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1
Mid term 2. 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:
Rt
Solution to y 0 (t) + g(t)y(t) = 0 is y(t) = y(0)e− 0 g(τ )dτ .
Z +∞
Z +∞
def 1
F(f )(ω) =
f (x)eiωx dx,
F −1 (f )(x) =
f (ω)e−iωx dω,
2π −∞
−∞
F(f (x − β))(ω) = F(f )(ω)eiωβ
r
2
x2
2α
α
π
1
−α|ω|
,
F(
)(ω)
=
e
,
F(e− 4α ) = e−αω .
F(e−α|x| ) =
2
2
2
2
πω +α
x +α
α
+∞
+∞
X
X
F S(f )(x) =
an cos(nπx/L) +
bn sin(nπx/L),
F(f ∗ g)(ω) = 2πF(f )(ω)F(g)(ω),
n=0
a0 =
1
2L
Z
L
f (x)dx,
−L
an =
1
L
Z
n=1
L
x
f (x) cos(nπ )dx,
L
−L
bn =
1
L
Z
(1)
(2)
(3)
(4)
(5)
L
x
f (x) sin(nπ )dx
L
−L
(6)
Question 1: Consider f : [−L, L] −→ R, f (x) = x2 . (a) Sketch the graph of the Fourier series of f .
F S(f ) is equal to the periodic extension of f (x) over R.
(b) For what values of x is F S(f ) equal to x2 ?
The periodic extension of f (x) = x2 over R is piecewise smooth and globally continuous. This means that
the Fourier series is equal to x2 over the entire interval [−L, +L].
(c) Is it possible to obtain F S(x) by differentiating 21 F S(f ) term by term?
Yes it is possible since the periodic extension of f (x) = x2 over R is continuous and piecewise smooth.
Question 2: Let f (x) = x, x ∈ [−L, L]. (a) Sketch the graph of the Fourier series of f .
The Fourier series is equal to the periodic extension of f , except at the points (2n + 1)L, n ∈ Z where it
is equal to 0 = 21 (1 − 1).
2
Mid-Term TEST, September 28, 2010
(b) Compute the coefficients of the Fourier series of f . (Hint:
Rb
a
R
Rb R
tg(t)dt = [t g]ba − a ( g)(t)dt).
f is odd, hence the cosine coefficients are zero. The sine coefficients bn are obtained by integration by
parts
Z
Z
nπx
nπx
1 L
1 L
x
L 1 L
x sin(
cos(
bn =
)dx = −
[x cos(nπ )]+L
)dx.
+
−L
L −L
L
L nπ
L
nπ L −L
L
L
L
As a result bn = −2 cos(nπ) nπ
= 2(−1)n+1 nπ
and
∞
F S(f )(x) =
2L X (−1)n+1
nπx
sin(
).
π 1
n
L
Question 3: Let L be a positive real number. Let V = span{1, cos(πt/L), sin(πt/L)} and consider
R
12
L
the norm kf kL2 = −L f (t)2 dt . (a) Compute the best approximation of 1 + t in V with respect
to the above norm.
We know from class that the truncated Fourier series
S1 (t) = a0 + a1 cos(πt/L) + b1 sin(πt/L)
is the best approximation. Now we compute a0 , a1 , a2
Z L
1
(1 + t)dt = 1,
2L −L
Z
1 L
(1 + t) cos(πt/L)dt = 0
a1 =
L −L
Z
Z
1 L
2L
1 L
L
b1 =
.
(1 + t) sin(πt/L)dt =
t sin(πt/L)dt = −2 cos(π) =
L −L
L −L
π
π
a0 =
As a result
S1 (t) = 1 +
2L
sin(πt/L)
π
(b) Compute the best approximation of 3 + 2 cos(πt/L) − 5 sin(πt/L) in V .
The function 3 + 2 cos(πt/L) − 5 sin(πt/L) is a member of V ; as a result, The best approximation is the
function itself.
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3
Question 4: 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).
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
4
Mid-Term TEST, September 28, 2010
Question 5: Solve ∂t u(x, t) − 2∂xx u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x) and u(±∞, t) = 0,
∂x u(±∞, t) = 0.
Applying the Fourier transform with respect to the x-variable:
∂t F(u)(ω, t) − 2(−iω)2 F(u)(ω, t) = 0.
This give the ODE
∂t F(u)(ω, t) + 2ω 2 F(u)(ω, t) = 0.
The solution is
2
F(u)(ω, t) = a(ω)e−2ω t .
The initial condition implies
F(u)(ω, t) = F(u0 )(ω)e
−2ω 2 t
r
=
x2
π
F(u0 )(ω)F(e− 8t )(ω, t).
2t
The convolution theorem gives
1
F(u)(ω, t) =
2π
As a result
r
u(x, t) =
1
8πt
r
x2
π
F(u0 ∗ e− 8t )(ω, t).
2t
Z
+∞
−∞
u0 (y)e−
(x−y)2
8t
dy.
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5
Question 6: 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.
This equation can be re-written using the convolution operator:
f ∗ f − 2(
x2
2
4
) ∗ f + 2π 2
= 0.
+1
x +4
We take the Fourier transform and use the convolution theorem (3) together with (4) 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
6
Mid-Term TEST, September 28, 2010
Question 7: Prove that if there exists a smooth solution to the Klein-Gordon equation then it is
unique: ∂tt u(x, t) − c2 ∂xx u(x, t) + m2 u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = f (x), ∂t u(x, 0) = g(x)
and u(±∞, t) = 0, ∂t u(±∞, t) = 0, ∂x u(±∞, t) = 0. (Hint: test with ∂t u(x, t) and use φφ0 = ( 21 φ2 )0 .)
Let u1 and u2 be two solutions. Then setting φ = u1 − u2 , we obtain that φ solves the homogeneous
problem ∂tt φ − c2 ∂xx φ + m2 φ = 0, x ∈ R, t > 0, with φ(x, 0) = φ(x), ∂t φ(x, 0) = 0 and φ(±∞, t) = 0,
∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0. Testing with ∂t φ(x, t) and integrating over R and using the property
∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0, we obtain
Z +∞
Z +∞
1
1
∂t ( φ2 )dx
∂xx φ∂t φdx + m2
∂t ( (∂t φ)2 )dx − c2
2
2
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞
1
1 2
= dt
∂x φ∂t ∂x φdx + m2 dt
(∂t φ)2 dx + c2
φ dx
2
2
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞
1
1 2
1
(∂t φ)2 dx + c2 dt
(∂x φ)2 dx + m2 dt
φ dx
= dt
−∞ 2
−∞ 2
−∞ 2
Z +∞
Z +∞
Z +∞
1
1 2
1
= dt
(∂t φ)2 dx + c2
(∂x φ)2 dx + m2
φ dx
−∞ 2
−∞ 2
−∞ 2
Z +∞
= dt
(∂t φ)2 + c2 (∂x φ)2 + m2 φ2 dx .
Z
0=
+∞
−∞
Introduce
Z
+∞
E(t) =
(∂t φ)2 + c2 (∂x φ)2 + m2 φ2 dx.
−∞
Then
dt E(t) = 0.
The fundamental Theorem of calculus gives
E(t) = E(0) = 0.
This means
Z
+∞
(∂t φ(x, t))2 + c2 (∂x φ(x, t))2 + m2 φ2 (x, t) dx = 0,
for all t ≥ 0.
−∞
This implies
0 = (∂t φ(x, t))2 + c2 (∂x φ(x, t))2 + m2 φ2 (x, t),
i.e. φ = 0, thereby proving the uniqueness.
for all t ≥ 0, x ∈ R,