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Contents
Acknowledgements
v
Preface
vii
1 The Abstract Problem
1
2 Examples
9
3 The Finite Element Method in its Simplest Form
29
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4 Examples of Finite Elements
35
5 General Properties of Finite Elements
53
6 Interpolation Theory in Sobolev Spaces
59
7 Applications to Second-Order Problems...
67
8 Numerical Integration
77
9 The Obstacle Problem
95
10 Conforming Finite Element Method for the Plate Problem 103
11 Non-Conforming Methods for the Plate Problem
113
ix
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1. The Abstract Problem
2
Definition 1.1. Let V be a normed linear space. A bilinear form a(·, ·)
on V is said to be V-elliptic if there exists a constant α > 0 such that for
all v ∈ V.
a(v, v) ≥ α||v||2 .
(1.4)
2
Theorem 1.1. Let V be a Banach space and K a closed convex subset
of V. Let a(·, ·) be V-elliptic. Then there exists a unique solution for the
problem (P).
Further this solution is characterised by the property:
(1.5)
a(u, v − u) ≥ f (v − u)
for all
v ∈ K.
Remark 1.1. The inequalities (1.5) are known as variational inequalities.
Proof. The V-ellipticity of a(·, ·) clearly implies that if a(v, v) = 0 then
v = 0. This together with the symmetry and bilinearity of a(·, ·) shows
that a(·, ·) defines an inner-product on V. Further the continuity and the
V-ellipticity of a(·, ·) shows that the norm
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v ∈ V → a(v, v) 2
(1.6)
defined by the inner-product is equivalent to the existing norm on V.
Thus V acquires the structure of a Hilbert space and we apply the Riesz
representation theorem to obtain the following: for all f ∈ V ′ , there
exists σ f ∈ V such that
(1.7)
for all v ∈ V.
f (v) = a(σ f, v)
The map σ : V ′ → V given by f 7→ σ f is linear. Now,
1
a(v, v) − f (v)
2
1
= a(v, v) − a(σ f, v)
2
J(v) =
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1. The Abstract Problem
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1
1
= a(v − σ f, v − σ f ) − a(σ f, σ f ).
2
2
3
The symmetry of a(·, ·) is essential in obtaining the last equality.
For a given f , since σ f is fixed, J is minimised if and only if a(v −
σ f, v − σ f ) is minimised. But this being the distance between v and
σ f , our knowledge of Hilbert space theory tells us that since K is a
closed convex subset, there exists a unique element u ∈ K such that this
minimum is obtained. This proves the existence and uniqueness of the
solution, which is merely the projection of σ f over K.
We know that this projection is characterised by the inequalities:
(1.8)
a(σ f − u, v − u) ≤ 0 for all
v ∈ K.
Geometrically, this means that the angle between the vectors (σ f −u)
and (v − u) is obtuse. See Fig. 1.1.
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Figure 1.1:
Thus, a(σ f, v − u) ≤ a(u, v − u) which by virtue of (1.7) is precisely
the relation (1.5). This completes the proof.
We can state the following
Corollary 1.1. (a) If K is a non-empty closed convex cone with vertex at origin 0, then the solution of (P) is characterised by:
a(u, v) ≥ f (v) for all v ∈ K
(1.9)
a(u, u) = f (u).
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1. The Abstract Problem
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(b) If K is a subspace of V then the solution is characterised by
(1.10)
a(u, v) = f (v)
for all
4
v ∈ K.
Remark 1.2. The relations (1.5), (1.9) and (1.10) are all called variational formulations of the problem (P).
Proof.
(a) If K is a cone with vertex at 0, then for u, v ∈ K, u + v ∈ K.
(cf. Fig. 1.2). If u is the solution to (P), then for all v ∈ K applying
(1.5) to (u + v) we get a(u, v) ≥ f (v) for all v ∈ K. In particular
this applies to u itself. Setting v = 0 in (1.5) we get −a(u, u) ≥
− f (u) which gives the reverse inequality necessary to complete
the proof of (1.9). Conversely, if (1.9) holds, we get (1.5) by just
subtracting one inequality from the other.
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Figure 1.2:
(b) Applying (a) to K, since any subspace is a cone with vertex at 0,
we get (b) immediately. For if v ∈ K, then −v ∈ K and applying
(1.9) both to v and −v we get (1.10).
This completes the proof.
Remark 1.3. The solution u of (P) corresponding to f ∈ V ′ (for a fixed
a(·, ·)) defines a map V ′ → V. Since this solution is the projection of
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