Let $u:[0,T]\times\mathbb{R} \mapsto \mathbb{R}_+$, $T>0$, $r>0$, $\mu>0$, $\sigma>0$ and $A>0$.

What is the solution to the following PDE: \begin{equation} u_t(t,x)+\mu x u_x(t,x)+\frac{\sigma^2x^2}{2}u_{xx}(t,x)-ru(t,x)+\frac{1}{A+x}=0 \end{equation} with the terminal condition $u(T,x)=0$.

By Feymann-Kac theorem, $u$ has the probabilistic representation $$u(t,x)=\mathbb{E}\left[\int_t^Te^{-r(u-t)}\frac{1}{A+S_u}du|S_t=x\right]$$ where $S$ follows the SDE $$dS_t=\mu S_tdt+\sigma S_tdW_t.$$ However, to find $u$ by first solving $\mathbb{E}\left[\frac{1}{A+S_u}\right]$ does not lead to an analytical solution because the random variable $\frac{1}{A+S_u}$ follows a logit normal distribution which is known for not having analytical expression for the mean.

  • $\begingroup$ There is a $r$ in the first equation. Is that a typo? Should it be $t$? $\endgroup$ May 18 at 11:23
  • $\begingroup$ Sorry, I forgot to declare $r>0$, now it is edited. $\endgroup$
    – Amira
    May 19 at 9:03
  • $\begingroup$ Function $1/(A+x)$ is not integrable. Is it proven that a solution exists? $\endgroup$
    – Andrew
    May 19 at 16:22
  • $\begingroup$ Since $S_t>0$ a.s. for all $t\in[0,T]$, then $0<\mathbb{E}[\frac{1}{A+S_u}]\leq \frac{1}{A}$, hence, it has finite moment. A solution $u$ should exist because the probabilistic representation is well-defined. $\endgroup$
    – Amira
    May 20 at 4:17
  • $\begingroup$ Based on a past question you asked, don't you rather mean $\frac{1}{A+\exp(\pm x)}$? As it stands, note that $\mathbb{E}\big(\frac{1}{A+S_u}\big)$ does not exist, since the integral does not converge $\endgroup$
    – charmd
    May 20 at 5:50


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