Question

Separate the wave equation in two-dimensional rectangular coordinates x, y. Consider a rectangular membrane, rigidly attache to supports along its sides, such that a ≤ x ≤ 0 and b ≤ y ≤ 0. Find the solution, including the specification of the characteristic frequencies of the membrane oscillations. In the case of a = b, show that two or more modes of vibration correspond to a single frequency

Answer 1

Consider a thin elastic membrane stretched tightly over a rectangular frame. Suppose the dimensions of the frame are a × b and that we keep the edges of the membrane fixed to the frame.

Perturbing the membrane from equilibrium results in some sort of vibration of the surface. Our goal is to mathematically model the vibrations of the membrane surface.

We let u(x, y,t) = deflection of membrane from equilibrium at position (x, y) and time t. For a fixed t, the surface z = u(x, y,t) gives the shape of the membrane at time t. Under ideal assumptions (e.g. uniform membrane density, uniform tension, no resistance to motion, small deflection, etc.) u(x.y.t), satisfies the two dimensional wave equation

(1)

for 0 < x < a, 0 < y < b.

the constant c has the units of velocity. It is given by c² = τ /ρ , where τ is the tension per unit length, and ρ is mass density.

we are keeping the edges of the membrane fixed is expressed by the boundary conditions

u(0, y,t) = u(a, y,t) = 0, 0 ≤ y ≤ b, t ≥ 0, u(x, 0,t) = u(x, b,t) = 0, 0 ≤ x ≤ a, t ≥ 0. (2)

We must also specify how the membrane is initially deformed and set into motion. This is done via the initial conditions

u(x, y, 0) = f (x, y), (x, y) ∈ R, ut(x, y, 0) = g(x, y), (x, y) ∈ R, where R = [0, a] × [0, b]. (3)

we shall use separation of variables to produce simple solutions to (1) and (2),

We seek nontrivial solutions of the form u(x, y,t) = X(x)Y(y)T(t). Plugging this into the wave equation (1) we get

XYT′′ = c² (X ′′YT + XY ′′T) .

If we divide both sides by c²XYT this becomes

Because the two sides are functions of different independent variables, they must be constant:

The first equality becomes T ′′ − c²AT = 0. The second can be rewritten as

Once again, the two sides involve unrelated variables, so both are constant:

If we let C = A − B these equations can be rewritten as X ′′ − BX = 0, Y ′′ − CY = 0.

The first boundary condition is 0 = u(0, y,t) = X(0)Y (y)T(t), 0 ≤ y ≤ b, t ≥ 0. Since we want nontrivial solutions only, we can cancel Y and T, yielding X(0) = 0. When we perform similar computations with the other three boundary conditions we also get X(a) = 0, Y (0) = Y (b) = 0. There are no boundary conditions on T.

We have already solved the two boundary value problems for X and Y . The nontrivial solutions are

Xm(x) = sin µmx, µm = mπ /a , m = 1, 2, 3, . . .

Y_n(y) = sin νny, νn = nπ /b , n = 1, 2, 3, . . .

with separation constants B = −µ_m² and C = −ν_n² .

Recall that T must satisfy T ′′ − c²AT = 0 with A = B + C = − µ_m²-ν_n² < 0. It follows that for any choice of m and n the general solution for T is T_mn(t) = B_mn cos λ_mnt + B*_mn sin λ_mnt,

where

.

These are the characteristic frequencies of the membrane.

Assembling our results, we find that for any pair m, n ≥ 1 we have the normal mode u_mn(x, y,t) = X_m(x)Y_n(y)T_mn(t) = sin µ_mx sin ν_ny (B_mn cos λ_mnt + B^∗_mn sin λ_mnt)

Also, for m=n, both of the normal modes reduces to same a same frequency.