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 vibration of the string in the xy-plane is determined by the one-dimensional wave equationMathworldPlanetmath

2ut2=c22ux2 (1)

satisfied by the ordinates  u(x,t)  of the points of the string with the abscissa x on the time   t(0). The boundary conditionsMathworldPlanetmath are thus

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

We suppose also the initial conditions

u(x, 0)=f(x),ut(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 equationMathworldPlanetmath (1) may be written

c2X′′X=T′′T. (2)

This is not possible unless both sides are equal to a same constant -k2 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′′=-(kc)2X,T′′=-k2T (3)

The solutions of these are, as is well known,

{X=C1coskxc+C2sinkxcT=D1coskt+D2sinkt (4)

with integration constants Ci and Di.

But if we had set both sides of (2) equal to  +k2, we had got the solution  T=D1ekt+D2e-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  C1=0,  and the one for X(p) that

C2sinkpc=0.

If one had  C2=0,  then X(x) were identically 0 which is naturally impossible.  So we must have

sinkpc=0,

which implies

kpc=nπ(n+).

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

k=nπcp(n=1, 2, 3,).

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

u=XT=C2sinnπpx[D1cosnπcpt+D2sinnπcpt]

or

u=[Ancosnπcpt+Bnsinnπcpt]sinnπpx

(n=1, 2, 3,) where An’s and Bn’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):=n=1(Ancosnπcpt+Bnsinnπcpt)sinnπpx (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

n=1Ansinnπpx=f(x),
n=1Bnnπcpsinnπpx=g(x)

on the interval[0,p].  But the left sides of these equations are the Fourier sine seriesMathworldPlanetmath of the functions f and g, and therefore we obtain the expressions for the coefficients:

An=2p0pf(x)sinnπxpdx,
Bn=2nπc0pg(x)sinnπxpdx.

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 stringPlanetmathPlanetmath
Related topic ExampleOfSolvingTheHeatEquation
Related topic EigenvalueProblem