A numerical method to obtain positive solution for classes of sublinear semipositone problems

Abstract

Using a numerical method based on sub- super-solution, we will show the existence of positive solution for the problem −Δu = λf(u(x)) for x ∈ Ω, with Dirichlet boundary condition.

Introduction

In this paper, we consider the semipositone problem

$$\left\{\begin{array}{cc}-\mathrm{\Delta }u(x)=\lambda f(u(x))\text{;}\hfill & x\in \Omega \text{,}\hfill \\ u(x)=0\text{;}\hfill & x\in \mathrm{\partial }\Omega \text{,}\hfill \end{array}\right.$$

where λ > 0 is a constant, Ω is a bounded region in Rⁿ with smooth boundary, and f is a smooth function such that f′(u) is bounded below,

$$f(0)<0\text{,}$$

and

$$\underset{u\to \infty }{\lim }\frac{f(u)}{u}=0\text{.}$$

We prove under some additional conditions the existence of a positive solution for λ ∈ I, where I is an interval close to the smallest eigenvalue of −Δ with Dirichlet boundary condition. It was shown that if λ is small then (1), (1.1), (1.2) has no positive solution. In fact, if λ is small, and if u_λ is a solution, then u_λ ⩽ 0 in Ω. Next we consider an anti-maximum principle by Clement and Peletier [3], from which we obtain the existence of a δ = δ(Ω) > 0 and a solution z_λ, positive in Ω, of

$$\left\{\begin{array}{cc}-\mathrm{\Delta }z-\lambda z=-1\text{;}\hfill & x\in \Omega \text{,}\hfill \\ z=0\text{;}\hfill & x\in \mathrm{\partial }\Omega \hfill \end{array}\right.$$

for λ ∈ (λ₁, λ₁ + δ), where λ₁ is the smallest eigenvalue of the problem

$$\left\{\begin{array}{cc}-\mathrm{\Delta }\varphi (x)=\lambda \varphi (x)\text{;}\hfill & x\in \Omega \text{,}\hfill \\ \varphi (x)=0\text{;}\hfill & x\in \mathrm{\partial }\Omega .\hfill \end{array}\right.$$

Let I = [α, γ], where α > λ₁ and γ < λ₁ + δ, and let

$$\sigma ≔\underset{\lambda \in I}{\max }\Vert z_{\lambda }\Vert \text{,}$$

where ∥·∥ denotes the supremum norm. Now assuming that there exists a m > 0 such that

$$f(x)⩾x-m\text{;}\forall x\in [0\text{,}m\gamma \sigma ]$$

(note here that (γσ > 1)) we prove that, (1), (1.1), (1.2) has a non-negative solution for λ ∈ I. Here the solution z_λ of (2) will be used to create a sub-solution, that is, a function ψ(x) such that

$$\left\{\begin{array}{cc}-\mathrm{\Delta }\psi (x)⩽\lambda f(\psi (x))\text{;}\hfill & x\in \Omega \text{,}\hfill \\ \psi (x)⩽0\text{;}\hfill & x\in \mathrm{\partial }\Omega .\hfill \end{array}\right.$$

Theorem 1

There exists λ₀ > 0 such that if 0 < λ < λ₀ and u is a solution of (1), (1.1), (1.2) then u ⩽ 0 in Ω.

The above theorem (proved in [1]) shows that if λ is small then (1), (1.1), (1.2) has no positive solution. In fact, if λ is small, and if u_λ is a solution, then u_λ ⩽ 0 in Ω.

Theorem 2

Let λ ∈ I and assume that f satisfies (1.1), (1.2), (1.4). Then (1) has a non-negative solution.

Theorem 2 also proved in [1] that we give a proof for the sake of completeness. Let ψ(x)≔(γm)z(x), where γ, m and z are as before. Then

$$-\mathrm{\Delta }\psi (x)=(\gamma m)\lambda z-1\text{;}x\in \Omega \text{,}\psi =0\text{;}x\in \mathrm{\partial }\Omega \text{.}$$

But

$$\lambda f(\psi (x))=\lambda f(\gamma \mathit{mz}(x))⩾\lambda \gamma \mathit{mz}(x)-m⩾\gamma m\lambda z-1=-\mathrm{\Delta }\psi (x)\text{.}$$

Thus ψ(x) is a sub-solution of (1) for λ ∈ I. Now let v(x) to be the unique positive solution of

$$\left\{\begin{array}{cc}-\mathrm{\Delta }v(x)=1\text{;}\hfill & x\in \Omega \text{,}\hfill \\ v(x)⩽0\text{;}\hfill & x\in \mathrm{\partial }\Omega \text{,}\hfill \end{array}\right.$$

and consider ϕ(x) = J v(x); J > 0 (to be chosen). Then ϕ satisfies

$$-\mathrm{\Delta }\varphi (x)=J\text{;}x\in \Omega \text{,}\varphi (x)=0\text{;}x\in \mathrm{\partial }\Omega \text{.}$$

But by (1.2), there exists a J₀ > 0 such that for J > J₀,

$$-\mathrm{\Delta }\varphi (x)=J⩾\lambda f(\mathit{Jv}(x))\text{;}x\in \Omega \text{,}$$

and thus ϕ(x) will be a super-solution for (1). Now choose J > J₀ large enough that

$$\varphi (x)=\mathit{Jv}(x)⩾(\gamma m)z(x)=\psi (x)\text{.}$$

This is clearly possible by Hopfs maximum principle, and hence by the well-known method of sub- and super-solutions, (1) has a non-negative solution u(x) such that γm z(x) ⩽ u(x) ⩽ J v(x), and Theorem 2 is proven. Here we give a simple example that satisfies (1.1), (1.2), (1.4). Consider

$$f(s)=m(s+1{)}^{\frac{1}{2}}-\frac{3m}{2}\text{;}m>0\text{.}$$

Then $f(0)=-\frac{m}{2}<0$ and ${\lim }_{s\to \infty }\frac{f(s)}{s}=0$. Thus (1.1), (1.2) are satisfied. Now let p > 0 be such that f(p) = p − m. That is,

$$\begin{array}{cc}\hfill & m(p+1{)}^{\frac{1}{2}}-\frac{3m}{2}=p-m\text{,}\hfill \\ \hfill & m^2(p+1)={\left\{p+\frac{m}{2}\right\}}^2\text{,}\hfill \\ \hfill & p^2+(m-m^2)p-\frac{3m^2}{4}=0\text{,}\hfill \end{array}$$

and

$$p=\frac{(m^2-m)+\sqrt{m^4-2m^3+m^2+3m^2}}{2}=\frac{(m^2-m)+m\sqrt{m(m-1{)}^2+3}}{2}\text{.}$$

Hence in order that (1.4) is satisfied, we must have

$$\frac{m^2-m+m\sqrt{(m-1{)}^2+3}}{2}⩾m(\gamma \alpha )\text{,}$$

that is,

$$(m-1)+\sqrt{(m-1{)}^2+3}⩾2(\gamma \alpha )$$

(Note here that γ and α are quantities that depend on only Ω). Clearly for a given γ and α, there exists a m₀ such that if m > m₀ then (1.7) is satisfied and equivalently (1.4) will be satisfied.

We investigate numerically positive solutions. Our numerical method is based on monotone iteration.

We say that u is a positive solution of (1) if u is a classical solution with u(x) ∈ I for all $x\in \overline{\Omega }$ and u(x) > 0 for all x ∈ Ω.