202 Hmnogeneous Difference Schemes
If cxk > 0 for k = 1, 2 ... , i - 1, then a; cxx. ' i > 0 and ex; > cx;_ 1 > 0.
In a similar way we can be pretty sure that f3x ' i < 0 and 0 < f3; < {3;_ 1.
2) cxN = {3 0 or cx(l) = {3(0). From the second Green forrnula
(ex, Af3) = ({3, A ex)+ aN (ex f3x - f3 cxx) N - a 1 (ex f3x - f3 cxx) 0
it follows immediately that cxN = {3 0 on account of conditions (10).
3) The determinant 6.; =a; (cxx,if3i -cx;f3x,i) =canst= cxN is pos-
itive for 0 < i < N. Applying the second Green's forrnula in the domain
{O < X; = ih < x; 0 = x} we obtain
; 0 -1
0 = L (cxA{3-{3Acx); h = a; 0 (ex f3x - {3 cxx); 0 - a 1 (ex f3x - f3 cx,:) 0
i=l
= -6.(x; 0 ) + {3(0) a 1 CX;; ' i = -6.(x; 0 ) + {3 0 •
Since X; 0 = x is an arbitrary node of the grid w h, we might have
6.(x) =canst = {3(0) = cx(l).
Other ideas are connected with a possibility of arranging the Green
function such as
for i < k,
( 11)
for i > k.
Whence it follows that Gok = GNk = 0.
The preceding representation will be proved if we succeed in showing
that the function specified by formulae (11) solves the equation A(i) G;k =
-O;k/h. Indeed, A(i) G;k = 0 for i -f k due to the facts that Acx; = 0 and
Af3; = 0. The expression A(i) G;k for i = k needs investigation:
(12)
1
(A(i) G;k)i=k = h 2 [ak+ 1 (cxk f3k+1 - cxk f3iJ
CXN
- ak (cxk f3k - cxk-1 f3k)] - dk Gkk ·
From the condition Ak+1 = ak+ 1 (ex k+ 1 {3k - ex k f3k+ 1 )/ h = ex N we
derive the expression ak+ 1 cxk f3k+ 1 = ak+ 1 cxk+ 1 {3k - hcxN and substitute it
into the right-hand side of formula (12). As a final result we get
(A(i)Gik)i=k = {3k (acxx)x k - _hl - dk{3k cxk = {3k Acxk - ~ = -~
CXN ' CXN CXN h h