d'Alembert and D. Bernoulli solutions of wave equation

Let's consider the d'Alembert's solution

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

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

$\displaystyle u'_t(x,\,0) \,:=\, g(x) \,\equiv\, 0.$ (2)

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

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

$\displaystyle f(y) \,=\, \sum_{n=1}^\infty A_n\sin\frac{n\pi y}{p}$   with$\displaystyle \;\;
A_n = \frac{2}{p}\int_0^pf(x)\sin\frac{n\pi x}{p}\,dx \quad (n = 1,\,2,\,\ldots)
$

Thus we may write

\begin{align*}\begin{cases}f(x\!-\!ct) = \sum_{n=1}^\infty A_n\sin\!\left(\frac{...
...ct}{p}+\cos\frac{n\pi x}{p}\sin\frac{n\pi ct}{p}\right). \end{cases}\end{align*}    

Adding these equations and dividing by 2 yield

$\displaystyle u(x,\,t) = \frac{1}{2}[f(x\!-\!ct)+f(x\!+\!ct)] = \sum_{n=1}^\infty A_n\cos\frac{n\pi ct}{p}\sin\frac{n\pi x}{p},$ (3)

which indeed is the solution of D. Bernoulli in the case $ g(x) \equiv 0$ .

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

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

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

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

Remark. In the case of quantum systems one has Schrödinger's wave equation whose solutions are different from the above.



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