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,

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2 Responses to Fourier… 4

  1. Nesha says:

    Nice derivation! It gives straightforward explanation about motivation for usage of change of variables. The question that may be interesting is where the motivation for usage of Fourier Transform to solve PDE comes from in the first step? It apparently must be obtained by some kind of trial and error experiments, and then, when it worked, it has been used for solution. Similar question can be asked for the solution of differential equation in frequency domain. Where that solution comes from? It’s a function guess that worked, in my view.

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