Partial differential equations and the finite element method 3-FEA1.md

This part is devoted to introduce the Galerkin method and its important special case, the Finite element method.

Consider the general framework

Let $V$be a Hilbert space, $a(., .): V\times V\rightarrow R$ a bilinear form and $l\in V'$. It is our task to find $u\in V$ such that
(1)

$$a(u,v)=l(v)$$

We assume that the bilinear form $a ( . , .)$ is bounded and V-elliptic, i.e., that there exist constants $C_b, C_{el} > 0$ such that

$$|a(u,v)|\leq C_b \parallel u\parallel_V\parallel v\parallel_V$$
and
$$a(u,v)\geq C_{el}\parallel v\parallel^2_V$$

Recall that the weak problem (1) has a unique solution by the Lax-Milgram lemma.

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The Galerkin method

As $V$ is infinite dim function space. The Galerkin method is based on a sequence of finite dimensional subspaces $\{V_n\}_{n=1}^{\infty}\subset V$, $V_n\subset V_{n+1}$, that fill the space $V$ in the limit. In each finite-dimensional space $V_n$, problem (1) is solved exactly. It can be shown that under suitable assumptions the sequence of the approximate solutions $\{u_n\}_{n=1}^{\infty}$ converges to the exact solution of problem (1).

Discrete problem

Find $u_n\in V_n$ such that
(2)

$$a(u_n,v)=l(v),v\in V_n$$

Lemma 1 (Unique solvability) Problem (2) has a unique solution $u_n\in V_n$,.

Proof: Recall Lax-Milgram lemma : Let $V$ be a Hilbert space, $a(., .): V\times V\rightarrow R$ a bounded V-elliptic bilinear form and $l\in V'$ Then there exists a unique solution to the problem

$$a(u,v)=l(v)$$ .

Suppose space $V$ has a finite basis $\{v_n\}_{n=1}^{N_n}$, then

$$u_n=\sum_{j=1}^{N_n} y_j v_j$$
where $y_j$s are unknown coefficients.
$$a(u_n,v_i)=\sum_{j=1}^{N_n} y_j a(v_j,v_i)=l(v_I)$$
Denote $S_n=\{a(v_j,v_i)\}_{i,j=1}^N$, $F_n=\{l(v_i)\}$, $Y_n=\{y_j\}$, then
(3) $$S_n Y_n=F_n$$ .

Lemma 2 (Positive definiteness of $S_n$) Let $V_n$ be a Hilbert space and $a(., .): V\times V\rightarrow R$a bilinear V-elliptic form. Then the stiffness matrix $S_n$ , of the discrete problem (3) is positive definite.

About the error $u-u_n$

Lemma 3 (Orthogonality of error for elliptic problems) Let $u\in V$ be the exact solution of the continuous problem (I ) and $u_n$ the exact solution of the discrete problem (3).
Then the error $e_n = u - u_n$, satisjies

$$a(u - u_n,v) = 0, \forall v\in V_n$$ .
Proof:Because $V_n\subset V$, then
$$a(u - u_n,v) =a(u,v)-a(u_n,v)=0.$$

Remark 1 (Geometrical interpretation) If the bilinear form $a ( . , .)$ is symmetric, it induces an energetic inner product, then

$$(e_n,v)_e = 0, \forall v \in V_n$$ ,

i.e., that the error of the Galerkin approximation $e_n = u - u_n$ is orthogonal to the Galerkin subspace $V_n$ in the energetic inner product. Hence the approximate solution u, E V, is an orthogonal projection of the exact solution $u$ onto the Galerkin subspace $V_n$ in the energetic inner product, and thus it is the nearest element in the space $V_n$ to the exact solution $u$ in the energy norm,

$$\parallel u-u_n\parallel_e=\inf_{v\in V_n}\parallel u-v\parallel_e$$

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FEA

Let $\Omega\subset R^d$ , where $d$ is the spatial dimension, be an open bounded set. If the Hilbert space $V$ consists of functions defined in $\Omega$ and the Galerkin subspaces $V_n$ comprise piecewise-polynomial functions, the Galerkin method is called the Finite element method (FEM).

Consider the model equation

$$-\nabla (a_1\nabla u)+a_0u=f,$$
where $f \in L^2(\Omega)$, in a bounded interval $\Omega = (a, b ) \subset R$, equipped with the homogeneous Dirichlet boundary conditions. At the beginning let $a_1$ and $a_0$ be constants and assume a simple load function of the form $f(x)=1$.

The Galerkin procedure assumes a sequence of finite-dimensional subspaces

$$V_1\subset V_2\subset\ldots\subset V$$

Consider a partition

$$a=x_0^{(n)}<x_1^{(n)}<\ldots<x_{M_n}^{(n)}=b$$
and define the finite element mesh
$$\mathcal {T}_n=\{K_1^{(n),K_2^{(n),\ldots,K_{M_n}^{(n)\}$$
The open intervals

$$K_i^{(n)=(x_{i-1}^{(n)},x_i^{(n)})$$ are called finite elements, and the value $$h(n)=max (x_{i}^{(n)}-x_{i-1}^{(n)})$$
is said to be the mesh diameter.

The piecewise linear basis functions $v_j$, satisfy $v_j(x_i)=\delta_{ij}$.

Then by (3), we can get $u_n$.