EXISTENCE OF ENTIRE EXPLOSIVE POSITIVE RADIAL SOLUTIONS OF QUASILINEAR ELLIPTIC SYSTEMS

have received much attention recently. See, for example, [5, 7, 9, 18, 26, 27, 28]. Problem (1.1) arises in the theory of quasiregular and quasiconformal mappings or in the study of non-Newtonian fluids. In the latter case, the pair (p,q) is a characteristic of the medium. Media with (p,q) > (2,2) are called dilatant fluids and those with (p,q) < (2,2) are called pseudoplastics. If (p,q)= (2,2), they are Newtonian fluids. When p = q = 2, system (1.1) becomes − u= f(u,v), x ∈RN, − v = g(u,v), x ∈RN, (1.2)

In this paper, we study the existence of entire explosive positive solutions of the system div |∇u| p−2 ∇u = m |x| v α , x ∈ R N , div |∇v| q−2 ∇v = n |x| u β , x ∈ R N , (1.4) where α > 0, β > 0, p > 1, and q > 1.As far as the author knows, there are no results that contain existence criteria of entire explosive positive solutions to the elliptic system (1.4).Motivated by this fact, we will study mainly this problem here.When p = q = 2, the related results have been obtained by [13].
Our theorem for existence extends the results [13].
in a domain D ⊂ R N and continuous on its boundary S. Then there exists a decreasing function g(R) determined by F(u) such that u(P ) ≤ g R(P ) . (2.13) Here P denotes a point in D and R(P ) denotes its distance from S. The function g(R) has the limits Proof.Each point P ∈ D can be the centre of a sphere of radius R(P ) which lies in D. Therefore, it suffices to prove the theorem in D as a sphere of radius R, and suppose that u is defined continuously on S. We define a function v in D and S as the solution of the problem (2.15) , where F(v) is the function occurring in condition (i) and θ is a constant, 0 < θ < β.Thus, from conditions (i) and (ii) and (2.15), we have Moreover, α is a positive constant which satisfies u ≤ α on S. (2.17) The existence and uniqueness of a positive solution v of (2.15) are assured because F 1 is a nondecreasing function.In fact, the existence can be obtained by the standard variational method and the uniqueness can be obtained by an idea similar to that in the proof of Lemma 2.1 (see [8]).From (2.To this end, we must examine v, the solution of (2.15).First we show that v must be a function of r only where r denotes the distance from the centre of the sphere.We can find a positive radial solution v(r ) of (2.15) by the variational method to the equivalent form of (2.15) in r : ) where Φ p (s) = |s| p−2 s.The uniqueness of the positive solution v of (2.15) implies that v is just the radial solution v(r ) of problem (2.21) and (2.22).Since v(0) is a monotonic increasing function of α, α is itself uniquely determined by v(0).Let v(0) = v 0 .As v 0 increases, α = v(R) also increases.We will show that α = v(R) becomes infinite for some finite value of v 0 .
This value of v 0 is denoted by lim α→∞ v 0 , which can be used to define the function g(R) in this lemma.
It is convenient to rewrite (2.21) in the form Integrating (2.23) from 0 to r yields From (2.24), we see that v ≥ 0. Therefore, v is a nondecreasing function, and we can obtain from (2.24) that Inserting (2.25) into (2.21),we have (2.27) We now multiply (2.27) by v and integrate from 0 to r to obtain that is, where (2.30) This implies Then we consider r as a function of v, and we have By condition (i), the integral in (2.32) converges as v becomes infinite when v 0 = 0.But then the integral also converges for any value of v 0 > 0. If we denote its limit by A(v 0 ), letting v → ∞, (2.32) yields where From (2.34), we see that, for each v 0 ,v becomes infinite at a finite value of r ∞ in the range indicated in (2.33).Therefore, r ∞ is a function of v 0 and is denoted by r ∞ (v 0 ).The function r ∞ (v 0 ) is continuous and nonincreasing.If it were increasing, then two solutions corresponding to different values of v 0 would have to be equal at some value of r .This is impossible because a solution of (2.21) with a prescribed value on the surface of a sphere is unique.Furthermore, the integral A(v 0 ) tends to +∞ as v 0 tends to −∞, and to zero as v 0 tends to +∞.Therefore, by (2.33), r ∞ (v 0 ) behaves in the same way.We now define g(R) := min{v 0 | r ∞ (v 0 ) = R}.This function is decreasing and satisfies (2.14), so it is the desired g(R) of Lemma 2.2.This completes the proof of Lemma 2.2.

Lemma 2.3. If f (u) is nondecreasing and satisfies Lemma 2.2, then in any bounded domain D there exists a solution of (2.1) which becomes infinite on S.
Proof.We note that, for any constant α and any domain D, there exists in D a solution u α of (2.1) which is equal to α on S provided that f (u) is nondecreasing (see [8]).Furthermore, at each point of D, u α increases with α.If f (u) satisfies Lemma 2.2(i), then Lemma 2.2 holds, and at each point P in D, all of the u α are bounded above.Thus, in every closed subdomain, u α converges uniformly to a limit u.This limit is also a solution of (2.1).As P approaches S, u(P ) increases infinitely since on S, u α = α becomes infinite.Thus, u is the desired solution and Lemma 2.3 is proved. (2.35) Proof.From Lemma 2.3, we have that for each k ∈ N, the boundary value problem has a positive solution.Furthermore, in R N .To prove our result, we only need to prove that (A) there exists dt, (2.38) where is the unique positive solution of the following problem: (2.41) Thus and from Lemma 2.1, we obtain g = v Lemma 2.5.Suppose that p −1 < α ≤ β, β > q−1, and p, q > 1.The problem has an entire explosive positive radial solution provided that the C(R N ) functions m(x), n(x) ≥ C > 0 and satisfy dt < ∞. (2.45) Proof.From Lemma 2.3, for each natural number k, let v k be a positive solution of the boundary value problem Again, by the maximum principle, we can show that To complete the proof, it is sufficient to show that there exists a function To do this, we first consider the equation (2.48) By Lemma 2.4, (2.48) has a positive solution u on R N such that u(x) → ∞ as |x| → ∞.We claim that w = u − 1 is a desired lower boundary for v k .Indeed, since Again, by the standard regularity argument for elliptic problems, it is a straightforward argument to prove that v is the desired solution of (2.43).By a similar argument, we can show that (2.44) has an entire explosive positive radial solution.
(b) We note first that, since g is radial, we get dt. (2.60) for all r ≥ 0. Thus we have dt = g(r ). (2.61) Thus we have u 1 ≤ g.Similar arguments will show, in sequence, that v 1 ≤ h, u 2 ≤ g,....

Main results
. By a modification of the method given in [13], we establish the following results.
If, in addition, the functions m and n satisfy then all entire positive radial solutions of (1.4) are large solutions.On the other hand, if m and n satisfy dt < ∞, then all entire positive radial solutions of (1.4) are bounded.
In case (i), we have that q − 1 ≥ 1.Using the inequality (1 . (3.16) Therefore, . (3.17) Letting Similarly, there exists a function d(r ) such that u k (r ) ≤ d(r ) for all k if α = p − 1.Thus the sequences {u k } and {v k } are bounded above on bounded sets and therefore converge.Let u(r ) = lim k→∞ u k (r ) and v(r ) = lim k→∞ v k (r ).By standard elliptic regularity theory, it can be shown that (u, v) is the desired solution of (1.4).
On the other hand, if the inequalities (3.3) hold, then lim r →∞ f (r ) < ∞ and lim r →∞ g(r ) < ∞ so that the estimates above providing upper bounds for the sequences {u k } and {v k } may be chosen independent of r so that the solution (u, v) is bounded above (and, in fact, any solution of (3.5) will be bounded when the inequalities (3.3) hold).
We now give our main theorems for the superlinear case, where p − 1 < α ≤ β, β ≥ q − 1, and 2 ≤ q ≤ p.We use the notation R + = [0, ∞) and define the set G as (u, v) is an entire radial solution of (1.4) .
(3.22) Theorem 3.2.There are infinitely many entire positive radial solutions of system (1.4) provided that the C(R N ) functions m and n satisfy (3.3).The set G is a closed bounded convex subset of R + × R + .Furthermore, the set G satisfies where the triangle T and the rectangle R are given by ) Proof.From Lemma 2.7, it is clear that [0,g(0)] × [0,h(0)] ⊂ G so that G is nonempty.We will show that G is a bounded, closed, and convex set and then prove relationship (3.23).
As a preliminary, note that, if (a, b) ∈ G, then any pair (a 0 ,b 0 ) for which 0 ≤ a 0 ≤ a and 0 ≤ b 0 ≤ b must be in G since the process used in Lemma 2.7 can be repeated with and v 0 = b, u 0 = a.Then, as in Lemma 2.7, the sequences {u k } and {v k } are monotonically increasing.Then, letting (U , V ) be the solution of (3.5) with central values (a, b), we can easily prove, since b 0 ≤ b, that v 0 ≤ V .Thus, u 1 ≤ U (since, also, a 0 ≤ a), and consequently v 1 ≤ V , and so on.Hence, we get u k ≤ U and v k ≤ V , and therefore, u ≤ U and v ≤ V , where (u, v) = lim k→∞ (u k ,v k ) is a solution of (1.4) (with central values (a 0 ,b 0 )).
To prove that G is bounded, assume that it is not.Therefore, since Without loss of generality, we assume that [0, ∞) × {0} ⊂ G. Let m(r ) = min{m(r ), n(r )} and let h be a positive radial solution of (See Lemma 2.3 for the proof of existence.)Let (u, v) be any solution, which exists by hypothesis, to (3.5) with a > h(0) and b = 0. Without loss of generality, we will assume that a ≥ 1.We now show that h ≤ u + v for all r ≥ 0 which, if proven, will contradict the fact that u + v exists for all r ≥ 0. Clearly, h(0) < a ≤ u(0) + v(0).Thus there exists > 0 such that h(r ) < u(r ) + v(r ) for all r ∈ [0, ).Let (3.27) We claim that R 0 = 1.Indeed, suppose that R 0 < 1.Since h(r ) < u(r )+v(r ) in [0,R 0 ), elementary estimates yield (3.28) This contradicts the fact that R 0 is a supremum.Thus, R 0 = 1, establishing the boundedness of the set G.
To prove that G is closed, we let (a 0 ,b 0 ) ∈ ∂G and show that (a 0 ,b 0 ) ∈ G. Let (u, v) be the solution of (3.5) which corresponds to a = a 0 and b = b 0 .Without loss of generality, we may assume that max{a 0 ,b 0 } > K = g(0), where the function g is given in Lemma 2.7.If max{a 0 ,b 0 } = a 0 , then K ≤ a 0 − 1/n for large n so that u n (r ) ≥ K for all r ≥ 0 and for all n sufficiently large where dt. (3.29) Thus we note that where m(r ) = min{1,K β−α } min{m(r ), n(r )}.Thus div ∇u n p−2 ∇u n ≥ m(r )v α n , div ∇v n q−2 ∇v n ≥ m(r )u α n . (3.31) Let h 1 (r ) and h 2 (r ) be positive solutions of where R 0 is an arbitrary positive real number.It is now easy to show by Lemma 2.1 that u n ≤ h 1 and v n ≤ h 2 ; thus . Since R 0 is arbitrary, the functions u and v exist on R N and are hence entire so that (a 0 ,b 0 ) ∈ G. On the other hand, if max{a 0 ,b 0 } = b 0 , then K ≤ b 0 − 1/n for large n so that v n ≥ K for all r ≥ 0 and for all sufficiently large n.Then u n (r ) ≥ K α f (r ), where 1) dt, and the proof continues as before with K replaced by K α f (r ).
To prove that G is convex, suppose that (a, b) ∈ G and (a, b) ∈ G. Let λ ∈ (0, 1), let (u, v) be the solution of (3.5), and let (U, V ) be the solution of (3.5) when (a, b) is replaced by (a, b).We need to prove that λ(a, b)+(1−λ)(a, b) ∈ G.To do this, we let {u n }, {v n }, {U n }, and {V n } be the increasing sequences of functions, as developed in Lemma 2.7, such that u n u, v n v, U n U, and V n V .Likewise, let {w n } and {z n } be the sequences developed again as in Lemma 2.7 corresponding to central values λa + (1 − λ)a and λb + (1 − λ)b, respectively.We also let z 0 = λb + (1 − λ)b.We will show that the increasing sequences {w n } and {z n } satisfy which, in turn, implies that {w n } and {z n } converge, and hence, their limits are entire, giving λ(a, b) dt.
(3.34) Since α > p − 1 ≥ 1, however, we know that (λc for any nonnegative numbers c and d.Applying this inequality, we get Using this result, we can prove similarly that z Continuing this process will produce (3.33).
To prove (3.23), it is clear that, since (A, 0) and (0,B) are in G and G is convex, the line x/A+y/B = 1 is in G. And, as noted earlier, if (a, b) ∈ G, then (x 0 ,y 0 ) ∈ G whenever 0 ≤ x 0 ≤ a and 0 ≤ y Since u is entire, we conclude that R = ∞ and lim r →∞ u(r ) = ∞.This completes the proof.