Problem
Regularized Problem
Semi-discrete Problem
Error Analysis of an Evolution Equation for
Microstructure
Richard Norton
Thursday 20/Friday 21 May 2010
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Error Analysis
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Consider the problem
∆ut + div(σ(∇u)) = 0
in Ω = (0, 1)2
u=0
on ∂Ω
u = u0
when t = 0
σ = DW where W is a double well potential such that
σ : R2 → R2 globally Lipschitz, and σ(p) · p ≥ c|p|2 − d for c > 0,
d ≥ 0.
For example, W (∇u) = 12 u21+1 (ux2 − 1)2 + 12 uy2 .
x
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
This problem is the H01 (Ω) gradient flow of I (u) :=
i.e.
ut = −∇I (u)
in H01 (Ω).
Error Analysis
R
Ω W (∇u)dx.
The direction chosen by the dynamics is the direction of steepest
descent.
In our example, the solution would like to satisfy ux = ±1, uy = 0.
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Semi-discrete Problem
Error Analysis
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Questions
I
What can we say about the long-time behaviour of solutions
and the appearance of microstructure?
I
Is microstructure a numerical artifact and can we approximate
the solution with FEM?
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
What can we prove?
Rewrite as
ut = F (u)
in H01 (Ω)
where F (u) := −∆−1 div(σ(∇u)). F : H01 (Ω) → H01 (Ω) is
Lipschitz
Standard theory and energy estimates (eg. Henry and Temam) ⇒
∃! solution
Z t
u(t) = u0 +
F (u(s))ds,
u∈
ut ∈
Z
0
1
∞
C ([0, ∞), H0 (Ω)) ∩ L ([0, ∞), H01 (Ω)),
C ((0, ∞), H01 (Ω)) ∩ L∞ ((0, ∞), H01 (Ω)) (both
Z
W (∇u(t))dx +
Ω
0
Lipschitz), and
t
k∇us (s)k2L2 (Ω) ds = const.
t ≥ 0,
⇒ ut ∈ L2 ((0, ∞), H01 (Ω)) and ut → 0 in H01 (Ω) as t → 0.
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
What can’t we prove?
I
No additional regularity or compactness (only H01 (Ω)) . This
limits what we can say about the long-time behaviour of
solutions. ku(t)kH 1 (Ω) ≤ C for all t ≥ 0 only implies weak
convergence.
I
Question
remains: Are all solutions minimising sequences for
R
W
(∇u(x))dx,
Ω
Ror are some solutions attracted to rest
points, for which Ω W (∇u(x))dx > 0?
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Numerics?
Can we approximate the solution with FEM?
Up to finite time it is possible to prove that uh → u in H01 (Ω) as
h → 0 but we do not get a rate of convergence (due to lack of
additional regularity).
Modify problem to create additional regularity...
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Regularized Problem
Consider the problem
∆ut − ∆2 u + div(σ(∇u)) = 0
in Ω = (0, 1)2
∆u = u = 0
u = u0 ∈
on ∂Ω
H01 (Ω)
Rewrite as
ut − ∆u = F (u)
F (u) := −∆−1 div(σ(∇u)).
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
in H01 (Ω)
at t = 0.
Error Analysis
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
In the gradient flow representation the additional term is bending
energy
Z
I (u) =
W (∇u) + (∆u)2 .
2
Ω
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Using same techniques as before we can prove similar results, e.g.
∃! solution u ∈ C ([0, ∞), H01 (Ω)) ∩ C ((0, ∞), V )
∆t
u(t) = e
Z
u0 +
t
e∆(t−s) F (u(s))ds
0
and . . .
(V := {v ∈ H01 (Ω) : ∆v ∈ H01 (Ω)}).
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
. . . it is also possible to prove that
kukH 1 ≤ C
t ∈ [0, ∞)
(
C −1/2 t −1/2 t ∈ (0, T ]
kukH 2 ≤
C −1/2
t ∈ (T , ∞)
(
Ct −1 t ∈ (0, T ]
kut kH 1 ≤
C
t ∈ (T , ∞)
... and other results eg. higher regularity, ∃ Lyapunov function,
ut → 0 in H 1 , ∃ compact attractor of finite dimension...
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Semi-discrete Problem
Vh ⊂ H01 (Ω). Define ∆h : Vh → Vh
(∆h uh , φh )L2 = −(∇uh , ∇φh )L2
∀uh , φh ∈ Vh .
Also define elliptic projection operator R = R(h) and L2
projection operator P = P(h) by
(∇(Ru − u), ∇φh )L2 = 0
∀φh ∈ Vh , u ∈ H01 (Ω)
(Pu − u, φh )L2 = 0
∀φh ∈ Vh , u ∈ H01 (Ω).
We have ∆h R = P∆. Assume
ku − RukL2 + hku − RukH 1 . hs kukH s
ku − PukL2 + hku − PukH 1 . hs kukH s
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
s = 1, 2.
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Apply the Galerkin method: Find uh ∈ C ([0, ∞), Vh ) such that
u = u0h := Ru0 at t = 0 and
uh,t − ∆h uh = Fh (uh )
in Vh for t > 0
where Fh (uh ) := RF (uh ) = −∆−1
h div(σ(∇uh ).
Same theory ⇒ ∃! solution uh such that
Z t
∆h t
uh (t) = e
u0h +
e∆h (t−s) Fh (uh (s))ds
0
Same regularity results as before (except H 2 norm replaced with
k∆h uh kL2 ).
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Error Analysis
To analyse the error we follow standard theory (eg. Larsson for
short time error), but pay particular attention to the dependence
on ,
to show that
kuh − ukH 1 . h−1/2 t −1/2
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
t ∈ (0, T ].
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Split the error into 2 parts
e = uh − u = uh − Ru + Ru
−u.
| {z } | {z }
θ(t)∈Vh
ρ(t)
ρ(t) is just the elliptic projection error,
kρ(t)kH 1 . hku(t)kH 2 . h−1/2 t −1/2
for t ∈ (0, T ].
θ(t) satisfies the following equation,
θt − ∆h θ = Fh (uh ) − Fh (u) + (P − R)(ut − F (u)).
∆h t
θh (t) = e
Z
θ0h +
t
h
i
e∆h (t−s) Fh (uh ) − Fh (u) + (P − R)(us − F (u)) ds
0
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
θ(t) = e∆h t θ0
Z t
+
e∆h (t−s) [Fh (uh (s)) − Fh (u(s))] ds
0
Z t
−
e∆h (t−s) (P − R)F (u(s))ds
+e
0
∆h t/2
Z
(P − R)u( 2t ) − e∆h t (P − R)u0
t
2
+
∆h e∆h (t−s) (P − R)u(s)ds
Z t0
+
e∆h (t−s) (P − R)us (s)ds
Error Analysis
(T1)
(T2)
(T3-4)
(T5)
(T6)
t
2
Now take k · kH 1 of each term separately and use our regularity
estimates to get the result.
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
kuh − u kH 1 . h−1/2 t −1/2
Error Analysis
t ∈ (0, T ].
If we choose h2−2δ < for some δ > 0 then
kuh − u kH 1 . hδ t −1/2
independent of .
Unfortunately we can only prove that up to finite time u → u in
H01 (Ω) as → 0. We do not have a rate of convergence for the
regularization error.
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Problem
Regularized Problem
Semi-discrete Problem
left: = 0.001, right: = 0.0001. h = 0.02.
Richard Norton
Error Analysis of an Evolution Equation for Microstructure
Error Analysis
Problem
Regularized Problem
Semi-discrete Problem
Error Analysis
Further Work
I
Long-time convergence result . . . convergence of attractors.
I
Error in L2 .
I
Regularization error? How does regularization change the
long-time behaviour of the PDE?
I
Existence of rest points for the original PDE.
Richard Norton
Error Analysis of an Evolution Equation for Microstructure