程序代写代做代考 Homework 6 Solutions

Homework 6 Solutions

Chapter 5, Exercise 9

If FR is defined as given, then one may check that F̂R(ξ) =
(
1− |ξ|

R

)
·1[−R,R](ξ).

This is easily derived from the result of Exercise 2 of the same chapter. Conse-
quently, we see by inversion that

(FR ∗ f)(x) =

R

̂(FR ∗ f)(ξ)e2πiξxdx

=


R
F̂R(ξ)f̂(ξ)e2πiξxdx

=

∫ R
−R

(
1−
|ξ|
R

)
f̂(ξ)e2πiξxdx.

To see that {FR} is a family of good kernels as R→∞, note that∫
R
|FR(x)|dx =


R
FR(x)dx = F̂R(0) = 1,

and using the bound sin2(πtR) ≤ 1, we also have that

R


{|x|>δ}

sin2(πtR)

(πtR)2
dt = 2R

∫ ∞
δ

sin2(πtR)

(πtR)2
dt ≤ 2R

∫ ∞
δ

1

(πtR)2
dt =

2

δπ2
R−1,

and the right side clearly tends to 0 as R→∞.

1

Chapter 5, Exercise 11

The fact that u is continuous on the closure of the upper half plane follows
immediately from Theorem 5.2.1(ii) in the same chapter.

To show that u(t, x)→ 0 as |x|+ t→∞, we apply the hint given to write

u(t, x) =


R
f(x−y)Ht(y)dy =


{|y|>|x|/2}

f(x−y)Ht(y)dy+

{|y|≤|x|/2}

f(x−y)Ht(y)dy.

Let’s call the terms on the right side as A,B, respectively. To bound A, note
that f is bounded (say |f | ≤ C


4π), and 1 < 2y/|x| if |y| > |x|/2, so

|A| ≤

{|y|>|x|/2}

|f(x− y)|Ht(y)dy


2C

t1/2

∫ ∞
|x|/2

e−y
2/4tdy


2C

t1/2

∫ ∞
|x|/2

2y

|x|
e−y

2/4tdy

=
8Ct1/2

|x|
e−x

2/16t.

To bound B, note that |x− y| ≥ 1
2
|x| if |y| ≤ 1

2
|x|, thus 1

1+(x−y)2 ≤
1

1+x2/4
. But

since f ∈ S, it follows that |f(x)| ≤ D/(1 + x2) for some D > 0. Thus,

|B| ≤

{|y|≤|x|/2}

|f(x− y)|Ht(y)dy


D

1 + x2/4


R
Ht(x)dx

=
D

1 + x2/4
.

On the other hand, |Ht(y)| ≤ t−1/2, and hence

|u(t, x)| ≤

R
|f(x− y)|Ht(y)dy ≤ t−1/2


R
|f(x)|dx =: Kt−1/2.

Thus, given � > 0, we choose T = T (�) large enough so that KT−1/2 < �, then we choose M = M(�) large enough so that D 1+x2/4 + 8CT 1/2 |x| e −x2/16T < � whenever |x| > M . Then it is true that |u(t, x)| < � whenever |x|+ t ≥M + T . 2 Chapter 5, Exercise 12 For notational convenience, we write H(t, x) for the heat kernel, and subscripts will always indicate partial derivatives throughout this problem. Let u(t, x) = xt−1H(t, x). Then it is clear that u(t, x) = −2Hx(t, x) if t > 0
and x ∈ R, thus we have

ut(t, x) = −2

∂t

∂x
H(t, x) = −2

∂x

∂t
H(t, x) = −2

∂3

∂x3
H(t, x) = uxx(t, x).

If x 6= 0, then limt→0 xt−3/2e−x
2/4t = 0; this follows immediately from the fact

that ex
2/4t ≥ 1

2
(x2/4t)2, which in turn follows from the fact that eu = 1 + u +

1
2
u2 + … ≥ 1

2
u2 for u ≥ 0. If x = 0, then obviously limt→0 xt−3/2e−x

2/4t = 0.
This proves that u(t, x)→ 0 as t→ 0 along lines parallel to the axis {x = 0}.

On the other hand, if we let (t, x) → 0 along the parabola x2 = 4t, then
we get u(t, x(t)) = (4π)−1/2(4t)1/2t−3/2e−1 = ct−1/2 → +∞ as t → 0, hence u
is not continuous at 0.

3

Chapter 5, Exercise 13

Suppose u is harmonic on the strip {(x, y) : −π
2
< y < π 2 }, and u extends continuously to the boundary, with u(x,−π/2) = u(x, π/2) = 0. Also assume that u vanishes at infinity. We want to show u = 0. We give two solutions. Solution 1: Given � > 0, we can choose M large enough so that |x| > M
implies |u(x, y)| ≤ �. Then the boundary condition on u and the maximum
principle imply that |u(x, y)| ≤ � whenever |x| ≤ M (the maximum principle
is an easy corollary of the mean value property, and says that any harmonic
function achieves its maximum on the boundary of a connected domain). Since
we can choose M as large as we’d like, this actually shows that |u(x, y)| < � everywhere on the strip. Since � is arbitrary, u = 0. Solution 2: Let’s define H = {(x, y) : x > 0}, and let us define a new func-
tion

v(x, y) = u
(1

2
log(x2 + y2), tan−1(y/x)

)
.

We claim that v is a harmonic function on H, and extends continuously to the
closure of H, where it is zero on the boundary and vanishes at ∞. If we can
prove this then we will be done, because then we would have v = 0 by theorem
5.2.7, from which it follows that u(x, y) = v(ex cos y, ex sin y) = 0.

To prove that v is harmonic, we write v in polar coordinates v(r, θ) = u(log r, θ),
from which we may compute the three quantities

vr =
1

r
ux(log r, θ), vrr =

1

r2
(uxx − ux)(log r, θ), vθθ = uyy(log r, θ).

Consequently we find using the formula for the laplacian in polar coordinates:

∆v = vrr +
1

r
vr +

1

r2
vθθ =

1

r2
(uxx + uyy) = 0,

so that v is harmonic. Showing that v does indeed satisfy the zero boundary
conditions is then an easy consequence of the boundary conditions on u.

Chapter 5, Exercise 14

Note from exercise 9 that F̂R(ξ) =
(
1 − |ξ|

R

)
· 1[−R,R](ξ). Hence by the Pois-

son summation formula, we find that

∞∑
n=−∞

FN (x+ n) =
∞∑

n=−∞
F̂N (n)e2πinx =

N∑
n=−N

(
1−
|n|
N

)
e2πinx.

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