Positive Solutions Of A Singular Dirichlet Problem To Nonlinear Elliptic Equations: Existence, Uniqueness And Boundary Behavior

In this paper,we consider the existence,unqueness,the boundary behavior of the positive solution of a singular Dirichlet problem to nonlinear elliptic equations-△u=q(x)g(u)+p(x)f(u),u>0,x∈Ω,u|(?)Ω=0,(1.1) Where Ω is a bounded domain with smooth boundary in RN. q,p satisfy(A1)q,p∈Cαloc(Ω),for some α∈(0,1),q is positive in Ω,p is non-negative in Ω；(A2)Poisson problem-△v1=q(x),v1>0,x∈Ω,v1|(?)Ω=0, and-△v2=p(x),v2>0,x∈Ω,v2|(?)Ω=0, have the unique solutions v1,v2∈C2+α(Ω)∩C(Ω). and g,f satisfy(g1)9∈C1((0,∞),(0,∞)), and g is decreasing on(0,∞)；(f1)f∈C∝loc.([0,∞),[0,∞)),f is increasing on [0,∞)(or(f01)f∈C1((0,∞),(0,∞)), f is decreasing on(0,∞))；(f2)There exists s0>0,satisfy f(s)/s+s0s decreasing on (0,∞),andWe have results as following:Theorem1If g satisfy(g1),f satisfy (f1)(or(f01))and(f2),the problem(1.1)has a unique solution u∈C2+α(Ω)∩C(Ω)only and only if(A1)and(A2)hold.On this condition,we have the following results:Theorem2Let f satisfy(f1),g satisfy (g2)there exists Cq>0,satisfy g satisfy(Q1) with where μ1=μ2=…=μj-1=1,μj>1,μi∈R for j+1≤j≤m. Then for the unique solution u of problem(1.1) where ψ1is the solution of the problem and ζ1=q1/μj-1,ζ2=q2/μj-1In particular,(ⅰ)when Cg=1,u verifies(ⅱ)when Cg<1and q1=q2=q0,u verifies where ζ01=q0/μj-1Theorem3If f satisfy(f1),q satisfy (Q2) and Ck+2Cg>2, then for the unique solution u of problem(1.1) where K(t):=∫t0k(s)dsκ,∈Λ,Λ denote the set of all positive monotonic functions in C1(0,δ0)∩L1(0,δ0)(δ0>0)which satisfy ζ3=q1/2(Ck+2Cg-2),ζ4=q2/2(Ck+2Cg-2) In particular.(ⅰ)when Cg=1,u verifies(ⅱ)when Cg <1and q1=q2=q0,u verifies where ζ02=q0/2(Ck+2Cg-2)Theorem4Let f satisfy (f01) and (f3)there exists Cf≠0,satisfy p satisfy(P1) If Ck+2Cg>2,Ck+2Cf>>2,(ⅰ) When then for the unique solution u of problem (1.1) where ζ3=q1+p0c0/2(Ck+2Cg-2),ζ4=q2+p0c0/2(Ck+2Cg-2)(ⅱ) when then for the unique solution u of problem (1.1) where ψ2is the solution of the problem and p1+qoc0/2(Ck+2Cf-2),ζ4=p2+q0c0/2(Ck+2Cf-2)...