Non-uniqueness of the solution of the equation for a plucked string I'm a bit confused about what is written in this PDF (in page 2). The author asserts that the differential equation $y'' +y = 0$ with boundary conditions $y(0)=0=y(\pi)$ has infinitely many solutions. Namely that $y(x) = A \sin(kx)$ is a solution for any $A\in\mathbb{R}$ and $k\in\mathbb{Z}$. Is he right? It seems to me that it works only for $k=1$ or $k=-1$.
Furthermore, for the more physically-oriented of you, this equation should represent "small oscillations of a plucked string", but I do not understand how it can be, because there is no dependence on time. I'd like to see where does this equation come from.
Thanks in advance for any response
 A: There is an error in the document. 
The solution to $y''+y=0$ with the boundary conditions $y(0) = y(\pi) = 0$ is unique up to a multiplicative factor, $y = A \sin x$. This is also not the wave equation. 
The partial differential equation describing the motion of a vibrating string is 
$$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2}$$
where $\psi = \psi(x,t)$ is the height of the string at position $x$ and time $t$, 
and where $c$ is the speed of the waves on the string.
($c = T/\rho$, where $T$ is the tension and $\rho$ is the linear density.) 
For convenience, set $c = 1$. 
Separate variables, $\psi(x,t) = y(x)T(t)$. 
We find 
$$\frac{y''}{y} = \frac{T''}{T}.$$
Since the LHS depends only on $x$ and the RHS only on $t$, each ratio must be equal to some constant (sometimes called the separation constant).
Call it $-k^2$. 
Thus, 
$$\begin{eqnarray*}
y'' + k^2 y &=& 0 \\
T'' + k^2 T &=& 0.
\end{eqnarray*}$$
The solutions $y(x)$ are of the form $y(x) = A \sin k x + B\cos k x$.
Impose the boundary conditions, $y(0) =y(\pi) = 0$. 
Thus, 
$$y(x) = A\sin n x$$
where $k = n = 1,2,\ldots$. 
This is the origin of our infinity of solutions. 
The solutions to the differential equation for $T$ will be of the form $T(t) = C \sin n t + D\cos n t$. 
(Notice the angular frequency is $k = n$, so the frequency of the $n$th solution is $n/(2\pi)$.)
Since we have not been given boundary conditions in time, the solution to the wave equation will be of the form
\begin{equation}
\psi(x,t) = \sum_{n=1}^\infty (a_n \sin n x \sin n t + b_n \sin n x \cos n t).\tag{1}
\end{equation}
There is an infinite tower of solutions. 
The $n=1$ solution is the fundamental mode of vibration of the string. 
The solutions for $n = 2, 3, \ldots$ correspond to the higher modes. 
Addendum: There is another way to see nonuniqueness. 
The general solution to the wave equation is 
$$\psi(x,t) = f(x-t) + g(x+t)$$
where $f$ and $g$ are arbitrary twice differentiable functions. 
($f(x-t)$ is a right-moving wave and $g(x+t)$ is left-moving.)
The boundary conditions imply
$$\begin{eqnarray*}
f(-t) + g(t) &=& 0 \\
f(\pi-t) + g(\pi + t) &=& 0. 
\end{eqnarray*}$$
Thus, the solution to the specified partial differential equation is
$$\psi(x,t) = f(x-t) - f(-x-t)$$
where $f(x)$ is any periodic function with period $2\pi$, $f(x+2\pi) = f(x)$. 
This solution corresponds, of course, to those functions attainable by the sum in equation (1). 
A: It must be that $y''+k^2 y=0$,(where $k$ is an integer). It might be a minor typo, I guess. Because the constant is arbitrary, there are infinitely many solutions. This type of equations appears in many linear PDEs. When we do separation of variables, we can get such kinds of several ODEs. Also in the heat equation like $u_t = u_{xx}$ with zero boundary condition, the steady-state is exactly what you are considering here. 
A: There is one other case where there might be infinitely many solutions satisfying the equation.  Consider the following:
$$ \left\{\begin{array}{l} y'' + y = 0\\
y(0) = y(\pi) = 0\end{array}\right.$$
This has infinitely many solutions, each of the form
$$y(x) = A \sin(x)$$
but there is no information allowing you to determine $A$.  Any value of $A$ would work, hence there are as many solutions as there are real numbers.  This would be an example of a differential equation with infinitely many solutions, but they have a unique shape (i.e. a constant times $\sin(x)$).
But we would not be able to claim that $y(x) = A\sin(kx)$ is a solution to the differential equation above, for any $A\in\mathbb{R}$ and any $k\in\mathbb{Z}$.  Indeed, choose $k=2$.  Then it is easy to verify that while $y(x) = A\sin(2x)$ satisfies the boundary value requirements (i.e. $y(0) = y(\pi) = 0$), this is not a solution to the original differential equation.  Hence the only argument against uniqueness is that the constant $A$ cannot be uniquely determined.
My guess is the author meant the differential equation to be $y'' + k^2y = 0$, which would allow solutions of the form $y = A\sin(kx)$, for any $A\in\mathbb{R}$, but $k$ would still be fixed.  So again, uniqueness would fail because there is no way to determine $A$.
