Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 50, pp. 1–11.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: 10.58997/ejde.2025.50

POSITIVE SOLUTIONS FOR GENERALIZED HARDY-HÉNON EQUATIONS

XIZHENG ZHANG, XIYOU CHENG, MEIHUA YANG

Abstract. This article concerns a generalized Hardy-Hénon equation and its associated Dirich-

let problem. We obtain upper and lower estimates for positive solutions, and establish the ex-

istence and nonexistence of positive solutions to both the equation and its associated Dirichlet
problem under certain parametric conditions.

1. Introduction

We consider the nonlinear elliptic equation

−∆u = [w(|x|)(1− |x|)]αup, x ∈ B1(0), (1.1)

and its corresponding Dirichlet problem

−∆u = [w(|x|)(1− |x|)]αup, x ∈ B1(0),

u = 0, |x| = 1,
(1.2)

where B1(0) ⊂ RN (N ≥ 3) is a ball of radius 1 centered at 0, p ̸= 1, α ∈ R and w ∈ C1([0, 1],R+
0 )

with R+
0 = (0,+∞). Equation (1.1) is usually called the generalized boundary Hardy-Hénon

equation because of the presence of weight function [w(|x|)(1− |x|)]α. In particular, when w ≡ 1,
equation (1.1) is the so-called boundary Hardy-Hénon equation [7]. Let us briefly recall some
relevant studies on the elliptic equation

−∆u = a(x)up, in Ω, (1.3)

where Ω ⊂ RN is a domain. When a ≡ 1, equation (1.3) is the Lane-Emden equation [4, 12].
When a(x) = |x|α and 0 ∈ Ω, equation (1.3) is called the Hardy-Hénon equation [7, 16]. For the
case α > −2 and p < (N + 2 + 2α)/(N − 2), Phan-Souplet [16] showed that equation (1.3) with
Ω = RN has no positive radial solutions. When α ≤ −2 and p > 1, Dancer-Du-Guo [8] proved that
the Hardy-Hénon equation (1.3) has no positive solutions in any domain Ω containing the origin.
For the case p < 0 and α > −2, Du-Guo [10] investigated the stable positive solutions of equation
(1.3). For the case a(x) = |x|α and Ω = B1(0), Cao-Peng-Yan [3] analyzed the profile of ground
state solutions and the existence of multi-peaked solutions with the Dirichlet boundary condition.
Du [9] established the existence, uniqueness and blow-up rate of large solutions of equation (1.3).
Cheng-Wei-Zhang [7] explored the estimates, existence and nonexistence of positive solutions to
equation (1.3) for the case a(x) = (1 − |x|)α and Ω = B1(0). For elliptic equations with the
Hardy potential, some profound results on the existence, nonexistence, and asymptotic behavior
of positive solutions were presented in [1, 2, 5, 6, 15] and the references therein.

The goal of this article is to establish the estimate and nonexistence of positive solutions to
(1.1) and to present the nonexistence, existence and uniqueness of positive solutions of (1.2). The
rest of this paper is organized as follows. We study positive solutions of (1.1) and (1.2) for the
case 1 < p < (N + 2)/(N − 2) in Section 2 and for the case p < 1 in Section 3, respectively.

2020 Mathematics Subject Classification. 35J60, 35B09, 35B45.

Key words and phrases. Hardy-Hénon equation; sub-super solution method; mountain pass theorem; singularity.
©2025. This work is licensed under a CC BY 4.0 license.
Submitted December 11, 2024. Published May 14, 2025.

1



2 X. ZHANG, X. CHENG, M. YANG EJDE-2025/50

2. Results for the case p > 1

Throughout this article, for simplicity, we denote maxt∈[0,1] |w′(t)| by |w′|∞, and denote Br(0)

by Br, and the closure of Br(0) by Br, for any r > 0. We start with the upper estimate of positive
solutions of equation (1.1).

Theorem 2.1. If 1 < p < (N +2)/(N − 2), for any positive solution u of (1.1) in B1 there exists
C = C(N, p, α, d, |w′|∞) > 0 such that

u(x) ≤ C[w(|x|)(1− |x|)]−
2+α
p−1 , x ∈ B1. (2.1)

For α ≤ −2, we obtain the following result.

Theorem 2.2. (i) If p > 1 and α ≤ −2, then (1.1) has no positive solutions with a positive
lower bound.

(ii) If p > 1 and α+ p+ 2 ≤ 0, then (1.1) has no positive solutions.
(iii) If 1 < p < (N + 2)/(N − 2) and α+ p+ 1 < 0, then (1.1) has no positive solutions.

For (1.2), when 1 < p < (N + 2)/(N − 2) and α > −2, we have the following result.

Theorem 2.3. If 1 < p < (N + 2)/(N − 2) and α > −2, then (1.2) has a positive solution.

To prove Theorems 2.1-2.3, we need the following two technical lemmas.

Lemma 2.4 ([16]). If N ≥ 3, 1 < p < (N + 2)/(N − 2), µ ∈ (0, 1] and a(x) ∈ Cµ(B1) satisfies

∥a∥Cµ(B1)
≤ C1 and a(x) ≥ C2, x ∈ B1,

for some constants C1, C2 > 0, then there exists C > 0 depending only on N, p, µ, C1, C2 such
that for any nonnegative classical solution u of

−∆u = a(x)up, x ∈ B1,

it holds

|u(x)|
p−1
2 + |∇u(x)|

p−1
p+1 ≤ C(1 +

1

1− |x|
), x ∈ B1. (2.2)

Lemma 2.5. If 1 < p < (N +2)/(N −2), then there exists C = C(N, p, α, d, |w′|∞) > 0 such that
any nonnegative solution u of (1.1) satisfies

u(x) ≤ C[w(|x|)(1− |x|)]−
2+α
p−1 and |∇u(x)| ≤ C[w(|x|)(1− |x|)]−

p+1+α
p−1 , x ∈ B1\B1/2. (2.3)

Proof. Let x0 ∈ B1. Then y := x0 + c(x0)x/2 ∈ B1 for all x ∈ B1, where c(x) = w(|x|)(1 − |x|).
Let U(x) = c(x0)

2+α
p−1 u(x0 + c(x0)x/2), x ∈ B1. Then U satisfies

−∆U = a(x;x0)U
p ∀x ∈ B1, where a(x;x0) =

c(y)α

4c(x0)α
.

For x0 ∈ B1, it follows that

c(y)

c(x0)
≥

d(1− |x0 +
w(|x0|)(1−|x0|)

2 x|)
1− |x0|

≥
d(1− |x0| − w(|x0|)(1−|x0|)

2 )

1− |x0|
≥ d

2
> 0

and

c(y)

c(x0)
≤

(1− |x0 +
w(|x0|)(1−|x0|)

2 x|)
d(1− |x0|)

≤
(1− |x0|+ w(|x0|)(1−|x0|)

2 )

d(1− |x0|)
≤ 3

2d
.

Thus, for x, x0 ∈ B1 it holds(d
2

)α

≤ 4a(x;x0) ≤
( 3

2d

)α

, as α ≥ 0,( 3

2d

)α

≤ 4a(x;x0) ≤
(d
2

)α

, as α < 0.



EJDE-2025/50 GENERALIZED HARDY-HÉNON EQUATIONS 3

We claim that ∥a(·;x0)∥C1(B1)
≤ C for x0 ∈ B1\B1/2, where C depends only on d, α and |w′|∞.

In fact, using

|Dia(x;x0)| =
∣∣∣α
8
(
c(y)

c(x0)
)α−1 x

i
0 +

w(|x0|)(1−|x0|)
2 xi

|x0 +
w(|x0|)(1−|x0|)

2 x|
[w′(|y|)(1− |y|)− w(|y|)]

∣∣∣,
where xi and xi

0 denote the i-th component of x and x0, for x ∈ B1 and x0 ∈ B1\B1/2, we have

|Dia(x;x0)| ≤

{
α
8 (|w

′|∞ + 1)
(

3
2d

)α−1
, as α ≥ 1,

|α|
8 (|w′|∞ + 1)

(
d
2

)α−1
, as α < 1.

In view of Lemma 2.4, we have

|U(x)|
p−1
2 + |∇U(x)|

p−1
p+1 ≤ C(1 +

1

1− |x|
), x ∈ B1.

Let x = 0. Then we deduce

|U(0)|
p−1
2 + |∇U(0)|

p−1
p+1 ≤ C,

U(0) = c(x0)
2+α
p−1 u(x0) = [w(|x0|)(1− |x0|)]

2+α
p−1 u(x0) ≥ 0.

So we have

U(0) + |∇U(0)| ≤ C,

which implies that for any x0 ∈ B1\B1/2 it holds

u(x0) ≤ C[w(|x0|)(1− |x0|)]−
2+α
p−1 ,

|∇u(x0)| ≤ C[w(|x0|)(1− |x0|)]−
p+α+1
p−1 .

By the arbitrariness of x0 ∈ B1\B1/2, inequality (2.3) follows. □

Proof of Theorem 2.1. On the one hand, by Lemma 2.5, there exists C1 = C1(N, p, α, d, |w′|∞) > 0
such that any positive solution u of equation (1.1) satisfies

u(x) ≤ C1[w(|x|)(1− |x|)]−
2+α
p−1 , x ∈ B1\B1/2.

On the other hand, we have 0 < d
2 ≤ w(|x|)(1 − |x|) ≤ 1 for x ∈ B1/2, which together with

Lemma 2.4 implies that there is C2 = C2(N, p, α, d, |w′|∞) > 0 such that u(x) ≤ C2 for x ∈ B1/2.

Therefore, there exists C = C(N, p, α, d, |w′|∞) > 0 such that u(x) ≤ C[w(|x|)(1 − |x|)]−
2+α
p−1 , for

x ∈ B1. □

Proof of Theorem 2.2. Assume that u ∈ C2(B1) is a positive solution of (1.1). Using spherical
coordinates to write u(x) = u(r, θ) with r = |x| and θ = x

|x| , we have

urr +
N − 1

r
ur +

1

r2
∆SN−1u = −[w(r)(1− r)]αup, r ∈ (0, 1). (2.4)

Let

ũ(r) =
1

|SN−1|

∫
SN−1

u(r, θ)dθ.

It follows from (2.4) that

ũrr +
N − 1

r
ũr = − [w(r)(1− r)]α

|SN−1|

∫
SN−1

u(r, θ)pdθ. (2.5)

Thus, we obtain

(rN−1ũ′(r))′ < 0, for all r ∈ (0, 1),

which implies that rN−1ũ′ is decreasing. It has a limit m ∈ [−∞,+∞) as r → 1−. In addition,
using Jensen’s inequality [19] for (2.5) yields

−(rN−1ũ′)′ ≥ [w(r)(1− r)]αrN−1ũp, r ∈ (0, 1). (2.6)

(i) Assume that u has a positive lower bound when α ≤ −2.



4 X. ZHANG, X. CHENG, M. YANG EJDE-2025/50

Case 1.. If m ≥ 0, then
rN−1ũ′(r) > m, r ∈ (0, 1).

So ũ′ > 0 holds for r ∈ (0, 1). Assume that ũ → n1 as r → 1−. Then there exist constants m1 > 0
and r1 ∈ (0, 1) such that ũ(r) > m1 for any r ∈ (r1, 1). From (2.6) it follows that

rN−1
1 ũ′(r1) = rN−1ũ′(r)−

∫ r

r1

(τN−1ũ′(τ))′ dτ

≥
∫ r

r1

[w(τ)(1− τ)]ατN−1ũ(τ)p dτ

≥ 2−prN−1
1 mp

1

∫ r

r1

(1− τ)α dτ, r ∈ (r1, 1).

Let r → 1−. In view of α ≤ −2, the above integral diverges to +∞, which is a contradiction.

Case 2. If m ∈ [−∞, 0), there exist r∗ > 0 and n2 > 0 such that

rN−1ũ′(r) < −n2, r ∈ (r∗, 1).

So there is n∗ ∈ (0, n2] such that ũ′(r) < −n∗ for r ∈ (r∗, 1). Noticing that u has a positive
lower bound. We assume that ũ(r) → n3 ∈ (0,+∞) as r → 1−, which together with the strictly
decreasing of ũ yields ũ(r) > n3 for r ∈ (r∗, 1). From equation (2.6) it follows that

−rN−1ũ′(r) ≥ −rN−1
∗ ũ′(r∗) + np

3

∫ r

r∗

[w(τ)(1− τ)]ατN−1 dτ

≥ np
3

∫ r

r∗

[w(τ)(1− τ)]ατN−1 dτ

≥ np
3r

N−1
∗

∫ r

r∗

(1− τ)α dτ, r ∈ (r∗, 1).

Then

ũ′(r) ≤ −np
3r

N−1
∗

∫ r

r∗

(1− τ)α dτ, r ∈ (r∗, 1).

Integrating the above inequality from r∗ to r leads to

ũ(r)− ũ(r∗) ≤ −np
3r

N−1
∗

∫ r

r∗

∫ t

r∗

(1− τ)α dτdt, r ∈ (r∗, 1).

Because α ≤ −2, the right-hand side diverges to −∞ as r → 1−, which yields a contradiction.
(ii) By an argument similar to Part (i), we can deduce a contradiction for Case 1. As for Case

2, we have
ũ′(r) < −n∗ for r ∈ (r∗, 1) and ũ(r) → n3 ∈ [0,+∞) as r → 1−.

If n3 ∈ (0,+∞), we can derive a contradiction analogous to Case 2 in the proof of Part (i).
Now, we suppose that n3 = 0. Using the differential mean value theorem leads to

ũ(r) ≥ n∗(1− r), r ∈ (r∗, 1). (2.7)

For any r ∈ (r∗, 1), from (2.6) it follows that

rN−1
∗ ũ′(r∗)− rN−1ũ′(r) ≥ np

∗

∫ r

r∗

[w(τ)]α(1− τ)α+pτN−1 dτ.

Hence, we obtain

−ũ′(r) ≥ np
∗r

1−N

∫ r

r∗

(1− τ)α+pτN−1 dτ

≥ np
∗r

N−1
∗

∫ r

r∗

(1− τ)α+p dτ

=
np
∗r

N−1
∗

α+ p+ 1
[(1− r∗)

α+p+1 − (1− r)α+p+1], r ∈ (r∗, 1).

Letting r → 1−, we can see a contradiction because α+ p+ 2 ≤ 0.



EJDE-2025/50 GENERALIZED HARDY-HÉNON EQUATIONS 5

(iii) The proof of Case 1 is similar to that of Case 1 in Part (i). For Case 2, the proof is
analogous to that of Part (ii) when n3 ∈ (0,+∞). Next, we only need to consider the case n3 = 0.
In view of 1 < p < (N + 2)/(N − 2), from Theorem 2.1 and (2.5)-(2.6) it follows that

[w(r)(1− r)]αrN−1ũp ≤ −(rN−1ũ′)′ ≤ CrN−1[w(r)(1− r)]α[w(r)(1− r)]−p(2+α)/(p−1),

where C > 0 is a positive constant. This together with (2.7) gives

C[w(r)(1− r)]−p(2+α)/(p−1) ≥ np
∗(1− r)p, r ∈ (r∗, 1).

That is,

(1− r)−p(α+p+1)/(p−1) ≥ np
∗

C(w(r))p(2+α)/(p−1)
≥ np

∗

C
> 0, r ∈ (r∗, 1).

In view of −p(α+ p+1)/(p− 1) > 0, we have (1− r)−p(α+p+1)/(p−1) → 0 as r → 1−, which yields
a contradiction. □

To establish the existence of positive solutions to (1.2), we start with a corresponding pertur-
bation problem. Applying the maximum principle and the regularity of elliptic equations [9, 14],
we can obtain the following lemma.

Lemma 2.6. If 1 < p < (N + 2)/(N − 2), α > −2, ϵ0 > 0 and ϵ ∈ (0, ϵ0], then there exists
C = C(N, p, α, ϵ0, d, |w′|∞) > 0 such that for any positive radial solution uϵ ∈ C2(B1) ∩C(B1) of

−∆u = [(w(|x|)(1 + ϵ− |x|)]αup, x ∈ B1,

u = 0, |x| = 1,
(2.8)

it holds
∥∇uϵ∥L∞(B1) + ∥uϵ∥L∞(B1) ≤ C. (2.9)

Proof. To show that ∥uϵ∥L∞(B1) ≤ C, we conversely suppose that there are a sequence of solutions
uk, ϵk ∈ (0, ϵ0] and Pk ∈ B1 such that

Mk = max
x∈B1

uk(x) = uk(Pk) → +∞, as k → ∞.

We claim that Pk = 0. Otherwise, if Pk ̸= 0, then by the symmetric property there exists Qk ∈ B1

such that |Pk| > |Qk| and uk achieves the local minimum at Qk. Thus, we have

0 ≥ −∆uk(Qk) = [(w(|Qk|)(1 + ϵk − |Qk|)]αuk(Qk)
p > 0,

which is a contradiction. Without loss of generality, we assume that ϵk → ϵ̃ ∈ [0, ϵ0]. Let

Uk(y) =
1

Mk
uk(M

−(p−1)/2
k y).

Then Uk satisfies

−∆Uk = [w(|M−(p−1)/2
k y|)(1 + ϵk − |M−(p−1)/2

k y|)]αUp
k ,

where 0 ≤ Uk ≤ 1 and Uk(0) = 1. According to the standard arguments of elliptic equations, we
can extract a subsequence of {Uk} converging to a function U in C2

loc(RN ), from which we derive
that

−∆U = [w(0)(1 + ϵ̃)]αUp in RN , and U(0) = 1.

This yields a contradiction with [13, Theorem 4.1].
To prove that ∥∇uϵ∥L∞(B1) ≤ C, we know that ∥uϵ∥L∞(B1) ≤ C for ϵ ∈ (0, ϵ0], and uϵ is

radially symmetric. For convenience, we denote uϵ(r) = uϵ(x) as |x| = r. For the case α ≥ 0, by
the regularity of elliptic equations, it is easy to see the desired result. For the case −2 < α < 0,
by way of contradiction, we suppose that there exist ϵk ∈ (0, ϵ0] and positive solution uk of (2.8)
with ϵ = ϵk such that ∥∇uk∥L∞(B1) → +∞ as k → ∞. From u′

k(0) = 0 and

−rN−1u′
k(r) =

∫ r

0

[w(τ)(1 + ϵk − τ)]ατN−1up
k(τ) dτ, r ∈ (0, 1],

we deduce that u′
k(r) < 0 for r ∈ (0, 1]. Let rk ∈ (0, 1] be the minimum point of u′

k. Using
the interior estimate of elliptic equations, we see that {rk} has a subsequence converging to 1.



6 X. ZHANG, X. CHENG, M. YANG EJDE-2025/50

Without loss of generality, assume that rk → 1 as k → ∞, then −u′
k(rk) → +∞ as k → ∞. From

(2.8) it follows that

−u′
k(r) = r1−N

∫ r

0

τN−1[w(τ)(1 + ϵk − τ)]αup
k(τ) dτ, r ∈ (0, 1),

where −2 < α < 0. By the differential mean value theorem, we have uk(r) ≤ (1 − r)|u′
k(rk)| for

r ∈ (0, 1), implying that

|u′
k(rk)| ≤ r1−N

k

∫ rk

0

τN−1[w(τ)(1− τ)]αup
k(τ) dτ

≤
∫ rk

0

[w(τ)(1− τ)]αup
k(τ) dτ

=

∫ rk

0

[w(τ)(1− τ)]αuk(τ)
1−ηuk(τ)

p+η−1 dτ

≤ Cp+η−1dα|u′
k(rk)|1−η

∫ rk

0

(1− τ)α+1−η dτ

for any given constant η ∈ (0,min{1, α+ 2}). Then we obtain

|u′
k(rk)| ≤ K|u′

k(rk)|1−η, for all k ∈ N and some K > 0,

which is a contradiction with |u′
k(rk)| → +∞ as k → ∞. □

Proof of Theorem 2.3. We need to consider two cases.

Case 1. When −2 < α ≤ 0, we consider the problem

−∆u = [w(|x|)(1 + n−1 − |x|)]α|u|p−1u, x ∈ B1,

u = 0, |x| = 1,
(2.10)

where n ∈ N. Define u+ = max{u, 0} and

Fn(u) =
1

2

∫
B1

|∇u|2dx− 1

p+ 1

∫
B1

[w(|x|)(1 + n−1 − |x|)]α(u+)p+1dx, u ∈ H1
0 (B1),

where H1
0 (B1) is equipped with the norm ∥u∥ = (

∫
B1

|∇u|2dx)1/2 for u ∈ H1
0 (B1). We now prove

that Fn has a radially symmetric critical point in H1
0 (B1). To this end, we choose the subspace

of H1
0 (B1) as

X =
{
u ∈ H1

0 (B1) : u is a radially symmetric function in B1

}
.

Clearly, for any fixed n, Fn satisfies the conditions of mountain pass lemma in X [18]. By the
theory of critical point on symmetric function spaces [18], Fn has a critical point un, which is
a radially symmetric function in H1

0 (B1). Thus un is a nontrivial nonnegative weak solution to
(2.10).

From the regularity and strong maximum principle [14], we have un ∈ C2(B1) ∩ C1(B1) and
un > 0. From Lemma 2.6, there is C > 0 such that ∥un∥C1(B1)

≤ C for all n ∈ N. By the

regularity of elliptic equations, un is bounded in C2+µ
loc (B1) with µ ∈ (0, 1). By the Arzela-Ascoli

theorem, we see un → u in C2
loc(B1). Hence, u ∈ C2(B1)∩C1(B1) is a radially symmetric solution

to (1.2).
We claim that u is a nontrivial solution. Without loss of generality, we suppose that un → u in

C1
loc(B1) as n → ∞. Otherwise, u ≡ 0. From un → u in C(B1), it follows that ∥un∥L∞(B1)

= o(1)

(as n → ∞). Using equations of un and un+1, we have

−∆(un+1 − un) = [w(|x|)]α[(1 + (n+ 1)−1 − |x|)αup
n+1 − (1 + n−1 − |x|)αup

n]

> [w(|x|)(1 + n−1 − |x|)]α(up
n+1 − up

n)

= [w(|x|)(1 + n−1 − |x|)]α(un+1 − un)χn(x), x ∈ B1,



EJDE-2025/50 GENERALIZED HARDY-HÉNON EQUATIONS 7

where ∥χn∥L∞(B1) = o(1) (as n → ∞), and thus

−∆(un+1 − un)− [w(|x|)(1 + n−1 − |x|)]αχn(x)(un+1 − un) > 0, x ∈ B1,

un+1 − un = 0, |x| = 1.
(2.11)

We denote by λ1[h(x), χ] the first eigenvalue of

−∆φ+ h(x)φ = λφ, x ∈ χ,

φ = 0, x ∈ ∂χ.
(2.12)

Let h(x) = −d2

4 [w(|x|)(1 + n−1 − |x|)]−2 and t(x) = − 1
4 (1 + n−1 − |x|)−2. From [11, Lemma 2.3]

it follows that
λ1[h(x), B1] ≥ λ1[t(x), B1] > 0.

In view of α > −2 and ∥χn∥L∞(B1) = o(1) (as n → ∞), for the sufficient large n we have

[w(|x|)(1 + n−1 − |x|)]αχn(x) ≤ −h(x) in B1.

Thus,
λ1[−(w(|x|)(1 + n−1 − |x|))αχn(x), B1] ≥ λ1[h(x), B1] > 0.

From (2.11) and the strong maximum principle, it follows that un+1(x) > un(x) for x ∈ B1 and
for the sufficient large n, which contradicts the fact that un → 0 in C(B1) as n → ∞. Therefore,
u ̸≡ 0 and u is a positive solution to (1.2).

Case 2. When α > 0, we define a functional F in H1
0 (B1) by

F (u) =
1

2

∫
B1

|∇u|2dx− 1

p+ 1

∫
B1

[w(|x|)(1− |x|)]α(u+)p+1dx.

By the mountain pass lemma [18], F has a positive critical point v ∈ H1
0 (B1). By 1 < p < N+2

N−2 and

α > 0, together with the regularity of elliptic equations [14], we obtain that v ∈ C2(B1)∩C1(B1)
is a positive solution to (1.2). □

3. Results for the case p < 1

In this section, we consider problems (1.1) and (1.2) when p < 1. Let us summarize our results
as follows.

Theorem 3.1. If 0 < p < 1, then there exists a constant C > 0 such that any positive solution u
to (1.1) satisfies

u(x) ≥ C[w(|x|)(1− |x|)]−
2+α
p−1 , x ∈ B1. (3.1)

Theorem 3.2. If 0 < p < 1 and 1 + p+ α < 0, then (1.2) has no positive solutions in C1(B1).

Theorem 3.3. If 0 < p < 1 and α ≤ −2, then (1.1) has no positive solutions in C1(B1).

Theorem 3.4. (i) If p < 1, and α > −2, then (1.2) has a positive classical solution. More-
over, if p < 0, the positive solution of (1.2) is unique.

(ii) If ζ ∈ C1(B1) is a nonnegative function, 0 < p < 1 and α ≥ 0, then the problem

−∆u = [w(|x|)(1− |x|)]αup, x ∈ B1,

u = ζ, |x| = 1,
(3.2)

has a unique positive solution in C2(B1) ∩ C1(B1).

To establish the lower estimates of positive solutions to (1.1), we need the following lemma.

Lemma 3.5 ([7]). If N ≥ 3, p < 1, µ ∈ (0, 1) and a ∈ Cµ(B1) satisfies a(x) ≥ C for x ∈ B1 and
some constant C > 0, then for any positive classical solution u to the problem

−∆u = a(x)up, x ∈ B1,

satisfies

|u(0)| ≥
( C

λ1(B1)

)1/(1−p)

, x ∈ B1,



8 X. ZHANG, X. CHENG, M. YANG EJDE-2025/50

where λ1(B1) is the first eigenvalue of −∆ with the Dirichlet boundary condition on B1.

Now, we are in a position to prove Theorem 3.1.

Proof of Theorem 3.1. Let x0 ∈ B1 and c(x) = w(|x|)(1− |x|). Then y := x0 + c(x0)x/2 ∈ B1 for
x ∈ B1. We define

U(x) = c(x0)
2+α
p−1 u(x0 + c(x0)x/2), x ∈ B1.

Then U satisfies

−∆U = a(x;x0)U
p, x ∈ B1, where a(x;x0) =

c(y)α

4c(x0)α
.

From the proof of Lemma 2.5 we know that, for all x, x0 ∈ B1,

4a(x;x0) ≥

{(
d
2

)α
, as α ≥ 0,(

3
2d

)α
, as α < 0.

Applying Lemma 3.5, we have U(0) ≥ C for some constant C > 0, i.e.,

u(x0) ≥ C[w(|x0|)(1− |x0|)]−
2+α
p−1 .

Because x0 ∈ B1 is arbitrary, we obtain

u(x) ≥ C[w(|x|)(1− |x|)]−
2+α
p−1 , x ∈ B1.

The proof is complete. □

Proof of Theorem 3.2. Assume that u ∈ C1(B1) is a positive solution of (1.2). According to
Theorem 3.1 and Hopf’s lemma [9, 14], there exist constants C1, C2 > 0 such that

C1[w(|x|)(1− |x|)]−
2+α
p−1 ≤ u(x) ≤ C2w(|x|)(1− |x|), x ∈ B1,

which together with − 2+α
p−1 < 1 yields a contradiction by letting |x| → 1−. □

Proof of Theorem 3.3. We suppose that (1.1) has a positive solution u. By Theorem 3.1, there
exists a positive constant C > 0 such that

u(x) ≥ C, x ∈ B1.

We denote

ũ(r) :=
1

|SN−1|

∫
SN−1

u(r, θ)dθ.

Then ũ satisfies
−(rN−1ũ′(r))′ ≥ Cp[w(r)(1− r)]αrN−1, r ∈ (0, 1).

Using the arguments as we did for the proof of Part (i) in Theorem 2.2, we can deduce a contra-
diction. □

Proof of Theorem 3.4. (i) For convenience, we separate the proof into two steps.

Step 1. Prove the existence of positive solutions to (1.2) with p < 1 and α > −2. We denote the
first eigenfunction and eigenvalue by φ1 and λ1(B1) of the problem

−∆φ = λφ, in B1,

φ = 0, on ∂B1.

Let β = 2+α
1−p and u = mφβ

1 . By Hopf’s lemma there exist constants c1, c2 > 0 such that

c1φ1(x) ≤ w(|x|)(1− |x|) ≤ c2φ1(x), x ∈ B1.

Then there exist a proper constant c > 0 and a sufficiently small constant m > 0 such that

−∆u− [w(|x|)(1− |x|)]αup

= −∆(mφβ
1 )− [w(|x|)(1− |x|)]α(mφβ

1 )
p

= −(mβφβ−1
1 ∆φ1 +mβ(β − 1)φβ−2

1 |∇φ1|2)− [w(|x|)(1− |x|)]αmpφpβ
1

≤ mβλ1(B1)φ
β
1 −mβ(β − 1)φβ−2

1 |∇φ1|2 − cαmpφpβ+α
1



EJDE-2025/50 GENERALIZED HARDY-HÉNON EQUATIONS 9

= mβφβ−2
1 [λ1(B1)φ

2
1 − (β − 1)|∇φ1|2 − cαβ−1mp−1] ≤ 0, x ∈ B1.

Let u = Kφρ
1 with the sufficiently large K > 0 and ρ ∈ (0,min{1, 2+α

1−p }]. Then we find that

−∆u− [w(|x|)(1− |x|)]αup

= −∆(Kφρ
1)− [w(|x|)(1− |x|)]α(Kφρ

1)
p

= −(Kρφρ−1
1 ∆φ1 +Kρ(ρ− 1)φρ−2

1 |∇φ1|2)− [w(|x|)(1− |x|)]αKpφρp
1

≥ Kρλ1(B1)φ
ρ
1 −Kρ(ρ− 1)φρ−2

1 |∇φ1|2 − cαKpφρp+α
1

= Kρφρ−2
1 [λ1(B1)φ

2
1 − (ρ− 1)|∇φ1|2 − cαρ−1Kp−1]

≥ 0, x ∈ B1,

where c > 0 is a constant.
By the sub-super solution method and the regularity of elliptic equations [9], equation (1.2) has

a positive solution in C2(B1) ∩ C1(B1).

Step 2. Show the uniqueness of positive solutions to (1.2) with p < 0 and α > −2. By the
estimates of positive solutions, there is a constant c∗ > 0 such that for any positive solution u to

(1.2), we have u(x) ≥ c∗φ
(2+α)/(1−p)
1 . Choose a positive constant m < c∗ in (i)-Step 1. Then there

is a minimal positive solution v∗ in [mφβ
1 ,Kφρ

1]. Assume that v is any positive solution to (1.2).

Then v(x) ≥ mφβ
1 and so min{v,Kφρ

1} is a supersolution to (1.2) and mφβ
1 ≤ min{v,Kφρ

1}, which
indicates that v∗ ≤ min{v,Kφρ

1}, in particular, v∗ ≤ v. Hence, v∗ is a minimal positive solution.
Now we claim that v∗ = v. Otherwise, there exists x0 ∈ B1 such that v∗(x0) − v(x0) =

minx∈B1
{v∗(x)− v(x)} < 0. Then

0 ≥ −∆(v∗ − v)(x0) = [w(|x0|)(1− |x0|)]α(v∗(x0)
p − v(x0)

p) > 0,

which obviously yields a contradiction.

(ii) To prove the existence of positive solutions to (3.2), from (i)-Step 1 we know that u = mφβ
1

satisfies

−∆u ≤ [w(|x|)(1− |x|)]αup, in B1,

u = 0 ≤ ζ, on ∂B1.

For any given constants δ > 0 and q ∈ (1, N+2
N−2 ), there is a positive solution uδ to the problem

−∆u = uq, in B1+δ,

u = 0, on ∂B1+δ.

We can take a sufficiently large M > 0 such that u := Muδ satisfies u ≥ u in B1 and

−∆u = Muq
δ ≥ [w(|x|)(1− |x|)]α(Muδ)

p = [w(|x|)(1− |x|)]αup, in B1,

u ≥ ζ, on ∂B1.

By the sub-super solution method and the standard arguments [9], (3.2) has a minimal positive

solution u∗ and a maximal positive solution u∗ in the interval [mφβ
1 ,Muδ].

To show the uniqueness of positive solution to (3.2), we assume that v is an arbitrary positive
solution to (3.2). From Theorem 3.1, there exists constant C > 0 such that

v(x) ≥ C[w(|x|)(1− |x|)]β , x ∈ B1.

Without loss of generality, we suppose that mφ1(x)
β ≤ C[w(|x|)(1 − |x|)]β for all x ∈ B1 and

some m > 0. Therefore, mφβ
1 and min{v, u} are a pair of subsolution and supersolution of (3.2).

Moreover, u∗(x) ≤ v(x) for all x ∈ B1. Because of the arbitrariness of v, we see that u∗ is a
minimal positive solution of (3.2).

Now, we prove that v = u∗. Choose a sufficiently large M > 0 such that Muδ(x) ≥ v(x) for
all x ∈ B1. Then, u∗(x) ≤ v(x) ≤ u∗(x) for all x ∈ B1. By α ≥ 0 and the regularity of elliptic
equations, u∗ and u∗ belong to C2(B1) ∩ C1(B1). We claim that u∗ = u∗. Otherwise, if u∗ ≥ u∗
and u∗ ̸≡ u∗, then by the strong maximum principle of the problem

−∆(u∗ − u∗) = [w(|x|)(1− |x|)]α[up
∗ − (u∗)p], x ∈ B1,



10 X. ZHANG, X. CHENG, M. YANG EJDE-2025/50

u∗ − u∗ = 0, |x| = 1,

we obtain u∗(x) > u∗(x) for all x ∈ B1. In addition, by Hopf’s lemma, ∂(u∗−u∗)
∂ν < 0 on ∂B1 with

ν being the exterior unit normal on ∂B1. Multiplying equation (3.2) with u = u∗ (resp. u = u∗)
by u∗ (resp. u∗), we obtain

−u∗∆u∗ = [w(|x|)(1− |x|)]α(u∗)
pu∗, x ∈ B1,

−u∗∆u∗ = [w(|x|)(1− |x|)]α(u∗)pu∗, x ∈ B1,
(3.3)

subtracting equations from each other in (3.3) and then integrating by parts over B1, we have

0 ≥
∫
∂B1

ζ
[∂u∗

∂ν
− ∂u∗

∂ν

]
=

∫
B1

[w(|x|)(1− |x|)]αu∗u
∗[up−1

∗ − (u∗)p−1] > 0,

which is a contradiction. □

Remark 3.6. In general, if w ∈ C1[0, 1] and d1 ≤ w ≤ d2 for some d2 ≥ d1 > 0, then Theorems

2.1-2.3 and 3.1-3.4 are still valid. In fact, we denote by ũ := d
α

p−1

2 u, w̃ = w
d2

and d̃ = d1

d2
. Then ũ

satisfies the equation

−∆ũ = [w̃(|x|)(1− |x|)]αũp, x ∈ B1(0), (3.4)

and u is a positive solution to (1.1) if and only if ũ is a positive solution of (3.4). It is obvious

that w̃ ∈ C1[0, 1] and 0 < d̃ ≤ w̃ ≤ 1. Then, it suffices to apply the mentioned results above to
equation (3.4).

Acknowledgments. This work is partially supported by the NSF of China (12371242).

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EJDE-2025/50 GENERALIZED HARDY-HÉNON EQUATIONS 11

Xizheng Zhang

School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, China

Email address: 220220934341@lzu.edu.cn

Xiyou Cheng (corresponding author)

School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, China
Email address: chengxy@lzu.edu.cn

Meihua Yang

School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074,
China

Email address: yangmeih@hust.edu.cn


	1. Introduction
	2. Results for the case p > 1
	3. Results for the case p < 1
	Acknowledgments

	References