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The University of British Columbia
MATH 257/316
Midterm Exam I – July 2015
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Signature
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This exam consists of 2 questions worth 50 marks. No notes nor calculators.
Question
Points
1
30
2
20
Total:
50
Score
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in examination questions.
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July 2015
Math 257/316 Midterm 1
Page 2 of 6
1. Consider the problem:
x2 y 00 + xy 0 + (2x − α2 )y = 0,
for x > 0,
where α is a non-negative constant. We are looking for the series solutions of the form
∞
X
φ(x, r) =
ak (r)xk+r .
k=0
(3 points)
(a) Explain that x = 0 is a regular singular point.
Since x2 y 00 + xy 0 + (2x − α2 )y = 0, then y 00 +
Answer.
p(x) =
1
2x − α2
and q(x) =
, then
x
x2
and x2 q(x) = 2x − α2 .
xp(x) = 1,
(8 points)
1 0 2x − α2
y = 0. Let
y +
x
x2
So x = 0 is a regular singular point.
(b) Find the recurrence relation of {ak (r)}∞
k=1 .
Answer.
Since φ(x, r) =
∞
X
ak (r)xk+r , then
k=0
φx (x, r) =
∞
X
(k + r)ak (r)xk+r−1 ,
and φxx (x, r) =
k=0
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 .
k=0
So we get
x2 φxx (x, r) + xφx (x, r) + (2x − α2 )φ(x, r)
∞
∞
∞
X
X
X
= x2
(k + r)(k + r − 1)ak (r)xk+r−2 + x
(k + r)ak (r)xk+r−1 + (2x − α2 )
ak (r)xk+r
=
=
k=0
∞
X
∞
X
k=0
∞
X
k=0
k=0
k=0
(k + r)(k + r − 1)ak (r)xk+r +
∞
X
k=0
(k + r)ak (r)xk+r + 2
ak (r)xk+r+1 − α2
ak (r)xk+r
k=0
∞
∞
∞
X
X
X
ak (r)xk+r
ak−1 (r)xk+r − α2
(k + r)ak (r)xk+r + 2
(k + r)(k + r − 1)ak (r)xk+r +
k=0
k=0
k=1
= r(r − 1)a0 (r) + ra0 (r) − α2 a0 (r)
∞
X
+
{[(k + r)(k + r − 1) + (k + r) − α2 ]ak (r) + 2ak−1 (r)}xk+r
k=1
2
∞
X
2
= (r − α )a0 (r) +
∞
X
{[(k + r)2 − α2 ]ak (r) + 2ak−1 (r)}ak (r).
k=1
Define {ak (r)}∞
k=1 be the following recurrence relation:
[(k + r)2 − α2 ]ak (r) + 2ak−1 (r) = 0,
for all k ≥ 1.
That is, we have
ak (r) =
−2
· ak−1 (r),
(k + r)2 − α2
for all k ≥ 1.
k=0
July 2015
(3 points)
Math 257/316 Midterm 1
Page 3 of 6
(c) Find the indicial equation and exponents at x = 0 (that is, the roots of the indicial
equation).
Answer.
By the computation in part (b), we know that the indicial equation is:
r2 − α2 = 0.
Then exponents at x = 0 are r1 = α and r2 = −α.
(5 points)
(d) Compute a1 (r), a2 (r), a3 (r) in terms of a0 (r) (You do not need to simplify). Find a
general pattern of {ak (r)}∞
k=1 .
Answer.
By the result of part (b), then
ak (r) =
−2
· ak−1 (r),
(k + r)2 − α2
for all k ≥ 1.
Then we have
a1 (r) =
a2 (r) =
=
a3 (r) =
=
−2
(1 + r)2 − α2
−2
(2 + r)2 − α2
−2
(2 + r)2 − α2
−2
(3 + r)2 − α2
−2
(3 + r)2 − α2
· a0 (r)
· a1 (r)
·
−2
· a0 (r)
(1 + r)2 − α2
· a2 (r)
·
−2
−2
·
· a0 (r).
2
2
(2 + r) − α (1 + r)2 − α2
In general, we have
ak (r) =
=
=
−2
(k + r)2 − α2
−2
(k + r)2 − α2
−2
(k + r)2 − α2
= a0 (r)
k
Y
i=1
k
· ak−1 (r)
−2
· ak−2 (r)
(k − 1 + r)2 − α2
−2
−2
·
·
· a0 (r)
(k − 1 + r)2 − α2 (1 + r)2 − α2
·
−2
(i + r)2 − α2
= (−2) a0 (r)
k
Y
i=1
1
,
(i + r)2 − α2
∀k ≥ 1.
July 2015
(5 points)
Math 257/316 Midterm 1
Page 4 of 6
1
(e) If α = , find the first two non-zero terms of two linearly independent series
4
solutions about x = 0.
1
1
1
Answer. If α = , by the result of part (c), then r1 = and r2 = − , which implies
4
4
4
1
that r1 − r2 = . Let a0 (r) = 1, by the computation in part (b), then
2
1
1
2
2
φ(x, r) = r −
a0 (r).
x φxx (x, r) + xφx (x, r) + 2x −
16
16
Then
X
∞
1
1
1
y1 (x) = φ x,
=
xk+ 4 ,
ak
4
4
k=0
X
∞
1
1
1
and y2 (x) = φ x, −
=
xk− 4
ak −
4
4
k=0
are two linearly independent solutions. By the result of part (d), we have
1
1
a0
= a0 −
=1
4
4
−2
1
4
a1
=
·1=−
1 2
1 2
4
3
1+ 4 − 4
1
−2
a1 −
=
2
2 · 1 = −4.
1
4
1−
− 1
4
(6 points)
4
(f) If α = 0, find the first two non-zero terms of two linearly independent series solutions
about x = 0.
Answer. If α = 0, by the result of part (c), then r1 = r2 = 0. Let a0 (r) = 1, by the
computation in part (b), then
x2 φxx (x, r) + xφx (x, r) + 2xφ(x, r) = r2 .
So we get
x2 φxxr (x, r) + xφxr (x, r) + 2xφr (x, r) = 2r.
Then
y1 (x) = φ(x, 0) =
∞
X
k
ak (0)x ,
and y2 (x) = φr (x, 0) =
k=0
∞
X
a0k (0)xk
+ ln x
k=0
∞
X
ak (0)xk
k=0
are two linearly independent solutions. By the result of part (d), we have a0 (r) = 1,
−2
−2
−2
a1 (r) =
and a2 (r) =
·
, which implies that
2
2
(r + 1)
(r + 2) (r + 1)2
a00 (r) = 0,
a01 (r) =
4
,
(r + 1)2
and a02 (r) =
4
−2
−2
4
·
+
·
.
3
2
2
(r + 2) (r + 1)
(r + 2) (r + 1)3
So we get
a0 (0) = 1,
a1 (0) = −2,
a01 (0) = 4,
and a02 (0) = −3.
July 2015
Math 257/316 Midterm 1
Page 5 of 6
(20 points) 2. Use the separation of variables to solve the following problem:


ut = 4uxx + 8u, 0 < x < 2, t > 0,




BC: ux (0, t) = 0, u(2, t) = 0,
t > 0,

π √

7π


x − 2 cos
x , 0 ≤ x ≤ 2.
 IC: u(x, 0) = cos
4
4
You must consider all cases for the separation constant.
Answer.
Let’s first look at the heat equation and the homogeneous boundary condition:
(
ut = 4uxx + 8u, 0 < x < 2, t > 0,
BC: ux (0, t) = 0,
u(2, t) = 0,
t > 0.
(1)
A separated solution to (1) 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 − 4uxx − 8u
= X(x)T 0 (t) − 4X 00 (x)T (t) − 8X(x)T (t)
= X(x)T 0 (t) − 4T (t)[X 00 (x) + 2X(x)].
So we get X(x)T 0 (t) = 4T (t)[X 00 (x) + 2X(x)], that is,
−
T 0 (t)
X 00 (x) + 2X(x)
=−
:= λ.
4T (t)
X(x)
X 00 (x) + 2X(x)
, then λt (x, t) = 0. By (2), then λx (x, t) = 0. So λ(x, t) is a
X(x)
constant. Since ux (0, t) = u(2, t) = 0, then X(x) satisfies
( 00
X + (λ + 2)X = 0,
(2)
Let λ(x, t) = −
X 0 (0) = X(2) = 0.
We are expected to find non-zero solutions to (3). Let X = erx be a solution to
X 00 + (λ + 2)X = 0, then
X 0 = rerx
X 00 = r2 erx
0 = X 00 + (λ + 2)X
= r2 erx + (λ + 2)erx
= erx (r2 + λ + 2).
(3)
July 2015
Math 257/316 Midterm 1
Page 6 of 6
a. If λ + 2 = 0, that is, r2 = 0. So r = 0, which implies that the general solution to
X 00 + (λ + 2)X = X 00 = 0 is X(x) = C1 + C2 x. Since X 0 (x) = C2 and X 0 (0) = 0, then
C2 = 0. Since X(2) = 0, that is, C1 + 2C2 = 0, then C1 = C2 = 0.
b. If λ + 2 > 0, that is, λ + 2 = µ2 for some µ > 0. So r2 + µ2 = 0, that is, r = ±µi, which
implies that the general solution to X 00 + (λ + 2)X = X 00 + µ2 X = 0 is
X(x) = C1 cos(µx) + C2 sin(µx). Since X 0 (x) = −C1 µ sin(µx) + C2 µ cos(µx) and
X 0 (0) = 0, then C2 µ = 0. Since µ > 0, then C2 = 0. Then X(x) = C1 cos(µx). Since
X(2) = 0, that is, C1 cos(2µ) = 0. In order to make X to be a non-zero solution, then we
π
(2n + 1)π
need cos(2µ) = 0, that is, 2µ = + nπ =
for some n ∈ N, that is,
2
2
2
(2n + 1)π
(2n + 1)π 2
2
λ=µ −2=
− 2. For λ =
− 2, then all solution to (3) are
4
4
(2n + 1)π
Xn (x) = C cos
x .
4
c. If λ + 2 < 0, that is, λ + 2 = −µ2 for some µ > 0. So r2 − µ2 = 0, that is, r = ±µ, which
implies that the general solution to X 00 + (λ + 2)X = X 00 − µ2 X = 0 is
X(x) = C1 eµx + C2 e−µx . Since X 0 (x) = C1 µeµx − C2 µe−µx and X 0 (0) = 0, µ > 0, then
C1 − C2 = 0. Since X(2) = 0, that is, C1 e2µ + C2 e−2µ = 0, then we have C1 = C2 = 0.
(2n + 1)π 2
In summary, (3) has non-zero solution if and only if λ =
− 2 for some
4
S
(2n + 1)π 2
− 2 in (3), all non-zero solutions to (3) are
n ∈ N {0}. When λ =
4
(2n + 1)π
(2n + 1)π 2
X(x) = C cos
x . When λ =
− 2, by (1), then
4
4
"
#
2
(2n + 1)2 π 2
(2n
+
1)π
0
0
0
0 = T (t) + 4λT (t) = T (t) + 4
− 2 T (t) = T (t) +
− 8 T (t). (4)
4
4
It’s easy to see that all solutions to (4) are given by:
(2n+1)2 π 2
−
−8 t
4
T (t) = Ce
.
S
In summary, for any n ∈ N {0}, we can find a non-zero solution to (1):
−
un (x, t) = e
Now since u(x, 0) = cos
(2n+1)2 π 2
−8
4
t
cos
(2n + 1)π
x .
4
π √
7π
x − 2 cos
x , then
4
4
u(x, t) = e
h 2
i
− π4 −8 t
h
i
2
π √
7π
− 49π
−8 t
4
cos
x − 2e
cos
x .
4
4