On the Cauchy Problem for Fokker–Planck–Kolmogorov Equations with Potential Terms on Arbitrary Domains

Abstract

We study the Cauchy problem for Fokker–Planck–Kolmogorov equations for finite measures with unbounded and degenerate coefficients. Sufficient conditions for the existence and uniqueness of solutions are given.

1 Introduction

In this paper we study the Cauchy problem for the Fokker–Planck–Kolmogorov equation

$$\begin{aligned} \partial _t\mu =\partial _{x_i}\partial _{x_j}\bigl (a^{ij}\mu \bigr )- \partial _{x_i}\bigl (b^i\mu \bigr )+c\mu , \quad \mu |_{t=0}=\nu \end{aligned}$$

(1.1)

with respect to measures on $D\times [0,T]$ for a domain $D\subset {\mathbb {R}}^d$; no boundary condition is imposed. Throughout the paper summation over all repeated indices is meant.

Equations of this type for transition probabilities of diffusion processes were first derived by Kolmogorov in his celebrated paper [18]. In the same paper, the question about the existence and uniqueness of probability solutions was posed (the case $c=0$). The classical works [2, 14, 27–29] deal with such equations with smooth coefficients having at most linear growth at infinity.

Equations with integrable and Sobolev coefficients in the class of bounded Borel measures have been intensively studied in the past decade. For the variational approach to (1.1) in the case of unit diffusion matrix, a gradient drift and $c=0$, see [1, 17]. The existence and uniqueness of solutions given by families of probability measures in the case where $c=0$, the diffusion matrix is nondegenerate and Sobolev regular and the drift is integrable have been studied in [4–6, 9]. The papers [13, 22, 24] deal with equations with degenerate diffusion matrices. In particular, the solvability of the Cauchy problem for equations with a degenerate Sobolev regular $A$ in the class of densities under certain growth restrictions on the lower order terms has been proved in [22]. The regularisation phenomenon of a multiplicative stochastic perturbation of Brownian type for transport equations (the case $A=0$) with non smooth drift coefficients was considered in [12]. Relations between the $L^1$- and $L^{\infty }$-uniqueness of semigroups, Liouville-type theorems and the uniqueness of $L^1$-solutions to the Cauchy problem for the Fokker–Planck–Kolmogorov equation have been studied in [21, 30]. In [11], the existence, uniqueness and behavior of nonnegative solutions of equation (1.1) with $b=c=0$ on ${\mathbb {R}}^{d+1}$ have been investigated. Nonnegative solutions of divergence type equations are considered in [16, 23].

Fokker–Planck–Kolmogorov equations are closely connected with the corresponding diffusion processes (see [18, 27]). It is well known that the transition probabilities of these processes satisfy such equations and in the case of globally bounded coefficients one can represent solutions of the Fokker–Planck–Kolmogorov equation as transition probabilities (see [13]). However, in the case of unbounded and nonsmooth coefficients associated diffusion processes are not unique and can explode. Moreover, in this case the Cauchy problem (1.1) can have several solutions even if the diffusion process is unique (see [9]). Note also that diffusion processes on arbitrary domains are investigated in [15], where, in particular, the existence and uniqueness of a diffusion process on $D=(-1, 1)$ with the generator

$$\begin{aligned} Lu(x)=2^{-1}\bigl |1-|x|\bigr |^{2\alpha }u''(x)+\bigl (\mathrm{tg}(-\pi x/2)+\mathrm{sgn} x\bigr )u'(x) \end{aligned}$$

is proved.

Nevertheless, in spite of the existing immense literature on the subject, one cannot say that Kolmogorov’s question is completely answered. For example, it is still unknown whether in the case of dimension $d=1$ or $d=2$ with $(a^{ij})=\mathrm{I}$ and smooth $b$ it can happen that there exist two different probability solutions to (1.1); for $d\ge 3$ such examples exist (see [9]).

Our paper is concerned with the following problems.

(I)  Existence of solutions under very weak local assumptions about the coefficients. For an arbitrary domain $D$ (not necessarily bounded, e.g., $D={\mathbb {R}}^d$) and a probability measure $\nu$ on $D$, we construct on $[0, T]$ a family of subprobability measures $\mu _t$ on $D$ such that $d\mu =d\mu _t\,dt$ solves (1.1) and the inequality

$$\begin{aligned} \mu _t(D)\le \nu (D)+\int _0^t\int _{D}c(x, s)\,\mu _s(dx)\,ds \end{aligned}$$

(1.2)

holds for almost all $t\in (0, T)$. Several existence results are presented here, for instance, such a solution exists if $A=(a^{ij})$ is locally nondegenerate, the coefficients $a^{ij}, b^i, c$ are locally bounded and $c\le 0$. It is important that no Sobolev regularity or global integrability of the coefficients is assumed. In order to construct such solutions we employ estimates from [7], and the main difficulty is to connect our weak solution in the interior of the domain with the initial measure $\nu$.

(II)   Conditions which ensure the equality in (1.2). If $c=0$, this means the conservation of mass: $\mu _t(D)=1$. It turns out that a very natural assumption (from the probabilistic point of view) is the existence of a Lyapunov function. We say that a function $V\in C^{2}(D)$ is a Lyapunov function for the differential operator

$$\begin{aligned} L=a^{ij}\partial _{x_i}\partial _{x_j}+b^i\partial _{x_i}+c \end{aligned}$$

if $V(x)\rightarrow +\infty$ as $x\rightarrow \partial D$ and $LV\le C_1+C_2V$. If we consider a diffusion process with the generator $L$ (assuming that the coefficients are sufficiently smooth), then a Lyapunov function ensures that the process does not explode.

(III)   Does the Cauchy problem (1.1) have a unique solution in our class? Recall that we consider subprobability measures $\mu _t$ which satisfy inequality (1.2), but not arbitrary solutions.

The reader is warned that no boundary condition is imposed even if $D\not ={\mathbb {R}}^d$!

Let us consider a very simple example: $D=(0, 1)$ and $\partial _t\mu =\partial _{xx}\mu$. It is obvious that there are several solutions, because no boundary condition is imposed. But if we replace $(0, 1)$ with the whole space, then, as shown by Widder [29], a nonnegative solution is unique without any restrictions on the behavior of $\mu$ at infinity. Hence, if we have no boundary conditions on $\partial D\times [0, T]$, the uniqueness depends on $D$ and $L$. Thus we have a question: how does it depend? It turns out that an answer can be given in terms of suitable Lyapunov functions. More precisely, if the coefficients are sufficiently smooth and there exists a Lyapunov function $V$ for the operator $L$, then the Cauchy problem has a unique solution. For instance, let $D=(0, 1), a(x)=x^2(1-x)^2$ and $b=c=0$. Then $V(x)=x^{-1}(1-x)^{-1}$ is a Lyapunov function for $L$ on $D$ and the Cauchy problem

$$\begin{aligned} \partial _t\mu =\partial _{xx}(x^2(1-x)^2\mu ), \quad \mu |_{t=0}=\nu \end{aligned}$$

has a unique solution. In order to prove uniqueness we apply some ideas from [9], where it is assumed that $D={\mathbb {R}}^d, c=0$ and $\mu _t(D)=1$. Here we obtain generalizations to the case of equations with potential terms on arbitrary domains. Our generalizations are not merely technical: we consider a new class of solutions. Moreover, an important difference is that we have managed to eliminate the assumption (which was crucial in the paper cited) that the Lyapunov functions involved are globally Lipschitzian.

Thus, we prove that under rather broad assumptions about the coefficients $a^{ij}, b^i$, and $c$, the existence of a Lyapunov function $V$ such that

$$\begin{aligned} V\in C^2(D), \quad \lim _{x\rightarrow \partial D}V(x)=+\infty , \quad LV\le C_1+C_2V \end{aligned}$$

ensures both existence and uniqueness of a solution to the Cauchy problem (1.1) given by a family of subprobability measures $\mu _t$ such that (1.2) holds. If $c=0$ and $\nu$ is a probability measure, then the measures $\mu _t$ are probability measures as well. In particular, if $D={\mathbb {R}}^d$ and $V(x)=|x|^2/2$, then for the existence and uniqueness it is sufficient to have a bound

$$\begin{aligned} \mathrm{tr}A(x, t)+(b(x, t), x)+|x|^2c(x, t)/2\le K+K|x|^2. \end{aligned}$$

In particular, in the case of equations with unbounded coefficients we partially answer the question posed by Kolmogorov [18] about the existence and uniqueness of solutions to the Cauchy problem (1.1). More precisely, Kolmogorov asked whether a probability solution to (1.1) is unique with the initial distribution that is Dirac’s measure; in addition, two other related questions of his were: under which conditions do nonnegative solutions exist and under which conditions are such solutions probabilistic and satisfying the Chapman–Kolmogorov equation?

We now proceed to the definitions and exact statements.

Let $T>0$ and let $D$ be an arbitrary open set in ${\mathbb {R}}^d$. We assume that along with the domain $D$ an increasing sequence of bounded open sets $D_k$ is given such that for every $k$ the closure $\overline{D_k}$ of $D_k$ is contained in $D_{k+1}$ and $\bigcup _{k=1}^{\infty }D_k=D$. For example, if $D={\mathbb {R}}^d$, then for $D_k$ the ball of radius $k$ centered at the origin can be taken.

We shall say that a locally finite Borel measure $\mu$ on the strip $D\times (0,T)$ is given by a family of Borel measures $(\mu _t)_{t\in (0, T)}$ if, for every Borel set $B\subset D$, the mapping $t\mapsto \mu _t(B)$ is measurable and for every function $u\in C^{\infty }_0(D\times (0, T))$ one has

$$\begin{aligned} \int _{D\times (0, T)}u\,d\mu =\int _0^T\int _{D}u(x, t)\,\mu _t(dx)\,dt. \end{aligned}$$

Obviously, the last identity extends to all functions of the form $fu$, where $u$ is as above and $f$ is $\mu$-integrable on every compact set in $D\times (0, T)$. For example, the transition probabilities $\mu _t(B)=P(x_t\in B)$ of a stochastic process $x_t$ in $D$ define a measure $\mu =\mu _t\,dt$ on $D\times (0, T)$.

Set

$$\begin{aligned} L\varphi =a^{ij}\partial _{x_i}\partial _{x_j}\varphi +b^i\partial _{x_i}\varphi +c\varphi , \end{aligned}$$

where $a^{ij}, b^i, c$ are Borel functions on $D\times [0, T]$ and $A=(a^{ij})$ is a symmetric non-negative definite matrix (called the diffusion matrix), i.e., $a^{ij}=a^{ji}, (A(x, t)y, y)\ge 0$ for all $(x, t)\in D\times [0, T]$ and all $y\in {\mathbb {R}}^d$. The mapping $b$ is called the drift coefficient and $c$ is called the potential.

Let $\nu$ be a probability measure on $D$. We shall say that a measure $\mu =(\mu _t)_{t\in (0, T)}$ given by a family of measures satisfies the Cauchy problem (1.1) if $a^{ij}, b^i$ and $c$ belong to $L^1(\overline{D_k}\times J, |\mu |)$ for each domain $D_k$ and each interval $J\subset (0, T)$ and for every function $\varphi \in C^{\infty }_0(D)$ the identity

$$\begin{aligned} \int _{D} \varphi d\mu _{t}-\int _{D} \varphi d\nu =\lim _{\varepsilon \rightarrow 0+} \int _{\varepsilon }^{t}\int _{D} L\varphi d\mu _{s}ds \end{aligned}$$

(1.3)

holds for a.e. $t\in (0, T)$. We note that in general the set of points $t$ for which (1.3) holds depends on $\varphi$. If the function $\displaystyle t\mapsto \int _{D}\varphi \,d\mu _t$ is continuous on $(0, T)$, then identity (1.3) holds for all $t\in [0, T)$. If one has the inclusion $L\varphi \in L^1(D\times [0, T])$, then

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+} \int _{\varepsilon }^{t}\int _{D} L\varphi d\mu _{s}ds= \int _{0}^{t}\int _{D}L\varphi d\mu _{s}ds. \end{aligned}$$

We shall also use another definition of a solution, which is, however, equivalent to the previous one (see [9]). Namely, the measure $\mu =(\mu _t)_{0<t<T}$ satisfies $\partial _t\mu =L^{*}\mu$ if

$$\begin{aligned} \int _0^T\int _{D}(\partial _tu+Lu)\,d\mu _t\,dt=0 \quad \forall u\in C^{\infty }_0(D\times (0, T)). \end{aligned}$$

The measure $\mu =(\mu _t)_{0<t<T}$ satisfies the initial condition $\mu |_{t=0}=\nu$ if, for each function $\varphi \in C^{\infty }_0(D)$, there exists a set $J_{\varphi }\subset (0,T)$ of full Lebesgue measure such that

$$\begin{aligned} \lim _{t\rightarrow 0, t\in J_{\varphi }}\int _{D}\varphi \,d\mu _t=\int _{D}\varphi \,d\nu . \end{aligned}$$

As before, if the function $\displaystyle t\mapsto \int _{D}\varphi \,d\mu _t$ is continuous on $(0, T)$, then $J_{\varphi }=(0, T)$.

We shall always assume that $c\le 0$. This assumption can obviously be replaced with $c\le c_0$ for some number $c_0$. Indeed, in order to remove $c_0$ it suffices to consider $e^{-c_0t}\mu _t$ in place of $\mu _t$.

We study the existence and uniqueness in the class ${\mathcal {M}}_{\nu }$ of measures $\mu$ given by families of nonnegative measures $(\mu _t)_{0<t<T}$ such that $\mu$ is a solution to the Cauchy problem (1.1), $c\in L^1(D\times (0, T), \mu )$ and inequality (1.2) holds for a.e. $t\in (0, T)$.

Let us recall some notation. Let $C^{\infty }_0(D)$ and $C^{\infty }_0(D\times (0, T))$ denote the classes of smooth compactly supported functions on $D$ and $D\times (0, T)$, respectively. Let $L^p(D)$ and $L^p(D\times [0, T])$ denote the spaces of functions integrable to power $p$ with respect to Lebesgue measure on the corresponding domains; the analogous classes with respect to the measure $\mu$ will be denoted by $L^p(D\times [0, T], \mu )$.

The class $L^r([0, T], L^p(D)), r,p\ge 1$, consists of the equivalence classes of measurable functions $u$ on $[0,T]\times D$ such that $t\mapsto \Vert u(\,\cdot \, ,t)\Vert _{L^p(D)}$ belongs to $L^r[0,T]$ and is equipped with its natural norm defined as the $L^r$-norm of the latter function.

Let $W^{1, p}(D)$ denote the Sobolev space of all functions $f\in L^p(D)$ which have weak derivatives $\partial _{x_i}f$ belonging to $L^p(D)$. We write $f\in L^p_{loc}(D)$ or $f\in W^{1, p}_{loc}(D)$ if $f\zeta \in L^p(D)$ or $f\zeta \in W^{1, p}(D)$ for every $\zeta \in C^{\infty }_0(D)$. The class $C^{2,1}(D\times (0,T))$ consists of functions on $D\times (0,T)$ twice continuously differentiable in $x$ and continuously differentiable in $t$; $C(D\times [0,T))$ is the class of continuous functions on $D\times [0,T)$.

Let $\nabla f$ denote the gradient of $f$ with respect to the space variable $x$, i.e.,

$$\begin{aligned} \nabla f(x, t)=(\partial _{x_1}f(x, t), \ldots , \partial _{x_d}f(x, t)). \end{aligned}$$

Existence results are discussed in Sect. 2 and uniqueness results are presented in Sect. 3.

2 Existence Results

In this section we prove several existence theorems under different assumptions about coefficients.

Theorem 2.1

Suppose that $c\le 0$, for each $k\in {\mathbb {N}}$ the coefficients $a^{ij},\, b^{i}$ and $c$ are bounded on $D_k\times [0, T]$ and there exist positive numbers  $m_k$ and $M_k$ such that the inequality

$$\begin{aligned} m_k|y|^{2}\le (A(x, t)y, y)\le M_k|y|^{2} \end{aligned}$$

holds for all $y\in {\mathbb {R}}^d$ and $(x, t)\in D_k\times [0, T]$. Then, for every probability measure $\nu$ on $D$, the set ${\mathcal {M}}_{\nu }$ is not empty.

Proof

We divide the proof into several steps.

1. We set $a^{ij}(x, t)=0, b^{i}(x, t)=0$ and $c(x, t)=0$ if $t\not \in [0, T]$ or if $t\in [0, T]$, but $x\not \in D$. Let $\omega$ be a function such that

$$\begin{aligned} \omega \in C^{\infty }_0({\mathbb {R}}^{d+1}), \quad \omega \ge 0, \quad \int _{{\mathbb {R}}^{d+1}}\omega (x, t)\,dx\,dt=1. \end{aligned}$$

Set $\omega _{\varepsilon }(x, t)=\varepsilon ^{-d-1}\omega (x\varepsilon ^{-1}, t\varepsilon ^{-1})$. Let $I_n$ be the indicator of $D_n\times [0, T]$. Let

$$\begin{aligned} a^{ij}_{n}=(a^{ij}I_n+\delta ^{ij}(1-I_n))*\omega _{1/n}, \quad b^i_n=(b^iI_n)*\omega _{1/n}, \quad \mathrm{and} \quad c_n=(cI_n)*\omega _{1/n}. \end{aligned}$$

It is obvious that for every fixed $n$ the functions $a^{ij}_n, b^i_n$ and $c_n$ are smooth and uniformly bounded along with all their derivatives. Moreover, $(A_n(x, t)y, y)\ge |y|^2\min \{m_n, 1\}$ for all $t, x$ and $y$; here $m_n$ is the number from the assumptions of the theorem corresponding to the set $D_n\times [0, T]$. Finally, for every $k$ and $p\ge 1$ the sequences $a^{ij}_n, b^i_n$, and $c_n$ converge in $L^p(D_k\times [0, T])$ to the functions $a^{ij}, b^i$ and $c$, respectively.

We extend the measure $\nu$ by zero outside of $D$ to a measure on the whole space ${\mathbb {R}}^d$. Let $\eta _{n}\in C_{0}^{\infty }(D)$ be a sequence of nonnegative functions such that $\eta _{n}\,dx$ are probability measures on ${\mathbb {R}}^d$ weakly convergent to $\nu$.

On ${\mathbb {R}}^d\times [0, T]$ we consider the Cauchy problem

$$\begin{aligned} \partial _{t}u_{n}=\partial _{x_{i}}\partial _{x_{j}}(a_{n}^{ij}u_{n}) -\partial _{x_{i}}(b_{n}^{i}u_{n}) +c_{n}u_{n},\quad u_{n}|_{t=0}=\eta _{n}. \end{aligned}$$

Let us rewrite it as

$$\begin{aligned} \partial _{t}u_{n}=a_{n}^{ij}\partial _{x_{i}x_{j}}u_{n} +\left( 2\partial _{x_{j}}a_{n}^{ij}-b_{n}^{i}\right) \partial _{x_{i}}u_{n} +\left( \partial _{x_{i}x_{j}}a_{n}^{ij} -\partial _{x_{i}}b_{n}^{i}+q_{n}\right) u_{n}, \quad u_{n}|_{t=0}=\eta _{n}. \end{aligned}$$

Here all the coefficients are smooth and bounded for each $n$. It is well-known (see [14], Chap. 1, §7, Theorem 12) that there exists a smooth bounded nonnegative classical solution $u_n\in L^{1}({\mathbb {R}}^{d}\times [0,T])$. Further, the classical solution $u_n$ is also a weak solution in sense of distributions, so for every function $\psi \in C_{0}^{\infty }({\mathbb {R}}^{d})$ one has

$$\begin{aligned} \int _{{\mathbb {R}}^d}\psi (x)u_{n}(x,t)\,dx= \int _{{\mathbb {R}}^d}\psi (x)\eta _{n}(x)\,dx +\int _{0}^{t}\int _{{\mathbb {R}}^d} L_n\psi (x, s) u_{n}(x,s)\,dx\,ds. \end{aligned}$$

(2.1)

Let $\zeta \in C_{0}^{\infty }({\mathbb {R}}^d)$ be such that $\zeta (x)=1$ if $|x|\le 1, \zeta (x)=0$ if $|x|>2, |\zeta |\le 1$ and $\zeta$ has two bounded derivatives. Let us plug $\psi (x)=\zeta (x/N)$ in equality (2.1) and let $N$ go to infinity. For every fixed $n$ the functions $a_{n}^{ij},\, b_{n}^{i},\, c_{n}$ are globally bounded, so the Lebesgue dominated theorem yields that

$$\begin{aligned} \int _{{\mathbb {R}}^d}u_{n}(x,t)\,dx =\int _{{\mathbb {R}}^d}\eta _{n}(x)\, dx+\int _{0}^{t}\int _{{\mathbb {R}}^d}c_{n}(x,s)u_{n}(x,s)\, dx\, ds. \end{aligned}$$

In particular, the measures $u_{n}(x,t)\, dx$ are subprobability measures for all $t\in [0, T]$. Taking into account that $\eta _n=0$ outside of $D$ and $c_n\le 0$, we obtain

$$\begin{aligned} \int _{D}u_{n}(x,t)\,dx\le \int _{D}\eta _{n}(x)\, dx+\int _{0}^{t}\int _{D}c_{n}(x,s)u_{n}(x,s)\, dx\, ds. \end{aligned}$$

(2.2)

2. We now choose a convergent subsequence in $\{u_{n}\}$.

We observe that for any fixed $k$ and all sufficiently large $n$ the bound $(A_n(x, t)y, y)\ge |y|^2\min \{m_{k+1}, 1\}$ holds for all $(x, t)\in D_k\times [0, T]$ and $y\in {\mathbb {R}}^d$. Indeed,

$$\begin{aligned} \left( a^{ij}I_n+(1-I_n)\delta ^{ij}\right) *\omega _{1/n}(x, t)= \left( a^{ij}I_{k+1}+(1-I_{k+1})\delta ^{ij}\right) *\omega _{1/n}(x, t) \end{aligned}$$

if $(x, t)\in D_k\times [0, T]$ and $n$ is large enough so that the support of the function $(y, \tau )\mapsto \omega _{1/n}(y-x, \tau -t)$ belongs to $D_{k+1}\times (-1, T+1)$. Similarly, for fixed $n$ (large enough) and $k$ one has

$$\begin{aligned}&\Vert a^{ij}_n\Vert _{L^{\infty }(D_k\times [0, T])}\le \Vert a\Vert _{L^{\infty }(D_{k+1}\times [0, T])}+1,\quad \Vert b^i_n\Vert _{L^{\infty }(D_k\times [0, T])}\le \Vert b^i\Vert _{L^{\infty }(D_{k+1}\times [0, T])},\\&\Vert c_n\Vert _{L^{\infty }(D_k\times [0, T])}\le \Vert c\Vert _{L^{\infty }(D_{k+1}\times [0, T])}. \end{aligned}$$

Due to [7], Corollary 3.2], for every $k>2$ one has

$$\begin{aligned} \int _{D_k\times [Tk^{-1}, T(1-k^{-1})]}u_{n}^{(d+1)/d}\, dx\, dt\le C_k, \end{aligned}$$

where $C_k$ depends only on $m_{k+1}, \Vert a\Vert _{L^{\infty }(D_{k+1}\times [0, T])}, \Vert b^i\Vert _{L^{\infty }(D_{k+1}\times [0, T])}, \Vert c\Vert _{L^{\infty }(D_{k+1}\times [0, T])}$ and does not depend on $n$.

Since the unit ball in $L^{(d+1)/d}$ is weakly compact, for every $k>2$ one can extract from $\{u_{n}\}$ a subsequence weakly convergent in $L^{(d+1)/d}(D_k\times [Tk^{-1}, T(1-k^{-1})])$. Without loss of generality, using the diagonal procedure, we can assume that $\{u_{n}\}$ converges weakly to a nonnegative function $u$ that belongs to $L^{(d+1)/d}(D_k\times [Tk^{-1}, T(1-k^{-1})])$ for every $k$.

3. As we have shown above, for every $k$ the coefficients $a_{n}^{ij}, b_{n}^{i}$ and $c_{n}$ are uniformly bounded (with respect to $n$) on $D_k\times [0, T]$. Let $\psi \in C^{\infty }_0(D)$. Then $\text{ supp }\,\psi \subset D_k$ for some $k$ and there exists a number $C(\psi )$ (independent of $n$) such that

$$\begin{aligned} \left| \int _{D}\psi (x)u_{n}(x,t)\, dx- \int _{D}\psi (x)u_{n}(x,s)\, dx\right|= & {} \left| \int _{s}^{t}\int _{D}\left( L_{n}\psi (x, \tau )\right) u_{n}(x, \tau )dx\,d\tau \right| \\&\le C(\psi )|t-s| \end{aligned}$$

for all $n$ and $s, t\in [0, T]$. Hence the functions

$$\begin{aligned} f_{n}(t):=\int _{D}\psi (x)u_{n}(x,t)\, dx \end{aligned}$$

are Lipschitzian with the constant $C(\psi )$ independent of $n$. Thus, this is a uniformly bounded and equicontinuous family of functions for every fixed $\psi$. By the Arzelà–Ascoli theorem there exists a subsequence uniformly convergent on $[0, T]$. We observe that in the space $L^{(d+1)/d}[0,T]$ the same subsequence converges to

$$\begin{aligned} f(t):=\int _{D}\psi (x)u(x,t)\, dx \end{aligned}$$

and, since the weak and uniform limits coincide a.e., every subsequence of $\{f_{n}\}$ (and thus the whole sequence as well) converges uniformly to the same Lipschitzian function $\widetilde{f}$ which coincides with $f$ on a set of full Lebesgue measure in $[0,T]$. Obviously, this set depends on $\psi$. Denote it by ${\mathbb {T}}(\psi )$.

4. Let us show that $u$ constructed at Step 2 is a solution to (1.1). Fix $\psi \in C_{0}^{\infty }({\mathbb {R}}^{d})$ with support in some $D_k$. We have $L\psi \in L^{\infty }(D_k\times [0, T])$ by the assumptions of the theorem. Moreover, there is a number $C_k$ such that $\sup _n\Vert L_n\psi \Vert _{L^{\infty }(D_k\times [0, T])}\le C_k$, and $L_n\psi$ converges to $L\psi$ in $L^p(D_k\times [0, T])$ for every $p\ge 1$. Take $t$ in the full measure set ${\mathbb {T}}(\psi )$; for every element of this set convergence

$$\begin{aligned} \int _D\psi (x)u_{n}(x,t)\, dx\rightarrow \int _D\psi (x)u(x,t)\, dx \end{aligned}$$

takes place. Let $0<s<t$. One has

$$\begin{aligned}&\left| \int _D\psi (x)u_{n}(x,t)\, dx- \int _D\psi (x)\eta _{n}(x)\, dx -\int _{s}^{t}\int _D u_{n}L_{n}\psi \, dx\, d\tau \right| \\&\qquad =\left| \int _D\psi (x)u_{n}(x,s)\, dx-\int _D\psi (x)\eta _{n}(x)\,dx\right| \le C(\psi )s, \end{aligned}$$

where $C(\psi )$ is independent of $n$ and $s$. Hence,

$$\begin{aligned} \left| \int _D\psi (x)u_{n}(x,t)\, dx- \int _D\psi (x)\eta _{n}(x)\, dx - \int _{s}^{t}\int _D u_nL_{n}\psi \, dx\, d\tau \right| \le C(\psi )s. \end{aligned}$$

(2.3)

We observe that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{s}^{t}\int _D L_{n}\psi (x, \tau )u_{n}(x,\tau )\, dx\, d\tau = \int _{s}^{t}\int _D uL\psi \, dx\, d\tau . \end{aligned}$$

Indeed,

$$\begin{aligned}&\left| \int _{s}^{t}\int _D u_n(x, \tau )L_{n}\psi (x, \tau )\, dx\, d\tau - \int _{s}^{t}\int _D u(x, \tau )L\psi (x, \tau )\, dx\, d\tau \right| \\&\quad \le \Vert L_n\psi -L\psi \Vert _{L^{d+1}(D_k\times [s, t])}\Vert u_n\Vert _{L^{(d+1)/d}(D_k\times [s, t])} \\&\qquad +\left| \int _{s}^{t}\int _D (L\psi (x, \tau ))u_{n}(x,\tau )\, dx\, d\tau - \int _{s}^{t}\int _D (L\psi (x, \tau ))u(x,\tau )\, dx\, d\tau \right| , \end{aligned}$$

where the first summand on the right-hand side tends to zero due to convergence of $L_n\psi$ to $L\psi$ and the uniform norm boundedness of $\{u_n\}$ shown above. The second summand tends to zero by the weak convergence of $\{u_n\}$ and the boundedness of $L\psi$. Thus, letting $n\rightarrow \infty$ in (2.3), we obtain

$$\begin{aligned} \left| \int _D\psi (x)u(x,t)\, dx- \int _D\psi (x)\,d\nu - \int _{s}^{t}\int _D uL\psi \, dx\, d\tau \right| \le C(\psi )s. \end{aligned}$$

Letting $s\rightarrow 0$, we arrive at the equality

$$\begin{aligned} \int _D\psi (x)u(x,t)\, dx=\int _D\psi (x)\,d\nu +\int _{0}^{t}\int _D uL\psi \, dx\, d\tau . \end{aligned}$$

Hence, the function $u$ is a nonnegative solution to the Cauchy problem (1.1).

5. Let us show that the measure $u(x,t)dx\, dt$ is a solution in the class ${\mathcal {M}}_{\nu }$. We recall that $c_{n}\le 0$ and $\eta _n\,dx$ are probability measures. By (2.2), for every function $\psi \in C_{0}^{\infty }(D), 0\le \psi \le 1$ the following inequality holds:

$$\begin{aligned} \int _D\psi (x)u_{n}(x,t)\, dx-\int _{0}^{t}\int _D\psi (x)c_{n}(x,s)u_{n}(x,s)\, dx\, ds\le 1. \end{aligned}$$

(2.4)

Let $\psi _{N}\in C^{\infty }_0(D), 0\le \psi _N\le 1$ and $\psi _N(x)=1$ if $x\in D_N$. Let $t$ also belong to the full measure set ${\mathbb {T}}=\bigcap _{N\in {\mathbb {N}}}\mathbb {T}(\psi _{N})$, i.e., for all $N\in {\mathbb {N}}$ the following convergence takes place:

$$\begin{aligned} \int _D\psi _{N}(x)u_{n}(x,t)\, dx\rightarrow \int _D\psi _{N}(x)u(x,t)\, dx. \end{aligned}$$

Plugging such $\psi _{N}$ and $t$ into (2.4) and letting $n\rightarrow \infty$, we obtain

$$\begin{aligned} \int _{D}\psi _{N}(x)u(x,t)\, dx-\int _{0}^{t}\int _{D}\psi _{N}(x)c(x,s)u(x,s)\, dx\, ds\le 1. \end{aligned}$$

Finally, letting $N\rightarrow \infty$ and applying Fatou’s lemma, we arrive at the required inequality

$$\begin{aligned} \int _D u(x,t)\, dx-\int _{0}^{t}\int _D c(x,s)u(x,s)\, dx\, ds\le 1=\nu (D). \end{aligned}$$

This completes the proof. $\square$

Remark 2.2

According to [7], Corollary 3.2], under the assumptions of Theorem 2.1, every solution $\mu$ in ${\mathcal {M}}_{\nu }$ is given by a density $\varrho \in L^{(d+1)/d}_{loc}(D\times (0, T))$ with respect to Lebesgue measure.

The assumptions of local boundedness can be weakened under the additional assumption that the elements of the matrix $A$ are Sobolev regular. The next theorem was proved in [4] in the case $c=0$ under the assumption of existence of a Lyapunov function. We give here a different and shorter proof and do not impose global restrictions on the coefficients.

Theorem 2.3

Let $p>d+2$. Assume that for every $k$ we have $a^{ij}(\, \cdot \,, t)\in W_{loc}^{1,p}(D_k)$,

$$\begin{aligned} \sup _{t\in (0, T)}\Vert a^{ij}(\,\cdot \,, t)\Vert _{W^{1, p}(D_k)}<\infty \end{aligned}$$

and $(A(x, t)y, y)\ge m_k|y|^2$ for all $(x, t)\in D_k\times [0, T], y\in {\mathbb {R}}^d$ and some $m_k>0$. Assume also that $b\in L^{p}(D_k\times [0, T])$ and $c\in L^{p/2}(D_k\times [0, T])$ for each number $k$. Then, for every probability measure $\nu$ on $D$, the set ${\mathcal {M}}_{\nu }$ is not empty.

Proof

1. Exactly as at Step 1 of the proof of the previous theorem, we construct sequences of smooth bounded functions $a^{ij}_n, b^i_n$ and $c_n$ such that for every domain $D_k$ one has

$$\begin{aligned}&\lim _{n\rightarrow \infty }\Vert a^{ij}_n- a^{ij}\Vert _{L^{p}(D_k\times [0, T])}=0,\\&\lim _{n\rightarrow \infty }\Vert b^i_n- b^i\Vert _{L^{p}(D_k\times [0, T])}=0,\\&\lim _{n\rightarrow \infty }\Vert c_n- c\Vert _{L^{p/2}(D_k\times [0, T])}=0, \end{aligned}$$

in particular, the norms $\Vert a^{ij}_n\Vert _{L^p(D_k\times [0, T])}, \Vert b^i\Vert _{L^p(D_k\times [0, T])}, \Vert c_n\Vert _{L^{p/2}(D_k\times [0, T])}$ are bounded uniformly in $n$. Moreover, $c_n\le 0$ and $(A_n(x, t)y, y)\ge \min \{m_{k}, 1\}$ for all $n>k$. We extend the measure $\nu$ by zero outside of $D$ to a measure on ${\mathbb {R}}^d$. Let $\eta _{n}\in C_{0}^{\infty }(D)$ be a sequence of nonnegative functions such that $\eta _{n}\,dx$ are probability measures on ${\mathbb {R}}^d$ weakly convergent to $\nu$.

Let $\{u_n\}$ be a smooth bounded solution of the Cauchy problem

$$\begin{aligned} \partial _{t}u_{n}=\partial _{x_{i}}\partial _{x_{j}}(a_{n}^{ij}u_{n})- \partial _{x_{i}}(b_{n}^{i}u_{n}) +c_{n}u_{n},\quad u_{n}|_{t=0}=\eta _{n}. \end{aligned}$$

Then $u_n(x, t)\,dx$ are subprobability measures for every $t$ and

$$\begin{aligned} \int _{{\mathbb {R}}^d}u_n(x, t)\,dx=\int _{{\mathbb {R}}^d}\eta _n(x)\,dx +\int _0^t\int _{{\mathbb {R}}^d}c_n(x, t)u_{n}(x, t)\,dx\,d\tau . \end{aligned}$$

2. We now pick a convergent subsequence in $\{u_{n}\}$. By [7], Corollary 3.9], for every $k>2$ the following estimate on the Hölder norm holds:

$$\begin{aligned} \Vert u_{n}\Vert _{C^{\alpha }(D_k\times [Tk^{-1}, T(1-k^{-1})])}\le C_k, \end{aligned}$$

where $\alpha \in (0, 1)$ and $C_k$ is independent of $n$. The Arzelà–Ascoli theorem, the diagonal method and a passage to a subsequence enable us to conclude that the sequence $\{u_{n}\}$ converges uniformly to some function $u$ on $D_k\times [Tk^{-1}, T(1-k^{-1})]$ for every $k$. It is obvious that $u$ is a nonnegative continuous function. Let us show that $u$ satisfies (1.1).

Let $\psi \in C_{0}^{\infty }(D)$. Then $\text{ supp }\,\psi \subset D_k$ for some $k$. The uniform convergence of $\{u_{n}\}$ immediately yields that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _D\psi (x)u_{n}(x,t)dx=\int _D\psi (x)u(x,t)dx \end{aligned}$$

for all $t\in (0, T)$. Now let $0<s<t<T$. We observe that

$$\begin{aligned}&\left| \int _s^t\int _{D}L_n\psi u_n\,dx\,d\tau -\int _s^t\int _{D}L\psi u\,dx\,d\tau \right| \\&\quad \le \Vert L_n\psi -L\psi \Vert _{L^1(D_k\times [s, t])}\Vert u_n\Vert _{L^{\infty }(D_k\times [s, t])} +\Vert L\psi \Vert _{L^1(D_k\times [s, t])}\Vert u- u_n\Vert _{L^{\infty }(D_k\times [s, t])}, \end{aligned}$$

where the first summand on the right-hand side tends to zero by convergence of $a^{ij}_n, b^i_n$ and $c_n$ to $a^{ij}, b^i$ and $c$ respectively. The uniform convergence of $\{u_n\}$ to $u$ yields convergence of the second summand to zero. Therefore,

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _s^t\int _{D}L_n\psi u_n\,dx\,d\tau =\int _s^t\int _{D}L\psi u\,dx\,d\tau . \end{aligned}$$

Thus, letting $n\rightarrow \infty$, we obtain

$$\begin{aligned} \int _D\psi (x)u(x, t)=\int _D\psi (x)u(x, s)\,dx+\int _s^t\int _DL\psi u\,dx\,d\tau \end{aligned}$$

for all $s, t\in (0, T)$.

3. Let us justify the limit as $s\rightarrow 0$.

Let $0<\tau <T$ and $y\in C_{0}^{\infty }(D)$. We extend the function $y$ by zero outside of $D$. Let also $w_{n,\tau }$ be the solution of the adjoint problem

$$\begin{aligned} \partial _{t}w_{n,\tau }+a_{n}^{ij}\partial _{x_{i}}\partial _{x_{j}}w_{n,\tau } +b_{n}^{i}\partial _{x_{i}}w_{n,\tau }+c_{n}w_{n,\tau }=0,\quad w_{n,\tau }|_{t=\tau }=y. \end{aligned}$$

Let $\zeta \in C_{0}^{\infty }({\mathbb {R}})$ be such that $\zeta (x)=1$ if $|x|\le 1, \zeta (x)=0$ if $|x|>2, |\zeta |\le 1$ and let $\zeta$ have two bounded derivatives. Multiplying the adjoint equation by $\zeta _Nu_n$, integrating by parts and then letting $N\rightarrow \infty$, we obtain

$$\begin{aligned} \int _{D} y(x)u_{n}(x,\tau )\, dx=\int _{D}w_{n,\tau }(x,0)\eta _{n}(x)\,dx, \end{aligned}$$

where in the last equality we have taken into account that $\text{ supp }\,y,\quad \text{ supp }\,\eta _n \subset D$. By [20], Part III,Theorem 10.1], for every ball $U\subset {\mathbb {R}}^d$ the estimate

$$\begin{aligned} \Vert w_{n,\tau }(x,0)-y(x)\Vert _{L^{\infty }(U)}\le C(U)\tau ^{\alpha } \end{aligned}$$

holds with $C$ and $\alpha$ independent of $n$. Thus,

$$\begin{aligned}&\left| \int _{D}y(x)u_n(x, \tau )\,dx- \int _{D}y\,d\nu \right| \\&\quad \le \int _{D}|w_{n, \tau }(x, 0)-y(x)|\eta _n(x)\,dx +\left| \int _{D}y(x)\eta _n(x)\,dx- \int _{D}y\,d\nu \right| \\&\qquad \le C\tau ^{\alpha }+\left| \int _{D}y(x)\eta _n(x)\,dx- \int _{D}y\,d\nu \right| . \end{aligned}$$

Letting $n\rightarrow \infty$, we obtain the estimate

$$\begin{aligned} \left| \int _{D}y(x)u(x, \tau )\,dx-\int _{D}y\,d\nu \right| \le C\tau ^{\alpha }, \end{aligned}$$

which yields that

$$\begin{aligned} \lim _{\tau \rightarrow 0}\int _{D}y(x)u(x, \tau )\,dx=\int _{D}y\,d\nu . \end{aligned}$$

Finally, the inequality

$$\begin{aligned} \int _{D} u(x,t)\, dx- \int _{0}^{t}\int _{D} c(x,s)u(x,s)\, dx\, ds\le 1=\nu (D) \end{aligned}$$

can be justified exactly in the same way as at Step 5 of the proof of Theorem 2.1. $\square$

Remark 2.4

By [7], Corollary 3.9], under the assumptions of Theorem 2.3, every solution $\mu$ in ${\mathcal {M}}_{\nu }$ is given by a locally Hölder continuous density $\varrho$ with respect to Lebesgue measure.

We now proceed to the case of a degenerate matrix $A$.

Theorem 2.5

Let the coefficients $a^{ij},\, b^{i}$ and $c$ be continuous in $x$, measurable in $t$ and bounded on $D_k\times [0, T]$ for every $k$. Assume that the diffusion matrix $A$ is symmetric and $(A(x, t)y, y)\ge 0$ for all $(x, t)\in D\times [0, T]$ and $y\in {\mathbb {R}}^d$. Then for every probability measure $\nu$ the set ${\mathcal {M}}_{\nu }$ is not empty.

Proof

We use the well-known method of “vanishing viscosity”. 1. Let us introduce the operator

$$\begin{aligned} L_{\varepsilon }:=\varepsilon \Delta +L \end{aligned}$$

for every $\varepsilon >0$ and consider the Cauchy problem

$$\begin{aligned} \partial _{t}\mu _{t}=L_{\varepsilon }^{*}\mu , \quad \mu |_{t=0}=\nu . \end{aligned}$$

(2.5)

Obviously, the hypotheses of Theorem 2.1 are fulfilled. Then, for every $n$ the Cauchy problem (2.5) with $\varepsilon =1/n$ has a solution $\mu ^n$ given by a family of subprobability measures $(\mu _{t}^{n})_{t\in (0,T)}$ on $D$ for which the following inequality holds:

$$\begin{aligned} \mu _t^n(D)\le \nu (D)+\int _0^t\int _{D}c(x, s)\mu _s^n(dx)\,ds. \end{aligned}$$

2. Next, we choose a convergent subsequence of solutions.

There exists a subsequence of indices $n_{l}$ such that the measures $\mu _{t}^{n_{l}}$ converge weakly on each compact set $\overline{D_k}$ for each $t\in [0,T]$. To prove this, it suffices to apply Prokhorov’s theorem for any fixed compact set $\overline{D_k}$ for a dense countable set $S\in [0,T]$ along with the diagonal method. Then, using again the diagonal method for compact sets $\overline{D_k}$, we can extract a subsequence of measures that converges on all compact sets. Let us show that the constructed subsequence is a Cauchy sequence for every $t\in [0,T]$. Let $t\in [0,T], s\in S$ and $\varphi \in C_{0}^{\infty }({\mathbb {R}}^{d})$. Since the coefficients of $L$ are bounded on cylinders, one has

$$\begin{aligned}&\left| \int _D\varphi d\mu _{t}^{n_{p}}-\int _D\varphi d\mu _{t}^{n_{k}}\right| \le \left| \int _D\varphi d\mu _{t}^{n_{p}}-\int _D\varphi d\mu _{s}^{n_{p}}\right| \\&\quad +\left| \int _D\varphi d\mu _{s}^{n_{p}}-\int _D\varphi d\mu _{s}^{n_{k}}\right| + \left| \int _D\varphi d\mu _{s}^{n_{k}}-\int _D\varphi d\mu _{t}^{n_{k}}\right| \\&\qquad \le 2C(\varphi )\cdot |t-s|+\left| \int _D\varphi d\mu _{s}^{n_{p}}-\int _D\varphi d\mu _{s}^{n_{k}}\right| . \end{aligned}$$

Given $\varepsilon >0$, we can choose $s$ close enough to $t$ to make the first summand less than $\varepsilon /2$. Since the sequence $\mu _{s}^{n_{l}}$ converges, there exists a number $N$ such that for all $p,k>N$ the second summand is less than $\varepsilon /2$. Therefore, it is proved that for every function $\varphi \in C_{0}^{\infty }(D)$ the sequence of integrals ${\displaystyle \int _D\varphi \, d\mu _{t}^{n_{l}}}$ is a Cauchy sequence, hence has a finite limit.

Fix a number $k$. Since we can chose a weakly convergent subsequence in each subsequence of the restrictions of $\mu _t^{n_l}$ to $\overline{D_k}$ and, by the previous step, such weakly convergent subsequences must have the same limit, we conclude that for every $t$ the sequence $\mu _{t}^{n_l}$ converges weakly to some subprobability measure $\mu _t$ on each $D_k$.

We observe that for every continuous function $f$ the mapping ${\displaystyle t\mapsto \int _{D_k} f\, d\mu _{t}}$ is Borel measurable on $[0,T]$ as a limit of measurable functions. Consider the class $\Phi$ of bounded Borel functions $\varphi$ on $D_k$ for which the mapping ${\displaystyle t\mapsto \int _{D_k}\varphi \, d\mu _{t}}$ is Borel measurable on $[0,\tau ]$. The set $\Phi$ contains the algebra of continuous bounded functions on $D_k$ and is closed with respect to uniform and monotone limits. By the monotone class theorem (see [3], Theorem 2.2.12]) the set $\Phi$ contains all bounded Borel functions on $D_k$. In particular, the mapping $t\mapsto \mu _{t}(B)$ is Borel measurable on $[0,T]$ for each Borel set $B\subset D_k$. Since $k$ was arbitrary, this is true for every Borel subset of $D$. Let $\mu$ be the measure given by the family $(\mu _{t})_{t\in [0,T]}$. Obviously, $\{\mu ^{n_{l}}\}$ converges weakly to $\mu$ on $D_k\times [0, T]$ for every $k$.

3. Passing to the subsequence constructed above, we conclude that the sequence $\mu ^n=(\mu _t^n)_{t\in (0, t)}$ is such that $\{\mu ^n\}$ converges weakly to $\mu =(\mu _t)_{t\in (0, T)}$ and $\{\mu _t^n\}$ converges weakly to $\mu _t$ for each $t$.

Let $\varphi \in C_{0}^{\infty }(D)$. For every $n$ one has

$$\begin{aligned} \int _{D}\varphi d\mu _{t}^{n}- \int _{D}\varphi d\mu _{0}= \int _{0}^{t}\int _{D}L_{1/n}\varphi d\mu _{s}^{n}ds. \end{aligned}$$

(2.6)

Since

$$\begin{aligned} \left| \int _{0}^{t}\int _{D} (L_{1/n}-L)\varphi d\mu _{s}^{n}ds\right| \le \frac{T}{n}\sup _{D} |\Delta \varphi |, \end{aligned}$$

by the weak convergence of $\mu ^{n}$ to $\mu$ we conclude that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\int _{0}^{t}\int _{D}L_{1/n}\varphi d\mu _{s}^{n}ds=\int _{0}^{t}\int _{D}L\varphi d\mu _{s}ds, \end{aligned}$$

and also for every $t$

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\int _{D}\varphi d\mu _{t}^{n}=\int _{D}\varphi d\mu _{t}. \end{aligned}$$

Hence one can let $n\rightarrow \infty$ in (2.6) and obtain

$$\begin{aligned} \int _D\varphi d\mu _{t}-\int _D\varphi d\mu _{0}=\int _{0}^{t}\int _D L\varphi d\mu _{s}ds. \end{aligned}$$

This means that $\mu$ is the required solution to the Cauchy problem (1.1). Finally, the inequality

$$\begin{aligned} \int _{D} u(x,t)\, dx-\int _{0}^{t}\int _{D} c(x,s)u(x,s)\, dx\, ds\le 1=\nu (D) \end{aligned}$$

is justified as at Step 5 of the proof of Theorem 2.1. $\square$

Remark 2.6

If we assume that the coefficients and the initial data are more regular, then one can construct a solution given by a density even in the case of a degenerate diffusion matrix. Such results were obtained for $D={\mathbb {R}}^d$ in [10, 13, 22]. These results can be extended to the case of an arbitrary domain $D$ as follows. Let $q\ge 1$. Suppose that for every $k$ the following assumptions are fulfilled: $b, c\in L^{q}(D_k\times [0, T]), a^{ij}(\, \cdot \,,t)\in W^{1, q}(D_k)$ and

$$\begin{aligned} \sup _{t\in [0, T]}\Vert a^{ij}\Vert _{W^{1, q}}(D_k)<\infty . \end{aligned}$$

Let $p=(q-1)/q$. Suppose that $(pc+(p-1)\mathrm{div} h)^{+}\in L^1([0, T], L^{\infty }(D))$, where $h^i=\partial _{x_j}a^{ij}-b^i.$ Let also $\varrho _0\in L^p(D)$ and $\nu =\varrho _0\,dx$. Then, there exists a solution $\mu \in {\mathcal {M}}_{\nu }$ given by a density $\varrho \in L^{\infty }([0, T], L^p(D))$.

The proof essentially repeats the reasoning from [10, 22]. We assume that the coefficients are smooth (if suffices to convolute the equation with a smooth kernel). Thus, the only problem is the unboundedness of the coefficients on $D$ and the degenerate diffusion matrix. Let $\psi _k\in C^{\infty }_0(D)$ and let $\psi _k(x)=1$ if $x\in D_k$. Let $p>1$. Consider the Cauchy problem $\partial _tu_k=L_k^{*}u_k, u_k|_{t=0}=u_0$, where

$$\begin{aligned}&L_k=(\psi _ka^{ij}+k^{-1}\delta ^{ij})\partial _{x_i}\partial _{x_j}+\psi _kb^i\partial _{x_i}+ p^{-1}(p-1)\\&\quad \bigl ((2\partial _{x_j}a^{ij}-b^i)\partial _{x_i}\psi _k+a^{ij}\partial _{x_i}\partial _{x_j}\psi _k\bigr ) +\psi _kc. \end{aligned}$$

Observe that there exists a solution $u_k$ of this Cauchy problem which is a smooth function and $u_k\in L^{\infty }([0, T], L^1(D))$. Then the inequality

$$\begin{aligned} \partial _t |u_k|^p\le \partial _{x_i}\partial _{x_j}((\psi _ka^{ij}+k^{-1}\delta ^{ij})|u_k|^p)- \partial _{x_i}(\psi _kb|u_k|^p)+\psi _k(pc+(p-1)\mathrm{div} h)|u_k|^p \end{aligned}$$

and Grownwall’s inequality immediately yield that

$$\begin{aligned} \sup _{t\in [0, T]}\int _{D}|u_k(x, t)|^p\,dx\le M \end{aligned}$$

with a constant $M$ independent of $k$. Furthermore, in the standard way one can extract a subsequence from $\{u_k\}$ which converges to a solution $\varrho$ of class $L^{\infty }([0, T], L^p(D))$.

To complete this section, we shall give conditions under which the identity

$$\begin{aligned} \mu _t(D)=\nu (D)+\int _0^t\int _{D}c(x, s)\,d\mu _s\,ds \end{aligned}$$

(2.7)

holds instead of an inequality (in particular, this means that if $c=0$ and $\nu$ is a probability measure, then $\mu _t$ is also probability measures).

Theorem 2.7

Let $\mu =(\mu _t)_{0<t<T}\in {\mathcal {M}}_{\nu }$ and $c\le 0$. Suppose that there exists a function $V$ such that $V\in C^{2, 1}(D\times (0, T))\bigcap C(D\times [0, T))$, for every interval $[\alpha , \beta ]\in (0, T)$ one has

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}\times [\alpha , \beta ]}V(x, t)=+\infty \end{aligned}$$

and, for some functions $K, H\in L^1((0, T))$ with $H\ge 0$, the following estimate holds:

$$\begin{aligned} \partial _tV(x, t)+LV(x, t)\le K(t)+H(t)V(x, t). \end{aligned}$$

Suppose also that $V(\, \cdot \,,0)\in L^1(\nu )$. Then, for a.e. $t\in (0, T)$, one has

$$\begin{aligned} \mu _t(D)=\nu (D)+\int _0^t\int _{D}c(x, s)\,d\mu _s\,ds \end{aligned}$$

and the estimate

$$\begin{aligned} \int _{D}V(x, t)\,d\mu _t\le Q(t)+R(t)\int _{D}V(x, 0)\,d\nu \end{aligned}$$

holds, where

$$\begin{aligned} R(t)=\exp \left( \int _0^tH(s)\,ds\right) , \quad Q(t)=R(t)\int _0^t\frac{K(s)}{R(s)}\,ds. \end{aligned}$$

Proof

Let $\zeta _N\in C^2([0, +\infty ))$ be such that $0\le \zeta '\le 1, \zeta ''\le 0$, where $\zeta _N(s)=s$ if $s\le N-1$ and $\zeta (s)=N$ if $s>N+1$. Let also $\eta \in C^{\infty }_0((0, T))$. For the function $u(x, t)=(\zeta _{N}(V(x, t))-N)\eta (t)$ and the solution $\mu =(\mu _t)_{0<t<T}\in {\mathcal {M}}_{\nu }$ we have

$$\begin{aligned} \int _0^T\int _D(\partial _tu+Lu)\,d\mu _t\,dt=0, \end{aligned}$$

which yields that

$$\begin{aligned} -\int _0^T\eta '(t)\int _{D}(\zeta _{N}(V(x, t))-N)\,d\mu _t\,dt=\int _0^T\eta (t)\int _{D}L(\zeta _{N}(V(x, t))-N)\,d\mu _t\,dt. \end{aligned}$$

Since $\eta$ is arbitrary,

$$\begin{aligned} \frac{d}{dt}\int _D(\zeta _{N}(V(x, t))-N)\,d\mu _t=\int _{D}L(\zeta _{N}(V(x, t))-N)\,d\mu _t. \end{aligned}$$

Hence there holds the identity

$$\begin{aligned}&\int _{D}\zeta _{N}(V(x, t))\,d\mu _t\!=\!\int _{D}\zeta _{N}(V(x, s))\,d\mu _s\!+\! \left( \mu _t(D)\!-\!\nu (D)-\int _0^t\int _{D}c(x, \tau )\,d\mu _{\tau }\,d\tau \right) N\\&\quad +\int _s^t\int _{D}\Bigl (\zeta _N'(V)(\partial _tV+LV)+\zeta _N''(V)|\sqrt{A}\nabla V|^2\Bigr )\,d\mu _{\tau }\,d\tau \\&\quad +\int _s^t\int _{D}c\left( \zeta _N(V)-\zeta '_N(V)V\right) \,d\mu _{\tau }\,d\tau . \end{aligned}$$

Observing that $z\zeta '_N(z)\le \zeta _N(z)$, we arrive at the estimate

$$\begin{aligned}&\int _{D}\zeta _N(V(x, t))\,d\mu _t\le \int _{D}\zeta _N(V(x, s))\,d\mu _s\\&\quad \!+ \left( \mu _t(D)\!-\!\nu (D){-}\int _0^t\int _{D}c(x, \tau )\,d\mu _{\tau }\,d\tau \right) N\!+\! \int _s^t K(\tau )\!+\!H(\tau )\!\int _{D}\zeta _N(V(x, \tau ))\,d\mu _{\tau }\,d\tau , \end{aligned}$$

Letting $s\rightarrow 0$, we obtain

$$\begin{aligned}&\int _{D}\zeta _N(V(x, t))\,d\mu _t\le \int _{D}\zeta _N(V(x, 0))\,d\nu +\! \left( \mu _t(D)\!-\!\nu (D)\!-\!\int _0^t\int _{D}c(x, \tau )\,d\mu _{\tau }\,d\tau \right) N\nonumber \\&\quad +\int _0^tK(\tau )+H(\tau )\int _{D}\zeta _N(V(x, \tau ))\,d\mu _{\tau }\,d\tau . \end{aligned}$$

(2.8)

Since

$$\begin{aligned} \mu _t(D)\le \nu (D)+\int _0^t\int _{D}c(x, s)\,d\mu _s\,ds, \end{aligned}$$

the last inequality can be rewritten in the following way:

$$\begin{aligned} \int _{D}\zeta _N(V(x, t))\,d\mu _t\le \int _{D}\zeta _N(V(x, 0))\,d\nu + \int _0^tK(\tau )+H(\tau )\int _{D}\zeta _N(V(x, \tau ))\,d\mu _{\tau }\,d\tau . \end{aligned}$$

Using Grownwall’s inequality, we have

$$\begin{aligned} \int _{D}\zeta _N(V(x, t))\,d\mu _t\le Q(t)+R(t)\int _{D}\zeta _N(V(x, 0))\,d\nu . \end{aligned}$$

Letting $N\rightarrow \infty$, we obtain the required estimate.

Moreover, if $\mu _t(D)<\nu (D)+\displaystyle \int _0^t\int _{D}c(x, s)\,d\mu _s\,ds$, then, letting $N\rightarrow \infty$ in (2.8), we obtain

$$\begin{aligned} \int _{D}V(x, t)\,d\mu _t-\int _{D}V(x, 0)\,d\nu - \int _0^tK(\tau )+H(\tau )\int _{D}V(x, \tau )\,d\mu _{\tau }\,d\tau =-\infty , \end{aligned}$$

which is impossible. Hence $\mu _t(D)=\nu (D)+\displaystyle \int _0^t\int _{D}c(x, s)\,d\mu _s\,ds$. The theorem is proven. $\square$

Observe that the assumption $V(\,\cdot \,,0)\in L^{1}(\nu )$ is not restrictive: if there is some Lyapunov function, then there is a Lyapunov function integrable with respect to the initial condition. More precisely, the following generalization of the lemma from [4] is true.

Proposition 2.8

Let $\mu =(\mu _{t})_{t\in [0,T]}$ be a solution to the Cauchy problem $\partial _{t}\mu =L^{*}\mu$ and $\mu |_{t=0}=\nu$, where $\nu$ is a probability measure on $D$. Suppose that there exists a nonnegative function $V$ such that $V\in C^{2, 1}(D\times (0, T))\bigcap C(D\times [0, T))$, for every interval $[\alpha , \beta ]\in (0, T)$ we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}\times [\alpha , \beta ]}V(x, t)=+\infty \end{aligned}$$

and for some functions $K, H\in L^1((0, T))$ with $H\ge 0$ the following estimate holds:

$$\begin{aligned} \partial _tV(x, t)+LV(x, t)\le K(t)+H(t)V(x, t). \end{aligned}$$

Then, there exists a nonnegative function $W\in C^{2, 1}(D\times (0, T))\bigcap C(D\times [0, T))$ such that for every interval $[\alpha , \beta ]\in (0, T)$ one has

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}\times [\alpha , \beta ]}W(x, t)=+\infty \end{aligned}$$

and $\partial _tW(x, t)+LW(x, t)\le K(t)+H(t)W$ holds and $W(x, 0)\in L^1(\nu )$.

Proof

We construct a nonnegative function $\theta \in C^{2}({\mathbb {R}})$ such that

$$\begin{aligned} \theta (0)=0,\quad \lim \limits _{r\rightarrow \infty }\theta (r)=+\infty , \quad 0\le \theta '(r)\le 1, \quad \theta ''(r)\le 0, \quad \theta (V(\, \cdot \,, 0))\in L^{1}(\nu ). \end{aligned}$$

To this end it suffices to find a function $\theta$ with the properties listed above and integrable with respect to the measure $\sigma =\nu \circ W^{-1}(\,\cdot \,, 0)$. Let us find an increasing sequence of numbers $z_{k}$ such that $z_{k+1}-z_{k}\ge z_{k}-z_{k-1}\ge 1$ and $\sigma ([z_{k},\infty ))\le 2^{-k}$. Let $\theta _0$ be a linear function on each interval $[z_{k}, z_{k+1}]$ with $\Theta _0(z_{k})=k-1$. We obtain a $\sigma$-integrable increasing concave function $\theta _{0}$. However, it does not belong to the class $C^{2}$. To improve this, we take for $\theta$ the function

$$\begin{aligned} \theta (z)=\int _{0}^{z}g(s)ds,\quad g\in C^{1}({\mathbb {R}}), \end{aligned}$$

where $g'(z)\le 0$ and $g(z)=\theta _{0}'(z)$ if $z\in (z_{k}, z_{k+1}-k^{-1})$. Further, taking into account that $\theta$ is concave and $c$ is nonpositive, we obtain

$$\begin{aligned} \partial _{t}\theta (V)\!+\!L\theta (V) \!=\!\theta '(V)\bigl (\partial _{t}V\!+\!LV\bigr ) \!+\!\theta ''(V)(A\nabla V,\nabla V)\!+\!c\bigl (\theta (V)\!-\!\theta '(V)V\bigr ) \le K\!+\!H\theta (V). \end{aligned}$$

Now $W:=\theta (V)$ is the required function. $\square$

Remark 2.9

Modifying a bit the reasoning above, one can get sufficient conditions for (2.7) of another type. Let $V\in C^{2}(D)$ and

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}}V(x)=+\infty . \end{aligned}$$

Let $\zeta \in C^{\infty }([0, +\infty )), \zeta (x)=1$ if $x<1$ and $\zeta (x)=0$ if $x>2$. Set $\varphi _N(x)=\zeta (V(x)/N)$. If $\mu \in {\mathcal {M}}_{\nu }$, then

$$\begin{aligned} \int _{D}\varphi _N\,d\mu _t=\int _{D}\varphi _N\,d\nu +\int _0^t\int _{D}[L_0\varphi _N+c\varphi _N]\,d\mu _s\,ds, \end{aligned}$$

where $L_0=a^{ij}\partial _{x_i}\partial _{x_j}+b^i\partial _{x_i}$. Observe that $\varphi _N\rightarrow 1$ if $N\rightarrow \infty$. Thus, (2.7) is ensured by the equality

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _0^t\int _{D}L_0\varphi _N\,d\mu _s\,ds=0. \end{aligned}$$

It is fulfilled, for example, if

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _0^T\int _{N\le V\le 2N}\Bigl (N^{-1}|L_0 V|+N^{-2}|\sqrt{A}\nabla V|^2\Bigr )\,d\mu =0. \end{aligned}$$

If $a^{ij}(\,\cdot \,t)\in W^{1, 1}_{loc}(D), \mu _t=\varrho (x, t)\,dx$ and $\varrho (\,\cdot \,, t)\in W^{1, 1}_{loc}(D)$, then (2.7) is ensured, for example, by

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _0^T\int _{N\le V\le 2N} \Bigl (N^{-1}|(b-\beta _{\mu })\nabla V|+N^{-2}|\sqrt{A}\nabla V|^2\Bigr )\,d\mu =0, \end{aligned}$$

where $\beta ^{i}_{\mu }=\sum _{j=1}^{d}\bigl (\partial _{x_j}a^{ij} +a^{ij}\varrho ^{-1}\partial _{x_j}\varrho \bigr )$.

We also note that (2.7) has a clear probabilistic sense: the diffusion process corresponding to the operator $L$ does not reach the boundary of the domain $D$ in finite time.

3 Uniqueness Results

In this section, we give some conditions sufficient for uniqueness and the resulting theorem on the existence and uniqueness for the Cauchy problem (1.1). We generalize the results of [9, 25, 26]. Note that some examples of non-uniqueness are given in [9]. Namely, for $d=4$ and $(a^{ij})=\mathrm{I}$, a smooth vector field $b$ on ${\mathbb {R}}^d$ independent of $t$ is constructed such that (1.1) with $c=0$ and some initial probability measure $\nu$ has two different solutions given by families of probability measures; a similar example can be constructed for any $d\ge 3$, but the cases $d=1$ and $d=2$ remain open; moreover, even in the case $d=4$ we have no such examples with Dirac’s initial distributions. Another example (even for $d=1$) shows that we can lose uniqueness if we drop condition (1.2), i.e., do not confine ourselves to subprobability solutions. Indeed, set $d=1, a=1$ and $b(x)=-2x(1+x^2)^{-1}-(1+x^2)\mathrm{arctg}\,x.$ Then $(x, b(x))\le 0$ and the Cauchy problem (1.1) with $\nu =(\pi (1+x^2))^{-1}\,dx$ has a probability solution. But there exists one more solution, which is given by the density $e^t(\pi (1+x^2))^{-1}$.

Since a proof of uniqueness is usually divided into the steps dealing with a local estimate and its application in some limiting procedure, we do not repeat this first step in details (which can be found in the cited papers).

Let the matrix $A(x, t)=(a^{ij}(x, t))_{1\le i, j\le d}$ be symmetric and satisfy the following condition:

(H1) for every $D_k\subset D$ there exist strictly positive numbers $m_k$ and $M_k$ such that the estimate

$$\begin{aligned} m_k|y|^2\le (A(x, t)y, y)\le M_k|y|^2 \end{aligned}$$

holds for all $y\in {\mathbb {R}}^d$ and all $(x, t)\in D\times (0, T)$.

Let us recall the definition of the functional class VMO.

Let $g$ be a bounded function on ${\mathbb {R}}^{d+1}$. Set

$$\begin{aligned} O(g, R)=\sup _{(x, t)\in {\mathbb {R}}^{d+1}}\sup _{r\le R}r^{-2}|U(x, r)|^{-2} \int _t^{t+r^2}\int \int _{y, z\in U(x, r)}|g(y, s)-g(z, s)|\,dy\,dz\,ds. \end{aligned}$$

If $\lim \limits _{R\rightarrow 0}O(g, R)=0$, then the function $g$ is said to belong to the class $VMO_x({\mathbb {R}}^{d+1})$.

If $g\in VMO_x({\mathbb {R}}^{d+1})$, then one can always assume that $O(g, R)\le w(R)$ for all $R>0$, where $w$ is a continuous function on $[0, +\infty )$ and $w(0)=0$.

Suppose that a function $g$ is defined on $D\times [0, T]$ and is bounded on $D_k\times [0, T]$ for every $k$. Let us extend $g$ by zero to all of ${\mathbb {R}}^{d+1}$. If for every function $\zeta \in C^{\infty }_0(D)$ the function $g\zeta$ belongs to the class $VMO_x({\mathbb {R}}^{d+1})$, then we say that $g$ belongs to the class $VMO_{x, loc}({\mathbb {R}}^d\times [0, T])$. Elliptic and parabolic equations with $VMO$ coefficients are considered in [19].

Set

$$\begin{aligned} L_0u=a^{ij}\partial _{x_i}\partial _{x_j}u+b^i\partial _{x_i}u. \end{aligned}$$

(3.1)

Theorem 3.1

Suppose that $a^{ij}\in VMO_{x, loc}(D\times [0, T])$ and that the matrix $A=(a^{ij})$ satisfies condition (H1) above. Suppose also that there is a function $V\in C^{2}(D)$ such that

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}}V(x)=+\infty . \end{aligned}$$

Then the set ${\mathbb {M}}_{\nu }$ consisting of all measures $\mu \in {\mathcal {M}}_{\nu }$ for which the functions $|L_0V|, |\sqrt{A}\nabla V|^2$ belong to $L^1(\mu , D_k\times [0, T])$ for every $k$ and

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _0^T\int _{N\le V\le 2N} \Bigl (N^{-1}|L_0 V|+N^{-2}|\sqrt{A}\nabla V|^2\Bigr )\,d\mu =0 \end{aligned}$$

contains at most one element.

Proof

The proof is similar to the proof of Theorem 3.1 from [9], but there are some new aspects concerning the domain $D$ and the term $c$. More precisely, the local part of the proof of Theorem 3.1 remains unchanged, but the global part of the proof requires some nontrivial additional considerations.

We start with the case of coefficients of class $\bigcap _{k\in {\mathbb {N}}}C^{\infty }(D_k\times [0, T])$.

1. Let $\varphi _N(x)=\eta (V(x)/N)$, where $\eta \in C^{\infty }_0([0, +\infty ))$ is a nonnegative function such that $\eta (z)=1$ if $0\le z\le 1$ and $\eta (z)=0$ if $z>2$. Moreover, $0\le \eta \le 1$ and there is a number $K>0$ such that the estimate $|\eta '(z)|^2\eta ^{-1}(z)\le K$ holds for all $x\in \text{ supp }\,\eta$. We observe that $\varphi _N\in C^{\infty }_0(D)$, in particular, there exists a positive integer $k_0=k_0(N)$ such that $\text{ supp }\,\varphi _N \subset D_{k_0}$.

We redefine the functions $a^{ij}, b^i, c$ outside of $D_{k_0}\times [0, T]$ so that the new functions are bounded along with all derivatives on ${\mathbb {R}}^d\times [0, T]$.

2. Let $\psi \in C^{\infty }_0(D), |\psi |\le 1$ and let $f$ be the solution of the adjoint Cauchy problem

$$\begin{aligned} \partial _sf+Lf=0, \quad f|_{s=t}=\psi . \end{aligned}$$

Suppose that $\sigma _1$ and $\sigma _2$ are two solutions in the set ${\mathbb {M}}_{\nu }$. Set $\mu =\sigma _1-\sigma _2$. Multiplying the equation $\partial _t\mu =L^{*}\mu$ by $f\varphi _N$ and integrating, we arrive at the equality

$$\begin{aligned} \int _{D}\psi \varphi _N\,d\mu _t= \int _0^t\int _{D}\Bigl (2(A\nabla \varphi _N, \nabla f)+fL_0\varphi _N\Bigr )\,d\mu _s\,ds. \end{aligned}$$

3. Set $\sigma =(\sigma _1+\sigma _2)/2$. Let us estimate the integral

$$\begin{aligned} \int _{0}^t\int _{D}\varphi _N|\sqrt{A}\nabla f|^2\,d\sigma _s\,ds. \end{aligned}$$

Multiplying the equation $\partial _t\sigma =L^{*}\sigma$ by $f^2\varphi _N$ and integrating, we arrive at the equality

$$\begin{aligned}&\int _{D}\psi ^2\varphi _N\,d\sigma _t- \int _{D}f^2(x, 0)\varphi _N(x)\,\nu (dx) =2\int _{0}^t\int _{D}|\sqrt{A}\nabla f|^2\varphi _N\,d\sigma _s\,ds\\&\quad +\,2\int _{0}^t\int _{D} \bigl [f(A\nabla f, \nabla \varphi _N) +f^2L_0\varphi _N\bigr ]\,d\sigma _s\,ds. \end{aligned}$$

Note that $0\le \varphi _N(x)\le 1$. By the maximum principle $|f(x, s)|\le \max _x|\psi (x)|\le 1$. Using the Cauchy inequality, we obtain the following estimate:

$$\begin{aligned} \int _{0}^t\int _{D}\varphi _N|\sqrt{A}\nabla f|^2\,d\sigma _s\,ds\le 2+\int _0^t\int _D\Bigl (|\sqrt{A}\nabla \varphi _N|^2\varphi _N^{-1}+|L_0 \varphi _N|\Bigr )\,d\sigma _s\,ds. \end{aligned}$$

Thus,

$$\begin{aligned} \int _{D}\psi \varphi _N\,d\mu _t\le (1+R_N)^{1/2}R_N^{1/2}+R_N, \end{aligned}$$

where

$$\begin{aligned} R_N=\int _{0}^T\int _{D} \Bigl (|L_0\varphi _N|+2\varphi _N^{-1}|\sqrt{A}\nabla \varphi _N|\Bigr )\,d\sigma _s\,ds. \end{aligned}$$

Finally, letting $N\rightarrow \infty$, we arrive at the estimate

$$\begin{aligned} \int _{D}\psi \,d\mu _t\le 0. \end{aligned}$$

Replacing $\psi$ by $-\psi$, we get the opposite inequality. Hence for a.e. $t\in [0, T]$ we have

$$\begin{aligned} \int _{D}\psi \,d\mu _t=0. \end{aligned}$$

Since $\psi$ was an arbitrary function in $C^{\infty }_0({\mathbb {R}}^d)$ with the only restriction $|\psi |\le 1$, we conclude that $\mu _t=0$ and hence $\sigma ^1=\sigma ^2$.

In the general case (without assumptions about the smoothness of coefficients) one has to solve the Cauchy problems $\partial _sf_n+L_nf_n=0, f_n|_{s=t}=\psi$, and apply a priori estimates from [7, 19] in order to get an analogous estimate, where the coefficients of the operator $L_n$ are smooth approximations of the coefficients of $L$ on $\text{ supp }\,\varphi _N$. $\square$

Example 3.2

The assumption of Theorem 3.1 is fulfilled for all measures $\mu \in {\mathcal {M}}_{\nu }$ if

$$\begin{aligned} (1+|V(x)|)^{-1}|L_0V(x, t)|+(1+|V(x)|)^{-2}|\sqrt{A}(x, t)\nabla V(x)|^2\le W(x, t), \end{aligned}$$

where $W\in C^{2, 1}(D\times (0, T))\bigcap C(D\times [0, T)), W(\, \cdot \,,0)\in L^1(\nu )$, for every interval $[\alpha , \beta ]$ from $(0, T)$ one has

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}\times [\alpha , \beta ]}W(x, t)=+\infty \end{aligned}$$

and for some functions $K, H\in L^1((0, T))$ with $H\ge 0$ the estimate

$$\begin{aligned} \partial _tW(x, t)+LW(x, t)\le K(t)+H(t)W(x, t) \end{aligned}$$

holds.

Indeed, by Theorem 2.7 we have $W\in L^1(\mu , D\times [0, T])$, hence

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _0^T\int _{N<V<2N}W\,d\mu =0. \end{aligned}$$

Let us note that the assumptions of Theorem 3.1 admit practically any growth of coefficients (the function $W$ from the example above may grow arbitrarily fast), but we impose restrictions on $|L_0V|$, i.e., we control the growth of $L_0V$ from both sides. Moreover, we have proved the uniqueness only in the class ${\mathbb {M}}_{\nu }$, but not for all measures from ${\mathcal {M}}_{\nu }$. It is possible to eliminate these constraints in the case of more regular coefficients.

We shall assume now that along with (H1) we have

  (H2)   for every positive integer $k$ there is a number $\Lambda _k>0$ such that

$$\begin{aligned} |a^{ij}(x, t)-a^{ij}(y, t)|\le \Lambda _k|x-y| \end{aligned}$$

for all $x, y\in D_k$ and $t\in [0, T]$.

Let us recall some facts from [7]. The assumption (H1) ensures the existence of a density $\varrho$ of the solution $\mu$ with respect to Lebesgue measure. Moreover, if along with (H1) and (H2) we have $b\in L^{p}_{loc}(D\times (0, T))$ and $c\in L^{p/2}_{loc}(D\times (0, T))$ for some $p>d+2$, then we can choose a version of $\varrho$ continuous on $D\times (0, T)$ such that for a.e. $t\in (0, T)$ the function $\varrho (\, \cdot \,, t)$ belongs to $W^{1, p}(U)$ for every closed ball $U\subset D$. Since for a.e. $t\in (0, T)$ the measure $\mu _t(dx)=\varrho (x, t)\,dx$ is a subprobability measure on $D$, Harnack’s inequality ensures that for every closed ball $U$ from $D$ and for every interval $J\subset (0, T)$ there exists a number $C>0$ such that $\varrho (x, t)\ge C$ for all $(x, t)\in U\times J$.

Below we deal with the continuous version of the density $\varrho$.

We recall that, for every measure $\mu$ given by a Sobolev density $\varrho$ with respect to Lebesgue measure, its logarithmic gradient $\beta _{\mu }$ with respect to the metric generated by the matrix $A$ is defined by the following formula:

$$\begin{aligned} \beta ^{i}_{\mu }=\sum _{j=1}^{d}\bigl (\partial _{x_j}a^{ij} +a^{ij}\varrho ^{-1}\partial _{x_j}\varrho \bigr ). \end{aligned}$$

Further in this section we assume that the coefficients $b$ and $c$ are locally integrable with respect to Lebesgue measure on $D\times (0, T)$ to powers $p$ and $p/2$, respectively, for some $p>d+2$ and that the conditions (H1) and (H2) are fulfilled.

Moreover, we consider only the set ${\mathbb {M}}_{\nu }$ of measures $\mu \in {\mathcal {M}}_{\nu }$ satisfying the condition

$$\begin{aligned} b\in L^2(\mu , D_k\times [0, T]) \quad \forall D_k. \end{aligned}$$

For example, the latter condition is fulfilled for the whole class ${\mathcal {M}}_{\nu }$ in the case of a drift bounded on $D_k\times [0, T]$ or in the case $b\in L^s(D_k\times [0, T])$ and $\mu =\varrho \,dx\,dt$ with $\varrho \in L^r(D_k\times [0, T])$, where $2/s+1/r=1$.

Suppose that there are two solutions of the Cauchy problem (1.1) in the class ${\mathbb {M}}_{\nu }$ given by densities $\sigma$ and $\varrho$ with respect to Lebesgue measure. Then these densities are continuous on $D\times (0, T)$. In addition, the functions $\sigma$ and $\varrho$ are strictly positive. Let $v(x, t)=\sigma (x, t)/\varrho (x, t)$. The function $v$ is continuous and positive on $D\times (0, T)$.

Lemma 3.3

Suppose that for a.e. $t\in (0, T)$ we have the estimates

$$\begin{aligned} \int _{D}\varrho (x, t)\,dx=\nu (D)+\int _0^t\int _{D}c(x, s)\varrho (x, s)\,dx\,ds \end{aligned}$$

and

$$\begin{aligned} \int _{D}\sigma (x, t)\,dx\le \nu (D)+\int _0^t\int _{D}c(x, s)\sigma (x, s)\,dx\,ds. \end{aligned}$$

Suppose also that for every $\lambda >0$ we have for a.e. $t\in (0, T)$

$$\begin{aligned} \int _{D}e^{\lambda (1-v(x, t))}\varrho (x, t)\,dx\le 1. \end{aligned}$$

(3.2)

Then $v\equiv 1$, i.e., $\sigma =\varrho$.

Proof

Let $t$ be such that $\varrho (\, \cdot \,, t)$ and $\sigma (\, \cdot \,, t)$ satisfy the listed conditions and (3.2) holds for every positive integer $\lambda$. We observe that the set of points $t$ for which this is not true is a set of zero Lebesgue measure. If there is a ball $U\subset D$ such that $v(x, t)\le 1-\delta$ for each $x\in U$ and some $\delta >0$, then

$$\begin{aligned} e^{\lambda \delta }\int _{U}\varrho \,dx\le \int _{U}e^{\lambda (1-v(x, t))}\varrho (x, t)\,dx\le 1. \end{aligned}$$

Letting $\lambda \rightarrow +\infty$, we obtain a contradiction. Hence, $v(x, t)\ge 1$ and $\sigma \ge \varrho$ for all $(x, t)\in D\times (0, T)$. Moreover,

$$\begin{aligned} \!\int _{D}\varrho (x, t)\,dx\le \!\int _{D}\sigma (x, t)\,dx, \quad \!\int _0^t\int _{D}|c(x, s)|\!\varrho (x, s)\,dx\,ds\le \int _0^t\int _{D}|c(x, s)|\!\sigma (x, s)\,dx\,ds. \end{aligned}$$

We observe that

$$\begin{aligned} \nu (D)= & {} \int _{D}\varrho (x, t)\,dx+\int _0^t\int _{D}|c(x, s)|\varrho (x, s)\,dx\,ds\\&\le \int _{D}\sigma (x, t)\,dx+\int _0^t\int _{D}|c(x, s)|\sigma (x, s)\,dx\,ds=\nu (D). \end{aligned}$$

Hence we have

$$\begin{aligned} \int _{D}\varrho (x, t)\,dx=\int _{D}\sigma (x, t)\,dx, \end{aligned}$$

which ensures that $v\equiv 1$. The lemma is proven.

The next lemma was proved in [25] when $D={\mathbb {R}}^d$ and $c=0$. Since the estimate is local (only on $\text{ supp }\,\psi$), there is no difference between arbitrary $D$ and ${\mathbb {R}}^d$. The addition of a new term $c$ does not create new difficulties and the reasoning is completely the same. In this lemma, it is possible to take $e^{\lambda (1-z)}$ and $e^{\lambda (1-z)}-e^{\lambda }$ for $f$.

Lemma 3.4

Let $\psi \in C^{\infty }_0(D), \psi \ge 0$ and $0<t<T$. Then the following estimate holds:

$$\begin{aligned} \int _{D}f(v(x, t))\varrho (x, t)\psi (x)\,dx\le f(1)\int _{D}\psi (x)\,d\nu + \int _0^t\int _{D}\varrho f(v)L\psi \,dx\,ds. \end{aligned}$$

(3.3)

If, in addition, $|b-\beta _{\mu }|\varrho \in L^1(D_k\times (0, T))$ for every $k$, then

$$\begin{aligned}&\int _{D}f(v(x, t))\varrho (x, t)\psi (x)\,dx\le f(1)\int _{D}\psi (x)\,d\nu \nonumber \\&\quad +\,\frac{1}{2}\int _0^t\int _{D}\varrho (A\nabla \psi , \nabla \psi )\psi ^{-1}|f'(v)|^2f''(v)^{-1}\,dx\,ds \nonumber \\&\qquad +\int _0^t\int _{D}f(v)(b-\beta _{\mu }, \nabla \psi )\varrho \,dx\,ds +\int _0^t\int _{D}f(v)c\psi \varrho \,dx\,ds. \end{aligned}$$

(3.4)

Applying Lemmas 3.3 and 3.4 we obtain the following result.

Theorem 3.5

Suppose that (H1) and (H2) are fulfilled and

$$\begin{aligned} b\in L^p_{loc}(D\times (0, T)), \quad c\in L^{p/2}_{loc}(D\times (0, T)) \end{aligned}$$

for some $p>d+2$. Assume also that there exists a function $V$ such that $V\in C^{2}(D)$ and

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}}V(x)=+\infty . \end{aligned}$$

Suppose that at least one of the following conditions is fulfilled:

(i) for some measure $\mu \in {\mathbb {M}}_{\nu }$ one has

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _0^T\int _{N\le V\le 2N} \Bigl (N^{-1}|L_0 V|+N^{-2}|\sqrt{A}\nabla V|^2\Bigr )\,d\mu =0, \end{aligned}$$

(ii) for some measure $\mu \in {\mathbb {M}}_{\nu }$ one has

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _0^T\int _{N\le V\le 2N} \Bigl (N^{-1}|(b-\beta _{\mu })\nabla V|+N^{-2}|\sqrt{A}\nabla V|^2\Bigr )\,d\mu =0, \end{aligned}$$

(iii) for some $K>0$ and all $(x, t)\in D\times (0, T)$ the inequality

$$\begin{aligned} LV(x, t)\le K+KV(x) \end{aligned}$$

holds. Then the class ${\mathbb {M}}_{\nu }$ consists of at most one element.

Proof

Let us prove (i). Let a measure $\mu \in {\mathbb {M}}_{\nu }$ satisfy the condition from (i) and have a density $\varrho$ with respect to Lebesgue measure. By Remark 2.9, equality (2.7) holds for $\mu$. Suppose that there is yet another measure in ${\mathbb {M}}_{\nu }$ given by a density $\sigma$. Set $v=\sigma /\varrho$. Let $\psi (x)=\zeta (V(x)/N)$, where $\zeta \in C^{\infty }_0({\mathbb {R}}^d)$ is a nonnegative function such that $\zeta (x)=1$ if $|x|\le 1$ and $\zeta (x)=0$ if $|x|>2$, and there exists a number $M>0$ such that for all $x\in \text{ supp }\,\zeta$ one has

$$\begin{aligned} |\zeta (x)|\le M, \quad |\nabla \zeta (x)|\le M, \quad |\nabla \zeta (x)|^2\zeta ^{- 1}(x)\le M. \end{aligned}$$

Let $f(z)=e^{\lambda (1-z)}$. It is clear that $|f(z)|\le e^{\lambda }$ if $z\ge 0$. Using (3.4) from Lemma 3.4 and the fact that $c\le 0$, we obtain

$$\begin{aligned}&\int _{D}e^{\lambda (1-v(x, t))}\varrho (x, t)\zeta (V(x)/N)\,dx\le \int _{D}\zeta (V(x)/N)\,d\nu \\&\quad +\,e^{\lambda }M\int _0^t\int _{N\le V\le 2N}N^{-1}|L_0 V|+N^{-2}|\sqrt{A}\nabla V|^2\,\varrho \,dx, \end{aligned}$$

where $L_0$ is defined by (3.1). Letting $N\rightarrow +\infty$ and using Lemma 3.3, we obtain the required assertion.

The case(ii) can be treated similarly. Let us now prove (iii).

Suppose that there are two measures in ${\mathbb {M}}_{\nu }$ given by densities $\sigma$ and $\varrho$ with respect to Lebesgue measure. According to Theorem 2.7, one has (2.7) for both measures. Let us set $v=\sigma /\varrho$ and consider the function $\psi (x)=\zeta (N^{-1}V(x))$, where a nonnegative function $\zeta \in C^{\infty }_0({\mathbb {R}})$ is such that $\zeta (0)=1, \zeta (z)=0$ if $|z|>1, 0\le \zeta \le 1$ and, moreover, $\zeta '(z)\le 0$ and $\zeta ''(z)\ge 0$ if $z>0$.

Let $f(z)=e^{\lambda (1-z)}-e^{\lambda }$. Then $f(z)\le 0$ and $|f(z)|\le 2e^{\lambda }$ if $z\ge 0$. Observe that

$$\begin{aligned} f(v)\zeta 'LV\le (K+KV)f(v)\zeta ' \end{aligned}$$

since $f(v)\zeta '\ge 0$. Using estimate (3.3) from Lemma 3.4, we obtain

$$\begin{aligned}&\int _{D}(e^{\lambda (1-v(x, t))}-e^{\lambda })\varrho (x, t)\zeta (N^{-1}V(x))\,dx\le (1-e^{\lambda })\int _{D}\zeta (N^{-1}V(x))\,d\nu \\&+\,2e^{\lambda }MN^{- 1}\int _0^t\int _{V<N}(K+KV)\varrho \,dx\,ds \!+\!\int _0^t\int _{D}\zeta (N^{- 1}V(x))f(v(x, s))c(x, s)\varrho (x, s)\,dx\,ds. \end{aligned}$$

We observe that

$$\begin{aligned} \lim _{N\rightarrow +\infty }N^{-1}\int _0^t\int _{V<N}(K+KV)\varrho \,dx\,ds=0. \end{aligned}$$

(3.5)

Indeed, let $\gamma \in (0, 1)$ and $N>\gamma ^{-1}$. Then

$$\begin{aligned} N^{- 1}\int _0^t\int _{V<N}(K+KV)\varrho \,dx\,ds\le \gamma \int _0^t\int _{V<\gamma N}\varrho \,dx\,ds +K\int _0^t\int _{\gamma N<V<N}\varrho \,dx\,ds. \end{aligned}$$

Hence we have

$$\begin{aligned} \lim _{N\rightarrow +\infty }N^{-1}\int _0^t\int _{V<N}(K+KV)\varrho \,dx\,ds\le \gamma \int _0^t\int _{D}\varrho \,dx\,ds. \end{aligned}$$

Letting $\gamma \rightarrow 0$, we obtain (3.5). Thus, letting $N\rightarrow +\infty$, we obtain

$$\begin{aligned} \int _{D}(e^{\lambda (1-v(x, t))}\!-\! e^{\lambda })\varrho (x, t)\,dx\le (1- e^{\lambda })\int _{D}\,d\nu \!+\! \int _0^t\int _{D}(e^{\lambda (1-v)}\!-\!e^{\lambda })c(x, s)\varrho (x, s)\,dx\,ds. \end{aligned}$$

Since $c\le 0$ and for a.e. $t\in (0, T)$ the identity

$$\begin{aligned} \int _{D}\varrho (x, t)\,dx=\nu (D)+\int _0^t\int _{D}c(x, s)\varrho (x, s)\,dx\,ds, \end{aligned}$$

holds, for such $t$ we have

$$\begin{aligned} \int _{D}e^{\lambda (1-v(x, t))}\varrho (x, t)\,dx\le 1. \end{aligned}$$

Using Lemma 3.3, we complete the proof. $\square$

A combination of Theorems 2.3 and 3.5 yields the following sufficient conditions for existence and uniqueness.

Theorem 3.6

Suppose that (H1) and (H2) hold and

$$\begin{aligned} c\in L^{p/2}_{loc}(D\times (0, T)), \quad b\in L^{\infty }(D_k\times [0, T]) \end{aligned}$$

for some $p>d+2$ and all $k$. Assume that there exists a function $V\in C^{2}(D)$ such that

$$\begin{aligned} \lim _{k\rightarrow \infty }\inf _{D_k\setminus D_{k-1}}V(x)=+\infty \end{aligned}$$

and for some number $K>0$ and all $(x, t)\in D\times (0, T)$ one has the inequality

$$\begin{aligned} LV(x, t)\le K+KV(x). \end{aligned}$$

Then the class ${\mathcal {M}}_{\nu }$, where $\nu$ is a probability measure on $D$, consists of exactly one element $\mu =(\mu _t)_{t\in (0, T)}$. Moreover, for a.e. $t$ the identity

$$\begin{aligned} \mu _t(D)=\nu (D)+\int _0^t\int _Dc(x, s)\,d\mu _s\,ds \end{aligned}$$

holds. In particular, if $c=0$, the measures $\mu _t$ are probabilities for a.e. $t$.

Remark 3.7

In Remark 2.6 we have discussed a construction of a solution given by a density $\varrho \in L^{\infty }([0, T], L^p(D))$ in the case of a degenerate diffusion matrix. Following [10, 22], one can find sufficient conditions for the uniqueness of a solution. Suppose that, in addition to the conditions from Remark 2.6, there exists a function $V\in C^{2}(D)$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\inf _{D_n\setminus D_{n-1}}V(x)=+\infty \end{aligned}$$

and for some number $K>0$ and all $(x, t)\in D\times (0, T)$ on has

$$\begin{aligned} L_0V(x, t)\ge -K- KV(x), \quad |\sqrt{A(x, t)}V(x)|\le KV(x), \end{aligned}$$

where, as above, $L_0 \psi =a^{ij}\partial _{x_i}\partial _{x_j}\psi +b^i\partial _{x_i}\psi$. Then a solution of class $L^{\infty }([0, T], L^p(D))$ is unique.

Indeed, one can show that, for any solution $\varrho$ to the Cauchy problem with zero initial condition and any $\psi \in C^{\infty }_0(D)$, the following inequality holds:

$$\begin{aligned} \int _D|\varrho (x, t)|^p\psi (x)\,dx\le \int _0^t\int _{D}\bigl [L_0\psi +\psi (pc+(p- 1)\mathrm{div} h)^{+}\bigr ]|\varrho |^p\,dx\,dt. \end{aligned}$$

Let $\psi _N(x)=\zeta (N^{-1}V(x)),$ where $\zeta \in C^{\infty }_0({\mathbb {R}})$ is a nonnegative function such that $\zeta (z)=1$ if $|z|<1$ and $\zeta (z)=0$ if $|z|>2, 0\le \zeta \le 1$, moreover, $\zeta '(z)\le 0$. We observe that for some number $C_1>0$ and all $(x, t)\in D\times (0, T)$ we have

$$\begin{aligned} L_0\psi (x, t)=N^{-1}\zeta '(N^{-1}V(x))L_0V(x, t) +N^{-2}\zeta ''(N^{-1}V(x))|\sqrt{A(x, t)}\nabla V(x)|^2\le K_1. \end{aligned}$$

Hence,

$$\begin{aligned}&\int _D|\varrho (x, t)|^p\psi _N(x)\,dx\le K_1\int _0^t\int _{N<V<2N}|\varrho |^p\,dx\,dt\\&\quad + \int _0^t\int _{D}\psi _N(pc+(p-1)\mathrm{div} h)^{+}|\varrho |^p\,dx\,dt. \end{aligned}$$

Letting $N\rightarrow \infty$, we arrive at the inequality

$$\begin{aligned} \int _D|\varrho (x, t)|^p\,dx\le \int _0^t\int _{D}(pc+(p-1)\mathrm{div} h)^{+}|\varrho |^p\,dx\,dt. \end{aligned}$$

Grownwall’s inequality yields that

$$\begin{aligned} \int _D|\varrho (x, t)|^p\,dx=0 \end{aligned}$$

and $\varrho \equiv 0$. This yields the uniqueness of a solution.

Remark 3.8

Let $c=0$, let $a^{ij}, b^i\in C(D)$, and let $\mathrm{det}A>0$. Assume also that there is a function $V\in C^{2}(D)$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\inf _{D_n\setminus D_{n-1}}V(x)=+\infty . \end{aligned}$$

Suppose that for some number $K>0$ and some number $n$ the estimate $LV(x)\le -KV$ holds for every $x\in D\setminus D_n$. Then, for every probability measure $\nu$, there exists a unique solution to the Cauchy problem $\mu =(\mu _t)_{t\in (0, +\infty )}$ given by probability measures $\mu _t$. Moreover, the solution is ergodic, i.e., the measures

$$\begin{aligned} \sigma _t(dx)=t^{-1}\int _0^t\mu _s(dx)\,ds \end{aligned}$$

converge weakly to a probability solution $\mu$ of the stationary equation $L^{*}\mu =0$ on $D$ as $t\rightarrow +\infty$.

The existence and uniqueness of a solution on $(0, +\infty )$ follow by the above theorems. The solution is a probability solution by the fact that $c=0$ and the existence of a Lyapunov function. Grownwall’s inequality and a reasoning similar to the proof of Theorem 2.7 yield that

$$\begin{aligned} \int _{D}V\,d\sigma _t=t^{- 1}\int _0^t\int _{D}V\,d\mu _s\,ds\le K_1, \end{aligned}$$

where $K_1$ does not depend on $t$. Hence, the family $\{\sigma _{t}\}$ is uniformly tight and each sequence $\{\sigma _{t_n}\}$ contains a subsequence weakly convergent to some measure $\sigma$. Obviously, $\sigma$ satisfies the equation $L^{*}\sigma =0$. By the uniqueness of a probability solution the whole subsequence $\sigma _{t_n}$ converges to it (the uniqueness for stationary equations follows from the existence of a Lyapunov function and in the case of an arbitrary domain $D$ can be justified in the same way as in [8] for $D={\mathbb {R}}^d$) .

Remark 3.9

Suppose that all conditions sufficient for the existence and uniqueness in the class ${\mathcal {M}}_{\nu }$ are fulfilled for $L$. Let $P(s, y, t, dx)$ denote the solution of the Cauchy problem with the initial condition $\mu |_{t=s}=\delta _y$. We observe that for every Borel set $B$ the mapping $(s, y)\mapsto P(s, y, t, B)$ is measurable as a limit of measurable mappings which correspond to the solutions $P_n$ of the approximating problems with smooth coefficients (in the existence theorem the solution is constructed in exactly the same way as above; the fact that the constructed solution is unique ensures convergence of the whole sequence and not only of some subsequence). Moreover, $P_n$ satisfy the Chapman–Kolmogorov equation (see [14])

$$\begin{aligned} P_n(s, y, t, B)=\int _{D}P_n(s, y, \tau , dy)P_n(\tau , y, t, B), \quad s<\tau <t. \end{aligned}$$

This ensures that $P$ also satisfies this equation.

To conclude, we consider some more examples. We begin with the example from [15] which has already been mentioned in the introduction.

Example 3.10

Let $\nu$ be a probability measure on $D=(-1, 1)$. Given $\alpha >0$, we consider the Cauchy problem

$$\begin{aligned} \partial _{t}\mu _{t}= \frac{1}{2}\partial _{xx}\left( \left| 1-\left| x\right| \right| ^{2\alpha }\mu \right) -\partial _{x}\left( \left( \mathrm{tg}\left( -\frac{\pi x}{2}\right) +\mathrm{sign}\, x\right) \mu \right) , \quad \mu _{0}=\nu . \end{aligned}$$

(3.6)

Note that the coefficients in the above equations are rather irregular. The drift coefficient is discontinuous at $x=0,\, x=2k+1$. Moreover, the diffusion coefficient does not satisfy the linear growth condition for $\alpha >1/2$ and is not Hölder continuous with exponent $1/2$ if $\alpha <1/2$.

Let us show that this Cauchy problem has a unique probability solution $\{\mu _{t}\} _{t\in [0,T]}$ in the domain $D=(-1,1)$ for every $T>0$.

We introduce the following exhaustion $\{D_{k}\} _{k\in {\mathbb {N}}}$ of the domain $D$:

$$\begin{aligned} D_{k}=\left( -1+2^{-k},\,1-2^{-k}\right) . \end{aligned}$$

Note that the diffusion coefficient is nondegenerate on each cylinder $D_{k}\times [0,T]$. Moreover, the local regularity assumptions from Theorem 3.6 are fulfilled. Consider the following Lyapunov function:

$$\begin{aligned} V(x)=\frac{2-x^{2}}{1-x^{2}}. \end{aligned}$$

Let us show that all the assumptions of Theorem 3.6 are fulfilled for this Lyapunov function.

1) $V\in C^{2}(D), V>0$ in $D$ and

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\inf \limits _{D_{k}\backslash D_{k-1}}V(x)= \lim \limits _{k\rightarrow \infty } \frac{2-\left( 1-2^{-k}\right) {}^{2}}{1-\left( 1-2^{-k}\right) {}^{2}}=+\infty . \end{aligned}$$

2) The following estimate is true:

$$\begin{aligned} LV(x)=2^{-1}\bigl |1- |x|\bigr |^{2\alpha }V''(x)+\bigl (\mathrm{tg}(-\pi x/2)+ \mathrm{sgn}x\bigr )V'(x)\le K_{1}\cdot V(x) \end{aligned}$$

for some constant $K_{1}>0$ and

$$\begin{aligned} \lim \limits _{|x|\rightarrow 1} \frac{LV(x)}{V(x)}=-\infty . \end{aligned}$$

(3.7)

Hence we can apply Theorem 3.6 and obtain that (3.6) has a unique subprobability solution. Furthermore, $c=0$ and thus for a.e. $t\in [0,T]$ the measure $\mu _{t}$ is a probability measure.

Furthermore, the solution is ergodic in the above sense. Indeed, due to (3.7), there exist a positive constant $K_{2}$ and a number $k\in {\mathbb {N}}$ such that on $D\backslash D_{k}$ one has $LV\le -K_{2}\cdot V$. According to Remark 3.8, this inequality ensures the weak convergence of measures

$$\begin{aligned} \sigma _{t}=\frac{1}{t}\int _{0}^{t}\mu _{s}ds \quad as \quad t\rightarrow +\infty \end{aligned}$$

to a probability measure $\sigma$ on $D$ which solves the stationary equation $L^{*}\sigma =0$.

Example 3.11

Let $\nu$ be a probability measure on $D={\mathbb {R}}^{d}$. Consider the Cauchy problem

$$\begin{aligned} \partial _{t}\mu _{t}=\partial _{x_i}\partial _{x_j}(a^{ij}\mu )- \partial _{x_i}(b^i\mu )+c\mu , \quad \mu |_{t=0}=\nu . \end{aligned}$$

Suppose that (H1) and (H2) are fulfilled and

$$\begin{aligned} c\in L^{p/2}_{loc}({\mathbb {R}}^d\times (0, T)), \quad b\in L^{\infty }(B(0, k)\times [0, T]) \end{aligned}$$

for some $p>d+2$ and all $k$, where $B(0, k)$ is the ball of radius $k$ centered at the origin. Let $V(x)=|x|^2/2$. Then the condition $LV\le K+KV$ takes the form

$$\begin{aligned} \mathrm{tr}A(x, t)+(b(x, t), x)+|x|^2c(x, t)/2\le K+K|x|^2 \end{aligned}$$

for all $(x,t)\in {\mathbb {R}}^d\times [0,T]$. If the latter inequality holds, then the set ${\mathcal {M}}_{\nu }$ consists of exactly one element.