Singular solutions of elliptic equations with iterated exponentials
Marius Ghergu, Olivier Goubet

TL;DR
This paper constructs positive singular solutions for a class of elliptic equations involving iterated exponentials, extending previous work to more complex nonlinearities and analyzing their behavior near singularities.
Contribution
It introduces a method to construct and analyze singular solutions for elliptic equations with iterated exponential nonlinearities, expanding the understanding of such equations.
Findings
Constructed positive singular solutions with prescribed behavior.
Extended previous results to iterated exponential nonlinearities.
Provided a framework for analyzing solutions near singularities.
Abstract
We construct positive singular solutions for the problem in (), on , having a prescribed behaviour around the origin. Our study extends the one in Y. Miyamoto [Y. Miyamoto, A limit equation and bifurcation diagrams of semilinear elliptic equations with general supercritical growth. J. Differential Equations \textbf{264} (2018), 2684--2707] for such nonlinearities. Our approach is then carried out to elliptic equations featuring iterated exponentials.
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Singular solutions of elliptic equations with iterated exponentials
Marius Ghergu
Institute of Mathematics “Simion Stoilow” of the Romanian Academy, P.O. Box 1-764, 014700 Bucharest, Romania
School of Mathematics and Statistics, University College Dublin, Belfield, Dublin 4, Ireland
and
Olivier Goubet
Laboratoire Amiénois de Mathématiques Fondamentales et Appliquée, (LAMFA UMR 7352 CNRS UPJV) 33 rue saint-Leu, Université de Picardie Jules Verne 80039 Amiens France
Abstract.
We construct positive singular solutions for the problem in (), on , having a prescribed behaviour around the origin. Our study extends the one in Y. Miyamoto [Y. Miyamoto, A limit equation and bifurcation diagrams of semilinear elliptic equations with general supercritical growth. J. Differential Equations 264 (2018), 2684–2707] for such nonlinearities. Our approach is then carried out to elliptic equations featuring iterated exponentials.
Key words and phrases:
Singular solutions, prescribed singularity, iterated exponentials
2010 Mathematics Subject Classification:
Primary 35J61, 35J75; Secondary 35B40
1. Introduction and the main results
Consider the problem
[TABLE]
where is the open unit ball, is a real number and
[TABLE]
The related problem, also known as the Gelfand problem, namely
[TABLE]
has been long investigated starting with J. Liouville since 1853 (see [16]). One particular feature of (1.2) is that for it has the explicit singular solution . Joseph and Lundgren [14] completely determined the structure of the radial solutions of (1.2) emphasizing the role of the singular solution in the global picture of the solution set to (1.2). Thanks to the standard Hardy inequality, the explicit singular solution is stable for all space dimensions . Further studies related to (1.2) are contained in [4, 11, 20, 17, 18, 21] and in the monograph [6]. Problems with exponential nonlinearities also appear in other contexts involving higher order operators [1, 2, 3, 7], -Laplace operators [8] or -Hessian operators [12, 13] or even systems of coupled equations [5, 10].
Returning to (1.1) we point out that such a problem does not possess an explicit singular solution. However, we are able to construct a radial singular solution with a prescribed behaviour around the origin. We prove:
Theorem 1.1**.**
There exists a unique such that (1.1) has a singular solution such that, as we have
[TABLE]
and
[TABLE]
Letting and using the Maclaurin series approximation we may re-write (1.3) as
[TABLE]
as .
The related problem
[TABLE]
was recently studied in [15]. It is proved in [15] that (1.6) has a singular solution that satisfies
[TABLE]
Also the Morse index of is infinite (resp finite) provided (resp. ).
We would like to point out that in [19] a positive radial singular solution of
[TABLE]
is constructed. Such a singular solution has the property that
[TABLE]
where
[TABLE]
We are able to show that the solution of (1.7) coincides with in a neighbourhood of the origin. Thus, we may further investigate the bifurcation problem
[TABLE]
By the classical result of Gidas, Ni and Nirenberg [9] all solutions of (1.10) are radially symmetric. Furthermore (see [19]) the solution set of (1.10) can be described as where and . Hence, the solution set is a curve emanating from . Using [19, Theorem 1.1, Corollary 1.2, Corollary 1.3] we have:
Theorem 1.2**.**
Let be the solution obtained in Theorem 1.1.
- (i)
If , then the Morse index of is finite; 2. (ii)
If , then the Morse index of is infinite. Furthermore:
- (ii1)
The curve has infinitely many turning points around . In particular problem (1.10) has infinitely many solutions for ; 2. (ii2)
The number of intersection points between and the singular solution tends to infinity as .
We also address in this article the similar problem with iterared exponential, that reads for and and
[TABLE]
For problem (1.11) we prove
Theorem 1.3**.**
Let and let be the iterated logarithm (). There exists a unique such that (1.11) has a singular solution such that, as we have, for ,
[TABLE]
and
[TABLE]
The next sections contain the proofs of the main results. Throughout this paper for any functions , defined in a neighbourhood of infinity, we use the notation (resp. as to indicate that is bounded (resp. tends to zero) as . A similar notation is used for . Also, the symbols , stand for generic positive constants whose values may be different on each occurence.
2. Proof of Theorem 1.1
Let be a radial solution of (1.1). Letting v(x)=u\big{(}\frac{x}{\sqrt{\lambda}}\big{)} we find
[TABLE]
Letting and we find that satisfies
[TABLE]
We next look for a solution of (2.2) in the form
[TABLE]
where
[TABLE]
Let us observe that, as , we have
[TABLE]
Letting f=\ln\big{(}2t+\varphi(t)\big{)} we have
[TABLE]
Also,
[TABLE]
Observe that
[TABLE]
so by (2.4) and the above calculations we find
[TABLE]
Using equation (2.3) we have
[TABLE]
so
[TABLE]
where
[TABLE]
Let also .
Using the first equation of (2.2) together with (2.3), (2.8)-(2.11) we deduce that satisfies
[TABLE]
where
[TABLE]
We shall show that equation (2.12) has a solution where is a real number and
[TABLE]
equipped with the norm . As in [15], we discuss in the following the case , the case being similar. We transform (2.12) into the fixed point equation
[TABLE]
where the integral operator is given by
[TABLE]
and . The existence of a solution to (2.14) will be derived by means of the contraction principle; to this end, for set
[TABLE]
Lemma 2.1**.**
There exist such that and is a contraction.
Proof.
[TABLE]
where . Let now . Then, for large, we estimate
[TABLE]
Thus, by taking and large enough we have
[TABLE]
Using this fact we have
[TABLE]
since . This shows that .
To prove that is a contraction, let . Then
[TABLE]
From (2.11) we have
[TABLE]
Using the Mean Value Theorem and (2.10) we estimate
[TABLE]
for some between and . Thus,
[TABLE]
Also, leting , by (2.9) and the Mean Value Theorem we estimate
[TABLE]
for some between and . By the Mean Value Theorem we further estimate
[TABLE]
where lies between and . Hence,
[TABLE]
Now, using the above estimates in (2.17) we deduce
[TABLE]
By taking now large enough it follows that is a contraction. ∎
We are now in a position to prove the result in Theorem 1.1. First, there exists a a solution of (2.14), that is, satisfies
[TABLE]
Thus, the function given by (2.3) is positive in a neighbourhood of infinity and satisfies
[TABLE]
We claim that vanish at some point . Otherwise, in and by the continuation principle satisfies (2.19) on the whole real line. We claim that this implies that is monotone increasing. Indeed, assuming the contrary, there would exist a minimum point at which and which contradicts (2.19). Hence, is monotone increasing and there exists .
Multiply in (2.19) by and using the fact that in increasing, we find
[TABLE]
Integrating in the above equality over the interval , , we find
[TABLE]
where . This implies further that which contradicts the fact that is finite. This shows that vanishes at some point and satisfies
[TABLE]
Letting and u^{*}(x)=v^{*}\big{(}\sqrt{\lambda^{*}}x\big{)} (where is the solution of (2.1) with ) we obtain that is a solution of (1.1) with which satisfies (1.3). Concerning the proof of the asymptotic behaviour in (1.5) we have
[TABLE]
as . Since
[TABLE]
we have
[TABLE]
Using (2.22)-(2.23) in (2.21) we obtain
[TABLE]
which proves (1.5).
We next focus on the expansion of the gradient around the origin.
Lemma 2.2**.**
The solution of (2.18) satisfies
[TABLE]
Proof.
From the proof of Lemma 2.1 we infer that
[TABLE]
Then is solution to the integral equation
[TABLE]
The result follows promptly. ∎
Now, from (2.3), (2.7) and (2.24) we have
[TABLE]
Recall that the singular solution is given by u^{*}\big{(}\frac{r}{\sqrt{\lambda^{*}}}\big{)}=v(r)=w(t) where and are solutions of (2.1) and (2.2) (with ). Since , from (2.25) we find
[TABLE]
3. Proof of Theorem 1.2
Let us recall that the solution of (1.7) is obtained in [19, Lemma 3.1] as
[TABLE]
Further, the unknowns and are found as a unique fixed point through a contraction mapping in the set
[TABLE]
From (3.1) we find
[TABLE]
Let now be the singular solution of (1.1) for constructed in Theorem 1.1. Let v^{*}(x)=u^{*}\big{(}\frac{x}{\sqrt{\lambda^{*}}}\big{)} which satisfies (2.1), that is the same equation as (1.7) in . Define
[TABLE]
In order to prove that in a neighbourhood of infinity it is enough to show that belongs to .
Letting where , we have that satisfies (2.2). Thus, (3.3) reads
[TABLE]
In order to conclude the proof it suffices to show that
[TABLE]
Then, for large we have so, by the uniqueness of the fixed point in a neighbourhood of infinity, that is in a neighbourhood of the origin and then we conclude from Theorem 1.1, Corollary 1.2, Corollary 1.3 from [19]. Let us now turn to (3.5).
From the definition of in (1.8) and L’Hospital’s rule we find
[TABLE]
Thus,
[TABLE]
Now, using (2.3) we find
[TABLE]
Since as it follows that
[TABLE]
which proves the first part of (3.5). For the second part, we first note that
[TABLE]
Thus, from the above arguments we find
[TABLE]
Using (2.3) and Lemma 2.2 we have
[TABLE]
We also have
[TABLE]
Since as , we find
[TABLE]
Combining (3.7), (3.8) and (3.9) we deduce which finishes our proof. ∎
4. Proof of Theorem 1.3
We follow the lines of the proof of Theorem 1.1. For , we plan to solve here the equation
[TABLE]
We look for a solution of (4.1) in the form
[TABLE]
where
[TABLE]
Here is the iterated logarithm function defined by . We then have to solve the equation
[TABLE]
where
[TABLE]
with
[TABLE]
Here is defined thanks to Taylor’s formula as
[TABLE]
The next result provides the estimates we need to construct our solution in the space defined in (2.16).
Lemma 4.1**.**
Let .
- (a)
For we have
[TABLE]
and
[TABLE] 2. (b)
Also
[TABLE] 3. (c)
For there holds
[TABLE]
Proof.
(a) The estimates (4.8)-(4.9) follow from the identities
[TABLE]
(b) We have
[TABLE]
From (4.9) we find
[TABLE]
Combining the above equality with (4.9) we find
[TABLE]
(c) The estimate (4.11) follows from the following computations
[TABLE]
∎
We seek a solution by Lemma 2.1 as in the proof of Theorem 1.1.
We begin with an upper bound for defined in (4.5). By Mean Value theorem we have that
[TABLE]
for some between and . Since is decreasing and is negative for large , from (4.16) and (4.9)-(4.10) we find
[TABLE]
We now bound defined in (4.4). First, by the estimates (4.8) we have
[TABLE]
Also, by the Mean Value Theorem and (4.8) we have
[TABLE]
for some between and . Hence, from (4.18)-(4.19) we find
[TABLE]
We now handle the nonlinear term defined in (4.6). Observe first that given in (4.6)-(4.7) can be written as
[TABLE]
Since and , a further adjustment of and Mean Value Theorem together with (4.8) yields
[TABLE]
Since is increasing, using (4.11) we deduce
[TABLE]
Let now and denote . Using that
[TABLE]
we have
[TABLE]
Since and are inverse each other, for we may write
[TABLE]
Thus
[TABLE]
By the Mean Value Theorem, (4.23) and (4.24), for some between and , we have
[TABLE]
Now, the second term in (4.25) is bounded from above by
[TABLE]
Let us observe first that from (4.22)-(4.24) and the Mean Value Theorem we have
[TABLE]
To bound from above the first term in (4.27) we have
[TABLE]
For , from (4.21) we have
[TABLE]
Then
[TABLE]
As before,
[TABLE]
Also, by the Mean Value Theorem, for some between and we have
[TABLE]
The last two estimates yield
[TABLE]
Finally, combining (4.25)-(4.30) we deduce
[TABLE]
Setting gives and we can appply the Lemma 2.1. We then have a unique solution of (4.3) in .
Proceeding as in the proof of Theorem 1.1, we have that there exists a such that and hence we have completed the proof of the existence and uniqueness result.
We next derive the asymptotic expansion (1.12). We begin with
[TABLE]
We have, using (4.9), that
[TABLE]
Then
[TABLE]
Finally, we use use the expansion (4.15) to deduce (1.12). We now prove the estimate for the gradient. Exactly as in the proof of Theorem 1.1, we have that . Then
[TABLE]
since . We use the expansion
[TABLE]
since . This completes the proof of Theorem 1.3, using that .
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