The Gaussian, in general, gives:
\begin{equation} e^{-\frac{ax^{2}}{2}} \end{equation}
which is a Bell-Shaped curve. It’s pretty darn important
solving heat equation without boundary
for general expression:
\begin{equation} \pdv{U}{t} = \alpha \pdv[2]{U}{x} \end{equation}
\begin{equation} U(t,x) = \frac{1}{\sqrt{4\pi \alpha t}}\int_{\mathbb{R}} f(y) e^{-\frac{(x-y)^{2}}{4\alpha t}} \dd{y} \end{equation}
where,
\begin{equation} \hat{U}(t,\lambda) = \hat{f}(\lambda)e^{-\alpha t \lambda^{2}} \end{equation}
\begin{equation} \hat{U}(t,\lambda) = \hat{f}(\lambda)e^{-\lambda^{2}(t)} \end{equation}
Heat Equation and Gaussian
\begin{equation} H(t,x) = \frac{1}{\sqrt{2\pi} t}e^{-\frac{x^{2}}{2t}} \end{equation}
You will note that \(H\) does satisfy the heat equation:
\begin{equation} \pdv{U}{t} = \pdv[2]{U}{x} \end{equation}
closed form solution
\begin{equation} U(t,x) = \frac{1}{\sqrt{2\pi} t} \int_{\mathbb{R}} f(y) e^{-\frac{(x-y)^{2}}{2t}} \dd{y} \end{equation}
this is exactly:
\begin{equation} \int_{\mathbb{R}}f(y) H(t,(x-y)) \dd{y} = \int_{\mathbb{R}}\frac{1}{\sqrt{2\pi} t}e^{-\frac{(x-y)^{2}}{2t}} f(y) \dd{y} \end{equation}
We can understand this when \(t \to 0\), where there is a single, narrow, area \(1\) band which we sweep across all of \(y\). Because its thin and \(1\), its basically \(f(x)\) at each \(y\).
solving Heat Equation without boundary
Consider the partial Fourier Transform on the \(x\) variable of the heat equation.
\begin{equation} U(t,x) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{ix\lambda} \hat{U} \qty(t,\lambda) \dd{\lambda} \end{equation}
Taking derivatives of this:
\begin{equation} \pdv{U}{t} (t,x) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{i\lambda x} \pdv{\hat{U}}{t} (t,\lambda) \dd{\lambda} \end{equation}
and:
\begin{equation} \pdv[2]{U}{x} = \frac{1}{2\pi} \int_{\mathbb{R}} \qty(-\lambda^{2}) e^{ix \lambda } \hat{U}(t,\lambda) \dd{\lambda} \end{equation}
Because these two are equal, it gives us that:
\begin{equation} \hat{U}(t,\lambda) = -\lambda^{2} \hat{U}(t,\lambda) \end{equation}
meaning:
\begin{equation} \hat{U}(t,\lambda) = a(\lambda)e^{-\lambda^{2}t} \end{equation}
Finally, at:
\begin{equation} \hat{U}(0,\lambda) = a(\lambda) = \hat{f}(\lambda) \end{equation}
We see that:
\begin{equation} \hat{U}(t,\lambda) = \hat{f}(\lambda)e^{-\lambda^{2}(t)} \end{equation}
To get our original function back, we need to inverse Fourier transform it:
\begin{equation} U(t,x) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{ix\lambda - \lambda^{2}t} \hat{f}(\lambda) \dd{\lambda} \end{equation}
Integrating Gaussian, more Generally
Let’s integrate:
\begin{equation} \int_{-\infty}^{\infty} e^{-\frac{{ax}^{2}}{2}} \dd{x} \end{equation}
Let’s replace: \(s = \sqrt{a} x\)
This gives us that (based on Integrating Gaussian):
\begin{equation} x = \sqrt{\frac{2\pi}{a}} \end{equation}
If we replace \(a\) by \(\frac{1}{t}\), we obtain:
\begin{equation} \frac{1}{\sqrt{2\pi}t} \int_{-\infty}^{\infty} e^{-\frac{x^{2}}{2t}} \dd{x} = 1 \end{equation}
by rescaling \(x(a)\) function above.
If \(t\) increases, you will see that this function diffuses from a single point at \(0\) and spreading out. Notice, that over the whole real line, no matter what the \(t\) is, you always end up with integral \(1\).
Integrating Gaussian
Let’s integrate:
\begin{equation} \int_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}} \dd{x} \end{equation}
computing this is funny:
\begin{equation} A \cdot A = \int_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}} \dd{x} \int_{-\infty}^{\infty} e^{-\frac{y^{2}}{2}} \dd{y} \end{equation}
We can think of this as a double integral:
\begin{equation} A \cdot A = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}} e^{-\frac{y^{2}}{2}} \dd{x} \dd{y} \end{equation}
meaning we get:
\begin{equation} A \cdot A = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^{2}+y^{2}}{2}} \dd{x} \dd{y} \end{equation}
Its polar time; recall:
\begin{equation} x^{2} + y^{2} = r^{2} \end{equation}
we can now go over this whole thing by converting into polar (notice the extra factor \(r\)):
\begin{equation} A \cdot A = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-\frac{r^{2}}{2}} r \dd{r} \dd{\theta} \end{equation}
very suddenly we can use u sub on \(r\) to obtain:
\begin{equation} 2\pi \int_{0}^{\infty} e^{-u} \dd{u} = 2\pi \cdot 1 = 2\pi \end{equation}
Meaning:
\begin{equation} A = \sqrt{2\pi} \end{equation}