Let \(B_t\) be a standard Brownian Motion starting from zero and define
\[ p(t,x) = \frac1{\sqrt{2\pi t}}e^{-\frac{x^2}{2t} } \]
Given any \(x \in \mathbf R \), define \(X_t=x + B_t\) . Of course \(X_t\) is just a Brownian Motion stating from \(x\) at time 0. Fixing a smooth, bounded, compactly supported function \(f:\mathbf R \rightarrow \mathbf R\), we define the function \(u(x,t)\) by
\[u(x,t) = \mathbf E_x f(X_t)\]
where we have decorated the expectation with the subscript \(x\) to remind us that we are starting from the point \(x\).
- Explain why \[ u(x,t) = \int_{\infty}^\infty f(y)p(t,x-y)dy\]
- Show by direct calculation using the formula from the previous question that for \(t>0\), \(u(x,t)\) satisfies the diffusion equation
\[ \frac{\partial u}{\partial t}= c\frac{\partial^2 u}{\partial x^2}\]
for some constant \(c\). (Find the correct \(c\) !) - Again using the formula from part 1), show that
\[ \lim_{t \rightarrow 0} u(t,x) = f(x)\]
and hence the initial condition for the diffusion equation is \(f\).