# solving the wave equation due to D. Bernoulli

A string has been strained between the points $(0,\mathrm{\hspace{0.17em}0})$ and $(p,\mathrm{\hspace{0.17em}0})$ of the $x$-axis. The vibration of the string in the $xy$-plane is determined by the one-dimensional wave equation^{}

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

satisfied by the ordinates $u(x,t)$ of the points of the string with the abscissa $x$ on the time $t\phantom{\rule{veryverythickmathspace}{0ex}}(\geqq 0)$.
The boundary conditions^{} are thus

$$u(0,t)=u(p,t)=0.$$ |

We suppose also the initial conditions

$$u(x,\mathrm{\hspace{0.17em}0})=f(x),{u}_{t}^{\prime}(x,\mathrm{\hspace{0.17em}0})=g(x)$$ |

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

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

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

The solutions of these are, as is well known,

$\{\begin{array}{cc}X={C}_{1}\mathrm{cos}\frac{kx}{c}+{C}_{2}\mathrm{sin}\frac{kx}{c}\hfill & \\ T={D}_{1}\mathrm{cos}kt+{D}_{2}\mathrm{sin}kt\hfill & \end{array}$ | (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

$${C}_{2}\mathrm{sin}\frac{kp}{c}=0.$$ |

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

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

which implies

$$\frac{kp}{c}=n\pi \mathit{\hspace{1em}}(n\in {\mathbb{Z}}_{+}).$$ |

This means that the only suitable values of $k$ satisfying the equations (3), the so-called eigenvalues, are

$$k=\frac{n\pi c}{p}\mathit{\hspace{1em}}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots}).$$ |

So we have infinitely many solutions of (1), the eigenfunctions

$$u=XT={C}_{2}\mathrm{sin}\frac{n\pi}{p}x\left[{D}_{1}\mathrm{cos}\frac{n\pi c}{p}t+{D}_{2}\mathrm{sin}\frac{n\pi c}{p}t\right]$$ |

or

$$u=\left[{A}_{n}\mathrm{cos}\frac{n\pi c}{p}t+{B}_{n}\mathrm{sin}\frac{n\pi c}{p}t\right]\mathrm{sin}\frac{n\pi}{p}x$$ |

$(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots})$ 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

$u(x,t):={\displaystyle \sum _{n=1}^{\mathrm{\infty}}}\left({A}_{n}\mathrm{cos}{\displaystyle \frac{n\pi c}{p}}t+{B}_{n}\mathrm{sin}{\displaystyle \frac{n\pi c}{p}}t\right)\mathrm{sin}{\displaystyle \frac{n\pi}{p}}x$ | (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

$$\sum _{n=1}^{\mathrm{\infty}}{A}_{n}\mathrm{sin}\frac{n\pi}{p}x=f(x),$$ |

$$\sum _{n=1}^{\mathrm{\infty}}{B}_{n}\frac{n\pi c}{p}\mathrm{sin}\frac{n\pi}{p}x=g(x)$$ |

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:

$${A}_{n}=\frac{2}{p}{\int}_{0}^{p}f(x)\mathrm{sin}\frac{n\pi x}{p}dx,$$ |

$${B}_{n}=\frac{2}{n\pi c}{\int}_{0}^{p}g(x)\mathrm{sin}\frac{n\pi x}{p}dx.$$ |

## References

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

Title | solving the wave equation due to D. Bernoulli |
---|---|

Canonical name | SolvingTheWaveEquationDueToDBernoulli |

Date of creation | 2013-03-22 16:31:41 |

Last modified on | 2013-03-22 16:31:41 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 12 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 35L05 |

Synonym | vibrating string^{} |

Related topic | ExampleOfSolvingTheHeatEquation |

Related topic | EigenvalueProblem |