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Existence of positive entire solutions of a semilinear elliptic problem with a gradient term
Nonlinear Analysis 71 (2009) 3113–3118
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Existence of positive entire solutions of a semilinear elliptic problem with a gradient term Hongtao Xue ∗ , Xigao Shao School of Science, Yantai Nanshan University, Yantai, Shandong 265713, PR China
article
abstract
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Article history: Received 14 May 2008 Accepted 26 January 2009
By a sub–supersolution argument and a perturbed argument, we show the existence of entire solutions to a semilinear elliptic problem −∆u + h(x)|∇ u|q = b(x)g (u), u > 0, α x ∈ RN , lim|x|→∞ u(x) = 0, where q ∈ (1, 2], b, h ∈ Cloc (RN ) for some α ∈ (0, 1), N 1 h(x) ≥ 0, b(x) > 0, ∀ x ∈ R , and g ∈ C ((0, ∞), (0, ∞)) which may be singular at 0. No monotonicity condition is imposed on the functions g (s) and g (s)/s. © 2009 Elsevier Ltd. All rights reserved.
MSC: 35J65 35B05 35O75 35R05 Keywords: Semilinear elliptic equations Entire solutions Existence
1. Introduction and the main results The purpose of this note is to investigate the existence of entire solutions to the following model problem
− ∆u + h(x)|∇ u|q = b(x)g (u),
u > 0, x ∈ RN ,
lim u(x) = 0,
|x|→∞
(1.1)
α where q ∈ (1, 2], h ∈ Cloc (RN ) for some α ∈ (0, 1) is non-negative in Ω , g satisfies
(g1 ) g ∈ C 1 ((0, ∞), (0, ∞)); (g2 ) lims→0+ g (s)/s = ∞; (g3 ) lims→∞ g (s)/s = 0; and b satisfies α (b1 ) b ∈ Cloc (RN ) and b(x) > 0, ∀ x ∈ RN ; (b2 ) the linear problem
− ∆u = b(x),
u > 0, x ∈ RN ,
lim u(x) = 0
|x|→∞
2+α has a unique solution w ∈ Cloc (RN ).
∗
Corresponding author. E-mail address: [email protected] (H. Xue).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.01.222
(1.2)
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H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
First, let us review the following model
− ∆u = b(x)g (u),
u > 0, x ∈ RN ,
lim u(x) = 0.
(1.3)
|x|→∞
Problem (1.3) arises from many branches of mathematics and applied mathematics. It was discussed and extended to more general problems in a number of works, for instance, [1–13]. When the equation is considered over a bounded smooth domain Ω instead of RN , the corresponding problem was studied, for example, in [14–25] and the references cited therein. For g (u) = u−γ with γ > 0, if b satisfies (b1 ) and the following condition (b3 )
R∞ 0
r φ(r )dr < ∞, where φ(r ) = max|x|=r b(x),
2+α Lair and Shaker [9] showed that problem (1.3) has a unique solution u ∈ Cloc (RN ). Later, Lair and Shaker [10] and Zhang [13] extended the above result to the more general g which satisfies (g1 ) and
(g4 ) g is non-increasing on (0, ∞) and lims→0+ g (s) = ∞. Cˇirstea and Rˇadulescu [3] also extended the above results to the more general g which satisfies (g1 ), (g2 ) and g (s)
(g5 ) s+s is decreasing on (0, ∞) for some s0 > 0; 0 (g6 ) g is bounded in a neighborhood of ∞. Recently, Dinu [26] further generalized the above results to the cases that (i) b satisfies (b1 ) and (b3 ); (ii) g satisfies (g1 ), (g3 ) and g (s) (g7 ) s is decreasing on (0, ∞); and lim|x|→∞ u(x) = l > 0 instead of lim|x|→∞ u(x) = 0 in problem (1.3); or the following cases that (i1 ) (b4 ) (ii2 ) (g8 )
b satisfies (b1 ) and R∞ r N −1 φ(r )dr < ∞, where N ≥ 3; 0 g satisfies (g1 )–(g3 ) and g is increasing on (0, ∞).
Afterwards, Goncalves and Santos [8] also generalized the above results to the case that g satisfies (g1 )–(g3 ) and (g7 ). Ye and Zhou [12, Theorem 4.2] showed that if g satisfies (g1 ) and is non-increasing on (0, ∞), b satisfies (b1 ), then problem (1.3) admits a solution if and only if b satisfies (b2 ). Moreover, if a solution of problem (1.3) exists, it is unique. Now let us return to problem (1.1). α When g (u) = u−γ with γ > 0, q ∈ (1, 2], h ∈ Cloc (RN ) is non-negative in RN , b satisfies (b1 ) and (b3 ), Dinu [26] showed 2+α α that problem (1.1) has a unique solution u ∈ Cloc (RN ). Recently, the author [27] showed that whenq ∈ (0, 1), h ∈ Cloc (RN ) for some α ∈ (0, 1), h(x) < 0, ∀ x ∈ RN , g satisfies (g1 )–(g3 ), and b satisfies (b3 ) instead of (b2 ), problem (1.1) have at least one solution. In this paper we continue to consider the existence of entire solutions to problem (1.1) for the functions g (s) and g (s)/s which do not have monotonicity. Our main result is summarized in the following theorem. α Theorem 1.1. Let q ∈ (1, 2], h ∈ Cloc (RN ) be non-negative in RN , and b satisfy (b1 ) and (b2 ). If g satisfies (g1 )–(g3 ), then 2+α problem (1.1) has at least one solution u ∈ Cloc (RN ).
Remark 1.1 ([27]). The condition (b3 ) implies (b2 ), but (b2 ) is invariant under translations. Remark 1.2. Some basic examples of the functions, which satisfy (g1 )–(g3 ), are (i) (ii) (iii) (iv)
u−γ + up + sin f (u) + 1, where γ > 0, p < 1 and f ∈ C 2 (R); γ e1/u + up + cos f (u) + 1, where γ > 0, p < 1 and f ∈ C 2 (R); −γ u ln−q1 (1 + u) + lnq2 (1 + u) + up + sin f (u) + 2 with f ∈ C 2 (R), γ > 0, p < 1, q2 > 0 and q1 > 0; u−γ + arctan f (u) + π with f ∈ C 2 (R) and γ > 0.
Remark 1.3. The technique of this paper in our proofs can be applied to the more general problem
−∆u + h(x)|∇ u|q = b(x)g (u) + a(x)f (u),
u > 0, x ∈ RN ,
lim u(x) = 0,
|x|→∞
α where q ∈ (1, 2], h ∈ Cloc (RN ) is non-negative in RN , a and b satisfy (b1 ) and (b2 ), g and f satisfy (g1 ), g + f satisfies (g2 )–(g3 ).
The paper is organized as follows. In Section 2 we give some preliminary considerations. Finally we show the existence of solutions to problem (1.1).
H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
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2. Preliminaries We first consider the following Dirichlet problem
− ∆u + h(x)|∇ u|q = b(x)g (u),
u > 0, x ∈ Ω , u|∂ Ω = 0,
(2.1)
where Ω is a bounded domain with smooth boundary in RN (N ≥ 1). Concerning the existence of the solutions to problem (2.1), refer to [28,29,20] and the references cited therein. For the convenience, we denote f (x, u, ∇ u) = b(x)g (u) − h(x)|∇ u|q . Now we introduce a sub–supersolution method with the boundary restriction. Definition 2.1. A function u ∈ C 2+α (Ω ) ∩ C (Ω ) is called a subsolution of problem (2.1) if
− ∆u ≤ f (x, u, ∇ u),
u > 0, x ∈ Ω , u|∂ Ω = 0.
(2.2)
Definition 2.2. A function u¯ ∈ C 2+α (Ω ) ∩ C (Ω ) is called a supersolution of problem (2.1) if
− ∆u¯ ≥ f (x, u¯ , ∇ u¯ ),
u¯ > 0, x ∈ Ω , u¯ |∂ Ω = 0.
(2.3)
Lemma 2.1 ([15, Lemma 3]). Let f (x, u, ξ ) satisfy the following two basic conditions: (D1 ) f (x, u, ξ ) is locally Hölder continuous in Ω × (0, ∞) × RN and continuously differentiable with respect to the variables u and ξ ; (D2 ) for any Ω1 ⊂⊂ Ω and any a, b ∈ (0, ∞)(a < b), there exists a corresponding constant C = C (Ω1 , a, b) > 0 such that ¯ 1 , ∀ u ∈ [a, b], ∀ ξ ∈ RN . |f (x, u, ξ )| ≤ C (1 + |ξ |2 ), ∀ x ∈ Ω If problem (2.1) has a supersolution u¯ and a subsolution u such that u ≤ u¯ in Ω , then problem (2.1) has at least one solution ¯ ) in the ordered interval [u, u¯ ]. u ∈ C 2+α (Ω ) ∩ C (Ω Lemma 2.2 (Existence, [8, Theorem 1.2 and Proof of Theorem 1.1]; uniqueness, [3, Section 2]). Let b satisfy (b1 ). If g satisfies (g1 )–(g3 ) and (g7 ), then the following problem
− ∆u = b(x)g (u),
u > 0, x ∈ Ω , u|∂ Ω = 0
¯)∩C has a unique solution u ∈ C (Ω
2+α
(2.4)
(Ω ).
Lemma 2.3. If g satisfies (g1 )–(g3 ), then there exists a function f¯1 such that (i) (ii) (iii) (iv)
f¯1 ∈ C 1 ((0, ∞), (0, ∞)); g (s)/s ≤ f¯1 (s), ∀ s > 0 and lims→0+ f¯1 (s) = ∞; f¯1 is non-increasing on (0, ∞); lims→∞ f¯1 (s) = 0.
Proof. By (g1 )–(g3 ), we can denote f¯ (s) = sup g (t )/t .
(2.5)
t ≥s>0
Observe that f¯ (s) ≥ g (t )/t ,
∀ s > 0 and t ≥ s;
and f¯ is non-increasing on (0, ∞). Moreover, lim f¯ (s) = ∞ and
s→0+
lim f¯ (s) = 0.
s→∞
Now we can assume f¯ ∈ C 1 (0, ∞). If not, we can replace it by 2 f¯1 (s) = s
Z
s s/2
f¯ (t )dt ,
s > 0.
Obviously, f¯ (s) ≤ f¯1 (s) ≤ f¯ (s/2),
∀ s > 0.
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H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
And, for s > 0, f¯10 (s) =
≤
2
s 2
s
f¯ (s) −
1
f¯ (s) −
1
2 2
f¯ (s/2)
s2
f¯ (s/2)
s
Z
2
−
s/2
2 s
−
s2 2
f¯ (t )dt
f¯ (s) =
1 s
(f¯ (s) − f¯ (s/2)) ≤ 0,
i.e., f¯1 ∈ C 1 ((0, ∞), (0, ∞)). The proof of Lemma 2.3 is completed.
¯ ). For the convenience, we denote |u|∞ = maxx∈Ω¯ |u(x)| whenever u ∈ C (Ω ¯ ), h(x) ≥ 0, b(x) > 0, ∀ x ∈ Ω ¯ . If g satisfies (g1 )–(g3 ), then problem (2.1) has at least Lemma 2.4. Let q ∈ (1, 2], b, h ∈ C α (Ω ¯ ) ∩ C 2+α (Ω ). one solution u ∈ C (Ω ¯ ) ∩ C 2+α (Ω ) be the first eigenfunction corresponding to the first eigenvalue λ1 of Proof. Let ψ1 ∈ C 1 (Ω − ∆u = λu,
u > 0, x ∈ Ω , u|∂ Ω = 0.
(2.6)
Let β = q/(q − 1). It follows by (g2 ) that there exists a positive constant δ1 ∈ (0, 1) such that g (s) s
≥
λ1 β + |h|∞ β q |∇ψ1 |q∞ , min b(x)
∀ s ∈ (0, δ1 ).
¯ x∈Ω
β
δ1
Let u = c1 ψ1 with c1 ∈ 0, min{1,
β |ψ1 |∞
} . Since c1q−1 < 1, we see that
β
β−2
−∆u + h(x)|∇ u|q = βλ1 c1 ψ1 − c1 β(β − 1)ψ1 β
q(β−1)
|∇ψ1 |2 + h(x)β q c1q ψ1 β
≤ min b(x)g (c1 ψ1 ) ≤ b(x)g (c1 ψ1 ) = b(x)g (u), ¯ x∈Ω
|∇ψ1 |q
x ∈ Ω,
β
i.e., u = c1 ψ1 is a subsolution to problem (2.1). To construct a supersolution, we see by Lemma 2.2 that the following problem
1 , − ∆u = b(x)u f¯1 (u) +
u > 0, x ∈ Ω , u|∂ Ω = 0,
u
(2.7)
¯ ) ∩ C 2+α (Ω ), which is a supersolution to problem (2.1). has a unique solution u¯ ∈ C (Ω Using the same maximum principle argument as the following proof of (3.2) in Section 3, we can obtain that u(x) ≤ ¯ ) in the ordered u¯ (x), x ∈ Ω . It follows by Lemma 2.1 that problem (2.1) has at least one solution u ∈ C 2+α (Ω ) ∩ C (Ω interval [u, u¯ ]. 2 Lemma 2.5. Let b satisfy (b1 ) and (b2 ). If g satisfies (g1 )–(g3 ) and f¯1 is in Lemma 2.3, then there exists a function v ∈ Cloc (RN ) satisfying
1 ¯ − ∆v ≥ b(x)v f1 (v) + , v
v(x) > 0, x ∈ RN , lim v(x) = 0. |x|→∞
(2.8)
Proof. By (g1 )–(g3 ), we define
Γ (t ) =
t
Z 0
s sf¯1 (s) + 1
ds,
t ≥ 0.
It follows by L’Hôspital’s rule that lim
t →∞
Γ (t ) t
t
= lim
t →∞
tf 1 (t ) + 1
= lim
t →∞
1 f 1 (t ) + t −1
= ∞.
Let w be the solution to problem (1.2) and c0 = maxRN w(x). Therefore, we see that there exists a positive constant c2 such that c0 c2 ≤ Γ (c2 ) =
c2
Z 0
s sf 1 (s) + 1
ds.
We now define a function v by
w(x) =
1 c2
v(x)
Z 0
s sf 1 (s) + 1
ds,
∀ x ∈ RN .
H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
3117
Then 0 < v(x) ≤ c2
lim v(x) = 0.
and
|x|→∞
Moreover, by Lemma 2.3, we obtain c2 b(x) = −c2 ∆w
= ≤
−∆v f 1 (v(x)) + (v(x))−1
−∆v
f 1 (v(x)) + (v(x))−1
d
1
+
dv
,
x ∈ RN ,
f 1 (v) + (v)−1
|∇v(x)|2
i.e.,
−∆v ≥ c2 b(x)(f¯1 (v) + v −1 ) ≥ b(x)v(f¯1 (v) + v −1 ),
x ∈ RN .
3. Proof of Theorem 1.1 Consider the perturbed problem
− ∆uk + h(x)|∇ uk |q = b(x)g (uk ),
uk > 0, x ∈ B(0, k), uk |∂ B(0,k) = 0,
(3.1)
where B(0, k) = {x ∈ R : |x| < k}, k = 1, 2, 3, . . . . It follows by Lemma 2.4 that problem (3.1) has one solution uk ∈ C 2+α (B(0, k)) ∩ C (B¯ (0, k)). Put N
uk ( x ) = 0 ,
∀ |x| > k.
Let v be as in Lemma 2.5, we assert that uk (x) ≤ v(x),
x ∈ RN , k = 1, 2, 3, . . . .
(3.2)
Assume the contrary, i.e., there exists a positive integer m and x0 ∈ B(0, m) such that v(x0 ) < um (x0 ), i.e., sup (ln(um (x)) − ln(v(x)))
x∈B(0,m)
exists and is positive in B(0, m). At the point, say x1 , we have
∇ (ln(um (x1 )) − ln(v(x1 ))) = 0 and ∆ (ln(um (x1 )) − ln(v(x1 ))) ≤ 0. On the other hand, it follows by the definition of f¯1 that g (um (x1 )) um ( x 1 )
≤ f¯1 (v(x1 )).
So, we see that
∆um (x1 ) ∆v(x1 ) |∇ um (x1 )|2 |∇v(x1 )|2 − − + 2 um ( x 1 ) v(x1 ) (um (x1 )) (v(x1 ))2 ∆um (x1 ) ∆v(x1 ) = − um ( x 1 ) v(x1 ) h(x1 )|∇ um (x1 )|α g (um (x1 )) 1 ≥ − b(x1 ) − (f¯1 (v(x1 )) + ) > 0, um ( x 1 ) um (x1 ) v(x1 )
∆ (ln(um (x1 )) − ln(v(x1 ))) =
which is a contradiction. Hence (3.2) holds. Now, we need to estimate {uk }. For any bounded C 2+α -smooth domain Ω 0 ⊂ RN , take Ω1 and Ω2 with C 2+α -smooth boundaries, and K1 large enough, such that
Ω 0 ⊂⊂ Ω1 ⊂⊂ Ω2 ⊂⊂ Bk ,
k ≥ K1 .
Note that uk (x) ≥ u(x) > 0,
∀ x ∈ B(0, K1 );
(3.3)
when B(0, K1 ) is the substitution for Ω in the proof of Lemma 2.4. Let
ρk (x) = b(x)g (uk (x)) − h(x)|∇ uk (x)|q ,
x ∈ B¯ (0, K1 ).
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H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
Since −∆uk (x) = ρk (x), x ∈ B(0, K1 ), by the interior estimate theorem of Ladyzenskaja and Ural’tseva [30, Theorem 3.1, p. 266], we get a positive constant C1 independent of k such that max |∇ uk (x)| ≤ C1 max uk (x) ≤ C1 max v(x), x∈B¯ (0,K1 )
¯2 x∈Ω
x∈B¯ (0,K1 )
∀ x ∈ B(0, K1 ),
(3.4)
p ¯ 2 . It follows that {ρk }∞ ¯ i.e., |∇ uk (x)| is uniformly bounded on Ω K1 is uniformly bounded on Ω2 and hence ρk ∈ L (Ω2 ) for any p > 1. Since −∆uk (x) = ρk (x), x ∈ Ω2 , we see by [31, Theorem 9.11] that there exists a positive constant C2 independent of k such that
kuk kW 2,p (Ω1 ) ≤ C2 kρk kLp (Ω2 ) + kuk kLp (Ω2 ) ,
∀ k ≥ K1 .
(3.5)
Taking p > N such that α < 1 − N /p and applying Sobolev’s embedding inequality, we see that {kuk kC 1+α (Ω¯ 1 ) }∞ K1 is
¯ 1 ) and {kρk kC α (Ω¯ ) }∞ uniformly bounded. Therefore ρk ∈ C α (Ω is uniformly bounded. It follows by Schauder’s interior 1 K1 estimate theorem (see [31, Chapter 1, p. 2]) that there exists a positive constant C3 independent of k such that kuk kC 2+α (Ω¯ 0 ) ≤ C3 kρk kC α (Ω¯ 1 ) + kuk kC (Ω¯ 1 ) ,
∀ k ≥ K1 ;
(3.6)
∞
i.e., {kuk kC 2+α (Ω¯ 0 ) }K1 is uniformly bounded. Using Ascoli–Arzela’s theorem and the diagonal sequential process, we see that 2 ¯ 0 2 ¯ 0 {uk }∞ K1 has a subsequence that converges uniformly in the C (Ω ) norm to a function u ∈ C (Ω ) and u satisfies
−∆u + h(x)|∇ u|q = b(x)g (u),
¯ 0. x∈Ω
By (3.3), we obtain that u > 0,
¯ 0. ∀x∈Ω
2+α ¯ 0 ). Since Ω 0 is arbitrary, we also see that u ∈ Cloc Applying Schauder’s regularity theorem we see that u ∈ C 2+α (Ω (RN ). It follows by (3.2) that lim|x|→∞ u(x) = 0. Thus, a standard bootstrap argument (with the same details as in [9]) shows that u is a classical solution to problem (1.1). The proof is finished.
Acknowledgment The authors would like to thank Professor Zhijun Zhang for valuable suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31]
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Existence of positive entire solutions of a semilinear elliptic problem with a gradient term Hongtao Xue ∗ , Xigao Shao School of Science, Yantai Nanshan University, Yantai, Shandong 265713, PR China
article
abstract
info
Article history: Received 14 May 2008 Accepted 26 January 2009
By a sub–supersolution argument and a perturbed argument, we show the existence of entire solutions to a semilinear elliptic problem −∆u + h(x)|∇ u|q = b(x)g (u), u > 0, α x ∈ RN , lim|x|→∞ u(x) = 0, where q ∈ (1, 2], b, h ∈ Cloc (RN ) for some α ∈ (0, 1), N 1 h(x) ≥ 0, b(x) > 0, ∀ x ∈ R , and g ∈ C ((0, ∞), (0, ∞)) which may be singular at 0. No monotonicity condition is imposed on the functions g (s) and g (s)/s. © 2009 Elsevier Ltd. All rights reserved.
MSC: 35J65 35B05 35O75 35R05 Keywords: Semilinear elliptic equations Entire solutions Existence
1. Introduction and the main results The purpose of this note is to investigate the existence of entire solutions to the following model problem
− ∆u + h(x)|∇ u|q = b(x)g (u),
u > 0, x ∈ RN ,
lim u(x) = 0,
|x|→∞
(1.1)
α where q ∈ (1, 2], h ∈ Cloc (RN ) for some α ∈ (0, 1) is non-negative in Ω , g satisfies
(g1 ) g ∈ C 1 ((0, ∞), (0, ∞)); (g2 ) lims→0+ g (s)/s = ∞; (g3 ) lims→∞ g (s)/s = 0; and b satisfies α (b1 ) b ∈ Cloc (RN ) and b(x) > 0, ∀ x ∈ RN ; (b2 ) the linear problem
− ∆u = b(x),
u > 0, x ∈ RN ,
lim u(x) = 0
|x|→∞
2+α has a unique solution w ∈ Cloc (RN ).
∗
Corresponding author. E-mail address: [email protected] (H. Xue).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.01.222
(1.2)
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H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
First, let us review the following model
− ∆u = b(x)g (u),
u > 0, x ∈ RN ,
lim u(x) = 0.
(1.3)
|x|→∞
Problem (1.3) arises from many branches of mathematics and applied mathematics. It was discussed and extended to more general problems in a number of works, for instance, [1–13]. When the equation is considered over a bounded smooth domain Ω instead of RN , the corresponding problem was studied, for example, in [14–25] and the references cited therein. For g (u) = u−γ with γ > 0, if b satisfies (b1 ) and the following condition (b3 )
R∞ 0
r φ(r )dr < ∞, where φ(r ) = max|x|=r b(x),
2+α Lair and Shaker [9] showed that problem (1.3) has a unique solution u ∈ Cloc (RN ). Later, Lair and Shaker [10] and Zhang [13] extended the above result to the more general g which satisfies (g1 ) and
(g4 ) g is non-increasing on (0, ∞) and lims→0+ g (s) = ∞. Cˇirstea and Rˇadulescu [3] also extended the above results to the more general g which satisfies (g1 ), (g2 ) and g (s)
(g5 ) s+s is decreasing on (0, ∞) for some s0 > 0; 0 (g6 ) g is bounded in a neighborhood of ∞. Recently, Dinu [26] further generalized the above results to the cases that (i) b satisfies (b1 ) and (b3 ); (ii) g satisfies (g1 ), (g3 ) and g (s) (g7 ) s is decreasing on (0, ∞); and lim|x|→∞ u(x) = l > 0 instead of lim|x|→∞ u(x) = 0 in problem (1.3); or the following cases that (i1 ) (b4 ) (ii2 ) (g8 )
b satisfies (b1 ) and R∞ r N −1 φ(r )dr < ∞, where N ≥ 3; 0 g satisfies (g1 )–(g3 ) and g is increasing on (0, ∞).
Afterwards, Goncalves and Santos [8] also generalized the above results to the case that g satisfies (g1 )–(g3 ) and (g7 ). Ye and Zhou [12, Theorem 4.2] showed that if g satisfies (g1 ) and is non-increasing on (0, ∞), b satisfies (b1 ), then problem (1.3) admits a solution if and only if b satisfies (b2 ). Moreover, if a solution of problem (1.3) exists, it is unique. Now let us return to problem (1.1). α When g (u) = u−γ with γ > 0, q ∈ (1, 2], h ∈ Cloc (RN ) is non-negative in RN , b satisfies (b1 ) and (b3 ), Dinu [26] showed 2+α α that problem (1.1) has a unique solution u ∈ Cloc (RN ). Recently, the author [27] showed that whenq ∈ (0, 1), h ∈ Cloc (RN ) for some α ∈ (0, 1), h(x) < 0, ∀ x ∈ RN , g satisfies (g1 )–(g3 ), and b satisfies (b3 ) instead of (b2 ), problem (1.1) have at least one solution. In this paper we continue to consider the existence of entire solutions to problem (1.1) for the functions g (s) and g (s)/s which do not have monotonicity. Our main result is summarized in the following theorem. α Theorem 1.1. Let q ∈ (1, 2], h ∈ Cloc (RN ) be non-negative in RN , and b satisfy (b1 ) and (b2 ). If g satisfies (g1 )–(g3 ), then 2+α problem (1.1) has at least one solution u ∈ Cloc (RN ).
Remark 1.1 ([27]). The condition (b3 ) implies (b2 ), but (b2 ) is invariant under translations. Remark 1.2. Some basic examples of the functions, which satisfy (g1 )–(g3 ), are (i) (ii) (iii) (iv)
u−γ + up + sin f (u) + 1, where γ > 0, p < 1 and f ∈ C 2 (R); γ e1/u + up + cos f (u) + 1, where γ > 0, p < 1 and f ∈ C 2 (R); −γ u ln−q1 (1 + u) + lnq2 (1 + u) + up + sin f (u) + 2 with f ∈ C 2 (R), γ > 0, p < 1, q2 > 0 and q1 > 0; u−γ + arctan f (u) + π with f ∈ C 2 (R) and γ > 0.
Remark 1.3. The technique of this paper in our proofs can be applied to the more general problem
−∆u + h(x)|∇ u|q = b(x)g (u) + a(x)f (u),
u > 0, x ∈ RN ,
lim u(x) = 0,
|x|→∞
α where q ∈ (1, 2], h ∈ Cloc (RN ) is non-negative in RN , a and b satisfy (b1 ) and (b2 ), g and f satisfy (g1 ), g + f satisfies (g2 )–(g3 ).
The paper is organized as follows. In Section 2 we give some preliminary considerations. Finally we show the existence of solutions to problem (1.1).
H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
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2. Preliminaries We first consider the following Dirichlet problem
− ∆u + h(x)|∇ u|q = b(x)g (u),
u > 0, x ∈ Ω , u|∂ Ω = 0,
(2.1)
where Ω is a bounded domain with smooth boundary in RN (N ≥ 1). Concerning the existence of the solutions to problem (2.1), refer to [28,29,20] and the references cited therein. For the convenience, we denote f (x, u, ∇ u) = b(x)g (u) − h(x)|∇ u|q . Now we introduce a sub–supersolution method with the boundary restriction. Definition 2.1. A function u ∈ C 2+α (Ω ) ∩ C (Ω ) is called a subsolution of problem (2.1) if
− ∆u ≤ f (x, u, ∇ u),
u > 0, x ∈ Ω , u|∂ Ω = 0.
(2.2)
Definition 2.2. A function u¯ ∈ C 2+α (Ω ) ∩ C (Ω ) is called a supersolution of problem (2.1) if
− ∆u¯ ≥ f (x, u¯ , ∇ u¯ ),
u¯ > 0, x ∈ Ω , u¯ |∂ Ω = 0.
(2.3)
Lemma 2.1 ([15, Lemma 3]). Let f (x, u, ξ ) satisfy the following two basic conditions: (D1 ) f (x, u, ξ ) is locally Hölder continuous in Ω × (0, ∞) × RN and continuously differentiable with respect to the variables u and ξ ; (D2 ) for any Ω1 ⊂⊂ Ω and any a, b ∈ (0, ∞)(a < b), there exists a corresponding constant C = C (Ω1 , a, b) > 0 such that ¯ 1 , ∀ u ∈ [a, b], ∀ ξ ∈ RN . |f (x, u, ξ )| ≤ C (1 + |ξ |2 ), ∀ x ∈ Ω If problem (2.1) has a supersolution u¯ and a subsolution u such that u ≤ u¯ in Ω , then problem (2.1) has at least one solution ¯ ) in the ordered interval [u, u¯ ]. u ∈ C 2+α (Ω ) ∩ C (Ω Lemma 2.2 (Existence, [8, Theorem 1.2 and Proof of Theorem 1.1]; uniqueness, [3, Section 2]). Let b satisfy (b1 ). If g satisfies (g1 )–(g3 ) and (g7 ), then the following problem
− ∆u = b(x)g (u),
u > 0, x ∈ Ω , u|∂ Ω = 0
¯)∩C has a unique solution u ∈ C (Ω
2+α
(2.4)
(Ω ).
Lemma 2.3. If g satisfies (g1 )–(g3 ), then there exists a function f¯1 such that (i) (ii) (iii) (iv)
f¯1 ∈ C 1 ((0, ∞), (0, ∞)); g (s)/s ≤ f¯1 (s), ∀ s > 0 and lims→0+ f¯1 (s) = ∞; f¯1 is non-increasing on (0, ∞); lims→∞ f¯1 (s) = 0.
Proof. By (g1 )–(g3 ), we can denote f¯ (s) = sup g (t )/t .
(2.5)
t ≥s>0
Observe that f¯ (s) ≥ g (t )/t ,
∀ s > 0 and t ≥ s;
and f¯ is non-increasing on (0, ∞). Moreover, lim f¯ (s) = ∞ and
s→0+
lim f¯ (s) = 0.
s→∞
Now we can assume f¯ ∈ C 1 (0, ∞). If not, we can replace it by 2 f¯1 (s) = s
Z
s s/2
f¯ (t )dt ,
s > 0.
Obviously, f¯ (s) ≤ f¯1 (s) ≤ f¯ (s/2),
∀ s > 0.
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H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
And, for s > 0, f¯10 (s) =
≤
2
s 2
s
f¯ (s) −
1
f¯ (s) −
1
2 2
f¯ (s/2)
s2
f¯ (s/2)
s
Z
2
−
s/2
2 s
−
s2 2
f¯ (t )dt
f¯ (s) =
1 s
(f¯ (s) − f¯ (s/2)) ≤ 0,
i.e., f¯1 ∈ C 1 ((0, ∞), (0, ∞)). The proof of Lemma 2.3 is completed.
¯ ). For the convenience, we denote |u|∞ = maxx∈Ω¯ |u(x)| whenever u ∈ C (Ω ¯ ), h(x) ≥ 0, b(x) > 0, ∀ x ∈ Ω ¯ . If g satisfies (g1 )–(g3 ), then problem (2.1) has at least Lemma 2.4. Let q ∈ (1, 2], b, h ∈ C α (Ω ¯ ) ∩ C 2+α (Ω ). one solution u ∈ C (Ω ¯ ) ∩ C 2+α (Ω ) be the first eigenfunction corresponding to the first eigenvalue λ1 of Proof. Let ψ1 ∈ C 1 (Ω − ∆u = λu,
u > 0, x ∈ Ω , u|∂ Ω = 0.
(2.6)
Let β = q/(q − 1). It follows by (g2 ) that there exists a positive constant δ1 ∈ (0, 1) such that g (s) s
≥
λ1 β + |h|∞ β q |∇ψ1 |q∞ , min b(x)
∀ s ∈ (0, δ1 ).
¯ x∈Ω
β
δ1
Let u = c1 ψ1 with c1 ∈ 0, min{1,
β |ψ1 |∞
} . Since c1q−1 < 1, we see that
β
β−2
−∆u + h(x)|∇ u|q = βλ1 c1 ψ1 − c1 β(β − 1)ψ1 β
q(β−1)
|∇ψ1 |2 + h(x)β q c1q ψ1 β
≤ min b(x)g (c1 ψ1 ) ≤ b(x)g (c1 ψ1 ) = b(x)g (u), ¯ x∈Ω
|∇ψ1 |q
x ∈ Ω,
β
i.e., u = c1 ψ1 is a subsolution to problem (2.1). To construct a supersolution, we see by Lemma 2.2 that the following problem
1 , − ∆u = b(x)u f¯1 (u) +
u > 0, x ∈ Ω , u|∂ Ω = 0,
u
(2.7)
¯ ) ∩ C 2+α (Ω ), which is a supersolution to problem (2.1). has a unique solution u¯ ∈ C (Ω Using the same maximum principle argument as the following proof of (3.2) in Section 3, we can obtain that u(x) ≤ ¯ ) in the ordered u¯ (x), x ∈ Ω . It follows by Lemma 2.1 that problem (2.1) has at least one solution u ∈ C 2+α (Ω ) ∩ C (Ω interval [u, u¯ ]. 2 Lemma 2.5. Let b satisfy (b1 ) and (b2 ). If g satisfies (g1 )–(g3 ) and f¯1 is in Lemma 2.3, then there exists a function v ∈ Cloc (RN ) satisfying
1 ¯ − ∆v ≥ b(x)v f1 (v) + , v
v(x) > 0, x ∈ RN , lim v(x) = 0. |x|→∞
(2.8)
Proof. By (g1 )–(g3 ), we define
Γ (t ) =
t
Z 0
s sf¯1 (s) + 1
ds,
t ≥ 0.
It follows by L’Hôspital’s rule that lim
t →∞
Γ (t ) t
t
= lim
t →∞
tf 1 (t ) + 1
= lim
t →∞
1 f 1 (t ) + t −1
= ∞.
Let w be the solution to problem (1.2) and c0 = maxRN w(x). Therefore, we see that there exists a positive constant c2 such that c0 c2 ≤ Γ (c2 ) =
c2
Z 0
s sf 1 (s) + 1
ds.
We now define a function v by
w(x) =
1 c2
v(x)
Z 0
s sf 1 (s) + 1
ds,
∀ x ∈ RN .
H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
3117
Then 0 < v(x) ≤ c2
lim v(x) = 0.
and
|x|→∞
Moreover, by Lemma 2.3, we obtain c2 b(x) = −c2 ∆w
= ≤
−∆v f 1 (v(x)) + (v(x))−1
−∆v
f 1 (v(x)) + (v(x))−1
d
1
+
dv
,
x ∈ RN ,
f 1 (v) + (v)−1
|∇v(x)|2
i.e.,
−∆v ≥ c2 b(x)(f¯1 (v) + v −1 ) ≥ b(x)v(f¯1 (v) + v −1 ),
x ∈ RN .
3. Proof of Theorem 1.1 Consider the perturbed problem
− ∆uk + h(x)|∇ uk |q = b(x)g (uk ),
uk > 0, x ∈ B(0, k), uk |∂ B(0,k) = 0,
(3.1)
where B(0, k) = {x ∈ R : |x| < k}, k = 1, 2, 3, . . . . It follows by Lemma 2.4 that problem (3.1) has one solution uk ∈ C 2+α (B(0, k)) ∩ C (B¯ (0, k)). Put N
uk ( x ) = 0 ,
∀ |x| > k.
Let v be as in Lemma 2.5, we assert that uk (x) ≤ v(x),
x ∈ RN , k = 1, 2, 3, . . . .
(3.2)
Assume the contrary, i.e., there exists a positive integer m and x0 ∈ B(0, m) such that v(x0 ) < um (x0 ), i.e., sup (ln(um (x)) − ln(v(x)))
x∈B(0,m)
exists and is positive in B(0, m). At the point, say x1 , we have
∇ (ln(um (x1 )) − ln(v(x1 ))) = 0 and ∆ (ln(um (x1 )) − ln(v(x1 ))) ≤ 0. On the other hand, it follows by the definition of f¯1 that g (um (x1 )) um ( x 1 )
≤ f¯1 (v(x1 )).
So, we see that
∆um (x1 ) ∆v(x1 ) |∇ um (x1 )|2 |∇v(x1 )|2 − − + 2 um ( x 1 ) v(x1 ) (um (x1 )) (v(x1 ))2 ∆um (x1 ) ∆v(x1 ) = − um ( x 1 ) v(x1 ) h(x1 )|∇ um (x1 )|α g (um (x1 )) 1 ≥ − b(x1 ) − (f¯1 (v(x1 )) + ) > 0, um ( x 1 ) um (x1 ) v(x1 )
∆ (ln(um (x1 )) − ln(v(x1 ))) =
which is a contradiction. Hence (3.2) holds. Now, we need to estimate {uk }. For any bounded C 2+α -smooth domain Ω 0 ⊂ RN , take Ω1 and Ω2 with C 2+α -smooth boundaries, and K1 large enough, such that
Ω 0 ⊂⊂ Ω1 ⊂⊂ Ω2 ⊂⊂ Bk ,
k ≥ K1 .
Note that uk (x) ≥ u(x) > 0,
∀ x ∈ B(0, K1 );
(3.3)
when B(0, K1 ) is the substitution for Ω in the proof of Lemma 2.4. Let
ρk (x) = b(x)g (uk (x)) − h(x)|∇ uk (x)|q ,
x ∈ B¯ (0, K1 ).
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H. Xue, X. Shao / Nonlinear Analysis 71 (2009) 3113–3118
Since −∆uk (x) = ρk (x), x ∈ B(0, K1 ), by the interior estimate theorem of Ladyzenskaja and Ural’tseva [30, Theorem 3.1, p. 266], we get a positive constant C1 independent of k such that max |∇ uk (x)| ≤ C1 max uk (x) ≤ C1 max v(x), x∈B¯ (0,K1 )
¯2 x∈Ω
x∈B¯ (0,K1 )
∀ x ∈ B(0, K1 ),
(3.4)
p ¯ 2 . It follows that {ρk }∞ ¯ i.e., |∇ uk (x)| is uniformly bounded on Ω K1 is uniformly bounded on Ω2 and hence ρk ∈ L (Ω2 ) for any p > 1. Since −∆uk (x) = ρk (x), x ∈ Ω2 , we see by [31, Theorem 9.11] that there exists a positive constant C2 independent of k such that
kuk kW 2,p (Ω1 ) ≤ C2 kρk kLp (Ω2 ) + kuk kLp (Ω2 ) ,
∀ k ≥ K1 .
(3.5)
Taking p > N such that α < 1 − N /p and applying Sobolev’s embedding inequality, we see that {kuk kC 1+α (Ω¯ 1 ) }∞ K1 is
¯ 1 ) and {kρk kC α (Ω¯ ) }∞ uniformly bounded. Therefore ρk ∈ C α (Ω is uniformly bounded. It follows by Schauder’s interior 1 K1 estimate theorem (see [31, Chapter 1, p. 2]) that there exists a positive constant C3 independent of k such that kuk kC 2+α (Ω¯ 0 ) ≤ C3 kρk kC α (Ω¯ 1 ) + kuk kC (Ω¯ 1 ) ,
∀ k ≥ K1 ;
(3.6)
∞
i.e., {kuk kC 2+α (Ω¯ 0 ) }K1 is uniformly bounded. Using Ascoli–Arzela’s theorem and the diagonal sequential process, we see that 2 ¯ 0 2 ¯ 0 {uk }∞ K1 has a subsequence that converges uniformly in the C (Ω ) norm to a function u ∈ C (Ω ) and u satisfies
−∆u + h(x)|∇ u|q = b(x)g (u),
¯ 0. x∈Ω
By (3.3), we obtain that u > 0,
¯ 0. ∀x∈Ω
2+α ¯ 0 ). Since Ω 0 is arbitrary, we also see that u ∈ Cloc Applying Schauder’s regularity theorem we see that u ∈ C 2+α (Ω (RN ). It follows by (3.2) that lim|x|→∞ u(x) = 0. Thus, a standard bootstrap argument (with the same details as in [9]) shows that u is a classical solution to problem (1.1). The proof is finished.
Acknowledgment The authors would like to thank Professor Zhijun Zhang for valuable suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31]
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