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}
Steps: recall we ended up at
\begin{equation} \hat{U} (t, \lambda) = \hat{f}(\lambda) e^{- t \frac{\lambda^{2}}{2}} \end{equation}
Let’s call:
\begin{equation} \hat{g}(\lambda) = e^{- t \frac{\lambda^{2}}{2}} \end{equation}
so we have:
\begin{equation} \hat{U} (t, \lambda) = \hat{f}(\lambda) \hat{g}(\lambda) \end{equation}
we can use convolution to figure U(t,x).
Recall that the Fourier transform of a Gaussian:
\begin{equation} \mathcal{F}\left(e^{-\frac{ax^{2}}{2}}\right) = \sqrt{\frac{2\pi}{a}}e^{-\frac{\lambda^{2}}{2a}} \end{equation}
Let’s first set:
\begin{equation} a = \frac{1}{t} \end{equation}
Which will give us that:
\begin{equation} g(x) = \frac{1}{\sqrt{2\pi t} } e^{-\frac{x^{2}}{2t}} \end{equation}
Meaning, with convolution:
\begin{equation} \mathcal{F}^{-1}(\hat{f} \hat{g}) = f * g \end{equation}
why does this make sense We are convolving a Gaussian against f(x). Meaning, at very small t , we are taking a very small window of size 1 against.
Heavyside function \begin{equation} f(x) = \begin{cases} 1, x\geq 0 \ 0, x<0 \end{cases} \end{equation}
This gives: if we split the room by x. Recall:
\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}
Given our f, this becomes:
\begin{equation} U (t, x) = \frac{1}{\sqrt{2\pi} t} \int_{0}^{\infty} e^{-\frac{(x-y)^{2}}{2t}} \dd{y} \end{equation}
If we change variables:
\begin{align} \frac{(x-y)^{2}}{2t} - \left( \frac{x}{\sqrt{2t}} - \frac{y}{\sqrt{2t}}\right)^{2} \end{align}
which means:
\begin{equation} z = \frac{y}{2\sqrt{t}} \end{equation}
\begin{equation} \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} e^{^{-\left(\frac{x}{\sqrt{2t}} - z\right)^{2}}} \dd{z} \end{equation}
and we will also apply:
\begin{equation} w = z - \frac{x}{\sqrt{2t}} \end{equation}
which will give:
\begin{equation} \frac{1}{\sqrt{\pi}} \int_{-\frac{x}{\sqrt{2t}}}^{\infty} e^{-w^{2}} \dd{w} \end{equation}
notice, as x increases, we are integrating more of a Gaussian, which will be exceedingly close to 1; as x decreases, we’ll get closer to 0. And also, t smoothed x out, which means as t increases the interface between 0 and 1 becomes smoother.
erf erf
convolution see convolution