The Fourier transform is defined for all integrable functions as

\begin{align}
\widehat{f}(z) = \int_{-\infty}^{\infty} f(x) e^{-ixz} dx, \quad z\in \mathbb{R} \tag{1}
\end{align}

The inverse Fourier transform
$$f(x) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \widehat{f}(x) e^{ixz} dx, \quad x\in \mathbb{R}\tag{2}$$

Theorem

If the shift of $f(x)$ is
$$Ef(x) = f(x+h)$$
then the Fourier transform of the shifted function is
$$\widehat{E f}(z) = e^{ihz}\widehat{f}(z)\tag{3}$$

If
$$Df(x) \equiv f'(x), \quad \text{ then } \quad \widehat{D f}(z) = iz f'(z) \tag{4}$$

the transform of the derivative is the derivative of the transform

This is a very important statement that deserves to be stated in words and with the mathematical nomenclature

the transform of the derivative $\widehat{D }(\cdot)$
is,
the derivative (achieved by multiplication by $iz$) of the transform $\widehat{f}(\cdot)$.

Each time we differentiate, we multiply by $iz$. What are we multiplying? The transform! In Fourier transform space, differentiation is a simple multiplication exercise.

Proof

1. Fourier transform of the shifted function

\begin{align}
\widehat{Ef}(x) & = \int_{-\infty}^{\infty} Ef(x) e^{-ixz} dx\\
& = \int_{-\infty}^{\infty} f(x + h) e^{-ixz} dx\\
\text{ Let }\qquad \qquad \qquad \qquad y &= x + h\\
\widehat{Ef}(x) & = \int_{-\infty}^{\infty} f(y) e^{-i(y \; – h)z} dy
\end{align}
where $e^{-iyz} \equiv e^{-ixz} $. Just a change of notation.

$$\therefore \qquad \widehat{Ef}(x) = e^{ihz}\widehat{f}(z)$$

 

2. Fourier transform of the derivative of $\widehat{f}(z)$

We defined the Fourier transform $(1)$ as
\begin{align}
\widehat{f}(z)& = \int_{-\infty}^{\infty} f(x) e^{-ixz} dx\\
\text{ the modulus, }\qquad \qquad \qquad \qquad |e^{-ixz}| &= \sqrt{\underbrace{\cos^2 (xz)}_{\text{real part}} + \underbrace{\sin^2 (xz)}_{\text{imaginary part}} } = \sqrt{1} = 1
\end{align}

Therefore the integral exists which implies that $\lim_{x \rightarrow \pm \infty} F(x) = 0$.
By differentiating both sides of $(1)$,
\begin{align}
\widehat{D f}(z) & = \int_{-\infty}^{\infty} f'(x) e^{-ixz} dx\\
&=\underbrace{\biggr[ f(x) e^{-ixz} \bigg]_{x = -\infty}^{x = \infty}}_{ = 0} + iz \int_{-\infty}^{\infty} f(x) e^{-ixz} dx
\end{align}

(by parts)

Application of the theorem

We can apply this theorem to the second central difference approximation
$$\dfrac{f(x+h) – 2f(x) + f(x-h) }{h^2} = g(z)$$

where $g(z)$ is the second central difference approximation comprised of shifted terms.
\begin{align}
\widehat{g}(z) &= \dfrac{1}{h^2}(e^{ihz} \;-2 + e^{-ihz}) \widehat{f}(z)\\
&= \dfrac{1}{h^2}( \cos {hz} + i\sin (hz) \; – 2 + \cos (-hz) + i\sin (-hz) ) \\
&= \dfrac{1}{h^2}( \cos {hz} + i\sin (hz) \; – 2 + \cos (hz) \; -\; i\sin (hz) )\\
&= \dfrac{1}{h^2}( 2\cos (hz) \; – 2 )
\end{align}

$$
\begin{array}{r|r}
S&A \\
\hline
T&C \\
\end{array}
$$

so instead of a partial differential equation, it is an ordinary differential equation which now needs to be sovled. Thinking of $z$ as fixed,

\begin{align}
\hat u(z,t) = \hat u(z, o) e^{-\frac{1}{2} z^2 t}
\end{align}