Non-existence of localized travelling waves with
non-zero speed in single reaction-diffusion equations
Yong Jung Kim
∗
Department of Mathematical Sciences
Korea Advanced Institute of Science and Technology
Daejon 305-701, Republic of Korea
Wei-Ming Ni
†
Center for Partial Differential Equations
East China Normal University
Minhang, Shanghai, 200241, P.R. China
and
Department of Mathematics, University of Minnesota
127 Vincent Hall, 206 Church St. S.E.,
Minneapolis, MN 55455 USA
Masaharu Taniguchi
‡
Department of Mathematical and Computing Sciences,
Tokyo Institute of Technology,
O-okayama 2-12-1-W8-38, Meguro-ku, Tokyo 152-8552 Japan
Abstract
Assume a single reaction-diffusion equation has zero as an asymptotically stable
stationary point. Then we prove that there exist no localized travelling waves with
non-zero speed. That is, if [lim inf |x|→∞ u(x), lim sup|x|→∞ u(x)] is included in an open
interval of zero that does not include other stationary points, then the speed has to
be zero or the travelling profile u has to be identically zero.
2010 Mathematics Subject Classification: 35C07, 35K57
Key words: travelling waves, travelling spots, reaction-diffusion equations
∗
[email protected]
[email protected]
‡
[email protected], This work was supported by Grant-in-Aid for Scientific Research
(C), 23540235, Japan Society for the Promotion of Science, 2011-2013.
†
1
1
Introduction
In this paper we consider the following equation
∂u
= ∆u + f (u)
∂t
u|t=0 = u0
in Rn , t > 0,
(1.1)
in Rn .
Here n ≥ 1 and u0 is a given bounded uniformly continuous function. The Laplacian ∆
def ∑n
2
2
1
stands for ∆ =
j=1 ∂ /∂xj . Here f ∈ C (R) satisfies f (0) = 0. In addition, we assume
either
f ′ (0) < 0
(1.2)
or
0 < lim inf
u→0
f (u)
f (u)
≤ lim sup
<∞
p
−u|u|
−u|u|p
u→0
(1.3)
for a positive constant p with
n−1−
2
< 0.
p
Typical examples are
(a) f (u) = −u + |u|q−1 u with q > 1;
(b) f (u) = −u(u − b0 )(u − 1) with 0 < b0 < 1;
(c) f (u) = u|u|(u − 1) with n = 1, 2.
We study a travelling wave solution v(x1 , . . . , xn −ct) with speed c. We write y = xn −ct
and v(x1 , . . . , xn−1 , y) satisfies
)
( n−1
2
∑ ∂2
∂
∂v
+ 2 v+c
+ f (v) = 0
(x1 , . . . , xn−1 , y) ∈ Rn .
2
∂xj
∂y
∂y
j=1
This equation is the profile equation for a travelling wave solution v(x1 , . . . xn−1 , xn − ct).
We write v(x1 , . . . , xn−1 , y) by u(x1 , . . . , xn ). Then we have
∆u + c
∂u
+ f (u) = 0
∂xn
in Rn .
(1.4)
In this paper we study this equation with
[lim inf u(x), lim sup u(x)] ⊂ (−a0 , b0 ).
|x|→∞
|x|→∞
Here we put
b0
def
=
sup{b ∈ (0, ∞) | f (s) < 0 for all s ∈ (0, b)} ∈ (0, ∞],
a0
def
sup{a′ ∈ (0, ∞) | f (s) > 0 for all s ∈ (−a′ , 0)} ∈ (0, ∞],
=
2
(1.5)
and have −a0 < 0 < b0 . If u satisfies lim|x|→∞ u(x) = 0, we have (1.5). If u satisfies (1.4)
and lim|x|→∞ u(x) = 0, we call u a localized travelling wave solution.
If c = 0, a localized travelling wave is a stationary solution. Many works have been
studied for the existence, uniqueness and the spherical symmetry of stationary solutions.
For the existence and the uniqueness of stationary solutions, see [14] , [3] , [4], [5], [16], [12]
and [19] for instance. For the spherical symmetry of positive solutions, see [7], [8] and [13]
for instance.
Localized travelling waves for systems of reaction-diffusion equations are studied in [15],
[10] and [6] for examples, and they are sometimes called travelling spots or travelling spikes.
For a single reaction-diffusion one might suppose that (1.1) is a gradient system for an
energy functional and thus a localized initial state with finite energy has to be a stationary
state or cannot keep its shape as time goes on. This intuitive idea suggests that there exist
no localized travelling wave solutions with speed c ̸= 0. The aim of this paper is to give a
simple proof of this non-existence.
The main assertion of this paper is as follows.
Theorem 1. Assume f ∈ C 1 (R) satisfies f (0) = 0. In addition assume f ′ (0) < 0 or (1.3).
If u ∈ L∞ (Rn ) ∩ C 2 (Rn ) satisfies (1.4) and (1.5), one has c = 0 or u ≡ 0.
Remark 1. If one assumes either
[lim inf u(x), lim sup u(x)] ⊂ [−a0 , b0 ) or [lim inf u(x), lim sup u(x)] ⊂ (−a0 , b0 ]
|x|→∞
|x|→∞
|x|→∞
|x|→∞
in stead of (1.5), Theorem 1 does not hold true. Indeed, we put n = 1, a0 = 1, b0 = 1, and
choose f with f ′ (−1) > 0,
f ′ (u) < f ′ (−1)
f (u) > 0,
for all u ∈ (−1, 0),
√
and have a monotone decreasing solution connecting 0 and −1 with any speed c ≥ 2 f ′ (−1)
to (1.4) by [11, 1, 2]. Thus the interval (−a0 , b0 ) in the condition (1.5) is maximal in
Theorem 1.
2
Proof of Theorem 1
We put
def
∫
F (u) =
u
f (s) ds.
0
def
def
For any R > 0 we use B(0; R) = {x ∈ Rn | |x| ≤ R} and ∂B(0; R) = {x ∈ Rn | |x| = R}.
We write the outward unit vector ν = (ν1 , . . . , νn ) at x ∈ ∂B(0; R) as
ν=
x
x
=
|x|
R
for x ∈ ∂B(0; R).
3
We denote ∂/∂xj by Dj and ∂ 2 /∂xi ∂xj by Dij for 1 ≤ i, j ≤ n. Under the assumption of
Theorem 1 we put
{
}
def
def
′
A = 1 + sup |u(x)|,
M = max 1, max |f (s)| ,
−A≤s≤A
x∈Rn
B1
def
=
lim inf u(x),
|x|→∞
def
B2 = lim sup u(x),
|x|→∞
and have −A ≤ B1 ≤ B2 ≤ A.
Lemma 1. If u ∈ L∞ (Rn ) ∩ C 2 (Rn ) satisfies (1.4), one has
(
)
∫
∫
∫
1
2
2
c
(Dn u) =
|∇u| − F (u) νn −
Dn u(∇u, ν).
2
B(0;R)
∂B(0;R)
∂B(0;R)
Proof. Using
div (Dn u∇u) = Dn u∆u +
n
∑
(2.6)
Djn uDj u
j=1
and
(
) ∑
1
Dn |∇u|2 =
Dj uDjn u,
2
j=1
n
we get
(
)
1
div (Dn u∇u) = Dn |∇u|2 + Dn u∆u.
2
Multiplying (1.4) by Dn u, we have
(
)
1
0 = div (Dn u∇u) − Dn |∇u|2 + c(Dn u)2 + Dn (F (u)) .
2
∫
∫
Applying the Gauss divergence formula B(0;R) div F = ∂B(0;R) (F, ν), we obtain
∫
∫
∫
∫
1
2
2
0=
Dn u(∇u, ν) −
|∇u| νn + c
(Dn u) +
F (u)νn .
2 ∂B(0;R)
∂B(0;R)
B(0;R)
∂B(0;R)
This completes the proof.
We take δ0 > 0 small enough such that we have
sf (s) < 0
if s ∈ [−δ0 , δ0 ]\{0}
and
1
|f (s)|
|f (s)|
− f ′ (0) ≤
inf
≤
sup
≤ −2f ′ (0)
s∈[−δ0 ,δ0 ]\{0} |s|
2
|s|
s∈[−δ0 ,δ0 ]\{0}
a≤
|f (s)|
|f (s)|
≤
sup
< +∞
1+p
1+p
s∈[−δ0 ,δ0 ]\{0} |s|
s∈[−δ0 ,δ0 ]\{0} |s|
inf
4
if f ′ (0) < 0,
if f satisfies (1.3).
Here a ∈ (0, ∞) is a constant. We put
1
def
α = − f ′ (0) > 0
2
if f ′ (0) < 0.
For any x0 ∈ [−δ0 , δ0 ] we define T (t; x0 ) by
T ′ (t; x0 ) = f (T (t; x0 ))
T (0; x0 ) = x0 .
t > 0,
Here ′ means d/dt. If x0 ̸= 0, we have
∫ T (t;x0 )
∫ x0
ds
ds
t=
=
.
f (s)
x0
T (t;x0 ) −f (s)
We put
1
min {1, a0 + B1 , b0 − B2 }
if a0 < ∞, b0 < ∞,

2


 1 min {1, b − B }
if a0 = ∞, b0 < ∞,
def
0
2
2
ε0 =
1

min {1, a0 + B1 }
if a0 < ∞, b0 = ∞,


 12
if a0 = ∞, b0 = ∞,
2
and have 0 < ε0 ≤ 1/2 and [B1 − ε, B2 + ε] ⊂ (−a0 , b0 ) for all ε ∈ (0, ε0 ). For all ε ∈ (0, ε0 )
there exists rε > 0 such that
B1 − ε ≤ u(x) ≤ B2 + ε
if |x| ≥ rε .
(2.7)
Lemma 2. Suppose f ′ (0) < 0. For x0 ∈ [−δ0 , δ0 ] one has
|T (t; x0 )| ≤ |x0 |e−αt
for all t ≥ 0.
One has
sup eαt max {|T (t; B1 − ε)|, |T (t; B2 + ε)|} < +∞.
t≥1
Proof. The assertion holds true for the case x0 = 0. For the former statement we give a
proof for the case x0 > 0, since the case x0 < 0 can be proved similarly. Using
1
1
≤
−f (s)
αs
for all s ∈ (0, δ0 ], we get
∫
x0
T (t;x0 )
ds
=t≤
−f (s)
∫
x0
T (t;x0 )
ds
.
αs
This gives
log T (t; x0 ) ≤ −αt + log x0
for all t ≥ 0. The latter statement follows from the former statement.
5
Lemma 3. Suppose f satisfies (1.3). For x0 ∈ [−δ0 , 0) ∪ (0, δ0 ] one has
(
)− 1
|T (t; x0 )| ≤ apt + |x0 |−p p
One has
for all t ≥ 0.
1
sup(apt) p max {|T (t; B1 − ε)|, |T (t; B2 + ε)|} < +∞.
t≥1
Proof. For the former statement we give a proof for the case x0 > 0, since the case x0 < 0
can be proved similarly. We have
∫ x0
ds
t≤
,
1+p
T (t;x0 ) as
which gives
)
1 (
T (t; x0 )−p − x−p
.
0
ap
The latter statement follows from the former statement.
t≤
We define a constant k > 0 as
{
1 + supt≥1 eαt max {|T (t; B1 − ε)|, |T (t; B2 + ε)|}
if f satisfies f ′ (0) < 0,
def
k =
1
1 + supt≥1 (apt) p max {|T (t; B1 − ε)|, |T (t; B2 + ε)|} if f satisfies (1.3).
We define a positive constant β as

{
}
1
1
1


,√
if f satisfies (1.2),
 min
def
4
{ max{1, |c|} α + M }
β =
1
1
1


,√
if f satisfies (1.3).
 min
4
max{1, |c|} 1 + M
Using a method in [18], we will prove the following a priori estimate.
Proposition 1. For any ε ∈ (0, ε0 ) let rε > 0 be as in (2.7). Assume that u ∈ L∞ (Rn ) ∩
C 2 (Rn ) satisfies (1.4) and (1.5). In the case of (1.2) one has
|u(x)| ≤ 2ke−αβ|x|
if
(2.8)
{
(
)}
√
1 rε (β|c| + n)
1
2A(rε )n
|x| ≥ max 1, ,
+
log
.
n
β
2αβ 2
αβ
k(πβ) 2
In the case of (1.3) one has
|u(x)| ≤ 2k (apβ|x|)− p
1
if
{
1 rε (β|c| +
|x| ≥ max 1, ,
β
2β 2
√
n)
6
1
+ log
β
(2.9)
(
1
2A(rε )n a p
n
1
k(πβ) 2 e p
)}
.
Proof. We only give a proof for the upper estimate on u, since the lower estimate on u can
be proved similarly.
def
We put w(x, t) = u(x) − T (t; B2 + ε) for x ∈ Rn and t > 0. Then we have
wt − ∆w − cDn w = f (u(x)) − f (T (t; B2 + ε))
for x ∈ Rn and t > 0. Thus we have
∫ 1
wt − ∆w − cDn w − w
f ′ (θu(x) + (1 − θ)T (t; B2 + ε)) dθ = 0
for all x ∈ Rn , t > 0,
0
w(x, 0) = u(x) − B2 − ε
for all x ∈ Rn .
We introduce w(x,
e t) by
∫ 1
w
et − ∆w
e − cDn w
e−w
e
f ′ (θu(x) + (1 − θ)T (t; B2 + ε)) dθ = 0
0
{
2A
if |x| ≤ rε ,
w(x,
e 0) =
0
if |x| > rε .
for all x ∈ Rn , t > 0,
By the maximum principle, we obtain
0 ≤ w(x,
e t)
w(x, t) ≤ w(x,
e t)
for all x ∈ Rn , t > 0,
for all x ∈ Rn , t > 0.
For the maximum principles see [17] or [9] for instance. We introduce W (x, t) by
Wt − ∆W − cDn W − M W = 0
for all x ∈ Rn , t > 0,
{
2A
if |x| ≤ rε ,
W (x, 0) =
0
if |x| > rε .
We have
w(x,
e t) ≤ W (x, t)
for all x ∈ Rn , t > 0,
and
u(x) ≤ T (t; B2 + ε) + W (x, t)
Using
for all x ∈ Rn , t > 0.
( ∑n−1
)
2
2
+
(x
+
ct)
x
1
n
j
j=1
K(x, t) = eM t
−
,
n exp
4t
(4πt) 2
def
∫
we have
K(x − y, t) dy.
0 < W (x, t) = 2A
B(0;rε )
Using B(0; rε ) ⊂ [−rε , rε ]n we have
∫
K(x − y, t) dy,
0 < W (x, t) < 2A
[−rε ,rε ]n
7
(2.10)
and thus
∫
0 < W (x, t) < 2Ae
Mt
xn +r
√ ε +ct
4t
xn −r
√ ε +ct
4t
n−1 ∫
∏
e−z
√ dz
π
j=1
2
xj +rε
√
4t
xj −rε
√
4t
e−z
√ dz
π
2
(2.11)
for all x ∈ Rn and all t > 0.
Then (2.10) and (2.11) give
u(x) ≤ T (t; B2 + ε) + 2Ag(x, t)
(2.12)
for all x ∈ Rn and all t > 0, where
∫
def
g(x, t) = eM t
x +r
xn + rε + ct
∫ j√ ε −z2
√
n−1
−z 2
∏
e
4t
4t e
xj − rε √π dz.
xn − rε + ct √π dz
√
√
j=1
4t
4t
Now we study g. Using
min{(xj + rε )2 , (xj − rε )2 } ≥ |xj |2 − 2rε |xj | + rε2 = (|xj | − rε )2 ,
we have
xj + rε
(
)
√
2
−z 2
e
(|x
|
−
r
)
r
j
ε
ε
0 < x −4t
rε √π dz ≤ √πt exp −
j
4t
√
4t
∫
if |xj | > rε ,
and
xj + r ε
√
2
e−z
rε
√
√
0 < x −4t
dz
≤
r
j
π
πt
√ ε
4t
Combining these estimates together, we obtain
∫
if |xj | ≤ rε .
xj + r ε
(
)
( 2)
√
2
e−z
rε
rε
(|xj | − rε )2
4t
0 < x − r √ dz ≤ √ exp
exp −
j
4t
4t
π
πt
√ ε
4t
∫
for all xj ∈ R and t > 0. Using
}
{
min (xn + ct + rε )2 , (xn + ct − rε )2 ≥ (|xn + ct| − rε )2 ,
we obtain
∫
xn + rε + ct
(
)
√
2
−z 2
(|x
+
ct|
−
r
)
e
r
n
ε
ε
4t
xn − rε + ct √π dz ≤ √πt exp −
4t
√
4t
8
if |xn + ct| > rε ,
and
xn + rε + ct
√
2
e−z
rε
4t
√
dz ≤ √
xn − rε + ct
π
πt
√
4t
Combining these two estimates together, we obtain
∫
if |xn + ct| ≤ rε .
xn + rε + ct
(
)
( 2)
√
2
−z 2
r
(|x
e
r
+
ct|
−
r
)
ε
√ dz ≤ √ ε exp ε exp − n
0 < x − r 4t
n
ε + ct
4t
4t
π
πt
√
4t
∫
for all xn ∈ R and t > 0. Now we obtain
0 < g(x, t) ≤ e
Mt
(rε )n
n
(πt) 2
(
exp
n(rε )2
4t
where
def
J(x, t) = (|xn + ct| − rε ) +
2
)
n−1
∑
(
J(x, t)
exp −
4t
)
,
(|xj | − rε )2 .
j=1
Now we get
J(x, t) = |x| + 2ctxn + c t − 2rε |xn + ct| − 2rε
2
2 2
n−1
∑
|xj | + n(rε )2 .
j=1
Now we set t = β|x| and have
J(x, β|x|) = |x| + 2βcxn |x| + β c |x| − 2rε
2
2 2
2
n−1
∑
|xj | − 2rε |xn + ct| + nrε2 .
j=1
Using
n
∑
|xj | ≤
√
n|x|,
j=1
we get
√
J(x, β|x|) ≥ (1 + β 2 c2 − 2β|c|)|x|2 − 2rε n|x| − 2rε β|c||x| + nrε2 ,
and thus
√
J(x, β|x|) ≥ (1 − β|c|)2 |x|2 − 2rε ( n + β|c|)|x| + nrε2 .
From the definition of β we have
(1 − β|c|)2 ≥
and
1
2
√
1
J(x, β|x|) ≥ |x|2 − 2rε (β|c| + n)|x| + nrε2 .
2
Thus we get
(rε )n
0 < g(x, β|x|) ≤
n exp
(πβ|x|) 2
(
rε (β|c| +
2β
9
√
n)
)
) )
( (
1
− M β |x| .
exp −
8β
From the definition of β we have
2αβ <
2β <
1
− Mβ
8β
1
− Mβ
8β
if f ′ (0) < 0,
if f satisfies (1.3).
In the case of f ′ (0) < 0 we have
u(x) ≤ 2ke−αβ|x|
if
αβ|x|
e
2A(rε )n
≥
n exp
k(πβ|x|) 2
(
rε (β|c| +
2β
√
n)
)
.
This gives (2.8).
In the case of (1.3) we have
(
√ )
2A(rε )n
rε (β|c| + n) −2β|x|
u(x) ≤ k (apβ|x|) +
e
n exp
2β
(πβ|x|) 2
)
(
(
√ )
1
rε (β|c| + n)
2A(rε )n
− p1
−β|x|
−β|x|
e
≤ k (apβ|x|)
1+
(apβ|x|) p e
n exp
2β
k(πβ) 2
(
(
)
√ )
rε (β|c| + n) ( a ) p1 −β|x|
2A(rε )n
− p1
≤ k (apβ|x|)
1+
e
n exp
2β
e
k(πβ) 2
− p1
if |x| ≥ 1. Here we used
( a ) p1
1
sup(apβr) p e−βr =
e
r>0
.
This gives (2.9) and completes the proof.
2,s
Let s ∈ (n, +∞) be arbitrarily given. If w ∈ Wloc
(B(0; 2)) ∩ Ls (B(0; 2)) and h ∈
s
L (B(0; 2)) satisfy
(−∆ − cDn )w = h
in B(0; 2),
one has
∥w∥W 2,s (B(0;1)) ≤ K0 ∥h∥Ls (B(0;2))
by the Schauder interior estimate, where K0 is a positive constant. See [9, Theorem 9.11]
for a general theory. Then W 2,s (B(0; 1)) ⊂ C 1 (B(0; 1)) and L∞ (B(0; 2)) ⊂ Ls (B(0; 2))
yield
∥w∥C 1 (B(0;1)) ≤ K∥h∥L∞ (B(0;2)) ,
where K is a positive constant. See also [9] for the Sobolev imbedding theorems.
def
Proof of Theorem 1. For any y ∈ Rn we put w(x) = u(x + y) and have
−∆w(x) − cDn w(x) = f (w(x))
10
for x ∈ B(y; 2).
The Schauder estimate gives
∥w∥C 1 (B(0;1)) ≤ K∥f (w)∥L∞ (B(0;2)) ,
which yields
|∇u(y)| ≤ K sup |f (u(y + x))|
for all y ∈ Rn .
|∇u(x)| ≤ KM sup |u(x + y)|
for all x ∈ Rn .
|x|≤2
Especially we get
|y|≤2
In the case of (1.2) we obtain
|∇u(x)| ≤ 2kKM e−αβ(|x|−2) ≤ 2kKM e2αβ e−αβ|x|
{
(
)}
√
1 rε (β|c| + n)
1
2A(rε )n
|x| ≥ max 3, ,
+
log
.
n
β
2αβ 2
αβ
k(πβ) 2
if
In the case of (1.3) we obtain
|∇u(x)| ≤ 2kKM 3 p (apβ|x|)− p
1
{
if
1 rε (β|c| +
|x| ≥ max 3, ,
β
2β 2
√
n)
1
+ log
β
1
(
1
2A(rε )n a p
1
n
)}
.
k(πβ) 2 e p
Using estimates on u and |∇u|, we get
∫
∂B(0;R)
(
)
∫
1
2
|∇u| − F (u) νn −
Dn u(∇u, ν)
2
∂B(0;R)
(
)
n
M
2π 2 n−1 3
2
2
max |∇u| +
max |u| .
≤ (n)R
2 ∂B(0;R)
2 ∂B(0;R)
Γ 2
Here Γ is the Gamma function. Recall that 2π n/2 /Γ (n/2) is the area of a unit ball in Rn .
If f ′ (0) < 0, we have
∫
)
(
∫
1
2
|∇u| − F (u) νn −
Dn u(∇u, ν) = 0
lim R→∞
2
∂B(0;R)
∂B(0;R)
from Proposition 1.
If f satisfies (1.3), we have
max |u| = O(R− p ),
max |∇u| = O(R− p )
1
∂B(0;R)
1
∂B(0;R)
as R → ∞ using Proposition 1. Thus we obtain
∫
(
)
∫
1
2
= O(Rn−1− p2 )
|∇u|
−
F
(u)
ν
−
D
u(∇u,
ν)
n
n
2
∂B(0;R)
∂B(0;R)
11
as R → ∞. Using
n−1−
2
<0
p
and sending R → ∞ in (2.6) in Lemma 1, we obtain
∫
c
(Dn u)2 = 0
Rn
for both cases. This equality gives c = 0 or Dn u ≡ 0. If Dn u ≡ 0, the condition
lim|x|→∞ u(x) = 0 implies u ≡ 0. Thus we find that c = 0 or u ≡ 0. This completes
the proof of Theorem 1.
3
Applications
Let n ≥ 3, 1 < q < (n + 2)/(n − 2) and f (u) = −u + |u|q−1 u. Equation (1.4) for c = 0 is
called the scalar field equation. It is well known that there exists a unique positive U of
class C 2 [0, ∞) satisfying
U ′′ (r) +
n−1 ′
U (r) + f (U (r)) = 0 for all r > 0,
r
U ′ (0) = 0, lim U (r) = 0.
r→∞
If u > 0 satisfies
∆u + f (u) = 0
x ∈ Rn ,
u(x) = U (|x − y|) for some y ∈ Rn from [7, 8]. From Theorem 1 in this paper one has c = 0
if u ̸≡ 0 satisfies (1.4) and (1.5) for some c ∈ R. Thus there exist no localized travelling
waves with non-zero speed for f (u) = −u + |u|q−1 u.
We give another example of f that satisfies (1.3). Let f (u) = u|u|(u − 1). If u > 0
satisfies
∆u + f (u) = 0
x ∈ Rn ,
u has to be spherically symmetric for some point in R2 from [13]. For an example of
spherically symmetric solution is as follows. For n = 2, the following equation
U ′′ (r) + 1r U ′ (r) + f (U (r)) = 0 for all r > 0,
lim U (r) = 0.
(3.13)
r→∞
has a solution
4
.
+2
From Theorem 1 one has c = 0 if n = 1, 2 and u ̸≡ 0 satisfies (1.4) and (1.5) for some c ∈ R.
Thus there exist no localized travelling waves with non-zero speed for f (u) = u|u|(u − 1)
when n = 1, 2. However our assumption does not hold true when p = 1 and n ≥ 3. Thus it
is an interesting open problem to prove the existence or non-existence of travelling waves
for f = u|u|(u − 1) when n ≥ 3 or more generally for f with (1.3) when n ≥ 1 + (2/p).
U (r) =
r2
12
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14