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)
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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)
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(3)
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where the partial derivatives have now become complete derivatives. Multiplying (3) by gives
(4)
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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)
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This has solution
(6)
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Plugging (5) back into (◇),
(7)
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which we can rewrite as
(8)
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since the left and right sides again must both be equal to a constant. We can now separate out the equation
(9)
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where we have defined a new constant satisfying
(10)
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Equations (◇) and (◇) have solutions
(11)
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(12)
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We now apply the boundary conditions to (11) and (12). The conditions and mean that
(13)
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Similarly, the conditions and give and , so and , where and are integers. Solving for the allowed values of and then gives
(14)
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Plugging (◇), (◇), (◇), (◇), and (14) back into (◇) gives the solution for particular values of and ,
(15)
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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)
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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)
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where is defined by combining (◇) and (◇) to yield
(18)
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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)
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where is the Kronecker delta. This can be demonstrated by direct integration. Let so in (◇), then
(20)
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Now use the trigonometric identity
(21)
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to write
(22)
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Note that for an integer , the following integral vanishes
(23)
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(24)
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(25)
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(26)
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since when is an integer. Therefore, when . However, does not vanish when , since
(27)
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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)
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Now plug in , set , and prime the indices to distinguish them from the and in (28),
(29)
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Making use of (◇) in (29),
(30)
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so the sums over and collapse to a single term
(31)
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Equating (30) and (31) and solving for then gives
(32)
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An analogous derivation gives the s as
(33)
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