Existence of positive radial solutions for a semipositone elliptic equation

J. Math. Anal. Appl. 484 (2020) 123735

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Existence of positive radial solutions for a semipositone elliptic equation Ryuji Kajikiya a,1 , Eunkyung Ko b,∗,2 a

Department of Mathematics, Faculty of Science and Engineering, Saga University, Saga, 840-8502, Japan b Major in Mathematics, Faculty of Basic Sciences, College of Natural Sciences, Keimyung University, Daegu, 42601, South Korea

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 3 April 2019 Available online 29 November 2019 Submitted by J. Shi Keywords: Semipositone Elliptic equation Radial solution Positive solution Variational method

In this paper, we study the existence of positive radial solutions for a semipositone elliptic equation −Δu = λf (u)

in Ω,

u=0

on ∂Ω,

where Ω is a ball or an annulus in RN with N ≥ 2, λ > 0 is a parameter and f is a continuous function satisfying the semipositone condition f (0) < 0. We give a weak and general sufficient condition on f for the existence of positive radial solutions for λ > 0 large. We show that the linearized operator at each solution has the nonnegative first eigenvalue. These solutions are obtained as minimizers of a Lagrangian functional. © 2019 Elsevier Inc. All rights reserved.

1. Introduction We study the existence of positive radial solutions for the semipositone elliptic equation −Δu = λf (u) in Ω,

u = 0 on ∂Ω,

(1.1)

where Ω is an annulus A or a unit ball B in RN with N ≥ 2, λ > 0 is a parameter and f (s) satisfies the semipositone condition f (0) < 0. Here the annulus A and the ball B are defined by A := {x ∈ RN : a < |x| < b},

B := {x ∈ RN : |x| < 1}.

* Corresponding author. E-mail addresses: [email protected] (R. Kajikiya), [email protected] (E. Ko). The first author was supported by JSPS KAKENHI Grant Number 16K05236. 2 The second author was supported by the National Research Foundation of Korea (NRF) Grant funded by the Korea Government (NRF-2017R1D1A1B03030681). 1

https://doi.org/10.1016/j.jmaa.2019.123735 0022-247X/© 2019 Elsevier Inc. All rights reserved.

2

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

Semipositone problems arise in various fields such as mechanical systems, design of suspension bridges, chemical reactions, astrophysics (thermal equilibrium of plasmas), combustion and management of natural resources (constant effort harvesting) (see [11,12,14]). As pointed out by P. L. Lions in [13], semipositone problems are mathematically very challenging, from the point of view of many important natural applications, in particular for positive solutions. The study of semipositone problems was formally introduced by Castro and Shivaji in [3]. Also, it has been shown in [16] that semipositone problems lead to symmetry breaking phenomena. In the same work, the authors proved that f (0) < 0 is a necessary condition for symmetry breaking of positive solutions in a ball in RN . Finding positive solutions to the problem (1.1) has been actively studied for a long time, and it is not an easy task in semipositone problems. The difficulty is due to the fact that solutions have to live in regions where the reaction term f (u) is negative as well as positive. However, many authors have established rich results for the equation (1.1) via sub-super solutions method, variational method, degree theory, fixed point theory, shooting method, reflection argument, maximum principles, bifurcation theory etc. The range of the parameter λ for which (1.1) has a positive solution is dramatically changed according to the behavior of f near infinity as well as the sign of f (0). When f satisfies a sublinear condition near infinity (i.e. lims→∞ f (s)/s = 0), it has been known that (1.1) has a positive solution for λ large under suitable assumptions on f . In [6] the method of sub-super solutions was employed to prove an existence result for λ large on the condition that f is bounded from below. Ambrosetti, Arcoya and Buffoni in [1], by means of bifurcation theory, established existence results when f (u) behaves like uβ , 0 ≤ β < 1, near infinity. In [8], for nonlinearity with falling zero, the variational methods were used to prove an existence results for λ large. When Ω is a ball in RN , for classes of sublinear nonlinearities bifurcation diagrams are quite well known. In [4] and [5], they established that there exist 0 < λ1 < λ2 such that for λ < λ1 no solution, for λ1 < λ ≤ λ2 exactly two solutions, and for λ > λ2 and λ = λ1 exactly one solution. Further, they proved that upper branch of the solution is stable and the lower branch including λ = λ1 is unstable. In [5], the authors studied differentiable increasing nonlinearities (e.g., f (u) = (u + 1)1/3 − 2), while in [4] differentiable nonlinearities with falling zero (e.g., f (u) = −u2 + u − 1) were discussed. The authors obtained the existence results by means of ODE techniques, such as shooting method (also see [17]). In [9], for continuous nonlinearities, the variational method is used to establish existence and nonexistence results. The authors proved that there exist 0 < λ1 ≤ λ2 such that there is no solution for λ < λ1 and at least one positive radial solution for λ > λ2 . When Ω is an annulus, the existence of positive solutions are studied in [2,7,10]. In [2], a superlinear semipositone equation is considered and nonnegative solutions are obtained. In [7,10], a system of p-Laplacian or a generalized p-Laplacian is studied and the existence and nonexistence of positive solutions are proved. The purpose of the present paper is to give a weak and general sufficient condition on f (s) for the existence of positive radial solutions for large λ and to study both on a ball and on an annulus in the same variational method. Furthermore, we prove that all eigenvalues of the linearized operator at the solution are nonnegative. Hence the solution is stable. We employ the variational method to establish the existence and nonexistence results. We here note that the constant zero u(x) ≡ 0 must not be a solution of (1.1) because f (0) = 0. Throughout the paper, we always assume that N ≥ 2. We consider positive radial solutions for (1.1). Then (1.1) with u = u(r), r = |x| is rewritten as u +

N −1  u + λf (u) = 0 in (a, b), r

(1.2)

where a = 0 and b = 1 when Ω is the ball B and 0 < a < b when Ω is the annulus A. The boundary condition is written as

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

3

u (0) = u(1) = 0 when Ω = B,

(1.3)

u(a) = u(b) = 0 when Ω = A.

(1.4)

We define s F (s) :=

f (t)dt.

(1.5)

0

In what follows, by (1.2) we mean (1.2) with (1.3) if Ω = B and (1.2) with (1.4) if Ω = A. We state our main results. We first consider the nonexistence of positive solutions under the condition, lim sup F (s)/s2 < ∞.

(1.6)

s→∞

Theorem 1.1. Let f (s) be a continuous function on [0, ∞) which satisfies f (0) < 0 and (1.6). Then (1.2) has no positive solutions for small λ > 0. It seems that the theorem above is valid for the radial solutions only. For a general domain Ω, we give another nonexistence result. Consider the next condition which is slightly stronger than (1.6). lim sup f (s)/s < ∞.

(1.7)

s→∞

Theorem 1.2. Let Ω be a smooth bounded domain in RN and let f be a continuous function on [0, ∞) satisfying f (0) < 0 and (1.7). Then for small λ > 0, (1.1) has no nonnegative solutions. We consider the existence of positive radial solutions. Some authors used the bifurcation theory, the approach from the ordinary differential equation or the sub-super solution method. We employ the variational method. To this end, we define 1 H0,r (Ω) := {u ∈ H01 (Ω) : u(x) is radial}, 1 H + := {u ∈ H0,r (Ω) : u(x) ≥ 0

in Ω},

(1.8)

where Ω is the annulus A or the ball B. We define F (s) by (1.5) and put b  I(u) = I(u, λ) :=

 1  2 u (r) − λF (u(r)) rN −1 dr, 2

(1.9)

a

where a = 0 and b = 1 when Ω = B and 0 < a < b when Ω = A. For a solution u of (1.2), we define the linearized operator L(u) at u by  L(u)v :=

 d N −1 d  − 2− − λf (u) v, dr r dr

with the boundary condition v  (0) = v(1) = 0 when Ω = B, v(a) = v(b) = 0 when Ω = A. We define

(1.10)

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

4

i(λ) := inf + I(u, λ) ≥ −∞. u∈H

(1.11)

To state the existence of positive radial solutions, we introduce the following assumption. Assumption 1.3. Let f (s) be a continuous function on [0, ∞) which satisfies f (0) < 0 (semipositone) and the following conditions: (i) (ii) (iii) (iv)

F (S) > 0 with some S > 0, where F (s) is defined by (1.5), if F (s) > 0, then f (s) > 0, lims→∞ F (s)/s2 = 0, if Ω = B, there exist constants p, C > 0 such that |f (s)| ≤ C(sp + 1) for s ≥ 0, where 1 < p < ∞ when N = 2 and 1 < p < (N + 2)/(N − 2) when N ≥ 3.

The condition (iv) means that f (s) has a subcritical growth order as s → ∞. When Ω = A, we do not need such a condition (see (2.1) later on). From Assumption 1.3, f (s) has a zero in (0, ∞). Denote the first (smallest) zero of f (s) by z. Then f (s) < 0

in [0, z),

f (z) = 0.

(1.12)

By Assumption 1.3 (i), we can define Z := sup{s > 0 : F (t) ≤ 0 in [0, s]}.

(1.13)

By Assumption 1.3 with (1.12), 0 < z < Z < S < ∞. We shall show that the combination of (i) and (ii) is equivalent to the following conditions: F (s) ≤ 0 in [0, Z],

F (Z) = 0,

(1.14)

F (s) > 0,

in (Z, ∞).

(1.15)

f (s) > 0

Suppose that (i) and (ii) hold. Define Z by (1.13). Then (1.14) obviously holds. By the definition of Z, there exists a sequence εn > 0 such that F (Z + εn ) > 0 and εn → 0. Fix n arbitrarily. Since F (Z + εn ) > 0, f (Z + εn ) is also positive by (ii) in Assumption 1.3. If f (s) has a zero larger than Z + εn , we choose the first zero r0 ∈ (Z + εn , ∞). Then F (r0 ) > 0 and so f (r0 ) > 0 by (ii) in Assumption 1.3. A contradiction occurs. Hence f (s) > 0 for s > Z + εn . Since n is arbitrary, f (s) is positive for all s > Z and F (s) as well. Thus (i) and (ii) imply (1.14) and (1.15). The reverse assertion can be easily proved. Hence (i) and (ii) are equivalent to (1.14) and (1.15). Another main result is as follows. Theorem 1.4. Suppose that Assumption 1.3 holds. Then for any λ > 0 fixed, I(u, λ) is bounded from below on u ∈ H + , i.e., i(λ) > −∞, where i(λ) is defined by (1.11). There exists a λ0 > 0 such that i(λ) = 0 when λ ∈ (0, λ0 ], i(λ) < 0 when λ > λ0 and limλ→∞ i(λ) = −∞. For λ ≥ λ0 , (1.2) has a positive solution uλ which satisfies I(uλ , λ) = inf+ I(v, λ)(= i(λ)). v∈H

If f belongs to C 1 [0, ∞), all the eigenvalues of the linearized operator L(uλ ) are nonnegative.

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

5

Costa, Tehrani and Yang [9] proved the similar result as in Theorem 1.4 for a ball. They used the Schwarz Symmetrization to prove that the nonnegative solution is strictly positive. Therefore their method does not seem to be applicable to an annulus case. Our result holds true for both a ball and an annulus. We give sufficient conditions for Assumption 1.3 (ii), (iii) and (iv). (ii) f (s) has a unique zero z in (0, ∞) and f (s) < 0 in [0, z), f (z) = 0 and f (s) > 0 in (z, ∞). (iii) lims→∞ f (s)/s = 0. In fact, (i) and (ii) imply that F has a unique zero Z such that z < Z and F (s) < 0 in (0, Z) and F (s) > 0 in (Z, ∞). This leads to (ii). The condition (iii) implies (iii) by L’Hospital’s rule and ensures (iv) clearly. Then the next corollary follows. Corollary 1.5. Let f (s) be a continuous function on [0, ∞) which satisfies f (0) < 0, (i), (ii) and (iii) . Then the same conclusion as in Theorem 1.4 is valid. We shall show that (i) and (iii) in Assumption 1.3 are necessary for Theorem 1.4. It is not difficult to verify the next result, however we shall prove it for the reader’s convenience. Proposition 1.6. Suppose that F (s) ≤ 0 for all s. Then (1.2) has no positive solutions. The proposition above shows that the condition (i) in Assumption 1.3 is necessary for the existence of positive solutions. The next result says that (iii) is necessary for the functional I to be bounded from below. Proposition 1.7. Suppose that F (s) > 0 for all large s. Assume that I(u, λ) is bounded from below with respect to u for fixed λ ∈ (λ1 , ∞) with some λ1 > 0, i.e., inf I(u, λ) > −∞

u∈H +

for λ ∈ (λ1 , ∞).

Then F (s)/s2 converges to 0 as s → ∞. To prove the main results, we organize this paper into four sections. In Section 2, we show that I(u, λ) is bounded from below and prove the existence of a minimizer of I in H + . In Section 3, we investigate the properties of positive solutions and prove that a minimizer is strictly positive in Ω. In Section 4, we prove the main theorems. 2. Minimizer In this section, we shall show that I is bounded from below and it attains its minimum at a certain point u ∈ H + . Throughout the paper, we suppose Assumption 1.3 unless otherwise stated. We define the norm u p with 1 ≤ p ≤ ∞ by ⎛ b ⎞1/p  u p := ⎝ |u(r)|p rN −1 dr⎠ ,

u ∞ := ess sup|u(r)|.

a

Therefore ⎛ b ⎞1/2  u 2 = ⎝ |u (r)|2 rN −1 dr⎠ a

6

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

1 becomes the norm of H0,r (Ω) because of the Poincaré inequality. We extend f (s) to the whole R by putting f (s) := f (0) for s < 0. Let F (s) be given by (1.5). By (iii), |F (s)| ≤ C(s2 + 1) for s ∈ R with some C > 0. 1 Then I(u, λ) given by (1.9) is well defined in the whole H0,r (Ω). It is a C 1 functional by the subcritical condition (iv) in Assumption 1.3 when Ω = B. Let Ω = A. Then the two norms

⎛ b ⎞1/2  ⎝ u (r)2 rN −1 dr⎠ , a

⎛ b ⎞1/2  ⎝ u (r)2 dr⎠ ,

(2.1)

a

1 are equivalent to each other because a > 0. Thus H0,r (A) = H01 (a, b). This space is embedded in C 1/2 [a, b], which is a set of Hölder continuous functions with exponent 1/2. Therefore I(u) is a C 1 functional and its derivative is computed as



b

I (u)v =

(u (r)v  (r) − λf (u)v)) rN −1 dr,

a 1 for u, v ∈ H0,r (Ω). In the next lemma, we show that I is bounded from below. 1 Lemma 2.1. For any λ > 0, I(u, λ) is bounded from below in H0,r (Ω).

Proof. Since we defined f (s) = f (0) for s < 0, F (s) = f (0)s for s < 0. Therefore (iii) remains valid as s → −∞, that is, lim F (s)/s2 = 0.

(2.2)

s→±∞

Fix λ > 0 arbitrarily. By (2.2), for any ε > 0, there exists an M = M (ε) > 0 such that |F (s)| ≤ εs2 + M

for s ∈ R.

(2.3)

We here choose ε > 0 so small that λε/μ1 ≤ 1/4, where μ1 is the first eigenvalue of −Δ in Ω with the 1 Dirichlet boundary condition. Hence it holds that μ1 u 22 ≤ u 22 for u ∈ H0,r (Ω). This inequality with (2.3) gives an estimate of I(u), 1  2 u 2 − λ(ε u 22 + M bN −1 (b − a)) 2   1 λε − u 22 − λM bN −1 (b − a) ≥ 2 μ1 1 ≥ u 22 − λM bN −1 (b − a) ≥ −λM bN −1 (b − a), 4

I(u) ≥

1 because λε/μ1 ≤ 1/4. Thus I(·) is bounded from below in H0,r (Ω). The proof is complete.

(2.4) 

By Lemma 2.1, i(λ) given by (1.11) is well defined and i(λ) > −∞. We call u a minimizer of I in H + if u ∈ H + and I(u, λ) = inf+ I(v, λ)(= i(λ)). v∈H

We shall look for a minimizer u of I and then we shall show that it becomes a positive solution of (1.2). The infimum of I is nonpositive because I(0, λ) = 0. Therefore, if the infimum is 0, the minimizer may be

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

7

equal to the constant zero u ≡ 0. Clearly, this is not a solution of (1.2). Therefore we are interested in the case where the infimum is negative. This case occurs for λ large, which will be proved in the next lemma. Lemma 2.2. Suppose that Assumption 1.3 holds. Then there exists a Λ > 0 such that i(λ) < 0 for λ > Λ. Moreover, limλ→∞ i(λ) = −∞. Proof. We shall show that there exists a function U (x) satisfying U ∈ H + , U (r) ≡ 0 and b

F (U (r))rN −1 dr > 0.

(2.5)

a

By Assumption 1.3, there exists a constant S > 0 such that F (S) > 0. Let 1 < q < ∞. Then there exists 1 a sequence un ∈ C 1 [a, b] ∩ H0,r (Ω) such that 0 ≤ un (r) ≤ S and un converges to S in Lq (a, b) and almost everywhere. The Lebesgue dominated convergence theorem shows that b F (un )r

N −1

b dr →

a

F (S)rN −1 dr > 0

a

b as n → ∞. Therefore a F (un )rN −1 dr > 0 for large n. Put U (r) := un (r), which satisfies (2.5). We choose Λ > 0 so large that b 

 1  2 U (r) − ΛF (U (r)) rN −1 dr < 0. 2

a

Then it follows readily that for λ > Λ, b  I(U, λ) =

 1 2 |U | − λF (U ) rN −1 dr 2

a

b  <

 1 2 |U | − ΛF (U ) rN −1 dr < 0. 2

a

Therefore i(λ) ≤ I(U, λ) < 0 for λ > Λ. In the inequality above, as λ → ∞, I(U, λ) diverges to −∞, and so i(λ) → −∞. The proof is complete.  We shall show that if i(λ0 ) < 0 at some point λ0 , then i(λ) is negative in [λ0 , ∞). Lemma 2.3. If i(λ) < 0, then i(μ) < 0 for μ > λ. Proof. Let i(λ) < 0. Then there exists a point u ∈ H + such that I(u, λ) < 0. Therefore 1 0 < u 22 < λ 2

b a

which shows that

F (u)rN −1 dr,

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

8

b

F (u)rN −1 dr > 0.

a

For μ ∈ (λ, ∞), we have 1 I(u, μ) = u 22 − μ 2

b

F (u)rN −1 dr

a

<

1  2 u 2 − λ 2

b

F (u)rN −1 dr = I(u, λ) < 0.

a

Consequently, i(μ) ≤ I(u, μ) < 0. The proof is complete.



We shall show that i(λ) = 0 for λ > 0 small under the assumption (1.6), which is weaker than (iii). Lemma 2.4. Suppose that the same assumption as in Theorem 1.1 holds. Then there exists a λ1 > 0 such that I(u, λ) ≥ 0 for u ∈ H + and λ ∈ (0, λ1 ). In particular, i(λ) = 0 in (0, λ1 ). Proof. Since f (0) < 0, F (s) ≤ −cs for small s > 0 with a constant c > 0. This inequality with (1.6) shows the existence of M > 0 such that F (s) ≤ M s2 for s ≥ 0. Then we have I(u, λ) ≥

1  2 u 2 − λM u 22 ≥ (1/2 − λM/μ1 ) u 22 ≥ 0, 2

provided that λ ∈ (0, μ1 /2M ), where μ1 is the first eigenvalue of −Δ in Ω with the Dirichlet boundary condition. Thus I(u, λ) ≥ 0 for u ∈ H + . Since I(0, λ) = 0, i(λ) = 0 hods for λ ∈ (0, μ1 /2M ). The proof is complete.  We define λ0 := inf{λ > 0 : i(λ) < 0}.

(2.6)

By Lemmas 2.2–2.4, λ0 is positive. Lemma 2.5. i(λ) = 0 for λ ≤ λ0 and i(λ) < 0 for λ > λ0 . Proof. Recall that i(λ) ≤ 0 for any λ because I(0, λ) = 0. Hence either i(λ) = 0 or i(λ) < 0 holds. By Lemma 2.3, i(λ) = 0 for λ < λ0 and i(λ) < 0 for λ > λ0 . If i(λ0 ) < 0, it can be proved easily that i(λ) < 0 for λ close to λ0 . Thus i(λ0 ) = 0. The proof is complete.  We show the existence of a minimizer of I in H + . Theorem 2.6. Let λ > λ0 . Then I(·, λ) has a minimizer in H + . Proof. Let λ > λ0 . Then i(λ) < 0 by Lemma 2.5. I is bounded from below and its infimum is negative. We put I0 := i(λ) = inf + I(u) < 0. u∈H

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

9

Choose a sequence un ∈ H + such that I(un ) converges to I0 . We shall show that a subsequence of un 1 converges. Substitute u = un in (2.4). Since I(un ) converges, un is bounded in H0,r (Ω). Hence a subsequence 1 of un (denoted by un again) converges to a certain limit u0 weakly in H0,r (Ω). We shall show that the convergence is strong. By the compact embedding, un converges to u0 strongly in L2 (Ω). By (2.3), there exists a constant C > 0 such that |F (s)| ≤ C(s2 + 1) for s ∈ R. By this inequality, the integral of F (un ) converges to that of F (u0 ) (see [15, p. 89, Proposition B.1]), i.e., b

F (un )rN −1 dr →

a

b

F (u0 )rN −1 dr

as n → ∞.

(2.7)

a

Using the definition of I(u), I(un ) → I0 and (2.7), we have 1  2 u = I(un ) + λ 2 n 2

b

F (un )rN −1 dr

a

b → I0 + λ

F (u0 )rN −1 dr

a

b ≤ I(u0 ) + λ

F (u0 )rN −1 dr =

1  2 u , 2 0 2

a

where we have used the fact that I0 is the infimum of I(u). This shows that lim sup un 2 ≤ u0 2 . n→∞

Since un weakly converges to u0 , it holds that u0 2 ≤ lim inf un 2 . n→∞

Combining the two inequalities above, we find that un 2 → u0 2 . Therefore un strongly converges to u0 . Hence we have I(u0 ) = I0 = inf + I(u) < 0. u∈H

Therefore u0 is a minimizer. The proof is complete.



We do not know whether a minimizer of I(·, λ) in H + is a critical point of I(·, λ). Indeed, for a minimizer u, we put g(t) := I(u + tv, λ) with any test function v ∈ H + . Since u is a minimizer, g(0) ≤ g(t) holds for t ≥ 0. Then it follows that g  (0) ≥ 0. If g(0) ≤ g(t) for t ∈ (−ε, ε),

(2.8)

with a small ε > 0, then g  (0) = 0 follows and u becomes a critical point of I. To prove (2.8), we shall show the strict positivity of u in the next section.

10

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

3. Properties of minimizers For a minimizer u given by Theorem 2.6, we shall show that u(r) > 0 in (a, b). To this end, in this section, we study the properties of positive radial solutions and minimizers of the functional I. For a positive solution u(r) of (1.2), we define the energy function E(r) by E(r) :=

1  2 u (r) + λF (u(r)). 2

(3.1)

N −1  2 u (r) ≤ 0. r

(3.2)

Then we have E  (r) = −

Although the next lemma is easily proved by using E(r), we shall show it for the completeness. Lemma 3.1. Let u(r) be a positive solution of (1.2). Then maxa≤x≤b u(x) > Z, E(r) > 0 in [a, b) and the following results hold. (i) Let Ω = B. Then E  (r) < 0 in (0, 1), u (r) < 0 in (0, 1) and hence the maximum of u(r) is attained at the origin r = 0 only. (ii) Let Ω = A. Then u(r) has a unique maximum point r0 and E  (r) < 0 in (a, b) \ {r0 }, u (r) > 0 in (a, r0 ), u (r0 ) = 0 and u (r) < 0 in (r0 , b). Proof. Let Ω = B. Since E(r) is nonincreasing by (3.2) and u ≡ 0 in (0, 1), we have E(0) > E(1). Since u (0) = u(1) = 0, we find that E(0) = λF (u(0)) > E(1) =

1  2 u (1) ≥ 0. 2

(3.3)

By (3.3), (1.14) and (1.15), u(0) > Z. Then f (u(0)) > 0 by (1.15). Since u (r)/r → u (0) as r → +0 by L’Hospital’s rule, we let r → +0 in (1.2) to obtain N u (0) + λf (u(0)) = 0. Since f (u(0)) > 0, u (0) is negative. This assertion with u (0) = 0 implies that u (r) < 0 for r > 0 small. We shall show that u (r) < 0 in (0, 1). Suppose to the contrary that u (r0 ) = 0 at some r0 ∈ (0, 1). We denote the smallest zero of u (r) by r0 again. Hence u (r) < 0 in (0, r0 ) and u (r0 ) = 0. Then u (r0 ) ≥ 0. Since E(r0 ) = λF (u(r0 )) > E(1) ≥ 0, we have u(r0 ) > Z and f (u(r0 )) > 0. Thus u (r0 ) ≥ 0, u (r0 ) = 0 and f (u(r0 )) > 0. This contradicts (1.2). Consequently, u (r) < 0 in (0, 1). This with (3.2) implies that E  (r) < 0 in (0, 1). Therefore E(r) is strictly decreasing and E(r) > E(1) ≥ 0 for r ∈ [0, 1). We consider the case where Ω = A. The proof is the same as in the case where Ω = B except for the uniqueness of a maximum point. We shall show it. Since E(a) = (1/2)u (a)2 > E(b) ≥ 0, we have u (a) > 0. Since u(a) = u(b) = 0, there exists a critical point of u(r). We choose the smallest critical point r0 , which is greater than a because u (a) > 0. Then u (r) > 0 in [a, r0 ) and u (r0 ) = 0. Since E(r0 ) = λF (u(r0 )) > E(b) ≥ 0, it holds that u(r0 ) > Z. Hence f (u(r0 )) > 0. By this inequality with u (r0 ) = 0, (1.2) implies that u (r0 ) < 0. Accordingly, u (r) < 0 for r slightly larger than r0 . In the same way as in the case Ω = B, we can prove that u (r) < 0 in (r0 , b). The proof is complete. 

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

11

We shall show the strict positivity of a minimizer. Before going to the proof, we shall mention the 1 continuity of u ∈ H0,r (Ω). As stated in (2.1), if Ω = A, then u ∈ C 1/2 [a, b]. In the same discussion as above, when Ω = B, u belongs to C 1/2 [ε, 1] with any ε ∈ (0, 1). In particular, u is continuous in (0, 1]. However, it 1 is not necessarily continuous at r = 0. Indeed, u(r) := r−α − 1 with α > 0 belongs to H0,r (B) when N is large enough. Therefore we have the next lemma. 1 Lemma 3.2. Let u ∈ H0,r (Ω). If Ω = A, then u belongs to C 1/2 [a, b]. If Ω = B, then u belongs to C 1/2 [ε, 1] with any ε ∈ (0, 1).

For a ≤ s < t ≤ b, we define t  I(u; s, t) =

 1  2 u (r) − λF (u(r)) rN −1 dr. 2

(3.4)

s

For any partition a = r0 < r1 < r2 < · · · < rn = b, it holds clearly that I(u, λ) =

n

I(u; ri−1 , ri ).

(3.5)

i=1

When λ > λ0 , i(λ) is negative by Lemma 2.5. Then I(·, λ) has a minimizer in H + by Theorem 2.6. The existence of a nontrivial minimizer for λ = λ0 , i.e., u ≡ 0 satisfying I(u, λ0 ) = i(λ0 ) = 0, will be proved later on. We investigate the properties of the minimizer. Lemma 3.3. Let λ0 ≤ λ < ∞ and let u be a minimizer of I(·, λ) in H + satisfying u ≡ 0. Then the following assertions hold. (i) u(r) > Z somewhere in (a, b). (ii) Suppose that there exist a ≤ r1 < r2 ≤ b such that u(r1 ) = u(r2 ) = 0 and u(r) > 0 in (r1 , r2 ). Then u(r3 ) > Z at some r3 ∈ (r1 , r2 ). In the lemma above, we suppose that λ0 ≤ λ < ∞ and u is a minimizer of I in H + satisfying u ≡ 0. When λ0 < λ, I(u, λ) = i(λ) < 0 and hence u ≡ 0 clearly holds. However, if λ = λ0 , then i(λ0 ) = 0 and a minimizer u may be equal to u ≡ 0. The assumption of the lemma rules out such a minimizer. Proof of Lemma 3.3. We first show that u(r) > Z somewhere in (a, b). If u(r) ≤ Z for all r ∈ (a, b), then F (u(r)) ≤ 0. Since u(r) ≡ 0 by assumption, u (r) also does not identically vanish. Hence I(u, λ) is positive. This contradicts the fact that I(u, λ) = i(λ) ≤ 0. Therefore u(r) > Z somewhere. We shall show the assertion (ii) of the lemma. Assume that u(r1 ) = u(r2 ) = 0 and u(r) > 0 in (r1 , r2 ). Suppose to the contrary that u(r) ≤ Z in (r1 , r2 ). Then F (u(r)) ≤ 0 in (r1 , r2 ). Since u (r) ≡ 0, we have r2  I(u; r1 , r2 ) =

 1  2 u (r) − λF (u(r)) rN −1 dr > 0, 2

r1

where I(u; r1 , r2 ) has been defined by (3.4). We define v(r) := u(r) in [a, b] \ (r1 , r2 ) and v(r) := 0 in (r1 , r2 ). Then v ∈ H + and I(v; r1 , r2 ) = 0,

I(v; a, r1 ) = I(u; a, r1 ),

I(v; r2 , b) = I(u; r2 , b),

12

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

which with (3.5) shows that I(v, λ) < I(u, λ). This contradicts the fact that u is a minimizer. Therefore u(r) > Z somewhere in (r1 , r2 ). The proof is complete.  In the next lemma, we show that a minimizer does not identically vanish near the boundary ∂Ω. Lemma 3.4. Let λ0 ≤ λ < ∞ and let u be a minimizer of I(·, λ) in H + satisfying u ≡ 0. Then there exists a sequence tn ∈ (a, b) such that u(tn ) > 0 and tn → b as n → ∞. If Ω = A, then there exists a sequence sn ∈ (a, b) such that u(sn ) > 0 and sn → a as n → ∞. Proof. Let Ω = B, a = 0 and b = 1. We shall show the existence of tn . Suppose to the contrary that such a sequence does not exist. Then u(r) ≡ 0 on [θ, 1] with a certain θ ∈ (0, 1). Define v(r) := u(θr) for r ∈ [0, 1]. Then v ∈ H + . Changing the variables s = θr, we compute 1  I(v, λ) =

 1  2 v (r) − λF (v) rN −1 dr 2

0

=θ

−N

θ 

 θ2  2 u (s) − λF (u) sN −1 ds 2

0

=θ

−N

1 

 θ2  2 u (s) − λF (u) sN −1 ds, 2

0

where we have used the fact that u ≡ 0 on [θ, 1] in the last identity. Put g(t) := P t−N +2 − Qt−N , 1 P := 2

1



2 N −1

u (r) r

1 dr,

Q := λ

0

F (u(r))rN −1 dr.

0

Observe that I(v) = g(θ) and I(u) = g(1) = P − Q. Since I(u) = i(λ) ≤ 0, we have P ≤ Q. Since u ≡ 0, P is positive. When N ≥ 3, we compute for t ∈ (0, 1], g  (t) = (−N + 2)P t−N +1 + N Qt−N −1 = N t−N −1 (Q − ((N − 2)/N )P t2 ) > N t−N −1 (Q − P ) ≥ 0. Even if N = 2, g  (t) is positive. Thus g(t) is increasing in (0, 1] and hence we have I(v) = g(θ) < g(1) = I(u), which contradicts u to be a minimizer. Consequently, u(tn ) > 0 with a sequence tn converging to 1. Let Ω = A. Let us show the existence of a sequence tn . Suppose to the contrary that u(r) ≡ 0 in [θb, b] with a certain θ ∈ (0, 1). Extend u(r) by putting u(r) ≡ 0 in R \ [a, b]. We define v(r) := u(θr) for r ∈ [a, b]. Then v(a) = v(b) =0 and v ∈ H + . The rest of the proof is the same as in the case Ω = B. Thus we obtain a sequence tn converging to b such that u(tn ) > 0. We shall show the existence of a sequence sn converging to a satisfying u(sn ) > 0 when Ω = A. Suppose to the contrary that u(r) ≡ 0 in [a, r0 ] with some r0 > a. By Lemma 3.3 (i), we can choose a point r1 such that r0 < r1 < b, 0 ≤ u(r) ≤ Z in [r0 , r1 ] and u(r1 ) = Z. We put a1 := a + r1 − r0 and v(r) := u(r − a + r0 )

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

13

in [a, a1 ], v(r) := Z in [a1 , r1 ] and v(r) := u(r) in [r1 , b]. Then v ∈ H + . Using the change of variables t = r − a + r0 , we have a1  I(v; a, a1 ) =

 1  2 v (r) − λF (v) rN −1 dr 2

a

r1  =

 1  2 u (t) − λF (u) (t + a − r0 )N −1 dt. 2

r0

Since F (u(t)) ≤ 0 and F (u(t)) ≡ 0 in [r0 , r1 ], the integrand is nonnegative. Hence the right hand side is strictly less than the integral, r1 

 1  2 u (t) − λF (u) tN −1 dt = I(u; r0 , r1 ). 2

r0

Accordingly, I(v; a, a1 ) < I(u; r0 , r1 ). Moreover, we have I(v; a1 , r1 ) = 0,

I(u; a, r0 ) = 0,

I(v; r1 , b) = I(u; r1 , b).

Therefore I(v; a, b) < I(u; a, b). A contradiction occurs. Hence there exists a sequence sn → a such that u(sn ) > 0. The proof is complete.  We shall show the strict positivity of a minimizer in the lemma below. Lemma 3.5. Let λ0 ≤ λ < ∞ and let u be a minimizer of I(·, λ) in H + satisfying u ≡ 0. Then u(r) > 0 in (0, 1) when Ω = B and u(r) > 0 in (a, b) when Ω = A. At the moment, we have not proved yet that a minimizer becomes a solution of (1.2). Therefore, we can not use the energy E(r) in a proof of the lemma. Proof. Let Ω = B, a = 0 and b = 1. We shall show that u(r) > 0 in (0, 1). Suppose to the contrary that u(r0 ) = 0 at some r0 ∈ (0, 1). By Lemma 3.4 and u(1) = 0, we have two points r1 , r2 such that 0 < r1 < r2 ≤ 1, u(r1 ) = u(r2 ) = 0 and u(r) > 0 in (r1 , r2 ). By Lemma 3.3 (ii), u(r) > Z at a certain r ∈ (r1 , r2 ). We divide the proof into two cases: Case 1. u(r) > Z at some points r ∈ (0, r1 ). Case 2. u(r) ≤ Z in (0, r1 ). Suppose that Case 1 occurs. Then there exist points t1 , t2 such that 0 < t1 < r1 < t2 , u(t1 ) = u(t2 ) = Z and u(r) < Z in (t1 , t2 ). We define w(r) := u(r) in (0, 1) \ [t1 , t2 ] and w(r) := Z in [t1 , t2 ]. Then w ∈ H + . Since F (Z) = 0, we have I(w; t1 , t2 ) = 0. However I(u; t1 , t2 ) > 0. Thus I(w) < I(u), which contradicts u to be a minimizer. Consequently, Case 1 leads to a contradiction. We deal with Case 2: u(r) ≤ Z in (0, r1 ). Define t2 by t2 := min{t ∈ [r1 , r2 ] : u(t) = Z} Then u(r) ≤ Z in (0, t2 ]. We define w(r) := Z in (0, t2 ] and w(r) := u(r) in [t2 , 1]. The same computation as in Case 1 is valid and we have a contradiction. Thus u(r) > 0 in (0, 1). Let Ω = A. Suppose that u(r0 ) = 0 at some r0 ∈ (a, b). By Lemmas 3.3 and 3.4, there exist r1 , r2 such that a < r1 < r0 < r2 < b and u(ri ) > Z with i = 1, 2. Then we can choose points t1 , t2 such that

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

14

r1 < t1 < r0 < t2 < r2 , u(t1 ) = u(t2 ) = Z and u(r) < Z in (t1 , t2 ). Define w(r) := Z in [t1 , t2 ] and w(r) := u(r) in [a, b] \ [t1 , t2 ]. Then we find a contradiction. The proof is complete.  1 Let Ω = B. Then u(r) ∈ H0,r (B) is not necessarily continuous at r = 0. By Lemma 3.5, a minimizer u(r) is positive. However, there exists a possibility that the limit inferior of u(r) as r → +0 may be equal to 0. We shall show that this behavior does not occur.

Lemma 3.6. Suppose the same assumption as in Lemma 3.5. Let Ω = B. Then lim inf r→+0 u(r) ≥ Z. Proof. Suppose that the lemma is false, i.e., lim inf u(r) < Z. r→+0

(3.6)

We shall show that there exists a sequence {tn } such that u(tn ) > Z,

tn > 0,

tn → 0.

(3.7)

Suppose to the contrary that such a sequence does not exist. Then u(r) ≤ Z in (0, ε) with a small ε > 0. Since u(r) > Z somewhere by Lemma 3.3 (i), there exists a point s1 ∈ (0, 1) such that u(s1 ) = Z and u(r) ≤ Z in (0, s1 ). By (3.6), u(r) ≡ Z in (0, s1 ) and u (r) ≡ 0. Hence I(u; 0, s1 ) > 0. We define v(r) := Z in (0, s1 ] and v(r) := u(r) in (s1 , 1). Then I(v, λ) < I(u, λ). This contradicts u to be a minimizer. Therefore there exists a sequence tn satisfying (3.7). By (3.6) and (3.7), there exist two points s1 , s2 such that 0 < s1 < s2 < 1, u(s1 ) = u(s2 ) = Z and u(r) < Z in (s1 , s2 ). We define w(r) := Z in (s1 , s2 ) and w(r) := u(r) in (0, 1) \ (s1 , s2 ). Then I(w, λ) < I(u, λ). A contradiction occurs. Consequently, lim inf r→+0 u(r) ≥ Z. The proof is complete.  We define ∞ C0,r (Ω) := {u ∈ C0∞ (Ω) : u(x) is radial}. ∞ Lemma 3.7. Let λ0 ≤ λ < ∞ and let u be a minimizer of I(·, λ) in H + satisfying u ≡ 0. Let v ∈ C0,r (Ω) be any test function. Then there exists an ε > 0 such that u(r) + tv(r) > 0 in (a, b) for |t| < ε. ∞ Proof. Let Ω = B. Let v ∈ C0,r (B) be any test function. Since v has a compact support in B, there exists a δ > 0 such that v(r) = 0 in [1 − δ, 1]. By Lemma 3.2, u(r) is continuous in (0, 1]. By Lemmas 3.5 and 3.6, u(r) > c in (0, 1 − δ) with some c > 0. Putting M := v ∞ , we have

u(r) + tv(r) ≥ c − |t|M > 0 for r ∈ (0, 1 − δ), |t| < c/M. It holds clearly that u(r) + tv(r) = u(r) > 0 in (1 − δ, 1). In the case where Ω = A, the argument above is still valid. The proof is complete.  4. Proof of the theorems In this section, we prove the theorems stated in Section 1. Theorem 4.1. Let λ > λ0 and let uλ be a minimizer of I in H + which is obtained in Theorem 2.6. Then it is a positive solution of (1.2) which belongs to C 2 [a, b] and satisfies I(uλ , λ) = i(λ) < 0.

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

15

Proof. Let u = uλ be a minimizer of I in H + . It is strictly positive in Ω by Lemmas 3.5 and 3.6. Let ∞ v ∈ C0,r (Ω) be any test function. By Lemma 3.7, u + tv ∈ H + for |t| < ε with a small ε > 0. Put g(t) := I(u + tv, λ). Then g(t) attains its minimum at t = 0 in (−ε, ε). Hence g  (0) = 0, i.e.,





b

0 = g (0) = I (u)v =

(u (r)v  (r) − λf (u)v) rN −1 dr.

a

Thus u is a weak solution of (1.2). When Ω = B, we use the bootstrap argument with the elliptic regularity theorem and (iv) of Assumption 1.3. Then it can be proved that u belongs to W 2,q (Ω) with any 1 ≤ q < ∞ When Ω = A, u is in C 1/2 [a, b] by Lemma 3.2. This result with the elliptic regularity theorem shows that u ∈ W 2,q (Ω) with any 1 ≤ q < ∞. Moreover, since (1.2) is an ODE and f (u(r)) ∈ C[a, b], we have u ∈ C 2 [a, b]. The proof is complete.  We shall show the existence of a solution at λ = λ0 . Theorem 4.2. For λ = λ0 , (1.2) has a positive solution u0 which belongs to C 2 [a, b] and satisfies I(u0 , λ0 ) = i(λ0 ) = 0. Proof. Let λn be a sequence such that λn > λ0 and it converges to λ0 . Let un be a minimizer obtained by Theorem 4.1 with λ = λn . Since I(un , λn ) < 0, we have 1  2 u < λn 2 n 2

b

F (un )rN −1 dt.

a

This inequality with (2.3) implies that 1  2 u ≤ λn ε un 22 + λn M C ≤ (λn ε/μ1 ) un 22 + λn M C, 2 n 2 with a constant C > 0 independent of n. Choose ε > 0 satisfying λ0 ε/μ1 < 1/4. Since λn → λ0 , un 2 is bounded. Since un is a solution of (1.2), by the elliptic regularity theorem, it is bounded in W 2,q (Ω) with any 1 ≤ q < ∞. This space is compactly embedded in C 1,θ (Ω) when q is large enough. Along a subsequence, un converges to a certain limit u0 in C 1,θ (Ω). Since un is a solution of (1.2), it satisfies b

(un v  − λn f (un )v)rN −1 dr = 0,

a ∞ for any v ∈ C0,r (Ω). Letting n → ∞, we have

b

(u0 v  − λ0 f (u0 )v)rN −1 dr = 0.

a

Thus u0 is a nonnegative weak solution of (1.2). Since (1.2) is an ordinary differential equation, u0 belongs to C 2 [a, b]. Letting n → ∞ in the inequality, I(un , λn ) ≤ I(v, λn )

for v ∈ H + ,

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

16

we have I(u0 , λ0 ) ≤ I(v, λ0 )

for v ∈ H + .

Thus u0 is a minimizer of I(·, λ0 ), i.e., I(u0 , λ0 ) = i(λ0 ) = 0. Since u0 is a solution of (1.2), it does not identically vanish. Then it is strictly positive in Ω by Lemmas 3.5 and 3.6. The proof is complete.  We give a proof of Theorem 1.4. Proof of Theorem 1.4. In Lemma 2.5, we have already proved that i(λ) = 0 for λ ≤ λ0 and i(λ) < 0 for λ > λ0 . By Lemma 2.2, limλ→∞ i(λ) = −∞. By Theorems 4.1 and 4.2, for λ ≥ λ0 , there exists a positive solution uλ of (1.2) such that I(uλ , λ) = i(λ). Let f belong to C 1 [0, ∞). We shall show the first eigenvalue of L(uλ ) is nonnegative. Fix λ ∈ [λ0 , ∞). Instead of uλ , we write u. The second derivative I  (u) is a bilinear form,

I  (u)vw =

b

(v  (r)w (r) − λf  (u)vw)rN −1 dr

1 for v, w ∈ H0,r (Ω).

a

Since u is a solution of (1.2), it belongs to C 2 [a, b] and therefore the integral above is finite. Let φ ∈ H + . Put g(t) := I(u + tφ) for t ≥ 0. Since g(t) attains its minimum at t = 0 and g  (0) = I  (u)φ = 0, we have g  (0) ≥ 0. This is written as 



b 2

g (0) = I (u)φ =

(φ (r)2 − λf  (u)φ2 )rN −1 dr ≥ 0,

a 1 for any φ ∈ H + . Let ψ ∈ H0,r (Ω) be any test function, which may change its sign. Put φ := |ψ|. Then φ ∈ H +    and φ 2 = ψ 2 . Hence I (u)ψ 2 = I  (u)φ2 ≥ 0. Since the first eigenfunction is radially symmetric, it is enough to compute I  (u)ψ 2 for a radial test function ψ. Then the first eigenvalue is nonnegative because of its variational characterization. The proof is complete. 

We prove Theorem 1.1, which is the nonexistence result for small λ > 0. Proof of Theorem 1.1. Since f (0) < 0 and (1.6), there exists an M > 0 such that F (s) ≤ M s2

for s ≥ 0.

(4.1)

Let Ω = A. Suppose to the contrary that (1.2) has a positive solution u with a small λ > 0. By Lemma 3.1, u has a unique critical point r0 ∈ (a, b) such that u (r) > 0 in (a, r0 ), u (r0 ) = 0, u (r) < 0 in (r0 , b) and u(r0 ) > Z. We use (3.2) to obtain −E  (r) =

N −1  2 N −1  2 u (r) ≥ u (r) . r b

Integrating both sides over [r0 , b], we have N −1 E(r0 ) − E(b) ≥ b

b r0

u (r)2 dr.

(4.2)

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

17

We use the Schwarz inequality to get b u(r0 ) = −

u (r)dr ≤



⎛ b ⎞1/2  b − r0 ⎝ u (r)2 dr⎠ ,

r0

r0

which is rewritten as b

u (r)2 dr ≥

r0

u(r0 )2 u(r0 )2 ≥ . b − r0 b−a

Combining this inequality with (4.2), we have E(r0 ) − E(b) ≥

N −1 u(r0 )2 , b(b − a)

which means that 1 N −1 λF (u(r0 )) − u (b)2 ≥ u(r0 )2 . 2 b(b − a) Using (4.1) in this inequality, we obtain  λM −

N −1 b(b − a)

 u(r0 )2 ≥

1  2 u (b) . 2

This is impossible if λ > 0 is small enough because u(r0 ) > 0. Therefore (1.2) has no positive solution for λ > 0 small. Let Ω = B. Use the same argument as in the case where Ω = A with r0 = 0 and b = 1 in (4.2). Then we have a contradiction. The proof is complete.  We prove Theorem 1.2, which is the nonexistence result for a general domain Ω with a small λ > 0. Proof of Theorem 1.2. Let Ω be a smooth bounded domain. Since f (0) < 0 and (1.7), there exists an M > 0 such that f (s) ≤ M s for s ≥ 0. Let u be a nonnegative solution of (1.1) with a small λ > 0. Let μ1 be the first eigenvalue of −Δ with the Dirichlet boundary condition and φ be the corresponding eigenfunction satisfying φ(x) > 0. We have −Δu = λf (u) ≤ λM u. Taking the L2 inner product of the both sides with φ, we obtain μ1 (u, φ)2 ≤ λM (u, φ)2 , where (u, φ)2 denotes the L2 (Ω) inner product. We here note that u ≥ 0 and u ≡ 0 because the constant zero must not be a solution of (1.1). If λM < μ1 , we have a contradiction. Therefore, when λ ∈ (0, μ1 /M ), (1.1) has no nonnegative solutions. The proof is complete.  We give a proof of Proposition 1.6.

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

18

Proof of Proposition 1.6. Suppose that F (s) ≤ 0 for all s > 0 and (1.2) has a positive solution u(r). Let E(r) be the energy defined by (3.1). Let Ω = B. By (3.3), we have a contradiction. Even if Ω = A, this method is valid.  We conclude this paper by proving Proposition 1.7. Proof of Proposition 1.7. It is enough to show that if lim sups→∞ F (s)/s2 > 0, then I(u, λ) is unbounded from below. Suppose that the limit superior is positive. Then there exist δ > 0 and a sequence sn such that F (sn )/s2n ≥ δ and sn → ∞. We first deal with the case where Ω = B. Define un (r) := sn for 0 ≤ r ≤ 1/2, un (r) := 2sn (1 − r) for r ∈ [1/2, 1]. Then un ∈ H + . We compute un 22

1 =

(−2sn )2 rN −1 dr ≤ 2s2n .

(4.3)

1/2

Since F (s) > 0 for large s > 0 by assumption, F (s) is bounded from below for s ≥ 0. Put −C := mins≥0 F (s). Then we have 1

F (un )rN −1 dr ≥ −C0 ,

1/2

with a constant C0 independent of n. Since F (sn ) ≥ δs2n , we obtain 1 F (un )r

N −1

1/2 dr ≥ F (sn )rN −1 dr − C0 ≥ c0 s2n − C0 ,

0

0

where c0 > 0 is a constant independent of n. From the inequality above with (4.3), we have

I(un , λ) =

(1/2) un 22

1 −λ

F (un )rN −1 dr ≤ s2n − λc0 s2n + λC0 .

0

We fix λ > 0 so large that 1 < λc0 and let n → ∞. Then I(un , λ) → −∞. Therefore I(·, λ) is unbounded from below. Let Ω = A. We put c := (2a + b)/3 and d := (a + 2b)/3 and define un (r) := 3sn (r − a)/(b − a) un (r) := sn un (r) := 3sn (b − r)/(b − a)

when a ≤ r ≤ c, when c ≤ r ≤ d, when d ≤ r ≤ b.

The rest of the proof is the same as in the case where Ω = B. The proof is complete.



References [1] A. Ambrosetti, D. Arcoya, B. Buffoni, Positive solutions for some semi-positone problems via bifurcation theory, Differential Integral Equations 7 (3–4) (1994) 655–663. [2] D. Arcoya, A. Zertiti, Existence and non-existence of radially symmetric non-negative solutions for a class of semi-positone problems in an annulus, Rend. Mat. Appl. (7) 14 (4) (1994) 625–646. [3] A. Castro, R. Shivaji, Nonnegative solutions for a class of nonpositone problems, Proc. Roy. Soc. Edinburgh Sect. A 108 (3–4) (1988) 291–302.

R. Kajikiya, E. Ko / J. Math. Anal. Appl. 484 (2020) 123735

19

[4] A. Castro, R. Shivaji, Positive solutions for a concave semipositone Dirichlet problem, Nonlinear Anal. 31 (1–2) (1998) 91–98. [5] A. Castro, G. Sudhasree, Uniqueness of stable and unstable positive solutions for semipositone problems, Nonlinear Anal. 22 (4) (1994) 425–429. [6] A. Castro, J.B. Garner, R. Shivaji, Existence results for classes of sublinear semipositone problems, Results Math. 23 (3–4) (1993) 214–220. [7] M. Chhetri, D.D. Hai, R. Shivaji, On positive solutions for classes of p-Laplacian semipositone systems, Discrete Contin. Dyn. Syst. 9 (4) (2003) 1063–1071. [8] P. Clément, G. Sweers, Existence and multiplicity results for a semilinear elliptic eigenvalue problem, Ann. Sc. Norm. Super. Pisa Cl. Sci. (4) 14 (1) (1987) 97–121. [9] D. Costa, H. Tehrani, J. Yang, On a variational approach to existence and multiplicity results for semipositone problems, Electron. J. Differential Equations (11) (2006), 10 pp. [10] D.D. Hai, R. Shivaji, Positive solutions for semi-positone systems in an annulus, Rocky Mountain J. Math. 29 (4) (1999) 1285–1299. [11] A.C. Lazer, P.J. McKenna, Existence, uniqueness and stability of oscillations in differential equations with asymmetric nonlinearities, Trans. Amer. Math. Soc. 315 (1989) 721–739. [12] A.C. Lazer, P.J. McKenna, Large-amplitude periodic oscillations in suspension bridges: some new connections with nonlinear analysis, SIAM Rev. 32 (4) (1990) 537–578. [13] P.L. Lions, On the existence of positive solutions of semilinear elliptic equations, SIAM Rev. 24 (4) (1982) 441–467. [14] S. Oruganti, J. Shi, R. Shivaji, Diffusive logistic equation with constant effort harvesting, I: steady states, Trans. Amer. Math. Soc. 354 (2002) 3601–3619. [15] P.H. Rabinowitz, Minimax Methods in Critical Point Theory with Applications to Differential Equations, CBMS Regional Conf. Ser. in Math., vol. 65, Amer. Math. Soc., Providence, 1986. [16] J.A. Smoller, A.G. Wasserman, Symmetry-breaking for solutions of semilinear elliptic equations with general boundary conditions, Comm. Math. Phys. 105 (3) (1986) 415–441. [17] J.A. Smoller, A.G. Wasserman, Existence of positive solutions for semilinear elliptic equations in general domains, Arch. Ration. Mech. Anal. 98 (3) (1987) 229–249.