d’Alembert and D. Bernoulli solutions of wave equation

Let’s consider the d’Alembert’s solution (http://planetmath.org/WaveEquation)

$u(x,t):={\displaystyle \frac{1}{2}}[f(x-ct)+f(x+ct)]+{\displaystyle \frac{1}{2c}}{\displaystyle {\int }_{x-ct}^{x+ct}}g(s)𝑑s$(1)

of the wave equation in one dimension in the special case when the other initial condition is

$u_t^{\prime }(x,\mathrm{\hspace{0.17em}0}):=g(x)\equiv \mathrm{\hspace{0.17em}0}.$(2)

We shall see that the solution is equivalent with the solution of D. Bernoulli.

We the given function $f$ to the Fourier sine series on the interval  $[0,p]$:

$$f(y)=\sum _{n=1}^{\mathrm{\infty }}{A}_n\sin \frac{n\pi y}{p}\mathit{\hspace{1em}}\text{with}{A}_n=\frac{2}{p}{\int }_0^{p}f(x)\sin \frac{n\pi x}{p}dx\mathit{\hspace{1em}}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots })$$

Thus we may write

$\{\begin{array}{cc}f(x-ct)={\sum }_{n=1}^{\mathrm{\infty }}{A}_n\sin \left(\frac{n\pi x}{p}-\frac{n\pi ct}{p}\right)={\sum }_{n=1}^{\mathrm{\infty }}{A}_n\left(\sin \frac{n\pi x}{p}\cos \frac{n\pi ct}{p}-\cos \frac{n\pi x}{p}\sin \frac{n\pi ct}{p}\right),\hfill & \\ f(x+ct)={\sum }_{n=1}^{\mathrm{\infty }}{A}_n\sin \left(\frac{n\pi x}{p}+\frac{n\pi ct}{p}\right)={\sum }_{n=1}^{\mathrm{\infty }}{A}_n\left(\sin \frac{n\pi x}{p}\cos \frac{n\pi ct}{p}+\cos \frac{n\pi x}{p}\sin \frac{n\pi ct}{p}\right).\hfill & \end{array}$

Adding these equations and dividing by 2 yield

$u(x,t)={\displaystyle \frac{1}{2}}[f(x-ct)+f(x+ct)]={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{A}_n\cos {\displaystyle \frac{n\pi ct}{p}}\sin {\displaystyle \frac{n\pi x}{p}},$(3)

which indeed is the solution of D. Bernoulli (http://planetmath.org/SolvingTheWaveEquationByDBernoulli) in the case  $g(x)\equiv 0$.

Note.  The solution (3) of the wave equation is especially in the special case where one has besides (2) the sine-formed initial condition

$u(x,\mathrm{\hspace{0.17em}0}):=f(x)\equiv \sin {\displaystyle \frac{\pi x}{p}}.$(4)

Then  $A_n=0$  for every $n$ except 1, and one obtains

$u(x,t)=\cos {\displaystyle \frac{\pi ct}{p}}\sin {\displaystyle \frac{\pi x}{p}}.$(5)

Remark. In the case of quantum systems one has Schrödinger’s wave equation (http://planetmath.org/SchrodingersWaveEquation)whose solutions are different from the above.