Math 602
6
43
Fourier transform
Here are some formulae that you may want to use:
Z +∞
Z +∞
def 1
iωx
−1
F(f )(ω) =
f (x)e dx,
F (f )(x) =
f (ω)e−iωx dω,
2π −∞
−∞
(3)
F(f ∗ g) = 2πF(f )F(g),
1
2α
α
F(e−α|x| ) =
,
F( 2
)(ω) = e−α|ω| ,
π ω 2 + α2
x + α2
F(f (x − β))(ω) = eiβω F(f )(ω),
2
ω2
1
F(e−αx ) = √
e− 4α
4πα
1
1
−ax
F(H(x)e
)(ω) =
,
H is the Heaviside function
2π a − iω
i
i
1
F(pv( ))(ω) = sign(ω) = (H(ω) − H(−ω))
x
2
2
1
π
1
2
def
F(sech(ax)) =
sech( ω),
sech(ax) =
= x
2a
2a
ch(x)
e + e−x
cos(a) − cos(b) = −2 sin( 12 (a + b)) sin( 21 (a − b))
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
Question 102: (i) Let f be an integrable function on (−∞, +∞). Prove that for all a, b ∈ R,
and for all ξ ∈ R, F([eibx f (ax)])(ξ) = a1 F(f )( ξ+b
a ).
Solution: The definition of the Fourier transform together with the change of variable ax 7−→ x0
implies
Z +∞
1
f (ax)eibx eiξx dx
F[eibx f (ax)])(ξ) =
2π −∞
Z +∞
1
=
f (ax)ei(b+ξ)x dx
2π −∞
Z +∞
(ξ+b) 0
1
1
=
f (x0 )ei a x dx0
2π −∞ a
= a1 F(f )( ξ+b
a ).
2
(ii) Let c be a positive real number. Compute the Fourier transform of f (x) = e−cx sin(bx).
√
Solution: Using the fact that sin(bx) = −i 21 (eibx − e−ibx ), and setting a = c we infer that
f (x) =
=
1 −(ax)2 ibx
(e − e−ibx )
2i e
1 −(ax)2 ibx
1 −(ax)2 i(−b)x
e − 2i
e
e
2i e
and using (i) and (7) we deduce
fˆ(ξ) =
2
− 14 ( ξ+b
1 1 √1
a )
2i a 4π (e
− e− 4 (
1
ξ−b 2
a )
fˆ(ξ) =
1
− 4c
(ξ+b)2
1 √1
2i 4πc (e
− e− 4c (ξ−b) ).
1
2
).
In conclusion
Question 103: (a) Compute the Fourier transform of the function f (x) defined by
(
1 if |x| ≤ 1
f (x) =
0 otherwise
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44
Solution: By definition
F(f )(ω) =
1
2π
Z
1
1 1 iω
(e − e−iω )
2π iω
1 2 sin(ω)
.
=
2π
ω
eiξω = =
−1
Hence
F(f )(ω) =
1 sin(ω)
.
π ω
(b) Find the inverse Fourier transform of g(ω) =
sin(ω)
ω .
Solution: Using (a) we deduce that g(ω) = πF(f )(ω), that is to say, F −1 (g)(x) = πF −1 (F(f ))(x).
Now, using the inverse Fourier transform, we deduce that F −1 (g)(x) = πf (x) at every point x where
f (x) is of class C 1 and F −1 (g)(x) = π2 (f (x− ) + f (x+ )) at discontinuity points of f . As a result:


π if |x| < 1
−1
F (g)(x) = π2 at |x| = 1


0 otherwise
Question 104: Find the inverse Fourier transform of
Solution: Since F(Sλ (x)) =
1 sin(λω
π
ω .
1 sin(λω)
,
π
ω
the inverse Forier transform theorem implies that


1 if |x| < λ
1 sin(λω)
−1
F
(x) = 21 if |x| = λ

π
ω

0 otherwise.
Question 105: Use the Fourier transform technique to solve the following PDE:
∂t u(x, t) + c∂x u(x, t) + γu(x, t) = 0,
for all x ∈ (−∞, +∞), t > 0, with u(x, 0) = u0 (x) for all x ∈ (−∞, +∞).
Solution: By taking the Fourier transform of the PDE, one obtains
∂t F(u) − iωcF(y) + γF(y) = 0.
The solution is
F(u)(ω, t) = c(ω)eiωct−γt .
The initial condition implies that c(ω) = F(u0 )(ω):
F(u)(ω, t) = F(u0 )(ω)eiωct e−γt .
The shift lemma in turn implies that
F(u)(ω, t) = F(u0 (x − ct))(ω)e−γt = F(u0 (x − ct)e−γt )(ω).
Applying the inverse Fourier transform gives:
u(x, t) = u0 (x − ct)e−γt .
Question 106: Solve by the Fourier transform technique the following equation: ∂xx φ(x) −
2∂x φ(x) + φ(x) = H(x)e−x , x ∈ (−∞, +∞), where H(x) is the Heaviside function. (Hint:
use the factorization iω 3 + ω 2 + iω + 1 = (1 + ω 2 )(1 + iω) and recall that F(f (x))(−ω) =
F(f (−x))(ω)).
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Solution: Applying the Fourier transform with respect to x gives
(−ω 2 + 2iω + 1)F(φ)(ω) = F(H(x)e−x )(ω) =
1
1
.
2π 1 − iω
where we used (8). Then, using the hint gives
1
1
1
1
=
2π (1 − iω)(−ω 2 + 2iω + 1)
2π iω 3 + ω 2 + iω + 1
1
1
1
=
.
2π 1 + ω 2 1 + iω
F(φ)(ω) =
We now use again (8) and (5) to obtain
F(φ)(ω) = π
1
1
1 1
= πF(e−|x| )(ω)F(H(x)e−x )(−ω).
π 1 + ω 2 2π 1 − i(−ω)
Now we use F(H(x)e−x )(−ω) = F(H(−x)ex )(ω) and we finally have
F(φ)(ω) = πF(e−|x| )(ω)F(H(−x)ex )(ω).
The Convolution Theorem (4) gives
F(φ)(ω) = π
1
F(e−|x| ∗ (H(−x)ex ))(ω).
2π
We obtain φ by using the inverse Fourier transform
φ(x) =
i.e.,
Z
1 −|x|
e
∗ (H(−x)ex ),
2
+∞
φ(x) =
−∞
1 −|x−y|
e
H(−y)ey dy
2
and recalling that H is the Heaviside function we finally have
(
Z
1 −x
e
1 0 y−|x−y|
φ(x) =
e
dy = 41
2 −∞
( 4 − x)ex
if x ≥ 0
if x ≤ 0.
Question 107: Use the Fourier transform technique to solve the following ODE y 00 (x) − y(x) =
f (x) for x ∈ (−∞, +∞), with y(±∞) = 0, where f is a function such that |f | is integrable over
R.
Solution: By taking the Fourier transform of the ODE, one obtains
−ω 2 F(y) − F(y) = F(f ).
That is
F(y) = −F(f )
1
.
1 + ω2
and the convolution Theorem, see (4), together with (5) gives
1
F(y) = −πF(f )F(e−|x| ) = − F(f ∗ e−|x| ).
2
Applying F −1 on both sides we obtain
1
1
y(x) = − f ∗ e−|x| = −
2
2
Z
∞
−∞
e−|x−z| f (z)dz
Math 602
That is
1
y(x) = −
2
Z
46
∞
e−|x−z| f (z)dz.
−∞
Question 108: Use the Fourier transform method to compute the solution of utt − a2 uxx = 0,
where x ∈ R and t ∈ (0, +∞), with u(0, x) = f (x) := sin2 (x) and ut (0, x) = 0 for all x ∈ R.
Solution: Take the Fourier transform in the x direction:
F(u)tt + ω 2 a2 F(u) = 0.
This is an ODE. The solution is
F(u)(t, ω) = c1 (ω) cos(ωat) + c2 (ω) sin(ωat).
The initial boundary conditions give
F(u)(0, ξ) = F(f )(ω) = c1 (ω)
and c2 (ω) = 0. Hence
F(t, ω) = F(f )(ω) cos(ωat) =
1
F(f )(ω)(eiaωt + e−iaωt ).
2
Using the shift lemma, we infer that
F(t, ω) =
1
(F(f (x − at))(ω) + F(f (x + at))(ω)).
2
Usin the inverse Fourier transform, we finaly conclude that
u(t, x) =
1
1
(f (x − at) + f (x + at)) = (sin2 (x + at) + sin2 (x − at)).
2
2
Note that this is the D’Alembert formula.
Question 109: Use the Fourier transform method to compute the solution of utt − a2 uxx = 0,
where x ∈ R and t ∈ (0, +∞), with u(0, x) = f (x) := cos2 (x) and ut (0, x) = 0 for all x ∈ R.
Solution: Take the Fourier transform in the x direction:
F(u)tt + ω 2 a2 F(u) = 0.
This is an ODE. The solution is
F(u)(t, ω) = c1 (ω) cos(ωat) + c2 (ω) sin(ωat).
The initial boundary conditions give
F(u)(0, ξ) = F(f )(ω) = c1 (ω)
and c2 (ω) = 0. Hence
F(t, ω) = F(f )(ω) cos(ωat) =
1
F(f )(ω)(eiaωt + e−iaωt ).
2
Using the shift lemma (i.e., formula (6)) we obtain
u(t, x) =
1
1
(f (x − at) + f (x + at)) = (cos2 (x + at) + cos2 (x − at)).
2
2
Note that this is the D’Alembert formula.
Question 110: Solve the integral equation: f (x) +
x ∈ (−∞, +∞).
1
2π
R +∞
f (y)
dy
−∞ (x−y)2 +1
=
1
x2 +4
+
1
x2 +1 ,
for all
Math 602
47
Solution: The equation can be re-written
f (x) +
1
1
1
1
f∗ 2
= 2
+
.
2π
x +1
x + 4 x2 + 1
We take the Fourier transform of the equation and we apply the Convolution Theorem (see (4))
F(f ) +
1
1
1
1
2πF( 2
)F(f ) = F( 2
) + F( 2
).
2π
x +1
x +4
x +1
Using (5), we obtain
1
1
1
F(f ) + e−|ω| F(f ) = e−2|ω| + e−|ω| ,
2
4
2
which gives
1
1
1
F(f )(1 + e−|ω| ) = e−|ω| ( e−|ω| + 1).
2
2
2
We then deduce
1 −|ω|
e
.
2
Taking the inverse Fourier transform, we finally obtain f (x) =
F(f ) =
1
x2 +1 .
Question 111: Solve the following integral equation (Hint: solution is short):
Z +∞
√ Z +∞ − y2
x2
e 2π f (x − y)dy = −2πe− 4π .
f (y)f (x − y)dy − 2 2
∀x ∈ R.
−∞
−∞
Solution: This equation can be re-written using the convolution operator:
√
x2
x2
f ∗ f − 2 2e− 2π ∗ f = −2πe− 4π .
We take the Fourier transform and use the convolution theorem (4) together with (7) to obtain
√
−ω 2 11
−ω 2 11
1
1
4
4
2π = −2π q
4π
e
e
2πF(f )2 − 2π2 2F(f ) q
1
1
4π 2π
4π 4π
F(f )2 − 2F(f )e−ω
2π
2
+ e−ω
(F(f ) − e
−ω 2 π
2
2
π
=0
)2 = 0.
This implies
F(f ) = e−ω
2π
2
.
Taking the inverse Fourier transform, we obtain
f (x) =
√
x2
2e− 2π .
Question 112: Solve the integral equation: f (x) +
(−∞, +∞).
3
2
R +∞
−∞
e−|x−y| f (y)dy = e−|x| , for all x ∈
Solution: The equation can be re-written
3
f (x) + e−|x| ∗ f = e−|x| .
2
We take the Fourier transform of the equation and apply the Convolution Theorem (see (4))
3
F(f ) + 2πF(e−|x| )F(f ) = F(e−|x| ).
2
Using (5), we obtain
F(f ) + 3π
1 1
1 1
F(f ) =
,
2
π1+ω
π 1 + ω2
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48
which gives
F(f )
ω2 + 4
1 1
=
.
1 + ω2
π 1 + ω2
We then deduce
1 1
1
= F(e−2|x| ).
2
π4+ω
2
Taking the inverse Fourier transform, we finally obtain f (x) = 21 e−2|x| .
R +∞
2
1 2
Question 113: Solve the following integral equation −∞ e−(x−y) g(y)dy = e− 2 x for all x ∈
(−∞, +∞), i.e., find the function g that solves the above equation.
F(f ) =
Solution: The left-hand side of the equation is a convolution; hence,
2
1
2
e−x ∗ g(x) = e− 2 x .
By taking the Fourier transform, we obtain
1 2
1 2
1
1
2π √ e− 4 ω Fg(ω) = √ e− 2 ω .
4π
2π
That gives
1 2
1
F(g)(ω) = √ e− 4 ω .
2π
By taking the inverse Fourier transform, we deduce
r
√
4π −x2
2 −x2
e
.
g(x) = √ e
=
π
2π
Question 114: Solve the integral equation:
How many solutions did you find?
R +∞
−∞
f (y)f (x−y)dy =
4
x2 +4 ,
for all x ∈ (−∞, +∞).
Solution: The equation can be re-written
f ∗f =
4
.
x2 + 4
We take the Fourier transform of the equation and apply the Convolution Theorem (see (4))
2πF(f )2 = F(
x2
4
)
+4
Using (5), we obtain
2πF(f )2 = e−2|ω| .
which gives
1
F(f ) = ± √ e−|ω| .
2π
Taking the inverse Fourier transform, we finally obtain
1
2
f (x) = ± √
.
2
2π x + 1
We found two solutions: a positive one and a negative one.
Question 115: Use the Fourier transform method to solve the equation ∂t u +
u(x, 0) = u0 (x), in the domain x ∈ (−∞, +∞) and t > 0.
Solution: We take the Fourier transform of the equation with respect to x
2t
∂x u)
1 + t2
2t
= ∂t F(u) +
F(∂x u)
1 + t2
2t
= ∂t F(u) − iω
F(u).
1 + t2
0 = ∂t F(u) + F(
2t
1+t2 ∂x u
= 0,
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49
This is a first-order linear ODE:
∂t F(u)
d
2t
= iω (log(1 + t2 ))
= iω
2
F(u)
1+t
dt
The solution is
2
F(u)(ω, t) = K(ω)eiω log(1+t ) .
Using the initial condition, we obtain
F(u0 )(ω) = F(u)(ω, 0) = K(ω).
The shift lemma (see formula (6)) implies
F(u)(ω, t) = F(u0 )(ω)eiω log(1+t
2
)
= F(u0 (x − log(1 + t2 ))),
Applying the inverse Fourier transform finally gives
u(x, t) = u0 (x − log(1 + t2 )).
Question 116: Solve the integral equation:
Z +∞ Z +∞ √
√ − y2
1
2 − (x−y)2
2π
f (y) − 2e
−
f (x−y)dy = −
e 2π dy,
2
2
1+y
−∞
−∞ 1 + y
∀x ∈ (−∞, +∞).
(Hint: there is an easy factorization after applying the Fourier transform.)
Solution: The equation can be re-written
f ∗ (f −
√
x2
2e− 2π −
√
x2
1
1
)=−
∗ 2e− 2π .
2
2
1+x
1+x
We take the Fourier transform of the equation and apply the Convolution Theorem (see (4))
√
√
x2
x2
1
1
2πF(f ) F(f ) − 2F(e− 2π ) − F(
)
= −2πF(
) 2F(e− 2π )
2
2
1+x
1+x
Solution 1: Using (5), (7) we obtain
√
x2
2F(e− 2π ) =
F(
√
2
− ω1
πω 2
1
2q
e 4 2π = e− 2
1
4π 2π
1
1
) = e−|ω| ,
2
1+x
2
which gives
πω 2
πω 2
1
1
F(f ) F(f ) − e− 2 − e−|ω| = − e−|ω| e− 2 .
2
2
This equation can also be re-written as follows
F(f )2 − F(f )e−
πω 2
2
πω 2
1
1
− F(f ) e−|ω| + e−|ω| e− 2 = 0,
2
2
and can be factorized as follows:
(F(f ) − e−
This means that either F(f ) = e−
finally obtain two solutions
πω 2
2
f (x) =
πω 2
2
1
)(F(f ) − e−|ω| ) = 0.
2
or F(f ) = 21 e−|ω| . Taking the inverse Fourier transform, we
√
x2
2e− 2π ,
or
f (x) =
1
.
1 + x2
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50
Solution 2: Another solution consists of remarking that the equation with the Fourier transform can
be rewritten as follows:
√
x2
F(f )2 − F( 2e− 2π )F(f ) − F(
√
x2
1
1
)F(f ) + F( 2e− 2π )F(
) = 0,
2
1+x
1 + x2
which can factorized as follows:
√
x2
(F(f ) − F( 2e− 2π ))(F(f ) − F(
1
)) = 0.
1 + x2
√
x2
1
The either F(f ) = F( 2e− 2π )) or (F(f ) = F( 1+x
2 ). The conclusion follows easily.
Question 117: Use the Fourier transform technique to solve ∂t u(x, t) + sin(t)∂x u(x, t) + (2 +
3t2 )u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x). (Use the shift lemma: F(f (x − β))(ω) =
R +∞
1
f (x)eiωx dx)
F(f )(ω)eiωβ and the definition F(f )(ω) := 2π
−∞
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + sin(t)(−iω)F(u)(ω, t) + (2 + 3t2 )F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= iω sin(t) − (2 + 3t2 ).
F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −iω(cos(t) − 1) − (2t + t3 ).
This implies
3
F(u)(ω, t) = F(u0 )(ω)e−iω(cos(t)−1) e−(2t+t ) .
Then the shift lemma gives
3
F(u)(ω, t) = F(u0 (x + cos(t) − 1)(ω)e−(2t+t ) .
This finally gives
3
u(x, t) = u0 (x + cos(t) − 1)e−(2t+t ) .
Question 118: Solve the following integral equation (Hint: x2 − 3xa + 2a2 = (x − a)(x − 2a)):
Z
+∞
√
y2
x2
(f (y) − 3 2e− 2π )f (x − y)dy = −4πe− 4π .
∀x ∈ R.
−∞
Solution: This equation can be re-written using the convolution operator:
√
x2
x2
f ∗ f − 3 2e− 2π ∗ f = −4πe− 4π .
We take the Fourier transform and use (7) to obtain
√
−ω 2 11
−ω 2 11
1
1
4
4
2π = −4π q
4π
2πF(f )2 − 2π3 2F(f ) q
e
e
1
1
4π 2π
4π 4π
F(f )2 − 3F(f )e−ω
(F(f ) − e
−ω 2 π
2
2π
2
+ 2e−ω
)(F(f ) − 2e
2
−ω 2 π
2
π
=0
) = 0.
This implies
either
F(f ) = e−ω
2π
2
,
or
F(f ) = 2e−ω
2π
2
.
Math 602
Taking the inverse Fourier transform, we obtain
√
x2
either f (x) = 2e− 2π ,
or
51
√
x2
f (x) = 2 2e− 2π .
Question 119: Use the Fourier transform technique to solve ∂t u(x, t)−∂xx u(x, t)+sin(t)∂x u(x, t)+
(2 + 3t2 )u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x). (Hint: use the definition F(f )(ω) :=
R +∞
2
ω2
1
1
iωx
dx), the result F(e−αx )(ω) = √4πα
e− 4α , the convolution theorem and the
2π −∞ f (x)e
shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ . Go slowly and give all the details.)
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + ω 2 F(u)(ω, t) + sin(t)(−iω)F(u)(ω, t) + (2 + 3t2 )F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= −ω 2 + iω sin(t) − (2 + 3t2 ).
F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −ω 2 t − iω(cos(t) − 1) − (2t + t3 ).
This implies
2
3
F(u)(ω, t) = F(u0 )(ω)e−ω t e−iω(cos(t)−1) e−(2t+t ) .
2
Using the result F(e−αx )(ω) =
ω2
√ 1 e− 4α
4πα
where α =
1
4t ,
this implies that
r
3
x2
π
F(u0 )(ω)F(e− 4t )(ω)e−iω(cos(t)−1) e−(2t+t )
t
r
3
x2
π
=
F(u0 ∗ e− 4t )(ω)e−iω(cos(t)−1) e−(2t+t ) .
t
F(u)(ω, t) =
x2
Then setting g = u0 ∗ e− 4t the convolution theorem followed by the shift lemma gives
r
3
1
π
F(u)(ω, t) =
F(g(x + cos(t) − 1))(ω)e−(2t+t ) .
2π t
This finally gives
r
r
Z +∞
(x+cos(t)−1−y)2
1
1
−(2t+t3 )
−(2t+t3 )
4t
u(x, t) =
g(x + cos(t) − 1)e
=e
u0 (y)e−
dy.
4πt
4πt −∞
Question 120: Consider the telegraph equation ∂tt u + 2α∂t u + α2 u − c2 ∂xx u = 0 with α ≥ 0,
u(x, 0) = 0, ∂t u(x, 0) = g(x), x ∈ R, t > 0 and boundary condition at infinity u(±∞, t) =
0. Solve the equation by the Fourier transform technique. (Hint: the solution to the ODE
φ00 (t) + 2αφ0 (t) + (α2 + λ2 )φ(t) = 0 is φ(t) = e−αt (a cos(λt) + b sin(λt))
Solution: Applying the Fourier transform with respect to x to the equation, we infer that
0 = ∂tt F(u)(ω, t) + 2α∂t F(u)(ω, t) + α2 F(u)(ω, t) − c2 (−iω)2 F(u)(ω, t)
= ∂tt F(u)(ω, t) + 2α∂t F(u)(ω, t) + (α2 + c2 ω 2 )F(u)(ω, t)
Using the hint, we deduce that
F(u)(ω, t) = e−αt (a(ω) cos(ωct) + b(ω) sin(ωct)).
The initial condition implies that a(ω) = 0 and F(g)(ω) = ωcb(ω); as a result, b(ω) = F(g)(ω)/(ωc)
and
sin(ωct)
.
F(u)(ω, t) = e−αt F(g)
ωc
Math 602
Then using (??), we have
F(u)(ω, t) =
52
π −αt
e F(g)F(Sct ).
c
The convolution theorem implies that
u(x, t) = e
−αt
1
1
g ∗ Sct = e−αt
2c
2c
Z
∞
g(y)Sct (x − y)dy.
−∞
Finally the definition of Sct implies that Sct (x − y) is equal to 1 if −ct < x − y < ct and is equal
zero otherwise, which finally means that
Z x+ct
−αt 1
g(y)dy.
u(x, t) = e
2c x−ct
Question 121: Use the Fourier transform technique to solve ∂t u(x, t)−∂xx u(x, t)+cos(t)∂x u(x, t)+
(1 + 2t)u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x). (Hint: use the definition F(f )(ω) :=
R +∞
2
ω2
1
1
iωx
dx), the result F(e−αx )(ω) = √4πα
e− 4α , the convolution theorem and the
2π −∞ f (x)e
shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ . Go slowly and give all the details.)
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + ω 2 F(u)(ω, t) + cos(t)(−iω)F(u)(ω, t) + (1 + 2t)F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= −ω 2 + iω cos(t) − (1 + 2t).
F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −ω 2 t + iω sin(t) − (t + t2 ).
This implies
2
2
F(u)(ω, t) = F(u0 )(ω)e−ω t eiω sin(t) e−(t+t ) .
2
Using the result F(e−αx )(ω) =
ω2
√ 1 e− 4α
4πα
where α =
1
4t ,
this implies that
r
2
x2
π
F(u0 )(ω)F(e− 4t )(ω)eiω sin(t) e−(t+t )
t
r
2
x2
π
=
F(u0 ∗ e− 4t )(ω)eiω sin(t) e−(t+t ) .
t
F(u)(ω, t) =
x2
Then setting g = u0 ∗ e− 4t the convolution theorem followed by the shift lemma gives
r
2
1
π
F(u)(ω, t) =
F(g(x − sin(t)))(ω)e−(t+t ) .
2π t
This finally gives
r
u(x, t) =
2
2
1
g(x − sin(t))e−(t+t ) = e−(t+t )
4πt
r
1
4πt
Z
+∞
u0 (y)e−
(x−sin(t)−y)2
4t
dy.
−∞
Question 122: Solve the following integral equation (Hint: (a + b)2 = a2 + 2ab + b2 ):
Z
+∞
Z
+∞
f (y)f (x − y)dy − 2
−∞
−∞
2
4
f (x − y)dy + 2π 2
=0
y2 + 1
x +4
∀x ∈ R.
Math 602
53
Solution: This equation can be re-written using the convolution operator:
f ∗ f − 2(
2
4
) ∗ f + 2π 2
= 0.
x2 + 1
x +4
We take the Fourier transform and use the convolution theorem (??) together with (??) to obtain
2πF(f )2 − 4πF(f )e−|ω| + 2πe−2|ω| = 0
F(f )2 − 2F(f )e−|ω| + e−2|ω| = 0
(F(f ) − e−|ω| )2 = 0
This implies
F(f ) = e−|ω| .
Taking the inverse Fourier transform, we obtain
f (x) =
x2
2
.
+1
Question 123: State the shift lemma (Do not prove).
Solution: Let f be an integrable function over R (in L1 (R)), and let β ∈ R. Using the definitions
above, the following holds:
F(f (x − β))(ω) = F(f )(ω)eiωβ ,
∀ω ∈ R.
Question 124: Use the Fourier transform technique to solve ∂t u(x, t)+t∂x u(x, t)+2u(x, t) = 0,
x ∈ R, t > 0, with u(x, 0) = u0 (x).
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + t(−iω)F(u)(ω, t) + 2F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= iωt − 2.
F(u)(ω, t)
Then applying the fundamental theorem of calculus we obtain
1
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = iω t2 − 2t.
2
This implies
1 2
F(u)(ω, t) = F(u0 )(ω)eiω 2 t e−2t .
Then the shift lemma gives
1
F(u)(ω, t) = F(u0 (x − t2 )(ω)e−2t .
2
This finally gives
1
u(x, t) = u0 (x − t2 )e−2t .
2
Question 125: Prove the shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ , ∀ω ∈ R.
Solution: Let f be an integrable function over R (in L1 (R)), and let β ∈ R. Using the definitions
above, the following holds:
Z +∞
1
F(f (x − β))(ω) =
f (x − β)eiωx dx
2π −∞
Math 602
54
Upon making the change of variable z = x − β, we obtain
Z +∞
Z +∞
1
1
f (z)eiω(z+β) dz = eiωβ
f (z)eiωz dz
F(f (x − β))(ω) =
2π −∞
2π −∞
= eiωβ F(f )(ω).
which proves the lemma.
Question 126: Use the Fourier transform technique to solve ∂t u(x, t) + cos(t)∂x u(x, t) +
sin(t)u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x).
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + cos(t)(−iω)F(u)(ω, t) + sin(t)F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= iω cos(t) − sin(t).
F(u)(ω, t)
Then applying the fundamental theorem of calculus, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = iω sin(t) + cos(t) − 1.
This implies
F(u)(ω, t) = F(u0 )(ω)eiω sin(t) ecos(t)−1 .
Then the shift lemma gives
F(u)(ω, t) = F(u0 (x − sin(t))(ω)ecos(t)−1 .
This finally gives
u(x, t) = u0 (x − sin(t))ecos(t)−1 .