J. Math. Anal. Appl. 424 (2015) 598–612

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Journal of Mathematical Analysis and Applications

www.elsevier.com/locate/jmaa

A three solution theorem for singular nonlinear elliptic boundary

value problems

R. Dhanya a, Eunkyung Ko b, R. Shivaji c,∗

a Department of Mathematics, IISc, Bangalore 560012, Indiab PDE and Functional Analysis Research Center, Department of Mathematical Sciences, Seoul National University, Seoul 151-747, South Koreac Department of Mathematics and Statistics, University of North Carolina at Greensboro, Greensboro, NC 27412, USA

a r t i c l e i n f o a b s t r a c t

Article history:Received 12 June 2014Available online 11 November 2014Submitted by J. Shi

Keywords:Three solutionsSingular nonlinear elliptic BVPs

We establish a three solution theorem for singular elliptic boundary value problems of the form −Δu = f(u)

uβ in Ω, u = 0 on ∂Ω where Ω is a bounded domain in RN , N ≥ 1 with a smooth boundary ∂Ω. Here f : [0, ∞) → [0, ∞) is a C1 function in [0, ∞) with f(0) > 0 and β ∈ (0, 1). In particular if there exist two pairs of sub-supersolutions (ψ1, φ1) and (ψ2, φ2) where ψ1 ≤ ψ2 ≤ φ1, ψ1 ≤ φ2 ≤ φ1 with ψ2 � φ2, and ψ2, φ2 are strict sub and supersolutions, then we establish existence of three solutions u1, u2 and u3 for the above boundary value problem such that u1 ∈ [ψ1, φ2], u2 ∈ [ψ2, φ1] and u3 ∈ [ψ1, φ1] \([ψ1, φ2] ∪ [ψ2, φ1]). Our results extend the work in [1] (and also in [17]) where such multiplicity results were discussed in the non-singular case β = 0. Further our results strengthen the multiplicity results in [12] and [13] for such singular problems (β = 0), namely, establish an additional third solution. We also establish a sub-super solution theorem for infinite semipositone problems (i.e. the case when f(0) < 0).

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

Consider the following singular elliptic boundary value problem:

(P )

⎧⎪⎪⎨⎪⎪⎩

−Δu = f(u)uβ

in Ω

u > 0 in Ω

u = 0 on ∂Ω.

We study the existence of three solutions in C2(Ω) ∩ C10 (Ω) for the above BVP where Ω is a bounded

domain in RN , N ≥ 1 with a smooth boundary ∂Ω and 0 < β < 1.

* Corresponding author.E-mail addresses: dhanya.tr@gmail.com (R. Dhanya), ekko1115@snu.ac.kr (E. Ko), shivaji@uncg.edu (R. Shivaji).

http://dx.doi.org/10.1016/j.jmaa.2014.11.0120022-247X/© 2014 Elsevier Inc. All rights reserved.

R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612 599

We make the following assumptions on f :

(H1) f : [0, ∞) → [0, ∞) is a C1 function with f(0) > 0.

We can rewrite our problem (P ) as below:

(Q)

⎧⎪⎪⎨⎪⎪⎩

−Δu− f(0)uβ

= g(u) in Ω

u > 0 in Ω

u = 0 on ∂Ω,

where g(u) = f(u)−f(0)uβ . By the mean value theorem g(t) = f ′(s)t1−β for some s ∈ (0, t). Since

lims→0 |f ′(s)| < ∞ and 0 < β < 1, we have g(0) = 0. Thus we can treat g as a continuous function on [0, ∞) with g(0) = 0.

We further assume:

(G1) There exists a non-negative constant k such that g(t) = g(t) + kt is increasing in [0, ∞).

Our aim is to establish a three solution theorem for (P ) using fixed point arguments (see [1] and [17] for similar arguments in the non-singular case, i.e. β = 0). The main difficulty here is due to the singular nature of the reaction term which prevents the operator associated with (P ) to be monotone. Once the singularity has been absorbed into the operator as in (Q) we can prove the monotonicity of the associated problem. We will in fact show that a certain operator associated with the new problem is strongly increasing as well as completely continuous and hence prove our main result (see Theorem 3.9). See [2] where the authors study an approximate non-singular boundary value problem by truncating the reaction terms near the origin, and later return to the original problem by studying the limit of the solutions to the approximate problem. See also [5] where given a sub-solution ψ > 0 in Ω and a super solution φ(≥ ψ) the authors first study the non singular problem −Δu = f(u)

uβ in Ω, u = ψ on ∂Ω where Ω � Ω and then return to the original problem by studying the limit of the solution u ∈ [ψ, φ] in Ω as dist(Ω, ∂Ω) → 0. Here we provide a direct approach to establish a theory of sub-super solutions to such singular problems (P ). See also [3,7,16,19,20] for related results for such singular boundary value problems.

This paper is divided into several sections. In Section 2 we discuss some of the required preliminary material. In Section 3 we state and prove several important results and use them to conclude the proof of our three solution theorem for the BVP (P ) in the case f(0) > 0. In Section 4 as an application we consider a singular coperative system. Section 5 deals with a sub-super solution theorem for the infinite semipositone case, f(0) < 0. Constants, functions and variables used in each section are independent to those used in previous sections unless specifically mentioned.

2. Preliminaries

We now recall some results from Amann [1]. Let e ∈ C2(Ω) denote the unique positive solution of

−Δe = 1 in Ω

e = 0 on ∂Ω.

Then e(x) > 0 in Ω, ∂e∂ν < 0 on ∂Ω and e(x) ≥ kd(x) for some constant k > 0, where d(x) = dist(x, ∂Ω)

denote the distance of x from the boundary of Ω. Ce(Ω) is the set of functions in u ∈ C0(Ω) such that

600 R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612

−te ≤ u ≤ te for some t > 0. Ce(Ω) equipped with ‖u‖e = inf{t > 0 : −te ≤ u ≤ te} is a Banach space. Also the following continuous embedding holds:

C10 (Ω) ↪→ Ce(Ω) ↪→ C0(Ω).

Further Ce(Ω) is an ordered Banach space (OBS) whose positive cone Pe = {u ∈ Ce(Ω) : u(x) ≥ 0} is normal and has non empty interior. In particular the interior of P 0

e consists of all those functions u ∈ C0(Ω)with t1e ≤ u ≤ t2e for some t1, t2 > 0.

Definition 2.1. Define the map Ag : C0(Ω) → C10 (Ω) as Ag(u) = w iff w is a weak solution of

(P1)

⎧⎪⎪⎨⎪⎪⎩

−Δw − f(0)wβ

= g(u) in Ω

w > 0 in Ω

w = 0 on ∂Ω.

By a weak solution w ∈ H10 (Ω) we mean

∫Ω

∇w∇ϕ− f(0)∫Ω

ϕ

wβ=

∫Ω

g(u)ϕ holds for all ϕ ∈ C∞c (Ω). (2.1)

Lemma 2.2 (Hardy–Sobolev inequality). For every ϕ ∈ H10 (Ω) there exists a constant C independent of ϕ

such that the following holds:

∫Ω

∣∣∣∣ ϕ

d(x)

∣∣∣∣2

≤ C

∫Ω

|∇ϕ|2.

Suppose ϕ, w ∈ H10 (Ω) and w is known to satisfy the estimate w(x) ≥ cd(x) for some c > 0. Then

Holder’s inequality together with the Hardy–Sobolev inequality will yield∣∣∣∣∫Ω

ϕ

wβ

∣∣∣∣ ≤ c1

∫Ω

|ϕ|d(x)β = c1

∫Ω

|ϕ|d(x) d(x)1−β ≤ c1

∣∣∣∣∣∣∣∣ ϕ

d(x)

∣∣∣∣∣∣∣∣L2

∥∥d(x)1−β∥∥L2 ≤ C∗‖∇ϕ‖L2(Ω)

for some C∗ > 0.Let w ∈ H1

0 (Ω) be a weak solution of (P1) satisfying the lower bound w(x) ≥ cd(x), then using standard density arguments we find∫

Ω

∇w∇ϕ− f(0)∫Ω

ϕ

wβ=

∫Ω

g(u)ϕ holds for all ϕ ∈ H10 (Ω). (2.2)

We extend the function f and g from R+ to R in a continuous manner by defining f(t) = f(0) and g(t) = g(0) for all t ≤ 0. W.l.o.g we will assume throughout this paper that g is increasing in R+(i.e. k = 0in (G1)). If not, we can study

(P2)

⎧⎪⎪⎨⎪⎪⎩

−Δw − f(0)wβ

+ kw = g(u) in Ω

w > 0 in Ω

w = 0 on ∂Ω

instead of (P1) and establish our results. In this case we define Ag(u) = w iff w is a weak solution of (P2).

R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612 601

Definition 2.3. We say a function u > 0 in Ω is a solution of (P ) if u ∈ C2(Ω) ∩ C10 (Ω), and u satisfies

the equation −Δu = f(u)uβ in the classical sense. We say u > 0 is a subsolution (or supersolution) if

u ∈ C2(Ω) ∩ C10 (Ω) and −Δu ≤ f(u)

uβ (or −Δu ≥ f(u)uβ ) pointwise in Ω.

3. Statements and proofs of main results

Proposition 3.1. A function u solves (P ) iff u is a fixed point of the map Ag.

Proof. If u ∈ C10 (Ω) solves (P ) in the classical sense, then clearly u is a fixed point of Ag(u). Conversely

suppose that Ag(u) = u for some u ∈ C10 (Ω). Since u(x) > 0 and g(u(x)) is locally Holder continuous

in Ω one can appeal to standard elliptic regularity theory to obtain u ∈ C2(Ω). Thus u solves (P ) iff u = Ag(u). �

Given a λ > 0, for β ∈ (0, 1) by [4]

(P0){

−Δξλ = λξ−βλ in Ω

ξλ = 0 on ∂Ω

has a unique solution ξλ ∈ C2(Ω) ∩ C10 (Ω) and there exist constants c1, c2 > 0 such that

c1d(x) ≤ ξλ(x) ≤ c2d(x) ∀x ∈ Ω.

Lemma 3.2. The map Ag is a well defined, monotone operator from C0(Ω) to C10 (Ω).

Proof. Given a u ∈ C0(Ω), we define the composition function v(x) = g(u(x)). Then v(x) ∈ C0(Ω) and non-negative. Let ξ be the unique solution of −Δξ = f(0)ξ−β , ξ = 0 on ∂Ω. Let w = ξ and w = ξ + Me, where M > ‖v‖∞. Then w and w are the sub and supersolution for the BVP

(P1)

⎧⎪⎪⎨⎪⎪⎩

−Δw − f(0)wβ

= g(u) in Ω

w > 0 in Ω

w = 0 on ∂Ω.

Define the convex set M = {z ∈ H10 (Ω) : w ≤ z ≤ w}. Define the energy functional

E(z) = 12

∫Ω

|∇z|2 − f(0)1 − β

∫Ω

z1−β −∫Ω

vz.

Clearly E : M → R is coercive and weakly lower semicontinuous on the weakly closed set M. Thus there exists a function w ∈ H1

0 (Ω) minimizing the functional of E defined on M. We make the following assertion and prove it via Perron’s method as described in Struwe [18].

Claim. w is a weak solution of (P1).

We need to show that∫Ω

∇w∇ϕ− f(0)∫Ω

ϕ

wβ=

∫Ω

vϕ holds for all ϕ ∈ C∞c (Ω).

Let ϕ ∈ C∞c (Ω), ε > 0 and vε = min{w, max{w, w + εϕ}}. Then vε = w + εϕ − ϕε + ϕε where

602 R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612

ϕε = max{0, w + εϕ− w} ≥ 0,

ϕε = −min{0, w + εϕ− w} ≥ 0.

Note that ϕε, ϕε ∈ H10 (Ω) and vε ∈ M. Now since M is convex, w + t(vε − w) ∈ M. Let J(z) =

∫Ωz1−β

where z ∈ M. For t > 0, consider the difference quotient

J(w + t(vε − w)) − J(w)t

=∫Ω

(w + t(vε − w))1−β − w1−β

t

= (1 − β)∫Ω

vε − w

(w + s(vε − w))β

for some s ∈ [0, t]. Now letting t → 0+, since vε−w(w+s(vε−w))β is bounded above by the L1 function |vε−w|

wβ , applying dominated convergence theorem we obtain,

limt→0+

J(w + t(vε − w)) − J(w)t

= (1 − β)∫Ω

vε − w

wβ.

Further, limt→0+E(w+t(vε−w))−E(w)

t exists which we denote by 〈DE(w), vε−w〉. Since w is a local minimizer of E on M we have

0 ≤⟨DE(w), vε − w

⟩=

∫Ω

∇w∇(vε − w) − f(0)∫Ω

vε − w

wβ−∫Ω

v(vε − w). (3.1)

Next for any given ϕ ∈ H10 (Ω), w−βϕ ∈ L1(Ω), and we denote

∫Ω∇w∇ϕ− f(0)

∫Ω

ϕwβ −

∫Ωvϕ by

〈DE(w), ϕ〉. Substituting for vε − w = εϕ − ϕε + ϕε in (3.1) we have

0 ≤⟨DE(w), vε − w

⟩= ε

⟨DE(w), ϕ

⟩−⟨DE(w), ϕε

⟩+

⟨DE(w), ϕε

⟩(3.2)

which implies

⟨DE(w), ϕ

⟩≥ 1

ε

[⟨DE(w), ϕε

⟩−⟨DE(w), ϕε

⟩]. (3.3)

Since w is a supersolution for (P1),

⟨DE(w), ϕε

⟩=

⟨DE(w), ϕε

⟩+

⟨DE(w) − DE(w), ϕε

⟩≥

⟨DE(w) − DE(w), ϕε

⟩=

∫Ωε

∇(w − w)∇ϕε − f(0)∫Ωε

ϕε

(1wβ

− 1wβ

)

where Ωε = {x ∈ Ω : w(x) + εϕ(x) > w(x) > w(x)}. Now

∫Ωε

∇(w − w)∇ϕε =∫Ωε

∣∣∇(w − w)∣∣2 + ε

∫Ωε

∇(w − w)∇ϕ ≥ ε

∫Ωε

∇(w − w)∇ϕ

and

R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612 603

∫Ωε

ϕε

(1wβ

− 1wβ

)=

∫Ωε

(w − w)(

1wβ

− 1wβ

)+ ε

∫Ωε

ϕ

(1wβ

− 1wβ

)≤ ε

∫Ωε

2|ϕ|wβ

.

Hence, we get

⟨DE(w), ϕε

⟩≥ ε

∫Ωε

∇(w − w)∇ϕ− f(0)ε∫Ωε

2|ϕ|wβ

.

Note that the Lebesgue measure of Ωε goes to 0 as ε → 0 and by the Hardy–Sobolev inequality 2|ϕ|wβ ∈ L1(Ω),

we obtain

⟨DE(w), ϕε

⟩≥ o(ε),

where o(ε)/ε → 0 as ε → 0. In a similar way we have

⟨DE(w), ϕε

⟩≤ o(ε).

Hence from (3.3), 〈DE(w), ϕ〉 ≥ 0. Reversing the sign of ϕ, we conclude∫Ω

∇w∇ϕ− f(0)∫Ω

ϕ

wβ=

∫Ω

vϕ holds for all ϕ ∈ C∞c (Ω)

and hence the claim is verified. Recall that the solution w(x) here obtained is bounded below by kd(x), for some positive constant k. Thanks to the regularity results of Gui–Lin [10], we conclude that w ∈ C1,γ

0 (Ω)for some γ ∈ (0, 1).

Next suppose u1 ≤ u2, then g(u1) ≤ g(u2). Now for i = 1, 2 let wi = Ag(ui) where

⎧⎪⎪⎪⎨⎪⎪⎪⎩

−Δwi −f(0)wδ

i

= g(ui) in Ω

wi > 0 in Ω

wi = 0 on ∂Ω.

Then we have

−Δ(w2 − w1) − f(0)(

1wβ

2− 1

wβ1

)= g(u2) − g(u1) ≥ 0. (3.4)

We know wi(x) ≥ ξ(x) ≥ cd(x) for i = 1, 2. Choosing test function ϕ = (w2 −w1)− = − min{w2 −w1, 0} in (2.2) we get

−∫Ω

∣∣∇(w2 − w1)−∣∣2 − f(0)

∫Ω

(1wβ

2− 1

wβ1

)(w2 − w1)− ≥ 0.

This implies that ‖(w2 −w1)−‖H10 (Ω) = 0, hence w1 ≤ w2. Since wi = Ag(ui) it is equivalent to saying that

the map Ag is monotone.Now if u1 = u2, we can show that ‖(w2 −w1)−‖H1

0 (Ω) = ‖(w1 −w2)−‖H10 (Ω) = 0, i.e. w1 = w2. Hence we

conclude that the map Ag is well defined. �Theorem 3.3. The map Ag : Ce(Ω) → Ce(Ω) is completely continuous.

604 R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612

Proof. For 0 < β < 1, following the lines of proof of Step 1, Lemma 4.2 of [6] we can show that Ag : C0(Ω) →C1,γ

0 (Ω) is continuous for some γ ∈ (0, 1). Since C1,γ0 (Ω) � C1

0 (Ω) we have Ag : C0(Ω) → C10 (Ω) is

completely continuous. Now recalling the continuous embedding

C10 (Ω) ↪→ Ce(Ω) ↪→ C0(Ω)

we conclude Ag : Ce(Ω) → Ce(Ω) is a completely continuous map. �Next important step to prove the three solution theorem is to show that the map Ag is strongly increasing,

i.e. if u1 ≤ u2 and u1 ≡ u2, then Ag(u2) −Ag(u1) ∈ P 0e . In other words, there exist constants t1, t2 > 0 such

as t1e(x) ≤ w2(x) − w1(x) ≤ t2e(x). First we prove a strong maximum principle with a singular coefficient a(x) and use that to conclude that the map Ag is strictly increasing.

Let us state the well known Hopf maximum principle.

Lemma 3.4. Let Ω be any open subset of RN satisfying the interior sphere condition. Assume that u ∈C2(Ω) ∩C1(Ω) and satisfies −Δu + a(x)u ≥ 0 for some a(x) ∈ C(Ω) with a(x) ≥ 0. Suppose at some point x0 ∈ ∂Ω,

0 ≤ u(x0) < u(x) holds ∀x ∈ Ω.

Then ∂u∂ν (x0) < 0.

The assumptions of strong maximum principle as stated in Evans [8] and Lin–Han [11] demand a(x) to be a non-negative continuous function in Ω. But a careful observation of the details of the proof tells us that the conclusion of the SMP is valid even if a(x) ∈ C(Ω), in particular a(x) can blow up to +∞ as xapproaches ∂Ω. For sake of completeness we repeat their arguments and prove the result here.

Proposition 3.5 (Strong maximum principle). Let u ∈ C2(Ω) ∩C10 (Ω) be nonnegative, a(x) ∈ C(Ω), a(x) ≥ 0

and −Δu + a(x)u ≥ 0 in Ω. Then either u(x) ≡ 0 or u(x) > 0 in Ω.

Proof. Assume that u ≡ 0 and let S = {x ∈ Ω : u(x) = 0} be non empty. Define V = {x ∈ Ω : u(x) > 0}and then by assumption V is nonempty. Choose a point y ∈ V satisfying dist(y, S) < dist(y, ∂Ω). Let Bdenote the largest possible ball with center at y and with interior contained in V. Then there exists at least one point x0 ∈ S ∩ ∂B. Note that B � Ω and hence a(x) ∈ C(B). By the Hopf lemma we conclude that ∂u∂ν (x0) < 0. But this is a contradiction since x0 is a local minimum of u in Ω and ∇u(x0) = 0. Thus the set S is empty. �

As before let wi = Ag(ui) for i = 1, 2 and recall Eq. (3.4) satisfied by (w2 − w1). By the mean value theorem, there exists a continuous function θ(x) ∈ (w1(x), w2(x)) and

−Δ(w2 − w1) + βf(0)θβ+1 (w2 − w1) = g(u2) − g(u1). (3.5)

If u1 ≤ u2, we have already seen that w1 ≤ w2. Now if a(x) = βf(0)θβ+1 , then a(x) ∈ C(Ω). Since g is strictly

increasing, by strong maximum principle we get Ag is strictly increasing, in other words

If u1 ≤ u2, u1 ≡ u2 then Ag(u1) < Ag(u2). (3.6)

In fact we establish a stronger result in the next theorem, namely Ag is strongly increasing. Here the idea of choosing the subsolution in the proof is adapted from Proposition A.1 of [15].

R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612 605

Theorem 3.6. The map Ag : Ce(Ω) → Ce(Ω) is strongly increasing.

Proof. Let u1 ≤ u2, wi = Ag(ui) and denote w = w2 − w1. By strong maximum principle we know that w > 0 in Ω. From Eq. (3.5), −Δw + βf(0)

θβ+1 w ≥ 0. Since w1(x) ≤ θ(x) ≤ w2(x), we get βf(0)θβ+1 ≤ c

d(x)β+1 for some constant c > 0. We fix this constant c throughout the proof of this theorem. Thus,

−Δw + c

d(x)β+1 w ≥ 0 in Ω. (3.7)

For any r > 0, define the open set Ωr = {x ∈ Ω : d(x, ∂Ω) < r}. Let

v(x) = e(x) + e(x)γ for some γ > 1 to be chosen later. (3.8)

Then an explicit calculation will yield −Δv(x) = 1 +γeγ−1−γ(γ−1)eγ−2|∇e|2. If we define α = infx∈∂Ω | ∂e∂ν |, then α > 0. Therefore we can choose r small enough so that |∇e(x)| > α

2 for all x ∈ Ωr. Now,

−Δv + c

d(x)β+1 v = 1 + γeγ−1 − γ(γ − 1)eγ−2|∇e|2 + c

d(x)β+1 v(x)

≤ C − αγ(γ − 1)2e2−γ

+ ce(x)d(x)β+1 + ce(x)γ

d(x)β+1

≤ C − c1γ(γ − 1)d(x)2−γ

+ c2d(x)β + c3

d(x)β+1−γ.

We find that if we choose a γ ∈ (1, 2 − β), then d(x)γ−2 is the dominating term in RHS of the previous expression. Thus for r small enough,

−Δv + c

d(x)β+1 v ≤ 0 in Ωr

v(x) > 0 in Ωr

v(x) = 0 on ∂Ω.

From (3.6) and (3.7) we obtain

−Δw + c

d(x)β+1 w ≥ 0 in Ωr

w(x) > 0 in Ωr

w(x) = 0 on ∂Ω.

Now choose a k > 0 so that kv − w < 0 on ∂Ωr − ∂Ω. Thus,

−Δ(kv − w) + c

d(x)β+1 (kv − w) ≤ 0 in Ωr

kv − w ≤ 0 on ∂Ωr.

Hence w ≥ k(e(x) + e(x)γ) in Ωr. Since infx∈Ωcrw > 0, there exists a constant t > 0 such that w(x) > te(x)

for all x ∈ Ω. Thus Ag defined on Ce(Ω) is strongly increasing. �Theorem 3.7 (Minimal and maximal solutions). Let ψ, φ be positive sub and supersolution of (P ) satisfying ψ ≤ φ. Then there exists a minimal as well as a maximal solution for (P ) in the ordered interval [ψ, φ].

606 R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612

Proof. Since ψ ∈ C10 (Ω) ∩ C2(Ω) is a sub solution of (P ), we have

−Δψ − f(0)ψβ

≤ g(ψ).

Let us define ψ be the unique solution of −Δψ − f(0)ψβ = g(ψ). In other words according to our notation

ψ = Ag(ψ). Note that

−Δ(ψ − ψ) − f(0)(

1ψβ

− 1ψβ

)≥ 0

and thus ψ ≥ ψ or Ag(ψ) ≥ ψ. Similarly φ being a super solution we show that Ag(φ) ≤ φ. Let E = Ce(Ω)and P be the positive cone of E. Then (E, P ) is an ordered Banach space. The map Ag : [ψ, φ] → E is an increasing completely continuous map from Theorems 3.2 and 3.3. Also ψ ≤ Ag(ψ) ≤ Ag(φ) ≤ φ. Thus by Corollary 6.2 of Amann [1] there exist the minimal and maximal fixed points for the map Ag in the interval [ψ, φ]. From Proposition 3.1 the minimal and maximal fixed points of the map Ag turn out to be the minimal and maximal solutions in the interval [φ, ψ]. �Lemma 3.8. (See Amann [1].) Let X be a retract of some Banach space and f : X → X be a completely continuous map. Suppose that X1 and X2 are disjoint retracts of X and let Uk, k = 1, 2 be open subsets of X such that Uk ⊂ Xk, k = 1, 2. Moreover, suppose that f(Xk) ⊂ Xk and that f has no fixed points on Xk \ Uk, k = 1, 2. Then f has at least three distinct fixed points x, x1, x2 with xk ∈ Xk, k = 1, 2 and x ∈ X \ (X1 ∪X2).

Our multiplicity result for the singular problem (P ) is based on the above lemma applied to the map Ag.

Theorem 3.9 (Three solution theorem). Suppose there exist two pairs of ordered sub and supersolutions (ψ1, φ1) and (ψ2, φ2) of (P ) with the property that ψ1 ≤ ψ2 ≤ φ1, ψ1 ≤ φ2 ≤ φ1 and ψ2 � φ2. Additionally assume that ψ2, φ2 are not solutions of (P ). Then there exist at least three solutions ui, i = 1, 2, 3 for (P )where u1 ∈ [ψ1, φ2], u2 ∈ [ψ2, φ1] and u3 ∈ [ψ1, φ1] \ ([ψ1, φ2] ∪ [ψ2, φ1]).

Proof. Recall from Proposition 3.1 that obtaining a solution for (P ) is equivalent to finding fixed points for the map Ag. Let X = [ψ1, φ1], X1 = [ψ1, φ2] and X2 = [ψ2, φ1]. Then X and X ′

ks being non empty closed convex subsets of Banach space Ce(Ω) are retracts of Ce(Ω). Note that X1 and X2 are disjoint subsets of X. The map Ag restricted to X has the following property:

Claim. Ag : X → X is completely continuous and strongly increasing.

Since Ag is increasing and ψ1, φ1 are ordered sub and supersolutions of (P ) we have

ψ1 ≤ Ag(ψ1) ≤ Ag(φ1) ≤ φ1

i.e. Ag(X) ⊂ X and it is also true that Ag(Xk) ⊂ Xk for k = 1, 2. Ag is completely continuous and strongly increasing because of Theorems 3.3 and 3.6 and hence the claim.

It is given that φ2 is a strict supersolution of (P ). Thus by strong maximum principle Ag(φ2) < φ2. Now by Corollary 6.2 of [1] Ag has a maximal fixed point u1 ∈ X1 and ψ1 ≤ u1 < φ2. Similarly since ψ2 is a strict subsolution Ag has a minimal fixed point u2 in X2 with ψ2 < u2 ≤ φ1.

R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612 607

Next notice that φ2 − u1 satisfies

−Δ(φ2 − u1) − f(0)(

1φβ

2− 1

uβ1

)≥ 0.

Repeating the ideas in the proof of Theorem 3.6 we find there exists a constant t1 > 0 such that φ2 − u1 ≥t1e(x). Similarly there exists a constant t2 > 0 such that u2 − ψ2 ≥ t2e(x). Define,

Bk = X ∩{z ∈ Ce(Ω) : ‖z − uk‖e < tk

}for k = 1, 2.

Then Bk ⊂ Xk and Bk is open in X, k = 1, 2. Hence Xk has non empty interior and let Uk be the largest open set in Xk containing uk such that Ag has no fixed point in Xk \ Uk for (k = 1, 2). Now the existence of the third fixed point for Ag easily follows from Lemma 3.8 and hence the problem (P ) has at least three solutions u1 ∈ [ψ1, φ2], u2 ∈ [ψ2, φ1] and u3 ∈ [ψ1, φ1] \ ([ψ1, φ2] ∪ [ψ2, φ1]). �Remark 3.10. We encourage the readers to see [12] and [13] where such singular boundary value problems were discussed. Under certain additional assumptions on f they established two pairs of sub-super solutions (ψ1, φ1) and (ψ2, φ2) where 0 ≤ ψ1 ≤ ψ2 ≤ φ1, 0 ≤ ψ1 ≤ φ2 ≤ φ1 with ψ2 � φ2 and ψ2, φ2 are strict sub and supersolutions respectively. However due to lack of such a three solution theorem for such singular problems they concluded the existence of only two positive solutions u1, u2 with u1 ∈ [ψ1, φ2] and u2 ∈ [ψ2, φ1]. Our result now provides a third positive solution u3 ∈ [ψ1, φ1] \ ([ψ1, φ2] ∪ [ψ2, φ1]).

4. Applications to a coperative system of singular boundary value problems

We consider the coperative system

(P3)

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−Δu = f1(v) + g1(u)uβ

in Ω

−Δv = f2(u) + g2(v)vβ

in Ω

u > 0, v > 0 in Ω

u = v = 0 on ∂Ω,

where gi(0) > 0 for i = 1, 2. We rewrite this system as

(P4)

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−Δu− g1(0)uβ

= f1(v) + g1(u) in Ω

−Δv − g2(0)vβ

= f2(u) + g2(v) in Ω

u > 0, v > 0 in Ω

u = v = 0 on ∂Ω,

where gi(t) = gi(t)−gi(0)tβ

for i = 1, 2. We further assume that for some constant k ≥ 0, the functions fi(t) +kt

and gi(t) are strictly increasing in [0, ∞). Again w.l.o.g we assume k = 0 and let Ai(u, v) = wi, i = 1, 2where

608 R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

−Δw1 −g1(0)wβ

1= f1(v) + g1(u) in Ω

−Δw2 −g2(0)wβ

2= f2(u) + g2(v) in Ω

w1 > 0, w2 > 0 in Ω

w1 = w2 = 0 on ∂Ω.

Let us define the map A : C0(Ω) × C0(Ω) → C10 (Ω) × C1

0 (Ω) as A(u, v) = (A1(u, v), A2(u, v)). Let P =(p1, p2), Q = (q1, q2) and P, Q ∈ C0(Ω) ×C0(Ω). We say P ≤ Q iff pi ≤ qi for i = 1, 2. Note that if (u, v) ≤(u, v) then Ai(u, v) ≤ Ai(u, v). Therefore, A : Ce(Ω) ×Ce(Ω) → Ce(Ω) ×Ce(Ω) is monotonically increasing. Since each component of the map A is completely continuous, we have A : Ce(Ω) ×Ce(Ω) → Ce(Ω) ×Ce(Ω)is completely continuous. Now suppose that there exist two pairs of sub and supersolution for (P3) given by (Ψ1, Φ1) and (Ψ2, Φ2) where Ψi = (ψi, ψi) and Φi = (φi, φi) for i = 1, 2. Also assume that 0 ≤ Ψ1 ≤Ψ2 ≤ Φ1, 0 ≤ Ψ1 ≤ Φ2 ≤ Φ1 with Ψ2 � Φ2, and Ψ2, Φ2 are strict sub and supersolutions respectively. Then there exist three solutions U1, U2 and U3 for the system such that U1 ∈ [Ψ1, Φ2], U2 ∈ [Ψ2, Φ1] and U3 ∈ [Ψ1, Φ1] \ ([Ψ1, Φ2] ∪ [Ψ2, Φ1]).

5. Infinite semipositone problems

Consider the following elliptic boundary value problem

(P5)

⎧⎪⎪⎨⎪⎪⎩

−Δu = f(u)uβ

in Ω

u > 0 in Ω

u = 0 on ∂Ω,

where 0 < β < 1 and f : [0, ∞) → R is a C1 map with f(0) < 0, and f(t)tβ

is strictly increasing. Also

we assume that there exists a constant β0 > 0 such that f(t)tβ

> 0 for t > β0. In the literature such kind of problems are referred to as infinite semipositone problems (see [14] and references therein). Clearly if u ∈ C2(Ω) is a solution of (P5) then supx∈Ω u(x) > β0, if not it would contradict the maximum principle.

In this section we aim to prove existence of a minimal and maximal solution for (P5) when there exists a pair of sub-super solutions (ψ, φ) which additionally satisfies the hypothesis c0d(x) ≤ ψ(x) ≤ φ(x) for some positive constant c0 and ‖ψ‖∞ > β0. Our proof relies on standard Lp estimates and Holder regularity results for singular boundary value problems which we state below.

Theorem 5.1 (Elliptic estimates). (See [9].) Given f(x) ∈ Lp(Ω) with 1 < p < ∞, the Dirichlet problem −Δu = f in Ω, u = 0 on ∂Ω has a unique weak solution u ∈ W 2,p(Ω) ∩ W 1,p

0 (Ω). Also there exists a constant C independent of u such that ‖u‖W 2,p(Ω) ≤ C‖f‖Lp(Ω).

Theorem 5.2. (See Proposition 3.4, Gui–Lin [10].) Let Ω be a bounded Lipchitz domain in Rn, and suppose that u ∈ C2(Ω) ∩ C(Ω) satisfies the following:

∣∣Δu(x)∣∣ ≤ Md(x)−β and

∣∣u(x)∣∣ ≤ Md(x)α

for some positive constants M , β, α. Then there exists some γ ∈ (0, 1) depending upon β and α such that ||u||C1,γ(Ω) ≤ C(M, α, β).

Now onwards we assume that (P5) has a subsolution ψ ∈ C10 (Ω) with ψ(x) ≥ c0d(x), c0 a positive

constant and ‖ψ‖∞ > β0. We define

R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612 609

C ={u ∈ C1

0 (Ω) : u(x) ≥ ψ(x)}. (5.1)

Consider a function u ∈ C and let g(u) = f(u)uβ . Then there exists a constant c1 = c1(u) such that |g(u)| ≤

c1d(x)β . Note that d(x)−β ∈ Lq(Ω) for all q ∈ (1, 1β ). Let w be the unique weak solution of −Δw = g(u) in Ω, w = 0 on ∂Ω which is now guaranteed to exist by Theorem 5.1. Since g is monotonically increasing and u ≥ ψ, we have

−Δw = g(u) ≥ g(ψ) ≥ −Δψ in Ω, and w = ψ = 0 on ∂Ω.

Now by the weak comparison principle we have w ≥ ψ. Also note that f(u)uβ ≤ C

ξβ1, where C is a positive

constant depending on u and ξ1 is the unique solution of −Δξ1 = ξ−β1 in Ω, ξ1 = 0 on ∂Ω. Therefore,

−Δw = g(u) ≤ C

ξβ1= −Δ(Cξ1) in Ω, and w = ξ1 = 0 on ∂Ω.

Again from the weak maximum principle and the previous lower bound of w we have

c0d(x) ≤ ψ(x) ≤ w(x) ≤ Cξ1(x) ≤ Cd(x) in Ω.

By the interior regularity w ∈ C2(Ω) and from the previous estimate w ∈ C0(Ω). Finally we notice that wsatisfies the assumptions of Theorem 5.2 and thus w ∈ C1,γ(Ω) for some γ ∈ (0, 1).

Definition 5.3. We define the map Ag : C → C as Ag(u) = w iff w is a solution of

−Δw = g(u) in Ω

w = 0 on ∂Ω.

Clearly from the previous discussion Ag is a well defined map from C to itself. Next we will show that the map Ag is completely continuous.

Lemma 5.4. Ag : C → C10 (Ω) is completely continuous.

Proof. First we will show that the map Ag : C → C1,γ0 (Ω) is continuous for some γ ∈ (0, 1). Fix u ∈ C, then

u(x) ≥ c0d(x). Choose a δ small enough so that ‖h‖C10(Ω) < δ and u(x) +h(x) ≥ c0

2 d(x) and ‖u +h‖∞ > β0. Let wh be the unique solution of

{−Δwh = g(u + h) in Ω

wh = 0 on ∂Ω

and w the weak solution of{

−Δw = g(u) in Ω

w = 0 on ∂Ω.

Note that g(u + h) → g(u) point wise in Ω as ‖h‖C10 (Ω) → 0. Now we fix a p ∈ (1, 1β ), then

∣∣g(u + h) − g(u)∣∣ ≤ |f(u + h)|

β+ |f(u)|

β≤ C1d(x)−β ∈ Lp(Ω).

(u + h) u

610 R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612

Thus by dominated convergence theorem ∫Ω|g(u +h) − g(u)|p → 0. Let us consider the difference (wh −w)

which satisfies {−Δ(wh − w) = g(u + h) − g(u) in Ω

wh − w = 0 on ∂Ω.

From the elliptic estimates we have

‖wh − w‖W 2,p(Ω) ≤ C∥∥g(u + h) − g(u)

∥∥Lp(Ω) → 0 as ‖h‖C1

0 (Ω) → 0,

i.e. wh → w a.e. (upto a subsequence) as ‖h‖C10(Ω) → 0. Also note that

∣∣Δ(wh − w)∣∣ ≤ C1

d(x)β and |wh − w| ≤ Cd(x) =⇒ ‖wh − w‖C1,γ(Ω) < ∞.

Since we already know that wh → w a.e. in Ω, by the Ascoli Arzela theorem wh → w in C1,γ(Ω) for all 0 < γ < γ. In other words, Ag(u + h) → Ag(u) in C1,γ(Ω) as ‖h‖C1

0 (Ω) → 0. Using the fact that C1,γ(Ω) � C1(Ω) and the compact embedding is continuous we conclude that Ag : C → C1

0 (Ω) is completely continuous. �Theorem 5.5 (Existence of minimal and maximal solutions). Let ψ, φ ∈ C1

0 (Ω) be an ordered pair of sub-supersolutions for the (P5). Additionally assume that there exists a constant c0 > 0 such that c0d(x) ≤ ψ(x) ≤ φ(x). Then there exists a minimal as well as a maximal solution for (P5) within the ordered interval [ψ, φ].

Proof. Let X = {u ∈ C10 (Ω) : ψ ≤ u ≤ φ}. Let Ag be the map as in Definition 5.3 and for u ∈ X, let

w = Ag(u). Since g is monotonically increasing and ψ, φ are sub and supersolutions we have

−Δψ ≤ g(ψ) ≤ −Δw = g(u) ≤ g(φ) ≤ −Δφ.

By the weak comparison principle, ψ ≤ w ≤ φ, or in other words Ag maps from X to X itself and by Lemma 5.4 Ag is a completely continuous map from X to itself. Clearly Ag is increasing as g is monotonically increasing. By Amann [1] exists a minimal and maximal fixed point for Ag in X which implies the (P5) has a minimal and a maximal solution in [ψ, φ]. �Example 5.6. Let f : [0, ∞) → R is a continuous monotonically increasing function with f(0) < 0 and f(t)tβ

> 0 for all t ≥ β0. Assume that there exists a positive constant A such that f(s) ≥ Asσ for s � 1. Then there exists a sub solution ψ ∈ C1

0 (Ω) for the boundary value problem −Δu = λ f(u)uβ in Ω, u|∂Ω = 0

satisfying the conditions ψ(x) ≥ c0d(x) and ‖ψ‖∞ > β0 for λ � 1. If in addition f satisfies lims→∞f(s)sβ+1 = 0

then the boundary value problem has a super solution φ(≥ ψ) and hence a positive solution for λ � 1.

The idea of construction of sub solution is quite similar to Lemma 5.2 of [14] and hence we will be brief. We define ψ = λr(φ + φ

21+β ), where φ is the first eigenfunction of −Δ, and r to be chosen later. We may

also assume that ‖φ‖∞ = 1. Then

−Δψ = λrφ−2β1+β

(λ1φ

1+3β1+β + 2λ1

1 + βφ2 − 2(1 − β)

(1 + β)2 |∇φ|2)

in Ω.

Let Ωr = {x ∈ Ω : dist(x, ∂Ω) ≤ r}. Fix a δ > 0 such that (λ1φ1+3β1+β + 2λ1

1+βφ2 − 2(1−β)

(1+β)2 |∇φ|2) < −m in Ωδ

for some m > 0. Now in Ωδ,

R. Dhanya et al. / J. Math. Anal. Appl. 424 (2015) 598–612 611

−Δψ ≤ λr

φ2β

1+β

(−m) ≤ −λr+rβ

ψβm ≤ λf(0)

ψβ.

The last inequality in the previous expression is possible if we choose λ � 1 and r > 21+β . Since f is assumed

to be monotonically increasing we have −Δψ ≤ λ f(ψ)ψβ .

Now in Ω \Ωδ we have φ > μ, hence φ + φ2

1+β > μ > 0. If we choose λ � 1 we have

f(ψ) > f(λrμ

)≥ Aλrσ(μ)σ. (5.2)

Thus in Ω \Ωσ,

−Δψ ≤ λrc0μ

where c0 =(λ1 + 2

1 + βλ1

)and μ = μ

2β1+β

≤ λ[Aλrσμσ](λr + λr 2

1+β )βif λ � 1 and r <

11 + β − σ

≤ λf(ψ)ψβ

in Ω \Ωσ (from (5.2)).

Thus −Δψ ≤ λ f(ψ)ψβ in Ω and ψ ≥ c0d(x) and ‖ψ‖∞ > β0 if λ � 1.

Remark 5.7. From the previous example we infer that it is quite natural to impose the additional assump-tion on subsolutions. The problems discussed in [14] can be dealt in a similar way with a subsolution bounded below by c0d(x) for some c0 > 0. Further, by doing so, we are also guaranteed on the existence of minimal/maximal solutions in [ψ, φ].

Acknowledgment

R. Dhanya would like to thank Prof. Prashanth and Prof. Giacomoni for introducing various techniques involved in the study of singular boundary value problems. Dhanya was Dr. D.S. Kothari Postdoctoral fellow (MA/12-13/2006) and was supported by NBHM travel grant (2/44(84)/2013/R&D-II/3743) when this work was being carried out. Eunkyung Ko gratefully acknowledges support from TIFR CAM. Dhanya and Eunkyung Ko would like to thank of UNC, Greensboro for their warm hospitality. This work was partially supported by a grant from the Simons Foundation (#317872) to Ratnasingham Shivaji.

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