# Derivation of the Fokker-Planck equation

Let

• $$b\in C^1(\mathbb R)$$ be Lipschitz continuous
• $$\sigma\in C^2(\mathbb R)$$ be Lipschitz continuous with $$\sigma(\mathbb R)\subseteq\mathbb R\setminus\left\{0\right\}$$ and $$\sigma''$$ being bounded
• $$Lf:=bf'+\frac12\sigma^2f''$$ for $$f\in C^2(\mathbb R)$$ and $$L^\ast g:=\frac12(\sigma^2g)''-(bg)'$$ for $$g\in C^2(\mathbb R)$$
• $$(\kappa_t)_{t\ge0}$$ be a semigroup of Markov kernels on $$(\mathbb R,\mathcal B(\mathbb R))$$ and $$\kappa_t f:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)\tag1$$ for bounded Borel measurable $$f:\mathbb R\to\mathbb R$$.

Assume $$(\kappa_t)_{t\ge0}$$ is a strongly continuous semigroup (via $$(1)$$) on $$C_0(\mathbb R)$$ (continuous functions on $$\mathbb R$$ vanishing at infinity equipped with the supremum norm) with infinitesimal generator $$(\mathcal D(A),A)$$ given by the restriction $$A$$ of $$L$$ to $$\mathcal D(A):=\left\{f\in C_0(\mathbb R)\cap C^2(\mathbb R):Lf\in C_0(\mathbb R)\right\}.$$ Moreover, assume there is a Borel measurable $$p_t:\mathbb R\times\mathbb R\to[0,\infty)$$ with $$\kappa_t(x,B)=\int_Bp_t(x,y)\:\lambda({\rm d}y)\;\;\;\text{for all }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\tag2,$$ where $$\lambda$$ denotes the Lebesgue measure, for all $$t\ge0$$ with

1. $$[0,\infty)\ni t\mapsto p_t(x,y)$$ is differentiable for all $$x,y\in\mathbb R$$ and $$[0,\infty)\times\mathbb R\ni(t,y)\mapsto\frac{\partial p}{\partial t}(t,x,y)\tag3$$ is locally bounded for all $$x\in\mathbb R$$
2. $$\mathbb R\ni y\mapsto p_t(x,y)$$ is twice continuously differentiable for all $$(t,x)\in[0,\infty)\times\mathbb R$$.

We can observe the following: Fix $$f\in C_c^2(\mathbb R)$$. Then, $$\frac{\kappa_{t+h}f-\kappa_tf}h\xrightarrow{h\to0}A(\kappa_tf)=\kappa_t(Af)\tag4,$$ where the convergence is with respect to the supremum norm, for all $$t\ge0$$. Fix $$(t,x)\in[0,\infty)\times\mathbb R$$. On the one hand, $$\frac{(\kappa_{t+h}f)(x)-(\kappa_tf)(x)}h\xrightarrow{h\to0}\int\frac{\partial p}{\partial t}(t,x,y)f(y)\:{\rm d}y\tag5$$ by 1. and the dominated convergence theorem. On the other hand, $$\left(\kappa_t\left(Af\right)\right)(x)=\int f(y)\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\:{\rm d}y\tag6$$ by 2. and integration by parts. Thus, $$\int\frac{\partial p}{\partial t}(t,x,y)f(y)\:{\rm d}y=\int f(y)\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\:{\rm d}y\tag7.$$ Are we able to conclude $$\frac{\partial p}{\partial t}(t,x,y)=\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\tag8$$ (since $$f$$ was arbitrary)?

Essentially your question is whether $$C_c^2(\mathbb{R})$$ is measure-determining and the answer is "yes".

Definition Let $$\mu$$ and $$\nu$$ be (non-negative) measures on a measurable space $$(X,\mathcal{A})$$. A family $$\mathcal{D} \subseteq L^1(\mu) \cap L^1(\nu)$$ is called determining if $$\left\{ \forall f \in \mathcal{D}: \quad \int f \, d \mu = \int f \, d\nu \right\} \implies \mu = \nu.$$

It is known that $$C_c^{\infty}(\mathbb{R}^n)$$ is determining for all measures which are finite on compact sets, see e.g. Theorem 17.12 in $$\{1\}$$. The idea is to use the fact that $$C_c^{\infty}(\mathbb{R}^n)$$ is dense in $$L^1(\mu)$$ for any measure $$\mu$$ which is finite on compacts. As a consequence of this, we get that $$\mathcal{D}=C_c^2(\mathbb{R}^n)$$ is determining, i.e.

$$\left\{\forall f \in C_c^2(\mathbb{R}^n): \quad \int f \, d\mu = \int f \, d\nu \right\} \implies \mu=\nu \tag{1}$$

for any (non-negative) measures $$\mu$$, $$\nu$$ which are finite on compacts. Alternatively, $$(1)$$ can be proved by noting that it suffices to show $$\mu(K)=\nu(K)$$ for any compact set $$K$$ and applying Urysohn's lemma to approximate the indicator function $$1_K$$ by (sufficiently) smooth functions.

In your case we are not dealing with non-negative measures but that's not much of a bother. If $$p$$, $$q$$ are two locally bounded functions such that $$\int f(x) p(x) \, dx = \int f(x) q(x) \, dx, \qquad f \in C_c^2(\mathbb{R}^n)$$ then $$\int f(x) (p^+(x)+q^-(x)) \, dx = \int f(x) (p^-(x)+q^+(x)) \, dx$$ where $$\pm$$ denotes the positive and negative part, respectively. Applying $$(1)$$ we get

$$\int_B (p^+(x)+q^-(x)) \, dx = \int_B (p^-(x)+q^+(x)) \, dx$$

for any Borel set $$B$$, i.e.

$$\int_B p(x) \, dx= \int q(x) \, dx.$$

Choosing $$B=\{p>q\}$$ and $$B=\{q this yields $$p=q$$ (Lebesgue)-almost everywhere. If $$p$$ and $$q$$ are continuous, then this, in turn, entails that $$p=q$$.

Hence, $$(7)$$ gives $$(8)$$ provided that $$\partial_t p(t,x,\cdot)$$ and $$L^* p_t(x,\cdot)$$ are locally bounded and continuous.

Reference

$$\{1\}$$ R. Schilling: Measures, integrals and martingales, Cambridge, 2nd edition.

• I've only got a copy of the 1st edition. The relevant chapter is completely new, right? Jan 25, 2019 at 17:30
• @0xbadf00d Seems so, yes. You can take a look at this question which gives two references (though not exactly for the statement which I mentioned above; since I don't have copies of the books I can't check what material they contain about determining families).
– saz
Jan 25, 2019 at 17:40
• Which version of Urysohn's lemma do you've got in mind? (Surely not the one from Wikipedia.) Jan 26, 2019 at 0:34
• @0xbadf00d For instance something like this
– saz
Jan 26, 2019 at 7:42