Singular solutions for semilinear elliptic equations with general supercritical growth

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Introduction and main results
We are concerned with singular radial solutions of the semilinear elliptic equation where Ω = {x ∈ N ∶ |x| < R} is a finite ball with N ≥ 3 and f ∈ C 2 [0, ∞) . In this study we consider solutions to the ordinary differential equation By a singular solution u * (r) of (1.1) we mean that u * (r) is a classical solution of (1.1) for 0 < r ≤ r 0 with some r 0 > 0 and it satisfies u * (r) → ∞ as r → 0.
It is well known that the singular solution plays an important role in the study of the solution structure of (1.1) in the supercritical case (see, e.g., [6,11,20,21,23]), and various properties of the singular solutions have been studied extensively. Let us recall some known results for the existence and uniqueness of the singular solution of (1.1). When f (u) = u p with p > N∕(N − 2) , (1.1) has the exact singular solution It was shown by Serrin-Zou [25,Proposition 3.1] that, if p > p S = (N + 2)∕(N − 2) , the singular solution of (1.1) is unique. In the case f (u) = e u , it was shown by Mignot-Puel [19] that is the unique singular solution of (1.1) when N ≥ 3 . Later, the singular radial solutions have been studied for various nonlinearities. For f (u) = u p + g(u) with lower order term g, we refer to [2,4,11,12,18,20,23,24]. For f (u) = e u + g(u) , see [21,24]. In [24]  Recently, the singular radial solutions have been constructed by [14] for f (u) = exp(u p ) with p > 0 , and by [9] for the cases f (u) = exp(e u ) and, more generally, Furthermore, the existence of the singular solution was obtained by [16,22] for a certain general class of supercritical nonlinearities. On the other hand, the uniqueness of the singular radial solution seems hard to obtain, and it was left open in [9,14,16,22].
In this paper, we will investigate qualitative properties of the singular solution to (1.1) with the following general hypotheses (f1) and (f2) on f.
(1.6) q = lim u→∞ F(u)f � (u). and hence 1∕p + 1∕q = 1 . The exponent q, defined by (1.5), was first introduced by Dupaigne and Farina [7] in the study of the stable solutions in N . Fujishima and Ioku [8] investigated the solvability of semilinear parabolic equations with the nonlinear term f satisfying (f1) and (1.6). It is easy to see that various examples of f satisfy (f1) and (f2) with some exponent q ≥ 1 . One can easily check that q = p∕(p − 1) > 1 if f (u) = u p with p > 1 , and that q = 1 if f (u) = e u . For example, if f (u) = u p (log(u + e)) with p > 1 and ∈ , then q = p∕(p − 1) . Hence, the principal term of f is not necessarily u p if q = p∕(p − 1) > 1 . The case q = 1 corresponds to a super-power case, and various examples of f were provided in [22,Sect. 2]. In particular, the case q = 1 includes the super-exponential nonlinearities, e.g., f (u) = exp(u p ) with p > 0 and (1.3) with n ≥ 1 . (See also Lemma 5.1 below).
Let q S denote the Hölder conjugate of the critical Sobolev exponent Then the supercritical case p > p S corresponds to the case q < q S . In this paper, we study the properties of singular solutions to (1.1) with a nonlinearity f satisfying (f1) and (f2) with q < q S . For > 0 , we denote by u(r, ) a regular solution of (1.1) satisfying u(0) = and u � (0) = 0 . We also show the convergence property of u(r, ) to the singular solution.
Theorem 1.1 Suppose that N ≥ 3 and (f 1) and (f 2) with q < q S hold. Then, there exists a unique singular solution u * (r) of (1.1) for 0 < r ≤ r 0 with some r 0 > 0 , and the regular solution u(r, ) satisfies Furthermore, the singular solution u * satisfies  [22]. In the proof of Theorem 1.1 we construct the singular solution in a completely different way, which will enable us to obtain the convergence property simultaneously. (ii) When f (u) = u p , it was shown in [3,5] [26].) Therefore, we cannot expect the uniqueness of the singular solution in the critical and subcritical cases q ≥ q S with N ≥ 3 and in the super-power case q = 1 with N = 1, 2. (iii) By Theorem 1.1, we can easily calculate an exact asymptotic expansion of the singular solution near the origin in the typical cases. We use here the notation f (u) ∼ g(u) means that f (u)∕g(u) → 1 as u → ∞ . In the case f (u) ∼ u p and Hospital's rule. Then q = p∕(p − 1) and F −1 (u) ∼ ((p − 1)u) −1∕(p−1) , and hence (1.8) can be written as where A is the constant defined in (1.2). Thus a principal term of the asymptotic expansion of the singular solution u * at r = 0 is given by (1.2). In the case f (u) ∼ e u and f � (u) ∼ e u , we obtain by a similar argument.
Let us consider the asymptotic expansion of the singular solution near the origin in the super-exponential case f (u) = exp(g(u)) . Define g n (u) with n ≥ 1 by In the cases f (u) = exp(g(u)) , where g(u) = u p with p > 0 and g(u) = g n (u) with n ≥ 1 , it was shown by [22] that (f1) and (f2) hold with q = 1 . (See also Lemma 5.1 below.) Then (1.1) has a unique singular solution u * , which satisfies (1.8) by Theorem 1.1. In particular, we obtain the following asymptotic expansion near the origin.
(i) In the case g(u) = u p with p > 0 , the singular solution u * of (1.1) satisfies (ii) In the case g(u) = g n (u) , defined by (1.9), with n ≥ 1 , the singular solution u * of (1.1) satisfies where log 1 u = log u and log n u = log(log n−1 u) for n ≥ 2 . In particular, if g(u) = exp u , then the singular solution u * of (1.1) satisfies Remark 1.2 In the case g(u) = u p , Kikuchi and Wei [14] constructed the singular solution u * satisfying (1.10) by using a contraction mapping principle. Also, in the case g(u) = g n (u) with n ≥ 1 , Ghergu and Goubet [9] constructed the singular solution which has a prescribed behavior around the origin. In our proof, we make use of the formulas (1.8) to derive the asymptotic expansions (1.10) and (1.11) in a unified way. u * (r) = Ar −2∕(p−1) (1 + o(1)) as r → 0, u * (r) = −2 log r + log 2(N − 2) + o(1) as r → 0 (1.9) g 1 (u) = exp u and g n (u) = exp(g n−1 (u)) for n ≥ 2.
As an application, we consider the nonlinear eigenvalue problem where B = {x ∈ N ∶ |x| < 1} with N ≥ 3 and > 0 is a parameter. We assume that f (u) > 0 for u ≥ 0 . By the symmetry result of Gidas, Ni and Nirenberg [10], every classical solution of (1.13) is radially symmetric about the origin. Hence, the problem (1.13) can be reduced to an ODE problem. Denote by {( ( ), v(r, ))} a solution of (1.13) with v(0, ) = > 0 . It is well known that the solution set of (1.13) can be described as the curve {( ( ), v(r, )) ∶ 0 < < ∞} . (see, e.g., [15,20,21]). In the cases f (u) = e u and f (u) = (u + 1) p , the bifurcation diagram of the problem (1.13) was completely characterized by Joseph-Lundgren [13]. In these cases, there is a special change of variables such that (1.13) can be transformed into an autonomous first order system. For a general nonlinearity, we cannot expect to find such a change of variables, but we obtain the following result as a corollary of Theorem 1.1. (i) The problem (1.13) has a unique radial singular solution ( * , v * ) , that is, there exists a unique * > 0 such that the problem (1.13) with = * has a singular solution v * , and the singular solution v * is a unique singular radial solution of (1.13) with = * . (ii) As → ∞ , the solution ( ( ), v(r, )) of (1.13) described above satisfies where ( * , v * ) is the singular radial solution in (i).

Remark 1.3
One of the open problems raised by [1] is whether for which (1.13) has a singular solution is unique or not. Corollary 1.2 (i) gives an affirmative answer to this question in the radially symmetric case.
Under the assumptions (f1) and (f2) the following quasi-scaling, which was introduced in [8], works well: It was found by [22] that the limit equation of (1.1) with respect to (1.14) as → 0 is and that (1.15) has the exact singular solution for r > 0.
In this paper we will find a singular solution to (1.1) of the form Then u * given by (1.16) solves (1.1) for 0 < r ≤ r 0 if and only if x satisfies where a = N + 2 − 4q , b = 2N − 4q and t 0 = − log r 0 . In the proof of Theorem 1.1, we will show that x(t) → 0 as t → ∞ if u * given by (1.16) is a singular solution of (1.1). We also show that (1.17) has a unique solution x satisfying x(t) → 0 as t → ∞ . The transformation (1.16) enables us to deal with a quite wide class of nonlinear terms. The paper is organized as follows. In Sect. 2, we give some preliminary results and in Sect. 3, we investigate the asymptotic behavior of the singular solution. In Sect. 4, we show the uniqueness and the existence of singular solution, and give the proof of Theorem 1.1. In Sect. 5, we consider the super-exponential cases, and give the proof of Corollaries 1.1 and 1.2.

Preliminaries
First we show the following lemma.
Furthermore, the exponent q in (1.5) satisfies q ≥ 1 and q is also given by (1.6).

Proof
If we assume that f � (u) is bounded as u → ∞ , then f (u) ≤ Cu with some constant C > 0 . This contradicts F(u) < ∞ . Then we obtain lim sup u→∞ f � (u) = ∞ . If we assume that lim inf u→∞ f � (u) < ∞ , then there exists a sequence u n → ∞ of local maximum of f � (u) such that f � (u n ) > 0 and f �� (u n ) = 0 . This contradicts (1.5). Thus we obtain f � (u) → ∞ as u → ∞ . By L'Hospital's rule, we have In the remaining part of this paper except the last section, we assume that (f 1) and (f 2) with q < q S hold. By Lemma 2.1 we may assume that by replacing u 0 in (f1).
For a solution u of (1.1), define x(t) by (1.17) and t = − log r.

Lemma 2.2 Let u be a solution of (1.1), and define x(t) by
Thus we obtain Then we obtain (2.3). In the case q > 1 , by a direct calculation, we obtain (2.5) and (2.6). ◻

Lemma 2.3
Let u be a positive solution of (1.1) satisfying u(r) ≥ u 0 for 0 < r ≤ r 0 . Then the following (i) and (ii) hold.
Integrating the above on (r, r 1 ] , we obtain Letting r → 0 , we obtain u(r) → −∞ . This is a contradiction. Thus we obtain where we used the fact that f(u) is increasing for u ≥ u 0 which follows from (2.1). Then it follows that Integrating the above on [ , r] with 0 < < r , and letting → 0 , we have F(u(r)) ≥ r 2 ∕(2N). ◻ Lemma 2.4 For any > 0 , there exists a constant C > 0 such that for sufficiently large u.
Proof From (1.5) there exists u 1 ≥ u 0 such that Then it follows that is nondecreasing for u ≥ u 1 , and hence we obtain Observe that Then f (u)F(u) q+ is nonincreasing in u ≥ u 1 , and hence , and letting n → ∞ , we obtain (2.11). ◻ Lemma 2.6 Let u be a singular solution of (1.1). Then Proof Assume by contradiction that Take q 0 ∈ (q, q S ) , and define z 0 (t) by Observe that for 0 < r ≤ r 2 . Then r ∕F(u(r)) is nondecreasing and bounded for 0 < r ≤ r 2 . From (2.22) we see that F(u) q 0 f (u) is also bounded for 0 < r ≤ r 2 . Thus we have (2.19) f � (u(r))F(u(r)) ≤ q 0 for 0 < r ≤ r 1 (2.21) d dr which implies that u � (r) = O(r 1− q 0 ) as r → 0 . Since q 0 < 1 , we have u � (r) → 0 as r → 0 , and hence lim r→0 u(r) < ∞ . This is a contradiction. Thus we obtain the conclusion. ◻

Asymptotic behavior of the singular solution
Let u be a singular solution of (1.1) satisfying u(r) ≥ u 0 for 0 < r ≤ r 0 . Define x(t) by (2.2) for t ≥ t 0 , where t 0 = − log r 0 . In this section, we will show the following proposition. To prove Proposition 3.1, we first show the following lemma.
First we consider the case (i). .
Proof By Lemma 2.6 and (2.2) we obtain lim sup t→∞ x(t) > −∞ . Since x(t) is monotone increasing or decreasing for sufficiently large t, we have lim t→∞ x(t) > −∞ . Then, by Lemma 3.1 (i), x(t) is bounded, and hence there exists a constant c ∈ such that x(t) → c as t → ∞ . We show that c = 0 . On the contrary, assume that c ≠ 0 . We first show that We consider the case x � (t) ≥ 0 for all t large. Since x(t) is bounded, we have Assume by contradiction that lim sup t→∞ x � (t) > 0 . Let t n → ∞ be a sequence of local minimum points of x � (t) . Then, from (3.2), we have

By (2.3) and the fact that f � (u)F(u) → q as t → ∞ , we obtain
On the other hand, This is a contradiction. In the case where x � (t) ≤ 0 for all t large, we get the contradiction by a similar argument. Thus (3.1) holds. Letting t → ∞ in (2.3), from (3.1) we obtain x �� (t) → −b(e c − 1) ≠ 0 as t → ∞ . This implies that |x � (t)| → ∞ as t → ∞ , which is a contradiction. Thus, we obtain x(t) → 0 as t → ∞ . ◻ Next, we consider the case (ii). First we show that x � (t) is bounded for t ≥ t 0 .
Since f � (u(r))F(u(r)) → 1 as t → ∞ , we have This contradicts Lemma 3.1 (i), since M > 0 is arbitrary. Thus, x � (t) is bounded in both cases q > 1 and q = 1 . ◻ In order to show that x(t) → 0 as t → ∞ in the case (ii), we consider the following ordinary differential equation where > 0 and p > 1 are constants, c ∈ C[t 0 , ∞) and G ∈ C[t 0 , ∞) . In (3.4), we assume that with some constant c * > 0 , and that G(t) → 0 as t → ∞ . We can state the following lemma.

Lemma 3.4 Let w ∈ C 2 [t 0 , ∞) be a bounded positive solution of (3.4). Assume that the sign of w � (t) changes infinitely many times as t → ∞ . Then w(t) → 1 as t → ∞.
Since the proof of Lemma 3.4 is rather complicated, we will give the proof after completing the proof of Proposition 3.1. By using Lemma 3.4, we will show the following lemma.
We show that H(t, z(t)), defined by (3.7), satisfies In fact, we observe that Lemmas 3.1 (i) and 3.3 imply that x(t) is bounded above and x � (t) is bounded. Since f � (u)F(u) − q → 0 as t → ∞ , we see that (3.9) holds. Then, applying Lemma 3.4 with

Proof of Lemma 3.4 Define and , respectively, by
Since w(t) is bounded above, we have 0 ≤ ≤ < ∞ . We will show that = = 1 by dividing the proof into three steps.
Step Since  which implies that ̃= . Then, from (3.17), we find that Step 3 We show that = = 1. From (3.15), we have ≤ 1 ≤ . Assume by contradiction that < 1 . Then > 1 . Put Let {s i } and { i } be the sequences appearing in (3.19), and let t = t i (w) is the inverse function of w = w(t) for s i < t < i . Put w � (t) = w � (t i (w)) = q i (w) . We will show that Take any ŵ ∈ +1 2 , 1 . From (3.19), there exists ŝ i satisfying s i <ŝ i < i and w(ŝ i ) =ŵ for sufficiently large i. Multiplying (3.4) by w � (t) , and integrating it over [s i ,ŝ i ] , we obtain
From w � (s i ) = 0 and (3.5), we obtain Then it follows that Note here that, from (3.15) and (3.18), we have

Proof of Theorem 1.1
We first show the uniqueness of the singular solution.
Theorem 4.1 There exists at most one singular solution of (1.1).
To show Theorem 4.1, we need the following lemma by [17,Lemma 4.2].

Lemma 4.1 Let y(t) be a solution of where A(t) and B(t) are continuous functions satisfying
If y(t) is bounded as t → ∞ , then y(t) ≡ 0.

Super-exponential nonlinearity
In this section we consider the case where f(u) has the form f (u) = exp(g(u)).
(ii) Since f (u) > 0 for all u ≥ 0 , u(r, ) has the first zero, which we denote by r 0 ( ) .