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Next: 2 or more Dimensions Up: Elliptic Equations Previous: Elliptic Equations

One Dimension

We start by considering the one dimensional Poisson's equation
\begin{displaymath}
\left.\d^2 V\over\d x^2\right. = f(x).
\end{displaymath} (3.2)

The 2nd derivative may be discretised in the usual way to give
\begin{displaymath}
V_{n-1} - 2 V_n + V_{n+1} = \delta x^2\cdot f_n
\end{displaymath} (3.3)

where we define $f_n = f(x = x_n = n\delta x)$.

The boundary conditions are usually of the form $V(x) = V_0$ at $x = x_0$ and $V(x) = V_{N+1}$ at $x = x_{N+1}$, although sometimes the condition is on the first derivative. Since $V_0$ and $V_{N+1}$ are both known the $n=1$ and $n=N$ equations (3.3) may be written as

$\displaystyle -2 V_1 + V_2$ $\textstyle =$ $\displaystyle \delta x^2\cdot f_1 - V_0$ (3.4)
$\displaystyle V_{N-1} - 2 V_N$ $\textstyle =$ $\displaystyle \delta x^2\cdot f_N - V_{N+1}.$ (3.5)

This may seem trivial but it maintains the convention that all the terms on the left contain unknowns and everything on the right is known. It also allows us to rewrite the (3.3) in matrix form as
\begin{displaymath}
\left[\begin{array}{llllllll}
-2 &1 & & & & & & \\
1 &-2 &1...
...ot f_{N-1}\\
\delta x^2\cdot f_N - V_{N+1}
\end{array}\right]
\end{displaymath} (3.6)

which is a simple matrix equation of the form
\begin{displaymath}
\bss{A}\bi{x} = \bi{b}
\end{displaymath} (3.7)

in which $\bss{A}$ is tridiagonal. Such equations can be solved by methods which we shall consider below. For the moment it should suffice to note that the tridiagonal form can be solved particularly efficiently and that functions for this purpose can be found in most libraries of numerical functions.

There are several points which are worthy of note.


next up previous
Next: 2 or more Dimensions Up: Elliptic Equations Previous: Elliptic Equations