I have the following PDE:
$$\begin{cases}
\left (-\partial_t + \mu x \partial_x + \dfrac{1}{2} \sigma^2 x^2 \partial_x^2 \right )u = -f(t, x)\\
u(0, x) = 0
\end{cases}$$
where $f(t, x) = x^2 \left ( (\mu + \sigma^2 ) e^{\mu t} – \left (\mu + \frac{1}{2}\sigma^2\right )\right )$.
Since $\mu x \partial_x + \dfrac{1}{2} \sigma^2 x^2 \partial_x^2$ is the infinitesimal generator of a Geometric Brownian Motion with drift $\mu$ and volatility $\sigma$, by Feynman-Kac the solution should be given by:
$$u(t, x) = \mathbb{E}^x \left (\int_0^t f(s, X_s)dx\right )$$
where $X$ has dynamics
$$dX_t = \mu X_t dt + \sigma X_t dW_t.$$
I know that the exact solution should be $u(t, x) = \frac{1}{2} x^2 \left ( e^{(2\mu + \sigma^2) t} – 2e^{\mu t} + 1\right )$. However, when I try to compute the solution using Feynman-Kac, I get:
$$u(t, x) = \int_0^t \mathbb{E}^x (f(s, X_s))ds = \int_0^t x^2 e^{(2\mu + \sigma^2) s}\left ( (\mu + \sigma^2 ) e^{\mu s} – \left (\mu + \frac{1}{2}\sigma^2\right )\right ) ds$$ which doesn't agree with the above solution.
What am I doing wrong?
Best Answer
Almost. It should be $u(t,x)=\Bbb E^x\left(\int_0^t f(t-s,X_s)\,ds\right)$.