# Existence results for classes of infinite semipositone problems

## Abstract

We consider the problem

${ − Δ p u = a u p − 1 − b u γ − 1 − c u α , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,$

where $Δ p u=div( | ∇ u | p − 2 ∇u)$, $p>1$, Ω is a smooth bounded domain in $R n$, $a>0$, $b>0$, $c≥0$, $γ>p$ and $α∈(0,1)$. Given a, b, γ and α, we establish the existence of a positive solution for small values of c. These results are also extended to corresponding exterior domain problems. Also, a bifurcation result for the case $c=0$ is presented.

## 1 Introduction

Consider the nonsingular boundary value problem:

${ − Δ u = a u − b u 2 − c h ( x ) , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,$
(1)

where Ω is a smooth bounded domain in $R n$, $a>0$, $b>0$, $c≥0$, $Δu=div(∇u)$ is the Laplacian of u and $h: Ω ¯ →R$ is a $C 1 ( Ω ¯ )$ function satisfying $h(x)≥0$ for $x∈Ω$, $h(x)≢0$, $max x ∈ Ω ¯ h(x)=1$ and $h(x)=0$ for $x∈∂Ω$. Existence of positive solutions of problem (1) was studied in . In particular, it was proved that given an $a> λ 1$ and $b>0$ there exists a $c ∗ (a,b,Ω)>0$ such that for $c< c ∗$ (1) has positive solutions. Here, $λ 1$ is the first eigenvalue of −Δ with Dirichlet boundary conditions. Nonexistence of a positive solution was also proved when $a≤ λ 1$. Later in , these results were extended to the case of the p-Laplacian operator, $Δ p$, where $Δ p u=div( | ∇ u | p − 2 ∇u)$, $p>1$. Boundary value problems of the form (1) are known as semipositone problems since the nonlinearity $f(s,x)=as−b s 2 −ch(x)$ satisfies $f(0,x)<0$ for some $x∈Ω$. See  for some existence results for semipositone problems.

In this paper, we study positive solutions to the singular boundary value problem:

${ − Δ p u = a u p − 1 − b u γ − 1 − c u α , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,$
(2)

where $Δ p u=div( | ∇ u | p − 2 ∇u)$, $p>1$, Ω is a smooth bounded domain in $R n$, $a>0$, $b>0$, $c≥0$, $α∈(0,1)$, $p>1$, and $γ>p$. In the literature, problems of the form (2) are referred to as infinite semipositone problems as the nonlinearity $f(s)= a s p − 1 − b s γ − 1 − c s α$ satisfies $lim s → 0 + f(s)=−∞$. One can refer to , and  for some recent existence results of infinite semipositone problems. We establish the following theorem.

Theorem 1.1 Given $a,b>0$, $γ>p$, and $α∈(0,1)$, there exists a constant $c 1 = c 1 (a,b,α,p,γ,Ω)>0$ such that for $c< c 1$, (2) has a positive solution.

Remark 1.1 In the nonsingular case ($α=0$), positive solutions exist only when $a> λ 1$ (the principal eigenvalue) (see [1, 2]). But in the singular case, we establish the existence of a positive solution for any given $a>0$.

Next, we study positive radial solutions to the problem:

(3)

where $Ω={x∈ R n ||x|> r 0 }$ is an exterior domain, $n>p$, $a>0$, $b>0$, $c≥0$, $α∈(0,1)$, $p>1$, $γ>p$ and $K:[ r 0 ,∞)→(0,∞)$ belongs to a class of continuous functions such that $lim r → ∞ K(r)=0$. By using the transformation: $r=|x|$ and $s= ( r r 0 ) − n + p p − 1$, we reduce (3) to the following boundary value problem:

${ − ( | u ′ | p − 2 u ′ ) ′ = h ( s ) ( a u p − 1 − b u γ − 1 − c u α ) , 0 < s < 1 , u ( 0 ) = u ( 1 ) = 0 ,$
(4)

where $h(s)= ( p − 1 n − p ) p r 0 p s − p ( n − 1 ) n − p K( r 0 s − ( p − 1 ) n − p )$. We assume:

($H 1$) $K∈C([ r 0 ,∞),(0,∞))$ and satisfies $K(r)< 1 r n + θ$ for $r≫1$, and for some θ such that $( n − p p − 1 )α<θ< n − p p − 1$.

With the condition ($H 1$), h satisfies: (5)

We note that if $θ≥ n − p p − 1$ then $h(s)$ is nonsingular at 0 and $h∈C([0,1],(0,∞))$. In this case, problem (4) can be studied using ideas in the proof of Theorem 1.1. Hence, our focus is on the case when $θ< n − p p − 1$ in which, h may be singular at 0. Note that in this case $h ˆ = inf s ∈ ( 0 , 1 ) h(s)>0$.

Remark 1.2 Note that $ρ+α<1$ since $θ>( n − p p − 1 )α$.

We then establish the following theorem.

Theorem 1.2 Given $a,b>0$, $γ>p$, $α∈(0,1)$, and assume ($H 1$) holds. Then there exists a constant $c 2 = c 2 (a,b,α,p,γ)>0$ such that for $c< c 2$, (3) has a positive radial solution.

Finally, we prove a bifurcation result for the problem

(6)

where Ω is a smooth bounded domain in $R n$, a is a positive parameter, $b,α>0$, $p>1+α$ and $γ>p$. We prove the following.

Theorem 1.3 The boundary value problem (6) has a branch of positive solutions bifurcating from the trivial branch of solutions $(a,0)$ at $(0,0)$ (as shown in Figure 1).

Our results are obtained via the method of sub-super solutions. By a subsolution of (2), we mean a function $ψ∈ W 1 , p (Ω)∩C( Ω ¯ )$ that satisfies

and by a supersolution we mean a function $Z∈ W 1 , p (Ω)∩C( Ω ¯ )$ that satisfies:

where . The following lemma was established in .

Lemma 1.4 (see [13, 18])

Let ψ be a subsolution of (2) and Z be a supersolution of (2) such that $ψ≤Z$ in Ω. Then (2) has a solution u such that $ψ≤u≤Z$ in Ω.

Finding a positive subsolution, ψ, for such infinite semipositone problems is quite challenging since we need to construct ψ in such a way that $lim x → ∂ Ω − Δ p ψ=−∞$ and $− Δ p ψ>0$ in a large part of the interior. In this paper, we achieve this by constructing subsolutions of the form $ψ=k ϕ 1 β$, where k is an appropriate positive constant, $β∈(1, p p − 1 )$ and $ϕ 1$ is the eigenfunction corresponding to the first eigenvalue of $− Δ p ϕ=λ | ϕ | p − 2 ϕ$ in Ω, $ϕ=0$ on Ω.

In Sections 2, 3, and 4, we provide proofs of our results. Section 5 is concerned with providing some exact bifurcation diagrams of positive solutions of (2) when $Ω=(0,1)$ and $p=2$.

## 2 Proof of Theorem 1.1

We first construct a subsolution. Consider the eigenvalue problem $− Δ p ϕ=λ | ϕ | p − 2 ϕ$ in Ω, $ϕ=0$ on Ω. Let $ϕ 1$ be an eigenfunction corresponding to the first eigenvalue $λ 1$ such that $ϕ 1 >0$ and $∥ ϕ 1 ∥ ∞ =1$. Also, let $δ,m,μ>0$ be such that $|∇ ϕ 1 |≥m$ in $Ω δ$ and $ϕ 1 ≥μ$ in $Ω− Ω δ$, where $Ω δ ={x∈Ω|d(x,∂Ω)≤δ}$. Let $β∈(1, p p − 1 + α )$ be fixed. Here, note that since $α∈(0,1)$, $p p − 1 + α >1$. Choose a $k>0$ such that $2b k γ − p + β p − 1 λ 1 k α ≤a$. Define $c 1 =min{ k p − 1 + α β p − 1 (β−1)(p−1) m p , 1 2 k p − 1 μ β ( p − 1 ) (a− β p − 1 λ 1 k α )}$. Note that $c 1 >0$ by the choice of k and β. Let $ψ=k ϕ 1 β$. Then

$− Δ p ψ= k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) − k p − 1 β p − 1 (β−1)(p−1) | ∇ ϕ 1 | p ϕ 1 p − β ( p − 1 ) .$

To prove ψ is a subsolution, we need to establish: (7)

in Ω if $c< c 1$. To achieve this, we split the term $k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 )$ into three, namely,

$k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) = a k p − 1 − α ϕ 1 β ( p − 1 − α ) − 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) ( a − k α ϕ 1 α β β p − 1 λ 1 ) − 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) ( a − k α ϕ 1 α β β p − 1 λ 1 ) .$

Now to prove (7) holds in Ω, it is enough to show the following three inequalities: (8) (9) (10)

From the choice of k, $−(a− β p − 1 λ 1 k α )≤−2b k γ − p$, hence,

$− 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) ( a − k α ϕ 1 α β β p − 1 λ 1 ) ≤ − b k γ − 1 − α ϕ 1 β ( p − 1 − α ) ≤ − b k γ − 1 − α ϕ 1 β ( γ − 1 − α ) .$
(11)

Using $ϕ 1 ≥μ$ in $Ω− Ω δ$ and $c< 1 2 k p − 1 μ β ( p − 1 ) (a− β p − 1 λ 1 k α )$

$− 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) ( a − k α ϕ 1 α β β p − 1 λ 1 ) ≤ − k p − 1 ϕ 1 β ( p − 1 ) ( a − k α λ 1 β p − 1 ) 2 k α ϕ 1 α β ≤ − c k α ϕ 1 α β .$
(12)

Finally, since $|∇ ϕ 1 |≥m$, in $Ω δ$, and $c< k p − 1 + α β p − 1 (β−1)(p−1) m p$,

$− k p − 1 β p − 1 ( β − 1 ) ( p − 1 ) | ∇ ϕ 1 | p ϕ 1 p − β ( p − 1 ) ≤ − k p − 1 + α β p − 1 ( β − 1 ) ( p − 1 ) m p k α ϕ 1 α β ϕ 1 p − β ( p − 1 ) − α β ≤ − c k α ϕ 1 α β ϕ 1 p − β ( p − 1 + α ) .$

Since $p−β(p−1+α)>0$,

$− k p − 1 β p − 1 (β−1)(p−1) | ∇ ϕ 1 | p ϕ 1 p − β ( p − 1 ) ≤ − c k α ϕ 1 α β .$
(13)

From (11), (12) and (13) we see that equation (7) holds in Ω, if $c< c 1$. Next, we construct a supersolution. Let e be the solution of $− Δ p e=1$ in $Ω,e=0$ on Ω. Choose $M ¯ >0$ such that $a u p − 1 − b u γ − 1 − c u α ≤ M ¯ p − 1$ $∀u>0$ and $M ¯ e≥ψ$. Define $Z= M ¯ e$. Then Z is a supersolution of (2). Thus, Theorem 1.1 is proven.

## 3 Proof of Theorem 1.2

We begin the proof by constructing a subsolution. Consider

$− ( | ϕ ′ | p − 2 ϕ ′ ) ′ = λ | ϕ | p − 2 ϕ , t ∈ ( 0 , 1 ) , ϕ ( 0 ) = ϕ ( 1 ) = 0 .$
(14)

Let $ϕ 1$ be an eigenfunction corresponding to the first eigenvalue of (14) such that $ϕ 1 >0$ and $∥ ϕ 1 ∥ ∞ =1$. Then there exist $d 1 >0$ such that $0< ϕ 1 (t)≤ d 1 t(1−t)$ for $t∈(0,1)$. Also, let $ϵ< ϵ 1$ and $m,μ>0$ be such that $| ϕ 1 ′ |≥m$ in $(0,ϵ]∪[1−ϵ,1)$ and $ϕ 1 ≥μ$ in $(ϵ,1−ϵ)$. Let $β∈(1, p − ρ p − 1 + α )$ be fixed and choose $k>0$ such that $2b k γ − p + β p − 1 λ 1 k α h ˆ ≤a$. Define $c 2 =min{ k p − 1 + α β p − 1 ( β − 1 ) ( p − 1 ) m p d 1 ρ , 1 2 k p − 1 μ β ( p − 1 ) (a− β p − 1 λ 1 k α h ˆ )}$. Then $c 2 >0$ by the choice of k and β. Let $ψ=k ϕ 1 β$. This implies that:

$− ( | ψ ′ | p − 2 ψ ′ ) ′ = k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) − k p − 1 β p − 1 (β−1)(p−1) | ϕ 1 ′ | p ϕ 1 p − β ( p − 1 ) .$

To prove ψ is a subsolution, we need to establish: (15)

Here, we note that the term $k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) = h ˆ k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) h ˆ$$h(t)(a k p − 1 − α ϕ 1 β ( p − 1 − α ) − 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) (a− k α ϕ 1 α β β p − 1 λ 1 h ˆ )− 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) (a− k α ϕ 1 α β β p − 1 λ 1 h ˆ ))$, where $h ˆ = inf s ∈ ( 0 , 1 ) h(s)$. Now to prove (15) holds in $(0,1)$, it is enough to show the following three inequalities: (16) (17) (18)

From the choice of k, $−(a− β p − 1 λ 1 k α h ˆ )≤−2b k γ − p$, hence,

$− 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) ( a − k α ϕ 1 α β β p − 1 λ 1 h ˆ ) ≤ − b k γ − 1 − α ϕ 1 β ( p − 1 − α ) ≤ − b k γ − 1 − α ϕ 1 β ( γ − 1 − α ) .$
(19)

Using $ϕ 1 ≥μ$ in $(ϵ,1−ϵ)$ and $c< 1 2 k p − 1 μ β ( p − 1 ) (a− β p − 1 λ 1 k α h ˆ )$

$− 1 2 k p − 1 − α ϕ 1 β ( p − 1 − α ) ( a − k α ϕ 1 α β β p − 1 λ 1 h ˆ ) ≤ − k p − 1 ϕ 1 β ( p − 1 ) ( a − k α λ 1 β p − 1 h ˆ ) 2 k α ϕ 1 α β ≤ − c k α ϕ 1 α β .$
(20)

Next, we prove (18) holds in $(0,ϵ]$. Since $| ϕ 1 ′ |≥m$, and $p−β(p−1)>αβ+ρ$

$− k p − 1 β p − 1 ( β − 1 ) ( p − 1 ) | ϕ 1 ′ | p ϕ 1 p − β ( p − 1 ) ≤ − k p − 1 + α β p − 1 ( β − 1 ) ( p − 1 ) m p k α ϕ 1 α β ϕ 1 ρ ≤ − k p − 1 + α β p − 1 ( β − 1 ) ( p − 1 ) m p k α ϕ 1 α β d 1 ρ t ρ .$

Since $h(t)≤ 1 t ρ$ in $(0,ϵ]$, and $c< k p − 1 + α β p − 1 ( β − 1 ) ( p − 1 ) m p d 1 ρ$,

$− k p − 1 β p − 1 (β−1)(p−1) | ϕ 1 ′ | p ϕ 1 p − β ( p − 1 ) ≤ − c h ( t ) k α ϕ 1 α β .$
(21)

Proving (18) holds in $[1−ϵ,1)$ is straightforward since h is not singular at $t=1$. Thus, from equations (19), (20) and (21), we see that (15) holds in $(0,1)$. Hence, ψ is a subsolution. Let $Z= M ¯ e$ where e satisfies $− ( | e ′ | p − 2 e ′ ) ′ =h(t)$ in $(0,1)$, $e(0)=e(1)=0$ and $M ¯$ is such that $a u p − 1 − b u γ − 1 − c u α ≤ M ¯ p − 1$ $∀u>0$ and $M ¯ e≥ψ$. Then Z is a supersolution of (4) and there exists a solution u of (4) such that $u∈[ψ,Z]$. Thus, Theorem 1.2 is proven.

## 4 Proof of Theorem 1.3

We first prove (6) has a positive solution for every $a>0$. We begin by constructing a subsolution. Let $ϕ 1$ be as in the proof of Theorem 1.1 (see Section 2). Let $β∈(1, p p − 1 )$, and choose a $k>0$ such that $b k γ − p + β p − 1 λ 1 k α ≤a$. Let $ψ=k ϕ 1 β$. Then

$− Δ p ψ= k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) − k p − 1 β p − 1 (β−1)(p−1) | ∇ ϕ 1 | p ϕ 1 p − β ( p − 1 ) .$

To prove ψ is a subsolution, we will establish:

$k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) ≤a k p − 1 − α ϕ 1 β ( p − 1 − α ) −b k γ − 1 − α ϕ 1 β ( γ − 1 − α )$
(22)

in Ω. To achieve this, we rewrite the term $k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 )$ as $k p − 1 β p − 1 λ 1 ϕ 1 β ( p − 1 ) =a k p − 1 − α ϕ 1 β ( p − 1 − α ) − k p − 1 − α ϕ 1 β ( p − 1 − α ) (a− k α ϕ 1 α β β p − 1 λ 1 )$. Now to prove (22) holds in Ω, it is enough to show $− k p − 1 − α ϕ 1 β ( p − 1 − α ) (a− k α ϕ 1 α β β p − 1 λ 1 )≤−b k γ − 1 − α ϕ 1 β ( γ − 1 − α )$. From the choice of k, $−(a− β p − 1 λ 1 k α )≤−b k γ − p$, hence,

$− k p − 1 − α ϕ 1 β ( p − 1 − α ) ( a − k α ϕ 1 α β β p − 1 λ 1 ) ≤ − b k γ − 1 − α ϕ 1 β ( p − 1 − α ) ≤ − b k γ − 1 − α ϕ 1 β ( γ − 1 − α ) .$

Thus, ψ is a subsolution. It is easy to see that $Z= ( a b ) 1 γ − p$ is a supersolution of (6). Since k, can be chosen small enough, $ψ≤Z$. Thus, (6) has a positive solution for every $a>0$. Also, all positive solutions are bounded above by Z. Hence, when a is close to 0, every positive solution of (6) approaches 0. Also, $u≡0$ is a solution for every a. This implies we have a branch of positive solutions bifurcating from the trivial branch of solutions $(a,0)$ at $(0,0)$.

## 5 Numerical results

Consider the boundary value problem

${ − u ″ ( x ) = a u − b u 2 − c u α , x ∈ ( 0 , 1 ) , u ( 0 ) = 0 = u ( 1 ) ,$
(23)

where $a,b>0$, $c≥0$ and $α∈(0,1)$. Using the quadrature method (see ), the bifurcation diagram of positive solutions of (23) is given by

$G(ρ,c)= ∫ 0 ρ d s [ 2 ( F ( ρ ) − F ( s ) ) ] = 1 2 ,$
(24)

where $F(s):= ∫ 0 s f(t)dt$ where $f(t)= a t − b t 2 − c t α$ and $ρ=u( 1 2 )= ∥ u ∥ ∞$. We plot the exact bifurcation diagram of positive solutions of (23) using Mathematica. Figure 2 shows bifurcation diagrams of positive solutions of (23) when $a=8$ ($< λ 1$) and $b=1$ for different values of α.

Bifurcation diagrams of positive solutions of (23) when $a=15$ ($> λ 1$) and $b=1$ for different values of α is shown in Figure 3.

Finally, we provide the exact bifurcation diagram for (6) when $p=2$, and $Ω=(0,1)$. Consider

${ − u ″ ( x ) = a u − b u 2 u α , x ∈ ( 0 , 1 ) , u ( 0 ) = 0 = u ( 1 ) ,$
(25)

where $a,b,α>0$. The bifurcation diagram of positive solutions of (25) is given by

$G ˜ (ρ,a)= ∫ 0 ρ d s [ 2 ( F ˜ ( ρ ) − F ˜ ( s ) ) ] = 1 2 ,$
(26)

where $F ˜ (s):= ∫ 0 s f ˜ (t)dt$ where $f ˜ (t)= a t − b t 2 t α$ and $ρ=u( 1 2 )= ∥ u ∥ ∞$. The bifurcation diagram of positive solutions of (25) as well as the trivial solution branch are shown in Figure 4 when $α=0.5$ and $b=1$.

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## Acknowledgements

EK Lee was supported by 2-year Research Grant of Pusan National University.

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Goddard, J., Lee, E.K., Sankar, L. et al. Existence results for classes of infinite semipositone problems. Bound Value Probl 2013, 97 (2013). https://doi.org/10.1186/1687-2770-2013-97 