I have a question about pg.10 from Stein and Shakarchi's fourier analysis.
We must now connect this result with our original problem, that is, the physical motion of a string. There, we imposed the restrictions $0 \leq x \leq \pi$, the initial shape of the string $u(x, 0) = f (x)$, and also the fact that the string has fixed end points, namely $u(0, t) = u(\pi, t) = 0$ for all $t$. To use the simple observation above, we first extend $f$ to all of $\mathbb{R}$ by making it odd on $[−\pi,\pi]$, and then periodic in $x$ of period $2\pi$, and similarly for $u(x, t)$, the solution of our problem. Then the extension $u$ solves the wave equation on all of $\mathbb{R}$, and $u(x, 0) = f(x)$ for all $x \in \mathbb{R}$. Therefore, $u(x, t) = F (x + t) + G(x − t)$, and setting $t = 0$ we find that $$F(x) + G(x) = f(x).$$ Since many choices of $F$ and $G$ will satisfy this identity, this suggests imposing another initial condition on $u$ (similar to the two initial conditions in the case of simple harmonic motion), namely the initial velocity of the string which we denote by $g(x)$: $$\frac{\partial u}{\partial t}(x,0) = g(x),$$ where of course $g(0) = g(\pi) = 0$. Again, we extend $g$ to $\mathbb{R}$ first by making it odd over $[−\pi,\pi]$, and then periodic of period $2\pi$. The two initial conditions of position and velocity now translate into the following system: $$F(x) + G(x) = f(x), \quad F'(x) − G'(x) = g(x).$$
It seemed okay to me when it first assumed $f(x)$ to be odd (although I'm not sure why this is allowed), when it assumed the velocity $g(x)$ to be also odd, I don't understand why this can be assumed. Just by assuming $f(x,t)=\sin(x+t)$ it tells me $g(x,t)$ has to be even.
Can I get more explanation on why these assumptions are valid?
Thank you.
The reason you want to make the position odd about $x=0$ is to ensure the extended function over $-\pi\leq x \leq \pi$ is smooth. Otherwise, being even while also having $f(0)=0$ would add kink to the function -- an unnecessary complication that will not help with a Fourier series approximation.
Why odd velocity? For a similar reason. We have fixed $g(0)=0$ for all $t$, so it would be physically odd to have both sides of the "extended" string rise up and down with the same velocity -- that that this couldn't happen but it is again imposing a kink that will kill smoothness of the $g(x)$.
Finally, the authors are also relying on some good ol' physical intuition here to guide the most sensible choice.