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Hopf boundary maximum principle violation for semilinear elliptic equations
Nonlinear Analysis 72 (2010) 3346–3355
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Hopf boundary maximum principle violation for semilinear elliptic equations Yavdat Il’yasov a,∗ , Youri Egorov b a
Institute of Mathematics, Ufa Scientific Center of RAS, Ufa, Russia
b
Université Paul Sabatier, Toulouse, France
article
info
Article history: Received 31 January 2009 Accepted 8 December 2009 MSC: 35J60 35J67 Keywords: Pohozaev’s identity Non-Lipschitz nonlinearity Variational approach
abstract We consider elliptic equations with non-Lipschitz nonlinearity
−∆u = λ|u|β−1 u − |u|α−1 u in a smooth bounded domain Ω ⊂ Rn , with Dirichlet boundary conditions; here 0 < α < β < 1. We prove the existence of a weak nonnegative solution which does not satisfy the Hopf boundary maximum principle, provided that λ is large enough and n > max{2, 2(1 + α)(1 + β)/(1 − α)(1 − β)}. © 2009 Elsevier Ltd. All rights reserved.
1. Introduction Let Ω be a bounded domain in Rn with a smooth boundary ∂ Ω , which is strictly star-shaped with respect to the origin in Rn . We consider the following problem:
−1u = λ|u|β−1 u − |u|α−1 u in Ω , u = 0 on ∂ Ω .
(1)
Here λ is a real parameter and 0 < α < β < 1, so the nonlinearity f (λ, u) := λ|u|β−1 u − |u|α−1 u on the right-hand side of (1) is non-Lipschitzean at zero. Our interest in this problem has been induced by [1], where in the case of dimension n = 1, Ω = (−1, 1), among other results, the authors showed that, for certain values λ > 0, Eq. (1) possesses nonnegative solutions u(x), x ∈ (−1, 1) with a special feature u(−1) = u(1) = 0,
u0 (−1) = u0 (1) = 0.
(2)
This result means that the Hopf boundary maximum principle is violated at the points x = −1, x = 1 and the uniqueness is lost for the initial value problem for (1) with u(−1) = u0 (−1) = 0, since u ≡ 0 also satisfies (1). Furthermore, it can easily be shown that the existence of such a solution with λ0 > 0 yields the existence of a continuum set of nonnegative solutions to this boundary value problem for any λ > λ0 . Note that when the nonlinearity f (λ, u) is a locally Lipschitz function such a phenomenon is impossible due to the uniqueness of the solution of the initial value problem and the Hopf boundary maximum principle. Observe that property (2) implies that the function u is also a weak solution of (1) on the
∗
Corresponding author. E-mail addresses: [email protected] (Y. Il’yasov), [email protected] (Y. Egorov).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.12.015
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
3347
whole line R. In the theory of integrable systems, the solutions of such type are known as ‘‘compactons’’: solitary waves with compact support [2,3]. In the PDE literature the term ‘‘free boundary solution’’ (see [1,4]) is also used for such type of solution. The purpose of the present paper is to find an answer to the following question: Can the Hopf boundary maximum principle be violated for semilinear elliptic equations such as (1) in higher dimensions n > 1, including non-radial cases? Let us state our main result. We consider a weak solution u ∈ H01 := H01 (Ω ), where H01 (Ω ) denotes the closure C0∞ (Ω ) in the standard Sobolev space H 1 (Ω ) with the norm k · k1 . We use the term compacton solution for a weak solution u ∈ C 1 (Ω ) ∂u of (1), which does not satisfy the Hopf boundary maximum principle, i.e. ∂νλ = 0 on ∂ Ω , where ν denotes the unit outward normal. In the opposite case the weak solution u ∈ C 1 (Ω ) of (1) is said to be a non-compacton solution. Our main result is the following.
Theorem 1. Let Ω be a bounded domain in Rn with smooth boundary, which is strictly star-shaped with respect to the origin. Assume that 0 < α < β < 1 and n > max{2, 2(1 + α)(1 + β)/(1 − α)(1 − β)}. Then there exists λ∗ > 0 such that, for every λ ≥ λ∗ , problem (1) has a compacton solution uλ ∈ C 1,κ (Ω ) for κ ∈ (0, 1), which is nonnegative in Ω . Moreover, the number of such solutions for λ > λ∗ is infinite. Furthermore, there exists λ ∈ (0, λ∗ ] such that problem (1) has no nontrivial solution for all λ ∈ [0, λ). The proof of the theorem relies on variational arguments. Furthermore, basic ingredients in the proof consist in using Pohozaev’s identity [5] corresponding to (1) and in applying a spectral analysis with respect to the fibering procedure [6] introduced in [7]. From [8] it follows that there exists λ∗∗ ≤ λ∗ such that for λ > λ∗∗ problem (1) has a second weak solution wλ ∈ C 1,µ (Ω ), λ < 0 holds on ∂ Ω . µ ∈ (0, 1), which is positive wλ > 0 in Ω and satisfies the Hopf boundary maximum principle, i.e. ∂w ∂ν Furthermore, this result in [8] does not require the above restrictions on n and the star-shaped property of Ω in Theorem 1. However, in the framework of our approach we are able to obtain a different result in this direction. In Section 5 (see Corollary 14) we derive from our variational problem that (1) possesses a weak non-compacton solution wλ which is a ground state for (1), i.e. 0 < Eλ (wλ ) ≤ Eλ (vλ ) for any solution vλ of (1), where Eλ is the Lagrangian associated with (1) (see below (3)). A result on the existence of nonnegative compacton solution, but for the different version of problem (1), has been obtained in [9]. The results in [1] have been extended in [4] both in allowing singular terms in the right-hand side of (1), i.e. −1 < α < β < 1 and to consider a quasilinear differential operator like the p-Laplacian. We also refer the reader to [4] for a deeper discussion of the problem and for further references. The paper is organized as follows. In Section 2, we apply a spectral analysis related to the fibering procedure [7] to introduce two spectral points Λ0 , Λ1 which play a basic role in the proof of the main result. In Section 3, we derive some important consequences from Pohozaev’s identity. In Section 4, we prove the existence of the solution to an auxiliary constrained minimization problem. In Section 5, we prove Theorem 1. In Section 6, we indicate briefly some open problems in connection with (1) and make a few remarks. Some technical results are proved in Appendices A and B. 2. Spectral analysis with respect to the fibering procedure In this section we apply a spectral analysis related to the fibering procedure [7] to introduce two spectral points which will play important roles in the proof of the main result. Throughout this section we assume that n > 2. Observe that problem (1) is the Euler–Lagrange equation associated with the functional Eλ (u) =
1 2
T ( u) − λ
1
β +1
B(u) +
1
α+1
A(u),
u ∈ H01 ,
(3)
where we use the notations T (u) =
Z
|∇ u| dx, 2
Ω
B(u) =
Z
β+1
| u| Ω
dx,
A(u) =
Z Ω
|u|α+1 dx.
Let u ∈ H01 . Consider the function eλ (t ) := Eλ (tu) defined for t ∈ R+ . Introduce the functionals Qλ (u) = e0λ (t )|t =1 , Lλ (u) = e00λ (t )|t =1 for u ∈ H01 . Then Qλ (u) = T (u) − λB(u) + A(u), Let u ∈
Lλ (u) = T (u) − λβ B(u) + α A(u).
H01
\ {0}. Following the spectral analysis [7], we solve the system 2 1+β 1+α Qλ (tu) = t T (u) − λt B(u) + t A(u) = 0 Eλ (tu) =
t2 2
T (u) − λ
t 1+β
1+β
B(u) +
t 1+α
1+α
A(u) = 0
(4)
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and find the corresponding solution 2(β − α)
t0 (u) =
A(u)
1 1−α
(1 + α)(1 − β) T (u)
,
(5)
α,β
λ0 (u) = c0 λ(u),
(6)
where α,β
c0
(1 − α)(1 + β) = (1 − β)(1 + α)
(1 + α)(1 − β) 2(β − α)
β−α 1−α
and β−α
1−β
A(u) 1−α T (u) 1−α
λ(u) =
B(u)
.
(7)
Thus with respect to the spectral analysis, we have the following spectral point:
Λ0 = inf λ0 (u).
(8)
H01 \{0}
Introduce the second point Λ1 . Let u ∈ H01 \ {0}. Consider now the following system:
Qλ (tu) = t 2 T (u) − λt 1+β B(u) + t 1+α A(u) = 0 Lλ (tu) = t 2 T (u) − λβ t 1+β B(u) + α t 1+α A(u) = 0
(9)
for t ∈ R+ , λ ∈ R+ . Solving this system, we find, as above, α,β
λ1 (u) = c1 λ(u),
(10)
where α,β
c1
=
1−α
1−β
1−β
β−α 1−α
β −α
.
(11)
Then we have
Λ1 = inf λ1 (u).
(12)
H01 \{0}
Proposition 2. 0 < Λ1 < Λ0 < +∞. α,β
Proof. By (6) and (10), the inequality Λ1 < Λ0 is equivalent to c1
(1 + β) (1 + α)
1+α
β−α 1−α
2
> 1 ⇐⇒
1+β
1−α
1+α
>
2
α,β
< c0 , which can also be written as follows:
β−α
1+α
.
β−α
Now, letting µ = 11−α > 0, η = 1+α > 0, we obtain +α 1
1
(1 + η) η > (1 + µ) µ . 1
Hence, since the function x 7→ (1 + x) x is decreasing on ]0, +∞[ and µ > η, we obtain the desired inequality Λ1 < Λ0 . It is clear that Λ0 < +∞. Let us show that 0 < Λ1 . Note that λ(u) is a zero-homogeneous function on H01 \ {0}. Therefore we may restrict the infimum in (12) to the set S := {v ∈ H01 : kvk1 = 1}. Set
γ =
(1 + α)(2∗ − 1 − β) , (2∗ − 1 − α)
p=
(1 + α) , γ
q=
2∗
(1 + β − γ )
,
(13)
where 2∗ = 2n/(n − 2). Then p, q > 1, 1/p + 1/q = 1 for n > 2, and by the Hölder inequality we have B(u) ≤
Z
∗
u2 dx Ω
1/q
· A(u)1/p .
(14)
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
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By Sobolev’s inequality,
Z
∗
∗
u2 dx ≤ C0 kuk21 = C0 < +∞
Ω
(15)
for u ∈ S, where C0 does not depend on u ∈ H01 . Hence, for any u ∈ S, we have
λ(u) =
A(u)(1−β) B(u)(1−α)
− (1−α) q
≥ C0
A(u)
− (1−α) q
p(1−β)−(1−α) p
= C0
∗
A(u)
2)(β−α) − (2(2− ∗ −1−α)
.
Since A(u) ≤ C1 < +∞ on S, where 0 < C1 < +∞ does not depend on u ∈ H01 , we see from (10) that Λ1 > 0.
3. Pohozaev’s identity We will need the following regularity result. Proposition 3. Assume that 0 < α < β < 1. Suppose that u ∈ H01 is a weak solution of (1). Then u ∈ C 1,κ (Ω ) for κ ∈ (0, 1). Proof. Let u ∈ H01 be a weak solution of (1). Since |f (λ, u)| < C (1 + |u|), u ∈ R with some C > 0, then (see Lemma B.3 in [10]) u ∈ Lq (Ω ) for any q ∈ [1, ∞). This implies that −1u = f (λ, u) ∈ Lq (Ω ) for any q ∈ [1, ∞). Thus, by the Caldéron–Zygmund inequality (see [11]), u ∈ H 2,q (Ω ) for any q ∈ [1, ∞), whence u ∈ C 1,κ (Ω ) for κ ∈ (0, 1) by the Sobolev embedding theorem. We will denote by Pλ the functional Pλ (u) :=
(n − 2) 2n
T (u) − λ
1
β +1
B(u) +
1
α+1
A(u)
defined for u ∈ H01 . Lemma 4. Suppose Ω is a smooth bounded domain in Rn . Let u be a weak solution of (1), u ∈ H01 . Then Pohozaev’s identity Pλ (u) +
1
Z
2n ∂ Ω
2 ∂u x · ν ds = 0 ∂ν
(16)
holds. Proof. Observe that by Proposition 3 we know that u ∈ H 2,q ∩ H01 for any q ∈ [1, ∞). We use the notations F (λ, z ) =
1 1 f (λ, τ ) dτ ≡ λ β+ B(z ) − α+ A(z ), ui = ∂∂xu , uij = ∂∂x xu and we adopt the summation convention on repeated indices. As 1 1 i i j in [5,12], multiplication of both parts of the equation by ui xi and integration by parts yields 2
Rz 0
Z Ω
f (λ, u)ui xi dx = −n
Z Ω
F (λ, u) dx +
Z ∂Ω
F (λ, u)x · ν ds
and
Z
Z
− Ω
ujj ui xi dx =
Ω
Z
uj (δij ui + xi uij ) dx −
|∇ u|2 dx −
= Ω
n 2
Z
Z ∂Ω
uj νj xi ui ds
|∇ u|2 dx − Ω
Thus, since u = 0 on the boundary, we obtain the result.
2 ∂u x · ν ds. 2 ∂ Ω ∂ν
1
Z
Let u ∈ H01 . Based on the ideas of spectral analysis [7], we consider the following system of equations:
Q (u) := T (u) − λB(u) + A(u) = 0 λ Lλ (u) := T (u) − λβ B(u) + α A(u) = 0 1 1 (n − 2) Pλ (u) := T ( u) − λ B(u) + A(u) = 0. 2n β +1 α+1
(17)
The computation of the corresponding determinant shows that this system is solvable if and only if
θ ≡ 2(1 + α)(1 + β) − n(1 − α)(1 − β) = 0.
(18)
Note that θ < 0 if and only if n > 2(1 + α)(1 + β)/(1 − α)(1 − β). Observe that the equation e0λ (t ) = 0, t > 0, has at most two roots, t 1 (u) := tλ1 (u) ∈ R+ and t 2 (u) := tλ2 (u) ∈ R+ , such that t 1 (u) ≤ t 2 (u), e00λ (t 1 (u)) ≤ 0 and e00λ (t 2 (u)) ≥ 0.
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Proposition 5. Assume that θ < 0, n > 2. If u ∈ H01 \ {0} and t > 0 are such that Qλ (tu) = 0 and Pλ (tu) ≤ 0, then Lλ (tu) > 0. Proof. Let u ∈ H01 \ {0} and t > 0 as in the assumption. Then T (u) = λt β−1 B(u) − t α−1 A(u) Lλ (tu) > 0 ⇔ λt β−α
(19)
(1 − β) B(u) > A(u). (1 − α)
(20)
Equality (19) implies that Pλ (t 1 (u)u) ≤ 0 holds if and only if
λt β−α
[2(1 + β) + n(1 − β)](1 + α) B(u) ≥ A(u). [2(1 + α) + n(1 − α)](1 + β)
(21)
Observe that the assumptions θ < 0, n > 2 imply that
[2(1 + β) + n(1 − β)](1 + α) (1 − β) < . [2(1 + α) + n(1 − α)](1 + β) (1 − α)
(22)
Thus (22) and (21) give
λt β−α
(1 − β) B(u) > A(u), (1 − α)
and therefore by (20) the proof is complete.
Corollary 6. If u0 is a compacton solution of (1), then Eλ (u0 ) > 0. Furthermore, if in addition θ < 0, n > 2, then Qλ (u0 ) = 0,
Pλ (u0 ) = 0,
Lλ (u0 ) > 0.
Proof. Observe that, if u0 is a compacton solution of (1), then by (16) we have Pλ (u0 ) = 0. Hence, using Eλ (u) = Pλ (u) + (1/2n)T (u), we get Eλ (u0 ) =
1 n
T (u0 ) > 0.
Note that Qλ (u0 ) = 0 if u0 is a solution of (1), and Pλ (u0 ) = 0 if in addition this solution is the compacton solution. Hence the assumptions θ < 0, n > 2 by Proposition 5 imply that Lλ (u0 ) > 0. 4. Constrained minimization problems Consider the following constrained minimization problem:
Eλ (u) → min Qλ (u) = 0.
(23)
We denote by Mλ := {w ∈ H01 \ {0} : Qλ (u) = 0} the admissible set of (23), and by Eˆλ := min{Eλ (u) : u ∈ Mλ } the minimal value in this problem. We say that (um ) is a minimizing sequence of (23) if Eλ (um ) → Eˆλ
as m → ∞ and um ∈ Mλ ,
m = 1, 2, . . . .
(24)
Proposition 7. If λ > Λ1 , then the set Mλ is not empty; meanwhile, the set Mλ is empty when λ < Λ1 . Proof. Let λ > Λ1 . Then by (8) there exists u ∈ H01 \ {0} such that Λ1 < λ(u) < λ and Lλ(u) (t (u)u) = 0, Qλ(u) (t (u)u) = 0. Hence, Qλ (t (u)u) < 0, since λ(u) < λ, and therefore there exists t > 0 such that Qλ (tu) = 0, i.e. tu ∈ Mλ . The proof of the second part of the proposition follows immediately from (8). From here it follows that Corollary 8. Eˆλ < +∞ for any λ > Λ1 . 4.1. Existence of the solution of (23) Lemma 9. Assume that n > 2. Then, for any λ > Λ1 , problem (23) has a solution u0 ∈ H01 \ {0}, i.e. Eλ (u0 ) = Eˆλ and u0 ∈ Mλ .
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
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Proof. Let λ > Λ1 . Then Mλ is not empty and there is a minimizing sequence (um ) of (23). Set tm ≥ 0 and vm ∈ H01 , m = 1, 2, . . . such that um = tm vm , kvm k1 = 1. Let us show that (tm ) is bounded. Observe that β−1 α−1 1 − λtm B(vm ) + tm A(vm ) = 0,
(25)
since Qλ (tm um ) = 0, m = 1, 2, . . . . Note that, since kvm k1 = 1, B(vm ), A(vm ) are bounded. Suppose that there exists a subsequence again denoted (tm ) such that tm → ∞ as m → +∞. Then the left-hand side of (25) tends to 1 as m → +∞. But this contradicts the assumption that Qλ (um ) = 0, m = 1, 2, . . . . Suppose now that there exist subsequences again denoted (tm ), (vm ) such that tm → 0 and/or vm + 0 weakly in H01 as m → +∞. β−1 α−1 Assume that tm A(vm ) → C as m → ∞, where 0 ≤ C < +∞. Then λtm B(vm ) → 1 + C as m → ∞. By (14), we have 1/p B(vm ) ≤ C0 · A(vm ) , where 0 < C0 < +∞ does not depend on m = 1, 2, . . . . Therefore β−1+ (1−α) p
β−1 β−1 tm B(vm ) ≤ C0 · tm A(vm )1/p = C0 · tm
(tmα−1 A(vm ))1/p .
(26)
Let us show that
β −1+
(1 − α) p
> 0.
(27)
(2∗ −1−α)
Substituting p = (2∗ −1−β) , we get
β −1+
(1 − α) p
=
(β − 1)(2∗ − 1 − α) + (1 − α)(2∗ − 1 − β) . (2∗ − 1 − α)
Since
(β − 1)(2∗ − 1 − α) + (1 − α)(2∗ − 1 − β) = 2∗ (β − α) − (1 + α)(β − 1) − (1 − α)(1 + β) = 2∗ (β − α) − 2(β − α) = (β − α)(2∗ − 2) > 0, β−1
we get (27). This implies that the right-hand side of (26) tends to zero, and therefore tm contradicts our assumption. α−1 Assume now that tm A(vm ) → +∞ as m → ∞. Then, by (25), we have A(vm ) β−α
λtm
B(vm )
→1
B(vm ) → 0 as m → ∞, which
(28)
as m → ∞. Using (14) and (15), we deduce that A(vm ) β−α
λtm
B(vm )
> C2
A(vm )(p−1)/p β−α
tm
= C2
(tm(α−1) A(vm ))(p−1)/p (p−1)(α−1)/p+β−α
tm
,
(29)
where 0 < C2 < +∞ does not depend on m = 1, 2, . . . . From (27), we have
(p − 1)(α − 1) p
+β −α =β −1+
1−α p
> 0.
This implies that the right-hand side of (29) tends to +∞, contrary to (28). Thus (um ) is bounded in H01 , and hence, by Sobolev’s embedding theorem, (um ) has a subsequence which converges weakly in H01 and strongly in Lp , 1 < p < 2∗ . Denoting this subsequence again by (um ), we get um → u0 weakly in H01 and strongly in Lp , 1 ≤ p < 2∗ for some u0 ∈ H01 . By the above, the sequences (tm ) and (vm ) are separated from zero, and
therefore u0 6= 0. Thus Eλ (u0 ) ≤ Eˆ λ and Qλ (u0 ) ≤ 0. If Qλ (u0 ) = 0, then u0 ∈ Mλ , and thus Eλ (u0 ) = Eˆ λ . Assume that Qλ (u0 ) < 0. Since n > 2, then there exists tλ2 (u0 ) > 0 such that Qλ (tλ2 (u0 )u0 ) = 0 and e00λ (t 2 (u0 )) ≥ 0. Therefore Eλ (tλ2 (u0 )u0 ) < Eλ (u0 ). Thus Eλ (tλ2 (u0 )u0 ) < Eλ (u0 ) ≤ Eˆ λ , and we get a contradiction. This completes the proof of Lemma 9. By (8), and using arguments as in the proof of Lemma 9, the following is deduced. Corollary 10. Assume that n > 2. If λ > Λ0 , then Eˆλ < 0. If Λ1 < λ < Λ0 , then 0 < Eˆλ < +∞, and if λ = Λ0 , then Eˆλ = 0.
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4.2. Existence of the solution of (1) Let λ > Λ1 ; then, by Lemma 9, there exists a solution u0 ∈ H01 \ {0} of (23). This implies that there exist Lagrange multipliers µ1 , µ2 such that
µ1 DEλ (u0 ) = µ2 DQλ (u0 ),
(30)
and |µ1 | + |µ2 | 6= 0. Proposition 11. Let θ < 0, n > 2, λ > Λ1 and u0 ∈ H01 be a solution of (23). Assume that Pλ (u0 ) ≤ 0. Then u0 is a weak nonnegative solution of (1). Proof. Note that, by Proposition 5, we have Lλ (u0 ) 6= 0, since θ < 0, n > 2, Qλ (u0 ) = 0, and by the assumption Pλ (u0 ) ≤ 0. From (23) and (30), we have 0 = µ1 Qλ (u0 ) = µ2 Lλ (u0 ). But Lλ (u0 ) 6= 0, and therefore µ2 = 0. Thus, by (30), we have DEλ (u0 ) = 0; that is, u0 is a weak solution of (1). Furthermore, since Eλ (|u0 |) = Eλ (u0 ), Qλ (|u0 |) = Qλ (u0 ) = 0, we may assume that u0 ≥ 0. This completes the proof. Corollary 12. Let θ < 0, n > 2, λ ≥ Λ0 , and u0 ∈ H01 be a solution of (23). Then Pλ (u0 ) < 0 and u0 is a weak solution of (1). Proof. Corollary 10 implies that Eˆ λ < 0 when λ > Λ0 , and Eˆ λ = 0 when λ = Λ0 . Hence, for any λ ≥ Λ0 , we have Eλ (u0 ) ≤ 0, and therefore the identity Eλ (u0 ) = Pλ (u0 )+(1/2n)T (u0 ) implies that Pλ (u0 ) < 0. Applying now Proposition 11, we complete the proof. 5. Proof of Theorem 1 Let us introduce Z := {λ > 0 : ∃uλ ∈ Mλ s.t. Eλ (uλ ) = Eˆ λ , Pλ (uλ ) < 0}.
(31)
Denote by Z the closure of Z . By assumption, n > 2(1 +α)(1 +β)/(1 −α)(1 −β) and n > 2. Hence Lemma 9 and Corollary 12 imply that Z is bounded below by Λ1 and [Λ0 , +∞) ⊂ Z , i.e. Z 6= ∅. Furthermore, Corollary 19 from Appendix A yields that Z \ Λ1 is an open set in R. Introduce
λ∗ := inf Z . Lemma 13. There exists a solution u∗ of (23) with λ = λ∗ . Furthermore, Λ1 < λ∗ and Pλ∗ (u∗ ) = 0. Proof. Since Z is an open set, λ∗ ∈ Z , and we can find a sequence λm ∈ Z , λm > λ∗ m = 1, 2, . . ., such that λm → λ∗ as m → ∞. By the definition of Z , for any m = 1, 2, . . . there exists solution uλm of (23) such that Pλm (uλm ) < 0. Lemma 20 from Appendix A yields the existence of the limit solution u∗ of (23) and the existence of a subsequence (again denoted by (uλm )) such that uλm → u∗ strongly in H 1 as λm → λ∗ . This yields that Pλ∗ (u∗ ) ≤ 0. Let us show that Λ1 < λ∗ . To obtain a contradiction, suppose that Λ1 = λ∗ . Then by the proof of Lemma 20 from Appendix A we have Λ1 = λ1 (u∗ ). This and (12) yield that u∗ is a critical point of λ(u), and therefore, by Proposition 21 from Appendix B, the function u∗ weakly satisfies (1) with λ = Λ0 . On the other hand, since Pλm (uλm ) < 0, m = 1, 2, . . . , by Proposition 11, uλm weakly satisfies (1) with λ = λm , m = 1, 2, . . . . From this, and since uλm → u∗ strongly in H 1 as λm → Λ1 , we derive that u∗ is a weak solution of (1) with λ = Λ1 . But Λ1 6= Λ0 , and we get a contradiction. Thus λ∗ ∈ (Λ1 , +∞). Suppose, contrary to our claim, that Pλ∗ (u∗ ) < 0. Then λ∗ ∈ Z . However, since Z is an open set, we get a contradiction. Thus Pλ∗ (u∗ ) = 0. Conclusion of the proof of Theorem 1 From Lemma 13 and Proposition 11, it follows that there exists a weak nonnegative solution u∗ ∈ H01 of (1) with λ = λ∗ . By Lemma 4, Pohozaev’s identity holds, and hence
R ∂ u∗ 2 ∂ν x · ν dx = 0, since Pλ∗ (u∗ ) = 0. By assumption, Ω is a strictly
u star-shaped domain with respect to the origin, i.e. x · ν > 0 on ∂ Ω . Therefore ∂∂ν = 0 on ∂ Ω , and thus u∗ is a compacton solution of (1) with λ = λ∗ . Let us now show that, for any λ > λ∗ , problem (1) has a compacton solution. Let σ > 1. Then Ωσ := {x ∈ Rn : x · σ ∈ Ω } ⊂ Ω , since Ω is a star-shaped domain with respect to the origin. Let us set uσ (x) = uλ∗ (x · σ ), x ∈ Ωσ , and uσ (x) = 0 in Ω \ Ωσ . Then the following identity ∗
−
1
σ2
1uσ = λ0 (uσ )β − (uσ )α
uσ weakly holds in Ωσ . Furthermore, since uσ = 0 and ∂∂ν = 0 on ∂ Ωσ , this identity weakly holds also in Ω . This implies 2 that the function wσ (x) = σ 1−α · uσ (x), which has a compact support in Ω , weakly satisfies problem (1) in Ω with
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355 2(β−α) 1−α
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1−α
· λ∗ . Hence, for any λ > λ∗ , setting σ = (λ/λ∗ ) 2(β−α) , we get the compacton solution wσ of (1). Furthermore, it is easily seen that the function wσ ,y (x) = wσ (x + y) is also a compacton solution of (1) for any y ∈ Rn , |y| < δ , where δ is a sufficiently small number. This implies that, for any λ > λ∗ , problem (1) has an infinite number of compacton solutions. Using Proposition 21 from Appendix B, similarly to in the proof of Lemma 13, we deduce that problem (1) with λ = Λ1 has no weak solution. Hence, and by Proposition 7, it follows that there exists λ ∈ (Λ1 , λ∗ ] such that problem (1) has no nontrivial solution for all λ ∈ [0, λ). This completes the proof. Now, following [13] we show that the solution of (1) obtained by the constrained minimization method (23) is a ground state. We recall that a solution wλ ∈ H01 of (1) is said to be a ground state if it has the property of minimizing the action (Lagrangian) among all solutions of (1) (see [13]), namely λ=σ
0 < Eλ (wλ ) ≤ Eλ (vλ ) for any solution vλ of (1). Corollary 14. Let Ω be a bounded domain in Rn with smooth boundary, which is strictly star-shaped with respect to the origin. Assume that 0 < α < β < 1, n > 2 and n > 2(1 + α)(1 + β)/(1 − α)(1 − β). Then, for every λ ∈ Z , problem (1) has a ground state wλ such that wλ ∈ C 1,µ (Ω ) for µ ∈ (0, 1) and wλ ≥ 0 in Ω . Furthermore, wλ is a non-compacton solution such ∂w that wλ ≥ 0 in Ω and ∂νλ < 0 on some subset U ⊆ ∂ Ω of positive (n − 1)-dimensional Lebesgue measure. Proof. Let λ ∈ Z and wλ be a solution of (23). Then we have P (wλ ) < 0. Hence, by Proposition 11, wλ is a weak nonnegative solution of (1). Now, let vλ denotes another solution of (1). Then Qλ (vλ ) = 0; that is, vλ belongs to the admissible set Mλ of minimization problem (23). Hence, since wλ solves (23), we have 0 < Eλ (wλ ) ≤ Eλ (vλ ); i.e. wλ is a ground state of (1).
R ∂wλ 2 ∂ν x · ν dx > 0. But Ω is a strictly star-shaped domain with respect to the origin; i.e. x · ν > 0 on ∂ Ω . Therefore there exists a subset U ⊆ ∂ Ω of positive (n − 1)-dimensional ∂w Lebesgue measure such that ∂νλ (s) < 0 for every s ∈ U. Since Pλ (wλ ) < 0, then Pohozaev’s identity implies that
Applying the same arguments to the case λ = λ∗ , we deduce the following. Corollary 15. The compacton solution u∗ is a ground state of (1) with λ = λ∗ . From Corollary 12, we have [Λ0 , +∞) ⊂ Z . Thus Corollary 14 implies that, for every λ ≥ Λ0 , problem (1) has a ground state wλ which is a non-compacton solution. By Corollary 12, we know that PΛ0 (uΛ0 ) < 0. This and Lemma 13 yield the following. Corollary 16. Λ1 < λ∗ < Λ0 . 6. Some additional remarks To conclude the paper, we would like to indicate briefly and informally some open problems in connection with (1) and make a few more remarks. In the present paper the number n = max{2, 2(1 +α)(1 +β)/(1 −α)(1 −β)} plays a crucial role. An open and important question is whether this number is a critical exponent for (1), i.e. if problem (1) for n ≤ n has no compacton solution. It follows from Theorem 1 that there is a point λ∗∗ ≥ λ such that, for λ ≥ λ∗∗ , problem (1) possesses nonnegative solutions, whereas if λ < λ∗∗ , then (1) has no nonnegative solutions. We conjecture that λ∗∗ = λ, and for all λ > λ∗∗ , problem (1) has at least two distinct nonnegative solutions such that one of them is a positive non-compacton solution. It is an open question whether λ∗∗ < λ∗ or λ∗∗ = λ∗ . If λ∗∗ < λ∗ , it seems reasonable to suppose that, for λ = λ∗∗ , similar to one-dimensional case in [1,4], there is a unique positive solution of (1) which is a non-compacton solution. On the other hand, if λ∗∗ = λ∗ , then by Theorem 1 we know that problem (1) for λ = λ∗∗ = λ∗ has a compacton solution. Let us mention that these both possibilities exhibit different structure for the set of nonnegative solutions from those in [1,4]. Furthermore, it is reasonable to suppose that in the above proof of Theorem 1 the sets Z and (λ∗ , +∞) coincide, and so, by Corollary 14, for all λ > λ∗ , there exists a positive non-compacton solution wλ of (1) which is a ground state. Furthermore, it is reasonable to conjugate that wλ satisfies the Hopf boundary maximum principle on the whole boundary uλ ∂ Ω ; i.e. ∂∂ν (s) < 0 for all s ∈ ∂ Ω . Note that if one considers the radial symmetric solutions of (1) in the ball BR , then Corollary 14 implies that this property holds for all λ ∈ Z . From our investigation we could not draw a conclusion about the relative position of both the non-compacton wλ and compacton uλ solutions. We point out that in [1,4] for the one-dimensional case of (1) it is shown that wλ > uλ . Acknowledgements The authors thank the referee for his useful suggestions and comments. The first author was partly supported by grants RFBR 08-01-00441-a and RFBR 08-01-97020-p-a.
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Appendix A Proposition 17. Let λ ∈ [Λ1 , +∞) and uλm be a sequence of solutions of (23), where λm → λ as m → +∞. Then there exist a subsequence (again denoted by (uλm )) and a limit point u¯ λ such that uλm → u¯ λ strongly in Lp (Ω ), where 1 ≤ p < 2∗ , and uλm + u¯ λ weakly in H01 as m → +∞. Furthermore, Eλ (¯uλ ) ≤ lim Eλm (uλm ),
(32)
Qλ (¯uλ ) ≤ 0.
(33)
m→∞
Proof. Let λ ∈ [Λ1 , +∞) and uλm be a sequence of solutions of (23), where λm → λ as m → +∞. Let tm ≥ 0 and vm ∈ H01 , m = 1, 2, . . . , be such that uλm = tm vm , kvm k1 = 1. As in the proof of Lemma 9, using (25), it is derived that (tm ) is bounded. This implies that the set (uλm ), m = 1, 2, . . ., is bounded in H01 . Hence by the Sobolev embedding theorem and by the Eberlein–Šmulian theorem we may assume that uλm → u¯ λ strongly in Lp (Ω ), where 1 ≤ p < 2∗ , and uλm + u¯ λ weakly in H01 as m → +∞ for some limit point u¯ λ . This yields that u¯ λ ∈ H01 is a weak nonnegative solution of (1) and (32), (33) hold. As in the proof of Lemma 9, using (25), it is derived that u¯ λ 6= 0. Lemma 18. Assume that λ ∈ Z , λm ∈ (Λ1 , +∞), m = 1, 2, . . . , λm → λ as m → +∞ and uλm is a sequence of solutions of (23). Then there exist a subsequence (again denoted by (uλm )) and a limit solution uλ of (23) such that uλm → uλ strongly in H 1 as m → +∞. Proof. Let λ ∈ Z . By Lemma 9, there exists a solution uλ of (23), i.e. uλ ∈ Mλ and Eˆ λ = Eλ (uλ ). Furthermore, Pλ (uλ ) < 0, since λ ∈ Z . This and Proposition 5 yield that Lλ (uλ ) > 0. Observe that
|Eλ (uλ ) − Eλm (uλ )| ≤
1
β +1
|λm − λ| · B(uλ ) < C1 |λ − λm |,
|Lλ (uλ ) − Lλm (uλ )| ≤ β|λm − λ| · B(uλ ) < C2 |λ − λm |,
(34) (35)
where C1 , C2 < +∞. Inequality (35) and Lλ (uλ ) > 0 yield that Lλm (uλ ) > 0 as m > N, where N is a sufficiently large number. Therefore there exists tλ2m (uλ ) > 0 such that Eλm (uλ ) ≥ Eλm (tλ2m (uλ )uλ )
if m > N .
(36)
Furthermore, we have Eλm (tλm (uλ )uλ ) ≥ Eλm (uλm ) since tλm (uλ )uλ ∈ Mλm . Thus Eλm (uλ ) ≥ Eλm (uλm ) if m > N. This and (34) give 2
2
Eλ (uλ ) + C1 |λ − λm | > Eλm (uλ ) ≥ Eλm (uλm ) as m > N . Thus Eˆ λ := Eλ (uλ ) ≥ limm→∞ Eλm (uλm ), and by (32) we obtain Eλ (¯uλ ) ≤ Eˆ λ . Assume that Qλ (¯uλ ) < 0. Then Qλ (tλ2 (¯uλ )¯uλ ) = 0, that is tλ2 (¯uλ )¯uλ ∈ Mλ and Eλ (tλ2 (¯uλ )¯uλ ) < Eλ (¯uλ ) ≤ Eˆ λ . Hence we get a contradiction. Therefore, by (33), we have Qλ (¯uλ ) = 0. This implies that Eλ (¯uλ ) = Eˆ λ . Hence, and by Proposition 17, we conclude that uλm → u¯ λ strongly in H 1 as m → +∞. The next corollary follows immediately from Lemma 18. Corollary 19. Z \ Λ1 is an open set in R. Lemma 20. Assume that λ ∈ Z , λm ∈ (Λ1 , +∞), λm > λ, m = 1, 2, . . . , λm ↓ λ as m → +∞ and uλm is a sequence of solutions of (23). Then there exist a subsequence (again denoted by (uλm )) and a limit solution uλ of (23) such that uλm → uλ strongly in H 1 as m → +∞. Proof. Assume that λ = Λ1 and Λ1 < λm , m = 1, 2, . . . . By (10), we see that Λ1 < λ1 (uλm ) ≤ λm , and therefore λ1 (uλm ) → Λ1 as m → ∞. Thus (uλm ) is a minimizing sequence of (12). By Proposition 17, there exist a subsequence (again denoted by (uλm )) and a limit point uλ such that uλm → uλ strongly in Lp (Ω ), where 1 ≤ p < 2∗ , and uλm + uλ weakly in H01 as m → +∞. Using this, we deduce that λ1 (uλ ) ≤ Λ1 . Since inequality λ1 (uλ ) < Λ1 is impossible, we conclude that λ1 (uλ ) = Λ1 and that uλm → uλ strongly in H 1 as m → +∞. Thus we get the proof when λ = Λ1 . Let now λ ∈ Z \ Λ1 . Since λ > Λ1 , then by Lemma 9 there exists a solution uλ of (23) and (34) holds if m > N, where N is a sufficiently large number. Since Qλ (uλ ) = 0 and λm > λ, then Qλm (uλ ) < 0. Note that Qλm (t · uλ ) < 0 if and only if t ∈ (tλ1 (uλ ), tλ2 (uλ )). This implies that Eλm (uλ ) > Eλm (tλ2m (uλ )uλ ) provided m > N (see (36)). Now, repeating the same arguments that have used after formula (36) in the proof of Lemma 18, we obtain the proof of the lemma.
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
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Appendix B Proposition 21. Let t0 (u) and λ(u) be functions on H01 \ {0} defined by (5) and (7), respectively. Suppose that w0 ∈ H01 \ {0} is a critical point of λ(u). Then t0 (w0 )w0 is a weak solution of (1) with λ = Λ0 . Proof. We first compute Dλ(u) =
2(β − α)
(1 − α)
β−1
1−β
·
A(u) 1−α T (u) 1−α B(u)
1−β
β−α
1−α 1−α ˙ (u) − (β + 1) · A(u) T (u) ˙ ( u) · DT · DB B(u)2
α−β
β−α
(1 − β)(1 + α) A(u) 1−α T (u) 1−α ˙ · · DA(u), 1−α B(u) R ˙ (u)(φ) = (∇ u, ∇φ)dx, where we use the notations DT Z Z ˙ (u)(φ) = |u|β−1 uφ dx, ˙ (u)(φ) = |u|α−1 uφ dx DB DA +
for u, φ ∈ H01 .
α,β
Let w0 ∈ H01 \ {0} be a critical point of λ(u). Then Dλ(w0 ) = 0. Hence, and using the identity c0 λ(w0 ) = Λ0 , we deduce that α,β
0 = Dc0 λ(w0 )
1−β (β + 1) α,β 2(β − α) A(w0 ) 1−α ˙ (w0 ) − Λ0 · DB ˙ (w0 ) · DT = · c0 B(w0 ) (1 − α)(β + 1) T (w0 ) α−β 1−α A (w ) ( 1 − β)( 1 + α) 0 α,β ˙ (w0 ) . + c0 · DA (1 − α)(β + 1) T (w0 ) Now, substituting u0 = t0 (w0 )w0 =
A(w0 ) 2(β−α) (1+α)(1−β) T (w0 )
1 1−α
· w0 into this equation and multiplying by t0 (w0 )β , we obtain
the desired equality:
˙ (u0 ) − Λ0 DB ˙ (u0 ) + DA ˙ (u0 ) = 0. DT References [1] J.I. Díaz, J. Hernández, Global bifurcation and continua of nonnegative solutions for a quasilinear elliptic problem, C. R. Acad. Sci., Paris 329 (1999) 587–592. [2] P. Rosenau, J.M. Hyman, Compactons: Solitons with finite wavelength, Phys. Rev. Lett. 70 (5) (1993) 564–567. [3] P. Rosenau, E. Kashdan, Compactification of nonlinear patterns and waves, Phys. Rev. Lett. 101 (2008) 264101. [4] J.I. Díaz, J. Hernández, F.J. Mancebo, Branches of positive and free boundary solutions for some singular quasilinear elliptic problems, J. Math. Anal. Appl. 352 (1) (2009) 449–474. [5] S.I. Pohozaev, Eigenfunctions of the equation 1u + λf (u) = 0, Soviet Math. Dokl. 5 (1965) 1408–1411. [6] S.I. Pohozaev, On the method of fibering a solution in nonlinear boundary value problems, Proc. Steklov Inst. Math. 192 (1990) 157–173. [7] Y.S. Il’yasov, Nonlocal investigations of bifurcations of solutions of nonlinear elliptic equations, Izv. Math. 66 (6) (2002) 1103–1130. [8] J. Hernández, F.J. Mancebo, J.M. Vega, Positive solutions for singular nonlinear elliptic equations, Proc. Roy. Soc. Edinburgh 137A (2007) 41–62. [9] Y. Haitao, Positive versus compact support solutions to a singular elliptic problem, J. Math. Anal. Appl. 319 (2) (2006) 830–840. [10] M. Struwe, Variational Methods, Application to Nonlinear Partial Differential Equations and Hamiltonian Systems, Springer-Verlag, Berlin, Heidelberg, New York, 1996. [11] D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, 2nd ed., in: Grundlehren, vol. 224, Springer, Berlin, Heidelberg, New York, Tokyo, 1983. [12] S.I. Pohozaev, On the eigenfunctions of quasilinear elliptic problems, Mat. Sb. 11 (2) (1970) 192–212. 171. [13] S. Coleman, V. Glazer, A. Martin, Action minima among solutions to a class of Euclidean scalar field equations, Comm. Math. Phys. 58 (1978) 211–221.
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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Hopf boundary maximum principle violation for semilinear elliptic equations Yavdat Il’yasov a,∗ , Youri Egorov b a
Institute of Mathematics, Ufa Scientific Center of RAS, Ufa, Russia
b
Université Paul Sabatier, Toulouse, France
article
info
Article history: Received 31 January 2009 Accepted 8 December 2009 MSC: 35J60 35J67 Keywords: Pohozaev’s identity Non-Lipschitz nonlinearity Variational approach
abstract We consider elliptic equations with non-Lipschitz nonlinearity
−∆u = λ|u|β−1 u − |u|α−1 u in a smooth bounded domain Ω ⊂ Rn , with Dirichlet boundary conditions; here 0 < α < β < 1. We prove the existence of a weak nonnegative solution which does not satisfy the Hopf boundary maximum principle, provided that λ is large enough and n > max{2, 2(1 + α)(1 + β)/(1 − α)(1 − β)}. © 2009 Elsevier Ltd. All rights reserved.
1. Introduction Let Ω be a bounded domain in Rn with a smooth boundary ∂ Ω , which is strictly star-shaped with respect to the origin in Rn . We consider the following problem:
−1u = λ|u|β−1 u − |u|α−1 u in Ω , u = 0 on ∂ Ω .
(1)
Here λ is a real parameter and 0 < α < β < 1, so the nonlinearity f (λ, u) := λ|u|β−1 u − |u|α−1 u on the right-hand side of (1) is non-Lipschitzean at zero. Our interest in this problem has been induced by [1], where in the case of dimension n = 1, Ω = (−1, 1), among other results, the authors showed that, for certain values λ > 0, Eq. (1) possesses nonnegative solutions u(x), x ∈ (−1, 1) with a special feature u(−1) = u(1) = 0,
u0 (−1) = u0 (1) = 0.
(2)
This result means that the Hopf boundary maximum principle is violated at the points x = −1, x = 1 and the uniqueness is lost for the initial value problem for (1) with u(−1) = u0 (−1) = 0, since u ≡ 0 also satisfies (1). Furthermore, it can easily be shown that the existence of such a solution with λ0 > 0 yields the existence of a continuum set of nonnegative solutions to this boundary value problem for any λ > λ0 . Note that when the nonlinearity f (λ, u) is a locally Lipschitz function such a phenomenon is impossible due to the uniqueness of the solution of the initial value problem and the Hopf boundary maximum principle. Observe that property (2) implies that the function u is also a weak solution of (1) on the
∗
Corresponding author. E-mail addresses: [email protected] (Y. Il’yasov), [email protected] (Y. Egorov).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.12.015
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
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whole line R. In the theory of integrable systems, the solutions of such type are known as ‘‘compactons’’: solitary waves with compact support [2,3]. In the PDE literature the term ‘‘free boundary solution’’ (see [1,4]) is also used for such type of solution. The purpose of the present paper is to find an answer to the following question: Can the Hopf boundary maximum principle be violated for semilinear elliptic equations such as (1) in higher dimensions n > 1, including non-radial cases? Let us state our main result. We consider a weak solution u ∈ H01 := H01 (Ω ), where H01 (Ω ) denotes the closure C0∞ (Ω ) in the standard Sobolev space H 1 (Ω ) with the norm k · k1 . We use the term compacton solution for a weak solution u ∈ C 1 (Ω ) ∂u of (1), which does not satisfy the Hopf boundary maximum principle, i.e. ∂νλ = 0 on ∂ Ω , where ν denotes the unit outward normal. In the opposite case the weak solution u ∈ C 1 (Ω ) of (1) is said to be a non-compacton solution. Our main result is the following.
Theorem 1. Let Ω be a bounded domain in Rn with smooth boundary, which is strictly star-shaped with respect to the origin. Assume that 0 < α < β < 1 and n > max{2, 2(1 + α)(1 + β)/(1 − α)(1 − β)}. Then there exists λ∗ > 0 such that, for every λ ≥ λ∗ , problem (1) has a compacton solution uλ ∈ C 1,κ (Ω ) for κ ∈ (0, 1), which is nonnegative in Ω . Moreover, the number of such solutions for λ > λ∗ is infinite. Furthermore, there exists λ ∈ (0, λ∗ ] such that problem (1) has no nontrivial solution for all λ ∈ [0, λ). The proof of the theorem relies on variational arguments. Furthermore, basic ingredients in the proof consist in using Pohozaev’s identity [5] corresponding to (1) and in applying a spectral analysis with respect to the fibering procedure [6] introduced in [7]. From [8] it follows that there exists λ∗∗ ≤ λ∗ such that for λ > λ∗∗ problem (1) has a second weak solution wλ ∈ C 1,µ (Ω ), λ < 0 holds on ∂ Ω . µ ∈ (0, 1), which is positive wλ > 0 in Ω and satisfies the Hopf boundary maximum principle, i.e. ∂w ∂ν Furthermore, this result in [8] does not require the above restrictions on n and the star-shaped property of Ω in Theorem 1. However, in the framework of our approach we are able to obtain a different result in this direction. In Section 5 (see Corollary 14) we derive from our variational problem that (1) possesses a weak non-compacton solution wλ which is a ground state for (1), i.e. 0 < Eλ (wλ ) ≤ Eλ (vλ ) for any solution vλ of (1), where Eλ is the Lagrangian associated with (1) (see below (3)). A result on the existence of nonnegative compacton solution, but for the different version of problem (1), has been obtained in [9]. The results in [1] have been extended in [4] both in allowing singular terms in the right-hand side of (1), i.e. −1 < α < β < 1 and to consider a quasilinear differential operator like the p-Laplacian. We also refer the reader to [4] for a deeper discussion of the problem and for further references. The paper is organized as follows. In Section 2, we apply a spectral analysis related to the fibering procedure [7] to introduce two spectral points Λ0 , Λ1 which play a basic role in the proof of the main result. In Section 3, we derive some important consequences from Pohozaev’s identity. In Section 4, we prove the existence of the solution to an auxiliary constrained minimization problem. In Section 5, we prove Theorem 1. In Section 6, we indicate briefly some open problems in connection with (1) and make a few remarks. Some technical results are proved in Appendices A and B. 2. Spectral analysis with respect to the fibering procedure In this section we apply a spectral analysis related to the fibering procedure [7] to introduce two spectral points which will play important roles in the proof of the main result. Throughout this section we assume that n > 2. Observe that problem (1) is the Euler–Lagrange equation associated with the functional Eλ (u) =
1 2
T ( u) − λ
1
β +1
B(u) +
1
α+1
A(u),
u ∈ H01 ,
(3)
where we use the notations T (u) =
Z
|∇ u| dx, 2
Ω
B(u) =
Z
β+1
| u| Ω
dx,
A(u) =
Z Ω
|u|α+1 dx.
Let u ∈ H01 . Consider the function eλ (t ) := Eλ (tu) defined for t ∈ R+ . Introduce the functionals Qλ (u) = e0λ (t )|t =1 , Lλ (u) = e00λ (t )|t =1 for u ∈ H01 . Then Qλ (u) = T (u) − λB(u) + A(u), Let u ∈
Lλ (u) = T (u) − λβ B(u) + α A(u).
H01
\ {0}. Following the spectral analysis [7], we solve the system 2 1+β 1+α Qλ (tu) = t T (u) − λt B(u) + t A(u) = 0 Eλ (tu) =
t2 2
T (u) − λ
t 1+β
1+β
B(u) +
t 1+α
1+α
A(u) = 0
(4)
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Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
and find the corresponding solution 2(β − α)
t0 (u) =
A(u)
1 1−α
(1 + α)(1 − β) T (u)
,
(5)
α,β
λ0 (u) = c0 λ(u),
(6)
where α,β
c0
(1 − α)(1 + β) = (1 − β)(1 + α)
(1 + α)(1 − β) 2(β − α)
β−α 1−α
and β−α
1−β
A(u) 1−α T (u) 1−α
λ(u) =
B(u)
.
(7)
Thus with respect to the spectral analysis, we have the following spectral point:
Λ0 = inf λ0 (u).
(8)
H01 \{0}
Introduce the second point Λ1 . Let u ∈ H01 \ {0}. Consider now the following system:
Qλ (tu) = t 2 T (u) − λt 1+β B(u) + t 1+α A(u) = 0 Lλ (tu) = t 2 T (u) − λβ t 1+β B(u) + α t 1+α A(u) = 0
(9)
for t ∈ R+ , λ ∈ R+ . Solving this system, we find, as above, α,β
λ1 (u) = c1 λ(u),
(10)
where α,β
c1
=
1−α
1−β
1−β
β−α 1−α
β −α
.
(11)
Then we have
Λ1 = inf λ1 (u).
(12)
H01 \{0}
Proposition 2. 0 < Λ1 < Λ0 < +∞. α,β
Proof. By (6) and (10), the inequality Λ1 < Λ0 is equivalent to c1
(1 + β) (1 + α)
1+α
β−α 1−α
2
> 1 ⇐⇒
1+β
1−α
1+α
>
2
α,β
< c0 , which can also be written as follows:
β−α
1+α
.
β−α
Now, letting µ = 11−α > 0, η = 1+α > 0, we obtain +α 1
1
(1 + η) η > (1 + µ) µ . 1
Hence, since the function x 7→ (1 + x) x is decreasing on ]0, +∞[ and µ > η, we obtain the desired inequality Λ1 < Λ0 . It is clear that Λ0 < +∞. Let us show that 0 < Λ1 . Note that λ(u) is a zero-homogeneous function on H01 \ {0}. Therefore we may restrict the infimum in (12) to the set S := {v ∈ H01 : kvk1 = 1}. Set
γ =
(1 + α)(2∗ − 1 − β) , (2∗ − 1 − α)
p=
(1 + α) , γ
q=
2∗
(1 + β − γ )
,
(13)
where 2∗ = 2n/(n − 2). Then p, q > 1, 1/p + 1/q = 1 for n > 2, and by the Hölder inequality we have B(u) ≤
Z
∗
u2 dx Ω
1/q
· A(u)1/p .
(14)
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3349
By Sobolev’s inequality,
Z
∗
∗
u2 dx ≤ C0 kuk21 = C0 < +∞
Ω
(15)
for u ∈ S, where C0 does not depend on u ∈ H01 . Hence, for any u ∈ S, we have
λ(u) =
A(u)(1−β) B(u)(1−α)
− (1−α) q
≥ C0
A(u)
− (1−α) q
p(1−β)−(1−α) p
= C0
∗
A(u)
2)(β−α) − (2(2− ∗ −1−α)
.
Since A(u) ≤ C1 < +∞ on S, where 0 < C1 < +∞ does not depend on u ∈ H01 , we see from (10) that Λ1 > 0.
3. Pohozaev’s identity We will need the following regularity result. Proposition 3. Assume that 0 < α < β < 1. Suppose that u ∈ H01 is a weak solution of (1). Then u ∈ C 1,κ (Ω ) for κ ∈ (0, 1). Proof. Let u ∈ H01 be a weak solution of (1). Since |f (λ, u)| < C (1 + |u|), u ∈ R with some C > 0, then (see Lemma B.3 in [10]) u ∈ Lq (Ω ) for any q ∈ [1, ∞). This implies that −1u = f (λ, u) ∈ Lq (Ω ) for any q ∈ [1, ∞). Thus, by the Caldéron–Zygmund inequality (see [11]), u ∈ H 2,q (Ω ) for any q ∈ [1, ∞), whence u ∈ C 1,κ (Ω ) for κ ∈ (0, 1) by the Sobolev embedding theorem. We will denote by Pλ the functional Pλ (u) :=
(n − 2) 2n
T (u) − λ
1
β +1
B(u) +
1
α+1
A(u)
defined for u ∈ H01 . Lemma 4. Suppose Ω is a smooth bounded domain in Rn . Let u be a weak solution of (1), u ∈ H01 . Then Pohozaev’s identity Pλ (u) +
1
Z
2n ∂ Ω
2 ∂u x · ν ds = 0 ∂ν
(16)
holds. Proof. Observe that by Proposition 3 we know that u ∈ H 2,q ∩ H01 for any q ∈ [1, ∞). We use the notations F (λ, z ) =
1 1 f (λ, τ ) dτ ≡ λ β+ B(z ) − α+ A(z ), ui = ∂∂xu , uij = ∂∂x xu and we adopt the summation convention on repeated indices. As 1 1 i i j in [5,12], multiplication of both parts of the equation by ui xi and integration by parts yields 2
Rz 0
Z Ω
f (λ, u)ui xi dx = −n
Z Ω
F (λ, u) dx +
Z ∂Ω
F (λ, u)x · ν ds
and
Z
Z
− Ω
ujj ui xi dx =
Ω
Z
uj (δij ui + xi uij ) dx −
|∇ u|2 dx −
= Ω
n 2
Z
Z ∂Ω
uj νj xi ui ds
|∇ u|2 dx − Ω
Thus, since u = 0 on the boundary, we obtain the result.
2 ∂u x · ν ds. 2 ∂ Ω ∂ν
1
Z
Let u ∈ H01 . Based on the ideas of spectral analysis [7], we consider the following system of equations:
Q (u) := T (u) − λB(u) + A(u) = 0 λ Lλ (u) := T (u) − λβ B(u) + α A(u) = 0 1 1 (n − 2) Pλ (u) := T ( u) − λ B(u) + A(u) = 0. 2n β +1 α+1
(17)
The computation of the corresponding determinant shows that this system is solvable if and only if
θ ≡ 2(1 + α)(1 + β) − n(1 − α)(1 − β) = 0.
(18)
Note that θ < 0 if and only if n > 2(1 + α)(1 + β)/(1 − α)(1 − β). Observe that the equation e0λ (t ) = 0, t > 0, has at most two roots, t 1 (u) := tλ1 (u) ∈ R+ and t 2 (u) := tλ2 (u) ∈ R+ , such that t 1 (u) ≤ t 2 (u), e00λ (t 1 (u)) ≤ 0 and e00λ (t 2 (u)) ≥ 0.
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Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
Proposition 5. Assume that θ < 0, n > 2. If u ∈ H01 \ {0} and t > 0 are such that Qλ (tu) = 0 and Pλ (tu) ≤ 0, then Lλ (tu) > 0. Proof. Let u ∈ H01 \ {0} and t > 0 as in the assumption. Then T (u) = λt β−1 B(u) − t α−1 A(u) Lλ (tu) > 0 ⇔ λt β−α
(19)
(1 − β) B(u) > A(u). (1 − α)
(20)
Equality (19) implies that Pλ (t 1 (u)u) ≤ 0 holds if and only if
λt β−α
[2(1 + β) + n(1 − β)](1 + α) B(u) ≥ A(u). [2(1 + α) + n(1 − α)](1 + β)
(21)
Observe that the assumptions θ < 0, n > 2 imply that
[2(1 + β) + n(1 − β)](1 + α) (1 − β) < . [2(1 + α) + n(1 − α)](1 + β) (1 − α)
(22)
Thus (22) and (21) give
λt β−α
(1 − β) B(u) > A(u), (1 − α)
and therefore by (20) the proof is complete.
Corollary 6. If u0 is a compacton solution of (1), then Eλ (u0 ) > 0. Furthermore, if in addition θ < 0, n > 2, then Qλ (u0 ) = 0,
Pλ (u0 ) = 0,
Lλ (u0 ) > 0.
Proof. Observe that, if u0 is a compacton solution of (1), then by (16) we have Pλ (u0 ) = 0. Hence, using Eλ (u) = Pλ (u) + (1/2n)T (u), we get Eλ (u0 ) =
1 n
T (u0 ) > 0.
Note that Qλ (u0 ) = 0 if u0 is a solution of (1), and Pλ (u0 ) = 0 if in addition this solution is the compacton solution. Hence the assumptions θ < 0, n > 2 by Proposition 5 imply that Lλ (u0 ) > 0. 4. Constrained minimization problems Consider the following constrained minimization problem:
Eλ (u) → min Qλ (u) = 0.
(23)
We denote by Mλ := {w ∈ H01 \ {0} : Qλ (u) = 0} the admissible set of (23), and by Eˆλ := min{Eλ (u) : u ∈ Mλ } the minimal value in this problem. We say that (um ) is a minimizing sequence of (23) if Eλ (um ) → Eˆλ
as m → ∞ and um ∈ Mλ ,
m = 1, 2, . . . .
(24)
Proposition 7. If λ > Λ1 , then the set Mλ is not empty; meanwhile, the set Mλ is empty when λ < Λ1 . Proof. Let λ > Λ1 . Then by (8) there exists u ∈ H01 \ {0} such that Λ1 < λ(u) < λ and Lλ(u) (t (u)u) = 0, Qλ(u) (t (u)u) = 0. Hence, Qλ (t (u)u) < 0, since λ(u) < λ, and therefore there exists t > 0 such that Qλ (tu) = 0, i.e. tu ∈ Mλ . The proof of the second part of the proposition follows immediately from (8). From here it follows that Corollary 8. Eˆλ < +∞ for any λ > Λ1 . 4.1. Existence of the solution of (23) Lemma 9. Assume that n > 2. Then, for any λ > Λ1 , problem (23) has a solution u0 ∈ H01 \ {0}, i.e. Eλ (u0 ) = Eˆλ and u0 ∈ Mλ .
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
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Proof. Let λ > Λ1 . Then Mλ is not empty and there is a minimizing sequence (um ) of (23). Set tm ≥ 0 and vm ∈ H01 , m = 1, 2, . . . such that um = tm vm , kvm k1 = 1. Let us show that (tm ) is bounded. Observe that β−1 α−1 1 − λtm B(vm ) + tm A(vm ) = 0,
(25)
since Qλ (tm um ) = 0, m = 1, 2, . . . . Note that, since kvm k1 = 1, B(vm ), A(vm ) are bounded. Suppose that there exists a subsequence again denoted (tm ) such that tm → ∞ as m → +∞. Then the left-hand side of (25) tends to 1 as m → +∞. But this contradicts the assumption that Qλ (um ) = 0, m = 1, 2, . . . . Suppose now that there exist subsequences again denoted (tm ), (vm ) such that tm → 0 and/or vm + 0 weakly in H01 as m → +∞. β−1 α−1 Assume that tm A(vm ) → C as m → ∞, where 0 ≤ C < +∞. Then λtm B(vm ) → 1 + C as m → ∞. By (14), we have 1/p B(vm ) ≤ C0 · A(vm ) , where 0 < C0 < +∞ does not depend on m = 1, 2, . . . . Therefore β−1+ (1−α) p
β−1 β−1 tm B(vm ) ≤ C0 · tm A(vm )1/p = C0 · tm
(tmα−1 A(vm ))1/p .
(26)
Let us show that
β −1+
(1 − α) p
> 0.
(27)
(2∗ −1−α)
Substituting p = (2∗ −1−β) , we get
β −1+
(1 − α) p
=
(β − 1)(2∗ − 1 − α) + (1 − α)(2∗ − 1 − β) . (2∗ − 1 − α)
Since
(β − 1)(2∗ − 1 − α) + (1 − α)(2∗ − 1 − β) = 2∗ (β − α) − (1 + α)(β − 1) − (1 − α)(1 + β) = 2∗ (β − α) − 2(β − α) = (β − α)(2∗ − 2) > 0, β−1
we get (27). This implies that the right-hand side of (26) tends to zero, and therefore tm contradicts our assumption. α−1 Assume now that tm A(vm ) → +∞ as m → ∞. Then, by (25), we have A(vm ) β−α
λtm
B(vm )
→1
B(vm ) → 0 as m → ∞, which
(28)
as m → ∞. Using (14) and (15), we deduce that A(vm ) β−α
λtm
B(vm )
> C2
A(vm )(p−1)/p β−α
tm
= C2
(tm(α−1) A(vm ))(p−1)/p (p−1)(α−1)/p+β−α
tm
,
(29)
where 0 < C2 < +∞ does not depend on m = 1, 2, . . . . From (27), we have
(p − 1)(α − 1) p
+β −α =β −1+
1−α p
> 0.
This implies that the right-hand side of (29) tends to +∞, contrary to (28). Thus (um ) is bounded in H01 , and hence, by Sobolev’s embedding theorem, (um ) has a subsequence which converges weakly in H01 and strongly in Lp , 1 < p < 2∗ . Denoting this subsequence again by (um ), we get um → u0 weakly in H01 and strongly in Lp , 1 ≤ p < 2∗ for some u0 ∈ H01 . By the above, the sequences (tm ) and (vm ) are separated from zero, and
therefore u0 6= 0. Thus Eλ (u0 ) ≤ Eˆ λ and Qλ (u0 ) ≤ 0. If Qλ (u0 ) = 0, then u0 ∈ Mλ , and thus Eλ (u0 ) = Eˆ λ . Assume that Qλ (u0 ) < 0. Since n > 2, then there exists tλ2 (u0 ) > 0 such that Qλ (tλ2 (u0 )u0 ) = 0 and e00λ (t 2 (u0 )) ≥ 0. Therefore Eλ (tλ2 (u0 )u0 ) < Eλ (u0 ). Thus Eλ (tλ2 (u0 )u0 ) < Eλ (u0 ) ≤ Eˆ λ , and we get a contradiction. This completes the proof of Lemma 9. By (8), and using arguments as in the proof of Lemma 9, the following is deduced. Corollary 10. Assume that n > 2. If λ > Λ0 , then Eˆλ < 0. If Λ1 < λ < Λ0 , then 0 < Eˆλ < +∞, and if λ = Λ0 , then Eˆλ = 0.
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4.2. Existence of the solution of (1) Let λ > Λ1 ; then, by Lemma 9, there exists a solution u0 ∈ H01 \ {0} of (23). This implies that there exist Lagrange multipliers µ1 , µ2 such that
µ1 DEλ (u0 ) = µ2 DQλ (u0 ),
(30)
and |µ1 | + |µ2 | 6= 0. Proposition 11. Let θ < 0, n > 2, λ > Λ1 and u0 ∈ H01 be a solution of (23). Assume that Pλ (u0 ) ≤ 0. Then u0 is a weak nonnegative solution of (1). Proof. Note that, by Proposition 5, we have Lλ (u0 ) 6= 0, since θ < 0, n > 2, Qλ (u0 ) = 0, and by the assumption Pλ (u0 ) ≤ 0. From (23) and (30), we have 0 = µ1 Qλ (u0 ) = µ2 Lλ (u0 ). But Lλ (u0 ) 6= 0, and therefore µ2 = 0. Thus, by (30), we have DEλ (u0 ) = 0; that is, u0 is a weak solution of (1). Furthermore, since Eλ (|u0 |) = Eλ (u0 ), Qλ (|u0 |) = Qλ (u0 ) = 0, we may assume that u0 ≥ 0. This completes the proof. Corollary 12. Let θ < 0, n > 2, λ ≥ Λ0 , and u0 ∈ H01 be a solution of (23). Then Pλ (u0 ) < 0 and u0 is a weak solution of (1). Proof. Corollary 10 implies that Eˆ λ < 0 when λ > Λ0 , and Eˆ λ = 0 when λ = Λ0 . Hence, for any λ ≥ Λ0 , we have Eλ (u0 ) ≤ 0, and therefore the identity Eλ (u0 ) = Pλ (u0 )+(1/2n)T (u0 ) implies that Pλ (u0 ) < 0. Applying now Proposition 11, we complete the proof. 5. Proof of Theorem 1 Let us introduce Z := {λ > 0 : ∃uλ ∈ Mλ s.t. Eλ (uλ ) = Eˆ λ , Pλ (uλ ) < 0}.
(31)
Denote by Z the closure of Z . By assumption, n > 2(1 +α)(1 +β)/(1 −α)(1 −β) and n > 2. Hence Lemma 9 and Corollary 12 imply that Z is bounded below by Λ1 and [Λ0 , +∞) ⊂ Z , i.e. Z 6= ∅. Furthermore, Corollary 19 from Appendix A yields that Z \ Λ1 is an open set in R. Introduce
λ∗ := inf Z . Lemma 13. There exists a solution u∗ of (23) with λ = λ∗ . Furthermore, Λ1 < λ∗ and Pλ∗ (u∗ ) = 0. Proof. Since Z is an open set, λ∗ ∈ Z , and we can find a sequence λm ∈ Z , λm > λ∗ m = 1, 2, . . ., such that λm → λ∗ as m → ∞. By the definition of Z , for any m = 1, 2, . . . there exists solution uλm of (23) such that Pλm (uλm ) < 0. Lemma 20 from Appendix A yields the existence of the limit solution u∗ of (23) and the existence of a subsequence (again denoted by (uλm )) such that uλm → u∗ strongly in H 1 as λm → λ∗ . This yields that Pλ∗ (u∗ ) ≤ 0. Let us show that Λ1 < λ∗ . To obtain a contradiction, suppose that Λ1 = λ∗ . Then by the proof of Lemma 20 from Appendix A we have Λ1 = λ1 (u∗ ). This and (12) yield that u∗ is a critical point of λ(u), and therefore, by Proposition 21 from Appendix B, the function u∗ weakly satisfies (1) with λ = Λ0 . On the other hand, since Pλm (uλm ) < 0, m = 1, 2, . . . , by Proposition 11, uλm weakly satisfies (1) with λ = λm , m = 1, 2, . . . . From this, and since uλm → u∗ strongly in H 1 as λm → Λ1 , we derive that u∗ is a weak solution of (1) with λ = Λ1 . But Λ1 6= Λ0 , and we get a contradiction. Thus λ∗ ∈ (Λ1 , +∞). Suppose, contrary to our claim, that Pλ∗ (u∗ ) < 0. Then λ∗ ∈ Z . However, since Z is an open set, we get a contradiction. Thus Pλ∗ (u∗ ) = 0. Conclusion of the proof of Theorem 1 From Lemma 13 and Proposition 11, it follows that there exists a weak nonnegative solution u∗ ∈ H01 of (1) with λ = λ∗ . By Lemma 4, Pohozaev’s identity holds, and hence
R ∂ u∗ 2 ∂ν x · ν dx = 0, since Pλ∗ (u∗ ) = 0. By assumption, Ω is a strictly
u star-shaped domain with respect to the origin, i.e. x · ν > 0 on ∂ Ω . Therefore ∂∂ν = 0 on ∂ Ω , and thus u∗ is a compacton solution of (1) with λ = λ∗ . Let us now show that, for any λ > λ∗ , problem (1) has a compacton solution. Let σ > 1. Then Ωσ := {x ∈ Rn : x · σ ∈ Ω } ⊂ Ω , since Ω is a star-shaped domain with respect to the origin. Let us set uσ (x) = uλ∗ (x · σ ), x ∈ Ωσ , and uσ (x) = 0 in Ω \ Ωσ . Then the following identity ∗
−
1
σ2
1uσ = λ0 (uσ )β − (uσ )α
uσ weakly holds in Ωσ . Furthermore, since uσ = 0 and ∂∂ν = 0 on ∂ Ωσ , this identity weakly holds also in Ω . This implies 2 that the function wσ (x) = σ 1−α · uσ (x), which has a compact support in Ω , weakly satisfies problem (1) in Ω with
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355 2(β−α) 1−α
3353
1−α
· λ∗ . Hence, for any λ > λ∗ , setting σ = (λ/λ∗ ) 2(β−α) , we get the compacton solution wσ of (1). Furthermore, it is easily seen that the function wσ ,y (x) = wσ (x + y) is also a compacton solution of (1) for any y ∈ Rn , |y| < δ , where δ is a sufficiently small number. This implies that, for any λ > λ∗ , problem (1) has an infinite number of compacton solutions. Using Proposition 21 from Appendix B, similarly to in the proof of Lemma 13, we deduce that problem (1) with λ = Λ1 has no weak solution. Hence, and by Proposition 7, it follows that there exists λ ∈ (Λ1 , λ∗ ] such that problem (1) has no nontrivial solution for all λ ∈ [0, λ). This completes the proof. Now, following [13] we show that the solution of (1) obtained by the constrained minimization method (23) is a ground state. We recall that a solution wλ ∈ H01 of (1) is said to be a ground state if it has the property of minimizing the action (Lagrangian) among all solutions of (1) (see [13]), namely λ=σ
0 < Eλ (wλ ) ≤ Eλ (vλ ) for any solution vλ of (1). Corollary 14. Let Ω be a bounded domain in Rn with smooth boundary, which is strictly star-shaped with respect to the origin. Assume that 0 < α < β < 1, n > 2 and n > 2(1 + α)(1 + β)/(1 − α)(1 − β). Then, for every λ ∈ Z , problem (1) has a ground state wλ such that wλ ∈ C 1,µ (Ω ) for µ ∈ (0, 1) and wλ ≥ 0 in Ω . Furthermore, wλ is a non-compacton solution such ∂w that wλ ≥ 0 in Ω and ∂νλ < 0 on some subset U ⊆ ∂ Ω of positive (n − 1)-dimensional Lebesgue measure. Proof. Let λ ∈ Z and wλ be a solution of (23). Then we have P (wλ ) < 0. Hence, by Proposition 11, wλ is a weak nonnegative solution of (1). Now, let vλ denotes another solution of (1). Then Qλ (vλ ) = 0; that is, vλ belongs to the admissible set Mλ of minimization problem (23). Hence, since wλ solves (23), we have 0 < Eλ (wλ ) ≤ Eλ (vλ ); i.e. wλ is a ground state of (1).
R ∂wλ 2 ∂ν x · ν dx > 0. But Ω is a strictly star-shaped domain with respect to the origin; i.e. x · ν > 0 on ∂ Ω . Therefore there exists a subset U ⊆ ∂ Ω of positive (n − 1)-dimensional ∂w Lebesgue measure such that ∂νλ (s) < 0 for every s ∈ U. Since Pλ (wλ ) < 0, then Pohozaev’s identity implies that
Applying the same arguments to the case λ = λ∗ , we deduce the following. Corollary 15. The compacton solution u∗ is a ground state of (1) with λ = λ∗ . From Corollary 12, we have [Λ0 , +∞) ⊂ Z . Thus Corollary 14 implies that, for every λ ≥ Λ0 , problem (1) has a ground state wλ which is a non-compacton solution. By Corollary 12, we know that PΛ0 (uΛ0 ) < 0. This and Lemma 13 yield the following. Corollary 16. Λ1 < λ∗ < Λ0 . 6. Some additional remarks To conclude the paper, we would like to indicate briefly and informally some open problems in connection with (1) and make a few more remarks. In the present paper the number n = max{2, 2(1 +α)(1 +β)/(1 −α)(1 −β)} plays a crucial role. An open and important question is whether this number is a critical exponent for (1), i.e. if problem (1) for n ≤ n has no compacton solution. It follows from Theorem 1 that there is a point λ∗∗ ≥ λ such that, for λ ≥ λ∗∗ , problem (1) possesses nonnegative solutions, whereas if λ < λ∗∗ , then (1) has no nonnegative solutions. We conjecture that λ∗∗ = λ, and for all λ > λ∗∗ , problem (1) has at least two distinct nonnegative solutions such that one of them is a positive non-compacton solution. It is an open question whether λ∗∗ < λ∗ or λ∗∗ = λ∗ . If λ∗∗ < λ∗ , it seems reasonable to suppose that, for λ = λ∗∗ , similar to one-dimensional case in [1,4], there is a unique positive solution of (1) which is a non-compacton solution. On the other hand, if λ∗∗ = λ∗ , then by Theorem 1 we know that problem (1) for λ = λ∗∗ = λ∗ has a compacton solution. Let us mention that these both possibilities exhibit different structure for the set of nonnegative solutions from those in [1,4]. Furthermore, it is reasonable to suppose that in the above proof of Theorem 1 the sets Z and (λ∗ , +∞) coincide, and so, by Corollary 14, for all λ > λ∗ , there exists a positive non-compacton solution wλ of (1) which is a ground state. Furthermore, it is reasonable to conjugate that wλ satisfies the Hopf boundary maximum principle on the whole boundary uλ ∂ Ω ; i.e. ∂∂ν (s) < 0 for all s ∈ ∂ Ω . Note that if one considers the radial symmetric solutions of (1) in the ball BR , then Corollary 14 implies that this property holds for all λ ∈ Z . From our investigation we could not draw a conclusion about the relative position of both the non-compacton wλ and compacton uλ solutions. We point out that in [1,4] for the one-dimensional case of (1) it is shown that wλ > uλ . Acknowledgements The authors thank the referee for his useful suggestions and comments. The first author was partly supported by grants RFBR 08-01-00441-a and RFBR 08-01-97020-p-a.
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Appendix A Proposition 17. Let λ ∈ [Λ1 , +∞) and uλm be a sequence of solutions of (23), where λm → λ as m → +∞. Then there exist a subsequence (again denoted by (uλm )) and a limit point u¯ λ such that uλm → u¯ λ strongly in Lp (Ω ), where 1 ≤ p < 2∗ , and uλm + u¯ λ weakly in H01 as m → +∞. Furthermore, Eλ (¯uλ ) ≤ lim Eλm (uλm ),
(32)
Qλ (¯uλ ) ≤ 0.
(33)
m→∞
Proof. Let λ ∈ [Λ1 , +∞) and uλm be a sequence of solutions of (23), where λm → λ as m → +∞. Let tm ≥ 0 and vm ∈ H01 , m = 1, 2, . . . , be such that uλm = tm vm , kvm k1 = 1. As in the proof of Lemma 9, using (25), it is derived that (tm ) is bounded. This implies that the set (uλm ), m = 1, 2, . . ., is bounded in H01 . Hence by the Sobolev embedding theorem and by the Eberlein–Šmulian theorem we may assume that uλm → u¯ λ strongly in Lp (Ω ), where 1 ≤ p < 2∗ , and uλm + u¯ λ weakly in H01 as m → +∞ for some limit point u¯ λ . This yields that u¯ λ ∈ H01 is a weak nonnegative solution of (1) and (32), (33) hold. As in the proof of Lemma 9, using (25), it is derived that u¯ λ 6= 0. Lemma 18. Assume that λ ∈ Z , λm ∈ (Λ1 , +∞), m = 1, 2, . . . , λm → λ as m → +∞ and uλm is a sequence of solutions of (23). Then there exist a subsequence (again denoted by (uλm )) and a limit solution uλ of (23) such that uλm → uλ strongly in H 1 as m → +∞. Proof. Let λ ∈ Z . By Lemma 9, there exists a solution uλ of (23), i.e. uλ ∈ Mλ and Eˆ λ = Eλ (uλ ). Furthermore, Pλ (uλ ) < 0, since λ ∈ Z . This and Proposition 5 yield that Lλ (uλ ) > 0. Observe that
|Eλ (uλ ) − Eλm (uλ )| ≤
1
β +1
|λm − λ| · B(uλ ) < C1 |λ − λm |,
|Lλ (uλ ) − Lλm (uλ )| ≤ β|λm − λ| · B(uλ ) < C2 |λ − λm |,
(34) (35)
where C1 , C2 < +∞. Inequality (35) and Lλ (uλ ) > 0 yield that Lλm (uλ ) > 0 as m > N, where N is a sufficiently large number. Therefore there exists tλ2m (uλ ) > 0 such that Eλm (uλ ) ≥ Eλm (tλ2m (uλ )uλ )
if m > N .
(36)
Furthermore, we have Eλm (tλm (uλ )uλ ) ≥ Eλm (uλm ) since tλm (uλ )uλ ∈ Mλm . Thus Eλm (uλ ) ≥ Eλm (uλm ) if m > N. This and (34) give 2
2
Eλ (uλ ) + C1 |λ − λm | > Eλm (uλ ) ≥ Eλm (uλm ) as m > N . Thus Eˆ λ := Eλ (uλ ) ≥ limm→∞ Eλm (uλm ), and by (32) we obtain Eλ (¯uλ ) ≤ Eˆ λ . Assume that Qλ (¯uλ ) < 0. Then Qλ (tλ2 (¯uλ )¯uλ ) = 0, that is tλ2 (¯uλ )¯uλ ∈ Mλ and Eλ (tλ2 (¯uλ )¯uλ ) < Eλ (¯uλ ) ≤ Eˆ λ . Hence we get a contradiction. Therefore, by (33), we have Qλ (¯uλ ) = 0. This implies that Eλ (¯uλ ) = Eˆ λ . Hence, and by Proposition 17, we conclude that uλm → u¯ λ strongly in H 1 as m → +∞. The next corollary follows immediately from Lemma 18. Corollary 19. Z \ Λ1 is an open set in R. Lemma 20. Assume that λ ∈ Z , λm ∈ (Λ1 , +∞), λm > λ, m = 1, 2, . . . , λm ↓ λ as m → +∞ and uλm is a sequence of solutions of (23). Then there exist a subsequence (again denoted by (uλm )) and a limit solution uλ of (23) such that uλm → uλ strongly in H 1 as m → +∞. Proof. Assume that λ = Λ1 and Λ1 < λm , m = 1, 2, . . . . By (10), we see that Λ1 < λ1 (uλm ) ≤ λm , and therefore λ1 (uλm ) → Λ1 as m → ∞. Thus (uλm ) is a minimizing sequence of (12). By Proposition 17, there exist a subsequence (again denoted by (uλm )) and a limit point uλ such that uλm → uλ strongly in Lp (Ω ), where 1 ≤ p < 2∗ , and uλm + uλ weakly in H01 as m → +∞. Using this, we deduce that λ1 (uλ ) ≤ Λ1 . Since inequality λ1 (uλ ) < Λ1 is impossible, we conclude that λ1 (uλ ) = Λ1 and that uλm → uλ strongly in H 1 as m → +∞. Thus we get the proof when λ = Λ1 . Let now λ ∈ Z \ Λ1 . Since λ > Λ1 , then by Lemma 9 there exists a solution uλ of (23) and (34) holds if m > N, where N is a sufficiently large number. Since Qλ (uλ ) = 0 and λm > λ, then Qλm (uλ ) < 0. Note that Qλm (t · uλ ) < 0 if and only if t ∈ (tλ1 (uλ ), tλ2 (uλ )). This implies that Eλm (uλ ) > Eλm (tλ2m (uλ )uλ ) provided m > N (see (36)). Now, repeating the same arguments that have used after formula (36) in the proof of Lemma 18, we obtain the proof of the lemma.
Y. Il’yasov, Y. Egorov / Nonlinear Analysis 72 (2010) 3346–3355
3355
Appendix B Proposition 21. Let t0 (u) and λ(u) be functions on H01 \ {0} defined by (5) and (7), respectively. Suppose that w0 ∈ H01 \ {0} is a critical point of λ(u). Then t0 (w0 )w0 is a weak solution of (1) with λ = Λ0 . Proof. We first compute Dλ(u) =
2(β − α)
(1 − α)
β−1
1−β
·
A(u) 1−α T (u) 1−α B(u)
1−β
β−α
1−α 1−α ˙ (u) − (β + 1) · A(u) T (u) ˙ ( u) · DT · DB B(u)2
α−β
β−α
(1 − β)(1 + α) A(u) 1−α T (u) 1−α ˙ · · DA(u), 1−α B(u) R ˙ (u)(φ) = (∇ u, ∇φ)dx, where we use the notations DT Z Z ˙ (u)(φ) = |u|β−1 uφ dx, ˙ (u)(φ) = |u|α−1 uφ dx DB DA +
for u, φ ∈ H01 .
α,β
Let w0 ∈ H01 \ {0} be a critical point of λ(u). Then Dλ(w0 ) = 0. Hence, and using the identity c0 λ(w0 ) = Λ0 , we deduce that α,β
0 = Dc0 λ(w0 )
1−β (β + 1) α,β 2(β − α) A(w0 ) 1−α ˙ (w0 ) − Λ0 · DB ˙ (w0 ) · DT = · c0 B(w0 ) (1 − α)(β + 1) T (w0 ) α−β 1−α A (w ) ( 1 − β)( 1 + α) 0 α,β ˙ (w0 ) . + c0 · DA (1 − α)(β + 1) T (w0 ) Now, substituting u0 = t0 (w0 )w0 =
A(w0 ) 2(β−α) (1+α)(1−β) T (w0 )
1 1−α
· w0 into this equation and multiplying by t0 (w0 )β , we obtain
the desired equality:
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