Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/Aug06 667

Solution 4a
The symmetric difference quotient is given by

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

Hence we have the following system equations that incorporates the initial conditions $$u(0)=u(1)=0 \!\,$$.

$$ \underbrace{ \frac{1}{h^2} \begin{bmatrix} 1 &    &        &        &          &     & \\ 1 & -2  &  1     &        &          &     & \\  &  1  & -2     & 1      &          &     & \\  &     & \ddots & \ddots & \ddots   &     & \\ &    &        &        &        1 & -2  & 1 \\   &     &        &        &          &     & 1 \end{bmatrix} }_{A} \underbrace{ \begin{bmatrix} u_0 \\ u_1 \\ u_2 \\ \vdots \\ \vdots \\ u_n \end{bmatrix} }_{U} = \underbrace{ \begin{bmatrix} 0 \\ f(u_1) \\ f(u_2) \\ \vdots\\ f(u_{n-1})\\ 0 \end{bmatrix} }_{f} \!\,$$

Solution 4b
$$ \underbrace{AU-f}_{F(U)}=0\!\,$$

Domain: $$ R^{N+1} \!\,$$

Range: $$ R^{N+1} \!\,$$

$$U \!\,$$ is a vector containing $$N+1 \!\,$$ approximations $$u_i \!\,$$ for the solution $$u \!\,$$ at $$x_i \!\,$$

Newton's Method
$$U^{k+1}=U^{k}-J(F(U^{(k)})^{-1}F(U^{(k)}) \!\,$$

where $$J(A) \!\,$$ denotes the Jacobian of a matrix $$A\!\,$$.

Specifically,

$$J(F(U^{(k)})=J(AU-F)=A-J(F) \!\,$$

Sufficient Condition
If $$J(F(U^{(k)})^{-1} \!\,$$ exists, then $$U^{(k)} \!\,$$ iterates are defined.

Convergence of sequence
We cannot decide if the sequence converges since Newton's method only guarantees local convergence.

In general, for local convergence of Newton's method we need:


 * $$F \!\,$$ differentriable


 * $$D(F(U^*)) \!\,$$ invertible


 * $$D(F) \!\,$$ Lipschitz


 * $$U^{(0)} \!\,$$ close to solution $$U^* \!\,$$

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

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

which after integrating by parts and plugging in initial conditions we have

$$\int_0^1 u'v'= \int_0^1 fv - \alpha u(1)v(1) \!\,$$

Let $$\{x_i\}_{i=0}^{N+1} \!\,$$ be the nodes of a uniform partition of $$ [0,1]\!\,$$ where $$x_0=0 \!\,$$ and $$x_i=ih \!\,$$.

Let $$\{\phi_i \}_{i=0}^{N+1} \!\,$$ be the standard "hat" functions defined as follows:

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

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

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

Also $$\phi_0=0 \!\,$$ since $$u(0)=0 \!\,$$

Then $$\{ \phi_i \}_{i=0}^{N+1} \!\,$$ forms a basis for the discrete space $$U_h=\{ u \in H^1[0,1] : u(0)=0, u \mbox{ piecewise linear } \} \!\,$$

Discrete Weak Formulation
Find $$u_h \in U_h \!\,$$ such that for all $$v \in U_h \!\,$$

$$\int_0^1 u_h' v_h' = \int_0^1 f v_h - \alpha u(1) v(1) \!\,$$

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

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

Also for $$j=1,\ldots,N+1 \!\,$$

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

In matrix form

$$ \begin{bmatrix} \frac{2}{h}  & -\frac{1}{h} &             &              && \\ -\frac{1}{h} & \frac{2}{h}  &-\frac{1}{h} &              && \\ & \ddots      & \ddots      & \ddots       && \\ &             &             &              &-\frac{1}{h} & \frac{2}{h}+\alpha \end{bmatrix} \begin{bmatrix} u_{h_1}\\ u_{h_2}\\ \vdots\\ u_{h_{N+1}} \end{bmatrix} = \begin{bmatrix} \int_0^1 f \phi_1 \\ \int_0^1 f \phi_2 \\ \vdots \\ \int_0^1 f\phi_{N+1} \end{bmatrix}

\!\,$$

Solution 6b
$$ \begin{align} y_{n+1} &=\sum_{j=0}^p a_j y_{n-j}+h\sum_{j=-1}^p b_j f(t_{n-j},y_{n-j}) \\ &=a_0y_n +a_1 y_{n-1}+\ldots+a_py_{n-p}+h [b_{-1}f(t_{n+1},y_{n+1})+b_0f(t_{n},y_n)+b_1f(t_{n-1},y_{n-1})+\ldots+b_pf(t_{n-p},y_{n-p})]

\end{align} \!\,$$

Conditions:

(i) $$\sum_{j=0}^p a_j=1 \!\,$$

$$\frac43 + -\frac{1}{3}=1 \!\,$$

(ii)$$\sum_{j=-1}^p b_j- \sum_{j=0}^p j a_j = 1\!\,$$

$$\frac23 - (0 \cdot \frac 43 + 1 \cdot -\frac13) = 1 \!\,$$

Problem 6c
The scheme satisfies the root condition but not the strong root condition since the roots are given by

$$\lambda=\frac{\frac43 \pm \sqrt{\frac{16}{9}-\frac43} }{2} \!\,$$

which implies $$\lambda_1 =1 \!\,$$ and $$\lambda_2=\frac13 \!\,$$