December 14, 2015
Lecturer Dmitri Zaitsev
Michaelmas Term 2015
Course 2E01 2015 (SF Engineers & MSISS & MEMS)
Sheet 9
Due: at the end of the tutorial
Exercise 1
Identify even or odd functions and find their Fourier series for the period −1 ≤ x ≤ 1:
(i) f (x) = 2x;
(ii) f (x) = −x2 .
Exercise 2
Use Fourier series to find a solution of the equation
y 00 (x) − y(x) = a(x),
where a(x) = 2x for −1 ≤ x ≤ 1.
Solution:
For the interval [−1, 1] and the odd function a(x) we calculate
Z
1
Z
bn =
a(x)sin(nπx) dx =
−1
=
1
(uv)|1−1
Z
2xsin(nπx) dx =
−1
Z
1
1
cos(nπx) 1
|−1 + 2
−
u v dx = −2x
nπ
−1
0
uv 0 dx
−1
Z
1
cos(nπx) dx
−1
4(−1)n
4
= − cos(nπ) + 0 = −
,
nπ
nπ
hence
a(x) =
X
bn sin(nπx) = −
4 X (−1)n
sin(nπx).
π
n
Next, for y we use general Fourier Series
y(x) = a0 +
∞
X
(an cos(nπx) + bn sin(nπx)),
n=1
and its second derivative
00
y (x) = −
∞
X
(n2 π 2 an cos(nπx) + n2 π 2 bn sin(nπx)),
n=1
and substitute in our differential equation. Identifying constant coefficients we obtain
−a0 = 0.
Identifying coefficients of cos(nπx) we have
n2 π 2 an − an = 0,
whence
an = 0.
Finally identifying coefficients of sin(nπx) we obtain
−(n2 π 2 bn − bn ) = −
4(−1)n
,
nπ
from where
bn =
4(−1)n
nπ(1 + n2 π 2 )
and hence
y(x) =
X
bn sin(nπx) = 4
X
4(−1)n
sin(nπx).
nπ(1 + n2 π 2 )
Exercise 3
Find the Fourier integral representation of the function
f (x) =
x if |x| < 1
0 if |x| > 1.
Solution:
Since the function is odd, we only need to compute the odd coefficient
Z
Z
1 1
1 1 0
B(w) =
x sinwx dx =
uv dx
π −1
π −1
Z 1
Z 1
1
1
−coswx x=1
coswx
x=1
0
=
(uv)|x=−1 −
u v dx =
(x
)|x=−1 +
dx
π
π
w
w
−1
−1
1 −2cosw sinwx x=1
1 −2cosw 2sinw
,
|
dx =
=
+
+
π
w
w2 x=−1
π
w
w2
and so the Fourier integral is
Z +∞
Z
2 +∞ sinw − wcosw
f (x) =
B(w)sinwx dw =
sinwx dw.
π 0
w2
0