Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/Aug09 667

Problem 4

Given the two-point boundary value problem $\{\begin{array}{c}-u^{″}+\alpha u=f(x),0\le x\le 1,\alpha >0\\ u^{\prime }(0)=A,\\ u^{\prime }(1)=B\end{array}$

Problem 4a

Set up the finite element approximation for this problem, based on piecewise linear elements in equidistant points. Determine the convergence rate in an appropriate norm

Solution 4a

Let $V=\{v\in H^1[0,1]\}$

Find $u\in V$ such that for all $v\in V$

$\underset{0}{\overset{1}{\int }}-u^{″}v+\alpha \underset{0}{\overset{1}{\int }}uv=\underset{0}{\overset{1}{\int }}fv$

or after integrating by parts and including initial conditions

$\underset{0}{\overset{1}{\int }}{u}^{\prime }{v}^{\prime }+\alpha \underset{0}{\overset{1}{\int }}uv=\underset{0}{\overset{1}{\int }}fv+Bv(1)-Av(0)$

Discrete Variational Form

$V_h=\{v\in H^1[0,1]:v=$ piecewise linear $\}$

$\{{\varphi }_i{\}}_{i=0}^{N+1}$ is basis for $V_h$;$h=\frac{1}{N+2}$

For $i=1,2,\dots ,N$

${\varphi }_i=\{\begin{array}{c}\frac{x-x_{i-1}}{h}\text{ for }x\in [x_{i-1},x_i]\\ \frac{x_{i+1}-x}{h}\text{ for }x\in [x_i,x_{i+1}]\\ 0\text{ otherwise}\end{array}$

${{\varphi }_i}^{\prime }=\{\begin{array}{c}\frac{1}{h}\text{ for }x\in [x_{i-1},x_i]\\ -\frac{1}{h}\text{ for }x\in [x_i,x_{i+1}]\\ 0\text{ otherwise}\end{array}$

$\int {\varphi }_i{\varphi }_j=\{\begin{array}{c}\frac{2}{3}h\text{ for }i=j\\ \frac{1}{6}h\text{ for }|i-j|=1\\ 0\text{ otherwise}\end{array}$

$\int {{\varphi }_i}^{\prime }{{\varphi }_j}^{\prime }=\{\begin{array}{c}\frac{2}{h}\text{ for }i=j\\ -\frac{1}{h}\text{ for }|i-j|=1\\ 0\text{ otherwise}\end{array}$

${\varphi }_0=\{\begin{array}{c}\frac{x_1-x}{h}\text{ for }x\in [x_0,x_1]\\ 0\text{ otherwise }\end{array}$

${\varphi }_{N+1}=\{\begin{array}{c}\frac{x-x_N}{h}\text{ for }x\in [x_N,x_{N+1}]\\ 0\text{ otherwise }\end{array}$

Find $u_h\in V_h$ such that for all $v\in V_h$

$\int {u_h}^{\prime }{v_h}^{\prime }+\alpha \underset{0}{\overset{1}{\int }}{u}_h{v}_h=\underset{0}{\overset{1}{\int }}f{v}_h+Bv(1)-Av(0)$

Since $\{{\varphi }_i{\}}_{i=0}^{N+1}$ forms a basis

$u_h=\sum _{i=0}^{N+1}{u}_{h_i}{\varphi }_i$

Therefore we have system of equations

For $j=0,1,\dots ,N+1$

$\sum _{i=0}^{N+1}{u}_{h_i}\underset{0}{\overset{1}{\int }}{{\varphi }_i}^{\prime }{{\varphi }_j}^{\prime }+\alpha \sum _{i=0}^{N+1}{u}_{h_i}\underset{0}{\overset{1}{\int }}{\varphi }_i{\varphi }_j=\underset{0}{\overset{1}{\int }}f{\varphi }_j+B{\varphi }_j(1)-A{\varphi }_j(0)$

$\left[\begin{array}{cccc}\alpha \frac{1}{3}h+\frac{1}{h}& \frac{\alpha }{6}h-\frac{1}{h}& & \\ \frac{\alpha }{6}-\frac{1}{h}& \alpha \frac{2}{3}h+\frac{2}{h}& \frac{\alpha }{6}h-\frac{1}{h}& \\ \ddots & \ddots & \ddots & \\ & \frac{\alpha }{6}h-\frac{1}{6}& \alpha \frac{2}{3}h+\frac{2}{h}& \frac{\alpha }{6}h-\frac{1}{h}\\ & & \frac{\alpha }{6}h-\frac{1}{h}& \alpha \frac{1}{3}h+\frac{1}{h}\end{array}\right]\left[\begin{array}{c}{u}_{h_0}\\ u_{h_1}\\ ⋮\\ u_{h_N}\\ u_{h_{N+1}}\end{array}\right]=\left[\begin{array}{c}\underset{0}{\overset{1}{\int }}f{\varphi }_0-A\\ \underset{0}{\overset{1}{\int }}f{\varphi }_1\\ ⋮\\ \underset{0}{\overset{1}{\int }}f{\varphi }_N\\ \underset{0}{\overset{1}{\int }}f{\varphi }_{N+1}+B\end{array}\right]$

Convergence Rate

In general terms, we can use Cea's Lemma to obtain

$\Vert u-u_h{\Vert }_1\le C\underset{v_h\in V_h}{\inf }\Vert u-v_h{\Vert }_1$

In particular, we can consider $v_h$ as the Lagrange interpolant, which we denote by $v_I$. Then,

$\Vert u-u_h{\Vert }_1\le C\Vert u-v_I{\Vert }_1\le Ch\Vert u{\Vert }_{H^2(0,1)}$.

It's easy to prove that the finite element solution is nodally exact. Then it coincides with the Lagrange interpolant, and we have the following punctual estimation:

$\Vert u(x)-u_h(x){\Vert }_{L^{\mathrm{\infty }}([x_{i-1},x_i])}\le \underset{x\in [x_{i-1},x_i]}{\max }\frac{u^{(2)}(x)}{(2)!}{\left(\frac{h}{2}\right)}^2$

Problem 4b

Explain whether $\alpha >0$ is necessary for the convergence in part (a).

Solution 4b

If $\alpha \ge 0$, then the stiffness matrix is diagonally dominant and hence solvable.