# Existence of Positive Bounded Solutions of Semilinear Elliptic Problems

Authors: Faten Toumi
Journal: International Journal of Differential Equations
 (2010)

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## Abstract

This paper is concerned with the existence of bounded positive solution for the semilinear elliptic problemΔu=λp(x)f(u) in Ω subject to some Dirichlet conditions, where Ω is a regular domain in ℝn (n≥3) with compact boundary. The nonlinearity f is nonnegative continuous and the potential p belongs to some Kato class K(Ω). So we prove the existence of a positive continuous solution depending on λ by
the use of a potential theory approach.


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## Body

## 1. Introduction

In this paper, we study the existence of positive bounded solution of semilinear elliptic problemPΔu=λp(x)f(u)inΩ,u∂Ω=αφ,lim⁡|x|→+∞u(x)=β(ifΩisunbounded),
where Ω is a 𝒞1,1-domain in ℝn(n≥3) with compact boundary, α and β are fixed nonnegative constants such that α+β>0, and β=0 when Ω is bounded. The parameter λ is nonnegative, and the function φ is nontrivial nonnegative and continuous on ∂Ω.Numerous works treated semilinear elliptic equations of the typeQΔu=F(x,u)inD,u(x)=Φ(x)on∂D,lim⁡|x|→+∞u(x)=b>0.
For the case of nonpositive function F, many results of existence of positive solutions are established in smooth domains or in ℝn, for instance, see [1–5] and the references therein.In the case whereF changes sign, many works can be cited, namely, the work of  Glover and McKenna [6], whose used techniques of probabilistic potential theory for solving semilinear elliptic and parabolic differential equations in ℝn. Ma and Song [7] adapted the same techniques of those of Glover and McKenna to elliptic equations in bounded domains. More precisely, the hypotheses in [6, 7] require in particular that F(x,u)=F(u) and on each compact, there is a positive constant A such that -Au≤F(u)≤0.In [8], Chen et al. used an implicit probabilistic representation together with Schauder's fixed point theorem to obtain positive solutions of the problem (Q). In fact, the authors considered a Lipschitz domain D in ℝn(n≥3), with compact boundary and imposed to the function F to satisfy on D×(0,b), b∈(0,+∞](1.1)-U(x)t≤F(x,t)≤V(x)f(t),
where f  is nonnegative Borel measurable function defined on (0,b) and the potentials U,V are nonnegative Green-tight functions in D.  In particular, the authors showed the existence of solutions of (Q) bounded below by a positive harmonic function.In [9], Athreya studied (Q) with the singular nonlinearity F(x,t)=g(t)≤max(1,t-α), 0<α<1, in a simply connected bounded C2-domain D in ℝn,n≥3. He showed the existence of solutions bounded below by a given positive harmonic function h0, under the boundary condition Φ≥(1+A)h0, where A is a constant depending on h0, α, and D.Recently, Hirata [20] gave a Chen-Williams-Zhao type theorem for more general regular domains D. More precisely, the author imposed to the function F to satisfy (1.2)|F(x,u)|≤V(x)u-μ,μ>0,
where the potential V belongs to a class of functions containing Green-tight ones. We remark that the class of functions introduced by Hirata coincides with the classical Kato class introduced for smooth domains in [10, 11].In this paper, we will considerF(x,u)=λp(x)f(u). We impose to the potential p  to be in a new Kato class K(Ω) (see Definition 1.1 below), which contains the Green-tight functions and the classical Kato class used by Hirata. More precisely, we will prove using potential theory's tools, the existence of positive solution for (P). Moreover, we will give global behaviour for the solution.So, in the remainder of this introduction, we will give some results related on potential theory, and we will prove others. In the second section, we will give the main theorem and some examples of applications.Let us recall thatℬ(Ω) is the set of Borel measurable functions in Ω and 𝒞0(Ω) is the set of continuous ones vanishing at ∂Ω∪{∞}. The exponent + means that only the nonnegative functions are considered.We denote byHΩφ the bounded continuous solution of the Dirichlet problem(1.3)Δu=0inΩ,u=φon∂Ω,lim⁡|x|→+∞u(x)=0,ifΩisunbounded,
where the function φ is nontrivial nonnegative continuous on ∂Ω. In the remainder of this paper, we denote h=1-HΩ1, and we remark that h=0 when Ω is bounded.Let us recall some notations and notions concerning essentially the potential theory.(i)
Forf∈ℬ+(Ω), we denote by Vf  the potential defined in Ω by
(1.4)Vf(x)=∫ΩG(x,y)f(y)dy,
where G is the Green function of the Laplace operator Δ on Ω with Dirichlet conditions.(ii)
We recall that iff∈Lloc1(Ω) and Vf∈Lloc1(Ω), then we have Δ(Vf)=-f  in Ω (in the sense of distributions), see [10, page 52].(iii)
Let(Xt,t>0) be the Brownian motion in ℝn and Px be the probability measure on the Brownian continuous paths starting at x. For p∈ℬ+(Ω), we define the kernel Vp by
(1.5)Vpf(x)=Ex(∫0τΩe-∫0tp(Xs)dsf(Xt)dt),
where Ex is the expectation on Px and τΩ=inf{t>0:Xt∉Ω}. If p∈ℬ+(Ω) is such that Vp<∞, the kernel Vp satisfies (see [10, 12])
(1.6)V=Vp+Vp(pV)=Vp+V(pVp).
So for, each u∈ℬ(Ω) such that V(p|u|)<∞, we have
(1.7)(u-Vp(pu))(u+V(pu))=(u+V(pu))(u-Vp(pu))=u.(iv)
We recall that a functionf:[0,∞)→ℝ is called completely monotone if (-1)nf(n)≥0, for each n∈ℕ. Moreover, if f is completely monotone on [0,∞), then by [13, Theorem  12a], there exists a nonnegative measure μ on [0,∞) such that
(1.8)f(t)=∫0∞exp⁡(-tx)dμ(x).
So, using this fact and the Hölder inequality, we deduce that if f is completely monotone from [0,∞) to (0,∞), then Log(f) is a convex function.(v)
Letf∈ℬ+(Ω) be such that Vf<∞. From (1.5), it is easy to see that for each x∈Ω, the function t→Vtqf(x) is completely monotone on [0,∞).Now, we recall some properties relating to the Kato classK(Ω).Definition 1.1 (see [14, 15]).
A Borel measurable functionp in Ω belongs to the class K(Ω) if p satisfies(1.9)lim⁡α→0(sup⁡x∈Ω∫Ω∩B(x,α)ρ(y)ρ(x)G(x,y)|p(y)|dy)=0,lim⁡M→∞⁡(sup⁡x∈Ω⁡∫Ω∩(|y|≥M)ρ(y)ρ(x)G(x,y)|q(y)|dy)=0(ifΩisunbounded),
where ρ(x)=min(1,δ(x)) and δ(x) is the Euclidean distance between x and ∂Ω.Remark 1.2.
WhenΩ is a bounded domain, then we can replace ρ(x) by δ(x) and the condition (1.9) is superfluous.Proposition 1.3 (See [14, 15]).
Letp be a nonnegative function in K(Ω). Then one has (i)
∥p∥Ω:=supx∈Ω∫Ωρ(y)/ρ(x)G(x,y)p(y)dy<∞,(ii)
the potentialVp∈𝒞0(Ω).Proposition 1.4 (see [16, 17]).
Letp be a nonnegative function belonging to K(Ω). Then, one has (i)
(1.10)αp=sup⁡x,y∈Ω∫ΩG(x,z)G(z,y)G(x,y)p(z)dz<∞,(ii)
for any nonnegative superharmonic functionv in Ω, one has
(1.11)∫ΩG(x,y)v(y)p(y)dy≤αpv(x),∀x∈D.Proposition 1.5.
Letv be a nonnegative superharmonic function in Ω and p be a nonnegative function in K(Ω). Then, for each x∈Ω such that 0<v(x)<∞, one has
(1.12)exp⁡(-αp)v(x)≤v(x)-Vp(pv)(x)≤v(x).Proof.
Letv be a nonnegative superharmonic function, then by [18, Theorem  2.1, page 164], there exists a sequence (vk)k of nonnegative measurable functions in Ω such that the sequence (Vvk)k defined in Ω by
(1.13)Vvk(x):=∫ΩG(x,y)vk(y)dy
increases to v.
Letx∈Ω such that 0<v(x)<∞. Then, there exists k0∈ℕ such that 0<Vvk(x)<∞, for k≥k0.
Now, for a fixedk≥k0, we consider the function κ(t)=Vtpvk(x).  Since the function κ is completely monotone on [0,∞), then log(κ) is convex on [0,∞). Therefore,
(1.14)κ(0)≤κ(1)exp⁡(-κ′(0)κ(0)),
which means
(1.15)Vvk(x)≤Vpvk(x)exp⁡(V(pVvk)(x)Vvk(x)).
Hence, it follows from Proposition 1.4 (i) that
(1.16)exp⁡(-αv)Vvk(x)≤Vpvk(x).
Consequently, from (1.6), we obtain that
(1.17)exp⁡(-αp)Vvk(x)≤Vvk(x)-Vp(pVvk(x))(x)≤Vvk(x).
By letting k→+∞, we deduce the result.


## 2. Main Result

In this section, we will give an existence result for the problem (P). Assume the following assumptions.(A1)
The functionp is nonnegative and belongs to K(Ω).(A2)
The functionf is a nonnegative, continuous on [0,+∞) and satisfies ∀c>0,∃a≥0 such that, ∀0≤s<t≤c, f(t)-f(s)≤a(t-s).(A3)
σ0=infx∈Ω¯(αHΩφ(x)+βh(x))/f(0)Vp(x)>0.Remark 2.1.
Letf be in C1([0,+∞)), then for a:=max(supt∈[0,c]f′(t),0), the function f satisfies (A2). In particular, if f is nonincreasing, then a=0 holds.Consider the functionθ:λ→λexp(λaαp), where αp is the constant associated to the potential p defined by (1.10). It is obvious to see that θ is bijective from [0,+∞) to [0,+∞).Theorem 2.2.
Assume that the hypotheses (A1)–(A3) are satisfied. Then, for each λ∈[0,θ-1(σ0)), the problem (P) has a positive continuous bounded solution satisfying
(2.1)(1-θ(λ)σ0)exp⁡(-λaαp)(αHΩφ+βh)≤u≤αHΩφ+βh.Remark 2.3.
We remark that iff satisfies the hypothesis  (A2) and f(0)=0, we take σ0=+∞, in this case for each λ∈ℝ+, the problem (P) has a positive bounded solution satisfying
(2.2)exp⁡(-λaαp)(αHΩφ+βh)≤u≤αHΩφ+βh.Now, let us give some examples of applications of the above theorem.Example 2.4.
Assume that(A1) is satisfied. Let μ≥1. Then, for each λ∈ℝ+, the following problem
(2.3)Δu=λp(x)uμinΩ,u(x)=αφ(x)on∂Ω,lim⁡|x|→+∞u(x)=β
admits a positive continuous bounded solution. Indeed, for each c>0, one verifies that for a=μcμ-1, the function f(t)=tμ satisfies (A2).Example 2.5.
Letμ≥0. Assume (A1) and (A3). Consider the following:
(2.4)Δu=λp(x)(1+u)-μinΩ,u(x)=αφ(x)on∂Ω,lim⁡|x|→+∞u(x)=β,u>0.
Then, the function f(t)=(1+t)-μ is in C1([0,+∞)) and decreasing. By Remark 2.1, the hypothesis (A2) is satisfied for a=0. So that for each λ∈[0,σ0), (2.4) has a positive continuous bounded solution satisfying
(2.5)(1-λσ0)(αHΩφ+βh)≤u≤αHΩφ+βh.Example 2.6.
LetΩ be a C1,1-bounded domain and suppose that the hypothesis (A2) is satisfied. Let g be a nonnegative function in Lq(Ω) such that q>n/2 and suppose that μ<1-n/q. Then,
(2.6)Δu=λg(x)δ(x)μf(u)inΩ,u(x)=αφ(x)on∂Ω
has a positive continuous solution.
Let us verify the assumptions(A1) and (A3). From [16, Proposition 2.3], the function p=g/δ(·)μ∈K(Ω), and so the hypothesis (A1) is satisfied. From [16, Proposition 2.7(iii)], there exists a constant c1>0 such that we have for each x∈Ω(2.7)Vp(x)≤c1δ(x).
Now, since φ is nontrivial continuous function at ∂Ω, then there exists c2>0, such that one has on Ω(2.8)HΩφ(x)≥c2‖φ‖1δ(x).
Thus, σ0=infx∈Ω¯(HΩφ(x)/f(0)Vp(x))>0 and so the assumption (A3) is satisfied.Example 2.7.
LetΩ=B(0,1)̅c be the exterior of the unit ball in ℝn(n≥3). Suppose that the hypothesis (A2) is satisfied. Let γ,μ∈ℝ such that γ<1<2<n<μ. Then,
(2.9)Δu=λ1|x|μ-γ(|x|-1)γf(u)inΩ,u(x)=αφ(x)in∂Ω,lim⁡|x|→+∞u(x)=β
has a positive continuous solution.
From [14], the function p(x)=1/|x|μ-γ(|x|-1)γ∈K(Ω) and so the assumption (A1) is satisfied. Moreover, from [14, Proposition  3.5], there exists a constant c1>0 such that one has
(2.10)Vp(x)≤c1|x|-1|x|n-1.
Now, from [19, page 258], there exists a constant c2>0 such that one has on Ω(2.11)αHΩφ(x)+βh(x)≥c2|x|-1|x|n-1.
Thus, σ0=infx∈Ω¯((αHΩφ(x)+βh(x))/f(0)Vp(x))≥c2/c1>0 and so the assumption (A3) is satisfied.In the next, we will give the proof of Theorem2.2.Proof.
Letp∈K+(Ω) and put w:=αHΩφ+βh. Let c=∥ω∥∞>0, then from (A2), there exists a≥0, such that the function ψ:t→at+f(0)-f(t) is a nondecreasing function on [0,c]. Let σ0 be the constant given by (A3), and let λ∈[0,θ-1(σ0)) where θ(λ):=λexp(λaαp). Put q:=λap. Consider the nonempty bounded convex set given by
(2.12)Λ:={u∈B+(Ω):(1-θ(λ)σ0)exp⁡(-λaαp)w≤u≤w}.
Let T be the operator defined on Λ by
(2.13)Tu:=w-Vq(qw)-λf(0)Vqp+λVq(pψ(u)).
We claim that the operator T maps Λ to itself. Indeed, by (A2) and using the monotony of the function ψ, we have for u∈Λ(2.14)Tu=w-Vq(λp(aw+f(0)))+λVq(pψ(u))=w-Vq(λp(ψ(w)+f(w)))+λVq(pψ(u))≤w-λVq(pf(w))+λVq(p(ψ(u)-ψ(w)))≤w.
On the other hand, by using Proposition 1.5 and (A3), we have
(2.15)Tu≥exp⁡(-λaαp)w-λf(0)Vqp+λVq(pψ(u))≥exp⁡(-λaαp)w-λf(0)Vp≥exp⁡(-λaαp)w-λσ0w=(1-θ(λ)σ0)exp⁡(-λaαp)w.
Hence, TΛ⊂Λ.
Next, we prove that the operatorT is nondecreasing on Λ. Let u1,u2∈Λ such that u1≤u2,then by hypothesis (A2), we obtain(2.16)Tu2-Tu1=λVqp(ψ(u2)-ψ(u1))≥0.
Now, consider the sequence (uk)k defined by
(2.17)u0=w-λaVq(pw)-λf(0)Vq(p),
and uk+1=Tuk for k∈ℕ.
It is obvious to see thatu0∈Λ and u1=Tu0≥u0. Thus, using the fact that Λ is invariant under T and the monotony of T, we deduce that(2.18)(1-θ(λ)σ0)exp⁡(-λaαp)w≤u0≤u1≤⋯≤uk≤w.
Hence, the sequence (uk)k converges to a function u∈Λ.
Therefore, from the monotone convergence theorem and the fact thatψ is continuous, the sequence (Tuk)k converges to Tu. So,(2.19)u:=w-Vq(qw)-λf(0)Vqp+λVq(pψ(u)),
or equivalently
(2.20)(u-Vq(qu))=(w-Vq(qw))-Vq(λpf(u)).
Applying the operator (I+V(q·)) to both sides of the above equality and using (1.6) and (1.7), we conclude that u satisfies
(2.21)u=αHΩφ+βh-λV(pf(u)).
Finally, let us verify that u is a solution of the problem (P). Using the fact that p∈K+(Ω) and f(u) is bounded on [0,c], we obtain pf(u)∈K+(Ω). So, Proposition 1.3 (ii) yields V(pf(u))∈𝒞0(Ω) which implies with the continuity of the harmonic continuous function w  that u is continuous on Ω. This completes the proof.



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