Partial Differential Equations - Modelling and ... - ResearchGate

170 B. Maury
b(u ε ,u ε )/ε is bounded, so that b(u ε ,u ε )goesto0withε. Consequently, it
holds 0 ≤ b(z,z) ≤ lim inf b(u ε ,u ε ) = 0, which implies z ∈ K, sothatz = u.
To establish the strong character of the convergence, we show that u ε
converges toward u for the norm associated to a(·, ·), which is equivalent
to the original norm. As u ε converges weakly to u for this scalar product
(a(u ε ,v) → a(u, v) for any v ∈ V ), it is sufficient to establish the convergence
of |u ε | a
= a(u ε ,u ε ) 1/2 towards |u| a
. Firstly |u| a
≤ lim inf |u ε | a
, and the other
inequality comes from (3):
1
2 a(uε ,u ε ) −〈ϕ, u ε 〉≤ 1 a(u, u) −〈ϕ, u〉,
2
so that lim sup |u ε | a
≤|u| a
.
The proposition does not say anything about the rate of convergence, and
it can be very poor, as the following example illustrates.
Example 1. Consider I =]0, 1[, V = H 1 (I), and the problem which consists
in minimizing the functional
J(v) = 1 ∫
|u ′ | 2 ,
2 I
over K = {v ∈ V | v(x) = 0 a.e. in O = ]0, 1/2[}. The solution to that
problem is obviously u =max{0, 2(x − 1/2)}. Now let us denote by u ε the
minimum of the penalized functional
J ε = 1 ∫
|u ′ | 2 + 1 ∫
|u| 2 ,
2 I 2ε O
The solution to the penalized problem can be computed exactly:
( )
( ( )
x√ε x√ε
u ε =k ε (x)sh
in ]0, 1/2[ with k ε (x)= sh + 1 ( )) −1 x√ε
2 √ ε ch ,
and u ε affine in ]1/2, 1[, continuous at 1/2. This makes it possible to estimate
|u ε − u|, which turns out to behave like ε 1/4 .
Yet, in many situations, convergence can be shown to be of order 1, given
some assumptions are fulfilled. Let us introduce ξ ∈ V ′ as the unique linear
functional such that
a(u, v)+〈ξ,v〉 = 〈ϕ, v〉 ∀v ∈ V. (4)
Before stating the first order convergence result, we show here that the penalty
method provides an approximation of ξ.
Proposition 2. Let ξ ε ∈ V ′ be defined by
v ∈ V ↦−→ 〈ξ ε ,v〉 = 1 ε b(uε ,v).
Then ξ ε converges (strongly) to ξ in V ′ , at least as fast as u ε converges to u.