The one-dimensional wave equation is given by
(1)
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In order to specify a wave, the equation is subject to boundary conditions
(2)
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(3)
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and initial conditions
(4)
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(5)
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The one-dimensional wave equation can be solved exactly by d'Alembert's solution, using a Fourier transform method, or via separation of variables.
d'Alembert devised his solution in 1746, and Euler subsequently expanded the method in 1748. Let
(6)
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(7)
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By the chain rule,
(8)
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(9)
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The wave equation then becomes
(10)
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Any solution of this equation is of the form
(11)
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where and are any functions. They represent two waveforms traveling in opposite directions, in the negative direction and in the positive direction.
The one-dimensional wave equation can also be solved by applying a Fourier transform to each side,
(12)
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which is given, with the help of the Fourier transform derivative identity, by
(13)
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where
(14)
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(15)
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This has solution
(16)
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Taking the inverse Fourier transform gives
(17)
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(18)
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(19)
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(20)
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where
(21)
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(22)
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This solution is still subject to all other initial and boundary conditions.
The one-dimensional wave equation can be solved by separation of variables using a trial solution
(23)
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This gives
(24)
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(25)
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So the solution for is
(26)
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Rewriting (25) gives
(27)
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so the solution for is
(28)
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where . Applying the boundary conditions to (◇) gives
(29)
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where is an integer. Plugging (◇), (◇) and (29) back in for in (◇) gives, for a particular value of ,
(30)
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(31)
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The initial condition then gives , so (31) becomes
(32)
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The general solution is a sum over all possible values of , so
(33)
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Using orthogonality of sines again,
(34)
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where is the Kronecker delta defined by
(35)
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gives
(36)
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(37)
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(38)
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so we have
(39)
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The computation of s for specific initial distortions is derived in the Fourier sine series section. We already have found that , so the equation of motion for the string (◇), with
(40)
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is
(41)
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where the coefficients are given by (◇).
A damped one-dimensional wave
(42)
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given boundary conditions
(43)
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(44)
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initial conditions
(45)
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(46)
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and the additional constraint
(47)
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can also be solved as a Fourier series.
(48)
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where
(49)
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(50)
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(51)
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