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Large and entire large solution for a quasilinear problem
Nonlinear Analysis 70 (2009) 1738–1745 www.elsevier.com/locate/na
Large and entire large solution for a quasilinear problem Dragos-Patru Covei ∗ Department of Mathematics, University Constantin Brancusi of Tg-Jiu, Bld. Republicii 1, 210152, Romania Received 18 October 2007; accepted 21 February 2008
Abstract In this article we give the existence results for the quasilinear elliptic problem ∆ p u = b(x) f (u) in Ω , u(x) → ∞ as x → ∂Ω , where Ω ⊆ R N is a domain. c 2008 Elsevier Ltd. All rights reserved.
MSC: 35J60 Keywords: Large solution; Entire solution
1. Introduction and the main results In this article, we are concerned with the existence of a blow-up solution for the quasilinear elliptic equation ∆ p u = b(x) f (u),
u > 0, in Ω , u(x)
→
∞,
as x→∂ Ω
(1.1)
where N > 2, 1 < p < N , ∆ p u is the p-Laplacian operator, Ω = R N or Ω is a domain in R N and b(x) is assumed to satisfy the following hypotheses, not necessary simultaneous: (b1) b is nonnegative, nontrivial in Ω and it has the property: for any x0 ∈ Ω satisfying b(x0 ) = 0, there exists a domain Ω0 such that x0 ∈ Ω0 , Ω 0 b Ω , and b(x) > 0 for all x ∈ ∂Ω0 ; (b2) b ∈ C 0,α (Ω ); 0,α (b3) For b(x) ∈ Cloc (R N ) and Φ(r ) = max|x|=r b(x) Z ∞ r 1/( p−1) Φ 1/( p−1) (r )dr < ∞ if 1 < p ≤ 2, Z0 ∞ ( p−2)N +1 r p−1 Φ(r )dr < ∞ if 2 ≤ p < ∞. 0
We assume that the nonlinearity f satisfies (f1) f ∈ C 1 (R), f 0 (s) ≥ 0 and f (s) > 0, ∀s ∈ R, or f ∈ C 1 [0, ∞), f (0) = 0, f 0 (s) ≥ 0 and f (s) > 0 for s > 0; ∗ Tel.: +40 0742269582.
E-mail address: [email protected]. c 2008 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2008.02.057
D.-P. Covei / Nonlinear Analysis 70 (2009) 1738–1745
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R +∞ (f2) α [ p · ( p − 1)−1 · F(s)]−1/ p ds < +∞, F 0 (s) = f (s), ∀α > 0. This problem has been considered by several authors in the case p = 2 and in the case of a bounded domain Ω . The first example of a boundary blow-up problem where Ω ⊂ R2 is a open regular bounded domain, f (u) = eu , b(x) = 1 has been already studied by Bieberbach [1] in 1916. In this case the problem plays an important role in the theory of Riemannian surfaces of constant negative curvatures and in the theory of automorphic functions. When Ω ⊂ R3 this problem arises in the study of the electrostatic potential in a glowing hollow metal body. Rademacher [11] in 1943, using the ideas of Bieberbach, extended this result to a bounded domain in R3 with C 2 boundary. Later, the problem of existence of solutions has been studied extensively by many other authors for a more general class of functions. Assumption (f2) has been first introduced in 1957 by Keller [6] and Osserman [14]. Loewner and Nirenberg in [8] have studied the problem of uniqueness and asymptotic behavior of solutions when b(x) = const > 0 and f (u) = u (N +2)/(N −2) , N > 2 with an arbitrary domain instead of just R N . They showed that this problem is motivated by a concrete problem arising in Riemannian Geometry. In particular, we mention the work of Matero [9] and the work of Mohammed [10] where it is proved the existence of solutions to the problem (1.1) when ∆ is replaced by the p-Laplacian. In this paper we will use the same technique as in [7,12,15] in order to prove the existence of large solutions of (1.1), that are solutions u satisfying u(x) → ∞ as x → ∂Ω (if Ω 6= R N ), or u(x) → ∞ as |x| → ∞ (if Ω = R N ). In the latter case the solution is called an entire large solution. For the study of a solution given in this paper we advise the reader to consult the reference [10]. Our main results are summarized in the following theorems. Theorem 1.1 (See also [5]). If b satisfies (b1), (b2) and Ω is a bounded C 2 open domain of R N then − ∆ p v = b(x),
v > 0,
in Ω ,
v=0
on ∂Ω ,
(1.2)
C 1,α (Ω ).
has a unique weak solution v ∈ Moreover, there exist constants l1 ≥ k1 > 0, such that k1 d(x) ≤ v(x) ≤ l1 d(x) on Ω , where d(x) denotes the distance from x ∈ Ω to the boundary ∂Ω . Theorem 1.2. Let f and b satisfy the assumptions (f1), (f2), (b1), (b2) and Ω be a bounded C 2 open domain of R N . Then, the problem (1.1) has a large solution. Theorem 1.3. Let f and b satisfy the assumptions (f1), (f2), (b1), (b3) and Ω = R N . Then, the problem (1.1) has an entire large solution. We establish some preliminary results that will be necessary later to prove our main theorems. 2. Preliminary results In this section, we will give some lemmas. The next lemma can be found in [12]. Lemma 2.1. Let Ω be a bounded open set of R N which satisfies the uniform external sphere condition, then there exists a sequence {Ωm }∞ 1 of open sets such that Ω m ⊂ Ωm+1 ⊂ Ω ,
∪∞ m=1 Ωm = Ω ,
and the boundary ∂Ωm is a smooth (C ∞ ) submanifold of dimension N − 1 for m ≥ 1. Lemma 2.2 (See [3]). If b satisfies (b1), (b2) and Ω is a bounded C 2 open domain of R N then the problem −∆ p u(x) = b(x)[u(x) + ]−1 ,
u > 0, x ∈ Ω ,
u |∂ Ω = 0,
(2.1)
has a unique solution u ∈ C 1,α (Ω ) for any ε > 0. Lemma 2.3. Suppose that (b1) and (b3) are satisfied. Then the problem ( p−2 0 −(r N −1 v 0 (r ) v (r ))0 = r N −1 Φ(r )(r =| x |) on R N , lim v(r ) = 0, v 0 (0) = 0, r →∞
has a unique positive bounded radially symmetric solution.
(2.2)
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Proof. Set I (r ) :=
r
Z
ξ 1−N
0
ξ
Z
σ N −1 Φ(σ )dσ
1/( p−1)
r > 0.
dξ,
0
Now we prove that Z Z +∞ 1−N K := ξ
ξ
σ
N −1
Φ(σ )dσ
1/( p−1) dξ,
0
0
which is, in fact, finite. 1 Case 1. 1 < p ≤ 2. Since in this case 1 ≤ p−1 < ∞, by the Hardy inequality (see [2, Theorem 2.1]), we have 1/( p−1) Z ξ 1/( p−1) Z ξ Z +∞ Z +∞ − N −1 σ N −1 Φ(σ )dσ dξ = ξ p−1 σ N −1 Φ(σ )dσ dξ ξ 1−N 0
0
0
0
−1 #1/( p−1) Z +∞ 1 N −1 − N −1 ≤ · −1 ξ p−1 (ξ ξ N −1 Φ(ξ ))1/( p−1) dξ p−1 p−1 0 1 Z +∞ p−1 1 ξ 1/( p−1) Φ 1/( p−1) (ξ )dξ < ∞. = N−p 0 "
1 Case 2. Let 2 ≤ p < +∞, so 1 ≤ p − 1, it follows that 1 ≥ p−1 > 0. Set Z ξ Z ξ σ N −1 Φ(σ )dσ ≤ 1 for ξ > 0, or σ N −1 Φ(σ )dσ > 1 for ξ > 0. 0
0
In the first case Z ξ 1/( p−1) N −1 σ Φ(σ )dσ ≤ 1, 0
so Z
r
ξ
1−N p−1
ξ
Z
0
σ N −1 Φ(σ )dσ
1/( p−1)
Z
r
dξ ≤
0
ξ
1−N p−1
dξ,
0
is finite as r → ∞ and N > p. In the second case, Z ξ 1/( p−1) Z ξ N −1 σ Φ(σ )dσ ≤ σ N −1 Φ(σ )dσ, 0
0
for ξ ≥ 0, so Z Z r 1−N ξ p−1 0
ξ
σ N −1 Φ(σ )dσ
1/( p−1)
Z dξ ≤
0
r
ξ
1−N p−1
0
Z
ξ
σ N −1 Φ(σ )dσ dξ.
0
Integration by parts gives Z r Z ξ Z Z ξ 1−N p − 1 r d p−N ξ p−1 σ N −1 Φ(σ )dσ dξ = − ξ p−1 σ N −1 Φ(σ )dσ dξ N − p dξ 0 0 0 0 Z r Z r ( p−2)N +1 p−N p−1 N −r p−1 = σ −1 Φ(σ )dσ + ξ p−1 Φ(ξ )dξ . N−p 0 0 Using L’ Hˆopital’s rule, we have Z r Z p−N lim −r p−1 σ N −1 Φ(σ )dσ + r →∞
0
= lim
−
Rr 0
ξ
( p−2)N +1 p−1
Φ(ξ )dξ
0
σ N −1 Φ(σ )dσ + r
N−p p−1
Rr 0
ξ
( p−2)N +1 p−1
Φ(ξ )dξ
N−p p−1
r →∞
Z = lim
r
r →∞ 0
r
ξ
( p−2)N +1 p−1
r Z Φ(ξ )dξ =
0
∞
ξ
( p−2)N +1 p−1
Φ(ξ )dξ < ∞.
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Now setting w(r ) := K − I (r ), it is easily shown that w is the unique positive bounded solution of the problem (2.2). The next Weak comparison principle can be found in [2]. Lemma 2.4. Let Ω be a bounded domain in R N , N ≥ 2 with smooth boundary ∂Ω and w1 , w2 ∈ W 1, p (Ω ) satisfy −∆ p w1 ≤ −∆ p w2
in Ω .
If w1 ≤ w2 on ∂Ω then w1 ≤ w2
in Ω .
Lemma 2.5. If f satisfies (f1) and (f2) then f (t) F(t)
p−1 p
(2.3)
→ ∞ as t → ∞.
Therefore, there exists γ > 0 such that Z ∞ ds < ∞. 1/( p−1) (s) f γ
(2.4)
Proof. We remark that limt→∞ F(t)/t p = +∞, since condition (f2) implies that Z t t 1 ds t− ≤ → 0, √ √ p p 2 F(t) F(s) t/2 as t → +∞. On the other hand, by (f1) we have f (t) F(t) ≥ p , t t p−1 which implies that
(2.5)
f (t) = ∞. t→+∞ t p−1 lim
Using again (2.5), we have 1 f (t) f (t) p ≥ → +∞, p−1 t p−1 F(t) p as t → +∞. The proof of (2.3) is complete. Now, from (2.3) there exists t1 > 0, such that f (t) ≥ F(t)
p−1 p
,
(2.6)
for all t ≥ t1 . But the relation (2.6) is equivalent to 1 1 ≤ √ . p f 1/( p−1) (s) F(s)
(2.7)
Integrating (2.7) from t1 to +∞ we have Z ∞ Z ∞ 1 1 dt ≤ dt < ∞. √ p 1/( p−1) (t) f F(t) t1 t1 The proof of (2.4) is complete for γ := t1 .
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In view of this we define Z +∞ ds , R(u) = 1/( p−1) (s) f u(x)
and w(x) = R(u(x)),
and then R : (0, ∞) → (0, ∞) is strictly decreasing. Then (1.1) is transformed into the following equivalent singular Dirichlet problem −∆ p w + g(w) |∇w| p = b(x), w > 0, in Ω , (2.8) w → 0, as x → ∂Ω , where g(w) =
f 0 (u) f ( p−2)/( p−1) (u)
=
f 0 (R −1 (w)) , f ( p−2)/( p−1) (R −1 (w))
and R −1 denotes the inverse function which is strictly decreasing in (0, ∞). In particular, when f (u) = eu , we have w = ( p − 1)e−u(x)/( p−1) , and −∆ p w + ( p − 1)
|∇w| p = b(x), w
w > 0, in Ω ,
w → 0, as x → ∂Ω ,
when f (u) = u γ and γ > p − 1, we have w(x) = C[u(x)]−C |∇w| p = b(x), w C := ( p − 1)/(γ − p + 1). −∆ p w + γ C
w > 0, in Ω ,
−1
and
w → 0, as x → ∂Ω ,
Lemma 2.6. Let Ω be an open bounded C 2 domain of R N . If (f1), (f2), (b1), (b2) are satisfied then any solution u of (1.1) satisfies R −1 (v(x)) ≤ u(x),
∀x ∈ Ω ,
where v(x) is the unique solution of the problem (1.2). In particular, when f (u) = eu , ( p − 1) ln[( p − 1)/v(x)] ≤ u(x),
∀x ∈ Ω ,
and when f (u) = u γ and γ > p − 1, [C/v(x)]C ≤ u(x),
∀x ∈ Ω .
Proof. We consider the problem (2.8). Let w be any solution of (2.8), we claim that Z ∞ ds w(x) = ≤ v(x), x ∈ Ω . 1/( p−1) (s) u(x) f Suppose, on the contrary, that {x ∈ Ω |w(x) > v(x)} 6= ∅. Since g(w) ≥ 0, we have −∆ p w + ∆ p v = −g(w) |∇w| p ≤ 0. On the other hand by w(x) = v(x) = 0 on ∂Ω , it follows by Lemma 2.4 that w(x) ≤ v(x), x ∈ Ω . This contradiction proves the lemma. The next result describes an interior regularity result for weak solutions to (1.1). 1, p
∞ (Ω ), p > 1 be a local weak solution of ∆ u = b(x, r ) in x ∈ Ω Lemma 2.7 (See [4,13]). Let u ∈ Wloc (Ω ) ∩ L loc p (open domain in R N ) where b(x, r ) is measurable in x ∈ Ω and continuous in r ∈ R such that |b(x, r )| ≤ γ on Ω × R and let Ω 0 be a subdomain of Ω such that Ω 0 ⊂ Ω . Given a subdomain compact K ⊂ Ω 0 , there exist positive constants C00 , C11 and α ∈ (0, 1), depending only upon N , p, γ , M = ess supΩ 0 |u| and dist(K , ∂Ω 0 ) such that k∇u(x)k∞,K ≤ C00 and x → ∇u(x) is locally H¨older continuous in Ω 0 i.e. u x (x) − u x (y) ≤ C11 |x − y|α , x, y ∈ Ω 0 , i = 1, 2, . . . , N . (2.9) i i
D.-P. Covei / Nonlinear Analysis 70 (2009) 1738–1745
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Lemma 2.8 (See also [12]). Under hypotheses (f1), (f2), (b1), (b2) and Ω is a bounded, connected, open subset of R N whose boundary ∂Ω is a smooth (C ∞ ) submanifold of dimension N − 1, (1.1) has a solution. Proof. For each m ∈ N , consider the boundary value problem ∆ p u = b(x) f (u), x ∈ Ω u(x) = m, x ∈ ∂Ω .
(2.10)
Since b ∈ C α (Ω )R and it satisfies (b1), we see that u m = m is an upper solution of (2.10). To construct a lower solution ∞ ds u m of (2.10), let u (x) f 1/( p−1) = w 1 (x), where w1 ∈ C 1,α (Ω ) is the unique solution of (s) 1 −∆ pZv = b(x), v(x) > 0, in Ω , ∞ ds > 0, on ∂Ω , v = 1/( p−1) f (s) 1 (which exists by the same argument as in Theorem 1.1). Then we see that u 1|∂ Ω = 1 ≤ m and u 1 = R −1 (w1 ) satisfies −∆ p w1 =
p ∆ p u1 f 0 (u ) − 2 1 ∇u 1 = b(x), f (u 1 ) f (u 1 )
x ∈ Ω,
which yields ∆ p u 1 ≥ b(x) f (u 1 ), x ∈ Ω . Thus u 1 is a lower solution of (2.10). By Lemma 2.4 we have u 1 ≤ m. By the lower and upper solution method it follows that (2.10) has a solution u m in the order interval [u 1 , m]. Furthermore, by the above lemmas, it is clear that 0 ≤ R −1 (w 1 ) = u 1 ≤ u 2 ≤ · · · ≤ u m ≤ u m+1 ≤ · · · ≤
in Ω .
(2.11)
Then the sequence {u m } is nondecreasing. Now let x0 be an arbitrary point in Ω . A straightforward computation shows that there is an open set O ⊂ Ω containing x0 and a positive constant, C O , such that u m (x) ≤ C O on O (using for example [10,12]). Moreover, limx→∂ Ω u(x) = ∞, where u(x) = limm→∞ u m (x) for x ∈ Ω . This concludes the basic details. Now we prove Theorems. 3. Proof of Theorem 1.1 Since, for any ε > 0, t → [t + ]−1 is a nonincreasing function in (0, ∞) and u ≤ kuk∞ it follows that b(x)[u(x) + ]−1 ≥ b(x)[ku(x)k∞ + ]−1 , and therefore −∆ p u = b(x)[u(x) + ]−1 ≥ b(x)[ku(x)k∞ + ]−1 , that is u = u[ku(x)k∞ + ]1/( p−1) is an upper solution to the problem (1.2). Remember u exists by Lemma 2.2. Using the fact that u = 0 is a lower solution to (1.2) and the upper and lower solution method we have that the problem (1.2) has a solution v ∈ C 1,α (Ω ), which is positive by the maximum principle. Moreover this solution is the unique solution to (1.2). For the last part of the theorem see [5]. 4. Proof of Theorem 1.2 Let us consider the problem ∆ p u = b(x) f (u), u > 0, in Ωm , u(x) → ∞, as x → ∂Ωm ,
(4.1)
0,α (Ω ), it where {Ωm }∞ 1 is as in Lemma 2.1 and our assumptions on b(x), for each component Ωm . Since b ∈ C follows that b ∈ C 0,α (Ω m ) and by Lemma 2.8, we see that (4.1) has a solution u m . It follows by Lemmas 2.1–2.8 that
0 < R −1 (v(x)) ≤ u m+1 ≤ u m (x) ≤ · · · ,
(4.2)
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for all x ∈ Ωm . If B is a compact subset of Ω , then there exists m 0 such that B ⊂ Ωm 0 and it follows by (4.2) that the −1 sequence {u m (x)}∞ m 0 is decreasing for x ∈ B and is bounded below by R (v(x)) on B. So u(x) = limm→∞ u m (x)
1,α exists for all x ∈ Ω . By the elliptic regularity theory u ∈ Cloc (Ω ) and ∆ p u(x) = b(x) f (u(x)), x ∈ Ω (see also the −1 argument used in [10]). By (4.2) and R (v(x)) → ∞ as x → ∂Ω , we see that u is a solution of (1.1) and the proof is complete.
5. Proof of Theorem 1.3 By Theorem 1.2, follows that the following problem ∆ p u = b(x) f (u), u > 0, in |x| < m, u(x) = ∞, as |x| = m, 1,α has a solution u m ∈ Cloc (Bm ) which satisfies
u 1 (x) ≥ u 2 (x) ≥ · · · ≥ u m (x) ≥ u m+1 (x),
in |x| < 1,
where Bm := {x| |x| < m, m ∈ N}. To prove our result, we need only to prove (A) R −1 (v(x)) ≤ u m (x), ∀x ∈ Bm , where v ∈ C 1 (R N ) is the unique solution of the problem (2.2). (B) u(x) → ∞ as |x| → ∞, where u := limm→∞ u m . Now, we observe that Lemma 2.5 allows us to define a positive valued smooth function on [0, ∞) by Z ∞ ds wm (x) = , ∀ |x| ≤ m. 1/( p−1) (s) f u m (x)
(5.1)
We claim that wm (x) ≤ v(x),
R −1 (v(x)) ≤ R −1 (wm (x)) = u m (x),
for any x ∈ Bm . Clearly, this inequality holds for |x| = m where wm = 0 and u m = ∞. We also observe that −∆ p wm (x) =
∆ p um ∆ p um f 0 (u m ) |∇u m | p ≤ − 2 = b(x) ≤ Φ(r ) = −∆ p v(r ). f (u m ) f (u m ) f (u m )
Since v(r ) > 0, for any x ∈ Bm , it follows by Lemma 2.4 that wm (x) ≤ v(r ), for any x ∈ Bm . Consequently, R −1 (v(r )) ≤ u(x) in R N and (A) is proved. Since lim|x|→∞ v(|x|) = 0 and lim|x|→∞ R −1 (v (r )) = ∞, it is clear that u(x) → ∞ as |x| → ∞. The proof is complete. Acknowledgments The author is grateful to the reviewer(s) for the attention with which this paper was read. References [1] L. Bieberbach, ∆u = eu und die automorphen funktionen, Math. Ann. 77 (1916) 173–212. [2] M.V. Borsuk, V. Kondratiev, Elliptic Boundary Value Problems of Second Order in Piecewise Smooth Domains, in: Mathematical Library, vol. 69, Elsevier, North-Holland, 2006, p. 531. [3] J.I. Diaz, J.E. Sa´a, Existence et unicite de solutions positives pour certaines equations elliptiques quasilineaires, CRAS. 305 (Serie I) (1987) 521–524. [4] E. DiBenedetto, C 1,α - local regularity of weak solutions of degenerate elliptic equations, Nonlinear Anal. 7 (1983) 827–850. [5] Z.M. Guo, J.R.L. Webb, Uniqueness of positive solutions for quasilinear elliptic equations when a parameter is large, in: Proc. R. Soc. Edinb., 124A, 1994, pp. 189–198. [6] J.B. Keller, On solution of ∆u = f (u), Commun. Pure Appl. Math. 10 (1957) 503–510. [7] A.V. Lair, A necessary and sufficient condition for existence of large solutions to semilinear elliptic equations, J. Math. Anal. Appl. 240 (1999) 205–218. [8] C. Loewner, L. Nirenberg, Partial differential equations invariant under conformal or projective transformations, in: Contributions to Analysis, Academic Press, New York, 1974, pp. 245–272.
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[9] J. Matero, Nonlinear elliptic problems with boundary blow-up, Uppsala Dissertations in Mathematics, 1997. [10] A. Mohammed, Existence and asymptotic behavior of blow-up solutions to weighted quasilinear equations, J. Math. Anal. Appl. 298 (2004) 621–637. [11] H. Rademacher, Finige besondere probleme partieller Differentialgleichungen, in: Die Differential und Integralgleichungen der Mechanick und Physik I, 2nd ed., Rosenberg, New York, 1943, pp. 838–845. [12] S. Tao, Z. Zhang, On the existence of explosive solutions for semilinear elliptic problems, Nonlinear Anal. 48 (2002) 1043–1050. [13] P. Tolksdorf, Regularity for a more general class of quasilinear elliptic equations, J. Differential Equations 51 (1984) 126–150. [14] R. Osserman, On the inequality ∆u ≥ f (u), Pacific J. Math. 7 (1957) 1641–1647. [15] Z. Zhang, A remark on the existence of explosive solutions for a class of semilinear elliptic equations, Nonlinear Anal. 41 (2000) 143–148.
Large and entire large solution for a quasilinear problem Dragos-Patru Covei ∗ Department of Mathematics, University Constantin Brancusi of Tg-Jiu, Bld. Republicii 1, 210152, Romania Received 18 October 2007; accepted 21 February 2008
Abstract In this article we give the existence results for the quasilinear elliptic problem ∆ p u = b(x) f (u) in Ω , u(x) → ∞ as x → ∂Ω , where Ω ⊆ R N is a domain. c 2008 Elsevier Ltd. All rights reserved.
MSC: 35J60 Keywords: Large solution; Entire solution
1. Introduction and the main results In this article, we are concerned with the existence of a blow-up solution for the quasilinear elliptic equation ∆ p u = b(x) f (u),
u > 0, in Ω , u(x)
→
∞,
as x→∂ Ω
(1.1)
where N > 2, 1 < p < N , ∆ p u is the p-Laplacian operator, Ω = R N or Ω is a domain in R N and b(x) is assumed to satisfy the following hypotheses, not necessary simultaneous: (b1) b is nonnegative, nontrivial in Ω and it has the property: for any x0 ∈ Ω satisfying b(x0 ) = 0, there exists a domain Ω0 such that x0 ∈ Ω0 , Ω 0 b Ω , and b(x) > 0 for all x ∈ ∂Ω0 ; (b2) b ∈ C 0,α (Ω ); 0,α (b3) For b(x) ∈ Cloc (R N ) and Φ(r ) = max|x|=r b(x) Z ∞ r 1/( p−1) Φ 1/( p−1) (r )dr < ∞ if 1 < p ≤ 2, Z0 ∞ ( p−2)N +1 r p−1 Φ(r )dr < ∞ if 2 ≤ p < ∞. 0
We assume that the nonlinearity f satisfies (f1) f ∈ C 1 (R), f 0 (s) ≥ 0 and f (s) > 0, ∀s ∈ R, or f ∈ C 1 [0, ∞), f (0) = 0, f 0 (s) ≥ 0 and f (s) > 0 for s > 0; ∗ Tel.: +40 0742269582.
E-mail address: [email protected]. c 2008 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2008.02.057
D.-P. Covei / Nonlinear Analysis 70 (2009) 1738–1745
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R +∞ (f2) α [ p · ( p − 1)−1 · F(s)]−1/ p ds < +∞, F 0 (s) = f (s), ∀α > 0. This problem has been considered by several authors in the case p = 2 and in the case of a bounded domain Ω . The first example of a boundary blow-up problem where Ω ⊂ R2 is a open regular bounded domain, f (u) = eu , b(x) = 1 has been already studied by Bieberbach [1] in 1916. In this case the problem plays an important role in the theory of Riemannian surfaces of constant negative curvatures and in the theory of automorphic functions. When Ω ⊂ R3 this problem arises in the study of the electrostatic potential in a glowing hollow metal body. Rademacher [11] in 1943, using the ideas of Bieberbach, extended this result to a bounded domain in R3 with C 2 boundary. Later, the problem of existence of solutions has been studied extensively by many other authors for a more general class of functions. Assumption (f2) has been first introduced in 1957 by Keller [6] and Osserman [14]. Loewner and Nirenberg in [8] have studied the problem of uniqueness and asymptotic behavior of solutions when b(x) = const > 0 and f (u) = u (N +2)/(N −2) , N > 2 with an arbitrary domain instead of just R N . They showed that this problem is motivated by a concrete problem arising in Riemannian Geometry. In particular, we mention the work of Matero [9] and the work of Mohammed [10] where it is proved the existence of solutions to the problem (1.1) when ∆ is replaced by the p-Laplacian. In this paper we will use the same technique as in [7,12,15] in order to prove the existence of large solutions of (1.1), that are solutions u satisfying u(x) → ∞ as x → ∂Ω (if Ω 6= R N ), or u(x) → ∞ as |x| → ∞ (if Ω = R N ). In the latter case the solution is called an entire large solution. For the study of a solution given in this paper we advise the reader to consult the reference [10]. Our main results are summarized in the following theorems. Theorem 1.1 (See also [5]). If b satisfies (b1), (b2) and Ω is a bounded C 2 open domain of R N then − ∆ p v = b(x),
v > 0,
in Ω ,
v=0
on ∂Ω ,
(1.2)
C 1,α (Ω ).
has a unique weak solution v ∈ Moreover, there exist constants l1 ≥ k1 > 0, such that k1 d(x) ≤ v(x) ≤ l1 d(x) on Ω , where d(x) denotes the distance from x ∈ Ω to the boundary ∂Ω . Theorem 1.2. Let f and b satisfy the assumptions (f1), (f2), (b1), (b2) and Ω be a bounded C 2 open domain of R N . Then, the problem (1.1) has a large solution. Theorem 1.3. Let f and b satisfy the assumptions (f1), (f2), (b1), (b3) and Ω = R N . Then, the problem (1.1) has an entire large solution. We establish some preliminary results that will be necessary later to prove our main theorems. 2. Preliminary results In this section, we will give some lemmas. The next lemma can be found in [12]. Lemma 2.1. Let Ω be a bounded open set of R N which satisfies the uniform external sphere condition, then there exists a sequence {Ωm }∞ 1 of open sets such that Ω m ⊂ Ωm+1 ⊂ Ω ,
∪∞ m=1 Ωm = Ω ,
and the boundary ∂Ωm is a smooth (C ∞ ) submanifold of dimension N − 1 for m ≥ 1. Lemma 2.2 (See [3]). If b satisfies (b1), (b2) and Ω is a bounded C 2 open domain of R N then the problem −∆ p u(x) = b(x)[u(x) + ]−1 ,
u > 0, x ∈ Ω ,
u |∂ Ω = 0,
(2.1)
has a unique solution u ∈ C 1,α (Ω ) for any ε > 0. Lemma 2.3. Suppose that (b1) and (b3) are satisfied. Then the problem ( p−2 0 −(r N −1 v 0 (r ) v (r ))0 = r N −1 Φ(r )(r =| x |) on R N , lim v(r ) = 0, v 0 (0) = 0, r →∞
has a unique positive bounded radially symmetric solution.
(2.2)
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Proof. Set I (r ) :=
r
Z
ξ 1−N
0
ξ
Z
σ N −1 Φ(σ )dσ
1/( p−1)
r > 0.
dξ,
0
Now we prove that Z Z +∞ 1−N K := ξ
ξ
σ
N −1
Φ(σ )dσ
1/( p−1) dξ,
0
0
which is, in fact, finite. 1 Case 1. 1 < p ≤ 2. Since in this case 1 ≤ p−1 < ∞, by the Hardy inequality (see [2, Theorem 2.1]), we have 1/( p−1) Z ξ 1/( p−1) Z ξ Z +∞ Z +∞ − N −1 σ N −1 Φ(σ )dσ dξ = ξ p−1 σ N −1 Φ(σ )dσ dξ ξ 1−N 0
0
0
0
−1 #1/( p−1) Z +∞ 1 N −1 − N −1 ≤ · −1 ξ p−1 (ξ ξ N −1 Φ(ξ ))1/( p−1) dξ p−1 p−1 0 1 Z +∞ p−1 1 ξ 1/( p−1) Φ 1/( p−1) (ξ )dξ < ∞. = N−p 0 "
1 Case 2. Let 2 ≤ p < +∞, so 1 ≤ p − 1, it follows that 1 ≥ p−1 > 0. Set Z ξ Z ξ σ N −1 Φ(σ )dσ ≤ 1 for ξ > 0, or σ N −1 Φ(σ )dσ > 1 for ξ > 0. 0
0
In the first case Z ξ 1/( p−1) N −1 σ Φ(σ )dσ ≤ 1, 0
so Z
r
ξ
1−N p−1
ξ
Z
0
σ N −1 Φ(σ )dσ
1/( p−1)
Z
r
dξ ≤
0
ξ
1−N p−1
dξ,
0
is finite as r → ∞ and N > p. In the second case, Z ξ 1/( p−1) Z ξ N −1 σ Φ(σ )dσ ≤ σ N −1 Φ(σ )dσ, 0
0
for ξ ≥ 0, so Z Z r 1−N ξ p−1 0
ξ
σ N −1 Φ(σ )dσ
1/( p−1)
Z dξ ≤
0
r
ξ
1−N p−1
0
Z
ξ
σ N −1 Φ(σ )dσ dξ.
0
Integration by parts gives Z r Z ξ Z Z ξ 1−N p − 1 r d p−N ξ p−1 σ N −1 Φ(σ )dσ dξ = − ξ p−1 σ N −1 Φ(σ )dσ dξ N − p dξ 0 0 0 0 Z r Z r ( p−2)N +1 p−N p−1 N −r p−1 = σ −1 Φ(σ )dσ + ξ p−1 Φ(ξ )dξ . N−p 0 0 Using L’ Hˆopital’s rule, we have Z r Z p−N lim −r p−1 σ N −1 Φ(σ )dσ + r →∞
0
= lim
−
Rr 0
ξ
( p−2)N +1 p−1
Φ(ξ )dξ
0
σ N −1 Φ(σ )dσ + r
N−p p−1
Rr 0
ξ
( p−2)N +1 p−1
Φ(ξ )dξ
N−p p−1
r →∞
Z = lim
r
r →∞ 0
r
ξ
( p−2)N +1 p−1
r Z Φ(ξ )dξ =
0
∞
ξ
( p−2)N +1 p−1
Φ(ξ )dξ < ∞.
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Now setting w(r ) := K − I (r ), it is easily shown that w is the unique positive bounded solution of the problem (2.2). The next Weak comparison principle can be found in [2]. Lemma 2.4. Let Ω be a bounded domain in R N , N ≥ 2 with smooth boundary ∂Ω and w1 , w2 ∈ W 1, p (Ω ) satisfy −∆ p w1 ≤ −∆ p w2
in Ω .
If w1 ≤ w2 on ∂Ω then w1 ≤ w2
in Ω .
Lemma 2.5. If f satisfies (f1) and (f2) then f (t) F(t)
p−1 p
(2.3)
→ ∞ as t → ∞.
Therefore, there exists γ > 0 such that Z ∞ ds < ∞. 1/( p−1) (s) f γ
(2.4)
Proof. We remark that limt→∞ F(t)/t p = +∞, since condition (f2) implies that Z t t 1 ds t− ≤ → 0, √ √ p p 2 F(t) F(s) t/2 as t → +∞. On the other hand, by (f1) we have f (t) F(t) ≥ p , t t p−1 which implies that
(2.5)
f (t) = ∞. t→+∞ t p−1 lim
Using again (2.5), we have 1 f (t) f (t) p ≥ → +∞, p−1 t p−1 F(t) p as t → +∞. The proof of (2.3) is complete. Now, from (2.3) there exists t1 > 0, such that f (t) ≥ F(t)
p−1 p
,
(2.6)
for all t ≥ t1 . But the relation (2.6) is equivalent to 1 1 ≤ √ . p f 1/( p−1) (s) F(s)
(2.7)
Integrating (2.7) from t1 to +∞ we have Z ∞ Z ∞ 1 1 dt ≤ dt < ∞. √ p 1/( p−1) (t) f F(t) t1 t1 The proof of (2.4) is complete for γ := t1 .
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In view of this we define Z +∞ ds , R(u) = 1/( p−1) (s) f u(x)
and w(x) = R(u(x)),
and then R : (0, ∞) → (0, ∞) is strictly decreasing. Then (1.1) is transformed into the following equivalent singular Dirichlet problem −∆ p w + g(w) |∇w| p = b(x), w > 0, in Ω , (2.8) w → 0, as x → ∂Ω , where g(w) =
f 0 (u) f ( p−2)/( p−1) (u)
=
f 0 (R −1 (w)) , f ( p−2)/( p−1) (R −1 (w))
and R −1 denotes the inverse function which is strictly decreasing in (0, ∞). In particular, when f (u) = eu , we have w = ( p − 1)e−u(x)/( p−1) , and −∆ p w + ( p − 1)
|∇w| p = b(x), w
w > 0, in Ω ,
w → 0, as x → ∂Ω ,
when f (u) = u γ and γ > p − 1, we have w(x) = C[u(x)]−C |∇w| p = b(x), w C := ( p − 1)/(γ − p + 1). −∆ p w + γ C
w > 0, in Ω ,
−1
and
w → 0, as x → ∂Ω ,
Lemma 2.6. Let Ω be an open bounded C 2 domain of R N . If (f1), (f2), (b1), (b2) are satisfied then any solution u of (1.1) satisfies R −1 (v(x)) ≤ u(x),
∀x ∈ Ω ,
where v(x) is the unique solution of the problem (1.2). In particular, when f (u) = eu , ( p − 1) ln[( p − 1)/v(x)] ≤ u(x),
∀x ∈ Ω ,
and when f (u) = u γ and γ > p − 1, [C/v(x)]C ≤ u(x),
∀x ∈ Ω .
Proof. We consider the problem (2.8). Let w be any solution of (2.8), we claim that Z ∞ ds w(x) = ≤ v(x), x ∈ Ω . 1/( p−1) (s) u(x) f Suppose, on the contrary, that {x ∈ Ω |w(x) > v(x)} 6= ∅. Since g(w) ≥ 0, we have −∆ p w + ∆ p v = −g(w) |∇w| p ≤ 0. On the other hand by w(x) = v(x) = 0 on ∂Ω , it follows by Lemma 2.4 that w(x) ≤ v(x), x ∈ Ω . This contradiction proves the lemma. The next result describes an interior regularity result for weak solutions to (1.1). 1, p
∞ (Ω ), p > 1 be a local weak solution of ∆ u = b(x, r ) in x ∈ Ω Lemma 2.7 (See [4,13]). Let u ∈ Wloc (Ω ) ∩ L loc p (open domain in R N ) where b(x, r ) is measurable in x ∈ Ω and continuous in r ∈ R such that |b(x, r )| ≤ γ on Ω × R and let Ω 0 be a subdomain of Ω such that Ω 0 ⊂ Ω . Given a subdomain compact K ⊂ Ω 0 , there exist positive constants C00 , C11 and α ∈ (0, 1), depending only upon N , p, γ , M = ess supΩ 0 |u| and dist(K , ∂Ω 0 ) such that k∇u(x)k∞,K ≤ C00 and x → ∇u(x) is locally H¨older continuous in Ω 0 i.e. u x (x) − u x (y) ≤ C11 |x − y|α , x, y ∈ Ω 0 , i = 1, 2, . . . , N . (2.9) i i
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Lemma 2.8 (See also [12]). Under hypotheses (f1), (f2), (b1), (b2) and Ω is a bounded, connected, open subset of R N whose boundary ∂Ω is a smooth (C ∞ ) submanifold of dimension N − 1, (1.1) has a solution. Proof. For each m ∈ N , consider the boundary value problem ∆ p u = b(x) f (u), x ∈ Ω u(x) = m, x ∈ ∂Ω .
(2.10)
Since b ∈ C α (Ω )R and it satisfies (b1), we see that u m = m is an upper solution of (2.10). To construct a lower solution ∞ ds u m of (2.10), let u (x) f 1/( p−1) = w 1 (x), where w1 ∈ C 1,α (Ω ) is the unique solution of (s) 1 −∆ pZv = b(x), v(x) > 0, in Ω , ∞ ds > 0, on ∂Ω , v = 1/( p−1) f (s) 1 (which exists by the same argument as in Theorem 1.1). Then we see that u 1|∂ Ω = 1 ≤ m and u 1 = R −1 (w1 ) satisfies −∆ p w1 =
p ∆ p u1 f 0 (u ) − 2 1 ∇u 1 = b(x), f (u 1 ) f (u 1 )
x ∈ Ω,
which yields ∆ p u 1 ≥ b(x) f (u 1 ), x ∈ Ω . Thus u 1 is a lower solution of (2.10). By Lemma 2.4 we have u 1 ≤ m. By the lower and upper solution method it follows that (2.10) has a solution u m in the order interval [u 1 , m]. Furthermore, by the above lemmas, it is clear that 0 ≤ R −1 (w 1 ) = u 1 ≤ u 2 ≤ · · · ≤ u m ≤ u m+1 ≤ · · · ≤
in Ω .
(2.11)
Then the sequence {u m } is nondecreasing. Now let x0 be an arbitrary point in Ω . A straightforward computation shows that there is an open set O ⊂ Ω containing x0 and a positive constant, C O , such that u m (x) ≤ C O on O (using for example [10,12]). Moreover, limx→∂ Ω u(x) = ∞, where u(x) = limm→∞ u m (x) for x ∈ Ω . This concludes the basic details. Now we prove Theorems. 3. Proof of Theorem 1.1 Since, for any ε > 0, t → [t + ]−1 is a nonincreasing function in (0, ∞) and u ≤ kuk∞ it follows that b(x)[u(x) + ]−1 ≥ b(x)[ku(x)k∞ + ]−1 , and therefore −∆ p u = b(x)[u(x) + ]−1 ≥ b(x)[ku(x)k∞ + ]−1 , that is u = u[ku(x)k∞ + ]1/( p−1) is an upper solution to the problem (1.2). Remember u exists by Lemma 2.2. Using the fact that u = 0 is a lower solution to (1.2) and the upper and lower solution method we have that the problem (1.2) has a solution v ∈ C 1,α (Ω ), which is positive by the maximum principle. Moreover this solution is the unique solution to (1.2). For the last part of the theorem see [5]. 4. Proof of Theorem 1.2 Let us consider the problem ∆ p u = b(x) f (u), u > 0, in Ωm , u(x) → ∞, as x → ∂Ωm ,
(4.1)
0,α (Ω ), it where {Ωm }∞ 1 is as in Lemma 2.1 and our assumptions on b(x), for each component Ωm . Since b ∈ C follows that b ∈ C 0,α (Ω m ) and by Lemma 2.8, we see that (4.1) has a solution u m . It follows by Lemmas 2.1–2.8 that
0 < R −1 (v(x)) ≤ u m+1 ≤ u m (x) ≤ · · · ,
(4.2)
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for all x ∈ Ωm . If B is a compact subset of Ω , then there exists m 0 such that B ⊂ Ωm 0 and it follows by (4.2) that the −1 sequence {u m (x)}∞ m 0 is decreasing for x ∈ B and is bounded below by R (v(x)) on B. So u(x) = limm→∞ u m (x)
1,α exists for all x ∈ Ω . By the elliptic regularity theory u ∈ Cloc (Ω ) and ∆ p u(x) = b(x) f (u(x)), x ∈ Ω (see also the −1 argument used in [10]). By (4.2) and R (v(x)) → ∞ as x → ∂Ω , we see that u is a solution of (1.1) and the proof is complete.
5. Proof of Theorem 1.3 By Theorem 1.2, follows that the following problem ∆ p u = b(x) f (u), u > 0, in |x| < m, u(x) = ∞, as |x| = m, 1,α has a solution u m ∈ Cloc (Bm ) which satisfies
u 1 (x) ≥ u 2 (x) ≥ · · · ≥ u m (x) ≥ u m+1 (x),
in |x| < 1,
where Bm := {x| |x| < m, m ∈ N}. To prove our result, we need only to prove (A) R −1 (v(x)) ≤ u m (x), ∀x ∈ Bm , where v ∈ C 1 (R N ) is the unique solution of the problem (2.2). (B) u(x) → ∞ as |x| → ∞, where u := limm→∞ u m . Now, we observe that Lemma 2.5 allows us to define a positive valued smooth function on [0, ∞) by Z ∞ ds wm (x) = , ∀ |x| ≤ m. 1/( p−1) (s) f u m (x)
(5.1)
We claim that wm (x) ≤ v(x),
R −1 (v(x)) ≤ R −1 (wm (x)) = u m (x),
for any x ∈ Bm . Clearly, this inequality holds for |x| = m where wm = 0 and u m = ∞. We also observe that −∆ p wm (x) =
∆ p um ∆ p um f 0 (u m ) |∇u m | p ≤ − 2 = b(x) ≤ Φ(r ) = −∆ p v(r ). f (u m ) f (u m ) f (u m )
Since v(r ) > 0, for any x ∈ Bm , it follows by Lemma 2.4 that wm (x) ≤ v(r ), for any x ∈ Bm . Consequently, R −1 (v(r )) ≤ u(x) in R N and (A) is proved. Since lim|x|→∞ v(|x|) = 0 and lim|x|→∞ R −1 (v (r )) = ∞, it is clear that u(x) → ∞ as |x| → ∞. The proof is complete. Acknowledgments The author is grateful to the reviewer(s) for the attention with which this paper was read. References [1] L. Bieberbach, ∆u = eu und die automorphen funktionen, Math. Ann. 77 (1916) 173–212. [2] M.V. Borsuk, V. Kondratiev, Elliptic Boundary Value Problems of Second Order in Piecewise Smooth Domains, in: Mathematical Library, vol. 69, Elsevier, North-Holland, 2006, p. 531. [3] J.I. Diaz, J.E. Sa´a, Existence et unicite de solutions positives pour certaines equations elliptiques quasilineaires, CRAS. 305 (Serie I) (1987) 521–524. [4] E. DiBenedetto, C 1,α - local regularity of weak solutions of degenerate elliptic equations, Nonlinear Anal. 7 (1983) 827–850. [5] Z.M. Guo, J.R.L. Webb, Uniqueness of positive solutions for quasilinear elliptic equations when a parameter is large, in: Proc. R. Soc. Edinb., 124A, 1994, pp. 189–198. [6] J.B. Keller, On solution of ∆u = f (u), Commun. Pure Appl. Math. 10 (1957) 503–510. [7] A.V. Lair, A necessary and sufficient condition for existence of large solutions to semilinear elliptic equations, J. Math. Anal. Appl. 240 (1999) 205–218. [8] C. Loewner, L. Nirenberg, Partial differential equations invariant under conformal or projective transformations, in: Contributions to Analysis, Academic Press, New York, 1974, pp. 245–272.
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