NONRADIAL LEAST ENERGY SOLUTIONS OF THE p -LAPLACE ELLIPTIC EQUATIONS

. We study the p -Laplace elliptic equations in the unit ball under the Dirichlet boundary condition. We call u a least energy solution if it is a minimizer of the Lagrangian functional on the Nehari manifold. A least energy solution becomes a positive solution. Assume that the nonlinear term is radial and it vanishes in | x | < a and it is positive in a < | x | < 1. We prove that if a is close enough to 1, then no least energy solution is radial. Therefore there exist both a positive radial solution and a positive nonradial solution.

1. Introduction. We study nonradial positive solutions for the p-Laplace elliptic equation − ∆ p u = f (x, u), u > 0 in B, u = 0 on ∂B, (1.1) where ∆ p u := div(|∇u| p−2 ∇u) is the p-Laplacian with p ≥ 2, B is the unit ball in R N and f (x, u) is a continuous function on B × [0, ∞) which is radially symmetric with respect to x and has a subcritical growth with respect to u. In order to use the second derivative of the Lagrangian functional, we suppose that p ≥ 2.
In the present paper, we shall show that no least energy solution is radial. To define a least energy solution, we first define the Lagrangian functional I(u) for (1. Denote the Fréchet derivative of I(u) by I (u), which is computed as Here W 1,p 0 (B) is the usual Sobolev space. We put We define the Nehari manifold N and the least energy We define a solution of (1.1) by u if u ∈ W 1,p 0 (B) and it satisfies (1.1) in the distribution sense. Therefore a function u in W 1,p 0 (B) is a solution of (1.1) if and only if I (u) = 0. Since any solution u fulfills J(u) = I (u)u = 0, it belongs to the Nehari manifold, i.e., any solution u of (1.1) satisfies (1.4) We call u a least energy solution if u ∈ N and I(u) = I 0 . Such a minimizer exists and becomes a positive solution of (1.1) after replacing u by −u if necessary. This claim will be proved in Theorem 2.3.
On the other hand, we define N r := {u ∈ N : u is radial}, I r := inf{I(u) : u ∈ N r }. (1.5) We call u a radial least energy solution if u ∈ N r and I(u) = I r . To avoid confusion, a usual least energy solution is called a global least energy solution.
Theorem 1.1 (Moore and Nehari [16, pp.32-33]). For a suitable a ∈ (0, 1) (a is close to 1), (1.6) has at least three positive solutions: the first one is even, the second one u(x) is non-even and the third one is the reflection u(−x).
Put N = 1 in (1.7). Then the weight function |x| λ with λ large enough is similar to h a (x) with a sufficiently close to 1, where h a (x) is defined after (1.6). Hence we expect that a noneven solution in Theorem 1.1 may be a least energy solution. Indeed, we have proved in [12,13] that no least energy solution of (1.6) is even when a is sufficiently close to 1. There are many contributions which have studied the Hénon equation ( [1,2,3,4,5,6,7,10,11,13,18,20]). The purpose of the present paper is to extend Theorem 1.1 to more general equation (1.1).
In our paper [14], we studied the equation where B is the unit ball in R N , 2 ≤ p < q < p * and f (x) is radial. Here p * is the critical exponent defined by We assume that with some a ∈ (0, 1). Here f + (x) := max{f (x), 0}. Since the nonlinear term f (x)u q−1 in (1.8) has homogeneity with degree q − 1, we can define the Rayleigh quotient Instead of the Lagrangian functional I(u), we use the Rayleigh quotient R(u) and define a least energy solution by a minimizer of R in N . Then we proved the next result.
is a radial function which satisfies (1.10) with a ∈ (0, 1) close enough to 1, then no global least energy solution of (1.8) is radial.
To prove the theorem above, we used the homogeneity of the nonlinear term in the paper [14], however (1.1) has no homogeneity. This makes our problem complicated and we cannot apply the method used in [14] to problem (1.1). Another purpose of the present paper is to extend the result in [14] to more general nonlinear term f (x, u).
This paper is organized into five sections. In Section 2, we state the main results and give an example of f (x, u). In Section 3, we show the existence of a global (or radial) least energy solution and prove that it becomes a positive solution of (1.1). In Section 4, we give an estimate of a positive radial solution. In Section 5, we show that no global least energy solution is radial.
2. Main results. In this section, we state main results.
Assumption 2.1. We assume that p ≥ 2 and f (x, u) is a continuous function on B × [0, ∞) which satisfies the following conditions.
(i) f (x, u) is a radial function of x.
Let I 0 and I r be defined by (1.3) and (1.5), respectively. We state the first main result, which ensures the existence of positive solutions for (1.1). Theorem 2.3. Let Assumption 2.1 hold. Then there exist a global least energy solution u and a radial least energy solution v, that is, u ∈ N and I(u) = I 0 , v ∈ N r and I(v) = I r . Moreover, u is a positive solution of (1.1) and v is a positive radial solution of (1.1) after replacing u by −u and v by −v if necessary.
In the theorem above, v is radial. However, we do not know whether u is radial or not. There is a possibility that u = v and I 0 = I r (see Remark 2.5). To guarantee that u is not radial, we need the assumption that a is close to 1, where a is the constant given in (iii) of Assumption 2.1. Recall that in Assumption 2.1, f (x, u) ≡ 0 in |x| ≤ a and f (x, u) > 0 in a < |x| < 1. The closeness of a to 1 makes the symmetry breaking. Indeed, we have the next theorem.
Theorem 2.4. There exists a constant a 0 ∈ (0, 1) such that if f (x, u) satisfies Assumption 2.1 with a ∈ (a 0 , 1), then I 0 < I r and hence no global least energy solution is radial. Therefore (1.1) has both a positive radial solution and a positive nonradial solution.
We supposed in Assumption 2.1 that f (x, u) is continuous in x for the sake of simplicity. Observing the proof of Theorem 2.4, we see that Theorem 2.4 is applicable to f (x, u) = h(x)u q with h(x) = 0 for |x| ≤ a and h(x) = 1 for a < |x| ≤ 1. Therefore our theorem is a generalization of the result by Moore and Nehari [16].
Remark 2.5. In Theorem 2.4, we suppose that a is sufficiently close to 1. This assumption is essential for the symmetry breaking of solutions. If a is close to 0, then Theorem 2.4 is not necessary valid. Indeed, put N = 1, p = 2 and f (x, u) = h a (x)u q and consider (1.6). Then we proved in our paper [14, Theorem 1.2] that if a > 0 is small enough, then (1.6) has a unique positive solution and it is even. Since N = 1, a radial solution means an even solution. In this case, it happens that u = v and I 0 = I r , where u and v are given in Theorem 2.3.

Least energy solution.
In this section, we shall show that a least energy solution exists and becomes a positive solution of (1.1). Hereafter we always suppose Assumption 2.1 and define an annulus A by where a is the constant given by (iii) of Assumption 2.1. We extend f (x, u) as an odd function by putting We first study the Nehari manifold N . Let F (x, u) be given by (1.2). We note that F (x, u) = 0 for |x| ≤ a and u ∈ R because f (x, u) = 0 for |x| ≤ a and u ∈ R. Let A be defined by (3.1) and let u ∈ W 1,p 0 (B). By the notation, u ≡ 0 in A, we mean that the set of x ∈ A satisfying u(x) = 0 has positive Lebesgue measure.
then there exists a unique λ > 0 such that λu ∈ N . Indeed, we have the next lemma.
Proof. Since f (x, u) = 0 for |x| ≤ a, the definition of J implies that for t > 0, Then the equation above is rewritten as We define If these assertions would be proved, then t −p J(tu) has a unique zero and the lemma follows. By (2.1), we have with some C > 0. Accordingly, as t → +0. Thus the first assertion in (3.2) holds. We shall show the second assertion in (3.2). For δ > 0, we define Since u ≡ 0 in A, we choose δ > 0 so small that |D| > 0. Here |D| denotes the Lebesgue measure of D. For ε > 0, we put We can choose ε > 0 so small that |E| > 0. Indeed, if |E| = 0 for any ε > 0, then the set of x ∈ D satisfying f (x, δ) > 0 has Lebesgue measure zero. This contradicts the fact that D has a positive Lebesgue measure. Therefore |E| > 0 for small ε > 0. We fix such an ε > 0. By (2.1), we have Hence we obtain x ∈ E and t > 1. Then L(t) is estimated as The proof is complete.
Lemma 3.1 ensures the next result.
By Lemma 3.1, we can define λ(u) as below.
We shall show that the global least energy I 0 is positive. Lemma 3.6. I 0 := inf u∈N I(u) > 0.

Proof. Inequality (3.3) is valid for all
Since p < q < r, for any ε > 0 there exists a C ε > 0 such that In what follows, we denote the L p (B) norm of u by u p . Combining (3.4), (3.5) and using the Sobolev embedding, we have ∇u p p ≤ ε u p p + C ε u r r ≤ εC 0 ∇u p p + C ε C 0 ∇u r p , where C 0 is an embedding constant independent of ε. Choose ε > 0 so small that εC 0 < 1/2, which shows that Hence there exists a constant c 0 > 0 such that u). Integrating this inequality with respect to u, we get f (x, u)u ≥ qF (x, u) for x ∈ A and u > 0. Since the both sides are even in u, this inequality is valid for all u ∈ R and x ∈ B. Using this inequality with (3.5), we estimate I(u) for u ∈ N as Combining the inequality above and (3.6), we get inf u∈N I(u) > 0. The proof is complete.
Let I(u) and J(u) be defined in Section 1. Furthermore, we define Since I(u) = K(u) for u ∈ N because of (3.5), we have Lemma 3.7. If J(u) < 0, then K(u) > I 0 .
In the next lemma, we shall show the existence of a global least energy solution. Our method remains valid for the existence of a radial least energy solution. Proof. Put I 0 := inf N I(u). Choose a minimizing sequence u n , i.e., u n ∈ N and I(u n ) → I 0 as n → ∞. Substituting u n in (3.7), we find that u n is bounded in W 1,p 0 (B). Hence it has a subsequence (denoted by u n again) converging to a weak limit u ∈ W 1,p 0 (B). By the compact embedding with (2.2), we get as n → ∞. Since u n ∈ N , it holds that K(u n ) = I(u n ) → I 0 as n → ∞. Then (3.8) shows that We here note that I 0 > 0 by Lemma 3.6. Since u n ∈ N , it holds that Since u n weakly converges to u in W 1,p 0 (B), we use the second convergence in (3.8) to get Thus J(u) ≤ 0. If J(u) < 0, then (3.9) contradicts Lemma 3.7. Hence J(u) = 0. If u = 0, then K(u) = 0, which contradicts (3.9). Therefore u = 0 and so u belongs to N . Since u ∈ N , I(u) is equal to K(u). By (3.9), I(u) = K(u) = I 0 and u ∈ N . The proof is complete.
To prove Theorem 2.3, we need the Lagrange multiplier rule. We shall show that a global least energy solution becomes a positive solution of (1.1). The method below is still valid for a radial least energy solution with the help of the principle of the symmetric criticality by Palais [17].
Proof of Theorem 2.3. By (2.1), we have (3.10) It is easy to see that for u, v ∈ W 1,p 0 (B). In Lemma 3.8, we have already proved the existence of a global least energy solution. Let v be a global least energy solution, i.e., v ∈ N and I(v) = I 0 . Then it satisfies (3.5), that is, Combining two equations above and using (3.10), we have Since v is a minimizer of I under the restriction J(v) = 0 and J (v) is surjective, we use Lemma 3.9 to obtain a Lagrange multiplier λ ∈ R which satisfies (3.12) or equivalently (3.11), λ is equal to 0. By (3.12) with λ = 0, v is a critical point of I. Since I(v) = I 0 > 0, v is a nontrivial solution of (1.1). Observe that |v| ∈ N and I(v) = I(|v|). Accordingly, |v| is also a minimizer of I on N . Hence w := |v| is a solution of (1.1), which satisfies We here need the strong maximum principle for the p-Laplacian. For its statement and the proof, we refer the readers to [19, p.34 It is easy to prove the next lemma, but it plays an important role in the proof of Theorem 2.4. By (4.2), u(0) = u(a) = u ∞ . Therefore Combining the two inequalities above, we obtain (4.3). The proof is complete.

5.
Positive nonradial solution. In this section, we shall prove Theorem 2.4. Let B(x, r) denote a ball centered at x with radius r > 0. We observe the inclusion, Since the left hand side is compact, we have a finite covering, f (x, u)uφ 2 dx, Proof. Let u be a positive radial solution. Put u(x) = 0 outside B. Let x i with 1 ≤ i ≤ M satisfy (5.1). Since u is radial and |x i | = 1, it holds that for any i when |y| = 1. Since f (x, u)u ≥ 0, it follows from (5.1) with the identity above that for any y satisfying |y| = 1. Since |z| = 1 and φ(x) = 1 in B(z, 1/2), the inequality above shows that Since f (x, u) = 0 for |x| ≤ a, the inequality above is reduced to provided that a ∈ (3/4, 1).
Hereafter ω N denotes the surface area of the unit sphere in R N . Then any radial f (x, u)udx.
Using this inequality with (5.5), we obtain (5.4). The proof is complete.
We employ a method developed in our paper [15]. Let φ be the function given by (5.2). For a radial least energy solution u, we define g(t, s) := I((1 + t)(1 + sφ)u) for t, s ∈ R. (5.7) We shall show that with some t, s ∈ R. If this claim would be proved, then This inequality shows that no global least energy solution is radial, which is the desired conclusion. To prove our claim, we investigate g(t, s) near (t, s) = (0, 0). By Assumption 2.1 and p ≥ 2, I(u) is twice continuously differentiable in the sense of the Fréchet derivative. The second derivative I (u) is a bilinear form, for u, v, w ∈ W 1,p 0 (B). Let u be a radial least energy solution. Define g(t, s) by (5.7). We compute the second derivative of g as below. Substituting t = s = 0 and using (5.8) and noting I (u) = 0, we have g t (0, 0) = 0, g s (0, 0) = 0, We shall show that g ts (0, 0) = 0. Since u is radial, it is even. However φ is odd. Accordingly, the integral of the second term on the right hand side of (5.11) vanishes, i.e., B f u (x, u)u 2 φdx = 0.
The first term on the right hand side of (5.11) is rewritten as because ∆ p u is radial and φu is odd. Therefore g ts (0, 0) = 0. Then the next lemma follows.
Let u be a radial least energy solution and define g(t, s) by (5.7). Let a be the constant given in (iii) of Assumption 2.1. We shall show that the closeness of a to 1 ensures that g tt (0, 0) and g ss (0, 0) are negative. Lemma 5.3. There exists a constant a 0 ∈ (0, 1) depending only on p, q and N such that if Assumption 2.1 holds with a ∈ (a 0 , 1), then g tt (0, 0) and g ss (0, 0) are negative.
Since C depends only on N and p, the constant a 0 is determined by p, q and N . For a ∈ (a 0 , 1), it holds that γ > 0 and g ss (0, 0) < 0. The proof is complete.
We conclude this paper by proving Theorem 2.4.
Proof of Theorem 2.4. Let a 0 be the constant given by Lemma 5.3. Suppose that Assumption 2.1 holds with a ∈ (a 0 , 1). Let u be a radial least energy solution.
The proof is complete.