Fourier Integral
qThe Fourier series of f on the interval (-p,p) is
f ( x) =
æ p
1 p
1 ¥ éæ p
np ö
np
np ö np ù
f (t )dt + å êçç ò f (t ) cos
tdt ÷÷ cos
x + çç ò f (t ) sin
tdt ÷÷ sin
xú
ò
p
2p
p n=1 ëè - p
p
p
p
p û
ø
è -p
ø
qIf we let an=np/p, Da=an+1-an=p/p. Then
Fourier Transform and Solving Partial
Differential Equations
f ( x) =
p
1 æ p
1 ¥ é p
ù
ç f (t )dt ö÷Da + å êæç ò- p f (t ) cos a ntdt ö÷ cos a n x + æç ò- p f (t ) sin a ntdt ö÷ sin a n x úDa
ø
ø
è
ø
2p è ò- p
p n =1 ëè
û
qIf we let p ® ¥, Da ® 0, the first term is zero and
the summation becomes an integral
f ( x) =
1
¥
éæ ¥ f (t ) cos atdt ö cos ax + æ ¥ f (t ) sin atdt ö sin ax ù da
ç
÷
ç ò-¥
÷
úû
ø
è
ø
p ò0 êëè ò-¥
This is called the Fourier integral of f on (-¥,¥)
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Fourier Cosine and Sine Integrals
Class Exercises
qThe Fourier integral of an even function on the
interval (-¥,¥) is the cosine integral.
f ( x) =
2
¥
A(a ) cos axda ,
pò
0
qFind Fourier integral representation of
x < -1
ì 0,
ï- 1, - 1 < x < 0
ï
f ( x) = í
ï 2, 0 < x < 1
ïî 0,
x >1
¥
A(a ) = ò f ( x) cos axdx
0
qThe Fourier integral of an odd function on the
interval (-¥,¥) is the sine integral.
f ( x) =
2
¥
B(a ) sin axda ,
pò
0
¥
B(a ) = ò f ( x) sin axdx
qFind Fourier cosine and sine function of
0
qThe complex or exponential form of Fourier
integral is
1 ¥
f ( x) =
C (a )e -iax da ,
2p ò-¥
Advanced Mech. Eng. Analysis (ME 230)
Younes Shabany
¥
f ( x) = xe -2 x , x > 0
C (a ) = ò f ( x)e dx
iax
-¥
157
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Fourier Transform
Transform of Derivative
qFourier integrals are used to define Fourier
transforms.
qFourier transform and inverse Fourier transform
qUsing integration by part and assuming f(x)®0 as
x® ±¥ it can be shown that
¥
F { f ( x)} = ò f ( x)eiax dx = F (a )
-¥
F -1{F (a )} =
¥
F { f ' ( x)} = ò f ' ( x)eiax dx
-¥
1 ¥
F (a )e -iax da = f ( x)
2p ò-¥
= f ( x ) e iax
qFourier sine transform and inverse Fourier sine
transform
¥
Fs { f ( x)} = ò f ( x) sin axdx = Fs (a )
0
Fs-1{Fs (a )} =
2
¥
p ò0
Fc { f ( x)} = ò f ( x) cos axdx = Fc (a )
0
Advanced Mech. Eng. Analysis (ME 230)
-1
c
F
2
¥
- ia ò f ( x)eiax dx
-¥
¥
-¥
= -iaF (a )
Fs (a ) sin axda = f ( x)
qAssuming f’(x)®0 as x® ±¥ it can be shown that
F { f ' ' ( x)} = -a 2 F (a )
¥
{Fc (a )} = ò0 Fc (a ) cosaxda = f ( x)
p
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Younes Shabany
-¥
= -ia ò f ( x)eiax dx
qFourier cosine transform and inverse Fourier
cosine transform
¥
¥
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Transform of Derivative
Transform of Derivative
qFourier cosine and sine of second derivative of a
function are given in terms of Fourier cosine and
sine of the function itself respectively.
qNote that Fourier cosine and sine of first derivative
of a function are given in terms of Fourier sine and
cosine of the function itself respectively.
¥
Fc { f ' ' ( x)} = ò f ' ' ( x) cos axdx
0
¥
Fc { f ' ( x)} = ò f ' ( x) cos axdx
¥
0
¥
= f ( x) cos ax 0 + a ò f ( x) sin axdx
= - f ' (0) + a é f ( x) sin ax 0 - a ò f ( x) cos axdxù
êë
úû
0
= - f ' (0) - a 2 Fc { f ( x)}
¥
0
= - f (0) + aFs { f ( x)}
¥
0
0
¥
¥
Younes Shabany
¥
= f ' ( x) sin ax 0 - a ò f ' ( x) cos axdx
¥
= f ( x) sin ax 0 - a ò f ( x) cos axdx
= -aFc { f ( x)}
¥
Fs { f ' ' ( x)} = ò f ' ' ( x) sin axdx
¥
Fs { f ' ( x)} = ò f ' ( x) sin axdx
Advanced Mech. Eng. Analysis (ME 230)
¥
¥
= f ' ( x) cos ax 0 + a ò f ' ( x) sin axdx
0
0
¥
¥
= -a é f ( x) cos ax 0 + a ò f ( x) sin axdxù
êë
úû
0
= af (0) - a 2 Fs { f ( x)}
0
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Fourier Transform and Solution of BVPs
Class Exercises
qSolve below BVP using Fourier transform
¶ 2u ¶u
=
, -¥ < x < ¥
¶x 2 ¶t
ìu
u ( x,0) = f ( x), f ( x) = í 0
î 0,
qSolve BVPs using Fourier transform.
k
k
x <1
x >1
2
u ( x,0) = e - x , - ¥ < x < ¥
qLet’s define Fourier transform of u as
Note that
¥
F {u ( x, t )} = ò u ( x, t )eiax dx = U (a , t )
dU
- ka U (a , t ) =
dt
Þ
2
¥
1
-¥
-1
BC : F {u ( x,0)} = ò f ( x)eiax dx = ò u0 e iax dx =u0
U (a , t ) = 2u0
sin a
a
Þ
{
2
2
}
2
F e- x / 4 p = 2 p pe - p a
-¥
ì ¶ 2u ü
ì ¶u ü
Eq : F ík 2 ý = F í ý
î ¶t þ
î ¶x þ
¶ 2u ¶u
= , - ¥ < x < ¥, t > 0
¶x 2 ¶t
dU
+ ka 2U (a , t ) = 0
dt
2
e i a - e - ia
sin a
= 2u0
ia
a
2
e - ka t
u ( x, t ) = F -1{U (a , t )} =
u0
¥
sin a
p ò-¥ a
Advanced Mech. Eng. Analysis (ME 230)
2
e - ka t e -iax da =
u0
¥
p ò-¥
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sin a cos ax
a
2
e - ka t da
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