Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 07, pp. 1–19.

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

DOI: 10.58997/ejde.2025.07

INFINITELY MANY SIGN-CHANGING SOLUTIONS FOR AN

ASYMPTOTICALLY LINEAR AND NONLOCAL

SCHRÖDINGER EQUATION

RUOWEN QIU, RENQING YOU, FUKUN ZHAO

Abstract. In this article, we consider the nonlocal schrödinger equation

−LKu+ V (x)u = f(x, u), x ∈ RN ,

where −LK is an integro-differential operator and V is coercive at infinity, and
f(x, u) is asymptotically linear for u at infinity. Combining minimax method

and invariant set of descending flow, we prove that the problem possesses

infinitely many sign-changing solutions.

1. Introduction

The existence and multiplicity of sign-changing solutions are interesting topics
in the studies of nonlinear elliptic equations. Recently, much more attention has
been paid to such topics of the following classical elliptic equations

−∆u = f(x, u), x ∈ Ω, (1.1)

where Ω ⊂ RN is a bounded domain with smooth boundary. In fact, there are
different ways to obtain sign-changing solutions of the (1.1). Via a variational
argument and a version of deformation lemma, Castro, Cossio and Neuberger [10]
proved that (1.1), on a bounded domain Ω, possesses a sign-changing solution which
changes sign only once. Dancer and Du [12] considered the equation

−∆u = u|u|p−1 + g(u) in Ω,

u = 0, on ∂Ω,

where 1 < p < 2∗ − 1, g : R → R is Lipschitz continuous and

lim sup
u→0

g(u)

u
< λ1 , lim sup

|u|→∞

g(u)

|u|p
= 0.

The authors showed that the above problem has at least one sign-changing solution,
besides a positive solution and a negative solution, and their method based on a
topological degree argument combined with an priori bound of solutions. Suppose
f ′(0) < λ2 and f is superlinear but subcritical at infinity, Bartsch and Wang [3]
showed that there exists a solution u1 of (1.1) which changes sign, and if u2 is a

2020 Mathematics Subject Classification. 35R11, 35A15, 35B28.
Key words and phrases. Sign-changing solution; integro-differential operator; invariant set;

variational method.
©2025. This work is licensed under a CC BY 4.0 license.
Submitted September 9, 2024. Published January 15, 2025.

1



2 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

second nontrivial solution then u2 > u1 (respectively, u2 < u1) implies that u2 is
positive (respectively, negative). They found that there is a critical point u1 whose
critical group Ck(J, u1) := Hk(J

c, Jc − {u1}) is not trivial for some k ≥ 2, then u1

can be never positive nor negative. In [6], by constructing invariant sets of descent
flow, Bartsch, Liu and Weth obtained a sign-changing solution with precisely two
nodal domains and infinitely many nodal solutions for the Schrödinger equation

−∆u+ V (x)u = f(x, u), x ∈ RN . (1.2)

In addition, there are many useful results about this equation, such as [4, 8, 38].
In almost the above-mentioned papers, the (AR) condition which introduced by
Ambrosetti and Rabinowitz [1] was imposed, i.e. for some µ > 2,

0 ≤ µF (x, u) ≡ µ

∫ u

0

f(x, s)ds ≤ f(x, u)u, ∀(x, u) ∈ RN × {R \ {0}}. (1.3)

We remark that inequality (1.3) is one of the main tools to prove the boundedness
of the PS sequence. By a simple calculation, (1.3) shows that f(x, u) must be
superlinear with respect to u at infinity, that is

lim
u→∞

f(x, u)

u
= +∞.

However, the study of many practical problems such as the self-trapping of an
electromagnetic wave, under some suitable assumptions, leads to some problems
related to (1.2), in which f(x, u) is asymptotically linear with respect to u at infinity,
the background and the results for some typical models can be found in [31, 32].

In the previus decades, some results about existence of positive solutions for
elliptic problems that are asymptotically linear at infinity have been obtained. In
[33], Stuart and Zhou obtained a positive radial solution by applying mountain pass
theorem, where the equation is radially symmetric and V is a constant. Liu, Su and
Weth [21] established the compactness of PS sequences for the associated energy
functional under general spectral-theoretic assumptions, and obtained existence of
three nontrivial solutions if the energy functional has a mountain pass geometry.
Asymptotically linear problems with steep potential well have been studied in [35,
36], in which multiple solutions were constructed without giving nodal information
about the solutions. Unlike with the superliear case, less was known about the sign-
changing solutions to the asymptotically linear case. Maia, Miyagaki and Soares
[22] investigated the problem

−∆u+ λu = f(u), x ∈ RN , (1.4)

where the nonlinearity f is asymptotically linear at infinity. The authors showed
the existence of a sign-changing solution of (1.4), which changes sign exactly once.

In this article, we are concerned with the existence and multiplicity of sign-
changing solutions of the nonlocal Schrödinger equation

−LKu+ V (x)u = f(x, u), x ∈ RN , (1.5)

where V (x) is a nonnegative potential function and f(x, u) is asymptotically linear
for u at infinity, and LK is an integro-differential operator defined as follow

LKu(x) :=

∫
RN

(u(x+ y) + u(x− y)− 2u(x))K(y)dy, x ∈ RN ,

and the kernel K which is a measurable function and satisfies the following assump-
tions:



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 3

(A1) there is θ > 0 and s ∈ (0, 1) such that K(x) ≥ θ|x|−(N+2s) for any x ∈
RN\{0};

(A2) mK ∈ L1(RN ), where m(x) := min{|x|2, 1}.
If K(x) = |x|−N−2s, then −LK change into the fractional Laplacian operator

(−∆)s, and when s → 1−, (−∆)s → −∆, for more details we refer to the readers
to [13, 24] and the references therein. From a physical point of view, the nonlocal
operators play a crucial role in describing several different physical phenomena,
such as in the anomalous diffusion [27, 34], in the fractional quantum mechanics
[25] and so on.

Different from the operator −∆, the integro-differential operator LK is nonlocal,
which brings us some difficulties in applying variational methods. We refer the
reader to [28], and [29] for the variational setting and the existence of nontrivial
solution of such problem settled on a bounded domain of RN . It is worth mentioning
that there also are many interesting results related to nonlocal elliptic equations
with integro-differential operators in books [9, 24]. These result most focus on the
existence and multiplicity of nontrivial solutions or positive solutions.

However, to the best of our knowledge, there are few results concerning the exis-
tence of sign-changing solution for the nonlocal schrödinger equation (1.5). When
K(x) = |x|−N−2s, Wang and Zhou [37] obtained a radial sign-changing solution
of a fractional Schrödinger equation. Moreover, Chang and Wang [11] consid-
ered a fractional Laplacian equation and obtained the existence and multiplicity of
sign-changing solutions via applying the Caffarelli-Silvestre extension method and
invariant sets of descending flow. In [14, 15], by combining constraint variational
method and quantitative deformation, the authors prove the the equation

−LKu = f(x, u), in Ω,

u = 0, in RN \ Ω

possesses one least energy sign-changing solution and infinitely many sign-changing
solutions. The case with potential function can be seen [16]. We mention that
the above results are heavily based on the nonlinearity term f is superlinear and
subcritical. Simultaneously, the problem settled on a bounded domain.

Recall that a solution of (1.5) is called sign-changing if u± ̸= 0, where

u+(x) = max
{
u(x), 0

}
and u−(x) = min

{
u(x), 0

}
.

To state our main results, we need the following assumptions on V (x) and f :

(A3) V ∈ C(RN ,R) satisfies V0 := infx∈RN V (x) > 0;
(A4) For each M > 0, there exists r > 0 such that

meas({x ∈ Br(y) : V (x) ≤ M}) → 0 as |y| → ∞,

where meas denotes for the Lebesgue measure, and BR(x) denotes an open
ball of RN centered at x and of radius R > 0, while we simply write BR

when x = 0.

We assume f satisfies the following assumptions:

(A5) f ∈ C(RN × R,R) and f(x, u) = o(|u|) as u → 0 uniformly in x;
(A6) There is a constant a ∈ (0,+∞) such that f(x, u)u−1 → a as u → ∞

uniformly in x and

a > inf σ(−LK + V (x)),



4 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

where σ(−LK + V (x)) denotes the spectrum of the operator −LK + V (x);
(A7) f(x, u)/|u| is a increasing function of u ̸= 0;
(A8) infx∈RN ,u∈R\{0} F (x, u) > 0;
(A9) limu→∞(f(x, u)u− 2F (x, u)) = +∞.

Theorem 1.1. If (A3)–(A9) hold, then (1.5) has infinitely many sign-changing
solutions.

Remark 1.2. Condition (A4), which is weaker than the coercive assumption:
V (x) → ∞ as |x| → ∞, was firstly introduced by Bartsch and Wang in [2] to
overcome the lack of compactness.

Remark 1.3. There are functions satisfying (A5)–(A9). For example, f(x, u) =
au3

1+u2 , where a is the constant can be found in (A6). By direct calculations, we know

F (x, u) = a[ 12u
2 − 1

2 ln(1 + u2)], it is easy to prove that function f(x, u) satisfies
the assumptions (A5)–(A9).

Notation. Throughout this paper, we denote by | · |p the usual norm of the space
Lp(RN ), 1 ≤ p < ∞. un ⇀ u and un → u mean the weak and strong convergence,
respectively, as n → ∞. Bρ = {u ∈ E, ||u|| < ρ}.

2. Preliminary lemmas

First, we shall introduce some notation. For any s ∈ (0, 1), we define

Xs(RN ) =
{
u : RN → R : u is Lebesgue measurable u ∈ L2(RN ) and the mapping

(x, y) 7→
(
u(x)− u(y)

)√
K(x− y) ∈ L2(RN × RN )

}
,

where the kernel K satisfies (A1) and (A2). The norm in Xs(RN ) is defined as

∥u∥Xs :=
(∫

RN

∫
RN

|u(x)− u(y)|2K(x− y) dx dy +

∫
RN

|u|2dx
)1/2

and
(
Xs(RN ), ∥ · ∥Xs

)
is a Hilbert space, we refer to [13, 29] for more properties of

Xs(RN ). Since there is a potential functional V (x) is involved in (1.5), we introduce
the following subspace E of Xs(RN )

E :=
{
u ∈ Xs(RN ) :

∫
RN

V (x)u2dx < +∞
}
,

which is a Hilbert space equipped with the inner product

(u, v) =

∫
RN

∫
RN

(u(x)− u(y))(v(x)− v(y))K(x− y) dx dy +

∫
RN

V (x)uv dx.

The norm on E induced by the above inner productis denoted by ∥u∥. We will look
for solutions of (1.5) in the space E. We say that u ∈ E is a weak solution of (1.5)
if ∫

RN

∫
RN

(u(x)− u(y))(v(x)− v(y))K(x− y) dx dy +

∫
RN

V (x)uv dx

=

∫
RN

f(x, u)v dx

for all v ∈ E. So the energy functional associated with (1.5) is

Ψ(u) =
1

2

∫
RN

∫
RN

(u(x)− u(y))2K(x− y) dx dy +
1

2

∫
RN

V (x)u2dx



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 5

−
∫
RN

F (x, u) dx, u ∈ E.

Under our assumptions, it is standard to check that Ψ ∈ C1(E,R) and for v ∈ E,
it holds

⟨Ψ′(u), v⟩ =
∫
RN

∫
RN

(u(x)− u(y))(v(x)− v(y))K(x− y) dx dy

+

∫
RN

V (x)uv dx−
∫
RN

f(x, u)v dx.

Next we prove some preliminary lemmas, which are crucial for proving our main
results. Firstly, to overcome the difficulties brought by the nonlocal feature of
operator LK , we need the following embedding result.

Theorem 2.1. If (A3), (A4) hold, then the embeddings E ↪→ Lp(RN ) are contin-
uous for p ∈ [2, 2∗s] and compact for p ∈ [2, 2∗s), where 2∗s = 2N

N−2s is the fractional
Sobolev critical exponent.

Proof. First, we show that E ↪→ Lp(RN ) for p ∈ [2, 2∗s]. In fact, by (A3) we know
that E ↪→ Xs(RN ) is continuous, from the [13, Theorem 6.5] and (A1) we obtain
Xs(RN ) ↪→ Lp(RN ) for p ∈ [2, 2∗s]. Then the conclusion follows.

Next we show that E ↪→ Lp(RN ) is compact for p ∈ [2, 2∗s). In fact, let {un} ⊂ E
be a bounded sequence of E, going if necessary to a subsequence we have un ⇀ u
in E and un → u in Lp

loc(RN ), p ∈ [2, 2∗s) and ∥un(x)∥+ ∥u(x)∥ ≤ C̄, where C̄ is a
positive constant. We first prove that un → u in L2(RN ), it suffices to prove that
|un|2 → |u|2.

Fix M > 0 and set AM (y) := {x ∈ RN : V (x) ≤ M} ∩ Br(y), where r > 0 is
given by (A4). When p = 2, un → u in L2

loc(RN ) implies that un → u in L2(BR)
for any R > 0. Now, we choose {yn} ⊂ RN such that RN ⊂

⋃∞
i=1 Br(yi) and each

x ∈ RN is covered by at most 2N balls. Denote the set CM (yi) := {x ∈ RN :
V (x) > M} ∩Br(yi), we have∫

RN\BR

|un(x)− u(x)|2 dx

≤
∞∑

|yi|≥R−r

∫
Br(yi)

|un(x)− u(x)|2 dx

=

∞∑
|yi|≥R−r

(∫
CM (yi)

|un(x)− u(x)|2 dx+

∫
AM (yi)

|un(x)− u(x)|2 dx
)
.

Using the definition of CM (yi) and Hölder inequality we obtain∫
CM (yi)

|un(x)− u(x)|2 dx ≤ 1

M

∫
Br(yi)

V (x)|un(x)− u(x)|2 dx,∫
AM (yi)

|un(x)− u(x)|2dx ≤
(∫

AM (yi)

|un(x)− u(x)|2tdx
)1/t(∫

AM (yi)

1t
′
)1/t′

= |un(x)− u(x)|2L2t(AM (yi))

(
measAM (yi)

)1/t′
,

where t ∈ (1, N
N−2s ) and

1
t +

1
t′ = 1. Hence,∫

RN\BR

|un(x)− u(x)|2dx



6 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

≤
∑

|yi|>R−r

( 1

M

∫
Br(yi)

V (x)|un(x)− u(x)|2dx

+ sup
|yi|>R−r

(
measAM (yi)

)1/t′

|un(x)− u(x)|2L2t(AM (yi))

)
≤ 2N

M

∫
RN\BR−2r

V (x)|un(x)− u(x)|2dx

+ 2NC2
2 sup
|yi|>R−r

(
measAM (yi)

)1/t′

∥un(x)− u(x)∥2

≤ 2N C̄2

M
+ 2N (C2C̄)2 sup

|yi|>R−r

(
measAM (yi)

)1/t′

,

where C2 > 0 is the embedding constant. Now, for any ε > 0 we choose M > 0 so
large that

2N+1C̄2

M
< ε. (2.1)

For fixed M > 0, there exists RM > 0 such that

2N+1(C2C̄)2 sup
|yi|>RM−r

(
measAM (yi)

)1/t′

< ε, (2.2)

since

sup
|yi|≥R−r

(
measAM (yi)

)1/t′ → 0 as R → ∞.

For such RM , by (2.1) and (2.2) we have∫
RN\BRM

|un(x)− u(x)|2dx ≤ ε.

Hence,∫
RN

|un(x)− u(x)|2dx

=

∫
B(0,RM )

|un(x)− u(x)|2dx+

∫
RN\BRM

|un(x)− u(x)|2dx ≤ 2ε,
(2.3)

this proves that |un|2 → |u|2 in L2(RN ). Finally, by the Interpolation inequality
we have (up to renaming C)

|un − u|p ≤ C|un − u|θ2 |un − u|1−θ
2∗s

≤ C|un − u|θ2||un − u||1−θ

≤ C|un − u|θ2(||un||+ ||u||)1−θ,

(2.4)

where 1
p = θ

2 + 1−θ
2∗s

and θ ∈ (0, 1). Hence the right hand of (2.4) is small enough,

therefore, un → u in Lp(RN ) for p ∈ (2, 2∗s). □

Next, we consider the eigenvalues problem. As in [5, 29], we have the following
results.

−LKu+ V u = λu, x ∈ RN . (2.5)

Proposition 2.2. Let s ∈ (0, 1), N > 2s, and K : RN \ {0} → (0,+∞) be a
function satisfying assumptions (A1) and (A2). Then



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 7

(1) (2.5) admits an eigenvalue λ1 that is positive, simple and that can be char-
acterized as follows:

λ1 = inf
u∈E, |u|2=1

{∫
RN

∫
RN

|u(x)− u(y)|2K(x− y) dx dy

+

∫
RN

V (x)u2(x) dx
}
.

(2.6)

(2) The set of the eigenvalues of (2.5) consists of a sequence {λk}k∈N with

0 < λ1 < λ2 ≤ · · · ≤ λk ≤ λk+1 ≤ · · ·
and λk → +∞ as k → +∞. Moreover, for each k ∈ N, the eigenvalues can
be characterized as

λk+1 = inf
u∈X⊥

k , |u|2=1

{∫
RN

∫
RN

|u(x)− u(y)|2K(x− y) dx dy

+

∫
RN

V (x)u2(x) dx
}
,

(2.7)

where Xk := span{e1, e2, · · · , ek}.
(3) The sequence {ek}k∈N of eigenfunctions corresponding to λk is an orthonor-

mal basis of L2(RN ) and an orthogonal basis of E.

We give some properties of f(x, u) and F (x, u) in the following lemma.

Lemma 2.3. (1) If (A5) and (A6) hold, then for any p ∈ (2, 2∗s) and ε > 0,
there exists Cε > 0 such that for all u ∈ R,

|f(x, u)| ≤ ε|u|+ Cε|u|p−1 |F (x, u)| ≤ εu2 + Cε|u|p.
(2) If (A7) hold, then for any (x, u) ∈ RN × (R \ {0}), we have

1

2
f(x, u)u− F (x, u) ≥ 0,

where F (x, u) =
∫ u

0
f(x, s)ds.

Proof. Conclusion (1) is easy, so we omit it here. It follows from (A7) that, for any
t ≥ 0, u ∈ R \ {0}, one has

1− t2

2
uf(x, u) + F (x, tu)− F (x, u) =

∫ 1

t

[
f(x, u)

u
− f(x, su)

su

]
su2ds ≥ 0. (2.8)

Taking t = 0 in (2.8), we obtain for any (x, u) ∈ RN × (R \ {0}),
1

2
f(x, u)u ≥ F (x, u). (2.9)

This completes the proof of conclusion (2). □

Under conditions (A5)–(A7), we will show that Ψ(u) has a mountain pass geom-
etry, see Lemma 2.4 and Lemma 2.5, where Lemma 2.4 can be directly obtained
from the embedding result Theorem of 2.1 and Lemma 2.3.

Lemma 2.4. Suppose A3)–(A6) hold. Then Ψ(u) = 1
2∥u∥

2+o(∥u∥2), ⟨Ψ′(u), u⟩ =
∥u∥2 + o(∥u∥2) as u → 0 in E.

Lemma 2.5 ([17]). Suppose (A3)–(A8) hold. Then there is a v ∈ E with v ̸= 0
such that Ψ(v) < 0.



8 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

Proof. By Proposition 2.2 we know that

inf σ(−LK + V (x))

= inf
u∈E, |u|2=1

{∫
RN

∫
RN

|u(x)− u(y)|2K(x− y) dx dy +

∫
RN

V (x)u2(x) dx
}
.

By (A6), we have an ũ ∈ E such that |ũ|2 = 1 and ∥ũ∥2 < a. Replacing ũ by |ũ|
(still renaming ũ ), we can suppose that ũ ≥ 0 a.e. on RN . To prove the Lemma,
it suffices to show that

lim
t→+∞

Ψ(tũ)

t2
< 0. (2.10)

First, we claim that

lim
t→+∞

∫
RN

F (x, tũ)

t2
dx =

1

2
a. (2.11)

To prove (2.11), without loss of generality we can assume that ũ is defined every-
where on RN and divide the argument into two situations: ũ(x) > 0 and ũ(x) = 0.
When ũ(x) > 0. By (f2), we obtain

lim
t→+∞

F (x, tũ)

t2
= lim

t→+∞

F (x, tũ)

(tũ)2
(ũ)2 =

1

2
a(ũ)2. (2.12)

When ũ(x) = 0, for all t > 0

F (x, tũ)

t2
= 0 =

1

2
a(ũ)2. (2.13)

In view of (2.12) and (2.13) we know that

lim
t→+∞

F (x, tũ)

t2
=

1

2
a(ũ)2 a.e. on RN . (2.14)

On the other hand, by (A6)–(A8), there exists a C > 0 such that

0 ≤ f(x, u)

u
≤ C for all u ∈ R \ {0},

and thus

0 ≤ F (x, u)

u2
≤ C

2
for all u ∈ R \ {0}.

Therefore,

0 ≤ F (x, tũ)

t2
≤ C

2
(ũ)2 for all u ∈ R \ {0}. (2.15)

Equations (2.14) and (2.15) allow us to apply Lebesgue dominated convergence
theorem to obtain

lim
t→+∞

∫
RN

F (x, tũ)

t2
dx =

a

2

∫
RN

(ũ)2dx =
a

2
,

that is claim (2.10). According to (2.10), we easily obtain that

lim
t→+∞

Ψ(tũ)

t2
=

1

2
∥ũ∥2 − lim

t→+∞

∫
RN

F (x, tũ)

t2
dx =

1

2

(
∥ũ∥2 − a

)
< 0,

so the Lemma is proved. □

Another difficulty that needs to be overcome is the lack of boundedness for
Palais-Smale sequences when the (AR) condition (1.3) is does not satisfy. As in
[33] our proof of the boundedness of {un} relies on the work of Lions [19, 20] on
the concentration compactness principle.



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 9

Lemma 2.6. If (A3)–(A9) hold, and {un} ⊂ N is a sequence such that {Ψ(un)}
is bounded, then {un} is bounded.

Proof. Assuming that the statement does not hold, then we can suppose that there
exists a subsequence again denoted by {un} such that {Ψ(un)} is bounded but
∥un∥ → ∞ when n → ∞. By the definition of N , for all u ∈ N we obtain that

Ψ(u) = Ψ(u)− 1

2
⟨Ψ′(u), u⟩

=

∫
RN

(1
2
f(x, u)− F (x, u)

)
dx ≥ 0.

(2.16)

Then, up to a subsequence, Ψ(un) → l ≥ 0 by (2.16). If l > 0, we define vn :=
2
√
lun

∥un∥ , and ∥vn∥ = 2
√
l. If l = 0, we define vn := un

∥un∥ , so that ∥vn∥ = 1.

Now, we prove the following Claim.
Claim: There exist r, d > 0 and a sequence {yn} ⊂ RN such that

lim inf
n→∞

∫
Br(yn)

v2ndx ≥ d > 0. (2.17)

If the claim is not true, then by the nonlocal type Lions lemma, vn → 0 in Lp
(
RN

)
,

where 2 < p < 2∗s. Using Lemma 2.3 we obtain∣∣ ∫
RN

F (x, vn) dx
∣∣ ≤ ε

∫
RN

v2ndx+ Cε

∫
RN

|vn|pdx,

since {vn} ⊂ E is bound and the arbitrariness of ε > 0 we have

lim
n→∞

∫
RN

F (x, vn) dx = 0.

When l = 0, we have

lim inf
n→∞

Ψ(vn) = lim inf
n→∞

(1
2
∥vn∥2 −

∫
RN

F (x, vn) dx
)
=

1

2
. (2.18)

To obtain a contradiction, we need the inequality

Ψ(tu) ≤ Ψ(u), for t ≥ 0 and u ∈ N . (2.19)

Indeed, let u ∈ N and define the function

ξ(t) :=
t2

2
f(x, u)u− F (x, tu), t ≥ 0,

and for any t > 0

ξ′(t) = tf(x, u)u− f(x, tu)u = tu2
(f(x, u)

u
− f(x, tu)

tu

)
.

By (A7), for every t > 0, we know ξ(t) ≤ ξ(1). Then, after integration on RN and
using that ⟨Ψ′(u), u⟩ = 0, we have

Ψ(tu) = Ψ(tu)− t2

2
⟨Ψ′(u), u⟩

=

∫
RN

( t2
2
f(x, u)u− F (x, tu)

)
dx

≤
∫
RN

(1
2
f(x, u)u− F (x, u)

)
dx

= Ψ(u).



10 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

This completes the proof of (2.19). On the other hand, taking t = 1
∥un∥ in (2.19),

we have

Ψ(vn) = Ψ(tun) ≤ Ψ(un) = l + on(1) = on(1),

which contradicts (2.18).
When l > 0,

lim inf
n→∞

Ψ(vn) = lim inf
n→∞

(1
2
∥vn∥2 −

∫
RN

F (x, vn) dx
)
= 2l,

and taking t = 2
√
l

∥un∥ in (2.19) we have

Ψ(vn) ≤ Ψ(un) = l + on(1),

getting the same contradiction, thus the Claim holds.
By the above claim we infer a contradiction in both cases: when {yn} is bounded

or unbounded. This will complete the proof.

Case 1. {yn} is bounded. Then there is r̃ > 0 such that {yn} ⊂ Br̃. By (2.17), we
have ∫

Br(yn)

v2ndx >
d

2
.

Thus we can choose r̂ > r + r̃ with Br(yn) ⊂ Br̂ and∫
Br̂

v2ndx >
d

2
.

Since {vn} is bounded in E, there exists a subsequence still denoted by {vn} such
that vn ⇀ v in E, vn → v in Lp(RN ) for 2 ≤ p < 2∗s, and vn(x) → v(x) a.e. on RN .
In particular, we have∫

Br̂

v2ndx →
∫
Br̂

v2dx and

∫
Br̂

v2dx ≥ d

2
> 0,

which implying that v ̸≡ 0. Thus there exists a set Ω ⊂ Br̂ with the meas(Ω) > 0
such that v(x) ̸= 0 for every x ∈ Ω. Hence for a fixed x ∈ Ω and the constant

t > 0, vn(x) =
tun(x)
∥un∥ ̸= 0 when n large enough which implied that un(x) ̸= 0. As

a consequence of ∥un∥ → ∞, |un(x)| → ∞. So |un(x)| → ∞ for every x ∈ Ω. Since

Ψ(un) =

∫
RN

(1
2
f(x, un)un − F (x, un)

)
dx ≥

∫
Ω

(1
2
f(x, un)un − F (x, un)

)
dx,

by (A9) and Fatou’s lemma, we have

lim inf
n→∞

Ψ(un) ≥
∫
Ω

lim inf
n→∞

(1
2
f(x, un)un − F (x, un)

)
dx = ∞,

which imply Ψ(un) → ∞, contradicting that Ψ(un) → l ∈ R.
Case 2. {yn} is unbounded. We set a new sequence ṽn(x) := vn(x − yn) and
∥ṽn∥ = ∥vn∥ is a constant. Thus, up to a subsequence ṽn ⇀ ṽ in E, ṽn → ṽ in
Lp

(
RN

)
for 2 ≤ p < 2∗s, and ṽn(x) → ṽ(x) a.e. on RN . From (2.17),

lim inf
n→∞

∫
Br(yn)

v2ndx = lim inf
n→∞

∫
Br

ṽ2ndx ≥ d

and hence ∫
Br

ṽ2dx ≥ d > 0,



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 11

implying that ṽ ̸≡ 0. Therefore, there exists a subset Λ ⊂ Br with positive measure
such that ṽ ̸≡ 0 for every x ∈ Λ. Similar to Case 1, |un(x + yn)| → ∞ for every
x ∈ Λ as n → ∞. So

Ψ(un) =

∫
RN

(1
2
f(x, un)un − F (x, un)

)
dx

≥
∫
Br(yn)

(1
2
f(x, un)un − F (x, un)

)
dx

=

∫
Br

(1
2
f(x, un(x+ yn))un(x+ yn)− F (x, un(x+ yn)

)
dx

≥
∫
Λ

(1
2
f(x, un(x+ yn))un(x+ yn)− F (x, un(x+ yn))

)
dx.

Thus, we have the same contradiction with Case 1. □

Lemma 2.7. If (A3)–(A4) hold, then Ψ satisfies Palais-Smale condition at any
level c > 0.

Proof. Let {un} ⊂ E be a (PS)c sequence of Ψ, that is Ψ(un) → c, and Ψ′(un) → 0.
By Lemma 2.6, we know that {un} is bounded in E. So we can assume that up to
a subsequence, there exists a u ∈ E such that

un ⇀ u in E,

un → u in Lp(RN ) for p ∈ [2, 2∗s).

Observe that

∥un − u∥2 = ⟨Ψ′(un)−Ψ′(u), un − u⟩+
∫
RN

(f(x, un)− f(x, u)) (un − u) dx.

It follows from the Hölder inequality and Lemma 2.3 that∣∣ ∫
RN

(f(x, un)− f(x, u))(un − u) dx
∣∣

≤
∫
RN

(|f(x, un)|+ |f(x, u)|)|un − u| dx

≤
∫
RN

(ε|un|+ ε|u|+ Cε|un|p−1 + Cε|u|p−1)|un − u| dx

≤ 4ε(|un|22 + |u|22)|un − u|2 + Cε(|un|p−1
p + |u|p−1

p )|un − u|p.

Thus we have verified that un → u in E, that is Ψ satisfies (PS)c condition. □

3. Proof of Theorem 1.1

We define an operator A : E → E as

Au := (−LKu+ V u)
−1 ◦ h(u), u ∈ E,

where h(u) := f(x, u). When u ∈ E fixed, we consider the functional

J(v) =
1

2

∫
RN

∫
RN

|v(x)− v(y)|2K(x− y) dx dy+
1

2

∫
RN

V (x)v2dx−
∫
RN

F (x, u) dx.

It is easy to prove that J ∈ C1(E,R) and coercive, bounded below and strictly con-
vex in E. Therefore, by [23, Theorem 1.1], J(v) admits a unique global minimizer



12 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

v = Au, and v = Au is the unique solution to the equation

−LKv + V (x)v = f(x, u), ∀u ∈ E

That is,∫
RN

∫
RN

(v(x)− v(y))(φ(x)− φ(y))K(x− y) dx dy +

∫
RN

V (x)vφ dx

=

∫
RN

f(x, u)φdx, ∀φ ∈ E.

(3.1)

Lemma 3.1. If (A3), (A4) hold, then the operator A satisfies:

(1) A is continuous and maps bounded sets into bounded sets.
(2) ⟨Ψ′(u), u−Au⟩ = ∥u−Au∥2
(3) ∥Ψ′(u)∥ ≤ ∥u−Au∥

Proof. (1) Let {un} ⊂ E such that un → u in E. Let vn := Aun and v := Au.
Then (3.1) implies that∫

RN

∫
RN

(vn(x)− vn(y)) (φ(x)− φ(y))K(x− y) dx dy +

∫
RN

V (x)vnφdx

=

∫
RN

f(x, un)φ(x) dx, φ ∈ E,

(3.2)

∫
RN

∫
RN

(v(x)− v(y))(φ(x)− φ(y))K(x− y) dx dy +

∫
RN

V (x)vφ dx

=

∫
RN

f(x, u)φ(x) dx, φ ∈ E.

(3.3)

In view of the Hölder inequality, Theorem 2.1 and (3.2), (3.3), we have

∥vn − v∥2

=

∫
RN

∫
RN

(vn(x)− vn(y)− v(x) + v(y))
2
K(x− y) dx dy +

∫
RN

V (x)|vn − v|2dx

=

∫
RN

f(x, un)vn dx+

∫
RN

f(x, u)vdx−
∫
RN

f(x, un)v dx−
∫
RN

f(x, u)vn dx

=

∫
RN

(f(x, un)− f(x, u)) (vn − v)dx

≤
(∫

RN

|vn − v|2
∗
sdx

) 1
2∗s
(∫

RN

|f(x, un)− f(x, u)|
2∗s

2∗s−1 dx
) 2∗s−1

2∗s

≤ C∥vn − v∥
(∫

RN

|f(x, un)− f(x, u)|
2∗s

2∗s−1 dx
) 2∗s−1

2∗s .

So,

∥vn − v∥ ≤ C
(∫

RN

|f(x, un)− f(x, u)|
2∗s

2∗s−1 dx
) 2∗s−1

2∗s .

It follows from the Lebesgue dominated convergence theorem that

lim
n→+∞

∫
RN

|f(x, un)− f(x, u)|
2∗s

2∗s−1 dx = 0,

hence,

∥vn − v∥ → 0, as n → +∞,



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 13

which implies that A is continuous on E. Next, we prove the boundedness of A. In
(3.1), we taking φ = Au ∈ E and combining Lemma 2.3 and the Hölder inequality,
we know

∥Au∥2

=

∫
RN

f(x, u)Audx

≤ C
(∫

RN

|Au∥u|dx+

∫
RN

|u|p−1|Au|dx
)

≤ C
(∫

RN

|u|2dx
)1/2(∫

RN

|Au|2dx
)1/2

+ C
(∫

RN

|u|pdx
) p−1

p
(∫

RN

|Au|pdx
)1/p

≤ C∥Au∥
(
∥u∥+ ∥u∥p−1

)
.

Therefore, ∥Au∥ ≤ C
(
∥u∥+ ∥u∥p−1

)
, which implies that A maps bounded sets into

bounded sets.
(2) Taking φ = u−Au ∈ E into (3.1), we have∫

RN

∫
RN

(Au(x)−Au(y))(u(x)−Au(x)− u(y) +Au(y))K(x− y) dx dy

+

∫
RN

V (x)Au(u−Au) dx

=

∫
RN

f(x, u)(u−Au) dx.

Hence

⟨Ψ′(u), u−Au⟩

=

∫
RN

∫
RN

(u(x)− u(y))(u(x)−Au(x)− u(y) +Au(y))K(x− y) dx dy

+

∫
RN

V (x)u(u−Au) dx−
∫
RN

f(x, u)(u−Au) dx

=

∫
RN

∫
RN

(u(x)−Au(x)− u(y) +Au(y))2K(x− y) dx dy

+

∫
RN

V (x)(u−Au)2dx

= ∥u−Au∥2.

(3) By the Hölder inequality, for any φ ∈ E, we obtain

|⟨Ψ′(u), φ⟩|

=
∣∣∣ ∫

RN

∫
RN

(u(x)− u(y))(φ(x)− φ(y))K(x− y) dx dy

+

∫
RN

V (x)uφdx−
∫
RN

f(x, u)φdx
∣∣∣

≤
∫
RN

∫
RN

|(u(x)−Au(x)− u(y) +Au(y))(φ(x)− φ(y))K(x− y)| dx dy

+

∫
RN

|V (x)(u−Au)φ| dx



14 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

≤
(∫

RN

∫
RN

|u(x)−Au(x)− u(y) +Au(y)|2K(x− y) dx dy
)1/2

×
(∫

RN

∫
RN

|φ(x)− φ(y)|2K(x− y) dx dy
)1/2

+
(∫

RN

V (x)|u−Au|2dx
)1/2(∫

RN

V (x)φ2dx
)1/2

≤
(∫

RN

∫
RN

|u(x)−Au(x)− u(y) +Au(y)|2K(x− y) dx dy

+

∫
RN

V (x)|u−Au|2dx
)1/2(∫

RN

∫
RN

|φ(x)− φ(y)|2K(x− y) dx dy

+

∫
RN

V (x)φ2dx
)1/2

= ∥u−Au∥∥φ∥,
which implies that ∥Ψ′(u)∥ ≤ ∥u−Au∥. □

As in [6], we consider the convex cones E+ := {u ∈ Xs : u ≥ 0} and E− := {u ∈
Xs : u ≤ 0}. For an arbitrary ε > 0, we define

D+
ε := {u ∈ Xs : dist(u,E+) < ε}, D−

ε := {u ∈ Xs : dist(u,E−) < ε},
where dist(u,E±) = infv∈E± ∥v − u∥.

Lemma 3.2. If (A3)–(A8) hold, then there exists ε0 > 0 such that for 0 < ε < ε0,
A(∂D+

ε ) ⊂ D+
ε , A(∂D−

ε ) ⊂ D−
ε .

Proof. Taking φ = v+ in (3.1), by the Hölder inequality and Lemma 2.3, we obtain
that

∥v+∥2 =

∫
RN

∫
RN

(
v+(x)− v+(y)

)2
K(x− y) dx dy +

∫
RN

V (x)
∣∣v+∣∣2 dx

≤
∫
RN

∫
RN

(
v+(x)− v+(y)

)2
K(x− y) dx dy +

∫
RN

V (x)vv+dx+
(
v+, v−

)
=

∫
RN

∫
RN

(v(x)− v(y))
(
v+(x)− v+(y)

)
K(x− y) dx dy +

∫
RN

V (x)vv+dx

=

∫
RN

f(x, u)v+dx

≤
∫
RN

f
(
x, u+

)
v+dx

≤
∫
RN

(
ε
∣∣u+

∣∣+ Cε

∣∣u+
∣∣p−1

)
v+dx

≤ ε
∣∣u+

∣∣
2

∣∣v+∣∣
2
+ Cε

∣∣u+
∣∣p−1

p

∣∣v+∣∣
p
.

Let u ∈ E and v = Au, by the Theorem 2.1, for any p ∈ [2, 2∗s], there exists Cp > 0
such that

|u±|p = inf
v∈E∓

|v − u|p ≤ Cp inf
v∈E∓

∥v − u∥ = Cp dist(u,E
∓). (3.4)

It is easy to know that dist(v,E−) ≤ ∥v+∥, combining with (3.4) we know there
exists C > 0 such that

dist(v,E−)∥v+∥ ≤ ∥v+∥2



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 15

≤ ε|u+|2|v+|2 + Cε|u+|p−1
p |v+|p

≤ C
(
εCε dist(u,E

−) + CεC
p−1
p (dist(u,E−))p−1

)
∥v+∥.

Therefore,

dist(Au,E−) ≤ C(ε dist(u,E−) + Cε(dist(u,E
−))p−1),

where C > 0 is different from the previous line. Then there exists ε0 > 0 such that,
for all ε ∈ (0, ε0) and u ∈ ∂D−

ε

dist(Au,E−) < ε.

In particular, we have A (∂D−
ε ) ⊂ D−

ε . A (∂D+
ε ) ⊂ D+

ε can be proved analogously.
□

By the Lemma 3.1, we only know that A is continuous. Next, we construct a
locally Lipschitz continuous operator B which inherits the properties of A. Similar
to [7, Lemma 2.1], we have the following lemma.

Lemma 3.3. There is a locally Lipschitz continuous odd operator B : E\K → E
satisfies the following properties:

(1) B (∂D−
ε ) ⊂ D−

ε , B (∂D+
ε ) ⊂ D+

ε ;
(2) 1

2∥u−Bu∥ ≤ ∥u−Au∥ ≤ 2∥u−Bu∥;
(3) ⟨Ψ′(u), u−Bu⟩ ≥ 1

2∥u−Au∥2;
(4) ∥Ψ′(u)∥ ≤ 2∥u−Bu∥;

where K = {u ∈ E | Ψ′(u) = 0}.

By a similar arguments as [15, Lemma 3.4], we have the following Lemma.

Lemma 3.4. Suppose that N is a symmetric closed neighborhood of Kc := {u ∈
E | Ψ′(u) = 0 , Ψ(u) = c}. Then there exists ε1 > 0 such that for 0 < ε < ε′ < ε1,
and a continuous map σ : [0, 1]× E → E satisfying:

(1) σ(0, u) = u,∀u ∈ E.
(2) σ(t, u) = u,∀t ∈ [0, 1],Ψ(u) /∈ [c− ε′, c+ ε′].
(3) σ(t,−u) = −σ(t, u),∀(t, u) ∈ [0, 1]× E.
(4) σ (1,Ψc+ε\N) ⊂ Ψc−ε.

(5) σ(t,D+
ε ) ⊂ D+

ε , σ(t,D
−
ε ) ⊂ D−

ε .

In particular, if N is a symmetric closed neighborhood of Kc\W , where W = D+
ε ∪

D−
ε , then there exists ε1 > 0 such that for 0 < ε < ε1 there will be a continuous

map η : E → E such that

(6) η(−u) = −η(u),∀u ∈ E.
(7) η|Ψc−2ε = id.
(8) η (Ψc+ε\(N ∪W )) ⊂ Ψc−ε.

(9) η(D+
ε ) ⊂ D+

ε , η(D
−
ε ) ⊂ D−

ε .

Proof of Theorem 1.1. Let λi, i = 1, 2, . . . be the ith eigenvalue of (2.5) and ei be
the eigenfunction corresponding to λi, Xj = span {e1, e2, · · · , ej}. Firstly, we define

M := {u ∈ E | 1
4
∥u∥2 >

∫
RN

F (x, u) dx} ∪Bρ,

where ρ > 0 such that

{u ∈ E | 1
4
∥u∥2 =

∫
RN

F (x, u) dx} ∩ ∂Bρ ̸= ∅.



16 R. QIU, R. YOU, F. ZHAO EJDE-2025/07

By Lemma 2.3, the interpolation inequality and the definition of λi, which is defined
in Proposition 2.2, for all u ∈ ∂M ∩ X⊥

j−1, there exists different constants C > 0
such that ∫

RN

F (x, u) dx ≤
∫
RN

(
ε|u|2 + Cε|u|p

)
dx

≤ C

∫
RN

|u|pdx

≤ C
(∫

RN

|u|2dx
)pθ/2(∫

RN

|u|2
∗
sdx

) p(1−θ)
2∗s

≤ Cλ
− pθ

2
j ∥u∥pθ∥u∥p(1−θ)

= Cλ
− pθ

2
j

(∫
RN

F (x, u) dx
)p/2

,

where θ ∈ (0, 1) satisfying 1
p = θ

2 + 1−θ
2∗s

, thus∫
RN

F (x, u) dx ≥ Cλ
pθ

p−2

j . (3.5)

By (3.5), for any u ∈ ∂M ∩X⊥
j−1, we have

Ψ(u) =
1

2
∥u∥2 −

∫
RN

F (x, u) dx =

∫
RN

F (x, u) dx ≥ Cλ
pθ

p−2

j .

Sincep pθ
p−2 > 0, we obtain

inf
u∈∂M∩X⊥

j−1

Ψ(u) ≥ Cλ
pθ

p−2

j → +∞ as j → +∞.

By similar arguments to those in Lemma 2.4, we can choose Rj large enough such
that Ψ(u) < 0 for u ∈ Xj \BRj

. Like in [18], we define

cj = inf
D∈Γj

sup
u∈D\W

Ψ(u),

where

Γj =
{
H

(
Xj+1 ∩BRj+1

)
: H ∈ C

(
Xj+1 ∩BRj+1

, E
)
, H is odd and

H
∣∣
Xj+1∩∂BRj+1

= id
}
.

Next, we assert that

(D\W ) ∩X⊥
j−1 ∩ ∂M ̸= ∅,∀D ∈ Γj , j ≥ 2.

In fact, by the definition of Γj , when D = H
(
Xj+1 ∩BRj+1

)
, we know H ∈

C
(
Xj+1 ∩BRj+1 , X

)
, H is odd and H|∂BRj+1

∩Xj+1 = id. Let Ô = {u ∈ Xj+1 ∩
BRj+1 | H(u) ∈ intM} and O be the connected component of Ô containing 0.
Clearly O is a bounded symmetric neighborhood of 0 in Xj+1 and O ∩ Xj+1 ∩
∂BRj+1

= ∅. By Borsuk’s theorem [30],

γ(∂O) = j + 1 and H(∂O) ⊂ ∂M,

where γ(∂O) denote the genus of ∂O, readers can learn more about the properties
of genus from [26, 30]. Now, we define I : W ∩ ∂M → R by

I(u) =

∫
RN

F
(
x, u+

)
dx−

∫
RN

F (x, u−) dx. (3.6)



EJDE-2025/07 TRAVELING WAVES FOR A CHEMOTAXIS MODEL 17

It is easy to see that I is an odd continuous map and 0 /∈ I(W ∩ ∂M). Indeed,
if 0 ∈ I(W ∩ ∂M), then there exists u ∈ W ∩ ∂M such that

∫
RN F (x, u+) dx =∫

RN F (x, u−) dx. When u ∈ W , we know∫
RN

F (x, u+) dx =

∫
RN

F (x, u−) dx ≤ Cε.

But when u ∈ ∂M , there exists C > 0 such that∫
RN

F (x, u) dx ≥ C > 0,

which is a contradiction when ε is small enough. As a consequence, γ(∂M∩W ) = 1.
Thus, γ((I(∂O)\W )∩∂M) ≥ j+1−1 = j, which is contradict to codim

(
X⊥

j−1

)
=

j − 1 < j. So H(∂O) \ W ∩ ∂M ∩ X⊥
j−1 ̸= ∅, since H(∂O) \ W ⊂ D \ W , then

the claim holds. To complete the proof, we only need to prove Kcj \W ̸= ∅, j ≥ 2.
Otherwise, it follows from Lemma 3.4 that there exists ε > 0 and an odd continuous
map η : E → E such that

η|Ψcj−2ε = id,

η(Ψcj+ε \W ) ⊂ Ψcj−ε,

η(D±
ε ) ⊂ D±

ε .

Hence, regarding the ε mentioned above, there exists D0 ∈ Γj such that

sup
u∈D0\W

Ψ(u) < cj + ε,

that is D0\W ⊂ Ψcj+ε. Let U := η (D0), it is easy to verify that U ∈ Γj and
cj ≤ supu∈U\W Ψ(u). Note that

U\W = η(D0)\W ⊂ (η(D0\W )∪η(W ))\W ⊂ η(D0\W )\W ⊂ η(Ψcj+ε\W ) ⊂ Ψcj−ε.

Thus

cj ≤ sup
u∈U\W

Ψ(u) ≤ cj − ε,

which is a contradiction. The proof is complete. □

Acknowledgments. This work was supported by the NSFC 12261107, and by the
Yunnan key Laboratory of Modern Analytical Mathematics and Applications.

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Ruowen Qiu
Department of Mathematics, Yunnan Normal University, Kunming, 650221, China

Email address: 1239814486@qq.com

Renqing You

Department of Mathematics, Yunnan Normal University, Kunming, 650221, China

Email address: 1768332868@qq.com

Fukun Zhao (corresponding author)

Department of Mathematics, Yunnan Normal University, Kunming, 650221, China
Email address: fukunzhao@163.com