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

= c² ⋅

(1)

satisfied by the ordinates u(x, t) of the points of the string with the abscissa x on the time moment t (≧ 0). The boundary conditions 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 equation (1) may be written

c² ⋅

(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′′ = −

²X, T′′ = −k²T

(3)

The solutions of these are, as is well known,

{ kx kx
X = C1 cos c + C2 sin c
T = D1 coskt + D2 sin kt

(4)

with integration constants C_i and D_i.

But if we had set both sides of (2) equal to +k², we had got the solution T = D₁e^kt + 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

kp- C2 sin c = 0.

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

kp- sin c = 0,

which implies

kp- c = n π (n ∈ ℤ+ ).

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

k = n-πc (n = 1, 2, 3, ...). p

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

[ ] nπ- nπc- n-πc u = XT = C2 sin p x D1 cos p t + D2 sin p t

[ ] nπc- nπc- n-π u = An cos p t + Bn sin p t sin p x

(n = 1, 2, 3, …) 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) := ∑ _n=1^∞ ( )
A cos n-πct + B sin n-πct
n p n p

sin

(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

∞ ∑ nπ- An sin p x = f(x), n=1

∞ ∑ n-πc nπ- Bn p sin p x = g(x ) n=1

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:

∫ p An = 2- f (x )sin n-πx dx, p 0 p

∫ p B = -2-- g(x)sin n-πx dx. n nπc 0 p

References

[1] K. V ais al a: Matematiikka IV. Hand-out Nr. 141. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).

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