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I Difference Equation Boundary Conditions0.

  1. Oct 10, 2016 #1
    This question is inspired by Gilbert Strang's Course on Computational Science and Engineering, MIT 18.085.

    Consider the three matrices
    Fixed-Fixed
    $$K=\begin{bmatrix}
    2 &-1 & 0 &0 \\
    -1&2 & -1 &0 \\
    0 & -1 &2 & -1 \\
    0 & 0 & -1 & 2 \\
    \end{bmatrix} $$
    Free-Fixed
    $$T=\begin{bmatrix}
    1 &-1 & 0 &0 \\

    -1&2 & -1 &0 \\

    0 & -1 &2 & -1 \\

    0 & 0 & -1 & 2 \\
    \end{bmatrix} $$
    Free-Free
    $$B=\begin{bmatrix}

    1 &-1 & 0 &0 \\

    -1&2 & -1 &0 \\

    0 & -1 &2 & -1 \\

    0 & 0 & -1 & 1 \\

    \end{bmatrix} $$
    $$ \mathbf{u} =\begin{bmatrix}
    u_1 \\ u_2 \\ u_3 \\ u_4
    \end{bmatrix} $$
    The problem is to solve ##(\text{scale})A\mathbf{u} = \mathbf{b} ## where##A=K,T, \text{ or } B ## and ##\mathbf{b} ## is (I presume) an arbitrary source vector. This should be the finite difference solution corresponding to
    $$\frac{\partial^2 }{\partial x^2} u(x) = b(x)$$
    subject to some boundary conditions.
    I what way do these matrices correspond the boundary conditions described when trying to solve the equation? I think continuous constraints may take the form of specific values or slopes of ##u## at the boundaries. How does that get translated into a finite difference? Also, could anyone explain his comment here?
     
  2. jcsd
  3. Oct 15, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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