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[parent] solving the wave equation due to D. Bernoulli (Example)

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).



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Cross-references: functions, differential equations, partial differential equation, velocity, position, boundary, wave equation

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