Existence of solution for a non-linear SDE

Question

Since $\exp(\cdot)$ is locally Lipschitz, the following SDE has a strong solution: $$ \mathrm{d}X_s=\exp(X_s) \, \mathrm{d}B_s,\quad X_0=1, $$ where $B$ is a standard Brownian motion. I wonder if the following expression holds: $$\mathbb{E}\int_0^T\exp(2X_s) \, \mathrm d s<\infty.$$

Answer 1

Let us show that
\begin{equation*} E\int_0^T e^{2X_t}\,dt=\infty \quad\text{for real }T\ge T_*:=e^{-2}/2. \tag{1}\label{1} \end{equation*}

Indeed, letting $Y_t:=e^{2X_t}$, we have $Y_0=e^2$ and, by Itô's lemma, for real $t\ge0$
\begin{equation*} dY_t=2e^{4X_t}dt+2e^{3X_t}dB_t= 2Y_t^2dt+2e^{3X_t}dB_t. \end{equation*} So, for real $t$ and $u$ such that $0\le t\le u$ and $m(t):=EY_t$ we have $m(0)=e^2$ and \begin{equation*} m(u)-m(t)=2\int_t^u EY_s^2ds\ge2\int_t^u m(s)^2ds. \end{equation*} So, on any interval $[0,T)$ where the function $m$ is finite, $m\ge m(0)=e^2>0$ and $m$ is a continuous increasing function. Moreover, then for any $t\in[0,T)$ \begin{equation*} m'_+(t):=\liminf_{u\downarrow t}\frac{m(u)-m(t)}{u-t}\ge 2m(t)^2. \end{equation*} So, for $h:=-1/m$ and any $t\in[0,T)$ \begin{equation*} h'_+(t)=\liminf_{u\downarrow t}\frac{h(u)-h(t)}{u-t} =\liminf_{u\downarrow t}\frac{m(u)-m(t)}{u-t}\frac1{m(u)m(t)}\ge2. \end{equation*} So, \begin{equation*} 0>h(T)\ge h(0)+2T=-e^{-2}+2T, \end{equation*} whence $T<T_*$. So, if $T\ge T_*$, then the increasing function $m$ is not finite on $[0,T)$. So, $m(t)=EY_t=Ee^{2X_t}=\infty$ for all $t$ in a left neighborhood of $T$ if $T\ge T_*$. So, we get \eqref{1}. $\quad\Box$