Institute for Analysis
Prof. Dr. Wolfgang Reichel
C.I.M.E. summer school Cetraro, June 19 – 23, 2017
”Geometry of PDE’s and related problems”
Symmetry properties for solutions of higher-order
elliptic boundary value problems (4th version of June 20, 2017)
Please let me know errors or typos: [email protected]
Goals of this lecture-series:
• Discussion of symmetry properties of solutions to (−∆)m u = f (x, u) either on Rn
or on balls with additional Dirichlet boundary conditions.
• Methods based on contraction mapping
• Methods based on the moving plane method
• An example of an overdetermined 4th order problem using Newton’s inequalities
1. Linear problems: weak solutions, eigenvalues, regularity, Green
functions
Suppose m ∈ N and Ω ⊂ Rn is a bounded domain. Here we consider the polyharmonic
boundary value problem
m−1
∂
∂
m
(1)
(−∆) u = f (x) in Ω, u =
u = 0 on ∂Ω.
u = ... =
∂ν
∂ν
To formulate solutions concepts for (1) we need to define some function spaces. The
classical Lp -spaces on a domain Ω ⊂ Rn are denoted by Lp (Ω) for 1 ≤ p ≤ ∞. For
derivatives of a function u : Ω → R we use the multi-index notation with a multi-index
α = (α1 , . . . , αn ) ∈ Nn0 to define
Dα u =
∂ |α|
u with |α| = α1 + . . . + αn .
∂xα11 · · · ∂xαnn
Let Cc∞ (Ω) be the space of test-functions. For 0 ≤ γ ≤ 1 and k ∈ N0 let
C k (Ω) := {u : Ω → R : Dα u exist and have continuous and bounded extensions to Ω
∀α with |α| ≤ k}
C0k (Ω) := {u ∈ C k (Ω) : Dα u|∂Ω = 0 ∀ α with |α| ≤ k}
C k,γ (Ω) := {u ∈ C k (Ω) : [Dα u]γ < ∞ ∀ α with |α| = k}
where [f ]γ = supx,y∈Ω,x6=y
|f (x)−f (y)|
.
|x−y|γ
The spaces are normed by
kukC k := max kDα uk∞ ,
|α|≤k
kukC k,γ := kukC k + max[Dα u]γ
|α|=k
1
Symmetry for higher-order elliptic problems
and C0k (Ω) is a closed subspace of C k (Ω). For functions u : Ω → R having weak derivatives
of order |α| we use the symbol Dα u : Ω → R to denote the weak derivative. For 1 ≤ p ≤ ∞
let
W k,p (Ω) = {u : Ω → R : Dα u ∈ Lp (Ω) ∀α with |α| ≤ k}
P
1/p
p
α
with norm kukW k,p :=
kD
uk
if 1 ≤ p < ∞ and kukW k,∞ := max|α|≤k kDα uk∞
p
L
|α|≤k
and
k·kW k,p
W0k,p (Ω) = Cc∞ (Ω)
.
For bounded domains Ω the spaces W0k,p (Ω) can be equipped with the equivalent norm
P
1/p
p
α
kukW k,p :=
if 1 ≤ p < ∞ and kukW k,∞ := max|α|=k kDα uk∞ . This
|α|=k kD ukLp
0
0
is a consequence of the Poincaré inequality kuk ≤ C(j, Ω, p)k∂j ukLp for u ∈ W01,p (Ω)
and j = 1, . . . , n. In the case p = 2 we define the Hilbert spaces H k (Ω) := W k,2 (Ω) and
H0k (Ω) := W0k,2 (Ω).
Lp
On the Hilbert space H0k (Ω) one can introduce yet another interesting equivalent norm.
Lemma 1. Let k ∈ N. For u ∈ H0k (Ω) define
r
∆u
if k = 2r,
k
D u=
r
∇∆ u if k = 2r + 1.
Then the scalar product
Z
hu, viH0k :=
Dk u Dk v dx,
u, v ∈ H0k (Ω)
Ω
1
generates via kukH0k := hu, uiH2 k a norm which is equivalent to k · kW k,2 .
0
0
R
P
Proof. Observe that hu, viH0k = |α|=k Ω Dα uDα v dx. By density it is enough to verify
this for u, v ∈ Cc∞ (Ω). The relation is then a consequence of several integrations by parts.
For illustration let us, e.g., consider the case k = 2:
n Z
XZ
X
∂2
∂2
α
α
D uD v dx =
u
v dx
∂x
∂x
∂x
∂x
i
j
i
j
Ω
Ω
i,j=1
|α|=2
n Z
X
∂2
∂2
=
u
v dx
∂xi ∂xi ∂xj ∂xj
i,j=1 Ω
Z
=
∆u∆v dx.
Ω
We consider three kinds of solutions: weak, strong and classical.
2
Symmetry for higher-order elliptic problems
Definition 2 (weak solution). Let f ∈ L2 (Ω). A function u ∈ H0m (Ω) is called a weak
solution of (1) if
Z
Z
m
m
D u D φ dx =
f φ dx for all φ ∈ H0m (Ω).
Ω
Ω
2
Lemma 3. For every f ∈ L (Ω) there exists a unique weak solution u ∈ H0m (Ω) of (1)
and the solution operator T : L2 (Ω) → L2 (Ω), T f := u is compact. It has a complete
L2 -orthonormal basis {φi : i ∈ N} of eigenfunctions with corresponding eigenvalues {λi :
i ∈ N}, where the eigenvalues λi are ordered
λ1 ≤ λ2 ≤ λ3 . . .
Within the eigenvalue sequence, values are repeated according to the multiplicity of the
eigenvalue.
Definition 4 (strong solution). Let f ∈ Lp (Ω) for some p ∈ (1, ∞). A function u ∈
W 2m,p (Ω) ∩ W0m,p (Ω) is called a strong solution of (1) provided the differential equation in
(1) holds for almost all x ∈ Ω.
The following result establishes existence and uniqueness of strong solutions to (1), cf.
Corollary 2.21 in [4].
Lemma 5. Let 1 < p < ∞ and assume ∂Ω ∈ C 2m . For every f ∈ Lp (Ω) there exists
a unique strong solution u ∈ W 2m,p (Ω) ∩ W0m,p (Ω) of (1) and the solution operator T :
Lp (Ω) → W 2m,p (Ω) ∩ W0m,p (Ω), T f := u is bounded.
Definition 6 (classical solution). Let f ∈ C(Ω). A function u ∈ C 2m (Ω) ∩ C0m−1 (Ω) is
called a classical solution of (1) provided the differential equation in (1) holds for all x ∈ Ω.
The corresponding existence and uniqueness result for classical solutions to (1) can be
found in Theorem 2.19 in [4].
Lemma 7. Assume ∂Ω ∈ C 2m,α . For every f ∈ C α (Ω) there exists a unique classical
solution u ∈ C 2m,α (Ω) ∩ C0m (Ω) of (1) and the solution operator T : C α (Ω) → C 2m,α (Ω) ∩
C0m (Ω), T f := u is bounded.
Definition 8 (positivity preserving property). The operator T from Lemma 3, Lemma 5
or Lemma 7 is called positivity preserving if f ≥ 0 implies u ≥ 0.
The fact that the solution operator for a ball, i.e., Ω = BR (P ) ⊂ Rn , is positivity preserving
is deduced from the following formula originally given by Boggio, cf. Section 2.6 in [4].
Definition 9 (Green function for balls). For x ∈ B1 (0) let x∗ = x/|x|2 and define for
x, y ∈ B1 (0)
Z |x| |x∗ −y|/|x−y|
2m−n
G(x, y) := km,n |x − y|
(t2 − 1)m−1 t1−n dt.
1
where km,n =
Γ(1+n/2)
.
nπ n/2 4m−1 Γ(m)2
3
Symmetry for higher-order elliptic problems
Lemma 10. Let Ω = B1 (0) ⊂ Rn . Then for 1 < p < ∞ and f ∈ Lp (B1 (0)) the solution
operator from Lemma 5 is given by
Z
u(x) = (T f )(x) =
G(x, y)f (y) dy, x ∈ B1 (0)
B1 (0)
and hence it is positivity preserving. The same is true for the solution operator of Lemma 3
if f ∈ L2 (B1 (0)) and for the solution operator of Lemma 7 if f ∈ C α (B1 (0)).
2. Symmetry, simplicity and positivity for first eigenfunctions
Recall from Lemma 3 that (−∆)m with Dirichlet boundary conditions on any bounded
domain Ω has a first eigenvalue
λ1 =
min
u∈H0m (Ω)\{0}
kuk2H0m
kuk2L2
.
Three natural questions arise:
(1) Is λ1 simple?
(2) If so, do the eigenfunctions corresponding to λ1 have one sign?
(3) If Ω is symmetric, do the eigenfunctions corresponding to λ1 have the same symmetry?
In this section we answer all three questions in a positive way when Ω = B1 (0). Let us
first sketch the usual way of answering question (1) in case m = 1. Let u be a Dirichlet
eigenfunction of −∆ corresponding to λ1 and decompose u = u+ + u− with the usual
pointwise definitions u+ = max{u, 0} and u− = min{u, 0}. In H01 (Ω) a (nontrivial) fact
is that u+ , u− ∈ H01 (Ω) and ∇u = ∇u+ χ{u>0} + ∇u− χ{u<0} and hence |∇|u|| = |∇u| a.e.
in Ω. Assume for contradiction that u is sign-changing, i.e., u+ , u− are both non-trivial.
Thus |u|, u+ , u− are also minimizers for λ1 . The fact that, e.g., u+ vanishes on a set of
positive measure then contradicts the strong minimum principle. For m ≥ 2 this argument
fails twice. First, u ∈ H0m (Ω) implies in general not that u+ belongs to H m (Ω). Second,
the maximum principle in general fails.
The way out for m ≥ 2 is to decompose u = u1 + u2 with u1 ≥ 0 a.e. in Ω and u1 ⊥H0m u2 .
This relies on an abstract construction due to Moreau [9].
Definition 11 (Dual cones). Let (H, h·, ·i) be a real Hilbert space. Let K ⊂ H be a cone,
i.e., v ∈ K implies tv ∈ K for all t ≥ 0. The dual cone K∗ is defined by
K∗ = {w ∈ H : hw, vi ≤ 0 for all v ∈ K}.
Theorem 12 ( [9]). Let (H, h·, ·i) be a real Hilbert space and let K ⊂ H be a closed convex
cone. Then for every u ∈ H there exist unique elements u1 ∈ K, u2 ∈ K∗ such that
u = u1 + u2 with u1 ⊥ u2 . Moreover for all u, ũ ∈ H
(2)
ku − ũk2 ≥ ku1 − ũ1 k2 + ku2 − ũ2 k2 .
4
Symmetry for higher-order elliptic problems
Proof. One defines u1 as the (unique) minimizer ku−u1 k = minv∈K ku−vk and u2 := u−u1 .
By the cone property and convexity of K we have u1 + tv ∈ K whenever t ≥ 0 and v ∈ K
so that
(3)
ku − u1 k2 ≤ ku − (u1 + tv)k2 for all t ≥ 0.
The quantity on the right-hand side of (3) has a minimum at t = 0 and thus
d
2
0 ≤ ku − (u1 + tv)k = −2hu − u1 , vi.
dt
t=0
∗
Hence hu2 , vi ≤ 0 for all v ∈ K so that u2 ∈ K . Moreover, setting v = u1 we see that (3)
holds for all t ≥ −1 and thus we can use
d
2
0 = ku − (1 + t)u1 )k = −hu − u1 , u1 i
dt
t=0
so that u1 ⊥ u2 . Let us finally prove (2). Note that for all u, ũ ∈ H we have
ku − ũk2 = ku1 − ũ1 k2 + ku2 − ũ2 k2 + 2hu1 − ũ1 , u2 − ũ2 i
= ku1 − ũ1 k2 + ku2 − ũ2 k2 − 2 hu1 , ũ2 i −2 hũ1 , u2 i
| {z }
| {z }
≤0
2
≤0
2
≥ ku1 − ũ1 k + ku2 − ũ2 k .
This inequality also implies uniqueness of u1 , u2 with the stated properties.
Let us briefly discuss the consequences of taking K = {v : v ≥ 0} where the functions v
are taken from the Hilbert spaces L2 (Ω), H01 (Ω), H02 (Ω). Clearly in the first case, where
H = L2 (Ω) we get K = {v ∈ L2 (Ω) : v ≥ 0} and K∗ = {w ∈ L2 (Ω) : w ≤ 0} so that
the splitting according to Theorem 12 yields u = u1 + u2 with u1 = Ru+ , u2 = u− . In the
second case let us equip H01 (Ω) with the scalar product hu, ũiH01 := Ω ∇u · ∇ũ dx. Then
K = {v ∈ H01 (Ω) : v ≥ 0} and K∗ = {w ∈ H01 (Ω) : w is weakly sub-harmonic}. Clearly,
by the maximum
principle, K∗ ⊂ −K but K∗ 6= −K. Finally, in the third case let us take
R
hu, ũiH02 := Ω ∆u · ∆ũ dx and K = {v ∈ H02 (Ω) : v ≥ 0}. Then K∗ = {w ∈ H02 (Ω) :
(−∆)2 w ≤ 0 weakly}. In this case it is in general not true that K∗ is a subset of −K. It
is however true for the case where Ω = B1 (0) as the next result shows.
H = H0m (B1 (0)) with hu, ũi :=
RLemmam 13.mLet Ω = B1 (0) and consider the Hilbert space
∗
D uD ũ dx. Let K = {v ∈ H : v ≥ 0}. Then K ⊂ −K and, more precisely, for
B1 (0)
every w ∈ K∗ we have w < 0 or w ≡ 0 in B1 (0)).
Proof. We only prove the statement K∗ ⊂ −K. The more precise formulation w < 0 or
w ≡ 0 for all w ∈ K∗ can be found in [4]. Let w ∈ K∗ and let φ ∈ L2 (B1 (0)) with φ ≥ 0
and consider the solution vφ ∈ H0m (B1 (0)) of
m−1
∂
∂
m
(−∆) vφ = φ in B1 (0), vφ =
vφ = . . . =
vφ = 0 on ∂B1 (0).
∂ν
∂ν
R
By Lemma 10, vφ ∈ K, so that 0 ≥ hw, vφ i = B1 (0) φw dx. Since this holds for all
φ ∈ L2 (B1 (0)) with φ ≥ 0 we conclude w ≤ 0.
5
Symmetry for higher-order elliptic problems
Finally, we are ready to prove simplicity, positivity and symmetry of first eigenfunctions.
Theorem 14. The first eigenvalue λ1 of (−∆)m , m ≥ 1 on B1 (0) with Dirichlet boundary
conditions is simple. Any corresponding eigenfunction is of one sign and radially symmetric.
Proof. Let u ∈ H0m (B1 (0)) be an eigenfunction corresponding to λR1 . Assume for contradiction that u is sign-changing. We use the scalar product hu, ũi := B1 (0) Dm uDm ũ dx and
split u = u1 + u2 according to Theorem 12 with K = {v ∈ H0m (B1 (0)) : v ≥ 0}. The
assumption that u is sign-changing implies u1 ∈ K \ {0} and u2 ∈ K∗ \ {0}. Lemma 13
implies that u2 < 0 in B1 (0). Therefore, if we set ũ := u1 − u2 , then ũ ∈ K and moreover
u2 − u1 ≤ u2 ≤ u1 + u2 < u1 − u2 a.e. in B1 (0).
Moreover, the first inequality is strict on a set of positive measure. Therefore ku1 +u2 kL2 <
ku1 − u2 kL2 . However, due to u1 ⊥H0m u2 we see that ku1 + u2 k2H0m = ku1 − u2 k2H0m . This
contradicts the minimality of λ1 . Therefore, any eigenfunction corresponding to λ1 does
not change sign and in particular it is either strictly positive (u ≥ 0 solving the eigenvalue
equation implies −u ∈ K∗ ) or strictly negative (nontrivial elements of K∗ are strictly
negative).
A standard argument allows now to deduce simplicity from positivity. Suppose that u, ũ
are linearly independent eigenfunctions corresponding to λ1 . Then w.l.o.g u, ũ are both
strictly positive. However, u − αũ for α ∈ R is also an eigenfunction for λ1 , and it is
sign-changing for suitable α > 0. This contradiction shows simplicity.
Finally, radial symmetry of the eigenfunction is a consequence of simplicity.
3. Symmetry for nonlinear problems by uniqueness and non-resonance
In this section we prove two statements that directly imply symmetry for solutions of the
nonlinear boundary value problem
m−1
∂
∂
m
(4)
(−∆) u = f (x, u) in B1 (0), u =
u = ... =
u = 0 on ∂B1 (0).
∂ν
∂ν
Recall the eigenvalue sequence λ1 < λ2 ≤ λ3 ≤ . . . for the polyharmonic operator (−∆)m
on the unit ball with Dirichlet boundary conditions.
Theorem 15 (Uniqueness). Let f : B1 (0) × R → R be a Carathéodory-function with the
following properties
(i) |x| = |y| ⇒ f (x, s) = f (y, s) for all s ∈ R,
(ii) f (·, 0) ∈ L2 (B1 (0)),
(iii) ∃L, > 0 such that
(∗)1
−L ≤
f (x, s) − f (x, t)
≤ λ1 − for all s, t ∈ R, s 6= t, x ∈ B1 (0)
s−t
6
Symmetry for higher-order elliptic problems
(∗)i
or ∃L, > 0, i ≥ 2 such that
f (x, s) − f (x, t)
≤ λi − for all s, t ∈ R, s 6= t, x ∈ B1 (0).
λi−1 + ≤
s−t
Then (4) has a unique weak solution u ∈ H0m (B1 (0)). This solution is necessarily radially
symmetric.
Proof. We give the proof only in case of (∗)i for i ≥ 2. The case (∗)1 differs only in a few
details. Clearly λi−1 < λi . The inequality
|f (x, s)| ≤ |f (x, 0)| + |s|λi
and assumption (ii) guarantee that u ∈ L2 (B1 (0)) implies f (·, u) ∈ L2 (B1 (0)). Let κ :=
λi−1 +λi
and set g(x, t) := f (x, t) − κt. Then (iii) implies
2
λi−1 − λi + 2
g(x, s) − g(x, t)
λi − λi−1 − 2
≤
≤
for all s, t ∈ R, s 6= t, x ∈ B1 (0).
2
s−t
2
Let Tκ : L2 (B1 (0)) → H0m (B1 (0)) be the solution operator mapping the right-hand side
h ∈ L2 (B1 (0)) to the solution w ∈ H0m (B1 (0)) of the problem
m−1
∂
∂
m
(−∆) w − κw = h in B1 (0), w =
w = ... =
w = 0 on ∂B1 (0).
∂ν
∂ν
(5)
Notice that Tκ is well-defined and continuous since κ is not a Dirichlet eigenvalue of (−∆)m .
The map Tκ : L2 (B1 (0)) → L2 (B1 (0)) is compact and its operator norm can be computed
as
kTκ hkL2
1
2
kTκ k = sup
=
=
.
min{|λj − κ| : j ∈ N}
λi − λi−1
h6=0 khkL2
Next, consider the operator
2
L (B1 (0)) → L2 (B1 (0)),
G :=
u 7→ g(·, u).
Then u ∈ H0m (B1 (0)) is a weak solution of (4) if and only if u = Tκ (G(u)), u ∈ L2 (B1 (0)).
Hence, it is enough to show that Tκ ◦ G is a contraction in L2 (B1 (0)), i.e.,
(5)
kTκ (G(u)) − Tκ (G(v))kL2 ≤ kTκ k kG(u) − G(v)kL2 ≤
2
λi − λi−1 − 2
ku − vkL2 .
λi − λi−1
2
|
{z
}
<1
By the contraction mapping principle this establishes existence and uniqueness. Notice that
if u(·) solves (4) then by (i) the function u(A·) also solves (4) for every orthogonal matrix
A ∈ O(n). By uniqueness we get u(x) = u(Ax), which establishes radial symmetry.
Infinitely many eigenvalues of the above sequence have radially symmetric eigenfunctions.
But there is also a sequence {νi }i∈N of eigenvalues corresponding to non-radially symmetric
eigenfunctions. The idea behind the next result (Theorem 20) goes back to [7] and can
be explained as follows: if the nonlinearity f in (4) is non-resonant w.r.t. the non-radial
Dirichlet spectrum of (−∆)m on Ω then all solutions of (4) must be radially symmetric. A
precise meaning of non-resonant will be given below in Theorem 20.
7
Symmetry for higher-order elliptic problems
Definition 16. Let us denote by L2rad (B1 (0)) the space of all radially symmetric functions
in L2 (B1 (0)), and let Y = L2rad (B1 (0))⊥ be its L2 -orthogonal complement. The operator
(−∆)m : H 2m (B1 (0))∩H0m (B1 (0))∩L2rad (B1 (0)) → L2rad (B1 (0)) is selfadjoint and has (simrad
ple) eigenvalues {λrad
i }i∈N with corresponding radially symmetric eigenfunctions {φi }i∈N
building an orthonormal basis of L2rad (B1 (0)). Let {µi }i∈N be the set of non-radial eigenvalues, i.e., eigenvalues for which there exists a least one eigenfunction belonging to Y . We
denote by {ψi }i∈N an ONB of Y consisting of non-radial eigenfunctions corresponding to
{µi }i∈N .
Remark 17.
by Theorem 14 and hence µ1 ≥ λ2 .
(i) Note that λ1 = λrad
1
: i ∈ N} and
(ii) L2 (B1 (0)) = L2rad (B1 (0)) ⊕ L2rad (B1 (0))⊥ with L2rad (B1 (0)) = span{φrad
i
2
⊥
Lrad (B1 (0)) = span{ψi : i ∈ N}.
(iii) We do not exclude {λrad
i }i∈N ∩ {µi }i∈N 6= ∅. In the intersection could be an eigenvalue which has both radial and non-radial eigenfunctions. Its eigenspace has an
ONB with elements from {φrad
i }i∈N as well as from {ψi }i∈N . According to the above
definition such an eigenvalue is called non-radial.
Lemma 18. Let h ∈ Y = L2rad (B1 (0))⊥ and κ 6∈ {µi }i∈N . Then, in the space H0m (B1 (0))∩Y
there exists a unique element w =: Tκ h solving
m−1
∂
∂
m
(6)
(−∆) w − κw = h in B1 (0), w =
w = ... =
w = 0 on ∂B1 (0).
∂ν
∂ν
Moreover,
kTκ k = sup
h6=0
1
kTκ hkL2
.
=
khkL2
min{|µj − κ| : j ∈ N}
Proof. Every element h ∈ Y can be expressed by the basis {ψi }i∈N , i.e., h =
and hence, by the non-resonance assumption on κ, we have that
X hh, ψi iL2
w=
ψi
µ
i−κ
i∈N
P
i∈N hh, ψi iL2 ψi
solves (6). Hence
kwk2L2 =
X hh, ψi i2 2
1
L
≤
khk2L2
2
2
(µ
min{|µ
i − κ)
j − κ| : j ∈ N}
i∈N
1
which shows that kTκ k ≤ min{|µj −κ|:j∈N}
. The fact that equality holds can be seen by
choosing h = ψj0 where j0 ∈ N is the index where |µj0 − κ| = min{|µj − κ| : j ∈ N}.
Lemma 19. Let φ ∈ H0m (B1 (0)) ∩ L2rad (B1 (0)) and ψ ∈ H0m (B1 (0)) ∩ Y . Then
Z
Dm φ Dm ψ dx = 0.
B1 (0)
8
Symmetry for higher-order elliptic problems
∞
Proof. By density it is enough to show the claim
when φ is a radial
function.
R
R C0 (B1 (0))
m
m
In this case integration by parts shows that B1 (0) D φ D ψ dx = B1 (0) (−∆)m φψ dx. The
latter vanishes because (−∆)m φ is radial and hence belongs to L2rad (B1 (0)).
Theorem 20 (Nonresonance w.r.t. non-radial spectrum). Let f : [0, ∞) × R → R be a
Carathéodory-function with the following properties
(i) |x| = |y| ⇒ f (x, s) = f (y, s),
(ii) f (·, 0) ∈ L2 (B1 (0)),
(iii) ∃L, > 0 such that
f (x, s) − f (x, t)
≤ µ1 − for all s, t ∈ R, s 6= t, x ∈ B1 (0)
s−t
or ∃L, > 0, i ≥ 2 such that
f (x, s) − f (x, t)
µi−1 + ≤
≤ µi − for all s, t ∈ R, s 6= t, x ∈ B1 (0).
s−t
−L ≤
(∗∗)1
(∗∗)i
Then every weak solution of (4) is radially symmetric.
Proof. As in the proof of Theorem 15 we only deal with the case of (∗∗)i with i ≥ 2.
Consider orthogonal projections P : L2 (B1 (0)) → L2rad (B1 (0)) and Q : L2 (B1 (0)) →
L2rad (B1 (0))⊥ . Let κ = µi−12+µi and set g(x, t) := f (x, t) − κt. If we split u ∈ H0m (B1 (0))
into u = v + w with v ∈ H0m (B1 (0)) ∩ L2rad (B1 (0)) and w ∈ H0m (B1 (0)) ∩ Y then (4) can
be split into two equations
m−1
∂
m
(−∆) v − κv = P g(x, v + w) in B1 (0), v = . . . =
(7)
v = 0 on ∂B1 (0),
∂ν
m−1
∂
m
(8) (−∆) w − κw = Qg(x, v + w) in B1 (0), w = . . . =
w = 0 on ∂B1 (0).
∂ν
In fact, (7) and (8) follow by Lemma 19 from the weak form of (4) and they are understood
as
Z
Z
m
m
(D v D φ − κvφ) dx =
g(x, v + w)φ dx for all φ ∈ H0m (B1 (0))) ∩ L2rad (B1 (0))
B1 (0)
and
Z
Ω
m
m
Z
(D w D ψ − κwψ) dx =
B1 (0)
g(x, v + w)ψ dx for all ψ ∈ H0m (B1 (0))) ∩ L2rad (B1 (0))⊥ ,
Ω
respectively. Using the solution operator Tκ : Y → Y from Lemma 18 and the operator
2
L (B1 (0)) → L2 (B1 (0)),
G :=
u 7→ g(·, u).
we see that (8) is equivalent to
(9)
w = (Tκ ◦ Q ◦ G)(v + w).
For every fixed v ∈ H0m (B1 (0)) ∩ L2rad (B1 (0)), we can show exactly like in Theorem 15 that
the map w 7→ (Tκ ◦ Q ◦ G)(v + w) is a contraction on Y . Hence (9) has a unique solution.
9
Symmetry for higher-order elliptic problems
Notice that w = 0 solves (9) since G(v) = f (v) − κv ∈ L2rad (B1 (0)) and so Q(G(v)) = 0.
Thus, all solutions u = v + w of (4) necessarily have w = 0 and are therefore radial.
4. Symmetry via moving plane method – an example from potential theory
Let us step back more than three centuries into the year 1686/1687 when Isaac Newton
published the first edition of Principia Mathematica. In this book he developed among
many other seminal ideas the concept of the gravitational potential u of a body Ω ⊂ R3
with mass-density ρ : Ω → [0, ∞). In nowadays mathematical language we write it in the
form
Z
1
ρ(y)
u(x) =
dy,
x ∈ R3 .
4π Ω |x − y|
If Ω is a ball BR (0) and if the mass density is a constant ρ > 0 then Newton computed

R2 |x|2 

ρ
−
, |x| ≤ R,

2
6
u(x) =
ρR3


,
|x| ≥ R.

3|x|
Two observations are important:
(i) Outside the ball the gravitational potential coincides with that of a single point
centered at the origin whose mass equals the mass of the entire ball.
(ii) The potential u is radially symmetric. In particular: u|∂BR (0) is constant.
The first observation (and its generalization to radially symmetric mass densities) allows
to reduce celestial mechanics of stars and planets to the interaction of point masses. The
second observation has an impact that one can check: the gravitational force F = −∇u
exerted upon a unit point mass by the body BR (0) is always normal to the surface and
has everywhere on the surface the same value. In other words: wherever you travel on the
surface of a ball-shaped planet and carry a unit mass with you it always weighs the same
and when you let it drop then it falls perpendicularly to the surface of the planet (the
surface being a level set of the gravitational potential). Observation (ii) also triggers the
following mathematical question.
Are there bodies Ω ⊂ R3 different from the ball with the property that the constantdensity gravitational potential is constant everywhere on ∂Ω?
Surprisingly, the question was answered mathematically rigorously only in 2000 – more
than three hundred years after Newton. In fact it was answered for the generalization of
the Newtonian potential to n dimensions.
10
Symmetry for higher-order elliptic problems
Theorem 21 (Fraenkel [3]). Let Ω ⊂ Rn be a bounded open set and let ωn be the surface
measure of the unit-sphere in Rn . Consider
Z

1
1


dy,
n = 2,
log

2π Ω
|x − y|
Z
u(x) =
1
1



dy, n ≥ 3.
(n − 2)ωn Ω |x − y|n−2
If u is constant on ∂Ω then Ω is a ball.
Note that no boundary-regularity of Ω is needed. The goal of this chapter is to prove
yet another generalization of Fraenkel’s theorem, cf. [10]. Unlike in Fraenkel’s theorem we
need to a-priori restrict the class of open sets.
Theorem 22. Let Ω ⊂ Rn be a bounded convex domain. For α ≥ 2 consider
 Z
1


dy, n = α,
log

|x − y|
Ω
Z
(10)
u(x) =
1



dy, n 6= α.
n−α
Ω |x − y|
If u is constant on ∂Ω then Ω is a ball.
R
Remark 23. 1. The potential Ω |x−y|1n−α dy is called a Riesz-potential with unit density.
2. Lu, Zhu [8] have found another version of Theorem 22, where instead of the convexity of
Ω they assumed ∂Ω ∈ C 1 . Moreover, in the follow-up paper [5] the analysis was extended
to the case of Bessel potentials.
It is easy to see that the converse of both Theorem 21 and Theorem 22 hold: suppose
Ω = BR (0) ⊂ Rn is a ball centered at the origin. Then u is radially symmetric and hence
u is constant on ∂Ω.
Let us begin with the crucial observation that ∂Ω = u−1 {β} where β := u|∂Ω .
Lemma 24. Let u, Ω be as in Theorem 22 and β := u|∂Ω . If n ≥ α then u > β in Ω and
C
C
u < β in Ω . If n < α then u < β in Ω and u > β in Ω .
Remark 25. In Fraenkel’s case α = 2 this lemma was proven under the sole assumption
Ω open, bounded. In the case α > 2 we use convexity of Ω to establish this result.
Proof. We begin with the remark that Riesz-potentials are C l (Rn ) for all 0 < l < α. We
assume w.l.o.g that α > 2 (α = 2 corresponds to Fraenkel’s theorem) so that we have
u ∈ C 2 (Rn ). Now let us restrict to the case 2 < α ≤ n. The remaining case α > n
works in a very similar way. Note that ∆|x|α−n = (α − n)(α − 2)|x|α−n−2 < 0 and
1
∆ log |x|
= (2 − n)|x|−2 < 0. Therefore u is superharmonic on Rn and hence inside Ω the
function u is larger than the value β of u on ∂Ω. It remains to consider u outside Ω. We
show that the convexity of Ω implies that u has no local extremum outside Ω. Since either
u(x) → 0 (for α < n) or u(x) → −∞ (for α = n) as |x| → ∞ the absence of local extrema
implies that u is smaller than β outside Ω and we are done. So let x ∈ Rn \ Ω. By the
11
Symmetry for higher-order elliptic problems
convexity of Ω we can separate x from Ω through a hyperplane, i.e., there exists a unit
vector e ∈ Rn and a point z0 ∈ Rn \ Ω such that
(y − z0 ) · e < 0 < (x − z0 ) · e for all y ∈ Ω.
In particular (x − y) · e > 0 for all y ∈ Ω. Since
Z
(x − y) · e
∇u(x) · e = cα,n
dy
n−α+1
Ω |x − y|
and the integrand is strictly positive we see that u has no local extremum outside Ω.
As a consequence, one can rewrite the definition of u as follows. Let fH : R → R be the
Heaviside-function, i.e., fH (t) := 1 for t > 0 and fH (t) := 0 for t ≤ 0. Thus, Lemma 24
can be phrased as
fH (u − β), n ≥ α,
χΩ =
fH (β − u), n < α.
And now one can rewrite the








(11)
u(x) =







potential u as
Z
1
fH (u(y) − β) dy, n = α,
log
|x − y|
n
ZR
fH (u(y) − β)
dy,
n > α,
|x − y|n−α
n
ZR
fH (β − u(y))
dy,
n < α,
n−α
Rn |x − y|
i.e., one can consider u to be the solution of the nonlinear integral equation (11). Notice
that the unknown domain Ω has disappeared from (11). The main idea of the proof is to
show radial symmetry of the solution u of (11) via the method of moving planes – suitably
adapted for nonlinear integral equations.
Before we prove the main result let us make some connections to polyharmonic boundary
value problems.
Definition 26 (Polyharmonic Green function in Rn ). Let x, y ∈ Rn . Define
G(x, y) := km,n |x − y|2m−n , km,n =
Γ( n2 − m)
22m π n/2 Γ(m)
if n > 2m
or
n < 2m and n odd
and
n
2m−n
G(x, y) := km,n |x − y|
(−1)m− 2
1
log
, km,n = 2m−1 n/2
|x − y|
2
π Γ(m)
if n ≤ 2m and n even.
Then G is the Green function of (−∆)m on Rn .
R
Remark 27. Let Ω be open, bounded and U (x) = Ω G(x, y) dy for x ∈ Rn . Then U ∈
C 2m (Ω) with (−∆)m U = 1 on Ω and U ∈ C 2m (Rn \Ω) and (−∆U ) = 0 on Rn \Ω. Moreover
U ∈ C l (Rn ) for 0 < l < 2m and (−∆)m U = χΩ in Rn in the distributional sense.
12
Symmetry for higher-order elliptic problems
Hence, if we assume that α = 2m then in the case n < 2m, n odd, or n ≥ 2m the potential
u from Theorem 22 weakly satisfies
(−∆)m u =
1
km,n
χΩ in Rn .
If we combine this with Lemma 24 then the above equation can be restated as
fH (u − β) in Rn , n ≥ 2m,
1
m
(12)
(−∆) u =
km,n fH (β − u) in Rn , n < 2m, n odd,
i.e., u (weakly) satisfies the semilinear polyharmonic equation (12) on Rn where the
reference to the unknown domain Ω has disappeared. Therefore, Theorem 22 can be seen
as an example, where radial symmetry for positive solutions of semilinear polyharmonic
equations can be shown by the moving plane method. However, as the moving plane
method for second order elliptic differential equations is based on the maximum principle
which is unavailable for higher order equations, the way to proceed for (12) is to adapt the
moving plane method for the nonlinear integral equation (11).
After this comment we can now begin with the moving-plane proof of Theorem 22.
4.1. Asymptotic expansion and the role of the barycenter. Let us define the
barycenter of Ω by
Z
1
B(Ω) :=
y dy.
vol Ω Ω
Lemma 28. Let u, Ω be as in Theorem 22. Then

 vol Ω log 1 − |x|−2 x · B(Ω) + h(x)
|x|
(13)
u(x) =

α−n
vol Ω |x|
+ (n − α)|x|α−n−2 x · B(Ω) + h(x)
if n = α,
if n 6= α
where h satisfies |h(x)| ≤ C|x|α−n−2 , |∇h(x)| ≤ C|x|α−n−3 for some constant C > 0.
Proof. Let n 6= α. A direct application of Taylor’s theorem to g(t) := |x − ty|α−n yields
(14)
|x − y|α−n = |x|α−n − (α − n)|x|α−n−2 x · y + k(x, y)
where there exists a constant C > 0 and a radius R0 > 0 such that
(15)
|k(x, y)| ≤ C|x|α−n−2 ,
|∇x k(x, y)| ≤ C|x|α−n−3
for all |x| ≥ R0 , y ∈ Ω.
Here R0 > 0 is chosen such that Ω ⊂ BRR0 (0). The claim of the lemma follows from
integrating (14). The estimate for h(x) := Ω k(x, y) dy follows from (15). The proof for
n = α is similar.
The lemma has the following consequence: if u, Ω are like in Theorem 22 then for arbitrary
q ∈ Rn the function u(·+q) and the domain Ω−q also satisfy the assumption of Theorem 22.
Hence, w.l.o.g, we may assume that B(Ω) = 0 and, as a consequence, that the middle term
in the asymptotic expansion (13) vanishes.
13
Symmetry for higher-order elliptic problems
4.2. The moving plane method. For a point x = (x1 , x0 ) ∈ Rn let xλ = (2λ − x1 , x0 )
be the reflection of x at the hyperplane Tλ := {x ∈ Rn : x1 = λ}. Hence |xλ |2 − |x|2 =
4λ(λ − x1 ). Also define the half-space Hλ := {x ∈ Rn : x1 < λ} and note that ∂Hλ = Tλ .
On H λ define the function
wλ (x) := u(x) − u(xλ ) if α ≤ n
and
wλ (x) := −u(x) + u(xλ ) if α > n.
We will show that the function wλ satisfies
(16)
wλ (x) > 0 in Hλ ,
∂wλ
(x) < 0 on Tλ
∂x1
for all λ > 0. By continuity this implies u(x1 , x0 ) ≥ u(−x1 , x0 ) for all x ∈ Rn , x1 ≤ 0.
The reverse inequalities also hold by repeating the moving plane argument with the −x1 direction. Hence u(−x1 , x0 ) = u(x1 , x0 ) for all x ∈ Rn and moreover u is strictly monotone
in the positive x1 -direction. Repeating the moving-plane argument with an arbitrary unitdirection instead of the x1 -direction one obtains that the function u is radially symmetric
with respect to the origin and moreover radially strictly monotone. Together with the fact
that ∂Ω is a level-surface of the function u this implies that Ω must be a ball centered at
the origin. Thus, Theorem 22 is proved if we show (16) for all values of λ > 0.
Let us now outline the prototypical moving-plane procedure how to prove (16) for all values
of λ > 0. It is done in 5 steps:
(i)
(ii)
(iii)
(iv)
(v)
for every λ > 0 one has wλ (x) > 0 if |x| sufficiently large, cf. Lemma 29
wλ > 0 in Hλ if λ is sufficiently large, cf. Lemma 30
wλ ≥ 0 and λ > 0 implies wλ > 0, cf. Lemma 31
the set J := {λ > 0 : wλ > 0 in Hλ } is open, cf. Lemma 32
J = (0, ∞), cf. Lemma 33
We state the precise form of the lemmas next and directly give their proofs.
Lemma 29. For every λ > 0 there exists a value R(λ) > 0 such that for all x ∈ Hλ
with |x| ≥ R(λ) we have wλ (x) > 0. The function R(λ) can be chosen such that R(λ) is
continuous and non-increasing in λ.
Proof. According to the value of α the proof needs to be divided into several cases. Here
we only do the case 2 ≤ α < n. Recall that |x| < |xλ | for x ∈ Hλ and λ > 0. With h being
the function from Lemma 28 we have
u(x) − u(xλ ) = vol Ω(|x|α−n − |xλ |α−n ) + h(x) − h(xλ )
Assume first that |xλ |2 ≤ 2|x|2 . By convexity of the function s 7→ s
|x|α−n − |xλ |α−n >
where C1 := (n − α)2
Hence
(17)
α−n
2
α−n
2
for s > 0 we have
n − α λ α−n−2
|x |
4λ(λ − x1 ) ≥ C1 |x|α−n−2 λ(λ − x1 )
2
. By Lemma 28 we have |h(x) − h(xλ )| ≤ 2C|x|α−n−3 (λ − x1 ).
u(x) − u(xλ ) > |x|α−n−3 (λ − x1 ) vol ΩC1 |x|λ − 2C .
14
Symmetry for higher-order elliptic problems
Next assume that |xλ |2 ≥ 2|x|2 . Then
|x|α−n − |xλ |α−n ≥ |x|α−n (1 − 2
α−n
2
) =: C2 |x|α−n ,
where C2 > 0. Again by Lemma 28 |h(x) − h(xλ )| ≤ 2C|x|α−n−2 . Thus
u(x) − u(xλ ) ≥ |x|α−n vol ΩC2 −
(18)
2C .
|x|2
Hence, the right-hand sides both in (17) and (18) are positive provided
(
)
r
2C
2C
|x| > R(λ) := max
,
.
vol ΩC1 λ
vol ΩC2
Lemma 30. There exists λ∗ > 0 such that for all λ > λ∗ we have wλ (x) > 0 in Hλ .
Proof. We give the proof only in the case 2 ≤ α < n. Let R(λ) be the function defined
in Lemma 29. Let c1 := min|x|≤R(1) u(x). Hence c1 > 0, and since u(x) decays to 0 as
|x| → ∞ there exists a value λ∗ ≥ 1 such that |x| ≥ λ∗ implies u(x) ≤ c1 /2. Let now
λ > λ∗ . Consider x ∈ Hλ with |x| > R(1). For such x we have |x| > R(λ) and hence
u(x) > u(xλ ) by Lemma 29. Now consider x ∈ Hλ with |x| ≤ R(1). Since |xλ | ≥ λ > λ∗
we find u(x) ≥ c1 > u(xλ ), and the claim is proved.
The next lemma can be seen as an equivalent to the strong minimum principle.
Lemma 31. Let λ > 0. If wλ ≥ 0 in Hλ then wλ > 0 in Hλ and
∂wλ
(x)
∂x1
< 0 on Tλ .
Proof. As usual we only treat 2 ≤ α < n. Let us first derive the two identities (20) and
(21). By the nonlinear integral equation (11)
Z
Z
Z
fH (u(y) − β)
(19)
. . . dy
u(x) =
dy =
. . . dy +
n−α
Rn |x − y|
Hλ
Rn \Hλ
Z
fH (u(y) − β) fH (u(y λ ) − β)
=
+
dy.
n−α
|x − y λ |n−α
Hλ |x − y|
Therefore
Z
(20)
wλ (x) =
λ
fH (u(y) − β) − fH (u(y ) − β)
Hλ
1
1
−
dy
|x − y|n−α |xλ − y|n−α
|
{z
}
=:(∗)
∂u
λ
and for x ∈ Tλ we have ∂w
(x) = 2 ∂x
(x) so that by (19)
∂x1
1
(21)
Z
1 ∂wλ
∂
1
∂
1
λ
dy.
(x) =
fH (u(y) − β)
+fH (u(y ) − β)
2 ∂x1
∂x1 |x − y|n−α
∂x1 |x − y λ |n−α
Hλ
|
{z
}
=:(∗∗)
15
Symmetry for higher-order elliptic problems
What do we need to conclude the statement of the lemma? Suppose we knew for all
x, y ∈ Hλ that (∗) > 0 and for all x ∈ Tλ , y ∈ Hλ that (∗∗) < 0 as well as
1
∂
1
= 0.
(22)
+
∂x1 |x − y|n−α |x − y λ |n−α
Then the statement of the lemma would follow like
this: First, wλ ≥ 0 and 6≡ 0 by
λ
Lemma 29. Therefore fH (u(y) − β) − fH (u(y ) − β) ≥ 0 on Hλ and actually positive on a
subset of Hλ of positive measure. Hence the integral in (20) is strictly positive for x ∈ Hλ .
The same observation together with (∗∗) < 0 allows to further estimate (21) for x ∈ Tλ
Z
1 ∂wλ
∂
1
1
λ
(x) <
fH (u(y ) − β)
+
dy = 0 by (22)
2 ∂x1
∂x1 |x − y|n−α |x − y λ |n−α
Hλ
and the claim would be shown. So what remains to be done?
Let us look at (∗): for x, y ∈ Hλ one has
|xλ − y|2 = 4 (λ − x1 ) (λ − y1 ) +|x − y|2
| {z } | {z }
>0
>0
which implies (∗) > 0 since n > α.
Let us look at (∗∗): for x ∈ Tλ and y ∈ Hλ one has
∂
x1 − y1
λ − y1
|x − y| =
=
>0
∂x1
|x − y|
|x − y|
which implies (∗∗) > 0 since n > α.
Finally, let us look at (22): for fixed y ∈ Hλ , x0 ∈ Rn−1 the function x1 7→
is even around x1 = λ. Therefore its derivative vanishes at x1 = λ.
1
+ |x−yλ1 |n−α
|x−y|n−α
Lemma 32. The set J ⊂ (0, ∞) is open.
Proof. Assume that J is not open, i.e., there exists λ0 ∈ J which is an accumulation
point of J C . Thus, there exist sequences λk → λ0 as k → ∞ and xk ∈ Hλk such that
wλk (xk ) ≤ 0. Let R(λ) be the function from Lemma 29. Clearly |xk | ≤ R(λ0 /2), because
|xk | > R(λ0 /2) ≥ R(λk ) for large k would imply wλk (xk ) > 0 by Lemma 29, which cannot
hold. Hence (by extracting a subsequence) we have xk → x0 ∈ H(λ0 ). Since wλ0 > 0 in
∂u
λ
Hλ0 we must have x0 ∈ Tλ0 . Thus, we find 2 ∂x
(x0 ) = ∂w
(x0 ) < 0 due to Lemma 31. This
∂x1
1
λk
contradicts 0 ≥ wλ (xk ) = u(xk ) − u(xk ) for large k.
Lemma 33. The set J = (0, ∞).
Proof. The set J is non-empty by Lemma 30. Let (µ, ∞) be the largest open interval
contained in J. Then wµ ≥ 0 in Hµ . Assume for contradiction that µ > 0. By Lemma 31
we see that wµ > 0 in Hµ so that µ ∈ J. A contradiction is reached since J is open by
Lemma 32.
16
Symmetry for higher-order elliptic problems
5. An example of symmetry in overdetermined 4th order problem
The most famous second-order overdetermined problem was first studied by Serrin [11] and
Weinberger [12]: let Ω ⊂ Rn be a bounded C 2 -domain and u ∈ C 2 (Ω) a solution of
−∆u = 1 in Ω,
u = 0 on ∂Ω,
∂u
= const. < 0 on ∂Ω.
∂ν
Then Serrin, and independently Weinberger, proved that Ω is a ball and u is radially
symmetric. Bennett [1] studied the following 4th order overdetermined problem:
(23)
∆2 u = −1 in Ω,
∂u
u =
= 0 on ∂Ω,
∂ν
∆u = const. = c on ∂Ω.
Using Weinberger’s method Bennett proved the following result.
Theorem 34. Let Ω bepa bounded C 4,α -domain and suppose u ∈ C 4 (Ω) solves (23). Then
Ω is a ball of radius |c| (n2 + 2n) and u is given by
−1 (n + 2)nc)2 nc 2
1
4
+ r +
r
2n
4
2
4(n + 2)
where r denotes the distance of x to the center of the ball.
Recently, another proof of Serrin’s result was given in [2] by Brandolini, Nitsch, Salani,
Trombetti. Instead of giving Bennett’s original proof, let us show in the case c > 0 how
some of the ideas of [2] can be used to give an alternative proof of Bennett’s result.
In [2] the authors are making use of the so-called Newton inequality, which can be stated
as follows. Let A be a symmetric
nP× n matrix
and S1 (A) := trace(A) and S2 (A) :=
P
1
1
2
2
2
2
(trace(A) − |A| ) = 2 ( j ajj ) − i,j aij be the first two elementary symmetric func2
tions of the eigenvalues of A. Then
n−1
S1 (A)2 .
2n
Equality holds if and only if A is a multiple of the identity matrix. This result and more
on Newton’s inequalities can be found in [6]. Let us denote by S2ij (A) := ∂a∂ij S2 (A) =
trace(A)δij − aij . An important identity for S2 (D2 φ) and φ ∈ C 3 (Ω) is as follows
!
n
n
X
X
∂
∂φ
2S2 (D2 φ) =
S2ij (D2 φ)
∂xi j=1
∂xj
i=1
= div (∆φ Id −D2 φ)∇φ .
17
S2 (A) ≤
Symmetry for higher-order elliptic problems
Therefore, Newton’s inequality applied to A = D2 φ reads as follows
n−1
n−1
2
(S1 (D2 φ))2 =
(∆φ)2 .
(24)
div (∆φ Id −D φ)∇φ = 2S2 (D2 φ) ≤
n
n
Proof of Theorem 34: (assuming c > 0:)
Step 1: u < 0 in Ω. Suppose for contradiction that somewhere in Ω the function u takes a
nonnegative value. Then u attains is maximum at a point q ∈ Ω. At this point we have
∆u(q) ≤ 0. Since ∆u|∂Ω = c > 0 the function ∆u attains is minimum over Ω at a point
p ∈ Ω. Hence ∆2 u(p) ≥ 0 in contradiction to (23).
Step 2: If u solves (23) then the following integral relation holds:
Z
Z
−n 2
c vol(Ω).
(25)
u dx = − (∆u)2 dx =
n+4
Ω
Ω
We start with a differential identity which holds for arbitrary functions φ, ψ ∈ C 3 (Ω):
1
div − xψ∆φ − (∇φ · ∇ψ)x + (∇φ · x)∇ψ + (∇ψ · x)∇φ
2
n
1
1
= − ψ∆φ − ψ ∇(∆φ) · x + (2 − n)∇φ · ∇ψ + (∇φ · x)∆ψ + (∇ψ · x)∆φ.
2
2
2
We choose φ = u and ψ = ∆u, use the boundary condition ∇u = 0, ∆u = c on ∂Ω and
integrate to obtain
Z
n
n 2
− (∆u)2 + (2 − n)∇u · ∇(∆u) + (∇u · x)∆2 u dx
− c vol(Ω) =
2
2
ZΩ
n
= ( − 2)(∆u)2 − (∇u · x) dx
2
ZΩ
n
= ( − 2)(∆u)2 + nu dx
Ω 2
Z
n
= −( + 2) (∆u)2 dx,
2
Ω
where for the last equality we have used the differential equation tested with u. This
establishes the claim.
Step 3: If u solves (23) then the following integral relation holds:
Z
Z
1
c2
2
(26)
∇uD (∆u)∇(∆u) dx = −
(∆u)2 + vol(Ω).
4 Ω
4
Ω
Since
1
1
div(∇u|∇∆u|2 ) = ∆u|∇∆u|2 + ∇uD2 (∆u)∇(∆u)
2
2
1
1
= div (∆u)2 ∇(∆u) − (∆u)2 ∆2 u + ∇uD2 (∆u)∇(∆u)
4
4
the claim follows by integration using ∆2 u = −1 in Ω and ∇u = 0, ∆u = c on ∂Ω.
18
Symmetry for higher-order elliptic problems
Step 4: If we apply Newton’s inequality (24) to φ = ∆u we obtain
n−1 2 2 n−1
(∆ u) =
.
div ∆2 u Id −D2 (∆u) ∇(∆u) = 2S2 (D2 ∆u) ≤
n
n
Multiplication with u and using u ≤ 0 we obtain after integration
Z
Z
n−1
2
∇u∇(∆u) + ∇uD (∆u)∇(∆u) dx ≥
u dx.
n
Ω
Ω
One further integration by parts and (26) yields
Z
Z
n−1
5
c2
2
−
(∆u) + vol(Ω) ≥
u dx.
4 Ω
4
n
Ω
But (25) tells us that in fact we have equality in the above inequality. Tracing this back to
Newton’s inequality we find that D2 ∆u is a multiple of the identity matrix, i.e., D2 ∆u =
− n1 δij in Ω. If a ∈ Ω denotes the point where A := ∆u(a) = maxΩ ∆u then
1
∆u(x) =
(A − |x − a|2 ).
2n
If we now consider
the
boundary
conditions
for ∆u then we see that Ω is a ball centered
√
at a with radius A − 2cn. The boundary condition for u and ∂ν u then yield the stated
form of u.
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