# Motion of string fixed at both ends

I was reading about the Fourier analysis from Waves by Frank S Crawford Jr. But I got trapped at the very beginning; this is the excerpt that troubled me:

# Motion of string fixed at both ends.

Suppose that for $t\lt 0$, we constrain the string to follow a prescribed shape $f(z)$ by means of some sort of template. Then at $t= 0,$ we let the spring go by suddenly removing the template. Thus at $t=0$ each part of the string has its displacement $\psi(z,0)$ equal to $f(z)$ and has velocity equal to $v(z,0)$ equal to zero. Now the $n$th term of the velocity is proportional to $\sin(\omega_nt+\phi_n),$ which reduces to $\sin\phi_n$ at $t= 0.$ Thus we can make $v(z,0)=0$ for all $z$ simply by setting the phase constant $\phi_n$ equal to either zero or to $\pi.$ However, the phase constant $\phi_1=\pi$ (for example) is just equivalent to a minus sign affixed to $A_1.$ Therefore, we can satisfy these initial conditions is we set all phase constants to zero but allow the amplitudes $A_1,A_2,$ etc., to be either positive or negative. Thus, we have, for $v(z,0)= 0,$

$$\psi(z,t)= A_1\sin k_1z\cos\omega_1t +A_2\sin k_2z\cos\omega_2t+\ldots,\tag{38}$$ and at $t=0$

$$\psi(z,0)= A_1\sin k_1z + A_2\sin k_2 z+\ldots, \tag{39}$$

# Fourier series for function with zeros at both ends.

Now, the function $f(z)$ can be a very general function of $z.$ The only condition we specified was that it was to constrain the string. Therefore, virtually all we require of $f(z)= 0$ at $z=0$ and $z=L.$ We also require that $f(z)$ not be "jagged" on a "small" scale, since our wavefunction $\psi(z,t)$ is supposed to be a slowly varying function of $z.$ Therefore, $f(z)$ must be reasonably smooth in order for us to be able to use it to constrain the string and still the string to obey the differential equation that we obtained in the "continuous" approximation. Thus, we've found that any reasonable function $f(z)$ that vanishes at $z=0\;\&\; L$ can be expanded in a series of the form of Eq. $(39)$. Equation $(39)$ is called a Fourier series[...]

Our function $f(z)$ was used to constrain the string, and therefore it was defined only between $z=0\;\&\;L.$ However, the functions $\sin k_1 z,~ \sin 2k_1 z,~\sin 3k_1 z,$ etc., that make up the infinite series of Eq.$(39)$ are defined for all $z$ from $-\infty$ to $+\infty.$[...]

What I've not conceived is

1. So, what is $f(z)?$ Crawford told all the story here solely of $f(z).$ What is it actually? I thought $\psi(z,t)$ is the wavefunction- the general state of the string which is just the superpositions of infinite modes as in Eq.$(38);$ where did $f(z)$ come from?? Crawford mentions that $f(z)$ is for $t\lt 0$. Then why is, at $t=0$, the string described by $f(z)$?? I'm not understanding that; first he mentioned that $f(z)$ is some sort of constraining the string at $t\lt 0$; then suddenly he wrote at $t= 0,$ the string is described by $f(z)$. Why is it so? Also, why should the velocity $v(z,t)$ be $0$ at $t= 0?$

2. Why should not $f(z)$ "jagg"? Why should it be "reasonably smooth"?

3. He mentioned about $\sin k_1 z,~ \sin 2k_1 z,~\sin 3k_1 z,$ "that make up in Eq.(39)"; but I'm not seeing them in Eq.$(39).$ Could anyone point out those in Eq.$(39)$? I'm just seeing one term containing $k_1$- only the first term.

• A hint is that $f(z)$ should always obey the boundary conditions. – ja72 Oct 28 '15 at 18:23
• $f(z)=\psi(z,0)$ is the initial condition... Oops... ACuriosMind took care of that. With regards to "smoothness", there is an entire mathematical discipline called "functional analysis", which tries to extend things like the Fourier transform to ever less well behaved functions and distributions, that, however, turns out to be a very difficult topic all in itself and it is of little interest to physics, which can almost always pretend that things are smooth enough (there are exceptions to that). – CuriousOne Oct 28 '15 at 18:25

1. $f$ is the initial condition for the string. If you have some differential equation (e.g. the wave equation) for the shape of the string varying in time, i.e. $\psi(x,t)$, you will require some initial condition to get a unique solution. $f$ is defined to be that initial condition, i.e. $f(z) = \psi(z,0)$. That the velocity is zero is part of what we suppose - if the string was fixed to the shape $f$ until $t=0$, it does not move in the instant $t=0$, which one might formally see by taking the left-derivative of $\psi(x,t)$ with $\psi(x,t) = f(x)$ for $t \leq 0$. Since $f$ and $\psi$ are "reasonably smooth" the first derivative - the velocity - exist and is equal to the left-derivative, hence it vanishes at $t=0$.
2. Since it is an initial condition for a differential equation, we must be able to take the derivatives of $f$ that occur in the equation in the first place. So "reasonably smooth" means "smooth enough to take all relevant derivatives".
3. $(39)$ is the infinite Fourier series $$\psi(x,0) = \sum_{n\in\mathbb{N}} A_n\sin(k_n x)$$ containing each of the functions $f_i(x) = \sin(k_i x),i\in\mathbb{N}$ as a summand. If the string has length $L$, then, by the general way Fourier series are constructed, we have $k_n = \frac{2\pi}{L}n$, and hence $k_n = nk_1$.
• Crawford wrote that $f$ is for $t\lt 0$; then why should it be for $t= 0$ also? – user36790 Oct 28 '15 at 18:30
• @user36790: By continuity. If $\psi(x,t) = f(x)$ for $t < 0$, then if $\psi$ is continuous, $\psi(x,0) = f(x)$ follows since we have $\psi(x,0) = \psi(x,-\lim_{n\to\infty}\frac{1}{n}) = \lim_{n\to\infty}\psi(x,-\frac{1}{n})= \lim_{n\to\infty} f(x) = f(x)$. – ACuriousMind Oct 28 '15 at 18:34
• +1; If it is continuous & of-course, wavefunction $\psi$ is continuous. But why is velocity zero at $t= 0?$ I think it is also due to continuity as prior to $t= 0,$ the string was set fixed by some template. Is it so? Doesn't 'reasonably smooth' & 'slowly varying' refer to the continuity of $\psi?$ – user36790 Oct 28 '15 at 18:51
• @user36790: Added something, but it's essentially the same argument as why $\psi(x,t) = f(x)$ at $t=0$, just by differentiability, not continuity. $\psi$ must be "reasonably smooth" (in both $x$ and $t$) to solve the wave equation, and thus $f$ is reasonably smooth, too, since $\psi(x,0) = f(x)$. – ACuriousMind Oct 28 '15 at 19:08