The one-dimensional wave equation is given by
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(1)
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In order to specify a wave, the equation is subject to boundary conditions
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(2)
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(3)
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and initial conditions
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(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
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(6)
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(7)
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By the chain rule,
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(8)
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(9)
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The wave equation then becomes
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(10)
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Any solution of this equation is of the form
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(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,
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(12)
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which is given, with the help of the Fourier transform derivative identity, by
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(13)
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where
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(14)
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(15)
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This has solution
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(16)
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Taking the inverse Fourier transform gives
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(17)
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(18)
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(19)
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(20)
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where
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(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
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(23)
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This gives
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(24)
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(25)
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So the solution for is
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(26)
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Rewriting (25) gives
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(27)
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so the solution for is
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(28)
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where .
Applying the boundary conditions
to (◇) gives
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(29)
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where
is an integer. Plugging (◇), (◇) and (29) back in for
in (◇) gives, for a particular value of
,
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(30)
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(31)
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The initial condition then gives
, so (31) becomes
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(32)
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The general solution is a sum over all possible values of , so
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(33)
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Using orthogonality of sines again,
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(34)
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where
is the Kronecker delta defined by
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(35)
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gives
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(36)
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(37)
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(38)
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so we have
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(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
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(40)
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is
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(41)
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where the coefficients are given by (◇).
A damped one-dimensional wave
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(42)
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given boundary conditions
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(43)
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(44)
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initial conditions
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(45)
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(46)
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and the additional constraint
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(47)
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can also be solved as a Fourier series.
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(48)
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where
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(49)
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(50)
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(51)
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