LARGE SOLUTIONS OF SEMILINEAR ELLIPTIC EQUATIONS WITH NONLINEAR GRADIENT TERMS

We show that large positive solutions exist for the equation (P±) :∆u±|∇u|q = p(x)uγ in Ω ⊆ RN(N ≥ 3) for appropriate choices of γ > 1,q > 0 in which the domain Ω is either bounded or equal to RN . The nonnegative function p is continuous and may vanish on large parts of Ω. If Ω = RN , then p must satisfy a decay condition as |x| →∞. For (P+), the decay condition is simply ∫∞ 0 tφ(t)dt <∞, where φ(t)=max|x|=t p(x). For (P−), we require that t2+βφ(t) be bounded above for some positive β. Furthermore, we show that the given conditions on γ and p are nearly optimal for equation (P+) in that no large solutions exist if either γ ≤ 1 or the function p has compact support in Ω.


Introduction.
We consider existence and nonexistence of large solutions of the equation ∆u ±|∇u| q = p(x)u γ (P ±) in which q and γ are positive constants, the function p is nonnegative, and the domain Ω is either bounded with smooth boundary or equal to R N .A solution u(x) of (P ±) is called a large solution if u → ∞ as x → ∂Ω.In the case Ω = R N , x → ∂Ω means |x| → ∞ and such solutions are called entire large solutions.Large solutions of semilinear elliptic equations have been studied for decades.Almost all such studies have dealt with equations of the form ∆u = g(x, u) (1.1) in which the function g takes various forms (see [1,2,6,8,10,12] and their references).
Except for [2] and [10], all of these have restricted their attention to bounded domains and functions g which are strictly positive when u > 0.
In [2], Bandle and Marcus proved the existence of large solutions when g(x, u) = p(x)u γ in which p is allowed to be zero at finitely many places in Ω. Lair and Shaker [10] proved the existence of large solutions in bounded domains and entire large solutions in R N for g(x, u) = p(x)f (u), allowing p to be zero on large parts of Ω.
A few authors considered large solutions of semilinear equations containing nonlinear gradient terms (see [1,7,11]).Motivation for the present study stems from the work of Bandle and Giarrusso [1] who developed existence and asymptotic behavior results for large solutions of ∆u ±|∇u| q = f (u) (1.2) on a bounded domain.Our goal here is to develop comprehensive (and nearly optimal) existence results for (P +) when Ω is either bounded or equal to R N , and at the same time develop somewhat comparable results for problem (P −).In particular, for Ω bounded, we show that problem (P +) has a positive large solution if γ and q satisfy γ > max{1,q}, (1.3) and p satisfies the following condition.
Condition (P ).p(x) ≥ 0, ∀x ∈ Ω; p(x) ∈ C( Ω); if z ∈ Ω and p(z) = 0, then there exists a domain D z containing z for which D z ⊆ Ω and p(x) > 0 for all x ∈ ∂D z .Thus, p is allowed to vanish on significant portion of Ω.Indeed, the function p could be zero on all of Ω except for a small (in measure) open set containing ∂Ω.For problem (P −) with Ω bounded, we do not need inequality (1.3) when we establish the existence of a nonnegative solution (see Theorem 4).
For Ω = R N , we prove the existence of a positive entire large solution if p also satisfies where φ(r )= max |x|=r p(x) (see Theorem 3).For problem (P −), we require the stronger decay condition that r 2+β φ(r ) be bounded above for some β > 0. In addition, restrictions are placed on the relationship between q and γ (see Theorem 5).Furthermore, for both the bounded and unbounded cases, we show that our conditions on γ, q, and p are nearly optimal for (P +) in that if γ ≤ 1 or p has compact support in Ω, then (P +) has no positive large solutions.
We also note that, among the many open problems related to (P ±), the existence of positive large solutions remains unproved for (P +) if 1 < γ ≤ q and for (P −) if the function p is required to satisfy the weaker decay condition (1.4).

Equation (P +).
In this section, we develop a critical boundedness result (Lemma 1), which will prove very useful in developing our existence theorem for bounded domains (Theorem 2).This result, in turn, provides the critical element for the existence proof when Ω = R N .Interestingly, Bandle and Giarrusso [1] had no boundedness result comparable to that of Lemma 1. Indeed, their proof of their main existence result simply assumes that such an upper bound exists.

Lemma 1. Let u n be a solution of the problem
(2.1) Proof.To prove that u n > 0 in Ω, without loss of generality, we let n = 1.We observe from the maximum principle that 0 ≤ u 1 ≤ 1.Furthermore, it is clear that, for any 0 < o < 1, any solution, say z, to the problem (which exists by [9,Thm. 8.3,p. 301]) satisfies z ≤ u 1 and 0 ≤ z ≤ o .Hence, if we show that z > 0 in Ω for some choice of o ∈ (0, 1), we will be done.To do this, let x o ∈ R N \ Ω.We assume, without loss of generality, that x o = 0. Let r = |x| and choose R o > 0 large so that Ω ⊆ B(0,R o ).
Choose M o > 0 so that p(x) < M o on Ω.Now, choose 0 < o < 1 so that Thus, we suppose that max(v − w) in B(0,R o ) is positive.In this case, the point where the maximum occurs must lie in Ω since Therefore, at the point where max(v − w) occurs, we have Now, let be a sufficiently small positive number so that B R+ ⊆ Ω and let v n be a solution of (2.6) A similar argument as above implies that v n > 0 in B R+ .By the maximum principle, it is clear that v n ≤ v n+1 ,n = 1, 2,... .By [4, Thm.A], it is easy to show that v n is a radial solution.Thus, v n satisfies (2.7) It is clear that v n (0) = 0 and v n (r ) ≥ 0 ∀n and ∀r .Equation (2.7) may be rewritten as which may be integrated to get (2.9) Thus, we have (2.11) Let v be a solution of (2.12) The existence of v is justified by [10,Thm. 1].By the maximum principle, v n ≤ v in B R+ for all n.Thus, v n is bounded above on B R by a constant which is independent of n.By (2.11), v n (r ) is also bounded above by a constant independent of n.Let K be an upper bound for both v n and v n on B R .
If we can find a function w n which satisfies where K o is a constant independent of n, then by the maximum principle, we have u n ≤ w n ≤ K o , and we will be done.
Let w n = cv λ n , where v n is a solution of (2.7), the constants c and λ, both independent of n, are determined later.Since to complete the proof, it suffices to find c and λ such that on B R+ for all n, for then we have (2.11), the left side of the above equals (2.16) Now, since γ > q, we can choose λ large so that (2.17) and we may find β > 0 such that v n (r ) ≥ β ∀n and ∀r .For the above choice of λ, choose the constant c ≥ 1 so that the following holds.
Theorem 2. Assume that (1.3) and condition (P ) hold.Suppose that Ω is a bounded domain in R N , N ≥ 3, with smooth boundary.Then, equation (P +) has a large positive solution in Ω.
Proof.By [9, Thm.8.3, p. 301], it is easy to prove that, for each k ∈ N, the boundary value problem has a unique positive classical solution.By the maximum principle it can be shown that We assume, without loss of generality, that x o = 0. Let r = |x|.Then, for some small > 0, v + ε/(1 + r ) has a negative minimum in Ω.At that minimum, we have In what follows, it is understood that the maximum principle is applied as above, where the factor ε/(1 + r ) is used whenever the function p is not strictly positive.
To complete the proof, it suffices to show the following: To prove (C1), we consider two cases.Case (a).p(x o ) > 0. Since p is continuous, there exists a ball B(x o ,r ) such that p(x) ≥ m o in B(x o ,r ) for some m o > 0. (C1) then follows easily from Lemma 1.
Case (b).p(x o ) = 0.By the hypothesis, there exists a domain Ω o ⊆ Ω such that x o ∈ Ω o and p(x) > 0 ∀x ∈ ∂Ω o .From Case (a) above, we know that ∀x ∈ ∂Ω o there exists a ball B(x, r x ) and a positive constant Since Ω is bounded (and hence Ω o is bounded), ∂Ω o is compact.Thus, there exists a finite number of such balls that cover The proof of (C2) is straightforward.For any L > 0 and any sequence x k → x ∈ ∂Ω, since v L+1 = L+1 on ∂Ω and is continuous, there is some To prove (C3), we let x o ∈ Ω and let B(x o ,r ) be the ball of radius r centered at x o such that it is contained in Ω.Let ψ be a C ∞ function which is equal to 1 on B(x o ,r /2) and zero off B(x o ,r ).
Let f (s) = 1/(1 + s).Multiplying both sides of equation (2.21) by ψ 2 f (v k ) and integrating over B(x o ,r ) yields (2.23) Integration by parts produces (2.24) Thus, we have where M r is an upper bound for v k on B(x o ,r ), k = 1, 2,..., is any positive number, and the constants M 1 and M 2 are independent of k.Hence, we get By the standard regularity argument (see [10]), we may find a number r 1 > 0 such that there is a subsequence of {v k } ∞  1 , which we still call {v k } ∞ 1 , that converges in C 1+α (B(x o ,r 1 )) for some positive number α < 1.
Let ψ be as before but with r replaced by r 1 .Now, we consider two cases regarding the regularity of the function p(x).
and is a solution of (P +).

Case 2. p(x) ∈ C(Ω). We have v k s−C(B(xo,r 1 ))v
→ and, consequently, That the Laplacian is a closed linear operator implies that v ∈ D(∆), ∆v = z.Since x o is arbitrary, we have that v is a classical solution of (P +).(1.3), (1.4), and condition (P ) hold.Then equation (P +) has a positive entire large solution.

Ω
Proof.By Theorem 2, for k = 1, 2,..., the boundary blow-up problem has a classical solution.By the maximum principle, it is clear that We claim that v is the desired solution.To prove this, we consider the related problem (2.29) It is shown in [10] that (2.29) has a unique positive solution for each k, and that for some w → ∞ as |x| → ∞.It follows easily from the maximum principle that v k ≥ u k for k = 1, 2 ... .Thus, v(x) → ∞ as |x| → ∞.By a similar argument as (C3) in Theorem 2, we have that v is the desired solution.

Ω is a bounded domain.
The following theorem is our main result for problem (P −) on a bounded domain.Much of the proof is similar to that of Theorem 2 and is, therefore, only outlined.Theorem 4. Suppose that γ > 1, q > 0, condition (P ) holds, and Ω is a bounded domain in R N , N ≥ 3, with smooth boundary.Then equation (P −) has a large nonnegative solution in Ω.
Proof.The only significant difference between this proof and that of Theorem 2 lies in obtaining an upper bound M for the sequence {v k } near x 0 .There, Lemma 1 is needed, but for (P −) the proof is much easier.Indeed, it is easy to prove that v k (x) ≤ w(x) for all x ∈ Ω, where w is a solution of (See [10,Thm. 1]). (2.31) Our main result for equation (P −) is the following theorem.
If there exist positive numbers β and R such that 0 ≤ p(x) ≤ Mr −2−β , whenever r ≡ |x| > R, then equation (P −) has a nonnegative entire large solution provided that max{γ, q} > 2 if q ≥ 1, and To prepare for proving this theorem, we now state and prove three lemmas.Lemma 6.Let M be any nonnegative number and β any positive number.Then, for R sufficiently large, there is a nonnegative solution of the problem Let α be a positive real number whose value will be made precise later.We have (2.33) Requiring that α 1 ≥ 0 and respectively.Thus, for q < 1, we require that (2.36) In this case, we take α = (1 − q)/(2 − q).For q > 2, we take γ > 1 and α ≤ min{β/(γ − 1), (q − 2)/(q − 1)}.Hence, we may choose R large so that L(r α ) ≥ 0. Consequently, w(r ) ≡ r α , where α is as defined above, is the desired solution.
(2.41) 1) , it is sufficient to have This is true since we can choose s large enough so that Hence, we have L(w) ≥ 0.
Lemma 8. Suppose that β > 0 and 0 ≤ p(x) ≤ 2 2+β Mr −2−β for large M. Let w be a nonnegative solution of Then, any solution of for any k > R.
Proof.Suppose that the conclusion is false.That is, suppose that (2.48) Hence, we get (2.49) Hence, we get which contradicts (2.49).Hence, w(x) We now prove Theorem 5.
Proof.By Theorem 4, we have that, for k = 1, 2,..., the boundary blow-up problem has a nonnegative classical solution.By the maximum principle, it is clear that We claim that v is the desired solution.To prove this, we need only to prove that v satisfies (P −) and that v → ∞ as |x| → ∞.
To prove the second statement, it suffices to find a nonnegative lower bound, say w(x), for the sequence {v k } ∞ 1 such that w → ∞ as |x| → ∞.Also, by another standard regularity argument (see [10]), we can show that v is smooth and is a classical solution of (P −).
If q > 2 or q < 1, let w be a solution of (2.44) which we know to exist by Lemma 6, where M is a constant.Then, by Lemma 11,v We conclude that v is a solution of (P −).In particular, if q > 2, then v ≥ w = r (q−2)/(q−1) .
Assume that 1 ≤ q ≤ 2. By Lemma 6, there is a solution, say u, to equation (2.37), where s > 2. Now, let w be a solution of (2.38).It is clear that w solves (2.44) and hence v ≥ w − M o , where M o is defined as in Lemma 8. Again, we get v → ∞ as |x| → ∞, and is a classical solution of (P −).
Proof.We first show that (1.4) implies that Indeed, for any R > 0, we have (3.3)Now, suppose that (P +) has a positive large solution u(x).Using technique similar to that described in [5], we define where v o (s N−1 r ) is the volume of the (N − 1)-dimensional sphere and σ r is the measure on the sphere.We have (By Jensen's Inequality) = φ(r ) ūγ (r ).
(3.5) Thus, we have Integrating the above inequality yields or equivalently, Thus, ū is bounded and hence cannot be a large solution.Consequently, u cannot be a large solution.
Combining Theorems 3 and 9, we get the following corollary.
Corollary 10.Assume that conditions (P ) and (1.4) hold.Let γ > q, then equation (P +) has a positive entire large solution if and only if γ > 1.
Proof.Suppose that (P +) has a positive large solution u(x) in Ω. Assume, without loss of generality, that 0 ∈ Ω.Let B be a ball of radius R centered at 0 and containing Ω such that ∂B ∩ ∂Ω = ∅.Let M = max x∈ Ω p(x).
Let v n be the unique solution of As shown in Lemma 1, v n is radially symmetric and thus satisfies Hence, we have  Corollary 12. Suppose that γ > q and Ω is a bounded domain in R N (N ≥ 3).Let p(x) satisfy condition (P ).Then equation (P +) has a positive large solution if and only if γ > 1.
4. Condition on p is nearly optimal.We have shown in Theorem 2 that if the nonnegative function p is such that each of its zero points is enclosed by a bounded surface of nonzero points, then equation (P +) has a large positive solution.In this section we show that, if the condition does not hold in the sense that p vanishes in an "outer ring" of the domain, then equation (P +) has no positive large solution.