Suppose I have the following nonlinear ODE for a function $X:\mathbb{R}\rightarrow\mathbb{R}$ such that \begin{equation} \frac{dX(t)}{dt}=k^{1/2}((X(t))^{2}(X(t)-1)^{1/2}) \end{equation} with initial data $X(0)=X_{o}=1$ and $k>0$. The formal solution is then \begin{equation} X(t)=X_{o}+\int_{0}^{t}((X(v))^{2}(X(v)-1)^{1/2}dv \end{equation} The ODE is readily integrated to give $ \bigg|\frac{(X(t)-1))^{1/2}}{X(t)}+\tan^{-1}(X(t)-1))^{1/2}\bigg|-\bigg| \frac{(X(0)-1))^{1/2}}{X(0)}-\tan^{-1}(X(0)-1))^{1/2}\bigg|=k^{1/2}t $. Since $X(0)=1$ and $\tan^{-1}(0)=0$ then \begin{equation} \bigg|\frac{(X(t)-1))^{1/2}}{X(t)}+\tan^{-1}(X(t)-1))^{1/2}\bigg|=k^{1/2}t \end{equation} When $X(t)=\infty$ then $\tan^{-1}(\infty)=\pi/2$ and $\frac{(X(t)-1))^{1/2}}{X(t)}=0$ so there is then a blowup at $t=t_{*}=\pi/2k^{1/2}$. This is equivalent to a parametric cyloid equation solution such that $X(t)=X(0)/[1+cos\varphi]$ and $t=[\varphi+sin\varphi]/2k^{1/2}$ so the blowup occurs at $\varphi=\pi$.
Suppose now the ODE is subject to multiplicative white noise perturbations $\widehat{\mathcal{W}}(t)$ such that \begin{equation} \frac{dX(t)}{dt}=k^{1/2}((X(t))^{2}(X(t)-1)^{1/2}).\widehat{\mathcal{W}}(t) \end{equation} giving a SDE or diffusion without drift \begin{equation} d\widehat{X}(t)=k^{1/2}((X(t))^{2}(X(t)-1)^{1/2}).d\widehat{\mathcal{B}}(t)= F[X(v)].d\widehat{\mathcal{B}}(v) \end{equation} and where $d\widehat{B}(t)=\widehat{W}(t)dt$ is the standard Wiener process or Brownian motion. The formal solution is then the Ito stochastic integral: \begin{equation} X(t)=X_{o}+k^{1/2}\int_{0}^{t}((X(v))^{2}(X(v)-1)^{1/2}).d\widehat{\mathcal{B}}(v) \end{equation} The expectation is then \begin{equation} \mathbf{E}[\widehat{X}(t)]=X_{o}+k^{1/2}\mathbf{E}\left[\int_{0}^{t}((X(v))^{2}(X(v)-1)^{1/2}). d\widehat{\mathcal{B}}(v)\right]=X_{o} \end{equation} So what I would like to know is: [1] Is is justified to say that there is at least a unique weak global solution?And that this stochastic integral is now a square-integrable martingale whose suprema is bounded and finite--and therefore blow-up free--for all $t\in\mathbb[0,\infty)$, or at least does not explode on $[0,t_{*}]$. Then $\mathbf{P}[|X(t)|=\infty]=0$. Ideally, I would like the blowup to be "dissipated" for all t>0. But to be square integrable one must have \begin{equation} \mathbf{E}[|\widehat{X}(t)|^{2}]=X_{o}^{2}+k\mathbf{E}\bigg[\bigg|\int_{0}^{t}((X(v))^{2}(X(v)-1)^{1/2}].d\mathcal{B}(v)\bigg|^{2}\bigg]<\infty \end{equation} which is tantamount to the volatility. But the Ito isometry requires that \begin{equation} \int_{0}^{t}(X(v))^{4}(X(v)-1)dv<\infty \end{equation} I get this to diverge at $t=t_{*}$ which is confusing since I would expect a finite and bounded volatility $\mathbf{E}[|X(t)|^{2}]<\infty$ if the mean is finite in that $\mathbf{E}[|\widehat{X}(t)|]<\infty$. I would also like square-integrability for the purpose of obtaining a strong solution.
[2]Also by the Englebert-Schmidt Theorem and Feller's Test for explosion, one expects such pure diffusions with no drift, of the form $d\widehat{X}(t)=F(X(t))d\widehat{\mathcal{B}}(t)$ to have non-exploding solutions even if the underlying ODE $dX(t)=F(X(t))dt$ does explode. Applying Fellers test I deduce that this SDE does not explode.
[3] Same questions for a stochastic Riccati differential equation? \begin{equation} d\widehat{X}(t)=-\alpha[(X(t))^{2}+\beta^{2}].d\widehat{\mathcal{B}}(t) \end{equation} with $X(0)<0$, and $\alpha,\beta>0$. Thanks.