II. FINITE DIFFERENCE METHOD 1 Difference formulae

h = 1/(n + 1) and i = 0, 1, ..., n + 1. Let us denote by ¯yi the values of
the approximated solution in the points xi. By discretizing y ′′ by means f
central differences, the following nonlinear system is got:
¯y◦ = 0, ¯yn+1 = 0
¡ ¯yi+1−2¯yi+¯yi−1
h 2
+ 1
2 (1 + xi + ¯yi) 3 = 0, (1 · i · n)
Eliminating the variables ¯y◦, ¯yn+1, the system can be written in the following
matrix form
with ¯y = (¯y1, ..., ¯yn) T and
A =
⎛
⎜
⎝
2 ¡1
¡1 2 ¡1
.
.
. . .
¡1 2 ¡1
¡1 2
A¯y + h 2 B(¯y) = 0 (11)
⎞
⎟
⎠
, B(¯y) = diag(f(xi, ¯yi)), i = 1, 2, ..., n
The nonlinear system (11) can be solved, for example, by Newton’s method.
Indeed, the Jacobian matrix of the system is given by
J = A + h 2 By
(12)
By virtue of the property fy ¸ 0, such a Jacobian turns out to be a matrix
with diagonal predominance and positive definite. Therefore the Newton’s
method is convergent if the initial values are chosen sufficiently near the
solution. The implementation of Newton’s method is the following. A
plausible initial point ¯y (◦) is chosen, the (r+1)¡th iterated vector is obtained
by solving the linear system
A + h 2
By
and setting
v = ¡A¯y (r) ¡ h 2 B(¯y (r) )
¯y (r+1) = ¯y (r) + v.
As it is easily verified, the exact solution of the continuous problem is y(x) =
¡ x ¡ 1.
2
2−x
By a program of the following type we obtain y (1) from an initial vector,
y (◦) = 0. By the same program we obtain y (2) from y (1) = 0... and so on:
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