Distributions - Part I
Wolfgang Stefani
LMU München
Hütte am 12.-15.12.2013
Motivation
We would like
δ(
x) : δ(
x) =
0 forx ,0∧
RRd
δ(
x)
dx=
1.This allowsR
Rd
δ(
x)ϕ(
x)
dx= ϕ(
0)
, forϕ ∈
C0∞(
Rd)
. Consider now:Z 1
−1
|
x|ϕ
0(
x)
dx= −
Z 1−1
sign
(
x)ϕ(
x)
dx,
Z 1
−1
sign
(
x)ϕ
0(
x)
dx=
Z 0−1
−ϕ
0(
x)
dx+
Z 10
ϕ
0(
x)
dx= −
2ϕ(
0).
Test functions - D(Ω)
Let
Ω ⊂
Rd andC0∞(Ω) = {ϕ ∈
C∞(Ω)|
supp(ϕ) ⊂⊂ Ω}
. Whereby supp(ϕ) = {
x∈ Ω|ϕ(
x)
,0}
.With the usual pointwise addition and skalar multiplication this is a vector space (overC).
We further define the following convergence:
lim
j→∞
ϕ
j= ϕ
inC0∞(Ω),
if•
∃
K⊂⊂ Ω ∀
j:
supp(ϕ
j) ⊂
K,•
∀α ∈
Nd:
Dαϕ
j→
Dαϕ
uniformly, i. e.limj→∞supx∈K
|
Dαϕ
j(
x) −
Dαϕ(
x)| =
0.Distributions - D
0(Ω)
The mapT
: D(Ω) →
Cis a distribution, if and only if it is a sequentially continuous linear form, i.e.(ϕ
j→ ϕ ⇒
T(ϕ
j) →
T(ϕ)) ∧
T(ϕ + λψ) =
T(ϕ) + λ
T(ψ).
Equivalently:
∀
K⊂⊂ Ω ∃
c,
Nconstant, such that:∀ϕ ∈ D(
K) : |
T(ϕ)| ≤
ckϕk
N,∞;K=
cP|α|≤Nsupx∈K
|∂
αϕ(
x)|
.If in the above a certainNsuffices for all compactK, then the smallest of those is called the order ofT.
Regular Distributions
Let
∅
,Ω ⊂
Rdopen,(
Rd, λ
d)
the Lebesgue measure space and f∈
Lloc1(Ω)
, then following is a Distribution of order 0:Tf
:
C0∞(Ω) →
C, ϕ 7→
Z
Ω
f
· ϕ
dλ
d.
= (
f, ϕ)
If we additionally let
α ∈
Nd, then(
DαTf)(ϕ) := (−
1)
|α|Tf(
Dαϕ) = (−
1)
|α|Z
Ω
f
·
Dαϕ
dλ
d is a Distribution, since forϕ
j→ ϕ
:|(
DαTf)(ϕ
j− ϕ)| = |
ZK
f
·
Dα(ϕ
j− ϕ)
dλ
d| ≤ k
Dα(ϕ
j− ϕ)k
∞;Kk
fk
1;K→
0.
Properties D
0(Ω)
By definition,
D
0permits aCvectorspace structure. Denoting T(ϕ) = (
T, ϕ)
, we get a bilinear form onD
0.Let nowa
∈
C∞(Ω)
andα, β ∈
Nd. With the following definitions we get:(
aT, ϕ) := (
T,
aϕ) ⇒
aT∈ D
0,(
DαT, ϕ) := (−
1)
|α|(
T,
Dαϕ) ⇒
DαT∈ D
0. And further:Di
(
aT) =
Di(
a)
T+
aDi(
T)
, Dα+βT=
Dα(
DβT) =
Dβ(
DαT)
.Localization
ForT
∈ D
0(Ω)
we define its restriction onΩ
0⊂ Ω
, by TΩ
0
(ϕ) =
T(ϕ) ∀ϕ ∈ D(Ω
0).
Using this we define the support ofT
∈ D
0(Ω)
: supp(
T) = {
x∈ Ω| ∀δ >
0 :TΩ∩Bδ(x),0
}.
For regular distributions, we have: supp
(
Tf) =
supp(
f)
. We also have the general implication forT∈ D
0 andϕ ∈ D
:supp
(
T) ∩
supp(ϕ) = ∅ ⇒ (
T, ϕ) ≡
0.
Convolution with functions
ForT
∈ D
0(Ω)
andψ ∈ D(Ω)
we define their convolution atx∈ Ω
, by(
T∗ ψ)(
x) :=
T(ψ(
x− ·)) = (
T, ψ(
x− ·)).
For a regular distributionTf we have:
(
Tf∗ ψ)(
x) =
RΩf
(
y)ψ(
x−
y)
dy.This convolution has the following properties:
• T
∗ ϕ ∈
C∞(
Rd)
,• supp
(
T∗ ϕ) ⊂
supp(
T) +
supp(ϕ)
,• Dα
(
T∗ ϕ) = (
DαT) ∗ ϕ =
T∗ (
Dαϕ)
.For
η ∈ D
as well we get further:(
T∗ η) ∗ ϕ =
T∗ (η ∗ ϕ)
.Convergence in D
0We define the following convergence on
D
0: limk→∞Tk
=
T:⇔ ∀ϕ ∈ D :
limk→∞
(
Tk, ϕ) = (
T, ϕ).
This convergence makes
D
0a complete space.With the mollifierJε
(
x) = ε
−dJ(ε
−1x) = ε
−dcexp(−
1−|ε1−1x|2)
1{|ε−1x|<1}we get the convergence:
T
∗
Jε→
T inD
0.
So the spaceC∞is dense in
D
0and analogously for the space with compact support.Delta Distribution
Leta
∈ Ω
and define,δ
0∈ D
0: δ
0(ϕ) = ϕ(
0) ∀ϕ ∈ D.
ZΩ
f
(
x)ϕ(
x)
dxApproximation via Dirac sequences
(
tk)
k⊂
L1(
Rd)
: 1.∀
x∈
Rd∀
k∈
N:
tk(
x) ≥
0,2.
∀
k∈
N:
RRdtk
(
x)
dx=
1, 3.∀ε >
0:
limk→∞RRd\Bε(0)tk
(
x)
dx=
0.For example:
tk
(
x) = √
1 2π
k−1exp
(−
x2
2k−1
)
or tk(
x) =
Jε=1 k(
x)
Radon Measures
Let
(
X, B, µ)
withX Hausdorff, such that•
µ
is locally finite, i.e.∀
x∈
X∃
Uopen: µ(
U) < ∞
,•
µ
is inner regular, i.e.∀
A∈ B : µ(
A) =
sup{µ(
K)|
K⊂
A:
K compact}
. LetM(Ω)
be the set of Radon Measures onΩ
andµ ∈
M(Ω)
, then(µ, ϕ) :=
Z
Ω
ϕ(
x)
dµ(
x)
, forϕ ∈ D,
defines a Distribution.In particular, the Dirac measure δ(A) =
1, 0∈A,
0, 0<A (A ⊂Rd) gives the
Delta Distribution:
(δ
0, ϕ) =
ZΩ
ϕ(
x)
dδ
0(
x) = ϕ(
0) = δ
0(ϕ)
forϕ ∈ D.
Distributions - Part II
Wolfgang Stefani
LMU München
Hütte am 12.-15.12.2013
Distributional Solution
LetT
∈ D
0,f∈ C(
Rd,
R)
andP(
D) =
X|α|≤k
aα
(
x) ∂
α∂
xα. T is a distributional solution toP(
D)
u=
f, if and only ifP
(
D)
T=
Tf⇔ (
P(
D)
T−
Tf, ϕ) =
0∀ϕ ∈ D.
For the Laplacian Operator
∆
onT we have:∆
T=
d
X
i=1
DiDiT
⇒ (∆
T, ϕ) = (
T, ∆ϕ),
forϕ ∈ D.
Ifg
∈
C2(
Rd,
R)
andT=
Tg we have the classical Green fromula Z(ϕ∆
g−
g∆ϕ)
dx=
0.
Fundamental Solution
ForP
(
D) =
X|α|≤k
aα
∂
α∂
xα, the solutionγ
ofP
(
D)
u(
x) =
0,
i.e. P(
D)γ = δ
0 is called a fundamental solution.The inhomogeneous solution to
P
(
D)
u(
x) =
f(
x)
is u(
x) = (γ ∗
f)(
x) =
Zγ(
x−
y)
f(
y)
dy.
Reminder
We have seen:
Z 1
−1
|
x|ϕ
0(
x)
dx= −
Z 1−1
sign
(
x)ϕ(
x)
dxZ 1
−1
sign
(
x)ϕ
0(
x)
dx= −
2ϕ(
0) = −
Z 1−1
2
δ
0(
x)ϕ(
x)
dx.
So:(
sign0, ϕ) = −(
sign, ϕ
0) = −(−
2δ
0, ϕ) ⇒
sign0=
2δ
0=
Z2
δ
0·
dxExample - I
For the Heaviside functionH
(
x) =
1
,
x≥
00
,
x<
0 we have:(
H0, ϕ) = −(
H, ϕ
0) = −
Z ∞0
1
ϕ
0(
x)
dx= ϕ(
0) = (δ
0, ϕ) ⇒ ∂
∂
xH(
x) = δ
0(
x)
So for the Laplacian∆
inR:∆γ = ∂
2∂ x
2γ =
∂
∂ x
∂
∂ x γ
= δ
0⇒ ∂
∂ x γ =
H(
x) +
c⇒ γ =
xH(
x) +
cx+
c0c=12,c0=0
⇒ γ(
x) =
1 2|
x|
Example - II
From classical theory inRd:
γ(
x) =
2π1 ln
(|
x|)
(d−2)ω1 d
|
x|
2−d and Z|x|≤c
|γ(
x)| =
Z c
0
|
ln(
r)|
r dr,
d=
2,
1 (d−2)
Z c
0
r2−drd−1
,
d≥
2 Then withR sufficiently large,(∆γ, ϕ) = (γ, ∆ϕ) =
Zγ(
x)∆ϕ(
x)
dx=
limε&0
Z
ε<|x|<R
γ(
x)∆ϕ(
x)
dx=
limx&0Jε
.
Using Green’s second identity:
Jε
=
Zε<|x|<R
ϕ∆γ
dx+
Z|x|=ε
γ ∂ϕ
∂ν − ϕ ∂γ
∂ν
dσ.
Example - II cont.
Since
∆γ =
0 for|
x| > ε
and for|
x| = ε
we have∂
∂ν = ∂
∂
r: Jε=
Z
|x|=ε
γ ∂ϕ
∂ν − ϕ ∂γ
∂ν
dσ =
Z
|x|=ε
γ(
x) ∂ϕ
∂
r− ϕ(
x)
1(2−d)
ω
d(
2−
d)|
x|
1−d dσ
=
1ω
dZ
|x|=ε
ε
2−d d−
2∂ϕ(
x)
∂
r− ϕ(
x)ε
1−d dσ
Using the mean value theorem with|
x0| = |
x00| = ε
:Jε
=
1ω
dε
2−d d−
2∂(
x0)
∂
r− ε
1−dϕ(
x00)
ω
dε
d−1= ε
d−
2∂ϕ(
x0)
∂
r+ ϕ(
x00).
As the derivative is finite for
ε &
0 we get:(∆γ, ϕ) =
limε&0Jε
= ϕ(
0) = (δ
0, ϕ).
The Schwartz Space S
The vector space of all rapidly decreasing functions
S(
Rd) := {φ ∈
C∞(
Rd) | ∀α, β ∈
Nd0:
supx∈Rd
|
xαDβφ(x)| < ∞}
is called Schwartz space. We have
D
(S
. he.g. exp
(−|
x|
2)
i Topology and convergence are induced by the seminormskφk
N=
supx∈Rd
|α|,|β|<Nmax
|
xαDβϕ(
x)| ⇔
pk,l(
φ) =
supx∈Rd
(|
x|
k+
1)
X|β|≤l
|
Dβφ(
x)|
and φk
→
φinS :⇔ ∀
N∈
N0: kφ
k−
φkN→
0.
Fourier Transform F
For someφ
∈ S
its Fourier transform is defined asF
φ(ξ) = (
2π)
−d2Z
Rd
e−ix·ξφ
(
x)
dx, ξ ∈
Rd.
which maps as
F : S → S
and is bijective and bicontinuous with inverse(F
−1φ)(x) = (
2π)
−d2ZRd
eix·ξφ(ξ)d
ξ, ϕ ∈ S.
Forφ
∈ S
we also getxαφ,
Dαφ, F
φ,
DαF
φ, F (
Dαφ) ∈ S
and DαF
φ= (−
i)
|α|F (
xαφ),ξ
αF
φ= (−
i)
|α|F (
Dαφ)So, in particular
F (
Dαφ) = (−
i)
−|α|ξ
αF
φTempered Distributions S
0The space
S
0contains all linear formsS →
C, which are sequentially continous, i. e.φk
→
φinS ⇒
T(φ
k) →
T(φ).
Since φk
→
φinD ⇒
φk→
φinS ⇒
T(φ
k) →
T(φ),
we haveT∈ S
0⇒
T∈ D
0. And byF
T(ϕ) :=
T(F ϕ)
we define the Fourier transform ofT∈ S
0.Example - III
The regular distributionTe is not tempered. Let
ϕ ∈ D(
R)
, thenψ
k(
x) :=
e−kϕ(
x−
k) →
0 inS(
R),
withψ
k∈ D(
R)
, butTe
(ψ
k) =
ZR
ex
ψ
k(
x)
dx=
ZR
exe−k
ϕ(
x−
k)
dx=
ZR
ey
ϕ(
y)
dy ,0=
Te(
0).
Example - IV.1
Consider the heat equation for distributions:
∂u∂tu
(
t,
x) = ∆
xu(
t,
x)
u(
t,
x) =
f(
t,
x)
i.e.(
∂t∂T, ϕ) = (∆
xT, ϕ) (
0, ∞) ×
Rd(
T, ϕ) = (
Tf, ϕ) {
t=
0} ×
Rd Determine its Fourier transform forx:F
x(∆
xT, ϕ) = F
x(
T, ∆
xϕ) = (
T, F
x(∆
xϕ)) = (
T, (−
i)
−2ξ
(0,2)F
xϕ)
Thus we get:
F
x( ∂
∂
tT, ϕ) = F
x(−
T, ∂
∂
tϕ) = (−
T, ∂
∂
tF
xϕ) =
( ∂
∂
tT, F
xϕ) = ∂
∂
t(
T, F
xϕ) = = (−|ξ|
! 2T, F
xϕ)
Example - IV.2
We have now the ordinary differential equation in
F
x(
T)
:∂t∂
F
x(
T, ϕ) = −|ξ|
2F
x(
T, ϕ) F
x(
T, ϕ) = F
x(
Tf, ϕ)
i.e.∂t∂
F
x(
T) = −|ξ|
2F
x(
T) (
0, ∞) ×
RdF
x(
T) = F (
Tf) {
t=
0} ×
Rd Thus we get:F
x(
T) =
e−t|ξ|2F (
Tf) ⇒ (
T, F
xϕ) =
e−t|ξ|2(
Tf, F ϕ)
⇒ F
ξ−1(
T, F ϕ) = F
ξ−1(
Tf,
e−t|ξ|2F ϕ) ⇒ (
T, ϕ) =
Tf
, F
ξ−1(
e−t|ξ|2F ϕ)
Example - IV.3
The convolution theorem forT1
∈ S
0 andT2∈ E
0:F (
T1∗
T2) = (
2π)
d2F (
T1)F (
T2)
With this we get:F
ξ−1(
e−t|ξ|2F
xϕ) = (
2π)
−d2F
ξ−1(
e−t|ξ|2) ∗ ϕ
FurtherF
ξ−1(
e−t|ξ|2) = (
2π)
−d2 Zeix·ξe−t|ξ|2 d
ξ =
ZRd
eix·ξ−t|ξ|2d
ξ
= (
2t)
−d2e−|x|2
4t
=
g(
t,
x)
Example - IV.4
So finally:
(
Tf,
g∗ ϕ) =
Zf
(
t,
x)(
2π)
−d2 Zg
(
t−
t0,
x−
x0)ϕ(
t0,
x0)
d(
t0,
x0)
d(
t,
x) =
Zϕ(
t0,
x0)(
2π)
−d2 Zf
(
t,
x)
g(
t−
t0,
x−
x0)
d(
t,
x)
d(
t0,
x0) =
Z(
2π)
−d2 Zf
(
t,
x)
g(
t−
t0,
x−
x0)
d(
t,
x)
ϕ(
t0,
x0)
d(
t0,
x0) =
Z
(
4π(
t−
t0))
−d2f(
t,
x)
e−|x−x0|2 4(t−t0) d
(
t,
x)
ϕ(
t0,
x0)
d(
t0,
x0) =
Z
(
4π(−
t0))
−d2 Zf
(
x)
e−|x−x0|2 4(−t0) d
(
x)
ϕ(
t0,
x0)
d(
t0,
x0).
Example - IV.5
The solution is the regular distributionTh with:
h
(
t,
x) = (
4π
t)
−d2 Zf
(
x)
e−|x−x0|2 4t d