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

Strong solutions of semilinear SPDEs with unbounded diffusion 1 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

SPDEs as infinite dimensional SDEs For a , b ∈ R the stochastic differential equation � dX t = aX t dt + bX t dW t X 0 = x ∈ R admits a unique solution (geometric Brownian motion), which satisfies moreover the equation � t X t = e at x + e a ( t − s ) bX s dW s 0 To consider the problem in infinite dimensions, do the replacements Real number a Linear (unbounded) operator A Exponential of real number a Semigroup ( S t ) t generated by A Real number b Nonlinear bounded operator B Brownian motion Cylindrical Brownian motion Florian Bechtold 2/ 21

Cylindrical Brownian motion and stochastic integration in infinite dimensions Formally, for a Hilbert space U with ONB ( e k ) k , set for a sequence ( β k ) k of independet Brownian motions as cylindrical Brownian motion ∞ � e k β k W t := ” t ” ∈ U k =1 Problem: This is not an element of L 2 (Ω , U ). Suppose for B ∈ L ( U , H ) and for an ONB ( f k ) k of H , there exists ( λ k ) k ⊂ R such that ∞ � Bu = λ k f k � u , e k � k Now if ( λ k ) k ∈ ℓ 2 , then B ∈ L 2 ( U , H ) and � t � � BdW s || 2 λ k f k β k t || 2 λ 2 E [ || H ] = E [ || H ] = k t < ∞ 0 k k Florian Bechtold 3/ 21

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 ( S t ) t ≤ T of operators and B : H → L 2 ( U , H ), let W be a cylindrical Wiener process on U . Then a progressively measurable process ( u t ) t ≤ T with values in H is called mild solution to du t = Au t dt + B ( u t ) dW t , u 0 = ξ ∈ H if P -almost surely for all t ≤ T � t u t = S t ξ + S t − s B ( u s ) dW s 0 � �� � =:( K ( u )) t holds as an equality in H . Florian Bechtold 4/ 21

The key tool: A maximal inequality Lemma (Maximal inequality for stochastic convolutions) Let H be Hilbert, u ∈ L p (Ω , L ∞ ([0 , T ] , H )) , let B : H → L 2 ( U , H ) , let W be a U-cylindrical Brownian motion, then for all contraction semi groups S on H one has � t S t − s B ( u s ) dW s || p || B ( u s ) || p H ] ≤ CT p / 2 E [sup E [sup || L 2 ( U , H ) ] . 0 t ≤ T t ≤ T Corollary (Existence and uniqueness of mild solutions) If B : H → L 2 ( U , H ) is Lipschitz and of linear growth, then K : L p (Ω , L ∞ ([0 , T ] , H )) → L p (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

Our problem It is natural to ask, if one can also consider unbounded diffusions, i.e. problems of the form � = ∆ udt + ( − ∆) δ/ 2 B ( u t ) dW t du = u 0 ∈ L p ( T N ) u (0) for δ ∈ [0 , 1). The corresponding mild formulation becomes � t ( − ∆) δ/ 2 S t − s B ( u s ) dW t u t = S t u 0 + 0 Can one replace the maximal inequality for stochastic convolutions with something more general, adapted to this context? Florian Bechtold 6/ 21

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 ( S t ) t ≤ T . Let B : H → L 2 ( U , H ) satisfy || B ( z ) || 2 L 2 ( U , H ) ≤ C (1 + || z || 2 H ) . Let W be a cylindrical Wiener process on U. Let δ ∈ [0 , 1) , T > 0 . 2 Then for p > 1 − δ and every progressively measurable u ∈ L p (Ω , L ∞ ([0 , T ] , H )) there holds � t ( − A ) δ/ 2 S t − s B ( u s ) dW s || p E [sup || H ] 0 t ≤ T p 2 (1 − δ ) E [sup || B ( u t ) || p ≤ C δ, p T L 2 ( U , H ) ] t ≤ T and the associated convolution process is continuous. Florian Bechtold 7/ 21

A maximal inequality and mild solutions for δ ∈ [0 , 1) Theorem (General existence theorem) Let H be Hilbert. Let B : H → L 2 ( U , H ) be Lipschitz continuous, satisfying the growth condition previously stated. Then for 2 δ ∈ [0 , 1) and p > 1 − δ the SPDE du t = ∆ u t dt + ( − ∆) δ/ 2 B ( u t ) dW t u (0) = u 0 ∈ H admits a unique mild solution, meaning there exists a unique progressively measurable process u ∈ L p (Ω , L ∞ ([0 , T ] , H )) such that � t ( − ∆) δ/ 2 S t − s B ( u s ) dW s u t = S t u 0 + 0 � �� � =:( K ( u )) t which is P -almost surely continuous in time. Florian Bechtold 8/ 21

A maximal inequality and mild solutions for δ ∈ [0 , 1) Idea of the proof: Use Banach fixed-point theorem in the space Z H := L p (Ω , L ∞ ([0 , T ] , H )) exploiting the maximal inequality � s ||K ( u ) − K ( v ) || p ( − ∆) δ/ 2 S s − r B ( u r ) − B ( v r ) dW r || p Z H = E [sup || H ] 0 s ≤ T ≤ C δ, p T p / 2(1 − δ ) E [sup || B ( u s ) − B ( v s ) || p L 2 ( U , H ) ] s ≤ t ≤ C δ, p T p / 2(1 − δ ) E [sup || u s − v s || p H ] s ≤ t = C δ, p T p / 2(1 − δ ) || u − v || p Z H For δ < 1, one can choose T sufficiently small for K to become a contraction. Florian Bechtold 9/ 21

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 ( e i ) e ≤ d we have � � d d � � B i ( u s ) dW i B ( u s ) dW s = = B i ( u s ) � dW s , e i � s i =1 i =1 The question thus becomes: under which conditions on the functions ( B i ) i ≤ d does the associated Nemytskii operator have the required properties, more precisely: Given the Hilbert space L 2 ( T N ), under what conditions on a real valued function B i : R → R is the operator B : L 2 ( T N ) → L 2 ( U , L 2 ( T N )) � � d � z → u → B i ( z ) � u , e i � i =1 Lipschitz and satisfying the growth condition? Florian Bechtold 10/ 21

Nemytskii operators and strong solutions for δ ∈ [0 , 1) Lemma (A Nemytskii operator type result for L 2 ( T N )) Let U be a d-dimensional Hilbert space with ONB ( e i ) i ≤ d . Let B 1 , . . . , B d ∈ C 1 ( R ; R ) be of bounded derivative and satisfy the growth condition d � | B i ( ξ ) | 2 ≤ C (1 + | ξ | 2 ) i =1 Then the associated Nemytskii operator is well defined, Lipschitz and satisfies the growth condition || B ( z ) || 2 L 2 ( U , L 2 ( T N )) ≤ C (1 + || z || 2 L 2 ( T N ) ) . Florian Bechtold 11/ 21

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

Thanks for your attention! Florian Bechtold 13/ 21

Starting point In her first publication, Hofmanova considered a class of problems comprising in particular � = ∆ udt + div F ( u ) dt + � d k =1 B k ( u ) dW k du t u (0) = u 0 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

Hofmanova’s setting The mild formulation corresponding to Hofmanova’s problem du = ∆ u t dt + div F ( u t ) dt + B ( u t ) dW t u 0 = ξ is given by � t � t u t = S t ξ + S t − s div F ( u s ) ds + S t − s B ( u s ) dW s 0 0 � t � t ( − ∆) 1 / 2 S t − s ( − ∆) − 1 / 2 div F ( u s ) ds + = S t ξ + S t − s B ( u s ) dW s 0 0 Key idea in Hofmanova’s paper to deal with Bochner integral: 1 || ( − ∆) 1 / 2 S t − s || ≤ C √ t − s ( − ∆) − 1 / 2 div F ( · ) is a bounded operator Use trinangle inequality, get control via time horizon T ( → think contraction for fixed point argument) Florian Bechtold 15/ 21

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 π ( t − s ) α − 1 ( s − σ ) − α ds = sin πα σ for α ∈ (0 , 1), one has due to Fubini’s theorem � t ( − A ) δ/ 2 S t − s B ( u s ) dW s 0 � t �� s � =sin πα ( t − s ) α − 1 ( − A ) δ/ 2 S t − s ( s − σ ) − α S s − σ B ( u σ ) dW σ ds π 0 0 � �� � =: Y s Moreover, use the inequality 1 || ( − A ) δ/ 2 S t − s || ≤ C ( t − s ) δ/ 2 Florian Bechtold 16/ 21

The limit case δ = 1: Heuristics What about the limit case δ = 1, i.e. problems of the form � = ∆ udt + ( − ∆) 1 / 2 � d k =1 B k ( u t ) dW k du t = u 0 ∈ L p ( T N ) u (0) Heuristically, looking at the problem in Fourier basis, the Laplacian makes us gain k 2 , the noise potentially lose k , yet due to Itô Isometry, one has to square, so we lose potentially in order k 2 . In order for the equation to still admit a solution, one needs to thus require that the non-linearities B k are "small" in the following sense. Florian Bechtold 17/ 21