# On Positive Radial Solutions for a Class of Elliptic Equations

Authors: Ying Wu; Guodong Han
Journal: The Scientific World Journal
 (2014)

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## Abstract

A class of elliptic boundary value problem in an exterior domain is considered under some conditions concerning the first eigenvalue of the relevant linear operator, where the variables of nonlinear termfs,u need not to be separated. Several new theorems on the existence and multiplicity of positive radial solutions are obtained by means of fixed point index theory. Our conclusions are essential improvements of the results in Lan and Webb (1998), Lee (1997), Mao and Xue (2002), Stańczy (2000), and Han and Wang (2006).


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## Body

## 1. Introduction

The existence and multiplicity of positive radial solution for the following elliptic boundary value problem(1)-Δu=f(|x|,u)for|x|>1,x∈ℝn,n⩾3,u=0for|x|=1,u⟶0as|x|⟶∞,
are considered in this paper, where f∈C((1,∞)×ℝ+,ℝ+) and ℝ+=[0,∞).In recent years, similar problems have been discussed by several authors; see [1–11] and references therein. The usual approaches include variational method [2, 6, 7, 9], topological method [4, 10, 11], and sub- and supersolution method [3, 5].In [11], by using the norm-type cone expansion and compression theorem, Stańczy proved that problem (1) has at least one positive radial solution under the following conditions:(B1)
for anyM>0, there exists a function pM∈C((1,∞),ℝ+) with ∫1∞‍s(1-s2-n)pM(s)ds<∞ such that
(2)0⩽f(s,u)⩽pM(s)forany(s,u)∈(1,+∞)×[0,M];(B2)
there exists a setB⊂(1,∞) of positive measure such that
(3)lim⁡u→∞f(s,u)u=∞uniformlywithrespecttos∈B;(B3)
there exists a functionp∈C((1,∞),ℝ+) with ∫1∞‍s(1-s2-n)p(s)ds<∞ such that
(4)lim⁡u→0+f(s,u)p(s)u=0uniformlywithrespecttos∈(1,∞).In a recent paper [4], replacing the conditions listed above by the weaker ones
(5)lim⁡⁡inf⁡u→∞min⁡s∈[c,d]f(s,u)u>ξ,limsup⁡u→0+f(s,u)p(s)u<η
uniformly with respect to  s∈(1,∞)  for suitable positive numbers  ξ  and  η, the authors proved that problem (1) still has at least one positive radial solution.In the present paper, we continue the study in [4]. Under some conditions concerning the first eigenvalues corresponding to the relevant linear operators, we improve the above positive numbers ξ and η by using the fixed point index. Furthermore, we obtain several existence theorems on multiple positive radial solutions of (1). Our results cover both sub- and superlinear problems. It seems to be difficult to utilize the norm-type cone expansion and compression theorem to prove our results.In the remainder of this section, we recall some facts on the fixed point index for completely continuous operators on a cone in the Banach space in order to prove our main results. Please refer to [12–14] for more details.LetE be a real Banach space and P a cone in E. The following lemma is a well-known result of the fixed point index theory, which will play an important role in the proof of our main results.Lemma 1 (see [12–14]).
LetΩ be a bounded open set in E with θ∈Ω, A:P∩Ω¯→P a completely continuous operator, where θ denotes the null element of E. Assume that A has no fixed point on P∩∂Ω.(i)
(Homotopy invariance) Ifu≠μAu for all μ∈[0,1] and u∈P∩∂Ω, then the fixed point index i(A,P∩Ω,P)=1;(ii)
(omitting a direction) if there exists an elementψ0∈P∖{θ} such that u≠Au+μψ0 for all u∈P∩∂Ω and μ⩾0, then i(A,P∩Ω,P)=0;(iii)
(cone expansion) if∥Au∥⩾∥u∥ for u∈P∩∂Ω, then i(A,P∩Ω,P)=0;(iv)
(additivity) supposeΩ1 is an open subset of Ω with θ∈Ω1 and u≠Au for u∈P∩∂Ω1; then
(6)i(A,P∩Ω,P)=i(A,P∩Ω1,P)+i(A,P∩(Ω∖Ω¯),P);(v)
ifi(A,P∩Ω,P)≠0, then A has at least one fixed point in P∩Ω.The paper is organized as follows. In Section2 we change problem (1) into a singular two-point boundary value problem and then investigate the existence and multiplicity of its positive solutions. And some examples are presented in Section 2. Several theorems on existence and multiplicity of positive radial solutions of problem (1) are established in Section 3.


## 2. Positive Solutions of Singular Two-Point Boundary Value Problems

Looking for radial solutionsu(x)=z(|x|) of (1), where z:ℝ+→ℝ, one can substitute v(t)=z((1-t)1/(2-n)) thus reducing (1) to the following singular two-point boundary value problem, which is singular at 1:
(7)-v′′=g(t,v),t∈(0,1),v(0)=v(1)=0,
where
(8)g(t,v)=1(n-2)2(1-t)(2n-2)/(2-n)f((1-t)1/(2-n),v).It is well known that the solution of (7) in C2[0,1] is equivalent to the solution of the following Hammerstein integral equation in C[0,1]:
(9)v(t)=∫01‍k(t,s)g(s,v(s))ds,t∈[0,1],
where Green's function
(10)k(t,s)={s(1-t),0⩽s⩽t⩽1,t(1-s),0⩽t⩽s⩽1.Define an operatorA:C[0,1]→C[0,1] as follows:
(11)(Av)(t)=∫01‍k(t,s)g(s,v(s))ds,t∈[0,1].
Then the solution v* of (9) in C[0,1] is equivalent to the fixed point of A in C[0,1].LetE=C[0,1] be our Banach space with the norm ∥v∥=max⁡t∈[0,1]|v(t)| for all v∈E, P={v∈E:v(t)⩾0fort∈[0,1]}, and Q={v∈P:min⁡t∈[a,b]v(t)⩾min⁡{a,1-b}∥v∥}, where [a,b]⊂(0,1). It is easy to show that P and Q are cones in E. Let Ωr={u∈E:∥u∥<r} be the open ball of radius r in E. Define a set H by
(12)H={∫01‍h∈C((0,1),ℝ+):h≢0,∫01‍t(1-t)h(t)dt<+∞}.Forh∈H, define an operator Th by
(13)(Thv)(t)=∫01‍k(t,s)h(s)v(s)dsforv∈E.Then we have the following lemma.Lemma 2.
For anyh∈H,(i)
Th:E→E is a completely continuous positive linear operator, and the spectral radius r(Th)≠0 and Th has a positive eigenfunction φ1h corresponding to its first eigenvalue λ1h=(r(Th))-1;(ii)
Th(P)⊂Q;(iii)
there existδ1,δ2>0, such that
(14)δ1k(t,s)⩽φ1h(s)⩽δ2k(s,s)fort,s∈[0,1];(iv)
define a functionalJh by Jh(v)=∫01‍h(t)φ1h(t)v(t)dt for v∈E; then Jh(Thv)=λ1h-1Jh(v) forv∈E;(v)
let(15)P0={v∈P:Jh(v)⩾λ1h-1δ1∥v∥};then P0 is a cone in E and Th(P)⊂P0, where δ1 is defined by (14).To prove Lemma2, we need the following lemmas.Lemma 3 (see [15]).
Suppose thatE is a Banach space, Tn:E→E(n=1,2,3,…) are completely continuous operators, T:E→E, and
(16)lim⁡n→+∞max⁡∥u∥<r∥Tnu-Tu∥=0,∀r>0;
then T is a completely continuous operator.Lemma 4 (see [16, 17]).
Suppose thatT:E→E is a completely continuous linear operator andT(P)⊂(P). If there exist ϕ∈E∖(-P) and a constantc>0 such that cTϕ⩾ϕ, then the spectral radiusr(T)≠0 and T has a postive eigenfunction corresponding to its first eigenvalue λ1=(r(T))-1.Proof of Lemma2.
(i) It follows from the definition ofH that
(17)|(Thv)(t)|⩽∫01‍k(t,s)h(s)|v(s)|ds⩽∥v∥·∫01‍k(s,s)h(s)ds<+∞foranyv∈E.
Hence, by Lebesgue’s dominated convergence theorem, it is easy to see thatTh:E→E. Obviously, Th(P)⊂P and Th is a linear operator; namely, Th is a positive linear operator. Next, we will show that Th is completely continuous. For any natural number n(n⩾2), let
(18)hn(t)={inf⁡t<s⩽1/nh(s),0⩽t⩽1n,h(t),1n⩽t⩽n-1n,inf⁡(n-1)/n⩽s<th(s),n-1n⩽t⩽1.
Thenhn:[0,1]→[0,+∞) is continuous and hn(t)⩽h(t) for all t∈(0,1). Let
(19)(Thnv)(t)=∫01‍k(t,s)hn(s)v(s)ds.
It is clear thatThn:E→E is completely continuous. For any r>0 and v∈Ωr, according to (18), (19), and the absolute continuity of integral, we have
(20)lim⁡n→∞∥Thnv-Thv∥=lim⁡n→∞max⁡t∈[0,1]|∫01‍k(t,s)(hn(s)-h(s))v(s)ds|⩽∥v∥lim⁡n→∞∫01‍k(s,s)(h(s)-hn(s))ds=∥v∥lim⁡n→∞∫e(n)‍k(s,s)(h(s)-hn(s))ds⩽∥v∥lim⁡n→∞∫e(n)‍k(s,s)h(s)ds=0,
where e(n)=[0,1/n]∪[(n-1)/n,1]. Therefore, by Lemma 3, Th:E→E is a completely continuous operator.
It is obvious that there existst1∈(0,1) such that k(t1,t1)h(t1)>0. Thus there is [a1,b1]⊂(0,1) such that t1∈(a1,b1) and k(t,s)h(s)>0 for all t,s∈[a1,b1]. Take ζ∈P such that ζ(t1)>0 and ζ(t)=0 for all t∉[a1,b1]. Then for t∈[a1,b1],
(21)(Thζ)(t)=∫01‍k(t,s)h(s)ζ(s)ds⩾∫a1b1‍k(t,s)h(s)ζ(s)ds>0.
So there exists a constantc>0 such that c(Thζ)(t)⩾ζ(t) for all t∈[0,1]. From Lemma 4, we have that the spectral radius r(Th)≠0 and Th has a positive eigenfunction corresponding to its first eigenvalue λ1h=(r(Th))-1.
(ii) To proveTh(P)⊂Q, we only need to show
(22)min⁡t∈[a,b](Thv)(t)⩾min⁡{a,1-b}∥Thv∥forv∈P.
In fact, for everyv∈P, from k(t,s)⩽k(s,s)=s(1-s) for t,s∈[0,1], we have
(23)(Thv)(t)=∫01‍k(t,s)h(s)v(s)ds⩽∫01‍s(1-s)h(s)v(s)dsfort∈[0,1],
so
(24)∥Thv∥⩽∫01‍s(1-s)h(s)v(s)ds∀v∈P.
Notice that, fort∈[a,b],
(25)k(t,s)={s(1-t)⩾s(1-b),s⩽t;t(1-s)⩾a(1-s),t⩽s;
thus,
(26)k(t,s)⩾min⁡⁡{a,1-b}s(1-s)for(t,s)∈[a,b]×[0,1].
It follows from (24) and (26) that, for all v∈P,
(27)(Thv)(t)=∫01‍k(t,s)h(s)v(s)ds⩾min⁡⁡{a,1-b}∫01‍s(1-s)h(s)v(s)ds⩾min⁡⁡{a,1-b}∥Thv∥fort∈[a,b].
So (22) holds; thus, Th maps P into Q.
(iii) Sinceφ1h is a positive eigenfunction of Th, we know from the maximum principle (see [18]) that φ1h(t)>0 for all t∈(0,1). Note that k(0,s)=k(1,s)≡0 for s∈[0,1]; we have φ1h(0)=φ1h(1)=0. This implies that φ1h′(0)>0 and φ1h′(1)<0 (see [18]). Define a function Φh on [0,1] by
(28)Φh(s)={φ1h′(0),s=0,φ1h(s)s(1-s),s∈(0,1),-φ1h′(1),s=1.
Then it is easy to see thatΦh is continuous on [0,1] and Φh(s)>0 for all s∈[0,1]. So, there exist δ1,δ2>0 such that δ1⩽Φh(s)⩽δ2 for all s∈[0,1]. Thus
(29)δ1k(t,s)⩽δ1s(1-s)⩽φ1h(s),⩽δ2s(1-s)=δ2k(s,s),
for all t,s∈[0,1].
(iv) From (14), Jh(v)=∫01‍h(t)φ1h(t)v(t)ds⩽δ2∫01‍t(1-t)h(t)v(t)dt<+∞ for all v∈E. So J:E→ℝ1 is well defined. For v∈E,
(30)Jh(Thv)=∫01‍h(t)φ1h(t)(∫01‍k(t,s)h(s)v(s)ds)dt=∫01‍h(s)v(s)(∫01‍k(s,t)h(t)φ1h(t)dt)ds=∫01‍h(s)v(s)(r1hφ1h(s))ds=λ1h-1Jh(v).
(v) It is easy to verify thatP0 is a cone in E. It follows from (14) and (30) that
(31)Jh(Thv)=λ1h-1∫01‍h(s)φ1h(s)v(s)ds⩾λ1h-1δ1∫01‍h(s)k(t,s)v(s)ds=λ1h-1δ1(Thv)(t)forv∈P.
SoJh(Thv)⩾λ1h-1δ1∥Thv∥ for all v∈P. The proof is completed.Denote(32)M1=(min⁡t∈[a,b]∫ab‍k(t,s)ds)-1,λ=(max⁡t∈[0,1]∫ab‍k(t,s)ds)-1.We list some conditions as follows which will be useful in this section.(H1)
g∈C((0,1)×ℝ+,ℝ+) and for any M>0 there exists a function hM∈H such that
(33)g(t,v)⩽hM(t)∀(t,v)∈(0,1)×[0,M];(H2)
there exists a functionh∈H such that
(34)limsup⁡v→0+g(t,v)h(t)v<λ1huniformlyw.r.t.t∈(0,1);(H3)
there exists a functionh∈H such that
(35)limsup⁡v→+∞g(t,v)h(t)v<λ1huniformlyw.r.t.t∈(0,1);(H4)
liminf⁡v→0+⁡min⁡t∈[a,b]g(t,v)/v>M1;(H5)
liminf⁡v→+∞min⁡t∈[a,b]g(t,v)/v>M1;(H6)
there exists a numberl>0 such that
(36)g(t,v)>λlfor(t,v)∈[a,b]×[min⁡{a,1-b}l,l];(H7)
there exists a functionh∈H such that
(37)liminf⁡v→0+g(t,v)h(t)v>λ1huniformlyw.r.t.t∈(0,1);(H8)
there existh∈H with h(t)≢0 for t∈[a,b] and q∈C(ℝ+,ℝ+) such that
(38)g(t,v)⩾h(t)q(v)∀(t,v)∈(0,1)×ℝ+,(39)liminf⁡v→+∞q(v)v>λ1h.Lemma 5.
Assume (H1) holds. Then A:Q→Q is a completely continuous operator.Proof.
The proof is similar to that of Lemma 3.1 in [4], so we only sketch it. Under (H1), A is well defined and for every v∈Q, Av is nonnegative and continuous on [0,1]. Note the property of k(t,s); it is easy to see that A(Q)⊂Q. (H1) and Lebesgue's dominated convergence theorem ensure the continuity of A. Finally, by using Ascoli-Arzela theorem, we can prove that A is completely continuous.Lemma 6.
Assume (H1) holds.(i)
If (H2) is satisfied, then i(A,Q∩Ωr,Q)=1 for sufficiently small positive number r;(ii)
if (H3) is satisfied, then i(A,Q∩ΩR,Q)=1 for sufficiently large positive number R;(iii)
if (H4) is satisfied, then i(A,Q∩Ωr,Q)=0 for sufficiently small positive number r;(iv)
if (H5) is satisfied, then i(A,Q∩ΩR,Q)=0 for sufficiently large positive number R;(v)
if (H6) is satisfied, then i(A,Q∩Ωl,Q)=0;(vi)
if (H7) is satisfied, then i(A,Q∩Ωr,Q)=0 for sufficiently small positive number r;(vii)
if (H8) are satisfied, then i(A,Q∩ΩR,Q)=0 for sufficiently large positive number R.Proof.
(i) By (H2), there exists r>0 such that
(40)g(t,v)⩽λ1hh(t)v∀(t,v)∈(0,1)×[0,r].
DefineShv=λ1hThv for v∈E; then Sh:E→E is a bounded linear operator with Sh(P)⊂Q and the spectral radial r(Sh)=1. For every v∈Q∩∂Ωr, it follows from (40) that
(41)(Av)(t)=∫01‍k(t,s)g(s,v(s))ds⩽λ1h∫01‍k(t,s)h(s)v(s)ds=λ1h(Thv)(t)=(Shv)(t)fort∈[0,1].
So(42)Av⩽Shv∀v∈Q∩∂Ωr.
If there existv1∈Q∩∂Ωr and μ1∈[0,1] such that v1=μ1Av1, then it is easy to see that μ1∈(0,1). Thus τ1=μ1-1>1 and τ1v1=Av1⩽Shv1. By induction, we have τ1nv1⩽Shnv1, n=1,2,…. Then τ1nv1=Shnv1⩽∥Sh∥∥v1∥ and taking the supremum on [0,1] gives τ1n⩽∥Shn∥. By the spectral radius formula, we have
(43)r(Sh)=lim⁡n→∞∥Shn∥n⩾τ1>1,
which is a contradiction. According to the homotopy property invariance of fixed point index, we have i(A,Q∩Ωr,Q)=1.
(ii) By (H3), there exist η>0 and ε0∈(0,1) such that
(44)g(t,v)⩽ε0λ1hh(t)v∀(t,v)∈(0,1)×[η,+∞).
From (H1), there is hη∈H such that g(t,v)⩽hη(t) for all (t,v)∈(0,1)×[0,η]. Hence,
(45)g(t,v)⩽ε0λ1hh(t)v+hη(t)∀(t,v)∈(0,1)×[0,∞).
DefineShv=ε0λ1hThv for v∈E; then Sh:E→E is a bounded linear operator and Sh(P)⊂Q. Let C1=∫01‍t(1-t)hη(t)dt<+∞. Set
(46)W={v∈Q:v=μAv,μ∈[0,1]}.
Next we prove thatW is bounded. For any v∈W, from (45), we have
(47)v(t)=μ(Av)(t)⩽(Av)(t)=∫01‍k(t,s)g(s,v(s))ds⩽ε0λ1h∫01‍k(t,s)h(s)ds+∫01‍k(t,s)hη(s)ds⩽ε0λ1h(Thv)(t)+C1=(Shv)(t)+C1,t∈[0,1].
Thus(48)((I-Sh)v)(t)⩽C1∀v∈W,t∈[0,1].
Sinceλ1h is the first eigenvalue of Th and ε0∈(0,1), the first eigenvalue of Sh, r(Sh)-1>1. Therefore, the inverse operator (I-Sh)-1 exists and
(49)(I-Sh)-1=I+Sh+Sh2+⋯+Shn+⋯.
It follows fromTh(P)⊂Q that (I-Sh)-1(P)⊂Q. Hence, we have from (48) that
(50)v(t)⩽(I-Sh)-1C1∀v∈W,t∈[0,1].
That is,W is bounded. Choose R>{η,sup⁡W}; then v≠μAv for all μ∈[0,1] and v∈Q∩ΩR. By the homotopy invariance property of fixed point index, we have i(A,Q∩ΩR,Q)=1.
(iii)–(v) have been proved in [4], so we skip it.
(vi) By (H7), there exists r>0 such that
(51)g(t,v)⩾λ1hh(t)v∀(t,v)∈(0,1)×[0,r].
For anyv∈Q∩Ωr, we have
(52)(Av)(t)⩾λ1h∫01‍k(t,s)h(s)v(s)ds=λ1h(Thv)(t)fort∈[0,1].
Without loss of generality, we can suppose thatA has no fixed point on Q∩∂Ωr. Suppose that there exist v1∈Q∩∂Ωr and μ1⩾0 such that v1=Av1+μ1φ1h. Then μ1>0 and v1=Av1+μ1φ1h⩾μ1φ1h. Let
(53)μ*=sup⁡{μ>0:v1⩾μφ1h}.
Thenμ*⩾μ1>0 and v1⩾μ*φ1h. Since Th is a positive linear operator, we have
(54)λ1hThv1⩾μ*λ1hThφ1h=μ*φ1h.
Hence, by (52) we have
(55)v1=Av1+μ1φ1h⩾λ1hThv1+μ1φ1h⩾μ*φ1h+μ1φ1h,
which contradicts the definition of μ*. Thus according to the property of omitting a direction for fixed point index, we have i(A,Q∩Ωr,Q)=0.
(vii) From (39), there exist η>0 and ε0∈(0,1) such that
(56)q(v)⩾λ1h(1+ε0)v∀v∈[η,+∞).
Sinceq is bounded on [0,η], there is a constant C2>0 such that
(57)q(v)⩾λ1h(1+ε0)v-C2∀v∈[0,η].
Thusq(v)⩾λ1h(1+ε0)v-C2 for all v∈[0,+∞). Hence, by (38) we have
(58)g(t,v)⩾λ1h(1+ε0)h(t)v-C2h(t)
for all (t,v)∈(0,1)×[0,+∞). Let C3=C2∫01‍h(t)φ1h(t)(∫01‍k(t,s)h(s)ds)dt; then C3>0 is a finite constant. Take
(59)R>C3(ε0min⁡{a,1-b}∫ab‍h(t)φ1h(t)dt)-1.
Suppose that there existv1∈Q∩∂ΩR and μ1⩾0 such that v1=Av1+μ1φ1h; then
(60)Jh(v1)=J(Av1)+μ1J(φ1h)⩾J(Av1)⩾∫01‍h(t)φ1h(t)(λ1h(1+ε0)∫01‍k(t,s)h(s)v1(s)dsmmmmmmmmmmmm-C2Th(1)∫01‍)dt=λ1h(1+ε0)Jh(Thv1)-C3=(1+ε0)Jh(v1)-C3.
Hence,(61)Jh(v1)⩽C3ε0-1.
On the other hand,(62)Jh(v1)=∫01‍h(t)φ1hv1(t)dt⩾∫ab‍h(t)φ1hv1(t)dt⩾Rmin⁡{a,1-b}∫ab‍h(t)φ1h(t)dt.
By the maximum principle,φ1h(t)>0 for all t∈(0,1). By h(t)≢0 for t∈[a,b], we have ∫ab‍h(t)φ1h(t)dt>0. Thus from (62) and (61), we have
(63)R⩽(min⁡{a,1-b}∫ab‍h(t)φ1h(t)dt)-1Jh(v1)⩽C3(ε0min⁡{a,1-b}∫ab‍h(t)φ1h(t)dt)-1.
This is a contradiction. So, by the property of omitting a direction for fixed point index, we havei(A,Q∩ΩR,Q)=0. The proof is completed.Now, we are ready to state our main results of this section.Theorem 7.
Assume that (H1), (H2), (H3), and (H6) hold; then the singular BVP (7) has at least two positive solutions.Proof.
According to Lemma6, we can choose sufficiently small positive number r and sufficiently large positive number R satisfying 0<r<l<R, i(A,P∩Ωr,P)=1, and i(A,P∩ΩR,P)=1. From i(A,P∩Ωl,P)=0 and the additivity property of the fix point index, we obtain
(64)i(A,P∩(Ωl∖Ω¯r),P)=0-1=-1,i(A,P∩(ΩR∖Ω¯l),P)=1-0=1.
Hence A has at least two fixed points, one in Ωl∖Ω¯r and another in ΩR∖Ω¯l. That is, the singular BVP (7) has at least two positive solutions. The proof is completed.Theorem 8.
If (H1) and one of the following conditions are satisfied, then the singular BVP (7) has at least one positive solution.(i)
(H2) and (H5) hold;(ii)
(H2) and (H6) hold;(iii)
(H2) and (H8) hold;(iv)
(H3) and (H4) hold;(v)
(H3) and (H6) hold;(vi)
(H3) and (H7) hold.Proof.
By the properties of the fixed point index, we only need to choose suitable positive numbersr and R. This completes the proof.Remark 9.
The following conditions are little stronger than (H1), (H2), and (H3), respectively. And they are somewhat easy to verify. In fact, (H1*) is a key condition in [19].(H1*)
g∈C((0,1)×ℝ+,ℝ+) and there exist h∈H and ψ∈C(ℝ+,ℝ+) such that g(t,v)⩽h(t)ψ(v) for all (t,v)∈(0,1)×ℝ+;(H2*)
limsup⁡v→0+ψ(v)/v<λ1h;(H3*)
limsup⁡v→∞ψ(v)/v<λ1h.
If (H1*) holds, then, for any M>0, let hM(t)=h(t)max⁡v∈[0,M]ψ(v); then we have g(t,v)⩽h(t)ψ(v)⩽hM(t) for all (t,v)∈(0,1)×[0,M] and hM∈H. Consequently, (H1) holds. From g(t,v)/h(t)v⩽h(t)ψ(v)/h(t)v=ψ(v)/v, it is easy to see that if (H2*) or (H3*) holds, then (H2) or (H3) holds, respectively.Remark 10.
In [11], Stańczy established a one-solution theorem for the singular problem (7). In addition to (H1), the key conditions imposed on the nonlinear term g are the following ones:(A2)
there exists a setA⊂(0,1) of positive measures such that
(65)lim⁡v→+∞g(t,v)v=∞uniformlyw.r.t.t∈A;(A3)
there exists a functionh∈H such that
(66)lim⁡v→0+g(t,v)h(t)v=0uniformlyw.r.t.t∈(0,1).
It is easy to see that (A2) is equivalent to the following condition:(A2′)
there exists a set[a,b]⊂(0,1) such that
(67)lim⁡v→+∞min⁡t∈[a,b]g(t,v)v=∞.
In a recent paper [4], Han and Wang improved the results in [11] by substituting the following conditions for the above ones:(HW5)liminf⁡v→∞min⁡t∈[a,b]⁡g(t,v)/v>M1;(HW2) there exists a functionh∈H such that
(68)limsup⁡v→0+g(t,v)h(t)v<mhuniformlyw.r.t.t∈(0,1),where mh=(max⁡t∈[0,1]∫01‍k(t,s)h(s)ds)-1 for h∈H.
Additionally, they obtained a twin-solution theorem for the singular BVP (7) by using (HW2), (H6), and(HW3) there exists a functionh∈H such that
(69)limsup⁡v→+∞g(t,v)h(t)v<mhuniformlyw.r.t.t∈(0,1).
Sincemh⩽λ1h, (H2) is an improvement of (HW2). In fact, without loss of generality, suppose φ1h, the positive eigenfunction corresponding to λ1h, satisfies ∥φ1h∥=φ1h(t0)=1; then
(70)1=φ1h(t0)=λ1h∫01‍k(t0,s)h(s)φ1h(s)ds⩽λ1h∫01‍k(t0,s)h(s)ds⩽λ1hmax⁡t∈[0,1]∫01‍k(t,s)h(s)ds.
Somh⩽λ1h and then (HW2) implies (H2). Therefore, Theorems 7 and 8 are essential improvements of Theorems 3.1, 3.2 in [4] and Theorem 2.2 in [11]. Furthermore, we give the new conditions (H7) and (H8). These improvements allow us to deal with more singular problems.Remark 11.
For the following singular BVP with general boundary conditions,(71)-v′′(t)=g(t,v(t))fort∈(0,1),αv(0)-βv′(0)=0,γv(1)+δv′(1)=0,
where α,β,γ, and δ⩾0 and ρ=γβ+αγ+αδ>0, our results still hold. In fact, the Green function
(72)G(t,s)={1ρ(γ+δ-γt)(β+αs),0⩽s⩽t⩽1,1ρ(γ+δ-γs)(β+αt),0⩽t⩽s⩽1,
has analogous properties as k(t,s) is defined in (10) (see [4, Remark 3.3]). Hence, Theorems 7, 8 can be generalized to the singular problem (71) without any essential difficulty. See [4, 19, 20] for details.For the following singular two-point BVP whose variables of nonlinear term are separated, the hypotheses and results will be more concise:(73)-v′′=h(t)g(v),t∈(0,1),v(0)=v(1)=0.Sinceh is a fixed function, Th, λ1h, and φ1h are confirmed exclusively. So we skip the subscript h in the following. Corresponding  to (H1)–(H8), we formulate the conditions for singular BVP (73):(H1′)
g∈C(ℝ+,ℝ+) and h∈H;(H2′)
limsup⁡v→0+g(v)/v<λ1;(H3′)
limsup⁡v→+∞g(v)/v<λ1;(H4′)
min⁡t∈[a,b]⁡h(t)>0 and liminf⁡v→0+g(v)/v>M1(min⁡t∈[a,b]h(t))-1;(H5′)
min⁡t∈[a,b]⁡h(t)>0 and liminf⁡v→+∞g(v)/v>M1(min⁡t∈[a,b]⁡h(t))-1;(H6′)
there exists a numberl>0 such that
(74)h(t)g(v)>λlfor(t,v)∈[a,b]×[min⁡{a,1-b}l,l];(H7′)
liminf⁡v→0+g(v)/v>λ1;(H8′)
liminf⁡v→+∞g(v)/v>λ1.Remark 12.
Taket1∈[a,b] with φ1(t1)=min⁡t∈[a,b]φ1(t). Then
(75)φ1(t1)=λ1∫01‍k(t1,s)h(s)φ1(s)ds⩾λ1∫ab‍k(t1,s)h(s)φ1(s)ds⩾λ1φ1(t1)∫ab‍k(t1,s)h(s)ds⩾λ1φ1(t1)min⁡t∈[a,b]∫ab‍k(t,s)h(s)ds⩾λ1φ1(t1)min⁡t∈[a,b]h(t)min⁡t∈[a,b]∫ab‍k(t,s)ds.
Notice thatφ1(t1),min⁡t∈[a,b]h(t), and min⁡t∈[a,b]∫ab‍k(t,s)ds>0; we have
(76)λ1⩽M1(min⁡t∈[a,b]h(t))-1.
So (H4′) implies (H7′) and (H5′) implies (H8′).Remark 13.
Observe that the condition(77)h(t)≢0fort∈[a,b]
is not contained in (H8′). In fact, the cone Q will be replaced by the cone P0 which is defined in (15) as we consider one solution by (H8′). See the proof of Theorem 15 in the following.Theorem 14.
Assume that (H1′), (H2′), (H3′), and (H6′) hold; then the singular BVP (73) has at least two positive solutions.Proof.
This theorem is a direct corollary of Theorem7.Theorem 15.
If (H1′)  and one of the following conditions are satisfied, then the singular BVP (7) has at least one positive solution.(i)
(H2′) and (H5′) hold;(ii)
(H2′) and (H6′) hold;(iii)
(H2′) and (H8′) hold;(iv)
(H3′) and (H4′) hold;(v)
(H3′) and (H6′) hold;(vi)
(H3′) and (H7′) hold.Proof.
(i), (ii), and (iv)–(vi) are direct corollaries of Theorem7. Next, we prove (iii). In contrast to (H8), (H8′) does not contain the condition that h≢0 for t∈[a,b].
(iii) SinceT(P)⊂P0 is completely continuous, we know that A(P)⊂P0 is also completely continuous. Similar to item (i) of Lemma 6, by (H2′), it is not difficult to prove i(A,P0∩Ωr,P0)=1 for sufficiently small positive number r. By (H8′), there exist η>0 and ε0∈(0,1) such that
(78)g(v)⩾λ1(1+ε0)v∀v∈[η,+∞).
Sinceg is bounded on [0,η], there is a constant C4>0 such that g(v)⩾λ1(1+ε0)v-C4 for all v∈[0,η]. Thus
(79)g(v)⩾λ1(1+ε0)v-C4∀v∈[0,+∞).
LetC5=C4∫01‍h(t)φ1(t)(∫01‍k(t,s)h(s)ds)dt; then C5>0 is a finite constant. Take
(80)R>max⁡{r,C5λ1δ1-1ε0-1}.
Suppose that there existv1∈P0∩∂ΩR and μ1⩾0 such that v1=Av1+μ1φ1; then from (79) we have
(81)J(v1)=J(Av1)+μ1J(φ1)⩾J(Av1)⩾∫01‍h(t)φ1(t)(λ1(1+ε0)∫01‍k(t,s)h(s)v1(s)dsmmmmmmmmm-C4T(1)∫01‍)dt=λ1(1+ε0)J(Tv1)-C5=(1+ε0)J(v1)-C5.
Hence,(82)J(v1)⩽C5ε0-1.
On the other hand, sincev1∈P0∩∂ΩR, we have
(83)J(v1)⩾λ1-1δ1∥v1∥=λ1-1δ1R.
Thus from (82) and (83), we have
(84)R⩽λ1δ1-1J(v1)⩽C5λ1δ1-1ε0-1.
This contradicts with (80). By the property of omitting a direction for fixed point index, we have i(A,P0∩ΩR,P0)=0. So i(A,P0∩(ΩR∖Ω¯r))=0-1=-1. Thus A has a fixed point in ΩR∖Ω¯r; that is, the singular BVP (73) has at least one positive solution. The proof is completed.Remark 16.
Lan and Webb have studied BVP (73) in [21]. Their key conditions are(i)
h∈L1(0,1), h⩾0 a.e. on [0,1] and g∈C(ℝ+,ℝ+);(ii)
One of the following conditions holds:
(h1)
limsup⁡v→0+g(v)/v<α and β<liminf⁡v→+∞g(v)/v;(h2)
limsup⁡v→+∞g(v)/v<α and β<liminf⁡v→0+g(v)/v, where α=mh and β=(min⁡t∈[a,b]∫ab‍k(t,s)h(s)ds)-1.
Under the above condition (i), we still can prove Lemma2 (see Theorem 2.1 in [21]). By (70) and (75), we have α⩽λ1h⩽β. Therefore, the items (iii) and (vi) of Theorem 15 extend the main results in [21] essentially. Furthermore, our conditions (H2′), (H3′), (H7′), and (H8′) cannot be improved anymore.At the end of this section, we present three simple examples to which our theorems can be applied, respectively. We choose[a,b]=[1/4,3/4].Example 17.
Let(85)g(t,v)={1t(1-t)(cvt+1),v∈[0,18l],1t(1-t)×[(88-c)(v-18l)t+18clt+1],v∈[18l,14l],1t(1-t)(11lt+1),v∈[14l,l],1t(1-t)[c(v-l)+11lt+1],v∈[l,+∞],
where c>0 is a constant. Obviously, g(t,v)⩽h(t)ψ(v) for all (t,v)∈(0,1)×ℝ+, where h(t)=t-1(1-t)-1 and
(86)ψ(v)={cv+2,v∈[0,18l],(88-c)(v-18l)+18cl+2,v∈[18l,14l],11l+2,v∈[14l,l],c(v-l)+11l+2,v∈[l,+∞].
Sinceλ=32/3<11, if lim⁡v→0+ψ(v)/v=c<λ1h and lim⁡v→+∞ψ(v)/v=c<λ1h, then g satisfies all the conditions of Theorem 7; thus, we infer that the singular BVP (7) has at least two positive solutions. Furthermore, if
(87)mh=(max⁡t∈[0,1]∫01‍k(t,s)h(s)ds)-1=(ln⁡2)-1≈1.443<c<λ1h,
then Theorem 3.1 in [4] is invalid for this example.Example 18.
Let(88)g(t,v)=11-tcv+12t(1-t)v1/2+11-t=1t(1-t)(tcv+12v1/2+t)⩽1t(1-t)(cv+12v1/2+1)≜h(t)ψ(v),
where 0<c<λ1h. Since lim⁡v→0+ψ(v)/v=+∞>λ1h and lim⁡v→+∞ψ(v)/v=c<λ1h, the item (iv) of Theorem 8 implies that singular BVP (7) has at least one positive solution.Example 19.
Let(89)g(t,v)=h(t)ψ(v)=tp-1(1-t)q-1∑i=1n‍civi,
where p,q∈(0,1), ai>0 for i=1,2,…,n, and a1<λ1h. Since lim⁡v→0+ψ(v)/v=a1<λ1h and lim⁡v→+∞ψ(v)/v=+∞>λ1h, the item (iii) of Theorem 15 ensures that the singular BVP (73) has at least one positive solution.


## 3. Positive Radial Solutions of Elliptic Boundary Value Problems

Define a set(90)K={∫1∞‍p∈C(1,∞,ℝ+):p≢0,∫1∞‍s(1-s2-n)p(s)ds<+∞}.Denotec=(1-a)1/(2-n) and d=(1-b)1/(2-n). For p∈K, let
(91)h(t)=1(n-2)2s2n-2p((1-t)1/(2-n)).Then we haveh∈H. As in (13) and Lemma 2, h confirms an operator Th and its first eigenvalue λ1h. To emphasize their relation with p, we use the notations hp, λ1hp, and φ1hp.According to (8), we formulate the following conditions which correspond to those in Section 2.(C1)
f∈C((1,∞)×ℝ+,ℝ+) and for any M>0 there exists a function pM∈K such that
(92)f(s,u)⩽pM(s)∀(s,u)∈(1,∞)×[0,M];(C2)
there exists a functionp∈K such that
(93)limsup⁡u→0+f(s,u)p(s)u<λ1hpuniformlyw.r.t.s∈(1,∞);(C3)
there exists a functionp∈K such that
(94)limsup⁡u→+∞f(s,u)p(s)u<λ1hpuniformlyw.r.t.s∈(1,∞);(C4)
liminf⁡u→0+min⁡s∈[c,d]⁡f(s,u)/u>c2-2n(n-2)2M1;(C5)
liminf⁡u→+∞min⁡s∈[c,d]⁡f(s,u)/u>c2-2n(n-2)2M1;(C6)
there exists a numberl>0 such that
(95)f(s,u)>(n-2)2λlfor(s,u)∈[c,d]×[min⁡{a,1-b}l,l];(C7)
there exists a functionp∈K such that
(96)liminf⁡u→0+f(s,u)p(s)u>λ1hpuniformlyw.r.t.s∈(1,∞);(C8)
there existp∈K with p(s)≢0 for s∈(c,d) and q∈C(ℝ+,ℝ+) such that
(97)f(s,u)⩾p(s)q(u)∀(s,u)∈(1,+∞)×ℝ+,liminf⁡u→+∞q(u)u>λ1hp.Now we are ready to state our main results for the elliptic BVP (1).Theorem 20.
If (C1), (C2), (C3), and (C6) are satisfied, then the elliptic BVP (1) has at least two positive radial solutions.Theorem 21.
If (C1) and one of the following conditions are satisfied, then the elliptic BVP (1) has at least one positive radial solution:(i)
(C2) and (C5) hold;(ii)
(C2) and (C6) hold;(iii)
(C2) and (C8) hold;(iv)
(C3) and (C4) hold;(v)
(C3) and (C6) hold;(vi)
(C3) and (C7) hold.Remark 22.
According to Remark10, we know
(98)mp=(n-2)(max⁡τ∈[0,1]∫1∞‍k(τ,1-s2-n)sn-1p(s)ds)-1<λ1hpforp∈K.
Therefore, Theorems20 and 21 are essential improvements of Theorems 4.1 and 4.2 in [4], respectively.Remark 23.
For the following BVP whose nonlinear terms are variable-separated, it is easy to give more concise theorems as Theorems14 and 15; we skip them here:
(99)-Δu=p(|x|)f(u)for|x|>1,x∈ℝn,n⩾3,u=0for|x|=1,u⟶0as|x|⟶+∞,
where p∈C((1,∞),ℝ+) and f∈C(ℝ+,ℝ+).Remark 24.
One can see that, through variables substitution, it is easy to give some examples like in Section3, so we skip it.



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