analysis of the dpg method for the poisson equation

52 Pages·2011·0.55 MB·English

ANALYSIS OF THE DPG METHOD
FOR THE POISSON EQUATION
a
Leszek Demkowicz
b
and Jay Gopalakrishnan
a
ICES, The University of Texas at Austin
b
Department of Mathematics, University of Florida
FE Rodeo, Texas A& M,
Februrary 26, 2011
Acknowledgments
Current and Past Support:
Department of Energy [National Nuclear Security Administration]
(Award Number [DE-FC52-08NA28615])
KAUST (collaborative research grant)
Boeing Company
Air Force
Feb 26, 2011 DPG Method 2 / 38
◮ Proof of well-posedness for the DPG formulation in
multidimensions.
◮ Numerical experiments.
Outline of Presentation
◮ Abstract B3 framework.
Feb 26, 2011 DPG Method 3 / 38
◮ Numerical experiments.
Outline of Presentation
◮ Abstract B3 framework.
◮ Proof of well-posedness for the DPG formulation in
multidimensions.
Feb 26, 2011 DPG Method 3 / 38
Outline of Presentation
◮ Abstract B3 framework.
◮ Proof of well-posedness for the DPG formulation in
multidimensions.
◮ Numerical experiments.
Feb 26, 2011 DPG Method 3 / 38
Petrov–Galerkin Method
with Optimal Test Functions
3
Abstract B Framework
(Repetitio Mater Studiorum Est)
Feb 26, 2011 DPG Method 4 / 38
Abstract Variational Problem
{
′
u ∈ U Bu = l B : U → V
⇔
b(u, v) = l(v) ∀v ∈ V < Bu, v >= b(u, v) ∀v ∈ V
where
◮ U, V are Hilbert spaces,
◮ b(u, v) is a continuous bilinear form on U × V ,
|b(u, v)| ≤ M‖u‖U ‖v‖V
that satisfies the inf-sup condition (⇔ B is bounded below),
inf sup |b(u, v)| =: γ > 0
‖u‖U=1 ‖v‖
V =1
◮ l ∈ V ′ represents the load and satisfies the compatibility
condition l(v) = 0, ∀v ∈ V0 where
V0 := {v ∈ V : b(u, v) = 0 ∀u ∈ U}
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Energy Norm
Banach Closed Range Theorem implies that there exists a unique
1
solution u that depends continuously upon the data, ‖u‖ ≤
γ‖l‖V ′.
The supremum in the inf-sup condition defines an equivalent,
problem-dependent energy (residual) norm,
‖u‖E := sup |b(u, v)| = ‖Bu‖V ′
‖v‖=1
For the energy norm, M = γ = 1. Recalling that the Riesz operator
′
is an isometry form V into V , we may characterize the energy norm
in an equivalent way as
‖u‖E = ‖vu‖V
where vu is the solution of the variational problem,
{
vu ∈ V
(vu, δv)V = b(u, δv) ∀δv ∈ V
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Optimal Test Functions
Select your favorite trial basis functions: ej, j = 1, . . . , N. For each
function ej, introduce a corresponding optimal test (basis) function
e¯j ∈ V that realizes the supremum,
|b(ej, e¯j)| = sup |b(ej, v)|
‖v‖V =1
i.e. it solves the variational problem,
{
e¯j ∈ V
(e¯j, δv)V = b(ej, δv) ∀δv ∈ V
Define the discrete test space as
V¯hp := span{e¯j, j = 1, . . . , N} ⊂ V . It follows from the construction
of the optimal test functions that the discrete inf-sup constant
inf sup |b(uhp, vhp)| = 1
‖uhp‖E=1 ‖v
hp‖=1
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The Best Approximation
Consequently, Babuˇska’s Theorem
M
‖u − uhp‖E ≤ inf ‖u − whp‖E
γhp whp∈Uhp
implies that
‖u − uhp‖E ≤ inf ‖u − whp‖E
whp∈Uhp
i.e., the method delivers the best approximation error in the energy
norm.
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