Math 6610 • Analysis of Numerical Methods I

{Heat Equation Stability:}

\partitem{a} Consider the initial value problem for the constant-coefficient
diffusion equation (with \(\beta>0\))
\[
  v_t=\beta v_{xx},\qquad t>0
\]
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 there are no boundary conditions here.)

\partitem{b} Consider the variable coefficient diffusion equation
\[
  v_t=(\beta(x)v_x)_x,\qquad 0<x<1,\quad t>0
\]
with Dirichlet boundary conditions
\[
  v(0,t)=0,\qquad v(1,t)=0
\]
and initial data \(v(x,0)=f(x)\). Assume that \(\beta(x)\ge\beta_0>0\), 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{Fact 1:} A real symmetric \(n\times n\) matrix \(A\) can be diagonalized by
an orthogonal similarity transformation, and \(A\)'s eigenvalues are real.

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

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

\textbf{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
\(\nabla r(x)=0\).