Genni Fragnelli1, Jerome A. Goldstein2, Rosa Maria Mininni1, and
2
Silvia Romanelli1
3
1Department of Mathematics, University of Bari Aldo Moro, Via E.
4
Orabona 4, 70125 Bari, Italy
5
2Department of Mathematical Sciences, University of Memphis, 373
6
Dunn Hall, Memphis, TN 38152-3240, USA
7
March 17, 2019
8
Abstract
9
We consider a differential operator of order 2n of the type Anu =
10
(−1)n(au(n))(n), where a(x) > 0 in [0, 1] \ {x
0} and a(x0) = 0. We
11
show that, for any n ∈ N, the operator −An generates a contractive
12
analytic semigroup of angle π/2 on L2(0, 1). Note that the domain of
13
Andepends on the type of degeneracy of a. Our theorems extend some
14
previous results in [3] where n = 1.
15
Keywords. Degenerate linear differential operators of order 2n, analytic
16
semigroups
17
1
Introduction
18
Recently, in [3] the authors considered second order differential operators
19
with coefficients having interior degeneracy and acting on L2(0, 1) or L2(0, 1)
20
with a weight. In particular, we stated that both the operators A1u := (au0)0
21
and A2u := au00 having interior degeneracy, in the sense of Definition 2.1
22
or Definition 3.1, are nonpositive and selfadjoint on L2(0, 1) or L21 a
(0, 1),
23
respectively, provided that they are equipped with suitable domains
con-24
taining Dirichlet boundary conditions and additional conditions (including
25
regularity of a) depending on the kind of degeneracy of the coefficient a.
26
Analogous results hold even when the boundary conditions are of Wentzell
27
type (see [4]). The case of the above operators with Dirichlet or Neumann
28
boundary conditions was studied in the framework of Carleman estimates
1
and controllability by G. Fragnelli and D. Mugnai in [7], [8], for Dirichlet
2
boundary conditions, and by I. Boutaayamou, G. Fragnelli and L. Maniar
3
in [1], for Neumann ones. Other related results and their applications to
4
optimal control can be found in [5], [6].
5
Here we focus on the extensions of the results obtained in [3] to the
6
case of differential operators of order 2n with interior degeneracy, where
7
the operator is in divergence form. We deal with the weak degeneracy
8
in Section 2 and with the strong degeneracy in Section 3. This paper is
9
dedicated to Gis`ele Ruiz Goldstein for celebrating her 60th birthday with
10
great admiration and friendship and for thanking her for so many years of
11
wonderful cooperation. We also hope she would like to join us in the future
12
steps of our research in this setting.
13
2
Weakly degenerate operators of order 2n
14
Let us consider the interval [0, 1] and a function a defined on it. Let us
15
introduce the notion of weak degeneracy.
16
Definition 2.1. A function a ∈ C[0, 1] is said weakly degenerate if there
17
exists x0 ∈ (0, 1) such that a(x0) = 0, a(x) > 0 in [0, 1] \ {x0}, and
1 a ∈
18
L1(0, 1). The operator A
nu := (au(n))(n) is called weakly degenerate if a is
19
weakly degenerate.
20
Example 2.1. The function a(x) = |x − x0|α is an example of weakly
21
degenerate function if 0 < α < 1.
22
Let a be weakly degenerate and introduce the following weighted spaces, for n ≥ 1.
Kan(0, 1) := {u ∈ Hn−1(0, 1) : un−1 is absolutely continuous in [0, 1], √
a u(n)∈ L2(0, 1), u(k)(j) = 0, j = 0, 1, k = 0, .., n − 1}
with the norm
23 kukn:=kuk2L2 + k √ au(n)k2L2 12 , and the space
Wan(0, 1) := {u ∈ Kan(0, 1) | au(n)∈ Hn(0, 1)}.
Here H0(0, 1) = L2(0, 1), u(0) = u.
Remark 1. For any n ∈ N, we have that
u ∈ Kan+1(0, 1) implies u0 ∈ Kn
a(0, 1), (2.1)
u ∈ Wan+1(0, 1) implies u0∈ Wan(0, 1). (2.2) We have the following Green’s formula.
1
Lemma 2.1. Assume that a ∈ C[0, 1] is weakly degenerate. Then for all
2
(u, v) ∈ Wan(0, 1) × Kan(0, 1), n ≥ 1, one has
3 Z 1 0 (au(n))(n)vdx = (−1)¯ n Z 1 0 a u(n)¯v(n)dx. (2.3)
Proof. It is sufficient to work with u, v real valued functions. We argue by
4
induction on n ≥ 1. By [3, Lemma 2.1], the assertion holds for n = 1.
5
Assume now that (2.3) is true for a certain n ∈ N and show that (2.3) holds
6
for (n + 1), following the idea of [3, Lemma 2.1]. We have to show that, for
7 all (u, v) ∈ Wan+1(0, 1) × Kan+1(0, 1), 8 Z 1 0 (au(n+1))(n+1)vdx = (−1)(n+1) Z 1 0 au(n+1)v(n+1)dx. (2.4)
First of all, let us prove that
9 Z 1 0 (au(n+1))(n+1)vdx = − Z 1 0 (au(n+1))(n)v0dx. (2.5)
Indeed, for any sufficiently small δ > 0 one has Z 1 0 (au(n+1))(n+1)vdx = Z x0−δ 0 (au(n+1))(n+1)vdx+ Z x0+δ x0−δ (au(n+1))(n+1)vdx + Z 1 x0+δ (au(n+1))(n+1)vdx. Hence Z 1 0
(au(n+1))(n+1)vdx = ((au(n+1))(n)v)(x0− δ) − ((au(n+1))(n)v)(0)
− Z x0−δ 0 (au(n+1))(n)v0dx + Z x0+δ x0−δ (au(n+1))(n+1)vdx + ((au(n+1))(n)v)(1) − ((au(n+1))(n)v)(x0+ δ)x − Z 1 x0+δ (au(n+1))(n)v0dx
= ((au(n+1))(n)v)(x0− δ) − Z x0−δ 0 (au(n+1))(n)v0dx + Z x0+δ x0−δ (au(n+1))(n+1)vdx − ((au(n+1))(n)v)(x0+ δ) − Z 1 x0+δ (au(n+1))(n)v0dx, (2.6)
since au(n+1)∈ Hn+1(0, 1) and v(0) = 0 = v(1). Now we prove that
lim δ→0+ Z x0−δ 0 (au(n+1))(n)v0dx = Z x0 0 (au(n+1))(n)v0dx lim δ→0+ Z 1 x0+δ (au(n+1))(n)v0dx = Z 1 x0 (au(n+1))(n)v0dx lim δ→0+ Z x0+δ x0−δ (au(n+1))(n+1)vdx = 0. (2.7)
To this end we observe that Z x0−δ 0 (au(n+1))(n)v0dx = Z x0 0 (au(n+1))(n)v0dx − Z x0 x0−δ (au(n+1))(n)v0dx, (2.8) Z 1 x0+δ (au(n+1))(n)v0dx = Z 1 x0 (au(n+1))(n)v0dx − Z x0+δ x0 (au(n+1))(n)v0dx. (2.9)
Moreover, (au(n+1))(n+1)v and (au(n+1))(n)v0 belong to L1(0, 1). Thus for any > 0, by the absolute continuity of the integral, there exists δ := δ() > 0 such that Z x0 x0−δ (au(n+1))(n)v0dx ≤ Z x0 x0−δ |(au(n+1))(n)v0|dx < , Z x0+δ x0−δ (au(n+1))(n+1)vdx ≤ Z x0+δ x0−δ |(au(n+1))(n+1)v|dx < , Z x0+δ x0 (au(n+1))(n)v0dx ≤ Z x0+δ x0 |(au(n+1))(n)v0|dx < .
Let us take such a δ in (2.6). Thus, being arbitrary, we obtain lim δ→0+ Z x0 x0−δ (au(n+1))(n)v0dx = lim δ→0+ Z x0+δ x0−δ (au(n+1))(n+1)vdx = lim δ→0+ Z x0+δ x0 (au(n+1))(n)v0dx = 0.
Clearly, (2.8), (2.9) imply that
lim δ→0+ Z x0−δ 0 (au(n+1))(n)v0dx = Z x0 0 (au(n+1))(n)v0dx lim δ→0+ Z 1 x0+δ (au(n+1))(n)v0dx = Z 1 x0 (au(n+1))(n)v0dx.
In order to prove the desired result, it is sufficient to verify that
1 lim δ→0+((au (n+1))(n)v)(x 0− δ) = lim δ→0+((au (n+1))(n)v)(x 0+ δ), (2.10)
but this follows in a simple way. Indeed (au(n+1))(n) ∈ H1(0, 1) and v ∈
Kan+1(0, 1), thus lim δ→0+((au (n+1))(n)v)(x 0− δ) = ((au(n+1))(n)v)(x0) (2.11) = lim δ→0+((au (n+1))(n)v)(x 0+ δ).
Hence, by (2.6),(2.7) and (2.10), we have
2 Z 1 0 (au(n+1))(n+1)vdx = − Z 1 0 (au(n+1))(n)v0dx,
and (2.5) is true. Now, according to (2.1), (2.2), (u0, v0) ∈ Wan(0, 1) × Kan(0, 1), thus we obtain Z 1 0 (au(n+1))(n+1)vdx = Z 1 0 (a(u0)(n))(n+1)vdx = − Z 1 0 (a(u0)(n))(n)v0dx.
The induction assumption implies that
Z 1 0 (a(u0)(n))(n)v0dx = (−1)n Z 1 0 (a(u0)(n))(v0)(n)dx
and, consequently, Z 1 0 (au(n+1))(n+1)vdx = − Z 1 0 (a(u0)(n))(n)v0dx = (−1)n+1 Z 1 0 (a(u0)(n))(v0)(n)dx = (−1)n+1 Z 1 0 au(n+1)v(n+1)dx.
This proves the assertion.
1
Lemma 2.1 enables us to show the following result.
2
Theorem 2.1. The operator eA1,n with domain D( eA1,n) = Wan(0, 1) defined
3
by eA1,nu := (−1)n(au(n))(n) is positive and selfadjoint on L2(0, 1). Hence
4
− eA1,ngenerates a contractive analytic semigroup of angle
π 2 on L 2(0, 1), for 5 any n ∈ N. 6
Proof. The case n = 1 was proved in [2]. Now observe that D( eAn) is dense
7
in L2(0, 1). It is also clear that eAnis positive and symmetric. We only need
8
to verify that (I + eAn) is surjective. Observe that Kan(0, 1) equipped with
9
the inner product
10
(u, v)n:=
Z 1 0
(u¯v + au(n)v¯(n))dx,
for any (u, v) ∈ Kan(0, 1) × Kan(0, 1), is a Hilbert space. Moreover,
11
Kan(0, 1) ,→ L2(0, 1) ,→ (Kan(0, 1))∗, (2.12)
where (Kan(0, 1))∗ is the dual space of Kan(0, 1) with respect to L2(0, 1).
12
Indeed, the embedding of Kan(0, 1) in L2(0, 1) is readily seen to be dense
13
and continuous. In addition, for any f ∈ Kan(0, 1) and ϕ ∈ L2(0, 1),
14 |hf, ϕiL2| = Z 1 0 f ¯ϕdx ≤ kf kL2kϕkL2 ≤ kf kKn akϕkL2.
Hence L2(0, 1) ,→ (Kan(0, 1))∗. Then (Kan(0, 1))∗is the completion of L2(0, 1) with respect to the norm of (Kan(0, 1))∗, i.e., kf k(Kn
a(0,1))∗ = sup{ R1 0 f ¯ϕdx :
kϕkKn
a(0,1) ≤ 1}. For f ∈ L
2(0, 1), define the functional F : Kn
a(0, 1) → R by F (v) := Z 1 0 f ¯vdx.
Since Kan(0, 1) ,→ L2(0, 1), we have that F ∈ (Kan(0, 1))∗. By Riesz’s
Theo-1
rem, there exists a unique u ∈ Kan(0, 1) such that for all v ∈ Kan(0, 1),
2 (u, v)n= Z 1 0 f ¯vdx. (2.13) In particular, since Cc∞(0, 1) ⊂ Kn
a(0, 1), (2.13) holds for all v ∈ Cc∞(0, 1),
3 i.e. 4 Z 1 0 au(n)¯v(n)dx = Z 1 0 (f − u)¯vdx. (2.14)
In other words, by (2.3), (2.14) is equivalent to
5 (−1)n Z 1 0 (au(n))(n)¯vdx = Z 1 0 (f − u)¯vdx
and we obtain that f − u = eA1,nu, hence (I + eA1,n)(u) = f .
6
3
Strongly degenerate operators of order 2n
7
In this section we will deal with the case of strong degeneracy, according
8
to the following definition.
9
Definition 3.1. A function a ∈ W1,∞(0, 1) is said strongly degenerate if
10
there exists x0∈ (0, 1) such that a(x0) = 0,
11
a(x) > 0 in [0, 1] \ {x0}, (3.1)
and 1 a 6∈ L
1(0, 1). The operator A
nu := (au(n))(n) is called strongly
degen-12
erate if a is strongly degenerate.
13
Example 3.1. The function a(x) = |x − x0|α is an example of a strongly
14
degenerate function if 1 ≤ α.
15
Now let us assume that a is strongly degenerate and introduce the fol-lowing weighted spaces for n ≥ 1,
Ks,an (0, 1) := {u ∈ Hn−1(0, 1) | u(n−1) is locally absolutely continuous in [0, x0) ∪ (x0, 1],
√
equipped with the norm 1 kukn:= kuk2L2+ k √ a u(n)k2L2 12 ,
and the space
Wan(0, 1) := {u ∈ Ks,an (0, 1)| au(n)∈ Hn(0, 1),
(au(k))(h)(x0) = 0, for all k, h = 0, .., n}.
Remark 2. For any n ∈ N, we have that
u ∈ Ks,an+1(0, 1) implies u0 ∈ Ks,an (0, 1), (3.2)
u ∈ Wan+1(0, 1) implies u0∈ Wn
a(0, 1). (3.3)
In order to state an analogous Green’s formula as in Lemma 2.1, we need
2
some preliminary results.
3
Proposition 3.1. Let n ≥ 1 and consider
4
Dn:= {u ∈ Hn−1(0, 1) | u(n−1) is locally absolutely continuous in [0, x0) ∪ (x0, 1],
au(k) ∈ H01(0, 1), k = 0, .., n − 1,
au(n)∈ Hn(0, 1), (au(k))(h)(x0) = 0, for all k, h = 0, .., n}.
Then Dn= Wan(0, 1).
5
Remark 3. Notice that if u ∈ Dn, then au(k) ∈ H01(0, 1), for k = 0, 1.., n−1.
6
This fact and the assumption (3.1), imply u(k)(0) = u(k)(1) = 0, for k =
7
0, .., n − 1.
8
Before proving Proposition 3.1, we note the following.
9
Lemma 3.1. For any u ∈ Dn, we have
10
| (au(k))(h)(x) | ≤ k(au(k))(h+1)k
L2p|x − x0| (3.4)
for all k = 0, .., n, h = 0, .., n − 1, x ∈ [0, 1].
11
Proof. Let u ∈ Dn. Fix k = 0, .., n. Since (au(k))(h)(x0) = 0, for any
h = 0, .., n − 1, one can deduce that
| (au(k))(h)(x) | = Z x x0 (au(k))(h+1)(s) ds ≤ k(au(k))(h+1)kL2p|x − x0|. 12
Proof of Proposition 3.1. We prove that Dn⊂ Wan(0, 1). Let u ∈ Dn. It is
sufficient to show√au(n)∈ L2(0, 1). As a first step, we observe the following
formula, Z 1 x h (au(n))(n)u¯i(s) ds = m X k=1 (−1)kh(au(n))(n−k)u¯(k−1)i(x) + (−1)n Z 1 x [a|u(n)|2](s)ds. (3.5)
It is sufficient to work with u real valued. We prove (3.5) by induction on n ≥ 1. Clearly, the assertion follows for n = 1. Assume now that (3.5) holds for a certain n ≥ 1 and show it for n + 1. To this end, let u ∈ Dn+1.
Thus u0∈ Dn. We evaluate Z 1 x [(au(n+1))(n+1)u](s)ds = h (au(n+1))(n)u(s) i1 x− Z 1 x h (au(n+1))(n)u0 i (s)ds (by Remark 3) = −h(au(n+1))(n)ui(x) − Z 1 x h (a(u0)(n))(n)u0i(s)ds (by (3.5) applied to u0) = −h(au(n+1))(n)ui(x) − " n X k=1 (−1)ka(u0)(n)(n−k)(u0)(k−1)(x) + (−1)n Z 1 x h a((u0)(n))2i(s)ds = − h (au(n+1))(n)u i (x) + n X k=1 (−1)k+1 (au(n+1))(n−k)u(k) (x) + (−1)n+1 Z 1 x h a(u(n+1))2 i (s)ds = − h (au(n+1))(n)u i (x) + n+1 X k=2 (−1)k (au(n+1))(n+1−k)u(k−1) (x) + (−1)n+1 Z 1 x h a(u(n+1))2 i (s)ds
= n+1 X k=1 (−1)k(au(n+1))(n+1−k)u(k−1)(x) + (−1)n+1 Z 1 x h a(u(n+1))2i(s)ds.
By induction it follows that (3.5) holds for any x ∈ [0, 1]. Let us now
1
show that, for fixed k = 1, .., n,
2 lim x→x+0 (au(n))(n−k)u(k−1) (x) = 0. (3.6) Indeed, if we define L = lim x→x+0 (au(n))(n−k)u(k−1)(x),
assume L 6= 0. Then there exists C > 0 such that (au(n))(n−k)u(k−1) (x) ≥ C for all x in a right neighborhood of x0, x 6= x0.
3
Thus, by (3.4) there exists C1 > 0 such that
|u(k−1)(x)| ≥ C |(au(n))(n−k)(x)| ≥ C1 k(au(n))(n−k+1)k L2 ·√ 1 x − x0 .
This implies that u(k−1) ∈ L/ 2(0, 1). Hence L = 0 and
4 lim x→x+0 n X k=1 (−1)k(au(n))(n−k)u(k−1)(x) = 0.
Similar arguments imply that
5
lim
x→x−0
(au(n))(n−k)u(k−1)(x) = 0 (3.7)
for fixed k = 1, .., n. Therefore, from (3.5), the limits (3.6) and (3.7) enable us to deduce that Z 1 x0 (au(n))(n)u(s) ds = (−1)n Z 1 x0 a(u(n))2(s) ds
and Z x 0 0 (au(n))(n)u (s) ds = (−1)n Z x0 0 a(u(n))2 (s) ds. Thus Z 1 0 (au(n))(n)u (s) ds = (−1)n Z 1 0 a(u(n))2 (s) ds.
Since (au(n))(n)u ∈ L1(0, 1), then √a u(n)∈ L2(0, 1) and D
n⊂ Wan(0, 1).
1
In order to prove that Wan(0, 1) ⊂ Dn, let u ∈ Wan(0, 1). We only need
2
to show that
3
au(k)∈ H01(0, 1), k = 0, .., n − 1. (3.8) For k = 0 we proceed as in [7]. We already know that (au)(j) = 0, j = 0, 1, and au ∈ L2(0, 1). Moreover, since a ∈ W1,∞(0, 1),
(au)0 = a0u + au0∈ L2(0, 1).
Thus for x < x0 one has
4
(au)(x) = Z x
0
(au)0(t)dt.
This implies that there exists limx→x−
0(au)(x) = (au)(x0) =
Rx0
0 (au)
0(t)dt =
5
L ∈ R. If L 6= 0, then there exists C > 0 such that
6
|(au)(x)| ≥ C
for all x in a neighborhood of x0, x 6= x0. Setting C1 :=
C2 max[0,1]a(x) > 0, 7 it follows that 8 |u2(x)| ≥ C 2 a2(x) ≥ C1 a(x)
for all x in a left neighborhood of x0, x 6= x0. But, since the operator is
9
strongly degenerate, 1 a 6∈ L
1(0, 1) thus u 6∈ L2(0, 1). Hence L = 0.
Analo-10
gously, starting from
11
(au)(x) = − Z 1
x
(au)0(t)dt for x > x0,
one can prove that limx→x+
0(au)(x) = (au)(x0) = 0 and thus
Rx0 0 (au) 0(t)dt = 12 0 = −Rx1 0(au)
0(t)dt. From this it also easily follows that (au)0 is the
distri-13
butional derivative of au, and so au ∈ H1 0(0, 1).
Assume that (3.8) holds for a certain k ∈ {0, .., n − 1} such that (k + 1) ∈ {0, .., n − 1}. We show that au(k+1)∈ H1 0(0, 1). Indeed, (au(k+1))(j) = 0, j = 0, 1 and au(k+1)∈ L2(0, 1). Moreover, (au(k+1))0= a0u(k+1)+ au(k+2)∈ L2(0, 1)
and proceeding as before, au(k+1) ∈ H1
0(0, 1). Then the assertion holds.
1
By induction the following result holds.
2
Lemma 3.2. For all (u, v) ∈ Wan(0, 1) × Ks,an (0, 1) one has
Z 1 0 (au(n))(n)v dx = (−1)n Z 1 0 au(n)v(n)dx.
Similar arguments as in Theorem 2.1 show the next result. The case
3
n = 1 was proved in [3].
4
Theorem 3.1. If a is strongly degenerate, then the operator eAnwith domain
5
D( eAn) = Wan(0, 1) defined by eAnu := (−1)n(au(n))(n)is positive and
selfad-6
joint on L2(0, 1). Hence − eAn generates a contractive analytic semigroup of
7 angle π 2 on L 2(0, 1), for any n ∈ N. 8
Acknowledgments
9G. Fragnelli, R.M. Mininni and S. Romanelli are members of Gruppo
10
Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni
11
(GNAMPA), Istituto Nazionale di Alta Matematica (INdAM). G. F. is
par-12
tially supported by the FFABR “Fondo per il finanziamento delle attivit`a
13
base di ricerca” 2017, by the INdAM- GNAMPA Project 2019
“Controlla-14
bilit`a di PDE in modelli fisici e in scienze della vita” and by PRIN 2017-2019
15
Qualitative and quantitative aspects of nonlinear PDEs.
16
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