Solution
According to the cooling formula:
\[\frac{\partial u}{\partial t} = C \cdot \frac{\partial^2 u}{\partial x^2}\]
Applying Fourier transform for spacial variable \(x\) on both sides:
\[\begin{split}\mathcal{F} (\frac{\partial u}{\partial t}) (s) & =
\int_{-\infty}^{\infty} e^{-2 \pi i s x}
\frac{\partial}{\partial t} u(x, t) dx
\\ & =
\frac{\partial}{\partial t}
\int_{-\infty}^{\infty} e^{-2 \pi i s x} u(x, t) dx
\\ & =
\frac{\partial}{\partial t} U(s, t)\end{split}\]
where \(U(s, t) = \mathcal{F} u (s, t)\)
\[ \begin{align}\begin{aligned}\begin{split}\mathcal{F} (C \cdot \frac{\partial^2 u}{\partial x^2}) (s) & =
C \cdot \int_{-\infty}^{\infty} e^{-2 \pi i s x}
\frac{\partial^2}{\partial x^2} u(x, t) dx
\\ & =
C \cdot (2 \pi i s)^2
\int_{-\infty}^{\infty} e^{-2 \pi i s x} u(x, t) dx
\\ & =
-4 C \pi^2 s^2 U(s, t)\end{split}\\\begin{split}\\\end{split}\end{aligned}\end{align} \]
Therefore we have a differential equation:
\[\frac{\partial}{\partial t} U(s, t) = -4 C \pi^2 s^2 U(s, t)\]
whose solution:
\[U(s, t) = F(s) \cdot G(s, t)\]
where \(F(s) = \mathcal{F} f(x)\) and \(G(s, t) = e^{-4 C \pi^2 s^2 t}\).
Notice that \(G(s, t)\) is a Gaussian function of variable \(s\), which can well be
denoted as a Fourier transform of another Gaussian function \(g(x, t)\) of
variable \(x\).
Suppose:
\[ \begin{align}\begin{aligned}g(x, t) = A(t) e^{(- \pi (a(t) \cdot x)^2)} = A(t) e^{- \pi a^2 (t) x^2}\\\begin{split}\\\end{split}\\\begin{split}\therefore
G(s, t) & =
(\mathcal{F} g) (s, t)
\\ & =
(\mathcal{F} (A(t) e^{- \pi a^2 (t) x^2})) (s)
\\ & =
A(t) \cdot \frac{1}{|a(t)|} e^{- \pi \frac{s^2}{a^2(t)}}
\\ & =
e^{-4 C \pi^2 s^2 t}\end{split}\end{aligned}\end{align} \]
By solving the system of equations:
\[\begin{split}\begin{cases}
-4 C \pi^2 t = \frac{\pi}{a^2 (t)}
\\
A(t) = |a(t)|
\end{cases}\end{split}\]
we have:
\[ \begin{align}\begin{aligned}A(t) = a(t) = \frac{1}{2 \sqrt{C \pi t}}\\\begin{split}\\\end{split}\\\therefore
g(x, t) = \frac{1}{2 \sqrt{C \pi t}} e^{- \frac{x^2}{4 C t}}\\\begin{split}\\\end{split}\\\begin{split}\therefore
U(s, t) & =
F(s) \cdot G(s, t)
\\ & =
\mathcal{F} f (s) \cdot \mathcal{F} g (s, t)
\\ & =
\mathcal{F} (f (x) * g(x, t))\end{split}\\\begin{split}\\\end{split}\\\begin{split}\therefore
u(x, t) & =
f(x) * g(x, t)
\\ & =
f(x) * \frac{1}{2 \sqrt{C \pi t}} e^{- \frac{x^2}{4 C t}}\end{split}\end{aligned}\end{align} \]
Back to Fourier Transform.