Kolmogorov equations in Hilbert spaces IV

Transcription

1 March 26, 2010

2 Other types of equations Let us consider the Burgers equation in = L 2 (0, 1) dx(t) = (AX(t) + b(x(t))dt + dw (t) X(0) = x, (19) where A = ξ 2, D(A) = 2 (0, 1) 0 1 (0, 1), b(x) = ξ 2 (x 2 ), D(b) = 0 1 (0, 1).

3 The corresponding Kolmogorov operator looks like N 0 ϕ(x) = Lϕ(x) + ξ (x 2 ), Dϕ(x), ϕ E A (). It is well known that equation (19) has a unique solution X(t, x). We denote by P t the corresponding transition semigroup P t ϕ(x) := E[ϕ(X(t, x))], ϕ C b (). It is also well known that P t has a unique invariant probability measure µ.

4 Theorem N 0 is essentially m dissipative in L 2 (, µ) and its closure coincide with the infinitesimal generator of P t. see DP A. Debussche, Potential Analysis 2007

5 Sketch of proof The proof is similar to the previous one. One introduces an approximating equation λϕ n Lϕ n b n (x), Dϕ n = f, with b n regular and looks for an estimate for b(x) 2 µ(dx) and Dϕ n by means of an estimate of DX n (t, x)h. These estimates are delicate, the basic one is ( E ( A) 1/2 D x X n (t, x)h 2) c( x L 4 + 1) 7 e ct (1 + t 7/8 ) h 2.

6 2D Navier Stokes equation We consider 2D NS equation in D = [0, 2π] 2 dx(t) = (AX(t) + b(x(t))dt + C dw (t) X(0) = x, (20) with periodic boundary conditions where = {x L 2 (D) : div x = 0} V ( 2 # (D))2, V = {x 1 # (D))2 : div x = 0} A = ν 0 P, b(x) = P(x x) P is the projector on divergence free vectors and C L() is symmetric, positive and of trace class.

7 It is well known that equation (23) has a unique solution X(t, x). Moreover the corresponding transition semigroup P t ϕ(x) := E[ϕ(X(t, x))], ϕ C b (). has a unique probability invariant measure µ under minimal assumptions on C, see M. airer, J. Mattingly, Ann. Math

8 Theorem N 0 is essentially m dissipative in and its closure coincides with the infinitesimal generator of P t. see V. Barbu, DP, A. Debussche, Rend. Acc. Lincei 04

9 Sketch of proof Again one needs to find an estimate for b(x) 2 µ(dx) and for Dϕ n. One finds for α < C 1 ( E DX n (t, x)h 2) e 2ωαt e α x 2 V h 2.

10 Final bibliographic comments Essential m dissipativity of Kolmogorov operators in spaces L p (, µ) where µ is an invariant measures (also called in the literature L p -uniqueness problem) has been extensively studied for several equation. We will just recall some papers. (i) For the 2D- Navier Stokes and Euler equations see S. Albeverio, A Cruzeiro, Commun. Math. Phys S. Albeverio, B. Ferrario, LNiM 1942, Springer, 2008.

11 (ii) For the stochastic quantization equation see S. Albeverio, M. Röckner, PTRF DP-L.Tubaro, PTRF (iii) Another method to study Kolmogorov operator for Burgers and Navier Stokes equations in L 1 (, µ) is presented in a forthcoming paper by Es-Sarhir-Stannat, JFA 2010

12 Introduction We are given a Kolmogorov operator Nϕ = Lϕ + b(t, ), D x ϕ, where L is the Ornstein Uhlenbeck operator consider before and b : [0, T ] D(b) is a nonlinear mapping. Given µ 0 P() and T > 0 we want to find (µ t ) t (0,T ] P() such that d u(t, x)µ t (dx) = (D t u(t, x) + Nu(t, x))µ t (dx), (1) dt for all u in some test functions space.

13 As test functions we choose the space E A ([0, T ] ), the linear span of all functions of the form u φ,h (t, x) = φ(t)e i x,h(t) with φ C 1 ([0, T ]; R), φ(t ) = 0 and h C 1 ([0, T ]; D(A )). Equation (1) is called Fokker Planck equation.

14 It is naturally related to the stochastic differential equation dx = (AX + b(t, X))dt + BdW (t), X(s) = x, s t. (2) Assume in fact that (2) has a unique solution X(t, x). Set P t ϕ(x) = E[ϕ(X(t, x))], ϕ C b () and denote by P t the transpose operator ϕ, P t F = P t ϕ, F.

15 Then it is easy to see that µ t = P t µ 0, t [0, T ] is a solution of (1). The fact that this is the unique solution it is less evident, it requires a proof which involves the consideration of the problem D t u(t, x) + Nu(t, x) = f (t, x) (3) u(t, x) = 0

16 The first part of this lecture is devoted to the case when b is regular so that the stochastic differential equation (2) can be solved. In the second part we consider b singular when we are not able to solve (2). We can show, under very weak assumptions, existence of a solution of (1), simply by introducing an approximating equation with a regular b α replacing b and then by showing tightness of the corresponding approximated solution (µ α t ) of (1). This method works under very general conditions, for instance for 3D-NS.

17 More difficult is the uniqueness. ere we show uniqueness provided the operator K := D t + N, defined on E A ([0, T ] ) has an m-dissipative extension to L 1 ([0, T ], µ), where µ(dt, dx) := µ t (dx)dt, for each solution (µ t ) of (1). All results described in this talk are included in a series of papers in collaboration with V. Bogachev and M. Röckner in Doklady Math 2008, JFA 2009, Ascona 2009, JEE (to appear), preprint 2010.

18 The regular case We use the following notations. separable ilbert space. P() set of all Borel probability measures in. P m () subset of µ P() such that x m µ(dx) <. A : D(A), infinitesimal generator of a strongly continuous semigroup e ta, B L(). W, -valued cilyndrical Wiener process defined on a filtered probability space (Ω, F, (F t ) t 0, P). b : [0, T ] D(b).

19 The stochastic equation We consider the stochastic differential equation dx = (AX + b(t, X))dt + BdW (t), X(s) = x, s t (3) which we shall write in the following (mild) form. t X(t, s, x) = e (t s)a x + e (t r)a b(r, X(r, x))dr + W A (t, s), s where W A (t, s) = t s e (t r)a BdW (r).

20 We shall make the following assumptions ypothesis 1 (i) The linear operator Q t = t 0 esa BB e sa ds, t 0, is of trace class for any t > 0. (ii) For all t > 0 we have e ta () Q 1/2 t () and Λ t = Q 1/2 t fulfills Λ t ct α, t > 0, for some α [1/2, 1). (iii) b C 1 ([0, T ]; Cb 1 (; )). e ta

21 Proposition Assume that ypothesis 1 holds. Then for any x, 0 s t, problem (3) has a unique mild solution X(, s, x) C W ([s, T ]; ). Moreover for any n N there exists C n > 0 such that E X(t, s, x) n C n (1 + x n ), s t [0, T ], n N. (4)

22 Itô s formula Proposition For any ϕ E A () we have t E[ϕ(X(t, s, x))] = ϕ(x) + E[Lϕ + b(r, ), Dϕ ](X(r, s, x))dr. s (5)

23 The transition evolution operator Let us define the transition evolution operator setting P s,t ϕ(x) := E[ϕ(X(t, s, x))], ϕ B b (). It is to check that P s,t is Feller, that is for all ϕ C b () and all 0 s < t T we have P s,t ϕ C b (). From Itô s formula it follows that d dt P s,tϕ(x) = P s,t L(t)ϕ(x), ϕ E A (). (6)

24 Definition For any 0 s t T and any x we denote by π s,t (x, ) the law of X(t, s, x). Then P s,t ϕ(x) = ϕ(y)π s,t (x, dy), ϕ B b ().

25 The transpose transition evolution operator P s,t We denote by C b () the topological dual of C b() and by, the duality between C b () and C b (). For any F C b () and any ϕ C b() we shall write F(ϕ) = ϕ, F. Moreover for any t 0 we denote by P s,t the transpose of P s,t defined as ϕ, P s,tf = P s,t ϕ, F, ϕ C b (), F C b ().

26 There is a natural imbedding of P() into Cb (). Namely µ P() C b (), µ F µ, where F µ (ϕ) = ϕ, F µ = ϕdµ, ϕ C b (). In the following we shall identify µ and F µ.

27 Proposition For any 0 s t T and µ P() we have P s,t µ P(). Proof. Let µ P(), t > 0 and set F = Ps,t µ, which means ϕ, F = P s,t ϕ, µ = P s,t ϕ(x)µ(dx), ϕ C b (). Set µ s,t (A) := A P s,t 1l A (x)µ(dx), A B(). Since P s,t acts on B b (), we see that µ s,t P(). Finally, it is clear that µ s,t = F s,t.

28 The backward Kolmogorov operator We define the backward Kolmogorov operator as (K u)(t, x) = d dt u(t, x) + Lu(t, x) + b(t, x), Du(t, x), (7) for all u E A ([0, T ] ). We recall that u E A ([0, T ] ) is the linear span of all functions of the form u φ,h (t, x) = φ(t)e i x,h(t) such that φ C 1 ([0, T ]; R), φ(t ) = 0. h C 1 ([0, T ]; D(A )).

29 We are going to show that K space is essentially m-dissipative in the C T ([0, T ]; C b ()) := {g C([0, T ]; C b ()) : g(t ) = 0} Let in fact λ R, f C T ([0, T ]; C b ()) and consider the equation λu K u = f. (8)

30 Equation (8) is equivalent to the system λu(t, x) d dt u(t, x) Lu(t, x) b(t, x), Du(t, x) = f (t, x) u(t, 0) = 0, which one can solve formally for any λ R obtaining u(t, x) = F(λ)f (t, x) := T t e λ(r t) (P t,r f (r, ))(x)dr. (9)

31 Then one can show that F( ) is the resolvent of an m-dissipative operator which is precisely the closure of K in C T ([0, T ]; C b ()). Notice that this is an identification problem similar to that seen in the second lecture because b is regular. We shall denote by K the closure of K. Then we can say that the space E A ([0, T ] ) is a core for K. Finally, it is important to notice that the spectrum of K is empty, in particular that 0 belongs to the resolvent set.

32 The Fokker Planck equation Let µ 0 P 1 (). We say that (µ t ) t (0,T ] P 1 () is a solution to the Fokker Planck equation d u(t, x)µ t (dx) = K u(t, x)µ t (dx), (11) dt if for each ϕ C b (), the mapping t ϕdµ t is measurable and equation (11) is fulfilled for all u E A ([0, T ] ).

33 Equivalent forms of the Fokker Planck equation By integrating both sides of (11) from 0 to t we obtain u(t, x)µ t (dx) = u(0, x)µ 0 (dx) t [ ] + (K u)(s, x)µ s (dx) ds. 0 (12) which is equivalent (because u(t, x) = 0) to T [ ] u(0, x)µ 0 (dx) = (K u)(s, x)µ s (dx) ds. (13) 0

34 Existence and uniqueness Theorem Assume that ypothesis 1 holds and let µ 0 P 1 (). Then there exists a unique solution (µ t ) t 0 P 1 () of the Fokker Planck equation (11).

35 Proof of existence Set µ t = P0,t µ 0, which implies ϕ(x)µ t (dx) = P 0,t ϕ(x)µ 0 (dx), ϕ C b (). Let now u E A ([0, T ] ). Then, recalling that P 0,t acts on C b,1 () and Itô s formula, we have d u(t, x)µ t (dx) = d P 0,t u(t, x)µ 0 (dx) dt dt = P 0,t (K u)(t, x)µ 0 (dx) = (K u)(t, x)µ t (dx). So, µ t is a solution of (11).

36 Uniqueness Let µ t, ν t be solutions of (11), equivalently of (13), with µ 0 = ν 0 and set ζ t = µ t ν t. Then we have T 0 dt K u(t, x)ζ t (dx) = 0, u E A ([0, T ] ). (14) Fix now any f C T ([0, T ]; ) and consider the solution u of equation K u = f

37 Assume first that u E A ([0, T ] ). Then by (14) we find T 0 dt f (t, x)ζ t (dx) = 0 (15) Since E A ([0, T ] ) is a core for K it follows that (15) holds for any f C T ([0, T ] ). In conclusion, ζ t = 0 by the arbitrariness of f.

38 The case when b is irregular We shall consider again the Fokker Planck equation T [ ] u(0, x)µ 0 (dx) = (K u)(s, x)µ s (dx) ds. (13) 0 where K is the Kolmogorov operator K u = D t u + Lu + b(t, ), D x u, u E A ([0, T ] ).

39 ypothesis 2 (i) (BB ) 1 L(). (ii) b : [0, T ] D(b). There exists a sequence (b α ) α (0,1] C 1 ([0, T ]; Cb 1 (; )) such that lim b α(t, x) = b(t, x), α 0 (t, x) [0, T ] D(b). We call X α and (µ α t ) t [0,T ] the solutions of equations (3) and (11) (with µ 0 P 1 () given), with b α replacing b. Then we set µ α (dx dt) = µ α t (dx)dt.

40 Main steps for existence One tries to show that (µ α ) α>0 is tight. We have lim α 0 [0,T ] K α u(t, x) µ α (dt dx) = [0,T ] K u(t, x) µ(dx), u E A ([0, T ] ). (15)

41 Then if µ is a weak limit point of (µ α ) α>0, letting α 0 in the identity T [ ] u(0, x)µ 0 (dx) = (K α u)(s, x)µ α s (dx) ds, 0 we have that µ is a solution to (13). We notice that to prove (15) it is enough to show that lim b α (t, x) µ α (dt, dx) = b(t, x) µ(dt, dx). α 0 [0,T ] [0,T ] (16)

42 Tightness of (µ α ) α>0 We need an estimate of the form T E X α (t, 0, x) 2 Y dt C x 2 (17) 0 where Y is a subspace of X compactly embedded in X. Assume that (17) is fulfilled, then recalling that µ α t = (P0,t α ) µ 0 and consequently ϕ(x)µ α t (dx) = P0,t α ϕ(x)µ 0(dx), ϕ C b,n ()

43 we have (provided x 2 X µ 0(dx) < ) T 0 x 2 Y µα t (dx)dt = T 0 dt E X α (t, 0, x) 2 Y µ 0(dx) CT x 2 µ 0(dx) C 1 This prove the tightness of the sequence (µ α ).

44 Now it remains to prove that lim b α (t, x) µ α (dt dx) = b(t, x) µ(dt dx). α 0 [0,T ] [0,T ] This fact depends on the different situations one is dealing with. Concrete examples can be founded in the papers Bogachev, DP, Röckner, Doklady 2007, JFA 2010

45 Uniqueness We again assume ypothesis 2 (i) (BB ) 1 L(). (ii) There exists a sequence (b α ) α (0,1] C 1 ([0, T ]; Cb 1 (; )) such that lim b α(t, x) = b(t, x), α 0 (t, x) [0, T ] D(b).

46 Theorem Let Λ be the set of all solutions µ(dt, dx) = µ t (dx)dt to the Fokker Planck equation (11) such that µ 0 P 1 (). there is C > 0 such that T 0 We have dt b α (t, x) 2 µ t (dx) C, α (0, 1]. T lim dt b(t, x) b α (t, x) 2 µ t (dx) = 0. (19) α 0 0 Then #(Λ) 1.

47 Proof In the first part of the proof we shall show that K is essentially m dissipative in L 1 ([0, T ] ; µ) for any µ Λ such that (18) and (19) are fulfilled. Using this result, we shall finally show the uniqueness using a simple argument of measure theory. We shall proceed in different steps.

48 Step 1. Let µ Λ. Then K is dissipative in L 1 ([0, T ] ; µ). In fact, for any u E ([0, T ] ) we have K (u 2 ) = 2u K u + B D x u 2. Now, replacing u with u 2, in the Fokker Planck equation (13) we have,

49 T 0 = ds K u(s, x) u(s, x) µ s (dx) T 0 T 0 ds B D x u 2 µ s (dx) 1 2 u 2 (s, x) µ 0 (dx) ds B D x u 2 µ s (dx), u E ([0, T ] ). (18) So, K is dissipative on L p ([0, T ], µ) for any p 1, where µ(dt, dx) = µ t (dx)dt.

50 In order to show that K is essentially m-dissipative on L 1 ([0, T ], µ) we introduce the approximating equation dx α (t) = [AX α (t) + b α (t, X α (t))]dt + BdW (t), X α (0) = x, s t (19) and the corresponding Kolmogorov equation where f C 1 ([0, T ] ) and λ R. λu α K α u α = f. (20)

51 Step 2 Estimate for the gradient of the approximating equation. There exists a constant C(λ, f 0 ) > 0 such that T 0 dt C(λ, f 0 ) 1 + D x u α (t, x) 2 µ t (dx) ( T 0 dt ) b α (t, x) b(t, x) 2 µ t (dx). (21)

52 To prove (21) we argue as before proving that u α belongs to the domain of K. Therefore we can write (20) as λu α K u α = f + b α b, D x u α. (22) Notice now that by the maximum principle we have u α 0 1 λ f 0. (23)

53 Then multiplying both sides of (22) by u α, integrating on µ t (dx)dt over [0, T ] and taking into account of (18) yields T λ dt uα(t, 2 x)µ t (dx) T 0 T 0 dt dt B D x u α (t, x) 2 µ t (dx)µ t (dx)ds f (t, x) u α (t, x) µ t (dx) T + dt 0 b α (t, x) b(t, x) D x u α (t, x) u α (t, x) µ t (dx). (24)

54 Taking into account (23) yields 1 2 T 0 dt B D x u α (t, x) 2 µ t (dx)µ t (dx) 1 λ f λ f T 0 dt b α (t, x) b(t, x) D x u α (t, x) µ t (dx) Now the claim follows, since (BB ) 1 is bounded, by a standard argument. (25)

55 Step 3. K is m-dissipative and his spectrum is empty. It is enough to show T lim dt b α b, D x u α µ t (dx) = 0. (26) α 0 0 This will imply that the range of λ K includes Cb 1 ([0, T ] ) so that it is dense in in L 1 ([0, T ], µ). Then the conclusion will follow from the Lumer Phillips theorem.

56 To prove (26) notice that by the ölder inequality we have T 0 T 0 b α b, D x u α dtµ t (dx) 2 T b α b 2 dtµ t (dx) D x u α (t, x) 2 dtµ t (dx). 0 Now the conclusion follows from (21).

57 Step 4 Conclusion Let µ (1) (dt dx) = µ (1) t (dx)dt, µ (2) (dt dx) = µ (2) t (dx)dt be two solutions of Fokker Planck equation (13). Then we have u(0, x)dµ 0 = K u dµ (1) so that = [0,T ] [0,T ] [0,T ] K u dµ (2), u D(K ). K u d(µ (1) µ (2) ) = 0 (27)

58 We cannot conclude by (27) that µ (1) = µ (2) because we do not know whether the range of K is dense in the space of integrable functions with respect to the signed measure µ (1) µ (2). We only know, by the previous steps, that K (D(K )) is dense both in L 2 ([0, T ] ; µ (1) ) and in L 2 ([0, T ] ; µ (2) ). So, we proceed as follows. We define µ := 1 2 (µ(1) + µ (2) ). Then µ is also a solution of the Fokker Planck equation (13).

59 Consequently K is essentially m dissipative in L 2 ([0, T ] ; µ) and that K (D(K )) is dense in L 2 ([0, T ] ; µ). Moreover, since µ (1) << µ, µ (2) << µ there exist densities σ 1 and σ 2 such that µ (1) (dt dx) = σ 1 (t, x)µ(dt dx), µ (2) (dt dx) = σ 2 (t, x(µ(dt dx) and as easily seen. 0 σ 1 2, 0 σ 2 2,

60 Finally, by [0,T ] K u (σ 1 σ 2 )dµ = 0. we deduce that σ 1 σ 2 = 0 because σ 1 σ 2 L ([0, T ], µ), and K (D(K )) is dense in L 2 ([0, T ], µ) So, µ (1) = µ (2).