Math 6610 • Analysis of Numerical Methods I

{Heat Equation Stability:}

\partitem{a} Consider the initial value problem for the constant-coefficient
diffusion equation
\[
  v_t=\beta v_{xx},\qquad t>0
\]
with initial data \(v(x,0)=f(x)\). A scheme for this problem is:
\[
  \frac{u_j^{n+1}-u_j^n}{k}
  =
  \frac{\beta}{h^2}
  \left\{u_{j-1}^{n+1}-2u_j^{n+1}+u_{j+1}^{n+1}\right\}.
\]
Analyze the 2-norm stability of this scheme. For which values of \(k>0\) and
\(h>0\) is the scheme stable? (Note that this there are no boundary conditions
here.)

\partitem{b} Consider the variable coefficient diffusion equation
\[
  v_t=(\beta v_x)_x,\qquad 0<x<1,\quad t>0
\]
with Dirichlet boundary conditions
\[
  v(0,t)=0,\qquad v(1,t)=0
\]
and initial data \(v(x,0)=f(x)\). Assume that \(\beta(x)\geq\beta_0>0\), and that
\(\beta(x)\) is smooth. Let \(\beta_{j+1/2}=\beta(x_{j+1/2})\). A scheme for this
problem is:
\[
  \frac{u_j^{n+1}-u_j^n}{k}
  =
  \frac{1}{h^2}
  \left\{
    \beta_{j-1/2}u_{j-1}^{n+1}
    -(\beta_{j-1/2}+\beta_{j+1/2})u_j^{n+1}
    +\beta_{j+1/2}u_{j+1}^{n+1}
  \right\}.
\]
Analyze the 2-norm stability of this scheme for solving this initial boundary
value problem. DO NOT NEGLECT THE FACT THAT THERE ARE BOUNDARY CONDITIONS!

\newpage

\textbf{8) Numerical Methods for ODEs:} Consider the Linear Multistep Method
\[
  y_{n+2}-\frac{4}{3}y_{n+1}+\frac{1}{3}y_n=\frac{2}{3}k f_{n+2}
\]
for solving an initial value problem \(y'=f(y,x)\), \(y(0)=\eta\). You may assume
that \(f\) is Lipschitz continuous with respect to \(y\) uniformly for all \(x\).

\partitem{a} Analyze the consistency, stability, accuracy, and convergence
properties of this method.

\partitem{b} Sketch a graph of the solution to the following initial value
problem.
\[
  y'=-10^8[y-\cos(x)]-\sin(x),\qquad y(0)=2.
\]
Would it be more reasonable to use this method or Euler's method for this
problem? What would you consider in choosing a timestep \(k\) for each of the
methods? Justify your answer.

\newpage

\fact{1} A real symmetric \(n\times n\) matrix \(A\) can be diagonalized by an
orthogonal similarity transformation, and \(A\)'s eigenvalues are real.

\fact{2} The \((N-1)\times(N-1)\) matrix \(M\) defined by
\[
  M=
  \begin{bmatrix}
    -2 & 1 & 0 & \cdots & 0 & 0\\
    1 & -2 & 1 & \ddots & \vdots & \vdots\\
    0 & 1 & -2 & \ddots & 0 & 0\\
    \vdots & \ddots & \ddots & \ddots & 1 & 0\\
    0 & \cdots & 0 & 1 & -2 & 1\\
    0 & \cdots & 0 & 0 & 1 & -2
  \end{bmatrix}
\]
has eigenvalues
\[
  \mu_l=-4\sin^2\left(\frac{\pi l}{2N}\right),
  \qquad l=1,2,\ldots,N-1.
\]

\fact{3} The \((N+1)\times(N+1)\) matrix:
\[
  \begin{bmatrix}
    -1 & 1 & 0 & \cdots & 0 & 0\\
    1 & -2 & 1 & \ddots & \vdots & \vdots\\
    0 & 1 & -2 & \ddots & 0 & 0\\
    \vdots & \ddots & \ddots & \ddots & 1 & 0\\
    0 & \cdots & 0 & 1 & -2 & 1\\
    0 & \cdots & 0 & 0 & 1 & -1
  \end{bmatrix}
\]
has eigenvalues
\[
  \mu_l=-4\sin^2\left(\frac{\pi l}{2(N+1)}\right),
  \qquad l=0,1,\ldots,N.
\]

\fact{4} For a real \(n\times n\) matrix \(A\), the Rayleigh quotient of a vector
\(x\in\R^n\) is the scalar
\[
  r(x)=\frac{x^T Ax}{x^T x}.
\]
The gradient of \(r(x)\) is
\[
  \nabla r(x)=\frac{2}{x^T x}(Ax-r(x)x).
\]
If \(x\) is an eigenvector of \(A\) then \(r(x)\) is the corresponding eigenvalue and
\(\nabla r(x)=0\).