Dynamic Systems and Applications 15 (2006) 133-146

MULTIPLICITY RESULTS FOR SEMIPOSITONE PROBLEMS ON BALLS

STEPHEN B. ROBINSON^{1} AND MATTHEW RUDD^{2}

1Department of Mathematics, Wake Forest University, Winston–Salem, NC 27109 USA. 2Department of Mathematics, University of Texas at Austin, Austin, TX 78712 USA.

Dedicated to John Baxley on the occasion of his retirement from Wake Forest University.

ABSTRACT. By carefully applying elementary techniques to a model problem, we obtain multiplicity results for semilinear semipositone problems posed on balls. The structural hypotheses on the nonlinearities considered are rather mild and, in particular, do not require any convexity or concavity. We thereby generalize some recent ODE results of Baxley, Henderson, Thompson, et al. to higher dimensions. AMS (MOS) Subject Classification. 34B15.

1. INTRODUCTION

There is a vast literature devoted to semilinear Dirichlet problems of the form

(1.1)  \{ \begin{matrix}−Δu & = & λf(u), & in & Ω, \\ u & = & 0, & on & ∂Ω, \end{matrix}

where Ω ⊂ ℝ^{N} is a bounded domain with smooth boundary, f : ℝ → ℝ is continuous, and λ is a real parameter. A standard problem is to describe the dependence of solutions of (1.1) on the parameter λ, and one finds that (1.1) has solution continua { (λ, u) } of various shapes according to the properties of the nonlinearity f and the underlying domain Ω. A class of problems which is particularly relevant to the present paper is that of S–shaped bifurcation curves: when f is sufficiently convex, for example, there is an interval Λ := (\underline{λ}, \overline{λ}) such that (1.1) has at least three solutions for any λ ∈ Λ ([5]).

In this paper, we study a different structural condition on f which guarantees the existence of at least three solutions of problem (1.1) when λ ≡ 1 and Ω is the unit ball. Specifically, the nonlinearity f in (1.1) will be modeled on the step function

Received November 13, 2004              1056-2176 $15.00 \begin{matrix}c \\ ◯ \end{matrix}Dynamic Publishers, Inc.

(1.2)  f̄(t) := \{ \begin{matrix}k, & for & t ≤ 1, \\ K, & for & t > 1, \end{matrix}

for constants k < 0 and K > 0. Theorem 4.1 states the conditions on f precisely, but we emphasize that we do not impose any convexity or concavity conditions on f (cf., e.g., [6],[7], where concavity assumptions are essential). The resulting problem is a semipositone (f(0) < 0) version of the problem considered in [8], in which f is strictly positive. The latter paper, in turn, was motivated by some recent work by Henderson and Thompson ([10],[11],[12]) and Baxley et al. ([3],[4]), which we now describe.

In [12], Henderson and Thompson proved the following result.

Theorem 1.1. Let 0 < a < b < \frac{c}{2}, and suppose that f : [0, ∞) → [0, ∞) is a continuous function satisfying

(i) f(t) < 8a, for 0 ≤ t ≤ a, (ii) f(t) ≥ 16b, for b ≤ t ≤ 2b, and (iii) f(t) ≤ 8c, for 0 ≤ t ≤ c.

Then the boundary–value problem

(1.3)  \{ \begin{matrix}u′′ + f(u) = 0, in (0, 1) , \\ u(0) = u(1) = 0 \end{matrix}

has at least three distinct symmetric nonnegative solutions.

The constants 8 and 16 appearing in this theorem are, in fact, sharp, as can be seen by constructing explicit solutions of (1.3) (see, e.g., the introduction of [8]). In [3] and [4], Baxley et al. have shown that, with appropriate modifications of these bounds on f, one can obtain any desired odd number of positive solutions of (1.3). The subsequent paper [8] considers the PDE analogue of this problem, i.e., (1.1) for a bounded domain Ω ⊂ ℝ^{N} with N ≥ 2 and λ = 1, and successfully generalizes Theorem 1.1 and its counterpart in [4] to higher dimensions.

As stated above, we are interested in (1.1) when the nonlinearity f is semipositone, f(0) < 0. Arndt and Robinson have investigated this problem in one dimension and have found conditions guaranteeing the existence of three solutions, at least one of which is positive ([2]). Under slightly more restrictive conditions, they conclude that two of the three solutions are positive. In the hopes of gaining insight into this problem on more general domains and in higher dimensions, we consider problem (1.1) when λ ≡ 1 and Ω ⊂ ℝ^{N} is the unit ball, for N ≥ 2. Since Ω is radially symmetric, we concentrate on radial solutions of (1.1), thereby reducing the equation above to the ODE

(1.4)  −u′′ + (\frac{1 − N}{r} ) u′ = f(u) ,

where u = u(r) and r = |x|. By the classical results of Gidas, Ni, and Nirenberg ([9]), any positive solution of (1.1) on the unit ball must be radial, but we will also consider nonpositive radial solutions of (1.1).

Since we are particularly interested in the possible relationship between k and K, we carefully analyze the model problem (1.1) with the step nonlinearity f̄ above. Section 2 treats dimensions N ≥ 3, and Section 3 then handles the problem in dimension N = 2. In either case, we will need to solve the elementary ordinary differential equation

(1.5)  −u′′ + (\frac{1 − N}{r} ) u′ = c ,

for a given constant c, and we find easily that the general solution of (1.5) is

(1.6)  u(r) = α + β r^{2−N} − (\frac{c}{2N} ) r^{2} ,

when N ≥ 3, while

(1.7)  u(r) = α + β ln r − (\frac{c}{4}) r^{2}

when N = 2. The constants α and β will be determined by the specifics of the boundary–value problems which arise.

Our methods in Sections 2 and 3 are entirely elementary and yield precise information on admissible pairs k and K for the model problems considered. As detailed in Section 4, these idealized problems produce lower solutions of (1.1), which we combine with other upper and lower solutions to obtain multiple solutions of (1.1).

2. THE MODEL PROBLEM ON A BALL, N ≥ 3

2.1. Explicit calculations. Let f̄ be the step function defined by (1.2). In this section, we will construct a bounded solution u of the model problem (1.1) for N ≥ 3, where u has the form

u(r) = \{ \begin{matrix}u_{1}(r), & for & r_{0} ≤ r ≤ 1 , \\ u_{2}(r), & for & 0 ≤ r ≤ r_{0} , \end{matrix}

for u_{1} ≤ 1, u_{2} ≥ 1, and r_{0} ∈ (0, 1). Since u is bounded, f̄(u) ∈ L^{2}(0, 1) and elliptic regularity guarantees that u ∈ H^{2}(0, 1) ⊂ C^{1}(0, 1). Consequently, u_{1} and u_{2} must meet smoothly: u_{1}(r_{0}) = u_{2}(r_{0}) and u′_{1}(r_{0}) = u′_{2}(r_{0}).

Since u_{1} ≤ 1 on (r_{0}, 1), f̄(u_{1}) = k and we have the boundary–value problem

(2.1)  \{ \begin{matrix}−u′′_{1} + (\frac{1 − N}{r} ) u′_{1} = k , \\ u_{1}(r_{0}) = 1, u_{1}(1) = 0 . \end{matrix}

The general solution of the ODE in (2.1) is

u_{1}(r) = α_{1} + β_{1} r^{2−N} − (\frac{k}{2N} ) r^{2} .

Since u_{1}(1) = 0, we must have

α_{1} + β_{1} = \frac{k}{2N} ,

so that u_{1} has the form

u_{1}(r) = α_{1} + (\frac{k}{2N} − α_{1}) r^{2−N} − (\frac{k}{2N} ) r^{2} .

As for u_{2}, we have the boundary–value problem

(2.2)  \{ \begin{matrix}−u′′_{2} + (\frac{1 − N}{r} ) u′_{2} = K , \\ u_{2}(r_{0}) = 1 . \end{matrix}

The solution of (2.2) has the form

u_{2}(r) = α_{2} − (\frac{K}{2N} ) r^{2} .

Note that we omit the r^{2−N} term from u_{2} since this term is singular at the origin.

Since

u_{1}(r_{0}) = 1 = u_{2}(r_{0}) ,

we have

α_{1} + (\frac{k}{2N} − α_{1}) r_{0}^{2−N} − (\frac{k}{2N} ) r_{0}^{2} = 1 = α_{2} − (\frac{K}{2N} ) r_{0}^{2} .

Solving for α_{1} and α_{2} yields

α_{1} = \frac{1}{1 − r_{0}^{2−N}} + (\frac{k}{2N} ) (\frac{r_{0}^{2} − r_{0}^{2−N}}{1 − r_{0}^{2−N}} ) < 0

and

α_{2} = 1 + (\frac{K}{2N} ) r_{0}^{2} > 0 .

We also have

β_{1} = \frac{1}{r_{0}^{2−N} − 1} + (\frac{k}{2N} ) (\frac{1 − r_{0}^{2}}{1 − r_{0}^{2−N}} ) > 0 .

We now impose the condition

u′_{1}(r_{0}) = u′_{2}(r_{0}) ,

MULTIPLICITY FOR SEMIPOSITONE PROBLEMS 137

which gives

(2 − N) β_{1}r_{0}^{1−N} − (\frac{k}{N} ) r_{0} = − (\frac{K}{N} ) r_{0} .

Substituting the expression above for β_{1} and simplifying, we obtain

(\frac{k − K}{N(2 − N)}) r_{0}^{N} + (\frac{2K − Nk}{2N(2 − N)}) r_{0}^{2} + 1 − \frac{k}{2N} = 0 .

Completing our construction of u thus rests on the existence of a root r_{0} ∈ (0, 1) of the polynomial

(2.3)  P (r) := (\frac{k − K}{N(2 − N)}) r^{N} + (\frac{2K − Nk}{2N(2 − N)}) r^{2} + 1 − \frac{k}{2N} .

Note that

P (0) = 1 − \frac{k}{2N} > 1 = P (1) ,

so we know that P is positive at the endpoints r = 0 and r = 1. Moreover, since

P′(r) = (\frac{k − K}{2 − N} ) r^{N−1} + (\frac{2K − Nk}{N(2 − N)}) r ,

we see that

P′(0) = 0 and P′(1) = \frac{K}{N} > 0 .

We need P to have a negative minimum inside the interval (0, 1). We see that P has two nonnegative critical points: r = 0 and the unique positive solution r_{∗} of

(\frac{k − K}{2 − N} ) r^{N−2} + (\frac{2K − Nk}{N(2 − N)}) = 0 , i.e.,

(2.4)  r_{∗}^{N−2} = \frac{2K − Nk}{N(K − k)} > 0 .

Since N > 2, k < 0, and K > 0, we find that

(2.5)  \frac{2}{N} < r_{∗}^{N−2} < 1 .

Note also that

P′′(r_{∗}) = \frac{2K − Nk}{N} > 0 ,

so that a minimum indeed occurs at r_{∗}.

Finally, we need P (r_{∗}) ≤ 0, which yields

(2.6) P (r_{∗}) = (\frac{k − K}{N(2 − N)}) (\frac{2K − Nk}{N(K − k)})^{N/(N−2)} +

(\frac{2K − Nk}{2N(2 − N)}) (\frac{2K − Nk}{N(K − k)})^{2/(N−2)} + 1 − \frac{k}{2N} ≤ 0 .

Let r_{∗} be any number such that (2.5) holds. We can then use (2.4) to relate k and K,

(2.7)  k = (\frac{Nr_{∗}^{N−2} − 2}{Nr_{∗}^{N−2} − N} ) K ,

and combining this with (2.6) yields the estimates

(2.8)  K ≥ \frac{2N(Nr_{∗}^{N−2} − N)}{(2 − N)r_{∗}^{N} + Nr_{∗}^{N−2} − 2}

and

(2.9)  k ≤ \frac{2N(Nr_{∗}^{N−2} − 2)}{(2 − N)r_{∗}^{N} + Nr_{∗}^{N−2} − 2} .

Again, the bounds (2.8) and (2.9) hold for any chosen r_{∗} ∈ (\frac{2}{N} , 1) and for any pair (k, K) which satisfy (2.7) for this choice of r_{∗}.

Having determined an admissible pair (k, K) corresponding to a chosen r_{∗} ∈ (\frac{2}{N} , 1), we know that P (r) either has the isolated root r_{∗} (if equality holds in both (2.8) and (2.9)) or two distinct roots r_{1} and r_{2} such that 0 < r_{1} < r_{∗} < r_{2} < 1. In the first case, our construction above yields one solution u of (1.1) for the step function f defined by (1.2); in the second case, we obtain two such solutions. In either case, it remains to check the sign of the resulting solution(s). Since

u′′_{1} = (2 − N) (1 − N) β_{1}r^{−N} − (\frac{k}{N} ) > 0 ,

it suffices to check the sign of u′_{1}(1): the solution u constructed from u_{1} and u_{2} will be positive on [0, 1] if and only if u′_{1}(1) ≤ 0.

Suppose that we choose k and K such that the polynomial P above has the unique root r_{∗}, for some r_{∗} ∈ (\frac{2}{N} , 1). By a direct computation, we have

u′_{1}(1) = (2 − N) [\frac{1}{r_{∗}^{2−N} − 1} + (\frac{k}{2N} ) (\frac{1 − r_{∗}^{2}}{1 − r_{∗}^{2−N}} )] − (\frac{k}{N} ) ,

and u′_{1}(1) ≤ 0 if and only if

(2 − N) [1 − (\frac{k}{2N} ) (1 − r_{∗}^{2})] − (\frac{k}{N} ) (r_{∗}^{2−N} − 1) ≤ 0 .

Simplifying this inequality yields

\frac{2N (N − 2) r_{∗}^{N−2}}{(2 − N) r_{∗}^{N} + Nr_{∗}^{N−2} − 2} ≤ k ,

which must be consistent with the precise value of k given by (2.9); thus, substituting this value of k, we must verify that

\frac{2N (N − 2) r_{∗}^{N−2}}{(2 − N) r_{∗}^{N} + Nr_{∗}^{N−2} − 2} ≤ \frac{2N(Nr_{∗}^{N−2} − 2)}{(2 − N)r_{∗}^{N} + Nr_{∗}^{N−2} − 2} .

Since both denominators are equal and negative, this last inequality holds because r_{∗}^{N−2} ≤ 1. In fact, r_{∗}^{N−2} < 1, and we conclude from these calculations that the strict inequality u′_{1}(1) < 0 holds.

For an arbitrary admissible pair (k, K), (2.4) determines the corresponding value of r_{∗}, and the distinct roots r_{1} and r_{2} of P (r) satisfy r_{1} < r_{∗} < r_{2}. Let (k_{∗}, K_{∗}) denote the admissible pair for which P (r) has the isolated root r_{∗}, and let u_{∗} denote Figure 1. Admissible pair boundary curves Γ_{N}, for dimensions N = 2, 3, 5, and 8. Admissible pairs (k, K) lie on and above the corresponding curve Γ_{N}.

the corresponding solution constructed above. Let u_{1} and u_{2} denote the solutions corresponding to the meeting points r_{1} and r_{2}, respectively. Since r_{2} > r_{∗}, it follows easily that u_{2} > u_{∗} on (0, 1), and we see that u_{2} is positive. The sign of u_{1}, on the other hand, may change on the interval (0, 1). Regardless, we find that there exists a positive solution of (1.1) for each admissible pair (k, K).

2.2. Summary. For any dimension N ≥ 3, define the curve Γ_{N} : (\frac{2}{N} , 1) → ℝ^{2} by

Γ_{N}(t) := ( \frac{2N(Nt^{N−2} − 2)}{(2 − N)t^{N} + Nt^{N−2} − 2} , \frac{2N(Nt^{N−2} − N)}{(2 − N)t^{N} + Nt^{N−2} − 2} ) .

For each pair (k, K) on or above the curve Γ_{N}, there exists a positive solution of the model problem (1.1) on the unit ball with the step nonlinearity (1.2).

Figure 1 illustrates the dependence of these curves on the dimension N, for dimensions 2 (covered in the next section), 3, 5, and 8.

3. THE MODEL PROBLEM ON A BALL, N = 2

3.1. Explicit calculations. We now consider the case in which Ω is the unit ball in ℝ^{2}, with f̄ defined by (1.2) as in the previous section. As before, we will construct a bounded solution u of (1.1) of the form

u(r) = \{ \begin{matrix}u_{1}(r), & for & r_{0} ≤ r ≤ 1 , \\ u_{2}(r), & for & 0 ≤ r ≤ r_{0} , \end{matrix}

where u_{1} ≤ 1, u_{2} ≥ 1, and u_{1} and u_{2} meet smoothly at r_{0} ∈ (0, 1).

Since u_{1} ≤ 1 on (r_{0}, 1), f̄(u_{1}) = k and we have the boundary–value problem

(3.1)  \{ \begin{matrix}−u′′_{1} − (\frac{1}{r}) u′_{1} = k , \\ u_{1}(r_{0}) = 1, u_{1}(1) = 0 . \end{matrix}

The general solution of the ODE in (3.1) is

u_{1}(r) = α_{1} + β_{1} ln r − (\frac{k}{4}) r^{2} .

Since u_{1}(1) = 0, we must have

α_{1} = \frac{k}{4} ,

so that u_{1} has the form

u_{1}(r) = \frac{k}{4} + β_{1} ln r − (\frac{k}{4}) r^{2} .

As for u_{2}, we have the boundary–value problem

(3.2)  \{ \begin{matrix}−u′′_{2} − (\frac{1}{r}) u′_{2} = K , \\ u_{2}(r_{0}) = 1 , \end{matrix}

whose solution has the form

u_{2}(r) = α_{2} − (\frac{K}{4} ) r^{2} .

Note that we omit the logarithmic term from u_{2} since this term is singular at the origin.

Since

u_{1}(r_{0}) = 1 = u_{2}(r_{0}) ,

we have

\frac{k}{4} + β_{1} ln r_{0} − (\frac{k}{4}) r_{0}^{2} = 1 = α_{2} − (\frac{K}{4} ) r_{0}^{2} .

MULTIPLICITY FOR SEMIPOSITONE PROBLEMS 141

Solving for β_{1} and α_{2} yields

β_{1} = \frac{4 − k + k r_{0}^{2}}{4 ln r_{0}}

and

α_{2} = 1 + (\frac{K}{4} ) r_{0}^{2} > 0 .

Imposing the condition

u′_{1}(r_{0}) = u′_{2}(r_{0}) ,

we successively obtain

\frac{β_{1}}{r_{0}} − (\frac{k}{2}) r_{0} = − (\frac{K}{2} ) r_{0}

and

\frac{4 − k + k r_{0}^{2}}{4r_{0} ln r_{0}} − (\frac{k}{2}) r_{0} = − (\frac{K}{2} ) r_{0} .

We must therefore find an r_{0} ∈ (0, 1) such that

4 − k + k r_{0}^{2} = 2 (k − K) r_{0}^{2} ln r_{0} , i.e.,

the function

P (r) := 4 − k + k r^{2} − 2 (k − K) r^{2} ln r

must vanish at some r_{0} ∈ (0, 1). Since

P′(r) = 2kr − 2 (k − K) (2r ln r + r) = 4 (K − k) r ln r + 2Kr ,

P has critical points at r = 0 and at the unique solution r_{∗} of

4 (K − k) ln r = −2K ,

so that

r_{∗} = exp \frac{K}{2 (k − K)} .

Since k − K < 0, we see that r_{∗} ∈ (0, 1). For any r_{∗} ∈ (0, 1) such that

ln r_{∗} = \frac{K}{2 (k − K)} ,

we have

(3.3)  k = ( \frac{2 ln r_{∗} + 1}{2 ln r_{∗}} ) K = (1 + \frac{1}{2 ln r_{∗}} ) K .

Since k < 0 and K > 0, we must have

1 + \frac{1}{2 ln r_{∗}} < 0 ,

from which we see that r_{∗} must satisfy

r_{∗} > \frac{1}{\sqrt{e}} .

Calculating directly, we find that P (0) = 4 − k > 0, P (1) = 4 > 0, and P′′(r_{∗}) = 4 (K − k) (ln r_{∗} + 1) + 2K = 8K − 4k > 0 .

Thus, P attains a minimum at r_{∗}. In fact,

P (r_{∗}) = 4 − k + (k − K) exp (\frac{K}{k − K} ) ,

and, using the relationship (3.3), we have

P (r_{∗}) = 4 + (\frac{r_{∗}^{2} − 1}{2 ln r_{∗}} − 1) K .

Since this minimum value must be nonpositive, the inequality

4 + (\frac{r_{∗}^{2} − 1}{2 ln r_{∗}} − 1) K ≤ 0

must hold; simplifying yields the lower bound

(3.4)  K ≥ \frac{8 ln r_{∗}}{1 + 2 ln r_{∗} − r_{∗}^{2}} ,

from which we also obtain the upper bound

(3.5)  k ≤ \frac{8 ln r_{∗} + 4}{1 + 2 ln r_{∗} − r_{∗}^{2}} .

Recall that the estimates (3.4) and (3.5) hold for any chosen r_{∗} ∈ ( \frac{1}{\sqrt{e}} , 1).

3.2. Summary. The boundary curve of admissible pairs (k, K) in dimension 2, Γ_{2} : ( \frac{1}{\sqrt{e}} , 1) → ℝ^{2}, is defined by

Γ_{2}(t) := (\frac{8 ln t + 4}{1 + 2 ln t − t^{2}} , \frac{8 ln t}{1 + 2 ln t − t^{2}} ) .

As t → \frac{1}{\sqrt{e}}, points on the curve Γ_{2} approach the point (0, \frac{4}{e}). For each pair (k, K) on or above the curve Γ_{2}, there exists a positive solution of the model problem (1.1) on the unit ball with the step nonlinearity (1.2).

4. A THREE–SOLUTIONS THEOREM

Let us begin by recalling the definitions of upper and lower solutions. For a general domain Ω ⊂ ℝ^{N} with smooth boundary ∂Ω, a function u ∈ W^{2,p}(Ω), for p > N, is an upper solution of (1.1) if

(4.1)  \{ \begin{matrix}Δu + f(u) ≤ 0, in Ω, \\ u ≥ 0, on ∂Ω . \end{matrix}

One defines a lower solution of (1.1) by reversing both of these inequalities. As in the previous sections, we henceforth let Ω denote the unit ball in ℝ^{N}, for N ≥ 2, and we focus on radial upper and lower solutions.

As above, Γ_{N} denotes the boundary curve of admissible pairs (k, K) for the model problem (1.1) with the step nonlinearity (1.2). For a fixed point (k, K) on Γ_{N}, the

MULTIPLICITY FOR SEMIPOSITONE PROBLEMS 143

construction in Sections 2 and 3 yields precisely one positive radial solution ψ of the corresponding model problem, and the constant

M := max_{x∈Ω} { ψ(x) }

satisfies M > 1. With this notation in place, we have our first multiplicity result.

Theorem 4.1. Let (k, K) be a point on Γ_{N}, let ψ be the positive radial solution of the corresponding model problem as above, and let M be its maximum value. Let a, b, and c be positive constants satisfying

0 < a < b < \frac{c}{K} ,

and let f : ℝ → ℝ be a continuous function such that

(i) ak < f(t) < 0, for t ≤ a,

(ii) f(t) > bK, for b ≤ t ≤ Mb, and

(iii) ak < f(t) < c for all t.

Then the Dirichlet problem (1.1) has at least three radial solutions, two of which do not change sign in Ω.

Proof. First, define the radial functions

\overline{u}_{1}(r) := 0 and \overline{u}_{2}(r) := \frac{c}{2N} (1 − r^{2}) .

Since f(0) < 0, it is easy to see that \overline{u}_{1} is an upper solution. On the other hand, since

−Δ\overline{u}_{2} = c in Ω and \overline{u}_{2} = 0 on ∂Ω ,

we have

Δ\overline{u}_{2} + f(\overline{u}_{2}) = −c + f(\overline{u}_{2}) < 0,

and we find that \overline{u}_{2} is a strict upper solution.

The first lower solution can be obtained similarly. Define \underline{u}_{1} by

\underline{u}_{1}(r) := \frac{ak}{2N} (1 − r^{2}) ,

so that

−Δ\underline{u}_{1} = ak in Ω and \underline{u}_{1} = 0 on ∂Ω .

It follows that \underline{u}_{1} is negative in Ω and, since

Δ\underline{u}_{1} + f(\underline{u}_{1}) = −ak + f(\underline{u}_{1}) > 0 ,

\underline{u}_{1} is a strict lower solution.

Now define \underline{u}_{2} := bψ. For points where \underline{u}_{2} ≤ b, we have

Δ\underline{u}_{2} + f(\underline{u}_{2}) = −kb + f(\underline{u}_{2}) ≥ k (a − b) ≥ 0 ,

while, for points where b ≤ \underline{u}_{2} ≤ bM, we have

Δ\underline{u}_{2} + f(\underline{u}_{2}) = −bK + f(\underline{u}_{2}) ≥ 0 .

Consequently, \underline{u}_{2} is a lower solution. Moreover, we know that \underline{u}_{2} is positive in Ω since ψ > 0 in Ω.

We therefore have two upper solutions and two lower solutions which clearly satisfy \underline{u}_{1} ≤ \overline{u}_{1} ≤ \underline{u}_{2} ≤ \overline{u}_{2}. Applying standard degree–theoretic arguments (cf. [1],[13]) shows that (1.1) has three radial solutions u_{1}, u_{2}, and u_{3} such that

\underline{u}_{1} ≤ u_{1} ≤ \overline{u}_{1} , \underline{u}_{2} ≤ u_{2} ≤ \overline{u}_{2} , and \underline{u}_{1} ≤ u_{3} ≤ \overline{u}_{2} ,

with u_{3}(x_{1}) < \underline{u}_{2}(x_{1}) and u_{3}(x_{2}) > \overline{u}_{1}(x_{2}) for some points x_{1} and x_{2}. These three solutions are thus distinct. Moreover, we see that u_{1} is negative and u_{2} is positive, completing the proof.

To obtain multiple positive solutions, we rely on the following observation, from which we directly obtain Corollary 4.3.

Lemma 4.2. Let k < 0 and c > 0 be fixed constants, and let u be a radially symmetric solution of

Δu + f(u) = 0 in Ω, u = 0 on ∂Ω ,

where −k ≤ f(u) ≤ 0 for u ≤ 1 and f(u) ≤ c for all u. There is a c^{∗}, depending only on k, such that if c ≤ c^{∗}, then u is either strictly negative or strictly positive in Ω, i.e. u cannot be a sign-changing solution.

Proof. Consider the case where u is sign-changing with u′(1) > 0, u(r) < 0 on some interval (r_{0}, 1), u is concave on (r_{1}, 1] with 0 < r_{1} < r_{0} < 1 with u(r_{1}) = 1, and u > 1 in [0, r_{1}) with u′(r) < 0 in (0, r_{1}) and u′(0) = 0.

For purposes of comparison, let \overline{u} be a radially symmetric function satisfying Δ\overline{u} = −k in the unit ball minus the origin, with \overline{u}(1) = 0 and \overline{u}′(1) = 0. Let r_{k} be such that \overline{u}(r_{k}) = 1, and note that r_{k} → 0 as k → 0^{−}. (We can solve for all of this explicitly as in the previous sections.)

A standard comparison shows that u ≤ \overline{u} in [r_{k}, 1], so 0 < r_{1} < r_{k}. Moreover, by the concavity of u we see that |u′(r_{1})| ≥ \frac{1}{1−r_{1}} , the absolute value of the slope of the line connecting (r_{1}, 1) and (1, 0). It follows via the Mean Value Theorem that u′′(r) = \frac{−1}{r_{1}(1−r_{1})} for some r in (0, r_{1}). However, if c < \frac{1}{r_{1}(1−r_{1})}, then a contradiction is reached when we try to substitute into

u′′ + \frac{N − 1}{r}u′ = −f(u) ≥ −c.

(Recall that u′(r) < 0.) Hence we can choose c^{∗} as the minimum of \frac{1}{r(1−r)} for r in (0, r_{k}]. This minimum is just \frac{1}{r_{k}(1−r_{k})} when r_{k} ≤ \frac{1}{2}, and thus becomes very large as k → 0^{−}.

Corollary 4.3. In addition to the hypotheses of Theorem 4.1, suppose that (k, K) is a point on Γ_{N} with k sufficiently close to zero to guarantee that c < c^{∗}. Then (1.1) has at least three radial solutions, two of which are positive and one of which is negative.

Proof. As in the proof of Theorem 4.1, we have a positive radial solution u_{1}, a negative radial solution u_{2}, and a third radial solution u_{3}. Since u_{3}(x_{2}) > \overline{u}_{1}(x_{2}) = 0 at some x_{2}, Lemma 4.2 guarantees that u_{3} is strictly positive.

In the case N = 1, Arndt and Robinson [2] have refined the result in Corollary 4.3 as follows: two of the three solutions are positive if either k > −8 or −24 < k < −8 and K < \frac{−16k}{k+8}.

REFERENCES

[1] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Review 18 (1976), 620–709.

[2] A. Arndt and S. B. Robinson, Multiple positive solutions for a nonlinear boundary value problem, preprint.

[3] J. V. Baxley and P. T. Carroll, Nonlinear boundary value problems with multiple positive solutions, Discrete Contin. Dyn. Syst. (2003), 83–90.

[4] J. V. Baxley and L. J. Haywood, Nonlinear boundary value problems with multiple solutions, Nonlinear Anal. 47 (2001), 1187–1198.

[5] K. J. Brown, M. M. A. Ibrahim, and R. Shivaji, S-shaped bifurcation curves, Nonlinear Anal. 5 (1981), 475–486.

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

[7] A. Castro and R. Shivaji, Positive solutions for a concave semipositone Dirichlet problem, Nonlinear Anal. 31 (1998), 91–98.

[8] P. Drábek and S. B. Robinson, Multiple positive solutions for elliptic boundary value problems, Rocky Mountain J. Math, to appear.

[9] B. Gidas, W. M. Ni, and L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phys. 68 (1979), 209–243.

[10] J. Henderson and H. B. Thompson, Existence of multiple solutions for second order boundary value problems, J. Differential Equations 166 (2000), 443–454.

[11] , Existence of multiple solutions for some n-th order boundary value problems, Comm. Appl. Nonlinear Anal. 7 (2000), 55–62.

[12] , Multiple symmetric positive solutions for a second order boundary value problem, Proc. Amer. Math. Soc. 128 (2000), 2373–2379.

146 S. B. ROBINSON AND M. RUDD

[13] R. Shivaji, A remark on the existence of three solutions via sub–super solutions, vol. 109 of Lecture Notes Pure Applied Math. 1987, pp. 561–566.