EXACT MULTIPLICITY OF POSITIVE SOLUTIONS

IN SEMIPOSITONE PROBLEMS WITH

CONCAVE-CONVEX TYPE NONLINEARITIES

Sudhasree Gadam and Joseph A. Iaia

Abstract. We study the existence, multiplicity, and stability of positive solutions to:

\begin{aligned}−u′′(x) & = λf(u(x)) for x ∈ (−1, 1), λ > 0, \\ u(−1) & = 0 = u(1), \end{aligned}

where f : [0, ∞) → ℝ is semipositone (f(0) < 0) and superlinear (lim_{t→∞} f(t)/t = ∞). We consider the case when the nonlinearity f is of concave-convex type having exactly one inflection point. We establish that f should be appropriately concave (by establishing conditions on f) to allow multiple positive solutions. For any λ > 0, we obtain the exact number of positive solutions as a function of f(t)/t and establish how the positive solution curves to the above problem change. Also, we give examples where our results apply. This work extends the work in [1] by giving a complete classification of positive solutions for concave-convex type nonlinearities.

1. Introduction

We study the positive solutions to the two point boundary value problem:

(1.1)  −u′′(x) = λf(u(x)) for x ∈ (−1, 1), λ > 0,
(1.2)  u(−1) = 0 = u(1),

where f : [0, ∞) → ℝ is a twice differentiable function such that:

(1.3)  f(0) < 0 (semipositone), lim_{t→∞} \frac{f(t)}{t} = ∞ (superlinear), and f has a unique positive zero β.

We define F by F (t) = ∫_{0}^{t} f(s) ds, and we observe that by (1.3):

(1.4)  F has a unique positive zero θ > β.

We also assume that f has exactly one inflection point t^{∗} with:

(1.5)  f′′(t) < 0 on (0, t^{∗}), f′′(t) > 0 on (t^{∗}, ∞), and t^{∗} > β.

Since (\frac{f(t)}{t})′ = \frac{tf′(t)−f(t)}{t^{2}} and (tf′(t) − f(t))′ = tf′′(t) with f(0) < 0, it follows from (1.5) that either:

(1.5)_{1}  (f(t)/t)′ ≥ 0 for all t > 0, or

(1.5)_{2}  (f(t)/t)′ > 0 for t ∈ (0, t_{1}) ∪ (t_{2}, ∞) and (f(t)/t)′ < 0 for t ∈ (t_{1}, t_{2})

for some t_{1}, t_{2} with 0 < t_{1} < t^{∗} < t_{2}.

EJQTDE, 2001 No. 4, p. 1 For future reference we define:

1991 Mathematics Subject Classification. Primary 34B15: Secondary 35J65.

Key words and phrases. Semipositone, concave, convex.

(1.6)  H(t) = F (t) − \frac{1}{2}tf(t)

and observe that:

(1.7)  H′(t) = −\frac{1}{2}t^{2}(f(t)/t)′.

Finally, for a positive solution of (1.1)-(1.2), we define:

ρ = sup_{(−1,1)} u(x).

We refer the reader to [2, 3] where the classification (1.5)_{1}, (1.5)_{2} helps in giving a complete description of positive solution curves for concave nonlinearities. In [7], Shi and Shivaji consider (1.5)_{2} and obtain a similar result to Theorem 1 section (2) with reasonably different methods from ours.

We also note that in [9], Wang considers the positone problem (f(0) > 0) with f initially convex and then concave. Finally, semipositone problems occur in several harvesting models (see [4]) and have been extensively studied in [1-3] and [5-8].

Our main results are:

Theorem 1.

(1) If f satisfies (1.3)-(1.5) and (1.5)_{1}, then there exists λ^{∗} with 0 < λ^{∗} < ∞ such that (1.1)-(1.2) has no positive solutions for λ > λ^{∗} and has a unique positive solution for λ ∈ (0, λ^{∗}] (see Fig. 1).

In addition, ρ ≡ ρ_{λ} is a decreasing function of λ with ρ_{λ} : (0, λ^{∗}] → [θ, ∞) such that ρ_{λ^{∗}} = θ and

lim_{λ→0^{+}} ρ_{λ} = +∞.

(2) If f satisfies (1.3)-(1.5), (1.5)_{2}, and H(t^{∗}) ≥ 0, then there exist λ_{1}, λ_{2}, λ^{∗} with 0 < λ_{1} < λ_{2} < ∞ and λ_{1} < λ^{∗} < ∞ such that (1.1)-(1.2) has no positive solutions for λ > max{λ_{2}, λ^{∗}} and has a unique positive solution for λ < λ_{1} while for λ = λ_{1} it has exactly two positive solutions. Also, ρ_{λ^{∗}} = θ and

lim_{λ→0^{+}} ρ_{λ} = +∞.

Subcase a: If λ_{2} ≤ λ^{∗} then for λ ∈ (λ_{1}, λ_{2}) (1.1)-(1.2) has exactly three positive solutions while for λ = λ_{2} it has exactly two positive solutions. Finally, if λ ∈ (λ_{2}, λ^{∗}] then (1.1)-(1.2) has exactly one positive solution (see Fig. 2A).

EJQTDE, 2001 No. 4, p. 2 Subcase b: If λ_{2} > λ^{∗} then for λ ∈ (λ_{1}, λ^{∗}] (1.1)-(1.2) has exactly three positive solutions while for λ ∈ (λ^{∗}, λ_{2}) (1.1)-(1.2) has exactly two positive solutions. Finally, for λ = λ_{2} the problem (1.1)-(1.2) has exactly one positive solution (see Fig. 2B).

This paper is organized as follows. In Section 2, we study the variations of the positive solutions with respect to the parameters λ and ρ. We prove Theorem 1 in Section 3. In Section 4 we give a family of examples which satisfies the hypotheses of Theorem 1.

2. First and Second Variations with respect to parameters

We first observe that any positive solution of (1.1)-(1.2) must be symmetric about the origin. To see this, let x_{0} ∈ (−1, 1) be the point at which u attains its maximum. Denote u(x_{0}) = ρ > 0. Thus u′(x_{0}) = 0 and it follows that u(x_{0}+x) and u(x_{0}−x) satisfy the differential equation (1.1) as well as the same initial conditions at x_{0}. Therefore, by uniqueness of solutions of initial value problems, we must have u(x_{0} + x) = u(x_{0} − x). So assuming without loss of generality that x_{0} ≥ 0, we see then that 0 = u(1) = u(2x_{0} − 1) and since u > 0 on (−1, 1), we must have 2x_{0} − 1 = −1 - i.e. x_{0} = 0 and thus u is symmetric about the origin.

With this result, for any ρ > 0 and any λ > 0 we define u(x, λ, ρ) to be the solution to the initial value EJQTDE, 2001 No. 4, p. 3 problem:

(2.1)  u′′(x) + λf(u(x)) = 0, λ > 0,
(2.2)  u(0) = ρ > 0, u′(0) = 0,

where ′ denotes differentiation with respect to x. Observing that u(−x, λ, ρ) also solves (2.1) and (2.2), it follows from the uniqueness of solutions of initial value problems that u(−x, λ, ρ) = u(x, λ, ρ). Thus we see that the set of positive solutions of (1.1)-(1.2) is precisely the set of solutions of (2.1)-(2.2) for which:

(2.3)  u(x, λ, ρ) > 0 for x ∈ (0, 1) and u(1, λ, ρ) = 0.

We now prove some elementary properties of positive solutions of (1.1)-(1.2) (and hence of (2.1)-(2.3) for some ρ > 0). Multiplying (2.1) by u′(x), integrating over (0, x), and using (2.2) yields:

(2.4)  \frac{1}{2}[u′(x)]^{2} + λF (u(x)) = λF (ρ).

Evaluating this at x = 1 gives:

(2.5)  0 ≤ \frac{1}{2}[u′(1)]^{2} = λF (ρ).

Since for ρ > 0 we have F (ρ) ≥ 0 if and only if ρ ≥ θ (by (1.4)), we see from (2.5) that:

(2.6)  positive solutions of (1.1)-(1.2) satisfy ρ ≥ θ, and
(2.7)  positive solutions of (1.1)-(1.2) satisfy u′(1) < 0 if ρ > θ and u′(1) = 0 if ρ = θ.

Also observe that if u is a positive solution to (2.1)-(2.3), then u′′(0) = −λf(ρ) < 0 (by (1.1), (1.3), and (2.6)) and therefore u′ < 0 on (0, ϵ) for some ϵ > 0. In fact u′(x) < 0 on (0, 1) for if u′(x_{1}) = 0 at some first x_{1} ∈ (0, 1) then 0 < u(x_{1}) < ρ while from (2.4) and (2.5) we have F (u(x_{1})) = F (ρ) ≥ 0. Thus by (1.4) β < θ ≤ u(x_{1}) < ρ. But this is impossible since F is increasing for x > β (by (1.3)) and thus:

(2.8)  positive solutions of (1.1)-(1.2) satisfy u′(x) < 0 on (0, 1).

Next we observe that u(xd, λ, ρ) and u(x, λd^{2}, ρ) satisfy the same initial value problem and so by uniqueness of solutions of initial value problems we have:

u(xd, λ, ρ) = u(x, λd^{2}, ρ).

After differentiating this with respect to d and setting d = 1, we obtain:

(2.9)  xu′(x, λ, ρ) = 2λ\frac{∂u}{∂λ}(x, λ, ρ).

Next let v denote the solution to the corresponding linearized problem of (1.1):

(2.10)  v′′(x) + λf′(u(x))v(x) = 0,
(2.11)  v(0) = 1, v′(0) = 0,

and let w denote the solution to the problem:

(2.12)  w′′(x) + λf′(u(x))w(x) + λf′′(u(x))v^{2}(x) = 0,
(2.13)  w(0) = 0, w′(0) = 0.

That is, v and w are the first and second derivatives of u with respect to ρ - i.e. v ≡ \frac{∂u}{∂ρ}(x, λ, ρ) and

w ≡ \frac{∂^{2}u}{∂ρ^{2}}(x, λ, ρ).

Now observe that by multiplying (2.10) by u′(x) and integrating on (0, x) we obtain:

(2.14)  u′(x)v′(x) + λf(u(x))v(x) = λf(ρ).

Similarly, multiplying (2.12) by u′(x) and integrating on (0, x) gives:

(2.15)  u′(x)w′(x) + λf(u(x))w(x) + v′^{2}(x) + λf′(u(x))v^{2}(x) = λf′(ρ).

Lemma 2.1. Suppose f satisfies (1.3). Let u(x, λ_{0}, ρ_{0}) be a positive solution to (1.1)-(1.2). Then v(x) ≡ \frac{∂u}{∂ρ}(x, λ_{0}, ρ_{0}) has at most one zero in [0, 1].

Proof. We first observe that if v(x_{0}) = 0 then v′(x_{0}) ≠ 0 for if v′(x_{0}) = 0 then by uniqueness of solutions of initial value problems, it follows that v ≡ 0. On the other hand, v(0) = 1 ≠ 0.

Now on to the proof of the lemma. Suppose by the way of contradiction that x_{1} and x_{2} are the first two consecutive zeros of v. Then by the remarks in the previous paragraph and since v(0) = 1, we have v′(x_{1}) < 0 and v′(x_{2}) > 0. Also by (2.14) it follows that u′(x_{2})v′(x_{2}) = λ_{0}f(ρ_{0}) and so we see that u′(x_{2}) and f(ρ_{0}) have the same sign. But since ρ_{0} ≥ θ (by (2.6)), it follows from (1.3)-(1.4) that f(ρ_{0}) > 0 and hence u′(x_{2}) > 0. But this contradicts (2.7)-(2.8). Hence, v(x) can have at most one zero on [0, 1]. □

Remark: Note that the above lemma does not rely on the concavity properties of f. □

Lemma 2.2. Suppose f satisfies (1.3)-(1.5). Let u(x, λ_{0}, ρ_{0}) be a positive solution to (1.1)-(1.2) with θ ≤ ρ_{0} ≤ t^{∗} and suppose also that v(1) = \frac{∂u}{∂ρ}(1, λ_{0}, ρ_{0}) = 0. Then w(1) = \frac{∂^{2}u}{∂ρ^{2}}(1, λ_{0}, ρ_{0}) > 0.

Proof. Recall that v ≡ \frac{∂u}{∂ρ} satisfies (2.10)-(2.11) and w ≡ \frac{∂^{2}u}{∂ρ^{2}} satisfies (2.12)-(2.13). Multiplying (2.10) by w and (2.12) by v, subtracting one from the other, integrating over (0, 1), and using v(1) = 0 we obtain:

(2.16)  w(1)v′(1) = ∫_{0}^{1} λ_{0}f′′(u(x))v^{3}(x) dx.

Since v(1) = 0, it follows from lemma 2.1 that we have v > 0 on [0, 1) and it also follows from the uniqueness of solutions to initial value problems that v′(1) < 0. Since θ ≤ ρ_{0} ≤ t^{∗} and u(x) is decreasing on (0,1) (by (2.8)), it follows that u(x) < ρ_{0} ≤ t^{∗} on (0, 1) and so by (1.5) we have f′′(u(x)) < 0 on (0, 1). These facts and (2.16) imply w(1) > 0. This proves the lemma. □

Lemma 2.3. If f satisfies (1.3)-(1.5), (1.5)_{2}, and H(t^{∗}) ≥ 0, then the function defined by J : [0, ∞) → ℝ, J(t) = f′(t)F (t) − \frac{1}{2}f^{2}(t) has exactly one positive zero, t^{∗∗}, and θ < t^{∗} < t^{∗∗} < t_{2}.

Proof. By (1.5), t^{∗} > β. Combining this with the fact that H(t^{∗}) ≥ 0 implies F (t^{∗}) ≥ \frac{1}{2}t^{∗}f(t^{∗}) > 0 (since t^{∗} > β) and so F (t^{∗}) > 0 which implies t^{∗} > θ (by (1.4)).

Next observe that J′(t) = f′′(t)F (t) so J is increasing on (0, θ) ∪ (t^{∗}, ∞) and decreasing on (θ, t^{∗}). Also, observe J(θ) < 0 so that J < 0 on [0, t^{∗}]. Hence J has at most one positive zero.

Also, J = f′H − fH′ hence J(t_{2}) = f′(t_{2})H(t_{2}) and f(t_{2}) = t_{2}f′(t_{2}) (by (1.5)_{2}). Since t_{2} > t^{∗} > β (by (1.5)_{2}), we have t_{2}f′(t_{2}) = f(t_{2}) > 0 and so J(t_{2}) > 0 because H has a maximum at t_{2} and so H(t_{2}) > H(t^{∗}) ≥ 0. Thus, J has exactly one positive zero, t^{∗∗}, and θ < t^{∗} < t^{∗∗} < t_{2}. This completes the proof of the lemma. □

Lemma 2.4. Suppose f satisfies (1.3)-(1.5) and (1.5)_{2}. Let u(x, λ_{0}, ρ_{0}) be a positive solution of (1.1)-(1.2) with ρ_{0} ≥ t^{∗∗} and suppose also that v(1) = \frac{∂u}{∂ρ}(1, λ_{0}, ρ_{0}) = 0. Then w(1) = \frac{∂^{2}u}{∂ρ^{2}}(1, λ_{0}, ρ_{0}) < 0.

Proof. We define:

E = v′^{2} + λ_{0}f′(u)v^{2}

and observe (by (2.10)) that:

E′ = λ_{0}f′′(u)u′v^{2}.

Since ρ_{0} ≥ t^{∗∗} > t^{∗}, examining the sign of E′ along with (1.5) and (2.8), we see that E is decreasing on (0, x^{∗}) and increasing on (x^{∗}, 1) where x^{∗} is the point at which u(x^{∗}) = t^{∗}.

Thus, E has exactly one local minimum and no local maxima on (0,1). Hence the maximum of E on [0,1] occurs either at x = 0 or x = 1.

Next, we see from lemma 2.3 that ρ_{0} ≥ t^{∗∗} implies J(ρ_{0}) ≥ 0. Using (2.4), (2.11), (2.14), and the fact that v(1) = 0, we obtain:

E(0) − E(1) = \frac{λ_{0}}{F (ρ_{0})}[f′(ρ_{0})F (ρ_{0}) − \frac{f^{2}(ρ_{0})}{2}] = \frac{λ_{0}}{F (ρ_{0})}J(ρ_{0}) ≥ 0.

EJQTDE, 2001 No. 4, p. 5 Thus, for x ∈ [0, 1] we have v′^{2} + λ_{0}f′(u)v^{2} = E(x) ≤ E(0) = λ_{0}f′(ρ_{0}). Hence, by (2.15):

u′w′ + λ_{0}f(u)w ≥ 0 on [0, 1].

Now solving (2.4) for u′, using (2.8) and substituting into the above inequality gives:

w′ − \sqrt{\frac{λ_{0}}{2}} \frac{f(u)}{\sqrt{F (ρ_{0}) − F (u)}}w ≤ 0 on (0, 1].

Multiplying by the appropriate integrating factor and then integrating on (ϵ, x) ⊂ (0, 1] for ϵ > 0 we have:

∫_{ϵ}^{x}(we^{−\sqrt{\frac{λ_{0}}{2}} ∫_{ϵ}^{x} \frac{f(u) dt}{\sqrt{F (ρ_{0})−F (u)}}} )′ ≤ 0.

Now, for ϵ small enough we have w(ϵ) < 0 because by (2.12)-(2.13) we have w(0) = 0, w′(0) = 0, and w′′(0) = −λ_{0}f′′(ρ_{0}) < 0 since ρ_{0} ≥ t^{∗∗} > t^{∗}. Therefore:

w(x)e^{−\sqrt{\frac{λ_{0}}{2}} ∫_{ϵ}^{x} \frac{f(u) dt}{\sqrt{F (ρ_{0})−F (u)}}} ≤ w(ϵ) < 0.

Hence w(x) < 0 on (ϵ, 1]. In particular, w(1) < 0. This completes the proof of the lemma. □

3. Proof of Theorem 1

We begin by rewriting (2.4), and we obtain:

\frac{−u′(x)}{\sqrt{2}\sqrt{F (ρ) − F (u(x))}} = \sqrt{λ} on (0, 1).

Thus, after integrating on (x, 1) and using u(1) = 0 we obtain:

(3.1)  \frac{1}{\sqrt{2}} ∫_{0}^{u(x)} \frac{dt}{\sqrt{F (ρ) − F (t)}} = \sqrt{λ}(1 − x).

Letting x → 0 gives:

(3.2)  \sqrt{λ} = \frac{1}{\sqrt{2}} ∫_{0}^{ρ} \frac{dt}{\sqrt{F (ρ) − F (t)}} ≡ G(ρ).

Thus, given a positive solution of (1.1)-(1.2) (and hence of (2.1)-(2.3) for some ρ ≥ θ), we see that λ and ρ are related by equation (3.2).

Conversely, given λ_{0} > 0, if there exists a ρ_{0} ∈ [θ, ∞) with G(ρ_{0}) = \sqrt{λ_{0}}, then we can obtain a positive solution of (1.1)-(1.2) as follows. Define K : [0, ρ_{0}] → ℝ by:

K(x) = \frac{1}{\sqrt{2}} ∫_{0}^{x} \frac{dt}{\sqrt{F (ρ_{0}) − F (t)}}.

Since ρ_{0} ≥ θ, it follows from (1.3)-(1.4) that 1/\sqrt{F (ρ_{0}) − F (t)} is integrable on [0, ρ_{0}]. Thus K is continuous on [0, ρ_{0}] while from (3.2) we have K(ρ_{0}) = G(ρ_{0}) = \sqrt{λ_{0}}. Also:

K′(x) = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{F (ρ) − F (x)}} > 0 on [0, ρ_{0}).

Thus K is continuous and increasing on [0, ρ_{0}] and so K has an inverse. In addition,

(K^{−1}(x))′ = \sqrt{2}\sqrt{F (ρ) − F (K^{−1}(x))}.

Taking a hint from (3.1) which says a positive solution of (1.1)-(1.2) satisfies K(u(x)) = \sqrt{λ}(1−x), we define

u(x) = K^{−1}(\sqrt{λ}_{0}(1 − x)).

It is then straightforward to show that u solves (2.1)-(2.3) with λ = λ_{0} and ρ = ρ_{0}. Thus, we see that the set of λ for which there is a positive solution of (1.1)-(1.2) is precisely those positive λ for which there is a solution - ρ - of G(ρ) = \sqrt{λ}. Therefore we now turn our attention to a study of the function G = \sqrt{λ} defined in (3.2). We begin by changing variables in (3.2) and obtain:

\sqrt{λ(ρ)} = G(ρ) = \frac{1}{\sqrt{2}} ∫_{0}^{1} \frac{ρ dv}{\sqrt{F (ρ) − F (ρv)}}

and from (1.3)-(1.4) it follows \sqrt{λ(ρ)} is a positive continuous function on [θ, ∞). Also, by (1.3)-(1.4):

\sqrt{λ(θ)} = G(θ) = \frac{1}{\sqrt{2}} ∫_{0}^{1} \frac{θ dv}{\sqrt{−F (θv)}} ≡ \sqrt{λ^{∗}} = finite, positive.

In addition, \sqrt{λ(ρ)} is differentiable over (θ, ∞) and:

(3.3)  \frac{λ′(ρ)}{2\sqrt{λ(ρ)}} = G′(ρ) = \frac{1}{\sqrt{2}} ∫_{0}^{1} \frac{H(ρ) − H(ρv)}{[F (ρ) − F (ρv)]^{3/2}} dv

where H is given by (1.6). Since u(x, λ(ρ), ρ) is a positive solution of (1.1)-(1.2), we also have:

u(1, λ(ρ), ρ) = 0.

Differentiating this with respect to ρ gives:

(3.4)  \frac{∂u}{∂λ}(1, λ(ρ), ρ)λ′(ρ) + \frac{∂u}{∂ρ}(1, λ(ρ), ρ) = 0.

We now show that lim_{ρ→θ^{+}} λ′(ρ) = −∞. We know from above that lim_{ρ→θ^{+}} λ(ρ) = λ(θ) = λ^{∗} is positive and finite. Also, lim_{ρ→θ^{+}} \frac{∂u}{∂λ}(1, λ(ρ), ρ) = lim_{ρ→θ^{+}} \frac{1}{2λ(ρ)}u′(1, λ(ρ), ρ) = \frac{1}{2λ(θ)}u′(1, λ(θ), θ) = 0 by (2.7) and (2.9). On the other hand, (2.7) and (2.14) imply lim_{ρ→θ^{+}} \frac{∂u}{∂ρ}(1, λ(ρ), ρ) = \frac{f(θ)}{f(0)} < 0. It now follows from (3.4) that:

(3.5)  lim_{ρ→θ^{+}} λ′(ρ) = −∞.

We claim now that λ′(ρ) < 0 for large ρ and lim_{ρ→∞} λ(ρ) = 0. Since H′ = \frac{1}{2}(f − tf′) < 0 for ρ large and H′′ = −\frac{1}{2}tf′′ < 0 for ρ > t^{∗}, it follows that lim_{ρ→∞} H(ρ) = −∞. Combining these facts, it follows that for large ρ we have H(ρ) < H(ρv) for all v ∈ (0, 1). Therefore, by (3.3)

(3.6)  λ′(ρ) < 0 for large ρ.

EJQTDE, 2001 No. 4, p. 7 Next, we rewrite \sqrt{λ} as:

\sqrt{λ(ρ)} = G(ρ) = \frac{1}{\sqrt{2}} ∫_{0}^{1/2} \frac{ρ dv}{\sqrt{F (ρ) − F (ρv)}} + \frac{1}{\sqrt{2}} ∫_{1/2}^{1} \frac{ρ dv}{\sqrt{F (ρ) − F (ρv)}}

From (1.5), f′′ > 0 for t > t^{∗} and from (1.3) f(t)/t → ∞ as t → ∞, thus f(= F′) and f′ are positive for large t and lim_{t→∞} F (t) = ∞. Therefore, for 0 < v < \frac{1}{2} and ρ large we have F (ρv) ≤ F (\frac{1}{2}ρ). And so by the mean value theorem:

F (ρ) − F (ρv) ≥ F (ρ) − F (\frac{1}{2}ρ) ≥ \frac{1}{2}ρf(\frac{1}{2}ρ).

Also for \frac{1}{2} < v < 1 and large ρ, we have again by the mean value theorem:

F (ρ) − F (ρv) ≥ ρf(\frac{1}{2}ρ)(1 − v).

Combining these estimates into the first and second integrals above respectively gives:

\sqrt{λ(ρ)} = G(ρ) ≤ \frac{1}{\sqrt{2}} ∫0^{\frac{1}{2}} \frac{ρ}{\sqrt{\frac{1}{2}ρf(\frac{1}{2}ρ)}} + \frac{1}{\sqrt{2}} ∫ 1_{\frac{1}{2}} \frac{ρ}{\sqrt{ρf(\frac{1}{2}ρ)}} \frac{1}{\sqrt{1 − v}} dv = \frac{3}{2}\sqrt{\frac{ρ}{f(\frac{1}{2}ρ)}}.

Thus, by the superlinearity of f - (1.3) - we see that

(3.7)  lim_{ρ→∞} λ(ρ) = 0.

Consequently, since λ(ρ) is continuous on [θ, ∞) and tends to 0 at infinity (by (3.7)), we see that λ(ρ) is a bounded function. Thus, (1.1)-(1.2) has no positive solutions for λ > max_{[θ,∞)} λ(ρ).

Case (1.5)_{1} : It remains to prove that λ′(ρ) < 0 for ρ ∈ (θ, ∞). From (1.6) we have H′(t) = \frac{1}{2}[f(t) − tf′(t)] and H′′(t) = −\frac{1}{2}tf′′(t). Since (1.5)_{1} holds we infer that H′(t) ≤ 0 (in fact, H′(t) = 0 for at most one value of t) and hence λ′(ρ) < 0 follows from (3.3). This together with that λ(ρ) is continuous on [θ, ∞) implies that λ(ρ) has an inverse, ρ_{λ} : (0, λ^{∗}] → [θ, ∞) and ρ′_{λ} < 0 on (θ, ∞) with ρ_{λ^{∗}} = θ and lim_{λ→0^{+}} ρ_{λ} = ∞. This completes the proof of Case (1.5)_{1}.

Case (1.5)_{2} : In view of (1.5)_{2} and (1.7) we have H′(t) < 0 on [0, t_{1}) ∪ (t_{2}, ∞) and H′(t) > 0 on (t_{1}, t_{2}). Thus for ρ ∈ (t^{∗}, t^{∗∗}) ⊂ (t_{1}, t_{2}) H is increasing and H(ρ) > H(t^{∗}) ≥ 0. Also, since H(0) = 0 and H is decreasing on (0, t_{1}), it follows that H(ρv) < H(ρ) for all v ∈ (0, 1) and all ρ ∈ (t^{∗}, t^{∗∗}). Hence by (3.3):

(3.8)  λ′(ρ) > 0 for ρ ∈ (t^{∗}, t^{∗∗}).

Combining this with (3.5) and (3.6) we see that λ(ρ) has at least one local minimum on (θ, t^{∗}) and at least one local maximum on (t^{∗∗}, ∞). To complete the proof of theorem 1 we will show that these are the only critical points of λ(ρ). First, suppose ρ_{0} ∈ (θ, t^{∗}) and λ′(ρ_{0}) = 0. From (3.4) we see \frac{∂u}{∂ρ}(1, λ(ρ_{0}), ρ_{0}) = 0.

From lemma 2.2 we see that \frac{∂^{2}u}{∂ρ^{2}}(1, λ(ρ_{0}), ρ_{0}) > 0. Differentiating (3.4) and evaluating at ρ_{0} gives:

(3.9)  \frac{∂u}{∂λ}(1, λ(ρ_{0}), ρ_{0})λ′′(ρ_{0}) + \frac{∂^{2}u}{∂ρ^{2}}(1, λ(ρ_{0}), ρ_{0}) = 0.

Since \frac{∂u}{∂λ}(1, λ(ρ_{0}), ρ_{0}) < 0 by (2.7) and (2.9), we see that λ′′(ρ_{0}) > 0. Hence, ρ_{0} must be a local minimum of λ(ρ). If there were a second critical point, ρ_{1} ∈ (θ, t^{∗}), of λ(ρ), the same argument shows that it too would be a local minimum of λ(ρ) and thus between ρ_{0} and ρ_{1} there would be a local maximum, ρ_{2}, with λ′′(ρ_{2}) > 0 but this is clearly impossible. Thus, ρ_{0} is the only critical point of λ(ρ) on (θ, t^{∗}). Similarly, suppose ρ_{0} ∈ (t^{∗∗}, ∞) and λ′(ρ_{0}) = 0. Then as before (3.4) implies \frac{∂u}{∂ρ}(1, λ(ρ_{0}), ρ_{0}) = 0. Now using lemma

2.4 we see that \frac{∂^{2}u}{∂ρ^{2}}(1, λ(ρ_{0}), ρ_{0}) < 0. And as above, using (3.9) we see that λ′′(ρ_{0}) < 0. Hence, ρ_{0} must be a local maximum of λ(ρ) and as above this is the only critical point of λ(ρ) on (t^{∗∗}, ∞). This completes the proof of theorem 1. □

EJQTDE, 2001 No. 4, p. 8

4. Examples

Consider f(t) = t^{3} −3At^{2} +6Bt−C where A, B, and C are positive. Then f is semipositone and superlinear. Also, f has exactly one inflection point at t^{∗} = A. We have f′(t) = 3t^{2} − 6At + 6B hence f′(t) ≥ 0 for all t if and only if 2B ≥ A^{2}. Thus if 2B ≥ A^{2}, f has exactly one zero β and since we have f(t^{∗}) = f(A) = −2A^{3} + 6AB − C, we see that t^{∗} > β if 6AB > 2A^{3} + C. Next, H(t) = F (t) − \frac{1}{2}tf(t) = −\frac{1}{4}t^{4} + \frac{A}{2}t^{3} − \frac{1}{2}Ct, H′(t) = −t^{3} + \frac{3A}{2}t^{2} − \frac{1}{2}C, and H′′(t) = −3t^{2} + 3At. Thus, H′ has exactly one local maximum at t^{∗} = A. If H′(A) > 0 then H′ has two zeros, while H′ ≤ 0 if H′(A) ≤ 0. Note that H′(A) > 0 if and only if A^{3} > C and H(t^{∗}) = H(A) ≥ 0 if and only if A^{3} ≥ 2C. Thus, (1.3)-(1.5) and (1.5)_{1} are satisfied if we choose positive A, B, C so that 6B > \frac{C}{A} + 2A^{2}, C ≥ A^{3} whereas (1.3)-(1.5) and (1.5)_{2} are satisfied if 6B > \frac{C}{A} + 2A^{2}, A^{3} ≥ 2C, and 2B ≥ A^{2}.

References

1. A. Castro and R. Shivaji, Non-negative solutions for a class of non-positone problems, Proc. Roy. Soc. Edinburgh. 108A (1988), 291–302.

2. A. Castro, S. Gadam and R. Shivaji, Positive solution curves of semipositone problems with concave nonlinearities, Proc. Roy. Soc. Edinburgh. 127A (1997), 921–934.

3. A. Castro, S. Gadam and R. Shivaji, Evolution of positive solution curves in semipositone problems with concave nonlinearities, to appear in Contemp. Math.

4. M. R. Myerscough, B. F. Gray, W. L. Hogarth, and J. Norbury, An analysis of an ordinary differential equations model for a two species predator-prey system with harvesting and stocking, J. Math. Biol. 30 (1992), 389–411.

5. A. Castro and S. Gadam, Uniqueness of stable and unstable positive solutions for semipositone problems, Nonlinear Analysis, TMA 22 No. 4 (1994), 425–429.

6. A. Castro, S. Gadam and R. Shivaji, Branches of radial solutions for semipositone problems, Jour. Diff. Eqns. 120 No. 1 (1995), 30–45.

7. J. Shi and R. Shivaji, Exact multiplicity of solutions for classes of semipositone problems with concave-convex type nonlinearity, Discrete and Continuous Dynamical Systems 7 no. 3 (2001), 559–571.

8. S-H. Wang, Positive solutions for a class of nonpositone problems with concave nonlinearities, Proc. Roy. Soc. Edinburgh 124 No. 3 (1994), 507–515.

9. S-H. Wang, On S-shaped bifurcation curves, Nonlinear Analysis, TMA 22 No. 12 (1994), 1475–1485.

Mathematics, ‘Yashodha’, JCR VI Cross, Chitradurga, India 577501

Dept. of Mathematics, University of North Texas, Denton, TX 76203, U.S.A.

E-mail address: iaia@unt.edu