1549301742-The_Theory_of_Difference_Schemes__Samarskii

62 Basic Concepts of the Theory of Difference Schemes

Since g 2 ( s) > 0, Lagrange's formula provides support for the representation

1

Vxx = v^11 (x + ()h) j g 2 (s) ds = v^11 (x + ()h) = v^11 (~),

-1

-l<()<l - - ,

where~ is the middle point of the segment [x-h, x+h]. The second formula

(10) can be derived in the same manner as before.

Below we follow these procedures in a step-by-step fashion: collect (14)

and (16), carry 2v(x) over to the left-hand side and divide the resulting

expression by h^2 • Simple algebra gives

where

{

( 1 + s )^3

g1(s) = 3

(1 - s)

for

for

I

-l<s<O - - ,

j g 4 (s) ds = ~

-I

Because g 4 (s) > 0 and v(^4 l(x) is continuous, applying Lagrange's formula

yields

h2

v-;t>X = v^11 (x) + - 12 v(^4 l(x + ()h) '

Remark 1 By exactly the same reasoning as before, we can establish

h2 h4

( 17) v-xx = v^11 (x) + - 12 v(^4 l(x) + - 360 v(^6 l(") <, , ~ = x + ()h,

if v(x) belongs to the class C(^6 l[x - h, x + h].

Remark 2 A similar result is still valid for the derivative

( 18) Vxxxx = h^1 4 [v( x + 2h) • - 4 v( x + h) + 6 v( x)

• 4 v(x - h) + v(x - 2h)] = v(^4 l(~),

where~ = x + ()h, 1()1 < 2, is the middle point of the segment [ x - 2h, x + 2h]
and v E C(^4 l[x - 2h, x + 2h]. This expression will be proved if we succeed
in showing that
2
Vxxxx -- 6 1 J g(s) V (4) (x +sh) ds,
-2