Traveling Wave Solutions of Allen-Cahn Equation with
Fractional Laplacians
Mingfeng Zhao
University of Connecticut, USA
October 14, 2012
Joint work with Professor
Mingfeng Zhao (UConn)
Changfeng Gui
Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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Outline
1
Traveling Wave Solutions
2
Fractional Laplacians
3
Traveling Wave Solutions with Boundary Reactions
4
Traveling Wave Solutions with Fractional Laplacians
5
The Main Result and Outline of the Proof
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Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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Traveling Wave Solutions
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Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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Traveling Wave Solutions
Assume f = −F 0 for F ∈ C 3 (R), consider
ut (x, t) − ∂xx u(x, t) = f (u),
∀x ∈ R, t > 0
(1.1)
The traveling wave solution u of (1.1) is of the form:
u(x, t) = φ(x − µt)
where µ is a constant. Then
−φ00 (x) − µφ0 (x) = f (φ(x)),
∀x ∈ R
(1.2)
Three types of traveling waves:
u is a train if φ is periodic.
u is a front if φ is bounded, nonconstant and monotone.
u is a pulse if φ is bounded, nonconstant and has same limit at ±∞.
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Traveling Fronts
In the case of traveling front, WLOG, assume µ ≥ 0 and φ is increasing and has
limits ±1 at ±∞,i.e.,

−φ00 (x) − µφ0 (x) = f (φ(x)), ∀x ∈ R



φ0 (x) > 0, ∀x ∈ R
(1.3)


 lim φ(x) = ±1.
x→±∞
Then
Z y
1 0
2
[φ (y )] + µ
[φ0 (x)]2 dx = F (φ(y )) − F (−1)
2
−∞
Z
µ [φ0 (x)]2 dx = F (1) − F (−1)
(1.4)
(1.5)
R
F (t) > F (−1),
∀t ∈ (−1, 1)
(1.6)
There are three interesting nonlinearities:
Fisher-KPP, Combustion and Bistable
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Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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Fisher-KPP Model
Fisher-KPP Model:
f (t) < 0, ∀t ∈ (−1, 1),
f 0 (−1) < 0
and f 0 (1) > 0
Figure: Fisher-KPP Model
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Combustion Model
Combustion Model:
There exists some t0 ∈ (−1, 1) such that
(
f (t) < 0, if t ∈ (−1, t0 )
and f 0 (−1) < 0
f (t) = 0, if t ∈ [t0 , 1]
Figure: Combustion Model
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Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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Bistable Model: Allen-Chan
Bistable Model: There exists some t0 ∈ (−1, 1) such that
(
f (t) < 0, if t ∈ (−1, t0 )
and f 0 (±1) < 0
f (t) > 0, if t ∈ (t0 , 1)
Figure: Bistable Model
In this case, F is called a double well potential.
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Double Well Potential with Equal Depths
If F (−1) = F (1), we say F has equal depths, f is balanced bistable.
Note: µ = 0.
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Double Well Potential with Non Equal Depths
If F (−1) < F (1), we say F has non equal depths, f is unbalanced bistable.
Note: µ > 0.
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Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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Classical Results

−φ00 (x) − µφ0 (x) = f (φ(x)),



φ0 (x) > 0, ∀x ∈ R


 lim φ(x) = ±1.
∀x ∈ R
(1.3)
x→±∞
Combustion Model:
There exists a unique pair (µ, φ) as the solution to (1.3).
Fisher-KPP Model:
There exists a minimal speed µ0 such that for any µ ≥ µ∗ , there is a unique φ such
that (µ, φ) is a solution to (1.3), where
s
p
f (t)
∗
sup
2 f 0 (1) ≤ µ ≤ 2
−1<t<1 t − 1
Bistable Model:
There exists a unique pair (µ, φ) as the solution to (1.3). Moreover, there exists
some c1 , c2 > 0 such that
1 − φ(x) ∼ e −c1 x , as x → ∞ and
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− 1 + φ(x) ∼ e c2 x , as x → −∞
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Fractional Laplacians
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Fractional Laplacians
For any 0 < s < 1, the fractional Laplacian (−∆)s arises from Lévy process, in
fact, (−∆)s serves as the infinitesimal generator of the 2s-stable symmetric Lévy
process.
Mathematically, for any 0 < s < 1 and f ∈ S(Rn ), the fractional Laplacian (−∆)s u
of u is defined as (−∆)s u = F −1 ((2π| · |)2s û), i.e.,
Z
(−∆)s u(x) =
e 2πix·ξ (2π|x|)2s û(y ) dy , ∀x ∈ Rn .
Rn
Γ n−2s
2
|x|2s−n .
Note: The fundamental solution of (−∆) is 2s n
2 π 2 Γ(s)
s
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Integral Form of (−∆)s
Using Fourier transform, we can see
Z
u(x) − u(y )
(−∆)s u(x) = C (n, s)P.V.
dy
(2.1)
n+2s
n
R |x − y |
Z
C (n, s)
u(x + y ) + u(x − y ) − 2u(x)
=−
dy
(2.2)
2
|y |n+2s
n
R
Z
u(x) − u(y ) − ∇u(x) · (x − y )1B1 (x) (y )
dy (2.3)
= C (n, s)
|x − y |n+2s
Rn
s22s Γ n+2s
2
where C (n, s) = n
.
π 2 Γ(1 − s)
Note: In (2.1), if 0 < s < 12 , the principal value is not needed, but if
the principal value MUST be needed.
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Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
October 14, 2012
1
2
≤ s < 1,
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Z
Proof: Let g (x) =
Rn
form, we obtain
f (x + y ) + f (x − y ) − 2f (x)
dy , by taking Fourier trans|y |n+2s
[e 2πiz·y + e −2πiz·y − 2] fˆ(z)
dy
|y |n+2s
Rn
Z
|e πiz·y − 1|2
−fˆ(z)
dy
|y |n+2s
Rn
z
Z
|e i 2|z| ·x − 1|2
2s ˆ
−(2π|z|) f (z)
dx
|x|n+2s
Rn
Z
ĝ (z)
=
=
=
Z
Note:
Rn
Let x = 2π|z|y .
z
|e i 2|z| ·x − 1|2
dx is a positive constant function on Rn \{0}.
|x|n+2s
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(−∆)s Looks Like −∆
The fractional Laplacian (−∆)s is an integral and nonlocal operator, but it shares
many similar properties with usual Laplacian −∆, such as, the strong maximum
principle, Harnack inequality, Poison kernel, potential theory, regularity theory,
etc....
Note: The strong maximum principle and Harnack inequality hold for (−∆)s in the
sense of global maximum principle and global Harnack inequality. For example, let
(−∆)s u(x) ≤ 0 in B1 (0) and assume for some x0 ∈ B1 (0), we have
u(x0 ) = sup u(x)
x∈Rn
Then u(x) ≡ u(x0 ) in Rn .
In fact, M. Kassmann(2007) constructed a function u ∈ C 2 (B1 (0)) such that

s

 (−∆) u(x) = 0, in B1 (0)
|u(x)| ≤ 1,
in Rn


u(x) > 0 = u(0), in B1 (0)\{0}.
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s-Harmonic Extension
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s-Poisson Kernel
s-Poisson Kernel:
For any 0 < s < 1, the s-Poisson kernel on Rn+1
is defined as
+
Γ n+2s
y 2s
Pys (x) = n 2
∀(x, y ) ∈ Rn+1
n+2s ,
+
π 2 Γ(s) [|x|2 + y 2 ] 2
Note: If s 6= 12 , then
Pys (x)6=
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Z
2s
e −2πy |ξ| e −2πiξ·x dξ
Rn
Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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Caffarelli-Silvestre’s s-Harmonic Extension
L. Caffarelli, L. Silvestre(2007):
Let 0 < s < 1, for any u ∈ C 2 (Rn ), let u(x, y ) = Pys ∗ u(x) for all (x, y ) ∈ Rn+1
+ .
n+1
2
Then u ∈ C (R+ ) and

div[y 1−2s ∇u(x, y )] = 0, ∀(x, y ) ∈ Rn+1

+




n
lim u(x, y ) = u(x), ∀x ∈ R
(2.4)
y &0

1−2s

2
Γ(1
−
s)


(−∆)s u(x), ∀x ∈ Rn
 − lim y 1−2s Dy u(x, y ) =
y &0
Γ(s)
Note: The operator div[y 1−2s ∇u(x, y )] is a degenerate and divergence operator
with A2 -weight |y |1−2s . E. Fabes, D. Jerison, C. Kenig, R. Serapioni (1982)
developed the whole machinery about equations involving the divergence operators
with A2 -weight.
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Proof: Because div [y 1−2s ∇Pys (x)] = 0 and Pys is an approximate identity, one can
derive div [y 1−2s ∇f¯(x, y )] = 0 and lim f¯(x, y ) = f (x). On the other hand, we
y &0
know
f¯(x, y ) − f (x)
− lim y 1−2s Dy f¯(x, y ) = −2s lim
y &0
y &0
y 2s
Z
n+2s
Γ
y 2s [f (z) − f (x)]
dz
= −2s n 2
lim y −2s
2
2 n+2s
π 2 Γ(s) y &0
Rn [|z − x| + y ] 2
Z
Γ n+2s
f (x + w ) + f (x − w ) − 2f (x)
2
= −s n
lim
dw
n+2s
π 2 Γ(s) y &0 Rn
[|w |2 + y 2 ] 2
21−2s Γ(1 − s)
(−∆)s f (x).
=
Γ(s)
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Traveling Wave Solutions with Boundary Reactions
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Traveling Wave Solutions with Boundary Reactions
Consider the following boundary reaction problem:

2

 ∆u(x, y ) + µu(x, y ) = 0, ∀(x, y ) ∈ R+


 ∂u(x, 0)
= f (u(x, 0)), ∀x ∈ R
∂n




 ux (x, 0) > 0, ∀x ∈ R and
lim u(x, 0) = ±1.
(3.1)
x→±∞
Combustion Model: L. Caffarelli, A. Mellet, Y. Sire (2011)
There exists a unique pair (µ, u) as the solution to (3.1). Moreover, there exists
some c > 0 such that
Z ∞
2
2
−1 + u(x, 0) ≤ √ √ e −z dz, as x → −∞.
π
c|x|
Fisher-KPP Model: Unknown!
Bistable Model: X. Cabré, N. Cónsul, J. Mandé (2010)
There exists a unique pair (µ, u) as the solution to (3.1).
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Traveling Wave Solutions with Fractional Laplacians
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Traveling Wave Solutions with Fractional Laplacians
Consider the following problem

s
0

 (−∆) u(x) − µu (x) = f (u(x)),

u 0 (x) > 0, ∀x ∈ R


 lim u(x) = ±1.
∀x ∈ R
(4.1)
x→±∞
Note: There are nonexistence and existence results for (4.1).
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Nonexistence Results
Fisher-KPP Model:
(
ut (x, t) + (−∆)s u(x, t) = f (u(x, t)),
|u(x, 0)| ≤ 1,
∀x ∈ Rn , t > 0
(4.2)
∀x ∈ Rn
R. Mancinelli, D. Vergni, A. Vulpiani (2003), D. Del-Castillo-Negrete,
B. Carrears, V. Lynch(2003):
They showed numerically that the front propagates exponentially fast in t.
X. Cabré, J. Roquejoffre(2011):
For any 0 < s < 1, the position of all level sets of solution to (4.2) moves
exponentially fast in t. In particular, there is no solution to (4.1).
Combustion Model: C. Gui, T. Huan(2012)
If 0 < s ≤ 12 , then there is no solution to (4.1).
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Existence Results

(−∆)s u(x) − µu 0 (x) = f (u(x)),



u 0 (x) > 0, ∀x ∈ R


 lim u(x) = ±1.
∀x ∈ R
(4.1)
x→±∞
Combustion Model: A. Mellet, J. Roquejoffre, Y. Sire(2011)
If 12 < s < 1, then there exists a unique pair (µ, u) as the solution to (4.1).
Moreover, there exists C > 0 such that
C
1 − u(x) ≤
, as x → ∞.
|x|2s−1
Balanced Bistable Model: X. Cabré, J. Solá-Morales(2005), X. Cabré, Y.
Sire (2010)
There exists a unique solution u to (4.1) with µ = 0. Moreover, we have
1 − u(x) ∼
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1
, as x → ∞ and
|x|2s
− 1 + u(x) ∼
1
, as x → −∞
|x|2s
Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
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De Giorgi Type Conjecture
Conjecture: Find critical dimension ns ∈ N such that for all n ≤ ns , f be balanced
bistable, and u be a bounded solution of the following problem:

(−∆)s u(x) = f (u(x)), ∀x ∈ Rn



Dn u(x) > 0, ∀xRn


 lim u(x 0 , xn ) = ±1, ∀x 0 ∈ Rn−1
xn →±∞
Then u is one dimensional, that is, u(x) = g (a · x + b) in Rn for some a ∈ Rn and
b ∈ R.
Note: X. Cabré, Solà-Morales(2005), E. Valdinoci, Y. Cinti(2009), X. Cabré,
Cinti(2012) affirmed ns ≥ 3 if 21 ≤ s < 1 if n ≤ 3 and ns ≥ 2 if 0 < s < 21
by using methods (Liouville Theorem and Energy Estimates) of proving De Giorgi
Conjecture for −∆.
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How about unbalanced bistable case?
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Main Result

(−∆)s u(x) − µu 0 (x) = f (u(x)),



u 0 (x) > 0, ∀x ∈ R


 lim u(x) = ±1.
∀x ∈ R
(4.1)
x→±∞
C. Gui, M. Zhao(2012):
Let F be any double well potential, then there exists a unique pair (µ, u) as the
solution to (4.1). Moreover, we have
1 − u(x) ∼
1
, as x → ∞ and
|x|2s
− 1 + u(x) ∼
1
, as x → −∞
|x|2s
Note: The continuation method.
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Continuation Method
P. Bates, P. Fife, X. Ren, X. Wang (1997):
Let JZ be a nonnegative symmetric kernel such that J(x), |x|J(x), J 0 (x) ∈ L1 (R)
with
J(x) dx = 1 and f be a bistable nonlinearity. Then there exists a unique
R
pair (µ, u) as the solution to the problem

−J ∗ u(x) + u(x) − µu 0 (x) = f (u(x)), ∀x ∈ R



u 0 (x) > 0, ∀x ∈ R


 lim u(x) = ±1.
x→±∞
Z
Note: −J ∗ u(x) + u(x) =
J(x − y )[u(x) − u(y )] dy .
R
R. Frank, E. Lenzmann (2010):
For all 0 < s < 1, the ground state solution of the following problem
(−∆)s Q + Q − Q α+1 = 0,
is unique provided 0 < α <
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4s
1−2s
for 0 < s <
1
2
∀x ∈ R
and 0 < α < ∞ for
Traveling Wave Solutions of Allen-Cahn Equation with Fractional Laplacians
1
2
≤ s < 1.
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Continuation Method
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 ). Consider
Fθ (t) = (1 − θ)F0 (t) + θF1 (t),
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∀θ ∈ [0, 1].
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Continuation Method
Let g be the solution of (4.1) with balanced double well potential F0 . Let uθ = g +v ,
where v (0) = 0, v ∈ C 2 (R) and lim v (x) = 0. Equivalently, we need to consider
v →±∞
the zero set of the mapping
S(θ, µ, v ) = −v + ((−∆)s + 1)−1 [v + µv 0 + fθ (v + g ) − (−∆)s g + µg 0 ]
Solution set
Σ = {θ ∈ [0, 1] : (4.1) has a solution with fθ }
I. Nondegeneracy of Linearized Equation
The first eigenfunction is simple for fractional Laplacians.
II: Boundedness of (µ, v )
Using elliptic estimate for (−∆)s , it suffices to show the boundedness of speeds µθ
for all θ ∈ [0, 1].
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Motivation in the ODE Case
(
−uθ00 (x) − µθ uθ0 (x) = fθ (uθ (x)),
uθ0 (x)
Then
> 0,
∀x ∈ R
and
∀x ∈ R
(5.1)
lim uθ (x) = ±1.
x→±∞
Z y
1 0
2
[u (y )] + µθ
[uθ0 (x)]2 dx = Fθ (uθ (y )) − Fθ (−1)
2 θ
−∞
Z
µθ [uθ0 (x)]2 dx = Fθ (1) − Fθ (−1)
(5.2)
(5.3)
R
WOLG, assume uθ (0) = t0 , then
1 0
[u (0)]2 ≥ Fθ (t0 ) − Fθ (1) ≥ F1 (t0 ) − F1 (1)
(5.4)
2 θ
0
0
0
00
Let y0 ∈ R be such that uθ (y0 ) = sup uθ (y ) ≥ uθ (0), then uθ (y0 ) = 0. Hence, we
y ∈R
get
µθ =
kf0 kC ([−1,1]) + kf1 kC ([−1,1])
−fθ (uθ (y0 ))
p
≤
,
uθ0 (y0 )
2[F1 (t0 ) − F1 (1)]
∀θ ∈ [0, 1]
Note: The key is that F1 (t0 ) > F1 (1).
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Upper Bounds of Speed
(
(−∆)s uθ00 (x) − µθ uθ0 (x) = fθ (uθ (x)),
uθ0 (x)
∀x ∈ R
> 0,
and
∀x ∈ R
(5.5)
lim uθ (x) = ±1.
x→±∞
Then
Z y
Z y
−
(−∆)s uθ00 (x)uθ0 (x) dx + µθ
[uθ0 (x)]2 dx = Fθ (uθ (y )) − Fθ (−1)
−∞
−∞
Z
µθ [uθ0 (x)]2 dx = Fθ (1) − Fθ (−1)
(5.6)
(5.7)
R
WOLG, assume uθ (0) = t0 , then
Z
0
−
(−∆)s uθ00 (x)uθ0 (x) dx ≥ Fθ (t0 ) − Fθ (1) ≥ F1 (t0 ) − F1 (1)
(5.8)
−∞
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Upper Bounds of Speed: 0 < s <
1
2
For any R > 0, we know that
Z
Z 1
2C (1, s) −2s
z
−(−∆)s uθ (y ) ≤
R
+ C1,s
uθ0 (y + tz) dzdt.
1+2s
s
|z|<R 0 |z|
By Cauchy-Schwarz’s inequality, we have
Z 0
4C (1, s) −2s C (1, s) 1−2s 0 2
R
+
R
kuθ kL2 (R)
−
(−∆)s uθ (y )uθ0 (y ) dy ≤
s
1 − 2s
−∞
4C (1, s) −2s C (1, s) 1−2s F1 (1) − F1 (−1)
≤
R
+
R
(5.9)
s
1 − 2s
µθ
By (5.8) and (5.9), 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
Let R be such that
Mingfeng Zhao (UConn)
4C (1, s) −2s
F1 (t0 ) − F1 (1)
R
=
, then
s
2
µθ ≤ C , ∀θ ∈ [0, 1]
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Upper Bounds of Speed: s =
1
2
Consider the functionZ
ρ(y ) = [|Dx uθ (x, y )|2 − |Dy uθ (x, y )|2 ] dx, ∀y ≥ 0
R
Z
Then
0
ρ (y ) = 2 Dy uθ (x, y )∆uθ (x, y ) dx ≡ 0, ∀y > 0.
R
Since ρ(y ) → 0 as y → ∞, then ρ(y ) ≡ 0 for all y ≥ 0, i.e.,
Z
Z
|Dx uθ (x, y )|2 dx =
|Dy uθ (x, y )|2 dx
R
(5.10)
R
By Cauchy-Schwarz’s inequality and (5.7), then
Z 0
Z 0
Z
s 00
0
0
−
(−∆) uθ (x)uθ (x) dx =
Dx uθ (x, 0)uθ (x) dx ≤
|uθ0 (x)|2 dx
−∞
−∞
R
F1 (1) − F1 (−1)
≤
µθ
(5.11)
By (5.8) and (5.11), then
µθ ≤
Mingfeng Zhao (UConn)
F1 (1) − F1 (−1)
,
F1 (t0 ) − F1 (1)
∀θ ∈ [0, 1]
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Upper Bounds of Speed:
For subcritical case, i.e.,
µθ as denominator of
1
2
1
2
<s<1
< s < 1, we will estimate get an upper bound involving
Z
0
−
(−∆)s uθ00 (x)uθ0 (x) dx
−∞
We also have
µθ ≤ C ,
Mingfeng Zhao (UConn)
∀θ ∈ [0, 1].
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Hamiltonian Identity
Z
It suffices to show
(−∆)s u(x)u 0 (x) dx = 0.
(5.12)
R
1
1B (0)c , go to the case of
|x|1+2s BFRW(1997), in fact, since J ∗ u(x) → ±1, as x → ±∞ and J is even, then
Z
Z
u 0 (x)J ∗ u(x) dx = − u(x)J0 ∗ u(x) dx
R
ZR Z
= −
u(x)u(y )J0 (x − y ) dxdy = 0
Approach I: For any > 0, let J (x) =
R
R
Approach II: Let u be the extension, consider
Z
1 ∞
[|u x (x, t)|2 − |u y (x, t)|2 ]t 1−2s dt
v (x) =
2 0
Z x
21−2s Γ(1 − s)
0
2
−
−µ
|u (t)| dx + F (u(x)) − F (−1)
Γ(s)
−∞
Then v 0 (x) ≡ 0 in R and v (x) → 0 as |x| → ∞.
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Uniqueness and Asymptotic Behavior
Uniqueness: Sliding method. Take (µ1 , u1 ) and (µ2 , u2 ) with µ1 ≥ µ2 and u1 (0) =
u2 (0). Consider
w t (x) = u1 (x + t) − u2 (x),
∀x ∈ R, ∀t ≥ 0.
Then

(−∆)s w t (x) − µ1 (w t )0 (x) + d t (x)w t (x) = (µ1 − µ2 )u20 (x) ≥ 0, ∀x ∈ R



 t
f (u1 (x)) − f (u2 (x))
d (x) := −
, ∀x ∈ R
(5.13)
u1 (x) − u2 (x)


t

 lim w (x) = 0
|x|→∞
For f , then there exists some τ ∈ (0, 1) such that f is decreasing on (−1, −τ ) and
(τ, 1). Take large X0 > 0 such that τ < |u1 (x)|, |u2 (x)| < 1 for |x| ≥ X0 . Take
large X1 > 0 such that u1 (x) ≥ max
max
x∈[−X0 ,X0 ]
u1 (x),
max
x∈[−X0 ,X0 ]
u2 (x) ,
∀x ≥ X1 .
Then for any t ≥ T0 = X0 + X1 , we have w t (x) > 0 in R.
Asymptotic Behavior: Compare with solutions in the balanced bistable case
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Open Question
Open Questions:
What will happen if we add more zeros in (−1, 1) of f in general for all of −∆,
(−∆)s and convolution operator?
Note: The above question is always true for the case of balanced bistable model
because of the variational method.
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Thank You!
Mingfeng Zhao (UConn)
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