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# Shifted Brownian Motion and a PDE

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 $$K(x,y,t)$$)