Homework 6 Solutions
Math 309, Fall 2015
10.2.18 The function is odd, and so it immediately follows that
a0 = 0 and am = 0 for all m.
Since f is 4-periodic, L = 2 and so our formula gives
Z
Z
mπx mπx 1 1
1 2
f (x) sin
x sin
dx =
dx.
bm =
2 −2
2
2 −1
2
A straightforward application of integration by parts yields
Z
mπx mπ mπ 1 1
2
4
x sin
dx = −
cos
+ 2 2 sin
.
2 −1
2
mπ
2
mπ
2
We thus conclude that the Fourier series for the given function is
∞ mπ mπ mπx X
2
4
−
cos
+ 2 2 sin
sin
mπ
2
mπ
2
2
m=0
10.2.21 The function is even, and so we immediately conclude that
bm = 0 for all m.
Since the function is again 4-periodic, we plug L = 2 into our formula for the
Fourier coefficients and find
Z
1 2 x2
4
a0 =
dx = .
2 −2 2
3
1
Similarly, integrating by parts twice, we find that
Z
mπx 1 2 x2
8
8
am =
cos
dx = 2 2 cos(mπ) = 2 2 (−1)m .
2 −2 2
2
π m
π m
The Fourier series for f is thus
∞
mπx 2 X 8
m
+
(−1)
cos
3 m=1 π 2 m2
2
10.2.27 For part (a), we use substitution. Namely, let u = x + T . We find
then that
Z a
Z T +a
Z T +a
g(x)dx =
g(u − T )du =
g(u)du,
0
T
T
where we have used the T -periodicity of g in the second equality. Since u is
just a variable of integration,
this proves the assertion in the hint. Now we
R T +a
write the integral 0 g(x)dx in two different ways:
Z
a
T +a
Z
Z
T +a
g(x)dx.
g(x)dx +
T
0
0
a
T
Z
g(x)dx =
g(x)dx =
g(x)dx +
0
T +a
Z
Since the hint tells us that the first integral on the left equals the second
integral on the right, we subtract both from this equation to find
Z T
Z T +a
g(x)dx.
g(x)dx =
a
0
For part (b), observe that there is some integer k such that a0 := a + kT ∈
[0, L]. Then setting u = x + kT , we have
Z
T +a
Z
a+kT +T
a0 +T
g(u − kT )du =
g(x)dx =
a
Z
g(u)du
a0
a+kT
where we have again used the T -periodicity of g. Since 0 ≤ a0 ≤ L, we simply
apply part (a) to conclude
Z
T +a
Z
T +a0
g(x)dx =
a
Z
g(x)dx =
a0
g(x)dx.
0
2
T
To show part (c), it follows directly from part (b) that
Z T +b
Z T
Z T +a
g(x)dx.
g(x)dx =
g(x)dx =
b
0
a
10.3.13 When ω 2 6= n2 , the solution to the homogeneous problem is
yhom (t) = c1 sin(ωt) + c2 cos(ωt),
and so a particular solution to this equation must have the form
yp (t) = a sin(nt) + b cos(nt).
Plugging this expression into the differential equation, we see that we must
have
a(ω 2 − n2 ) = 1
b(ω 2 − n2 ) = 0.
Therefore, we must have a = (ω 2 − n2 )−1 and b = 0. Thus, in this case, the
general solution to the nonhomogeneous problem is
y(t) = c1 sin(ωt) + c2 cos(ωt) + (ω 2 − n2 )−1 sin(nt).
Plugging in the initial values, we find
y(0) = 0 ⇒ c2 = 0,
y 0 (0) = 0 ⇒ c1 =
−n
.
ω(ω 2 − n2 )
One may thus express the solution to this initial value problem as the book
does, namely
ω sin(nt) − n sin(ωt)
y(t) =
ω(ω 2 − n2 )
Our approach is the same when n2 = ω 2 , except now, when finding a particular solution to the equation, we must assume it has the form
yp (t) = at sin(nt) + bt cos(nt).
Observe that since we are solving a scalar equation, it would be extraneous
to include terms of the form “c sin(nt) + d cos(nt)” since these can obviously
3
be absorbed into the general solution. Plugging the above particular solution
into the equation, we find b = −1/2n and a = 0. The general solution is thus
y(t) = c1 sin(nt) + c2 cos(nt) −
1
t cos(nt).
2n
Again, the initial values give us
y(0) = 0 ⇒ c2 = 0 y 0 (0) = 0 ⇒ nc1 −
1
1
= 0 ⇒ c1 = 2 .
2n
2n
The solution to this initial value problem is thus
y(t) =
sin(nt) − nt cos(nt)
2n2
10.3.14 Observe that if ω 6= n for any integer n, the solution to the problem
y 00 + ω 2 y =
N
X
bn sin(nt),
y(0) = 0, y 0 (0) = 0
n=1
is simply
y(t) =
N
X
bn
n=1
ω sin(nt) − n sin(ωt)
.
ω(ω 2 − n2 )
This follows directly from part (a) because the sum on the right hand side of
the differential equation is finite. Simply taking N → ∞, we conclude that
the formal solution to the problem
y 00 + ω 2 y =
∞
X
bn sin(nt),
y(0) = 0, y 0 (0) = 0
n=1
is given by
y(t) =
∞
X
n=1
bn
ω sin(nt) − n sin(ωt)
ω(ω 2 − n2 )
The reasoning is precisely the same when ω = m ∈ Z. The formal solution
is an infinite sum exactly as appears in the box above except that the term
corresponding to n = m is replaced by
bm
sin(mt) − mt cos(mt)
.
2m2
4
Your textbook thus expresses this solution as


∞
sin(mt) − mt cos(mt)
 X m sin(nt) − n sin(mt) 
y(t) = 
bm
 + bm
2
2
m(m − n )
2m2
m=1
n6=m
10.3.15 Though the problem does not state this explicitly, we assume ω 2 ∈
/ Z.
Since f (t) is clearly an odd function, its Fourier series is equal to its Fourier
sine series. Our usual formula for Fourier coefficients and straightforward
integration yields
(
Z π
4/mπ if m even
1
1
f (t) sin(mt)dt =
(2 − 2 cos(mπ)) =
.
bm =
π −π
mπ
0 if m odd
We now simply plug these values of bm into the right hand side of the differential equation in the previous problem, and thus into its solution:
y(t) =
4 ω sin(mt) − m sin(ωt)
mπ
ω(ω 2 − m2 )
m odd
X
m>0
4 X 2
1
1
2 −1
=
(ω − m )
sin(mt) − sin(ωt)
π m odd
m
ω
m>0
10.4.31 Setting u = −x, we have
Z L
Z
Z 0
Z L
f (x)dx =
f (x)dx =
f (x)dx +
−L
−L
0
0
Z
f (−u)(−du) +
Now, using the fact that f is an odd function, we have
Z 0
Z 0
Z L
f (−u)(−du) =
f (u)du = −
f (u)du.
L
L
0
Since u is just a variable of integration, we conclude that
Z
L
f (x)dx = 0
−L
5
f (x)dx.
0
L
L
10.4.36 Define f (x) to the 4-periodic “triangular wave”, i.e.
(
−x
if − 2 ≤ x ≤ 0
f (x) =
, f (x + 4) = f (x).
x
if 0 ≤ x ≤ 2
We know from Example 1 in Section 10.2 that the Fourier series for this
function is given by
∞
1
(2n − 1)πx
8 X
.
cos
1− 2
π n=1 (2n − 1)2
2
Moreover, since the given function is continuous, the Fourier series convergence theorem implies that the Fourier series converges to f (x) for all x.
Next, observe that f (0) = 0. Simply plugging this into the Fourier series,
since all the cosines become 1, we conclude that
∞
∞
X
8 X
1
1
π2
0=1− 2
⇒
=
π n=1 (2n − 1)2
(2n − 1)2
8
n=1
The sum in the box above is easily seen to be the same as the sum in the
problem, where we note that our summation starts at n = 1 and the book’s
starts at n = 0.
6