On vibrating string and differentiating infinite sum

In summary, the discussion on vibrating strings involves the principles of wave mechanics, where the behavior of a string under tension can be analyzed through harmonic frequencies and modes of vibration. The mathematical representation often includes infinite series that describe the sum of sine and cosine functions, leading to a deeper understanding of the string's motion. Differentiating such infinite sums can reveal important characteristics of the waveforms, providing insights into the amplitude, frequency, and overall dynamics of the vibrating string system. This interplay between physical phenomena and mathematical analysis underscores the fundamental principles of wave theory.
  • #1
psie
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TL;DR Summary
I'm reading Vretblad's Fourier Analysis and its Applications. In a section on some basic applications of Fourier series, the author discusses the vibrating string in one example. I'm confused about a technical detail concerning differentiation of infinite sums.
Consider a homogeneous vibrating string of length ##\pi## fixed at both endpoints. The deviation from equilibrium is denoted ##u(x,t)## and the vibrations are assumed to be small so that they are at right angle to the ##x##-axis; gravitation is disregarded. The problem can be formulated as

\begin{align} u_{tt}&= u_{xx},\quad &0<x<\pi, \ t>0; \\ u(0,t)&=u(\pi,t)=0,\quad &t>0; \\ u(x,0)&=f(x),\quad &0<x<\pi; \\ u_t(x,0)&=g(x),\quad &0<x<\pi. \end{align}

I'm quoting now from the book:

The usual attempt ##u(x,t)=X(x)T(t)## this time leads up to this set of coupled problems:
$$\begin{cases}
X''(x)+\lambda X(x)=0, \\
X(0)=X(\pi)=0;
\end{cases}\qquad T''(t)+\lambda T(t)=0.$$ The ##X## problem is familiar by now; it has nontrivial solutions exactly for ##\lambda=n^2## (##n=1,2,3,\ldots##), viz., multiples of ##X_n(x)=\sin nx##. For these values of ##\lambda##, the ##T## problem is solved by ##T_n(t)=a_n\cos{nt}+b_n\sin{nt}.## Because of homogeneity we obtain the following solutions of the subproblem ##(1),(2)##: $$u(x,t)=\sum_{n=1}^\infty X_n(x)T_n(t)=\sum_{n=1}^\infty (a_n\cos{nt}+b_n\sin{nt})\sin nx.\tag 5$$ Letting ##t=0## in order to investigate ##(3)##, we get $$f(x)=u(x,0)=\sum_{n=1}^\infty a_n\sin{nx}.$$ Termwise differentiation with respect to ##t## and then substitution of ##t=0## gives for the second initial condition ##(4)## that $$g(x)=u_t(x,0)=\sum_{n=1}^\infty nb_n\sin nx.$$ Thus if we choose ##a_n## to be the sine coefficients of (the odd extension of) ##f##, and choose ##b_n## so that ##nb_n## are the corresponding coefficients of ##g##, then the series ##(5)## ought to represent the wanted solution.

Why can we differentiate ##(5)## termwise and why does it equal the differentiated function? A priori, we don't know the coefficients ##a_n## and ##b_n##, therefor I don't see how we can determine in what sense the series converges...
 
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  • #2
The assumption - because we reject any solution which dones't satisfy this as unphysical - is that the coefficients decay sufficiently fast for the solution to be at least twice differentiable in time (so that every point on the string has a defined velocity and acceleration). That entitles us to differentiate term by term.
 
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  • #3
pasmith said:
The assumption - because we reject any solution which dones't satisfy this as unphysical - is that the coefficients decay sufficiently fast for the solution to be at least twice differentiable in time (so that every point on the string has a defined velocity and acceleration). That entitles us to differentiate term by term.
Ok.

There is an exercise in this section of the book about the so-called plucked string, i.e. a point on the string is pulled from its resting position and then released with no initial speed. The mathematical formulation of the problem is ##(1)## to ##(4)## above, however, with $$f(x)=ax, \ 0\leq x\leq \frac12 \pi\quad f(x)=a(\pi-x), \ \frac12\pi\leq x\leq \pi$$ and ##g(x)=0##. The solution to that problem is $$u(x,t)=\frac{4a}{\pi}\sum_{n=1}^\infty \frac{(-1)^n}{(2k+1)^2}\cos(2k+1)t\sin(2k+1)x.$$ If we differentiate ##u## termwise with respect to ##t## or ##x## once, I don't think the differentiated series converges, right? To what degree is that a problem? Is ##u## only a formal solution to the exercise then? The author does remark that the model described above is unphysical and that with distributions as derivatives, the mathematical troubles go away.
 
  • #4
psie said:
The solution to that problem is$$u(x,t)=\frac{4a}{\pi}\sum_{n=1}^\infty \frac{(-1)^n}{(2k+1)^2}\cos(2k+1)t\sin(2k+1)x.$$
Doesn't look right. perhaps ##\displaystyle\sum_{n=0}^\infty## and not ##k## but ##n## ?

##\ ##
 
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  • #5
Yes, thanks, those were typos.

To make sense of the exercise above and the solution given, I guess one has to view it as a weak solution to the problem ##(1)## to ##(4)##.
 
  • #6
psie said:
Yes, thanks, those were typos.
but the brackets were a genuine omission. So we have$$
u(x,t)=\frac{4a}{\pi}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\; \cos\bigl( (2n+1)t\bigr )\; \sin\bigl((2n+1)x\bigr )\ ,$$which agrees with what we see e.g. here.

And you are worried about the convergence of the summation for ##u_t## $$
u_t(x,t)=\ldots \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}\; \sin\bigl( (2n+1)t\bigr ) \; \sin\bigl((2n+1)x\bigr )\ $$
As a physicist I don't even give that a second thought, but I can imagine that a mathematician needs convincing :wink:. Wasn't there a handy rule that exploits the ##(-1)^n## in combination with ##|\sin| \le 1## ?

That way the 'weak solution' can promote to 'the solution' :smile:

Note: as a physicist I know that there is a lot of approximation going on in this:
E.g. the summation for ##f(x)## must terminate: the string diameter is finite.
There will be damping with different factors for different ##n##
etc, etc

##\ ##
 
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  • #7
BvU said:
And you are worried about the convergence of the summation for ##u_t## $$
u_t(x,t)=\ldots \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}\; \sin\bigl( (2n+1)t\bigr ) \; \sin\bigl((2n+1)x\bigr )\ $$
Indeed, I'm worried about that equality holds here, because if you have $$s(x)=\sum u_n(x),$$ then first, ##s(x)## needs to converge pointwise and second, ##\sum u_n'(x)## needs to converge uniformly in order for ##s'(x)=\sum u_n'(x)##. There is the so-called M-test, although I'm not sure if it works here...
 
  • #8
You can use the method of characteristics to sidestep the need to justify differentiating a series with unknown cofficients term by term with respect to time and instead justify integrating a known series (the odd extension of the continuous function [itex]g[/itex]) term by term with respect to [itex]x[/itex].
 
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