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By Stephen R. Bernfeld

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A superfunction may be defined similarly by re1 ' 2 versing the respective inequalities. x E We have immediately the following r e s u l t . 1. 1) XI1 where f E C[J X R xR,R]. 1) on J . Proof: Let Then to 6 Jo sufficiently small. = f(t,X,X') Cp and h > 0, k > 0, h - E C(2)[[t0-k,t0+h],R]. 2 t h e BVP - X" = f(t,X,X'), i s a lower solution of - (to+ h ) Cp' ( t o k) h+k -> X' (to + h ) X' cp is a - (to k) h+k = for some t o- k < s < to + h. the continuity of 47 f together 1. 2, implies that f ( s , x ( s ) , x t ( s ) ) --3 f ( t O , x ( t O ) , x t ( t O ) ) as h + k + 0.

2), we then a r r i v e a t the contradiction - D-a' ( t ) D-B' ( t ) > 0 which establishes t h e claim to E [a,s). 2) t o conclude a ' ( t ) that (to) > p(b). on 6= b [t0,6]. This which implies The l a s t conclusion, i n i t s turn, leads t o the contradiction a l ( b ) -< p ' ( b ) 17 because of t h e second 1. METHODS INVOLVING DIFFERENTIAL INEQUALITIES inequality i n ( i v ) . lation a(a) > p(a) i s impossible. a ( t ) 5 B(t) a ( & ) > p(a) Thus a(b) If we assume t h a t > i s impossible. 1.

U swe only need t o We s h a l l only prove t h a t x(t) 5 @ ( t ) on J . The arguments are e s s e n t i a l l y t h e same f o r t h e case a(t) x(t). Assume, i f possible, t h a t - x(t) > B(t) f o r some t E J . Then x ( t ) p ( t ) has a positive maximum a t a point 0 to E J Hence it follows t h a t x ' ( t , ) = @ ' ( t o ) , I x ' ( t O ) ( c c . and x"(to) = F ( t O , x ( t O ) , x '( t o ) ) 19 1. METHODS INVOLVING DIFFERENTIAL INEQUALITIES and therefore, we arrive at - which i s impossible a t a maximum of x ( t ) B(t).

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