×

Close

- Finite Element Methods - FEM
- Note
**Jawaharlal Nehru Technological University Anantapur (JNTU) College of Engineering (CEP), Pulivendula, Pulivendula, Andhra Pradesh, India - JNTUACEP**- 11 Topics
**110 Views**- 4 Offline Downloads
- Uploaded 1 year ago

Touch here to read

Page-3

- The A bstract problem - ( 5 - 8 )
- Examples - ( 9 - 28 )
- The finite element method inits simplest form - ( 29 - 35 )
- Examples of finite elements - ( 36 - 52 )
- General properties of finite elements - ( 53 - 59 )
- Interpolation theory in solvable spaces - ( 60 - 66 )
- Applications to second-order problems over polygonal domains - ( 67 - 76 )
- Numerical integration - ( 77 - 94 )
- The obstacle problem - ( 95 - 102 )
- Conforming finite element method for the plate problem - ( 103 - 112 )
- Non- conforming methods for the plate problem - ( 113 - 128 )

Topic:

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

1. The Abstract Problem 5 σ f on K, it follows that the above map is linear if and only if K is a subspace. The problems associated with variational inequalities are therefore nonlinear in general. Exercise 1.1. Let V be as in Theorem 1.1. For f1 , f2 ∈ V ′ , let u1 , u2 5 be the corresponding solutions of (P). If || · ||∗ denotes the norm in V ′ , prove that 1 ||u1 − u2 || ≤ || f1 − f2 ||∗ . α Remark 1.4. The above exercise shows, in particular, the continuous dependence of the solution of f , in the sense described above. This together with the existence and uniqueness establishes that the problem (P) is “well-posed” in the terminology of partial differential equations. Exercise 1.2. If V is a normed linear space, K a given convex subset of V and J : V → R any functional which is once differentiable everywhere, then (i) if u ∈ K is such that J(u) = Min J(v), u satisfies, J ′ (u)(v − u) ≥ 0 v∈K for all v ∈ K. (ii) Conversely, if u ∈ K such that J ′ (u)(v − u) ≥ 0 for all v ∈ K, and J is everywhere twice differentiable with J ′′ satisfying J ′′ (v)(w, w) ≥ α||w||2 , for all v, w ∈ K and some α ≥ 0, then J(u) = Min J(v). v∈K Exercise 1.3 (1 ). Apply the previous exercise to the functional 1 a(v, v) = f (v) 2 with a(·, ·) and f as in Theorem 1.1. If K is a subspace of V, show that J ′ (u)(v) = 0 for all v ∈ K. In particular if K = V, J ′ (u) = 0. J(v) = It was essentially the symmetry of the bilinear form which provided the Hilbert space structure in Theorem 1.1. We now drop the symmetry assumption on a(·, ·) but we assume V to be a Hilbert space. In addition we assume that K = V. Theorem 1.2 (LAX-MILGRAM LEMMA). Let V be a Hilbert space. 6 1 Exercises (1.2)(i) and 1.3 together give relations (1.5)

1. The Abstract Problem 6 a(·, ·) a continuous, bilinear, V-elliptic form, f ∈ V ′ . If (P) is the problem: to find u ∈ V such that for all v ∈ V, (1.11) a(u, v) = f (v), then (P) has a unique solution in V 1 . Proof. Since a(·, ·) is continuous and V-elliptic, there are constants M, α > 0 such that |a(u, v)| ≤ M||u|| ||v||, (1.12) a(v, v) ≥ α||v||2 , for all u, v ∈ V. Fix any u ∈ V. Then the map v 7→ a(u, v) is continuous and linear. Let us denote it by Au ∈ V ′ . Thus we have a map A : V → V ′ defined by u 7→ Au. (1.13) || Au ||∗ = sup v∈V v,0 | Au(v)| |a(u, v)| = sup ≤ M||u||. ||v|| ||v|| v∈V v,0 Thus A is continuous and ||A|| ≤ M. We are required to solve the equation (1.14) Au = f. Let τ be the Riesz isometry, τ : V ′ → V so that (1.15) f (v) = ((τ f, v)), where ((·, ·)) denotes the inner product in V. Then, Au = f if and only if τ Au = τ f or equivalently. u = u − ρ(τ Au −τ f ), (1.16) 7 where ρ > 0 is a constant to be specified. We choose ρ such that g : V → V is a contraction map, where g is defined by (1.17) 1 g(v) = v − ρ(τAv − τ f ) cf. Corollary 1.1(b). for v ∈ V.

## Leave your Comments