Previous theorems established that there is zero probability of a  diffusion blowup such that $\mathbf{P}[\psi(t)=\infty]=0$ for all $t\in[t_{\epsilon},\infty)=\mathbf{Y}\bigcup\mathbf{Z}\subset\mathbf{R}^{+}$ where

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

and $t=t_{*}$ is the blowup time for the unperturbed system. From the perspective of consistency, the same conclusion should also follow from a Fokker-Planck equation, or Kolmogorov’s second equation, which is essentially an equation for conservation of probability. For a pure diffusion with non drift, the FP equation will reduce to a diffusion equation for the underlying probability density function.

Let $\psi(t)$ be a generic diffusion satisfying athe nonlinear SDE

$d\widehat{\psi}(T)=U[\psi(t)]dt +V[\psi(t)]\circ d\widehat{\mathcal{B}}(t)$

with $U:\mathbf{R}\times \mathbf{R}\rightarrow \mathbf{R}$ and $V:\mathbf{R}\times \mathbf{R}\rightarrow \mathbf{R}$

some initial data $\psi(0),\widehat{\mathcal{B}}(0)=0.$ with $\psi(t)\in [\psi(0),\infty)$. A pure drift-free diffusion is then

$d\widehat{\psi}(T)=V[\psi(t)]\circ d\widehat{\mathcal{B}}(t)=\lim_{U\rightarrow 0}U[\psi(t)]dt +V[\psi(t)]\circ d\widehat{\mathcal{B}}(t)$

The corresponding nonlinear Fokker-Planck equation for a probability density $\mathscr{P}[\psi(T),t]$ is (refs)

$\partial_{t}\mathscr{P} [\psi(T),t]=-\partial_{\psi} V[\psi(t)]+\frac{1}{2}V[\psi(t)] \partial_{\psi} V[\psi(t)]]\mathscr{P}[\psi(t),t]+\frac{1}{2}\partial_{\psi}\partial_{\psi}[V[\psi(t)]^{2} \mathscr{P}[\psi(t),t]$

Taking the Ito interpretation

$\widehat{\psi}(t+\delta t)-\widehat{\psi}(T)=U[\psi(t)]\delta t +V[\psi(t)]{\displaystyle \int_{t}^{t+\delta t}}d\mathcal{B}(t')$

The Fokker-Planck equation now becomes (ref)

$\partial_{t}\mathscr{P}[\psi(t),t]=\partial_{\psi}U[\psi(t),t]\mathscr{P}[\psi(t),t]+\frac{1}{2}\partial_{\psi}\partial_{\psi}[V[\psi(t)]^{2}\mathscr{P}[\psi(t),t]$

If $U[\psi(t)=0$ then this is a nonlinear diffusion equation for $\mathscr{P}[\psi(t),t]$

$\partial_{t}\mathscr{P}[\psi(t),t]=+\frac{1}{2} \partial_{\psi}\partial_{\psi}[V[\psi(t)]^{2}\mathscr{P}[\psi(t),t]$

The general solution of a nonlinear Fokker-Planck equation cannot usually be found but a stationary solution is possible in the infinite-time relaxation limit such that

$\partial_{t}\mathscr{P}[\psi(t),t]=0$ so that

$G\mathscr{P}[\psi(t)]=\partial_{\psi} V[\psi(t),t]\mathscr{P}[\psi(t]+\frac{1}{2} \partial_{\psi}\partial_{\psi}[V[\psi(t)]^{-2}\mathscr{P}[D(t]=$

where G is the generator of the diffusion as in (-). The stationary solution at $t=\infty$ for the probability density is

$\mathscr{P}[\psi(t),t=\infty]= \bigg|\frac{C}{(V[\psi(t)]^{2}}\bigg| exp\bigg(2{\displaystyle \int_{\psi_{0}}}^{\psi(t)}\frac{U[\bar{\psi}(t)]}{(\Psi[\bar{\psi}(t)])^{2}}d\bar{\psi}(t)\bigg)$

where $latex C$ is a constant. The normalised solution is

${\displaystyle \int_{\psi(0)}^{\infty}}\mathscr{P}[\psi(t)]d\psi(t)={\displaystyle \int_{\psi_{0}}^{\infty}} \bigg|\frac{C}{(V[\psi(t)]^{2}}\bigg| exp\bigg(2{\displaystyle \int_{\psi_{0}}^{\psi(t)}}\frac{U[\bar{\psi}(t)]}{(V[\bar{\psi}(t)])^{2}}d\bar{\psi}(t)\bigg)=1$

so that

$\mathscr{P}[\psi(t),t=\infty]=$ $\frac{\bigg|\frac{1}{(V[\psi(t),t=\infty]^{2}}\bigg| exp\bigg(2{\displaystyle \int_{\psi_{0}}^{\psi(t)}}\frac{iU[\bar{\psi}(t)]} {(V[\bar{\psi}(t)])^{2}}d \bar{\psi}(t)\bigg)}{{\displaystyle \int_{\psi_{0}}^{\infty}} \bigg|\frac{1}{(V[\psi(t)]^{2}} exp\bigg(2{\displaystyle \int_{\psi_{0}}^{\psi(t)}}\frac{U[\bar{\psi}(t)]}{(V[\bar{\psi}(t)])^{2}} d \bar{\psi}(t)\bigg)}$

Integrating over the probability density gives

$\mathbf{P}[\widehat{\psi}(t),t=\infty]={\displaystyle \int_{\psi(0)}^{\psi(t)}}\mathscr{P}[\psi(t),t=\infty]dD(t)=$ ${\displaystyle \int_{\psi(0)}^{\psi(t)}}\frac{\bigg|\frac{1}{(V[\psi(t)]^{2}}\bigg| exp\bigg(2{\displaystyle \int_{\psi_{0}}^{\psi(t)}}\frac{U[\bar{\psi}(t)]} {(V[\bar{\psi}(t)])^{2}}d\bar{\psi}(t)\bigg)}{{\displaystyle \int_{\psi_{0}}^{\infty}} \bigg|\frac{1}{(V[\psi(t)]^{2}}\bigg| exp\bigg(2{\displaystyle \int_{\psi_{0}}^{\psi(t)}}\frac{U[\bar{\psi}(t)]}{(V[\bar{\psi}(t)])^{2}} d\bar{\psi}(t)\bigg)} d\psi(t)$

The probability that $\psi(t)<\infty$ at $t=\infty$ is then

$\mathbf{P}[\widehat{\psi}(t)<\infty,t=\infty]={\displaystyle \int_{\psi(0)}^{\infty}}\mathscr{P}[\psi(t),t=\infty]d\psi(t)=$ ${\displaystyle \int_{\psi(0)}^{\infty}}\frac{\bigg|\frac{1}{(V[\psi(t)]^{2}}\bigg| exp\bigg(2{\displaystyle \int_{\psi_{0}}^{\psi(t)}}\frac{U[\bar{\psi}(t)]} {(V[\bar{\psi}(t)])^{2}}d\bar{\psi}(t)\bigg)}{{\displaystyle \int_{\psi_{0}}^{\infty}} \bigg|\frac{1}{(V[\psi(t)]^{2}}\bigg| exp\bigg(2{\displaystyle \int_{\psi_{0}}^{\psi(t)}}\frac{U[\bar{\psi}(t)]}{(V[\bar{\psi}(t)])^{2}} d\bar{\psi}(t)\bigg)} d\psi(t)=1$

For a pure diffusion with $V[\psi(t)]=0$ the solution reduces to

$\mathbf{P}[\widehat{\psi}(t)<\infty]=\lim_{\psi(t)\rightarrow\infty}{\displaystyle \int_{\psi(0)}^{\psi(t)}}\mathscr{P}[\psi(t),t=\infty]=$ $\lim_{\psi(t)\rightarrow\infty}\frac{\displaystyle{\int_{\psi(0)}^{\psi(t)}}\bigg|\frac{1}{(V[\psi(t)]^{2}}\bigg|}{{\displaystyle \int_{\psi_{0}}^{\infty}} \bigg|\frac{1}{(V[\psi(t)]^{2}}\bigg|}=1$T

Technical details of the FP equation are found in (refs.)

Given the SDE $d\widehat{\psi}(t)=V[\psi(t)]\otimes d\widehat{\mathcal{B}}(t)$ defined for all $t\in\mathbf{Y}\bigcup\mathbf{Z}$ with initial matter density $\psi_{\epsilon}$ the Fokker-Planck diffusion equation now becomes (ref)

$\partial_{t}\mathscr{P}[\psi(t),t]=+\frac{1}{2} \partial_{\partial}\partial_{\psi}((\psi(t))^{4}(\psi(t)-1))\mathscr{P}[\psi(t),t]$

For a stationary solution in the limit as $t\rightarrow\infty$

$G\mathbf{\mathscr{P}}[\psi(t),t=\infty]$ = $\frac{1}{2}\partial_{\psi}\partial_{\psi}((\psi(t))^{4}(\psi(t)-1))\mathbf{\mathscr{P}}[\psi(t),t=\infty]=0$

The probability that the diffusion $\psi(t)$ is finite at $t=\infty$ is then

$\mathbf{P}[\psi(T)<\infty,t=\infty]=\lim_{\psi(t)\rightarrow\infty} {\displaystyle \int_{\psi_{\epsilon}}^{\psi(t)}}\mathscr{P}[\psi(t),t=\infty]d\psi(t)$

$= \lim_{\psi(t)\rightarrow\infty}\frac{{\displaystyle \int_{\psi_{\epsilon}}^{\psi(t)}}\bigg|\frac{1}{(\psi(t))^{4}(\psi(t)-1)}\bigg|} {\int_{\psi_{\epsilon }}^{\infty} \bigg|\frac{1}{(\psi(t))^{4}(\psi(t)-1)}\bigg|}=1$

This result can also be computed explicitly. Given the SDE $d\widehat{\psi}(t)=V[\psi(t)]\circ d\widehat{\mathcal{B}}(t)$ defined for all
$t\in\mathbf{Y}\bigcup\mathbf{Z}$ with initial matter density $\psi_{\epsilon}$ the Fokker-Planck diffusion equation is as before(ref)

$\partial_{t}\mathscr{P}[[\psi(t),t]$ $=+\frac{1}{2} \partial_{\psi}\partial_{\psi}\psi^{4}(t)(\psi(t)-1))\mathbf{\mathscr{P}}[\psi(t),t]$

For a stationary solution in the limit as $t\rightarrow\infty$

$G\mathscr{P}[\psi(t),t=\infty]=\frac{1}{2}\partial_{\psi}\partial_{\psi}\psi^{4}(t))(\psi(t)-1))\mathscr{P}[\psi(t),t=\infty]=0$

The stationary solution for the Fokker-Planck probability density is

$\mathscr{P}[\psi(t),t=\infty]=$  $\bigg(\frac{6(\psi_{\epsilon})^{3}} {6\psi_{\epsilon}^{3} ln(\psi_{\epsilon}^{-1}-1)-6\psi_{\epsilon}^{3}-2\psi_{\epsilon}-3\psi_{\epsilon}^{4}}\bigg) \frac{1}{(\psi(t))^{4}(\psi(t)-1)}$

The probability that the diffusion $\widehat{\psi}(t)$ is finite at $\infty$ is now given by (-) as

$\mathbf{P}[\psi(t)<\infty,t=\infty]={\displaystyle\int_{\psi_{\epsilon}}^{\infty}}\mathscr{P}[\psi(t),t=\infty]$  $\bigg(\frac{6\psi_{\epsilon}^{3}}{6\psi_{\epsilon}^{3} ln(\psi_{\epsilon}^{-1}-1)-6\psi_{\epsilon}^{3}- \psi_{\epsilon}-3\psi_{\epsilon}^{4}}\bigg) {\displaystyle \int_{\psi_{\epsilon}}^{\infty}}\frac{d\psi(t)}{(\psi(t))^{4}(\psi(t)-1)}=1$

Performing the integral i (-)

$\mathbf{P}[\psi(t)<\infty,t=\infty]={\displaystyle \int_{\psi_{\epsilon}}^{\infty}}\mathscr{P}[\psi(t),t=\infty]$= $\bigg(\frac{6\psi_{\epsilon}^{3}}{6\psi_{\epsilon}^{3} ln(\psi_{\epsilon}^{-1}-1)-6\psi_{\epsilon}^{3}- 2\psi_{\epsilon}-3\psi_{\epsilon}^{4}} \bigg)$

$\times\bigg(\frac{1}{3(\psi(t))^{3}}+\frac{1}{2(\psi(t))^{2}}+\frac{1}{\psi(t)}+ ln (1-(\psi(t))^{-1})\bigg)_{\psi(t)=\infty}$  $-\bigg(\frac{6\psi_{\epsilon}^{3}}{6\psi_{\epsilon}^{3} ln(\psi_{\epsilon}^{-1}-1)-6\psi_{\epsilon}^{3}- 2\psi_{\epsilon}-3\psi_{\epsilon}^{4}}\bigg) \bigg(\frac{1}{3(D_{\epsilon})^{3}} -\frac{1}{2(\psi_{\epsilon})^{2}}-\frac{1}{\psi_{\epsilon}}-ln (1-\psi_{\epsilon}^{-1})\bigg)$

The first term vanishes at $\psi(t)=\infty$ since $ln(1)=0$ so that (-) reduces to

$\mathbf{P}[\psi(t)<\infty,t=\infty]={\displaystyle \int_{\psi_{\epsilon}}^{\infty}}\mathscr{P}[\psi(t),t=\infty]$

$= -\bigg(\frac{6\psi_{\epsilon}^{3}}{6\psi_{\epsilon}^{3} ln(\psi_{\epsilon}^{-1}-1)-6\psi_{\epsilon}^{3}- 2\psi_{\epsilon}-3\psi_{\epsilon}^{4}}\bigg)$

$\times\bigg(\frac{1}{3(\psi_{\epsilon})^{3}} -\frac{1}{2(\psi_{\epsilon})^{2}}-\frac{1}{\psi_{\epsilon}}-ln (1-\psi_{\epsilon}^{-1})\bigg)$ $=-\bigg(\frac{6\psi_{\epsilon}^{3}}{6\psi_{\epsilon}^{3} ln(\psi_{\epsilon}^{-1}-1)-6\psi_{\epsilon}^{3}- 2\psi_{\epsilon}-3\psi_{\epsilon}^{4}}\bigg)\times-\bigg(\frac{6\psi_{\epsilon}^{3} ln(\psi_{\epsilon}^{-1}-1)-6\psi_{\epsilon}^{3}- 2\psi_{\epsilon}-3\psi_{\epsilon}^{4}}{6\psi_{\epsilon}^{3}}\bigg)=1$

Hence, the diffusion remains finite in the infinite-time relaxation limit.