# Ngô Quốc Anh

## January 22, 2010

### Finite element method: A brief introduction via an 2-point boundary value problem of laplacian equation in 1D

Filed under: Các Bài Tập Nhỏ, Giải tích 9 (MA5265), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 17:05

Followed by this topic where I discussed how to use Finite Difference Method in numerical analysis. Today, we will study a so-called Finite Element Method, an improved Finite Difference Method. We first consider the following problem

$-u''(x)=f(x)$ where $0 \leq x \leq 1$ with boundary conditions $u(0)=0=u(1)$.

The above problem is usually called 2-point boundary value problem of laplacian equation in 1D.

We first consider such problem by using Finite Difference Method. The idea of using the Finite Difference Method is to split equally the interval $[0,1]$ into $n$ pieces by $x_i=\frac{i}{n}$ where $i=\overline{0,n}$. We then approximate $u''$ by using the following

$\displaystyle u''({x_i}) = \frac{{{u_{i - 1}} - 2{u_i} + {u_{i + 1}}}}{{{h^2}}}, \quad i=\overline{1,n-1}$

where $h:=\frac{1}{n}$. Therefore, we have the following system of linear equations

$\displaystyle\frac{{ - 1}}{{{h^2}}}\left( {\begin{array}{*{20}{c}} { - 2} & 1 & {} & {} & {}\\ 1 & { - 2} & 1 & {} & {} \\ {} & {} & {} & {} & {}\\ {} & {} & {} &\ddots& 1\\ {} & {} & {} & 1 & 2 \\ \end{array} } \right) = \left( {\begin{array}{*{20}{c}} {f({x_1})}\\ {f({x_1})}\\ \vdots \\ {f({x_{n - 1}})}\\ \end{array} } \right).$

The idea of the Finite Element Method is to split the interval $[0,1]$ into $n$ pieces by $x_i$ where $i=\overline{0,n}$ but not necessarily equality. In order to deal with the problem, we shall look for weak solutions. In the sense of the distribution, $u$ is called a weak solution if

$\displaystyle\int_0^1 {\nabla u\nabla vdx} = \int_0^1 {f(x)vdx} ,\quad\forall v \in H_0^1([0,1])$.

Under some conditions, for example, if $f \in L^2([0,1])$, the existence of weak solution is well-understood via the Lax-Milgram lemma.

Let $V_h$ be a collection of piecewise linear map $g$ in $[0,1]$ so that $g$ is linear on every intervals $[x_{i-1},x_i]$ where $i=\overline{1,n}$. We also suppose that $g$ is such that $g(0)=g(1)=0$. Clearly the space $V_h$ is finite dimension with a basis $\{\phi_i\}_{i=1}^{n-1}$ constructed as follows

$\displaystyle {\phi _i}(x) = \left\{ \begin{gathered}1,\quad x = {x_i}, \hfill \\ 0 , \quad x \ne {x_i}. \hfill \\ \end{gathered}\right.$

It is easy to see that $\phi_i \in H_0^1([0,1])$.

We will find $u$ as a linear combination of these functions $\phi_i$ denoted by $u_h$, to be exact, we will approximate $u$ by $u_h$ such that at $u(x_i)=u_h(x_i)$ for every $i=\overline{1,n-1}$. Precisely, let

$\displaystyle {u_h}(x) = \sum\limits_{i = 1}^{n - 1} {{\xi _i}{\phi _i}(x)}$

we then have

${\xi _i}=\displaystyle u({x_i}), \quad i = \overline {1,n - 1}$.

From the equation defining the notion of weak solution we replace $v$ by $\phi _i$ and $u$ by $u_h$, we now get

$\displaystyle\int_0^1 {\left( {\sum\limits_{j= 1}^{n - 1} {u({x_j}){{\phi '}_j}(x)} } \right){{\phi '}_i}(x)dx} = \int_0^1 {f(x){\phi _i}(x)dx}$

which yields

$\displaystyle\sum\limits_{j = 1}^{n - 1} {u({x_j})\int_0^1 {{{\phi '}_j}(x){{\phi '}_i}(x)dx} } = \int_0^1 {f(x){\phi _i}(x)dx}$.

The above identity can be rewritten as a linear system $Ax=b$ as the following

$\displaystyle\left( {\begin{array}{*{20}{c}} {({\phi' _1},{\phi' _1})} & {({\phi '_2},{\phi '_1})} &\cdots& {} & {} &\cdots& {({\phi' _{n - 1}},{\phi' _1})}\\{({\phi '_1},{\phi' _2})} & {({\phi' _2},{\phi' _2})} & \cdots & {} & {} & {} & \vdots\\ \vdots&\vdots&\ddots& {} & {} & {} & {}\\{} & {} & {} & {} & {} & {} & {}\\{} & {} & {} & {} & {} & {} & {}\\ \vdots&\vdots& {} & {} & {} &\ddots&\vdots \\{({\phi '_1},{\phi' _{n - 1}})} & {({\phi' _2},{\phi' _{n - 1}})} &\cdots& {} & {} &\cdots& {({\phi' _{n - 1}},{\phi' _{n - 1}})}\\ \end{array} } \right)\left( {\begin{array}{*{20}{c}} {u({x_1})}\\{u({x_2})}\\ \vdots \\ \vdots \\ \vdots \\{u({x_{n - 2}})}\\{u({x_{n - 1}})}\\ \end{array} } \right) = \left( {\begin{array}{*{20}{c}} {(f,{\phi _1})}\\{(f,{\phi _2})}\\ \vdots \\ \vdots \\ \vdots \\{(f,{\phi _{n - 2}})}\\{(f,{\phi _{n - 1}})}\\ \end{array} } \right)$

where we use the following notation

$\displaystyle (f,g) = \int_0^1 {f(x)g(x)dx}$.