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# 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)