Assignment 3
Due date: July 28, 2015
1. Solve x2 y 00 − 2xy 0 + 2y = 0, y(1) = −1, y(2) = 1 if the solution exists.
Answer.
Let y = xr be a solution to x2 y 00 − 2xy 0 + 2y = 0, then
y 0 = rxr−1
y 00 = r(r − 1)xr−2
0 = x2 y 00 − 2xy 0 + 2y
= x2 · r(r − 1)xr−2 − 2x · rxr−1 + 2xr
= r(r − 1)xr − 2rxr + 2xr
= xr [r(r − 1) − 2r + 2]
= xr [r2 − 3r + 2].
Then r2 − 3r + 2 = 0, that is, r1 = 1 and r2 = 2. So the general solution to x2 y 00 − 2xy 0 + 2y = 0 is:
y = C1 x + C2 x 2 .
Since y(1) = −1 and y(2) = 1, then
C1 + C2 = −1
and
2C1 + 4C2 = 1.
Then we get
5
C1 = − ,
2
and C2 =
3
2
So the solution to x2 y 00 − 2xy 0 + 2y = 0, y(1) = −1, y(2) = 1 is:
5
3
y = − x + x2 .
2
2
2. Solve x2 y 00 + 5xy 0 + (4 + π 2 )y = ln x, y(1) = 0, y(e) = 0 if the solution exists. (Hint: Use the change of variable
x = et .)
Answer.
Let x = et (if we assume x > 0), by the chain rule, we have
dy
dt
=
=
d2 y
dt2
=
=
=
dy dx
dy
·
= et ·
dx dt
dx
dy
x·
dx
dy
d2 y dx
et ·
+ et · 2 ·
dx
dx dt
dy
d2 y
et ·
+ e2t · 2
dx
dx
dy
d2 y
x·
+ x2 · 2 .
dx
dx
By the product rule
That is, we have
x·
dy
dy
=
,
dx
dt
d2 y
dy
d2 y
d2 y dy
=
−
x
·
− .
=
dx2
dt2
dx
dt2
dt
and x2 ·
So x2 y 00 + 5xy 0 + (4 + π 2 )y = ln x is equivalent to:
dy
d2 y dy
dy
d2 y
2
−
+ 4 + (4 + π 2 )y = t.
+
5
+
(4
+
π
)y
=
dt2
dt
dt
dt2
dt
d2 y
dy
+ 4 + (4 + π 2 )y = 0, then
2
dt
dt
Let y = ert be a solution to
dy
dt
d2 y
dt2
= rert
= r2 ert
d2 y
dy
+ 4 + (4 + π 2 )y
2
dt
dt
= r2 ert + 4rert + (4 + π 2 )rt
e
0 =
= ert (r2 + 4r + 4 + π 2 ).
Then r2 + 4r + 4 + π 2 = 0, that is,
r1 = −2 + πi,
and r2 = −2 − πi.
Notice that
er1 t = e(−2+πi)t = e−2t · eiπt = e−2t [cos(πt) + i sin(πt)].
Then the general solution to
d2 y
dy
+ 4 + (4 + π 2 )y = 0 is:
dt2
dt
y = C1 e−2t cos(πt) + C2 e−2t sin(πt).
By using the method of undetermined coefficients, let yp (t) = at + b be a particular solution to
(4 + π 2 )y = t, then
4a + (4 + π 2 )(at + b) = t.
So we get
(4 + π 2 )a = 1,
and
4a + (4 + π 2 )b = 0.
So we get
a=
1
,
4 + π2
and b = −
Page 2
4
.
(4 + π 2 )2
dy
d2 y
+4 +
dt2
dt
So the general solution to
d2 y
dy
+ 4 + (4 + π 2 )y = t is:
2
dt
dt
y = C1 e−2t cos(πt) + C2 e−2t sin(πt) +
4
t
−
.
2
4+π
(4 + π 2 )2
Since t = ln x, then the general solution to x2 y 00 + 5xy 0 + (4 + π 2 )y = ln x is:
y = C1 x−2 cos(π ln x) + C2 x−2 sin(π ln x) +
ln x
4
−
.
2
4+π
(4 + π 2 )2
Since y(1) = 0 and y(e), then
C1 −
4
= 0,
(4 + π 2 )2
and
− C1 e−2 +
1
4
−
= 0.
2
4+π
(4 + π 2 )2
So we get
4
1
4
2
C1 =
,
=e
−
(4 + π 2 )2
4 + π 2 (4 + π 2 )2
which is absurd. So we do not have solution.
3. Find the eigenvalues and eigenfunctions of y 00 + λy = 0, y 0 (0) = y 0 (π) = 0.
Answer.
Let y = erx be a solution to y 00 + λy = 0, then
y 0 = rerx
y 00 = r2 erx
0 = X 00 + λX
= r2 erx + λerx
= erx (r2 + λ).
a. If λ = 0, that is, r2 = 0. So r = 0, which implies that the general solution to y 00 + λy = y 00 = 0 is
y(x) = C1 + C2 x. Since y 0 (x) = C2 , and y 0 (0) = y 0 (π) = 0, then C2 = 0, which implies that all solutions
are
y(x) = C.
b. If λ < 0, that is, λ = µ2 for some µ > 0. So r2 + µ2 = 0, that is, r = ±µi, which implies that
the general solution to y 00 + λy = y 00 + µ2 y = 0 is y(x) = C1 cos(µx) + C2 sin(µx). Since y 0 (x) =
−C1 µ sin(µx) + C2 µ cos(µx) and y 0 (0) = y 0 (π) = 0, then
C2 µ = −C1 µ sin(µπ) + C2 µ cos(µπ) = 0.
Since µ > 0, then
C1 sin(µπ) = 0,
and C2 = 0.
In order to make y to be a non-zero solution, then we need sin(πµ) = 0, that is, πµ = nπ for some n ∈ N,
that is, λ = µ2 = n2 . For λ = n2 , then all solution are:
Xn (x) = C cos(nx).
Page 3
c. If λ > 0, that is, λ = −µ2 for some µ > 0. So r2 − µ2 = 0, that is, r = ±µ, which implies that the
general solution to y 00 + λy = y 00 − µ2 y = 0 is y(x) = C1 eµx + C2 e−µx . Since y 0 (x) = C1 µeµx − C2 µe−µx
and y 0 (0) = y 0 (π) = 0, then
C1 − C2 = C1 µeµπ − C2 µe−µπ = 0.
Since µ > 0, then C1 = C2 = 0.
S
In summary, all eigenvalues of y 00 + λy = 0, y 0 (0) = y 0 (π) = 0 are: λ = n2 with n ∈ N {0}, and for eigenvalue
λ = n2 , the eigenfunctions are y = C cos(nx).
4. Find the eigenvalues and eigenfunctions of x2 y 00 − xy 0 + λy = 0, y 0 (1) = y 0 (L) = 0, where L > 1.
Answer.
Let y = xr be a solution to x2 y 00 − xy 0 + λy = 0, then
y 0 = rxr−1
y 00 = r(r − 1)xr−2
0 = x2 y 00 − 2xy 0 + 2y
= x2 · r(r − 1)xr−2 − x · rxr−1 + λxr
= r(r − 1)xr − rxr + λxr
= xr [r(r − 1) − r + λ]
= xr [r2 − 2r + λ].
Then
r=
2±
√
√
4 − 4λ
= 1 ± 1 − λ.
2
• If λ = 1, that is, r1 = r2 = 1, which implies that the general solution to x2 y 00 − xy 0 + λy = 0 is:
y = C1 x + C2 x ln x.
Then y 0 (x) = C1 + C2 ln x + C2 . Since y 0 (1) = y 0 (L) = 0, then C1 + C2 = C1 + C2 + C2 ln L = 0, then
C2 = 0. So the solution is:
y = C1 .
• If λ > 1, that is, λ = µ2 + 1 for some µ > 0, then r = 1 ± µi, which implies that the general solution to
x2 y 00 − xy 0 + λy = 0 is:
y = C1 x cos(µ ln x) + C2 x sin(µ ln x).
Then y 0 (x) = C1 cos(µ ln x) − C1 sin(µ ln x) + C2 sin(µ ln x) + C2 cos(µ ln x). Since y 0 (1) = y 0 (L) = 0,
then C1 + C2 = C1 cos(µ ln L) − C1 sin(µ ln L) + C2 sin(µ ln L) + C2 cos(µ ln L) = 0, that is, µC2 = 0 and
µC1
µC2
−
sin(µ ln L) +
cos(µ ln L) = 0. Since µ > 0, so C2 == −C1 and −2C1 sin(µ ln L) = 0. We are
L
L
looking for non-zero solutions, then µ ln L = nπ for n ∈ N, that is,
ln L 2
2
λ=µ +1=
+ 1.
nπ
And solutions are:
ln L
ln L
y = C1 x cos
ln x − x sin
ln x .
nπ
nπ
Page 4
• If λ < 1, that is, λ = 1 − µ2 for some µ > 0, then r = 1 ± µ, which implies that the general solution to
x2 y 00 − xy 0 + λy = 0 is:
y = C1 x1+µ + C2 x1−µ .
Then y 0 (x) = (1 + µ)C1 xµ + (1 − µ)C2 x−µ . Since y 0 (1) = y 0 (L) = 0, then (1 + µ)C1 + (1 − µ)C2 =
(1 + µ)C1 Lµ + (1 − µ)C2 L−µ = 0. Since µ > 0 and L > 1, then C1 = C2 = 0.
ln L 2
In summary, eigenvalues are 1, and
+ 1, eigenfunctions are
nπ
ln L
ln L
y = C, and y = C x cos
ln x − x sin
ln x , respectively.
nπ
nπ
5. If f (x) = L − x for 0 < x < 2L, and if f (x + 2L) = f (x), find a formula for f (x) in the interval −L < x < 0.
Answer. For any −L < x < 0, then L < x + 2L < 2L. Since f (x + 2L) = f (x), then
f (x) = f (x + 2L) = L − (x + 2L) = −x − L.
So f (x) = −x − L for −L < x < 0.
1, −L ≤ x < 0,
, f (x + 2L) = f (x).
6. Find the Fourier series for f (x) =
0, 0 ≤ x < L
Answer.
Let the Fourier series of f be:
∞
nπ nπ i
a0 X h
an cos
+
x + bn sin
x .
2
L
L
n=1
Then
an =
=
=
bn =
=
=
So we get
1
L
Z
L
nπ x dx
L
−L
Z
nπ 1 0
cos
x dx
L −L
L
1, if n = 0,
0, if n ≥ 1
Z L
nπ 1
f (x) sin
x dx
L −L
L
Z
nπ 1 0
sin
x dx
L −L
L
(−1)n − 1
.
nπ
f (x) cos
∞
∞
nπ 1 X (−1)n − 1
2X 1
(2k + 1)π
f (x) = +
sin
x =1−
sin
x .
2
nπ
L
π
2k + 1
L
n=1
k=0
Page 5
7. Find the Fourier series for f (x) =
Answer.
x + L, −L ≤ x ≤ 0,
, f (x + 2L) = f (x).
L,
0<x<L
Let the Fourier series of f be:
∞
nπ i
nπ a0 X h
+
x + bn sin
x .
an cos
2
L
L
n=1
Then
an =
=
=
bn =
=
=
So we get
Z
L
nπ x dx
L
−L
Z
Z
nπ nπ 1 0
1 L
(x + L) cos
L cos
x dx +
x dx
L −L
L
L 0
L

3L


,
if n = 0,

2
n

 L[1 − (−1) ] , if n ≥ 1

n2 π 2
Z L
nπ 1
f (x) sin
x dx
L −L
L
Z
Z
nπ nπ 1 0
1 L
x dx +
x
(x + L) sin
L sin
L −L
L
L 0
L
L(−1)n+1
.
nπ
1
L
f (x) cos
∞
f (x) =
3L 2L X
1
+ 2
cos
4
π
(2k + 1)2
∞
nπ (2k + 1)π
L X (−1)n+1
x +
sin
x .
L
π
n
L
n=1
k=0
8. Solve the problem:


ut = uxx , 0 < x < 40, t > 0,
BC : u(0, t) = u(40, t) = 0, t > 0,

IC : u(x, 0) = 50, 0 < x < 40.
Answer.
Recall that the “general solution” to
u(x, t) =
∞
X
ut = uxx , 0 < x < 40, t > 0,
is:
BC : u(0, t) = u(40, t) = 0, t > 0.
cn e−
n2 π 2
t
402
n=1
Let the Fourier sine series of 50 is:
50 =
∞
X
bn sin
n=1
Page 6
sin
nπ x .
40
nπ x .
40
Then
40
Z
nπ x dx
40
0
1 − (−1)n
= 100 ·
.
nπ
2
40
bn =
Then
50 sin
∞
∞
nπ 200 X
1
100 X 1 − (−1)n
(2k + 1)π
50 =
sin
x =
sin
x .
π
n
40
π
2k + 1
40
n=1
k=0
So the solution to the given heat conduction problem is:
∞
(2k+1)2 π 2
200 X 1
u(x, t) =
e− 402 t sin
π
2k + 1
k=0
(2k + 1)π
x .
40
9. Solve the problem:


ut = uxx , 0 < x < 40, t > 0,
BC : u(0, t) = u(40, t) = 0, t > 0,

IC : u(x, 0) = x, 0 < x < 40.
Answer.
ut = uxx , 0 < x < 40, t > 0,
is:
BC : u(0, t) = u(40, t) = 0, t > 0.
Recall that the “general solution” to
u(x, t) =
∞
X
cn e−
n2 π 2
t
402
n=1
Let the Fourier sine series of 50 is:
x=
∞
X
bn sin
n=1
sin
nπ x .
40
nπ x .
40
Then
bn =
2
40
Z
= 80 ·
Then
x=
40
x sin
0
nπ x dx
40
(−1)n+1
.
nπ
∞
nπ 80 X (−1)n+1
sin
x .
π
n
40
n=1
So the solution to the given heat conduction problem is:
∞
u(x, t) =
nπ 80 X (−1)n+1 − n2 π22 t
e 40 sin
x .
π
n
40
n=1
Page 7
10. Find the Fourier series for the extended function of f (x) =
Answer.
L + x, −L ≤ x < 0,
.
L − x, 0 ≤ x < L
Let the Fourier series of f be:
∞
nπ i
nπ a0 X h
+
x + bn sin
x .
an cos
2
L
L
n=1
Then
an =
=
=
bn =
=
=
Z
L
nπ x dx
L
−L
Z
Z
nπ nπ 1 0
1 L
(x + L) cos
x dx +
(L − x) cos
x dx
L −L
L
L 0
L

if n = 0,
 L,
n
 2L[1 − (−1) ] , if n ≥ 1
n2 π 2
Z L
nπ 1
f (x) sin
x dx
L −L
L
Z
Z
nπ nπ 1 0
1 L
(x + L) sin
x dx +
(L − x) sin
x
L −L
L
L 0
L
0.
1
L
f (x) cos
So we get
∞
L X 2L[1 − (−1)n ]
f (x) = +
cos
2
n2 π 2
n=1
(2k + 1)π
x
L
∞
1
L 4L X
= + 2
cos
2
π
(2k + 1)2
k=0
(2k + 1)π
x .
L
11. Solve the problem:


ut = uxx , 0 < x < 40, t > 0,
BC : ux (0, t) = ux (40, t) = 0, t > 0,

IC : u(x, 0) = 50, 0 < x < 40.
Answer.
Recall that the “general solution” to
u(x, t) =
∞
X
ut = uxx , 0 < x < 40, t > 0,
is:
BC : ux (0, t) = ux (40, t) = 0, t > 0.
cn e−
n2 π 2
t
402
cos
n=1
So the solution to the given heat conduction problem is:
u(x, t) = 50.
Page 8
nπ x .
40
12. Solve the problem:


ut = uxx , 0 < x < 40, t > 0,
BC : ux (0, t) = u(40, t) = 0, t > 0,

IC : u(x, 0) = x, 0 < x < 40.
Answer.
Recall that the “general solution” to
u(x, t) =
∞
X
ut = uxx , 0 < x < 40, t > 0,
is:
BC : ux (0, t) = u(40, t) = 0, t > 0.
(2n+1)2 π 2
t
4·402
−
cn e
cos
n=0
Let
x=
∞
X
an cos
n=0
(2n + 1)π
x .
2 · 40
(2n + 1)π
x .
2 · 40
Then
an =
=
Then
Z 40
2
(2n + 1)π
x cos
x dx
40 0
2 · 40
160[(−1)n (2n + 1)π − 2]
(2n + 1)2 π 2
∞
x=
160 X (−1)n (2n + 1)π − 2
cos
π2
(2n + 1)2
n=0
(2n + 1)π
x .
80
So the solution to the given heat conduction problem is:
∞
160 X − (2n+1)22 π2 (−1)n (2n + 1)π − 2
4·40
u(x, t) = 2
e
cos
π
(2n + 1)2
n=0
(2n + 1)π
x ..
80
13. Solve the problem:


ut = uxx , 0 < x < 40, t > 0,
BC : u(0, t) = ux (40, t) = 0, t > 0,

IC : u(x, 0) = x, 0 < x < 40.
Answer.
Recall that the “general solution” to
u(x, t) =
∞
X
−
cn e
ut = uxx , 0 < x < 40, t > 0,
is:
BC : u(0, t) = ux (40, t) = 0, t > 0.
(2n+1)2 π 2
t
4·402
n=0
Page 9
sin
(2n + 1)π
x .
2 · 40
Let
x=
∞
X
n=0
bn sin
(2n + 1)π
x .
2 · 40
Then
Z 40
(2n + 1)π
2
x sin
x dx
40 0
2 · 40
320(−1)n
.
(2n + 1)2 π 2
bn =
=
Then
∞
320 X (−1)n
x= 2
sin
π
(2n + 1)2
n=0
(2n + 1)π
x .
80
So the solution to the given heat conduction problem is:
∞
u(x, t) =
320 X − (2n+1)22 π2 (−1)n
4·40
e
sin
π2
(2n + 1)2
n=0
Page 10
(2n + 1)π
x ..
80