Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/Aug07 667

Problem 4a
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Fixed point iteration
Equation $$(1) \!\,$$ is conveniently in fixed point iteration form.

$$y_{n}^{j+1}=\underbrace{\alpha h f(t_n,y_n^j) + g_{n-1}}_{\phi(y_n^{j})} \!\,$$

Notice that the right hand side is only a function of $$ y_n^j \!\,$$ since $$\alpha, h, t_n, g_{n-1} \!\,$$ are fixed when solving for the fixed point $$y_n^* \!\,$$ where $$\phi(y_n^*)=y_n^* \!\,$$

Also note that $$j \!\,$$ is the fixed point iteration index.

Conditions for local convergence
The fixed point iteration will converge when the norm of the Jacobian of $$\phi \!\,$$ is less than 1 i.e.

$$\| D(\phi) \| < 1 \!\,$$

Since $$ \|D(\phi)\| = \| \alpha h D(f) \| \!\,$$, we equivalently have the condition

$$\|D(f)\| < \frac{1}{\alpha h} \!\,$$

Newton iteration
The Newton iteration solves $$\psi(y)=0 \!\,$$ and the iteration is given by

$$y_{i+1}=y_i - D(\psi(y_i))^{-1}\psi(y_i) \!\,$$

Let $$\psi(y)= y-\phi(y) \!\,$$

Conditions for local convergence
If $$D(\psi(y))^{-1} \!\,$$ exists, i.e. $$ D(\psi(y)) \!\,$$ is invertible or equivalently non-singular, then local convergence is guaranteed.

Note that

$$D(\psi(y))= I - D(\phi(y)) \!\,$$

Conditions for quadratic convergence
If $$D(\psi(y)) \!\,$$ is Lipschitz, then we have quadratic convergence and $$ \psi(y) \!\,$$ is twice continuously differentiable in a neighborhood of the root

Trapezoid method (Implicit, Adams-Moulton)
$$ y_{n+1} = y_n + \tfrac12 h \big( f(t_{n+1},y_{n+1}) + f(t_n,y_n) \big) \!\,$$

Define Local Truncation Error
The local truncation error is given as follows:


 * $$\mbox{error}= (y(t_{n+1}) - y(t_n)) - \tfrac12 h \big( f(t_{n+1},y(t_{n+1})) + f(t_n,y_n) \big)\!\,$$

Find Local Truncation Error Using Taylor Expansion
Note that $$y_i \approx y(t_i) \!\,$$. The uniform step size is $$h \!\,$$. Hence,

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

Therefore, the given equation may be written as

$$y(t_i+h)-y(t_i) \approx \frac12 h (y'(t_i+h)+y'(t_n)) \!\,$$

Expand Left Hand Side
Expanding about $$ t_n \!\, $$, we get

$$

\begin{array}{|c|c|c|c|} \mbox{Order} & y(t_n+h) & -y(t_n) & \Sigma \\ \hline & & & \\ 0& y(t_n) & -y(t_n) & 0 \\ & & & \\ 1& y'(t_n)h & 0 & y'(t_n)h \\ & & & \\ 2& \frac{1}{2!}y(t_n)h^2 & 0 & \frac{1}{2}y(t_n)h^2 \\ & & & \\ 3& \frac{1}{3!}y(t_n)h^3 & 0 & \frac{1}{6}y(t_n)h^3 \\ & & & \\ 4& \mathcal{O}(h^4) & 0 & \mathcal{O}(h^4) \\ & & & \\ \hline \end{array}

$$

Expand Right Hand side
Also expanding about $$ t_n \!\,$$ gives

$$ \begin{array}{|c|c|c|c|} \mbox{Order }h & \frac12 hy'(t_n+h) & \frac12 h y'(t_n) & \Sigma \\ \hline &&&\\ 0& 0 &0 & 0 \\ &&&\\ 1& \frac12 h \cdot y'(t_n) & \frac12 hy'(t_n) & y'(t_n)h \\ &&&\\ 2& \frac12 h \cdot y(t_n) h & 0 & \frac12 y(t_n)h^2 \\ &&&\\ 3& \frac12 h \cdot \frac{y}{2}(t_n)h^2 & 0 & \frac{y}{4}(t_n)h^3\\ &&&\\ 4& \mathcal{O}(h^4)& \mathcal{O}(h^4) & \mathcal{O}(h^4)\\ \hline \end{array}

$$

Calculate local truncation error
Since the order 3 terms of $$h \!\,$$ do not agree ($$\frac{y(t_n)}{6}h^3 \neq \frac{y(t_n)}{4}h^3 \!\,$$), the error is of order $$h^2 \!\,$$.

Solution 5b
We need to show that $$y(t_{n+1}) - 2y(t_n) + y(t_{n-1}) + hy'(t_{n-1}) - hy'(t_{n}) = \mathcal{O}(h^3) \!\,$$

Again, note that $$\begin{align} t_{n+1} &= t_{n} + h \\ t_{n-1} &= t_n - h \end{align}\!\,$$

$$ \begin{array}{|c|c|c|c|c|c|c|} \mbox{Order of }h & y(t_{n}+h) & -2y(t_n) & y(t_{n}-h) & hy'(t_{n}-h) & -hy'(t_{n}) & \Sigma \\ \hline 0 & y(t_n)              & -2y(t_n) & y(t_n)                & 0                    & 0        & 0 \\ &&&&&&\\ 1 & y'(t_n)h            &  0       & -y'(t_n)h             & y'(t_n)h             & -y'(t_n) & 0 \\ &&&&&&\\ 2 & \frac12 y(t_n)h^2 &  0       & \frac12 y(t_n)h^2   & -y''(t_n)h^2         & 0        & 0 \\ &&&&&&\\ 3 & \frac16 y(t_n)h^3 & 0       & -\frac16 y(t_n)h^3 & \frac12 y(t_n)h^3 & 0        & \frac12 y(t_n)h^3 \\ &&&&&&\\ \hline \end{array} $$

So this method is also consistency order 2.

Solution 5c
The trapezoid is stable because its satisfies the root condition. (The root of the characteristic equation is 1 and has a simple root)

The second method is not stable because the characteristic equation has a double root of 1.

Both the trapezoid method and second method are consistent with order $$h^2 \!\,$$

Note that convergence occurs if and only the method is both stable and consistent.

Therefore, the trapezoid method converges in general but the second method does not. mkmkmkmlmklml

Weak Form
Integrating by parts gives, for all $$v \in V \!\,$$

$$\int u'v' + b \int uv = \int fv \!\,$$

Specifically,

$$\int u'u_h' + b \int uu_h= \int fu_h \!\,$$

Discretized Form (Finite Element Formulation)
Similarly, the finite element formulation is find $$ u_h \in V_h\!\,$$ such that for all $$ v_h \in V_h \!\,$$

$$\int u_h'v_h'+ b \int u_h v_h = \int fv_h \!\,$$

Specifically,

$$\int u_h'u_h'+ b \int u_h u_h = \int fu_h \!\,$$

Equate Both Sides and Apply Inequalities
$$ \begin{align} \int_0^1 u_h'u_h' + b\int_0^1 u_hu_h &= \int_0^1 u'u_h' + b\int_0^1 uu_h \\ &\leq |u|_1 |u_h|_1 + b \|u\|_0 \|u_h\|_0 \mbox{  Cauchy-Schwartz} \\ &\leq |u|_1|u_h|_1+b \frac12 |u|_1 \frac12 |u_h|_1 \mbox{  Poincare} \\ &= |u_h|_1 (1+\frac{b}{4}) |u|_1 \end{align} $$

Hence we have,

$$ \begin{align} \end{align} $$
 * u_h|_1^2 \leq |u_h|_1^2 + b \|u_h\|_0^2 &\leq |u_h|_1 (1+\frac{b}{4}) |u|_1 \\
 * u_h|_1^2 &\leq |u_h|_1 (1+\frac{b}{4}) |u|_1 \\
 * u_h|_1 &\leq (1+\frac{b}{4}) |u|_1

Prove equality
We have for all $$v \in H_0^1(0,1) \!\,$$

$$a(u,v)=\int u'v'dx+b\int uv dx=\int f v \!\,$$

Specifically, for all $$v_h \in V_h\!\,$$

$$a(u,v_h)=\int u' v_h' dx + b \int u v_h dx = \int f v_h \quad \quad (1)\!\,$$

The discrete form of the energy scalar product is for all $$ u_h,v_h \in V_h\!\,$$

$$a(u_h,v_h)=\int u_h' v_h'dx + b \int u_h v_h dx = \int f v_h \quad \quad (2) \!\,$$

Subtracting equation (2) from equation (1), we have

$$a(u-u_h,v_h)=0 \!\,$$

Let $$R_h(I_hu)=\overline{u_h} \in V_h \!\,$$. Note that by hypothesis $$ I_hu \in V_h\!\,$$. Then,

$$a(I_hu-\overline{u_h},I_hu-\overline{u_h})=0 \!\,$$

By ellipticity,

$$0=a(I_hu-\overline{u_h},I_hu-\overline{u_h}) \geq \alpha \| I_hu-\overline{u_h}\|^2\!\,$$

which implies

$$I_hu-\overline{u_h} = 0\!\,$$

i.e.

$$I_hu=\overline{u_h} \!\,$$

Deduce inequality
Hence we have

$$R_h(u-I_hu)= u_h-I_hu \!\,$$

Arguing as we did in part (a), we have

$$|u_h-I_hu|_1 \leq \Lambda |u-I_hu|_1 \!\,$$

Show inequality
$$ \begin{align} & \leq |u-I_hu|_1 + |I_hu - u_h|_1 \\ & \leq |u-I_hu|_1 + \Lambda|u-I_h u|_1 \\ & = (1+\Lambda)|u-I_h u|_1 \end{align} $$
 * u-u_h|_1 & = |u-I_hu + I_hu - u_h|_1 \\

Bound Right Hand Side
For $$x \in [x_{i-1}, x_{i}] \!\,$$, Newton's polynomial interpolation error gives for some $$\xi_i \in [x_{i-1}, x_{i}] \!\,$$

$$ \begin{align} &=\int_0^1 \left( \left[ \frac{u^{(2)}(\xi_i)}{2}(x-x_i)(x-x_{i-1})\right]' \right)^2 \\ &=\left(\frac{u^{(2)}(\xi_i)}{2}\right)^2\int_0^1 [ \underbrace{(x-x_{i-1})}_{\leq h}\underbrace{(x-x_i)}_{\leq h}] dx \\ &\leq (u^{(2)}(\xi_i)h)^2
 * u-I_h u|_1^2 &= \left| \frac{u^{(2)}(\xi_i)}{2}(x-x_i)(x-x_{i-1}) \right|_1^2 dx \\

\end{align} \!\,$$

Therefore the error on the entire interval is given by

$$ \begin{align} &\leq \max_{0 < \xi < 1} (u^{(2)}(\xi)h)^2 \cdot n \end{align} \!\,$$
 * u-I_hu|_1^2 &\leq \sum_{i=1}^n (u^{(2)}(\xi_i)h)^2 \\

which implies

$$ |u-I_h u|_1 \leq \max_{0 < \xi < 1} u^{(2)}(\xi_i) \sqrt{n} h \!\,$$

$$u \!\,$$ needs to be twice differentiable.