Solutions with radial symmetry for a semilinear elliptic system with weights

Solutions with radial symmetry for a semilinear elliptic system with weights

Accepted Manuscript Solutions with radial symmetry for a semilinear elliptic system with weights Dragos-Patru Covei PII: DOI: Reference: S0893-9659...

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Accepted Manuscript Solutions with radial symmetry for a semilinear elliptic system with weights

Dragos-Patru Covei

PII: DOI: Reference:

S0893-9659(17)30282-3 http://dx.doi.org/10.1016/j.aml.2017.09.003 AML 5331

To appear in:

Applied Mathematics Letters

Received date : 31 July 2017 Revised date : 6 September 2017 Accepted date : 6 September 2017 Please cite this article as: D.-P. Covei, Solutions with radial symmetry for a semilinear elliptic system with weights, Appl. Math. Lett. (2017), http://dx.doi.org/10.1016/j.aml.2017.09.003 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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Solutions with radial symmetry for a semilinear elliptic system with weights Dragos-Patru Covei Department of Applied Mathematics The Bucharest University of Economic Studies Piata Romana, 1st district, postal code: 010374, postal office: 22, Romania

Abstract In this paper, we provide sufficient conditions for the boundedness and unboundedness of the entire solutions for a semilinear elliptic system of the following type   ∆u = p (|x|) f (u, v) , x ∈ RN , 1  ∆v = p2 (|x|) g (u) , x ∈ RN ,

N ≥ 3.

Here p1 , f , p2 and g are continuous functions satisfying certain properties. Furthermore, we study the case

where the system is not of a variational type. Our results are obtained by a straightforward application of the Arzela–Ascoli theorem. Keywords: Entire solution; Large solution; Elliptic system. 2010 MSC: 35B08, 35B09, 35J25, 35J47, 35J99

1. Introduction and the main results The aim of this paper is to investigate the existence of radial solutions for the following semilinear elliptic system

  ∆u = p1 (x) f (u, v) , x ∈ RN ,  ∆v = p (x) g (u) , x ∈ RN , 2

N ≥ 3.

(1)

We impose the following conditions on the weight functions p1 , p2 and to the nonlinearities f , g : (P1) p1 , p2 : [0, ∞) → [0, ∞) are spherically symmetric continuous functions (i.e., for r = |x| we have p1 (x) = p1 (r) and p2 (x) = p2 (r)); 5

(C1) f : [0, ∞) × [0, ∞) → [0, ∞) and g : [0, ∞) → [0, ∞) are continuous, increasing, g (s) > 0 whenever s > 0; (C2) for fixed parameters a, b ∈ (0, ∞) there exist continuous functions h : [0, ∞) × [0, ∞) → [0, ∞) and ϕ : [0, ∞) → [0, ∞) such that: h is nondecreasing and there exists C ∈ (0, ∞) such that f (t1 , t2 · s2 ) ≤ Ch (t1 , t2 ) · ϕ (s2 ) , ∀ t1 ≥ 0, ∀ s2 ≥ 1, ∀ t2 ≥ M · g (a) ,

(2)

where M = b/g (a) if b > g (a) or M = 1 if b ≤ g (a). Preprint submitted to Applied Mathematics Letters

September 6, 2017

The motivation for our study stems from many recent works: Alves-Holanda [1], Figueiredo-Yang [5], Garc´ıa-Meli´ an [6, 7], Lair-Mohammed [14], Singh [21] and Li-Zhang-Zhang [17]. We give a quick review of the related literature. The pioneering work on large solutions for elliptic systems of the form (1) seems to be due to Lair and Wood [16]. However, in the case when there exists a Hamiltonian function F : Ω × D ⊆ Rn × R2 → R+ such that p1 (x) f (u, v) = Fu (x, u, v) and p2 (x) g (u) = Fv (x, u, v), problem (1) has a variational structure and thus under some suitable hypotheses on F one can study it by variational methods as in [1]. However, in a more general case, problem (1) is not of variational type. The study of the system

  ∆u = p1 (x) uα v β , x ∈ Ω,  ∆v = p (x) uγ v η , x ∈ Ω, 2

Ω ⊂ RN ,

has been done in Garc´ıa-Meli´ an [6, 7] and Lair-Mohammed [14]. The work of Li-Zhang-Zhang [17] have

10

established some new results for the following special case of (1)   ∆u = p (|x|) f (v) , x ∈ RN , 1 (N ≥ 3).  ∆v = p2 (|x|) g (u) , x ∈ RN ,

(3)

In this article we consider the system studied in Li-Zhang-Zhang [17] within a more general framework

which include a general class of nonlinearities. Our results complement the work of many authors [2, 3, 4, 5, 8, 9, 10, 11, 12, 18, 20, 21, 22]. In order to facilitate the presentation of our main results we introduce some notations Z r 1 dt, H (∞) = lim H (s) , H (r) = s→∞ a h (t, M g (t))   Z r Z s Z t Z z P (r) = s1−N tN −1 p1 (t) ϕ 1 + z 1−N lN −1 p2 (l) dldz dtds, 0 0 Z0 r Z0 z  1−N N −1 −1 z Q (r) = CP (t) dtdz, Q (∞) = lim Q (r) , t p2 (t) g H r→∞ 0 0   Z r Z y Z t Z z P (r) = y 1−N tN −1 p1 (t) f a, b + g (a) z 1−N lN −1 p2 (l) dldz) dtdy, 0 0 0 0   Z r Z y Z t Z z 1−N N −1 1−N N −1 y t p2 (t) g a + f (a, b) z s p1 (s) dsdz dtdy, Q (r) = 0

Q (∞) =

0

0

0

lim Q (r) , P (∞) = lim P (r) , P (∞) = lim P (r) .

r→∞

r→∞

r→∞

The following are our main results. 15

Theorem 1. Assume that H (∞) = ∞ and (P1) holds. Furthermore, if f satisfies the hypotheses (C1) and

(C2), then the system (1) has positively radial solution (u, v) ∈ C 2 ([0, ∞)) × C 2 ([0, ∞)) , and:

20

1.)

If P (∞) < ∞ and Q (∞) < ∞ then lim|x|→∞ u (|x|) < ∞ and lim|x|→∞ v (|x|) < ∞.

2.)

If P (∞) = ∞ and Q (∞) = ∞ then lim|x|→∞ u (|x|) = ∞ and lim|x|→∞ v (|x|) = ∞.

3.)

If P (∞) < ∞ and Q (∞) = ∞ then lim|x|→∞ u (|x|) < ∞ and lim|x|→∞ v (|x|) = ∞.

4.)

If P (∞) = ∞ and Q (∞) < ∞ then lim|x|→∞ u (|x|) = ∞ and lim|x|→∞ v (|x|) < ∞. 2

Theorem 2. Assume that (P1), (C1) and (C2) hold true. It follows that: i.)

If P (∞) < H (∞) < ∞ and Q (∞) < ∞ are satisfied, then the system (1) has positively bounded

radial solution (u, v) ∈ C 2 ([0, ∞)) × C 2 ([0, ∞)) which satisfies

ii.)

 a + P (r) ≤ u (r) ≤ H −1 CP (r) and b + Q (r) ≤ v (r) ≤ b + Q (r) .

If P (∞) < H (∞) < ∞ and Q (∞) = ∞, then the system (1) has positively radial solution (u, v) ∈

C 2 ([0, ∞)) × C 2 ([0, ∞)) which satisfies limr→∞ u (r) < ∞ and limr→∞ v (r) = ∞.

Next we provide examples in which our hypotheses are satisfied. They are inspired from [15]. 25

Example 3. Let a = b = 1, f (u, v) = u1/4 v 1/16 , g (u) = u4 , p1 (r) = that for C = 1, M = 1, h (t1 , t2 ) = f (t1 , t2 ) and ϕ (s2 ) =

1/16 s2

1 1+r 2

and p2 (r) =

r 1+r 2 .

Let us notice

the hypotheses (P1), (C1) and (C2) hold

true. Moreover, H (r) P (∞) Q (∞)

  √ 1 = 2 r − 1 → ∞ as r → ∞, g H −1 CP (r) = ( P (r) + 1)8 ≥ 1 2 Z ∞ Z ∞ Z y 1 1−N N −1 rp1 (r) dr = ∞, ≥ y t p1 (t) dtdy = N −2 0 Z0 ∞ Z0 y Z ∞ 1 ≥ y 1−N tN −1 p2 (t) dtdy = rp2 (r) dr = ∞, N −2 0 0 0

and thus by Theorem 1 case 2.) we get that the system (1) has positively radial solution (u, v) such that lim|x|→∞ u (|x|) = ∞ and lim|x|→∞ v (|x|) = ∞. 30

Example 4. For a = b = 1, f (u, v) = u1/4 v 1/16 , g (u) = u4 , p1 (r) =

r r 6 +1

and p2 (r) =

r (1+r)5

the 1/16

hypotheses (P1), (C1) and (C2) hold true for C = 1, M = 1, h (t1 , t2 ) = f (t1 , t2 ) and ϕ (s2 ) = s2

.

Moreover, H (r) P (∞) Q (∞)

  √ 1 r − 1 → ∞ as r → ∞, g H −1 CP (r) ≤ ( sup P (r) + 1)8 , 2 r≥0 Z ∞ Z ∞ 1 1 ≤ ϕ(1 + rp2 (r) dr) rp1 (r) dr < ∞ , N −2 0 N −2 0 Z ∞ Z ∞ π 1 1 ≤ [ ϕ(1 + rp2 (r) dr) + 1]8 rp2 (r) dr < ∞ , 12 (N − 2) N −2 0 N −2 0

= 2

and hence by Theorem 1 case 1.) we obtain that the system (1) has positively radial solution (u, v) such that lim|x|→∞ u (|x|) < ∞ and lim|x|→∞ v (|x|) < ∞. 35

1 (1+r 2 )5

Example 5. We consider a = b = 1, f (u, v) = u1/4 v 1/16 , g (u) = u4 , p1 (r) = √ 4 1/16

Then, for C = 1, M = 1, h (t1 , t2 ) = f (t1 , t2 ) and ϕ (s2 ) = s2

3

and p2 (r) =

√ 1 . 1+r 2

the hypotheses (P1), (C1) and (C2) hold

true. In addition, using [15, p.106] H (r) P (r) Q (r)

 √ r − 1 → ∞ as r → ∞, Z r/2 Z s Z s 1 1 ≥ sp1 (s) ϕ(s2−N [1 − ( )N −2 ] tN −3 lp2 (l) dldt)ds → ∞ as r → ∞, N −2 2 0 0 0 Z r Z y ≥ y 1−N tN −1 p2 (t) dtdy → ∞ as r → ∞, = 2

0

0

and thus by Theorem 1 case 2.) we see that the system (1) has positively radial solution (u, v) such that lim|x|→∞ u (|x|) = ∞ and lim|x|→∞ v (|x|) = ∞. 40

Example 6. Let a = b = 1, f (u, v) = u1/4 v 1/16 , g (u) = u4 , p1 (r) = for C = 1, M = 1, h (t1 , t2 ) = f (t1 , t2 ) and ϕ (s2 ) =

1/16 s2

√ 1 1+r 2

and p2 (r) =

1 . (1+r 2 )4

Then,

the hypotheses (P1), (C1) and (C2) hold true. In

addition, using [15, p.106]  √ r − 1 → ∞ as r → ∞,

H (r)

=

2

P (r)

<

Q (r)



∞ as r → ∞, Z r Z y y 1−N tN −1 p2 (t) dtdy → ∞ as r → ∞, 0

0

and thus by Theorem 1 case 3.) we get that the system (1) has positively radial solution (u, v) such that lim|x|→∞ u (|x|) < ∞ and lim|x|→∞ v (|x|) = ∞. 45

Remark 7 (see [23]). When

R1

1 dt 0 h(t,Mg(t))

= ∞, one can see that there is a > 0 sufficiently small such

that P (∞) < H (∞) < ∞ holds, provided P (∞) < ∞ and H (∞) < ∞, where a is given as in C2). Remark 8. Our assumption (C2) is considered in Krasnosel’skii and Rutickii [13] (see also Rao and Ren [19]), where it is paid special attention.

2. Proofs of the main results We start by noticing that radial solutions of the system of ODE  ′   rN −1 u′ (r) = rN −1 p1 (r) f (u (r) , v (r)) , 0 ≤ r < ∞,   ′ rN −1 v ′ (r) = rN −1 p2 (r) g (u (r)) , 0 ≤ r < ∞,     u (0) = a, v (0) = b, u′ (0) = 0, v ′ (0) = b,

(4)

are also solutions of system (1). Let us rewrite the system (1) as a system of integral equations as follows   u (r) = a + R r t1−N R t sN −1 p (s) f (u (s) , v (s)) dsdt, 1 0 0 (5) R R r t  v (r) = b + t1−N sN −1 p (s) g (u (s)) dsdt. 0

2

0

4

50

We define the sequences {un }n≥0 and {vn }n≥0 on [0, ∞) iteratively by:    u = u (0) = a, v0 = v (0) = b,   0 Rr Rt un (r) = a + 0 t1−N 0 sN −1 p1 (s) f (un−1 (s) , vn−1 (s)) dsdt, r ≥ 0,     v (r) = b + R r t1−N R t sN −1 p (s) g (u (s)) dsdt, r ≥ 0. n 2 n 0 0

(6)

We prove that the sequences {un }n≥0 and {vn }n≥0 are nondecreasing on [0, ∞). Indeed, let us consider Z r Z t u1 (r) = a + t1−N sN −1 p1 (s) f (u0 (s) , v0 (s)) dsdt 0 0 Z r Z t = a+ t1−N sN −1 p1 (s) f (a, b) dsdt 0 0 Z r Z t 1−N ≤ a+ t sN −1 p1 (s) f (u1 (s) , v1 (s)) dsdt = u2 (r) . 0

0

This proves that u1 (r) ≤ u2 (r) which in turn leads to v1 (r) ≤ v2 (r). A mathematical induction argument

applied to (6) yields un (r) ≤ un+1 (r) for any n ∈ N,and vn (r) ≤ vn+1 (r) for any n ∈ N.

Next we show that the non-decreasing sequences {un } and {vn } are bounded above on any bounded interval. Indeed, by the monotonicity of {un }n≥0 and {vn }n≥0 we get 

55

′ ′ rN −1 (un (r))  N −1 ′ r (vn (r))′

=

rN −1 p1 (r) f (un−1 (r) , vn−1 (r)) ≤ rN −1 p1 (r) f (un (r) , vn (r)) ,

(7)

=

rN −1 p2 (r) g (un (r)) .

(8)

By integrating (7) from 0 to r and by taking into account (2), we find out that Z r ′ (un (r)) ≤ r1−N tN −1 p1 (t) f (un (t) , vn (t)) dt 0 Z r Z t Z z = r1−N tN −1 p1 (t) f (un (t) , b + z 1−N sN −1 p2 (s) g (un (s)) dsdz)dt 0 0 0 Z t Z r Z z 1−N N −1 ≤ r z 1−N t p1 (t) f (un (t) , b + g (un (t)) sN −1 p2 (s) dsdz)dt 0 0 0 Z t Z z Z r b 1−N 1−N N −1 + z sN −1 p2 (s) dsdz))dt (9) ≤ r t p1 (t) f (un (t) , g (un (t)) ( g (u (t)) n 0 0 0 Z t Z z Z r b + z 1−N sN −1 p2 (s) dsdz))dt ≤ r1−N tN −1 p1 (t) f (un (t) , g (un (t)) ( g (a) 0 0 0 Z r Z t Z z 1−N N −1 1−N ≤ r t p1 (t) f (un (t) , M g (un (t)) (1 + z sN −1 p2 (s) dsdz))dt 0 0 0 Z r Z t Z z ≤ Ch (un (r) , M g (un (r))) r1−N tN −1 p1 (t) ϕ(1 + z 1−N sN −1 p2 (s) dsdz)dt. 0

0

0

Then by using the inequality (9) we see that (un (r))′ ≤ Cr1−N h (un (r) , M g (un (r))) By integrating (10), from 0 to r, we get Z un (r) a

Z

r

tN −1 p1 (t) ϕ(1 +

0

Z

t

z 1−N

0

1 dt ≤ CP (r) , h (t, M g (t)) 5

Z

0

z

lN −1 p2 (l) dldz)dt.

(10)

which is the same as H (un (r)) ≤ CP (r) .

(11)

Now we can easily see that H is a bijection with the inverse function H −1 strictly increasing on [0, ∞). By combining this with the previous inequality, we get  un (r) ≤ H −1 CP (r) .

(12)

Rr Substitution of (12) into (vn (r))′ = r1−N 0 tN −1 p2 (t) g (un (t)) dt we get Z r Z r  (vn (r))′ = r1−N tN −1 p2 (t) g (un (t)) dt ≤ r1−N tN −1 p2 (t) g H −1 CP (t) dt. 0

(13)

0

By integrating this over the interval [0, r] one gets Z r Z z  vn (r) ≤ b + z 1−N tN −1 p2 (t) g H −1 CP (t) dtdz = b + Q (r) . 0

(14)

0

To summarize we have found upper bounds for {un }n≥0 and {vn }n≥0 which depend on r. Now let us

complete the proof of Theorems 1-2. Proof of Theorem 1 completed: We claim that the sequences {un }n≥0 and {vn }n≥0 are bounded and ′

60



equicontinuous on [0, c0 ] for arbitrary c0 > 0. Indeed, since (un (r)) ≥ 0 and (vn (r)) ≥ 0 it follows that  un (r) ≤ un (c0 ) ≤ C1 and vn (r) ≤ vn (c0 ) ≤ C2 . Here C1 = H −1 CP (c0 ) and C2 = b + Q (c0 ) are positive constants. We have proved that {un }n≥0 and {vn }n≥0 are bounded on [0, c0 ] for arbitrary c0 > 0. By using

this fact in (9) and (13) we show that the same is true for (un (r))′ and (vn (r))′ . Indeed, for any r ≥ 0, Z r Z r (un (r))′ = r1−N tN −1 p1 (t) f (un−1 (t) , vn−1 (t)) dt ≤ r1−N rN −1 p1 (t) f (un (t) , vn (t)) dt 0 0 Z r Z r ≤ kp1 k∞ f (un (t) , vn (t)) dt ≤ kp1 k∞ f1 (C1 , C2 ) dt 0



0

kp1 k∞ f (C1 , C2 ) c0 on [0, c0 ] ,

and, similar arguments show that ′

(vn (r))

= r

1−N

Z

0

r

t

N −1

p2 (t) g (vn (t)) dt ≤ r

1−N

≤ kp2 k∞ g (C2 ) c0 on [0, c0 ] .

g (C2 ) r

N −1

kp2 k∞

Z

r

dt

0

To finish the proof it is left to establish that {un }n≥0 and {vn }n≥0 are equicontinuous functions on [0, c0 ] 65

for arbitrary c0 > 0. Let ε1 , ε2 > 0 be arbitrary constants. To verify the equicontinuity on [0, c0 ] we apply the mean value theorem to get ′ |un (x) − un (y)| = (un (ξ1 )) |x − y| ≤ kp1 k∞ f (C1 , C2 ) c0 |x − y| , ′ |vn (x) − vn (y)| = (vn (ξ2 )) |x − y| ≤ kp2 k g (C2 ) c0 |x − y| , ∞

for all n ∈ N and all x, y ∈ [0, c0 ] .

6

So it suffices to take δ1 =

ε1 kp1 k∞ f (C1 ,C2 )c0

and δ2 =

ε2 g(C2 )kp2 k∞ c0 ,

to see that {un }n≥0 and {vn }n≥0 are

equicontinuous functionson [0, c0 ]. Therefore, since both {un }n≥0 and {vn }n≥0 are bounded and equicontin70

uous functions on [0, c0 ] it follows from the Arzela–Ascoli theorem that there exists a function u ∈ C ([0, c0 ]) and a subsequence N1 of N∗ with un (r) converging uniformly to u on [0, c0 ] as n → ∞ through N1 .

On the other hand, {vn }n≥1 is a uniformly bounded sequence, so it follows from the Arzela–Ascoli theorem that there exists a function v ∈ C ([0, c0 ]) and a subsequence N2 of N∗ with vn (r) converging uniformly to v on [0, c0 ] as n → ∞ through N2 . Thus {(un (r) , vn (r))}n∈N2 converges uniformly on [0, c0 ] to 75

(u, v) ∈ C ([0, c0 ]) × C ([0, c0 ]) through N2 (see L¨ u-O’Regan-Agarwal [18]). We also point out that, since {un }n≥0 and {vn }n≥0 are non-decreasing on [0, ∞) we see that {(un , vn )}n≥0 converges uniformly to (u, v) on [0, c0 ] × [0, c0 ]. The solution constructed in this way is a radially symmetric solution of the system (1). Since the radial solutions of (1) are solutions of the ordinary differential equations system (5) it follows that the radial solutions of (1) with u (0) = a, v (0) = b satisfy: Z r Z y 1−N u (r) = a + y tN −1 p1 (t) f (u (t) , v (t)) dtdy, r ≥ 0, 0 0 Z r Z y y 1−N v (r) = b + tN −1 p2 (t) g (u (t)) dtdy, r ≥ 0. 0

80

(15) (16)

0

Next we prove that all four Cases 1., 2., 3. and 4. hold true. Case 1.): Assume P (∞) < ∞ and Q (∞) < ∞. Proceeding as in the proof of (12) and (14), with the integral equations (15) and (16), one gets the estimates  u (r) ≤ H −1 CP (∞) < ∞ and v (r) ≤ b + Q (∞) < ∞ for all r ≥ 0.

In other words, we get that (u, v) is bounded. We next consider: Case 2.): In the case P (∞) = Q (∞) = ∞, we observe that u (r)

Z

r

Z

t

sN −1 p1 (s) f (u (s) , v (s)) dsdt Z r Z y Z t Z z 1−N N −1 1−N = a+ y t p1 (t) f (u (s) , b + z sN −1 p2 (s) g (u (s)) dsdz)dtdy 0 0 0 0 Z r Z y Z t Z z 1−N N −1 1−N ≥ a+ y t p1 (t) f (u (s) , b + g (a) z lN −1 p2 (l) dldz)dtdy 0 0 0 0 Z r Z y Z t Z z 1−N N −1 1−N ≥ a+ y t p1 (t) f (a, b + g (a) z lN −1 p2 (l) dldz))dtdy = a+

t

1−N

0

0

0

0

0

(17)

0

= a + P (r) .

Proceeding as above we get that Z r Z y 1−N y v (r) = b + tN −1 p2 (t) g (u (t)) dtdy 0 0  Z r Z t Z y Z 1−N 1−N N −1 y z ≥ b+ t p2 (t) g a + f (a, b) 0

0

0

7

0

z

s

N −1



p1 (s) dsdz dtdy = b + Q (r) .

Consequently, by taking limits we get entire large solutions limr→∞ u (r) = limr→∞ v (r) = ∞. This com85

pletes the proof. Case 3.): In the spirit of Case 1. and Case 2. above, we have  u (r) ≤ H −1 CP (∞) < ∞, v (r) ≥ b + Q (r) .

Therefore, if P (∞) < ∞ and Q (∞) = ∞ we get limr→∞ u (r) < ∞ and limr→∞ v (r) = ∞, as was to be shown. In order, to complete the proof of Theorem 1 it remains to proceed to the Case 4.): In this case, the idea is to mimic the proof of the Case 3. We observe that u (r) ≥ a + P (r) , v (r) ≤ b + Q (r) .

(18)

Then the claim of the Theorem follows by letting r → ∞ in (18). 90

Proof of Theorem 2: i.) It follows from (11) and the conditions of the theorem that H (un (r)) ≤ CP (∞) < CH (∞) < ∞, vn (r) ≤ b + Q (∞) < ∞.  On the other hand, since H −1 is strictly increasing on [0, ∞), we find out that un (r) ≤ H −1 CP (∞) < ∞,

and then the non-decreasing sequences {un }n≥0 and {vn }n≥0 are bounded above for all r ≥ 0 and all n. We n→∞

use this observation to conclude that (un (r) , vn (r)) → (u (r) , v (r)) , and thus the limiting functions u and v are positive entire bounded radial solutions of the system (1). We now turn to the case 95

ii.): The proof is similar and is therefore omitted.

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