Clearly, H(£0) = 0, H(£) > 0 if £ € (0,£o), and H(£) < 0 if £ > Suppose ip(ro) > £o ■

0 < -V"(ro) = - V(ro) = H(4>(r0)) < 0, ro which is impossible. Thus,

In this case we find from (16) and (18) that o > -V"(n) = -V{ri) - ^-V(n) = AV(ri) - /(Ha - ri)^(n)

= A^(ri) - /(i?2 - r0)Vp(n) + [/(J*2 - r0) - f(R2 - ri)]Vp(ri)

As ip' < 0 in (ro,ri), we have that ip(ri) < V,(ro)> and, hence, (19) implies < Thus,

Ji(V>(ri))>0. Moreover, since — ^o > R2 — r\, we find from (1) that f(R2-r0)>f(R2-ri)

and, therefore,

0>-<p"(r1)>H(^(n))>0, which is impossible. Consequently, (17) must be satisfied. Subsequently, for each e € (0, R2 — Rm), we set a__ — Rm > ^

and consider the function ip£ defined through

$e(r) = V'min (ae(r - Rm) + Rm) , Rm < T < R2 ~ £ . This function satisfies ii'e{Rm) = aEip'min(Rm) = 0, lim i>e(r) = lim Vmin(p) = oo.

rlR2-e ptR2

Moreover, setting p := Rm + ae(r - Rm), Rm<r<R2-£, we find that, for each r £ (Rm, R2 — £),

Thus, taking into account that

A > o , Vmin(p) > 0 > ae > 1 , ~ = --- < aE , it is apparent that

>A Mr)-a2J(R2-r)r£(r), because r < p implies f(R2~r)>f(R2-p).

The fact that r < p follows readily from the definition of p, since r > R„ and ae > 1. Therefore, for each r € (Rm, R2 — e), we have that

-#'(r) - 1^i>'£(r) > A&(r) ~ a2J(R2 - r)#(r) and, consequently, the function

:= ar*Mr), r 6 (Rm, ifc - e), provides us with a supersolution of

Thus, as an easy consequence of the strong maximum principle, it is apparent that, for each r € (Rm, R2 ~ z), af1 Mr) = af1^min (ae(r - Rm) + Rm) > t/WxM . (20)

As (20) holds true for any e G (0, i?2 ~ Rm)i passing to the limit as e j 0 gives max (r), re [R m, R2). This shows the uniqueness in case (15) and concludes the proof.

3. A pivotal one-dimensional problem

In this section we study the problem r u" = fup , t > 0 ,

3.1. Existence and uniqueness

The main result of this section is the following.

Theorem 3.1. Suppose p > 1 and f G C[0,oo) satisfies (1). Then, (21) possesses a unique solution.

Proof. Subsequently, for each M > 0 and 6 > 0, we consider the auxiliary problem

Clearly, (u, u) = (0, M) provides us with an ordered sub-supersolution pair of (22). Hence, the problem possesses a solution u G [0,M]. Necessarily, u(t) > 0 for each t G [0,6), since u(t) > 0 for sufficiently small t > 0, because u(0) = M, and, for each t0 G (0,6), the pair (u, v) = (0,0) is the unique solution of u' = v, v' = fup, u(t0) = v(t0) = 0.

Thus, u(t) > 0 for each t G [0,6), since u > 0. Note that u is decreasing, since u"(t) = f(t)up(t) > 0 for each t G (0,6]. Moreover, the solution is unique. Indeed, if u\ ^ u2 are two solutions of (22) with u\ (to) > u2{to) for some i0 g (0,6), then there exist t\ G [0, to) and t2 G (to, 6] such that ui(tj) = u2(tj), j G {1,2}, ui(t)>u2(t), t G (ti,t2).

Pick tm G (ti,t2) such that ui(tm) - u2(tm) = max {ui(t) - u2(t)} . te[ti,i2]

Then, o > (til - U2)"(tm) = /(im)[«?(im) - ul{tm)} > 0 , which is impossible. Therefore, the solution of (22) is unique. Subsequently, we will denote it by U[M,b]- Note that U[m,6] < M. Now, we claim that

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