An introduction to the theory of Regularity Structures

Transcription

1 An introduction to the theory of Regularity Structures Lorenzo Zambotti (LPMA, Univ. Paris 6)

2 The KPZ equation Formally, the KPZ equation is written t u = 2 x u + ( x u) 2 + ξ, t 0, x S 1 := R/2πZ. ξ is a random external force. It is not a function but a Schwartz distribution (like the derivative of a continuous non-differentiable function). This equation was introduced in 1986 by physicists as a model for the fluctuations of a randomly growing interface and later recognised as an universal object, believed to be the limit of many discrete models in statistical mechanics. u is expected to be similar to a Brownian motion in x, to x u is not a function but a Schwartz distribution and its square is ill-defined.

3 Regularization of the noise and stability Let us consider a smooth ξ ε that converges to ξ in some sense as ε 0. t u ε = 2 x u ε + ( x u ε ) 2 + ξ ε. We are interested in the following stability problem: find topologies on ξ ε and u ε such that 1. (Probabilistic step) ξ ε converges to the space-time white noise as ε 0 2. (Analytic step) ξ u is continuous.

4 Analytic step: Products of random distributions The analytic step involves a treatment of products of random Schwartz distributions, like the term ( x u) 2. The theory says that under certain conditions on the SPDE (a subcriticality assumption in Hairer s terminology, physicists would say that the theory is super-renormalisable) u ε = F (g 1 (ξ ε ),..., g k (ξ ε )) where F is a continuous functional and {g 1,..., g k } are explicit polynomial functionals of the smooth random function ξ ε. For instance, if we add the term ( x u) 3 to the equation, it is not subcritical anymore and the theory fails to apply.

5 The probabilistic step: Renormalisation and convergence In the probabilistic step we need to prove that {g j (ξ ε )} converge in probability. However the {g j ( )} are not continuous with respect to the very weak topology we try to impose on ξ. Convergence needs to be proved on each term separately. In fact, some of the {g j (ξ ε )} fail to converge. This is a structural problem. So what? Some of the g j (ξ ε ) need to be modified (renormalised): we want to find {ĝ ε j (ξ ε)} which converges still retaining some link with {g j (ξ ε )}. If we can do this, û ε = (ĝ ε 1 (ξ ε),..., ĝ ε K (ξ ε)) converges to a renormalised solution.

6 Renormalisation of KPZ The probabilistic step contains an additional fundamental element: the renormalisation procedure. For instance, in the case of the KPZ equation, the solution u ε of t u ε = 2 x u ε + ( x u ε ) 2 + ξ ε does not converge as ε 0. It is necessary to define t û ε = 2 x ûε + ( x û ε ) 2 C ε + ξ ε where C ε + in order to have a convergent function û ε as ε 0. The term ( x û ε ) 2 C ε converges to the renormalised square of x u.

7 References A theory of regularity structures, M. Hairer Introduction to Regularity Structures, M. Hairer Singular stochastic PDEs, M. Hairer

8 Variables, multi-indices, monomials, derivatives We use variables for d 1 x, y, z R d, k, i N d. For a multi-index k N d we define For k N d we define monomials k := k k d. x k = x k 1 1 xk d d with homogeneity k := k k d. For k N d and ϕ C (R d ) we set k ϕ(x) := k 1 x k 1 1 k d x k ϕ(x). d d With these notations, in what follows one can (almost) forget that d is not necessarily equal to 1.

9 Abstract Monomials We introduce symbols X 1,..., X d and X 0 := 1, X k := X k 1 1 Xk d d, k Nd and the evaluation operators Π x X k (y) = (y x) k. Note that Π x X k (y) C y x k, x, y R d.

10 Hölder functions and abstract Taylor series By definition, f C γ, γ R + \ N, iff f (y) = i γ i f (x) i! We associate to f the abstract Taylor series (y x) i + r(x, y), r(x, y) C y x γ. F(x) = i γ i f (x) i! This is equivalent to x ( i f (x), i γ). X i. Then : Π x F(x)(y) = i γ f (x) = Π x F(x)(x), i f (x) i! (y x) i (reconstruction) trivial in this case.

11 Γ operators The function f (y) := (y z) k has the abstract Taylor series F(x) = i k and in particular F(z) = X k. We define ( ) k (x z) k i X i = (X + x z) k i Γ xz X k = (X + x z) k = i k ( ) k (x z) k i X i. i Γ xz is a rule to transform a Taylor sum around z in one around x. This definition satisfies the simple properties Γ xy Γ yz = Γ xz, Γ xx = Id, Π z = Π x Γ xz, Γ xz X k X k < k, Γ xz X k X k i C x z k i.

12 Hölder functions To f C γ, γ R + \ N, we associate Then F(x) Γ xz F(z) = F(x) = i γ X i i! i γ i f (x) i! i f (x) X i. j γ i i+j f (z) j! and in particular f C γ iff i f C γ i for all i γ, i.e. F(x) Γ xz F(z) i C x z γ i. (x z) j

13 Hölder functions To f C γ, γ R + \ N, we associate Then F(x) Γ xz F(z) = F(x) = i γ X i i! i γ i f (x) i! i f (x) X i. j γ i i+j f (z) j! and in particular f C γ iff i f C γ i for all i γ, i.e. F(x) Γ xz F(z) i C x z γ i. (x z) j We want to add to this classical framework generalised monomials: these will be random (Schwartz) distributions.

14 Schwartz distributions The Schwartz space or space of rapidly decreasing functions on R d is S(R d ) := {ϕ C (R d ) : p α,β (ϕ) < +, α, β N d } p α,β (ϕ) := sup x R d x α β ϕ(x), α, β N d. A tempered (or Schwartz) distribution is a linear functional T : S(R d ) R such that there exist a constant C and α 1, β 1,..., α k, β k N d s.t. T(ϕ) C k p αi,β i (ϕ), ϕ S(R d ). i=1 We write T S (R d ). Examples: If f L p (R d ), p 1, then T(ϕ) := ϕ f dx For any finite measure µ on R d, T(ϕ) := ϕ dµ For any finite measure µ on R d and α N d, T(ϕ) := α ϕ dµ Linear combinations of the above examples

15 Generalised monomials We consider a class of symbols T {X k, k 0} representing generalised monomials. We associate to each τ T a degree or homogeneity : τ T, τ R, with the assumption that X k = k = k k d. We want to endow T with operators (Π x, Γ xz ) such that: 1. Π x associates to each symbol τ T a (random) distribution on R d whose local regularity around each point x is no worse than the homogeneity τ ; informally: Π x τ(y) C y x τ, x, y R d 2. Γ xz is a rule to transform an abstract Taylor sum around z in one around x.

16 General models A model of T is given by a couple (Π x, Γ xz ) such that for all x, Π x : T S (R d ) and for all ϕ C c (R d ) where ϕ x,δ (y) := δ d ϕ Π x τ(ϕ x,δ ) Cδ τ, ( ) y x δ, δ > 0. for all x, y, z, Γ xz : T T is such that Γ xx = Id, Γ xy Γ yz = Γ xz and Γ xz τ τ < τ, Γ xz τ τ α C z x τ α, α < τ. for all x, z: Π z = Π x Γ xz. Very important: (Π x, Γ xz ) is in general random (measurable with respect to ξ or ξ ε ), T is fixed and deterministic.

17 A 1-dimensional example Let us add a symbol W to T. Let (B t ) t R be a two-sided Brownian motion. Then a possible choice for Π x and Γ xz is Π x W(y) = B y B x, Γ xz W = W + (B x B z )1. W = 1 ε =: α, 2 Π xw(y) = B y B x C y x α Π x Γ xz W(y) = Π x (W + (B x B z )1)(y) = B y B x + B x B z = B y B z = Π z W(y) Γ xz W W 0 = B x B z C x z α 0

18 A 1-dimensional example Let us add yet another symbol Ξ to T. A possible choice for Π x and Γ xz is for ϕ C c (R) Π x Ξ(ϕ) = R ϕ(y) db y = ϕ (y) (B y B x ) dy, Γ xz Ξ = Ξ. R The homogeneity of Ξ is Ξ = α 1 = 1 2 ε. It is easy to show that the required assumptions are satisfied: ( ) Π x Ξ(ϕ x,δ ) = y x (B y B x ) δ 2 ϕ dy δ Cδ 1 2 ε R

19 Hölder functions For γ > 0 we say that F D γ if F takes values in the linear span of the symbols with homogeneity < γ and for all α < γ F(x) Γ xz F(z) α C x z γ α where α is the norm of the projection onto the span of the symbols with homogeneity equal to α. (Note that α is not restricted to N anymore!). The inspiration comes from Gubinelli s theory of controlled rough paths.

20 A 1-dimensional example Let σ : R R be smooth. Then σ(b x ) σ(b z ) C B x B z C x z α. This means F(x) = σ(b x )1 defines a function F D α, since F(x) Γ xz F(z) F(x) Γ xz F(z) 0 C x z α 0 Can we go beyond α?

21 A 1-dimensional example By the Ito formula x σ(b x ) = σ(b z ) + σ (B u ) db u + 1 x σ (B u ) du. z 2 z By a (nontrivial) result similar to the Kolmogorov criterion and due to M. Gubinelli x (σ (B u ) σ (B z )) db u C x z 2α. z Then σ(b x ) σ(b z ) σ (B z )(B x B z ) C x z 2α

22 A 1-dimensional example Then let us set G(x) = σ(b x )1 + σ (B x )W. Then G(x) Γ xz G(z) = = [ σ(b x ) σ(b z ) σ (B z )(B x B z ) ] 1 + (σ (B x ) σ (B z ))W and G D 2α : G(x) Γ xz G(z) β C x z 2α β, β {0, α}. Note that: 1. F is a truncation of G 2. If ξ C β with β 2α, then H(x) = (σ(b x ) + ξ x )1 + σ (B x )W defines a function H D 2α

23 Hölder functions In the general case, for γ > 0 we say that F D γ if F takes values in the linear span of the symbols with homogeneity < γ and for all α < γ F(x) Γ xz F(z) α C x z γ α where α is the norm of the projection onto the span of the symbols with homogeneity equal to α. (Note that α is not restricted to N anymore!). Given a Taylor expansion F(x) around each point x, can we find a function/distribution f which has this expansion up to a remainder? It f C γ, γ R + \ N, and F(x) = i f (x) i γ i! X i, then F D γ and f (x) = Π x F(x)(x). But in general Π x F(x)( ) is a distribution and we can not compute it at x.

24 The reconstruction theorem Reconstruction Theorem. For all γ > 0 and F D γ there exists a unique distribution RF on R d such that or, more precisely, such that RF(y) Π x F(x)(y) C x y γ RF(ϕ x,δ ) Π x F(x)(ϕ x,δ ) Cδ γ. Again for all ϕ C c (R d ), ϕ x,δ (y) := δ d ϕ ( ) y x δ, δ > 0. Π x F(x)( ) being a distribution, we can not set RF(x) = Π x F(x)(x). The proof is based on Wavelets Analysis, like for Paraproducts in Massimiliano s course.

25 How to define a x db x Let us consider a continuos process a x, x R, and a deterministic ϕ C c (R). Can we define R ϕ(x) a x db x pathwise? Idea: use the reconstruction theorem! If we define F(x) := a x Ξ then Π x F(x)(y) = a x db y and formally RF(x) = Π x F(x)(x) = a x db x. But we need F D γ with γ > 0.

26 How to define a x db x F(x) := a x Ξ, F D γ with γ > 0 depends on F(x) Γ xz F(z) = a x Ξ Γ xz a z Ξ = (a x a z )Ξ satisfying F(x) Γ xz F(z) i C x z γ i for all i γ i.e. a x a z C x z γ+1 α ( Ξ = α 1 = 1/2 ε). Then F D γ with γ > 0 if a C γ+1 α with γ + 1 α > 1/2 + ε. This is in fact a Theorem due to Young (1936). However this is unsatisfactory since it rules out SDEs like X t = X 0 + t 0 σ(x s) db s.

27 The symbol for B y db y Let us add another symbol ΞW to T. Π x ΞW(ϕ) = ϕ(y)(b y B x ) db y (Ito integral) R 1 = R 2 ((B y B x ) 2 (y x)) ϕ (y) dy Γ xz ΞW = Ξ(W + (B x B z )1) = ΞW + (B x B z )Ξ. Π x Γ xz ΞW = ϕ(y)(b y B x ) db y + ϕ(y)(b x B z ) db y ΞW := Ξ + W = 2α 1 = 2ε. R Π x ΞW(ϕ x,δ ) = R Cδ 2ε 1 2 ((B y B x ) 2 (y x)) δ 2 ϕ R ( ) y x dy δ

28 How to define a x db x Let us now try F(x) := a x Ξ + b x ΞW, F(x) Γ xz F(z) = (a x a z b z (B x B z ))Ξ + (b x b z )ΞW and we must have F(x) Γ xz F(z) i C x z γ i i.e. a x a z b z (B x B z ) C x z γ+1 α ( Ξ = α 1) b x b z C x z γ+1 2α ( ΞW = 2α 1)

29 How to define a x db x If we set A(x) := a x 1 + b x W, then A(x) Γ xz A(z) = (a x a z b z (B x B z ))1 + (b x b z )W we know from the previous slide that a x a z b z (B x B z ) C x z γ+1 α ( 1 = 0) b x b z C x z γ+1 2α ( W = α) and this means A D γ+1 α. Then A D γ+1 α F = AΞ D γ. Note that Ξ = α 1.

30 How to define a x db x Let us fix α ]1/3, 1/2[ and set γ = 3α 1, so that γ + 1 α = 2α. We assume that A(x) := a x 1 + b x W belongs to D 2α. Moreover x H(x) := Ξ belongs to D since H(x) Γ xz H(z) 0. Note that RA = a, RH = db. We have shown that A H(x) = F(x) = a x Ξ + b x ΞW belongs to D 3α 1 with 3α 1 > 0. The reconstruction theorem yields RF =: a y db y such that for all x ϕ x,δ a db a x ϕ x,δ db b x ϕ x,δ (B B x ) db Cδ3α 1. R R R This is a result due to M. Gubinelli (2004).

31 An exercise Let γ > β. For all U D γ with U(x) = u τ (x)τ τ <γ set Q β U(x) = τ <β u τ (x)τ and show that Q β U Dβ. If moreover γ > β > 0 then show that RU = RQ β U.

32 Another exercise Suppose that 0 < α < γ, U D γ and U takes values in the linear span of {X k, k} V with τ α, τ V. Let V(x) be the projection of U(x) on the linear span of {X k, k}. Then show that 1. V D α, 2. the function x u(x) := U(x), 1 is in C α 3. RU = RV = u.

33 Some notations: the heat kernel For x = (x 1,..., x d ) R d we define the heat kernel G : R d R G(x) = 1 (xd >0) 1 (2πx d ) d 1 2 exp Given k = (k 1,..., k d ) N d we define ( x x2 d 1 2x d ). G (k) (x) = k 1 x k 1 1 k d x k G(x). d d

34 Parabolic scaling The heat kernel has a very important scaling property: G(δx 1,..., δx d 1, δ 2 x d ) = 1 δ G(x 1,..., x d 1, x d ), δ > 0 which is associated with the scaling s := (1,..., 1, 2). This motivates the following definitions: x y s := x 1 y x d 1 y d 1 + x d y d 1/2, k s := k k d 1 + 2k d, k N d.

35 Regularity Structures A regularity structure T = (A, T, G) consists of the following elements: An index set A R such that 0 A, A is bounded below and A has no finite accumulation point A graded vector space T = α A T α, with each T α a Banach space. Furthermore T 0 R1. A structure group G of linear operators acting on T such that Γ G, α A, a T α Γa a β<α T β. Furthermore Γ1 = 1 for all Γ G.

36 Models A model of T = (A, T, G) with scaling s is given by a couple (Π x, Γ xz ) such that for all x, Π x : T S (R d ) and for all ϕ C c (R d ) where ϕ x,δ (y) := δ d 1 ϕ ( y1 x 1 δ Π x τ(ϕ x,δ ) Cδ l, τ T l,..., y d 1 x d 1 δ, y d x d ), δ > 0. δ 2 a map Γ : R d R d G such that for all x, y, z, Γ xx = Id, Γ xy Γ yz = Γ xz and Γ xz τ τ α C z x l α s, τ T l. for all x, z: Π z = Π x Γ xz.

37 The reconstruction theorem For γ > 0 we say that F D γ if F : R d α<γ T α and for all α < γ F(x) Γ xz F(z) Tα C x z γ α s where Tα is the norm of the projection onto T α. Reconstruction Theorem. For all γ > 0 and F D γ there exists a unique RF S (R d ) such that or, more precisely, such that RF(y) Π x F(x)(y) C x y γ s RF(ϕ x,δ ) Π x F(x)(ϕ x,δ ) Cδ γ.

38 The mild formulation Now, how can we use this formalism to solve SPDEs? Let us go back to the regularized (unrenormalised) KPZ t u ε = 2 x u ε + ( x u ε ) 2 + f (u ε ) + ξ ε. with an additional non-linear term, for f : R R smooth. Here everything is classical, and the solution is given by the fixed point of the mild formulation ) u ε (t, x) = G t s (x y) (( x u ε ) 2 + f (u ε ) + ξ ε (s, y) ds dy [0,t] S 1 (we assume for simplicity that u ε (0, ) 0).

39 Lift ot the equation Now we want to consider the equation ) u ε = G (( x u ε ) 2 + f (u ε ) + ξ ε, build a regularity structure T = (A, T, G) with a model (Π ε, Γ ε ), and find a U ε D γ = D γ (Π ε, Γ ε ) such that RU ε = u ε, U ε = K(DU ε DU ε + F(U ε ) + Ξ) K : D γ D γ+2, R(KV) = G RV D : D γ D γ 1, R(DV) = x RV F : D γ D γ, R(FV) = f (RV) and finally there is the

40 How to build the regularity structure (A, T, G) We shall need Ξ T. We also use the notation Ξ = Let us write U 0 = 0 and (forgetting the non-linear term f (u)) U n+1 = I(DU n DU n + Ξ). Then U 1 = I(Ξ) = ( ) U 2 = I (DI(Ξ)) 2 + Ξ = I ( + ) = +, U 3 = I( ) In this way we generate iteratively a list of symbols (trees):...

41 Abstract Monomials We define the following family F of symbols: 1, X 1,..., X d, Ξ F if τ 1,..., τ n F then τ 1 τ n F (commutative and associative product) if τ F then I(τ) F and I k (τ) F (formal convolution with the heat kernel differentiated k N d times), with I(X k ) = 0 Examples: I(Ξ), X n ΞI k (Ξ), I((I 1 (Ξ)) 2 ), Ξ n To a symbol τ we associate a real number τ s called its homogeneity: Ξ s < d + 1 2, X 1 s = = X d 1 s = 1, X d s = 2, 1 = 0 τ 1 τ n s = τ 1 s + + τ n s, I k (τ) s = τ s + 2 k s. However F is far too large. We do not need Ξ n for n 2.

42 Sub-criticality assumption We consider a SPDE t u = u + F(u, u, ξ) on R d. In the formal expression for F: we replace ξ by Ξ, with Ξ = α < 0 if α + 1 < 0 then we associate to i u a symbol P i, P i = α + 1, i = 1,..., d 1 if α + 2 < 0 then we associate to u a symbol U, U = α + 2. We assume that 1. The resulting expression is polynomial in the symbols. 2. Terms containing Ξ do not contain other symbols, and all other monomials have homogeneity > α. Then we say that the SPDE is sub-critical.

43 Examples of sub-critical equations KPZ in 1 space-dim. (d = 2). Here α = Ξ = 3 2 δ. t u = u + F(u, u, ξ), F(u, p, ξ) = p 2 + ξ. Then the formal replacement gives F(U, P, Ξ) = P 2 + Ξ, P = α + 1, P 2 = 2α + 2 > α since α > 2. If we modify the equation t u = u + F(u, u, ξ), F(u, p, ξ) = p 3 + ξ. Then the formal replacement gives F(U, P, Ξ) = P 3 + Ξ, P = α + 1, P 3 = 3α + 3 < α since α < 3 2. Then this class of equations is sub-critical for F(u, p, ξ) = p 2 + ξ and critical or super-critical for F(u, p, ξ) = p k + ξ, k 3.

44 Examples of sub-critical equations Parabolic Anderson model. Here the space-dim. is 2 (d = 3) and ξ is only a function of space. Therefore α = 1 δ. The equation is t u = u + F(u, ξ), F(uξ) = uξ. Then the formal replacement gives F(u, Ξ) = uξ, α + 2 > 0. Another important example is the Φ 4 3 model: space-dim. is 3 (d = 4), α = 5 2 δ, t u = u + F(u, ξ), F(u, ξ) = u 3 + ξ. Then the formal replacement gives F(U, Ξ) = U 3 + Ξ, U = α + 2 < 0, U 3 = 3α + 6 > α since α < 3 2. Then this class of equations is sub-critical for F(u, ξ) = u 4 + ξ and critical or super-critical for F(u, ξ) = u k + ξ, k 5.

45 Construction of the regularity structure Now we define a set M F of monomials in the above symbols: Ξ m U n P k M F if F contains a term ξ m u n (Du) k with m m 1, n n, k k. For KPZ: F(u, p, ξ) = p 2 + ξ, M F = {Ξ, P, P 2 }. For PAM: F(u, p, ξ) = uξ, For Φ 4 3 : F(u, p, ξ) = u3 + ξ, M F = {U, UΞ, Ξ}. M F = {U, U, U 3, Ξ}.

46 Construction of the regularity structure We set W 0 = U 0 = P0 i = and, given A, B F, we also write AB := {τ τ : τ A, τ B}. Then, we define the sets W n, U n and Pn i for n > 0 recursively by W n = W n 1 Q(U n 1, P n 1, Ξ), Q M F U n = {X k } { } I(τ) : τ W n, Pn i = {X k } { } I i (τ) : τ W n, and finally F F := n 0(W n U n ).

47 Construction of the regularity structure Lemma The set {τ F F : τ s γ} is finite for all γ R iff the SPDE is sub-critical. Then we set A := { τ s : τ F F }, T γ := {λ 1 τ λ k τ k : τ i s = γ}. By the Lemma each T γ is finite-dimensional (if non-empty). The group structure G will be defined below.

48 Another example: Φ 4 3 We have seen the family of symbols (trees) of KPZ. For Φ 4 3 we have with the notation Ξ =, I(Ξ) =, I(Ξ) 3 =, I(Ξ)I(I(Ξ) 3 ) =, the list of symbols (trees) T =,,,,,,, X i, 1,,,....

49 The Π x operators We fix a (continuous) path of ξ ε and define recursively (continuous) generalized monomials Π ε xτ around the base point x Π ε xx i (y) = (y i x i ), Π ε xξ(y) = ξ ε (y), Π ε x(τ 1 τ n )(y) = n Π ε xτ i (y), Π ε xi k (τ)(y) = (G (k) Π ε xτ)(y) (y x)i (G (i+k) Π ε i! xτ)(x) i s< I k (τ) s Then we can interpret analytically τ s i=1 Π ε xτ(y) C ε y x τ s s.

50 The Γ operators We have the recursive definition of Γ ε xz : T T Γ ε xzx i = X i + (x i z i ), Γ ε xzξ = Ξ, Γ ε xz τ i = i i Γ ε xzτ i Γ ε xzi k (τ) = I k (Γ ε xzτ) j s< τ s+2 k On can check again the compatibility condition Π ε z = Π ε xγ ε xz and the properties Γ ε xx = Id, Γ ε xyγ ε yz = Γ ε xz, (Π ε xi k+j (Γ ε (X + x z)j xzτ))(z) j! Γ ε xzτ τ s < τ s, Γ ε xzτ τ l C x z τ l s, l < τ s.

51 Examples Let for instance d = 2. Then α = Ξ s = 3 2 δ, δ > 0. Π ε xi(ξ)(y) = (G ξ ε )(y) (G ξ ε )(x) The homogeneity is I(Ξ) s = 1 2 δ. Π ε xi(ξi(ξ))(y) = G (ξ ε (G ξ ε ))(y) G (ξ ε (G ξ ε ))(x) and the homogeneity is I(ΞI(Ξ)) s = 1 δ.

52 Examples Setting h := G (ξ ε (G (ξ ε (G ξ ε ))), Π ε xi(ξi(ξi(ξ)))(y) = h(y) h(x) x1 h(x) (y 1 x 1 ) and the homogeneity is 3 2 δ. Π x I(X 1 Ξ)(y) = = (G(y z) G(x z) x1 G(x z) (y 1 x 1 )) (z 1 x 1 ) ξ ε (z) dz and the homogeneity is I(XΞ) s = 3 2 δ.

53 Examples For τ = I(ΞI(ΞI(Ξ))) the homogeneity is 3 2 δ and (dropping ε) Π ε xi(ξi(ξi(ξ)))(y) = G (ξg (ξg ξ))(y) G (ξg (ξg ξ))(x) + (y 1 x 1 )( x1 G ξ(x)(g (ξg ξ)(x) G ξ(x) 2 )) + (y 1 x 1 ) x1 G (ξg ξ)(x)g ξ(x) (y 1 x 1 ) x1 G (ξg (ξg ξ))(x) G ξ(x)g (ξg ξ)(y) + G ξ(x) 2 G ξ(y) G (ξg ξ)(x)g ξ(y) G ξ(x) 3 + 2G ξ(x)g (ξg ξ)(x)

54 Examples For instance Γ ε xzi(ξ) = I(Ξ) + (G ξ ε (x) G ξ ε (z))1 and indeed setting h := (G ξ ε ) Π ε xγ ε xzi(ξ)(y) =h(y) h(x) + h(x) h(z) = h(y) h(z) =Π ε zi(ξ)(y). Another example: Γ ε xzi(ξi(ξ)) =(G (ξ ε (G ξ ε ))(x) G (ξ ε (G ξ ε ))(z) +(G ξ ε (z) G ξ ε (x))g ξ ε (z))1 + (G ξ ε (x) G ξ ε (z)) I(Ξ) + I(ΞI(Ξ))

55 Symbols with negative homogeneity At the beginning we said that we would find, under a subcriticality assumption on the SPDE, a representation u ε = F (g 1 (ξ ε ),..., g k (ξ ε )) where F is a continuous functional and {g 1,..., g k } are explicit polynomial functionals of the smooth random function ξ ε. Well, the functionals {g i (ξ ε )} are given by {Π ε τ, τ F F, τ s < 0}. Each of these (finitely many) terms must be proved to converge separately. If this is done, convergence for all other symbols (trees) follows automatically by the theory. However, before converging these terms must be renormalised.

56 Classical multiplication Consider f i C γ i, i = 1, 2, with γ > 0 and f i (y) = j γ i a j i (x) (y x)j + r i (x, y), r i (x, y) C(x) x y γ i. Pointwise multiplication gives, setting γ = γ 1 γ 2, f 1 f 2 (y) = a j1 1 (x) a j 2 2 (x) (y x) j 1+j 2 j 1 γ 1 j 2 γ 2 + r 1 (x, y)(f 2 (y) r 2 (x, y)) + r 2 (x, y)(f 1 (y) r 1 (x, y)) + r 1 r 2 (x, y) = j γ b j (x) (y x) j + r(x, y), with r(x, y) C(x) x y γ. After all, if f 1 and f 2 are in C 1 then f 1 f 2 is not necessarily in C 2!

57 Multiplication of modelled distributions We say that F D γ η if F D γ and F(x) = N a j (x) τ j, τ j F F, τ j s η. j=1 Consider F i D γ i η i, γ i > 0, i = 1, 2. By the reconstruction theorem, f i := RF i S (R d ) satisfies N i f i (y) = a j i (x) Π xτ j i (y) + r i(x, y) j=1 where r i (x, y) C(x) x y γ i s is a remainder. Question: Can we define the product of f 1 and f 2?

58 Multiplication of Taylor sums For F i D γ i η i and f i := RF i S (R d ), j = 1, 2, N i F i (x) = a j i (x) τ j i, f i(y) = j=1 N i j=1 a j i (x) Π xτ j i (y) + r i(x, y), r i (x, y) C(x) x y γ i s. Formal pointwise multiplication yields ( f 1 f 2 )(y) = N 1 N 2 j 1 =1 j 2 =1 a j 1 1 (x)a j 2 2 (x) Π x τ j 1 1 (y)π xτ j 2 2 (y) + r 1 (x, y)π x F 2 (x)(y) + r 2 (x, y)π x F 1 (x)(y) + r 1 r 2 (x, y).

59 Multiplication of Taylor sums ( f 1 f 2 )(y) = N 1 N 2 j 1 =1 j 2 =1 a j 1 1 (x)a j 2 2 (x) Π x τ j 1 1 (y)π xτ j 2 2 (y) + r 1 (x, y)π x F 2 (x)(y) + r 2 (x, y)π x F 1 (x)(y) + r 1 r 2 (x, y). Now r 1 (x, y)π x F 2 (x)(y) x y γ 1+η 2 s. Therefore, setting γ = (γ 1 + η 2 ) (γ 2 + η 1 ), we have ( f 1 f 2 )(y) = with r(x, y) C(x) x y γ s. a 1 i (x)a 2 i (x) Π x τ j1 1 (y)π xτ j 2 2 (y) + r(x, y) τi 1 + τ i 2 <γ However Π x τ j 1 1 (y) and Π xτ j 2 2 (y) are distributions, so in general Π x τ j 1 1 (y)π xτ j 2 2 (y) is ill-defined.

60 Multiplication of Taylor sums This suggests the following definition: (F 1 F 2 )(x) = a 1 i (x)a 2 i (x) τi 1 τi 2 τi 1 s+ τ i 2 s<γ and, by the reconstruction theorem, if F 1 F 2 D γ and γ > 0 then f 1 f 2 := R(F 1 F 2 ) satisfies ( f 1 f 2 )(y) = a 1 i (x)a 2 i (x) Π x (τ j1 1 τ j 2 2 )(y) + r(x, y) τi 1 s+ τ i 2 s<γ and can be defined as the product of f 1 and f 2. If Π x τ 1 and Π x τ 2 are genuine distributions, their pointwise product is ill-defined in general and therefore Π x can fail to be multiplicative. In the applications to SPDEs, it will be necessary to renormalise some of these products.

61 An abstract integration operator We want to construct an operator K : D γ D γ+2, γ > 0, γ / N, s.t. R(KV) = G (RV). This is given by the following formula: if V = τ V τ (x) τ then KV(x) = V τ (x) I(τ) τ + X k ) G (k) (x y) (RV(dy) Π ε k! xq R d k s 2 V(dy) k s<γ+2 where Q η is the projection onto β<η T β.

62 An abstract differentiation operator We want to construct an operator D i : D γ D γ 1, γ > 0, s.t. R(D i V) = i (RV). Let us set first D i τ: D i 1 = 0, D i X j = δ ij, D i I k (τ) = I k+δi (τ), D i (ττ) = τd i τ + τd i τ. Then we see that Π ε xd i τ = i Π ε xτ and D i : D γ D γ 1 is constructed by linearity.

63 The lift of KPZ We want to formulate the PDE in 1 space-dim. as a fixed point in D γ t h = h + ( x h) 2 + ξ ε H = K((DH) 2 + Ξ). What is the right γ? Note that H is a linear combination of powers of X and of I(τ) with τ s 3 2 δ. Then DH Dγ 1 1 δ. Now 2 (DH) 2 D γ δ so that we need γ > δ in order for the r.h.s. of the equation to be well defined. We choose γ = δ.

64 The lift of KPZ We expect H = h 1 + a + b + c X 1 + d + e D δ. Then and DH = a + b + c 1 + d + e D δ (DH) 2 = a 2 + 2ab + 2ac + b 2 + 2bc + c 2 1 D δ so that K((DH) 2 + Ξ) = G (( h) 2 + ξ ε ) a 2 + G (( h) 2 + ξ ε a 2 Π ε ) X 1 + 2ab + 2ac D δ

65 The lift of KPZ We obtain that a = b = 1, c = h, d = 2, e = 2h, i.e. H = h h X h = h 1 + I(Ξ) + I((I 1 (Ξ)) 2 ) + h X 1 + 2I(I 1 (Ξ)(I 1 (Ξ)) 2 ) + 2h I(I 1 (Ξ)). The theory shows that this fixed point in D δ has (locally in time) a unique solution, which converges if Π ε τ converges for all τ s < δ. However, as we have already announced several times, the unrenormalised family Π ε τ fails to converge.

66 White noise Let (X, A, m) be a σ-finite measure space with atomless measure m. A white noise on (X, A, m) is a map W : H := L 2 (X, m) L 2 (Ω, P), where (Ω, F, P) is a probability space, such that W(h) N (0, h 2 ) E(W(h 1 )W(h 2 )) = h 1, h 2. We can assume w.l.o.g. that F = σ(w(h), h H). Then a classical result says that there exists a natural isometry between L 2 (Ω, P) and the symmetric Fock space H = H k, k 0 H k = L 2 sym (X k, m k ) = H sk H k is the k-th homogeneous Wiener chaos, and j k H j the k-th inhomogeneous Wiener chaos.

67 Multiplication of Wiener chaoses We denote by I k the natural isometry between H k and L 2 sym(x k, m k ). If f L 2 (X k ) then we denote by f s its projection onto the symmetric subspace L 2 sym(x k ) and I k (f ) := I k (f s ). Then a formula (which we do not make more precise here) states that I l (f )I m (g) j l+m with explicit projections on each homogeneous chaos. H j

68 An example: white noise on a point If (X, m) is a point (0) with m = δ 0, then L 2 (X, m) = R and L 2 (Ω, P) = L 2 (R, N (0, 1)), H k = RHk, where (H k ) k are the Hermite polynomials: H 0 = 1, H n+1 (x) = 2xH n (x) H n(x). For instance H 1 (x) = x, H 2 (x) = x 2 1, H 3 (x) = x 3 3x, H 4 (x) = x 4 6x and the formula for the product is for m n H m (x) H n (x) = 2 n n! n i=0 ( ) m Hm n+2i (x) n i 2 i. i! The Wick product is defined by H n H m = H n+m.

69 White noise in R d Let now (X, m) be R d with the Lebesgue measure. We denote W(h) =: h(y) ξ( dy). R d Then ξ is a Gaussian field with the covariance function E(ξ(y)ξ(x)) = δ(x y). It can be seen with the help of the Kolmogorov criterion that x W(1 [0,x] ) is a.s. continuous, where [0, x] = [0, x 1 ] [0, x d ]. Then it can be seen that for h S(R d ) W(h) = ( 1) d R d d h x 1 x d W(1 [0,x] ) dx and therefore ξ is a well-defined random Schwartz distribution.

70 Regularised white noise in R d We set now ξ ε (x) := W(ρ ε (x )), ( ρ ε (y) := ε d 1 y1 ρ ε,..., y d 1 ε, y ) d ε 2 and ρ is a smooth mollifier. Recall: Π ε xx i (y) = (y i x i ), Π ε xξ(y) = ξ ε (y), Π ε x(τ 1 τ n )(y) = n Π ε xτ i (y), Π ε xi k (τ)(y) = (G (k) Π ε xτ)(y) (y x)i (G (i+k) Π ε i! xτ)(x) i s< I k (τ) s By the previous discussion, Π ε τ is a r.v. in the k-th inhomogeneous Wiener chaos, with k the number of occurrences of the symbol Ξ in τ. i=1

71 The Π ε operator Let us set Π ε X i (y) = y i, Π ε Ξ(y) = ξ ε (y), Π ε (τ 1 τ n )(y) = n Π ε τ i (y), i=1 Π ε I k (τ)(y) = (G (k) Π ε τ)(y). This operator is the "stationary" version of Π ε x. Again, Π ε τ is a r.v. in the k-th inhomogeneous Wiener chaos, with k the number of occurrences of the symbol Ξ in τ.

72 Divergence of the Π ε operator Let us consider for instance τ =. Then Π ε (ϕ) = ϕ(y)(π ε ) 2 (y) dy = R 2 ϕ(y)(g ξ ε ) 2 (y) dy R 2 which belongs to H 0 H 2. It is easy to see that E(Π ε (ϕ)) = ϕ(y)(g ρ ε ρ ε G )(0) dy 1 R 2 ε. Instead we have that Var(Π ε (ϕ)) remains bounded.

73 Renormalisation of Π ε We now introduce a linear M ε : T T and ˆΠ ε := Π ε M ε. However M ε can not be arbitrary and must respect the structure we have built. In particular, we expect ˆΠ ε to have the following recursive construction: ˆΠ ε X i (y) = y i, ˆΠε Ξ(y) = ξ ε (y), n ˆΠ ε (τ 1 τ n )(y) = ˆΠ ε τ i (y) ˆΠ ε [L ε (τ 1 τ n )] (y), i=1 ˆΠ ε I k (τ)(y) = (G (k) ˆΠ ε τ)(y). We are modifying the products. This imposes several restrictions on L ε and M ε.

74 The ˆΠ x operators We then define ˆΠ ε xx i (y) = (y i x i ), ˆΠε x Ξ(y) = ξ ε (y), ˆΠ ε x(τ 1 τ n )(y) = n ˆΠ ε xτ i (y) ˆΠ ε x [L ε (τ 1 τ n )] (y), i=1 ˆΠ ε xi k (τ)(y) = (G (k) ˆΠ ε xτ)(y) (y x)i (G (i+k) i! ˆΠ ε xτ)(x) i s< I k (τ) s We have in general ˆΠ ε xτ(y) Π ε xm ε τ(y), ˆΠε x τ(x) = Π ε xm ε τ(x).

75 Renormalisation of KPZ These matrices are of the form M = exp( 3 i=0 C i L i ), where the generators L i are determined by the following contraction rules: L 0 : 1, L 1 : 1, L 2 : 1 L 3 : 1. This should be understood in the sense that if τ is an arbitrary formal expression, then L 0 τ is the sum of all formal expressions obtained from τ by performing a substitution of the type 1. For example, one has L 0 = 2, L 0 = 2 +, etc.

76 Renormalisation of KPZ Theorem Let M ε = exp( 3 i=0 C ε i L i) be as above and let (ˆΠ ε, ˆΓ ε ) be the corresponding renormalised model. Let furthermore H be the solution to the lifted KPZ H = K((DH) 2 + Ξ) with respect to this model. Then, the function ĥ ε = RH solves the equation t ĥ ε = 2 x ĥε + ( x ĥ ε ) 2 4C ε 0 x ĥ ε + ξ ε (C ε 1 + C ε 2 + 4C ε 3).

77 Proof DH = + + h h D δ (DH) 2 +Ξ = Ξ h h +4h +(h ) 2 1. We recall the rules L 0 = L 1 = L 2 = L 3 = 1, L 0 = 2, L 0 = 2, L 0 = 2 +. Then M ε DH = DH 4C0 ε, (M εdh) 2 = (DH) 2 8C0 ε ( M ε (DH) 2 + Ξ ) = (DH) 2 + Ξ C0( ε h 1 ) C1 ε C2 ε 4C3 ε = (M ε DH) 2 + Ξ 4C0 ε M ε DH (C1 ε + C2 ε + 4C3) ε. Then ĥ ε = G (( x ĥ ε ) 2 + ξ ε 4C0 ε xĥε (C1 ε + Cε 2 + 4Cε 3 )).

78 Convergence of the model Theorem. If we choose C ε 0 = 0, C ε 1 = c 1 ε, Cε 2 = 4c 2 log ε + c 3, C ε 3 = c 2 log ε + c 4. then the renormalised model (ˆΠ ε, ˆΓ ε ) converges. In the limit, the solution is parametrised by c 3 + 4c 4. This is a general fact: there is a renormalisation group R such that M ε R. If M ε and N ε are different renormalisation maps, then M ε Nε 1 converges to R R and the two limit solution differ by the action of R. The group R acts on the set of renormalised solutions.