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On the KellerOsserman conjecture in one dimensional case
Boundary Value Problems volumeÂ 2016, ArticleÂ number:Â 162 (2016)
Abstract
A sufficient and necessary condition for existence of solution for the boundary blowup problem in one dimensional case is obtained. This problem can be seen as the KellerOsserman conjecture, which comes from the study on elliptic equations.
1 Introduction
The boundary blowup problem
where Î© is a bounded domain in \(R^{N}\) (\(N\geq2\)), arises in many fields, such as the theory of automorphic functions and Riemann surfaces of constant negative curvature, the study of the electric potential in a glowing hollow metal body, etc. In 1916, this type of problem was first studied by Bieberbach (see [1]). A very famous result is the following.
Theorem A
Assume that \(f(0)=0\), \(f^{\prime}(t)\) is continuous and \(f^{\prime}(t)\geq0\) for all \(t\geq0\), then (1.1) has a solution if and only if
The condition (1.2) plays an important role in the study of the boundary blowup problem, it was first proposed by Keller [2] and Osserman [3], thus lately, this condition was called the KellerOsserman condition. Also see [4] and [5]. For later investigations on boundary blowup problems we refer to [6â€“10] etc. An interesting problem is the following.
KellerOsserman conjecture
When \(N=1\), the conclusion of TheoremÂ A is also true.
For the KellerOsserman conjecture, see Anuradha et al. (see [11]) who first considered the following autonomous two point boundary value problem:
Using the quadratic method, they gave some necessary, sufficient conditions for the existence of nonnegative solutions. As they said, â€˜The gap between the class of functions that satisfy the necessary condition but not the sufficient condition is quite small.â€™ In their paper, they gave some examples to illustrate their conclusions.
Remark 1
Anuradha et al. [11], Remarks 3.1 and 3.2, showed that, to problem (1.3):

(i)
for \(f(u)=u(\ln u+1)\), there are no solutions,

(ii)
for \(f(u)=u [ 2 ( \ln u ) ^{3}+3(\ln u)^{2} ] \), there exist solutions for some \(\lambda>0\).
It is easy to verify that \(f(u)=u(\ln u+1)\) does not satisfy the KellerOsserman condition (1.2), but \(f(u)=u [ 2 ( \ln u ) ^{3}+3(\ln u)^{2} ] \) satisfies this KellerOsserman condition.
Recently, Wang [12] used the same method as Anuradhaâ€™s and generalized the results of Anuradhaâ€™s; he obtained a more suitable condition for the existence of solutions for (1.3). The gap between the class of functions that satisfy the necessary condition but not the sufficient condition becomes smaller than that of Anuradhaâ€™s, but we still have a distance to KellerOssermanâ€™s conjecture. Later Zhang [13] considered this problem again. In this paper, we want to study problem (1.3) again and solve this conjecture.
In this paper, we investigate the following boundary blowup problem:
where \(f(t)\geq0\), \(f(t)\) is continuous and monotone increasing for all \(t\geq0\). For convenience, we set \(u(10)=+\infty\) if \(\lim_{t\rightarrow 1^{}}u(t)=+\infty\), for short, denote \(u(1)\).
First we introduce the definition of regularly varying which can be found in [14].
Definition 1
A positive measurable function R defined on \([D,+\infty)\), for some \(D>0\), is called regularly varying (at infinity) with index \(q\in \mathbb{R} \), written \(R\in \mathbb{R} _{q}\), if for all \(\xi>0\),
When the index of regular variation q is zero, we say that the function is slowly varying.
Remark

(a)
Any function \(R\in \mathbb{R} _{q}\) can be written in terms of a slowly varying function. Indeed, set \(R(u)=u^{q}L(u)\), then L varies slowly.

(b)
For any \(m>0\), \(u^{m}L(u)\rightarrow+\infty\), \(u^{m}L(u)\rightarrow0\) as \(u\rightarrow+\infty\).
Here come our main results.
Theorem 1
If \(f(0)=0\), then equation (1.4) has a positive solution if and only if the KellerOsserman condition (1.2) holds true.
Theorem 2
If \(f(0)>0\), then equation (1.4) has a positive solution if and only if the KellerOsserman condition (1.2) and \(\max _{a\geq0}g(a)\geq\frac{\sqrt{2}}{2}\) hold true, where the function g is defined in (2.3).
Theorem 3
Assume that \(f(0)=0\) and \(f(t)\in \mathbb{R} _{q}\), \(q>1\). Then the solution \(u(t)\) of (1.4) satisfies
where
Theorem 4
Assume that \(f(0)=0\), \(\frac{f(u)}{u}\) is increasing on (\(0,+\infty\)) and \(f(t)\in \mathbb{R} _{q}\), \(q>1\). Then equation (1.4) has only one positive solution.
From TheoremÂ 1, we can easily obtain the following.
Corollary 1
If \(f(0)=0\), then for any \(\lambda>0\), equation (1.3) has a positive solution if and only if the KellerOsserman condition (1.2) holds true.
CorollaryÂ 1 generalizes TheoremÂ 3.1, TheoremÂ 3.2 and TheoremÂ 3.6 in [11], also generalizing TheoremÂ 2.1 and TheoremÂ 2.2 in [12].
2 The proof of main results
In order to prove our main results, we need some lemmas, presented below.
Lemma 1
If \(u(t)\) is a solution of (1.4), then \(u(1t)\) is also a solution of (1.4).
The proof of this lemma is trivial, we omit it here.
Lemma 2
If \(u(t)\) is a solution of (1.4), then there exists only one point \(t_{0}\in(0,1)\) such that \(u^{\prime}(t_{0})=0\), in fact, \(t_{0}=\frac{1}{2}\).
The proof of this lemma can be obtained by the generalized Rolle theorem. \(t_{0}=\frac{1}{2}\) can be obtained easily by LemmaÂ 1.
Lemma 3
If \(u(t)\in C^{2}[\frac{1}{2},1)\) is a solution of the following problem:
then
is a solution of (1.4), in the contrary case, the conclusion also holds true.
Proof
The proof is trivial, we omit it.â€ƒâ–¡
Lemma 4
Assume that the KellerOsserman condition holds true. If \(f(0)>0\), then the function
is well defined on the interval \([0,+\infty)\) and satisfies
If \(f(0)=0\), then the function \(g(a)\) is well defined on the interval \((0,+\infty)\) and satisfies
Proof
Notice that \(f(s)\) is increasing, thus
Therefore
set
where
By the KellerOsserman condition, we know that \(J(a,b)<+\infty\).
If \(f(0)>0\), then
which means that the function \(g(a)\) is well defined on the interval \([0,+\infty)\).
If \(f(0)=0\), notice that \(f^{\prime}(0)\) exists, thus there exist some positive constants c and k such that \(f(t)\leq kt\), \(c< b\) for \(t\in [0,c]\), therefore
which means that the function \(g(a)\) is well defined on the interval \((0,+\infty)\) (equation (2.4) holds for any \(a>0\)) and \(\lim_{a\rightarrow 0^{+}}g(a)=+\infty\).
From the KellerOsserman condition (1.2) and noticing that \(( \int_{0}^{t}f(s)\,ds ) ^{1/2}\) is monotone decreasing, we have
Hence
therefore
by (2.4), we get
Note that
then
therefore
â€ƒâ–¡
Lemma 5
If \(f(0)=0\), then equation (2.1) has a solution if and only if the KellerOsserman condition holds true.
Proof
Sufficiency. Conversely, suppose that there is no solution \(u(t)\). Given any real number a, let \(u(t)\) be a solution of the following problem:
Its maximal existence interval is \([\frac{1}{2},r_{a})\), where \(r_{a}\) such that \(\lim_{t\rightarrow r_{a}^{}}u(t)=+\infty\) and \(\frac{1}{2}< r_{a}\) and \(r_{a}\neq1\). Notice that (2.6) implies that
here we used the inequality \(u^{\prime}(t)\geq0\) for \(t\in[ \frac{1}{2},r_{a})\). From (2.7), we have
By LemmaÂ 4 and the zero theorem, we know that there exists \(a_{0}>0\) such that
then (2.8) implies that \(r_{a_{0}}=1\), which is a contradiction.
Necessity. Assume that
then for any \(a>0\), we have \(g(a)=+\infty\), by (2.8), we know that equation (2.1) has no solution, which is a contradiction.â€ƒâ–¡
From (2.8) and LemmaÂ 4, we can easily obtain the following.
Lemma 6
If \(f(0)>0\), then equation (2.1) has a solution if and only if the KellerOsserman condition and \(\max_{a\geq0}g(a)\geq \frac{\sqrt{2}}{2}\) are all satisfied.
Proof of TheoremÂ 1
From LemmaÂ 3 and LemmaÂ 5, the conclusion of TheoremÂ 1 is obvious.â€ƒâ–¡
Proof of TheoremÂ 2
3 Asymptotical stability
In this section, we want to study the behavior of large solutions near the boundary and the uniqueness of the solution. We start with the following comparison theorem.
Lemma 7
(Comparison theorem)
Assume that \(f(t)\geq 0\), \(f(t): \mathbb{R} \rightarrow \mathbb{R} \) is continuous and monotone increasing. Let \(u(t),v(t)\in C^{2}[\frac {1}{2},1)\) such that
then \(u(t)\geq v(t)\) for all \(t\in[\frac{1}{2},1]\).
Proof
Assume that it is false, then there exists some \(t_{0}< t_{1}\) in \([\frac{1}{2},1)\) such that \(u(t_{0})< v(t_{0})\), \(u^{\prime }(t_{0})=v^{\prime}(t_{0})\), \(u(t)\leq v(t)\) and \(u^{\prime }(t)\geq v^{\prime}(t)\) for \(t\in[ t_{0},t_{1}]\). Let \(w(t)=u(t)v(t)\), \(t\in [ t_{0},t_{1}]\), then (3.1) and (3.2) imply that
Multiplying \(w^{\prime}(t)\) to both sides of (3.3), we have
integrating from \(t_{0}\) to t (\(t\in[ t_{0},t_{1}] \)), we get
which is a contradiction.â€ƒâ–¡
Remark
In LemmaÂ 7, if we replace \(\frac{1}{2}\) and 1 by any other real numbers a and b (\(a< b\)), then the conclusion is also true.
Corollary 2
(Comparison theorem)
Assume that \(f: \mathbb{R} \rightarrow \mathbb{R} \) is continuous and monotone increasing with \(f(0)=0\). Let \(u(t)\in C^{2}[\frac{1}{2},1)\) such that
then \(u(t)\geq0\) for all \(t\in[\frac{1}{2},1]\).
Proof
In LemmaÂ 7, let \(v(t)\equiv0\), then we can immediately get the conclusion.â€ƒâ–¡
Proof of TheoremÂ 3
First, we claim that if \(h\in \mathbb{R} _{p}\) is continuous and nonnegative on \([ a,+\infty ) \), here \(p<1\), then \(h\in L^{1}( [ a,+\infty ) )\). To see this, suppose that \(h\notin L^{1}( [ a,+\infty ) )\), then \(\lim_{s\rightarrow +\infty}H(s)=\int_{a}^{+\infty}h(s)\,ds=\infty\), by the Lâ€™Hospital rule, we have
Notice that H is increasing, it follows that \(\xi\rightarrow\xi^{p+1}\) is increasing as well, therefore \(p\geq1\), which contradicts \(p<1\). \(f(t)\in \mathbb{R} _{q}\) implies \(F(t)\in \mathbb{R} _{q+1}\) (by the Lâ€™Hospital rule), hence \(F^{1/2}(t)\in R_{\frac{q+1}{2}}\). Notice that \(q>1\) implies \(\frac{q+1}{2}<1\), then the above claim shows that \(F^{1/2}(t)\in L^{1}( [ a,+\infty ) )\), that is, the KellerOsserman condition (1.2) holds true, then from TheoremÂ 1, we know that the solution of (1.4) exists, by CorollaryÂ 2, these solutions (this solution) \(u(t)\) satisfy (satisfies) \(u(t)\geq0\). Then (2.6) implies that
From (3.4), we have
thus
for any boundary blowup solution \(u(t)\) of (1.4), we can easily obtain
Then (3.5) implies that
therefore
where \(\beta(t)\) satisfies
In order to prove (1.5), we only need to prove
From (3.6), we see, for any \(\varepsilon>0\), that there exists some \(\delta <1/2\) such that
Notice that
thus
which implies that
In the following, we prove that \(\phi(t)\in \mathbb{R} _{2/(1q)}\) (at zero) provided that \(f(t)\in \mathbb{R} _{q}\). In fact, if \(f(t)\in \mathbb{R} _{q}\), then
which implies that
therefore
thus
So
letting \(\varepsilon\rightarrow0^{+}\), and then by (3.8), (3.7) holds true.â€ƒâ–¡
Proof of TheoremÂ 4
Let \(u(t)\), \(v(t)\) be the solutions of (1.4), by (1.5), we have
also notice that \(f(t)\in \mathbb{R} _{q}\), \(q>1\). Then for any \(\varepsilon>0\), there exists \(\delta>0\) such that
set \(\overline{u}(t)=(1+\varepsilon)v(t)\), then
We claim that for any \(t\in[\frac{1}{2},1)\), \(u(t)\leq\overline {u}(t)\). Notice that
from LemmaÂ 7, the above claim is obvious. Similarly, we can prove that for any \(t\in[\frac{1}{2},1)\), \(u(t)\geq\underline{u}(t)\), where \(\underline{u}(t)=(1\varepsilon)\) \(v(t)\). Passing to the limit \(\varepsilon \rightarrow0^{+}\), we conclude that \(u(t)\equiv v(t)\).â€ƒâ–¡
Remark
From the proof of TheoremÂ 3 and TheoremÂ 4, we can easily see that the condition \(\frac{f(u)}{u}\) is increasing on \((0,+\infty)\) and \(f(t)\in \mathbb{R} _{q}\), \(q>1\) can be replaced by the following.

(H)
There exists \(q>1\) such that for any \(\varepsilon>0\) sufficiently small, the function f satisfies
$$ f \bigl[ (1+\varepsilon)s \bigr] \geq(1+\varepsilon)^{q}f(s)\quad \mbox{and}\quad f \bigl[ (1\varepsilon)s \bigr] \leq(1 \varepsilon)^{q}f(s),\quad s\in (0,+\infty). $$
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Acknowledgements
The authors thank the reviewers for their insightful and detailed comments. This work is supported by NSF of China (11201213, 11371183), NSF of Shandong Province (ZR2015AM026, ZR2013AM004) and the Project of Shandong Provincial Higher Educational Science and Technology (J15LI07).
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Wang, LL., Fan, YH. On the KellerOsserman conjecture in one dimensional case. Bound Value Probl 2016, 162 (2016). https://doi.org/10.1186/s1366101606677
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DOI: https://doi.org/10.1186/s1366101606677