Chapter 1
The Structure of Finite Element Method
The finite element method is a numerical computational method for differential equations and
partial differential equations. In order to solve the general field problem by using finite element
method, it must pass through the following processes:
1 Find the variational formulation associated with original field problem.
2 Establish finite element subspace. For example, select the element type and associated
phase functions.
3 Establish element stiffness matrix, element column and assemble global stiffness matrixfull
column.
4 Treatment of the boundary conditions and solving of the system of finite element equations.
5 Come back to the real world.
In this book, the first four processes will be systematic formulations in the first chapter till
third chapter.
1.1 Galerkin Variational Principle and Ritz Variational Principle
As an example, we consider the linear elliptic boundary value problem of two dimension,
1.1.1
where, Ω is a connected domain in R2, .Ω = Γ1 ∪ Γ2 is a piecewise smooth boundary. Let
n denote the unit outward normal vector to .Ω defined almost everywhere on .Ω. px, y ∈
C1Ω, px, y ≥ p0 0, σx, y ∈ C0Ω and σx, y ≥ 0.
Throughout this chapter we make notation: C0Ω = the set of all continuous function in
an open subset in Rn. CkΩ = the set of functions v ∈ C0Ω, whose derivatives of order k,
exist and are continuous;
where α = α1, ? ? ? , αn, |α| = α1 + ? ? ? + αn.
Assume that ux, y ∈ C2Ω satisfies 1.1.1 in Ω and on .Ω, the function ux, y is called
classical solution of problem 1.1.1.
Next, we consider weak solution of 1.1.1. Define the norm
1.1.2
Sobolev space H1Ω is a closure of C∞Ω, under the norm 1.1.2 with the inner product
1.1.3 H1Ω is a Hilbert space which is called one order Sobolev space. Let
C∞0 Ω = {v : v is an infinite differentiable function and support of v . Ω},
H10 Ω = the closure of C∞0 Ω under the norm1.1.2,
it is equivalent to
H10 Ω = {v : v ∈ H1Ω, v|.Ω = 0}.
In addition, let
C∞# Ω = {v : v ∈ C∞Ω, v|Γ1 = 0},
V Ω = closure ofC∞# Ω under the norm1.1.2,
which is equivalent to
V = {v : v ∈ H1Ω, v|Γ1 = 0}.
It is clear that V is a Hilbert space with inner product 1.1.3. Furthermore,
H10 Ω . V . H1Ω.
Let us introduce bilinear functional
1.1.4
In 1.1.4, fixed u, then Bu, v is a linear functional of v, while v is fixed, it is a linear functional
of u. In other words, suppose α1, α2, β1, β2 are arbitrary constants, then
Bα1u1 + α2u2, β1v1 + β2v2 =α1β1Bu1, v1 + α1β2Bu1, v2
+ α2β1Bu2, v1 + α2β2Bu2, v2, .u1, u2, v1, v2 ∈ H1Ω.
It is clear that 1.1.4 satisfies
1 Symmetry,
Bu, v = Bv, u. 1.1.5
2 The continuity in V × V , i.e., there exists a constant M 0, such that
|Bu, v| M u 1,Ω v 1,Ω, .u, v ∈ V. 1.1.6
3 Coerciveness in V , i.e., there exists constant γ 0, such that
Bu, u γ u 2 1,Ω, .u ∈ V. 1.1.7
Of course,
is a continuous linear functional in v.
The Galerkin Variational Formulation for 1.1.1: Find u ∈ V , such that
Bu, v = fv, .v ∈ V. 1.1.8
A solution u satisfying 1.1.8 is called a weak solution of 1.1.1. The space V is called
admissible space or trial space. On the other hand, 1.1.8 must be satisfied for every v ∈ V ,
therefore, V is called test function space. If trial and test space for the variational problem are
the same Hilbert V , in this case, V is called energy space.
Owing to the boundary condition on Γ2 is contained in the variational problem 1.1.8, the
boundary condition on Γ2 is called nature boundary condition, while the boundary condition
on Γ1 is called essential boundary condition.
The following proposition gives the relationship between classical solution and weak solution
of 1.1.1.
Proposition 1.1 Suppose u ∈ C2Ω. If u is a classical solution of 1.1.1, then, u is
the weak solution of 1.1.1. Otherwise, if u is a weak solution of 1.1.1, then u is a classical
solution of 1.1.1.
Proof Assume that u ∈ C2Ω is a classical solution of 1.1.1, .v ∈ V , multiplying both
sides of 1.1.1 by v and integrating
Applying Gauss theorem
In view of v ∈ V , and u satisfying boundary condition we have
Bu, v = fv, .v ∈ V,
i.e., u satisfies 1.1.7.
Conversely, let u ∈ C2Ω be a solution of 1.1.8, owing to v|Γ1 = 0, we obtain
Substituting above equity into 1.1.8 leads to
By the arbitrary of v ∈ V , it yields that u is a classical solution of 1.1.1. The proof is
complete.
The following Lax-Milgram theorem guarantees the existence of the Garlerkin variational
problem 1.2.8.
Theorem 1.1Lax-Milgram Theorem Let V be a Hilbert space, Bu, v is a bilinear
functional in V × V and satisfies:
Symmetry Bu, v = Bv, u, .u, v ∈ V. 1.1.9
Continuity There exists a positive constant M independent of u, v, such that
|Bu, v| M u v , .u, v ∈ V. 1.1.10
Coerciveness There exists a constant
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