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

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?

• Your edit would make sense if you had an infinite string and required it to oscillate only between two points without these points being physically fixed. This however is not your setup, so you get a series, not an integral. All three answers are correct. – safesphere May 22 at 5:58
• Forget about periodicity. Does the series converge to the function on the interval $[0,L]$? If yes, that's the only thing that matters. There is a very good reason to write the solution as a Fourier series when there are second partial derivatives of cartesian coordinates involved. – Felipe May 22 at 6:40

## 4 Answers

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.

• but the geometry of the string is not that of a circle. It's weird that if we idenfy x=0 and x=L, no physics is lost. – mithusengupta123 May 22 at 10:40
• @mithusengupta123 It’s because you’re applying circular boundary conditions. If you didn’t, then indeed something would have been lost. – knzhou May 22 at 16:55
• But my concern is exactly that. The geometry is not circular. How can you get away with using a circular geometry without losing physics? It's like pretending the geometry to be circular when it really is not. – mithusengupta123 May 23 at 5:04
• @mithusengupta123 If you want, just forget entirely that I said the word "circle". The Fourier series applies to functions defined on the interval $[0, a]$ such that $f(0) = f(a)$. Special cases of this include: periodic functions with period $a$, functions defined on a circle of circumference $a$, and a string of length $a$ with boundary conditions $f(0) = f(a)$, but these cases are all independent of each other. – knzhou May 23 at 5:38
• @mithusengupta123 Certainly, you can also take the perspective that the Fourier series is mainly for periodic functions (since that's the most common application). Then you would say that functions on the circle, and strings with boundary conditions $f(0) = f(a)$ can be corresponded with periodic functions. It's all just slightly different ways of saying the same thing. – knzhou May 23 at 19:38

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.

• I know that we can do a periodic extension. But this procedure is redefining the function in such a way that it becomes periodic. But note that the actual situation (the t=0 configuration between 0 to L) is not periodic at all. – mithusengupta123 May 22 at 2:13
• @mithusengupta123 The periodic extension is essentially just a formality. If the Fourier series expansion works for the periodic extension, then any particular period of the periodic extension can be described using that Fourier series. The aperiodic function is just one period of the periodic extension. – probably_someone May 22 at 2:25

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 $$[0,a]$$

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.

• This does not answer the question. Forget about $y(x,t)$. Note that in order to derive the expression for $y(x,t)$ you need to assume the Fourier expansion of $y(x,0)$, which is not periodic in space. That was my problem. The string does not extend beyond 0 and L. on either side. So you cannot say that $y(x,0)=y(x+L,0)$. – mithusengupta123 May 23 at 4:50
• That is the point. It is not periodic within $0 \ldots L$, but it is periodic in $-\infty \ldots +\infty$, with period of $2L$. The mathematical representation of reality is periodic, even if reality isn't. And that is good enough as it allows us to make future predictions. – John Alexiou May 23 at 5:11