53.2.1 Sectorial Operators And Analytic Semigroups
With the theory of functions of a complex variable, it is time to consider the notion of
analytic semigroups. These are better than continuous semigroups. I am mostly following
the presentation in Henry [?]. In what follows H will be a Banach space unless specified
to be a Hilbert space.
Definition 53.2.1Let ϕ < π∕2 and for a ∈ ℝ, let S_{aϕ}denote the sector in the complexplane
{z ∈ ℂ∖{a} : |arg(z − a)| ≤ π − ϕ}
This sector is as shown below.
PICT
A closed, densely defined linear operator, A is calledsectorial if for some sector asdescribed above, it follows that for all λ ∈ S_{aϕ},
(λI − A)−1 ∈ ℒ (H,H )
and for some M it satisfies
|| ||
||||(λI − A )−1|||| ≤-M----
|λ − a|
To begin with it is interesting to have a perturbation theorem for sectorial operators.
First note that for λ ∈ S_{aϕ},
A (λI − A )− 1 = − I + λ(λI − A )−1
Proposition 53.2.2Suppose A is a sectorial operator as defined above so it is a denselydefined closed operator on D
(A )
⊆ H which satisfies
|| ||
||||A(λI − A)−1|||| ≤ C (53.2.3)
(53.2.3)
whenever
|λ|
,λ ∈ S_{aϕ}, is sufficiently large and suppose B is a densely defined closedoperator such that D
(B)
⊇ D
(A )
and for all x ∈ D
(A)
,
||Bx|| ≤ ε||Ax ||+ K ||x|| (53.2.4)
(53.2.4)
where εC < 1. Then A + B is also sectorial.
Proof: I need to consider
(λI − (A+ B ))
^{−1}. This equals
( ( −1) )−1
I − B (λI − A) (λI − A) . (53.2.5)
(53.2.5)
The issue is whether this makes any sense for all λ ∈ S_{bϕ} for some b ∈ ℝ. Let b > a be
very large so that if λ ∈ S_{bϕ}, then 53.2.3 holds. Then from 53.2.4, it follows that for
||x||
≤ 1,
|| || || || || ||
||||B (λI − A )−1 x|||| ≤ ε||||A(λI − A)−1x||||+ K ||||(λI − A )−1x||||
≤ εC + K ∕|λ − a|
and so if b is made still larger, it follows this is less than r < 1 for all
||x||
≤ 1. Therefore,
for such b,
( )
I − B (λI − A )−1 − 1
exists and so for such b, the expression in 53.2.5 makes sense and equals
( )−1
(λI − A)−1 I − B (λI − A)−1
and furthermore,
|| ||
||||(λI − A )− 1(I − B (λI − A )− 1) −1||||≤-M----1---≤ -M-′--
|| || |λ − a|1− r |λ − b|
by adjusting the constants because
-M----|λ-− b|
|λ − a|1− r
is bounded for λ ∈ S_{bϕ}. This proves the proposition.
It is an interesting proposition because when you have compact embeddings, such
inequalities tend to hold.
Definition 53.2.3Let ε > 0 and for a sectorial operator as defined above, let thecontour γ_{ε,ϕ}be as shown next where the orientation is also as shown by the arrow.
PICT
The little circle has radius ε in the above contour.
Definition 53.2.4For t ∈ S_{0}
(ϕ+π∕2)
^{0}the open sector shown in the followingpicture,
PICT
define
1 ∫
S (t) ≡--- eλt(λI − A)−1dλ (53.2.6)
2πi γε,ϕ
(53.2.6)
where ε is chosen such that t is a positive distance from the set of points included in γ_{ε,ϕ}.The situation is described by the following picture which shows S_{0}
(ϕ+ π∕2)
^{0}and S_{0ϕ}. Notehow the dotted line is at right angles to the solid line.
PICT
Also define S
(0)
≡ I. It isn’t necessary that ε be small, just that γ_{ε,ϕ}not contain t.This is because the integrand in 53.2.6is analytic.
Then it is necessary to show the above definition is well defined.
Lemma 53.2.5The above definition is well defined for t ∈ S_{0}
(ϕ+ π∕2)
^{0}. Also there is aconstant, M_{r}such that
||S(t)|| ≤ M
r
for every t ∈ S_{0}
(ϕ+π∕2)
^{0}such that
|argt|
≤ r <
(π )
2 − ϕ
.
Proof: In the definition of S
(t)
one can take ε = 1∕
|t|
. Then on the little circle
which is part of γ_{ε,ϕ} the contour integral equals
1 ∫ π−ϕ i(θ+arg(t)) ( 1 ) −1 1
2π- ee |t|eiθ − A |t|eiθdθ
ϕ−π
and by assumption the norm of the integrand is no larger than
M 1
1∕|t||t|
and so the norm of this integral is dominated by
M ∫ π−ϕ M
2π- dθ = 2π-(2π − 2ϕ) ≤ M
ϕ−π
which is independent of t.
Now consider the part of the contour used to define S
(t)
which is the top line
segment. This equals
∫
-1- ∞ eywt(ywI − A )−1wdy
2πi 1∕|t|
where w is a fixed complex number of unit length which gives a direction for
motion along this segment, arg
∫ ∞
1-- exp(− c(r)|t|y) M-dy
2π∫1∕|t| y
1-- ∞ M-|t|-1
= 2π 1 exp(− c (r)x) x |t|dx
1 ∫ ∞ M
= 2π- exp(− c (r)x) x-dx
1
where c
(r)
< 0 independent of
|arg(t)|
≤ r. A similar estimate holds for the integral on
the bottom segment. Thus for
|arg(t)|
≤ r,
||S (t)||
is bounded. This proves the
Lemma.
Also note that if the contour is shifted to the right slightly, the integral over the
shifted contour, γ_{ε,ϕ}^{′} coincides with the integral over γ_{ε,ϕ} thanks to the Cauchy integral
formula and an approximation argument involving truncating the infinite contours and
joining them at either end. Also note that in particular,
||S(t)||
is bounded for all
positive real t. The following is the main result.
Theorem 53.2.6Let A be a sectorial operator as defined in Definition 53.2.1forthe sector S_{0ϕ}.
Then S
(t)
given above in 53.2.6is analytic for t ∈ S_{0(ϕ+ π∕2)
}^{0}.
For any x ∈ H and t > 0, then for n a positive integer,
S (n)(t) x = AnS (t)x
S is a semigroup on the open sector, S_{0(ϕ+π∕2)
}^{0}. That is, for all t,s ∈ S_{0(ϕ+π∕2)
}^{0},
S (t +s) = S(t)S(s)
t → S
(t)
x is continuous at t = 0 for all x ∈ H.
For some constants M,N such that if t is positive and real,
||S(t)|| ≤ M
N
||AS (t)|| ≤--
t
Proof: Consider the first claim. This follows right away from the formula.
∫
S(t) ≡-1- eλt(λI − A)−1dλ
2πi γε,ϕ
The estimates for uniform convergence do not change for small changes in t and so the
formula can be differentiated with respect to the complex variable t using the dominated
convergence theorem to obtain
′ -1-∫ λt −1
S (t) ≡ 2πi γ λe (λI − A) dλ
ε,ϕ
∫ ( )
= -1- eλt I + A (λI − A)−1 dλ
2πi γε,ϕ
1 ∫ λt − 1
= 2πi e A (λI − A ) dλ
γε,ϕ
because of the Cauchy integral theorem and an approximation result. Now approximating
the infinite contour with a finite one and then the integral with Riemann sums, one can
use the fact A is closed to take A out of the integral and write
( 1 ∫ )
S′(t) = A --- eλt(λI − A)−1 dλ = AS (t)
2πi γε,ϕ
To get the higher derivatives, note S
(t)
has infinitely many derivatives due to t being a
complex variable. Therefore,
′ ′
S′′(t) = lim S-(t-+-h)−-S-(t)-= lim A S(t+-h)-− S-(t)
h→0 h h→0 h
and
S (t+ h)− S(t)
------h-------→ AS (t)
and so since A is closed, AS
(t)
∈ D
(A )
and
A2S (t)
Continuing this way yields the claims 1.) and 2.). Note this also implies S
(t)
x ∈ D
(A)
for each t ∈ S_{0(ϕ+π∕2)
}^{0}.
Next consider the semigroup property. Let s,t ∈ S_{0}
(ϕ+ π∕2)
^{0} and let ε be sufficiently
small that γ_{ε,ϕ} is at a positive distance from both s and t. As described above let γ_{ε,ϕ}^{′}
denote the contour shifted slightly to the right, still at a positive distance from t.
Then
( 1 )2 ∫ ∫ λt −1 μs −1
S (t)S (s) = 2πi ′ e (λI − A ) e (μI − A ) dμdλ
γε,ϕ γε,ϕ
At this point note that
−1 −1 −1( − 1 −1)
(λI − A) (μI − A) = (μ − λ) (λI − A ) − (μI − A) .
Then substituting this in the integrals above, it equals