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Homesolving the wave equation due to D. Bernoulli

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# 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

$\displaystyle\frac{\partial^{2}u}{\partial t^{2}}=c^{2}\cdot\frac{\partial^{2}% u}{\partial x^{2}}$ | (1) |

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

$u(x,\,0)=f(x),\quad u_{t}^{{\prime}}(x,\,0)=g(x)$ |

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

$\displaystyle c^{2}\cdot\frac{X^{{\prime\prime}}}{X}=\frac{T^{{\prime\prime}}}% {T}.$ | (2) |

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:

$\displaystyle X^{{\prime\prime}}=-\left(\frac{k}{c}\right)^{2}X,\quad T^{{% \prime\prime}}=-k^{2}T$ | (3) |

The solutions of these are, as is well known,

$\displaystyle\begin{cases}X=C_{1}\cos\frac{kx}{c}+C_{2}\sin\frac{kx}{c}\\ T=D_{1}\cos{kt}+D_{2}\sin{kt}\\ \end{cases}$ | (4) |

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

$C_{2}\sin\frac{kp}{c}=0.$ |

If one had $C_{2}=0$, then $X(x)$ were identically 0 which is naturally impossible. So we must have

$\sin\frac{kp}{c}=0,$ |

which implies

$\frac{kp}{c}=n\pi\quad(n\in\mathbb{Z}_{+}).$ |

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

$\displaystyle u(x,\,t):=\sum_{{n=1}}^{\infty}\left(A_{n}\cos\frac{n\pi c}{p}t+% B_{n}\sin\frac{n\pi c}{p}t\right)\sin\frac{n\pi}{p}x$ | (5) |

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:

$A_{n}=\frac{2}{p}\int_{{0}}^{p}\!f(x)\sin\frac{n\pi x}{p}\,dx,$ |

$B_{n}=\frac{2}{n\pi c}\int_{{0}}^{p}\!g(x)\sin\frac{n\pi x}{p}\,dx.$ |

# References

- 1 K. Väisälä: Matematiikka IV. Hand-out Nr. 141. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).

## Mathematics Subject Classification

35L05*no label found*

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