Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
Changfeng Gui and Mingfeng Zhao
Department of Mathematics, University of Connecticut
Traveling Fronts with Fractional Laplaicans
The Continuation Method
Upper Bounds of Speed when 0 < s <
Let 0 < s < 1 and f = −F ′ for some F ∈ C 3(R), consider

su(x) − µu′(x) = f (u(x)), ∀x ∈ R

(−∆)


u′(x) > 0, ∀x ∈ R


 lim u(x) = ±1.
Let F1 be any fixed double well potential with F1(−1) < F1(1) and f1(t0) = 0 for some t0 ∈
(−1, 1). Take any double well potential F0 with F0(−1) = F0(1) = F1(−1), f1(t0) = 0 and
F0(t0) = F1(t0), define
For any R > 0, we know that
(1)
Fθ (t) = (1 − θ)F0(t) + θF1(t) and
fθ (t) = −Fθ′ (t),
(−∆)su(x) =
s22sΓ 1+2s
2
1
2
π Γ(1 − s)
P.V.
Z
u(x) − u(y)
dy,
1+2s
R |x − y|
Z 1
|z|<R 0 |z|
z
′ (y + tz) dzdt.
u
1+2s θ
By Cauchy-Schwarz’s inequality and (3), we have
Z 0
4C(1, s) −2s C(1, s) 1−2s ′ 2
s
′
(−∆) uθ (y)uθ (y) dy ≤
−
R
+
R
kuθ kL2(R)
s
1 − 2s
−∞
4C(1, s) −2s C(1, s) 1−2s F1(1) − F1(−1)
≤
R
+
R
s
1 − 2s
µθ
Where µ is a constant and
s
∀θ ∈ [0, 1].
x→±∞
2C(1, s) −2s
s
−(−∆) uθ (y) ≤
R
+ C1,s
Z
1
2
∀x ∈ R
Problem: Is there any pair (µ, u) as the solution to (1) ?
By (3), we know that
4C(1, s) −2s C(1, s) 1−2s F1(1) − F1(−1)
R
+
R
F1(t0) − F1(1) ≤
s
1 − 2s
µθ
Main Result
Unbalanced Bistable: Gui, Zhao(2012)
Let 0 < s < 1, f ′(±1) < 0, f (±1) = 0, and there exists some t0 ∈ (−1, 1) such that
f (t) < 0, ∀t ∈ (−1, t0),
and f (t) > 0, ∀t ∈ (t0, 1)
Then there exists a unique pair (µ, u) with µ ≥ 0 as the solution to (1) up to translation.
Moreover, we have
1
1 − u(x) ∼ 2s , as x → ∞ and
|x|
1
− 1 + u(x) ∼ 2s , as x → −∞
|x|
Note: The continuation method.
Balanced Bistable: Cabré, Solá-Morales(2005), Cabré, Sire (2010)
Let 0 < s < 1 and µ = 0, then (1) has a unique solution up to translation iff
and F (t) > F (−1) = F (1), ∀t ∈ (−1, 1)
Moreover, if f ′(±1) < 0, then
1
1 − u(x) ∼ 2s , as x → ∞ and
|x|
1
− 1 + u(x) ∼ 2s , as x → −∞
|x|
Combustion: Mellet, Roquejoffre, Sire(2011)
Let 21 < s < 1, f ′(−1) < 0 and there exists some t0 ∈ (−1, 1) such that
f (t) < 0, ∀t ∈ (−1, t0),
and f (t) = 0, ∀t ∈ [t0, 1]
Then there exists a unique pair (µ, u) with µ > 0 as the solution to (1) up to translation.
Moreover, there exists some C > 0 such that
C
1 − u(x) ≤ 2s−1 , as x → ∞
|x|
f (t) < 0, ∀t ∈ (−1, 1),
f ′(1) > 0
Then (1) has no solution.
Combustion: Gui, Huan(2012)
Let 0 < s ≤ 12 , f ′(−1) < 0 and there exists some t0 ∈ (−1, 1) such that
f (t) < 0, ∀t ∈ (−1, t0),
Upper Bounds of Speed when s =
and f (t) = 0, ∀t ∈ [t0, 1]
1
2
Let uθ be the harmonic extension in R2+ of uθ and
Z
ρ(y) = [|Dxuθ (x, y)|2 − |Dy uθ (x, y)|2] dx,
∀y ≥ 0
R
Then
Let µθ be the solution of the following problem:

′′(x) − µ u′ (x) = f (u (x)),
−u

θ θ
θ θ
θ

u′θ (x) > 0, ∀x ∈ R

 lim uθ (x) = ±1.
∀x ∈ R
R
x→±∞
Then
1 ′
[uθ (y)]2 + µθ
[u′θ (x)]2 dx = Fθ (uθ (y)) − Fθ (−1)
2
Z −∞
R
R
Dy uθ (x, y)∆uθ (x, y) dx ≡ 0,
R
R
−∞
≤
[u′θ (x)]2 dx = Fθ (1) − Fθ (−1)
−∞
F1(1) − F1(−1)
By (3), then
Let y0 ∈ R be such that u′θ (y0) = sup u′θ (y) ≥ u′θ (0), then u′′θ (y0) = 0. Hence, we get
y∈R
Upper Bounds of Speed when
F1(1) − F1(−1)
µθ ≤
,
F1(t0) − F1(1)
1
2
∀θ ∈ [0, 1]
<s<1
We can get an upper bound involving µθ as denominator of
Z 0
−
(−∆)suθ (x)u′θ (x) dx
∀θ ∈ [0, 1]
−∞
Discussion
Let (µθ , uθ ) be a solution of (1) with fθ , then
Z y
Z y
[u′θ (x)]2 dx = Fθ (uθ (y)) − Fθ (−1)
−
(−∆)suθ (x)u′θ (x) dx + µθ
−∞
−∞
Z
µθ [u′θ (x)]2 dx = Fθ (1) − Fθ (−1)
• What will happen if we add more zeros in (−1, 1) of f ?
• Let J be a nonnegative symmetric kernel such that J(x), J ′(x) ∈ L1(R) with
(2)
R
WOLG, assume uθ (0) = t0, then
Z 0
(−∆)suθ (x)u′θ (x) dx ≥ Fθ (t0) − Fθ (1) ≥ F1(t0) − F1(1)
−
−∞
R
µθ
WOLG, assume uθ (0) = t0, then
1 ′
[uθ (0)]2 ≥ Fθ (t0) − Fθ (1) ≥ F1(t0) − F1(1)
2
−fθ (uθ (y0)) kf0kC([−1,1]) + kf1kC([−1,1])
p
≤
,
µθ =
′
uθ (y0)
2[F1(t0) − F1(1)]
∀y > 0.
By Cauchy-Schwarz’s inequality and (2), then
Z 0
Z 0
Z
−
(−∆)suθ (x)u′θ (x) dx =
Dxuθ (x, 0)u′θ (x) dx ≤
|u′θ (x)|2 dx
Z y
µθ
ρ′(y) = 2
Z
Since ρ(y) → 0 as y → ∞, then ρ(y) ≡ 0 for all y ≥ 0, in particular,
Z
Z
Z
|Dy uθ (x, 0)|2 dx
|u′θ (x)|2 dx =
|Dxuθ (x, 0)|2 dx =
Hamiltonian Identity
Fisher-KPP: Cabré, Roquejoffre(2011)
Let 0 < s < 1 and f satisfies
f ′(−1) < 0and
4C(1, s) −2s F1(t0) − F1(1)
Let R be such that
R
=
, then
s
2
µθ ≤ C, ∀θ ∈ [0, 1]
Note: The key is that F1(t0) > F1(1).
Nonexistence Results
Then (1) has no solution.
= {θ ∈ [0, 1] : (1) has a solution (µθ , uθ ) with fθ }
X
• The balanced bistable result implies 0 ∈
.
X
• The openness of
follows from the nondegeneracy of linearized equation, since u′θ (x) > 0
in R which is the first eigenfunction.
X
• The closeness of
follows from the uniform boundedness of speed µθ for all θ ∈ [0, 1].
Motivation from the ODE Case
Existence Results
f (±1) = 0,
Consider the solution set
X
J(x) dx = 1
R
and f be a combustion or Fisher-KPP nonlinearity. Do we have solution to the problem

′(x) = f (u(x)), ∀x ∈ R

−J
∗
u(x)
+
u(x)
−
µu

u′(x) > 0, ∀x ∈ R

 lim u(x) = ±1.
x→±∞
(3)
Z
• Other problems related with traveling waves.