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- 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:
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...
\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...