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Received: 14 February 2023 Revised: 2 July 2023 Accepted: 13 September 2023
DOI: 10.1002/mma.9692
RESEARCH ARTICLE
Boundary controllability for a degenerate beam equation
Alessandro Camasta1Genni Fragnelli2
1Mathematics Department, University of
Bari Aldo Moro, Bari, Italy
2Department of Ecology and Biology,
Tuscia University, Viterbo, Italy
Correspondence
Genni Fragnelli, Department of Ecology
and Biology, Tuscia University, Largo
dell'Università, 01100 Viterbo, Italy.
Email: genni.fragnelli@unitus.it
Communicated by: B. Harrach
Funding information
FFABR "Fondo per il finanziamento delle
attività base di ricerca" 2017; INdAM
GNAMPA Project, Grant/Award Number:
codiceCUP_E53C22001930001; Horizon
Europe Seeds STEPs;
Horizon_EU_DM737_project 2022 COPS
at Tuscia University; PRIN 2017-2019
"Qualitative and quantitative aspects of
nonlinear PDEs"; Project "Mathematical
models for interacting dynamics on
networks (MAT-DYN-NET) CA18232"
The paper deals with the controllability of a degenerate beam equation. In par-
ticular, we assume that the left end of the beam is fixed, while a suitable control
𝑓acts on the right end of it. As a first step, we prove the existence of a solution for
the homogeneous problem, then we prove some estimates on its energy. Thanks
to them, we prove an observability inequality, and using the notion of solution
by transposition, we prove that the initial problem is null controllable.
KEYWORDS
boundary observability, degenerate beam equation, fourth-order operator, null controllability
MSC CLASSIFICATION
35L80, 93B05, 93B07
1INTRODUCTION
We consider a boundary controllability problem for a system modeling the bending vibrations of a degenerate beam of
length L=1. Denote by uthe deflection of the beam and assume that the left end of the beam is fixed, while a suitable
shear force 𝑓is exerted on the right end of the beam; thus, the motion describing beam bending is given by the following
problem
utt(t,x)+Au(t,x)=0,(t,x)∈QT,
u(t,0)=0,ux(t,0)=0,t∈(0,T),
u(t,1)=0,ux(t,1)=𝑓(t),t∈(0,T),
u(0,x)=u0(x),ut(0,x)=u1(x),x∈(0,1),
(1.1)
where QT∶= (0,T)×(0,1),T>0, and Au ∶= auxxxx .
An equation similar to the one of (1.1) can be found, for example, in models that describe the vibrations of a bridge.
Indeed, a suspension bridge may be seen as a beam of given length L, with hinged ends and whose downward deflection
is measured by a function u(t,x)subject to three forces. These forces can be summarized as the stays holding the bridge
up as nonlinear springs with spring constant k, the constant weight per unit length of the bridge Wpushing it down, and
the external forcing term 𝑓(t,x). This leads to the equation
utt +𝛾uxxxx =−ku++W+𝑓(t,x),
This is an open access article under the terms of the Creative Commons Attribution-NonCommercial-NoDerivs License, which permits use and distribution in any medium,
provided the original work is properly cited, the use is non-commercial and no modifications or adaptations are made.
© 2023 The Authors. Mathematical Methods in the Applied Sciences published by John Wiley & Sons Ltd.
Math. Meth. Appl. Sci. 2024;47:907–927. wileyonlinelibrary.com/journal/mma 907
CAMASTA and FRAGNELLI
where 𝛾is a physical constant depending on the beam, on Young's modulus and on the second moment of inertia. If 𝛾is
a function that depends on the variable xand the external function acts only on the boundary, then we have exactly the
equation of (1.1) (see, e.g., Camasta & Fragnelli [1] for other applications of (1.1)).
The novelty of this paper is that a∶[0,1]→Ris such that a(0)=0anda(x)>0forallx∈(0,1].Ifthereexistsa
boundary function 𝑓that drives the solution of (1.1) to 0 at a given time T>0, in the sense that
u(T,x)=ut(T,x)=0
for all x∈(0,1), then the problem is said null controllable.
Boundary exact controllability on linear beam problems has been studied for many years by a lot of authors; see, for
example, earlier studies [2–9] and the references therein. For quasilinear beams or nonlinear beams, we refer to Yao and
Weiss [10] and previous research [11, 12], respectively.
In all the previous papers, the equation is always nondegenerate. The first results on boundary controllability for degen-
erate problems can be found in earlier work [13, 14] and in [15]. In particular, in [15] considers the equation in divergence
form
utt −(x𝛼ux)x=0
for 𝛼∈(0,1), and the control acts in the degeneracy point x=0. In the same period, Alabau-Boussouira et al. [13]
consider the equation
utt −(a(x)ux)x=0,(1.2)
where a∼xK,K>0. In this case, the authors establish observability inequalities when K<2; if K2, a negative
result is given. We remark that in Gueye [15], the observability inequality, and hence null controllability, is obtained via
spectral analysis, while in Alabau-Boussouira et al. [13] via suitable energy estimates. In Boutaayamou et al. [14], the
same problem of (1.2) in nondivergence case with a drift term is considered. Clearly, the presence of the operator in
nondivergence form as well as the presence of a drift term leads the authors to use different spaces with respect to the ones
in Alabau-Boussouira et al. [13] or in Gueye [15] and gives rise to some new difficulties. However, thanks to some suitable
assumptions on the drift term, the authors prove some estimates on the energy that are crucial to prove an observability
inequality and hence null controllability for the initial problem.
As far as we know, this is the first paper where the boundary controllability for a degenerate beam equation is considered.
For the function a, we consider two cases: the weakly degenerate case and the strongly degenerate one. More precisely,
we have the following definitions:
Definition 1.1. A function ais weakly degenerate at 0,(WD)for short, if a∈[0,1]∩1(0,1],a(0)=0, a>0on
(0,1]and if
K∶= sup
x∈(0,1]
xa′(x)
a(x),(1.3)
then K∈(0,1).
Definition 1.2. A function ais strongly degenerate at 0,(SD)for short, if a∈1[0,1],a(0)=0, a>0on(0,1]and
in (1.3) we have K∈[1,2).
We underline that, contrary to degenerate wave equations for which null controllability fails if K2, for degenerate
beam equations null controllability is an open problem if K2 is an open problem; indeed, the assumption K<2is
essential in this paper just for technical reasons.
Clearly, the presence of the degenerate operator Au ∶= auxxxx leads us to use different spaces with respect to the ones
in the previous papers [13, 14] or [15], and in these new spaces, we prove an estimate similar to the following one
E𝑦(0)C
T
0
𝑦2
xx(t,1)dt,
where 𝑦and E𝑦are the solution and the energy of the homogeneous adjoint problem associated to (1.1), respectively, and C
is a strictly positive constant. Then, thanks to the introduction of the solutions by transposition for (1.1), we prove that (1.1)
is null controllable for Tsufficiently large. Actually, in Komornik [16], null controllability is proved for nondegenerate
beam equations also for arbitrary short times. We expect that this result still holds for degenerate beam equations, but it
will be the subject of a forthcoming paper.
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CAMASTA and FRAGNELLI
We underline also that, as for wave equations or for parabolic models, in this paper we consider the same boundary
conditions in the weakly and in the strongly degenerate case since here we consider the operator in nondivergence form.
Different boundary conditions are considered if the operator is in divergence form; see Camasta and Fragnelli [17]. Observe
that one can rewrite the operator in divergence form using the operator in nondivergence form in the following way
(a(x)𝑦xx(t,x))xx =a′′ (x)𝑦xx (t,x)+2a′(x)𝑦xxx (t,x)+a(x)𝑦xxxx (t,x).
But in order to apply the results of this paper to the same problem in divergence form, one has to make stronger assump-
tions on the function a. For this reason, the null controllability for a degenerate beam equation in divergence form is
studied in Camasta and Fragnelli [17].
The paper is organized in the following way: in Section 2, we consider the homogeneous problem associated to (1.1)
and we prove that this problem is well-posed in the sense of Theorem 2.2; in Section 3, we consider the energy associated
to it and we prove two estimates on the energy from below and from above. In Section 4, thanks to these estimates and to
the boundary observability (see Corollary 4.1), we prove that the original problem has a unique solution by transposition
and this solution is null controllable. Section 5 presents some open problems. The paper ends with the Appendix where
we give two proofs to make the article self-contained.
We underline that in this paper, Cdenotes universal positive constants which are allowed to vary from line to line.
2WELL-POSEDNESS FOR THE PROBLEM WITH HOMOGENEOUS
DIRICHLET BOUNDARY CONDITIONS
In this section, we study the well-posedness of the following degenerate hyperbolic problem with Dirichlet boundary
conditions
𝑦tt(t,x)+a𝑦xxxxx (t,x)=0,(t,x)∈(0,+∞) × (0,1),
𝑦(t,0)=𝑦(t,1)=0,t∈(0,+∞),
𝑦x(t,0)=𝑦x(t,1)=0,t∈(0,+∞),
𝑦(0,x)=𝑦0
T(x),x∈(0,1),
𝑦t(0,x)=𝑦1
T(x),x∈(0,1).
(2.5)
We underline the fact that the choice of denoting initial data with T-dependence is connected to the approach for null
controllability used in the next sections.
As in earlier work [1, 18] or [19], let us consider the following weighted Hilbert spaces:
L2
1
a
(0,1)∶=
u∈L2(0,1)∶
1
0
u2
adx <+∞
and
Hi
1
a
(0,1)∶=L2
1
a
(0,1)∩Hi
0(0,1),i=1,2,
with the related norms
u2
L2
1
a
(0,1)∶=
1
0
u2
adx ∀u∈L2
1
a
(0,1)
and
u2
Hi1
a
(0,1)∶= u2
L2
1
a
(0,1)+
i
k=1u(k)2
L2(0,1)∀u∈Hi
1
a
(0,1),
i=1,2, respectively. We recall that Hi
0(0,1)∶={u∈Hi(0,1)∶u(k)(𝑗)=0,𝑗 =0,1,k=0,…,i−1},withu(0)=uand
i=1,2. Observe that for all u∈Hi
1
a
(0,1), using the fact that u(k)(𝑗)=0forallk=0,…,i−1and𝑗=0,1, it is easy to
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CAMASTA and FRAGNELLI
prove that u2
Hi1
a
(0,1)is equivalent to the following one
u2
i∶= u2
L2
1
a
(0,1)+u(i)2
L2(0,1),i=1,2.
If i=1, the previous assertion is clearly true.
Moreover, under an additional assumption on a, one can prove that the previous norms are equivalent to the following
one
ui,∼∶= u(i)L2(0,1),i=1,2.
Indeed, assume the following.
Hypothesis 2.1. The function a ∶[0,1]→Ris continuous in [0,1],a(0)=0,a >0on (0,1], and there exists K ∈(0,2)
such that the function
x→ xK
a(x)(2.6)
is nondecreasing near x =0.
Observe that if ais weakly or strongly degenerate, then (1.3) implies that the function
x→ x𝛾
a(x)
is nondecreasing in (0,1]for all 𝛾K; in particular, (2.6) holds globally. Moreover,
lim
x→0
x𝛾
a(x)=0 (2.7)
for all 𝛾>K. The properties above will play a central role in the next sections.
Thanks to Hypothesis 2.1, one can prove the following equivalence.
Proposition 2.1. Assume Hypothesis 2.1. Then for all u ∈Hi
1
a
(0,1)the norms uHi1
a
(0,1),uiand ui,∼,i=1,2,are
equivalent.
Proof. By Cannarsa et al. [20, Proposition 2.6], one has that there exists C>0suchthat
1
0
u2
adx C
1
0
(u′)2dx,
for all u∈L2
1
a
(0,1)∩H1
0(0,1). Thus, the thesis follows immediately if i=1.
Now, assume i=2. Proceeding as for i=1 and applying the classical Hardy inequality (see, e.g., Fragnelli & Mugnai
[21]) to z∶= u′(observe that z∈H1
0(0,1)), we have
1
0
u2
adx C
1
0
(u′)2dx C
1
0
z2
x2dx 4C
1
0
(z′)2dx =4C
1
0
(u′′)2dx
and the thesis follows. □
Hence, assuming Hypothesis 2.1 in the rest of the paper, we can use indifferently ·ior ui,∼in place of ·Hi1
a
(0,1),
i=1,2.
Using the previous spaces, it is possible to define the operator (A,D(A)) by
Au ∶= au′′′′ for all u∈D(A)∶=u∈H2
1
a
(0,1)∶au′′′′ ∈L2
1
a
(0,1).
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CAMASTA and FRAGNELLI
Moreover,
Au,vL2
1
a
(0,1)=
1
0
u′′v′′ dx,
that is,
1
0
u′′′′vdx =
1
0
u′′v′′ dx (2.8)
for all (u,v)∈D(A)×H2
1
a
(0,1)(see Camasta & Fragnelli [18, Proposition 2.1]).
Another important Hilbert space, related to the well-posedness of (2.5), is the following one
0∶= H2
1
a
(0,1)×L2
1
a
(0,1),
endowed with the inner product
(u,v),(
u,
v)0∶=
1
0
u′′
u′′dx +
1
0
v
v
adx
and with the norm
(u,v)2
0∶=
1
0
(u′′)2dx +
1
0
v2
adx
for every (u,v),(
u,
v)∈0. Then, consider the matrix operator ∶D()⊂0→0given by
∶= 0Id
−A0,D()∶=D(A)×H2
1
a
(0,1).
Using this operator, we rewrite (2.5) as a Cauchy problem. Indeed, setting
(t)∶=𝑦(t)
𝑦t(t)and 0∶= 𝑦0
T
𝑦1
T,
one has that (2.5) can be formulated as .
(t)=(t),t0,
(0)=0.(2.9)
Theorem 2.1. Assume Hypothesis 2.1. Then the operator (,D()) is nonpositive with dense domain and generates a
contraction semigroup (S(t))t0.
Proof. According to Engel and Nagel [22, Corollary 3.20], it is sufficient to prove that ∶D()→0is dissipative
and −is surjective, where ∶= Id 0
0Id .
is dissipative: Take (u,v)∈D().Then(u,v)∈D(A)×H2
1
a
(0,1)and so (2.8) holds. Hence,
(u,v),(u,v)0=(v,−Au),(u,v)0
=
1
0
u′′v′′ dx −
1
0
vAu 1
adx =0.
By Engel and Nagel [22, Chapter 2.3], the operator is dissipative.
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CAMASTA and FRAGNELLI
−is surjective: Take (𝑓,g)∈0=H2
1
a
(0,1)×L2
1
a
(0,1).Wehavetoprovethatthereexists(u,v)∈D()such that
(−)u
v=𝑓
g⇐⇒ v=u−𝑓,
Au +u=𝑓+g.(2.10)
Thus, define F∶H2
1
a
(0,1)→Ras
F(z)=
1
0
(𝑓+g)z1
adx,
for all z∈H2
1
a
(0,1).Obviously,F∈H2
1
a
(0,1)∗
,beingH2
1
a
(0,1)∗
the dual space of H2
1
a
(0,1)with respect to the
pivot space L2
1
a
(0,1). Now, introduce the bilinear form L∶H2
1
a
(0,1)×H2
1
a
(0,1)→Rgiven by
L(u,z)∶=
1
0
uz 1
adx +
1
0
u′′z′′ dx
for all u,z∈H2
1
a
(0,1). Clearly, thanks to the equivalence of the norms given before, L(u,z)is coercive. Moreover, L(u,z)
is continuous. Indeed, for all u,z∈H2
1
a
(0,1),wehave
L(u,z)uL2
1
a
(0,1)zL2
1
a
(0,1)+u′′ L2(0,1)z′′L2(0,1),
and the conclusion follows again by the equivalence of the norms.
As a consequence, by the Lax–Milgram Theorem, there exists a unique solution u∈H2
1
a
(0,1)of
L(u,z)=F(z)for all z∈H2
1
a
(0,1),
namely,
1
0
uz 1
adx +
1
0
u′′z′′ dx =
1
0
(𝑓+g)z1
adx (2.11)
for all z∈H2
1
a
(0,1).
Now, take v∶= u−𝑓;thenv∈H2
1
a
(0,1).Wewillprovethat(u,v)∈D()and solves (2.10). To begin with, (4.38)
holds for every z∈∞
c(0,1).Thus, we have
1
0
u′′z′′ dx =
1
0
(𝑓+g−u)z1
adx
for every z∈∞
c(0,1).Hence, (u′′)′′ =(𝑓+g−u)1
aa.e. in (0,1),thatis,Au =𝑓+g−ua.e. in (0,1). Thus, as in Camasta
and Fragnelli [18, Theorem 2.1], u∈D(A); hence, (u,v)∈D()and u+Au =𝑓+g. Recalling that v=u−𝑓,we
have that (u,v)solves (2.10). □
Now, if 0∈0,then(t)=S(t)0is the mild solution of (2.9). Also, if 0∈D(), then the solution is classical and
the equation in (2.5) holds for all t0. Hence, by Bensoussan et al. [23, Propositions 3.1 and 3.3], one has the following
theorem.
Theorem 2.2. Assume Hypothesis 2.1. If 𝑦0
T,𝑦
1
T∈0, then there exists a unique mild solution
𝑦∈1[0,+∞); L2
1
a
(0,1)∩[0,+∞); H2
1
a
(0,1)
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CAMASTA and FRAGNELLI
of (2.5) which depends continuously on the initial data 𝑦0
T,𝑦
1
T∈0. Moreover, if 𝑦0
T,𝑦
1
T∈D(), then the solution 𝑦
is classical in the sense that
𝑦∈2[0,+∞); L2
1
a
(0,1)∩1[0,+∞); H2
1
a
(0,1)∩([0,+∞); D(A))
and the equation of (2.5) holds for all t 0.
Remark 1.
1. Due to the reversibility (in time) of the equation, solutions exist with the same regularity also for t<0.
2. Observe that the proofs of Theorems 2.1 and 2.2 are independent of (2.6).
3ENERGY ESTIMATES
In this section, we prove some estimates of the energy associated to the solution of (2.5). To this aim, we give the next
definition.
Definition 3.1. Let 𝑦be a mild solution of (2.5) and consider its energy given by the continuous function defined as
E𝑦(t)∶= 1
2
1
0𝑦2
t(t,x)
a(x)+𝑦2
xx(t,x)dx ∀t0.
The definition above guarantees that the classical conservation of the energy still holds also in this degenerate situation.
Theorem 3.1. Assume Hypothesis 2.1 and let 𝑦be a mild solution of (2.5). Then
E𝑦(t)=E𝑦(0)∀t0.(3.12)
Proof. First of all, suppose that 𝑦is a classical solution. Then multiplying the equation
𝑦tt +A𝑦=0
by 𝑦t
a, integrating over (0,1)and using the formula of integration by parts (2.8), one has
0=1
2
1
0𝑦2
t
at
dx +
1
0
𝑦xxxx 𝑦tdx =1
2
d
dt
1
0𝑦2
t
a+𝑦2
xxdx
=d
dt E𝑦(t).
Consequently, the energy E𝑦associated to 𝑦is constant.
If 𝑦is a mild solution, we approximate the initial data with more regular ones, obtaining associated classical
solutions for which (3.12) holds. Thanks to the usual estimates, we can pass to the limit and obtain the thesis. □
In the next results, we establish some inequalities for the energy from above and from below; these inequalities will
be used in the next section to establish the controllability result. First of all, we start with the following theorem, whose
proof is based on the next lemma.
Lemma 3.1. Assume Hypothesis 2.1.
1. If 𝑦∈H1
1
a
(0,1),thenlim
x→0
x
a𝑦2(x)=0.
2. Assume a (SD) at 0.If𝑦∈D(A),then𝑦′′ ∈W1,1(0,1).
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CAMASTA and FRAGNELLI
The previous results are proved in Boutaayamou et al. [14, Lemma 3.2.5] and Camasta and Fragnelli [1, Proposition
3.2], respectively; anyway, we rewrite their proof in the Appendix to make the paper self-contained.
Theorem 3.2. Assume a (WD) or (SD) at 0. If 𝑦is a classical solution of (2.5), then 𝑦xx(·,1)∈L2(0,T)for any T >0and
1
2
T
0
𝑦2
xx(t,1)dt =
1
0𝑦tx2
a𝑦xt=T
t=0
dx +1
2
QT
x
a𝑦2
t2−xa′
adx dt +3
QT
x𝑦2
xxdx dt ,(3.13)
Proof. Multiply the equation in (2.5) by x2𝑦x
aand integrate over QT. Integrating by parts, we obtain
0=
QT
𝑦tt
x2𝑦x
adx dt +
QT
x2𝑦x𝑦xxxxdx dt
=
1
0𝑦t
x2𝑦x
at=T
t=0
dx −
QT
𝑦tx2
a𝑦xtdx dt +
QT
x2𝑦x𝑦xxxxdx dt
=
1
0𝑦t
x2𝑦x
at=T
t=0
dx −
QT
1
2
x2
a(𝑦2
t)xdx dt +
QT
x2𝑦x𝑦xxxxdx dt
=
1
0𝑦t
x2𝑦x
at=T
t=0
dx −1
2
T
0x2
a𝑦2
tx=1
x=0
dt +1
2
QTx2
a′
𝑦2
tdx dt
+
QT
x2𝑦x𝑦xxxxdx dt .
(3.14)
Now, x2
a′
=2xa−x2a′
a2=x
a2−xa′
a. Hence, (3.14) reads
1
0𝑦tx2
a𝑦xt=T
t=0
dx −1
2
T
0x2
a𝑦2
tx=1
x=0
dt +1
2
QT
x
a𝑦2
t2−xa′
adx dt +
QT
x2𝑦x𝑦xxxxdx dt =0.(3.15)
Furthermore, by the regularity of the solution, 𝑦t∈H2
1
a
(0,1)⊂H1
1
a
(0,1),thus,
lim
x→0
x2
a(x)𝑦2
t(t,x)=0
by Lemma 3.1; therefore, by the boundary conditions of 𝑦,onehas 1
a(1)𝑦2
t(t,1)=0. Now, consider the term
QTx2𝑦x𝑦xxxxdx dt , which is well-defined; indeed, using the fact that x2
ais nondecreasing, we have that there exists a
positive constant Csuch that
x21
a
𝑦xa𝑦xxxx
C𝑦xa𝑦xxxx .
By hypothesis, 𝑦xand a𝑦xxxx belong to L2(0,1); thus, by the Hölder inequality, x2𝑦x𝑦xxxx ∈L1(0,1).
Let 𝛿>0 and write
QT
x2𝑦x𝑦xxxxdx dt =
T
0
𝛿
0
x2𝑦x𝑦xxxxdx dt +
T
0
1
𝛿
x2𝑦x𝑦xxxxdx dt .(3.16)
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CAMASTA and FRAGNELLI
Obviously, by the absolute continuity of the integral,
lim
𝛿→0
T
0
𝛿
0
x2𝑦x𝑦xxxxdx dt =0.(3.17)
Now, we will estimate the second term in (3.16). By definition of D(A),settingI∶= (𝛿, 1],wehave𝑦xxxx ∈L2(I);
thus, 𝑦∈H4(I)by Camasta and Fragnelli [18, Lemma 2.1]. Hence, we can integrate by parts
T
0
1
𝛿
x2𝑦x𝑦xxxxdx dt =
T
0x2𝑦x𝑦xxxx=1
x=𝛿dt −
T
0
1
𝛿
(x2𝑦x)x𝑦xxxdx dt
=
T
0x2𝑦x𝑦xxxx=1
x=𝛿dt −
T
0
1
𝛿
(2x𝑦x+x2𝑦xx)𝑦xxx dx dt
=
T
0x2𝑦x𝑦xxxx=1
x=𝛿dt −
T
0
1
𝛿
2x𝑦x𝑦xxxdx dt −1
2
T
0
1
𝛿
x2𝑦2
xxxdx dt
=
T
0x2𝑦x𝑦xxxx=1
x=𝛿dt −
T
0
[2x𝑦x𝑦xx]x=1
x=𝛿dt +2
T
0
1
𝛿
(x𝑦x)x𝑦xxdx dt
−1
2
T
0x2𝑦2
xxx=1
x=𝛿dt +1
2
T
0
1
𝛿
2x𝑦2
xxdx dt
=
T
0x2𝑦x𝑦xxxx=1
x=𝛿dt −
T
0
[2x𝑦x𝑦xx]x=1
x=𝛿dt −1
2
T
0x2𝑦2
xxx=1
x=𝛿dt
+2
T
0
1
𝛿
𝑦x𝑦xxdx dt +2
T
0
1
𝛿
x𝑦2
xxdx dt +
T
0
1
𝛿
x𝑦2
xxdx dt
=
T
0x2𝑦x𝑦xxxx=1
x=𝛿dt −
T
0
[2x𝑦x𝑦xx]x=1
x=𝛿dt −1
2
T
0x2𝑦2
xxx=1
x=𝛿dt
+
T
0
1
𝛿𝑦2
xxdx dt +3
T
0
1
𝛿
x𝑦2
xxdx dt
=
T
0x2𝑦x𝑦xxxx=1
x=𝛿dt −
T
0
[2x𝑦x𝑦xx]x=1
x=𝛿dt −1
2
T
0x2𝑦2
xxx=1
x=𝛿dt
+
T
0𝑦2
xx=1
x=𝛿dt +3
T
0
1
𝛿
x𝑦2
xxdx dt .
(3.18)
But x𝑦2
xx ∈L1(0,1)and using the absolute continuity of the integral, we obtain
lim
𝛿→0
T
0
1
𝛿
x𝑦2
xxdx dt =
T
0
1
0
x𝑦2
xxdx dt .
915
CAMASTA and FRAGNELLI
Now, we evaluate the boundary terms that appear in (3.18). To this aim, observe that thanks to the boundary conditions
of 𝑦,
x2𝑦x𝑦xxxx=1
x=𝛿−[2x𝑦x𝑦xx ]x=1
x=𝛿−1
2x2𝑦2
xxx=1
x=𝛿+𝑦2
xx=1
x=𝛿=−𝛿2𝑦x(t,𝛿)𝑦xxx (t,𝛿)
+2𝛿𝑦x(t,𝛿)𝑦xx (t,𝛿)− 1
2𝑦2
xx(t,1)+ 1
2𝛿2𝑦2
xx(t,𝛿)−𝑦2
x(t,𝛿).
Hence, we have to estimate the following quantities:
𝛿2𝑦x(t,𝛿)𝑦xxx (t,𝛿),
𝛿𝑦x(t,𝛿)𝑦xx (t,𝛿),
𝛿2𝑦2
xx(t,𝛿),
𝑦2
x(t,𝛿)
as 𝛿goes to 0. Naturally, since 𝑦∈H2
0(0,1),𝑦xis a continuous function. This implies that
lim
𝛿→0𝑦2
x(t,𝛿)=𝑦2
x(t,0)=0.(3.19)
Thanks to Lemma 3.1,
lim
𝛿→0𝛿2𝑦2
xx(t,𝛿)=0= lim
𝛿→0𝛿𝑦x(t,𝛿)𝑦xx (t,𝛿).
It remains to prove that
∃ lim
𝛿→0𝛿2𝑦x(t,𝛿)𝑦xxx (t,𝛿)=0.(3.20)
By (3.19), it is sufficient to prove that ∃ lim
𝛿→0𝛿𝑦xxx(t,𝛿)∈R.Tothisaim,werewrite𝛿𝑦xxx(t,𝛿)as
𝛿𝑦xxx(t,𝛿)=𝑦xxx(t,1)−
1
𝛿
(x𝑦xxx(t,x))xdx =𝑦xxx (t,1)−
1
𝛿
𝑦xxx(t,x)dx −
1
𝛿
x𝑦xxxx(t,x)dx.(3.21)
Note that x𝑦xxxx (t,x)=a(x)𝑦xxxx (t,x)x
a(x)∈L2(0,1)⊆L1(0,1)(indeed, a𝑦xxxx ∈L2(0,1)and x
a(x)∈L∞(0,1),
thanks to (2.6)). Hence, by the absolute continuity of the integral lim
𝛿→0
1
𝛿
x𝑦xxxx(x)dx =
1
0
x𝑦xxxx(x)dx.On the other
hand, 1
𝛿
𝑦xxx(t,x)dx =
1
𝛿
𝑦xxx(t,1)−
1
x
𝑦xxxx(t,s)ds
dx
=(1−𝛿)𝑦xxx(t,1)−
1
𝛿
1
x
𝑦xxxx(t,s)ds dx.
Now, we estimate the last term in the previous equation
1
𝛿
1
x
𝑦xxxx(t,s)ds dx =
1
𝛿
s
𝛿
𝑦xxxx(t,s)dxds =
1
𝛿
𝑦xxxx(t,s)(s−𝛿)ds
=
1
𝛿
s𝑦xxxx(t,s)ds −𝛿
1
𝛿
𝑦xxxx(t,s)ds.
(3.22)
916
CAMASTA and FRAGNELLI
As before, lim
𝛿→0
1
𝛿
s𝑦xxxx(t,s)ds =
1
0
s𝑦xxxx(t,s)ds. Moreover, as far as the second term in the last member of (3.22) is
concerned, we have
0<𝛿
1
𝛿𝑦xxxx(t,s)ds =𝛿
1
𝛿a(s)𝑦xxxx(t,s)
a(s)
ds
𝛿
1
𝛿
1
a(s)ds
1
2
a𝑦xxxx
L2(0,1)
=𝛿1−K
2
1
𝛿
𝛿K
a(s)ds
1
2
a𝑦xxxx
L2(0,1)
𝛿1−K
2
1
𝛿
sK
a(s)ds
1
2
a𝑦xxxx
L2(0,1)
C𝛿1−K
2(1−𝛿)1
2
a𝑦xxxx
L2(0,1),
for a positive constant C. Thus, since K<2, it follows that lim
𝛿→0𝛿
1
𝛿
𝑦xxxx(t,s)ds =0. Consequently,
lim
𝛿→0
1
𝛿
1
x
𝑦xxxx(t,s)ds dx = lim
𝛿→0
1
𝛿
𝑦xxxx(t,s)(s−𝛿)ds =
1
0
s𝑦xxxx(t,s)ds
As a consequence, coming back to (3.21),
∃ lim
𝛿→0𝛿𝑦xxx(t,𝛿)∈R,
and, in particular, (3.20) is proved.
Thus, by (3.18), one has
lim
𝛿→0
T
0
1
𝛿
x2𝑦x𝑦xxxxdx dt =−
1
2
T
0
𝑦2
xx(t,1)dt +3
T
0
1
0
x𝑦2
xxdxdt.
By the previous equality, (3.15)–(3.17), the thesis follows. □
As a consequence of the previous equality on 1
2T
0𝑦2
xx(t,1)dt, we have the next estimate from below on the energy.
Theorem 3.3. Assume a (WD) or (SD) at 0. If 𝑦is a mild solution of (2.5), then
T
0
𝑦2
xx(t,1)dt 12T+4max4
a(1),1E𝑦(0)(3.23)
Proof. As a first step, assume that 𝑦is a classical solution of (2.5); thus, (3.13) holds. Now, set z(t,x)∶=𝑦x(t,x);since
𝑦x(t,0)=0, by the classical Hardy inequality, for any T>0weobtain
1
0
𝑦2
xdx =
1
0
z2dx =
1
0
z2
x2x2dx 1
0
z2
x2dx 4
1
0
z2
xdx =4
1
0
𝑦2
xxdx.(3.24)
917
CAMASTA and FRAGNELLI
Thus, applying (2.6), one has
1
0
x2𝑦x(𝜏,x)𝑦t(𝜏, x)
a(x)dx
1
2
1
0
x4
a(x)𝑦2
x(𝜏,x)dx +1
2
1
0
𝑦2
t(𝜏,x)
a(x)dx
1
2a(1)
1
0
𝑦2
x(𝜏,x)dx +1
2
1
0
𝑦2
t(𝜏,x)
a(x)dx
2
a(1)
1
0
𝑦2
xx(𝜏, x)dx +1
2
1
0
𝑦2
t(𝜏,x)
a(x)dx
for all 𝜏∈[0,T]. By Theorem 3.1, we get
1
0x2𝑦x(𝜏,x)𝑦t(𝜏, x)
a(x)𝜏=T
𝜏=0
dx
2
a(1)
1
0
𝑦2
xx(T,x)dx +1
2
1
0
𝑦2
t(T,x)
a(x)dx
+2
a(1)
1
0
𝑦2
xx(0,x)dx +1
2
1
0
𝑦2
t(0,x)
a(x)dx
max 4
a(1),1(E𝑦(T)+E𝑦(0))
=2max4
a(1),1E𝑦(0).
(3.25)
Moreover, using the fact that xa′Ka,wefind
QT
x
a𝑦2
t2−xa′
adx dt
QT
x
a𝑦2
t(2+K)dx dt
(2+K)
QT
𝑦2
t
adx dt.
(3.26)
Clearly,
QT
x𝑦2
xxdx dt
QT
𝑦2
xxdx dt (3.27)
and from (3.13), (3.25), (3.26), and (3.27), we get (3.23) if 𝑦is a classical solution of (2.5). Now, let 𝑦be the mild solution
associated to the initial data (𝑦0,𝑦
1)∈0. Then, consider a sequence {(𝑦n
0,𝑦
n
1)}n∈N⊂D()that converges to (𝑦0,𝑦
1)
and let 𝑦nbe the classical solution of (2.5) associated to (𝑦n
0,𝑦
n
1). Clearly, 𝑦nsatisfies (3.23); then, we can pass to the
limit and conclude. □
Now, we will prove an inequality on the energy from above. To this aim, we need on
T
0
𝑦2
xx(t,1)dt an equality different
from (3.13).
Theorem 3.4. Assume a (WD) or (SD) at 0. If 𝑦is a classical solution of (2.5), then 𝑦xx(·,1)∈L2(0,T)for any T >0and
1
2
T
0
𝑦2
xx(t,1)dt =
1
0x𝑦t𝑦x
at=T
t=0dx +1
2
QT
𝑦2
t
a1−xa′
adx dt +3
2
QT
𝑦2
xxdx dt ,(3.28)
918
CAMASTA and FRAGNELLI
Proof. Multiplying the equation in (2.5) by x𝑦x
aand integrating over QT,weobtain
0=
1
0x𝑦x𝑦t
at=T
t=0dx −
QT
1
2
x
a(𝑦2
t)xdx dt +
QT
x𝑦x𝑦xxxxdx dt
=
1
0x𝑦x𝑦t
at=T
t=0dx −1
2
T
0x
a𝑦2
tx=1
x=0dt +1
2
QTx
a′
𝑦2
tdx dt +
QT
x𝑦x𝑦xxxxdx dt .
(3.29)
Now, x
a′
=a−xa′
a2=1
a1−xa′
a. Hence, (3.29) reads
1
0x𝑦x𝑦t
at=T
t=0dx −1
2
T
0x
a𝑦2
tx=1
x=0dt +1
2
QT
𝑦2
t
a1−xa′
adx dt +
QT
x𝑦x𝑦xxxxdx dt =0.(3.30)
As before,
lim
x→0
x
a(x)𝑦2
t(t,x)=0
and 1
a(1)𝑦2
t(t,1)=0, so that
T
0x
a𝑦2
tx=1
x=0dt =0. In addition, the term QTx𝑦x𝑦xxxxdx dt is well-defined since x𝑦x𝑦xxxx =
x
a𝑦xa𝑦xxxx ∈L1(0,1).Thus,wetake𝛿>0, and as in the proof of Theorem 3.2, we rewrite
QT
x𝑦x𝑦xxxxdx dt =
T
0
𝛿
0
x𝑦x𝑦xxxxdx dt +
T
0
1
𝛿
x𝑦x𝑦xxxxdx dt .
Since x𝑦x𝑦xxxx ∈L1(0,1),wehavelim
𝛿→0
T
0
𝛿
0
x𝑦x𝑦xxxxdx dt =0.Moreover, integrating by parts the second term of the
previous equality and thanks to the boundary conditions on 𝑦,wehave
T
0
1
𝛿
x𝑦x𝑦xxxxdx dt =
T
0
[x𝑦x𝑦xxx]x=1
x=𝛿dt −
T
0
1
𝛿
(x𝑦x)x𝑦xxxdx dt
=−
T
0
𝛿𝑦x(t,𝛿)𝑦xxx (t,𝛿)dt −
T
0
1
𝛿
𝑦x𝑦xxxdx dt −1
2
T
0
1
𝛿
x𝑦2
xxxdx dt
=−
T
0
𝛿𝑦x(t,𝛿)𝑦xxx (t,𝛿)dt −
T
0
[𝑦x𝑦xx]x=1
x=𝛿dt +
T
0
1
𝛿
𝑦2
xxdx dt −1
2
T
0
[x𝑦2
xx]x=1
x=𝛿dt +1
2
T
0
1
𝛿
𝑦2
xxdx dt
=−
T
0
𝛿𝑦x(t,𝛿)𝑦xxx (t,𝛿)dt +
T
0
𝑦x(t,𝛿)𝑦xx (t,𝛿)dt −1
2
T
0
𝑦2
xx(t,1)dt
+1
2
T
0
𝛿𝑦2
xx(t,𝛿)dt +3
2
T
0
1
𝛿
𝑦2
xxdx dt .
Proceeding as in the proof of Theorem 3.2, one can prove that ∃lim
𝛿→0𝛿𝑦xxx(t,𝛿)∈R; hence,
lim
𝛿→0𝛿𝑦x(t,𝛿)𝑦xxx (t,𝛿)=0.
919
CAMASTA and FRAGNELLI
Moreover, by Lemma 3.1, we get
lim
𝛿→0𝑦x(t,𝛿)𝑦xx (t,𝛿)=0,lim
𝛿→0𝛿𝑦2
xx(t,𝛿)=0.
Hence,
lim
𝛿→0
T
0
1
𝛿
x𝑦x𝑦xxxxdx dt =−
1
2
T
0
𝑦2
xx(t,1)dt +3
2
QT
𝑦2
xxdx dt .
Coming back to (3.30), it follows that
1
0x𝑦x𝑦t
at=T
t=0dx +1
2
QT
𝑦2
t
a1−xa′
adx dt −1
2
T
0
𝑦2
xx(t,1)dt +3
2
QT
𝑦2
xxdx dt =0
and (3.28) holds. □
Thanks to (3.28), we can prove the following estimate on the energy from above.
Theorem 3.5. Assume a (WD) or (SD) at 0. If 𝑦is a mild solution of (2.5), then
T
0
𝑦2
xx(t,1)dt T(2−K)−4max1,4
a(1),4K
a(1)E𝑦(0)
for any T >0.
Proof. Multiplying the equation in (2.5) by −K𝑦
2aand integrating over QT,wehave
0=−
K
2
QT
𝑦tt𝑦
adx dt −K
2
QT
𝑦𝑦xxxxdx dt =−
K
2
1
0𝑦𝑦t
at=T
t=0dx +K
2
QT
𝑦2
t
adx dt −K
2
QT
𝑦2
xxdx dt ,
thanks to (2.8). Summing the previous equality to (3.28) multiplied by 2 and using the degeneracy condition (1.3), we
have
T
0
𝑦2
xx(t,1)dt =2
1
0x𝑦t𝑦x
at=T
t=0dx +
QT
𝑦2
t
a1−xa′
adx dt +3
QT
𝑦2
xxdx dt
−K
2
1
0𝑦𝑦t
at=T
t=0dx +K
2
QT
𝑦2
t
adx dt −K
2
QT
𝑦2
xxdx dt
=2
1
0x𝑦t𝑦x
at=T
t=0dx −K
2
1
0𝑦𝑦t
at=T
t=0dx
+
QT
𝑦2
t
a1−xa′
a+K
2dx dt +3−K
2
QT
𝑦2
xxdx dt
2
1
0x𝑦t𝑦x
at=T
t=0dx −K
2
1
0𝑦𝑦t
at=T
t=0dx
+1−K
2
QT
𝑦2
t
adx dt +1−K
2
QT
𝑦2
xxdx dt
=2
1
0x𝑦t𝑦x
at=T
t=0dx −K
2
1
0𝑦𝑦t
at=T
t=0dx +(2−K)TE𝑦(0).
920
CAMASTA and FRAGNELLI
Now, we analyze the boundary terms that appear in the previous relation. By (3.25),
2
1
0x𝑦x(𝜏,x)𝑦t(𝜏, x)
a(x)𝜏=T
𝜏=0
dx
4max4
a(1),1E𝑦(0).
Furthermore, by (2.6),
𝑦(𝜏,x)𝑦t(𝜏, x)
a(x)1
2
𝑦2
t(𝜏,x)
a(x)+1
2a(1)
𝑦2(𝜏,x)
x2
for all 𝜏∈[0,T]; in particular, by Theorem 3.1 and (3.24), we have
1
0
𝑦(𝜏,x)𝑦t(𝜏, x)
a(x)dx
1
2
1
0
𝑦2
t(𝜏,x)
a(x)dx +1
2a(1)
1
0
𝑦2(𝜏,x)
x2dx
1
2
1
0
𝑦2
t(𝜏,x)
a(x)dx +4
2a(1)
1
0
𝑦2
x(𝜏,x)dx
1
2
1
0
𝑦2
t(𝜏,x)
a(x)dx +16
2a(1)
1
0
𝑦2
xx(𝜏, x)dx
max 1,16
a(1)E𝑦(0),
for all 𝜏∈[0,T]. Hence,
K
2
1
0𝑦𝑦t
a𝜏=T
𝜏=0dx
Kmax 1,16
a(1)E𝑦(0),
and the thesis follows if 𝑦is a classical solution. If 𝑦is a mild solution, then we can proceed as in Theorem 3.3. □
4BOUNDARY OBSERVABILITY AND NULL CONTROLLABILITY
Inspired by Alabau-Boussouira et al. [13], we give the next definition.
Definition 4.1. Problem (2.5) is said to be observable in time T>0 via the second derivative at x=1ifthereexists
a constant C>0 such that for any 𝑦0
T,𝑦
1
T∈0, the classical solution 𝑦of (2.5) satisfies
CE𝑦(0)T
0
𝑦2
xx(t,1)dt.(4.31)
Moreover, any constant satisfying (4.31) is called observability constant for (2.5) in time T.
Setting
CT∶= sup {C>0∶Csatisfies (4.31)},
we have that problem (2.5) is observable if and only if
CT=inf
(𝑦0
T,𝑦1
T)(0,0)
T
0𝑦2
xx(t,1)dt
E𝑦(0)>0.
921
CAMASTA and FRAGNELLI
The inverse of CT,thatis,cT∶= 1
CT
, is called the cost of observability (or the cost of control)intimeT.
Theorem 3.5 admits the following straightforward corollary.
Corollary 4.1. Assume that a is (WD) or (SD) at 0. If
T>4
2−Kmax 1,4
a(1),4K
a(1),
then (2.5) is observable in time T. Moreover,
T(2−K)−4max1,4
a(1),4K
a(1)CT.
In the following, we will study the problem of null controllability for (1.1). As a first step, we give the definition of a
solution for (1.1) by transposition, which permits low regularity on the notion of solution itself. Precisely, we have the
following:
Definition 4.2. Let 𝑓∈L2
loc[0,+∞) and (u0,u1)∈L2
1
a
(0,1)×H2
1
a
(0,1)∗
.Wesaythatuis a solution by transposition
of (1.1) if
u∈1[0,+∞); H2
1
a
(0,1)∗∩[0,+∞); L2
1
a
(0,1)
and for all T>0,
ut(T),v0
TH2
1
a
(0,1)∗
,H2
1
a
(0,1)
−
1
0
1
au(T)v1
Tdx =u1,v(0)H2
1
a
(0,1)∗
,H2
1
a
(0,1)
−
1
0
1
au0vt(0,x)dx −
T
0
𝑓(t)vxx(t,1)dt
(4.32)
for all v0
T,v1
T∈0,wherevsolves the backward problem
vtt(t,x)+a(x)vxxxx(t,x)=0,(t,x)∈(0,+∞) × (0,1),
v(t,0)=0,vx(t,0)=0,t>0,
v(t,1)=0,vx(t,1)=0,t>0,
v(T,x)=v0
T(x),vt(T,x)=v1
T(x),x∈(0,1).
(4.33)
Observe that, by Theorem 2.2, there exists a unique mild solution of (4.33) in [T,+∞).Now,setting𝑦(t,x)∶=v(T−t,x),
one has that 𝑦satisfies (2.5) with 𝑦0
T(x)=v0
T(x)and 𝑦1
T(x)=−v1
T(x). Hence, we can apply Theorem 2.2 to (2.5) obtaining
that there exists a unique mild solution 𝑦of (2.5) in [0,+∞). In particular, there exists a unique mild solution vof (4.33)
in [0,T]. Thus, we can conclude that there exists a unique mild solution
v∈1[0,+∞); L2
1
a
(0,1)∩[0,+∞); H2
1
a
(0,1)
of (4.33) in [0,+∞) which depends continuously on the initial data VT∶= v0
T,v1
T∈0.
By Theorem 3.1, the energy is preserved in our setting, as well, so that the method of transposition done in
Alabau-Boussouira et al. [13] continues to hold thanks to (3.23). Therefore, there exists a unique solution by transposi-
tion u∈1[0,+∞); H−2
1
a
(0,1)∩[0,+∞); L2
1
a
(0,1)of (1.1), that is, a solution of (4.32). To prove this fact, consider
922
CAMASTA and FRAGNELLI
the functional ∶0→Rgiven by
v0
T,v1
T=u1,v(0)H2
1
a
(0,1)∗
,H2
1
a
(0,1)
−
1
0
1
au0vt(0,x)dx −
T
0
𝑓(t)vxx(t,1)dt,(4.34)
for all T>0, where vsolves (4.33). Clearly, is linear. Moreover, it is continuous. Indeed, for all T>0, we have
v0
T,v1
Tu1H2
1
a
(0,1)∗v(0)H2
1
a
(0,1)+u0L2
1
a
(0,1)vt(0)L2
1
a
(0,1)+𝑓L2(0,T)
T
0
v2
xx(t,1)dt
1
2
.
By (3.23) and Theorem 3.1, there exists a positive constant Csuch that
T
0
v2
xx(t,1)dt CEv(0)=CEv(T)= C
2
1
0v2
t(T,x)
a(x)+v2
xx(T,x)dx =C
2
v0
T,v1
T
0
.
Hence,
v0
T,v1
Tu1H2
1
a
(0,1)∗v(0)H2
1
a
(0,1)+u0L2
1
a
(0,1)vt(0)L2
1
a
(0,1)+𝑓L2(0,T)
v0
T,v1
T
0
.
Using again Theorem 3.1, we have vt(0)L2
1
a
(0,1)Ev(T);thus,vt(0)L2
1
a
(0,1)C
v0
T,v1
T
0
. Analogously, thanks to
Proposition 2.1, there exists C>0suchthatv(0)H2
1
a
(0,1)C
v0
T,v1
T
0
.Thus, we can conclude that there exists C>0
so that
v0
T,v1
TC
v0
T,v1
T
0
,
that is, is continuous.
Being ∈(0)∗=H2
1
a
(0,1)∗
×L2
1
a
(0,1),we can use the Riesz Theorem obtaining that for any T>0, there exists a
unique
u0
T,
u1
T∈(0)∗such that
v0
T,v1
T=
u0
T,
u1
T,v0
T,v1
T(0)∗,0=
u0
T,v0
TH2
1
a
(0,1)∗
,H2
1
a
(0,1)
+
1
0
u1
Tv1
T
adx.(4.35)
Moreover,
u0
T,
u1
Tdepend continuously on T, so there exists a unique u∈[0,+∞); L2
1
a
(0,1)∩1[0,+∞); H−2
1
a
(0,1)
such that u(T)=−
u1
Tand ut(T)=
u0
T. By (4.34) and (4.35), we can conclude that uis the unique solution by transposition
of (1.1).
Now, we are ready to examine null controllability. To this aim, consider the bilinear form Λ∶0×0→Rdefined as
Λ(VT,WT)∶=
T
0
vxx(t,1)wxx (t,1)dt,
where vand ware the solutions of (4.33) associated to the data VT∶= v0
T,v1
Tand WT∶= w0
T,w1
T, respectively. The
following lemma holds.
Lemma 4.1. Assume a (WD) or (SD) at 0. The bilinear form Λis continuous and coercive.
923
CAMASTA and FRAGNELLI
Proof. By Theorem 3.1, Evand Eware constant in time, and due to (3.23), one has that Λis continuous. Indeed, by
Holder's inequality and (3.23),
Λ(VT,WT)T
0vxx(t,1)wxx (t,1)dt
T
0
v2
xx(t,1)dt
1
2
T
0
w2
xx(t,1)dt
1
2
CE
1
2
v(T)E
1
2
w(T)=C
1
0
(v1
T)2(x)
adx +
1
0
[(v0
T)xx]2(x)dx
1
2
·
1
0
(w1
T)2(x)
adx +
1
0
[(w0
T)xx]2(x)dx
1
2
=C(v(T),vt(T))0(w(T),wt(T))0=CVT0WT0
for a positive constant Cindependent of (VT,WT)∈0×0.
In a similar way one can prove that Λis coercive. Indeed, by Theorem 3.5, one immediately has that there exists
C>0suchthat
Λ(VT,VT)=
T
0
v2
xx(t,1)dt CTEv(0)=CTEv(T)CVT2
0
for all VT∈0.□
Function Λis used to prove the null controllability property for the original problem (1.1). To this aim, let us start
defining T0as the lower bound found in Corollary 4.1, that is,
T0∶= 4
2−Kmax 1,4
a(1),4K
a(1).(4.36)
Theorem 4.1. Assume a (WD) or (SD) at 0. Then, for all T >T0and for every (u0,u1)∈L2
1
a
(0,1)×H−2
1
a
(0,1),thereexists
a control 𝑓∈L2(0,T)such that the solution of (1.1) satisfies
u(T,x)=ut(T,x)=0∀x∈(0,1).(4.37)
Proof. Consider the map ∶0→Rgiven by
(VT)∶=u1,v(0)H2
1
a
(0,1)∗
,H2
1
a
(0,1)
−
1
0
u0vt(0,x)
adx,
where vis the solution of (4.33) associated to the initial data VT∶= v0
T,v1
T∈0. Clearly, is continuous and linear
and, thanks to Lemma 4.1, we can apply the Lax–Milgram Theorem. Thus, there exists a unique VT∈0such that
Λ(VT,WT)=(WT)∀WT∈0.
Set 𝑓(t)∶=vxx(t,1),wherevis the unique solution of (4.33) associated to VT.Then
T
0
𝑓(t)wxx(t,1)dt =
T
0
vxx(t,1)wxx (t,1)dt =ΛVT,WT=(WT)=u1,w(0)H2
1
a
(0,1)∗
,H2
1
a
(0,1)
−
1
0
u0wt(0,x)
adx
(4.38)
for all WT∈0.
924
CAMASTA and FRAGNELLI
Finally, denote by uthe solution by transposition of (1.1) associated to the function 𝑓introduced above. We have
that
T
0
𝑓(t)wxx(t,1)dt =−
ut(T),w0
TH2
1
a
(0,1)∗
,H2
1
a
(0,1)
+
1
0
u(T)w1
T
adx +u1,w(0)H2
1
a
(0,1)∗
,H2
1
a
(0,1)
−
1
0
u0wt(0,x)
adx.
(4.39)
Combining (4.38) and (4.39), it follows that
ut(T),w0
TH2
1
a
(0,1)∗
,H2
1
a
(0,1)
−
1
0
u(T)w1
T
adx =0
for all w0
T,w1
T∈0. Hence, we have (4.37). □
5CONCLUSIONS AND OPEN PROBLEMS
In this paper, we have proved that if the function ais weakly or strongly degenerate and T>T0,whereT0is given in (4.36),
then (1.1) is boundary null controllable, that is,
u(T,x)=ut(T,x)=0
for all x∈(0,1). However, in contrast to wave equations, nondegenerate beam equations can be generically controlled
in arbitrary short times since there is no finite speed of propagation (see Komornik [16]). Hence, one would expect to
obtain null controllability for (1.1) also for short times using the same technique proposed in Komornik [16], where
nondegenerate variable coefficients are considered.
Another open problem is to prove null controllability for (1.1) or to show that it fails when K2. Indeed, for degenerate
wave equations, we know that, if K2, null controllability fails; on the other hand, for degenerate beam equations, we
do not know anything in this case: the assumption K<2 is made here only for technical reasons.
AUTHOR CONTRIBUTIONS
Alessandro Camasta: validation, writing—review and editing, and investigation. Genni Fragnelli: investigation,
validation, writing—review and editing, supervision, and methodology.
ACKNOWLEDGEMENTS
A. Camasta and G. Fragnelli are members of the Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Appli-
cazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM) and a member of .They are partially supported
by the project Mathematical models for interacting dynamics on networks (MAT-DYN-NET) CA18232,byINdAM GNAMPA
Project, CUP E53C22001930001 and by the PRIN 2017-2019 qualitative and quantitative aspects of nonlinear PDEs.
G. Fragnelli is also supported by FFABR Fondo per il finanziamento delle attività base di ricerca 2017 and by the
HORIZON−EU−DM737 project 2022 COntrollability of PDEs in the Applied Sciences (COPS) at Tuscia University. The
authors thank the two referees for their comments and the Project Horizon Europe Sees STEPS: STEerability and
controllability of PDEs in Agricultural and Physical models.
CONFLICT OF INTEREST STATEMENT
This work does not have any conflicts of interest.
ORCID
Genni Fragnelli https://orcid.org/0000-0002-5436-7006
925
CAMASTA and FRAGNELLI
REFERENCES
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(2017), 2052–2087.
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SIAM J. Control Optim. 61 (2023), 1934–1954.
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Wiley & Sons, Ltd., Chichester, 1994.
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AUTHOR BIOGRAPHIES
Alessandro Camasta is a Ph.D student at the Department of Mathematics of the University of Bari Aldo Moro. His
principal mathematical interests are Stability and Controllability of PDEs.
Genni Fragnelli received her Ph.D in 2002 from the University of Tuebingen, Germany. Currently, she is an Associate
Professor at the Department of Ecology and Biology of the Tuscia University. Her principal mathematical interests are
PDEs, Control Theory, and Semigroup Theory.
How to cite this article: A. Camasta and G. Fragnelli, Boundary controllability for a degenerate beam equation,
Math. Meth. Appl. Sci. 47 (2024), 907–927, DOI 10.1002/mma.9692.
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CAMASTA and FRAGNELLI
APPENDIX A:
Proof of Lemma 3.1.1. If K<1, where Kis the constant of Hypothesis 2.1, then the assertion follows immediately
by (2.7) with 𝛾=1. Thus, assume K1. Set z(x)∶= x
a(x)𝑦2(x).Thenz∈L1(0,1). Indeed,
1
0
x
a𝑦2dx 1
0
𝑦2
adx.
Moreover, z′=𝑦2
a+2x𝑦𝑦′
a−a′x
a2𝑦2;thus, for a suitable 𝜀>0 given by Hypothesis 2.1,
𝜀
0z′dx 1
0
𝑦2
adx +2
𝜀
0
x2(𝑦′)2
adx
1
2
1
0
𝑦2
adx
1
2
+K
1
0
𝑦2
adx
(1+K)
1
0
𝑦2
adx +2𝜀
a(𝜀)
1
0
(𝑦′)2dx
1
2
1
0
𝑦2
𝜎dx
1
2
.
This is enough to conclude that z∈W1,1(0,1),andthus,thereexistslim
x→0z(x)=L∈R.IfL0, sufficiently close
to x=0 we would have that 𝑦2(x)
aL
2x∉L1(0,1),while 𝑦2
a∈L1(0,1).
Proof of Lemma 3.1.2. In order to prove the lemma, fixed 𝑦∈D(A), we will establish that 𝑦′′ is absolutely continuous
in [0,1]. To this aim, let 𝛿>0. Clearly,
𝑦′′(𝛿)=𝑦′′ (1)−
1
𝛿
𝑦′′′(x)dx;(A1)
thus,
𝑦′′(𝛿)=𝑦′′ (1)−
1
𝛿
𝑦′′′(1)−
1
x
𝑦′′′′(s)ds
dx =𝑦′′(1)−(1−𝛿)𝑦′′′(1)+
1
𝛿
1
x
𝑦′′′′(s)ds
dx
=𝑦′′(1)−𝑦′′′ (1)+𝛿𝑦′′′(1)+
1
𝛿
s
𝛿
𝑦′′′′(s)dx ds =𝑦′′ (1)−𝑦′′′(1)+𝛿𝑦′′′ (1)+
1
𝛿
𝑦′′′′(s)(s−𝛿)ds.
(A2)
Trivially, lim
𝛿→0𝛿𝑦′′′(1)=0 and, proceeding as in Theorem 3.2, we have
lim
𝛿→0
1
𝛿
𝑦′′′′(s)(s−𝛿)ds =
1
0
s𝑦′′′′(t,s)ds.
Thus, if we pass to the limit as 𝛿→0 in (A2), we conclude that
∃ lim
𝛿→0𝑦′′(𝛿)=𝑦′′ (1)−𝑦′′′(1)+
1
0
s𝑦′′′′(s)ds ∈R.
By continuity, it is possible to define 𝑦′′(0)∶=lim𝛿→0𝑦′′ (𝛿).In particular, by (A1),
1
0
𝑦′′′(x)dx =𝑦′′ (1)−𝑦′′(0)
and the thesis follows.
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