MAT-INF4300
Appendices
A.3 Notation
Notation for derivatives. Assume u : U 7→ R, x ∈ U.
Provided that the limit exists,
u(x + hei ) − u(x)
∂u
(x) = lim
.
h→0
∂xi
h
The usual shorthand version.
∂u
∂2u
∂3u
u xi =
,
u xi xj =
,
u xi xj xk =
∂xi
∂xi xj
∂xi xj xk
The multiindex notation consists of a vector of the form α = (α1 , . . . , αn ),
where each component αi is a nonnegative integer. This vector is called the
multiindex, and has order
|α| = α1 + · · · + αn .
Given a multiindex α we define
D α u(x) :=
∂ |α| u(x)
= ∂xα11 . . . ∂xαnn u
α
α
n
n
∂xn . . . ∂xn
If k is a nonnegative integer,
D k u(x) := D α u(x) | |α| = k
is the set of all partial derivatives of order k. By ordering the partial
derivatives, we can also regard D k u(x) as a point in Rn .
X
1/2
.
|D k u| =
|D α u|2
|α|=k
If in the special case k = 1 we have the elements of Du being a vector.
Du = ux1 , . . . , uxn = gradient vector.
If k = 2 we regard the elements of D 2 u as being arranged in a matrix:
 ∂2u
2u 
. . . ∂x∂1 ∂x
∂x21
n


2
.
.
D u=
 = Hessian matrix
.
∂2u
∂xn x1
∆u =
n
X
i=1
...
∂2u
∂x2n
uxixi = tr D 2 u = Laplacian of u.
1
B.2. Elementary Inequalities
*a. Cauchy’s inequality
ab ≤
a2 b2
+
2
2
(a, b ∈ R)
Proof
0 ≤ (a − b)2 = a2 − 2ab + b2
a2 b2
2ab ≤ a + b =⇒ ab ≤
+
2
2
2
2
*b. Cauchy’s inequality with ǫ
ab ≤ ǫa2 +
b2
4ǫ
(a, b > 0, ǫ > 0).
Proof. Using the regular Cauchy’s inequality.
√ b 2ǫa2
b2
b2
√
2ǫa
ab =
+
= ǫa2 +
≤
2
2(2ǫ)
4ǫ
2ǫ
c. Young’s inequality. Let 1 < p, q < ∞ and
ap bq
+
ab ≤
p
q
1
p
+
1
q
= 1. Then
(a, b > 0).
Proof. The mapping x 7→ ex is convex, and consequently
1
p+ 1
q
ab = elog a+log b = e p log a
log bq
1
ap bq
1
p
q
+
≤ elog a + elog b =
p
q
p
q
*d. Young’s inequality with ǫ.
ab ≤ ǫap + C(ǫ)bq
(a, b > 0, ǫ > 0)
for C(ǫ) = (ǫp)−q/p q −1 .
Proof. Using Young’s inequality.
ab = (ǫp)1/p a
ǫpap
+
=
p
bq
(ǫp)q/p
q
b
(ǫp)1/p
≤
p
(ǫp)1/p a
p
+
b
(ǫp)1/p
q
q
= ǫap + bq (ǫp)−q/p q −1 = ǫap + C(ǫ)bq
2
e. Hölder’s inequality. Assuming 1 < p, q < ∞ and
u ∈ Lp (U) and v ∈ Lq (U), we have
Z
|uv|dx ≤ kukLp (U ) kvkLq (U )
1
p
+
1
q
= 1. Then, if
U
Proof. By homogeneity, we may assume kukLp = kvkLq = 1. Then by Young’s
inequality, for 1 < p, q < ∞,
Z
Z
Z
1
1
p
|u| dx +
|v|q dx = 1 = kukLp kvkLq
|uv|dx ≤
p
q
U
U
U
f. Minowski’s inequality. Assume 1 ≤ p ≤ ∞ and u, v ∈ Lp (U). Then,
ku + vkLp (U ) ≤ kukLp (U ) + kvkLp (U )
Proof.
ku +
Z
U
|u+v|
So,
p−1
vkpLp (U )
=
(|u|+|v|)dx ≤
Z
p
U
|u + v| dx =
Z
Z
U
|u + v|p−1 |u + v|dx ≤
Z
1/p Z
1/p p−1
p
p
p
|u| dx
+
|v| dx
|u+v| dx
p
U
U
p
p
= ku + vkp−1
Lp (U ) kukL (U ) + kvkL (U )
p
p
ku + vkpLp (U ) ≤ ku + vkp−1
Lp (U ) kukL (U ) + kvkL (U )
ku + vkLP (U ) ≤ kukLp (U ) + kvkLp (U )
U
=⇒
*g. General Hölder inequality. Let 1 ≤ p1 , . . . , pm ≤ ∞, with
and assume uk ∈ Lpk (U) for k = 1, . . . , m. Then
Z
U
|u1 . . . um |dx ≤
m
Y
k=1
P
1
i pi
=1
kuk kLpk (U)
Proof.
Induction using Hölder’s inequality.
3
h. Interpolation inequality for Lp -norms. Assume 1 ≤ s ≤ r ≤ t ≤ ∞
and
1
θ (1 − θ)
= +
.
r
s
t
Suppose also u ∈ Ls (U) ∩ Lt (U). Then u ∈ Lr (U), and
kukLr (U ) ≤ kukθLs (U ) kuk1−θ
Lt (U ) .
= 1.
Proof. We can use Hölder’s inequality since θrs + (1−θ)r
t
Z
Z
r
|u| dx =
|u|θr |u|(1−θ)r dx
U
≤
Z
U
|u|
U
s
θr θr
θrs Z
(1−θ)r
t
t
dx
|u|(1−θ)r (1−θ)r
.
U
i. Cauchy-Schwarz inequality.
(x, y ∈ Rn ).
|xy| ≤ |x||y|
Proof. Let ǫ > 0 and note
0 ≤ |x ± ǫy|2 = |x|2 ± 2ǫxy + ǫ2 |y|2 =⇒
±xy ≤
Provided y 6= 0 we set ǫ =
±xy ≤
|x|
.
|y|
So
1 2 ǫ 2
|x| + |y|
2ǫ
2
1
2ǫ
=
|y|
2|x|
and
ǫ
2
=
|x|
.
2|y|
We get,
|y| 2
|x| 2 |x||y| |x||y|
|x| +
|y| =
+
= |x||y|
2|x|
2|y|
2
2
|xy| ≤ |x||y|
4
C.2 Gauss-Green Theorem
In this section we assume U is a bounded, open subset of Rn , and ∂U is C 1 .
Gauss-Green Theorem. Suppose u ∈ C 1 Ū . Then
Z
Z
uxi dx =
uν i dS
(i = 1, . . . , n).
U
∂U
The integral of the partial derivative of u over xi over the vector x in U, is
equal to the surface integral of the antiderivative u times the unit vector ν i
on the boundary ∂U. This result is not proven, but just accepted as a fact.
Integration by parts formula. Let u, v ∈ C 1 (Ū ). Then
Z
Z
Z
uxi vdx = − uvxi dx +
uvν i dS
(i = 1, . . . , n)
U
U
∂U
Proof
First we apply the Gauss-Green with uv instead of u
Z
Z
(uv)xi dx =
uvν i dS.
U
∂U
Then we observe that (uv)xi = uxi v + uvxi by the product rule of
differentiation, so we get:
Z
Z
Z
Z
(uv)xi dx = uxi v + uvxi dx =
uxi vdx +
uvxi dx
U
U
U
Setting both these equal each other, we get the identity. Green’s formulas
For u, v ∈ C 2 (U ).
(i)
Z
U
△udx =
Z
∂U
∂u
dS
∂ν
Proof
We use Gauss-Green’s theorem with uxi instead of u and arrive at the
equality:
Z
Z
uxixi dx =
uxi ν i dS.
U
∂U
Summing over i = 1, . . . , n, we get
n Z
n Z
X
X
uxixi dx =
i=1
U
i=1
5
uxi ν i dS
∂U
From elementary calculus;
Z X
n
uxixi dx =
U i=1
Z
n
X
uxi ν i dS.
∂U i=1
In the integral on the left side, we recognise the expression of the Laplacian.
On the right side we have the definition of the expression for the dot product
of Du · ν which is the definition of ∂u
. Hence, we have
∂ν
Z
Z
∂u
dS.
△udx =
U
∂U ∂ν
(ii)
Z
U
Dv · Dudx = −
Z
Z
u△vdx +
U
∂U
∂v
udS
∂ν
Proof
Using the integration by parts formula with vxi instead of v.
Z
Z
Z
uxi vxi dx = − uvxi xi dx +
uvxi ν i dS
U
U
∂U
Summing from i = 1 to n and taking the integral over the entire sum, and
factor out u from the sums on the right side.
Z X
n
U i=1
uxi vxi dx = −
Z
U
u
n
X
vxi xi dx +
i=1
Z
u
∂U
n
X
vxi ν i dS
i=1
On the left side, we see we have the dot product of the gradients. In the first
term on the right side we have the Laplacian of v and in the last term we get
∂v
the defining expression for ∂ν
. Finally:
Z
Z
Z
∂v
Du · Dvdx = − u△vdx +
u dS
U
U
∂U ∂ν
(iii)
Z
U
u△v − v△udx =
6
Z
u
∂U
∂v
∂u
−v
∂ν
∂v
Proof
Using identity (ii) and shifting the terms:
Z
Z
Z
Du · Dvdx = − u△vdx +
U
Z
U
u△vdx =
U
Z
u
∂U
u
∂U
∂v
dS −
∂ν
Z
U
∂v
dS
∂ν
Du · Dvdx
And by using (ii) again with u and v interchanged and rewriting we get
Z
Z
Z
∂u
v△udx =
v dS −
Dv · Dudx
U
∂U ∂ν
U
Subtracting the left and right sides yields
Z
Z
Z
Z
Z
Z
∂v
∂u
u△vdx − v△udx =
u dS − Du
v dS + Dv
Dudx
· Dvdx −
· U
U
U
U
∂U ∂ν
∂U ∂ν
We collect the terms in the integrals, and end up with
Z
Z
∂v
∂u
u△v − v△udx =
u
− v dS
∂ν
U
∂U ∂ν
Polar Coordinates
(i) Let f : Rn 7→ R be continuous and summable. Then
Z
Z ∞Z
f dx =
dS dr
Rn
0
B(x0 ,r)
for each point x0 ∈ Rn . The integral of the function f over the entire space
is the same as taking the surface integral over the ball with radius r from 0
to ∞.
(ii) In particular
d
dr
Z
B(x0 ,r)
Z
f dx =
f dS
∂B(x0 ,r)
for each r > 0. Differentiating the integral over the ball with radius r gives
us the surface integral over the boundary for the same ball.
7
C.7. Uniform convergence
Suppose that {fk }∞
k=1 is a sequence of real-valued functions in R such that
|fk (x)| ≤ M
k = 1, . . . , x ∈ Rn
for some constant M. Every function in the sequence is bounded by the
constant M over the entire domain. We also need the sequence of fk ’s to
be uniformly equicontinuous. Then there exists a subsequence {fkj }∞
j=1 ⊂
{fk }∞
and
a
continuous
function
f
such
that
k=1
fkj → f
uniformly on compact subsets of Rn .
To say {fk } is uniformly equicontinuous means that for each ε > 0, there
exists a δ > 0 such that |x − y| < δ implies fk (x) − fk (y)| < ε for x, y ∈ Rn
for k = 1, . . .. (All the fk ’s are continuous).
D.1 Banach Spaces
We let X denote a real linear space.
Definition
A mapping k.k : X 7→ [0, ∞) is called a norm if:
(i) ku + vk ≤ k|uk + kvk for all u, v ∈ X. (Triangle inequality).
(ii) kλuk = |λ|kuk for all u ∈ X, λ ∈ R.
(iii) kuk = 0 ⇐⇒ u = 0.
Hereafter we assume X is a normed, linear space.
Definition
We say a sequence {uk }∞
k=1 ⊂ X converges to u ∈ X, written uk → u, if
lim kuk − uk = 0.
k→∞
Definitions
(i) A sequence {uk }∞
k=1 ⊂ X is called a Cauchy sequence if for each
ε > 0 there exists N > 0 such that
kuk − ul k < ε for all k, l ≥ N.
(ii) X is complete if each Cauchy sequence in X converges; that is,
∞
whenever {uk }∞
k=1 is a Cauchy sequence, there exists u ∈ X such that {uk }k=1
converges to u.
(iii) A Banach space X is a complete, normed linear space.
8
Definition
We say X is separable if X contains a countable, dense subset.
Examples
(i) Lp spaces. Assume U is an open subset of Rn , and 1 ≤ p ≤ ∞. If
f : U 7→ R is measurable, we define
R
1/p
p
|f
|
dx
if 1 ≤ p < ∞
U
kf kLp (U ) :=
ess supU |f |
if p = ∞.
We define Lp (U) to be the linear space of all measurable functions f : U 7→ R
for which kf kLp (U ) < ∞. Then Lp (U) is a Banach space, provided we identify
two functions which agree a.e.
Hölder spaces.
(iii) Sobolev spaces.
D.2 Hilbert Spaces
Let H be a real linear space.
Definition
A mapping (·, ·) : H × H 7→ R is called an inner product if
(i) (u, v) = (v, u) for all u, v ∈ H
(ii) the mapping u 7→ (u, v) is linear for each v ∈ H.
(iii) (u, u) ≥ 0 for all u ∈ H.
(iv) (u, u) = 0 if, and only if, u = 0.
Notation
If (·, ·) is an inner product, the associated norm is
kuk := (u, u)1/2
(u ∈ H)
(1)
The Cauchy-Schwarz inequality states
|(u, v)| ≤ kukkvk (u, v ∈ H).
(2)
This inequality was proved in appendix B.2. Using (2) we easily verify (1)
defines a norm on H.
Definition
A Hilbert space H is a Banach space endowed with an inner product which
generates the norm.
9
Examples
(a) The space L2 (U) is a Hilbert space, with
Z
(f, g) =
f gdx.
U
1
(b) The Sobolev space H (U) is a Hilbert space, with
Z
(f, g) =
f g + Df · Dgdx.
U
Definitions
(i) Two elements u, v ∈ H are orthogonal if (u, v) = 0.
(ii) A countable basis {wk }∞
k=1 ⊂ H is called orthonormal if
(wk , wl ) = 0 (k, l = 1, . . . ; k 6= l)
kwk k = 1
(k = 1, . . .).
If u ∈ H and {wk }∞
k=1 ⊂ H is an orthonormal basis, we can write
u=
∞
X
(u, wk )wk ,
k=1
where the series converges in H. In addition,
2
kuk =
∞
X
(u, wk )2 .
k=1
Definition
If S is a subspace of H, S ⊥ = {u ∈ H | (u, v) = 0} for all v ∈ S is the
subspace orthogonal to S.
D.3 Bounded Linear Operators
Definition
(i) A bounded linear operator u∗ : X 7→ R is called a bounded linear
functional on X.
(ii) We write X ∗ to denote the collection of all bounded linear
functionals on X; X ∗ is the dual space of X.
Theorem (Riesz Representation Theorem)
H ∗ can be canonically identified with H; more precisely, for each u∗ ∈ H ∗
there exists a unique element u ∈ H such that
hu∗ , vi = (u, v) for all v ∈ H.
(This mapping is a linear isomorphism of H ∗ onto H).
10
E.1 Lebesgue Measure
Lebesgue measure provides a way of describing the ”size” or ”volume” of
certain subsets of Rn .
Definition
A collection of M of subsets of Rn is called a σ-algebra if
(i) ∅, Rn ∈ M
(ii) A ∈ M =⇒ Ac ∈ M.
∞
∞
(iii) If {Ak }∞
k=1 ⊂ M, then ∪k=1 Ak , ∩k=1 Ak ∈ M
Theorem (Existence of Lebesgue -measures and -sets
There exists a σ-algebra M of subsets of Rn and a mapping
| · | : M 7→ [0, +∞]
with the following properties:
(i) Every open subset of Rn , and thus every closed subset of Rn , belong
to M.
B.
(ii) If B is a ball in Rn , then |B| equals the n-dimensional volume of
(iii) If {Ak }∞
k=1 ⊂ M and the sets are pairwise disjoint, then
∞
∞
X
[
|Ak |
(countable additivity)
Ak =
k=1
k=1
(iv) If A ⊆ B, where B ∈ M and |B| = 0, then A ∈ M and |A| = 0.
Notation
The sets in M are called Lebesgue measurable sets and | · | is the ndimensional Lebesgue measure.
If some property holds everywhere on Rn except for a measurable set
with Lebesgue measure zero, we say the property holds Almost everywhere,
abbreviated ”a.e”.
Remarks
From the previous information we see that |A| equals the volume of any set
A with piecewise smooth boundary. We see that |∅| = 0 and that if the sets
in (iii) are not disjoint, we have a ≤ instead of equality (subadditivity).
11
E.2 Measurable functions and integration
Definition
Let f : Rn 7→ R. We say f is a measurable function if
f −1 (U) ∈ M
for each open subset U ∈ R.
Note in particular that if f is continuous, then f is measurable. The sum
and product of two measurable functions are measurable and for a sequence
{fk }of measurable functions, lim sup fk and lim inf fk , are also measurable.
Theorem (Ergoroff’s Theorem)
let {fk }, f be measurable functions and
fk → f
a.e on A,
where A ⊂ Rn is measurable, |A| < ∞. Then for each ε > 0, there exists a
measurable subset E ⊂ A such that
(i) |A − E| ≤ ε.
(ii) fk → f uniformly on E.
Now if f is a nonnegative, measurable function, it is possible, by an
approximation of f with simple functions, to define the Lebesgue integral
Z
f dx.
Rn
This agrees with the usual integral if f is continuous or Riemann integrable.
If f is measurable, but not necessarily nonnegative, we define
Z
Z
Z
+
f dx =
f dx −
f − dx,
Rn
Rn
Rn
provided at least one of the terms on the right hand side is finite. In this case
we say f is integrable.
Definition
A measurable function f is summable if
Z
|f |dx < ∞.
Rn
12
Remark
Note carefully the terminology: a measurable function is integrable if it has
an integral (which may be infinite) and is summable if this integral is finite.
Notation
If the real-valued function f is measurable, we define the essential supremum
of f to be
ess sup f := inf µ ∈ R |{f > µ} = 0 .
E.3 Convergence theorems for integrals
The Lebesgue theory of integration is especially useful since it provides the
following powerful convergence theorems.
Theorem (Fatou’s Lemma)
Assume the functions {fk } are nonnegative and measurable. Then,
Z
Z
fk dx.
lim inf fk dx ≤ lim inf
Rn
k→∞
k→∞
Rn
Theorem (Monotone Convergence Theorem)
Assume the functions {fk } are measurable, with
0 ≤ f1 ≤ f2 ≤ · · · ≤ fk ≤ fk+1 ≤ . . . ,
then
Z
lim fk dx = lim
Rn k→∞
k→∞
Z
fk dx.
Rn
Theorem (Dominated Convergence Theorem)
Assume the functions {fk } are integrable and fk → f a.e. Suppose also
|fk | ≤ g a.e for some summable function g. Then
Z
Z
fk dx →
f dx.
Rn
Rn
13
—————————————————————-
24/08-2010
Normal ODE’s are of the form
′
y (t) = f t, y(t)
.
y(0) = c
PDE’s have the form
Υ t, x1 , . . . , xn , u, ut, ux1 , . . . = 0
| {z }
ind. var.
where u = u(t, x1 , . . . , xn ), a function of several variables.
Lipschitz continuity is whenever |f (x)−f (y)| ≤ L|x−y| for some constant L.
We write f ∈ C k (Ω) for some function f : Ω 7→ R which means f, f ′ , . . . , f (k)
exist and f (k) is continuous.
As opposed to ODE, there are usually no general solutions for PDE’s.
Recall Taylor’s theorem for a series expansion.
(x − a)n (n)
f (a).
n!
f (x) = f (a) + (x − a)f ′ (a)0 . . . +
Proving using the fundamental theorem of calculus.
Z x
Z
′
f (t) = f (x) − f (a) =⇒ f (x) = f (a) +
a
x
f ′ (t)
a
and by integration-by-parts:
′
′
= f (a) + xf (x) − af (a) −
= f (a) +
Z
x
tf ′′ (t)dt
a
x
′′
a
where
Z
Z
′
′
f (t)dt + xf (a) − af (a) −
x
a
Z
f ′′ (t)dt = x f ′ (x) − f ′ (a)
′
= f (a) + (x − a)f (a) +
Z
a
tf ′′ (t)dt
a
x
(x − t)f ′′ (t)dt
and by expanding forever, we eventually arrive at (*).
14
x
(*)
The n-dimensional version.
f (x) = f (a) +
Z
|α|≤K
1
D α f (a)(x − a)α
α!
α
Where the one dimensional notation would be D α f = ∂∂xαf and the multi|α| f
dimensional notation means D α = ∂xα1∂,...,∂x
αn which can also be written as
n
1
α1
αn
∂x1 , . . . , ∂xn f .
Here we have the vectors |α = (α1 , . . . , αn ) ∈ Rn , x = (x1 , . . . , xn ) ∈ Rn
the scalar |α| = α1 + · · · + αn and the cross-faculty α! = α1 ! . . . αn !.
When we set |α| = K for some positive, integer constant K this means
all combinations α1 + · · · + αn = K. Notation wise the three vectors
(x − a)α = (x1 − a1 )α1 . . . (xn − an )αn .
Proof
Fix some x ∈ Rn . Join a and x by u(t) = a + t(x − a), reducing the multidimensional to the 1-dimensional case, so we can use Taylor’s expansion. We
have performed a linear interpolation, so if this was for n = 2, we would have
for two points in the plane:
The function f u(t) : R 7→ R with f u(1) = f (x) and f u(0) = f (a).
Now we use the 1-dimensional Taylor’s form.
n
X
1 dk f
u(t)
f (x) = f u(1) = f (a) +
k! dtk t=0
k=1
X K · D α f u(t) t=0 (x − a)α
= f (a) +
α
|α|≤K
X K D α f (a)(x − a)α
= f (a) +
α
|α|≤K
Gauss-Green formula
For some open set U ⊆ Rn and some function u : U 7→ R, where u ∈ C 1 Ū
∂u
where we have the multi-dimensional u = u(x1 , . . . , xn ) and uxi = ∂x
i
Z
Z
uxi dx =
uν i ds
U
∂U
15
where ∂U denotes the boundary of U. ν = (ν 1 , . . . , ν n ) is the outward norm
from the boundary:
We use ds because the integral over ∂U has a different measure than the first
integral.
Green’s Formula (Important)
P
For a Laplacian: △u = ni=1 uxixi (with no cross derivatives) we have, for
u, v ∈ C 2 (Ū
Z
Z
∂u
ds
(i) △udx =
U
∂U ∂ν
.
=
Du
·
ν
(dot
product)
and
Du
=
u
,
.
.
.
,
u
where ∂u
x
x
n
1
∂ν
This can be proved with a 1-dimensional Gauss-Green theorem:
Z
Z
ux dx =
u · νds.
U
∂U
Verifying Green’s formula (i) using Gauss-Green’s theorem with u instead of
ux
Z
Z
ux ux =
(ux · ν)ds.
U
(ii)
Z
U
∂U
DuDvdx = −
since (uv)xi = uvxi + uxi v.
(iii)
Z
U
Z
u△vdx +
U
(u△v − v△u)dx =
Z
∂U
Z
∂U
u
∂v
∂u ds
−v
∂ν
∂ν
Convolution
Let f, g be measurable functions.
(f ∗ g)(x) =
Z
Rn
f (x − y)g(y)dy
assuming that the integrals exist.
16
∂v
ds
∂ν
Properties
(i) f ∗ g = g ∗ f
which follows from change of variables.
(ii) f ∗ (g ∗ h) = (f ∗ g) ∗ h
(iii) supp(f ∗ g) ⊂ supp(f ) ∗ supp(g)
where supp = support. So supp(f ) = {x ∈ Rn : f (x) 6= 0} and x 6∈
supp(f ) ⇒ f (x) = 0: the region of the function that is not 0.
Proof of (iii)
Assume x ∈ supp(f ∗ g). Assume for contradiction x 6∈ supp(f ) + supp(g).
This implies
∀y ∈ supp(g),
x − y 6∈ supp(f )
But
(f ∗ g)(x) =
Z
Rn
f (x − y)g(y)dy =
Z
supp(g)
f (x − y)g(y)dy
or else it would be zero. So x − y 6∈ supp(f ) = 0 which means f (x − y) = 0
and the integral is 0. Contradiction. Thus,
x ∈ supp(f ∗ g) =⇒ x ∈ supp(f ) + supp(g).
We consider the function
η(x) =
C exp
1
|x|2 −1
0,
, |x| ≤ 1
|x| > 1
R
so supp(y) = B(0, 1): the closed unit ball. We choose C such that R⋉ ηdx =
R
1, η ε (x) = ε1n η xε . For this function supp(η ε ) = B(0, ε). We have Rn η ε dx =
1.
The support is decreasing as ε → 0; the function is higher and narrower,
but the area remains 1.
17
ε↓0
We have the Dirac delta: η ε (x) R= δ(x). This is a measure where δ(x) = 0
except x = 0 where δ(0) = ∞ and δ(x)dx = 1.
We introduce the regularization of f .
Z
ε
f (x − y)ηε (y)dy
f = f ∗ nε =
| {z }
Rn
∈C ∞
where f ε : is a smooth, well behaved function. Some properties:
(i) f ε ∈ C ∞ (U)
(ii) f ε → f
a.e as ε ↓ 0.
If f is “not smooth” the derivative does not exist, so we use the convolution
to obtain a more manageable function. In fact this is the standard way of
dealing with non-smooth functions while working with PDE’s.
—————————————————————-
27/08-2010
We consider an elliptic
PDE, the Laplace equation △u = 0. We have x ∈ Rn ,
P
n
u : Rn 7→ R, △u = i=1 uxi xi .
Another important elliptic PDE is the Poisson equation △u = f .
Definition of distribution
D(Ω) = φ : φ ∈ Cc∞ (Ω)
the space of test functions.
∂k u
Cc∞ (Ω) = u :
; exists ∀k supp(u) is compact
k
∂x
|{z}
∞ deriv.
The distribution T : D(Ω) 7→ R and satisfies T (φn ) → 0 if φn → 0.
If T1 = T2 ⇔ T1 (φ) = T2 (φ) ∀φ ∈ D(Ω).
If (the first is differentiated) T1′ = T2 ⇒ (T2 , φ) = −(T1 , φ′ ) ∀φ ∈ D(Ω),
in the sense of distribution.
(1) measure B(0, r) = α(n)r n : the volume of unit ball in Rn .
(2)
Z
—
r↓0
∂B(0,r)
g(x)ds(x) −→ g(0)
18
where the special integral sign is the average of the integral:
Z
Z
1
—f (x)dx =
f (x)dx.
α(n)r n B(0,r)
We will now look closer at the Laplacian equation, △u = 0 in Rn . What u
satisfies this equation?
This equation has a radial equation, which is a strong property and
benefitial.pIt means u(x) = u(|x|), so if x = (x1 , . . . , xn ), we can instead
use |x| = x21 + . . . + x2n . We denote u(x) = u(|x|) = v(r) where r = |x|, the
radius. We can then consider △v = 0
By the chain rule uxi = v ′ (r) xri :
u(x1 , . . . , xn ) = v(r) =⇒
∂r ∂
∂
∂u
=
v(r) =
v(r)
u xi =
∂xi
∂xi
∂r
∂xi
The double derivative (product rule):
u xi xi
x2i
1 x2i
′
= v (r) ·
+ v (r)
− 3
r
r
r
′′
(n − 1) ′
v (r) = 0
r
We hve not reduced the PDE to a second degree ODE.
△u = 0 =⇒ v ′′ (r) +
Solution:
c
log(v ′ ) = (1 − n) log r + log c =⇒ v ′ = n−1
r
b log r + c n = 2
v(r) =
b
+c n≥3
r n−2
for some constants b and c.
Claim
Choose constants b, c such that −△x u = δ0 (Dirac distr.) in the sense of
distribution. Then
Z
Z
− △x ugdx = g(x)δ0 = g(0).
The Laplacian equation △u(0) = 0 since |x| = r → 0 means the solutions
tend to infinity.
19
Fundamental solution for the Laplacian.
△x u = δ0 . This is the case in the solution we found v(r). The fundamental
solution is:
1
− 2π
log |x|
n=2
φ(x) =
1
1
·
n
≥3
n(n−2)α(n) |x|n−2
We will prove the claim −△x φ = δ0 where φ is the fundamental solution. We
set △φ = 0 in Rn \{0} so we can find solutions.
Z
△x φ = δ0 =⇒ −
△x φg(x)dx = g(0) ∀g ∈ D(Ω)
(⋆)
Rn
For φ we associate a distribution Fφ .
Z
(Fφ , g) =
φ(x)g(x)dx
Rn
∀g ∈ D(Ω) = D(Rn ).
The g is a smooth, test function. We are interested in (F△φ , g) =
and using integration by parts:
Z
Z
Z
′′
′ ′
f gdx = − f g dx = f g ′′ dx
so we get
=
We continue.
Z
Rn
φ△gdx = Fφ , △g
F△φ , g =
Z
R⋉
△φgdx
φ(x)△g(x)
Rn
We split up this integral in two parts.
Z
Z
=
φ(x)△g(x) +
B(0,δ)
R
φ(x)△g(x) := I + J
Rn \B(0,δ)
Solving I:
There are two situations, n = 2 and n ≥ 3. First we calculate for n = 2.
Z
Z
1
1
log |x| △g(x)dx ≤ k△gkL∞ log |x|dx
2π {z }
B(0,δ) 2π
B(0,δ) |
φ(x)
Going to polar coordinates.
Z 2π Z δ
Z δ
≤ C
log |r|rdrdθ ≤ C log |r|rdr ≤ C log |δ|δ 2
0
0
0
20
where C log |δ|δ 2 → 0 as δ → 0.
Now we consider the case n ≥ 3
Z
1
1
· n−2
△g(x)dx
B(0,δ) n(n − 2)α(n) |x|
Z δ Z
≤ k△gkL∞ 1
ds(y)dr n−2
0
B(0,δ) |y|
Z
Z δ
Z δ
1
1
ds(y)dr
≤
C
nα(n)r n−1 dr
≤C
n−2
n−2
r
r
B(0,δ)
0
0
Z δ
Cnα(n) 2
δ → 0 as δ → 0
= Cnα(n)
rdr =
2
0
Thus, I → 0 as δ → 0.
We direct our attention to the other integral, J.
Z
φ(x)△g(x)dx
Rn \B(0,δ)
Z
∂g
∂φ
g(x)ds(x)+
φ(x) ds(x)
=
∂ν
∂(Rn \B(0,δ)
Rn \B(0,δ)
∂(Rn \B(0,δ) ∂ν
R
R
∂v
= 0 − J1 + J2 where Ω (u△v − v△u)dx = Ω (u ∂ν
− v ∂u
). Useful facts are
∂ν
∂φ
x
x
1
that ∇x φ(x) = − nα(n)|x|n , ν = − |x| and ∂ν = △φ · ν = nα(n)|x|
n−1 (scalar
product).
Z
1
J1 = −
g(x)ds(x)
n−1
∂(B(0,δ) nα(n)|x|
Z
Z
1
g(x)ds(x) = − —
g(x)ds(x) → −g(0) as δ ↓ 0
=−
nα(n)δ n−1 ∂(B(0,δ)
∂B(0,δ)
Z
Z
△φ(x)g(x)dx−
Z
J2 = Z
∂g
∂g
φ(x) ds(x) ≤ k kL∞
|φ(x)|dx
∂ν
∂ν
∂B(0,δ)
∂B(0,δ)
Again we must consider the two cases n = 2 and n ≥ 3. First n = 2.
Z
log |x|ds(x) ≤ C log |δ|(2πδ) ≤ Cδ log |δ| → 0 as δ ↓ 0
∂B(0,δ)
Similar for n ≥ 3 which means J2 → 0 as δ ↓ 0. So, finally, I + J = g(0). We let δ → 0 to cover the whole space.
−△x φ = δ0
21
and remember △φ 6= 0 ∀x ∈ Rn .
Z
v(x) =
φ(x − y)f (y)dy = φ ∗ f
Rn
△x v(x) =
Z
Rn
△x φ(x − y)f (y)dy =
Applying −△x φ = δ0 , we get
=−
Z
Z
Rn
△yφ(x − y)f (y)dy
δ{x} f (y)dy = −f (x)
so −△u = f which is the Poisson equation. If you know φ, the fundamental
solution, you know v(x) which satisfies −△u = f .
Theorem
Assume f ∈ C 2 (Rn ),
u(x) =
where −△x φ = δ0 , then
Z
Rn
φ(x − y)f (y)dy
(i) u ∈ C 2 (Rn )
(ii) −△u = f
where (i) follows from f ∈ C 2 , since for φ ∗ f , vx = φx ∗ f = φ ∗ fx .
—————————————————————-
31/08-2010
Recap
Z
—
Z
—
1
u(x)dx =
α(n)r n
B(x,r)
Z
1
u(y)ds(y) =
nα(n)r n
∂B(x,r)
u(x)dx.
B(x,r)
Z
u(y)ds(y)
∂B(x,r)
Harmonic Functions
We say that u : Rn 7→ R is harmonic if △u = 0.
Mean value theorem
Let Ω ⊂ Rn . If u ∈ C 2 (Ω) is harmonic, then
Z
Z
u(x) = —
u(y)ds(y) = —
u(y)dy ∀B(x, r) ⊂ Ω.
∂B(x,r)
B(x,r)
22
Proof (First equality)
For r > 0
Z
φ(r) = —
u(y)ds(y).
∂B(x,r)
For r = 0, φ(r) = u(x), since limr→0+ φ(r) = u(x). Claim: φ is constant,
which means φ′ (r) = 0.
Z
φ(r) = —
u(y)ds(y); y = x + rz
∂B(x,r)
Z
=—
u(x + rz)ds(z)
∂B(0,1)
Z
φ (r) = —
′
Z
=—
By definition. Since
∂u
∂ν
∂B(0,1)
∂B(x,r)
∇u(x + rz)zds(z)
∇u(y)
y−x
ds(y)
r
= ∇u · ν
Z
=—
∂u
ds(y)
∂B(x,r) ∂ν
Z
∂u
1
ds(y)
=
n−1
nα(n)r
∂B(x,r) ∂ν
Integration by parts.
1
=
nα(n)r n−1
Z
B(x,r)
∇(∇u)dy
We recognise this as the Laplacian.
Z
1
=
△u(y)dy = 0
nα(n)r n−1 B(x,r)
The equality follows since u is harmonic, so △u = 0. Thus φ(r) = φ(0) =
u(x).
Proving the other equality.
Z
u(x) = —
u(y)dy.
B(x,r)
23
Begin by using a very common step.
Z
Z r Z
u(y)dy =
B(x,r)
=
Z
0
0
∂B(x,s)
hZ
r
n−1
nα(n)s
—
∂B(x,s)
u(y)ds(y) dr
i
u(y)ds(y) dr
The part inside the brackets is u(x) by the first part.
Z r
=
nα(n)sn−1 u(x)dr = α(n)u(x)r n
0
Thus,
Z
—
u(y)dy = u(x)
B(x,r)
This theorem has a converse.
Theorem
If u ∈ C 2 (Ω) satisfies
Z
u(x) = —
u(y)ds(y),
∂B(x,r)
∀B(x, r) ⊂ Ω,
then u is harmonic, i.e △u = 0.
Proof
Assume
Z
φ(r) = —
u(y)ds(y) = u(x)
∂B(x,r)
′
is a constant so φ (r) = 0 and
Z
r
φ (r) = —△u(y)dy.
n
′
Let △u 6= 0 for contradiction. Then △u > 0 ot △u < 0 in B(x, r), then
φ′ (r) > 0 or φ′ (r) < 0. But φ′ (r) = 0 so △u = 0. Contradiction.
Maximum Principle
Let Ω ⊂ Rn be open and bounded and assume u ∈ C 2 (Ω)∩C(Ω̄) is harmonic.
(1) max u(x) = max u(x)
Ω̄
∂Ω
(2) If Ω is connected and ∃x0 ∈ Ω such that u(x0 ) = max u(x)
Ω̄
If both these conditions are satisfied, u is a constant in Ω. This theorem has
applications to harmonic functions.
24
Proof
We note that condition 2 implies condition 1, so we only have to prove
condition 2.
We assume that x0 ∈ Ω such that M = u(x0 ) = max u(x) for some
Ω̄
constant M. This satisfies the mean value theorem, so
Z
M =—
u(y)dy = u(x0 ) = max u(x) ≤ M.
Ω̄
B(x,r)
The equality holds if u(y) = M, ∀y ∈ B(x0 , r). This means we have u(y)
included in B(x0 , r), which means it is constant there.
The set A = {x : u(x) = M} is, by connectedness both open and closed,
so A = Ω. By using −u instead of u the maximum principle will become the
minimum principle with the same theorem except we use ’min’ instead of
’max’.
We have shown how to solve (the boundary value problem)
−△u = f in Ω
u=g
on ∂Ω
Can we say if this solution is unique?
We assume u, v are two solutions and we claim u = v. We set
w = u−v
w
e = v−u
which means w, w
e both satisfy
△w = 0 in Ω
w = 0 on ∂Ω
We can then apply the maximum principle
max |u − v| = max |u − v| = 0 ⇒ u = v
∂Ω
Ω
We use the maximum principle to prove unique solutions for e.g parabolic
equations.
Theorem Smoothness of Harmonic Functions
Let Ω ⊆ Rn be open and bounded. If u ∈ C(Ω) satisfies
Z
u(x) = —
u(y)ds(y),
∀B(x, r) ⊂ Ω
∂B(x,r)
25
then, u ∈ C ∞ (Ω), we have a smooth function. In particular, u harmonic
implies it is smooth.
A recap. For
1/(|x|2 −1)
Ce
|x| < 1
η(x) =
0
|x| ≥ 1
R
we choose the constant C such that Rn η(x)dx = 1 and supp(η) ⊂ B(0, 1).
Now
1 x
ηε (x) = n η
ε
ε
R
with Rn ηε (x)dx = 1 (the approximation function) and supp(ηε ) ⊂ B(0, ε).
Proof
We introduce
uε (x) =
Z
Ω
Then,
ηε (x − y)u(y)dy = ηε ∗ f.
(i) uε ∈ C ∞ (Ω)
(ii) uε (x) = u(x).
We prove these to prove the theorem. Claim (i) is a smooth C ∞ by the
convolution property.
Z
(uε )x = (ηε )x (x − y)u(y)d
Ω
(uε )xx =
Z
Ω
(ηε )xx (x − y)u(y)d
so uε ∈ C ∞ because ηε ∈ C ∞ .
Because of this we only need to show (ii) to complete the proof.
Z
uε (x) =
ηε (x − y)u(y)dy
B(x,ε)
By def. of ηε .
Z
|x − y| 1
u(y)dy
= n
η
ε B(x,ε)
ε
Z Z
r 1 ε
u(y)ds(y) dr
η
= n
ε 0
ε
∂B(x,r)
Z
Z
1 ε r
= n
u(y)ds(y)dr
η
ε 0
ε ∂B(x,r)
26
1
= n
ε
Z
0
ε
Z
r n−1
nα(n)r
—
u(y)ds(u)dr
η
ε
∂B(x,r)
Now we can use the smoothness property.
Z
1 ε r = n
nα(n)r n−1 u(x)dr
η
ε 0
ε
Z
Z
1 ε r
= u(x) n
ds(y)dr
η
ε 0
ε ∂B(0,r)
Combining the two integrals back to one.
Z
|y| 1
= u(x) n
dy
η
ε B(0,ε)
ε
Z
= u(x)
ηε (y)dy = u(x)
B(0,ε)
|
{z
}
=1
So u(x) = uε (x) and the smoothness is proved: u(x) ∈ C ∞ (Ω).
—————————————————————-
03/09-2010
We remember that u is harmonic implies that u satisfies MVP, the mean
value property, and the reverse: if u ∈ C 2 and u satisfies MVP that implies
that u is harmonic.
The Max principle states that
max u = max u
∂U
U
which is used for proof of a unique solution.
We also have the result that u harmonic ⇒ u ∈ C ∞ (U).
Liouville’s Theorem
Let u : Rn 7→ R be harmonic and bounded. Then u is constant.
27
Proof
We fix some x0 ∈ Rn , so
Z
u(x0 ) = —
u(y)dy.
B(x0 ,r)
From a previous result, u harmonic ⇒ u ∈ C ∞ (R⋉). Since △u = 0, then
△uxi = 0 for all i = 1, . . . , n, and furthermore uxi is itself harmonic. This
means, by the MVP,
Z
Z
1
uxi (x0 ) = —
uxi (y)dy =
ux (y)dy
α(n)r n B(x0 ,r) i
B(x0 ,r)
Integration by parts,
1
=
α(n)r n
Z
uνi dS(y).
∂B(x0 ,r)
1
|uxi (x0 )| ≤ kνkL∞
α(n)r n
Using that u is bounded,
≤ CkukL∞ (Rn
Z
∂B(x0 ,r)
|u|dS(y)
1
n
· α(n)nr n−1 = C ∀r
n
α(n)r
r
R
since ∂B(x0 ,r) dS(y) = α(n)nr n−1 . If we take the limit, r → ∞, then
uxi (x0 ) ≡ 0 ∀x0 ∈ Rn , which is true only when u is constant. Representation formula
let f ∈ Cc2 (Rn ), and let n ≥ 3. Then every bounded solution of −△u = f ,
x ∈ Rn , has the form
Z
u(x) =
φ(x − y)f (y)dy + C
Rn
where C is some constant and φ is the fundamental solution of the Laplacian.
Proof
First: why is this not true when n = 2? Because for n = 2,
φ(x) = −
1
log |x| → ∞ as |x| → ∞
2π
so the integral can be unbounded. However, for n ≤ 3, we have
φ(x) =
K
|x|n−2
K=
28
1
n(n − 2)α(n)
which is not bounded (but that is not obvious).
we have already proved that
Z
u(x) =
φ(x − y)f (y)dy
Rn
is a solution for the Poisson equation. We must now show that this solution
is bounded, i.e show that:
Z
1
|u(x)| = K
f (y)dy n−2
Rn |x − y|
Z
Z
1
1
≤ K
f (y)dy+|K
f (y)dy ≤ |I1 +I2 | ≤ C
n−2
n−2
B(x,ε) |x − y|
Rn \B(x,ε) |x − y|
for some upper bound C.
First we see that I2 is bounded. This is because f ∈ Cc2 (Rn ), that is, f
has compact support. That means that outside some specified region f will
be 0. Hence the integral is not taken over the entire set.
Other parts blow up when x → y, but we have removed the ε-ball around
x, so y cannot be arbitrarily close to x. This means it is finite. From these
observations, I2 is bounded.
I1 has the form
Z rZ
1
I1 =
dsdr
n−2
0
∂B(x,r) r
which we have already solved. From this we know that I1 is bounded. Since
both I1 and I2 are bounded, then |I1 + I2 | ≤ C, so
Z
u(x) =
φ(x − y)f (y)dy
Rn
is a bounded solution of −△u = f . Let u and u
e be two bounded solutions. For w = u − u
e, we have −△w = 0
which means w is constant (Liouville’s theorem), and this in turn means that
u=u
e0C. Any two solutions are just shifted by a constant.
Harnack’s Inequality
(For V ⊂⊂ U we mean V ⊂ V ⊂ U, where V is a compact set).
For each connected, open set V ⊂⊂ U, ∃C, C > 0 depending on V , such
that
sup u ≤ C inf u
v
V
for all non-negative, harmonic functions u in U.
29
Proof
We set
1
r = dist(V, ∂U) = inf{dist(V, ∂U)}
4
and choose x, y such that |x − y| ≤ r.
Z
Z
1
1
udz = n u(y)
u(x) = —
udz ≥ n —
2
2
B(y,r)
B(x,2r)
This is true for any x, y, so
=⇒ 2n u(y) ≥ u(x) ≥
1
u(y)
2n
V connected and V compact means that
∃{Bi }N
i=1 ⇒ u(x) ≥
1
2nN
u(y)
And u(x) ≥ 21n u(y) means u(y) ≥ 21n u(z) so, naturally, u(x) ≥ 21n u(y) ≥
2n
12 u(z). Since we have compactness, we have a finite number of “balls” that
cover the entire set: u(x) ≥ 2nN1u(y) .
This theorem says we have a sense of uniformity. The different values
don’t differ more than a constat factor C.
Green’s Function
We need Green’s function to classify all solutions of
−△u = f in Ω
u=g
on ∂Ω
Let us first consider the one-dimensional version.
30
−u′′ (x) = f (x) in Ω
u(0) = u(1) = 0 on ∂Ω
What is u? We set the “fundamental theorem”:
Z x
u(x) = C1 +
u′ (y)dy
0
′
u (y) = C2 +
Z
y
′′
0
u (z)dz = C2 −
Z
=⇒ u(x) = C1 + C2 x −
We define F (y) :=
Ry
0
0
f (z)dz, so
Z tZ
0
y
f (z)dz =
0
Z
Use integration by parts,
Z x
Z
x
′
yF (y)dy = xF (x) −
yF (y) 0 −
0
Z
x
y
Z
y
f (z)dz
0
f (z)dzdy.
0
x
F (y)dy
0
x
yf (y)dy =
0
Z
0
x
(x − y)f (y)dy.
Using the boundary condition u(0) = u(1) = 0,
Z 1
C1 = 0, C2 =
(1 − y)f (y)dy
0
so
u(x) = x
Green’s function
Z
1
0
(1 − y)f (y)dy −
Z
0
x
(x − y)f (y)dy.
y(1 − x) 0 ≤ y ≤ x
G(x, y) =
x(1 − y) x ≤ y ≤ 1
So,
u(x) = x
Z
1
0
(1 − y)f (y)dy −
Z
x
0
(x − y)f (y)dy =
Some basic properties are
• G is continuous.
• G(x, y) = G(y, x) (symmetric).
31
Z
0
1
G(x, y)f (y)dy.
• Writing the solution in a more convenient way.
Z
u(x) =
φ(x − y)f (y)dy + C
Rn
solves −△u = f .
Green’s function is the “fundamental solution” to
−△u = f in Ω
u=g
on ∂Ω
For later usage: Green’s formula:
Z
Z
u△v − v△udy =
Ω
u
∂Ω
∂u
∂u
− v dS(y).
∂ν
∂ν
Multi-dimensional case
Let U ⊆ Rn be open and bounded.
−△u = f in U
u=g
on ∂U
Let u ∈ C 2 (U ) be any function. Fix x ∈ U such that B(x, ε) ⊂ U. We apply
Green’s formula: Vε = U\B(e, ε) to u(y) and φ(y − x).
Z
Z
∂φ
∂u
u(y) (y−x)−φ(y−x) (y)dS(y)
u(y)△φ(y−x)−φ(y−x)△u(y)dy =
∂ν
∂ν
∂Vε
Vε
For the boundary ∂Vε we have to consider both the inner and outer boundary
as shown in the picture:
We have seen before that:
Z
∂u
φ(y − x) dS(y) → 0 as ε ↓ 0
∂ν
B(x,ε)
In addition, we have also already seen that
Z
∂φ
u(y)
(y
−
x)d?
→ u(x) as ε ↓ 0
∂ν
∂B(x,ε)
32
(These are the inner boundaries in the two terms we get after Green’s
function).
We know that φ(x) = 0 if x 6= 0. The terms u(y)△φ(y − x) = 0 since
△φ(y − x) = 0 when y 6= x and this is the case since we have removed the
ball B(x, ε) around x.
As ε ↓ 0
Z Z
∂u
u(x) =
φ(y − x)
dS(y) −
φ(y − x)△u(y)dy.
∂ν
∂U
U
and Vε → U when ε → 0.
From this expression we have a problem with the term
not know this.
∂U
∂ν
since we do
Correction function
Fix x, φx − φx (y) such that △φx = 0 in U, φx = φ(y − x) on ∂U.
.
(We need this to find a way around the problem with ∂U
∂ν
Apply Green’s formula to φx and u (using △φx = 0 on U).
Z
Z ∂u ∂φx
x
dS(y)
− φx (y)
− φ (y)△u(y)dy =
u(y)
∂ν
∂ν
U
∂U
We use that φx = φ(y − x) on ∂U.
Z ∂φx
∂u =
u(y)
dS(y)
− φ(y − x)
∂ν
∂ν
∂U
Implementing the corrector function:
G(x, y) := φ(y − x) − φx (y) for x 6= y
Z
Z
∂G
u(x) = − G(x, y)△u(y)dy −
u(y)
(x, y)dS(y)
∂ν
U
∂U
= △yG(x, y) · ν. We have another expression that does not include
where ∂G
∂ν
∂u
(which is unknown on the boundary).
∂ν
Finally we get,
Z
Z
∂G
dS(y)
u(x) =
f (y)G(x, y)dy −
g(y)
∂ν
U
∂U
33
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07/09-2010
For u ∈ C 2 (U)
−△u = f in U
u=g
on ∂U
the solution has the form
Z
∂G
+
u(x) −
g(y)
∂ν
∂U
Z
f (y)G(x, y)dy.
U
We have G(x, y) = φ(x − y) − φx (y) for
△y φx (y) = 0
U
x
φ (y) = φ(y − x) y ∈ ∂U
Theorem
Green’s function is symmetric: G(x, y) = G(y, x).
Proof
We set v(z) = G(x, z), and w(z) = G(y, z). We want to check if G(x, y) =
v(y) = w(x) = G(y, x). We have:
△v = △w = 0
since x 6= z and y 6= z for the
first and second
equality, respectively.
We consider the set U\ B(x, ε) ∪ B(y, ε) = V depicted in the following
image.
0=
Z
0=
0=
Z
Z
V
v△w − w△vdx
v
∂w
∂v
− w dS
∂ν
∂ν
v
∂v
∂w
− w dS
∂ν
∂ν
∂B(y,ε)
∂B(x,ε)
34
Z
=
V
Z
DvDw − DwDvdx
w
B(x,ε)
Z
∂v
∂w
dS =
− v
∂ν
∂ν
| {z }
Vanishes
v
∂B(y,ε)
∂w
∂v
− w dS
∂ν
∂ν
We look closer at the problematic part:
Z ∂w 1
≤ C n−2 εn−1 → 0 when ε → 0.
v
∂ν
ε
As we have shown before:
lim
ε→0
Similarly,
Z
w
Z
v
∂B(x,ε)
∂B(y,ε)
∂φ
dS = w(x).
∂ν
∂w
→ v(y).
∂ν
We now know that y 7→ G(x, y) is harmonic, and x 7→ G(y, x) is harmonic,
as well as G itself.
Green’s function for a half space
We will end up with a general solution for the Laplacian problem. Green’s
function for R+
n (like the upper plane for n = 2 or the positive octant in
n = 3. Formally we write Rn+ = {x ∈ Rn |xn > 0}.
For the half-space we get the boundary problem:
−△u = f x ∈ Rn+
u=g
x ∈ ∂Rn+
for which we can find the general solution. To construct this solution we solve
the auxiliary problem:
−△y φx = 0
Rn+
x
φ (y) = φ(x − y) Rn+
For the vector x = (x1 , . . . , xn ) we define x
e = (x1 , . . . , −xn ), which is a
reflection of x as shown in the following image.
35
Claim: φx (y) = φ(y − x
e). This is harmonic. Consider y ∈ ∂Rn+ . φ is constant
on the surface: φ(x − y) = φ(|x − y|):
Since φ is a function of the magnitude, and not the point, φ(x−y) = φ(y −e
x).
A candidate for the solution is
Z
∂G
(y)g(y)dS(y)
u(x) = −
∂ν
Rn
+
and we have
and
G(x, y) = φ(x − y) − φ(y − x
e)
∂G
∂G
=−
∂ν
∂yn
∂G
xn − yn
1
−xn − yn n
=−
−
|
∂ν
nα(n) |x − y|n
|e
x−y
and using that |e
x − y|n = |x − y|n on the boundary,
==
1
∂xn
·
.
nα(n) |x − y|n
The candidate is on the form
∂xn
u(x) =
nα(n)
Z
R⋉−1
g(y)
dy
|x − y|n
and apparently we have convolution on the kernel, and we introduce the
Poisson kernel:
Z
=
K(x, y)g(y)dy
Rn−1
which is a candidate, not a solution.
36
Theorem
We assume g ∈ C 1 (Rn−1 ) ∩ L∞ (Rn−1 ). Then
(i) u ∈ C ∞ (Rn−1 ) ∩ L∞ (Rn+ )
(ii) △u = 0 in R+
n
(iii) For x ∈ Rn+ , we have limx→xo u(x) = g(x0 ) for x0 ∈ Rn−1 . (When you
approach a point on the boundary from inside, you get the right path).
Proof
For x ∈ Rn+ , y 7→ K(x, y) is C ∞ (∂Rn+ ). K is a smooth function on the
boundary as long as x is not on the boundary. Thus, u ∈ C ∞ (Rn+ .
The integral of K, the kernel, is exactly 1.
Z
Z
Z ∞
1
2xn
2xn
(n − 1)α(n − 1)r n−2
K(x, y)dy =
dy
=
dr
n
nα(n) Rn−1 |x − y|n
nα(n) 0
(x2n + r 2 ) 2
Rn−1
!
Z
2xn (n − 1)α(n − 1) ∞ 1
r n−2
dr
=
nα(n)
xn 1 + ( xrn )2
0
Substitution, s =
r
,
xn
dr = xn ds
x
x
2
n
n (n − 1)α(n − 1)
=
n
nα(n)xn
1
,
cos2 θ
Second substitution, s = tan θ, ds =
2(n − 1)α(n − 1)
⇒
nα(n)
Z
π
2
∞
sn−2
xn−2
n
0
(1 + s2 ) 2
1 + s2 =
1
.
cos2 θ
sinn−2 θ
1
n
cos
θ
dθ
cosn−2 θ
cos2 θ
0
2(n − 1)α(n − 1)
=
nα(n)
Z
Z
π
2
sinn−2 θdθ = 1
0
The step showing that this equals 1 is a proof by induction. It is quite easily
verified for n = 2, as α(n) is the volume/area/length of the unit circle in Rn :
2α(1)
2α(2)
Z
π
2
1dθ =
0
2(2) π = 1.
2π 2
Since g is bounded by assumption and the integral of the kernel is 1, we have
Z
u(x) =
K(x, y)g(y)dy ≤ kgkL∞ .
Rn−1
37
By definition y 7→ K(x, y) is harmonic, and
K=
∂G
∂φx
−
.
∂yn
∂yn
Since K(x, y) = K(y, x), it follows that x 7→ K(y, x) is harmonic as well.
Using that g is continuous |y − x0 | ≤ δ ⇒ |g(y) − g(x0 )| < ε.
δ
0
We choose some x ∈ R+
n such that |x − x | ≤ 2 . We then verify (iii) in the
proof by considering
Z
0
|u(x) − g(x )| =
K(x, y) g(y) − g(x0 ) dy Splitting the integral in two parts.
Z
Z
0 ≤ I +J =
K(x, y) g(y) −g(x ) dy +
|x0 −y|≤ε
|x0 −y|>ε
K(x, y)g(y) −g(x0)dy
The integral of the kernel is 1, and by continuity, |g(y) − g(x0 )| ≤ ε, so I ≤ ε,
so it is arbitrarily small.
xn
What happens when y is far away from x0 ? Since K(x, y) = |x−y|
n , when
|y − x| is large, K becomes smaller. Furthermore,
|y −x0 | = |y −x+x−x0 | ≤ |y −x|+|x−x0 | < |y −x|+
δ
1
≤ |y −x0 |+|y −x|.
2
2
In short |y − x0 | ≤ 21 |y − x0 | + |y − x| so 21 |y − x0 | ≤ |y − x|, and we have
removed the δ-dependence. Since g is bounded, we can assume the ’worst
case scenario’ and set |g(y) − g(x0 )| ≤ 2kgkL∞ , so
Z
1
2n+2 kgkL∞ xn
dy → 0
J ≤ 2kgkL∞
K(x, y)dy ≤
nα(n)
|y − x0 |n
|x0 −y|>ε
as xn → 0+ . Both parts of the integral tend to 0, so the difference between
u(x) and g(x0 ) becomes arbitrarily small, so u(x) → g(x0 ) as x → x0 . 38
—————————————————————-
10/09-2010
Theorem (Poisson’s formula for a ball)
Assume g ∈ C(∂B(0, r)) and define u by
Z
r 2 − |x|2
g(y)
u(x) =
dS(y)
nα(n)r ∂B(0,r) |x − y|n
then,
(i) u ∈ C ∞ (B(0, r)
(ii) △u = 0 in B(0, r)
(iii) For x ∈ B(0, r), limx→x0 u(x) = g(x0 ) for all x0 ∈ ∂B(0, r).
Proof
If we can prove (ii), (i) follows since all harmonic functions are C ∞ . We see
that ∀x the mapping y 7→ G(x, y) is harmonic when y 6= x, and by symmetry
x 7→ G(x, y) is harmonic. Hence,
x 7→
Since G is harmonic,
P ∂G
∂yi
△x
X ∂G
∂yi
yi = −K(x, y).
is harmonic.
X ∂G
u(x) =
∂yi
Z
yi =
X ∂
△x G yi
∂yi
K(x, y)g(y)dS =⇒
∂B(0,r)
△x u(x) =
Z
∂B(0,r)
△x K(x, y)g(y)dS = 0
which proves that (ii) u is harmonic and it follows that (i) u ∈ C ∞ B(0, r) .
From the expression for u(x), we have the Kernel,
Z
r 2 − |x|2
1
K(x, y) =
.
nα(n)r ∂B(0,r) |x − y|n
From the maximum principle, we know that the boundary problem:
−△u = f in B(0, r)
u=g
on ∂B(0, r)
39
has a unique solution. From assumption, g ∈ C ∂B(0, r) , which means that
for y, x ∈ ∂B(0, r),
|y − x0 | < δ =⇒ |g(y) − g(x0 )| < ε.
Having established this, we examine
Z
0
0
K(x, y) g(y) − g(x ) dS |u(x) − g(x )| = ∂B(0,r)
We split the integral in two:
Z
Z
0 ≤
K(x, y) g(y)−g(x ) dS+
∂B(x,δ)∩∂B(x0 ,r)
∂B(x,δ)\∂B(x0 ,r)
K(x, y)g(y)−g(x0)dS
which we call I1 + I2 , and they are the surface integrals in the image:
By continuity we have |g(y) − g(x0 )| < ε, and the surface integral of K(x, y)
is 1, so
Z
<ε+
K(x, y)g(y) − g(x0 )dS
∂B(x,δ)\∂B(0,r)
Taking a closer look at the boundary:
And, as on page 39. in the text book, we get |y − x| ≥ 21 |y − x0 |, and when
we return to the integral, we get
Z
r 2 − |x|2
4
∞ dS → 0
<ε+
·
kgk
K
|y − x0 |
∂B(x,δ)\∂B(x0 ,r) nα(n)r
as x → x0 , which implies |x| → r, and it follows that r 2 − |x|2 → 0. 40
Energy Methods
Again we specify the Poisson boundary problem and denote it (P).
−△u = f in B(0, r)
(P)
u=g
on ∂B(0, r)
We assume ∂U is C 1 and u ∈ C 2 U .
To prove that the solution is unique: assume u, v are solutions and set
w = v − u.
−△w = 0 in U
w=0
on ∂U
Z
Z
Z
∂w
Green’s formula
0=
w△w
==
w
dS −
|Dw|2dx
∂ν
U
∂U
U
= 0, the first term vanishes, and we are left with |Dw| = 0 in U.
Since ∂w
∂ν
There is no gradient so w must be flat: w(y) = w(x) for x ∈ U and y ∈ ∂U
We know from this that there is at most one solution (there is of course
the possibility that there is no solution).
We have A which is a class of functions, and we define it as
o
n
2
A = u|u ∈ C (C), u = g on ∂U
or in other words, any function taking the values we want.
Z
1
I u :=
|Du|2 − f udu
2
U
Theorem We have the equivalence
u solves (P) ⇐⇒ I[u] ≤ I[w] ∀w ∈ A.
In physics this is the “least energy” or “minimized energy property”.
Proof
Implications from left to right. We assume u solves (P) and w ∈ A.
Z
Z
0 = (−△u − f )(u − w)dx =
Du · (Du − Dw) − f u + f wdx
U
U
=
Z
U
|Du|2 − DuDw − f u + f wdx
We use a very important, elementary inequality. It is obvious that 12 (a −
b)2 ≥ 0, which is true both for numbers and vectors. Multiplying (or dot
41
multiplying) and shifting terms yields 12 a2 + 12 b2 ≥ ab. using this, we get
−DuDw ≥ − 21 (|Du|2 + |Dw|2):
Z
1
1
≥
|Du|2 − |Du|2 − |Dw|2 − f u + f wdx
2
2
U
Z
Z
1
1
2
|Du| − f udx −
|Dw|2 − f wdx = I[u] − I[w].
=
U 2
U 2
In summary 0 ≥ I[u] − I[w] which leaves us with I[u] ≤ I[w].
For the converse we assume we have a minimiser. For a small number τ and
v ∈ C0∞ (U), then u + τ v ∈ A, we define i(τ ) := I[u + τ v], for i(0) ≤ i(τ ) for
τ 6= 0 and i′′ (0) = 0.
Z
i
2
d h 1 ′
i (τ ) =
Du + τ Dv − f (u + τ v) dx
U dt 2
Z
i
τ2
d h 1 Du|2 + τ Du · Dv + |Dv|2 − f u − τ f v dx
=
2
U dt 2
We now have a polynomial, which is easy to differentiate.
Z
=
DuDv + τ |Dv|2 − f vdx
U
′
0 = i (0) =
Z
U
Integration by parts.
Z
DuvdS +
Z Z
U
∂U
DuDv − f vdx
U
−△u · v − f vdx
Using that v = 0 on the boundary,
Z
= (−△u − f )vdx = 0.
U
This is quite strong, as the only assumption we made was that v ∈ C0∞ (U),
so
Z
wvdx = 0∀v ∈ C0∞ (U) =⇒ w = 0.
U
(The trick is that v ≈ w).
42
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14/10-2010
43
—————————————————————-
17/10-2010
44
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21/10-2010
Weak Derivative
We say v = D α u in the weak sense if for u, v ∈ L1loc (U) we have
Z
Z
α
|α|
vφdx
∀φ ∈ Cc∞ (U).
u · D φdx = (−1)
U
U
Sobolev Space
The definition of the Sobolev space is that if u ∈ W k,p(U), then D α u exists
in the weak sense for all |α| ≤ k.
We write kukW k,p(U ) < ∞, and we define
X
kukW k,p (U ) =
|α|≤k
kukW k,p(U ) =
kD α ukpLp (U )
X
|α|≤k
p1
1 ≤ p < ∞.
kD α ukL∞ (U )
p = ∞.
As we have shown earlier, k · kW k,p (U ) is a norm.
Example
Let U = B(0, 1) ⊂ R and let u(x) = |x|−a . For what values of a > 0, n, p is
u ∈ W 1,p (U)?
We recall the definition of the weak derivative, which for k = 1
amounts to: u ∈ W k,p (U) which means D α u ∈ Lp (U) for all |α| ≤ 1. By
the definition of Sobolev norms,
kukpW 1,p (U )
=
X
|α|≤1
kD
α
ukpLp (U ) k
=
| {z }
Z
p
B(0,1)
|u| dx =
Z
B(0,1)
We utilize the coarea formula.
Z 1Z
Z tZ
−ap
|x| dS dr =
0
+
α=0
Calculating the first term:
kukpLp (U )
=
kukpLp (U )
∂B(0,r)
0
45
n
X
|i=1
kuxi kpLp (U )
{z
α=1
|x|−ap dx
∂B(0,r)
r −ap dSdr
}
We use that |∂B(0, r)| = cr n−1 , so
Z 1
=C
r −ap+n−1dr.
0
This is finite if −ap + n − 1 > −1 which is equivalent to saying ap < n.
We move on to the sum of the first derivatives, and start by calculating
what it is.
X a
−ax
−
uxi = ∂xi |x|−a = ∂xxi
x2i 2 = a+2
|x|
which implies that |Du(x)| = |x|aa+1 . We now calculate the Lp -norm for this
gradient.
Z
Z
Coarea
p
|Du(x)| dx = a
|x|−(a+1)p dx ==
B(0,1)
a
Z 1Z
0
B(0,1)
r
−(a+1)p
dSdr = aC
∂B(0,r)
Z
1
r −(a+1)p+n−1 dr.
0
This is finite if −(a + 1)p + n − 1 > −1 which is equivalent to (a + 1)p < n.
Thus, u, Du are Lp (U) if (n + 1)p < n (as this is a stricter condition than
the first).
(If u, Du ∈ Lp (U), then by definition they are in W 1,p (U).
(2) Let φ ∈ Cc∞ (U) and fix ε > 0. Define Vε = U\B(0, ε).
<BILDE>
Note that Vε → U as ε → 0. Using integration by parts;
Z
Z
Z
uφν i dSx .
uxi φdx +
uφxi dx = −
∂Vε
Vε
Vε
Vε has two boundaries: an inner and an outer boundary: ∂Vε = ∂U ∪∂B(0, ε),
but by compact support φ(∂U) = 0, so the surface integral of the outer
boundary vanishes.
Z
Z
uxi φdx +
uφν idSx .
=−
∂B(0,ε)
Vε
We estimate the second term.
Z
Z
i
≤
uφν
dS
x
∂B(0,ε)
∂B(0,ε)
46
|x|−a |φ|dSx ≤
(since ∂B(0, ε) ⇔ |x| = ε)
Z
kφkL∞
ε−a dSx ≤ Cε−a+n−1 → 0 as ε → 0, if a + 1 ≤ n.
∂B(0,ε)
Moreover, u, Du ∈ L1 (U) if a + 1 < n. Taking the limit ε → 0.
Z
Z
Z
uxi φdx + lim
uφν i dSx
uφxi dx = − lim
lim
ε→0
ε→0
Vε
ε→0
Vε
∂B(0,ε=
(we can put the limit inside the integral)
Z
Z
Z
Z
α
uφxi dx = − uxi φdx + 0 =⇒
uD φdx = − D α uφdx
U
U
U
U
for (a + 1) < n. Thus,
• Du exists in the weak sense if a + 1 < n.
• u, Du ∈ Lp if (a + 1)p < n.
and based on this we know that u ∈ W 1,p (U) if (a + 1)p < n or a <
n−p
.
p
Theorem: Properties of Weak Derivatives
Assume u, v ∈ W k,p (U), |α| ≤ k. Then
(i) D α u ∈ W k−|α|,p(U) and D β (D α u) = D α (D β u) = D α+β u. for all
α, β where |α| + |β| ≤ k.
(ii) For each λ, µ ∈ R, λu + µv ∈ W k,p (U). and D α (λu + µv) =
λD u + µD α v for all |α| ≤ k.
α
(iii) If V ⊂ U, then u ∈ W k,p (V ).
P
β≤α
(iv)
ξ ∈ Cc∞ (U) then ξu
βIf α−β
α
α!
D ξD
u where αβ = β!(α−β)!
.
β
∈
W k,p (U) and D α (ξu)
Proof
(i) Fix φ ∈ Cc∞ (U), then D β φ ∈ Cc∞ (U) is also a test function.
Z
Z
Z
α
β
|α|
|α|
α
β
uD α+β φdx.
uD (D φdx) = (−1)
D u D φ dx = (−1)
|{z}
U
U
U
=
Inf. diff.
(The partial derivatives of φ is the normal derivative, not the weak kind).
We know that u is k-times differentiable and |α + β| ≤ k.
Z
Z
α+β
β
|α+β|
α
D α+β uφdx =⇒
D
uφdx = (−1)
= (−1) (−1)
U
U
47
Z
α
β
|β|
D D φdx = (−1)
Z Z
U
U
D α+β uφdx =⇒
U
D β (D α u) = D α+β u in the weak sense.
For the other claim.
kD α ukpW k−|α|,p (U ) =
X
|β|≤k−|α|
kD β (D α u)kpLp (U ) =
X
|α+β≤k
kD α+β ukpLp (U ) = kukW k,p (U ) .
(ii) Fix φ ∈ Cc∞ (U).
Z
Z
Z
α
α
(λu + µv)D φdx = λ uD φdx + µ vD αφdx =
U
U
|α|
(−1)
(−1)|α|
Z
Z
U
α
λD uφdx + (−1)
|α|
Z
µD α vφdx =
U
U
(λD α u + µD α v)φdx =⇒ D α (λu + µv) = λD α u + µD α v.
U
Clearly, kλu + µvkW k,p(U ) ≤ |λ|kukW k,p(U ) + |µ|kvkW k,p(U ) , where we used that
k · kW k,p (U ) is a norm.
(iii) Let φ ∈ Cc∞ (V ), then φ ∈ Cc∞ (U).
Z
Z
α
|α|
D α uφdx
uD φdx = (−1)
U
U
φ: only defined for V (??).
Z
Z
α
|α|
D α uφdx.
uD φdx = (−1)
V
V
Justifies that u has weak derivatives. Also, clearly,
kukW k,p(V ) ≤ kukW k,p (U ) .
(iv) We use an induction argument on |α|. First assume that |α| = 1, then
for any φ ∈ Cc∞ (U):
Z
Z
α
(ξu)D φdx =
u
ξD α φ
dx
| {z }
U
U
D α (ξφ)−D α (ξ)φ
=
Z
α
U
uD (ξφ)dx −
Z
U
α
uD ξφdx = −
48
Z
U
α
D u · ξφdx −
Z
U
uD α ξ · φdx =
−
Z
(D α ξu + ξDα u) φdx,
U
and by definition of the weak derivative:
X α β
Dξ D α−β u
D (ξu) = D ξ · u + ξD u =
β
β≤α
α
α
α
(and we have assumed |α| = 1). We assume this holds for l < k and the
formula is valid for all |α| ≤ l. Choose α with |α| = l + 1 which implies
α = β + γ where |β| = l and γ = 1. Claim: the formula is valid for all such
α.
Z
α
ξuD φdx =
U
Z
β
γ
|β|
ξuD (D φ)dx = (−1)
|{z}
U
∈Cc∞
Z
D β (ξu)D αφdx
U
and by the induction assumption,
Z X β
|β|
D β ξDβ−σ uD γ φdx.
= (−1)
U σ≤β σ
Our goal now is to try and transfer the α outside..
Z X β
|β|+|γ|
D γ (D σ ξDβ−σ u)φdx
= (−1)
σ
U σ≤β
Since |γ| = 1 we can use the formula from the induction step.


Z X β  γ+σ β−σ
D ξD u + D β ξDβ+γ−σu  φdx
= (−1)|α|
|
{z
} |
{z
}
σ
U
σ≤β
(I)
X β σ≤β
σ
D
γ+σ
ξD
I
β−σ
II
X
ρ=σ+γ
u ==
γ≤ρ≤β+γ
β
D ρ ξDβ+γ−ρu
ρ−γ
Starts from order 1, so |γ| = 1,
X β D ρ ξDβ+γ−ρ u + D β+γ ξu = IRes.
ρ−γ
γ≤ρ≤β+γ
49
(II)
X β X β σ
β+γ−σ
β+γ−σ
D σ ξDβ+γ−σ = IIRes
D ξD
u = ξD
u+
σ
σ
γ≤σ≤β
σ≤β
Adding these together.
X β β D σ ξDβ+γ−σ u + ξDβ+γ−σ u + D β+γ ξu.
+
IRes + IIRes =
σ
σ−γ
γ≤σ≤β
and using
β
σ−γ
+
β
σ
β+γ
σ
=
, we get
X β + γ D γ ξDβ+γ−σ u + ξDβ+γ u + D β+γ ξu
=
σ
γ≤σ≤β
X α
X β + γ γ
β+γ−σ
D α ξDα−σ u
D ξD
u=
=
σ
σ
σ≤γ
σ≤β+γ
Therefore,
Z
ξDα φdx = (−1)|α|
Z
U
U
!
X α
D α ξDα−γ uφ dx
σ
σ≤α
which, finally, implies,
X α
D σ ξDα−σ u.
D (ξu) =
σ
σ≤α
α
Theorem Sobolev Spaces as function spaces
For each k = 1, . . . and 1 ≤ p ≤ ∞, the Sobolev space is a Banach space.
Proof
We know that k · kW k,p (U ) is a norm, so it only remains to prove that
the Sobolev space is complete, i.e for every Cauchy sequence in W k,p (U)
converges to some function in the same space W k,p (U).
Assume that {um } is a Caucy sequence in W k,p(U). Then {D α um } is
Cauchy in Lp (U). Since
kum − un kW k,p (U ) → 0 as l, m → ∞
by definition we have
kD α um − D α ul kLp (U ) → 0 as l, m → ∞.
50
By completeness of Lp (U), D α um → uα in Lp (U). In particular, take
α = (0, . . . , 0), then um → u(0,...,0) in Lp (U). We define u = u(0,...,0) and claim
that uα = D α u. If we can show this, we have shown that D α um → D α u ⇒
um → u in W k,p(U) for |α| ≤ k. To show this, we first fix a φ ∈ Cc∞ (U).
Then
Z
Z
α
uD φdx = lim
um D α φdx.
R
R
m→∞
U
R
U
(where u = lim um = lim um ). The functions un have weak derivatives
up to order k, so
Z
Z
α
|α|
|α|
uα φdx =⇒
= (−1) lim
D um φdx = (−1)
n→∞
U
U
D α u = uα for all |α| ≤ k, hence D α um → D α u in Lp (U), which implies
um → u in W k,p (U), since by defintion
X
kum − ukW k,p (U ) =
kD α um − D α ukpLp (U ) .
In conclusion: W k,p (U) is a Banach space.
51
—————————————————————-
24/10-2010
52
—————————————————————-
28/09-2010
Smooth Approximation
Recall
•
•
∃ηε ∈
Cc∞
B(0, ε) ,
Z
ηε dx = 1.
Rn
Uε = x ∈ U | dist(x, ∂U) > ε
Theorem - Local Approximation by Smooth functions
For some integer k ≥ 0 and 1 ≤ p < ∞, let u ∈ W k,p(U). Set uε = ηε ∗ u in
Uε . Then,
(i)
uε ∈ Cc∞ (Uε ) for each ε > 0
(ii)
k,p
uε −→ u in Wloc
(U).
(We showed a related result for Lp spaces, but this theorem states it is also
true for Sobolev spaces).
Proof
Proving (i).
k=0
u ∈ W k,p (U) =⇒ u ∈ Lp (U) =⇒ uε ∈ C ∞ (Uε )
where the last implication follows from the properties of mollifiers.
Proving (ii). We begin proving point (ii) with verifying the following
claim. For all |α| ≤ k,
α ε
Dαu
D
| {zu} = ηε ∗ |{z}
Norm. der
Weak der.
The left side derivatives are normal, partial derivatives since uε is a smooth
function, and the derivatives on the right side are the weak derivatives as
defined for Sobolev spaces.
Proof of claim
By the definition of mollifiers.
ε
u (x) =
Z
U
ηε (x − y)u(y)dy,
53
so
α ε
D u (x) =
Z
U
Dxα ηε (x − y)u(y)dy
where we have, by the definition of weak derivatives (?),
Dxα ηε (x − y) = (−1)|α| Dyα ηε (x − y)
and plugging this in the first expression,
Z
α ε
|α|
Dyα ηε (x − y)u(y)dy.
D u (x) = (−1)
U
But for some fixed x ∈ Uε , ηε (x − y) ∈ Cc∞ (U), so
Z
α
|α|
|α|
ηε (x − y)D u(y)dy
(−1)
= (−1)
U
We get (−1)2|α| = 1, and we are left with the desired result:
Z
ηε (x − y)D αu(y)dy = ηε ∗ D α u(x).
U
By assumption and construction of Sobolev spaces,
D α u(x) ∈ Lp (Uε )
hence,
By definition,
D α uε −→ D α u in Lp (V ), V ⊂⊂ U.
keε − ukpW k,p(V ) =
X
|α|≤k
kD α uε − D α ukpLp (V ) → 0
k,p
as ε → 0, so we can conclude uε → u in Wloc
(U). Theorem - Global approx. by smooth functions.
Assume U is bouned and suppose u ∈ W k,p(U), 1 ≤ p < ∞. Then there
exists functions
um ∈ C ∞ (U) ∩ W k,p(U)
such that um → u as m → ∞ in W k,p(U).
Proof (Difficult)
(1) Define
1
Ui = x ∈ U | dist(x, ∂U) >
i
for i = 1, 2, . . ., so Ui ⊂ Ui+1 .
54
We write Vi = Ui+3 − U i , and we get the sets as indicated by the image.
S
We choose V0 ⊂⊂ U such that U = ∞
i=0 Vi , e.g V0 = U3 .
(2) Let {ξi }∞
i=0 be a smooth partition of unity subordinate to the open sets
∞
{Vi }i=0 , that is, suppose
0 ≤ ξi P
≤ 1, ξi ∈ Cc∞ (Vi ), i ∈ N
∞
i=0 ξi = 1 on U.
(This is a theorem from Topology).
Now, if u ∈ W k,p(U), then ξi u ∈ W k,p (U) (we have already proved this), and
supp(ξi u)⊂ Vi .
(3) Define U i = ηεi ∗ (ξi u) for each εi > 0, for i = 0, 1, . . ..
Note the following: fro sufficiently small εi > 0, then supp(U i ) ⊂ Wi ,
i = 1, 2, . . . where Wi = U+4 − U . The support of the product ξi u is the
support of of the sum of the sets each of them have support. Then, for
ξi > 0,
δ
kui − ξi ukW k,p (U ) ≤ i+1
(*)
2
for some small δ > 0, i = 0, 1, . . .. By the local approximation theorem,
kui − ξiuk converges to 0.
(4) Define
v=
∞
X
ui
i=0
for a fixed x ∈ U. For each V ⊂⊂ U, there are at most finitely many non-zero
terms in the sum. For example, if x ∈ W1 , then ui (x) = 0 for i ≥ 4, and if
x ∈ Wj then ui(x) for |i − j| ≥ 4. Because of this the sum on the set V will
be finite, and it follows that the full sum is finite.
Using the finite sum and that ui ∈ C ∞ (U), the function v is smooth.
We will use the following:
u=1·u=
55
∞
X
i=1
ξi u.
We want to approximate v by y. For V ⊂⊂ U,
kv − ukW k,p (V ) ≤
∞
X
i=0
i
ku − ξi ukW k,p (V )
∞
X
δ
≤
=δ
i+1
2
i=0
The first inequality is the triangle inequality, and the second one follows from
(*).
kv − ukW k,p (U ) = sup kv − ukW k,p (V ) ≤ δ.
V ⊂⊂U
We have proved theorem:
v ∈ C ∞ (U) ∩ W k,p (U).
Theorem - Global approx. by smooth functions up to the boundary.
Assume U is bounded and ∂U ∈ C 1 . Suppose u ∈ W k,p (U) for 1 ≤ p < ∞,
then ∃um ∈ C ∞ (U ) such that um → u in W k,p(U).
Proof
(1) Fix x0 ∈ ∂U. Since ∂U ∈ C 1 , ∃r > 0 and a C 1 function γ : Rn−1 7→ R
such that
U ∩ B(x0 , r) = x ∈ B(x0 , r) | xn > γ(x′ )
where x′ = (x1 , . . . , xn−1 ). (This follows since the boundary is C 1 , see
appendix). We call V = U ∩ B(x0 , 2r ). See image.
(2) Now define
xε = x + λεen ,
(x ∈ V, ε > 0, λ > 0)
and en is (0, . . . , 0, 1). We only add λ to the last component.
56
Claim
B(xε , ε) ⊂ U ∩ B(x0 , r) ⇔ y ∈ B(xε , ε) ⇒ y ∈ B(x0 , r)
for yn < γ(y ′ ) and y ′ = (y1 , . . . , yn−1). To verify this claim: we can write
y ∈ B(xε , ε) as y = xε + εw for |w| = 1.
|y − x0 | = |xε + εw − x0 | = |x + λεen + εw − x0 | ≤ |x − x0 | + ε|λ + 1| < r
since |x − x0 | < 2r and ε|λ + 1| <
within the ball. We have
r
2
for sufficietly small ε, thus we know y is

w1
 .. 


ε
y = x + εw = x + λεen + εw) = x + ε(λen + w = x + ε  . 
 wn − 1 
λ + wn

so,
  

y1
x1 + εw1
  

..
y =  ...  = 

.
yn
xn + ε(λ + wn )
where we note that yn = xn + ε(λ + wn ). We can approximate γ(y ′ ) =
γ(x′ + εw ′ ) ≤ γ(x′ ) + εM for some M > 0, since
Z ε
Z ε
d
′
′
′
′
′
(∂(x +ετ )dτ = γ(x )+
∇γ ·wdτ ≤ γ(x′ )+εM.
γ(x +εw ) = γ(x )+
0
0 dτ
Since x ∈ V we also have γ(x′ ) < xn so we can conclude that γ(y ′) < xn +εM.
From earlier we have
yn = xn + ε(λ + wn ) ⇒ xn = yn − ε(λ + wn ) ⇒
γ(y ′) = yn − ε(λ + wn )0εM = yn − ε(λ + wn − M).
If we choose λ > 0 so large that λ + wn > M, we get yn > γ(y ′ ). Claim
shown.
(3) We define uε (x) := u(xε ), for x ∈ V , and we write v ε = ηε ∗ uε . We do
this to get away from the boundary since the mollification vanishes at the
boundary due to the support. We mollify at the point xε instead, which is
well within the ball.
Claim: v ε ∈ C ∞ (V ).
57
We will verify this. By definition of mollifiers;
Z
ε
v (x) = ηε ∗ uε (x) =
ηε (x − y)uε(y)dy
B(x,ε)
and by our definition: uε (x) = u(xε ), we get
Z
Z
ε
=
ηε (x − y)u(y )dy =
ηε (x − y)u(y + λεen )dy
B(x,ε)
B(x,ε)
Change of variable, z = y + λεen ,
Z
Z
=
ηε (x + λεen − z)u(z)dz =
B(x+λεen ,ε)
B(xε ,ε)
ηε (xε − y)u(y)dz = ηε ∗ u(xε ).
So, we have shown v ε (x) = ηε ∗ u(xε ). Now x ∈ V implies xε ∈ W ⊂⊂ U
which again implies v ε ∈ C ∞ (V ). Claim shown.
(4) Claim:
v ε → u in W k,p(V ).
We look at all α, where |α| ≤ k.
kD α v ε − D α ukLp (V ) ≤ kD α v ε − D α uε kLp (V ) + kD α uε − D α ukLp (V )
The first term tends to 0 as ε → 0 since uε is the mollification, and the second
term tends to 0 as ε → 0 by the local approximation theorem.
We need that Lp is continuous under translation, that is:
Z
|f (y) − f (x)|p dx −→ 0 as y → x.
We have uε = eε (x) = u(xε ) = u(x + λεen ) which goes to u(x) as ε → 0, and
since Lp is continuous under translation,
kD α uε − D α ukLp (V ) → 0
as ε → 0.
(5) Since ∂U is compact, we can find finitely many xi0 ∈ ∂U, ri > 0
and corresponding sets Vi = U ∩ B(xi0 , r2i ) and functions vi ∈ C ∞ (V i ) for
i = 1, . . . , N. such that
N
[
ri
∂U ⊂
B(xi0 , )
2
i=1
58
(that is, we can cover the entire boundary with a finite amount of balls Vi ),
and for each such ball Vi the function vi satisfies:
kvi − ukW k,p (Vi ) ≤ δ.
(**)
(For each Vi we aruge in the exact same way as we did for V , only repeating
it N times).
S
∞
We choose V0 ⊂⊂ U such that U ⊂ N
i=0 Vi and select v0 ∈ C (V 0 )
satisfying
kv0 − ukW k,p (V0 ) ≤ δ.
(***)
(which we can do by a previous theorem).
(6) Let {ξi }N
i=0 be a smooth partition of unity on U, subordinate to the sets
ri N
V0 , B(xi0 , ) i=1 .
2
Define
v=
We write u =
PN
i=0
i=0 ξi u.
α
N
X
ξi vi =⇒ v ∈ C ∞ (U ).
For all |α| ≤ k
α
kD v − D ukLp (U ) ≤
N
X
i=0
kD α (ξi vi ) − D α (ξi u)kLp (U )
which follows by the triangle inequality.
=
N
X
i=0
kD α ξi (vi − u) kLp (U )
where we can write
α
D ξi (vi − u) =
so we have
X β β≤α
α
D β ξi − D α−β (vi − w)
N X X
β
β
α−β
D
ξ
−
D
(v
−
u)
=
i
i
α
i=0 β≤α
Lp (U )
≤
N X X
β
i=0 β≤α |
α
kDξαi kL∞ (U ) · kD α−β (vi − u)kLp (U )
|
{z
}
{z
}
≤kvi −uk k,p
W
≤C∈R
59
(U )
≤C
N
X
i=0
kvi − ukW k,p(U ) ≤ CNδ.
In the last inequality we used (**) and (***), and since δ can be arbitrarily
small, this means v → u in W k,p(U). 60
—————————————————————-
01/10-2010
2.5.8
We have the boundary conditions
△u = 0 in Rn+
u = g on ∂Rn+
Assume g ∈ L∞ (bounded) and g(x) = |x| for x ∈ ∂Rn+ and |x| ≤ 1. Show
that the gradient Du is not bounded near x = 0.
The solution u(x) si given by Poisson’s formula for a half space, so we have
Z
g(y)
2xn
dS.
(x ∈ Rn+ )
u(x) =
nα(n) ∂Rn+ |x − y|n
We are given a hint, which is to estimate (en denotes the n’th unit vector:
en = (0, . . . , 0, 1) ∈ Rn )
u(λen ) − u(0)
.
λ
Clearly u(0) = 0.
Z
g(y)
2λ
dS
(x ∈ Rn+ ).
u(λen ) =
nα(n) ∂Rn+ |λen − y|n
We have: y ∈ ∂Rn+ ⇔ y = (y1 , . . . , yn−1, 0) ⇒ λen − y = (−y1 , . . . , −yn−1 , λ).
2
|λen − y| = y12 + y22 + . . . + yn−1
+ λ2
n
21
= |y| + λ2
21
so |λen − y|n = (λ2 + |y|2) 2 . Returning to the function, we have
Z
2λ
g(y)
u(λen ) =
n dS.
2
nα(n) ∂Rn+ (λ + |y|2) 2
61
It follows that
Z
u(λen ) − u(0)
g(y)
2
= u(λen ) =
n dS.
2
λ
nα(n) ∂Rn+ (λ + |y|2) 2
By a definition from Evans
2
=
nα(n)
Z
Rn−1
(λ2
g(y)
n−1
y.
n d
2
+ |y| ) 2
Now we split the integral in two parts, and use what we know about u.
Z
Z
2
|y|
g(y)
n−1
n−1
=
y+
y
n d
n d
2
2
nα(n) |y|≤1 (λ2 + |y|2) 2
|y|>1 (λ + |y| ) 2
for y ∈ Rn−1 . However, now we see that, from the first integral, as λ → 0,
(λ2
1
|y|
.
n →
2
2
|y|n−1
+ |y| )
If we integrate this limit (which we can due to monotone convergence
theorem), using the coarea formula,
Z
Z 1Z
1
1
n−1
d
y
=
dS
dt =
n−1
n−1
y≤1 |y|
0
|y|=r r
Z 1
Z 1
1
1
n−2
Cr dr = C
dr = +∞.
n−1
0 r
0 r
What we have calculated is the derivative of xn , and
uxn = ∞ =⇒ Du unbounded.
5.4 Extensions
Extensions of W 1,p (U) to W 1,p (Rn ).
Assume U is bounded and ∂U ∈ C 1 . Choose V ⊃⊃ U for a bounded V .
Then there exists a bounded linear operator E : W 1,p (U) → W 1,p (Rn ) such
that for each u ∈ W 1,p (U):
(i) Eu = u a.e in U.
(ii) supp(Eu) ⊂ V .
(iii) kEukW 1,p (Rn ) ≤ CkukW 1,p (U ) for C ∈ R and C only depends on
n,p and U.
Definition: Eu is called the extension of u to Rn .
Proof
Fix x0 ∈ ∂U and assume first that ∂U is flat near x0 , lying in the plane
{xn = 0}.
62
For r ≥ 0, let
B + = B(x0 , r) ∩ {xn ≥ 0}
B − = B(x0 , r) ∩ {xn ≤ 0}
where B + ⊂ U , and B − ⊂ Rn \U.
We make the temporary assumption u ∈ C 1 U, and then we define
u(x)
x ∈ B+
u(x) :=
−3u(x1 , . . . , xn−1 , −xn ) + 4u(x1 , . . . , xn−1 , − x2n ) x ∈ B −
so for some x ∈ B − , we get u(x) ∈ B + (a higher order reflection).
(1) Claim: u ∈ C 1 (B) (where B = B + ∪ B − ).
To prove this claim, we write
u+ = uB+ .
u− = uB− ,
It suffices to show that
D α u− {xn =0}
= D α u+ {xn =0} ,
∀|α| ≤ 1.
If we can show this, the claim follows. To show it, we first observe that
−
u+ = u− on {xn = 0}. (−3u(0) + 4u(0) = u(0)), and u+
xi = uxi on {xn = 0}
for i = 1, . . . , n − 1, and for i = n we verify this directly:
xn
u−
)
xi (x) = u B − = 3uxn (x1 , . . . , xn−1 , −xn ) − 2uxn (x1 , . . . , xn−1 , −
2
so, we have
+ u−
=
u
=
u
x
n {xn =0}
xn {xn =0}
xn {xn =0}
63
and therefore the claim follows.
(2) Claim:
kukW 1,p (B) ≤ CkukW 1,p (B+ ) .
By definition,
kukW 1,p (B) =
where,
and
Z
Z
B−
X
kalpha≤1
p
B
kD
|u| dx =
−
p
α
ukpLp (B)
Z
|u (x)| dx =
=
Z
p
B
|u| dx +
p
+
| u (x) | dx +
B + | {z }
u(x)
Z
B−
n Z
X
i=1
B
|uxi (x)|p dx,
|u− (x)|p dx
Z
− 3u(x′ , xn ) + 4u(x′ , − xn p dx.
2
B−
Using the inequality |a + b|p ≤ C(|a|p + |b|p ), we get
Z
Z
xn p
′
p
′
C
|u(x , −xn )| dx +
|u(x , − )| dx
2
B−
B−
and using a change of variable,
≤C
Z
B+
|u(x)|p dx.
So, we have established that
Z
Z
p
|u(x)| dx ≤ C
B+
|u(x)|p dx
B+
|uxi (x)|p dx,
B
and by a similar argument,
Z
Z
p
|uxi (x)| dx ≤ C
B
and when we combine these, we get
kukW 1,p (B) ≤ CkukW 1,p (B+ )
which proves the claim.
(3) Consider the situation where ∂U is not flat near x0 . In this case we can
find a C 1 mappnig Φ with inverse Ψ such that Φ straightens out the boundary
∂U near x0 .
64
yi = xi , i = 1, . . . , n − 1
y = Φ(x)
yn = xn − γ(x1 , . . . , xn−1 )
xi = yi , i = 1, . . . , n − 1
x = Ψ(y).
xn = yn + γ(y1 , . . . , yn−1 )
In the ”new” domain we inscribe a ball around y0 inside the transformed ball
and call each part B + and B − . In addition we write
u′ (y) = u (Ψ(y)) ,
x = Ψ(y), y = Ψ(x).
With the same reasoning as we used with the assumption of a flat boundary,
we can extend u′ to u′ defined on B such that u′ ∈ C 1 (B). Moreover, we can
also use the inequality,
ku′ kW 1,p (B) ≤ Cku′ kW 1,p (B+ ) .
By definition,
u(x) = u′ (Φ(x)) =⇒ uxi = u′yi (Φ(x)) + u′yn (Φ(x))γxi
for i = 1, . . . , n − 1, and for i = n
uxn (x) = u′xn (Φ(x)).
Let W = Ψ(B) ⇔ B = Ψ(W ).
Then, by change of variable and our previous definitions;
kukW 1,p (W ) ≤ ku′ kW 1,p (B)
and by (*)
≤ Cku′ kW 1,p (W + ) ≤ CkukW 1,p (W + ) ≤ CkukW 1,p (U )
65
(*)
so,
kukW 1,p (W ) ≤ CkukW 1,p (U )
(**)
(4) Since ∂U is compact ∃{xi0 }N
i=1 ⊂ ∂U (a set of points on the boundary),
N
and open sets {Wi }i=1 and extensions ui of u to Wi (i = 1, . . . , N) such that
N
∂U ⊂ ∪N
i=1 Wi . We choose W0 ⊂⊂ U such that U ⊂ ∪i=0 Wi .
P
N
N
We let {ξ}i=0 be a partition of unity wrt {Wi }i=0 . We write u = N
i=0 ξi ui ,
N
where u0 = u. Note that supp(U ⊂ ∪i=0 Wi . Then
kukW 1,p (Rn ) ≤ C
N
X
i=0
(∗∗)
kui kW 1,p (Wi ) ≤ C
In short, we havae established that
N
X
i=0
kukW 1,p (U ) = CNkukW 1,p (U ) .
n
kuk1,p
W (R ) ≤ CkukW 1,p (U ) .
(I)
(5) Write Eu = u, with u 7→ Eu being linear (follows from the definition).
The construction so far is for u ∈ C 1 (U ). Now assume u ∈ W 1,p (U). Then
∃um ∈ C ∞ (U ) such that
kum − ukW 1,p (U ) → 0 as n → ∞.
Thus,
(I)
kEum − Eun kW 1,p (Rn ) ≤ Ckum − ul kW 1,p (U ) → 0 as m, l → ∞.
1,p
which means {Eum }∞
(Rn ), hence it converges
m=1 is a Cauchy sequence in W
(since Sobolev spaces are complete) to some u = Eu in W 1,p (Rn ). So, we
know, since um ∈ C ∞ (smooth)
kEum kW 1,p (Rn ) ≤ Ckum kW 1,p (W )
↓
for all u ∈ W 1,p (U). ↓
as m → ∞
kEukW 1,p (Rn ) ≤ CkukW 1,p (U )
5.5 Traces
Theorem - Trace Theorem
Assume U is bounded and ∂U ∈ C 1 . Then there exists a bounded, linear
operator T : W 1,p (U) 7→ Lp (∂U) such that
(i) T u = u
if u ∈ C(U ) ∩ W 1,p (U).
∂W
(ii) kT ukLp (∂U ) ≤ CkukW 1,p (U ) for all u ∈ W 1,p (U).
Definition: T u is the trace of u on ∂U.
66
Proof
(1) Assume u ∈ C 1 (U), and assume x0 ∈ ∂U and ∂U is flat near x0 . Let
b = B(x0 , r ) (the smaller ball contained in B). We also
B = B(x0 , r), B
2
b We define Γ = B
b ∩ ∂U, and
introduce ξ ∈ Cc∞ (B), ξ ≥ 0 and ξ = 1 on B.
′
n−1
x = (x1 , . . . , xn−1 ) ∈ R
= {xn = 0}.
We take the integral of u over Γ, and since ξ = 1 on Γ we can write:
Z
Z
Z
p
′
p
′
|u| dx = ξ|u| dx ≤
ξ|u|pdx′
Γ
Γ
{xn =0}
(since {xn = 0} is a bigger domain than Γ). Using Green’s formula, we get
Z
Z
Z
p
p−1
p
pξ|u| sgn(u)uxn dx
ξxn |u| dx +
=−
(ξ|u| )xn dx = −
B+
B+
B+
≤C
Z
B+
p
|u| + p|u|
p−1
|uxn |dx
and we take a closer look at p|u|p−1|uxn |. Using Young’s inequality: ab ≤
q
ap
+ bq for 1/p + 1/q = 1, we get
p
|u|p−1|uxn | ≤
|uxn |p |u|(p−1)q
|uxn |p p − 1 p
+
=
+
|u|
p
q
p
p
since 1/p + 1/q = 1 ⇒ 1/q = (p − 1)/p ⇒ q = p/(p − 1), so we get
Z
Z
Z
p
p
′
p
|u|p +|Du|pdx ≤ CkukpW 1,p (B+ )
|u| dx ≤ C
|u| + |uxn | dx ≤ C
Γ
B+
B+
where C only depends on ξ and p.
67
—————————————————————-
05/10-2010
Theorem from last time
Assume U is bounded with ∂U ∈ C 1 . Then there exists a bounded, linear
operator
T : W 1,p (U) 7→ K p (∂U)
such that
(i) T u = u∂U if u ∈ C(U) ∩ W 1,p (U)
(ii) kT ukLp (∂U ) ≤ CkukW 1,p (U ) for all u ∈ W 1,p (U) for some constant
C = C(n, p, U).
Proof
Assume first u ∈ C 1 (U), and x0 ∈ ∂U and assume ∂U is flat near x0 . As we
had last time:
Z
Γ
p
′
|u| dx ≤ C
Z
B+
|u|p + |Du|p dx
(I)
The second step is to assume ∂U is not flat near x0 , in which case we can
simply straighten out ∂U near x0 with functions Φ and Ψ:
Applying (I) and change of variables, we get
Z
Z
p
|u| dS ≤
|u|p + |Du|pdx
Γ
U
for Γ ⊂ ∂U containing x0 .
68
(II)
Since the boundary is compact we can find points {x0 }N
i=1 ⊂ ∂U, Γi ⊂ ∂U
where i = 1, . . . , N such that ∂U = ∪N
Γ
and
by
(II),
which
applies to each
i=1 i
i,
Z
Z
p
|u| dS ≤
|u|p + |Du|p dx =⇒
Γi
Then
U
kukLp (Γi ) ≤ CkukW 1,p (U ) .
kukLp (∂U ) ≤
N
X
i=1
i = (1, . . . , N)
kukLp (Γi ) ≤ C · NkukW 1,p (U )
which is the desired estimate. We write T u = u∂U .
kT ukLp (∂U ) ≤ CkukW 1,p (U ) ,
u ∈ C 1 (U).
(III)
Next, assume u ∈ W 1,p (U) and we show that the inequality is true for this
as well. By the approximation theorem ∃um ∈ C ∞ (U) such that um → u in
W 1,p (U) (IV). Then, since the functions are smooth:
(III)
kT um − T ul kLp (∂U ) ≤ Ckim − ul kW 1,p (U ) −→ 0 by (IV)
as m, l → ∞. Thus, {T um } is Cauchy in Lp (∂U), so T um → u∗ := T u in
Lp (∂U). We combine this with (IV).
(III)
kT um kLp (∂U ) ≤ Ckum kW 1,p (U )
↓
(III)
↓ as m → ∞
kT ukLp (∂U ) ≤ CkukW 1,p (U )
Assume in addition that u ∈ C)U, then um → u uniformly on U , which again
means T um → T u uniformly on U .
T um = um ∂U =⇒ T u = u∂U
Recall:
W01,p (U) is the closure of Cc∞ (U) in W 1,p (U), that is
u ∈ W01,p (U) ⇐⇒ ∃um ∈ Cc∞ (U) ⇒ um → u in W 1,p (U).
Theorem
Assume U is bounded and ∂U ∈ C 1 . Suppose u ∈ W 1,p (U). Then
u ∈ W01,p (U) ⇐⇒ T u = 0 on ∂U.
(where we call T u the trace of u).
Proof : Evans page 261.
69
5.6 - Sobolev Inequalities
Assume 1 ≤ p < n. Motivation: we want to establish an estimate of the form
kukLq (Rn ) ≤ CkDukLp (Rn )
(I)
for all u ∈ Cc∞ (Rn ), and some constant C > 0 which is independent of the
function u and 1 ≤ q < ∞.
Question: Assuming that the estimate (I) holds, what is the relation
beween p,q and the dimension n?
We choose u ∈ Cc∞ (U) where u 6= 0. We define uλ (x) = u(λx) for
λ > 0, x ∈ Rn . Then by (I),
kuλ kLq (Rn ) ≤ CkDuλ kLp (Rn )
But,
Z
Z
1
|uλ(x)| dx =
|u(λx)| dx = n
λ
Rn
Rn
q
q
Z
Rn
(II)
|u(y)|q dy.
where we made the substitution y = λx, dy = λn dx, which yields dx =
(1/λn )dy. Also,
Z
Z
Z
Z
λp
p
p
p
p
|Dux (x)| dx =
|Du(λx)| dx = n
|Du(y)|pdy.
|D(u(λx))| dx = λ
λ
n
n
n
n
R
R
R
R
where we used the same substitution. Now, by (II)
n
λ− q kukLq (Rn ) ≤ Cλ
p−n
p
kDukLp (Rn )
which is equivalent to,
n
n
kukLq (Rn ) ≤ Cλ1− p + q kDukLp (Rn )
We note that if 1 − np + nq > 0, then by letting λ → 0 we get 0 on the right
hand side, which means we get a contraction. If 1 − np + nq < 0 we get a
contraction by letting λ → ∞.
Both these scenarios must be untrue since we have assumed u 6= 0. So,
for (I) to hold we must have
1−
n n
1
1 1
np
+ = 0 ⇐⇒
= −
⇐⇒ q =
.
p
q
q
p n
n−p
Definition: A q satisfying this is called the Sobolev conjugate of p and is
denoted p∗ .
70
Theorem. The Gagliardo-Nirenberg-Sobolev Inequality
np
(the Sobolev conjugate), then there exists
Assume 1 ≤ p < n and p∗ = n−p
a constant C > 0 depending only on p and n such that
kukLp∗ (Rn ) ≤ CkDukLp (Rn )
for all u ∈ Cc∞ (Rn ).
Proof
Case p = 1: Since supp(U) is compact,
Z xi
u(x) =
uxi (x1 , . . . , yi , . . . , xn )dyi
(i = 1, . . . , n)
−∞
(u vanishes before infinity because it is compact), so
Z ∞
|u(x)| ≤
uxi (x1 , . . . , yi, . . . , xn )dyi
(i = 1, . . . , n)
−∞
which implies,
n
|u(x)| ≤
n Z
Y
∞
−∞
i=1
|uxi (x1 , . . . , yi, . . . , xn )|
and taking the 1/(n − 1)’th power on both sides:
|u(x)|
n
n−1
≤
n Z
Y
i=1
∞
−∞
|uxi (x1 , . . . , yi , . . . , xn )|
1
n−1
We use the shorthand notation: uxi (x) = uxi (x1 , . . . , yi, . . . , xn ), so we write
n
n−1
|u(x)|
≤
n Z
Y
i=1
∞
−∞
|uxi (x)|
1
n−1
=
Z
∞
−∞
|ux1 (x)| dyi
1
n−1
n Z
Y
i=2
∞
−∞
|uxi (x)|
1
n−1
We take the integral on both sides wrt x1 .
Z
∞
−∞
|u(x)|
n
n−1
dx1 ≤
Z
∞
−∞
|ux1 (x)| dyi
1 Z
n−1
∞
n Z
Y
−∞ i=2
∞
−∞
|uxi (x)|
1
n−1
dx1
(*)
We want to estimate the integral on the far right and do so using the general
Hölder inequality.
Z
f1 . . . fn dx ≤
Z
f1p1
71
p1
1
...
Z
f1pn
p1
n
if
Pn
1
i=1 pi
= 1, or the shorthand version:
Z Y
n
i=1
We set
fi =
Z
n Z
Y
fi dx ≤
i=1
∞
−∞
fipi
|uxi (x)|dyi
1
n−1
p1
i
.
pi = n − 1
,
Pn 1
1
since
i=2 pi = n−1 (n − 1) = 1, which means we can use the Hölder
inequality. We get
1
1
n−1
n−1
Z ∞Y
n Z ∞ Z ∞
n Z ∞
Y
dx1 ≤
|uxi (x)|dyidx1
|uxi (x)|dyi
−∞ i=2
−∞
−∞
i=2
−∞
which implies, when we use the inequality on (*),
Z
∞
−∞
|u(x)|
n
n−1
dx1 ≤
Z
∞
−∞
|ux1 (x)| dyi
1
n−1
n Z
Y
∞
−∞
i=2
We now integrate both sides wrt to x2 , define k :=
ZZ
∞
k
−∞
|u(x)| dx1 dx2 ≤
Z Z
∞
−∞
|uxi (x)dx1 dx2
k Z
Z
∞
−∞
|
This implies,
I=
where
f1n−1
I≤
Z Z
=
Z
∞
n
Y
−∞ i=1, i6=2
"Z
∞
−∞
|ux1 (x)|dyi
∞
−∞
|ux1 (x)|dyi dx2
This implies,
ZZ
∞
−∞
|u(x)|k dx1 dx2 ≤
Z Z
∞
−∞
|ux1 (x)dyi dx2
k
−∞
Z
n Z Z Z
Y
=
fin−1
72
i=3
|
−∞
|uxi (x)|dyidx1
{z
}
k
dx2
fi , 3≤i≤n
1
n−1
−∞
|ux1 (x)|dyi.
∞
−∞
∞
∞
|uxi (x)|dx1 dx2 dyi
∞
−∞
k Y
n ZZ
∞
#n−1
1
n−1
|uxi (x)|dyi dx1
1
n−1
1
n−1
f1
−∞
k Z Z
−∞
|ux1 (x)| dyi
{z
}
i=1, i6=2
i=3
∞
∞
Y Z
fi dx2 ≤
Z
|ux1 (x)|dx1 dx2
k
k !
n ZZZ
Y
i=3
.
∞
−∞
|uxi (x)|dx1 dx2 dyi
!k
Continuing in this way wrt x3 , . . . , xn we will eventually arrive at the
inequality:
Z
...
Z
∞
−∞
|u(x)|
n
n−1
n Z
Y
dx1 . . . dxn ≤
...
Z
∞
−∞
i=1
|uxi (x)|dx1 . . . dyi . . . dxn
1
n−1
But now we simply use |uxi (x)| ≤ |Du(x)|, which is true for i = 1, . . . , n.
Using that the integral from −∞ to ∞ n times is the integral over Rn we get
Z
Rn
|u(x)dx|
=
n
n−1
n Z
Y
Z
and if we take the
n−1
n
∞
|Du(x)|dx1 . . . dyi . . . dxn
−∞
i=1
Rn
|u(x)dx|
Rn
...
Z
1
n−1
Z
|Du(x)|dx
=
Rn
i=1
So we have:
≤
n Z
Y
n
n−1
≤
Z
Rn
1
n−1
n
n−1
|Du(x)|dx
n
n−1
|Du(x)|dx
power on each side, we get
Z
Rn
|u(x)dx|
n
n−1
n−1
n
≤
Z
Rn
|Du(x)|d
and this is equivalent to
n
≤ CkDukL1(Rn ) .
kukL n−1
(Rn )
np
We have verified this for p = 1, and note that p∗ = n−p
so when p = 1 we
n
∗
have p = n−1 which is what we have on the left side.
We now generalise this. For p = 1 we just established that
Z
Rn
|u(x)|
n−1
n
n
n−1
≤C
Z
Rn
|Du(x)|dx.
(**)
The case 1 < p < n. We apply the inequality (**) to v = |u|γ , where
γ > 1 is to be chosen later.
Z
Rn
|u(x)|
γn
n−1
n−1
n
≤C
≤C
Z
Rn
Z
|u|
Rn
γ
|D(|u| )|dx = C
(γ−1)p
p−1
p−1
Z
p
Rn
73
Z
Rn
|Du|p
γ|u|γ−1 |Du|dx
p1
We have
p−1
p
+
Z
1
p
Rn
= 1, so Hölder’s inequality is okay. The expression implies,
γn
n−1
|u(x)|
n−1
n
≤C
Z
Rn
|u|
(γ−1)p
p−1
p−1
p
kDukLp (Rn )
where we have what we need on the right hand side. Now we choose γ such
that
(γ − 1)p
n
p
−p
γn
=
⇐⇒ γ(
−
)=
n−1
p−1
n−1 p−1
p−1
−p
np − n − np + p
p−n
n−1
⇐⇒ γ
=
⇐⇒ γ
= −p ⇐⇒ γ = p
>1
(n − 1)(p − 1)
p−1
n−1
n−p
So,
So, setting
γn
n−1
γn
p
(n−1)
n
np
=
·
=
= p∗ .
n−1
n−p (n − 1)
n−p
= p∗ in the previous expression, means
Z
Rn
But,
n−1
− p−1
n
p
|u| dx
≤ CkDukLp (Rn )
p∗
np − p − np + n
n−p
1
n−1 p−1
−
=
=
= ∗
n
p
np
np
p
which implies, and finally concludes the proof,
kukLp∗ (Rn ) ≤ CkDukLp (Rn )
74
∀u ∈ Cc 1 (Rn )
—————————————————————-
12/10-2010
Theorem: Morrey’s Inequality
For n < p ≤ ∞ we have
kukC 0,γ (Rn ) ≤ CkukW 1,p (U )
∀u ∈ Cc1 (R)
where γ = 1 − np .
Proof
Z
—
B(x,r)
|u(y) − u(x)|dy ≤ C
Z
B(x,r)
|Du(y)|
|x − y|n−1
kukL∞ (Rn ) ≤ CkukW 1,p (Rn )
Now let W = B(x, r) ∩ B(y, r) for any x, y ∈ Rn and |x − y| = r.
75
(1)
(2)