solving the wave equation due to D. Bernoulli

A string has been strained between the points  $(0,\mathrm{\hspace{0.17em}0})$  and  $(p,\mathrm{\hspace{0.17em}0})$  of the $x$-axis.  The vibration of the string in the $xy$-plane is determined by the one-dimensional wave equation

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

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

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

We suppose also the initial conditions

$$u(x,\mathrm{\hspace{0.17em}0})=f(x),u_t^{\prime }(x,\mathrm{\hspace{0.17em}0})=g(x)$$

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

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

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

$c^2\cdot {\displaystyle \frac{X^{\prime \prime }}{X}}={\displaystyle \frac{T^{\prime \prime }}{T}}.$ (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:

$X^{\prime \prime }=-{\left({\displaystyle \frac{k}{c}}\right)}^{2}X,T^{\prime \prime }=-k^{2}T$ (3)

The solutions of these are, as is well known,

$\{\begin{array}{cc}X=C_1\cos \frac{kx}{c}+C_2\sin \frac{kx}{c}\hfill & \\ T=D_1\cos kt+D_2\sin kt\hfill & \end{array}$ (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  $T=D_1{e}^{kt}+D_2{e}^{-kt}$  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

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

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

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

which implies

$$\frac{kp}{c}=n\pi \mathit{\hspace{1em}}(n\in {\mathbb{Z}}_{+}).$$

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

$$k=\frac{n\pi c}{p}\mathit{\hspace{1em}}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots }).$$

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

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

or

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

$(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots })$ 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

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

$$\sum _{n=1}^{\mathrm{\infty }}{A}_n\sin \frac{n\pi }{p}x=f(x),$$

$$\sum _{n=1}^{\mathrm{\infty }}{B}_n\frac{n\pi c}{p}\sin \frac{n\pi }{p}x=g(x)$$

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:

$$A_n=\frac{2}{p}{\int }_0^{p}f(x)\sin \frac{n\pi x}{p}dx,$$

$$B_n=\frac{2}{n\pi c}{\int }_0^{p}g(x)\sin \frac{n\pi x}{p}dx.$$

Title	solving the wave equation due to D. Bernoulli
Canonical name	SolvingTheWaveEquationDueToDBernoulli
Date of creation	2013-03-22 16:31:41
Last modified on	2013-03-22 16:31:41
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	12
Author	pahio (2872)
Entry type	Example
Classification	msc 35L05
Synonym	vibrating string
Related topic	ExampleOfSolvingTheHeatEquation
Related topic	EigenvalueProblem