# 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

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It's probably a typo. The differential equation should be $y''+\lambda y=0$, where $\lambda$ is a constant. –  Jim Belk Apr 15 '12 at 15:58
So the $\lambda$ is a constant to be defined after solving the equation? Could you please link me something that explains where this equation comes from? –  Fiat Lux Apr 15 '12 at 17:49
You should look for an introductory source on differential equations (which the notes you linked to are not). Alternatively, it may work better to read a sophomore-level physics book on mechanics (e.g. Barger and Olsson). –  Jim Belk Apr 15 '12 at 18:33

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 $$\psi(x,t) = \sum_{n=1}^\infty (a_n \sin n x \sin n t + b_n \sin n x \cos n t).\tag{1}$$ 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).

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Thanks for your beautiful exposition. If I understand well the $n$ in the solution of $y(x)$ originates from the value of the separation constant. But why should it be an integer? –  Fiat Lux Apr 15 '12 at 21:08
@FiatLux: You're very welcome. The solution $\cos k x$ is killed by the boundary condition $y(0) = B = 0$. $k = n$ is forced on us by the boundary condition $y(\pi) = A\sin k \pi = 0$, since the zeros of the sine function are at $n\pi$ for integer $n$. –  user26872 Apr 15 '12 at 21:18
Thanks for your addendum, but now I have two more questions: Why could you separate variables $\psi (x,t) =y(x)T(t)$ in the first step without losing solutions? Why the most general solution, a function of time and space that is periodic in space, is represented by this sum? For the second question a reference would be enough for me, I know this is connected to the Fourier transform but the fact there is also time involved confuses me a bit... –  Fiat Lux Apr 17 '12 at 23:09
@Fiat Lux: You're welcome. Separation of variables is equivalent to an eigenfunction expansion (the series in (1)). The eigenfunctions form a complete set, so we don't lose any solutions. The keyword for your second question is "Fourier series." After some study you will find that the coefficients in the Fourier series in (1) can be chosen so that the series converges almost everywhere to any function of the form given in the addendum. I recommend first studying one-dimensional Fourier series to get a handle on this. Here is a good place to start. –  user26872 Apr 18 '12 at 0:02

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.

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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 $A$.

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