Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/Jan07 667

Solution 4a
Note the following identity


 * $$F(x) - F(y) = \int_0^1 F'(sx+(1-s)y)(x-y)ds \qquad (*) \!\,$$

The error $$e_{k+1} \!\,$$ is given by

$$ \begin{align} e_{k+1} &= x_{k+1} - x_* \\ &= \underbrace{x_k - x_*}_{e_k} - B_k^{-1} f(x_k) \\ &= e_k - B_k^{-1} ( f(x_k) - \underbrace{f(x_*)}_{0}  )  \\ &= B_k^{-1} \left\{ (B_k - f'(x_*))e_k - \int_0^1 \left [ f'(sx_k+(1-s)x_*) - f'(x_*) \right] e_k ds \right \} \\

\end{align}

$$

Solution 4b
Assume $$B \!\,$$ is invertible, $$\|B -f'(x_*)\| \!\,$$ is bounded, and $$f' \!\,$$ is Lipschitz.

$$ \begin{align}

\| e_{k+1} \| &\leq \| B^{-1} \| \left \{ \|B-f'(x_*)\| \|e_k\| - \int_0^1 \underbrace{\|f'(sx_k+(1-s)x_*)-f'(x_*)\|}_{\leq L\|sx_k+(1-s)x_*-x_* \|} \|e_k\| ds \right \} \\

&\leq \| B^{-1} \| \left \{ \|B-f'(x_*)\| \|e_k\| - \int_0^1 L \|e_k\|^2 s ds \right \} \\

&\leq \| B^{-1} \| \left \{ \|B-f'(x_*)\| - \frac{L}{2} \|e_k\| \right \} \|e_k\|

\end{align} $$

This implies local superlinear convergence.

Super linear convergence
$$\frac{\|e_{k+1}\|}{\|e_k\|} \rightarrow 0 \!\,$$ as $$k \rightarrow \infty \!\,$$

Find condition for super linear convergence
From part(b)

$$\frac{\|e_{k+1}\|}{\|e_k\|} \leq \|B_k^{-1}\| \{ \|B_k-f'(x_*)\| - \frac{L}{2} \|e_k\| \} \!\,$$

Since $$ \| e_k \| \rightarrow 0 \mbox{ as } k \rightarrow \infty \!\,$$, if

$$ \|B_k-f'(x_*)\| \rightarrow 0 \!\,$$

as $$k \rightarrow 0 \!\,$$, we have super linear convergence i.e.

$$ B_k=f'(x_k) \mbox{ Newton Iteration form}\!\,$$

Solution 5a
The implicit method can be viewed as a fix point iteration:

$$y_{i+1}=\underbrace{y_i+\frac{h}{2}(f(t_i,y_i)+f(t_{i+1},y_{i+1}))}_{g(y_{i+1})} \!\,$$

We want $$\left|\frac{\partial g(y_{i+1})}{\partial y_{i+1}} \right|< 1\!\,$$

which implies

$$h < \frac{2}{\left|\frac{\partial f(t_{i+1},y_{i+1})}{\partial y_{i+1}}\right|} \!\,$$

Solution 5b
Re-writing (1) and replacing $$y_k \mbox{ with } y(t_k) \mbox{ and } f(t_k,y_k) \mbox{ with } y'(t_k) \!\,$$ we have a formula for consistency of order p:

$$ y(t_{i+1}) - y(t_i) - {h \over 2} y'(t_i) - {h \over 2} y(t_{i+1}) = \mathcal{O}(h^{p+1}) \!\,$$

For uniform stepsize h


 * $$t_{i+1} = t_i +h \!\,$$

Expanding in Taylor Series about $$t_{i} \!\,$$ gives

Solution 6a
Using Taylor Expansions, we can approximate the second derivative as follows

$$u''(x)=\frac{u(x+h)-2u(x)+u(x-h)}{h^2} \!\,$$

We can eliminate two equations from the n+2 equations by substituting the initial conditions $$u(0)=u_0, u(1)=u_1 \!\,$$ into the equations for $$U_1 \!\,$$ and $$U_n \!\,$$ respectively.

We then have the system

$$ \underbrace{ \begin{bmatrix} \frac{2a}{h^2}+b & -\frac{a}{h^2}  &                &        &        &        &\\ -\frac{a}{h^2}  &\frac{2a}{h^2}+b  & -\frac{a}{h^2} &        &        &        &\\ &-\frac{a}{h^2}   &\frac{2a}{h^2}+b & -\frac{a}{h^2} &        &    \\ &                 & \ddots          & \ddots         & \ddots &     \\ &                 &                &        &  -\frac{a}{h^2}       & \frac{2a}{h^2}+b \end{bmatrix} }_A \underbrace{ \begin{bmatrix} U_1 \\ U_2 \\ U_3 \\ \vdots \\ U_n \end{bmatrix} }_U = \underbrace{ \begin{bmatrix} f(x_1)+\frac{a}{h^2}u_0 \\ f(x_2)\\ f(x_3)\\ \vdots\\ f(x_n)+\frac{a}{h^2}u_1 \end{bmatrix} }_F \!\,$$

$$A \!\,$$ is nonsingular since it is diagonally dominant.

Local Truncation Error
$$ LTE = \frac{u(x_i+h) + u(x_i -h) - 2u(x_i)}{h^2} - u''(x_i) = \frac{u^{(4)}(\xi)}{12}h^2 \!\,$$

Bound for Local Truncation Error
$$\frac{1}{12} \|u^{(4)}\|_\infty h^2 \!\,$$

Derive Bound for Max Error
Let $$L=-a\partial_{xx} + b \!\,$$, $$L_h=A \!\,$$, and $$R_h \!\,$$ is the local truncation error.

Then

$$ \begin{align} Lu &= f \\ L_h u &= f + R_h \\ L_h U &= f \end{align} \!\,$$

Subtracting the two last equations gives

$$ L_h(u-U)=R_h\!\,$$

Hence,

$$\|u-U\|_\infty \leq \|L_h^{-1}\| \|R_h\| \!\,$$

$$\|u-U\|_\infty \leq C h^2 \!\,$$, that is the error has order 2.

Solution 6c
Note that $$A \!\,$$ is a $$M \!\,$$ matrix and hence the discrete maximum principle applies. (See January 05 667 test for definition of $$M \!\,$$ matrix)

Discrete Maximum Principle
$$ AU=F \!\,$$

If $$F \leq 0 \!\,$$, then $$U \leq 0 \!\,$$.

Specifically let $$F=f_1-f_2 \!\,$$, then $$U_1 - U_2 \leq 0 \!\,$$ which implies $$U_1 \leq U_2 \!\,$$