1.
Introduction
In this paper, we consider the Lane-Emden equation
and the equation
The structures of the positive solutions of (1.1) and (1.2) have been studied intensively in the last several years. When n=3, (1.1) arises in the stellar structure in astrophysics. When n=4, (1.1) is relevant to the famous Yang-Mills equations. When n=2, (1.2) is an interesting problem in differential geometry and is known as the "Prescribing Gaussian Curvature" problem.
For Eq (1.1), the Sobolev exponent
plays a central role in the solvability question. In the subcritical case 1<p<ps(n), it was established by Gidas and Spruck in their celebrated work [1] that (1.1) has no positive solution. If p=(n+2)/(n−2), then (1.1) is a special case of the Yamabe problem in conformal geometry. In [2], using the asymptotic symmetry technique, Caffarelli, Gidas and Spruck were able to classify all the positive solutions of (1.1) for n≥3. They showed that any positive solutions of (1.1) can be written in the form
where λ>0 and x0 is some point in Rn. In [3], Chen and Li proved the same result for (1.1) by applying the moving plane method. In n=2, Eq (1.2) is also classified in [3] under the additional assumption that
It is proved in [3] that if u is a solution of (1.2) such that (1.3) holds, then
for some λ>0 and some point x0∈R2.
In the supercritical case p>ps(n), it is more difficult to classify the positive solutions of (1.1). The first result in this direction was given by Zou in [4]. It was proved in [4] that if ps(n)<p<ps(n−1) and if u is a positive solution of (1.1) with algebraic decay rate 2/(p−1) at infinity, then u is radially symmetric about some point x0∈Rn. In [5], Guo generalized Zou's result to p≥ps(n−1) by assuming that
Moreover, it is showed in [5] that (1.4) is a necessary and sufficient condition for a positive solution of (1.1) to be radially symmetric about some point. The analogous result for second order equation (1.2) is considered in [6]. It is proved in [6] that if n≥4 and if u∈C2(Rn) is an entire solution of (1.2), then u is radially symmetric about some point x0∈Rn if and only if
If we focus on radial solutions, then the structure of positive solutions of (1.1) has been completely classified in [7]. They showed that for any a>0, (1.1) admits a unique positive radial solution u=ua(r) with ua(0)=a. Moreover, no two positive radial solutions of (1.1) can intersect each other when p>pJL(n), where pJL(n) is the exponent given by
Another important topic is the classification of stable solutions. In general, a solution of the semilinear equation
with f be a Lipschitz function is called stable if
One of the most interesting questions concerning stable solutions is the following De Giorgi's conjecture.
Conjecture: Let u be a bounded solution of the equation
such that ∂u∂xn>0. Then the level sets of u are hyperplanes, at least if n≤8.
De Giorgi's conjecture was proved in dimension n=2 by Ghoussoub and Gui in [8]. For n=3, this is proved by Ambrosio and Cabré in [9]. Savin proved in [10] that for 4≤n≤8, the above conjecture is true under the additional limit condition that
For n>9, a counterexample is constructed in [11]. The conjecture is still open for dimensions 4≤n≤8 without the additional assumption (1.5).
For Eq (1.1), there are also some results concerning stable solutions. In [12], Liouville type results for solutions with finite Morse index were established. By making a delicate use of the classical Moser iteration method, Farina was able to classify finite Morse index solutions in his seminal paper [13]. It was proved in [13] that if u∈C2(Rn) is a stable solution of (1.1) with 1<p<pJL(n), then u≡0. Moreover, (1.1) admits a smooth positive, bounded, stable and radial solution for n≥11,p>pJL(n). Actually, it was showed in [13] that the radial solutions considered in [7] are stable when n≥11,p>pJL(n). The results in [13] also have a lot of generalizations, we refer to [14,15,16,17,18,19]. As for the classification of the stable solutions of (1.2), it was proved in [20] that for 1≤n≤9, there is no stable solution u∈C2(Rn) of (1.2).
In spite of the above mentioned results, there are some intriguing problems which are still open. In [21], the authors proposed the following conjecture, which is a natural extension of De Giorgi type conjecture:
Conjecture: Let n≥11,pJL(n)<p<pJL(n−1), then all stable solutions to (1.1) must be radially symmetric around some point.
Remark 1.1. When p>pJL(n−1), (1.1) has a positive stable solution which is not radially symmetric. Indeed, let u be a positive radial stable solution of the equation
for p>pJL(n−1) (see [13]), then u can also be viewed as a stable solution of the equation
But it is obvious that this solution is not radially symmetric in Rn.
Our first objective in this paper is to give some partial results toward the above conjecture. To state our results in a more precise way, let us first introduce some new exponents. Let γ be a constant such that
we define
Let
and let
We consider
as a function with respect to the variable p. Let S be the set
and let p∗ be the infimum of S. We define
With the help of these numbers, we can give the statement of our first result.
Theorem 1.2. Let n≥11,pJL(n)<p<pcs(n), where pcs(n) is defined in (1.10). Let u be a positive stable solution of (1.1) such that u is even symmetric with respect to the planes {xi=0},i=1,2,⋯n, then u is radially symmetric with respect to the origin.
In Theorem 1.2, we need the assumption that u is a positive solution of (1.1). But under suitable conditions, we can show that stable solutions of (1.1) do not change sign. For our purpose, we use p1(n)≤p2(n)≤p3(n) to denote the three numbers such that
We define
Theorem 1.3. Let n≥11,pJL(n)<p<psi(n), where psi(n) is defined in (1.12). Let u be a stable solution of Eq (1.1), then u does not change sign. If u is a axially symmetric stable solution of (1.1), then u does not change sign when pJL(n)<p<pJL(n−1).
Remark 1.4. By using MATLAB, we can give some examples for pcs(n) and psi(n).
For Eq (1.2), we have the following result.
Theorem 1.5. Let u be a smooth stable solution of Eq (1.2) for n=10, then u is radially symmetric with respect to some point in Rn.
Remark 1.6. If n≥11, then Eq (1.2) has a smooth stable solution which is not radially symmetric. For more discussions, we refer to [22].
The rest of the paper will be organized as follows. In section 2, we consider rigidity results on the unit sphere for some second order equations. Rigidity results on compact manifolds have been considered by several authors, see for instance [1,23,24,25,26,27]. We point out that in the above papers, the proof of the rigidity results depends heavily on the classical Bochner formula. In section 3, we first use a monotonicity formula to study the qualitative properties of solutions. Then, by combing the rigidity results and the qualitative properties of solutions, we can verify the assumption of Theorem 1.1 in [5] and obtain the symmetry properties of stable solutions. In section 4, we give the prove of Theorem 1.5.
Notation. In some situations, we will write a point x∈Rn as x=(r,θ), where (r,θ) is the spherical coordinates and Sn−1⊂Rn is the unit sphere. In the rest of the paper, c will denote a positive constant which may vary from line to line.
2.
A second order equation on the unit sphere
In this section, we consider the equation
where
and ΔSn−1 is the Laplace-Beltrimi operator on the unit sphere. In the rest of this section, we will always assume that n≥11. The main result in this section is the following.
Theorem 2.1. Let γ be a constant such that
If pJL(n)<p<pcs(n) with pcs(n) be the number defined in (1.9) and if ϕ is a positive solution of (2.1) such that (2.2) holds. Suppose
where Φi,i=1,2,⋯,n are the eigenfunctions of the operator −ΔSn−1 corresponding to the eigenvalue n−1, then ϕ is a constant solution of Eq (2.1). If {ϕ−β1p−1} has at least three connected components, then the same result holds.
In order to prove Theorem 2.1, we need several lemmas.
Lemma 2.2. Let pJL(n)<p<pJL(n−1) and let ϕ∈H1(Sn−1) be a weak solution of (2.1) such that
for every ψ∈H1(Sn−1), then ϕ∈C2(Sn−1).
Proof. We take ψ=|ϕ|γ−12ϕ into (2.2), where γ is a positive constant which will be chosen later. Then
Multiplying the both sides of (2.1) by |ϕ|γ−1ϕ and integrating over Sn−1, we can get that
(2.4) is equivalent to
By combining (2.3) and (2.5) together, we can obtain that
It is easy to check that
when 1≤γ<2p+2√p(p−1)−1. By applying Hölder's inequality, we have
It follows from (2.7) that ∫Sn−1|ϕ|p+γdθ is finite. By formula (5.10) in [13], we know that there exists a constant γ such that (p+γ)/(p−1)>(n−1)/2. Therefore, |ϕ|p−1∈Lq(Sn−1) for some q>(n−1)/2. The standard regularity results in [28] imply that ϕ∈C2(Sn−1).
Lemma 2.3. Let pJL(n)<p<pJL(n−1) and let ϕ∈C2(Sn−1) be a positive solution of (2.1) such that (2.2) holds, then
where γ is the constant used in the proof of Lemma 2.2 and α(p,γ,n) is the constant defined in (1.6).
Proof. Let θ0∈Sn−1 be a point such that ϕ(θ0)=‖ϕ‖L∞(Sn−1)=η. By taking suitable orthogonal transformation, we may assume that θ0 is the south pole. Let us introduce the following coordinates on Sn−1,
where ξ∈[0,π),ξ1∈[0,2π),ξk∈[0,π) for k=2,3,⋯n−2. The coordinate of the point θ0 is given by (0,0,⋯,0). By (2.1), we know that ϕ satisfies the equation
where Sn−2 is the unit sphere in Rn−1 and ΔSn−2 is the Laplace -Beltrami operator on Sn−2. We define
where ωn−2 is the area of Sn−2. It follows from (2.9) that ˆϕ satisfies
By the Jensen's inequality, we can get that
Let ξ1 be the first point such that ˆϕ(ξ1)=η2. It follows from (2.11) that ˆϕ is strictly decreasing in (0,ξ1). We will focus on the case ξ1<π2 since the case ξ1≥π2 can be dealt with similarly. Let γ be the constant used in the proof of Lemma 2.2. By (2.10), we can obtain that
This implies ξ1≥η1−p2. By the above analysis, we can get that
By Lemma 2.2, we know that
We get from (2.12) and (2.13) that
Hence (2.8) holds.
Lemma 2.4. Let ϕ be a positive solution of (2.1) such that
where Φi,i=1,2,⋯,n are the eigenfunctions of the operator −ΔSn−1 corresponding to the eigenvalue n−1, then
Proof. We define
where ¯ϕ is given by
Then ˜ϕ satisfies the equation
Multiplying the both sides of (2.17) by ˜ϕ and using integration by part, we can get that
By (2.15) and the definition of ˜ϕ, we know that
By (2.18) and the Poincaré's inequality, we have
If ˜ϕ≠0, then
It follows that
Hence (2.16) holds.
Lemma 2.5. If ϕ is a positive solution of (2.1) such that {ϕ−β1p−1≠0} has at least three connected components, then
Proof. The equation (2.1) can be written as
Since {ϕ−β1p−1≠0} has at least three connected components, then there is a connected component Ω0 of {ϕ−β1p−1≠0} such that the area of Ω0 is less than 13ωn−1. let 1Ω0 be the function defined by
Multiplying the both sides of (2.22) by (ϕ−β1p−1)1Ω0 and using integration by part, we can get that
Let λ1(Ω0) be the first eigenvalue of the eigenvalue problem
By (2.23) and the mean value theorem, we can get that
It follows from (2.24) that
Since the area of Ω0 is less than 13ωn−1, where ωn−1 is the area of the unit sphere in Rn. By using Schwartz symmetrization, we can get that
It follows from (2.25) and (2.26) that
Hence (2.21) holds.
Remark 2.6. We notice that
then
Lemma 2.7. Let ¯p be a constant such that pJL(n)<¯p<pJL(n−1). There exists a positive constant c such that if ϕ∈C2(Sn−1) is a nonconstant solution of (2.1) for pJL(n)<p<¯p, then
Proof. Suppose (2.28) does not hold, then there exists a sequence {ϕm} such that ϕm satisfies
and
Since −ϕm is also a solution of (2.29), without loss of generality, we can assume that
It follows from the proof of Lemma 2.2 that ∫Sn−1ϕ2mdθ remains bounded. So (2.30) implies
By (2.8) and (2.32), we can get that there exist two constants p0 and c0 such that
Moreover, c0 is a constant solution of (2.1) for p=p0. Therefore,
We get from (2.31) that
Therefore,
It follows from (2.34) that c0 is not zero. Let
then limm→+∞ψm=0 and ψm satisfies the equation
It is easy to verify that
for some positive constant c independent of m. We define
then vm satisfies
Since
and
By standard elliptic estimates, we know that there exists a nontrivial function v∞ such that vm→v∞ in H1(Sn−1). Moreover, v∞ satisfies the equation
Then we deduce that v∞ is a nontrivial eigenfunction of −ΔSn−1 corresponding to the eigenvalue (p0−1)β0. On the other hand, it is easy to see that
and pJL(n)>(n+1)/(n−3) when n≥11. Therefore, (p0−1)β0 can not be an eigenvalue of −ΔSn−1. By combining these two facts together, we obtain a contradiction. Next, we can give some estimates about the constant c in Lemma 2.7.
Lemma 2.8. Let pJL(n)<p<pJL(n−1) and let ϕ be a positive solution of (2.1) such that
then the constant c in Lemma 2.7 can be estimated by cs(p,n), where cs(p,n) is defined by (1.8).
Proof. Multiplying the both sides of (2.1) by ϕ and integrating over Sn−1, we can get that
We take ψ=ϕ into (2.2), then
By (2.38) and (2.39), we can get that
By the Poincaré's inequality, we know that
It follows from (2.40) and (2.41) that
In order to estimate the constant c in Lemma 2.7, we need to give a lower bound for ∫Sn−1|∇Sn−1ϕ|2dθ. Since we have assumed that
then there exists a point θ0 such that ϕ(θ0)=β(p,n). By taking suitable orthogonal transformation, we may assume that θ0 is the south pole. We use the coordinates used in the proof of Lemma 2.2. By (2.1), we know that ϕ satisfies Eq (2.9). We define
then ˆϕ satisfies (2.10) and (2.11). Let ξ1 be the first point such that
We know from (2.11) that
We will assume that ξ1<π2 since the case ξ1<π2 can be dealt with similarly. By (2.10), we can get that
We deduce that
Let
By (2.1) and the Jensen's inequality, we can get that ¯ϕ≤β1p−1. Therefore,
It follows from the Poincaré's inequality that
Therefore,
By the above analysis, we know that (1.8) holds.
Lemma 2.9. Let ϕ be a positive solution of (2.1) such that (2.2) holds. If
then ϕ is a constant when pJL(n)<p<pcs(n), where pcs(n) is defined by (1.9).
Proof. By (2.38) and (2.39), we have
Let ϕ be a nonconstant solution of (2.1) satisfying (2.2), we know from Lemma 2.7 that ϕ satisfies (2.28). By combining (2.28) and (2.45) together, we can get that
It follows from (2.46) that
when pJL(n)<p<pcs(n). Since we have assumed that ϕ is a nonconstant solution of (2.1), this is a contradiction.
Proof of Theorem 2.1:. This result follows from Lemma 2.4, Lemma 2.5 and Lemma 2.9.
Remark 2.10. In this section, we always assume that ϕ is positive. But we can prove that if ϕ is a solution of (2.1) depends only on the variable ξ, then ϕ does not change sign. The proof of this fact will be given in the appendix.
Remark 2.11. It is proved in [29] that if n≥4 and (n+1)/(n−3)<p<pJL(n−1), then (2.1) has a nonconstant positive solution.
Remark 2.12. By Lemma 1 in [30], we have the following Hardy type inequality,
The equation (2.1) has a singular solution which is given by
Suppose ϕ∗ satisfies (2.2), then
If p=pJL(n−1), then
Let us define
then
If p>(n−1)/(n−5), then g′(p)<0. Therefore, the singular solution ϕ∗ satisfies (2.2) if p≥pJL(n−1).
3.
Qualitative properties of stable solutions
In this section, we give the proof of Theorem 1.2 and Theorem 1.3.
Lemma 3.1. Let n≥11,pJL(n)<p<psi(n), where psi(n) is defined in (1.12). If ϕ is a nontrivial solution of (2.1) such that (2.2) holds, then ϕ does not change sign.
Proof. We assume that ϕ change sign. Without loss of generality, we can assume that there exists a connected component Ω1 of {ϕ>0} such that λ1(Ω1)≥n−1, where λ1(Ω1) is the first eigenvalue of the eigenvalue problem
Multiplying the both sides of (2.1) by ϕ and integrating over Ω1, we can get that
We take ψ=u1Ω1 into (2.2), where 1Ω1 is the function defined by
Then
By (3.1) and (3.2), we know that
It follows that if pJL(n)<p<psi(n), then ϕ vanishes identically on Ω1. Since we have assumed that ϕ>0 on Ω1, this is a contradiction.
Proof of Theorem 1.3: We consider the transform
Since u satisfies (1.1), then w is a bounded solution of the equation
We set
By (3.4), we get that
By the estimates in [19], we can get that ∂tw,∂ttw,|∇Sn−1| are uniformly bounded. Integrating (3.6) from −s to s, we find
for some constant c independent of s. Let s tend to +∞ in (3.7), then
Similar to the proof of Theorem 1.4 in [1], we can obtain that
For any sequence {tk} such that tk→∞ as k→∞, we consider the translation of w defined by wk(t,θ)=w(t+tk,θ). Then there exist a subsequence {wlk(t,θ)} and a function w∞(t,θ) such that wlk(t,θ)→w∞(t,θ) in C2([−1,1]×Sn−1). By (3.8) and the dominated convergence theorem, we know that there exists a function ϕ(θ) such that w∞(t,θ)=ϕ(θ). Moreover, ϕ is a solution of (2.1) such that (2.2) holds. If ϕ=0, then limt→+∞E(w)(t)=0. But we also have limt→−∞E(w)(t)=0 since u is regular at the origin. It follows easily that w≡0. Since we have assumed that u is a nontrivial solution, this is a contradiction. Therefore ϕ is not zero. If ϕ≠0, we know from Lemma 3.1 that ϕ does not change sign. Suppose there exist two sequences {tk} and {˜tk} such that
and
then {u≠0} has a bounded connected component. Without loss of generality, we can assume there exists a bounded connected component Ω− such that u<0 on Ω−. Then u satisfies the equation
Since u is a stable solution of (1.1), then L=Δ+p|u|p−1 satisfies the refined maximum principle (see [31]). Since
we get from the refined maximum principle that u≥0 on Ω−. In view of the definition of Ω−, we get a contradiction. By the above arguments, we know that there exits a positive constant R0 such that u doesn't change sign on Rn∖BR0. By applying the refined maximum principle again, we know that u does not change sign.
If u is axially symmetric, then the proof is essentially the same as the arguments above. The only difference is that we need to use remark 2.10 rather than Lemma 3.1 to show that there exits a positive constant R0 such that u doesn't change sign on Rn∖BR0.
Proof of Theorem 1.2. Let n≥11,pJL(n)<p<pcs(n) and let u be a positive stable solution of (1.1). Let us consider the transform
Since u satisfies (1.1), then w is a bounded solution of Eq (3.4). Let {tk} be a sequence such that tk→∞ as k→∞. Similar to the arguments used in the proof of Theorem 1.3, we know that there exists a function ϕ∈H1(Sn−1) such that
Moreover, ϕ is a solution of (2.1) such that (2.2) holds. By Lemma 2.2, we have ϕ∈C2(Sn−1). If u is even symmetric with respect to the {xi=0},i=1,2,⋯n, then
Since pJL(n)<p<pcs(n), we know from Theorem 2.1 that ϕ is a constant function. In particular, we have
Since the sequence {tk} can be arbitrary, we conclude that
Since p>pJL(n), then p>n/(n−4). By Theorem 4.4 in [5], we can get that
where
and
Moreover, for any integer τ≥0, we have ν(r,θ) satisfies
uniformly in Cτ(Sn−1), where V equals either zero or a first eigenfunctions of the operator −ΔSn−1. Since we have obtained the asymptotic expansion (3.11) which is good enough to apply the moving plane method, then the rest of the proof is essentially the same as the proof of Theorem 1.1 in [4].
4.
The proof of Theorem 1.5
In this section, we give the proof of Theorem 1.5, the proof is mainly based on the following observation.
Proposition 4.1. Let n=10 and let u be a smooth stable solution of Eq (1.2), then
In order to prove Proposition 4.1, we first recall a monotonicity formula.
Lemma 4.2. If u is a solution of the equation (1.2), then
where
Moreover, if u is a smooth stable solution of (1.1), then
Proof. The proof of (4.2) follows from a scaling argument which is similar to the proof Proposition 5.1 in [32]. The proof of (4.3) follows easily from the capacity estimates in [33].
With the help of Lemma 4.2, we can give the proof of Proposition 4.1.
proof of Proposition 4.1. The proof of Proposition 4.1 will consist of the following four steps.
Step 1: Let {λk} be a sequence such that limk→+∞λk=+∞. For any λk, we define uλk(x)=u(λkx)+2ln(λk). It is easy to check that uλk(x) is also a stable solution of (1.1). By the capacity estimates (see for instance [33]), we know that uλk→u∞ for some function u∞∈H1loc(Rn). Moreover, u∞ is a stable solution of (1.1).
Step 2: For any 0<R1<R2<+∞, by Lemma 4.2,
By the scaling invariance of E, we have
We use Lemma 4.2 again, then
Therefore,
It follows that there exists a function ϕ∈H1(Sn−1) such that u∞=ϕ−2ln(r). Moreover, ϕ satisfies the equation
Step 3: For every δ>0, we choose a function ηδ∈C∞0((δ2,2δ)) such that ηδ≡1 in (δ,1δ), and r|η′δ(r)|≤4. For every ψ∈H1(Sn−1), we define ψδ=r−n−22ψ(θ)ηδ(r). For every ψ∈H1(Sn−1), we define ψδ=r−n−22ψ(θ)ηδ(r). Since u∞ is stable, we have
Therefore, ϕ satisfies
for every ψ∈H1(Sn−1).
Step 4: We take ψ=eϕ2 into (4.9), then
Multiplying the both sides of (4.8) by eϕ and using integration by part, we have
If n=10, then (n−2)2/4=2(n−2). By (4.10) and (4.11), we can get that
It follows from (4.12) that ϕ=ln(16) is a constant. Since {λk} can be arbitrary, we can obtain that proposition 4.1 holds.
Proof of Theorem 1.5. It follows from proposition 4.1 and Theorem 1.3 in [6].
Appendix 1: A Liouville type result
In this appendix, we prove the claim in remark 2.10. The proof is based on the the following result.
Proposition 4.3. Let p≥n+1n−3 and (p−1)μ≥n−1. If ϕ is a solution of the equation
where Br⊂Rn−1 is a ball and 0<r<1, then ϕ=0.
Proof. Multiplying the both sides of (4.13) by (21+|x|2)n−1ϕ and using integration by part, we can get that
Multiplying the both sides of (4.13) by (21+|x|2)n−1(x⋅∇ϕ) and using integration by part, we can get that
where
It follows that
Multiplying the both sides of (4.13) by x⋅∇(21+|x|2)n−1ϕ and using integration by part, we can get that
By some computations, we can get that
By (4.16) and (4.17), we have
We combine (4.14), (4.15) and (4.18) in the following way:
then
If p≥n+1n−3 and (p−1)μ≥(n−1), then the left hand side of the last identity will become non-positive, therefore, Eq (4.13) has only trivial solution.
Corollary 4.4. If p≥n+1n−3 and if ϕ is a nontrivial solution of Eq (2.1) depends only on the variable ξ, here we use the coordinates in the proof of Lemma 2.3, then ϕ does not change sign.
Proof. If ϕ change sign, then there exists 0<r<1 such that (4.13) has a nontrivial solution, this is a contradiction.
Acknowledgments
The research of J. Wei is partially supported by NSERC of Canada.
Conflict of interest
We declare that we have no financial and personal relationships with other people or organizations that can inappropriately influence our work.