Wave Equation--Rectangle

To find the motion of a rectangular membrane with sides of length  and  (in the absence of gravity), use the two-dimensional wave equation

(1)

where  is the vertical displacement of a point on the membrane at position () and time . Use separation of variables to look for solutions of the form

(2)

Plugging (2) into (1) gives

(3)

where the partial derivatives have now become complete derivatives. Multiplying (3) by  gives

(4)

The left and right sides must both be equal to a constant, so we can separate the equation by writing the right side as

(5)

This has solution

(6)

Plugging (5) back into (◇),

(7)

which we can rewrite as

(8)

since the left and right sides again must both be equal to a constant. We can now separate out the  equation

(9)

where we have defined a new constant  satisfying

(10)

Equations (◇) and (◇) have solutions

(11)

(12)

We now apply the boundary conditions to (11) and (12). The conditions  and  mean that

(13)

Similarly, the conditions  and  give  and , so and , where  and  are integers. Solving for the allowed values of  and  then gives

(14)

Plugging (◇), (◇), (◇), (◇), and (14) back into (◇) gives the solution for particular values of  and ,

(15)

Lumping the constants together by writing  (we can do this since  is a function of  and , so can be written as ) and , we obtain

(16)

Plots of the spatial part for modes are illustrated above.

The general solution is a sum over all possible values of  and , so the final solution is

(17)

where  is defined by combining (◇) and (◇) to yield

(18)

Given the initial conditions  and , we can compute the s and s explicitly. To accomplish this, we make use of the orthogonality of the sine function in the form

(19)

where  is the Kronecker delta. This can be demonstrated by direct integration. Let  so  in (◇), then

(20)

Now use the trigonometric identity

(21)

to write

(22)

Note that for an integer , the following integral vanishes

			(23)
			(24)
			(25)
			(26)

since  when  is an integer. Therefore,  when . However,  does not vanish when , since

(27)

We therefore have that , so we have derived (◇). Now we multiply  by two sine terms and integrate between 0 and  and between 0 and ,

(28)

Now plug in , set , and prime the indices to distinguish them from the  and  in (28),

(29)

Making use of (◇) in (29),

(30)

so the sums over  and  collapse to a single term

(31)

Equating (30) and (31) and solving for  then gives

(32)

An analogous derivation gives the s as

(33)