TW364
Tutorial #5 Solutions
22 September 2016
Problem 1:
Z
∞
(a) C(α) =
f (x)eiαx dx =
−∞
(b)
Z
∞
e(iα−1)x dx =
0
1
1 − iα
=⇒ f (x) =
1
2π
Z
∞
−∞
1
e−iαx dα
1 − iα
1
1
α
1 + iα
=
+i
and e−iαx = cos(αx) − i sin(αx)
=
1 − iα
1 + α2
1 + α2
1 + α2
Z ∞h
i
1
α ih
1
+
i
cos(αx)
−
i
sin(αx)
dα
Hence f (x) =
2π −∞ 1 + α2
1 + α2
Z ∞h
1
cos(αx) i sin(αx) iα cos(αx) α sin(αx) i
=
−
+
+
dα
2π −∞ 1 + α2
1 + α2
1 + α2
1 + α2
Z
1 ∞ cos(αx) + α sin(αx)
=
dα [terms 1 and 4 in the line above are even in α, and the others odd]
π 0
1 + α2
Problem 2:
(a) We note that f is even, so we determine the cosine integral:
Z ∞
Z ∞
πα sin(πα) + cos(πα) − 1
A(α) =
f (x) cos(αx) dx =
x cos(αx) dx =
α2
0
0
Z ∞
πα sin(πα) + cos(πα) − 1
2
cos(αx) dα
f (x) =
π 0
α2
(b) We note that f is odd, so we determine the sine integral:
Z ∞
Z ∞
sin(πα) − πα cos(πα)
B(α) =
f (x) sin(αx) dx =
x sin(αx) dx =
α2
0
0
Z
2 ∞ sin(πα) − πα cos(πα)
f (x) =
sin(αx) dα
π 0
α2
Problem 3:
n ∂ 2 u o d2 U
n ∂2u o
Let U (α, t) = F{u(x, t)}. Then F
=
and
F
a2 2 = −a2 α2 U .
∂t2
dt2
∂x
d2 U
= −a2 α2 U =⇒ U (α, t) = c1 cos(aαt) + c2 sin(aαt).
dt2
Taking the Fourier transform of the initial conditions yields U (α, 0) = F (α) and Ut (α, 0) = G(α).
We solve the ODE
U (α, 0) = F (α) =⇒ c1 · 1 + c2 · 0 = F (α) =⇒ c1 = F (α)
Ut (α, 0) = G(α) =⇒ −aαc1 · 0 + aαc2 · 1 = G(α) =⇒ c2 =
Thus U (α, t) = F (α) cos(aαt) +
G(α)
aα
G(α)
sin(aαt).
aα
We take the inverse Fourier transform: u(x, t) =
1
2π
Z
∞
G(α)
F (α) cos(aαt) +
sin(aαt) e−iαx dα.
aα
−∞
1
Problem 4:
∂2u ∂2u
+ 2 = 0, uy (x, 0) = 0,
∂x2
∂y
n ∂ 2 u o d2 U
n ∂2u o
= −α2 U and F
=
.
Let U (α, y) = F{u(x, y)}. Then F
2
∂x
∂y 2
dy 2
The boundary value problem to be solved:
We solve the ODE
d2 U
= α2 U
dy 2
2
u(x, 1) = e−x .
=⇒ U (α, y) = c1 cosh(αy) + c2 sinh(αy).
2
Taking the Fourier transform of the initial conditions yields Uy (α, 0) = 0 and U (α, 1) = F{e−x } =
√
πe−
α2
4
.
Uy (α, 0) = 0 =⇒ αc1 · 0 + αc2 · 1 = 0 =⇒ c2 = 0
U (α, 1) =
√
πe−
α2
4
=⇒ c1 cosh α + 0 =
√
Thus U (α, y) =
√
√
πe−
α2
4
=⇒ c1 =
α2
πe− 4
cosh α
α2
πe− 4
cosh(αy).
cosh α
1
We take the inverse Fourier transform: u(x, y) = √
2 π
Z
∞
−∞
√
α2
πe− 4
cosh(αy)e−iαx dα.
cosh α
Note: this integral can be simplified by writing e−iαx in the form cos(αx) + i sin(αx), and splitting the integrand
into even and odd parts.
2