Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/Aug05 667

Solution 4a
$$x_{k+1} = x_k - \frac{x_k - x_{k-1}}{f(x_k) - f(x_{k-1})}f(x_k) \!\,$$

Solution 4b
If $$f''(x) \!\,$$ is bounded and $$ x_0 \!\,$$ is close to $$ x_* \!\,$$, then the secant method has convergence order $$\phi = \frac{1+ \sqrt{5}}{2} \approx 1.6180339887 \!\,$$ (the Golden ratio).

A partial proof of this can be found here

Solution 5a
$$y(t_{n+1})-y(t_n) = \int_{t_n}^{t_{n+1}}y' = \int_{t_n}^{t_{n+1}} f(t,y) \approx \frac{h}{2}( f(t_{n+1},y_{n+1})+ f(t_n,y_n) ) \!\,$$

Solution 5b
Letting $$f(t,y) = \lambda y \!\,$$, we have

$$y_{n+1}=y_n+\frac{h}{2}\lambda y_{n+1}+\frac{h}{2} \lambda y_n \!\,$$

If we let $$h \lambda =z \!\,$$, and rearrange the equation we have

$$y_{n+1}= \underbrace{\left(  \frac{1+\frac{z}{2}}{1-\frac{z}{2}}\right)}_M y_n \!\,$$

We require $$M <1 \!\,$$. This is true if $$\lambda \!\,$$ is a negative real number.

Solution 5
We now want instead

$$\left\| \frac{1+\frac{h}{2}A}{1-\frac{h}{2}A} \right\| <1 \!\,$$

i.e.

$$ \|1+ \frac{h}{2}A \| < \| 1- \frac{h}{2} A \| \!\,$$

or (since $$A\!\,$$ is symmetric)

$$\rho(I+\frac{h}{2}Q \Lambda Q^T) < \rho(I-\frac{h}{2} Q \Lambda Q^T) \!\,$$

or (since multiplying by orthogonal matrices does not affect the norm)

$$\rho(I+\frac{h}{2}\Lambda ) < \rho(I-\frac{h}{2} \Lambda) \!\,$$

or (by definition)

$$\max_i \{ 1+\frac{h}{2}\lambda_i\} < \max_i \{ 1 - \frac{h}{2}\lambda_i \} \!\,$$

If $$A \!\,$$ is negative definite (all its eigenvalues are negative), the above inequality holds.

Variational Form
Derive the variational form by multiplying by test function $$v \in H \!\,$$ and integrating from 0 to 1. Use integration by parts and substitute initial conditions to then have:

Find $$u \in H = \{ v \in H^1(0,1) \} \!\,$$ such that for all $$ v \in H\!\,$$

$$\underbrace{\int_0^1 u'v' + \int_0^1 uv}_{a(u,v)} = \underbrace{\int_0^1 v}_{F(v)} \!\,$$

Relationship between (1) and (2)
(2) is an equivalent formulation of (1) but it does not involve second derivatives.

Existence of Unique Solution
By the Lax-Milgram theorem, we have the existence of a unique solution.


 * bilinear form continuous/bounded: $$ a(u,v) \leq C \|u\|_1 \|v\|_1 \!\,$$


 * bilinear form coercive: $$ a(u,u) \geq 1 \cdot \|u\|_1^2 \!\,$$


 * functional bounded: $$ F(v) \leq \|v\|_1 \!\,$$

$$ \begin{align} \int v &= \int 1 \cdot v \\ &\leq \|1\| \|v\|_0 \\ &\leq \|v\|_1 \end{align} \!\,$$

Solution 6b
Define hat functions $$\{ \phi_i \}_{i=1}^N \!\,$$ as basis of the discrete space. Note that $$\phi_0 \!\,$$ and $$\phi_n \!\,$$ have only half the support as the other basis functions. Using this basis we have

$$ \underbrace{ \begin{bmatrix} \frac{h}{3}+\frac{1}{h} & \frac{h}{6}-\frac{1}{h}   &                       & & & & \\ \frac{h}{6}-\frac{1}{h} & \frac{2}{3}h + \frac{2}{h} &\frac{h}{6}-\frac{1}{h}& & & & \\ & \ddots                    & \ddots                &\ddots &&& \\ &                           &                       &       &&\frac{h}{6}-\frac{1}{h}&\frac{h}{3}+\frac{1}{h} \end{bmatrix} }_A \underbrace{ \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix} }_U = \underbrace{ \begin{bmatrix} \frac{h}{2} \\ h \\ \vdots \\ \frac{h}{2} \end{bmatrix} }_F \!\,$$

Observe that $$A \!\,$$ is symmetric. It is positive definite by Gergoshin's theorem. The solution $$U \!\,$$ is unique since $$A \!\,$$ is diagonally dominant.

Solution 6c
If $$v \in V_1 \!\,$$ then it is also in $$V_2 \!\,$$ since $$ P_2 \!\,$$ is a refinement of $$P_1 \!\,$$. In other words, since $$ V_1\!\,$$ is piecewise linear over each intervals, it is also piecewise linear over a refinement of its interval.

Solution 6
From orthogonality of error, we have

$$a(u-u_h, v_h)=0 \!\,$$ for all $$v \in V_h \!\,$$

Specifically,

$$a(u-u_2, u_1-u_2)=0 \!\,$$

Then

$$\underbrace{a(u-u_1,u-u_1)}_{\|u-u_1\|_{H^1}^2}+2\underbrace{a(u-u_2,u_1-u_2)}_0=\underbrace{a(u-u_2,u-u_2)}_{\|u-u_2\|_{H^1}^2}+\underbrace{a(u_1-u_2,u_1-u_2)}_{\|u_1-u_2\|_{H^1}^2} \!\,$$