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

Solution 4a
Let $$V= \{v \in H^1[0,1] \} \!\,$$

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

$$\int_0^1 -u''v + \alpha \int_0^1 uv = \int_0^1 fv \!\,$$

or after integrating by parts and including initial conditions

$$ \int_0^1 u'v' + \alpha \int_0^1 uv = \int_0^1 fv + Bv(1)-Av(0)\!\,$$

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

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

For $$i=1,2,\ldots,N \!\,$$

$$\phi_i = \begin{cases} \frac{x-x_{i-1}}{h} \mbox{ for } x \in [x_{i-1}, x_i ] \\ \frac{x_{i+1}-x}{h} \mbox{ for } x \in [x_i, x_{i+1}] \\ 0 \mbox{ otherwise} \end{cases} \!\,$$

$$ \phi_i'= \begin{cases} \frac1h \mbox{ for } x \in [x_{i-1},x_i] \\ -\frac1h \mbox{ for } x \in [x_i, x_{i+1}] \\ 0 \mbox{ otherwise} \end{cases} \!\,$$

$$ \int \phi_i \phi_j = \begin{cases} \frac23 h \mbox{ for } i=j \\ \frac16 h \mbox{ for } |i-j| =1 \\ 0 \mbox{ otherwise} \end{cases} \!\,$$

$$ \int \phi_i' \phi_j' = \begin{cases} \frac2h \mbox{ for } i=j \\ -\frac1h \mbox{ for } |i-j|=1 \\ 0 \mbox{ otherwise} \end{cases} \!\,$$

$$ \phi_0 = \begin{cases} \frac{x_1-x}{h} \mbox{ for } x \in [x_0,x_1] \\ 0 \mbox{ otherwise } \end{cases} \!\,$$

$$ \phi_{N+1} = \begin{cases} \frac{x-x_N}{h} \mbox{ for } x\in [x_N, x_{N+1}] \\ 0 \mbox{ otherwise } \end{cases} \!\,$$

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

$$\int u_h' v_h' + \alpha \int_0^1 u_h v_h = \int_0^1 f v_h + Bv(1)-Av(0) \!\,$$

Since $$\{ \phi_i \}_{i=0}^{N+1} \!\,$$ forms a basis

$$u_h = \sum_{i=0}^{N+1}u_{h_i} \phi_i \!\,$$

Therefore we have system of equations

For $$j=0,1,\ldots, N+1 \!\,$$

$$\sum_{i=0}^{N+1} u_{h_i}\int_0^1 \phi_i' \phi_j' + \alpha \sum_{i=0}^{N+1} u_{h_i} \int_0^1 \phi_i \phi_j = \int_0^1 f \phi_j + B\phi_j(1) -A \phi_j(0) \!\,$$

$$ \begin{bmatrix} \alpha \frac13 h + \frac1h & \frac{\alpha}{6} h - \frac{1}{h} & & \\ \frac{\alpha}{6} -\frac1h & \alpha\frac23 h + \frac{2}{h} & \frac{\alpha}{6}h - \frac{1}{h} & \\ \ddots&\ddots & \ddots& \\ & \frac{\alpha}{6} h -\frac16 & \alpha \frac23 h + \frac2h & \frac{\alpha}{6} h - \frac1h \\ &                             &       \frac{\alpha}{6} h -\frac1h & \alpha \frac13 h + \frac1h \end{bmatrix}

\begin{bmatrix} u_{h_0}\\ u_{h_1}\\ \vdots \\ u_{h_N}\\ u_{h_{N+1}} \end{bmatrix} = \begin{bmatrix} \int_0^1 f \phi_0 -A \\ \int_0^1 f \phi_1 \\ \vdots \\ \int_0^1 f \phi_N \\ \int_0^1 f \phi_{N+1} +B \end{bmatrix}

\!\,$$

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

$$ \| u - u_h\|_1 \leq C\inf_{v_h \in V_h} \| u - v_h \|_1   $$

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

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

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:

$$ \|u(x) - u_h(x)\|_{L^{\infty}([x_{i-1},x_i])} \leq \max_{x \in [x_{i-1},x_i]} \frac{u^{(2)}(x)}{(2)!} \left(\frac{h}{2}\right)^2 $$

Solution 4b
If $$\alpha \geq 0 \!\,$$, then the stiffness matrix is diagonally dominant and hence solvable.