Given the generator $G=\frac{1}{2} k^{1/2} D(t)^{2}(D(t)-1)^{1/2}$ for the pure density function diffusion $\widehat{D}(t)$ on the partition

$\mathbf{X}\bigcup \mathbf{Y}\bigcup \mathbf{Z}=[0,t_{\epsilon}] \bigcup [t_{\epsilon},t_{*}] \bigcup [t_{*},\infty)$

where

$d \widehat{D}(t) = k^{1/2} D^{2}(t)(D(t)-1)^{1/2} \otimes d\widehat{\mathcal{B}}(t)$

$\widehat{D}(t) = D_{\epsilon}+k^{1/2}{\displaystyle \int_{t_{\epsilon}}^{t}} D^{2}(v)(D(v)-1)^{1/2} \otimes d\widehat{\mathcal{B}}(v)$

for initial data $D(t_{\epsilon})$. It is possible to define and find a Lyapunov function $\mathcal{L}y[D]$ related to boundedness and non-explosion of the density function diffusion(refs). A $C^{2}$ function $\mathcal{L}_{y}:\mathbf{R}\rightarrow\mathbf{R}$ is a Lyapunov function if

1. $\mathcal{L}y[D]\ge 0$.
2 $lim_{D\rightarrow\infty}\|\mathcal{L}y[D]\|=\infty$
3. Given the generator $G$ for the diffusion,$\exists$ constants $c,Q$ such that
$G \mathcal{L}y[D]\le c\mathcal{L}y[D]+Q$

Given the generator $G=\tfrac{1}{2}(D(t))^{2}(D(t)-1)\partial_{D}\partial_{D}$ for the matter density diffusion $\widehat{D}(t)$ for all $t\in\mathbf{Y}\bigcup\mathbf{Z}\equiv[t_{\epsilon},\infty)$ suppose there exists a Lyapunov function $\mathbf{\mathcal{L}y}[D]$ then the SDE does not explode and the density diffusion $\widehat{D}(t)$ is finite and bounded for all $t\in\mathbf{Y}\bigcup\mathbf{Z}\equiv[t_{\epsilon},\infty)$
Proof:
For $N,M\in\mathbb{Z}$, Define the hitting times
$\mathcal{T}^{N}=\inf\lbrace t>t_{\epsilon}:\mathbf{\mathcal{L}y}[\widehat{D}(t)]\ge N\rbrace,~~ T^{M}=\inf\lbrace t>t_{\epsilon}:\widehat{D}(t)\ge M\rbrace$

Now Ito’s formula is applied to $\mathbf{\mathcal{L}y}[D(t\wedge \mathcal{T}^{N}\wedge T^{M}]$
$\mathbf{\mathcal{L}y}[D(t\wedge \mathcal{T}^{N}\wedge T^{M}]= [D_{\epsilon}]+{\displaystyle \int_{t_{\epsilon}}^{\mathbf{\mathcal{L}y}[D(t\wedge \mathcal{T}^{N}\wedge T^{M}}}) G\mathcal{L}y[D(v)]dv$+

$\sum_{\xi=1}^{M} {\displaystyle \int_{0}^{\mathcal{L}y[D(t\wedge \mathcal{T}^{N}\wedge T^{M}}}\mathcal{L}y[D(v)]\Psi_{\xi}[D(s)]\circ d\widehat{\mathcal{B}}^{\xi}(s)\mathcal{L}y[D(v)][(D(v))^{2}(D(v)-1))^{1/2}]_{\xi} d\widehat{\mathcal{B}}^{\xi}(s)$

Since $\mathbf{\mathcal{L}y}$ and $\Psi[D(t)]$ are continuous and bounded on a ball $\mathbb{B}$ of radius $M$, the local martingale is a martingale. For the generator G of the pure diffusion $\widehat{D}(t)$

Since $\mathcal{L}y$ and $\Psi[D(t)]$ are continuous and bounded on a ball $\mathbb{B}$ os radius $M$, the local martingale is a martingale. For the generator G of the pure diffusion $\widehat{D}(t)$
$\mathcal{L}y[D(t\wedge \mathcal{T}^{N}\wedge T^{M}]= [D_{\epsilon}]+k^{1/2}{\displaystyle \int_{t_{\epsilon}}^{\mathcal{L}y[D(t\wedge \mathcal{T}^{N}\wedge T^{M}}}[(D(v))^{2}(D(v)-1)]\partial_{D}\partial_{D}\mathcal{L}y[D(v)]dv$+$k^{1/2} \sum_{\xi=1}^{M}$  ${\displaystyle \int_{0}^{\mathcal{L}y[D(t\wedge\mathcal{T}^{N}\wedge T^{M}}}$

Taking the expectation $\mathbf{E}\lbrace...\rbrace$ this reduces to
$\mathbf{E} (\mathcal{L}y$  $[D(t\wedge \mathcal{T}^{N}\wedge T^{M}] )$

= $[D_{\epsilon}] + k^{1/2}{\displaystyle \int_{t_{\epsilon}}^{t}}$ $\mathbf{E} ( [(D(v))^{2} (D(v)-1)^{1/2}]$ $\partial_{D} \partial_{D} \mathcal{L}y [D(v)]dv$ $) U[v<\mathcal{T}^{N}\wedge T^{M}]$ $<$ ${\displaystyle \int_{t_{\epsilon}}^{t}}$ $\mathbf{E}([[C\mathbf{\mathcal{L}}y[D]+Q]dv) U[v<\mathcal{T}^{N}\wedge T^{M}]\le\mathcal{L}y[D_{\epsilon}]$

+ $Qt +{\displaystyle \int_{\epsilon}^{t}}\mathbf{E}(\mathcal{L}y[D(s\wedge \mathcal{T}^{N} \wedge T^{M}]dv$

From Gronwall’s inequality it follows that
$\mathbf{E}\lbrace \mathcal{L}y[D(t\wedge \mathcal{T}^{N}\wedge T^{M}]\rbrace < (\mathcal{L}y[D_{\epsilon}]+ Qt ) exp(Ct)$

Letting $M\rightarrow\infty$ and observing that $\widehat{D}(t)$ does not explode before comoving time $\mathcal{T}^{N}$

$\mathbf{E}( \mathcal{L} y$ $[D(t \wedge \mathcal{T}^{N}])$ <$(\mathcal{L}y [D_{\epsilon}]+ Qt ) exp(Ct)$

But
$\mathbf{E} ( \mathcal{L}y$  $D[t\wedge\mathcal{T}^{n}])$= $\mathbf{E} \mathcal{L} y [\widehat{D}(t)]$  $U[t < \mathcal{T}^{N}] + N \mathbf{P} [t \ge \mathcal{T}^N)$

Now
$N \mathbf{P} [t \ge \mathcal{T}^N)] <= (\mathcal{L}y [D_{\epsilon}]+ Qt ) exp(Ct)$
so that
$\mathbf{P} [t \ge \mathcal{T}^{N}] \le \frac{1}{N} (\mathcal{L} [D_{\epsilon}]+ Qt ) exp(Ct)$
taking the limit $N\rightarrow\infty$ or $\mathcal{T}^{N} \rightarrow\infty$ gives
$\mathbf{P}[t\ge\mathcal{T}^{\infty}]=0$ or $\mathbf{P}[t\le\mathcal{T}^{\infty}]=1$ and so $t$ is always less than $\mathcal{T}^{\infty}$