PlanetPhysics/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 [[../WaveEquation/|wave equation]]

$\begin{array}{c}\frac{{\partial }^{2}u}{\partial t^2}=c^2\cdot \frac{{\partial }^{2}u}{\partial x^2}\end{array}$

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 [[../PiecewiseLinear/|boundary]] conditions are thus $u(0,t)=u(p,t)=0.$ We suppose also the initial conditions $u(x,0)=f(x),{u_t}^{\prime }(x,0)=g(x)$ which give the initial [[../Position/|position]] of the string and the initial [[../Velocity/|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 [[../DifferentialEquations/|partial differential equation]] (1) may be written

$\begin{array}{c}{c}^2\cdot \frac{X^{″}}{X}=\frac{T^{″}}{T}.\end{array}$

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 [[../DifferentialEquations/|differential equations]] of second order:

Failed to parse (unknown function "\begin{matrix}"): {\displaystyle \begin{matrix} X'' = -\left(\frac{k}\right)^2 X,\quad T'' = -k^2T \end{matrix}}

The solutions of these are, as is well known,

Failed to parse (unknown function "\begin{cases}"): {\displaystyle \begin{matrix} \begin{cases} X = C_1\cos\frac{kx}+C_2\sin\frac{kx}\\ T = D_1\cos{kt}+D_2\sin{kt}\\ \end{cases} \end{matrix}}

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 $C_2\sin \frac{kp}{=}0.$ If one had\, $C_2=0$,\, then $X(x)$ were identically 0 which is naturally impossible.\, So we must have $\sin \frac{kp}{=}0,$ which implies $\frac{kp}{=}n\pi (n\in {\mathbb{Z}}_{+}).$ This means that the only suitable values of $k$ satisfying the equations (3), the so-called eigenvalues, are $k=\frac{n\pi c}{p}(n=1,2,3,\dots ).$ So we have infinitely many solutions of (1), the eigenfunctions $u=XT=C_2\sin \frac{n\pi }{p}x[D_1\cos \frac{n\pi c}{p}t+D_2\sin \frac{n\pi c}{p}t]$ or $u=[A_n\cos \frac{n\pi c}{p}t+B_n\sin \frac{n\pi c}{p}t]\sin \frac{n\pi }{p}x$ $(n=1,2,3,\dots )$ where $A_n$'s and $B_n$'s are for the time being arbitrary constants.\, Each of these [[../Bijective/|functions]] satisfy the boundary conditions.\, Because of the linearity of (1), also their sum series

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

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: $A_n=\frac{2}{p}\underset{0}{\overset{p}{\int }}f(x)\sin \frac{n\pi x}{p}dx,$ $B_n=\frac{2}{n\pi c}\underset{0}{\overset{p}{\int }}g(x)\sin \frac{n\pi x}{p}dx.$