Fourier Transform

heat equation on the entire line

\begin{equation} \pdv{u}{t} = \frac{1}{2} \pdv[2]{u}{x} \end{equation}

We can try to find a:

\begin{equation} U(0,x) = f(x) \end{equation}

if we write:

\begin{equation} \hat{U}(t,\lambda) = \int e^{-i x \lambda} U(t,x) \dd{x} \end{equation}

which means we can write, with initial condtions:

\begin{equation} \hat{U} (t, \lambda) = \hat{f}(\lambda) e^{- t \frac{\lambda^{2}}{2}} \end{equation}

We want to reach a close form:

\begin{equation} U (t, x) = \frac{1}{\sqrt{2\pi} t} \int_{-\infty}^{\infty} f(y) e^{-\frac{(x-y)^{2}}{2t}} \dd{y} \end{equation}

This is the staging file for the midterm sheet, which I don’t usually publicise.

We have:

\begin{equation} \pdv[2]{u}{x} + \pdv[2]{u}{y} = 0 \end{equation}

Ignoring the boundary conditions when \(u(0,y)\), we know that we have Dirichlet boundaries in \(y\). This gives:

\begin{equation} u(x,0) = u(x,\pi) = 0 \end{equation}

Assuming our solution takes on the shape of \(u=X(x)Y(y)\), we obtain:

\begin{equation} X’’(x)Y(y) + Y’’(y)X(x) = 0 \end{equation}

by plugging in derivatives of that assumption; meaning:

\begin{equation} X’’(x)Y(y) = -Y’’(y)X(x) \end{equation}

This gives rise to:

\begin{align} \frac{X’’(x)}{X(x)} = -\frac{Y’’(y)}{Y(y)} = c \end{align}