Partial Differential Equations (Week 5)
Distribution Theory I and Laplace’s Equation
Gustav Holzegel
March 2, 2014
1
Distributions I – the basics
1.1
Introduction
We have touched the idea of distributional solutions when we discussed Burger’s
equation, and also when we discussed Holmgren’s theorem. The key in proving
the latter was the observation that for a linear partial differential operator P =
P
α
n
|α|≤m aα (x) D and a classical solution of P u = w in Ω (open subset of R )
with Dβ u = 0 on ∂Ω for |β| ≤ m − 1, we have the integration by parts formula
Z X
Z
|α|
(−1) Dα (aα (x) v) · u dx
(1)
w · v dx =
Ω
Ω |α|≤m
valid for all v ∈ C0∞ (Ω). The formula (1) makes sense for u merely continuous
or even u ∈ L1loc (Ω) and it is natural to declare u ∈ L1loc (Ω) a “weak” or a
“distributional” solution of P u = w if it satisfies (1) for all v ∈ C0∞ (Ω).
Generally, with any (real- or) complex valued function f : Ω → C, continuous
on Ω ⊂ Rn open (or even u ∈ L1loc (Ω)), we can associate its integrals against
test functions by defining
Z
f (x) φ (x) dx
(2)
f [φ] :=
Ω
for φ ∈ D :=
C0∞
(Ω). Note that
• f [φ] is a linear functional on the space of test functions D
• f [φ] is well defined for f ∈ L1loc (Ω)
• the definition supports the idea of “smeared averages” from physics: If f
is an observable like a velocity or a temperature, you will never be able
to determine its value at a point but only averaged over a small interval
(finite detector size).
• If f is continuous, then f [φ] determines f uniquely (so in some sense we
don’t lose anything by considering f [φ] instead of f )
1
• One can differentiate f in the sense of distributions by defining
Dk f [φ] := −f [Dk φ]
which agrees with the usual derivative if f ∈ C 1 by the standard integration by parts formula. Therefore, linear partial differential operators act
naturally on the functionals f [φ].
1.2
The space D ′ (Ω)
We broaden our view further and consider general linear functionals on the
space D of test-functions (of which those arising by integration against an L1
function, the f [φ] above, are a particular example). We shall introduce a notion
of continuity of such functionals below (which is desirable if we would like to
keep the interpretation as physical observables). This notion of continuity ist
most easily formulated via sequential continuity.
Definition 1.1. We say that φn ∈ D converges to φ in D if
• there is a compact set K such that all φn vanish outside K
• there is a φ ∈ D such that for all α ∈ Nd we have ∂ α φn → ∂ α φ uniformly
in x.
Definition 1.2. A distribution is a linear functional ℓ : D (Ω) → C, which is
continuous in the sense that if φn converges to φ in D, the ℓ (φn ) → ℓ (φ). The
vectorspace of distributions in denoted D′ (Ω).
Example 1.3. Each continuous (or L1loc ) function generates a distribution via
(2). Such distributions are called regular distributions. Note every distribution
is regular, as the next example shows.
not regular. Indeed, the formula
RExample 1.4. The distribution δξ [φ] = φ (ξ) is
dx g (x) φ (x) = φ (ξ) would imply that g ∈ L1loc vanishes everywhere (modulo a
set of measure 0). This example also makes it intuitive to talk about the support
of a distribution: If f [φ] = g [φ] for all φ with support in ω ⊂ Ω, we’ll say that
the two distributions agree in ω.
The above notion of continuity may be cumbersome to check in practical
applications. However, we have the following
Proposition 1.5. The function ℓ : D (Ω) → C belong to D′ (Ω) if and only if
for every compact subset K ⊂ Ω there is an integer n (K, ℓ) and a c ∈ R such
that for all φ ∈ D (Ω) with support in K we have
X
max |∂ α φ|
(3)
with kφkC n =
| ℓ [φ] | ≤ ckφkC n
|α|≤n
2
x
Proof. The “if” follows immediately from the estimate (3). For “only if” suppose
that (3) was violated for some compact set K. Then we can find for this K a
sequence φn with kφn kC n = 1 and |ℓ [φn ] | ≥ n (otherwise the estimate (3) would
hold with c = N ). But then ψn = n−1/2 φn is a sequence converging to zero in
D, while |ℓ [ψn ] | ≥ n1/2 does not go to zero. Contradiction.
If there is a c such that (3) holds, ℓ is said to be of order n on K. IF ℓ is of
order n on every compact subset K ⊂ Ω, the ℓ is of order n on Ω.
Example 1.6. Any regular distribution (Example 1.3) is of order 0. The Dirac
delta of Example 1.4 is also of order zero.
Example 1.7. The principal value distribution
Z
Z
φ (x)
φ (x)
ℓ [φ] := lim
= P.V.
ǫ→0 |x|>ǫ
x
x
is a distribution of order 1 (near 0 at least; away from zero it is order 0). The
proof is an exercise. Hint: Taylor-expand φ near 0 and use the symmetry of the
integral.
1.3
Distributional Derivatives
We can define the distributional derivative Dk f as the distribution
Dk f [φ] = −f [Dk φ]
or more generally Dα f [φ] = (−1)|α| f [Dα φ] .
(4)
You should check that this indeed defines a distribution.
Exercise 1.8. Compute Dk δξ [φ].
Therefore, we can apply a linear operator P of order m to a distribution
u [φ] via
P u [φ] = u P t φ .
Below we will be particularly interested in distributional solutions of
P u = δξ .
(5)
A distribution u satisfying (5) is called a fundamental solution with pole ξ for
the operator P .
1.4
Relation with weak derivatives
If f is a regular distribution, i.e.
f [φ] =
Z
φ (x) f (x) dx
Ω
3
for some f ∈ L1loc (Ω) then it may be that the distributional derivative is again
a regular distribution. In other words, there could be a g ∈ L1loc such that
Z
Z
Dk f [φ] := − Dk φ (x) f (x) dx = φ (x) g (x) dx
holds for any φ in D. In this case, we say that f has g = Dk f as its weak
derivative. Using an argument similar to one already used above, it is easy to
show that the weak derivative, if it exists, is unique.
To see that not every function (≡ regular distribution) has a weak derivative
consider the example of the step function H : R → R defined as
1 for x ≥ 0
H (x) =
(6)
0 for x < 0 .
This is clearly in L1loc but the distributional derivative is easily seen to be the
delta distribution in view of the following computation:
Z ∞
Z ∞
Dx H [φ] = −
Dx φ (x) H (x) dx =
−Dx φ (x) H (x) = φ (0) .
−∞
0
Using the notion of a weak derivatives one can define various notions of “weak
solutions” to a PDE, which will typically require some number of weak derivatives to exist.
1.5
Convergence of Distributions
Definition 1.9. A sequence of distributions ℓn ∈ D′ (Ω) converges to ℓ ∈ D′ (Ω)
if and only if for every test function φ ∈ D (Ω) we have
ℓn [φ] → ℓ [φ]
with the usual notion of convergence in C. We will write ℓn ⇀ ℓ to denote
this convergence and say that ℓn convergences “weakly” or “in the sense of
distributions” to ℓ.
Exercise 1.10. Show that the sequence of (regular) distributions n2 einx converges weakly to zero as n → ∞.
R
Exercise 1.11. Let j ∈ D Rd with Rd j (x) dd x = 1. Define jǫ (x) = ǫ−d j xǫ .
Show that jǫ ⇀ δ0 .
The previous example is remarkable as it show that the non-regular delta
distribution can be approximated by function in D. In fact, any element in
D′ can be approximated in this way: The space D is dense in D′ . This can
be used to extend (uniquely) the usual operations of calculus (differentiation,
translation, convolution) to D′ , which is very useful for PDE.
When we return to distributions next time, we will define the Schwartz
space of test functions D (Ω) ⊂ S (Ω), which are functions decaying faster than
any polynomial near infinity. This is the natural space to discuss the Fourier
transform as it can be shown to be a bijection S → S....
4
2
Laplace’s equation
We start now with the analysis of one of the most important PDEs of mathematics and physics, the Laplace equation
∆u :=
d
X
∂i2 u = 0 ,
(7)
i=1
or, more generally, the Poisson equation, ∆u = f , which has a prescribed inhomogeneity f on the right hand side.1 Equation (7) appears naturally in
electrostatics, complex analysis and many other areas. We have already seen
that this PDE is elliptic. Our approach will be to understand the behavior of
solutions to (7) in great detail first (which is doable in view of the highly symmetric form of the operator) before turning to more general elliptic operators
and equations. In summary, our goals are
• “Solve” ∆u = f . What formulation is well-posed and which one’s are
ill-posed? What are the conditions on f ?
• What is the regularity of u? How does it depend on f ?
• What happens for more general elliptic operators Lu = f ?
I will follow very closely Fritz John’s book, Chapter 4, for the first part.
2.1
Uniqueness for Dirichlet and Neumann problem
Let Ω ⊂ Rd be an open bounded connected region of Rd whose boundary is
sufficiently regular for Stokes theorem to apply. Recall the Green’s identities
(which are special
cases of our formula for the transpose of an operator), valid
for u, v ∈ C 2 Ω̄ :
Z
Z X
Z
du
v dS ,
vxi uxi dx +
v∆u = −
(8)
dn
∂Ω
Ω i
Ω
Z
Ω
v∆u =
Z
u∆v +
Ω
Z
∂Ω
du
dv
v
dS ,
−u
dn
dn
(9)
P
d
and recall that dn
f = i ξ i ∂i f means differentiating in the direction of the
outward unit normal ξ to the boundary ∂Ω. Applying (9) with v = 1 we find
Z
Z
du
dS ,
(10)
∆udx =
dn
Ω
∂Ω
1 In applications, f could be a charge-distribution whose electrostatic potential u is to be
determined.
5
while applying (8) with v = u produces
Z
Z
Z X
2
u∆u =
(∂i u) +
Ω
∂Ω
Ω
i
u
du
dS .
dn
(11)
From (11) we immediately obtain a uniqueness statement for Poisson’s equation.
Consider the following problems
∆u = f in Ω
(12)
u = g on ∂Ω
∆u = f in Ω
du
dn = g on ∂Ω
(13)
for prescribed (say smooth for the moment) functions f and g. Problem (12) is
called the Dirichlet problem, problem (13) the Neumann problem for Poisson’s
equation.
Suppose we have two solutions
∆u1 = f and ∆u2 = f of the Dirichlet
problem with u1 , u2 ∈ C 2 Ω̄ , then their difference satisfies Laplace’s equation,
∆ (u1 − u2 ) = 0 with u1 − u2 = 0 on the boundary. By (11)
X
2
(∂i [u1 − u2 ]) = 0 ,
i
and hence u1 −u2 = c must be constant in Ω. But since
u1 −u2 = 0 on ∂Ω, we can
conclude u1 = u2 and hence the uniqueness of C 2 Ω̄ solutions to the Dirichlet
problem (12). For the Neumann problem (13) there remains ambiguity up to
a constant. Note, that there is an obvious constraint for existence of solutions
to the Neumann problem given by (10) which produces a non-trivial relation
between f and g.
2.2
The fundamental solution
The Laplacian is spherically symmetric. By this we mean that if u (x) is a
solution of ∆u = 0 then v (x) = u (Rx), with R a rotation in Rd , satisfies
∆v = 0 (Exercise). There is also invariance under translations and dilations. In
two-dimension, the Laplacian is also invariant under spherical inversion:
Exercise 2.1. Prove that ∆u = 0 for u = u (x1 , . . . , xd ) implies that
x
2−d
∆ |x|
u
=0
|x|2
in the domain of u. Conclude that in two dimensions the Laplacian is invariant
under conformal transformations.
6
In view of the spherical symmetry of the Laplace operator, one may try to
find spherically symmetric solutions of ∆u = 0. Setting u = ψ (r) and expressing
the Laplacian in spherical coordinates one obtains the ODE
ψ ′′ (r) +
d−1 ′
ψ (r) = 0
r
which is readily solved as ψ ′ = Cr1−d for some constant C and
C 2−d
for d > 2
2−d r
ψ (r) =
C log r
for d = 2
(14)
We can still add a non-trivial constant to ψ (which we suppress). The function
(14) solves ∆ψ = 0 for r 6= 0 by construction but
has a singularity for r = 0.
Note that ψ (r) and also ψ ′ (r) are still in L1loc Rd but that ψ ′′ (r) fails (barely).
Note also that in view of the translation invariance we could have introduced
polar coordinates around any point. We will now prove
Lemma 2.2. Let Ω ⊂ Rd open, bounded and connected and ξ ∈ Ω. Then for
any u ∈ C 2 Ω̄ and ξ ∈ Ω we have the identity
Z
Z dK (x, ξ)
du
dSx (15)
(x) − u (x)
u (ξ) =
K (x, ξ) ∆udx −
K (x, ξ)
dnx
dnx
Ω
∂Ω
with K (x, ξ) = ψ (|x − ξ|) and the constant C in ψ chosen such that C −1 = ωd
equals the surface area of the unit-sphere in d − 1 dimensions.
Corollary 2.3. The function K (x, ξ) is a fundamental solution with pole ξ for
the Laplace operator in Ω in that
∆v = δξ
holds in Ω the sense of distributions.
Proof. Indeed, by the identity of the Lemma,
Z
K (x, ξ) ∆udx = u (ξ)
Ω
for all test functions u ∈ C0∞ (Ω) ⊂ C 2 Ω̄ .
Proof of Lemma 2.2. Consider the region Ωρ = Ω \ B (ξ, ρ) obtained by cutting
out a small ball from Ω around ξ.
∂Ω
Ω
ξ
S (ξ, ρ)
7
Then, for u ∈ C 2 Ω̄ we have from (9) applied in Ωρ (where v is harmonic) the
identity
Z Z
Z
du
dv
dv
du
v∆u =
v
−u
−u
dS +
v
dS
(16)
dn
dn
dn
dn
∂Ω
Ωρ
S(ξ,ρ)
d
where dn
denotes the inward normal on S (ξ, ρ) and the outward normal on ∂Ω.
We claim that the boundary term on S (ξ, ρ) converges to u (ξ) in the limit as
ρ → 0. To see this, note that since v is radial around ξ we have for d > 2
Z
Z
Z
du du v
=
=
(ρ)
(ρ)
∆u
v
v
dn
dn
S(ξ,ρ)
S(ξ,ρ)
B(ξ,ρ)
C 2−d d−1
ρ ρ ωd | max ∆u (ξ) |
(17)
≤
2−d
B(ξ,ρ)
where we have used (8), the explicit form of v and recall the notation ωd for the
surface area of the d − 1-dimensional unit-sphere. (For d = 2 we would obtain
log ρ · ρ instead of ρ.) Clearly, the right hand side goes to zero as ρ → 0. On
the other hand, by a similar computation, we have
Z
Z
dv
= Cρ1−d
u
u = u ξ̃ Cωd
(18)
S(ξ,ρ) dn
S(ξ,ρ)
for some ξ̃ ∈ S (ξ, ρ) by the mean value theorem. By the continuity of u, the
expression converges to Cωd u (ξ) in the limit as ρ → 0. Choosing C = ωd−1
produces the identity claimed in the Lemma.
The formula (15) is extremely useful, as it expresses the value of u at any
point in Ω by the value of ∆u in Ω (which is prescribed in Poission’s equation)
and the values of u and its normal derivative on the boundary. However, we
already know that the solution of Poisson’s equation is already unique by prescribing either u on the boundary or its normal derivative. In other
words,
once u is prescribed on the boundary, there can be at most one C 2 Ω̄ solution,
whose normal derivative would then also be uniquely determined and can no
longer be freely specified. This tells us that the Cauchy-problem for Laplace’s
equation ∆u = 0 (i.e. specifying the function and its normal derivative on ∂Ω)
is not solvable in general. Of course we know it is solvable for analytic data in
a small neighborhood of a point on the boundary by the Cauchy Kovalevskaya
theorem. However, this does not mean that one can solve the problem in the
analytic class in a full neighborhood of all of ∂Ω (and indeed one cannot in
general!).
2.3
Regularity of harmonic functions
The formula (15) provides important insights into the regularity of harmonic
functions. The latter satisfy (recall this formula was derived for u ∈ C 2 Ω̄ )
Z dK (x, ξ)
du
dSx .
(19)
(x) − u (x)
K (x, ξ)
u (ξ) = −
dnx
dnx
∂Ω
8
We can easily show that u is C ∞ at any
ξ ∈Ω. Namely, given ξ we find a small
ball B (ξ, ρ) around ξ such that dist ξ̃, ∂Ω ≥ δ > 0 for all ξ̃ ∈ B (ξ, ρ). Then
we apply formula (19) in that ball B (ξ, ρ). Differentiating u (ξ) we can interchange the derivative with the integral as the integrand and all its derivatives
are uniformly bounded on ∂Ω. It follows that u is C 1 in the ball B (ξ, ρ) and
hence u ∈ C 1 (Ω). By the same argument one sees u ∈ C n (Ω) for any n. For
this argument, we have
only used
the formula (19) in small balls inside Ω, on
which u is always C 2 B (ξ, ρ) provided u is C 2 (Ω). We summarize this as
Lemma 2.4. Any u ∈ C 2 (Ω) which solves ∆u = 0 in Ω is actually C ∞ (Ω).
In fact, one can show something stronger:
Lemma 2.5. Any u like in the previous Lemma is actually analytic in Ω.
The proof is an exercise. It only requires suitably extending the formula
(19) into a complex neighborhood. Fixing a ξ it is not hard to see that (19)
defines a complex analytic function one the disc |z − ξ| < δ for suitable small
δ. Alternatively, you can construct the power series for u directly and show
it converges using the estimates obtained from the mean value theorem (see
section ... and Evans, Chapter 2.2).
The two previous Lemmata give yet another proof of the fact that the
Cauchy-problem for the Laplace equation cannot be solved in general. Assume
that you specified Cauchy data for u on xd = 0 in Rd
∆u = 0
xd = 0
and that you were able to solve Laplace’s equation in a small ball around the
origin. By the Lemmas, you know that the solution has to be C ∞ (in fact,
analytic) inside that ball. But this implies that your Cauchy-data would have
to have been analytic in the first place! It follows that you cannot solve the
Cauchy problem for Laplace’s equation except if the data are analytic. This is
very similar to the equation ut + iux = 0 we discussed previously!
2.4
Mean value formulas
Let us return once more to the formula (15). It is clear that the function
G (x, ξ) = K (x, ξ) + w (x)
(20)
for w ∈ C 2 Ω̄ with ∆w = 0 in Ω is another fundamental solution with pole ξ
and that the formula
Z Z
dG (x, ξ)
du
dSx (21)
(x) − u (x)
G (x, ξ)
G (x, ξ) ∆udx −
u (ξ) =
dnx
dnx
∂Ω
Ω
9
is valid. (Simply observe
that the right hand side is zero if G is replaced by the
harmonic w ∈ C 2 Ω̄ in view of (9).)
For instance, taking for Ω a ball B (ξ, ρ) centered around ξ and
G (x, ξ) = K (x, ξ) − ψ (ρ) = ψ (|x − ξ|) − ψ (ρ) ,
(22)
dG
we have that G (x, ξ) = 0 for x ∈ ∂Ω and, in view of dn
|∂Ω = ω1d ρ1−d , the
x
identity
Z
Z
1
(ψ (|x − ξ|) − ψ (ρ)) ∆u (x) dx +
u (ξ) =
u (x) dSx
ωd ρd−1 |x−ξ|=ρ
|x−ξ|<ρ
(23)
expressing the values of u at the center of a ball in terms of its values on the
boundary and a volume integral involving ∆u (which is typically prescribed).
Noting that ψ (|x − ξ|) − ψ (ρ) < 0 (recall ψ (|x − ξ|) is monotone and gets large
and negative for small arguments), we immediately produce
Lemma 2.6. If u ∈ C 2 in B (ξ, ρ) and ∆u ≥ 0 in the ball, then
Z
1
u (ξ) ≤
u (x) dSx .
ωd ρd−1 |x−ξ|=ρ
(24)
Note that for harmonic functions this is the well-known mean value property:
The value of a harmonic function at a point equals its average value over a sphere
surrounding that point.
Definition 2.7. We call u subharmonic in Ω if (24) holds for any point ξ ∈ Ω
and sufficiently small balls B (ξ, ρ).
By Lemma 2.6, functions with ∆u ≥ 0 in Ω are subharmonic. Conversely, a
subharmonic function which is also C 2 satisfies ∆u ≥ 0. This follows directly
from (23). It particular, a C 2 function which has the mean value property is
harmonic.2
2.5
Poisson’s formula
The goal of this subsection is to prove Poisson’s formula,
Z
u (ξ) = ∆ξ
K (x, ξ) u (x) dx
(25)
Ω
valid for u ∈ C 2 Ω̄ and ξ ∈ Ω. The “quick and dirty” prove of (25) would be
to pull the Laplacian through the integral and use that ∆ξ K = δξ .
To do it carefully, we first let u ∈ C02 (Ω). Then formula (15) reduces to
Z
Z
K (x, ξ) ∆x u (x) dx
(26)
K (x, ξ) ∆x u (x) dx =
u (ξ) =
Rd
Ω
2 This
remains true assuming only that u is C 0 .
10
since the boundary terms vanish by the assumption of compact support. The
domain can be increased to all of Rd for the same reason. Then
Z
Z
Z
ψ (|y|) ∆y u (y + ξ) dy
ψ (|x − ξ|) ∆x u (x) dx =
K (x, ξ) ∆x u (x) dx =
d
Rd
Rd
ZR
Z
ψ (|y|) ∆ξ u (y + ξ) dy = ∆ξ
ψ (|y|) u (y + ξ) dy
=
Rd
Rd
Z
Z
= ∆ξ
ψ (|x − ξ|) u (x) dx = ∆ξ
K (x, ξ) u (x) dx ,
Rd
Ω
C02
which proves the desired formula
for u ∈
(Ω) and any ξ ∈ Ω. Now for a
given ξ, a general u ∈ C 2 Ω̄ can be decomposed into a part which is supported
away from the boundary and a part which is supported away from a small ball
around ξ. More precisely, let ξ ∈ Ω be given and choose a small ball b ⊂ Ω
around ξ. Choose also a larger ball B such that b̄ ⊂ B ⊂ Ω and a cut-off
function χ ∈ C02 (Ω) which is equal to 1 in B. Then u ∈ C 2 Ω̄ can be written
as
u = u1 + u2 = χ · u + (1 − χ) u
with u1 ∈ C02 (Ω) and u2 supported away from B. For u2 we have
Z
Z
∆ξ
K (x, ξ) u2 (x) dx = ∆ξ
K (x, ξ) u2 (x) dx = 0
Ω
Ω\B
since K (x, ξ) is in C 2 Ω \ B and harmonic for any ξ ∈ b. This concludes the
proof.
The relevance of Poisson’s formula is that for a given u ∈ C 2 Ω̄ (for instance, a charge distribution), the expression
Z
K (x, ξ) u (x) dx
w (ξ) =
Ω
provides a special solution to Poisson’s equation ∆ξ w = u (whose negative
gradient is the electric field). We can in fact consider the above w (x) as a
function on all of Rd . We know that w is C 2 inside Ω and3 satisfies Poisson’s
equation. It’s not hard to see that w is harmonic (hence analytic) at points
ξ∈
/ Ω̄, and that for points on ∂Ω, w may fail to be C 2 .
2.6
Maximum Principles
The previous considerations relied heavily on the exact form of the Laplacian.
The ideas considered in this section, on the other hand, carry over to more
general (even non-linear!) elliptic equations. More about this in the exercises.
3 This is clear for u supported away from ξ.
On the other hand, for u of compact
support
away from the
R
R boundary ∂Ω one has,Rsimilar to the above computation, w 1(ξ) =
K (x, ξ) u (x) dx = Rd ψ (|x − ξ|) u (x) dx = Rd ψ (|y|) u (y + ξ) dy. Since ψ is in Lloc , w
Ω
can be differentiated (at least) as many times as u.
11
Proposition 2.8. (Weak maximum principle) Let u ∈ C 2 (Ω) ∩ C 0 Ω̄ and
∆u ≥ 0 in Ω. Then
max u = max u
∂Ω
Ω̄
Proof. Since u ∈ C 0 Ω̄ the expressions are well-defined. For ∆u > 0 the
conclusion would follow immediately since at an interior maximum we would
need both ∂x2 u ≤ 0 and ∂y2 u ≤ 0 and hence ∆u ≤ 0. For the general case, we
consider the auxiliary function
v (x) = u (x) + ǫ|x|2
which satisfies ∆v = ∆u + 2ǫn > 0 in Ω for any ǫ > 0. Therefore
max u + ǫ|x|2 = max u + ǫ|x|2
∂Ω
Ω̄
max u + ǫ min |x|2 ≤ max u + ǫ max |x|2
Ω̄
∂Ω
Ω̄
∂Ω
and since this holds for any ǫ > 0, maxΩ̄ u ≤ max∂Ω u. Since the reverse
inequality is trivial, we obtain the desired equality.
For u being harmonic, the same argument leads to the equality for the
minimum and in view of |u| = max (u, −u) we also have
Corollary 2.9. Let u ∈ C 2 (Ω) ∩ C 0 Ω̄ and ∆u = 0 in Ω. Then
max |u| = max |u|
∂Ω
Ω̄
Corollary 2.10. Let u ∈ C 2 (Ω) ∩ C 0 Ω̄ and ∆u = 0. Then u = 0 on ∂Ω
implies u = 0 in all of Ω.
Corollary 2.10 improves our uniqueness statement about the Dirichlet prob
lem for Poisson’s equation (12) from Section 2.1, which required u ∈ C 2 Ω̄ .
2
Now we
know there can only be one solution to this problem in u ∈ C (Ω) ∩
0
C Ω̄ .
Proposition 2.8 still allows the maximum of u to be attained in the interior
(of course in addition to it being attained on the boundary). The following
“strong” maximum principle asserts that this can only happen if u is actually
constant:
Proposition 2.11. Let u ∈ C 2 (Ω) and ∆u ≥ 0 in Ω. Then either u is constant
or
u (ξ) < sup u
Ω
for all ξ ∈ Ω.
Corollary 2.12. If u ∈ C 2 (Ω) ∩ C 0 Ω̄ is non-constant, then
u (ξ) < max u
∂Ω
12
Proof. The continuity implies supΩ u = maxΩ̄ u = max∂Ω u, the last step being
Proposition 2.8.
Proof of Proposition 2.11. Set M = sup u and decompose Ω = Ω1 ∪ Ω2 as
Ω1 = { ξ ∈ Ω | u (ξ) = M } and Ω2 = { ξ ∈ Ω | u (ξ) < M }
(27)
These sets are obviously disjoint. If we can show they are both open, then one
of them must be empty (a connected set cannot be written as the union of two
open sets, by definition). Now Ω2 is open by the continuity of u in Ω. To see
that Ω1 is open, fix ξ ∈ Ω1 and consider the mean value inequality (24) in a
small ball around ξ contained in Ω
Z
Z
[u (x) − u (ξ)] dSx
u (x) dSx − ωd ρn−1 u (ξ) =
0≤
|x−ξ|=ρ
|x−ξ|=ρ
Since u (ξ) = M and u (x) ≤ M for all x, the integrand is non-positive. This is
only consistent if u (x) = M holds in the ball, establishing that Ω1 is open.
2.7
Existence of solutions: Green’s function for a ball
So far we have been talking about the uniqueness of solutions and about their
properties but not about whether they actually
exist! Let us return to equation
(21) which we know holds for any u ∈ C 2 Ω̄ . The claim is the following: If we
would be able to find a G (x, ξ) (and hence a w (x)) such that G (x, ξ) = 0 holds
for all of x ∈ ∂Ω (independently of ξ ∈ Ω), then we would have a formal solution
of the Dirichlet problem (12) expressed purely in terms of the “data” f and g.
We say formal because formula (21) hinges on u ∈ C 2 Ω̄ and it is not clear
a-priori whether a solution of (12) has to be C 2 up to the boundary. However,
once one has found the desired G (x, ξ), one can directly check whether the u (ξ)
thus defined satisfies (12) and what regularity properties it has.
We will carry out this program for Ω being a ball of radius a, which wlog
(translation invariance) we put at the origin. Given a ξ ∈ B (0, a) we define it’s
dual point (cf. Exercise 2.1) to be
ξ⋆ =
a2
ξ
|ξ|2
⋆
The point of this definition is that the quotient rr of r = |x − ξ|, the distance
from a point x to ξ, and r⋆ = |x − ξ ⋆ |, the distance from a point x to ξ ⋆ , is
constant for x being any point on the boundary of the ball S (0, a):
r⋆
a
=
r
|ξ|
for x ∈ ∂Ω. This is verified by a quick computation. It follows that, for d > 2,
if we define
2−d
|ξ|
G (x, ξ) = K (x, ξ) −
K (x, ξ ⋆ ) ,
a
13
then G (x, ξ) = 0 for x ∈ ∂Ω. Moreover,
for any fixed ξ ∈ Ω, we have ξ ⋆ ∈
/ Ω̄
⋆
2
and hence K (x, ξ ) is both C Ω̄ in x and harmonic (recall this is required
of the w in (20)!) Note also that for ξ → 0 we recover the G used in the mean
value formula (22).
Plugging this into (21) we obtain the formula
Z
1 a2 − |ξ|2
H (x, ξ) gdSx
where H (x, ξ) =
u (ξ) =
(28)
aωd |x − ξ|d
|x|=a
valid for |ξ| < a as a natural candidate to solve the Dirichlet problem “∆u
=0
in B (0, a) and u = g on S (0, a)”. (In fact, we know that if u ∈ C 2 Ω̄ then
this has to be the solution.) The following Proposition checks this directly.
Proposition 2.13. Let g be continuous for |x| = a. Then the function
g (ξ)
for |ξ| = a
R
u (ξ) =
H (x, ξ) g (x) dSx for |ξ| < a
|x|=a
(29)
is continuous in the closed ball B (0, a) and C ∞ and harmonic in the inside.
By our improved uniqueness theorem
(Corollary 2.10) we know this is the
unique solution in C 2 (Ω) ∩ C 0 Ω̄ of the desired Dirichlet problem.
Proof. Next time.
3
Exercises
1. Problem on mollification (see Section 4)
2. Problem on convolution (see Section 4)
3. Show that the distributional derivative of log |x| is P.V. x1 .
4. (Fritz John, 3.6 (3)) Show that the function
1 for x1 > ξ1 , x2 > ξ2
u (x1 , x2 ) =
0
for all other x1 , x2
(30)
defines a fundamental solution with pole (ξ1 , ξ2 ) of the operator L =
∂2
∂x1 ∂x2 in the x1 x2 -plane.
5. At various stages we have used the interchange of differentiation and integration. The following theorem covers these situations. You can prove
it using the dominant convergence theorem.
Suppose that I ⊂ R is open and that f : I × Ω → R, (t, x) 7→ f (t, x) with
Ω ⊂ Rn is a function which is in L1 (Ω) for all t, differentiable in t for all
∂
x with the derivative satisfying | ∂t
f (t, x) | ≤ g (x) for all (t, x) with g a
1
function in L (Ω). Then
Z
Z
d
∂f
(t, x) dx < ∞ .
f (t, x) dx =
dt Ω
Ω ∂t
14
6. Let ∆u = 0 for u ∈ C 2 (Ω) ∩ C 0 Ω̄ and g ≥ 0 on ∂Ω. If g > 0 at one
point of ∂Ω, then u > 0 in Ω.
7. (Fritz John, 4.2 (1)) Let Ω denote the unbounded set |x| > 1. Let u ∈
C 2 Ω̄ , ∆u = 0 in Ω and lim|x|→∞ u (x) = 0. Show that
max |u| = max |u| .
∂Ω
Ω̄
8. (Fritz John, 4.2 (2)) Let u ∈ C 2 (Ω) ∩ C 0 Ω̄ be a solution of
∆u +
n
X
ak (x) uxk + c (x) u = 0
k=1
where c (x) < 0 in Ω. Show that u = 0 on ∂Ω implies u = 0 in Ω.
9. (Fritz John, 4.2 (3)) Prove the weak maximum principle for solutions of
the two-dimensional elliptic equation
Lu = auxx + 2buxy + cuyy + 2dux + 2euy = 0
where a, b, c, d, e are continuous functions of x and y in Ω̄ and ac − b2 > 0
(ellipticity) as well as a > 0 hold. HINT: Prove
it first under the stricti
h
2
condition Lu > 0, then use u+ǫv for v = exp M (x − x0 ) + M (y − y0 )
with appropriate M , x0 , y0 .
4
4.1
2
Additional Material
Distributions (see Appendix of Rauch’s book)
This problem will take you in a rather informal way through the main ideas
of extending various operations of calculus from test functions to distributions.
The key is Proposition 2 in the Appendix of Rauch’s book, which we repeat
below. Remember we already have a notion of convergence in D, Definition 1.1.
Proposition 4.1. Suppose L : D (Ω1 ) → D (Ω2 ) is a linear, sequentially continuous map. Suppose in addition that there is a linear, sequentially continuous
map Lt : D (Ω2 ) → D (Ω1 ) which is the transpose of L in the sense that
Z
Z
φ · Lt ψ holds for all φ ∈ D (Ω1 ) , ψ ∈ D (Ω2 ) .
Lφ · ψ =
Ω2
Ω1
Then the operator L extends to a sequentially continuous map of D′ (Ω1 ) →
D′ (Ω2 ) given by
L (ℓ) [ψ] = ℓ Lt ψ
for all ℓ ∈ D′ (Ω1 ) , ψ ∈ D (Ω2 ) .
You can find the (very easy) proof in Rauch’s book or do it yourself. This
simple proposition allows us to define the following operations on distributions
15
• multiplication of a distribution with a C ∞ function f ∈ C ∞ (Ω). Since on
test functions the transpose is itself (multiplication by f ), we have
(f · ℓ) [ψ] = ℓ [f · ψ]
• translation of a distribution (say Ω = Rd so that we don’t have to keep
track of domains). Since (τy f ) (x) = f (x − y) on test functions has transpose τ−y we have τy (ℓ) [ψ] = ℓ [τ−y ψ] on distributions.
• reflection of a distribution: R (ℓ) [ψ] = ℓ [Rψ], as the transpose of (Rf ) (x) =
f (−x) on test functions is itself.
• derivative of a distribution (we already did that!)
• convolution of a distribution with a C ∞ function (see below)
The remarkable point of convolving a distribution with a smooth function is
that the result is actually a smooth function. You will prove this below. This
is very useful and can be usedto show that we can approximate any element in
D′ Rd by elements in D Rd .
Let Ω = Rd and f ∈ D Rd . For g ∈ D (Rn ), the convolution of g with f is
defined as
Z
f (x − y) g (y) dy = (g ⋆ f ) (x) .
(31)
(f ⋆ g) (x) :=
Rd
Exercise 4.2. Show that (on test functions) the transpose of convolution with
f is convolution with Rf
Therefore we define
(f ⋆ ℓ) [ψ] = ℓ [Rf ⋆ ψ]
(32)
Exercise 4.3. Compute f ⋆ δ0 .
We can now show that the convolution (32) is actually smooth. This is
suggested interpreting the convolution (31) itself as the regular distribution
τ−x Rf acting on a test function g:
(f ⋆ g) (x) = τ−x Rf [g] = g [τ−x Rf ] .
Now the right hand side (which is a complex-valued function of x) actually makes
sense for distributional g providing an alternative definition of the convolution
of a distribution with a smooth function. Fortunately, the two “definitions”
agree:
Exercise 4.4. With ℓ ∈ D′ Rd and f ∈ D Rd given, show that
• the function x 7→ ℓ [τ−x Rf ] is C ∞ Rd
16
• ℓ [Rf ⋆ ψ] = ℓ [τ−x Rf ] · ψ holds for all test functions (where the right hand
side is to be understood as integrating the C ∞ function again the test
function ψ,)
The second statement is precisely the statement that the two definitions
agree as distributions. You can prove it expressing the right hand side in terms
of Riemann sums....
Exercise 4.5.
4.2
Mollification (see Appendix of Evans)
Let Ω ⊂ Rn be open. The goal is to approximate functions in L1loc (Ω) or C 0 (Ω)
by functions which are smooth. (Obviously, this can be very convenient in
arguments involving lots of integration by parts.)
ǫ
Ωǫ
We write Ωǫ ⊂ Ω for
Ωǫ := {x ∈ Ω | dist (x, ∂Ω) > ǫ}
Let η ∈ C ∞ (Rn ) be the function (“standard mollifier”)
(
|x| < 1
C exp |x|21−1
η (x) =
0
|x| ≥ 1
R
with C chosen such that Rn η dx = 1. For ǫ > 0 we define
ηǫ (x) :=
1 x
η
ǫn
ǫ
(33)
(34)
RNote that ηǫ is smooth, supported on the closed unit ball B (0, ǫ) and that
ηǫ (x) dx = 1.
Definition 4.6. The mollification of an L1loc (Ω) function f : Ω → R is defined
as
f ǫ := ηǫ ⋆ f
in Ωǫ .
(35)
We claim that on Ωǫ , the function f ǫ is smooth and approximates f in an
appropriate sense as ǫ → 0:
Exercise 4.7. Show that
17
1. f ǫ ∈ C ∞ (Ωǫ )
2. f ǫ → f almost everywhere as ǫ → 0
3. If f ∈ C 0 (Ω) (=continuous), then f ǫ → f uniformly on compact subsets
of Ω
4. If f ∈ Lploc (Ω) with 1 ≤ p < ∞, then f ǫ → f in Lploc (Ω)
Here is a guideline:
1. Write out the difference quotients
f ǫ (x + hei ) + f ǫ (x)
h
with h sufficiently small so that x + hei is still in Ωǫ . Then use uniform
convergence to interchange limit and integral.
2. Start with Lebesgue’s Differentiation theorem
Z
1
lim
|f (x) − f (y) |dy = 0 ,
r→0 vol (B (x, r)) B(x,r)
which holds at almost every x for f in L1 , to estimate |f ǫ (x) − f (x) |.
3. Follows from 2. using uniform continuity on compact subsets.
4. Apply Hölder to the definition of f ǫ to estalish
kf ǫ kLp (U) ≤ kf kLp(V )
(36)
for U ⊂⊂ V ⊂⊂ Ω. To show that f ǫ → f in U ⊂⊂ V ⊂⊂ Ω, suppose
δ > 0 is given. Choose a continuous g ∈ C 0 (V ) such that
kf − gkLp(V ) < δ
which is possible by density. Then show that kf ǫ − f kLp(U) ≤ 3δ for ǫ
sufficiently small using the triangle inequality and (36).
18