1549301742-The_Theory_of_Difference_Schemes__Samarskii

700 Methods for Solving Grid Equations

Current explanations need certain clarification by having recourse to

one simplest example devoted to the governing equation

u"=-f(x), O<x<l, u(O)=u(l)=O,

and associated three-point scheme

(72) Ay = -/(:1:), .e E wh, y(O) = 0, y(l) = 0,

which is constructed on the grid

so that at two inner points

since y 0 = y 3 = 0. For clarity only, we take h 1 < h and 2h + h 1 = 1.

The next step is to find the eigenvalues of the operator A. By defini-

tion, Ay + >.y = 0 or, what amounts to the same,

!_ ( Y2 - Y1 _ !!2.) + ).. = O

h h h 1 Yi '

Having completed the eli111ination of y 2 , we derive the quadratic equation

related to μ = >.h^2 :

(73) t μ^2 - (1+3t)μ+2 + t = 0,

with the roots

1 + 3t ± J(l + 3t)2 - 4t (2 + t)

μ(1,2)= 2t ,

P(l,'2)

)..(1 ,~ 'J) = --h2

and the characteristic relation between >..(1) = >.min = b and >..( 2 ) = >.max =
~:

(j 2t(2+t)

17= ~ = --------------,================

~ (1+3t)2 - 2t (2 + t) + (1+3t) J(l + 3t)^2 - 4t (2 + t)

vVhence it follows that fort = 1, that is, on an equidistant grid r/ = 1/3,
while 17 ~ 2t = 2h 1 /h for small ratios t = h 1 /h ~ l; making worse the