Abstract
In this paper, we consider a class of semilinear elliptic equation with critical exponent and -1 growth. By using the critical point theory for nonsmooth functionals, two positive solutions are obtained. Moreover, the symmetry and monotonicity properties of the solutions are proved by the moving plane method. Our results improve the corresponding results in the literature.
1 Introduction and main result
Consider the multiplicity of positive solutions for the following singular elliptic equation with critical growth
where B⊂ℝN (N=3) is the unit ball, μ is a positive constant. Problem (1.1) has a variational structure given by the functional
for
According to Theorem 1 of [1], we know that problem (1.2) has a unique positive solution
if and only if t ∈ (0, 1).
The following singular elliptic problem has been extensively considered
where Ω⊂ℝN(N=3) is a bounded domain with smooth boundary ∂Ω, 0 < p ≤ 2⋆−1 and γ>0. For examples, [2, 3, 4, 5, 6, 7, 8, 9, 10, 11] studied the case of 0<γ<1 for problem (1.4). Particularly, by the variational method and Nehari method, Sun, Wu and Long investigated the multiplicity of positive solutions for the singular elliptic problem for the first time in [8]. And, Yang [11] discussed problem (1.4) with p=2⋆−1 for the first time, and obtained two positive solutions by using the variational method and sub-supersolution method. For the case γ>1, problem (1.4) is considered by [1,12,13].
To our best knowledge, problem (1.4) with γ=1 is only investigated by [4], and two positive solutions are obtained when 1 < p<2⋆−1. A nature question is whether there exist positive solutions for problem (1.4) with γ=1 and p=2⋆−1. In the present note, we give a positive answer by the critical point theory for nonsmooth functionals, and obtain two positive solutions for problem (1.1). Moreover, based on the moving plane technique, we study the symmetry and monotonicity properties of positive solutions to problem (1.1).
In order to study problem (1.1), we define f:B × ℝ → [0, +∞) by
Consider the following auxiliary problem
Problem (1.5) has a variational structure given by the functional
where
Now our main result is as follows.
Theorem 1.1
There exists µ⋆>0 such that for 0<μ<µ⋆, problem (1.1) has two positive radially symmetric solutions. Moreover, the solutions are monotone decreasing about the origin.
2 Proof of Theorem 1.1
We divide two parts to prove Theorem 1.1. First, by using the critical point theory for nonsmooth functionals, we prove that problem (1.1) has at least two positive solutions. Then, we prove that the solution of problem (1.1) is radially symmetric and monotone decreasing about the origin by the moving plane method.
2.1 Existence of Two Positive Solutions
By the definition of F, the following statement is valid.
Lemma 2.1
Assume that {un} is bounded in
Proof
When u < wμ, one has
From the above information, one has
From [1], we have
By the dominated convergence theorem, (2.1) holds. The proof is complete. □
We now recall some concepts adapted from critical point theory for nonsmooth functionals. Let (X, d) be a complete metric space,
The extended real number ∣Df∣(u) is called the weak slope of f at u, see [14,15].
A sequence {un} of X is called Palais-Smale sequence of the functional f, if ∣Df∣(un) → 0 as n → ∞ and f(un) is bounded. We say that u ∈ X is a critical point of f if ∣Df∣(u)=0. Since u → ∣Df∣(u) is lower semicontinuous, any accumulation point of a (PS) sequence is clearly a critical point of f.
Since looking for positive solutions of problem (1.1), we consider the functional J as defined on the closed positive cone P of
P is a complete metric space and J is a continuous functional on P. Then we have the following conclusion.
Lemma 2.2
Assume that u ∈ P and ∣DJ∣(u)<+∞, then for any v ∈ P there holds
Proof
Similar to the proof of Lemma 3.1 in [15]. Let ∣DJ∣(u) < c,
where U is a neighborhood of u. Then ∣∣σ(z, t) − z∣∣=t, combining with (2.4), there exists a pair (z, t) ∈ U × [0, δ] such that
Consequently, we assume that there exist sequences {un}⊂ P and {tn}⊂[0, +∞), such that un → u, tn → 0+, and
that is,
where
Set
and
Notice that
where ξn ∈ (un−snun, un+sn(v − un)), which implies that ξn → u(un → u) as sn → 0+. Note that F(x, t) is increasing in t, then I1,n≥0 for all n. Applying Fatou’s Lemma to I1,n, we obtain
for v ∈ P. For I2,n, by the differential mean value theorem, we have
From the above information, one has
for every v ∈ P. Since ∣DJ∣(u) < c is arbitrary, this leads us to the proof of Lemma 2.2. □
Lemma 2.3
(i)Assume that u is a critical point of J, then u is a weak solution of problem (1.5), that is, for all
Moreover, u=wμ a.e. in B.
(ii)If u is a critical point of J, then u is a positive solution of problem (1.1).
Proof
(i) Let u be a critical point of J. By Lemma 2.2, for
Since ∇ u(x)=0 for a.e. x ∈ B with u(x)=0 and meas{x ∈ B∣ u(x)+sφ(x) < 0, u(x) > 0} → 0 as s → 0, we have
Therefore
as s → 0. We obtain
By the arbitrariness of the sign of φ, we can deduce that (2.7) holds. Thus, u is a weak solution of problem (1.5).
Next, we prove that u=wμ a.e. in B. Choosing φ=(u − wμ)− in (2.7), one has
Notice that
Hence, it follows from (2.8) and (2.9) that
which implies that ∣(u − wμ)−∣=0, that is, u(x)=wμ(x) a.e. in B. Thus, u ∈ P.
(ii) This follows from (i). The proof of Lemma 2.3 is completed. □
Let S be the best Sobolev constant, namely
Lemma 2.4
Assume that 0<μ<1 and
Proof
Let
Taking v=2un ∈ P in (2.11), we have
for some positive constant c. Therefore, {un} is bounded in
It follows from (2.12) and (2.13) that
Since f(x, un)un≤1 for all n, by the dominated convergence theorem, one gets
Set wn=un−u and limn → ∞∣wn∣=l > 0, by Brézis-Lieb’s lemma, Lemma 2.1 and (2.14), we have
On the other hand, for φ ∈ P, taking v=un+φ in (2.11) and letting n → ∞, by Fatou’s lemma, we obtain
Denote
Taking the limit n → ∞ first, then T → ∞, one gets
It follows from (2.16) and (2.17) that
Therefore, similar to the proof of (i) in Lemma 2.3, for all
In particular, one has
It follows from (2.15) and (2.18) that
Consequently, it follows from (2.10) that
where
which contradicts the above inequality. Therefore, l=0, which implies that un → u in
Now, we show that the functional J satisfies the Mountain-pass lemma.
Lemma 2.5
There exist constants r, ρ, μ0 > 0 such that the functional J satisfies the following conclusions:
(i)
(ii) There exists
Proof
(i) By (2.2) and (2.3), one has
We see that there exist constants ρ, r, Λ0 > 0, such that
So we obatin J (tu)<0 for all u+≠0 and t small enough. Therefore, for ∣u∣ small enough, one has
(ii) For every
According to Lemma 2.1, similar to [16], we can easy obtain that d is attained at some u⋆ ∈ Bρ. According to Lemma 2.3 and Lemma 2.5 (i), we obtain the following result.
Theorem 2.6
For 0<μ<μ0, problem (1.1) has a positive solution u⋆ with J(u⋆)=d < 0.
Denote
Then, the infimum (2.10) is attained at U(x). Choosing
Lemma 2.7
There holds
Proof
From [17], we have
where c > 0 is a constant. Since J(u⋆+tφε) → −∞ as t → ∞, J(u⋆)<0, according to Lemma 2.5 (i), we can assume that there exist t1, t2 > 0 such that supt≥0J(u⋆+tφε)=supt∈[t1, t2]J(u⋆+tφε).
Since u⋆ is a positive solution of problem (1.1), we have
Moreover, according to u⋆>wμ, we deduce that
Consequently, one has
for all t≥0. Therefore, from (2.22), (2.23) and the following inequality
we have
For t=0, let
Then, we have
such that g′(t) > 0 for 0 < t<tmax and g′(t) < 0 for t > tmax. Moreover, g(tmax)=maxt≥0g(t). By a standard regularity argument, one has u⋆ ∈ C1(B, ℝ+) and there exists a positive constant C (independent of x) such that u⋆<C. Consequently, it follows from (2.21)-(2.24) that
Let
for all μ ∈ (0, μ1). This leads us to the proof of Lemma 2.7. □
Theorem 2.8
There exists µ⋆>0 such that for 0<μ<µ⋆, problem (1.1) has at least two positive solutions.
Proof
Let 0<μ<µ⋆=min{μ0, μ1}. By Lemma 2.4 and Lemma 2.7, J satisfies the Palais-Smale condition at the level c. By Lemma 2.5, there exists a Palais-Smale sequence {un} such that ∣DJ∣(un) → 0, J(un) → c as n → ∞. Up to a subsequence, un → v⋆ in
2.2 Radially Symmetric and Monotone Decreasing Solution
In this part, we shall prove that every positive solution u of problem (1.1) is radially symmetric and monotone decreasing about the origin. Let
Therefore, it is sufficient to prove that v is radially symmetric and monotone decreasing about the origin. First, we introduce some notations.
Choose any direction to be the x1 direction. Let
be the moving plane, and the set
be the region to the left of the plane. We use the standard notation
for the reflection of x about the plane Tλ.
We denote v(xλ)=vλ(x), then compare the values of v(x) and vλ(x), let
Otherwise, after a direct calculation, we derive that wλ(xλ)=−wλ(x), hence it is said to be anti-symmetric.
We carry out the proof in two steps. To begin with, we show that for λ sufficiently close to −1, we have
This provides a starting point to move the plane Tλ.
Next, we move the plane Tλ along the x1 direction to the right as long as inequality (2.26) holds. The plane Tλ will eventually stop at some limiting position λ0, where
then we are able to claim that
The symmetry and monotonicity of solution v about Tλ follow naturally from the proof. Also, because of the arbitrariness of the x1 direction, we conclude that v must be radially symmetric and monotone decreasing about the origin.
Step 1. We show that for λ sufficiently close to −1, we have
If not, we denote there exists some point x0 ∈ Σλ, such that
By problem (2.25) and the Mean Value Theorem, we can easy derive that
where ξλ(x0), ηλ(x0) are in between vλ(x0) and v(x0), and the last inequality is due to the negative point x0 of wλ(x). That is
we have
Then we obtain that
where
Since v ∈ C∞(B) ∩ L∞(B) (see [4]), v(x0) is bounded. We can see that c(x0)>0 provided μ enough small.
On the other hand, at the minmum point, we have that
which contradicts with (2.29). Therefore (2.27) hods. This completes the preparation for the moving of planes.
Step 2. Inequality (2.27) provides a starting point, from which we move the plane Tλ toward the right as long as (2.27) holds to its limiting position to show that v is monotone decreasing about the origin. More precisely, we define
We will show that
If not, then λ0 < 0, we show that the plane can be moved further right to cause a contradiction with the definition of λ0. More precisely, there exists a small ε>0 such that for all λ ∈ (λ0, λ0+ε) such that
which contradicts the definition of λ0. Hence, (2.30) must be true.
Now we need to prove (2.31) is true. First, by the definition of λ0, we have
We claim that there exists a constant C0 such that
Due to the continuity of
Denote Ωλ=Σλ\Σλ0−δ this is a narrow region, we need to prove that wλ satisfies the conditions of narrow region principle. From the (2.28), we know that, for any x ∈ Ωλ,
where ξλ(x), ηλ(x) are in between vλ(x) and v(x). Consequently, one has
where
then we derive that
Combining with (2.32), we have
This (2.31) is true. Therefore, we must have
Further more, we have
If we move the plane from λ is sufficiently close to 1, then we move the plane Tλ along the x1 direction to the left. By a similar argument, we can derive
then we have
We drive that v(x) is symmetric about the plane T0. Moreover, the arbitrariness of the x1 direction leads to the radial symmetry of v(x) about the origin, so does u. The monotonicity comes directly from the argument. This completes the proof.
Acknowledgements
The authors express their gratitude to the reviewer for careful reading and helpful suggestions which led to an improvement of the original manuscript.
Supported by Science and Technology Foundation of Guizhou Province(KJ[2019]1163); Fundamental Research Funds of China West Normal University(18B015); Innovative Research Team of China West Normal University(CXTD2018-8).
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© 2021 Chun-Yu Lei and Jia-Feng Liao, published by De Gruyter
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