Let \(f \in C_0^2(\mathbf R^n)\) and \(\alpha(x)=(\alpha_1(x),\dots,\alpha_n(x))\) with \(\alpha_i \in C_0^2(\mathbf R^n)\) be given functions and consider the partial differential equations

\begin{align*}

\frac{\partial u}{\partial t} &= \sum_{i=1}^n \alpha_i(x)

\frac{\partial u}{\partial x_i} + \frac{1}{2} \frac{\partial^2

u}{\partial x_i^2} \ \text{ for } t >0 \text{ and }x\in \mathbf R^n \\

u(0,x)&=f(x) \ \text{ for } \ x \in \mathbf R^n

\end{align*}

Use the Girsonov theorem to show that the unique bounded solution \(u(t,x)\) of this equation can be expressed by

\begin{align*}

u(t,x) = \mathbf E_x \left[ \exp\left(\int_0^t \alpha(B(s))\cdot dB(s) –

\frac{1}{2}\int_0^t |\alpha(B(s))|^2 ds \right)f(B(t))\right]

\end{align*}

where \(\mathbf E_x\) is the with respect to \(\mathbf P_x\) when the Brownian Motion starts at \(x\). (Note there maybe a sign error in the above exponential term. Use what ever sign is right.) For the remainder, assume that \(\alpha\)

is a fixed constant \(\alpha_0\). Now using what you know about the distribution of \(B_t\) write the solution to the above equation as an integral kernel integrated against \(f(x)\). (In other words, write \(u(t,x)\) so that your your friends who don’t know any probability might understand it. ie \(u(t,x)=\int K(x,y,t)f(y)dy\) for some \(K(x,y,t)\))