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Corrigendum to “Bifurcation and admissible solutions for the Hessian equation” [J. Funct. Anal. 273 (2017) 3200–3240]
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[m1L; v1.254; Prn:13/03/2019; 11:46] P.1 (1-3) Journal of Functional Analysis ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Functional Analysis www.elsevier.com/locate/jfa
Corrigendum
Corrigendum to “Bifurcation and admissible solutions for the Hessian equation” [J. Funct. Anal. 273 (2017) 3200–3240] ✩ Guowei Dai School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, PR China
a r t i c l e
i n f o
Article history: Received 7 February 2019 Accepted 22 February 2019 Available online xxxx Communicated by H. Brezis
a b s t r a c t The uniqueness obtained in [1] may not right if f (s)/sk is only decreasing for s > 0. We correct this error by requiring f (s)/sk is “strictly” decreasing for s > 0. © 2019 Elsevier Inc. All rights reserved.
MSC: 34A12 34C23 35J60 Keywords: Uniqueness Strict monotonicity Bifurcation
DOI of original article: https://doi.org/10.1016/j.jfa.2017.08.001. Research supported by NNSF of China (No. 11871129) and Xinghai Youqing funds from Dalian University of Technology. E-mail address: [email protected]. ✩
https://doi.org/10.1016/j.jfa.2019.02.020 0022-1236/© 2019 Elsevier Inc. All rights reserved.
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2
1. Main results Consider the following k-Hessian equation
Sk D2 u = λk f (−u) u=0
in B, on ∂B.
(1.1)
Global bifurcation result is established in [1]. However, the proof of Theorem 7.2 in [1] is incorrect. Another stronger condition must be added to the hypothesis: Theorem 7.2. Let f ∈ C 1 (R+ , R+ ) such that f (s)/sk is strictly decreasing for s > 0. Then (i) if f0 ∈ (0, +∞) and f∞ = 0, problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (λ1 /f0 , +∞) and has only the trivial radially symmetric solution for any λ ∈ (0, λ1 /f0 ]; (ii) if f0 ∈ (0, +∞) and f∞ ∈ (0, +∞) with f∞ = f0 , problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (λ1 /f0 , λ1 /f∞ ) and has only the trivial radially symmetric solution for any λ ∈ (0, λ1 /f0 ] ∪ [λ1 /f∞ , +∞); (iii) if f0 = +∞ and f∞ ∈ (0, +∞), problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (0, λ1 /f∞ ) and has only the trivial radially symmetric solution for any λ ∈ [λ1 /f∞ , +∞]; (iv) if f0 = +∞ and f∞ = 0, problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (0, +∞). Proof. We only give the proof of (i) since the arguments of (ii)–(iv) are completely similar. The existence of a connected curve C bifurcating (λ1 /f0 , +∞) has been obtained in [1]. And uλ with uλ = −vλ is decreasing with respect to λ for any (λ, vλ ) ∈ C. Furthermore, since vλ is differentiable with respect to λ (as a consequence of Implicit duλ λ Function Theorem), one has that dv dλ ≥ 0. Therefore, we have dλ ≤ 0. It suffices to show the uniqueness of radially symmetric k-admissible of problem (1.1). For λ ∈ (λ1 /f0 , +∞), assume that there exist two solutions v1 and v2 with v1 ∈ C. If v2 ∈ / C, it has been shown in [1] that v2 = v1 . If v2 ∈ C, since C is a connected curve, there exists t ∈ [0, +∞) such that v2 = tv1 . Substituting v1 and v2 into problem (1.3) of [1], we have that
k
rN −k (−v1 )
= λk
N N −1 r f (v1 ) , r ∈ (0, 1) k CN
and N k tk rN −k (−v1 ) = λk k rN −1 f (tv1 ) , r ∈ (0, 1). CN
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It follows that f (tv1 ) f (v1 ) = . tk v1k v1k The strict monotonicity of f (s)/sk implies that t = 1. So, we still get v2 = v1 .
2
Note that the uniqueness may not right if the strict monotonicity of f (s)/sk is removed. For example, consider
Sk D2 u = λk (−u)k (1 + h(u)) u=0
in B, on ∂B,
where h(s) =
0 s2 − d2
if |s| ≤ d, if |s| > d
for some d > 0. For this example, it is not hard to see that C is vertical the trivial solution axes in a finite distance. So, it possesses infinitely many solutions for λ = λ1 . Here we also point out that there exists a sign error in the Picone-Hessian identity [1, identity (2.2)]. The corrected form should be
k uk+1 rN −k (−u2 ) N −k k 1 − u1 r (−u1 ) = uk2
k+1 k −u2 −u1 u2 k+1 N −k k+1 k wu1 + r +k + (k + 1)u1 u1 (−u1 ) . u2 u2 References [1] G. Dai, Bifurcation and admissible solutions for the Hessian equation, J. Funct. Anal. 273 (2017) 3200–3240.
AID:8199 /ERR
[m1L; v1.254; Prn:13/03/2019; 11:46] P.1 (1-3) Journal of Functional Analysis ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Functional Analysis www.elsevier.com/locate/jfa
Corrigendum
Corrigendum to “Bifurcation and admissible solutions for the Hessian equation” [J. Funct. Anal. 273 (2017) 3200–3240] ✩ Guowei Dai School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, PR China
a r t i c l e
i n f o
Article history: Received 7 February 2019 Accepted 22 February 2019 Available online xxxx Communicated by H. Brezis
a b s t r a c t The uniqueness obtained in [1] may not right if f (s)/sk is only decreasing for s > 0. We correct this error by requiring f (s)/sk is “strictly” decreasing for s > 0. © 2019 Elsevier Inc. All rights reserved.
MSC: 34A12 34C23 35J60 Keywords: Uniqueness Strict monotonicity Bifurcation
DOI of original article: https://doi.org/10.1016/j.jfa.2017.08.001. Research supported by NNSF of China (No. 11871129) and Xinghai Youqing funds from Dalian University of Technology. E-mail address: [email protected]. ✩
https://doi.org/10.1016/j.jfa.2019.02.020 0022-1236/© 2019 Elsevier Inc. All rights reserved.
JID:YJFAN
AID:8199 /ERR
[m1L; v1.254; Prn:13/03/2019; 11:46] P.2 (1-3)
G. Dai / Journal of Functional Analysis ••• (••••) •••–•••
2
1. Main results Consider the following k-Hessian equation
Sk D2 u = λk f (−u) u=0
in B, on ∂B.
(1.1)
Global bifurcation result is established in [1]. However, the proof of Theorem 7.2 in [1] is incorrect. Another stronger condition must be added to the hypothesis: Theorem 7.2. Let f ∈ C 1 (R+ , R+ ) such that f (s)/sk is strictly decreasing for s > 0. Then (i) if f0 ∈ (0, +∞) and f∞ = 0, problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (λ1 /f0 , +∞) and has only the trivial radially symmetric solution for any λ ∈ (0, λ1 /f0 ]; (ii) if f0 ∈ (0, +∞) and f∞ ∈ (0, +∞) with f∞ = f0 , problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (λ1 /f0 , λ1 /f∞ ) and has only the trivial radially symmetric solution for any λ ∈ (0, λ1 /f0 ] ∪ [λ1 /f∞ , +∞); (iii) if f0 = +∞ and f∞ ∈ (0, +∞), problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (0, λ1 /f∞ ) and has only the trivial radially symmetric solution for any λ ∈ [λ1 /f∞ , +∞]; (iv) if f0 = +∞ and f∞ = 0, problem (1.1) has a unique radially symmetric k-admissible solution for any λ ∈ (0, +∞). Proof. We only give the proof of (i) since the arguments of (ii)–(iv) are completely similar. The existence of a connected curve C bifurcating (λ1 /f0 , +∞) has been obtained in [1]. And uλ with uλ = −vλ is decreasing with respect to λ for any (λ, vλ ) ∈ C. Furthermore, since vλ is differentiable with respect to λ (as a consequence of Implicit duλ λ Function Theorem), one has that dv dλ ≥ 0. Therefore, we have dλ ≤ 0. It suffices to show the uniqueness of radially symmetric k-admissible of problem (1.1). For λ ∈ (λ1 /f0 , +∞), assume that there exist two solutions v1 and v2 with v1 ∈ C. If v2 ∈ / C, it has been shown in [1] that v2 = v1 . If v2 ∈ C, since C is a connected curve, there exists t ∈ [0, +∞) such that v2 = tv1 . Substituting v1 and v2 into problem (1.3) of [1], we have that
k
rN −k (−v1 )
= λk
N N −1 r f (v1 ) , r ∈ (0, 1) k CN
and N k tk rN −k (−v1 ) = λk k rN −1 f (tv1 ) , r ∈ (0, 1). CN
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[m1L; v1.254; Prn:13/03/2019; 11:46] P.3 (1-3)
G. Dai / Journal of Functional Analysis ••• (••••) •••–•••
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It follows that f (tv1 ) f (v1 ) = . tk v1k v1k The strict monotonicity of f (s)/sk implies that t = 1. So, we still get v2 = v1 .
2
Note that the uniqueness may not right if the strict monotonicity of f (s)/sk is removed. For example, consider
Sk D2 u = λk (−u)k (1 + h(u)) u=0
in B, on ∂B,
where h(s) =
0 s2 − d2
if |s| ≤ d, if |s| > d
for some d > 0. For this example, it is not hard to see that C is vertical the trivial solution axes in a finite distance. So, it possesses infinitely many solutions for λ = λ1 . Here we also point out that there exists a sign error in the Picone-Hessian identity [1, identity (2.2)]. The corrected form should be
k uk+1 rN −k (−u2 ) N −k k 1 − u1 r (−u1 ) = uk2
k+1 k −u2 −u1 u2 k+1 N −k k+1 k wu1 + r +k + (k + 1)u1 u1 (−u1 ) . u2 u2 References [1] G. Dai, Bifurcation and admissible solutions for the Hessian equation, J. Funct. Anal. 273 (2017) 3200–3240.