Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/Jan06 667

Solution 4a
Newton's method:

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

Solution 4b
From the picture, notice that if $$x_k < x_* \!\,$$, then after one step $$x_{k+1} \!\,$$ will be greater than $$x_* \!\,$$. This is because from hypothesis, the function is always increasing and concave up.

Then without loss of generality assume $$x_k > x_* \!\,$$.

Subtracting $$x_* \!\,$$ from both sides of Newton's method gives an expression for the relationship between consecutive errors.

$$ \underbrace{x_{k+1}- x_*}_{e_{k+1}} = \underbrace{x_k - x_*}_{e_k} - \frac{f(x_k)}{f'(x_k)} \qquad (*) \!\,$$

Expanding $$f(x_k) \!\,$$ around $$f(x_*) \!\,$$ using Taylor expansion gives

$$f(x_k)= \underbrace{f(x_*)}_0 + f'(\xi) \underbrace{(x_k- x_*)}_{e_k} \!\,$$

where $$\xi \in [x_*, x_k] \!\,$$

Substituting this expression into (*), we have

$$e_{k+1}=(1-\underbrace{\frac{f'(\xi)}{f'(x_k)}}_M)e_k \!\,$$

Since $$f'(x) > 0 \!\,$$ and is always increasing (from hypothesis), $$ M \!\,$$ is a positive number less than 1. Therefore the error decreases as $$ k\!\,$$ increases which implies the method always converges.

Define Suitable Subspace
$$V_N = \{ v \in H_0^1 [0,1] : v \mbox{ piecewise linear} \} \!\,$$

which has a basis the hat functions $$\{\phi_i\}_{i=1}^{N-1} \!\,$$ defined as follows:

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

How to Determine Coefficients
The discrete weak variational form is given as such:

Find $$u_N \in V_N \!\,$$ such that for all $$v \in V_N \!\,$$

$$\underbrace{\int_0^1 u_N'v_N'}_{a(u_N,v_N)} = \underbrace{\int_0^1 f v_N}_{F(v_N)} \!\,$$

Since we have a basis $$\{\phi_i\}_{i=1}^{N-1} \!\,$$, we have a system of equations (that can be expressed in matrix form):

For $$j=1,2,\ldots,N-1 \!\,$$

$$\sum_{i=1}^{N-1} u_i \int_0^1 \phi_i \phi_j = \int_0^1 f \phi_j \!\,$$

Existence of Unique Solution
The existence of a unique solution follows from Lax-Milgram.

Note the following:


 * bilinear form continuous (bounded) e.g. $$a(u_N,v_N) \leq \|u_N\|_1 \|v_N\|_1 \!\,$$

$$ \begin{align} a(u,v) &= \int u'v' \\ &\leq |u|_1 |v|_1 \mbox{ Cauchy Schwartz in L2 }\\ &\leq \|u\|_1 \|v\|_1 \mbox{ dominance of spaces } \end{align} \!\,$$


 * bilinear form coercive e.g. $$a(u_N,u_N) \geq C \|u_N\|_1^2 \!\,$$

$$ \begin{align} a(v,v) &= \int v'^2 \\ &= |v|_1^2 \\ &\geq C\|v\|_1^2 \mbox{ Poincare inequality} \end{align} \!\,$$

Poincare Inequality: $$\|v\|_0 \leq \|v\|_1 \leq C |v|_1 \!\,$$


 * $$F(v_N) \leq C\|v_N\|_1 \!\,$$

$$ \begin{align} \int fv &\leq \|f\|_0 \|v\|_0 \\ &\leq \|f\|_1 \|v\|_1 \end{align} \!\,$$

Solution 5b

 * $$ a(u,v)=a(v,u) \!\,$$


 * $$ a(\alpha u, v) = \int \alpha u' v' = \alpha \int u'v'= \alpha a(u,v) \!\,$$


 * $$a(u+w,v) = \int (u'+w')v' = \int u'v' + \int w'v' = a(u,v)+a(w,v) \!\,$$


 * $$ a(u,u) = \int u'^2 \neq 0 \mbox { if } u \neq 0 \!\,$$

Solution 5c
$$ \phi_1 = \begin{cases} 2x &\mbox{ for } x \in [0, \frac12] \\ 2-2x &\mbox{ for } x \in [\frac12, 1] \end{cases} \!\,$$

$$ \phi_2= \begin{cases} 8x              & \mbox { for } x \in [0,\frac14 ] \\ -8(x-\frac12)   &\mbox{ for } x \in [\frac14, \frac12 ] \\ 0               &\mbox{ otherwise } \end{cases} \!\,$$

$$ \phi_3= \begin{cases} 8(x-\frac12)              & \mbox { for } x \in [\frac12,\frac34 ] \\ -8(x-1)   &\mbox{ for } x \in [\frac14, 1 ] \\ 0               &\mbox{ otherwise } \end{cases} \!\,$$

Define a new hat function on each new pair of adjoining subintervals. The hat functions should have all have the same height as the previous basis's hat functions.

Solution 5d
For our system in (a), this system yields a diagonal matrix.

Solution 6a
$$

\begin{array}{|c|c|c|c|c|c|c|c|} \mbox{Order} & y(t+h) & - y(t) & -\frac{h}{2} y'(t) & -\frac{h}{2} y'(t+h) & -\frac{h^2}{12} y(t) & \frac{h^2}{12} y(t+h) & \Sigma \\ \hline & & & & & & &\\ 0 & y(t)     & -y(t) & 0                  &          0       & 0 & 0 & 0 \\ 1 & y'(t)h   &  0    & -\frac{h}{2} y'(t) &-\frac{h}{2}y'(t) & 0 & 0 & 0  \\ 2 & \frac{y(t)h^2}{2} &0 & 0 & -\frac{h}{2} y(t)h & -\frac{h^2}{12}y(t) & \frac{h^2}{12} y(t) & 0 \\ 3 & \frac{y(t)h^3}{6}& 0 & 0 & -\frac{h}{2}\frac{y(t)}{2}h^2 & 0 & \frac{h^2}{12}y'''(t)h & 0\\ 4& y'(t)\frac{h^4}{24} & 0 &0 & -\frac{h}{2}y'(t)\frac{h^3}{6} & 0 & \frac{h^2}{12}y(t)\frac{h^2}{2} & 0 \\ 5& O(h^5) & 0 &0 & O(h^5) & 0 & O(h^5) & O(h^5) \\ \hline \end{array}

$$

Solution 6b
$$y_n'' = \frac{d}{dx} \lambda y + \lambda y \frac{d}{dy}\lambda y = \lambda^2 y \!\,$$

$$y_{n+1}=y_n +\frac{h}{2}\lambda y_n +\frac{h}{2} \lambda y_{n+1} + \frac{h^2}{12}[ \lambda^2 y_n - \lambda^2 y_{n+1} ] \!\,$$

Letting $$z=h \lambda \!\,$$ and rearranging terms gives

$$y_{n+1}=\underbrace{ \left (\frac{1+\frac{z}{2}+\frac{z^2}{12}}{1-\frac{z}{2}+\frac{z^2}{12}} \right)}_M y_n \!\,$$

If $$z \!\,$$ is a negative real number, then $$M < 1 \!\,$$