HOMEWORK 8
SHUANGLIN SHAO
1. Section 5.3. # 3.
Proof. By the exercise #2 in Section 4.2, we have
∞
X
(2n + 1)πct
(2n + 1)πct
(2n + 1)πx
An cos
+ Bn sin
.
u(x, t) =
sin
2l
2l
2l
n=0
Since ut (x, 0) = 0, we have
∞ X
(2n + 1)πc
(2n + 1)πx
0=
Bn
sin
.
2l
2l
n=0
Thus Bn = 0.
To compute An , n = 0, 1, 2, · · · , fix m.
Z l
Z l
∞
X
(2m + 1)πx
(2n + 1)πx
(2m + 1)πx
x sin
dx =
An
sin
sin
dx.
2l
2l
2l
0
0
n=0
We first compute A0 . Set m = 0. By integration by parts,
2
Z l 2l
πx 0
2l
x − cos
dx =
.
π
2l
π
0
For the right hand side,
Z l
x sin
0
(2n + 1)πx
πx
l
sin
dx = .
2l
2l
2
So
A0 =
8l
.
π2
For m ≥ 1. By integration by parts, the left hand side is
2
Z l
(2m + 1)πx
2l
x sin
dx = (−1)m+1
.
2l
(2m + 1)π
0
1
For the right hand side,
Z l
1
(n + m + 1)πx
(n − m)πx
(− ) cos
RHS =
dx
− cos
2
l
l
0
Z 1 l
(n + m + 1)πx
(n − m)πx
=−
cos
dx
− cos
2 0
l
l
= 0,
when n 6= m. When n = m, it is equal to l/2. So for m = 0, 1, 2, · · · ,
Am =
8(−1)m+1 l
.
(2m + 1)2 π 2
Therefore
u(x, t) =
∞
X
8(−1)n+1 l
(2n + 1)πx
(2n + 1)πct
sin
cos
.
2
2
(2n + 1) π
2l
2l
n=0
2. Section 5.3. # 4.
Proof. Part (a).
linear equations:
Let v(x, t) = u(x, t) − U . Then we have the system of


= kvxx ,
vt
v(0, t) = 0, vx (0, t) = 0,


v(x, 0) = −U.
By (17) in Section 4.1,
v(x, t) =
∞
X
nπ 2
kt
l
)
sin
An sin
nπx
.
l
An e−(
n=1
nπx
.
l
By the initial condition v(x, 0) = −U ,
−U =
∞
X
n=1
Fix m. We have
Z l
Z l
∞
X
mπx
nπx
mπx
(−U ) sin
dx =
An
sin
sin
dx.
l
l
l
0
0
n=1
Thus
Am =
Therefore
v(x, t) =
∞
X
2U
n=1
nπ
2U
((−1)m − 1) .
mπ
((−1)n − 1) e−(
2
nπ 2
kt
l
)
sin
nπx
.
l
Part (b). By the L’Hospital rule,
lim
x→∞
e−x
1
x2
2
2x
1
x2
= 0.
2 = lim
2 = lim
x
x
x→∞ 2xe
x→∞ ex2
x→∞ e
= lim
Thus there exists a constant C > 0 such that for 0 < x < ∞,
C
.
x2
2
e−x ≤
So for t > 0,
e−(
nπ 2
kt
l
)
Cl2
.
n2 π 2 kt
≤
Therefore
|v(x, t)| ≤
∞
X
C
,
n3
n=1
which is convergent. Therefore for 0 < x < l, t > 0,
∞
X
C
+ U.
|u(x, t)| ≤ |v(x, t)| + U ≤
n3
n=1
3. Section 5.3. # 6.
Proof. We solve the eigenvalue problem
−X 0 (x) = λX(x), 0 < x < l.
Integrating with the factor eλx ,
0
eλx X 0 (x) + λeλx X(x) = eλx X(x) = 0.
Integrating from 0 to 1,
eλx X(x)|10 = eλ X(1) − X(0) = X(1) eλ − 1 = 0.
So
λn = 2niπ, n ∈ Z.
Therefore
Xn (x) = X(0)e−2inπx , 0 < x < 1.
The set {Xn }n∈Z is orthogonal: for n, m ∈ Z and n 6= m,
Z 1
Z 1
2
Xn (x)Xm (x)dx = X (0)
e2(m−n)iπx dx = 0.
0
0
3
4. Section 5.3. # 8.
Proof. We know that Xj , j = 1, 2, satisfy the same boundary condition at
x = a and x = b:
Xj0 (a) − a0 Xj (a) = 0,
Xj0 (b) + al Xj (b) = 0.
Then
−X10 X2 + X1 X20 |ba
= −X10 (b)X2 (b) + X1 (b)X20 (b) − −X10 (a)X2 (a) + X1 (a)X20 (a)
= (al X1 (b)X2 (b) − al X1 (b)X2 (b)) − (−a0 X1 (a)X2 (a) + a0 X1 (a)X2 (a))
= 0.
5. Section 5.3. # 9.
Proof. We need to show that for any two eigenfunctions X1 and X2 satisfy
0 = X10 X2 − X1 X20 |ba
= X10 (b)X2 (b) − X1 (b)X20 (b) − X10 (a)X2 (a) − X1 (a)X20 (a)
= (δα − βγ) X10 (a)X2 (a) − X1 (a)X20 (a)
by the conditions
Xj (b) = αXj (a) + βXj0 (a),
Xj0 (b) = γXj (a) + δXj0 (a),
for j = 1, 2. So the equation implies that, the boundary conditions are
symmetric if and only if
δα − βγ = 0.
6. Section 5.3. # 10.
Proof. We show that for i < j,
(Zi , Zj ) = 0
by the mathematical induction.
Part (a). If i = 1 and j = 2, we prove that
(Z1 , Z2 ) = 0
4
by establishing (Y1 , Y2 ) = 0:
0 = (X2 − (X2 , Z1 )Z1 , X1 )
= (X2 , X1 ) − (X2 , Z1 )(Z1 , X1 )
X1
X1
= (X2 , X1 ) − (X2 ,
)(
, X1 )
kX1 k kX1 k
= (X2 , X1 ) − (X2 , X1 )
= 0.
Step 2.
Suppose that the set {Z1 , Z2 , · · · , Zn } is mutually orthogonal.
We prove that Zn+1 is orthogonal to each vector in {Z1 , Z2 , · · · , Zn }. We
n+1
know that Zn+1 = kYYn+1
k and
Yn+1 = Xn+1 − (Xn+1 , Zn )Zn − (Xn+1 , Zn−1 )Zn−1 − · · · − (Xn+1 , Z1 )Z1
Let 1 ≤ i ≤ n.
(Yn+1 , Zi ) = (Xn+1 , Zi ) − (Xn+1 , Zi )(Zi , Zi )
= (Xn+1 , Zi ) − (Xn+1 , Zi )
=0
since (Zi , Zj ) = 0 for i 6= j, and kZi k = 1. This implies that {Z1 , Z2 , · · · , Zn+1 }
is mutually orthogonal. So {Z1 , Z2 , · · · , Zn , · · · } is mutually orthogonal.
Part (b). Let X1 and X2 denote the two functions.
Z π
Z π
kX1 k2 =
X12 (x)dx =
(cos x + cos 2x)2 dx = π.
0
0
So
Z1 =
cos x + 2 cos 2x
√
.
π
To obtain Z2 ,
Z π
1
(3 cos x − 4 cos 2x) (cos x + cos 2x) dx
(X2 , Z1 ) = √
π 0
√
π
=−
.
2
Therefore
21
7 cos x 7 cos 2x
Y2 = X2 −(X2 , Z1 )Z1 = 3 cos x−4 cos 2x+ (cos x+cos 2x) =
−
.
2
2
2
Thus
cos x − cos 2x
√
Z2 =
π
Therefore we get an orthogonal pair Z1 , Z2 .
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7. Section 5.3. # 11.
Proof. (a). If f satisfies the Dirichlet condition, f (a) = f (b) = 0. Then
f (x)f 0 (x)|ba = 0.
If f satisfies the Neumann condition, f 0 (a) = f 0 (b) = 0. Then
f (x)f 0 (x)|ba = 0.
If f satisfies the periodic boundary condition, f (a) = f (b) and f 0 (a) = f 0 (b).
Then
f (x)f 0 (x)|ba = f (b)f 0 (b) − f (a)f 0 (a) = 0.
If f satisfies the Robin boundary conditions,
f (a) − a0 f 0 (a) = 0,
f (b) + al f 0 (b) = 0.
Then
f (x)f 0 (x)|ba = f (b)f 0 (b) − f (a)f 0 (a) = −al f 0 (b)
2
− a0 f 0 (a)
2
≤ 0.
8. Section 5.3. # 15.
Proof. Suppose that the eigenvalue problem
X (4) (x) = λX(x).
From the boundary conditions X(0) = X(1) = X 00 (0) = X 00 (1) = 0,
Z l
λ
(X(x))2 dx
0
Z
=
l
X (4) (x)X(x)dx
0
= X (3) X(x)|l0 −
l
Z
X (3) X 0 (x)dx
0
00
= −X (x)X
0
(x)|l0
Z
l
−
X 0 (x)
2
dx
0
Z
=−
l
X 0 (x)
2
dx ≤ 0.
0
Therefore λ ≤ 0.
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9. Section 5.4. # 1.
Proof. Part (a). Since
the interval −1 < x < 1.
P∞
n 2n
n=0 (−1) x ,
it converges to
1
1+x2
point-wise in
Part (b).
It does not converge uniformly in the interval −1 < x < 1
because at x = 1,
∞
X
n 2n
(−1) x
n=1
=
∞
X
(−1)n
n=1
which diverges because 1/(1 + x2 ) = 1/2.
Part (c). Fix N ∈ N. We have
∞
N
∞
X
X
x2(N +1)
X
(−1)n x2n = .
(−1)n x2n =
(−1)n x2n −
1 + x2
n=1
n=1
n=N +1
We evaluate the difference in the integral,
∞
2
N
X
X
n 2n
n 2n (−1)
x
−
(−1)
x
dx
−1 n=1
n=1
Z 1 4(N +1)
x
dx
=
(1
+ x2 )2
−1
Z 1 4(N +1)
x
=2
dx
(1
+ x2 )2
0
Z 1
x4N +4 dx
≤2
0
Z 1
0
2
≤
x4N +5 dx
4N + 5 0
2
=
4N + 5
Z
1
which goes to zero as N goes to infinity. Therefore the series
converges in the L2 sense.
P∞
n 2n
n=0 (−1) x
10. Section 5.4. # 2.
Proof. If the series converges to f uniformly, the series converges to f pointwise.
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Next we prove that it converges to f in the L2 sense: for any > 0, there
exists N ∈ N such that for any m ≥ N ,
√
m
X
fn (x) − f (x) ≤ √
.
b−a
n=1
2
Z b X
Z b
m
1dx = .
fn (x) − f (x) dx ≤
b−a a
a n=1
This proves that the series converges to f in the L2 sense.
11. Section 5.4. # 3.
Proof. Part (a). For x ∈ R, if x = 12 , then fn ( 12 ) = 0, which converges to
zero as n goes to infinity.
For x < 12 , there exists N such that
1
1
−x> .
2
N
Thus for n ≥ N , fn (x) = 0, which goes to zero as n goes to infinity.
For x > 12 , the proof is similar. So fn (x) → 0 as n goes to infinity.
Part (b). Take γn = n2 . If the convergence is uniform, it would imply the
L1 convergence. Hence
Z
Z 1/2
2
2γn
|fn (x)|dx = 2
|fn (x)|dx = γn =
= 2n,
n
n
1/2−1/n
which goes to infinity as n goes to infinity.
Part (c). By the previous computation,
Z
2n2/3
2
= 1/3 → 0,
|fn (x)|2 dx =
n
n
as n goes to infinity. Thus fn → 0 in the L2 sense.
Part (d).
because
If γn = n, then fn does not converge to zero in the L2 sense
Z
|fn |2 dx = 2.
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12. Section 5.4. # 4.
Proof. Since gn ( 14 ) = 1,
1
gn ( ) → 1, as n → ∞.
4
Thus the sequence gn fails to converge to zero point-wise.
For n ≥ 100, the interval [ 14 −
are disjoint. So we compute
Z
1 1
,
n2 4
+
|gn (x)|2 dx =
1
)
n2
and the interval [ 34 −
Proof. By (6) in Section 5.1, on the interval (0, 4),
∞
A0 X
nπx
+
An cos
.
2
4
n=1
For m = 0,
Z
0
4
1
φ(x)dx =
2
Z
1
Z
0dx +
0
3
1dx
= 1.
1
For m = 1,
Z
Z
πx
1 3
πx
1 3
dx =
dx
φ(x) cos
cos
2 1
4
2 1
4
Z 1 3 4
πx 0
=
sin
dx
2 1
π
4
2
πx 3
= sin
|
π 4 1
2
3π
π
=
sin
− sin
π
4
4
= 0.
A1 =
9
1
)
n2
13. Section 5.4. # 5.
2
A0 =
4
+
2
4
2
+ 2 = 2
2
n
n
n
which goes to zero as n goes to infinity.
φ(x) =
1 3
,
n2 4
For m = 2,
Z
Z
1 3
2πx
2πx
1 3
A2 =
φ(x) cos
cos
dx =
dx
2 1
4
2 1
4
Z 1 3 4
2πx 0
=
dx
sin
2 1
2π
4
1
πx 3
= sin
|
π 2 1
3π
1
π
sin
=
− sin
π
2
2
2
=− .
π
For m = 3,
Z
Z
1 3
3πx
1 3
3πx
φ(x) cos
dx =
cos
dx
2 1
4
2 1
4
Z 1 3 4
3πx 0
=
sin
dx
2 1
3π
4
2
3πx 3
=
sin
|
3π 4 1
2
9π
3π
=
sin
− sin
3π
4
4
= 0.
A3 =
14. Section 5.4. # 12.
Proof. By (1) in Section 5.1,
f (x) =
∞
X
An sin
n=1
10
nπx
,
l
where
Am
2
=
l
Z
l
f (x) sin
0
Z
mπx
dx
l
l
mπx
x sin
dx
l
0
Z
l
−2 l
mπx 0
x
=
cos
dx
l 0
mπ
l
Z l
mπx l
mπx
−2
=
x cos
| −
cos
dx
mπ
l 0
l
0
Z l
l
−2
mπx 0
m
(−1) l −
dx
=
sin
mπ
mπ 0
l
2(−1)m+1 l
=
.
mπ
By the Parseval’s equality,
Z l
Z l
∞
X
l3
2
2
|An |
|Xn (x)| dx =
x2 dx = ,
3
0
0
2
l
=
n=1
where
Z
l
Z
l
nπx
dx
sin2
l
0
Z 2πx
1 l
1 − cos
dx
=
2 0
l
l
= .
2
2
|Xn (x)| dx =
0
Therefore
∞
X
m=1
Thus
2l3
l3
=
.
(mπ)2
3
∞
X
π2
1
=
.
6
m2
m=1
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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