By Michael P Ryan

ISBN-10: 0387057412

ISBN-13: 9780387057415

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Since u" (0) = 0, this implies that if a is small enough, then u" < 0 in a right-neighbourhood of the origin. Let (0, a) be the largest interval on which u" < O. Then u(x) < ax and u'(x) < a on (0, a). 4). 1). Here Ie R+ is the maximal interval of the form (0, &) on which ~(a) is continuous, and hence u'(~(a), a) = O. It remains to find a point on C+ where u'" = O. By a sequence of reflections, the solution on [0, ~] can then be continued to a periodic solution of the SBS equation, with period 4~.

20) 1By Hk (R) we denote the space of functions in L 2 (R) whose first k generalised derivatives also belong to L 2 (R)[A]. 34 1. Introduction Another route to periodic solutions was followed by Mark Peletier [PMal], who investigated the phenomenon of sequential buckling in elastic rods (see Hunt et al. [Hul). 21) F(U)} dx, subject to the constraint J(U) = ~ { (u')2dx =)... 22) 2 JR Here)" > 0 represents the prescribed value of the total shortening of the rod. The potential function F is given by F(s) = is2 - is4 + ~s6, with 01.

2) that u'u'" :::: > I~I (u')2 - ~{(U2 ~(u')2 _ 1 2 4E 4 1)2 - 4E} if 0 < u < 1. 2c) and u" > 0 and u' > 0 as long as 0 < u < 1. 7) holds, then u' cannot have a first zero ~ such that u(~) < 1. 7) cannot hold, so the assertion follows. (b) We assume, to the contrary, that ~(a*) = 00. Then 0 < u ~ c and u' > 0 on R+, so that u(x) tends to a limit, say e, as x -+ 00, and e E (0, c]. This leads to a contradiction. We leave the details as an exercise. 2. 2. (c) Suppose that u(~(a*), a*) < c. 5), u" < 0 at ~(a*) so that by the Implicit Function Theorem, a* cannot be the supremum, a contradiction.

### Hamiltonian cosmology-new by Michael P Ryan

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