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首页partial differential equations (by Lawrance C. Evans)答案
Graduate studies in Mathematics part one: representation for solutons part two: theory for partial differential equations part three: theory for nonlinear partial differential equations
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Authors: Joe Benson, Denis Bashkirov, Minsu Kim, Helen Li, Alex Csar
Evans PDE Solutions, Chapter 2
Joe: 1, 2,11; Denis: 4, 6, 14, 18; Minsu: 2,3, 15; Helen: 5,8,13,17. Alex:10, 16
Problem 1. Write down an explicit formula for a function u solving the initial-value problem
(
u
t
+ b · Du + cu = 0 on R
n
× (0, ∞)
u = g on R
n
× {t = 0}
Here c ∈ R and b ∈ R
n
are constants.
Sol: Fix x and t, and consider z(s) := u(x + bs, t + s)
Then
˙z(s) = b · Du + u
t
= −cu(x + bs, t + s)
= −cz(s)
Therefore, z(s) = De
−cs
, for some constant D. We can solve for D by letting s = −t. Then,
z(−t) = u(x − bt, 0)
= g(x − bt)
= De
ct
i.e. D = g(x − bt)e
−ct
Thus, u(x + bs, t + s) = g(x − bt)e
−c(t+s)
and so when s = 0, we get u(x, t) = g(x − bt)e
−ct
.
Problem 2. Prove that Laplace’s equation ∆u = 0 is rotation invariant; that is, if O is an orthogonal
n × n matrix and we define
v(x) := u(Ox) (x ∈ R)
then ∆v = 0.
Solution:
Let y := Ox, and write O = (a
i j
). Thus,
v(x) = u(Ox)
= u(y)
where y
j
=
P
n
i=1
a
ji
x
i
. This then gives that
∂v
∂x
i
=
n
X
j=1
∂u
∂y
j
∂y
j
∂x
i
=
n
X
j=1
∂u
∂y
j
a
ji
1
2
Thus,
∂v
∂x
1
.
.
.
∂v
∂x
n
=
a
11
. . . a
n1
.
.
.
.
.
.
a
1n
. . . a
nn
∂u
∂y
1
.
.
.
∂u
∂y
n
= O
T
∂u
∂y
1
.
.
.
∂u
∂y
n
D
x
· v = O
T
D
y
· u
Now,
∆v = D
x
v · D
x
v
= (O
T
D
y
u) · (O
T
D
y
u)
= (O
T
D
y
u)
T
O
T
D
y
u
= (D
y
u)
T
(O
T
)
T
O
T
D
y
u
= (D
y
u)
T
OO
T
D
y
u
= (D
y
u)
T
D
y
u because O is orthogonal
= (D
y
u) · (D
y
u)
= ∆u(y)
= 0
Problem 3. Modify the proof of the mean value formulas to show for n ≥ 3 that
u(0) =
1
nα(n)r
n−1
Z
∂B(0,r)
gdS +
1
n(n − 2)α(n)
Z
B(0,r)
1
|x|
n−2
−
1
r
n−2
f dx,
provided
−∆u = f in B
0
(0, r)
u = g on ∂B(0, r).
Solution: Set
φ(t) =
1
nα(n)t
n−1
Z
∂B(0,t)
u(y)dS (y), 0 ≤ t < r,
and
φ(r) =
1
nα(n)r
n−1
Z
∂B(0,r)
u(y)dS (y) =
1
nα(n)r
n−1
Z
∂B(0,r)
gdS.
Then,
φ0(t) =
t
n
1
α(n)t
n
Z
B(0,t)
∆u(y)dy
=
t
n
1
α(n)t
n
Z
B(0,t)
−f dy
=
−1
α(n)t
n−1
Z
B(0,t)
f dy.
(See the proof of Thm2)
3
Let > 0 be given.
(1) φ() = φ(r) −
Z
r
φ0(t)dt =
1
nα(n)r
n−1
Z
∂B(0,r)
gdS −
Z
r
φ0(t)dt.
Using integration by parts, we compute
−
Z
r
φ0(t)dt =
Z
r
1
nα(n)t
n−1
Z
B(0,t)
f dydt
=
1
nα(n)
Z
r
1
t
n−1
Z
B(0,t)
f dydt
=
1
nα(n)
1
2 − n
1
t
n−2
Z
B(0,t)
f dy
r
−
Z
r
1
2 − n
1
t
n−2
Z
∂B(0,t)
f dS dt
=
1
n(n − 2)α(n)
Z
r
1
t
n−2
Z
∂B(0,t)
f dS dt −
1
r
n−2
Z
B(0,r)
f dy +
1
n−2
Z
B(0,)
f dy
=:
1
n(n − 2)α(n)
I −
1
r
n−2
Z
B(0,r)
f dy + J
.
Observe that
J :
1
n−2
Z
B(0,)
f dy ≤ C ·
2
, for some constant C > 0
and
Z
B(0,)
1
|x|
n−2
f (x)dx =
Z
r
0
dt
Z
∂B(0,t)
1
t
n−2
f dS .
As → 0, I + J →
R
B(0,)
1
|x|
n−2
f (x)dx. Thus,
lim
→0
−
Z
r
φ0(t)dt =
1
n(n − 2)α(n)
Z
B(0,r)
1
|x|
n−2
f (x)dx −
1
r
n−2
Z
B(0,r)
f dy
=
1
n(n − 2)α(n)
Z
B(0,r)
1
|x|
n−2
−
1
r
n−2
f dx.
Therefore, letting → 0, we have from (1)
u(0) = φ(0) =
1
nα(n)r
n−1
Z
∂B(0,r)
gdS +
1
n(n − 2)α(n)
Z
B(0,r)
1
|x|
n−2
−
1
r
n−2
f dx.
Problem 4. We say v ∈ C
2
(
¯
U) is subharmonic if
−∆v ≤ 0 in U.
(a) Prove for subharmonic v that
v(x) ≤
?
B(x,r)
v dy for all B(x, r) ⊂ U.
(b) Prove that therefore max
¯
U
v = max
∂U
v.
(c) Let φ : R → R be smooth and convex. Assume u is harmonic and v := φ(u). Prove v is
subharmonic.
4
(d) Prove v := |Du|
2
is subharmonic, whenever u is harmonic.
Solution.
(a) As in the proof of Theorem 2, set φ(r) :=
>
∂B(x,r)
v dS (y) and obtain
φ
0
(r) =
r
n
?
B(x,r)
∆v(y)dy ≥ 0.
For 0 < < r,
Z
r
φ
0
(s)ds = φ(r) − φ() ≥ 0.
Hence, φ(r) ≥ lim
→0
φ() = v(x). Therefore,
?
B(x,r)
v dy =
1
α(n)r
n
Z
B(x,r)
v dy =
1
α(n)r
n
Z
r
0
Z
∂B(x,s)
v(z) dS (z)
!
ds
=
1
α(n)r
n
Z
r
0
nα(n)s
n−1
φ(s) ds ≥
1
r
n
Z
r
0
ns
n−1
v(x) ds = v(x)
(b) We assume that U ⊂ R
n
is open and bounded. For a moment, we assume also that U is
connected. Suppose that x
0
∈ U is such a point that v(x
0
) = M := max
¯
U
v. Then for
0 < r < dist(x
0
, ∂U),
M = v(x
0
) ≤
?
B(x
0
,r)
v dy ≤ M.
Due to continuity of v, an equality holds only if v ≡ M within B(x
0
, r). Therefore, the set
u
−1
({M}) ∩ U = {x ∈ U|u(x) = M} is both open and relatively closed in U. By the connect-
edness of U, v is constant within the set U. Hence, it is constant within
¯
U and we conclude
that max
¯
U
v = max
∂U
v.
Now let {U
i
|i ∈ I} be the connected components of U. Pick any x ∈ U and find j ∈ I
such that x ∈ U
j
. We obtain
v(x) ≤ max
¯
U
j
v = max
∂U
j
v ≤ max
∂U
v
and conclude that max
¯
U
v = max
∂U
v.
(c) For x = (x
1
, ..., x
n
) ∈ U and 1 ≤ i, j ≤ n,
∂
2
v
∂x
i
∂x
j
(x) =
∂
2
∂x
i
∂x
j
φ(u(x)) = φ
00
(u(x)) ·
∂u
∂x
i
(x) ·
∂u
∂x
j
(x) + φ
0
(u(x)) ·
∂
2
u
∂x
i
∂x
j
(x).
Since φ is convex, then φ
00
(x) ≥ 0 for any x ∈ R. Recall that u is harmonic and obtain
∆v = φ
00
(u) ·
n
X
i=1
∂u
∂x
i
!
2
+ ∆u = φ
00
(u) ·
n
X
i=1
∂u
∂x
i
!
2
≥ 0.
(d) We set v := |Du|
2
=
n
P
k=1
∂u
∂x
k
2
. For x = (x
1
, ..., x
n
) ∈ U and 1 ≤ i, j ≤ n,
∂
2
v
∂x
i
∂x
j
(x) = 2
n
X
k=1
"
∂
2
u
∂x
i
∂x
k
(x) ·
∂
2
u
∂x
i
∂x
j
(x) +
∂u
∂x
k
(x) ·
∂
3
u
∂x
i
∂x
j
∂x
k
(x)
#
.
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