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2 Part II
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15CCM211A and 6CCM211B Outline Solutions 2 Part II
Part II
(a) Make the change of variable = b y.
(b) We have fromdf
dx b f(x) = (x)
that
(f )() b f() = (),or, integrating by parts in the first term
i f() b f() = ()
using the fact that f(x) 0 as |x| 0 where? important!On the other hand, we have
() =
0
eix ebxdx =1
b i .
Hence
f() =1
b2 + 2.
Therefore, using Fourier inversion,
f(x) =1
2pi
eixb2 + 2
d = 12beb|x|,
where we use part (a) for the last equality.
Check: show with this f that dfdx b f(x) = e b x for positive x > 0, and is
zero for negative x < 0. At x = 0 the function is discontinuous and so
f is only continuous but not differentiable at zero. (So f is not a classical
solution at x = 0, only a distributional solution (see he web notes) but we
will ignore this se3 3rd and 4th year courses for details.)
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