φ0
2
dx dt.
0
0
These equalities give
(h(x)φ
−
−
1
2
Z
1
2
Z
Z
0
Z Z
3 T L
2
h
dx dt +
p(x)h0 (x)φ00 dx dt
2 0 0
0
0
Z TZ L
Z L
1
2
2
p0 (x)h(x)φ00 dx dt −
(p(x)h00 (x))0 φ0 dx dt
2 0 0
0
T
0
Z
1
+
2
, φt )|T0
T
T
Z
L
0
(x)φ2t
2
p(x)h(x)φ00 |L
0 dt
0
T Z L
=
0
m1 (x)φ2 + m2 (x)φ0
2
dx dt.
0
Next we use another multiplier h0 (x)φ. Integrating by part on [0, T ] × [0, L], we
obtain
Z
L
T
Z
0
0
φtt h (x)φ dx dt = (h
0
Z
(x)φ, φt )|T0
0
h0 (x)φ2t dx dt.
0
(p(x)φ00 )00 h0 (x)φ dx dt
0
T
Z
L
T
Z
0
Z
−
0
L
T
Z
Z
L
p(x)h0 (x)(φ00 )2 dx dt +
=
0
0
T Z L
Z
−
0
Z
0
1
2
Z
0
T
Z
L
(p(x)h000 (x))00 φ2 dx dt
0
2
p(x)h000 (x) + (p(x)h00 (x))0 φ0 dx dt ,
0
T
Z
L
b1 (x)φ − (b2 (x)φ)0 + (b3 (x)φ)00 h0 (x)φ dx dt
0
Z
T
Z
=
0
L
m3 (x)φ2 + m4 (x)φ0
2
dx dt,
0
where functions mi (x)(i = 1, 2, 3, 4) are the function of b1 (x), b2 (x), b3 (x), h(x)
and their derivatives. Considering the fact that functions mi (x)(i = 1, 2, 3, 4) are
all lower order term, we omit specific functions form here.
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BOUNDARY CONTROLLABILITY
By using the above equalities, we have
Z
1 T
2
p(x)h(x)φ00 |L
0 dt
2 0
T Z T Z L
1
h0 (x)φ2t dx dt
= h(x)φ0 − h0 (x)φ, φt +
2
0
0
0
Z Z
1 T L
0
00 2
+
p(x)h (x)φ dx dt
2 0 0
Z Z
1 T L
2
+
(p(x)h0 (x) − p0 (x)h(x))φ00 dx dt
2 0 0
Z TZ L
2
[d1 (x)φ2 + d2 (x)φ0 ] dx dt,
−
13
(3.17)
0
0
Where functions d1 (x), d2 (x) are the function of b1 (x), b2 (x), b3 (x), h(x) and their
derivatives.
Let h(x) be the solution of the problem
b
h + 1,
p
h(0) = 0.
hx =
(3.18)
where b(x) = max(px , 0), e.g.,
h(x) = e
Rx
0
b
p dx
Z
x
e−
Rs
b
dτ
0 p
ds,
0 < x < L.
0
Therefore, if we introduce
T
1
Y = h(x)φ0 − h0 (x)φ, φt .
2
0
From (3.17), we obtain
Z
Z T
Z TZ L
1 T
2
2
p(L)h(L)φ00 (L, t)dt ≥
E(t)dt − |Y | − C
(φ2 + φ0 ) dx dt. (3.19)
2 0
0
0
0
For |Y |, we have
Z
L
Z
L
2
+ Cε
(φ2 + φ0 )dx|T0
0
0
≤ ε E(T ) + E(0) + Cε kφ(T )k2H 1 + kφ(0)k2H 1 .
|Y | ≤ ε
2
φt dx|T0
Using the inequality in Lemma 3.2, we compute
Z T
Z T
N (T )
,
E(t)dt ≥
e−ct dt E(0) − T N (T ) ≥ kT [E(0) + E(T )] −
2
0
0
RT
−cT
where kT = ( 0 e−ct dt) e 2 and E(T ) ≥ e−cT E(0) − N (T ).
Using inequalities (3.19)-(3.20) and (3.15), we obtain
Z
1 T
2
p(L)h(L)φ00 (L, t)dt
2 0
Z TZ L
N (T )
2
≥ kT [E(0) + E(T )] −
−C
(φ2 + φ0 ) dx dt
2
0
0
(3.20)
14
X.-M. CAO
−ε
n
EJDE-2015/239
o
E(T ) + E(0) + Cε kφ(T )k2H 1 + kφ(0)k2H 1
≥ (kT − ε)[E(0) + E(T )] − CT kφk2L∞ ([0,T ];H 1 (0,L))
≥ (kT − ε)[E(0) + e−cT E(0) − N (T )] − CT kφk2L∞ ([0,T ];H 1 (0,L)) .
Choosing ε > 0 small enough, such that kT − ε > 0, we obtain
Z T
2
αE(0) ≤
p(L)h(L)φ00 (L)dt + CT kφk2L∞ ([0,T ];H 1 (0,L)) ,
(3.21)
0
where α = 2(kT − ε)(1 + e−CT ).
By using a compactness and uniqueness argument similar to Lagnese [11], we
can easily prove that there exist constant C > 0,such that
Z T
2
2
kφkL∞ (0,T ;H 1 (0,L)) ≤ C
φ00 (L, t)dt.
0
So the proof is complete.
Therefore, Λ defines an isomorphism from H02 (0, L) × L2 (0, L) onto H −2 (0, L) ×
L (0, L). For ψ0 ∈ L2 (0, L), ψ1 ∈ H −2 (0, L) given , we solve
2
Λ(φ0 , φ1 ) = (ψ1 , −ψ0 ).
Then we define φ as the corresponding solution of (3.6) and we take
u(t) = φ00 (L, t) ∈ L2 (0, T ),
which is the control driving the system to rest at time t = T .
Let us now introduce a new norm
Z T
2
k(φ0 , φ1 )k2new =
p(L)φ00 (L, t)dt.
0
It follows from (3.15) that for T > 0, k(φ0 , φ1 )knew is a norm which is equivalent
to the H02 (0, L) × L2 (0, L) norm.
However, the above control strategy only gives distributed control functions because solutions (η, ηt ) of the controlled system (3.7) are only in L2 × H −2 no matter
(φ0 , φ1 ) are smooth or not.
3.2. Smooth control. Smooth control has been considered by Lasiecka and Triggiani [13, 14], Tataru [23]. Here we shall modify the above control strategy to obtain
smooth control to meet the need of Theorem 3.1 by induction on the order of the
space.
Firstly, we define operator B by
Bu = −(p(x)u00 )00 + b1 (x)u − (b2 (x)u)0 + (b3 (x)u)00 ,
∀u ∈ H 4 (0, L).
(3.22)
Let T > 0 be given. We assume that z ∈ C ∞ (−∞, ∞) is such that 0 ≤ z(t) ≤ 1
with
(
0, t ≥ T
z(t) =
(3.23)
1, t ≤ 0
EJDE-2015/239
BOUNDARY CONTROLLABILITY
15
For (φ0 , φ1 ) ∈ (H 4 (0, L) ∩ H02 (0, L)) × H02 (0, L) given, we solve (3.6) and then,
instead of (3.7), we solve the problem
ηtt + (p(x)η 00 )00 = b1 η + b2 η 0 + b3 η 00 ,
η(x, T ) = 0,
η(0, t) = 0,
ηt (x, T ) = 0,
0
η (0, t) = 0,
x ∈ (0, L), t ∈ (0, T ),
x ∈ (0, L),
t ∈ (0, T ),
η 0 (L, t) = z(t)φ00 (L, t),
η(L, t) = 0,
(3.24)
t ∈ (0, T ),
Let Λ be given by (3.8) where η is the solution of (3.24) this time. It is easy to
check that, for any (φ0 , φ1 ), (ϕ0 , ϕ1 ) ∈ H02 (0, L) × L2 (0, L),
Z T
z(t)p(L)φ00 (L, t)ϕ00 (L, t)dt,
(3.25)
hΛ(φ0 , φ1 ), (ϕ0 , ϕ1 )iL2 ×L2 =
0
where φ and ϕ are solutions of (3.6) with initial data (φ0 , φ1 ) and (ϕ0 , ϕ1 ), respectively.
We shall show that problem (3.24) provides smooth controls to Theorem 3.1 by
the following lemma.
Lemma 3.4. Let Λ be given by (3.8) where η is the solution of (3.24). Then there
are constants c1 > 0 and c2 > 0 such that
c1 k(φ0 , φ1 )kH 4 (0,L)×H 2 (0,L) ≤ kΛ(φ0 , φ1 )kL2 (0,L)×H 2 (0,L)
(3.26)
≤ c2 k(φ0 , φ1 )kH 4 (0,L)×H 2 (0,L) ,
for all (φ0 , φ1 ) ∈ (H 4 (0, L) ∩ H02 (0, L)) × H02 (0, L),
Proof. From the previous argument,
c1 k(φ0 , φ1 )kH 2 (0,L)×L2 (0,L) ≤ kΛ(φ0 , φ1 )kH −2 (0,L)×L2 (0,L)
≤ c2 k(φ0 , φ1 )kH 2 (0,L)×L2 (0,L)
(3.27)
holds. Let (φ0 , φ1 ) ∈ (H 4 (0, L) ∩ H02 (0, L)) × H02 (0, L) be given, Suppose that φ is
the solution of problem (3.6) corresponding to the initial data (φ0 , φ1 ).
For (ϕ0 , ϕ1 ) ∈ (H 6 (0, L) ∩ H02 (0, L)) × (H 4 (0, L) ∩ H02 (0, L)), let ϕ is the solution
of (3.6) corresponding to the initial data (ϕ0 , ϕ1 ). Then ϕt and ϕtt are the solutions
of (3.6) corresponding to the initial data (ϕ1 , Bϕ0 ) and (Bϕ0 , Bϕ1 ), respectively.
Using the initial data (φ0 , φ1 ) and (Bϕ0 , Bϕ1 ) in the formula (3.25), we obtain
Z T
(η(0), Bϕ1 ) − (ηt (0), Bϕ0 ) = −
z(t)p(L)φ00 (L, t)ϕ00tt (L, t)dt.
(3.28)
0
On the one hand, integrating by parts with respect to the variable t on [0, T ], we
obtain
Z T
−
z(t)p(L)φ00 (L, t)ϕ00tt (L, t)dt
0
=
p(L)φ000 (L)ϕ001 (L)
Z
+ p(L)
T
zt (t)φ00 (L, t)ϕ00t (L, t)dt
0
Z
+ p(L)
T
z(t)φ00t (L, t)ϕ00t (L, t)dt
0
On the other hand, using (3.22),
(η(0), Bϕ1 )
16
X.-M. CAO
L
Z
η(0)(p(x)ϕ001 )00 dx +
=−
L
Z
EJDE-2015/239
η(0) b1 (x)ϕ1 − (b2 (x)ϕ1 )0 + (b3 (x)ϕ1 )00 dx
0
0
= p(L)φ000 (L)ϕ001 (L) −
Z
L
(p(x)η 00 (0))00 ϕ1 dx
0
L
Z
+
b1 (x)η(0) + b2 (x)η 0 (0) + b3 (x)η 00 (0) ϕ1 dx
0
then
(η(0), Bϕ1 ) − (ηt (0), Bϕ0 )
Z L
00
00
(p(x)η 00 (0))00 ϕ1 dx
= p(L)φ0 (L)ϕ1 (L) −
0
L
Z
+
b1 (x)η(0) + b2 (x)η 0 (0) + b3 (x)η 00 (0) ϕ1 dx − (ηt (0), Bϕ0 ).
0
So we have the identity
I(φ, ϕ) = (B ∗ η(0), ϕ1 ) − (ηt (0), Bϕ0 ),
(3.29)
where
Z
I(φ, ϕ) = p(L)
T
zt (t)φ
00
(L, t)ϕ00t (L, t)dt
Z
+ p(L)
0
T
z(t)φ00t (L, t)ϕ00t (L, t)dt,
0
B ∗ η(0) = −(p(x)η 00 (0))00 + b1 (x)η(0) + b2 (x)η 0 (0) + b3 (x)η 00 (0).
Since (H 6 (0, L) ∩ H02 (0, L)) × (H 4 (0, L) ∩ H02 (0, L)) is dense in (H 4 (0, L) ∩
× H02 (0, L), the identity (3.29) is true for all (ϕ0 , ϕ1 ) ∈ (H 4 (0, L) ∩
× H02 (0, L). Letting ϕ0 = 0 in (3.29), we obtain
H02 (0, L))
H02 (0, L))
I(φ, ϕ) = (B ∗ η(0), ϕ1 ),
(3.30)
for ϕ1 ∈ H02 (0, L) where ϕ is the solution of (3.6) for the initial data (0, ϕ1 ).
Moreover, by inequality (3.15), we have the estimate
I(φ, ϕ)|
T
Z
zt (t)φ00 (L, t)ϕ00t (L, t)dt + p(L)
= p(L)
0
≤c
hZ
T
z(t)φ00t (L, t)ϕ00t (L, t)dt
0
T
p(L)(φ00 (L, t))2 dt +
Z
0
×
Z
T
i1/2
p(L)(φ00t (L, t))2 dt
0
Z
T
(3.31)
1/2
p(L)(ϕ00t (L, t))2 dt
0
≤ c E(φ0 , φ1 ) + E(φ1 , Bφ0 )
1/2 Z
T
1/2
p(L)(ϕ00t (L, t))2 dt
0
≤ c kφ0 k24 + kφ1 k22
1/2
kϕ1 k2 .
In terms of (3.30)-(3.31), we obtain
kB ∗ η(0)k−2 ≤ sup
B ∗ η(0), ϕ1 ≤ c(kφ0 k24 + kφ1 k22 )1/2 .
kϕ1 k2 =1
(3.32)
EJDE-2015/239
BOUNDARY CONTROLLABILITY
17
Now using the ellipticity of the operator B ∗ and from (3.32), we have
kη(0)k2 ≤ c kη(0)k0 + kB ∗ η(0)k−2
≤ c kη(0)k0 + (kφ0 k24 + kφ1 k22 )1/2
≤ c(kφ0 k24 + kφ1 k22 )1/2 .
where kη(0)k0 ≤ c(kφ0 k22 + kφ1 k20 )1/2 is used. A similar argument yields
kηt (0)k0 ≤ c(kφ0 k24 + kφ1 k22 )1/2 ,
4
(3.33)
H02 (0, L)), ϕ1
after we let ϕ0 ∈ (H (0, L) ∩
= 0 in (3.29).
Next, let us prove the left hand side of inequality (3.26). We set φ0 = ϕ0 and
φ1 = ϕ1 in (3.29),
Z T1
Z T
p(L)(φ00t (L, t))2 dt − cε
p(L)(φ00 (L, t))2 dt
I(φ, φ) ≥
0
0
T
Z
p(L)(φ00t (L, t))2 ]dt
−ε
0
Z
T1
≥
(3.34)
p(L)(φ00t (L, t))2 dt − cε E(φ0 , φ1 ) − εE(φ1 , Bφ0 )
0
≥ C1 E(φ1 , Bφ0 ) − cε (kφ0 k22 + kφ1 k20 ).
Since
(B ∗ η(0), φ1 )
Z L
Z
=−
p(x)η 00 (0)φ001 dx +
0
L
b1 (x)η(0) + b2 (x)η 0 (0) + b3 (x)η 00 (0) φ1 dx,
0
we have
(B ∗ η(0), φ1 ) − (ηt (0), Bφ0 )
≤ c(kη(0)k2 kφ1 k2 + kηt (0)k0 kBφ0 k0 )
1/2 1/2
≤ c kη(0)k22 + kηt (0)k20
kφ1 k22 + kBφ0 k20
.
(3.35)
Using (3.34)-(3.35) and the induction assumption
c(kφ0 k22 + kφ1 k20 ) ≤ kη(0)k20 + kηt (0)k2−2 ,
it follows that
kφ0 k24 + kφ1 k22 ≤ cE(φ1 , Bφ0 ) + c kφ0 k22 + kφ1 k20
≤ c kη(0)k22 + kηt (0)k20 + c kφ0 k22 + kφ1 k20
≤ c kη(0)k22 + kηt (0)k20 .
The proof is complete.
A similar argument can used for establishing the inequality
c1 k(φ0 , φ1 )kH 6 (0,L)×H 4 (0,L) ≤ kΛ(φ0 , φ1 )kH 2 (0,L)×H 4 (0,L)
≤ c2 k(φ0 , φ1 )kH 6 (0,L)×H 4 (0,L) .
for all (φ0 , φ1 ) ∈ (H 6 (0, L) ∩ H02 (0, L)) × (H 4 (0, L) ∩ H02 (0, L)).
(3.36)
18
X.-M. CAO
EJDE-2015/239
Proof of Theorem 3.1. It follows that the operator
Λ : (H 6 (0, L) ∩ H02 (0, L)) × (H 4 (0, L) ∩ H02 (0, L)) → V (0, L) × (H 4 (0, L) ∩ V (0, L))
is surjective. Let (ψ0 , ψ1 ) ∈ (H 4 (0, L) ∩ V (0, L)) × V (0, L) be given, then there is
(φ0 , φ1 ) ∈ (H 6 (0, L) ∩ H02 (0, L)) × (H 4 (0, L) ∩ H02 (0, L)) such that the control
u(t) = z(t)φ00 (L, t)
which drives system (3.4) to rest at the time T , where φ is the solution of (3.6)
with the initial data (φ0 , φ1 ).
Since φt , φtt is the solution of (3.6) with the initial data (φ1 , Bφ0 ) and (Bφ0 , Bφ1 )
respectively, we conclude that φ00t (L, t),φ00tt (L, t) ∈ L2 (0, T ) from Lemma 3.3. Then
u ∈ H 2 (0, T ).
If we change the boundary control condition into φ(0, t) = 0, φ00 (0, t) = 0,
φ(L, t) = 0, φ00 (L, t) = u(t) or others, the methods in this article still be applicable.
Acknowledgements. This research is supported by the National Science Foundation of China, grants no. 61104129, no. 11201272 and no. 11171195. I want
to thank Professor Peng-Fei Yao for his valuable guidance. The author would like
to thank the editor and the reviewer for their valuable comments and constructive
suggestions which help to improve the presentation of this article.
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Xiao-Min Cao
School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China
E-mail address: caoxm@sxu.edu.cn