# I Difference Equation Boundary Conditions0.

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1. Oct 10, 2016

### MisterX

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. Oct 15, 2016