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}

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