[PPT] - Strong solutions of semilinear SPDEs with unbounded diffusion 1 PowerPoint Presentation

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Strong solutions of semilinear SPDEs with unbounded diffusion1

Florian Bechtold

LPSM – Sorbonne Université

Les probabilités de demain 14/06/2019 Université Paris Diderot

1This project has received funding from the European Union’s Horizon 2020 research and innovation programme under the Marie Skłodowska-Curie grant agreement No 754362. Florian Bechtold 1/ 21

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SPDEs as infinite dimensional SDEs

For a, b ∈ R the stochastic differential equation

dXt

= aXtdt + bXtdWt X0 = x ∈ R admits a unique solution (geometric Brownian motion), which satisfies moreover the equation Xt = eatx +

t

ea(t−s)bXsdWs To consider the problem in infinite dimensions, do the replacements Real number a Linear (unbounded) operator A Exponential of real number a Semigroup (St)t generated by A Real number b Nonlinear bounded operator B Brownian motion Cylindrical Brownian motion

Florian Bechtold 2/ 21

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Cylindrical Brownian motion and stochastic integration in infinite dimensions

Formally, for a Hilbert space U with ONB (ek)k, set for a sequence (βk)k of independet Brownian motions as cylindrical Brownian motion Wt := ”

∞

k=1

ekβk

t ” ∈ U

Problem: This is not an element of L2(Ω, U). Suppose for B ∈ L(U, H) and for an ONB (fk)k of H, there exists (λk)k ⊂ R such that Bu =

∞

k

λkfku, ek Now if (λk)k ∈ ℓ2, then B ∈ L2(U, H) and E[||

t

BdWs||2

H] = E[||

k

λkfkβk

t ||2 H] =

k

λ2

kt < ∞

Florian Bechtold 3/ 21

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The mild formulation for semilinear SPDEs

Definition (Mild solution) Let H be Hilbert, A : D(A) ⊂ H → H be generator of a strongly continuous semigroup (St)t≤T of operators and B : H → L2(U, H), let W be a cylindrical Wiener process on U. Then a progressively measurable process (ut)t≤T with values in H is called mild solution to dut = Autdt + B(ut)dWt, u0 = ξ ∈ H if P-almost surely for all t ≤ T ut = Stξ +

t

St−sB(us)dWs

=:(K(u))t

holds as an equality in H.

Florian Bechtold 4/ 21

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The key tool: A maximal inequality

Lemma (Maximal inequality for stochastic convolutions) Let H be Hilbert, u ∈ Lp(Ω, L∞([0, T], H)), let B : H → L2(U, H), let W be a U-cylindrical Brownian motion, then for all contraction semi groups S on H one has E[sup

t≤T

||

t

St−sB(us)dWs||p

H] ≤ CT p/2E[sup t≤T

||B(us)||p

L2(U,H)].

Corollary (Existence and uniqueness of mild solutions) If B : H → L2(U, H) is Lipschitz and of linear growth, then K : Lp(Ω, L∞([0, T], H)) → Lp(Ω, L∞([0, T], H)) admits a unique fixed point for T sufficiently small, i.e. the corresponding SPDE admits a unique mild solution.

Florian Bechtold 5/ 21

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Our problem

It is natural to ask, if one can also consider unbounded diffusions, i.e. problems of the form

du

= ∆udt + (−∆)δ/2B(ut)dWt u(0) = u0 ∈ Lp(TN) for δ ∈ [0, 1). The corresponding mild formulation becomes ut = Stu0 +

t

(−∆)δ/2St−sB(us)dWt Can one replace the maximal inequality for stochastic convolutions with something more general, adapted to this context?

Florian Bechtold 6/ 21

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A maximal inequality and mild solutions for δ ∈ [0, 1)

Theorem (A maximal inequality) Let H be Hilbert. Let A : D(A) ⊂ H → H be generator of a strongly continuous contraction semi-group of operators (St)t≤T. Let B : H → L2(U, H) satisfy ||B(z)||2

L2(U,H) ≤ C(1 + ||z||2 H).

Let W be a cylindrical Wiener process on U. Let δ ∈ [0, 1), T > 0. Then for p >

2 1−δ and every progressively measurable

u ∈ Lp(Ω, L∞([0, T], H)) there holds E[sup

t≤T

||

t

(−A)δ/2St−sB(us)dWs||p

H]

≤ Cδ,pT

p 2 (1−δ)E[sup

t≤T

||B(ut)||p

L2(U,H)]

and the associated convolution process is continuous.

Florian Bechtold 7/ 21

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A maximal inequality and mild solutions for δ ∈ [0, 1)

Theorem (General existence theorem) Let H be Hilbert. Let B : H → L2(U, H) be Lipschitz continuous, satisfying the growth condition previously stated. Then for δ ∈ [0, 1) and p >

2 1−δ the SPDE

dut = ∆utdt + (−∆)δ/2B(ut)dWt u(0) = u0 ∈ H admits a unique mild solution, meaning there exists a unique progressively measurable process u ∈ Lp(Ω, L∞([0, T], H)) such that ut = Stu0 +

t

(−∆)δ/2St−sB(us)dWs

=:(K(u))t

which is P-almost surely continuous in time.

Florian Bechtold 8/ 21

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A maximal inequality and mild solutions for δ ∈ [0, 1)

Idea of the proof: Use Banach fixed-point theorem in the space ZH := Lp(Ω, L∞([0, T], H)) exploiting the maximal inequality ||K(u) − K(v)||p

ZH = E[sup s≤T

||

s

(−∆)δ/2Ss−rB(ur) − B(vr)dWr||p

H]

≤ Cδ,pT p/2(1−δ)E[sup

s≤t

||B(us) − B(vs)||p

L2(U,H)]

≤ Cδ,pT p/2(1−δ)E[sup

s≤t

||us − vs||p

H]

= Cδ,pT p/2(1−δ)||u − v||p

ZH

For δ < 1, one can choose T sufficiently small for K to become a contraction.

Florian Bechtold 9/ 21

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Nemytskii operators and strong solutions for δ ∈ [0, 1)

While the previous theorem works for infinite dimensional noise, we restrict ourselves in the following to finite dimension noise, i.e. U is of finite dimension, for some basis (ei)e≤d we have B(us)dWs =

d

i=1

Bi(us)dW i

s

=

d

i=1

Bi(us)dWs, ei

The question thus becomes: under which conditions on the

functions (Bi)i≤d does the associated Nemytskii operator have the required properties, more precisely: Given the Hilbert space L2(TN), under what conditions on a real valued function Bi : R → R is the operator B : L2(TN) → L2(U, L2(TN)) z →

u →

d

i=1

Bi(z)u, ei

Lipschitz and satisfying the growth condition?

Florian Bechtold 10/ 21

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Nemytskii operators and strong solutions for δ ∈ [0, 1)

Lemma (A Nemytskii operator type result for L2(TN)) Let U be a d-dimensional Hilbert space with ONB (ei)i≤d. Let B1, . . . , Bd ∈ C1(R; R) be of bounded derivative and satisfy the growth condition

d

i=1

|Bi(ξ)|2 ≤ C(1 + |ξ|2) Then the associated Nemytskii operator is well defined, Lipschitz and satisfies the growth condition ||B(z)||2

L2(U,L2(TN)) ≤ C(1 + ||z||2 L2(TN)).

Florian Bechtold 11/ 21

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Nemytskii operators and strong solutions for δ ∈ [0, 1)

Problem in passing from Lebesgue to Sobolev spaces: Nemytskii operators lose Lipschitz continuity. Yet, the growth condition can be preserved, which is sufficient Picard iterations uniformly bounded (geometric series for T small) Extraction of a weak-*-convergent subsequence (in Sobolev space) Argue via uniqueness of limit

Florian Bechtold 12/ 21

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Thanks for your attention!

Florian Bechtold 13/ 21

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Starting point

In her first publication, Hofmanova considered a class of problems comprising in particular

du

= ∆udt + divF(u)dt + d

k=1 Bk(u)dW k t

u(0) = u0 By using the mild formulation, she can Control the divergence by the smoothing of the semigroup Establish regularity in space of the unique solution thanks to the Sobolev embedding theorem (requires the theory of stochastic integration in 2-smooth Banach spaces)

Florian Bechtold 14/ 21

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Hofmanova’s setting

The mild formulation corresponding to Hofmanova’s problem du = ∆utdt + divF(ut)dt + B(ut)dWt u0 = ξ is given by ut = Stξ +

t

St−sdivF(us)ds +

t

St−sB(us)dWs = Stξ +

t

(−∆)1/2St−s(−∆)−1/2divF(us)ds +

t

St−sB(us)dWs Key idea in Hofmanova’s paper to deal with Bochner integral: ||(−∆)1/2St−s|| ≤ C

1 √t−s

(−∆)−1/2divF(·) is a bounded operator Use trinangle inequality, get control via time horizon T (→ think contraction for fixed point argument)

Florian Bechtold 15/ 21

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A maximal inequality and mild solutions for δ ∈ [0, 1)

Idea of the proof: Use factorization method by Da Prato, Zabczyk, Kwapien exploiting the identity

t

σ

(t − s)α−1(s − σ)−αds = π sin πα for α ∈ (0, 1), one has due to Fubini’s theorem

t

(−A)δ/2St−sB(us)dWs =sin πα π

t

(t − s)α−1(−A)δ/2St−s

s

(s − σ)−αSs−σB(uσ)dWσ

=:Ys

ds Moreover, use the inequality ||(−A)δ/2St−s|| ≤ C 1 (t − s)δ/2

Florian Bechtold 16/ 21

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The limit case δ = 1: Heuristics

What about the limit case δ = 1, i.e. problems of the form

du

= ∆udt + (−∆)1/2 d

k=1 Bk(ut)dW k t

u(0) = u0 ∈ Lp(TN) Heuristically, looking at the problem in Fourier basis, the Laplacian makes us gain k2, the noise potentially lose k, yet due to Itô Isometry, one has to square, so we lose potentially in order k2. In

rder for the equation to still admit a solution, one needs to thus

require that the non-linearities Bk are "small" in the following sense.

Florian Bechtold 17/ 21

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The limit case δ = 1: Uniform bounds

In particular, (uδ)δ ⊂ L2(Ω × [0, T], TN), i.e. using Itô’s formula ||ut||2

L2(TN) = ||u0||2 L2(TN) − 2

t

TN |∇u|2dxds + Martingale

+

t

d

i=1
TN |(−∆)δ/2Bi(us)|2dxds

Lemma Let (Bk)k≤d be as before and satisfy additionally

d

k=1

||(−∆)δ/2Bk(u)||2

L2(TN)

≤ C||∇u||2

L2(TN)

for C < 2. Then (uδ)δ is uniformly bounded in L2(Ω × [0, T] × TN)

Florian Bechtold 18/ 21

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The limit case δ = 1: Using Banach-Alaoglu

By Banach-Alaoglu, we conclude the existence of a weakly convergent subsequence in the reflexsive Banach space L2(Ω × [0, T] × TN). In what sense is now the original equation solved?

Florian Bechtold 19/ 21

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The limit case δ = 1: Weak-mild solutions

Definition (Weak-mild solutions) We call u ∈ L2(Ω × [0, T] × TN) weak-mild solution to the problem du = ∆udt + (−∆)1/2B(ut)dWt u(0) = u0 (1) if P-almost surely one has for all ξ ∈ D((−∆)1/2) ut, ξ = Stu0, ξ +

t

(−∆)1/2ξ, St−sB(us)dWs Lemma The weak limit u ∈ L2(Ω × [0, T] × TN) obtained above satisfies (1) in the weak-mild sense given that B is of linear growth.

Florian Bechtold 20/ 21

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The limit case δ = 1: An application

In particular, one is able to apply the machinery presented to

du