SMA 5878 Functional Analysis II€¦ · Semigroups and Their Generators To beter explain this remark we consider rst the case A 2L(X). In this case, the semigroup t 7!T(t) is the

Semigroups and Their Generators

SMA 5878 Functional Analysis II

Alexandre Nolasco de Carvalho

Departamento de MatematicaInstituto de Ciencias Matematicas and de Computacao

Universidade de Sao Paulo

May 02, 2018

Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II



In this chapter we present some basic facts of the theory ofsemigroups of bounded linear operators which are essential for theunderstanding of the techniques of resolution of parabolic andhyperbolic semilinear PDEs.

Most of the exposition will be centered on the characterization ofthe generators of semigroups of bounded linear operatorssince, in applications of the theory, in general, we know thedifferential equation and do not know the solution operator.



The study of semigroups of bounded linear operators is associatedto the study of linear Cauchy problems of the form

d

dtx(t) = Ax(t)

x(0) = x0

(1)

where A : D(A) ⊂ X → X is linear (in general unbounded).

The semigorup {T (t) : t ≥ 0} is the solution operator of (1), thatis, given x0 ∈ X , t 7→ T (t)x0 is the solution (in some sense) of (1).



To beter explain this remark we consider first the case A ∈ L(X ).In this case, the semigroup t 7→ T (t) is the solution operator (inthe usual sense) of the problem

d

dtT (t) = AT (t)

T (0) = B ∈ L(X ).(2)

with B = I . This solution will be denoted by T (t) =: etA.

Let us show that there exists a unique solution for (2) and that thesemigroup properties are satisfied. This follows from the Banachcontraction principle that we state next.



LemmaLet X be a complete metric space with metric dX : X × X → R+

and a function F :X→X such that dX (F n(x),F n(y)) ≤ k dX (x , y)for some positive integer n and k < 1 (F n is a contraction). Then,there exists a unique x ∈ X such that F (x) = x . The elementx ∈ X is called a fixed point of F .



Now we will seek for solutions for (2) which are functions in{U(·) ∈ C ([0, τ ], L(X )) ∩ C 1((0, τ ], L(X )) : U(0) = B} thatverify (2). Let K = {U(·) ∈ C ([0, τ ],L(X )) : U(0) = B} anddefine the map F : K → K by

F (U)(t) = B +

∫ t

0AU(s)ds

and observe that a solution of (2) is a fixed point of F in K andthat is a fixed point of F is a solution of (2). Note that K is acomplete metric space with the metric induced by the norm ofC ([0, τ ],L(X )).



We wish to show that there exists a positive integer n such that F n

is a contraction. In fact:

‖F (U)(t)− F (V )(t)‖ ≤∣∣∣∣∫ t

0‖AU(s)− AV (s)‖ds

∣∣∣∣≤ |t|‖A‖ sup

t∈[0,τ ]‖U(t)− V (t)‖

≤ τ‖A‖ supt∈[0,τ ]

‖U(t)− V (t)‖



Suppose that, for t ∈ [0, τ ],

‖F n−1U(t)− F n−1V (t)‖ ≤ |t|n−1‖A‖n−1

(n − 1)!sup

t∈[0,τ ]‖U(t)− V (t)‖,

then

‖F n(U)(t)− F n(V )(t)‖ ≤∣∣∣∣∫ t

0‖AF n−1U(s)− AF n−1V (s)‖ds

∣∣∣∣≤ |t|

n‖A‖n

n!sup

t∈[0,τ ]‖U(t)− V (t)‖

≤ |τ |n‖A‖n

n!sup

t∈[0,τ ]‖U(t)− V (t)‖.



Noting that |τ |n‖A‖nn! → 0 as n→∞, we have that there is a

positive integer n0 such that F n0 is a contraction and it followsfrom the Banach Contraction Principle that there is a unique fixedpoint for F . It is easy to see that this fixed point is a continuouslydifferentiable function and that it satisfies (2).



Since the above reasoning holds for all τ ∈ R we obtain that thesolutions of (2) are globally defined.

Now let us verify that the semigroup property is satisfied for thesolution T (t) of (2) with B = I .

Note that U(t) = T (t + s) and V (t) = T (t)T (s) are solutions of(2) satisfying U(0) = V (0) = T (s).

It follows from the uniqueness of solutions thatT (t + s) = T (t)T (s). Thus, {T (t) : t ∈ R} is a uniformlycontinuous group of bounded linear operators.



Clearly, we will be interested in more general situations since, inseveral applications, the operator A is not bounded.

Reciprocally, given a semigroup of bounded linear operators we canassociate it to a differential equation through the followingdefinition.



Definicoes e resultados basicos

DefinitionA semigroup of bounded linear operators in X is a family{T (t) : t ≥ 0} ⊂ L(X ) such that

(i) T (0) = IX ,

(ii) T (t + s) = T (t)T (s), for all t, s ≥ 0.

If besides that

(iii) ‖T (t)− IX‖L(X )t→0+−→ 0, we say that the semigroup is

uniformly continuous.

(iv) ‖T (t)x − x‖Xt→0+−→ 0, for each x ∈ X , we say that the

semigroup is strongly continuous.



DefinitionIf {T (t), t ≥ 0} ⊂ L(X ) is a strongly continuous semigroup ofbounded linear operators, its infinitesimal generator is theoperator defined by A : D(A) ⊂ X → X , where

D(A) =

{x ∈ X : lim

t→0+

T (t)x − x

texists

},

Ax = limt→0+

T (t)x − x

t, ∀ x ∈ D(A).



Example

Let A ∈ L(X ) and define eAt :=∞∑n=0

Antn

n! . Then {eAt : t ∈ R} is a

uniformly continuous group of bounded linear operators withgenerator A satisfying ‖eA t‖ ≤ e |t|‖A‖.

The series∞∑n=0

Antn

n! converges absolutely, uniformly in compact

subsets of R. In fact, since ‖An‖ ≤ ‖A‖n, we have that

‖eAt‖ ≤∞∑n=0

∥∥∥∥Antn

n!

∥∥∥∥ ≤ ∞∑n=0

(|t| ‖A‖)n

n!= e |t| ‖A‖, t ∈ R and



∞∑n=1

∥∥∥∥ Antn−1

(n − 1)!

∥∥∥∥ ≤ ‖A‖ ∞∑n=0

(|t| ‖A‖)n

n!= ‖A‖e |t| ‖A‖, t ∈ R.

Therefored

dteAt = AeAt , t ∈ R.

Also‖eAt − I‖ ≤ |t|‖A‖e |t|‖A‖ → 0

as t → 0. It follows that {T (t) : t ∈ R} is the unique solution ofx = Ax with x(0) = I . The result now follows from the previousconsiderations.



The following result very useful to obtain properties of regularity ofsemigroups.

Lemma (2)

Let φ be a continuous function which is right differentiable in theinterval [a, b). If D+φ is continuous in [a, b) then, φ iscontinuously differentiable in [a, b).

Proof: Exercise.



Every strongly continuous semigroup posses an exponential boundthat is given in the following theorem.

Theorem (1)

Suppose that {T (t), t ≥ 0} ⊂ L(X ) is a strongly continuoussemigroup. Then, there exists M ≥ 1 and β ∈ R such that

‖T (t)‖L(X ) ≤ Meβ t , ∀t ≥ 0.

For any ` > 0 we can choose β ≥ 1` log ‖T (`)‖L(X ) and then

choose M.



Proof: First note that there is an η > 0 such thatsupt∈[0,η] ‖T (t)‖<∞. This is a consequence of the fact that, for

each sequence {tn}n∈N em (0,∞) with tnn→∞−→ 0+, {T (tn)x}n∈N is

bounded for each x ∈ X and, from the Uniform BoundednessPrinciple, {‖T (tn)‖L(X )}n∈N is bounded.



Choose ` > 0 such that sup{‖T (t)‖L(X ), 0 ≤ t ≤ `}=M<∞ and

let β ≥ 1` log{‖T (`)‖L(X )} that is ‖T (`)‖L(X ) ≤ eβ` and then

‖T (n`+ t)‖ = ‖T (`)nT (t)‖ ≤ ‖T (`)‖n‖T (t)‖ ≤ Meβn`

≤ Me |β|`eβ(n`+t), 0 ≤ t ≤ `; n = 0, 1, 2 · · ·

and the result follows.



The theorem that follows characterizes completely the uniformlycontinuous semigroups of bounded linear operators through itsgenerators.

TheoremGiven a strongly continous semigroup {T (t), t ≥ 0} ⊂ L(X ), thefollowing statements are equivalent:

(a) The semigroup is uniformly continuous,

(b) The infinitesimal generator is defined in all of X ,

(c) For some A in L(X ), T (t) = et A.



Proof: If T (t) = et A for some A ∈ L(X ) the other statementswere proved in Example 1.

If the infinitesimal generator of {T (t) : t ≥ 0} is globally defined

then,{∥∥∥T (t)x−x

t

∥∥∥X

}0≤t≤1

is bounded for each x and from the

Uniform Boundedness Principle

{∥∥∥T (t)−It

∥∥∥L(X )

}0≤t≤1

is bounded

and therefore T (t)→ I in L(X ) as t → 0+.

It is enough to prove that, if T (t)t→0+−→ I in L(X ), there is

A ∈ L(X ) with T (t) = eAt .



Supposing that T (t)→ I in L(X ) as t → 0+, there is a δ > 0such that ‖T (t)− I‖L(X ) ≤ 1/2, 0 ≤ t ≤ δ. So, for t > 0,

‖T (t + h)− T (t)‖L(X ) = ‖(T (h)− I )T (t)‖L(X ) → 0,

‖T (t)− T (t − h)‖L(X ) = ‖(T (h)− I )T (t − h)‖L(X ) → 0

as h→ 0+, since ‖T (t)‖L(X ) is bounded in [0, δ].



Therefore t → T (t) : R+ → L(X ) is continuous and the integral∫ t

0T (s)ds is well defined. Besides that,

∥∥∥∥1

δ

∫ δ

0T (s)ds − I

∥∥∥∥L(X )

≤ 1/2

and therefore

(∫ δ

0T (s)ds

)−1∈ L(X ). Define

A = (T (δ)− I )

(∫ δ

0T (s)ds

)−1.



For each h > 0,

h−1(T (h)−I )∫ δ

0T (s)ds = h−1

{∫ δ+h

hT (s)ds −

∫ δ

0T (s)ds

}= h−1

∫ δ+h

δT (s)ds − h−1

∫ h

0T (s)ds

h→0+−→ T (δ)− I .

Hence h−1(T (h)− I )h→0+−→ A and h−1(T (t + h)− T (t)) =

T (t)T (h)−Ih = T (h)−I

h T (t)h→0+−→ T (t)A = AT (t).



Thus t → T (t) has a right derivative

d+

dtT (t) = T (t)A = AT (t)

which is continuous for t ≥ 0.

It follows from Lemma 2 that t 7→ T (t) is continuouslydifferentiable and, from the uniqueness of solutions for the problemx = Ax with x(0) = I , consequently, T (t) = eAt , t ≥ 0.

In view of this theorem the theory of semigroups concentrates inthe study of strongly continuous semigroups and their generators.



Our next result collects some important facts about stronglycontinuous semigroups of bounded linear operators that will befrequently used in the rest of this chapter.



TheoremLet {T (t)} be a strongly continuous semigroup. Then,

1. For each x ∈ X , t → T (t)x is continuous for t ≥ 0.

2. t → ‖T (t)‖L(X ) is lower semicontinuous and thereforemeasurable.

3. If A is the generator of T (t), then, A is densely defined andclosed. For x ∈ D(A), t 7→ T (t)x is continuouslydifferentiable and

d

dtT (t)x = AT (t)x = T (t)Ax , t > 0.

4. ∩m≥1D(Am) is dense in X .

5. For Reλ > β (β from Theorem 1), λ ∈ ρ(A) and

(λ− A)−1x =

∫ ∞0

e−λtT (t)xdt, ∀x ∈ X



Proof: 1. The continuity of t 7→ T (t)x is a consequence ofTheorem 1 and from the fact that, t > 0 and x ∈ X ,

‖T (t + h)x − T (t)x‖X = ‖(T (h)− I )T (t)x‖Xh→0+−→ 0,

‖T (t)x − T (t − h)x‖X ≤ ‖T (t − h)‖L(X )‖T (h)x − x‖Xh→0+−→ 0.



2. Let us show that {t ≥ 0 : ‖T (t)‖L(X ) > b} is open in [0,∞) foreach b which implies the statement.

But ‖T (t0)‖L(X ) > b implies that there exists x ∈ X with‖x‖X = 1 such that ‖T (t0)x‖ > b.

It follows from 1. that ‖T (t)x‖ > b for all t sufficiently close to t0,so ‖T (t)‖L(X ) > b for t in a neighborhood of t0 and the resultfollows.



3. Let x ∈ X and for ε > 0, xε = 1ε

∫ ε

0T (t)x dt then, xε → x as

ε→ 0+ and, for h > 0,

h−1(T (h)xε − xε) =1

εh

{∫ ε+h

εT (t)x dt −

∫ h

0T (t)x dt

}h→0+−→ 1

ε(T (ε)x − x).



So xε ∈ D(A). It will be an immediate consequence of 5. that A isclosed since (λ− A)−1 ∈ L(X ).

If x ∈ D(A) it is clear that

d+

dtT (t)x = lim

h→0+

1

h{T (t + h)x − T (t)x} = AT (t)x = T (t)Ax

is continuous and any function with right derivative continuous iscontinuously differentiable.



4. Let φ : R→ R be a function in C∞(R) and φ(t) = 0 in aneighborhood of t = 0 and also for t sufficiently large, let x ∈ X

and f =

∫ ∞0φ(t)T (t)x dt.

It follows easily from

h−1(T (h)f − f ) = h−1∫ ∞h

(φ(t − h)− φ(t))T (t)x dt

that f ∈ D(A) and that Af = −∫ ∞0φ′(t)T (t)x dt.



Since −φ′ satisfies the same conditions that φ,

Amf = (−1)m∫ ∞0φ(m)(t)T (t)x dt

for all m ≥ 1 e f ∈ ∩m≥1D(Am).



To show that such set of points is dense in X , choose φ above

satisfying also

∫ ∞0φ(t)dt = 1 then, if

fn =

∫ ∞0

nφ(nt)T (t)xdt =

∫ ∞0φ(s)T (s/n)xds, n = 1, 2, 3, · · · ,

we have that fn ∈ ∩m≥1D(Am) and fn → x as n→∞.



5. Define R(λ) ∈ L(X ) by

R(λ)x =

∫ ∞0

e−λtT (t)xdt.

Note that ‖R(λ)‖≤ MReλ−β for all λ ∈ C such that Reλ>β and

‖T (t)‖L(X ) ≤ Meβt .



Let x ∈ X and h > 0

h−1(T (h)− I )R(λ)x = R(λ)T (h)x − x

h

= h−1[∫ ∞

he−λt+λhT (t)x −

∫ ∞0

e−λtT (t)x

]= h−1

[−∫ h

0eλ(h−t)T (t)x +

∫ ∞0

(eλh − 1)e−λtT (t)x

]h→0+−→ −x + λR(λ)x .

(3)



Therefore R(λ)x ∈ D(A) and (λ− A)R(λ)x = x , and λ− A isonto. Also, if x ∈ D(A) then, integrating by parts,R(λ)Ax = λR(λ)x − x = AR(λ)x .

It follows that (λ−A)R(λ)x = x = R(λ)(λ−A)x for all x ∈ D(A)and λ−A is also injective. Hence (λ−A) is a bijection from D(A)onto X with bounded inverse R(λ) and the proof is complete.



TheoremLet {T (t), t ≥ 0} and {S(t), t ≥ 0} strongly continuoussemigroups with generators A and B respectively. If A = B thenT (t) = S(t), t ≥ 0.



Proof: Let x ∈ D(A) = D(B). From Theorem 3 it follows easilythat the function s 7→ T (t − s)S(s)x is differentiable and that

d

dsT (t − s)S(s)x = −AT (t − s)S(s)x + T (t − s)BS(s)x

= −T (t − s)AS(s)x + T (t − s)BS(s)x = 0.

Therefore s 7→ T (t − s)S(s)x is constant and in particular itsvalues at s = 0 and s = t are the same, that is, T (t)x = S(t)x .

This holds for all x ∈ D(A) and since D(A) is dense in X andS(t), T (t) are bounded, T (t)x = S(t)x for all x ∈ X .


Documents

SMA 5878 Functional Analysis II€¦ · Semigroups and Their Generators To beter explain this remark we consider rst the case A 2L(X). In this case, the semigroup t 7!T(t) is the