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 vibration of the string in the $xy$-plane is determined by the one-dimensional wave equation

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

 $u(x,\,0)=f(x),\quad u_{t}^{\prime}(x,\,0)=g(x)$

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

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

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

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

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

The solutions of these are, as is well known,

 $\displaystyle\begin{cases}X=C_{1}\cos\frac{kx}{c}+C_{2}\sin\frac{kx}{c}\\ T=D_{1}\cos{kt}+D_{2}\sin{kt}\\ \end{cases}$ (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

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

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

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

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

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

$(n=1,\,2,\,3,\,\ldots)$ 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

 $\sum_{n=1}^{\infty}A_{n}\sin\frac{n\pi}{p}x=f(x),$
 $\sum_{n=1}^{\infty}B_{n}\frac{n\pi c}{p}\sin\frac{n\pi}{p}x=g(x)$

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:

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

References

• 1 K. Väisälä: Matematiikka IV.  Hand-out Nr. 141. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).
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