Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift

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148 W. Desch and S.-O. Londen

Proof. First let u

=0sothatu(t)=S

α,1

. We will apply Lemma 5.4 with

β =1.Noticethata

(t) = 1, so that v(t)=S

α,1

−1

ζ<δ

. To prove Part (1),

put ζ = δ

= 0 and take any η<1. By Lemma 5.4 (3) we have, pointwise for all

ω ∈ Ω,

u(t, ω)=A



(t − s)u(s, ω) ds + u

Moreover, with these parameters, v(t)=u(t), and (5.10) holds. By Lemma 5.4,

Part (4) and (5.11), u(·,ω) admits a fractional time derivative of order η with

values in L

(B; R), such that the following estimate holds:

D

u(t, ω)

(B;R)

≤ Mt

−η

u

(ω)

(B;R)

Taking convolution with t

η−1

/Γ(η), we obtain with a suitable constant M

inde-

pendent of t,

u(t, ω)

(B;R)

≤ M

u

(ω)

(B;R)

Integrating with respect to ω we obtain

u(t)

(Ω×B;R)

≤ M

u



(Ω×B;R)

Thus u ∈ L

∞

([0,T]; L

(Ω × B; R)).

To prove Part (2), take again β = 1 and let (5.12) hold. Then (5.10) (with

instead of δ) holds a fortiori. Let v(t)=S

α,1

− 1

ζ<δ

.Fixω ∈ Ω. By

Lemma 5.4 (4), v(·,ω) has a fractional time derivative of order η which satisﬁes

an estimate

D

v(t, ω)

D((−A)

)

≤ Mt

αδ

−η−αζ

u

(ω)

D((−A)

)

Since, by (5.12), αδ

− η − αζ > −

, the estimate above implies

D

v(·,ω)

([0,T ];D((−A)

))

≤ M

u

(ω)

D((−A)

)

Integrating again with respect to Ω, we obtain Part (2).

The case u

=0,u

=0,u(t)=S

α,2

is treated similarly with β =2.Notice

that a

(t)=t.Intheendwecancombinebothcases. 

Lemma 5.7 (Contribution of f ). Let A satisfy Hypothesis 3.1,let(Ω, F, P) be

a probability space, p ∈ [2, ∞).Letα ∈ (0, 2), γ>0, 0 <T<∞,andf ∈

([0,T] × Ω; L

(B; R)).

1) For almost t ∈ [0,T], the following integral exists in L

(B; R),pointwisefor

almost all ω ∈ Ω,aswellasinL

(Ω; L

(B; R)):

u(t, ω)=



α,γ

(t − s)f(s, ω) ds. (5.14)

Stochastic Integral Equations 149

Moreover, u ∈ L

([0,T] ×Ω; L

(B; R)),andforalmostallω ∈ Ω and almost

all t ∈ [0,T],



(t −s)u(s, ω) ds ∈D(A),

u(t, ω)=A



(t −s)u(s, ω) ds +



(t −s)f(s, ω) ds.

2) Suppose, in addition, that f ∈ L

([0,T] × Ω; D((−A)



)) with some  ∈ [0, 1],

let η ∈ (−1, 1), ζ ∈ [0, 1] be such that

η + αζ < γ + α. (5.15)

Put

v(t)=



u(t) if ζ ≥ ,

u(t) −



(t −s)f(s) ds if ζ<.

Then, if η>0, the function t → v(t) ∈ L

(Ω; D((−A)

)) has a fractional

derivative of order η in L

([0,T] × Ω; D((−A)

)).Ifη<0, the function

t → v(t) ∈ L

(Ω; L

(B; R)) has a fractional integral of order −η with values

in L

(Ω; D((−A)

)).Ifη =0, we deﬁne D

v = v. In either case there exists

a constant M

T,Leb

dependent on A, T, p, α, γ, , ζ, η such that

D

v

([0,T ]×Ω;D((−A)

))

≤ M

T,Leb

f

([0,T ]×Ω;D((−A)



))

Moreover, the constant M

T,Leb

can be made arbitrarily small by taking the

time interval [0,T] suﬃciently short.

Proof. The function t →



(t − s)

γ−1

f(s)

(Ω×B,R)

ds is the convolution of an

-function and an L

-function, therefore it is in L

([0,T], R). From (5.11) with

δ = ζ = η =0weobtainS

α,γ

(t)

(B;R)→L

(B;R)

≤ Mt

γ−1

. Consequently, the

integral

u(t)=



α,γ

(t −s)f(s) ds

exists as an integral in L

(Ω × B; R) for a.e. t,andu ∈ L

([0,T] × Ω,L

(B,R)).

By standard arguments the integral (5.14) exists also in L

(B; R) for a.e. ω ∈ Ω

and a.e. t ∈ [0,T]. Now (5.9) implies (almost everywhere in Ω and [0,T])

u(t) −



(t − s)f(s, ω) ds =



α,γ

(t −s)f(s, ω) −a

(t −s)f(s, ω)] ds



t−s

(σ)S

α,γ

(t −s −σ)f (s, ω) dσ] ds.

We use the closedness of A and interchange the order of integrals to obtain

u(t) −



(t −s)f(s, ω) ds = A



(σ)u(t − σ, ω) dσ.

This proves Part (1) of the lemma.

150 W. Desch and S.-O. Londen

To prove Part (2), let η,ζ,  be such that (5.15) holds. For shorthand put

V (t)x =



α,γ

x if  ≤ ζ,

α,γ

(t)x −a

x]else.

From (5.11) with β replaced by γ,andδ replaced by ,wehave

V (t)x

D((−A)

)

≤ Mt

(γ+α)−(η+αζ)−1

x

D((−A)



)

We obtain by a straightforward convolution argument that





V (t − s)f(s) ds

([0,T ]×Ω;D((−A)

))

ds ≤ M

t,Leb

f

(Ω,D((−A)



))

with

T,Leb

= M



(γ+α)−(η+αζ)−1

dt.

Clearly, M

T,Leb

converges to 0 as T → 0. All we have to show is that in fact

v(t)=



V (t − s)f(s) ds.

We treat the case η>0, >ζ, the other cases are done similarly. The deﬁnition

of V (t)x yields



(s)V (t −s)xdx= S

α,γ

(t)x − a

(t)x.

Fubini’s Theorem implies



(s)



t−s

V (t − s −σ)f (σ) dσ ds =



t−σ

(s)V (t −σ − s)f (σ) ds dσ



α,γ

(t − σ) −a

(t −σ)] f (σ) dσ = v(t).

Thus v(t), considered as a function with values in D((−A)

), admits a fractional

derivative of order η which is V ∗ f . 

The following lemma is the key to estimate the contribution of the stochastic

integral:

Lemma 5.8. Let A satisfy Hypothesis 3.1, p ∈ [2, ∞).Letα ∈ (0, 2), β>

ζ ∈ [0, 1] and η ∈ (−1, 1), such that (5.3) holds, i.e.,

+ η + αζ < β.LetT>0.

Then there exists a constant

T,Ito

> 0 depending on A, p, T, α, β, η, ζ such that

for all h ∈ L

([0,T]; L

(B; l

)),







|(−A)

α,β

(t − s)h(s, x)|



dΛ(x) dt

≤

T,Ito



|h(s, x)|

dΛ(x) ds.

Stochastic Integral Equations 151

Moreover, the constant

T,Ito

can be made arbitrarily small by taking the time

interval [0,T] suﬃciently short.

Proof. Write V (t):=(−A)

α,β

(t) and notice that by (5.11) (with δ =0),

V (t)

(B,l

)→L

(B,l

)

≤ Mt

β−(η+αζ)−1

First assume that p>2. Notice that

and

p−2

are conjugate exponents. Take

f :[0,T] × B → R

such that



p−2

(t, x) dΛ(x) dt =1.Weestimate



f(t, x)



|V (t −s)h(s, x)|

ds dΛ(x) dt



f(t, x) |V (t − s)h(s, x)|

dΛ(x) ds dt

≤



f(t, x)

p−2

dΛ(x)

p−2



|V (t − s)h(s, x)|

dΛ(x)

ds dt

≤



f(t, ·)

p−2

(B;R)



V (t − s)

(B;l

)→L

(B;l

)

h(s, ·)

(B;l

)

ds dt

≤



f(t, ·)

p−2

(B;R)

p−2



V (t − s)

(B;l

)→L

(B;l

)

h(s, ·)

(B;l

)

ds|

Thus







|V (t −s)h(s, x)|



dΛ(x) dt

≤



V (t − s)

(B;l

)→L

(B;l

)

h(s, ·)

(B;l

)

ds|

(5.16)

For p = 2, the estimate (5.16) is obvious. In either case, we obtain (by estimating

the convolution with respect to s)







|V (t − s)h(s, x)|



dΛ(x) dt

≤



V (s)

(B;l

)→L

(B;l

)



h(s, .)

(B;l

)

≤ M



2(β−(η+αζ)−1)

ds h

([0,T ];L

(B;l

))

152 W. Desch and S.-O. Londen

By (5.3) we infer that s

2(β−(η+αζ)−1)

is integrable on [0,T]sothat

T,Ito



2(β−(η+αζ)−1)

is ﬁnite and converges to 0 as T → 0+. 

Lemma 5.9 (Contribution of g). Let A satisfy Hypothesis 3.1, and let the probability

space (Ω, F, P) and the Wiener processes w

be as in Hypothesis 3.2.LetT>0,

2 ≤ p<∞, α ∈ (0, 2),andβ>

.Letg ∈ L

([0,T] × Ω; L

(B; l

)) and {g

} be

a sequence approximating g in the sense of Lemma 4.1, where the values of g

are

in D(A) ∩ L

(B; R) ∩ L

∞

(B; R).Leta

g be given by Deﬁnition 4.5.Forj ∈ N

put

(t)=

k=1



α,β

(t −s)g

(s) dw

1) The limit u(t) = lim

j→∞

(t) exists in L

([0,T] × Ω; L

(B; R)). Moreover,

for almost all t ∈ [0,T] and almost all ω ∈ Ω,

u(t, ω)=A



(t − s)u(s, ω) ds +(a

g)(t, ω).

2) Suppose 0 ≤ ζ ≤ 1 and η ∈ (−1, 1) are such that

η + αζ +

<β. (5.17)

Then, if η>0, the function u :[0,T] → L

(Ω, D((−A)

)) has a fractional

derivative of order η.Ifη<0,thenu :[0,T] → L

(Ω; L

(Ω, R)) has a

fractional integral of order −η with values in L

(Ω; D((−A)

)).Ifη =0,we

denote D

u = u. In either case there exists a constant M

T,Ito

dependent on

A, p, α, β, η, ζ such that

D

u

([0,T ]×Ω;D((−A)

))

≤ M

T,Ito

g

([0,T ]×Ω;L

(B;l

))

Moreover, the constant M

T,Ito

can be made arbitrarily small by choosing the

time interval [0,T] suﬃciently short.

Proof. First, let h ∈ L

([0,T] × Ω; L

(B; l

)) be of the elementary structure like

the g

in Lemma 4.1. Evidently, the following integral exists

∞

k=1



α,β

(t − s)h

(s) dw

∞

k=1

∞

i=1



i−1

α,β

(t −s)h

where τ

are suitable stopping times, h

∈D(A)∩L

(B; R) ∩L

∞

(B; R), and both

sums are in fact only ﬁnite sums. For η ∈ (−1, 1), ζ ∈ [0, 1], satisfying (5.17),

Stochastic Integral Equations 153

put V (t)x =(−A)

α,β

(t)x. We apply Lemma 4.3 and integrate for t ∈ [0,T].

Subsequently we apply Lemma 5.8:



∞

k=1



V (t − s)h

(s) dw

(B;R)

dP(ω) dt (5.18)

≤ M







|[V (t − s)h(s, ω)](x)|



dP(ω) dΛ(x) dt

≤ M

T,Ito

h

([0,T ]×Ω;L

(B;l

))

In particular, with ζ = η =0,andh = g

−g

,wehave

u

− u



([0,T ]×Ω;L

(B;R))

≤ Mg

−g



([0,T ]×Ω;L

(B;l

))

so that {u

} is a Cauchy sequence in L

([0,T] × Ω; L

(B; R)) and has a limit

u. Without loss of generality, taking a subsequence, if necessary, we may assume

that u

converges also pointwise for almost all t ∈ [0,T].Againweusethesimple

structure of g

,inparticularthatg

(t, ω) ∈D(A). From (5.9) and the Stochastic

Fubini Theorem we obtain



α,β

(t − σ)g

(σ, ω) − a

(t − σ)g

(σ, ω)] dw



t−σ

(s)S

α,β

(t −σ − s)g

(σ, ω) ds] dw

= A



t−σ

(s)S

α,β

(t − σ − s)g

(σ, ω) ds dw

= A



(s)



t−s

α,β

(t − σ − s)g

(σ, ω) dw

ds.

Taking the sum over k =1,...,j we obtain

(t, ω) − a

g

(t, ω)=A



(s)u

(t −s, ω) ds.

Taking limits for j →∞(pointwise a.e. in [0,T]), and using the closedness of A

we have for almost all t ∈ [0,T]

u(t, ω) −a

g(t, ω)=A



(s)u(t −s, ω) ds.

Thus Part (1) of the lemma is proved.

To prove Part (2), let η ∈ (−1, 1) and ζ ∈ [0, 1] satisfy (5.17). With V (t)x =

(−A)

α,β

x and h = g

, we obtain from (5.18),



k=1

V (t − s)g

(s) dw

([0,T ]×Ω;L

(B;R))

≤ M

T,Ito

g



([0,T ]×Ω;L

(B;l

))

154 W. Desch and S.-O. Londen

with a suitable constant M

T,Ito

whichconvergesto0asT → 0. We have to show

that in fact

k=1



V (t − s)g

(s) dw

=(−A)

(t)

First let η>0. By deﬁnition we know that



(s)V (t − s)xds =(−A)

α,β

(t)x.

Taking integrals and using the Stochastic Fubini Theorem, we obtain

(−A)

(t)=

k=1



(−A)

α,β

(t − σ)g

(σ, ω) dw

k=1



t−σ

(s)V (t − σ − s)g

(σ, ω) ds dw



(s)

k=1



t−s

V (t −s −σ)g

(σ, ω) dw

Thus (−A)

has a fractional derivative of order η which is Vg

.Takingthe

limit for j →∞we infer that D

(−A)

u = Vg.Nowletη<0. Similarly as

above, the Stochastic Fubini Theorem yields

k=1



V (t − σ)g

(σ) dw

=(−A)



−η

(s)u

(t −s) ds.

Again we take the limit for j →∞and use the closedness of A, to see that the

fractional integral D

u takes values in D((−A)

)with(−A)

u = Vg. 

6. The semilinear equation

This section is devoted to the proof of Theorems 3.6, 3.7, 3.11, and Corollary 3.8.

Lemma 6.1. Let (Ω, F, P) be a probability space, let A satisfy Hypothesis 3.1,and

let F and G satisfy Hypotheses 3.4 and 3.5.Wedeﬁnetheoperators

: L

([0,T] × Ω; D((−A)

)) → L

([0,T] × Ω; D((−A)



)),

: L

([0,T] × Ω; D((−A)

)) → L

([0,T] × Ω; L

(B; l

))

v](t, ω):=F (t, ω, v(t)), [N

v](t, ω):=G(t, ω, v(t)).

(1) Then N

and N

are well deﬁned and Lipschitz continuous with Lipschitz

constants M

, M

, respectively.

Stochastic Integral Equations 155

(2) Let F

satisfy Hypotheses 3.4, 3.5,and3.9.Letv ∈ L

([0,T] ×

Ω; D((−A)

)).Then

N

v −N

v

([0,T ]×Ω;D((−A)



))

≤μ

ΔF



([0,T ]×Ω;R)

N

v −N

v

([0,T ]×Ω;L

(B;l

))

≤μ

ΔG



([0,T ]×Ω;R)

Here the constants M

, M

and the functions μ

ΔF

and μ

ΔG

are as in Hypothe-

ses 3.4, 3.5,and3.9.

Proof. These are straightforward estimates. 

Lemma 6.2. Let the assumptions of Theorem 3.6 hold, in addition assume that

≤ θ, δ

≤ θ.Forv ∈ L

([0,T] × Ω; D((−A)

)) let T

[F,G,u

]

v :[0,T] × Ω →

(B; R) be the unique solution u of

u(t, ω)=A



(t −s)u(s, ω) ds + u

+ tu

∞

k=1



(t − s)G

(s, ω, v(s)) dw



(t −s)F (s, ω, v(s)) ds.

in the sense of Proposition 5.1.

(1) Then T

[F,G,u

]

is well deﬁned as a nonlinear operator from L

([0,T] ×

Ω; D((−A)

)) into L

([0,T]×Ω; D((−A)

)). Moreover, T

[F,G,u

]

is globally

Lipschitz continuous with a Lipschitz constant dependent on A, p, T , α, β,

γ, , θ, M

, M

(2) There exists an equivalent norm on L

([0,T] × Ω; D((−A)

)), such that the

Lipschitz constant of T

[F,G,u

]

is smaller than 1. This norm depends on

T,p,A,α,β,γ,θ,,M

(3) There exists a constant M, depending on A, T, p, M

,α,β,,θ,δ

,δ

such that the following Lipschitz estimate holds:

If F

satisfy Hypotheses 3.4, 3.5,and3.9,ifu

0,1

, u

0,2

are in

(Ω; D((−A)

)) and u

1,1

1,2

∈ L

(Ω; D((−A)

)), measurable with respect

to F

, then for any v ∈ L

([0,T] × Ω; D((−A)

)) we have

T

0,1

1,1

]

v −T

0,2

1,2

]

v

([0,T ]×Ω;D((−A)

))

≤ M

u

0,1

− u

0,2



(Ω;D((−A)

))

+ u

1,1

−u

1,2



(Ω;D((−A)

))

+ μ

ΔF



([0,T ]×Ω;R)

+ μ

ΔG



([0,T ]×Ω;R)

(4) T

[F,G,u

]

has a unique ﬁxed point u

[F,G,u

]

∈ L

([0,T] × Ω; D((−A)

)).

Moreover, there exists a constant M dependent on A, p, T , M

, M

, α, β,

, θ, δ

, δ

such that the following Lipschitz estimate holds:

156 W. Desch and S.-O. Londen

If u

i,j

, F

, G

are as in (3),then

u

0,1

1,1

]

−u

0,2

1,2

]



([0,T ]×Ω;D((−A)

))

≤ M

u

0,1

− u

0,2



(Ω;D((−A)

))

+ u

1,1

−u

1,2



(Ω;D((−A)

))

+ μ

ΔF



([0,T ]×Ω;R)

+ μ

ΔG



([0,T ]×Ω;R)

Proof. We recall Proposition 5.2 with η =0andζ replaced by θ. Notice that the

conditions (5.2), and (5.3), (5.4), (5.5) are satisﬁed. Let u solve

u = Aa

∗ u + a

∗ f + a

g+ u

+ tu

Notice that with the present choice of coeﬃcients the function v in (5.6) is simply

u.Thusu ∈ L

([0,T] × Ω; D((−A)

)) with

u(t)

([0,T ]×Ω;D((−A)

))

(6.1)

≤ M

init

u



(Ω;D((−A)

))

+ u



(Ω;D((−A)

))

+ M

T,Leb

f

([0,T ]×Ω;D((−A)



))

+ M

T,Ito

g

([0,T ]×Ω;L

(B,l

))

Given v ∈ L

([0,T] × Ω; D((−A)

)), we put f = N

v and g = N

v as in

Lemma 6.1. Then f ∈ L

([0,T] × Ω; D((−A)



)) and g ∈ L

([0,T] × Ω; L

(B; l

)).

Thus, by (6.1), u = T

[F,G,u

]

v ∈ L

([0,T]×Ω; D((−A)

)). In particular for v =0

we have

T

[F,G,u

]

(0)

([0,T ]×Ω;D((−A)

))

≤ M

init

u



(Ω,D((−A)

))

+ u



(Ω;D((−A)

))

+ M

T,Leb

F,0

+ M

T,Ito

G,0

We could immediately get a Lipschitz estimate for T

[F,G,u

]

by (6.1), but we will

get a better (contraction) estimate in an equivalent norm below.

To prove (2), we recall from Proposition 5.2 that M

T,Leb

and M

T,Ito

can be

taken arbitrarily small, if the time intervals are suﬃciently short. In particular,

there exists m ∈ N such that

T/m,Leb

+ M

T/m,Ito

With some κ>0 to be speciﬁed below, we deﬁne for v ∈ L

([0,T]×Ω; D((−A)

)),

|||v||| :=

q=1



Tq/m

T (q−1)/m



v(t, ω)

D((−A)

)

dP(ω) dt

1/p

For q =1,...,m we put

(t, ω, v):=I

(q−1)T/m≤t<qT/m

(t) F (t, ω, v(t)),

(t, ω, v):=I

(q−1)T/m≤t<qT/m

(t) G(t, ω, v(t)).

If v, ˜v ∈ L

([0,T] × Ω; D((−A)

)), then

[F,G,u

]

v −T

[F,G,u

]

˜v =

q=1

Stochastic Integral Equations 157

where w

solves

= Aa

∗w

+ a

∗ [F

(v) − F

(˜v)] + a

 [G

(v) − G

(˜v)].

Now w

=0on[0,

T (q−1)

]. Lemma 6.1(1) and (6.1) imply





Tq/m

T (q−1)/m



w

(t, ω)

D((−A)

)

dP(ω) dt



1/p

≤ M

T/m,Leb





Tq/m

T (q−1)/m



v(t, ω) − ˜v(t, ω)

D((−A)

)

dP(ω) dt



1/p

+ M

T/m,Ito





Tq/m

T (q−1)/m



v(t, ω) − ˜v(t, ω)

D((−A)

)

dP(ω) dt



1/p

≤





Tq/m

T (q−1)/m



v(t, ω) − ˜v(t, ω)

D((−A)

)

dP(ω) dt



1/p

On the intervals

(r−1)T

with r>qwe have the estimate





Tr/m

T (r−1)/m



w

(t, ω)

D((−A)

)

dP(ω) dt



1/p

≤





w

(t, ω)

D((−A)

)

dP(ω) dt



1/p

≤ M





Tq/m

T (q−1)/m



v(t, ω) − ˜v(t, ω)

D((−A)

)

dP(ω) dt



1/p

with M = M

T,Leb

+ M

T,Ito

.Wechooseκ ∈ (0, 1) suﬃciently small, such

that M

∞

r=1

. We have therefore

|||w

||| =

r=q



Tr/m

T (r−1)/m



w

(t, ω)

D((−A)

)

dP(ω) dt

1/p

≤

+ M

r=q+1

r−q



Tq/m

T (q−1)/m

v(t, ω) − ˜v(t, ω)

D((−A)

)

dP(ω) dt

1/p

≤



Tq/m

T (q−1)/m

v(t, ω) − ˜v(t, ω)

D((−A)

)

dP(ω) dt

1/p

Summing for q =1,...,m we obtain

|||T

[F,G,u

]

v −T

[F,G,u

]

˜v||| ≤

q=1

|||w

|||

≤

q=1



Tq/m

T (q−1)/m

v(t, ω) − ˜v(t, ω)

D((−A)

)

dP(ω) dt

1/p

|||v − ˜v|||.

Part (3) is a straightforward application of (6.1) and Lemma 6.1 (2).