J.evol.equ. 1 (2001) 243 – 276 1424–3199/01/030243 – 34 $ 1.50 + 0.20/0 © Birkh¨auser Verlag, Basel, 2001
On a class of degenerate elliptic operators arising from Fleming-Viot processes Sandra Cerrai and Philippe ClÉment Dedicated to the memory of Ralph Phillips
Abstract. We are dealing with the solvability of an elliptic problem related to a class of degenerate second order operators which arise from the theory of Fleming-Viot processes in population genetics. In the one dimensional case the problem is solved in the space of continuous functions. In higher dimension we study the problem in L2 spaces with respect to an explicit measure which, under suitable assumptions, can be taken invariant and symmetrizing for the operators. We prove the existence and uniqueness of weak solutions and we show that the closure of the operator in such L2 spaces generates an analytic C0 -semigroup.
1. Introduction The aim of this paper is to study the solvability of the degenerate elliptic problem ω(x)|x, Dϕ(x)i = f (x), λϕ(x) − γ (x) Tr [C(x)D 2 ϕ(x)] − hω(x) − |e
x ∈ S, (1.1)
where h·, ·i denotes the usual inner product in R , λ is any positive constant, S is the simplex of Rd consisting of all x ∈ [0, 1]d such that x1 + · · · + xd ≤ 1, C(x) is the d × d matrix of components d
ci,j (x) = xi (δij − xj ),
i, j = 1, . . . , d,
which degenerates on the boundary of S, ω(x) is the vector of Rd of components ω(x) is the vector of Rd+1 of components ω0 (x), ω1 (x), . . . , ωd (x). ω1 (x), . . . , ωd (x) and e P Notice that here and in what follows we denote |e ω(x)| = di=0 ωi (x). The functions γ and ω0 , ω1 , . . . , ωd are supposed to be continuous and strictly positive on S. The operator ω(x)|x, Dϕ(x)i, Lϕ(x) = γ (x) Tr [C(x)D 2 ϕ(x)] + hω(x) − |e
x ∈ S,
(1.2)
arises in the theory of Fleming-Viot processes as the generator of a Markov C0 -semigroup defined on C(S), the space of continuous functions on S. Fleming-Viot processes are Received December 4, 2000; accepted December 9, 2000. 2000 Mathematics Subject Classification: 60J35, 60K35, 92D15. Key words: Fleming-Viot process, degenerate elliptic problems, generation of C0 -semigroups.
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measure-valued processes that describe the genetic evolution of a population and are defined as the limit in distribution of empirical processes associated with suitable sequences of Markov chains. The operator (1.2) is the generator corresponding to some diffusion model in population dynamics in which each individual is of some type and the type space E is given by a finite number d + 1 of elements. In this case the state space is ( ) d X xi = 1 , 1E = x ∈ [0, 1] d+1 ; i=0
where xi denotes the proportion of the population that is of type i. Moreover, the first order term corresponds to a mutation operator A which depends on x ∈ 1E and which is given by Z Ax f (j ) = (f (j ) − f (i)) p(x, di), j ∈ E, x ∈ 1E , E
where f : E → R, and p (x, {i}) = ωi (x), for any i ∈ E. There are models in which the number of different types is infinite, like in the Ohta and Kimura’s model, where E = Z. By using a different approach it is also possible to construct models in which E is an arbitrary complete separable metric space, like for example R or [0, 1]Z+ . In this case Fleming and Viot in [6] had proposed to topologize E and replace 1E by P(E), the space of probability measure on E, with the topology of the weak convergence. For a complete description of such models we refer to the papers by Ethier and Kurtz [5] and Dawson and March [2]. It is known that if the domain of the operator L defined by (1.2) is C 2 (S), then it is dissipative in C(S) and hence, as densely defined, is closable in C(S) with dissipative closure. By the Lumer-Phillips theorem its closure L¯ is the generator of C0 -semigroup of contractions if and only if the range of λ I − L is dense in C(S), for some λ > 0. This is the reason we are led to study the solvability of the problem (1.1). When the coefficients γ and ωi are constant, it is easy to show that the image of I − L contains a dense subset of C(S) and the generation of a semigroup follows (see [12] for the case of a finite type space and [2], [5] for the case of general type spaces E). When the coefficients are not constant the situation is more delicate. However, in [3bis], by using an argument based on the Trotter product formula, it is proved that L¯ generates a non-negative C0 -semigroup of contractions on C(S). In Section 3 we consider the problem (1.1) when the population is divided into only two different types, that is when d = 1 and S = [0, 1]. In this case by assuming that the functions γ and ωi are continuous on [0, 1] and satisfy some H¨older condition at the boundary points, we show that C 2 [0, 1] is a core for the operator L with domain (1.3) D = ϕ ∈ C 1 [0, 1] ∩ C 2 (0, 1) : lim x(1 − x)ϕ 00 (x) = 0 . t→0+ ,1−
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Thus, by applying a recent result by Metafune [8] based on some earlier results of Angenent [1], we can extend the result proved in [3bis] and we can conclude that L¯ is the generator of a C0 -semigroup of contractions on C[0, 1], which is positive and analytic and its domain coincides with the set D in (1.3). In the section 4 we study the problem (1.1) in a L2 context. Introduce a measure ν on S which is absolutely continuous with respect to the Lebesgue measure and for which we give an explicit formula of the density. It is important to notice that when γ and ωi fulfill suitable assumptions, the measure ν can be chosen invariant and even symmetrizing and when γ and ωi are constant, it coincides with a Dirichlet distribution D(q) on S, of some parameters qi related to the coefficients (see e.g. [9] and [12]). Notice that in general the measure ν is neither invariant nor symmetrizing. Nevertheless, we are able to prove an integration by parts formula which allows to show that the operator L is closable in L2 (S, ν) and its closure L¯ generates a C0 -semigroup. Such a formula also provides a characterization of the domain of L¯ in L2 (S, ν) in terms of appropriate Sobolev spaces, and a factorization of the symmetric part of L¯ in terms of its square root (which coincides with L¯ when there exists a symmetrizing invariant measure). Moreover, due to the Lax-Milgram theorem we can define an operator A in L2 (S, ν) which is the generator of an analytic semigroup in L2 (S, ν) with dense domain, which ¯ Therefore, as L¯ generates a C0 -semigroup, A coincides with L¯ and extends the operator L. we have generation of analytic semigroup. 2. Notations and preliminary results We are here concerned with the operator ω(x)|x, Dϕ(x)i, Lϕ(x) = γ (x) Tr [C(x)D 2 ϕ(x)] + hω(x) − |e where C(x) is the matrix of components cij (x) defined by cij (x) = xi (δij − xj ),
1 ≤ i, j ≤ d
and S is the simplex of Rd defined by ) ( d X xi ≤ 1 . S = x ∈ [0, 1] d ; i=1
We recall that in what follows we shall denote |e ω(x)| =
d X
ωi (x).
i=0 ◦
Moreover, we shall denote by S the interior of S.
x ∈ S,
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HYPOTHESIS 2.1. The functions γ and ωi , i = 0, 1, . . . , d, are continuous and strictly positive on S. Due to the previous condition, the operator L can be written as MA, with Mϕ(x) = γ (x)ϕ(x), Aϕ(x) = Tr [C(x)D 2 ϕ(x)] +
1 hω(x) − |e ω(x)|x, Dϕ(x)i. γ (x)
(2.1)
We have the following general result. LEMMA 2.2. Let C(X) be the Banach space of continuous functions on a compact set X, equipped with the supremum norm k · k0 and let A be a m-dissipative operator in C(X). Moreover, let B be the operator defined by D(B) = D(A),
B = MA,
where Mu(x) = m(x)u(x),
x ∈ X,
for some continuous and strictly positive function m defined in X. Then, if B is dissipative B is m-dissipative. Proof. Let f be any function in C(X). We have to prove that there exists u ∈ D(A) such that u − MAu = f . Set λ0 = min
x∈ X
1 , m(x)
p(x) = λ0 −
1 ≤0 m(x)
and set Cu = pu, for any u ∈ C(X). Then C is m-dissipative and bounded, so that also A + C is m-dissipative (for a proof of this fact see for example [10]-Chapter 3, Corollary 3.3). Hence there exists one and only one u ∈ D(A) such that λ0 u − (Au + pu) = f/m, and this is equivalent to u = MAu + f. ¨ Whence due to the previous lemma, once one proves that L¯ is dissipative, it follows that ¯ This is the reason why in the m-dissipativity of L¯ is equivalent to m-dissipativity of A. what follows, without any loss of generality, we can write the operator L in the form σ (x)|x, Dϕ(x)i, Lϕ(x) = Tr [C(x)D 2 ϕ(x)] + hσ (x) − |e
x ∈ S,
(2.2)
where σi (x) = ωi (x)/γ (x), for any x ∈ S and i = 0, 1, . . . , d. Notice that all σi are continuous and strictly positive on S for each i, so that the Hypothesis 2.1 is fulfilled by σi as well. Thus, next step is showing that L¯ is dissipative in C(S). To this purpose, we first recall some well known facts concerning the matrix C(x), x ∈ S.
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LEMMA 2.3. For any h ∈ Rd , h 6= 0, it holds hC(x)h, hi ≥ 0,
◦
if x ∈ S and hC(x)h, hi > 0,
if x ∈ S .
(2.3)
Proof. We have d X
hC(x)h, hi =
xi (δij − xj )hi hj =
i,j =1 d X
=
xi h2i
−
i=1
d X
!2 xi hi
d X
xi h2i −
i=1
d X
xi xj hi hj
i,j =1
.
i=1
If we take x in S, it holds d X
!2 ≤
xi hi
i=1
d X
d X
xi h2i
i=1
xi ≤
i=1
d X
xi h2i ,
i=1
and then hC(x)h, hi ≥
d X
xi h2i −
i=1
d X
xi h2i = 0.
i=1
◦
Furthermore, if x ∈ S , then x1 + · · · + xd < 1, so that the strict inequality holds in (2.3). ¨ LEMMA 2.4. For any x ∈ S it holds det C(x) =
d Y
xi ,
i=0
where x0 = 1 −
Pd
m=1 xm .
Thus det C(x) = 0 if and only if x ∈ ∂S.
Proof. For any x ∈ S we have
· · · −x1 xd x1 (1 − x1 ) −x1 x2 −x2 x1 x2 (1 − x2 ) · · · −x2 xd C(x) = .. .. .. .. . . . . −xd x2 · · · xd (1 − xd ) −xd x1
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By easy calculations 1 − x1 −1 −1 · · · 1 − x1 −x · · · −x 1 1 −x2 1 0 ··· d d −x2 1 − x2 · · · Y Y −x2 −x3 0 1 ··· xi . xi det C(x) = .. .. .. = .. . . .. .. . . . .. i=1 i=1 .. . . −x −x · · · 1 − x d d d −x 0 0 ··· d
1
−1 0 0 .. .
and developing along the first row we get det C(x) = =
d Y
xi 1 − x1 +
d X
i=1
i=2
d Y
d X
xi 1 − x1 −
i=1
! −(−1)
i+1
! =
xi
i=2
i
(−1) (−xi )
d Y
xi .
i=0
¨ LEMMA 2.5. If we denote 3(x) = diag {x1 , . . . , xd } and x ⊗ x = xx t , we have C(x) = 3(x) − x ⊗ x,
x ∈ S.
(2.4) ◦
Moreover, C(x) is invertible for any x ∈ S and it holds C −1 (x) = 3−1 (x) +
e⊗e , x0
◦
x ∈ S,
(2.5)
where e is the vector of Rd having all components equal 1. Proof. For any h, k ∈ Rd and x ∈ S we have hh, C(x)ki =
d X i=1
=
d X
hi
d X
xi (δij − xj )kj
j =1
xi hi ki −
i=1
d X
xi hi xj kj = hh, 3(x)ki − hh, (x ⊗ x)ki,
i, j =1 ◦
so that (2.4) follows. Concerning C −1 (x), let us solve for any x ∈ S and v ∈ Rd the equation 3(x)u − hx, ui x = v. If we set λ = hx, ui, we have xi ui − λxi = vi , for any i = 1, . . . , d and then ui =
vi + λxi vi = + λ. xi xi
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This implies that λ=
d X
xi ui =
i=1
d X
vi + λ
i=1
d X i=1
so that λ = hv, ei/x0 . Therefore C −1 (x)v = u = 3−1 (x) v +
xi =
d X
vi + λ(1 − x0 ),
i=1
hv, ei e e⊗e = 3−1 (x) + v. x0 x0 ¨
Next proposition is the main result in order to have the dissipativity of the operator L. PROPOSITION 2.6. Let ϕ ∈ C 2 (S) and let x¯ ∈ S be a point where ϕ achieves its minimum. Then hω(x) ¯ − |e ω(x)| ¯ x, ¯ Dϕ(x)i ¯ ≥ 0.
¯ ≥ 0, Tr [C(x)D ¯ 2 ϕ(x)]
Proof. It will be convenient to rewrite the operator L in y-coordinates y0 (x) = 1 −
d X
xm ,
yi (x) = xi , 1 ≤ i ≤ d.
m=1
In this case, for any x ∈ S we have that y(x) belongs to the set d X yj = 1 . 1 = y ∈ [0, 1] d+1 ; j =0
If ψ ∈ C 2 (1) and y = y(x), for some x ∈ S, we define ϕ(x1 , . . . , xd ) = ψ(y0 (x), y1 (x), . . . , yd (x)). Then d ∂ ∂ ∂ϕ(x) X ∂ψ(y) ∂yj = =− − ψ(y) ∂xi ∂yj ∂xi ∂y0 ∂yi j =0
and ∂ 2 ϕ(x) = ∂xi ∂xj =
∂ ∂ − ∂y0 ∂yi
∂ ∂ − ∂y0 ∂yj
ψ(y)
∂ 2 ψ(y) ∂ 2 ψ(y) ∂ 2 ψ(y) ∂ 2 ψ(y) + − − . ∂yi ∂yj ∂y0 ∂yj ∂y0 ∂yi ∂y02
(2.6)
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Therefore we get d X
Tr [C(x)D 2 ϕ(x)] =
yi (δij − yj )
i,j =1
−
d X ∂ 2 ψ(y) ∂ 2 ψ(y) − yi (δij − yj ) ∂yi ∂yj ∂yi ∂y0 i,j =1
d X
yi (δij − yj )
i,j =1 d X
=
i,j =1
yi (δij − yj )
i,j =1
−
d X
d X ∂ 2 ψ(y) ∂ 2 ψ(y) + yi (δij − yj ) ∂yj ∂y0 ∂y02
yj y0
j =1
∂ 2 ψ(y) ∂yi ∂yj
−
d X
yi y0
i=1
∂ 2 ψ(y) ∂yi ∂y0
∂ 2 ψ(y) ∂ 2 ψ(y) + y0 (1 − y0 ) , ∂yj ∂y0 ∂y02
so that Tr [C(x)D 2 ϕ(x)] =
d X
yi (x)(δij − yj (x))
i,j =0
∂ 2 ψ(y(x)) . ∂yi ∂yj
(2.7)
Similarly, it is possible to show that d X
∂ψ(y(x)) ∂ϕ(x) X = ω(x)|yi (x)) . (ωi (x) − |e ∂xi ∂yi d
ω(x)|xi ) (ωi (x) − |e
i=1
i=0
Now, if x¯ is a point in S where ϕ has its minimum, let us denote by y¯ the corresponding point in 1 where ψ achieves its minimum. We first suppose that y¯ is an extremal point, that is all coordinates y¯i ’s are zero but one y¯i0 = 1. It is easy to check that in this case the matrix of components y¯i (δij − y¯j ), 0 ≤ i, j ≤ d, is identically zero, indeed the coefficients can be non zero only if i = j = i0 and in this case we have y¯i0 (1 − y¯i0 ) = 0. Thus, due to (2.7) we have ¯ = 0. Tr [C(x)D ¯ 2 ϕ(x)] Now, we consider the case when y¯i < 1 for each i = 0, 1, . . . , d. If we denote I (y) ¯ = { i ∈ {0, 1, . . . , d} ; y¯i ∈ (0, 1)}, we have that y¯ is an interior point of the set ¯ { y ∈ 1 ; yi = 0, i ∈ I c (y)}. ¯ i,j ∈ I (y) It follows that the submatrix [Dij2 ψ(y)] ¯ is nonnegative definite and then d X i,j =0
y¯i (δij − y¯j )
∂ 2 ψ(y) ¯ = ∂yi ∂yj
X i,j ∈ I (y) ¯
y¯i (δij − y¯j )
∂ 2 ψ(y) ¯ ≥ 0. ∂yi ∂yj
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Due to (2.7) this implies the first inequality in (2.6). Concerning the second inequality, we observe that for any θ ∈ S it holds hθ − x, ¯ Dϕ(x)i ¯ ≥ 0. Actually, since S is convex, the function [0, 1] → S,
t 7 → ϕ((1 − t)x¯ + tθ )
is in C 1 (S) and attains its minimum at t = 0. This implies that hθ − x, ¯ Dϕ(x)i ¯ =
d+ ϕ((1 − t)x¯ + tθ )|t=0 ≥ 0. dt
(2.8)
Next, we remark that ω(x) ¯ hω(x) − x, ¯ Dϕ(x) ¯ . ¯ − |e ω(x)| ¯ x, ¯ Dϕ(x)i ¯ = |e ω(x)| ¯ |e ω(x)| ¯
Thus, since |e ω(x)| ¯ > 0,
1 ω(x) ¯ ∈ S, |e ω(x)| ¯ ¨
from (2.8) we conclude that the second inequality in (2.6) holds.
As an immediate consequence of the previous proposition we have the following maximum principle for the operator L. COROLLARY 2.7. Let m1 and m2 be two nonnegative functions in C(S) and let λ > 0 and ϕ ∈ C 2 (S) be such that ω(x)|x, Dϕ(x)i ≥ 0, λϕ(x) − m1 (x)Tr [C(x)D 2 ϕ(x)] − m2 (x) hω(x) − |e
x ∈ S.
Then ϕ(x) ≥ 0, for any x ∈ S. In particular, the operator L : D(L) = C 2 (S) → C(S) defined by ω(x)|x, Dϕ(x)i, Lϕ(x) = γ (x) Tr [C(x)D 2 ϕ(x)] + hω(x) − |e
x ∈ S,
is dissipative and closable in C(S), with dissipative closure. Moreover, if for any i = 0, 1, . . . , d σi (x) =
ωi (x) ≡ σ¯ i , γ (x)
x ∈ S,
L¯ generates a C0 -semigroup of positive contractions in C(S).
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Proof. The first part of the corollary immediately follows from the Proposition 2.6. As L 1 = 0, this implies the dissipativity of L in C(S) and, since D(L) is dense in C(S), L is closable with dissipative closure. Now, thanks to the Lemma 2.2, if we show that, under the assumption that all σi are constant, the closure of the operator A introduced in (2.1) is m-dissipative, then L¯ is mdissipative as well and, due to the Lumer-Phillips Theorem, L¯ is the generator of a C0 semigroup of contractions in C(S). Since A¯ is dissipative, we have only to show that ¯ = C(S), for some λ > 0, or, equivalently, Range(λI − A) is dense in Range(λI − A) C(S), for some λ > 0. As we are assuming all σi constant, the operator λI − A maps the subset Pk (S) of polynomials on S of degree less or equal to k into itself. Moreover, the operator λI − A : Pk (S) → Pk (S) is injective and then, as dim(Pk (S)) < ∞, it is also surjective. This implies that [ Pk (S) = P(S), Range(λI − A) ⊃ k≥0
so that Range (λI − A) is dense in C(S). The positivity of the semigroup follows from the fact that λϕ(x) − Lϕ(x) ≥ 0, on S H⇒ ϕ(x) ≥ 0 on S. ¨ Notice that in [3bis] the same result is also proved when the coefficients ωi are not constant. But in this case the proof is more delicate, as the argument based on polynomials fails. 3. The one-dimensional case In the 1-d case, the operator L is Lϕ(x) = γ (x) x(1 − x) ϕ 00 (x) + (ω1 (x) − (ω0 (x) + ω1 (x)) x) ϕ 0 (x),
x ∈ [0, 1].
For simplicity we shall assume that for some θ ∈ (0, 1) ω1 (x) =
θ ω0 (x), 1−θ
x ∈ S,
so that the operator L: DL = C 2 (S) → C(S) can be rewritten as Lϕ(x) = γ (x) x(1 − x) ϕ 00 (x) +
ω0 (x) (θ − x)ϕ 0 (x), 1−θ
x ∈ [0, 1].
Besides to the Hypothesis 2.1, in what follows we shall assume that the functions γ and ω0 fulfill the following conditions.
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HYPOTHESIS 3.1. If we denote σ (x) = it holds Z 1 0
ω0 (x) (θ − x), γ (x)
x ∈ [0, 1],
|σ (x) − σ (0)| |σ (x) − σ (1)| + x 1−x
dx < ∞.
(3.1)
Then the following results holds. THEOREM 3.2. Assume the Hypotheses 2.1 and 3.1. Then L is closable and its closure L¯ is the infinitesimal generator of a C0 -semigroup of contractions on C[0, 1], which is positive and analytic of angle π/2. Moreover 1 2 00 ¯ D(L) = ϕ ∈ C [0, 1] ∩ C (0, 1) ; lim x(1 − x)ϕ (x) = 0 . x→0+ ,1−
Before proceeding with the proof of the theorem, we recall a result proved by Metafune in [8]. THEOREM 3.3. Let m be a continuous and strictly positive function on [0, 1] and let b be any continuous function which satisfies (3.1) and such that b(0) > 0 and b(1) < 0. Set 1 2 00 D(A) = ϕ ∈ C [0, 1] ∩ C (0, 1) ; lim x(1 − x)ϕ (x) = 0 x→0+ ,1−
Aϕ(x) = m(x)(x(1 − x)ϕ 00 (x) + b(x)ϕ 0 (x)),
x ∈ (0, 1).
Then A generates a C0 -semigroup of contractions on C [0, 1] which is positive and analytic. Proof of Theorem 3.2. We introduce the following operator e = ϕ ∈ C 1 [0, 1] ∩ C 2 (0, 1) ; lim x(1 − x)ϕ 00 (x) = 0 , D(L) x→0+ , 1−
e Lϕ(x) = Lϕ(x)
x ∈ (0, 1).
e : D(L) e ⊂ C [0, 1] → C [0, 1] is m-dissipative, Due to Metafune’s theorem, the operator L densely defined and generates a positive analytic semigroup of angle π/2. Therefore, it is e hence L¯ ⊆ L, e as the graph norms e Clearly C 2 [0, 1] ⊂ D(L), enough to show that L¯ = L. 2 e e of D(L) and D(L) coincide on C [0, 1] and L is closed. The opposite inclusion follows e once one proves that C 2 [0, 1] is a core for L.
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e Our aim is to show that there exists a sequence {ϕn } ⊂ C 2 [0, 1] such We fix ϕ ∈ D(L). that lim ϕn = ϕ and
n→+∞
e1 ϕn = L e1 ϕ, lim L
n→+∞
in C [0, 1],
(3.2)
where e1 ϕ(x) = γ (x) x(1 − x)ϕ 00 (x) + 1 (θ − x)ϕ 0 (x), L ω0 (x) 1−θ
x ∈ S.
As e1 ϕ(x), e Lϕ(x) = ω0 (x)L e this immediately implies that C 2 (S) is a core for D(L). 1 e1 ϕ in C [0, 1] and a We fix a sequence {fn } ⊂ C [0, 1] which converges to f := ϕ − L sequence {ηn } in C 2 [0, 1], which converges to γ /ω0 in C [0, 1], such that l = inf min ηn (x) > 0.
(3.3)
n∈N x∈ S
e1 , with γ /ω0 replaced by ηn . As the functions e1,n the operator defined as L We denote by L ηn can be chosen in such a way that 1/ηn fulfills the Hypothesis 3.1, for each n ∈ N there exists a unique solution ϕn to the problem e1,n ϕ + fn . ϕ=L
(3.4)
We claim that ϕn ∈ C 2 [0, 1] and (3.2) holds. STEP 1. We prove that ϕn ∈ C 2 [0, 1], for each n ∈ N. Since ηn and fn are in C 1 [0, 1], we can consider the problem 2−θ ψ = αn ψ 00 + (β + αn0 )ψ 0 + fn0 , 1−θ where αn (x) = ηn (x) x(1 − x),
β(x) =
1 (θ − x). 1−θ
Notice that formally we get such a problem by differentiating each side of (3.4) and by setting ψ = ϕ 0 . Moreover, it can be written as 2−θ ψ(x) = ηn (x) (x(1 − x)ψ 00 (x) + bn (x)ψ 0 (x)) + fn0 (x), 1−θ where 1 bn (x) = ηn (x)
1 0 (θ − x) + ηn (x) x(1 − x) + (1 − 2x). 1−θ
(3.5)
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Now, since the function bn fulfills the conditions assumed for b in the Theorem 3.3, it e By setting follows that the equation (3.5) has a unique solution ψn in D(L). Z x ψn (t) dt χn (x) = 0
and by integrating each side of (3.5) on [0, 1], it follows e1,n χn + fn − fn (0), χn = αn χn00 + βχn0 + fn − fn (0) = L
x ∈ [0, 1].
e ⊂ By uniqueness this implies that ϕn = χn + fn (0) and then ϕn0 = ψn . As ψn ∈ D(L) 1 2 C [0, 1], we have that ϕn ∈ C [0, 1]. STEP 2. We prove an a priori estimate for the derivative of ϕn . More precisely we show that there exists a constant c > 0, independent of n ∈ N, such that for every ϕ ∈ C 2 [0, 1] e1,n ϕk0 ), kϕ 0 k0 ≤ c (kϕk0 + kL
n ∈ N.
(3.6)
For x ∈ [0, 1/2] we have θ L1,n ϕ(x) ϕ 0 (x) = xϕ 00 (x) + ηn (0)(1 − θ ) ηn (x)(1 − x) β(x) θ − ϕ 0 (x) = g(x). + ηn (0)(1 − θ ) ηn (x)(1 − x) If we set cn = θ/ηn (0)(1 − θ ) − 1, by multiplying each side by x cn , we get (x cn +1 ϕ 0 (x))0 = x cn g(x),
x ∈ [0, 1/2].
Hence, since cn + 1 > 0 and ϕ 0 ∈ C [0, 1/2], for any x ∈ (0, 1/2] we easily get Z x 1 0 t cn g(t) dt. ϕ (x) = c +1 xn 0 Thus, due to (3.3), for any δ ≤ 1/2 and x ∈ [0, δ] 1 sup |g(t)| cn + 1 t∈ [0,δ] β(t) 0 θ 1 e 2 sup |L1,n ϕ(t)| + − |ϕ (t)| . ≤ cn + 1 t∈ [0,δ] l ηn (0)(1 − θ ) ηn (t)
|ϕ 0 (x)| ≤
Due to the continuity of ηn and β, recalling that cn + 1 = θ/ηn (0)(1 − θ ) ≥ c, for some positive constant c, we can find δ1 ≤ 1/2 such that sup |ϕ 0 (t)| ≤ t∈ [0,δ1 ]
1 cl
e1,n ϕ(t)|. sup |L t∈ [0,δ1 ]
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Similarly we can find δ2 ≤ 1/2 such that sup
|ϕ 0 (t)| ≤
t∈ [1−δ2 ,1]
1 cl
sup
e1,n ϕ(t)|. |L
t∈ [1−δ2 ,1]
Moreover, from standard estimates for the regular Sturm-Liouville problem, there exists c > 0 such that |ϕ 0 (t)| ≤ c
sup t∈ [δ1 ,1−δ2 ]
sup
e1,n ϕ(t)| + |ϕ(t)|), (|L
t∈ [δ1 ,1−δ2 ]
and this completes the proof. e1 ) = D(L) e the unique STEP 3. Finally, we can prove (3.2). We denote by ψn ∈ D(L solution of the problem e1 ψ + fn , ψ =L
(3.7)
e1,n ϕn + fn , by easy calculations we have for any n ∈ N. Recalling that ϕn = L e1 ϕn + fn + hn , ϕn = L with hn (x) = (ηn (x) − η(x)) x(1 − x)ϕn00 (x). e1 we get Then, as ψn is the solution of the problem (3.7), due to the dissipativity of L kϕn − ψn k0 ≤ khn k0 . Observe that
1 1 0 e L1,n ϕn (x) − (θ − x)ϕn (x) hn (x) = (ηn (x) − η(x)) ηn (x) 1−θ 1 1 ϕn (x) − fn (x) − (θ − x)ϕn0 (x) . = (ηn (x) − η(x)) ηn (x) 1−θ
Hence khn k0 ≤
1 kηn − ηk0 (kϕn k0 + kfn k0 + c kϕn0 k0 ). l
e1 ϕn k0 ≤ kfn k0 + kϕn ko and kϕn k0 ≤ kfn k0 ≤ c we have Recalling (3.6), as kL e1,n ϕn k0 ) ≤ c kηn − ηk0 . khn k0 ≤ c kηn − ηk0 sup (kfn k0 + kL n∈ N
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e1 (ϕ −ψn )+f −fn , This implies that hn converges to 0 in C [0, 1]. Moreover, as ϕ −ψn = L we have kϕ − ψn k0 ≤ kf − fn k0 → 0, as n goes to infinity, so that kϕ − ϕn k0 ≤ kϕn − ψn k0 + kϕ − ψn k0 ≤ khn k0 + kf − fn k0 → 0, as n goes to infinity. Finally, we can conclude, as we have e1 ϕn = lim ϕn − fn − hn = ϕ − f = L e1 ϕ. lim L
n→+∞
n→+∞
¨ REMARK 3.4. It appears that in dimension d = 1 the domain of the closure of L is neither ϕ ∈ C [0, 1] ∩ C 2 (0, 1) ; lim Lϕ(x) exists , maximal domain x→0+ ,1−
nor ϕ ∈ C [0, 1] ∩ C (0, 1) ; 2
lim
x→0+ ,1−
Lϕ(x) = 0 ,
Ventcel’s boundary conditions.
4. The multidimensional case in L2 spaces In this section we study the operator L in the space of square integrable functions, with respect to a suitable measure ν. Such a measure is chosen absolutely continuous with respect to the Lebesgue measure and an explicit formula for its density ρ is given. Moreover, under suitable conditions on the coefficients σi , the measure ν can be taken invariant and symmetrizing for the operator L. For each i = 0, 1, . . . , d, let αi be a continuous function on S such that α¯ i := min αi (x) > −1.
(4.1)
x∈ S
◦
For any x ∈ S we define the function ρα (x) by setting ρα (x) =
d Y i=0
α (x) xi i
= exp
d X i=0
! αi (x) log xi ,
(4.2)
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where we recall x0 = 1 − Z ρα (x) dx = S
Z Y d S i=0
Pd
i=1 xi .
α (x) xi i dx
If α¯ i is as in (4.1), we have ≤
Z Y d S i=0
xiα¯ i
dx ≤
d Z Y
1
t α¯ i dt =
i=0 0
d Y i=0
1 , α¯ i + 1
so that ρα ∈ L1 (S, dx). Hence, as ρα (x) > 0, we can introduce the probability measure να on S defined by να (dx) = hρα i−1 ρα (x) dx =
Z
−1 ρα (y) dy
ρα (x) dx.
S
The measure να is a generalization of the Dirichlet distribution with parameters αi (x) − 1, which depend on x ∈ S. Our aim is to study the operator L in the Hilbert space H = L2 (S, να ), endowed with the inner product Z Z hϕ, ψiH = ϕ(x)ψ(x) dνα (x) = hρα i−1 ϕ(x)ψ(x) ρα (x) dx, S
S
and the corresponding norm k · kH . In what follows we shall assume that the following assumptions hold. HYPOTHESIS 4.1. If b(x) is the vector of components bi (x) =
1 − σi (x) 1 − σ0 (x) − , xi x0
◦
x ∈ S,
(4.3)
for each i = 0, 1, . . . , d there exists a function αi ∈ C 1 (S) which fulfills (4.1), such that hC(D log ρα + b), D log ρα + bi ∈ L∞ (S).
(4.4)
4.1. An integration by parts formula The first important step is given by the following identity. THEOREM 4.2. Assume that the Hypotheses 2.1 and 4.1 hold. Then, if L is the operator defined by (2.2), for any ϕ ∈ C 2 (S) and ψ ∈ C 1 (S) it holds Z Z hC Dϕ, Dψi dνα + hC Dϕ, D log ρα + bi ψ dνα = − hLϕ, ψiH . (4.5) S
S
Proof. For any δ ∈ (0, 1/(d + 1)) we define Sδ = { x ∈ S ; x ∈ [δ, 1 − δ]d , x0 ∈ [δ, 1 − δ]}.
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Notice that since we have taken δ ∈ (0, 1/(d + 1)), the definition above is meaningful and [ ◦ S1 . S= n>d+1
n
Moreover, it is immediate to check that ∂Sδ =
d [
{x ∈ S ; xi = δ} =
i=0
d [
∂Sδ,i .
i=0
This implies that if we denote by nk the exterior normal to ∂Sδ,k , we have nk (x) = −ek ,
1 n0 (x) = √ (1, . . . , 1)t . d
1 ≤ k ≤ d,
Now, for any δ < 1/(d + 1) we have Z
Z hCDϕ, Dψi dνα = Sδ
d X
cij (x)Di ϕ(x)Dj ψ(x) dνα = Iδ,0 − Iδ,1 − Iδ,2 ,
Sδ i, j =1
where Iδ,0 = hρα i−1 Iδ,1 = hρα i−1 Iδ,2 = hρα i−1
Z
d X
Sδ i, j =1 Z X d Sδ i, j =1 Z X d
Dj (ρα cij Di ϕ ψ) dx cij Dij ϕ ψρα dx Di ϕ ψDj (cij ρα ) dx.
Sδ i, j =1
By the divergence theorem we have Iδ,0 = hρa i−1
d Z X
nj cij ρα Di ϕ ψ dσ
i, j =1 ∂Sδ
= hρα i−1
d Z d X X k=0 i, j =1 ∂Sδ,k
nkj cij ρα Di ϕ ψ dσ,
where nj denotes the j -th component of the exterior normal to ∂Sδ and njk denotes the j -th component of the exterior normal to ∂Sδ,k . Our aim is to show that lim Iδ,0 = 0.
δ→0+
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To this purpose it is sufficient to show that for each k = 0, . . . , d and i = 1, . . . , d Z
d X
lim
δ→0+ ∂Sδ,k j =1
njk cij ρα Di ϕ ψ dσ = 0.
(4.6)
If x ∈ ∂Sδ,k , for some 1 ≤ k ≤ d, we have d X
cij (x)njk (x)ρα (x) = −
j =1
d X
cij (x)δkj ρα (x) = −cik (x)ρα (x)
j =1
= −xi (δik − xk )
d Y
α (x)
xh h
= −δik
h=0
d Y
α (x)+δhi
xh h
+
h=0
d Y
α (x)+δhi +δhk
xh h
.
h=0
Now, if x ∈ ∂Sδ,k we have that xk converges to 0, as δ goes to 0. Thus, recalling that ϕ ∈ C 2 (S) and ψ ∈ C 1 (S), from the dominated convergence theorem it immediately follows (4.6). Next, if x ∈ ∂Sδ,0 we have d X
d d Y 1 X α (x) cij (x)nj0 (x)ρα (x) = √ xi (δij − xj ) xh h d j =1 j =1 h=0 d d d 1 Y αh (x)+δhi 1 Y αh (x)+δhi 1 Y αh (x)+δhi +δh0 =√ xh −√ xh (1 − x0 ) = √ xh , d h=0 d h=0 d h=0
and then by arguing as above (4.6) follows also for k = 0. Concerning Iδ,1 we have Z Tr [CD 2 ϕ] ψ dνα . Iδ,1 = Sδ
Finally, for Iδ,2 we have Z Iδ,2
= Z
d X
Dj (cij ρα )ψ Di ϕ dx
Sδ i,j =1
hCDϕ, D log ρα i ψ dνα +
= Sδ
Z
d X
Di ϕDj cij ψ dνα .
Sδ i,j =1
For each i = 1, . . . , d we have d X j =1
Dj cij =
d X j =1
Dj (xj (δij − xi )) = 1 − (d + 1)xi .
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Hence, we get Z Z hCDϕ, D log ρα i ψ dνα + hDϕ, e − (d + 1)xi ψ dνα . Iδ,2 = Sδ
Sδ
This implies that Z Z hCDϕ, Dψi dνα = Iδ,0 − hLϕ, ψiH − hCDϕ, D log ρα i ψ dνα Sδ
Sδ
Z
hCDϕ, C −1 (e − σ + (|e σ | − (d + 1))x)iψ dνα .
− Sδ
Thanks to (2.5) we have σ (x)| − (d + 1))x) C −1 (x) (e − σ (x) + (|e e⊗e (e + (|e σ (x)| − (d + 1)) x − σ (x)), = 3−1 (x) + x0 so that σ (x)| − (d + 1))x) C −1 (x) (e − σ (x) + (|e σ (x)| − (d + 1)) e − 3−1 (x)σ (x) = 3−1 (x)e + (|e +
hσ (x), ei 1 − x0 d e + (|e σ (x)| − (d + 1)) e− e. x0 x0 x0
Rearranging all terms we easily get σ (x)| − (d + 1))x) = b(x), C −1 (x) (e − σ (x) + (|e where b(x) is the vector defined in (4.3), so that, for any δ ∈ (0, 1/(d + 1) we have Z Z hCDϕ, Dψi dνα + hCDϕ, D log ρα + bi ψ dνα = Iδ,0 − hLϕ, ψiH . Sδ
Sδ
Thus, thanks to the Hypothesis 4.1, if we take the limit as δ goes to zero, (4.5) follows. ¨ The integration by parts formula (4.5) has the following important consequence. THEOREM 4.3. Under the Hypotheses 2.1 and 4.1, the operator L is closable and its closure L¯ generates a C0 -semigroup on L2 (S, να ).
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Proof. Thanks to (4.5), if ϕ ∈ D(L) we have Z Z hLϕ, ϕiH = − hCDϕ, Dϕi dνα − hCDϕ, D log ρα + bi ϕ dνα . S
S
By using first the H¨older inequality and then the Young inequality, thanks to (4.4) we get Z hCDϕ, D log ρα + bi ϕ dνα S
Z hCDϕ, Dϕi1/2 hC (D log ρα + b) , D log ρα + bi1/2 ϕ dνα
≤ S
Z
1/2 hCDϕ, Dϕi dνα
≤ Hα
Z kϕkH ≤
S
1 hCDϕ, Dϕi dνα + Hα2 kϕk2H , 4 S
where Hα2 = suphC(x)(D log ρα (x) + b(x)), D log ρα (x) + b(x)i. ◦
(4.7)
x ∈S
This implies that hLϕ, ϕiH ≤
1 2 H kϕk2H . 4 α
Thus, by setting c0 = Hα2 /4, it follows that L − c 0 is dissipative. Now, as D(L) is dense in H , we can conclude that L − c 0 is closable and its closure L − c0 = L − c0 is dissipative. Now, as proved in [3bis], the closure of L in C(S) generates a C0 -semigroup of contractions. Then, due to the Lumer-Phillips theorem the image of λ − L is dense in C(S). As C(S) is dense in L2 (S, να ), this implies that the image of λ − L is dense in L2 (S, να ) and then, by ¨ using again the Lumer-Phillips theorem, L¯ generates a C0 -semigroup. 4.2. Existence of an invariant measure Under stronger conditions on the coefficients σi , due to the integration by parts formula (4.5) it is possible to exhibit an invariant measure for the operator L which is even symmetrizing. HYPOTHESIS 4.4. For each i = 0, 1, . . . , d, there exists σi ∈ C([0, 1]) such that σi (x) = σi (xi ),
x ∈ S.
THEOREM 4.5. Assume the Hypotheses 2.1 and 4.4. If we define for any i = 0, 1, . . . , d R 1 1−σi (s) ds s , t ∈ (0, 1), (4.8) αi (t) = t log t
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we have that αi can be continuously extended to [0, 1] and αi (t) > −1. Moreover, if we define ρα (x) =
d Y
α (xi )
xi i
,
◦
x ∈ S,
i=0
the probability measure να (dx) = hρα i−1 ρα dx is invariant and symmetrizing for the operator L. Proof. For each i = 0, 1, . . . , d, the function αi is well defined and continuous in (0, 1). Thanks to the theorem of De L’Hopital, we have R 1 1−σi (s) ds s = σi (0) − 1 lim t + log t t→0 and similarly R 1 1−σi (s) lim
t→1−
t
s
ds
log t
= σi (1) − 1.
Thus αi can be continuously extended at t = 0 and t = 1. Moreover it is immediate to check that αi (t) > −1, for any t ∈ [0, 1]. Next, if we show that D log ρα + b = 0, due to (4.5) we have that Z hCDϕ, Dψi dνα = − hLϕ, ψiH , S
and hence να is invariant and symmetrizing. For any j = 1, . . . , d we have Dj log ρα =
d X
Dj (αk (x) log xk ) =
k=0
d d (αj (xj ) log xj ) − (α0 (x0 ) log x0 ) dxj dx0
and then, since for any i = 0, . . . , d σi (xi ) − 1 d , (αi (xi ) log xi ) = dxi xi we immediately have that D log ρα + b = 0.
¨
REMARK 4.6. By taking the formula (4.8) for αi as a starting point, we can construct an example of functions αi which fulfill the condition (4.4). Assume that for any i = 0, 1, . . . , d σi (x) = si (xi )fi (x) + σ¯ i ,
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for some si ∈ C([0, 1]), fi ∈ C 1 (S) and σ¯ i > 0. We define R 1 1−σi (xi (s)) ds ◦ x s , x ∈ S, αi (x) = i log xi where (xi (s))j = xj (1 − δij ) + sδij . As seen in the proof of the previous theorem, αi can be extended as a continuous function on S. For any i, j = 1, . . . , d we have Z 1 σj (x) − 1 si (s) ds (1 − δij ) δij − Dj fi (x) Dj (αi log xi ) (x) = xj s xi and Dj (α0 log x0 ) (x) =
1 − σ0 (x) − Dj f0 (x) x0
Z
1
x0
s0 (s) ds. s
If we show that d X 1 − σj (x) 1 − σ0 (x) Dj (αk (x) log xk ) + − sup < ∞, xj x0 ◦ x∈ S k=0
we immediately have (4.2) Therefore, we need to assume that X Z 1 d si (s) ds < ∞. Dj fi (x) sup s ◦ xi x∈ S i=0 i6=j This is satisfied for example if we assume that si (s)/s ∈ L1 (0, 1) or if the functions fi are constant in a neighborhood of the boundary of S. 4.3. The variational formulation For any ϕ, ψ ∈ C 1 (S), we define the semi-inner product Z hhϕ, ψii1,C = hCDϕ, Dψi dνα
(4.9)
S
with the corresponding semi-norm [·]1,C . In what follows we shall denote by WC1,2 (S, να ) the completion of C 1 (S) with respect to the norm k · k1,C induced by the inner product hϕ, ψi1,C = hϕ, ψiH + hhϕ, ψii1,C . Similarly, for any ϕ, ψ ∈ C 2 (S) we define the semi-inner product Z hhϕ, ψii2,C = Tr [(CD 2 ϕ)(CD 2 ψ)] dνα , S
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with the corresponding semi-norm [·]2,C . We shall denote by WC2,2 (S, να ) the completion of C 2 (S) with respect to the norm k · k2,C induced by the inner product hϕ, ψi2,C = hϕ, ψi1,C + hhϕ, ψii2,C . From the definition we are giving for WC1,2 (S, να ) and WC2,2 (S, να ), we can imbed WC1,2 (S, να ) in L2 (S, να ) and WC2,2 (S, να ) in WC1,2 (S, να ), with continuous embeddings. Clearly, we have kϕkH ≤ kϕk0 ,
ϕ ∈ C(S)
(4.10)
ϕ ∈ C i (S).
(4.11)
and for i = 1, 2 kϕki,C ≤ kϕki ,
Thus, since C 1 (S) is dense in C(S), from (4.10) it follows that WC1,2 (S, να ) is dense in L2 (S, να ). Similarly, since C 2 (S) is dense in C 1 (S), from (4.11) it follows that C 2 (S) is dense in WC1,2 (S, να ), hence WC2,2 (S, να ) is dense in L2 (S, να ). In the sequel we shall denote by V the Hilbert space WC1,2 (S, να ), and by k · kV , [·]V , h·, ·iV and hh·, ·iiV respectively the norm, the seminorm, the inner product and the semi inner product in WC1,2 (S, να ), as defined above. For any ϕ, ψ ∈ C 1 (S) we define the bilinear form Z Z (4.12) aα (ϕ, ψ) = hCDϕ, Dψi dνα + hCDϕ, D log ρα + bi ψ dνα . S
S
As V is the completion of C 1 (S) with respect to the norm induced by the inner product Z Z hϕ, ψiV = ϕ ψ dνα + hCDϕ, ψi dνα , S
S
it is immediate to check that aα can be extended as a bilinear form on V × V . Moreover, due to (4.5), if ϕ ∈ D(L) and ψ ∈ C 1 (S) we have aα (ϕ, ψ) = − hLϕ, ψiH . LEMMA 4.7. Under the Hypotheses 2.1 and 4.1, the bilinear form aα : V × V → R is continuous, that is there exists a constant c such that |aα (ϕ, ψ)| ≤ c kϕkV kψkV ,
ϕ, ψ ∈ V .
(4.13)
Moreover, for any δ < 1 there exists cδ ∈ R such that |aα (ϕ, ϕ)| ≥ δ kϕk2V − cδ kϕk2H ,
ϕ ∈ V.
(4.14)
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Proof. We first prove (4.13). From (4.12) and (4.9) we have Z aα (ϕ, ψ) = hhϕ, ψiiV + hCDϕ, D log ρα + bi ψ dνα . S
Thus, thanks to (4.4) and to the H¨older inequality, we have |aα (ϕ, ψ)| ≤ kϕkV kψkV + Hα [ϕ]V kψkH , which easily yields (4.13). Now, let us prove (4.14). We have Z aα (ϕ, ϕ) = [ϕ]2V + hCDϕ, D log ρα + bi ϕ dνα . S
If we fix δ < 1, we have that (1 − δ)/2 > 0 and then, by using the H¨older inequality and the Young inequality, we get Z 2 hCDϕ, D log ρα + bi ϕ dνα ≤ 1 − δ [ϕ]2 + Hα kϕk2 . V H 2 2(1 − δ) S Hence it follows aα (ϕ, ϕ) ≥
1+δ 2 Hα2 [ϕ]V − kϕk2H . 2 2(1 − δ)
Thanks to the Young inequality, for any > 0 we have 1 kϕk2H [ϕ]2V = (kϕkV − kϕkH )2 ≥ (1 − ) kϕk2V + 1 − and then, if we choose = (1 − δ)/(1 + δ), by easy calculations we obtain aα (ϕ, ϕ) ≥ δ kϕk2V −
Hα2 + 2δ(1 + δ) kϕk2H . 2(1 − δ)
This yields (4.14), with cδ =
Hα2 + 2δ(1 + δ) 2(1 − δ)
(4.15) ¨
Now, due to (4.13), for any fixed ψ ∈ V the linear mapping V → R, ϕ 7 → aα (ϕ, ψ), is continuous and then there exists some fψ ∈ V ? such that aα (ϕ, ψ) = fψ (ϕ),
ϕ ∈ V.
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This allows us to define the operator A : D(A) ⊂ H → H in the classical way D(A) = {ψ ∈ V : fψ ∈ H },
Aψ = fψ .
(4.16)
For any δ < 1 we define the operator Aδ : D(A) ⊂ H → H,
Aδ = A + cδ I,
where cδ is the constant introduced in (4.15). By using the Lemma 4.8 and the LaxMilgram theorem, it is possible to show that Range( Aδ ) = H and 0 ∈ ρ(Aδ ). Moreover D(Aδ ) = D(A) is dense in V (and hence in H ) and −Aδ is a maximal dissipative operator which generates an analytic semigroup (for the proofs of these facts see for example [13]Section 2.2 and Theorem 3.6.1). Thus we have the following result. PROPOSITION 4.8. Under the Hypotheses 2.1 and 4.1 the C0 -semigroup generated by ¯ ⊇ (Hα2 /2, +∞), where Hα is the constant introduced in (4.7). L¯ is analytic and ρ(L) Proof. For any δ < 1 we have A = Aδ − cδ I and then it immediately follows that D(−A) is dense in H and −A generates an analytic semigroup in H . Moreover, as −Aδ is dissipative and 0 ∈ ρ(−Aδ ), we have ρ(−Aδ ) ⊇ [0, +∞) so that ρ(−A) ⊇ [cδ , +∞). Now, as our arguments work for any δ < 1 and cδ ↓ Hα2 /2, as δ goes to zero, we conclude that ρ(−A) ⊇ (Hα2 /2, +∞). Finally, let us prove that L ⊆ −A. We have already seen that if ϕ ∈ D(L), then for any ψ ∈ C 1 (S) aα (ϕ, ψ) = − hLϕ, ψiH and, since C 1 (S) is dense in V , the same relation holds for any ψ ∈ V . Recalling the definition of A, this implies that ϕ ∈ D(A) and Aϕ = −Lϕ, so that L ⊆ −A. As A is a closed operator, this yields L ⊆ −A. Moreover, since L¯ generates a C0 -semigroup, it is the only extension of L and then L¯ = −A. ¨ An important consequence of the previous proposition is the existence of a unique solution for the variational problem associated with the operator L. THEOREM 4.9. Under the Hypotheses 2.1 and 4.1, for any f ∈ H and λ > Hα2 /2 ¯ to the problem there exists a unique solution ϕ ∈ D(L) Z Z λ hϕ, ψiH + hCDϕ, Dψi dνα + hCDϕ, D log ρα + bi ψ dνα = hf, ψiH , S
S
for any ψ ∈ V . Furthermore, if ϕ ∈ D(L), then it is a solution of (1.1), that is λϕ − Lϕ = f.
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4.4. Characterization of D(L) In this subsection we are giving a characterization of D(L) in H . Moreover, in the case the coefficients σi fulfill the Hypothesis 4.4 (so that there exists an invariant measure να which is symmetrizing), we give a factorization of L which provides a characterization of the domain of the square root of −L. The crucial step is the following identity. PROPOSITION 4.10. Under the Hypotheses 2.1 and 4.1, for any ϕ ∈ C 2 (S) and λ ∈ R it holds Z Z σ |) hCDϕ, Dϕi dνα + hD log ρα + b, CD 2 ϕ (CDϕ)idνα [ϕ]22,C + (λ + |e Z S ZS σ |i dνα − Tr [(CDϕ ⊗ Dϕ)Dσ ] dνα + h(CDϕ ⊗ Dϕ) x, D|e S S Z = kLϕk2H − λ hLϕ, ϕiH − hCDϕ, D log ρα + bi (λϕ − Lϕ) dν. (4.17) S
Proof. If we prove (4.17) for ϕ ∈ C 3 (S), then the case of ϕ ∈ C 2 (S) follows by a standard approximation argument. Thus, let us fix ϕ ∈ C 3 (S) and λ ∈ R and let us define f = λϕ − Lϕ, that is σ (x)|x, Dϕ(x)i = f (x), λϕ(x) − Tr [C(x)D 2 ϕ(x)] − hσ (x) − |e
x ∈ S.
If we derive each side with respect to xj , we get λDj ϕ(x) − Tr [C(x)D 2 Dj ϕ(x)] − hσ (x) − |e σ (x)|x, DDj ϕ(x)i σ (x)|x − |e σ (x)|Dj x, Dϕ(x)i − Tr [Dj C(x)D 2 ϕ(x)] − hDj σ (x) − Dj |e = Dj f (x). where Dj = ∂/∂xj . If we multiply each side by cij (x)Di ϕ(x), by integrating with respect to να and by taking the sum over i and j , it follows Z Z λ Dj ϕ cij Di ϕ dνα − Tr [CD 2 Dj ϕ]cij Di ϕ dνα S S Z Z − hσ − |e σ |x, DDj ϕicij Di ϕ dνα − Tr [Dj CD 2 ϕ]cij Di ϕ dνα H S Z Z − hDj σ − Dj |e σ |x − |e σ |Dj x, Dϕicij Di ϕ dνα = Dj f cij Di ϕ dνα (4.18) S
S
(here and in what follows, any time we have the same index twice we are taking the sum on that index).
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Now let us calculate each term in (4.18). We have Z Z λ Dj ϕ cij Di ϕ dνα = λ hCDϕ, Dϕi dνα . S
S
By using (4.5), it holds Z Z σ |x, DDj ϕ(x)icij Di ϕ dνα − Tr [CD 2 Dj ϕ]cij Di ϕ dνα − hσ − |e S
Z
Z
=−
LDj ϕ cij Di ϕ dνα =
Z
hCDDj ϕ, D(cij Di ϕ)i dνα
S
+
H
S
Z
hCDDj ϕ, D log ρα + bi cij Di ϕ dνα = S
Z +
hCDDj ϕ, Dcij iDi ϕ dνα S
Z hCDDj ϕ, DDi ϕicij dνα +
hCDDj ϕ, D log ρα + bi cij Di ϕ dνα
S
S
= I1 + I2 + I3 . By some calculations it is possible to show that Z I2 = Tr [(CD 2 ϕ)2 ] dνα S
and
Z
I3 =
hD log ρα + b, CD 2 ϕ CDϕi dνα , S
so that Z Z σ |x, DDj ϕicij Di ϕ dνα − Tr [CD 2 Dj ϕ]cij Di ϕ dνα − hσ − |e S
H
Z =
Z hCDDj ϕ, Dcij iDi ϕ dνα +
S
Tr [(CD 2 ϕ)2 ] dνα S
Z +
hD log ρα + b, CD 2 ϕ CDϕi dνα . S
Next step is proving the following identity LEMMA 4.11. It holds Z Z hCDDj ϕ, Dcij iDi ϕ dνα − Tr [Dj CD 2 ϕ]cij Di ϕ dνα = 0. S
S
(4.19)
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Proof. It is not difficult to check that the left hand side in (4.19) is given by Z Di ϕ Dkj ϕ(chk Dh cij − cih Dh ckj ) dνα . S
Thus, if we show that for any i, j, k = 1, . . . , d d X
I (x) =
chk (x)Dh cij (x) − cih (x)Dh ckj (x) = 0,
x ∈ S,
h=1
we immediately get (4.19). Recalling that ckj (x) = xk (δkj − xj ), we have Dh ckj (x) = δhk (δkj − xj ) − δhj xk , and this implies that I (x) = δij cik (x)(1 − 2xi ) − (cik (x)xj + cj k (x)xi )(1 − δij ) −δkj cik (x)(1 − 2xk ) + (cik (x)xj + cij (x)xk )(1 − δkj ). If i = j = k, we immediately have that I (x) = 0. If i = j and k 6 = i, j we have I (x) = −cik (x)xi + cik (x) + cii (x)xk = xi2 xk − xi xk + xi (1 − xi )xk = 0. If k = j and i 6 = j we have I (x) = −cik (x)(1 − 2xk ) − xk cik (x) − xi ckk (x) = −xi xk2 + xi xk − xi xk (1 − xk ) = 0. Finally, if i 6 = j 6 = k we have I (x) = −xi cj k (x) + cij (x)xk = xi xj xk − xi xj xk = 0, so that I (x) = 0, for any x ∈ S and i, j, k = 1, . . . , d.
¨
Concerning the remaining terms in (4.18), if we denote by {ej } the standard orthonormal basis in Rd , we have Z Z − hDj σ, Dϕicij Di ϕ dνα = − hDσ ej , DϕihCDϕ, ej i dνα S
S
Z =−
Z h(CDϕ ⊗ Dϕ) Dσ ej , ej i dνα = −
S
and
Tr [(CDϕ ⊗ Dϕ) Dσ ] dνα S
Z Dj |e σ |hx, Dϕicij Di ϕ dνα S
Z =
Z hCDϕ, D|e σ |ihx, Dϕi dνα =
S
h(CDϕ ⊗ Dϕ) x, D|e σ |i dνα . S
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Moreover, Z Z |e σ |hDj x, Dϕicij Di ϕ dνα = |e σ |hCDϕ, Dϕi dνα . S
S
Finally, as Z
Z Dj f cij Di ϕ dνα = S
hDf, CDϕidνα , S
by using (4.5) we have Z
Z Dj f cij Di ϕ dνα = −hf, LϕiH − S
hCDϕ, D log ρα + bi f dνα S
Z = kLϕk2H − λhϕ, LϕiH −
hCDϕ, D log ρα + bi (λϕ − Lϕ) dνα . S
Therefore, collecting all terms in (4.18) we get (4.17).
¨
The identity (4.17) proved in the previous proposition allows us to give a characterization of D(L). THEOREM 4.12. The domain of L in H = L2 (S, να ) equals WC2,2 (S, να ), with equivalence of norms. Proof. In view of the definition of WC2,2 (S, να ) and L, it suffices to prove that there exists two constants c1 , c2 > 0 such that for any ϕ ∈ C 2 (S) c1 kϕkD(L) ≤ kϕk2,C ≤ c2 kϕkD(L) .
(4.20)
Due to (4.17) we have Z σ |) hCDϕ, Dϕi dνα kLϕk2H = [ϕ]22,C + (λ + |e S Z Z + hD log ρα + b, CD 2 ϕ (CDϕ)i dνα − Tr [(CDϕ ⊗ Dϕ) Dσ ] dνα S ZS + h(CDϕ ⊗ Dϕ) x, D|e σ |i dνα +λhLϕ, ϕiH ZS + hCDϕ, D log ρα + bi (λϕ − Lϕ) dνα . S
(4.21)
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Then, thanks to the Young inequality and to (4.4), after some calculations we have 1 σ k0 + 2 Hα2 ) [ϕ]21,C kLϕk2H ≤ [ϕ]22,C + (λ + ke 2 Z + |hD log ρα + b, CD 2 ϕ (CDϕ)i|dνα S
5 + λ2 kϕk2H + 4
Z
Z |Tr [(CDϕ ⊗ Dϕ) Dσ ]| dνα +
S
|h(CDϕ ⊗ Dϕ) x, D|e σ |i| dνα . S
For any A, B ∈ L(Rd ) with A ≥ 0, it holds kABkL(Rd ) ≤ |Tr [AB]| ≤ Tr [A] kBkL(Rd ) ,
(4.22)
and for any u, v ∈ Rd Tr [u ⊗ v] = hu, vi. Therefore we have Z Z |Tr [(CDϕ ⊗ Dϕ) Dσ ]| dνα ≤ kDσ k0 Tr [CDϕ ⊗ Dϕ] dνα S S Z = kDσ k0 hCDϕ, Dϕi dνα = kDσ k0 [ϕ]21,C .
(4.23)
S
Moreover, we have Z Z |h(CDϕ ⊗ Dϕ) x, D|e σ |i| dνα ≤ kDσ k0 kCDϕ ⊗ DϕkL(Rd ) dνα S
S
= kDσ k0 [ϕ]21,C .
(4.24)
Finally, Z |hD log ρα + b, CD 2 ϕ (CDϕ)i| dνα S
Z ≤ Hα
h(C 1/2 D 2 ϕC 1/2 )2 C 1/2 Dϕ, C 1/2 Dϕi1/2 dνα S
and then, by using once more the Young inequality, from (4.22) it follows Z H2 1 |hD log ρα + b, CD 2 ϕ (CDϕ)i| dνα ≤ α [ϕ]21,C + [ϕ]22,C . 2 2 S Then, collecting all terms we have 5Hα2 5 + 2kDσ k0 [ϕ]21,C + λ2 kϕk2H . σ k0 + kLϕk2H ≤ 3[ϕ]22,C + 2 λ + ke 2 2
(4.25)
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This implies that WC2,2 (S, να ) ⊂ D(L) and there exists a constant c1 > 0 such that the first inequality in (4.20) holds. On the other hand, by using (4.21), (4.23), (4.24) and (4.25) and the Young inequality, we have kLϕk2H ≥ [ϕ]22,C + λ [ϕ]21,C −
Hα2 1 [ϕ]21,C − [ϕ]22,C − 2kDσ k0 [ϕ]21,C 2 2
1 λ2 λ2 1 − kLϕk2H − kϕk2H − Hα2 [ϕ]21,C − kϕk2H − kLϕk2H , 2 2 2 2 so that 1 2 3 2 1 2 kLϕkH ≥ [ϕ]2,C + λ − Hα − 2kDσ k0 [ϕ]21,C − λ2 kϕk2H . 2 2 2 Thus, if we fix 3 λ > Hα2 + 2kDσ k0 , 2 we can conclude that D(L) ⊆ WC2,2 (S, να ) and there exists a constant c2 > 0 such that the second inequality in (4.20) holds. ¨ Next, we define the operator Ls as D(Ls ) = D(L) = C 2 (S), Ls ϕ = Lϕ + hCDϕ, D log ρα + bi, ϕ ∈ D(L).
(4.26)
Due to (4.5), for any ϕ, ψ ∈ D(L) we have Z Z Z Ls ϕ ψ dνα = − hCDϕ, Dψi dνα = ϕ Ls ψ dνα , S
S
s
so that Ls can be considered the symmetric part of L. Moreover, we introduce the space L2C (S, να ), defined as the set of Borel measurable functions 8 : S → Rd such that Z hC(x)8(x), 8(x)i dνα (x) < ∞. S
In view of (2.3), the space L2C (S, να ) becomes a Hilbert space if equipped with the inner product Z h8, 9iC = hC(x)8(x), 9(x)i dνα (x). S
We shall denote by k · kC the corresponding norm in L2C (S, να ). Finally, we introduce the gradient operator from L2 (S, να ) into L2C (S, να ) as follows D(DC ) = C 1 (S),
DC ϕ(x) = (D1 ϕ(x), . . . , Dd ϕ(x))t ,
We have the following result.
x ∈ S.
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PROPOSITION 4.13. Assume that the Hypotheses 2.1 and 4.1 hold. Then 1. the operator DC : C 1 (S) ⊂ L2 (S, να ) → L2C (S, να ) is closable and densely defined; 2. if D is the closure of DC , then D(D) = WC1,2 (S, να ); 3. if D? denotes the adjoint of D from L2C (S, να ) into L2 (S, να ), then −2L¯ s ⊆ D? D. Proof. We already know that C 1 (S) is dense in L2 (S, να ). We show that DC is closable. Let {ϕn } be a sequence in C 1 (S) converging to 0 in L2 (S, να ) and such that lim DC ϕn = 8,
n→+∞
that is
in L2C (S, να ),
Z
lim
n→+∞ S
hC(Dϕn − 8), Dϕn − 8i dνα = 0.
(4.27)
Since the integral in (4.27) is non negative, for any δ > 0 small enough we have Z hC(Dϕn − 8), Dϕn − 8i dνα = 0, lim n→+∞ S δ
where Sδ is the set introduced in the proof of the Proposition 4.2 Sδ = { x ∈ S : dist (x, ∂S) ≥ δ } . Thus, since there exists kδ > 0 such that inf hC(x)h, hi ≥ kδ |h|2 ,
x∈ Sδ
we have lim
h ∈ Rd
Z
n→+∞ S δ
|Dϕn − 8|2 dνα = 0.
As there exists a constant ρδ > 0 such that ρα (x) ≥ ρδ , for x ∈ Sδ , we have that in L2 (S) limn→+∞ ϕn = 0, limn→+∞ Dϕn = 8, in L2 (Sδ , Rd ). This implies that the restriction of 8 to Sδ is 0, a.s. for any δ > 0, so that 8 = 0 on S, a.s. and then DC is closable. The domain of its closure D coincides with WC1,2 (S, να ), as the graph norm of D equals the norm of WC1,2 (S, να ) on C 1 (S). Since the operator D is densely defined, the adjoint of D, D? : D(D? ) ⊂ L2C (S, να ) → L2 (S, να ), is well defined and since D is closed, from [7]-Chapter 5, Theorem 3.24, it follows that the operator D? D, with domain D(D? D) = ϕ ∈ D(D) : Dϕ ∈ D(D? ) ,
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is positive and self-adjoint in L2 (S, να ), D((D? D)1/2 ) = D(D) and D(D? D) is a core for D(D). Next, from the integration by parts formula (4.5), for any ϕ ∈ C 2 (S) and ψ ∈ C 1 (S) we have hDϕ, DψiC = −2hLs ϕ, ψiH . As C 1 (S) is dense in WC1,2 (S, να ) and the graph norm of D equals the norm in WC1,2 (S, να ), the same identity holds for ψ ∈ WC1,2 (S, να ). Moreover, as WC2,2 (S, να ) continuously embeds into WC1,2 (S, να ), thanks to the Theorem 4.12 we can extend the identity above to all ϕ ∈ WC2,2 (S, να ) and we have hDϕ, DψiC = −2hL¯ s ϕ, ψiH ,
ϕ ∈ WC2,2 (S, να ),
ψ ∈ WC1,2 (S, να ).
Finally, we prove that −2L¯ s ⊆ D? D. Let ϕ ∈ WC2,2 (S, να ) = D(L¯ s ) ⊂ WC1,2 (S, να ) = D(D) and set 8 = Dϕ ∈ L2C (S, να ). For any ψ ∈ D(D) we have |h8, DψiC | = 2|hL¯ s ϕ, ψiH | ≤ 2 kL¯ s ϕkH kψkH , and hence 8 ∈ D(D? ) and hD? 8, ψiH = h8, DψiC . Therefore ϕ ∈ D(D? D) and hD? Dϕ, ψiH = −2hL¯ s ϕ, ψiH . Since D(D) is dense in L2 (S, να ), it follows that D? Dϕ = −2L¯ s ϕ and −2L¯ s ⊆ D? D. ¨ In the case there exists a symmetrizing measure, the previous results allows us to give a ¯ 1/2 ). factorization of −L¯ and a characterization of D((−L) COROLLARY 4.14. Assume the Hypotheses 2.1 and 4.4. Then −2L¯ = D? D and ¯ 1/2 ) = W 1,2 (S, να ). D((−L) C Proof. Since in this case −2L¯ and D? D are both self-adjoint and −2L¯ s ⊆ D? D, we have that they coincide. ¨ Acknowledgment The authors would like to thank Professors D. Dawson, A. Greven and F. Den Hollander for bringing the problem to their attention and Professor G. Da Prato for giving them many useful suggestions. They also would like to thank Professor G. Metafune and Professor M. Roeckner for letting them know the papers [3bis] and [12], respectively.
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Sandra Cerrai Dip. Matematica per le Decisioni Universit`a di Firenze Via C. Lombroso 6/17 I-50134 Firenze Italy e-mail:
[email protected] Philippe Cl´ement Dept. of Applied Mathematical Analysis Technische Universiteit Delft Mekelweg 4 2628 CD Delft The Netherlands e-mail:
[email protected]
