1. Introduction
This paper is devoted to studying the stability of a beam-type degenerate equation with a small singular perturbation through a linear boundary feedback. To be more precise, we consider the following problem:
where
and
are non-negative constants and the function
a is such that
and
for all
. In particular, for the function
a, we consider two types of degeneracy according to the following definitions:
Definition 1. The function g is weakly degenerate, (WD) for short, at 0 if g ∈ is such that on , and ifthen . Definition 2. The function g is strongly degenerate, (SD) for short, at 0 if g ∈ is such that on and in (2) we have . Roughly speaking, when , it is (WD) if and (SD) if .
Problems similar to (
1) are considered in several papers (see, for example, [
1,
2,
3,
4,
5,
6,
7]). In particular, in [
3,
5], the following Euler–Bernoulli beam equation is considered:
with clamped conditions at the left end
and with dissipative conditions at the right end
Here,
y is the vertical displacement,
is the velocity,
is the rotation,
is the angular velocity,
m is the mass density per unit length,
is the flexural rigidity coefficient,
is the bending moment, and
is the shear. In particular, the boundary conditions (
5) mean that the shear is proportional to the velocity and the bending moment is negatively proportional to the angular moment. Observe that if we consider
in (
1), then we have boundary conditions analogous to those in (
5). Thus, the dissipative conditions at 1 are not surprising. We remark that the conditions
are necessary to study the well-posedness of the problem and to prove equivalence among all the norms introduced in this paper and that are crucial to obtain the stability result.
The qualitative behavior of (
3)–(
5) is studied in [
4], where it is proved that if
and
, the energy
of the vibration of the beam decays exponentially in a uniform way
for some
.
Observe that in all the references above, the equation is non-degenerate; however, there are some papers where the equation is degenerate in the sense that a
degenerate damping appears in the equation of (
3) (see, for example, [
8,
9,
10]). The first paper where the equation is degenerate in the sense that the fourth-order operator degenerates in a point as in (
1) is [
11]. However, to the best of our knowledge, [
12] is the first paper where the
stability for (
1) with
is considered. On the other hand, for a degenerate wave-equation, we refer to [
13] (see also the arxiv version of 2015) for a problem in divergence form and to [
14] for a problem in non-divergence form.
A position-dependent restoring force is introduced using the modified Euler–Bernoulli equation that includes a coefficient-dependent drift term. High stiffness, concentrated forces, or material discontinuities can be modeled by the term
, which produces a single behavior at
, suggesting a highly localized effect. Additionally, it denotes pre-stress or non-homogeneous stiffness, which applies to beams on uneven elastic foundations. Furthermore, if
is a distance function, the term affects tension in pre-stressed structures by acting like an inverse square law, similar to electrostatic or gravitational potentials. Additionally, the equation resembles singular potential quantum wave equations, which cause localized resonance effects in structural dynamics. All things considered, this formulation captures strong localized effects, pre-stress changes, and non-uniform limitations that are pertinent to practical physics and engineering (see [
15]).
As far as we know, for beam-type equations simultaneously admitting degeneracy and singularity, only controllability problems have been faced (see the recent paper [
16]), while nothing has been undertaken for stability. For this reason, in this paper, we focus on such a problem, proving that (
1) permits boundary stabilization, provided that the singular term has a small coefficient (see Theorem 2 below). Hence, we may regard this result as a natural continuation of [
12] and a perturbation of the related one in [
16]. Clearly, the presence of the singular term
introduces several difficulties with respect to [
12], which let us treat only the case of a function
d with weak degeneracy, according to the definition above. For a stability result for a degenerate/singular wave equation, we refer to [
17].
Strategy method. In order to prove (
6) for (
1), we use a multiplier method. In particular, after defining the energy associated with the problem, we prove an estimate on it using a multiplier method (see Proposition 6). Obviously, the presence of a degenerate fourth-order operator brings more difficulties with respect to the ones for the second-order case. These difficulties are related to some new terms that we have to face; for example, we have to estimate from above
for every
using the energy associated with the original problem. This is carried out in Proposition 7 thanks to a suitable fourth-order variational problem (see Proposition 3). Thanks to the estimates proved in Propositions 6 and 7 and using a result given in [
18], we prove the main result of the paper, i.e. Theorem 2.
This paper is organized as follows: In
Section 2, we give the functional setting and some preliminary results that we will use in the rest of the paper, together with the existence of solutions. In
Section 3, we introduce the energy associated with a solution for the problem and we show that it decays exponentially as time diverges. In particular, we prove that if
is small and
are not too degenerate (in the sense of Definitions 1 or 2), the energy satisfies (
6), as in [
4] for the non-degenerate and non-singular case. The last section is devoted to the conclusions and to some open problems.
2. Preliminary Results and Well-Posedness
In this section, we introduce the functional setting needed to treat (
1). However, here, our assumptions are more general than those required to obtain the stability result in the next section.
We start by assuming a very modest requirement.
Hypothesis 1. The functions are such that
, on ,
there are such that the functionsandare non-decreasing in a right neighborhood of .
It is clear that, if Hypothesis 1 holds, then
for all
, and
for all
.
Let us state that if
a is (WD) or (SD), then (
2) implies that (
7) holds on the whole domain
analogously for
d.
In order to treat (
1), let us introduce the following Hilbert spaces with the related inner products and norms given by the following:
for all
;
and
,
. In addition to the previous ones, we introduce the following important Hilbert spaces:
with the previous inner products
and norms
,
. Now, consider the scalar product
for all
, which induces the norm
. Observe that, if
a is continuous,
and (
7) is satisfied, then the norms
,
and
are equivalent in
. Here,
(see, e.g., [
11]). Clearly, if
, the previous equivalence is obviously satisfied. Indeed,
and
coincide and, for ([
19], Proposition 2.6), one proves that there is
such that
for all
. Let
Now, assume
and fix
. Proceeding as for
and applying the classical Hardy’s inequality to
(observing that
), we have
Hence,
and
are equivalent in
(actually, they are equivalent in
, see, e.g., [
11]). Moreover, by the previous inequality,
and the thesis follows. In particular,
for all
and
for all
(see ([
12], Proposition 2.1)).
As in ([
16], Proposition 2.3), one can prove the next result
Proposition 1. Assume Hypothesis 1 and take such that . If , then for , there is a positive constant such that Let
As in ([
20], Chapter V), we assume the next hypothesis:
Hypothesis 2. The constant is such that and Observe that the case
is already considered in [
12]. Thus, it is not restrictive to assume
.
Moreover, if
, we can take
such that
Hence, as a consequence of Proposition 1, one has the next estimate (see ([
16], Proposition 2.4)).
Proposition 2. Assume Hypothesis 1 and . If , then Under Hypotheses 1 and 2, one can consider in
also the product
which induces the norm
By Propositions 1 and 2, one can prove the following equivalence:
Corollary 1. Assume Hypotheses 1 and 2 and . Then, the norms , , , and are equivalent in .
In order to study the well-posedness of (
1), we introduce the operator
A by
, for all
, where the next Gauss–Green formula holds
for all
(see [
12]). Moreover, consider
where
Observe that if
and
, one proves that
; hence,
if and only if
, i.e.,
(for more details, we refer to [
16]). For this reason, in the following, we assume the next assumption:
Hypothesis 3. Assume Hypothesis 1 and .
Under this assumption, it is clear that d cannot be (SD). On the other hand, a can be (SD), but in this case, has to be very small.
Finally, to prove the well-posedness of (
1), we need to introduce the last Hilbert space
with inner product and norm given by
and
for every
, where
, and the matrix operator
given by
with domain
Thanks to (
19), one can prove the next theorem that contains the main properties of the operator
. Since the proof is similar to the one of [
21] or [
22], we omit it.
Theorem 1. Assume a (WD) or (SD). Then, the operator is non-positive with a dense domain and generates a contraction semigroup .
Thanks to the previous theorem, one has the next result, which can be proved as in ([
16], Theorem 2.7).
Theorem 2. Hypotheses 2 and 3 hold. If , then there is a unique mild solutionof (1), which depends continuously on the initial data. In addition, if , then the solution y is classical in the sense thatand the equation of (1) holds for all . Remark 1. Due to the reversibility in time of the equation, solutions exist with the same regularity for . We will use this fact in the proof of the controllability result by considering a backward problem whose final time data will be transformed in initial data: this is the reason for the notation of the initial data in problem (1). The last important result of this section is given by the next proposition. Let us start with
Hypothesis 4. Assume a (WD) or (SD), d (WD) with , with and .
Proposition 3. Assume Hypothesis 4 and definefor all . Then, the norms and are equivalent in . Moreover, for every , the variational problemadmits a unique solution , which satisfies the estimateswhereIn addition, and solves Proof. As a first step, observe that for all
, one has
and
for all
. Thus,
and
are equivalent. Indeed, for all
, if
, one proves immediately that
If
, by Proposition 2, one has
for all
, and so
for all
. In conclusion, we have
Now, we prove that there is
such that
for all
. Clearly, (
25) and (
24) imply
and
, respectively; hence, if
, one proves immediately that
; if
, by (
15), then
In any case, the claim holds.
Now, consider the bilinear and symmetric form
such that
As in [
22] or in [
17], one can easily prove that
is coercive and continuous. Now, consider the linear functional
with
and
. Clearly,
is continuous and linear. Thus, by the Lax-Milgram Theorem, there is a unique solution
of
for all
. In particular,
Concerning the other estimates, by (
24)–(
26) and (
31), we have
thus,
Moreover, by the equivalence of the norms in
, Proposition 2, and (
13), one has
where
is as in (
22). Thus, by (
32),
Now, we will prove that
z belongs to
and solves (
23). To this end, consider (
30) again; clearly, it holds for every
, so that
Thus,
a.e. in
(see, e.g., ([
23], Lemma 1.2.1)) and so
a.e. in
in particular
; this implies that
.
Now, coming back to (
30) and using (
19) and the fact that
, we have
for all
. Thus,
and
that is,
z solves (
23). □
3. Energy Estimates and Exponential Stability
In this section, we prove the main result of this paper. In particular, proving some estimates of the energy associated with (
1), we obtain the exponential stability.
To begin with, we give the next definition:
Definition 3. For a mild solution y of (1), we define its energy as the continuous function Recalling that
, one proves that if
y is a mild solution and if
, then
on the other hand, thanks to Equations (24)–(26), for all
and
,
where
is as in (
22). Thus, we have
where
Observe that if , being and (recall that ), Analogously for .
As in ([
21], Thoerem 3.1), it is possible to prove that the energy is a non-increasing function.
Theorem 1. Assume Hypothesis 4 and let y be a classical solution of (1). Then, the energy is non-increasing. In particular, Actually, one can prove that the previous monotonicity result also holds under weaker assumptions on the functions a and d.
Proposition 4. Assume Hypothesis 4. For the fixed , if y is a classical solution of (1), thenfor every . Here, . Proof. Since some computations are similar to the ones of ([
21], Proposition 4.7), we sketch them. Fix
. Multiplying the equation in (
1) by
and integrating over
, we have
As in [
21], one proves that
Hence, it remains to compute
. As in [
17], one proves
By ([
17], Lemma 1)
and the thesis follows. □
As a consequence of the previous equality, we have the next relation:
Proposition 5. Assume Hypothesis 4 and fix . If y is a classical solution of (1), then for every , we havewhere Proof. Let
y be a classical solution of (
1) and fix
. Multiplying the equation in (
1) by
, integrating by parts over
, and using (
19), we obtain
Obviously, all the previous integrals make sense, and by multiplying (
38) by
, one has
By summing (
35) and (
39) and using the boundary conditions at 1, we obtain the thesis. □
An immediate consequence of (
37) is the next result. However, to prove it, we assume an additional hypothesis on functions
a and
d.
Hypothesis 5. Assume a (WD) or (SD), d (WD) with , with and .
Observe that this hypothesis is more restrictive than Hypothesis 4; indeed, in Hypothesis 5, we exclude the case . In fact, as we can see already from the next result, the condition is important for the technique used in the following proposition:
Proposition 6. Assume Hypothesis 5, fix and let y be a classical solution of (1). Then, for every and for all , one provesif , andif . Here, and .
Proof. By assumption, we can take
; thus,
Now, we distinguish between the cases
and
.
Case .
In this case, the distributed terms in (
37) can be estimated from below in the following way:
Now, we estimate the boundary terms in (
37) from above. First of all, consider the integral
for all
. Using the fact that
, together with the classical Hardy inequality (
12) and proceeding as in ([
12], Proposition 3.3), one proves
Hence, by Proposition 2,
for all
. Hence, since the energy is non-increasing,
Now, based on (
34) and the fact that
, we have
Obviously,
and
Furthermore, recalling that
,
Hence, by (
37), (
41)–(
47) and Theorem 1, we have
Hence,
and the thesis follows.
Case . In this case, based on the definition of energy and (
15), one proves
hence,
Moreover, by (
37) and (
40), one has
where (B.T.) is the boundary terms in (
37). Now, by (
48),
Proceeding as for the case
and using the fact that
, one can estimate the boundary terms in the following way:
Hence,
and the thesis follows. □
In the next proposition, we will find an estimate from above for
To this end, set
and
Proposition 7. Assume Hypothesis 5 and fix . If y is a classical solution of (1), then for every and for every , we havewhere Proof. Set
,
, where
, and let
be the unique solution of
for all
By Proposition 3,
for all
t and solves
By (
21), we also have
where
is defined in (
22). Moreover, if
and
, then by Proposition 2 if
, one proves
and
On the other hand, if
and
, then by (
24), (
25) and (
29), it results in
In every case, for all considered
, we have
and
Finally, observe that for all considered
and all
, we have
Indeed, consider, first of all,
, then
If
and
is as in (
18), we obtain (
27), which implies that
in particular,
for all
.
Now, multiplying the equation in (
1) by
and integrating over
, we have
Hence, (
56) reads
On the other hand, multiplying the equation in (
51) by
and integrating over
, we have
By (
19), we obtain
Substituting in (
57), using the fact that
,
,
,
and proceeding as in [
12], we have
Then,
Thus, in order to estimate
, we have to consider the four terms in the previous equality.
So, by (
21), (
34), and Theorem 1, we have, for all
,
By Theorem 1, we have
Moreover, for any
, we have
by (
53). In a similar way, using (
54), it is possible to find the next estimate
Therefore, by summing (
60) and (
61) and applying Theorem 1, we obtain
Finally, we estimate the integral
. To this end, consider again the problem (
23) and differentiate with respect to
t. Thus,
Clearly,
satisfies (
52), in particular
and
Thus, by (
55) and the previous estiamte, for
, we find
Coming back to (
58) and using (
59), (
62), and (
63), we find
Hence, for every
,
and the thesis follows. □
As a consequence of Propositions 6 and 7, we can formulate the main result of the paper, whose proof is based on ([
18], Theorem 8.1).
Theorem 2. Assume Hypothesis 5, fix and if , then . Let y be a mild solution for (1). Then, for all and for all whereandHere, ν is defined as in (50). Proof. As a first step, consider
y a classical solution of (
1) and
. Take
Then, based on the definition of
and Propositions 6 and 7, we have
This implies
Hence, we can apply ([
18], Theorem 8.1) with
and (
64) holds.
Now, consider
. By Propositions 6 and 7, we have
Hence,
Recalling that
where
, one proves that
Hence, again by ([
18], Theorem 8.1) with
, (
64) holds.
If
y is the mild solution for the problem, we can proceed as in [
22], obtaining the thesis. □