Generalized Wentzell boundary conditions for
second order operators with interior degeneracy
Genni Fragnelli∗,
Gis`ele Ruiz Goldstein†, Jerome A. Goldstein†, Rosa Maria Mininni∗, Silvia Romanelli∗
Abstract
We consider operators in divergence form, A1u = (au0)0, and in
nondivergence form, A2u = au00, provided that the coefficient a
va-nishes in an interior point of the space domain. Characterizing the domain of the operators, we prove that, under suitable assumptions, the operators A1 and A2, equipped with general Wentzell boundary
conditions, are nonpositive and selfadjoint on spaces of L2 type.
Keywords. Second order operators in divergence and nondivergence form; Interior degeneracy; Generalized Wentzell boundary conditions.
AMS subject classifications. 47D06, 35K65, 47B25, 47N20
1
Introduction
It is well known that degenerate parabolic equations are widely used as mathematical models in the applied sciences to describe the evolution in time of a given system. For this reason, in recent years an increasing interest has been devoted to the study of second order differential degenerate operators in divergence or in nondivergence form. A wide non exhaustive description of them can be found in [9, Introduction] and in the references therein. In particular, operators of the type A1u := (au0)0, A2u := au00 with suitable
domains involving different boundary conditions, arise in a natural way in several contexts as aeronautics (Crocco equation), physics (boundary layer
∗
Department of Mathematics, University of Bari Aldo Moro, Via E.Orabona 4, 70125 Bari, Italy (genni.fragnelli@uniba.it, rosamaria.mininni@uniba.it, sil-via.romanelli@uniba.it)
†
Department of Mathematical Sciences, University of Memphis, 373 Dunn Hall, Mem-phis, TN 38152-3240, USA (ggoldste@memphis.edu, jgoldste@memphis.edu)
models), genetics (Wright-Fisher and Fleming-Viot models), mathematical finance (Black-Merton-Scholes models).
Here the novelty is that we deal with existence and regularity of solutions of Cauchy problems associated with parabolic equations having coefficients which degenerate in the interior of the spatial domain and satisfy general Wentzell boundary conditions in spaces of L2-type.
To our best knowledge, [16] is the first paper treating the existence of a solution for the Cauchy problem associated to a parabolic equation which degenerates in the interior of the spatial domain in the space L2(0, 1), while in [9] both the degenerate operators A1 and A2 in the space L2(0, 1), with
or without weight, were examined. In particular, the authors proved that both the operators are nonpositive and selfadjoint, hence they generate co-sine families and, as a consequence, analytic semigroups, provided that the coefficient a vanishes in an interior point of the spatial domain and Dirichlet boundary conditions hold.
These results are complemented in [12] and in [13], where other aspects of the associated parabolic problem, such as Carleman estimates and null controllability, are considered. We refer to the recent paper [4] for the anal-ogous results under Neumann boundary conditions.
For this kind of equations, in addition to well posedness, controllability and Carleman estimates, other features have been studied in several pa-pers, obtaining substantial progresses. Among these papa-pers, we cite [3], [5], [10] and [11], where the authors obtain results concerning inverse problems, stability, optimal existence and regularity theory also in higher dimension.
On the other hand, the importance of the study of the operators A1
and A2 equipped with general Wentzell boundary conditions ((GWBC)
for shortness) in spaces of Lp-type recently received a lot of attention, after the new directions opened in [8].
Indeed, it is worth to mention that, in the case of heat equations, (GWBC) allow to take into account the action of heat sources on the boundary (see [14]).
Additional motivation for the study of evolution equations with (GWBC) comes from their possible interpretation as evolution equations with dynam-ical boundary conditions. For a general view on the role of Wentzell bound-ary conditions we refer to [7], while for a physical interpretation of them see [14]. In this framework, up to now, operators with interior degeneracy were never considered. Thus we intend to fill this gap according to the ideas of [9] by using Hilbert spaces depending on the boundary conditions introduced in [8]. Indeed in the main theorems (Sections 3-5) we will prove that, under suitable assumptions, the operators A1 and A2 equipped with (GWBC)
are nonpositive and selfadjoint in suitable spaces of L2-type. It is worth noting that in the present work we deal with real function spaces, but the assertions can be easily extended to the complex case.
2
Basic assumptions and preliminary results
In the following we will introduce the notions of weak and strong degeneracy for a real-valued function a defined on the interval [0, 1].
Accordingly, we will define suitable weighted spaces and prove some formulas of Green type. These results will play a key role for the study of the operators A1 and A2 considered in Sections 3-5.
Definition 2.1. A function a ∈ C[0, 1] is said to be weakly degenerate if there exists x0 ∈ (0, 1) such that a(x0) = 0, a(x) > 0 in [0, 1] \ {x0}, and
1 a ∈ L
1(0, 1).
Example 2.1. We can take a(x) = |x − x0|α, 0 < α < 1, as an example of
a weakly degenerate function.
For any weakly degenerate a ∈ C[0, 1], let us introduce the following weighted spaces: Ha1(0, 1) := {u ∈ L2(0, 1) | u absolutely continuous in [0, 1], √ au0∈ L2(0, 1)}, Ha2(0, 1) := {u ∈ Ha1(0, 1)| au0 ∈ H1(0, 1)}, L21 a (0, 1) := u ∈ L2(0, 1) | Z 1 0 u2 a dx < ∞ , H11 a (0, 1) := L21 a (0, 1) ∩ H1(0, 1), H21 a (0, 1) := n u ∈ H11 a (0, 1) u0 ∈ H1(0, 1) o ,
endowed with the respective norms defined by
kuk2H1 a(0,1):= kuk 2 L2(0,1)+ k √ au0k2L2(0,1), for all u ∈ Ha1(0, 1), kuk2H2 a(0,1):= kuk 2 H1 a(0,1)+ k(au 0 )0k2L2(0,1), for all u ∈ Ha2(0, 1), kuk2L2 1 a (0,1):= Z 1 0 u2 a dx, for all u ∈ L 2 1 a (0, 1),
kuk2H1 1 a (0,1) := kuk 2 L2 1 a (0,1)+ ku 0k2 L2(0,1), for all u ∈ H11 a (0, 1), kuk2H2 1 a (0,1) := kuk 2 H1 1 a (0,1)+ kau 00k2 L2 1 a (0,1), for all u ∈ H 2 1 a (0, 1).
Similar arguments as in [9, Corollary 3.1] lead to the following result.
Proposition 2.1. If a is weakly degenerate, then the spaces H11 a
(0, 1) and H1(0, 1) coincide algebraically and their norms are equivalent. Hence C∞[0, 1] is dense in H11
a
(0, 1).
The following Green formulae are analogous to those proved in [9].
Lemma 2.1. If a is weakly degenerate, then
(i) for all (u, v) ∈ Ha2(0, 1) × Ha1(0, 1):
Z 1 0 (au0)0vdx = [au0v]x=1x=0− Z 1 0 au0v0dx. (2.1)
(ii) for all (u, v) ∈ H21 a (0, 1) × H11 a (0, 1): Z 1 0 u00vdx = [u0v]x=1x=0− Z 1 0 u0v0dx. (2.2)
The proof of Lemma 2.1 is given in the Appendix.
Now let us introduce another notion of interior degeneracy.
Definition 2.2. A function a ∈ W1,∞(0, 1) is called strongly degenerate if there exists x0 ∈ (0, 1) such that a(x0) = 0, a(x) > 0 in [0, 1] \ {x0}, and
1 a 6∈ L
1(0, 1).
Example 2.2. We can take a(x) = |x − x0|α, α ≥ 1, as an example of a
strongly degenerate function.
For any strongly degenerate a ∈ W1,∞(0, 1), let us introduce the corre-sponding weighted spaces
Ha1(0, 1) := {u ∈ L2(0, 1) | u locally absolutely continuous in [0, x0)∪(x0, 1],
√
au0 ∈ L2(0, 1)},
and consider the spaces L21 a (0, 1), H11 a (0, 1) and H21 a (0, 1) introduced in the weakly degenerate case.
Since in this situation a function u ∈ Ha2(0, 1) is locally absolutely con-tinuous in [0, 1] \ {x0} and not necessarily absolutely continuous in [0, 1] as
for the weakly degenerate case, the equality (2.1) is not true a priori. Now let us provide some useful results.
Proposition 2.2. Let a be strongly degenerate and define
X := {u ∈ L2(0, 1) | u locally absolutely continuous in [0, 1] \ {x0},
√ au0 ∈ L2(0, 1), au ∈ H1(0, 1) and (au)(x0) = 0}, Z := {u ∈ X | au0 ∈ H1(0, 1), (au0)(x0) = 0}. Then (i) Ha1(0, 1) = X (ii) Ha2(0, 1) = Z.
Proof. (i) Of course X ⊆ Ha1(0, 1). Conversely, let us take u ∈ Ha1(0, 1). Proceeding as in [12], we prove that au is continuous at x0, in particular
(au)(x0) = 0 that is, u ∈ X. To this aim, observe that, by the assumption on
u, au0 ∈ L2(0, 1) and, since a ∈ W1,∞(0, 1), then (au)0 = a0u+au0 ∈ L2(0, 1).
Therefore for 0 ≤ x < x0, one has
(au)(x) = Z x
0
(au)0(t)dt + (au)(0)
(observe that (au)(0) exists since au is continuous away from x0). This
implies that there exists
lim x→x−0 (au)(x) = Z x0 0 (au)0(t)dt + (au)(0) = L ∈ R. If L 6= 0, then there exists C > 0 such that
|(au)(x)| ≥ C
for all x in a left neighborhood of x0, x 6= x0. Thus setting C1:=
C2 max[0,1]a(x) > 0, it follows that |u2(x)| ≥ C2 a2(x) ≥ C1 a(x),
for all x in a left neighborhood of x0, x 6= x0. But, since a is strongly
degenerate, 1 a 6∈ L
1(0, 1) thus u 6∈ L2(0, 1). Hence L = 0. Analogously, one
can prove that limx→x+
0(au)(x) = 0 and thus au is continuous at x0 and
(au)(x0) = 0. Therefore, it easily follows that (au)0 is the distributional
derivative of au, and so au ∈ H1(0, 1), i.e. u ∈ X.
(ii) The inclusion Z ⊆ Ha2(0, 1) is obvious. Conversely, if u ∈ Ha2(0, 1), then we only need to show that (au0)(x0) = 0. Similar arguments as in the proof
of (i) imply that lim
x→x−0
(au0)(x) = L ∈ R. If L 6= 0, then there exists C > 0 such that
|(au0)(x)| ≥ C
for all x in a left neighborhood of x0, x 6= x0. It follows that
|(√au0)(x)| ≥ C pa(x),
for all x in a left neighborhood of x0, x 6= x0. Hence |(a(u0)2)(x)| ≥ C
2
a(x) for
all x in a left neighborhood of x0, x 6= x0. But a is strongly degenerate, so
1 a 6∈ L
1(0, 1) and thus √au0 6∈ L2(0, 1). Hence L = 0. Analogously, one can
prove that lim
x→x+0
(au0)(x) = 0 and thus au0 can be extended by continuity at x0 setting (au0)(x0) = 0.
We point out that Proposition 2.2 is based on the following
Lemma 2.2. (see [9, Lemma 2.5]) If a is strongly degenerate, then for all u ∈ Z we have that
|(au)(x)| ≤ k(au)0kL2(0,1)p|x − x0|,
and
|(au0)(x)| ≤ k(au0)0kL2(0,1)p|x − x0|, (2.3)
for all x ∈ [0, 1].
As for the weakly degenerate case and using the previous characteriza-tion, we can prove the following Green’s formulae. (See Appendix.)
Lemma 2.3. If a is strongly degenerate, then (i) for all (u, v) ∈ Ha2(0, 1) × Ha1(0, 1) one has
Z 1 0 (au0)0vdx = [au0v]x=1x=0− Z 1 0 au0v0dx.
(ii) for all (u, v) ∈ H21 a (0, 1) × H11 a (0, 1): Z 1 0 u00v dx = [u0v]x=1x=0− Z 1 0 u0v0dx.
Similar arguments as in [9, Propositions 3.6, 3.8] allow to characterize the spaces H11
a
(0, 1) and H21 a
(0, 1).
Let us make the following additional assumption on a. Hypothesis 2.1 There exists a positive constant K such that
1 a(x) ≤ K |x − x0|2 , for all x ∈ [0, 1] \ {x0}. (e.g. a(x) = |x − x0|α, 1 ≤ α ≤ 2.)
Proposition 2.3. If a is strongly degenerate and satisfies Hypothesis 2.1 then (i) H11 a (0, 1) = {u ∈ H11 a (0, 1) | u(x0) = 0},
and the norms kukH1 1 a (0,1) and R1 0(u 0)2dx 1 2 are equivalent. (ii) H21 a (0, 1) = {u ∈ H11 a (0, 1) | au00∈ L21 a (0, 1), au0 ∈ H1(0, 1), u(x0) = (au0)(x0) = 0}.
3
Operators in divergence form with (GWBC):
the weakly degenerate case
Let us fix βj, γj ∈ R, j = 0, 1, such that βj > 0 and γj ≥ 0, j = 0, 1.
Consider a weakly or strongly degenerate function a, define the operator in divergence form A1u = (au0)0 equipped with
(GWBC) A1u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1, (3.1)
and the space
associated with (GWBC) (see [8]). Here
dµ := dx |(0,1)⊗adS β |{0,1},
dx denotes the Lebesgue measure on (0, 1), β = (β0, β1), and adSβ denotes
the natural (Dirac) measure dS on {0, 1} with weight βa. More precisely, Xµ
is a Hilbert space with respect to the inner product given by
(f, g)Xµ= Z 1 0 f (x)g(x) dx + a(1)f (1)g(1) β1 + a(0)f (0)g(0) β0 ,
where f, g ∈ Xµ are written as (f χ(0,1), (f (0), f (1))), (g χ(0,1), (g(0), g(1))),
and χ(0,1) is the characteristic function of the interval (0, 1).
Hence Xµ is equipped with the norm defined by
kf k2 Xµ= Z 1 0 |f (x)|2dx + a(1)|f (1)|2 β1 + a(0)|f (0)| 2 β0 ,
for any f ∈ Xµ, provided that f is written as (f χ(0,1), (f (0), f (1))).
We now define the weighted spaces
b Ha1(0, 1) := {u ∈ Xµ | u absolutely continuous in [0, 1], √ au0 ∈ Xµ} and b Ha2(0, 1) := {u ∈ bHa1(0, 1)| au0 ∈ H1(0, 1)},
endowed with the norms defined, respectively, by kuk2 b H1 a(0,1) := kuk2Xµ+ k√au0k2 Xµ, for all u ∈ bH 1 a(0, 1) and kuk2 b H2 a(0,1) := kuk2 b H1 a(0,1) + k(au0)0k2X µ, for all u ∈ bH 2 a(0, 1).
Remark 3.1. Let us observe that bH1
a(0, 1) ⊂ Ha1(0, 1) and bHa2(0, 1) ⊂
Ha2(0, 1). It follows that for any (u, v) ∈ bHa2(0, 1) × bHa1(0, 1) the Green formula in Lemma 2.1(i) holds. Moreover C∞[0, 1] ⊂ bHa2(0, 1).
Further, let us define the domain of A1 to be the following subspace of
Xµ:
D(A1) = {u ∈ bHa2(0, 1)|A1u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1}.
Theorem 3.1. If a is weakly degenerate, then the operator A1 with domain
D(A1) is nonpositive and selfadjoint on Xµ.
Proof. Observe that D(A1) is dense in Xµ. Indeed, if we introduce the space
W1,a(0, 1) = {u ∈ bHa2(0, 1)| supp {u} compact, supp {u} ⊂ [0, 1] \ {x0},
A1u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1}, (3.2)
endowed with the norm
kuk2W 1,a(0,1) := kuk 2 b H2 a(0,1), it is clear that W1,a(0, 1) ⊂ D(A1) ⊂ Xµ.
Moreover, W1,a(0, 1) is dense in Xµ by using the following argument: take
u ∈ Xµ and define un:= ξnu, n ≥ max {x40,1−x4 0}, where
ξn= 0, in [x0−n1, x0+n1], 1, in [0, x0−n2] ∪ [x0+n2, 1], n(x0− x) − 1, in (x0−n2, x0−n1), n(x − x0) − 1, in (x0+n1, x0+n2). (3.3)
It follows that un → u in Xµ as n → ∞. Hence D(A1) is dense in Xµ.
In order to show that A1 is nonpositive and selfadjoint it suffices to prove
that A1 is symmetric, nonpositive and (I − A1)(D(A1)) = Xµ (see e.g. [1,
Theorem B.14] or [15]). A1 is symmetric.
By using Lemma 2.1(i) and (GWBC), for any u, v ∈ D(A1), one has
hA1u, viXµ = Z 1 0 v(au0)0dx + (au 0)0(0)a(0)v(0) β0 +(au 0)0(1)a(1)v(1) β1 = [au0v]x=1x=0− Z 1 0 au0v0dx + (β0u0(0) − γ0u(0)) a(0)v(0) β0 − (β1u0(1) + γ1u(1)) a(1)v(1) β1 = − Z 1 0 au0v0dx − γ0 β0 u(0)a(0)v(0) − γ1 β1 u(1)a(1)v(1) = hu, A1viXµ. A1 is nonpositive.
For any u ∈ D(A1), according to the previous calculations, one has hA1u, uiXµ = − Z 1 0 a|u0|2dx − γ0 β0 a(0)|u(0)|2− γ1 β1 a(1)|u(1)|2≤ 0. I − A1 is surjective.
Observe that bHa1(0, 1) is a Hilbert space with respect to the inner product
(u, v) b H1 a(0,1) = Z 1 0 (uv + au0v0) dx +a(0)u(0)v(0) β0 +a(1)u(1)v(1) β1 , (3.4)
for any u, v ∈ bHa1(0, 1). Moreover, we have that
b
Ha1(0, 1) ,→ Xµ,→ ( bHa1(0, 1)) ∗
, where ( bH1
a(0, 1))∗ is the dual space of bHa1(0, 1) with respect to Xµ. Let
f ∈ Xµ and define F : bHa1(0, 1) 7→ R such that
F (v) = Z 1 0 f v dx +a(0)f (0)v(0) β0 +a(1)f (1)v(1) β1 .
From bHa1(0, 1) ,→ Xµ, it follows that F ∈ ( bHa1(0, 1))∗. As a consequence,
by Riesz’s Theorem, there exists a unique u ∈ bHa1(0, 1) such that for any v ∈ bHa1(0, 1) (u, v) b H1 a(0,1)= F (v) = Z 1 0 f v dx +a(0)f (0)v(0) β0 +a(1)f (1)v(1) β1 .
From (3.4), the above equality means that Z 1
0
au0v0dx = Z 1
0
(f − u)v dx + a(0)(f − u)(0)v(0) β0 +a(1)(f − u)(1)v(1) β1 , (3.5) for all v ∈ bHa1(0, 1).
In particular, if we denote by Cc∞(0, 1) the space of all C∞(0, 1) functions with compact support in (0, 1), from Cc∞(0, 1) ⊂ bH2
a(0, 1), it follows that
(3.5) holds for all v ∈ Cc∞(0, 1). This implies Z 1 0 au0v0dx = Z 1 0 (f − u)v dx, v ∈ Cc∞(0, 1) . (3.6)
Thus the weak derivative (au0)0 in this context exists and (au0)0 ∈ L2(0, 1).
Hence u ∈ D(A1) and
Thanks to Theorem 3.1, one has that the problem ut− A1u = h(t, x), (t, x) ∈ QT := (0, T ) × (0, 1), (GWBC), u(0, x) = u0(x), x ∈ (0, 1), (3.7)
is wellposed in the sense of Theorem 3.2 below. But first we recall the following definition.
Definition 3.1. Assume that u0∈ Xµ and h ∈ L2(0, T ; Xµ). A function u
is said to be a weak solution of (3.7) if
u ∈ C([0, T ]; Xµ) ∩ L2(0, T ; bHa1(0, 1)) and satisfies Z 1 0 u(T, x)ϕ(T, x)dx − Z 1 0 u0(x)ϕ(0, x)dx − Z T 0 Z 1 0 ϕt(t, x)u(t, x)dxdt +a(1)u(T, 1)ϕ(T, 1) β1 − a(1)u0(1)ϕ(0, 1) β1 −a(1) β1 Z T 0 u(t, 1)ϕt(t, 1)dt +a(0)u(T, 0)ϕ(T, 0) β0 − a(0)u0(0)ϕ(0, 0) β0 −a(0) β0 Z T 0 u(t, 0)ϕt(t, 0)dt = − Z T 0 Z 1 0 auxϕxdxdt − γ1 β1 Z T 0 a(1)u(t, 1)ϕ(t, 1)dt − γ0 β0 Z T 0 a(0)u(t, 0)ϕ(t, 0)dt + Z T 0 Z 1 0 h(t, x)ϕ(t, x)dxdt + Z T 0 a(1)h(t, 1)ϕ(t, 1) β1 dt + Z T 0 a(0)h(t, 0)ϕ(t, 0) β0 dt, for all ϕ ∈ H1(0, T ; Xµ) ∩ L2(0, T ; bHa1(0, 1)).
Theorem 3.2. Assume that a is weakly degenerate. Then for all h ∈ L2(0, T ; Xµ) and u0 ∈ Xµ, there exists a unique weak solution u of (3.7)
such that sup t∈[0,T ] ku(t)k2Xµ ≤ Chku0k2Xµ+ khk 2 L2(Q T) i , (3.8)
where C > 0 is a suitable constant. Moreover, if u0 ∈ D(A1) , then u ∈
C([0, T ]; D(A1)) ∩ C1([0, T ]; Xµ) and the following inequality holds
sup t∈[0,T ] ku(t)k2Xµ+ Z T 0 ku(t)k2 b H1 a(0,1)dt ≤ K h kux(·, 0)k2L2(0,T ) + kux(·, 1)k2L2(0,T )+ ku0k2Xµ+ khk 2 L2(Q T) i , (3.9)
Proof. The assertion concerning the assumption u0 ∈ Xµand the regularity
of the solution u when u0 ∈ D(A1) is a consequence of the results in [2] and
of [6, Lemma 4.1.5 and Proposition 4.3.9]. We only need to prove (3.9). Now let us fix u0 ∈ D(A1) and consider that the corresponding weak solution u
is in C([0, T ]; D(A1)) ∩ C1([0, T ]; Xµ). In the differential equation of (3.7)
take the inner product in Xµ of each term by u(t), for any t ∈ (0, T ). The
result is 1 2 d dtku(t)k 2 Xµ+ k √ a ux(t)k2Xµ− (a(1)ux(t, 1))2 β1 − (a(0)ux(t, 0)) 2 β0 +γ0 β0 a(0)u2(t, 0) + γ1 β1 a(1)u2(t, 1) ≤ 1 2ku(t)k 2 Xµ+ 1 2kh(t)k 2 Xµ,
and the regularity of u(t) implies that also ux(·, 0) and ux(·, 1) are in L2(0, T ).
Hence we deduce that d dtku(t)k 2 Xµ ≤ d dtku(t)k 2 Xµ+ 2k √ a ux(t)k2Xµ ≤ 2 (a(1)ux(t, 1)) 2 β1 +(a(0)ux(t, 0)) 2 β0 + ku(t)k2Xµ+ kh(t)k2Xµ. (3.10)
By Gronwall’s Lemma, for any t ∈ [0, T ] one has
ku(t)k2X µ ≤ e T Z T 0 u2x(t, 1) dt ·2 a 2(1) β1 + Z T 0 u2x(t, 0) dt ·2 a 2(0) β0 +ku0k2Xµ+ khk 2 L2(Q T) i . (3.11)
Thus, there exists a positive constant eC such that
sup t∈[0,T ] ku(t)k2X µ ≤ eC h kux(·, 1)k2L2(0,T )+ kux(·, 0)k2L2(0,T ) +ku0k2Xµ+ khk 2 L2(Q T) i . (3.12)
Observe that, by integrating the second inequality of (3.10) over (0, T ) and using (3.12), we have Z T 0 k√a ux(t)k2Xµdt ≤ bC h kux(·, 1)k2L2(0,T )+ kux(·, 0)k2L2(0,T ) +ku0k2Xµ+ khk 2 L2(Q T) i , (3.13)
for a suitable positive constant bC. We can conclude that there exists a positive constant K such that
sup t∈[0,T ] ku(t)k2 Xµ+ Z T 0 ku(t)k2 b H1 a(0,1) dt ≤ Khkux(·, 0)k2L2(0,T ) + kux(·, 1)k2L2(0,T )+ ku0k2Xµ+ khk 2 L2(Q T) i , (3.14)
where K depends on T, a, β. Then the assertion follows.
4
Operators in divergence form with (GWBC):
the strongly degenerate case
Now let us assume that a is strongly degenerate and consider the operator (A1, bD(A1)), where A1 is defined as in the previous Section and bD(A1) is
obtained by replacing, in the definition of D(A1), the space bHa2(0, 1) by the
following
b
Ha s2 (0, 1) := {u ∈ bHa s1 (0, 1)| au0 ∈ H1(0, 1)}. Here
b
Ha s1 (0, 1) := {u ∈ Xµ| u locally absolutely continuous in [0, 1] \ {x0},
√
a u0 ∈ Xµ}.
In analogy with Proposition 2.2 one has the following
Proposition 4.1. Let a be strongly degenerate and define
b
X := {u ∈ Xµ| u locally absolutely continuous in [0, 1] \ {x0},
√ au0 ∈ Xµ, au ∈ H1(0, 1) and (au)(x0) = 0}, b Z := {u ∈ bX | au0 ∈ H1(0, 1), (au0)(x0) = 0}. Then b Ha s1 (0, 1) = bX and Hba s2 (0, 1) = bZ.
Remark 4.1. Let us observe that bHa s1 (0, 1) ⊂ Ha1(0, 1) and bHa s2 (0, 1) ⊂ H2
a(0, 1). It follows that for any (u, v) ∈ bHa s2 (0, 1) × bHa s1 (0, 1) the Green
formula in Lemma 2.1(i) holds. Moreover C∞[0, 1] ⊂ bHa s2 (0, 1).
Theorem 4.1. If a is strongly degenerate, then the operator (A1, bD(A1)) is
nonpositive and selfadjoint on Xµ.
Proof. Let us introduce the space
W1,as(0, 1) = {u ∈ bHas2 (0, 1)| supp {u} compact, supp {u} ⊂ [0, 1] \ {x0},
A1u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1},
and observe that
W1,as(0, 1) ⊂ bD(A1) ⊂ Xµ.
Then, analogous arguments as in Theorem 3.1 imply that W1,as(0, 1) is
dense in Xµand, hence, bD(A1) is dense in Xµ. Moreover, as a consequence
of Lemma 2.3 (i) and (GWBC), by arguing as in Theorem 3.1, one can show that A1 is symmetric and nonpositive. In order to prove that I − A1
is surjective, observe that W1,as(0, 1) is a Hilbert space with respect to the
inner product (u, v)W 1,as = Z 1 0 (uv + au0v0)dx +a(0)u(0)v(0) β0 +a(1)u(1))v(1) β1 ,
for any u, v ∈ W1,as(0, 1). Notice that
W1,as(0, 1) ,→ Xµ,→ (W1,as(0, 1))∗,
where (W1,as(0, 1))∗ is the dual space of W1,as(0, 1)) with respect to Xµ. Let
f ∈ Xµ and define F : W1,as(0, 1) 7→ R such that
F (v) = Z 1 0 f v dx +a(0)f (0)v(0) β0 +a(1)f (1)v(1) β1 .
From W1,as(0, 1) ,→ Xµ it follows that F ∈ (W1,as(0, 1))∗. Hence, by the
Riesz’s Theorem, there exists a unique u ∈ W1,as(0, 1) such that for any
v ∈ W1,as(0, 1) we have (u, v)W1,as = F (v) = Z 1 0 f v dx + a(0)f (0)v(0) β0 +a(1)f (1))v(1) β1 .
The above equality means that
Z 1
0
au0v0dx = Z 1
0
(f − u)v dx + a(0)(f − u)(0)v(0) β0
+a(1)(f − u)(1))v(1) β1
for all v ∈ W1,as(0, 1).
Let us denote by Cc∞((0, 1) \ {x0}) the space of all C∞(0, 1) functions
that vanish in a neighborhood of x0, with compact support in (0, 1) \ {x0}.
From Cc∞((0, 1) \ {x0}) ⊂ W1,as(0, 1), it follows that (4.1) holds for all
v ∈ Cc∞((0, 1) \ {x0}). This implies Z 1 0 au0v0dx = Z 1 0 (f − u)v dx
for all v ∈ Cc∞((0, 1) \ {x0}). Thus the weak derivative (au0)0 in this context
exists and (au0)0∈ L2(0, 1). Hence, u ∈ bD(A
1) and u − A1u = f.
Thus the analogous of Theorem 3.2 holds for (3.7) in the case a strongly degenerate. In addition we have a characterization of bD(A1).
Proposition 4.2. Let
D := {u ∈ Xµ | u locally absolutely continuous in [0, 1] \ {x0},
au ∈ H1(0, 1), au0∈ H1(0, 1), (au)(x0) = (au0)(x0) = 0
and A1u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1}.
Then
b
D(A1) = D.
Proof. D ⊆ bD(A1) :
Let u ∈ D. It is sufficient to prove that √au0 ∈ Xµ. Since au0 ∈ H1(0, 1)
and u is locally absolutely continuous in [0, 1] \ {x0}, the terms [au0u](1) and
[au0u](x), for x ∈ (x0, 1], are indeed well defined, so we have
Z 1
x
[(au0)0u](s)ds = [au0u]1x− Z 1
x
(a(u0)2)(s)ds.
Thus
(au0u)(x) = (au0u)(1) − Z 1 x [(au0)0u](s)ds − Z 1 x (a(u0)2)(s)ds.
Since u ∈ D, then (au0)0u ∈ L1(0, 1). Hence there exists
lim
x→x+0
since no integrability is known about a(u0)2 and such a limit could be −∞. If L 6= 0, there exists C > 0 such that
|(au0u)(x)| ≥ C
for all x in a right neighborhood of x0, x 6= x0. Thus by (2.3) there exists
C1 > 0 such that |u(x)| ≥ C |(au0)(x)| ≥ C1 √ x − x0 ,
for all x in a right neighborhood of x0, x 6= x0. This implies that u 6∈ L2(0, 1)
and thus u 6∈ Xµ. Hence L = 0 and
Z 1
x0
[(au0)0u](s)ds = [au0u](1) − Z 1
x0
(a(u0)2)(s)ds. (4.2)
If x ∈ [0, x0), proceeding as before, it follows that
Z x0
0
[(au0)0u](s)ds = −[au0u](0) − Z x0
0
(a(u0)2)(s)ds. (4.3)
By (4.2) and (4.3), it follows that Z 1
0
[(au0)0u](s)ds = [au0u]x=1x=0− Z 1
0
(a(u0)2)(s)ds.
Since (au0)0u ∈ L1(0, 1), then√au0 ∈ L2(0, 1). Thus √au0 ∈ X
µ and hence,
D ⊆ bD(A1).
b
D(A1) ⊆ D :
Let u ∈ bD(A1). By Remark 4.1 and Proposition 2.2 we know that au ∈
H1(0, 1) and (au)(x0) = 0. Thus it is sufficient to prove that (au0)(x0) = 0.
Toward this end, as in [9], observe that, since au0 ∈ H1(0, 1), there exists
L ∈ R such that lim
x→x0
(au0)(x) = (au0)(x0) = L. If L 6= 0, there exists C > 0
such that
|(au0)(x)| ≥ C, for all x in a neighborhood of x0, x 6= x0. Thus
|(a(u0)2)(x)| ≥ C
2
a(x),
for all x in a neighborhood of x0, x 6= x0. This implies that
√
au0 6∈ L2(0, 1)
and thus √au0 6∈ Xµ. Hence L = 0 and so (au0)(x0) = 0.
We point out the fact that the condition 1 a 6∈ L
1(0, 1) is crucial to prove
5
Operators in non divergence form with (GWBC)
Let us fix βj, γj ∈ R such that βj > 0, γj ≥ 0, j = 0, 1. Consider a weakly
or strongly degenerate function a and define the operator in nondivergence form A2u = au00 equipped with the general Wentzell boundary conditions
(GWBC) A2u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1 (5.1)
and the space
Yµ:= L21 a
([0, 1], dµ),
where the space L21 a
(0, 1) has been defined in Section 2, and
dµ := dx
a |(0,1)⊗ dS
β |{0,1}.
As usual, dx denotes the Lebesgue measure on (0, 1), β = (β0, β1), and dSβ
denotes the natural (Dirac) measure dS on {0, 1} with weight 1β. Thus Yµ
is a Hilbert space with the inner product given by
(f, g)Yµ = Z 1 0 f (x)g(x) a dx + f (1)g(1) β1 +f (0)g(0) β0 ,
where f, g ∈ Yµ, are written as (f χ(0,1), (f (0), f (1))), (g χ(0,1), (g(0), g(1))).
Hence Yµ is equipped with the norm defined by
kf k2Y µ = Z 1 0 |f (x)|2 a dx + |f (1)|2 β1 +|f (0)| 2 β0 ,
for any f ∈ Yµ, which is written as (f χ(0.1), (f (0), f (1))).
Let us now introduce the following spaces
e H11 a (0, 1) := Yµ∩ H1(0, 1), e H21 a (0, 1) :=nu ∈ eH11 a (0, 1) u0 ∈ H1(0, 1) o ,
endowed, respectively, with the associated norms
kuk2 e H1 1 a (0,1) := kuk 2 Yµ+ ku 0k2 L2(0,1), for all u ∈ eH11 a (0, 1), and kuk2 e H2 1 a (0,1) := kuk 2 e H1 1 a (0,1)+ kau 00k2 Yµ, for all u ∈ eH 2 1 a (0, 1).
Remark 5.1. Let us observe that eH11 a (0, 1) ⊂ H11 a (0, 1) and eH21 a (0, 1) ⊂ H21 a
(0, 1). It follows that for any (u, v) ∈ eH21 a
(0, 1) × eH11 a
(0, 1) the Green formula in Lemma 2.1 (ii) holds.
Let us define the domain of A2 as follows
D(A2) := {u ∈ eH21 a
(0, 1)| A2u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1}.
As a consequence of the results in Section 2, one has the next result.
Theorem 5.1. If a is weakly degenerate, then the operator (A2, D(A2)) is
selfadjoint and nonpositive on Yµ.
Proof. Since Cc∞(0, 1) ⊂ D(A2), the domain D(A2) is dense in Yµ. In order
to show that A2 is nonpositive and selfadjoint it suffices to prove that A2
is symmetric, nonpositive and (I − A2)(D(A2)) = Yµ (see e.g. [1, Theorem
B.14] or [15]). A2 is symmetric.
By Lemma 2.1(ii), for any u, v ∈ D(A2), one has
hA2u, viYµ = Z 1 0 au00v a dx + a(0)u00(0)v(0) β0 + a(1)u 00(1)v(1) β1 = [u0v]x=1x=0− Z 1 0 u0v0dx + (β0u0(0) − γ0u(0)) v(0) β0 − (β1u0(1) + γ1u(1)) v(1) β1 = − Z 1 0 u0v0dx − γ0 β0 u(0)v(0) − γ1 β1 u(1)v(1) = hu, A2viYµ. A2 is nonpositive.
For any u ∈ D(A2), according to the previous calculations, one has
hA2u, uiYµ = − Z 1 0 |u0|2dx − γ0 β0 |u(0)|2− γ1 β1 |u(1)|2 ≤ 0. I − A2 is surjective.
Further, if a is strongly degenerate analogous results as in Proposition 2.3 hold, provided that one replaces the spaces H11
a (0, 1) and H21 a (0, 1) by the spaces eH11 a (0, 1) and eH21 a
(0, 1), respectively. Thus one can deduce the following.
Theorem 5.2. If a is strongly degenerate and Hypothesis 2.1 is satisfied, then the operator (A2, D(A2)) is selfadjoint and nonpositive on Yµ.
Proof. First, by using similar arguments as in [9, Proposition 3.8] one can show that D(A2) := {u ∈ eH11 a (0, 1) | au00∈ L21 a (0, 1), au0 ∈ H1(0, 1), u(x0) = (au0)(x0) = 0,
and A2u(j) + (−1)j+1βju0(j) + γju(j) = 0, j = 0, 1}.
Now let us introduce the subspace
W1,1
a(0, 1) = {v ∈ H
2(0, 1)∩L2
1 a
(0, 1) | supp {v} compact, supp {v} ⊂ [0, 1]\{x0},
A2v(j) + (−1)j+1βjv0(j) + γjv(j) = 0, j = 0, 1},
endowed with the norm kvk2 W1, 1 a(0,1) := kvk2 e H2 1 a (0,1). It is evident that W1,1 a(0, 1) ⊂ D(A2) ⊂ Yµ. Moreover, W1,1
a(0, 1) is dense in Yµ by using the following argument: take
v ∈ Yµ and define vn := ξnv, n ≥ max {x40,1−x4 0}, where ξn is defined in
(3.3). It is clear that vn→ v in Yµ as n → ∞. Hence D(A2) is dense in Yµ.
Similar arguments as in the proof of Theorem 5.1 show that A2 is a
symmetric and nonpositive operator. Let us show that I − A2 is surjective,
i.e. (I − A2)D(A2) = Yµ. First of all, observe that W1,1
a(0, 1) is equipped
with the inner product
(u, v)1 = Z 1 0 uv a + u 0 v0 dx + u(1)v(1) β1 +u(0)v(0) β0 , (5.2) for any u, v ∈ W1,1
a(0, 1). Moreover, it follows that
W1,1
a(0, 1) ,→ Yµ,→ (W1, 1 a(0, 1))
where (W1,1 a(0, 1))
∗ is the dual space with respect to Y
µ. Let f ∈ Yµ and define F : W1,1 a(0, 1) 7→ R as follows F (v) = Z 1 0 f v a dx + f (1)v(1) β1 +f (0)v(0) β0 . Since W1,1 a(0, 1) ,→ Yµ, then F ∈ (W1, 1 a(0, 1)) ∗. As a consequence, by
Riesz’s Theorem, there exists a unique u ∈ W1,1
a(0, 1) such that for any
v ∈ W1,1 a(0, 1) (u, v)1 = Z 1 0 f v a dx + f (1)v(1) β1 +f (0)v(0) β0 .
From (5.2), the above equality is equivalent to write Z 1 0 u0v0dx = Z 1 0 (f − u)v a dx + (f − u)(1)v(1) β1 +(f − u)(0)v(0) β0 , (5.3) for all v ∈ W1,1 a(0, 1).
As in Section 4, let us denote by Cc∞((0, 1) \ {x0}) the space of C∞(0, 1)
functions that vanish in a neighborhood of x0, with compact support in
(0, 1) \ {x0}. Since Cc∞((0, 1) \ {x0}) ⊂ W1,1a(0, 1), (5.3) holds for all v ∈
Cc∞((0, 1) \ {x0}), i.e. Z 1 0 u0v0dx = Z 1 0 (f − u)v a dx, v ∈ C ∞ c ((0, 1) \ {x0}) . (5.4)
Thus the weak derivative u00 in this context exists and au00 ∈ L2
1 a
(0, 1). Hence, u ∈ D(A2), and, according to (5.4) and Lemma 2.3 (ii), we have
that
u − A2u = f.
As a consequence of Theorems 5.1 and 5.2, one has that A2 is the
in-finitesimal generator of a strongly continuous semigroup on Yµ. Hence, the
problem ut− A2u = h(t, x), (t, x) ∈ QT, (GWBC), u(0, x) = u0(x), x ∈ (0, 1), (5.5)
is well-posed in the sense of evolution operator theory. In particular, the following theorem holds.
Theorem 5.3. Assume that a is weakly degenerate (resp. strongly degen-erate and the Hypothesis 2.1 is satisfied). Then for all h ∈ L2(0, T ; Yµ) and
u0 ∈ Yµ, there exists a unique weak solution u ∈ C([0, T ]; Yµ)∩L2(0, T ; eH11 a (0, 1)) of (5.5) such that sup t∈[0,T ] ku(t)k2 Yµ ≤ C h ku0k2 Yµ+ khk 2 L2(Q T) i , (5.6)
where the constant C in (5.6) depends on T, a, β, but is independent of uo
and h. Moreover, if u0 ∈ D(A2), then u ∈ C([0, T ]; D(A2)) ∩ C1([0, T ]; Yµ).
Appendix
In the following we will give the proofs of Lemmas 2.1 and 2.3.
Proof of Lemma 2.1 (i). The proof is analogous to that of [9, Lemma 2.1], but we present it for the readers’ convenience. Let (u, v) ∈ Ha2(0, 1) × H1
a(0, 1). For any sufficiently small δ > 0 one has
Z 1 0 (au0)0vdx = Z x0−δ 0 (au0)0vdx + Z x0+δ x0−δ (au0)0vdx + Z 1 x0+δ (au0)0vdx = (au0v)(x0− δ) − (au0v)(0) − Z x0−δ 0 au0v0dx + Z x0+δ x0−δ (au0)0vdx + (au0v)(1) − (au0v)(x0+ δ) − Z 1 x0+δ au0v0dx = [au0v]x=1x=0+ (au0v)(x0− δ) − Z x0−δ 0 au0v0dx + Z x0+δ x0−δ (au0)0vdx − (au0v)(x0+ δ) − Z 1 x0+δ au0v0dx, (A-1) since au0 ∈ H1(0, 1). Now we prove that
lim δ→0 Z x0−δ 0 au0v0dx = Z x0 0 au0v0dx, lim δ→0 Z 1 x0+δ au0v0dx = Z 1 x0 au0v0dx and lim δ→0 Z x0+δ x0−δ (au0)0vx = 0. (A-2)
Toward this end, observe that Z x0−δ 0 au0v0dx = Z x0 0 au0v0dx − Z x0 x0−δ au0v0dx (A-3) and Z 1 x0+δ au0v0dx = Z 1 x0 au0v0dx − Z x0+δ x0 au0v0dx. (A-4)
Moreover, (au0)0v and au0v0 ∈ 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 au0v0dx ≤ Z x0 x0−δ |au0v0|dx < , Z x0+δ x0−δ (au0)0vdx ≤ Z x0+δ x0−δ |(au0)0v|dx < , Z x0+δ x0 au0v0dx ≤ Z x0+δ x0 |au0v0|dx < .
Now take such a δ in (A-1). Thus being arbitrary,
lim δ→0 Z x0 x0−δ au0v0dx = lim δ→0 Z x0+δ x0−δ (au0)0vdx = lim δ→0 Z x0+δ x0 au0v0dx = 0.
The previous equalities and (A-3), (A-4) imply
lim δ→0 Z x0−δ 0 au0v0dx = Z x0 0 au0v0dx lim δ→0 Z 1 x0+δ au0v0dx = Z 1 x0 au0v0dx. (A-5)
In order to obtain the desired result it is sufficient to prove that lim δ→0(au 0 v)(x0− δ) = lim δ→0(au 0 v)(x0+ δ). Since au0 ∈ H1(0, 1) and v ∈ H1 a(0, 1), lim δ→0(au 0v)(x 0− δ) = (au0v)(x0) = lim δ→0(au 0v)(x 0+ δ). (A-6)
Thus by (A-1), (A-2), (A-5) and (A-6), it follows that Z 1 0 (au0)0vdx = [au0v]x=1x=0− Z 1 0 au0v0dx.
Proof of Lemma 2.1 (ii). It is trivial, since (u, v) ∈ H21 a
(0, 1)×H11 a
(0, 1).
Proof of Lemma 2.3 (i). Let (u, v) ∈ Ha2(0, 1) × Ha1(0, 1). As for the weak case, one can prove that, for any δ > 0,
Z 1
0
(au0)0vdx = [au0v]x=1x=0+ (au0v)(x0− δ) −
Z x0−δ 0 au0v0dx + Z x0+δ x0−δ (au0)0vdx − (au0v)(x0+ δ) − Z 1 x0+δ au0v0dx. (A-7) Moreover lim δ→0 Z x0−δ 0 au0v0dx = Z x0 0 au0v0dx, lim δ→0 Z 1 x0+δ au0v0dx = Z 1 x0 au0v0dx (A-8) and lim δ→0 Z x0+δ x0−δ (au0)0vdx = 0. (A-9)
In order to obtain the desired result it is sufficient to prove that lim δ→0(au 0 v)(x0− δ) = lim δ→0(au 0 v)(x0+ δ). (A-10)
First of all, observe that, since au0 ∈ H1(0, 1) and v is locally absolutely
continuous on [0, x0) ∪ (x0, 1], the terms (au0v)(0) and (au0v)(1) are indeed
well defined. Moreover,
(au0v)(x0− δ) = (au0v)(0) + Z x0−δ 0 ((au0)0v)(s)ds + Z x0−δ 0 (au0v0)(s)ds and (au0v)(x0+ δ) = (au0v)(1) − Z 1 x0+δ ((au0)0v)(s)ds − Z 1 x0+δ (au0v0)(s)ds.
Since (au0)0, v ∈ L2(0, 1) and√au0, √av0 ∈ L2(0, 1), by H¨older’s inequality,
(au0)0v ∈ L1(0, 1) and au0v0 ∈ L1(0, 1). Thus there exist L
1, L2 ∈ R such that lim δ→0(au 0v)(x 0− δ) = (au0v)(0) + lim δ→0 Z x0−δ 0 ((au0)0v)(s) ds + lim δ→0 Z x0−δ 0 (au0v0)(s) ds = (au0v)(0) + Z x0 0 ((au0)0v)(s)ds + Z x0 0 (au0v0)(s) ds = L1
and lim δ→0(au 0v)(x 0+ δ) = (au0v)(1) − lim δ→0 Z 1 x0+δ ((au0)0v)(s) ds − lim δ→0 Z 1 x0+δ (au0v0)(s) ds = (au0v)(1) − Z 1 x0 ((au0)0v)(s)ds − Z 1 x0 (au0v0)(s)ds = L2.
If L1 6= 0, then there exists C > 0 such that
|(au0v)(x)| ≥ C
for all x in a left neighborhood of x0, x 6= x0. Thus by (2.3),
|v(x)| ≥ C |(au0)(x)| ≥
C1
√ x0− x
for all x in a left neighborhood of x0, x 6= x0, and for a suitable positive
constant C1. This implies that v 6∈ L2(0, 1). Hence L1 = 0. Analogously,
one can prove that L2 = 0. Thus (A-10) holds. In particular,
lim δ→0(au 0v)(x 0− δ) = lim δ→0(au 0v)(x 0+ δ) = 0
and the desired result follows.
Proof of Lemma 2.3 (ii). It is trivial, since (u, v) ∈ H21 a
(0, 1)×H11 a
(0, 1).
Acknowledgements Genni Fragnelli, Rosa Maria Mininni and Silvia Ro-manelli are members of the Gruppo Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM).
The research of Genni Fragnelli is partially supported by the research project “Sistemi con operatori irregolari” of the GNAMPA-INdAM.
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