Assignment 2
Due date: July 21, 2015
For Problem 1, 2, 3, 4, 5 and 6, you only need to three of them.
1. For the differential equation 2x2 y 00 + (3x − x2 )y 0 − y = 0 for x > 0, find the first three terms of a non-zero
series solution at x = 0, satisfying lim y(x) = 0.
x&0
Answer.
Since 2x2 y 00 + (3x − x2 )y 0 − y = 0, then
y 00 +
3x − x2 0
1
y − 2 y = 0.
2
2x
2x
That is,
y 00 +
Let p(x) =
3−x 0
1
y − 2 y = 0.
2x
2x
1
3−x
and q(x) = − 2 , then
2x
2x
xp(x) =
3−x
3 1
= − x,
2
2 2
1
and x2 q(x) = − .
2
So x = 0 is a regular singular point to the differential equation 2x2 y 00 + (3x − x2 )y 0 − y = 0, and the indicial
equation is:
3
1
1
1
0 = r(r − 1) + p0 r + q0 = r(r − 1) + r − = r2 + r − .
2
2
2
2
So the exponents at x = 0 are:
1
r1 = ,
2
Let φ(x, r) = xr
∞
X
k=0
ak (r)xk =
∞
X
and r2 = −1.
ak (r)xk+r for some {ak (r)}∞
k=0 which will be determined later, then
k=0
φx (x, r) =
∞
X
(k + r)ak (r)xk+r−1
k=0
φxx (x, r) =
∞
X
k=0
(k + r)(k + r − 1)ak (r)xk+r−2 .
So we have
2x2 φxx (x, r) + (3x − x2 )φx (x, r) − φ(x, r)
∞
∞
∞
X
X
X
= 2x2 ·
(k + r)(k + r − 1)ak (r)xk+r−2 + (3x − x2 )
(k + r)ak (r)xk+r−1 −
ak (r)xk+r
k=0
=
=
∞
X
k=0
∞
X
2(k + r)(k + r − 1)ak (r)xk+r +
(k + r)ak (r)xk+r+1 −
k=0
k=0
k=0
∞
X
∞
X
∞
X
2(k + r)(k + r − 1)ak (r)xk+r +
k=0
=
3(k + r)ak (r)xk+r −
k=0
∞
X
∞
X
2(n + r)(n + r − 1)an (r)xn+r +
n=0
3(k + r)ak (r)xk+r −
k=0
∞
X
n=0
= 2r(r − 1)a0 (r) + 3rao (r) − a0 (r) +
∞
X
ak (r)xk+r
k=0
(n + r − 1)an−1 (r)xn+r −
n=1
∞
X
3(n + r)an (r)xn+r −
∞
X
∞
X
ak (r)xk+r
k=0
∞
X
(n + r − 1)an−1 (r)xn+r −
n=1
an (r)xn+r
n=0
[2(n + r)(n + r − 1)an (r) + 3(n + r)an (r) − (n + r − 1)an−1 (r)
n=1
−an (r)]xn+r
2
= (2r − 2r + 3r − 1)a0 (r) +
∞
X
{[2(n + r)(n + r − 1) + 3(n + r) − 1]an (r) − (n + r − 1)an−1 (r)}xn+r
n=1
= (2r2 + r − 1)a0 (r) +
∞
X
{[2(n + r)(n + r − 1) + 3(n + r) − 1]an (r) − (n + r − 1)an−1 (r)}xn+r .
n=1
Let a0 (r) = 1, and define the sequence {an (r)}∞
n=1 be the following recurrence relation:
[2(n + r)(n + r − 1) + 3(n + r) − 1]an (r) − (n + r − 1)an−1 (r) = 0,
That is,
an (r) =
=
n+r−1
an−1 (r)
2(n + r)(n + r − 1) + 3(n + r) − 1
n+r−1
an−1 (r), ∀n ≥ 1.
(n + r)(2n + 2r + 1) − 1
So we get
2x2 φxx (x, r) + (3x − x2 )φx (x, r) − φ(x, r) = 2r2 + r − 1.
Page 2
∀n ≥ 1.
∞
1 X
1
1
2
=x
xk , then lim y(x) = 0 and
Let y(x) = φ x,
ak
x&0
2
2
k=0
1
a0
= 1
2
1 + 12 − 1
1
1
=
a1
· a0
1
1
2
2
(1 + 2 )(2 · 1 + 2 · 2 + 1) − 1
1
=
10
2 + 12 − 1
1
1
a2
=
· a1
1
1
2
2
(2 + 2 )(2 · 2 + 2 · 2 + 1) − 1
3
=
.
280
2. Consider the differential equation 2(x − 1)y 00 + y 0 + y = 0 for x > 1.
(a) Show that x = 1 is a regular singular point. Find the indicial equation and the exponents at x = 1.
Answer.
Since 2(x − 1)y 00 + y 0 + y = 0, then
y 00 +
Let p(x) =
1
1
y0 +
y = 0.
2(x − 1)
2(x − 1)
1
1
and q(x) =
, then
2(x − 1)
2(x − 1)
1
(x − 1)p(x) = ,
2
So x = 1 is a regular singular point and p0 =
and
1
2
(x − 1)2 q(x) =
x−1
.
2
and q0 = 0. Then the indicial equation is:
1
1
0 = r(r − 1) + p0 r + q0 = r(r − 1) + r = r2 − r.
2
2
Then the exponents at x = 1 are r1 =
1
2
and r2 = 0.
(b) Find the first two non-zero terms of two linearly independent series solutions about x = 1.
Answer.
r
Let φ(x, r) = (x − 1)
∞
X
k
ak (r)(x − 1) =
k=0
∞
X
ak (r)(x − 1)k+r for some {ak (r)}∞
k=0 which will
k=0
be determined later, then
φx (x, r) =
∞
X
(k + r)ak (r)(x − 1)k+r−1
k=0
φxx (x, r) =
∞
X
(k + r)(k + r − 1)ak (r)(x − 1)k+r−2 .
k=0
Page 3
Then we have
2(x − 1)φxx (x, r) + φx (x, r) + φ(x, r)
∞
∞
∞
X
X
X
= 2(x − 1)
(k + r)(k + r − 1)ak (r)(x − 1)k+r−2 +
(k + r)ak (r)(x − 1)k+r−1 +
ak (r)(x − 1)k+r
k=0
=
=
∞
X
k=0
2(k + r)(k + r − 1)ak (r)(x − 1)k+r−1 +
∞
X
k=0
(k + r)ak (r)(x − 1)k+r−1 +
∞
X
k=0
k=0
k=0
∞
X
∞
X
∞
X
2(k + r)(k + r − 1)ak (r)(x − 1)k+r−1 +
k=0
(k + r)ak (r)(x − 1)k+r−1 +
k=0
r−1
r−1
ak (r)(x − 1)k+r
ak−1 (r)(x − 1)k+r−1
k=1
= 2r(r − 1)a0 (r)(x − 1)
+ ra0 (r)(x − 1)
∞
X
+
[2(k + r)(k + r − 1)ak (r) + (k + r)ak (r) + ak−1 (r)](x − 1)k+r−1
k=1
= [2r(r − 1) + r]a0 (r)(x − 1)r−1 +
∞
X
{(k + r)[2(k + r − 1) + 1]ak (r) + ak−1 (r)}(x − 1)k+r−1
k=1
= [2r2 − r]a0 (r)(x − 1)r−1 +
∞
X
[(k + r)(2k + 2r − 1)ak (r) + ak−1 (r)](x − 1)k+r−1 .
k=1
Let a0 (r) = 0, define {ak (r)}∞
k=1 be the following recurrence relation:
(k + r)(2k + 2r − 1)ak (r) + ak−1 (r) = 0,
for all k ≥ 1.
Then we get
ak (r) = −
ak−1 (r)
,
(k + r)(2k + 2r − 1)
for all k ≥ 1.
So we have
2(x − 1)φxx (x, r) + φx (x, r) + φ(x, r) = (2r2 − r)(x − 1)r−1 .
Let
X
∞
1
1
1
y1 (x) = φ x,
=
ak
(x − 1)k+ 2 ,
2
2
k=0
and y2 (x) = φ(x, 0) =
∞
X
ak (0)(x − 1)k .
k=0
Since a0 (r) = 1, then y1 (x) and y2 (x) are two linearly independent solutions. Notice that
1
a0
= a0 (0) = 1
2
a0 12
1
a1
= −
2
1 + 12 2 · 1 + 2 · 12 − 1
1
= −
3
a0 (0)
a1 (0) = −
(1 + 0)(2 · 1 + 2 · 0 − 1)
= −1.
Page 4
(c) Find the first two non-zero terms of the series solution about x = 1, satisfying lim y(x) = 1 and
x&1
√
lim x − 1y 0 (x) = 1.
x&1
Answer.
By the result of part (b), then there are two constants C1 and C2 such that
y(x) = C1 y1 (x) + C2 y2 (x)
1
3
1
2
2
= C1 (x − 1) − (x − 1) + · · · + C2 1 − (x − 1) + a2 (0)(x − 1)2 + · · · .
3
Since lim y(x) = 1, then C2 = 1. Since
x&1
0
y (x) = C1
1
1
3 1
− 12
2
(x − 1) − · (x − 1) + · · · + C2 [−1 + 2a2 (x − 1) + · · · ] .
2
2 3
Then
√
0
x − 1y (x) = C1
√
i
h √
1
1
1 3 1
+1
− · (x − 1) 2 + · · · + C2 − x − 1 + 2a2 (x − 1)1+ 2 + · · · .
2 2 3
C1
= 1, that is, C1 = 2. So we get
2
1
3
1
2
2
y(x) = 2y1 (x) + y2 (x) = 2 (x − 1) − (x − 1) + · · · + 1 − (x − 1) + a2 (0)(x − 1)2 + · · · .
3
Since lim
x&1
x − 1y 0 (x) = 1, then
3. Consider the Laguerre differential equation:
xy 00 + (1 − x)y 0 + λy = 0,
x > 0,
where λ is a real constant.
(a) Show that x = 0 is a regular singular point. Find the indicial equation and the exponents at x = 0.
Answer.
Since xy 00 + (1 − x)y 0 + λy = 0, then
y 00 +
Let p(x) =
1 − x 0 λx
y +
= 0.
x
y
1−x
λ
and q(x) = , then
x
x
xp(x) = 1 − x,
and x2 q(x) = λx.
So x = 0 is a regular singular point, p0 = 1 and q0 = 0. So the indicial equation is:
0 = r(r − 1) + p0 r + q0 = r(r − 1) + r = r2 − r + r = r2 .
Then the exponents at x = 0 are r1 = r2 = 0.
Page 5
(b) Find the general series solutions near x = 0.
Answer.
Let φ(x, r) = x
r
∞
X
k
ak (r)x =
k=0
∞
X
ak (r)xk+r for some {ak (r)}∞
k=0 which will be determined
k=0
later, then
φx (x, r) =
∞
X
(k + r)ak (r)xk+r−1
k=0
φxx (x, r) =
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 .
k=0
Then
xφxx (x, r) + (1 − x)φx (x, r) + λφ(x, r)
∞
∞
∞
X
X
X
= x
(k + r)(k + r − 1)ak (r)xk+r−2 + (1 − x)
(k + r)ak (r)xk+r−1 + λ
ak (r)xk+r
=
=
k=0
∞
X
∞
X
k=0
∞
X
k=0
∞
X
k=0
∞
X
k=0
∞
X
k=0
k=1
(k + r)(k + r − 1)ak (r)xk+r−1 +
(k + r)(k + r − 1)ak (r)xk+r−1 +
k=0
+λ
∞
X
k=0
(k + r)ak (r)xk+r−1 −
(k + r)ak (r)xk+r−1 −
(k + r)ak (r)xk+r + λ
(k − 1 + r)ak−1 (r)xk+r−1
ak−1 (r)xk+r−1
k=1
k=1
r−1
= r a0 (r)x
+
∞
X
[(k + r)(k + r − 1 + 1)ak (r) − (k − 1 + r − λ)ak−1 (r)]xk+r−1
k=1
∞
X
= r2 a0 (r) +
[(k + r)2 ak (r) − (k − 1 + r − λ)ak−1 (r)]xk+r−1 .
k=1
Let a0 (r) = 1, define {ak (r)}∞
k=1 be the following recurrence relation:
(k + r)2 ak (r) − (k − 1 + r − λ)ak−1 (r) = 0,
∀k ≥ 1.
That is,
ak (r) =
k−1+r−λ
· ak−1 (r),
(k + r)2
Page 6
ak (r)xk+r
k=0
= r(r − 1)a0 (r)xr−1 + ra0 (r)xr−1
∞
X
+
[(k + r)(k + r − 1)ak (r) + (k + r)ak (r) − (k − 1 + r)ak−1 (r) + λak−1 (r)]xk+r−1
2
∞
X
∀k ≥ 1.
That is, we have
ak (r) =
=
=
k−1+r−λ
(k + r)2
k−1+r−λ
(k + r)2
k−1+r−λ
(k + r)2
= a0 (r)
k−1−1+r−λ
· ak−1−1 (r)
(k − 1 + r)2
k−1−1+r−λ
1−1+r−λ
·
· ··· ·
a0 (r)
(k − 1 + r)2
(1 + r)2
·
k
Y
i−1+r−λ
i=1
=
· ak−1 (r)
(i + r)2
k
Y
i−1−λ+r
i=1
(i + r)2
,
for all k ≥ 1.
Then
xφxx (x, r) + (1 − x)φx (x, r) + λφ(x, r) = r2 .
Let y1 (x) = φ(x, 0) = 1 +
∞
X
k=1
∞
k
X
Y
i−1−λ
ak (0)xk =
i2
k=0
(1)
!
xk , then y1 (x) is a solution. Notice that
i=1
a0 (0) = 1
1−1−λ+0
a1 (0) =
(1 + 0)2
= −λ
2−1+0−λ
a2 (0) =
· (−λ)
(2 + 0)2
1−λ
(−λ)
=
4
λ2 − λ
.
=
4
By (1), then we get
xφxxr (x, r) + (1 − x)φxr (x, r) + λφr (x, r) = 2r.
Let y2 (x) = φr (x, 0) =
∞
X
a0k (0)xk +
k=0
∞
X
ak (0)xk ln x, then y2 (x) is also a solution. Since a0 (r) = 1,
k=0
1−1−λ+r
r−λ
2−1+r−λ
1+r−λ
r−λ
a1 (r) =
=
, and a2 (r) =
· a1 (r) =
·
, then
2
2
2
2
(1 + r)
(r + 1)
(2 + r)
(r + 2)
(r + 1)2
a00 (0) = 0,
a01 (0) = 1 + 2λ,
Page 7
and a02 (0) =
1 + λ − 3λ2
.
4
Then general solution is :
y(x) = C1 y1 (x) + C2 y2 (x)
λ2 − λ 2
x + ···
= C1 1 − λx +
4
1 + λ − 3λ2 2
λ2 − λ 2
+C2 (1 + 2λ)x +
x + · · · + ln x 1 − λx +
x + ···
.
4
4
(c) Show that if λ is a non-negative integer n, then there is a polynomial solution of degree n.
Answer.
By the result of part (b), then
∞
k
X
Y
i−1−λ
y1 (x) = 1 +
i2
k=1
!
xk ,
i=1
is a solution.
When λ = 0, it’s easy to see that y = 1 is a solution. Now when λ = n for some positive integer, then
k
Y
i−1−λ
i=1
i2
k
Y
i−1−n
=
i2
i=1
= 0,
for all k ≥ n + 1.
On the other hand, we know that
an (0) =
n
Y
i−1−λ
i=1
i2
=
n
Y
i−1−n
i=1
i2
= (−1)n
n
Y
n+1−i
i=1
i2
= (−1)n ·
n!
(−1)n
=
.
(n!)2
n!
So y1 (x) is a polynomial solution of degree n.
(d) Find a polynomial solution for each of the case λ = n = 0, 1, 2, 3.
Answer.
By the result of part (c), then
• When λ = 0, y = 1 is a polynomial solution.
• When λ = 1, y = 1 − x is a polynomial solution.
1−1−2
1
1
• When λ = 2, y = 1 +
x + x2 = 1 − 2x + x2 is a polynomial solution.
12
2!
2
1−1−3
1−1−3 2−1−3 2 1 3
3
1
• When λ = 3, y = 1+
x+
·
x − x = 1−3x− x2 − x3 is a polynomial
2
2
2
1
1
2
3!
2
6
solution.
4. Consider the differential equation:
x(x − 1)y 00 + 6x2 y 0 + 3y = 0,
x > 0,
(a) Show that x = 0 is a regular singular point. Find the indicial equation and the exponents at x = 0.
Page 8
Answer.
Since x(x − 1)y 00 + 6x2 y 0 + 3y = 0, then
y 00 +
Let p(x) =
3
6x2
y0 +
y = 0.
x(x − 1)
x(x − 1)
6x2
6x
3
=
and q(x) =
, then
x(x − 1)
x−1
x(x − 1)
6x2
1
= −6x2 ·
x−1
1−x
∞
X
= −6x2
xk
xp(x) =
k=0
= −6
= −6
∞
X
k=0
∞
X
xk+2
xk
for all |x| < 1
k=2
1
3x
= −3x ·
x−1
1−x
∞
X
xk
= −3x ·
x2 q(x) =
k=0
= −3
= −3
∞
X
k=0
∞
X
xk+1
xk ,
for all |x| < 1.
k=1
So x = 0 is a regular singular point, and p0 = q0 = 0. So the indicial equation is:
0 = r(r − 1) + p0 r + q0 = r(r − 1) = r2 − r.
Then the exponents at x = 0 are r1 = 1 and r2 = 0.
(b) Find the general series solutions near x = 0. What would you expect the radius of convergence of the
series solution be?
Answer.
r
Let φ(x, r) = x
∞
X
k
ak (r)x =
k=0
∞
X
ak (r)xk+r for some {ak (r)}∞
k=0 which will be determined
k=0
later, then
φx (x, r) =
∞
X
(k + r)ak (r)xk+r−1
k=0
φxx (x, r) =
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 .
k=0
Page 9
Then
x(x − 1)φxx (x, r) + 6x2 φx (x, r) + 3φ(x, r)
∞
∞
∞
X
X
X
= (x2 − x)
(k + r)(k + r − 1)ak (r)xk+r−2 + 6x2
(k + r)ak (r)xk+r−1 + 3
ak (r)xk+r
k=0
k=0
k=0
∞
∞
∞
X
X
X
k+r
k+r−1
=
(k + r)(k + r − 1)ak (r)x
−
(k + r)(k + r − 1)ak (r)x
+
6(k + r)ak (r)xk+r+1
k=0
∞
X
+
k=0
k=0
3ak (r)xk+r
k=0
∞
∞
X
X
k−1+r
=
(k − 1 + r)(k − 1 + r − 1)ak−1 (r)x
−
(k + r)(k + r − 1)ak (r)xk+r−1
k=1
∞
X
+
=
k=0
6(k − 2 + r)ak−2 (r)xk−2+r+1 +
k=2
∞
X
∞
X
3ak−1 (r)xk−1+r
k=1
(k + r − 1)(k + r − 2)ak−1 (r)xk+r−1 −
k=1
∞
X
+
∞
X
(k + r)(k + r − 1)ak (r)xk+r−1
k=0
6(k + r − 2)ak−2 (r)xk+r−1 +
k=2
∞
X
3ak−1 (r)xk+r−1
k=1
= r(r − 1)a0 (r)xr − r(r − 1)a0 (r)xr−1 − (1 + r)ra1 (r)xr
∞
X
+
[(k + r − 1)(k + r − 2)ak−1 (r) − (k + r)(k + r − 1)ak (r) + 6(k + r − 2)ak−2 (r) + 3ak−1 (r)] xk+r−1 +
k=2
= −r(r − 1)a0 (r)xr−1 + [r(r − 1)a0 (r) − r(r + 1)a1 (r) + 3a0 (r)]xr
∞
X
+
{−(k + r)(k + r − 1)ak (r) + [(k + r − 1)(k + r − 2) + 3]ak−1 (r) + 6(k + r − 2)ak−2 (r)} xk+r−1
k=2
Define the sequence {ak (r)}∞
k=1 by the following recurrence relation:
r(r − 1)a0 (r) − r(r + 1)a1 (r) + 3a0 (r) = 0,
−(k + r)(k + r − 1)ak (r) + [(k + r − 1)(k + r − 2) + 3]ak−1 (r) + 6(k + r − 2)ak−2 (r) = 0,
∀k ≥ 2.
That is, we have
a1 (r) =
ak (r) =
r(r − 1) + 3
a0 (r)
r(r − 1)
[(k + r − 1)(k + r − 2) + 3]ak−1 (r) + 6(k + r − 2)ak−2
,
(k + r)(k + r − 1)
∀k ≥ 2.
Let a0 (r) = 1, it’s see that all ak (1)’s are well defined, let y1 (x) = φ(x, 1), then y1 (x) is a solution. Let
∂ a0 (r) = r, and y2 (x) =
(rφ(x, r)) is another solution. The general solution is:
∂r r=0
y(x) = C1 y1 (x) + C2 y2 (x).
Page 10
Since xp(x) = −6
∞
X
k
2
x and x q(x) = −3
∞
X
k=2
xk for all |x| < 1, then the radius of convergence of the
k=1
general series solution would be at least 1.
9
5. Find the general series solution near x = 0 to x y + xy + x −
4
2 00
Since x2 y 00 + xy 0 + x2 −
Answer.
9
4
0
2
y = 0.
y = 0, then
1 0 x2 − 49
y + y +
y = 0.
x
x2
00
x2 − 49
1
9
and q(x) =
, then x = 0 is a singular point. Notice that xp(x) = 1 and x2 q(x) = x2 − ,
2
x
x
4
both are analytic at x = 0, then x = 0 is a regular singular point. So p0 = 1 and q0 = − 94 , which implies that
the indicial equation is:
9
9
0 = r(r − 1) + p0 r + q0 = r(r − 1) + r − = r2 − .
4
4
Let p(x) =
So the exponents at x = 0 are r1 =
3
2
and r2 = − 23 . Let ψ(x, r) := xr
∞
X
ak (r)xk =
k=0
∞
X
ak (r)xk+r for some
k=0
differentiable functions ak (r), for all x > 0, then
ψx (x, r) =
ψxx (x, r) =
∞
X
k=0
∞
X
(k + r)ak (r)xk+r−1
(k + r)(k + r − 1)ak (r)xk+r−2 .
k=0
So we get
9
x ψxx (x, r) + xψx (x, r) + x −
ψ(x, r)
4
∞
∞
∞
X
X
9 X
2
k+r−2
k+r−1
2
= x
(k + r)(k + r − 1)ak (r)x
+x
(k + r)ak (r)x
+ x −
ak (r)xk+r
4
2
=
2
k=0
∞
X
∞
X
k=0
k=0
(k + r)(k + r − 1)ak (r)xk+r +
k=0
(k + r)ak (r)xk+r +
∞
X
k=0
ak (r)xk+r+2 −
k=0
∞
X
9
4
ak (r)xk+r
k=0
∞ ∞
X
X
9
=
(k + r)(k + r − 1) + k + r −
ak (r)xk+r +
ak (r)xk+r+2
4
k=0
k=0
∞
∞ X
X
9
=
(k + r)2 −
ak (r)xk+r +
ak−2 (r)xk+r
4
k=2
k=0
∞ X
9
9
9
2
r
2
1+r
2
=
r −
a0 (r)x + (1 + r) −
a1 (r)x
+
(k + r) −
ak (r) + ak−2 (r) xk+r .
4
4
4
k=2
Page 11
Let a1 (r) = 0, define the sequence {ak (r)}∞
k=1 by the recurrence relation:
9
(k + r)2 −
ak (r) + ak−2 (r) = 0, for all k ≥ 2.
4
That is,
ak (r) =
−1
(k + r)2 −
9
4
· ak−2 (r),
for all k ≥ 1.
Then
a2k+1 (r) = 0,
and a2k (r) = (−1)k a0 (r)
k
Y
1
,
2− 9
(2i
+
r)
4
i=1
for all k ≥ 0.
So we get
ψ(x, r) = x
r
∞
X
a2k (r)x
2k
r
= a0 (r)x + a0 (r)x
k=0
9
x2 ψxx (x, r) + xψx (x, r) + x2 −
4
r
∞
X
k
(−1)
k=1
9
2
ψ(x, r) =
r −
a0 (r)xr .
4
k
Y
1
(2i + r)2 −
i=1
!
9
4
x2k (2)
(3)
3
3 2 9
2
. By
For any 1 ≤ i ≤ k, then (2i + ) − = 4i + 3i > 0, which implies that we can define φ1 (x) = φ x,
2
4
2
(3), then φ1 (x) is a solution . Notice that
!
∞
k
Y
3
3 X
1
k
φ1 (x) = x 2 + x 2
x2k
(−1)
9
3 2
)
−
(2i
+
2
4
i=1
k=1
"
!
#
∞
k
X
Y
3
1
= x2 1 +
(−1)k
x2k .
(2i) · (2i + 3)
i=1
k=1
Let’s find another solution φ2 (x): Since r1 − r2 = 3, but a3 (r) = 0 which is well-defined. For any 2 ≤ i ≤ k,
3 2 9
we have 2i −
− = 4i2 − 6i = i(2i − 3) ≥ 2 · (2 · 2 − 3) = 2 > 0. Let a0 (r) = 1, then
2
4
a2 (r) = (−1)a0 (r) ·
=
1
(2 · 1 + r)2 −
−1
r+
r + 12
7
2
a2k (r) = (−1)k+1 · a2 (r) ·
k
Y
1
(2i + r)2 −
i=2
k
=
9
4
9
4
Y
(−1)k
1
·
.
7
1
r + 2 r + 2 i=2 (2i + r)2 − 49
Page 12
3
Hence all a2k (r)’s are defined near r =
Let φ2 (x) = φ x, − , since a0 (r) = 1, by (3), then φ2 (x) is a
2
solution which is linearly independent with φ1 (x). Notice that
3
φ2 (x) = φ x, −
2
"
!
#
∞
k
X
Y
3
1
= x− 2 1 +
x2k
(−1)k
3 2
9
−4
i=1 2i − 2
k=1
"
!
#
∞
k
X
Y
3
1
= x− 2 1 +
x2k .
(−1)k
(2i)(2i − 3)
− 32 .
i=1
k=1
So the general solution is:
y(x) = C1 φ1 (x) + C2 φ2 (x)
( "
∞
X
3
= C1 x 2 1 +
(−1)k
+C2
"
x
− 32
1+
1
(2i) · (2i + 3)
i=1
k=1
(
k
Y
∞
X
(−1)k
k
Y
i=1
k=1
1
(2i)(2i − 3)
6. Consider the differential equation:
x3 y 00 + αxy 0 + βy = 0,
x > 0,
where α and β are real constants and α 6= 0.
(a) Show that x = 0 is an irregular singular point.
Answer.
Since x3 y 00 + αxy 0 + βy = 0, then
y 00 +
Let p(x) =
α 0
β
y + 3 y = 0.
2
x
x
α
β
and q(x) = 3 , then
2
x
x
xp(x) =
α
,
x
and x2 q(x) =
So x = 0 is irregular.
(b) Find a solution of the from
∞
X
an xr+n with a0 = 1.
n=0
Page 13
!
β
.
x
#)
x
2k
!
#)
x2k
.
Answer.
Let φ(x, r) =
∞
X
ak (r)xk+r with a0 (r) = 1, and {ak (r)}∞
k=1 will be determined later, then
k=0
φx (x, r) =
∞
X
(k + r)ak (r)xk+r−1
k=0
φxx (x, r) =
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 .
k=0
Then we have
x3 φxx (x, r) + αxφx (x, r) + βφ(x, r)
∞
∞
∞
X
X
X
k+r+1
k+r
=
(k + r)(k + r − 1)ak (r)x
+α
(k + r)ak (r)x
+β
ak (r)xk+r
k=0
=
∞
X
k=0
(k − 1 + r)(k − 1 + r − 1)ak−1 (r)x
k+r
k=0
+α
k=1
=
∞
X
∞
X
(k + r)ak (r)x
k+r
∞
X
+β
k=0
(k + r − 1)(k + r − 2)ak−1 (r)xk+r + α
k=1
∞
X
k=0
(k + r)ak (r)xk+r + β
k=0
= αra0 (r)xr + βa0 (r)xr +
∞
X
ak (r)xk+r
∞
X
ak (r)xk+r
k=0
[(k + r − 1)(k + r − 2)ak−1 (r) + α(k + r)ak (r) + βak (r)]xk+r
k=1
∞
X
= (αr + β)xr + +
[(k + r − 1)(k + r − 2)ak−1 (r) + (α(k + r) + β)ak (r)]xk+r .
k=1
Then αr + β = 0. Let {ak (r)}∞
k=1 be defined by the following recurrence relation:
(k + r − 1)(k + r − 2)ak−1 (r) + (α(k + r) + β)ak (r) = 0,
for all k ≥ 1.
That is, we get
β
r=− ,
α
ak (r) = −
and
(k + r − 1)(k + r − 2)
· ak−1 (r),
α(k + r) + β
for all k ≥ 1.
β
Since r = − , then
α
α(k + r) + β = αk + αr + β = αk 6= 0, for all k ≥ 1.
β
β
So we know that φ x, −
is well defined. In summary, we know that y(x) = φ x, −
is a solution.
α
α
Notice that
ak (r) =
k
Y
(−1)(i + r − 1)(i + r − 2)
α(i + r) + β
i=1
k
= (−1)
k
Y
(i + r − 1)(i + r − 2)
i=1
αi + αr + β
Page 14
,
for all k ≥ 1.
So we get
β
β
k
i
−
+
1
i
−
+
2
Y
α
α
β
= (−1)k
ak −
α
αi
i=1
k β
(−1)k Y
β
i− +1
=
i− +2 ,
α
α
αk · k!
for all k ≥ 1.
i=1
7. Solve the transport equation with the initial condition:
ut + 3ux = 0, x ∈ R, t > 0.
u(x, 0) = cos(x), x ∈ R.
Answer.
For each x ∈ R, since u(x, 0) = cos(x), then we need to find the characteristic curve through the
point (0, x, cos(x)), that is, we should solve:
 0
t (s) = 1


 0
x (s) = 3
u0 (s) = 0,



t(0) = 0, x(0) = x, u(0) = cos(x).
So we get
t(s) = s,
x(s) = x + 3s,
and u(s) = cos(x).
Notice that u(s) = u(x(s), t(s)), then
u(x + 3t, t) = cos(x).
Since x ∈ R is arbitrary, then the solution is:
u(x, t) = cos(x − 3t) for x ∈ R and t ≥ 0.
8. Solve the transport equation with the initial condition:
ut + 2ux = ex+t , x ∈ R, t > 0.
u(x, 0) = sin(x), x ∈ R.
Answer.
For each x ∈ R, since u(x, 0) = cos(x), then we need to find the characteristic curve through the
point (0, x, cos(x)), that is, we should solve:
 0

 t 0(s) = 1

x (s) = 2
u0 (s) = ex(s)+t ,



t(0) = 0, x(0) = x, u(0) = sin(x).
Page 15
So we get
t(s) = s,
x(s) = x + 2s,
1
and u(s) = [ex+3s − ex ] + sin(x).
3
Notice that u(s) = u(x(s), t(s)), then
1
u(x + 2t, t) = [ex+3t − ex ] + sin(x).
3
Since x ∈ R is arbitrary, then the solution is:
1
u(x, t) = [ex+t − ex−2t ] + sin(x − 2t) for x ∈ R and t ≥ 0.
3
9. Determine whether the method of separation of variables can be used to replace the given PDEs by a pair of
ordinary differential equation. If so, find the equations.
(a) xuxx + ut = 0.
Answer.
Let u(x, t) = X(x)T (t), then
ux = X 0 (x)T (t),
uxx = X 00 (x)T (t),
and ut = X(x)T 0 (t).
Then
0 = xuxx + ut
= xX 00 (x)T (t) + X(x)T 0 (t).
So we get
−
xX 00 (x)
T 0 (t)
=
= λ.
X(x)
T (t)
So we can use separation of variables, and the ODEs are:
−xX 00 (x) − λX(x) = 0,
and T 0 (t) − λT (t) = 0.
(b) uxx + uxt + ut = 0.
Answer.
Let u(x, t) = X(x)T (t), then
ux = X 0 (x)T (t),
uxx = X 00 (x)T (t),
ut = X(x)T 0 (t),
and uxt = X 0 (x)T 0 (t)
Then
0 = uxx + uxt + ut
= X 00 (x)T (t) + X 0 (x)T 0 (t) + X(x)T 0 (t)
So we get
−
T 0 (t)
X 00 (x)
=
= λ.
X 0 (x) + X(x)
T (t)
So we can use separation of variables, and the ODEs are:
−xX 00 (x) − λ[X 0 (x) + X(x)] = 0,
Page 16
and T 0 (t) − λT (t) = 0.
(c) uxx + (x + y)uyy = 0.
Answer.
Let u(x, y) = X(x)Y (y), then
ux = X 0 (x)Y (y),
uxx = X 00 (x)Y (y),
uy = X(x)Y 0 (y),
and uyy = X(x)Y 00 (y).
Then
0 = uxx + (x + y)uyy
= X 00 (x)Y (y) + (x + y)X(x)Y 00 (y).
So we get
−
X 00 (x)
Y 00 (y)
=
.
(x + y)X(x)
Y (y)
So we can not use separation of variables.
(d) [p(x)ux ]x − r(x)utt = 0.
Answer.
Let u(x, t) = X(x)T (t), then
ux = X 0 (x)T (t),
uxx = X 00 (x)T (t),
ut = X(x)T 0 (t),
and utt = X(x)T 00 (t)
Then
0 = [p(x)ux ]x − r(x)utt
= [p(x)X 0 (x)T (t)]x − r(x)X(x)T 00 (t)
= T (t)[p(x)X 0 (x)]x − r(x)X(x)T 00 (t).
So we get
T 00 (t)
[p(x)X 0 (x)]x
=
= λ.
r(x)X(x)
T (t)
So we can use separation of variables, and the ODEs are:
[p(x)X 0 (x)]x − λr(x)X(x) = 0,
and T 00 (t) − λT (t) = 0.
(e) uxx + uyy + xu = 0.
Answer.
Let u(x, y) = X(x)Y (y), then
ux = X 0 (x)Y (y),
uxx = X 00 (x)Y (y),
and uyy = X(x)Y 00 (y).
Then
0 = uxx + uyy + xu
= X 00 (x)Y (y) + X(x)Y 00 (y) + xX(x)Y (y)
= [X 00 (x) + xX(x)]Y (y) + X(x)Y 00 (y).
Page 17
So we get
−
X 00 (x) + xX(x)
Y 00 (y)
=
= λ.
X(x)
Y (y)
So we can use separation of variables, and the ODEs are:
X 00 (x) + xX(x) + λX(x) = 0,
and Y 00 (y) − λY (y) = 0.
10. Solve y 00 + y = 0, y(0) = 0 and y 0 (π) = 1 if the solution exists.
Answer.
Let y = erx to be a solution to y 00 + y = 0, then
y 0 = rerx
y 00 = r2 erx
0 = y 00 + y
= r2 erx + erx
= erx (r2 + 1).
So r2 + 1 = 0, that is, r = ±i. Then the general solution to y 00 + y = 0 is:
y = C1 cos(x) + C2 sin(x).
Since y(0) = 0, that is, C1 = 0. Since y 0 (x) = −C1 sin(x) + C2 cos(x) and y 0 (π) = 1, that is, C2 cos(π) = 1. So
C2 = −1. Then the solution is:
y = − sin(x).
11. Solve y 00 + y = 0, y 0 (0) = 1 and y(π) = 0 if the solution exists.
Answer.
Let y = erx to be a solution to y 00 + y = 0, then
y 0 = rerx
y 00 = r2 erx
0 = y 00 + y
= r2 erx + erx
= erx (r2 + 1).
So r2 + 1 = 0, that is, r = ±i. Then the general solution to y 00 + y = 0 is:
y = C1 cos(x) + C2 sin(x).
Since y(π) = 0, that is, C1 = 0. Since y 0 (x) = −C1 sin(x) + C2 cos(x) and y 0 (0) = 1, that is, C2 cos(0) = 1. So
C2 = 1. Then the solution is:
y = sin(x).
Page 18
12. Use separation of variables to find

PDE:



BC:



IC:
Answer.
the solution to the heat conduction problem:
100uxx = ut ,
0 < x < 1, t > 0,
u(0, t) = 0, u(1, t) = 0,
t > 0,
u(x, 0) = sin(2πx) − sin(5πx),
0 ≤ x ≤ 1.
Let’s first look at the PDE and the homogeneous boundary condition:
(
ut = 100uxx , 0 < x < 1, t > 0,
BC: u(0, t) = 0,
u(1, t) = 0,
t > 0.
(4)
A separated solution to (4) is of the form u(x, t) = X(x)T (t), then
ut (x, t) = X(x)T 0 (t)
ux (x, t) = X 0 (x)T (t)
uxx (x, t) = X 00 (x)T (t)
0 = ut − 100uxx
= X(x)T 0 (t) − 100X 00 (x)T (t).
So we get
−
X 00 (x)
T 0 (t)
=−
:= λ.
100T (t)
X(x)
(5)
X 00 (x)
, then λt (x, t) = 0. By (5), then λx (x, t) = 0. So λ(x, t) is a constant, which is called
X(x)
the separation constant. Since u(0, t) = u(1, t) = 0, then X(x) satisfies
( 00
X + λX = 0,
(6)
X(0) = X(1) = 0.
Let λ(x, t) = −
We are expected to find non-zero solutions to (6). Let X = erx be a solution to X 00 + λX = 0, then X 0 = rerx ,
X 00 = r2 erx , and
0 = X 00 + λX = r2 erx + λerx
= erx (r2 + λ).
Then r2 + λ = 0. We have three cases:
a. If λ = 0, that is, r2 = 0. So r = 0, which implies that the general solution to X 00 + λX = X 00 = 0 is
X(x) = C1 + C2 x. Since X(0) = X(1) = 0, then
C1 = C1 + C2 L = 0.
Then C1 = C2 = 0.
Page 19
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 X 00 + λX = X 00 + µ2 X = 0 is X(x) = C1 cos(µx) + C2 sin(µx). Since X(0) = X(1) = 0, then
C1 = C1 cos(µ) + C2 sin(µ) = 0.
Then
C1 = 0,
and C2 sin(µ) = 0.
In order to make X to be a non-zero solution, then we need sin(µ) = 0, that is, µ = nπ for some n ∈ N,
that is, λ = µ2 = n2 π 2 . For λ = n2 π 2 , then all solution to (6) are
Xn (x) = C sin(nπx).
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 X 00 + λX = X 00 − µ2 X = 0 is X(x) = C1 eµx + C2 e−µx . Since X(0) = X(1) = 0, then
C1 + C2 = C1 eµ + C2 e−µ = 0.
Since µ > 0, then C1 = C2 = 0.
In summary, (6) has non-zero solution if and only if λ = n2 π 2 for some n ∈ N. When λ = n2 π 2 in (6), all
non-zero solutions to (6) are X(x) = C sin(nπ).
When λ = n2 π 2 , by (5), then
0 = T 0 (t) + 100λT (t)
= T 0 (t) + 100n2 π 2 T (t)
(7)
It’s easy to see that all solutions to (7) are given by:
2 π2 t
T (t) = Ce−100n
.
In summary, for any n ∈ N, we can find a non-zero solution to (4):
2 π2 t
un (x, t) = e−100n
sin(nπx).
Since u(x, 0) = sin(2πx) − sin(5πx), then the solution to the initial value problem is given by:
2
2
u(x, t) = e−400π t sin(2πx) − e−2500π t sin(5πx).
13. Use separation of variables

PDE:




BC:



 IC:
to find the solution to the heat conduction problem:
uxx = 4ut + u,
0 < x < 2, t > 0,
u(0, t) = 0, u(2, t) = 0,
t > 0,
πx u(x, 0) = 2 sin
− sin(πx) + 4 sin(2πx),
2
Page 20
0 ≤ x ≤ 2.
Answer.
Let’s first look at the PDE and the homogeneous boundary condition:
(
uxx = 4ut + u, 0 < x < 2, t > 0,
BC: u(0, t) = 0,
u(2, t) = 0,
t > 0.
(8)
A separated solution to (8) is of the form u(x, t) = X(x)T (t), then
ut (x, t) = X(x)T 0 (t)
ux (x, t) = X 0 (x)T (t)
uxx (x, t) = X 00 (x)T (t)
0 = 4ut + u − uxx
= 4X(x)T 0 (t) + X(x)T (t) − X 00 (x)T (t)
= 4X(x)T 0 (t) + [X(x) − X 00 (x)]T (t).
So we get [X 00 (x) − X(x)]T (t) = 4X(x)T 0 (t), that is,
−
X 00 (x) − X(x)
4T 0 (t)
=−
:= λ.
T (t)
X(x)
(9)
X 00 (x) − X(x)
, then λt (x, t) = 0. By (9), then λx (x, t) = 0. So λ(x, t) is a constant, which is
X(x)
called the separation constant. Since u(0, t) = u(2, t) = 0, then X(x) satisfies
( 00
X + (λ − 1)X = 0,
(10)
X(0) = X(2) = 0.
Let λ(x, t) = −
We are expected to find non-zero solutions to (10). Let X = erx be a solution to X 00 +λX = 0, then X 0 = rerx ,
X 00 = r2 erx , and
0 = X 00 + (λ − 1)X = r2 erx + (λ − 1)erx
= erx (r2 + λ − 1).
Then r2 + λ − 1 = 0. We have three cases:
a. If λ − 1 = 0, that is, r2 = 0. So r = 0, which implies that the general solution to X 00 + (λ − 1)X = X 00 = 0
is X(x) = C1 + C2 x. Since X(0) = X(2) = 0, then
C1 = C1 + C2 2 = 0.
Then C1 = C2 = 0.
b. If λ − 1 > 0, that is, λ − 1 = µ2 for some µ > 0. So r2 + µ2 = 0, that is, r = ±µi, which implies
that the general solution to X 00 + (λ − 1)X = X 00 + µ2 X = 0 is X(x) = C1 cos(µx) + C2 sin(µx). Since
X(0) = X(2) = 0, then
C1 = C1 cos(2µ) + C2 sin(2µ) = 0.
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Then
C1 = 0,
and C2 sin(2µ) = 0.
In order to make X to be a non-zero solution, then we need sin(2µ) = 0, that is, 2µ = nπ for some n ∈ N,
nπ 2
nπ 2
that is, λ = µ2 − 1 =
− 1. For λ =
− 1, then all solution to (10) are
2
2
nπx Xn (x) = C sin
.
2
c. If λ − 1 < 0, that is, λ − 1 = −µ2 for some µ > 0. So r2 − µ2 = 0, that is, r = ±µ, which implies that the
general solution to X 00 + (λ + 1)X = X 00 − µ2 X = 0 is X(x) = C1 eµx + C2 e−µx . Since X(0) = X(2) = 0,
then
C1 + C2 = C1 e2µ + C2 e−2µ = 0.
Since µ > 0, then C1 = C2 = 0.
nπ 2
nπ 2
In summary, (10) has non-zero solution if and only if λ =
− 1 for some n ∈ N. When λ =
−1
2
nπ 2
x .
in (10), all non-zero solutions to (10) are X(x) = C sin
2
nπ 2
When λ =
− 1, by (9), then
2
0 = 4T 0 (t) + λT (t)
nπ 2
0
− 1 T (t)
= 4T (t) +
2
(11)
It’s easy to see that all solutions to (11) are given by:
− 14
h
T (t) = Ce
2
( nπ
2 )
i
−1 t
.
In summary, for any n ∈ N, we can find a non-zero solution to (8):
− 41
un (x, t) = e
Since u(x, 0) = 2 sin
πx 2
h
2
( nπ
2 )
i
−1 t
sin
nπ 2
.
− sin(πx) + 4 sin(2πx), then the solution to the initial value problem is given by:
u(x, t) = 2e
− 14
h
( π2 )
2
i
−1 t
sin
πx 2
1
− e− 4 [π
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]t sin(πx) + 4e− 14 [4π2 −1]t sin(2πx).
2 −1