Existence and Uniqueness Results for Time-Inhomogeneous Time-Change Equations and Fokker–Planck Equations

Abstract

We prove existence and uniqueness of solutions to Fokker–Planck equations associated with Markov operators multiplicatively perturbed by degenerate time-inhomogeneous coefficients. Precise conditions on the time-inhomogeneous coefficients are given. In particular, we do not necessarily require the coefficients to be either globally bounded or bounded away from zero. The approach is based on constructing random time-changes and studying related martingale problems for Markov processes with values in locally compact, complete and separable metric spaces.

Appendix A: Auxiliary Results for the Construction of Time-Changes

In order to ensure the existence of the time-change $\tau$ as defined by the random differential equation (3.1), we used the following lemma concerning so-called Carathéodory differential equations.

Lemma A.1

Let $t_0> 0$ and consider the Carathéodory differential equation

$$\begin{aligned} \mathcal {T}(t) = \int _0^t \gamma (\mathcal {T}(s),s)\,\mathrm {d}s, \end{aligned}$$

(A.1)

where $\gamma :[0,t_0] \times [0,\infty ) \rightarrow [0,\infty )$ and the integral is understood in the Lebesgue sense. For some $S \in (0,t_0]$ and $T>0$, suppose that $\gamma (r,\cdot )$ is measurable for each $r \in [0,S]$, $\gamma (0,\cdot )$ is integrable on [0, T] and there exists an integrable function $f:[0,T]\rightarrow [0,\infty )$ such that

$$\begin{aligned} |\gamma (r,t) - \gamma (s,t) | \le f(t) |r -s| \quad \mathrm{for~all~} r,s\in [0,S],\, t \in [0,T]. \end{aligned}$$

Then, there exists a unique absolutely continuous function $\mathcal {T}:I \rightarrow [0,S]$ satisfying (A.1) for some interval $I \subset [0,\infty )$, where either there exists $T_0 \in (0,T]$ such that we may take $I =[0,T_0]$ and we have $\mathcal {T}(T_0) = S$ or we may take $I =[0,T]$ and have $\mathcal {T}(t) < S$ for all $t \le T$.

Proof

Since

$$\begin{aligned} |\gamma (r,t)| \le |\gamma (0,t)| + |f(t)| S\quad \text {for all}\quad r \in [0,S] \end{aligned}$$

and the right-hand side is integrable on [0, T], $\gamma$ satisfies the Carathéodory conditions in [16, Chap. 1] and thus [16, Chap. 1, Thm. 1] guarantees the existence of a solution $\mathcal {T}$ on an interval $[0,T_0]$ for some $T_0 >0$. The solution $\mathcal {T}$ can be extended either to the whole interval [0, T] provided $\mathcal {T}(t)\le S$ for all $t\in [0,T]$ or to the interval $[0,T_0]$ for some $T_0 \in (0,T]$ with $\mathcal {T}(T_0) = S$ (see e.g. [16, Chap. 1, Thm. 4]). Uniqueness of the solution follows by [16, Chap. 1, Thm. 2]. $\square$

Next, we provide a condition that is useful in verifying regularity of H (see Definition 2.4) needed for the existence of the time-change in Lemma 3.1 and the uniqueness in Lemma A.3 below.

Proposition A.2

Let $\mathcal {D} \subset C_0(E)$ dense in $C_0(E)$ and $\mathcal {A}:\mathcal {D} \rightarrow C_0(E)$ be linear. Suppose M is a solution on $(\varOmega ,\mathcal {F},\mathbb {P})$ to the RCLL-martingale problem for $(\mathcal {A},\mu _0)$, for some $\mu _0 \in \mathcal {P}(E)$. Denote by P the law on $D_E[0,\infty )$ of M. Then, any $H \in \mathcal {D}$ with $H \ge 0$ is regular for P.

Proof

Define $\rho$ as in (3.2) and recall that, by Definition 2.4, (3.7) and (3.8) have to be verified. Set

$$\begin{aligned} \rho _0 := \inf \left\{ s \in [0,\infty ) : H(M_s) = 0 \right\} . \end{aligned}$$

Since H is continuous and M is RCLL, $H(M_{\rho })=0$ on $\{\rho < \infty \}$ and $\rho _0 \le \rho$, $\mathbb {P}$-a.s. In particular, $\rho _0 = \rho$ on $\{\rho _0 = \infty \}$, and if $\{\rho _0 < \infty \}$ is a $\mathbb {P}$-null set, this already establishes the claim. Otherwise, the probability measure $\tilde{\mathbb {P}}(\,\cdot \,):=\mathbb {P}(\,\cdot \, |\{\rho _0 <\infty \})$ is well-defined, $\rho _0 < \infty$, $\tilde{\mathbb {P}}$-a.s. and to prove the proposition we only need to show $\tilde{\mathbb {P}}(\rho _0 \ge \rho )=1$. To do so, on $\{\rho _0 < \infty \}$ define for any $t \ge 0$ the random time

$$\begin{aligned} \tau (t):= \inf \left\{ s \in [0,\infty ) :\int _0^s H(M_{u+\rho _0})^{-1} \,\mathrm {d}u \ge t \right\} . \end{aligned}$$

Since $H(M_{\rho _0})=0$ and $\rho _0 \le \rho$, $\tilde{\mathbb {P}}$-a.s., it suffices to establish that $\tilde{\mathbb {P}}$-a.s. for any $t \ge 0$, $\tau (t)=0$.

For the proof of the last statement one proceeds as follows: Since H is bounded, $H(M_\rho )=0$ on $\{ \rho < \infty \}$, $\tilde{\mathbb {P}}$-a.s., and by footnote 2, Lemma 3.1 can be applied to the RCLL process $(M_{u+\rho _0})_{u \ge 0}$ on $(\varOmega ,\mathcal {F},\tilde{\mathbb {P}})$ with $\tilde{\sigma }=1$ and $\sigma =H$. This yields $\tilde{\mathbb {P}}$-a.s.,

$$\begin{aligned} \tau (t)=\int _0^t H(M_{\tau (u)+\rho _0})\,\mathrm {d}u, \quad t \ge 0, \end{aligned}$$

(A.2)

and $\tau (t) < \infty$ for any $t \ge 0$. Denote by $(\mathcal {F}_t)_{t \ge 0}$ the $\mathbb {P}$-usual augmentation of the filtration generated by M. Then, $\rho$ and $\rho _0$ (possibly modified on a $\mathbb {P}$-null set, see [13, Chap. 4, Cor. 3.13]) are $(\mathcal {F}_t)_{t \ge 0}$-stopping times and thus

$$\begin{aligned} \{ \tau (t)+\rho _0 \le s \} = \{ \rho _0 \le s \} \cap \left( \left\{ \int _{\rho _0}^{s} H(M_{u})^{-1} \,\mathrm {d}u \ge t \right\} \cup \{\rho \le s - \rho _0 \} \right) \in \mathcal {F}_s \end{aligned}$$

shows that also $\tau (t)+\rho _0$ is a stopping time. By assumption on H, M and $\mathcal {A}$ the process

$$\begin{aligned} N_t := H(M_t) - H(M_0) - \int _0^t \mathcal {A} H (M_s) \,\mathrm {d}s, \quad t \ge 0, \end{aligned}$$

is an $(\mathcal {F}^M_t)_{t\ge 0}$-martingale and thus, by [23, Lem. II.67.10], also an $(\mathcal {F}_t)_{t \ge 0}$-martingale. By the optional sampling theorem, for any $r \ge 0$, $\mathbb {P}$-a.s.,

$$\begin{aligned} \mathbb {E}[N_{(\tau (t)+\rho _0)\wedge r} | \mathcal {F}_{\rho _0 \wedge r}] =N_{\rho _0 \wedge r} \end{aligned}$$

or equivalently

$$\begin{aligned} \mathbb {E}[H(M_{(\tau (t)+\rho _0)\wedge r})| \mathcal {F}_{\rho _0 \wedge r}] =H(M_{\rho _0 \wedge r}) + \mathbb {E}\left[ \left. \int _{\rho _0 \wedge r}^{(\tau (t)+\rho _0)\wedge r} \mathcal {A} H(M_u) \,\mathrm {d}u\right| \mathcal {F}_{\rho _0 \wedge r}\right] . \end{aligned}$$

Multiplying by $\mathbb {1}_{\{ \rho _0 \le r \}}$, using $\{\rho _0 \le r\} \in \mathcal {F}_{\rho _0 \wedge r}$ and taking expectations gives

$$\begin{aligned} \mathbb {E}[H(M_{(\tau (t)+\rho _0)\wedge r})\mathbb {1}_{\{ \rho _0 \le r \}}]= & {} \mathbb {E}[H(M_{\rho _0 \wedge r})\mathbb {1}_{\{ \rho _0 \le r \}}]\\&+\,\mathbb {E}\left[ \int _{\rho _0 \wedge r}^{(\tau (t)+\rho _0)\wedge r} \mathcal {A} H(M_u) \,\mathrm {d}u \mathbb {1}_{\{ \rho _0 \le r \}}\right] . \end{aligned}$$

By assumption $\mathcal {D} \subset C_0(E)$ is dense and thus separating (see [13, Chap. 3, Ex. 11]) in the terminology of [13, Chap. 3, Sec. 4]. Therefore, by quasi-left continuity, [13, Chap. 4, Thm. 3.12],

$$\begin{aligned} \lim _{r \rightarrow \infty } M_{(\tau (t)+\rho _0)\wedge r}\mathbb {1}_{\{ \rho _0 \le r \}} = M_{\tau (t)+\rho _0}\mathbb {1}_{\{ \rho _0 <\infty \}} ,\quad \mathbb {P}\text {-a.s.}, \end{aligned}$$

and so using dominated convergence, boundedness and non-negativity of H, $H(M_{\rho _0})=0$ on $\{\rho < \infty \}$ and setting $C:=\Vert \mathcal {A} H\Vert$, one estimates

$$\begin{aligned} \mathbb {E}[H(M_{\tau (t)+\rho _0})\mathbb {1}_{\{ \rho _0< \infty \}}]= & {} \lim _{r \rightarrow \infty } \mathbb {E}[H(M_{(\tau (t)+\rho _0)\wedge r}) \mathbb {1}_{\{ \rho _0 \le r \}}] \nonumber \\= & {} \lim _{r \rightarrow \infty } \left| \mathbb {E}\left[ \int _{\rho _0}^{(\tau (t)+\rho _0)\wedge r} \mathcal {A} H(M_u) \,\mathrm {d}u \mathbb {1}_{\{ \rho _0 \le r \}}\right] \right| \nonumber \\\le & {} C \mathbb {E}[\tau (t)\mathbb {1}_{\{ \rho _0 < \infty \}}]. \end{aligned}$$

(A.3)

Using (A.2) and Tonelli’s theorem for the first and (A.3) for the second equality yields

$$\begin{aligned} \mathbb {E}[\tau (t)\mathbb {1}_{\{ \rho _0< \infty \}}]=\int _0^t \mathbb {E}[H(M_{\tau (u)+\rho _0})\mathbb {1}_{\{ \rho _0< \infty \}}] \,\mathrm {d}u \le C \int _0^t \mathbb {E}[\tau (u)\mathbb {1}_{\{ \rho _0 < \infty \}}] \,\mathrm {d}u \end{aligned}$$

(A.4)

and so Gronwall’s lemma implies that the left-hand side of (A.4) is 0 for any $t \ge 0$. But this implies that $\tilde{\mathbb {P}}$-a.s., $\tau (t) = 0$ for all $t \ge 0$ as desired. $\square$

1.1 A.1. Pathwise Uniqueness

To verify that the random times $(\tau (t))_{t\in [0,t_0]}$ solving the differential equation (3.1) are indeed stopping times with respect to the filtration generated by the process M, we show pathwise uniqueness of the time-changed Markov process $X_t:=M_{\tau (t)}$ for $t\in [0,t_0]$.

Lemma A.3

Let $\sigma$ and M be given as in Lemma 3.1. $(\tau (t))_{t \in [0,t_0]}$ is the family of random times from Lemma 3.1 with $\tau (t): = \tau (t_0)$ for $t >t_0$ and the time-changed process X is given by $X _{t} :=M_{\tau (t)}$ for $t \ge 0$. Suppose M is $(\mathcal {F}_t)$-adapted. Then, the following holds:

• (i) The time-changed process X has RCLL sample paths, $\mathbb {P}$-a.s.

• (ii) Any RCLL process $\tilde{X}$ satisfying

  $$\begin{aligned} \tilde{X}_t = M_{\int _0^t \sigma (u,\tilde{X}_u) \,\mathrm {d}u},\quad t\in [0,\infty ),\, \mathbb {P}\text {-}a.s., \end{aligned}$$

  (A.5)

  is indistinguishable from X.

• (iii) The random times $(\tau (t))_{t \in [0,t_0]}$ are $(\mathcal {F}_t)$-stopping times.

Proof

• (i) Since M has RCLL sample paths and $\tau$ is non-decreasing and absolutely continuous by Lemma 3.1, the time-changed process $X_{}$ has RCLL sample paths.

• (ii) Let $\tilde{X}$ be an RCLL process satisfying equation (A.5). Define the random time

  $$\begin{aligned} \tilde{\rho } := t_0\wedge \inf \big \{ t \ge 0 : H(\tilde{X}_t) = 0 \big \} \end{aligned}$$

  and set

  $$\begin{aligned} \tilde{\tau }(s) := \int _0^s \sigma (u,\tilde{X}_u ) \,\mathrm {d}u, \quad s\in [0,\infty ). \end{aligned}$$

  Notice that the integral is well-defined since $\sigma$ is bounded on compacts and $\tilde{X}$ is RCLL. Since $X_t=M_{\tau (t)}$ and $\tilde{X}_t= M_{\tilde{\tau }(t)}$, to verify that X and $\tilde{X}$ are indistinguishable, it is sufficient to show that $\tau (t)= \tilde{\tau }(t)$ for every $t\in [0,\infty )$, $\mathbb {P}$-a.s.

  By [21, Lem. 3.31] $\tilde{\tau }$ is absolutely continuous with weak derivative $\tilde{\tau }'(u) = \sigma (u, \tilde{X}_u )$ for $u \in [0,\infty )$ and invertible on $[0,\tilde{\rho }\wedge t_0]$ by the definition of $\tilde{\rho }$. The inverse of $\tilde{\tau }$ is denoted by $\tilde{\mathcal {T}}$ with domain $[0,\tilde{\tau }(\tilde{\rho }\wedge t_0)]$. Because $\tilde{\mathcal {T}}$ is also strictly increasing and absolutely continuous, the chain rule (see [21, Thm. 3.44]) gives

  $$\begin{aligned} 1 = \frac{\,\mathrm {d}}{\,\mathrm {d}t} \tilde{\tau } (\tilde{\mathcal {T}}(t)) =\sigma (\tilde{\mathcal {T}}(t), \tilde{X}_{\tilde{\mathcal {T}}(t)})\frac{\,\mathrm {d}}{\,\mathrm {d}t}\tilde{\mathcal {T}}(t)\quad \text {for almost all } t \in {[}0,\tilde{\tau }(\tilde{\rho }\wedge t_0)]. \end{aligned}$$

  Combining this with fundamental theorem of calculus (see [21, Thm. 3.30]), one has that $\tilde{\mathcal {T}}$ satisfies the integral equation

  $$\begin{aligned} \tilde{\mathcal {T}}(t) = \int _0^t \sigma (\tilde{\mathcal {T}}(s), \tilde{X}_{\tilde{\mathcal {T}}(s)})^{-1} \,\mathrm {d}s,\quad t \in {[}0,\tilde{\tau }(\tilde{\rho }\wedge t_0)]. \end{aligned}$$

  Moreover, notice that $M_t= M_{\tilde{\tau }(\tilde{\mathcal {T}}(t))}=\tilde{X}_{\tilde{\mathcal {T}}(t)}$ for $t \in [0,\tilde{\tau }(\tilde{\rho }\wedge t_0)]$. Therefore, $\mathcal {T}(t) = \tilde{\mathcal {T}}(t)$ for $t \in [0,\tilde{\tau }(\tilde{\rho }\wedge t_0)\wedge \tau (t_0)]$ since the solution to this equation is unique on $[0,\tau (t_0)]$, see (3.4). Furthermore, we have $\tilde{\tau }(\tilde{\rho }\wedge t_0) \le \rho \wedge \tilde{\tau } (t_0)$ since $t < \tilde{\tau }(\tilde{\rho }\wedge t_0)$ implies $\tilde{\mathcal {T}}(t) < \tilde{\rho }\wedge t_0$ and thus $t <\rho \wedge \tilde{\tau } (t_0)$, where we recall $\rho$ from (3.2) and that (3.7), (3.8) holds. In conclusion, $\mathcal {T}(t) =\tilde{\mathcal {T}}(t)$ for $t \in [0,\tilde{\tau }(\tilde{\rho }\wedge t_0)]$, which leads to $\tilde{\tau }(s) = \tau (s)$ for $s \in [0,\tilde{\rho }\wedge t_0]$.

  To see $\tilde{\tau }(s) = \tau (s)$ for $s>\tilde{\rho }\wedge t_0$, we first observe that $1/(H(M_s)\vee \varepsilon )$ is bounded for every $\varepsilon >0$ and $\tilde{\sigma }$ is bounded on compacts by Assumption 2.6. Applying a change of variables ([21, Cor. 3.57]) and using monotone convergence gives

  $$\begin{aligned} C t \ge \lim _{\varepsilon \rightarrow 0} \int _0^t \frac{\sigma (s,\tilde{X}_s)}{H(\tilde{X}_s) \vee \varepsilon } \,\mathrm {d}s = \lim _{\varepsilon \rightarrow 0} \int _0^{\tilde{\tau }(t)} \frac{1}{H(M_s) \vee \varepsilon } \,\mathrm {d}s = \int _0^{\tilde{\tau }(t)} \frac{1}{H(M_s)} \,\mathrm {d}s, \end{aligned}$$

  which ensures $\tilde{\tau }(t) \le \rho$ for all $t \ge 0$. Assuming $\tilde{\rho }< t_0$, there exist $\{t_n\}_{n \in \mathbb {N}} \subset [\tilde{\rho },t_0]$ with $t_n \downarrow \tilde{\rho }$ and $H(\tilde{X}_{t_n}) = 0$ and so $\rho \le \tilde{\tau }(\tilde{\rho })$ by (A.5) and (3.7), (3.8). In this case, $\tilde{\tau }(t) = \tilde{\tau }(\tilde{\rho }) = \rho =\tau (t)$ for all $t \ge \tilde{\rho }$. Assuming $\tilde{\rho } \ge t_0$, we have $\tilde{\tau }(t) = \tilde{\tau }(t_0)$ for all $t \ge t_0$ due to $\sigma (t,\cdot ) = 0$ for $t > t_0$ and in particular $\tilde{\tau }(t) = \tilde{\tau }(t_0)=\tau (t)$ for $t \ge t_0$.

• (iii) In order to apply a result from [13], we consider the two-dimensional process $Y_t:=(t,M_t)$ and the time-changed process $(t, X_t)$ for $t\in [0,T]$. Hence, [13, Chap. 6, Thm. 2.2 (b)] implies that $\tau (t)$ is a stopping time with respect to the usual augmentation of the filtration generated by M, and thus also an $(\mathcal {F}_t)$-stopping time, where we keep in mind that the first component of Y generates a trivial filtration.

$\square$

Corollary A.4

Let $\sigma$, M and $\tilde{X}$ be given as in Lemma A.3 and denote by P the law (on $D_E[0,\infty )$) of M under $\mathbb {P}$. Then, the law of $\tilde{X}$ under $\mathbb {P}$ is uniquely determined by P and $\sigma$.

Proof

By Lemma A.3(ii), the law of $\tilde{X}$ is identical to the law of X under $\mathbb {P}$. To show explicitly that the latter is uniquely determined by P and $\sigma$, one proceeds as follows: Let $n \in \mathbb {N}$, $t_1,\ldots ,t_n \in [0,\infty )$, $B_1,\ldots ,B_n \in \mathcal {B}(E)$ and let $\pi _1:D_E[0,\infty )\times D_E[0,\infty ) \rightarrow D_E[0,\infty )$ be the projection map on the first component. We have seen in the Proof of Lemma A.3 that there exist a unique solution to the time-change equation for $\mathbb {P}$-a.e. sample path $M(\omega )$. Hence, as in the proof of [13, Chap. 6, Lem. 2.1], the map

$$\begin{aligned} \gamma :D_E[0,\infty )\times D_E[0,\infty ) \rightarrow D_E[0,\infty ), \quad \gamma (M,X):= M_{\int _0^{\cdot } \sigma (u,X_u)\,\mathrm {d}u}, \end{aligned}$$

is Borel measurable and the set

$$\begin{aligned} C := \{ (m,x) \in D_E[0,\infty ) \times D_E[0,\infty ) : \gamma (m,x)=x, x_{t_1} \in B_1, \ldots , x_{t_n} \in B_n\} \end{aligned}$$

is in $\mathcal {B}(D_E[0,\infty )^2)$. Then, [13, Appendix 11, Thm. 11.3] implies that $\pi _1 C$ is in the P-completion of $\mathcal {B}(D_E[0,\infty ))$ and thus

$$\begin{aligned} \mathbb {P}(X_{t_1}\in B_1,\ldots ,X_{t_n} \in B_n) = P(\pi _1 C) \end{aligned}$$

is indeed uniquely determined by P and $\sigma$. $\square$