Korn’s inequality
6 marca 2013
Note 1: Text below is written using the Einstein summation convention.
Note 2: Due to the mechanical interpretation of the problem and for simplicity we work in R3 , but all the results and techniques translate without problems
to Rd .
We return to study of a problem stated by Miss Zaremba last week. We shall
concentrate on the static case.
1
Classical formulation


Au = f
ui = Ui

σij nj = Fi
in Ω
on ΓU
on ΓF
(1)
As a reminder:
Au = −(aijkh εkh (u))
(2)
σij
(3)
= aijkh εkh (u)
1
(ui,j + uj,i )
εij (u) =
2
2
(4)
Variational formulation
Last week’s results (or simple calculation) yields the following variational problem equivalent to the classical one:
Z
a(u, v − u) = (f, v − u) +
F (v − u)dΓ
(5)
ΓF
∀v
vi = Ui
on ΓU
We still need to choose suitable space of functions. The natural choice is
V = {v : v = (v1 , v2 , v3 ), vi ∈ H 1 (Ω)} = (H 1 (Ω))3
which is a Hilbert space for the scalar product
Z
(u, v) = (ui , vi )H 1 (Ω) = (ui vi + ui,j vi,j )dx.
Ω
1
Then, we can define the set of functions admissable as v in (5).
Uad = {v ∈ V : vi = Ui on ΓU }
In order to avoid problems with ”compatibility” at the interface between ΓF
and ΓU we shall assume that
F ∈ (L2 (ΓF ))3 .
To summarize, we are looking for a solution of a problem of finding a function
u ∈ Uad satisfying
Z
a(u, v − u) = (f, v − u) +
F (v − u)dΓ
∀v ∈ Uad
ΓF
3
Korn’s inequality
We would like to prove existence of a solution. In order to do that we’d like to
establish whether the form a(u, v) is coercive. This can be proved using Korn’s
inequality.
Twierdzenie 1 (Korn’s inequality). Let Ω be a bounded set with regular boundary1 . There exists a constant C > 0 (dependent on Ω) such that
Z
Z
εij (v)εij (v)dx +
vi vi dx ­ ckvk2V
∀v ∈ V.
(6)
Ω
Ω
Let us note that the converse inequality is trivial - the left hand side invloves
only certain combinations of first derivatives (namely vi,j + vj,i ), while the right
hand side has all of them.
So, the statement is equivalent to saying that
Z
Z
εij (v)εij (v)dx +
Ω
1/2
vi vi dx
Ω
is a norm in V equivalent to k · kV .
Theorem 1 is a consequence of a different theorem.
Twierdzenie 2. Let Ω be an open set with regular boundary. Let v be a
distribution on Ω such that
v ∈ H −1 (Ω),
v,i ∈ H −1 (Ω)
Then
v ∈ L2 (Ω).
First let us prove theorem 1.
1 The
result is also valid for certain nonbounded sets.
2
∀i.
(7)
Dowód. Let E be the space of v ∈ (L2 (Ω))3 such that
εij (v) ∈ L2 (Ω)
∀i, j;
E is a Hilbert space for the norm
Z
Z
εij (v)εij (v)dx +
Ω
1/2
.
vi vi dx
Ω
Using the definition of εij we establisht that
∂
∂
∂
∂ 2 vi
=
εik (v) +
εij (v) −
εjk (v)
∂xj ∂xk
∂xj
∂xk
∂xi
If v ∈ E then εij ∈ L2 (Ω), therefore
∂εij (v)
∈ H −1 (Ω)
∂xk
so, using the established equality
∂ 2 vi
∈ H −1 (Ω)
∂xj ∂xk
∀i, j, k.
Applying theorem 2 to vi,k we see that
vi,k ∈ L2 (Ω)
∀i, k.
Therefore v ∈ (H 1 (Ω))3 . We have the algebraic equality E = (H 1 (Ω))3 2 .
Since the injection (H 1 (Ω))3 → E is continuous, we apply the closed graph
theorem (or, more precisely the bounded inverse theorem) to get an isomorphism
of Banach spaces.
Now we will prove theorem 2.
Dowód. We introduce the space
X(Ω) = {v ∈ H −1 (Ω) :
v,i ∈ H −1 (Ω)
∀i}
which is Hilbert for the norm
kvk2H −1 (Ω)
+
n
X
!1/2
kv,i k2H −1 (Ω)
i=1
We have to show that
X(Ω) = L2 (Ω)
The proof consists of several steps.
2I
mean that they are isomorphic as vector spaces.
3
(8)
1. Relation (8) is true for Ω = R3 . Indeed, using the definition of H −1 through
Fourier transform the assumptions are equivalent to
(1 + |ξ|2 )−1/2 v̂ ∈ L2 (R3ξ ),
(1 + |ξ|2 )−1/2 ξi v̂ ∈ L2 (R3ξ ),
therefore
Z
1 + |ξ|
2 −1
1+
Ω
n
X
!
ξ
2
|v|2 dξ < ∞
i=1
i.e.
v ∈ L2 (Ω).
2. It is sufficient to prove (8) for a half-space
Ω = {x : x3 > 0}.
Indeed, let α0 , α1 , ·, αN be such that
α0 ∈ D(Ω),
α1 ∈ D(Ω),
, i = 1, . . . , N,
N
X
αi = 1,
i=1
and supp(αi ) ⊂ local chart defining Γ.
Let’s note that ϕ ∈ D(Ω), v → ϕv maps X(Ω) to itself. Also, we can write
PN
v = i=0 αi v. We can consider α0 v as an element of X(R3 ) (simply extending with 0) thus, using step one α0 v ∈ L2 (Ω). We will have that result
if we show that αi v ∈ L2 (Ω). If we assume that Γ is a once continuously
differentiable manifold of dimension n − 1, then the image of αi v is in the
space X(Ω) with Ω = {x : x3 > 0}, which ends the proof of this step.
3. We introduce
H01 (0, ∞; L2 (R2 )) =
ϕ|ϕ,
dϕ
∈ L2 (0, ∞; L2 (R2 )), ϕ(x0 , 0) = 0 ,
dx3
where x0 = (x1 , x2 ).
H −1 (0, ∞; L2 (R2 )) = dual of H01 (0, ∞; L2 (R2 ))
where L2 (R2 ) is identified with its dual,
Y (Ω) = {v| v, dv/dx3 ∈ H −1 (0, ∞; L2 (R2 ))}
We will now show that Y (Ω) is dense in X(Ω). Indeed, let ρm be a regularizing sequence for D(R2x0 )3 .
For v ∈ X(Ω) we define vm = v( x0 ) ∗ ρm , or more precisely
Z
vm (x) =
v(x0 − y 0 , x3 )ρm (y 0 )dy 0 .
R2
3ρ
Then, as m → ∞, we have vm → v in X(Ω). Note that vm is in particular
contained in Y (Ω).
R
2
m
is a sequence of functions ­ 0 of D(R ),
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ρm = 1, suppρm ⊂ B(0, m ), m → 0
4. D(Ω) is dense in X(Ω) It suffices to show density in Y (Ω). As a simple
consequence of the Hahn-Banach theorem we only need to show, that for
every functional M which is zero on D(Ω) is zero on a whole space Y (Ω).
Let v → M (v) be a linear form continuous on Y (Ω) and therefore of the
form
Z ∞
M (v) =
[(f, v) + (g, dv/dx3 )]dx3 ,
0
f, g ∈ H01 (0, ∞; L2 (R2x0 ))
Assume that M = 0 on D(Ω). If f˜, g̃ denote the continuations of f and g
as 0 for x3 < 0, the assumptions is equivalent to
dg̃
=0
f˜ −
dx3
therefore dg̃/dx3 ∈ H 1 (−∞, +∞, L2 (R2x0 ), but then
Z ∞
Z ∞
(g, dv/dx3 )dx3 = −
(dg/dx3 , v)dx3
0
∀v ∈ Y (Ω)
0
so M = 0 on Y (Ω).
5. For v ∈ D(Ω) we set
P v(x) =
v(x) if x3 > 0
a1 v(x0 , −x3 ) + a2 v(x0 , −2x3 )
where a1 + a2 = 1, a1 + a2 /2 = −1. We verify that v → P v is continuous
in D(Ω), provided with the topology induced by X(Ω) → X(R3 ).
Using that, we see from the previous point that we can extend P to a linear,
continuous mapping of X(Ω) → X(R3 ) and such that P restricted to Ω
equals to v. Then, for v ∈ X(Ω) P v ∈ X(R3 ), therefore as a consequence
of step one P v ∈ L2 (R3 ) therefore v ∈ L2 (Ω).
It remains to prove the continuity of P .
∂u
x3 > 0
∂
∂x3
Pv =
∂v
∂v
0
0
−a
∂x3
1 ∂x3 (x , −x3 ) − 2a2 ∂x3 (x , −2x3 ),
x3 < 0
Setting ∂v/∂x3 = w, we introduce
w(x) if x3 > 0
Qw =
−a1 w(x0 , −x3 ) − 2a2 w(x0 , −2x3 )
We need to prove that P (resp. Q) is continuous on D(Ω), with the topology induced by H −1 (Ω) → H −1 (R3 ), thus by transposition, that t P (resp.
5
t
Q) is continuous as a functional H 1 (R3 ) → H01 (Ω). Now, for H 1 (R3 ), we
have:
t
1
P ϕ(x) = ϕ(x) + a1 ϕ(x0 , −x3 ) + a2 ϕ(x0 , −x/2),
2
t
Qϕ(x) = ϕ(x) − a1 ϕ(x0 , −x3 ) − a2 ϕ(x0 , −x3 /2)
Then t P ϕ(x0 , 0) =t Qϕ(x0 , 0) = 0, from which the result follows.
4
Proving coercivity
Assumptions as in theorem 1. Let ΓU ⊂ Γ and ΓU has positive measure. Let
V0 = {v|v ∈ (H 1 (Ω))3 ,
v = 0 on ΓU }
Then there exists α0 > 0 such that
a(v, v) ­ α0 kvk2V
∀ ∈ V0
Dowód. In general
a(v, v) = 0 ⇔ v ∈ R
where
R = {v|v(x) = a + b × xa, b ∈ R3 }
. Because ΓU has positive measure
v ∈ R ∩ V0 ⇒ v = 0
and consequently a(v, v) = 0, v ∈ V0 ⇔ v = 0.
Thus we see that a(v, v) is a norm on V0 and we have to show that it is
equivalent to || · ||V . To simplify notation, we set
Z
ε(v) =
εij (v)εij (v)dx.
Ω
We now from the previous lecture that
a(v, v) ­ αε(v),
so, we only need to find a c0 > 0 such that
Z
2
2
ε(v) ­ c0 |v| , |v| =
vi vi dx,
∀v ∈ V0 .
Ω
If we replace v by v|v|−1 we can assume that |v| = 1; then we have to prove that
ε(v) ­ c0 . We argue by contraditcion. There exists a sequence vn ∈ V0 with:
|vn | = 1,
ε(vn ) → 0.
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From Korn’s inequality we then have ||vn ||V ¬ const; then we can select a
subsequence such that
vn * v weakly in V.
But then
lim inf ε(vn ) ­ ε(v),
therefore ε(v) = 0, therefore v = 0. Using Sobolev embedding theorem we choose
a subsequence vn → 0 strongly in L2 (Ω)3 which is a contradiction.
7