RADIAL SOLUTION OF ASYMPTOTICALLY LINEAR ELLIPTIC EQUATION WITH MIXED BOUNDARY VALUE IN ANNULAR DOMAIN∗

In this paper, we study nonlinear elliptic equation with mixed boundary value condition in annular domain. It is assumed that the nonlinearity is asymptotically linear and depends on the derivative term. Some results on the existence of solution are established by nonlinear analysis methods.


Introduction
In this paper, we study the following nonlinear elliptic equation with gradient term in annular domain where B i := {x ∈ R n : |x| < i}, i = 1, 2, n > 2, f : R × R × R → R. f = f (r, s, ξ) is continuous and C 1 with respect to (s, ξ). ∂/∂ν denotes the outward normal derivative.
Because of the wide interest in mathematics and applied mathematics, the existence of solution of elliptic equation in annular domains has been investigated by many authors, see [1, 2, 6, 8-10, 13-19, 24-26, 30] and the references cited therein. When f (r, s, ξ) = s p , the equation is known as Lane-Emden equation. In [18], Ni and Nussbaum established numerous results concerning the uniqueness and nonuniqueness for positive radial solution, when the domain is a ball or an annulus. When the domain Ω is star-shaped and f (r, s, ξ) = s n+2 n−2 , the well-known Pohozaev identity implies that the problem has no solution (see [20]). Brezis and Nirenberg [4] proved that the perturbation of lower-term can reverse this situation. If Ω is an annulus, Pohozaev theorem does not work any more since the annulus is not a starshaped domain. Therefore, it is possible that the constraints for the growth of f can be removed. Provided f (r, s, ξ) = −s + s 2N +1 , Coffman [5] pointed out that there are many rotationally nonequivalent positive solutions and the number of these solutions is unbounded as r → +∞. The case f (r, s, ξ) = g(r)h(s) was considered by Lin [15] and the case f (r, s, ξ) = λk(r)g(s) was studied by Wang [24]. Uniqueness of solutions was also studied when f (r, s, ξ) = f (s) (see [17]) or f (r, s, ξ) = s p + s q (see [30]). Recently, Dong and Wei [9] studied the existence of radial solution for elliptic equation with Dirichlet boundary value condition.
However, all of the above papers are devoted to the superlinear problems. In this paper, we focus on the asymptotically linear equation with mixed boundary value condition. There are some known papers related to asymptotically linear problems, such as [12,21] for second order elliptic equation, [29] for non-local elliptic equation, [27] for fourth-order elliptic equation and so on. For Sturm-Liouville equation involving p-Laplacian with mixed boundary condition, we refer to [22].
To state our main results, we introduce the following assumptions: is uniformly bounded and f (r, 0, ξ) = 0; (F1) There exists k ∈ Z + , and two continuous functions α(r), α(r), such that either of the following holds uniformly for (r, s, ξ) ∈ [1, 2] × R × R: The first main result of this paper is given as follows. Remark 1.1. We give a concrete example to illustrate the above result. Let n = 3. Consider the following boundary value problem: where 0 < |ε| < 1 2 , k ∈ Z + , h is C 1 continuous and there exist constants m 1 , m 2 > 0 such that 0 < m 1 < h(ζ) < m 2 for any ζ ∈ R. It is easy to know (F0) is satisfied. Besides, ε < 0 and ε > 0 correspond to the case i and ii of (F1), respectively. Theorem 1.1 implies that the above problem has at least one radial solution.
The approaches of the present paper are based on some methods of nonlinear analysis. We derive an equivalent ordinary differential equation for (M), and then, deal with the corresponding ordinary differential equation. Some fixed point theorems are used and some iterative methods are also introduced to overcome the difficulty caused by the gradient term. Schauder' s fixed point theorem is essential to the proof of Theorem 1.1. Meanwhile, Mountain pass theorem and iterative technique are applied to prove Theorem 1.2.
This paper is organized as follows. In Section 2, we derive an equivalent ordinary differential equation and introduce some function spaces. Section 3 is devoted to proving Theorem 1.1. We first study the special problem provided that the nonlinearity does not contain the gradient term. Then, the general case involving gradient term is considered. To prove the second main theorem, we apply Mountain pass theorem to establish existence of solutions for the non-gradient problem in Section 4. Finally, the proof of Theorem 1.2 is given in Section 5.

Preliminaries and Equivalent ODE
For x = (x 1 , · · · , x n ) ∈ R n , denote r = |x|. Then Hence, the elliptic problem (M) is equivalent to the following second order differential equation with mixed boundary condition Let t = t(r), which will be determined later. Then (2.1) implies To make the gradient term in the left of (2.2) vanish, we choose t(r) such that and Therefore, It follows from (2.6) that ). Since , g satisfies the following conditions: (G1) There exists k ∈ Z + and two continuous functions β(t) and β(t), such that either of the following holds uniformly for (t, s, η) ∈ [0, 1] × R × R: ii.
(G5) g satisfies local Lipschitz condition: there exist constants L and K, such that where ρ 1 , ρ 2 are positive constants, related tō ρ 1 ,ρ 2 in (F5), which will be determined later in Lemma 4.7. Moreover, Now we introduce the working spaces. Define I := (0, 1). Let C 1 (I) be the space of continuously differentiable functions in I, and By a standard argument, we know that the eigenvalue problem It is well known that is an equivalent norm in H 1 (I). Notice that λ 1 = π 2 4 , and the corresponding eigenfunction of λ 1 is denoted by ϕ 1 , which is positive in I. Moreover, from 3. Proof of Theorem 1.1 In this section, we first consider some special cases, where the nonlinearities do not contain the gradient terms. The similar argument can also be found in [28]. Consider the following problem: The following comparison theorem is known as Theorem of Strum-Picone (see [23]), which describes the location of the zero points of nontrivial solution.
Then problem    −x = P (t)x, only has the trivial solution x(t) ≡ 0.
To prove the lemma we make some extensions bȳ Hence, it is easy to check that For Case 1, we notice that y(t) = sin(k− 1 2 )πt is a solution of y +(k− 1 2 ) 2 π 2 y = 0. Now we comparex(t) with y(t) by Lemma 3.2. Obviously, the zeros of Hence, y(t) has 2k zero points in [0, 2], which impliesx(t) has at least 2k − 1 zero points in (0, 2) . Sincex(0) =x(2) = 0, we know thatx(t) has at least 2k + 1 zero points in [0, 2] . Denote byt 1 , · · · ,t 2k+1 the zero points ofx(t) in [0, 2] such that 0 =t 1 t 2 · · · t 2k+1 = 2. For any successive zerost i ,t i+1 ∈ [0, 2], Lemma 3.2 implies that Then, we get which leads to a contradiction. For Case 2, we notice that y(t) = sin kπt is a solution of y + k 2 π 2 y = 0. Then the zeros of y(t) in [0, 2] are t i = i k , 0 i 2k, i ∈ Z. Hence, y(t) has 2k + 1 zero points in [0, 2], which impliesx(t) has at least 2k zero points in (0, 2). Sincē x(0) =x(2) = 0, we know thatx(t) has at least 2k + 2 zero points in [0, 2]. Denotē t 1 , · · · ,t 2k+2 as the zero points ofx(t) in [0, 2] such that 0 =t 1 t 2 · · · t 2k+2 = 2. For any successive zerost i ,t i+1 ∈ [0, 2], Lemma 3.2 implies that Then, we get which is also a contradiction. Remark 3.1. It should be pointed out that this result is quite different from the result for Dirichlet problem. For Dirichlet problem, it is well-known that when P (t) locates between two successive eigenvalues, the linear equation only has trivial solution. This difference is mainly owing to the fact that Dirichlet condition ensures the right endpoint is also a zero point of the solution, so more accurate estimate about the distance between two zero points can be obtained. However, the same argument can not be applied to mixed boundary value problem, because the right endpoint 1 is not a zero point of the solution any more. We use the extension to treat the problem as a Dirichlet problem in [0, 2]. Actually, if the assumptions in the lemma are replaced by (k − 1 2 ) 2 π 2 P (t) (k + 1 2 ) 2 π 2 , P (t) may cross an eigenvalue of Dirichlet problem in [0, 2], which can not ensure the result. For example, if k = 1, π 2 4 P (t) 9π 2 4 , then P (t) may cross π 2 , which is an eigenvalue of Dirichlet problem in [0, 2].
The following lemma ensures the existence and uniqueness for problem (3.1).
Lemma 3.4. There exist two continuous functions β(t) and β(t), such that either of the following holds uniformly: Then the equation (3.1) has a unique solution.
To prove the above lemma, we first show the following results. Proof. Assume that u 1 (t), u 2 (t) are the solutions of (3.1), namely, and u(0) = u (1) = 0. According to Lemma 3.3, we know u ≡ 0. Now, we consider the existence of solutions for equation (3.1). Rewrite equation (3.1) in the following form: For any u ∈ C 1 M (I), from Lemma 3.3 we know that the linear boundary value problem has a unique solution v ∈ C 1 M (I). Define operator P : is the unique solution of equation (3.3). Then the existence of solution is equivalent to the existence of fixed point of P in C 1 M (I). Lemma 3.6. The operator P is continuous.
Proof. For any sequence {u n } ⊂ C 1 M (I) satisfying u n → u 0 as n → ∞, let v n = P u n , then we have {ω n } and {ω n } are uniformly bounded and equicontinuous sequence of functions. By Ascoli-Arzelà Theorem, {ω n } and {ω n } contain a uniformly convergent subsequence respectively (for convenience we also use the same notation), such that ω n −→ω 0 , ω n −→ϕ.
Let n → ∞, from the above we obtain By the uniqueness we know v 0 = P u 0 , which completes the proof.
Lemma 3.7. P is a compact operator.
Proof. For any bounded set S ⊂ C 1 M (I), we claim that P (S) is bounded in C 1 M (I). Otherwise, by an analogous manner as the proof of Lemma 3.6 we will get a contradiction. For every u ∈ S, v = P u is defined by (3.3). Since u , h u are all bounded, then v < ∞. Then we conclude that {v }, {v} are bounded and equicontinuous. By Ascoli-Arzelà Theorem, P is a compact operator.   in L 2 (I). Obviously, where β(t), β(t) satisfies either i or ii in Lemma 3.4. Let k → ∞, from (3.7) and (3.8), for a.e. t ∈ I, Hence, ω 0 ≡ 0, which is a contradiction to ω 0 = 1.
Proof of Lemma 3.4. The uniqueness is given in Lemma 3.5. To obtain the existence, assume D = {u ∈ C 1 M (I), u K + 1}, where K is given in Lemma 3.8. The continuity and compactness are established in Lemma 3.6 and Lemma 3.7, respectively. By Schauder's fixed point theorem, the operator P : D → D has at least one fixed point.
Then, for any v ∈ C 1 M (I), problem (M) v has a unique nontrivial solution u v . Proof. The proof can be obtained by Lemma 3.4. Proof. Assume that there exists {v n } such that u vn → ∞. Then Denote ω n = u vn / u vn and then ||ω n || = 1. Since (G0) holds, the second term in the above equation is bounded. Then a similar argument can be obtained, as the proof of Lemma 3.8.

Variational method
In this section, we consider (M) ODE by means of variational methods. In fact, the problem (M) ODE is non-variational because of the influence of the gradient term. We first study auxiliary problem (M) v . For any fixed v ∈ C 1 Then the weak solutions are equivalent to the critical points of the Euler-Lagrange functional J v : where G(t, u, η) := u 0 g(t, s, η)ds.
Define the corresponding functional J ± v : H 1 M (I) → R as follows: . Let u be a critical point of J ± v , which implies that u is a weak solution of (4.1). Furthermore, by the weak maximum principle it follows that u 0( 0) in I. Thus u is also a solution of problem (M) v . Hence, a nontrivial critical point of is actually a positive (negative) solution of (M) v .
From (4.2) we obtain Proof. From (G2)-(G4), we can take ε 0 > 0, C 0 > 0, τ > 2 such that Then Poincaré inequality and Sobolev inequality imply where C s is the Sobolev constant. Choosing u H 1 = ρ small enough, we obtain  Setting ϕ = u n and using Hölder inequality we have We claim that u n L 2 is bounded. Otherwise, passing to a subsequence, u n 2 L 2 → +∞, as n → +∞.
From the above two cases, for all t ∈ I, lim inf n→+∞ g + (t, u n , v ) u n L 2 ( π 2 4 + δ)ω. which follows that ω ≡ 0. This conclusion contradicts with ω n L 2 = 1, so u n L 2 is bounded. Then, from (4.8) we know that {u n } is bounded in H 1 M (I).
From (G5) we know that L < π 2 /4 and k := 2Kπ/(π 2 − 4L) satisfying k ∈ (0, 1). It can be easily seen that {u n } ⊂ H 1 M (I) is a Cauchy sequence, which implies that there exists u * ∈ H 1 M (I) such that u * ∈ T (u * ). According to the regularity theory, it follows that u * ∈ C 2 (I), which is actually a classical solution. Finally, from Lemma 4.5 we know that ||u * || H 1 c 0 , which means that u * is nontrivial.