Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 39, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF ENTIRE SOLUTIONS FOR SEMILINEAR
ELLIPTIC SYSTEMS UNDER THE KELLER-OSSERMAN
CONDITION
ZHIJUN ZHANG, YONGXIU SHI, YANXING XUE
Abstract. Under the Keller-Osserman condition on f + g, we show the existence and nonexistence of entire solutions for the semilinear elliptic system
∆u = p(x)f (v), ∆v = q(x)g(u), x ∈ RN , where p, q : RN → [0, ∞) are
continuous functions.
1. Introduction
The purpose of this paper is to investigate the existence and nonexistence of
entire solutions to the semilinear elliptic system
x ∈ RN (N ≥ 3),
∆u = p(x)f (v),
∆v = q(x)g(u),
x ∈ RN .
(1.1)
By an entire large solution (u, v), we mean a pair of functions u, v ∈ C 2 (RN ) that
satisfies (1.1) and
lim u(x) = lim v(x) = +∞.
(1.2)
|x|→∞
|x|→∞
In this article, we assume that p, q, f and g satisfy the following hypotheses:
(H1) p, q : RN → [0, ∞) and f, g : [0, ∞) → [0, ∞) are continuous and nontrivial;
(H2) f and g are nondecreasing on [0, ∞) and f (t) > 0, g(t) > 0 for all t > 0;
(H3) H(∞) := limr→∞ H(r) = ∞,
where
Z r
dt
p
H(r) :=
, r ≥ a > 0,
(1.3)
2(F (t) + G(t))
a
Z t
Z t
F (t) :=
f (s)ds, G(t) :=
g(s)ds.
(1.4)
0
We see that
0
1
H 0 (r) = p
> 0,
2(F (r) + G(r))
∀r > a
2000 Mathematics Subject Classification. 35J55, 35J60, 35J65.
Key words and phrases. Semilinear elliptic systems; entire solutions; existence.
c
2011
Texas State University - San Marcos.
Submitted January 22, 2011. Published March 9, 2011.
Supported by grants 10671169 from NNSF of China and 2009ZRB01795 from NNSF of
Shandong Province.
1
2
Z. ZHANG, Y. SHI, Y. XUE
EJDE-2011/39
and H has the inverse function H −1 on [a, ∞). Denote
φ1 (r) := max p(x),
φ2 (r) := min p(x),
ψ1 (r) := max q(x),
ψ2 (r) := min q(x).
|x|=r
|x|=r
|x|=r
(1.5)
|x|=r
First we review the single elliptic equation
∆u = p(x)f (u),
x ∈ RN .
(1.6)
For p ≡ 1 on RN and f satisfying (H1) and (H2), Keller-Osserman [8, 15] first
supplied the necessary and sufficient condition
Z ∞
dt
p
=∞
(1.7)
2F (t)
1
for the existence of entire radial large solutions to (1.6). For the weight p(x) = p(|x|)
and f (u) = uα with α ∈ (0, 1], Lair and Wood [10] proved that (1.6) has a nonnegation entire radial large solution if and only if
Z ∞
rp(r)dr = ∞.
(1.8)
0
Recently, Lair [11] obtained the following results.
Lemma 1.1. Let f and b satisfy (H1) and (H2) with f (0) = 0. Suppose
(i) (1.7) holds;
R∞
(ii) there exists a positive constant ε such that 0 r1+ε φ1 (r)dr < ∞,
(iii) r2N −2 φ1 (r) is nondecreasing near ∞.
Then (1.6) has one nonnegative nontrivial entire bounded solution. If, on the other
hand, p satisfies
Z ∞
rφ2 (r)dr = ∞
0
and (iii) holds, then (1.6) has no nonnegative nontrivial entire bounded solution.
Lemma 1.2. Let f and b satisfy (H1) and (H2) with f (0) = 0 and p(x) = p(|x|).
Suppose (1.7) holds. Then (1.6) has one nonnegative nontrivial entire solution.
Suppose further that (iii) and (1.8) hold, then any nonnegative nontrivial entire
solution of (1.6) is large. Conversely, if (1.6) has a nonnegative nontrivial entire
large solution, then p satisfies
Z ∞
r1+ε φ1 (r)dr = ∞, ∀ε > 0.
0
For more works, see for example [1, 2, 4, 9, 10, 11, 18, 20, 21, 22] and the
references therein.
Now let us return to (1.1).
When p(x) = p(|x|), q(x) = q(|x|), f (v) = v α , g(u) = uγ , and 0 < α ≤ γ, Lair
and Wood [12] considered the existence and nonexistence of entire positive radial
solutions to system (1.1). Moreover, when 0 < α ≤ 1 and 0 ≤ γ ≤ 1, Lair [13]
showed that (1.1) has a nonnegative entire radial large solution if and only if p and
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EXISTENCE OF ENTIRE SOLUTIONS
3
q satisfy both of the following conditions
Z ∞
Z t
α
2−N
tp(t) t
sN −3 Q1 (s)ds dt = ∞,
0
0
Z ∞
Z t
γ
tq(t) t2−N
sN −3 P1 (s)ds dt = ∞,
0
(1.9)
(1.10)
0
where
Z
P1 (r) =
r
Z
τ p(τ )dτ,
Q1 (r) =
0
r
τ q(τ )dτ.
0
Ghanmi, Mâagli, Rădulescu and Zeddini [5] generalized the results in [12] to the
case when f and g are satisfy the condition that: For all c > 0, there exists Lc > 0
such that for all s1 , s2 ∈ [c, ∞),
|f (s2 ) − f (s1 )| + |g(s2 ) − g(s1 )| ≤ Lc |s2 − s1 |.
(1.11)
Recently, the authors in [14] showed the existence of entire positive radial large
solutions for (1.1) under the condition
Z ∞
ds
= ∞.
(1.12)
f
(s)
+ g(s)
1
For related works, see [3, 4, 5, 16, 19, 21, 22, 23] and the references therein.
In this paper, we extend some of the existence results for entire positive solutions
in Keller [8], Osserman [15] and Lair [11] to (1.1). Our main results are as the
following.
Theorem 1.3. Under the hypotheses (H1)–(H3). Suppose that
(H4) r2N −2 φ1 (r) + ψ1 (r) is nondecreasing for large r;
(H5) there exists a positive constant ε such that
Z ∞
r1+ε φ1 (r) + ψ1 (r) dr < ∞,
0
then (1.1) has a positive entire bounded solution (u, v).
From Theorem 1.3, we have the following corollaries for the spherically symmetric
case p(x) = p(|x|) and q(x) = q(|x|).
Corollary 1.4. Under hypotheses (H1)–(H3), (1.1) has one positive solution (u, v).
Suppose furthermore that
(H6) P (∞) = Q(∞) = ∞, where
Z r
Z t
1−N
P (∞) := lim P (r), P (r) :=
t
sN −1 p(s)ds dt, r ≥ 0,
r→∞
0
0
Z r
Z t
Q(∞) := lim Q(r), Q(r) :=
t1−N
sN −1 q(s)ds dt, r ≥ 0.
r→∞
0
0
Then every positive radial entire solution (u, v) of (1.1) is large and satisfies
u(r) ≥ u(0) + f (v(0))P (r),
v(r) ≥ v(0) + g(u(0))Q(r),
∀r ≥ 0.
Corollary 1.5. Assume (H1)–(H4). If (1.1) has a non-negative radial entire large
solution, then
Z ∞
r1+ε p(r) + q(r) dr = ∞, ∀ε > 0.
(1.13)
0
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Z. ZHANG, Y. SHI, Y. XUE
EJDE-2011/39
Corollary 1.6. Under hypotheses (H1)–(H3), (1.1) has no radial entire large solutions if p + q satisfies one of the following conditions:
(i) p(r) + q(r) ≤ Cr2−2N
for large r;
(ii) r2N −2 p(r) + q(r) is nondecreasing near ∞ and
Z ∞p
p(r) + q(r)dr < ∞;
0
(iii)
R∞p
0
Λ(r)dr < ∞, where
Λ(r) = max p(t) + q(t) ,
r ≥ 0.
(1.14)
t∈[0,r]
Theorem 1.7. Under hypotheses (H1)–(H3), (1.1) has no radial entire large solutions if p + q satisifes
p(r) + q(r)
p(r) + q(r)
0 < lim inf
≤ lim sup
< ∞, β < −2.
(1.15)
r→∞
rβ
rβ
r→∞
Remark 1.8. By (H1) and (H2), we see that (H3) implies
Z ∞
Z ∞
ds
ds
p
p
=
= ∞.
F (s)
G(s)
a
a
R∞
Remark 1.9. By [10], we see that P (∞) = ∞ if and only if 0 rp(r)dr = ∞.
R∞
R∞
Remark 1.10. By [9], we see that if 1 √dt < ∞, then 1 fdt
(t) < ∞. In other
F (t)
R ∞ dt
R ∞ dt
R ∞ dt
words, if 1 f (t) = ∞, then 1 √
= ∞. Conversely, if 1 √
= ∞, then
F (t)
F (t)
R ∞ dt
= ∞ does not hold. For example,
1 f (t)
2σ−1
2σ
f (t) = 2(1 + t)(ln(t + 1)
ln(t + 1) + σ , F (t) = (t + 1)2 ln(t + 1) ,
R∞
where σ > 0. We can see that 1 fdt
(t) = ∞ if and only if σ ∈ (0, 1/2] and
R ∞ dt
√
= ∞ if and only if σ ∈ (0, 1].
1
F (t)
2. Proof of main theorems
Proof of Theorem 1.3. Suppose (H4) holds. We will show that (1.1) has a solution
by finding a supersolution, (ū, v̄) and a subsolution, (u, v), for which u ≤ ū and
v ≤ v̄. To do this, we first prove the existence of (u, v) to (1.1) by considering the
system of the integral equations
Z r
Z t
u(r) = β +
t1−N
sN −1 φ1 (s)f (v(s))ds dt, r ≥ 0,
0
0
(2.1)
Z r
Z t
1−N
N −1
v(r) = β +
t
s
ψ1 (s)g(u(s))ds dt, r ≥ 0,
0
0
where β ≥ a > 0, a is in (1.3). Let {v m }m≥0 and {um }m≥1 be the sequences of
positive continuous functions defined on [0, ∞) by
v 0 (r) = β,
r
Z
t1−N
um (r) = β +
t
Z
0
sN −1 φ1 (s)f (v m−1 (s))ds dt,
r ≥ 0,
0
Z
v m (t) = β +
0
r
t1−N
Z
0
t
sN −1 ψ1 (s)g(um (s))ds dt,
r ≥ 0.
(2.2)
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EXISTENCE OF ENTIRE SOLUTIONS
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Obviously, for all r ≥ 0 and m ∈ N, um (r) ≥ β, v m (r) ≥ β and v 0 ≤ v 1 . (H2 )
yields u1 (r) ≤ u2 (r) for all r ≥ 0, then v 1 (r) ≤ v 2 (r) for all r ≥ 0. By the same
argument, we obtain that the sequences {um (r)} and {v m (r)} are increasing with
respect to m for r ∈ [0, ∞). Moreover, for each r > 0,
Z r
0
1−N
um (r) = r
sN −1 φ1 (s)f (v m−1 (s))ds ≥ 0,
0
Z r
v 0m (r) = r1−N
sN −1 ψ1 (s)g(um (s))ds ≥ 0
0
and
0 0
rN −1 um (r) + v m (r)
= rN −1 φ1 (r)f (v m−1 (r)) + ψ1 (r)g(um (r))
≤ rN −1 φ1 (r) + ψ1 (r) f (v m (r) + um (r)) + g(v m (r) + um (r)) .
Let
Λ(r) = max φ1 (t) + ψ1 (t) ,
r ≥ 0.
t∈[0,r]
0
Multiplying this by 2rN −1 um (r) + v m (r) and integrate on [0, r], we obtain
0 2
rN −1 um (r) + v m (r)
Z r
≤2
t2(N −1) φ1 (t) + ψ1 (t)
0
0
f (v m (t) + um (t)) + g(v m (t) + um (t)) um (t) + v m (t) dt
Z um (r)+vm (r)
≤ 2r2(N −1) Λ(r)
f (σ) + g(σ) dσ
2β
≤ 2r
2(N −1)
Λ(r) F (um (r) + v m (r)) + G(um (r) + v m (r)) ,
and
0 p
1/2
. (2.3)
um (r) + v m (r) ≤ 2Λ(r) F (v m (r) + um (r)) + G(v m (r) + um (r))
Thus
Z
r
√
0
Z
u0m (t) + v 0m (t)
1/2 dt
2 F (um (t) + v m (t)) + G(um (t) + v m (t))
um (r)+v m (r)
=
dτ
p
2(F (τ ) + G(τ ))
Z
= H(um (r) + v m (r)) − H(2β) ≤
2β
r
p
M (t)dt.
0
Since H −1 is increasing on [0, ∞), we have
Z
um (r) + v m (r) ≤ H −1 H(2β) +
r
p
M (t)dt ,
∀r ≥ 0.
(2.4)
0
It follows by (H3) and (2.2) that the sequences {um } and {v m } are bounded and
equi-continuous on [0, c0 ] for arbitrary c0 > 0. By Arzela-Ascoli theorem, {um }
6
Z. ZHANG, Y. SHI, Y. XUE
EJDE-2011/39
and {v m } have subsequences converging uniformly to u and v on [0, c0 ]. By the
arbitrariness of c0 > 0, we see that (u, v) is a positive entire solution of
∆u = φ1 (r)f (v) ≥ p(x)f (v),
x ∈ RN ,
∆v = ψ1 (r)g(u) ≥ q(x)g(u),
x ∈ RN ;
(2.5)
i.e., (u, v) is a positive entire subsolution of (1.1).
Next we prove that (u, v) is bounded. Since (u, v) satisfies
0
rN −1 u0 (r) = rN −1 φ1 (r)f (v),
(2.6)
0
rN −1 v 0 (r) = rN −1 ψ1 (r)g(u).
(2.7)
2N −2
φ1 (r) + ψ1 (r) is nondecreasing on [R, ∞) and
Choose R > 0 so that r
u(r) > 0,
0
v(r) > 0,
∀r ≥ R.
0
Now, since u (r) ≥ 0 and v (r) ≥ 0 for r ≥ 0, and (H2) holds, multiplying (2.6)
and (2.7) by rN −1 u0 (r) and rN −1 v 0 (r), respectively, and integrating from 0 to r,
we have
Z r
2
2
rN −1 u0 (r) ≤ RN −1 u0 (R) + 2
t2(N −1) p(t)f (v(t))u0 (t)dt
R
Z
r d
2(N −1)
≤ C + 2r
φ1 (r) + ψ1 (r)
F (v(t) + u(t))dt
R dt
2(N −1)
≤ C + 2r
φ1 (r) + ψ1 (r) F (v(r) + u(r)),
and
2
rN −1 v 0 (r) ≤ C + 2r2(N −1) φ1 (r) + ψ1 (r) G(v(r) + u(r)),
2
for r > R, where C = RN −1 u0 (R) + v 0 (R)) , which yields
u0 (r) + v 0 (r)
p
√
1/2
≤ 2Cr−(N −1) + 2(φ1 (r) + ψ1 (r)) G(u(r) + v(r)) + F (v(r) + u(r))
,
and
d
dr
≤
Z
u(r)+v(r)
u(R)+v(R)
√
dτ
q
2 F (τ ) + G(τ )
−1/2 p
Cr1−N G(u(r) + v(r)) + F (v(r) + u(r))
+ φ1 (r) + ψ1 (r).
Integrating the above inequality and using the facts that
G(u(r) + v(r)) + F (v(r) + u(r)) ≥ G(u(R) + v(R)) + F (v(R) + u(R)) = C1 ,
for all r ≥ R, and
q
p
φ1 (r) + ψ1 (r) ≤ 2r1+ε φ1 (r) + ψ1 (r) r−1−ε ≤ r1+ε φ1 (r) + ψ1 (r) + r−(1+ε)
for ε > 0, we have
Z r
s1+ε φ1 (s) + ψ1 (s) ds + (εRε )−1
H(u(r) + v(r)) ≤ H(u(R) + v(R)) +
R
q
+ CC1−1 (N RN )−1 .
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EXISTENCE OF ENTIRE SOLUTIONS
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Letting r → ∞, we find that (u, v) is bounded since φ1 + ψ1 satisfies (H5) and f + g
satisfies (H3). Thus, Since (u, v) is nondecreasing, we have
lim u(r) = M1 > 0,
r→∞
lim v(r) = M2 > 0.
r→∞
In the same way, we can see that the system
ū(0) = v̄(0) = max{M1 , M2 }, ū0 (r) = v̄ 0 (r) = 0,
N −1 0
ū (r) = φ2 (r)f (v̄(r)), r > 0,
∆ū(x) = ū00 (r) +
(2.8)
r
N
−
1
∆v̄(x) = v̄ 00 (r) +
v̄ 0 (r) = ψ2 (r)g(ū(r)), r > 0
r
has a bounded solution (ū, v̄) which is a supersolution for (1.1). It is also clear that
ū(r) ≥ M1 ≥ u(r),
v̄(r) ≥ M2 ≥ v(r),
∀r ≥ 0.
Hence the standard super-sub solution principle (see [17, 7]) implies that (1.1) has
a bounded solution (u, v) such that u(x) ≤ u(x) ≤ ū(x) and v(x) ≤ v(x) ≤ v̄(x) on
RN . This completes the proof.
Proof of Theorem 1.7. We follow the arguments in ([6, Theorem 4.3] and [22, Theorem 3.4]) for studying the nonexistence of entire radial large solutions to (1.6).
Let
Z ∞
a(r) = rθ
t p(t) + q(t) dt, r ≥ 0.
(2.9)
r
By (1.15), there exist R0 > 0, C2 > C1 > 0 such that
C1 rβ ≤ p(r) + q(r) ≤ C2 rβ ,
r ≥ R0 ,
so
a0 (r) = θrθ−1
Z
∞
t p(t) + q(t) dt − rθ+1 p(r) + q(r)
r
= −r
β+θ+1
C1 −
C2 θ <0
−β − 2
provided θ ∈ 0, C1 C2−1 (−β − 2) ; i.e., a is decreasing in [R0 , ∞). Define
Z ∞
b(r) =
t p(t) + q(t) dt, r ≥ 0.
(2.10)
r
Now suppose that (1.1) has a radial entire large solution (u, v) with u(r) > 0
and v(r) > 0 for all r ≥ R, then for r ≥ R0
Z r
τ N −2 1
u(r) + v(r) = u(0) + v(0) +
1−
τ p(τ )f (v(τ ))
N −2 0
r
+ q(τ )g(u(τ )) dτ,
Z r
1
τ N −2 ≤ u(0) + v(0) +
1−
τ p(τ ) + q(τ )
N −2 0
r
× f (v(τ ) + u(τ )) + g(u(τ ) + v(τ )) dτ
Z r
C
τ N −2 =C+
1−
τ p(τ ) + q(τ )
N − 2 R0
r
× f (v(τ ) + u(τ )) + g(u(τ ) + v(τ )) dτ.
8
Z. ZHANG, Y. SHI, Y. XUE
EJDE-2011/39
Let τ = b−1 (s), w = (u + v) ◦ b−1 . By the monotonicity of b and a = rθ b in [R0 , ∞),
θ
t b−1 (t) is increasing in (0, t0 ], where t0 = b(R0 ), and
1 − rα ≤ Cα (1 − r),
∀r ∈ [0, 1] and and fixed α > 0,
(2.11)
we obtain, for t ∈ (0, t0 ],
t0
b−1 (s) N −2 f (w(s)) + g(w(s)) ds
−1
b (t)
t
Z t0 1
t (N −2)/θ ≤C+
1−
f (w(s)) + g(w(s)) ds
N −2 t
s
Z t0 1
t
≤C+
1−
f (w(s)) + g(w(s)) ds = z(t).
N −2 t
s
w(t) = C +
1
N −2
Z
1−
It is easy to see that z 0 (t) ≤ 0 for t ∈ (0, t0 ] and
C f (w(t)) + g(w(t))
C f (z(t)) + g(z(t))
z 00 (t) =
≤
,
t
t
which yields
Z t0
z 02 (t0 ) − z 02 (t) = 2
z 00 (s)z 0 (s)ds
t
Z t0
f (z(s)) + g(z(s)) z 0 (s)
ds
≥ 2C
s
t
Z t0
2C
≥
f (z(s)) + g(z(s)) z 0 (s)ds
t t
2C
F (z(t0 )) + G(z(t0 )) − F (z(t)) − G(z(t)) .
=
t
Since limt→0 w(t) = ∞, so is F (z(t)) + G(z(t)). We obtain, for 0 < t < t1 small
enough,
C F (z(t)) + G(z(t))
z 02 (t) ≤
,
t
and
C
z 0 (t)
−√ ≤ p
≤ 0.
t
F (z(t)) + G(z(t))
Integrating from t to t1 and letting t → 0, we obtain
Z t1
Z ∞
√
dσ
dt
p
√ = 2C t1 < ∞.
≤C
t
F (σ) + G(σ)
z(t1 )
0
This is a contradiction. The proof is completed.
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Zhijun Zhang
School of Mathematics and Information Science, Yantai University, Yantai, Shandong,
264005, China
E-mail address: zhangzj@ytu.edu.cn
Yongxiu Shi
School of Mathematics and Information Science, Yantai University, Yantai, Shandong,
264005, China
E-mail address: syxiu0926@126.com
Yanxing Xue
School of Mathematics and Information Science, Yantai University, Yantai, Shandong,
264005, China
E-mail address: xiaoxue19870626@163.com