Fokker–Planck Equations in Hilbert Spaces

Abstract

This paper includes in an unified way several results about existence and uniqueness of solutions of Fokker–Planck equations from (Bogachev et al., J Funct Anal, 256:1269–1298, 2009) [2], (Bogachev et al., J Evol Equ, 10(3):487–509, 2010) [3], (Bogachev et al., Partial Differ Equ, 36:925–939, 2011) [4] and (Bogachev et al., Bull London Math Soc 39:631–640, 2007) [1], using probabilistic methods. Several applications are provided including Burgers and 2D-Navier–Stokes equations perturbed by noise. Some of these applications were also studied by a different analytic approach in (Bogachev et al., J Differ Equ, 259(8):3854–3873, 2015) [5], (Bogachev et al., Ann Sc Norm Super Pisa Cl Sci 14(3):983–1023, 2015) [6], (Da Prato et al., Commun Math Stat, 1(3):281–304, 2013) [11].

1 Introduction and Setting of the Problem

We are given a separable Hilbert space H with norm $|\cdot |$ and inner product $\langle \cdot , \cdot \rangle .$ We denote by $\mathcal B(H)$ the set of all Borel subsets of H and by $\mathcal P(H)$ the set of all Borel probability measures on H. If $\psi :H\rightarrow [0,+\infty ]$ is convex and lower semi-continuous we denote by $\mathcal P_\psi (H)$ the set of all elements $\zeta \in \mathcal P(H)$ such that $\int _H \psi (x)\zeta (dx)<\infty .$ Clearly $\mathcal P_\psi (H)$ is a convex subset of $\mathcal P(H)$. We shall assume throughout the paper the following

Hypothesis 1

(i) $A:D(A)\subset H\rightarrow H$ is self-adjoint and there exists $\omega >0$ such that $\langle Ax,x \rangle \le -\omega |x|^2,\;\forall \;x\in D(A).$

(ii) There exists $\epsilon _0\in (0,1/2)$ such that $(-A)^{-1+2\epsilon _0}$ is of trace class.

(iii) $C:H\rightarrow H$ is bounded, symmetric and nonnegative.

(iv) $F:D(F)\subset [0,T]\times H\rightarrow H$ is Borel, whereas D(F) is a Borel subset of $[0,T]\times H$.

We have taken A and C time independent and A self-adjoint for the sake of simplicity, but more general cases (including: A infinitesimal generator of a $C_0$ semigroup and time dependent and equations with multiplicative noise) could be considered without important changes in the proofs.

For any $\beta \in (0,1)$ we shall use the notation $\Vert x\Vert _{\beta }:=|(-A)^\beta x|,\;\forall \;x\in D((-A)^\beta ).$ We recall that for any $\beta >0$ the embedding $D((-A)^\beta )\subset H$ is compact and that there exists $k_\beta >0$ such that

$$\begin{aligned} \Vert (-A)^\beta e^{tA}\Vert \le k_\beta t^{-\beta },\quad \forall \;t>0. \end{aligned}$$

(1)

A probability kernel $[0,T]\rightarrow \mathcal P(H),\; t\mapsto \mu _t,$ is a mapping $[0,T]\rightarrow \mathbb R,\; t\mapsto \mu _t(I)$ such that is measurable for any $I\in \mathcal B(H)$. We identify a probability kernel $(\mu _t)_{t\in [0,T]}$ with the measure $\underline{\mu }(dx\,dt)=\mu _t(dx)dt$ on $([0,T]\times H, \mathcal B([0,T])\times H))$ defined as

$$ \underline{\mu }([a,b]\times I)=\int _a^b\mu _t(I)dt,\quad \forall \;a,b\in \mathbb R,\;I\in \mathcal B(H). $$

Let us introduce the space of exponential functions; they will play the role of test functions. For any $h\in H$ we set $\varphi _h=e^{i\langle h,x \rangle },\; x\in H.$ Moreover, we denote by $\mathcal E_A(H)$ the linear span of the real parts of functions $\varphi _h$ such that $h\in D(A)$ and by $\mathcal E_A([0,T]\times H)$ the linear span of all functions of the form $u(t,x)=g(t)\varphi (x),\;(t,x)\in [0,T]\times H,$ where $g\in C^1([0,T]), \, g(T)=0$ and $\varphi \in \mathcal E_A(H)$. Finally, we define the Kolmogorov operator $K_F$, by setting

$$\begin{aligned} K_Fu(t,x)=D_tu(t,x)+\frac{1}{2}\,\text{ Tr }\;[CD^2_x u(t,x)]+ \langle x, AD_x u(t,x)\rangle + \langle F(t,x),D_x u(t,x)\rangle , \end{aligned}$$

(2)

for any $u\in \mathcal E_A([0,T]\times H)$ and $(t,x)\in [0,T]\times H.$

We consider the following problem. Given $\zeta \in \mathcal P(H)$, we want to find a probability kernel $\underline{\mu }=(\mu _t)_{t\in [0,T]}$ such that $\mu _0=\zeta$ and for all $u\in \mathcal E_A([0,T]\times H)$, $K_F u$ is $\underline{\mu }$-integrable and it results

$$\begin{aligned} \int _{H_T} K_F u(t,x)\,\mu _t(dx) dt=-\int _Hu(0,x)\,\zeta (dx), \end{aligned}$$

(3)

where $H_T=[0,T]\times H.$ $\underline{\mu }$ will be called a solution of the Fokker–Planck equation (3).

Let us describe the content of the paper. Section 2 is devoted to some preliminaries about the Ornstein–Uhlenbeck semigroup, following [2]. In Sect. 3 we present some existence result for the Fokker–Planck equation (3) under additional assumptions, Hypotheses 2 or 3. The first one is closely related to the paper [3], whereas the slightly different second one allows us to consider more general nonlinearities as Burgers and 2D-Navier–Stokes equations. Finally, Sect. 4 is devoted to uniqueness. Here we start from a basic rank condition, introduced in [1] and distinguish two cases: (i) C is invertible and $C^{-1}$ is bounded and (ii) C is of trace class. We note that an important consequence of uniqueness is that in this case the Chapman–Kolmogorov equation is well posed, see [3] for a thorough discussion of this fact.

We end this section with some notations used in what follows. By $C_b(H)$ we mean the space of all real continuous and bounded mappings $\varphi :H\rightarrow \mathbb R$ endowed with the sup norm $\Vert \cdot \Vert _{0}$. Moreover, $C^1_b(H)$ is the subspace of $C_b(H)$ of all continuously differentiable functions with bounded derivatives. Finally, $B_b(H)$ is the space of all real Borel mappings $\varphi :H\rightarrow \mathbb R$ endowed with the sup norm $\Vert \cdot \Vert _{0}$.

2 Preliminaries on the Ornstein–Uhlenbeck Semigroup

It is well known (see e.g. [8] or [13]) that under Hypothesis 1 the equation

$$\begin{aligned} \left\{ \begin{array}{l} dZ=AZ dt+\sqrt{C}\,dW(t),\quad t\ge 0\ \\ X(s)=x,\quad s\le t, \end{array}\right. \end{aligned}$$

(4)

has a unique mild solution given by

$$\begin{aligned} Z(t,s,x)=e^{(t-s)A}x+ W_A(t,s), \end{aligned}$$

(5)

where the stochastic convolution $W_A(t,s)$ is defined as

$$\begin{aligned} W_A(t,s)=\int _s^t e^{(t-r)A}\sqrt{C}\,dW(r),\quad \forall \;0\le s\le t. \end{aligned}$$

(6)

The following lemma will be used later

Lemma 1

For all $\beta \in [0,\epsilon _0]$ and $0\le s\le t\le T$ we have

$$\begin{aligned} \mathbb E|(-A)^\beta \,W_A(t,s)|^2\le \Vert C\Vert \,\text{ Tr }\,[(-A)^{-1+2\beta }]=:C(\beta ) \end{aligned}$$

(7)

Proof

We have in fact

$$\begin{aligned} \mathbb E|(-A)^\beta W_A(t,s)|^2&=\text{ Tr }\,\left[ C\int _s^t(-A)^{2\beta }e^{2rA}dr\right] \\&\le \frac{1}{2}\,\Vert C\Vert \, \text{ Tr }\, [(-A)^{2\beta -1}(e^{2sA}-e^{2tA})] \le \Vert C\Vert \, \text{ Tr }\, [(-A)^{2\beta -1}. \end{aligned}$$

$\square$

Let us consider the transition evolution operator $R_{s,t}$, corresponding to Eq. (4), which acts, in particular, on $C_{b,1}(H)$, the space of all real continuous mappings $\varphi$ in H such that $\sup _{x\in H}\,\tfrac{|\varphi (x)|}{1+|x|}<\infty$. For $0\le s\le t\le T$ we have

$$\begin{aligned} R_{s,t}\varphi (x):=\displaystyle \int _H\varphi (y)N_{e^{(t-s)A}x,Q_{t-s}}(dy)= \mathbb E[\varphi (Z(t,s,x))],\quad \forall \;\varphi \in C_{b,1}(H). \end{aligned}$$

(8)

Here $N_{e^{(t-s)A}x,Q_{t-s}}$ represents the Gaussian measure in H with mean $e^{(t-s)A}x$ and covariance

$$ Q_{t-s}=\int _0^{t-s} e^{rA}Ce^{rA}dr. $$

See e.g. [8]. Let us define a semigroup $S^{0,1}_\tau ,\,\tau \ge 0,$ in the space

$$C_T([0,T];C_{b,1}(H))=\{u\in C([0,T];C_{b,1}(H)):\;u(T,x)=0,\forall \;x\in H \},$$

by setting

$$\begin{aligned} (S^{0,1}_\tau u)(t,x)&= (R_{t,t+\tau }u(t+\tau ,\cdot )(x)\,{\mathbbm {1}}_{[0,T-\tau ]}(t) \nonumber \\&=\mathbb E[u(t+\tau ,Z(t+\tau ,t,x))]\,{\mathbbm {1}}_{[0,T-\tau ]}(t), \quad \forall \,u\in C_T([0,T];C_{b,1}(H)). \end{aligned}$$

(9)

Note that $S^{0,1}_\tau =0,\; \forall \;\tau \ge T.$ Let denote by $\mathcal K^1_0$¹ the infinitesimal generator of $S^{0,1}_\tau$, defined through its resolvent following [7] (see (10) below). The resolvent set of $\mathcal K^1_0$ coincides with $\mathbb R$ and its resolvent is given, for all $f\in C_T([0,T];C_{b,1}(H))$ and all $\lambda \in \mathbb R$, by

$$\begin{aligned} \displaystyle R(\lambda ,\mathcal K^1_0)f(t,x)&=\int _0^\infty e^{-\lambda \tau }S^{0,1}_\tau f(t,x)d\tau \displaystyle \nonumber \\&=\int _t^{T} e^{-\lambda (r-t)}\mathbb E[f(Z(r,t,x))]dr, \quad \forall \;(t,x)\in [0,T]\times H. \end{aligned}$$

(10)

$\mathcal K^1_0$ can also be defined as follows, following [14]. We say that a function $u\in C_T([0,T];C_{b,1}(H))$ belongs to the domain of $\mathcal K^1_0$ if there exists a function $f\in C_T([0,T];C_{b,1}(H))$ such that

(i)

$\displaystyle \lim _{\tau \rightarrow 0}\frac{1}{\tau }\;(S^{0,1}_\tau u(t,x)-u(t,x))=f(t,x),\quad \forall \;(t,x)\in [0,T]\times H.$

(ii) There exists $C_u>0$ such that

$$ \left| \frac{1}{\tau }\;(S^{0,1}_\tau u(t,x)-u(t,x))\right| \le C_u,\quad \forall \;(t,x)\in [0,T]\times H,\,\forall \;\tau \in [0,1]. $$

We set $\mathcal K^1_0 u=f$ and call $\mathcal K^1_0$ the infinitesimal generator of $S_\tau$ on $C_T([0,T];C_{b,1}(H))$. It is not difficult to check that $\mathcal E_A([0,T]\times H)\subset D(\mathcal K^1_0)$ and that for all $u\in \mathcal E_A([0,T]\times H)$ it results $\mathcal K^1_0 \,u=K_0\,u$. So, the abstract operator $\mathcal K^1_0$ is an extension of the differential Kolmogorov operator $K_0$. If, in particular, $u(t,x)=e^{i\langle x,h \rangle }\,g(t),$ $t\in [0,T],$ $x\in H,$ where $h\in D(A),$ $g\in C^1([0,T])$ and $g(T)=0$, we have

$$ K_0u(t,x)=[g'(t)-(\tfrac{1}{2}\,|C^{1/2}h|^2+ i\langle x, Ah\rangle )g(t)]e^{i\langle x,h \rangle }. $$

So, $K_0u$ has a linear growth in x; this is the reason for introducing the space $C_{b,1}(H)$ and for requiring that $h\in D(A)$.

One can show that functions from $C_T([0,T];C_{b,1}(H))$ can be approximated point-wise by sequences (more precisely multi-sequences) of elements of $\mathcal E_A([0,T]\times H).$ To describe this approximation, we shall use the following notation

$$ \lim _{n_1\rightarrow \infty }\lim _{n_2\rightarrow \infty }\lim _{n_3\rightarrow \infty }u_{n_1,n_2,n_3}=\lim _{\mathbf {n}\rightarrow \infty }u_{\mathbf {n}}. $$

The following three propositions were proven in [2], see also [9].

Proposition 2

For all $u\in C_T([0,T];C_{b,1}(H))$ there exists $(u_{\mathbf {n}})\subset \mathcal E_A([0,T]\times H)$ such that

(i) $\displaystyle \lim _{\mathbf {n}\rightarrow \infty }u_{\mathbf {n}}(t,x)=u(t,x),\quad \forall \;(t,x)\in [0,T]\times H.$

(ii) $|u_{\mathbf {n}}(t,x)|\le \Vert u\Vert _{C_T([0,T];C_{b,1}(H))}(1+|x|),\quad \forall \;(t,x)\in [0,T]\times H.$

Proposition 3

For any $u\in D(\mathcal K^1_0)$ there exists $(u_{\mathbf {n}})\subset \mathcal E_A([0,T]\times H)$ and $C>0$ such that

(i) $\displaystyle \lim _{\mathbf {n}\rightarrow \infty }u_{\mathbf {n}}(t,x)=u(t,x),\quad \forall \;(t,x)\in [0,T]\times H.$

(ii) $\displaystyle \lim _{\mathbf {n}\rightarrow \infty }K_0\,u_{\mathbf {n}}(t, x)=\mathcal K^1_0\,u(t, x),\quad \forall \;(t,x)\in [0,T]\times H.$

(iii) $\displaystyle |u_{\mathbf {n}}(x)|+| K_0u_{\mathbf {n}}(t, x)|\le C(1+|x|),\quad \forall \;(t,x)\in [0,T]\times H$

We say that $\mathcal E_A([0,T]\times H)$ is a core for $\mathcal K^1_0$.

Proposition 4

Let $f, D_xf\in C_T([0,T];C_{b,1}(H;H))$, $\lambda \in \mathbb R$ and let $u=R(\lambda ,\mathcal K^1_0)f$. Then there exists $u_{\mathbf {n}}\subset \mathcal E_A([0,T]\times H)$ and $C>0$ such that

(i) $\displaystyle \lim _{\mathbf {n}\rightarrow \infty }u_{\mathbf {n}}(t,x)=u(t,x),\quad \forall \;(t,x)\in [0,T]\times H.$

(ii) $\displaystyle \lim _{\mathbf {n}\rightarrow \infty }K_0\,u_{\mathbf {n}}(t,x)=\mathcal K^1_0 u(t,x),\quad \forall \;(t,x)\in [0,T]\times H.$

(iii) $\displaystyle \lim _{\mathbf {n}\rightarrow \infty }D_xu_{\mathbf {n}}(t,x)=D_xu(t,x),\quad \forall \;(t,x)\in [0,T]\times H.$

(iv) $\displaystyle |u_{\mathbf {n}}(t,x)|+|K_0u_{\mathbf {n}}(t,x)|+|D_x\varphi _{\mathbf {n}}(t,x)|\le C(1+|x|), \forall \,(t,x)\in [0,T]{\times } H.$

3 Existence

3.1 Basic Assumptions

We shall first introduce a suitable approximation $F_\alpha$ of F, $\alpha \in (0,1]$, such that the problem

$$\begin{aligned} \left\{ \begin{array}{l} dX_\alpha =(AX_\alpha +F_\alpha (t,X_\alpha ))dt+\sqrt{C}\,dW(t),\quad t\in [0,T],\\ X_\alpha (s)=x,\quad 0\le s\le t\le T \end{array}\right. \end{aligned}$$

(11)

has a unique mild solution $X_\alpha (t,s,x)$, that is

$$\begin{aligned} X_\alpha (t,s,x)=e^{(t-s)A}x+\int _s^te^{(t-r)A}F_\alpha (r, X_\alpha (r,s,x))dr+W_A(t,s). \end{aligned}$$

(12)

More precisely, we shall assume that

Hypothesis 2

For any $\alpha \in (0,1]$ there exists a mapping $F_\alpha :[0,T]\times H\rightarrow H, (t,x)\rightarrow F_\alpha (t,x)$ with the following properties:

(i) $F_\alpha$ is continuous and bounded and Eq. (11) has a unique mild solution.

(ii) There exists a convex and lower semicontinuous mapping $V:H\rightarrow [1,+\infty ]$ such that

$$\begin{aligned} |F_\alpha (t,x)|^2\le |F(t,x)|^2\le V(x),\quad \forall \;(t,x)\in [0,T]\times H, \end{aligned}$$

(13)

and

$$\begin{aligned} |F(t,x)-F_\alpha (t,x)|\le \alpha \,V(x) ,\quad \forall \;(t,x)\in [0,T]\times H,\;\alpha \in (0,1]. \end{aligned}$$

(14)

(iii) There exists $M>0$ such that

$$\begin{aligned} P^\alpha _{s,t}\,V(x)\le MV(x),\quad \forall \;0\le s\le t\le T,\, \alpha \in (0,1],\,x\in H, \end{aligned}$$

(15)

where $P^\alpha$ is the transition evolution operator, $P^\alpha _{s,t}\,\varphi (x)=\mathbb E[\varphi (X_\alpha (t,s,x))]$, $0\le s\le t\le T,$ $\varphi \in B_b(H).$

Now, fix $\zeta \in \mathcal P(H);$ then for any $t\in (0,T]$ and any $\alpha \in (0,1]$ we define a probability measure $\mu _t^\alpha$ in H such that (we take here $s=0$ for simplicity)

$$\begin{aligned} \int _H\varphi (x)\mu _t^\alpha (dx)&=\int _H P^\alpha _{0,t}\,\varphi (x)\zeta (dx)\nonumber \\&= \int _H \mathbb E[\varphi (X_\alpha (t,0,x))]\zeta (dx),\quad \forall \;\varphi \in \mathcal E_A(H) \end{aligned}$$

(16)

and set $\underline{\mu }^\alpha (dt\,dx)=\mu _t^\alpha (dx)dt.$ Let us deduce some consequences of Hypothesis 2 assuming that $\int _HV(x)\,\zeta (dx)<\infty$. First from (15) and (16) we find

$$\begin{aligned} \int _H V(x)\,\mu ^\alpha _t(dx)=\int _HP_{0,t}V(x)\,\zeta (dx)\le M\int _HV(x)\,\zeta (dx) \end{aligned}$$

(17)

and so, from (13)

$$\begin{aligned} \int _H |F_\alpha (t,x)|^2\,\mu ^\alpha _t(dx)\le \int _H |F(t,x)|^2\,\mu ^\alpha _t(dx)\le M\int _HV(x)\,\zeta (dx). \end{aligned}$$

(18)

Other useful estimates are provided by the following lemma.

Lemma 5

Assume that Hypotheses 1 and 2 are fulfilled. Let $x\in H$, $0<s\le t\le T$ and $\alpha \in (0,1]$; then $X_\alpha (t,s,x)\in D((-A)^{\epsilon _0})$, $\mathbb P$-a.s.² Moreover there exists $C>0$ such that

$$\begin{aligned} P^\alpha _{s,t}(|x|^2)=\mathbb E[| X_\alpha (t,s,x)|^2]\le C (1+|x|^2+ V(x)) \end{aligned}$$

(19)

and

$$\begin{aligned} P^\alpha _{s,t}(\Vert x\Vert _{\epsilon _0})=\mathbb E[|(-A)^{\epsilon _0} X_\alpha (t,s,x)|]\le Ct-s)^{-\epsilon _0}(1+|x|^2+ V(x)). \end{aligned}$$

(20)

Proof

Let us write $X_\alpha (t,s,x)=X_\alpha (t)$ for short. Then by (12) and Hölder’s inequality we have

$$ |X_\alpha (t)|^2\le 3 |x|^2+3T\int _s^t|F_\alpha (r,X_\alpha (r))|^2dr+3|W_A(t,s)|^2. $$

Now, taking expectation, recalling (7) and invoking (13) we obtain

$$ \mathbb E|X_\alpha (t)|^2\le 3 |x|^2+3T\int _s^t\mathbb E[V_\alpha (X_\alpha (r)]dr+3C(0). $$

Finally, taking into account (15), yields

$$ \mathbb E|X_\alpha (t)|^2\le 3 |x|^2+3T^2MV(x)+3C(0), $$

which implies (19) with a suitable $C>0$. To prove (20) write

$$ (-A)^{\epsilon _0} X_\alpha (t)=(-A)^{\epsilon _0} e^{(t-s)A}x+\int _s^t(-A)^{\epsilon _0} e^{(t-r)A}F_\alpha (r,X_\alpha (r))dr\ +(-A)^{\epsilon _0} W_A(t,s), $$

from which, recalling (1) and using that $|F_\alpha (r,X_\alpha (r))|\le 1+|F_\alpha (r,X_\alpha (r))|^2$, we have

$$\begin{aligned} |(-A)^{\epsilon _0} X_\alpha (t)|&\le \kappa _{\epsilon _0} \,(t-s)^{-\epsilon _0}|x|+\kappa _{\epsilon _0}\int _s^t(t-r)^{-\epsilon _0}(1+|F_\alpha (r,X_\alpha (r))|^2)dr\\&\quad +|(-A)^{\epsilon _0} W_A(t,s)|. \end{aligned}$$

Taking expectation and proceeding as before, we obtain

$$\begin{aligned} \displaystyle \mathbb E[|(-A)^{\epsilon _0} X_\alpha (t)|]&\le \displaystyle \kappa _{\epsilon _0} \,(t-s)^{-\epsilon _0}|x|+\kappa _{\epsilon _0}\int _s^t(t-r)^{-\epsilon _0}(1+V(X_\alpha (r))dr\\&\quad +(\mathbb E|(-A)^{\epsilon _0} W_A(t,s)|^2)^{1/2}\\&\le \displaystyle \kappa _{\epsilon _0}\,(t-s)^{-\epsilon _0} |x|^2+\kappa _{\epsilon _0}\,\frac{T^{1-\epsilon _0}}{1-\epsilon _0}(1+MV_\alpha (x))+C^{1/2}(\epsilon _0) \end{aligned}$$

which easily yields (20). $\square$

In the following we set

$$\begin{aligned} \psi (x):=1+|x|^2+V(x),\quad \forall \;x\in H. \end{aligned}$$

(21)

$\psi$ is convex and lower semicontinuous.

Corollary 6

Assume that Hypotheses 1 and 2 are fulfilled and let $\zeta \in \mathcal P_\psi (H)$. Then we have

$$\begin{aligned} \int _H|x|^2\,\mu _t^\alpha (dx)\le C \int _H \psi (x) \zeta (dx),\quad \forall \;\alpha \in (0,1] \end{aligned}$$

(22)

and

$$\begin{aligned} \int _H\Vert x\Vert _{\epsilon _0}\,\mu _t^\alpha (dx)\le C t^{-\epsilon _0}\int _H \psi (x) \zeta (dx),\quad \forall \;\alpha \in (0,1]. \end{aligned}$$

(23)

Proof

The proof follows by integrating both sides of (19) and (20) (with $s=0$) with respect to $\zeta$ over H. $\square$

We notice now that by Itô’s formula the measure $\underline{\mu }^\alpha (dt\,dx)=\mu ^\alpha _t(dx)$ solves the approximating equation

$$\begin{aligned} \int _{H_T} K_{F_\alpha } u(t,x)\mu _t^\alpha (dx)\,dt=-\int _H u(0,x)\, \zeta (dx) ,\quad \forall \;u\in \mathcal E_A([0,T]\times H), \end{aligned}$$

(24)

where $K_{F_\alpha }$ is the Kolmogorov operator

$$\begin{aligned} K_{F_\alpha }u =D_tu +\frac{1}{2}\,\text{ Tr }\;[CD^2_x u ]+ \langle x, AD_x u \rangle + \langle F_\alpha ,D_x u \rangle ,\quad \forall \,u\in \mathcal E_A([0,T]\times H). \end{aligned}$$

(25)

We are going to construct a sequence $(\mu ^{\alpha _h}_t)$ of measures weakly convergent to a measure $\mu _t$ for all $t\in [0,T]$ and finally, we shall pass to the limit as $h\rightarrow \infty$ in (24) (with $\alpha _h$ replacing $\alpha$) and prove that $\underline{\mu }(dx\,dt)=\mu _t(dx)\,dt$ is a solution of the Fokker–Planck equation (3).

3.2 Tightness

Proposition 7

Assume that $\zeta \in \mathcal P_{\psi }(H)$. Then $(\mu _t^\alpha )_{\alpha \in (0,1]}$ is tight for all $t\in (0,T]$ and we have

$$\begin{aligned} \mu _t^\alpha (\Vert x\Vert _{\epsilon _0}\ge R)\le \frac{Ct^{-\epsilon _0}}{R} \int _H \psi (x) \zeta (dx),\quad \forall \;R>0. \end{aligned}$$

(26)

Proof

Thanks to (23), we have for all $R>0$

$$ \begin{array}{l} \displaystyle \mu _t^\alpha (\Vert x\Vert _{\epsilon _0}\ge R)=\int _{\Vert x\Vert _{\epsilon _0}\ge R}\mu _t^\alpha (dx)\le \frac{1}{R}\int _H\Vert x\Vert _{\epsilon _0}\,\mu _t^\alpha (dx)\le \frac{Ct^{-\epsilon _0}}{R} \int _H \psi (x) \zeta (dx). \end{array} $$

The conclusion follows from the arbitrariness of R because balls in the topology of $D((-A)^{\epsilon _0})$ are compact in H. $\square$

Theorem 8

Assume that Hypotheses 1 and 2 are fulfilled and let $\zeta \in \mathcal P_{\psi }(H)$.³ Then there is a probability kernel $\underline{\mu }(dt\,dx)=\mu _t(dx)\,dt$ solving (3) and such that $\mu _t\in \mathcal P_\psi$ for all $t\in [0,T]$, where $\psi$ is defined by (21).

Proof

Since for any $t\in (0,T]$, $(\mu ^\alpha _t)_{\alpha \in (0,1]}$ is tight, there exists a sequence $\alpha _{h}(t)\rightarrow 0$ and $\mu _t\in \mathcal P(H)$ such that

$$ \lim _{h\rightarrow \infty }\int _{H}\varphi \,d\mu ^{\alpha _{h}(t)}_t=\int _{H}\varphi \,d\mu _t,\quad \forall \;\varphi \in C_b(H)\,t\in [0,T]. $$

By a diagonal extraction argument we can find a sequence $(\alpha _h)\rightarrow 0$ and a family of measures $(\mu _t)_{t\in \mathbb Q}$ such that⁴

$$\begin{aligned} \lim _{h\rightarrow \infty }\int _{H}\varphi \,d\mu ^{\alpha _h}_t=\int _{H}\varphi \,d\mu _t,\quad \forall \;\varphi \in C_b(H),\quad \forall \;t\in [0,T]\cap \mathbb Q. \end{aligned}$$

(27)

Now we are going to extend the family $(\mu ^{\alpha _h}_t)_{t\in \mathbb Q}$ to the whole interval (0, T] in such a way that the extension is a solution of (3). We shall proceed in three steps.

Step 1. For each $\varphi \in \mathcal E_A(H)$ there exists $C(\varphi )>0$ such that for all $\alpha \in (0,1]$ we have

$$\begin{aligned} \left| \int _H\varphi \,d\mu ^\alpha _t-\int _H\varphi \,d\mu ^\alpha _s\right| \le C(\varphi ) (t-s)\int _H\psi \,d\zeta ,\quad 0\le s\le t\le T. \end{aligned}$$

(28)

In fact by Itô’s formula we have

$$\begin{aligned} \mathbb E[\varphi (X_\alpha (t,0,x))]-\mathbb E[\varphi (X_\alpha (s,0,x))]=\mathbb E\int _s^t(L_r^\alpha \varphi )(X_\alpha (r,0,x))dr, \end{aligned}$$

(29)

where $L_t^\alpha ,\,t\in [0,T],$ is the Kolmogorov operator

$$ L_t^\alpha \varphi =\tfrac{1}{2}\,\text{ Tr }\,[CD^2\varphi ]+\langle x, AD\varphi \rangle +\langle F_\alpha (t,x), D\varphi \rangle ,\quad \forall \varphi \in \mathcal E_A(H). $$

Now, taking into account (16), write

$$\begin{aligned} \begin{array}{lll} \displaystyle \int _H\varphi \,d\mu ^\alpha _t-\int _H\varphi \,d\mu ^\alpha _s&{}=&{}\displaystyle \int _H(\mathbb E[\varphi (X_\alpha (t,0,x))]-\mathbb E[\varphi (X_\alpha (s,0,x))])\zeta (dx)\\ \\ &{}=&{}\displaystyle \mathbb E\int _H\int _s^t ( L_t^\alpha \varphi )(X(r,0,x))\,\zeta (dx)\,dr. \end{array} \end{aligned}$$

(30)

Let $C_1(\varphi )>0$ be such that

$$\begin{aligned}|( L_t^\alpha \varphi )(x)|&\le C_1(\varphi )(1+|x|+|F_\alpha (t,x)|)\\&\le 3C_1(\varphi )(1+|x|^2+|F_\alpha (t,x)|^2),\quad \forall \;x\in H. \end{aligned}$$

Therefore from (30), taking into account (15), (13) and (22), we find

$$\begin{aligned} \displaystyle \left| \int _H\varphi \,d\mu ^\alpha _t-\int _H\varphi \,d\mu ^\alpha _s\right|&\le \displaystyle 3C_1(\varphi )\int _H\int _s^t (1+ \mathbb E|X_\alpha (t,0,x))|^2\\&\quad +\mathbb E|F_\alpha (X_\alpha (t,0,x))|^2)\zeta (dx)dr\\&\le \displaystyle 3C_1(\varphi )(t-s)\int _H(1+C((1+|x|^2+V(x))\\&\quad +MV(x))\zeta (dx). \end{aligned}$$

So, (28) follows with a suitable constant $C(\varphi )$.

Step 2. Construction of $\mu _t$ for all $t\in (0,T]$.

Let $t_0\in (0,T]\setminus \mathbb Q$ and let $(t_j)$ be a sequence in $(0,T]\cap \mathbb Q$ convergent to $t_0$. The sequence $(\mu _{t_j})$ is tight by Proposition 7. Let us choose a limit point of $(\mu _{t_j})$ which we call $\mu _{t_0}$, that is (for a subsequence which we still call $(t_j)$)

$$\begin{aligned} \lim _{j\rightarrow \infty }\int _H\varphi \,d\mu _{t_j}=\int _H\varphi \,d\mu _{t_0},\quad \forall \;\varphi \in C_b(H). \end{aligned}$$

(31)

Claim. We have

$$\begin{aligned} \lim _{h\rightarrow \infty }\int _H\varphi \,d\mu ^{\alpha _h}_{t_0}=\int _H\varphi \,d\mu _{t_0},\quad \forall \;\varphi \in C_b(H). \end{aligned}$$

(32)

To prove the claim it is enough to show that

$$\begin{aligned} \lim _{h\rightarrow \infty }\int _H\varphi \,d\mu ^{\alpha _h}_{t_0}=\int _H\varphi \,d\mu _{t_0},\quad \forall \;\varphi \in \mathcal E_A(H). \end{aligned}$$

(33)

In fact, assume that we have proved (33). Since $(\mu ^{\alpha _h}_{t_0})$ is tight by (26), there exists a subsequence $(\mu ^{\alpha _{h_k}}_{t_0})$ weakly convergent to a probability measure $\nu$, that is such that

$$ \lim _{k\rightarrow \infty }\int _H\varphi \,d\mu ^{\alpha _{h_k}}_{t_0}=\int _H\varphi \,d\nu ,\quad \forall \;\varphi \in C_b(H). $$

On the other hand, by (33) it follows that

$$ \lim _{k\rightarrow \infty }\int _H\varphi \,d\mu ^{\alpha _{h_k}}_{t_0}=\int _H\varphi \,d\mu _{t_0},\quad \forall \;\varphi \in \mathcal E_A(H) $$

so that

$$ \int _H\varphi \,d\nu =\int _H\varphi \,d\mu _{t_0},\quad \forall \;\varphi \in \mathcal E_A(H), $$

which implies $\nu =\mu _{t_0}$ because $\mathcal E_A(H)$ is dense in $L^1(H,\nu )$.

Now we can prove the Claim. Let $\varphi \in \mathcal E_A(H)$. Then we have for any $j\in \mathbb N$

$$\begin{aligned} \displaystyle \left| \int _H\varphi \,d\mu _{t_0}-\int _H\varphi \,d\mu ^{\alpha _h}_{t_0}\right|&\le \left| \int _H\varphi \,d\mu _{t_0}-\int _H\varphi \,d\mu _{t_j}\right| \displaystyle \\&\quad +\left| \int _H\varphi \,d\mu _{t_j}-\int _H\varphi \,d\mu ^{\alpha _h}_{t_j}\right| +\left| \int _H\varphi \,d\mu ^{\alpha _h}_{t_j}-\int _H\varphi \,d\mu ^{\alpha _h}_{t_0}\right| . \end{aligned}$$

Taking into account (28), yields

$$\begin{aligned} \displaystyle \left| \int _H\varphi \,d\mu _{t_0}-\int _H\varphi \,d\mu ^{\alpha _h}_{t_0}\right|&\le \displaystyle \left| \int _H\varphi \,d\mu _{t_0}-\int _H\varphi \,d\mu _{t_j}\right| \nonumber \\&\quad \displaystyle +\left| \int _H\varphi \,d\mu _{t_j}-\int _H\varphi \,d\mu ^{\alpha _h}_{t_j}\right| \nonumber \\&\quad +C(\varphi )|t_j-t_0|\int _H\,\psi \,d\zeta =:J_1+J_2+J_3. \end{aligned}$$

(34)

Given $\delta >0$ there is $j_\delta \in \mathbb N$ such that $J_1+J_3<\delta .$ Therefore by (34) we have

$$ \left| \int _H\varphi \,d\mu _{t_0}-\int _H\varphi \,d\mu ^{\alpha _h}_{t_0}\right| \le \left| \int _H\varphi \,d\mu _{t_{j_\delta }}-\int _H\varphi \,d\mu ^{\alpha _h}_{t_{j_\delta }}\right| +\delta . $$

Now the conclusion follows from (27) and the arbitrariness of $\delta$.

Step 3. Conclusion

Let $(\mu _t)_{t\in [0,T]}$ be the family defined in Step 2. We are going to prove that for all $u\in \mathcal E_A([0,T]\times H)$

$$\begin{aligned} \lim _{h\rightarrow \infty } \int _{H_T} K_{F_{\alpha _h}}\,u(t,x)\,\mu ^{\alpha _h}_t(dx) dt =\int _{H_T} K_F\,u(t,x) \,\mu _t(dx) dt. \end{aligned}$$

(35)

This will imply

$$\begin{aligned} \int _{H_T} K_F u(t,x) \,\mu _t(dx) dt=-\int _H u(0,x)\zeta (dx), \end{aligned}$$

(36)

so that $(\mu _t)_{t\in [0,T]}$ is a solution of (3). To prove (35) is enough to show that

$$\begin{aligned} \lim _{h\rightarrow \infty } \int _{H_T} \langle AD_xu(t,x),x\rangle \,\mu ^{\alpha _h}_t(dx) dt =\int _{H_T} \langle AD_xu(t,x),x\rangle \,\mu _t(dx) dt \end{aligned}$$

(37)

and

$$\begin{aligned} \lim _{h\rightarrow \infty } \int _{H_T} \langle D_xu(t,x),F_{\alpha _h}(t,x)\rangle \,\mu ^{\alpha _h}_t(dx) dt =\int _{H_T} \langle D_xu(t,x),F(t,x)\rangle \,\mu _t(dx) dt. \end{aligned}$$

(38)

Let us prove (37). Recalling that $D_xu(t,x)\in D(A)$, that Au is bounded and that $|x|\,|\langle AD_xu(t,x),x\rangle |\le \Vert ADu\Vert _0\,|x|^2,$ we have for any $\epsilon >0$, thanks to (19),

$$ \begin{array}{l} \displaystyle \left| \int _{H_T} \langle AD_xu(t,x),x\rangle \,\mu ^{\alpha _h}_t(dx) dt -\int _{H_T} \langle AD_xu(t,x),x\rangle \,\mu _t(dx) dt \right| \le \epsilon \left| \int _{H_T} \frac{|x|\langle AD_xu(t,x),x\rangle }{1+\epsilon |x|}\,\mu ^{\alpha _h}_t(dx)dt\right| \\ \\ \displaystyle +\left| \int _{H_T} \frac{\langle AD_xu(t,x),x\rangle }{1+\epsilon |x|}\,\mu ^{\alpha _h}_t(dx) dt -\int _{H_T} \frac{\langle AD_xu(t,x),x\rangle }{1+\epsilon |x|}\,\mu _t(dx) dt \right| +\epsilon \left| \int _{H_T} \frac{|x|\langle AD_xu(t,x),x\rangle }{1+\epsilon |x|}\,\mu _t(dx)dt\right| \\ \\ \displaystyle \le 2 \epsilon C \Vert ADu\Vert _\infty \int _{H_T} \psi (x)\,\zeta (x)\,(dx)+\left| \int _{H_T} \frac{\langle AD_xu(t,x),x\rangle }{1+\epsilon |x|}\,\mu ^{\alpha _h}_t(dx) dt -\int _{H_T} \frac{\langle AD_xu(t,x),x\rangle }{1+\epsilon |x|}\,\mu _t(dx) dt \right| \end{array}. $$

Now (37) follows letting first $h\rightarrow \infty$ and then $\epsilon \rightarrow 0$.

Next let us prove (38). Set $g_h(t,x):= \langle Du(t,x),F_{\alpha _h}(t,x)\rangle$ and $g(t,x):= \langle Du(t,x),F(t,x)\rangle ,$ so that (38) is equivalent to

$$\begin{aligned} \lim _{h\rightarrow \infty } \int _{H_T} g_h(t,x)\,\mu ^{\alpha _h}_t(dx) dt =\int _{H_T}g(t,x)\,\mu _t(dx) dt. \end{aligned}$$

(39)

Notice that by (14) we have

$$\begin{aligned} |g(t,x)-g_h(t,x)|\le \alpha _h \Vert Du\Vert _0\,(1+V(x)). \end{aligned}$$

(40)

Now write

$$\begin{aligned} \left| \int _{H_T} g\,d\mu _t\,dt- \int _{H_T}g_h\,d\mu ^{\alpha _h}_t\,dt \right|&\le \left| \int _{H_T} g\,d\mu _t\,dt- \int _{H_T}g\,d\mu ^{\alpha _h}_t \right| \\&\quad +\left| \int _{H_T} (g-g_h)\,d\mu ^{\alpha _h}_t\,dt \right| =:I_{h}+J_h. \end{aligned}$$

As far as $I_h$ is concerned, we have for any $j\in \mathbb N$,

$$\begin{aligned} \displaystyle I_h&:=\left| \int _{H_T} g\,d\mu _t\,dt- \int _{H_T}g\,d\mu _t^{\alpha _h}\,dt \right| \le \left| \int _{H_T}(g-g_j)\,d\mu _t\,dt \right| \\&\quad \displaystyle + \left| \int _{H_T}g_j\,d\mu _t\,dt- \int _Hg_j\,d\mu _t^{\alpha _h}\,dt \right| + \left| \int _{H_T}(g-g_j)\,d\mu _t^{\alpha _h}\,dt \right| . \end{aligned}$$

Therefore, taking into account (40), yields

$$ \begin{array}{l} \displaystyle I_h\le 2\alpha _j M\Vert Du\Vert _0\int _{H_T}(1+V(x))\,d\mu _t\,dt+ \left| \int _{H_T}g_j\,d\mu _t\,dt- \int _{H_T}g_j\,d\mu _t^ {\alpha _h}\,dt \right| . \end{array} $$

Now, given $\epsilon >0$, choose $j_0$ such that $2\alpha _{j_0} \Vert Du\Vert _0\int _{H_T}(1+V(x))\,d\mu _t\,dt<\frac{\epsilon }{2}.$ Then

$$ I_h\le \frac{\epsilon }{2}+ \left| \int _{H_T}g_{j_0}\,d\mu _t\,dt- \int _{H_T}g_{j_0}\,d\mu _t^ {\alpha _h}\,dt \right| , $$

so that $\lim _{h\rightarrow \infty }I_h=0,$ for the arbitrariness of $\epsilon$. Finally, as far as $J_h$ is concerned, we have

$$ J_h\le \alpha _hM\Vert Du\Vert _0\int _H(1+V(x))\,\mu _t^ {\alpha _h}(dx)\,dt\rightarrow 0, $$

as $h\rightarrow \infty$. The proof is complete. $\square$

Let us present now some examples.

Example 9

(Tr $C<\infty$) We assume here that Tr $C<\infty$, $F:[0,T]\times H\rightarrow H$ is continuous and there exist $k>0$ and $N\in \mathbb N$ such that

$$\begin{aligned} \langle F(t,x),x \rangle \le k,\quad |F(t,x)|\le k(1+|x|^{2N_1}),\quad \forall \;(t,x)\in [0,T]\times H. \end{aligned}$$

(41)

Then we set

$$\begin{aligned} F_\alpha (t,x)=\frac{F(t,x)}{1+\alpha |F(t,x)|},\quad \forall \;(t,x)\in [0,T]\times H, \end{aligned}$$

(42)

so that

$$\begin{aligned} |F(t,x)-F_\alpha (t,x)|\le \alpha |F(t,x)|^2,\quad \forall \;(t,x)\in [0,T]\times H, \end{aligned}$$

(43)

Let $X_\alpha$ be the solution to (11). Then by Itô’ formula and (41), for any $m\in \mathbb N$ there exists $a_m>0$ such that

$$\begin{aligned} \mathbb E[|X_\alpha (t,s,x)|^{2m}]\le e^{-2m\omega (t-s)}|x|^{2m}+a_m,\quad \forall \;(t,x)\in [0,T]\times H,\,\alpha \in (0,1]. \end{aligned}$$

(44)

Therefore Hypothesis 2 is fulfilled with $V(x)=C(1+|x|^{4N}),\;x\in H$.

Example 10

($C=I$) We assume here that $C=I$, $F:[0,T]\times H\rightarrow H$ is continuous and there exist $k >0$, $N_1,\,N\in \mathbb N$ such that for all $(t,x)\in [0,T]\times H$ we have

$$\begin{aligned} \langle F(t,x+z),x \rangle \le k(1+|z|^{2N_1}),\quad \forall \;z\in H\quad ,\quad |F(t,x)|\le k(1+|x|^{2N}). \end{aligned}$$

(45)

Then we define $F_\alpha$ by (42). Let $X_\alpha$ be the solution to (11). Set $Y_\alpha (t,s,x):=X_\alpha (t,s,x)-W_A(t,s),$ so that

$$\begin{aligned} \frac{d}{dt}\,Y_\alpha (t,s,x)=AY_\alpha (t,s,x)+F_\alpha (t,Y_\alpha (t,s,x)). \end{aligned}$$

(46)

Now, setting $Y(t)=Y_\alpha (t,s,x)$, it follows by a simple computation that for any $m\in \mathbb N$ we have

$$\begin{aligned} \frac{1}{2m}\,\frac{d}{dt}\,|Y(t)|^{2m}&=-|(-A)^{1/2}Y(t)|^2\,|Y(t)|^{2m-2}+|Y(t)|^{2m-2} \langle Y(t)+W_A(t,s),Y(t)\rangle \\&\le k(1+|W_A(t,s)|^{2N_2}). \end{aligned}$$

Recalling [8, Proposition 4.3] it follows that there is $C(T)>0$ such that $\frac{d}{dt}\,\mathbb E|Y(t)|^{2m}\le C(T),$ which implies

$$\begin{aligned} \mathbb E|Y_\alpha (t,s,x)|^{2m}\le |x|^{2m}+C(T),\quad \forall \;x\in H. \end{aligned}$$

(47)

Therefore Hypothesis 2 is fulfilled with $V(x)=C(1+|x|^{4N})$.

Example 11

(Reaction–diffusion equations) Here we take $H=L^2(0,1)$, $Ax=D^2_\xi$ for all $x\in D(A)=H^2(0,1)\cap H^1_0(0,1)$ and $F(t,x)(\xi )=\sum _{i=0}^Na_i(t)(x(\xi ))^i,\; t\in [0,T],\;\xi \in [0,1],\,x\in L^2(0,1),$ where $N\in \mathbb N$ is odd and greater than 1, $a_i\in C([0,T])$, $i=1,\ldots ,N$, and $a_N(t)<0$ for all $t\in [0,T].$ Then Hypothesis 1 is fulfilled. Moreover, setting

$$\begin{aligned} F_\alpha (t,x)=\frac{F(t,x)}{1+\alpha |F(t,x)|},\quad (t,x)\in [0,T]\times [0,1], \end{aligned}$$

(48)

we have

$$\begin{aligned} |F_\alpha (t,x)|_{L^2(0,1)}\le C\left( 1+|x|^{2N}_{L^{2N}(0,1)}\right) \end{aligned}$$

(49)

and

$$\begin{aligned} |F(t,x)-F_\alpha (t,x)|_{L^2(0,1)}\le C\alpha |F(t,x)|_{L^2(0,1)}^2\le C_1\left( 1+|x|^{4N}_{L^{2N}(0,1)}\right) . \end{aligned}$$

(50)

Now Hypothesis 2(i)(ii) are fulfilled with $V(x)=C\left( 1+|x|^{4N}_{L^{2N}(0,1)}\right) .$ Finally, Hypothesis 2(iii) is fulfilled as well, see [8, Theorem 4.8] and [3, p. 505] where a more general example is also presented. Then Theorem 8 applies.

3.3 Other Assumptions

In this section we set $G(t,x)=(-A)^{-1/2}F(t,x)$, $J_\alpha =(1-\alpha A)^{-1}$, $\alpha \in (0,1].$

Hypothesis 3

We set $F_\alpha (t,x)=(-A)^{1/2}J_\alpha G_\alpha (t,x)$ and assume that:

(i) $G_\alpha$ is continuous and bounded. and the mild equation

$$\begin{aligned} X_\alpha (t,s,x)=e^{(t-s)A}x+\int _s^t(-A)^{1/2}e^{(t-r)A}J_\alpha G _\alpha (r, X_\alpha (r,s,x))dr+W_A(t,s), \end{aligned}$$

(51)

has a unique solution.

(ii) There exists a convex and lower semicontinuous mapping $V:H\rightarrow [1,+\infty ]$ such that

$$\begin{aligned} |G_\alpha (t,x)|^2\le |G(t,x)|^2\le V(x) ,\quad \forall \;(t,x)\in [0,T]\times H, \end{aligned}$$

(52)

and

$$\begin{aligned} |G(t,x)-G_\alpha (t,x)|\le \alpha \,V(x) ,\quad \forall \;(t,x)\in [0,T]\times H. \end{aligned}$$

(53)

(iii) There exists $M\ge 0$ such that

$$\begin{aligned} P^\alpha _{0,t}|x|^2\le M|x|^2,\quad P^\alpha _{0,t}V(x)\le MV(x),\quad \forall \;t\in [0,T],\, \alpha \in (0,1], \end{aligned}$$

(54)

where $P^\alpha$ is the transition evolution operator $P^\alpha _{s,t}\varphi (x)=\mathbb E[\varphi (X_\alpha (t,s,x))],\; 0\le s\le T,\;\varphi \in B_b(H).$

(Note the difference between Hypothesis 2(iii) and Hypothesis 3(iii); the reason is that from (51) we are not able to estimate $\mathbb E|X_\alpha (t,s,x)|^2$ independently of $\alpha$.)

Now fix $\zeta \in \mathcal P(H)$ and for any $t\in (0,T]$ and any $\alpha \in (0,1]$ define the probability measure $\mu _t^\alpha$ in H setting as in Sect. 3.1

$$\begin{aligned} \int _H\varphi (x)\mu _t^\alpha (dx)=\int _H P^\alpha _{0,t}\,\varphi (x)\zeta (dx)= \int _H \mathbb E[\varphi (X_\alpha (t,0,x))]\zeta (dx),\quad \forall \;\varphi \in \mathcal E_A(H) \end{aligned}$$

(55)

and $\underline{\mu }^\alpha (dt\, dx)=\mu _t^\alpha (dx)dt.$

Let us deduce some estimates from Hypothesis 3. First from (54) we have

$$\begin{aligned} \int _H |x|^2\,\mu ^\alpha _t(dx)=\int _HP_{0,t}|x|^2\,\zeta (dx)\le M\int _H|x|^2\,\zeta (dx) \end{aligned}$$

(56)

and

$$ \int _H V(x)\,\mu ^\alpha _t(dx)=\int _H P_{0,t}V(x)\,\zeta (dx)\le M\int _HV(x)\,\zeta (dx) $$

and so, from (52)

$$\begin{aligned} \int _H |F(t,x)|^2\,\mu ^\alpha _t(dx)\le M\int _HV(x)\,\zeta (dx). \end{aligned}$$

(57)

Lemma 12

Assume that Hypotheses 1 and 3 are fulfilled and let $x\in H$, $0<s\le t\le T$ and $\alpha \in (0,1]$. Then $X_\alpha (t,s,x)\in D((-A)^{\epsilon _0})$, $\mathbb P$-a.s. and there exists $C>0$ such that

$$\begin{aligned} \mathbb E[|(-A)^{\epsilon _0} X_\alpha (t,s,x)|]\le Ct-s)^{-\epsilon _0}(1+|x|^2+ V(x)). \end{aligned}$$

(58)

Proof

Writing $X_\alpha (t,s,x)=X_\alpha (t)$ we have

$$\begin{aligned} (-A)^{\epsilon _0} X_\alpha (t)&=(-A)^{\epsilon _0} e^{(t-s)A}x+\int _s^t (-A)^{\epsilon _0+1/2}J_\alpha e^{(t-r)A}G_\alpha (r,X_\alpha (r))dr \nonumber \\&\quad +(-A)^{\epsilon _0} W_A(t,s). \end{aligned}$$

(59)

Therefore, since $\Vert J_\alpha \Vert \le 1$ and using that $|G_\alpha (r,X_\alpha (r))|\le 1+ |G_\alpha (r,X_\alpha (r))|^2$, we have

$$\begin{aligned} \displaystyle |(-A)^{\epsilon _0} X_\alpha (t)|&\le \kappa _ {\epsilon _0}(t-s)^{-\epsilon _0}\,|x|+\kappa _{(\epsilon _0+1/2)}\int _s^t(t-r)^{-\epsilon _0-1/2}(1+ |G_\alpha (r,X_\alpha (r))|^2)dr \\&\quad +|(-A)^{\epsilon _0} W_A(t,s)|. \end{aligned}$$

Taking expectation, yields

$$\begin{aligned} \mathbb E|(-A)^{\epsilon _0} X_\alpha (t)|&\le \kappa _ {\epsilon _0}(t-s)^{-\epsilon _0}\,|x|+\kappa _{(\epsilon _0+1/2)}\int _s^t(t-r)^{-\epsilon _0-1/2}(1+ \mathbb E|G_\alpha (r,X_\alpha (r))|^2)dr\\&\quad +(\mathbb E|(-A)^{\epsilon _0} W_A(t,s)|^2)^{1/2}. \end{aligned}$$

Consequently, taking into account (7) and (52) we find

$$\begin{aligned} \mathbb E|(-A)^{\epsilon _0} X_\alpha (t)|&\le \kappa _ {\epsilon _0}(t-s)^{-\epsilon _0}\,|x|+\kappa _{(\epsilon _0+1/2)}\int _s^t(t-r)^{-\epsilon _0-1/2}(1+ MV(x))dr \\&\quad +(\mathbb E|(-A)^{\epsilon _0} W_A(t,s)|^2)^{1/2} \end{aligned}$$

and (58) follows. $\square$

In the following we shall set as before

$$\begin{aligned} \psi (x):=1+|x|^2+V(x),\quad \forall \;x\in H. \end{aligned}$$

(60)

Now, integrating (58) with respect to $\zeta$ we obtain

Corollary 13

Assume that Hypotheses 1 and 3 are fulfilled. Then for all $\zeta \in \mathcal P_\psi (H)$, $t\in (0,T]$, $x\in H$ and $\alpha \in (0,1]$ we have

$$\begin{aligned} \int _H\Vert x\Vert _{\epsilon _0}\,\mu _t^\alpha (dx)\le C t^{- \epsilon _0}\int _H \psi (x) \zeta (dx). \end{aligned}$$

(61)

Now by Itô’s formula we see that $\underline{\mu }^\alpha (dt\,dx)=\mu _t^\alpha (dx)dt$ solves the approximating Fokker–Planck equation (24) where $K_{F_\alpha }$ is the Kolmogorov operator (25), moreover the tightness of $\mu _t^\alpha$ follows as in the proof of Proposition 7.

Proposition 14

Assume that $\zeta \in \mathcal P_{\psi }(H)$. Then $(\mu _t^\alpha )_{\alpha \in (0,1]}$ is tight for all $t\in (0,T]$ and we have

$$\begin{aligned} \mu _t^\alpha (\Vert x\Vert _{\epsilon _0}\ge R)\le \frac{Ct^{-\epsilon _0}}{R} \int _H \psi (x) \zeta (dx),\quad \forall \;R>0. \end{aligned}$$

(62)

Theorem 15

Assume that Hypotheses 1 and 3 are fulfilled and let $\zeta \in \mathcal P_{\psi }(H)$.Then there is a probability kernel $\underline{\mu }(dt\,dx)=\mu _t(dx)\,dt$ solving (3) and such that $\mu _t\in \mathcal P_\psi$ for all $t\in [0,T]$, where $\psi$ is defined by (60).

Proof

We proceed as in the proof of Theorem 8. The Step 1 is similar: we construct a sequence $(\alpha _h)\rightarrow 0$ and a family of measures $(\mu _t)_{t\in \mathbb Q}$ such that

$$\begin{aligned} \lim _{h\rightarrow \infty }\int _{H}\varphi \,d\mu ^{\alpha _h}_t=\int _{H}\varphi \,d\mu _t,\quad \forall \;\varphi \in C_b(H),\quad \forall \;t\in [0,T]\cap \mathbb Q. \end{aligned}$$

(63)

Then we extend the family $(\mu ^{\alpha _h}_t)_{t\in \mathbb Q}$ to the whole interval (0, T] in such a way that the extension is a solution of (3) proving that

Step 2. For each $\varphi \in \mathcal E_A(H)$ there exists $C(\varphi )>0$ such that for all $\alpha \in (0,1]$ we have

$$\begin{aligned} \left| \int _H\varphi \,d\mu ^\alpha _t-\int _H\varphi \,d\mu ^\alpha _s\right| \le C(\varphi ) (t-s)\int _H\psi \,d\zeta ,\quad 0\le s\le t\le T. \end{aligned}$$

(64)

In fact by Itô’s formula we have

$$\begin{aligned} \mathbb E[\varphi (X_\alpha (t,0,x))]-\mathbb E[\varphi (X_\alpha (s,0,x))]=\mathbb E\int _s^t(L_r^\alpha \varphi )(X_\alpha (r,0,x))dr, \end{aligned}$$

(65)

where $L_t^\alpha ,\,t\in [0,T],$ is the Kolmogorov operator

$$ L_t^\alpha \varphi =\tfrac{1}{2}\,\text{ Tr }\,[CD^2\varphi ]+\langle x, AD\varphi \rangle +\langle G_\alpha (t,x), (-A)^{1/2}J_\alpha D\varphi \rangle ,\quad \forall \varphi \in \mathcal E_A(H). $$

Now we write

$$\begin{aligned} \begin{array}{lll} \displaystyle \int _H\varphi \,d\mu ^\alpha _t-\int _H\varphi \,d\mu ^\alpha _s= & {} \displaystyle \mathbb E\int _H\int _s^t ( L_t^\alpha \varphi )(X(r,0,x))\,d\zeta \,dr \end{array} \end{aligned}$$

(66)

and choose $C_1(\varphi )>0$ such that

$$ \begin{array}{l} |( L_t^\alpha \varphi )(x)|\le \tfrac{1}{2}\,\Vert \text{ Tr }\,[CD^2\varphi ]\Vert _0+|x|\, \Vert AD\varphi \Vert _0+|G_\alpha (t,x)|\,\Vert (-A)^{1/2}J_\alpha D\varphi \Vert _0\\ \\ \le C_1(\varphi )(1+|x|+|G_\alpha (t,x)|)\le 3C_1(\varphi )(1+|x|^2+|G_\alpha (t,x)|^2),\quad \forall \;x\in H. \end{array} $$

Therefore (64) follows arguing as before as well as Step 2. We go now to the Step 3. We have still to show statements (37) and (38); the proof of (37) is exactly the same, whereas for the proof of (38) we have just to write

$$ \langle D_xu(t,x), F_{\alpha _h}(t,x) \rangle = \langle (-A)^{1/2}J_{\alpha _h} D_xu(t,x), G_{\alpha _h}(t,x) \rangle $$

and notice that $(-A)^{1/2}D_xu$ is bounded. Then the proof ends as before. $\square$

Example 16

(Burgers type equations) We take here $H=L^2(0,1)$, C of trace class and denote by A the realisation of the Laplace operator in [0, 1] with periodic boundary conditions: $Ax=D^2_\xi ,\quad D(A)=H^2_\#(0,1),$ where

$$H^2_\#(0,1)=\{x\in H^2(0,1):\,x(0)=x(1),\,D_\xi x(0)=D_\xi x(1)\}.$$

Moreover we take $F(t,x)(\xi )=D_\xi (g(t, x(\xi )),\;\;x\in H^1_\#(0,1)$, where $g:[0,T]\times \mathbb R\rightarrow \mathbb R$ is continuous. Then Hypothesis 1 is fulfilled. Set $G(t,x)(\xi )=g(t, x(\xi ))$, so that $F(t,x)(\xi )=(-A)^{1/2}G(t,x),\;\forall \;x\in H^1_\#(0,1).$ Set moreover for $\alpha \in (0,1]$

$$\begin{aligned} F_\alpha (t,x)(\xi )=(-A)^{1/2} J_\alpha G_\alpha (t,x)=(-A)^{1/2}J_\alpha \left( \frac{G(t,x)}{1+\alpha |G(t,x)|} \right) ,\quad \forall \;x\in H^1_\#(0,1). \end{aligned}$$

(67)

To check Hypothesis 3(i)(ii) we need suitable estimates of the solution $X_\alpha (t,s,x)$ of Eq. (11). For any $m\in \mathbb N$ set $\varphi _m(x)=\frac{1}{2m}\int _0^1|x(\xi )|^{2m}d\xi .$ By Itô’s formula we have

$$\begin{aligned} \displaystyle d\varphi _m(X(t))&=(2m-1)\langle X(t)^{2m-1},AX(t)+F_\alpha (t,X(t))dt+\sqrt{C}\,dW(t)\rangle \displaystyle \nonumber \\&\quad +\frac{1}{2}\,(2m-1)\sum _{h=1}^\infty c_h\int _0^1X(t)^{2m-2}e_h^2\,d\xi \;dt, \end{aligned}$$

(68)

where $(e_h)$ is an orthonormal basis in H and $Ce_h=c_he_h,\;h\in \mathbb N$. Therefore

$$\begin{aligned} \displaystyle \frac{d}{dt}\,\mathbb E[\varphi _m(X(t))]&\le (2m-1)\mathbb E\langle X(t)^{2m-1},AX(t)+F_\alpha (t,X(t))\rangle \displaystyle \nonumber \\&\quad +\frac{1}{2}\,(2m-1)\text{ Tr }\,C\,\mathbb E\int _0^1X(t)^{2m-2}\,d\xi \end{aligned}$$

(69)

In particular, we have $\displaystyle \frac{d}{dt}\,\mathbb E|X(t)|^2\le -2\omega \mathbb E|X(t)|^2+\text{ Tr }\,C,$ so that

$$\begin{aligned} \mathbb E|X(t)|^2\le e^{-2\omega t}|x|^2+\frac{\text{ Tr }\,C}{2\omega } \end{aligned}$$

(70)

Moreover by (69)

$$\begin{aligned} \begin{array}{l} \displaystyle \frac{1}{4}\,\frac{d}{dt}\,\mathbb E[|X(t)|_{L^4(0,1)}^4]= 3\mathbb E\langle X(t)^3,AX(t)+D_\xi F(t,X(t))\rangle \displaystyle +\frac{3}{2}\,\text{ Tr }\,C\,\mathbb E[|X(t)|^{2}]. \end{array} \end{aligned}$$

(71)

But

$$ \langle X(t)^3,AX(t)\rangle =-3\int _0^1 X(t)^2(D_\xi X(t)^2d\xi \le 0 $$

and

$$ \langle X(t)^3,D_\xi F(t,X(t))\rangle =-3\int _0^1D_\xi X(t)\,X(t)^2g(t,X(t))d\xi =0. $$

So,

$$\begin{aligned} \frac{1}{4}\,\frac{d}{dt}\,\mathbb E\left[ |X(t)|_{L^4(0,1)}^4\right] \le \frac{3}{2}\,\text{ Tr }\,C\,\mathbb E[|X(t)|^{2}] \end{aligned}$$

(72)

and taking into account (70) we have

$$\displaystyle \frac{d}{dt}\,\mathbb E\left[ |X(t)|_{L^4(0,1)}^4\right] \le 6\,\text{ Tr }\,C\,\left( e^{-2\omega t}|x|^2+\frac{\text{ Tr }\,C}{2\omega }\right) .$$

It follows that

$$\begin{aligned} \mathbb E\left[ |X(t)|_{L^4(0,1)}^4\right] \le |x|^4+ 6T\,\text{ Tr }\,C\,\left( |x|^2+\frac{\text{ Tr }\,C}{2\omega }\right) \end{aligned}$$

(73)

Iterating this procedure we find that there exists $C_m(T)>0$ such that $\mathbb E[|X(t)|_{L^{2m}(0,1)}^{2m}]\le C_m(T)(1+ |x|^{2m}).$ Therefore Hypothesis 3(i)(ii)(iii) is fulfilled with $V(x)=C(1+|x|^4_{L^4(0,1)}).$

Example 17

(Burgers equation perturbed by white noise) We take here $H\,{=}\,L^2(0,1),$ $A\,{=}\,Ax\,{=}\,D^2_\xi ,\quad D(A)\,{=}\,H^2_\#(0,1),$ and $F(t,x)(\xi )=D_\xi (x^2(\xi )),\;\forall \;x\in H^1_\#(0,1)$.⁵ Then Hypothesis 1 is fulfilled. Set $G(t,x)(\xi )= x^2(\xi ),\;\forall \;x\in H^1_\#(0,1),$ so that $F(t,x)(\xi )=(-A)^{1/2}G(t,x),\;\forall \;x\in H^1_\#(0,1).$ Set moreover for $\alpha \in (0,1]$

$$\begin{aligned} F_\alpha (t,x)(\xi )=(-A)^{1/2}J_\alpha G_\alpha (t,x)=(-A)^{1/2}J_\alpha \left( \frac{x^2}{1+\alpha x^2} \right) ,\quad \forall \;x\in H^1_\#(0,1). \end{aligned}$$

(74)

Then Hypothesis 3(i)(ii) is fulfilled with $V(x)=C(1+|x|^4_{L^4(0,1)}).$ Finally, Hypothesis 3(iii) follows from [10, Proposition 2.2].⁶

Example 18

(2D-Navier–Stokes equation perturbed by coloured noise) Let us consider the space $L_\#^2$ of all real $2\pi$-periodic functions in the real variables $\xi _1$ and $\xi _2,$ which are measurable and square integrable on $[0,2\pi ]\times [0,2\pi ]$ endowed with the usual scalar product $\langle \cdot ,\cdot \rangle$ and norm $|\cdot |$. We denote by $(L_\#^2)^2$ the space consisting of all pairs $x=\left( \begin{array}{c} x^1\\ x^2 \end{array}\right)$ of elements of $L_\#^2$ endowed with the inner product $\displaystyle \langle x,y \rangle =\int _\mathcal O [x^1(\xi )y^1(\xi )+x^2(\xi )y^2(\xi )]d\xi ,\; x,y\in (L_\#^2)^2.$ Moreover, for any $x\in (L_\#^2)^2$ we set $|x|= \langle x,x \rangle ^\frac{1}{2}.$ (We shall consider everywhere also the corresponding complexified spaces). Let $f_{h,k}=\left( \begin{array}{c} e_h\\ e_k \end{array} \right) ,\;h,k\in \mathbb Z^2,$ be the complete orthonormal system of $(L_\#^2)^2,$ where $e_k(\xi )=\tfrac{1}{2\pi }\;e^{i\langle k,\xi \rangle },\; k=(k_1,k_2)\in \mathbb Z^2,\;\xi =(\xi _1,\xi _2),$ and $\langle k,\xi \rangle =k_1\xi _1+k_2\xi _2$. Then we shall denote by H the closed subspace of $(L_\#^2)^2$ of all divergence free vectors, that is: $H=\text{ linear } \text{ span }\;\{g_k:\;k\in \mathbb Z^2\},$ where

$$ g_k=\frac{k^\perp }{|k|}\; e_k=\left( \begin{array}{c} -\frac{k_2}{|k|}\; e_k\\ \\ \frac{k_1}{|k|}\; e_k \end{array} \right) ,\quad \quad k^\perp =\left( \begin{array}{c} -k_2\\ k_1 \end{array} \right) ,\quad k\in \mathbb Z^2. $$

Note in fact that $\text {div}\;g_k(\xi )=\frac{i}{|k|}\;e_k(-k_1k_2+k_1k_2)=0,\; k\in \mathbb Z^2.$ Any element $x=\left( \begin{array}{c} x^1\\ x^2 \end{array} \right)$ of H can be developed as a Fourier series $\displaystyle x=\sum _{k\in \mathbb Z^2}x_k g_k,$ where

$$\begin{aligned} \displaystyle x_k=\langle x,g_k \rangle _2&=-\frac{k_2}{|k|}\;\langle x^1,e_k \rangle _2+\frac{k_1}{|k|}\; \langle x^2,e_k \rangle _2 \displaystyle \nonumber \\&= \frac{1}{2\pi |k|}\;\int _\mathcal O[-k_2x^1(\eta )+k_1x^2(\eta )]e^{-i\langle k,\eta \rangle }d\eta . \end{aligned}$$

(75)

Moreover, we shall denote by $L^p_\#$, $p\ge 1$, the subspace of $(L_\#^p)^2$ of all divergence free vectors and by $|\cdot |_{p}$ the norm in $L^p_\#$. Let us define the Stokes operator $A:D(A)\rightarrow H$ setting $Ax=\mathcal P(\varDelta _\xi x-x),\; x\in D(A)=H_\#^{2},$ A is self-adjoint and $A g_k=-(1+|k|^2)g_k,\;k\in \mathbb Z^2.$ We also define the non linear operator F setting

$$ F(x)=\mathcal P(D_\xi x\cdot x),\quad x\in H^1_\#. $$

For any $\sigma > 0$ we have

$$ H^{2\sigma }_\#:=\left\{ x\in H:\;\Vert (-A)^\sigma x\Vert ^2=\sum _{k\in \mathbb Z^2}(1+|k|^2)^{\sigma /2}|x_k|^2\right\} . $$

see [8, Chap. 6] for details. If $x\in H^{1}_\#$, taking into account that div $x=0$, we have

$$\begin{aligned} F(x)=\begin{pmatrix} x_1D_{\xi _1}x_1+x_2D_{\xi _2}x_1\\ x_1D_{\xi _1}x_2+x_2D_{\xi _2}x_2 \end{pmatrix}= \begin{pmatrix} D_{\xi _1}(x^2_1)+D_{\xi _2}(x_1x_2)\\ D_{\xi _1}(x_1x_2)+D_{\xi _2}(x^2_2). \end{pmatrix} \end{aligned}$$

(76)

Set $G(x)=(-A)^{1/2}F(x),$ and moreover for $\alpha \in (0,1]$

$$ F_\alpha (x)= \begin{pmatrix} D_{\xi _1}(\tfrac{x^2_1}{1+\alpha (x^2_1+x^2_2)})+D_{\xi _2}(\tfrac{x_1x_2}{1+\alpha (x^2_1+x^2_2)})\\ D_{\xi _1}(\tfrac{x_1x_2}{1+\alpha (x^2_1+x^2_2)})+D_{\xi _2}(\tfrac{x^2_2}{1+\alpha (x^2_1+x^2_2)}) \end{pmatrix} $$

and $G_\alpha (x)=(-A)^{-1/2} F_\alpha (x).$ Then for $h\in H^{1}_\#$ and $\alpha \in (0,1]$ we have

$$ \begin{array}{l} \langle F_\alpha (x),h \rangle = \left\langle \begin{pmatrix} D_{\xi _1}(\tfrac{x^2_1}{1+\alpha (x^2_1+x^2_2)})+D_{\xi _2}(\tfrac{x_1x_2}{1+\alpha (x^2_1+x^2_2)})\\ D_{\xi _1}(\tfrac{x_1x_2}{1+\alpha (x^2_1+x^2_2)})+D_{\xi _2}(\tfrac{x^2_2}{1+\alpha (x^2_1+x^2_2)}) \end{pmatrix},\begin{pmatrix} h_1\\ h_2 \end{pmatrix}\right\rangle \\ \\ =-\langle \tfrac{x^2_1}{1+\alpha (x^2_1+x^2_2)},D_{\xi _1}h_1 \rangle -\langle \tfrac{x_1x_2}{1+\alpha (x^2_1+x^2_2)}, D_{\xi _2}h_1 +D_{\xi _1}h_2 \rangle -\langle \tfrac{x^2_2}{1+\alpha (x^2_1+x^2_2)},D_{\xi _2}h_2 \rangle \end{array} $$

Finally, assume that Tr $[CA^{-1]}<\infty$. then it is easy to see that, thanks to [8, Lemma 6.7], Hypothesis 3 is fulfilled with $G(x)=(1+|x|^4_{L^4_\#})$.

4 Uniqueness

4.1 The Rank Condition

We start with the following crucial result, proved in [1].

Theorem 19

Let $\psi :H\rightarrow [0,+\infty ]$ be convex and lover semi-continuous and let $\zeta \in \mathcal P_\psi (H)$. Assume that for any solution $\underline{\mu }=(\mu _t)_{t\in [0,T]}$ of (3) such that $\mu _0=\zeta$ the following statement holds

$$\begin{aligned} K_F(\mathcal E_A([0,T]\times H))\;\text{ is } \text{ dense } \text{ in }\;L^1([0,T]\times H, \underline{\mu }). \end{aligned}$$

(77)

Then Eq. (3) has at most one solution.

The condition (77) is called the rank condition.

Proof

Assume that $\underline{\mu }_1 =(\mu _{1,t})$ and $\underline{\mu }_2 =(\mu _{2,t})$ are solution of (3) such that $\mu _{1,0}=\mu _{2,0}=\zeta \in \mathcal P_\psi (H)$ and set

$$\underline{\lambda }:=\frac{1}{2}\;(\underline{\mu }_1+\underline{\mu }_2). $$

Then $\lambda _0=\zeta$ and $\underline{\mu }_1\ll \underline{\lambda },\quad \underline{\mu }_2\ll \underline{\lambda }.$ Denote by $\rho _1$ and $\rho _2$ the respective densities

$$ \rho _1:=\frac{d\underline{\mu }_1}{d\underline{\lambda }},\quad \rho _2:=\frac{d\underline{\mu }_2}{d\underline{\lambda }}. $$

We claim that

$$\begin{aligned} 0\le \rho _1(t,x)\le 2,\quad 0\le \rho _2(t,x)\le 2,\quad \underline{\lambda }\;\text{-a.s. },\quad \forall \;(t,x)\in [0,T]\times H. \end{aligned}$$

(78)

Let us show for instance that $0\le \rho _1(t,x)\le 2$. In fact, for any $A\in \mathcal B([0,T]\times H)$ we have $\underline{\mu }_2(A)=2\underline{\lambda }(A)-\underline{\mu }_1(A),$ so that $\underline{\mu }_2(A)=\int _A(2-\rho _1)d\underline{\lambda }$, which implies that $0\le \rho _1(t,x)\le 2$, $\underline{\lambda }$-a.e. by the arbitrariness of A.

Now for any $u\in \mathcal E_A([0,T]\times H)$ we have (recall that $H_T=[0,T]\times H$))

$$ \int _{H_T} K_F u\,(d\underline{\mu }_1-d\underline{\mu }_2)=\int _{H_T} K_F u\,(\rho _1-\rho _2)d\underline{\lambda }=0. $$

Since $\rho _1-\rho _2\in L^\infty ([0,T]\times H;\underline{\lambda }),$ by the rank condition it follows that $\rho _1=\rho _2$. $\square$

A useful remark for dealing with the rank condition is that for any solution $\underline{\mu }$ to Eq. (3) such that $\mu _0\in \mathcal P_\psi (H)$, $K_F$ is dissipative in $L^1([0,T]\times H;\underline{\mu })$, as the next proposition shows.

Proposition 20

Let $F:D(F)\subset [0,T]\times H\rightarrow H$ be Borel and $\zeta \in \mathcal P(H)$. Assume that $\underline{\mu }$ is a solution to (3). Then $K_F$ is dissipative in $L^p([0,T]\times H;\underline{\mu })$ for all $p\ge 1$.

Proof

First we show that for any $u\in \mathcal E_A([0,T]\times H)$ we have

$$\begin{aligned} \displaystyle \int _{H_T} K_F u(t,x)\;u(t,x)\;\mu _t(dx)\,dt \displaystyle&= -\frac{1}{2}\int _{H_T}|C^{1/2}D_x u(t,x)|^2\;\mu _t(dx)\,dt\nonumber \\&\quad -\frac{1}{2}\int _{ H}u^2(0,x)\;\zeta (dx). \end{aligned}$$

(79)

In fact if $u\in \mathcal E_A([0,T]\times H)$ we have $u^2\in \mathcal E_A([0,T]\times H)$ as well and

$$ D_t(u^2)=2uD_tu,\quad D_x(u^2)=2uD_xu,\quad D^2_x(u^2)=2D_xu\otimes D_xu+2uD^2_xu. $$

It follows that

$$\begin{aligned} K_F(u^2(t,x))=2u(t,x)\; K_F u(t,x)+|C^{1/2}D_xu(t,x)|^2. \end{aligned}$$

(80)

Integrating both sides of (80) with respect to $\underline{\mu }$ over $[0,T]\times H$, yields (79). Now, by (79) we have

$$ \int _{H_T}K_F\,u(t,x)\;u(t,x)\;\mu _t(dx)\,dt\le 0,\quad \forall \;u\in \mathcal E_A([0,T]\times H), $$

which implies the dissipativity of $K_F$ in $L^2([0,T]\times H,\underline{\mu })$. The case $p\ne 2$ follows from standard arguments about diffusions operators, see [12]. $\square$

Remark 21

Assume that $\underline{\mu }$ is a solution to (3). By Proposition 20 it follows that $K_F$ is closable in $L^1([0,T]\times H,\underline{\mu })$. We shall denote by $\overline{K_F}$ its closure. Clearly, if $\psi :H\rightarrow [0,+\infty ]$ is convex lover semi-continuous and

$$ \overline{K_F}\,(D(\overline{K_F}))\,\text{ is } \text{ dense } \text{ in }\,L^1([0,T]\times H,\underline{\mu })\;\text{ for } \text{ all }\; \underline{\mu }\; \text{ such } \text{ that } \;\mu _0\in \mathcal P_\psi (H),$$

then the rank condition (77) is fulfilled.

4.2 The Semigroup Associated to a Non Autonomous Problem

We assume here that Hypothesis 2 (resp. Hypothesis 3) is fulfilled. As usual in studying non autonomous systems, it is convenient to consider, besides (11) (resp. (51)), a problem in the unknowns $(H(\tau ),y(\tau ))$ (intended in the mild sense.)⁷

$$\begin{aligned} \left\{ \begin{array}{l} dH(\tau )=AH(\tau )d\tau +F_\alpha (y(\tau ),H(\tau ))d\tau +\sqrt{C}\,W(\tau ),\quad \tau \ge 0,\\ dy(\tau )=d\tau ,\quad \tau \ge 0 \\ H(0)=x,\quad y(0)=t. \end{array}\right. \end{aligned}$$

(81)

(resp. a similar problem for (51)). To solve problem (81) we set $y(\tau ,t)=t+\tau$, so that (81) reduces to

$$\begin{aligned} \left\{ \begin{array}{l} dH(\tau )=AH(\tau )d\tau +F_\alpha (t+\tau ,H(\tau ))d\tau +\sqrt{C}\,dW(\tau )\\ \\ H(0)=x. \end{array}\right. \end{aligned}$$

(82)

We denote by $H(\tau )=H(\tau ,t,x)$ the mild solution of (81) which, however, is only defined for $\tau \in [0,T-t]$.

Let us define a semigroup $S^{F_\alpha }_\tau ,\,\tau \ge 0,$ in the space $C_T([0,T]\times H)=\{u\in C([0,T]\times H):\;u(T,x)=0,\forall \;x\in H \}$ by setting

$$ (S^{F_\alpha }_\tau u)(t,x)= \mathbb E[u(t+\tau ,H(\tau + t,x))]\,{\mathbbm {1}}_{[0,T-\tau ]}(t), \;u\in C_T([0,T]\times H). $$

Notice that $S^{F_\alpha }_\tau =0,\; \forall \;\tau \ge T.$ Since the law of $H(\tau ,t,x)$ coincides, as easily seen, with that of $X_\alpha (t+\tau ,t,x)$, for any $x\in H$ and any $\tau$ such that $t+\tau \le T$, it results

$$\begin{aligned} (S^{F_\alpha }_\tau u)(t,x)&= (P^\alpha _{t,t+\tau }u(t+\tau ,\cdot )(x)\,{\mathbbm {1}}_{[0,T-\tau ]}(t)\nonumber \\&=\mathbb E[u(t+\tau ,X_\alpha (t+\tau ,t,x))]\,{\mathbbm {1}}_{[0,T-\tau ]}(t), \quad \forall \,u\in C_T([0,T]\times H). \end{aligned}$$

(83)

Let denote by $\mathcal K_{F_\alpha }$ the infinitesimal generator of $S^{F_\alpha }_\tau$, defined through its resolvent (see (84) below). Then the resolvent set of $\mathcal K_{F_\alpha }$ coincides with $\mathbb R$ and its resolvent is given, for all $f\in C_T([0,T]\times H)$ and all $\lambda \in \mathbb R$, by

$$\begin{aligned} \displaystyle R(\lambda ,\mathcal K_{F_\alpha })f(t,x)&=\int _0^\infty e^{-\lambda \tau }S^{F_\alpha }_\tau f(t,x)d\tau \nonumber \\&=\int _t^{T} e^{-\lambda (r-t)}\mathbb E[f(X_\alpha (r,t,x))]dr, \quad (t,x)\in [0,T]\times H. \end{aligned}$$

(84)

$\mathcal K_{F_\alpha }$ can also be defined as follows (following [14]). We say that a function $u\in C_T([0,T]\times H)$ belongs to the domain of $\mathcal K_{F_\alpha }$ if there exists a function $f\in C_T([0,T]\times H)$ such that

(i)

$\displaystyle \lim _{\tau \rightarrow 0}\frac{1}{\tau }\;(S^{F_\alpha }_\tau u(t,x)-u(t,x))=f(t,x),\quad \forall \;(t,x)\in [0,T]\times H.$

(ii) There exists $M_u>0$ such that

$$ \left| \frac{1}{\tau }\;(S^{F_\alpha }_\tau u(t,x)-u(t,x)))\right| \le M_u,\quad \forall \;(t,x)\in [0,T]\times H,\,\tau \in [0,1]. $$

We set $\mathcal K_{F_\alpha }u=f$ and call $\mathcal K_{F_\alpha }$ the infinitesimal generator of $S_\tau$ on $C_T([0,T]\times [0,H])$. As easily seen, the abstract operator $\mathcal K_{F_\alpha }$ is an extension of the differential Kolmogorov operator $K_{F_\alpha }$ defined by (2) (with $F_\alpha$ replacing F).

4.3 The Case When $C^{-1}$ is Bounded

Theorem 22

Assume, besides Hypotheses 1 and 2 that there exists $M_1>0$ such that

$$\begin{aligned} P_{0,t}V^2(x)\le M_1V^2(x),\quad \forall \;x\in H. \end{aligned}$$

(85)

Let moreover $\zeta \in \mathcal P_{\psi +V^2}(H)$, where $\psi$ is defined by (21). Then the Fokker–Planck equation (3) has a unique solution $\underline{\mu }=(\mu _t)_{t\in [0,T]}$ and $\mu _t\in \mathcal P_{\psi +V^2}(H)$ for any $t\ge 0$.

Proof

Let $f\in \mathcal E_A([0,T]\times H))$ and consider the approximating equation

$$\begin{aligned} \mathcal K_{F_\alpha }\, u_\alpha =f. \end{aligned}$$

(86)

Thanks to (84) Eq. (86) has a unique solution $u_\alpha$ given by

$$\begin{aligned} u_\alpha (t,x)=-\int _0^{T-t} \mathbb E[f(t+\tau ,X_\alpha (t+\tau ,t,x))]d\tau ,\quad (t,x)\in [0,T]\times H \end{aligned}$$

(87)

and therefore

$$\begin{aligned} |u_\alpha (t,x)|\le T\Vert f\Vert _0,\quad \forall \;(t,x)\in [0,T]\times H,\, \alpha \in (0,1]. \end{aligned}$$

(88)

Step 1. $u_\alpha \in D(\mathcal K^1_0)\cap C^1_b(H)$ and it results

$$\begin{aligned} \mathcal K_{F_\alpha }u_\alpha = \mathcal K^1_0\,u_\alpha +\langle F_\alpha (t,x),D_xu_\alpha \rangle =f. \end{aligned}$$

(89)

Fix in fact $t\in [0,T]$ and $h>0$ such that $t+h\le T$. Then, recalling (12) we write

$$\begin{aligned} X_\alpha (t+h,t,x)=Z(t+h,t,x)+g_\alpha (t+h,t,x), \end{aligned}$$

(90)

where

$$ Z(t+h,t,x)=e^{hA}x+\int _t^{t+h}e^{(t+h-r)A}\sqrt{C}\,\;dW(r) $$

and

$$ g_\alpha (t+h,t,x)=\int _t^{t+h}e^{(t+h-r)A}F_\alpha (r,X(r,t,x))dr. $$

Then by (9) and (83) we have

$$ S^{0,1}_hu_\alpha (t,x)=\mathbb E[u_\alpha (t+h,Z(t+h,t,x))] $$

and

$$\begin{aligned} S^{F_\alpha }_hu_\alpha (t,x)&=\mathbb E[u_\alpha (t+h,X(t+h,t,x))] \\&=\mathbb E[u_\alpha [t+h,\!Z(t+h,t,x)\!+\!g_\alpha (t+h,t,x))]. \end{aligned}$$

Consequently,

$$\begin{aligned} S^{F_\alpha }_hu_\alpha (t,x)&=S^{0,1}_hu_\alpha (t,x) \displaystyle +\mathbb E\int _0^1\langle D_xu_\alpha (t+h,Z(t+h,t,x)\\&\quad +\xi g_\alpha (t+h,t,x)),g_\alpha (t+h,t,x)\rangle \,d\xi , \end{aligned}$$

which implies

$$ \begin{array}{l} \displaystyle \frac{1}{h}(S^{F_\alpha }_hu_\alpha (t,x)-u_\alpha (t,x))=\frac{1}{h}(S^{0,1}_hu_\alpha (t,x)-u_\alpha (t,x))\\ \\ \displaystyle +\frac{1}{h}\mathbb E\int _0^1\langle D_xu_\alpha (t+h,Z(t+h,t,x)+\xi g_\alpha (t+h,t,x)),g_\alpha (t+h,t,x)]\,d\xi . \end{array} $$

Now, letting $h\rightarrow 0$ and taking into account that $\lim _{h\rightarrow 0}\frac{1}{h}\;g(t+h,t,x)=F_\alpha (t,x),$ yields (89).

Notice that from Step 1 does not follow that $u_\alpha \in D(K_F)= \mathcal E_A([0,T]\times H$), but we are going to show in the next step that $u_\alpha$ belongs to the domain of the closure $\overline{ K_F}$ of $K_F.$ (Recall Remark 21).

Step 2. $u_\alpha \in D(\overline{ K_F})$ and we have

$$\begin{aligned} \overline{ K_F}u_\alpha =\mathcal K^1_0u_\alpha +\langle F(t,x) ,D_xu_\alpha \rangle =f+\langle F(t,x)-F_\alpha (t,x),D_xu_\alpha \rangle . \end{aligned}$$

(91)

By Proposition 4 there is a multi-sequence $(u_{\alpha ,\mathbf {j}})$ in $\mathcal E_A([0,T]\times H)$ such that for all $(t,x)\in [0,T]\times H$

• (i) $\displaystyle \lim _{\mathbf {j}\rightarrow \infty }u_{\alpha ,\mathbf {j}}(t,x)= u_\alpha (t,x),$

• (ii) $\displaystyle \lim _{\mathbf {j}\rightarrow \infty }K_0 u_{\alpha ,\mathbf {j}}(t,x)= \mathcal K^1_0u_\alpha (t,x),$

• (iii) $\displaystyle \lim _{\mathbf {j}\rightarrow \infty }D_xu_{\alpha ,\mathbf {j}}(t,x)= D_xu_\alpha (t,x),$

• (iv) $\displaystyle |u_{\alpha ,\mathbf {j}}(t,x)|+| K_0\, u_{\alpha ,\mathbf {j}}(t,x)|+|D_xu_{\alpha ,\mathbf {j}}(t,x)|\le C(1+|x|),\quad x\in H .$

Then we have

$$ \begin{array}{lll} \displaystyle \lim _{\mathbf {j}\rightarrow \infty }K_{F} u_{\alpha ,\mathbf {j}}(t,x)&{}=&{}\displaystyle \lim _{\mathbf {j}\rightarrow \infty }(K_0u_{\alpha ,\mathbf {j}}(t,x)+\langle D_xu_{\alpha ,\mathbf {j}}(t,x), F(t,x)\rangle )\\ &{}=&{}\displaystyle \mathcal K^1_0u_{\alpha }(t,x)+\langle D_xu_{\alpha }(t,x),F(t,x)\rangle ,\quad \forall \;(t,x)\in [0,T]\times H. \end{array} $$

Since

$$\begin{aligned} |K_{F}\, u_{\alpha ,\mathbf {j}}(t,x)| \le C(1+|x|)+ C(1+|x|)|F(t,x)|,\quad \forall \;(t,x)\in [0,T]\times H, \end{aligned}$$

(92)

we have by Hypothesis 2

$$\begin{aligned} \displaystyle \int _{H_T}|K_{F}\, u_{\alpha ,\mathbf {j}}(t,x)|\,d\mu _t\,dt&\le \int _{H_T}[C(1+|x|)+ C(1+|x|)|F(t,x)|]\,d\mu _t\,dt \displaystyle \\&\le C'\int _{H_T} (1+|x|)V(x)\mu _t(dx)\,dt. \end{aligned}$$

By the dominated convergence theorem it follows that $u_\alpha \in D(\overline{K_F})$ and

$$ \lim _{\mathbf {j}\rightarrow \infty }K_{F} u_{\alpha ,\mathbf {j}}= \overline{ K_{F}}\,u_\alpha =\mathcal K^1_0u_\alpha +\langle F(t,x) ,D_xu_\alpha \rangle \quad \text{ in }\;L^1([0,T]\times H;\underline{\mu }). $$

Therefore $u_\alpha \in D(\overline{ K_F})$ and (91) is fulfilled .

Step 3. There is $C(T)>0$ such that

$$\begin{aligned} \int _{H_T}|D_xu_\alpha |^2\,d\underline{\mu }\le C(T)\Vert f\Vert _0^2\left( 1+\int _{H_T}|F-F_\alpha |^2\,d\underline{\mu }\right) . \end{aligned}$$

(93)

Multiplying both sides of Eq. (91) by $u_\alpha$ and integrating in $\underline{\mu }$ over $H_T$, yields

$$ \begin{array}{l} \displaystyle \int _{H_T}\overline{K_F}u_\alpha \,u_\alpha \,d\underline{\mu }\displaystyle =\int _{H_T}f\,u_\alpha \,d\underline{\mu }+ \int _{H_T}\langle F-F_\alpha ,D_xu_\alpha \rangle \,u_\alpha \,d\underline{\mu }. \end{array} $$

Taking into account (79), we obtain

$$\begin{aligned} \displaystyle \frac{1}{2}\int _{H_T}|C^{1/2}D_x u_\alpha |^2\,d\underline{\mu }&\le -\int _{H_T}\overline{K_F}u_\alpha \,u_\alpha \,d\underline{\mu }\displaystyle \\&=- \int _{H_T}f\,u_\alpha \,d\underline{\mu }+ \int _{H_T}\langle F_\alpha -F,D_xu_\alpha \rangle \,u_\alpha \,d\underline{\mu }. \end{aligned}$$

Now, recalling (88), we obtain

$$\begin{aligned} \begin{array}{lll} \displaystyle \frac{1}{2}\int _{H_T}|C^{1/2}D_x u_\alpha |^2\,d\underline{\mu }&{}\le &{}\displaystyle T^2\Vert f\Vert _0^2+T\Vert f\Vert _0 \int _{H_T}|F_\alpha -F|\,|D_xu_\alpha |\,d\underline{\mu }\\ \\ &{}\le &{}\displaystyle T^2\Vert f\Vert _0^2+T\Vert f\Vert _0\Vert C^{-1/2}\Vert \int _{H_T}|F_\alpha -F|\,|C^{1/2}D_xu_\alpha |\,d\underline{\mu }. \end{array} \end{aligned}$$

(94)

Now the conclusion follows by standard arguments.

Step 4.

$$ \overline{ K_F} (D( K_F))\;\text{ is } \text{ dense } \text{ in }\; L^1([0,T]\times H,\underline{\mu }). $$

In view of (91) it is enough to show that

$$\begin{aligned} \lim _{\alpha \rightarrow 0} \int _{H_T}|\langle F_\alpha -F,D_xu_\alpha \rangle |\,d\underline{\mu }=0. \end{aligned}$$

(95)

We have in fact by Hölder’s inequality, taking into account (85), (88) and (94) that,

$$ \begin{array}{l} \displaystyle \left( \int _{H_T}|\langle F_\alpha -F,D_xu_\alpha \rangle |\,d\underline{\mu }\right) ^2 \le \int _{H_T}| F_\alpha -F|^2\,d\underline{\mu }\; \int _{H_T}|D_xu_\alpha |^2\,d\underline{\mu }\\ \\ \displaystyle \le C(T)\Vert f\Vert _0^2\left( 1+\int _{H_T}|F-F_\alpha |^2\,d\underline{\mu }\right) \int _{H_T}| F_\alpha -F|^2\,d\underline{\mu }\\ \\ \displaystyle \le C(T)\Vert f\Vert _0^2\left( 1+\alpha ^2M_1\int _{H_T}V^2(x)\,d\underline{\mu }\right) \alpha ^2M_1\int _{H_T}V^2(x)\,d\underline{\mu }\rightarrow 0 \end{array} $$

as $\alpha \rightarrow 0$. The proof is complete. $\square$

Example 23

We continue here Example 10 using notations and assumptions there. As we have seen, Hypothesis 3 is fulfilled in this case. To apply Theorem 22 it remains to check that (85) is fulfilled. This clearly holds by (47), thus Theorem 22 applies.

Now we assume Hypotheses 1 and 3. Then, to repeat the proof of Theorem 22 we should find an estimate for $|(-A)^{1/2}D_xu_\alpha |^2$ rather than for $|D_xu_\alpha |^2$ which seems to be difficult. So, we shall assume (95) and, arguing as before, we can prove

Theorem 24

Assume, besides Hypotheses 1 and 3 that

$$\begin{aligned} \lim _{\alpha \rightarrow 0} \int _{H_T}|\langle F_\alpha -F,D_xu_\alpha \rangle |\,d\underline{\mu }=0. \end{aligned}$$

(96)

Let moreover $\zeta \in \mathcal P_{\psi }(H)$, where $\psi$ is defined by (21). Then the Fokker–Planck equation (3) has a unique solution $\underline{\mu }=(\mu _t)_{t\in [0,T]}$ and $\mu _t\in \mathcal P_{\psi }(H)$ for any $t\ge 0$.

Example 25

(Burgers equation) We consider here the setting of Example 17. Then statement (96) follows from [10, Lemma 4.1] and so Theorem 24 applies.

4.4 The Case When Tr $C<\infty$

Trying to repeat the proof of Theorem 22, whereas Steps 1 and 2 can be repeated without anyproblems, there is a difficulty for the proof of step 3 which requires $C^{-1}\in L(H)$. The key point is again to prove the statement (95).

Theorem 26

Assume, besides Hypotheses 1 and 3 that

$$\begin{aligned} \lim _{\alpha \rightarrow 0} \int _{H_T}|\langle F_\alpha -F,D_xu_\alpha \rangle |\,d\underline{\mu }=0. \end{aligned}$$

(97)

Then there is a unique solution $\underline{\mu }$ of the Fokker–Planck equation (3).

Example 27

Consider the 2D-Navier–Stokes equation from Example 18 and assume that Tr$[(-A)C]<\infty$. Then (97) is fulfilled by [8, Eq. 5.4.8]. So, Theorem 26 applies.