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.)