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1
Quiz 5 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with
no justification will not be graded. Here are some formulae that you may want to use:
1
F(f )(ω) =
2π
def
F(e−α|x| ) =
Z
+∞
f (x)e
iωx
dx,
F
−1
Z
−∞
1
α
,
2
π ω + α2
+∞
(f )(x) =
−∞
F(
x2
2α
)(ω) = e−α|ω| ,
+ α2
f (ω)e−iωx dω,
r
2
x2
π
F(e− 4α ) = e−αω .
α
(1)
(2)
Question 1: State the convolution theorem (Do not prove).
Let f and g be two integrable functions over R (in L1 (R)). Using the definitions above, the following holds:
F(f ∗ g)(ω) = 2πF(f )(ω)F(g)(ω),
∀ω ∈ R.
Question 2: (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 ).
The definition of the Fourier transform together with the change of variable ax 7−→ x0 implies
Z +∞
1
F[eibx f (ax)])(ξ) =
f (ax)eibx eiξx dx
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
Quiz 5, February 20, 2014
Question 3: Prove the shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ , ∀ω ∈ R.
Let f be an integrable function over R (in L1 (R)), and let β ∈ R. Using the definitions above, the following holds:
1
F(f (x − β))(ω) =
2π
Z
+∞
f (x − β)eiωx dx
−∞
Upon making the change of variable z = x − β, we obtain
Z +∞
Z +∞
1
iω(z+β)
iωβ 1
f (z)e
dz = e
f (z)eiωz dz
F(f (x − β))(ω) =
2π −∞
2π −∞
= eiωβ F(f )(ω).
which proves the lemma.
Question 4: 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).
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 .