Plucked string at $t=0$ and its Fourier decomposition

Question

Decomposing a function into a Fourier series is possible for periodic functions. Fourier transform, on the other hand, is used for aperiodic functions. How can we use Fourier series to analyse the initial configuration of a plucked string at $t=0$?

Edit The existing answer talks about periodic extension which I am aware of. To me, the periodic extension is a redefinition of an aperiodic function in such a way so as to make it periodic. We pretend that it is periodic while in the real problem it's not. For example, in the situation I described, the configuration of the string at $t=0$, between $x=0$ to $x=L$, is not repeated in space. Here, periodic extension is something we demand by brute force.

Why is there is no difference between an actual periodic function i.e., a periodically repeated pattern in space (for example, density in a crystal lattice) and that which is repeated by brute force?

Answer 1

Actually, you've gotten the issue backwards!

You complain that the Fourier series is illegitimate because we don't have an "actual" periodic function, so we repeat the function by "brute force". But that's not the right way to look at it. The Fourier series properly represents functions defined on the circle, i.e. functions $f(x)$ for $x \in [0, a]$ with $f(0) = f(a)$. That's true for the plucked string, because of the boundary conditions, so the Fourier series requires no modification of the function at all.

But isn't the "right" way to handle this a Fourier transform? No, neither practically or philosophically. Fourier transforms apply to functions defined on the real line, and the string is not defined on the real line, because it doesn't even exist outside the interval $[0, a]$. To apply a Fourier transform, you have to extend the definition of the function by brute force, while you don't have to do that for the Fourier series. Worse, the Fourier transform coefficients will be more complicated (as you can directly see from their definition), which is a consequence of this unnatural extension.

Answer 2

For any aperiodic function $f(x)$ defined on a finite interval $[0,a]$, we can compute the Fourier series of its periodic extension over the real numbers:

$$f_p(x)=f(x)\mod{a}$$

Unlike $f$, $f_p$ is periodic, and therefore there's no problem in constructing the Fourier transform.

Answer 3

We have the general solution $$y(x,t) = \sum_{n=0}^\infty sin(\frac{n\pi x}{a})(b_ncos(\frac{n\pi ct}{a})+c_nsin(\frac{n\pi ct}{a}))$$ For a string constrained to $y(0,t) = y(a,t) = 0$. We can then take the derivative of this with respect to time; $$y_t(x,t) = \sum_{n=0}^\infty \frac{n\pi c}{a}sin(\frac{n\pi x}{a})(-b_nsin(\frac{n\pi ct}{a})+c_ncos(\frac{n\pi ct}{a}))$$ Setting $t = 0$ in the first case removes the $c_n$ terms and so we can perform a cosine fourier series (although I presume that the plucking of the string means it is zero so $y(x,0) = 0$). Again with $t=0$, we remove now the $b_n$ terms, allowing us to perform another cosine fourier series to determine the other terms. This works as when one performs a fourier series, you don't do the integral over zero to infinity anyway, you do it over the periodic interval, so clearly in this case you have to do it over $[0,a]$

Answer 4

I am not sure what your main issue here is. Any function describing the deflection of a string for $x = 0 \ldots L$ can be assumed to be periodic as it can repeat itself for all other intervals, such as $x = 2L \ldots 3L$.

Specifically, a plucked string at $x_p$ has the form

$$ y(x,0) = Y_0 \begin{cases} \tfrac{x}{x_p} & 0 \leq x < x_p \\ 1-\tfrac{x-x_p}{L-x_p} & x_p \leq x \leq L \end{cases} \tag{1} $$

has the Fourier transform as

$$ y(x,t) = \sum_{i=1}^\infty Y_0 \tfrac{2 L^2}{i^2 \pi^2 x_p (L-x_p)} \sin \left( \tfrac{i \pi x_p}{L} \right) \sin\left( \tfrac{i \pi x}{L} \right) \cos \left( \tfrac{i \pi c t}{L} \right) $$

In plotting $y(x,0)/Y_0$ for $L=10$ and $x_p=3$ you see the periodicity for every 20 units of $x$. Below is the sum with 9, 27, 49, and 144 terms.