Consider the following boundary value problem:
$u^4=f,$ $in\ (0,1),$
$u(0)=u^′(0)=u(1)=u^′(1)=0$,
The weak formulation of the problem is
$u ∈ V,\ ∫_0^1u^"v^"dx=\int_0^1f\ v \ dx \ ∀ v ∈ V,$
where $V = H_0^2(0, 1)$. If we choose the finite element space $V_h$ to be a
subspace of $V$ , then any function in $V_h$ must be $C^1$ continuous.Suppose
$V_h$ consists of piecewise polynomials of degree less than or equal to $p$. The requirement that a finite element function be $C^1$ is equivalent to the $C^1$ continuity of the function across the interior nodal points $\{x_i\},i=1,…,N-1$ ,which places $2 (N−1)$ constraints. Additionally,the Dirichlet boundary conditions impose $4$ constraints.Hence,
$dim (V_h) = (p + 1)N − 2 (N − 1) − 4 = (p − 1)N − 2.$ Now it is evident that the polynomial degree p must be at least 2.However, with $p=2$, we cannot construct basis functions with small supports.Thus we should choose $p$ to be at least $3$. For $p=3$, our finite element space is taken to be
$V_h = \{v_h ∈ C^1(I) |\ v_h|I_i ∈ P_3(I_i),\ 1 ≤ i ≤ N,\ v_h(x) = v^′_h(x) = 0,\ at\ x = 0, 1\}.$It is then possible to construct basis functions with small supports using interpolation conditions of the function and its first derivative at the interior nodes $\{x_i\},i=1,…,N-1.$More precisely, associated with each interior node $x_i$, there are two basis functions $φ_i$ and $ψ_i$ satisfying the interpolation conditions
$φ_i(x_j) = δ_{ij},\ φ^′_i(x_j)=0,$
$ψ_i(x_j)=0,\ ψ^′_i(x_j) = δ_{ij} .$
A more plausible approach to constructing the basis functions is to use the
reference element technique.To this end,choose $I_0 = [0, 1]$ as the
reference element.Then the mapping $F_i : I_0 → I_i,F_i(ξ) = x_{i−1}+h_iξ.$ is a bijection between $I_0$ and $I_i$.Over the reference element $I_0$, we construct cubic functions $Φ_0, Φ_1, Ψ_0$, and $Ψ_1$ satisfying the interpolation conditions
$Φ_0(0) = 1, Φ_0(1) = 0, Φ^′_0(0) = 0, Φ^′_0(1) = 0,$
$Φ_1(0) = 0, Φ_1(1) = 1, Φ^′_1(0) = 0, Φ^′_1(1) = 0,$
$Ψ_0(0) = 0, Ψ_0(1) = 0, Ψ^′_0(0) = 1, Ψ^′_0(1) = 0,$
$Ψ_1(0) = 0, Ψ_1(1) = 0, Ψ^′_1(0) = 0, Ψ^′_1(1) = 1.$
we find these functions:
$Φ_0(ξ) = (1+2ξ) (1 − ξ)^2,
Φ_1(ξ) =(3 − 2ξ) ξ^2,
Ψ_0(ξ)=ξ (1 − ξ)^2,
Ψ1(ξ) = −(1 − ξ)ξ^2.$These functions, defined on the reference element, are called shape functions. With the shape functions, it is an easy matter to construct the basis functions with the aid of the mapping functions $\{F_i\},i=1,…,N-1$ .We have
$φ_i(x)=\begin{cases}
Φ_1(F^{-1}_i (x)),\ x ∈ I_i, \\
Φ_0(F^{−1}_{i+1}(x)),\ x ∈ I_{i+1} \\
0, \ otherwise \\
\end{cases}$ and $ψ_i(x)=\begin{cases}
h_iΨ_1(F^{−1}_i(x)),\ x ∈ I_i, \\
h_{i+1}Ψ_0(F^−1_{i+1}(x)),\ x ∈ I_{i+1} \\
0, \ otherwise \\
\end{cases}.$the computations of the stiffness
matrix and the load are done on the reference element.
In view of the above, I would like to know:
1) Why are two basic functions defined?
2) How are the stiffness matrix and the load vector calculated for solving the finite element system?
3) Given the two basic functions, what is the approximate answer?
please guide me.
Best Answer
The function $\phi$ is a hat function with flat derivatives at the nodes. The function $\psi$ is a "wavelet" with value zero at the nodes and derivative $1$ at the middle node. You need both to get the freedom of arbitrary values and derivatives at the nodes.
This happens as usual, using all the basis functions. Set $u=\sum a_i\phi_i+\sum b_i\psi_i$ and cycle $v$ through the set of $\phi_i$, $\psi_i$. That is, for the equation $B(u,v)=F(v)$, $B(u,v)=\int_0^1u''v''dx$, $F(v)=\int_0^1fvdx$, set up the system \begin{align} \sum a_iB(\phi_i,\phi_j)+\sum b_iB(\psi_i,\phi_j)&=F(\phi_j)\\ \sum a_iB(\phi_i,\psi_j)+\sum b_iB(\psi_i,\psi_j)&=F(\psi_j) \end{align} The system matrix in this form alternating $\phi$ and $\psi$ should still be a banded diagonal matrix, if the step sizes are equal then it has a periodic pattern.
Then after establishing the linear system for the coefficients, solve it and compose the solution from it.