---
title: "SU-MATH53 MAR112024"
source: https://www.jemoka.com/posts/kbhsu_math53_mar112024/
date: 2024-03-11
---

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&rsquo;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}\qty(e^{-\frac{ax^{2}}{2}}) = \sqrt{\frac{2\pi}{a}}e^{-\frac{\lambda^{2}}{2a}} \end{equation}
Let&rsquo;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&lt;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} - \qty( \frac{x}{\sqrt{2t}} - \frac{y}{\sqrt{2t}})^{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^{^{-\qty(\frac{x}{\sqrt{2t}} - z)^{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&rsquo;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
