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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


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

Marius Ghergu / Sunghan Kim / Henrik Shahgholian
Published Online: 2018-06-14 | DOI: https://doi.org/10.1515/anona-2017-0261

Abstract

We study the semilinear elliptic equation

-Δu=uα|logu|βin B1{0},

where B1n, with n3, nn-2<α<n+2n-2 and -<β<. Our main result establishes that the nonnegative solution uC2(B1{0}) of the above equation either has a removable singularity at the origin or it behaves like

u(x)=A(1+o(1))|x|-2α-1(log1|x|)-βα-1as x0,

with A=[(2α-1)1-β(n-2-2α-1)]1α-1.

Keywords: Singular solutions; asymptotic behavior; log-type nonlinearity

MSC 2010: 35J61; 35J75; 35B40; 35C20

1 Introduction

Let n3 and B1 be the unit open ball in n. This paper is concerned with the behavior of nonnegative solutions of

-Δu=uα|logu|βin B1{0},(1.1)

where α and β are real numbers satisfying

nn-2<α<n+2n-2and-<β<.(1.2)

We say that u is a nonnegative solution of (1.1) if uC2(B1{0}) is nonnegative and satisfies (1.1) pointwise. In addition, we say that a nonnegative solution u of (1.1) is singular if u is unbounded in any punctured ball Br{0}, with 0<r<1.

The case β=0 in (1.1) is by now well understood; in their pioneering work [4], Gidas and Spruck established a series of results that completely characterize the asymptotic behavior of local solutions of (1.1) (with β=0). The main goal of this paper is to obtain similar results for (1.1) when the exponents α and β are in the range given by (1.2).

Our main result is the following.

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:

  • (i)

    either u has a removable singularity at the origin,

  • (ii)

    or u is a singular solution and satisfies

    u(x)=(A+o(1))|x|-2α-1(log1|x|)-βα-1as x0,(1.3)

    where

    A=[(2α-1)1-β(n-2-2α-1)]1α-1.(1.4)

For β=0, we recover the result in [4, Theorem 1.3]. Let us note that in the case β=0, the approach in [4] relies to a large extend on the properties of the scaling function uλ(x)=λ2α-1u(λx) (λ>0). Thus, if u is a solution of (1.1) (with β=0), then so is uλ. A similar scaling is not available to us in case β0 due to the presence of the logaritmic term in (1.1). In turn, we shall take advantage of the result in [1, Theorem 1.1] which allows us to derive that singular solutions of (1.1) are asymptotically radial. The exact asymptotic behavior (1.3) is further deduced by looking at the corresponding ODE of the scaled function |x|2α-1(log1|x|)βα-1u(x) in polar coordinates.

The asymptotic behavior of nonnegative singular solutions has been studied in various settings. In addition to the classical results [4] and [1], Korevaar et al. [6] derived the improved asymptotic behavior of the nonnegative singular solutions of -Δu=un+2n-2 by a more geometric approach. Meanwhile, C. Li [7] extended the result on the asymptotic radial symmetry of singular solutions of -Δu=g(u) for a more general g(u) considered in [1]. Recently, the asymptotic radial symmetry has been achieved for other operators, such as conformally invariant fully nonlinear equations [5, 8], fractional equations [2], and fractional p-laplacian equations [3].

This paper extends the classical argument in [4] and [1] to a log-type nonlinearity. One of the key observations is that from the asymptotic radial symmetry achieved in [1] for nonnegative solutions of -Δu=g(u), one can obtain an optimal asymptotic upper bound for g(u)u. Hence, we are left with preserving the optimality by transforming g(u)u to u under a suitable inverse mapping.

This observation indeed allows us to consider a more general class of equations of the type

-Δu=uαf(u)in B1{0},

where f is a slowly varying function at infinity, under some additional assumptions. A typical example is

f(u)=|log(k1)u|β1|log(k2)u|β2|log(km)u|βm,

where ki are positive integers, βi are real numbers and log(k)u=log(log(k-1)u) for k2 with log(1)u=logu. However, we shall not specify the additional assumptions for the nonlinearity f as they turn out to involve technical and cumbersome computations. Hence, we present the argument only with f(u)=|logu|β in order to simplify the presentation.

Throughout the paper, we shall write f(x)=O(g(x)) if |f(x)|C|g(x)| uniformly in x, where C>0 depends at most on n, α and β. We shall also use the notation f(x)=o(g(x)) as x0 to denote that |f(x)||g(x)|0 as x0.

2 Asymptotic behavior around a non-removable singularity

Let u¯(r) denote the spherical average of u on the ball of radius r, that is,

u¯(r)=Bru𝑑σ.

The following result is a slight modification of [1, Theorem 1.1].

Theorem 2.1.

Let u be a nonnegative solution of

-Δu=g(u)in B1{0},

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:

  • (i)

    g(t) is nondecreasing in t,

  • (ii)

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

  • (iii)

    g(t)ctp for some pnn-2 and c>0.

Then

u(x)=(1+O(|x|))u¯(|x|)as x0.(2.1)

The original result in [1, Theorem 1.1] requires condition (i) above to be satisfied for all t>0, but a careful analysis of its proof shows that this condition is enough to hold in a neighborhood of infinity.

It is not hard to see that g(t)=tα|logt|β fulfills conditions (i)–(iii) in Theorem 2.1. Moreover, it follows from [1, Lemma 2.1] that uα|logu|βL1(B1) and u is a distribution solution of (1.1) in B1, i.e., for any ηCc(B1), we have

-B1uΔηdx=B1uα|logu|βηdx.

The next lemma provides an asymptotic upper bound for u¯.

Lemma 2.2.

We have

u¯(r)=O(r-2α-1(log1r)-βα-1)(2.2)

and

u¯(r)=O(r-α+1α-1(log1r)-βα-1)(2.3)

as r0.

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 uα|logu|βL1(B1), and thus from the divergence theorem and (1.1), we deduce that

-u¯(r)=crn-1Bruα|logu|βdx.(2.4)

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 r0, since the case u¯(r)=O(1) already satisfies (2.2). Under this assumption, we have u¯(rk) for some rk0. Then the monotonicity of u¯ implies that u¯(r) as r0.

Taking r small enough, and using (2.1) and the fact that ssα(logs)β is increasing for large s, we deduce that

-u¯(r)cru¯α(r)(logu¯(r))β.

Hence, from the assumption u¯(r) as r0 and the fact u¯(r)<0, it follows that

u¯(r)dssα(logs)β=-0ru¯(r)dru¯α(r)(logu¯(r))βcr2.

Note that for any sufficiently large s satisfying 2|β|(α-1)logs, we have

-1α-1dds(1sα-1(logs)β)=(1-β(α-1)logs)1sα(logs)β12sα(logs)β,

whence we may proceed from the integral above as

1u¯α-1(r)(logu¯(r))βcr2

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

u¯α-1(r)(logu¯(r))β=O(r-2)as r0.(2.5)

Setting w(s) to be the inverse function1 of ses, we know that sw(s) is the inverse function of slogs. Since tα-1(logt)β=(cslogs)β, with s=tα-1β, we deduce from (2.5) and the choice of w that

u¯(r)=O(r-2α-1w(r-2β)-βα-1)as r0.

However, since logs-loglogsw(s)logs for sufficiently large s, we arrive at (2.2). ∎

Let us next define

ψ(t,θ)=r2α-1(log1r)βα-1u(r,θ),(2.6)

with t=-logr and θ𝕊n-1.

Lemma 2.3.

We have

ψtt+Δθψ+aψt-bψ+ζβψα=0(2.7)

for large t>1 and θ𝕊n-1, where

a(t)=4α-1-n+2-2β(α-1)t,(2.8)b(t)=((n-2)-2α-1+β(α-1)t)(2α-1-β(α-1)t)-β(α-1)t2(2.9)

and

ζ(t,θ)=2α-1-βα-1logtt+logψ(t,θ)t.(2.10)

Proof.

We take r0>0 small enough such that logu>0 in Br0, and set t0=-logr0. In what follows, we take tt0 and 0<rr0, unless stated otherwise. For notational convenience, let us write

ϕ(r)=r2α-1(log1r)βα-1,

so that ψ(t,θ)=ϕ(r)u(r,θ). Since t=-rr and tt=rr+r2rr, we have

ψtt+Δθψ=r2ϕΔu+(2rϕ-(n-2)ϕ)rur+(rϕ+r2ϕ′′)u,

where the left and right side are evaluated in (t,θ) and, respectively, in (r,θ), and by ϕ and ϕ′′, we denoted dϕdr and, respectively, d2ϕdr2. Setting

η(r)=2α-1+β(α-1)logr,

we observe that rϕ=ηϕ and r2ϕ=(η2-η+rη)ϕ, and therefore

ψtt+Δθψ=-r2ϕuα(logu)β+(2η-n+2)rurϕ+(η2+rη)ϕu=-r2ϕuα(logu)β-(2η-n+2)ψt+((n-2)η-η2+rη)ψ,

where we used the fact that ψt=-rϕu-rϕur=-ηψ-rϕur and ψ=ϕu in deriving the second identity.

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

a(t)=2η(r)-n+2(2.11)

and

b(t)=(n-2)η(r)-η2(r)+rη(r).(2.12)

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

logu(r,θ)=2tα-1-βα-1logt+logψ(t,θ),

from which we may also deduce that

ζ(t,θ)=logu(r,θ)log1r.(2.13)

One may also notice from (2.6) that

r2ϕ(r)u(r,θ)=t-βψα(t,θ).(2.14)

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

Let us define

ψ¯(t)=𝕊n-1ψ(t,θ)𝑑θ=r2α-1(log1r)βα-1u¯(r)(2.15)

and

ζ¯(t)=2α-1-βα-1logtt+logψ¯(t)t.(2.16)

Averaging (2.7) over 𝕊n-1, we obtain

ψ¯′′+aψ¯-bψ¯+ζ¯βψ¯α+𝕊n-1(ψαζβ-ψ¯αζ¯β)𝑑θ=0(2.17)

for large t.

Lemma 2.4.

We have

ψ(t,θ)-ψ¯(t)=ψ¯(t)O(e-t),(2.18)|t(ψ(t,θ)-ψ¯(t))|+|θ(ψ(t,θ)-ψ¯(t,θ))|=ψ¯(t)O(e-t),(2.19)ψ¯(t)=O(1)andψ¯(t)=O(1)(2.20)

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

ψ(t,θ)-ψ¯(t)=r-2α-1(log1r)-βα-1(u(r,θ)-u¯(r)),

(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 Ar the annulus B2rB¯r/2. From (2.1), we have

-Δ(u-u¯)=u¯α|logu¯|βO(r)in Ar as r0.

Therefore, it follows from the interior gradient estimates that

|(u-u¯)|C(1ru-u¯L(Ar)+r2u¯α|logu¯|βL(Ar))C(u¯L(Ar)+r2u¯α|logu¯|βL(Ar))on Br.

We regard (1.1) as -Δu=m(x)u in B1{0}, where m=uα-1|logu|β. In view of (2.5) and (2.1), we have that m(x)=O(|x|-2), so the Harnack inequality implies

supAruCinfAru.

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

|(u-u¯)|C(u¯+r2u¯α|logu¯|β)Cu¯on Br.(2.21)

Since

ψ(t,θ)-ψ¯(t)=r2α-1(log1r)βα-1(u(r,θ)-u¯(r)),

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

Corollary 2.5.

We have

𝕊n-1(ψα(t,θ)ζβ(t,θ)-ψ¯α(t)ζ¯β(t))𝑑θ=O(e-t)as t.(2.22)

Proof.

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

ζ(t,θ)-ζ¯(t)=1tlogψ(t,θ)ψ¯(t)

for any t>0 and θ𝕊n-1. Due to (2.18), we have

ζ(t,θ)-ζ¯(t)=1tlog(1+O(e-t))=O(e-tt)as t.

Using the above estimate together with Lemma 2.4, we have

ψα(t,θ)ζβ(t,θ)-ψ¯α(t)ζ¯β(t)=(ψα(t,θ)-ψ¯α(t))ζβ(t,θ)+ψ¯α(t)(ζβ(t,θ)-ζ¯β(t))=(ψ(t,θ)-ψ¯(t))O(1)+(ζ(t,θ)-ζ¯(t))O(1)=O(e-t) as t.

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

Lemma 2.6.

We have either

limtψ¯(t)=0

or

limtψ¯(t)=A,(2.23)

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

12[ψ¯2]tT+tTaψ¯2𝑑s-12tTb(ψ¯2)𝑑s+1α+1tTζ¯β(ψ¯α+1)𝑑s+O(e-t)=0.

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

tTb(ψ¯2)𝑑s=[bψ¯2]tT-tTbψ¯2𝑑s=O(1+tTdss2)=O(1).(2.24)

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

tTζ¯β(ψ¯α+1)𝑑s=O(1+tTlogss2𝑑s)=O(1).(2.25)

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

tTψ¯2𝑑s=O(1).(2.26)

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

limtψ¯(t)=0.(2.27)

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

tT(ψ¯′′)2𝑑s+12tTa(ψ¯2)𝑑s-tTbψ¯ψ¯′′𝑑s+tTζ¯βψ¯αψ¯′′𝑑s+O(e-t)=0,(2.28)

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

tTa(ψ¯2)𝑑s=[aψ¯2]tT-tTaψ¯2𝑑s=O(1).(2.29)

On the other hand, since ψ¯ψ¯′′=12(ψ¯2)′′-ψ¯2, a further integration by parts produces

tTbψ¯ψ¯′′𝑑s=12[b(ψ¯2)]tT-tT(12b(ψ¯2)+bψ¯2)𝑑s=O(1),(2.30)

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

tTζ¯βψ¯αψ¯′′𝑑s=O(1).(2.31)

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

tT(ψ¯′′)2𝑑s=O(1).(2.32)

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

limtψ¯′′(t)=0.(2.33)

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

limtb(t)=2α-1(n-2-2α-1),

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

limtζ(t)=2α-1.

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=tj, after passing to the limit in (2.17), with t=tj, we obtain, from (2.22), (2.27) and (2.33), that

2α-1(n-2-2α-1)ψ¯0-(2α-1)βψ¯0α=0.

Thus, in view of (1.4), we have

ψ¯0=0orψ¯0=A.(2.34)

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 tj and tj such that ψ¯(tj)0 and ψ¯(tj)A, then by the intermediate value theorem, there must exist some other tj′′ such that ψ¯(tj′′)A2, which violates (2.34). Thus, the proof is completed. ∎

We are now in a position to prove Theorem 1.1.

Proof of Theorem 1.1.

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

u¯(r)=A(1+o(1))r-2α-1(log1r)-βα-1as r0.

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

limtψ¯(t)=0.

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

ψ¯′′+(a0+o(1))ψ¯-(b0+o(1))ψ¯+ζ¯βψ¯α+𝕊n-1(ζβψα-ζ¯βψ¯α)𝑑θ=0,

with

a0=4α-1-n+2andb0=2α-1(n-2-2α-1).

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

λ2+a0λ-b0=0.

Since a02+4b0=(n-2)2, the root λ is

λ=-12(a0+a02+4b0)=-2α-1.

Therefore, we have

ψ¯(t)=O(e-2tα-1)as t.

In view of (2.15), we obtain

u¯(r)=(log1r)-βα-1O(1)as r0.(2.35)

Now if β>0, then we deduce from (2.35) that u¯(r)0 as r0, from which, combined with (2.1), it follows that u(x)0 as x0. 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

B1uq𝑑xCB1(log1|x|)-βqα-1𝑑xC01rn-1(log1r)-βqα-1𝑑rC

for each q1, for some constant C>0 depending on n, α, β and q. Therefore, uα|logu|βLp(B1) for any p1, and in particular for p>n. This implies that ΔuLp(B1) for p>n, so uC1,α(B1/2) for α=1-np, proving again that the origin is a removable singularity. Thus, the proof of Theorem 1.1 (i) is completed. ∎

Acknowledgements

This work was initiated in June 2017 when M. Ghergu was visiting the Royal Institute of Technology (KTH) in Stocholm. The invitation and hospitality of the Department of Mathematics in KTH in greatly acknowledged.

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Footnotes

  • 1

    w is known as the Lambert W-function. 

About the article

Received: 2017-11-16

Accepted: 2017-11-21

Published Online: 2018-06-14


Funding Source: National Research Foundation of Korea

Award identifier / Grant number: NRF-2014-Fostering Core Leaders of the Future Basi

Funding Source: Swedish Foundation for International Cooperation in Research and Higher Education

Award identifier / Grant number: CH2015-6380

S. Kim has been supported by National Research Foundation of Korea (NRF) grant funded by the Korean government (NRF-2014-Fostering Core Leaders of the Future Basic Science Program). H. Shahgholian has been supported in part by Swedish Research Council.


Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 995–1003, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2017-0261.

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