1549301742-The_Theory_of_Difference_Schemes__Samarskii

The maximum principle 261

Since y(P) > y(Q) for all Q E Patt'(P), we find by virtue of the relations

D(P) > 0 and y(P) > 0 that

£ y(P) = D(P) y(P)+ L B(P, Q) (y(P)-y(Q)) > D(P) y(P) > o.

QEPatt'(P)

From such reasoning it seems clear that only the case £ y(P) = 0

needs investigation. As formula (7) shows, it is possible only if D( P) = 0

and y(Q) = y(P) for all Q E Patt'(P).

We now take the node P1 E Patt'(P) at which y(P 1 ) = y(P) = M 0

and adopt those ideas. Since y( P) -::/- const on the grid w an~ the grid is

connected, there exists a sequence of nodes P1, P2, ... , Pm, P, satisfying

conditions ( 4), such that

but

Then

y(Pm) = y(P) =Mo, y(P) < Mo,

PE Patt'(Pm).

£ y(Pm) > D(Pm) y(Pm) + B(Pm, P) (y(Pm) - y(P))

B(Pm, P) (y(P) - y(P)) > 0,

meaning P Pm and justifying the first assertion of the theorem. The

second assertion will be reduced to the first one once we replace y( P) by

-y(P).

Corollary 1 Let conditions (2) and (4) hold and let a grid function y(P)

defined on w + I /;>e nonnegative on the boundary, that is, y( P) > 0 for

P E / and£ y(P) > 0 on w. Then y(P) is nonnegative on w + /, that is,

y(P) > 0 for all PE w + / But ify(P) < 0 on I and £y(P) < 0 on w,

then y( P) < 0 on w + / ·

Proof Let£ y(P) > 0 on wand y(P) > 0 on/. Suppose that y(P 0 ) < 0 at
least at one inner node Po E w. Then y( P) should take its minimal negative
value inside the grid w, what is impossible on account of Theorem 1, since
y( P) -::/- const on the grid w (y( P 0 ) < 0, YI,, > 0). The second assertion of
the theorem can be proved in a similar manner.