solving the wave equation due to D. Bernoulli

A string has been strained between the points $ (0,\,0)$ and $ (p,\,0)$ of the $ x$-axis. The transversal vibration of the string in the $ xy$-plane is determined by the one-dimensional wave equation

$\displaystyle \frac{\partial^2u}{\partial t^2} = c^2\cdot\frac{\partial^2u}{\partial x^2}$ (1)

satisfied by the ordinates $ u(x,\,t)$ of the points of the string with the abscissa $ x$ on the time moment $ t\,(\geqq 0)$. The boundary conditions are thus

$\displaystyle u(0,\,t) = u(p,\,t) = 0.$

We suppose also the initial conditions

$\displaystyle u(x,\,0) = f(x),\quad u_t'(x,\,0) = g(x)$

which give the initial position of the string and the initial velocity of the points of the string.

For trying to separate the variables, set

$\displaystyle u(x,\,t) := X(x)T(t).$

The boundary conditions are then $ X(0) = X(p) = 0$, and the partial differential equation (1) may be written

$\displaystyle c^2\cdot\frac{X''}{X} = \frac{T''}{T}.$ (2)

This is not possible unless both sides are equal to a same constant $ -k^2$ where $ k$ is positive; we soon justify why the constant must be negative. Thus (2) splits into two ordinary linear differential equations of second order:

$\displaystyle X'' = -\left(\frac{k}{c}\right)^2 X,\quad T'' = -k^2T$ (3)

The solutions of these are, as is well known,

\begin{align*}\begin{cases}X = C_1\cos\frac{kx}{c}+C_2\sin\frac{kx}{c}\\ T = D_1\cos{kt}+D_2\sin{kt}\\ \end{cases}\end{align*} (4)

with integration constants $ C_i$ and $ D_i$.

But if we had set both sides of (2) equal to $ +k^2$, we had got the solution $ T = D_1e^{kt}+D_2e^{-kt}$ which can not present a vibration. Equally impossible would be that $ k = 0$.

Now the boundary condition for $ X(0)$ shows in (4) that $ C_1 = 0$, and the one for $ X(p)$ that

$\displaystyle C_2\sin\frac{kp}{c} = 0.$

If one had $ C_2 = 0$, then $ X(x)$ were identically 0 which is naturally impossible. So we must have

$\displaystyle \sin\frac{kp}{c} = 0,$

which implies

$\displaystyle \frac{kp}{c} = n\pi \quad (n \in \mathbb{Z}_+).$

This means that the only suitable values of $ k$ satisfying the equations (3), the so-called eigenvalues, are

$\displaystyle k = \frac{n\pi c}{p} \quad (n = 1,\,2,\,3,\,\ldots).$

So we have infinitely many solutions of (1), the eigenfunctions

$\displaystyle u = XT = C_2\sin\frac{n\pi}{p}x
\left[D_1\cos\frac{n\pi c}{p}t+D_2\sin\frac{n\pi c}{p}t\right]$

or

$\displaystyle u = \left[A_n\cos\frac{n\pi c}{p}t+B_n\sin\frac{n\pi c}{p}t\right]
\sin\frac{n\pi}{p}x$

$ (n = 1,\,2,\,3,\,\ldots)$ where $ A_n$'s and $ B_n$'s are for the time being arbitrary constants. Each of these functions satisfy the boundary conditions. Because of the linearity of (1), also their sum series

$\displaystyle u(x,\,t) := \sum_{n=1}^\infty\left(A_n\cos\frac{n\pi c}{p}t+B_n\sin\frac{n\pi c}{p}t\right)\sin\frac{n\pi}{p}x$ (5)

is a solution of (1), provided it converges. It fulfils the boundary conditions, too. In order to also the initial conditions would be fulfilled, one must have

$\displaystyle \sum_{n=1}^\infty A_n\sin\frac{n\pi}{p}x = f(x),$

$\displaystyle \sum_{n=1}^\infty B_n\frac{n\pi c}{p}\sin\frac{n\pi}{p}x = g(x)$

on the interval $ [0,\,p]$. But the left sides of these equations are the Fourier sine series of the functions $ f$ and $ g$, and therefore we obtain the expressions for the coefficients:

$\displaystyle A_n = \frac{2}{p}\int_{0}^p\!f(x)\sin\frac{n\pi x}{p}\,dx,$

$\displaystyle B_n = \frac{2}{n\pi c}\int_{0}^p\!g(x)\sin\frac{n\pi x}{p}\,dx.$

Bibliography

1
K. V AIS AL A: Matematiikka IV. Hand-out Nr. 141.    Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).



Contributors to this entry (in most recent order):

As of this snapshot date, this entry was owned by bloftin.