## Fourier… 4

Note:  This is not really a strict mathematical discussion, but a quasi-math.

My post about Fourier thing got a nice interesting comment about a week ago, which reminded me that I have not settled the theme yet, and that I suspended the topic even before mentioning Fourier transform.  Actually the post was not really about Fourier, but a complaint about a sound engineering textbook that discusses wave equations a lot.  So, let me get back to the main trail.

The 1D wave equation of my interest looked like

$\displaystyle \dfrac{\partial^2 u}{\partial t^2} = c^2 \dfrac{\partial^2 u}{\partial x^2}$.

And I was about to take Fourier transform of this equation, with respect to $x$.  So let us just do that.  The left hand side becomes

$\displaystyle \int_{-\infty}^{\infty} \dfrac {\partial^2 u}{\partial t^2} e^{-i \omega x} dx = \dfrac {\partial^2}{\partial t^2} \int_{-\infty}^{\infty} u e^{-i \omega x} dx = \dfrac {\partial^2}{\partial t^2} \hat{u}$

(here $\hat{u}$ denotes Fourier transform of $u$),  and F.T. of the right hand side is

$\displaystyle c^2 \int_{-\infty}^{\infty} \dfrac {\partial^2 u}{\partial x^2} e^{-i \omega x} dx$.

Here let me rely on a wishful thinking again that $u(x, t)$ vanishes when $x$ tends to  $\pm \infty$.  Then Fourier transform of 1st derivative of a function $u$ is

$\displaystyle \widehat {\dfrac{\partial u}{\partial x}} = \int_{-\infty}^{\infty} \dfrac {\partial u}{\partial x} e^{-i \omega x} dx = u e^{-i \omega x}] _{-\infty}^{\infty} + i \omega \int_{-\infty}^{\infty} u e^{-i \omega x} dx = i \omega \hat {u}$,

(the term $u e^{-i \omega x}] _{-\infty}^{\infty}$ vanishes).  Likewise, F.T. of the 2nd derivative is

$\displaystyle \widehat {\dfrac{\partial^2 u}{\partial x^2}} = - \omega^2 \hat {u}$.

So F.T. of the right hand side of the wave equation becomes

$\displaystyle -(\omega c)^2 \hat{u}$.

Since both sides should be equal, we get another differential equation in frequency domain

$\displaystyle \dfrac {\partial^2}{\partial t^2} \hat{u} = -(\omega c)^2 \hat{u}$.

This differential equation is actually only about $t$, and can be solved fairly easily.  Its solution is given by:

$\displaystyle \widehat{u(\omega, t)} = F(\omega) e^{-i \omega c t} + G(\omega) e^{i \omega c t}$,

where $F$ and $G$ are arbitrary functions of $\omega$, independent of $t$.

Then inverse-transform of this will give me the solution to the original wave equation.  Computing the inverse gives:

$\displaystyle u(x, t) = \dfrac{1}{2 \pi} (\int_{-\infty}^{\infty} F(\omega) e^{-i \omega c t} e^{i \omega x}d \omega + \int_{-\infty}^{\infty} G(\omega) e^{i \omega c t} e^{i \omega x}d \omega)$

$\displaystyle = \dfrac{1}{2 \pi} (\int_{-\infty}^{\infty} F(\omega) e^{i \omega (x - c t)} d \omega + \int_{-\infty}^{\infty} G(\omega) e^{-i \omega (x + c t)} d \omega)$.

Well here we note the $x - ct$ and $x + ct$ came up again, that were the sources of change of variable in the earlier post, this time in a natural way by a simple computation.

Okay, though the integral above might look scary, it can reach back to the definition of inverse Fourier transform by the same change of variables:

$\displaystyle X_1 = x - ct$,

$\displaystyle X_2 = x + ct$.

By this, the integral becomes (no complaints this time):

$\displaystyle u = \dfrac{1}{2 \pi}[\int_{-\infty}^{\infty} F(\omega) e^{i \omega X_1} d \omega + \int_{-\infty}^{\infty} G(\omega) e^{-i \omega X_2} d \omega]$.

Since the 2 terms are just definition of the inverse-transform of $F$ and $G$ respectively, letting them be $f$ and $g$ finally we get

$\displaystyle u(x, t) = f(X_1) + g(X_2)$

$\displaystyle = f(x - ct) + g(x + ct)$.

Now I got a bit happier than before…  But wait a second,