EXISTENCE OF STATIONARY SOLUTIONS FOR
SOME INTEGRO-DIFFERENTIAL EQUATIONS
WITH ANOMALOUS DIFFUSION
Vitali Vougalter1 , Vitaly Volpert2
1
2
Department of Mathematics, University of Toronto
Toronto, Ontario, M5S 2E4, Canada
e-mail: vitali@math.toronto.edu
Institute Camille Jordan, UMR 5208 CNRS, University Lyon 1
Villeurbanne, 69622, France
e-mail: volpert@math.univ-lyon1.fr
Abstract: The article deals with the existence of solutions of an integro-differential
equation arising in population dynamics in the case of anomalous diffusion involving the negative Laplace operator raised to a certain fractional power. The proof of
existence of solutions is based on a fixed point technique. Solvability conditions
for non-Fredholm elliptic operators in unbounded domains along with the Sobolev
inequality for a fractional Laplacian are being used.
AMS Subject Classification: 35J05, 35P30, 47F05
Key words: integro-differential equations, non Fredholm operators, Sobolev spaces
1. Introduction
The present article is devoted to the studies of the existence of stationary solutions
of the following integro-differential equation
∫
3
∂u
1
s
= −D(−∆) u +
<s<
(1.1)
K(x − y)g(u(y, t))dy + f (x),
∂t
4
4
R3
appearing in cell population dynamics. The space variable x corresponds to the
cell genotype, u(x, t) denotes the cell density as a function of their genotype and
time. The right side of this equation describes the evolution of cell density by means
of cell proliferation, mutations and cell influx. Here the anomalous diffusion term
is correspondent to the change of genotype via small random mutations, and the
nonlocal term describes large mutations. In this context g(u) stands for the rate
1
of cell birth which depends on u (density dependent proliferation), and the function
K(x−y) gives the proportion of newly born cells which change their genotype from
y to x. Let us assume that it is dependent on the distance between the genotypes.
Finally, the last term in the right side of this equation designates the influx of cells
for different genotypes.
1
3
The operator (−∆)s , < s < in equation (1.1) represents a particular case
4
4
of the anomalous diffusion actively studied in relation with various applications in
plasma physics and turbulence [13], [14], surface diffusion [15], [16], semiconductors [17] and so on. The physical meaning of the anomalous diffusion is that
the random process occurs with longer jumps in comparison with normal diffusion.
The moments of jump length distribution is finite in the case of normal diffusion, but
1
3
this is not the case for superdiffusion. The operator (−∆)s , < s < is defined
4
4
by means of the spectral calculus. A similar problem in the case of the standard
Laplace operator in the diffusion term was studied recently in [28].
Let us set D = 1 and establish the existence of solutions of the problem
∫
3
1
s
−(−∆) u +
<s< .
(1.2)
K(x − y)g(u(y))dy + f (x) = 0,
4
4
R3
1
The particular case of this equation when s = was treated recently in [29]. We
2
consider the case when the linear part of this operator does not satisfy the Fredholm
property. Therefore, conventional methods of nonlinear analysis may not be applicable. Let us use solvability conditions for non Fredholm operators along with the
method of contraction mappings.
Consider the equation
−∆u + V (x)u − au = f,
(1.3)
where u ∈ E = H 2 (Rd ) and f ∈ F = L2 (Rd ), d ∈ N, a is a constant and the
scalar potential function V (x) is either zero identically or converges to 0 at infinity.
For a ≥ 0, the essential spectrum of the operator A : E → F corresponding to the
left side of equation (1.3) contains the origin. As a consequence, this operator fails
to satisfy the Fredholm property. Its image is not closed, for d > 1 the dimension
of its kernel and the codimension of its image are not finite. The present work
is devoted to the studies of some properties of the operators of this kind. Note that
elliptic problems with non Fredholm operators were studied actively in recent years.
Approaches in weighted Sobolev and Hölder spaces were developed in [2], [3],
[4], [5], [6]. The non Fredholm Schrödinger type operators were treated with the
methods of the spectral and the scattering theory in [18], [21], [23]. The Laplace
operator with drift from the point of view of non Fredholm operators was studied
in [22] and linearized Cahn-Hilliard problems in [24] and [26]. Nonlinear non
Fredholm elliptic equations were treated in [25] and [27]. Significant applications
2
to the theory of reaction-diffusion equations were developed in [8], [9]. Operators
without Fredholm property arise also when studying wave systems with an infinite
number of localized traveling waves (see [1]). Particularly, when a = 0 the operator
A is Fredholm in some properly chosen weighted spaces (see [2], [3], [4], [5], [6]).
However, the case of a ̸= 0 is considerably different and the approach developed in
these works cannot be used. Front propagation problems with anomalous diffusion
were studied actively in recent years (see e.g. [30], [31]).
We set K(x) = εK(x) with ε ≥ 0 and suppose that the assumption below is
satisfied.
1
3
Assumption 1. Consider < s < . Let f (x) : R3 → R be nontrivial, such that
4
4
f (x) ∈ L1 (R3 ) and (−∆)1−s f (x) ∈ L2 (R3 ). Assume also that K(x) : R3 → R
and K(x) ∈ L1 (R3 ). Additionally, (−∆)1−s K(x) ∈ L2 (R3 ), such that
Q := ∥(−∆)1−s K(x)∥L2 (R3 ) > 0.
Let us choose the space dimension d = 3, which is related to the solvability
conditions for the linear Poisson type problem (3.22) established in Lemma 5. From
the point of view of applications, the space dimension is not limited to d = 3 since
the space variable is correspondent to cell genotype but not to the usual physical
space.
We use the Sobolev inequality for the fractional Laplacian (see e.g. Lemma 2.2
of [10], also [11])
1
3
<1−s<
(1.4)
4
4
along with the assumption above and the standard interpolation argument, which
yields
f (x) ∈ L2 (R3 ).
∥f (x)∥
6
L 4s−1 (R3 )
≤ cs ∥(−∆)1−s f (x)∥L2 (R3 ) ,
Let us use the Sobolev spaces
H 2s (R3 ) := {u(x) : R3 → R | u(x) ∈ L2 (R3 ), (−∆)s u ∈ L2 (R3 )},
0<s≤1
equipped with the norm
∥u∥2H 2s (R3 ) := ∥u∥2L2 (R3 ) + ∥(−∆)s u∥2L2 (R3 ) .
(1.5)
The standard Sobolev embedding tells that
∥u∥L∞ (R3 ) ≤ ce ∥u∥H 2 (R3 ) ,
(1.6)
where ce > 0 is the constant of the embedding. When the nonnegative parameter
ε = 0, we obtain the linear Poisson type equation (3.22). By virtue of Lemma 5
below along with Assumption 1 equation (3.22) has a unique solution
u0 (x) ∈ H 2s (R3 ),
3
3
1
<s< ,
4
4
such that no orthogonality conditions are required. By means of Lemma 5, when
3
≤ s < 1, a specific orthogonality relation (3.24) is needed to be able to solve
4
1
problem (3.22) in H 2s (R3 ). On the other hand, we need s > to be able to use the
4
Sobolev type inequality (1.4). By means of Assumption 1, using that
−∆u(x) = (−∆)1−s f (x) ∈ L2 (R3 ),
we have for the unique solution of the linear equation (3.22) that u0 (x) ∈ H 2 (R3 ).
We seek the resulting solution of the nonlinear problem (1.2) as
u(x) = u0 (x) + up (x).
Clearly, we arrive at the perturbative equation
∫
s
(−∆) up = ε
K(x − y)g(u0 (y) + up (y))dy,
R3
(1.7)
1
3
<s< .
4
4
(1.8)
We introduce a closed ball in the Sobolev space
Bρ := {u(x) ∈ H 2 (R3 ) | ∥u∥H 2 (R3 ) ≤ ρ},
0 < ρ ≤ 1.
(1.9)
Let us seek the solution of equation (1.8) as the fixed point of the auxiliary nonlinear
problem
∫
3
1
s
(−∆) u = ε
<s<
(1.10)
K(x − y)g(u0 (y) + v(y))dy,
4
4
R3
in ball (1.9). For a given function v(y) this is an equation with respect to u(x).
The left side of (1.10) contains the non Fredholm operator (−∆)s : H 2s (R3 ) →
L2 (R3 ). Its essential spectrum fills the nonnegative semi-axis [0, +∞). Therefore,
this operator has no bounded inverse. The similar situation appeared in articles [25]
and [27] but as distinct from the present work, the problems studied there required
orthogonality relations. The fixed point technique was used in [20] to estimate
the perturbation to the standing solitary wave of the Nonlinear Schrödinger (NLS)
equation when either the external potential or the nonlinear term in the NLS were
perturbed but the Schrödinger operator involved in the nonlinear problem there had
the Fredholm property (see Assumption 1 of [20], also [7]). We define the interval
on the real line
I := [−ce ∥u0 ∥H 2 (R3 ) − ce , ce ∥u0 ∥H 2 (R3 ) + ce ].
(1.11)
Let us make the following assumption on the nonlinear part of problem (1.2).
Assumption 2. Let g(z) : R → R, such that g(0) = 0 and g ′ (0) = 0. It is also
assumed that g(z) ∈ C2 (R), such that
a2 := supz∈I |g ′′ (z)| > 0.
4
Clearly, a1 := supz∈I |g ′ (z)| > 0 as well, otherwise the function g(z) will be constant on the interval I and then a2 = 0. For instance, g(z) = z 2 obviously satisfies
the assumption above.
We introduce the operator Tg , such that u = Tg v, where u is a solution of
equation (1.10). Our main proposition is as follows.
Theorem 3. Let Assumptions 1 and 2 hold. Then problem (1.10) defines the map
Tg : Bρ → Bρ , which is a strict contraction for all 0 < ε < ε∗ for some ε∗ > 0.
The unique fixed point up (x) of this map Tg is the only solution of equation (1.8) in
Bρ .
Evidently, the resulting solution of problem (1.2) given by (1.7) will be nontrivial because the source term f (x) is nontrivial and g(0) = 0 according to our
assumptions. Let us make use of the following trivial lemma.
Lemma 4. For R ∈ (0, +∞) consider the function
β
1
3
,
< s < , α, β > 0.
4s
R
4
4
(
) 31
4βs
It achieves the minimal value at R∗ =
, which is given by
α(3 − 4s)
φ(R) := αR3−4s +
φ(R∗ ) = 3(3 − 4s) 3 −1 (4s)− 3 α 3 β 1− 3 .
4s
4s
4s
4s
Let us proceed to the proof of our main result.
2. The existence of the perturbed solution
Proof of Theorem 3. We choose arbitrarily v(x) ∈ Bρ and denote the term involved
in the integral expression in the right side of problem (1.10) as
G(x) := g(u0 (x) + v(x)).
Let us apply the standard Fourier transform (3.25) to both sides of equation (1.10),
which yields
b G(p)
b
3 K(p)
.
u
b(p) = ε(2π) 2
|p|2s
Thus for the norm we have
∫
∥u∥2L2 (R3 )
3 2
= (2π) ε
R3
5
2 b
b
|K(p)|
|G(p)|2
dp.
|p|4s
(2.12)
As distinct from articles [25] and [27] involving the standard Laplacian operator
in the diffusion term, here we do not try to control the norm
K(p)
b
.
2s |p| ∞ 3
L (R )
We estimate the right side of (2.12) using inequality (3.26) with R > 0 as
∫
2 b
2 b
b
b
|K(p)|
|G(p)|2
|K(p)|
|G(p)|2
3 2
dp
+
(2π)
ε
dp ≤
|p|4s
|p|4s
|p|>R
∫
3 2
(2π) ε
|p|≤R
{
≤ε
2
∥K∥2L1 (R3 )
}
1
R3−4s
1
2
2
∥G(x)∥L1 (R3 )
+
∥G(x)∥L2 (R3 ) .
2π 2
3 − 4s R4s
(2.13)
Due to the fact that v(x) ∈ Bρ , we get
∥u0 + v∥L2 (R3 ) ≤ ∥u0 ∥H 2 (R3 ) + 1.
The Sobolev embedding (1.6) gives us
|u0 + v| ≤ ce ∥u0 ∥H 2 (R3 ) + ce .
∫
u0 +v
By means of the formula G(x) =
g ′ (z)dz with the interval I defined in
0
(1.11), we obtain
|G(x)| ≤ supz∈I |g ′ (z)||u0 + v| = a1 |u0 + v|.
Hence
∥G(x)∥L2 (R3 ) ≤ a1 ∥u0 + v∥L2 (R3 ) ≤ a1 (∥u0 ∥H 2 (R3 ) + 1).
∫ u0 +v [ ∫ y
]
Evidently, G(x) =
dy
g ′′ (z)dz . This yields
0
0
a2
1
|G(x)| ≤ supz∈I |g ′′ (z)||u0 + v|2 = |u0 + v|2 ,
2
2
a2
a2
∥u0 + v∥2L2 (R3 ) ≤ (∥u0 ∥H 2 (R3 ) + 1)2 .
(2.14)
2
2
Hence we arrive at the estimate from above for the right side of (2.13) as
}
{
2
2
a
a
1
2
ε2 ∥K∥2L1 (R3 ) (∥u0 ∥H 2 (R3 ) + 1)2
(∥u0 ∥H 2 (R3 ) + 1)2 R3−4s + 4s ,
8π 2 (3 − 4s)
R
∥G(x)∥L1 (R3 ) ≤
6
where R ∈ (0, +∞). Lemma 4 gives us the minimal value of the expression above.
Therefore,
8s
∥u∥2L2 (R3 )
≤ε
2
∥K∥2L1 (R3 ) (∥u0 ∥H 2 (R3 )
+ 1)
2+ 8s
3
2− 8s
3
3a23 a1
4s
8s
8s
(3 − 4s)s 3 π 3 24s+ 3
.
(2.15)
Clearly, (1.10) yields
∫
−∆u = ε(−∆)
1−s
R3
K(x − y)G(y)dy.
By virtue of (3.26) along with (2.14) we arrive at
∥∆u∥2L2 (R3 ) ≤ ε2 ∥G∥2L1 (R3 ) Q2 ≤ ε2
a22
(∥u0 ∥H 2 (R3 ) + 1)4 Q2 .
4
(2.16)
Therefore, by means of the definition of the norm (1.5) along with inequalities
(2.15) and (2.16) we derive the upper bound for ∥u∥H 2 (R3 ) as
ε(∥u0 ∥H 2 (R3 ) + 1)2 a2 ×
[
(
× ∥K∥2L1 (R3 )
a2 (∥u0 ∥
+ 1)
H 2 (R3 )
) 8s3 −2
2
3
4s
3
8s
3
4s+ 8s
3
(3 − 4s)s π 2
a1
+
Q
4
] 12
≤ρ
for all ε > 0 sufficiently small. Therefore, u(x) ∈ Bρ as well. If for a certain
v(x) ∈ Bρ there exist two solutions u1,2 (x) ∈ Bρ of equation (1.10), their difference
w(x) := u1 (x) − u2 (x) ∈ L2 (R3 ) solves
(−∆)s w = 0.
Because the operator (−∆)s considered in the whole space does not possess nontrivial square integrable zero modes, w(x) = 0 a.e. in R3 . Hence, equation (1.10)
defines a map Tg : Bρ → Bρ for all ε > 0 small enough.
Our goal is to establish that this map is a strict contraction. Let us choose arbitrarily v1,2 (x) ∈ Bρ . The argument above gives us u1,2 = Tg v1,2 ∈ Bρ as well. By
means of (1.10)
∫
s
(−∆) u1 = ε
K(x − y)g(u0 (y) + v1 (y))dy,
(2.17)
R3
∫
s
(−∆) u2 = ε
R3
K(x − y)g(u0 (y) + v2 (y))dy,
3
1
< s < . We define
4
4
G1 (x) := g(u0 (x) + v1 (x)),
G2 (x) := g(u0 (x) + v2 (x))
7
(2.18)
and apply the standard Fourier transform (3.25) to both sides of equations (2.17)
and (2.18). This yields
3
ub1 (p) = ε(2π) 2
b G
c1 (p)
K(p)
,
|p|2s
3
ub2 (p) = ε(2π) 2
b G
c2 (p)
K(p)
.
|p|2s
Obviously,
∫
∥u1 −
u2 ∥2L2 (R3 )
2
3
= ε (2π)
R3
2 c
b
c2 (p)|2
|K(p)|
|G1 (p) − G
dp.
|p|4s
Evidently, it can be estimated from above by virtue of (3.26) by
{
}
3−4s
1
R
1
ε2 ∥K∥2L1 (R3 )
∥G1 (x) − G2 (x)∥2L1 (R3 )
+ ∥G1 (x) − G2 (x)∥2L2 (R3 ) 4s ,
2π 2
3 − 4s
R
where R ∈ (0, +∞). We use the identity
∫
u0 +v1
G1 (x) − G2 (x) =
g ′ (z)dz.
u0 +v2
Hence
|G1 (x) − G2 (x)| ≤ supz∈I |g ′ (z)||v1 − v2 | = a1 |v1 − v2 |.
Thus
∥G1 (x) − G2 (x)∥L2 (R3 ) ≤ a1 ∥v1 − v2 ∥L2 (R3 ) ≤ a1 ∥v1 − v2 ∥H 2 (R3 ) .
Clearly,
∫
[∫
u0 +v1
G1 (x) − G2 (x) =
y
dy
u0 +v2
]
g ′′ (z)dz .
0
Let us obtain the estimate from above for G1 (x) − G2 (x) in the absolute value as
1
a2
supz∈I |g ′′ (z)||(v1 − v2 )(2u0 + v1 + v2 )| = |(v1 − v2 )(2u0 + v1 + v2 )|.
2
2
The Schwarz inequality yields the upper bound for the norm ∥G1 (x)−G2 (x)∥L1 (R3 )
as
a2
∥v1 −v2 ∥L2 (R3 ) ∥2u0 +v1 +v2 ∥L2 (R3 ) ≤ a2 ∥v1 −v2 ∥H 2 (R3 ) (∥u0 ∥H 2 (R3 ) +1). (2.19)
2
Thus we arrive at the estimate from above for the norm ∥u1 (x) − u2 (x)∥2L2 (R3 ) given
by
ε2 ∥K∥2L1 (R3 ) ∥v1 − v2 ∥2H 2 (R3 )
{a
2
2
2π
(∥u0 ∥H 2 (R3 ) + 1)2
2
8
a1 2 }
R3−4s
+ 4s .
3 − 4s R
Lemma 4 allows us to minimize the expression above over R ∈ (0, +∞) to obtain
the upper bound for ∥u1 (x) − u2 (x)∥2L2 (R3 ) as
ε2 ∥K∥2L1 (R3 ) ∥v1 − v2 ∥2H 2 (R3 )
[
2− 8s
3
3a1
4s
(3 − 4s)24s s 3
a2 (∥u0 ∥H 2 (R3 ) + 1)
π
] 8s3
.
(2.20)
Identities (2.17) and (2.18) give us
∫
(−∆)(u1 − u2 ) = ε(−∆)
1−s
R3
K(x − y)[G1 (y) − G2 (y)]dy.
By means of inequalities (3.26) and (2.19) we arrtive at
∥∆(u1 − u2 )∥2L2 (R3 ) ≤ ε2 ∥(−∆)1−s K∥2L2 (R3 ) ∥G1 − G2 ∥2L1 (R3 ) ≤
≤ ε2 Q2 a22 ∥v1 − v2 ∥2H 2 (R3 ) (∥u0 ∥H 2 (R3 ) + 1)2 .
(2.21)
By virtue of (2.20) and (2.21) the norm ∥u1 − u2 ∥H 2 (R3 ) is estimated from above by
the expression εa2 (∥u0 ∥H 2 (R3 ) + 1)×
{
×
[
3∥K∥2L1 (R3 )
(3 −
4s)24s s
4s
3
π
8s
3
a2 (∥u0 ∥H 2 (R3 ) + 1)
a1
] 8s3 −2
} 12
+ Q2
∥v1 − v2 ∥H 2 (R3 ) .
This yields that the map Tg : Bρ → Bρ defined by equation (1.10) is a strict contraction for all values of ε > 0 small enough. Its unique fixed point up (x) is the
only solution of problem (1.8) in the ball Bρ . The resulting u(x) ∈ H 2 (R3 ) given
by (1.7) is a solution of equation (1.2).
3. Auxiliary results
Let us obtain the solvability conditions for the following linear Poisson type
equation with a square integrable right side
(−∆)s u = f (x),
x ∈ R3 ,
0 < s < 1.
(3.22)
We designate the inner product as
∫
(f (x), g(x))L2 (R3 ) :=
f (x)ḡ(x)dx,
(3.23)
R3
with a slight abuse of notations when the functions involved in (3.23) are not square
integrable, like for instance the ones present in orthogonality conditions (3.24) and
(3.34) below. Indeed, if f (x) ∈ L1 (R3 ) and g(x) ∈ L∞ (R3 ) , then the integral in
the right side of (3.23) is well defined. Our technical statement is as follows.
9
Lemma 5. Let f (x) ∈ L2 (R3 ).
1) When 0 < s < 34 and additionally f (x) ∈ L1 (R3 ), problem (3.22) has a
unique solution u(x) ∈ H 2s (R3 ).
2) When 34 ≤ s < 1 and in addition |x|f (x) ∈ L1 (R3 ), equation (3.22) admits
a unique solution u(x) ∈ H 2s (R3 ) if and only if the orthogonality condition
(f (x), 1)L2 (R3 ) = 0
(3.24)
holds.
Proof. Let us first note that by virtue of the norm definition (1.5) along with the
square integrability of the right side of (3.22), it would be sufficient to establish the
solvability of equation (3.22) in L2 (R3 ). The solution u(x) ∈ L2 (R3 ) will clearly
belong to H 2s (R3 ), 0 < s < 1 as well.
We prove the uniqueness of solutions for problem (3.22). Suppose u1,2 (x) ∈
H 2s (R3 ) both solve (3.22). Then their difference w(x) := u1 (x) − u2 (x) satisfies
the homogeneous equation
(−∆)s w = 0.
Since the operator (−∆)s in R3 does not possess nontrivial square integrable zero
modes, w(x) vanishes a.e. in the whole space. We will use the standard Fourier
transform
∫
1
b
f (p) :=
f (x)e−ipx dx.
(3.25)
3
3
2
(2π) R
Obviously, we have the inequality
∥fb(p)∥L∞ (R3 ) ≤
1
3
(2π) 2
∥f (x)∥L1 (R3 ) .
(3.26)
We apply (3.25) to both sides of problem (3.22). This yields
u
b(p) =
fb(p)
fb(p)
3 | |p|≤1} +
χ
χ{p∈R3 | |p|>1} ,
{p∈R
|p|2s
|p|2s
(3.27)
where χA denotes the characteristic function of a set A ⊆ R3 . Clearly for all
0 < s < 1 the second term in the right side of (3.27) is square integrable due to the
estimate
∫
|fb(p)|2
χ{p∈R3 | |p|>1} dp ≤ ∥f ∥2L2 (R3 ) < ∞.
4s
|p|
3
R
To prove the square integrability of the first term in the right side of (3.27) when
0 < s < 34 , we use estimate (3.26), which yields
∫
R3
∥f (x)∥2L1 (R3 )
|fb(p)|2
3
χ
dp
≤
< ∞,
{p∈R | |p|≤1}
|p|4s
2π 2 (3 − 4s)
10
which completes the proof of part 1) of the lemma.
To study the solvability of equation (3.22) when 34 ≤ s < 1, we use the expansion
∫ |p| b
∂ f (s, ω)
fb(p) = fb(0) +
ds.
∂s
0
Here and below ω denotes the angle variables on the sphere. This enables us to
write the first term in the right side of (3.27) as
fb(0)
χ{p∈R3 | |p|≤1} +
|p|2s
∫ |p|
0
∂ fb(s,ω)
ds
∂s
χ{p∈R3 | |p|≤1} .
2s
|p|
(3.28)
Definition (3.25) yields
∂ fb(|p|, ω) 1
∥|x|f (x)∥L1 (R3 ) < ∞
≤
∂|p| (2π) 32
by means of one of our assumptions. Thus,
∫ |p| b
∂ f (s,ω)
ds
1
0
1−2s
∂s
χ{p∈R3 | |p|≤1} ≤
χ{p∈R3 | |p|≤1} ∈ L2 (R3 ).
3 ∥|x|f ∥L1 (R3 ) |p|
2s
(2π) 2
|p|
fb(0)
χ{p∈R3 | |p|≤1} ∈ L2 (R3 ) if and only if fb(0) = 0,
2s
|p|
which yields orthogonality condidition (3.24) in case 2) of the lemma.
The remaining term in (3.28)
Note that for the lower values of the power of the negative Laplacian 0 < s < 34
under the assumptions stated above no orthogonality relations are needed to solve
the linear Poisson type equation (3.22) in H 2s (R3 ).
We prove that one can incorporate a shallow, short-range potential into the linear
Poisson type equation considered above and generalize the result of Lemma 5. Let
us consider the equation with a square integrable right side
(−∆ + V (x))s u = f (x),
x ∈ R3 ,
0<s<1
(3.29)
with the operator (−∆ + V (x))s defined by means of the spectral calculus. Under
our assumptions the operator −∆ + V (x) on L2 (R3 ) is nonnegative as discussed
below.
Assumption 6. The potential function V (x) : R3 → R satisfies the estimate
|V (x)| ≤
C
1 + |x|3.5+ε
11
with some ε > 0 and x ∈ R3 a.e. such that
1
8
2
1 9
4 9 (4π)− 3 ∥V ∥L9 ∞ (R3 ) ∥V ∥ 9 4 3 < 1
L 3 (R )
8
and
√
cHLS ∥V ∥L 23 (R3 ) < 4π.
(3.30)
This is analogous to Assumption 1.1 of [21] under which by means of Lemma
2.3 of [21] our Schrödinger operator −∆ + V (x) is self-adjoint and unitarily equivalent to −∆ on L2 (R3 ) via the wave operators. Hence the essential spectrum of
(−∆ + V (x))s on L2 (R3 ) fills the nonnegative semi-axis [0, +∞). Therefore, such
operator does not have a bounded inverse and therefore it fails to satisfy the Fredholm property. Here C denotes a finite, positive constant and cHLS the constant in
the Hardy-Littlewood-Sobolev inequality
∫ ∫
3
f1 (x)f1 (y)
dxdy
≤ cHLS ∥f1 ∥2 32 3 , f1 ∈ L 2 (R3 )
2
L (R )
R3 R3 |x − y|
given on p.98 of [12]. The functions of the continuous spectrum of our Schrödinger
operator satisfy
(−∆ + V (x))φk (x) = k 2 φk (x),
k ∈ R3 ,
in the integral formulation the Lippmann-Schwinger equation (see e.g. p.98 of [19])
∫
ei|k||x−y|
eikx
1
φk (x) =
(V φk )(y)dy
(3.31)
−
3
4π R3 |x − y|
(2π) 2
and the orthogonality relations
(φk (x), φq (x))L2 (R3 ) = δ(k − q),
k, q ∈ R3 .
They form a complete system in L2 (R3 ). When the wave vector k vanishes, we deal
with the function φ0 (x) used in orthogonality relation (3.34) below. We denote by
tilde the generalized Fourier transform with respect to these functions, such that
f˜(k) := (f (x), φk (x))L2 (R3 ) ,
k ∈ R3 .
(3.32)
The integral operator involved in the right side of equation (3.31) is
∫
1
ei|k||x−y|
(Qφ)(x) := −
(V φ)(y)dy, φ ∈ L∞ (R3 ).
4π R3 |x − y|
We consider Q : L∞ (R3 ) → L∞ (R3 ). By virtue of Lemma 2.1 of [21] under
Assumption 6 above on the scalar potential we have ∥Q∥∞ < 1. Furthermore, this
12
norm is bounded above by I(V ), which denotes the left side of the first inequality
in (3.30). Corollary 2.2 of [21] yields the estimate
|f˜(k)| ≤
1
1
∥f ∥L1 (R3 ) .
(2π) 1 − I(V )
3
2
(3.33)
We have the following proposition.
Lemma 7. Let the potential V (x) satisfy Assumption 6 and f (x) ∈ L2 (R3 ).
a) Let 0 < s < 34 and additionally f (x) ∈ L1 (R3 ). Then equation (3.29) admits
a unique solution u(x) ∈ L2 (R3 ).
b) Let 34 ≤ s < 1 and in addition |x|f (x) ∈ L1 (R3 ). Then problem (3.29) has a
unique solution u(x) ∈ L2 (R3 ) if and only if the orthogonality condition
(f (x), φ0 (x))L2 (R3 ) = 0
(3.34)
holds.
Proof. We first suppose that equation (3.29) admits two solutions u1,2 (x) ∈
L2 (R3 ). Then their difference w(x) := u1 (x) − u2 (x) ∈ L2 (R3 ) satisfies the homogeneous equation
(−∆ + V (x))s w(x) = 0,
which cannot have nontrivial square integrable solutions due to the fact that our
self-adjoint operator −∆ + V (x) is unitarily equivalent to −∆ on L2 (R3 ). Hence
w(x) = 0 a.e. in R3 .
We apply the generalized Fourier transform (3.32) to both sides of problem
(3.29), which yields
ũ(k) =
f˜(k)
f˜(k)
3 | |k|≤1} +
χ
χ{k∈R3 | |k|>1} .
{k∈R
|k|2s
|k|2s
(3.35)
The second term in the right side of (3.35) is square integrable for all 0 < s < 1,
since
∫
|f˜(k)|2
χ{k∈R3 | |k|>1} dk ≤ ∥f ∥2L2 (R3 ) < ∞.
4s
R3 |k|
We use inequality (3.33) to obtain the upper bound on the square of the L2 norm of
the first term in the right side of (3.35) as
∫
R3
1
|f˜(k)|2
1
χ{k∈R3 | |k|≤1} dk ≤ 2
∥f (x)∥2L1 (R3 ) < ∞
4s
|k|
2π (3 − 4s) (1 − I(V ))2
for 0 < s < 34 , which completes the proof of part a) of the lemma.
13
To study the solvability of equation (3.29) when
∫
|k|
f˜(k) = f˜(0) +
0
3
4
≤ s < 1, we use the identity
∂ f˜(s, ω)
ds,
∂s
which enables us to express the first term in the right side of (3.35) as
f˜(0)
χ{k∈R3 | |k|≤1} +
|k|2s
∫ |k|
0
∂ f˜(s,ω)
ds
∂s
χ{k∈R3 | |k|≤1} .
2s
|k|
(3.36)
Note that ∇k f˜(k) ∈ L∞ (R3 ) by virtue of Lemma 2.4 of [21]. Hence the second
term in (3.36) can be estimated from above in the absolute value by
∥∇k f˜(k)∥L∞ (R3 ) |k|1−2s χ{k∈R3 | |k|≤1} ∈ L2 (R3 ),
3
≤ s < 1.
4
f˜(0)
χ{k∈R3 | |k|≤1} ∈ L2 (R3 ) if and only if f˜(0) van2s
|k|
ishes, which gives us orthogonality relation (3.34) in case 2) of the lemma.
The remaining term in (3.36)
Note that in case a) of the lemma above, when our Schrödinger operator is
raised to a lower power 0 < s < 34 , under the given assumptions no orthogonality
conditions are needed for solving equation (3.29) in L2 (R3 ). The particular case of
Lemma 7, when s = 21 was treated recently in [29].
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