Let's consider the d'Alembert's solution
![$\displaystyle u(x,\,t) \,:=\, \frac{1}{2}[f(x\!-\!ct)+f(x\!+\!ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds$ $\displaystyle u(x,\,t) \,:=\, \frac{1}{2}[f(x\!-\!ct)+f(x\!+\!ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds$](http://images.physicslibrary.org/cache/objects/661/l2h/img1.png) |
(1) |
of the wave equation in one dimension in the special case when the other initial condition is
 |
(2) |
We shall see that the solution is equivalent with the solution of D. Bernoulli.
We expand the given function to the Fourier sine series on the interval :
 with 
Thus we may write
Adding these equations and dividing by 2 yield
![$\displaystyle u(x,\,t) = \frac{1}{2}[f(x\!-\!ct)+f(x\!+\!ct)] = \sum_{n=1}^\infty A_n\cos\frac{n\pi ct}{p}\sin\frac{n\pi x}{p},$ $\displaystyle u(x,\,t) = \frac{1}{2}[f(x\!-\!ct)+f(x\!+\!ct)] = \sum_{n=1}^\infty A_n\cos\frac{n\pi ct}{p}\sin\frac{n\pi x}{p},$](http://images.physicslibrary.org/cache/objects/661/l2h/img8.png) |
(3) |
which indeed is the solution of D. Bernoulli in the case
.
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
 |
(4) |
Then for every except 1, and one obtains
 |
(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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