Contents Acknowledgements v Preface vii 1 The Abstract Problem 1 2 Examples 9 3 The Finite Element Method in its Simplest Form 29 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 ix 113
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 1 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) f (v) = a(σ f, v) for all v ∈ 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) =
1. The Abstract Problem 3 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. 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).
1. The Abstract Problem 4 (b) If K is a subspace of V then the solution is characterised by (1.10) a(u, v) = f (v) for all 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. 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 4