Writing (possible since the eigenvectors form a basis)

for ω2 6= ωn2 , n = 1, 2, . . . , N . The solution with initial values

We observe that there is a beat whenever the driving frequency ω is close to a normal mode of oscillation ωn. Compare this discussion with that of Boyce & DiPrima [4].

6.8Vibrating Membrane

6.8.1Helmholtz Equation

In the previous section we discussed the vibrating string. Recall that we have a string of unstretched length L that is tied down at ends 0 and L. If u = u(x; t) denotes the vertical

displacement of the string at position x, 0 ≤ x ≤ L, at time t, then we showed that for small displacements u satisfies the one-dimensional wave equation

∂2u 1 ∂2u ∂x2 − v2 ∂t2 = 0

where v2 = T /µ, T equals the tension in string and µ is the density of the string. We solved this equation subject to the boundary conditions u(0, t) = u(L, t) = 0 for all t and with initial conditions u(x, 0) = f (x) and ∂u∂t (x, 0) = g(x) where f and g are given.

Now we imagine a uniform, flexible membrane, of mass ρ per unit area, stretched under a uniform tension T per unit length over a region Ω in the plane whose boundary ∂Ω is a smooth curve (with a possible exception of a finite number of corners).

We now let U = U (x, y; t) denote the vertical displacement of the membrane at position (x, y) Ω at time t from its equilibrium position. We again assume that the membrane is tied down at the boundary; that is1

U (x, y; t) = 0 for (x, y) ∂Ω.

The motion of U = U (x, y; t) is governed by the two-dimensional wave equation:

introduce

∂∂

=∂x2 + ∂y2

the wave equation takes the form

1 ∂2U

U − v2 ∂t2 = 0.

We proceed as before an look for solutions of (6.24) in which the variables separate

U (x, y; t) = u(x, y)T (t).

Substituting this into (6.24), dividing by uT gives

The right-hand side depends only upon t where as the left-hand side depends only upon x, y. Thus for the two sides to be equal they must equal the same constant. Call this constant −k2. Thus we have the two equations

1In one dimension Ω = (0, L) and the boundary of Ω consists of the two points 0 and L.

The di erential equation for T has our well-known solutions

eiωt and e−iωt.

The second equation (6.25), called the Helmholtz equation, is a partial di erential equation for u = u(x, y). We wish to solve this subject to the boundary condition

u(x, y) = 0 for (x, y) ∂Ω.

6.8.2Rectangular Membrane

Consider the rectangular domain

For this rectangular domain the Helmholtz equation can be solved by the method of separation of variables. If one assumes a solution of the form (variables x and y separate)

u(x, y) = X(x)Y (y)

then the problem is reduced to two one-dimensional problems. It is an exercise to show that the allowed frequencies are

6.8.3 Circular Membrane: The Drum

We now consider the circular domain

Ω = (x, y) : x2 + y2 < a2

so the boundary of Ω is the circle ∂Ω : x2 + y2 = a2. Even though the variables separate in the Cartesian coordinates x and y, this is of no use since the boundary is circular and we would not be able to apply the BC u = 0 on the circular boundary ∂Ω. Since the domain is circular it is natural to introduce polar coordinates

x = r cos θ, y = r sin θ.

It is an exercise in the chain rule to show that in polar coordinates the 2D Laplacian is

and hence, the Helmholtz equation in polar coodinates is

We write u = u(r, θ).2

2Possible point of confusion: We wrote u = u(x, y) so really our new function of r and θ is u(r cos θ, r sin θ). Technically we should give this function of r and θ a new name but that would be rather pedantic.