By Robert E. 1941- Gaines

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U' (to) grad VI ' (to) Thus 45 and EIII • u" (to) = H '(to + grad Vl " '(to ~< 0. "(to If we let y = x'(t0) and note that x"(t0) = f(t0, x(t0), x'(t0)) we have a contradiction to ( ~ 8 ) . Remarks. 2 apply here. 3 in terms of curva- tures. Suppose C = (t, x(t)) is a solution trajectory in GI which "touches" ~G l at (to, xo). By i), ii), and iii), grad Vl • T I = v~ l = N = 0. Let + IIx' (to)ll ~ ~ra d VI IIgrad Vlll ~ p r o j e c t i o n of C onto ) = CI ~ t h e plane of N and T (d(t) • T) T + (d(t) • N) N + (to, x0) 46 _ ( i n t e r s e c t i o n of Vl(t,x) = 0) C ~ - ~ w i t h the plane of N and T / = ~ T + ~(G)N + (to, x0) ~here d(t) = (t-to, x(t)-x(t0)) and VI(~ T + S N + (to, xD)) = 0.

Ix,(0)I = h(x(0)) = (x'(t)) 2 - and (h(x(t))) Ix'(T) I = h(x(T)). Consider 2 Since C(t) < O, C(O) = O, and c(T) = 0 we must have ~'(0) < O, and o'(T) > 0. But O'(t) = 2x'(t) x"(t) - 2h(x(t)) h'(x(t)) x'(t) ~ ' ( o ) = 2x'(O) [ ~ f ( O , x ( O ) , x ' ( O ) ) + ~ ( p x ' ( O ) l ) ] o'(m) = 2x'(T) The two expressions [tf(T,x(T),x'(m)) on the right we again have a contradiction, b) Behavior are nonzero + @(lx'(m)l)]. and of the same sign. Thus, and we conclude on the Kernel of L. Ker L O ~ x 9 ~.

1). If x(t) is a solution (b, x(b), x'(b)) E G~ then x ~ ~ where = {x Proof. Suppose x E ~ : x e C1[a,b], (t,x(t), x'(t)) E G for t E [a,b]}. I) on [a,b]. Then (t,x(t),x'(t)) C G on [a,b] and (to, x(t0), x'(t0)) E 8G for some to. 47 Since (a,x(a),x'(a)), (b,x(b),x'(b)) @ G, to E (a,b). 4. If (t0~ x(t0)) ~ 9Gl which contradicts. 2 to obtain a contradiction. Remarks. 6 are useful for determining "candidates" for sets applying the Cbr~Luuatiom Theorem. several technical problems ~ in However, given such a ~ we are left with : i) We must show that solutions x(t) to our boundary value problem satisfy (a,x(a),x'(a)), (b,x(b),x'(b)) E G.

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