On vibrating string and differentiating infinite sum

[psie]

TL;DR

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...

 

[pasmith]

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.

 

[psie]

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.

 

[BvU]

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 $n$ ?

$\$

 

[psie]

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 $(4)$.

 

[BvU]

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 . 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'

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

$\$

 

[psie]

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...

 

[pasmith]

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 g) term by term with respect to x.