Mathematical Methods for Physicists: A concise introduction - Site Map

FOURIER SERIES AND INTEGRALS
Solution: Let us denote the Fourier transforms of y…x†; f …x† and r…x† by
Y…!†; F…!†; and R…!† respectively. Taking the Fourier transform of both sides
of the given integral equation, we have by the convolution theorem
Y…!† ˆF…!†‡Y…!†R…!† or Y…!† ˆ F…!†
1 R…!† :
Calculations of Fourier transforms
Fourier transforms can often be used to transform a di€erential equation which is
dicult to solve into a simpler equation that can be solved relatively easy. In
order to use the transform methods to solve ®rst- and second-order di€erential
equations, the transforms of ®rst- and second-order derivatives are needed. By
taking the Fourier transform with respect to the variable x, we can show that
…a† F
@u
ˆ iF…u†;
@x
!
9>
…b† F
@2 u
=
@x 2 ˆ 2 F…u†;
…4:60†
…c† F
@u
ˆ @
@t @t F…u†: > ;
Proof: (a) By de®nition we have
F
@u Z 1
@u
ˆ
@x 1 @x eix dx;
p
where the factor 1=
2 has been dropped. Using integration by parts, we obtain
F
@u Z 1
@u
ˆ
@x 1 @x eix dx
1 Z 1
ˆ ue ix ‡ i ue ix dx
1
ˆ iF…u†:
1
(b) Let u ˆ @v=@x in (a), then
!
F
@2 v
@x 2 ˆ iF
@v
ˆ…i† 2 F…v†:
@x
Now if we formally replace v by u we have
F
!
@2 u
@x 2 ˆ 2 F…u†;
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