# 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 wave equation
   ${\begin{matrix}{\frac {\partial ^{2}u}{\partial t^{2}}}=c^{2}\cdot {\frac {\partial ^{2}u}{\partial x^{2}}}\end{matrix}}$
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 $u(0,\,t)=u(p,\,t)=0.$ We suppose also the initial conditions $u(x,\,0)=f(x),\quad u_{t}'(x,\,0)=g(x)$ 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 $u(x,\,t):=X(x)T(t).$ The boundary conditions are then\, $X(0)=X(p)=0$,\, and the partial differential equation (1) may be written
   ${\begin{matrix}c^{2}\cdot {\frac {X''}{X}}={\frac {T''}{T}}.\end{matrix}}$
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:
   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 ${\frac {kp}{=}}n\pi \quad (n\in \mathbb {Z} _{+}).$ 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
   ${\begin{matrix}u(x,\,t):=\sum _{n=1}^{\infty }\left(A_{n}\cos {\frac {n\pi c}{p}}t+B_{n}\sin {\frac {n\pi c}{p}}t\right)\sin {\frac {n\pi }{p}}x\end{matrix}}$
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: $A_{n}={\frac {2}{p}}\int _{0}^{p}\!f(x)\sin {\frac {n\pi x}{p}}\,dx,$ $B_{n}={\frac {2}{n\pi c}}\int _{0}^{p}\!g(x)\sin {\frac {n\pi x}{p}}\,dx.$