ON POSITIVE SOLUTION TO A SECOND ORDER ELLIPTIC EQUATION WITH A SINGULAR NONLINEARITY

. We consider a second order elliptic equation with measurable bounded coeﬃcients where σ > 0, s is any real number, and Ω ⊂ R n , n ≥ 3 is a bounded domain, which contains the origin O . The aim of this paper is to establish existence, nonexistence and behavior of positive weak solutions near the isolated singularity O .

1. Introduction. Semi-linear elliptic equations has been a subject of many researches lately, and it is well known that properties of equation depend essentially on the character of nonlinear terms. This paper is devoted to a second order elliptic equation where Ω ⊂ R n , n ≥ 3 is a bounded domain, which contains the origin O. We assume σ > 0, s ∈ R, all coefficients to be measurable bounded functions, 0 < C 1 < p(x) < C 2 , a ij (x) = a ji (x), and the uniformly elliptic condition is satisfied: where λ > 0, ∀ζ ∈ R n a.e. in Ω.
It is easy to see that a nonlinear term becomes singular at O when s is negative, and also as u → 0. We will consider only positive weak solutions of the equation, which we define below. Definition 1.1. We say that a function u(x) ∈ W 1 2,loc (Ω\{O}), such that u(x) −1 ∈ L ∞,loc (Ω \ {O}), and u(x) > 0 a. e. is a weak solution to (1.1) if ∀δ > 0 and The main purpose of this work is to find existence and nonexistence conditions for a positive weak solution near an isolated singular point, and also to study the behavior of the solution in the neighborhood of this point.
The model equation can be written in the form It was considered earlier by some authors from the point of view of existence of entire positive solution (see, for instance, [1], [2], the list is not complete). In particular, for n ≥ 3, the existence of a solution u(x) ∈ C 2,α loc (R n ) such that was proved under the conditions Their methods of proofs used essentially the presents of the Laplacian in the equation.
Recently an existence and nonexistence problem was studied in [4] for a general equation (1.1). Solutions of the equation were considered in an outer domain, and the results of [4] correspond with those we present here near an isolated singular point. It can be verified by using a generalization of Kelvin's transformation for (1.1) [7]. However the methods of proofs are different. Also we study a behavior of a positive solution near the singularity.
In the following text we denote Lu ≡ (a ij (x)u xi ) xj .
Proof. First we make a replacement This follows from Keller-Osserman's type of estimate proved in [3] for s = 0, which can be extended without a problem for any real s.
and C 0 = α γ 0 . Therefore the lemma is proved. with There is a singularity of the solution at O.
where the constant C > 0 does not depend on u.

Proof.
Without loss of generality we suppose that B 1 (0) ⊂ Ω. Let us denote m ρ = inf |x|=ρ u(x), and rewrite the equation as the following The last estimate follows from Lemma 2.1. Then from Harnack inequality we have where the positive constant K doesn't depend on ρ.
Let Γ(x) be a fundamental solution, such that Such Γ(x) exists and satisfies the inequalities with some positive constants C 1 and C 2 [7]. Then Suppose ρ < 1/2, and fix now |x| = 1/2, then where the constant C 3 does not depend on ρ. Using again Harnack inequality we complete the proof of the lemma: Remark 2.2. From Harnack inequality it follows that m ρ = inf |x|=ρ u(x) and M ρ = sup |x|=ρ u(x) must have the same rate of growth as ρ → 0.
If there exits a solution u(x) then u → ∞ as |x| → 0. Integrating the equation (1.1) in D = {x : 1 < u < M } by part, we get The first term in the right hand side is fixed, the second is equivalent to a divergent integral as M → ∞, and then the right hand side is negative while the left hand side is positive, which is impossible. Therefore there is no solution, and the theorem is proved.
Then Γ(x) is a subsolution for (1.1) Let us consider a linear equation where α > 0 will be chosen later. This equation possesses a solution Γ α (x) such that [8]. Then using the estimate 2 − n < s+2 1+σ we obtain Hence, Γ α is a super-solution for (1.1), and therefore exists a positive solution u in Finally, C 1 |x| 2−n ≤ u(x) ≤ K α |x| 2−n , and taking K 1 = C 1 , and K 2 = K α we complete the proof.
4. Properties of solutions in the case 2 + s ≥ 0.
Theorem 4.1. Let s + 2 > 0, u(x) is a positive solution to (1.1) in Ω \ {O}. Then either u is a solution of (1.1) in Ω and u(0) > 0 or Proof. It follows from Theorem 3.1 that there exists a solution u(x) such that The function v 1 is a subsolution for the equation (1.1) Let's consider v 2 : There is a solution v 2 > 0 and v 2 ∈ C(B 1 (0)), since 2 + s > 0 [5]. From (4.1) and (4.2) we obtain So, v 2 is a supersolution for (1.1) And therefore, there exists a solution v > 0 of the problem

From (1.1) and (4.3) we have
There is a fundamental solution Γ(x) [6]: Then, since u(x) = o(|x| 2−n ) as x → 0, ∀ǫ > 0 there exists a ball B δ (0), such that Hence, taking in the account that we conclude from the maximum principle that Taking now ǫ → 0 we obtain that w(x 0 ) = 0, which leads to a contradiction. Then we have w ≡ 0 in B 1 \ {O}, and can put u(0) = v(0), where v(0) > 0. The theorem is proved.
Hence u ≥ C 0 , and v(x) can be considered as a solution of the equation where G(x, y) is the Green function for our problem. It is known that G(x, y) < 0 and there are constants C 1 , C 2 > 0 such that After the scaling y = 2 k x we get the same equation with respect to y, and then by applying Lemma 4.1 we get that u(x) > C 0 + δ x ∈ B 1 (0) \ {O}.
Continuing the procedure, we obtain u(x) > C 0 + 2 k δ for |x| = 2 −k (4.4) does not imply u(x) → ∞ as |x| → 0. However, the desired contradiction is clear since u was assumed to be bounded while from (3.4) it follows the contrary. Theorem 4.2 is proved.