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A Short Course on Duality, Adjoint Operators, Green’sFunctions, andA PosterioriError Analysis
Donald J. Estep
estep@math.colostate.edu
Department of Mathematics
Colorado State University
Fort Collins, CO 80523
Duality, Adjoints, Green’s Functions – p. 1/304
Outline of this course
This course has four parts:
• Basic background and theory of the concepts of duality,adjoint operators, and Green’s functions
• Development of an approach to a posteriori error analysisand adaptive error control based on a generalization of theGreen’s function
• An application of these ideas to using the effective domainof influence in elliptic problems to form a decomposition of asolution
• Extension of the ideas to nonlinear problems.
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Duality, Adjoint Operators, and Green’s Functions
Duality, Adjoints, Green’s Functions – p. 3/304
The Green’s Function
In the classic problem, u solves
−∆u = f, x ∈ Ω,
u = 0, x ∈ ∂Ω,
where Ω is a domain in Rd with boundary ∂Ω.
The Green’s function φ satisfies
−∆φ(y;x) = δy(x), x ∈ Ω,
φ(y;x) = 0, x ∈ ∂Ω,
δy is the delta function at a point y ∈ Ω.
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The Green’s Function
Function representation formula:
u(y) =
∫
Ωδy(x)u(x) dx =
∫
Ω−∆φ(y;x)u(x) dx
=
∫
Ωφ(y;x) · −∆u(x) dx =
∫
Ωφ(y;x)f(x) dx.
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Background in linear algebra
X is a vector space with norm ‖ ‖ over the real numbers
An important property of the spaces depends on the notion of aCauchy sequence:
DefinitionA sequence xn in X is a Cauchy sequence if we can makethe distance between elements in the sequence arbitrarily smallby restricting the indices to be large.
More precisely, for every ε > 0 there is an N such that‖xn − xm‖ < ε for all n,m > N .
This is a computable criterion for checking convergence.
Duality, Adjoints, Green’s Functions – p. 6/304
Background in linear algebra
ExampleConsider the sequence 1/n∞n=1 in [0, 1]. It is a Cauchysequence since
∣
∣
∣
∣
1
n−
1
m
∣
∣
∣
∣
=
∣
∣
∣
∣
m− n
mn
∣
∣
∣
∣
≤ 2maxm,n
mn=
2
minm,n
can be made arbitrarily small by taking m and n large.
It converges to 0 which is in [0, 1].
Sequences that converge are Cauchy sequences. Cauchysequences do not necessarily converge.
ExampleThe sequence 1/n∞n=1 is a Cauchy sequence in (0, 1) but doesnot converge in (0, 1).
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Background in linear algebra
Spaces in which Cauchy sequences converge are greatlypreferred.
DefinitionA Banach space is a vector space with a norm such that everyCauchy sequence converges to a limit in the space.
We also say the space is complete .
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Background in linear algebra
ExampleRn with the norms defined for x = (x1, · · · , xn)
>,
‖x‖1 = |x1| + · · · + |xn|
‖x‖2 =(
|x1|2 + · · · + |xn|
2)1/2
‖x‖∞ = max |xi|.
are all Banach spaces.
We use ‖ ‖ = ‖ ‖2 unless noted otherwise.
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Background in linear algebra
Spaces of functions:
DefinitionFor an interval [a, b], the space of continuous functions isdenoted C([a, b]), where we take the maximum norm
‖f‖ = maxa≤x≤b
|f(x)|.
C1([a, b]) denotes the space of functions that have continuousfirst derivatives on [a, b], where we use the norm
‖f‖ = maxa≤x≤b
|f(x)| + maxa≤x≤b
|f ′(x)|.
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Background in linear algebra
Spaces of functions:
DefinitionFor 1 ≤ p ≤ ∞,
Lp(Ω) = f : f is measurable on Ω and ‖f‖p <∞
where for 1 ≤ p <∞,
‖f‖p =
(∫
Ω‖f‖p dx
)1/p
, 1 ≤ p <∞, and ‖f‖∞ = ess supΩ‖f‖.
L2 is particularly important.
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Background in linear algebra
TheoremThe Lp spaces and C([a, b]) are Banach spaces.
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Background in linear algebra
Consider two vector spaces X and Y with norms ‖ ‖X and ‖ ‖Y
DefinitionA map or operator L from X to Y is a rule or association thatassigns to each x in X a unique element y in Y .
DefinitionA map L : X → Y is linear if L(αx1 + βx2) = αL(x1) + βL(x2)for all numbers α, β and x1, x2 in X.
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Background in linear algebra
ExampleEvery linear map from R
m to Rn is obtained by multiplying
vectors in Rm by a n×m matrix, i.e., they have the form Ax,
where A is a n×m matrix.
ExampleDifferentiation is a linear map from C1([a, b]) to C([a, b]).
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Background in linear algebra
The maps we consider also have to behave continuously.
DefinitionA map L : X → Y is continuous if for every sequence xn inX that converges to a limit x in X, i.e., xn → x, we haveL(xn) → L(x).
ExampleLinear maps from R
m to Rn are continuous.
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Background in linear algebra
There is an equivalent property for linear maps.
DefinitionA linear map L : X → Y is bounded if there is a constant C > 0such that ‖Lx‖Y ≤ C‖x‖X for all x in X.
TheoremA linear map between normed vector spaces is continuous ifand only if it is bounded.
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Background in linear algebra
The set of all linear transformations between two vector spacesX and Y is a vector space.
DefinitionIf X and Y are normed vector spaces, we use L(X,Y ) to denotethe vector space of all bounded linear maps from X to Y .L(X,Y ) is a normed vector space under the operator norm
‖L‖ = sup‖x‖X=1
‖Lx‖Y = supx 6=0
‖Lx‖Y‖x‖X
. (1)
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Background in linear algebra
ExampleIf the linear transformation L is given by the n× n matrix A, then
‖L‖1 = ‖A‖1 = max1≤j≤n
n∑
i=1
|aij |
‖L‖2 = ‖A‖2 =√
σ(ATA)
‖L‖∞ = ‖A‖∞ = max1≤i≤n
n∑
j=1
|aij |
where σ(A>A) is the spectral radius of A>A.
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Background in linear algebra
TheoremIf X and Y are normed vector spaces and Y is complete, thenL(X,Y ) is complete.
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Linear functionals and dual spaces
The concept of duality starts with linear functionals.
DefinitionA linear functional on a vector space X is a linear map from Xto R.
ExampleLet v in R
n be fixed. The map F (x) = v · x = (x, v) is a linearfunctional on R
n.
ExampleConsider C([a, b]). Both I(f) =
∫ ba f(x) dx and F (f) = f(y) for
a ≤ y ≤ b are linear functionals.
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Linear functionals and dual spaces
A linear functional provides a “low dimensional snapshot” of avector.
ExampleIn the example above with F (x) = (x, v), consider v = ei, the ith
standard basis function. Then F (x) = xi where x = (x1, · · · , xn).
We can take v = (1, 1, · · · , 1)/n and compute the average of thecomponents of a given input vector.
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Linear functionals and dual spaces
Exampleδy gives a linear functional on sufficiently smooth, real valuedfunctions via
F (u) = u(y) =
∫
Ωδy(x)u(x) dx.
Another linear functional is the average value of an integrablefunction,
F (u) =1
vol. of Ω
∫
Ωu(x) dx.
We can view the formulas defining the Fourier coefficients of afunction as a set of linear functionals.
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Linear functionals and dual spaces
We are interested in the continuous linear functionals.
DefinitionIf X is a normed vector space, the space L(X,R) of boundedlinear functionals on X is called the dual space of or on or toX, and is denoted by X∗.
The dual space is a normed vector space under the dual normdefined for y ∈ X∗ as
‖y‖X∗ = supx∈X
‖x‖X=1
|y(x)| = supx∈Xx 6=0
|y(x)|
‖x‖.
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Linear functionals and dual spaces
ExampleConsider X = R
n. Every vector v in Rn is associated with a
linear functional Fv(·) = (·, v). This functional is clear boundedsince |(x, v)| ≤ ‖v‖ ‖x‖ (The “C” in the definition is ‖v‖).
A classic result in linear algebra is that all linear functionals onRn have this form, i.e., we can make the identification
(Rn)∗ ' Rn.
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Linear functionals and dual spaces
ExampleFor C([a, b]), consider I(f) =
∫ ba f(x) dx. It is easy to compute
‖I‖C([a,b])∗ = supf∈C([a,b])max |f |=1
∣
∣
∣
∣
∫ b
af(x) dx
∣
∣
∣
∣
by looking at a picture.
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Linear functionals and dual spaces
1
-1
ab
possible functions
Computing the dual norm of the integration functional.
The maximum value for I(f) is clearly given by f = 1 or f = −1,and we get ‖I‖C([a,b])∗ = b− a.
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Linear functionals and dual spaces
ExampleRecall Hölder’s inequality for f ∈ Lp(Ω) and g ∈ Lq(Ω) with1
p+
1
q= 1 for 1 ≤ p, q ≤ ∞ is
‖fg‖L1(Ω) ≤ ‖f‖Lp(Ω)‖g‖Lq(Ω).
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Linear functionals and dual spaces
Each g in Lq(Ω) is associated with a bounded linear functional
on Lp(Ω) when1
p+
1
q= 1 and 1 ≤ p, q ≤ ∞ by
F (f) =
∫
Ωg(x)f(x) dx.
We can “identify” (Lp)∗ with Lq when 1 < p, q <∞, The casesp = 1, q = ∞ and p = ∞, q = 1 are trickier.
The case L2 is special in that we can identify (L2)∗ with L2.
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Linear functionals and dual spaces
The dual space is the collection of “reasonable” possiblesamples.
An important characteristic of a dual space is how much we canreveal about a vector by considering samples in the dual space.
ExampleBy considering the set of n functionals corresponding to takingthe inner product with e1, · · · , en, we can “reconstruct” anygiven vector in R
n by looking at the functional values.
The question of whether or not we can “recover” a vector ucompletely by computing sufficiently many linear functionalsdepends heavily on properties of the underlying space.In practice, we will often be content with one or just a few“snapshots”.
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Linear functionals and dual spaces
We might wonder about the number of bounded linearfunctionals that exist on an arbitrary normed vector space.
The celebrated Hahn-Banach theorem says there is a greatabundance of them.
Let x0 6= 0 be a fixed element of Banach space X.
The set X0 = αx0 : α ∈ R forms a vector subspace of X.
The linear functional F (αx0) = α is defined on X0 and isbounded since |F (αx0)| = |α| = ‖αx0‖/‖x‖.
So, we have found a bounded linear functional on a subspace ofX.
Can we extend this to be defined on all of X?
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Linear functionals and dual spaces
Issues:• What does it mean to extend?• Does the norm increase upon extension?• What if there is an infinite number of extensions involved?
TheoremHahn-Banach Let X be a Banach space and X0 a subspace ofX. Suppose that F0(x) is a bounded linear functional defined onX0. There is a linear functional F defined on X such thatF (x) = F0(x) for x in X0 and ‖F‖ = ‖F0‖.
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Linear functionals and dual spaces
Dual spaces are connected to distribution theory and weakconvergence.
The dual space can be better behaved than the original normedvector space.
TheoremIf X is a normed vector space over R, then X∗ is a Banachspace (whether or not X is a Banach space).
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Linear functionals and dual spaces
Useful notation:
DefinitionIf x is in X and y is in X∗, we denote the value
y(x) =< x, y > .
This is called the bracket notation .
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Linear functionals and dual spaces
The norms on X and its dual X∗ are closely related. Recall thatif y ∈ X∗, then
‖y‖X∗ = supx∈X
‖x‖X=1
|y(x)| = supx∈Xx 6=0
|y(x)|
‖x‖.
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Linear functionals and dual spaces
DefinitionThe generalized Cauchy inequality is
|< x, y >| ≤ ‖x‖X ‖y‖X∗ , x ∈ X, y ∈ X∗.
TheoremIf X is a Banach space, then
‖x‖X = supy∈X∗
y 6=0
|y(x)|
‖y‖X∗
= supy∈X∗
‖y‖X∗=1
|y(x)|
for all x in X.
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Hilbert spaces and duality
ExampleRn with the standard Euclidean norm ‖ ‖ = ‖ ‖2 can be identified
with its dual space.
L2 can be identified with its dual space.
Both of these spaces are Hilbert spaces.
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Hilbert spaces and duality
One way to get a normed vector space is to place an innerproduct (i.e., dot product) on the space.
TheoremIf X has an inner product (x, y), then it is a normed vector spacewith norm ‖x‖ = (x, x)1/2 for x in X.
DefinitionA vector space with an inner product that is a Banach spacewith respect to the associated norm is called a Hilbert space.
ExampleRn with ‖ ‖ = ‖ ‖2 and L2 are both Hilbert spaces.
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Hilbert spaces and duality
If X is a Hilbert space with inner product (x, y), then each y ∈ Xdetermines a linear functional Fy(x) =< x, y >= (x, y) for x in X.
This functional is bounded by Cauchy’s inequality,|(x, y)| ≤ ‖x‖ ‖y‖.
The Riesz Representation theorem says this is the only kind oflinear functional on a Hilbert space.
TheoremRiesz Representation For every bounded linear functional Fon a Hilbert space X, there is a unique element y in X such that
F (x) = (x, y), all x ∈ X and ‖y‖X∗ = supx∈Xx 6=0
|F (x)|
‖x‖.
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Hilbert spaces and duality
A Hilbert space is isometric to its dual space.
DefinitionTwo normed vector spaces X and Y are isometric if there is alinear 1-1 and onto map L : X → Y such that ‖L(x)‖Y = ‖x‖Xfor all x in X.
We often replace the bracket notation and the generalizedCauchy inequality by the inner product and the “real” Cauchyinequality without comment.
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Hilbert spaces and duality
The Sobolev spaces are Hilbert spaces based on L2.
ExampleFor k = 1, 2, 3, · · · , we define Hk(Ω) to be the distributionfunctions in L2(Ω) whose partial derivatives of order k and lessare also distributions in L2(Ω).
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Hilbert spaces and duality
Definitionindex notation: For α = (α1, · · · , αn) with integer coefficients,we define
D =
(
∂
∂x1, · · · ,
∂
∂xn
)
, Dα =∂|α|
∂xα1
1 · · · ∂xαnn
with |α| = α1 + · · · + αn.
Hk(Ω) = u,Dαu ∈ L2(Ω), |α| ≤ k. The inner products andnorms are
(u, v)k =∑
|α|≤k
(Dαu,Dαv), ‖u‖k = (u, u)1/2k .
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Hilbert spaces and duality
The Riesz Representation theorem says that every linearfunctional on Hk has the form (u, v)k for some fixed v in Hk.
Defining the dual space to Hk runs into subtle difficulties due toa collision with the requirements for using distribution theory.
We define the dual spaces to a subspace.
Hk0 (Ω) =
u ∈ Hk(Ω) : u =∂u
∂n= · · · =
∂k−1u
∂nk−1= 0 on ∂Ω
,
where ∂/∂n denotes the normal derivative on the boundary ∂Ω.
We let H−k(Ω) is the set of all linear functionals on Hk0 (Ω).
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Adjoint operators - definition
We now explain how a linear transformation between twonormed vector spaces X and Y is naturally associated withanother linear transformation between Y ∗ and X∗.
This is the infamous adjoint operator.
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Adjoint operators - definition
Suppose L ∈ L(X,Y ) is a bounded linear transformation.
For each y∗ ∈ Y ∗,
y∗ L(x) = y∗(L(x)) =< Lx, y∗ >
assigns a number to each x ∈ X, hence defines a functionalF (x). F (x) is clearly linear.
It is also bounded since
|F (x)| = |y∗(L(x))| ≤ ‖y∗‖Y ∗‖L(x)‖Y ≤ ‖y∗‖Y ∗‖L‖ ‖x‖X .
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Adjoint operators - definition
There is an x∗ ∈ X∗ such that y∗(L(x)) = x∗(x) for all x ∈ X.
x∗ is unique.
Thus, to each y∗ ∈ Y ∗, we have assigned a unique x∗ ∈ X∗ andthus have defined a linear transformation L∗ : Y ∗ → X∗.
We can write these relations as
y∗(L(x)) = L∗y∗(x)
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Adjoint operators - definition
Using the bracket notation,
< L(x), y∗ >=< x,L∗(y∗) > x ∈ X, y∗ ∈ Y ∗.
DefinitionThis is called the bilinear identity .
It defines the adjoint operator L∗ : Y ∗ → X∗ associated to abounded linear transformation L : X → Y .
Note that we have defined the adjoint transformation viasampling by elements in the dual space.
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Adjoint operators - definition
ExampleLet X = R
m and Y = Rn, where we take the standard inner
product and norm.
By the Riesz Representation theorem, the bilinear identity forL ∈ L(Rm,Rn) reads
(Lx, y) = (x, L∗y), x ∈ Rm, y ∈ R
n.
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Adjoint operators - definition
There is a unique n×m matrix A so that if y = L(x) = Ax where
A =
a11 · · · a1m
......
an1 · · · anm
, y =
y1
...yn
, x =
x1
...xm
and
yi =m∑
j=1
aijxj , 1 ≤ i ≤ n.
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Adjoint operators - definition
For a linear functional y∗ = (y∗1, · · · , y∗n)
> ∈ Y ∗
L∗y∗(x) = y∗(L(x)) =
(y∗1, · · · , y∗n),
∑mj=1 a1jxj
...∑m
j=1 anjxj
=
m∑
j=1
y∗1a1jxj + · · ·m∑
j=1
y∗nanjxj
=m∑
j=1
(
n∑
i=1
y∗i aij)
xj
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Adjoint operators - definition
L∗(y∗) is given by the inner product with y = (y1, · · · , ym)>
where
yj =
n∑
i=1
y∗i aij .
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Adjoint operators - definition
The matrix A∗ of L∗ is
A∗ =
a∗11 · · · a∗1n...
...a∗m1 · · · a∗mn
=
a11 a21 · · · an1
......
a1m a2m · · · anm
= A>.
The bilinear identity is
y>Ax = x>A>y
using the fact that (x, y) = (y, x).
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Adjoint operators - definition
The following theorem is crucial.TheoremL∗ ∈ L(Y ∗, X∗) and ‖L∗‖ = ‖L‖.
proof
|L∗y∗(x)| ≤ ‖y∗‖Y ∗‖L‖‖x‖X .
Therefore,
‖L∗y∗‖X∗ = supx 6=0
|L∗y∗(x)|
‖x‖≤ ‖y∗‖Y ∗‖L‖,
which implies that L∗ ∈ L(Y ∗, X∗) and ‖L∗‖ ≤ ‖L‖.
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Adjoint operators - definition
To show the reverse inequality, we prove that
‖Lx‖Y ≤ ‖L∗‖ ‖x‖X , x ∈ X.
The bilinear identity implies that
|y∗(Lx)| ≤ ‖L∗y∗‖X∗‖x‖X ≤ ‖L∗‖‖y∗‖Y ∗‖x‖X
and
supy∗ 6=0
|y∗(Lx)|
‖y∗‖Y ∗
≤ ‖L∗‖ ‖x‖X , x ∈ X.
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Adjoint operators - definition
Properties of the adjoint operator:
TheoremLet X, Y , and Z be normed linear spaces. Then,
0∗ = 0
(L1 + L2)∗ = L∗
1 + L∗2, all L1, L2 ∈ L(X,Y )
(αL)∗ = αL∗, all α ∈ R, L ∈ L(X,Y )
If L2 ∈ L(X,Y ) and L1 ∈ L(Y,Z), then L1L2 ∈ L(X,Z),(L1L2)
∗ ∈ L(Z∗, X∗), and
(L1L2)∗ = L∗
2L∗1.
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Adjoint operators - definition
We are often deal with linear operators that are not defined onthe entire space.
ExampleConsider D = d/dx on X = C([0, 1]). This linear map is onlydefined on the subspace C1([0, 1]).
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Adjoint operators - definition
DefinitionLet X and Y be normed vector spaces. A map L that assigns toeach x in a subset D(L) of X a unique element y in Y is called amap or operator with domain D(L).
L is linear if (1) D(L) is a vector subspace of X and (2)L(αx1 + βx2) = αL(x1) + βL(x2) for all α, β ∈ R andx1, x2 ∈ D(L).
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Adjoint operators - definition
We define the dual of a linear operator by examining its behavioron its domain.
We wantL∗y∗(x) = y∗(Lx) all x ∈ D(L).
We say that y∗ ∈ D(L∗) if there is an x∗ ∈ X∗ such that
x∗(x) = y∗(Lx), all x ∈ D(L).
The existence of x∗ is no longer automatic.
When x∗ exists, we define L∗y∗ = x∗.
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Adjoint operators - definition
For this to work, x∗ must be unique.
In other words, x∗(x) = 0 for all x ∈ D(L) should imply x∗ = 0.
This depends on the “size” of D(L).
DefinitionA subspace A of a normed linear space X is dense if everypoint in X is either in A or the limit of a sequence of points in A.
ExampleThe rational numbers are dense in the real numbers and thepolynomials are dense in C([a, b]).
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Adjoint operators - definition
We can define L∗ for a linear operator L : X → Y provided D(L)is dense in X.
We define D(L∗) to be those y∗ ∈ Y ∗ for which there is anx∗ ∈ X∗ with x∗(x) = y∗(Lx) for all x ∈ X.
This x∗ is unique and L∗y∗ = x∗.
The Hahn-Banach theorem implies that if there is a C such that|y∗(Lx)| ≤ C‖x‖ for all x ∈ D(L), then y∗ ∈ D(L∗).
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Adjoint operators - motivation
Common problem: Given normed vector spaces X and Y , anoperator L(X,Y ), and b ∈ Y , find x ∈ X such that
Lx = b.
We explain the role of the adjoint in solving this kind of problems.
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Adjoint operators - motivation
DefinitionThe set of b for which there is a solution is called the range ,R(L), of L.
The set of x for which L(x) = 0 is called the null space , N (L),of L.
0 is always in N (L).
If it is the only element in N (L), then Lx = b can have at mostone solution.
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Adjoint operators - motivation
TheoremN (L) is a subspace of X and R(L) is a subspace of Y .
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Adjoint operators - motivation
If y ∈ R(L), there is an x with Lx = y.
For y∗ ∈ Y ∗,y∗(Lx) = y∗(y).
By the definition of the adjoint,
L∗y∗(x) = y∗(y).
If y∗ ∈ N (L∗), then y∗(y) = 0.
A necessary condition that y ∈ R(L) is that y∗(y) = 0 for ally∗ ∈ N (L∗).
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Adjoint operators - motivation
Is this sufficient?
We require just one condition.
DefinitionA subset A of a normed vector space X is closed if everysequence xn in A that has a limit in X has its limit in A.
TheoremLet X and Y be normed linear spaces and L ∈ L(X,Y ). Anecessary condition that y ∈ R(L) is y∗(y) = 0 for ally∗ ∈ N (L∗). This is a sufficient condition if R is closed in Y .
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Adjoint operators - motivation
ExampleSuppose that L ∈ L(X,Y ) is associated with the n×m matrixA, i.e., L(x) = Ax.
The necessary and sufficient condition for the solvability ofAx = b is that b is orthogonal to all linearly independentsolutions of AT y = 0.
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Adjoint operators - motivation
ExampleIn the case X is a Hilbert space and L ∈ L(X,Y ), thennecessarily R(L∗) ⊂ N (L)⊥, where S⊥ is the subspace ofvectors that are orthogonal to a subspace S.
If R(L∗) is “large”, then N (L)⊥ must be “large” and N (L) mustbe “small”.
The existence of sufficiently many solutions of thehomogeneous adjoint equation implies there is at most onesolution of Lx = b for a given b.
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Adjoint operators - motivation
ExampleThe setting of general partial differential equations:
TheoremHolmgren Uniqueness The generalized initial value problemconsisting of the equation
L(u) =∑
|α|≤m
Aα(x)Dαu = f(x), x ∈ Rn,
where Aα and f are analytic functions, together with the data
Dβu(x) = gβ(x), |β| ≤ m− 1, x ∈ S
given on an analytic noncharacteristic surface S, has at mostone solution in a neighborhood of S.
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Adjoint operators - motivation
The adjoint has an important role in the solution of a generaln×m system Ax = b, where A is a n×m matrix, x ∈ R
m, andb ∈ R
n.
The reason to dwell on such problems is that differentialoperators do not tend to be “square”.
Exampley′′ = d2y/dx2 requires two boundary conditions to define aproblem for the associated differential equation that has aunique solution.
We may want to study the differential operator without anyboundary conditions, or more or less than two conditions.
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Adjoint operators - motivation
ExampleDivergence,
divu =∂u1
∂x1+ · · · +
∂un∂xn
,
associates a scalar with a given vector function, and theassociated differential equation is very “under-determined”.
The gradient,
gradu =
(
∂u1
∂x1, · · · ,
∂un∂xn
)
,
associates a vector field with a given scalar function, and theassociated equations are very “over-determined”.
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Adjoint operators - motivation
Consider the n×m system Ax = b, where A is a n×m matrix,x ∈ R
m, and b ∈ Rn.
We enlarge the system by adding the adjoint m× n systemA>y = c, where y and c are independent of x and b.
The new problem is an (n+m) × (n+m) symmetric problem
Sz = d,
with
z =
(
y
x
)
, d =
(
b
c
)
, S =
(
0 A
A> 0
)
.
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Adjoint operators - motivation
Since S is symmetric, it is diagonalizable with eigenvaluessatisfying
Av = λu
A∗u = λvor
AA>u = λ2u
A>Av = λ2v
The eigenvalues of S are the singular values of A and last twoequations give the left and right singular vectors of A.
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Adjoint operators - motivation
Consequences:• The theorem on the adjoint condition for solvability falls out
right away.• It yields a “natural” definition of a solution or gives
conditions for a solution to exist in the over-determined andunder-determined cases.
• It also gives a way of determining the condition of thesolution process.
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Adjoint operators - motivation
One interesting observation is that there is a reciprocalrelationship between the degree of over/under-determination ofthe original system and the adjoint system.
The more over-determined the original system, the moreunder-determined the adjoint system, and so forth.
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Adjoint operators - motivation
ExampleConsider 2x1 + x2 = 4, where L : R
2 → R. L∗ : R → R2 is given
by L∗ =
(
2
1
)
. The extended system is
0 2 1
2 0 0
1 0 0
y1
x1
x2
=
4
c1
c2
,
from which we see that 2c1 = c2 is required in order to have asolution.
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Adjoint operators - motivation
ExampleOn the other hand, if the problem is
2x1 + x2 = 4
x2 = 3,
with L : R2 → R
2, then there is a unique solution. The extendedsystem is
0 0 2 1
0 0 0 1
2 0 0 0
1 1 0 0
y1
y2
x1
x2
=
c1
c2
4
3
,
where we can specify any values for c1, c2.
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Adjoint operators - motivation
In the under-determined case, we can eliminate the deficiencyby posing the method of solution
AA>y = b
x = A>y.or
L(L∗(y)) = b
x = L∗(y).
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Adjoint operators - motivation
This works for differential operators
ExampleConsider the under-determined problem
divF = ρ.
The adjoint to div is -grad modulo boundary conditions.
If F = gradu, where u is subject to the boundary conditionu(′′∞′′) = 0, then we obtain the “square”, well-determinedproblem
div gradu = ∆u = −ρ,
which has a unique solution because of the boundary condition.
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Adjoint operators - motivation
ExampleConsider
Ax = b
A>φ = ei
where A is a n× n invertible matrix.
There are no constraints on the data for the adjoint problem.
We have specified the ith standard basis vector of Rn.
xi = (x, ei) = (x,A>φ) = (Ax, φ) = (b, φ).
y is the discrete Green’s vector associated to Ax = b.
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Adjoint operators - computation
The computation of the adjoint to a general differential operatoris not easy.
A tedious, formal approach:
1. Take the problem for the linear operator, including anyboundary and/or initial conditions imposed with theoperator, and discretize it.
2. Extract the matrix from the resulting discrete system, andcompute its transpose.
3. Let the discretization parameter tend to its limit anddetermine the differential operator that is approached by thetransposed matrix.
Determination of the adjoint of a differential operator is heavilyinfluenced by the boundary and/or initial conditions, as well asthe underlying spaces.
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Adjoint operators - computation
ExampleA standard difference approximation of
−u′′(x) = f(x), 0 < x < 1,
u(0) = 0, u(1) = 0,
yields the matrix
1
h2
2 −1 0
−1 2 −1. . . . . . . . .
−1 2 −1
0 −1 2
,
The differential operator is self-adjoint.
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Adjoint operators - computation
ExampleIf we change the boundary conditions to
−u′′(x) = f(x), 0 < x < 1,
u(0) = 0, u′(0) = 0,
we get a triangular matrix after discretization.
The adjoint corresponds to a problem
−v′′(x) = g(x), 0 < x < 1,
v(1) = 0, v′(1) = 0,
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Adjoint operators - computation
Example
u′ = f(x), 0 < x < 1,
u(0) = 0⇒
−v′ = g(x), 1 > x > 0,
v(1) = 0,
Example
−u′′ = f(x), 0 < x < 1,
no boundary conditions
⇒
−v′′ = g(x), 0 < x < 1,
v(0) = v′(0) = v(1) = v′(1) = 0.
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Adjoint operators - computation
In the case of L2, we can use the bilinear identity
< Lu, v∗ >=< u,L∗v∗ > all u ∈ X, v∗ ∈ Y ∗,
which is
(Lu, v) = (u, L∗v) all suitable u, v ∈ L2(Ω).
DefinitionWe say that we are evaluating the bilinear identity when wecompute
< Lu, v∗ > − < u,L∗v∗ >= (Lu, v) − (u, L∗v)
for some suitable functions u and v.
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Adjoint operators - computation
• Since we are considering differential operators, these willnot be defined on all of L2(Ω), but only a subset ofsufficiently smooth functions.
• The L2 inner product involves integration over the domain,and we can interpret the process producing the bilinearidentity as a succession of integration by parts. This yieldsterms that involve the integrals over the boundary of Ω ofthe functions involved as well as certain derivatives.
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Adjoint operators - computation
Computing the adjoint using the bilinear identity proceeds in twostages.
First, we compute a formal adjoint neglecting all boundary termsby assuming that the functions involved have compact supportinside Ω.
DefinitionA function on a domain Ω has compact support in Ω if itvanishes identically outside a bounded set contained inside Ω.
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Adjoint operators - computation
1. Take the differential operator applied to a smooth functionwith compact support and multiply by a smooth test functionwith compact support
2. Integrate over the domain
3. Integrate by parts to move all derivatives onto the testfunction while ignoring boundary terms.
Functions that have compact support are identically zeroanywhere near the boundary and any boundary terms arisingfrom integration by parts will vanish.
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Adjoint operators - computation
DefinitionLet L be a differential operator. The formal adjoint L∗ is thedifferential operator that satisfies
(Lu, v) = (u, L∗v)
(∫
ΩLu · v dx =
∫
Ωu · L∗v dx
)
for all sufficiently smooth u and v with compact support in Ω.
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Adjoint operators - computation
ExampleConsider
Lu(x) = −d
dx
(
a(x)d
dxu(x)
)
+d
dx(b(x)u(x))
on [0, 1]. Integration by parts neglecting boundary terms gives
−
∫ 1
0
d
dx
(
a(x)d
dxu(x)
)
v(x) dx
=
∫ 1
0a(x)
d
dxu(x)
d
dxv(x) dx− a(x)
d
dxu(x)v(x)
∣
∣
∣
∣
1
0
= −
∫ 1
0u(x)
d
dx
(
a(x)d
dxv(x)
)
dx+ u(x)a(x)d
dxv(x)
∣
∣
∣
∣
1
0
,
Duality, Adjoints, Green’s Functions – p. 88/304
Adjoint operators - computation
∫ 1
0
d
dx(b(x)u(x))v(x) dx = −
∫ 1
0u(x)b(x)
d
dxv(x) dx+b(x)u(x)v(x)
∣
∣
∣
∣
1
0
,
where all of the boundary terms vanish.
Therefore,
L∗v = −d
dx
(
a(x)d
dxv(x)
)
− b(x)d
dx(v(x)).
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Adjoint operators - computation
In higher space dimensions, we use the divergence theorem.
ExampleA general linear second order differential operator L in R
n canbe written
L(u) =n∑
i=1
n∑
j=1
aij∂2u
∂xi∂xj+
n∑
i=1
bi∂u
∂xi+ cu,
where aij, bi, and c are functions of x1, x2, · · · , xn. Then,
L∗(u) =
n∑
i=1
n∑
j=1
∂2(aijv)
∂xi∂xj−
n∑
i=1
∂(biv)
∂xi+ cv.
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Adjoint operators - computation
It can be verified directly that
vL(u) − uL∗(v) =
n∑
i=1
∂pi∂xi
,
where
pi =
n∑
j=1
(
aijv∂u
∂xj− u
∂(aijv)
∂xj
)
+ biuv.
The divergence theorem yields∫
Ω(vL(u) − uL∗(v)) dx =
∫
∂Ωp · nds = 0,
where p = (p1, · · · , pn) and n is the outward normal in ∂Ω.
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Adjoint operators - computation
A typical term,
va11∂2u
∂x21
= va11∂
∂x1
(
∂u
∂x1
)
=∂
∂x1
(
va11∂u
∂x1
)
−∂(a11v)
∂x1
∂u
∂x1
=∂
∂x1
(
va11∂u
∂x1
)
−∂
∂x1
(
u∂(a11v)
∂x1
)
+ u∂2(a11v)
∂x21
yielding
va11∂2u
∂x21
− u∂2(a11v)
∂x21
=∂
∂x1
(
a11v∂u
∂x1− u
∂(a11v)
∂x1
)
.
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Adjoint operators - computation
ExampleLet L be a differential operator of order 2p of the form
Lu =∑
|α|,|β|≤p
(−1)|α|Dα(
aαβ(x)Dβu)
,
thenL∗v =
∑
|α|,|β|≤p
(−1)|α|Dα(
aβα(x)Dβv)
,
and L is elliptic if and only if L∗ is elliptic.
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Adjoint operators - computation
Some special cases.
grad∗ = −div
div∗ = −grad
curl∗ = curl
Lu =∑
|α|≤p
aα(x)Dαu
thenL∗v =
∑
|α|≤p
(−1)|α|Dα(aα(x)v(x)).
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Adjoint operators - computation
ExampleIf
Lu = ρutt −∇ · (a∇u) + bu
then L∗ = L.
IfLu = ut −∇ · (a∇u) + bu
thenL∗v = −vt −∇ · (a∇v) + bv
where time runs “backwards”.
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Adjoint operators - computation
This procedure also works for systems
ExampleLet
L(~u) = A~ux +B~uy + C~u,
where A,B, and C are n× n matrices, then
L∗(~v) = (−A>~v)x − (B>~v)y + C>v,
so that~v>L~u− ~u>L∗~v = ∇ · (~v>A~u,~v>B~u).
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Adjoint operators - computation
In the second stage of computing the adjoint, we deal withboundary conditions by removing the assumption that thefunctions involved have compact support.
Integration by parts yields additional terms involving integralsover the boundary of the functions involved and their derivatives.
We want to determine boundary conditions such that the bilinearidentity still holds, e.g., such that any boundary terms that arisevanish.
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Adjoint operators - computation
It turns out that the form of the boundary conditions imposed inthe problem for L are important, but the values given for theseconditions are not.
If the boundary conditions are not homogeneous, we make themso for the purpose of determining the adjoint.
DefinitionThe adjoint boundary conditions posed on the formal adjointoperator are the minimal conditions required to make theboundary terms that appear when evaluating the bilinear identityfor general smooth functions vanish.
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Adjoint operators - computation
Some of the boundary terms that appear when evaluating thebilinear identity will vanish because of the boundary conditionsimposed in the original problem.
The point of this assumption is to make the formal adjoint serveas the true adjoint by pairing it with the correct boundaryconditions.
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Adjoint operators - computation
This definition can be made completely precise. Issues involved:
• Placing conditions on the differential operator L so thatevaluating the bilinear identity for general smooth functionsresults in expressions involving only values on the boundary.
• Making precise the meaning of “minimal conditions” neededfor the adjoint problem, and proving these always exist.
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Adjoint operators - computation
ExampleConsider Newton’s equation of motion s′′(x) = f(x) with x =“time”, normalized with mass 1. First, suppose we assumes(0) = s′(0) = 0, and 0 < x < 1. We have
s′′v − sv′′ =d
dx(vs′ − sv′)
and∫ 1
0(s′′v − sv′′) dx = (vs′ − sv′)
∣
∣
1
0.
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Adjoint operators - computation
The boundary conditions imply the contributions at x = 0 vanish,while at x = 1 we have
v(1)s′(1) − v′(1)s(1).
To insure this vanishes, we must have v(1) = v′(1) = 0. Theseare the adjoint boundary conditions.
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Adjoint operators - computation
ExampleSuppose instead we impose conditions such that at the massreturns to the origin with zero speed at x = 1.
This gives four boundary conditionss(0) = s′(0) = s(1) = s′(1) = 0 on the original problem.
In this situation, all of the boundary terms are zero and noboundary conditions will be imposed on the adjoint.
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Adjoint operators - computation
Solving the over-determined problem.
ExampleBased on the discussion above, we require the data f to beorthogonal to the solution of the adjoint problem v′′ = 0, which isv = a+ bx.
Hence, f must be orthogonal in L2(0, 1) to 1 and x.
Assume for example that f(x) = a+ bx+ cx2. It is easy to seethat (f, 1) = 0 and (f, x) = 0 forces a = c/6 and b = −c.
Choosing c = 1 for example, means that f(x) = 1/6 − x+ x2.
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Adjoint operators - computation
We solve the forward problem by integrating twice and using theboundary conditions at x = 0 to get a formula for the solution.
The resulting solution satisfies the conditions at x = 1 as well.
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Adjoint operators - computation
ExampleSince
∫
Ω(u∆v − v∆u) dx =
∫
∂Ω
(
u∂v
∂n− v
∂u
∂n
)
ds,
the Dirichlet and Neumann boundary value problems for theLaplacian are their own adjoints.
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Adjoint operators - computation
ExampleLet Ω ⊂ R
2 be bounded with a smooth boundary and let s =arclength along the boundary. Consider
−∆u = f, x ∈ Ω,∂u∂n + ∂u
∂s = 0, x ∈ ∂Ω.
Since∫
Ω(u∆v − v∆u) dx =
∫
∂Ω
(
u
(
∂v
∂n−∂v
∂s
)
− v
(
∂u
∂n+∂u
∂s
))
ds,
the adjoint problem is
−∆v = g, x ∈ Ω,∂v∂n − ∂v
∂s = 0, x ∈ ∂Ω.
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Green’s functions
For simplicity, we consider
Lu = f, x ∈ Ω,
suitable b.c. and i.v., x ∈ ∂Ω,
where L is a linear differential operator.
We specify the correct boundary and/or initial conditions sothere is a unique solution.
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Green’s functions
DefinitionThe Green’s function satisfies
L∗φ(y, x) = δy(x), x ∈ Ω,
adjoint b.c. and i.v., x ∈ ∂Ω,
where L∗ is the formal adjoint of L.
We are solving the extended system,
Lu = f, x ∈ Ω,
suitable b.c. and i.v., x ∈ ∂Ω,
L∗φ(y, x) = δy(x), x ∈ Ω,
adjoint b.c. and i.v., x ∈ ∂Ω.
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Green’s functions
Based on the bilinear identity, we obtain a representationformula for the value of the solution of the original problem at apoint y ∈ Ω,
u(y) = (u, δy) = (y, L∗φ) = (Lu, φ) = (f, φ).
The imposition of the adjoint boundary conditions is key here.
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Green’s functions
ExampleFor
−∆u = f, x ∈ Ω,
u = 0, x ∈ ∂Ω,
φ solves
−∆φ(y;x) = δy(x), x ∈ Ω,
φ(y;x) = 0, x ∈ ∂Ω,
and the bilinear identity reads
u(y) =
∫
Ωf(x)φ(y;x) dx.
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Green’s functions
Duality, Adjoints, Green’s Functions – p. 112/304
Green’s functions
ExampleFor
ut − ∆u = f, x ∈ Ω, 0 < t ≤ T,
u(t, x) = 0, x∂Ω, 0 < t ≤ T,
u(0, x) = 0, x ∈ Ω,
φ solves
−φt − ∆φ = δ(s,y)(t, x), x ∈ Ω, T > t ≥ 0,
φ(t, x) = 0, x∂Ω, T > t ≥ 0,
φ(T, x) = 0, x ∈ Ω,
and the bilinear identity reads
u(s, y) =
∫ T
0
∫
Ωf(t, x)φ(s, y; t, x) dxdt.
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Green’s functions
Motivations:• The Green’s function is defined without reference to the
particular data f that determines a particular solution. TheGreen’s function depends on the operator and its properties.
• The Green’s function generally has special propertiesarising from the properties of the delta function, such aslocalized support and symmetry.
• The Green’s function gives a “low-dimensional” snapshot ofthe solution.
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Green’s functions
ExampleThe Green’s function for the Dirichlet problem for the LaplacianL = −∆ on the ball Ω of radius r centered at the origin in R
3 is
φ(y;x) =1
4π×
|y − x|−1 − r|y|−1∣
∣
r2y|y|2 − x
∣
∣
−1, y 6= 0,
|x|−1 − r−1, y = 0,
where |x| denotes the Euclidean norm of x
The formula for the disk of radius r is
φ(y;x) =1
2π×
ln
(
|y|∣
∣r2y
|y|2−x∣
∣
r|y−x|
)
, y 6= 0,
ln(
r|x|
)
, y = 0.
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Green’s functions
Important issues: Existence, uniqueness, and smoothness ofthe Green’s function.
Generally, everything goes as for the original problem exceptthat the Green’s function may not be very smooth or may evenbe undefined at a point.
The point of evaluation y must lie inside the domain Ω. TheGreen’s function often behaves badly in the limit of yapproaching the boundary.
By the way, the theory also extends to over-determinedproblems.
However, if the original problem is under-determined, thisapproach fails.
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Green’s functions
Many expositions of the Green’s function avoid the adjoint.
When the original problem has a unique solution and the originaloperator is the adjoint to the adjoint operator, the Green’sfunction φ for the original problem and the Green’s function forthe adjoint problem φ∗ are related via
φ(y;x) = φ∗(x; y).
This is known as the reciprocity theorem.
This makes it possible to introduce Green’s functions for somekinds of problems without talking about the adjoint.
This depends on the dual space of the dual space of the originalspace being identifiable with the original space.
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Green’s functions
Including nonhomogeneous conditions is really a minor issuethat usually “solves itself” without trouble.
We carry out the analysis using the Green’s function for thehomogeneous problem and some additional integrals involvingdata and the Green’s function will appear.
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Green’s functions
ExampleSuppose the problem is
−∆u = f, x ∈ Ω,
u = g, x ∈ ∂Ω.
We define the Green’s function as for the homogeneousproblem, i.e.,
−∆φ(y;x) = δy(x), x ∈ Ω,
φ(y;x) = 0, x ∈ ∂Ω.
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Green’s functions
Evaluating the bilinear identity yields
∫
Ω(u∆φ− φ∆u) dx =
∫
∂Ω
(
u∂φ
∂n− φ
∂u
∂n
)
ds =
∫
∂Ωu∂φ
∂nds.
This yields
u(y) =
∫
Ωf(x)φ(y;x) dx−
∫
∂Ωg(s)
∂φ(y; s)
∂nds.
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Green’s functions
Different representations are possible.
ExampleFor
−∆u = f, x ∈ Ω,
u = 0, x ∈ ∂Ω,
We can define the Green’s function as the solution of
−∆φ(y;x) = 0, x ∈ Ω,
φ(y;x) = δy(x), x ∈ ∂Ω.
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Green’s functions
Evaluating the bilinear identity yields
∫
Ω(u∆φ− φ∆u) dx =
∫
∂Ω
(
u∂φ
∂n− φ
∂u
∂n
)
ds
= −
∫
∂Ωφ∂u
∂nds = −
∫
∂Ωδy∂u
∂nds = −
∂u
∂n(y).
This gives the value of the normal derivative of u at a point y onthe boundary,
∂u
∂n(y) = −
∫
Ωφ(y;x)f(x) dx.
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Green’s functions
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A PosterioriError Analysis and Adaptive Error Control
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A generalization of the Green’s function
The Green’s function yields
u(y) = (u, δy) = (φ(y;x), f(x)).
The value of a function at a point is an example of a linearfunctional.
There are frequently other quantities of interest that can beexpressed as functionals.
The Riesz Representation theorem suggests these functionalshave the form (u, ψ) for a suitable distribution ψ.
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A generalization of the Green’s function
Some useful choices of ψ include:
• We can construct ψ = δc to get the average value∮
c e(s) ds
of the error over a curve c in Rn, n = 2, 3, and ψ = δs to get
the average value of the error over a plane surface s in R3.
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A generalization of the Green’s function
This choice requires a certain smoothness, which is given by
TheoremSobolev If s > k + n/2, then there is a constant C such that forf ∈ Hs(Rn),
max|α|≤k
supx∈Rn
|Dα(x)| ≤ C‖f‖s.
This implies that the derivatives of f of order k and less arecontinuous.
The Sobolev theorem shows that if k ≤ n and s > k/2,restricting a function in Hs to a submanifold of (co)dimension kis well-defined.
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A generalization of the Green’s function
Some useful choices of ψ include:
• We can construct ψ = δc to get the average value∮
c e(s) ds
of the error over a curve c in Rn, n = 2, 3, and ψ = δs to get
the average value of the error over a plane surface s in R3.
• We can obtain errors in derivatives using dipoles in a similarway.
• ψ = χω/|ω| gives the error in the average value over asubset ω ⊂ Ω, where χω is the characteristic function of ω.
• Choosing ψ to be the residual of the finite elementapproximation sometimes yields the energy norm of theerror.
• ψ = χωe/‖e‖ω, where e is the error of the finite elementdiscretization, gives the L2(ω) norm of the error.
Only some of these data ψ have spatially local support.
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A generalization of the Green’s function
We consider
Lu = f, x ∈ Ω,
suitable b.c. and i.v., x ∈ ∂Ω,
where L is a linear differential operator.
The boundary and/or initial conditions should insure there is aunique solution.
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A generalization of the Green’s function
DefinitionThe generalized Green’s function corresponding to thequantity of interest (u, ψ) satisfies
L∗φ(y, x) = ψ(x), x ∈ Ω,
adjoint b.c. and i.v., x ∈ ∂Ω,
where L∗ is the formal adjoint of L.
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Discretization by the finite element method
We concentrate on
Lu = f, x ∈ Ω,
u = 0, x ∈ ∂Ω,
where
L(D,x)u = −∇ · a(x)∇u+ b(x) · ∇u+ c(x)u(x),
with u : Rn → R, a is a n× n matrix function of x, b is a n-vector
function of x, and c is a function of x.
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Discretization by the finite element method
We assume• Ω ⊂ R
n, n = 2, 3, is a smooth or polygonal domain
• a = (aij), where ai,j are continuous in Ω for 1 ≤ i, j ≤ n andthere is a a0 > 0 such that v>av ≥ a0 for all v ∈ R
n \ 0 andx ∈ Ω
• b = (bi) where bi is continuous in Ω
• c and f are continuous in Ω
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Discretization by the finite element method
The associated variational formulation reads
Find u ∈ H10 (Ω) such that
A(u, v) = (a∇u,∇v)+(b·∇u, v)+(cu, v) = (f, v) for all v ∈ H10 (Ω).
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Discretization by the finite element method
We construct a piecewise polygonal approximation of ∂Ω whosenodes lie on ∂Ω and which is contained inside Ω.
This forms the boundary of a convex polygonal domain Ωh.
Th denotes a simplex triangulation of Ωh that is locallyquasi-uniform.
hK denotes the length of the longest edge of K ∈ Th and h isthe mesh function with h(x) = hK for x ∈ K.
We also use h to denote maxK hK .
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Discretization by the finite element method
U = 0
Th
K
hK
Discretization of a domain Ω with curved boundaries.
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Discretization by the finite element method
Vh denotes the space of functions that are• continuous on Ω
• piecewise linear on Ωh with respect to Th• zero on the boundary ∂Ωh and Ω \ Ωh
Vh ⊂ H10 (Ω), and for smooth functions, the error of interpolation
into Vh is O(h2) in ‖ ‖
DefinitionThe finite element method is:
Compute U ∈ Vh such that A(U, v) = (f, v) for all v ∈ Vh.
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An a posteriori analysis for an algebraic equation
Estimate the error e = x−X of a numerical solution X of
Ax = b
where b, x ∈ Rn and A is a n× n matrix.
The computable residual error of X is
R = AX − b
Goal: Find a relation between the residual and the error.
The residual can be small even if the error is large.
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An a posteriori analysis for an algebraic equation
Since the residual error of the true solution is zero
Ae = −R.
To obtain an estimate on a quantity of interest, we introduce thegeneralized Green’s vector solving the adjoint problem
A>φ = ψ,
where ψ is any unit vector.
This gives an a posteriori estimate
|(e, ψ)| = |(e,A>φ)| = |(Ae, φ)| = |(R,φ)|.
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An a posteriori analysis for an algebraic equation
A posteriori error bound
|(e, ψ)| ≤ ‖φ‖ ‖R‖.
Definition‖φ‖ is called the stability factor
∣
∣
∣
∣
(
e
‖x ‖, ψ
)∣
∣
∣
∣
≤ cond ψ(A)‖R‖
‖b‖,
where the “weak” condition number is
cond ψ(A) = ‖φ‖ ‖A‖ = ‖A−>ψ‖ ‖A‖
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An a posteriori analysis for a finite element method
Goal: Estimate the error in a quantity of interest computed fromthe finite element solution U .
We use a generalized Green’s function φ solving the adjointproblem corresponding to a special choice of data ψ.
The error in the desired information may not have much to dowith the error in some global norm.
It may be computationally infeasible as well as veryinefficient to attempt to control the error in a globalnorm when all that is desired is accuracy in somequantities of interest.
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An a posteriori analysis for a finite element method
We assume that the information can be represented as (u, ψ).
DefinitionThe generalized Green’s function φ solves the weak adjointproblem : Find φ ∈ H1
0 (Ω) such that
A∗(v, φ) = (∇v, a∇φ)−(v,div (bφ))+(v, cφ) = (v, ψ) for all v ∈ H10 (Ω),
corresponding to the adjoint problem L∗(D,x)φ = ψ.
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An a posteriori analysis for a finite element method
(e, ψ) = (∇e, a∇φ) − (e,div (bφ)) + (e, cφ)
= (a∇e,∇φ) + (b · ∇e, φ) + (ce, φ)
= (a∇u,∇φ) + (b · ∇u, φ) + (cu, φ)
− (a∇U,∇φ) − (b · ∇U, φ) − (cU, φ)
= (f, φ) − (a∇U,∇φ) − (b · ∇U, φ) − (cU, φ).
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An a posteriori analysis for a finite element method
Definitionπhφ denotes an approximation of φ in Vh
TheoremThe error in the quantity of interest computed from the finiteelement solution satisfies the error representation ,
(e, ψ) = (f, φ−πhφ)−(a∇U,∇(φ−πhφ))−(b·∇U, φ−πhφ)−(cU, φ−πhφ),
where φ is the generalized Green’s function corresponding todata ψ.
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An a posteriori analysis for a finite element method
We use the error representation by approximating φ using arelatively high order finite element method.
Good results are obtained using the space V 2h
DefinitionThe approximate generalized Green’s function satisfies:Compute Φ ∈ V 2
h such that
A∗(v,Φ) = (∇v, a∇Φ)−(v,div (bΦ))+(v, cΦ) = (v, ψ) for all v ∈ V 2h .
The approximate error representation is
(e, ψ) ≈ (f,Φ − πhΦ) − (a∇U,∇(Φ − πhΦ))
− (b · ∇U,Φ − πhΦ) − (cU,Φ − πhΦ).
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An a posteriori analysis for a finite element method
Examples are computed using FETkLab.• Runs under MATLAB• Solves general nonlinear elliptic systems on general
domains in two space dimensions• Implements the a posteriori error estimate for 16
simultaneous adjoint data ψi• Uses bisection or red-green quadrisection to refine
elements• Only those elements whose element indicators are larger
than the mean plus one standard deviation of all of theelement indicators in that level are refined.
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An a posteriori analysis for a finite element method
ExampleAn elliptic problem on the unit square Ω = (0, 1) × (0, 1)
−∆u = 200 sin(10πx) sin(10πy), (x, y) ∈ Ω,
u(x, y) = 0, (x, y) ∈ ∂Ω
The solution is highly oscillatory
u(x, y) = sin(10πx) sin(10πy).
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An a posteriori analysis for a finite element method
-1
1
0u
xy
Highly oscillatory solution.
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An a posteriori analysis for a finite element method
We estimate the error in the average value by choosing ψ ≡ 1.
The generalized Green’s function is approximated on the samemesh using a piecewise quadratic finite element function.
We plot the ratios error/estimate for a wide variety of meshes.
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An a posteriori analysis for a finite element method
102
103
104
Number of elements
0.0
0.4
0.8
1.2
err
or/
esti
mate
0.0 0.1 1.0 10.0 100.0
percent error
0
1
2
3
erro
r/es
tim
ate
Plots of the error/estimate ratio.
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Adaptive error control
Goal of adaptive error control: Generate a mesh with a relativelysmall number of elements such that for a given tolerance TOLand data ψ,
|(e, ψ)| ≤ TOL.
This cannot be verified in practice.
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Adaptive error control
Rewrite the error representation elementwise:
(e, ψ) ≈∑
K∈Th
∫
K
(
(f−b·∇U−cU)(Φ−πhΦ)−a∇U ·∇(Φ−πhΦ))
dx.
DefinitionThe mesh acceptance criterion is∣
∣
∣
∣
∣
∑
K∈Th
∫
K
(
(f − b · ∇U − cU)(Φ − πhΦ) − a∇U · ∇(Φ − πhΦ))
dx
∣
∣
∣
∣
∣
≤ TOL
If this is satisfied, then the numerical solution is deemedacceptable.
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Adaptive error control
When the mesh criterion is not satisfied, we have to “enrich” thediscretization.
The cancellation among the contributions from each elementmake the enrichment decision problematic.
There is currently no theory or practical method foraccommodating cancellation of errors in an adaptiveerror control in a way that achieves true optimality ofefficiency.
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Adaptive error control
A standard approach is to formulate the discretization selectionproblem as an optimization problem.
This requires an estimate consisting of a sum over elements ofpositive quantities.
Inserting norms gives
|(e, ψ)| ≤∑
K∈Th
∫
K
∣
∣(f−b·∇U−cU)(Φ−πhΦ)−a∇U ·∇(Φ−πhΦ)∣
∣ dx.
If the acceptance criterion fails, the mesh is refined to achieve
∑
K∈Th
∫
K
∣
∣(f−b·∇U−cU)(Φ−πhΦ)−a∇U ·∇(Φ−πhΦ)∣
∣ dx ≤ TOL.
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Adaptive error control
The calculus of variations yields“Principle of Equidistribution” The element contributions ona nearly optimal mesh are roughly equal across the elements.
DefinitionTwo element acceptance criteria for the element indicatorsare
maxK
∣
∣(f − b · ∇U − cU)(Φ − πhΦ) − a∇U · ∇(Φ − πhΦ)∣
∣ .TOL|Ω|
,
or∫
K
∣
∣(f − b · ∇U − cU)(Φ− πhΦ)− a∇U · ∇(Φ− πhΦ)∣
∣ dx .TOLM
,
where M is the number of elements in Th.
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Adaptive error control
Computing a mesh using these criteria is usually performed by a“compute-estimate-mark-refine” adaptive strategy.
We start with a coarse mesh then successively refine somefraction of the elements on which the element acceptancecriterion fails.
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Adaptive error control
Example
−∆u =4800π
(
1 − 400((x− .5)2 + (y − .5)2))
e−400((x−.5)2+(y−.5)2),
(x, y) ∈ Ω,
u(x, y) = 0, (x, y) ∈ ∂Ω,
f is an approximation of a delta function at (.5, .5).
We control the error in the average value using a tolerance of.05%.
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Adaptive error control
0
350 0.08
0
Solution Generalized Green’s Function
-30
-20
-10
0
Mesh Element Contributions
Results for the initial mesh.
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Adaptive error control
0
140
0
0.08
Solution Generalized Green’s Function
-30
-20
-10
0
Mesh Element Contributions
Results after 10 iterations.
Duality, Adjoints, Green’s Functions – p. 158/304
Adaptive error control
0 50 100 150 200
percent error
0.90
0.95
1.00
1.05
1.10
erro
r/es
tim
ate
The error/estimate ratio.
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Further analysis on the a posteriori error estimate
We derive a more common form of the estimate for the simpleproblem with L(u) = −∆u.
The error representation formula is
(e, ψ) =
∫
Ωf(φ− πhφ) dx−
∫
Ω∇U · ∇(φ− πhφ) dx.
We break up the second integral on the right as∫
Ω∇U · ∇(φ− πhφ) dx =
∑
K∈Th
∫
K∇U · ∇(φ− πhφ) dx.
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Further analysis on the a posteriori error estimate
Using Green’s formula,
∫
K∇U · ∇(φ− πhφ) dx = −
∫
K∆U(φ− πhφ) dx
+
∫
∂K∇U · n∂K(φ− πhφ) ds,
where the last term is a line integral and n∂K denotes theoutward normal to ∂K.
Summing over all of the elements, the boundary integrals givetwo contributions from each element edge, computed inopposite directions.
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Further analysis on the a posteriori error estimate
If K1,K2 ∈ Th share a common edge σ1 ⊂ ∂K1 = σ2 ⊂ ∂K2,then
∫
σ1
∇U |K1· nσ1
(φ− πhφ) ds+
∫
σ2
∇U |K2· nσ2
(φ− πhφ) ds
=
∫
σ1
∇U |K1· nσ1
(φ− πhφ) ds
−
∫
σ1
∇U |K2· nσ1
(φ− πhφ) ds
= −
∫
σ1
[∇U ] · nσ1(φ− πhφ) ds,
where [U ] = ∇U |K2−∇U |K1
denotes the “jump” in ∇U acrossσ1 in the direction of the normal n∂K1
.
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Further analysis on the a posteriori error estimate
An alternate error representation,:
(e, ψ) = −∑
K∈Th
(∫
K(∆U + f)(φ− πhφ) dx
−1
2
∫
∂K[∇U ] · n∂K(φ− πhφ) ds
)
.
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Further analysis on the a posteriori error estimate
DefinitionThe residual and corresponding adjoint or dual weights are
RK =
(
‖∆U + f‖K‖h−1/2[∇U ]‖∂K/2
)
, WK =
(
‖φ− πhφ‖K‖h1/2(φ− πhφ)‖∂K
)
.
TheoremThe error of the finite element approximation is bounded by
|(e, ψ)| ≤∑
K∈Th
RK · WK .
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Further analysis on the a posteriori error estimate
We derive a priori bounds on the residual and dual weights.
There is a constant C independent of the mesh such that
RK ≤ C|K|1/2,
where |K| denote the area of K ∈ Th.
The bound on the first component follows from‖∆U + f‖K = ‖f‖K ≤ maxΩ |f | × |K|1/2.
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Further analysis on the a posteriori error estimate
For the second component, consider an integral over thecommon edge σ between two elements K1 and K2,
‖[∇U ]‖σ = ‖∇U |K2−∇U |K1
‖σ ≤ ‖∇U |K2−∇u|σ‖σ+‖∇u|σ−∇U |K1
‖σ.
By a trace inequality, the standard energy norm convergenceresult, and a standard elliptic regularity result,
‖∇U |Ki−∇u|σ‖σ ≤ ‖∇U −∇u‖
1/2Ki
‖∇U −∇u‖1/21,Ki
≤ C‖hu‖1/22,Ki
‖u‖1/22,Ki
≤ C‖h1/2f‖Ki,
for i = 1, 2. The local quasi-uniformity of the mesh implies
1
2‖h−1/2[∇U ]‖∂K ≤ Cmax
Ω|f | × |K|1/2.
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Further analysis on the a posteriori error estimate
Clearly, the convergence of the Galerkin approximation isstrongly influenced by the dual weights φ− πhφ, i.e. by theapproximation properties of Vh and the smoothness of φ.
This reflects the importance of the cancellation of errors inherentto the Galerkin method.
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The Effective Domain of Influence and SolutionDecomposition
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The Effective Domain of Influence
A characteristic property of elliptic partial differential equations isa global domain of influence.
A local perturbation of data near one point affects the solutionthroughout the domain of the problem.
Elliptic problems also have the property that the strength of theeffect of a localized perturbation on a solution decayssignificantly with the distance from the support of theperturbation, at least in some directions.
It turns out that this property also has profound consequencesfor the numerical solution of elliptic problems.
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The Effective Domain of Influence
ExampleConsider the Green’s function for the Dirichlet problem for theLaplacian −∆u = f on the ball Ω of radius r centered at theorigin in R
3.
If the data f is perturbed by a smooth function δf , theperturbation in the value of the solution δu(y) is given by
δu(y) =
∫
Ωφ(y;x)δf(x) dx, y ∈ Ω.
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The Effective Domain of Influence
If δf has compact support supp(δf) ⊂ Ω,
|y − x| ≤
∣
∣
∣
∣
r2y
|y|2− x
∣
∣
∣
∣
, x ∈ supp(δf), y ∈ Ω \ supp(δf).
We conclude that
|δu(y)| ≤max |δf | × volume of supp(δf) ×
(
1 + r|y|
)
4π × the distance from y to supp(δf).
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Poisson’s equation in a disk
Ω denotes the disk of radius r centered at the origin in R2
Consider the Dirichlet problem for the Laplacian L = −∆u = f .
Suppose that ω is a small region contained in Ω located wellaway from ∂Ω.
We wish to estimate the error e = u− U in the energy norm‖e‖1,ω in ω.
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Poisson’s equation in a disk
Ω
ω
Duality, Adjoints, Green’s Functions – p. 173/304
Poisson’s equation in a disk
We evaluate the norm weakly
‖e‖1,ω = supψ∈H−1(ω)‖ψ‖−1,ω=1
(e, ψ).
The supremum is achieved for some ψ ∈ H−1(ω).
We extend ψ to H−1(Ω) by setting it to zero in Ω \ ω.
To discuss orders of convergence, we use the a posterioribound.
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Poisson’s equation in a disk
G(x; y) denotes the Green’s function for the Laplacian on Ω, so
φ(x) =
∫
ΩG(x; y)ψ(y) dy =
∫
ωG(x; y)ψ(y) dy.
Recall,
G(x; y) =1
2π×
ln
(
|x|∣
∣ r2x
|x|2−y∣
∣
r|x−y|
)
, x 6= 0,
ln(
r|y|
)
, x = 0.
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Poisson’s equation in a disk
Case 1: For y ∈ ω, G(x; y) is a smooth function of x for x ∈ Ω \ ωand so is φ.
We assume that δ > 0 satisfies• It is small enough that ωδ = x ∈ Ω : dist (x, ω) ≤ δ is
contained in Ω
• It is large enough that for K ⊂ Ω \ ωδ, the union N (K) of Kand the elements bordering K does not intersect ω
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Poisson’s equation in a disk
∂Ω
∂ω
∂ωδ
For K ⊂ Ω \ ωδ, we let πh be the Lagrange nodal interpolant withrespect to Th, so that
‖φ− πhφ‖K ≤ C∑
|α|=2
‖h2Dαφ‖K .
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Poisson’s equation in a disk
Case 2: φ is not so smooth in ω.
For K ∩ ωδ 6= ∅, πh is the Scott-Zhang interpolant, for which
‖φ− πhφ‖K ≤ C|hφ|1,N (K),
for a mesh-independent constant C.
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Poisson’s equation in a disk
The second component of WK is bounded similarly after using atrace theorem,
‖h1/2(φ− πhφ)‖∂K ≤ ‖φ− πhφ‖1/2N (K)‖h(φ− πhφ)‖
1/21,N (K),
and the local quasi-uniformity of the mesh.
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Poisson’s equation in a disk
We conclude,TheoremFor any δ > 0 small enough that ωδ ⊂ Ω but large enough thatN (K) ∩ ω = ∅ for K ⊂ Ω \ ωδ, there is a constant C such that
‖e‖1,ω ≤∑
K⊂Ω\ωδ
∑
|α|=2
C‖h2Dαφ‖K |K|1/2+∑
K∩ωδ 6=∅
C|hφ|1,N (K)|K|1/2.
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Poisson’s equation in a disk
We estimate the quantities in the a posteriori bound.
To handle the first sum, we use
‖Dαxφ‖
2K =
∫
K
(∫
ωDαxG(x, y)ψ(y) dy
)2
dx
≤
∫
K‖Dα
xG(x, ·)‖21,ω‖ψ‖
2−1,ω dx
=∑
|β|=1
∫
K
∫
ω|Dα
xDβyG(x, y)|2 dydx+
∫
K
∫
ω|Dα
xG(x, y)|2 dydx.
Duality, Adjoints, Green’s Functions – p. 181/304
Poisson’s equation in a disk
There is a constant C such that
|DαxD
βyG(x, y)| ≤
C
|x− y|2, x 6= y ∈ Ω, |α| = 2, |β| ≤ 1.
We conclude there is a constant C independent of the meshsuch that for K ⊂ Ω \ ωδ,
‖φ− πhφ‖K ≤Ch2
K
dist (K,ω)2|K|1/2.
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Poisson’s equation in a disk
To handle the second sum on the right, we use the basic stabilityestimate,
‖φ‖1,Ω ≤ ‖ψ‖−1,Ω = ‖ψ‖−1,ω = 1.
If we assume a uniform (small) size hK = h for elements suchthat K ∩ ωδ 6= ∅, we obtain
∑
K∩ωδ 6=∅
ChK |φ|1,N (K) ≤ Ch‖φ‖1,Ω = Ch≤C
|ωδ|
∑
K∩ωδ 6=∅
h|K|.
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Poisson’s equation in a disk
We concludeTheoremFor any δ > 0 small enough that ωδ ⊂ Ω but large enough thatN (K) ∩ ω = ∅ for K ⊂ Ω \ ωδ, there is a constant C such that
‖e‖1,ω ≤∑
K⊂Ω\ωδ
Ch2K
dist (K,ω)2|K| +
∑
K∩ωδ 6=∅
Ch|K|.
Duality, Adjoints, Green’s Functions – p. 184/304
Poisson’s equation in a disk
We apply the “Principle of Equidistribution”
The element indicators are Ch2K/dist (K,ω)2 respectively Ch,
and in an optimal mesh,
h2K
dist (K,ω)2≈ h or hK ≈ h1/2 × dist (K,ω), K ⊂ Ω \ ωδ.
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Poisson’s equation in a disk
The decay of influence inherent to the Laplacian on the diskmeans that away from the region ω where we estimate the norm,we can choose elements asymptotically larger than the elementsize used in ωδ.
Moreover, the elements can increase in size as the distance toωδ increases
ωδ is an the effective domain of influence for the error in theenergy norm in ω.
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Poisson’s equation in a disk
DefinitionAn effective domain of influence corresponding to the data ψis the region ωψ in which the corresponding elements must besignificantly smaller in size than the elements used in thecomplement Ω \ ωψ in order to satisfy the adaptive goal.
Equivalently, if Th comprises uniformly sized elements, then theeffective domain of influence comprises those elements onwhich the element indicators, are substantially larger than thosein the complement.
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A decomposition of the solution
Often, the goal of solving a differential equation is to computeseveral pieces of information.
The problem of computing multiple quantities of interest alsoarises naturally when the data ψ for the adjoint problem does nothave spatially localized support (averages and norms).
We cannot expect a significant localization effect from the decayof influence when the support of the data for the adjoint problemis not spatially localized.
If the data ψ has the property that the adjoint weight φ− πhφ hasa more-or-less uniform size throughout Ω, then the degree ofnon-uniformity in an adapted mesh depends largely on thespatial variation of the residual.
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A decomposition of the solution
We use a partition of unity to “localize” a problem in whichsupp (ψ) does not have local support.
DefinitionSuppose that Ωi
Ni=1 is a finite open cover of Ω. A Lipschitz
partition of unity subordinate to Ωi is a collection of functionspi
Ni=1 with the properties
supp (pi) ⊂ Ωi, 1 ≤ i ≤ N,N∑
i=1
pi(x) = 1, x ∈ Ω,
pi is continuous on Ω and differentiable on Ωi, 1 ≤ i ≤ N,
‖pi‖L∞(Ω) ≤ C and ‖∇pi‖L∞(Ωi) ≤ C/diam (Ωi), 1 ≤ i ≤ N,
where C is a constant and diam (Ωi) is the diameter of Ωi.
Duality, Adjoints, Green’s Functions – p. 189/304
A decomposition of the solution
Corresponding to a partition of unity pi, ψ ≡∑N
i=1 ψpi,
DefinitionThe quantities (U,ψpi) corresponding to the data ψi = ψpiare called the localized information corresponding to thepartition of unity.
We consider the problem of estimating the error in the localizedinformation for 1 ≤ i ≤ N .
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A decomposition of the solution
We obtain a finite element solution via:
Compute Ui ∈ Vi such that A(Ui, v) = (f, v) for all v ∈ Vi,
where Vi is a space of continuous, piecewise linear functions ona locally quasi-uniform simplex triangulation Ti of Ω obtained bylocal refinement of an initial coarse triangulation T0 of Ω.
Vi is globally defined and the “localized” problem is solvedover the entire domain
We hope that this will require a locally refined mesh because thecorresponding data has localized support.
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A decomposition of the solution
A partition of unity approximation in the sense of Babuška andMelenk uses Ui = χiUi, 1 ≤ i ≤ N , where χi is the characteristicfunction of Ωi.
The local approximation Ui is in the local finite element spaceVi = χiVi.
DefinitionThe partition of unity approximation is defined byUp =
∑Ni=1 Uipi, which is in the partition of unity finite element
space
Vp =
N∑
i=1
Vipi =
N∑
i=1
vipi : vi ∈ Vi
.
Duality, Adjoints, Green’s Functions – p. 192/304
A decomposition of the solution
The partition of unity approximation recovers the fullconvergence properties of an approximation of the originalsolution.
Note
Up =
N∑
i=1
Uipi =
N∑
i=1
χiUipi ≡N∑
i=1
Uipi.
The values of Ui or Ui outside of Ωi are immaterial in forming theglobal partition of unity approximation.
Duality, Adjoints, Green’s Functions – p. 193/304
A decomposition of the solution
We use the generalized Green’s function satisfying the adjointproblem:
Find φi ∈ H10 (Ω) such that A∗(v, φi) = (v, ψi) for all v ∈ H1
0 (Ω).
We expand the global error
(u− Up, ψ) =N∑
i=1
(
(u− Ui)pi, ψ)
.
We estimate each summand on the right as
(
(u− Ui)pi, ψ)
= (u− Ui, ψi) = A∗(u− Ui, φi)
= (f, φi) − (a∇Ui,∇φi) − (b · ∇Ui, φi) − (cUi, φi).
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A decomposition of the solution
Letting πiφi denote an approximation of φi in Vi,
TheoremThe error of the partition of unity finite element solution Upsatisfies the error representation,
(u− Up, ψ) =N∑
i=1
(
(f, φi − πiφi) − (a∇Ui,∇(φi − πiφi))
− (b · ∇Ui, φi − πiφi) − (cUi, φi − πiφi))
.
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A decomposition of the solution
DefinitionThe approximate generalized Green’s functions solve:Compute Φi ∈ V 2
i such that
A∗(v,Φi) = (v, ψi) for all v ∈ V 2i , 1 ≤ i ≤ N,
where V 2i is the space of continuous, piecewise quadratic
functions with respect to Ti.
The approximate error representations are
(u− Ui, ψi) ≈ (f,Φi − πiΦi) − (a∇Ui,∇(Φi − πiΦi))
− (b · ∇Ui,Φi − πiΦi) − (cUi,Φi − πiΦi).
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A decomposition of the solution
If the localized error satisfies
∣
∣
(
u− Ui, ψi)∣
∣ ≤TOLN
, 1 ≤ i ≤ N,
then |(u− Up, ψ)| ≤ TOL.
This justifies treating the N “localized” problems independentlyin terms of mesh refinement.
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Computation of multiple quantities of interest
We present an algorithm for computing multiple quantities ofinterest efficiently using knowledge of the effective domains ofinfluence of the corresponding Green’s functions.
We assume that the information is specified as (U,ψi)Ni=1 for a
set of N functions ψiNi=1.
These data might arise as particular goals or via localizationthrough a partition of unity.
We assume that the goal is to compute the informationassociated to ψi so that the error is smaller than a toleranceTOLi for 1 ≤ i ≤ N .
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Computation of multiple quantities of interest
Two approaches:
Approach 1: A Global ComputationFind one triangulation such that the corresponding finiteelement solution satisfies |(e, ψi)| ≤ TOLi, for 1 ≤ i ≤ N .
Approach 2: A Decomposed ComputationFind N independent triangulations and finite elementsolutions Ui so that the errors satisfy |(ei, ψi)| ≤ TOLi, for1 ≤ i ≤ N .
The Global Computation can be implemented with astraightforward modification of the standard adaptive strategy inwhich the N corresponding mesh acceptance criteria arechecked on each element and if any of the N criteria fail, theelement is marked for refinement.
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Computation of multiple quantities of interest
If the correlation, i.e., overlap, between the effective domains ofinfluence associated to the N data ψi is relatively small andthe effective domains of influence are relatively small subsets ofΩ, then each individual solution in the DecomposedComputation will require significantly fewer elements than thesolution in the Global Computation to achieve the desiredaccuracy.
This can yield significant computational advantage in terms oflowering the maximum memory requirement to solve theproblem.
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Computation of multiple quantities of interest
Decreasing the maximum memory required to solve a problemcan be significant in at least two situations.
• If the individual solutions in the Decomposed Computationare computed in parallel and they require significantly fewerelements than the Global Computation, we can expect tosee significant speedup.
• In an environment with limited memory capabilities,decomposing a Global Computation requiring a largenumber of elements into a set of significantly smallercomputations can greatly increase the accuracy that can beachieved and/or decrease the time required.
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Computation of multiple quantities of interest
If the effective domains of influence associated to the N dataψ have relatively large intersections, then the individualsolutions in the Decomposed Computation will require roughlythe same number of elements as the solution for the GlobalComputation.
In this case, there is little to be gained in using the DecomposedComputation.
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Computation of multiple quantities of interest
In general, some of the N effective domains of influenceassociated to data ψi in the Decomposed Computation willcorrelate significantly and the rest will have low correlation.
We can optimize the use of resources by combiningcomputations for data whose associated domains of influencehave significant correlation and treating the rest independently.
Duality, Adjoints, Green’s Functions – p. 203/304
Computation of multiple quantities of interest
Determining the Solution Decomposition
1. Discretize Ω by an initial coarse triangulation T0 andcompute an initial finite element solution U0.
2. Estimate the error in each quantity (U0, ψi) by solving the Napproximate adjoint problems and computing the estimate.
3. Using the element indicators associated to identify theeffective domains of influence for the data ψi in terms ofthe mesh T0 and significant correlations between theeffective domains of influence.
4. Decide on the number of approximate solutions to becomputed and the subset of information to be computedfrom each solution.
5. Compute the approximate solutions independently usingadaptive error control aimed at computing the specifiedquantity or quantities of interest accurately.
Duality, Adjoints, Green’s Functions – p. 204/304
Identifying significant correlations
The key issue in the proposed algorithm is identifying theeffective domains of influence and recognizing significantcorrelation, or overlap, between different effective domains ofinfluence.
Mesh refinement decisions are based on the sizes of theelement indicators
Ei|K = maxK
∣
∣(f − b · ∇Ui− cUi)(Φi− πiΦi)− a∇Ui · ∇(Φi− πiΦi)∣
∣
or
Ei|K =
∫
K
∣
∣(f−b ·∇Ui−cUi)(Φi−πiΦi)−a∇Ui ·∇(Φi−πiΦi)∣
∣ dx,
Duality, Adjoints, Green’s Functions – p. 205/304
Identifying significant correlations
DefinitionWe let Ei(x) denote the piecewise constant element errorindicator function associated to data ψi with Ei(x) ≡ Ei|K forK ∈ T0.
Identifying the effective domain of influence means finding a setof elements on which the element error indicators aresignificantly larger than on the complement.
Identifying significant correlation between the effective domainsof influence of two data entails showing that the effectivedomains of influence have a significant number of elements incommon.
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Identifying significant correlations
Pattern matching uses
DefinitionThe (cross-)correlation of two functions f ∈ Lp(Ω) andg ∈ Lq(Ω), defined as:
(f g)(y) =
∫
Ωf(x)g(y + x) dx,
which is an L1(Ω) function.
The correlation indicator c(f, g) is a functional of thecorrelation function (f g).
We treat the element error indicator functions Ei as signalfunctions.
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Identifying significant correlations
Discounting translation or rotation, correlation of Ei and Ejreduces to the L2-inner-product:
(Ei Ej)(0) =
∫
ΩEi(x)Ej(x) dx = (Ei, Ej)Ω.
DefinitionThe correlation indicator c(Ei, Ej) is
c(Ei, Ej) = |(Ei Ej)(0)| = (Ei, Ej)Ω.
Duality, Adjoints, Green’s Functions – p. 208/304
Identifying significant correlations
DefinitionWe say that the effective domain of influence associated to ψi issignificantly correlated to the domain of influence associatedto ψj if two conditions hold.
Duality, Adjoints, Green’s Functions – p. 209/304
Identifying significant correlations
DefinitionWe say that the effective domain of influence associated to ψi issignificantly correlated to the domain of influence associatedto ψj if two conditions hold.
Condition 1The correlation of Ei and Ej is larger than a fixed fraction of thenorm of Ej , or
Correlation Ratio 1 =c(Ei, Ej)
‖Ej‖2≥ γ1,
for some fixed 0 ≤ γ1 ≤ 1.
This means that the projection of Ei onto Ej is sufficiently large.
Duality, Adjoints, Green’s Functions – p. 210/304
Identifying significant correlations
Element
Ele
men
t In
dic
ator
Element
Ele
men
t In
dic
ator
Element
Ele
men
t In
dic
ator
Ei
Ej Ei
Ej Ei
Ej
Three examples of significant correlation of Ei with Ej . Plottedare the element indicator functions Ei(x), Ej(x) versus the
element number.
Duality, Adjoints, Green’s Functions – p. 211/304
Identifying significant correlations
DefinitionWe say that the effective domain of influence associated to ψi issignificantly correlated to the domain of influence associatedto ψj if two conditions hold.
Condition 2The component of Ej orthogonal to Ei is smaller than a fixedfraction of the norm of Ej , or
Correlation Ratio 2 =
∥
∥
∥Ej −c(Ej ,Ei)‖Ei‖2 Ei
∥
∥
∥
‖Ej‖≤ γ2,
for some fixed 0 ≤ γ2 ≤ 1.
This corrects for the difficulties that arise when Ei is much largerthan Ej .
Duality, Adjoints, Green’s Functions – p. 212/304
Identifying significant correlations
Element
Ele
men
t In
dic
ato
r
Ej
Ei
An example in which the second condition fails.
Duality, Adjoints, Green’s Functions – p. 213/304
Identifying significant correlations
1 9 18 27 36
Element Number
0
1
2
3E1
E2
E3
E4
E5
E6
E7
E8
E9
Plots of nine element indicator functions Ei versus the elementnumber.
Duality, Adjoints, Green’s Functions – p. 214/304
Identifying significant correlations
With γ1 = .9 and γ2 = .7:
E1 with E8 E4 with none E7 with noneE2 with E6, E7 E5 with E2, E6 E8 with noneE3 with E1, E8 E6 with none E9 with none
We emphasize that the initial identification of significantcorrelation between effective domains of influence ofvarious Green’s functions in a computation is carriedout on a coarse initial partition of the domain and henceis relatively inexpensive.
Duality, Adjoints, Green’s Functions – p. 215/304
Examples
We solve elliptic problems using adaptive mesh refinement toachieve accuracy in a set of quantities of interest using a GlobalComputation and a Decomposed Computation.
The results suggest that the individual solutions in theDecomposed Computation require significantly fewer elementsto achieve the desired accuracy than the Global Computation ina variety of situations.
Duality, Adjoints, Green’s Functions – p. 216/304
Examples
As a relatively universal measure of the gain from using theDecomposed Computation, we use
DefinitionThe Final Element Ratio is the number of elements in the finalmesh refinement level required to achieve accuracy in thespecified quantities of interest in the Global Computation to themaximum number of elements in the final mesh refinementlevels for the individual computations in the DecomposedComputation.
Generally, we expect the gain in efficiency to scalesuper-linearly with the Final Element Ratio.
We compute the Final Element Ratio using solutions that arehave roughly the same accuracy.
Duality, Adjoints, Green’s Functions – p. 217/304
Examples
Example 1
We solve
− 110π2 ∆u(x) = sin(πx) sin(πy), (x, y) ∈ Ω,
u(x, y) = 0, (x, y) ∈ ∂Ω,
on the Ω = [0, 8] × [0, 8].
The solution is u(x, y) = 5 sin(πx) sin(πy).
Goal: Control the error in the average value of u by choosingψ ≡ 1/|Ω| = 1/64.
Duality, Adjoints, Green’s Functions – p. 218/304
Examples
Global Computation: Mesh is adapted the error in the averagevalue of u is smaller than the error tolerance of 5%.
Initial mesh is 10 × 10 elements and after five refinement levels,we end up with 3505 elements, achieving an error of .022.
Duality, Adjoints, Green’s Functions – p. 219/304
Examples
Initial Mesh Final Mesh
Initial and final meshes.
Duality, Adjoints, Green’s Functions – p. 220/304
Examples
Numerical solutions.
Duality, Adjoints, Green’s Functions – p. 221/304
Examples
Estimates, errors, and error/estimate ratios:
Level Elements Estimate Error Ratio1 100 .1567 .1534 .9786
2 211 .1157 .1224 1.058
3 585 .3063 .3078 1.005
4 1309 .1159 .1166 1.006
5 3505 .02163 .02148 .9975
Duality, Adjoints, Green’s Functions – p. 222/304
Examples
2 4
31
1
2 3
4
5
6 7
8
13
14 15
16
9
10 11
12
Partitions for the decomposition.
Duality, Adjoints, Green’s Functions – p. 223/304
Examples
The partition of unity yields four data ψ1, ψ2, ψ3, ψ4.
We compute the four localized approximations U1, · · · , U4using the same initial mesh as before.
The Correlation Ratios indicates that all four localized solutionsshould be computed independently.
Duality, Adjoints, Green’s Functions – p. 224/304
Examples
We obtain acceptable results using the tolerance of 5%.
Data Level Elements Estimateψ1 3 618 .01242
ψ2 3 575 −.0009109
ψ3 3 618 .01242
ψ4 3 575 −.0009109
This yields a partition of unity solution Up with accuracy .023.
Using the Decomposed Computation yields a Final ElementRatio of ≈ 5.7.
Duality, Adjoints, Green’s Functions – p. 225/304
Examples
Final Mesh for U1^
Final Mesh for U2^
Two of the final meshes.
Duality, Adjoints, Green’s Functions – p. 226/304
Examples
The generalized Green’s functions for the global average errorand the localized data ψ2.
Duality, Adjoints, Green’s Functions – p. 227/304
Examples
A Decomposed Computation using a partition of unity on 16equal-sized regions yields significant correlations:
E2 with E3 E5 with E8 E10 with E9 E13 with E14
E4 with E3 E7 with E8 E12 with E9 E15 with E14
This computation yields a Final Element Ratio of ≈ 2.6
Duality, Adjoints, Green’s Functions – p. 228/304
Examples
Example 2
We solve
−∇ ·(
(1.1 + sin(πx) sin(πy))∇u(x, y))
= −3 cos2(πx) + 4 cos2(πx) cos2(πx)
+2.2 sin(πx) sin(πy) + 2 − 3 cos2(πy), (x, y) ∈ Ω,
u(x, y) = 0, (x, y) ∈ ∂Ω,
where Ω = [0, 2] × [0, 2].
The exact solution is u(x, y) = sin(πx) sin(πy).
Duality, Adjoints, Green’s Functions – p. 229/304
Examples
We compute the average error using ψ1 ≡ 1/4 and four pointvalues using ψ2 ≈ δ(.5,.5), ψ3 ≈ δ(.5,1.5), ψ4 ≈ δ(1.5,1.5), andψ5 ≈ δ(1.5,.5).
We use
δ(cx,cy) =400
πe−400((x−cx)2+(y−cy)2)
to approximate the delta function δ(cx,cy).
Duality, Adjoints, Green’s Functions – p. 230/304
Examples
Global Computation: We use a tolerance of 2% starting with an8 × 8 mesh.
ψ1 ψ2
Lev. Elt’s1 64
2 201
3 763
4 2917
Est. Err. Rat..035 .035 1.0
.0088 .0089 1.0
.0027 .0027 1.0
.00044 .00044 1.0
Est. Err. Rat..090 .29 3.3
.042 .082 1.9
.020 .020 .99
.0050 .00504 1.0
Duality, Adjoints, Green’s Functions – p. 231/304
Examples
Global Computation: We use a tolerance of 2% starting with an8 × 8 mesh.
ψ3
Lev. Elt’s1 64
2 201
3 763
4 2917
Est. Err. Rat..24 .022 .091
.0024 .014 6.0
.0020 .0020 1.0
.0049 .00504 1.0
Duality, Adjoints, Green’s Functions – p. 232/304
Examples
Initial Mesh Final Mesh
Initial and final meshes. Final mesh has 2917 elements.
Duality, Adjoints, Green’s Functions – p. 233/304
Examples
Decomposed Computation: Use data ψ1, · · · , ψ5independently using tolerances of 2%.
We vary the initial meshes; using 7 × 7 for U1; 9 × 9 for U2 andU4; and 12 × 12 for U3 and U5.
Data Level Elements Estimateψ1 3 409 −.0004699
ψ2 4 1037 −.007870
ψ3 2 281 −.005571
ψ4 4 1037 −.007870
ψ5 2 281 −.005571
The Final Element Ratio is ≈ 2.8.
Duality, Adjoints, Green’s Functions – p. 234/304
Examples
Final Mesh for U1^
Final Mesh for U2^
Final Mesh for U3^
Final meshes for U1, U2, U3.
Duality, Adjoints, Green’s Functions – p. 235/304
Examples
Example 3
We solve
−∆u = 16(y − y2 + x− x2) (x, y) ∈ Ω,
u(x, y) = 0, (x, y) ∈ ∂Ω,
where Ω = [0, 1] × [0, 1].
The exact solution is u(x, y) = 8x(1 − x)y(1 − y).
Duality, Adjoints, Green’s Functions – p. 236/304
Examples
We estimate the error in the average value of u.
Since the domain is small and the solution and the generalizedGreen’s function are very smooth, the gain from decomposingthe solution is greatly reduced.
Beginning with a 4 × 4 mesh and using a tolerance of 1%, weobtain a sufficiently accurate solution using a GlobalComputation after five refinements.
The final mesh uses 885 elements and produces an error of.0008699.
Duality, Adjoints, Green’s Functions – p. 237/304
Examples
If we partition the domain using four equal regions there are nosubstantial correlations.
Computing the four solutions independently in the DecomposedComputation yields a Final Element Ratio of around 1.5.
Duality, Adjoints, Green’s Functions – p. 238/304
Examples
If we use sixteen equal regions, there are a number ofsubstantial correlations.
Correlation Ratio 1 for E1 on E2 = .98,Correlation Ratio 2 for E1 on E2 = .44,Correlation Ratio 1 for E2 on E1 = .82,Correlation Ratio 2 for E2 on E1 = .44.
Duality, Adjoints, Green’s Functions – p. 239/304
Examples
Computing U1 corresponding to the localized data ψ1 using atolerance of 1% requires 367 elements.
Computing U2 requires a mesh 494 elements.
Combining these two computations by using data equal to thesum of the two partition functions for the regions Ω1 and Ω2
requires 496 elements.
We gain almost nothing by computing U1 and U2 independently
Duality, Adjoints, Green’s Functions – p. 240/304
Examples
Final Mesh for U1^
Final Mesh for U2^
Final Mesh for "U1+U2"^ ^
Final meshes.
Duality, Adjoints, Green’s Functions – p. 241/304
Examples
We consider the error in the average value and the point valuesat (.25, .25) and (.5, .5).
We use a partition of unity decomposition for the error in theaverage to get data ψ1, · · · , ψ4. We let ψ5 ≈ δ(.25,.25) andψ6 ≈ δ(.5,.5).
We compare the correlation indicators on initial meshes rangingfrom 16 to 144 or 400 uniformly sized elements.
Duality, Adjoints, Green’s Functions – p. 242/304
Examples
0 50 100 150
Number of Elements
0.0
0.5
1.0
Corr. Rat. 1: E1 on E2
Corr. Rat. 2: E1 on E2
Corr. Rat. 1: E2 on E1
Corr. Rat. 2: E2 on E1
0 50 100 150
Number of Elements
0.0
0.5
1.0
Corr. Rat. 1: E1 on E3
Corr. Rat. 2: E1 on E3
Corr. Rat. 1: E3 on E1
Corr. Rat. 2: E3 on E1
Correlation Ratios versus number of elements.
Duality, Adjoints, Green’s Functions – p. 243/304
Examples
0 100 200 300 400
Number of Elements
0
5
10
Corr. Rat. 1: E1 on E5
Corr. Rat. 2: E1 on E5
Corr. Rat. 1: E5 on E1
Corr. Rat. 2: E5 on E1
0 100 200 300 400
Number of Elements
0
2
4
Corr. Rat. 1: E1 on E6
Corr. Rat. 2: E1 on E6
Corr. Rat. 1: E6 on E1
Corr. Rat. 2: E6 on E1
Correlation Ratios versus number of elements.
Duality, Adjoints, Green’s Functions – p. 244/304
Examples
0 50 100 150
Number of Elements
0.0
0.6
1.2
Corr. Rat. 1: E5 on E6
Corr. Rat. 2: E5 on E6
Corr. Rat. 1: E6 on E5
Corr. Rat. 2: E6 on E5
0 100 200 300 400
Number of Elements
0
1
2
3Corr. Rat. 1: E3 on E5
Corr. Rat. 2: E3 on E5
Corr. Rat. 1: E5 on E3
Corr. Rat. 2: E5 on E3
Correlation Ratios versus number of elements.
Duality, Adjoints, Green’s Functions – p. 245/304
Examples
In general, we find that all Correlation Ratios converge to a limitas the number of elements increases.
Generally, the second Correlation Ratio varies relatively little asthe mesh density increases for all data.
The first Correlation Ratio between data representing a partitionof unity decomposition also varies relatively little.
The first Correlation Ratio varies quite a bit on coarse mesheswhen one of the data is an approximate delta function.
Duality, Adjoints, Green’s Functions – p. 246/304
Examples
The determination that two effective domains of influence arenot closely correlated is relatively robust with respect to thedensity of the mesh.
There is a mild tendency to combine computations that are moreefficiently treated independently if the correlation indicators arecomputed on very coarse meshes.
Duality, Adjoints, Green’s Functions – p. 247/304
Examples
Example 4We solve
−∇ ·((
.05 + tanh(
10(x− 5)2 + 10(y − 1)2))
∇u)
+
(
−100
0
)
· ∇u = 1, (x, y) ∈ Ω,
u(x, y) = 0, (x, y) ∈ ∂Ω,
where Ω = [0, 10] × [0, 2].
Duality, Adjoints, Green’s Functions – p. 248/304
Examples
2
1
0
10
x
y
Dif
fusi
on
Plot of the diffusion coefficient.
Duality, Adjoints, Green’s Functions – p. 249/304
Examples
We use an initial mesh of 80 elements and an error tolerance ofTOL = .04%.
Level Elements Estimate1 80 −.0005919
2 193 −.001595
3 394 −.0009039
4 828 −.0003820
5 1809 −.0001070
6 3849 −.00004073
7 9380 −.00001715
8 23989 −.000007553
Duality, Adjoints, Green’s Functions – p. 250/304
Examples
Final Mesh.
Duality, Adjoints, Green’s Functions – p. 251/304
Examples
Pe=1000
Pe=.1
Plots of the meshes for the original problem with Pe = 1000 anda problem with Pe = .1.
Duality, Adjoints, Green’s Functions – p. 252/304
Examples
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
Domains for the partition of unity.
Duality, Adjoints, Green’s Functions – p. 253/304
Examples
Significant Correlations:
E3 with E4 E6 with E7 E7 with E6 E9 with E8 E10 with E8, E9
E13 with E14 E16 with E17 E17 with E16 E19 with E18 E20 with E18, E19
Note, there are no significant correlations in the “cross-wind”direction.
Duality, Adjoints, Green’s Functions – p. 254/304
Examples
Solutions are symmetric across y = 1.
Data TOL Level Elements Estimateψ1 .04% 7 7334 −6.927 × 10−7
ψ2 .04% 7 8409 −5.986 × 10−7
ψ3 .04% 7 7839 −5.189 × 10−7
ψ4 .04% 7 7177 −5.306 × 10−7
ψ5 .04% 7 7301 −4.008 × 10−7
ψ6 .02% 7 6613 −2.471 × 10−7
ψ7 .02% 7 4396 −2.938 × 10−7
ψ8 .02% 7 4248 −1.656 × 10−7
ψ9 .02% 7 3506 −1.221 × 10−7
ψ10 .02% 7 1963 −5.550 × 10−8
The Final Element Ratio is ≈ 2.9.Duality, Adjoints, Green’s Functions – p. 255/304
Examples
Final Mesh for U1^
^
Final Mesh for U5^
Plots of the final meshes for the localized solutions U1 and U5.Duality, Adjoints, Green’s Functions – p. 256/304
Examples
Final Mesh for U9^
Plots of the final mesh for the localized solution U9.
Duality, Adjoints, Green’s Functions – p. 257/304
Examples
2
1
0 02
46
8100
2
4
6
x 10 -4
y
x2
1
0 02
46
8100
2
4
6
x 10 -4
y
x
Plots of the generalized Green’s functions corresponding to ψ11
(left) and ψ19 (right).
Duality, Adjoints, Green’s Functions – p. 258/304
Examples
The effective domains of influence may not be spatiallycompactly-shaped.
Duality, Adjoints, Green’s Functions – p. 259/304
Examples
We combine some of the localized computations by solving forlocalized solutions corresponding to summing the two of thepartition of unity data.
Data TOL Level Elements Estimateψ3 + ψ4 .04% 7 8330 −9.8884 × 10−7
ψ6 + ψ7 .02% 7 5951 −5.897 × 10−7
ψ8 + ψ9 .02% 7 4406 −3.486 × 10−7
ψ9 + ψ10 .02% 7 3202 −2.243 × 10−7
The solutions for ψ3 + ψ4 and ψ8 + ψ9 use a few more elementsthan required for either of the original localized solutions.
The solutions for ψ6 + ψ7 and ψ9 + ψ10 use less than themaximum required for the individual localized solutions.
Duality, Adjoints, Green’s Functions – p. 260/304
Examples
Example 5
We solve
− 1π2 ∆u = 2 + 4e−5((x−.5)2+(y−2.5)2), (x, y) ∈ Ω,
u(x, y) = 0, (x, y) ∈ ∂Ω,
where Ω is the “square annulus” Ω = [0, 3] × [0, 3] \ [1, 2] × [1, 2].
The initial mesh has 48 elements and the error tolerance is 1%.
Duality, Adjoints, Green’s Functions – p. 261/304
Examples
Level Elements Estimate1 48 −5.168
2 125 −1.584
3 380 −.6879
4 894 −.3029
5 2075 −.1435
Duality, Adjoints, Green’s Functions – p. 262/304
Examples
Initial Mesh Final Mesh
Initial and Final Meshes.
Duality, Adjoints, Green’s Functions – p. 263/304
Examples
12
3 0
1
2
3
0
1
2
3
4
5
6
y
x
12
3 0
1
2
3
0
0.4
0.8
1.2
1.6
y
x
Solution and Generalized Green’s Function on the Final Mesh.
Duality, Adjoints, Green’s Functions – p. 264/304
Examples
1 2 3
4
567
8
The partition of unity has 8 square subdomains.
Duality, Adjoints, Green’s Functions – p. 265/304
Examples
There are no significant correlations.
Data Level Elements Estimateψ1 5 1082 −.01935
ψ2 5 1101 −.01399
ψ3 5 1144 −.01540
ψ4 5 1107 −.01360
Duality, Adjoints, Green’s Functions – p. 266/304
Examples
There are no significant correlations.
Data Level Elements Estimateψ5 5 1104 −.01436
ψ6 5 1110 −.01587
ψ7 5 1074 −.02529
ψ8 5 1098 −.01660
Final Element Ratio ≈ 1.8.
Duality, Adjoints, Green’s Functions – p. 267/304
Examples
Final Mesh for U3^
Final Mesh for U4^
Final meshes.
Duality, Adjoints, Green’s Functions – p. 268/304
Examples
Final Mesh for U6^
Final Mesh for U7^
Final meshes.
Duality, Adjoints, Green’s Functions – p. 269/304
Examples
0
12
3 0
1
2
3
1
y
x
12
3 0
1
2
3
0
0.2
0.6
1
1.4
y
x
Plots of some Generalized Green’s Functions.
Duality, Adjoints, Green’s Functions – p. 270/304
Nonlinear Problems
Duality, Adjoints, Green’s Functions – p. 271/304
A nonlinear algebraic equation
The problem is to estimate the error e = x−X of a numericalsolution X of a system of nonlinear algebraic equations,
f(x) = b
where the data b, nonlinearity f , and the solution x all have thesame dimension.
The residual error isR = f(X) − b
which immediately gives
f(x) − f(X) = −R.
The error is related to the residual through a nonlinear equation.
Duality, Adjoints, Green’s Functions – p. 272/304
A nonlinear algebraic equation
The mean value theorem for integrals:
f(x) − f(X) =
∫ 1
0f ′(sx+ (1 − s)X) (x−X) ds
where f ′ is the Jacobian matrix of f .
We get
Ae = −R
with
A =
∫ 1
0f ′(sx+ (1 − s)X) ds.
This is the linear problem obtained by linearizing f around anaverage of x and −X.
Duality, Adjoints, Green’s Functions – p. 273/304
A nonlinear algebraic equation
We obtain
|(e, ψ)| = |(e, A>φ)| = |(Ae, φ)| = |(R,φ)|.
To obtain a computable adjoint problem, we can try to replace xby X in the definition of A,
A→ A =
∫ 1
0f ′(sX + (1 − s)X) ds = f ′(X),
But, what is the effect on the estimate?
Duality, Adjoints, Green’s Functions – p. 274/304
Defining the adjoint to a nonlinear operator
We assume that the Banach spaces X and Y are Sobolevspaces and use ( , ) for the L2 inner product, and so forth.
We define the adjoint for a specific kind of nonlinear operator.
We assume f is a nonlinear map from X into Y , where thedomain D(f) is a convex set.
DefinitionA subset A of a vector space is convex if for any a, b ∈ A, theset of points on the “line segment” joining a and b, i.e.,sa+ (1 − s)b| 0 ≤ s ≤ 1 is contained in A.
Duality, Adjoints, Green’s Functions – p. 275/304
Defining the adjoint to a nonlinear operator
We choose u inside D(f) and define
F (e) = f(u+ e) − f(u),
where we think of e as representing an “error”, i.e., e = U − u.
The domain of F is
D(F ) = v ∈ X| v + u ∈ D(f).
We assume that D(F ) is independent of e and dense in X.
Note that 0 ∈ D(F ) and F (0) = 0.
We also assume that D(F ) is a vector subspace of X.
Duality, Adjoints, Green’s Functions – p. 276/304
Defining the adjoint to a nonlinear operator
Two reasons to work with functions of this form:• This is the kind of nonlinearity that arises when estimating
the error of a numerical solution or studying the effects ofperturbations.
• Nonlinear problems typically do not enjoy the globalsolvability that characterizes linear problems, only a localsolvability.
Duality, Adjoints, Green’s Functions – p. 277/304
Defining the adjoint to a nonlinear operator
The first definition is based on the bilinear identity.
DefinitionAn operator A∗(e) with domain D(A∗) ⊂ Y ∗ and range in X∗ isan adjoint operator corresponding to F if
(F (e), w) = (e,A∗(e)w) for all e ∈ D(F ), w ∈ D(A∗).
This is an adjoint operator associated with F , not the adjointoperator to F .
Duality, Adjoints, Green’s Functions – p. 278/304
Defining the adjoint to a nonlinear operator
ExampleSuppose that F can be represented as F (e) = A(e)e, whereA(e) is a linear operator with D(F ) ⊂ D(A).
For a fixed e ∈ D(F ), define the adjoint of A satisfying
(A(e)w, v) = (w,A∗(e)v) for all w ∈ D(A), v ∈ D(A∗).
Substituting w = e shows this defines an adjoint of F as well.
If there are several such linear operators A, then there will beseveral different possible adjoints.
Duality, Adjoints, Green’s Functions – p. 279/304
Defining the adjoint to a nonlinear operator
ExampleLet (t, x) ∈ Ω = (0, 1) × (0, 1), with X = X∗ = Y = Y ∗ = L2
denoting the space of periodic functions in t and x, with periodequal to 1.
Consider a periodic problem
F (e) =∂e
∂t+ e
∂e
∂x+ ae = f
where a > 0 is a constant and the domain of F is the set ofcontinuously differentiable functions.
Duality, Adjoints, Green’s Functions – p. 280/304
Defining the adjoint to a nonlinear operator
We can write F (e) = Ai(e)e where
A1(e)v =∂v
∂t+ e
∂v
∂x+ av
A2(e)v =∂v
∂t+
(
a+∂e
∂x
)
v
A3(e)v =∂v
∂t+
1
2
∂(ev)
∂x+ av.
Duality, Adjoints, Green’s Functions – p. 281/304
Defining the adjoint to a nonlinear operator
The adjoints are
A∗1(e)w = −
∂w
∂t−∂(ew)
∂x+ aw
A∗2(e)w = −
∂w
∂t+
(
a+∂e
∂x
)
w
A∗3(e)w = −
∂w
∂t−e
2
∂w
∂x+ aw.
Duality, Adjoints, Green’s Functions – p. 282/304
Defining the adjoint to a nonlinear operator
We base the second definition of an adjoint on the integral meanvalue theorem.
We assume that the original nonlinearity is Frechetdifferentiable.
The integral mean value theorem states
f(U) = f(u) +
∫ 1
0f ′(u+ se) ds e
where e = U − u and f ′ is the Frechet derivative of f .
Duality, Adjoints, Green’s Functions – p. 283/304
Defining the adjoint to a nonlinear operator
We rewrite this as
F (e) = f(U) − f(u) = A(e)e
with
A(e) =
∫ 1
0f ′(u+ se) ds.
Note that we can apply the integral mean value theorem to F :
A(e) =
∫ 1
0F ′(se) ds.
To be precise, we should discuss the smoothness of F .
Duality, Adjoints, Green’s Functions – p. 284/304
Defining the adjoint to a nonlinear operator
DefinitionFor a fixed e, the adjoint operator A∗(e), defined in the usual wayfor the linear operator A(e), is said to be an adjoint for F .
ExampleContinuing the previous example,
F ′(e)v =∂v
∂t+ e
∂v
∂x+
(
a+∂e
∂x
)
v.
After some technical analysis of the domains of the operatorsinvolved,
A∗(e)w = −∂w
∂t−e
2
∂w
∂x+ aw.
This coincides with the third adjoint computed above.
Duality, Adjoints, Green’s Functions – p. 285/304
Analysis of a space-time finite element method
We study a system of D reaction-diffusion equations consistingof d parabolic equations and D − d ordinary equations for theRD valued function u = (ui):
ui −∇ · (εi(u, x, t)∇ui) = fi(u, x, t), (x, t) ∈ Ω × R+, 1 ≤ i ≤ D,
ui(x, t) = 0, (x, t) ∈ ∂Ω × R+, 1 ≤ i ≤ d,
u(x, 0) = u0(x), x ∈ Ω,
where there is a constant ε > 0 such that
εi(u, x, t) ≥ ε for 1 ≤ i ≤ d and εi(u, x, t) ≡ 0 for the rest.
Duality, Adjoints, Green’s Functions – p. 286/304
Analysis of a space-time finite element method
Assumptions:
• Ω is a convex polygonal domain in R2 with boundary ∂Ω
• ε = (εi) and f = (fi) have smooth second derivatives
ui denotes the time derivative of ui
We write εi(u, x, t) = εi(u) and f(u, x, t) = f(u).
Duality, Adjoints, Green’s Functions – p. 287/304
Analysis of a space-time finite element method
We use up and uo to denote the parts of u associated to theparabolic and ordinary differential equations respectively.
upi = ui for 1 ≤ i ≤ d and upi = 0 for d < i ≤ D
uo = u− up
The presence of ordinary differential equations in the systemhas strong consequences for the smoothness of solutions.
Duality, Adjoints, Green’s Functions – p. 288/304
Analysis of a space-time finite element method
Discretization of the space-time domain:
We partition [0,∞) as 0 = t0 < t1 < t2 < · · · < tn < . . . , denotingeach time interval by In = (tn−1, tn] and time step bykn = tn − tn−1.
To each interval In, we associate a triangulation Tn of Ωarranged so the union of the elements in Tn is Ω.
Duality, Adjoints, Green’s Functions – p. 289/304
Analysis of a space-time finite element method
Space
Tim
e
Ω
A space-time discretization.
Duality, Adjoints, Green’s Functions – p. 290/304
Analysis of a space-time finite element method
As usual, we assume the intersection of any two elements in aspace triangulation is either a common edge, node, or is empty.
We also assume that the smallest angle of any triangle in atriangulation is bounded below by a fixed constant
hn denotes the piecewise constant mesh functionhn|K = diam(K) for K ∈ Tn and h denotes the global meshfunction, where h|In
= hn.
k denotes the piecewise constant function that is kn on In.
Duality, Adjoints, Green’s Functions – p. 291/304
Analysis of a space-time finite element method
The approximations are polynomials in time and piecewisepolynomials in space on each space-time “slab” Sn = Ω × In.
Vn ⊂ (H10 (Ω))d × (H1(Ω))D−d denotes the space of piecewise
linear continuous vector-valued functions v(x) ∈ RD defined on
Tn, where the first d components of v are zero on ∂Ω.
We define
W qn =
w(x, t) : w(x, t) =
q∑
j=0
tjvj(x), vj ∈ Vn, (x, t) ∈ Sn
.
W q denotes the space of functions defined on the space-timedomain Ω × R
+ such that v|Sn∈W q
n for n ≥ 1.
Duality, Adjoints, Green’s Functions – p. 292/304
Analysis of a space-time finite element method
Functions in W q are generally discontinuous across the discretetime levels.
[w]n = w+n − w−
n , where w±n = lims→tn± w(s), denotes the jump
across tn
Pn is the L2 projection onto Vn, i.e., Pn : L2(Ω) → Vn is definedby (Pnv, w) = (v, w) for all w ∈ Vn. P is defined by P = Pn onSn.
πn : L2(In) → Pq(In) denotes the L2 projection onto thepiecewise polynomial functions in time, where Pq(In) is thespace of polynomials of degree q or less on In. π is defined byπ = πn on Sn.
Duality, Adjoints, Green’s Functions – p. 293/304
Analysis of a space-time finite element method
DefinitionThe continuous Galerkin cG(q) approximation U ∈W q
satisfies U−0 = P0u0 and for n ≥ 1, the Galerkin orthogonality
relation
∫ tn
tn−1
(
(Ui, vi) + (εi(U)∇Ui,∇vi))
dt =
∫ tn
tn−1
(fi(U), vi) dt
for all v ∈W q−1n , 1 ≤ i ≤ D,
U+n−1 = PnU
−n−1.
U is continuous across time nodes over which there is no meshchange.
Duality, Adjoints, Green’s Functions – p. 294/304
Analysis of a space-time finite element method
DefinitionThe discontinuous Galerkin dG(q) approximation U ∈W q
satisfies U−0 = P0u0 and for n ≥ 1,
∫ tn
tn−1
(
(Ui, vi) + (εi(U)∇Ui,∇vi))
dt+(
[Ui]n−1, v+i
)
=
∫ tn
tn−1
(fi(U), vi) dt
for all v ∈W qn , 1 ≤ i ≤ D.
Duality, Adjoints, Green’s Functions – p. 295/304
Analysis of a space-time finite element method
ExampleWe discretize the scalar problem
u− ∆u = f(u), (x, t) ∈ Ω × R+,
u(x, t) = 0, (x, t) ∈ ∂Ω × R+,
u(x, 0) = u0(x), x ∈ Ω,
using the dG(0) method.
We let ~Un denote the Mn vector of nodal values with respect tothe nodal basis ηn,i
Mn
i=1 for Vn on In.
Duality, Adjoints, Green’s Functions – p. 296/304
Analysis of a space-time finite element method
Bn :(
Bn)
ij=(
ηn,i, ηn,j)
for 1 ≤ i, j ≤Mn and
Bn,n−1 :(
Bn,n−1
)
ij=(
ηn,i, ηn−1,j
)
for 1 ≤ i ≤Mn,
1 ≤ j ≤Mn−1 denotes the mass matrices
An :(
An)
ij=(
∇ηn,i,∇ηn,j)
denotes the stiffness matrices.
Un satisfies
(
Bn + knAn)
~Un − ~F (U−n )kn = Bn,n−1
~Un−1, n ≥ 1,
where (~F (U−n ))i = (f(U−
n ), ηn,i).
Duality, Adjoints, Green’s Functions – p. 297/304
Analysis of a space-time finite element method
Use of quadrature yields standard difference schemes
ExampleIf Mn is constant and the lumped mass quadrature is used toevaluate the coefficients of Bn and Bn,n−1 = Bn, then theresulting set of equations for the dG(0) approximation is thesame as the equations for the nodal values of the backwardEuler difference scheme.
The dG(0) method is related to the backward Euler method, thecG(1) method is related to the Crank-Nicolson scheme, and thedG(1) method is related to the third order sub-diagonal Padédifference scheme.
Duality, Adjoints, Green’s Functions – p. 298/304
Analysis of a space-time finite element method
Under general assumptions, the cG(q) and dG(q) have order ofaccuracy q + 1 in time and 2 in space at any point.
In addition, they enjoy a superconvergence property in time attime nodes. The dG(q) method will have order of accuracy2q + 1 in time and the cG(q) method will have order 2q in time attime nodes for sufficiently smooth solutions.
The discontinuous Galerkin method has stability properties thatmake it well suited for the solution of dissipative problems.
The continuous Galerkin method is “energy” preserving and iswell-suited for problems with a conserved quantity.
Duality, Adjoints, Green’s Functions – p. 299/304
Analysis of a space-time finite element method
We use the second definition of the adjoint, so use
εi = εi(u, U) =
∫ 1
0εi(
us+ U(1 − s))
ds,
βij = βij(u, U) =
∫ 1
0
∂εj∂ui
(
us+ U(1 − s))∇(uis+ Ui(1 − s))
ds,
fij = fij(u, U) =
∫ 1
0
∂fj∂ui
(us+ U(1 − s))
ds.
Typically, ε and f are piecewise continuous with respect to t andcontinuous, H1 functions in space while β is discontinuous intime and space.
Duality, Adjoints, Green’s Functions – p. 300/304
Analysis of a space-time finite element method
The adjoint problem is
−φi −∇ ·(
εi∇φi)
+∑D
j=1 βji · ∇φj −∑D
j=1 fijφj = ψi,
(x, t) ∈ Ω × (tn, 0], 1 ≤ i ≤ D,
φi (x, t) = 0, (x, t) ∈ ∂Ω × (tn, 0], 1 ≤ i ≤ d,
φ(x, tn) = 0, x ∈ Ω,
ExampleIn the case of the scalar problem with constant diffusion, theadjoint problem is
−φ− ε∆φ− fφ = ψ, (x, t) ∈ Ω × (tn, 0],
φ(x, t) = 0, (x, t) ∈ ∂Ω × (tn, 0],
φ(x, tn) = 0, x ∈ Ω.
Duality, Adjoints, Green’s Functions – p. 301/304
Analysis of a space-time finite element method
ExampleIn the case of one parabolic equation with nonlinear diffusioncoupled to one ordinary differential equation, the dual problem is
−φ1 −∇ · ε1∇φ1 + β11∇φ1 − f11φ1 − f12φ2 = ψ1,
−φ2 + β12∇φ1 − f21φ1 − f22φ2 = ψ2,
φ1(x, t) = 0,
φ(x, tn) = 0.
Duality, Adjoints, Green’s Functions – p. 302/304
Analysis of a space-time finite element method
For the cG method, we obtain
∫ tn
0(e, ψ) dt = (e+(0), φ(0))
+
∫ tn
0
(
(U , πPφ−φ)+(ε(U)∇U,∇(πPφ−φ))−(f(U), πPφ−φ))
dt.
For the dG method, we obtain
∫ tn
0(e, ψ) dt = (e−(0), φ(0)) +
n∑
j=1
(
[U ]j−1, (πPφ− φ)+j−1
)
+
∫ tn
0
(
(U , πPφ−φ)+(ε(U)∇U,∇(πPφ−φ))−(f(U), πPφ−φ))
dt.
Duality, Adjoints, Green’s Functions – p. 303/304
Collaborators
Sean Eastman, Colorado State UniversityMike Holst, UCSDClaes Johnson, ChalmersMats Larson, Umea UniversityDuane Mikulencak, Georgia TechDavid Neckels, Colorado State UniversityTim Wildey, Colorado State UniversityRoy Williams, Caltech
Duality, Adjoints, Green’s Functions – p. 304/304
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