Fourier Transform Examples
Steven Bellenot
November 5, 2007
1
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Formula Sheet
F [f (x)] = fb(w) or simply F [f ] = fb
F −1 [fb(w)] = f (x) or simply F −1 [fb] = f
Z ∞
1
F [f (x)](w) = fb(w) = √
f (x)e−iwx dx
2π −∞
Z ∞
1
F −1 [fb(w)](x) = √
fb(w)eiwx dw
2π −∞
Z ∞
1
u(x, t)e−iwx dx
F [u(x, t)](w, t) = u
b(w, t) = √
2π −∞
Z ∞
1
F −1 [b
u(w, t)](x, t) = √
u
b(w, t)eiwx dw
2π −∞
F [af (x) + bg(x)](w) = afb(w) + bb
g (w)
(9)
(10)
(11)
(12)
(13)
F [f 0 (x)](w) = iwfb(w)
F [f 00 (x)](w) = −w2 fb(w)
∂
u(x, t)](w, t) = iwb
u(w, t)
∂x
∂2
F [ 2 u(x, t)](w, t) = −w2 u
b(w, t)
∂x
∂
∂
b(w, t)
F [ u(x, t)](w, t) = u
∂t
∂t
2
2
∂
∂
F [ 2 u(x, t)](w, t) = 2 u
b(w, t)
∂t
Z∂t
F[
∞
(14)
[f ∗ g](x) =
f (w)g(x − w) dw = [g ∗ f ](x) =
−∞
√
(15)
F [f ∗ g] =
(16)
f (x − a) = F −1 [e−iwa fb(w)]
(17)
(18)
(19)
Z
∞
−∞
2π fbgb
1
−w2
F [exp (−ax2 )] = √ exp(
)
4a
2a
eiwa − e−iwa
sin wa =
2i
eiwa + e−iwa
cos wa =
2
1
f (x − w)g(w) dw
2
Formula Justifications
Equations (1), (3) and (5) readly say the same thing, (3) being the usual definition. (Warning, not all
textbooks define the these transforms the same way.) Equations (2), (4) and (6) are the respective inverse
transforms.
What kind of functions is the Fourier transform defined for? Clearly if f (x) is real, continuous and zero
R∞
outside an interval of the form [−M, M ], then fb is defined as the improper integral −∞ reduces to the
RM
proper integral −M . If f (x) decays fast enough as x → ∞ and x → −∞, then fb(w) is also defined. However
there are much larger collections of objects for which the transform can be defined. For example, if δ(x)
√
b
is the Dirac delta function, then δ(w)
= 1/ √2π the constant function. Also one can see that the inverse
transform of δ(w) is the constant function 1/ 2π.
Equation (7) follows because the integral is linear, the inverse transform is also linear.
Equation (8) follows from integrating by parts, using u = e−iwx and dv = f 0 (x) dx and the fact that f (x)
decays as x → ∞ and x → −∞.
Z ∞
Z ∞
∞
f 0 (x)e−iwx dx = f (x)e−iwx x=−∞ −
−f (x)iweiwx dx = (0 − 0) + iwfb(w)
−∞
−∞
Equation (9) is just (8) applied twice. And (10) and (11) are just restatements with more variables.
Equation (12) requires going back to the definition of the limit.
Z ∞
u(x, t + ∆t) − u(x, t) −iwx
u(x, t + ∆t) − u(x, t)
=
e
dx
F
∆t
∆t
−∞
u
b(w, t + ∆t) − u
b(w, t)
∂
→ u
b(w, t)
∆t
∂t
One now takes limits of both sides. We need to know that the fourier transform is continuous with this kind
of limit, which is true, but beyond our scope to show. Equation (13) is (12) done twice.
Equation (14) says f ∗ g = g ∗ f and this is done by substitution; use u = x − w; du = −dw; w = x − u;
u = ∞ when w = −∞ and u = −∞ when w = ∞ to obtain
Z ∞
Z −∞
Z ∞
f (w)g(x − w) dw =
f (x − u)g(u) (−du) =
f (x − u)g(u) du
=
w=−∞
u=∞
u=−∞
which used the negative sign to change the order of integration.
Equation (15) uses
Z ∞
Z ∞ Z ∞
(f ∗ g)e−iwx dx =
f (s)g(x − s) dse−iwx dx
x=−∞
x=−∞
=
Z
∞
Z
x=−∞
=
Z
∞
s=−∞
f (s)g(x − s)e−iwx ds dx
s=−∞
Z
s=−∞
∞
∞
f (s)g(x − s)e−iwx dx ds
x=−∞
Note that we have interchanged the order of integration, now we let u = x − s, x = u + s, du = ds, u = ±∞
when x = ±∞
Z ∞ Z ∞
=
f (s)g(u)e−iw(u+s) du ds
s=−∞
=
Z
∞
s=−∞
u=−∞
−iws
f (s)e
ds
Z
∞
g(u)e−iwu du
u=−∞
since
the u terms are constant as the √
integral
So the inital expression is
√
√ with respect to ds is concerned.
√
2πF [f ∗ g] and the end epression is 2π fb 2πb
g which is where the 2π factor comes from.
To show equation (16) we compute F [f (x − a)] and substitute u = x − a; x = u + a; dx = du :
Z ∞
Z ∞
Z ∞
f (x − a)e−iwx dx =
f (u)e−iw(u+a) du = e−iwa
f (u)e−iwu du = e−iwa fb(w)
x=−∞
u=−∞
u=−∞
For the bell shaped curves, equation (17) is done in earlier editions of the textbook. We repeat the
calculation for reference only
Z ∞
1
exp(−ax2 − iwx) dx
F [exp(−ax2 )] = √
2π −∞
2 2 !
Z ∞
√
iw
1
iw
√
exp −
ax + √
=√
+
dx
2 a
2 a
2π −∞
Z ∞
2 !
√
1
iw
w2
= √ exp −
exp −
ax + √
dx
4a
2 a
2π
−∞
p
We claim that the integral above has value I = πa . First we do the substitution
v=
so that dv =
√
adx and hence
Z
Z
iw
ax + √
2 a
∞
dv
exp(−v 2 ) √
a
−∞
I=
The result follows since
√
∞
exp(−v 2 ) dv =
√
π
−∞
comes from Calculus 3.
Finally (18) and (19) are from Euler’s eiθ = cos θ + i sin θ.
3
Solution Examples
• Solve 2ux + 3ut = 0;
u(x, 0) = f (x) using Fourier Transforms.
Take the Fourier Transform of both equations. The initial condition gives
and the PDE gives
u
b(w, 0) = fb(w)
2(iwb
u(w, t)) + 3
Which is basically an ODE in t, we can write it as
and which has the solution
∂
u
b(w, t) = 0
∂t
∂
2
u
b(w, t) = − iwb
u(w, t)
∂t
3
u
b(w, t) = A(w)e−2iwt/3
and the initial condition above implies A(w) = fb(w)
u
b(w, t) = fb(w)e−2iwt/3
We are now ready to inverse Fourier Transform and equation (16) above, with a = 2t/3, says that
u(x, t) = f (x − 2t/3)
• Solve 2tux + 3ut = 0;
u(x, 0) = f (x) using Fourier Transforms.
Take the Fourier Transform of both equations. The initial condition gives
u
b(w, 0) = fb(w)
and the PDE gives
2t(iwb
u(w, t)) + 3
Which is basically an ODE in t, we can write it as
∂
u
b(w, t) = 0
∂t
2
∂
u
b(w, t) = − iwtb
u(w, t)
∂t
3
and which has the solution (separate variables)
2
u
b(w, t) = A(w)e−iwt
/3
and the initial condition above implies A(w) = fb(w)
2
u
b(w, t) = fb(w)e−iwt /3
We are now ready to inverse Fourier Transform and equation (16) above, with a = t2 /3, says that
u(x, t) = f (x − t2 /3)
• Solve the heat equation c2 uxx = ut ;
u(x, 0) = f (x)
Take the Fourier Transform of both equations. The initial condition gives
u
b(w, 0) = fb(w)
and the PDE gives
c2 (−w2 u
b(w, t)) =
Which is basically an ODE in t, we can write it as
Which has the solution
∂
u
b(w, t)
∂t
∂
u
b(w, t) = −c2 w2 u
b(w, t)
∂t
2
u
b(w, t) = A(w)e−c
w2 t
and the initial condition above implies A(w) = fb(w)
2 2
u
b(w, t) = fb(w)e−c w t
We are now ready to inverse Fourier Transform: First use (17) with
1
1
= c2 t or a = 2
4a
4c t
to note that
√
2 2
2
x2
√ F exp(− 2 = e−c w t
4c t
2c t
So that by the convolution equation (15)
u(x, t) = f (x) ∗
1
√
2c πt
x2
exp − 2
4c t
• Solve the wave equation c2 uxx = utt ;
u(x, 0) = f (x) and ut (x, 0) = g(x)
Take the Fourier Transform of both equations. The initial condition gives
and the PDE gives
u
b(w, 0) = fb(w)
∂
ubt (w, 0) =
u
b(x, t)
= gb(w)
∂t
t=0
c2 (−w2 u
b(w, t)) =
Which is basically an ODE in t, we can write it as
∂2
u
b(w, t)
∂t2
∂2
u
b(w, t) + c2 w2 u
b(w, t) = 0
∂t2
Which has the solution, and derivative
u
b(w, t) = A(w) cos cwt + B(w) sin cwt
∂
u
b(w, t) = −cwA(w) sin cwt + cwB(w) cos cwt
∂t
so the first initial condition gives A(w) = fb(w) and the second gives cwB(w) = gb(w) make the solution
u
b(w, t) = fb(w) cos cwt +
Lets first look at
gb(w) sin cwt
w
c
eiwct + e−iwct
2
1 b
=
f (w)eicwt + fb(w)e−icwt
2
Applying equation (16) with a = −ct and with a = ct yields
fb(w) cos cwt = fb(w)
1
F −1 [fb(w) cos cwt] = (f (x + ct) + f (x − ct))
2
The second piece
gb(w) sin cwt
gb(w) sin cwt
=
w
c
iw −ic
and now the first factor looks like an integral, as a derivative with respect to x would cancel the iw in
bottom. Define
Z x
h(x) =
g(s) ds
s=0
By fundamental theorem of calculus
h0 (x) = g(x)
and by (8)
So
gb(w) = iwb
h(w)
icwt
gb(w) sin cwt b
e
− e−icwt
1
= h(w)
w
c
2i
−ic
1 b
=
h(w)eicwt − b
h(w)e−icwt
2c
Applying equation (16) with a = −ct and with a = ct yields
1
1
gb(w) sin cwt] =
(h(x + ct) − h(x − ct))
wc
2c
Z x+ct
Z x+ct
Z x+ct
1
1
g(s) ds −
g(s) ds =
g(s) ds
=
2c
2c x−ct
0
0
Putting both pieces together we get D’Alembert’s solution
Z x+ct
1
1
u(x, t) = (f (x − ct) + f (x + ct)) +
g(s) ds
2
2c x−ct
F −1 [
(The careful reader will notice that there might be a problem finding the fourier transform of h(x) due
to likelyhood of limx→±∞ h(x) 6= 0. But that is a story for another day.)
• Solve uxx + uyy = 0 on infinite strip (−∞, ∞) × [0, 1] with boundary conditions u(x, 0) = 0 and
u(x, 1) = f (x).
Take the Fourier Transform of all equations. The boundary conditons yield
u
b(w, 0) = 0
and the PDE gives
u
b(w, 1) = fb(w)
∂2
u
b(w, y) = 0
∂y 2
Which is basically an ODE in y, with a solution of the form
−w2 u
b(w, y) +
u
b(w, y) = A(w) cosh wy + B(w) sinh wy
The y = 0 condition implies A(w) = 0 and the y = 1 implies
fb(w)
sinh w
sinh wy
u
b(w, y) = fb(w)
sinh w
B(w) =
We get the solution
1
u(x, y) = √
2π
Z
∞
sinh wy iwx
fb(w)
e
dw
sinh w
w=−∞
• Solve ux + ut = 0; u(x, 0) = f (x) (Old Homework Problem) Take the Fourier Transform of both
equations. The initial condition gives
u
b(w, 0) = fb(w)
and the PDE gives
∂
u
b(w, t) = 0
∂t
Which is basically an ODE in t, we can write it as
iwb
u(w, t) +
and which has the solution
∂
u
b(w, t) = −iwb
u(w, t)
∂t
u
b(w, t) = A(w)e−iwt
and the initial condition above implies A(w) = fb(w)
u
b(w, t) = fb(w)e−iwt
We are now ready to inverse Fourier Transform and equation (16) above, with a = t, says that
u(x, t) = f (x − t)
• Solve ux + ut + u = 0;
u(x, 0) = f (x) (Old Homework Problem)
Take the Fourier Transform of both equations. The initial condition gives
and the PDE gives
u
b(w, 0) = fb(w)
∂
u
b(w, t) + u
b(w, t) = 0
∂t
Which is basically an ODE in t, we can write it as
iwb
u(w, t)) +
∂
u
b(w, t) = (−iw − 1)b
u(w, t)
∂t
and which has the solution
u
b(w, t) = A(w)e(−iw−1)t
and the initial condition above implies A(w) = fb(w)
u
b(w, t) = e−t fb(w)e−iwt
We are now ready to inverse Fourier Transform and equation (16) above, with a = t, says that
u(x, t) = e−t f (x − t)