P

where we have used £" = f£p > 0. By Theorem 3.2, at r = R the inequality (81) reduces to

-(l + e)>-(l + e)p, which is satisfied because 1 + e > 1 and p > 1. By continuity, there exists 6 — ¿(e) > 0, such that (81) holds for each r £ (R — 6,R}. By choosing a sufficiently large A, (81) is satisfied in [0,ii], since p > 1 and £ is positive and bounded away from zero in compact intervals of (0,oo). This shows that i])E is a supersolution of (80) for sufficiently large A.

Now, we will show that, for each sufficiently small e > 0, there exists C < 0 such that

is a non-negative subsolution of (80). Indeed, by reversing the inequality (81), it is obvious that ip is a subsolution of (80) if in the region where

C + ( 1-e) (^)2W~r)> 0 the following inequality is satisfied

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