Fourier… 3

Posted on November 8, 2010 by azumih

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A few years ago, I was with some students to read a textbook about audio engineering. I did not like the book in a few ways. One of the reasons is about treatment of wave equation.

1D wave equation looks something like this:

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

After presenting the equation, the book immediately declares its solution is given by

$\displaystyle u(x, t) = f(x -ct) + g(x + ct),$

for arbitrary function $f$ and $g$ , without any derivation, explanation no nothing. The solution comes down like an oracle. I still don’t like the way it is given.

This solution is usually called “d’Alembert’s Solution“. And in case his name chimes in, we can see a slightly better derivation.

The derivation goes like following. Introducing new variables

$\displaystyle \xi = x -ct$ ,

$\displaystyle \eta = x + ct$ ,

the PDE becomes

$\displaystyle \dfrac{\partial^2 u}{\partial \xi \partial \eta} = 0$ .

Integrating this with respect to $\xi$ gives

$\displaystyle \dfrac{\partial u}{\partial \eta} = G(\eta)$ ,

(G: arbitrary function of $\eta$ , independent of $\xi$ , since differentiating G with respect to $\xi$ gives $0$ ).

Then integrating above with respect to $\eta$ gives

$\displaystyle u = f(\xi) + g(\eta)$ ,

(f: arbitrary function of $\xi$ , and $g(\eta) = \int G(\eta) d \eta$ ). By substituting the variables back, we reach the solution

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

Okay. There is a bit of explanation above. It is better, only very slightly though, than nothing. But still $\xi$ and $\eta$ come down out of blue sky. It is unclear how this change of variables invented.

I recently saw Fourier comes for rescue, which is like

lll

Fourier… 2

Posted on September 30, 2010 by azumih

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.

Fourier, Laplace…

Posted on September 25, 2010 by azumih

Again this is quasi-mathematical imprecise thoughts about blah blah transforms.

I saw a book that has both Fourier and Laplace transforms in its title. The book says that Laplace transform is a generalisation of Fourier transform, in something like this way…

Let’s say $x(t)$ is a function that has such and such nice properties. Then there is another function $X(\omega)$ such that

$\displaystyle X(\omega) = \int_{-\infty}^\infty x(t) e^{-i \omega t}dt$ ,

which is called Fourier transform of $x(t)$ .

Then letting $i \omega = s$ , we have

$\displaystyle X(s) = \int_{-\infty}^\infty x(t) e^{-st}dt$ ,

which is called Lapalce trasnform. —I think $X(\omega)$ should have been $X(-i \omega)$ to make the things more straight.

Okay. The point is taken. $s$ is simpler than $i\omega$ . I must admit Laplace transform is more general. This seemed to be all spoken about relation between Fourier and Laplace transforms in the book.

But wait a second. Usually Laplace transform is introduced to us as a means for solving initial value problems of differential equations, isn’t it? I thought Fourier transform also should be talked about as a similar method for solving IVPs, to make relationship between these trasnforms clearer…

0909

und sofort ins unendliche

Tag Archives: Differential Equation

Fourier… 3

Fourier… 2

Fourier, Laplace…