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Diffusion and Brownian motion
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 \(f\).
A PDE example
Observe that for \(k=0,1,\dots\)
\[\phi_k(x) = \sin(\pi k x/2) \]
form an orthonormal basis of function for \(L^2([0,2]\) with \(\phi(0)=\phi(2)=0\). Here the inner-product of two functions in \(f,g \in L^2([0,2]\) is
\[\langle f,g\rangle =\int_0^2 f(x)g(x) dx\]
Define the operator \(L\) acting on a function \(\phi(x)\) by
\[L\phi(x)=\frac12 \frac{\partial^2 \phi}{\partial^2x}(x) – 5 \phi(x)\]
To solve the equation
\[ \frac{\partial u}{\partial t}(x,t) = (L u)(x,t) \]
with
\[ u(0,t)=u(2,t)=0 \qquad\text{and}\qquad u(x,0)=F(x) \]
assume that \(u(x,t)\) takes the from
\[u(x,t)=\sum_{k=0}^\infty a_k(t) \phi_k(x)\]
Find the equations for the \(a_k\) and solve then find an expression for \(u(x,t)\).