## Fourier… 2

Let me resume the Fourier thing.

Back and forth transform pair are:

$\displaystyle f(x) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i\omega x} d\omega$,

$\displaystyle \hat{f}(\omega) = \int_{-\infty}^{\infty} f(x) e^{-i\omega x} dx$.

Since I was concerning about solving ODE using Fourier transform in a similar manner as Lapalce’s, I want to see what the Fourier transform of derivatives (with respect to $x$) of $f(x)$ look like.  For the 1st derivative, I get:

$\displaystyle\hat{f\prime}(\omega) = \int_{-\infty}^{\infty} f\prime(x) e^{-i\omega x} dx = [f(x) e^{-i\omega x}]_{-\infty}^{\infty} -\int_{-\infty}^{\infty} f(x) (e^{-i\omega x})\prime dx$.

by using integration by parts.  If $f(x)$ vanishes at infinity ($\lim_{x \to \pm \infty}f(x) = 0$), then the 1st term becomes 0, so I get:

$\displaystyle\hat{f\prime}(\omega) = -\int_{-\infty}^{\infty} f(x) (e^{-i\omega x})\prime dx = i\omega \int_{-\infty}^{\infty} f(x) e^{-i\omega x} dx = i\omega \hat{f}(\omega)$

Now higher derivatives are easy.  Fourier transform of the 2nd derivative, for example, is:

$\displaystyle \hat{f\prime\prime}(\omega) = -\omega^2 \hat{f}(\omega)$.

Okay.  Let me pick a 2nd order ODE:

$\displaystyle u\prime\prime(x) + au\prime(x) + bu(x) = v(x)$.

Taking Fourier transform of both sides, I get:

$\displaystyle -\omega^2 \hat{u} + i a \omega \hat{u} + b\hat{u} = \hat{v}$,

$\displaystyle \hat{u}(\omega) = \dfrac{\hat{v}(\omega)}{-\omega^2 + i a \omega + b}$.

Finally I reach $f(x)$ by inverse-transforming both sides:

$\displaystyle u(x) = \dfrac{1}{2\pi}\int_{-\infty}^{\infty} \dfrac{\hat{v}(\omega)}{-\omega^2 + i a \omega + b} e^{i\omega x} d\omega$.

mmm… but is the integral on the right hand side looks so clumsy…  It is a double-integral actually.