Exact behavior around isolated singularity for semilinear elliptic equations with a log-type nonlinearity

Theorem 1.1.

Assume α and β satisfy (1.2) and let u be a nonnegative solution of (1.1). Then the following alternative holds:

1. either u has a removable singularity at the origin,

2. or u is a singular solution and satisfies

   (1.3) u ⁢ ( x ) = ( A + o ⁢ ( 1 ) ) ⁢ | x | - 2 α - 1 ⁢ ( log ⁡ 1 | x | ) - β α - 1   as  ⁢ x → 0 ,

   where

   (1.4) A = [ ( 2 α - 1 ) 1 - β ⁢ ( n - 2 - 2 α - 1 ) ] 1 α - 1 .

Theorem 2.1.

Let u be a nonnegative solution of

with an isolated singularity at the origin. Suppose that g ⁢ ( t ) is a locally Lipschitz function, which in a neighborhood of infinity satisfies the conditions below:

1. g ⁢ ( t ) is nondecreasing in t,

2. t - n + 2 n - 2 ⁢ g ⁢ ( t ) is nonincreasing,

3. g ⁢ ( t ) ≥ c ⁢ t p for some p ≥ n n - 2 and c > 0 .

Then

Lemma 2.2.

We have

and

as r → 0 .

Proof.

Throughout this proof, c > 0 depends at most on n, α and β, and may differ from one line to another. As mentioned earlier, we have u α | log u | β ∈ L 1 ( B 1 ) , and thus from the divergence theorem and (1.1), we deduce that

In particular, u ¯ ⁢ ( r ) is monotone decreasing in r. Moreover, if (2.2) holds, then one may easily derive (2.3) from (2.4) and (2.1).

Henceforth, we shall prove (2.2). Especially, we shall assume that u ¯ ⁢ ( r ) ≠ O ⁢ ( 1 ) as r → 0 , since the case u ¯ ⁢ ( r ) = O ⁢ ( 1 ) already satisfies (2.2). Under this assumption, we have u ¯ ⁢ ( r k ) → ∞ for some r k → 0 . Then the monotonicity of u ¯ implies that u ¯ ⁢ ( r ) → ∞ as r → 0 .

Taking r small enough, and using (2.1) and the fact that s ↦ s α ⁢ ( log ⁡ s ) β is increasing for large s, we deduce that

Hence, from the assumption u ¯ ⁢ ( r ) → ∞ as r → 0 and the fact u ¯ ′ ⁢ ( r ) < 0 , it follows that

Note that for any sufficiently large s satisfying 2 ⁢ | β | ≤ ( α - 1 ) ⁢ log ⁡ s , we have

whence we may proceed from the integral above as

for sufficiently small r > 0 . Thus, we arrive at

Setting w ⁢ ( s ) to be the inverse function^[1] of s ⁢ e s , we know that s w ⁢ ( s ) is the inverse function of s ⁢ log ⁡ s . Since t α - 1 ⁢ ( log ⁡ t ) β = ( c ⁢ s ⁢ log ⁡ s ) β , with s = t α - 1 β , we deduce from (2.5) and the choice of w that

However, since log ⁡ s - log ⁡ log ⁡ s ≤ w ⁢ ( s ) ≤ log ⁡ s for sufficiently large s, we arrive at (2.2). ∎

Lemma 2.3.

We have

for large t > 1 and θ ∈ 𝕊 n - 1 , where

and

Proof.

We take r 0 > 0 small enough such that log ⁡ u > 0 in B r 0 , and set t 0 = - log ⁡ r 0 . In what follows, we take t ≥ t 0 and 0 < r ≤ r 0 , unless stated otherwise. For notational convenience, let us write

so that ψ ⁢ ( t , θ ) = ϕ ⁢ ( r ) ⁢ u ⁢ ( r , θ ) . Since ∂ t = - r ⁢ ∂ r and ∂ t ⁢ t = r ⁢ ∂ r + r 2 ⁢ ∂ r ⁢ r , we have

where the left and right side are evaluated in ( t , θ ) and, respectively, in ( r , θ ) , and by ϕ ′ and ϕ ′′ , we denoted d ⁢ ϕ d ⁢ r and, respectively, d 2 ⁢ ϕ d ⁢ r 2 . Setting

we observe that r ⁢ ϕ ′ = η ⁢ ϕ and r 2 ⁢ ϕ ′ = ( η 2 - η + r ⁢ η ′ ) ⁢ ϕ , and therefore

where we used the fact that ψ t = - r ⁢ ϕ ′ ⁢ u - r ⁢ ϕ ⁢ u r = - η ⁢ ψ - r ⁢ ϕ ⁢ u r and ψ = ϕ ⁢ u in deriving the second identity.

In view of (2.8) and (2.9), it is not hard to check that

and

On the other hand, we know from (2.6) that

from which we may also deduce that

One may also notice from (2.6) that

Hence, inserting (2.11)–(2.14), we arrive at equation (2.7), which completes the proof. ∎

Lemma 2.4.

We have

as t → ∞ .

Proof.

In this proof, C > 0 will depend on n only and may differ from one line to another. The estimates in (2.20) follow immediately from (2.15), (2.2) and (2.3). Moreover, since

(2.18) can be easily deduced from (2.1). Thus, we are only left with proving (2.19).

For notational convenience, let u ¯ ⁢ ( x ) denote u ¯ ⁢ ( | x | ) . Also let us denote by A r the annulus B 2 ⁢ r ∖ B ¯ r / 2 . From (2.1), we have

Therefore, it follows from the interior gradient estimates that

We regard (1.1) as - Δ ⁢ u = m ⁢ ( x ) ⁢ u in B 1 ∖ { 0 } , where m = u α - 1 | log u | β . In view of (2.5) and (2.1), we have that m ⁢ ( x ) = O ⁢ ( | x | - 2 ) , so the Harnack inequality implies

Using this observation along with (2.1), (2.5) and the above gradient estimate, we find

Since

(2.19) follows from (2.21), (2.2) and (2.1). ∎

Corollary 2.5.

We have

Proof.

From (2.10) and (2.16), we know that

for any t > 0 and θ ∈ 𝕊 n - 1 . Due to (2.18), we have

Using the above estimate together with Lemma 2.4, we have

An integration over 𝕊 n - 1 in the above estimate, will lead us to (2.22). ∎

Lemma 2.6.

We have either

or

with A given by (1.4).

Proof.

Let ψ and ψ ¯ be defined by (2.6) and (2.15), respectively. Multiplying (2.17) by ψ ¯ ′ and integrating it over [ t , T ] , from (2.18)–(2.20) and (2.22), we find

By (2.9) and (2.20), we have b ⁢ ( t ) = O ⁢ ( 1 ) and b ′ ⁢ ( t ) = O ⁢ ( t - 2 ) , which leads us to

Similarly, from (2.10) and (2.20), we find ζ ¯ ⁢ ( t ) = O ⁢ ( 1 ) and ζ ¯ ′ ⁢ ( t ) = O ⁢ ( t - 2 ⁢ log ⁡ t ) , from which it follows that

Since α is chosen as in (1.2), we know from (2.8) that a ⁢ ( t ) is positive and bounded away from zero for all large t > 1 . Thus, (2.24), (2.25) and (2.17) yield

In view of (2.17), it follows from (2.20) and (2.22) that ψ ¯ ′′ ⁢ ( t ) = O ⁢ ( 1 ) , and hence ψ ¯ ′ ⁣ 2 ⁢ ( t ) is uniformly Lipschitz for large t > 1 . Hence, we deduce that

Now we multiply (2.17) by ψ ¯ ′′ and integrate it over [ t , T ] , which leads us to

due to (2.18)–(2.20) and (2.22), as before. Note that from (2.8) we have a ⁢ ( t ) = O ⁢ ( 1 ) and a ′ ⁢ ( t ) = O ⁢ ( t - 2 ) as t → ∞ . Hence, from (2.20) and (2.26), we derive

On the other hand, since ψ ¯ ⁢ ψ ¯ ′′ = 1 2 ⁢ ( ψ ¯ 2 ) ′′ - ψ ¯ ′ ⁣ 2 , a further integration by parts produces

where the second equality can be deduced analogously to the derivation of (2.24). Similarly, we also observe that

Due to (2.29), (2.30) and (2.31), (2.28) leads us to

Differentiating (2.17) with respect to t, we deduce, from (2.20), (2.22) and ψ ¯ ′′ ⁢ ( t ) = O ⁢ ( 1 ) , that ψ ¯ ′′′ ⁢ ( t ) = O ⁢ ( 1 ) . Therefore, ψ ¯ ′′ ⁣ 2 ⁢ ( t ) is uniformly Lipschitz for large t > 1 , from which combined with (2.32), we obtain

To this end, we shall pass to the limit in (2.17) with t → ∞ . Note that from (2.9) we have

while, from (2.16) and (2.20), it follows that

Although we do not know yet if ψ ¯ ⁢ ( t ) converges as t → ∞ , we still know from (2.20) that it converges along a subsequence. Denoting by ψ ¯ 0 a limit value of ψ ¯ ⁢ ( t ) along a subsequence, say t = t j → ∞ , after passing to the limit in (2.17), with t = t j , we obtain, from (2.22), (2.27) and (2.33), that

Thus, in view of (1.4), we have

Now the continuity of ψ ¯ implies that ψ ¯ ⁢ ( t ) converges as t → ∞ (without extracting any subsequence) either to 0 or A. If there are two distinct sequences t j → ∞ and t j ′ → ∞ such that ψ ¯ ⁢ ( t j ) → 0 and ψ ¯ ⁢ ( t j ′ ) → A , then by the intermediate value theorem, there must exist some other t j ′′ → ∞ such that ψ ¯ ⁢ ( t j ′′ ) → A 2 , which violates (2.34). Thus, the proof is completed. ∎

Proof of Theorem 1.1.

If (2.23) is true, then, in view of (2.15), we observe that

Hence, from (2.1), we derive (1.3) and (1.4), which establishes the proof for Theorem 1.1 (ii).

Henceforth, let us suppose that

The rest of the argument follows closely that of the proof in [1, Theorem 1.3].

In view of (2.8) and (2.9), we may rephrase (2.17) as

with

Thus, the decay of ψ ¯ ⁢ ( t ) is determined by the negative root of

Since a 0 2 + 4 ⁢ b 0 = ( n - 2 ) 2 , the root λ is

Therefore, we have

In view of (2.15), we obtain

Now if β > 0 , then we deduce from (2.35) that u ¯ ⁢ ( r ) → 0 as r → 0 , from which, combined with (2.1), it follows that u ⁢ ( x ) → 0 as x → 0 . Hence, the origin is a removable singularity. Similarly, if β = 0 , then (2.35) implies that u ¯ ⁢ ( r ) = O ⁢ ( 1 ) , and thus the origin is again a removable singularity.

Hence, we are only left with the case β < 0 . Since u ⁢ ( x ) = ( 1 + O ⁢ ( r ) ) ⁢ u ¯ ⁢ ( r ) , (2.35) implies that

for each q ≥ 1 , for some constant C > 0 depending on n, α, β and q. Therefore, u α | log u | β ∈ L p ( B 1 ) for any p ≥ 1 , and in particular for p > n . This implies that Δ ⁢ u ∈ L p ⁢ ( B 1 ) for p > n , so u ∈ C 1 , α ⁢ ( B 1 / 2 ) for α = 1 - n p , proving again that the origin is a removable singularity. Thus, the proof of Theorem 1.1 (i) is completed. ∎