First, the following lemma establishes the oscillation properties of the diffusion (martingale) through a subinterval or slab. The expected number of ‘upcrossings’ of the density function diffusion $\widehat{D}(t)$ across a subinterval or “slab” $\mathfrak{S}_{ab}=[D_{a},D_{b}] \subset [D_{\epsilon},\infty]$
is finite for all $t\in\mathbf{X}\bigcup\mathbf{Y}$ and the number of upcrossings on a semi-infinite interval $\mathbf{\mathfrak{G}}_{a \infty}=[D_{a},\infty]$ is always zero. This is essentially a Doob upcrossing inequality. Let $\widehat{D}(t)$ be the density function diffusion or martingale solution of the SDE (-) for initial data $\mathscr{I}_{\epsilon}=[t=t_{\epsilon},D_{\epsilon}<0,\widehat{\mathcal{B}}_{\epsilon}=0]$ so that $\widehat{D}(t)\in[|D_{\epsilon}|,\infty)$. Let $[|D_{\alpha}|,|D_{\beta}|]\subset[|D_{\epsilon}|,\infty)\subset\mathbf{R}^{+}$ be a finite interval with $|D_{\alpha}|<|D_{\beta}|$. Now let $\mathcal{U}(\alpha,\beta:t)$ denote the number of ‘upcrossings’ of the interval or slab $\mathcal{S}_{\alpha\beta}=[|D_{\alpha}|,|D_{\beta}|]$, which is the number of times that the density function diffusion $|\widehat{D}(t)|$ has passed from below $|D_{\alpha}|$ to above $D_{\beta}|$ at some $t\in\mathbf{Y}\bigcup\mathbf{Z}$. Then if $\widehat{D}(t)$ is a martingale

[1] The number of upcrossings through the slab $\mathfrak{S}$ is finite for all $t\in\mathbf{Y}\bigcup\mathbf{Z}$

$\mathbf{E}(\mathbf{\mathcal{U}}(a \rightleftarrows b; t)~\le~\frac{1}{D_{b}-D_{a}} \mathbf{E}([|\widehat{D}(t)|]]+|D_{a}|)$

$= \frac{1}{D_{b}-D_{a}}\mathbf{E}|D_{\epsilon}| +\frac{k^{1/2}}{D_{b}-D_{a}} \mathbf{E}({\displaystyle \int_{t_{\epsilon}}^{t}}|(D(v))^{2} (D(v)-1))^{1/2} \otimes d\widehat{\mathcal{B}}(v)+D_{a}) < \infty$

[2] The limit $|D_{\beta}| \rightarrow \infty$ gives a semi-infinite
interval or slab $[|D_{\alpha}|,\infty)$ so that the number of upcrossings from below $D_{a}$ to infinity at any $t\in\mathbf{Y}\bigcup\mathbf{Z}$, is zero

$\mathbf{E}(\mathbf{\mathcal{U}}(a,\infty :t)]$ $<=lim_{|D_{b}| \rightarrow \infty}$ $\frac{1}{D_{b}-D_{a}}(\mathbf{E}(\widehat{D}(t)|=0$

so that for the initial data, the density diffusion $\widehat{D}(t)$ never blows up for any finite $t\in\mathbf{Y}\bigcup\mathbf{Z}$. Define the ‘hitting times’ ${S}_{n}=\inf\lbrace t>{T}_{n-1}:|D(t)|\le |D_{a}|\rbrace$

${T}_{n}=\inf\lbrace t>{S}_{n}:|D(t)|\ge |D_{b}|\rbrace$

with $T_{0}=0$ and $n\in\mathbf{N}$. Construct a stochastic process

$\widehat{Z}(t) = \sum_{n}(|\widehat{D}(t \wedge {T}_{n})|-|\widehat{D}(t \wedge {S}_{n})|)$

Beginning with the first time $T_{1}$ that $|D(t)| \le |D_{a}|$, the process
$\widehat{Z}(t)$ evolves via increments of $\widehat{D}(s)$ until the the first time ${T}_{1}\ge|D_{b}|$. The process repeats if again the martingale diffusion $|\widehat{D}(t)|$ falls below $D_{a}|$ or $|\widehat{D}(t)|<= |D_{a}|$ at time $\mathcal{S}_{2}$ and so on. All terms in the summation vanish for sufficiently large $n$ that $\mathcal{S}_{n}>t$ so that there are at most $\mathbf{\mathcal{U}}(a\rightleftarrows\|t)$ terms that are non-zero. Then

$\widehat{Z}(t)=\sum_{n}\big(\widehat{D}(t\wedge{T}_{n})-\widehat{D} (t\wedge{S}_{n})\big)$

$\ge (D_{B}-D_{A}) \mathbf{\mathcal{U}}(a\rightleftarrows |t)(\widehat{D}(t)-|D_{a})$

so that

$\mathbf{E}(\mathbf{\widehat{Z}}(t)D) \ge (D_{\beta}-D_{\alpha})\mathbf{E} (\mathbf{\mathcal{U}}(a \rightleftarrows~B;t) )$

$+\mathbf{E}(D(s)-D_{\alpha})$ $\ge (D_{\beta}-D_{\alpha})\mathbf{E} \mathbf{\mathcal{U}}(a\rightleftarrows b; ~t)-\mathbf{E}(D(s)- D_{\alpha})$

$\ge (D_{\beta}-D_{\alpha})\mathbf{E}\mathbf{\mathcal{U}}(a\rightleftarrows b; ~t) \mathbf{E}( D(t)|-D_{\alpha})$

Since $\widehat{Z}(t)$ is also a martingale then

$\mathbf{E}(\widehat{Z}(t))=\mathbf{E}\langle \widehat{Z}_{o})$. It follows that

\$\mathbf{E}(\mathbf{\mathcal{U}}(a\rightleftarrows b|t)~<=~\frac{1}{D_{\beta}-D_{\alpha}} (\mathbf{E}(|\widehat{D}(t)|)+|D_{\alpha} ))$

Taking the limit as $|D_{b}| \rightarrow \infty$.

$\mathbf{E}(\mathbf{\mathcal{U}}(\alpha, \infty:t))= lim_{D_{\beta}, \infty}\mathbf{E}(\mathbf{\mathcal{U}}(a\rightleftarrows b:t)]$

$~\le~lim_{D_{\beta}\rightarrow\infty} \frac{1}{D_{\beta}-D_{\alpha}}(\mathbf {E}\langle \big|\widehat{D}(t)|\rangle+|D_{\alpha}\big|\big)=0$

This then gives $\mathbf{E}(\mathbf{\mathcal{U}}(a\rightarrow \infty;~ t)=0$ for any $t\in\mathbf{Y}\bigcup\mathbf{Z}$ so a blowup never occurs.